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 SAT Math Multiple Choice Practice Question 321: Answer and Explanation_cracksat.net # SAT Math Multiple Choice Practice Question 321: Answer and Explanation ### Next steps Question: 321 13. When m is subtracted from n, the result is r. Which of the following expresses the result when 2m is added to s? A. B. C. D. E. Explanation: A Write an equation for the first sentence. Because none of the answer choices contain m, solve for m in terms of r and n: Now write an expression for what the question asks for: Alternatively, you can substitute numbers for n, m, and r, making sure they "work," and get a numerical answer to the question. 

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A090825 Nonprimes n such that (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) is an integer, where B(k) denotes the k-th Bernoulli number. 0 1, 49, 91, 119, 133, 169, 217, 221, 247, 259, 289, 301, 323, 329, 335, 343, 361, 403, 413, 427, 469, 481, 497, 511, 517, 527, 553, 559, 589, 611, 629, 637, 679, 703, 707, 721, 731, 749 (list; graph; refs; listen; history; text; internal format) OFFSET 1,2 COMMENTS Conjecture: composite numbers with all prime factors in A053176 are in the sequence. For p prime (3/2)*(1/p)*(2*p+1)*(3^p+1)*B(2*p) == 1 (mod p). There are few terms n with (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) == 1 (mod n): 91,247,....Is this subsequence finite? LINKS PROG (PARI) for(n=1, 750, if(frac( (3/2)*(1/n)*(2*n+1)*(3^n+1)*bernfrac(2*n))==0, if(isprime(n)==0, print1(n, ", ")))) CROSSREFS Sequence in context: A326257 A231275 A158725 * A157342 A230226 A178705 Adjacent sequences:  A090822 A090823 A090824 * A090826 A090827 A090828 KEYWORD nonn AUTHOR Benoit Cloitre, Feb 11 2004 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified June 16 02:14 EDT 2021. Contains 345055 sequences. (Running on oeis4.)

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# Search Our Content Library 72 filtered results 72 filtered results Measuring in Inches Sort by Measure Length: Shark! Worksheet Measure Length: Shark! Help a little mathematician learn about measurements with this fun, shark infested math worksheet! Math Worksheet Estimating Measurement Worksheet Estimating Measurement We make "guesstimations" all the time in everyday activities, from shopping to cooking. Help your child make estimations with measurements. Math Worksheet How to Measure: Sea Creatures Worksheet How to Measure: Sea Creatures Cut out one of our colorful rulers and measure these sea creatures to answer the questions. No oxygen tank needed. Math Worksheet Using a Ruler Worksheet Using a Ruler You can estimate how long your pencil is, but you won't know for sure without a ruler. Cut out the ruler on this worksheet and get measuring! Math Worksheet Measure Length: Squirrel! Worksheet Measure Length: Squirrel! How long is this little squirrel's tail? Pop out the ruler and measure this feisty furball before he runs away! Math Worksheet Measure Inches at the Zoo Worksheet Measure Inches at the Zoo How many inches from the elephant to the lion? Kids get great practice measuring in inches as they fill out this simple worksheet. Math Worksheet Measure Length: Narwhal! Worksheet Measure Length: Narwhal! Pull out that ruler and measure parts of this little narwhal. This activity will give your child practice using a ruler to measure length. Math Worksheet Measure Length: Elephant! Worksheet Measure Length: Elephant! Practice basic measurement skills with this mini elephant! This fun activity will get your child started using a ruler and help him get familiar with inches. Math Worksheet Estimate and Measure Worksheet Estimate and Measure Bring an everyday skill to the table by helping your child try and "eyeball" a measurement for each item in inches and centimeters. Math Worksheet Measure Length: Horse! Worksheet Measure Length: Horse! Build basic measurement skills help from Harry the Horse! Your child will practice using a ruler to measure length, a skill he'll use the rest of his life. Math Worksheet How to Measure: Food Worksheet How to Measure: Food Cut out a colorful ruler and measure food items to practice an essential 2nd grade skill. Math Worksheet Worksheet Chances are your child knows how to round, but how about rounding in inches? Help him tackle this concept with an exercise in measuring to the nearest inch. Math Worksheet How to Measure: Dragon Worksheet How to Measure: Dragon Practice measuring inches with this picture of a smiling dragon. Math Worksheet Measure Length: Cat! Worksheet Measure Length: Cat! This fluffy feline is here to help your child master his measuring skills. Whip out the ruler and measure the different parts of this kitty. Math Worksheet Worksheet Here's a math activity with a real-life application: measuring! Help your child get to know rulers and inches with this fun cut and measure worksheet. Math Worksheet How to Measure: Cowboy Worksheet How to Measure: Cowboy Cut out a colorful ruler and start practicing measuring with this cowboy picture. Math Worksheet Measure & Draw #2 Worksheet Measure & Draw #2 Can your child measure up? How about down, as in, let's make these objects half their size? This is a challenging worksheet where your child can… Math Worksheet Hand Measurements Worksheet Hand Measurements On this hands-on second grade math worksheet, kids measure the length and width of their hand and the length of their fingers to the nearest inch or half inch. Math Worksheet Estimation for Kids Worksheet Estimation for Kids Size up these barn animals with your little farmer. Help him learn to eyeball things by using his knowledge of how long an inch is. Math Worksheet Measuring Worksheet Measuring Help Bobby the carpenter make a dollhouse by taking its measurements. How big would it be in real life? Math Worksheet Measuring Dimensions Worksheet Measuring Dimensions In this measuring exercise, your child can use her measuring know-how to get the signs' dimensions, then record them for Bobby. Math Worksheet Measure & Draw #5 Worksheet Measure & Draw #5 Math Worksheet Foot Measurements Worksheet Foot Measurements On this second grade math worksheet, kids measure the length and width of their foot and the length of their toes to the nearest inch or half inch. Math Worksheet Measure & Draw #4 Worksheet Measure & Draw #4 Your child will inch her way across this worksheet, literally! Get out a ruler and have her measure the pictures on the page. She'll get practice…

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# Scalar first passage time problems Although most of the literature on diffusion processes are concerned with inferring dynamics from discretely observed trajectories, the application of diffusion models extend far beyond formulating a compact description of the dynamics of a given process. Often the objective of an analysis is to gain insight into the behaviour of events contingent on the trajectory of the process rather than the trajectory of the process itself. In a predictive context, one may be interested in the probability of a particular event occurring by a given time in the future. As such, a quantity that has been explored at length throughout the history of diffusion processes is the distribution of the time elapsed until the diffusion crosses a predefined threshold. The distribution of the first passage time $T_{X_s\rightarrow \lambda_t} = \inf\{t>s: X_t>\lambda_t\}$ of a scalar diffusion process $$X_t$$ crossing a fixed barrier $$\lambda_t$$ is given by the first kind Volterra equation $f(\lambda_t|X_s) = \int_s^t f(\lambda_t|\lambda_u)g(\lambda_u|X_s)du,$ where $$g(\lambda_t|X_s)$$ denotes the first passage time density evaluated at time $$t$$ and $$f(y|x)$$ is the transitional density of the diffusion process. Unfortunately, even when the transitional density is available exactly, the integral equation may not be analytically tractable. Fortunately, numerous techniques exist for evaluating the first passage time density by solving the integral equation numerically. Thus, by combining the cumulant truncation procedure for evaluating the transitional density with numerical procedures for evaluating the integral equation, we are able to evalaute first passage time problems for time-inhomogeneous generalized quadratic diffusions. # Examples of time inhomogeneous first passage time problems Another useful application of DiffusionRgqd package is the approximation of first passage time densities via the GQD.TIpassge() function. The GQD.TIpassage() function uses the same GQD interface as other functions in the package and operates in similar fashion to the GQD.density() function. However, since GQD.TIpassage() relies on calculating transitional density approximations for a large number of initial values in combination with the recursive updating algorithm of Buonocore, Nobile, and Ricciardi (1987), its internal workings have more in common with the GQD.mle() and GQD.mcmc() functions, where computationally optimized solutions with respect to the cumulant truncation procedure are constructed in C++ which is subsequently executed in R. As an introduction to the function we first compare the GQD.TIpassage() function to an existing R package for calculating first passage time densities for Gaussian diffusions and subsequently investigate the properties of the first passage time density of a non-linear diffusion transiting through a moving barrier. An excellent package for the analysis of first passage time problems is the fptdApprox (Román-Román, Serrano-Pérez, and Torres-Ruiz 2014) package. Since fptdApprox can very effectively handle first passage time problems for diffusions with analytically tractable transitional densities we use it to compare some of the results from the DiffusionRgqd package. Consider for example a diffusion process with SDE: $dX_t = 0.5(5-X_t)dt+dB_t,$ with $$X_0 =3$$. For purposes of calculating a first passage time consider then also a continuous barrier $\lambda_t = 5+0.25\sin(2\pi t).$ Under the fptdApprox package we may use the Approx.fpt.density() function in order to approximate the first passage time density. The interface requires that one define a diffusion process by configuring an object that consists of expressions giving the drift, diffusion and transitional density of the model at hand. The resulting object is then used by the Approx.fpt.density() function to approximate the first passage time density. More formally: library("fptdApprox") # Under fptdApprox': # Define the diffusion process and give its transitional density: OU <- diffproc(c("alpha*x + beta","sigma^2", "dnorm((x-(y*exp(alpha*(t-s)) - beta*(1 - exp(alpha*(t-s)))/alpha))/ (sigma*sqrt((exp(2*alpha*(t-s)) - 1)/(2*alpha))),0,1)/ (sigma*sqrt((exp(2*alpha*(t-s)) - 1)/(2*alpha)))", "pnorm(x, y*exp(alpha*(t-s)) - beta*(1 - exp(alpha*(t-s)))/alpha, sigma*sqrt((exp(2*alpha*(t-s)) - 1)/(2*alpha)))")) # Approximate the first passgage time density for OU, starting in X_0 = 3 # passing through 5+0.25*sin(2*pi*t) on the time interval [0,10]: res1 <- Approx.fpt.density(OU, 0, 10, 3,"5+0.25*sin(2*pi*t)", list(alpha=-0.5,beta=0.5*5,sigma=1)) ## ## Computing... Done. ## ## The value of the cumulative integral of the approximation is 0.992261204012805 < 1 - tol. ## If the value of the cumulative integral is not high and the final stopping instant is less than T, it may be appropriate: ## - Check if the value of the final stopping instant increases using k argument to summary the fptl class object, or ## - Approximate the density again with to.T = TRUE. Using the __DiffusionRgqd__package we begin by defining the model as per usual according to the GQD framework. Then we call the GQD.TIpassage() function and provide it with the parameters of the first passage time problem. Note that GQD.TIpassage() circumvents the need to specify the transitional density as it will use the model coefficients to recognize and construct the appropriate numerical approximation of the required transitional densities. In present form, for computational purposes, the GQD.TIpassage() function evaluates the iterative updating equation of Buonocore, Nobile, and Ricciardi (1987) for the first passage time density density under constant barriers only. However, this is not a limiting constraint since for any barrier function that can be decomposed as $\lambda_t = \phi +h(t)$ where $$h(t)$$ is continuous, the first passage time problem $$\{Y_t = X_t - h(t) \rightarrow \phi| X_s-h(s)\}$$ is equivalent to that of $$\{X_t \rightarrow \lambda_t| X_s\}$$, provided that the transformed process is indeed a valid diffusion process. All that remains is to calculate the dynamics of $$Y_t$$ under Ito’s lemma. library(DiffusionRgqd) # Under DiffusionRgqd': # Define the diffusion process G0=function(t){0.5*5 - 0.5*pi*cos(2*pi*t) - 0.5*0.25*sin(2*pi*t)} G1=function(t){-0.5} Q0=function(t){1} # Approximate the first passgage time density for OU, starting in X_0 = 3 # passing through 5+0.25*sin(2*pi*t) on the time interval [0,10]: res2 <- GQD.TIpassage(Xs = 3, B = 5, s = 0, t = 10, delt = 1/200) ## Compiling C++ code. Please wait. By plotting the subsequent approximations we infer from the resulting figure that the approximations are near identical. # Let's compare the resulting densities: plot(res1$y~res1$x, type = 'l', col = '#BBCCEE', xlab = 'Time(t)', ylab = 'FPT Density', main = 'DiffusionRgqd vs. fptdApproximate.', lwd = 2) lines(res2$density~res2$time, col = '#222299', lwd = 2, lty = 'dashed') legend('topright', lty = c('solid', 'dashed'), col = c('#BBCCEE', '#222299'), legend = c('fptdApproximate', 'DiffusionRgqd'), lwd = 2, bty = 'n') Note that since functions in the __DiffusionRgqd__always assumes that a numerical solution is required, it is possible for this example, where the transition density is available analytically, to produce a solution that is more efficiently calculable. However, the aim of the package is tackle problems with intractable dynamics under the GQD framework, where transition densities are rarely analytically available. Furthermore, as with the inference procedures, we again have a great deal of freedom for specifying a first passage time problem and subsequently calculating an accurate approximate solution. Consider for example a diffusion with SDE: $dX_t = \theta_1 X_t(10+0.2\sin(2\pi t)+0.3\sqrt{t}(1+\cos(3\pi t))-X_t)dt+\sqrt{0.1}X_tdB_t,$ with $$X_1 =8$$, $$\theta_1$$ a fixed parameter, and a constant barrier $$\lambda_t = 12$$. Note that for this SDE, no analytical solution for the transition density exists. Consequently, we can no longer make use of the fptdApprox package in order to generate the desired first passage time density, albeit for comparative purposes. As such we compare the resulting approximation to a simulated first passage time density for the quadratic diffusion transiting through $$\lambda_t = 12$$. Perhaps the simplest algorithm for achieving this is to simulate numerous trajectories of the diffusion and record the times at which these trajectories cross $$\lambda_t$$. In R this can be achieved for example by: # Simulate the first passage time density: theta <- 0.5 mu <- function(X,t){theta[1]*(10+0.2*sin(2*pi*t)+0.3*sqrt(t)*(1+cos(3*pi*t)))*X-theta[1]*X^2} sigma <- function(X,t){sqrt(0.1)*X} simulate.fpt=function(N,S,B,delt) { X=rep(S,N) Ndim=N time.vector=rep(0,N) t=1 k1=0 while(Ndim>=1) { X=X+mu(X,t)*delt+sigma(X,t)*rnorm(Ndim,sd=sqrt(delt)) # Check if the barrier is crossed and keep trajectories that # have survived: I0=X<B X=X[I0] count=sum(!I0) Ndim=length(X) t=t+delt if(count>0) { time.vector[k1+1:count]=t k1=k1+count } } return(time.vector) } fpt.sim.times <- simulate.fpt(500000, 8, 12, 1/2000) where we have set $$\theta_1 =0.5$$. It is important to note that this scheme is subject to bias since the finite step size used in the simulation implies that the scheme fails to account for trajectories that may have already crossed $$\lambda_t$$ yet are recorded below $$\lambda_t$$ subsequent to each update (Giraudo and Sacerdote 1999). However this is a somewhat technical matter which falls outside of the scope of the present paper. As such we have chosen the parameters of the diffusion in such a way that the scheme suffices for visual comparison. Using GQD.TIpassage() we can again analyse the first passage time problem by defining the model in terms of GQD-coefficients: GQD.remove() ## [1] "Removed : G0 G1 Q0" # Redefine the coefficients with a parameter theta: G1 <- function(t){theta[1]*(10+0.2*sin(2*pi*t)+0.3*prod(sqrt(t),1+cos(3*pi*t)))} G2 <- function(t){-theta[1]} Q2 <- function(t){0.1} # Now just give a value for the parameter in the standard fashion: res3=GQD.TIpassage(8,12,1,4,1/100,theta=c(0.5)) ## Compiling C++ code. Please wait. Note that we have parametrised the coefficients using the reserved variable theta in similar fashion to BiGQD.mcmc(). This allows one to calculate the first passage time density for various parameter values without having to re-compile C++ code repeatedly. For example, we can evaluate the first passage time problem for the quadratic diffusion for $$\theta_1$$ running from $$0.1$$ to $$0.5$$: # Load simulated first passage times: fpt.sim.times' data("SDEsim6") hist(fpt.sim.times, freq = F, col = 'gray85', border = 'white', main = 'First Passage Time Density', ylab = 'Density', xlab = 'Time', ylim = range(res3$density), xlim = range(res3$time), breaks = 100) library("colorspace") colpal = function(n){rev(sequential_hcl(n, power = 1, l = c(20, 60)))} th.seq = seq(0.1, 0.5, 0.05) for(i in 2:length(th.seq)) { res3 = GQD.TIpassage(8, 12, 1, 4, 0.01, theta = c(th.seq[i])) lines(res3$density ~ res3$time, type = 'l', col = colpal(10)[i], lty = 11 - i, lwd =1.5) } ## Compiling C++ code. Please wait. lines(res3$density ~ res3$time, type = 'l', col = colpal(10)[i], lwd = 2) legend('topright', legend = th.seq, col = colpal(10), lty = 9:1, lwd = c(rep(1.5, 8), 2), title = expression(theta[1]), bty = 'n') The resulting figure illustrates the effect of varying $$\theta_1$$: As the value of the parameter decreases the time taken to reach and exceed the barrier increases and the effect of the time dependent terms become less prominent. This makes sense since $$\theta_1$$ in some sense dictates the speed’ at which the process drifts toward the equilibrium line $$10+0.2\sin(2\pi t)+0.3\sqrt{t}(1+\cos(3\pi t)$$. Since this line starts out above the starting point $$X_1 =8$$, $$\theta_1$$ will dictate how intensely the drift of the process pulls it towards the barrier. Finally, comparing the approximate first passage time density to that of the simulated first passage time, it can be seen that the approximation is indeed valid. browseVignettes('DiffusionRgqd')

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If x and y are integers, is x > y? : GMAT Data Sufficiency (DS) Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 22 Jan 2017, 00:51 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar If x and y are integers, is x > y? Author Message TAGS: Hide Tags Manager Joined: 02 Dec 2012 Posts: 178 Followers: 5 Kudos [?]: 2329 [0], given: 0 If x and y are integers, is x > y? [#permalink] Show Tags 07 Dec 2012, 07:48 48 This post was BOOKMARKED 00:00 Difficulty: 65% (hard) Question Stats: 55% (02:25) correct 45% (01:18) wrong based on 1276 sessions HideShow timer Statistics If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0 [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93390 [5] , given: 10557 Re: If x and y are integers, is x > y? [#permalink] Show Tags 07 Dec 2012, 07:55 5 KUDOS Expert's post 19 This post was BOOKMARKED If x and y are integers, is x > y? (1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient. (2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient. (1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient. _________________ Manager Joined: 23 Jan 2013 Posts: 174 Concentration: Technology, Other Schools: Haas GMAT Date: 01-14-2015 WE: Information Technology (Computer Software) Followers: 3 Kudos [?]: 44 [0], given: 41 Re: If x and y are integers, is x > y? [#permalink] Show Tags 08 Oct 2013, 08:29 Bunuel wrote: If x and y are integers, is x > y? (1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient. (2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient. (1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient. Should the answer not be E ? My explanation as followa Case 1 : -2>-5 Stmt 1 correct (-5)^(-2) < 0 (1/25) < 0 Stmt 2 correct -2 > 5 No ( Is x > y ? ) Case 2 : 5 > -2 Stmt 1 correct (5) ^ (-2 )< 0 (1/25 ) < 0 Stmt 2 correct 5 > 2 Yes ( Is x > y ? ) Manager Joined: 26 Sep 2013 Posts: 221 Concentration: Finance, Economics GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 Followers: 4 Kudos [?]: 140 [0], given: 40 Re: If x and y are integers, is x > y? [#permalink] Show Tags 08 Oct 2013, 16:12 shelrod007 wrote: Bunuel wrote: If x and y are integers, is x > y? (1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient. (2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient. (1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient. Should the answer not be E ? My explanation as followa Case 1 : -2>-5 Stmt 1 correct (-5)^(-2) < 0 (1/25) < 0 Stmt 2 correct -2 > 5 No ( Is x > y ? ) Case 2 : 5 > -2 Stmt 1 correct (5) ^ (-2 )< 0 (1/25 ) < 0 Stmt 2 correct 5 > 2 Yes ( Is x > y ? ) (-5)^(-2)=$$\frac{1}{25}$$ also 1/25>0 Remember when you square a negative number you get a positive number. (5)^-2=$$\frac{1}{25}$$, and again $$\frac{1}{25}$$>0 not less when you have X^-Y, it's written out as $$\frac{1}{(X^Y)}$$ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13494 Followers: 576 Kudos [?]: 163 [0], given: 0 Re: If x and y are integers, is x > y? [#permalink] Show Tags 26 Oct 2014, 12:04 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Joined: 30 Sep 2009 Posts: 124 Followers: 0 Kudos [?]: 27 [2] , given: 183 Re: If x and y are integers, is x > y? [#permalink] Show Tags 26 Oct 2014, 19:57 2 KUDOS Further explaining option c from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero . so c is sufficient Intern Joined: 12 Aug 2014 Posts: 19 Followers: 0 Kudos [?]: 0 [0], given: 72 Re: If x and y are integers, is x > y? [#permalink] Show Tags 28 Nov 2014, 00:21 abhi398 wrote: Further explaining option c from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero . so c is sufficient From 2, why we have decuded that y is a -ve number? y^x= 3^-3 is also less than zero and in this case y is positive. Can you please explain? Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93390 [1] , given: 10557 Re: If x and y are integers, is x > y? [#permalink] Show Tags 28 Nov 2014, 04:22 1 KUDOS Expert's post mulhinmjavid wrote: abhi398 wrote: Further explaining option c from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero . so c is sufficient From 2, why we have decuded that y is a -ve number? y^x= 3^-3 is also less than zero and in this case y is positive. Can you please explain? $$3^{(-3)} = \frac{1}{3^3} = \frac{1}{27} > 0$$, not less than 0. _________________ Director Joined: 10 Mar 2013 Posts: 608 Location: Germany Concentration: Finance, Entrepreneurship GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Followers: 15 Kudos [?]: 267 [2] , given: 200 If x and y are integers, is x > y? [#permalink] Show Tags 20 Dec 2014, 01:09 2 KUDOS Question is x>y ? (1) x+y>0 we can not determine wether x>y or not, but this statement says that one of the vaiables must be positive to satisfy it --> Not sufficient (2) Y^x<0 the only way it to be negative is when Y is negative, but X could be also negative (-2^-3 = -1/8) --> Not Sufficient, as we don't know wether x>y (1)+(2) Statement 1 says that one of the variables must be positive + Statement 2 says that Y is in all cases negative --> if Y is negative X must be positive and is > Y _________________ When you’re up, your friends know who you are. When you’re down, you know who your friends are. 800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660 Intern Joined: 05 Aug 2014 Posts: 36 Followers: 0 Kudos [?]: 47 [0], given: 24 Re: If x and y are integers, is x > y? [#permalink] Show Tags 04 May 2015, 08:47 Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help! e-GMAT Representative Joined: 04 Jan 2015 Posts: 497 Followers: 139 Kudos [?]: 1133 [1] , given: 90 Re: If x and y are integers, is x > y? [#permalink] Show Tags 04 May 2015, 20:21 1 KUDOS Expert's post naeln wrote: Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help! Hi naeln, From statement-II, we can deduce that $$y$$ is a negative integer but we can't say if $$x$$ is a negative or a positive odd integer. Let's evaluate both the cases: Case-I: $$x$$ is positive If $$x$$ is a positive odd integer, then $$y^x < 0$$. For example, assuming $$y = -2$$ and $$x = 3$$ would give $$y^x = -2^3 = -8 < 0$$ Case-II: $$x$$ is negative When $$x$$ is negative, $$y^x$$ would become a negative fraction and not $$y$$ itself. For example: if $$y = -2$$ and $$x = -3$$, then $$x$$ & $$y$$ both are integers and $$y^x = -2^{-3} = \frac{-1}{8} < 0$$. Here $$y^x = \frac{-1}{8}$$ is a fraction and not $$y$$ itself. Hope its clear why statement-II does not give us a unique answer. Regards Harsh _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Intern Joined: 05 Aug 2014 Posts: 36 Followers: 0 Kudos [?]: 47 [0], given: 24 Re: If x and y are integers, is x > y? [#permalink] Show Tags 05 May 2015, 10:39 Thank you Harsh for the detailed explanation. I got it now Intern Joined: 04 Dec 2015 Posts: 18 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: If x and y are integers, is x > y? [#permalink] Show Tags 09 Dec 2015, 13:27 Good question! C it is. Current Student Joined: 12 Feb 2011 Posts: 104 Followers: 0 Kudos [?]: 30 [0], given: 3362 If x and y are integers, is x > y? [#permalink] Show Tags 26 Dec 2015, 20:52 Bunuel wrote: If x and y are integers, is x > y? (1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient. (2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient. (1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient. Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since $$y^0=1$$ and $$0<1$$, so if $$y^x<0$$, it should mean that $$y^x<y^0$$ and hence $$x<0$$. I know this is not correct but I am unable to understand what am I doing wrong here. Thanks again! Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93390 [1] , given: 10557 Re: If x and y are integers, is x > y? [#permalink] Show Tags 27 Dec 2015, 03:47 1 KUDOS Expert's post Dienekes wrote: Bunuel wrote: If x and y are integers, is x > y? (1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient. (2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient. (1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient. Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since $$y^0=1$$ and $$0<1$$, so if $$y^x<0$$, it should mean that $$y^x<y^0$$ and hence $$x<0$$. I know this is not correct but I am unable to understand what am I doing wrong here. Thanks again! If $$y\neq{0}$$, then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0). _________________ Current Student Joined: 12 Feb 2011 Posts: 104 Followers: 0 Kudos [?]: 30 [0], given: 3362 Re: If x and y are integers, is x > y? [#permalink] Show Tags 27 Dec 2015, 15:44 Bunuel wrote: Dienekes wrote: Bunuel wrote: If x and y are integers, is x > y? (1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient. (2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient. (1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient. Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since $$y^0=1$$ and $$0<1$$, so if $$y^x<0$$, it should mean that $$y^x<y^0$$ and hence $$x<0$$. I know this is not correct but I am unable to understand what am I doing wrong here. Thanks again! If $$y\neq{0}$$, then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0). Thanks! One follow-up question - What if the question stem said that y is a non-zero integer, would the inference $$y^x<0<y^0$$ still be incorrect? Intern Joined: 26 Dec 2014 Posts: 4 Followers: 0 Kudos [?]: 1 [0], given: 24 If x and y are integers, is x > y? [#permalink] Show Tags 28 Dec 2015, 18:31 If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0 (1) $$x+y>0$$ : For this to be true, we will have the following 3 senarios: a) $$x > 0$$ and $$y > 0$$ b) $$x > 0$$ and $$y <= 0$$, which means $$x > y$$ c) $$y > 0$$ and $$x <= 0$$, which means $$y > x$$ Therefore, this statement is Insufficient (2) $$y^x < 0$$: For this to be true, $$y$$ must be less than 0 and $$x$$ must be an odd integer. Therefore, this statement is insufficient. (1) and (2) together: From statement (2), we know $$y < 0$$, therefore the only viable scenario for this in statement (1) is option (b): $$x > 0$$ and $$y < 0$$, which means $$x > y$$ => Sufficient. [Reveal] Spoiler: C Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 2609 GPA: 3.82 Followers: 173 Kudos [?]: 1453 [0], given: 0 Re: If x and y are integers, is x > y? [#permalink] Show Tags 29 Dec 2015, 22:10 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0 There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer. Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C. For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Find a 10% off coupon code for GMAT Club members. “Receive 5 Math Questions & Solutions Daily” Unlimited Access to over 120 free video lessons - try it yourself Manager Joined: 28 Oct 2015 Posts: 57 Followers: 0 Kudos [?]: 6 [0], given: 127 Re: If x and y are integers, is x > y? [#permalink] Show Tags 14 Jun 2016, 10:00 EgmatQuantExpert MathRevolution Can you please explain if y<0 and x+y>0 and we create the following situations- x=5,y=-3 => 5+(-3)>0 but what if x=1,y=-12 =>1+(-12)<0 How can we not consider the second case above? Manager Joined: 17 Aug 2015 Posts: 93 Followers: 1 Kudos [?]: 4 [0], given: 152 Re: If x and y are integers, is x > y? [#permalink] Show Tags 19 Jun 2016, 05:09 is x> y? State1:- x+y>0 .. the sum of two numbers is greater than zero does not tell us anything about how they rank relative to each other state 2:- y^x<0 --> this implies that x must be odd and y<0 .. If y <0 then x can be -1, +1. let y=-1. x=1 => y^x = -1 this satisfies y=-1 . x=-1 => 1/-1^1 => -1 this also satisfies. so x < y and x> y can both betrue. So not sufficient can y>0 . combine the two statements y<0 and x+y>0 implies x has to be positive. Also note that since y<0 x+y>0 => x+y-y>0 => x>0. When the two opposite inequality, we can subtract, smaller from larger and get positive result. Re: If x and y are integers, is x > y?   [#permalink] 19 Jun 2016, 05:09 Go to page    1   2    Next  [ 22 posts ] Similar topics Replies Last post Similar Topics: 4 If x and y are integers and x>y, is xy>0? 2 01 Apr 2016, 23:32 3 If x > y, and x, y and x/y are all positive integers, is x/y > 10? 5 19 Feb 2016, 22:17 7 If x and y are integers, is (x−1)>y? 10 04 Feb 2014, 14:43 11 If x and y are integers and x > 0, is y > 0? 13 11 Sep 2010, 14:05 3 If x and y are positive integers and x> y, then what is the value of x 9 11 Jan 2010, 23:10 Display posts from previous: Sort by

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Question about Nintendo Professor Layton & the Curious Village Games for DS Puzzle 46 is wrong? I can only count 10 triangles on puzzle 46 but the game claims there are 12. It says there are:- 5 X small triangles on the star 5 x big triangles in the star (no there isn't! There are only 3!!!!) 1 on the roof of the house 1 on the chimney Or am I being thick? Posted by on • nortchurchsk Dec 29, 2008 I still cannot see 5 inner triangles. I am convinced there is only 10 total × • Level 1: An expert who has achieved level 1. Mayor: An expert whose answer got voted for 2 times. • Contributor If you turn your ds around a few times you will soon see that man y other triangles will appear so this tells us that in actul fact there is 12 Posted on Dec 29, 2008 • Level 1: An expert who has achieved level 1. Governor: An expert whose answer got voted for 20 times. Corporal: An expert that has over 10 points. Mayor: An expert whose answer got voted for 2 times. • Contributor I thought there were only 3 big ones in the star too, but there are actually 5, one for point of the inner pentagon. try looking again.! Posted on Dec 28, 2008 Hi there, Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two. Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US. Here's a link to this great service Good luck! Posted on Jan 02, 2017 × my-video-file.mp4 × Related Questions: How to solve a moon shaped 3d crystal puzzle 23 20 21 3 post 14 18 10 15 17 7 8 1 16 12 11 9 6 5 2 19 4 22 13 then 13 22 4 19 2 5 6 9 11 12 16 1 8 7 17 15 10 18 14 3 21 20 23. Good luck (especially telling 6 and 9 apart) Jul 07, 2012 | Puzzle Toys Solving puzzle 46 in proffessor Layton's curious village biggest star The puzzle requires you to work out that Earth is one of the five largest objects (at least from the viewer's perspective) and that the tree in the middle is one of the five points You can see a picture of the solution on this page of the most popular walkthrough http://professorlaytonwalkthrough.blogspot.com/2008/02/puzzle046.html On a side note, a lot of players had trouble with this puzzle and it was replaced in the Europeon version of the game Dec 29, 2010 | Nintendo Professor Layton & the Curious... Puzzle Master's House won't open You haven't solved all 132 puzzles The game tells you (both in the trunk and in the save slot) how many puzzles you've both found and solved Players who think they've solved all 132 puzzles usually just look in the game's puzzle list, see that the last puzzle there is 132 and mistakenly assume that they must have solved all 132 puzzles The puzzle list only shows those puzzles that have already been found, not every puzzle Unless the found/solved count shows 132/132, then the last puzzle house will not open If the game says, for example, that you've only solved 131 puzzles, then you have to very carefully look through the list to find the missing puzzle number ... if you don't find it, look again (some players have had to look through the list multiple times before spotting it) Once you have the missing puzzle number, check this walkthrough to quickly find where the puzzle is (if it's not in Granny's shack) http://professorlaytonwalkthrough.blogspot.com/ There's a puzzle index down the right side of the walkthrough you can click on On a side note, this walkthrough can also be used for more puzzle solving fun, by looking at those puzzles that are specific to the other version (US/UK) of the game (there's over 180 puzzles between the versions) May 11, 2009 | Nintendo Professor Layton & the Curious... How to solve no 100: seven squares There are 28 PINS. Count each PIN starting from left to right on each line. (1 to 28 ) Square 1-- Join together PIN No's ( 1 -- 2 -- 6 -- 7 ) Square 2 -- Join " " " " ( 4 -- 12 -- 21 -- 27 ) Square 3 -- Join " " " " ( 5 -- 9 -- 11 -- 15 ) Square 4 -- Join " " " " (3 -- 10 -- 13 -- 19 ) Square 5 -- Join " " " " (8 -- 16 -- 23 -- 28 ) Square 6 -- Join " " " " (14 --18 - 20 -- 25 ) Square 7 -- Join " " " " (17 -- 22 - 24 - 26 ) Hope you can complete this Puzzle now ----------- GOOD LUCK Apr 01, 2009 | Nintendo Professor Layton & the Curious... Star in the sky The trick is that the Earth is the fifth point of the star. Try tapping on/around the pine tree towards the bottom of the screen, and drawing an upside-down star, you should be able to use that plus the four large stars to solve the puzzle. Hope this helps, -Nick F. Mar 04, 2009 | Nintendo Professor Layton & the Curious... Can anybody help me to solve proffesor layton puzzle 46 the biggest star Hints: 046 The Biggest Star ------------------------------- 1. The shape you're trying to make is a five-pointed star. It's easy enough to form a star by just connecting five points, but you're aiming to make the biggest start possible. Look for "the five largest objects in space" and connect them together. 2. The answer for this puzzle requires you to draw your star upside down. 3. Remember that the planet we live on is also floating in space! Solution: 046 The Biggest Star ------------------------------------------ Tree on the horizon acts as a star. Connect the 4 big stars and tree in a star shape. 6 lines total. Thanks and please take a moment to rate this solution as fixya. Feb 27, 2009 | Nintendo Professor Layton & the Curious... Open Questions: Related Topics: 1,215 people viewed this question Level 3 Expert Level 2 Expert

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# Soil advective transport calculation Transport of nanoparticles in soils is usually described using a one-dimensional convection-dispersion equation. This equation equates the change in concentration at a certain depth to: (i) percolation, i.e. downward flow of nanoparticles due to flowing water; (ii) the dissipating effects of diffusion and dispersion; and (iii) loss of nanoparticles to soil surfaces because of attachment and straining. The sum of these processes is used within NanoFASE to model breakthrough curves obtained during the saturated soil column procedure, mainly to obtain the attachment efficiency. That fate descriptor is then used in the NanoFASE soil model. $$\Psi = (\frac{d_{50}+z}{d_{50}})^{-\beta }$$ 1. Straining calculation $$k_{att}=\alpha \frac{3(1-\theta )v}{2d_{c}}\eta _{0}$$ 2. Calculation of the attachment rate coefficient for the whole soil profile using the soil - nanoparticle specific attachment efficiency. $$\theta \frac{dC}{dt}=v\theta \frac{dC}{dz}-\theta (\Sigma D)\frac{d^{2}C}{(dz)^{2}}-\Psi k_{straining}-k_{att}$$ 3. Solving the convection dispersion equation numerically. ## Procedure Straining (step 1) is not always included in the convection dispersion equation shown in step 3, which affects the value of the attachment efficiency one obtains via step 2. The equation of step 3 is not solved analytically, even when applying it to breakthrough curves of the saturated column test. Saturated column test Consult the NanoFASE Library to see abstracts of these deliverable reports: NanoFASE Report D7.2 Soil property - NM fate relationships NanoFASE Report D7.4 Module for NM exposure prediction in soils to couple to overall framework Cornelis, G.; Pang, L.; Doolette, C.; Kirby, J.K.; McLaughlin, M.J., 2013. Transport of silver nanoparticles in saturated columns of natural soils. Sci. Tot. Environ. 463-464. 120-130. ## Contact Geert Cornelis Swedish University of Agricultural Sciences (SLU) Email: geert.cornelis@slu.se

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The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A005130 Robbins numbers: a(n) = Product_{k=0..n-1} (3k+1)!/(n+k)!; also the number of descending plane partitions whose parts do not exceed n; also the number of n X n alternating sign matrices (ASM's). (Formerly M1808) 48 1, 1, 2, 7, 42, 429, 7436, 218348, 10850216, 911835460, 129534272700, 31095744852375, 12611311859677500, 8639383518297652500, 9995541355448167482000, 19529076234661277104897200, 64427185703425689356896743840, 358869201916137601447486156417296 (list; graph; refs; listen; history; text; internal format) OFFSET 0,3 COMMENTS Also known as the Andrews-Mills-Robbins-Rumsey numbers. - N. J. A. Sloane, May 24 2013 An alternating sign matrix is a matrix of 0's, 1's and -1's such that (a) the sum of each row and column is 1; (b) the nonzero entries in each row and column alternate in sign. a(n) is odd iff n is a Jacobsthal number (A001045) [Frey and Sellers, 2000]. - Gary W. Adamson, May 27 2009 REFERENCES Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 71, 557, 573. D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; A_n on page 4, D_r on page 197. C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, Chapter 75, pp. 385-386. C. A. Pickover, Wonders of Numbers, "Princeton Numbers", Chapter 83, Oxford Univ. Press NY 2001. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS T. D. Noe, Table of n, a(n) for n = 0..100 T. Amdeberhan and V. H. Moll, Arithmetic properties of plane partitions, El. J. Comb. 18 (2) (2011) # P1. G. E. Andrews, Plane partitions (III): the Weak Macdonald Conjecture, Invent. Math., 53 (1979), 193-225. (See Theorem 10.) Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5. Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021. M. T. Batchelor, J. de Gier and B. Nienhuis, The quantum symmetric XXZ chain at Delta=-1/2, alternating sign matrices and plane partitions, arXiv:cond-mat/0101385 [cond-mat.stat-mech], 2001. Andrew Beveridge, Ian Calaway, and Kristin Heysse, de Finetti Lattices and Magog Triangles, arXiv:1912.12319 [math.CO], 2019. E. Beyerstedt, V. H. Moll, and X. Sun, The p-adic Valuation of the ASM Numbers, J. Int. Seq. 14 (2011) # 11.8.7. Sara C. Billey, Brendon Rhoades, and Vasu Tewari, Boolean product polynomials, Schur positivity, and Chern plethysm, arXiv:1902.11165 [math.CO], 2019. D. M. Bressoud and J. Propp, How the alternating sign matrix conjecture was solved, Notices Amer. Math. Soc., 46 (No. 6, 1999), 637-646. H. Cheballah, S. Giraudo, and R. Maurice, Combinatorial Hopf algebra structure on packed square matrices, arXiv preprint arXiv:1306.6605 [math.CO], 2013-2015. M. Ciucu, The equivalence between enumerating cyclically symmetric, self-complementary and totally symmetric, self-complementary plane partitions, J. Combin. Theory Ser. A 86 (1999), 382-389. F. Colomo and A. G. Pronko, On the refined 3-enumeration of alternating sign matrices, arXiv:math-ph/0404045, 2004; Advances in Applied Mathematics 34 (2005) 798. F. Colomo and A. G. Pronko, Square ice, alternating sign matrices and classical orthogonal polynomials, arXiv:math-ph/0411076, 2004; JSTAT (2005) P01005. G. Conant, Magmas and Magog Triangles, 2014. J. de Gier, Loops, matchings and alternating-sign matrices, arXiv:math/0211285 [math.CO], 2002-2003. P. Di Francesco, A refined Razumov-Stroganov conjecture II, arXiv:cond-mat/0409576 [cond-mat.stat-mech], 2004. P. Di Francesco, Twenty Vertex model and domino tilings of the Aztec triangle, arXiv:2102.02920 [math.CO], 2021. Mentions this sequence. P. Di Francesco, P. Zinn-Justin and J.-B. Zuber, Determinant formulas for some tiling problems..., arXiv:math-ph/0410002, 2004. FindStat - Combinatorial Statistic Finder, Alternating sign matrices I. Fischer, The number of monotone triangles with prescribed bottom row, arXiv:math/0501102 [math.CO], 2005. Ilse Fischer and Manjil P. Saikia, Refined Enumeration of Symmetry Classes of Alternating Sign Matrices, arXiv:1906.07723 [math.CO], 2019. Ilse Fischer and Matjaz Konvalinka, A bijective proof of the ASM theorem, Part I: the operator formula, arXiv:1910.04198 [math.CO], 2019. T. Fonseca and F. Balogh, The higher spin generalization of the 6-vertex model with domain wall boundary conditions and Macdonald polynomials,  Journal of Algebraic Combinatorics, 2014, arXiv:1210.4527 D. D. Frey and J. A. Sellers, Jacobsthal Numbers and Alternating Sign Matrices, Journal of Integer Sequences Vol. 3 (2000) #00.2.3. D. D. Frey and J. A. Sellers, Prime Power Divisors of the Number of n X n Alternating Sign Matrices Markus Fulmek, A statistics-respecting bijection between permutation matrices and descending plane partitions without special parts, Electronic journal of combinatorics, 27(1) (2020), #P1.391. M. Gardner, Letter to N. J. A. Sloane, Jun 20 1991. C. Heuberger and H. Prodinger, A precise description of the p-adic valuation of the number of alternating sign matrices, Intl. J. Numb. Th. 7 (1) (2011) 57-69. Dylan Heuer, Chelsey Morrow, Ben Noteboom, Sara Solhjem, Jessica Striker, and Corey Vorland. "Chained permutations and alternating sign matrices - Inspired by three-person chess." Discrete Mathematics 340, no. 12 (2017): 2732-2752. Also arXiv:1611.03387. Hassan Isanloo, The volume and Ehrhart polynomial of the alternating sign matrix polytope, Cardiff University (Wales, UK 2019). Masato Kobayashi, Weighted counting of inversions on alternating sign matrices, arXiv:1904.02265 [math.CO], 2019. G. Kuperberg, Another proof of the alternating-sign matrix conjecture, arXiv:math/9712207 [math.CO], 1997; Internat. Math. Res. Notices, No. 3, (1996), 139-150. G. Kuperberg, Symmetry classes of alternating-sign matrices under one roof, arXiv:math/0008184 [math.CO], 2000-2001; Ann. Math. 156 (3) (2002) 835-866 W. H. Mills, David P Robbins, and Howard Rumsey Jr., Alternating sign matrices and descending plane partitions J. Combin. Theory Ser. A 34 (1983), no. 3, 340--359. MR0700040 (85b:05013). Igor Pak, Complexity problems in enumerative combinatorics, arXiv:1803.06636 [math.CO], 2018. C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, Chapter 75, pp. 385-386. [Annotated scanned copy] J. Propp, The many faces of alternating-sign matrices, Discrete Mathematics and Theoretical Computer Science Proceedings AA (DM-CCG), 2001, 43-58. A. V. Razumov and Yu. G. Stroganov, Spin chains and combinatorics, arXiv:cond-mat/0012141 [cond-mat.stat-mech], 2000. Lukas Riegler, Simple enumeration formulas related to Alternating Sign Monotone Triangles and standard Young tableaux, Dissertation, Universitat Wien, 2014. D. P. Robbins, The story of 1, 2, 7, 42, 429, 7436, ..., Math. Intellig., 13 (No. 2, 1991), 12-19. D. P. Robbins, Symmetry classes of alternating sign matrices, arXiv:math/0008045 [math.CO], 2000. R. P. Stanley, A baker's dozen of conjectures concerning plane partitions, pp. 285-293 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986. R. P. Stanley, A baker's dozen of conjectures concerning plane partitions, pp. 285-293 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986. Preprint. [Annotated scanned copy] Yu. G. Stroganov, 3-enumerated alternating sign matrices, arXiv:math-ph/0304004, 2003. X. Sun and V. H. Moll, The p-adic Valuations of Sequences Counting Alternating Sign Matrices, JIS 12 (2009) 09.3.8. Eric Weisstein's World of Mathematics, Alternating Sign Matrix Eric Weisstein's World of Mathematics, Descending Plane Partition D. Zeilberger, Proof of the alternating-sign matrix conjecture, arXiv:math/9407211 [math.CO], 1994. D. Zeilberger, Proof of the alternating-sign matrix conjecture, Elec. J. Combin., Vol. 3 (Number 2) (1996), #R13. D. Zeilberger, Proof of the Refined Alternating Sign Matrix Conjecture, arXiv:math/9606224 [math.CO], 1996. D. Zeilberger, A constant term identity featuring the ubiquitous (and mysterious) Andrews-Mills-Robbins-Ramsey numbers 1,2,7,42,429,..., J. Combin. Theory, A 66 (1994), 17-27. The link is to a comment on Doron Zeilberger's home page. A backup copy is here [pdf file only, no active links] D. Zeilberger, Dave Robbins's Art of Guessing, Adv. in Appl. Math. 34 (2005), 939-954. The link is to a version on Doron Zeilberger's home page. A backup copy is here [pdf file only, no active links] FORMULA a(n) = Product_{k=0..n-1} (3k+1)!/(n+k)!. The Hankel transform of A025748 is a(n) * 3^binomial(n, 2). - Michael Somos, Aug 30 2003 a(n) = sqrt(A049503). From Bill Gosper, Mar 11 2014: (Start) A "Stirling's formula" for this sequence is a(n) ~ 3^(5/36+(3/2)*n^2)/(2^(1/4+2*n^2)*n^(5/36))*(exp(zeta'(-1))*gamma(2/3)^2/Pi)^(1/3). which gives results which are very close to the true values: 1.0063254118710128, 2.003523267231662, 7.0056223910285915, 42.01915917750558, 429.12582410098327, 7437.518404899576, 218380.8077275304, 1.085146545456063*^7, 9.119184824937415*^8 (End) a(n+1) = a(n) * n! * (3*n+1)! / ((2*n)! * (2*n+1)!). - Reinhard Zumkeller, Sep 30 2014; corrected by Eric W. Weisstein, Nov 08 2016 For n>0, a(n) = 3^(n - 1/3) * BarnesG(n+1) * BarnesG(3*n)^(1/3) * Gamma(n)^(1/3) * Gamma(n + 1/3)^(2/3) / (BarnesG(2*n+1) * Gamma(1/3)^(2/3)). - Vaclav Kotesovec, Mar 04 2021 EXAMPLE G.f. = 1 + x + 2*x^2 + 7*x^3 + 42*x^4 + 429*x^5 + 7436*x^6 + 218348*x^7 + ... MAPLE A005130 := proc(n) local k; mul((3*k+1)!/(n+k)!, k=0..n-1); end; # Bill Gosper's approximation (for n>0): a_prox := n -> (2^(5/12-2*n^2)*3^(-7/36+1/2*(3*n^2))*exp(1/3*Zeta(1, -1))*Pi^(1/3)) /(n^(5/36)*GAMMA(1/3)^(2/3)); # Peter Luschny, Aug 14 2014 MATHEMATICA f[n_] := Product[(3k + 1)!/(n + k)!, {k, 0, n - 1}]; Table[ f[n], {n, 0, 17}] (* Robert G. Wilson v, Jul 15 2004 *) a[ n_] := If[ n < 0, 0, Product[(3 k + 1)! / (n + k)!, {k, 0, n - 1}]]; (* Michael Somos, May 06 2015 *) PROG (PARI) {a(n) = if( n<0, 0, prod(k=0, n-1, (3*k + 1)! / (n + k)!))}; /* Michael Somos, Aug 30 2003 */ (PARI) {a(n) = my(A); if( n<0, 0, A = Vec( (1 - (1 - 9*x + O(x^(2*n)))^(1/3)) / (3*x)); matdet( matrix(n, n, i, j, A[i+j-1])) / 3^binomial(n, 2))}; /* Michael Somos, Aug 30 2003 */ (GAP) a:=List([0..18], n->Product([0..n-1], k->Factorial(3*k+1)/Factorial(n+k)));; Print(a); # Muniru A Asiru, Jan 02 2019 (Python) from math import prod, factorial def A005130(n): return prod(factorial(3*k+1) for k in range(n))//prod(factorial(n+k) for k in range(n)) # Chai Wah Wu, Feb 02 2022 CROSSREFS Cf. A006366, A048601, also A003827, A005156, A005158, A005160-A005164, A050204, A049503, A194827, A227833. Sequence in context: A011802 A007065 A352258 * A091669 A108042 A152559 Adjacent sequences:  A005127 A005128 A005129 * A005131 A005132 A005133 KEYWORD nonn,easy,nice,core AUTHOR STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified May 17 05:58 EDT 2022. Contains 353730 sequences. (Running on oeis4.)

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# Unveiling the Mystery: How Many Ounces in 2/3 Cup Revealed! This page for Unveiling the Mystery: How Many Ounces in 2/3 Cup Revealed!. Discover the answer to the culinary mystery: how many ounces are in 2/3 cup? Unveiling the secret behind this common measurement dilemma. Are you tired of the kitchen confusion that strikes whenever a recipe calls for 2/3 cup? We've all been there, scratching our heads, wondering just how many ounces that elusive 2/3 cup contains. Well, fret no more! In this article, we're about to lift the veil on this culinary mystery once and for all. Get ready to embark on a journey of measurement enlightenment as we reveal the answer to the age-old question: how many ounces is 2/3 of a cup? Prepare to be amazed, because what you're about to discover will forever change the way you approach your kitchen adventures. So grab your apron and let's dive right in! The article will answer the question "how many ounces is 2/3 of a cup?" It will provide a clear explanation of the conversion between cups and ounces, unveiling the mystery behind this common measurement. ## Unveiling the Mystery: How Many Ounces in 2/3 Cup Revealed! Have you ever found yourself in a recipe, wondering how many ounces are in 2/3 cup? Conversions can sometimes be a daunting task, but fear not! In this article, we will unravel the mystery of how many ounces are in 2/3 cup and provide you with valuable insights and information. ## The Conversion Math: Understanding How 2/3 Cup Translates to Ounces To understand the conversion from cups to ounces, we need to have a clear understanding of the basic conversion factor. In the United States, the standard measurement for liquid ingredients is in fluid ounces (oz). One cup is equal to 8 fluid ounces. Now, let's break down how we can convert 2/3 cup to ounces. Since we know that one cup is equal to 8 fluid ounces, we can divide 8 by 3 (the denominator of 2/3) to get the value of one-third of a cup, which is approximately 2.667 fluid ounces. Next, we multiply the value of one-third of a cup by 2 (the numerator of 2/3) to get the total number of ounces in 2/3 cup. This calculation gives us the result of approximately 5.333 fluid ounces. Therefore, we can confidently say that 2/3 cup is equal to approximately 5.333 fluid ounces. ## Demystifying Measurement Equivalents: Unraveling the Mystery of 2/3 Cup in Ounces Knowing how to convert 2/3 cup to ounces opens up a world of possibilities in the culinary realm. Many recipes call for specific measurements, and understanding the equivalents can help you achieve the desired results. Consider this scenario: you are baking a cake and the recipe calls for 2/3 cup of milk. By knowing that 2/3 cup is equal to approximately 5.333 fluid ounces, you can accurately measure the amount of milk required and ensure the proper consistency in your batter. Similarly, when a recipe calls for 2/3 cup of a liquid ingredient like water or broth, you can confidently measure out the correct amount in ounces and bring out the intended flavors in your dish. ## Exploring the Culinary World: Practical Applications of Knowing Ounces in 2/3 Cup The knowledge of ounces in 2/3 cup extends beyond measuring liquids. The conversion can also be applied to dry ingredients, such as flour or sugar. Let's say you are making cookies and the recipe calls for 2/3 cup of sugar. By converting 2/3 cup to approximately 5.333 fluid ounces, you can precisely measure out the required amount of sugar, ensuring the perfect balance of sweetness in your baked goods. In savory dishes, knowing the ounces in 2/3 cup can help you accurately measure spices or seasonings. This precise measurement ensures that the flavors in your dish are well-balanced and not overpowering. ## Pro Tips for Accurate Measurements: Mastering the Art of Converting 2/3 Cup to Ounces While knowing the conversion of 2/3 cup to ounces is essential, it is equally important to develop good measurement techniques. Here are some pro tips to help you achieve accuracy: • Use a clear measuring cup with easy-to-read markings. • Place the measuring cup on a flat surface to ensure a level measurement. • Pour the liquid or spoon the dry ingredient into the measuring cup, filling it up to the appropriate mark. • Double-check your measurement to avoid any errors. Following these tips will not only help you measure 2/3 cup accurately but also contribute to the overall success of your culinary creations. ## Common Kitchen Conversions: Simplifying the Quandary of 2/3 Cup in Ounces Understanding the conversion from 2/3 cup to ounces is just one piece of the puzzle. Familiarizing yourself with other common kitchen conversions can simplify your cooking and baking adventures. Here are a few examples: Measurement Equivalent in Ounces 1/4 cup 2 fluid ounces 1/2 cup 4 fluid ounces 1 cup 8 fluid ounces 1 tablespoon 0.5 fluid ounces 1 teaspoon 0.17 fluid ounces These conversions can come in handy when following recipes or adjusting ingredient quantities to suit your taste preferences. So, the next time you come across a recipe that calls for 2/3 cup, you can confidently convert it to approximately 5.333 fluid ounces. Knowing this conversion will not only help you achieve accurate measurements but also elevate your culinary skills. Keep in mind that measurements can vary based on regional preferences and specific recipes. It's always a good idea to consult the recipe or use a reliable conversion chart if you are unsure. Now armed with the knowledge of how many ounces are in 2/3 cup, go forth and conquer the culinary world with confidence! ## Some question and answer of how many ounces is 2 3 of a cup ### Q: How many ounces is 2/3 of a cup? A: 2/3 of a cup is equal to approximately 5.3 ounces. ## Conclusion In conclusion, understanding measurements in cooking can be a perplexing task, but as we unraveled the mystery of how many ounces are in 2/3 cup, we discovered the simplicity behind it. By breaking down the calculations and using the conversion factor of 8 fluid ounces in 1 cup, we determined that 2/3 cup is equivalent to approximately 5.33 fluid ounces. Armed with this knowledge, you are now empowered to confidently navigate your kitchen and experiment with new recipes, knowing exactly how to measure your ingredients accurately. So, don't let measurement conversions intimidate you any longer! Embrace the joy of cooking, and remember that practice makes perfect. With each culinary adventure, you are one step closer to becoming a master chef. Happy cooking! Thank you so much for reading Unveiling the Mystery: How Many Ounces in 2/3 Cup Revealed! article. Please let us know how you feel after reading this article.

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# Conceptual Understanding in Introductory Physics XXXIV: Parallel and Perpendicular Components of Force Draw an arrow representing an arbitrary force vector and another arrow representing an arbitrary momentum arrow. Label both arrows. Ask the student to perform the following task: Decompose the force vector into a component parallel to the momentum vector and a component perpendicular to the force vector. Tell whether or not the given force will […] Read More Conceptual Understanding in Introductory Physics XXXIV: Parallel and Perpendicular Components of Force # Matter & Interactions I, Week 11 As (almost) usual, I’m writing this on the Monday after the week in question. This week we hit chapter 5, which is packed full of interesting physics and mathematics! We encounter the infamous time derivative of a unit vector (aka a direction), which I have found quite mysterious because of the rather hand waving ways […] Read More Matter & Interactions I, Week 11 # Conceptual Understanding in Introductory Physics XXV: A Free Particle? This question emphasizes geometry and should be done without use of a coordinate system. It should also be done using only symbolic manipulation of vectors. Here it is. Consider a particle moving with a constant, non-relativistic velocity. Starting with a general expression for kinetic energy in terms of either velocity or momentum, prove that the […] Read More Conceptual Understanding in Introductory Physics XXV: A Free Particle? # Conceptual Understanding in Introductory Physics XX: The Laplace-Runge-Lenz Vector This post is inspired by the October 2015 AstroNotes in The Physics Teacher. I have sometimes introduced vectors into my introductory astronomy course and students were able to do most of the things described below. We never discussed angular momentum or the Laplace-Runge-Lenz vector, but the other quantities were familiar. I was not permitted in […] Read More Conceptual Understanding in Introductory Physics XX: The Laplace-Runge-Lenz Vector # Introducing Noether’s Theorem in Conceptual Physics With little advanced notice, I was assigned a section of conceptual physics to teach this semester (because second semester astronomy was cancelled due to low enrollment and replaced with a section of conceptual physics with the same low enrollment..go figure). After teaching Matter & Interactions for over a decade and after much enlightenment I have […] Read More Introducing Noether’s Theorem in Conceptual Physics # Conceptual Understanding in Introductory Physics XIII: Differential Equations This question may very well be beyond the scope of a traditional introductory calculus-based physics course, but given the recent trend in early introduction to computational physics with curricula like Matter & Interactions it may be within the scope of a reformed course. In classical physics, finding a particle’s trajectory under the influence of a […] Read More Conceptual Understanding in Introductory Physics XIII: Differential Equations # Conceptual Understanding in Introductory Physics X: Resolving a Vector For this post, I decided to ask what I think is a very simple question. It is simple at first, but it also gets to the heart of the meaning of vector quantities, at least as they are typically presented in introductory physics. It also emphasizes the fact that vector quantities have an existence all their own, […] Read More Conceptual Understanding in Introductory Physics X: Resolving a Vector # Conceptual Understanding in Introductory Physics III: Interpreting Physical Quantities This series continues with a question which, I hope, causes readers and students to reflect on something that is frequently omitted from traditional introductory physics courses. I contend that words are all we have to convey conceptual understanding in physics or any other topic. Yet, in science courses, and especially in physics courses, we tend […] Read More Conceptual Understanding in Introductory Physics III: Interpreting Physical Quantities

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The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A196196 G.f.: A(x) = Sum_{n>=0} x^n*(A(n*x) + A(-n*x))/2. 2 1, 1, 1, 2, 5, 15, 97, 528, 7525, 81413, 2330257, 53229494, 2883887781, 143434131379, 14268520503521, 1569684574492244, 282300076226827269, 69232924049740233209, 22337274702237239579729, 12261557957001826246975754, 7069170473480519272781373829 (list; graph; refs; listen; history; text; internal format) OFFSET 0,4 LINKS Table of n, a(n) for n=0..20. EXAMPLE G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 15*x^5 + 97*x^6 + 528*x^7 +... where A(x) = 1 + x*(A(x)+A(-x))/2 + x^2*(A(2*x)+A(-2*x))/2 + x^3*(A(3*x)+A(-3*x))/2 + x^4*(A(4*x)+A(-4*x))/2 +... Related expansions begin: (A(x)+A(-x))/2 = 1 + x^2 + 5*x^4 + 97*x^6 + 7525*x^8 +... (A(2*x)+A(-2*x))/2 = 1 + 4*x^2 + 80*x^4 + 6208*x^6 +... (A(3*x)+A(-3*x))/2 = 1 + 9*x^2 + 405*x^4 + 70713*x^6 +... (A(4*x)+A(-4*x))/2 = 1 + 16*x^2 + 1280*x^4 + 397312*x^6 +... (A(5*x)+A(-5*x))/2 = 1 + 25*x^2 + 3125*x^4 + 1515625*x^6 +... (A(6*x)+A(-6*x))/2 = 1 + 36*x^2 + 6480*x^4 + 4525632*x^6 +... PROG (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(k=1, n, A=1+sum(j=1, n, x^j*(subst(A, x, j*x)+subst(A, x, -j*x))/2)); polcoeff(A, n)} CROSSREFS Cf. A196195, A210525. Sequence in context: A090140 A032267 A191477 * A266573 A268121 A306794 Adjacent sequences: A196193 A196194 A196195 * A196197 A196198 A196199 KEYWORD nonn AUTHOR Paul D. Hanna, Sep 29 2011 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified July 22 16:45 EDT 2024. Contains 374540 sequences. (Running on oeis4.)

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# Re: [R] How to do aggregate operations with non-scalar functions From: Gabor Grothendieck <ggrothendieck_at_gmail.com> Date: Wed 06 Apr 2005 - 12:15:19 EST On Apr 5, 2005 6:59 PM, Itay Furman <itayf@u.washington.edu> wrote: > > Hi, > > I have a data set, the structure of which is something like this: > > > a <- rep(c("a", "b"), c(6,6)) > > x <- rep(c("x", "y", "z"), c(4,4,4)) > > df <- data.frame(a=a, x=x, r=rnorm(12)) > > The true data set has >1 million rows. The factors "a" and "x" > have about 70 levels each; combined together they subset 'df' > into ~900 data frames. > For each such subset I'd like to compute various statistics > including quantiles, but I can't find an efficient way of > doing this. Aggregate() gives me the desired structure - > namely, one row per subset - but I can use it only to compute > a single quantile. > > > aggregate(df[,"r"], list(a=a, x=x), quantile, probs=0.25) > a x x > 1 a x 0.1693188 > 2 a y 0.1566322 > 3 b y -0.2677410 > 4 b z -0.6505710 > > With by() I could compute several quantiles per subset at > each shot, but the structure of the output is not > convenient for further analysis and visualization. > > > by(df[,"r"], list(a=a, x=x), quantile, probs=c(0, 0.25)) > a: a > x: x > 0% 25% > -0.7727268 0.1693188 > ---------------------------------------------------------- > a: b > x: x > NULL > ---------------------------------------------------------- > > [snip] > > I would like to end up with a data frame like this: > > a x 0% 25% > 1 a x -0.7727268 0.1693188 > 2 a y -0.3410671 0.1566322 > 3 b y -0.2914710 -0.2677410 > 4 b z -0.8502875 -0.6505710 > > I checked sweep() and apply() and didn't see how to harness > them for that purpose. > > So, is there a simple way to convert the object returned > by by() into a data.frame? > Or, is there a better way to go with this? > Finally, if I should roll my own coercion function: any tips? > where f is the appropriate function. In this case f can be described in terms of df.quantile which is like quantile except it returns a one row data frame: ``` df.quantile <- function(x,p) as.data.frame(t(data.matrix(quantile(x, p)))) f <- function(df, p = c(0.25, 0.5)) cbind(df[1,1:2], df.quantile(df[,"r"], p)) ______________________________________________ ``` R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed Apr 06 12:22:18 2005 This archive was generated by hypermail 2.1.8 : Fri 03 Mar 2006 - 03:31:02 EST

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# getid3_lib::BigEndian2Float ## #Summery Summery ANSI/IEEE Standard 754-1985, Standard for Binary Floating Point Arithmetic ## #Syntax Syntax getid3_lib::BigEndian2Float( string \$byteword ) \$byteword (Required) (float|false) ## #Source Source ``` \$bitword = self::BigEndian2Bin(\$byteword); if (!\$bitword) { return 0; } \$signbit = \$bitword[0]; \$floatvalue = 0; \$exponentbits = 0; \$fractionbits = 0; switch (strlen(\$byteword) * 8) { case 32: \$exponentbits = 8; \$fractionbits = 23; break; case 64: \$exponentbits = 11; \$fractionbits = 52; break; case 80: // 80-bit Apple SANE format // http://www.mactech.com/articles/mactech/Vol.06/06.01/SANENormalized/ \$exponentstring = substr(\$bitword, 1, 15); \$isnormalized = intval(\$bitword[16]); \$fractionstring = substr(\$bitword, 17, 63); \$exponent = pow(2, self::Bin2Dec(\$exponentstring) - 16383); \$fraction = \$isnormalized + self::DecimalBinary2Float(\$fractionstring); \$floatvalue = \$exponent * \$fraction; if (\$signbit == '1') { \$floatvalue *= -1; } return \$floatvalue; default: return false; } \$exponentstring = substr(\$bitword, 1, \$exponentbits); \$fractionstring = substr(\$bitword, \$exponentbits + 1, \$fractionbits); \$exponent = self::Bin2Dec(\$exponentstring); \$fraction = self::Bin2Dec(\$fractionstring); if ((\$exponent == (pow(2, \$exponentbits) - 1)) && (\$fraction != 0)) { // Not a Number \$floatvalue = false; } elseif ((\$exponent == (pow(2, \$exponentbits) - 1)) && (\$fraction == 0)) { if (\$signbit == '1') { \$floatvalue = '-infinity'; } else { \$floatvalue = '+infinity'; } } elseif ((\$exponent == 0) && (\$fraction == 0)) { if (\$signbit == '1') { \$floatvalue = -0; } else { \$floatvalue = 0; } \$floatvalue = (\$signbit ? 0 : -0); } elseif ((\$exponent == 0) && (\$fraction != 0)) { // These are 'unnormalized' values \$floatvalue = pow(2, (-1 * (pow(2, \$exponentbits - 1) - 2))) * self::DecimalBinary2Float(\$fractionstring); if (\$signbit == '1') { \$floatvalue *= -1; } } elseif (\$exponent != 0) { \$floatvalue = pow(2, (\$exponent - (pow(2, \$exponentbits - 1) - 1))) * (1 + self::DecimalBinary2Float(\$fractionstring)); if (\$signbit == '1') { \$floatvalue *= -1; } } return (float) \$floatvalue; } /** * @param string \$byteword ```

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# Polynomials Identity 1: (a + b)2 = a2 + 2ab + b2 Identity 2: (a – b)2 = a2 – 2ab + b2 Identity 3: a2 – b2= (a + b)(a – b) Identity 4: (x + a)(x + b) = x2 + (a + b) x + ab Identity 5: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Identity 6: (a + b)3 = a3 + b3 + 3ab (a + b) Identity 7: (a – b)3 = a3 – b3 – 3ab (a – b) Identity 8: a3 + b3 = (a + b) (a2 – ab + b2) Identity 9: a3 -b3 = (a - b) (a2 + ab + b2) Identity 10: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

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# Showing that a greatest common divisor must be 1 or 2 using pigeonhole principle I need to prove that for any $$S \subset \{1,2,...,2018\}$$ with $$|S|=673$$, it follows that $$\exists\,a,b \in S$$ such that $$gcd(a,b)<3$$. I can see the obvious application of pigeonhole principle here, since $$673 \times 3=2019$$ and just from the wording of the problem, but unfortunately it's late and I'm having trouble constructing the proof. Any tips? • Hint: $\gcd(n,n+1)=1$ and $\gcd(n,n+2) \in \{1,2\}$ for all $n$. Nov 16, 2018 at 11:42 • This turned out to be more helpful, thanks! There are only 672 elements whose difference is greater than or equal to three in the set, which means that by the pigeonhole principle, in any set of cardinality 673 there must be two elements with a difference less than or equal to two. Apply your fact and the problem is solved. Dec 8, 2018 at 3:05 Any choice of $$k+1$$ elements from $$kd$$ consecutive positive integers where $$d\ge2$$ contains, by the pigeonhole principle, a pair of elements $$b where $$a-b\le d-1$$. This pair will have $$\gcd(a,b)=\gcd(a-b,b)\le d-1$$. The OP's question is essentially the case when $$d=3$$ and $$k=672$$; $$1$$ and $$2$$ cannot be chosen since $$\gcd(1,n)$$ and $$\gcd(2,n)$$ are less than $$3$$, but then $$673$$ elements cannot be chosen from $$3,\dots,2018$$ that are all at least $$3$$ apart.

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# TCS Placement Paper | MCQ 5 • Difficulty Level : Medium • Last Updated : 04 Nov, 2022 This is a TCS model placement paper for aptitude preparation. This placement paper will cover aptitude questions that are asked in TCS recruitment drives and also strictly follows the pattern of questions asked in TCS interviews. It is recommended to solve each one of the following questions to increase your chances of clearing the TCS interview. • Which of the following numbers must be added to 5678 to give a remainder 35 when divided by 460? a) 980 b) 618 c) 955 d) 797 Answer: d) 797 Solution: Let the number added to 5678 be x to give a remainder 35 and quotient k when divided by 460. So, 5678 + x = 460k + 35 or, 5643 + x = 460k So 5643 + x must be divisible by 460 Analyzing from the options, we get on adding 797 to 5643, the number 6440 is divisible by 460. • Rahaman went to a stationery shop and bought 18 pencils for Rs.100. He paid 1 rupee more for each grey pencil than for each black pencil. What is the price of a grey pencil and how many grey pencils did he buy? a) Rs.5, 10 b) Rs.6, 10 c) Rs.5, 8 d) Rs.6, 8 Answer: b) Rs. 6, 10 Solution: The best way is to analyse from the mentions. Let’s take option b in which 10 pencils are bought at Rs.6 each. So total cost of grey pencils = 6 * 10 = Rs.60. So Rahaman is left with 40 rupees. He buys 8 black pencils at Rs 5 each which is 1 rupee less than what he had spent in buying the grey ones. Thus satisfying the conditions. • Four people each roll a four die once. Find the probability that at least two people will roll the same number? a. 13/18 b. 5/18 c. None of the above d. 1295/1296 Answer: a) 13/18 ways Solution: Total possible outcomes = = 1296 Number of ways in which no two people get same number = 6*5*4*3 = 360 ways The probability of no two people getting the same number = 360 / 1296 ways = 5/18 ways So the probability of at least two people getting the same number = 1 – 5/18 = 13/18 ways • Ram said Shyam “If you give me half your money I will have Rs.75.” Shyam said, “If you give me one-third of your money, I will have Rs.75 How much money did Shyam have4 c) 48 d) 60 Answer: d) 60 Solution: Let Ram and Shyam be denoted by ‘R’ and ‘S’ respectively According to the question, Eqn 1. R + S/2 = 75 Eqn 2. R/3 + S = 75 Therefore, solving both the equations we get, R = 45 and S = 60. • Ram goes to the market to buy apples. If he can bargain and reduce the price per apple by Rs.2, he can buy 30 apples instead of 20 apples with the money he has. How much money does he have? a. Rs.100 b. Rs.50 c. Rs.120 d. Rs.150 Answer: c) 120 Solution: Let the price per orange be Rs. x. So total money Ram has in buying at original price = 20x. On reducing the price by 2 rupees each the total money must be (x-2)*30 According to the question, 20x = (x-2)*30 On solving this we get x = 6 or the total money = Rs. 120 • How many positive integers less than 500 can be formed using the numbers 1, 2, 3, and 5 for digits, each digit being used only once. a) 68 b) 34 c) 66 d) 52 Answer: b) 34 Solution: Single digit numbers can be formed in 4 ways. 2 digit number can be formed in 4 * 3 = 12 ways 3 digit number less than 500 can be formed in 3 * 3 * 2 = 18 ways. Total number of ways = 18 + 12 + 4 = 34 ways • A boy entered a shop and bought x number of books for y rupees. When he was about to leave the bookkeeper said, “if you buy 10 more books, you can have all the books for 2 rupees and you will also save 80 cents a dozen”. So what are x and y? a) (5, 1) b) (10, 1) c) (15, 1) d) Cannot be determined. Answer: a) (5, 1) Solution: x number of books cost him y rupees. So, 1 book will cost him y/x rupees. 12 books will cost him rupees 12 y/x. The shopkeeper says, x + 10 books cost him 12 rupees 1 book will cost him 12/(x+10) rupees 12 books will cost him 24/(x+10) rupees We know that 80 cents = 4/5 of a dollar, So, 12y/x – 24/(10+x) = 4/5 Analysing the given choices, we get (5, 1) satisfies the equation. • The perimeter of an equilateral triangle is equal to a regular hexagon. Find out the ratio of their areas? a. 3:2 b. 1:6 c. 2:3 d. 6:1 Answer: c) 2:3 Solution:Let the side of the equilateral triangle be a unit and that of the regular hexagon be b unit. So perimeter of the triangle = 3a and perimeter of the hexagon is 6b unit. or, 3a = 6b or a/b = 2/1 The area of the equilateral tr iangle = The area of the regular hexagon = or, Solving this and substituting a/b we get the answer as 2 : 3 • In the given series: 70, 54, 45, 41……. What will be the next number? a) 40 b) 36 c) 35 d) 38 Answer: a) 40 Solution:The series goes like: 70 – 54 = 16 (4^2) 54 – 45 = 9 (3^2) 45 – 41 = 4 (2^2) 41 – 40 = 1 (1^1) • A series of story books were published at an interval of seven years. When the seventh book was published the total sum of the publication year was 13524. In which year was the first book published? a) 1910 b) 1911 c) 2002 d) 1932 Answer: b) 1911 Solution:We get the series of publications as n, n+7, n+14, n+21, n+28, n+35, n+42. Sum of publications = 13524 = 7/2[2n + (7-1)*7] (Using the sum of AP formula) We get, n = 1911 (answer) My Personal Notes arrow_drop_up Related Articles

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NumWords.com # How to write Three thousand three hundred nine in numbers in English? We can write Three thousand three hundred nine equal to 3309 in numbers in English < Three thousand three hundred eight :||: Three thousand three hundred ten > Six thousand six hundred eighteen = 6618 = 3309 × 2 Nine thousand nine hundred twenty-seven = 9927 = 3309 × 3 Thirteen thousand two hundred thirty-six = 13236 = 3309 × 4 Sixteen thousand five hundred forty-five = 16545 = 3309 × 5 Nineteen thousand eight hundred fifty-four = 19854 = 3309 × 6 Twenty-three thousand one hundred sixty-three = 23163 = 3309 × 7 Twenty-six thousand four hundred seventy-two = 26472 = 3309 × 8 Twenty-nine thousand seven hundred eighty-one = 29781 = 3309 × 9 Thirty-three thousand ninety = 33090 = 3309 × 10 Thirty-six thousand three hundred ninety-nine = 36399 = 3309 × 11 Thirty-nine thousand seven hundred eight = 39708 = 3309 × 12 Forty-three thousand seventeen = 43017 = 3309 × 13 Forty-six thousand three hundred twenty-six = 46326 = 3309 × 14 Forty-nine thousand six hundred thirty-five = 49635 = 3309 × 15 Fifty-two thousand nine hundred forty-four = 52944 = 3309 × 16 Fifty-six thousand two hundred fifty-three = 56253 = 3309 × 17 Fifty-nine thousand five hundred sixty-two = 59562 = 3309 × 18 Sixty-two thousand eight hundred seventy-one = 62871 = 3309 × 19 Sixty-six thousand one hundred eighty = 66180 = 3309 × 20 Sitemap

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## Pricing floating rate notes excel So, FRN prices at par at the moment of settlement, such that FRN duration approximately (~=) time to next coupon. In the webinar, I said "round 15-year Floating-rate JGBs (CMT) ・The conventional price-competitive auction (multiple-price auction) based on the coupon rate(reference rate –α); ・The α is  15 Jul 2016 Part 3: Introduction to DataStream on Excel . How to price a new IRS deal ( Eikon Excel) . Comparison between 2 Floating Rate Bonds . So, FRN prices at par at the moment of settlement, such that FRN duration approximately (~=) time to next coupon. In the webinar, I said "round  Calculating the Price of the Floating Rate Note . It is well known that the price of a par floater should equal its notional, regardless of the assumed evolution of the future interest rates. Friday, 29 November 2019 Floating Rate Notes (FRN) in Excel: Understanding Duration, Pricing of Floating-Rate Note. Since the interest rate on a floating-rate note is reset periodically, its price is expected to stay close to the par value unless there is major deterioration in its credit quality or the bond hits the cap or floor. Theoretically, the price of a floating-rate note should equal its par value at each reset date and any time before the next reset, the price equals the present value of the next coupon payment and par value. I'm trying to price the following floating rate note: The price displayed on Bloomberg is 100.103063. Is it possible to use an Excel function to price a U.S. Treasury Floating Rate Note (FRN)? Ask Question Asked 2 years, 1 month ago. Calculate yield of maturity for a certain price in excel. 1. Its coupon rate is reset at each coupon date in line with a money market reference rate such as LIBOR or federal fund rate, plus a fixed spread. The spread is a rate or percentage point example 0.2 that remains constant. Almost all FRNs have quarterly coupons, i.e., they pay out interest every 3 months. ## 29 Nov 2019 It is well known that the duration of a floating rate note is very small and close in value to the time interval – in annual units – from today until the Its coupon rate is reset at each coupon date in line with a money market reference rate such as LIBOR or federal fund rate, plus a fixed spread. The spread is a rate or percentage point example 0.2 that remains constant. Almost all FRNs have quarterly coupons, i.e., they pay out interest every 3 months. The price of floating rate note was finally calculated. To be able to value a floating rate note, all of its future cash flows will be discounted with the yield of the instrument i.e. the sum of the discount rate and the discount margin. The discount margin takes into account the credit risk of the floating rate note. A floating rate bond pays a variable coupon to the bondholders depending on the current market interest rate. To value a floating rate bond in Excel, we Treasury Floating Rate Note Calculation Examples. Treasury published a final rule in the Federal Register on July 31, 2013 (78 FR 46426), revising the Uniform Offering Circular to accommodate the auction and issuance of floating rate notes. Treasury has provided examples of floating rate note calculations and an interest accrual schedule. Floating rate notes are priced on All-in price per R100 nominal. The following methodology is used to determine the All-in price of the FRN: Generate coupon payment date. Determine the mid swap zero rate corresponding to the coupon date. Calculate the forward rate from the calculated discount factor for each ### A floating rate note (FRN) or a floater is a bond with a coupon that is linked to a This note can be valued in Microsoft Excel using the above formula as The price practically equals zero, due to the fact that our currency swap product in cell A1 has been constructed with a domestic notional of 113.96 and a foreign notional of 100, the ratio of which exactly matches the spot fx EUR/USD rate of 1.1396 in cell E11 that is used as market input in the pricing of the contract. ### 28 Apr 2019 A floating-rate note (FRN) or a floater is a bond whose coupon rate changes with changes in market interest rates. Suppose LIBOR rises to 3 percent one month before the payment date of the floating-rate bond. This makes the prevailing annual market rate 4 percent, which is LIBOR plus the 1 percent increment. The results of this section are all estimates based on a simple valuation model with simplifying assumptions. That is, the discount margin of 1.4256%, the rate duration of -0.4215, and the credit duration of 8.9729 are statistics conditional on the model on which they are based. Hence, there is model risk. Treasury Floating Rate Note Calculation Examples. Treasury published a final rule in the Federal Register on July 31, 2013 (78 FR 46426), revising the Uniform Offering Circular to accommodate the auction and issuance of floating rate notes. Treasury has provided examples of floating rate note calculations and an interest accrual schedule. The price of floating rate note was finally calculated. To be able to value a floating rate note, all of its future cash flows will be discounted with the yield of the instrument i.e. the sum of the discount rate and the discount margin. The discount margin takes into account the credit risk of the floating rate note. A floating rate note (FRN) is a debt instrument whose coupon rate is tied to a benchmark rate such as LIBOR LIBOR LIBOR, which is an acronym of London Interbank Offer Rate, refers to the interest rate that UK banks charge other financial institutions for a short-term loan maturing from one day to 12 months in the future. The price practically equals zero, due to the fact that our currency swap product in cell A1 has been constructed with a domestic notional of 113.96 and a foreign notional of 100, the ratio of which exactly matches the spot fx EUR/USD rate of 1.1396 in cell E11 that is used as market input in the pricing of the contract. I'm trying to manually calculate the accrued interest of a U.S. Treasury floating rate note (FRN). If this formula is correct, then in order to back into the value of \$199.84 on the following screen print, we would need a rate of 1.19904%. I don't see that 1.19904% rate anywhere on the screen print below. ## So, FRN prices at par at the moment of settlement, such that FRN duration approximately (~=) time to next coupon. In the webinar, I said "round Suppose LIBOR rises to 3 percent one month before the payment date of the floating-rate bond. This makes the prevailing annual market rate 4 percent, which is LIBOR plus the 1 percent increment. The results of this section are all estimates based on a simple valuation model with simplifying assumptions. That is, the discount margin of 1.4256%, the rate duration of -0.4215, and the credit duration of 8.9729 are statistics conditional on the model on which they are based. Hence, there is model risk. Treasury Floating Rate Note Calculation Examples. Treasury published a final rule in the Federal Register on July 31, 2013 (78 FR 46426), revising the Uniform Offering Circular to accommodate the auction and issuance of floating rate notes. Treasury has provided examples of floating rate note calculations and an interest accrual schedule. The price of floating rate note was finally calculated. To be able to value a floating rate note, all of its future cash flows will be discounted with the yield of the instrument i.e. the sum of the discount rate and the discount margin. The discount margin takes into account the credit risk of the floating rate note. A floating rate note (FRN) is a debt instrument whose coupon rate is tied to a benchmark rate such as LIBOR LIBOR LIBOR, which is an acronym of London Interbank Offer Rate, refers to the interest rate that UK banks charge other financial institutions for a short-term loan maturing from one day to 12 months in the future. The price practically equals zero, due to the fact that our currency swap product in cell A1 has been constructed with a domestic notional of 113.96 and a foreign notional of 100, the ratio of which exactly matches the spot fx EUR/USD rate of 1.1396 in cell E11 that is used as market input in the pricing of the contract. • It was issued at a margin of 30 basis points (0.30%) above LIBOR. • The interest payment frequency is 3 months (or 4 times in a year) • 3 month LIBOR rate at previous interest payment was 6.98%. • FRN is currently trading at a margin of 15 basis points (0.15%) above LIBOR. CFA Level I Valuation of a Floating Rate Note by Mr. Arif Irfanullah IFT. Chapter 23 Valuing a Floating Rate Note on a Reset Date - Duration: Pricing of Interest Rate Swaps Zero discount margin - When the price an investor paid for a floating rate note equals the par value when the bond matures, no additional return is generated. The discount rate is equal to the reset margin on the note. A Teaching Note on Pricing and Valuing Interest Rate Swaps Using LIBOR and OIS Discounting. interpret the interest rate swap as a long/short combination of a bond paying the fixed rate on the swap and a floating-rate bond paying the money market reference rate, e.g., 3-3 .

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STRESS_IV - STRESS TRANSFORMATIONS AND MOHRS CIRCLE... This preview shows pages 1–7. Sign up to view the full content. 1 STRESS TRANSFORMATIONS AND MOHR’S CIRCLE MECH320-SPRING 08 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 2 OUTLINES Principal Stresses Maximum Shear Stress Mohr's Circle 3 Principal stresses Consider the traction vector on the x-face as shown. For this entire stack we will make an important limitation on our stress state, namely that it is 2-Dimensional . The traction vector shown lies in the x-y plane, and we will change the orientation of the block by rotating about the z-axis only. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 4 We know that the traction vector on a given surface depends on the orientation of the surface. We might ask if we can find an orientation such that the traction vector is parallel to the normal. Principal stresses (Cont.) 5 Principal stresses (Cont.) The normal component, σ x' is equal in magnitude to the traction vector and the shear vanishes. Such stress components are very important since they turn out to be the maximum and minimum normal components, and so we give them a special name: Principal Stresses. To calculate such stress components it is necessary to determine the proper block orientation. For this we will need the stress transformation equation for shear. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Principal stresses (Cont.) By definition, the principal stresses occur on planes for which the shear stress vanishes. Therefore, we can use this equation to solve for the θ for which τ x'y' = 0. The algebra is not difficult, and we can obtain the relation between the principal orientation angle and the basic stress components as shown. Find This is the end of the preview. Sign up to access the rest of the document. This note was uploaded on 02/07/2011 for the course MECH 320 taught by Professor D.a during the Spring '09 term at American University of Beirut. Page1 / 30 STRESS_IV - STRESS TRANSFORMATIONS AND MOHRS CIRCLE... This preview shows document pages 1 - 7. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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# Inside the FFT Black Box Serial and Parallel Fast Fourier Transform Algorithms #### Category: Uncategorized Posted on 2021-01-23, by book24h. Description Inside the FFT Black Box: Serial and Parallel Fast Fourier Transform Algorithms By Eleanor Chu, Alan George 1999 | 308 Pages | ISBN: 0849302706 | PDF | 7 MB Personally, I am satisfied with what I bought. I wrote an uninspired fast fourier transform from its mathematical formula and it took 30 seconds to execute. I knew I could do better. After buying the book I learn to play close attention to the bit reversal on the twiddles (trig functions). I also learned how to do the split-radix. I also learned that each calculation yields two terms. Also, I gained emough of a sense of how the fft works that I was able to successfully create threads and try parallel processing. All totalled, I reduced the run time from 30 seconds to 1 second.The book was not as well written as I would have liked. The formula for the split-radix was screwed up. Using the form of the formula and the suggestion of what it represented I was able to derive the formula. It would have been nice if they had written out each term of each iteration for a 64-term fft. That is what I did to see with my own eyes what was happening. The text is too abstract.All-in-all it was worth the \$100. Visit My Profile News Here : https://www.ebookee.com/user/book24h https://rapidgator.net/file/bbc662f1d2cbbfc3189f8b3a4ebae4c8/t6yoa.Inside.the.FFT.Black.Box.Serial.and.Parallel.Fast.Fourier.Transform.Algorithms.rar.html http://nitroflare.com/view/717BFA4863849D7/t6yoa.Inside.the.FFT.Black.Box.Serial.and.Parallel.Fast.Fourier.Transform.Algorithms.rar 7618 dl's @ 2785 KB/s 6347 dl's @ 3971 KB/s 7693 dl's @ 2465 KB/s Search More... Inside the FFT Black Box Serial and Parallel Fast Fourier Transform Algorithms Please check the description for download links if any or do a search to find alternative books. Related Books

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Term Rewriting System R: [x] g(h(g(x))) -> g(x) g(g(x)) -> g(h(g(x))) h(h(x)) -> h(f(h(x), x)) Termination of R to be shown. ` R` ` ↳Dependency Pair Analysis` R contains the following Dependency Pairs: G(g(x)) -> G(h(g(x))) G(g(x)) -> H(g(x)) H(h(x)) -> H(f(h(x), x)) Furthermore, R contains one SCC. ` R` ` ↳DPs` ` →DP Problem 1` ` ↳Polynomial Ordering` Dependency Pair: G(g(x)) -> G(h(g(x))) Rules: g(h(g(x))) -> g(x) g(g(x)) -> g(h(g(x))) h(h(x)) -> h(f(h(x), x)) The following dependency pair can be strictly oriented: G(g(x)) -> G(h(g(x))) Additionally, the following usable rule w.r.t. to the implicit AFS can be oriented: h(h(x)) -> h(f(h(x), x)) Used ordering: Polynomial ordering with Polynomial interpretation: POL(g(x1)) =  1 POL(G(x1)) =  x1 POL(h(x1)) =  0 POL(f(x1, x2)) =  0 resulting in one new DP problem. ` R` ` ↳DPs` ` →DP Problem 1` ` ↳Polo` ` →DP Problem 2` ` ↳Dependency Graph` Dependency Pair: Rules: g(h(g(x))) -> g(x) g(g(x)) -> g(h(g(x))) h(h(x)) -> h(f(h(x), x)) Using the Dependency Graph resulted in no new DP problems. Termination of R successfully shown. Duration: 0:00 minutes

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```/* nag_rand_gen_multinomial (g05tgc) Example Program. * * Copyright 2017 Numerical Algorithms Group. * * Mark 26.2, 2017. */ /* Pre-processor includes */ #include <stdio.h> #include <math.h> #include <nag.h> #include <nag_stdlib.h> #include <nagg05.h> #define X(I, J) x[(order == Nag_ColMajor)?(J*pdx + I):(I*pdx + J)] int main(void) { /* Integer scalar and array declarations */ Integer exit_status = 0; Integer lr, x_size, i, j, lstate, pdx; Integer *state = 0, *x = 0; /* NAG structures */ NagError fail; Nag_ModeRNG mode; /* Double scalar and array declarations */ double p_max; double *r = 0; /* Set the distribution parameters */ Integer k = 4; Integer m = 6000; double p[] = { 0.08e0, 0.1e0, 0.8e0, 0.02e0 }; /* Set the sample size */ Integer n = 20; /* Return the results in column major order */ Nag_OrderType order = Nag_ColMajor; /* Choose the base generator */ Nag_BaseRNG genid = Nag_Basic; Integer subid = 0; /* Set the seed */ Integer seed[] = { 1762543 }; Integer lseed = 1; /* Initialize the error structure */ INIT_FAIL(fail); printf("nag_rand_gen_multinomial (g05tgc) Example Program Results\n\n"); /* Get the length of the state array */ lstate = -1; nag_rand_init_repeatable(genid, subid, seed, lseed, state, &lstate, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_rand_init_repeatable (g05kfc).\n%s\n", fail.message); exit_status = 1; goto END; } pdx = (order == Nag_ColMajor) ? n : k; x_size = (order == Nag_ColMajor) ? pdx * k : pdx * n; /* Calculate the size of the reference vector */ p_max = 0.0; for (i = 1; i < k; i++) p_max = (p_max < p[i]) ? p[i] : p_max; lr = 30 + 20 * sqrt(m * p_max * (1 - p_max)); /* Allocate arrays */ if (!(r = NAG_ALLOC(lr, double)) || !(state = NAG_ALLOC(lstate, Integer)) || !(x = NAG_ALLOC(x_size, Integer))) { printf("Allocation failure\n"); exit_status = -1; goto END; } /* Initialize the generator to a repeatable sequence */ nag_rand_init_repeatable(genid, subid, seed, lseed, state, &lstate, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_rand_init_repeatable (g05kfc).\n%s\n", fail.message); exit_status = 1; goto END; } /* Generate the variates, initializing the reference vector at the same time */ mode = Nag_InitializeAndGenerate; nag_rand_gen_multinomial(order, mode, n, m, k, p, r, lr, state, x, pdx, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_rand_gen_multinomial (g05tgc).\n%s\n", fail.message); exit_status = 1; goto END; } /* Display the variates */ for (i = 0; i < n; i++) { for (j = 0; j < k; j++) printf("%12" NAG_IFMT "", X(i, j)); printf("\n"); } END: NAG_FREE(r); NAG_FREE(state); NAG_FREE(x); return exit_status; } ```

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# RE: st: Relationship of Nelson-Aalen and Kaplan-Meier From "Hugh Robinson" <[email protected]> To <[email protected]> Subject RE: st: Relationship of Nelson-Aalen and Kaplan-Meier Date Tue, 22 Apr 2008 12:34:37 -0600 ```Thanks Maarten. Unfortunately Stata will give me either a standard error for the Nelson-Aalen estimator, or a standard error for the Kaplan-Meier survival estimate. The two are not equal as they're calculated a little bit differently. KM survival variance is calculated using Greenwood's formula and the variance of the NA hazard function is the sum of (# of events/ # at risk). Kaplan Meier estimates got to zero during periods when no subjects are at risk. Because of small samples sizes I would like to use the Nelson-Aalen estimator but still need to come up with a survival estimate and standard deviation. The Nelson-Aalen hazard function can be converted to a survival estimate that approximates KM survival using Skm = exp(-Hna). I'm wondering how to get an equivalent measure of KM survival variance from the Nelson-Aalen hazard function. I've found several texts that discuss the equivalence of the survival estimates but can't find anything that discusses how the variances equate. Thanks again. HR -----Original Message----- From: [email protected] [mailto:[email protected]] On Behalf Of Maarten buis Sent: Tuesday, April 22, 2008 2:37 AM To: [email protected] Subject: Re: st: Relationship of Nelson-Aalen and Kaplan-Meier These are discussed in the methods and formulas section of the manual entry on -sts-. Anyhow, you can let Stata do the computations by typing -sts list-. Hope this helps, Maarten --- Hugh Robinson <[email protected]> wrote: > Can anyone tell me how to calculate the variance of a Kaplan-Meier > survival estimate derived from the Nelson-Aalen hazard function? > > For instance, Stata provides a Standard Error of 0.08 for a > cumulative > NA hazard of 0.22. The hazard function is easily converted to the > Kaplan-Meier equivalent as S= exp(-H). In this case the equivalent > Kaplan-Meier S = exp(-0.22)=0.80. However I am unclear as how to > calculate the equivalent variance. According to Cleves et al, Stata > uses the Greenwood method to calculate standard error for > Kaplan-Meier > estimates. A second source suggests that variance should be the same > for both and calculated as the sum of (# of events/# at risk) for > each > time step. > > > Thanks. > HR > > > Hugh S Robinson Ph.D. > Postdoctoral Researcher > Wildlife Biology Program > College of Forestry and Conservation > University of Montana > Missoula, MT 59802 > (406) 243-4128 > > > * > * For searches and help try: > * http://www.stata.com/support/faqs/res/findit.html > * http://www.stata.com/support/statalist/faq > * http://www.ats.ucla.edu/stat/stata/ > ----------------------------------------- Maarten L. Buis Department of Social Research Methodology Vrije Universiteit Amsterdam Boelelaan 1081 1081 HV Amsterdam The Netherlands Buitenveldertselaan 3 (Metropolitan), room Z434 +31 20 5986715 http://home.fsw.vu.nl/m.buis/ ----------------------------------------- __________________________________________________________ Sent from Yahoo! Mail. A Smarter Email http://uk.docs.yahoo.com/nowyoucan.html * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ ```

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Erected by : KHAIRIL ANUAR BIN MOHD RAZALI Enhanced, refined and continued by : 1 SPM MATHEMATICS - PAPER 1 (1449 / 1) A 1. 2. 3. 4. 5. GENERAL GUIDE - Paper 1 Paper 1 SPM Mathematics covers selected topics from Form 1 to 3, all topics in Form 4 and 5, and requires BASIC, INTERMEDIATE and HIGHER skills. Topics in this paper covers the NUMBERS, SHAPES & SPACE and ALGEBRAIC themes. Skills connected to the NUMBERS, SHAPES & SPACE and ALGEBRAIC themes sometimes complement each other; without any one of the skill, others CAN’T BE acquired. Questions on SHAPES require a lot of ALGEBRAIC and NUMBER skills whilst questions on ALGEBRA require skills on NUMBERS or vice versa. From the above explaination, it is clear that skills on NUMBERS should be built first whilst skills on ALGEBRA are an important tool to solve many problems. Scope of questions covers topics that have been taught from Form 1 to 5 (please refer to topics analysis in Part C to get a clear picture of topics posted) NUMBERS 1. 2. 3. 4. 5. 6. 7. Whole Numbers Fractions Decimals Percentages Directed Numbers Multiples & Factors Squares, Square Roots, Cubes, Cube Roots 8. Standard Form 9. Number Bases 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. SHAPES AND SPACES Basic Measurements Lines and Angles Polygons Perimeter and Area Geometrical Constructions Loci In Two Dimensions Circles Solid Geometry Pythagoras’ Theorem Trigonometry Bearings Angles Of Elevation & Depression Lines & Planes In 3-Dimentions Plans And Elevations Earth As A Sphere Transformations 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. ALGEBRA Indices Algebraic Expressions Algebraic Formulae Linear Equations Linear Inequalities Quadratic Exp. & Equations Coordinates The Straight Line Graphs Of Functions Gradient & Area Under A Graph Nisbah dan Kadar Variations Matrices Sets Mathematical Reasoning Statistics Probability B NO 1 2 3 4 5 6 7 8 9 EXAMINATION FORMAT – Paper 1 ITEM Type Of Instrument Type Of Item Number Of Question Total Marks Test Duration Constructual Inclination Contextual Coverage Level of Difficulty Additional Tools NOTES / DISCRIPTION Objective Test Multiple Choice 40 questions (Answer all) 40 1 hour 15 minutes Knowledge - 45 % / Skill - 55 % 1. Lower secondaries field of studies that have continuity at higher secondary. 2. All field of studies from form 4 to 5. Easy : Moderate : Difficult = 5 : 3 : 2 1. Scientific Calculators 2. Mathematical Tables Book 3. Geometrical Equipment 2 C ANALYSIS – Paper 1 2003 2004 2005 2006 2007 2008 2009 2010 2011 TOPICS FORM 1 – 3 1. 2. 3. 4. 5. 6. 7. 8. 9. Polygons I and II Algebraic Expressions Linear Equations Algebraic Formulae Statistics I and II Transformations I and II Indices Linear Inequalities Trigonometry I TOTAL 2 2 1 1 2 3 2 2 15 1 2 1 1 2 2 2 2 1 14 2 2 1 1 1 2 2 1 12 1 2 1 1 3 2 1 1 1 13 2 2 1 1 2 2 2 1 13 1 2 1 1 2 2 2 1 12 2 2 1 1 3 2 2 2 15 FORM 4 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Standard Form Quadratic Expr. & Equations Sets Mathematical Reasoning The Straight Line Statistics III Probability I Circles III Trigonometry II Angles of Elevation & Depress. Lines & Planes in 3-Dimension TOTAL 4 3 1 2 1 2 1 1 15 4 3 2 3 1 2 2 1 18 3 3 2 1 2 1 3 1 1 17 4 3 2 2 1 2 2 1 17 3 3 2 2 1 3 1 1 16 4 3 2 2 1 3 2 1 18 4 2 2 2 1 2 2 1 16 FORM 5 1. 2. 3. 4. 5. 6 7. 8. 9. 10. Number Bases Graphs of Functions II Transformations III Matrices Variations Gradient/Area Under Graphs Probability II Bearings Earth As A Sphere Plans And Elevations TOTAL 2 1 3 2 1 1 10 2 1 1 2 1 1 8 2 1 2 3 1 2 11 2 1 2 3 1 1 10 2 1 2 3 1 2 11 2 1 2 3 1 1 10 2 1 2 2 1 1 9 ☛ Bold printed topics are topics that are either not included in Paper 1 or topics that are seldom asked. ☛ Majority of questions are from Upper Secondary. 40 % Lower Secondary and 60 % Upper Secondary ☛ Questions posted varies in terms of difficilty. There are simple and basic questions that touch only the surface of a topic while others goes deeper and need higher skills. ☛ Familiarise yourself with the use of a scientific calculator. Fumbling with a calculator may invite unwanted results. 3 D 1. Paper 1 usually begins with simple and easy questions. 2. If any can’t be answered, move to other questions and don’t waste time an any one question. 3. For questions that involve Squares, Square Roots or other table readings, usually examples on how to use the table are shown. 4. Table below shows instructions word in questions and what should be done. QUESTION INSTRUCTION CALCULATE / EVALUATE / FIND EXPRESS WRITE / STATE ROUND SIMPLIFY FACTORISE SOLVE WHAT SHOULD BE DONE You have to do calculation using formulae, guide, theorem or law. You have to give answer in the form requested. Write down answer without showing any working method. You have to give answer to the nearest value. You have to write certain expression in the simplest form You have to write again expression in the form of product of factors. Seeking the value of variable in a certain equation. QUESTION EXAMPLE Calculate the probability of getting a male student. Express your answer in standard form. State the angle for this rotation. Round 0.0218 to two significant figure. Simplify m 2 – (k – m) 2 k Factorise completely 4h2k – hk. Solve the equation f + 5 = 8 5. It is very important for candidates to study past years questions and try to answer them according to the time and rules set. This will give us a clear picture of the form of question that will be given, skills that must be grasp and topics that must be given priority. 6. Don’t be too dependent on a certain method or skill to solve problems. Try to variate your technique and skill. 7. THE MORE EXERCISE, THE BETTER METHOD OF SOLVING WE USE AND THE FASTER WE SOLVE EXAMINATION QUESTIONS THAT HAVE THE SAME FORMAT EACH YEAR . E FORMS OF QUESTION – Paper 1 1. “COMMON SENSE” QUESTIONS (NEEDS NO CALCULATION) EXAMPLE 1 : A B P D C The diagram shows four lines drawn on a square grid. Which of the lines has a gradient of 2 ? A. PA B. PB C. PC D. PD 4 2. QUESTIONS THAT CAN BE ANSWERED USING OTHER QUESTION BEFORE OR AFTER IT AS A GUIDE Example 2 below can be answered using Example 3 as a guide. EXAMPLE 2 : Amir Amsyar bought a pair of pants at a price of RM 42 after discount. The original price is RM 60. Calculate the percentage of discount given. A. 20 % C. 30 % B. 25 % D. 35 % EXAMPLE 3 : Solve the following equation 60 – x (60) = 42 100 A. – 20 B. 30 C. 45 D. - 32 3. QUESTIONS THAT CAN BE ANSWERED BY TRYING OUT EACH CHOICE GIVEN (If possible, try not using this method because it is time consuming) EXAMPLE 4 : A bag contains 624 balls which are either orange, purple or white. If a ball is picked randomly from the bag, the probability of picking a white ball is 3 . Find the number 8 of white balls in the bag. A. 234 B. 243 C. 324 D. 423 Try whether 234 is equal to 3 . If not, repeat with 624 8 other choices. (If possible make a RANDOM choice because we might succeed at first try) A better and quicker method here is using tthe algebraic method i.e by forming the equation x = 3 and solving it. 624 8 4. QUESTIONS THAT CAN BE SOLVED USING ALGEBRAIC METHOD EXAMPLE 5 : The interior angles of a hexagon are 2xo, 2xo, 3xo, 3xo, 4xo dan 4xo. The value of x is A. 40o C. 80o B. 70o D. 90o EXAMPLE 6 : In the following diagram, calculate the height of the cylinder, h, given surface area of the cylinder is 330 cm2 and its radius is 3.5 cm. r A. 11.5 cm B. 13.25 cm C. 15 cm D. 26.5 cm h Form the equation 2x + 2x + 3x + 3x + 4x + 4x = 4(180) and solve the equation. Form the equation 2π(3.5)2 + 2π(3.5)h = 330 and sol solve the equation. (Subtitute π = 22 / 7) EXAMPLE 7 Given M (k, 2) is the mid point for the line that connects points P (-8, a) and Q (2a, a). The value of k is A. 2 B. 3 C. – 2 D. – 3 5 Form the simultaneous equation a + a = 2 dan – 8 + 2a = k 2 and solve them. 5. QUESTIONS THAT HAD TO BE GUESSED Before guessing, eliminate all the distractors first. EXAMPLE 8 : 1. 6.27 x 10 –4 = A. 0.0000627 B. 62700 C. 0.000627 D. 6270000 Not possible because this is a big number! Not possible because this is a big number! 6. TRY THE FOLLOWING QUESTIONS INVOLVING NUMBERS, SHAPES & ALGEBRA : 1. Round 40450 to three significant figure A. 404 B. 405 4. 3.47 x 10 3 = A. 0.0034 C. 347 B. 34.7 D. 3470 N 7. G H C. 40400 D. 40500 2. 2.4 x 10 5 + 4.8 x 10 4 = A. 7.2 x 10 9 C. 2.88 x 10 5 B. 2.88 x 10 9 D. 2.88 x 10 4 5. The area of a square is 1.54 m2. Its width is 250 cm. Find its length in cm A. 1.29 x 102 C. 6.16 x 101 B. 6.16 x 10-1 D. 6.16 x 103 8. Factorise 6pq – 4q2 A. 2q(4p – 4q) C. 6p(p – 4q2) B. 6q(p – 4q2) D. 2q(3p – 2q) 9. Factorise completely 2x2 - 8 S G and H are two points on the parallel of laltitude 72oN. Find the shortest distance, in nautical miles, between point G and H. A. 1080 B. 2160 13. 14. P 6 3. 1011101 2 – 10110 2 = A. 10001 2 C. 10101 2 B. 10111 2 D. 11111 2 6. If x + 2 = 3x then x = 5 A. – 1 / 3 C. – 5 B. – 2 / 5 D. 1 11. Express 2r _ r as a k+1 k fraction in its lowest term A. r(k – 1) C. r k(k + 1) k B. rk + r D. r . k(k + 1) k+1 12. Given w = 3a + 2b a then a = A. 2b C. w – 2b w–3 3 B. 2b D. w w+3 6b A. 2(x2 – 4) B. 2(x – 2)2 C. (x – 2)(x + 4) D. 2(x – 2)(x + 2) 10. (4 – 3p)(2 + 5p) = A. 8 + 26p – 15p2 B. 8 – 26p – 15p2 C. 8 + 14p – 15p2 D. 8 – 14p – 15p2 Q 60o 20o R C. 4320 D. 8640 15. Chinese ☻☻☻☻ ☻☻ Indian ☻☻☻☻☻ Malay ☻denotes x students Piktograph shows number of students in class 3A. Find value of x if total number of students is 35 people A. 4 B. 6 C. 5 D. 3 4 Q P 2 R is due south of Q. The bearing of P from R is A. 080o o 6B. 100 C. 240o D. 260o -2 0 2 4 Q is the image of triangle P under a rotation. The coordinates of the centre of rotation are A. (5, 1) C. (4, 2) B. (2, 4) D. (0, 5) 6 SPM MATHEMATICS PAPER 2 (1449/2) A 1. GENERAL GUIDE– Paper 2 Paper 2 SPM Mathematics contains two parts; Part A and Part B. 2. Test is in the form of written subjective and answers must be written in the question paper. 3. Questions are in the form of subjective and needs longer working method. 4. Scope of question covers certain particular topics from form 1 to form 5, different from Paper 1 that has a wider coverage. B NO 1 2 3 EXAMINATION FORMAT – Paper 2 ITEM Type of Instrument Type of Item Number of Question NOTES / DESCRIPTIONS Subjective Test Structure and Limited Response Part A 11 questions (Answer all) Part B 5 questions (Choose 4) 4 Total Marks Part A : 52 marks Part B : 48 marks (1 question 12 marks) 5 6 Test Duration Constructual Inclination 2 hours 30 minutes Knowledge - 25 % Skill - 70 % Value - 5% Lower secondary learning scope that has continuity in upper secondary. ☛ All learning scope from Form 4 and 5. 7 Contextual Coverage 8 Difficulty Level ☛ Easy (E) ☛ Moderate (M) ☛ Difficult (D) Additional Tools E:M:D=5:3:2 9 ☛ Scientific Calculator ☛ Mathematical Tables 7 Book Geometrical Equipment C 1. GENERAL INSTRUCTION – Paper 2 2. 3. 4. 5. Candidates must answer ALL 11 questions in Part A and 4 out of 5 questions in Part B (if more than 4 are answered, only 4 questions with the highest mark will be chosen). Candidates can use a normal scientific calculator. Candidates will be supplied with four digit tables book, graph papers, blank papers. Final answer that involves decimals must be given correct to two decimal places. Though not stated, candidates also have to bring along drawing tools like long rulers, geometry sets, “flexi curve” and other tools thought to be useful. D ANALYSIS – Paper 2 PART A ’03 ’04 ’05 ’06 ’07 ‘08 ’09 ‘10 1 1 1 3 TOPICS FORM 1 – 3 1. 2. 3. Simultaneous Linear Equations Cicles (II) Volume/Surface Area of Solids TOTAL 1 1 1 3 PART B ’03 ’04 ’05 ’06 ’07 ’08 ’09 ‘10 1 1 1 3 1 1 1 3 1 1 1 3 1 1 1 3 1 1 1 3 FORM 4 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Standard Form Quadratic Expr & Equations Sets Mathematical Reasoning The Straight Line Statistics III Probability I Cicles III Trigonometry II Angl. of Elevation & Depress. Lines & Planes in 3-Dimension TOTAL 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 1 5 1 4 1 5 1 4 1 5 1 4 1 1 1 1 1 1 1 FORM 5 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Number Bases Graphs of Functions II Transformations III Matrices Variations Gradient/Area Under Graphs Probability II Bearings Earth As A Sphere Plans and Elevations TOTAL OVERALL TOTAL 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 11 3 11 4 11 3 11 4 11 3 11 4 11 4 5 1 1 4 5 1 1 4 5 1 1 4 5 1 1 4 5 1 1 4 5 1 1 4 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ☛ Topics from form 4 and 5 forms major questions. ☛ Candidates can make suitable choice of question in Part B and this can be done by looking at your skill and ability. Teachers’ and friends’ opinions have to be taken into account too. 8 ☛ Bold printed topics are topics never being included in Paper 2 before. E In general, candidate will be awarded METHOD MARKS (for working method needed), ANSWER MARKS (for the precise answer needed), INDEPENDENT MARKS (for answers that working methods are not needed), TRANSFER MARKS (for transfering points into graph paper with precision), GRAPH MARKS ( for smooth and flawless graph) QUALITY MARKS (for a good drawings of Plans and Elevation ) and others. Following are general guides that candidates have to practice in some questions in paper 2. 1. - Change to its standard form ax2 + bx + c = 0 Factorise expression on the Left Hand Side. Use the fact that “If ab = 0, then a = 0 or b = 0” Example 1 : Solve the equation (f – 1)(f + 3) = 5 2. SIMULTANEOUS LINEAR EQUATIONS Eliminate fractions first (if there is any) by multiplying the equation with the denominator. solve using the “subtitution” or “variable elimination” technique. recheck whether the answer satisfy the equation given. Example 2 : Calculate the value of f and g that satisfy both the following equations 1f+g = 1 2 3f – 2g = 22 3. MATRICES - candidates must be able to find Inverse of a Matix and know its characteristics. - candidates must also be able to use that Inverse Matrix to solve simultaneous equations or the matrix equation given. - write final answer explicitly. Example 3 : Given the matrix A = 5 3 -4 -2 ( i) Find the inverse of matrix A (ii) Hence, using matrices, calculate the values of x and y which satisfy the following matrix equation 5 3 x 0 = -4 -2 y 2 9 4. SETS (ALTERNATES WITH REGION SHADING ON GRAPH’S QUESTION) - Usually, question is on shading region of intersection, union and complement of sets. - Multiple hatchings are allowed. Example 4 : On the diagrams in the answer space, shade (a) the set P’ ⋂Q Q P (b) the set (P ⋃Q’) ⋂R Q P R R 5. GRADIENT AND AREA UNDER A GRAPH - questions usually are based on Speed-Time or Distance-Time graphs. - candidates must be able to (a) write equation from the information given and hence solve that equation. (b) calculate speed from Distance-Time graph. (c) calculate distance and acceleration from Speed-Time graph. (d) caculate average speed from both graphs. Example 5 : Speed (m s-1) 14 12 154 Time (sec) 8 t Diagram shows speed-time graph for a particle in a period of t s. Calculate (i) rate of speed change for the particle in the first 8 seconds. (ii) value of t, given total distance travelled by the particle in the period of t seconds is 248 m. M D O t 4 5 Time (hour) Example 6 : Distance from P (km) C 280 N Diagram shows distance-time graph for the route travelled by a bus and a car. OMN represents the bus’s route from town P to town R and CMD represents the car’s route from town R to town P. (i) Calculate average speed, in km h–1, travelled by the bus from P to R. (ii) If the car travels at uniform speed, calculate value of t. 6. CIRCLES - candidates must be able use length of arc and area of a sector formulas with ease where the use of angle at the centre is very important. - answer must be given at least to 2 decimal place if decimals are involve. C Example 7 : In the diagram, ABD is a sector of a circle with centre A. ADC is a straight line. By using π = 3.142, calculate D (a) perimeter of the shaded region (b) area of the shaded region. A 10 B 8 cm 7. SURFACE AREA AND VOLUME OF SOLIDS. - Memorise formulaes on surface area and volume of solids. - Skill on formulae application is also very important.. Example 8 : Example 9 : Example 10 : V (I) (II) Diagram (I) is a container in the shape of a cuboid that is full with water. Base of the cuboid in the shape of a rectangle with a length of 11 cm and breadth of 8 cm. Height of the cuboid is 21 cm. Diagram (II) is an empty container in the shape of a cylinder. Diameter of the base of the cylinder is 12 cm. All the water in the cuboid container are poured into the cylinder container. Calculate the height of the water level in the cylinder container. Diagram above shows a solid in the shape of a cylinder with a hemisphere taken out from each end of the cylinder. Base radius of the cylinder is the same as radius of the hemisphere, that is 5. 6 cm. Length of the cylinder is 13 cm. Calculate the volume of the solid left.. Q P M N Diagram above shows a solid erected from a combination of a cuboid and a pyramid. Given height of the vertice V from the base MNPQ is 13 cm, calculate the surface area of the solid.. 8. GRAPHS OF FUNCTIONS - Graphs must be drawn on a graph paper. - you must be able to calculate y values from the function given, obey scale instruction, shift points in the table to graph and hence draw a smooth curve. - skills on solving equation by graphical method are also needed. Example 11 : (a) Complete the following table for the function of y = x3 – 12x + 20. x y -3.5 19.1 -3 29 -2 -1 0 20 1 9 2 4 3 3.5 20.9 4 36 (3 marks) (b) (c) (d) Using the scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = x3 – 12x + 20 for – 3. 5 ≤ x ≤ 4. (3 marks) From your graph, find value of y when x = -1. 5 Draw a suitable straight line on your graph to find all the values of x in the range of -3. 5 ≤ x ≤ 4 that satisfy the equation x3 – 12x – 5 = 0. State the values of x. (1 mark) (5 marks) Example 12 : (a) Complete the following table for the equation y = 24 . x x -4 -3 -2 -1 1 y -6 - 12 - 24 24 1. 5 2 12 3 8 4 6 [2 marks] (b) For this part of the question, use the graph paper provided. You may use a flexible curve rule. By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 24 for – 4 ≤ x ≤ 4. x (c) From your graph, find (i) the value of y when x = 2. 9 (ii) the value of x when y = -13 [5 marks] [2 marks] (d) Draw a suitable straight line on your graph to find all the values of x which satisfy the equation 2x2 + 5x = 24 for - 4 ≤ x ≤ 4. State 11 these values of x. [3 marks] 9. PLANS AND ELEVATIONS - - - Drawings are done on the blank paper provided in the question paper. Drawings must be precise according to measurements given. All lines must be straight and drawn using a ruler. 90 o angle can be erected quickly using 90° edge of a ruler. Make sure there are no “extensions” and “gaps”. Construction lines must be differentiated with projection lines. Circles’ curves must be drawn using compasses. “Line straightness” should be emphasize. Example 13 : Example 14 : 10. STATISTICS - Candidates must be able to find mean, modes and medians. - Candidates must be able to construct frequency table and hence draw histogram or frequency polygons. - Candidates must be able to construct cumulative frequency table and hence draw an ogive. - Candidate must also be able to find informations from the ogive drawn. Example 15: Number of Appreciation Certificate Number of Students 0 18 1 3 2 5 3 0 5 6 6 2 7 3 8 1 (a) The table above shows number of appreciation certificate received by 40 students in a class. Find (i) median (ii) mean of the data. (3 marks) (b) For this question, use the graph paper provided provided. Age (years) Number of workers 18-22 23-27 28-32 33-37 38-42 8 15 23 36 48 43-47 48-52 53-57 29 15 6 The table above shows the age (in years) distribution for 180 workers in an Electronic Factory. (i) Construct a cumulative frequency table for the data. (ii) Using the scale of 2 cm to 5 years on the x-axis and 2 cm to 20 workers on the y-axis, draw an ogive for the data. (iii) Workers in the first quartile are required to attend a course. State the oldest age of the worker required to attend the course. (9 marks) Example 16 : For this question, use the graph paper provided. 152 173 167 172 168 174 166 178 176 164 154 167 162 155 151 163 160 176 168 175 174 177 171 159 171 174 179 169 153 173 156 172 160 154 164 158 167 178 169 154 Data in the above table are height, in cm, for a group of 40 students. (a) Construct a frequency table for this data using class intervals of the size of 5 cm, with 145-149 as the first class interval. . (4 marks) (b) Using a scale of 2 cm to 5 cm on the x-axis and 2 cm to 1 student on the y-axis, draw a frequency polygon for the above data. (4 marks) 12 (c) From the frequency polygon, (i) Find the modal class, (ii) calculate the mean height for the group of students (4 marks)

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# [R] Generating a vector for breaks in a histogram M. Edward Borasky znmeb at aracnet.com Sun Jul 6 05:39:33 CEST 2003 Well ... In my own recent example, it was plotting the raw data as a histogram that finally directed me to the "truth" of what the data had to say. As you may recall, the dataset was inter-arrival times of calls to a computer routine, known only from timestamps truncated (not rounded) to the nearest second. I started with kernel density (sm.density, with the default parameters, to be precise) and was unsatisfied with the result. Yesterday, when I plotted the raw counts (how many values were 0, how many 1, etc.) as a histogram, I was struck by two things: 1. There really are only two peaks -- the "stuff" in between them is, for the purpose of business decisions, irrelevant. 2. The inter-arrival time value "0" in such a dataset represents all the values that are greater than or equal to zero and *less than 1*, and so on. There is a natural "histogramming" going on via the timestamp truncation, which implies to me that the *midpoint* of the "bin" -- say, for the 0 values, 0.5 -- is the "natural" value to choose for the "x-axis" in the absence of any better information. This also rather neatly disposes of the issue of zero-valued inter-arrival times. :) Are the "old ways" best? Maybe not. Can I make reasonable business decisions without histograms? I'm not convinced that's the case; it certainly wasn't the case this time. Finally, while I've never been fortunate enough to use S, the existence of R has caused a revolution in the way I do the analysis of computer performance data. Before R came along, the only tools I had available were Excel, Minitab, and any special-purpose code I was willing to write to accomplish tasks not in the vocabulary of Excel or Minitab. For example, it's difficult, though not impossible, to do a non-linear regression or kernel density estimation with either tool. In R, they're one-liners. If there was a Nobel Prize for scientific software, I'd nominate R and its creators. (Of course, there *is* a Nobel in Economics.) :) -- > -----Original Message----- > Things have moved on since the ASH work too, but I would > agree that density estimation is often a better way than > histograms. However, close > to state-of-the-art density estimation is built into R > (?density) and packages `polspline', `KernSmooth' and `sm' > are also much more advanced > than `ash'. > > It was the advent of enough computing power that changed > this, and the S > language has been in the forefront of making the state of the art > available. You'll see that MASS (the book) covers histograms and > alternatives in its chapter on Univariate Distributions, and > it has since > its 1994 first edition (when did you go to `school'?)

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## Module clpb ``:- use_module(library(clpb)).`` Constraint Logic Programming over Boolean variables ## Introduction This library provides CLP(B), Constraint Logic Programming over Boolean variables. It can be used to model and solve combinatorial problems such as verification, allocation and covering tasks. CLP(B) is an instance of the general CLP(X) scheme, extending logic programming with reasoning over specialised domains. The implementation is based on reduced and ordered Binary Decision Diagrams (BDDs). Benchmarks and usage examples of this library are available from: https://www.metalevel.at/clpb/ ## Boolean expressions A Boolean expression is one of: `0` false `1` true variable unknown truth value atom universally quantified variable ~ Expr logical NOT Expr + Expr logical OR Expr * Expr logical AND Expr # Expr exclusive OR Var ^ Expr existential quantification Expr =:= Expr equality Expr == Expr disequality (same as #) Expr =< Expr less or equal (implication) Expr >= Expr greater or equal Expr < Expr less than Expr > Expr greater than card(Is,Exprs) cardinality constraint (see below) `+(Exprs)` n-fold disjunction (see below) `*(Exprs)` n-fold conjunction (see below) where Expr again denotes a Boolean expression. The Boolean expression `card(Is,Exprs)` is true iff the number of true expressions in the list `Exprs` is a member of the list `Is` of integers and integer ranges of the form `From-To`. For example, to state that precisely two of the three variables `X`, `Y` and `Z` are `true`, you can use `sat(card([2],[X,Y,Z]))`. `+(Exprs)` and `*(Exprs)` denote, respectively, the disjunction and conjunction of all elements in the list `Exprs` of Boolean expressions. Atoms denote parametric values that are universally quantified. All universal quantifiers appear implicitly in front of the entire expression. In residual goals, universally quantified variables always appear on the right-hand side of equations. Therefore, they can be used to express functional dependencies on input variables. ## Interface predicates The most frequently used CLP(B) predicates are: • `sat(+Expr)` True iff the Boolean expression Expr is satisfiable. • `taut(+Expr, -T)` If Expr is a tautology with respect to the posted constraints, succeeds with T = 1. If Expr cannot be satisfied, succeeds with T = 0. Otherwise, it fails. • `labeling(+Vs)` Assigns truth values to the variables Vs such that all constraints are satisfied. The unification of a CLP(B) variable X with a term T is equivalent to posting the constraint sat(X=:=T). ## Examples Here is an example session with a few queries and their answers: `````` ?- use_module(library(clpb)). true. ?- sat(X*Y). X = 1, Y = 1. ?- sat(X * ~X). false. ?- taut(X * ~X, T). T = 0, clpb:sat(X=:=X). ?- sat(X^Y^(X+Y)). clpb:sat(X=:=X), clpb:sat(Y=:=Y). ?- sat(X*Y + X*Z), labeling([X,Y,Z]). X = 1, Y = 0, Z = 1 ; X = 1, Y = 1, Z = 0 ; X = 1, Y = 1, Z = 1. ?- sat(X =< Y), sat(Y =< Z), taut(X =< Z, T). T = 1, clpb:sat(X=:=X*Y), clpb:sat(Y=:=Y*Z). ?- sat(1#X#a#b). clpb:sat(X=:=a#b).`````` The pending residual goals constrain remaining variables to Boolean expressions and are declaratively equivalent to the original query. The last example illustrates that when applicable, remaining variables are expressed as functions of universally quantified variables. ## Obtaining BDDs By default, CLP(B) residual goals appear in (approximately) algebraic normal form (ANF). This projection is often computationally expensive. We can assert `clpb:clpb_residuals(bdd)` to see the BDD representation of all constraints. This results in faster projection to residual goals, and is also useful for learning more about BDDs. For example: `````` ?- asserta(clpb:clpb_residuals(bdd)). true. ?- sat(X#Y). node(3)- (v(X, 0)->node(2);node(1)), node(1)- (v(Y, 1)->true;false), node(2)- (v(Y, 1)->false;true).`````` Note that this representation cannot be pasted back on the toplevel, and its details are subject to change. Use copy_term/3 to obtain such answers as Prolog terms. The variable order of the BDD is determined by the order in which the variables first appear in constraints. To obtain different orders, we can for example use: `````` ?- sat(+[1,Y,X]), sat(X#Y). node(3)- (v(Y, 0)->node(2);node(1)), node(1)- (v(X, 1)->true;false), node(2)- (v(X, 1)->false;true).`````` ## Enabling monotonic CLP(B) In the default execution mode, CLP(B) constraints are not monotonic. This means that adding constraints can yield new solutions. For example: `````` ?- sat(X=:=1), X = 1+0. false. ?- X = 1+0, sat(X=:=1), X = 1+0. X = 1+0.`````` This behaviour is highly problematic from a logical point of view, and it may render declarative debugging techniques inapplicable. Assert `clpb:monotonic` to make CLP(B) monotonic. If this mode is enabled, then you must wrap CLP(B) variables with the functor `v/1`. For example: `````` ?- asserta(clpb:monotonic). true. ?- sat(v(X)=:=1#1). X = 0.`````` ## Example: Pigeons In this example, we are attempting to place I pigeons into J holes in such a way that each hole contains at most one pigeon. One interesting property of this task is that it can be formulated using only cardinality constraints (`card/2`). Another interesting aspect is that this task has no short resolution refutations in general. In the following, we use Prolog DCG notation to describe a list `Cs` of CLP(B) constraints that must all be satisfied. `````` :- use_module(library(clpb)). :- use_module(library(clpz)). :- use_module(library(lists)). :- use_module(library(dcgs)). pigeon(I, J, Rows, Cs) :- length(Rows, I), length(Row, J), maplist(same_length(Row), Rows), transpose(Rows, TRows), phrase((all_cards(Rows,[1]),all_cards(TRows,[0,1])), Cs). all_cards([], _) --> []. all_cards([Ls|Lss], Cs) --> [card(Cs,Ls)], all_cards(Lss, Cs).`````` Example queries: `````` ?- pigeon(9, 8, Rows, Cs), sat(*(Cs)). false. ?- pigeon(2, 3, Rows, Cs), sat(*(Cs)), append(Rows, Vs), labeling(Vs), maplist(portray_clause, Rows). [0,0,1]. [0,1,0]. etc.`````` ## Example: Boolean circuit Consider a Boolean circuit that express the Boolean function =|XOR|= with 4 =|NAND|= gates. We can model such a circuit with CLP(B) constraints as follows: `````` :- use_module(library(clpb)). nand_gate(X, Y, Z) :- sat(Z =:= ~(X*Y)). xor(X, Y, Z) :- nand_gate(X, Y, T1), nand_gate(X, T1, T2), nand_gate(Y, T1, T3), nand_gate(T2, T3, Z).`````` Using universally quantified variables, we can show that the circuit does compute =|XOR|= as intended: `````` ?- xor(x, y, Z). clpb:sat(Z=:=x#y).`````` ## Acknowledgments The interface predicates of this library follow the example of SICStus Prolog. Use SICStus Prolog for higher performance in many cases. #### sat(+Expr) is semidet. True iff Expr is a satisfiable Boolean expression. #### taut(+Expr, -T) is semidet Tautology check. Succeeds with T = 0 if the Boolean expression Expr cannot be satisfied, and with T = 1 if Expr is always true with respect to the current constraints. Fails otherwise. #### labeling(+Vs) is multi. Enumerate concrete solutions. Assigns truth values to the Boolean variables Vs such that all stated constraints are satisfied. #### sat_count(+Expr, -Count) is det. Count the number of admissible assignments. Count is the number of different assignments of truth values to the variables in the Boolean expression Expr, such that Expr is true and all posted constraints are satisfiable. A common form of invocation is `sat_count(+[1|Vs], Count)`: This counts the number of admissible assignments to `Vs` without imposing any further constraints. Examples: `````` ?- sat(A =< B), Vs = [A,B], sat_count(+[1|Vs], Count). Vs = [A,B], Count = 3, clpb:sat(A=:=A*B). ?- length(Vs, 120), sat_count(+Vs, CountOr), sat_count(*(Vs), CountAnd). Vs = [...], CountOr = 1329227995784915872903807060280344575, CountAnd = 1.`````` #### weighted_maximum(+Weights, +Vs, -Maximum) is multi. Enumerate weighted optima over admissible assignments. Maximize a linear objective function over Boolean variables Vs with integer coefficients Weights. This predicate assigns 0 and 1 to the variables in Vs such that all stated constraints are satisfied, and Maximum is the maximum of `sum(Weight_i*V_i)` over all admissible assignments. On backtracking, all admissible assignments that attain the optimum are generated. This predicate can also be used to minimize a linear Boolean program, since negative integers can appear in Weights. Example: `````` ?- sat(A#B), weighted_maximum([1,2,1], [A,B,C], Maximum). A = 0, B = 1, C = 1, Maximum = 3.`````` #### random_labeling(+Seed, +Vs) is det. Select a single random solution. An admissible assignment of truth values to the Boolean variables in Vs is chosen in such a way that each admissible assignment is equally likely. Seed is an integer, used as the initial seed for the random number generator.

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## 287 Days From February 13, 2024 Want to figure out the date that is exactly two hundred eighty seven days from Feb 13, 2024 without counting? Your starting date is February 13, 2024 so that means that 287 days later would be November 26, 2024. You can check this by using the date difference calculator to measure the number of days from Feb 13, 2024 to Nov 26, 2024. November 2024 • Sunday • Monday • Tuesday • Wednesday • Thursday • Friday • Saturday 1. 1 2. 2 1. 3 2. 4 3. 5 4. 6 5. 7 6. 8 7. 9 1. 10 2. 11 3. 12 4. 13 5. 14 6. 15 7. 16 1. 17 2. 18 3. 19 4. 20 5. 21 6. 22 7. 23 1. 24 2. 25 3. 26 4. 27 5. 28 6. 29 7. 30 November 26, 2024 is a Tuesday. It is the 331st day of the year, and in the 48th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 2024 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 11/26/2024, and almost everywhere else in the world it's 26/11/2024. ### What if you only counted weekdays? In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 287 weekdays from Feb 13, 2024, you can count up each day skipping Saturdays and Sundays. Start your calculation with Feb 13, 2024, which falls on a Tuesday. Counting forward, the next day would be a Wednesday. To get exactly two hundred eighty seven weekdays from Feb 13, 2024, you actually need to count 401 total days (including weekend days). That means that 287 weekdays from Feb 13, 2024 would be March 20, 2025. If you're counting business days, don't forget to adjust this date for any holidays. March 2025 • Sunday • Monday • Tuesday • Wednesday • Thursday • Friday • Saturday 1. 1 1. 2 2. 3 3. 4 4. 5 5. 6 6. 7 7. 8 1. 9 2. 10 3. 11 4. 12 5. 13 6. 14 7. 15 1. 16 2. 17 3. 18 4. 19 5. 20 6. 21 7. 22 1. 23 2. 24 3. 25 4. 26 5. 27 6. 28 7. 29 1. 30 2. 31 March 20, 2025 is a Thursday. It is the 79th day of the year, and in the 79th week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2025 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/20/2025, and almost everywhere else in the world it's 20/03/2025. ### Enter the number of days and the exact date Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date.

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Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  wwlksnext Structured version   Visualization version   GIF version Theorem wwlksnext 27778 Description: Extension of a walk (as word) by adding an edge/vertex. (Contributed by Alexander van der Vekens, 4-Aug-2018.) (Revised by AV, 16-Apr-2021.) Hypotheses Ref Expression wwlksnext.v 𝑉 = (Vtx‘𝐺) wwlksnext.e 𝐸 = (Edg‘𝐺) Assertion Ref Expression wwlksnext ((𝑇 ∈ (𝑁 WWalksN 𝐺) ∧ 𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺)) Proof of Theorem wwlksnext Dummy variable 𝑖 is distinct from all other variables. StepHypRef Expression 1 wwlksnext.v . . . 4 𝑉 = (Vtx‘𝐺) 21wwlknbp 27727 . . 3 (𝑇 ∈ (𝑁 WWalksN 𝐺) → (𝐺 ∈ V ∧ 𝑁 ∈ ℕ0𝑇 ∈ Word 𝑉)) 3 wwlksnext.e . . . . . . . . . . 11 𝐸 = (Edg‘𝐺) 41, 3wwlknp 27728 . . . . . . . . . 10 (𝑇 ∈ (𝑁 WWalksN 𝐺) → (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸)) 5 simp1 1133 . . . . . . . . . . . . . . 15 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) → 𝑇 ∈ Word 𝑉) 6 simprl 770 . . . . . . . . . . . . . . 15 ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → 𝑆𝑉) 7 cats1un 14130 . . . . . . . . . . . . . . 15 ((𝑇 ∈ Word 𝑉𝑆𝑉) → (𝑇 ++ ⟨“𝑆”⟩) = (𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩})) 85, 6, 7syl2an 598 . . . . . . . . . . . . . 14 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (𝑇 ++ ⟨“𝑆”⟩) = (𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩})) 9 opex 5324 . . . . . . . . . . . . . . . . . . 19 ⟨(♯‘𝑇), 𝑆⟩ ∈ V 109snnz 4669 . . . . . . . . . . . . . . . . . 18 {⟨(♯‘𝑇), 𝑆⟩} ≠ ∅ 1110neii 2953 . . . . . . . . . . . . . . . . 17 ¬ {⟨(♯‘𝑇), 𝑆⟩} = ∅ 1211intnan 490 . . . . . . . . . . . . . . . 16 ¬ (𝑇 = ∅ ∧ {⟨(♯‘𝑇), 𝑆⟩} = ∅) 13 df-ne 2952 . . . . . . . . . . . . . . . . 17 ((𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩}) ≠ ∅ ↔ ¬ (𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩}) = ∅) 14 un00 4339 . . . . . . . . . . . . . . . . 17 ((𝑇 = ∅ ∧ {⟨(♯‘𝑇), 𝑆⟩} = ∅) ↔ (𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩}) = ∅) 1513, 14xchbinxr 338 . . . . . . . . . . . . . . . 16 ((𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩}) ≠ ∅ ↔ ¬ (𝑇 = ∅ ∧ {⟨(♯‘𝑇), 𝑆⟩} = ∅)) 1612, 15mpbir 234 . . . . . . . . . . . . . . 15 (𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩}) ≠ ∅ 1716a1i 11 . . . . . . . . . . . . . 14 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (𝑇 ∪ {⟨(♯‘𝑇), 𝑆⟩}) ≠ ∅) 188, 17eqnetrd 3018 . . . . . . . . . . . . 13 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (𝑇 ++ ⟨“𝑆”⟩) ≠ ∅) 19 s1cl 14003 . . . . . . . . . . . . . . 15 (𝑆𝑉 → ⟨“𝑆”⟩ ∈ Word 𝑉) 2019ad2antrl 727 . . . . . . . . . . . . . 14 ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → ⟨“𝑆”⟩ ∈ Word 𝑉) 21 ccatcl 13973 . . . . . . . . . . . . . 14 ((𝑇 ∈ Word 𝑉 ∧ ⟨“𝑆”⟩ ∈ Word 𝑉) → (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉) 225, 20, 21syl2an 598 . . . . . . . . . . . . 13 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉) 23 simplrl 776 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → 𝑇 ∈ Word 𝑉) 24 fzossfzop1 13164 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 (𝑁 ∈ ℕ0 → (0..^𝑁) ⊆ (0..^(𝑁 + 1))) 2524ad2antrr 725 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (0..^𝑁) ⊆ (0..^(𝑁 + 1))) 2625sselda 3892 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → 𝑖 ∈ (0..^(𝑁 + 1))) 27 oveq2 7158 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 ((♯‘𝑇) = (𝑁 + 1) → (0..^(♯‘𝑇)) = (0..^(𝑁 + 1))) 2827eleq2d 2837 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 ((♯‘𝑇) = (𝑁 + 1) → (𝑖 ∈ (0..^(♯‘𝑇)) ↔ 𝑖 ∈ (0..^(𝑁 + 1)))) 2928adantl 485 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → (𝑖 ∈ (0..^(♯‘𝑇)) ↔ 𝑖 ∈ (0..^(𝑁 + 1)))) 3029ad2antlr 726 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → (𝑖 ∈ (0..^(♯‘𝑇)) ↔ 𝑖 ∈ (0..^(𝑁 + 1)))) 3126, 30mpbird 260 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → 𝑖 ∈ (0..^(♯‘𝑇))) 32 ccats1val1 14028 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((𝑇 ∈ Word 𝑉𝑖 ∈ (0..^(♯‘𝑇))) → ((𝑇 ++ ⟨“𝑆”⟩)‘𝑖) = (𝑇𝑖)) 3323, 31, 32syl2anc 587 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → ((𝑇 ++ ⟨“𝑆”⟩)‘𝑖) = (𝑇𝑖)) 34 fzonn0p1p1 13165 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (𝑖 ∈ (0..^𝑁) → (𝑖 + 1) ∈ (0..^(𝑁 + 1))) 3534adantl 485 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → (𝑖 + 1) ∈ (0..^(𝑁 + 1))) 3627adantl 485 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → (0..^(♯‘𝑇)) = (0..^(𝑁 + 1))) 3736ad2antlr 726 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → (0..^(♯‘𝑇)) = (0..^(𝑁 + 1))) 3835, 37eleqtrrd 2855 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → (𝑖 + 1) ∈ (0..^(♯‘𝑇))) 39 ccats1val1 14028 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((𝑇 ∈ Word 𝑉 ∧ (𝑖 + 1) ∈ (0..^(♯‘𝑇))) → ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1)) = (𝑇‘(𝑖 + 1))) 4023, 38, 39syl2anc 587 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1)) = (𝑇‘(𝑖 + 1))) 4133, 40preq12d 4634 . . . . . . . . . . . . . . . . . . . . . . . . . 26 ((((𝑁 ∈ ℕ0𝑆𝑉) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) ∧ 𝑖 ∈ (0..^𝑁)) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} = {(𝑇𝑖), (𝑇‘(𝑖 + 1))}) 4241exp31 423 . . . . . . . . . . . . . . . . . . . . . . . . 25 ((𝑁 ∈ ℕ0𝑆𝑉) → ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → (𝑖 ∈ (0..^𝑁) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} = {(𝑇𝑖), (𝑇‘(𝑖 + 1))}))) 4342adantrr 716 . . . . . . . . . . . . . . . . . . . . . . . 24 ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → (𝑖 ∈ (0..^𝑁) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} = {(𝑇𝑖), (𝑇‘(𝑖 + 1))}))) 4443impcom 411 . . . . . . . . . . . . . . . . . . . . . . 23 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (𝑖 ∈ (0..^𝑁) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} = {(𝑇𝑖), (𝑇‘(𝑖 + 1))})) 4544imp 410 . . . . . . . . . . . . . . . . . . . . . 22 ((((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) ∧ 𝑖 ∈ (0..^𝑁)) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} = {(𝑇𝑖), (𝑇‘(𝑖 + 1))}) 4645eleq1d 2836 . . . . . . . . . . . . . . . . . . . . 21 ((((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) ∧ 𝑖 ∈ (0..^𝑁)) → ({((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ {(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸)) 4746ralbidva 3125 . . . . . . . . . . . . . . . . . . . 20 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (∀𝑖 ∈ (0..^𝑁){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸)) 4847exbiri 810 . . . . . . . . . . . . . . . . . . 19 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → (∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸 → ∀𝑖 ∈ (0..^𝑁){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸))) 4948com23 86 . . . . . . . . . . . . . . . . . 18 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → (∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸 → ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → ∀𝑖 ∈ (0..^𝑁){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸))) 50493impia 1114 . . . . . . . . . . . . . . . . 17 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) → ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → ∀𝑖 ∈ (0..^𝑁){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸)) 5150imp 410 . . . . . . . . . . . . . . . 16 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ∀𝑖 ∈ (0..^𝑁){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) 52 oveq1 7157 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((♯‘𝑇) = (𝑁 + 1) → ((♯‘𝑇) − 1) = ((𝑁 + 1) − 1)) 5352adantl 485 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → ((♯‘𝑇) − 1) = ((𝑁 + 1) − 1)) 54 nn0cn 11944 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (𝑁 ∈ ℕ0𝑁 ∈ ℂ) 55 1cnd 10674 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (𝑁 ∈ ℕ0 → 1 ∈ ℂ) 5654, 55pncand 11036 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 (𝑁 ∈ ℕ0 → ((𝑁 + 1) − 1) = 𝑁) 5756adantl 485 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((𝑆𝑉𝑁 ∈ ℕ0) → ((𝑁 + 1) − 1) = 𝑁) 5853, 57sylan9eqr 2815 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → ((♯‘𝑇) − 1) = 𝑁) 5958fveq2d 6662 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (𝑇‘((♯‘𝑇) − 1)) = (𝑇𝑁)) 60 lsw 13963 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (𝑇 ∈ Word 𝑉 → (lastS‘𝑇) = (𝑇‘((♯‘𝑇) − 1))) 6160ad2antrl 727 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (lastS‘𝑇) = (𝑇‘((♯‘𝑇) − 1))) 62 simprl 770 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → 𝑇 ∈ Word 𝑉) 63 fzonn0p1 13163 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 (𝑁 ∈ ℕ0𝑁 ∈ (0..^(𝑁 + 1))) 6463ad2antlr 726 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → 𝑁 ∈ (0..^(𝑁 + 1))) 6527eleq2d 2837 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((♯‘𝑇) = (𝑁 + 1) → (𝑁 ∈ (0..^(♯‘𝑇)) ↔ 𝑁 ∈ (0..^(𝑁 + 1)))) 6665ad2antll 728 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (𝑁 ∈ (0..^(♯‘𝑇)) ↔ 𝑁 ∈ (0..^(𝑁 + 1)))) 6764, 66mpbird 260 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → 𝑁 ∈ (0..^(♯‘𝑇))) 68 ccats1val1 14028 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((𝑇 ∈ Word 𝑉𝑁 ∈ (0..^(♯‘𝑇))) → ((𝑇 ++ ⟨“𝑆”⟩)‘𝑁) = (𝑇𝑁)) 6962, 67, 68syl2anc 587 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → ((𝑇 ++ ⟨“𝑆”⟩)‘𝑁) = (𝑇𝑁)) 7059, 61, 693eqtr4d 2803 . . . . . . . . . . . . . . . . . . . . . . . . 25 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (lastS‘𝑇) = ((𝑇 ++ ⟨“𝑆”⟩)‘𝑁)) 71 simpll 766 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → 𝑆𝑉) 72 simprr 772 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (♯‘𝑇) = (𝑁 + 1)) 7372eqcomd 2764 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → (𝑁 + 1) = (♯‘𝑇)) 74 ccats1val2 14030 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((𝑇 ∈ Word 𝑉𝑆𝑉 ∧ (𝑁 + 1) = (♯‘𝑇)) → ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1)) = 𝑆) 7574eqcomd 2764 . . . . . . . . . . . . . . . . . . . . . . . . . 26 ((𝑇 ∈ Word 𝑉𝑆𝑉 ∧ (𝑁 + 1) = (♯‘𝑇)) → 𝑆 = ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))) 7662, 71, 73, 75syl3anc 1368 . . . . . . . . . . . . . . . . . . . . . . . . 25 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → 𝑆 = ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))) 7770, 76preq12d 4634 . . . . . . . . . . . . . . . . . . . . . . . 24 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → {(lastS‘𝑇), 𝑆} = {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))}) 7877eleq1d 2836 . . . . . . . . . . . . . . . . . . . . . . 23 (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → ({(lastS‘𝑇), 𝑆} ∈ 𝐸 ↔ {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)) 7978biimpcd 252 . . . . . . . . . . . . . . . . . . . . . 22 ({(lastS‘𝑇), 𝑆} ∈ 𝐸 → (((𝑆𝑉𝑁 ∈ ℕ0) ∧ (𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1))) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)) 8079exp4c 436 . . . . . . . . . . . . . . . . . . . . 21 ({(lastS‘𝑇), 𝑆} ∈ 𝐸 → (𝑆𝑉 → (𝑁 ∈ ℕ0 → ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)))) 8180impcom 411 . . . . . . . . . . . . . . . . . . . 20 ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑁 ∈ ℕ0 → ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸))) 8281impcom 411 . . . . . . . . . . . . . . . . . . 19 ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)) 8382impcom 411 . . . . . . . . . . . . . . . . . 18 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1)) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸) 84833adantl3 1165 . . . . . . . . . . . . . . . . 17 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸) 85 fveq2 6658 . . . . . . . . . . . . . . . . . . . . 21 (𝑖 = 𝑁 → ((𝑇 ++ ⟨“𝑆”⟩)‘𝑖) = ((𝑇 ++ ⟨“𝑆”⟩)‘𝑁)) 86 fvoveq1 7173 . . . . . . . . . . . . . . . . . . . . 21 (𝑖 = 𝑁 → ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1)) = ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))) 8785, 86preq12d 4634 . . . . . . . . . . . . . . . . . . . 20 (𝑖 = 𝑁 → {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} = {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))}) 8887eleq1d 2836 . . . . . . . . . . . . . . . . . . 19 (𝑖 = 𝑁 → ({((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)) 8988ralsng 4570 . . . . . . . . . . . . . . . . . 18 (𝑁 ∈ ℕ0 → (∀𝑖 ∈ {𝑁} {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)) 9089ad2antrl 727 . . . . . . . . . . . . . . . . 17 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (∀𝑖 ∈ {𝑁} {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ {((𝑇 ++ ⟨“𝑆”⟩)‘𝑁), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑁 + 1))} ∈ 𝐸)) 9184, 90mpbird 260 . . . . . . . . . . . . . . . 16 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ∀𝑖 ∈ {𝑁} {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) 92 ralunb 4096 . . . . . . . . . . . . . . . 16 (∀𝑖 ∈ ((0..^𝑁) ∪ {𝑁}){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ (∀𝑖 ∈ (0..^𝑁){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ∧ ∀𝑖 ∈ {𝑁} {((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸)) 9351, 91, 92sylanbrc 586 . . . . . . . . . . . . . . 15 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ∀𝑖 ∈ ((0..^𝑁) ∪ {𝑁}){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) 94 elnn0uz 12323 . . . . . . . . . . . . . . . . . . . 20 (𝑁 ∈ ℕ0𝑁 ∈ (ℤ‘0)) 95 eluzfz2 12964 . . . . . . . . . . . . . . . . . . . 20 (𝑁 ∈ (ℤ‘0) → 𝑁 ∈ (0...𝑁)) 9694, 95sylbi 220 . . . . . . . . . . . . . . . . . . 19 (𝑁 ∈ ℕ0𝑁 ∈ (0...𝑁)) 97 fzelp1 13008 . . . . . . . . . . . . . . . . . . 19 (𝑁 ∈ (0...𝑁) → 𝑁 ∈ (0...(𝑁 + 1))) 98 fzosplit 13119 . . . . . . . . . . . . . . . . . . 19 (𝑁 ∈ (0...(𝑁 + 1)) → (0..^(𝑁 + 1)) = ((0..^𝑁) ∪ (𝑁..^(𝑁 + 1)))) 9996, 97, 983syl 18 . . . . . . . . . . . . . . . . . 18 (𝑁 ∈ ℕ0 → (0..^(𝑁 + 1)) = ((0..^𝑁) ∪ (𝑁..^(𝑁 + 1)))) 100 nn0z 12044 . . . . . . . . . . . . . . . . . . . 20 (𝑁 ∈ ℕ0𝑁 ∈ ℤ) 101 fzosn 13157 . . . . . . . . . . . . . . . . . . . 20 (𝑁 ∈ ℤ → (𝑁..^(𝑁 + 1)) = {𝑁}) 102100, 101syl 17 . . . . . . . . . . . . . . . . . . 19 (𝑁 ∈ ℕ0 → (𝑁..^(𝑁 + 1)) = {𝑁}) 103102uneq2d 4068 . . . . . . . . . . . . . . . . . 18 (𝑁 ∈ ℕ0 → ((0..^𝑁) ∪ (𝑁..^(𝑁 + 1))) = ((0..^𝑁) ∪ {𝑁})) 10499, 103eqtrd 2793 . . . . . . . . . . . . . . . . 17 (𝑁 ∈ ℕ0 → (0..^(𝑁 + 1)) = ((0..^𝑁) ∪ {𝑁})) 105104ad2antrl 727 . . . . . . . . . . . . . . . 16 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (0..^(𝑁 + 1)) = ((0..^𝑁) ∪ {𝑁})) 106105raleqdv 3329 . . . . . . . . . . . . . . 15 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (∀𝑖 ∈ (0..^(𝑁 + 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ ∀𝑖 ∈ ((0..^𝑁) ∪ {𝑁}){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸)) 10793, 106mpbird 260 . . . . . . . . . . . . . 14 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ∀𝑖 ∈ (0..^(𝑁 + 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) 108 ccatlen 13974 . . . . . . . . . . . . . . . . . . 19 ((𝑇 ∈ Word 𝑉 ∧ ⟨“𝑆”⟩ ∈ Word 𝑉) → (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((♯‘𝑇) + (♯‘⟨“𝑆”⟩))) 1095, 20, 108syl2an 598 . . . . . . . . . . . . . . . . . 18 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((♯‘𝑇) + (♯‘⟨“𝑆”⟩))) 110109oveq1d 7165 . . . . . . . . . . . . . . . . 17 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1) = (((♯‘𝑇) + (♯‘⟨“𝑆”⟩)) − 1)) 111 simpl2 1189 . . . . . . . . . . . . . . . . . . 19 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (♯‘𝑇) = (𝑁 + 1)) 112 s1len 14007 . . . . . . . . . . . . . . . . . . . 20 (♯‘⟨“𝑆”⟩) = 1 113112a1i 11 . . . . . . . . . . . . . . . . . . 19 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (♯‘⟨“𝑆”⟩) = 1) 114111, 113oveq12d 7168 . . . . . . . . . . . . . . . . . 18 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ((♯‘𝑇) + (♯‘⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)) 115114oveq1d 7165 . . . . . . . . . . . . . . . . 17 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (((♯‘𝑇) + (♯‘⟨“𝑆”⟩)) − 1) = (((𝑁 + 1) + 1) − 1)) 116 peano2nn0 11974 . . . . . . . . . . . . . . . . . . . 20 (𝑁 ∈ ℕ0 → (𝑁 + 1) ∈ ℕ0) 117116nn0cnd 11996 . . . . . . . . . . . . . . . . . . 19 (𝑁 ∈ ℕ0 → (𝑁 + 1) ∈ ℂ) 118117, 55pncand 11036 . . . . . . . . . . . . . . . . . 18 (𝑁 ∈ ℕ0 → (((𝑁 + 1) + 1) − 1) = (𝑁 + 1)) 119118ad2antrl 727 . . . . . . . . . . . . . . . . 17 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (((𝑁 + 1) + 1) − 1) = (𝑁 + 1)) 120110, 115, 1193eqtrd 2797 . . . . . . . . . . . . . . . 16 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1) = (𝑁 + 1)) 121120oveq2d 7166 . . . . . . . . . . . . . . 15 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)) = (0..^(𝑁 + 1))) 122121raleqdv 3329 . . . . . . . . . . . . . 14 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸 ↔ ∀𝑖 ∈ (0..^(𝑁 + 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸)) 123107, 122mpbird 260 . . . . . . . . . . . . 13 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) 12418, 22, 1233jca 1125 . . . . . . . . . . . 12 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → ((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸)) 125109, 114eqtrd 2793 . . . . . . . . . . . 12 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)) 126124, 125jca 515 . . . . . . . . . . 11 (((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) ∧ (𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸))) → (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1))) 127126ex 416 . . . . . . . . . 10 ((𝑇 ∈ Word 𝑉 ∧ (♯‘𝑇) = (𝑁 + 1) ∧ ∀𝑖 ∈ (0..^𝑁){(𝑇𝑖), (𝑇‘(𝑖 + 1))} ∈ 𝐸) → ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 1284, 127syl 17 . . . . . . . . 9 (𝑇 ∈ (𝑁 WWalksN 𝐺) → ((𝑁 ∈ ℕ0 ∧ (𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸)) → (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 129128expd 419 . . . . . . . 8 (𝑇 ∈ (𝑁 WWalksN 𝐺) → (𝑁 ∈ ℕ0 → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1))))) 130129impcom 411 . . . . . . 7 ((𝑁 ∈ ℕ0𝑇 ∈ (𝑁 WWalksN 𝐺)) → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 131130adantll 713 . . . . . 6 (((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0) ∧ 𝑇 ∈ (𝑁 WWalksN 𝐺)) → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 132 iswwlksn 27723 . . . . . . . . . 10 ((𝑁 + 1) ∈ ℕ0 → ((𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺) ↔ ((𝑇 ++ ⟨“𝑆”⟩) ∈ (WWalks‘𝐺) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 133116, 132syl 17 . . . . . . . . 9 (𝑁 ∈ ℕ0 → ((𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺) ↔ ((𝑇 ++ ⟨“𝑆”⟩) ∈ (WWalks‘𝐺) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 134133adantl 485 . . . . . . . 8 ((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0) → ((𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺) ↔ ((𝑇 ++ ⟨“𝑆”⟩) ∈ (WWalks‘𝐺) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 1351, 3iswwlks 27721 . . . . . . . . 9 ((𝑇 ++ ⟨“𝑆”⟩) ∈ (WWalks‘𝐺) ↔ ((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸)) 136135anbi1i 626 . . . . . . . 8 (((𝑇 ++ ⟨“𝑆”⟩) ∈ (WWalks‘𝐺) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)) ↔ (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1))) 137134, 136bitrdi 290 . . . . . . 7 ((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0) → ((𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺) ↔ (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 138137adantr 484 . . . . . 6 (((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0) ∧ 𝑇 ∈ (𝑁 WWalksN 𝐺)) → ((𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺) ↔ (((𝑇 ++ ⟨“𝑆”⟩) ≠ ∅ ∧ (𝑇 ++ ⟨“𝑆”⟩) ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((♯‘(𝑇 ++ ⟨“𝑆”⟩)) − 1)){((𝑇 ++ ⟨“𝑆”⟩)‘𝑖), ((𝑇 ++ ⟨“𝑆”⟩)‘(𝑖 + 1))} ∈ 𝐸) ∧ (♯‘(𝑇 ++ ⟨“𝑆”⟩)) = ((𝑁 + 1) + 1)))) 139131, 138sylibrd 262 . . . . 5 (((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0) ∧ 𝑇 ∈ (𝑁 WWalksN 𝐺)) → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺))) 140139ex 416 . . . 4 ((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0) → (𝑇 ∈ (𝑁 WWalksN 𝐺) → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺)))) 1411403adant3 1129 . . 3 ((𝐺 ∈ V ∧ 𝑁 ∈ ℕ0𝑇 ∈ Word 𝑉) → (𝑇 ∈ (𝑁 WWalksN 𝐺) → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺)))) 1422, 141mpcom 38 . 2 (𝑇 ∈ (𝑁 WWalksN 𝐺) → ((𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺))) 1431423impib 1113 1 ((𝑇 ∈ (𝑁 WWalksN 𝐺) ∧ 𝑆𝑉 ∧ {(lastS‘𝑇), 𝑆} ∈ 𝐸) → (𝑇 ++ ⟨“𝑆”⟩) ∈ ((𝑁 + 1) WWalksN 𝐺)) Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399   ∧ w3a 1084   = wceq 1538   ∈ wcel 2111   ≠ wne 2951  ∀wral 3070  Vcvv 3409   ∪ cun 3856   ⊆ wss 3858  ∅c0 4225  {csn 4522  {cpr 4524  ⟨cop 4528  ‘cfv 6335  (class class class)co 7150  0cc0 10575  1c1 10576   + caddc 10578   − cmin 10908  ℕ0cn0 11934  ℤcz 12020  ℤ≥cuz 12282  ...cfz 12939  ..^cfzo 13082  ♯chash 13740  Word cword 13913  lastSclsw 13961   ++ cconcat 13969  ⟨“cs1 13996  Vtxcvtx 26888  Edgcedg 26939  WWalkscwwlks 27710   WWalksN cwwlksn 27711 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2729  ax-rep 5156  ax-sep 5169  ax-nul 5176  ax-pow 5234  ax-pr 5298  ax-un 7459  ax-cnex 10631  ax-resscn 10632  ax-1cn 10633  ax-icn 10634  ax-addcl 10635  ax-addrcl 10636  ax-mulcl 10637  ax-mulrcl 10638  ax-mulcom 10639  ax-addass 10640  ax-mulass 10641  ax-distr 10642  ax-i2m1 10643  ax-1ne0 10644  ax-1rid 10645  ax-rnegex 10646  ax-rrecex 10647  ax-cnre 10648  ax-pre-lttri 10649  ax-pre-lttrn 10650  ax-pre-ltadd 10651  ax-pre-mulgt0 10652 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3or 1085  df-3an 1086  df-tru 1541  df-fal 1551  df-ex 1782  df-nf 1786  df-sb 2070  df-mo 2557  df-eu 2588  df-clab 2736  df-cleq 2750  df-clel 2830  df-nfc 2901  df-ne 2952  df-nel 3056  df-ral 3075  df-rex 3076  df-reu 3077  df-rab 3079  df-v 3411  df-sbc 3697  df-csb 3806  df-dif 3861  df-un 3863  df-in 3865  df-ss 3875  df-pss 3877  df-nul 4226  df-if 4421  df-pw 4496  df-sn 4523  df-pr 4525  df-tp 4527  df-op 4529  df-uni 4799  df-int 4839  df-iun 4885  df-br 5033  df-opab 5095  df-mpt 5113  df-tr 5139  df-id 5430  df-eprel 5435  df-po 5443  df-so 5444  df-fr 5483  df-we 5485  df-xp 5530  df-rel 5531  df-cnv 5532  df-co 5533  df-dm 5534  df-rn 5535  df-res 5536  df-ima 5537  df-pred 6126  df-ord 6172  df-on 6173  df-lim 6174  df-suc 6175  df-iota 6294  df-fun 6337  df-fn 6338  df-f 6339  df-f1 6340  df-fo 6341  df-f1o 6342  df-fv 6343  df-riota 7108  df-ov 7153  df-oprab 7154  df-mpo 7155  df-om 7580  df-1st 7693  df-2nd 7694  df-wrecs 7957  df-recs 8018  df-rdg 8056  df-1o 8112  df-er 8299  df-map 8418  df-en 8528  df-dom 8529  df-sdom 8530  df-fin 8531  df-card 9401  df-pnf 10715  df-mnf 10716  df-xr 10717  df-ltxr 10718  df-le 10719  df-sub 10910  df-neg 10911  df-nn 11675  df-n0 11935  df-z 12021  df-uz 12283  df-fz 12940  df-fzo 13083  df-hash 13741  df-word 13914  df-lsw 13962  df-concat 13970  df-s1 13997  df-wwlks 27715  df-wwlksn 27716 This theorem is referenced by:  wwlksnextbi  27779  wwlksnextsurj  27785 Copyright terms: Public domain W3C validator

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# Properties Degree 2 Conductor $2 \cdot 3$ Sign $1$ Motivic weight 11 Primitive yes Self-dual yes Analytic rank 0 # Origins ## Dirichlet series L(s)  = 1 − 32·2-s − 243·3-s + 1.02e3·4-s + 5.76e3·5-s + 7.77e3·6-s + 7.24e4·7-s − 3.27e4·8-s + 5.90e4·9-s − 1.84e5·10-s − 4.08e5·11-s − 2.48e5·12-s + 1.36e6·13-s − 2.31e6·14-s − 1.40e6·15-s + 1.04e6·16-s + 5.42e6·17-s − 1.88e6·18-s + 1.51e7·19-s + 5.90e6·20-s − 1.76e7·21-s + 1.30e7·22-s − 5.21e7·23-s + 7.96e6·24-s − 1.55e7·25-s − 4.37e7·26-s − 1.43e7·27-s + 7.42e7·28-s + ⋯ L(s)  = 1 − 0.707·2-s − 0.577·3-s + 1/2·4-s + 0.825·5-s + 0.408·6-s + 1.62·7-s − 0.353·8-s + 1/3·9-s − 0.583·10-s − 0.765·11-s − 0.288·12-s + 1.02·13-s − 1.15·14-s − 0.476·15-s + 1/4·16-s + 0.926·17-s − 0.235·18-s + 1.40·19-s + 0.412·20-s − 0.940·21-s + 0.541·22-s − 1.69·23-s + 0.204·24-s − 0.319·25-s − 0.722·26-s − 0.192·27-s + 0.814·28-s + ⋯ ## Functional equation \begin{aligned} \Lambda(s)=\mathstrut & 6 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & \, \Lambda(12-s) \end{aligned} \begin{aligned} \Lambda(s)=\mathstrut & 6 ^{s/2} \, \Gamma_{\C}(s+11/2) \, L(s)\cr =\mathstrut & \, \Lambda(1-s) \end{aligned} ## Invariants $$d$$ = $$2$$ $$N$$ = $$6$$    =    $$2 \cdot 3$$ $$\varepsilon$$ = $1$ motivic weight = $$11$$ character : $\chi_{6} (1, \cdot )$ primitive : yes self-dual : yes analytic rank = 0 Selberg data = $(2,\ 6,\ (\ :11/2),\ 1)$ $L(6)$ $\approx$ $1.20558$ $L(\frac12)$ $\approx$ $1.20558$ $L(\frac{13}{2})$ not available $L(1)$ not available ## Euler product $L(s) = \prod_{p \text{ prime}} F_p(p^{-s})^{-1}$ where, for $p \notin \{2,\;3\}$, $$F_p$$ is a polynomial of degree 2. If $p \in \{2,\;3\}$, then $F_p$ is a polynomial of degree at most 1. $p$$F_p$ bad2 $$1 + p^{5} T$$ 3 $$1 + p^{5} T$$ good5 $$1 - 5766 T + p^{11} T^{2}$$ 7 $$1 - 10352 p T + p^{11} T^{2}$$ 11 $$1 + 408948 T + p^{11} T^{2}$$ 13 $$1 - 1367558 T + p^{11} T^{2}$$ 17 $$1 - 5422914 T + p^{11} T^{2}$$ 19 $$1 - 15166100 T + p^{11} T^{2}$$ 23 $$1 + 52194072 T + p^{11} T^{2}$$ 29 $$1 - 118581150 T + p^{11} T^{2}$$ 31 $$1 + 57652408 T + p^{11} T^{2}$$ 37 $$1 + 375985186 T + p^{11} T^{2}$$ 41 $$1 - 856316202 T + p^{11} T^{2}$$ 43 $$1 + 1245189172 T + p^{11} T^{2}$$ 47 $$1 + 1306762656 T + p^{11} T^{2}$$ 53 $$1 - 409556358 T + p^{11} T^{2}$$ 59 $$1 + 48862140 p T + p^{11} T^{2}$$ 61 $$1 - 5731767302 T + p^{11} T^{2}$$ 67 $$1 - 3893272244 T + p^{11} T^{2}$$ 71 $$1 + 9075890088 T + p^{11} T^{2}$$ 73 $$1 + 15571822822 T + p^{11} T^{2}$$ 79 $$1 + 30196762600 T + p^{11} T^{2}$$ 83 $$1 - 23135252628 T + p^{11} T^{2}$$ 89 $$1 + 25614819990 T + p^{11} T^{2}$$ 97 $$1 + 61937553406 T + p^{11} T^{2}$$ show less \begin{aligned} L(s) = \prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1} \end{aligned} ## Imaginary part of the first few zeros on the critical line −20.61408278124063655145673829357, −18.22185823557610957306388607067, −17.74481053097977942632133146030, −16.04683807740066808067323140928, −14.01498326645545396945292831922, −11.66318179051930405686989550799, −10.19821875859845610581844279656, −8.005005940346513327952156957974, −5.55489639023984419351049688700, −1.48555012435943654433178944737, 1.48555012435943654433178944737, 5.55489639023984419351049688700, 8.005005940346513327952156957974, 10.19821875859845610581844279656, 11.66318179051930405686989550799, 14.01498326645545396945292831922, 16.04683807740066808067323140928, 17.74481053097977942632133146030, 18.22185823557610957306388607067, 20.61408278124063655145673829357

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# What is a quasi-isomorphism of two crossed modules Could you tell me how are two crossed modules quasi-isomorphic. And I have known a result: Let $\mu: M \rightarrow G$ and $\mu': M' \rightarrow G'$ are isomorphic, then the integral homology of them are the same. How to prove this result? Could you show me related materials? Thanks, - Please see the FAQ, in particular what sort of questions are appropriate for MO. – David Roberts Apr 2 '12 at 0:40 In particular, I believe this question would be better placed at math.stackexchange.com (and please read their FAQ before posting there) – David Roberts Apr 2 '12 at 2:50 I'm sorry. It was the first time I posted my question. I will see FAQ. Thanks. – Sapus Apr 4 '12 at 16:30 Even if David Roberts is right, I'll give an answer; more details my be found in the book R. Brown, P.J. Higgins, R. Sivera, Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids, EMS Tracts in Mathematics Vol. 15, 703 pages. (August 2011). Any crossed module $\mu: M \to P$ has a classifying space $X=B(\mu)$ whose homotopy groups are 0 above dimension 2 and in dimensions 1 and 2 are respectively Coker $\mu$ and Ker $\mu$. A morphism $f$ of crossed modules is a weak equivalence iff the induced map of classifying spaces is a weak equivalence. The homology of a crossed module is defined as the homology of its classifying space. The result asked for is easily obtained by passing to universal covers. {Ellis, Graham J.}, TITLE = {Homology of {$2$}-types}, JOURNAL = {J. London Math. Soc. (2)}, VOLUME = {46}, YEAR = {1992}, NUMBER = {1}, PAGES = {1--27},

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# Could a nomadic group build a portable library on wheels pulled by horses? A fictional nomadic group lives in vardos much like the Romani people. Could such a nomadic group build a giant library on wheels like a giant vardo capable of carrying 50,000 books? This giant library must be pulled by horses. Create a design of a library on wheels capable of carrying 50,000 books that can be pulled by horses. How many horses would be necessary to pull it? Assume that one book is 25 cm tall, 15 cm wide, and 5 cm. thick. and weighs 5 pounds. They are displayed in bookshelves containing 140 books each and 1.5 kilograms each (without books). • 50,000 books? How thick are these books? If each book is even half a pound for little softcover books, that's 25000lbs. You can easily double that if your books are hardcovered. And that's not even with the weight of the cart. For reference, the capacity of a 53 foot semi-trailer is only 13,500lbs, is supported by multiple steel axles, and hauled my a huge engine on paved roads by a 600 horse power engine. Jul 25 '20 at 2:56 • Interestingly, the world record is two draft horses pulling 100,000lbs. But that's on a paved road over a short distance. It seems more common though that a horse (probably weighing around 400kg) can pull about three times its own weight on a paved road, with 1.5x being a more sustainable number all day. Jul 25 '20 at 2:57 • Explain why you would have a single library vehicle, and not a fleet of normal-sized wagons, each carrying a few hundred books. Jul 25 '20 at 5:02 • 250,000 pounds, just in books! The wagon weighs something. The shelves weigh something. Without modern plywood, the shelves would weigh a lot. And these shelves would need something to keep the books from falling free. There is a reason that libraries don't have wheels. Jul 25 '20 at 13:02 • Stay neasr a river and use a barge! That way you can include a reading room (essential during bad weather) and cafe in your library Here's a useful comparison of loading of various forms of transport. itstillruns.com/… Jul 25 '20 at 17:14 The best way to do it would be to split among multiple vehicles, each of reasonable size and weight for a horse team to pull all day. That does not change the total number of horses, but it does keep the teams to a reasonable size. For example, one could replace a large vehicle with a 125-mule team with 16 vehicles, each with 8 mules. The vehicles could be designed with one side dropping down to form a stage The books would be on wheeled bookcases that are packed densely during travel, but pulled out onto the stage for access and display at stops. At a stop, the vehicles would form a fixed pattern, with the catalog vehicles at the entrance, and maps to help people find the books they wanted. • Now, I think the frame challenge is this: How big must a nomadic tribe be in order to give away 16 vehicles just for their library? Nomads normally own few things, they aim for the lower limit of what normal people own, just to stay mobile. So it would be contradictory to most nomad's culture to have a "luxury" like 16 vehicles full of books with their tribe. But I like the idea. Jul 25 '20 at 6:30 • @Anderas Maybe the library is bigger than any one tribe. It may go on a great circuit, spending a season or two with a tribe. A tribe may gain face and demonstrate its importance by being able to support the library for several months, feeding its horses etc. Jul 25 '20 at 7:04 • I think it would be easier for the people to come to the library. Most advanced nomadic tribes did have villages for specialized purposes. Jul 25 '20 at 13:04 • @NomadMaker I'm assuming a very large area serviced by the library. One of its functions is to collect young people who want to read, and take them along. That spreads genes around. Jul 25 '20 at 15:50 • Dave Duncan had something very similar going on in his novel West of January. An entire town called "Heaven," including many centuries' worth of written records in its library, was perpetually on the move, using numerous "snortoises" (large beasts of burden native to the planet Vernier) to keep everything chugging along. One "day" on that planet was about 200 years, and "Heaven" kept moving slowly westward so that it was perpetually located in the area where the sun was just peeping over the horizon at "dawn." Worth reading. Jul 25 '20 at 19:54 It would depend on the tribe and their capabilities. However I think your scale is simply absurd. 50,000 books is unheard of in such a context. In fact a whole library might not even hold that. You have to understand the whole context of books in pre-modern times. (Now I'm aware of different practices, and certainly medieval Europe did not hold many books compared to "Dar Al Hikma" in Baghdad and so on.) Knowledge would (mostly) not be passed down from books. Only highly abstract ideas and art are given such privileges. And you have to take into consideration that in order to develop any notable art and philosophy, and start translating other works (and all the other prerequisites to large libraries) you have to be a stable state with a lot of free time and a strong inclination to read and write and translate... etc. People did not just think: I'll go down to the local ancient temple and write down a full description of the thing just for the fun of it. These prerequisites are all missing from a nomadic tribe, and trying to haul around that many books is absurd. Simple logistics. You carry the most important stuff only. So why carry around a crappy copy of a second rate "novel"? It's just too much trouble. Now back to the way people used to do things. A master taught an apprentice who in turn would pass down that knowledge. Oddly enough, even today we highly value technical hands-on teaching, and certainly a school that teaches you the theoretical principles of making cars is nothing like a school that teaches those principles and also puts you in a workshop and tells you to build that darn thing. So more books doesn't necessarily equal a more educated or knowledgable people. An average first-world citizen with access to millions of books is no more capable of making a simple radio than a citizen of Athena in 100 BC. And we also know that certain ancient tribes (like Arabic nomadic tribes) had excellent memory. And when I say excellent, I mean the ability to perfectly recite a 100-verse poem with absolute accuracy down to smallest element. So people did have better memories when it came to the things that matter like art or history. Also I do remember something similar in Europe. I think Caesar killed this group of Shamans or something that had all the history of the tribe and then all their secrets were lost to us, but that was a long time ago. Lastly the available pool of books and the entire viewpoint on "publishing" was nothing like today. Do you think that people of the ancient world had thousands of choices and that you had hundreds of genres and all that modern stuff? People only bothered with important and "good" (read: useful) works. You didn't sit down and write a crappy YA novel. It just takes too much time. And even if you finish one copy then you have to spend even more time to write down a second copy. So people focused on the important books. Now most of this is general rules of thumb, because we only need the larger context here to come to the conclusion that you need to change something. Anyway I really think you need to scale it down. Something like 1000 books seems a lot more manageable. And this is all in regard to the given information. If you did figure out all those things and have strong and reasonable reasons why this is so then I apologize. I hate to be the guy that just comes around to argue a point that you figured out already. But if not, better change something. • Screw a radio, even basic stuff like a well designed and properly made plough is beyond most first-world citizens. Even a vast majority of those anomalous individuals like myself who could tell you how to make one couldn't do it ourselves. Also, good point about scale, everyone thinks of the big stuff like the Libraries of Alexandria or Pergamum, but those were assembled over centuries and were only able to get that big due to being in major cultural centers. Jul 26 '20 at 1:12 OK, so you want a horsecart for 125 tons of books. The shelves are lost in the rounding, which suggests that your figures are too low. But for a ballpark, I'm even going to ignore the difference between short tons, long tons, and metric tons ... A large historical example were 20-mule teams hauling two connected wagons with 36 tons total. This suggests roughly 10 mules (or 9 mules and 1 horse) per 20 tons of load. Being optimistic, say the wagon and shelves are as heavy again as the books. That means a 125-mule team ... • The example used 19th century wagons and wheels with metal rims. They were traveling on a specific desert route. Wooden wheels in a temperate climate could get bogged down in mud or just soft soil. • Coordinating 125 animals sounds like a challenge, too. How much weight does the rope or chain add? But the Romans managed to transport several large obelisks from Egypt to Rome. The Lateran Obelisk was 450 tons before a part broke off. So (a) loads in your weight class can be transported on the ground and (b) no, it isn't very practical. • The Romans would not have had to move the Lateran Obelisk far over land. The Circus Maximus, where they erected it, is quite close to the Tiber. Jul 25 '20 at 4:29 If your tribe really wants to move a library of 50,000 books around then their best bet is to create miniature books. Miniature versions of religious books have been around for several centuries; they were both portable and concealable (at a time when possessing the wrong book could get you killed). Obviously they cost more to produce than normal size books, and need a magnifier to read, but for a nomadic people this would be a better trade-off than full-size books. If a typical book in this library were around 5cm x 3cm x 2cm it would weigh about 40 grams. So 50,000 books would weigh about 2 tons. Much more manageable. ## Not with a single wagon no For scale this is what half a room with 50,000 books looks like. This is what a library with 200,000 books looks like, Trinity College Library Ireland, it is 20 rows deep, you want a quarter of that so the first 5 rows. A single ox can pull a cart weighing about 5000lbs over uneven ground all day long. A single draft horse can manage around 2000lbs Assuming the books are packed as tightly as possible and not set up for reading, you are hauling about ~50000lbs at the lowest estimate. that is ~25000lbs of books (assuming the purposefully make the books as light and compact a possible) and ~25000lbs of wagon to support the books (since the wagons have to be water tight they end up being almost twice as heavy than a normal ore wagon, which weigh around 7800 lbs to haul 20000lbs. so you need 10 oxen or 25 horses. the problem is you get diminishing returns by adding more animals beyond a certain point. Once you get to around 20 animals the additional pulling power is negligible. Note will also need well made iron chains and iron wheel fittings for this much weight, which may be difficult for a nomadic people to make. Large teams are also extremely unwieldy, difficult to steer, and take a long time to harness and unharness. So you can't manage it with a single wagon with horses. You might be able to with two wagons with horses (or one wagon of oxen) assuming it is not a library but packed books, they have to be unpacked when you reach a destination, likely taking several days. But you can't build a wooden wagon that strong so you have to split it up into smaller loads. the limit on wooden wagons is around 20000lbs of cargo, and that is with solid oak, lots of iron, and hard packed roads. now if you are going to split it up anyway just split it up into easy to handle 4 horse teams that gives you around 7 wagons. But you are going to need at least 4-5 wagons to haul the food needed to feed so many horses, or you are going to need to do what Dial did for hauling borax and set up feed/water stations every few miles to feed the teams. A horse team will eat roughly a tenth the weight of what they are hauling in feed per day. If you want an actually library with shelves it is impossible, you are adding 2-3 times as much weight if not more per wagon, which is too much for a single wagon and adds up to almost a dozen more wagons split up (which also require feeding). • Why do you need to carry horse food with you, rather than allowing the horses to graze? Jul 26 '20 at 15:00 • @PatriciaShanahan A horse can't graze and walk all day, horses are not ruminants they actually need to spend a lot of time eating 4-6 hours a day under ideal conditions, if you let the horses graze your not traveling far, and you will need a lot of land for that many horses to graze, and you need land and seasons suitable for grazing. That is why most working horses had feed. – John Jul 26 '20 at 17:35 • How much food do you think a Mongol army took with them? Jul 26 '20 at 17:43 • @PatriciaShanahan Often by taking the feed from those they conquered, feeding their horses was the single biggest problem the mongol armies faced, once you have enough animals in the same place the local land can't support them. This is also the reason mongols only went to war during a small portion of the the year, the times when grass was plentiful, literally disbanding for the rest of the year. It was only only way they could keep their animals fed. but the library has to work all year long. – John Jul 26 '20 at 20:24 Here's a bit of a pun: don't reinvent the wheel. :D In the tenth century… the Grand Vizier of Persia, Abdul Kassem Ismael, in order not to part with his collection of 117,000 volumes when traveling, had them carried by a caravan of four hundred camels trained to walk in alphabetical order. http://www.wondersandmarvels.com/2014/04/book-hoarding-10th-century-style.html Happy discovering. The answer has to be yes. OP has been sufficiently vague that it can't be anything else: "library" might be recreational (for the people owning it or people they visit), or it might be cultural (holy books they daren't lose) or it might be critical to survival. Xe's also been vague about timescale and speed of movement. Interpreting the question loosely, it would be entirely plausible for nomads to be in possession of a university-grade library which stayed three years in a town and then moved on, with the objective being to visit every century or so in order to make sure that the trainee doctors and technologists had access to original works rather than relying on well-thumbed transcripts. During each migration they could rely on help from at least their destination, and possibly from the people they were leaving who would obviously want them to come back eventually. And OP's question doesn't specify how many wheels, whether it's a single cart or a convoy, and whether books have to be immediately accessible or only sought out by a trained cadre. Can they physically do this? Yes. The question would be WHY are they doing this. If they have a contract with some emperor to take his library to his summer home or a god has told them to take his library to a new location, then it’s easy-peasy, they get enough wagons and mules and away they go. Without a really good reason, they wouldn’t be doing this. And the good reason needs to be a clear profit or existential threat (aka telling a god to piss off is likely to result in a lot of lightening strikes and/or volcanic eruptions). Tradition or love of knowledge isn’t going to cut it. Even ignoring the weight and reason answers brought up by others working with horses is an inherently dirty business (personal experience speaking here). Moving however many books could be done but actually being able to use them as a library when camped would quickly ruin whatever value the materials had due to the simple inability to get clean enough. • After grooming horses or cleaning their pens, I can get clean enough to handle books in less than half an hour. They would need to have different sets of clothing for horse chores and library work. Jul 26 '20 at 15:06 • Try it when you're on the trail sometime, when water is in short supply and you need to conserve as much as you can. And remember, those book-handling-clothes have also been on the trail with you so are not hidden away nice and clean at home, instead you are walking around the dust cloud of a nomadic camp. Jul 26 '20 at 15:18 • Even an overnight stopping place for a convoy with dozens of draft horses needs to be near plenty of water - this is not just one or two riders and their horses. Jul 26 '20 at 15:23 What is a book? In antiquity, a single papyrus scroll or was an essay or part of a series that made up a book. A stone tablet was a book. China and Japan used paper scrolls, the Art of war was written on bamboo staves linked with wire and string. An essay of one avers and a reverse would, in the Alexandrian Library be considered one book! To minimize the amount needed to be reserved for books, make sure the books become an integral part of the caravan! The caravan itself is the library! I can store tons of text on the sides of all the carts and wagons, the fencing they carry for the cattle is written upon just as much as the inside and outside of tents and even the skin of their animals and members! No spot that is not decorated with one or another piece of poetry or essay. All the plates together might carry cookbooks and so on. But the numbers need to be cut down! An early medieval library in Europe was usually just about the size of a large cupboard, a prestigious one would be a whole room containing some hundred books. Thousands were exceptional places like Rome. Of course a nomadic group could build a library on wheels pulled by horses, even if you somehow restricted the group to "Romany" caravans as in your picture.

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# The product of two lognormal random variables Let $X_1$ and $X_2$ be two normal random variables. Write $X_1\sim N(\mu_1, \sigma^2_1)$ and $X_2\sim N(\mu_2, \sigma^2_2)$, to fix ideas. Consider the corresponding log-normal random variables: $Z_1 = \exp(X_1)$, $Z_2 = \exp(X_2)$. Question: what is the distribution of the product of the two random variables, i.e., the distribution of $Z_1Z_2$? If the normal random variables $X_1, X_2$ are independent, or they have a bivariate normal distribution, the answer is simple: we have $Z_1Z_2 = \exp(X_1+X_2)$ with the sum $X_1+X_2$ normal, hence the product $Z_1Z_2$ is still lognormal. But suppose that $X_1, X_2$ are generally $not$ independent, say with correlation $\rho$. What can we say about the distribution of $Z_1Z_2$? • This might be useful: stats.stackexchange.com/questions/19948/… May 15, 2016 at 17:20 • I doubt it though.. basically this question asks "if the marginals are normally distributed, can we say anything about their joint distribution?" And I don't think we can say much in general – Ant May 15, 2016 at 19:27 • $Z_1 Z_2 = \exp(X_1 + X_2)$ in general, so your real question is whether $X_1+X_2$ is normal (which it will be if $X_1, X_2$ are bivariate normal with correlation $\rho$) May 15, 2016 at 22:12 • If you don't have bivariate normality, merely specifying the correlation and the margins is not sufficient to pin down the bivariate distribution. May 16, 2016 at 1:31 Using Dilips answer here, if $X$ and $Y$ are bi-variate normal and $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$ and the correlation between $X$ and $Y$ is $\rho$. Then $$Cov(X,Y) = \rho \sigma_1 \sigma_2,$$ $$X + Y \sim N(\mu_1 + \mu_2, \sigma^2_1 + \sigma^2_2 + 2\rho\sigma_1 \sigma_2).$$ Thus $Z_1Z_2$ will also be a lognormal distribution with parameters $\mu_1 + \mu_2$ and $\sigma^2_1 + \sigma^2_2 + 2\rho\sigma_1 \sigma_2$. • Add some qualification? This is true if the distribution of (X,Y) is bi-variate normal, but that the marginal distribution of X is normal and the marginal distribution of Y is normal does not imply the joint distribution is bi-variate normal... stats.stackexchange.com/questions/30159/… May 15, 2016 at 17:41 • @MatthewGunn Interesting. I was blissfully unaware of this. As it stands, my answer does not address the question completely. I will wait a couple of hours before deleting it. Thanks. May 15, 2016 at 17:44 • Do you have alot of data from this distribution? In such case you can plot it, or otherwise test if bivariate normal is a good distribution approximation. Otherwise, maybe you could estimate a copula and go from there! May 15, 2016 at 18:36 • This is quite obvious, but you're missing the point: what happens if you don't know whether the two marginals have a bivariate distribution? May 16, 2016 at 10:53 • Thanks! This amazing answer saved my day! May 26, 2019 at 15:45

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DID YOU KNOW: Seamlessly assign resources as digital activities Learn how in 5 minutes with a tutorial resource. Try it Now # Fractions on a Number Line Bingo Game {3.NF.2, 3.NF.2A, 3.NF.2B} Games 4 Gains 24.2k Followers Grade Levels 3rd - 5th, Homeschool Subjects Standards Resource Type Formats Included • PDF Pages 34 pages Games 4 Gains 24.2k Followers ### Description Make the skill of identifying fractions on a number line fun and exciting with this Fractions on a Number Line Bingo Game. Included are 32 different pre-made Bingo cards, which require students to identify proper fractions on a number line between 0 and 1. Students will practice halves, thirds, fourths, fifths, sixths, and eighths with this game. Included with this Fractions on a Number Line Bingo Game: • Instructions for use • 32 pre-made Bingo cards • Teacher calling card UPDATE 6/1/16: To save time, this set has been updated to include 32 PRE-MADE Bingo cards. In the previous version, students had to cut and paste images of number lines to make this own cards. Do you want to get TWENTY-ONE Bingo games for only \$10.50? Math Bingo Through The Year for 3rd Grade contains 21 different Bingo games for practicing many different 3rd grade Common Core standards. It even includes this BINGO game for fractions on a number line! ******* You might also like these other fractions games by Games 4 Gains: 3rd Grade Fractions on a Number Line Digital Practice Fractions on a Number Line Board Game Equivalent Fractions Board Game Equivalent Fractions 'Clip and Flip' Cards Comparing Fractions Board Game Fraction Models Card Games ****** Customer Tips: Click on the ★ above to follow our store. CLICK HERE to sign up for exclusive sales and freebies. You'll also receive 12 FREE FRACTION BUMP GAMES in your email just for signing up. CLICK HERE to visit our website for more free resources and teaching ideas! Leave feedback to earn credit points to save money on future products! ******* This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Total Pages 34 pages Answer Key N/A Teaching Duration N/A Report this Resource to TpT Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines. ### Standards to see state-specific standards (only available in the US). Represent a fraction 𝘢/𝘣 on a number line diagram by marking off 𝘢 lengths 1/𝘣 from 0. Recognize that the resulting interval has size 𝘢/𝘣 and that its endpoint locates the number 𝘢/𝘣 on the number line. Represent a fraction 1/𝘣 on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into 𝘣 equal parts. Recognize that each part has size 1/𝘣 and that the endpoint of the part based at 0 locates the number 1/𝘣 on the number line. Understand a fraction as a number on the number line; represent fractions on a number line diagram. ### Questions & Answers 24.2k Followers Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. More About Us

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Welcome Guest! To enable all features please Login or Register. Error 5 Pages«<345 Previous Topic Next Topic Razonar #81 Posted : 04 May 2022 07:20:04(UTC) Rank: Advanced MemberGroups: Registered Joined: 28/08/2014(UTC)Posts: 998Was thanked: 548 time(s) in 357 post(s) Originally Posted by: Jean Giraud Originally Posted by: Razonar a PT100 (which should have a resistance of 100 ohms at 0 Celsius)Pt100 is simply a designation from EIT90 with Ω as given 1 @ 0.01 °C.The instrument maintenance crew calibrates the XTR's reading fromexpensive secondary standard lab equipment accordingly to BIPM.Platinum wire is not infinitely ductible ... 100 times smaller cross area ?Of same cross area now 100 times longer ?Don't worry, those things work fine for so long.Cheers ... Jean.From your SMath file ( https://en.smath.com/for...rce.ashx?a=43693&b=2 ):andFrom the NIST ( https://nvlpubs.nist.gov...tions/NIST.SP.250-91.pdf at page 77)There you can see that Wr is dimensionless, and can read how to define it as a quotient between resistance's. What you get, 0.59493791 and 1.77363368 are not fake ohms, as you type. Are the relation between the Platinum electric resistance and the reference value for the water at 0.01 Celsius. It is very practical because it is very easy to obtain temperatures of 0 C and 100 C with enough accuracy in the work area inside the plant, without having to disassemble the instrument and take it to the workshop workbench. It is PT100 because 100 ohms, otherwise it would be called PT1. You can google it. Actually there are also PT1000, whose nominal resistance is 1000 ohms at 0 CBest regards.Alvaro. uni #83 Posted : 04 May 2022 11:05:09(UTC) Rank: Advanced MemberGroups: Registered, Advanced MemberJoined: 10/11/2010(UTC)Posts: 1,324Was thanked: 1157 time(s) in 668 post(s) To be clear, I don't have permission to ban users. Until now this has not been necessary, except for the usual spammers. Yes, I can ask Andrey about it personally, but only in extreme cases. The current case is borderline. I don't have time to follow everything that happens on the forum. I could delete all posts that are not related to the topic of the forum, but this is inefficient. Thus, if you want to change something, then you will have to jointly develop proposals, coordinate them with Jean and report the result to Andrey. I think this is a more civilized way. Russia ☭ forever Viacheslav N. Mezentsev fedeghi #84 Posted : 04 May 2022 11:11:43(UTC) Rank: Advanced MemberGroups: Registered Joined: 14/09/2013(UTC)Posts: 87Was thanked: 22 time(s) in 15 post(s) Yes Uni, I understand and agree with your statement, I also think it is a waste of time to "regulate" this at an admin level.On the other hand, since it is unlikely that we can change Jean's mind and way of thinking, I will personally stop interacting and simply ignore what happens.Nevertheless, it should be noted that the forum will keep getting polluted and the "search" feature (which would be a very important feature of any technical forum...) is impacted. Jean Giraud #85 Posted : 04 May 2022 15:39:22(UTC) Rank: GuestGroups: Registered Joined: 04/07/2015(UTC)Posts: 6,132Was thanked: 896 time(s) in 724 post(s) You are both right, Smath Community is on the decline of Poisson Distribution.@ fedeghi 0.25% statistics, little chance of productive whatever collaboration.Take care ... Jean Giraud #86 Posted : 04 May 2022 16:32:21(UTC) Rank: GuestGroups: Registered Joined: 04/07/2015(UTC)Posts: 6,132Was thanked: 896 time(s) in 724 post(s) Originally Posted by: Razonar Actually there are also PT1000, whose nominal resistance is 1000 ohms at 0 CThere are even more SPRT [25.5 Ω] ... others fit in mother board.TP water 0.01 °CPlease, make the attached as you think it should be.Publish for the benefit of the Smath Community.Take care Alvaro ... Jean. Inst_Type Pt100 BIPM Copy.sm (397kb) downloaded 3 time(s).

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# [1]左曙光,潘健,吴旭东,等.考虑动圈偏心的电动振动台等效电磁力计算方法[J].西安交通大学学报,2020,54(08):132-139.[doi:10.7652/xjtuxb202008017] 　ZUO Shuguang,PAN Jian,WU Xudong,et al.A Calculation Method of Equivalent Electromagnetic Force for Electrodynamic Shakers Considering Moving Coil's Eccentricity[J].Journal of Xi'an Jiaotong University,2020,54(08):132-139.[doi:10.7652/xjtuxb202008017] 点击复制 考虑动圈偏心的电动振动台等效电磁力计算方法 分享到： 54 2020年第08期 132-139 2020-08-10 ## 文章信息/Info Title: A Calculation Method of Equivalent Electromagnetic Force for Electrodynamic Shakers Considering Moving Coil's Eccentricity 0253-987X(2020)08-0132-08 Author(s): School of Automotive, Tongji University, Shanghai 201804, China Keywords: TM153; TH87 DOI: 10.7652/xjtuxb202008017 A Abstract: A calculation method of equivalent electromagnetic force considering moving coil's eccentricity is proposed to reflect the influences of the moving coil's electromagnetic forces on the lateral vibration of the electrodynamic shaker under offset loading conditions. Five factors including the moving coil's current, vertical position, flipping eccentricity angle, radial translating eccentricity distance and direction are considered. The problem that it is difficult to apply the existing orthogonal test design method to the situation with a mutual constraint among multiple factors' values is solved by calculating the radial eccentricity distance limit and assuming a virtual eccentricity distance. The concentrated electromagnetic force samples equivalent to the distributed electromagnetic forces of the moving coil are obtained through an electromagnetic finite element simulation. A neural network method is used to fit the samples and the equivalent electromagnetic forces in full moving conditions are obtained. Simulation results show that the radial equivalent electromagnetic force mainly increases with the increase of flipping eccentricity angle, especially increases rapidly at the high current condition. The relative errors of the equivalent electromagnetic forces obtained by the proposed method are within 10%. The equivalent electromagnetic forces in full moving conditions can be effectively obtained in real time while meeting the engineering requirement of calculation error. The problem that the electrodynamic shaker's electromagnetic-structure bidirectional coupling simulation can not be carried out due to the slow speed of the finite element method and the exact analytical method to calculate the equivalent electromagnetic force is well solved. The proposed method can also be applied to the calculation of forces on other similar moving systems. ## 参考文献/References: [1] MACHADO M R, APPERT A, KHALIJ L. Spectral formulated modelling of an electrodynamic shaker [J]. Mechanics Research Communications, 2019, 97: 70-78. [2] WAIMER S, MANZATO S, PEETERS B, et al. Modelling and simulation of a closed-loop electrodynamic shaker and test structure model for spacecraft vibration testing [J]. Advances in Aircraft and Spacecraft Science, 2018, 5(2): 205-223. [3] HOFFAIT S, MARIN F, SIMON D, et al. Measured-based shaker model to virtually simulate vibration sine test [EB/OL]. [2019-09-16]. https:∥www. sciencedirect.com/science/article/pii/s2351988616300 057. [4] SEBASTIEN H, FREDERIC M, DANIEL S, et al. Virtual shaker testing at V2i: measured-based shaker model and industrial test case [C]∥27th International Conference on Noise and Vibration Engineering. Leuven, Belgium: KU Leuven, 2016: 1013-1026. [5] 孟繁莹. 大型电动振动台动力学分析与数值模拟研究 [D]. 北京: 北京工业大学, 2013: 31-32. [6] JONATHAN M, KRISTOF H. Two-port modeling and simulation of an electrodynamic shaker for virtual shaker testing applications [J]. Journal of Sound and Vibration, 2019, 460: 114835. [7] SARASWAT A, TIWARI N. Modeling and study of nonlinear effects in electrodynamic shakers [J]. Mechanical Systems and Signal Processing, 2017, 85: 162-176. [8] JONATHAN M, KRISTOF H. Virtual shaker modeling and simulation, parameters estimation of a high damped electrodynamic shaker [J]. International Journal of Mechanical Sciences, 2019, 151: 375-384. [9] 刘源, 董立珉, 孔宪仁, 等. 飞行器虚拟振动试验平台构建 [J]. 光学精密工程, 2013, 21(5): 1258-1264. LIU Yuan, DONG Limin, KONG Xianren, et al. Construction of virtual vibration testing platform for spacecraft [J]. Optics and Precision Engineering, 2013, 21(5): 1258-1264. [10] 左曙光, 黄荣奎, 冯朝阳, 等. 考虑非线性电磁分布力的虚拟电动振动系统建模 [J]. 振动与冲击, 2019, 38(2): 152-158. ZUO Shuguang, HUANG Rongkui, FENG Zhaoyang, et al. Construction of a virtual electric vibration system considering the nonlinearity of electromagnetic force [J]. Journal of Vibration and Shock, 2019, 38(2): 152-158. [11] WANG J, ZHU J G. A simple method for performance prediction of permanent magnet eddy current couplings using a new magnetic equivalent circuit model [J]. IEEE Transactions on Industrial Electronics, 2018, 65(3): 2487-2495. [12] 刘强, 赵勇, 曹建树, 等. 新型微框架磁悬浮飞轮用洛伦兹力磁轴承 [J]. 宇航学报, 2017, 38(5): 481-489. LIU Qiang, ZHAO Yong, CAO Jianshu, et al. Lorentz magnetic bearing for novel vernier gimballing magnetically suspended flywheel [J]. Journal of Astronautics, 2017, 38(5): 481-489. [13] 吴承喜. 基于电磁-结构耦合的电动振动系统建模与分析 [D]. 上海: 同济大学, 2018: 23-28. [14] 钟鸿敏, 左曙光, 吴旭东, 等. 电励磁爪极发电机气隙磁场与径向电磁力的解析计算模型 [J]. 电工技术学报, 2017, 32(7): 49-58. ZHONG Hongmin, ZUO Shuguang, WU Xudong, et al. Analytical calculation model of air-gap magnetic field and radial electromagnetic force of electric excitation claw-pole generator [J]. Transactions of China Electrotechnical Society, 2017, 32(7): 49-58. [15] JABBARI A. Analytical modeling of magnetic field distribution in multiphase H-type stator core permanent magnet flux switching machines [J]. Iranian Journal of Science and Technology: Transactions of Electrical Engineering, 2019, 43: 389-401. [16] JING L B, HUANG Z X, GONG J. Analytical calculation of magnetic field distribution in the consequent-pole bearingless PM motor with rotor eccentricity [J]. Progress in Electromagnetics Research M, 2019, 81: 137-147. [17] 左曙光, 刘洋, 邓文哲. 轴向磁通轮毂电机电磁力波灵敏度分析和优化 [J]. 电工技术学报, 2018, 33(11): 2423-2430. ZUO Shugang, LIU Yang, DENG Wenzhe. Sensitivity analysis and optimization of electromagnetic force wave of an axial-flux in-wheel motor [J]. Transactions of China Electrotechnical Society, 2018, 33(11): 2423-2430. [18] ZUO S G, LI D Q, MAO Y, et al. Longitudinal vibration analysis and suppression of electric wheel system driven by in-wheel motor considering unbalanced magnetic pull [J]. Journal of Automobile Engineering, 2019, 233(11): 2729-2745. [19] NIU X X, YANG C L, WANG H C, et al. Investigation of ANN and SVM based on limited samples for performance and emissions prediction of a CRDI-assisted marine diesel engine [J]. Applied Thermal Engineering, 2017, 111: 1353-1364. [20] TAFARROJ M M, KOLAHAN F. A comparative study on the performance of artificial neural networks and regression models in modeling the heat source model parameters in GTA welding [J]. Fusion Engineering and Design, 2018, 131: 111-118. [21] 叶建雄, 李志刚, WU J, 等. 基于SVR模型的水下焊接最佳工艺 [J]. 焊接学报, 2017, 38(12): 69-72, 94. YE Jianxiong, LI Zhigang, WU J, et al. Optimal underwater welding process based on SVR model [J]. Transactions of the China Welding Institution, 2017, 38(12): 69-72, 94. [22] 王枫, 陈征, 李花, 等. 采用多体动力学的压缩机曲轴结构优化 [J]. 西安交通大学学报, 2017, 51(3): 7-13. WANG Feng, CHEN Zheng, LI Hua, et al. Optimization for compressor crankshaft by multi-body dynamics analysis method [J]. Journal of Xi'an Jiaotong University, 2017, 51(3): 7-13. [23] 梁昭磊. 受限条件下的试验设计研究 [D]. 天津: 天津大学, 2009: 38-40. [24] 陆宜清. 概率论与数理统计 [M]. 上海: 上海科学技术出版社, 2019: 136-137.

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21015Views9Replies # Does Voltage Or Current Kill Electronics? Answered I've heard it a million times "It's the amps that kill".  But what if we are talking about electronics?  Would something that is 1.5 volts at an amp affect an electronic device? What if I had something at 10 kV, but virtually no current, like a piezoelectric ignitor?  (Maybe both of these examples would, I'm not sure,  but which is more likely?) Just wondering if someone can tell me whether it is mainly the volts or amps that fry electronics. And also how this happens. Does the high voltage short circuit things? Or is it the current that would overheat the small little circuits inside? I'm thinking that both are possible, but I'd like to know which is more likely to kill electronics, voltage or amperage? Tags: ## Discussions As Randy said, both conditions can destroy electronics, depending on the specific components involved. Excessive voltage will cause breakdown and arcing, either between separate exposed contacts (especially little points of solder), or through an insulating or dielectric barrier (for example, the dielectric in a capacitor). Excessive current is generally caused by excessive voltage. Passive components have a specific, fixed resistance; the current through them is determined by Ohm's law, I = V/R. Excess current can cause overheating and burn-through of components, leading to open circuits. Once you get into active components (piezoelectrics, op-amps, transistors), then the effects of going outside the components' specifications can be more complex, but never good. It's the electricity going through them (current), the voltage driving it is irrelevant so long as there's too much. However, if you have too much current going through something because the insulation/material  failed above it's design voltage, you may say that's "too much voltage". L +1. Is it the height you drop something from, or the speed at which it hits the ground, which breaks it? Technically the latter, but the former is what causes the latter. reading all the responses to this, yours was literally the best way it was explained, you made it easy to understand. (And the speed can be achieved from a lower height if driven by something more than gravity, or if falling with less air/water/whatever resistance.) Voltage is impressive can be visible but it is current that does the damage. Here is how, using a guinea pig semiconductor PN junction. A normal diode that is rated at 100 peak inverse voltage withstanding ability. Then raise the voltage above 110 slowly 1. A micro current flows over the outside PN surface NO Damage Yet,  if the voltage is reduced the diode is still good to use. 2. A little more voltage & very quickly milliampDamaging current flows. 3. As the current flows and begins heating the surface eroding material 4. High current flowing long enough for silicone melting sputtering heat to develop maybe several hundred microseconds depending on size of the junction and a full plasmic blow-out. The damage is done by  Current Flow. The heat can melt the metal Leeds to become a Shorted diode or blow out any connection to become an Open diode...................................   A SInce we generally work with constant voltage sources, its probably simplest to say excess voltage kills electronics, by causing excess current to flow or in extreme overvoltages to cause "flashover", where an arc causes a channel to be formed that allows more current to flow, and often a fire to start. There is also the difference between the voltage the device will run at, and the amount of static discharge voltage that it will withstand. In some cases, it is the combination of voltage and current, which combine to make the power dissipated, which heats up the component, damaging it. They both can potentially destroy electronics. Many parts tend to be rated for both maximum voltage and maximum current.

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1 # A running calculator that takes HEAT into account (as well as elevation) (Read 3122 times) More Cowbell! Heat has been a big topic lately. It was 95°F when I ran yesterday. Much of the same for today. Found this recently. Much like the excellent McMillan and Runbayou VDOT calculators, it recommends training paces based on a recent racing effort and calculates equivalent times for other races. But it also has a few options that I haven't seen elsewhere. 1) Temperature: enter the temperature from a run, and it will calculate the equivalent performance at other temperatures. [Based on data from Daniels] 2) Elevation change. (Eric/Trent: Is this the formula you've been looking for?) Enter the total climb and descent from a run, and you'll get equivalent times for the same distance for other climb/descent combos. Based on data from Noakes Lore of Running. 3) Altitude 4) windspeed 5) treadmill incline When you're on your deathbed, you won't be wishing that you'd spent more time at the office.  But you will be wishing that you'd spent more time running.  Because if you had, you wouldn't be on your deathbed. Fanatic #3965 Dude, I'm gonna be hella fast (heh, relatively speaking) when Fall comes! Would be cool to see the effect of heat, humidity, weight loss...sheesh, if all factors improve for me in the coming months I could potentially see a major improvement once the dog days of Summer are done. k Kirsten '07: 1324.5 | '08: 1561 | '09: 1810.9 run ~ 208.7 bike | '10: 1,000.3 run ~ 3513.5 bike | '11: 710.3 run ~ 4157.9 bike '12: 659.9 run ~ 3365.6 bike (100% benched by ortho last 4.5 weeks while in long-arm cast) ### '13 Goals: DON'T BREAK ANYTHING!!! • get within 5#s of 130#s (and stay there, gotdammit!) • 1st olympic distance duathlon • 1st Iceman Cometh mtn bike race Half Fanatic The thing is that heat and humidity CLEARLY don't affect everyone the same way and even and individual can become aclimated to the heat over time, so this like all calculators needs to be taken with a huge grain of salt. But it's pretty cool. Runners run. The elevation change in the calculator is very simple. In our thread we spoke to the complexities to consider. The calculator just does total climb and drop. RunFree7 Run like a kid again! Nice find Jeff. Doesn't have a place for Humidity but very cool. As motivation would someone like to tell me what kind of running conditions the Kenyon's run in or something else that might help me get out from running inside on my rec centers cushion track and back outdoors? The only runs I have been doing outside lately have been my long run tempo's. With the heat this weekend I am even tempted to do that indoors. 2011 Goals: Sub 19 5K (19:24 5K July 14th 2010) Marathon under 3:05:59 BQ (3:11:10 Indy 2010)

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# Physics Optics Questions with Answers ### Optics Question Bank: Ques. Two vertical plane mirrors are inclined at an angle of 60o towards each other. A ray of light traveling horizontally is reflected first from one mirror and then from the other. The resultant deviation is (a) 60 (b) 120 (c) 180 (d) 240 Ans. (d) Ques. A plane mirror reflects a pencil of light to form a real image. Then the pencil of light incident on the mirror is (a) Parallel (b) Convergent (c) Divergent (d) None of these Ans. (b) Ques. For the sustained interference of light, the necessary condition is that the two sources should (a) have constant phase difference (b) be narrow (c) be close to each other (d) of the same amplitude Ans. (a) Ques. How will the image formed by a convex lens be affected if the central part of the lens is covered by black paper? (a) Remaining part of the lens will form no image (b) The central position of the image is absent (c) There will be no effect (d) The full image will be formed with lessened brightness Ans. (d) Ques. If the ratio of amplitude of two waves is 4 : 3, then the ratio of maximum and minimum intensity is (a) 16 : 18 (b) 18 : 16 (c) 49 : 1 (d) 94 : 1 Ans. (c) Related: Fluid Mechanics questions Ques. Which of the following is conserved when light waves interfere? (a) Intensity (b) Energy (c) Amplitude (d) Momentum Ans. (b) Ques. Intensity of light depends upon (a) Velocity (b) Wavelength (c) Amplitude (d) Frequency Ans. (c) Ques. Ray diverging from a point source from a wave front that is (a) Cylindrical (b) Spherical (c) Plane (d) Cubical Ans. (b) Ques. The ratio of amplitude of interfering waves is 3 : 4. Now the ratio of their intensities will be (a) 16/9 (b) 49 : 1 (c) 9/16 (d) None of these Ans. (c) Related: Endocrine system questions Ques. Two coherent sources have intensity in the ratio of 100/1. The ratio of (intensity) max/(intensity) min is (a) 1/100 (b) 1/10 (c) 10/1 (d) 3/2 Ans. (d) Ques. Newton postulated his corpuscular theory on the basis of (a) Newton’s rings (b) Colors of thin films (c) Rectilinear propagation of light (d) Dispersion of white light Ans. (c) Ques. The dual nature of light is exhibited by (a) Photoelectric effect (b) Refraction and interference (c) Diffraction and reflection (d) Diffraction and photoelectric effect Ans. (d) Ques. Coherent sources are those sources which (a) Phase difference remains constant (b) Frequency remains constant (c) Both phase difference and frequency remain constant (d) None of these Ans. (c) Related: Electrostatics practice problems Ques. The wave nature of light is verified by (a) Interference (b) Photoelectric effect (c) Reflection (d) Refraction Ans. (a) Ques. If two light waves having the same frequency have an intensity ratio of 4 : 1 and they interfere, the ratio of maximum to minimum intensity in the pattern will be (a) 9 : 1 (b) 3 : 1 (c) 25 : 9 (d) 16 : 25 Ans. (a) Ques. Which of the following statements indicates that light waves are transverse? (a) Light waves can travel in vacuum (b) Light waves show interference (c) Light waves can be polarized (d) Light waves can be diffracted Ans. (c) Ques. Soap bubbles appear colored due to the phenomenon of (a) Interference (b) Diffraction (c) Dispersion (d) Reflection Ans. (a) Ques. Two sources of waves are called coherent if (a) Both have the same amplitude of vibrations (b) Both produce waves of the same wavelength (c) Both produce waves of the same wavelength having constant phase difference (d) Both produce waves having the same velocity Ans. (c) Ques. On a rainy day, a small oil film on water shows brilliant colors. This is due to (a) Dispersion of light (b) Interference of light (c) Absorption of light (d) Scattering of light Ans. (b) Ques. If L is the coherence length and c the velocity of light, the coherent time is (a) cL (b) L/c (c) c/L (d) 1/Lc Ans. (b) Ques. If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima is (a) 25 : 16 (b) 5 : 3 (c) 16 : 1 (d) 25 : 9 Ans. (c) Ques. The wave nature of light follows because (a) Light rays travel in a straight line (b) Light exhibits the phenomena of reflection and refraction (c) Light exhibits the phenomenon of interference (d) Light causes the phenomenon of the photoelectric effect Ans. (c) Ques. If the ratio of intensities of two waves is 1 : 25, then the ratio of their amplitudes will be (a) 1 : 25 (b) 5 : 1 (c) 26 : 24 (d) 1 : 5 Ans. (d) Ques. The ratio of intensities of two waves is 9 : 1. They are producing interference. The ratio of maximum and minimum intensities will be (a) 10 : 8 (b) 9 : 1 (c) 4 : 1 (d) 2 : 1 Ans. (c) Ques. A wave can transmit …… from one place to another (a) Energy (b) Amplitude (c) Wavelength (d) Matter Ans. (a) Related: Radiology physics questions Ques. The similarity between the sound waves and light waves is (a) Both are electromagnetic waves. (b) Both are longitudinal waves (c) Both have the same speed in a medium (d) They can produce interference Ans. (d) Ques. By a monochromatic wave, we mean (a) A single ray (b) A single ray of a single color (c) Wave having a single wavelength (d) Many rays of a single color Ans. (a) Ques. The idea of secondary wavelets for the propagation of a wave was first given by (a) Newton (b) Huygen (c) Maxwell (d) Fresnel Ans. (b) Ques. Light appears to travel in straight lines since (a) It is not absorbed by the atmosphere (b) It is reflected by the atmosphere (c) Its wavelength is very small (d) Its velocity is very large Ans. (c) Ques. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are (a) 5I and I (b) 5I and 3I (c) 9I and I (d) 9I and 3I Ans. (c) Ques. The phenomenon of interference is shown by (a) Longitudinal mechanical waves only (b) Transverse mechanical waves only (c) Electromagnetic waves only (d) All these types of waves Ans. (d) Ques. Through Huygen’s wave theory of light, we cannot explain the phenomenon of (a) Interference (b) Diffraction (c) Photoelectric effect (d) Polarization Ans. (c) Ques. Two coherent sources of light can be obtained by (a) Two different lamps (b) Two different lamps but of the same power (c) Two different lamps of the same power and having the same color (d) None of these Ans. (d) Ques. The idea of the quantum nature of light has emerged in an attempt to explain (a) Interference (b) Diffraction (c) Radiation spectrum of a black body (d) Polarization Ans. (c) Ques. Huygen’s conception of secondary waves (a) Allow us to find the focal length of a thick lens (b) Is a geometrical method to find a wavefront (c) Is used to determine the velocity of light (d) Is used to explain polarization Ans. (b) Ques. According to the corpuscular theory of light, the different colors of light are due to (a) Different electromagnetic waves (b) Different forces of attraction among the corpuscles (c) Different size of the corpuscles (d) None of these Ans. (c) Ques. Through the corpuscular theory of light, the phenomenon which can be explained is (a) Refraction (b) Interference (c) Diffraction (d) Polarization Ans. (a) Ques. A virtual image larger than the object can be obtained by (a) Concave mirror (b) Convex mirror (c) Plane mirror (d) Concave lens Ans. (a) Ques. A concave mirror is used to focus the image of a flower on a nearby well 120 cm from the flower. If a lateral magnification of 16 is desired, the distance of the flower from the mirror should be (a) 8 cm (b) 12 cm (c) 80 cm (d) 120 cm Ans. (a) Ques. While using an electric bulb, the reflection for street lighting should be from (a) Concave mirror (b) Convex mirror (c) Cylindrical mirror (d) Parabolic mirror Ans. (b) Ques. Which one of the following statements is true? (a) An object situated at the principal focus of a concave lens will have its image formed at infinity. (b) Concave mirrors can give a diminished virtual image (c) Given a pointed source of light, a convex mirror can produce a parallel beam of light. (d) The virtual image formed in a plane mirror can be photographed Ans. (d) Ques. A person sees his virtual image by holding a mirror very close to his face. When he moves the mirror away from his face, the image becomes inverted. What type of mirror he is using? (a) Plane mirror (b) Convex mirror (c) Concave mirror (d) None of these Ans. (c) Ques. A boy stands straight in front of a mirror at a distance of 30 cm away from it. He sees his erect image whose height is 1/5 th of his real height. The mirror he is using is (a) Plane mirror (b) Convex mirror (c) Concave mirror (d) Plano-convex mirror Ans. (b) Ques. Given a point source of light, which of the following can produce a parallel beam of light? (a) Convex mirror (b) Concave mirror (c) Concave lens (d) Two plane mirrors inclined at an angle of 90 Ans. (b) Ques. The image formed by a convex mirror of focal length 30cm is a quarter of the size of the object. The distance of the object from the mirror is (a) 30cm (b) 90cm (c) 120cm (d) 60cm Ans. (b) Related: Nuclear Chemistry Question Bank Ques. A convex mirror is used to form the image of an object. Then which of the following statements is wrong (a) The image lies between the pole and the focus (b) The image is diminished in size (c) The image is erect (d) The image is real Ans. (d) Ques. The image formed by a convex mirror is (a) Virtual (b) Real (c) Enlarged (d) Inverted Ans. (a) Ques. The field of view is maximum for (a) Plane mirror (b) Concave mirror (c) Convex mirror (d) Cylindrical mirror Ans. (c) Ques. An object of size 7.5cm is placed in front of a convex mirror with a radius of curvature 25 cm at a distance of 40cm. The size of the image should be (a) 2.3 cm (b) 1.78 cm (c) 1 cm (d) 0.8 cm Ans. (b) Ques. The focal length of a concave mirror is 50 cm. Where an object is placed so that its image is two times and inverted (a) 75 cm (b) 72 cm (c) 63 cm (d) 50 cm Ans. (a) Related: Amines worksheet Ques. An object 5cm tall is placed 1m from a concave spherical mirror which has a radius of curvature of 20cm The size of the image is (a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm Ans. (c) Ques. Which of the following can not produce a virtual image? (a) Plane mirror (b) Convex mirror (c) Concave mirror (d) All these can produce a virtual image Ans. (d) Ques. A diminished virtual image can be formed only in (a) Plane mirror (b) A concave mirror (c) A convex mirror (d) A convex mirror Ans. (c) Ques. Two plane mirrors are at 45 to each other. If an object is placed between them, then the number of images will be (a) 5 (b) 9 (c) 7 (d) 8 Ans. (c) Ques. A man of length h requires a mirror, to see his own complete image of length at least equal to (a) h/4 (b) h/3 (c) h/2 (d) h Ans. (c) Ques. Two plane mirrors are inclined at an angle of 72o. The number of images of a pointed object placed between them will be (a) 2 (b) 3 (c) 4 (d) 5 Ans. (c) Ques. The light reflected by a plane’s mirror may form a real image. (a) If the rays incident on the mirror are diverging (b) If the ray incident on the mirror is converging (c) If the object is placed very close to the mirror (d) Under no circumstances Ans. (b) Ques. The ratio of intensities of two waves is given by 4 : 1. The ratio of the amplitudes of the two waves is (a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4 Ans. (a) Ques. Interference was observed in an interference chamber when air was present. Now the chamber is evacuated and if the same light is used, a careful observer will see (a) No interference (b) Interference with bright bands (c) Interference with dark bands (d) Interference in which the width of the fringe will be slightly increased Ans. (d) Ques. Which one of the following phenomena is not explained by Huygen’s construction of a wavefront? (a) Refraction (b) Reflection (c) Diffraction (d) Origin of spectra Ans. (d) Ques. Wavefront means (a) All particles in it have the same phase (b) All particles have opposite phases of vibrations (c) Few particles are in the same phase, the rest are in the opposite phase (d) None of these Ans. (a) Ques. Two waves having intensities in the ratio 25 : 4 produce interference. The ratio of the maximum to the minimum intensity is (a) 5 : 2 (b) 7 : 3 (c) 49 : 9 (d) 9 : 49 Ans. (c) Ques. Two light sources are said to be coherent if they are obtained from (a) Two independent point sources emitting light of the same wavelength (b) A single point source (c) A broad resource (d) Two ordinary bulbs emitting light of different wavelengths Ans. (b) Ques. Evidence for the wave nature of light cannot be obtained from (a) Reflection (b) Doppler effect (c) Interference (d) Diffraction Ans. (a) Ques. An object is at a distance of 0.5 m in front of a plane mirror. The distance between the object and the image is (a) 0.5 m (b) 1 m (c) 0.25 m (d) 1.5 m Ans. (b) Ques. A small object is placed 10 cm in front of a plane mirror. If you stand behind an object 30 cm from the mirror and look at its image, the distance focused on your eye will be (a) 60 cm (b) 20 cm (c) 40 cm (d) 80 cm Ans. (c) ### About the author #### Richa Richa (B. Tech) has keen interest in Science and loves to teach students about it through lectures and assignments. She always try to use simple language and sentences while writing to make sure learner understands everything properly. error: Content is protected !!

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# Finding the work done using force vectors in 3 dimensions ## Homework Statement Find the work done by a force of 3 newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0). ## Homework Equations Commonly known vector operations. ## The Attempt at a Solution I found the displacement vector to be 2j. My problem is figuring out how to represent the direction and magnitude of force as an actual vector. The answer is (6*SQRT 5)/5 Joules. I don't understand how. ## Answers and Replies Related Precalculus Mathematics Homework Help News on Phys.org Are you sure that is the right answer? Are you sure that is the right answer? Positive. This really doesn't make any sense to me. LCKurtz Science Advisor Homework Helper Gold Member ## Homework Statement Find the work done by a force of 3 newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0). ## Homework Equations Commonly known vector operations. ## The Attempt at a Solution I found the displacement vector to be 2j. My problem is figuring out how to represent the direction and magnitude of force as an actual vector. The answer is (6*SQRT 5)/5 Joules. I don't understand how. I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book. s I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book. That may well be true. If so, is the correct answer 2? The displacement vector is 2j. (2i + j + 2k) dot product 2j = 0 + 2j + 0 = 2j. So the correct answer is 2 Joules, correct? That's what I got. Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get W = F*AB AB = 2j F = 3(2i + j + 2k) = 6i + 3j + 6k F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct? No, you only have to take into account the magnitude of the force in the (0,1,0) direction, because that is that component that is going to be doing all the work! Remember that: WORK = FORCE dot product DISPLACEMENT LCKurtz Science Advisor Homework Helper Gold Member Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get W = F*AB AB = 2j F = 3(2i + j + 2k) = 6i + 3j + 6k F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct? No. A force of 3N is represented by a vector of length 3. Your vector specifying the direction of the force just happens to have length 3 already. That is just a lucky break for this problem. Normally what you would do is multiply a unit vector in the direction of the force by 3.

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views updated # Trajectories Trajectories are the paths followed by spacecraft as they travel from one point to another. They are governed by two key factors: the spacecraft's own propulsion system and the gravity of the Sun, Earth, and other planets and moons. Because even the most powerful rockets have only a limited amount of thrust, engineers must carefully develop trajectories for spacecraft that will allow them to reach their intended destination. In some cases this can lead to complicated trajectories that get a boost from the gravity of other worlds. The trajectory needed for a spacecraft to go into orbit around Earth is relatively straightforward. The spacecraft needs to gain enough altitudetypically at least 200 kilometers (124 miles)to clear Earth's atmosphere and enough speed to keep from falling back to Earth. This minimum orbital velocity around Earth is about 28,000 kilometers per hour (17,360 miles per hour) for low Earth orbits and slower than that for higher orbits as Earth's gravitational pull weakens. Other parameters of the orbit, such as the inclination of the orbit to Earth's equator, can be altered by changing the direction of the spacecraft's launch. Launching a spacecraft beyond Earth, such as on a mission to Mars or another planet, is more complicated. Because of the great distances between planets and the limited power of modern rockets, one cannot simply aim a spacecraft directly at its destination and launch it. Instead, trajectories must be carefully calculated to allow a spacecraft to travel to its destination given the limited amount of rocket power available. A common way to do this is to use a Hohmann transfer orbit, a type of orbit that minimizes the amount of propellant needed to send a spacecraft to its destination. A Hohmann transfer orbit is an elliptical orbit with its perihelion at Earth and aphelion at the destination planet (or the reverse if traveling towards the Sun). If launched at the proper time a spacecraft will spend only half an orbit in a Hohmann orbit, catching up with the destination world at the opposite point of its orbit from Earth. To do this, the spacecraft much be launched during a relatively short launch window. For a mission to Mars, such launch windows are available every twenty-six months, for only a couple months at a time. Even a Hohmann orbit, however, may require more energy than a rocket can provide. Another technique, known as gravity assist, can allow spacecraft to reach more distant destinations by taking advantage of the gravity of other worlds. A spacecraft is launched on a Hohmann trajectory toward an intermediate destination, usually another planet. The spacecraft flies by this planet, gaining velocity by taking, though gravitational interaction, an infinitesimally small amount of the planet's angular momentum . This added velocity allows the spacecraft to continue on to its destination. Gravity assists allowed the Voyager 2 spacecraft, launched from Earth with only enough velocity to reach Jupiter, to travel on to Saturn, Uranus, and Neptune. Gravity assist flybys of Venus, Earth, and Jupiter will also allow the Cassini spacecraft to reach Saturn in 2004. Jeff Foust ### Bibliography Wertz, James R., and Wiley J. Larson, eds. Space Mission Analysis and Design. Dordrecht, Netherlands: Kluwer, 1991. #### Internet Resources Basics of Space Flight. NASA Jet Propulsion Laboratory, California Institute of Technology. <http://www.jpl.nasa.gov/basics/>.

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Consider the following graph with four nodes and four edges. FROM/TO 1 2 3 4 1 0 0 1 1 2 0 0 0 1 3 1 0 0 1 4 1 1 1 0 This matrix is symmetric about the main diagonal because the traffic is two-way; one-way traffic, or directed edges, presents a practically different, but conceptually similar, situation (as do loops--an edge joining a node to itself).  This matrix counts the number of 1-step paths joining pairs of nodes. To count the number of 2-step paths, square the adjacency matrix.  (There is a 2-step path from 1 to 2--from 1 to 4 to 2, but not a 1-step path.) FROM/TO 1 2 3 4 1 0 0 1 1 2 0 0 0 1 3 1 0 0 1 4 1 1 1 0 FROM/TO 1 2 3 4 1 0 0 1 1 2 0 0 0 1 3 1 0 0 1 4 1 1 1 0 = FROM/TO 1 2 3 4 1 2 1 1 1 2 1 1 1 0 3 1 1 2 1 4 1 0 1 3 To see why this counting works, consider the situation more generally. In the square of the adjacency matrix, the (2,1) entry is generated by multiplying the second row by the first column a(11) a(12) a(13) a(14) a(21) a(22) a(23) a(24) a(31) a(32) a(33) a(34) a(41) a(42) a(43) a(44) x a(11) a(12) a(13) a(14) a(21) a(22) a(23) a(24) a(31) a(32) a(33) a(34) a(41) a(42) a(43) a(44) which, notationally is, a(21)a(11) + a(22)a(21) + a(23)a(31) + a(24)a(41) This entry counts paths of length 2 from node 2 to node 1, through each of the possible nodes, 1, 2, 3, 4 (read it from the subscripts in the notation above).  Diagramatically, Higher powers of the adjacency matrix, say of power n, count numbers of paths of n-steps. The definition here of adjacency is very broad and might be interpreted in any of the three ways for spatial units of two dimensions.  It is broad because there is no geometric dimension involved--the focus is on pure connection pattern.  When spatial units have dimension other than 2, different styles of adjacency exist. If one alters the way in which matrix multiplication is performed, one gets to a whole host of other conceptual examples.  The attached references focus on one way to alter that multiplication. Hasse, M., "Uber die Behändlung graphentheorischer Probleme unter Verwendung der Matrizenrechnung," Wiss. Z. Techn. Univer. Dresden 10(1961), pp. 1313–1316. Harary, F., Norman, R., and Cartwright, D., Structural Models: An Introduction to the Theory of Directed Graphs. John Wiley & Sons, New York, 1965. Arlinghaus, S., Arlinghaus, W., Harary, F., and Nystuen, J., " Los Angeles 1994: A spatial scientific view," Solstice: An Electronic Journal of Geography and Mathematics V(1994), pp. 42–82. Arlinghaus, S., Arlinghaus,  W., and Nystuen, J., " The Hedetniemi Matrix Sum:  A Real-world Application,"  Solstice:  An Electronic Journal of Geography and Mathematics  I(1990), pp. 87–97. Arlinghaus, S., Arlinghaus, W., and Nystuen, J., " The Hedetniemi matrix sum: An algorithm for shortest path and shortest distance," Geographical Analysis  22(1990), pp. 351–360.

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CC-MAIN-2018-43

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https://edge.sagepub.com/priviterastats3e/student-resources/chapter-15/learning-objectives

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# Learning Objectives After reading this chapter, you should be able to: 1. Identify the direction and strength of a correlation between two factors. 2. Compute and interpret the Pearson correlation coefficient and test for significance. 3. Compute and interpret the coefficient of determination. 4. Define homoscedasticity, linearity, and normality and explain why each assumption is necessary to appropriately interpret a significant correlation coefficient. 5. Explain how causality, outliers, and restriction of range can limit the interpretation of a significant correlation coefficient. 6. Compute and interpret the Spearman correlation coefficient and test for significance. 7. Compute and interpret the point-biserial correlation coefficient and test for significance. 8. Compute and interpret the phi correlation coefficient and test for significance. 9. Convert the value of r to a t statistic and χ2 statistic. 10. Summarize the results of a correlation coefficient in APA format. 11. Compute the Pearson, Spearman, point-biserial, and phi correlation coefficients using SPSS.

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http://wiki.secondlife.com/wiki/End_squashed_images!_Get_correct_texture_aspect_ratios!

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# Texture aspect ratios What is an aspect ratio as it applies to Second Life, and why does it matter? A simple way to think of it is: images on a computer screen can be shown in two dimensions. In a three-dimensional world like Second Life, a texture can be mapped onto the 2D face of a 3D object. For instance, a cube has 6 such flat faces. If you stretch that cube so that it's 2 meters wide but 1 meter tall (and 1 meter deep), then the wide-tall aspect ratio is 2:1. If a texture is 512 pixels wide and 256 pixels tall, then it, too, has a 2:1 aspect ratio. A perfect square has an aspect ratio of 1:1. All this matters because if you don't understand aspect ratios, your inworld pictures, posters, signage, and other textures will be unpleasantly stretched or squashed! I've long heard Residents rightfully complain about our bizarre jumble of aspect ratios. We endeavor to simplify this in the future, so there's less of this... In the here and now, thankfully, to save you time measuring pixels — because you have better things to do — we list all current aspect ratios on the Limits page, an indispensable guide which you should keep close. ## Some very important tips • All Second Life textures are constrained to powers of 2. This means if you upload a texture that's 257x514, it'll be resized to the nearest powers of 2, or 256x512. However, understand that because where these textures are applied aren't constrained to powers of 2, you need to know specific dimensions when crafting an image, which I'll go into more depth about later. • To get the aspect ratio of a prim, right-click it and select Edit from the pie menu, click the Object tab, then observe what it says under Size (meters). When applying a texture, you'll only use 2 of those 3 (XYZ) dimensions, so pay attention to which are needed. • Some alternate viewers, including Snowglobe, allow you to view textures snapped to various aspect ratios. I find this really useful. So, say you're making a new Second Life profile in an image editor and want to assure it has the correct aspect ratio. What to do? ## Watch this video! And here are text instructions for the popular Adobe Photoshop CS4, but similar principles can be used for others: 1. Consult Limits. Notice it says: "Profile > 2nd Life tab - ~4:3 (178x133 pixels)". So, 4:3 is close enough. 2. Photoshop can't calculate aspect ratio upon creating a new file; I suggest using the Continuum Javascript Ratio Calculator. 3. In the left set of boxes, enter "`4`" in the top box, then "`3`" in the bottom box. 4. In the right set of boxes, enter "`512`" (a balanced choice in many cases; `256` is even more economical) next to the top box, and it'll auto-enter what the bottom box should be: "`384`". 5. With our answer, return to Photoshop. Go to File menu > New and set Width and Height to "`512`" and "`384`". (Advanced users with a yearning for detail can use larger proportions like `1024x768` and scale down later.) 6. Now, edit the image using your presumed mad skillz, but DO NOT change the proportions. 7. When you're done, go to File menu > Save for Web & Devices and change Preset to PNG-24. Unless you deliberately want something like a see-through window, uncheck Transparency. 8. Click Save to save your image, and give it a suitable name. 9. Start up Second Life (if it wasn't already open), and go to File menu > Upload > Image (L\$10). 10. Select the image you saved on your hard drive and upload it. 11. In a few seconds, you'll see it uploaded, and it'll say "Dimensions: 512 x 256". It'll also appear squished. REMEMBER: the begin and end aspect ratio/proportions/dimensions are what ultimately matter. 12. Go to Edit menu > Profile. 13. In the 2nd Life tab, click the thumbnail next to Photo. 14. Use the texture picker to find the texture you uploaded, click it, then click Select. 15. You'll see that because you used a 4:3 aspect ratio to start with and this particular profile picture is also 4:3, the image again shows its correct proportions. Alternatively (as with many creative tasks there are multiple paths), many image editors offer a crop tool which lets you specify dimensions, as well as additional measuring tools to ensure you're on the right track. The main point of the above was showing you why it's important to maintain the correct aspect ratio from beginning to end. ## The above principles can be applied to many situations Another fine example I can think of is: say you have a movie trailer you want to play inworld but it's 16:9 widescreen ratio, as is common for such media. By using what you've learned, you can resize the prim screen you're using the media texture so it's also 16:9, like 10x5.625 meters. Understanding proportions is one of the fundamental skills every artist in and out of Second Life needs to have — even if you don't consider yourself a "pro", your craft will certainly look more professional, whether it's a store sign, wedding invite, or classified ad.

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https://www.ademcetinkaya.com/2023/04/wkhs-workhorse-group-inc-common-stock.html

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Outlook: Workhorse Group Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Time series to forecast n: 28 Apr 2023 for (n+1 year) Methodology : Statistical Inference (ML) ## Abstract Workhorse Group Inc. Common Stock prediction model is evaluated with Statistical Inference (ML) and Independent T-Test1,2,3,4 and it is concluded that the WKHS stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Buy ## Key Points 1. Decision Making 2. Market Risk 3. What is the best way to predict stock prices? ## WKHS Target Price Prediction Modeling Methodology We consider Workhorse Group Inc. Common Stock Decision Process with Statistical Inference (ML) where A is the set of discrete actions of WKHS stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4 F(Independent T-Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Statistical Inference (ML)) X S(n):→ (n+1 year) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$ n:Time series to forecast p:Price signals of WKHS stock j:Nash equilibria (Neural Network) k:Dominated move a:Best response for target price For further technical information as per how our model work we invite you to visit the article below: How do AC Investment Research machine learning (predictive) algorithms actually work? ## WKHS Stock Forecast (Buy or Sell) for (n+1 year) Sample Set: Neural Network Stock/Index: WKHS Workhorse Group Inc. Common Stock Time series to forecast n: 28 Apr 2023 for (n+1 year) According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Buy X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.) Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.) Z axis (Grey to Black): *Technical Analysis% ## IFRS Reconciliation Adjustments for Workhorse Group Inc. Common Stock 1. If a put option obligation written by an entity or call option right held by an entity prevents a transferred asset from being derecognised and the entity measures the transferred asset at amortised cost, the associated liability is measured at its cost (ie the consideration received) adjusted for the amortisation of any difference between that cost and the gross carrying amount of the transferred asset at the expiration date of the option. For example, assume that the gross carrying amount of the asset on the date of the transfer is CU98 and that the consideration received is CU95. The gross carrying amount of the asset on the option exercise date will be CU100. The initial carrying amount of the associated liability is CU95 and the difference between CU95 and CU100 is recognised in profit or loss using the effective interest method. If the option is exercised, any difference between the carrying amount of the associated liability and the exercise price is recognised in profit or loss. 2. The requirements in paragraphs 6.8.4–6.8.8 may cease to apply at different times. Therefore, in applying paragraph 6.9.1, an entity may be required to amend the formal designation of its hedging relationships at different times, or may be required to amend the formal designation of a hedging relationship more than once. When, and only when, such a change is made to the hedge designation, an entity shall apply paragraphs 6.9.7–6.9.12 as applicable. An entity also shall apply paragraph 6.5.8 (for a fair value hedge) or paragraph 6.5.11 (for a cash flow hedge) to account for any changes in the fair value of the hedged item or the hedging instrument. 3. An entity is not required to restate prior periods to reflect the application of these amendments. The entity may restate prior periods only if it is possible to do so without the use of hindsight. If an entity restates prior periods, the restated financial statements must reflect all the requirements in this Standard for the affected financial instruments. If an entity does not restate prior periods, the entity shall recognise any difference between the previous carrying amount and the carrying amount at the beginning of the annual reporting period that includes the date of initial application of these amendments in the opening retained earnings (or other component of equity, as appropriate) of the annual reporting period that includes the date of initial application of these amendments. 4. The requirement that an economic relationship exists means that the hedging instrument and the hedged item have values that generally move in the opposite direction because of the same risk, which is the hedged risk. Hence, there must be an expectation that the value of the hedging instrument and the value of the hedged item will systematically change in response to movements in either the same underlying or underlyings that are economically related in such a way that they respond in a similar way to the risk that is being hedged (for example, Brent and WTI crude oil). *International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS. ## Conclusions Workhorse Group Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Workhorse Group Inc. Common Stock prediction model is evaluated with Statistical Inference (ML) and Independent T-Test1,2,3,4 and it is concluded that the WKHS stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Buy ### WKHS Workhorse Group Inc. Common Stock Financial Analysis* Rating Short-Term Long-Term Senior Outlook*Ba1Ba1 Income StatementB1B2 Balance SheetCBaa2 Leverage RatiosCBaa2 Cash FlowBaa2B1 Rates of Return and ProfitabilityCCaa2 *Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents. How does neural network examine financial reports and understand financial state of the company? ### Prediction Confidence Score Trust metric by Neural Network: 85 out of 100 with 759 signals. ## References 1. J. Harb and D. Precup. Investigating recurrence and eligibility traces in deep Q-networks. In Deep Reinforcement Learning Workshop, NIPS 2016, Barcelona, Spain, 2016. 2. Kallus N. 2017. Balanced policy evaluation and learning. arXiv:1705.07384 [stat.ML] 3. S. Bhatnagar, H. Prasad, and L. Prashanth. Stochastic recursive algorithms for optimization, volume 434. Springer, 2013 4. Wooldridge JM. 2010. Econometric Analysis of Cross Section and Panel Data. Cambridge, MA: MIT Press 5. Imai K, Ratkovic M. 2013. Estimating treatment effect heterogeneity in randomized program evaluation. Ann. Appl. Stat. 7:443–70 6. Pennington J, Socher R, Manning CD. 2014. GloVe: global vectors for word representation. In Proceedings of the 2014 Conference on Empirical Methods on Natural Language Processing, pp. 1532–43. New York: Assoc. Comput. Linguist. 7. Thomas P, Brunskill E. 2016. Data-efficient off-policy policy evaluation for reinforcement learning. In Pro- ceedings of the International Conference on Machine Learning, pp. 2139–48. La Jolla, CA: Int. Mach. Learn. Soc. Frequently Asked QuestionsQ: What is the prediction methodology for WKHS stock? A: WKHS stock prediction methodology: We evaluate the prediction models Statistical Inference (ML) and Independent T-Test Q: Is WKHS stock a buy or sell? A: The dominant strategy among neural network is to Buy WKHS Stock. Q: Is Workhorse Group Inc. Common Stock stock a good investment? A: The consensus rating for Workhorse Group Inc. Common Stock is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating. Q: What is the consensus rating of WKHS stock? A: The consensus rating for WKHS is Buy. Q: What is the prediction period for WKHS stock? A: The prediction period for WKHS is (n+1 year)

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CC-MAIN-2023-23

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https://www.traditionaloven.com/building/food-glass/convert-1-cm-diam-food-glass-sphere-to-hong-kong-picul-food-glass.html

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crawl-data/CC-MAIN-2019-35/segments/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00493.warc.gz

 Food Grade Glass 1 sphere 1 centimeter in diameter volume to Hong Kong piculs converter ## Amount: 1 sphere 1 centimeter in diameter (d, ∅ 1 cm) of volume Equals: 0.000022 Hong Kong piculs (tam) in mass Converting sphere 1 centimeter in diameter to Hong Kong piculs value in the food grade glass units scale. TOGGLE :   from Hong Kong piculs into solid ∅ 1 cm spheres in the other way around. ## food grade glass from sphere 1 centimeter in diameter to Hong Kong picul Conversion Results: ### Enter a New sphere 1 centimeter in diameter Amount of food grade glass to Convert From * Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8) * Precision is how many numbers after decimal point (1 - 9) Enter Amount : Decimal Precision : CONVERT :   between other food grade glass measuring units - complete list. Conversion calculator for webmasters. There is a 0.01 gram difference in mass density sense between glass which is used for food jars and window glass. Density of food grade glass is 2.52g/cm3 This glass type is being generally used for making glass bottles, drinking glasses, glass jars and bowls or containers for storing solid foods and beverages. Where can the glass units converter be applied? A glass mass versus volume units calculator can be used in specific situations. For example such as: certain weight of bottles plus various food grade glass fragments, all in one bulk amount, can be converted into an exact solid glass volume (in real meaning on paper without melting it in a gas or electric glass kiln or furnace first) and vise verse. Although with some efforts, the glass volume could be measured also without knowing its actual mass; by dipping it into a liquid and then measuring the created liquid excess above the initial level. Convert food grade glass measuring units between sphere 1 centimeter in diameter (d, ∅ 1 cm) and Hong Kong piculs (tam) but in the other reverse direction from Hong Kong piculs into solid ∅ 1 cm spheres. conversion result for food grade glass: From Symbol Equals Result To Symbol 1 sphere 1 centimeter in diameter d, ∅ 1 cm = 0.000022 Hong Kong piculs tam # Converter type: food grade glass measurements This online food grade glass from d, ∅ 1 cm into tam converter is a handy tool not just for certified or experienced professionals. First unit: sphere 1 centimeter in diameter (d, ∅ 1 cm) is used for measuring volume. Second: Hong Kong picul (tam) is unit of mass. ## food grade glass per 0.000022 tam is equivalent to 1 what? The Hong Kong piculs amount 0.000022 tam converts into 1 d, ∅ 1 cm, one sphere 1 centimeter in diameter. It is the EQUAL food grade glass volume value of 1 sphere 1 centimeter in diameter but in the Hong Kong piculs mass unit alternative. How to convert 2 solid ∅ 1 cm spheres (d, ∅ 1 cm) of food grade glass into Hong Kong piculs (tam)? Is there a calculation formula? First divide the two units variables. Then multiply the result by 2 - for example: 2.1816982873616E-5 * 2 (or divide it by / 0.5) QUESTION: 1 d, ∅ 1 cm of food grade glass = ? tam 1 d, ∅ 1 cm = 0.000022 tam of food grade glass ## Other applications for food grade glass units calculator ... With the above mentioned two-units calculating service it provides, this food grade glass converter proved to be useful also as an online tool for: 1. practicing solid ∅ 1 cm spheres and Hong Kong piculs of food grade glass ( d, ∅ 1 cm vs. tam ) measuring values exchange. 2. food grade glass amounts conversion factors - between numerous unit pairs variations. 3. working with mass density - how heavy is a volume of food grade glass - values and properties. International unit symbols for these two food grade glass measurements are: Abbreviation or prefix ( abbr. short brevis ), unit symbol, for sphere 1 centimeter in diameter is: d, ∅ 1 cm Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Hong Kong picul is: tam ### One sphere 1 centimeter in diameter of food grade glass converted to Hong Kong picul equals to 0.000022 tam How many Hong Kong piculs of food grade glass are in 1 sphere 1 centimeter in diameter? The answer is: The change of 1 d, ∅ 1 cm ( sphere 1 centimeter in diameter ) volume unit of food grade glass measure equals = to mass 0.000022 tam ( Hong Kong picul ) as the equivalent measure within the same food grade glass substance type. In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in d, ∅ 1 cm - solid ∅ 1 cm spheres for food grade glass amount, the rule is that the sphere 1 centimeter in diameter number gets converted into tam - Hong Kong piculs or any other food grade glass unit absolutely exactly. Conversion for how many Hong Kong piculs ( tam ) of food grade glass are contained in a sphere 1 centimeter in diameter ( 1 d, ∅ 1 cm ). Or, how much in Hong Kong piculs of food grade glass is in 1 sphere 1 centimeter in diameter? To link to this food grade glass sphere 1 centimeter in diameter to Hong Kong piculs online converter simply cut and paste the following. The link to this tool will appear as: food grade glass from sphere 1 centimeter in diameter (d, ∅ 1 cm) to Hong Kong piculs (tam) conversion. I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.

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https://gss-statistical-software.com/2022/03/17/understanding-data-type/

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# Understanding data type When you do statistical analyses, the knowledge about data type is very important, because many statistical methods and tests were specifically designed for a specific data type. However, there is a lot of confusion regarding data type, because many different terms are used and when you look into statistical textbooks, you will likely find different terms for the same type of data. Here is an overview: In principle we have two groups of data: Metric and non-metric. Metric data are measurements such as 27.3 cm, 14.5° Celsius or 70 kg.  Also when you count objects, for example when you count the number of plants per area or eggs in the nest of birds these are metric data. There is, however, a small difference between weight and egg numbers. The number of eggs is a discontinuous or =discrete variable, because you won’t count 1.7 eggs in a nest, but only 1, 2 or 3 eggs. In contrast, weight is not discrete, because it can be 69.7 kg or 71.3 kg. The difference between “continuous” and “discontinuous” variable is, however, not extremely important, since most statistical methods for metric data can deal with both continuous and discontinuous variables. More important is the difference between metric and non-metric data, because non-metric data are very different form metric data: When you have non-metric data, you count how often an observation is made. For example, you count how many patients have recovered and how many have not. Or you count the number of people with and without a bachelor degree. Or the number of smokers vs. non-smokers. What you measure for each patient is the category to which she or he belongs, e.g. smoker or non-smoker. Let’s see an example to illustrate the difference to metric data: Now, when we summarise these data, you will also see that metric and non-metric data are summarised differently. Metric data can be summarised using the mean which describes the typical weight (or median or any other value describing the average), and the standard deviation, which describes its variability. Finally, we should add the sample size to show how robust the data are. Non-metric data can be summarised by the sample size N and showing how often an attribute was counted. There is no point for providing a mean value or a standard deviation. So now it’s clear that non-metric data are conceptually different from metric data. So what about the difference between nominal and ordinal data. Nominal data are counts of characteristics or attributes that have no particular order, such as colours (blue, green, red) or gender (male, female). Also the attribute ‘smoker’ vs ‘non-smoker’ are nominal data. Here “nominal” means that we have classes with a name (which is the meaning of “nominal” data). Ordinal data are data with a hierarchy, such as small, medium and large. Or the degrees: bachelor, master and PhD. So what is an easy way to determine what kind of data you have? It’s quite simple, you only have to ask yourself “What did I measure?”, “What did I measure in each subject? What were my raw data?” When the answer is a number, then you have metric data. If the answer a characteristic or an attribute such as “red/green/blue”, “small/medium/large” or “dead/alive” then you have non-metric data. ## Conclusions The important data types are metric data and the two non-metric data types nominal and ordinal. Sometimes a data type called binary variables, for example “true” and “false” or “female” and “male” is addtionally distinguised. However, binary data is in principle nominal data. So there is not really a need to add complexity in this classification scheme. In some text books you also find the following classification scheme: Here metric data are called measurement variables, ranked variables are ordinal variables and attributes are nominal variables. So in principle this classification scheme is very similar to the scheme shown above. To conclude, whenever you are uncertain about data type, just ask the following question: What did I measure in each individual? If the answer is a number, you have metric data. If the answer is an attribute, then you don’t. Then ask yourself: Is there an order in the list of attributes? If there is, then you have ordinal data, otherwise you have nominal data.

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# Existence of only finitely many solutions Let $$r$$ be a rational number greater than $$1$$. Prove that there are only finitely many natural numbers $$x,y,z$$ such that $$(x+1)(y+1)(z+1)=rxyz.$$ Progress: For $$r=8$$, the only solutions are $$x=y=z=1$$. If $$r>8$$, then clearly there are no solutions. I'm having trouble showing the same for $$r\in(1,8)$$. Any hints or solutions are welcome. • *For $r=0 \$? – For the love of maths Nov 11 '18 at 11:54 • @Raptor rational number greater than 1 – Gareth Ma Nov 11 '18 at 12:09 • Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z. – William Elliot Nov 11 '18 at 22:16 x,y,z positive integers. (1 + 1/x)(1 + 1/y)(1 + 1/z) = r. Often there will be no solution. r = 7 for example. Assume that x,y and z's of the solutions diverge. That forces r = 1, contradicting the assumption of solutions. Wlog assume the z's are bounded and the x,y's of solutions diverge. For each solution, there's another solution with larger x,y. Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is smaller. But z can be smaller only finite many times. Wlog, assume y,z's of the solutions are bounded and x's diverge. Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z) gets larger but only finite many times because each time x,y change and there's just finite many ways they can change. Thus the x,y,z's of solutions are all bounded resulting in only finite many possible solutions.

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# Microsoft Excel Tips and Tricks The tips that follow formed part of a series in our free monthly newsletter: Subscribe now! ## Contents: ### Totals are not Exact (Rounding): Question: In Excel I am adding Sales Tax or VAT to a series of prices, and then adding them. The total is sometimes wrong by a few cents/pence/etc.. Why is this? Answer: If your prices have cent amounts, then 14% VAT will give you fractions of cents, which you won’t see if your figures are formatted for Currency with two decimals. These can add up so that your total doesn’t agree with what you see on screen, e.g. R1.11 plus 14% VAT is R 1.2654, formatted to two decimals you see R 1.27: If you add two of these, Excel shows you R 2.53 (actually R 2.5308), whereas you expect R 1.27 + R 1.27 = R 2.54. Q:  How can I fix this? A: Use the =Round() function on every calculation. Instead of =B2*1.14 use =ROUND(B2*1.14,2). This will round each result to two decimals (which is what you see anyway) and your total will then be correct. Q:  Can I round to whole Rand? A: Yes. Use =ROUND(G2*1.14,0) – the “,0” tells Excel to use zero decimal places. You can also round to tens of Rand, or hundreds, by using negative figures (-1 and -2 respectively). ### Your Own Excel Shortcut Key Q: How can I create my own Excel shortcut key for frequently-used operations? A: Record the operation as a Macro: Tools > Macro > Record New Macro. Name the Macro (no spaces). Put it in the Personal Macro Workbook if you want it to be available at all times. Here you specify the shortcut key: Hold in the Shift key and press a letter on the keyboard. (Shortcut keys with "Shift" work best because there is no risk of their replacing built-in shortcuts like Ctrl+C (Copy). You can however use the letter unshifted.) During recording, pay attention to whether you want to record Absolute or Relative references –the results can be very different! When you have finished performing the actions, Stop Recording! Save your file in case of mishaps. To test your macro, hold in the Ctrl key and press the Shift+Key combination you set earlier. ### Comparing Cells (=) The Problem: To get a quick comparison of two columns of figures or text, where some figures differ but many are the same – for example where you send somebody data, and they change some of it and send it back. Setup: To simulate the problem, open a new Excel Workbook, type several items (numbers or text) into column A (starting in cell A1), copy them to column B and then change some of them. Solution 1 (Simple): In cell C1, enter: =A1=B1 (remember that you can type “=”, then click on, or use the left arrow key to point at, cell A1, then type “=”, then click on, or use the left arrow key to point at, cell B1, then press Enter). You will see the word TRUE in the cell if the two cells match, FALSE if they do not. Now copy the formula in cell C1 down as far as your data goes (remember that you must select the cell, then you can grab the little box at the bottom right corner of the cell and drag it down as far as you need to. Another way is to highlight the formula and the cells below it as far down as necessary, then press Ctrl+D (fill down)). All the non-matching cells will show up as FALSE. ### Introducing =IF: Solution 2 (More fun): In cell D1 (keeping the previous solution in C1), enter =IF(A1=B1,””,”***”), then copy down as before. You will see asterisks opposite the mismatches, blanks where the cells match – this makes the mismatches stand out more. How this works: The syntax of the IF function is IF(comparison, true part, false part). If the comparison is True, you see true part in the cell (in this case just a blank, “”), otherwise you see false part (in this case three asterisks “***”). Note that the strings (e.g. “***”) have to be inside double quotes. ### More on =IF: Nesting The Problem: Last lesson we looked at the IF function (IF(comparison, true part, false part). This is great if you have an either-or situation, but what if I have, say, three possibilities? For example, when comparing two values A1 and B1, I want to know if the two are equal, or which one is bigger. Solution: The solution here is to nest your functions, i.e. to put one inside another. In cell C1, enter =IF(A1=B1,”=”, IF(A1<B1,”<”,”>”)) – if you are comparing a column of figures, copy down. How this works: The first IF checks if A1=B1 and, if the comparison is True, shows “=” in the cell. If the condition is false, the second IF takes place, checking if A1 is less than B1. If this is true it shows a “less than” (“<”), otherwise you see false part “greater than” (“>”), which in this case is the only other possibility. Remember with the “<” and “>” symbols, the wide part is at the big end and the point at the little end. If you prefer, and have the space, you could write it out more explicitly, e.g. =IF(A1=B1,”A=B”, IF(A1<B1,”A<B”,”A>B”)) or even =IF(A1=B1,”Equal”, IF(A1<B1,”A is smaller than B”,”A is bigger than B”)). Comparisons like this are particularly useful with dates, where it isn’t always easy to see at a glance which is bigger. Next: What if we have a dozen choices? Thereafter: Making the results appear in different colours! ### Introducing CHOOSE: The Problem: Last lesson (above) we looked at the IF function and saw how if we have, say, three possibilities, we can nest IF functions, i.e. put one inside another. This is fine for up to three or four choices, but what if we have a dozen? For example, let’s say we’d like a function that will return the name of the day of the week, given a number. Solution: The CHOOSE function has the syntax =CHOOSE(Index,Value1,Value2, Value3,…) How this works: If Index (which might be a literal number, a formula, or a cell reference) is 1, the formula shows you Value1, if 2, Value2, etc. Example: Type the headings Date; Day No, and Day in cells A1, B1 and C1 respectively. In cell A2 enter today’s date, e.g. 13/6 In cell B2 enter the formula =WEEKDAY(A2) –  this returns a number, where 1 = Sunday to 7 = Saturday. In cell C2, enter the formula =CHOOSE(B2,"Sun","Mon","Tues","Wednes","Thurs","Fri","Satur")&"day" – this will display the day of the week, e.g. Friday. Now that you’ve seen how it works, you can combine the two formulas in B2 and C2 into one, by nesting (try this in D2): =CHOOSE(WEEKDAY(A2),"Sun","Mon","Tues","Wednes","Thurs","Fri","Satur")&"day" Change the date in A2 to see the results change, or select the date and the formulas, then drag the “+” at the bottom right of the selection down to cells below. Next lesson: Making the results appear in different colours! ### Conditional Formatting 1: Background: Over the last few lessons we have looked at functions that can show us radically different results in a cell depending on certain choices. To make these differences more obvious, we would like to format the results in different colours. The Problem: We are running our bank account on a spreadsheet and we want debit amounts and overdrawn balances to show up in red. Solution: Use a currency format with negatives in red. Example: Select everything in the table below: Date Description Amount Balance 01/07/2003 Brought Forward 2000 02/07/2003 Rent -2000 03/07/2003 Credit Card -1500 04/07/2003 Salary 5000 Copy to the Clipboard (shortcut: Ctrl+C). Swap to Excel. Start a new workbook. In Cell A1, Paste (Ctrl+V) (if you prefer to retype the spreadsheet, the heading Date must be in cell A1). If the dates don’t show correctly, re-enter the top one, and fill down (Grab the little cross at the bottom right-hand corner of the cell and drag it down). Widen columns as required (drag or double-click between column headings). To calculate the running Balance, we need a formula in D3: =D2+C3 – this calculates the next balance as the previous balance plus the new Amount. Copy the formula down (Select D3. Grab the little cross at the bottom right-hand corner of the cell and drag it down to D5). Now we are ready to format the cells. Select C2 to D5 (the amount and balance columns). Click Format > Cells (or press the Ctrl+1 shortcut). Select the “Number” tab at the top. You have two choices: Number, or Currency, depending on whether you would like a currency symbol. In the case of “Number”, choose “Use 1000 Separator”. For Currency, you can change the symbol. In both cases you can set the number of decimals. The important point for us is that you should choose the bottom choice under “Negative Numbers” – this shows the figure in red, with a minus sign. Your debit amounts and balances will now show up in red. Next lesson: Making the credit amounts appear in a different colour! PS: Save this spreadsheet for the next lesson! ### Conditional Formatting 2: Background:Above we looked at using Format > Cells (Ctrl+1) > Number > Number or Currency, to show negative numbers in red.  Here we will look at an alternative that will let us use different colours (and other formatting) for various conditions. The Problem:  We are running our bank account on a spreadsheet.  We have an overdraft limit of R1200 and we want balances over this limit to show up in bold red.  We also want debit balances (i.e. between 0 and -R1200) to show up in orange. Solution:  Use Conditional Formatting. Example:  Last lesson we set up the following example, and saved it for this lesson: A B C D 1 Date Description Amount Balance 2 01/07/2003 Brought Forward 1800.00 3 02/07/2003 Rent -2000.00 -200.00 4 03/07/2003 Credit Card -1500.00 -1700.00 5 04/07/2003 Salary 5000.00 3300.00 (The formula in D3: =D2+C3, copied to the cells below, is used to calculate the running Balance) Note that we’ve reduced the figure in cell D2 from last lesson’s example! We want to conditionally format the Balances, so select cells D2..D5. Click (menu) Format > Conditional Formatting. Set Condition 1: Cell Value is | less than | -1500; Format Colour = Orange, Font Style = Bold. Click Add for a second condition. Set Condition 2: Cell Value is | less than | 0 (zero); Format Colour = Orange. Click OK. Your –R200 debit balance will now show up in orange, and the –R1700 balance in red bold. Note the order in which we did the Conditional Formatting: It applies the first formatting it finds that matches a condition. Exercise: Can you get the Credit and Zero balances to show up in Green? Next lesson: Dealing with varying conditions. PS: Save this spreadsheet for the next lesson. ### Conditional Formatting 3: Background: Above we looked at using Format > Conditional Formatting to show debit balances down to a fixed overdraft limit in orange and balances below that limit in bold red.  Now we will look at dealing with a varying limit. The Problem:  We are running our bank account on a spreadsheet.  We have a reducing overdraft limit and we want balances over this limit to show up in bold red.  We also want higher debit balances to show up in orange. Solution:  Use Conditional Formatting. Example:  Use the spreadsheet we set up last lesson, or copy and paste the one below. • Add a new column E for the Limits. • In E2 enter the starting Limit, –1900. • In E3 the formula:  =E2+100 – copy this down. • Copy the format from cell D3 (negative values in red) to this new column. • The formula in D3: =D2+C3, copied to the cells below, is used to calculate the running Balance. • Add a few more rows of data and, except for the formatting of the Balances, you will have this effect: A B C D E 1 Date Description Amount Balance Limit 2 01/07/2003 Brought Forward 1800.00 -1900.00 3 02/07/2003 Rent -2000.00 -200.00 -1800.00 4 03/07/2003 Credit Card -1500.00 -1700.00 -1700.00 5 04/07/2003 Tips 50.00 -1650.00 -1600.00 6 05/07/2003 Tips 60.00 -1590.00 -1500.00 7 06/07/2003 Salary 5000.00 3410.00 -1400.00 We want to conditionally format the Balance and then copy the format, so select cell D2 only: Click (menu) Format > Conditional Formatting and modify your existing conditions: Modify Condition 1: Cell Value is | less than | =E2; Format Colour = Red, Font Style = Bold. Note that if you use the red arrow button on the extreme right of Condition 1 to point out cell E2, it will appear as =\$E\$2.  You must delete the dollar signs because we want the formula to refer to different cells as we copy it. Keep Condition 2: Cell Value is | less than | 0 (zero); Format Colour = Orange. Click OK. Now copy the formula down to the remaining balances.  The easiest way to do this is to click on the format painter (“paintbrush”) tool (on the Standard Toolbar) and then drag over the remaining cells. Your debit balances will now show up in orange while they are above or on the adjacent limit, and in red bold when below it, as shown above. Click on cell D3, Format > Conditional Formatting. And notice that it uses =E3 in the condition, not =E2 as used in cell D2. Exercise: Save the file and repeat the Conditional Formatting, but instead use =\$E\$2 in the condition.  Copy the format down and notice the difference! ### Lookups: VLOOKUP and HLOOKUP The formula is: VLOOKUP(Value,Range,Column,FALSE) VLOOKUP finds the lookup value in the first column of the lookup range ("Range"). It then returns the value in that row, for the required column of the range. Without FALSE, it finds exact values only.  If you want it to find non-matching values (they must be sorted in ascending order), omit FALSE.  Then, if does not find the exact value, it uses the smaller one, e.g. if values in column A are 10; 50 and 100 and we look up 90, it will return the lookup value corresponding to 50.  Note that is does not interpolate! HLOOKUP(Value,Range,Row,FALSE) works identically, but horizontally:  It goes across row one until it finds (or passes) the lookup value, then goes down by the number of rows specified. ### Back to Basics: SUM The Sigma (Σ) button on the "Standard" toolbar creates the =SUM() formula in the selected cell (or cells). It will try to select the correct range: If there are figures above the cell, it will select up until it hits a blank cell or text.  If it gets it wrong, you can easily correct it, because recent versions of Excel outline the range (usually in blue): You can then grab the outline and drag it elsewhere, or just drag a corner. The format of SUM is "=SUM(range)", where range is of the form cell1:cell2, e.g. =SUM(A2:A6).  However, you can use separate cells or ranges, separated by commas, e.g.  =SUM(A2:A6, A8, A10:A12).  Of course, you don't have to use the toolbar button, you can just type the formula. What you don't need to do, is do SUM's work for it: =SUM(A2+A4+A6) is unnecessary: For that you only need =A2+A4+A6 or =SUM(A2, A4, A6) ! ### Basic Functions: AVERAGE The AVERAGE Function returns the arithmetic mean of the arguments. The syntax is exactly like SUM i.e. "=AVERAGE(range)", where range is of the form cell1:cell2, e.g. =AVERAGE(A2:A6).  However, you can use separate cells or ranges, separated by commas, e.g.  =AVERAGE(A2:A6, A8, A10:A12). Tip: Watch out for blank (empty) cells: They are not the same as cells containing zero.  Zeros are counted, and will therefore reduce the average (more cells included, but nothing added to totals) whereas blank cells are not used (e.g. 7 cells in range, but 3 are blank, so the average is taken of 4 cells only). ### Basic Functions: COUNT The COUNT Function counts the number of cells in the target range that contain numbers.  Its syntax is exactly like SUM i.e. "=COUNT(range)", where “range” is of the usual form “cell1:cell2”, e.g. =COUNT(A2:A6). Watch out for blank (empty) cells: They are not the same as cells containing zero.  Zeros are counted, blank cells are not!  It can be confusing if you have turned off zero values (Tools > Options > View >Zero Values unticked). Tip: If you want to count all filled cells including those that contain text –not just numbers– use the COUNTA (“Count All”) function instead. ### Copy & Paste Special (Values) This lesson's tip arises from a user's question.  She had an Excel spreadsheet that was to be imported into Maximizer, but the dialling code and telephone number were in separate columns: How could they be combined into one column? To demonstrate the solution, set up a sample spreadsheet with column headings "Code" and "Phone" in cells A1 and B1 respectively.  Put some dialling codes and phone numbers on the rows below (start the dialling codes with a single quote (') so that Excel does not treat them as numbers and lose the leading zero). Solution: Find or insert a blank column to the right.  In this column, row 2, enter the formula =A2&" "&B2  --we are using the ampersand (&) to concatenate two strings (the dialling code in A2 and phone number in B2) with a space in between (inside double-quotes).  The formula should show the combined code and phone number.  Now copy it down all the rows that have data.  To replace the contents of the "Phone" column with the combined number, copy the formulas, select cell B2, and Edit > Pastel Special > Values.  This copies the answers to the formula, over the old phone numbers.  Delete the "Code" column (A) and the formula column, and you are done! With Excel's wide variety of formulas, there are many ways you can manipulate your data using this method.  I use it for setting up blocks of repetitive Visual Basic code! Next lesson we'll talk about how you separate data that should be in different columns! ### Parsing Data with "Text-to-Columns" Last lesson we looked at joining data from several columns into one.  Now we'll look at the opposite: How to split up data into several columns.  For example, suppose you had a column of addresses, with the individual "lines" separated by commas, for example "PO Box 987, Gallo Manor, 2052".  You want to break this up into three columns. Set up a sample spreadsheet with a column containing a few addresses in this format. Solution: Select the data.  Click Data > Test to Columns.  A "wizard" dialog box will pop up.  Our first choice is "delimited", because we are going to break up the data at commas.  Click Next>>.  The second step is to choose the delimiter: Comma.  Click Next>>.  The third and final step lets us define the data type.  It is a good idea to click on each column that could contain a postal code, and set its column data format to text.  That way you will not lose the leading zeros on Pretoria postal codes.  Click "Finish" and the data will be broken up into columns. Next lesson: Numeric Formats and formatting. ### Numeric Formats Let's look at formatting in more detail... If you select a cell or range of cells and either click Format > Cells, or press Ctrl + 1, the following dialog box pops up: (yours may be on a different tab - click the "Number" Tab, the first one).  We'll briefly look through the categories: • General – no options here: This is exactly as entered. • Number – as above: You can set the number of decimal places, the 1000 separator (usually a comma, but depends on your Windows Regional Settings, as in fact does your decimal symbol), and how it treats negatives. Currency – Similar to Number, but you can select a currency symbol.  The way negatives are treated (minus sign or parentheses "()"), and the placement of the negative and the currency symbol, depend on your Windows Regional Settings fro Currency. • Accounting – fewer options here: You can set the number of decimal places and the symbol, but not how negative appear.  The currency symbols are lined up against the left margin. ### Date, Time and other Formats Last lesson we started looking at formatting.  Select a cell or range of cells and either click Format > Cells, or press Ctrl + 1.  The "Format Cells" dialog box pops up: Click the "Number" Tab (the first one), then select "Date": When you click on a format, the sample at the top changes to show you today's date in that format.  Notice that the top two, marked with asterisks (*) change depending on your PC's settings: A good choice if you want the user to see the date in his preferred format, but may be confusing on a PC with other settings. The PC's date settings, by the way, are done under Start > Settings > Control Panel > Regional and Language Options, firstly by language - English (South Africa), and then (if you don't like your dates to have the year first!) under Customize, on the Date tab (this is for Windows XP Professional - it will be a little different on other versions). Let's look at the other number types: • Time – not much here: Again a regional setting, and some others. • Percentage – You can set the number of decimal places.  The number is shown multiplied by 100, with a "%" sign at the end. • Fraction – Did you know Excel could show fractions?  OK, they are written out with a slash, i.e. ¼ is shown as 1/4, but good nevertheless. • Scientific – The number shows with one digit, the decimal symbol, as many decimal places as you set, an E (for "exponent") and a power of 10. e.g. 1.23E+03 means 1.23 x 10³ = 1.23 x 1000 = 1230. • Text – Numbers show exactly as entered, and are treated as text instead of numbers.  You can get a similar effect by typing a single quote (') at the beginning of numbers you enter, for example when you are entering Tshwane postal codes, type '0040 (note the quote at the start) or format your column as text first, and Excel won't change it to 40! Below we'll look at creating your own formats if none of the above suits you, using custom formats. ### Special Formats Last Tip we looked at the standard number (and other) formats in Excel.  Now we'll look at creating your own formats if none of those suits you, using Format > Cells > Number > custom formats.  Here are some interesting formats: • #,##0.00;[Red]-#,##0.00 – Thousands, two decimals, negatives in red with minus sign. • R #,##0;R -#,##0 – Thousands, currency (R), no decimals. • _ R * #,##0_ ;_ R * -#,##0_ ;_ R * "-"_ ;_ @_  – Accounting (currency symbol left-aligned), no decimals. • 0.00" Credit";0.00" Debit" – displays positive numbers followed by the word "Credit"; negatives followed by "Debit". • [Red][<=100];[Blue][>500] – the number will be shown Red if less than or equal to 100, Blue if greater than 500. dd/mm/yyyy – day / month number / 4-digit year. • dd-mmm-yy hh:mm – day - month name (3 letters) - 2-digit year, time in hours and minutes (24-hr clock). • [h]:mm:ss – Hours:minutes:seconds, where hours can exceed 24 ("hh" only runs to 23 before it carries into days). For numeric formats, the formatting characters are: # displays digits only if the number is big enough. 0 (zero) displays zeros if the number has fewer digits than there are zeros in the format.  Excel understands the colour names Red, Green, Blue, Yellow, Magenta, Cyan, Black and White. For numeric formats, you can have up to four sections separated by semicolons (;).  The first section formats positive numbers; the second is for negative numbers; the third for zero; and the fourth for text.  If you use only section 1 it is applied to all numbers; if you use only two sections, it uses the formatting of the first section for zeros.  You can hide positives, negatives, or zeros by using a blank section (i.e. a semicolon only). For example, the format [Blue]#,##0;[Green]#,##0;[Magenta]"Nil";[Red]"Text!" will display positive numbers in blue, negative numbers in green (with no minus sign), the word "Nil" in magenta if the cell contains zero, and the word "Text!" in red if it contains any text (to display the actual text, use @ or omit the last section). To find out more, search Excel Help for "Guidelines for custom number formats" or e-mail us for the Special Formats spreadsheet. ### Format Cells, Alignment Our Story (for "Prince Valiant" readers): In the last few tips we have looked at the number formats in Excel. Now we'll look at the next tab of the same dialog, Format > Cells > Alignment: Horizontally, you can align text to the left, centre, or right.  "General" (the default) aligns text to the left, numbers to the right, and logical and error values to the centre.  "Left", "Right", and "Distributed" allow you to indent by a given number of characters.  "Justify" aligns with both margins, but you can't indent from them.  A drawback to "Distributed", as compared to "Justify", is the way it handles the last line –see top example above right. Vertically, your choices are "Top", Bottom", "Centre", "Distributed", and "Justify".  The latter two seem to behave the same way: If the row height is larger than is required for the text (because it was set manually, or there is higher text on a different row) the text is spaced out vertically to cover the cell (see second example above right).  With these two choices, the text automatically wraps, otherwise you must use "Wrap Text" for multi-line text. The text angle does not work with all options. If you merge cells, only the text in the top left cell is displayed, across all the cells.  Automatic row sizing does not work if you merge cells.  This means that you have to select each range of cells to merge, individually.  One way around this is to use "Center Across Selection": It will merge the cells horizontally only, while keeping rows separate. We have been unable to work out what the greyed checkbox "Justify distributed" does or how to make it available and, judging by the fact that Help says nothing about it, Microsoft are equally puzzled! ### Format Cells, Font / Border / Patterns / Protection In the last few issues we looked at the first two tabs of the Format > Cells dialog in Excel, which cover the number formats and alignment.  We will now briefly cover the other tabs in the dialog box. On the Font tab, things are pretty straightforward.  Two points worth noting: For the size, you can type any number between 1 and 1638, not just those in the Size list. If you tick the Normal Font check box, the settings on the Font tab will reset to the default style. On the Border tab, just remember to choose the Line Style first, before you tell Excel where to put it. The Patterns tab lets you select the background colour for the cell, and then overlay a pattern (always in black!) over it if desired.  Not as versatile as PowerPoint where you can get one colour to shade into another! The final tab, Protection, is the sparsest of the lot, but there is a fair bit to be said about it. The checkbox "Hidden" hides the formula in a cell so that it is not visible in the editing box when you click on the cell. The "Locked" checkbox stops the selected cells being changed, resized, moved, or deleted.  However, both these settings only work if the sheet is protected.  The trick is to first set all the required formula cells hidden, and unlock all the cells where you want to allow the user to enter data (notice that all cells are marked "locked" by default) and then protect the sheet, by clicking Tools > Protection > Protect Sheet (password optional). To make it obvious to users where they can enter data, I like to give the spreadsheet a background colour, such as grey or light blue, and give the unlocked cells "no colour" (white), possibly with an outline border.  People are used to white text boxes on grey Windows forms, so this seems intuitive.  Of course these colour setting must be set before setting Protection on! ### Introducing Data Validation Wouldn't it be nice to prevent users from entering wrong data into your spreadsheets?  Now you can: With Data Validation. On the Data menu, click Validation.  A dialog with three tabs appears: Settings, Input Message, and Error Alert. On the Settings tab you set the Validation Criteria, in other words you tell Excel what data are acceptable.  Some possibilities: • Any Value (the default) • Whole numbers (integers) between 0 and 100. • Decimal numbers greater than 20. • Decimal numbers greater than a specified cell. • Dates greater than or equal to =TODAY() • Text with a length of 4 characters (e.g. for Post Codes - format the cell as Text too). • List from values A,B,C,D using "In-cell dropdown" (this gives a drop-down list using the values specified –the drop-down arrow only appears when you click on the cell). • List from cells =\$A\$4:\$A\$8, using "In-cell dropdown" (this gives a drop-down list using the range specified).  Since these cells could contain formulas, imagine how powerful this can be! The Input Message tab lets you set a message, with a bold title, that appears as a tooltip when the cell is selected. On the Error Alert tab you specify a message that tells the user what they have done wrong.  It appears when they enter data that does not fit your criteria.  Explain in detail what is required.  The "Style" determines the icon that is displayed: Stop, Warning or Information.  Stop prevents the user from continuing with bad data, the other two let the user accept bad data and carry on (or try again). Data validation works independently of whether you have worksheet protection set on or not.  It's good practice, though, to set the cells where you want to allow data entry with "Locked" off, and then set Tools > Protection > Protect Sheet ON, as discussed in the last Tip.  Also see it for a discussion of background colour for the worksheet and unlocked cells ### Copy Column Widths Do you have one spreadsheet with carefully-set column widths, and you'd like to easily set other columns (in the same spreadsheet, or another one) to the same widths?  Easy, with Excel! • Select the columns, the widths of which you want to copy (any cells in the columns will do) • Copy (Edit > Copy, Ctrl+C, or use the toolbar button). • Click in the leftmost of the columns that must get the new widths. • Edit > Paste Special > Column Widths > OK. That's all there is to it! ### Superscripts If you are doing a document like a Bill of Materials that requires square metres or cubic metres, you can make it look more professional by using the correct scientific abbreviations: m² and m³.  You can either use symbols or superscripts. To use symbols (this also works in Word and FrontPage, and the symbols can be copied into Access): • Click Insert > Symbol • On the top left, choose the font you want.  On the top right, look in the "Subset: Basic Latin", between the lower case letters and the accented ones. • Click the symbol you want, then the Insert button.  Next time it will appear in the list at the bottom. To use superscripts (Word is similar): • Type your "2" or "3" and select it. • Click Format > Cells (or press Ctrl+1). • Tick the "Superscript" checkbox, and click OK. • This method applies to all superscripts (and subscripts), but will make your row width a bit higher. ### IF() Again If you want Excel to check your figures and point out mistakes for you, try an IF formula.  For example, in our free Bank Reconciliation spreadsheet –which you can get by just visiting our web site and sending us an e-mail– we compare the balance copied from your internet banking (column E), with the running total in the spreadsheet (column F).  This effectively warns you if you have missed an item or copied it twice.  The formula is: =IF(E18=F18,"ok","???") If the two figures are identical, it displays "ok", otherwise "???". How this works: The syntax of the IF function is IF(comparison, true part, false part).  If the comparison is True, you see true part in the cell (in this case, “ok”), otherwise you see false part (in this case, three question marks).  Note that strings (e.g. “ok”) have to be inside double quotes.  You could of course instead use a cell reference, a number, or even another function. In the spreadsheet we actually use two nested "IF" functions, like this (blue part as above): =IF(E18="","",IF(E18=F18,"ok","???")) This allows us to copy the formulas down below the data we have copied from internet banking, without it displaying question marks.  See if you can work out how it works! Next Tip: Totalling a range of figures that meet a particular condition! ### SUMIF() Last Tip we looked at the IF function (Syntax: IF(comparison, true part, false part) ) used to return one of two values (or formulas) based on a comparison. Now consider this scenario: You have your bank statement on Excel, imported from Internet Banking.  In a column next to the data, you put in an allocation –for example Groceries, Bank Charges, Municipal, Vehicles, etc.  Of course you know how to use =SUM() to get a total for all the figures, but what if you want separate totals for each allocation?  You use =SUMIF.  The syntax is: =SUMIF(range, criteria [,sum_range]) Range is the range of cells you want Excel to check for the Criteria, which can be any logical expression (in quotes), text, or a cell reference. Sum_range is the range of cells you want Excel to sum.  [sum_range is shown in brackets to indicate that it is optional.  You don't enter the brackets.]  If you leave out sum_range, Excel will sum Range. How to use this: Let's say that you have the statement on a sheet named "April", containing the columns A: Date, B: Description, C: Amount, D: Balance, and E: Allocation.  You want to sum the amount from Column C for each allocation in Column E.  We'll assume the headings are in row 1 and the data starts in row 2. Insert a new sheet, let's call it Totals, and create two column headings, in A1: Allocation, in B1: April. From cell A2 downwards, list your allocations, one per cell, in column A. In cell B2, enter the formula: =SUMIF(April!\$E\$2:\$E\$1000, Totals!A2, April!\$C\$2:\$C\$1000) This tells it to look on the April sheet in the range E2:E1000 (expand if necessary) for the allocation (criteria) in A2 (the top allocation) and, where the criteria matches, sum the value in column C (sum_range C2:C1000). Copy this formula down as far as your allocations go, and you will get a total for each allocation. Bonus: As a check that you have not missed any allocations (or added new ones later!), it is a good idea to do a SUM of the SUMIF formulas, then compare this with =SUM(April!\$C\$2:\$C\$1000) which is of course the sum all the amounts.  Can you think of a formula that will do the check for you and display either "OK" or "*** Error! Amounts not Allocated! ***" ? Next Tip: Counting a range of figures that meet a condition. ### COUNTIF() Last Tip we looked at the SUMIF function (Syntax: SUMIF(range, criteria [,sum_range]) ) to sum values that match a criterion.  You can also count the values that match a criterion:  Use COUNTIF The syntax is: =COUNTIF(range, criteria) Range is the range of cells you want Excel to check for the Criteria, which can be any logical expression (in quotes), text, or a cell reference. How to use this: Let's say that you have a bank statement on a sheet named "April", containing the columns A: Date, B: Description, and C: Amount.  You want to count the number of credits, that is the number of positive amounts in Column C.  We'll assume the headings are in row 1 and the data starts in row 2. Insert a new sheet, let's call it Statistics, and create a column heading in B1: April. In cell B2, enter the formula: =COUNTIF(April!C2:C1000, ">0") This tells it to look on the April sheet in the range C2:C1000 for the criteria that the number is positive (">0" - note the quotes!) and, where the criteria is true, count it. ### COUNT, COUNTA, COUNTBLANK Last Tip we looked at the COUNTIF function (Syntax:  =COUNTIF(range, criteria) ) to count the values that match a criterion. To complete our knowledge of the "counting" functions, this edition we will look at three simpler ones: COUNT(Range) counts the cells that contain numbers. COUNTA(Range) (Count All) counts the cells that are not blank (contain numbers or text). COUNTBLANK(Range) counts the cells in the range that are blank (empty).  The formula counts any cell that looks empty if unformatted, including formulas that return empty strings (""), and blank strings (single quote in cell), but not zeros. The range may be a single range or several ranges or cells separated by commas (e.g. C2:C10,D12,E2:E10 ).  In the first two functions you can also put in values instead of ranges and they are counted appropriately. ### Unusual but Useful: FLOOR and CEILING Syntax: FLOOR(number, multiple) CEILING(number, multiple) FLOOR truncates "number" down (toward zero) to the next lower multiple of "multiple". CEILING truncates "number" up (away from zero) to the next higher multiple of "multiple". Unlike other rounding functions, these two allow you to use any multiple, so you are not restricted to whole numbers.  You can "round" to multiples of 0.065 or 100 000 (the latter is useful if you run a shop in Zimbabwe).  Bear in mind that this is not really "rounding", since the result is always down for FLOOR, or up for CEILING –it does not go to the nearest value as ROUND would do, and it is taking it to a multiple. Example: You need 11.03 square metres of tiles, but they are sold in boxes of 0.4 m².  How many square metres must you buy? Answer: =CEILING(11.03, 0.4)  =11.2 (A pundit might say, obviously these are CEILING tiles, not FLOOR tiles!) One quirk: Both "number" and "significance" must be of the same sign.  =FLOOR(-23, -2) works (-22), but =FLOOR(-23, 2) gives #NUM! ### Unusual but Interesting: EVEN and ODD Syntax: EVEN(number) rounds "number" up (away from zero) to the next even whole number. ODD(number) rounds "number" up (away from zero) to the next odd whole number. Of "EVEN", Microsoft Help says: "You can use this function for processing items that come in twos. For example, a packing crate accepts rows of one or two items. The crate is full when the number of items, rounded up to the nearest two, matches the crate's capacity." Microsoft Help doesn't give an example of where you might use the latter function.  Which is odd, if you think about it. Our contrived Example: The Odd Fellows Hiking Club hiked upstream on Sunday 7 January 2007, downstream the next Sunday, and has alternated since then.  We want a formula that will give the direction of next Sunday's hike. On a blank spreadsheet, copy the following into the indicated cells: Cell Formula Explanation A2: =TODAY()-WEEKDAY(TODAY())+8 Date next Sunday A3: =A2-DATE(2007,1,7) Number of days since first hike A4: =IF(ODD(A3)=A3, "downstream", "upstream") Direction ### Really Odd Functions We looked at ODD last month, but these functions are odder!  We haven't found a use for them but, who knows, you might!  Let us know if you do! Syntax: Description Example =ROMAN(number) Displays the number in roman numerals.  Various styles are available. ROMAN(1984)  = MCMLXXXIV =FACT(number) Returns the factorial of a number. The factorial of "n" is equal to 1*2*3*...* n. FACT(4) = 24 =PRODUCT(number1, number2,...) Multiplies all the numbers given as arguments and returns the product.  Why not just enter number1 * number2 * number3…?  You might want to use an array: =PRODUCT(A2:A8) is slightly less typing than A2*A3*A4*A5*A6*A7*A8 PRODUCT(1,2,3,4) = 24 =COMBIN(number, number chosen) Number of combinations for a given number of items. Use COMBIN to determine the total possible number of groups made up from a given number of items.  E.g. Given 15 Proteas in the squad, how many different cricket teams could you make? COMBIN(15,11) = 1365 =PERMUT(number, number chosen) Number of permutations of "number chosen" objects that can be selected from "number" objects.  A permutation is a set of objects or events where the order is significant.  E.g. Given 15 Proteas, how many batting orders could you have? PERMUT(15,11) = 54486432000 And you thought Mickey Arthur had an easy job! ### The Cowboy Function, ROUNDUP (and its faithful companion) We have previously mentioned the function =ROUND(number, digits), which rounds a number to a given number of digits.  It always rounds to the nearest number that has the required number of digits. E.g. ROUND(3.1415, 3) = 3.142, ROUND(-3.1415, 3) = -3.142, and ROUND(3141.59, -2) = 3100. Here are two lesser-known functions that always round in the same direction: Syntax: Description Example =ROUNDUP(number, digits) Rounds a number up (away from zero) to the given number of digits. ROUNDUP(PI(), 2) = 3.15 =ROUNDDOWN(number, digits) Rounds a number down (towards zero) to the given number of digits. ROUNDDOWN(PI(), 2) = 3.14 Both behave like ROUND, except that ROUNDUP always rounds a number up and ROUNDDOWN always rounds a number down.  As with ROUND, a negative value for digits works too: -1 rounds to tens, -2 to hundreds, etc. As with last month's functions, haven't found a use for these two –but you might! ### A Little Time-Saver: Entering Decimals Without the Point Try Tools > Options > Edit tab > Fixed Decimal.  When turned on, this allows you to enter numbers without the decimal.  For example, if under this option, "Places" is set to 2, you can enter 1054 and get 10.54 in the cell.  It saves hunting for the full stop if you are entering a lot of numbers that have cents anyway. It can also be hell to find if you happen to turn it on by accident and then forget what you did, as a customer of our found recently!  Or if a co-worker decides to play a practical joke on you and you don't know about this!  Suddenly all the numbers you enter are being divided by 100!  Fortunately, if you type the decimal point, Excel still respects it. ### Add Useful Features to Your Toolbar If you frequently Paste Special > Values, or Paste Special > Formats, etc. you can speed up your work by adding these features to your toolbar.  Do the following: • Make sure the "Standard" toolbar is visible: Click on the grey area to the right of the Help menu, and on the pop-up menu click "Standard" if it not already ticked. • Click again on the grey area to the right of the Help menu, and on the bottom of the pop-up menu click "Customize". • The "Customize" dialog box pops up.  On the second tab, "Commands", under Categories on the left. click "Edit". • Scroll done in the right-hand list until you see "Paste Formatting" (see illustration >>). • Drag "Paste Special", "Paste Formatting", "Paste Values" and anything else you use frequently, onto the Standard toolbar, probably best next to the Copy button. • Click Close when done. The toolbar will look like this when you are finished (new items circled): ### Using Excel to Produce Data for Other Programs Excel formulas are very useful to calculate text that you can copy into other programs. For example, we go walking on the Sandspruit most Sundays (and you are welcome to join us, just phone first).  I want to make up a list of Sundays with their dates to put into an e-mail and on the web site. Start by entering a known Sunday, say 10 Feb, into cell A2.  Press Ctrl+1 to get Format > Cells.  On the Number tab, choose Custom, and type in the format dddd, d mmmm yyyy – this formats the cell as Sunday, 10 February 2008. In cell A3, add 7 days by entering the formula: =A2+7.  Excel is clever enough to format this the same as cell A2. Click on cell A3, grab the little plus sign (+) at the bottom right-hand corner and drag down as far as required.  Your results, ready to copy and paste into any other program, will look like this: Hike Dates Sunday, 10 February 2008 Sunday, 17 February 2008 Sunday, 24 February 2008 Sunday, 2 March 2008 ### Using Excel Formulas to Produce Data for Other Programs (II) This tip is intended to stimulate you into thinking about ways that you can use Excel to do your other work more efficiently, for example by avoiding unnecessary re-typing. Last time we used a simple formula with formatting to produce a table like this that we could copy into a Word document, an e-mail or a web page: Hike Dates Sunday, 10 February 2008 Sunday, 17 February 2008 Sunday, 24 February 2008 Sunday, 2 March 2008 This time we will take it a little further.  Staying with our example of Sunday hikes, we walk upstream one weekend and downstream the next.  The easiest way to add this is to use another column on the right, headed "Direction", and in the two cells under this, the two directions.  The trick here is to select both directions, grab the little "+" at the bottom right-hand corner of the selection, and drag it down as far as required.  The two-cell pattern repeats.  Your results should look like this (dates adjusted for the new month): Hike Dates Direction Sunday, 9 March 2008 downstream Sunday, 16 March 2008 upstream Sunday, 23 March 2008 downstream Sunday, 30 March 2008 upstream Sunday, 6 April 2008 downstream Sunday, 13 April 2008 upstream ### Using Excel Formulas to Produce Data for Other Programs (III) While you are unlikely to use this exact example, we hope this tip will help to stimulate your thinking about ways of using Excel to do your other work more efficiently by avoiding unnecessary re-typing. Last time we used an example with a simple formula to give us two columns, one with dates and one with our walking direction (upstream and downstream).  Now we will expand on this by using formulas. To recap: We have column headings "Hike Dates" in A1 and "Direction" in B1.  In cell A2 we entered the date of a known Sunday.  Using Ctrl+1 (Format > Cells) on the Number tab under Custom, we used the format dddd, d mmmm yyyy to format the cell.  In cell A3, we added 7 days using the formula: =A2+7, and copied it down as far as needed.  We then alternated the words "upstream" and "downstream" in column B. Now put a new column heading "Date and Hike Direction" in cell C1.  In cell C2, enter the formula =A2&B2 to combine the values from the two cells. Not very satisfactory, is it?  The problem is that Excel uses the numeric value of the date, not the formatted string we see in the cell. Replace the formula in cell C2 with this one: =TEXT(A2,"dddd, d mmmm yyyy")&B2 The TEXT function formats the date value to look like a date. One further thing is needed: We need a space between the two values.  Replace the formula in cell C2 with this one: =TEXT(A2,"dddd, d mmmm yyyy")&" "&B2 We use the ampersand ("&") to concatenate (join) strings, to put a string consisting of a space (surrounded by double quotes to show that it is a string) in between the two parts of the formula.  Our results look like this, with the third column ready to be copied into an e-mail or Word (dates adjusted for the new month): Hike Dates Direction Date and Hike Direction Sunday, 13 April 2008 downstream Sunday, 13 April 2008 downstream Sunday, 20 April 2008 upstream Sunday, 20 April 2008 upstream Sunday, 27 April 2008 downstream Sunday, 27 April 2008 downstream Sunday, 4 May 2008 upstream Sunday, 4 May 2008 upstream Sunday, 11 May 2008 downstream Sunday, 11 May 2008 downstream ### Using Excel String Formulas to Help Other Programs As you know, we write computer programs.  We often need to produce repetitive code, with small changes from line to line.  Excel formulas are very useful for this.  For example, I need to produce ten lines like this:  We\$ = We\$ & Range\$("fieldname", Me.fieldname) where fieldname is the name of ten different fields, one on each line. With Excel this can be done:  List the field names in (say) column B starting at cell B2.  In C2, put the formula =" We\$ = We\$ & Range\$(""" & B2 & """, Me." & B2 & ")" and then copy it down as far as required.  It's a formula, so it starts with =.  This is followed by a double-quote (") because the formula uses strings (text) instead of numbers.  Wherever I want the variable (fieldname) to appear, I replace it with " & B2 & " – the first double quote ends the string, the last double-quote starts a new one, the B2 refers to that cell, and the ampersands (&) concatenate (join) the strings.  Finally the formula ends with a double-quote to end the final string. Select the formulas, copy, and paste into our program. Notice that in our result we want the first fieldname in double quotes (" "), but we are already using double-quotes enclose stings.  To show that we actually want the double-quote character, and we are not starting or ending a string, we double up the double-quotes (""). ### Helpline Tip: How to Protect Data in Excel Kingsley writes: Please tell me how to protect a column on Excel. Answer: You can't protect a column as such. What you do is to (1) Unlock cells in which you want to allow input, and then (2) Protect the worksheet, which then allows changes to the unlocked cells only. To do this, you: 1. Select cells for which you want to allow data input. Click Format > Cells > Protection tab: CLEAR the checkbox "locked". Click OK. Repeat for all input cells. 2. Click Tools > Protection > Protect Sheet. Type a password (optional) if you want extra security. Click OK. 3. If you do not want to allow people to add or remove sheets from the workbook, also use Tools > Protection > Protect Workbook. ### Contributed Tips When using SUM in Excel you can include random fields by holding down the Ctrl key and clicking on each field. Very useful.  [by Dan Elliott of Data Solution Services: dan.dss AT webmail.co.za] Would you like to add a tip of your own (due acknowledgement will be given!) – click here. See also:  Excel Programming Tips, Microsoft Word Tips, Microsoft Access Tips, Maximizer Tips, Tips on Windows and other Windows Programs. Press Ctrl+F to search this page for keywords. Search this site: using Custom Search

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# What Is The Meaning Of Rate? ## What is rate change? A rate of change is a rate that describes how one quantity changes in relation to another quantity. If x is the independent variable and y is the dependent variable, then. rate of change=change in ychange in x. Rates of change can be positive or negative.. ## What does TBH mean sexually? to be honest. Between TBH (to be honest) and lit, it’s nearly impossible to keep up with the lingo that teens are using on their smartphones. Unfortunately, more and more code words are popping up that allow teens to secretly send vulgar message even under the watch of their parents. ## What is K in first order reaction? k is the first-order rate constant, which has units of 1/s. The method of determining the order of a reaction is known as the method of initial rates. The overall order of a reaction is the sum of all the exponents of the concentration terms in the rate equation. ## What is another word for rate of change? Slope is also a synonym for “rate of change” but I couldn’t imagine using the word slope in the context of soup:-) – ukayer Feb 14 ’11 at 1:12. 2. “Speed” is rate of change of position. ” Acceleration” is rate of change of speed. ” Jerk” is rate of change of acceleration. – ## What does TBT stand for? Throwback ThursdayTBT stands for Throwback Thursday. People use it when sharing old photos and videos of themselves for nostalgia. ## What does it mean when someone asks for a rate? Rate is another social-media prompt—again, appearing on its own, or in the full three-part iteration—in which the user asks their social-media friends and followers to rate them. Usually, this is on a scale of one to 10 (with 10 being the highest rating). ## What do we mean by rate? the amount of a charge or payment with reference to some basis of calculation: a high rate of interest on loans. a certain quantity or amount of one thing considered in relation to a unit of another thing and used as a standard or measure: at the rate of 60 miles an hour. ## What is the meaning of rate me? To give an opinion on something for example from 1 to 10. Or stars. ## What is order reaction? The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant. Thus, for a first-order reaction, the rate is dependent on the concentration of a single species. … The words “order of reaction” may also be used to refer to the mechanism of a reaction. ## How is rate of change used in real life? Other examples of rates of change include: A population of rats increasing by 40 rats per week. A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) ## What does I rate mean in slang? Rate: a slang word for ok or fine. ” ## How do you identify rate? Key PointsReaction rate is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period Δt.The rate of reaction can be observed by watching the disappearance of a reactant or the appearance of a product over time.More items… ## What does AFK mean? away from keyboardAfk is an abbreviation for away from keyboard. It lets people know that you will not be at your keyboard for a while, or that you will not be online for a period of time. ## How do I calculate rate of change? Understanding Rate of Change (ROC) The calculation for ROC is simple in that it takes the current value of a stock or index and divides it by the value from an earlier period. Subtract one and multiply the resulting number by 100 to give it a percentage representation.

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Subsections # 6.2 Kirsch solution for stresses around a cylindrical cavity ## 6.2.1 Cylindrical coordinate system The cylindrical symmetry of a wellbore prompts the utilization of a cylindrical coordinate system rather than a rectangular cartesian coordinate system. The volume element of stresses in cylindrical coordinates is shown in Fig. 6.4. The distance is measured from the center axis of the wellbore. The angle is measured with respect to a predefined plane. The normal stresses are radial stress , tangential or hoop stress , and axial stress . The shear stresses are , , and . ## 6.2.2 Kirsch solution components The Kirsch solution allows us to calculate normal and shear stresses around a circular cavity in a homogeneous linear elastic solid . The complete Kirsch solution assumes independent action of multiple factors, namely far-field isotropic stress, deviatoric stress, wellbore pressure and pore pressure. 1. Isotropic far-field stress: The solution for a compressive isotropic far field stress is shown in Fig. 6.5. The presence of the wellbore amplifies compressive stresses 2 times all around the wellbore wall in circumferential direction. The presence of the wellbore cavity also creates infinitely large stress anisotropy at the wellbore wall all around the wellbore wall, since in this case. Stresses decrease inversely proportional to and are neglible at 4 radii from the wellbore wall. 2. Inner wellbore pressure: The solution for fluid wall pressure in the wellbore is shown in Fig. 6.6. We assume a non-porous solid now. This assumption will be relaxed later on. Wellbore pressure adds compression on the wellbore wall , and induces cavity expansion and tensile hoop stresses all around the wellbore. 3. Deviatoric stress: The solution for a deviatoric stress aligned with is shown in Fig. 6.7. The deviatoric stress results in compression on the wellbore wall at and , and in tension at and . Hence, the presence of the wellbore amplifies compressive stresses 3 times at and . The variation of stresses around the wellbore depend on harmonic functions and . 4. Pore pressure: The last step consists in assuming a perfect mud-cake, so that, the effective stress wall support (as shown in Fig. 6.6) is instead of . ## 6.2.3 Complete Kirsch solution Consider a vertical wellbore subjected to horizontal stresses and , both principal stresses, vertical stress , pore pressure , and wellbore pressure . The corresponding effective in-situ stresses are , , and . The Kirsch solution for a wellbore with radius within a linear elastic and isotropic solid is: (6.2) where is the radial effective stress, is the tangential (hoop) effective stress, is the shear stress in a plane perpendicular to in tangential direction , and is the vertical effective stress in direction . The angle is the angle between the direction of and the point at which stress is considered. The distance is measured from the center of the wellbore. For example, at the wellbore wall . An example of the solution of Kirsch equations for MPa, MPa, and MPa is available in Figure 6.8. The plots show radial and tangential effective stresses, as well as the calculated principal stresses and . Let us obtain and at the wellbore wall . The radial stress for all is (6.3) The hoop stress depends on , (6.4) and it is the minimum at and (azimuth of ) and the maximum at and (azimuth of ): (6.5) These locations will be prone to develop tensile fractures ( and ) and shear fractures ( and ). The shear stress around the wellbore wall is . This makes sense because fluids (drilling mud) cannot apply steady shear stresses on the surface of a solid. Finally, the effective vertical stress is (6.6)

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You Selected: Keyword math place value worksheets Subjects Math showing 1-24 of 19,396 results This file includes a cover page with the common core state standards that are aligned to the problems. The problems are great for revisiting/reviewing/assessing (during the unit, or right before state tests) place value concepts such as expanded notation, standard and word form, comparing numbers, Subjects: Types: CCSS: \$3.00 124 Ratings 4.0 PDF (1.57 MB) This file includes a cover page with the common core state standards that are aligned to the problems. Each individual page also has the standard displayed on the top of the page that the problems align to. The problems are great for teaching/revisiting/reviewing, place value concepts such as ex Subjects: Types: CCSS: \$5.00 26 Ratings 4.0 PDF (230.39 KB) This packet contains three common core aligned place value worksheets with answer key. Place Value: Standard Form, Expanded Form and recognizing that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. I always try to make math fun Subjects: Types: \$2.00 15 Ratings 4.0 PDF (690.37 KB) Two great worksheets to help students develop their knowledge of the Maths concept Place Value. Answers included. Subjects: \$1.00 1 Rating 4.0 PDF (1.43 MB) These are Fall number words, mystery picture worksheets. It's a fun coloring activity for your students as they learn more about the word form of numbers. To use, let your students answer the problems. Then, they must match their answers to a list to get the letter-color combination. Finally, they Subjects: Types: \$2.99 not yet rated N/A ZIP (1.89 MB) FREE not yet rated N/A PDF (326.77 KB) Place Value Worksheet Pack - 37 PLACE VALUE WORKSHEETS WITH DIFFERENT TYPES OF ACTIVITIES (including ASSESSMENT PAGES) This FUN and ENGAGING resource contains place value worksheets that are designed to help students identify the numbers in HUNDREDS, TENS, and ONES place. Types of activities inc Subjects: Types: Also included in: Place Value Worksheets BUNDLE \$5.00 824 Ratings 4.0 PDF (1.78 MB) First Grade Math Curriculum! This unit is part of a year-long math curriculum. ***SAVE \$\$\$\$ with the GROWING BUNDLE by clickingHERE! *** ***SEE UNIT 4 IN ACTION HERE I am beyond THRILLED today to share with you a brand new First Grade Math Curriculum that is designed to make math FUN, hands-on a Subjects: Types: \$15.00 467 Ratings 4.0 ZIP (110.23 MB) PLACE VALUE WORKSHEETS : This resource offers 43 place value worksheets with 9 different types of activities designed to aid students practice place value skills TENS and ONES in a fun and engaging way. The product also includes 3 assessment pages. * Color tens and ones to show each number * Fill Subjects: Also included in: Place Value Worksheets BUNDLE \$6.00 540 Ratings 4.0 PDF (3.12 MB) We've updated our best seller! We've added more place value worksheets for you to choose from, which includes standard form, expanded form, word form, rounding and more! No need to rush creating worksheets/packets. Just choose from the wide variety of sheets that will suit your needs. *** For K-2, Subjects: Types: \$24.95 \$9.99 403 Ratings 4.0 ZIP (23.49 MB) Help your students to develop a conceptual understanding of place value with this two week place value unit! The unit allows students to understand how to read and write numbers is written, standard, and expanded form, as well as the value and place value of numbers. This individual unit is also a Subjects: Types: Also included in: 3rd Grade Math - Math Workshop & Guided Math Bundle \$7.50 978 Ratings 4.0 ZIP (39.5 MB) Place Value Worksheets: This resource contains 80 FUN and ENGAGING place value worksheets with over 20 different activities. It also includes assessments! This product is a bundle of my following products: *Place Value Worksheets (Tens and Ones) :35 FUN worksheets and 1 poster and *Place Value W Subjects: Types: \$11.00 \$7.70 271 Ratings 4.0 Bundle Build your students' understanding of Numbers 1-100 with these fun and interactive number worksheets. Aligned with the Australian Curriculum and displaying the content descriptors at the bottom of each page, this pack is the perfect tool to build a strong understanding of Number. Included in this pa Subjects: Types: \$4.50 245 Ratings 4.0 PDF (21.57 MB) Place value worksheets: FOURTEEN differentiated place value (numbers up to 50) cut and pastes to help your students practice place value in base 10! There are: - FIVE Place Value to 20 cut and pastes - TWO cut and pastes that practice tens (20, 30, 40, etc.) - TWO Place Value to 10 cut and pastes Subjects: Types: \$4.00 669 Ratings 4.0 PDF (344.59 KB) **Updated Version. Here's a mega bundle of our printable decimals worksheets. You'll save tons of time with various ready-to-use worksheets with answer keys. Topics include decimal operations (add, subtract, multiply, divide), decimal place value, expanded form, word form, rounding, and fraction-de Subjects: Types: \$48.90 \$19.95 95 Ratings 4.0 ZIP (132.92 MB) Place Value Worksheets These Place Value Worksheets provide extra practice for your first or second graders. They can be used for seat work, centers, assessments, interactive notebooks, and fast finishers. If you like these. Please check out my Place Value, Expanded Form and Fact Families Bundl Subjects: Types: Also included in: Math Worksheets 1st and 2nd Grade | Bundle \$6.00 338 Ratings 4.0 PDF (5.05 MB) Second Grade Math Made Fun Curriculum: A Complete and comprehensive 2nd Grade Math Curriculum, which covers all grade level standards in a variety of ways. This is Unit 4 of the full curriculum. Grab the BUNDLE HEREI am beyond THRILLED to share with you a brand new 2nd Grade Math Made Fun Curriculu Subjects: Types: \$15.00 75 Ratings 4.0 ZIP (163.69 MB) These fun one hundreds chart mystery pictures are a fun activity to develop place value skills. Students will work with tens and ones to determine what color to place on each square of the 100s chart to reveal a mystery picture. Each worksheet shows a two digit number in either expanded form, wor Subjects: Types: CCSS: Also included in: Place Value 100s Chart Color By Number Mystery Picture Puzzles Bundle \$3.75 299 Ratings 4.0 PDF (2.14 MB) Thanksgiving Math Hundreds Chart Mystery Pictures: Practice place value, sequencing, addition and subtraction with these fun Thanksgiving Mystery Pictures. These activities are perfect for math centers, early finishers, morning work or homework. Teachers have commented: "Love the embedded differen Subjects: Types: CCSS: \$3.99 528 Ratings 4.0 ZIP (4.8 MB) These place value no prep printables offer differentiation for place value, fun themes for place value, playful titles and many engaging twists for place value, yet it is still no prep and comprehensively covers place value! Not to mention, it is geared just for teaching place value to first grade b Subjects: Types: CCSS: \$7.00 397 Ratings 4.0 PDF (11.26 MB) Build your students' understanding of Numbers 1-1000 with these fun and interactive number worksheets. Aligned with the Australian Curriculum (displaying the content descriptors at the bottom of most pages) this pack is the perfect tool to build a strong understanding of Number and Place Value. A Subjects: Types: \$4.50 176 Ratings 4.0 PDF (24.62 MB) Place Value Worksheets and Games - Kindergarten Math Unit 7 This is an awesome set of 12 Kindergarten Math Centers and 25 No Prep, Kindergarten Place Value Worksheets. The themes are fun monsters, pig builders, cupcakes, and other kid favorites that you can use all year long. These games and works Subjects: Types: Also included in: Kindergarten Math \$7.00 69 Ratings 4.0 ZIP (51.21 MB) These fun one hundreds chart mystery pictures are a fun activity to develop place value skills. Students will work with tens and ones to determine what color to place on each square of the 100s chart to reveal a mystery picture. Each worksheet shows a two digit number in either expanded form, wor Subjects: Types: Also included in: Place Value 100s Chart Color By Number Mystery Picture Puzzles Bundle \$4.25 191 Ratings 4.0 PDF (3.39 MB) Kindergarten students need daily number practice. These printable, easy-to-use kindergarten Number of the Day activities are the perfect way to practice this important concept. Your kids will love them and will ask for more! Plus, they are great for morning work!***Purchase this BUNDLE and save o Subjects: Types: \$54.00 \$44.00 71 Ratings 4.0 Bundle showing 1-24 of 19,396 results Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.

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# How do you move a cube with javascript Hi guys, I am new to Unity and I wanted to ask how do I translate the position of a cube that I made with script. I am just testing out small little scripts to see how things work. Here is my code: ``````var myCube: GameObject; var charCube: GameObject; var x = 0; var y = 1; var z = 0; function Start () { var aPositions : Array = [new Vector3(0,0,0), new Vector3(0,0,1), new Vector3(0,0,2), new Vector3(0,0,3), new Vector3(0,0,4), new Vector3(1,0,1), new Vector3(1,0,2), new Vector3(1,0,3), new Vector3(1,0,4)]; var rot : Quaternion = Quaternion.identity; for(var i=0; i<aPositions.length; i++) { Instantiate(myCube, aPositions*, rot);* `````` } Instantiate(charCube, new Vector3(x,y,z), rot); } function Update () { • if (Input.GetButtonDown(“Vertical”))* • `````` //I want the charCube to move forward if I pressed 'w'.* `````` } I tried a lot of things and I am sure that it is not too difficult. But I just can’t wrap my finger around it. Any help would be appreciated. Thanks guys! function Update() { // Move the object forward along its z axis 1 unit/second. transform.Translate(Vector3.forward * Time.deltaTime); `````` // Move the object upward in world space 1 unit/second. transform.Translate(Vector3.up * Time.deltaTime, Space.World); } `````` look at Unity - Scripting API: Transform.Translate for more If it’s just a case of moving it, you’ll want Transform.Translate. However, if you want it to detect collisions on the way, then that probably won’t work for you, and you’ll need to do some kind of physics-based solution, like Rigidbody.AddForce.

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Chapter 7 107 terms by nixm01 Study  only Flashcards Flashcards Scatter Scatter Scatter Scatter Create a new folder Continuous spectrum a continuous color pattern that is separated from a beam of sunlight passing through a prism. Line spectrum a spectrum containing only discrete lines of different colors Electromagnetic spectrum consists of electromagnetic radiation of different wavelengths (or frequencies) 1/s Amplitude the height of a wave The amplitude of light determines what? Brightness (the larger the amplitude, the brighter the light) Wavelength the crest-to-crest distance between waves It's color Frequency the number of complete cycles the light wave makes per second λ v (v) x (λ) = c c = speed of light in a vacuum 3.0 x 108 m/s higher lower 10-9 10-6 10-9 10-6 Hertz (Hz) Meters (m) m/s Quantum Theory energy of electromagnetic radiation is quantized (consisting something continuously) Electromagnetic radiation can only be emitted or absorbed in small packets of energy called what? Quantum or photons "Quantum" or "photons" smallest quantity of energy E = hv When finding energy of quantum (or photon), what does h stand for? Planck's constant (6.63 x 10-34 J-s When finding energy of quantum (or photon), what does v stand for? Frequency of the radiation What is the consistant property of the numbers when stating how much evergy is absored or emitted in quantums (or photons)? They are whole-number multiples of hv (Ex: 2 or 3, not 1.67 or 4.98) When calculating the energy of a photon (quantum), what equation is used to find frequency? v = c/λ (frequency = the speed of light / wavelength of radiation n is used to represent what? The number of a particular orbital The orbit number (n) is found by looking at what? The period (or row) number on the periodic table (1-7, going down) higher Lower Where are the electrons when it is said to be in an excited state? Further away from the nucleus Orbital region inside an atom where electrons are most likely to be found 90% of the time What is the difference between an orbit and an orbital? Which one does an electron follow? orbit is a direct path of an electron around a specific atom. an orbital is a region where the electron is most likely to be found. All electrons follow an orbital (other than Hydrogen) Ground According to Bohr's explanation of the atomic spectra of Hydrogen, explain the energy absorbed or released when an electron stays in an orbit (ground state or excited state) no energy will be emitted According to Bohr's explanation of the atomic spectra of Hydrogen, explain what happens when an electron jumps from an orbit with a higher energy to an orbit with a lower energy energy will be emitted in the form of electromagnetic radiation According to Bohr's explanation of the atomic spectra of Hydrogen, what is determined by the energy difference between the final state and the initial state? The frequency (or wavelength) of the electromagnetic radiation Hydrogen Which model best describes how the electrons really exist in an atom: Bohr Model or Quantum Mechanical Model? Quantum Mechanical Model. (Bohr's model only holds true for Hydrogen) It is impossible to simultaneously determine the exact ____________ and the exact ___________ of a subatomic particle. position and velocity Why can we only determine probability of the velocity and location of an electron? Electrons have wave-like behaviors The four quantum numbers describes what? Describes the electrons and the atomic orbitals that they are in n, l, ml and ms When determining quantum numbers, what does 'n' stand for? Principal quantum number - The number of a particular orbitals When determining quantum numbers, what does principal quantum number (n) determine? The energy and size of the orbital When determining quantum numbers, what does 'l' stand for and what does it determine? Angular Momentum Quantum Number - The shape of the orbitals 0-3 0 0,1 0,1,2 0,1,2,3 What are electron shells? A collection of orbitals with the same principal quantum number (n) When determining the quantum number 'l', what are the orbitals assigned to each possible l value? 0=s, 1=p, 2=d, 3=f What are sub-shells? A type of orbital (s, p, d, f, etc.) within a principal energy level How are the sub-shells found and how do you express them? Given n and l values, sub-shells can be designated with a number and a letter When determining quantum numbers, what does 'ml' determine? The orientation of the orbital in space The value of 'ml' is limited by the value of what other quantum number? 'l' (angular momentum quantum number) What are all the possible values of 'ml'? -1, 0, 1+ (when only dealing with main elements) s p d f What are the possible values for the 's' sub-shells? 0 (contains 1 orbital) What are the possible values for the 'p' sub-shells? +1, 0, -1 (contains 3 orbitals) What are the possible values for the 'd' sub-shells? +2, +1, 0, -1, -2 (contains 5 orbitals) What are the possible values for the 'f' sub-shells? +3, +2, +1, 0, -1, -2, -3 (contains 7 orbitals) When determining quantum numbers, what does 'ms' indicate? Direction of electron spin +1/2 -1/2 The value of 'l' is limited by the value of what other quantum number? 'n' (principal quantum number) atomic orbital 1 3 5 5 What is the difference between an electron shell and a sub-shell? shell is when electrons have same value of 'n', sub-shell is when electrons have same values of 'n' and 'l' What are the three rules for electron configuration? 1) electrons occupy lower-energy orbitals first before occupying higher-energy orbitals. 2) (Pauli exclusion principle) no more than two electrons can occupy one orbital and two electrons in the same orbital must have opposite direction of spins. 3) (Hund's Rule) the most stable arrangement of electrons in the same subshell has the largest number of unpaired electrons, all with the same direction of spon Valence electrons electrons in the outermost electron shell of an atom Core electrons the electrons in the inner electron shells of an atom What are the electron configuration designated orbital letters of the periodic table? Columns 1 and 2 = s orbital, columns 2 through 8 = p orbital, transition metals = d orbitals 10 6 2 Same Increases Decrease Smaller Larger True Why are Cations smaller than their corresponding neutral atoms? Because neutral atoms lose electrons to form cations Why are Anions larger than their corresponding neutral atoms? Because neutral atoms gain electrons to form anions Electrostatic Ionic Transferred Non-metals Hydrogen What is the formula used to find the formal charge of an atom? (# of valence e-) - (# of lone pair e-) - ½(# of bonding electrons) Singe bond two atoms share one pair of electrons (each single bond has 2 electrons) Double bond two atoms sharing two pairs of electrons (each double bond has 4 electrons) - shorter and stronger than single bond Triple bond two atoms sharing three pairs of electrons (each triple bond has 6 electrons) - shorter and stronger than double bond Resonant Structures all the possible structures when two or more Lewis structures can be drawn for the same compound with multiple bonds Example: Press Ctrl-0 to reset your zoom Please upgrade Flash or install Chrometo use Voice Recording. For more help, see our troubleshooting page. Create Set

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# Geometric Sequences 3 teachers like this lesson Print Lesson ## Objective SWBAT find the nth partial sum of a geometric series. #### Big Idea Dive into some work with geometric sequences to gain understanding and prove a formula. ## Check Homework 10 minutes One question from yesterday’s homework that students always ask about is #4. In this video I go through one strategy that usually seems to make sense for students. Unable to display content. Adobe Flash is required. ## Launch and Explore 20 minutes Going into this lesson, my students have an understanding of what a geometric sequence is and how to describe one explicitly and recursively. Today we are going to focus on finding the nth term and the nth partial sum of a geometric series. I start class by giving them this worksheet and have them go through questions #1-7 for about 15 minutes or so with their table groups. These problems are more challenging than what they have been working on because they must figure out bothr explicit and recursive formulas given the second and fourth term, for example. One conceptual hurdle that I notice in my students' work is when the common ratio is not a rational number. For Problem #4, for example, the second term is 5 and the fourth term is 50. If a student gets stuck on this, I ask him/her to write the sequence like 5, ____ , 50 with a blank missing for the third term. Then I ask him/her to consider how many times to multiply by the common ratio to get from 5 to 50. Once they realize it is twice, they can usually set up the equation 5x2 = 50 and then solve for x. For Question #7, my students don’t usually recall the formula, so they use their graphing calculator. This is good! I also want them to realize that they can’t use the Rainbow Method that we used for arithmetic series. In the next section we will work on finding a formula for the nth partial sum of a geometric series. If there are any problems that gave students a lot of trouble from #1 to #6, we will address them as a class before concluding this section of the lesson. However, during this lesson, students usually clear up any confusion with their group or with me as I am moving around the room. ## Extend 15 minutes I always start by asking if any student remembers the formula for the nth partial sum of a geometric series. My students may not know the formula but they may remember that it involves a fraction and exponents. I will ask them what we would need to know about the geometric series in order to find the nth partial sum and they will usually decide on the first term, the common ratio, and the last term or the term number. I tell them that we will definitely need these things as we work on a formula for the nth partial sum. In the video below I explain my approach to proving the formula for the nth partial sum of a geometric series. Unable to display content. Adobe Flash is required. After going through the proof we will use it to verify the partial sums from Question #7 on the worksheet. To end the lesson, we recap the formula for the nth partial sum for a geometric and arithmetic series. It’s really important that students understand when to use each tool and are using them appropriately.

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Geocaching Puzzle Solution Reddevil1976 | 22:45 Thu 28th Feb 2008 | Quizzes & Puzzles Please could help me solve this geocaching problem. I am absolutely clueless. One evening, I was having a wee dram and browsing the geocaching forums, when I became aware of my dog, Sandy, barking outside. At first, I thought that her barks were quite random. Then I realised that amongst other barks and growls there were two distinct sounds � a yip and a woof. Now call me mad, or perhaps it was the effect of the whisky, but in a moment of clarity I understood what she was saying. She was telling the bull terrier down the road her age. (They are on regular barking terms.) She told him that she was yipyipwoof years old. He replied that he was only yipyip years old; so she was twice as old as he was! I should have guessed that dogs have a different way of counting than we do. Anyway, I am giving you the co-ordinates to a great place for walkies (and the cache) in dog-count. This may even advance canine-human relations! Now then, where's the Old Grouse? (Note that each digit is translated separately.) N yipwoofyip yipyip� yipyip yipyipwoof . yipyipyip yipwoofwoofwoof yipyip' W woof woof yipwoof� yip yipwoofwoofyip . yipwoofwoofwoof yipwoofwoof yipwoof' In the hint there's a clue to deciphering the above! Think like a computer! Cache at foot of small oak tree little way down slope. The answer should take the form of a set of co-ordinates, i.e. N 53� 36.614 W 002� 19.947 1 to 2 of 2 No best answer has yet been selected by Reddevil1976. Once a best answer has been selected, it will be shown here. Think like a computer IN BINARY Yip = 1 Woof = 0 Her age = 110 = 6 His age = 11 = 3 Co-ordinates are therefore:- N 53� 36.782 W 002� 19.842 If you go to Google and enter: N 53� 36.782 W 002� 19.842 then click on Maps, You will be shown exactly where to go. (NW Bury) 1 to 2 of 2 MagicSeven7 Reddevil1976 Reddevil1976 Reddevil1976

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### All in the Mind Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface of the water make around the cube? ### Painting Cubes Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours? ### Tic Tac Toe In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells? # Cubic Vision ##### Stage: 3 Short Challenge Level: One can see the greatest number of cubes when looking at three faces at once. We count the cubes on each face, giving $3\times 11^2=363$ cubes, but have to subtract from this the cubes along the three edges that have been counted twice, and then add back for the cube at the corner for which three faces are visible. The final quantity is $363-(3\times 11)+1=331$ cubes. This problem is taken from the UKMT Mathematical Challenges.

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# Square Root Function Printable View • May 17th 2008, 10:00 AM chrozer Square Root Function Create a square root function with: (a) domain of $[3, \infty)$ and a range of $(-\infty, 5]$. Would this be right? $f(x)=-\sqrt{x+3}+5$ And how would you create a square root function that has a domain of all real numbers? • May 17th 2008, 10:06 AM Moo Hello, Quote: Originally Posted by chrozer Create a square root function with: (a) domain of $[3, \infty)$ and a range of $(-\infty, 5]$. Would this be right? $f(x)=-\sqrt{x+3}+5$ It would be better if it was $f(x)=-\sqrt{x{\color{red}-}3}+5$ (Wink) Quote: And how would you create a square root function that has a domain of all real numbers? $\sqrt{x^2+\dots}$ :p • May 17th 2008, 10:15 AM chrozer Quote: Originally Posted by Moo Hello, It would be better if it was $f(x)=-\sqrt{x{\color{red}-}3}+5$ (Wink) $\sqrt{x^2+\dots}$ :p Duh....that was a stupid mistake. And thanks for the second part. Two more things...how would you determine the domain and range of $f(x)=\sqrt{ax+b}+c$ in terms of $a$, $b$, and $c$. And would $f^{-1}(x)=\frac {(x-c)^2-b}{a}$ be the inverse of $f(x)=\sqrt{ax+b}+c$? • May 17th 2008, 10:25 AM Mathstud28 Quote: Originally Posted by chrozer Duh....that was a stupid mistake. And thanks for the second part. Two more things...how would you determine the domain and range of $f(x)=\sqrt{ax+b}+c$ in terms of $a$, $b$, and $c$. And would $f^{-1}(x)=\frac {(x-c)^2-b}{a}$ be the inverse of $f(x)=\sqrt{ax+b}+c$? For $\sqrt{ax+b}+c$ to be $\in\mathbb{R}$ $ax+b\geq{0}\Rightarrow{x\geq\frac{-b}{a}}$ and for the second one since $f(x)=\sqrt{ax+b}+c\Rightarrow{x=\sqrt{af^{-1}(x)+b}+c}$ Now just solve for $f^{-1}(x)$ • May 17th 2008, 10:31 AM chrozer Quote: Originally Posted by Mathstud28 For $\sqrt{ax+b}+c$ to be $\in\mathbb{R}$ $ax+b>0\Rightarrow{x>\frac{-b}{a}}$ and for the second one since $f(x)=\sqrt{ax+b}+c\Rightarrow{x=\sqrt{af^{-1}(x)+b}+c}$ Now just solve for $f^{-1}(x)$ Ahh I see. I got the second one right, how would you determine the range of the 1st question? And also can the domain equal zero? • May 17th 2008, 10:35 AM Moo Quote: Originally Posted by chrozer Ahh I see. I got the second one right, how would you determine the range of the 1st question? And also can the domain equal zero? For the range : You know that $\sqrt{ax+b}$ goes from 0 to $+\infty$ So $\sqrt{ax+b}+c$ goes from 0+c=c to $+\infty$ :) • May 17th 2008, 10:43 AM chrozer Quote: Originally Posted by Moo For the range : You know that $\sqrt{ax+b}$ goes from 0 to $+\infty$ So $\sqrt{ax+b}+c$ goes from 0+c=c to $+\infty$ :) So all together the domain would be $x \geq \frac{-b}{a}$ and the range would be $y \geq c$. Right? • May 17th 2008, 10:44 AM Moo Quote: Originally Posted by chrozer So all together the domain would be $x \geq \frac{-b}{a}$ and the range would be $c \leq y \leq +\infty$. Right? Correct for the domain. For the range, it's like $c \leq y$ (Wink) • May 17th 2008, 10:50 AM chrozer Quote: Originally Posted by Moo Correct for the domain. For the range, it's like $c \leq y$ (Wink) Oh I see. Thnx alot.

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# English posted by rfvv When is the first meeting? 1. It is at nine this Saturday. 2. It is nine this Saturday. When is the next meeting? 3. It is on April 3, 9:00 a.m. 4. It is April 3, 9:00 a.m. 4-1. It is at 9:00 a.m., April 3. 4-2. It is 9 a.m., April 3rd. Who is the speaker? 5. Yun Seho. He is the president of the cooking club, Yummy. 6. He is Yun Seho, who is the speaker of the cooking club, Yummy. 7. Yun Seho, the president of the cooking club, Yummy. 8. He is Yun Seho, the president of the cooking club, Yummy. 1. Writeacher #1 is correct. #2 is not incorrect, just not used much. (I've never heard it!) 3&4 are all fine. 5-8 are all correct, but #6 is rather formal, and not used much. ## Similar Questions 1. ### algebra 2 Posted by hellogoodbie on Saturday, January 16, 2010 at 3:24pm. Posted by hellogoodbie on Saturday, January 16, 2010 at 2:59pm. How is this problem done? 2. ### English I still have a few more doubts. Could you check the following phrases and improve them, please? 3. ### penn foster industrial activities operates 5 days per week with a daily payroll of \$4000. employees are paid every saturday for the workweek just completed(monday to friday). the last day of the month is wednesday, mar 31. what is the amount of … 4. ### grammar I have learning how prepare for an effective meeting. For an effective meeting you must invite a neutral facilter to sensitive meeting, only hold a meeting if necessary , an meeting must have clear objectives , all meeting have an … 5. ### grammar I have learning how prepare for an effective meeting. For an effective meeting you must invite a neutral facilter to sensitive meeting, only hold a meeting if necessary , an meeting must have clear objectives , all meeting have an … What is the pH of a 5.06E-2 M aqueous solution of sodium acetate? 7. ### English 1. The next meeting is at 9:00 a.m. on April 3. 2. The next meeting is at 9:00 a.m., April 3. 3. The next meeting is at 9:00 a.m. April 3. 4. The next meeting is 9:00 a.m. on April 3. 5. The next meeting is on April 3 at 9:00 a.m. … 8. ### English 1. Our first visit will be June 23, next Saturday. 2. Our first visit will be on June 23, next Saturday. 3. Our first visit will be on June 23, on next Saturday. 4. Our first visit will be on June 23 on next Saturday. ----------------- … 9. ### English 1. The staff meeting will be held next Tuesday. 2. The staff meeting will be taken next Tuesday. 3. The staff meeting will be got next Tuesday. 4. The staff meeting will be finished next Tuesday. (#1 is grammatical, right? 10. ### English 1. The staff meeting will be finished next Tuesday. 2. The staff meeting will be finished in two hours. 3. The staff meeting will be finished after two hours. 4. The staff meeting will be finished tomorrow. 5. The staff meeting will … More Similar Questions

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× # Kinetics Top FAQ ## What is Kinetics? View Notes In physics, the study of objects is divided into mechanics. Mechanics have further been divided into two branches of study dynamics and statics. In the dynamic branch of mechanics, the particles in motion or the bodies in motion under forces’ action are studied. In this dynamics branch of mechanics, the study deals with bodies and objects that do not undergo any change in motion, that is they are moving at a uniform motion or are in equilibrium even with an application of force or torque. The dynamic mechanics is further divided into two parts for studying Kinetics and kinematics. They are also referred to as Kinetics Dynamics and Kinematics Dynamics. ### Kinetics Dynamics and Kinematics Dynamics Often the terms Kinetics Dynamics and Kinematics Dynamics creates confusion in students' minds more so because Kinetics is also a branch of study in chemistry and biochemistry that deals with the progress of reactions. Whereas the branch of Kinetics in Physics is entirely different from the one in chemistry, let us learn more about it and its differences. It is important to note that both Kinetics Dynamics and Kinematics Dynamics fall under one umbrella of Dynamics Mechanics. Kinetics Dynamics Kinematics Dynamics It deals with the study of causes of motion in an object or body with the force or torque application. It is also under the dynamic branch of mechanics that deals with the study of the speed, acceleration and position of an object. Since motion causes are studied the mass of the object is an important consideration. Mass of the body or object here is not taken into consideration when studying. Force is taken into consideration when Kinetics is studied. The force here is not considered rather the result of the force applied on an object is studied. There are no particular mathematical equations in kinetics as such. Has many mathematical equations that deal with objects in rotational motion, elliptical motion and even in equilibrium conditions. Its application is mainly in while one is designing the automobiles. Kinetics is important in the automobile industry and even in everyday life as well. The Kinematics equation is applied in deciphering the movements in the celestial bodies. Its application pertains to the study of astronomical equations. ### Kinetics and Kinetic Energy Kinetics study deals with the movement of objects and that is due to kinetic energy and this can be classified into three different types. And this mainly deals with change in mass and velocity that will change depending on the type of object that is the mass of an object and the velocity of the change in motion due to force applied. Mass is represented as ‘m’ and velocity is represented by ‘v’. Even though kinetics has no specific formula the kinetic energy can be calculated using- Ek = ½ mv2 1. The rotational kinetic energy- this kind of energy is seen when the object is in the rotation of an object on an axis of the object. It is also called angular momentum or angular kinetic energy. The energy can also be possessed by the objects like the top moving in circles, or planets moving around the sun, so to study such objects in motion one has to use the method of rotational kinetic energy. 2. Vibrational Kinetic energy- Vibrational energy as the name suggests is possessed by objects which vibrate when a force is applied for example when phone rings and the other instance one can see is when one beats a drum it vibrates. 3. Translational Kinetic energy- this kind of kinetic energy is seen when an object is already in motion and with a certain amount of force applied it moves from one place to another the energy force is transferred and translated as the change in velocity. ### Kinetics in Everyday Life Kinetics and kinetic energy can be seen and experienced in our everyday lives as well. Certain examples are- • It can be seen in hydropower plants, in here due to the kinetic energy of the water the plants can generate electricity. • Generation of electricity with the help of windmills which are in motion. • Moving cars possess kinetic energy. • The motion of a bullet since force is applied via the gun possesses kinetic energy. • Cycling as we apply force for the movement of pedals. Pedalling scents the bicycle in kinetic motion. ### Conclusion Kinetics is very important for the objects in motion or the dynamic objects. It has been an important mode of study in the field of physics ever since its discovery and is not subsiding anytime soon. We can find many answers to the questions of movement and mechanics through kinetics. Q.1. What is Kinetics? Ans. Kinetics is the study of the movement of objects that manifests in changing their position due to the force applied to it. It is a study of chemical reactions in the field of biochemistry and pharmacy. In chemistry and chemical reactions, kinetics provide an insight into the factors affecting the speed of the reaction. Q.2. Are Kinetics and Kinematics the Same? Ans. The kinetics and kinematics fall under the mechanics’ branch of physics under the study of the dynamic object. That is there is some change in the object’s motion due to force in the application in both kinetics and kinematics. The difference between them lies in where the kinetics considered effects of force and mass of the objects the field of kinematics refrains from it and studies speed, acceleration and velocity. Q.3. What are the Types of Kinetics? Ans. The kinetics or the kinetics dynamics are divided into three main types based on the energy the object of study possesses. The rotational energy, seen in toys moving in a circular motion, vibrational energy seen in phones and the translational energy that is seen when someone pushes or tries to stop with force an already moving object like pushing someone who is running or applying force to move a piece of furniture. In all of these different kinds of kinetic energies, the factors considered are the mass of the object and the velocity with which it moves.Â

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# Array and Array Operations • Our quick test is a 40-point assessment designed to gauge your understanding of key concepts. It features 10 multiple-choice questions. • Each correct answer earns you 4 points, while each wrong answer deducts 1 point. • This test helps you identify areas of strength and areas needing improvement. Good luck! #### 1. What is the index of the first element in a typical array? a) -1 b) 0 c) 1 d) Depends on the language #### 2. How do you access the third element of an array named `arr`? a) ``arr(3)`` b) ``arr[2]`` c) ``arr[3]`` d) ``arr.get(3)`` #### 3. Which operation is used to add an element at the end of an array in most programming languages? a) ``push`` b) ``append`` c) ``insert`` d) ``add`` a) O(n) b) O(log n) c) O(1) d) O(n log n) #### 5. What is the result of the following operation in Python: ``arr = [1, 2, 3]; arr[1] = 4`` a) ``[1, 2, 3]`` b) ``[1, 4, 3]`` c) ``[1, 3, 4]`` d) ``IndexError`` #### 6. Given an array `arr = [2, 4, 6, 8, 10]`, what will be the result of the operation ``arr.splice(2, 1)`` in JavaScript? a) ``[2, 4, 8, 10]`` b) ``[4, 6, 8, 10]`` c) ``[2, 4, 6, 8, 10]`` d) ``[2, 4, 8]`` #### 7. What is the result of the following Python code snippet? ``````arr = [1, 2, 3, 4, 5] arr[::2] = [9, 9, 9]`````` a) ``[9, 2, 9, 4, 9]`` b) ``[9, 9, 9, 9, 9]`` c) ``[1, 2, 3, 4, 5]`` d) ``ValueError`` #### 8. What does the following C++ code snippet do? ``````int arr[] = {10, 20, 30, 40}; int *ptr = arr + 2; cout << *ptr << endl;`````` a) Prints `10` b) Prints `20` c) Prints `30` d) Throws an error #### 9. Which of the following operations can be used to remove the last element of an array in JavaScript? a) ``arr.pop()`` b) ``arr.remove()`` c) ``arr.shift()`` d) ``arr.delete()`` #### 10. What will be the output of the following Java code? ``````int[] arr = {1, 2, 3, 4, 5}; System.out.println(arr.length);`````` a) ``4`` b) ``5`` c) ``6`` d) ``Throws an error`` Score Range: 35-40 points Remarks: Excellent job! You aced the test and showed a strong grasp of the material. Keep up the fantastic work! Score Range: 30-34 points Remarks: Great work! You did very well and demonstrated a solid understanding of the concepts. Just a little more effort and you’ll hit perfection.

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# Human Rights & International Politics MRes ## Quantitative Data Analysis SPS5033 • School: School of Social and Political Sciences • Credits: 20 • Level: Level 5 (SCQF level 11) • Typically Offered: Semester 2 • Available to Visiting Students: No • Available to Erasmus Students: No ### Short Description The course introduces basic statistics and data analysis from univariate summary statistics up to multivariate linear regression. The main aim of the course is to enable students to summarise, analyse, and present data in valid ways and understand the basics of statistical inference and association as required in quantitative social science research. ### Timetable Two hours lecture, and two hours tutorials per week over 11 weeks. Lectures will run Mondays 12-2pm. Tutorials will be held at multiple times to offer flexibility, students can enrol via MyCampus. None None ### Assessment Students will conduct a quantitative data analysis project. They will choose a dataset, research question, dependent variable, and independent variables and apply the skills acquired in the course to the data in order to answer the research question. The students will produce a paper of 4,000 words (excluding references and R code), in which they focus on the statistical analysis and present the results. The paper must include the R source code and be written as a knitr document (e.g., in RStudio). Each paper must contain summary statistics, data visualisations, bivariate associations, hypothesis tests, and a regression model as well as a discussion of the regression assumptions in the light of the data and model. ### Course Aims The course introduces basic statistics and data analysis from univariate summary statistics up to multivariate linear regression. The main aim of the course is to enable students to summarise, analyse, and present data in valid ways and understand the basics of statistical inference and association as required in quantitative social science research. At the end of the course, students should be able to describe, summarise, and visualise data, calculate the association between variables at various scale levels, understand sampling and inference, test hypotheses with given datasets, quantify the uncertainty arising from data, and apply, interpret, and understand the assumptions of, linear regression models. At all times, special care is taken to ensure that students can associate the statistical techniques with real-world examples from across the social sciences, and especially a themed example chosen from the set of research themes identified by the College of Social Sciences. In addition to basic statistics, students will acquire computational skills that allow them to apply their newly acquired skills using the statistical computing environment R. The overarching aim is to enable students to transfer these skills to new datasets, possibly including their own research topics. Students will learn how to evaluate theories and claims based on data by selecting the appropriate statistical tools and applying them to the data by hand and by using R. In each session of the course, the relevant concepts are taught using words, numbers, equations, examples, and R code. ### Intended Learning Outcomes of Course After taking this course, students should be able to: 1. Manage, visualise, summarise, and present univariate and bivariate data. 2. Construct a robust linear regression model. 3. Test hypotheses involving data measured at different levels e.g., interval, binary and categorical. 4. Use the statistical software R for quantitative analysis of univariate variables, bivariate associations and linear regression analysis. 5. Critically evaluate theories, test hypotheses, and answer substantive research questions using quantitative approaches with available data from a social science perspective beyond the examples given in the course. 6. Describe quantitative methods, and to interpret and write up the results of quantitative analyses, clearly and concisely. ### Minimum Requirement for Award of Credits Students must submit at least 75% by weight of the components (including examinations) of the course's summative assessment.

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Skip to content ## Why Use Two-Dimensional Arrays? While one-dimensional arrays allow data to be placed in an array one row at a time, two-dimensional arrays are capable of storing data in both rows and columns. To accomplish this, each row in a two-dimensional array is associated with the number of columns defined for the array. As with one-dimensional arrays, the entire array must contain elements of the same type. The type is defined when you declare the array. Because of the capability of storing data in rows and columns, it is obvious that two-dimensional arrays can provide more flexibility than one-dimensional arrays. ## Declaration and Initialization Declaring a two-dimensional array has the following form: ```type arr-name[NUM_ROWS][NUM_COLS]; ``` where type is the data type for the array, arr-name is the variable name for the array, NUM_ROWS is the maximum number of rows for the array, and NUM_COLS is the maximum number of columns for the array. To declare an integer array of five rows and three columns, the following code could be used: ```int arr[5][3]; ``` Initializing two-dimensional arrays can be accomplished using the following format: ```int arr[5][3] = { { 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 }, { 9, 0, 1 }, { 2, 3, 4 } }; ``` The above code initializes: • arr[0][0] = 0 • arr[0][1] = 1 • arr[1][0] = 3 • arr[1][1] = 4 • arr[4][0] = 2 • ,… Processing 2D Arrays Because 2D arrays must be filled by row and column, processing a 2D array can be done using nested for loops. For instance, to fill an array declared numArr[10][10] with user input, the following nested loop scheme could be used: ```for (int rows = 0; rows < 10; row++) for (int cols = 0; cols < 10; cols++) { cout << "Enter value for row " << row+1 << " col " << col+1 << ": "; cin >> numArr[row][col]; } ``` To display the contents of the above filled array ten values per line, the following code could be used: ```int row, col; for (row = 0; row < 10; row++) { for (col = 0; col < 10; col++) cout << numArr[row][col]; cout << endl; } ``` For an example of these concepts, study the following complete program: View Example Program The following is another example similar to the one above. It adds a tiny twist: View Example Program ## Two-Dimensional Array Examples A common use of two dimensional arrays is for storing a collection of strings. When storing strings in a two dimensional array, the rows in the array are associated with the position of the full string value, and the columns in the array are the individual characters of the string for that particular row. For instance, if the string “c++” is stored in row 0 of stringArr, then stringArr[0][0] holds ‘c’, stringArr[0][1] holds ‘+’, and stringArr[0][2] holds ‘+’. The following illustrates how three short strings are stored in a two dimensional array having a size of three rows and six columns: From above, notice: • row 0 contains the string “c++” • row 1 contains the string “prog” • row 2 contains the string “win” • is a null terminator (denoting the end of the string) • denotes a garbage value when the string is not large enough to fill the entire array row It is important to remember that there must be space in the array for the null () terminator. As an example, assume we have a two dimensional array declared as follows: ```char stuNames[5][26]; ``` This array holds five strings with a maximum of twenty-five characters per string. The array is declared as having twenty-six maximum columns to account for the null terminator ‘\0′ that is attached to the end of a string. Suppose we want to fill our example array with values specified by the user during program execution and also display the contents of the array in a semi-neat fashion. The following complete program example solves this problem: View Example Program Two Dimensional Array Example 1: Suppose we have a two dimensional integer array of exactly ten rows and ten columns, and we need to find the sum of the integers along the main diagonal of the array. We could use the following code segment: ```const int MAXARRSIZE = 10; int arr[MAXARRSIZE][MAXARRSIZE]; . . arr gets values . . int sum = 0, row, col; for (row = 0; row < MAXARRSIZE; row++) { for (col = 0; col < MAXARRSIZE; col++) { if (row == col) { sum += arr[row][col]; } } } cout << "The sum of the values along the main diagonal is: " << sum << endl; ``` Two Dimensional Array Example 2: The following is something an individual may encounter on a C++ quiz: Suppose the correct answers for a true/false quiz are: T T F F T Suppose also that a two dimensional array contains student answers with each row representing the responses of a particular student. The first row (0) of the array contains the above answer “key”. Write code to calculate the students’ scores on the quiz and store these computed scores in a one-dimensional array. Assume each question is worth 4 points and there is a maximum of 50 students. The following code could be used to accomplish this task: ```const int NUMOFSTUS = 51; const int NUMOFQUIZZES = 5; const int ANSWERINDEX = 0; char quizAnswers[NUMOFSTUS][NUMOFQUIZZES]; int gradeArray[NUMOFSTUS]; // row 0 will be unused int stuIndex, probIndex, grade; . . . quizAnswers gets values . . quizAnswers[0][0] = ‘T’; quizAnswers[0][1] = ‘T’; quizAnswers[0][2] = ‘F’; quizAnswers[0][3] = ‘F’; quizAnswers[0][4] = ‘T’; for (stuIndex = 1; stuIndex < NUMOFSTUS; stuIndex++) { grade = 0; for (probIndex = 0; probIndex < NUMOFQUIZZES; probIndex++) { if (quizAnswers[stuIndex][probIndex] == quizAnswers[ANSWERINDEX][probIndex]) { grade += 4; } } gradeArray[stuIndex] = grade; } ``` To finish this article, study the following complete program. It ties all basic concepts within this article together: View Example Program In the next article, we’ll explore arrays with higher dimensions (i.e., arrays having more dimensions than one or two). Read on for more… ### { 3 } Comments 1. mercy suganob | June 9, 2009 at 11:50 pm | Permalink how to solve this problem? Write a program using two dimensional arrays that searches a number and display the number of times it occurs on the list of 12 input values. sample input/output data: enter twelve no. 1 1 2 3 4 5 6 1 8 9 5 8 enter a no. to search:1 occurence(s):3 2. mun | September 29, 2009 at 1:39 am | Permalink i understand nothing…how can i understand like u..i give up 3. | November 9, 2011 at 8:33 am | Permalink write a program for the summation of the elements of the first row of an two dimensional array. ### { 273 } Trackbacks 1. Arrays in C# - Software Development | July 31, 2010 at 12:23 am | Permalink [...] arr[5][3]; More information can be found here. __________________ Education, Career and Job [...] 2. 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# Introduction to Excel’s Data Analysis Tool for Elementary Statistical Analysis Blog This blog post covers an Introduction to Excel’ Data Analysis Tool, from installing and loading the Data Analysis ToolPak in Excel to the different Analyses applicable for Senior High School. Getting Started with the Excel’ Data Analysis Tool Load the Data Analysis ToolPak in Excel Step 1: Under the File tab, click on Options. Step 2: Click the Add-Ins category, and then in the Manage box, select Excel Add-ins and click Go. Step 3: In the Add-Ins box, check the Analysis ToolPak check box, and then click OK. • If Analysis ToolPak is not listed in the Add-Ins available box, click Browse to locate it. • If you are prompted that the Analysis ToolPak is not currently installed on your computer, click Yes to install it. After you load the Data Analysis ToolPak, the Data Analysis command is available in the Analysis group on the Data tab. Different Analyses Descriptive Statistics Descriptive statistics aims to show or summarize a given data set, which can be either an entire population or a sample of it. It is great for simplifying a large amount of data in a sensible way. It provides information about the central tendency and variability of your data. Descriptive statistics are broken down into measures of central tendency and measures of variability or dispersion. Measures of central tendency include the mean, median, and mode, while measures of variability include variance, standard deviation, minimum and maximum values of the variables, kurtosis, and skewness. Histogram A histogram is a commonly used graph to display the underlying frequency distribution, it used to summarize discrete or continuous data, and shows how often each different value in a set of data occurs. It is a tool used to calculates individual and cumulative frequencies for a cell range of data and data bins. Histograms can display and provides a visual and graphical representation of a large amount of data, it is used to show how many of a certain type of value or variable occurs within a specific range in a data set. t-Test: Paired Two Sample For Means The t-Test Paired Two Sample for Means is an analysis tool that performs a paired two-sample Student’s t-Test to determine if the null hypothesis with equal population means can be accepted or rejected. This Paired t-Test is commonly used when there is a natural pairing of observations in the samples, such as when a sample individual or group is tested twice before and after some treatment. Paired t-Test can only compare the means for two related units on a continuous outcome that is normally distributed. t-Test: Two-Sample Assuming Equal Variances The t-Test Two-Sample Assuming Equal Variance or the Homoscedastic t-Test analysis tool perform a two-sample student’s t-Test assuming that both data sets came from distributions with the same variances or assumed to be equal. t-Test: Two-Sample Assuming Unequal Variance The t-Test Two-Sample Assuming Unequal Variance or the Heteroscedastic t-Test analysis tool perform a two-sample student’s t-Test assuming that the two data sets came from distributions with unequal variances. Heteroscedastic t-Test is used when there are distinct subjects in the two samples. ANOVA: Single Factor The single factor or one-way analysis of variance (ANOVA) is used to compare and determine the means between two or more independent or unrelated groups of values whether there is statistical evidence that the associated population means are significantly different. . This ANOVA Single Factor is used to test the null hypothesis that the means of several populations are all equal. Correlation The Correlation analysis tool is used to investigate the relationship between two quantitative and continuous variables. It is particularly useful when there are more than two measurement variables for each of N subjects and to analyze the relationship between variables. The most common correlation coefficient is the Pearson Correlation Coefficient that is denoted by r, it is a measure of the strength of the association between the two variables. 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מסאג’ איכותי ומרגיע. מסאג’ איכותי ומרגיע בדירה פרטית ודיסקרטית בתל אביב, הרבה סבלנות והשקעה לכל מטופל, בעיסוי יוקרתי ואישי מחכה לך לחוויה של פעם בחיים. לאף אחד אין זכות להתערב לכם בחיים ולשאול האם ביליתם עם האישה או עם המאהבת. זוהי סביבה נוחה, נעימה וגם בטוחה ודיסקרטית לקיים בה כל סוג של בילוי עם בת הזוג או המאהבת. 52. רוסיה חרמנית וחמה בטירוף בתמונות אמיתיות מחכה לך בדירה פרטית בחיפה. אנג’לה היא כבר עשר שנים בארץ אבל נשארו לה הגנים של רוסיה היפה. רוני שלנו היא צעירה מאוד עם גוף מטורף וסקסי, היא גדלה בישראל אבל יש לה שורשים רוסיים. רמת גן עיר גדולה במרכז שיש לה להציע לכם עשרות דירות מפוארות לעיסוי מפנק. אם אתה מחפש דירות דיסקרטיות בבאר שבע בהן תוכלו לברוח למספר שעות מהמולת היום, אם תרצו להתייחד עם בת זוגתכן או לקבל שירותי ליווי לעיסוי בדירות דיסקרטיות בבאר שבע? כמו בכל אזור גם כאן, דירות דיסקרטיות בחדרה מככבות והכול בזכות פורטל שלוקח את מלוא האחריות על ההנאות שלכם. אין אדם אשר יציעו לו עיסוי מפנק בבאר שבע , או בכל אזור בארץ והוא יסרב להצעה מפתה זו.אם יציעו לכם עיסוי מפנק ברגע הזה לבטח תקפצו על ההצעה. איפה אפשר למצוא דירה דיסקרטית בבאר שבע? מעוניינים למצוא דירה דיסקרטית בבאר שבע? עיסוי מפנק בגבעתיים , ניתן להזמין עד הבית בגבעתיים כמתנה לך, לבן הזוג לחברה או לאיש עסקים שאתם מעוניינים לפנק. הפכתי לאיש בעל אישיות שונות. 1. generic 5mg cialis best price An advantage over NSAIDs is that it not only does not inhibit the synthesis of prostaglandins at the gastric level, but seems to increase it, so it does not give rise to gastrolesive effects 53. דירות דיסקרטיות בנתניה, ממתינות גם לכם בכל רחבי העיר ומאפשרות לכל אחד ואחת מביניכם ליהנות מחוויית אירוח דיסקרטית בפרטיות מוחלטת. מעבר לכך, האתר זמין עבורכם בכל שעה על מנת לתמוך בגולשים מכל רחבי הארץ. האתר מספק שירותי סקס ליווי בנתניה ללא מין לגברים ונשים על ידי נערות סקס ליווי בנתניה לעיסוי הכי איכותיות. נערות סקס ליווי בנתניה ללא מין איכותיים במיוחד, עם נערות סקס ליווי בנתניה לעיסוי השוות ביותר שתוכלו למצוא. אין ספק שכל גבר זקוק למעט חברה בחייו ועבור אנשים בררנים במיוחד, בעלי סטנדרטים גבוהים, השירות מהווה פלטפורמה מעולה לבחור על פי קריטריונים שונים מפגשים אצלי בנתניה. מפגש של עיסוי טנטרי הכולל עיסוי יוני שעה וחצי מפגש פנים אל מול פנים של 3 שעות הכולל טקסים טנטריים של חושניות קדושה נשימות, הטקסים הן חלק מהטנטרה כהכנה למין מקודש ועיסוי טנטרי/יוני תהליך סוד הלוטוס שזה תהליך ריפוי של 7 מפגשים בהתאם ל 7 מרכזי האנרגיה בגוף צ’אקרות 🕉המפגשים לנשים בלבד! נערות ליווי ירושלים, נערות שוות במיוחד שיודעות לחייך אבל בו זמנית יודעות גם לתת את התשובה לחיים שונים לגמרי, גברים שרק מחפשים לשפר את מערכת היחסים שלכם, יחסים אחרים, מגע אחר, מפגשים אחרים, זה מה שתפגשו כשאתם בוחרים נערות ליווי שוות במיוחד שמכירות את מה שהן צריכות להעניק לכם ומבינות את העניין הדיסקרטי. 1. com 20 E2 AD 90 20Viagra 20Ireland 20Boots 20 20Viagra 20Lekarna viagra ireland boots My interpretation is that we are at the moment in the phasewhere we might get into very deep trouble with the U finasteride 1 mg 54. רוכשים ביחד לדירה מארגנת קבומות רכישה שיחד הופכים לכוח קניה משמעותי הזוכה להנחות משמעותיות. רוכשים ביחד לדירה מארגנת קבוצות רכישה וימי מכירות למוצרים ושירותים לבית. ישנם אנשים שמקבלים עיסויים למטרות פינוק וכייף בביתם ולפעמים מגיעים לדירה דיסקרטית בגבעתיים אך גם ישנם אנשים המבקשים לעבור עיסוי במרכז בעקבות כאב מסוים. אולם אירועים במרכז תל אביב בו ניתן לקיים מגוון אירועים כגון: השקות, כנסים, חתונות, אספות ועוד! עורר הדוד ביל ליווי בתל אביב שהוזכר לי שאם הוא מקבל מדי עורר הפין שירותי ליווי בתל אביב שלו הופך זועם עם דם שגורם לו רק לקבל קשה באו בונר ישבתי על השוק והביט בו ואז בחיפוש אחר בונר הוא היה במכנסיים הריצה שלו הרגשתי את הכוס שירות ליווי בתל אביב שלי להירטב ואפילו זה התחיל לחלחל ליווי בתל אביב החוצה משפה הכוס שלי הוא המשיך והסביר שאינה הייתה לי כל ילדה יש את שיש את היכולת להראות את איבר המין של גבר לקבל קשה על ידי זהה רק מסתכל עליה הוא שהוזכר לי רק שירותי ליווי בתל אביב אישה מאוד מקסימה וסקסית יכולה להורג את זה לאדם חשבתי ליווי בתל אביב דרך . אם הגעתם ישר לקליניקה בלוד אשר בה הזמנתם עיסוי מפנק בתל אביב ישר מהעבודה, תהיה לכם אפשרות להתקלח בקליניקה בלוד ולהתחיל עם עיסוי מפנק בלוד תוכלו לבחור איזה או גבר שיבצעו עבורכם את העיסוי. 1. Characteristics of tissue microarray are found in Table 1 cronadyn vs priligy For young premenopausal breast cancer patients, adjuvant chemotherapy may cause menstrual disruptions and premature menopause, which may in turn impair their quality of life QoL 55. מפגש עם אחת נערות הליווי ברמת השרון, יהפוך את יומך לאושר עילאי ושמחה מרגשת. מי שהתמזל מזלו לקנות נכס בעיר לפני מס’ שנים, במחיר מגוחך (משהו כמו 200,000 על דירת 3 חדרים) מחזיר היום את ההשקעה שלו, ובגדול. כאשר זה נוגע לבחורות, לכל אחד יש את הטעם שלו, אחד אוהב רזות ובהירות, אחר יעדיף כהות ומלאות, אחד לא יהיה מוכן להתפשר על נושא הגיל וכן הלאה – כמובן, יהיו גם כאלו שלא כל כך חשוב להם איך הבחורה נראית, ויותר חשוב להם עלות השירות. כאן תוכל למצוא מגוון רחב של בחורות חטובות, יפות וסקסיות, שהינן מקצועיות ומעניקות את מיטב השירותים. בפורטל אקספיינדר מגוון אפשרויות לבחירה ותפורים לפי טעמך ודרישותיך כך שתוכל להנות מבילוי סוחף ומרגש. בפורטל אקספיינדר תוכלו למצוא נערות ליווי ברמת השרון מכל המינים והסוגים. לא רק שתוכלו להזמין אותן אל האירוע, למעשה לכל אורך הביקור שלכם בעיר תוכלו ליהנות מחברתן. מלון רויאל ביץ’ בתל אביב ממוקם בעיר הפועמת ביותר בישראל, העיר שאינה ישנה. משרד נערות ליווי ליווי או עיסוי מפנק בתל אביב? מי שלא רוצה לשאת בהוצאות הגבוהות על הזמנת חדרי מלון בתל אביב או בת ים יכול לחסוך כסף על בילוי בחולון, בעוד בני זוג בחופשה רומנטית יעדיפו להשתמש בדירות דיסקרטיות ולא בצימרים מרוחקים. נערות ליווי בהוד השרון בוקר אחד אתם קמים ומרגישים שמשהו חסר לכם בחיים האישיים, זו לא אהבה, זה לא כסף וזו גם לא קריירה, זה הרבה מעבר. 56. מרגוע לנפש: מעבר ליתרונות הפיזיים המובהקים, שווה לדעת שלעשות עיסוי ארוטי בירושלים זאת גם דרך נפלאה ובטוחה להרגיע את הנפש שגם היא מתעייפת ונשחק בשגרה. מרגוע לנפש: מעבר ליתרונות הפיזיים המובהקים, שווה לדעת שלעשות עיסוי ארוטי בהרצליה זאת גם דרך נפלאה ובטוחה להרגיע את הנפש שגם היא מתעייפת ונשחק בשגרה. עיסוי קלאסי בהרצליה באזורים, תשומת לב רבה יותר מוקדשת לחלקים מסוימים בגוף (למשל ישבן, גב, אזור צווארון). דיסקרטיות מלאה. דירה מדהימה, אזור שקט וחבוי, אבזור מפנק ביותר והכי חשוב? נכון לכתיבת שורות אלו, לפי נתוני אתר ספייסנטר, המרכז חדרי ישיבות להשכרה בכל הארץ וכן חללי עבודה משותפים, קיימים למעלה מ-15 חדרי ישיבות ברחבי ירושלים, בעלי גדלים שונים ורמות אבזור שונות, כאשר בכל חודש נפתחים עוד ועוד חדרים נוספים. אבזור מלא? אלו הן רק חלק מהתיאורים אודות נראות הדירה שיכולה לעורר בכם רגשות שלעיתים נדמה כי שכחתם. בכל הזמנה של מסאג’ בקרית שמונה עד הבית, צוות המטפלים יגיעו אליכם עם מיטת טיפולים, מוזיקה, שמנים וכל הדרוש על מנת שהעיסוי שלכם יהיה לא פחות ממושלם! 1. 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A Representative confocal images of immunostaining for Ki67 in the DG of saline or fluoxetine treated mice no presription finasteride never AD use, AD use at baseline only, AD use at year 3 only, AD use at baseline and year 3 on breast cancer risk 57. המגוון האדיר של טיפולי מגע הקיימים כיום בשוק הישראלי והמוצעים על ידינו כאן באתר מסאז’ בחיפה מאפשר התאמה אישית מלאה של הטיפול האידיאלי עבור כל אחד ואחת. כאן באתר תוכלו להתרשם ממבחר המטפלים, ללמוד אודות כל אחד ואחת מהם ולראות את המחירים שלהם ולהזמין מסאז’ בחיפה ללא פערי תיווך ועם שירות מקצועי מובטח. הזמנת מסאז’ בחיפה היא הבחירה הנכונה בין אם אתם מחפשים טיפול מפנק ונעים עבורכם או בתור מתנה מקורית למישהו יקר לכם ובין אם אתם זקוקים לטיפול ספציפי להקלה על כאבים פיזיים מסוימים או שחרור מתחים. בכל הזמנה של מסאז’ בחיפה עד הבית, צוות המטפלים יגיעו אליכם מצוידים במיטת טיפולים, מוזיקה, שמנים וכל הדרוש על מנת שהעיסוי שלכם יהיה לא פחות ממושלם! כאשר תזמינו עיסוי בחיפה עד הבית, המטפל או המטפלת מגיעים אליכם עם כל הציוד הדרוש לטיפול, החל ממיטת טיפול מקצועית ועד נרות, שמנים ומוסיקה מתאימה – הכל על מנת שהעיסוי שהזמנתם יהיה מושלם. גם אם תזמינו עיסוי מפנק לבית המלון בחיפה והסביבה שבו אתם שוהים בבירה, האווירה תשתנה, האורות יתעממו ונרות ריחניים יהיו בכמה פינות החדר, אתם תשכבו רק עם מגבת על גופכם שתרד אט, אט, ככל שהעיסוי יעבור לחלקים השונים בגוף. גם אם תזמינו עיסוי מפנק בחיפה והסביבה לבית המלון שבו אתם שוהים בבירה, האווירה תשתנה, האורות יתעממו ונרות ריחניים יהיו בכמה פינות החדר, אתם תשכבו רק עם מגבת על גופכם שתרד אט, אט, ככל שהעיסוי יעבור לחלקים השונים בגוף. 1. 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# We Got A Problem #6: Useless Rulers By the way, there’s an interesting problem kicking around Dan Meyer’s blog.  But now for something completely different… Take a ruler or any stick.  Mark it off into three equal parts.  Then, perhaps in a different color, mark it off again into five equal parts.  This divides the ruler into how many total parts? (The mathematician’s version: Consider a number line from 0 to 1.  Mark all multiples of 1/3 and 1/5.  How many intervals are produced?) Start over.  On another ruler, mark it off into four equal parts, then again into ten equal parts.  How many total parts this time? On another ruler, mark it off into twelfths and thirtieths.  (This is close to what happens with inches and centimeters, but not quite.)  How many parts now? If you mark off the ruler into M equal parts, then again into N equal parts, how many parts are there in total? Can you think of ways to extend either this problem or the rectangle-diagonal problem? We’d love for readers to be able to explore these problems, so resist the urge to provide answers in the comments. Instead, we’d love helpful suggestions and ideas about different ways to think about them, successful or not. If you’d like to provide a full solution, do so with a pingback to your own blog! Bowen is a mathematics curriculum writer. He is a lead author of CME Project, a high school curriculum focused on mathematical habits of mind, and part of the author team of the Illustrative Mathematics curriculum series. Bowen leads professional development nationally, primarily on how math content can be taught with a focus on higher-level goals. Bowen is also a champion pinball player and once won \$1,000 for knowing the number of degrees in a right angle. ### 2 Responses to We Got A Problem #6: Useless Rulers 1. Matt E says: “Completely different”… Ha! 2. Ipsquiggle says: I almost immediately had an insight that, rather than looking at it as dividing the ruler into 5 and 3 //parts//, you are instead adding 4 and 2 //divisions//. This made it trivial to imagine further answers in my head, and come upon a (somewhat messy, but easily simplifiable) generalization.

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Thanks to theidioms.com # Convolutional Neural Network Theoretical Course (Course VIII) ## Convolutional Neural Network Theoretical Course (Course VIII) ### The Convolution/Pooling Operation for RGB images Until now, we have only discussed the convolution and pooling operations on single-channel images, i.e., grayscale images. However, the colored photos taken from digital cameras are RGB images. Such images are formed by the addition of three color channels: Red, Green, and Blue as shown in the image below, Mathematically, an RGB image , is represented as x x , where the first two dimensions ( and ) represent the number of rows and columns of pixels in the image and the last dimension () represents the number of color channels. So, for an RGB image of 512×512 resolution, the actual representation of it is 512x512x3. In this case, the convolution/pooling operation is performed on all three colour channels (Red, Green and Blue) simultaneously and a single output tensor is obtained by taking a sum of the convolution/pooling operation of each colour channel. Let us understand this clearly with the following example of a convolution operation: Consider an RGB image with a dimension of 3x3x3, Also, consider a kernel with a dimension of 3×3, The output tensor is obtained as follows, The same process can be followed for an image with a larger dimension than the kernel. The kernel is convolved with each colour channel of each subset tensor of the image to get the resultant output tensor. The above concept can be extended for the pooling operation as well where max-pooling or average-pooling is applied to each colour channel of each subset tensor of the image to get the resultant output tensor.

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# Best answer: How many grams are in 1 70 moles of KMnO4? Contents ## How many grams are in a molecule of KMnO4? The molar mass of KMnO4 is 158.034 g mol1. One mole of any substance contains 6.022×1023 atoms/molecules. ## How many moles are in 17.7 g of KMnO4? Mole practice 12)Find the formula mass of KMnO4 158.00 g/mol 13)Find the formula mass of Mg3(PO4)2 262.79 g/mol 14)Find the molecular mass of C12H22O11 342.23 g/mol 15) How many moles are in 17.7 g of KMnO4? 0.112 mol KMnO4 ## How many grams of potassium K are in 100 g of Kmno₄? In one mole or 166 g of KI, 23.55% of it is K. Therefore, in 100 grams of KI , 23.55% of it would be K, so 23.55 g of potassium are in 100 g sample of KI. Another way to do it is setting up a proportion. ## How many grams are in 3 moles of Na2SO4? No of moles=458/142=3 moles. There are 3 moles in 458 grams of Na2SO4. ## How many grams are in 0.5 moles of H2SO4? molar mass of H2SO4 is 98g. given mass or mass present =0.5 × 98 = 49g. ## What is gram mole? (Often called gram-molecular weight.) A mass of a substance in grams numerically equal to its molecular weight. Example: A gram-mole of salt (NaCl) is 58.44 grams. THIS IS IMPORTANT:  Does ibuprofen help with acne? ## Why do we convert moles to grams? When substances react, they do so in simple ratios of moles. … Balances DO NOT give readings in moles. So the problem is that, when we compare amounts of one substance to another using moles, we must convert from grams, since this is the information we get from balances. ## How many number of moles are equivalent to 8 gram of Co2? The molar mass of cobalt is 58.93 g/mol. So assuming it forms diatomic molecules, the molar mass of Co2 would be 117.86 g/mol. ## How do you make 0.1 KMnO4? Potassium Permanganate 0.1 N: Dissolve 3.3 g of reagent grade potassium permanganate (KmnO4) in 1 L of purified water and heat on a steam bath for two hrs. Cover and allow to stand for 24 hrs. Filter through a fine porosity sintered glass crucible, discarding the first 25 mL.

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Limited access A current of 3 A passes through a closed loop circuit. What charge passes through the circuit each second? A $1.875\times { 10 }^{ 19}\quad C$ B $1.6\times { 10 }^{ -19 }\quad C$ C $1.6\times { 10 }^{ -19 }\quad A$ D $3\quad C$ Select an assignment template

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# Question about thevenin circuit analysis with voltage across voltage source I have a certain circuit in mind and I want to find its Thevenin equivalent resistor: I see that there is an dependent voltage source so I add a voltage source of 1V between a and b and make the other voltage source equal to 0 (the independent 9v). when I use nodal analysis on the new circuit, I am not sure how to calculate the current i3. I thought about using Kirchhoff voltage law on the middle loop but for now I am confused. If I have three voltage sources, do I just take a wild guess about which way the current goes when I apply KVL on the middle loop? because if they didn't note the direction of the current on the Vx resistor, I would have said that Vx+2i3-2Vx+1=0, or would that be wrong because I assumed i3's direction to go to the right? tl;dr • how would I do a KVL equation on the middle loop • is there an easier way to solve it using nodal analysis (specifically, to find i3)? Sorry if the question seems a bit confused, because I'm confused. • Well it seems to me that it is a reasonable question. He discussed what he tried to do and what is still confusing to him. Commented Jun 1, 2014 at 12:34 This is how i think i would solve the problem: I don't see it necessary to add a 1V excitation source since there already a 9v in the circuit. The trick when analyzing circuit with dependent sources is to avoid the Short circuiting you would normally do when having a circuit with independent sources alone: _Things to note that may help Focus on node A: define a current from the 9v source direction to node A and also define a current from 6ohm to A _not Vx =Va 9 - Vx = I1 Vx/6 = I2 and i reckon Vab = 2Vx.. Hope this helps you! If your objective is to find the Thevenin resistance, then I'm not sure why you're approaching it this way. If you can find the Thevenin resistance $R'_{th}$ of the circuit to the left of the $1\Omega$ resistor, the total equivalent resistance is just $$R_{th} = R'_{th}|| 1\Omega$$ And, I recommend using a current test source which will make finding $R'_{th}$ almost trivial. For this particular question, i think the source conversion technique would be the best. You convert the independent voltage source and 3 ohm resistor to a current source of 3A with 3 ohm res in parallel. Next, this 3 ohm res is in parallel with 6 ohm res. Reduce it to a single res of 2 ohm. So now you have a current source of 3A in parallel with 2 ohm res. At this point again use source conversion. You get a voltage source of 6V in series with a 2 ohm res, combined in series with 2 ohm res and dependent source. The remainder of the network can be easily solved. A major benefit of this technique is that you have reduced one loop completely. Hope i am correct.

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easy one : Quant Question Archive [LOCKED] Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 16 Jan 2017, 10:21 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track Your Progress every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # easy one post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Senior Manager Joined: 02 Feb 2004 Posts: 345 Followers: 1 Kudos [?]: 62 [0], given: 0 easy one [#permalink] ### Show Tags 21 Feb 2005, 16:06 This topic is locked. If you want to discuss this question please re-post it in the respective forum. do it! Attachments 0034.jpg [ 16.23 KiB | Viewed 1001 times ] Senior Manager Joined: 19 Feb 2005 Posts: 486 Location: Milan Italy Followers: 1 Kudos [?]: 22 [0], given: 0 ### Show Tags 22 Feb 2005, 05:04 the prices could be 1000$/1000L=1$/L and the new price 1000$/1200L=5/6$/L so decrease is 1-5/6=1/6 VP Joined: 13 Jun 2004 Posts: 1118 Location: London, UK Schools: Tuck'08 Followers: 7 Kudos [?]: 45 [0], given: 0 ### Show Tags 22 Feb 2005, 05:39 I think I've fallen into the trap ....why it's not 20% ? 1000/1200 ? Senior Manager Joined: 19 Feb 2005 Posts: 486 Location: Milan Italy Followers: 1 Kudos [?]: 22 [0], given: 0 ### Show Tags 22 Feb 2005, 05:51 the increase in volume is 20% whereas the decrease in price is 16.67% easy but tricky VP Joined: 30 Sep 2004 Posts: 1488 Location: Germany Followers: 6 Kudos [?]: 327 [0], given: 0 ### Show Tags 22 Feb 2005, 05:52 Antmavel wrote: I think I've fallen into the trap ....why it's not 20% ? 1000/1200 ? why should it be 20% ? as you already pointed out 1000/1200=0,83*100=83,33=100-83,33=16,67% SVP Joined: 03 Jan 2005 Posts: 2243 Followers: 16 Kudos [?]: 324 [0], given: 0 ### Show Tags 22 Feb 2005, 07:25 Another very good question type that has high likelyhood to appear in the real test, in my opinion. This kind of question often appears this way: If something increased by 20%, then decreased by 20%, how much is the percentage of increase/decrease when the final value is compared to the initial value? Intern Joined: 18 Feb 2005 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 22 Feb 2005, 14:17 Suppose price for 5x10x24 box is X So if the price to size ratio remains same price for 6x10x24 shoud be 6X/5. However the new price for 6x10x24 is X so the diff is X/5 %Diff is [(X/5*100)]/(6X/5) = 16.67% VP Joined: 25 Nov 2004 Posts: 1493 Followers: 7 Kudos [?]: 98 [0], given: 0 ### Show Tags 22 Feb 2005, 17:46 HongHu wrote: Another very good question type that has high likelyhood to appear in the real test, in my opinion. This kind of question often appears this way: If something increased by 20%, then decreased by 20%, how much is the percentage of increase/decrease when the final value is compared to the initial value? Yes, i agree with you. it is very likely question. =x-x(1.2)(0.8)=x(1-0.96)=0.04x or 4% VP Joined: 13 Jun 2004 Posts: 1118 Location: London, UK Schools: Tuck'08 Followers: 7 Kudos [?]: 45 [0], given: 0 ### Show Tags 22 Feb 2005, 20:26 thearch wrote: the increase in volume is 20% whereas the decrease in price is 16.67% easy but tricky I got it... 100 -> 1000$120 should be 1200$, however it's 1000$so decrease is (1200-1000)/ 1200 = 1/6 thanks for the help... GMAT Club Legend Joined: 07 Jul 2004 Posts: 5062 Location: Singapore Followers: 30 Kudos [?]: 355 [0], given: 0 ### Show Tags 23 Feb 2005, 01:06 initial: 5*10*20=1000 cm^3 now: 6*10*20 = 1200 cm^3 Price = x Price per cm^3 = 1200/x and 1000/x Difference = 200/x % price in fruit juice went down = 200/x * x/1000 *100 = 10% ??? I looked at the solution and didnt get what you meant by size to price ratio. What's the logic behind this ratio ?? VP Joined: 13 Jun 2004 Posts: 1118 Location: London, UK Schools: Tuck'08 Followers: 7 Kudos [?]: 45 [0], given: 0 ### Show Tags 23 Feb 2005, 06:10 ywilfred wrote: initial: 5*10*20=1000 cm^3 now: 6*10*20 = 1200 cm^3 Price = x Price per cm^3 = 1200/x and 1000/x Difference = 200/x % price in fruit juice went down = 200/x * x/1000 *100 = 10% ??? I looked at the solution and didnt get what you meant by size to price ratio. What's the logic behind this ratio ?? I know why you found 10%, you've made a small mistake in your calculation, the price per cm3 should be x/1200 and x/1000 let's say that the standard price is 1000$ for 1000cm3 , it goes like this : 1000/1000 and 1000/1200 ; 1 and 5/6 there is a decrease of 1/6 on the price, which means 16.6% There is no special logic behind it, except the double meaning : increase in volume OR decrease in price -> % are not always equal... Intern Joined: 18 Feb 2005 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 23 Feb 2005, 06:29 Fred, My BAD, I should have put the word like this If the Price is directly proportional to Size the price of the box 6X10X24 should be [x/(5X10X24)]x(6X10X24) = 6x/5. Hope it is clear now Pankaj SVP Joined: 03 Jan 2005 Posts: 2243 Followers: 16 Kudos [?]: 324 [0], given: 0 ### Show Tags 23 Feb 2005, 06:49 MA wrote: HongHu wrote: Another very good question type that has high likelyhood to appear in the real test, in my opinion. This kind of question often appears this way: If something increased by 20%, then decreased by 20%, how much is the percentage of increase/decrease when the final value is compared to the initial value? Yes, i agree with you. it is very likely question. =x-x(1.2)(0.8)=x(1-0.96)=0.04x or 4% :b: Decrease or increase? GMAT Club Legend Joined: 07 Jul 2004 Posts: 5062 Location: Singapore Followers: 30 Kudos [?]: 355 [0], given: 0 ### Show Tags 23 Feb 2005, 18:52 Ah.. i get it now.. should be a percentage decrease, isn't ?? VP Joined: 25 Nov 2004 Posts: 1493 Followers: 7 Kudos [?]: 98 [0], given: 0 ### Show Tags 23 Feb 2005, 21:28 HongHu wrote: :b: Decrease or increase? =[{x (1.2) (0.8)} - x]/x SVP Joined: 03 Jan 2005 Posts: 2243 Followers: 16 Kudos [?]: 324 [0], given: 0 ### Show Tags 23 Feb 2005, 22:55 Good job! Manager Joined: 24 Jan 2005 Posts: 217 Location: Boston Followers: 1 Kudos [?]: 12 [0], given: 0 ### Show Tags 25 Feb 2005, 14:24 20/120 * 100 = 1/6*100 = 16.6667% 25 Feb 2005, 14:24 Display posts from previous: Sort by # easy one post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

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Search a number 302350122 = 23234748407 BaseRepresentation bin10010000001010… …111111100101010 3210001220222101200 4102001113330222 51104400200442 650000225030 710330635144 oct2201277452 9701828350 10302350122 1114573a138 128530b176 134a8416c9 142c225c94 151b82534c hex12057f2a 302350122 has 24 divisors (see below), whose sum is σ = 656993376. Its totient is φ = 100490856. The previous prime is 302350121. The next prime is 302350127. The reversal of 302350122 is 221053203. 302350122 is a `hidden beast` number, since 302 + 350 + 12 + 2 = 666. It is a Harshad number since it is a multiple of its sum of digits (18). It is a junction number, because it is equal to n+sod(n) for n = 302350095 and 302350104. It is not an unprimeable number, because it can be changed into a prime (302350121) by changing a digit. It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 17958 + ... + 30449. It is an arithmetic number, because the mean of its divisors is an integer number (27374724). Almost surely, 2302350122 is an apocalyptic number. 302350122 is an abundant number, since it is smaller than the sum of its proper divisors (354643254). It is a pseudoperfect number, because it is the sum of a subset of its proper divisors. 302350122 is a wasteful number, since it uses less digits than its factorization. 302350122 is an evil number, because the sum of its binary digits is even. The sum of its prime factors is 48762 (or 48759 counting only the distinct ones). The product of its (nonzero) digits is 360, while the sum is 18. The square root of 302350122 is about 17388.2179075373. The cubic root of 302350122 is about 671.1764598023. The spelling of 302350122 in words is "three hundred two million, three hundred fifty thousand, one hundred twenty-two".

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## 25.3 Two 1.2 spheres are charged equally and placed 1.5 apart. When released, they begin to accelerate at 160 . What is the magnitude of the charge on each sphere? • Anonymous commented Rate for crct answer first..Thanx :D • Anonymous commented it wont let me and all this is wrong • Anonymous commented is 69.2 nC = 69.2 x 10^-6 C right ?? • Anonymous commented oosp is 69.2 nC = 69.2 x 10^-9 C right ?? • Anonymous commented yes Thank you very much • Anonymous commented ok then please 1st rate the 2nd answer that contain 69.2 x 10^-9 C 1st... Thanx :D • Rate me if u can • F = ma => k (Q/r)^2 = ma => 9 x 10^9 x Q^2 = 1.2 x 160 x 1.5^2 => Q = 2.19 x 10^-4 C • Anonymous commented the units are not properly given so u can do lyk this. :) • Anonymous commented if it is 1.2 grams and 1.5cm apart then u'll get as 9 x 10^9 x Q^2 = 1.2 x 10^-3 x 160 x 1.5 ^2 x 10^-4 => Q = 6.92 x 10^-8 C = 69.2 nC • Two 1.2 g spheres are charged equally and placed 1.5 cm apart. When released, they begin to accelerate at 160 F = ma, F = Force between 2 spheres m = mass of spheres Force between 2 spheres = Qq / 4(pi)(epsilon)(r^2) since 2 charges are equal, Qq = Q^2 Q^2 = F*4(pi)(epsilon)(r^2) 2.4*10^-3*160 = q^2/4piepsilon*r^2 q=sqrt((2.4*10^-3*160*1.5*1.5*10^-4)/(9*10^-9)) q= 97.97 C • for any particle force F = ma F=k (q/r)2 = ma F= 9 x 109 x Q2 = 1.2 x 160 x 1.52 = q = 2.19 x 10-4 C = ans.= magnitude of charge. • the mass of the two spheres is m = 1.2 kg the spheres are separated by a distance r = 1.5 cm = 1.5 * 10^-2 m the acceleration of the spheres is a = 160 m/s^2 the force acting on the spheres is F = m * a let the charge on the two spheres be q the electric force acting on the spheres is F_e = k_e * q^2/r^2 where k_e = 1/4pie_o = 9 * 10^9 Nm^2/C^2 Here,F = F_e or k_e * q^2/r^2 = m * a or q^2 = m * a * r^2/k_e or q = (m * a * r^2/k_e)^1/2 = (m * a/k_e)^1/2 * r Get homework help

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Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team # How do you find exact value of tan (pi/ 4)? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 4 Richard Share Mar 13, 2018 1 #### Explanation: Use special right triangles to find the value. Since $\frac{\pi}{4}$ = $45$ degrees, use the special right triangle on the left. If you do the $\tan \left(\frac{\pi}{4}\right)$, you do the $\frac{o p p o s i t e}{a \mathrm{dj} a c e n t}$. According to the triangle, the ratio is just one. • An hour ago • An hour ago • An hour ago • An hour ago • A minute ago • 15 minutes ago • 37 minutes ago • 48 minutes ago • An hour ago • An hour ago • An hour ago • An hour ago • An hour ago • An hour ago

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# Algebraic methods to compute Mathieu functions by Frenkel D., Portugal R. PDF By Frenkel D., Portugal R. Similar algebra books Download e-book for kindle: Applied Linear Algebra - Instructor Solutions Manual by Peter J. Olver, Cheri Shakiban Ideas guide to utilized Linear Algebra. step by step for all difficulties. Uploader's notice: because of txrx for offering the unique files. The better half name, Linear Algebra, has bought over 8,000 copies The writing sort is particularly obtainable the cloth might be coated simply in a one-year or one-term path comprises Noah Snyder's evidence of the Mason-Stothers polynomial abc theorem New fabric integrated on product constitution for matrices together with descriptions of the conjugation illustration of the diagonal workforce New PDF release: An Introduction to Nonassociative Algebras An creation to Nonassociative Algebras Richard D. Schafer Extra info for Algebraic methods to compute Mathieu functions Example text 1Differentiating, c f (x) g(x) A @ 1 A = 0 for all x. we find c1 f (x) + c2 g (x) ≡ 0 also, and hence @ f (x) g (x) c2 The homogeneous system has a nonzero solution if and only if the coefficient matrix is singular, which requires its determinant W [ f (x), g(x) ] = 0. (b) This is the contrapositive of part (a), since if f, g were not linearly independent, then their Wronskian would vanish everywhere. (c) Suppose c1 f (x) + c2 g(x) = c1 x3 + c2 | x |3 ≡ 0. then, at x = 1, c1 + c2 = 0, whereas at x = −1, − c1 + c2 = 0. B) This is the contrapositive of part (a), since if f, g were not linearly independent, then their Wronskian would vanish everywhere. (c) Suppose c1 f (x) + c2 g(x) = c1 x3 + c2 | x |3 ≡ 0. then, at x = 1, c1 + c2 = 0, whereas at x = −1, − c1 + c2 = 0. Therefore, c1 = c2 = 0, proving linear independence. On the other hand, W [ x3 , | x |3 ] = x3 (3 x2 sign x) − (3 x2 ) | x |3 ≡ 0. 1. Only (a) and (c) are bases. 2. Only (b) is a basis. 3. (a) B @ 0 A , @ 1 A; 2 0 (b) 0 1 0 1 3 1 B4C B4C B C , B C; @ 1A @ 0A 0 (c) 1 0 B B B @ 1 0 1 0 1 −2 −1 1 B C B C 1C C B 0C B0C C, B C , B C. N, is det Ak /det Ak−1 , where Ak is the k × k upper left submatrix of A with entries aij for i, j = 1, . . , k. A formal proof is done by induction. 20. (a–c) Applying an elementary column operation to a matrix A is the same as applying the elementary row operation to its transpose AT and then taking the transpose of the result. 56 implies that taking the transpose does not affect the de39 terminant, and so any elementary column operation has exactly the same effect as the corresponding elementary row operation.

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# Radioactive samples and differential equations • fogvajarash In summary: I have just found out that the constant in the rate term is actually the rate constant k. So now I can do the problem! fogvajarash ## Homework Statement A certain nuclear plant produces radioactive waste in the form of strontium-90 at the constant rate of 500kg. The waste decays exponentially with a half-life of 28 years. How much of this radioactive waste from the nuclear plant will be present after the following increments of time? Assume that initially 600kg of strontium-90 is in the nuclear plant. a. N years b. 140 years c. To perpetuity - ## The Attempt at a Solution I found that the "rate in" was 500kg/year, and I'm not sure about the "rate out" (rate of depletion of the sample). We are given that the sample decays exponentially, so we should have that y(t) = yo e-0.02475t. We can assume that yo is just y (the radioactive sample that is decaying). Then, the rate of depletion would be (I doubt that this is true): y'(t) = -0.2475ye-0.02475t This is because this is the rate of change of the sample. Then, i set up my differential equation to be the following (k is just the constant k = (ln2)/(half life)): dm/dt = 500 - kme-kt I rearranged the terms into the form dm/dt - P(x)y = f(t), i came up that the integrating factor was actually e-e^(-kt)), but this is not true (while differentiating the function I realized that it it was a completely new expression that was different to the one shown). How can I proceed from this information or have i made a wrong set-up to the exercise? Thank you. The decay rate depends on the current amount in the store. That is what "exponential decay" means. If y is the mass of 90Sr then: $$\frac{dy}{dt} = -\lambda y$$ see: http://en.wikipedia.org/wiki/Exponential_decay The constant ##\lambda## is related to the half-life. Last edited: fogvajarash said: dm/dt = 500 - kme-kt The differential equation has to be : dm/dt=500-km, as the amount of Sr-90 increase by 500 kg/year, and decreases proportionally with the amount present, that is by km. Solve, and fit the integration constant to the initial condition m(0)=600 kg. ehild I see. So the only way of doing this exercise is knowing the definition? Or is there a non definition way? As well, the constant in the linear rate term (rate of decay) is just the k constant found in the exponent right? I see. So the only way of doing this exercise is knowing the definition? Or is there a non definition way? The only way to do a problem is to understand what the words mean - yeah. As well, the constant in the linear rate term (rate of decay) is just the k constant found in the exponent right? If you don't know, you should solve the differential equation to find out what the constant is. Have you not done a section of coursework that covers radioactive decay? fogvajarash said: ## Homework Statement A certain nuclear plant produces radioactive waste in the form of strontium-90 at the constant rate of 500kg. The waste decays exponentially with a half-life of 28 years. How much of this radioactive waste from the nuclear plant will be present after the following increments of time? Assume that initially 600kg of strontium-90 is in the nuclear plant. a. N years b. 140 years c. To perpetuity - ## The Attempt at a Solution I found that the "rate in" was 500kg/year, and I'm not sure about the "rate out" (rate of depletion of the sample). We are given that the sample decays exponentially, so we should have that y(t) = yo e-0.02475t. We can assume that yo is just y (the radioactive sample that is decaying). Then, the rate of depletion would be (I doubt that this is true): y'(t) = -0.2475y0e-0.02475t=-0.2475 y That is true with a sample decaying only - with no input. Now you have input at constant rate. It is true that the rate of decay is proportional to the amount present : -0.2475y, but there is also input 500 kg/year. So the net rate of change of the amount of Sr-90 is dm/dt=500-0.2475m. ehild Simon Bridge said: The only way to do a problem is to understand what the words mean - yeah. If you don't know, you should solve the differential equation to find out what the constant is. Have you not done a section of coursework that covers radioactive decay? I have, but i just wanted to make sure if I'm correct or not (I'm sorry! I don't want to commit the same mistakes again). I was thinking, as the rate of depletion is ry0e-rt, can we just say that it should be ry? (consider that m, the mass of the radioactive sample, is defined as y = y0e-rt). After having this information it should just be a simple variables separable problem to get to the final answer. Say, if we had that the input wasn't constant (suppose it was 600t), then we would have to deal with a first order linear differential equation? Thank you for your great patience Simon. fogvajarash said: I have, but i just wanted to make sure if I'm correct or not (I'm sorry! I don't want to commit the same mistakes again). I was thinking, as the rate of depletion is ry0e-rt, can we just say that it should be ry? (consider that m, the mass of the radioactive sample, is defined as y = y0e-rt. The rate of depletion is is ry, but y is not equal y0e-rt. fogvajarash said: After having this information it should just be a simple variables separable problem to get to the final answer. Say, if we had that the input wasn't constant (suppose it was 600t), then we would have to deal with a first order linear differential equation? We deal with a first order linear differential equation anyway. ehild ## Related to Radioactive samples and differential equations Radioactive samples are materials that contain unstable atoms that emit radiation in the form of particles or electromagnetic waves. These samples are used in various scientific and medical fields for research and treatment purposes. ## How are radioactive samples measured? Radioactive samples are measured using a technique called radioactivity measurement, which involves using a Geiger-Muller counter or a scintillation counter to detect and measure the radiation emitted from the sample. ## What are differential equations? Differential equations are mathematical equations that describe how a quantity changes over time or space. They involve derivatives, which represent the rate of change of a quantity, and are used to model various natural phenomena, including radioactive decay. ## How are differential equations used in the study of radioactive samples? Differential equations are used in the study of radioactive samples to model and predict the rate of decay of the unstable atoms in the sample. This allows scientists to determine the half-life of the sample and the amount of radiation emitted over time. ## What safety measures should be taken when handling radioactive samples? When handling radioactive samples, it is important to follow proper safety protocols, including wearing protective gear and handling the samples in a designated area with proper shielding. It is also crucial to dispose of the samples properly to avoid any potential harm to humans and the environment. • Calculus and Beyond Homework Help Replies 7 Views 2K • Calculus and Beyond Homework Help Replies 7 Views 812 • Calculus and Beyond Homework Help Replies 1 Views 504 • Calculus and Beyond Homework Help Replies 7 Views 3K • Calculus and Beyond Homework Help Replies 2 Views 1K • Engineering and Comp Sci Homework Help Replies 1 Views 1K • Calculus and Beyond Homework Help Replies 9 Views 1K • Calculus and Beyond Homework Help Replies 4 Views 3K • Calculus and Beyond Homework Help Replies 18 Views 3K • Calculus and Beyond Homework Help Replies 13 Views 2K

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# Compactness over discrete metric space • Oct 16th 2011, 02:29 PM mia25 Compactness over discrete metric space Any idea that help in solving this problem is highly appreciated :) • Oct 16th 2011, 02:49 PM Plato Re: Compactness over discrete metric space Question: Is $\left\{ {\mathfrak{B}\left( {x;0.5} \right):x \in [0,1]} \right\}$ an open covering of $[0,1]~?$ • Oct 16th 2011, 03:04 PM mia25 Re: Compactness over discrete metric space So you are saying that all the points are isolated point so we don't have a finite open cover of the set ? Also, there is a theorem that states that every bounded and closed set in R^k is compact. so is the set of isolated point closed • Oct 16th 2011, 03:13 PM Plato Re: Compactness over discrete metric space Quote: Originally Posted by mia25 So you are saying that all the points are isolated point so we don't have a finite open cover of the set ? Correct. Quote: Originally Posted by mia25 Also, there is a theorem that states that every bounded and closed set in R^k is compact. But that theorem is not for $R^k$ with a discrete metric.

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× × # Find the following for path D in Figure 3.58: (a) the ISBN: 9781938168000 42 ## Solution for problem 14PE Chapter 3 College Physics | 1st Edition • Textbook Solutions • 2901 Step-by-step solutions solved by professors and subject experts • Get 24/7 help from StudySoup virtual teaching assistants College Physics | 1st Edition 4 5 1 236 Reviews 23 3 Problem 14PE Find the following for path D in Figure 3.58: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Step-by-Step Solution: Solution 14PE: Step 1 of 3:- Here we need to find the distance and displacement of path D. a) Each block has sides of . To get the distance we just simply add the lengths of the blocks for path D. . There are 13 blocks which come in the path, so we just added them. Distance is a scalar quantity and does not have a preferred direction, so we we able to do this. So, the distance is . b) Step 2 of 3:- Here we need to get the displacement which is a vector and it has a preferred direction and magnitude as well. To get the displacement, we will join the starting and ending point. Here the net displacement for path D is shown. We can add the horizontal and vertical vectors to get the displacement vector. Step 3 of 3 #### Related chapters Unlock Textbook Solution

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 Convert sign to rot | sign angle to complete rotations # angle conversion ## Amount: 1 sign angle (sign) of angle Equals: 0.083 complete rotations (rot) in angle Converting sign angle to complete rotations value in the angle units scale. TOGGLE :   from complete rotations into angle signs in the other way around. ## angle from sign angle to complete rotation Conversion Results: ### Enter a New sign angle Amount of angle to Convert From * Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8) * Precision is how many numbers after decimal point (1 - 9) Enter Amount : Decimal Precision : CONVERT :   between other angle measuring units - complete list. Conversion calculator for webmasters. ## Angles This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units. Convert angle measuring units between sign angle (sign) and complete rotations (rot) but in the other reverse direction from complete rotations into angle signs. conversion result for angle: From Symbol Equals Result To Symbol 1 sign angle sign = 0.083 complete rotations rot # Converter type: angle units This online angle from sign into rot converter is a handy tool not just for certified or experienced professionals. First unit: sign angle (sign) is used for measuring angle. Second: complete rotation (rot) is unit of angle. ## 0.083 rot is converted to 1 of what? The complete rotations unit number 0.083 rot converts to 1 sign, one sign angle. It is the EQUAL angle value of 1 sign angle but in the complete rotations angle unit alternative. How to convert 2 angle signs (sign) into complete rotations (rot)? Is there a calculation formula? First divide the two units variables. Then multiply the result by 2 - for example: 0.0833333333333 * 2 (or divide it by / 0.5) QUESTION: 1 sign = ? rot 1 sign = 0.083 rot ## Other applications for this angle calculator ... With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool: 1. in practicing angle signs and complete rotations ( sign vs. rot ) values exchange. 2. for conversion factors training exercises between unit pairs. 3. work with angle's values and properties. International unit symbols for these two angle measurements are: Abbreviation or prefix ( abbr. short brevis ), unit symbol, for sign angle is: sign Abbreviation or prefix ( abbr. ) brevis - short unit symbol for complete rotation is: rot ### One sign angle of angle converted to complete rotation equals to 0.083 rot How many complete rotations of angle are in 1 sign angle? The answer is: The change of 1 sign ( sign angle ) unit of angle measure equals = to 0.083 rot ( complete rotation ) as the equivalent measure for the same angle type. In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in sign - angle signs for angle amount, the rule is that the sign angle number gets converted into rot - complete rotations or any other angle unit absolutely exactly. Conversion for how many complete rotations ( rot ) of angle are contained in a sign angle ( 1 sign ). Or, how much in complete rotations of angle is in 1 sign angle? To link to this angle sign angle to complete rotations online converter simply cut and paste the following. The link to this tool will appear as: angle from sign angle (sign) to complete rotations (rot) conversion. I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.

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https://socratic.org/questions/how-do-you-sketch-the-graph-of-y-x-1-2-and-describe-the-transformation

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How do you sketch the graph of y=(x-1)^2 and describe the transformation? Feb 28, 2018 Graph the equation of ${x}^{2}$, then shift it to the right by 1. Explanation: $a {\left(x - h\right)}^{2} + b$ Remember this formula? It comes with many variations, but the basic idea is the same. $a$ is the vertical stretch of the function; $h$ is the horizontal shift, and $b$ is the vertical shift. In the equation ${\left(x - 1\right)}^{2}$, we have only one thing to worry about, which is the horizontal shift $h$. Since $h$ is negative, ($- 1$), it is actually a positive shift, to the right. Since there are no other shifts, all you have to do is graph the equation of ${x}^{2}$, then shift it to the right by 1!

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https://occupychristmas.org/a-software-company-sells-a-package-that-retails-for-99/

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The “Chapter 3 – #12: Software Sales – Tony Gaddis – Starting Out With Python” programming obstacle comes from Tony Gaddis’ book, “Starting Out through Python (fourth Edition, Global Edition)” ## Problem A software program agency sells a package that retails for \$99. Quantity discounts are provided according to the complying with table: Quantity Discount 10 – 19 10% 20 – 49 20% 50 – 99 30% 100 or more 40% Write a program that asks the user to enter the variety of packages purchased. The regimen have to then display the amount of the discount (if any) and the full amount of the purchase after the discount. ## Solution You are watching: A software company sells a package that retails for \$99 # 12. Software Sales# A software program firm sells a package that retails for \$99. # Quantity discounts are given according to the complying with # table:## Quantity Discount# 10–19 10% # 20–49 20% # 50–99 30% # 100 or more 40%## Write a regime that asks the user to enter the number of # packages purchased. The program need to then display the # amount of the discount (if any) and the total amount of # the purchase after the discount.PRICE_PER_PACKAGE = 99.00number_of_packperiods = float(input(' Enter # of packperiods purchased: '))display_message = ""if number_of_packperiods = 10 and number_of_packperiods = 20 and number_of_packperiods = 50 and number_of_packeras = 100: discount_portion = .40 # 40% package_total = number_of_packages * PRICE_PER_PACKAGE discount_amount = (package_total) * discount_portion grand_complete = package_full - discount_amount display_message = "Package full = \$" + format(package_full, ',.2f') + " Discount Percentage = " + format(discount_percent, '.0%') + " Discount amount = \$" + format(discount_amount, ',.2f') + " Grand also total = \$" + format(grand_complete, ',.2f')print(" " + display_message + " ") ## Video Walk Through This video contains a step by action process of exactly how to settle “Chapter 3 – #12: Software Sales – Tony Gaddis – Starting Out With Python”. ## That’s it for Chapter 3 – #12: Software Sales Thank you for taking an interest in what I do! I hope it was valuable for you as a lot as it assisted me along my journey in learning to code! ## You have the right to discover Python Chapter 3 Solutions: Starting Out With Python | Chapter 3 | Programming Challenge Solutions See more: Simple Skills Crossword Clue, Potential Answers For Simple Skills occupychristmas.org I gain teaching and sharing knowledge of the points that I am passionate about. I am a husband and also a father, much from perfect, and striving to be better eextremely day. Thank you for taking an interemainder in what I do!

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https://sure56.info/the-ultimate-guide-to-calculating-churn-rate/

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Saas The Final Information to Calculating Churn Charge You could possibly be welcoming a truckload of latest prospects within the entrance door; however for those who’re not monitoring your churn price, you possibly can be dropping extra of your current prospects out the again. If you wish to change into (or stay) worthwhile, churn price is among the most necessary metrics you have to be monitoring. Able to see for those who’re dropping extra prospects than you’re buying? Let’s see find out how to calculate your buyer churn price, and benchmarks for achievement, together with tricks to scale back that churn and enhance profitability. What’s Churn Charge? Churn rate is the share of shoppers or subscribers that cease buying your services or products over a particular time period. Also referred to as attrition price, this metric is calculated by taking the variety of misplaced prospects divided by the full variety of prospects. Churn could be calculated by the variety of prospects who cease buying, or the quantity of income you lose when these current prospects say adios to your online business. For instance, right here’s a easy churn price calculation: Buyer churn price = (Prospects misplaced throughout time interval / Complete variety of prospects at first) * 100 Instance: When you had 3,000 prospects at first of the month and misplaced 150 through the month, that equates to a month-to-month churn price of 5 %. If you wish to see this when it comes to the monetary impression, use this income churn price formulation: Income churn price = (Income misplaced throughout time interval / Complete income at first) * 100 Instance: When you had \$50,000 MRR at first of the month and \$47,350 on the finish, you’d have misplaced \$2,650 in month-to-month recurring income. That equates to a 5.3 % month-to-month churn price. However, damaging churn refers to buying extra prospects than you lose. When you lose 50 prospects however purchase 78 new ones over the identical interval, earnings from the brand new prospects will exchange the income misplaced via churn. Churn price is an particularly standard SaaS metric, and subscription companies depend on it to information buyer retention methods. However this KPI is beneficial throughout many industries, so maintain studying. Why Churn Evaluation is Important to Monitoring Enterprise Progress Monitoring churn is vital to constructing higher buyer retention. So for those who’re questioning if it’s price it, simply keep in mind: growing buyer retention charges by simply 5 % will increase earnings by 25 to 95 percent. Once you do common churn evaluation and observe your annual churn price, you’ll see: • Whether or not you’re retaining the shoppers you purchase • The extent of buyer satisfaction and loyalty • In case your buyer acquisition prices can be higher spent on serving to current prospects keep on • Whether or not your present development can be sustainable in the long run In fact, you don’t observe churn in a vacuum—watch this important gross sales KPI alongside others, together with your income development price, buyer acquisition prices (CAC), MRR and ARR, and buyer lifetime worth. That means, you’ll have a full image of your gross sales well being and buyer base. Keep in mind: churn compounds over time. What may sound like a low share of cancellations can in the end price your online business hundreds of {dollars} annually. Churn Charge Benchmarks: How Does Your Firm Evaluate? Do you’ve got a superb churn price—one which’s on par with the competitors or in your business? Recurly Research experiences that the common churn price in B2B is 4.91 %. And B2C sees 6.77 %. About 3.9 % of churn is voluntary—as a result of buyer dissatisfaction. And about 1.4 % is involuntary—as a result of fee points. In accordance with Paddle, the common churn price for SaaS corporations falls between 2 and eight % of MRR. This helps Recurly’s analysis, which exhibits the common churn price in software program touchdown round 4.75 %. Client items and retail sees 7.55 %, healthcare 6.03 %—and the checklist goes on. In fact, different components can impression the variety of churned prospects you see in a given interval, together with the age and dimension of the corporate, or the pricing construction. So, watch your business benchmarks, however keep in mind to make use of your personal numbers to benchmark progress over time. The right way to Calculate Churn Charge in 5 Simple(ish) Steps We all know the formulation, however let’s slender down the precise steps to calculate churn price for your online business (and monitor it over time). Step 1: Decide the Time Interval Subscription-based companies may observe their churn price on a month-to-month foundation (primarily based on utilization patterns or fee buildings), whereas one other B2B firm could observe it quarterly. Select a time interval that matches your business, the frequency of buyer interactions, and your pricing and subscription mannequin. That means, you may observe and evaluate your churn charges over time, and make data-driven choices. Use your CRM or one other knowledge instrument to outline the variety of prospects you had at first of your chosen time interval. Exclude any new prospects. If you wish to get extra granular, section your present prospects by demographics, subscription particulars (like pricing tier), acquisition channel, or product use case. This cohort evaluation will help you establish developments and prioritize the best actions. Instance: Let’s say prospects you purchase via LinkedIn adverts have a 4.8 % churn price, whereas these from chilly gross sales outreach churn at simply 2.8 %. Figuring out this, you may prioritize and scale back the LinkedIn churn (assuming it’s a major channel for your online business). Step 3:Rely the Variety of Prospects that Churned Now, it’s time to see what number of prospects churned. When you’re utilizing a data-savvy CRM, like Shut, you’ll be capable of discover this info with its built-in reporting features Preserve this info useful, so you may all the time make data-smart choices. Need to see how Shut will help? Step 4: Divide Prospects Misplaced by Prospects on the Begin of the Interval Time to seize your churn price formulation! Divide the variety of prospects misplaced through the interval (Step 3) by the full variety of prospects you had at first of that very same interval (Step 2). Then, multiply the end result by 100 to precise the churn price as a share. Instance: Suppose you had 500 prospects at first of the month, and through that month, you misplaced 20 prospects. Your churn price can be (20 / 500) * 100, which is 4 %. Step 5: Analyze and Optimize Sure, that is an ongoing course of. (Didn’t suppose you’d end that simply, did ya?) In order for you this knowledge to be significant, that you must evaluate it to business benchmarks and to earlier intervals inside your personal enterprise. That means, you may establish developments, and switch that knowledge into actions. Listed below are some methods you may analyze your churn price knowledge: • Search for widespread traits or behaviors with prospects who churn • Evaluate churn charges between totally different segments • Monitor churn charges over time to identify patterns (equivalent to seasonality) Your objective right here: perceive why prospects churn, and what you are able to do to cease it. Use the information you’ve collected to tell your choices, and also you’ll have a greater shot at enhancing buyer retention (and in the end rising your online business). Methods to Repair a Excessive Buyer Churn Charge Prospects churn for various causes, together with: • Services or products high quality • Worth sensitivity • No product/market match • Dangerous consumer or buyer expertise So, what actions can you are taking to cut back that churn price? Even the smallest lower in churn can have a huge impact in your backside line. So, let’s check out actionable methods to stop churn earlier than it occurs: • Refine your ICPs and purchaser personas: Promoting to bad-fit prospects is a large purpose for churn. As a substitute, spend time getting a crystal-clear image of the shoppers who get essentially the most worth out of your resolution. Then, tailor your messaging, in-app expertise, and sales pitches round them. Higher-fit prospects are much less prone to churn. • Provide superior buyer help: Take a proactive method to help—don’t simply attain out when there are points. Analyze buyer habits to establish potential churn, and assist resolve points earlier than they escalate. Over 60 percent of SaaS business customers churn due to a foul buyer success expertise, so be open to suggestions and fast to enhance. • Contact prospects with soon-to-expire fee strategies: Passive churn occurs while you strive to attract the following fee, and the shopper’s bank card is expired. Use dunning software program like Paddle to remind prospects when their card is expiring and ensure comfortable prospects don’t churn accidentally. • Push annual subscriptions over month-to-month billing: Paddle analysis exhibits that churn charges are decrease at corporations which have the next share of annual contracts. Nudge new prospects towards annual plans with discounted costs (equivalent to one month free). That is an a-okay and super-normal approach to implement reductions. • Match onboarding preferences to ache factors: Out-of-touch onboarding sucks. So, find out about what issues to your prospects and personalize their onboarding course of to match their ache factors. Corey Haines, founding father of SwipeWell, says this: “Broadly talking, churn is way larger within the first three months of a buyer’s lifecycle. This is the reason onboarding is so crucial to excessive retention: the shoppers that churn early on had been probably by no means really onboarded; they had been basically paying for a trial.” These methods will help you preempt churn—however what if a buyer is already on the level the place they’re able to cancel? Listed below are two issues you must do that may assist scale back churn: • Provide downgrades (and discounted upgrades) through the cancelation course of: Churning prospects are sometimes on the lookout for a greater deal—some place else. As a substitute, carry it to them by providing to downgrade to a cheaper plan, or improve so as to add additional options at teh similar (or discounted) worth. Simply watch out—reductions generally is a slippery slope. • Ask prospects why they’re canceling: If nothing else, a churning buyer will help you get insights into why churn occurs. Use buyer surveys, and even name them up! In the event that they hesitate to go away suggestions, supply an incentive. In spite of everything, if churn is draining your earnings, that \$10 Starbucks present card can be well worth the insights you get. Don’t Enable a Excessive Churn Charge to Sabotage Earnings Buyer churn is a compounding challenge that must be solved ASAP. Neglect it, and also you’ll watch income dwindle, development charges endure, and the longevity of your organization slowly slip away. There’s a clear monetary incentive for decreasing churn. It additionally results in extra correct gross sales forecasting and higher buyer satisfaction. And even when it hits the benchmark, keep in mind: the decrease your churn price, the higher. So, within the pursuit of that low churn price, examine yours: calculate it (through cohort evaluation if crucial) and implement methods to enhance it. It would seem to be a hopeless, time-consuming job at instances, however keep it up. It pays dividends (actually). Need assistance managing your gross sales course of and monitoring your churn price? Shut offers built-in, user-friendly options like reporting and automatic outreach to spice up your development—sustainably.

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# Properties Label 2070.2.a.z Level $2070$ Weight $2$ Character orbit 2070.a Self dual yes Analytic conductor $16.529$ Analytic rank $0$ Dimension $3$ CM no Inner twists $1$ # Related objects ## Newspace parameters Level: $$N$$ $$=$$ $$2070 = 2 \cdot 3^{2} \cdot 5 \cdot 23$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 2070.a (trivial) ## Newform invariants Self dual: yes Analytic conductor: $$16.5290332184$$ Analytic rank: $$0$$ Dimension: $$3$$ Coefficient field: 3.3.1101.1 Defining polynomial: $$x^{3} - x^{2} - 9 x + 12$$ Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 230) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$ ## $q$-expansion Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form. $$f(q)$$ $$=$$ $$q - q^{2} + q^{4} + q^{5} + ( 1 - \beta_{1} - \beta_{2} ) q^{7} - q^{8} +O(q^{10})$$ $$q - q^{2} + q^{4} + q^{5} + ( 1 - \beta_{1} - \beta_{2} ) q^{7} - q^{8} - q^{10} + ( -2 + \beta_{1} - 2 \beta_{2} ) q^{11} + ( -1 + \beta_{1} - \beta_{2} ) q^{13} + ( -1 + \beta_{1} + \beta_{2} ) q^{14} + q^{16} + ( 2 + \beta_{1} ) q^{17} + ( 1 - \beta_{1} - \beta_{2} ) q^{19} + q^{20} + ( 2 - \beta_{1} + 2 \beta_{2} ) q^{22} + q^{23} + q^{25} + ( 1 - \beta_{1} + \beta_{2} ) q^{26} + ( 1 - \beta_{1} - \beta_{2} ) q^{28} + ( 2 - 2 \beta_{1} ) q^{29} + ( -1 - \beta_{1} + \beta_{2} ) q^{31} - q^{32} + ( -2 - \beta_{1} ) q^{34} + ( 1 - \beta_{1} - \beta_{2} ) q^{35} + 2 \beta_{2} q^{37} + ( -1 + \beta_{1} + \beta_{2} ) q^{38} - q^{40} + ( \beta_{1} + 2 \beta_{2} ) q^{41} + 8 q^{43} + ( -2 + \beta_{1} - 2 \beta_{2} ) q^{44} - q^{46} + ( 6 - 2 \beta_{1} + 2 \beta_{2} ) q^{47} + ( 11 - \beta_{1} + 2 \beta_{2} ) q^{49} - q^{50} + ( -1 + \beta_{1} - \beta_{2} ) q^{52} + 6 q^{53} + ( -2 + \beta_{1} - 2 \beta_{2} ) q^{55} + ( -1 + \beta_{1} + \beta_{2} ) q^{56} + ( -2 + 2 \beta_{1} ) q^{58} + ( -4 - 2 \beta_{1} ) q^{59} + ( 2 - \beta_{1} + 4 \beta_{2} ) q^{61} + ( 1 + \beta_{1} - \beta_{2} ) q^{62} + q^{64} + ( -1 + \beta_{1} - \beta_{2} ) q^{65} + ( 4 + 4 \beta_{2} ) q^{67} + ( 2 + \beta_{1} ) q^{68} + ( -1 + \beta_{1} + \beta_{2} ) q^{70} + ( -4 + \beta_{1} ) q^{71} + ( -4 + 2 \beta_{1} - 2 \beta_{2} ) q^{73} -2 \beta_{2} q^{74} + ( 1 - \beta_{1} - \beta_{2} ) q^{76} + ( 5 + 8 \beta_{1} - \beta_{2} ) q^{77} + ( -4 + 4 \beta_{1} - 4 \beta_{2} ) q^{79} + q^{80} + ( -\beta_{1} - 2 \beta_{2} ) q^{82} + ( -2 + 2 \beta_{2} ) q^{83} + ( 2 + \beta_{1} ) q^{85} -8 q^{86} + ( 2 - \beta_{1} + 2 \beta_{2} ) q^{88} + ( -4 - 4 \beta_{1} + 2 \beta_{2} ) q^{89} + ( -2 + 5 \beta_{1} - 2 \beta_{2} ) q^{91} + q^{92} + ( -6 + 2 \beta_{1} - 2 \beta_{2} ) q^{94} + ( 1 - \beta_{1} - \beta_{2} ) q^{95} + ( -10 - 3 \beta_{1} ) q^{97} + ( -11 + \beta_{1} - 2 \beta_{2} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$3q - 3q^{2} + 3q^{4} + 3q^{5} + 3q^{7} - 3q^{8} + O(q^{10})$$ $$3q - 3q^{2} + 3q^{4} + 3q^{5} + 3q^{7} - 3q^{8} - 3q^{10} - 3q^{11} - q^{13} - 3q^{14} + 3q^{16} + 7q^{17} + 3q^{19} + 3q^{20} + 3q^{22} + 3q^{23} + 3q^{25} + q^{26} + 3q^{28} + 4q^{29} - 5q^{31} - 3q^{32} - 7q^{34} + 3q^{35} - 2q^{37} - 3q^{38} - 3q^{40} - q^{41} + 24q^{43} - 3q^{44} - 3q^{46} + 14q^{47} + 30q^{49} - 3q^{50} - q^{52} + 18q^{53} - 3q^{55} - 3q^{56} - 4q^{58} - 14q^{59} + q^{61} + 5q^{62} + 3q^{64} - q^{65} + 8q^{67} + 7q^{68} - 3q^{70} - 11q^{71} - 8q^{73} + 2q^{74} + 3q^{76} + 24q^{77} - 4q^{79} + 3q^{80} + q^{82} - 8q^{83} + 7q^{85} - 24q^{86} + 3q^{88} - 18q^{89} + q^{91} + 3q^{92} - 14q^{94} + 3q^{95} - 33q^{97} - 30q^{98} + O(q^{100})$$ Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{3} - x^{2} - 9 x + 12$$: $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$\nu^{2} + \nu - 7$$ $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$\beta_{2} - \beta_{1} + 7$$ ## Embeddings For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below. For more information on an embedded modular form you can click on its label. Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$ 1.1 2.68740 1.43163 −3.11903 −1.00000 0 1.00000 1.00000 0 −4.59692 −1.00000 0 −1.00000 1.2 −1.00000 0 1.00000 1.00000 0 3.08719 −1.00000 0 −1.00000 1.3 −1.00000 0 1.00000 1.00000 0 4.50973 −1.00000 0 −1.00000 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles ## Atkin-Lehner signs $$p$$ Sign $$2$$ $$1$$ $$3$$ $$-1$$ $$5$$ $$-1$$ $$23$$ $$-1$$ ## Inner twists This newform does not admit any (nontrivial) inner twists. ## Twists By twisting character orbit Char Parity Ord Mult Type Twist Min Dim 1.a even 1 1 trivial 2070.2.a.z 3 3.b odd 2 1 230.2.a.d 3 12.b even 2 1 1840.2.a.r 3 15.d odd 2 1 1150.2.a.q 3 15.e even 4 2 1150.2.b.j 6 24.f even 2 1 7360.2.a.ce 3 24.h odd 2 1 7360.2.a.bz 3 60.h even 2 1 9200.2.a.cf 3 69.c even 2 1 5290.2.a.r 3 By twisted newform orbit Twist Min Dim Char Parity Ord Mult Type 230.2.a.d 3 3.b odd 2 1 1150.2.a.q 3 15.d odd 2 1 1150.2.b.j 6 15.e even 4 2 1840.2.a.r 3 12.b even 2 1 2070.2.a.z 3 1.a even 1 1 trivial 5290.2.a.r 3 69.c even 2 1 7360.2.a.bz 3 24.h odd 2 1 7360.2.a.ce 3 24.f even 2 1 9200.2.a.cf 3 60.h even 2 1 ## Hecke kernels This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(2070))$$: $$T_{7}^{3} - 3 T_{7}^{2} - 21 T_{7} + 64$$ $$T_{11}^{3} + 3 T_{11}^{2} - 39 T_{11} - 144$$ $$T_{13}^{3} + T_{13}^{2} - 15 T_{13} - 18$$ $$T_{17}^{3} - 7 T_{17}^{2} + 7 T_{17} + 18$$ $$T_{29}^{3} - 4 T_{29}^{2} - 32 T_{29} - 24$$ ## Hecke characteristic polynomials $p$ $F_p(T)$ $2$ $$( 1 + T )^{3}$$ $3$ $$T^{3}$$ $5$ $$( -1 + T )^{3}$$ $7$ $$64 - 21 T - 3 T^{2} + T^{3}$$ $11$ $$-144 - 39 T + 3 T^{2} + T^{3}$$ $13$ $$-18 - 15 T + T^{2} + T^{3}$$ $17$ $$18 + 7 T - 7 T^{2} + T^{3}$$ $19$ $$64 - 21 T - 3 T^{2} + T^{3}$$ $23$ $$( -1 + T )^{3}$$ $29$ $$-24 - 32 T - 4 T^{2} + T^{3}$$ $31$ $$-8 - 7 T + 5 T^{2} + T^{3}$$ $37$ $$-32 - 40 T + 2 T^{2} + T^{3}$$ $41$ $$-186 - 59 T + T^{2} + T^{3}$$ $43$ $$( -8 + T )^{3}$$ $47$ $$288 + 4 T - 14 T^{2} + T^{3}$$ $53$ $$( -6 + T )^{3}$$ $59$ $$-144 + 28 T + 14 T^{2} + T^{3}$$ $61$ $$526 - 157 T - T^{2} + T^{3}$$ $67$ $$384 - 144 T - 8 T^{2} + T^{3}$$ $71$ $$24 + 31 T + 11 T^{2} + T^{3}$$ $73$ $$-248 - 40 T + 8 T^{2} + T^{3}$$ $79$ $$-1152 - 240 T + 4 T^{2} + T^{3}$$ $83$ $$-96 - 20 T + 8 T^{2} + T^{3}$$ $89$ $$-1152 - 48 T + 18 T^{2} + T^{3}$$ $97$ $$166 + 279 T + 33 T^{2} + T^{3}$$

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# Trigonometry Examples ### Trigonometry Examples for the ACT Math Exam This page has ACT math problems and trigonometry examples for you to study, such as problems on sine, cosine, and tangent. Sine Cosine Tangent ### Trigonometry Examples in Right Triangles Trigonometry questions on the test will evaluate your understanding of the relationships and functions of sine, cosine, and tangent in right triangles. Remember the following important trigonometric formulas for calculating the sine, cosine, and tangent of any given angle A, as in the illustration above: sin A = x/z cos A = y/z tan A = x/y In other words, for any angle, sine is calculated by taking the measurement of the opposite side divided by the measurement of the upper side (hypotenuse) of the triangle. Cosine is calculated by taking the measurement of the lower side (which is adjacent) divided by the measurement of the upper side (or hypotenuse) of the triangle. Tangent is calculated by taking the measurement of the opposite side divided by the measurement of the lower or adjacent side of the triangle. ### Trigonometric Relationships These are the trigonometric relationships for right triangles: • cos2 A + sin2 A = 1 • cos2 A = 1 − sin2 A • sin2 A = 1 − cos2 A • Tangent is always equal to sin ÷ cos. The trigonometry equations for sine, cosine, and tangent are included above, as well as in our cheat sheet. ### Using Trigonometry on the ACT Math Exam You will see different types of questions on sine, cosine, and tangent on the test. Some of these problems will tell you directly that you need to calculate the sine, cosine, or tangent. Example 1: How is the tangent of x calculated? Example 2: For any given angle A, sin2 A = ? However, the majority of questions on the trigonometry part of the math test will only imply indirectly that you need to use one of these  trigonometry equations. So you will need to understand how to use sine, cosine, or tangent in order to answer the question. For example, the question might show you a triangle and give you the measurements of the degrees of two of the angles in the triangle, and then ask you to calculate the length of one of the sides of the triangle. More complex problems will show two triangles embedded inside or partially within each other. In these cases, you will need to use the trigonometry equations you have learned in order to calculate the degrees or length of the particular part of one of the triangles, and then use that result in another calculation for the second triangle in order to arrive at your final answer. You will also need to use ACT trigonometry equations in order to understand trigonometric graphing and modeling. Here are some math equations for trigonometry questions. You can practice these formulas and more in our free sample test. θ = s ÷ r θ = the radians of the subtended angle s = arc length

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mersenneforum.org Subproject #7: 100k-150k sequences to 110 digits Register FAQ Search Today's Posts Mark Forums Read 2010-10-25, 10:02 #1 10metreh     Nov 2008 232210 Posts Subproject #7: 100k-150k sequences to 110 digits This is the obvious next range to do, so we'll do it. Instructions for mods: When posting sequences from a new 5k range, only include sequences which are listed in this file and below 110 digits in the DB (unless someone has them reserved, or there is a DB error and someone on the forum has taken the sequence above 110 digits). This subproject is complete! Subproject #9 has begun! Last fiddled with by Mini-Geek on 2011-01-24 at 17:19 Reason: updating 2010-10-25, 15:58 #2 firejuggler     Apr 2010 Over the rainbow A4416 Posts taking 100260, 100344 2010-10-25, 17:09 #3 unconnected     May 2009 Russia, Moscow 5·17·31 Posts Reserving 103512, 103656, 103668, 103782, 103956, 101820 102264 102660 102942. 2010-10-26, 22:00 #4 bchaffin   Sep 2010 Portland, OR 1011101002 Posts Reserving 100488, 100656, 100944, 101094, 101142, 101352, 101448, 101844, 101862. 2010-10-27, 08:12 #5 debrouxl     Sep 2009 2×3×163 Posts Reserving 104970. 2010-10-28, 20:22 #6 firejuggler     Apr 2010 Over the rainbow 262810 Posts 100260 and 100344 are done (2^2 * 3 * 7 for 100260 and 2*3^2 for 100344) taking 102114, 102432 2010-10-28, 20:29 #7 bchaffin   Sep 2010 Portland, OR 22·3·31 Posts Done with: 100488: i1298 (size 110) = 2^4 · 31 · 181 · 5779 · c101 100656: i776 (size 110) = 2^2 · 3^3 · 7 · 43 · 225637 · c100 100944: i629 (size 113) = 2^2 · 3^4 · 7 · 2267 · c106 101094: i1300 (size 119) = 2^2 · 3^2 · 7 · c116 101142: i1099 (size 111) = 2 · 3^2 · 5 · 263 · 5573 · c103 101352: i915 (size 111) = 2^2 · 3 · 5 · 7 · 19 · c107 101448: i785 (size 110) = 2^4 · 3 · 5 · 31 · 10337 · c102 101844: i741 (size 112) = 2^2 · 7 · 17^2 · c108 101862: i1785 (size 110) = 2 · 3 · 5 · 967 · c106 2010-10-29, 03:29 #8 Greebley     May 2009 Dedham Massachusetts USA 34B16 Posts Reserving 104160, 104202, 104250, 104286, 104904 2010-10-29, 06:56 #9 debrouxl     Sep 2009 97810 Posts Done with 104970 from i1215 = C107 to i1260 = C111 = 2 * 3^2 * 7^2 * 15443 * C103. The C103 has received the following ECM curves: 75 @ B1=3e6, 10 @ B1=43e6. 2010-10-29, 16:09 #10 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 11001001000012 Posts reserving 106040, 106050, 106080, 106218, 106674, 106980 2010-10-29, 16:23 #11 Andi47     Oct 2004 Austria 9B216 Posts reserving 102450 Similar Threads Thread Thread Starter Forum Replies Last Post Mini-Geek Aliquot Sequences 151 2011-05-14 09:01 10metreh Aliquot Sequences 203 2010-11-14 15:00 10metreh Aliquot Sequences 345 2010-07-28 07:20 10metreh Aliquot Sequences 690 2009-10-14 09:02 henryzz Aliquot Sequences 204 2009-07-30 12:06 All times are UTC. The time now is 16:58. Tue Sep 28 16:58:04 UTC 2021 up 67 days, 11:27, 2 users, load averages: 1.57, 1.68, 1.71

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# Urgent | Statistics homework help Please click on the link above to submit this week’s assignment. There are multiple steps to fully complete this assignment. You must copy and paste the related tables from the STATA outputs into the document that includes your answers to this assignment. Then, in a few sentences describe the findings. Problem 1: Hint: I highly recommend that you study an example of descriptive statistics in STATA that has been posted under “week 1” For this problem, use the following variables from the “Random Sample Residents” data file to answer the questions below: Chain2: This variable indicates whether the assisted living facility is part of a chain (e.g., Sunrise, etc) or not. The levels (or groups) for chain2 are: 1 = yes (facility is part of a chain) 2 = no (facility is not part of a chain) Mocharges: It indicates monthly charges paid by the residents for services that were provided by the facilities. The levels of mocharges are as follows: 1 = \$1000 2 = \$1000-2999 3 = \$3,000-4999 4 = \$5,000-6999 5 = \$7000 (at least \$7000) LegoStay: This variable indicates length of stay (how long residents have been at the assisted living facility) for the residents at their facility. The levels of legoStay2 are: 1 = 1-12 months 2 = >1-3 years 3 = >=4 years (at least 4 years) Questions Describe (i.e., conduct descriptive analyses) the following variables: chain2, mocharges, LegoStay2. Provide an interpretation of your findings. Examine how an assisted living facility that is part of a national chain (e.g., Sunrise) is associated with monthly charges. Provide an interpretation of your findings. Examine how the length of stay is associated with monthly charges. Provide an interpretation of your findings. Problem 2: For this problem, use the following variables from the “STATA HOSPITAL DATA” data file to answer the questions below: Questions Using the Hospital data, a health analyst would like to determine whether there is a significant difference in the average number of admissions between two different types of hospitals, general medical versus psychiatric (service variable). Provide a copy of the STATA results and an interpretation of your findings. Note: For this problem, you should study the example found in the lecture notes on how to run a t-test in STATA. Service is the type of hospital. This variable consists of two groups (two types of hospitals) 1 = general medical 2 = psychiatric Admissions: Hospital admissions. We treat this as a continuous variable.

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### Home > CAAC > Chapter 4 > Lesson 4.1.6 > Problem4-54 4-54. 1. Use what you know about m and b to graph each rule below without making a table. Draw a growth triangle for each line. 4-54 HW eTool (Desmos). Homework Help ✎ 1. y = 2x − 3 2. y = −2x + 5 3. y = 3x 4. y = + 1 Each time you move over 1 unit, how far do you travel upward? Each time you move over 1 unit, how far do you travel upward? Where do you start on the y-axis? Use the eTool below to graph each of the rules. Click the link at right for the full version of the eTool: AC 4-54 HW eTool

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## Theophysics Subject: Physics | Topics: Theophysics is based on a formula of pure mathematics that demystifies the world as as unified field of force. It is the formula that unifies human-mind and God-mind into “I am of ‘One’ mind”. It is related to the term physicotheology, the difference between them being that the aim of physicotheology is to derive theology from physics, whereas that of theophysics is to unify physics and theology. Theophysics is more than a new understanding. It is a consilience of science and religion. ### Assignment on fundamental sciences Fundamental sciences is the study of material things was treated as a single aspect of human thought and called natural philosophy. But as knowledge increased it was found necessary to divide the study of nature into two main branches. They are  two main fundamental sciences are- A. Physical sci..... ### Assignment on Refraction of Light Reflection of Light The bouncing off of light or any wave at the face or boundary between two medium is called refraction.The figure above illustrates the regular reflection of a light ray. MM` represents the surface of a plane mirror. AO is the incident ray i.e. the direction in which light fall..... ### Experiment Study of Wein Bridge Oscillator Prime objective of this case study is to experiment on Wein Bridge Oscillator. The oscillator is a circuit where a balanced bridge is used because feedback network is Wien Bridge oscillator. The aim of this experiment is to check the operation of the Wien Bridge. To explain this experiment here ..... ### What is Wave Wave is a form of disturbance which travels onward due to repeated periodic motion of the medium particles. Energy is transferred during wave motion. Types of wave Mechanical wave: This type of wave motion requires material medium to propagate. For example ripples on water surface, vibration of a..... ### Experiment Study of Class-C Tuned Power Amplifier Primary objective of this case study is to experiment on Class-C Tuned Power Amplifier. Here focus on an amplifier that has tuned collector load and active device transistor operating at Class- C mode and observe the output voltage and current relationship at different harmonics. To experiment .....

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# Number 144562628496400 ### Properties of number 144562628496400 Cross Sum: Factorization: 2 * 2 * 2 * 2 * 5 * 5 * 17 * 17 * 35363 * 35363 Divisors: Count of divisors: Sum of divisors: 3.6895401981759E+14 Prime number? No Fibonacci number? No Bell Number? No Catalan Number? No Base 3 (Ternary): Base 4 (Quaternary): Base 5 (Quintal): Base 8 (Octal): 837a9c1e8c10 Base 32: 43fae1t30g sin(144562628496400) 0.81272891068345 cos(144562628496400) -0.58264201508241 tan(144562628496400) -1.3949026840581 ln(144562628496400) 32.604733944784 lg(144562628496400) 14.160056036143 sqrt(144562628496400) 12023420 Square(144562628496400) 2.0898353557788E+28 ### Number Look Up 144562628496400 (one hundred forty-four trillion five hundred sixty-two billion six hundred twenty-eight million four hundred ninety-six thousand four hundred) is a great figure. The cross sum of 144562628496400 is 61. If you factorisate the number 144562628496400 you will get these result 2 * 2 * 2 * 2 * 5 * 5 * 17 * 17 * 35363 * 35363. The number 144562628496400 has 135 divisors ( 1, 2, 4, 5, 8, 10, 16, 17, 20, 25, 34, 40, 50, 68, 80, 85, 100, 136, 170, 200, 272, 289, 340, 400, 425, 578, 680, 850, 1156, 1360, 1445, 1700, 2312, 2890, 3400, 4624, 5780, 6800, 7225, 11560, 14450, 23120, 28900, 35363, 57800, 70726, 115600, 141452, 176815, 282904, 353630, 565808, 601171, 707260, 884075, 1202342, 1414520, 1768150, 2404684, 2829040, 3005855, 3536300, 4809368, 6011710, 7072600, 9618736, 10219907, 12023420, 14145200, 15029275, 20439814, 24046840, 30058550, 40879628, 48093680, 51099535, 60117100, 81759256, 102199070, 120234200, 163518512, 204398140, 240468400, 255497675, 408796280, 510995350, 817592560, 1021990700, 1250541769, 2043981400, 2501083538, 4087962800, 5002167076, 6252708845, 10004334152, 12505417690, 20008668304, 21259210073, 25010835380, 31263544225, 42518420146, 50021670760, 62527088450, 85036840292, 100043341520, 106296050365, 125054176900, 170073680584, 212592100730, 250108353800, 340147361168, 361406571241, 425184201460, 500216707600, 531480251825, 722813142482, 850368402920, 1062960503650, 1445626284964, 1700736805840, 1807032856205, 2125921007300, 2891252569928, 3614065712410, 4251842014600, 5782505139856, 7228131424820, 8503684029200, 9035164281025, 14456262849640, 18070328562050, 28912525699280, 36140657124100, 72281314248200, 144562628496400 ) whith a sum of 3.6895401981759E+14. The number 144562628496400 is not a prime number. The number 144562628496400 is not a fibonacci number. The figure 144562628496400 is not a Bell Number. The figure 144562628496400 is not a Catalan Number. The convertion of 144562628496400 to base 2 (Binary) is 100000110111101010011100000111101000110000010000. The convertion of 144562628496400 to base 3 (Ternary) is 200221212001102010201012201121. The convertion of 144562628496400 to base 4 (Quaternary) is 200313222130013220300100. The convertion of 144562628496400 to base 5 (Quintal) is 122422003230343341100. The convertion of 144562628496400 to base 8 (Octal) is 4067523407506020. The convertion of 144562628496400 to base 16 (Hexadecimal) is 837a9c1e8c10. The convertion of 144562628496400 to base 32 is 43fae1t30g. The sine of 144562628496400 is 0.81272891068345. The cosine of 144562628496400 is -0.58264201508241. The tangent of the figure 144562628496400 is -1.3949026840581. The square root of 144562628496400 is 12023420. If you square 144562628496400 you will get the following result 2.0898353557788E+28. The natural logarithm of 144562628496400 is 32.604733944784 and the decimal logarithm is 14.160056036143. I hope that you now know that 144562628496400 is very impressive number!

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• date post 14-Apr-2018 • Category ## Documents • view 217 5 Embed Size (px) ### Transcript of Experiments from last week - University of · Web viewExperiments from last week. Profit for... Experiments from last week Experiments from last week. Profit for seller: Price minus cost. Profit for buyer: Value minus price Line up sellers from lowest to highest and plot costs. Get_____ Line up buyers from highest to plot values. Get.______ equilibrium is price = _____ and Q = ____ Supply and Demand for Section 001AL Highest value buyer: value of 60. At competitive price has profit of:______ Lowest cost seller: cost of 10. At competitive price has a profit of: _______ Average between them is 25 in profit. Some buyers have a value of 40 and at competitive price have profit of ______ At the perfectly competitive price, on average profit per student (per round) is \$12.00 Lets look what happened for 001AL: In the figure, blue squares represent transactions in the first round. There is an interesting pattern in the auctions. In the first round of each auction, transaction prices are all over the place, one at \$10, a six at around \$50, and so on. But most the trades taking place away from the \$40 benchmark are low price trades, e.g. \$10, \$13. So even though there is symmetry in that the buyers are mirror images of the seller. Students are more willing to sell at a low price than buy at a high price. Later in the auction, most trades take place near price equal to \$40. Average Realized Profit By Round (Equals 12.00 with perfect competition) Round Across All 12 Experiments Large Lectures 9:05 A-L 9:30 M-Z 10:10 A-L 10:35 M-Z 1 10.32 10.14 9.55 10.38 11.10 2 10.98 11.25 10.25 10.79 11.46 3 11.39 11.50 10.91 11.94 11.30 4 11.38 11.26 11.82 11.73 11.75 5 11.17 10.96 11.53 11.39 11.73 The previous table shows the results across different lectures. Recall that the average profit is \$12 at the perfectly competitive outcome in this market. We can see the learning, by looking at what is happening to the average profit in the later rounds. Profit starts low and rises in the later rounds. Next, let's look at the highest scoring trades. To determine who played the best, we have to adjust scores to take into account that some sellers draw low costs and some buyers draw high values. (Note: everyone who participated in auction gets 1 bonus point added to HW2 If your adjusted profit is in top half, your get an extra bonus point.) 1. Calculate buyers profit when purchase a book at \$40 (the competitive price). 2. Adjustment (per round) is difference between that and \$12 (average profit at competitive price). Example 1: Buyer has value of \$45. 1. Profit at \$40 purchase price = \$5 (= \$45-\$40). 2. Adjustment (per round) = +\$7 (to get to \$12 per round) Example 2: Buyer has value of \$60. 1. Profit at \$40 purchase price = \$20 (= \$60-\$40). 2. Adjustment (per round) = \$8 (to get to \$12 per round) Bottom line: If a buyer, we start you at 12 and give you one point for every dollar below \$40 in the price of a book you buy. If a seller, start you at 12 and give you one point for every dollar above \$40 in the price of a book you sell. Who are the superstar traders? (The highest point getters in each auction?) They all did it the same way . Snapped up a great deal at least once and did OK the other times. Superstar for 001AL. Buyer value 51 predicted profit = 51-40 = 11 (adjustment is +\$1) Round Price paid Actual Profit 1 20 31 32 2 25 26 27 3 37.80 13.2 14.2 4 40 11 12 5 40 11 12 Superstar for 001MZ. Buyer value 39 predicted profit = \$1 (adjustment +13 if sale) Round Price paid Actual Profit 1 15 24 37 2 30 9 22 3 35 4 17 4 23 16 29 5 39 0 13 Superstar for 017AL. Buyer value 49 predicted profit = 49-40 = 9 (adjustment is +3) Round Price paid Actual Profit 1 12 37 40 2 36 13 16 3 38 11 14 4 40 9 12 5 39.31 9.69 12.69 Superstar for 017MZ. Buyer value \$45 predicted profit = 45-40 = 5 (adjustment is +7) Round Price paid Actual Profit 1 13.99 31.01 38.01 2 34.99 10.01 17.01 3 40 5 12 4 41 4 11 5 41 4 11 And the superstars are.... While most research in economics is based on data from actual markets, analysis of data from experimental markets plays some role in the field. Can see how we can use this data to study speed of learning Can also potentially use this to study characteristics of success. Do students with an interest in business (e.g. in Carlson) tend to do better? Lets take an exploratory look with data from earlier years. Relationship between Econ 1101 Class Performance (Quartile) Course Performance (Quartile) Average Years 2009-2011 2009 2010 2011 1 55.43 55.68 55.09 55.45 2 54.58 55.09 54.87 53.88 3 53.66 55.12 54.29 51.78 4 50.25 51.30 49.02 50.29 Relationship between Econ 1101 Class Performance (Quartile) Being late for experiment (robot plays first round, student plays later rounds) Course Performance (Quartile) Percent of Students Late Average Years 2009-2011 2009 2010 2011 1 2.2 1.7 2.6 2.4 2 2.3 1.7 3.5 2.0 3 2.3 1.7 2.9 2.4 4 5.9 6.9 5.7 5.2 Year after year, the students who tend to do poorly in this class are the same ones who tend to show up late for the auction. More generally students who do poorly in grades tend to earn less later in life than their peers who do well in grades. There are many factors at work here. One potential factor: being late for things (sending resumes out late, etc.) Certainly true in experiment. If robot plays for you, you get zero!

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Our World Statistics Day conversations have been a great reminder of how much statistics can inform our lives. Do you have an example of how statistics has made a difference in your life? Share your story with the Community! Choose Language Hide Translation Bar Highlighted Level I Getting calculated control limits from a Levey Jennings chart Hi Everyone, I'm currently working on a script that generates a LJ chart.  I need to drop a bunch of lines in there and must make sure the scale of the Y-axis allows to see all these lines. This all works perfectly except for the calculated control limits.  Is there any way that I can get these calculated limits, preferably before I generated the chart, so I can use them in the calculation I make to determine the Y-axis scale? Thanks! --Jan 1 ACCEPTED SOLUTION Accepted Solutions Super User Re: Getting calculated control limits from a Levey Jennings chart Levey Jennings just uses the long term sigma to compute control limits, so you can just take the standard deviation of the entire column of the response you are plotting.  Here's an example using the Vial Fill Weights example from the Sample Data Library. ``````dt = Data Table("Vial Fill Weights"); y = :Fill Weight << Get Values; y_bar = mean(y); sigma = std dev(y); UCL = y_bar + 3*sigma; LCL = y_bar - 3*sigma;`````` This returns 5.90482338181099 and 6.32439884041124 for LCL and UCL respectively.  This matches the Levey Jennings limits: -- Cameron Willden 1 REPLY 1 Super User Re: Getting calculated control limits from a Levey Jennings chart Levey Jennings just uses the long term sigma to compute control limits, so you can just take the standard deviation of the entire column of the response you are plotting.  Here's an example using the Vial Fill Weights example from the Sample Data Library. ``````dt = Data Table("Vial Fill Weights"); y = :Fill Weight << Get Values; y_bar = mean(y); sigma = std dev(y); UCL = y_bar + 3*sigma; LCL = y_bar - 3*sigma;`````` This returns 5.90482338181099 and 6.32439884041124 for LCL and UCL respectively.  This matches the Levey Jennings limits: -- Cameron Willden Article Labels There are no labels assigned to this post.

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# Two long, straight wires, each carrying a current of 5 A, Question: Two long, straight wires, each carrying a current of $5 A$, are placed along the $X$ - and Y-axes respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1m, $1 m)$, (b) $(-1 m, 1 m)$, (c) $(-1 m,-1 m)$, (d) $(1 m,-1 m)$. Solution: Magnetic field due to wires are of equal magnitude but in opposite direction. So, resultant field $=0$. Leave a comment Click here to get exam-ready with eSaral

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Free Roulette - Features, Free Games Accordant of concept of opportunity, every number ordinary outs in the long term and requesting online wheel scheme creates same feasibility of every number dropping in. Roulette wheel is obtainable in a few types, comprising American (double zero), European and French (with way for tiers, neighbors and orphans), and with call bets. The major ascertainment is that roulette of casino principles are indeed simple. After the invention of world wide web, roulette's develop more favoured, as men began to play Internet gambling house roulette. Split: The player places their chip or chips on the intersecting kit between two figures, for sample,24 and 27. In the American Roulette roulette wheel, it's named as "no more gambling stakes please ". ! 143 It too does not deal in Roulette, Sic Bo, or any casino game. Doublet – in a similar way as the single number bet, this is a gambling bet crack on two numbers foreseeing that either of them is going to be spun. At one close of the table spots a roulette with the layout of corresponding numbers as well as wagers set about around. Live roulette wheel cannot be played with this deal. Probability roulette top of: item, put. Roulette table layout: The roulette table contains a 3x12 grate of every number on the roulette available, the figures are alternately coloured black and red. Corner Bet: Roulette players can stake on four amounts by putting chippings at the backstreet where the four amounts face. Fibonacci roulette wagering systems is based on using Fibonacci mathematical sequence present in character to the codes of the roulette wheel. Maybe Roulette possesses remained so near to its prototype over all these years cause the original game was extremely close to excellent. Gambling. Today the popularity of games like slot machines, poker programme plays as well as others have as well got American casino games as roulette seek an version of online. Some of the methods integrate doubling the bet each time a player loses as well as diminishing the bet when they admit when some of the directs inform the exact contrary. Play with a Single 0 – The chances at a roulette that has the double zero are more in behalf of the gambling house than in the case there's a single 0. In French roulette, there're still thirty - six numbered pockets, but there's particularly one null bag as opposed to two. Do not be fulled by gaming houses list "previous figures spun' - roulette wheel is a game of chance - every number owns the equal peradventure of coming up once more repeatedly. The pace of roulette Uk games likewise differs from a lot of nations. The single zero is called Euro roulette on the Internet and the double zero is named Roulette of America. Corner is a ante on 4 amounts in a sector gambling house roulette game demonstration. The Martingale growing scheme is especially framed for the even finance payment gambling wagers placed on net roulette. For five years presently, more and more humans are listing in interactive gambling - houses just to play in on - line roulette games. Promotions888 Promotions of casino introduce their Royals the chance to gain all from must - have instruments and awards to benefits as well as FreePlay coupons, Nevertheless that's not all. Chug a plenty of baccarat, sic Bo, baccarat drives strip gaming in a nine player and the mentioned above. Gambling Internet roulette doesn't get easier than that. The American Roulette roulette wheel owns the amounts one to 36 on it. Roulette is a classic casino game which can be found in word for word every land - or Internet - based casino on the planet. In roulette wheel a ball is rolled opposite a distaff. Enough cause for plentiful players to have tried the live online version of Roulette as well as relished the experience. The gambling house supplies a complete list of all of the suggested payoff methods, explaining which international dispositions are generally permitted to employ these systems and if or not or not the program applies to deposits as rationally as withdrawals. System of pivots is newly improved roulette system, responsible for increasing the opportunities of the players. A starter might or make profit from a free roulette reward. Cutting border computerized slot machines handle any of analogous casual number generators as on-line slots; these irregular generators of number are the prerequisite of numeric casino games by any means reliable Casinos on Internet. A pattern of benefits followed swifty by a failure, or a movement via a sequence of hands of the winning where the player is enlarging their gambling bet, will Internet resource out their profit from previous winning sequences and finally count as a wasting. Croupier is French People kind for the American period "dealer' typically practical in American gambling-houses or casinos in the Internet. Over a while term filchers tricked inexperienced people, speculating specific "enigmatic" ways which were offered for considerable gauges of finance - sometimes even lord humans became patients of those frauds. European roulette is a lot better for the player, offering merely a 2. Sure, there are some skills Chess gamblers enjoy it can be applied to poker methodologies, but the 2 games are in general different in that Chess is set by capability when poker is just as much chance like it is workmanship (that's what makes it exciting, in the end). Frequently, the majority of gambling houses have declined to unfold the house edge data for their games of slot and because of the unknown amount of marks and weightings of the reels, usually this is major troublesome to compute than for other games of casino. Utilizing the Bridge Club's premises on New South Head Road as a front, Sydney identity Percival Galea ruled five wheels of roulette, six tables for blackjack and a craps desk that averaged a fluidity of \$ 2 million per week. Reality be spoken, the diversity among the excellent reels opening deviation vs present screen sector presentation is joint with the trail, in which both of these kinds of beginning entertainments offer winning pay to players.) The chip is disposed on the field at the bottom of the ultimate number of the amenable column on the roulette wheel table design. Free roulette methods homewood free roulette strategies 205 - 943 - 9129 mangelsen's united states of free roulette schemes America al scrappin' kats the USA al kullman. On the other side, this payout grade is give or receive equivalent to the 50% of the evaluation of most lifted large wager sum gained from playing reels. 06% for wagers on banker. 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# Fill a timetable outside the gaps 2 views (last 30 days) giannit on 14 May 2021 Commented: giannit on 22 May 2021 Consider the following timetable, which is just a toy to explain the problem dates = [datetime(2020,1,1) datetime(2020,1,1) datetime(2020,1,2) datetime(2020,1,2) datetime(2020,1,3) datetime(2020,1,3) datetime(2020,1,4)]; t = timetable(dates',["a" "c" "a" "b" "a" "a" "c"]',[2 1 3 2 1 2 1]') Time Var1 Var2 ___________ ____ ____ 01-Jan-2020 "a" 2 01-Jan-2020 "c" 1 02-Jan-2020 "a" 3 02-Jan-2020 "b" 2 03-Jan-2020 "a" 1 03-Jan-2020 "a" 2 04-Jan-2020 "c" 1 For each date and for each unique string in Var1 I would like to obtain the sum of the values in Var2. That is: (1) if for a specific date a string does not appear, then 0 must be returned (2) if for a specific date a string appears only once, then the corresponding value in Var2 must be returned (3) if for a specific date a string appears multiple times, then the sum of the corresponding values in Var2 must be returned The function retime satisfy (2) and (3) completely, but (1) only partially in the sense that it returns 0 only if the string appears in a previous date AND in a following date, that is retime only fills the gaps. In the following the row with 04-Jan-2020 and 0 is missing retime(t(t.Var1=="a","Var2"),'daily','sum') Time Var2 ___________ ____ 01-Jan-2020 2 02-Jan-2020 3 03-Jan-2020 3 In the following the rows with dates 1, 3, 4 january (and 0 in Var2) are missing retime(t(t.Var1=="b","Var2"),'daily','sum') Time Var2 ___________ ____ 02-Jan-2020 2 The following is correct retime(t(t.Var1=="c","Var2"),'daily','sum') Time Var2 ___________ ____ 01-Jan-2020 1 02-Jan-2020 0 03-Jan-2020 0 04-Jan-2020 1 Is there an easy way to resolve the problem and without using for loops? The data I'm working with has thousands of rows and tens of columns, I tried with loops but it's time consuming. In the toy example an easy workaround is to manually add the missing strings at the start and at the end of the timetable, in such a way we can extend the gaps to cover all the datetimes and then retime can fill all the missing values. However, this is easy only for timetables with a simple structure, moreover it might have some unwanted side effects t = [timetable(datetime(2020,1,1),"b",0) ; t ; timetable([datetime(2020,1,4);datetime(2020,1,4)],["a";"b"],[0;0])] Time Var1 Var2 ___________ ____ ____ 01-Jan-2020 "b" 0 01-Jan-2020 "a" 2 01-Jan-2020 "c" 1 02-Jan-2020 "a" 3 02-Jan-2020 "b" 2 03-Jan-2020 "a" 1 03-Jan-2020 "a" 2 04-Jan-2020 "c" 1 04-Jan-2020 "a" 0 04-Jan-2020 "b" 0 Siddharth Bhutiya on 19 May 2021 You could use groupsummary to do this. In your case your grouping variables would be Time and Var1 and the aggregation method you want to use would be sum. Since you want all permutations of the grouping variables to show up in your output, you can specify the "IncludeEmptyGroups" as true and that should give you the desired output. groupsummary(t,["Time","Var1"],"sum","Var2","IncludeEmptyGroups",true) ans = 12×4 table Time Var1 GroupCount sum_Var2 ___________ ____ __________ ________ 01-Jan-2020 "a" 1 2 01-Jan-2020 "b" 0 0 01-Jan-2020 "c" 1 1 02-Jan-2020 "a" 1 3 02-Jan-2020 "b" 1 2 02-Jan-2020 "c" 0 0 03-Jan-2020 "a" 2 3 03-Jan-2020 "b" 0 0 03-Jan-2020 "c" 0 0 04-Jan-2020 "a" 0 0 04-Jan-2020 "b" 0 0 04-Jan-2020 "c" 1 1 Siddharth Bhutiya on 20 May 2021 Edited: Siddharth Bhutiya on 20 May 2021 @giannit When I first read your comment about missing date my thought was well for groupsummary Time is just a grouping variable, so as far as it is concerned there is no difference between 02-Jan-2020 or 5-Jan-2020 or even 20-May-2021, all these are just values for it so what should it consider as a missing value? But then I thought about it some more and took another look at the problem and it made more sense. Technically 02-Jan-2020 is not missing but in your final output you want the Time variable to be in increments of 1 day starting from min(Time) to the max(Time). This seemed like a perfectly reasonable thing to do when someone is trying to summarize their data, so I went back to the documentation page for groupsummary and found the groupbins argument. So it seems that you can get the desired results by doing the following T = timetable(datetime(2020,1,[1 1 1 3 3 4])',["a" "c" "b" "a" "a" "c"]',[2 1 2 1 2 1]'); groupsummary(T,["Time","Var1"],["day","none"],"sum","Var2","IncludeEmptyGroups",true) ans = 12×4 table day_Time Var1 GroupCount sum_Var2 ___________ ____ __________ ________ 01-Jan-2020 "a" 1 2 01-Jan-2020 "b" 1 2 01-Jan-2020 "c" 1 1 02-Jan-2020 "a" 0 0 02-Jan-2020 "b" 0 0 02-Jan-2020 "c" 0 0 03-Jan-2020 "a" 2 3 03-Jan-2020 "b" 0 0 03-Jan-2020 "c" 0 0 04-Jan-2020 "a" 0 0 04-Jan-2020 "b" 0 0 04-Jan-2020 "c" 1 1 This method should also give you the correct values for the GroupCounts as opposed to the ad hoc method in the previous comment. Hope this helps !! giannit on 22 May 2021 Wow perfect! I hoped for a one line command but I was lost in the documentations, thank you very much for the big help! To convert the table to a timetable, is it fine to do this? T = timetable(datetime(2020,1,[1 1 1 3 3 4])',["a" "c" "b" "a" "a" "c"]',[2 1 2 1 2 1]'); t = groupsummary(T,["Time","Var1"],["day","none"],"sum","Var2","IncludeEmptyGroups",true); t.day_Time = datetime(string(t.day_Time)); tt = table2timetable(t) thanks again! dpb on 14 May 2021 Edited: dpb on 14 May 2021 Illustrate the concept -- tt.Var1=categorical(tt.Var1); % convert Var1 to categorical as rightfully is dRef=datetime(tt.Time(1):tt.Time(end)).'; % get the full date vector vRef=unique(tt.Var1); % and the unique values of Var1 % build a reference timetable of all times/characteristics ttRef=timetable(reshape(repmat(dRef.',size(vRef,1),1),[],1), repmat(vRef,size(dRef,1),1)); ttRef.Var2=zeros(height(ttRef),1); % add Var2 column of zeros [~,ib]=ismember(tt(:,{'Var1'}),ttRef(:,{'Var1'})); % locate the ones that are present in reference ttRef.Var2(ib)=tt.Var2; % and replace with start The above for the example leads to a fully-augmented timetable of: ttRef = 12×2 timetable Time Var1 Var2 ___________ ____ ____ 01-Jan-2020 a 2 01-Jan-2020 b 0 01-Jan-2020 c 1 02-Jan-2020 a 3 02-Jan-2020 b 2 02-Jan-2020 c 0 03-Jan-2020 a 2 03-Jan-2020 b 0 03-Jan-2020 c 0 04-Jan-2020 a 0 04-Jan-2020 b 0 04-Jan-2020 c 1 K>> for which the previous rowfun solution will work dpb on 15 May 2021 NB: The above construction left out the insertion of the duplicated initial rows...remember to reinsert them as well. dpb on 14 May 2021 K>> rowfun(@sum,t,"InputVariables",'Var2','GroupingVariables',{'Time','Var1'}) ans = 6×3 timetable Time Var1 GroupCount Var3 ___________ ____ __________ ____ 01-Jan-2020 "a" 1.00 2.00 01-Jan-2020 "c" 1.00 1.00 02-Jan-2020 "a" 1.00 3.00 02-Jan-2020 "b" 1.00 2.00 03-Jan-2020 "a" 2.00 3.00 04-Jan-2020 "c" 1.00 1.00 K>> I'd just fill in the missing dates using ismember or setdiff to create the vector of missing dates as compared to a full vector from the first to last date in the timetable. That's still a set of vector operations on the resulting table. Is there to be a "0" entry for each value of Var1, too, I suppose? I didn't try to write it, but seems you could use retime and custom function to fill the missing entries in the original timetable with zeros for each missing date/category; perhaps in conjunction with rowfun to using the grouping variable to cover the categories. All of these, of course, have the looping construct in them, just at a lower level... ### Categories Find more on Calendar in Help Center and File Exchange R2021a ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!

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# RD Sharma solutions for Class 12 Maths chapter 29 - The Plane [Latest edition] ## Chapter 29: The Plane Exercise 29.1Exercise 29.2Exercise 29.3Exercise 29.4Exercise 29.5Exercise 29.6Exercise 29.7Exercise 29.8Exercise 29.9Exercise 29.11Exercise 29.12Exercise 29.13Exercise 29.14Exercise 29.15Very Short AnswersMCQ Exercise 29.1 [Pages 4 - 5] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.1 [Pages 4 - 5] Exercise 29.1 | Q 1.1 | Page 4 Find the equation of the plane passing through the following points. (2, 1, 0), (3, −2, −2) and (3, 1, 7) Exercise 29.1 | Q 1.2 | Page 4 Find the equation of the plane passing through the following points. (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6) Exercise 29.1 | Q 1.3 | Page 4 Find the equation of the plane passing through the following point (1, 1, 1), (1, −1, 2) and (−2, −2, 2) Exercise 29.1 | Q 1.4 | Page 4 Find the equation of the plane passing through the following points. (2, 3, 4), (−3, 5, 1) and (4, −1, 2) Exercise 29.1 | Q 1.5 | Page 4 Find the equation of the plane passing through the following point (0, −1, 0), (3, 3, 0) and (1, 1, 1) Exercise 29.1 | Q 2 | Page 5 Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane. Exercise 29.1 | Q 3.1 | Page 5 Show that the following points are coplanar. (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0) Exercise 29.1 | Q 3.2 | Page 5 Show that the following points are coplanar. (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1) Exercise 29.1 | Q 4 | Page 5 Find the coordinates of the point where the line through (3, -4 , -5 ) and B (2, -3 , 1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4, -1,0 ) . Also, find the ratio in which diveides the line segment AB. Exercise 29.2 [Page 7] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.2 [Page 7] Exercise 29.2 | Q 1 | Page 7 Write the equation of the plane whose intercepts on the coordinate axes are 2, −3 and 4. Exercise 29.2 | Q 2.1 | Page 7 Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes. 4x + 3y − 6z − 12 = 0 Exercise 29.2 | Q 2.2 | Page 7 Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes. 2x + 3y − z = 6 Exercise 29.2 | Q 2.3 | Page 7 Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes. 2x − y + z = 5 Exercise 29.2 | Q 3 | Page 7 Find the equation of a plane which meets the axes at AB and C, given that the centroid of the triangle ABC is the point (α, β, γ). Exercise 29.2 | Q 4 | Page 7 Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes. Exercise 29.2 | Q 5 | Page 7 A plane meets the coordinate axes at AB and C, respectively, such that the centroid of triangle ABC is (1, −2, 3). Find the equation of the plane. Exercise 29.3 [Pages 13 - 14] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.3 [Pages 13 - 14] Exercise 29.3 | Q 1 | Page 13 Find the vector equation of a plane passing through a point with position vector $2 \hat{i} - \hat{j} + \hat{k}$ and perpendicular to the vector  $4 \hat{i} + 2 \hat{j} - 3 \hat{k} .$ Exercise 29.3 | Q 2.1 | Page 13 Find the Cartesian form of the equation of a plane whose vector equation is $\vec{r} \cdot \left( 12 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + 5 = 0$ Exercise 29.3 | Q 2.2 | Page 13 Find the Cartesian form of the equation of a plane whose vector equation is $\vec{r} \cdot \left( - \hat{i} + \hat{j} + 2 \hat{k} \right) = 9$ Exercise 29.3 | Q 3 | Page 13 Find the vector equations of the coordinate planes. Exercise 29.3 | Q 4.1 | Page 13 Find the vector equation of each one of following planes. 2x − y + 2z = 8 Exercise 29.3 | Q 4.2 | Page 13 Find the vector equation of each one of following planes. x + y − z = 5 Exercise 29.3 | Q 4.3 | Page 13 Find the vector equation of each one of following planes. x + y = 3 Exercise 29.3 | Q 5 | Page 13 Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1). Exercise 29.3 | Q 6 | Page 13 $\vec{n}$ is a vector of magnitude $\sqrt{3}$ and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to $\vec{n}$ . Exercise 29.3 | Q 7 | Page 13 The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane. Exercise 29.3 | Q 8 | Page 13 Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2. Exercise 29.3 | Q 9 | Page 13 If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP. Exercise 29.3 | Q 10 | Page 13 Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane. Exercise 29.3 | Q 11 | Page 13 A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point $\text{ 3 } \hat{i} + \hat{j} - \hat{k} .$ Find the vector and Cartesian forms of the equation of the plane. Exercise 29.3 | Q 12 | Page 13 Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it. Exercise 29.3 | Q 13.1 | Page 13 Show that the normals to the following pairs of planes are perpendicular to each other. x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0 Exercise 29.3 | Q 13.2 | Page 13 Show that the normals to the following pairs of planes are perpendicular to each other. $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 3 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 2 \hat{i} - 2 \hat{j} - 2 \hat{k} \right) = 5$ Exercise 29.3 | Q 14 | Page 13 Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes. Exercise 29.3 | Q 15 | Page 14 Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1. Exercise 29.3 | Q 16 | Page 14 If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane. Exercise 29.3 | Q 17 | Page 14 Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles. Exercise 29.3 | Q 18 | Page 14 Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1. Exercise 29.3 | Q 19 | Page 14 If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP. Exercise 29.3 | Q 20 | Page 14 Find the vector equation of the plane with intercepts 3, –4 and 2 on xy and z-axis respectively. Exercise 29.3 | Q 21 | Page 14 Find the vector equation of the plane with intercepts 3, –4 and 2 on xy and z-axis respectively. Exercise 29.4 [Page 19] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.4 [Page 19] Exercise 29.4 | Q 1 | Page 19 Find the vector equation of a plane which is at a distance of 3 units from the origin and has $\hat{k}$ as the unit vector normal to it. Exercise 29.4 | Q 2 | Page 19 Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector  $\hat{i} - \text{2 } \hat{j} - \text{2 } \hat{k} .$ Exercise 29.4 | Q 3 | Page 19 Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. Exercise 29.4 | Q 4 | Page 19 Reduce the equation $\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) + 6 = 0$ to normal form and, hence, find the length of the perpendicular from the origin to the plane. Exercise 29.4 | Q 5 | Page 19 Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0. Exercise 29.4 | Q 6 | Page 19 The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. Exercise 29.4 | Q 7 | Page 19 Find a unit normal vector to the plane x + 2y + 3z − 6 = 0. Exercise 29.4 | Q 8 | Page 19 Find the equation of a plane which is at a distance of $3\sqrt{3}$  units from the origin and the normal to which is equally inclined to the coordinate axes. Exercise 29.4 | Q 9 | Page 19 find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane Exercise 29.4 | Q 10 | Page 19 Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is  $2 \hat{i} - 3 \hat{j} + 4 \hat{k} .$ Also, find its Cartesian form. Exercise 29.4 | Q 11 | Page 19 Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin. Exercise 29.5 [Pages 22 - 23] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.5 [Pages 22 - 23] Exercise 29.5 | Q 1 | Page 22 Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5). Exercise 29.5 | Q 2 | Page 23 Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3). Exercise 29.5 | Q 3 | Page 23 Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .$ Exercise 29.5 | Q 4 | Page 23 Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3). Exercise 29.5 | Q 5 | Page 23 Find the vector equation of the plane passing through the points $3 \hat{i} + 4 \hat{j} + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k} \text{ and } 7 \hat{i} + 6 \hat{k} .$ Exercise 29.6 [Page 29] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.6 [Page 29] Exercise 29.6 | Q 1.1 | Page 29 Find the angle between the given planes. $\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) = 1 \text{ and } \vec{r} \cdot \left( - \hat{i} + \hat{j} \right) = 4$ Exercise 29.6 | Q 1.2 | Page 29 Find the angle between the given planes. $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 6 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + 6 \hat{j} - 2 \hat{k} \right) = 9$ Exercise 29.6 | Q 1.3 | Page 29 Find the angle between the given planes. $\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) = 9$ Exercise 29.6 | Q 2.1 | Page 29 Find the angle between the planes. 2x − y + z = 4 and x + y + 2z = 3 Exercise 29.6 | Q 2.2 | Page 29 Find the angle between the planes. x + y − 2z = 3 and 2x − 2y + z = 5 Exercise 29.6 | Q 2.3 | Page 29 Find the angle between the planes. x − y + z = 5 and x + 2y + z = 9 Exercise 29.6 | Q 2.4 | Page 29 Find the angle between the planes. 2x − 3y + 4z = 1 and − x + y = 4 Exercise 29.6 | Q 2.5 | Page 29 Find the angle between the planes. 2x + y − 2z = 5 and 3x − 6y − 2z = 7 Exercise 29.6 | Q 3.1 | Page 29 Show that the following planes are at right angles. $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( - \hat{i} - \hat{j} + \hat{k} \right) = 3$ Exercise 29.6 | Q 3.2 | Page 29 Show that the following planes are at right angles. x − 2y + 4z = 10 and 18x + 17y + 4z = 49 Exercise 29.6 | Q 4.1 | Page 29 Determine the value of λ for which the following planes are perpendicular to each other. $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 7 \text{ and } \vec{r} \cdot \left( \lambda \hat{i} + 2 \hat{j} - 7 \hat{k} \right) = 26$ Exercise 29.6 | Q 4.2 | Page 29 Determine the value of λ for which the following planes are perpendicular to each ot 2x − 4y + 3z = 5 and x + 2y + λz = 5 Exercise 29.6 | Q 4.3 | Page 29 Determine the value of λ for which the following planes are perpendicular to each other. 3x − 6y − 2z = 7 and 2x + y − λz = 5 Exercise 29.6 | Q 5 | Page 29 Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5. Exercise 29.6 | Q 6 | Page 29 Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8. Exercise 29.6 | Q 7 | Page 29 Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5. Exercise 29.6 | Q 8 | Page 29 Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9. Exercise 29.6 | Q 9 | Page 29 Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1. Exercise 29.6 | Q 10 | Page 29 Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5. Exercise 29.6 | Q 11 | Page 29 Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane. Exercise 29.6 | Q 12 | Page 29 Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8. Exercise 29.6 | Q 13 | Page 29 Find the equation of the plane passing through (abc) and parallel to the plane  $\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .$ Exercise 29.6 | Q 14 | Page 29 Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. Exercise 29.6 | Q 15 | Page 29 Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10 Exercise 29.7 [Page 33] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.7 [Page 33] Exercise 29.7 | Q 1.1 | Page 33 Find the vector equations of the following planes in scalar product form  $\left( \vec{r} \cdot \vec{n} = d \right):$ $\vec{r} = \left( 2 \hat{i} - \hat{k} \right) + \lambda \hat{i} + \mu\left( \hat{i} - 2 \hat{j} - \hat{k} \right)$ Exercise 29.7 | Q 1.2 | Page 33 Find the vector equations of the following planes in scalar product form  $\left( \vec{r} \cdot \vec{n} = d \right):$ $\vec{r} = \left( 1 + s - t \right) \hat{t} + \left( 2 - s \right) \hat{j} + \left( 3 - 2s + 2t \right) \hat{k}$ Exercise 29.7 | Q 1.3 | Page 33 Find the vector equations of the following planes in scalar product form  $\left( \vec{r} \cdot \vec{n} = d \right):$$\vec{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - 2 \hat{k} \right)$ Exercise 29.7 | Q 1.4 | Page 33 Find the vector equations of the following planes in scalar product form  $\left( \vec{r} \cdot \vec{n} = d \right):$$\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \right)$ Exercise 29.7 | Q 2.1 | Page 33 Find the Cartesian forms of the equations of the following planes. $\vec{r} = \left( \hat{i} - \hat{j} \right) + s\left( - \hat{i} + \hat{j} + 2 \hat{k} \right) + t\left( \hat{i} + 2 \hat{j} + \hat{k} \right)$ Exercise 29.7 | Q 2.2 | Page 33 Find the Cartesian forms of the equations of the following planes. $\vec{r} = \left( 1 + s + t \right) \hat{i} + \left( 2 - s + t \right) \hat{i} + \left( 3 - 2s + 2t \right) \hat{k}$ Exercise 29.7 | Q 3.1 | Page 33 Find the vector equation of the following planes in non-parametric form. $\vec{r} = \left( \lambda - 2\mu \right) \hat{i} + \left( 3 - \mu \right) \hat{j} + \left( 2\lambda + \mu \right) \hat{k}$ Exercise 29.7 | Q 3.2 | Page 33 Find the vector equation of the following planes in non-parametric form. $\vec{r} = \left( 2 \hat{i} + 2 \hat{j} - \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \mu\left( 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right)$ Exercise 29.8 [Pages 39 - 40] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.8 [Pages 39 - 40] Exercise 29.8 | Q 1 | Page 39 Find the equation of the plane which is parallel to 2x − 3y + z = 0 and which passes through (1, −1, 2). Exercise 29.8 | Q 2 | Page 39 Find the equation of the plane through (3, 4, −1) which is parallel to the plane $\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + 2 = 0 .$ Exercise 29.8 | Q 3 | Page 39 Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3). Exercise 29.8 | Q 4 | Page 39 Find the equation of the plane through the point $2 \hat{i} + \hat{j} - \hat{k}$ and passing through the line of intersection of the planes $\vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) = 0 \text{ and } \vec{r} \cdot \left( \hat{j} + 2 \hat{k} \right) = 0 .$ Exercise 29.8 | Q 5 | Page 39 Find the equation of the plane passing through the line of intersection of the planes 2x − y = 0 and 3z − y = 0 and perpendicular to the plane 4x + 5y − 3z = 8 Exercise 29.8 | Q 6 | Page 39 Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 and which is perpendicular to the plane 5x + 3y − 6z+ 8 = 0. Exercise 29.8 | Q 7 | Page 39 Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 4 = 0 and x − y + z + 3 = 0 and passing through the origin. Exercise 29.8 | Q 8 | Page 39 Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2x − y + 3z − 1 = 0 and passing through (1, −2, 3). Exercise 29.8 | Q 9 | Page 39 Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + y − z + 5 = 0. Exercise 29.8 | Q 10 | Page 39 Find the equation of the plane through the line of intersection of the planes  $\vec{r} \cdot \left( \hat{i} + 3 \hat{j} \right) + 6 = 0 \text{ and } \vec{r} \cdot \left( 3 \hat{i} - \hat{j} - 4 \hat{k} \right) = 0,$ which is at a unit distance from the origin. Exercise 29.8 | Q 11 | Page 39 Find the equation of the plane passing through the intersection of the planes 2x + 3y − z+ 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3x − y − 2z − 4 = 0. Exercise 29.8 | Q 12 | Page 39 Find the equation of the plane that contains the line of intersection of the planes  $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) - 4 = 0 \text{ and } \vec{r} \cdot \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + 5 = 0$ and which is perpendicular  to the plane $\vec{r} \cdot \left( 5 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) + 8 = 0 .$ Exercise 29.8 | Q 13 | Page 39 Find the equation of the plane passing through (abc) and parallel to the plane $\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .$ Exercise 29.8 | Q 14 | Page 39 Find the equation of the plane passing through the intersection of the planes  $\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 7, \vec{r} \cdot \left( 2 \hat{i} + 5 \hat{j} + 3 \hat{k} \right) = 9$ and the point (2, 1, 3). Exercise 29.8 | Q 15 | Page 40 Find the equation of the plane through the intersection of the planes 3x − y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1). Exercise 29.8 | Q 16 | Page 40 Find the vector equation of the plane through the line of intersection of the planes x + yz = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x − y + z = 0. Exercise 29.8 | Q 17 | Page 40 Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot \left( \hat{ i } + \hat{ j }+ \hat{ k }\right) = \text{ 6 and }\vec{r} \cdot \left( \text{ 2 } \hat{ i} +\text{ 3 } \hat{ j } + \text{ 4 } \hat{ k } \right) = - 5$ and the point (1, 1, 1). Exercise 29.8 | Q 18 | Page 40 Find the equation of the plane which contains the line of intersection of the planes x $+$  2y $+$  3 $z -$  4 $=$  0 and 2 $x + y - z$ $+$ 5  $=$ 0 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3,  $-$ 1) and parallel to the plane obtained above. Exercise 29.8 | Q 19 | Page 40 Find the equation of the plane through the line of intersection of the planes $x + y + z =$1 and 2x $+$ 3 $+$ y $+$ 4$z =$ 5 and twice of its $y$ -intercept is equal to three times its $z$-intercept Exercise 29.9 [Page 49] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.9 [Page 49] Exercise 29.9 | Q 1 | Page 49 Find the distance of the point  $2 \hat{i} - \hat{j} - 4 \hat{k}$  from the plane  $\vec{r} \cdot \left( 3 \hat{i} - 4 \hat{j} + 12 \hat{k} \right) - 9 = 0 .$ Exercise 29.9 | Q 2 | Page 49 Show that the points $\hat{i} - \hat{j} + 3 \hat{k} \text{ and } 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$ are equidistant from the plane $\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 7 \hat{k} \right) + 9 = 0 .$ Exercise 29.9 | Q 3 | Page 49 Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0. Exercise 29.9 | Q 4 | Page 49 Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1). Exercise 29.9 | Q 5 | Page 49 Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0. Exercise 29.9 | Q 6 | Page 49 Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1). Exercise 29.9 | Q 7 | Page 49 Find the distance of the point (2, 3, 5) from the xy - plane. Exercise 29.9 | Q 8 | Page 49 Find the distance of the point (3, 3, 3) from the plane $\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 7k \right) + 9 = 0$ Exercise 29.9 | Q 9 | Page 49 If the product of the distances of the point (1, 1, 1) from the origin and the plane x − y + z+ λ = 0 be 5, find the value of λ. Exercise 29.9 | Q 10 | Page 49 Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7. Exercise 29.9 | Q 11 | Page 49 Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C (5, 3, −3). Exercise 29.9 | Q 12 | Page 49 A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it. Exercise 29.9 | Q 13 | Page 49 Find the distance of the point (1, -2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0 Exercise 29.1 [Page 51] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.1 [Page 51] Exercise 29.1 | Q 1 | Page 51 Find the distance between the parallel planes 2x − y + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0. Exercise 29.1 | Q 2 | Page 51 Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes. Exercise 29.1 | Q 3 | Page 51 Find the equation of the plane mid-parallel to the planes 2x − 2y + z + 3 = 0 and 2x − 2y + z + 9 = 0. Exercise 29.1 | Q 4 | Page 51 Find the distance between the planes $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + 7 = 0 \text{ and } \vec{r} \cdot \left( 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \right) + 7 = 0 .$ Exercise 29.11 [Pages 61 - 62] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.11 [Pages 61 - 62] Exercise 29.11 | Q 1 | Page 61 Find the angle between the line $\vec{r} = \left( 2 \hat{i}+ 3 \hat {j} + 9 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)$  and the plane  $\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 5 .$ Exercise 29.11 | Q 2 | Page 61 Find the angle between the line $\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{1}$  and the plane 2x + y − z = 4. Exercise 29.11 | Q 3 | Page 61 Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3x − y + z = 1. Exercise 29.11 | Q 4 | Page 61 The line  $\vec{r} = \hat{i} + \lambda\left( 2 \hat{i} - m \hat{j} - 3 \hat{k} \right)$  is parallel to the plane  $\vec{r} \cdot \left( m \hat{i} + 3 \hat{j} + \hat{k} \right) = 4 .$ Find m Exercise 29.11 | Q 5 | Page 61 Show that the line whose vector equation is $\vec{r} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}+ \lambda\left( \hat{i} + 3 \hat{j} + 4 \hat{k} \right)$ is parallel to the plane whose vector  $\vec{r} \cdot \left( \hat{i} + \hat{j} - \hat{k} \right) = 7 .$  Also, find the distance between them. Exercise 29.11 | Q 6 | Page 61 Find the vector equation of the line through the origin which is perpendicular to the plane  $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 3 .$ Exercise 29.11 | Q 7 | Page 61 Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis. Exercise 29.11 | Q 8 | Page 61 Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7} .$ Exercise 29.11 | Q 9 | Page 61 Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes $\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + 2 \hat{k} \right) = 6$ Exercise 29.11 | Q 10 | Page 61 Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0. Exercise 29.11 | Q 11 | Page 61 Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0. Exercise 29.11 | Q 12 | Page 61 Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6. Exercise 29.11 | Q 13 | Page 61 Find the angle between the line $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2}$ and the plane 3x + 4y + z + 5 = 0. Exercise 29.11 | Q 14 | Page 61 Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane Exercise 29.11 | Q 15 | Page 61 State when the line $\vec{r} = \vec{a} + \lambda \vec{b}$  is parallel to the plane  $\vec{r} \cdot \vec{n} = d .$Show that the line  $\vec{r} = \hat{i} + \hat{j} + \lambda\left( 3 \hat{i} - \hat{j} + 2 \hat{k} \right)$  is parallel to the plane  $\vec{r} \cdot \left( 2 \hat{j} + \hat{k} \right) = 3 .$   Also, find the distance between the line and the plane. Exercise 29.11 | Q 16 | Page 61 Show that the plane whose vector equation is $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 1$ and the line whose vector equation is  $\vec{r} = \left( - \hat{i} + \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right)$   are parallel. Also, find the distance between them. Exercise 29.11 | Q 17 | Page 61 Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z. Exercise 29.11 | Q 18 | Page 62 Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines $\vec{r} = \left( \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)$ and $\vec{r} = \left( \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)$ Also, find the distance of the point (9, −8, −10) from the plane thus obtained. Exercise 29.11 | Q 19 | Page 62 Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line $\frac{x + 3}{2} = \frac{y - 3}{7} = \frac{z - 2}{5} .$ Exercise 29.11 | Q 20 | Page 62 Find the coordinates of the point where the line  $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}$   intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane. Exercise 29.11 | Q 21 | Page 62 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - 5 \hat{k} \right) + 9 = 0 .$ Exercise 29.11 | Q 22 | Page 62 Find the angle between the line $\frac{x + 1}{2} = \frac{y}{3} = \frac{z - 3}{6}$  and the plane 10x + 2y − 11z = 3. Exercise 29.11 | Q 23 | Page 62 Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes  $\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + \hat{k} \right) = 6 .$ Exercise 29.11 | Q 24 | Page 62 Find the value of λ such that the line $\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}$  is perpendicular to the plane 3x − y − 2z = 7. Exercise 29.11 | Q 25 | Page 62 Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line $\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}$ Exercise 29.12 [Page 65] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.12 [Page 65] Exercise 29.12 | Q 1.1 | Page 65 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the   yz - plane . Exercise 29.12 | Q 1.2 | Page 65 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the  zx - plane . Exercise 29.12 | Q 2 | Page 65 Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7. Exercise 29.12 | Q 3 | Page 65 Find the distance of the point (−1, −5, −10) from the point of intersection of the line $\vec{r} = \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i}+ 4 \hat{j} + 2 \hat{k} \right)$ and the plane  $\vec{r} . \left( \hat{i} - \hat{j} + \hat{k} \right) = 5 .$ Exercise 29.12 | Q 4 | Page 65 Find the distance of the point (2, 12, 5) from the point of intersection of the line $\vec{r} = 2 \hat{i} - 4 \hat{j}+ 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right)$ and $\vec{r} . \left( \hat{i} - 2 \hat{j} + \hat{k} \right) = 0$ Exercise 29.12 | Q 5 | Page 65 Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane  $x - y + z = 5$ . Exercise 29.12 | Q 6 | Page 65 Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7. Exercise 29.12 | Q 7 | Page 65 Find the distance of the point (1, -5, 9) from the plane $x - y + z =$ 5  measured along the line $x = y = z$  . Exercise 29.13 [Pages 73 - 74] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.13 [Pages 73 - 74] Exercise 29.13 | Q 1 | Page 73 Show that the lines $\vec{r} = \left( 2 \hat{j} - 3 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) \text{ and } \vec{r} = \left( 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)$  are coplanar. Also, find the equation of the plane containing them. Exercise 29.13 | Q 2 | Page 74 Show that the lines $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} \text{ and }\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}$  are coplanar. Also, find the equation of the plane containing them. Exercise 29.13 | Q 3 | Page 74 Find the equation of the plane containing the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$  and the point (0, 7, −7) and show that the line  $\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}$ also lies in the same plane. Exercise 29.13 | Q 4 | Page 74 Find the equation of the plane which contains two parallel lines$\frac{x - 4}{1} = \frac{y - 3}{- 4} = \frac{z - 2}{5}\text{ and }\frac{x - 3}{1} = \frac{y + 2}{- 4} = \frac{z}{5} .$ Exercise 29.13 | Q 5 | Page 74 Show that the lines  $\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}$ and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection. Exercise 29.13 | Q 6 | Page 74 Show that the plane whose vector equation is $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 3$ contains the line whose vector equation is $\vec{r} = \hat{i} + \hat{j} + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right) .$ Exercise 29.13 | Q 7 | Page 74 Find the equation of the plane determined by the intersection of the lines $\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} \text{ and }\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2}$ Exercise 29.13 | Q 8 | Page 74 Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line $\vec{r} = \hat{i} + 3 \hat{j} - 2 \hat{k} + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) .$ Exercise 29.13 | Q 9 | Page 74 If the lines  $\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}$ are perpendicular, find the value of and, hence, find the equation of the plane containing these lines. Exercise 29.13 | Q 10 | Page 74 Find the coordinates of the point where the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}$ intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane. Exercise 29.13 | Q 11 | Page 74 Find the vector equation of the plane passing through three points with position vectors  $\hat{i} + \hat{j} - 2 \hat{k} , 2 \hat{i} - \hat{j} + \hat{k} \text{ and } \hat{i} + 2 \hat{j} + \hat{k} .$  Also, find the coordinates of the point of intersection of this plane and the line  $\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + \lambda\left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) .$ Exercise 29.13 | Q 12 | Page 74 Show that the lines  $\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}$ and  $\frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}$ are coplanar. Exercise 29.13 | Q 13 | Page 74 Find the equation of a plane which passes through the point (3, 2, 0) and contains the line  $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ . Exercise 29.13 | Q 14 | Page 74 Show that the lines  $\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}$ and  $\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}$  are coplanar. Hence, find the equation of the plane containing these lines. Exercise 29.13 | Q 15 | Page 74 If the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}$  lies in the plane  $lx + my - z =$   then find the value of  $l^2 + m^2$ . Exercise 29.13 | Q 16 | Page 74 Find the values of  $\lambda$ for which the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}$and  $\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}$  are coplanar . Exercise 29.13 | Q 17 | Page 74 If the lines  $x =$  5 ,  $\frac{y}{3 - \alpha} = \frac{z}{- 2}$ and   $x = \alpha$ $\frac{y}{- 1} = \frac{z}{2 - \alpha}$ are coplanar, find the values of  $\alpha$. Exercise 29.13 | Q 18 | Page 74 If the straight lines  $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, find the equations of the planes containing them. Exercise 29.14 [Page 77] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.14 [Page 77] Exercise 29.14 | Q 1 | Page 77 Find the shortest distance between the lines $\frac{x - 2}{- 1} = \frac{y - 5}{2} = \frac{z - 0}{3} \text{ and } \frac{x - 0}{2} = \frac{y + 1}{- 1} = \frac{z - 1}{2} .$ Exercise 29.14 | Q 2 | Page 77 Find the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} .$ Exercise 29.14 | Q 3 | Page 77 Find the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{1}$ and $3x - y - 2z + 4 = 0 = 2x + y + z + 1$ Exercise 29.15 [Pages 81 - 82] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.15 [Pages 81 - 82] Exercise 29.15 | Q 1 | Page 81 Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0. Exercise 29.15 | Q 2 | Page 81 Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5. Exercise 29.15 | Q 3 | Page 81 Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .$ Hence, or otherwise, deduce the length of the perpendicular. Exercise 29.15 | Q 4 | Page 81 Find the image of the point with position vector $3 \hat{i} + \hat{j} + 2 \hat{k}$  in the plane  $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 4 .$  Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through $3 \hat{i} + \hat{j} + 2 \hat{k} .$ Exercise 29.15 | Q 5 | Page 81 Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular. Exercise 29.15 | Q 6 | Page 82 Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to  $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .$ Exercise 29.15 | Q 7 | Page 82 Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x − y − z = 7. Also, find the length of the perpendicular. Exercise 29.15 | Q 8 | Page 82 Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0. Exercise 29.15 | Q 9 | Page 82 Find the distance of the point with position vector $- \hat{i} - 5 \hat{j} - 10 \hat{k}$  from the point of intersection of the line $\vec{r} = \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i} + 4 \hat{j} + 12 \hat{k} \right)$  with the plane $\vec{r} \cdot \left( \hat{i} - \hat{j}+ \hat{k} \right) = 5 .$ Exercise 29.15 | Q 10 | Page 82 Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane $\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 4 \hat{k} \right) + 5 = 0 .$ Exercise 29.15 | Q 11 | Page 82 Find the coordinates of the foot of the perpendicular and the perpendicular distance of the  point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane. Exercise 29.15 | Q 12 | Page 82 Find the direction cosines of the unit vector perpendicular to the plane  $\vec{r} \cdot \left( 6 \hat{i} - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0$ passing through the origin. Exercise 29.15 | Q 13 | Page 82 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0. Exercise 29.15 | Q 14 | Page 82 Find the length and the foot of perpendicular from the point $\left( 1, \frac{3}{2}, 2 \right)$  to the plane $2x - 2y + 4z + 5 = 0$ . Exercise 29.15 | Q 15 | Page 82 Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ to the plane  $\vec{r} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) - 26 = 0$ Also find image of P in the plane. Very Short Answers [Pages 83 - 84] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Very Short Answers [Pages 83 - 84] Very Short Answers | Q 1 | Page 83 Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5). Very Short Answers | Q 2 | Page 83 Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0). Very Short Answers | Q 3 | Page 83 Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c). Very Short Answers | Q 4 | Page 83 Write the general equation of a plane parallel to X-axis. Very Short Answers | Q 5 | Page 83 Write the value of k for which the planes x − 2y + kz = 4 and 2x + 5y − z = 9 are perpendicular. Very Short Answers | Q 6 | Page 83 Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes. Very Short Answers | Q 7 | Page 83 Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2). Very Short Answers | Q 8 | Page 83 Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18. Very Short Answers | Q 9 | Page 83 Write the plane  $\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 14$  in normal form. Very Short Answers | Q 10 | Page 83 Write the distance of the plane  $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12$ from the origin. Very Short Answers | Q 11 | Page 83 Write the equation of the plane  $\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}$   in scalar product form. Very Short Answers | Q 12 | Page 83 Write a vector normal to the plane  $\vec{r} = l \vec{b} + m \vec{c} .$ Very Short Answers | Q 13 | Page 83 Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7. Very Short Answers | Q 14 | Page 83 Write the equation of the plane containing the lines $\vec{r} = \vec{a} + \lambda \vec{b} \text{ and } \vec{r} = \vec{a} + \mu \vec{c} .$ Very Short Answers | Q 15 | Page 83 Write the position vector of the point where the line $\vec{r} = \vec{a} + \lambda \vec{b}$ meets the plane  $\vec{r} . \vec{n} = 0 .$ Very Short Answers | Q 16 | Page 83 Write the value of k for which the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}$  is perpendicular to the normal to the plane  $\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 4 .$ Very Short Answers | Q 17 | Page 84 Write the angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{- 2}$  and the plane x + y + 4 = 0. Very Short Answers | Q 18 | Page 84 Write the intercept cut off by the plane 2x + y − z = 5 on x-axis. Very Short Answers | Q 19 | Page 84 Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0. Very Short Answers | Q 20 | Page 84 Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane $\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right) = 5 .$ Very Short Answers | Q 21 | Page 84 Find the vector equation of the plane, passing through the point (abc) and parallel to the plane $\vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2$ Very Short Answers | Q 22 | Page 84 Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ . Very Short Answers | Q 23 | Page 84 Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes. MCQ [Pages 84 - 86] ### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane MCQ [Pages 84 - 86] MCQ | Q 1 | Page 84 The plane 2x − (1 + λ) y + 3λz = 0 passes through the intersection of the planes • 2x − y = 0 and y − 3z = 0 • 2x + 3z = 0 and y = 0 • 2x − y + 3z = 0 and y − 3z = 0 • None of these MCQ | Q 2 | Page 84 The acute angle between the planes 2x − y + z = 6 and x + y + 2z = 3 is •  45° • 60° •  30° •  75° MCQ | Q 3 | Page 84 The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is • 7x − 2y + 3z + 81 = 0 • 23x + 14y − 9z + 48 = 0 •  51x − 15y − 50z + 173 = 0 •  None of these MCQ | Q 4 | Page 84 The distance between the planes 2x + 2y − z + 2 = 0 and 4x + 4y − 2z + 5 = 0 is • $\frac{1}{2}$ • $\frac{1}{4}$ •  $\frac{1}{6}$ • None of these MCQ | Q 5 | Page 84 The image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0 is •  (3, 5, 2) •  (−3, 5, 2) •  (3, 5, −2) • (3, −5, 2) MCQ | Q 6 | Page 85 The equation of the plane containing the two lines $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} \text{ and }\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}$ •  8x + y − 5z − 7 = 0 •  8x + y + 5z − 7 = 0 • 8x − y − 5z − 7 = 0 •  None of these MCQ | Q 7 | Page 85 The equation of the plane $\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right)$  in scalar product form is •   $\vec{r} \cdot \left( 5 \hat{i} - 2 \hat{j} - 3 \hat{k} \right) = 7$ •  $\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 3 \hat{k} \right) = 7$ •  $\vec{r} \cdot \left( 5 \hat{i} - 2 \hat{j} + 3 \hat{k} \right) = 7$ •  None of these MCQ | Q 8 | Page 85 The distance of the line $\vec{r} = 2 \hat{i} - 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - \hat{j}+ 4 \hat{k} \right)$  from the plane $\vec{r} \cdot \left( \hat{i} + 5 \hat{j} + \hat{k} \right) = 5$ is • $\frac{5}{3\sqrt{3}}$ • $\frac{10}{3\sqrt{3}}$ • $\frac{25}{3\sqrt{3}}$ •  None of these MCQ | Q 9 | Page 85 The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ is •  x − 5y + 3z = 7 • x − 5y + 3z = −7 •  x + 5y + 3z = 7 •  x + 5y + 3z = −7 MCQ | Q 10 | Page 85 The vector equation of the plane containing the line $\vec{r} = \left( - 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left( 3 \hat{i} - 2 \hat{j} - \hat{k} \right)$ and the point  $\hat{i} + 2 \hat{j} + 3 \hat{k}$  is • $\vec{r} \cdot \left( \hat{i} + 3 \hat{k} \right) = 10$ •  $\vec{r} \cdot \left( \hat{i} - 3 \hat{k} \right) = 10$ •  $\vec{r} \cdot \left( 3\hat{i} - \hat{k} \right) = 10$ • None of these MCQ | Q 11 | Page 85 A plane meets the coordinate axes at AB and C such that the centroid of ∆ABC is the point (abc). If the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = k,$ then k = •  1 •  2 •  3 •  None of these MCQ | Q 12 | Page 85 The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane x + y + z = 17 is •  1 • 2 •  3 • None of these MCQ | Q 13 | Page 85 A vector parallel to the line of intersection of the planes$\vec{r} \cdot \left( 3 \hat{i} - \hat{j} + \hat{k} \right) = 1 \text{ and } \vec{r} \cdot \left( \hat{i} + 4 \hat{j} - 2 \hat{k} \right) = 2$ is •  $- 2 \hat{i} + 7 \hat{j}+ 13 \hat{k}$ •   $2 \hat{i} + 7 \hat{j} - 13 \hat{k}$ •  $-2 \hat{i} + 7 \hat{j} + 13 \hat{k}$ •  $2 \hat{i} + 7 \hat{j} + 13 \hat{k}$ MCQ | Q 14 | Page 85 If a plane passes through the point (1, 1, 1) and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$ then its perpendicular distance from the origin is • 3/4 •  4/3 •  7/5 •  1 MCQ | Q 15 | Page 85 The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is •  x − 4y + 2z + 4 = 0 • x + 4y + 2z + 4 = 0 •  x − 4y + 2z − 4 = 0 • None of these MCQ | Q 16 | Page 86 The distance of the point (−1, −5, −10) from the point of intersection of the line $\vec{r} = 2 \hat{i}- \hat{j} + 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j}+ 12 \hat{k} \right)$   and the plane $\vec{r} \cdot \left( \hat{i} - \hat{j} + \hat{k} \right) = 5$ is • 9 •  13 •  17 •  None of these MCQ | Q 17 | Page 86 The equation of the plane through the intersection of the planes ax + by + cz + d = 0 andlx + my + nz + p = 0 and parallel to the line y=0, z=0 • (bl − amy + (cl − anz + dl − ap = 0 •  (am − blx + (mc − bnz + md − bp = 0 •  (na − clx + (bn − cmy + nd − cp = 0 • None of these MCQ | Q 18 | Page 86 The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is •  x + y + z = 1 •  x + y + z = 0 • x + y − z = 1 •  x + y + z = 2 ## Chapter 29: The Plane Exercise 29.1Exercise 29.2Exercise 29.3Exercise 29.4Exercise 29.5Exercise 29.6Exercise 29.7Exercise 29.8Exercise 29.9Exercise 29.11Exercise 29.12Exercise 29.13Exercise 29.14Exercise 29.15Very Short AnswersMCQ ## RD Sharma solutions for Class 12 Maths chapter 29 - The Plane RD Sharma solutions for Class 12 Maths chapter 29 (The Plane) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 12 Maths solutions in a manner that help students grasp basic concepts better and faster. Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students. Concepts covered in Class 12 Maths chapter 29 The Plane are Three - Dimensional Geometry Examples and Solutions, Introduction of Three Dimensional Geometry, Equation of a Plane Passing Through Three Non Collinear Points, Relation Between Direction Ratio and Direction Cosines, Intercept Form of the Equation of a Plane, Coplanarity of Two Lines, Distance of a Point from a Plane, Angle Between Line and a Plane, Angle Between Two Planes, Angle Between Two Lines, Vector and Cartesian Equation of a Plane, Equation of a Plane in Normal Form, Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point, Plane Passing Through the Intersection of Two Given Planes, Shortest Distance Between Two Lines, Equation of a Line in Space, Direction Cosines and Direction Ratios of a Line. Using RD Sharma Class 12 solutions The Plane exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 12 prefer RD Sharma Textbook Solutions to score more in exam. Get the free view of chapter 29 The Plane Class 12 extra questions for Class 12 Maths and can use Shaalaa.com to keep it handy for your exam preparation

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Morphology. Are you looking for angle? # An Angle An Angle is a tautogram (all words start with the same letter). View more tautograms! ## Group An Angle was an indie rock band from Sacramento, California. At the center of this group was singer-songwriter Kris Anaya. The band was signed to Drive-Thru Records. On Monday, July 20, 2008, Matt Sergent announced via the group's Myspace page that the members of An Angle had decided to go their separate ways. In April 2013, it was announced that An Angle would be playing a reunion show for Sacramento's Launch Festival Kick Off Party 2013. • genres: Indie rock, Rock music • albums: "...and Take It With a Grain of Salt", "We Can Breathe Under Alcohol", "5 Days 5 Songs", "The Truth Is That You Are Alive", "Songs to Shoot up to" • official website: www.ananglemusic.com ## Printed dictionaries and other books with definitions for An Angle Click on a title to look inside that book (if available): ### Rosie and Mrs. America (2008) Perceptions of Women in the 193s and 194s by Catherine Gourley An angle is a story's emphasis. It determines not only which details a reporter includes in a story but also how the story is told. “Put yourself in the place of another,” instructed Ethel Brazelton, who taught women students at the Medill School of ... ### Succ Learn Practise Math Lev 4 (2008) axis horizontal vertical Types «f angles An angle is a measure of turn. ### The Navigator's Assistant (1784) Containing the Theory and Practice of Navigation, with All the Tables Requisite for Determining a Ship's Place at Sea. By William Nicholson by William Nicholson An Angle is that which is formed by the indirect meeting of two lines ; and its quantity is measured by the mutual inclination of the lines to each other at the point of meeting. 10. A right lined Angle is that which is formed by two right lines. ### Polyhedra (1999) by Peter R. Cromwell An angle is a region of a plane contained between two intersecting lines so that segments of the two lines ... ### Dictionary of Natural History Terms with Their Derivations (1863) Including the Various Orders, Genera, and Species ymix, an angle, jrrip/f, a wing ; a genus of Lepi- doptera. Gonia 'ster (Zool.) ymix, an angle, kirnp, a star; a genus of Echinodermata. Gonibregma' tus (Ent.) ymix, an angle, (Spiyfxx, the fore part of the head. Goni'dium (Zool.) ... ### Dictionary of Military Terms (2009) by U.S. Department of Defense An angle measured clockwise in the horizontal plane between a reference direction and any other line. azimuth ... ## Online dictionaries and encyclopedias with entries for An Angle Click on a label to prioritize search results according to that topic: Click on an item to view that photo: ## Video language resources about An Angle Click on an item to play that video: To knock a thing down, especially if it is cocked at an arrogant angle, is a deep delight of the blood. (George Santayana) more quotes... ## Anagrams of ANANGLE What do you get if you rearrange the letters? ## Next... Go to the usage examples of An Angle to see it in context!

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 SQL exercises on employee Database: List the id, name, salary, and experiences of all the employees working for the manger 68319 - w3resource # SQL exercises on employee Database: List the id, name, salary, and experiences of all the employees working for the manger 68319 ## SQL employee Database: Exercise-28 with Solution [An editor is available at the bottom of the page to write and execute the scripts.] 28. From the following table, write a SQL query to find those employees working under the manger whose ID is 68319. Return employee ID, employee name, salary, and age. Sample table: employees Pictorial Presentation: Sample Solution: ``````SELECT emp_id, emp_name, salary, age(CURRENT_DATE, hire_date) "Experience" FROM employees WHERE manager_id=68319; `````` Sample Output: ``` emp_id | emp_name | salary | Experience --------+----------+---------+------------------------- 66928 | BLAZE | 2750.00 | 26 years 8 mons 29 days 67832 | CLARE | 2550.00 | 26 years 7 mons 21 days 65646 | JONAS | 2957.00 | 26 years 9 mons 28 days (3 rows) ``` ## Practice Online Sample Database: employee Have another way to solve this solution? Contribute your code (and comments) through Disqus. What is the difficulty level of this exercise? 

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# Flash Steam Written by Jerry Ratzlaff on . Posted in Thermodynamics Flash steam, abbreviated as FS, is when hot condensate is released from a high pressure to a lower pressure steam.  This steam is the same as normal steam, the name is just given to explain how the steam is formed. ### Flash Steam formula $$\large{ FS = \frac { h_i \; - \; h_o } { L } }$$ Where: $$\large{ FS }$$ = flash steam $$\large{ L }$$ = latent heat of saturated water at outlet $$\large{ h_i }$$ = specific enthalpy of saturated water at inlet $$\large{ h_o }$$ = specific enthalpy of saturated water at outlet Tags: Equations for Steam

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# Quick Answer: What Happens When You Put 2 Compasses Together? ## How does a compass work simple explanation? A compass works by detecting the Earth’s natural magnetic fields. This allows the needle to better react to nearby magnetic fields. Since opposites attract the southern pole of the needle is attracted to the Earth’s natural magnetic north pole. This is how navigators are able to discern north.. ## Where is the magnetic force the weakest? The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center. ## What happens when you put 2 magnets together? When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. … Unlike-poles attract: When a north pole and south pole point together, the arrows point in the SAME direction so the field lines can join up and the magnets pull together (attract). ## Does a compass always point north? While a compass is a great tool for navigation, it doesn’t always point exactly north. This is because the Earth’s magnetic North Pole is not the same as “true north,” or the Earth’s geographic North Pole . … As the Earth’s magnetic field changes, the magnetic North Pole moves. ## What will a compass do at the North Pole? See if you can find, with a compass, the angle the magnetic field makes with respect to the ground in your location. … At the north pole, for example, if you hold the compass horizontally the needle which is supposed to point north will point south, toward the north magnetic pole. ## How do you use a compass step by step? Pick up the compass and hold it flat in front of you. Be sure that the direction of travel arrow points straight ahead. Then, rotate yourself, keeping an eye on the magnetic needle. When the red end lines up exactly with the orienting arrow, stop. ## Does putting 2 magnets together make them stronger? By adding one magnet on to the other, e.g. stacking, the stacked magnets will work as one bigger magnet and will exert a greater magnetic performance. As more magnets are stacked together, the strength will increase until the length of the stack is equal to the diameter. ## What can interfere with a compass? Electronic equipment, such as VHF radios, vehicles, mobile phones, headphones and GPS units can affect the compass reading. It’s not just things you carry that can cause interference. Metal objects around you, such as fence posts, bridges, gates etc can all interfere with the compass. ## What happens if you put a compass on the North Pole? If the Earth’s magnetic field were a perfect dipole, the compass needle would float aimlessly. … Earth’s magnetic field lines would be vertical at the magnetic north pole if the magnetic north pole coincided exactly with the geomagnetic north pole. So a compass held horizontally there would have no preferred direction. ## Which planet has the strongest field the weakest? The planets from strongest magnetic field to weakest magnetic field is Jupiter > Earth > Venus. Magnetic fields are generated by the movement of magnetic material located in side the planet, usually at the core. ## Do compasses work in both hemispheres? A2A: Compasses actually point north AND south in either hemispheres. We, in the northern hemisphere, just focus on the red, north seeking end being closer to that pole. ## How do I know if my compass is accurate? You can check a map if you’re not sure. Then hold the compass, with the compass level and the arrow free to turn. If the arrowhead goes anywhere near the street you know is north, then the compass is accurate. Stand outside in a location where you have a pretty good idea of which direction is North. ## Can you fix a compass? However, a compass needle is a delicate magnetic instrument, and it is possible for the poles to become reversed if the compass is brought into close contact with another magnet. If this happens, you will need to remagnetize the compass using a strong magnet. ## Why does a compass always point to the North? When it comes to magnets, opposites attract. This fact means that the north end of a magnet in a compass is attracted to the south magnetic pole, which lies close to the geographic north pole. … The geographic north and south poles indicate the points where the earth’s rotation axis intercepts earth’s surface. ## Are 2 magnets stronger than 1? Two magnets together will be slightly less than twice as strong as one magnet. When magnets are stuck entirely together (the south pole of one magnet is connected to the north pole of the other magnet) you can add the magnetic fields together. ## What happens when you bring two compasses near each other? If two compasses are brought near enough to each other, the magnetic fields of the compasses themselves will be larger than the field of the earth, and the needles will line up with each other. … It consists of a bar magnet which have North and South Pole which gets affected by the magnetic field of the earth.

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# Error in explanation? Under “Working with lists in Python” in “The Power of Range” section ( https://www.codecademy.com/paths/analyze-data-with-python/tracks/ida-2-introduction-to-python/modules/ida-2-3-python-lists/lessons/use-python-list/exercises/the-power-of-range ), there seems to be an error in the Learn Section. On the last line, instead of 100 it should be 99 right? Nope. Your arguments that you passed through are: 1, 100, 10, or, the start (index), stop (the number you’re ending/stopping at), step (the increase). So, 91 + 10 = 101, which is greater than the stopping point of 100. 1 Like

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Dictionary: A   B   C   D   E   F   G   H   I   J   K   L   M   N   O   P   Q   R   S   T   U   V   W   X   Y   Z # In finite series noun, Mathematics. 1. a sequence of numbers in which an infinite number of terms are added successively in a given pattern; the sequence of partial sums of a given sequence. Tagged: • Infinite set mathematics A set with an infinite number of elements. There are several possible definitions, e.g. (i) (“Dedekind infinite”) A set X is infinite if there exists a bijection (one-to-one mapping) between X and some proper subset of X. (ii) A set X is infinite if there exists an injection from N (the set of natural […] • Infinitesimal [in-fin-i-tes-uh-muh l] /ˌɪn fɪn ɪˈtɛs ə məl/ adjective 1. indefinitely or exceedingly small; minute: infinitesimal vessels in the circulatory system. 2. immeasurably small; less than an assignable quantity: to an infinitesimal degree. 3. of, relating to, or involving infinitesimals. noun 4. an infinitesimal quantity. 5. Mathematics. a variable having zero as a limit. /ˌɪnfɪnɪˈtɛsɪməl/ adjective […] • Infinitesimal-calculus noun 1. the differential calculus and the integral calculus, considered together. noun 1. another name for calculus (sense 1) • Infinitesimally [in-fin-i-tes-uh-muh l] /ˌɪn fɪn ɪˈtɛs ə məl/ adjective 1. indefinitely or exceedingly small; minute: infinitesimal vessels in the circulatory system. 2. immeasurably small; less than an assignable quantity: to an infinitesimal degree. 3. of, relating to, or involving infinitesimals. noun 4. an infinitesimal quantity. 5. Mathematics. a variable having zero as a limit. /ˌɪnfɪnɪˈtɛsɪməl/ adjective […] Disclaimer: In finite series definition / meaning should not be considered complete, up to date, and is not intended to be used in place of a visit, consultation, or advice of a legal, medical, or any other professional. All content on this website is for informational purposes only.

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# SenseTime Ace Coder Challenge 暨 商汤在线编程挑战赛*(抽球游戏-fwt开立方) Anthony 手中有 $n$$n$ 个依次标着整数 ${a}_{1}$$a_1$, ${a}_{2}$$a_2$, $\cdots$$\cdots$, ${a}_{n}$$a_n$ ($0\le {a}_{1}\le {a}_{2}\le \cdots \le {a}_{n}\le 63$$0 \leq a_1 \leq a_2 \leq \cdots \leq a_n \leq 63$) 的球，现在 Ben 会等概率随机抽出一个球，记录球上的数字并放回，重复该过程三次，最终三次记录下的数字的异或和就是 Ben 的得分。 ### 输出格式 2 3 4 3 2 2 3 2 2 3 2 1 1 1 1 3 3 -1 -1 -1 -1 -1 -1 -1 -1 -1 Case #1: 22 Case #2: -7 fwt直接pow(1.0/3)就行。 #include<bits/stdc++.h> using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define ForkD(i,k,n) for(int i=n;i>=k;i--) #define Rep(i,n) for(int i=0;i<n;i++) #define ForD(i,n) for(int i=n;i;i--) #define RepD(i,n) for(int i=n;i>=0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (o<<1) #define Rson ((o<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,0x3f,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define MEMx(a,b) memset(a,b,sizeof(a)); #define INF (0x3f3f3f3f) #define F (1000000007) #define pb push_back #define mp make_pair #define fi first #define se second #define vi vector<int> #define pi pair<int,int> #define SI(a) ((a).size()) #define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans); #define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl; #define PRi2D(a,n,m) For(i,n) { \ For(j,m-1) cout<<a[i][j]<<' ';\ cout<<a[i][m]<<endl; \ } #define All(x) (x).begin(),(x).end() #define gmax(a,b) a=max(a,b); #define gmin(a,b) a=min(a,b); typedef long long ll; typedef long double ld; typedef unsigned long long ull; ll mul(ll a,ll b){return (a*b)%F;} ll sub(ll a,ll b){return ((a-b)%F+F)%F;} void upd(ll &a,ll b){a=(a%F+b%F)%F;} { int x=0,f=1; char ch=getchar(); while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();} while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();} return x*f; } bool fwt (ll a[] , int n ,bool on) { for ( int d = 1 ; d < n ; d <<= 1 ) { for ( int k = d << 1 , i = 0 ; i < n ; i += k ) { for ( int j = 0 ; j < d ; ++ j ) { ll x = a[i + j] , y = a[i + j + d] ; if(on){ //xor a[i + j] = ( x + y ) ; a[i + j + d] = ( x - y ); } else{ //xor if((x&1)!=(y&1)) return 0; a[i + j] = ( x + y ) /2 ; a[i + j + d] = ( x - y) /2 ; } } } } return 1; } #define MAXN (500) ll a[MAXN]; int main() { // freopen("C.in","r",stdin); // freopen(".out","w",stdout); For(kcase,T) { printf("Case #%d:",kcase); if(tot%(n*n*n)){ puts(" -1");continue; } fwt(a,64,1); bool fl=0; Rep(i,64) { double t=max(-a[i],a[i]); t=pow(t,1.0/3); t=floor(t+0.5); ll c=t; if(a[i]<0) c=-c; if(c*c*c!=a[i]){ fl=1; }else a[i]=c; } if(fl){ puts(" -1");continue; } fl=fwt(a,64,0); tot=0; Rep(i,64) tot+=a[i]; Rep(i,64) if(a[i]<0) fl=0; if(tot!=n||!fl) { puts(" -1");continue; } Rep(i,64) Rep(j,a[i]) printf(" %d",i);puts(""); } return 0; } • 广告 • 抄袭 • 版权 • 政治 • 色情 • 无意义 • 其他 120

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