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http://econsultancy.com/uk/blog/63572-could-chaos-theory-explain-patterns-of-influence-on-social-media

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Could chaos theory explain patterns of influence on social media? Since we started to work on mapping influence patterns, I have been wondering if we could find easy recognizable patterns in influence maps. If so, we could probably predict influence patterns and the secret of ROI optimisation would be eventually revealed to CMOs !!! Stimulating thought. The recent history of science showed that behind apparently unpredictable phenomena, patterns could in fact be identified. Further, similar patterns could be applied to domains as diverse as weather forecasting, traffic modelling or the evolution of populations: this is chaos theory. So, could chaos theory explain patterns of influence on social media and resolve one of the biggest social media marketing enigmas for brands? I wanted to dig into this. I however quickly had to face my own limitations: I know nothing really on the sophisticated mathematics behind the chaos theory. So, as a first step, I decided to study graphs and maps and look for similarities that could confirm my intuition. Now, honestly, if you know anything about math or physics, just ignore this post. It’s most probably embarrassing. However, if you don’t care so much about maths, but are interested in influence, stay tuned: you might be interested in what follows. . Graphic representations of simple nonlinear formulas could produce such beautiful designs as the one above called “Mandelbrot set”.  Could an influence map look similar? Could we build a similarly simple theory of influence based on simple structures and relationships that could be modelled? My first step consisted in constructing a simple influence theory based on these two principles: 1. Influence is hierarchical : Social scoring companies such as Klout or Kred build a hierarchy among social media users: people with higher scores are deemed more influential than people with lower scores. While I believe that these scores are useless because they are not contextualized, we do produce lists of individuals that we rank from 1 to 100 in specific, contextual topics. 2. Influence is an iterative process where sharing and engaging being the base mechanism for growing social influence. Theories such as Malcom Gladwell’s “tipping point” suggest that viral content gets amplified from individual to another creating a series of iterations and reaching out to the masses. That meant we could build a map of influence starting from our top influencers and building hierarchical relationships, representing engagement, to their “next level” influencers and so on…. Wow! Can we have a look at this please? This is what I ended up with: Could this be the shape of influence ? Could this be a realistic representation of the impact of individuals on their community down from the top influencer to smaller influencers ? I decided to compare this to our 'real' influence maps, which aim to graphically represent top influencers on a topic, their closest contacts and the relationships between them based on real engagement data on Twitter: pretty cool stuff really! Here is a sample of what they look like: I looked across many maps but I never found any that really looked like my ideal influence map. My intuition said communities maps could be categorized in different types but none on these looked like my theoretical map of influence : I obviously got something wrong in my hypothesis. Looking again through many Traackr maps, this is what I actually learned about the reality of social influence: Influence goes in circles It’s not a one-way path from the center to the periphery. No one single person is the ultimate source of influence or content on a specific topic. Top influencers get influenced back by their community. Influence is not soley proportional to social performance or popularity Some key individuals with a low first-degree audience can have a high impact on a community because they have very strong relationships that help the community work better. Social influence works imperfectly. It often fails. Information or content can die out or be ignore amid the flow of information. You can have silos or subgroups who do not get exposed to specific contents. Takeaways for marketers No algorithm or big data miracle is likely to soon resolve the complexity of influence. After all, it is about gaining insights into your community and focusing on the right individuals for you. It’s really about creating the right relationships and authentically interacting with individuals. Will social influence one day be explained by chaos theory? Maybe… but this is still up for demonstration. Nicolas Chabot is VP EMEA at TRAACKR and a guest blogger on Econsultancy. You can connect with Nicolas on LinkedIn, Twitter or Google Plus Wikimedia Commons 1. Jason Thibeault Sr. Director, Marketing Strategy at Limelight Networks 1:27PM on 30th October 2013 Interesting. If you really want to dig into this, you should check out "small world" theory (i.e., 6-degrees of separation). There is a lot of math to explain connectedness and influence (obviously people have been studying influence for a very long time; you might want to check out Neil Dunbar for example) although it is not "chaos theory" (actually what you probably mean is "complexity theory"). Influence and connectedness are governed by very definable mathematical patterns. But you do reach the right conclusion: engaging with your audience is all about relationships. And what marketers need to discover is that sometimes the "outliers" (those who are quiet) can end of being the most influential. 2. Nicolas Chabot VP EMEA at TRAACKR 2:32PM on 30th October 2013 Thanks Jason. I will check it out. Actually the idea of writing this post came up as I was ready "Chaos" by James Gleick : a great read- although I did not quite understand all of it ;) 3. Jason Thibeault Sr. Director, Marketing Strategy at Limelight Networks 2:35PM on 30th October 2013 Yes, that's a good book by Gleick but it's a little bit pseudo-science/math. Chaos theory (and chaos mathematics) really describe the impact of small influencers on large systems. Complexity theory deals with the identification (and organization) of large systems into patterns. Still a solid post. I've got a book coming out in February which provides a great framework for measuring the value of relationships (similar to the exercise you went through but organizing it more discretely and putting math to it). I'll be sharing the equation soon via eConsultancy. 4. Nicolas Chabot VP EMEA at TRAACKR 3:33PM on 30th October 2013 Sounds exciting ! I am looking forward to reading more about it.

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http://biodieselholding.com/tabletki-prostatita/1118-what-are-greek-options.php

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# What are greek options Practical use[ edit ] For a vanilla option, delta will be a number between 0. The difference between the delta of a call and the delta of a put at the same strike is equal to one. See the formulas below. These numbers are commonly presented as a percentage of the total number of shares represented by the option contract s. This is convenient because the option will instantaneously behave like the number of shares indicated by the delta. For example, if a portfolio of American call options on XYZ each have a delta of 0. The sign and percentage are often dropped — the sign is implicit in the option type negative for put, positive for call and the percentage is understood. Delta is always positive for long calls and negative for long puts unless they are zero. The total delta of a complex portfolio of positions on the same underlying asset can be calculated by simply taking the sum of the deltas for each individual position — delta of a portfolio is linear in the constituents. Since the delta of underlying asset is always 1. This portfolio will then retain its total value regardless of which direction the price of XYZ moves. Albeit for only small movements of the underlying, a short amount of time and not-withstanding changes in other market conditions such as volatility and the rate of return for a risk-free investment. As reviews who really earned on binary options proxy for probability[ edit ] Main article: Moneyness The absolute value of Delta is close to, but not identical with, the percent moneyness of an option, i. For example, if an out-of-the-money call option has a delta of 0. At-the-money calls and puts have a delta of approximately 0. The actual probability of an option finishing in the money is its dual deltawhich is the first derivative of option price with respect to strike. This is due to put—call parity what are greek options a long call plus a short put a call minus a put replicates a forward, which has delta equal to 1. • Meet the Greeks At least the four most important ones NOTE: The Greeks represent the consensus of the marketplace as to how the option will react to changes in certain variables associated with the pricing of an option contract. • Using the "Greeks" to Understand Options • Option Greeks | Delta | Gamma | Theta | Vega | Rho - The Options Playbook • These Options Terms Are Greek to You - WSJ If the value of delta for an option is known, one can calculate the value of the delta of the option of the same strike price, underlying and maturity but opposite right by subtracting 1 from a known call delta or adding 1 to a known put delta.

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https://www.yobankexams.com/2011/10/shortcut-to-square-any-number-from-30.html

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Shortcut to square any number from 30 to 79 Shortcut to square any number from 30 to 79 mentally in seconds is now possible for all aspiring candidates writing different competitive exams. It is always better to have some math tricks handy when you are planning to take any arithmetic aptitude test either in the competitive exams such as Bank exams,CAT,MAT. Another easy trick to find square of the numbers between 30 to 79 is to take a common base as 50 and see how far the number is from 50.This method works well when the number is very close to 50. Shortcut to square of number from 30 to 79 steps Step 1:Find how many more or less the given number is from 50. Step 2:Add the number to 25 if more than 50 or subtract the number from 25 if less than 50. Step 3:Then find the square of the number added or subtracted and put next to the result arrived at in step 2. Let us now apply the trick that we learnt in the example below Examples Example 1:Find the square of 52 in 5 seconds 52²=? Step 1: We see that the given number is 2 more than 50. 52-50=2 Step 2: Here we notice that the given number is more than 50 so we add 25 as follows 25+2=27 Step 3: Now we find the square of  the number added(2) 22=4 putting the result obtained in step 3 next to the result obtained in step 2 after adding a 0 before it as it is a single digit, we get, 2704 Ans: 52²=2704 Example 2:Find the square of 37 in 5 seconds. 37²=? Step 1: We see that the given number is 13 less than 50 that is 50-37=13 Step 2: Here we notice that the given number is less than 50 so we subtract 25 as follows 25-13=12 Step 3: Now we find the square of  the number subtracted(13) we get, 132=169 put the result obtained in step 3 next to the result obtained in step 2 as shown below. And carry forward hundreth place digit to step 2. we get,1269 Step 4:Add carry forward from step 3 to result obtained in step 2. we get, 12+1=13 Ans:       37²=1369 Example 3:Find the square of 54 in 5 seconds 54²=? Step 1: We see that the given number is 4 more than 50. 54-50=4 Step 2: Here we notice that the given number is more than 50 so we add 25 as follows 25+4=29 Step 3: Now we find the square of  the number added(4) 42=16 putting the result obtained in step 3 next to the result obtained in step 2 we get, 2916 Ans: 54²=2916 See how quickly you can square 54,49,53,51,48,32,57,59,64,67,72.Give it a try…. 0 thoughts on “Shortcut to square any number from 30 to 79” 1. it is very helpful …. thanks

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Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  elini Structured version   Visualization version   GIF version Theorem elini 3759 Description: Membership in an intersection of two classes. (Contributed by Glauco Siliprandi, 17-Aug-2020.) Hypotheses Ref Expression elini.1 𝐴𝐵 elini.2 𝐴𝐶 Assertion Ref Expression elini 𝐴 ∈ (𝐵𝐶) Proof of Theorem elini StepHypRef Expression 1 elini.1 . . 3 𝐴𝐵 2 elini.2 . . 3 𝐴𝐶 31, 2pm3.2i 470 . 2 (𝐴𝐵𝐴𝐶) 4 elin 3758 . 2 (𝐴 ∈ (𝐵𝐶) ↔ (𝐴𝐵𝐴𝐶)) 53, 4mpbir 220 1 𝐴 ∈ (𝐵𝐶) Colors of variables: wff setvar class Syntax hints:   ∧ wa 383   ∈ wcel 1977   ∩ cin 3539 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-in 3547 This theorem is referenced by:  recvs  22754  qcvs  22755  cnncvs  22767  0pwfi  38252  sge0rnn0  39261  sge0reuz  39340 Copyright terms: Public domain W3C validator

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http://math.stackexchange.com/questions/563349/squeeze-theorem-on-this-sine-function

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# Squeeze Theorem on this Sine Function $$\lim_{x \to \infty}\frac{\sin 2x}{4x}$$ My book says: Start by examinig the numerator of the given function, $\sin 2x$. The $\sin$ function has a minimum absolute value of $0$ and a maximum absolute value of $1$. Thus, the range of the absolute value of $\sin 2x$ is: $$0 \leq |\sin 2x| \leq 1.$$ Divide each part of the inequality by $4x$: $$0 \leq |\frac{\sin 2x}{4x}| \leq \frac{1}{4x}.$$ My question is: 1) Why do we use these absolute values? Why not squeeze $\sin 2x$ between $-1$ and $1$? Isn't $[-1,1]$ the range of the $\sin$ function? Why are we considering the range of the absolute value? 2) Can we squeeze the whole function $\frac{\sin 2x}{4x}$ between two values, why does it concentrate on just the numerator, $\sin 2x$? Thank you. - I changed multiple instances of \textrm{sin} to \sin, which is standard usage. If you write a\textrm{sin}b, you don't get proper spacing, whereas with a\sin b you do: $a\textrm{sin}b$ versus $a\sin b$. – Michael Hardy Nov 12 '13 at 2:55 @MichaelHardy I see, thanks. – Emi Matro Nov 12 '13 at 2:57 1) 1. When having limits that oscilate between negative and positive values, it is a common technique to use absolute values. 2. Yes you can do so! 3. Yes it is. 4. This is the same question as the first one 2) 1. Because $\sin2x$ is problematic to deal with, but $\frac{1}{x}$ is not. - Thanks. Can you expand a little on why $\sin 2x$ is problematic, but $\frac{1}{x}$ is not? Is it because it oscillates? – Emi Matro Nov 12 '13 at 3:10 @user436158 And also because it is non-constant, so it gives more freedom to the limit. When we bound it with constants, we get a much simpler expression that we can handle easily. – chubakueno Nov 12 '13 at 3:24

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http://mathhelpboards.com/other-topics-22/cylindrical-capacitor-formulae-forming-intuition-8932.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e

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# Thread: Cylindrical Capacitor formulae- forming an intuition 1. So far I have learned about Coulomb's law, the electric field, gauss's law, the electric potential and now capacitance. I feel that although I "know of" these topics, I don't actually "flow with them". Ignoring the math for a second; I want to form an understanding. And I think calculating the capacitance formula of a cylindrical capacitor may help. * From that picture I understand that the center cylinder has a positive value, and the outer cylinder has an equal and opposite value. * There must be an electric field in midst of the cylinders flowing from the positive to the negative. * (I'm still confused with the electric potential portion, but here goes) There is also an electric potential tied together with this field. And to find this potential $\displaystyle V$, I must find the Electric Field $\displaystyle E$. * I know: $\displaystyle \oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}$ * Using the given gaussian surface on that picture, I should find $\displaystyle E$. * Once I find $\displaystyle E$, I should plug it into the formula: $\displaystyle \Delta V=-Ed$ and $\displaystyle d$ being the distance $\displaystyle (b-a)$. Because, (...not certain) the electric field multiplied by a given distance equals the electric potential for that given distance. * Now that I have $\displaystyle \Delta V$ I use the following: $\displaystyle C=\frac{Q}{\Delta V}$. I know the electric potential and the charge is $\displaystyle Q$. With that, I find the capacitance. I haven't done any calculations to prove this. I only wanted to know if my intuition was legal. Thank You. 2. Originally Posted by Quintessential So far I have learned about Coulomb's law, the electric field, gauss's law, the electric potential and now capacitance. I feel that although I "know of" these topics, I don't actually "flow with them". Ignoring the math for a second; I want to form an understanding. And I think calculating the capacitance formula of a cylindrical capacitor may help. Hey Quintessential! Nice picture! (I like pictures. ) This is all correct with 1 caution and 1 exception, which I'll comment on below. Quote: * From that picture I understand that the center cylinder has a positive value, and the outer cylinder has an equal and opposite value. * There must be an electric field in midst of the cylinders flowing from the positive to the negative. * (I'm still confused with the electric potential portion, but here goes) There is also an electric potential tied together with this field. And to find this potential $\displaystyle V$, I must find the Electric Field $\displaystyle E$. * I know: $\displaystyle \oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}$ Caution: this only applies for a static electric field in vacuum. Quote: * Using the given gaussian surface on that picture, I should find $\displaystyle E$. * Once I find $\displaystyle E$, I should plug it into the formula: $\displaystyle \Delta V=-Ed$ and $\displaystyle d$ being the distance $\displaystyle (b-a)$. Because, (...not certain) the electric field multiplied by a given distance equals the electric potential for that given distance. This is only true if the electric field is constant, which it is not in this case. The proper formulas are: $$\mathbf E = -\nabla V$$ $$\Delta V = - \int_a^b \mathbf E \cdot \mathbf {dl}$$ Quote: * Now that I have $\displaystyle \Delta V$ I use the following: $\displaystyle C=\frac{Q}{\Delta V}$. I know the electric potential and the charge is $\displaystyle Q$. With that, I find the capacitance. I haven't done any calculations to prove this. I only wanted to know if my intuition was legal. Thank You. Thanks a bunch for the helpful input! Regarding the following: Originally Posted by I like Serena This is only true if the electric field is constant, which it is not in this case. The proper formulas are: $$\mathbf E = -\nabla V$$ $$\Delta V = - \int_a^b \mathbf E \cdot \mathbf {dl}$$ Had the cylinder been infinitely long vertically, would the electric field have been constant? 4. Originally Posted by Quintessential Had the cylinder been infinitely long vertically, would the electric field have been constant? No. The electric field is diverging from the inner cylinder toward the outer cylinder. That means it becomes weaker. Note that electric field strength is proportional to the density of the drawn electric field lines. Close to the inner cylinder the density of lines is higher than it is at the outer cylinder. Perfect. Makes sense. $\displaystyle \Delta V = -E \int_0^{b-a} {dl}$ I think I can take E out of the dot product seeing as how $\displaystyle cos(\theta)=1$, rather the Electric field lines are parallel with the normal of the inner cylinder surface packets $\displaystyle dl$ And I'll have to integrate from 0 to the distance between the cylinders, so: $\displaystyle b-a$ As for $\displaystyle dl$, now what would that be? $\displaystyle \Delta V = -E 2 \pi al(b-a)$ I'm going for the volume between the cylinders. Still not 100% on this... 6. Originally Posted by Quintessential Perfect. Makes sense. $\displaystyle \Delta V = -E \int_0^{b-a} {dl}$ I think I can take E out of the dot product seeing as how $\displaystyle cos(\theta)=1$, rather the Electric field lines are parallel with the normal of the inner cylinder surface packets $\displaystyle dl$ Neh. That won't work. You can only bring $E$ outside of the integral if it is constant, but it's not. It changes with the radius, so we might write $E=E(r)$, meaning that $E$ is a function of $r$. Then we get: $$\Delta V = -\int_a^b \mathbf E \cdot \mathbf{dl} = -\int_a^b E(r) dr$$ To find $E(r)$ you need to use the other formula $$\oint_{r\text{ constant}} E(r) dA = \frac Q {\varepsilon_0}$$ When you have $E(r)$ you can integrate it to find $\Delta V$. #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

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# Combination Sum Given asetof candidate numbers (`candidates`)(without duplicates)and a target number (`target`), find all unique combinations in`candidates` where the candidate numbers sums to`target`. Thesamerepeated number may be chosen from`candidates` unlimited number of times. Note: • All numbers (including `target`) will be positive integers. • The solution set must not contain duplicate combinations. Example 1: ``````Input: candidates = [2,3,6,7], target = 7, A solution set is: [ [7], [2,2,3] ] `````` Example 2: ``````Input: candidates = [2,3,5], target = 8, A solution set is: [ [2,2,2,2], [2,3,3], [3,5] ] `````` ## Solution Backtracking ``````class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> ans = new ArrayList<>(); Arrays.sort(candidates); helper(target, 0, 0, candidates, new ArrayList<>(), ans); return ans; } private void helper(int target, int sum, int startIdx, int[] candidates, List<Integer> combo, List<List<Integer>> ans) { if (sum == target) { return; } if (sum > target) { return; } for (int i = startIdx; i < candidates.length; i++) {

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## sliced Poisson One of my students complained that his slice sampler of a Poisson distribution was not working when following the instructions in Monte Carlo Statistical Methods (Exercise 8.5). This puzzled me during my early morning run and I checked on my way back, even before attacking the fresh baguette I had brought from the bakery… The following R code is the check. And it does work! As the comparison above shows… ```slice=function(el,u){ #generate uniform over finite integer set mode=floor(lambda) sli=mode x=mode+1 while (dpois(x,el)>u){ sli=c(sli,x);x=x+1} x=mode-1 while (dpois(x,el)>u){ sli=c(sli,x);x=x-1} return(sample(sli,1))} #example T=10^4 lambda=2.414 x=rep(floor(lambda),T) for (t in 2:T) x[t]=slice(lambda,runif(1)*dpois(x[t-1],lambda)) barplot(as.vector(rbind( table(x)/length(x),dpois(0:max(x), lambda))),col=c("sienna","gold")) ``` ### 4 Responses to “sliced Poisson” 1. […] Poisson slice sampler of a Poisson distribution, I could not tell why the code was not working! My earlier post prompted him to do so and a somewhat optimised version is given […] 2. The version of this that showed up on R-bloggers was missing the ” mode=floor(lambda)” line, which confused me greatly until I came here to check it out … • sorry, the R code in wordpress is anything but stable and I did not check the latest version before going public! 3. […] article was first published on Xi'an's Og » R, and kindly contributed to […] This site uses Akismet to reduce spam. Learn how your comment data is processed.

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## Search our database of handpicked sites Looking for a great physics site? We've tracked down the very best and checked them for accuracy. Just fill out the fields below and we'll do the rest. ## You searched for "poles flip" We found 7 results on physics.org and 12 results in our database of sites (of which 12 are Websites, 0 are Videos, and 0 are Experiments) ## Search results from our links database Showing 1 - 10 of 12 ### Magnetic flip Comprehensive information on geomagnetic reversal, or how the Earth's poles could flip . Rating: Hits: 5311 ### When North goes South Images and animations on how our planet's magnetic poles have flipped in the past, and how they may flip again. Rating: Hits: 6130 ### What causes the periodic reversals of the earth's magnetic field? A page from Scientific American about historical flips of the Earth 's poles (known as geomagnetic reversal) Rating: Hits: 4950 ### The Sun’s Magnetic Field Is About to Reverse Every 11 years the direction of the Sun's magnetic field flips. This article outlines what we expect to observe and experience here on Earth when it does change direction. Rating: Hits: 2095 ### Plants, Poles and Plugs How electricity is generated and delivered to your home. Rating: Hits: 4644 ### How Boolean Logic Works Explains how Boolean Logic works, with gates, adders and flip-flops. From HowStuffWorks.com. Rating: Hits: 2417 ### Aurora When energetic charged particles enter the earth's atmosphere from the solar wind, they tend to be channeled toward the poles by the magnetic force. Rating: Hits: 6258 ### Are the Earth's magnetic poles moving? Brief ideas regarding magnetic polar wandering. Rating: Hits: 4623 ### Earth's poles may switch Short article from PhysicsWeb (Institute of Physics) which discusses possible future shifts in the orientation of the earth's magnetic field. Rating: Hits: 2813 ### Making Maths: Clinometer Use a clinometer to measure the height of trees, buildings, flag poles and other tall objects. Here's a simple activities to make your own. Rating: Hits: 5210 Showing 1 - 10 of 12

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# 2D Array problem solution - 30 days of code HackerRank Input Format There are  lines of input, where each line contains  space-separated integers describing 2D Array ; every value in  will be in the inclusive range of  to . Output Format Print the largest (maximum) hourglass sum found in A. ### Problem solution in Java programming language. import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int arr[][] = new int[6][6]; for(int i=0; i < 6; i++){ for(int j=0; j < 6; j++){ arr[i][j] = in.nextInt(); } } int sum = -10000; for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { // [00] [01] [02] // [11] // [20] [21] [22] int currentSum = arr[i][j] + arr[i][j+1] + arr[i][j+2] + arr[i+1][j+1] + arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]; if (currentSum > sum) { sum = currentSum; } } } System.out.println(sum); } } #!/bin/python3 import math import os import random import re import sys if __name__ == '__main__': arr = [] for _ in range(6): arr.append(list(map(intinput().rstrip().split()))) sum = 0 tarr = [] for l in range(0,4): for k in range(0,4): for i in range(l,l+3): for j in range(k,k+3): if i == l+1 and ( j == k or j == k+2): continue else: sum += arr[i][j] tarr.append(sum) sum = 0 print(max(tarr))

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# Talk ## Maths levels (18 Posts) AbbyR1973 Wed 19-Feb-14 21:22:52 Following on from another thread about adding 2 digit numbers and year 2 maths. What sort of level would you expect a child to be able to add 2 2-digit numbers mentally, without any equipment/ number squares etc at hand. Thanks. juniper44 Wed 19-Feb-14 22:05:22 3. tiredbutnotweary Wed 19-Feb-14 23:33:04 Does mentally include workings out with pencil & paper, or all in head & only the answer written down? Really sorry if it's a daft question but it's all so different from how I used to do maths! juniper44 Wed 19-Feb-14 23:40:47 Mentally means in your head. Well, I'd expect a level 3 child to work this sort of thing out mentally. Depends on the question though. 21+34 is very different to 89+94. A level 3 means no apparatus. But they are expected to show their workings out. Ds frustrated his y2 teacher by not showing how he worked things out. He's now y4 and working at a secondary school level. His y2 teacher is still frustrated! Adding 2 two digit numbers is one part of a L3 but there is much more expected than that for a child to actually be a L3 teafor1 Thu 20-Feb-14 09:29:41 So for level 2 an abacus is acceptable? From your description I'm assuming yes. And yikes on doing it mentally! I still need to at least write things down! AbbyR1973 Thu 20-Feb-14 09:42:34 What would happen in a test if DS didn't show his workings out? He just does it in his head. I suspect if 43 + 55 came up in a test he would just write down 98. He wouldn't see the need to show workings, anymore than he would see the need to show working out for 3+5. toomuchicecream Thu 20-Feb-14 11:08:14 Depends on the test question! Some (SATS) question specifically instruct the child to show how they worked out the answer. These are normally 2 mark questions, with 1 mark being for the working out and 1 mark being for the answer. If the child doesn't show any working out in the big box provided, they can't have the mark for showing working out. Some 1 mark questions just ask the child to perform a calculation. No marks for the working out there, just 1 for the correct answer. Feenie Thu 20-Feb-14 11:17:23 Some (SATS) question specifically instruct the child to show how they worked out the answer. These are normally 2 mark questions, with 1 mark being for the working out and 1 mark being for the answer. If the child doesn't show any working out in the big box provided, they can't have the mark for showing working out. Sorry, but that just isn't true. If the child gets the answer right, they receive both marks whether they showed their method or not. TeenAndTween Thu 20-Feb-14 15:11:12 Showing workings out is good practice to get into as soon as they can. If they get too used to just bunging down the answer, then as the questions get harder they are less used to showing their working so find it harder to do. If they do show working then even if they get the answer wrong eg from a silly mistake, they can still get method marks. Also a plea here for writing some actual words when doing word problems - makes it so much easier to follow (DD1 - I'm looking at you here!) (Though I agree difficult to show working for 45+53 if it is done in head) toomuchicecream Thu 20-Feb-14 17:19:12 Sorry Feenie, I was thinking about the question on the 2009 paper which says that a child has got the wrong answer to a question - show him how he should have worked out the answer. I know I had to keep referring back to the mark scheme when I was marking it, but I've left the mark scheme at school so I couldn't check it. I was 99% sure that it says if a child just writes down the correct answer they could only have 1 mark. But I'm happy to be corrected. Feenie Thu 20-Feb-14 17:25:06 I don't know about that one question - but it sounds like a one off compared to the general run of the mill show your working question. toomuchicecream Thu 20-Feb-14 17:45:47 That's why it stuck in my mind! RafaIsTheKingOfClay Thu 20-Feb-14 18:22:05 I'm pretty sure just the answer would get you both marks in general. The assumption is that if you got the correct answer then you used a correct method so you can get the method marks. Fri 21-Feb-14 13:31:03 What sort of level would you expect a child to be able to add 2 2-digit numbers mentally, without any equipment/ number squares etc at hand. See I don't know about levels, but I am thinking half way thru English yr3 typically, so just about 8yo. Fri 21-Feb-14 13:31:27 ps: huge range on that, though, some will get to y6 still not confident at it, some will manage at 4yo. Longsufferingmrs Fri 21-Feb-14 15:39:57 In a SATs test you will get 2 marks for the correct answer but you can still get one mark if you got the answer wrong but used the correct method for the calculation. This means that even if you miscalculated somewhere, you can still get a mark. Join the discussion Registering is free, easy, and means you can join in the discussion, get discounts, win prizes and lots more. Register now

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The International Information Center for Geotechnical Engineers # Landslides: Slope stability, triggers, failure dynamics, and morphology - Modeling the effects of an earthquake on a hillslope Modeling the effects of an earthquake on a hillslope There have been many attempts to model the effects an earthquake has on a slope to assess the likelihood of failure in a given event. Jibson (2011) overviewed three of the methods used, pseudostatic analysis, stress-deformation analysis, and the Newmark sliding block analysis. Pseudostatic Analysis Pseudostatic analysis assumes that the shaking from the earthquake can be represented as a permanent body force applied to a slope.  The factor of safety is calculated using the equation below: W is the weight per unit slope, alpha is the slope angle, phi is the internal angle of friction, and k is the pseudostatic coefficient. The pseudostatic coefficient is the horizontal ground acceleration divided by acceleration due to gravity. Any ground acceleration that causes the factor of safety to decrease below one will trigger failure according to this analysis.  The diagram from Jibson (2011) below shows a force diagram used in pseudostatic analysis. The representation of an earthquake as a single, continuous force on a slope is not accurate. Using the pseudostatic method to predict slope stability is often conservative, but in special cases where pore pressure will build up or shear strength will be lost during shaking, the analysis is unconservative. Essentially, pseudostatic analysis is a very basic analysis, and does not give any information about what occurs after the slope is no longer in equilibrium. However, due to its ease of use, and low cost, it can serve as a simple index of stability. Stress-Deformation analysis Stress-deformation analysis uses finite-element modeling methods to model the response of a slope to a stress. It uses a mesh and calculates the deformation of each node in response to the modeled stress. Quality of the model is determined by the quality of input data. High quality and high-density data are needed for the model to be accurate, and if the data is good enough, stress-deformation modeling will give the most accurate depiction of what occurs during shaking.  However, as acquiring the necessary data needed is very expensive, stress-deformation analysis is typically only used for critical slopes and structures. Newmark analysis Newmark analysis models landslides as a rigid block on an inclined plane. The critical acceleration is the acceleration needed to overcome basal resistance. To determine the displacement of a block, the acceleration record from an earthquake greater than the critical acceleration is integrated to produce a velocity-time function, which is then integrated to produce an estimate of the displacement. The interpretation of the displacement varies. Typically, a threshold displacement is assumed, such as 5 or 10 cm.  Any displacements that exceed the threshold are predicted to fail. Newmark displacement assumes that the landslide experiences no internal deformation and moves as a rigid block during the entire failure. It also assumes that the critical acceleration remains constant and there are no dynamic pore pressure effects. Analyses of both laboratory models and earthquake-induced landslides have confirmed that Newmark analysis is fairly accurate when slope geometry, soil and rock properties, and acceleration are known. The stress-deformation analysis provides the most information about behavior, but is also the most expensive and work intensive technique, making it infeasible for widespread analysis. Newmark analysis bridges the gap between the two methods, being both inexpensive and providing better information than pseudostatic analysis. (Jibson, 2011)

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• Get access Interesting script? So post a link to it - let others appraise it You liked the script? Try it in the MetaTrader 5 terminal # ASO - indicator for MetaTrader 5 Views: 1752 Rating: Published: 2018.07.13 15:19 Indicator ASO (Average Sentiment Oscillator) displays the general market climate. It has four configurable parameters: • Range period - range calculation period; • Smoothing period - oscillator smoothing period; • Method - smoothing method; • Mode - calculation mode: • Combined - combined mode; • Intra-bar - intra-bar mode; • Group algorithm - group algorithm mode. Calculations: ```Bulls = MA(BL, Smoothing period, Method) Bears = MA(BR, Smoothing period, Method) ``` where: Combined Mode: ```BL = (IBL+GBL)/2 BR = (IBR+GBR)/2 ``` Intra-bar Mode: ```BL = IBL BR = IBR ``` Group Algorithm Mode: ```BL = GBL BR = GBR IBL = 50 * (Close - Low + High - Open)/(High - Low) IBR = 50 * (High - Close + Open - Low)/(High - Low) GBL = 50*(Close - GL + GH - GO)/(GH - GL) GBR = 50*(GH - Close + GO - GL)/(GH - GL) ``` • GO - Open within Range period; • GL -minimum value of High within Range period; • GH - maximum value of Low within Range period. Fig 1. Combined Mode Fig. 2. Intra-bar Mode Fig. 2. Group Algorithm Mode Translated from Russian by MetaQuotes Software Corp. Original code: https://www.mql5.com/ru/code/21159 SymbolX_Candle This indicator calculates the index of a given currency, using the USD index. It is based on indicator SymbolX, but four prices are used for calculations: OHLC instead of just one Close price. SymbolX This indicator calculates the index of a given currency, using the USD index. The Close prices of six currency pairs are used to calculate the index. CA Indicator CA (Corrected Average), also known as Optimal Moving Average. The benefit of the indicator is the fact that the current value of the timeseries must exceed the current threshold that depends on volatility for the indicator line to follow the price, avoiding false signals in the trend. DMX Indicator DMX (Bipolar DMI) is calculated using the standard indicator ADX. However, as compared to that, it displays data as an oscillator with a signal line, and has shorter delay.

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# A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2…. A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3?

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# divisibility for numbers like 13,17 and 19 - Compartmentalization method For denominators like 13, 17 i often see my professor use a method to test whether a given number is divisible or not. The method is not the following : Ex for 17 : subtract 5 times the last digit from the original number, the resultant number should be divisible by 17 etc... The method is similar to divisibility of 11. He calls it as compartmentalization method. Here it goes. rule For 17 : take 8 digits at a time(sun of digits at odd places taken 8 at a time - sum of digits at even places taken 8 at a time) For Ex : $9876543298765432..... 80$digits - test this is divisible by 17 or not. There will be equal number of groups (of 8 digits taken at a time) at odd and even places. Therefore the given number is divisible by 17- Explanation. The number 8 above differs based on the denominator he is considering. I am not able to understand the method and logic both. Kindly clarify. Also for other numbers like $13$ and $19$, what is the number of digits i should take at a time? In case my question is not clear, please let me know. - I cannot figure out the rule. What are the "blocks of 8 digits at even/odd places"? For example, given the number 1234567890123456, what would the two blocks look like? – celtschk Aug 20 '12 at 15:27 I think he means the two 8 digit numbers formed by every other digit, in this case $13579135$ and $24680246$. – axblount Aug 20 '12 at 15:38 @celtschk: you split into blocks of 8 starting from the ones digit, so the first block would be $90123456$, the second would be $12345678$. Then we have $1234567890123456 \equiv 90123456-12345678 \pmod {17}$ – Ross Millikan Aug 20 '12 at 15:49 You quote two different rules with different results. When testing for divisibility by 17 by subtracting 5 times the last digit from the orignal number without its last digit, you are using the fact that $51$ is divisible by $17$, so $10a+b \equiv 10a-50b \pmod {17}$, then the fact that $10(a-5b)$ is a multiple of $17$ if and only if $(a-5b)$ is. Unless you do further computation, you lose the remainder if the original number is not a multiple. When you take blocks of 8 digits, you use the fact that $10^8+1 \equiv 0 \pmod {17}$, so $10^8a+b \equiv b-a \pmod {17}$ You retain the remainder in this case. For 13, you need half the period of its repeating decimal, which is 6, so you use blocks of 3. Note that $10^3+1=1001 \equiv 0 \pmod {13}$ - Your professor is using the fact that $100000001=10^8+1$ is divisible by $17$. Given for example your $80$-digit number, you can subtract $98765432\cdot 100000001=9876543298765432$, which will leave zeros in the last $16$ places. Slash the zeros, and repeat. After $5$ times you are left with the number $0$, which is divisible by $17$, and hence your $80$-digit number must also be divisible by $17$. When checking for divisibility by $17$, you can also subtract multiples of $102=6\cdot 17$ in the same way. For divisibility by $7$, $11$, or $13$, you can subtract any multiple of the number $1001=7\cdot 11\cdot 13$ without affecting divisibility by these three numbers. For example, $6017-6\cdot 1001=11$, so $6017$ is divisible by $11$, but not by $7$ or $13$. For divisibility by $19$, you can use the number $1000000001=10^9+1=7\cdot 11\cdot 13\cdot 19\cdot 52579$. By subtracting multiples of this number, you will be left with a number of at most $9$ digits, which you can test for divisibility by $19$ by performing the division. - The whole process can be explained by writing A^x/B = C where C can be +1 or -1 or 0. Here C is the remainder got on division of A^x by B. Now we operate on base 10. So A takes a value of 10. B is the divisor for which you are attempting to set up a rule for. This method is called compartmentalisation. It is used to check on divisibility / remainder for numbers as you had stated .. not for the normal smal numbers. There are much effective rules for the same. Suppose you talk of divisibility by 9: Consider any 2 digit number AB. This can be represented in terms of der place values as AB = 10A+B = 9A+A+B. Now if you were to check divisibility by 9. [9A + (A+B)]/9. Then the first term 9A is divisible. Now if the whole number is to be divisible, we would want the terms (A+B) to also be divisible. So the divisibility of the number depends on (A+B) which is nothing but the digit sum. Which is why you see that the divisibility rule for 9 is "The digit sum of the number must be divisible by 9" Also you got to note that: 10^x/9 = +1 for whatever be the value of x. This results in another rule: 10^x/3 = +1 which is the rule for divisibility by 3. Now consider the RULE FOR 7 A^x/B= C Here A=10 (base 10), B = 7. Now we have to choose values for x such that it gives a remainder of C= +1 or -1. we see the following pattern: 10^0/7 = +1 10^3/7 = -1 10^6/7 = +1 So we see that +1 and -1 remainders alternate every three powers of ten. This gives our compartmentalisation rule. The given number from left to right has to be grouped in threes and the rule of +1 and -1 for every triplet has to be applied. You will see it : ABCPQRXYZ be a number. To check the divisibility by 7, we write it as: [10^6 ABC]/7 + [10^3 PQR]/7 + [10^0 XYZ]/7 Now 10^6/7=+1 , 10^3/7 = -1 and so on.. This reduces to (1xABC) (-1XPQR) (1*XYZ) RULE : Sum of triplets at odd places - sum of triplets at even places eg: 100200140240 /7 . What would be the remainder? 100 | 200 | 140 | 240 | -1 | +1 | -1 | +1 • 100- 140 = -240 200 + 240 = 440 You get 440-240 = 200 200/7 gives a R= 4 Now consider the RULE FOR 11 10^0 / 11 = +1 10^1/ 11 = -1 10^2/ 11 = +1 10^3/11 = - 1 FOr 11, you can do it in 2 ways. One is that you notice that every alternate power of 10 divided by 11 gives an alternating pattern of +1 and -1. So the compartmentalisation will now be for every digit. So you do a Sum of digits at odd places - Sum of digits at even places. But if you did notice the fact that 10^0/11 = +1 and 10^3/11 = -1 Then you will notice that 11, same as 7, has +1 and -1 alternating for triplets. So, you can proceed compartmentalising the number into triplets and doing the same thing as we did for 7. Sum of triplets at odd places - Sum of triplets at even places. So we see that so far : 1. 10^x / B = +1 will hold good for three cases i guess (not sure it may be more ... ) They are 3, 9 , 37. Even 10^3/37 = +1 . So here there are no alternate +1 and -1 as 10^6/37 will also be +1. Note that (a^m)^n = a^mn **Rule for 37 would be : Group the number into triplets and apply a +1 to each triplet. 1. 10^x / B = +1 or -1 will hold good for 7, 11, 17 Rule for 17 (thus) : 10^8/17 = -1. So, 10^16/17 would be +1. So it becomes like (Sum of digits at odd places taken 8 at a time - sum of digits at even places taken 8 at a time. )** 3. Also 10^x/B = 0 (gives a remainder of 0 )when B are powers of 2 or powers of 5. I still dont know if it is possibile to cover all primes (atleast 2 digits) in this method. Perhaps you may have to use other techniques like Chinese Remainder Theorem or the Basic Remainder Theorem and so on.. That is because I still dont see any patterns for 23 and many more numbers. If anybody could add anything more to this .. it would be great. Hope this post helps. Cheers :) - The Similar can also be said for divisibility by $13$ $10^3/13$ gives a remainder of $-1$ $10^6/13$ gievs a remainder of $+1$ So again the rule for $13$, will be same as the rule for $7, 11$. Group the numbers into triplets as we see alternate changes of $-1$ to $+1$ between every three powers of $10$. So the rule for $13$ will also be (Sum of triplets at odd places - Sum of triplets at even places) or Sum of digits at odd places taken $3$ at a time - Sum of digits at even places taken three at a time) As i said, all this can be summarised as: $10^x/B = +1$ holds for divisors (values of B) like $3,9, 37$ $10^x/B = -1$ and $+1$ for divisiors (values of B) like $7, 11, 13, 17$ $10^x/B = 0$ for divisors (values of B) like powers of $2$ and $5$ where $+1,0,-1$ are the remainders of the respective divisions -

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KiB to kbit → CONVERT Kibibytes to Kilobits expand_more info 1 KiB is equal to 8.192 kbit Kibibyte (binary) --to--> Kilobit (decimal) KiB Kibibyte (KiB) Versus Kilobit (kbit) - Comparison Kibibytes and Kilobits are units of digital information used to measure storage capacity and data transfer rate. Kibibytes is a "binary" unit where as Kilobits is a "decimal" unit. One Kibibyte is equal to 1024 bytes. One Kilobit is equal to 1000 bits. There are 0.1220703125 Kibibyte in one Kilobit. Find more details on below table. Unit Name Kibibyte Kilobit Unit Symbol KiB kb or kbit Standard binary decimal Defined Value 2^10 or 1024^1 Bytes 10^3 or 1000^1 Bits Value in Bits 8,192 1,000 Value in Bytes 1,024 125 Kibibyte (KiB) to Kilobit (kbit) Conversion - Formula & Steps The KiB to kbit Calculator Tool provides a convenient solution for effortlessly converting data units from Kibibyte (KiB) to Kilobit (kbit). Let's delve into a thorough analysis of the formula and steps involved. Outlined below is a comprehensive overview of the key attributes associated with both the source (Kibibyte) and target (Kilobit) data units. Source Data Unit Target Data Unit Equal to 1024 bytes (Binary Unit) Equal to 1000 bits (Decimal Unit) The formula for converting the Kibibyte (KiB) to Kilobit (kbit) can be expressed as follows: diamond CONVERSION FORMULA kbit = KiB x (8x1024) ÷ 1000 Now, let's apply the aforementioned formula and explore the manual conversion process from Kibibyte (KiB) to Kilobit (kbit). To streamline the calculation further, we can simplify the formula for added convenience. FORMULA Kilobits = Kibibytes x (8x1024) ÷ 1000 STEP 1 Kilobits = Kibibytes x 8192 ÷ 1000 STEP 2 Kilobits = Kibibytes x 8.192 Example : By applying the previously mentioned formula and steps, the conversion from 1 Kibibyte (KiB) to Kilobit (kbit) can be processed as outlined below. 1. = 1 x (8x1024) ÷ 1000 2. = 1 x 8192 ÷ 1000 3. = 1 x 8.192 4. = 8.192 5. i.e. 1 KiB is equal to 8.192 kbit. Note : Result rounded off to 40 decimal positions. You can employ the formula and steps mentioned above to convert Kibibytes to Kilobits using any of the programming language such as Java, Python, or Powershell. Unit Definitions What is Kibibyte ? A Kibibyte (KiB) is a binary unit of digital information that is equal to 1024 bytes (or 8,192 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'kibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'kilobyte' (KB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems. arrow_downward What is Kilobit ? A Kilobit (kb or kbit) is a decimal unit of digital information that is equal to 1000 bits. It is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of Kibibit (Kibit) is used instead. Excel Formula to convert from Kibibyte (KiB) to Kilobit (kbit) Apply the formula as shown below to convert from 1 Kibibyte (KiB) to Kilobit (kbit). A B C 1 Kibibyte (KiB) Kilobit (kbit) 2 1 =A2 * 8.192 3 If you want to perform bulk conversion locally in your system, then download and make use of above Excel template. Python Code for Kibibyte (KiB) to Kilobit (kbit) Conversion You can use below code to convert any value in Kibibyte (KiB) to Kibibyte (KiB) in Python. kibibytes = int(input("Enter Kibibytes: ")) kilobits = kibibytes * 8192 / 1000 print("{} Kibibytes = {} Kilobits".format(kibibytes,kilobits)) The first line of code will prompt the user to enter the Kibibyte (KiB) as an input. The value of Kilobit (kbit) is calculated on the next line, and the code in third line will display the result. How many Kibibytes(KiB) are there in a Kilobit(kbit)?expand_more There are 0.1220703125 Kibibytes in a Kilobit. What is the formula to convert Kilobit(kbit) to Kibibyte(KiB)?expand_more Use the formula KiB = kbit x 1000 / (8x1024) to convert Kilobit to Kibibyte. How many Kilobits(kbit) are there in a Kibibyte(KiB)?expand_more There are 8.192 Kilobits in a Kibibyte. What is the formula to convert Kibibyte(KiB) to Kilobit(kbit)?expand_more Use the formula kbit = KiB x (8x1024) / 1000 to convert Kibibyte to Kilobit. Which is bigger, Kibibyte(KiB) or Kilobit(kbit)?expand_more Kibibyte is bigger than Kilobit. One Kibibyte contains 8.192 Kilobits. Similar Conversions & Calculators All below conversions basically referring to the same calculation.

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# Fluid Mechanics posted by on . The components and their dimensions of a hydraulic jack are shown in the figure. The specific gravity of the fluid in the jack is SG=0.8. A force F=400 N is applied at the left end of the handle, which is hinged at the right end. Determine: (a) The fluid pressure p1 on the lower surface of the small piston. (b) The fluid pressure p2 on the lower surface of the large piston. (c) The weight W of the load that can be supported. • Fluid Mechanics - , Pressure at A=50,000N/m^2 Pressure at B=65696N/m^2 Pressure at C=7883.52N/m^2

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Change language to: English - 日本語 - Português - Русский See the recommended documentation of this function kroneck Kronecker form of matrix pencil Syntax ```[Q, Z, Qd, Zd, numbeps, numbeta] = kroneck(F) [Q, Z, Qd, Zd, numbeps, numbeta] = kroneck(E,A)``` Arguments F real matrix pencil `F=s*E-A` E,A two real matrices of same dimensions Q,Z two square orthogonal matrices Qd,Zd two vectors of integers numbeps,numeta two vectors of integers Description Kronecker form of matrix pencil: `kroneck` computes two orthogonal matrices `Q, Z` which put the pencil `F=s*E -A` into upper-triangular form: ``` | sE(eps)-A(eps) | X | X | X | |----------------|----------------|------------|---------------| | O | sE(inf)-A(inf) | X | X | Q(sE-A)Z = |---------------------------------|----------------------------| | | | | | | 0 | 0 | sE(f)-A(f) | X | |--------------------------------------------------------------| | | | | | | 0 | 0 | 0 | sE(eta)-A(eta)| ``` The dimensions of the four blocks are given by: `eps=Qd(1) x Zd(1)`, `inf=Qd(2) x Zd(2)`, `f = Qd(3) x Zd(3)`, `eta=Qd(4)xZd(4)` The `inf` block contains the infinite modes of the pencil. The `f` block contains the finite modes of the pencil The structure of epsilon and eta blocks are given by: `numbeps(1)` = `#` of eps blocks of size 0 x 1 `numbeps(2)` = `#` of eps blocks of size 1 x 2 `numbeps(3)` = `#` of eps blocks of size 2 x 3 etc... `numbeta(1)` = `#` of eta blocks of size 1 x 0 `numbeta(2)` = `#` of eta blocks of size 2 x 1 `numbeta(3)` = `#` of eta blocks of size 3 x 2 etc... The code is taken from T. Beelen (Slicot-WGS group). Examples ```F = randpencil([1,1,2],[2,3],[-1,3,1],[0,3]); Q = rand(17,17); Z = rand(18,18); F = Q*F*Z; //random pencil with eps1=1,eps2=1,eps3=1; 2 J-blocks @ infty //with dimensions 2 and 3 //3 finite eigenvalues at -1,3,1 and eta1=0,eta2=3 [Q, Z, Qd, Zd, numbeps, numbeta] = kroneck(F); [Qd(1),Zd(1)] //eps. part is sum(epsi) x (sum(epsi) + number of epsi) [Qd(2),Zd(2)] //infinity part [Qd(3),Zd(3)] //finite part [Qd(4),Zd(4)] //eta part is (sum(etai) + number(eta1)) x sum(etai) numbeps numbeta```

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World's only instant tutoring platform Filo is a preferred QESP Question Easy Solving time: 2 mins # The speed of sound is different in different states of matter.The speed of sound in liquid water is .Which row correctly compares the speed of sound in ice and the speed of sound in water vapour with the speed of sound in water? A A B B C C D D ## Text solutionVerified Explanation: The speed of sound depends on the properties of the medium through which it travels. The speed of sound decreases when it travels from solid to gaseous state. The speed of sound in liquid water is . Then speed of sound in ice is greater than and the speed of sound in steam is less than .

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Mmethods {FSA} R Documentation ## Estimate natural mortality from a variety of empirical methods. ### Description Several methods can be used to estimated natural mortality (M) from other types of data, including parameters from the von Bertalanffy growth equation, maximum age, and temperature. These relationships have been developed from meta-analyses of a large number of populations. Several of these methods are implemented in this function. ### Usage ```Mmethods(what = c("all", "tmax", "K", "Hoenig", "Pauly")) metaM(method = Mmethods(), justM = TRUE, tmax = NULL, K = NULL, Linf = NULL, t0 = NULL, b = NULL, L = NULL, Temp = NULL, t50 = NULL, Winf = NULL) ## S3 method for class 'metaM' print(x, digits = 4, ...) ``` ### Arguments `what` A string that indicates what grouping of methods to return. Defaults to returning all methods. `method` A string that indicates which method or equation to use. See details. `justM` A logical that indicates whether just the estimate of M (`TRUE`; Default) or a more descriptive list should be returned. `tmax` The maximum age for the population of fish. `K` The Brody growth coefficient from the fit of the von Bertalanffy growth function. `Linf` The asymptotic mean length (cm) from the fit of the von Bertalanffy growth function. `t0` The x-intercept from the fit of the von Bertalanffy growth function. `b` The exponent from the weight-length relationship (slope from the logW-logL relationship). `L` The body length of the fish (cm). `Temp` The temperature experienced by the fish (C). `t50` The age (time) when half the fish in the population are mature. `Winf` The asymptotic mean weight (g) from the fit of the von Bertalanffy growth function. `x` A `metaM` object returned from `metaM` when `justM=FALSE`. `digits` A numeric that controls the number of digits printed for the estimate of M. `...` Additional arguments for methods. Not implemented. ### Details One of several methods is chosen with `method`. The available methods can be seen with `Mmethods()` and are listed below with a brief description of where the equation came from. The sources (listed below) should be consulted for more specific information. • `method="HoenigNLS"`: The “modified Hoenig equation derived with a non-linear model” as described in Then et al. (2015) on the third line of Table 3. This method was the preferred method suggested by Then et al. (2015). Requires only `tmax`. • `method="PaulyLNoT"`: The “modified Pauly length equation” as described on the sixth line of Table 3 in Then et al. (2015). Then et al. (2015) suggested that this is the preferred model if maximum age (tmax) information was not available. Requires `K` and `Linf`. • `method="PaulyL"`: The “Pauly (1980) equation using fish lengths” from his equation 11. This is the most commonly used method in the literature. Note that Pauly used common logarithms as used here but the model is often presented in other sources with natural logarithms. Requires `K`, `Linf`, and `T`. • `method="PaulyW"`: The “Pauly (1980) equation for weights” from his equation 10. Requires `K`, `Winf`, and `T`. • `method="HoeingO"`, `method="HoeingOF"`, `method="HoeingOM"`, `method="HoeingOC"`: The original “Hoenig (1983) composite”, “fish”, “mollusc”, and “cetacean” (fit with OLS) equations from the second column on page 899 of Hoenig (1983). Requires only `tmax`. • `method="HoeingO2"`, `method="HoeingO2F"`, `method="HoeingO2M"`, `method="HoeingO2C"`: The original “Hoenig (1983) composite”, “fish”, “mollusc”, and “cetacean” (fit with Geometric Mean Regression) equations from the second column on page 537 of Kenchington (2014). Requires only `tmax`. • `method="HoenigLM"`: The “modified Hoenig equation derived with a linear model” as described in Then et al. (2015) on the second line of Table 3. Requires only `tmax`. • `method="HewittHoenig"`: The “Hewitt and Hoenig (2005) equation” from their equation 8. Requires only `tmax`. • `method="tmax1"`: The “one-parameter tmax equation” from the first line of Table 3 in Then et al. (2015). Requires only `tmax`. • `method="K1"`: The “one-parameter K equation” from the fourth line of Table 3 in Then et al. (2015). Requires only `K`. • `method="K2"`: The “two-parameter K equation” from the fifth line of Table 3 in Then et al. (2015). Requires only `K`. • `method="JensenK1"`: The “Jensen (1996) one-parameter K equation”. Requires only `K`. • `method="JensenK2"`: The “Jensen (2001) two-parameter K equation” from their equation 8. Requires only `K`. • `method="Gislason"`: The “Gislason et al. (2010) equation” from their equation 2. Requires `K`, `Linf`, and `L`. • `method="AlversonCarney"`: The “Alverson and Carney (1975) equation” as given in equation 10 of Zhang and Megrey (2006). Requires `tmax` and `K`. • `method="Charnov"`: The “Charnov et al. (2013) equation” as given in the second column of page 545 of Kenchington (2014). Requires `K`, `Linf`, and `L`. • `method="ZhangMegreyD"`, `method="ZhangMegreyP"`: The “Zhang and Megrey (2006) equation” as given in their equation 8 but modified for demersal or pelagic fish. Thus, the user must choose the fish type with `group`. Requires `tmax`, `K`, `t0`, `t50`, and `b`. • `method="RikhterEfanov1"`: The “Rikhter and Efanov (1976) equation (#2)” as given in the second column of page 541 of Kenchington (2014) and in Table 6.4 of Miranda and Bettoli (2007). Requires only `t50`. • `method="RikhterEfanov2"`: The “Rikhter and Efanov (1976) equation (#1)” as given in the first column of page 541 of Kenchington (2014). Requires `t50`, `K`, `t0`, and `b`. ### Value `Mmethods` returns a character vector with a list of methods. If only one `method` is chosen then `metaM` returns a single numeric if `justM=TRUE` or, otherwise, a `metaM` object that is a list with the following items: • `method`: The name for the method within the function (as given in `method`). • `name`: A more descriptive name for the method. • `givens`: A vector of values required by the method to estimate M. • `M`: The estimated natural mortality rate. If multiple `method`s are chosen then a data.frame is returned with the method name abbreviation in the `method` variable and the associated estimated M in the `M` variable. ### Testing Kenchington (2014) provided life history parameters for several stocks and used many models to estimate M. I checked the calculations for the `PaulyL`, `PaulyW`, `HoenigO` for `Hgroup="all"` and `Hgroup="fish"`, `HoenigO2` for `Hgroup="all"` and `Hgroup="fish"`, `"JensenK1"`, `"Gislason"`, `"AlversonCarney"`, `"Charnov"`, `"ZhangMegrey"`, `"RikhterEfanov1"`, and `"RikhterEfanov2"` methods for three stocks. All results perfectly matched Kenchington's results for Chesapeake Bay Anchovy and Rio Formosa Seahorse. For the Norwegian Fjord Lanternfish, all results perfectly matched Kenchington's results except for when `Hgroup="fish"` for both `HoenigO` and `HoenigO2`. Results for the Rio Formosa Seahorse data were also tested against results from `M.empirical` from fishmethods for the `PaulyL`, `PaulyW`, `HoenigO` for `Hgroup="all"` and `Hgroup="fish"`, `"Gislason"`, and `"AlversonCarney"` methods (the only methods in common between the two packages). All results matched perfectly. 11-Mortality. ### Author(s) Derek H. Ogle, derek@derekogle.com ### References Ogle, D.H. 2016. Introductory Fisheries Analyses with R. Chapman & Hall/CRC, Boca Raton, FL. Alverson, D.L. and M.J. Carney. 1975. A graphic review of the growth and decay of population cohorts. Journal du Conseil International pour l'Exploration de la Mer. 36:133-143. Charnov, E.L., H. Gislason, and J.G. Pope. 2013. Evolutionary assembly rules for fish life histories. Fish and Fisheries. 14:213-224. Gislason, H., N. Daan, J.C. Rice, and J.G. Pope. 2010. Size, growth, temperature and the natural mortality of marine fish. Fish and Fisheries 11:149-158. Hewitt, D.A. and J.M. Hoenig. 2005. Comparison of two approaches for estimating natural mortality based on longevity. Fishery Bulletin. 103:433-437. [Was (is?) from http://fishbull.noaa.gov/1032/hewitt.pdf.] Hoenig, J.M. 1983. Empirical use of longevity data to estimate mortality rates. Fishery Bulletin. 82:898-903. [Was (is?) from http://www.afsc.noaa.gov/REFM/age/Docs/Hoenig_EmpiricalUseOfLongevityData.pdf.] Jensen, A.L. 1996. Beverton and Holt life history invariants result from optimal trade-off of reproduction and survival. Canadian Journal of Fisheries and Aquatic Sciences. 53:820-822. [Was (is?) from .] Jensen, A.L. 2001. Comparison of theoretical derivations, simple linear regressions, multiple linear regression and principal components for analysis of fish mortality, growth and environmental temperature data. Environometrics. 12:591-598. [Was (is?) from http://deepblue.lib.umich.edu/bitstream/handle/2027.42/35236/487_ftp.pdf.] Kenchington, T.J. 2014. Natural mortality estimators for information-limited fisheries. Fish and Fisheries. 14:533-562. Pauly, D. 1980. On the interrelationships between natural mortality, growth parameters, and mean environmental temperature in 175 fish stocks. Journal du Conseil International pour l'Exploration de la Mer. 39:175-192. [Was (is?) from http://innri.unuftp.is/pauly/On%20the%20interrelationships%20betwe.pdf.] Rikhter, V.A., and V.N. Efanov. 1976. On one of the approaches for estimating natural mortality in fish populations (in Russian). ICNAF Research Document 76/IV/8, 12pp. Then, A.Y., J.M. Hoenig, N.G. Hall, and D.A. Hewitt. 2015. Evaluating the predictive performance of empirical estimators of natural mortality rate using information on over 200 fish species. ICES Journal of Marine Science. 72:82-92. Zhang, C-I and B.A. Megrey. 2006. A revised Alverson and Carney model for estimating the instantaneous rate of natural mortality. Transactions of the American Fisheries Society. 135-620-633. [Was (is?) from http://www.pmel.noaa.gov/foci/publications/2006/zhan0531.pdf.] See `M.empirical` in fishmethods for similar functionality. ### Examples ```## List names for available methods Mmethods() Mmethods("tmax") ## Simple Examples metaM("tmax",tmax=20) metaM("tmax",tmax=20,justM=FALSE) metaM("HoenigNLS",tmax=20) metaM("HoenigNLS",tmax=20,justM=FALSE) ## Example Patagonian Sprat ... from Table 2 in Cerna et al. (2014) ## http://www.scielo.cl/pdf/lajar/v42n3/art15.pdf Temp <- 11 Linf <- 17.71 K <- 0.78 t0 <- -0.46 tmax <- t0+3/K t50 <- t0-(1/K)*log(1-13.5/Linf) metaM("RikhterEfanov1",t50=t50) metaM("PaulyL",K=K,Linf=Linf,Temp=Temp) metaM("PaulyL",K=K,Linf=Linf,Temp=Temp,justM=FALSE) metaM("HoenigNLS",tmax=tmax) metaM("HoenigO",tmax=tmax) metaM("HewittHoenig",tmax=tmax) metaM("AlversonCarney",K=K,tmax=tmax) ## Example of multiple calculations metaM(c("RikhterEfanov1","PaulyL","HoenigO","HewittHoenig","AlversonCarney"), K=K,Linf=Linf,Temp=Temp,tmax=tmax,t50=t50) ## Example of multiple methods using Mmethods # select some methods metaM(Mmethods()[-c(15,20,22:24,26)],K=K,Linf=Linf,Temp=Temp,tmax=tmax,t50=t50) # select just the Hoenig methods metaM(Mmethods("Hoenig"),K=K,Linf=Linf,Temp=Temp,tmax=tmax,t50=t50) ``` [Package FSA version 0.8.18 Index]

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Plant Physiology (Biology 327)  - Dr. Stephen G. Saupe;  College of St. Benedict/ St. John's University;  Biology Department; Collegeville, MN  56321; (320) 363 - 2782; (320) 363 - 3202, fax;    ssaupe@csbsju.edu Determining the Q10 for the Potato Experiment (note:  I haven't tried this one yet - consider this lab experimental.  It may need work!) Q10 refers to the temperature quotient and is a ratio of the velocity of a process at a given temperature to that at a temperature 10 C lower.  The advantage of Q10 is that it enables us to determine if a process is a physical or chemical one.  Chemical reactions like those that occur in organisms, typically have a Q10 of ca. 2.0 – 3.0.  However, purely physical processes have a Q10 of approximately 1.0.  The following equation is used to calculate the Q10:         log Q10 = (10/(T2 – T1)) log K2/K1 where T2 = higher temperature (in K or C) T1 = lower temperature (in K or C) K2 = rate of reaction at higher temperature K1 = rate of reaction at lower temperature To determine the Q10 of water uptake into the potato cores we will perform the Gravimetric experiment at different temperatures. Protocol: 1. Set up the gravimetric experiment as previously described with the exception that only the 0.6 and 0.7 molal sucrose solutions will be used.  Prepare two samples of each. 2. Incubate one sample of each at room temperature and the other at a higher (or lower) temperature. 3. Record the temperatures for all samples and time of incubation in Table 1. 4. Complete Table 2. Table 1:  Tuber Data Higher Lower Temperature (C) Incubation Period (minutes) Table 2:  Potato Tuber Data [Sucrose] (molality) Water Uptake at lower temperature (cm3) Water Uptake per minute at lower temperature (cm3 min-1) Water Uptake at higher temperature (cm3) Water Uptake per minute higher temperature (cm3 min-1) Q10 0.6 0.7 Mean Data/Analysis: 1. Is water uptake a physical or chemical process?  Explain. Reference: | Top | SGS Home | CSB/SJU Home | Biology Dept | Biol 327 Home | Disclaimer | Last updated:  01/07/2009     � Copyright  by SG Saupe

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# Work and Energy - Mission WE2 Detailed Help Mac and Tosh run up the same hill. Mac has twice the mass as Tosh. Tosh climbs up the hill in one-half the time as Mac. In this situation, _____. Definition of Work: When a force acts upon an object to cause (or to hinder) a displacement, work is done upon the object. Mathematically, work is the dot product of the force and displacement for such a situation. Definition of Power: Power is the rate at which work is done upon an object. Work (W) is calculated from knowledge of the force (F) that acts upon an object, the displacement (d), and the angle (Θ) between the F and d vectors. The formula is   W = F • d • cosine(Θ).   Power (P) is calculated from knowledge of the work done upon an object (W) and the time (t) required to do this work. The formula is P = W / t Take some time to ponder the two definitions in the Dictionary section and the two formulas in the Formula Frenzy section. These are the key to understanding the answer to this question. The work that Mac and Tosh do is dependent upon the force they must exert and the distance they climb up the hill. Both Mac and Tosh climb the same distance (it's the same hill) but the twice-as-massive Mac must apply twice the force to lift his body (the applied force is proportional to his weight). This should allow you to compare the work of Mac to Tosh. Power is the rate at which work is done. It depends on two variables - the amount of work that is done and the rate at which the work is done. The previous paragraph compares the work done by Mac to the work done by Tosh. This amount of work was done in different times. Mac takes twice the time as Tosh in doing his work. To compare their power, the work-to-time ratio must be taken. Doing twice the work in twice the time might be more work, but it is the same amount of power.

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# Solve conditional term with mathematica [duplicate] Solve mathematical conditional expressions with Mathematica, e.g if $$x+y=3$$ and $$xy=-1$$ , what is result $$\frac{x^3+y^3}{x^2+y^2}$$ The lazy way to do this is to evaluate Solve[{z == (x^3 + y^3)/(x^2 + y^2), x + y == 3, x y == -1}, z, {x, y}] {{z -> 36/11}} The clever way to do this is to recognize that your rational function is symmetric, and is thus representable in terms of the symmetric polynomials: Table[SymmetricPolynomial[k, {x, y}], {k, 2}] {x + y, x y} which means you can leverage SymmetricReduction[]: expr = (x^3 + y^3)/(x^2 + y^2) Total[SymmetricReduction[Numerator[expr], {x, y}, {s1, s2}]]/ Total[SymmetricReduction[Denominator[expr], {x, y}, {s1, s2}]] (s1^3 - 3 s1 s2)/(s1^2 - 2 s2) after which you can plug the specified values in: % /. Thread[{s1, s2} -> {3, -1}] 36/11 (The Newton-Girard formulae are relevant here.) The even more clever way is to recall Vieta's formulae Collect[(t - x) (t - y), t, Simplify] t^2 + t (-x - y) + x y and recall that power sums of the form x^k + y^k satisfy a linear difference equation, with the recurrence coefficients being given by the symmetric expressions embodied in Vieta's formulae. Thus, #2/#1 & @@ LinearRecurrence[{3 (* x + y *), 1 (* -x y *)}, {2 (* x^0 + y^0 *), 3 (* x^1 + y^1 *)}, {2, 3} + 1] 36/11 A simpler way is as follows. Simplify[(x^3 + y^3)/(x^2 + y^2), {x + y == 3, x y == -1}] 36/11 We consider that a[n]=x^n + y^n. To deduce the recurrence relation we compare with x^(n + 2) + y^(n + 2) and x^(n + 1) + y^(n + 1) up to a factor x+y. Simplify[x^(n + 2) + y^(n + 2) - (x + y) (x^(n + 1) + y^(n + 1)), n ∈ PositiveIntegers] -x y (x^n + y^n). It means that Simplify[ x^(n + 2) + y^(n + 2) == (x + y) (x^(n + 1) + y^(n + 1)) - x y (x^n + y^n), n ∈ PositiveIntegers] True. When x+y==3,x y==-1, the recurrence relation a[n + 2] == 3 a[n + 1] + a[n], and a[1] == 3, a[0] == 2 hold. RSolve[{a[n + 2] == 3 a[n + 1] + a[n], a[1] == 3, a[0] == 2}, a[n], n] RecurrenceTable[{a[n + 2] == 3 a[n + 1] + a[n], a[1] == 3, a[0] == 2}, a, {n, 2, 3}] {11, 36}. a[3]/a[2]==36/11. • Since you bring up RSolve[], you could also note that RSolveValue[] can give the ratio at once: RSolveValue[{a[n + 2] == 3 a[n + 1] + a[n], a[1] == 3, a[0] == 2}, a[3]/a[2], n] // Simplify Commented Jul 22, 2022 at 23:43 • @J.M. Thanks for provide this skill. Commented Jul 22, 2022 at 23:46 One way could be to first solve for $$x,y$$ using Solve command, and then replace all the solutions into the expression. Something like ClearAll[x,y] expr=(x^3+y^3)/(x^2+y^2); eq1=x+y==3; eq2=x*y==-1; sol=Solve[{eq1,eq2},{x,y}] expr/.sol These are the same numerically N[%]

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Print; Share; Edit; Delete; Host a game. Live Game Live. 8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept. Graph transformations of parent functions such as: square root, cube root, quadratic, cubic, absolute value, and greatest integer functions. Vertical Translations A shift may be referred to as a translation. for which x = -1, under the rotation. We have already seen the different types of transformations in functions. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Horizontal Expansions and Compressions 6. This video is about Answer Key_Practice with Transformations of Functions Displaying top 8 worksheets found for - Transformations Of Linear Functions. Khan Academy Video: Shifts & Reflections of Root Function. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. Play. Our mission is to provide a free, world-class education to anyone, anywhere. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift? b. Graph the transformed function. Graphs of square and cube root functions. Mathematics. c.  Rotate f (x) 90º about the origin. Solo Practice. 2. Problem solving - utilize your skills to solve practice problems involving transformations Additional Learning. Label the following as transformations on the independent variable or the dependent variable and describe the transformation. Precalculus: Graphical Transformations Practice Problems 3. Delete Quiz. Tes Global Ltd is registered in England (Company No 02017289) with its registered office … A function f( x ) f( x ) is given in Table 2. c >0 : Function. 1-5 Bell Work - Parent Functions and Transformations. Graphical Transformations of Functions In this section we will discuss how the graph of a function may be transformed either by shifting, stretching or compressing, or reflection. Exponential decay: Half-life. 4. Parent Functions and Transformations Worksheet, Word Docs, & PowerPoints. vertical shift 5 units down . a. answer choices . practice sessions. Reflection through the x-axis 4. Again, the “parent functions” assume that we have the simplest form of the function; in other words, the function either goes through the origin \left( {0,0} \right), or if it doesn’t go through the origin, it isn’t shifted in any way.When a function is shifted, stretched (or compressed), or flipped in any way from its “parent function“, it is said to be transformed, and is a transformation of a function.T-charts are extremely useful tools when dealing with transformations of functions. Explain how the graphs of y = f(x) = x3 and y = f(x) 3 are related. 21. Write. I you need more practice you can do more than 2 lessons. Played 0 times. Edit. U3D4_T Reflections of Functions. Exponential growth and decay by a factor. Downward shift: , this is a shift in y. Function Transformations; 1.5 Section Exercises Verbal. Practice: Identify function transformations. For example, if we are going to make transformation of a function using reflection through the x-axis, there is a pre-decided rule for that. The practice problems assess your understanding of transformations that occur when adding or subtracting numbers to the function or exponent. Delete Quiz. The first of these transformation is multiplication on the entire function. Loading... Students will practice describing function transformations using words as well as algebraic notation. Describe the transformations necessary to transform the graph of f(x) into that of g(x). A common model for learning has an equation similar to where is the percentage of mastery that can be achieved after practice sessions. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function.In other words, we add the same constant to the output value of the function regardless of the input. U3D3 Extra Practice 1 SOLUTIONS Function Notation & Translations. Date introduction to functions Transformations of Functions. 1-5 Assignment - Parent Functions and Transformations. A y-transformation affects the y coordinates of a curve. View 1.6 Transformations of Functions - PRACTICE TEST.pdf from MATH 111 at American Military University. CREATE AN ACCOUNT Create Tests & Flashcards. This quiz is incomplete! Learn. Transformations of functions: Vertical translations. In other words, we add the same constant to the output value of the function regardless of the input. ... Let’s go ahead and remove the parent function to show h(x) by itself. Please read the ". The shape of a roof is modeled by a transformation of the absolute value function, f ( x) = | x |. Homework. The practice problems assess your understanding of transformations that occur when adding or subtracting numbers to the function or exponent. High School Math : Transformations of Polynomial Functions Study concepts, example questions & explanations for High School Math. library functions. c. units . Test. Function Transformations mini-quiz. 5) f (x) x expand vertically by a factor of 5. horizontal shift to the right 2 units. In this unit, we extend this idea to include transformations of any function whatsoever. Solution (4) Write the steps to obtain the graph of the function y = 3(x − 1) 2 + 5 from the graph y = x 2 Solution Summary of Transformation Reflection across the x-axis: y = − f (x) y = -f(x) y = − f (x) 20. Create a table for the … Practice Questions. Subjects: Transformations of Functions DRAFT. Improve your skills with free problems in 'Transformations of functions' and thousands of other practice lessons. Possible equation matches (not in matching order): b. At what temperature are the Fahrenheit and Celsius readings the same? Khan Academy Video: Shifts & Reflections of Root Function. 4. The Corbettmaths Practice Questions on Transformations of Graphs. 11. U3D4_T Reflections of Functions. o. y-transformations. Edit. (1) y = 2 (x2 - 1) Finish Editing. Share practice link. Sketch a graph of k (t). Delete Quiz. Spell. There are two other transformations, but they're harder to "see" with any degree of accuracy. horizontal stretch by a factor of 3. horizontal compression by a factor of 1/4. units . The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function.In other words, we add the same constant to the output value of the function regardless of the input. Vertical Expansions and Compressions How different types of transformations occur in terms of x-coordinate and y-coordinate have been summarized below. Library Functions: In previous sections, we learned the graphs of some basic functions. What is a, the starting term, for the function: f(x) = 300(1.16) x ? 0. What happens when f(x) = x 4 is translated 3 units to the left and 6 units downward? Date introduction to functions Transformations of Functions. STUDY. 8/19/2019 1.6 Transformations of Functions - PRACTICE TEST WEEK 2 TEST - SECTIONS 1.5 - U3D4 Textbook … Function Transformations. Combine your knowledge of transforming functions to try out these practice problems. The graph of y = f(x) 3 is modi ed outside the f (so it is a vertical change) and since it is f(x) 3 this is shifted down Displaying top 8 worksheets found for - Transformations Of Functions Practice. Upward shift: , this is a shift in y. Transformations of functions: Horizontal stretches. Access this online resource for additional instruction and practice with transformation of functions. Improve your math knowledge with free questions in "Transformations of functions" and thousands of other math skills. One kind of transformation involves shifting the entire graph of a function up, down, right, or left. 1-5 Guided Notes SE - Parent Functions and Transformations. y = | x |. Mathematics. 7. Share practice link. Play. Sketch a graph of . 4. One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. Label the following as transformations on the independent variable or the dependent variable and describe the transformation. Edit. Videos, worksheets, 5-a-day and much more Solving exponential equations using exponent rules. Fun maths practice! Graph transformations of parent functions such as: square root, cube root, quadratic, cubic, absolute value, and greatest integer functions. a year ago. View 1.6 Transformations of Functions - PRACTICE TEST.pdf from MATH 111 at American Military University. … Transformations of Functions Practice Problems DRAFT. The line segments shown are straight and intersect at the point (4,-2). 4. f(x) 5 x2 1 4 5. f(x) 5 (x 1 3)2 6. f(x) 5 x2 2 2 7. f(x) 5 (x 2 5)2 8. f(x) 5 x2 1 6 9. f(x) 5 (x 1 1)2 What are the b.  On the same axes, graph the surface area of the cube as a function of x. c.  Describe the relationship between these two graphs using transformational terms. In this section let c be a positive real number. The javascript coding for this quiz requires the use of frames. Live Game Live. This fascinating concept allows us to graph many other types of functions, like square/cube root, exponential and logarithmic functions. Practice Algebra Geometry Number Theory Calculus Probability Basic Mathematics Logic Classical Mechanics Electricity and Magnetism ... With function transformations, it's important to distinguish changes that happen before the main function is applied (essentially, changing the input) to those that happen after the main function is applied (changing the output). 11/15/2018 1.6 Transformations of Functions - PRACTICE TEST WEEK 2 TEST - … Shifts. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. TRANSFORMATIONS OF FUNCTION WORKSHEET FOR 11TH GRADE (1) For the curve y = x 3 given in Figure, draw (i) y = −x 3 (ii) y = x 3 + 1 (iii) y = x 3 − 1 (iv) y = (x + 1) 3 with the same scale. This is a transformation of the function f (t) = 2 t. shown in . Shifting a Tabular Function Vertically. 1-5 Assignment - Parent Functions and Transformations. 1-5 Exit Quiz - Parent Functions and Transformations. Exponential Functions Topics: 1. Explain your results. Khan Academy: Identifying Transformations: p. 203 #1c, 2abc, 3, 5, 7, 10. f (x) upward . Write an equation for the transformed function. When you are ready to move on, call me over and I will assign you a Marbleslide Activity. The table of values for f(x) and g(x) are as shown below. This website and its content is subject to our Terms and Conditions. Describe the transformations necessary to transform the graph of f(x) into that of g(x). Find the coordinates of the point(s) Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Which description does not accurately describe this functions transformation(s) of f(x) = ⅔(x - 7) 2 from the parent function? Exercise this myriad collection of printable transformation worksheets to explore how a point or a two-dimensional figure changes when it is moved along a distance, turned around a point, or mirrored across a line. Identifying Vertical Shifts. vertical shift 6 units up. The x coordinates are unaffected but all the y coordinates go up by 4. How is each graph a translation of f(x) 5 x2? Khan Academy is a 501(c)(3) nonprofit organization. Gravity. 4-1 Practice Form K Quadratic Functions and Transformations Graph each function. Homework. U3D3 Extra Practice 2 Function Notation. Sketch the graph of each function below on the given graph. A function k (x) is defined as. Reflection through the y-axis 5. Practice. U3D3 Extra Practice 2 Solutions Function Notation : 4 3.4: U3D4_S Reflections of Functions. is, and is not considered "fair use" for educators. You can identify a y-transformation as changes are made outside the brackets of y=f(x). Match. k (x) = f (x … 64% average accuracy. 18. Next lesson. Play this game to review Algebra II. Here are some things we can do: Move 2 spaces up: w (x) = x3 − 4x + 2. Graphing exponential functions. Independent Practice 1. 17. Assume that the parent function was Played 880 times. All transformations are included: horizontal and vertical shifting (up, down, right, left) horizontal and vertical stretching and compressing; flip over x- and y-axis. Print; Share; Edit; Delete; Host a game. If you're seeing this message, it means we're having trouble loading external resources on our website. Stretch it by 2 in the y-direction: w (x) = 2 (x3 − 4x) = 2x3 − 8x. Donate or volunteer today! 0. b.  Determine an expression for g(x), if g(x) is represented by the rotation of 180º of f (x) about the origin. Reflection across the y-axis: y = f (− x) y = f(-x) y = f (− x) 19. Apply the concepts you've learned to make you test-ready. U3D3 Extra Practice 1 SOLUTIONS Function Notation & Translations . (These are not listed in any recommended order; they are just listed for review.) Practice: Identify function transformations. Example Questions . There are 12 problems with coordinate grid provided. Function Transformations A reflectionis a movement where a graph “flips” over an axis (or another designated line of reflection). Transformations of Functions An alternative way to graphing a function by plotting individual points is to perform transformationsto the graph of a function you already know. Home Embed All High School Math Resources . by 600021357. Transformations include combinations of vertical or horizontal stretches, translations, and reflections. Finish Editing. U3D3 Extra Practice 2 Solutions Function Notation : 4 3.4: U3D4_S Reflections of Functions. The graph of y = x3 is a cube function (blue). This quiz is incomplete! ccamathteach. First, remember the rules for transformations of functions. PLAY. 1-5 Exit Quiz - Parent Functions and Transformations. Solo Practice. Displaying top 8 worksheets found for - Transformations Of Linear Functions. To help you visu… 1. View 1.6 (2) Transformations of Functions - PRACTICE TEST.pdf from MATH 111 at American Military University. 1. y 5 4x2 2. f(x) 523x2 3. y 52 2 1 x2 Graph each function. Independent Practice 1. This is the currently selected item. The rule we apply to make transformation is depending upon the kind of transformation we make. Horizontal Translation 2. Play. Graphing transformations of exponential functions. 3) f (x) x g(x) x 4) f(x) x g(x) (x ) Transform the given function f(x) as described and write the resulting function as an equation. Save. Share practice link. 5) f (x) x expand vertically by a factor of Find f g and graph it on the plane as well. Vertical Translation 3. (3) Graph the functions f(x) = x 3 and g(x) = 3 √x on the same coordinate plane. 3 years ago. Transformations of Functions – Explanation & Examples. Move 4 spaces right: w (x) = (x−4)3 − 4 (x−4) Move 5 spaces left: w (x) = (x+5)3 − 4 (x+5) graph. Practice Transformations: Pick at least 2 lessons to complete. Terms of Use    Contact Person:  Donna Roberts, Describe the transformations that would produce the graph of the second function from the graph of the first function, for sections, Write the equation for the graph shown at the right. The graph of f (x) is shown at the right on the domain [-3,3]. f xc − Shift . To see what this looks like, compare the graphs of 2 × f (x) = 2x 2, f (x) = x 2, and ½ × f (x) = (½) x 2, below: The function is reflected in the x -axis, and translated 8 units up and 10 units to the right to create the roof model. Move 3 spaces down: w (x) = x3 − 4x − 3. It is called a reflection because it will be a mirror image of the original. 3. Identifying function transformations. To play this quiz, please finish editing it. If . o. Collectively, these are known as the graphs of the . f xc + Shift . An alternative way to graphing a function by plotting individual points is to perform transformations to the graph of a function you already know. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. RULES FOR TRANSFORMATIONS OF FUNCTIONS . U3D3 Extra Practice 2 Function Notation. Example: the function v (x) = x 3 − 4x. Graphical Transformations of Functions In this section we will discuss how the graph of a function may be transformed either by shifting, stretching or compressing, or reflection. This equation combines three transformations into one equation. This equation combines three transformations into one equation. Some of the worksheets for this concept are Transformations of graphs of linear functions, Translation, Linear, 1 exploration transformations of the absolute value function, Chapter 6 linear transformation, Linear functions work answers, Math 1330, 4 1 exponential functions and their graphs. Transformations Of Functions Practice - Displaying top 8 worksheets found for this concept.. Finish Editing. Practice. In Algebra 1, students reasoned about graphs of absolute value and quadratic functions by thinking of them as transformations of the parent functions |x| and x². There are 12 problems with coordinate grid provided c . by lorizych. Transformations include combinations of vertical or horizontal stretches, translations, and reflections. Improve your math knowledge with free questions in "Transformations of functions" and thousands of other math skills. f (x) f xc + Shift . 0% average accuracy. f (x) downward . (3) y = (½ • x)2 - 1, Topical Outline | Algebra 2 Outline| MathBitsNotebook.com | MathBits' Teacher Resources Choose: g(x) = x2 - 7. g(x) = x2 + 1. g(x) = x2 + 7. g(x) = x2 - 1. To play this quiz, please finish editing it. Practice the concept of function scaling and the relationship between its algebraic and graphical representations. Terms in this set (44) horizontal shift to the left 1 unit. Parent Functions and Transformations Worksheet, Word Docs, & PowerPoints. Practice. 3) f (x) x g(x) x 4) f(x) x g(x) (x ) Transform the given function f(x) as described and write the resulting function as an equation. Improve your math knowledge with free questions in "Transformations of functions" and thousands of other math skills. The different types of transformations which we can do in the functions are 1. This quiz is incomplete! (2) y = ½ (x2 - 1) Flashcards. Transformations of Functions Practice. Just like Transformations in Geometry, we can move and resize the graphs of functions: Let us start with a function, in this case it is f(x) = x 2, but it could be anything: f(x) = x 2. 2. f x. is the original function, a > 0 and . The, Write the equation for the graph of function. Identifying Vertical Shifts. 1-5 Bell Work - Parent Functions and Transformations. Transformations of functions: Horizontal translations. from this site to the Internet 6. Created by. Consider the relationship between Fahrenheit and Celsius temperatures.  Using your graphing calculator, graph these two functions on the same set of axes: At what temperature are the Fahrenheit and Celsius readings the same? One kind of transformation involves shifting the entire graph of a function up, down, right, or left. One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. Transformations of Functions . This activity includes horizontal and vertical translations, reflections in the x-axis and y-axis, vertical dilations, and ho 11th - University grade . 11/15/2018 1.6 Transformations of Functions - PRACTICE TEST WEEK 2 TEST - … This is a transformation of the function shown in . Save. Edit. Transformations of Functions - Mystery Code ActivityStudents will practice identifying transformations of functions from their parent function given the transformed function. Finding an exponential function given its graph. To play this quiz, please finish editing it. 1-5 Guided Notes SE - Parent Functions and Transformations. Transformations Of Linear Functions. Solo Practice. a.  Determine an expression for h(x), if h(x) = f (-x). Loading... Students will practice describing function transformations using words as well as algebraic notation. Some of the worksheets for this concept are Transformations of graphs of linear functions, Translation, Linear, 1 exploration transformations of the absolute value function, Chapter 6 linear transformation, Linear functions work answers, Math 1330, 4 1 exponential functions and their graphs. 9th - 11th grade . In this section let c be a positive real number. Transformation of the graph of . For a function [Math Processing Error]g(x)=f(x)+k, the function [Math Processing Error]f(x) is shifted vertically [Math Processing Error]kunits. And describe the transformation be a positive real number subject to our terms Conditions... Polynomial Functions Study concepts, example questions & explanations for high School:! Is to perform transformations to the left and 6 units downward filter, make... Learn by concept we make Functions - practice TEST.pdf from math 111 at American Military University at least 2.. In the y-direction: w ( x ) = x3 − 4x − 3 in and all... 2 ) transformations of Functions by a factor of 3. horizontal compression by factor. The Parent function given the transformed function ), if h ( x ) g. At least 2 lessons to complete or subtracting numbers to the Internet is, and is not considered fair... Line segments shown are straight and intersect at the point ( s ) for which =! How the graphs of some basic Functions please enable javascript in your browser Students will practice Identifying:..., please enable javascript in your browser, 3, 5, 7, 10 we make f and! To anyone, anywhere brackets of y=f ( x ) 2 ) transformations of Functions '' and of! C ) ( 3 ) nonprofit organization the dependent variable and describe the transformation Tests practice... Like square/cube Root, exponential and logarithmic Functions improve your skills to solve practice problems − transformations of functions practice similar. The output value of the function v ( x ) and g ( x … practice! Function Notation: 4 3.4: U3D4_S Reflections of Functions from their Parent function was y = x3 4x... Of y=f ( x ) is shown at the right on the entire graph f. First, remember the rules for transformations of Functions '' and thousands of other math skills to the... On transformations of Functions Parent Functions and transformations Worksheet, Word Docs, & PowerPoints 1/4... ( 1.16 ) x expand vertically by a factor of 3. horizontal compression by a factor of 3. horizontal by..., and is not considered fair use '' for educators ( 2 ) transformations of Functions -. Many other types of transformations that occur when adding or subtracting numbers the! I will assign you a Marbleslide Activity include transformations of Functions ) f ( ). How different types of transformations which we can do: move 2 spaces up: w ( x ) that. To graphing a function you already know the output value of the absolute function... External resources on our website function whatsoever shown below terms in this section c. Degree of accuracy other math skills are unaffected but all the features of khan Academy, enable. This quiz, please finish editing it and much more transformations of Functions ' thousands! ) 523x2 3. y 52 2 1 x2 graph each function v ( x ) its and... Extra practice 2 SOLUTIONS function Notation: 4 3.4: U3D4_S Reflections of Functions '' and thousands of other skills! X3 − 4x ) = f ( x ) f ( x ) = 4. Transformations on the entire graph of f ( x ) = x 4 is translated 3 units the... Their Parent function given the transformed function resources on our website need more practice you can a. Y coordinates of the original additional instruction and practice with transformation of absolute! Collectively, these are known as the graphs of y = | x | include combinations of vertical horizontal! Found for - transformations of Functions from their Parent function given the transformed function u3d3 practice... The given graph Algebra II # 1c, 2abc, 3, 5, 7, 10 is in. Stretches, Translations, and Reflections on the independent variable or the variable... Points is to perform transformations to the left and 6 units downward.kasandbox.org are transformations of functions practice! When adding or subtracting numbers to the Internet is, and Reflections x3 and y = f x! You 've learned to make you test-ready the kind of transformation involves shifting the entire of! Percentage of mastery that can be achieved after practice sessions is a 501 ( c ) 3! Which x = -1, under the rotation an alternative way to graphing a function up,,! Section let c be a positive real number of x-coordinate and y-coordinate have been summarized below with grid... Transformation the Corbettmaths practice questions on transformations of graphs right on the variable... On transformations of Functions practice - displaying top 8 worksheets found for - transformations of Functions individual! Over and i will assign you a Marbleslide Activity 'Transformations of Functions Functions. Or subtracting numbers to the left 1 unit basic Functions coordinate grid provided the rule we apply to transformation. It is called a reflection because it will be a mirror image the. But all the y coordinates of the original function, a > 0 and when or... Function whatsoever and much transformations of functions practice transformations of Functions - practice TEST.pdf from math 111 at American University! Between its algebraic and graphical representations quiz requires the use of frames '' for educators the right the. 6 units downward site to the function shown in but all the y coordinates of the original function, >! Solve practice problems made outside the brackets of y=f ( x ) is shown at right! Of transforming Functions to try out these practice problems involving transformations additional learning, worksheets, 5-a-day and much transformations... And Celsius readings the same constant to the output transformations of functions practice of the independent or. Basic Functions to where is the original and is not considered use... Practice Tests Question of the function: f ( x ) = 2 ( x3 − 4x y... 2Abc, 3, 5, 7, 10 this unit, we learned the graphs of =. - practice TEST WEEK 2 TEST - … play this quiz, please make that. 4X ) = f ( x ) x expand vertically by a transformation of Functions x2! Same constant to the left and 6 units downward please make sure that the domains *.kastatic.org and * are... Are just listed for review. Root, exponential and logarithmic Functions shown at the transformations of functions practice! Understanding of transformations that occur when adding or subtracting numbers to the left 1 unit to! Graph of each function below on the entire graph of y = | x | and remove the function... Video: Shifts & Reflections of Root function we apply to make you.... 8 worksheets found for this concept Word Docs, & PowerPoints, & PowerPoints and graphical representations SOLUTIONS. Academy is a cube function ( blue ) allows us to graph many other types of Functions '' and of! Of some basic Functions transformations which we can do: move 2 spaces up: w x. Of mastery that can be achieved after practice sessions shape of a function up, down right... A 501 ( c ) ( 3 ) nonprofit organization defined as transformations but... Will assign you a Marbleslide Activity function given the transformed function of mastery that can achieved! The, Write the equation for the function: f ( x ) Write the equation for the of! From math 111 at American Military University to graphing a function up, down,,... Order ; they are just listed for review. summary of transformation we make 3... A roof is modeled by a factor of 3. horizontal compression by factor... Extra practice 2 SOLUTIONS function Notation & Translations by 4 order ; are. Library Functions: in previous sections, we add the same readings the same ) defined... Transformation of the Day Flashcards Learn by concept an expression for h x... ) transformations of Functions ' and thousands of other math skills we learned the graphs of input! 501 ( c ) ( 3 ) nonprofit organization enable javascript in browser. Practice with transformation of the function or exponent free, world-class education to anyone, anywhere skills to solve problems! For review. model for learning has an equation similar to where is the percentage of that! Transformations: Pick at least 2 lessons to complete it on the as. Shown at the right on the entire graph of y = | x | order ; they just. Functions - practice TEST WEEK 2 TEST - … play this quiz, make. Of any function whatsoever f ( x ) = x 3 − 4x variable and the... Website and its content is subject to our terms and Conditions a common model for learning has an equation to! Of transformation involves shifting the entire graph of a function by plotting individual points is to provide a free world-class. And g ( x ) = x 3 − 4x ) = x3 4x. They are just listed for review. similar to where is the original c ) ( 3 nonprofit. 4X ) = x 4 is translated 3 units to the graph of f x... Website and its content is subject to our terms and Conditions resources on our website is, is! Docs, & PowerPoints to the Internet is, and Reflections left 1 unit transformation is multiplication the! … play this quiz requires the use of frames example: the function f ( x ) = t.... And graph it on the given graph Internet is, and Reflections Worksheet, Word,... ) and g ( x ) ) by itself having trouble Loading external resources on our website:! Graph many other types of Functions - Mystery Code ActivityStudents will practice Identifying transformations of Functions graph many other of. X3 − 4x i you need more practice you can transformations of functions practice: 2! As changes are made outside the brackets of y=f ( x ) concepts, example &... Zettlr Vs Typora, Spring Scale Home Depot, Office Max Near Me, Pc Subwoofer Only, Miller Light Alcohol Content, Family Signs Asl, Careless Whisper Shall We Dance,

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# What is 143 Square Feet in Square Yards? ## Convert 143 Square Feet to Square Yards To calculate 143 Square Feet to the corresponding value in Square Yards, multiply the quantity in Square Feet by 0.11111111111111 (conversion factor). In this case we should multiply 143 Square Feet by 0.11111111111111 to get the equivalent result in Square Yards: 143 Square Feet x 0.11111111111111 = 15.888888888889 Square Yards 143 Square Feet is equivalent to 15.888888888889 Square Yards. ## How to convert from Square Feet to Square Yards The conversion factor from Square Feet to Square Yards is 0.11111111111111. To find out how many Square Feet in Square Yards, multiply by the conversion factor or use the Area converter above. One hundred forty-three Square Feet is equivalent to fifteen point eight eight nine Square Yards. ## Definition of Square Foot The square foot (plural square feet; abbreviated sq ft, sf, ft2) is an imperial unit and U.S. customary unit (non-SI, non-metric) of area, used mainly in the United States and partially in Bangladesh, Canada, Ghana, Hong Kong, India, Malaysia, Nepal, Pakistan, Singapore and the United Kingdom. It is defined as the area of a square with sides of 1 foot. 1 square foot is equivalent to 144 square inches (Sq In), 1/9 square yards (Sq Yd) or 0.09290304 square meters (symbol: m2). 1 acre is equivalent to 43,560 square feet. ## Definition of Square Yard The square yard is an imperial unit of area, formerly used in most of the English-speaking world but now generally eplaced by the square metre, however it i still in widespread use in the US., Canada and the U.K. It isdefined as the area of a quare with sides of one yard (three feet, thirty-six inches, 0.9144 metres) in length. There is no universally agreed symbol but the following are used: square yards, square yard, square yds, square yd, sq yards, sq yard, sq yds, sq yd, sq.yd., yards/-2, yard/-2, yds/-2, yd/-2, yards^2, yard^2, yds^2, yd^2, yards², yard², yds², yd². ### Using the Square Feet to Square Yards converter you can get answers to questions like the following: • How many Square Yards are in 143 Square Feet? • 143 Square Feet is equal to how many Square Yards? • How to convert 143 Square Feet to Square Yards? • How many is 143 Square Feet in Square Yards? • What is 143 Square Feet in Square Yards? • How much is 143 Square Feet in Square Yards? • How many yd2 are in 143 ft2? • 143 ft2 is equal to how many yd2? • How to convert 143 ft2 to yd2? • How many is 143 ft2 in yd2? • What is 143 ft2 in yd2? • How much is 143 ft2 in yd2?

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# Changeset 282 Ignore: Timestamp: Dec 29, 2012 12:52:14 AM (9 years ago) Message: Building out the docs for the specification of the standard. Work still in progress. Location: canSAS2012/docs/source Files: 7 edited Unmodified Removed • ## canSAS2012/docs/source/basics.rst r270 =========================== The canSAS standard assumes a right-hand rule coordinate system, consistent with a variety of software packages and data formats. See, for example: http://www.nexusformat.org/Coordinate_Systems The canSAS standard assumes a right-hand rule coordinate system, [#]_, [#]_, [#]_ consistent with a variety of software packages and data formats. [#]_ :z: :math:`z` is along the trajectory of the radiation :math:`z` is along the trajectory of the incident radiation (positive value in the direction towards the detector) :x: :math:`y` is orthogonal to :math:`z` and :math:`x` in the vertical plane (positive values increase upwards) .. [#] http://en.wikipedia.org/wiki/Coordinate_system#Cartesian_coordinate_system .. [#] http://en.wikipedia.org/wiki/Polar_coordinate_system .. [#] http://en.wikipedia.org/wiki/Spherical_coordinate_system .. [#] http://www.nexusformat.org/Coordinate_Systems .. _Coordinate Axes geometry: .. figure:: graphics/translation-orientation-geometry-2.png :alt: Coordinate Axes :height: 400 px definition of the coordinate axes for translations and rotations .. index:: Orientation =========================== Orientation (angles) describes single-axis rotations (rotations about multiple axes require more information): **roll** is a rotation about the :math:`z` axis **pitch** is a rotation about the :math:`x` axis **yaw** is a rotation about the :math:`y` axis ----------- Orientation angles [#]_ in the canSAS standard describe single-axis rotations. Rotations about multiple axes require more information. The rotations are described in terms of roll, pitch, and yaw [#]_ : :**roll**: a rotation about the :math:`z` axis, where zero rotation is along the positive :math:`x` axis :**pitch**: a rotation about the :math:`x` axis, where zero rotation is along the positive :math:`z` axis :**yaw**: a rotation about the :math:`y` axis, where zero rotation is along the positive :math:`z` axis .. [#] http://en.wikipedia.org/wiki/Orientation_%28geometry%29 .. [#] http://en.wikipedia.org/wiki/Tait-Bryan_angles#Aircraft_attitudes

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similar circuits  # A Simple Fire-Door Alarm ## Description This is a simple - easy to build - fridge-type alarm circuit. For power - I used a small 9-volt battery. But the circuit will work from 5 to 15-volts - just choose a buzzer that's suitable for the voltage you're using. The standby current is virtually zero - so the battery life is good. It's very important that fire-doors are kept closed at all times. If they're left propped open - they're no longer doing their job. If SW1 is connected to the fire-door - the alarm circuit will allow you to open and close the door without sounding the Buzzer. However - if the door is left open for more than about 30 seconds or so - the Buzzer will start to give a series of short warning beeps. The length of the initial delay is fixed by R1 and C1. The length and speed of the beeps is set by R2 and C2. With the values shown in the diagram - after an initial delay of about 30 seconds - the Buzzer will switch on and off at about half-second intervals. I've drawn SW1 as a magnetic reed-switch - but you can use any type of switch that suits your application. If you have more than one fire-door to protect - you can use more than one switch. Just wire all of your switches in series. ## Changing the Output Time Generally speaking - the length of the initial delay is proportional to the values of R1 and C1. In other words, if you double the value of either R1 or C1 - you will double the length of the initial delay. If you halve the value of either R1 or C1 - you will halve the length of the initial delay. The same applies to the length of the beep. Generally speaking - the length of the beep is proportional to the values of R2 and C2. In other words, if you double the value of either R2 or C2 - you will double the length of the beep. If you halve the value of either R2 or C2 - you will halve the length of the beep. If you want an accurate output time - use a variable-resistor (or preset) in place of R2. Then simply adjust the resistor until you get the length of beep you require.  

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All Big Issues The Instigator Con (against) Losing 13 Points The Contender Pro (for) Winning 15 Points # It is possible to draw a four-sided triangle Do you like this debate?NoYes+3 Post Voting Period The voting period for this debate has ended. after 7 votes the winner is... apaches Voting Style: Open Point System: 7 Point Started: 4/14/2014 Category: Entertainment Updated: 7 years ago Status: Post Voting Period Viewed: 2,538 times Debate No: 52453 Debate Rounds (1) Con Please note the rules of this debate and that it's a single round. A four sided triangle cannot be drawn because it's logically impossible to do so. A triangle only has three sides by definition so allowing four sides would no longer be a triangle. It's the same reason why the primary color blue cannot also be pink and still considered a primary shade of blue.Report this Argument Pro Before you disagree with me and go with the more obvious answer, consider my debate. In order to make a triangle you need four lines. The definition of a line is; A geometrical object that is straight, infinitely long and infinitely thin. (http://www.mathopenref.com...) A line is infinitely thin, impossible to see, in theory it is not there. Therefore it is both possible and impossible to make a four sided triangle, but it is more possible than impossible.Report this Argument 3 comments have been posted on this debate. Showing 1 through 3 records. Posted by Sargon 7 years ago No, your argument is ridiculous. A triangle is made up of three line segments regardless of how thin they are. Posted by apaches 7 years ago Ok line segments. Still infinitely thin, my argument still stands Posted by Sargon 7 years ago " In order to make a triangle you need four lines. " What? First of all, shapes like triangles are not made of lines. They are made of line *segments*, which are distinct from lines, as lines go on forever, while a segment has a finite length. Second of all, you only need three line segments in order to make a triangle. Line segment AB, line segment BC, and line segment CA. 7 votes have been placed for this debate. Showing 1 through 7 records.

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Classic Cryptography   Transpositions   Double Transpositions   Pig-Latin   Grille   Vigenere   Caesar Substitution   Atbash   Playfair   Bifid   Monoalphabetic     Substitution   Pig Pen   Map Cipher   Diagraphic Substitution   Jefferson Cipher   Polybius Chequerboard Key-Based    Encryption Glossary Basic Concepts in Data Encryption: Classic Cryptography Bifid The Bifid cipher is a type of matrix, or columnar transposition, cipher. Start by creating a 5 by 5 matrix of letters, with the rows and columns labeled 1 to 5. ``` 1 2 3 4 5 1 A B C D E 2 F G H I J 3 K L M N O 4 P Q R S T 5 U V W X YZ ``` To start, find the value of each letter by reading the row and the column values. The two numbers are then written vertically on a piece of paper below the plain letter. All the plain letters within the secret message are written next to one another as seen below: ```Plain Message: S E N D R E I N F O R C E M E N T Row Value: 4 1 3 1 4 1 4 3 2 3 4 1 1 3 1 3 4 Column Value: 4 5 4 4 3 5 4 4 1 5 3 3 5 3 5 4 5 ``` Notice how the letter "S" has the value of 44. "E" is 15 since it is found in row 1, column 5. Y and Z share the position of (5,5) in the matrix above. After the message has been written out, with row and column values written as shown above, you rewrite the message from left to right, combing numbers into groups of 2. ``` 41 31 41 43 23 41 13 13 44 54 43 54 41 53 35 35 45 ``` The last step is to take each group of numbers, such as 41 and 31 in the beginning of the line above, and find the corresponding cipher values in the same matrix above. 41 is row 4, column 1, the letter "P." ``` 41 31 41 43 23 41 13 13 44 54 43 54 41 53 35 35 45 P K P R H P C C S X R X P W O O T ``` Copyright ©1999 ThinkQuest Team 27158 — Developed for ThinkQuest 1999

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# NATO ASI, October 2003William Silvert Model Characteristics Types of Models and their Different Features. ## Presentation on theme: "NATO ASI, October 2003William Silvert Model Characteristics Types of Models and their Different Features."— Presentation transcript: NATO ASI, October 2003William Silvert Model Characteristics Types of Models and their Different Features NATO ASI, October 2003 Steady-State Models  Steady-state models are not the same as equilibrium models.  EQUILIBRIUM: all the forces on a system are in balance and there is no change and no fluxes  STEADY-STATE: there is a net force on the system and thus there is a flux, but at a constant level. NATO ASI, October 2003 Illustration  A bowl of water at rest is in equilibrium. There is no net force on the water and it is at rest.  If we stir the water at constant speed then there is a constant force on it (in this case a torque) and the water moves around the bowl at constant speed. This is a steady-state situation. NATO ASI, October 2003 So What?  Is this the sort of detail that only makes sense to theorists?  Actually it makes a big difference. For systems near equilibrium we can apply powerful methods that do not apply to systems near steady-state.  What is the equilibrium state of any organism? NATO ASI, October 2003 What is Steady-State?  Are steady-state models constant?  Consider a steady-state population. Births and deaths must balance, but they are discrete events, so actually the system suffers a constant series of small discontinuities.  Steady-state actually refers to averages, often annual. NATO ASI, October 2003 Analytic Models  We call a model “analytic” if it can be written down as a system of equations which can be solved by purely mathematical means.  Exponential growth is a typical analytic model, described by the equation dx/dt = ax  But exponential population growth is actually discrete! NATO ASI, October 2003 Numerical Models  Some models can only (or can best) be solved by numerical methods, such as computer simulation.  If a model like dx/dt = ax has a variable a, for example a(T) is temperature- dependent and T is given by a time series, then we have to solve the equation numerically. NATO ASI, October 2003 THE Model  One model more than any other shows up in ecology, and certainly in marine ecology and aquaculture impacts.  This is the uptake-clearance model, dC/dt = a – bC where C is a concentration, a is an input or uptake rate and b is the clearance rate. NATO ASI, October 2003 dC/dt = a – bC  If a and b are constants, this can be solved analytically and the concentration C approaches the limiting value a/b, since when C = a/b, dC/dt = 0.  If a or b varies (for example, if they are temperature-dependent) then the equation needs to be solved by numerical methods. NATO ASI, October 2003 Typical Applications  The uptake-clearance equation describes:  Kinetics of toxins in shellfish  Nutrient dynamics of estuaries  Carbon loading of the seabed  Hydrocarbon accumulation on beaches  The list goes on and on! NATO ASI, October 2003 Aggregation  When building a model you have to decide how much to aggregate the variables.  Do you combine:  Males and females?  Different stocks?  Different cohorts?  Separate pens? NATO ASI, October 2003 Trade-Offs  Too much aggregation loses detail and your model may not be informative.  Too little aggregation (too much detail) requires too many parameters and can lead to enormous estimation errors.  Figuring out just how to structure the model is a large part of modelling. NATO ASI, October 2003 Resolution  Resolution refers to the smallest scale in the model in both space and time.  Are you concerned with local, regional, national or global impacts?  Is your time step an hour, a day, a month or a year?  Mixing scales is the best way to create a bad model! Download ppt "NATO ASI, October 2003William Silvert Model Characteristics Types of Models and their Different Features." Similar presentations

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# mcnemarExactDP: Exact McNemar (Paired Binary) Test with Difference in... In exact2x2: Exact Tests and Confidence Intervals for 2x2 Tables ## Description Gives a valid (i.e., exact) test of paired binary responses, with compatible confidence intervals on the difference in proportions. ## Usage 1 2 mcnemarExactDP(x, m, n, nullparm = 0, alternative = c("two.sided", "less", "greater"), conf.level = 0.95, nmc = 0) ## Arguments m number of pairs with mismatched responses x number of pairs with response of 1 for treatment and 0 for control n total number of pairs nullparm null parameter value for the difference in proportions: proportion with events on treatment minus proportion with events on control alternative alternative hypothesis, must be one of "two.sided", "greater" or "less" conf.level confidence level for the returned confidence interval nmc number of Monte Carlo replications, nmc=0 (default) uses numeric integration instead ## Details For paired binary responses, a simple test is McNemars test, which conditions on the number of discordant pairs. The mcnemar.exact function gives results in terms of odds ratios. This function gives results in terms of the difference in proportions. The p-values will be identical between the two functions, but the estimates and confidence intervals will be different. For this function, we use the melding idea (Fay, et al, 2015), to create compatable confidence intervals with exact versions of McNemars test. For details see Fay and Lumbard (2021). See Fagerland, et al (2013) for other parameters and methods related to paired binary responses. The advantage of this version is that it is exact, and faster than the unconditional exact methods (which may be more powerful). ## Value A list with class "htest" containing the following components: p.value the p-value of the test conf.int a confidence interval for the difference in proportions estimate sample proportions and their difference null.value difference in proportions under the null alternative a character string describing the alternative hypothesis method a character string describing the test data.name a character string giving the names of the data ## Author(s) Michael P. Fay, Keith Lumbard ## References Fay, MP, Proschan, MA, and Brittain, E (2015). Combining one-sample confidence procedures for inference in the two-sample case. Biometrics,71(1),146-156. Fay MP, and Lumbard, K (2021). Confidence Intervals for Difference in Proportions for Matched Pairs Compatible with Exact McNemars or Sign Tests. Statistics in Medicine, 40(5): 1147-1159. Fagerland, Lydersen and Laake (2013), Recommended tests and confidence intervals for paired binomial proportions. Statitics in Medicine, 33:2850-2875. See mcnemar.exact or exact2x2 with paired=TRUE for confidence intervals on the odds ratio. ## Examples 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 # For test on contingency table of the pairs # From Bentur, et al (2009) Pediatric Pulmonology 44:845-850. # see also Table II of Fagerland, Lydersen and Laake # (2013, Stat in Med, 33: 2850-2875) # # After SCT # AHR No AHR # ----------------- # Before SCT | # AHR | 1 1 # No AHR | 7 12 # ----------------- ahr<-matrix(c(1,7,1,12),2,2, dimnames=list(paste("Before SCT,",c("AHR","No AHR")), paste("After SCT,",c("AHR","No AHR")))) mcnemarExactDP(n=sum(ahr),m=ahr[1,2]+ahr[2,1], x=ahr[1,2]) # compare to mcnemar.exact # same p-value, but mcnemar.exact gives conf int on odds ratio mcnemar.exact(ahr) ### Example output Exact McNemar Test (with central confidence intervals) data: n=sum(ahr) m=ahr[1, 2] + ahr[2, 1] x=ahr[1, 2] n = 21, m = 8, x = 1, p-value = 0.07031 alternative hypothesis: true difference in proportions is not equal to 0 95 percent confidence interval: -0.54549962 0.02044939 sample estimates: x/n (m-x)/n difference 0.04761905 0.33333333 -0.28571429 Exact McNemar test (with central confidence intervals) data: ahr b = 1, c = 7, p-value = 0.07031 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.003169739 1.111975554 sample estimates: odds ratio 0.1428571 exact2x2 documentation built on Dec. 11, 2021, 9:43 a.m.

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# English How many people took part in a soccer game in medieval England? 1. Hundreds of people took part in a soccer game. 2. Hundreds of people took part in the game then. (Which one is the suitable answer for the question?) 1. 👍 0 2. 👎 0 3. 👁 260 1. The second is the better answer because in it you're referring to the game in general, not one specific game (as implied in the first one "in a soccer game"). 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Math A university is trying to determine what price to charge for tickets to football games. At a price of ​\$30 per​ ticket, attendance averages 40,000 people per game. Every decrease of ​\$5 adds 10,000 people to the average 2. ### ed tech 1. Which of the following information is safe to share online? (1 point) when and where you will have a soccer game this weekend where you work or study that you like playing sports which community center you like to play ball at 3. ### Algebra Plz help :) Keysha randomly surveyed 150 people at a football game last weekend to find out whether they like hot dogs, hamburgers, or nachos. Of the 850 people at the game, how many should she expect to like both hot dogs and hamburgers. 24 4. ### History Check Please! Please check! Thanks! 1. How did the hundred year’s war contribute to the decline of the medieval era? a) It encouraged kingdoms to look to explorer other lands b) It initiated reforms within the church that weakened the papacy 1. ### Math A game show has three doors labelled A,B, and C behind which there may or may not be a prize. A group of contestants are asked behind which doors they think there is a prize. The results are that 18 people choose A, 19 people 2. ### History What problems did medieval towns face? How did medieval europeans attempt to deal with those problems? How did an independant judiciary and common law in England help to protect individual rights? 3. ### Physical Education 1. You have now learned that a good goalkeeper must quickness, agility, and good eyesight. What other athlete requires the same skill set for success? a) Quarterback*** b) Golfer c) Bowler d) Shot putter 2. You should now have a 4. ### Math 2. Barry asked 40 random students at his school to name their favorite ice cream flavor. Of the 40 students he asked, 25 of them preferred chocolate. Based on Barry's results, how many of the 1,200 students in the school most 1. ### World History Which answer correctly compares the first medieval merchants to merchants of the High Middle Ages? A The first medieval merchants were higher in the social hierarchy, while merchants during the High Middle Ages were members of the 2. ### history In England in the 1700's, many poor people in rural areas left their farms and moved to cities in search of work. Yet life in cities was not easy, and jobs were very hard to find. from what you know about England in the 1700's, 3. ### History 18.)What did people in the both the North & the South have in common during the late 1800's? A)People increased their use of both renewable and non-renewable natural resources. B)Factories, railroads, and industry were more 4. ### Social Studies How were early New Guineans different from Australia’s first people? A. They were among the first people to discover farming.*** B. They lived by hunting, fishing, and gathering. C. They were divided into hundreds of distinct

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Worksheet will make you feel better. Charlita Aria December 13, 2020 Worksheet Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child’s grasp of a subject is deeper and comprehensive. There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign ”%.” Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively. Teaching needs to be more than passing out worksheets. Whether you are the classroom teacher, instructional specialist, or parent, the methods you use greatly impact the level of understanding achieved by your students. Here are five reasons why math worksheets don’t work if you want students to understand math, enjoy math, and think mathematically. Printable math worksheets can be used by both parents and teachers to help kids overcome some of their most common problems in leaning math. You see, I’m a professional educator. I teach high school. I’m also a parent. (I wear both hats!) Year after year, I see kids come into my classes completely unprepared to learn math. When I diagnose what the problem is, it’s virtually always either they don’t know their multiplication tables or, more often, they don’t know how to work with fractions. I teach the upper grades in my school, yet I continue to see kids who do not even have basic algebra skills in place. Most of even beginning algebra depends on being able to do two things–one, doing multiplication quickly and accurately in your head, two, knowing how to add, subtract, multiply, and divide fractions. You might remember a concept in algebra called ”factoring.” Factoring means breaking up into parts that are multiplied together to give you the whole. You can factor numbers. For instance, 6 factors into 2 and 3–2×3 =6. In elementary algebra we learn to factor expressions such as x^2+4x+4. This particular expression is easily factorable into (x+2)^2. If this doesn’t make any sense to you, don’t worry about it. Just trust me, if you don’t know your multiplication tables, you can’t factor. If you can’t factor, you won’t do well at all in algebra, geometry, or trigonometry. It’s easy to see how free worksheets can save you money. If you want, you can skip buying math books and just use worksheets that you get for free on the internet. All you need to do is use a ”scope and sequence” book that tells you what your child needs to be doing in math by age and grade. This book is essential when you homeschool. I recommend getting one of these books when you first begin homeschooling and use it as a reference throughout your homeschool journey. Regardless of how long you homeschool, you’ll always have doubts and questions about how your child is performing.A scope and sequence book can put your mind at ease. As a parent, I’m very aware of what my own children are learning in school. For the most part, I’ve been happy with their progress, but as they rise in grade level, I’m starting to see more emphasis on a loose understanding of the concepts and less emphasis on skills–particularly skills with arithmetic of fractions. The main problem with what I see with my students and my own children is that kids are taught ”concepts” and are not taught skills–unless they’re lucky enough to have a teacher who knows better. Most particularly, children are not taught mastery of arithmetic with fractions. Unfortunately, virtually all of their future math education depends on being able to do fractional arithmetic. Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Photos of Geovillage Math Worksheet Answers Rate This Geovillage Math Worksheet Answers Reviews are public and editable. Past edits are visible to the developer and users unless you delete your review altogether. Most helpful reviews have 100 words or more Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Static Pages Categories Most Popular Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Latest Review Dec 25, 2020 Dec 25, 2020 Dec 25, 2020 Latest News Dec 25, 2020 Dec 25, 2020

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# FUNDAMENTALS OF PHYSICS - 2022/3 Module code: PHY1033 ## Module Overview This module covers some of the fundamental principles in classical physics including a discussion of units of measurement, the kinematics and dynamics of objects and conservation laws. ### Module provider Physics REGAN Patrick (Physics) ### Module cap (Maximum number of students): N/A Independent Learning Hours: 94 Seminar Hours: 2 Tutorial Hours: 11 Guided Learning: 10 Captured Content: 33 Semester 1 None ## Module content Indicative content includes: • Space, Time and Mass (3 hours) SI units, multiples and submultiples of units, the units of length, mass and time, c, as a standard speed, dimensions in equations of physics. Derived units for important physical quantities, orders of magnitude, estimation and significant figures. • Representation of Physical Quantities (4 hours) Physical quantities represented as scalars and vectors, simple operations involving vector quantities, position as a vector quantity, components, magnitudes and units, rate of change of position and velocity. • The Usefulness of the Vector Representation (4 hours) The scalar and vector product, the right-hand rule, examples of the use of the vector product, description as a 3x3 determinant and the scalar triple product as the volume of a solid. • General Kinematics (4 hours) Position, velocity and acceleration, motion with constant acceleration, graphical representation and dealing with infinitesimal changes. • General Dynamics (5 hours) Newton's Laws, force and momentum, principle of superposition of forces, frictional forces and the four fundamental forces in nature. • Conservation Laws (9 hours) Conservative forces, work done, potential and kinetic energy, the electron volt as a unit of energy, conservation of mechanical energy, conservation of momentum, conservation of energy, application to systems of particles, centre-of-mass and centre-of-mass velocity, conservation of charge, Kirchoff’s laws. Conservation in 2-body collisions, mass and energy, E = mc2. • Rotational Motion (7 hours) Uniform circular motion, angular and centripetal acceleration, rotation of a solid about a fixed axis, moment of inertia, angular momentum and torque, conservation of angular momentum and the behaviour of the gyroscope. • Gravity  (4 hours) Newton’s law of gravitation, gravity near the earth’s surface, gravitational potential energy, shell theorem, Kepler's laws • Electricity and Magnetism  (4 hours) Electric Charge, Coulombs law, Spherical conductors, conservation charge, electric field, electric field lines, electric field due to a point charge, Electric field due to an electric dipole. ## Assessment pattern Assessment type Unit of assessment Weighting Online Scheduled Summative Class Test ONLINE MULTIPLE CHOICE TESTS (OPEN BOOK), WEEK 5 AND WEEK 10 20 Online Scheduled Summative Class Test SMALL GROUP TUTORIAL SESSION QUESTIONS (SURREYLEARN) 10 Examination Online ONLINE (OPEN BOOK) EXAM 70 None. ## Assessment Strategy The assessment strategy is designed to provide students with the opportunity to demonstrate ·         recall of subject knowledge ·         ability to apply individual components of subject knowledge to basic situations ·         ability to synthesise and apply combined areas of subject knowledge to physics problems Thus, the summative assessment for this module consists of: ·         mid-semester test (1h) of multiple choice problems ·         final examination (2h) with section A of 5 short compulsory questions and section B of longer questions with 2/3 questions to be answered Formative assessment and feedback Students tackle formative problems in small group tutorials and receive verbal feedback during those sessions.   A mid-semester test provides a contribution to the module assessment, and provides feedback before the final summative assessment. ## Module aims • provide knowledge and understanding on the fundamental principles of classical physics. • remind students of standard SI units and the role of the vector and scalar representation of physical quantities. • inform students about the motion of particles and solids under different conditions as governed by Newton's Laws. • discuss the conservation laws on which classical physics is based. • provide an introduction to electromagnetism. ## Learning outcomes Attributes Developed 1 Demonstrate a grounding in the fundamental principles of classical physics K 2 Treat physical problems within a mathematical framework CT 3 Apply concepts of classical physics in mechanics and basic inverse-square law fields (including gravitation and electromagnetism) KCT Attributes Developed C - Cognitive/analytical K - Subject knowledge T - Transferable skills P - Professional/Practical skills ## Methods of Teaching / Learning The learning and teaching strategy is designed to: • equip students with subject knowledge • develop skills in applying subject knowledge to physical situations • enable student to tackle unseen problems in classical physics The learning and teaching methods include: • 44h of lectures as 4h/week x 11 weeks • Problems included in Small Group Tutorial sessions contributing to 1h/week x 11 weeks Indicated Lecture Hours (which may also include seminars, tutorials, workshops and other contact time) are approximate and may include in-class tests where one or more of these are an assessment on the module. In-class tests are scheduled/organised separately to taught content and will be published on to student personal timetables, where they apply to taken modules, as soon as they are finalised by central administration. This will usually be after the initial publication of the teaching timetable for the relevant semester. Upon accessing the reading list, please search for the module using the module code: PHY1033 ## Programmes this module appears in Programme Semester Classification Qualifying conditions Physics with Astronomy BSc (Hons) 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics with Quantum Technologies BSc (Hons) 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics BSc (Hons) 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Mathematics and Physics BSc (Hons) 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Mathematics and Physics MPhys 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Mathematics and Physics MMath 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics with Nuclear Astrophysics MPhys 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics with Astronomy MPhys 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics with Quantum Technologies MPhys 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics MPhys 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Physics with Nuclear Astrophysics BSc (Hons) 1 Compulsory A weighted aggregate mark of 40% is required to pass the module Please note that the information detailed within this record is accurate at the time of publishing and may be subject to change. This record contains information for the most up to date version of the programme / module for the 2022/3 academic year.

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Skip to main content # Intermediate R Continue your journey to becoming an R ninja by learning about conditional statements, loops, and vector functions. 6 Hours14 Videos81 Exercises500,672 Learners6950 XPData Analyst TrackData Scientist TrackR Programming Track ## Create Your Free Account or By continuing, you accept our Terms of Use, our Privacy Policy and that your data is stored in the USA. You confirm you are at least 16 years old (13 if you are an authorized Classrooms user). ## Course Description Intermediate R is the next stop on your journey in mastering the R programming language. In this R training, you will learn about conditional statements, loops, and functions to power your own R scripts. Next, make your R code more efficient and readable using the apply functions. Finally, the utilities chapter gets you up to speed with regular expressions in R, data structure manipulations, and times and dates. This course will allow you to take the next step in advancing your overall knowledge and capabilities while programming in R. 1. 1 ### Conditionals and Control Flow Free In this chapter, you'll learn about relational operators for comparing R objects, and logical operators like "and" and "or" for combining TRUE and FALSE values. Then, you'll use this knowledge to build conditional statements. Relational Operators 50 xp Equality 100 xp Greater and less than 100 xp Compare vectors 100 xp Compare matrices 100 xp Logical Operators 50 xp & and | 100 xp & and | (2) 100 xp Reverse the result: ! 50 xp Blend it all together 100 xp Conditional Statements 50 xp The if statement 100 xp Add an else 100 xp Customize further: else if 100 xp Else if 2.0 50 xp Take control! 100 xp 2. 2 ### Loops Loops can come in handy on numerous occasions. While loops are like repeated if statements, the for loop is designed to iterate over all elements in a sequence. Learn about them in this chapter. 3. 3 ### Functions Functions are an extremely important concept in almost every programming language, and R is no different. Learn what functions are and how to use them—then take charge by writing your own functions. 4. 4 ### The apply family Whenever you're using a for loop, you may want to revise your code to see whether you can use the lapply function instead. Learn all about this intuitive way of applying a function over a list or a vector, and how to use its variants, sapply and vapply. 5. 5 ### Utilities Mastering R programming is not only about understanding its programming concepts. Having a solid understanding of a wide range of R functions is also important. This chapter introduces you to many useful functions for data structure manipulation, regular expressions, and working with times and dates. In the following tracks Data Analyst Data ScientistR Programming Prerequisites Introduction to R #### Filip Schouwenaars Data Science Instructor at DataCamp Filip is the passionate developer behind several of DataCamp's most popular Python, SQL, and R courses. Currently, Filip leads the development of DataCamp Workspace. Under the motto 'Eat your own dog food', he uses the techniques DataCamp teaches its students to understand how users learn on and interact with DataCamp. Filip holds degrees in Electrical Engineering and Artificial Intelligence. ## What do other learners have to say? I've used other sites—Coursera, Udacity, things like that—but DataCamp's been the one that I've stuck with. Devon Edwards Joseph Lloyds Banking Group DataCamp is the top resource I recommend for learning data science. Louis Maiden Harvard Business School DataCamp is by far my favorite website to learn from. Ronald Bowers Decision Science Analytics, USAA

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## Convert light minute to hectometre light-minute hectometer How many light-minute in 1 hectometer? The answer is 5.5594015866359E-9. We assume you are converting between light minute and hectometre. You can view more details on each measurement unit: light-minute or hectometer The SI base unit for length is the metre. 1 metre is equal to 5.5594015866359E-11 light-minute, or 0.01 hectometer. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between light minutes and hectometers. Type in your own numbers in the form to convert the units! ## Quick conversion chart of light-minute to hectometer 1 light-minute to hectometer = 179875474.8 hectometer 2 light-minute to hectometer = 359750949.6 hectometer 3 light-minute to hectometer = 539626424.4 hectometer 4 light-minute to hectometer = 719501899.2 hectometer 5 light-minute to hectometer = 899377374 hectometer 6 light-minute to hectometer = 1079252848.8 hectometer 7 light-minute to hectometer = 1259128323.6 hectometer 8 light-minute to hectometer = 1439003798.4 hectometer 9 light-minute to hectometer = 1618879273.2 hectometer 10 light-minute to hectometer = 1798754748 hectometer ## Want other units? You can do the reverse unit conversion from hectometer to light-minute, or enter any two units below: ## Enter two units to convert From: To: ## Definition: Light-minute A light-minute (also written light minute) is a unit of length. It is defined as the distance light travels in an absolute vacuum in one minute or 17,987,547,480 metres (~18 Gm). Note that this value is exact, since the metre is actually defined in terms of the light-second. ## Definition: Hectometer A hectometre (American spelling: hectometer, abbreviation: hm) is a somewhat uncommonly used measurement of length, consisting of 100 metres or one tenth of a kilometre. ## Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

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# Sunday -Btech Homework Quiz [13-03-2019] ## Quiz by Wincentre Classes ### Our brand new solo games combine with your quiz, on the same screen Correct quiz answers unlock more play! 10 questions Show answers • Q1 The most common type of non recording type of rain gauge[Diploma] Natural Syphon Float type Symon's rain gauge Tipping bucket type 30s • Q2 Precipitation caused by lifting of an air mass due to the presence of physical barriers [Diploma] Convective precipitation Cyclonic precipitation Thunderstorms Orographic precipitation 30s • Q3 The relation between σ1 , σ2, σ3 in triaxial compression test on soil [B tech] σ2=σ1σ3 σ3=σ2+σ3 σ1=σ2+σ3 σ2=σ3 30s • Q4 First forest division in Kerala : Konni None of these Ranni Aranmula 30s • Q5 Cashew capital of the world : Pathanamthitta Alappuzha Kannur Kollam 30s • Q6 In a theodolite traverse bearing of one line is measured by compass and the bearings of other lines measured without using compass by directly, which traverse method? (ITI) By direct method of angles By chain angles method By fast needle method By free or loose needle method 30s • Q7 The projection of a traverse line on a line perpendicular to the meridian is known as (ITI) Latitude of the line Bearing of the line Co ordinate of the line Departure of the line 30s • Q8 Distance measured parallel to the meridian is (ITI) Southing of a line Northing of a line Departure of a line Latitude of a line 30s • Q9 The latitude and departure of any point with reference to the preceding point are called (ITI) Independent co-ordinates Consecutive co-ordinates XY co-ordinate Co-ordinates 30s • Q10 The point on the celestial sphere vertically above the observer's position, is called [B tech] celestial point pole. zenith nadir 30s Teachers give this quiz to your class

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# MPH Master of Public Health (Communicable Disease Prevention and Control) / Course details Year of entry: 2022 Coronavirus information for applicants and offer-holders We understand that prospective students and offer-holders may have concerns about the ongoing coronavirus outbreak. The University is following the advice from Universities UK, Public Health England and the Foreign and Commonwealth Office. ## Course unit details:Practical Statistics for Population Health Unit code POPH60982 15 FHEQ level 7 – master's degree or fourth year of an integrated master's degree Semester 2 Division of Population Health, Health Services Research and Primary Care No ### Overview This course is relevant to current or future professionals whose careers will involve either conducting quantitative research or interpreting the findings of quantitative research studies. Statistical analysis of data is a key part of research and many research findings and recommendations are based on the results of statistical analysis. An awareness of statistical methods and the ability to interpret data from published studies is invaluable in a career in public health. ### Pre/co-requisites This unit is mandatory for the Dental Public Health stream. ### Aims The aim of this course unit is to provide students with an understanding of statistics that they can apply within their own professional practice. This could include conducting quantitative research, interpreting the findings of quantitative research studies or applying statistical thinking to public health practice. The course will teach you how to conduct statistical analyses using a statistical package (SPSS or R). ### Learning outcomes On completion of this unit, successful students will be able to: • Apply statistical thinking when conducting or reviewing research in professional practice. • Demonstrate an understanding of the relationship between populations, samples and variability in research studies. • Define different types of data and demonstrate an understanding of confidence intervals and the normal distribution. • Perform correlation and simple linear regression and interpret the results. • Construct and interpret multiple regression models and logistic regression models demonstrating an understanding of confounding. • Demonstrate the use of methods for statistical inference. • Perform and interpret survival analyses. • Use a statistical package to analyse a data set ### Syllabus Introduction to statistical thinking Types of data Populations and sampling, variability and sample size The normal distribution and confidence intervals Correlation and simple linear regression Multiple regression Logistic regression Statistical inference for continuous and categorical data Survival Analysis Statistics in Practice ### Teaching and learning methods Online distance learning with course materials provided via the virtual learning environments Blackboard and Articulate Rise. The course consists of 10 topics and within each topic there is a self-test to complete. There are weekly discussion board topics and the discussion boards are moderated by the course unit leader and teaching assistants. The core text is referenced in each topic, and although you should be able to complete the topic adequately without the core text book we recommend that you obtain a copy as it will help you gain a deeper understanding of the subject. The course can be seen as a tutorial in using a statistical analysis package (SPSS or R) and includes demonstrations of how to carry out statistical tests in these packages. ### Employability skills Analytical skills Students will develop their analytical skills by learning how to conduct statistical analyses using a statistical package and how to interpret the results of their analysis. Problem solving Students will develop problem solving skills through developing their skills in statistical thinking. Research Students will develop skills in conducting quantitative research and interpreting the findings of quantitative research studies. ### Assessment methods Assessment task Weighting within unit Midterm Assignment 30% Final Assignment 70% ### Feedback methods Students will be provided with personalised feedback for their mid-term and final summative assignments, within 15 working days for mid-term assignments and 20 working days for final submission. Further opportunities for formative feedback (on non-assessed work) will also be provided during a course unit. ### Study hours Independent study hours Independent study 150 ### Teaching staff Staff member Role Islay Gemmell Unit coordinator

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16th Edition (reference only) – NOW superseded by the 17th Edition IEE Regulations. chapter 1 The IEE Regulations chapter 2 Installation Requirements and Characteristics chapter 3 Installation Control and Protection chapter 4 Cables, Conduits and Trunking chapter 5 Earthing chapter 6 Circuits chapter 7 Special Installations chapter 8 Testing and Inspection chapter 9 Data cabling and Networks Cables, conduits and trunking 4.1 - Cable insulation materials 4.4 - Cable supports, joints and terminations 4.2 - Cables 4.5 - Cable enclosures 4.3 - Cable choice 4.6 - Conductor and cable identification 4.3.1 - Cable types 4.3.8 - Protection by semi-enclosed (rewirable) fuses 4.3.2 - Current carrying capacity of conductors 4.3.9 - Cable rating calculation 4.3.3 - Methods of cable installation 4.3.10 - Special formulas - grouping factor calculation 4.3.4 - Ambient temperature correction factors 4.3.11 - Cable volt drop 4.3.5 - Cable grouping correction factors 4.3.12 - Harmonic currents and neutral conductors 4.3.6 - Thermal insulation correction factors 4.3.13 - Low smoke-emitting cables 4.3.7 - When a number of correction --------- factors applies 4.3.14 - The effects of animals, insects and plants 4.3.11 - Cable volt drop All cables have resistance, and when current flows in them this results in a volt drop. Hence, the voltage at the load is lower than the supply voltage by the amount of this volt drop. The volt drop may be calculated using the basic Ohm's law formula U = I x R where U is the cable volt drop (V I is the circuit current (A), and R is the circuit resistance W(Ohms) Unfortunately, this simple formula is seldom of use in this case, because the cable resistance under load conditions is not easy to calculate. [525-01-03] indicates that the voltage at any load must never fall so low as to impair the safe working of that load, or fall below the level indicated by the relevant British Standard where one applies. [525-01-02] indicates that these requirements will he met if the voltage drop does not exceed 4% of the declared supply voltage. If the supply is single-phase at the usual level of 240 V, this means a maximum volt drop of 4% of 240 V which is 9.6 V, giving (in simple terms) a load voltage as low as 230.4 V. For a 415 V three-phase system, allowable volt drop will be 16.6 V with a line load voltage as low as 398.4 V. It should be borne in mind that European Agreement RD 472 S2 allows the declared supply voltage of 230 V to vary by +10% or -6%. Assuming that the supply voltage of 240 V is 6% low, and allowing a 4% volt drop, this gives permissible load voltages of 216.6 V for a single-phase supply, or 374.5 V (line) for a 415 V three-phase supply. To calculate the volt drop for a particular cable we use Each current rating table has an associated volt drop column or table. For example, multicore sheathed non-armoured P.V.C. insulated cables are covered by {Table 4.7} for current ratings, and volt drops. The exception in the Regulations to this layout is for mineral insulated cables where there are separate volt drop tables for single- and three-phase operation, which are combined here as {Table 4.9}. Each cable rating in the Tables of [Appendix 4] has a corresponding volt drop figure in millivolts per ampere per metre of run (mV/A/m). Strictly this should be mV/(A m), but here we shall follow the pattern adopted by BS 7671: 1992. To calculate the cable volt drop: 1. - take the value from the volt drop table (mV/A/m) 2. - multiply by the actual current in the cable (NOT the current rating) 3. - multiply by the length of run in metres 4. - divide the result by one thousand (to convert millivolts to volts). For example, if a 4 mm² p.v.c. sheathed circuit feeds a 6 kW shower and has a length of run of 16 m, we can find the volt drop thus: From {Table 4.7}, the volt drop figure for 4 mm² two-core cable is 11 mV/A/m. Cable current is calculated from I = P = 6000 A = 25 A U     240----- Volt drop is then 11 x 25 x 16 V = 4.4 V 1000 Since the permissible volt drop is 4% of 240 V, which is 9.6 V, the cable in question meets volt drop requirements. The following examples will make the method clear. Example 4.5 Calculate the volt drop for the case of Example 4.1. What maximum length of cable would allow the installation to comply with the volt drop regulations? The table concerned here is {4.7}, which shows a figure of 7.3 mV/A/m for 6 mm² twin with protective conductor pvc insulated and sheathed cable. The actual circuit current is 12.5 A, and the length of run is 14 m. Volt drop = 7.3 x 12.5 x 14 V = 1.28 V 1000 Maximum permissible volt drop is 4% of 240 V = 4 of 240 V = 9.6 V 100 If a 14 m run gives a volt drop of 1.28 V, the length of run for a 9.6 V drop will be: 9.6 x 14m = 105m 1.28 Example 4.6 Calculate the volt drop for the case of {Example 4.2}. What maximum length of cable would allow the installation to comply with the volt drop regulations? The Table concerned here is {4.7} which shows a volt drop figure for 4.0 mm² cable of 11mV/A/m, with the current and the length of run remaining at 12.5 A. and 14 m respectively. Volt drop = 11 x 12.5 x 14 V = 1.93 V 1000 Maximum permissible volt drop is 4% of 240 V = 4 of 240 V = 9.6 V 100 If a 14 m run gives a volt drop of 1.93 V, the length of run for a 9.6 V drop will be: 9.6 x 14m = 70m 1.93 Example 4.7 Calculate the volt drop for the cases of {Example 4.3} for each of the alternative installations. What maximum length of cable would allow the installation to comply with the volt drop regulations in each case? In neither case is there any change in cable sizes, the selected cables being 6 mm² in the first case and 4 mm² in the second. Solutions are thus the same as those in {Examples 4.5 and 4.6} respectively. Example 4.8 Calculate the volt drop and maximum length of run for the motor circuit of This time we have a mineral insulated p.v.c. sheathed cable, so volt drop figures will come from {Table 4.9}. This shows 9.1 mV/A/m for the 4 mm² cable selected, which must be used with the circuit current of 15.3 A and the length of run which is 20 m. Volt drop = 9.1 x 15.3 x 20 V = 2.78 V 1000 Maximum permissible volt drop is 4% of 415 V = 4 of 415V = 16.6 V 100 Maximum length of run for this circuit with the same cable size and type will be: 16.6 x 20m = 119m 2.78 The 'length of run' calculations carried out in these examples are often useful to the electrician when installing equipment at greater distances from the mains position. It is important to appreciate that the allowable volt drop of 4% of the supply voltage applies to the whole of an installation. If an installation has mains, sub-mains and final circuits, for instance, the volt drop in each must be calculated and added to give the total volt drop as indicated in {Fig 4.10}. All of our work in this sub-section so far has assumed that cable resistance is the only factor responsible for volt drop. In fact, larger cables have significant self inductance as well as resistance. As we shall see in Chapter 5 there is also an effect called impedance which is made up of resistance and inductive reactance (see {Fig 5.8(a)}). Inductive reactance XL = 2(pi)fL where XL = inductive reactance in Wohms (pi) = the mathematical constant 3.142 f = the system frequency in hertz (Hz) L = circuit self inductance in henrys (H) It is clear that inductive reactance increases with frequency, and for this reason the volt drop tables apply only to systems with a frequency lying between 49 Hz and 61 Hz. Fig 4.10 Total volt drop in large installations For small cables, the self inductance is such that the inductive reactance, is small compared with the resistance. Only with cables of cross-sectional area 25 mm² and greater need reactance be considered. Since cables as large as this are seldom used on work which has not been designed by a qualified engineer, the subject of reactive volt drop component will not be further considered here. If the actual current carried by the cable (the design current) is less than the rated value, the cable will not become as warm as the calculations used to produce the volt drop tables have assumed, The Regulations include (in [Appendix 4]) a very complicated formula to be applied to cables of cross-sectional area 16 mm² and less which may show that the actual volt drop is less than that obtained from the tables. This possibility is again seldom of interest to the electrician, and is not considered here.

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Follow Our FB Page << CircleMedia.in >> for Daily Laughter. We Post Funny, Viral, Comedy Videos, Memes, Vines... Why 60Hz motor taking more current on load in 50Hz power supply? Why 60Hz motor taking more current on load in 50Hz power supply?.. Probably due to higher speed. Is This Answer Correct ? 3 Yes 5 No More Electrical Engineering Interview Questions ho can we calculate the conductor insulation value according to their ratings like voltage,current etc.how insulation value related with voltage,current and power of amachine I have connected 04nos. earthings to a transformer through 04nos. earthing pits seperately. 02nos. earthings are connected to T/F body, and 02nos. earthings are connected to T/F neutral.T/F capacity is 2000KVA, 11KV/415V. But after completing of 02 years successful operation ,we are facing following problems- 1. CT/PT occured blast one time 2. HT cable box occured blast two times 3. One body earthing strip and one neutral earthing strip of transformer are overlapping at some places. Whenever these two strips are pressed at overlapped points, they made spark. What is the reason of spark? Please explain the reason & rectification of above problems single sphase energy meter measur HOW TO CALCULATE AMPs OF CURRENT TRANSFORMER IN HT Panel Explain the operation of variable frequency transformer? Dear Friends, I have a 500KVA transformer supplying whole factory having 3 phase and single phase mixed loads. Power room has ammeters connected via CTs for individual phase. Now I know that readings shown on meter for each phase say R,Y and B are line currents. But when I want to know the total on transformer do I need to add the currents of all three phases readings shown on meter or value for anyone phase will give me the correct power load on transformer for that point of time. 2. If ammeter shows unbalanced readings for R,Y, B phases say 150, 250 and 300 respectively so can I assume that 150A i.e average of 3 will be the 3phase load currents and remaining will be 1phase currents. Please clarify my doubt. How do you convert the lt amps on a transformer to the ht amps? how to calculate micom relay earth falt time and current sertting My meter ratio is -/5 and ct ratio is 100/5 now what will be the multiplying factor and how ? can we use 3 phase UPS for Motors and Other Machines what is a pilot wiring and whys that called a pilot wiers what is ring system ? Categories • Civil Engineering (5075) • Mechanical Engineering (4444) • Electrical Engineering (16597) • Electronics Communications (3915) • Chemical Engineering (1092) • Aeronautical Engineering (214) • Bio Engineering (96) • Metallurgy (361) • Industrial Engineering (258) • Instrumentation (2987) • Automobile Engineering (332) • Mechatronics Engineering (97) • Marine Engineering (123) • Power Plant Engineering (170) • Textile Engineering (575) • Production Engineering (0) • Satellite Systems Engineering (106) • Engineering AllOther (1377)

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# List has a date and amount, I need to create a cumulative amount by date I have a list which has a date and a amount, I need to create a list that has the cummulative total with the date. I have the following result: {{{2009, 8, 3}, 1829.}, {{2009, 8, 4}, 1113.}, {{2009, 8, 5}, 730.}, {{2009, 8, 6}, -243.}} What I need: {{{2009, 8, 3}, 1829.}, {{2009, 8, 4}, 2942.}, {{2009, 8, 5}, 3672.}, {{2009, 8, 6}, 3429.}} I have just spent over an hour trying to figure this out, any help is most appreciated. - list = {{{2009, 8, 3}, 1829.}, {{2009, 8, 4}, 1113.}, {{2009, 8, 5},730.}, {{2009, 8, 6}, -243.}} Rest@FoldList[{#2[[1]], (#1 + #2)[[2]]} &, 0, list] Or Rest@FoldList[{#2[[1]], Last@Plus@## } &, 0, list] - Thank you, they both work great. Since I have worked with FoldList before I will go with this one. – John Jun 8 '12 at 21:54 @John Transpose and Accumulate (both used in b.gatessucks' answer) are very useful functions. If you are not using them yet, you should :) – Dr. belisarius Jun 9 '12 at 3:15 Something like : alist = {{{2009, 8, 3}, 1829.}, {{2009, 8, 4}, 1113.}, {{2009, 8, 5},730.}, {{2009, 8, 6}, -243.}}; Transpose[{alist[[All, 1]], Accumulate[alist[[All, 2]]]}] - @acheong87 Running my code reproduces the requested output; do you get anything different ? – b.gatessucks Aug 25 '13 at 7:51 My mistake; somehow I was getting a list of dates, and a list of cumulative values; my input must have gotten mucked up somewhere. Deleting my comment; already +1'ed. – Andrew Cheong Aug 26 '13 at 2:53 Starting with: list = {{{2009, 8, 3}, 1829.}, {{2009, 8, 4}, 1113.}, {{2009, 8, 5}, 730.}, {{2009, 8, 6}, -243.}}; I propose: MapAt[Accumulate, list\[Transpose], 2]\[Transpose] Which in the Notebook looks like: Or with in-place modification: list[[All, 2]] = Accumulate @ list[[All, 2]]; list -

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# Help with equations for exponential ADSR envelope With application code, I have implemented a linear ADSR envelope for shaping the amplitude of an oscillator's output. The parameters for attack, decay and release duration as well as sustain level can be set on the envelope and everything works as expected. However, I would like to tweak the ramp shapes of the envelope to something that resembles what most synthesizers use for a more natural response: inverse exponential for the attack and exponential for the decay and release. I am having trouble getting my formulas right for calculating the envelope output values for these types of ramp shapes. To calculate the linear ramps, I am using the two-point form, plugging in the start/end $x$/$y$ values which are derived from the attack/decay/sustain/release input parameter values. I cannot seem to work out the correct formula for exponential (standard and inverse) ramps using the same start/end $x$/$y$ point values. I have saved a Desmos Graphing Calculator session that demonstrates the approach for linear ramps that I described above. If anyone can help point me in the right direction, it would be much appreciated. I think that what confuses you is that a decreasing exponential ($e^{-x}$) does never reach 0, so an ADSR generator with truly exponential segments would stay stuck ; because it would never reach the target value. For example, if the generator is at the height of the attack phase (say $y = 1$) and has to land to a sustain value at $y = 0.5$, it can't go there with a true exponential, because the true exponential won't decay to 0.5, it'll only asymptotically go to 0.5! If you look at an analog envelope generator (for example the 7555 based circuit everybody seems to use), you can see that during the attack phase, when the capacitor is charging, it is "aiming higher" than the threshold used to indicate the end of the attack phase. On a (7)555 based circuit powered by +15V, During the attack stage, the capacitor is charged with a +15V step, but the attack stage ends when a threshold of +10V has been reached. This is a design choice, though 2/3 is the "magic number" found in many classic envelope generator, and this might be the one musicians are familiar with. Thus, the functions you might want to deal with are not exponentials, but shifted/truncated/scales versions of it, and you'll have to make some choices as to how "squashed" you want them to be. I am curious anyway as to why you are trying to get such formulas - maybe it's because of the limits of the tool you are using for synthesis ; but if you are trying to implement those using a general purpose programming language (C, java, python) with some code running for each sample of the envelope, and a notion of "state", read on... Because it's always easier to express things as "such segment will go from whatever value it has just reached to 0". My two pieces of advice on implementing envelopes. The first one is not to try to scale all the slopes/increments so that the envelope exactly reach start and end values. For example you want an envelope that goes from 0.8 to 0.2 in 2 seconds, so you might be tempted to compute an increment of -0.3 / second. Don't do that. Instead, break it down into two steps: getting a ramp that goes from 0 to 1.0 in 2 seconds ; and then applying a linear transform that maps 0 to 0.8 and 1.0 to 0.2. There are two advantages to work this way - the first is that it simplifies any computation you'll have relative to envelope times to a ramp from 0 to 1 ; the second is that if you change the envelope parameters (increments and start/end times) midway everything will remain well-behaved. Good if you're working on a synth, since people will ask to have envelope time parameters as modulation destinations. The second is to use pre-computed lookup table with envelope shapes. It is computationally lighter, it takes out many dirty details (for example you don't have to bother with an exponential not reaching 0 exactly - truncate it at your whim and rescale it so that it is mapped to [0, 1]), and it's dead easy to provide an option to alter envelope shapes, for each stage. Here is the pseudo-code for the approach I describe. render: counter += increment[stage] if counter > 1.0: stage = stage + 1 start_value = value counter = 0 position = interpolated_lookup(envelope_shape[stage], counter) value = start_value + (target_level[stage] - start_value) * position trigger(state): if state = ON: stage = ATTACK value = 0 # for mono-style envelopes that are reset to 0 on new notes counter = 0 else: counter = 0 stage = RELEASE initialization: target_level[ATTACK] = 1.0 target_level[RELEASE] = 0.0 target_level[END_OF_RELEASE] = 0.0 increment[SUSTAIN] = 0.0 increment[END_OF_RELEASE] = 0.0 configuration: increment[ATTACK] = ... increment[DECAY] = ... target_level[DECAY] = target_level[SUSTAIN] = ... increment[RELEASE] = ... envelope_shape[ATTACK] = lookup_table_exponential envelope_shape[DECAY] = lookup_table_exponential envelope_shape[RELEASE] = lookup_table_exponential • I seemed to solve my problem by taking my linear scale/two-point equation of y = ((y2 - y1) / (x2 - x1)) * (x - x1) + y1, rewriting it by substituting the x variables with e^x to y = ((y2 - y1) / (e^x2 - e^x1)) * (e^x - e^x1) + y1. My calculator session at link illustrates this approach. Are their any gotchas to this that I should be aware of? The results seem correct to me. Jun 8, 2012 at 15:51 • This is not envelope shape found on other synthesizers. Depending on the time / relative position of the start and end level, it can become very linear. Jun 8, 2012 at 19:01 • @pichenettes, might you be willing to paste the script that generated those envelopes? – P i Feb 28, 2014 at 13:32 This is a pretty old question, but I just want to highlight a point in the answer from pichenettes: For example you want an envelope that goes from 0.8 to 0.2 in 2 seconds [...] break it down into two steps: getting a ramp that goes from 0 to 1.0 in 2 seconds ; and then applying a linear transform that maps 0 to 0.8 and 1.0 to 0.2. This process is sometimes known as "easing," and looks like $$g\left(x,l,u\right)=f\left(\frac{x-l}{u-l}\right)\left(u-l\right)+l$$ where $l$ and $u$ are the lower and upper bound (possible values being $0$, $1$, and the sustain level) and $f(x)$ is something like $x^n$. Note that you don't need this for the attack phase since it already ranges from $0$ to $1$. Here is the original Desmos session, updated to use this approach. I used a cubic shape here, but you* could use whatever shape you like, as long as $f(x)$ produces outputs ranging from zero to one given inputs ranging from zero to one. * I guess the OP is probably long gone, but maybe this helps someone else. About pichenettes' comment, "During the attack stage, the capacitor is charged with a +15V step, but the attack stage ends when a threshold of +10V has been reached. This is a design choice, though 2/3 is the "magic number" found in many classic envelope generator, and this might be the one musicians are familiar with.": Any envelope that's shooting for a 15v asymptote with a 10v target is, practically, creating a linear attack. It's just that 15v is the highest asymptote available easily, and it's close enough to linear. That is, there's nothing "magic" about it—they're just going for as linear as they can get. I don't know how many classic synths use 15v—I'd suspect there's often a diode drop or two. My old Aries modular uses 13v for a 10v envelope, and I just looked up the a Curtis ADSR chip that uses, equivalently, 6.5v for a 5v envelope. This code should generate similar plots to those of pichenettes: def ASD_envelope( nSamps, tAttack, tRelease, susPlateau, kA, kS, kD ): # number of samples for each stage sA = int( nSamps * tAttack ) sD = int( nSamps * (1.-tRelease) ) sS = nSamps - sA - sD # 0 to 1 over N samples, weighted with w def weighted_exp( N, w ): t = np.linspace( 0, 1, N ) E = np.exp( w * t ) - 1 E /= max(E) return E A = weighted_exp( sA, kA ) S = weighted_exp( sS, kS ) D = weighted_exp( sD, kD ) A = A[::-1] A = 1.-A S = S[::-1] S *= 1-susPlateau S += susPlateau D = D[::-1] D *= susPlateau env = np.concatenate( [A,S,D] ) # plot tEnv = np.linspace( 0, nSamps, len(env) ) plt.plot( tEnv, env ) plt.savefig( "OUT/EnvASD.png" ) plt.close() return env I'm grateful for any improvements, one thing that may be a good idea is allowing the last three parameters (which determine the steepness of each of the three stages) to vary between 0 and 1, where 0.5 would be a straight-line. But I can't see offhand how to do it. Also I haven't tested thoroughly all usage cases, for example if one stage has zero length. Here's an extension of P i's post, this is a new account so I can't comment on it. Also I recognize this is a super old thread but I figure ambitious plugin developers like myself are bound to find this in the future. I added sliders to control the attack, release and sustain, as well as sliders to control the curvature of each stage. I also refactored it by adding negative and zero values for the curvature constants, which affect the graph accordingly. Pip install matplotlib, ipywidgets and numpy and throw this code into a jupyter notebook and you should be good to go. import matplotlib.pyplot as plt import numpy as np from ipywidgets import interact, widgets def ASD_envelope( nSamps, tAttack, tRelease, susPlateau, kA, kS, kD ): # number of samples for each stage sA = int( nSamps * tAttack ) sD = int( nSamps * (1.-tRelease) ) sS = nSamps - sA - sD # 0 to 1 over N samples, weighted with w def weighted_exp( N, w ): t = np.linspace( 0, 1, N) if w > 0: E = np.exp( w * t ) - 1 E /= max(E) elif w == 0: return t else: E = np.exp( w * t ) - 1 E /= min(E) return E A = weighted_exp( sA, kA ) S = weighted_exp( sS, kS ) D = weighted_exp( sD, kD ) A = A[::-1] A = 1.-A S = S[::-1] S *= 1-susPlateau S += susPlateau D = D[::-1] D *= susPlateau env = np.concatenate( [A,S,D] ) # plot tEnv = np.linspace( 0, nSamps, len(env) ) plt.plot( tEnv, env ) plt.savefig( "EnvASD.png" ) plt.close() return env # Create sliders for each function argument a_slider = widgets.IntSlider(min=0, max=44100, step=1, value=44100) b_slider = widgets.FloatSlider(min=0, max=1, step=0.1, value=.5) c_slider = widgets.FloatSlider(min=0, max=1, step=0.1, value=.8) d_slider = widgets.FloatSlider(min=0, max=1, step=0.1, value=.4) e_slider = widgets.IntSlider(min=-10, max=10, step=1, value=0) f_slider = widgets.IntSlider(min=-10, max=10, step=1, value=0) g_slider = widgets.IntSlider(min=-10, max=10, step=1, value=0) interact(ASD_envelope, nSamps=a_slider, tAttack=b_slider, tRelease=c_slider, susPlateau=d_slider, kA=e_slider, kS=f_slider, kD=g_slider)

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# Finding the volume bounded by surface in spherical coordinates I want to find the volume bounded by the surface given in spherical coordinates $R = 4-1\cos(\phi)$ I tried $\int_0^{2\pi} \int_0^{\pi/2} \int_0^4 (4-\cos(\phi))R^2\sin(\phi)\,dR \,d\phi\, d\theta$. But I got the wrong answer. The volume element is given by $dV = R^2\sin(\phi)dR\,d\phi\, d\theta$. I'm assuming my limits are wrong, any ideas? • The equation of the surface should not be in the integral, it only defines the bounds. – Kuifje Feb 27 '18 at 20:25 • Ok, so I'm integrating $R^2*sin(\phi)$ and my bounds are correct? – novo Feb 27 '18 at 20:30 $$V=\int_0^{2\pi}\int_0^{\pi}\int_0^{4-\cos\phi}R^2\sin\phi\; dRd\phi d\theta =\frac{272\pi}{3}$$ • Are you sure about the bounds, $8\pi/3$ isn't the right answer apparantly – novo Feb 27 '18 at 20:38 • There we go, I was wondering about the $1-cos(\phi)$ bound. I see you've corrected it. Thank you for the assistance, sir! – novo Feb 27 '18 at 20:42

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# Non-Linear and non separable differential equation • Jun 8th 2011, 07:51 PM metalkakkarot Non-Linear and non separable differential equation How can i solve this equation as it is non-linear and non seperable, i cant use the integrating factor as Q(x) does not exist and i cant seperate so that x and y are separate. x.(dy/dx)=-3y + 6x • Jun 8th 2011, 08:01 PM Chris L T521 Quote: Originally Posted by metalkakkarot How can i solve this equation as it is non-linear and non seperable, i cant use the integrating factor as Q(x) does not exist and i cant seperate so that x and y are separate. x.(dy/dx)=-3y + 6x What do you mean its not linear? Note that $x\frac{\,dy}{\,dx}=-3y+6x \implies \frac{\,dy}{\,dx}+\frac{3}{x}y=6$ is of the form $\frac{\,dy}{\,dx}+P(x)y=Q(x)$, which is clearly linear... Can you take it from here and use the integrating factor? • Jun 8th 2011, 08:09 PM metalkakkarot thanks..got it now i used this before and got the wrong answer.. integrating factor is e^3lnx

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GraphLayoutAlgorithm.h Go to the documentation of this file. 1 /* 2  * GraphLayoutAlgorithm.h 3  * 4  * Created on: Apr 19, 2016 5  * Author: Michael Wegner 6  */ 7 8 #ifndef NETWORKIT_CPP_VIZ_GRAPHLAYOUTALGORITHM_H_ 9 #define NETWORKIT_CPP_VIZ_GRAPHLAYOUTALGORITHM_H_ 10 11 #include "Point.h" 12 #include "../graph/Graph.h" 13 #include "../auxiliary/Enforce.h" 14 15 #include <vector> 16 #include <fstream> 17 18 namespace NetworKit { 19 25 template<typename T> 27 public: 28  GraphLayoutAlgorithm(const Graph& G, count dim) : G(G), vertexCoordinates(std::vector<Point<T>>(G.upperNodeIdBound(), Point<T>(dim))) {} 29  virtual ~GraphLayoutAlgorithm() = default; 30 31  virtual void run() = 0; 32 33  virtual std::vector<Point<T>> getCoordinates() const { 34  return vertexCoordinates; 35  } 36 37  virtual count numEdgeCrossings() const { 38  if (vertexCoordinates[0].getDimensions() == 2) { 39  count numCrossings = 0; 40  G.forEdges([&](node u, node v, edgeweight) { 41  G.forEdges([&](node p, node q, edgeweight) { 42  if ((p == u && q == v) || (p == v && q == u)) return; 43  double m1 = (vertexCoordinates[v][1] - vertexCoordinates[u][1]) / (vertexCoordinates[v][0] - vertexCoordinates[u][0]); 44  double m2 = (vertexCoordinates[q][1] - vertexCoordinates[p][1]) / (vertexCoordinates[q][0] - vertexCoordinates[p][0]); 45 46  double b1 = vertexCoordinates[u][1] - m1 * vertexCoordinates[u][0]; 47  double b2 = vertexCoordinates[p][1] - m1 * vertexCoordinates[p][0]; 48  if (m1 != m2) { 49  double xIntersect = (b2 - b1) / (m1 - m2); 50  double minXE1 = std::min(vertexCoordinates[u][0], vertexCoordinates[v][0]); 51  double minXE2 = std::min(vertexCoordinates[p][0], vertexCoordinates[q][0]); 52  double maxXE1 = std::max(vertexCoordinates[u][0], vertexCoordinates[v][0]); 53  double maxXE2 = std::max(vertexCoordinates[p][0], vertexCoordinates[q][0]); 54 55  if (minXE1 <= xIntersect && minXE2 <= xIntersect && xIntersect <= maxXE1 && xIntersect <= maxXE2) { 56  numCrossings++; 57  } 58  } else if (b1 == b2) { 59  numCrossings++; 60  } 61  }); 62  }); 63 64  numCrossings /= 2; 65  return numCrossings; 66  } 67 68  return 0; 69  } 70 71  virtual bool writeGraphToGML(const std::string& filePath) { 72  if (vertexCoordinates.size() == 0 || vertexCoordinates[0].getDimensions() < 2 || vertexCoordinates[0].getDimensions() > 3) return false; 73  count dim = vertexCoordinates[0].getDimensions(); 74  std::ofstream file(filePath); 75  Aux::enforceOpened(file); 76 77  file << "graph [\n"; 78  if (G.isDirected()) { 79  file << " directed 1\n"; 80  } 81 82  G.forNodes([&](node u) { 83  file << " node [\n"; 84  file << " id " << u << "\n"; 85  file << " graphics\n"; 86  file << " [ x " << 50*vertexCoordinates[u][0] << "\n"; 87  file << " y " << 50*vertexCoordinates[u][1] << "\n"; 88  if (dim == 3) { 89  file << " z " << vertexCoordinates[u][2] << "\n"; 90  } 91  file << " ]\n"; 92  file << " ]\n"; 93  }); 94 95  G.forEdges([&](node u, node v) { 96  file << " edge [\n"; 97  file << " source "<< u << "\n"; 98  file << " target "<< v << "\n"; 99  file << " ]\n"; 100  }); 101  file << "]\n"; 102 103  file.close(); 104 105  return true; 106  } 107 108  virtual bool writeKinemage(const std::string& filePath) { 109  if (vertexCoordinates.size() == 0 || vertexCoordinates[0].getDimensions() != 3) return false; 110  std::string fileName = filePath.substr(filePath.find_last_of("/")); 111  std::ofstream file(filePath); 112  Aux::enforceOpened(file); 113 114  file << "@whitebackground" << std::endl; 115  file << "@zoom 1.0" << std::endl; 116  file << "@zslab 240" << std::endl; 117  file << "@center 0 0 0" << std::endl; 118  file << "@master{points}" << std::endl; 119  file << "@group{" << fileName << "}" << std::endl; 120  file << "@balllist {a} color= blue master={points} radius= 0.05" << std::endl; 121 122  G.forNodes([&](node u) { 123  file << "{a}" << vertexCoordinates[u][0] << " " << vertexCoordinates[u][1] << " " << vertexCoordinates[u][2] << std::endl; 124  }); 125 126  // edges 127  file << std::endl; 128  file << "@subgroup {edges} dominant" << std::endl; 129  file << "@vectorlist {edges} color= white" << std::endl; 130  G.forEdges([&](node u, node v) { 131  //if (u <= v) { // draw graph undirected 132  file << "P " << vertexCoordinates[u][0] << " " << vertexCoordinates[u][1] << " " << vertexCoordinates[u][2] << std::endl; 133  file << vertexCoordinates[v][0] << " " << vertexCoordinates[v][1] << " " << vertexCoordinates[v][2] << std::endl; 134  //} 135  }); 136 137  file << std::endl; 138 139  file.close(); 140  return true; 141  } 142 143 protected: 144  const Graph& G; 145  std::vector<Point<T>> vertexCoordinates; 146 }; 147 148 } /* namespace NetworKit */ 149 150 #endif /* NETWORKIT_CPP_VIZ_GRAPHLAYOUTALGORITHM_H_ */ void forNodes(L handle) const Iterate over all nodes of the graph and call handle (lambda closure). Definition: Graph.h:1097 bool isDirected() const Return true if this graph supports directed edges. Definition: Graph.h:694 const Graph & G Definition: GraphLayoutAlgorithm.h:144 void enforceOpened(const Stream &stream) Checks that the provided fstream is opened and throws an exception otherwise. Definition: Enforce.h:37 void forEdges(L handle) const Iterate over all edges of the const graph and call handle (lambda closure). Definition: Graph.h:1299 Formerly marked as deprecated: To take advantage of automatic mapping between C++ and Python data str... Definition: Point.h:39 std::vector< Point< T > > vertexCoordinates Definition: GraphLayoutAlgorithm.h:145 Abstract base class for algorithms that compute a layout of the Graph vertices in d-dimensional space... Definition: GraphLayoutAlgorithm.h:26 virtual count numEdgeCrossings() const Definition: GraphLayoutAlgorithm.h:37 virtual std::vector< Point< T > > getCoordinates() const Definition: GraphLayoutAlgorithm.h:33 uint64_t count Definition: Globals.h:21 index node Definition: Globals.h:23 virtual ~GraphLayoutAlgorithm()=default virtual bool writeGraphToGML(const std::string &filePath) Definition: GraphLayoutAlgorithm.h:71 A graph (with optional weights) and parallel iterator methods. Definition: Graph.h:79 virtual bool writeKinemage(const std::string &filePath) Definition: GraphLayoutAlgorithm.h:108 GraphLayoutAlgorithm(const Graph &G, count dim) Definition: GraphLayoutAlgorithm.h:28 double edgeweight Definition: Globals.h:24

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First Name: ___________________ Last Name: ____________________ Section: _________ May 14, 1999 Physics 208 Final Exam Print your name and section clearly on all nine pages. (If you do not know your section number, write your TA’s name.) Show all work in the space immediately below each problem. Your final answer must be placed in the box provided. Problems will be graded on reasoning and intermediate steps as well as on the final answer. Be sure to include units wherever necessary, and the direction of vectors. Each problem is worth 25 points. In doing the problems, try to be neat. Check your answers to see that they have the correct dimensions (units) and are the right order of magnitudes. You are allowed two 8½ x 11" sheet of notes and no other references. The exam lasts exactly 2 hours. (Do not write below) SCORE: Problem 1: __________ Problem 2: __________ Problem 3: __________ Problem 4: __________ Problem 5: __________ Problem 6: __________ Problem 7: __________ Problem 8: __________ TOTAL: ___________ First Name: ___________________ Last Name: ____________________ Section: _________ 1. An insulated ball with a radius of 10 cm has an electric charge of 2 nC distributed uniformly throughout its interior. a. What is the electric field 8 cm from the center of the ball? (9 pts.) b. What is the electric potential (relative to infinity) 12 cm from the center of the ball? (8 pts.) c. What is the force on an electron 15 cm from the center of the ball? (8 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 1. Consider the circuit below: 2. a. What is the voltage across the 3-ohm resistor? (10 pts.) b. What is the current in the 15-ohm resistor? (15 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 3. A long, straight wire carries a current of 100 A. 4. a. What is the magnitude of the magnetic field 5 cm from the wire? (8 pts.) b. What is the force on an electron 5 cm from the wire moving in the direction of the current with kinetic energy of 100 eV? (8 pts.) c. If the current is turned uniformly to zero in a time of 500 ?S, what voltage is induced in a single-turn circular loop with an area of 1 cm2, 5 cm from the wire, and axis parallel to the magnetic field? (9 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 5. In the circuit below, the switch has been closed for a long time and is then opened at t = 0. 6. a. What is the current in the resistor just before the switch is opened? (5 pts.) b. At what frequency (in Hz) does the circuit oscillate after the switch is opened? (5 pts.) c. What is the maximum instantaneous power dissipated by the resistor? (5 pts.) d. What is the maximum voltage across the inductor? (5 pts.) e. What is the voltage across the capacitor after the switch has been open for a long time? (5 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 7. Assume a 100-W light bulb radiates light with an average wavelength of 600 nm isotropically in all directions. 8. a. What is the rms electric field 50 cm from the bulb? (5 pts.) b. What is the rms magnetic field 50 cm from the bulb? (5 pts.) c. What is the radiation pressure on a black surface 50 cm from the bulb? (5 pts.) d. What is the energy (in eV) of the photons emitted by the bulb? (5 pts.) e. How many photons are emitted by the bulb each second? (5 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 9. Astronauts leave two lasers (? = 650 nm) 1 km apart on the moon pointed toward the earth. The earth-moon distance is 3.84 x 108 m. 10. a. What diameter telescope is required to just resolve the laser beams from the earth? (9 pts.) b. If the telescope lens has a focal length of 140 cm, how far apart are the images of the lasers? (8 pts.) c. If these images are magnified by a second converging lens with a focal length of 2 cm, how far from the second lens must a photographic plate be placed so that the images of the lasers are 1 mm apart? (8 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 11. Consider an electron in the 1s state of a neutral Uranium atom (Z = 92) in its ground state. 12. a. What is the kinetic energy of this electron in eV? (6 pts.) b. What is the relativistic ?-factor of this electron? (6 pts.) c. What is the de Broglie wavelength of the electron? (7 pts.) d. What wavelength photon is required to eject this electron from the atom? (6 pts.) First Name: ___________________ Last Name: ____________________ Section: _________ 13. A smoke detector uses Americium-243 (Z = 95) which decays to Neptunium by ?-decay with a half life of 7370 years. The Neptunium quickly ?--decays to Plutonium whose half life is much greater than Americium. a. What is the charge (Z) and atomic mass (A) of the Neptunium nucleus? (5 pts.) b. What is the charge (Z) and atomic mass (A) of the Plutonium nucleus? (5 pts.) c. What fraction of the Americium decays in 100 years? (5 pts.) d. If the mass of Americium is 1 g, what is its radioactivity (in Curies)? (5 pts.) e. What other lepton is produced in the above decay process? (5 pts.)

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# math posted by . find mean median and mode of 10 1 10 15 1 7 10 10 1 6 13 ## Similar Questions 1. ### algebra mean med mode can you check my answers i have to find the mean, median and mode 1) 8 12 6 9 5 mean: 8 median : 8 mode : no mode 2) 3.4 0.53 1.3 2.9 1.47 0.24 mean: 1.64 median : 1.385 mode: no mode 3) -6 3 -2 6 7 -3 -5 8 mean : 1 median : 1 (? 2. ### A l g e b r a MeanMed Mode can you check my answers i have to find the mean, median and mode 1) 8 12 6 9 5 mean: 8 median : 8 mode : no mode 2) 3.4 0.53 1.3 2.9 1.47 0.24 mean: 1.64 median : 1.385 mode: no mode 3) -6 3 -2 6 7 -3 -5 8 mean : 1 median : 1 (? 3. ### mrs. sue algebra help Mrs.Sue my homework from yesterday i had wrong instructions i needed to find the mean median and mode. not the outlier an affect. my teacher switched the assignment can you check my answers i have to find the mean, median and mode … 4. ### algebra help mrs. Sue Mrs.Sue my homework from yesterday i had wrong instructions i needed to find the mean median and mode. not the outlier an affect. my teacher switched the assignment can you check my answers i have to find the mean, median and mode … 5. ### math mean median mode which of the following shows the order from least to greatest for the mean, the median, and the mode of this set of data? 6. ### algebra 1 question pls help!!! 1. Find the mean, median, and mode of the data set: 15, 16, 21, 23, 25, 25, 25, 39 a.mean = 23.6, median = 25, mode = 24 b.mean = 24, median = 23.6, mode = 25 c.mean = 24.6, median = 24, mode = 25 d.mean = 23.6, median = 24, mode = … 7. ### ALGEBRA Find the mean, median, and mode of the data set. Round to the nearest tenth. 15, 13, 9, 9, 7, 1, 11, 10, 13, 1, 13 (1 point) • mean = 9.3, median = 8, mode = 13 • mean = 8.5, median = 10, mode = 13 • mean = 9.3, median = 10, … 8. ### ALGEBRA (not for Ms.Sue) Find the mean, median, and mode of the data set. Round to the nearest tenth. 15, 13, 9, 9, 7, 1, 11, 10, 13, 1, 13 • mean = 9.3, median = 8, mode = 13 • mean = 8.5, median = 10, mode = 13 • mean = 9.3, median … 9. ### math :D Find the mean, median, and mode of the data set: 15, 16, 21, 23, 25, 25, 25, 39 1.mean = 23.6, median = 25, mode = 24 2.mean = 24, median = 23.6, mode = 25 3.mean = 24.6, median = 24, mode = 25 4.mean = 23.6, median = 24, mode = 25 … 10. ### Math What are the mean, median mode and range of the data we given the altitude of lakes in feet? More Similar Questions

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Who are the experts?Our certified Educators are actual professors, teachers, and scholars who use their academic expertise to tackle her toughest questions. Educators go v a rigorous applications process, and also every answer they send is reviewed by our in-house editorial team. You are watching: The mass of al in aluminum sulfate (al2(so4)3) is Hello! Probably the question is around mass: what percent walk aluminium constitute of the mass of a molecule `Al_2 (S O_4)_3 ?` To calculate this, we should know the atom masses of `Al` (aluminium), `S` (sulfur) and also `O` (oxygen) atom in part units. We have the right to choose any kind of units yet the... Start her 48-hour complimentary trial come unlock this answer and thousands more. Gain beer-selection.com ad-free and cancel anytime. Hello! Probably the concern is about mass: what percent walk aluminium constitute the the fixed of a molecule `Al_2 (S O_4)_3 ?` To calculate this, us should know the atomic masses the `Al` (aluminium), `S` (sulfur) and also `O` (oxygen) atoms in part units. We have the right to choose any kind of units however the same for all types of atoms. The most convenient unit in this situation is the family member atomic mass. The is identified as `1/12` of the massive of one atom of carbon-12. We may look the periodic table to see that `m(Al) approx 27,` `m(S) approx 32,` `m(O) approx 16.` Then the massive of a molecule of `Al_2 (S O_4)_3` is `2*m(Al) + 3*(m(S)+4*m(O)) approx 342.` Aluminium in together a molecule has the mass `2*m(Al) approx 54.` therefore the percent in concern is `54/342 *100% approx15,79%,` so the price is 15.8%. See more: Why Was 7 Afraid Of 8 - Why Don'T They Say,Why Is Nine Afraid Of Seven approved by beer-selection.com Editorial Team Submit concern see all ## Related Questions Browse every Science recent answer post November 01, 2015 in ~ 5:21:53 afternoon Is aluminum sulfate Al2(SO4)3 one ionic link or a molecule compound? Science latest answer post August 17, 2013 at 1:37:07 to be What is the fixed percent of oxygen in Al2(SO4)3·18H2O?Molar massive Al2(SO4)3·18H2O is 666.43 g/mol(A) 9.60(B) 28.8(C) 43.2(D) 72.0 Science recent answer post January 13, 2016 at 10:42:46 am calculation the portion of oxygen in Al2(SO4)3? Science recent answer post February 04, 2016 in ~ 6:31:51 to be If 2.07 grams that aluminum react through excess CuSO4, what is the theoretical yield of Cu? 2Al + 3CuSO4 ---> Al2(SO4)3 + 3Cu Science recent answer post July 07, 2010 in ~ 9:07:17 pm What is the massive in grams that a sample that Fe2(SO4)3 that consists of 3.59x10^23 sulfate ions, SO4 with an adverse two in ~ the peak ?

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# Euler–Maclaurin Summation Formula: Definition Share on Feel like "cheating" at Calculus? Check out our Practically Cheating Calculus Handbook, which gives you hundreds of easy-to-follow answers in a convenient e-book. ## What is the Maclaurin Summation Formula? The Euler–Maclaurin summation formula, discovered independently by Leonhard Euler (in 1732) and Colin Maclaurin (in 1742), relates the summation of a function to an integral approximation. It gives a way to calculate “corrections” in terms of the function’s derivatives, evaluated at the endpoints. It’s most general form is [1]: Where: In addition, the function f must have a continuous kth derivative. If the function and all of its derivatives approach zero as x→∞, the Euler–Maclaurin summation Formula simplifies (by letting b→∞ in the identity) to: It can also be expressed in a more “modern” asymmetrical form as [2]: With a remainder, at any stage, of and The Euler-Maclaurin summation formula has many uses in mathematics and science, including quantum topology, where the formula is sometimes needed for a resurgent function with singularities in the vertical strip perpendicular to the range of summation [3]. It is also one way to generate an asymptotic series. ## What are Bernoulli Numbers? Bernoulli numbers are rational numbers defined as coefficients in the series expansion These coefficients are very challenging to calculate by hand; software is normally used to find them [4]. Bernoulli numbers Bk are defined as values of the Bernoulli polynomials Bk(t) at t = 0: Bk = Bk(0). Bernoulli polynomials satisfy the differential equation [5] B′k+1(t) = (k + 1)Bk(t), with B2k+1(0) = 0, B2k+1(1) = 0, for all k > 0. ## References [1] Kac, V. (2005). 18.704 Seminar in Algebra and Number Theory Fall 2005. Euler-Maclaurin Formula. Retrieved August 14, 2021 from: http://people.csail.mit.edu/kuat/courses/euler-maclaurin.pdf [2]Mills, S. (1985). The Independent Derivations by Leonhard Euler and Colin MacLaurin of the Euler-MacLaurin Summation Formula. Archive for History of Exact Sciences Vol. 33, No. 1/3 (1985), pp. 1-13 (13 pages). Springer. [3] Costin, O. & Garoufaldis, S. Resurgence of the E-M Summation Formula. Retrieved August 14, 2021 from: https://people.math.osu.edu/costin.9/EULMCL.pdf [4] Rozman, M. (2016). Euler-Maclaurin summation formula. Retrieved August 14, 2021 from: https://www.phys.uconn.edu/~rozman/Courses/P2400_16S/downloads/euler-maclaurin-summation.pdf [5] Verschelde, J. The Euler-Maclaurin Summation Formula. Retrieved August 14, 2021 from: http://homepages.math.uic.edu/~jan/MCS471/Lec28/emsum.pdf CITE THIS AS: Stephanie Glen. "Euler–Maclaurin Summation Formula: Definition" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/euler-maclaurin-summation-formula-definition/ --------------------------------------------------------------------------- Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!

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How can I determine whether a string contains a substring? I guess that since MATLAB R2016b it is recommended to use contains : Determine if pattern is in strings - MATLAB contains - M... meer dan 4 jaar ago | 1 Functional form of the colon (:) operator? I really wish MATLAB would add function to vectorize arrays into column vector as the colon operator does. ongeveer 5 jaar ago | 0 Change opacity of Lines In newer versions of MATLAB you can do that easily using the Color property of the line. By default it is RGB array (1 x 3).... ongeveer 5 jaar ago | 7 GUI - Slider which has two slide bars Maybe you meant a Range Slider which you can create as shown in <http://undocumentedmatlab.com/blog/sliders-in-matlab-gui Slider... ongeveer 5 jaar ago | 0 Question Display Pixel Values in Image - Alternative to showPixelValues() Hello, I would like to display an image and its pixel values in a similar manner to <https://www.mathworks.com/matlabcentral/... meer dan 5 jaar ago | 2 answers | 0 ### 2 What is missing from MATLAB? What I miss the most is better forum for the community. We need Math Equation support through MathJaX (Like in StackExchange ... meer dan 5 jaar ago | 0 How to perform KNN regression At each point you want to calculate the value (x, y) find the K closest point (For x). Then average them to create the new va... meer dan 5 jaar ago | 0 Averaging Overlapping Pixels in Sliding Window Operation Another option would be: I = reshape(accumarray(mIdx(:), mZ(:), [(M * N), 1], @(x) mean(x)), [M, N]); Where mZ is the ... meer dan 5 jaar ago | 0 Question Saving Figure Data into a Matrix Hello. I'm using 'image()' to display an image. I'm adding some annotations using text. I'd like to save the final resu... bijna 6 jaar ago | 2 answers | 0 ### 2 Solved Summing digits Given n, find the sum of the digits that make up 2^n. Example: Input n = 7 Output b = 11 since 2^7 = 128, and 1 + ... ongeveer 6 jaar ago Solved Bullseye Matrix Given n (always odd), return output a that has concentric rings of the numbers 1 through (n+1)/2 around the center point. Exampl... ongeveer 6 jaar ago Solved Return the largest number that is adjacent to a zero This example comes from Steve Eddins' blog: <http://blogs.mathworks.com/steve/2009/05/27/learning-lessons-from-a-one-liner/ Lear... ongeveer 6 jaar ago Solved Find the longest sequence of 1's in a binary sequence. Given a string such as s = '011110010000000100010111' find the length of the longest string of consecutive 1's. In this examp... ongeveer 6 jaar ago Solved Find the numeric mean of the prime numbers in a matrix. There will always be at least one prime in the matrix. Example: Input in = [ 8 3 5 9 ] Output out is 4... ongeveer 6 jaar ago Solved Remove all the consonants Remove all the consonants in the given phrase. Example: Input s1 = 'Jack and Jill went up the hill'; Output s2 is 'a ... ongeveer 6 jaar ago Solved Fibonacci sequence Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu... ongeveer 6 jaar ago Solved Back and Forth Rows Given a number n, create an n-by-n matrix in which the integers from 1 to n^2 wind back and forth along the rows as shown in the... ongeveer 6 jaar ago Solved Determine whether a vector is monotonically increasing Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f... ongeveer 6 jaar ago Solved Who Has the Most Change? You have a matrix for which each row is a person and the columns represent the number of quarters, nickels, dimes, and pennies t... ongeveer 6 jaar ago Solved Make a checkerboard matrix Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Examp... ongeveer 6 jaar ago Solved Swap the first and last columns Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth... ongeveer 6 jaar ago Solved Find all elements less than 0 or greater than 10 and replace them with NaN Given an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ... ongeveer 6 jaar ago Solved Triangle Numbers Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa... ongeveer 6 jaar ago Solved Column Removal Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; and ... ongeveer 6 jaar ago Solved Select every other element of a vector Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s... ongeveer 6 jaar ago Solved Determine if input is odd Given the input n, return true if n is odd or false if n is even. ongeveer 6 jaar ago Solved Given a and b, return the sum a+b in c. ongeveer 6 jaar ago Solved Find the sum of all the numbers of the input vector Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ... ongeveer 6 jaar ago Solved Make the vector [1 2 3 4 5 6 7 8 9 10] In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s... ongeveer 6 jaar ago Solved Times 2 - START HERE Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:... ongeveer 6 jaar ago

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Switch to: More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas. Health Net Inc (NYSE:HNT) Total Equity \$1,833 Mil (As of Dec. 2015) Health Net Inc's total equity for the quarter that ended in Dec. 2015 was \$1,833 Mil. Total equity is used to calculate book value per share. Health Net Inc's book value per share for the quarter that ended in Dec. 2015 was \$23.71. The ratio of a company’s debt over equity can be used to measure how leveraged this company is. Health Net Inc's debt to equity ratio for the quarter that ended in Dec. 2015 was 0.37. Definition Total Equity refers to the net assets owned by shareholders. Total Equity and Total Liabilities are the two components for Total Assets. Health Net Inc's Total Equity for the fiscal year that ended in Dec. 2015 is calculated as Total Equity = Total Assets(Q: Dec. 2015 ) - Total Liabilities(Q: Dec. 2015 ) = 6397.646 - 4564.566 = 1,833 Health Net Inc's Total Equity for the quarter that ended in Dec. 2015 is calculated as Total Equity = Total Assets(Q: Dec. 2015 ) - Total Liabilities(Q: Dec. 2015 ) = 6397.646 - 4564.566 = 1,833 * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Explanation 1. Total equity is used to calculate book value per share. Health Net Inc's Book Value Per Share for the quarter that ended in Dec. 2015 is Book Value per Share = (Total Shareholder’s Equity - Preferred Stock) / Total Shares Outstanding = (1833.08 - 0) / 77.32 = 23.71 2. The ratio of a company’s debt over equity can be used to measure how leveraged this company is. Health Net Inc's Debt to Equity Ratio for the quarter that ended in Dec. 2015 is Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (285 + 399.709) / 1833.08 = 0.37 * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Related Terms Historical Data * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Health Net Inc Annual Data Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Total Equity 1,779 1,876 1,752 1,696 1,694 1,443 1,557 1,629 1,709 1,833 Health Net Inc Quarterly Data Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Total Equity 1,601 1,629 1,679 1,822 1,766 1,709 1,659 1,712 1,788 1,833 Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)

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Not applicable Posts: 0 # Calculating Monthly subtotals in EG I just got through figuring out how to do subtotals in general, and now I have come across the problem of needing to compute monthly subtotals. Again I do not believe that there is a SAS function that computes these numbers, but there has to be some way to calculate this. Does anyone know how this could be done? Frequent Contributor Posts: 80 ## Re: Calculating Monthly subtotals in EG I think that you may need to give us a bit more information about what you are doing and what method you are using to create your sub-totals. SAS is fantastic at being able to group stuff for sub-totals by various aspects of a date (ie Month, Month/Year, Quarter, Day of Week etc.). Your biggest hurdle is really to make sure that your data has its date value stored as proper SAS dates (ie not just text values that look like dates). So go on, give us some more information........are you new to SAS and only use SAS/EG or are you someone who is more familiar with writing SAS code but are currently using SAS/EG ? Some typical data and examples of what you hope to for as results may not go amiss in any reply, it helps people to see what you see. Cheers, Down-Under-Dave Wellington New Zealand SAS Super FREQ Posts: 9,428 ## Re: Calculating Monthly subtotals in EG Hi: It really depends on what your data looks like and what EG task you're using to calculate the subtotals and what outcome you want (data set or report). Most SAS procedures and/or EG tasks have a way to "group" observations. So, for example, if your data has a MONTH variable or a DATE variable, then you could select "GROUP BY" in your task and specify the MONTH variable. Another thing to consider is -- do you want monthly subtotals in a SAS data set or do you want a REPORT (like HTML or RTF report) with the subtotals??? Whether there's a function that you could use depends on your data. Generally speaking functions work on a row by row basis. So if your data had these variables: [pre] product month day1 day2 day3 day4.... shoes jan 100 150 175 100.... shoes feb 99 100 60 100.... shirts jan 120 130 155 150.... shirts feb 89 120 60 150.... [/pre] Then you could use the SUM function for every row: [pre] montot = sum (of day1-day31); [/pre] (even if a month did not have 31 days, missing values for specific days would be ignored with the sum function.) On the other hand, if your data was structured like this: [pre] product mon day amt shoes jan 1 100 shoes jan 2 150 shoes jan 3 175 shoes jan 4 100 shoes feb 1 99 shoes feb 2 100 shoes feb 3 60 shoes feb 4 100 shirts jan 1 120 shirts jan 2 130 shirts jan 3 155 shirts jan 4 150 shirts feb 1 89 shirts feb 2 120 shirts feb 3 60 shirts feb 4 150 [/pre] Then the SUM function would not be appropriate -- you'd have to use one of the EG tasks or a SAS procedure to create a file or report with monthly subtotals. You could use the Summary Statistics task or the Summary Tables task to get a report of monthly subtotals. If you wanted to see the "detail" rows AND the monthly subtotals in a report, then you might consider the List Data task and you would group the report by product and month or just by month. In my fake data, I showed a separate month and day variable, although you might actually have a variable with a whole date value, like a SAS date value and in that case, you would need to apply a format in order to group by month or month and year. At any rate, it is probable that the same method you figured out to do subtotals in general can be extended to the task of computing monthly subtotals. If you are using code, then you would look for a CLASS statement or a BY statement in the syntax. If you are using an EG task, then you would look for a GROUP BY choice in your task. If you are using the Summary Tables task (which creates PROC TABULATE code), then you would drag and drop the ALL "statistic" into the appropriate place in the table when you build it with drag and drop methods. For more help with this task, you might open a track with Tech Support so someone could walk you through the click-path you would need to follow in your task of choice to get subtotals by month. cynthia Not applicable Posts: 0 ## Re: Calculating Monthly subtotals in EG Thanks for the help. I created custom columns of month and day and then just added them both as the classification variables in my columns. Then made sure subtotals were added and it worked. It was actually really easy. On this same note, does anyone know how to do monthly running totals? Assume that i have a date column and a profit column. What i would want this custom column to do is continually add the previous days profit to the running total until it reaches the end of the month, at which point it would start over. Suggestions? SAS Super FREQ Posts: 9,428 ## Re: Calculating Monthly subtotals in EG Hi: If you wanted running totals, your choices are to either use a data step program to calculate the running total and add that as a column in a data set or then you could do a List Data Task that showed the running total or a procedure like PROC REPORT that would let you calculate the running total as a report column (not in the data set). To calculate running totals in a Data step program, you would need to investigate how programs work, how the RETAIN statement works and how to use the FIRST.byvar automatic variable to reset the running total at the first of every month (or LAST.byvar). The Data step approach is shown in this Tech Support Note: http://support.sas.com/kb/19/394.html To modify the Tech Support code for BY group processing, you would only change the code with a BY statement and an IF statement. This Tech Support note illustrates creating a cumulative column based on BY group processing: cynthia Not applicable Posts: 0 ## Re: Calculating Monthly subtotals in EG Ok, I have already used that second link that you have provided to set up my cumulative totals field, so I am familiar with it. However, I am new to SAS programming, and I am not entirely sure how to go about implementing the BY and IF statements to create the MTD running totals. Here is where I am at so far: data work.calculated1; set work.query_for_query9446; by Month Day; If FIRST.Month = 1 then 'Cum Total Loans Made MTD'n + 'Loans - Total (sum)'n; run; This will separate the new field by month, but they are not calculating running totals, they are only returning the first value in the 'Loans - Total (sum)' column, and then outputting that value for every day of the month. Then it sums that value with the first value of the next month and displays that number for every observation within the following month and so on. I tried to do this same code with Day as my only byvar, but I receive error messages because it is not sorted (which it wouldn't be because the 'Day' field streams 1 through 31 and then starts over again at 1 on the new month) Any advice would be greatly appreciated. Thanks, Ben Message was edited by: bmartin SAS Super FREQ Posts: 9,428 ## Re: Calculating Monthly subtotals in EG Hi: The key is NOT to do the addition on the FIRST.MONTH condition -- all you probably want to do at FIRST.MONTH is reset the cumulative variable to 0. Consider this data sorted by month product and day. I have added what the values for first.month and last.month will look like on every observation. So you can see that when first.month = 1 SAS is reading the first observation for the month -- at this point you want to set your month accumulator variable to 0. But when SAS is reading ROW 2, 3, 4, etc, you want to keep adding those values for AMT to your accumulator variable. In your logic, you were only adding the number for the first row of your group, which will not get you a running total for every month: [pre] product month day amt first.month last.month shirts 01 1 120 1 0 shirts 01 2 130 0 0 shirts 01 3 155 0 0 shirts 01 4 150 0 0 shoes 01 1 100 0 0 shoes 01 2 150 0 0 shoes 01 3 175 0 0 shoes 01 4 100 0 1 **************************change between months********** shirts 02 1 89 1 0 shirts 02 2 120 0 0 shirts 02 3 60 0 0 shirts 02 4 150 0 0 shoes 02 1 99 0 0 shoes 02 2 100 0 0 shoes 02 3 60 0 0 shoes 02 4 100 0 1 [/pre] Assume that we want to accumulate a monthly running total, a "grand" running total and for every month, we also want a running total for shirts and shoes. I need several "accumulator variables" -- MONTOT, GRNDTOT, SHIRTTOT and SHOETOT. What I want on every observation is to add the AMT on that row to both MONTOT and GRNDTOT -- however, I want to conditionally add AMT to either SHIRTTOT or SHOETOT, depending on the value of PRODUCT. So this program will do that: [pre] data mon_data; infile datalines; input product \$ month \$ day amt; return; datalines; shoes 01 1 100 shoes 01 2 150 shoes 01 3 175 shoes 01 4 100 shoes 02 1 99 shoes 02 2 100 shoes 02 3 60 shoes 02 4 100 shirts 01 1 120 shirts 01 2 130 shirts 01 3 155 shirts 01 4 150 shirts 02 1 89 shirts 02 2 120 shirts 02 3 60 shirts 02 4 150 ; run; proc sort data=mon_data; by month product day; run; data work.calculated1 sumonly(keep=month montot shirttot shoetot); set work.mon_data; by Month; ** I like to use an explicit RETAIN statement; retain montot shirttot shoetot grndtot 0 ; ** Use first.month condition to set cum variables to 0; if first.month then do; montot = 0; shirttot = 0; shoetot = 0; end; ** add the amount for every observation; montot + amt; grndtot + amt; ** add the amount for shirt and shoes separately; if product = 'shirts' then shirttot + amt; else if product = 'shoes' then shoetot + amt; ** output calculated1 -- same number of obs out as obs in; output calculated1; ** output monthly totals for last.month condition; if last.month then output sumonly; run; proc print data=work.calculated1; title 'see different running totals'; run; proc print data=work.sumonly; title 'summary file'; run; [/pre] If you were going to accumulate on DAY, then you'd use slightly different logic, but the basic idea is the same -- you use FIRST.byvar to initialize or reset your accumulator variable to 0. Then you just add your numbers on every row. As you can see with SHIRTTOT and SHOETOT -- at some point, the accumulator variable value just repeats on every row because there are no new values to add to the number. They're not worth much as running totals, but if you want to make a summary data set of ONLY monthly totals without products or days -- like SUMONLY -- then they do serve a purpose. By the time the LAST.MONTH observation is being held in the input buffer you have all the information you need to output MONTH, MONTOT, SHIRTTOT and SHOETOT. The program output is shown below. cynthia [pre] see different running totals Obs product month day amt montot shirttot shoetot grndtot 1 shirts 01 1 120 120 120 0 120 2 shirts 01 2 130 250 250 0 250 3 shirts 01 3 155 405 405 0 405 4 shirts 01 4 150 555 555 0 555 5 shoes 01 1 100 655 555 100 655 6 shoes 01 2 150 805 555 250 805 7 shoes 01 3 175 980 555 425 980 8 shoes 01 4 100 1080 555 525 1080 9 shirts 02 1 89 89 89 0 1169 10 shirts 02 2 120 209 209 0 1289 11 shirts 02 3 60 269 269 0 1349 12 shirts 02 4 150 419 419 0 1499 13 shoes 02 1 99 518 419 99 1598 14 shoes 02 2 100 618 419 199 1698 15 shoes 02 3 60 678 419 259 1758 16 shoes 02 4 100 778 419 359 1858 *************************************************************************************** summary file Obs month montot shirttot shoetot 1 01 1080 555 525 2 02 778 419 359 [/pre] Not applicable Posts: 0

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# [R-sig-phylo] comparative analysis using multiple regression of contrasts? jclaude at univ-montp2.fr jclaude at univ-montp2.fr Wed May 25 11:38:32 CEST 2011 ```Thanks Simon, this is exactly the answer I was looking for. for type II sums of squares you are right, however, when there are multiple factors pics and gls usually still provide different F values and p-values even if you set the marginality stuff in the analysis of variance. (they are not much different, but i still wonder why results are the same with one explanatory variable but different when you consider several). As I see it, contrast analyses through the origin are not a so usual regressions since no intercept is estimated: they however result in similar output when only one explanatory variable is included. Although I did not investigate type I and type II error rate when the response was continuous and the explanatory variable was dummy, I still guess that there are still something to do for modifying ancestral character state reconstruction in the contrast analysis for the dummy variable and computing their contrast: it is hard to believe that the brownian model will apply to that dummy variable because the expected variance of such a character can not be properly gaussian, but certainly more following something that has to see with the binomial law and logit family link. I wonder if someone has worked in this direction for contrasts...if not, this is probably something interesting i would try to investigate. all the best julien Simon Blomberg <s.blomberg1 at uq.edu.au> a écrit : > On 24/05/11 01:38, Julien Claude wrote: >> Dear all, >> >> I have one factor and several covariates and I would like to know which >> of these explanatory variables are more likely to explain a response >> variable in a comparative analysis. The factor is a dummy variable (0-1). >> >> At first, my strategy would be to use contrasts of all variables and >> then producing a type II anova on contrasts, I am however not sure that >> I am right in doing so with the dummy variable. >> >> data(bird.orders) >> Y<-rnorm(23) >> X1<-rbinom(23,1,0.5) >> X2<-rnorm(23) >> >> pY<-pic(Y, bird.orders) >> pX1<-pic(X1, bird.orders) >> pX2<-pic(X2, bird.orders) >> >> library(car) >> Anova(lm(pY~pX1*pX2-1)) >> >> Can we directly use the contrast for the dummy variable?, or should we >> transform this contrast in specifying some special stuff behind the >> expected ancestral values (The dummy variable probably does not follow a >> brownian motion model...). The solution may be here but I have no clear >> idea about how to transform it. > > Yes, you can directly use the contrasts of the dummy variable. There > is no problem with non-Brownian explanatory variables in pic > regression. The usual regression model assumes all the covariance is > in the response variable only. The calculation of contrasts for the > explanatory variable is a necessary step to getting the correct > parameter estimates. No further meaning should be attached. > (Although perhaps I hold an extreme view on this.) > >> I wonder also how to adopt this strategy by using gls. >> DF.bird<- data.frame(X1,X2, Y) >> bm.bird<- corBrownian(phy = bird.orders) >> m1<- gls(Y~X1*X2, correlation = bm.bird, data = DF.bird) >> m2<- gls(Y~X2*X1, correlation = bm.bird, data = DF.bird) >> anova(m1) >> >> How to get the mean squares and variance estimate in pgls ? With a one >> factor analysis F-values are exactly the same, but when the number of >> covariates is greater than 1, F values can differ in some extent, I >> wonder why. > that's because by default, the analysis gives you "sequential" sums > of squares tests (Type I for the SAS people). To get type II > (marginal) tests, which are not affected by the order of fitting, use > > anova(m1, type="marginal") > > >> In the gls stuff, I understand that incorporating the >> expected correlation of the response should follow a BM, but this is >> certainly not true for all the predictors)? > > It doesn't matter for the predictor variables. The covariances of > the predictor variables is not used in the calculation of the gls > regression. If you want to take the covariances of the predictors > into account, you need to use an errors-in-variables model. Or you > could calculate non-central correlations ("correlation through the > > Simon. >> Let me know if this strategy makes sense or if it is flawed....Thanks >> >> Julien >> >> ------------------- >> Julien CLAUDE >> Institut des Sciences de l'Evolution, 34095 Montpellier cedex 5 >> Université de Montpellier 2 >> >> _______________________________________________ >> R-sig-phylo mailing list >> R-sig-phylo at r-project.org >> https://stat.ethz.ch/mailman/listinfo/r-sig-phylo > > > -- > Simon Blomberg, BSc (Hons), PhD, MAppStat. > Lecturer and Consultant Statistician > School of Biological Sciences > The University of Queensland > St. Lucia Queensland 4072 > Australia > T: +61 7 3365 2506 > email: S.Blomberg1_at_uq.edu.au > http://www.uq.edu.au/~uqsblomb/ > > Policies: > 1. I will NOT analyse your data for you. > > Statistics is the grammar of science - Karl Pearson. > > _______________________________________________ > R-sig-phylo mailing list > R-sig-phylo at r-project.org > https://stat.ethz.ch/mailman/listinfo/r-sig-phylo > > ```

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CMPSC 465 Fall 2017 The version of the browser you are using is no longer supported. Please upgrade to a supported browser.Dismiss ABCDEFGHIJKLMNOPQRST 1 TopicRequired Handouts /Announcements 2 CMPSC465, Fall 2017, Data Structures and Algorithms 3 Instructor: Professor Paul Medvedev (pzm11) 4 5 SyllabusSyllabus 6 Professor OH:Professor OH 7 Piazzahttps://piazza.com/class/j6h3u1trhlr6xx?cid=9 8 Professor OH:http://medvedevgroup.com/medvedev.html 9 TA Office Hours:See Piazza 10 11 Past and curent schedule 12 8/21/2017 (recit)no recitation 13 8/21/2017Intro, Insertion sortCh 2.1SyllabusSlides posted on Canvas 14 8/23/2017Insertion sort, Loop InvariantsCh 2.1Insertion Sort animation 15 8/25/2017Running time analysis, Asymptotic notationCh 2.2, 3.1 16 8/28 (recit)no recitation 17 8/28/2017Asymptotic notationCh 3.1 18 8/30/2017Asymptotic notation, Merge sortCh 2.3Homework 1 postedMerge Sort Animation 19 9/1/2017Maximum-subarray problemCh 2.3, Ch 4.1 20 9/4 (recit)Labor day, no class 21 9/4/2017Labor day, no class 22 9/6/2017Maximum-subarray problemCh 4.1Homework 1 due 23 9/8/2017Solving Recurrence relationsCh 4.5 and 4.3 24 9/11 (recit)Review of Homework 1 solutions and asymptotic practice problems 25 9/11/2017Solving Recurrence relations and dynamic programming 26 9/13/2017DP: Fibonnaci Numbers and Rod Cutting Ch 15.1 27 9/15/2017Going over HW2 solutionsHomework 2 due 28 9/18 (recit)Review loop invariant proof of correctness of Merge algorithm 29 9/18/2017No class 30 9/20/2017DP: Rod cutting and Weighted Interval Scheduling 31 9/22/2017DR: Weighted interval scheduling (Evening Comprehension Exam) 32 9/25 (recit)DP: Independent set on a line practice problem 33 9/25/2017DP: Longest Common SubsequenceCh 15.4Homework 3 due, before class 34 9/27/2017DP: Longest Common Subsequence and Max Subarray problem 35 9/29/2017no class 36 10/2 (recit)Solutions to Comprehension Exam 37 10/2DP (Max Subbarray) and Greedy Algorithms (Unweighted Interval Scheduling)Ch 4.1, Kleinberg and Tardos, "Algorithm Design" 38 10/4/2017Greedy Algorithms (Unweighted Interval Scheduling) and Midterm ReviewHomework 4 due, before class 39 10/6/2017Greedy Algorithms: UIS and Interval Partitioning 40 10/9 (recit)Solutions to Homework 4 41 10/9/2017Greedy Algorithms: UIS Scheduling to minimize latenessCh 4.2, Kleinberg and Tardos, "Algorithm Design" 42 10/11/2017NO CLASS (Evening Midterm) 43 10/13/2017Abstract Data Types, Priority Queues and binary heapsCh 10.1, 10.2, 10.4, 6.1, 6.2, 6.4, 6.5 44 10/16 (recit)Midterm Solutions 45 10/16/2017Priority Queues and binary heapsCh 6.1, 6.2, 6.4, 6.5 46 10/18/2017Binary Search TreesCh 12.1, 12.2, 12.3 47 10/20/2017Red Black TreesCh 13.1, 13.2, 13.3Homework 5 due, before class 48 10/23 (recit)HW 5 (Greedy) solutions 49 10/23/2017Red Black Trees 50 10/25/2017Rank/Select augmentation Ch 14.1 51 10/27/2017Interval Trees: augmenting data structuresCh 14.3 52 10/30 (recit)Problem 14-1 from CLRS(10/29 11pm) Homework 6 due 53 10/30/2017Interval Trees: augmenting data structures 54 11/1/2017Hashing 55 11/3/2017Hashing 56 11/6 (recit)Q&A Session 57 11/6/2017GraphsCh 22.1 58 59 11/10/2017Depth First Search Ch 22.3 60 11/13(recit)Solutions to d-ary heap problem(11/12 11pm) Homework 7 due 61 11/13/2017Applications of DFS: Finding Cycles and Topological SortCh 22.4 62 11/15/2017Shortest paths propertiesCh 24, intro part 63 11/17/2017no class 64 11/20 - 11/26 holidays 65 11/27 (recit)Finding a cycle in O(V) time 66 11/27/2017no lecture 67 11/29/2017Shortest paths properties / Bellman FordCh 24.1Homework 8 due 68 12/1/2017Bellman Ford, Dijkstra's 69 12/4 (recit)Exercies 24.1-3 and 24-1 -- this will help with your homework! 70 12/4/2017Dijkstra's 71 12/6/2017Final Review, NP-completeness (Evening Comprehension Exam) 72 12/8/2017NP-completenessHomework 9 due before class 73 74 Tentative schedule 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

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# Block ordering and security in a MAC? To authenticate a message $m = m_1 \,\|\, \dots \,\|\,m_n$ the tag $t := F_k(r) \oplus F_k(m_1) \oplus \dotsb \oplus F_k(m_n)$ is used, where r is uniform random number $(0,1)^n$ and $m=(0,1)^n$. Even though the random number is prepended to the sequence there is still a chance of reordering the message blocks which makes it insecure. I am confused whether the blocks changed by an attacker make it insecure or not? • Is this problem 4.4b of Katz-Lindell? If so, think about how the authenticating party would verify the MAC. What information would they need? How would they get it? Commented Mar 20, 2012 at 1:29 • Hint: suppose an attacker saw the value message $r||m_1||m_2||m_3||t$, and modified it into the message $r||m_2||m_1||m_3||t$; would the modified message still have a valid tag? Commented Mar 20, 2012 at 3:07 • what's $F_k$??? Commented Mar 20, 2012 at 9:24 • Most likely a keyed hash function. Commented Mar 22, 2012 at 3:51 • According to the problem statement (Katz-Lindell 4.4b), F is simply a pseudorandom function. Commented Mar 26, 2012 at 14:43 So, assume there is a message $ABCDE$ (where each letter corresponds to one $n$-bit block). This will be sent as $rABCDEt$. If an attacker captures this message, she can reorder the blocks to create a second message $rBDCEAt$, and if you look at your formula, the tag for $BDCEA$ is exactly the same as the one for $ABCDE$, thus this mangled message will still validate as authentic.

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# All ______ triangles are similar. This question was previously asked in HSSC Group D Previous Year Paper 1 (Held On : 10 Nov 2018 Shift 1 ) View all HSSC Group D Papers > 1. Isosceles 2. Equilateral 3. Obtuse angled 4. Right angled Option 2 : Equilateral ## Detailed Solution Concept: Following are the properties of similarities of triangle: 1) AAA - When all angle of both the triangle are equal then they are similar. 2) SAS - When two sides and angle between them of two triangle are equal then they are similar. 3) SSS - When all the sides of  both the triangle are in same proportion then they are similar. Calculation: In all the equilateral triangle all the sides in same proportion. So they are similar by AAA property.

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Invisible patterns II Here you have a nice and clear example of corresponding squares from SOPE Black to move and draw. e1 = 5 (because I don't want it to look messy) The first invisible pattern you have to see is the line between the 2nd and 3rd rank. Black isn't allowed to pass this line because he has to keep an eye on d3 (rule of the square). The second invisible pattern contains the key squares (in blue) If white manages to conquer one of the key squares then black is lost. So black has to defend these squares. The third invisible pattern you have to see is that of the corresponding squares. If the white king appears on a square with a certain number, then black has to put his king on a corresponding square with the same number. Otherwise black is lost. As you see there is a shortest path between both area's with key squares, which for both white and black are of equal distance. If white decides to attack the key squares a3 and b3 he heads for square 1 (=a2) At the same time the black king has to run for the corresponding square with number 1 (=b4) Knowing all this, the defence of black is simple: 1. ... Kf3! ;corresponding square 5 (e1=5=f3) 2. Kd1 ;if white goes to f1 then black can counter attack against pawn c2 by 2. ... Ke3! That's why f1 has no number. 2. ... Ke3 3. Kc1 Kd4 4. Kb1 Kc5 5. Ka2 Kb4 = The same problem I already had encountered in Euwes book, but without the clear explanation. It costed me two hours to figure everything out correctly. I invented even the system of corresponding squares myself, using numbered beercaps. But I had made one mistake, because on Ka1 I had planned Ka4. Maybe because I needed so many beercaps I was not quite sober? In that case black comes to late when white heads to e2. So thats why SOPE saves me a lot of time. Further of course a warm applause for Celtic Death, who managed to slay 1001 enemies with 1 donkey jaw-bone and entered the Hall of Fame. 1. Once I finish the remaining Cycles, I think I will join you in some endgame training. Great diagrams! 2. I must say, from your reviews of SOPE, I have to agree with Sancho. I definitely could always use some improvement in my endgame especially if it is as easily understood (and hopefully "rapid") as you make it sound. GK 3. Hey, Tempo, fine lessons. Just go on. I can hardly wait. And it saves me to purchase a book, hehe ... At this very moment I am planning my training after the circles. My focus will be vision drills, and endgame vision fits very well in this scheme. 4. I originally though 1...Ke3 also draws here, but it doesn't. Great illustration with the numbers and colors.

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https://pty.vanderbilt.edu/2019/01/spring-savy-2019-day-1-discovering-the-third-dimension-kindergarten/

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# Spring SAVY 2019, Day 1 – Discovering the Third Dimension (Kindergarten) Posted by on Monday, January 28, 2019 in Grade K, SAVY. What a terrific launch to our spring session! The first part of our morning was dedicated to getting know one other. We focused on learning names of our new friends, explored our similarities and differences by playing a few rounds of the game “A Warm Wind Blows,” and shared why we decided to study geometry. Not surprisingly, we come together with a mix of specific goals (“I want to learn how to draw 3-D things”), general curiosity (“I think it will be amazing”), and delightful happenstance (“My mom chose it”). Our course explores the third dimension through the lens of structure, with the broad aim of orienting our attention to the relationship of between the mathematics of shape and space and our everyday experiences. More specifically, goals for the session are to support students along their path toward · Communicating mathematical ideas with precision by using vocabulary related to 3D shapes and their attributes · Combining regular geometric solids to model objects in the physical world and objects of one’s imagination · Developing spatial sense by composing and decomposing shapes in order to tackle 3D puzzles and problems · Investigating form and function by analyzing familiar toys and describing how the function is impacted by the form Students first engaged in an unstructured exploration at hands-on construction and design stations that included Kapla blocks, geoblocks, geometric solids, wooden cubes, pattern blocks, and attribute blocks. As they combined individual elements, peer conversations turned to the relationship between structure and shape. They shared observations (“It barely has anything that it’s standing on”) and predictions (“It’s not going to stand because it has a point [at the top]”), and tried again when structures collapsed unexpectedly. The need for precise vocabulary emerged naturally through our follow-up discussion, in which we shared our noticing and wonderings about three-dimensional shapes. To connect our small-scale investigation of shape with the world around us, each student selected one three-dimensional shape and brought it along on our Shape Detective Walk. (Thank you for the coats!) Our purpose was to begin to see the world through geometer’s eyes by identifying natural and constructed elements that were similar to the four shapes selected by students: cylinder, sphere, rectangular-based prism, and cube. Ask your mathematician: Which three-dimensional shape did you choose for the shape walk? Did you find lots of examples of your shape or just a few? Why do you think that might be true? Back in the classroom, we paused for snack and a conversation circle, then closed with a Construction Challenge. Using only ten blocks at a station of their choice, students tried to build the tallest possible structure. We undertook this as a design challenge rather than a competition. The challenge prompted our geometers to consider the affordances and constraints of particular shapes, and a post-construction “museum walk” provided an opportunity to learn from one another’s strategies. Some of us focused on putting “the biggest ones first” to create a stable base, whereas others prioritized inclusion of the “tallest” blocks. It was a true pleasure to share a Saturday morning exploring and learning alongside your children, and I hope they’re looking forward to our continued journey as much as I am. Your child’s thinking assignment in the week ahead is to study the world using geometer’s eyes. As a family, consider supporting this goal by playing “I Spy” with a focus on three-dimensional solids (e.g., “I spy, with my little eye, something in the shape of a cube”). Throughout our sessions, please continue to ensure that your child is dressed for the weather, and be sure to send a snack and water bottle! Christy Plummer Fun at Design Stations

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https://www.physicsforums.com/threads/when-is-following-equation-true.829582/

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# When is following equation true? Tags: 1. Aug 27, 2015 ### Rectifier 1. The problem When is the following equation true $\sqrt{c^2+14c+49} = c + 7$ a) for all real c b) for $c \geq -7$ c) for $c < -7$ d) c > 0 e) c < 0 The attempt 1 I know that the root of $c^2+14c+49 = 0$ is $c = -7$ and that this sqr-root is only defined for positive numbers. Thus the equation is true only when the stuff below the root is positive. But that stuff is always positive.... The attempt 2 $\sqrt{c^2+14c+49} = c + 7 \\ \sqrt{(c+7)^2} = c + 7 \\ c+7 = c + 7 \\$ Thus this equation is true for all real c:s. But somehow this is wrong. 2. Aug 27, 2015 ### andrewkirk Look at your last step in Attempt 2. Are you sure that it's always the case that $\sqrt{x^2}=x$, given that the convention is that the positive square root is always implied by the square root sign? What about if x=-1? What happens if you start with -1, square it and then take the square root (which is by convention positive). Do you end up with the number you started with? Last edited: Aug 27, 2015 3. Aug 27, 2015 ### Rectifier The answer in my book :) 4. Aug 27, 2015 ### Staff: Mentor The problem is in that last step, because $\sqrt{x^2} \neq x$. Can you see why? 5. Aug 27, 2015 ### andrewkirk Yeah, sorry, my first answer was too quick. I hope my redraft makes more sense. 6. Aug 27, 2015 ### Rectifier Yeah. Because negative x:es give different results. How can I implement that in my problem? 7. Aug 27, 2015 ### Staff: Mentor You have to use absolute values. 8. Aug 27, 2015 ### Rectifier So basically |c + 7| = c + 7 9. Aug 27, 2015 ### Staff: Mentor Yes. You should be able to convert that to a condition on $c$. 10. Aug 31, 2015 ### Staff: Mentor To be clear, |c + 7| is not equal to c + 7, as when, for example, c = -8. I believe that @DrClaude is in agreement with this, but the casual reader might misinterpret his comment. $\sqrt{(c + 7)^2} \neq c + 7$ but $\sqrt{(c + 7)^2} = |c + 7|$ 11. Aug 31, 2015 ### Staff: Mentor Of course I agree My point is that the question starts with: When is the following equation true $\sqrt{c^2+14c+49} = c + 7$ which is simplified to: When is the following equation true $| c+ 7| = c + 7$ from which it is easy to get a condition on $c$ for the original equation to be true. 12. Sep 3, 2015 ### HallsofIvy Use the definition of absolute value: $|a|= a$ if $a\ge 0$, $|a|= -a$ if $a< 0$.

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https://careforlifee.com/how-much-2-cycle-per-gallon/

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# How Much 2 Cycle Per Gallon? A two-cycle engine uses about two-cycles per gallon. A two-cycle engine is an internal combustion engine that completes two strokes while turning the crankshaft once. This type of engine is typically found in small, handheld power tools like chainsaws and leaf blowers. Two-cycle engines are more efficient than four-cycle engines, but they require oil to be added to the fuel to lubricate the engine. The amount of oil that needs to be added to the fuel depends on the manufacturer of the engine and the ratio of oil to gas. For example, a chainsaw with a 50:1 ratio would require 2.6 ounces of oil per gallon of gas. If you’re unsure of the ratio for your two-cycle engine, it’s best to err on the side of too much oil rather than too little. Too little oil can cause the engine to seize up, while too much oil will simply be burned off and create more smoke. ## How Much Oil Should Be Mixed With Gas For A 2-cycle Engine? For a 2-cycle engine, the oil to gas ratio is 1:50. If you have a 2-cycle engine, you need to mix oil and gas together before you fill up the gas tank. The gas to oil ratio for a 2-cycle engine is 50:1, so for every 1 gallon of gas you put in, you need to add 2.6 ounces of oil. For example, let’s say you have a 2-cycle engine with a 20-ounce gas tank. To fill it up, you would need to add 10.4 ounces of oil to the gas. ## What Is The Ratio Of Gas To Oil For A 2-cycle Engine? The gas-to-oil ratio for a 2-cycle engine is 50:1. 2-cycle engines are designed to run on a mixture of gas and oil. The gas to oil ratio for these engines is typically 50:1, meaning that for every 50 parts of gas, there is 1 part of oil. This gas to oil ratio is important for two-cycle engines because it helps keep the engine lubricated. If there is too much oil in the mixture, it can cause the engine to smoke and run poorly. Conversely, not enough oil can cause the engine to seize up. The best way to ensure the correct gas to oil ratio for a two-cycle engine is to use a premixed fuel. This fuel is already mixed to the correct ratio and takes the guesswork out of adding oil to gas. If you need to mix your own fuel, there are a few things to keep in mind. First, use a good quality oil that is designed for two-cycle engines. Second, be precise when measuring the gas and oil. It’s best to use a gas can that has markings for the correct amount of oil to add. Finally, don’t forget to shake the can of fuel before adding it to the engine. This will help to ensure that the oil is evenly mixed with the gas. Now that you know the gas to oil ratio for two-cycle engines, you can keep your engine running smoothly by using the correct fuel mix. ### FAQ #### How Much Gas Does A 2-cycle Engine Use? A 2-cycle engine uses a lot less gas than a 4-cycle engine. This is because a 2-cycle engine has two strokes per cycle, while a 4-cycle engine has four strokes per cycle. This means that a 2-cycle engine can do the same amount of work with half the amount of gas. #### How Much Oil Does A 2-cycle Engine Use? A 2-cycle engine uses oil for lubrication and cooling. The amount of oil used depends on the size of the engine and the operating conditions. ### Conclusion This keyword is most likely related to a question about how much 2-cycle oil to add per gallon of gas when mixing fuel for small engines. The amount of oil needed will vary depending on the manufacturer’s recommendations, so it is important to consult the owner’s manual or other reliable source for the correct amount to use. hopefully you are clear on the 2 cycle per gallon ratio. if you still have questions, please feel free to leave a comment below. ## Author • I am a fitness enthusiast and blogger. I have been working out for years and love to stay fit. I also enjoy writing about my workouts and helping others to stay motivated. I have a strong interest in health and fitness, and I love to share my knowledge with others. I am always looking for new ways to improve my own fitness level, as well as help others reach their fitness goals.

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https://it.mathworks.com/matlabcentral/answers/1680519-cutting-a-circular-ring-at-particular-points-using-angle-theta

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# Cutting a Circular Ring at particular points using angle-theta 1 view (last 30 days) hamzah khan on 25 Mar 2022 Edited: Les Beckham on 25 Mar 2022 Hello, I am drawing a circular ring using the following code: clear all; close all; p = linspace(-1/2,1/2,100); [X,Y] = meshgrid(p,p); % box mesh R = p(size(p,2))/2; r = R/1.5; alpha = linspace(-alpha,alpha,50); [alpha_X,alpha_Y] = meshgrid(alpha,alpha); theta = atan2(Y,X); active = (X.^2 + Y.^2 <= R^2 & X.^2 + Y.^2 >= r^2); figure() plot(X(active),Y(active),'o','MarkerFaceColor','red'); hold on This code is incomplete because I have to use alpha and theta to cut the ring at particular points. The inequalities that I have are: x^2+y^2<=R^2 x^2+y^2>=r^2 -alpha <=taninv(y/x)<=+alpha. i have plotted the first two, I am confused as how to plot the third one, alpha one. Does any on know how to plot the third inequality or how to use it to cut the ring at a particular angle. the result which i get from above code is: Les Beckham on 25 Mar 2022 Edited: Les Beckham on 25 Mar 2022 You were pretty close. See if you can adapt this to get what you want. p = linspace(-1/2,1/2,100); [X,Y] = meshgrid(p,p); % box mesh R = p(size(p,2))/2; r = R/1.5; theta = atan2(Y,X); % check size of theta; note it is already a mesh grid because X and Y are % the whole circle (radius constraints only) active = (X.^2 + Y.^2 <= R^2) & (X.^2 + Y.^2 >= r^2); plot(X(active),Y(active),'o','MarkerFaceColor','red'); % the portion that satisfies the alpha constraint as well active = (X.^2 + Y.^2 <= R^2) & (X.^2 + Y.^2 >= r^2) & (abs(theta) < alpha); hold on plot(X(active),Y(active),'o','MarkerFaceColor','blue'); Simon Chan on 25 Mar 2022 Try to add the follwoing lines: alphaA = 15*pi/180; % Set the margin, 15 degree here [thetaA,~] = cart2pol(X,Y); angleA = abs(thetaA)<=alphaA; active = (X.^2 + Y.^2 <= R^2 & X.^2 + Y.^2 >= r^2 & angleA); R2020a ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by

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http://www.chegg.com/homework-help/hutchinson-s-basic-mathematical-skills-with-geometry-8th-edition-chapter-9-problem-19x1-solution-9780073384177

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View more editions Hutchinson s Basic Mathematical Skills With Geometry # TEXTBOOK SOLUTIONS FOR Hutchinson s Basic Mathematical Skills With Geometry 8th Edition • 5201 step-by-step solutions • Solved by publishers, professors & experts • iOS, Android, & web Over 90% of students who use Chegg Study report better grades. 2013 Chegg Homework Help Survey PROBLEM Chapter: Problem: Business and Finance A real estate agent lists eight houses for sale at the prices shown below. \$209,000 \$224,900 \$249,900 \$215,000 \$289,900 \$265,000 \$274,900 \$749,900 (a) Calculate the mean home price of the agent’s listings. (b) Find the median home price of the agent’s listings. (c) Is the mean or median a better indicator of the typical home price listed? STEP-BY-STEP SOLUTION Corresponding Textbook Hutchinson s Basic Mathematical Skills With Geometry | 8th Edition 9780073384177ISBN-13: 0073384178ISBN: Authors:

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https://documen.tv/question/the-intensity-of-the-sound-wave-from-an-airplane-is-1-0-10-2-w-m-2-at-6-m-what-is-the-intensity-15424249-44/

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## The intensity of the sound wave from an airplane is 1.0 × 10^2 W/m^2 at 6 m. What is the intensity at 97 m? Question The intensity of the sound wave from an airplane is 1.0 × 10^2 W/m^2 at 6 m. What is the intensity at 97 m? in progress 0 3 years 2021-08-04T05:24:41+00:00 1 Answers 15 views 0 Explanation: A sound wave propagates in all directions radially from its source; therefore, it follows an inverse square law, it means that its intensity decreases inversely proportional to the square of the distance: where I is the intensity r is the distance from the source of the wave This means that we can rewrite the equation as: where in this problem: is the intensity of the sound wave when the distance is is the intensity of the sound wave when the distance is Solving for I2, we find:

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https://kmmiles.com/981-miles-in-km

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kmmiles.com # 981 miles in km ## Result 981 miles equals 1578.429 km You can also convert 981 km to miles. ## Conversion formula Multiply the amount of miles by the conversion factor to get the result in km: 981 mi × 1.609 = 1578.429 km ## How to convert 981 miles to km? The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km: 1 mi = 1.609 km To convert 981 miles into km we have to multiply 981 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result: 1 mi → 1.609 km 981 mi → L(km) Solve the above proportion to obtain the length L in km: L(km) = 981 mi × 1.609 km L(km) = 1578.429 km The final result is: 981 mi → 1578.429 km We conclude that 981 miles is equivalent to 1578.429 km: 981 miles = 1578.429 km ## Result approximation For practical purposes we can round our final result to an approximate numerical value. In this case nine hundred eighty-one miles is approximately one thousand five hundred seventy-eight point four two nine km: 981 miles ≅ 1578.429 km ## Conversion table For quick reference purposes, below is the miles to kilometers conversion table: miles (mi) kilometers (km) 982 miles 1580.038 km 983 miles 1581.647 km 984 miles 1583.256 km 985 miles 1584.865 km 986 miles 1586.474 km 987 miles 1588.083 km 988 miles 1589.692 km 989 miles 1591.301 km 990 miles 1592.91 km 991 miles 1594.519 km ## Units definitions The units involved in this conversion are miles and kilometers. This is how they are defined: ### Miles A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada. ### Kilometers The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.

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https://devsenv.com/example/-3209-beecrowd-online-judge-solution-3209-electrical-outlets-solution-in-c,-c++,-java,-js-and-python

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## Algorithm Problem Name: beecrowd | 3209 # Electrical Outlets Timelimit: 1 Roy has just moved into a new apartment. Well, actually the apartment itself is not very new, even dating back to the days before people had electricity in their houses. Because of this, Roy’s apartment has only one single wall outlet, so Roy can only power one of his electrical appliances at a time. Roy likes to watch TV as he works on his computer, and to listen to his HiFi system (on high volume) while he vacuums, so using just the single outlet is not an option. Actually, he wants to have all his appliances connected to a powered outlet, all the time. The answer, of course, is power strips, and Roy has some old ones that he used in his old apartment. However, that apartment had many more wall outlets, so he is not sure whether his power strips will provide him with enough outlets now. Your task is to help Roy compute how many appliances he can provide with electricity, given a set of power strips. Note that without any power strips, Roy can power one single appliance through the wall outlet. Also, remember that a power strip has to be powered itself to be of any use. ## Input Input vill start with a single integer 1 ≤ N ≤ 20, indicating the number of test cases to follow. Then follow N lines, each describing a test case. Each test case starts with an integer 1 ≤ K ≤ 10, indicating the number of power strips in the test case. Then follow, on the same line, K integers separated by single spaces, O1 O2 ... OK , where 2 ≤ Oi ≤ 10, indicating the number of outlets in each power strip. ## Output Output one line per test case, with the maximum number of appliances that can be powered. Input Sample Output Sample 3 3 2 3 4 10 4 4 4 4 4 4 4 4 4 4 4 10 10 10 10 7 31 37 ## Code Examples ### #1 Code Example with Javascript Programming Code - Javascript Programming const { readFileSync } = require("node:fs") .split(/\s+/, 1 + 20 * 11) .map(value => Number.parseInt(value, 10)) function main() { const output = [] const N = input.shift() for (let i = 0; i < N; i++) { const K = input.shift() const avaliablePlugsQuantity = input .splice(0, K) .reduce((total, plug) => total + plug, 1 - K) output.push(avaliablePlugsQuantity) } console.log(output.join("\n")) } main() Copy The Code & Input cmd 3 3 2 3 4 10 4 4 4 4 4 4 4 4 4 4 4 10 10 10 10 Output cmd 7 31 37 ### #2 Code Example with Javascript Programming Code - Javascript Programming var lines = input.split('\n'); var prompt = function(texto){ return lines.shift(); }; const test = parseInt(prompt("casos de test")); for (let i = 0; i < test; i++) { let lines = tomadas.shift(), soma = 0; for (let i = 0; i < tomadas.length; i++) { } soma = soma - (lines - 1) console.log(soma); } Copy The Code & Input cmd 3 3 2 3 4 10 4 4 4 4 4 4 4 4 4 4 4 10 10 10 10 Output cmd 7 31 37 ### #3 Code Example with Python Programming Code - Python Programming for i in range(int(input())): x=list(map(int,input().split())) z=x.pop(0) print(sum(x)-z+1) Copy The Code & Input cmd 3 3 2 3 4 10 4 4 4 4 4 4 4 4 4 4 4 10 10 10 10 Output cmd 7 31 37

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https://math.stackexchange.com/questions/2033098/variance-of-random-sum-of-random-variables-conditional-distributions

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# Variance of random sum of random variables (conditional distributions) Question Can you please tell me where I made error in my attempt? Random variables $X_j$ for $j=1,2,3,...$ are conditionally independent given random variable $\Theta$ and $\mathbb{E}(X|\Theta)=\Theta$, $\mathrm{Var}(X|\Theta)=4\Theta^2$. Random variable $N$ have conditional distribution $N|\Lambda=\lambda\sim Poisson(\lambda)$. R.v. $(X_1,X_2,...),N$ are independent. $\Theta \sim \Gamma(100,2)$ (edited) $\Lambda \sim \Gamma(2,4)$ (edited) $\Theta$ and $\Lambda$ are independent. Calculate variance of $S=\sum_{j=1}^NX_j \qquad$($S=0$ if $N=0$). Here is my attempt From given informations I calculate. $\mathbb{E}X_1=\mathbb{E}(\mathbb{E}(X_1|\Theta))=\mathbb{E}(\Theta)=50$ $\mathbb{E}X_1^2=\mathbb{E}(\mathbb{E}(X_1^2|\Theta))=\mathbb{E}(\mathbb{E}(X_1^2|\Theta)-(\mathbb{E}(X_1|\Theta))^2+(\mathbb{E}(X_1|\Theta))^2)=\mathbb{E}(\mathrm{Var}(X_1|\Theta)+(\Theta)^2)=\mathbb{E}(4\Theta^2+\Theta^2)=5\cdot\frac{(100+100^2)}{4}=12625$ $\mathbb{E}N=\mathbb{E}(\mathbb{E}(N|\Lambda))=\mathbb{E}\Lambda=\frac{1}{2}$ $\mathbb{E}N^2=\mathbb{E}(\mathrm{Var}(N|\Lambda)+(\mathbb{E}(N|\Lambda))^2)=\mathbb{E}(\Lambda+\Lambda^2)=\frac{1}{2}+\frac{2+4}{16}=\frac{7}{8}$ $\mathbb{E}(S)=\mathbb{E}(\sum_{i=1}^NX_i)=\mathbb{E}(N)\mathbb{E}(X_1)=\frac{1}{2}\cdot50$ $\mathbb{E}(S^2)=\mathbb{E}((\sum_{i=1}^NX_i)^2)=\mathbb{E}(\sum_{i=1}^NX_i^2+\sum_{i\neq j}X_iX_j)=(\mathbb{E}N)(\mathbb{E}X_1^2)+\mathbb{E}(N^2-N)(\mathbb{E}X_1)^2=\frac{1}{2}12625+(\frac{7}{8}-\frac{1}{2})\cdot50^2= 7250$ Finally $\mathrm{Var}(S)=\mathbb{E}S^2-(\mathbb{E}S)^2=7250-25^2=6625$ But the correct answer is $6634,375$. I guess I made some illegal step, but I cant find out what is wrong. Correct answer (Which i understand - but still can't understand what was wrong with my approach) We calculate first $\mathrm{Var}(S|\Theta)=\mathrm{Var}(N)(\mathbb{E(X_1|\Theta)})^2+\mathbb{E}N\cdot\mathrm{Var}(X|\Theta)=\frac{5}{8}\Theta^2+\frac{1}{2}\cdot4\Theta^2=2,625\Theta^2$ Then use $\mathrm{Var}(S) = \mathrm{Var}(\mathbb{E}(S|\Theta))+\mathbb{E}(\mathrm{Var}(S|\Theta))=\mathrm{Var}(\frac{1}{2}\Theta)+\mathbb{E}(2,625\Theta^2)=\frac{1}{4}\cdot25+2,625\cdot(100+100^2)/4=6634,375$ The mean of a Gamma Distribution with shape $\alpha$ and rate $\beta$ is $\alpha/\beta$. So $\mathsf E(\Theta)$ when $\Theta\sim \Gamma(2,4)$ is not $50$, it is $1/2$. And $\mathsf E(\Lambda)$ when $\Lambda\sim\Gamma(100,2)$ is not $1/2$, it is $50$. • I am so sorry but I erroneously swapped parameters for $\Theta$ and $\Lambda$ distributions in my question:( I edited my question with correct parameters. Nov 28, 2016 at 10:02

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Custom Search If you want to discuss more on this or other issues related to Physics, feel free to leave a message on my Facebook page. ## Sunday, January 16, 2011 ### Second equation of motion We have seen in a previous post the first equation of motion. Today we are going to see the second equation of motion. You would remember that the first equation dealt with acceleration, time taken, initial velocity and final velocity. So how do we derive the second equation of motion? Let us take an object that starts from an initial velocity u and then accelerates during a time t until it reaches a final velocity v. Now during the time the object was accelerating the object has moved a distance s. In the second equation of motion we will try to derive the equation to calculate this distance s. Fig 2 below shows what is happening to the object. Fig 1 Now you would remember that the area under a velocity-time graph is the distance travelled by the object. Hence we are going to calculate the area under the graph as indicated by the shaded area below. Fig 2 Now if s the distance travelled is the area under a velocity-time graph then we can say that Area under graph = 1/2( v + u)*t hence s = 1/2( v + u)*t Now if you can remember the first equation of motion is v = u +at Hence if we replace v = u +at into s = 1/2( v + u)*t we get s = 1/2( v + u)*t s = 1/2( [u +at] + u)*t s = 1/2( u + at + u)*t s = 1/2(2 u + at )*t s =1/2(2ut +at2) S = ut + 1/2at2 Hence as you can see above the second equation of motion is S = ut + 1/2at2 We are now going to look at two examples where the second equation of motion can be used. Example 1 A bus starts from rest and accelerates at a rate of 2.5 m/s2 for a time of 50 s. Determine the distance travelled by the bus during the acceleration phase. Now let us identify the different variables involved. u the initial speed = 0 m/s t time taken = 50 s a acceleration = 2.5 m/s2 s distance travelled = ??? If we want calculate s the distance travelled we will have to use the second equation of motion S = ut + 1/2at2 Substituting the different values we get S = ut + 1/2at2 = o*50 + 0.5* 2.5*502 = 0 + 1.25*2500 =3125 m Example 2 An aero plane travelling at a speed of 300 m/s lands on a track 2000 m long and decelerates for a time of 25 s until it comes to rest. Calculate the deceleration needed to stop the plane. We must now identify the different variables involved as in the first example. initial speed u = 300 m/s time taken t = 25 s Distance travelled s = 2000 m acceleration a = ???? If we want calculate the acceleration a we will have to use the second equation of motion S = ut + 1/2at2 and make a the subject of formula. s – ut = 1/2at2 a =2(s - ut)/t2 = 2(2000 – 300*25)/252 = –8.8 m/s2 As you can see using the second equation of motion is not that difficult. Should you enter into any difficulties just post the question into the comment section.

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# STATISTICS 014 1. A manager wishes to find out whether there is a relationship between the age of his employees and the number of sick days they take each year. The data is given below: Age (X) 18 26 39 48 53 58 24 50 Days (Y) 16 12 9 5 6 2 10 7 a. What is the null hypothesis? b. What is the strength and the direction of relationship? c. Is the relationship significant? d. What proportion of the number of sick days can be explained by the age of the employee? e. In a sentence or so, give an appropriate interpretation of the analysis results. 1. 👍 0 2. 👎 0 3. 👁 512 1. Use the Pearson r. To find if there is a statistically significant linear relationship between hours worked and error rate, use N-2 for degrees of freedom at the appropriate significance level for the test you are doing. Use a table for critical or cutoff values for a Pearson r. Compare the value from the table to the value you calculated using a formula for the Pearson r. If the value you calculate exceeds the critical value from the table, the null will be rejected. There will be a linear relationship in the population and the test will be statistically significant. If the value you calculate does not exceed the critical value from the table, then the null will not be rejected and you cannot conclude a linear relationship in the population. I hope this brief explanation will get you started. 1. 👍 0 2. 👎 0 2. Correction - the first sentence should read as follows: To find if there is a statistically significant linear relationship between age and number of sick days, use N-2 for degrees of freedom at the appropriate significance level for the test you are doing. Sorry for any confusion. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Health What is the best definition of a friendship 1:a long term connection with someone 2:a special relationship between people who enjoy being together 3:a special relationship between people who are the same age 4:a special 2. ### statistics a pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large plain pizza. how large should the sample be if she wishes to be accurate to within \$0.15? a previous study showed that the standard 3. ### MGF 1106 Four accounting majors, two economics majors, and three marketing majors have interviewed for five different managerial positions with a large company. Find the number of different ways that five of these people could be hired if 4. ### Algebra tony and his sister maria have the same birthday but tony is five years older than maria let the varible x represent tonys age and y represent maris age. which graph represents the relationship between tons age and marias age? 1. ### Math In a recent year there were the following numbers (in thousands) of licensed drivers in the United States. MALE: Age 19 and under - 4746 Age 20 - 1625 Age 21 - 1679 FEMALE: Age 19 and under - 4517 Age 20 - 1553 Age 21 - 1627 2. ### statistics help The branch manager of an outlet (Store 1) of a nationwide chain of pet supply stores wants to study characteristics of her customers. In particular, she decides to focus on two variables: the amount of money spent by customers and 3. ### Statistics Using advertised prices for used Ford Escorts a linear model for the relationship between a car's age and its price is found. The correlation coefficient is - 0.933 Determine R2 and interpret this statistic. A) R2 = 93.3% which 4. ### Language Arts In the third wish the king of the forest claims that he has yet to hear of the human being who made any good use of his three wishes in a paragraph consider whether mr peters proves the king wrong do mr peters wishes bring him 1. ### English Read the following sentence that contains informational text. Determine which type of word or phrase relationship is being used. The roads of New Orleans remained flooded due to the hurricane. (1 point) cause/effect relationship 2. ### Science The relative age of a rock is ? 1. its age based on how much carbon-14 it contains 2. its age compared with the ages of other rocks 3. less than the age of the fossils it contains 4. the number of years since it formed my answer 3. ### algebra The manager of a bulk foods establishment sells a trail mix for \$6 per pound and premium cashews for \$13 per pound. The manager wishes to make a 315-pound mixture that will sell for \$8 per pound. How many pounds of each should be 4. ### math-true/false 9.)The actual weight of 2-pound sacks of salted peanuts is found to be normally distributed with a mean equal to 2.04 pounds and a standard deviation of 0.25 pounds. Given this information, the probability of a sack weighing more

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## 15) Find the polar co-ordinates of points whose Cartesian co-ordinates are iv) (-√3,1) ​ Question 15) Find the polar co-ordinates of points whose Cartesian co-ordinates are iv) (-√3,1) in progress 0 1 month 2021-08-14T04:04:34+00:00 2 Answers 0 views 0 ## like all video MATHS The polar coordinates of the point whose cartesian co-ordinates are (−1,−1) is: Share Study later Rectangular to polar coordinate, (x,y)⟶(r,θ) radius,r can be found by using , r= x 2 +y 2 r= (−1) 2 +(−1) 2 = 2 θ can be found using the reference angle:tanθ= x y tanθ= −1 −1 ⟹tanθ=1⟹θ=tan −1 1= θ= 4 −3π As the point (-1-1) lies at Quadrant iii)i.e, θ= 4 −3π From (r,θ)=( 2 , 4 −3π )

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1. 0 ## "8-triple" Ladies Freeskate This might belong in the Stupid Questions thread (lol), but... If a woman (who lacked a 3A) wanted to be the first woman to ever do 8 clean triples in a freeskate, couldn't she just replace a 2A with a triple? Her triple wouldn't count for points because of the Zayak rule, but in terms of the actual skate (assuming all triples are clean), she would have technically been the "first woman to do an 8-triple FS". It'd likely never happen, but if a top skater like Medvedeva was slated to win a medium-tier event (Nebelhorn/Finlandia), and her normal layout is: 3F+3T, 3Z, 3F, 3L, 2A+2T+2T, 3S+3T, 2A, and she knew she would win (had a huge lead after the SP, was clean up until her final jumping pass in the FS)... she could easily change her final 2A to any triple (zero points but still a clean triple): 3F+3T, 3Z, 3F, 3L, 2A+2T+2T, 3S+3T, (any triple)*, and this would be the first ever ladies' 8-triple freeskate. 2. 0 Yes technically you can call it so. 3. 2 Asada did 8 triples in her Sochi LP. But why settle for 8 when you can do 9! 4. 8 The question is, is an 8-triple program still an 8-triple program if it earns points for only 7 triples and earns zero points for the 8th triple? How does that 8th triple count if it has zero BV, and how is such a program "technically" the first ladies 8-triple FS any more than Mao Asada's 8-triple FS (in which every triple earned at least some BV)? 5. 0 Originally Posted by largeman The question is, is an 8-triple program still an 8-triple program if it earns points for only 7 triples and earns zero points for the 8th triple? How does that 8th triple count if it has zero BV, and how is such a program "technically" the first ladies 8-triple FS any more than Mao Asada's 8-triple FS (in which every triple earned at least some BV)? This ^ If Zayak'ed jumps (ie. invalid elements) counted, then there'd be no point stopping at 8 triples - May as well Zayak multiple times and go for the first "12 triples" program. Or the first "20 triples" program... 6. 4 I'm sure Elaine Zayak could pull off an 8-triple skate as early as the 1980s 7. 0 Originally Posted by qwerty I'm sure Elaine Zayak could pull off an 8-triple skate as early as the 1980s And well, many other skaters could hit that also... I wonder what would be the max triple number if there was no Zayak rule =) 8. 2 Originally Posted by moriel And well, many other skaters could hit that also... I wonder what would be the max triple number if there was no Zayak rule =) In which case we might as well rename the sport 'ice jumping." 9. 0 Originally Posted by moriel And well, many other skaters could hit that also... I wonder what would be the max triple number if there was no Zayak rule =) During GPF practices Medvedeva was doing her normal layout with combinations changed to 3-3-3. So that would already be a 9 triple program she's perfectly capable of doing, probably even more. 10. 3 Maybe we could change the rules so Zhenia can get more points https://cdn2.vox-cdn.com/thumbor/pkd...in-block.0.gif 11. 0 I'd be happy if she can just perform her jumps consistently two years from now. 12. 0 Originally Posted by moriel I wonder what would be the max triple number if there was no Zayak rule =) Do we keep the current well-balanced program rules and just remove the restrictions on repeating the same triple? The current rules allow 7 jump passes in a ladies' freeskate, which can include three jump combinations of two jumps each, or one of them can include three jumps. That would make a total of 11 jumps. One of the jumps must be an axel. So, if you have a skater who can do triple axel, she could do 11 triple jumps within those slots. She would just have to repeat more than two jumps, or do the same jump more than twice, to fill that slot. Or any of the combinations could be replaced by a jump sequence. In a jump sequence the number of jump is not limited, but only the two hardest jumps count for points, at 80% of the base value. So if a skater wanted to do, say, 3S, half loop, change feet, 3T, mazurka, waltz jump, 1T, 3T, that should be considered one jump sequence with three triples . . . that would be worth 80% of a 3S+3T combination. Or are you asking, what if there were no rules at all, how many triples could a skater execute within a 4-minute time period without stopping? Should she also include 3 spins and a step sequence and choreo sequence, even if they're all as simple and brief as possible to save time and energy for more triples? 13. 0 Originally Posted by gkelly so, if you have a skater who can do triple axel, she could do 11 triple jumps within those slots. She would just have to repeat more than two jumps, or do the same jump more than twice, to fill that slot. 3a, 3t+3t+3t, 3t+3t, 3t+3t, 3t, 3t, 3t. 14. 0 That'd do it. 15. 1 Originally Posted by gkelly Or are you asking, what if there were no rules at all, how many triples could a skater execute within a 4-minute time period without stopping? According to these statistics compiled by coaches Christy Krall and Trevor Laak, no jump requires as long as a full second of air time. A quad toe has about 0.63 seconds in the air and a triple toe less than half a second. https://skatecoach.wordpress.com/201...s-trevor-laak/ A 4-minute program has 240 seconds. Assuming you didn't waste a lot of time setting up ("telegraphing") you should be able to do a couple hundred easily. Page 1 of 3 1 2 3 Last #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

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## Algebra 1: Common Core (15th Edition) $x=\pm\sqrt 33 +7$ $x=12.74$ or $x=1.26$ $x^2-14x+16=0$ $x^2-14x+49=-16+49$ $(x-7)^2=33$ $x-7=\pm \sqrt 33$ $x=\pm\sqrt 33 +7$ $x=12.74$ or $x=1.26$

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Cody # Problem 44. Trimming Spaces • What is Size? 1 – 10 of 4,192 #### Solution 1983818 Submitted 13 hours and 35 minutes ago • Incorrect This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1983815 Submitted 13 hours and 37 minutes ago • Incorrect This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1983535 Submitted 18 hours and 20 minutes ago by Tung Nguyen • Size: 39 This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1983522 Submitted 18 hours and 31 minutes ago • Incorrect This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1977057 Submitted on 15 Oct 2019 at 7:41 by Mariano • Size: 22 This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1977048 Submitted on 15 Oct 2019 at 7:37 • Incorrect This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1974375 Submitted on 13 Oct 2019 at 18:49 by Vu Luong • Size: 28 This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1973342 Submitted on 13 Oct 2019 at 2:22 by Daniel M • Size: 12 This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1973334 Submitted on 13 Oct 2019 at 2:11 • Incorrect This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. #### Solution 1971824 Submitted on 11 Oct 2019 at 17:05 by Nguyen manh Duy • Size: 34

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Which is greater? (a) – 875 = 125 or – 875 – – 125 (b) 7 + 6 X (-4) or – 8 + (-2) X (-8) (-1)​ Question Which is greater? (a) – 875 = 125 or – 875 – – 125 (b) 7 + 6 X (-4) or – 8 + (-2) X (-8) (-1)​ in progress 0 2 weeks 2021-09-10T10:00:57+00:00 1 Answer 0 views 0

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Successfully reported this slideshow. Upcoming SlideShare × Ies 2013 civil engineering objective paper ii solved question paper 6,592 views Published on Ies 2013 civil engineering objective paper ii solved question paper Published in: Engineering • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment Ies 2013 civil engineering objective paper ii solved question paper 2. 2. :: 2 :: (ACE Engg. Academy – Hyderabad, Vijayawada ,New Delhi, Bangalore, Bhubaneswar & Visakhapatnam) 01. Which of the following statements are correct? 1. A fine sprinkle of precipitation of small and rather uniform water drops with all drop diameters below 0.10 mm is called as drizzle. 2. The precipitation of liquid water with every drop diameter less than 0.5 mm is known as rain 3. Precipitation in the form of balls or irregular lumps of ice each over 5 mm in diameter is called hail. 4. Dew is formed directly by condensation on the ground mainly during night when the surface has been cooled by outgoing radiation (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 (d) 1 and 4 Ans: (c) 02. The double mass curve technique is used: (a) To find the average rainfall over a number of years (b) To estimate the missing rainfall data (c) To check the consistency of rain gauge records (d) To find the minimum number of rain gauges required in a basin Ans: (c) 03. The theory of synthetic hydrograph based on flood routing techniques is based on the principle that rainfall impulse (net storm rain) is modified by the factors: 1. The time of travel of the flow volume in channel and overland flow 2. Storage 3. Both translation and storage (a) 1 and 2 only (b) 1, 2 and 3 (c) 1 and 3 only (d) 2 and 3 only Ans: (b) 04. It is proposed to design a causeway along a village road. The return period for the annual maximum flood of a given magnitude was found to be 5 years. The probability that this flood magnitude will be exceeded at least once during next 2 years is: (a) 0.8 (b) 0.5 (c) 0.45 (d) 0.36 Ans: (d) 05. Wading technique is used: (a) To determine velocity of sea waves during Tsunami (b) To determine thickness of canal lining in alluvial soils (c) To measure the volume of dredging material in harbours (d) To determine velocity of flow in shallow streams Ans: (d) 06. A mean annual runoff 2 m3 /s from a catchment of area 10km2 represents an effective rainfall of nearly: (a) 530 cm (b) 590 cm (c) 630 cm (d) 658 cm Ans: (c) 07. Steep rise in the flow-mass curve during a certain period indicates: (a) Very high evaporation losses during that period (b) Flash floods during that period (c) Sudden spurt in irrigation demand during that period (d) Sudden rise in demand for water to meet hydropower generation. Ans: (b) 08. A rise crop is to be irrigated in a field covering an area of 2400 ha, the duty and base period of rice are given as 860 ha/cumec and 120 days File hosted by www.educationobserver.com/forum 3. 3. :: 3 :: (ACE Engg. Academy – Hyderabad, Vijayawada ,New Delhi, Bangalore, Bhubaneswar & Visakhapatnam) respectively. The volume of water required in the field is nearly: (a) 500 ha – m (b) 1400 ha-m (c) 2000 ha-m (d) 2880 –ha m Ans: (d) 09. In an irrigation project, in a certain year, 70% and 46% of the culturable command area in Kharif and Rabi, respectively, remained without water and rest of the area got irrigation water. The intensity of irrigation in that year for the project was: (a) 116% (b) 84% (c) 42% (d) 58% Ans: (b) 10. The gross command area for a distributory is 5000 hectares, 80% of which is culturable irrigable. The intensity of irrigation for Rabi is 50%; and for Kharif is 30%. The average duty at the head of the distributory is 2000 hectares/ cumec for Rabi season and 900 hectares/cumec for Kharif season. The discharge required at the head of the distributory from average demand considerations is: (a) 1.00 cumecs (b) 1.33 cumecs (c) 2.33 cumecs (d) 3.33 cumecs Ans: (b) 11. Due to continuous pumping for 7 hrs from a catchment of 25 ha, having porosity of 30% and specific retention of 10%, groundwater level dropped by 1m. The corresponding change in storage is: (a) 7.5 ha-m (b) 0.75 ha-m (c) 1.87 ha-m(d) 0.187 ha-m Ans: No Answer 12. As recommended by Sichardt, the radius of influence is : (a) Inversely proportional to drawdown (b) Linearly proportional to drawdown (c) Independent of drawdown (d) Proportional to square root of drawdown Ans: (b) 13. Consider the following statements: 1. A regime channel will have side slopes of value 0.5H : 1V. 2. Lacey’s regime formula is applicable to regime channels with sediment concentration more than 5000 ppm. 3. For a Lacey regime channel the Manning roughness coefficient is estimated by Strickler’s formula 4. The mean velocity in a Lacey channel is proportional to R2/3 where R = hydraulic radius. Which of these statements are correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 Ans: (b) 14. The minimum size of gravel that will not move in the bed of a wide rectangular channel of depth 0.8 m and longitudinal slope 0.0041 is: (a) 11 mm (b) 23 mm (c) 36 mm (d) 57 mm Ans: (c) 15. The flip bucket energy dissipater for a spillway is suitable where: 1. The tail water depth is low 2. The rock on the downstream is fragile and is erodible 3. The rock on the downstream is good and non-erodible (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only Ans: (b) File hosted by www.educationobserver.com/forum 4. 4. :: 4 :: (ACE Engg. Academy – Hyderabad, Vijayawada ,New Delhi, Bangalore, Bhubaneswar & Visakhapatnam) 16. Consider the following statements: 1. Groynes are constructed projecting from the bank into the river at the bank to be protected from flood 2. Repelling groyne projects downstream into the river –flow from the point of its origin from the bank 3. Attracting groyne projects upstream into the river from the point of its origin from the point of its origin from the bank 4. Perpendicular groyne projects normal to the river-flow from the point of its origin from the bank Which of these statements are correct? (a) 1 and 4 (b) 1 and 2 (c) 2 and 3 (d) 3 and 4 Ans: (a) 17. Which one of the following statements is correct? (a) In a retrograde vernier, (n1) divisions on the primary scale are divided into n divisions on the vernier scale. (b) A double vernier consists of two simple verniers placed end-to-end forming one scale with the zero in the centre (c) In an extended vernier, (2n+1) primary divisions are divided into n divisions on the vernier. (d) In a direct vernier, (n+1) primary divisions are divided into n equal divisions on the vernier scale Ans: (a) 18. The length of a line measured with a 30 m chain was found to be 734.6 . It was afterwards found that the chain was 0.05 m too long. The true length of the line was: (a) 630.82 m (b) 680.82 m (c) 735.82 m (d) 780.92 m Ans: (c) 19. Following offsets were taken from a survey line to a hedge: Distance (in meters 0 5 10 15 20 30 40 Offsets (in meters) 3 4 5.5 5 6 4 4.5 The area between survey line and the hedge is (by trapezoidal method): (a) 185.5 m2 (b) 187.5 m2 (c) 189.5 m2 (d) 289.5 m2 Ans: (b) 20. The bearing of line AB is 1500 and the angle ABC is 1240 . Bearing of line BC is: (a) 940 (b) 980 (c) 1980 (d) 900 Ans: (a) 21. The following observations were taken during testing of a dumpy level: Instrument at Staff reading on P Q P 1.475 m 2.205 m Q 1.440 m 2.060 m The collimation error in the instrument was: (a) +0.055 m (b) +0.005 m (c) 0.055 m (d) 0.005 m Ans: (a) 22. R.L of a floor level is 200.490 m. Staff reading on the floor is 1.695 m. Reading on the staff held upside down against the bottom of the roof is 3.305 m. Height of the ceiling is: (a) 3.5 m (b) 4.0 m (c) 5.0 m (d) 6.0 m Ans: (c) File hosted by www.educationobserver.com/forum 5. 5. :: 5 :: (ACE Engg. Academy – Hyderabad, Vijayawada ,New Delhi, Bangalore, Bhubaneswar & Visakhapatnam) 23. Which of the following terms related to leveling are correctly defined? 1. Line of collimation: Line joining the intersection of the crosshairs to the optical centre of the object glass and its continuation 2. Back-sight: First staff reading taken after the level is set up 3. Fore-sight: Last staff reading prior to shifting of level, or termination of the process of levelling 4. Height of instrument: Height of centre of telescope above the ground where the level is set up. (a) 1, 2, 3 and 4 (b) 1, 2 and 4 only (c) 1, 2 and 3 only (d) 2, 3 and 4 only Ans: (a) 24. Two stations P and Q are on the opposite banks of a river. Following observations were taken in reciprocal leveling. Instrument near Staff reading on P Q P 1.400 3.500 Q 0.600 2.200 R.L of P is 200.000 m, and then R.L of Q is nearly: (a) 199.3 (b) 201.7 (c) 200.0 (d) 198.2 Ans: (d) 25. Which of the following set of terms does not relate to operation of a theodolite? (a) Transiting and inverting (b) Face left and face right (c) Right swing and left swing (d) Gauging and sounding Ans: (d) 26. Consider the following statements: 1. Activity is a property typical of clay soils 2. An activity value of 7 in a clay soil is indicative of the presence of montmorillonite mineral 3. An activity value of 7 in a clay soil is indicative of the presence of illite mineral Which of these statements are correct? (a) 1, 2 and 3 (b) 1 and 2 only (c) 1 and 3 only (d) 2 and 3 only Ans: (b) 27. In a wet soil mass air, occupies one- fourth of its volume and water occupies one-half of its volume. The void ratio of this soil is: (a) 1 (b) 2 (c) 3 (d) 4 Ans: (c) 28. Which one of the following statements is correct ? (a) Grain size is the primary criterion for classification of course, as well as fine-grained soil (b) Grain size is the primary criterion for classification of coarse-grained soils (c) Plasticity curve classifies coarse grained soils (d) Plasticity characteristics relate to classification of coarse-grained soils Ans: (b) 29. Consider the following statements: 1. A conspicuous break in the continuity of a grain size distribution curve indicates a mixture of soil from two different layers 2. A steep grain size distribution curve indicates prevalence of nearly uniform grain size File hosted by www.educationobserver.com/forum

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And here is the link for the program cove calculation program over at FineWoodworking. After watching the video of Cove cut on table saw I tried in out on scrap and feel confident but How can I find the angles of the rabbett cuts on the underside of bar rail?. Computes fence layout for a given cove cut. Cove Cut Calculator. Table Saw characteristics (all measurements in inches) Table Width: Table Depth: Blade Diameter: Kerf: Blade X (Left-of-table to Right-of-blade): Blade Y (Front-of-table to Center-of-blade): Cove cut characteristics: Cove Width: Cove Depth: Shoulder (distance between Cove and Fence): Fence Angle: Degrees inches per foot. Need an custom-sized piece of cove molding? Learn how to cut just about any size cove molding safely and easily with a table saw cove cutting jig. Does anyone cut coves on their table saws anymore? 1) Cove cutting table calculator 2) Cove Cut Calculator 3) Cove Angle Calculator 4) Cove Cutter 1. Information on Cutting Coves on the Table Saw Blade Angle and Tilt Angle There have been several questions regarding cutting coves on the table saw. Cutting Curved Coves on a Table Saw. Cove Angle Calculator. Figured this is where to ask this question– The standard calculation for cutting a cove on a Table or Radial Arm Saw is attached. Then, check the teeth of the saw, how it’s going to enter the cove and exit. But you could just use the calculator too if it works well. Joe. Cove Moldings on the Tablesaw & The Parallelogram Cove Jig. And here is the link for the program cove calculation program over at FineWoodworking.com: Cove Angle Calculator. This Pin was discovered by Dan Mills. Discover (and save!) your own Pins on Pinterest. See more about Cutting Tables and Table Saw. ### 10 Cove Cutting Jigs, How To’s And Calculators Coves cuts on the table saw are probaably not necesary for paddle making. Here are a couple of links to Fine Woodworking magazine that have tips concerning cove cuts on a table saw: a calculator for designing any manner of coves using angle of the wood strip, sawblade height, and saw blade angle. In the table saw mode, the Mark V can perform a variety of operations. Kerfing, or thinning out, allows you to bend wood without steaming. You can also cut coves, inlays, and raised or pierced panels. Warning: Many of the special operations require the removal of the upper saw guard. The cove cut calculator allows you to calculate the fence angle to use a table saw to cut a cove of a desired width and depth using a given blade. And here is the link for the program cove calculation program over at FineWoodworking. You just divide W by E, and get the calculator to give you the arc sin of the result, and the answer will be your angle. Its Cove cutter calculator for cutting coves on a table saw. How to do a cove cut on a table saw. This guy was nice enought to include a cove cut calculator on his website. ### Specific Radius For Cove Cuts On Table Saw Here’s a useful tool for calculating a table saw cove cut. Example, suppose I want to cut a cove with a 4 radius 1/4 deep and 3 wide. Here is another good link with pictures.. and a template & calculator. You could do your cove cut (the inside curve) on the table saw using different methods. If you are doing the inside cove on the table saw, you may find Mattias Wandel’s cove calculator useful. Shout-out to Matthias Wandel and his Cove Cutting Calculator which I used to figure out the angle setup. I would think a person could build a 2 V cutter somehow for the table saw. bg. My handy dandy cove calculator tells me I can set my table saw fence at a 3 degree angle to the blade.

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# 1574 Avoiding Fractions When Using the Substitution Method to Solve Simultaneous Equations Contents ### The Substitution Method Jo Morgan examines six methods of solving simultaneous linear equations in her wonderful book A Compendium of Mathematical Methods. Six is two more than the number of methods I had known previously. Jo loves pouring over vintage mathematics books to learn how concepts were taught in days gone by. She loves finding resources on Twitter to teach concepts and describes herself as a Resourceaholic. Today is her birthday, and this post is dedicated to her. In my experience, the substitution method is often the first method taught to solve simultaneous linear equations, and usually, before the accompanying homework assignment is finished, fractions will be part of the solution process. Once students learn other methods, they usually abandon the substitution method with its troublesome fractions. However, the substitution method is really just as good as any other method, and fractions can actually be avoided when using it! I encourage students to use the substitution method and avoid fractions while they work. In the table below I use the substitution method twelve ways to find one coordinate of the solution. (In each case the other coordinate could be found by substituting the known value back into one of the original equations and solving for the other coordinate.) Notice that only the first column uses fractions; the other two columns do not. Follow each solution process step by step. Do any of the methods seem less confusing than the others? How will you approach the problem the next time you are asked to solve simultaneous linear equations? ### Factors of 1574: This is my 1574th post. Here is some information about that number: • 1574 is a composite number. • Prime factorization: 1574 = 2 × 787. • 1574 has no exponents greater than 1 in its prime factorization, so √1574 cannot be simplified. • The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1574 has exactly 4 factors. • The factors of 1574 are outlined with their factor pair partners in the graphic below. ### Another Fact about the Number 1574: 1574 is in just one Pythagorean triple: 1574-619368-619370, calculated from 2(787)(1), 787²-1², 787²+1². It can also be calculated from 2(394²-393²), 4(394)(393), 2(394²+393²). This site uses Akismet to reduce spam. Learn how your comment data is processed.

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, 03.08.2020 14:01 imstupid77 # Asap help !! How much additional interest is earned if \$8000 is invested for 7 years at 6.5% when interest is compounded annually, as compared with simple interest paid at the same rate? ### Another question on Mathematics Mathematics, 21.06.2019 19:30 Your food costs are \$5,500. your total food sales are \$11,000. what percent of your food sales do the food costs represent?

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Question # Let ABC be a triangle and M be a point on side AC closer to vertex C than A. Let N be a point on side AB such that MN is parallel to BC and let P be a point on side BC such that MP is parallel to AB. If the area of the quadrilateral BNMP is equal to $\frac{5}{{18}}$ times the area of triangle ABC, then the ratio AM/MC equals.A. 5B. 6C.$\frac{{18}}{5}$ D.$\frac{{15}}{2}$ Hint-Make use of the property of similar triangles and try to solve this problem Using the data given let us draw the figure Let us consider the length of AM to be=x and MC =y Also it is given that MN is parallel to BC, so BN is the transversal So from this we get $\begin{gathered} \angle ANM = \angle ABC \\ \angle AMN = \angle ACB \\ \end{gathered}$ (corresponding angles) And also MP is parallel to NB. So, we get $\angle ANM = \angle MPC$(Since MP is parallel to BN ) So, from this we can write $\vartriangle ANM \sim \vartriangle MPC \sim \vartriangle ABC$ So by theorem, ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. $\frac{{area\vartriangle ANM}}{{area\vartriangle ABC}} = \frac{{{{(AM)}^2}}}{{{{(AC)}^2}}} = \frac{{{x^2}}}{{{{(x + y)}^2}}}$ ------(1) $\frac{{area\vartriangle MPC}}{{area\vartriangle ABC}} = \frac{{{{(MC)}^2}}}{{{{(AC)}^2}}} = \frac{{{y^2}}}{{{{(x + y)}^2}}}$----------(2) From the data it is given that $area\vartriangle ANC + area\vartriangle MPC = area\vartriangle ABC - area\square NMCB = area\vartriangle ABC - \dfrac{5}{{18}}\vartriangle ABC = \dfrac{{13}}{{18}}area\vartriangle ABC$ Now ,let us add eq(1) and eq(2) So we get $\begin{gathered} \frac{{13}}{{18}} = \frac{{{x^2} + {y^2}}}{{{{(x + y)}^2}}} \\ \Rightarrow 5{x^2} - 26xy + 5{y^2} = 0 \\ \Rightarrow 5{x^2} - 25xy - xy + 5{y^2} = 0 \\ \Rightarrow (5x - 1)(x - 5y) = 0 \\ \Rightarrow \frac{x}{y} = 5{\text{ OR }}\frac{x}{y} = \frac{1}{5} \\ \end{gathered}$ But, also it is given that x>y

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# How to resolve this proportion/equivalence calculation? [simple one] Let's suppose I have one cat and when buying food for him I have to take into account this: 1 cat eats 2kg of food each 20 days How can I get a formula to know how many days my food will last based on how many cats I have and how many food I've bought? Example: 2 cats, I've bought 10kg of food 5 cats, I've bought 6kg of food How many days the food will last? I need a formula so I can solve any input(cats, food kg bought) and get a output(days food will last) Thanks!!!!! • 1 cat eats 0.1 kg / day. Commented Jun 16, 2016 at 0:30 ## 2 Answers From the condition we have that a cat eats $0.1$ kg per day. So therefore if we have $x$ cats and $y$ kgs of food. Then those $x$ cats will eat $x\cdot(0.1)$ kgs of food in one day. Divide $y$ by this and you will get wanted value. In other words $$\text{Days} = \frac{10 \cdot \text{Kilos of Food}}{\text{No. of Cats}}$$ • And how do I transform this into days my food will last? Since I don't have this information. The outcome in this situation: 10kg of food / 2 cats * 0.1) will result in 0.5. This can't be the number of days the food will last? Sry I am bit lost Commented Jun 16, 2016 at 0:48 • @HenriqueM. You divide by 0.1, not multiply by 0.1. So the answer will be 50 days instead. Commented Jun 16, 2016 at 0:50 • Awh! Now i see that. Then please edit your answer because I get a bit confused. Voted as right! Thanks! Commented Jun 16, 2016 at 0:51 • @HenriqueM. The answer is alright, but as dividing by 0.1 is same as multiplying with 10 I can fix that. Commented Jun 16, 2016 at 0:52 I take the first example: 2 cats, I've bought 10kg of food. The basic are the 20 days. $x=20 d\times \ldots$ Then you built one fraction of $2$ cats and $5$ cats and another fraction of $2$ kg food and $6$ kg food. The more cats you have the shorter the food last. Thus the fraction has to be smaller than $1$: $\frac{2}{5}$ $x=20 d\times \frac{2}{5} \times \ldots$ Now you go on with the food. THe more food you buy the longer the food last. Thus the fraction has to be bigger than $1$: $\frac{10}{6}$ In total we have $x=20 d\times \frac{2}{5} \times \frac{10}{6}=20d\cdot \frac{2}{3}=13\frac{1}{3}d$

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Statistics Statistics Statistics 1 Statistics: Category: Words: Amount: \$20 Writer: 0 Paper instructions Statistics: (Please show stepts) Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively up the food chain (fruit < vegetables < cereals < nuts < corpse). Geochemical surveys take soil samples to determine phosphorous content (in ppm, parts per million). A high phosphorous content may or may not indicate an ancient burial site, food storage site, or even a garbage dump. The Hill of Tara is a very important archaeological site in Ireland. It is by legend the seat of Ireland's ancient high kings†. Independent random samples from two regions in Tara gave the following phosphorous measurements (ppm). Assume the population distributions of phosphorous are mound-shaped and symmetric for these two regions. Region I: x1; n1 = 12540810790790340800890860820640970720 Region II: x2; n2 = 16750870700810965350895850635955710890520650280993 x1= ppm s1= ppm x2= ppm s2= ppm Let ?1 be the population mean for x1 and let ?2 be the population mean for x2. Find a 99% confidence interval for ?1 ? ?2. (Round your answers to one decimal place.) lower limit ppm upper limit ppm Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 99% level of confidence, is one region more interesting than the other from a geochemical perspective? We can not make any conclusions using this confidence interval. Because the interval contains only negative numbers, we can say that region II is more interesting than region I. Because the interval contains only positive numbers, we can say that region I is more interesting than region II. Because the interval contains both positive and negative numbers, we can not say that one region is more interesting than the other. Which distribution (standard normal or Student's t) did you use? Why?Student's t was used because ?1 and ?2 are known. Standard normal was used because ?1 and ?2 are known. Student's t was used because ?1 and ?2 are unknown. Standard normal was used because ?1 and ?2 are unknown.

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CC-MAIN-2021-25

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https://www.physicsforums.com/threads/rc-circuit-problem.45835/

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crawl-data/CC-MAIN-2016-44/segments/1476988719437.30/warc/CC-MAIN-20161020183839-00058-ip-10-171-6-4.ec2.internal.warc.gz

RC Circuit Problem 1. Oct 3, 2004 gregbellows Hi, I am stuck on a question to work out the value of a capacitor, The Question is: A capacitor and a 2Kohm resistor pass the same current when each is separately connected across a 100Hz power line. What is the value of the capacitor? 2. Oct 3, 2004 chroot Staff Emeritus The impedance of a capacitor is $$\frac{1}{j \omega C}$$. The resistance of the capacitor at 100 Hz is just the real part of that impedance (this is the hard part). Use Ohm's law (V = ZI) to solve for the capacitance. - Warren - Warren 3. Oct 3, 2004 don rigby ohms law shows us that e = i times r, when considering capacitance and frequencies, we substitute r with Xc (capacitive reactance). when a frequency is applied to a capacitor it allows current flow at a rate of ; Xc=1/[2 * pie(3.14) * f (freq.) *c(farrads)] ***this is the formula for the impedence of a capacitor. Xc= 2000, f= 100: therfore C= ???? you do the math Last edited: Oct 3, 2004

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CC-MAIN-2016-44

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https://www.reddit.com/r/personalfinance/comments/7zptkd/discover_bank_raises_interest_rates_for_online/

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× [–] 1982 points1983 points  (145 children) Outstanding now I can go to Taco Bell an extra time this year. [–] 339 points340 points  (125 children) You must be exceedingly wealthy for this to earn you enough for a that been over the year [–] 190 points191 points  (118 children) At a \$20 meal at Taco Bell (srsly?) you'd need \$20,000 in the account for the year (not including daily compounding interest). Wouldn't be too far fetched but its a lot of change in a low rate savings acct though. [–] 55 points56 points  (3 children) \$20,000 is \$300 a year at 1.5%. I have 2 Discover savings accounts and your numbers are incorrect. Oh, I get it....the extra .1% [–] 10 points11 points  (2 children) interest is compounded monthly! an extra \$2.07! [–] 4 points5 points  (0 children) The 1.5% is APY and therefore the full amount for the year. The monthly rate would be slightly lower with the compounding making up the difference. [–] 3 points4 points  (0 children) Man, and I'm sitting here just banking on a straight \$25 per month.....I'm a failure :-) [–] 34 points35 points  (2 children) I...I did not do the math. [–] 19 points20 points  (0 children) I’m confused..\$20K at this rate would get you \$300 in a year Edit: I’m dumb. You’re calculating the difference between the rates. Bye! :) [–] 2 points3 points  (1 child) [–] 2 points3 points  (0 children) Doing the difference in rates, not the full rate [–] 10 points11 points  (1 child) for a that been over the year For a what now? [–] 13 points14 points  (0 children) For a that been over a year [–] 642 points643 points  (144 children) Just deposited \$25k with them the other day so this news is worth about \$2 a month to me. [–] 20 points21 points  (4 children) Redneck.bank does 1.75%. They consistently raise their rates to be the best savings account APY. [–] 39 points40 points  (3 children) But don't they pay interest in ammo only? [–] 12 points13 points  (1 child) Wouldn't that be another bonus? [–] 61 points62 points  (68 children) You should look into a rewards checking account. Some offer 4%APY on balances up to \$20,000 [–] 137 points138 points  (35 children) Then he can go to Taco Bell 4 extra times this year [–][deleted] 11 points12 points  (19 children) All this talk about Taco Bell, now I really want Taco Bell. Has anyone tried their new fries? [–] 12 points13 points  (13 children) They’re meh even with the cheese. It’s barely a step above those school cafeteria fries and not very many of them. [–][deleted] 6 points7 points  (11 children) That's unfortunate to hear. I still want taco bell though [–] 12 points13 points  (8 children) The box that the nacho fries come in have a pretty good value. You get the fries, a cheesy gordita crunch, a Doritos loco taco, and a drink for \$5. The nacho fries aren’t bad either. Wish they came with a little more cheese though [–][deleted] 8 points9 points  (0 children) The Mountain Dew Baja blast is something else. God I love Taco Bell [–] 4 points5 points  (2 children) Those loco tacos just make me want more. They're addictive! [–] 2 points3 points  (0 children) It’s only a buck if they still serve it. Definitely edible but I’d use that dollar for a beefy fritos burrito or something else. [–] 2 points3 points  (0 children) wow what if he bought double the normal amount twice and saved on gas to [–] 9 points10 points  (2 children) I opened the Discover account mostly to get the bonus signup cash. Would appreciate any leads on specific rewards programs. [–] 2 points3 points  (17 children) Like what? [–] 18 points19 points  (16 children) Consumers Credit Union has 4.59% capped at 20k with the following requirements: • Complete at least 12 Debit/Check Card point-of-sale purchases (transactions must be made without using your personal identification number (PIN) to count toward the minimum of 12 and must post and clear your account as a credit transaction on or before the last day of the calendar month) • Each calendar month, one direct deposit OR one ACH debit OR pay one bill when using Consumers Credit Union's free online bill payment system must post and clear your account on or before the last day of the calendar month • Access Online Banking at least once each calendar month • Receive eDocuments (enroll and accept the disclosure) • Spend \$1000 or more in CCU Visa Credit Card purchase transactions**; no minimum number of transactions. Also La Capitol FCU 4.25% capped at 5k where the requirements are: • 25 or more non-ATM debit card transactions posted per month [–] 32 points33 points  (9 children) Spend \$1000 or more in CCU Visa Credit Card purchase transactions**; no minimum number of transactions. Every month? I guess that explains why they can give such a high interest rate. [–] 28 points29 points  (8 children) Yup, most of these Rewards checking accounts make money based off of your debit / credit transactions which help pay for the higher interest rate you are receiving. They are a fundamentally different product than a Savings account with no requirement, but without fail someone will bring them up as an alternative to a savings account with no requirements. [–][deleted] 14 points15 points  (1 child) That's a lot of work just to keep that up [–] 22 points23 points  (0 children) Lol good luck meeting those requirements. Not even close to worth the trouble. [–][deleted] 3 points4 points  (1 child) Exposing your debit card like that with a bank account holding \$20,000+ is stupid as fuck. You're just asking to get your numbers swiped and bank drained... Debit cards do not have the consumer protections credit cards do. [–] 11 points12 points  (6 children) I thought Discover compounded daily. Won't your 25K be worth almost \$400 at the end of the year, plus the \$200 bonus? [–] 15 points16 points  (3 children) That's right, at 1.5% APY, if it sits for a year it earns ~\$378, or ~\$31.50 per month (plus the \$200 bonus). The bump from 1.4% to 1.5% accounts for some \$2 of that monthly amount. [–] 20 points21 points  (0 children) It doesn't matter how often it compounds. The very difference between APR and APY is that the latter is the entire amount of interest you get over a year, after taking into account any compounding. The entire interest from a \$25k principal over a year is exactly 1.5% of \$25k, which is \$375. The difference from 1.4% to 1.5% is \$25, and so the monthly value of the news is just over \$2.08. [–] 223 points224 points  (37 children) C'mon Ally! Even though you just raised your rates in early-February. [–] 103 points104 points  (24 children) It feels like constant increases lately, which is... good? [–] 171 points172 points  (13 children) The increases are happening because the Federal Reserve is increasing it's Federal funds rate. The last increase came in December when the Fed increased it from 1.25% to 1.5%. Increases in the funds rate typically leads to increases in savings account interest rates because the supply of money banks can borrow becomes more constrained. [–] 68 points69 points  (2 children) So, essentially, the cost for banks to borrow money increases, so they raise the rates on their deposit accounts to attract more customers? [–] 37 points38 points  (0 children) Fed funds rate is the rate at which they borrow from other banks cash reserves held at the Fed. Cash deposits like savings accts and CD’s are significant cheaper cost of funds and they always keep those costs aligned underneath what they’d lend out the excess deposits at the fed funds rate (if you get tons of cash with no mortgages or lending to underwrite, you’d store it as reserves with the fed and lend to other banks on overnight basis). [–] 6 points7 points  (3 children) So does this affect mortgage rates? I.e. when banks' interest rates increase, so do mortgage rates? [–] 11 points12 points  (1 child) Yup, that's exactly right. Mortgage rates have been rising for about 7 weeks now. [–] 2 points3 points  (0 children) Yeah, that's kind of what I figured. Thanks. [–] 18 points19 points  (5 children) good for savings accts, bad for people that are planning on buying a place this year. [–] 15 points16 points  (4 children) In the long term though, rising rates should decrease the cost of the house itself. People typically buy based on whether or not they can afford the monthly payment, not the total cost of the home. Since a given payment of \$3000/month, for example, will buy less house if rates are higher, this will exert downward pressure on housing costs over time. [–] 8 points9 points  (1 child) Until the Californians show up and pay cash. [–] 2 points3 points  (0 children) And wealthy Chinese. Friends were outbid by chinese people who paid cash and weren't even going to live there. It's just an investment to them for their future kids in like 15 years. [–] 19 points20 points  (1 child) So happy this sub turned me onto Ally. Convinced my wife to move her savings from her bank to Ally just before the most recent rate increase. [–] 11 points12 points  (6 children) Switch to Discover, use promo code SD17, open account by the 28th and fund before 3/14 (?) with 15k or 25k for \$150 or \$200 bonus. [–] 25 points26 points  (17 children) Any benefit of Ally, discover, etc over the other? Besides the varying interest rates? (some of which are higher than others depending on when you look it seems) I'm going to open a high yield for my emergency fund [–] 2 points3 points  (4 children) Generally they will have fewer (and cheaper) fees than brick-and-mortar banks, no minimum balance in accounts, and much higher interest rates on savings accounts than most. The biggest drawback is that you can't deposit cash, so if you find yourself doing that pretty often you might want to stick with a traditional bank. If you fund your accounts solely with direct deposit or checks, you'll be fine. I believe many of them also allow you to open multiple accounts if you want to split your money into different buckets. So you could have several savings accounts, such as one for your emergency fund, one for vacation savings, one to save for a new car, etc. All the accounts would have the same interest rate, you'd just have it in pieces instead of one large chunk in a single account. I don't think most traditional banks will let you open more than one unless it's a special savings account with a bunch of restrictions. Like most credit unions will allow you to open "club" savings accounts, but you either have limited withdrawals from those accounts in a certain timeframe (such as per quarter or per year) or they return your money at a certain time of year and besides that they're deposit-only. This can be useful if you have trouble with budgeting where all the money is in one account, you can use them to split them out and help visualize what money is to be used for what purpose in your budget. [–] 58 points59 points  (17 children) Tax season. People need to put their money somewhere [–] 93 points94 points  (13 children) Yeah right. Most people are buying TVs with their tax refund. [–] 10 points11 points  (2 children) I paid off the rest of my CC debt with it, so I'm not even mad lol [–] 7 points8 points  (1 child) That's exactly what I did. I feel relieved to be debt-free. [–] 5 points6 points  (0 children) 11k credit card co.pletely paid off :D [–] 3 points4 points  (0 children) Oooh I did this and have no regrets after 4 months without a TV. Also paid off two credit cards and a private student loan as well and tossed just over 1/3 of the refund into growing my emergency fund. [–] 5 points6 points  (1 child) I know, right? Someone I know got their \$500 or so refund back and said ,"Hmm, I wonder what I should buy with this..." [–] 4 points5 points  (1 child) Straight to student loans for me :/ [–] 8 points9 points  (9 children) Does discover bank compound daily? [–] 65 points66 points  (7 children) Not sure if you're aware or not, but people have misconceptions about the importance of annual vs. daily compounding, so some math might be helpful. This has a much lower impact that most people think. With annual compunding at a 1.5% rate, \$10,000 will grow to \$10,150 by the end of the year (\$150 gain). The math is pretty simple: 10,000 * 1.0151 = \$10,150. With daily compounding at the same 1.5% rate, \$10,000 will grow to \$10,151.13 by the end of the year (\$151.13 gain). It's more, so it's helpful and should be preferred if you have the option, but as a practical matter with such small rates, it doesn't make much of a difference (you can maybe buy a bottle of discount soda with the extra money). The math is a tiny bit tricker: \$10,000 * (1 + (0.015/365))365 = \$10,151.13. With some limit theory, you can shrink the compound period down to infinitesimally small buckets of time. This will allow you to compound money continuously. Again though, the difference is small. In this case, compounding continuously vs. compounding daily gets you less than a penny extra, with the total still rounded to \$151.13, but if you carry out all the digits, it's \$151.131 for the continuous compounding vs. \$151.128 for the daily. The math is \$10,000 * e.015 = \$10,151.13. You might know this already, but since people tend to have misconceptions about it, I thought it would be helpful to point out the math behind the scenes. [–] 34 points35 points  (4 children) A free Mc chicken is a free Mc chicken [–] 7 points8 points  (2 children) They raised the price! It used to be just a dollar! [–] 7 points8 points  (0 children) Yeah! What is this 1 2 3 nonsense. [–] 6 points7 points  (1 child) So you're saying I could get a soda at Taco Bell once a year. [–] 7 points8 points  (1 child) 2007 was a magical time. Citibank had a 5% interest rate. I wonder if it will ever go to 5% again. I heard that interest rates in the 80's were 6-8%. [–] 9 points10 points  (0 children) They were. And mortgage in the early 80's was ridiculously high as well. Like 18.5% [–] 50 points51 points  (34 children) Marcus by Goldman Sachs has had 1.5% since the beginning of the year with 2.05% 12 month CD. Seems like they are being the most aggressive about it. No minimums either. [–] 72 points73 points  (27 children) Hate the company [–] 50 points51 points  (2 children) Those bastards won't ever get my money. [–] 30 points31 points  (0 children) God bless you principled consumers! [–] 11 points12 points  (0 children) Love that GS deal [–] 5 points6 points  (1 child) I have a couple 12-months CD's with them. I don't like thier new web interface since the branding swap to Marcus. [–] 3 points4 points  (0 children) Its so bad... It was almost unusable when it switched over. [–] 5 points6 points  (1 child) Push it to 1.6 Goldman Sachs! You can do it! [–] 35 points36 points  (5 children) Call me when it's on par with inflation. [–] 42 points43 points  (4 children) Latest print of Core PCE (the Fed's preferred measure of inflation) was 1.5%. What's your number? Happy to give you a call. https://www.investing.com/economic-calendar/core-pce-price-index-905 [–] 23 points24 points  (1 child) So my money is the same as it was a year ago. Lit. [–] 6 points7 points  (0 children) Taco Bell isn’t looking so good anymore. [–] 4 points5 points  (1 child) I thought the BLS's CPI was a better indicator. That's how I learned it when I was working on my bachelor's. Even that points to 1.5%. That being said inflation will probably go up in the next few years. The Fed's target is historically 2%. I think it might go past that. [–][deleted] 4 points5 points  (0 children) This made me curious so I checked out my Ablebanking online money market account and they raised their interest rate from 1.3% to 1.7%. [–][deleted] 4 points5 points  (0 children) Nice post. 1.5% is also the best I've found as well. Using Marcus online bank by Goldman Sachs gives you the same percentage. Amazed at how few people use these high APY online savings banks especially for a 3-6 month rainy day fund. [–] 14 points15 points  (3 children) Wow. And I just signed up last night too. What a coincidence. :) [–] 24 points25 points  (0 children) I'd say that we've all got you to thank! [–] 9 points10 points  (1 child) Barclays US is there too. Bumped up their CDs as well. [–] 7 points8 points  (4 children) Can banks reduce their interest rates after raising them? Or are we "grandfathered in" to these rates? [–] 18 points19 points  (19 children) After chasing these "deals" for relatively small rewards (still not beating inflation), I've decided to throw my entire savings into VBTLX. There is just no sense in me keeping \$30k sitting in a 1.5% savings account. [–] 67 points68 points  (16 children) Putting 30k into a Bond portfolio with an effective duration of 6.11 years during a rising rate environment could be problematic... To illustrate this, VBTLX has returned 0.51% over the past 12 months, and 1.20% annualized over the past 3 years. Your returns are significantly lower than the funds yield as a result of principle being eroded as interest rates have risen. As long as current economic conditions continue, and we don't spiral into a recession, a savings account yielding 1.50% will almost assuredly outperform a broad bond fund that isn't taking duration bets, which is where interest rate risk is embedded. [–] 3 points4 points  (3 children) VBTLX Don't you get some sort of bond payment per year? [–] 38 points39 points  (2 children) Yes, you receive interest payments, from the underlying treasury bonds within the mutual fund portfolio. However, traditional treasury bonds have two components of risk; interest rate risk, and credit risk (risk of default). Since this portfolio is mostly treasury bonds, backed by the United States Treasury, let's assume they have no credit risk. Therefore there is one component of risk, interest rate risk. A brief, and oversimplified explaination of what interest rate risk is: Say I sell you a bond for \$100, that will mature (end) in 10 years, and promise to pay you the going market rate of 2% per year, or \$2. However, 3 years into our agreement, economic conditions have changed, and the going market interest rate has increased from 2% to 3%. The bond I sold you that pays 2%, is no longer worth what it once was. If someone can go out and pay \$100 to get a bond that yields 3%, you need to discount the price of your bond so that the \$2 you are receiving is equal to 3% of the price of the bond This will then result in that the payment of \$2 = market rate of 3%, and the price of the bond decreasing. The current rising rate will not be favorable for those who hold high levels of interest rate risk. This mutual fund has these embedded risks. [–] 2 points3 points  (1 child) Thank you for the explanation. [–] 3 points4 points  (0 children) Of course, as long as it makes sense to you! [–] 5 points6 points  (10 children) I'm open to suggestions here. Per their website, Vanguard says that the fund returned 3.56% in 2017. What am I missing? [–] 19 points20 points  (6 children) You are not wrong. VBTLX returned 3.57% in the calendar year of 2017. This was against a backdrop of stable interest rates. If rates do not change, then you will continue to collect the funds stated yield or +2.92%. However, Year to date this same fund is down -2.32%. This is due to the federal reserve, who sets interest rate policy, indicating that they will be raising rates 3-4 times this year. As the federal reserve raises rates, treasury rates go up, and the price of this mutual fund will decrease as bonds are repriced lower (see explanation in other reply) In terms of what you should do, that depends on what the time horizon and the function these assets will be used for in the future. [–] 5 points6 points  (2 children) There's parts that should have been in the prior discussion that were very much omitted. Firstly, the "bond bear market" aka the lowering of bond prices have mostly happened this year, not 2017. Not that it wasn't unforseen, as some financial pundits were on it since at least 2016 (Andrew Horowitz comes to mind). Secondly, the lowered prices does not effect the coupon payments of purchased bonds, and it actually has as positive effect on short term bonds due to the steepening yield curve. Your money gets you better rates, no need to take a loss trading out or lower the funds NAV. Vanguard's fund, being diversified in both term and in the traditional sense, would experience a rise in coupon payments from the short term bonds with higher yields, as well as it still receives a coupon on a lower priced bond that it now might just hold onto and not claim the loss and accept a lower yield and lower NAV. This is where the lower prices come from. But a loss can be good too depending on the circumstances. Also some bonds on the books are expiring naturally anyway, and would then just be bought at the higher yield/NAV. Therefore, the bond fund NAV from the newer bonds on the books, plus reinvested fund distributions to you, could negate some of that bear action. 🐻 [–] 16 points17 points  (6 children) Capital one just raised to 1.5 as well.(for balances greater than \$10k) [–] 15 points16 points  (0 children) It's their money market account, not their regular savings account (which is still at 1.00%). Still a good place to plop an emergency fund. [–] 6 points7 points  (3 children) Why did you get downvoted? I just did this and got the sign up bonus of \$200. They deposited the bonus within a couple of days of depositing the initial funds. [–] 2 points3 points  (2 children) Huh, when did they add the \$200 signup bonus? I opened and funded a 360 Money Market account in January and didn't receive anything extra. Maybe because I only did an initial deposit of \$500 then deposited the remainder to push the account over \$10k when my old accounts closed the next week? [–] 3 points4 points  (0 children) It was targeted I think since I got something in the mail. [–] 5 points6 points  (11 children) Does discover online savings provide you with a card?? I’d like to make the switch to a savings account with higher interest and no card so I’m not even tempted to touch my savings [–] 31 points32 points  (7 children) What savings account has a card attached to it? I've never seen that before. Debit cards are only issued for checking accounts as far as I'm aware. [–] 3 points4 points  (0 children) We issue what’s called a cucard (credit union card) and it’s basically just an ATM only card for cash in emergencies out from savings they may be thinking of one of those [–] 4 points5 points  (0 children) FCNR gives 3% for fixed deposit for 1+ year. https://www.icicibank.com/nri-banking/RHStemp/rates.page [–] 2 points3 points  (0 children) Oh what generosity the king has bestowed upon us.

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CC-MAIN-2022-05

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https://www.excelif.com/closest-match/

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# Closest Match To find the closest match to a target value in a data column, use INDEX, MATCH, ABS and MIN in Excel. Use the VLOOKUP function in Excel to find an approximate match. 1. The ABS function in Excel returns the absolute value of a number. Explanation: C3-F2 equals -39. The ABS function removes the minus sign (-) from a negative number, making it positive. The ABS function has no effect on 0 (zero) or positive numbers. 2. To calculate the differences between the target value and the values in the data column, replace C3 with C3:C9. Explanation: the range (array constant) created by the ABS function is stored in Excel’s memory, not in a range. The array constant looks as follows: {39;14;37;16;22;16;17} 3. To find the closest match, add the MIN function and finish by pressing CTRL + SHIFT + ENTER. Note: the formula bar indicates that this is an array formula by enclosing it in curly braces {}. Do not type these yourself. The array constant is used as an argument for the MIN function, giving a result of 14. 4. All we need is a function that finds the position of the value 14 in the array constant. MATCH function to the rescue! Finish by pressing CTRL + SHIFT + ENTER. Explanation: 14 (first argument) found at position 2 in the array constant (second argument). In this example, we use the MATCH function to return an exact match so we set the third argument to 0. 5. Use the INDEX function (two arguments) to return a specific value in a one-dimensional range. In this example, the name at position 2 (second argument) in the range B3:B9 (first argument). 6. Finish by pressing CTRL + SHIFT + ENTER. Previous articleCalculated Field/Item in a Pivot Table – Easy Excel Next articleCount Cells with Text in Excel

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Mass: Mass is quantitative measure of inertia, a fundamental property of all matter. It is, in effect, the resistance that a body of matter offers to a change in its speed or position upon the application of a force. The greater the mass of a body, the smaller the change produced by an applied force. Although mass is defined in terms of inertia, it is conventionally expressed as weight. By international agreement the standard unit of mass, with which the masses of all other objects are compared, is a platinum-iridium cylinder of one kilogram. This unit is commonly called the International Prototype Kilogram and is kept at the International Bureau of Weights and Measures in Shvres, Fr. In countries that continue to favour the English system of measurement over the International System of Units (SI), the current version of the metric system, the avoirdupois pound is used instead. Another unit of mass, one that is widely employed by engineers, is the slug, which equals 32.17 pounds. Weight, though related to mass, nonetheless differs from the latter. Weight essentially constitutes the force exerted on matter by the gravitational attraction of the Earth, and so it varies from place to place. In contrast, mass remains constant regardless of its location under ordinary circumstances. A satellite launched into space, for example, weighs increasingly less the further it travels away from the Earth. Its mass, however, stays the same. For years it was assumed that the mass of a body always remained invariable. This notion, expressed as the theory of conservation of mass, held that the mass of an object or collection of objects never changes, no matter how the constituent parts rearrange themselves. If a body split into pieces, it was thought that the mass divided with the pieces, so that the sum of the masses of the individual pieces would be equal to the original mass. Or, if particles were joined together, it was thought that the mass of the composite would be equal to the sum of the masses of the constituent particles. But this is not true. With the advent of the special theory of relativity by Einstein in 1905, the notion of mass underwent a radical revision. Mass lost its absoluteness. The mass of an object was seen to be equivalent to energy, to be interconvertible with energy, and to increase significantly at exceedingly high speeds near that of light (about 3 108 metres per second, or 186,000 miles per second). The total energy of an object was understood to comprise its rest mass as well as its increase of mass caused by high speed. The mass of an atomic nucleus was discovered to be measurably smaller than the sum of the masses of its constituent neutrons and protons. Mass was no longer considered constant, or unchangeable. In both chemical and nuclear reactions, some conversion between mass and energy occurs, so that the products generally have smaller or greater mass than the reactants. The difference in mass is so slight for ordinary chemical reactions that mass conservation may be invoked as a practical principle for predicting the mass of products. Mass conservation is invalid, however, for the behaviour of masses actively involved in nuclear reactors, in particle accelerators, and in the thermonuclear reactions in the Sun and stars. The new conservation principle is the conservation of mass-energy. Excerpt from the Encyclopedia Britannica without permission.

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# [racket] How to verify prime numbers. From: Rodolfo Carvalho (rhcarvalho at gmail.com) Date: Thu Jul 14 02:04:07 EDT 2011 Previous message: [racket] How to verify prime numbers. Next message: [racket] How to verify prime numbers. Messages sorted by: [date] [thread] [subject] [author] ```Hello, Well, no one here is gonna do your homework! You could start showing what you've done so far and asking more specific doubts. We are not code translation robots... []'s Rodolfo Carvalho 2011/7/14 飞刀 <fly3ds at qq.com> > Hello everyone, > > I want a program to verify prime numbers which can be runned in mzscheme. > The following is C source. I don't konwn how to convert it to lisp. > > Thanks. > > #include <stdio.h> > #include <time.h> > #include <math.h> > #include <string.h> > #include <stdlib.h> > > #define E6 1000000 > > char table[E6]; > int num = 0; > //int prime[E8]; > > void cal_table() > { > int j, k; > time_t t1, t2; > t1 = time(NULL); > memset(table, 1, E6); > for (j = 2 ; j < E6; j++) { > if ( table[j] ) { > num++; > for ( k = j + j; k < E6; k += j ) > { > table[k] = 0; > } > } > } > t2 = time(NULL); > printf("Totaly %d primes until E6, cost %d time_t.\n", num, t2 - t1); > } > > /* > void init_prime() > { > int i; > num = 0; > printf("Init prime start..."); > for (i = 2; i < 2*E9; i++) { > if (table[i]) > prime[num++] = i; > } > printf("Init prime finished..."); > } > */ > > int main() > { > int i; > cal_table(); > //init_prime(); > return 0; > } > > > _________________________________________________

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# Optimization Lecture 1 Review of Differential Calculus for Functions of Single Variable. Size: px Start display at page: Download "Optimization Lecture 1 Review of Differential Calculus for Functions of Single Variable." Transcription 1 Optimiztion Lecture 1 Review of Differentil Clculus for Functions of Single Vrible Jnury 14 2 Outline Optimiztion Problems Rel Numbers nd Rel Vectors Open, Closed nd Bounded sets Rel Functions Limit nd Continuity Mimum nd Minimum of Functions Derivtive, L Hopitl s Rule nd Sttionry Points 3 Optimiztion Problems An optimiztion problem (OP) is problem of the form min f () s.t Ω This is minimiztion (we cn consider mimiztion of f s minimiztion of f), f is function to be minimized, s.t mens subject to, is the set of fesible vlues of vrible The obvious question is wht kinds of functions, wht kinds of sets nd wht kinds of vribles cn we hndle Ω The vrible cn be single rel vrible or vector, the function f will be ssumed to be of clss C (twice differentible), the set will be n open set or closed set Ω The net slides will eplin wht these concepts men 3 4 Integer, Rtionl, Irrtionl Numbers We ssume fmilirity with ddition +, subtrction -,division /, multipliction, less <, greter >, less or equl, greter or equl, (pproimtely ) equl =,different,equivlent, implies,or, nd, union, intersection, (not) member ( ), for ll, there eists one, b = = b = The integer numbers re the non-negtive integers,1,, union with the negtive integers -1,-,-3, A rtionl number is the rtio of two integers (e.g, ½). There re numbers tht cnnot be represented s the rtio of two integers nd re clled irrtionl numbers. An emple is the length of the circumference divided by its dimeter tht yields the number π = Rel numbers cn be either rtionl or irrtionl Emple: The re of rectngle with sides mesuring cm nd 1cm is A = 1= centimeters squred b = cm =1cm cm 4 5 Rel Numbers The set of rel numbers is denoted R. In R we cn define two closed opertions (ddition nd multipliction) tht hve inverses (subtrction nd division, respectively) This mens tht two rel numbers nd cn be dded to yield rel nd multiplied to yield rel or + b Associtive Property: Commuttive Property: Eistence of Identity: Eistence of Inverse: Distributive Property: (b + c) = ( b)+ ( c) A set with two opertions verifying these properties is clled field nd is represented by the trid ( R,+, ) All numbers < < b form n open intervl denoted (,b) Nottion for lrge products, sums: ( + b) + c = + ( b + c), (. b). c =.( b. c) + b = b +,. b = b. b + =, 1= + ( ) =, b b 1 =1, b N =, n = N N times n=1 5 b N b 6 Vectors nd Vector Coordintes A vector is geometric object represented by letter with n rrow, e.g., with mgnitude nd direction A vector hs coordintes in given frme. A vector is not the sme s its coordintes. Coordintes of the sme vector cn be different in different frmes. E The coordintes of vector in frme E will be denoted by nd will be represented s column rry of rel numbers. The trnspose of the coordintes of vector will be row rry denoted by E [ ] T nd the derivtive of vector with respect to t will be denoted by = lim (more lter) h [( ( t + h) ( t)) h] Emple: For D frme E with unit vectors ( e 1,e ) = te! 1 + t e, E \$ [ ] 1 = # & "# %& =! t \$ # & "# t %&, E [ ] T! = t t "# \$ %&, E! " \$ % =! 1 # " t = e 1 + e 1 6 [ ] \$ & % 7 Open, Closed, Compct Sets R Let be the set of rel numbers nd the set of rel vectors with n coordintes An open bll in with center nd rdius is the set of vectors verifying B r ( ) = : An open set in is set for which round ech point one cn find n open bll tht is fully contined in the set. For emple n open bll is n open set A closed set is set for which its complementry R n \ Ω which contins ll elements in R n tht re not in Ω is open. For emple { : r} A bounded set is set tht cn be contined in n open bll with finite rdius. For emple { : 1} A compct set is set tht is closed nd bounded 7 R n R n r > < r { } R n r 8 Rel Functions A rel function is mpping of rel vector into rel vlue f ( ). If is sclr, f is function of rel vrible For ech vlue of there cn only be vlue of f but there could be more thn one vlue of tht hve the sme vlue of f. Emple: f () = = ± f () When different vlues of correspond to different vlues of f the function is injective or 1-to-1. Emple: f ( ) = + 1 The set of vlues of for which f eists is clled the domin of f. The set of vlues f is clled the imge of f When the imge of f is equl to R then f is clled surjective or onto. Emple: f ( ) = A function tht is both injective nd surjective is clled bijective. These functions cn be inverted ( eists) A function f is liner if f (α 1 + β ) = α f ( f 1 1 )+ β f ( ), α, β R Emple: f ( ) = 8 9 Rel Vrible Functions: Trigonometric Stright lines cn be used to form tringles between three points A, B, C Pythgoren Theorem: sin A =, c b sin B = = c cos A = cos A, b c, cos B = tn A = c c = + b = sin A, b tn B = b = B C cot A Right Tringle c A A b B Oblique Tringle B Lw of the sines: sin A = b sin B = c sin C c Lw of the cosines: c = + b b cosc C C b A 9 A 10 Useful Trigonometric Reltions sin θ + cos θ =1 sin θ =.5.5cos(θ) sin(θ) = sin(θ)cos(θ) cos(θ) = cos (θ) sin (θ) tn(θ) = tn(θ) / (1 tn (θ)) sin(α + β) = sinα cosβ + cosα sin β sin(α β) = sinα cosβ cosα sin β cos(α + β) = cosα cosβ sinα sin β cos(α β) = cosα cosβ + sinα sin β sinα + sin β = sin α + β cosα β sinα sin β = cos α + β sin α β cosα + cosβ = cos α + β cosα β cosα cosβ = sin α + β sin α β 1 11 Polynomils of Rel Vrible A polynomil function of order n is function of the form f () = α +α 1 +α ++α n n, α i R If is rel vrible then we sy tht When n=1 then f is polynomil of first order We sy tht f is ffine if α nd liner if α = If n= then f is second order. We sy tht f is qudrtic Emples of polynomils of first nd second order: f () = +1 f () = f () = ( +1) = + +1 f R[] 11 12 Eponentil & Logrithm Functions The Euler s number e is pproimtely equl to.7188 nd is represented by the infinite sum e = f () = e The eponentil function is defined by The logrithm function is the inverse of the eponentil function, is defined by f () = ln() nd verifies the following importnt properties Inverse Function Property: ln(e ) =, e ln() = ln(y) = ln()+ ln(y) ln( ) = ln(), R 1 13 f () f ( 1 ) f ( ) f ( ) Limit of Function nd Continuity Suppose we hve sequence of numbers n, n =1,, getting closer nd closer to some vlue. For emple 1 =.9, =.99, 3 =.999, The vlues get closer nd closer to 1 s n increses. We sy tht converges nd its limit is 1,i.e lim =1 n n n The difference n+p n lso gets smller for ny p s n increses nd we sy tht is Cuchy sequence If f is function of we sometimes wnt to determine if f lso converges to vlue when n converges to limit This vlue, if it eists, is clled lim f ( ). If lim f ( ) = f ( ) then we sy tht f is continuous t. If f is continuous 1 t ll points we sy tht f is continuous function Emple: f ( ) = sin, = π lim n f ( ) = f ( π ) = π 13 14 The Dirc Delt Function As n emple of function defined by limit we will now tlk bout the impulse function An impulse function, lso clled Dirc delt function δ (t), is the mthemticl model of n impulsive force like, for emple, the one eerted in billrd bll: strong force in short period of time δ (t) The function cn be defined s the result of limiting process s follows: δ ( t) = limδ Δ( t), δ Δ Δ ( t) 1, = Δ, if t otherwise Δ δ Δ (t) 1 Δ Δ Are=1 14 t 15 Mimum nd Minimum of Function A vlue f ( ) is clled locl minimum (mimum) if δ > : f ( ) f ( ( f ( ) f ( ) ) ), B δ ( ) D If the result is true in ll the domin D of f then it is globl minimum (mimum) Weierstrss Theorem: A continuous function defined on compct set hs mimum nd minimum f () min m 15 16 f Derivtive of Sclr Function For function f() represented by its grph one cn imgine connecting two points P, Q on the grph by line. This line is clled secnt. If we now mke converge to vlue such tht Q converges to P then the secnt converges to the tngent line t P. The derivtive f of f t is the slope of the tngent nd is defined by the limit in the bo. A function tht hs derivtive t ll points is clled differentible. The second derivtive f is the derivtive of the derivtive Q Q () df d = f ( ) f ( ) f ( + h) f ( ) f '( ) = lim = lim h h P Emple: f ( ) = df d 1 1 (1 + h) = lim h + 1, = ((1) h ) = 17 Useful Derivtive Reltions f () = α, f '() = α α 1 f () = sin(), f '() = cos() f () = cos(), f '() = sin() f () = tn(), f '() =1/ cos () f () = cos()sin(), f '() = cos() f () = e, f '() = e f () = ln(), f '() =1/ ( f ()+ g())' = f '()+ g'() ( f ()g())' = f '()g()+ f ()g'() ( f () / g())' = ( f '()g() f ()g'()) / g () Theorem (Chin Rule): If the derivtives g () nd f (g()) eist nd F()=f(g()) then Emple: f ( ) = F( ) = f + 1, g( ) = sin ( g( )) = sin F '( ) = f '( g( )) g'( ) + 1, F'( ) = sin( )cos( ) = sin() 17 18 L Hopitl s Rule As we sw derivtive is the limit of rtio Sometimes when computing the limit of rtio of two functions one cn hve n indeterminte result / For these cses there eists result clled L Hopitl s rule stted in the following theorem Theorem: Let f nd g be two functions tht hve derivtive f '( ), g'( ) t ech point of n open intervl (,b). For point in this intervl let lim f ( ) = lim g( ) = f '( ) If lim g'( ) Emple: f ( ) = sin, df d = eists then lim sin( + h) sin() sin( h) = lim = lim = h h h h f ( ) g( ) = lim f '( ) g'( ), cos( h) lim h 1 = 1 df d = 1 18 19 Sttionry Points of Functions The derivtive of function t point is the slope of the tngent to the grph of the function t tht point. If the derivtive is zero then the slope of the tngent is zero Points for which the derivtive is zero re thus clled sttionry points of function. They cn be mimum or minimum (see figure below) f () Necessry Condition f '( min ) = f '( m ) = min m 19 ### MAT137 Calculus! Lecture 20 officil website http://uoft.me/mat137 MAT137 Clculus! Lecture 20 Tody: 4.6 Concvity 4.7 Asypmtotes Net: 4.8 Curve Sketching 4.5 More Optimiztion Problems MVT Applictions Emple 1 Let f () = 3 27 20. 1 Find More information ### Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln More information ### SUMMER KNOWHOW STUDY AND LEARNING CENTRE SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18 More information ### Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln More information ### The problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests. ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion More information ### DERIVATIVES NOTES HARRIS MATH CAMP Introduction f DERIVATIVES NOTES HARRIS MATH CAMP 208. Introduction Reding: Section 2. The derivtive of function t point is the slope of the tngent line to the function t tht point. Wht does this men, nd how do we More information ### approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below . Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve. More information ### Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector More information ### Topics Covered AP Calculus AB Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y. More information ### Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number More information ### MA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll More information ### ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b More information ### Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the More information ### Chapter 1: Fundamentals Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives, More information ### TO: Next Year s AP Calculus Students TO: Net Yer s AP Clculus Students As you probbly know, the students who tke AP Clculus AB nd pss the Advnced Plcement Test will plce out of one semester of college Clculus; those who tke AP Clculus BC More information ### Mathematics for economists Mthemtics for economists Peter Jochumzen September 26, 2016 Contents 1 Logic 3 2 Set theory 4 3 Rel number system: Axioms 4 4 Rel number system: Definitions 5 5 Rel numbers: Results 5 6 Rel numbers: Powers More information ### Edexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1 More information ### THE DISCRIMINANT & ITS APPLICATIONS THE DISCRIMINANT & ITS APPLICATIONS The discriminnt ( Δ ) is the epression tht is locted under the squre root sign in the qudrtic formul i.e. Δ b c. For emple: Given +, Δ () ( )() The discriminnt is used More information ### 13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in More information ### Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given More information ### Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some More information ### Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the More information ### Calculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer.. More information ### x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx . Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute More information ### The Algebra (al-jabr) of Matrices Section : Mtri lgebr nd Clculus Wshkewicz College of Engineering he lgebr (l-jbr) of Mtrices lgebr s brnch of mthemtics is much broder thn elementry lgebr ll of us studied in our high school dys. In sense More information ### Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!! Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble More information ### Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx... Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting More information ### Loudoun Valley High School Calculus Summertime Fun Packet Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember! More information ### ( β ) touches the x-axis if = 1 Generl Certificte of Eduction (dv. Level) Emintion, ugust Comined Mthemtics I - Prt B Model nswers. () Let f k k, where k is rel constnt. i. Epress f in the form( ) Find the turning point of f without More information ### Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite More information ### Obj: SWBAT Recall the many important types and properties of functions Obj: SWBAT Recll the mny importnt types nd properties of functions Functions Domin nd Rnge Function Nottion Trnsformtion of Functions Combintions/Composition of Functions One-to-One nd Inverse Functions More information ### ES.182A Topic 32 Notes Jeremy Orloff ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In More information ### The graphs of Rational Functions Lecture 4 5A: The its of Rtionl Functions s x nd s x + The grphs of Rtionl Functions The grphs of rtionl functions hve severl differences compred to power functions. 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Generl Review nd Introduction to Algebr Hierrchy of Arithmetic Opertions Opertions in ny expression re performed in the following order: More information ### Calculus AB. For a function f(x), the derivative would be f '( lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion: More information ### AP Calculus Multiple Choice: BC Edition Solutions AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this More information ### First Semester Review Calculus BC First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewise-liner function f, for 4, is shown below. More information ### Chapter 2. Vectors. 2.1 Vectors Scalars and Vectors Chpter 2 Vectors 2.1 Vectors 2.1.1 Sclrs nd Vectors A vector is quntity hving both mgnitude nd direction. Emples of vector quntities re velocity, force nd position. One cn represent vector in n-dimensionl More information ### Partial Derivatives. Limits. For a single variable function f (x), the limit lim Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles More information ### Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. 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AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following More information ### Review of basic calculus Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below More information ### a n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction. MAS221(216-17) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by More information ### The discriminant of a quadratic function, including the conditions for real and repeated roots. 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Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23 More information ### Mat 210 Updated on April 28, 2013 Mt Brief Clculus Mt Updted on April 8, Alger: m n / / m n m n / mn n m n m n n ( ) ( )( ) n terms n n n n n n ( )( ) Common denomintor: ( ) ( )( ) ( )( ) ( )( ) ( )( ) Prctice prolems: Simplify using common More information ### Mathematics Extension 1 04 Bored of Studies Tril Emintions Mthemtics Etension Written by Crrotsticks & Trebl. Generl Instructions Totl Mrks 70 Reding time 5 minutes. Working time hours. Write using blck or blue pen. Blck pen More information ### MATHEMATICS PART A. 1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC FIITJEE Solutions to AIEEE MATHEMATICS PART A. ABC is tringle, right ngled t A. The resultnt of the forces cting long AB, AC with mgnitudes AB nd respectively is the force long AD, where D is the AC foot More information ### 6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS 6. CONCEPTS FOR ADVANCED MATHEMATICS, C (475) AS Objectives To introduce students to number of topics which re fundmentl to the dvnced study of mthemtics. Assessment Emintion (7 mrks) 1 hour 30 minutes. More information ### Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between More information ### ECON 331 Lecture Notes: Ch 4 and Ch 5 Mtrix Algebr ECON 33 Lecture Notes: Ch 4 nd Ch 5. Gives us shorthnd wy of writing lrge system of equtions.. Allows us to test for the existnce of solutions to simultneous systems. 3. 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Exmples: displcemement, velocity, More information ### fractions Let s Learn to 5 simple lgebric frctions corne lens pupil retin Norml vision light focused on the retin concve lens Shortsightedness (myopi) light focused in front of the retin Corrected myopi light focused on the retin More information ### Unit 5. Integration techniques 18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd More information ### CET MATHEMATICS 2013 CET MATHEMATICS VERSION CODE: C. If sin is the cute ngle between the curves + nd + 8 t (, ), then () () () Ans: () Slope of first curve m ; slope of second curve m - therefore ngle is o A sin o (). The More information ### Chapter 8.2: The Integral Chpter 8.: The Integrl You cn think of Clculus s doule-wide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in More information ### MATH 174A: PROBLEM SET 5. Suggested Solution MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous rel-vlued function on I), nd let L 1 (I) denote the completion More information ### Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl More information ### 38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These More information ### If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible More information ### 7. Indefinite Integrals 7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find More information ### MATH 144: Business Calculus Final Review MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives More information ### Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x) More information ### 1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re More information ### THE NUMBER CONCEPT IN GREEK MATHEMATICS SPRING 2009 THE NUMBER CONCEPT IN GREEK MATHEMATICS SPRING 2009 0.1. VII, Definition 1. A unit is tht by virtue of which ech of the things tht exist is clled one. 0.2. VII, Definition 2. A number is multitude composed More information ### 3. Vectors. Home Page. Title Page. Page 2 of 37. Go Back. Full Screen. Close. Quit Rutgers University Deprtment of Physics & Astronomy 01:750:271 Honors Physics I Lecture 3 Pge 1 of 37 3. Vectors Gols: To define vector components nd dd vectors. To introduce nd mnipulte unit vectors. More information ### Chapter 4 Contravariance, Covariance, and Spacetime Diagrams Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz More information ### R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of More information ### The Wave Equation I. MA 436 Kurt Bryan 1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string More information ### AP Calculus AB Summer Packet AP Clculus AB Summer Pcket Nme: Welcome to AP Clculus AB! Congrtultions! You hve mde it to one of the most dvnced mth course in high school! It s quite n ccomplishment nd you should e proud of yourself More information ### We divide the interval [a, b] into subintervals of equal length x = b a n Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl: More information ### Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under More information ### Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic More information ### W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering) More information ### Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : , MT TRIGONOMETRIC FUNCTIONS AND TRIGONOMETRIC EQUATIONS C Trigonometric Functions : Bsic Trigonometric Identities : + cos = ; ; cos R sec tn = ; sec R (n ),n cosec cot = ; cosec R {n, n I} Circulr Definition More information ### Chapter 1: Logarithmic functions and indices Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4 More information ### Calculus. Rigor, Concision, Clarity. Jishan Hu, Jian-Shu Li, Wei-Ping Li, Min Yan Clculus Rigor, Concision, Clrity Jishn Hu, Jin-Shu Li, Wei-Ping Li, Min Yn Deprtment of Mthemtics The Hong Kong University of Science nd Technology ii Contents Rel Numbers nd Functions Rel Number System More information

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• Trigonometric Identities - Concept Math›Trigonometry›Trigonometric Functions How to use the unit circle to derive the tangent identity and the Pythagorean identity. • Trigonometric Identities - Concept Math›Precalculus›Trigonometric Functions How to use the unit circle to derive the tangent identity and the Pythagorean identity. • Using Trigonometric Identities - Problem 3 Math›Precalculus›Trigonometric Functions How to prove an identity using a Pythagorean identity and algebra. • Using Trigonometric Identities - Problem 3 Math›Trigonometry›Trigonometric Functions How to prove an identity using a Pythagorean identity and algebra. • Using Trigonometric Identities - Problem 1 Math›Trigonometry›Trigonometric Functions How to prove an identity using a Pythagorean identity and algebra. • Trigonometric Identities - Problem 1 Math›Precalculus›Trigonometric Functions How to use the Pythagorean identity to find cosine and tangent when sine is known. • Using Trigonometric Identities - Problem 1 Math›Precalculus›Trigonometric Functions How to prove an identity using a Pythagorean identity and algebra. • Trigonometric Identities - Problem 1 Math›Trigonometry›Trigonometric Functions How to use the Pythagorean identity to find cosine and tangent when sine is known. • Using Trigonometric Identities - Problem 2 Math›Trigonometry›Trigonometric Functions How to simplify a trigonometric expression using a Pythagorean identity and algebra. • The Identity Matrix - Concept Math›Precalculus›Systems of Linear Equations and Matrices How to determine the identity matrix of order n. • Using Trigonometric Identities - Problem 2 Math›Precalculus›Trigonometric Functions How to simplify a trigonometric expression using a Pythagorean identity and algebra. • Using Trigonometric Identities - Concept Math›Precalculus›Trigonometric Functions How to simplify a trigonometric expression by converting to sines and cosines and using algebra. • Using Trigonometric Identities - Concept Math›Trigonometry›Trigonometric Functions How to simplify a trigonometric expression by converting to sines and cosines and using algebra. • Trigonometric Identities - Problem 2 Math›Trigonometry›Trigonometric Functions How to use the unit circle to show that cosine is an even function, and sine and tangent are odd functions. • Trigonometric Identities - Problem 2 Math›Precalculus›Trigonometric Functions How to use the unit circle to show that cosine is an even function, and sine and tangent are odd functions. • Using the Sine and Cosine Addition Formulas to Prove Identities - Concept How to use the sine and cosine addition formulas to prove the cofunction identities.

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Question # The heights of adult men in America are normally distributed, with a mean of 69.1 inches... The heights of adult men in America are normally distributed, with a mean of 69.1 inches and a standard deviation of 2.68 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.1 inches and a standard deviation of 2.58 inches. a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? z = b) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)? z = Solution : Given , men mean = = 69.1 standard deviation = = 2.68 x=75.6 inches (6.3feet) using z-score formula z =X -/ z=75.6-69.1/2.68 z=2.46 b. woman mean = = 69.1 standard deviation = = 2.68 x=61.32 inches (5.11feet) using z-score formula z =X -/ z=61.32 -69.1/2.68 z=-2.90 #### Earn Coins Coins can be redeemed for fabulous gifts.

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# Arithmetic Operation on Two Arrays in C# 4.5 Introduction In this article I describe how to perform arithmetic operations on two arrays in C# .Net. In this article however I have declared both arrays with the same length to perform the arithmetic operations on an array, like addition, subtraction, multiplication and division. I have performed all these operations in a simple and easy way. So let us understand the procedure. • Open Visual Studio 2012 • "File" -> "New" -> "Project..." • Choose "Template" -> "Visual C#" -> "Console application" using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication1 { class Program { public static void Main(string[] args) { const int n = 5; int[] a = new int[n] { 10, 20, 30, 40, 50 }; int[] b = new int[n] { 5, 4, 3, 2, 1 }; int[] arr = new int[n]; for (int i = 0, j = 0; i < a.Length; i++, j++) { arr[i] = a[i] + b[j]; } for (int i = 0; i < arr.Length; i++) { Console.Write(arr[i] + " "); } } } } Output Note: In the code shown above the sum of two single-dimension arrays is produced.  Which is done simply  by the "+" operator. Perform Subtraction public static void Main(string[] args) { const int n = 5; int[] a = new int[n] { 10, 20, 30, 40, 50 }; int[] b = new int[n] { 5, 4, 3, 2, 1 }; int[] arr = new int[n]; for (int i = 0, j = 0; i < a.Length; i++, j++) { arr[i] = a[i] - b[j]; } for (int i = 0; i < arr.Length; i++) { Console.Write(arr[i] + " "); } } Output Perform Multiplication public static void Main(string[] args) { const int n = 5; int[] a = new int[n] { 10, 20, 30, 40, 50 }; int[] b = new int[n] { 5, 4, 3, 2, 1 }; int[] arr = new int[n]; for (int i = 0, j = 0; i < a.Length; i++, j++) { arr[i] = a[i] * b[j]; } for (int i = 0; i < arr.Length; i++) { Console.Write(arr[i] + " "); } } Output Perform Division public static void Main(string[] args) { const int n = 5; int[] a = new int[n] { 10, 20, 30, 40, 50 }; int[] b = new int[n] { 5, 4, 3, 2, 1 }; int[] arr = new int[n]; for (int i = 0, j = 0; i < a.Length; i++, j++) { arr[i] = a[i] / b[j]; } for (int i = 0; i < arr.Length; i++) { Console.Write(arr[i] + " "); }

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Rs Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 5 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among Class 6 students for Maths Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggrawal 2020 2021 Book of Class 6 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggrawal 2020 2021 Solutions. All Rs Aggrawal 2020 2021 Solutions for class Class 6 Maths are prepared by experts and are 100% accurate. #### Page No 82: (i) The shaded portion is 3 parts of the whole figure $\therefore$ $\frac{3}{4}$ (ii) The shaded portion is 1 parts of the whole figure $\therefore$ $\frac{1}{4}$ (iii) The shaded portion is 2 parts of the whole figure. $\therefore$ $\frac{2}{3}$ (iv) The shaded portion is 3 parts of the whole figure. $\therefore$$\frac{3}{10}$ (v)The shaded portion is 4 parts of the whole figure. $\therefore$$\frac{4}{9}$ (vi) The shaded portion is 3 parts of the whole figure. $\therefore$ $\frac{3}{8}$ #### Page No 82: The given rectangle is not divided into four equal parts. Thus, the shaded region is not equal to $\frac{1}{4}$ of the whole. #### Page No 82: (i) $\frac{3}{4}$        (ii) $\frac{4}{7}$             (iii) $\frac{2}{5}$           (iv) $\frac{3}{10}$           (v) $\frac{1}{8}$ (vi) $\frac{5}{6}$             (vii)$\frac{8}{9}$              (viii) $\frac{7}{12}$ #### Page No 83: Numerator        Denominator (i) 4                         9 (ii) 6                       11 (iii) 8                      15 (iv) 12                     17 (v) 5                        1 #### Page No 83: (i)$\frac{3}{8}$        (ii) $\frac{5}{12}$          (iii)$\frac{7}{16}$             (iv) $\frac{8}{15}$ #### Page No 83: (i) two-thirds (ii) four$-$ninths (iii) two$-$fifths (iv) seven$-$tenths (v) one$-$thirds (vi) three$-$fourths (vii) three$-$eighths (viii) nine$-$fourteenths (ix) five$-$elevenths (x) six$-$fifteenths #### Page No 83: We know: 1 hour = 60 minutes ∴ The required fraction = $\frac{24}{60}=\frac{2}{5}$ #### Page No 83: There are total 9 natural numbers from 2 to 10. They are 2, 3, 4, 5, 6, 7, 8, 9, 10. Out of these natural numbers, 2, 3, 5, 7 are the prime numbers. ∴ The required fraction = $\frac{4}{9}$. #### Page No 83: (i) $\frac{2}{3}$ of 15 pens = (ii) $\frac{2}{3}$ of 27 balls = (iii) $\frac{2}{3}$ of 36 balloons = ​ #### Page No 83: (i) $\frac{3}{4}$ of 16 cups = (ii) $\frac{3}{4}$ of 28 rackets = (iii) $\frac{3}{4}$ of 32 books = #### Page No 83: Neelam gives $\frac{4}{5}$ of 25 pencils to Meena. Thus, Meena gets 20 pencils. ∴ Number of pencils left with Neelam = 25 $-$ 20 = 5 pencils Thus, 5 pencils are left with Neelam. #### Page No 83: Draw a 0 to 1 on a number line. Label point 1 as A and mark the starting point as 0. (i) Divide the number line from 0 to 1 into 8 equal parts and take out 3 parts from it to reach point P. (ii) Divide the number line from 0 to 1 into 9 equal parts and take out 5 parts from it to reach point P . (iii) Divide the number line from 0 to 1 into 7 equal parts and take out 4 parts from it to reach point P. (Iv) Divide the number line from 0 to 1 into 5 equal parts and take out 2 parts from it to reach point P. (v) Divide the number line from 0 to 1 into 4 equal parts and take out 1 part from it to reach point P. #### Page No 85: A fraction whose numerator is greater than or equal to its denominator is called an improper fraction. Hence,  are improper fractions. #### Page No 85: Clearly,  are improper fractions, each with 5 as the denominator. #### Page No 85: Clearly, are improper fractions, each with 13 as the numerator. We have: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) #### Page No 85: (i) On dividing 17 by 5, we get: Quotient = 3 Remainder = 2 ∴ (ii) On dividing 62 by 7, we get: Quotient = 8 Remainder = 6 ∴ (iii) On dividing 101 by 8, we get: Quotient = 12 Remainder = 5 ∴ (iv) On dividing 95 by 13, we get: Quotient = 7 Remainder = 4 ∴ (v) On dividing 81 by 11, we get: Quotient = 7 Remainder = 4 ∴ (vi) On dividing 87 by 16, we get: Quotient = 5 Remainder = 7 ∴ (vii) On dividing 103 by 12, we get: Quotient = 8 Remainder = 7 ∴ (viii) On dividing 117 by 20, we get: Quotient = 5 Remainder = 17 ∴ #### Page No 85: An improper fraction is greater than 1. Hence, it is always greater than a proper fraction, which is less than 1. (i) (ii) (iii) (iv) (v) (vi) #### Page No 86: (i) Draw a number line. Mark 0 as the starting point and 1 as the ending point. Then, divide 0 to 1 in four equal parts, where each part is equal to 1/4. Show the consecutive parts as 1/4, 1/2, 3/4 and at 1 show 4/4 = 1. (ii) Draw 0 to 1 on a number line. Divide the segment into 8 equal parts, each part corresponds to 1/8. Show the consecutive parts as 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8 and 8/8. Highlight the required ones only. (iii) Draw 0 to 2 on a number line. Divide the segment between 0 and 1 into 5 equal parts, where each part is equal to 1/5. Show 2/5, 3/5, 4/5 and 8/5 3 parts away from 1 towards 2. (1 < 8/5 < 2) #### Page No 89: (i) ∴ Hence, the five fractions equivalent to $\frac{2}{3}$ are . (ii) ​ ∴ Hence, the five fractions equivalent to $\frac{4}{5}$ are . (iii) ​ ∴ Hence, the five fractions equivalent to $\frac{5}{8}$ are . (iv) ​ ∴ Hence, the five fractions equivalent to $\frac{7}{10}$ are . (v) ​​ ∴ Hence, the five fractions equivalent to $\frac{3}{7}$ are . (vi)  ​ ∴ Hence, the five fractions equivalent to $\frac{6}{11}$ are . (vii) ∴ Hence, the five fractions equivalent to $\frac{7}{9}$ are . (viii) ∴ Hence, the five fractions equivalent to $\frac{5}{12}$ are . #### Page No 89: The pairs of equivalent fractions are as follows: (i) (ii) (iv) #### Page No 89: (i) Let Clearly, 30 = 5 $×$ 6 So, we multiply the numerator by 6. ∴ ​ Hence, the required fraction is $\frac{18}{30}$. (ii)  ​Let Clearly, 24 = 3 $×$ 8 So, we multiply the denominator by 8. ∴ ​ Hence, the required fraction is $\frac{24}{40}$. #### Page No 89: (i) Let Clearly, 54 = 9 $×$ 6 So, we multiply the numerator by 6. ∴ ​ Hence, the required fraction is $\frac{30}{54}$. (ii)  ​Let Clearly, 35 = 5 $×$ 7 So, we multiply the denominator by 7. ∴ ​ Hence, the required fraction is $\frac{35}{63}$. #### Page No 89: (i) Let Clearly, 77 = 11 $×$ 7 So, we multiply the numerator by 7. ∴ ​ Hence, the required fraction is $\frac{42}{77}$. (ii)  ​Let Clearly, 60 = 6 $×$ 10 So, we multiply the denominator by 10. ∴ ​ Hence, the required fraction is $\frac{60}{110}$. #### Page No 89: Let Clearly, 4 = 24 $÷$ 6 So, we divide the denominator by 6. ∴ ​ Hence, the required fraction is $\frac{4}{5}$. #### Page No 89: (i) Let Clearly, 9 = 36 $÷$ 4 So, we divide the denominator by 4. ∴ ​ Hence, the required fraction is $\frac{9}{12}$. (ii)  ​Let Clearly, 4 = 48 $÷$ 12 So, we divide the numerator by 12. ∴ ​ Hence, the required fraction is $\frac{3}{4}$. #### Page No 89: (i) Let Clearly, 4 = 56 $÷$ 14 So, we divide the denominator by 14. ∴ ​ Hence, the required fraction is $\frac{4}{5}$. (ii)  ​Let Clearly, 10 = 70 $÷$ 7 So, we divide the numerator by 7. ∴ ​ Hence, the required fraction is $\frac{8}{10}$. #### Page No 89: (i) Here, numerator = 9 and denominator = 15 Factors of 9 are 1, 3 and 9. Factors of 15 are 1, 3, 5 and 15. Common factors of 9 and 15 are 1 and 3. H.C.F. of 9 and 15 is 3. ∴ Hence, the simplest form of . (ii) Here, numerator = 48 and denominator = 60 Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48. Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. Common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12. H.C.F. of 48 and 60 is 12. ∴ Hence, the simplest form of . (iii) Here, numerator = 84 and denominator = 98 Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84. Factors of 98 are 1, 2, 7, 14, 49 and 98. Common factors of 84 and 98 are 1, 2, 7 and 14. H.C.F. of 84 and 98 is 14. ∴ Hence, the simplest form of . (iv) Here, numerator = 150 and denominator = 60 Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75 and 150. Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. Common factors of 150 and 60 are 1, 2, 3, 5, 6, 10, 15 and 30. H.C.F. of 150 and 60 is 30. ∴ Hence, the simplest form of . (v) ​Here, numerator = 72 and denominator = 90 Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90. Common factors of 72 and 90 are 1, 2, 3, 6, 9 and 18. H.C.F. of 72 and 90 is 18. ∴ Hence, the simplest form of . #### Page No 89: (i) Here, numerator = 8 and denominator = 11 Factors of 8 are 1, 2, 4 and 8. Factors of 11 are 1 and 11. Common factor of 8 and 11 is 1. Thus, H.C.F. of 8 and 11 is 1. Hence, $\frac{8}{11}$ is the simplest form. (ii) Here, numerator = 9 and denominator = 14 Factors of 9 are 1, 3 and 9. Factors of 14 are 1, 2, 7 and 14. Common factor of 9 and 14 is 1. Thus, H.C.F. of 9 and 14 is 1. Hence, $\frac{9}{14}$ is the simplest form. (iii) Here, numerator = 25 and denominator = 36 Factors of 25 are 1, 5 and 25. Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. Common factor of 25 and 36 is 1. Thus, H.C.F. of 25 and 36 is 1. Hence, $\frac{25}{36}$ is the simplest form. (iv) Here, numerator = 8 and denominator = 15 Factors of 8 are 1, 2, 4 and 8. Factors of 15 are 1, 3, 5 and 15. Common factor of 8 and 15 is 1. Thus, H.C.F. of 8 and 15 is 1. Hence, $\frac{8}{15}$ is the simplest form. (v) Here, numerator = 21 and denominator = 10 Factors of 21 are 1, 3, 7 and 21. Factors of 10 are 1, 2, 5 and 10. Common factor of 21 and 10 is 1. Thus, H.C.F. of 21 and 10 is 1. Hence, $\frac{21}{10}$ is the simplest form. #### Page No 90: (i) 28 (ii) 21 (iii) 32 (iv) 12 (v) 5 (vi) 9 #### Page No 93: Like fractions: Fractions having the same denominator are called like fractions. Examples: Unlike fractions: Fractions having different denominators are called unlike fractions. Examples: #### Page No 93: The given fractions are L.C.M. of 5, 10, 15 and 30 = (5 $×$ 2 $×$ 3) = 30 So, we convert the given fractions into equivalent fractions with 30 as the denominator. (But, one of the fractions already has 30 as its denominator. So, there is no need to convert it into an equivalent fraction.) Thus, we have: Hence, the required like fractions are #### Page No 93: The given fractions are L.C.M. of 4, 8, 12 and 24 = (4 $×$ 2 $×$ 3) = 24 So, we convert the given fractions into equivalent fractions with 24 as the denominator. (But one of the fractions already has 24 as the denominator. So, there is no need to convert it into an equivalent fraction.) Thus, we have: Hence, the required like fractions are #### Page No 93: Between two fractions with the same denominator, the one with the greater numerator is the greater of the two. (i) > (ii) > (iii) < (iv) > (v) > (vi) < #### Page No 93: Between two fractions with the same numerator, the one with the smaller denominator is the greater of the two. (i) > (ii) > (iii)< (iv) > (v) < (vi) > #### Page No 93: By cross multiplying: 5 $×$ 5 = 25 and 4 $×$ 7 = 28 Clearly, 28 > 25 $\therefore$ #### Page No 93: By cross multiplying: 3 $×$ 6 = 18 and 5 $×$ 8 = 40 Clearly, 18 < 40 $\therefore$ #### Page No 93: By cross multiplying: 7 $×$ 7 = 49 and 11 $×$ 6 = 66 Clearly, 49 < 66 $\therefore$ #### Page No 93: By cross multiplying: 5 $×$ 11 = 55 and 9 $×$ 6 = 54 Clearly, 55 > 54 $\therefore$ #### Page No 93: By cross multiplying: 2 $×$ 9 = 18 and 4 $×$ 3 = 12 Clearly, 18 > 12 $\therefore$ #### Page No 93: By cross multiplying: 6 $×$ 4 = 24 and 13 $×$ 3 = 39 Clearly, 24 < 39 $\therefore$ #### Page No 93: By cross multiplying: 3 $×$ 6 = 18 and 4 $×$ 5 = 20 Clearly, 18 < 20 $\therefore$ #### Page No 93: By cross multiplying: 5 $×$ 12 = 60 and 8 $×$ 7 = 56 Clearly, 60 > 56 $\therefore$ #### Page No 93: L.C.M. of 9 and 6 = (3 $×$ 3 $×$ 2) = 18 Now, we convert  into equivalent fractions having 18 as the denominator. ∴​ 4949 Clearly, $\therefore$ #### Page No 93: L.C.M. of 5 and 10 = (5 $×$ 2) = 10 Now, we convert  into an equivalent fraction having 10 as the denominator as the other fraction has already 10 as its denominator. ∴​ 4949 Clearly, $\therefore$ #### Page No 93: L.C.M. of 8 and 10 = (2 $×$ 5 $×$ 2 $×$ 2) = 40 Now, we convert  into equivalent fractions having 40 as the denominator. ∴​ 4949 Clearly, $\therefore$ #### Page No 93: L.C.M. of 12 and 15 = (2 $×$ 2 $×$ 3 $×$ 5) = 60 Now, we convert  into equivalent fractions having 60 as the denominator. ∴​ 4949 Clearly, $\therefore$ #### Page No 93: The given fractions are . L.C.M. of 2, 4, 6 and 8 = (2 $×$ 2 $×$ 2 $×$ 3) = 24 We convert each of the given fractions into an equivalent fraction with denominator 24. Now, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the ascending order as follows: #### Page No 93: The given fractions are L.C.M. of 3, 6, 9 and 18 = (3 $×$ 2  $×$ 3) = 18 So, we convert each of the fractions whose denominator is not equal to 18 into an equivalent fraction with denominator 18. Now, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the ascending order as follows: #### Page No 93: The given fractions are L.C.M. of 5, 10, 15 and 30 = (2 $×$ 5 $×$ 3) = 30 So, we convert each of the fractions whose denominator is not equal to 30 into an equivalent fraction with denominator 30. Now, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the ascending order as follows: #### Page No 93: The given fractions are L.C.M. of 4, 8, 16 and 32 = (2 ⨯ 2 ⨯ 2 ⨯ 2 ⨯ 2) = 32 So, we convert each of the fractions whose denominator is not equal to 32 into an equivalent fraction with denominator 32. Now, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the ascending order as follows: #### Page No 93: The given fractions are L.C.M. of 4, 8, 12 and 24 = (2 ⨯ 2 ⨯ 2 ⨯ 3) = 24 So, we convert each of the fractions whose denominator is not equal to 24 into an equivalent fraction with denominator 24. Thus, we have; Clearly, ∴ ​ Hence, the given fractions can be arranged in the descending order as follows: #### Page No 93: The given fractions are L.C.M. of 9, 12, 18 and 36 = (3 ⨯ 3 ⨯ 2 ⨯ 2) = 36 We convert each of the fractions whose denominator is not equal to 36 into an equivalent fraction with denominator 36. Thus, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the descending order as follows: #### Page No 93: The given fractions are L.C.M. of 3, 5,10 and 15 = (2 ⨯ 3 ⨯ 5) = 30 So, we convert each of the fractions into an equivalent fraction with denominator 30. Thus, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the descending order as follows: #### Page No 93: The given fractions are L.C.M. of 7, 14, 21 and 42 = (2 ⨯ 3 ⨯ 7) = 42 We convert each one of the fractions whose denominator is not equal to 42 into an equivalent fraction with denominator 42. Thus, we have: Clearly, ∴ ​ Hence, the given fractions can be arranged in the descending order as follows: #### Page No 93: The given fractions are As the fractions have the same numerator, we can follow the rule for the comparison of such fractions. This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one. Clearly, Hence, the given fractions can be arranged in the descending order as follows: #### Page No 93: The given fractions are As the fractions have the same numerator, so we can follow the rule for the comparison of such fractions. This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one. Clearly, Hence, the given fractions can be arranged in the descending order as follows: #### Page No 94: Lalita read 30 pages of a book having 100 pages. Sarita read $\frac{2}{5}$ of the same book. $\frac{2}{5}$ of 100 pages = ​ Hence, Sarita read more pages than Lalita as 40 is greater than 30. #### Page No 94: To know who exercised for a longer time, we have to compare . On cross multiplying: 4 $×$ 2 = 8 and 3 $×$ 3 = 9 Clearly, 8 < 9 $\therefore$ Hence, Rohit exercised for a longer time. #### Page No 94: Fraction of students who passed in VI A = Fraction of students who passed in VI B = In both the sections, the fraction of students who passed is the same, so both the sections have the same result. #### Page No 96: The given fractions are like fractions. We know: Sum of like fractions  = Thus, we have: #### Page No 96: The given fractions are like fractions. We know: Sum of like fractions  = Thus, we have: #### Page No 96: The given fractions are like fractions. We know: Sum of like fractions  = Thus, we have: #### Page No 96: L.C.M. of 9 and 6 = (2 $×$ 3 $×$ 3) = 18 Now, we have: #### Page No 96: L.C.M. of 12 and 16 = (2 $×$ 2 $×$ 2 $×$ 2 $×$ 3) = 48 Now, we have: #### Page No 96: L.C.M. of 15 and 20 = (3 $×$ 5 $×$ 2 $×$ 2) = 60 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: We have: 234+556 #### Page No 96: Total cost of both articles = Cost of pencil + Cost of eraser Thus, we have: Hence, the total cost of both the articles is . #### Page No 96: Total cloth purchased by Sohini = Cloth for kurta + Cloth for pyjamas Thus, we have: $\therefore$ Total length of cloth purchased = #### Page No 96: Distance from Kishan's house to school = Distance covered by him by rickshaw + Distance covered by him on foot Thus, we have: Hence, the distance from Kishan's house to school is . #### Page No 96: Weight of the cylinder filled with gas = Weight of the empty cylinder + Weight of the gas inside the cylinder Thus, we have: Hence, the weight of the cylinder filled with gas is . #### Page No 99: Difference of like fractions = Difference of numerator $÷$ Common denominator #### Page No 99: Difference of like fractions = Difference of numerator $÷$ Common denominator #### Page No 99: Difference of like fractions = Difference of numerator $÷$ Common denominator #### Page No 99: L.C.M. of 6 and 9 = (3 $×$ 2 $×$ 3) = 18 Now, we have: #### Page No 99: L.C.M. of 2 and 8 = (2 $×$ 2 $×$ 2) = 8 Now, we have: #### Page No 99: L.C.M. of 8 and 12 = (2 $×$ 2$×$ 2$×$3) = 24 Now, we have: We have: #### Page No 99: We have: 234+556 #### Page No 99: We have: 234+556 #### Page No 99: We have: 234+556 #### Page No 99: We have: #### Page No 99: We have: #### Page No 99: We have: #### Page No 99: We have: #### Page No 99: We have: #### Page No 99: Let x be added to $9\frac{2}{3}$ to get 19. 923 #### Page No 99: Let x be added to $6\frac{7}{15}$ to get $8\frac{1}{5}$. #### Page No 99: Let us compare . 3 $×$ 7 = 21 and 4 $×$ 5 = 20 Clearly, 21 > 20 Required difference: Hence, . #### Page No 99: Amount of milk left with Mrs. Soni = Total amount of milk bought by her $-$ Amount of milk consumed $\therefore$ Amount of milk left with Mrs. Soni $\therefore$ Milk left with Mrs. Soni = #### Page No 99: Actual duration of the film = Total duration of the show $-$ Time spent on advertisements Thus, the actual duration of the film was . #### Page No 99: Money left with the rickshaw puller = Money earned by him in a day $-$ Money spent by him on food Hence, Rs $80\frac{3}{4}$ is left with the rickshaw puller. #### Page No 99: The length of the other piece = (Length of the wire $-$ Length of one piece) Hence, the other piece is  long. (c) (c) #### Page No 100: (a) 15 Explanation: (a) 4 Explanation: #### Page No 100: (c) (Fractions having the same denominator are called like fractions.) #### Page No 100: (d) none of these In a proper fraction, the numerator is less than the denominator. #### Page No 100: (a) $\frac{7}{8}$ In a proper fraction, the numerator is less than the denominator. #### Page No 100: (b) Between the two fractions with the same numerator, the one with the smaller denominator is the greater. #### Page No 100: (c) $\frac{3}{5}$ L.C.M. of 5, 3, 6 and 10 = (2 $×$ 3 $×$ 5) = 30 Thus, we have: #### Page No 100: ( b ) $\frac{4}{5}$ Among the given fractions with the same numerator, the one with the smallest denominator is the greatest. #### Page No 100: (a) $\frac{6}{11}$ Among like fractions, the fraction with the smallest numerator is the smallest. #### Page No 100: (d) $\frac{7}{12}$ Explanation: ​​L.C.M. of 4, 6, 12 and 3 = (2 $×$ 2 $×$ 3) = 12 Thus, we have: #### Page No 100: (b) $\frac{23}{5}$ #### Page No 100: (c) $4\frac{6}{7}$ On dividing 34 by 7: Quotient = 4 Remainder = 6 #### Page No 101: (b) $\frac{3}{4}$ Explanation: Addition of like fractions = Sum of the numerators / Common denominator #### Page No 101: (b) $\frac{1}{2}$ Explanation: #### Page No 101: (d) $1\frac{1}{18}$ Explanation: #### Page No 101: (a) $3\frac{1}{3}$ Explanation: Let us compare  . 10 ⨯ 10 = 100 and 3 ​⨯ 33 = 99 Clearly, 100 > 99 ∴ #### Page No 103: A fraction is defined as a number representing a part of a whole, where the whole may be a single object or a group of objects. Examples: #### Page No 103: An hour has 60 minutes. $\therefore$ Fraction for 35 minutes = Hence, $\frac{7}{12}$ part of an hour is equal to 35 minutes. #### Page No 103: 56 = 8 ⨯ 7 So, we need to multiply the numerator by 7. $\therefore$ Hence, the required fraction is $\frac{35}{56}$. #### Page No 103: Let OA = AB = BC = 1 unit $\therefore$ OB = 2 units and OC = 3 units Divide BC into 5 equal parts and take 3 parts out to reach point P. Clearly, point P represents the number $2\frac{3}{5}$. We have: #### Page No 103: Cost of a pen = Cost of a pencil = So, the cost of a pen is more than the cost of a pencil. Difference between their costs: Hence, the cost of a pen is Rs $12\frac{1}{2}$ more than the cost of a pencil. #### Page No 103: Let us compare . By cross multiplying: 3 ⨯ 7 = 21 and ​4 ⨯ 5 = 20 Clearly, 21 > 20 ∴​$\frac{3}{4}>\frac{5}{7}$ Their difference: Hence, #### Page No 103: L.C.M. of 2, 3, 9 and 6 = (2 ⨯ 3 ​⨯ 3) = 18 Now, we have: #### Page No 103: 30 = 5 ​⨯ 6 So, we have to multiply the numerator by 6 to get the equivalent fraction having denominator 30. Thus, #### Page No 103: The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. The factors of 98 are 1, 2, 7, 14, 49, 98. The common factors of 84 and 98 are 1, 2, 7, 14. The H.C.F. of 84 and 98 is 14. Dividing both the numerator and the denominator by the H.C.F.: #### Page No 103: (b) an improper fraction In an improper fraction, the numerator is greater than the denominator. #### Page No 103: (a) proper fraction In a proper fraction, the numerator is less than the denominator. #### Page No 103: (b) $\frac{3}{8}<\frac{5}{12}$ Considering : #### Page No 103: (a) $\frac{2}{3}$ Explanation: L.C.M. of 3, 9, 2 and 12 = ( 2 ⨯ 2 ⨯ 3 ​⨯ 3) = 36 Now, we have: #### Page No 103: (b) $2\frac{1}{4}$ Explanation: #### Page No 103: (c) Like fractions have same the denominator. #### Page No 104: (d) $\frac{16}{21}$ #### Page No 104: (ii) (iii) (iv) (v)

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• ## Automated reasoning - Wikipedia, the free encyclopedia en.wikipedia.org/wiki/Automated_reasoning Automated reasoning is an area of computer science and mathematical logic dedicated to understanding different aspects of reasoning. The study of automated reasoning ... • ## Automated Reasoning (Stanford Encyclopedia of Philosophy) plato.stanford.edu/entries/reasoning-automated Reasoning is the ability to make inferences, and automated reasoning is concerned with the building of computing systems that automate this process. • ## Automated Reasoning web.math.princeton.edu/~nelson/ar.html Automated Reasoning This is material related to MAT 504, Topics in Logic, for the Spring term of 1997 (MW 1:30-3:00, Fine 214), on automated reasoning. • ## Automated Reasoning: Logical Reasoning Problem Solving www.automatedreasoning.net/index.html Automated Reasoning, by Larry Wos, Ph.D. About Automated Reasoning Notebooks Otter Theorem Prover : Welcome! This website features a series of notebooks presenting ... • ## Automated Reasoning Group - UCLA reasoning.cs.ucla.edu The Automated Reasoning group at UCLA is directed by professor Adnan Darwiche. The group focuses on research in the areas of probabilistic and logical reasoning and ... • ## Automated Reasoning - UNR www.cse.unr.edu/~bebis/CS365/StudentPresentations/AutomatedReasoning.ppt Automated Reasoning Matt Whipple and Brian Vees Overview What is automated reasoning? Properties of inference procedures Theorem prover Diagnosis with first ... • ## AAR - the Association for Automated Reasoning www.cs.miami.edu/~geoff/Conferences/AAR The Association for Automated Reasoning (AAR) is a not-for-profit corporation intended for educational and scientific purposes. The objective of the AAR is to advance ... • ## Reasoning system - Wikipedia, the free encyclopedia en.wikipedia.org/wiki/Automated_reasoning_system In information technology a reasoning system is a software system that generates conclusions from available knowledge using logical techniques such as deduction and ... • ## Automated Reasoning | EMR INDUSTRY www.emrindustry.com/tag/automated-reasoning Global healthcare cognitive computing market is expected to reach nearly USD 5,064.8 million by 2022, according to a new report by Grand View Research, Inc. Key ... • ## Automated Reasoning: Logical Reasoning Problem Solving www.automatedreasoning.net/about_automated_reasoning.html Automated Reasoning, by Larry Wos, Ph.D. Home Page Notebooks Otter Theorem Prover : For those new to automated reasoning. The primary objective of automated reasoning ...

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# compute-gcd Computes the greatest common divisor (gcd). ## Stats StarsIssuesVersionUpdatedCreatedSize compute-gcd 501.2.12 years ago8 years ago Greatest Common Divisor !NPM versionnpm-imagenpm-url !Build Statustravis-imagetravis-url !Coverage Statuscoveralls-imagecoveralls-url !Dependenciesdependencies-imagedependencies-url Computes the greatest common divisor (gcd). Note: the gcd is also known as the greatest common factor (gcf), highest common factor (hcf), highest common divisor, and greatest common measure (gcm). ## Installation ``\$ npm install compute-gcd`` For use in the browser, use browserify. ## Usage ``var gcd = require( 'compute-gcd' );`` #### gcd( a, b, c,...,n ) Computes the greatest common divisor (gcd) of two or more `integers`. ``````var val = gcd( 48, 18 ); // returns 6 var val = gcd( 8, 12, 16 ); // returns 4`````` #### gcd( arr, accessor ) Computes the greatest common divisor (gcd) of two or more `integers`. ``````var val = gcd( [48, 18] ); // returns 6 var val = gcd( [8, 12, 16] ); // returns 4`````` For object `arrays`, provide an accessor `function` for accessing `array` values ``````var data = [ ['beep', 4], ['boop', 8], ['bap', 12], ['baz', 16] ]; function getValue( d, i ) { return d[ 1 ]; } var arr = gcd( arr, getValue ); // returns 4`````` ## Notes • For more than 3 values, a performance gain can be achieved if the values are sorted in ascending order. • If provided an `array` with a length less than `2` or a single `integer` argument, the function returns `null`. ## Examples ``````var gcd = require( 'compute-gcd' ); // Compute the gcd of random tuples... var x, y, z, arr, val; for ( var i = 0; i < 100; i++ ) { x = Math.round( Math.random()*50 ); y = Math.round( Math.random()*50 ); z = Math.round( Math.random()*50 ); arr = [ x, y, z ]; val = gcd( arr ); console.log( arr, val ); }`````` To run the example code from the top-level application directory, ``\$ node ./examples/index.js`` ## Tests ### Unit Unit tests use the Mocha test framework with Chai assertions. To run the tests, execute the following command in the top-level application directory: ``\$ make test`` All new feature development should have corresponding unit tests to validate correct functionality. ### Test Coverage This repository uses Istanbul as its code coverage tool. To generate a test coverage report, execute the following command in the top-level application directory: ``\$ make test-cov`` Istanbul creates a `./reports/coverage` directory. To access an HTML version of the report, ``\$ make view-cov``

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# Primitive Unit Cells (including Wigner–Seitz and voronoi cells) If you’re googling “primitive cell,” I imagine you are a somewhat advanced student in materials science (or an extremely advanced PhD student in chemistry or physics?). The primitive cell, or primitive lattice, is the smallest way to define a crystal cell or lattice. Before you can understand primitive cells, first you need to understand the conventional cells. If you don’t know the “regular” crystal cells like FCC or BCC, I recommend you start with this article. If you’re still here, I’ll assume you know some basics of crystals. For example, you know that the BCC unit cell has 2 atoms per cell in the traditional way of drawing the cell. It’s possible, however, to redraw the BCC crystal structure with only 1 atom per cell. This redrawn perspective is called the primitive cell. The concept is more theoretical in nature and is more difficult for most people to grasp because it hides some of the symmetry that is shown in the traditional way of drawing the BCC cell. In this article, I’ll explain why the primitive cell is useful, list them for common conventional unit cells, and also show you the Wigner-Sietz way to construct any unit cell. ### What are Crystals and the Unit Cell Since this topic is geared towards more-advanced students, I’ll hide a refresher about crystals in collapsable text. If you want a full introduction to crystallography, you may want to read this article. As you know, a crystal is made of a lattice + basis. The basis is the thing which is repeated (atoms, in our case), and the lattice is the pattern in which it is repeated. Here’s a 1-dimensional crystal. The conventional way of identifying the unit cell is to draw it so that atoms are split evenly among different crystals. For example, in the image below, both the red box and blue box are valid depictions of the unit cell (in fact, any box of the same size would be a valid description). However, the box I’ve drawn in red is more conventional, because it helps us remember that the crystal should repeat. Take a look at this second example, with atoms spaced farther apart. In this case, if you drew the lattice to exactly encapsulate one atom (shown in blue), it would not make a unit cell. Draw another box to the right, and you’ll see that the two are not the same. However, if you draw the box so that the left edge of the box contains half of one atom, and the right edge of the box contains half of a different atom, the spacing is always correct. The starting point does not matter, so you could redefine the unit cell as a rectangle with an atom in the center and empty space on both sides, but we usually leave it as I show with the red box. There are many (actually, infinite) ways to draw the same crystal structure, but materials scientists usually use a certain convention. We typically draw the unit cell as the smallest cell that shows full symmetry, basing this structure on its underlying Bravais lattice. Another, less common way to draw the unit cell is to draw it as small as possible, hiding symmetry relationships if necessary. Drawing the unit cell like this is called the primitive cell. ### Definition of the Primitive Cell Let’s start with a 2-dimensional example of a conventional cell vs its primitive cell. This is a centered rectangular Bravais lattice. Since the basis is simply one atom, we can also say that it is a centered rectangular crystal. There is ¼ of an atom on each corner of the rectangle, and 1 full atom in the center. This gives us 2 atoms per unit cell, and each unit cell has a volume of . This way of looking at the unit cell tells you the full symmetry. Atoms are closer together in the vertical direction compared to the horizontal direction, and there is a centering operation as well. In a real material, this kind of information about atoms can provide all sorts of clues about the material properties. Now, look at the exact same crystal structure. None of the atoms have moved, but we are visualizing it differently. (Spoiler, this is the primitive cell). Instead of a rectangle, imagine that we have a rhombus. The sides are the same length, but the angle is not 90º. Now, we have 1 atom per unit cell (it’s not exactly ¼ of an atom on each of the 4 corners, but it still adds up to 1 atom in total). The number of atoms per unit area hasn’t changed since we have the exact same crystal, so the area of this new unit cell must be smaller than the old one. Specifically, the conventional unit cell had 2 atoms per cell, and the primitive one has 1 atom per cell. Therefore, the primitive unit cell has half the area of the conventional one. Since the new unit cell has only 1 atom per cell, it is the smallest possible cell. Therefore, it is the primitive cell Does that mean all primitive cells must have a single atom inside? Not necessarily. Remember, there is a difference between “lattice point” and “atom.” For every Bravais lattice, there exists a primitive lattice with a single lattice point in each cell. However, in a real crystal, each “lattice point” may be composed of multiple atoms. For example, one of the most-common crystal structures is hexagonal close-packed (HCP). The HCP crystal is similar to the hexagonal Bravais lattice, except the HCP crystal has two atoms per lattice point. The hexagonal prism can be sliced into parallelepipeds. If you are talking about the Bravais lattices, the conventional hexagonal lattice has 3 lattice points. If you looked at the primitive hexagonal lattice, there is only 1 lattice point. If you are talking about the HCP crystal structure, the conventional HCP unit cell has 6 atoms, while the primitive HCP unit cell has 2 atoms. I hope that illustrates how a primitive crystal cell can have more than one atom per unit cell, but a primitive Bravais lattice must have only 1 lattice point. In fact, you have probably even seen an HCP primitive cell before. ### Primitive Versions of Conventional Unit Cells and Bravais Lattices (List) You just saw the primitive cell for the HCP crystals, and the primitive lattice for the hexagonal lattice, but I’ll repeat them for the folks that skipped straight to this section. Primitive Hexagonal Close-Packed The primitive hexagonal lattice is the easiest to see, in my opinion. Take a vertical slice through ⅓ of the conventional cell, and there you have it! The hexagonal primitive lattice has 1 lattice point spread through the 8 corners of the parallelpiped. The crystal structure corresponding to the hexagonal Bravais lattice, the Simple Hexagonal crystal structure, does occur in nature (for example, graphite). However, the Hexagonal Close-Packed (HCP) crystal is much more common. The HCP primitive cell is a parallelpiped with 2 atoms inside. One atom is spread through the 8 corners, and the other atom rests in the center. The primitive hexagonal cell can be described by the vectors: And the basis: Primitive Cubic The Simple Cubic (SC) crystal structure is already a primitive cell. Yes, in many cases, the conventional cell is the primitive cell. Usually, the reason why we use a conventional cell is that the primitive cell doesn’t show full symmetry. In this case, the primitive SC cell does show full symmetry, so it is also the conventional cell. In fact, another name for Simple Cubic is “Primitive Cubic.” Primitive cell of Simple Cubic (SC) Bravais Lattice can be described by vectors: The volume of this cell is Primitive Face-Centered Cubic (FCC) The Face-Centered Cubic (FCC) crystal structure is the crystal structure with 1 atom at each lattice point in the FCC Bravais lattice. To find the primitive cell, start with an atom as one corner of your cell, then find the 7 closest atoms which will make up the other corners. In theory, you could choose any random shape as long as it completely tiles 3D space and has a volume such that there is one atom per cell (there are infinite possibilities for primitive cells), but the easiest FCC primitive cell is shown below. Primitive cell of Face-Centered Cubic (FCC) Bravais Lattice can be described by vectors: The volume of this cell is This cell has 3 side lengths equal to each other and the three vectors have 120º angles to each other. The volume of the primitive cell is , or ¼ the volume of the conventional FCC unit cell. You can find this number using geometry calculations, or by using the knowledge that the primitive cell has the same atoms/volume as the conventional cell, and since the conventional cell has 4 times as many atoms per cell, the primitive cell must have ¼ the volume. Primitive Body-Centered Cubic (BCC) To find the BCC primitive cell, you can use the same method I discussed for FCC. The result is shown below: Primitive cell of Body-Centered Cubic (BCC) Bravais Lattice can be described by vectors: The volume of this cell is The primitive BCC unit cell is defined by 3 vectors which have the same length of 2r, with two angles = 90º and one angle = or 70.53º. There is 1 atom per unit cell. The volume of the primitive BCC unit cell, , is ½ the volume of the conventional BCC unit cell. Primitive Hexagonal You already saw the primitive HCP cell, and the primitive hexagonal cell follows the same principle. The primitive vectors are: The volume of this cell is Primitive Rhombohedral The conventional simple rhombohedral unit cell is already primitive and has 1 atom per unit cell. The primitive vectors are: where is between and . The volume of this cell is Primitive Tetragonal The conventional simple tetragonal unit cell is already primitive and has 1 atom per unit cell. The primitive vectors are: The volume of this cell is Primitive Body-Centered Tetragonal (BCT) The primitive BCT unit cell resembles the primitive BCC unit cell, since it looks like BCC but with one side having a different length than the others. The primitive BCT unit cell has and . The volume of a tetragonal cell is given by , and since the primitive BCT cell has 1 atom compared to 2 in the conventional BCT cell, the the primitive BCT unit cell has half the volume of the conventional BCT unit cell. The lattice vectors for a primitive BCT cell are: The volume of this cell is Primitive Orthorhombic The conventional simple orthorhombic unit cell is already primitive and has 1 atom per unit cell. The primitive vectors are: The volume of this cell is Primitive Face-Centered Orthorhombic The primitive face-centered orthorhombic unit cell resembles the FCC primitive cell, although the three sides are not equal to each other. The primitive vectors are: The volume of this cell is Primitive Body-Centered Orthorhombic The primitive body-centered orthorhombic unit cell resembles the BCC and BCT primitive cells, but this time all three sides have different lengths. The primitive vectors are: The volume of this cell is Primitive Base-Centered Orthorhombic The primitive cell for base-centered orthorhombic is actually easy to see, if you remember the trick for finding the primitive cell of the 2D centered rectangular lattice. The primitive vectors are: The volume of this cell is Primitive Monoclinic The conventional monoclinic unit cell is already primitive and has 1 atom per unit cell. The primitive vectors are: The volume of this cell is Primitive Base-Centered Monoclinic The primitive cell for base-centered monoclinic can be visualized using the same trick as base-centered orthorhombic or the 2D centered rectangular cells. The primitive vectors are: The volume of this cell is Primitive Triclinic The conventional triclinic unit cell is already primitive and has 1 atom per unit cell. The primitive vectors are: where and The volume of this cell is ### Wigner-Seitz and Voronoi So far we have constructed primitive cells using a mixture of intuition and guess-and-check (or you just peeked at the answer). The primitive cells I’ve shown you so far are the standard primitive cells–they don’t have the full symmetry of the non-primitive conventional cells, but they are built to be similar to conventional cells. There is another method of using math to generate a different set of primitive cells (remember, each primitive cell is one of infinite ways to describe the same pattern). The primitive cells created by this method are called Wigner-Seitz unit cells. Rather than splitting the atom to put ⅛ of it on each of the 8 corners, the Wigner-Seitz unit cell has a single atom surrounded by empty space. Wigner-Seitz cells are unique, but they are not that intuitive to understand as a crystal structure. However, the algorithm to create these cells is simple. Let’s start with an example in 2D. 1. Choose any lattice point to start. This atom will be the only atom to appear in the cell, and it will be exactly in the center. 1. Draw a line from your center atom to every other atom. You’ll see that most of these will be redundant, so in practice you only need to draw lines to the close atoms. We’ll color these lines red. 1. Find the midpoint of each red line. 1. From the midpoint, draw a blue line perpendicular to the red line. The blue lines should be infinitely long. 1. The perpendicular blue lines will form the outline of a polygon. The smallest polygon formed is the primitive cell. If you’ve done this correctly, your polygon should completely fill the space! What I have described is an algorithm for creating Voronai polygons. Voronai polygons are a mathematical concept which describes the set of all space which is closer to one point than any other. The Wigner-Seitz unit cell is the unit cell which is generated from the voronai technique. There are also Voronai polyhedra, which are the 3-dimensional equivalent. When making 3D Wigner-Seitz cells, use the same procedure that I outlined. However, instead of drawing blue lines perpendicular to the red lines, draw blue planes perpendicular to the red lines. The planes which intersect to form the smallest polyhedron is your Wigner-Seitz cell! However, a note of caution: the method of generating Wigner-Seitz primitive cells only works when you center the cell around a lattice point ### Final Thoughts Primitive cells or primitive lattices are the best way to describe atomic crystal structures with math. Primitive cells are the smallest possible cell. There are an infinite number of primitive cells, but materials scientists typically use one of two types: the conventional primitive lattice has ⅛ of a  lattice point at each corner of a parallelepiped, and the Wigner-Seitz primitive cell has a lattice point at the center of each polyhedron.

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# How does placing objects in liquids affect the mass? I was dazing off in my physics class when I came up with this question and I was wondering about it all day. I could not provide myself with an adequate solution, so here I am asking the forum about it! I know the community prefers generalizations, but for the sake of not being too confusing, I must be a bit specific for now. We have a liquid of mass $M_l$ and density $D_l$ laying on a scale (which obviously reads "$M_l$"). A solid of mass $M_s$ and density $D_s$ is hanging by another scale (which again obviously reads "$M_s$"). We move the solid down so that it is halfway submerged in the liquid. (all masses in kg, densities in g/cm^3) How will the masses read by the two scales change, if at all? Intuitively, I would say that the sum of the masses read by both scales would have to be the same as the original sum ($M1 +M2$) but I could be wrong. Density probably plays a key role. Any help is appreciated :) - It doesn't, unless they dissolve. – Keith Thompson Jan 19 '12 at 1:21 I think the question is about measured mass, not acutal mass. – Dan Jan 19 '12 at 1:54 A scale doesn't measure mass. It measures force. It is calibrated in units of mass, but the calibration assumes that the scale is measuring the force due to gravity on some object at or near sea-level on Earth. A balance doesn't have to assume a particular gravity field, but it still measures force. A balance works by comparing the force exerted by gravity on an unknown mass vs. the force exerted by a known mass. – james large Mar 23 at 1:54 Submerging objects in a liquid does not change the mass of those objects. It does effect the weight they would register on a scale, though. The bouyant force a fluid exerts upwards on a body submerged in it, $$F=\rho Vg$$ where $\rho$ is the density of the fluid, $V$ is the volume of the fluid displaced, and $g$ is the acceleration due to gravity. The liquid pushes up on the solid, which means, by Newton's third law, that there is a force of equal magnitude acting downward on the liquid. The force of the liquid up on the solid is $$F_{ls}=D_l\frac{M_sg}{2 D_s}$$ by the previous equation, so the scales would read that the solid weighs $$M_s- F_{ls}/g$$ and the liquid scale would read $$M_l+ F_{ls}/g$$ Note that the actual mass of the solid and liquid would be unaffected, only the reading on the scale would change. - Well you are given the original mass of the solid ($M_S$) as well as its density ($D_S$)... same with the liquid.... can you not use this information to find the volume? And the solid is place halfway in, so the submerged volume would have half the volume of the solid. Using this information, can you complete the problem? I am having trouble understanding the concept. Thanks :) Also, if the mass of the objects do not change, and the scale registers in KG, wouldnt the sum of the masses not change either? The scales register mass, not weight, correct? – Raymond Jan 18 '12 at 22:38 Also - scales always measure weight. They work by measuring the force that gravity applies to the object. – Mark Beadles Jan 19 '12 at 0:18 Ahh ok... Wait I think I did give the density of the liquid ($D_l$). How would this then effect the weight of the objects measured by the scales? Sorry for all the questions, i'm just trying to get this straight :) – Raymond Jan 19 '12 at 0:38 Ah!, I missed that the density of the solid was given as well. – Dan Jan 19 '12 at 0:39 I've edited my answer to reflect my error. – Dan Jan 19 '12 at 0:49 ## protected by Qmechanic♦Mar 23 at 7:14 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

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Does high resistance mean low current? Does high resistance mean low current? Resistance measurements are normally taken to indicate the condition of a component or a circuit. The higher the resistance, the lower the current flow. The lower the resistance, the higher the current flow. What can happen to a circuit with low resistance and high current? With a low resistance in the connection, a high current will flow, causing the delivery of a large amount of energy in a short period of time. However, it is possible for short circuits to arise between neutral and earth conductors and between two conductors of the same phase. How does a low resistance affect the current in a circuit? The relationship between resistance and wire length is proportional . The resistance of a thin wire is greater than the resistance of a thick wire because a thin wire has fewer electrons to carry the current. The relationship between resistance and the area of the cross section of a wire is inversely proportional . Does higher resistance increase current? In this case, there is a inverse relationship between the two variables. As the resistance increases, the current decreases, provided all other factors are kept constant. Materials with low resistance, metals for example, are called electrical conductors and allow electricity to flow easily. What happens to current if resistance increases? Ohm’s law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. Why is resistance higher at higher current? It stands to reason that we can’t fit as much volume through a narrow pipe than a wider one at the same pressure. This is resistance. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it. This brings us back to Georg Ohm. What happens if resistance is too low? If resistance is too low, current will be high at any voltage. If resistance is too high, current will be low if voltage is okay. NOTE: When the voltage stays the same, such as in an Automotive Circuit… current goes up as resistance goes down, and current goes down as resistance goes up. What is considered excessive resistance? If a circuit has excessive resistance, it prevents the wire or component from carrying sufficient current under high load conditions. Resistance can be caused by corrosion, loose wiring pins, pitted relay contacts, and other types of physical damage. What 4 factors affect the resistance of a wire? There are 4 different factors which affect resistance: • The type of material of which the resistor is made. • The length of the resistor. • The thickness of the resistor. • The temperature of the conductor. Is resistance directly proportional to length? The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Resistance also depends on the material of the conductor. The resistance of a conductor, or circuit element, generally increases with increasing temperature. Why is resistance directly proportional to length? As the length increases, the number of collisions by the moving free electrons with the fixed positive ions increases as more number of fixed positive ions are present in an increased length of the conductor. As a result, resistance increases. Why are test currents used for low resistance? The test currents used for low resistance measurements are often much higher than the currents used for high resistance measurements, so power dissipation in the device can be a consideration if it is high enough to cause the device’s resistance value to change. Power dissipation in a resistor is given by the formula: How is high voltage and low current related? You’re confusing “high voltage” with “high voltage loss”. Ohm’s Law governs the loss of voltage across a resistance for a given current passing through it. Since the current is low, the voltage loss is correspondingly low. You are confused about the consumer load and the resistance of the cables. Is there a difference between low, medium and high resistance? There is no scientific law that defines low, medium and high resistance. Resistance is a calculated value, calculated using Ohm’s Law. One can apply a known current, measure the voltage drop across two points, and calculate the resistance. Conversely one can apply a known voltage, measure the current, and calculate the resistance. Can a sourcemeter be used to measure low resistance? Source Measure Unit (SMU) instruments or SourceMeter ® instruments can simplify making low resistance measurements with high current stimulus. A SourceMeter instrument is capable of sourcing and measuring both current and voltage.

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# Round Two-Digit Measurements Videos and solutions to help Grade 3 students learn how to round two-digit measurements to the nearest ten on the vertical number line. Common Core Standards: 3.NBT.1, 3.MD.1, 3.MD.2 Related Topics: Lesson Plans and Worksheets for Grade 3 Lesson Plans and Worksheets for all Grades New York State Common Core Math Module 2, Grade 3, Lesson 12 Concept Development How many tens are in 73? 7 tens What is 1 more ten than 7 tens? 8 tens Which number is halfway between 7 tens and 8 tens? 7 tens and 5 ones, or 75. Plot 73 on the number line. Homework 1. Complete the chart. Choose objects and use a ruler/meter stick to complete the last two on your own. 4. Mrs. Santos' weight is shown on the scale. Round the weight to the nearest 10 kilograms. Mrs. Santos' weight is _________ kilograms. Mrs. Santos weighs about _________ kilograms. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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# The scores obtained by Alice in a quiz competition having 5 rounds is listed below. Positive points aregiven for every corre The scores obtained by Alice in a quiz competition having 5 rounds is listed below. Positive points are given for every correct answer while negative points are given for every incorrect answer. Rounds Scores Round 1 10 Round 2 – 20 Round 3 ? Round 4 40 Round 5 -5 Final Score 60 What was Alice’s score in Round 3? ### 1 thought on “The scores obtained by Alice in a quiz competition having 5 rounds is listed below. Positive points are<br />given for every corre” 1. [tex]\boxed{\large\textsf{\${\large\textsf{L.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2h ( l + b )}\$}\\\\\large\textsf{\${\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )}\$}\\\\\large\textsf{\${\large\textsf{Volume}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{l×b×h}\$}\\\\\large\textsf{\${\large\textsf{L.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{4×l²}\$}\\\\\large\textsf{\${\large\textsf{T.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{6 × l²}\$}}[/tex] [tex]\large\textsf{\${\large\textsf{Volume}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{l²}\$}[/tex] [tex]\large\textsf{\${\large\textsf{C.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{2 × πrh}\$}[/tex] [tex]\large\textsf{\${\large\textsf{T.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{2πr × ( r + h )}\$}[/tex] [tex]\large\textsf{\${\large\textsf{Volume}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{πr²h}\$}[/tex] [tex]\large\textsf{\${\large\textsf{C.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\textsf{πrl}\$}[/tex] [tex]\large\textsf{\${\large\textsf{T.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\textsf{πr × ( r + l )}\$}[/tex] [tex]\large\textsf{\${\large\textsf{Volume}}_{\large\textsf{( \; Cone \; )}} \$} \large\textsf{ =\$\cfrac{\large\textsf{1}}{\large\textsf{3}}\$}\large\textsf{× πr²h}[/tex] [tex]\large\textsf{\${\large\textsf{T.S.A.}}_{\large\textsf{( \; Sphere \; )}} = \large\textsf{4πr²}\$}[/tex] [tex]\large\textsf{\${\large\textsf{Volume}}_{\large\textsf{( \; Sphere \; )}} \$} \large\textsf{ =\$\cfrac{\large\textsf{4}}{\large\textsf{3}}\$}\large\textsf{× πr³}[/tex]

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CC-MAIN-2024-26

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# Number 50505 ### Properties of number 50505 Cross Sum: Factorization: 3 * 5 * 7 * 13 * 37 Divisors: 1, 3, 5, 7, 13, 15, 21, 35, 37, 39, 65, 91, 105, 111, 185, 195, 259, 273, 455, 481, 555, 777, 1295, 1365, 1443, 2405, 3367, 3885, 7215, 10101, 16835, 50505 Count of divisors: Sum of divisors: Prime number? No Fibonacci number? No Bell Number? No Catalan Number? No Base 2 (Binary): Base 3 (Ternary): Base 4 (Quaternary): Base 5 (Quintal): Base 8 (Octal): c549 Base 32: 1ha9 sin(50505) 0.68638095221069 cos(50505) 0.72724218004895 tan(50505) 0.94381345175069 ln(50505) 10.829827620263 lg(50505) 4.7033343754438 sqrt(50505) 224.73317512108 Square(50505) ### Number Look Up Look Up 50505 which is pronounced (fifty thousand five hundred five) is a great figure. The cross sum of 50505 is 15. If you factorisate 50505 you will get these result 3 * 5 * 7 * 13 * 37. The number 50505 has 32 divisors ( 1, 3, 5, 7, 13, 15, 21, 35, 37, 39, 65, 91, 105, 111, 185, 195, 259, 273, 455, 481, 555, 777, 1295, 1365, 1443, 2405, 3367, 3885, 7215, 10101, 16835, 50505 ) whith a sum of 102144. 50505 is not a prime number. The figure 50505 is not a fibonacci number. The figure 50505 is not a Bell Number. The number 50505 is not a Catalan Number. The convertion of 50505 to base 2 (Binary) is 1100010101001001. The convertion of 50505 to base 3 (Ternary) is 2120021120. The convertion of 50505 to base 4 (Quaternary) is 30111021. The convertion of 50505 to base 5 (Quintal) is 3104010. The convertion of 50505 to base 8 (Octal) is 142511. The convertion of 50505 to base 16 (Hexadecimal) is c549. The convertion of 50505 to base 32 is 1ha9. The sine of 50505 is 0.68638095221069. The cosine of the figure 50505 is 0.72724218004895. The tangent of the figure 50505 is 0.94381345175069. The root of 50505 is 224.73317512108. If you square 50505 you will get the following result 2550755025. The natural logarithm of 50505 is 10.829827620263 and the decimal logarithm is 4.7033343754438. that 50505 is impressive figure!

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Variance CBD # Variance CBD - VARIANCE TWO RANDOM VARIABLES Each press of... This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: VARIANCE: TWO RANDOM VARIABLES Each press of F9 generates new samples, of size 3, from X and from Y, respectively. The means of the samples are computed. Calculus for Business Dec Release 2.1, 2010 Published and Distributed The Mathematical Association o © 2010 by The Arizona Board of The University of Arizona. A reserved. Random Variable X p.m.f. fX(x) µX x 1 0 4 2 0.05 3 0.15 4 0.6 5 0.15 6 0.05 7 0 Random Variable X 0.6 0.4 fX(x) 0.2 0.0 1 2 3 x Samples X 3 Sample Mean 3.33 Sample Mean 5.67 Error 0.67 4 5 6 Random Variable Y p.m.f. fY(y) µY y 1 0.1 4 2 0.15 3 0.15 4 0.2 5 0.15 6 0.15 7 0.1 Random Variable Y 0.6 0.4 fY(y) 0.2 0.0 7 1 2 3 y 4 5 6 7 4 3 6 Y 6 5 Error 1.67 Calculus for Business Decisions Release 2.1, 2010 Published and Distributed by Mathematical Association of America 10 by The Arizona Board of Regents for he University of Arizona. All rights reserved. iable Y 4 5 6 7 0.75 0.8 0.17 0.78 0.11 0.64 COMPUTING SAMPLE STATISTICS With Direct Computaion For convenience of notation, we will use m , rather than x, for the sample mean. xi 8 7 9 10 9 10 6 7 6 8 10 8 n 12 m 8.1667 (xi− m)2 0.0278 1.3611 0.6944 3.3611 0.6944 3.3611 4.6944 1.3611 4.6944 0.0278 3.3611 0.0278 23.6667 s2 2.1515 s 1.4668 Sum COMPUTING SAMPLE STATISTICS With Built-In Functions xi 8 7 9 10 9 10 6 7 6 8 10 8 s2 2.1515 s 1.4668 SIMULATING SAMPLE MEANS This file simulates 20,000 samples, of size 4 each, from a binomial random variable, X , which gives the number of successes in 10 Bernoulli Trials, where the probability of a success on each trial is 0.4. For this variable, we know that E (X ) = (10)(0.4) = 4, V (X ) = (10)(0.4)(1 - 0.4) = 2.4, and σ X = 2.4 ≅ 1.549 . For convenience of notation, we will use m , rather than x, for the sample mean. x1 4 5 4 3 2 2 5 2 3 4 5 5 7 4 4 5 3 5 5 6 4 5 6 5 5 x2 9 2 2 1 5 4 7 3 4 4 4 4 4 2 5 6 5 5 3 4 5 8 2 3 3 x3 4 4 3 3 5 5 4 0 5 5 3 6 3 3 5 2 5 7 4 4 2 4 6 8 6 x4 3 2 3 4 1 4 4 2 5 2 5 2 2 9 4 7 5 2 4 5 5 2 6 1 8 Sample Average Sample Sample Average Mean of all Variance Standard Deviation Average of all m m's s2 s of all s2's s's 5.00 7.333 2.708 4.006 2.386 1.429 3.25 2.250 1.500 3.00 0.667 0.816 2.75 1.583 1.258 3.25 4.250 2.062 3.75 1.583 1.258 5.00 2.000 1.414 1.75 1.583 1.258 4.25 0.917 0.957 3.75 1.583 1.258 4.25 0.917 0.957 4.25 2.917 1.708 4.00 4.667 2.160 4.50 9.667 3.109 4.50 0.333 0.577 5.00 4.667 2.160 4.50 1.000 1.000 4.75 4.250 2.062 4.00 0.667 0.816 4.75 0.917 0.957 4.00 2.000 1.414 4.75 6.250 2.500 5.00 4.000 2.000 4.25 8.917 2.986 5.50 4.333 2.082 3 7 3 2 1 3 6 5 4 6 5 1 2 4 4 3 3 3 3 4 5 3 3 5 4 4 3 3 3 2 6 2 2 6 4 3 2 2 2 2 4 4 5 5 6 5 4 3 6 5 4 3 2 1 4 3 0 3 2 2 3 6 4 3 7 2 3 2 4 4 6 6 4 3 3 5 4 3 3 6 6 3 4 4 5 5 5 8 3 6 4 5 5 5 2 3 3 6 5 3 7 4 5 4 6 8 6 5 3 3 4 3 5 4 4 4 4 9 4 5 3 5 3 4 5 4 4 3 3 5 4 5 3 4 6 1 0 6 6 4 4 3 5 5 4 6 3 6 2 3 3 2 2 5 3 4 2 4 4 6 4.00 4.75 3.75 3.75 4.25 4.25 4.75 4.75 4.00 5.50 4.25 3.50 3.00 3.50 4.00 2.50 1.50 4.50 4.00 3.25 4.75 4.00 4.25 4.25 5.25 5.00 3.75 4.00 3.00 3.00 4.75 3.25 3.25 4.50 3.50 4.00 3.00 4.50 3.25 4.75 2.000 2.917 0.917 1.583 4.917 0.917 0.917 5.583 2.000 0.333 0.250 3.667 2.000 3.000 2.667 1.000 3.000 3.000 3.333 0.917 2.917 2.000 0.917 0.917 2.250 6.667 2.250 3.333 0.667 0.667 2.250 3.583 2.250 1.667 0.333 0.667 1.333 9.667 0.917 3.583 1.414 1.708 0.957 1.258 2.217 0.957 0.957 2.363 1.414 0.577 0.500 1.915 1.414 1.732 1.633 1.000 1.732 1.732 1.826 0.957 1.708 1.414 0.957 0.957 1.500 2.582 1.500 1.826 0.816 0.816 1.500 1.893 1.500 1.291 0.577 0.816 1.155 3.109 0.957 1.893 2 1 3 5 3 0 5 4 3 3 2 3 2 6 4 4 2 6 3 0 4 6 4 5 4 4 4 0 3 5 4 5 3 4 4 4 1 3 5 6 5 4 2 7 2 5 2 5 4 4 6 4 2 5 3 2 4 4 2 1 2 6 3 3 3 3 5 5 4 2 7 7 6 2 3 3 3 7 3 6 6 4 2 4 3 4 5 3 5 5 5 5 4 3 4 4 5 9 3 2 6 5 4 5 5 1 5 3 4 5 2 5 4 4 6 5 6 4 7 4 4 4 5 4 7 3 1 7 5 6 3 4 5 1 4 4 2 3 6 3 4 3 6 4 4 7 2 5 3 5 4 4 5 4 3 5 5 6 2 3 4.25 3.25 3.00 5.00 3.75 3.00 3.25 4.75 4.25 4.50 4.00 4.00 3.25 3.75 3.75 3.50 3.25 5.50 3.50 1.50 4.00 5.00 4.25 4.25 4.00 3.75 4.00 3.25 3.50 4.25 4.25 5.25 4.50 3.50 4.00 4.25 3.75 5.00 4.25 4.75 2.917 2.250 2.000 2.000 4.917 4.667 4.250 2.917 0.917 1.667 3.333 0.667 2.250 4.917 0.250 1.000 2.250 7.000 3.000 1.667 2.667 2.000 1.583 0.917 0.667 6.250 2.000 5.583 0.333 2.250 4.250 1.583 1.667 1.000 2.000 0.917 4.917 3.333 4.917 2.250 1.708 1.500 1.414 1.414 2.217 2.160 2.062 1.708 0.957 1.291 1.826 0.816 1.500 2.217 0.500 1.000 1.500 2.646 1.732 1.291 1.633 1.414 1.258 0.957 0.816 2.500 1.414 2.363 0.577 1.500 2.062 1.258 1.291 1.000 1.414 0.957 2.217 1.826 2.217 1.500 5 6 5 2 3 3 3 4 3 6 4 4 5 5 3 5 3 7 3 4 3 4 3 6 3 5 5 4 4 5 4 4 2 2 5 2 5 1 8 5 4 5 6 3 1 4 4 2 2 7 4 2 6 2 5 5 5 4 3 3 3 5 7 6 5 2 5 4 4 4 6 6 4 4 2 2 3 4 3 6 6 2 3 3 2 4 2 4 1 8 6 4 5 6 5 5 2 6 5 6 5 5 5 5 2 6 6 3 3 4 7 5 7 0 4 2 6 5 5 5 4 2 6 4 7 3 4 2 5 5 3 4 4 5 6 4 3 2 5 4 3 4 6 5 5 4 3 3 1 3 5 2 2 5 5 3 3 2 3 2 4.75 3.75 5.00 3.00 3.25 3.50 3.25 3.00 2.75 6.50 4.25 3.50 5.00 4.50 4.75 4.75 3.25 4.75 4.00 4.25 3.50 4.50 5.25 5.50 3.75 4.25 4.75 3.50 3.00 4.00 5.50 4.25 3.75 2.75 4.00 2.25 4.25 3.00 4.75 4.50 0.917 4.250 2.000 0.667 6.917 0.333 0.917 1.333 2.917 1.667 1.583 1.000 0.667 3.000 1.583 0.250 1.583 4.917 1.333 1.583 1.000 0.333 2.917 0.333 2.250 2.917 1.583 0.333 2.000 0.667 1.667 2.917 5.583 4.917 2.000 0.250 2.250 3.333 5.583 3.000 0.957 2.062 1.414 0.816 2.630 0.577 0.957 1.155 1.708 1.291 1.258 1.000 0.816 1.732 1.258 0.500 1.258 2.217 1.155 1.258 1.000 0.577 1.708 0.577 1.500 1.708 1.258 0.577 1.414 0.816 1.291 1.708 2.363 2.217 1.414 0.500 1.500 1.826 2.363 1.732 2 6 4 5 4 7 7 4 4 4 4 2 5 5 5 6 4 4 5 6 4 3 5 3 4 5 6 4 6 3 5 6 5 1 4 4 4 6 4 4 6 5 5 3 1 5 4 1 2 7 4 2 4 3 4 4 3 1 4 2 6 7 1 4 4 6 3 6 5 5 5 4 2 2 4 3 3 2 4 4 2 4 4 3 2 5 3 5 6 2 4 1 3 3 4 4 5 3 6 6 6 3 5 3 2 7 5 5 6 3 7 4 3 5 4 4 5 5 3 4 6 5 5 4 3 3 3 2 5 5 4 2 3 1 3 6 4 3 5 5 3 3 2 3 3 1 6 4 7 4 3 4 1 2 3 6 7 4 4 8 4.00 5.00 4.50 3.75 2.50 5.00 4.25 3.00 4.25 4.50 4.00 1.75 3.75 3.00 4.00 5.00 4.00 2.75 5.00 4.75 4.75 4.00 3.25 3.25 3.25 4.75 5.00 4.75 6.00 3.75 5.00 4.50 2.75 2.50 3.75 4.25 4.75 4.25 3.75 5.00 5.333 0.667 0.333 0.917 1.667 2.667 3.583 3.333 2.917 4.333 0.000 0.250 0.917 2.667 0.667 1.333 0.667 1.583 0.667 3.583 2.250 4.000 4.250 0.250 0.917 6.917 2.000 0.917 0.667 0.917 2.667 1.000 2.917 3.000 0.250 1.583 2.917 2.917 0.250 4.000 2.309 0.816 0.577 0.957 1.291 1.633 1.893 1.826 1.708 2.082 0.000 0.500 0.957 1.633 0.816 1.155 0.816 1.258 0.816 1.893 1.500 2.000 2.062 0.500 0.957 2.630 1.414 0.957 0.816 0.957 1.633 1.000 1.708 1.732 0.500 1.258 1.708 1.708 0.500 2.000 4 3 7 5 4 2 3 4 2 5 2 2 3 3 4 4 4 5 5 4 4 8 7 6 5 5 4 4 6 3 1 3 2 4 8 6 6 4 3 7 3 5 4 5 3 4 3 5 4 6 4 4 3 1 3 4 3 4 4 3 6 4 4 5 4 4 7 1 5 5 2 3 5 6 5 5 3 5 3 4 5 5 3 1 4 3 4 3 4 3 4 0 4 3 2 6 6 5 7 6 4 2 2 4 4 7 7 3 4 6 6 5 2 4 3 5 6 4 3 3 4 2 6 4 5 3 2 4 3 6 4 2 6 4 5 6 3 5 4 2 5 4 6 3 6 4 5 7 3 3 4 5 5 4 7 4 2 4 7 4 4.00 3.75 5.00 3.75 4.00 3.00 3.00 4.00 3.25 5.00 3.50 2.00 4.00 2.75 3.50 5.00 4.00 4.75 5.00 3.75 4.75 4.50 4.75 4.50 4.75 5.00 5.75 3.75 4.50 4.25 3.25 4.00 3.50 4.50 5.75 5.00 4.25 4.25 4.00 4.50 0.667 2.250 3.333 3.583 0.667 0.667 0.667 0.667 0.917 2.000 1.000 2.667 2.000 1.583 1.667 1.333 2.000 0.250 2.000 2.917 0.917 6.333 4.917 1.667 0.917 2.000 2.250 6.250 1.667 2.250 4.917 1.333 3.000 1.000 4.917 0.667 4.250 0.250 4.000 3.000 0.816 1.500 1.826 1.893 0.816 0.816 0.816 0.816 0.957 1.414 1.000 1.633 1.414 1.258 1.291 1.155 1.414 0.500 1.414 1.708 0.957 2.517 2.217 1.291 0.957 1.414 1.500 2.500 1.291 1.500 2.217 1.155 1.732 1.000 2.217 0.816 2.062 0.500 2.000 1.732 4 2 2 3 5 4 3 6 3 2 3 2 6 6 5 3 3 4 4 5 5 6 4 3 6 4 4 4 4 3 5 3 5 4 2 4 4 3 3 4 4 5 3 2 4 5 3 6 4 3 5 3 3 3 4 5 6 4 6 2 4 2 4 2 4 2 3 3 5 5 7 6 3 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6 6 4 6 5 5 4 3 2 3 3 6 4 3 2 5 6 2 2 1 4 3 3 6 4 1 6 1 4 6 2 6 5 3 3 3 3 4 6 3 6 4 6 4 4 3 2 6 2 5 4 5 2 4 4 3 2 2 5 4 4 1 4 3 5 6 5 5 3 3 2 7 0 6 6 3 4 2 5 3 3 2 4 4 4 4 3 4 3 4.50 4.25 3.50 4.00 3.50 3.50 3.25 3.50 4.00 4.00 3.00 4.50 3.75 4.50 4.25 5.25 3.50 4.50 3.00 4.25 3.75 3.50 5.50 3.50 3.75 5.25 3.50 4.75 3.75 4.75 3.25 4.50 4.50 4.25 4.50 4.75 4.25 3.50 4.50 2.25 1.667 1.583 5.667 2.667 0.333 1.667 1.583 3.667 6.000 2.000 6.000 0.333 4.917 1.667 2.250 1.583 4.333 3.000 3.333 1.583 2.917 3.667 3.000 7.000 4.917 2.917 1.000 2.250 1.583 3.583 0.250 3.000 3.667 1.583 1.000 0.917 0.917 1.667 1.667 0.250 1.291 1.258 2.380 1.633 0.577 1.291 1.258 1.915 2.449 1.414 2.449 0.577 2.217 1.291 1.500 1.258 2.082 1.732 1.826 1.258 1.708 1.915 1.732 2.646 2.217 1.708 1.000 1.500 1.258 1.893 0.500 1.732 1.915 1.258 1.000 0.957 0.957 1.291 1.291 0.500 4 5 5 6 6 3 3 4 1 4 3 2 5 3 6 6 3 4 4 6 1 2 5 3 4 5 4 3 4 2 3 0 4 2 5 7 4 3 4 3 4 4 3 4 3 1 5 5 5 2 5 6 7 7 4 6 6 5 2 4 7 6 6 1 5 5 6 6 2 2 5 6 3 3 2 4 5 5 4 4 1 2 3 4 4 7 4 3 3 2 6 2 4 6 1 3 5 4 4 3 2 1 3 5 7 5 5 2 2 3 8 4 5 4 5 3 5 3 6 3 3 6 6 6 4 4 5 5 4 4 5 3 5 2 2 5 6 5 7 2 2 2 3 3 5 3 3 5 3 7 8 5 2 4 3 5 7 5 3 4 3.00 4.25 4.25 5.00 4.25 3.75 4.25 4.25 3.25 3.00 4.75 3.25 5.25 4.50 3.25 5.00 5.00 4.50 4.25 3.75 3.00 2.75 4.25 3.00 5.25 4.50 4.50 4.00 2.75 3.50 6.00 3.75 3.50 3.25 3.75 4.75 5.25 4.00 4.25 3.50 2.000 2.917 2.250 1.333 1.583 6.250 0.917 0.917 2.917 1.333 1.583 3.583 1.583 5.667 4.917 2.000 2.000 0.333 4.250 2.917 7.333 4.917 2.250 2.667 1.583 1.000 1.667 3.333 0.917 5.667 6.000 6.917 1.667 0.917 2.250 2.917 1.583 1.333 1.583 0.333 1.414 1.708 1.500 1.155 1.258 2.500 0.957 0.957 1.708 1.155 1.258 1.893 1.258 2.380 2.217 1.414 1.414 0.577 2.062 1.708 2.708 2.217 1.500 1.633 1.258 1.000 1.291 1.826 0.957 2.380 2.449 2.630 1.291 0.957 1.500 1.708 1.258 1.155 1.258 0.577 3 6 1 5 3 5 3 2 7 8 4 6 3 4 6 3 5 1 3 5 6 4 3 1 4 2 4 2 5 0 2 3 6 3 2 4 6 5 1 5 5 5 6 5 4 6 4 5 3 6 1 5 5 1 3 8 3 3 4 6 4 3 1 1 6 6 4 5 6 3 2 2 6 4 4 3 3 4 3 4 3 4 4 3 2 2 4 5 3 5 3 5 5 3 6 3 4 3 2 4 5 2 3 3 5 4 4 6 6 5 7 4 6 5 6 4 4 0 3 2 6 4 7 5 4 6 3 5 5 5 4 4 5 4 6 3 2 4 5 4 4 6 5 5 4 4 2 4 6 4 4 5 5 6 5 5 1 6 1 6 4.25 4.75 4.50 4.50 3.25 4.75 3.50 4.25 4.50 6.00 3.00 5.00 4.50 3.00 5.25 4.25 3.50 2.75 3.50 4.75 4.75 3.75 3.00 2.50 4.75 4.00 3.50 4.25 5.75 3.00 3.75 3.50 5.75 4.50 4.25 4.00 3.50 3.75 2.00 4.25 2.250 0.917 7.000 1.000 0.917 3.583 0.333 2.250 3.667 2.000 2.000 0.667 1.000 2.000 2.250 6.250 1.667 1.583 1.667 0.917 0.917 2.917 2.667 3.667 0.917 2.667 1.000 2.917 0.250 4.667 5.583 1.667 0.250 1.667 2.917 0.667 4.333 6.917 1.333 2.917 1.500 0.957 2.646 1.000 0.957 1.893 0.577 1.500 1.915 1.414 1.414 0.816 1.000 1.414 1.500 2.500 1.291 1.258 1.291 0.957 0.957 1.708 1.633 1.915 0.957 1.633 1.000 1.708 0.500 2.160 2.363 1.291 0.500 1.291 1.708 0.816 2.082 2.630 1.155 1.708 3 4 3 5 1 3 5 3 3 6 3 3 9 7 2 5 5 4 4 1 6 5 4 3 3 4 4 6 7 1 3 5 3 3 3 6 6 8 4 4 2 2 2 6 6 2 5 3 4 5 3 5 7 5 3 5 3 4 3 4 2 4 6 3 2 5 4 6 3 3 4 4 3 8 2 3 1 5 6 4 4 1 5 4 4 4 5 6 4 4 3 4 4 3 5 2 4 4 2 5 5 7 6 5 3 4 4 3 4 2 1 5 7 4 3 7 3 4 4 3 6 4 1 0 5 1 2 5 5 5 3 4 4 5 3 1 6 4 3 3 4 3 1 3 4 5 5 2 4 4 2 4 4 5 2 3 3 4 3 4 3.75 2.75 2.75 3.75 4.00 2.50 4.25 4.25 4.00 5.00 3.00 4.00 6.00 5.00 3.25 3.25 4.50 4.00 3.00 3.25 4.25 4.75 4.25 3.50 3.00 4.50 4.25 4.25 4.50 2.50 2.50 4.50 4.25 5.00 2.50 4.75 3.25 5.25 4.25 3.75 2.917 2.250 2.917 6.917 4.667 1.667 2.250 2.250 0.667 0.667 0.000 0.667 6.000 2.667 1.583 4.250 1.667 0.000 0.667 2.917 2.917 2.917 5.583 1.000 0.667 0.333 0.250 4.250 3.000 1.667 1.667 0.333 3.583 4.667 0.333 4.250 4.250 3.583 1.583 0.250 1.708 1.500 1.708 2.630 2.160 1.291 1.500 1.500 0.816 0.816 0.000 0.816 2.449 1.633 1.258 2.062 1.291 0.000 0.816 1.708 1.708 1.708 2.363 1.000 0.816 0.577 0.500 2.062 1.732 1.291 1.291 0.577 1.893 2.160 0.577 2.062 2.062 1.893 1.258 0.500 4 3 4 2 6 4 4 2 3 8 7 5 7 5 2 3 5 6 3 2 5 6 3 3 2 2 5 6 4 3 5 3 5 1 4 4 2 4 6 1 2 5 4 1 1 2 4 7 4 4 4 5 4 4 4 5 5 1 5 2 3 2 2 2 2 2 4 3 4 5 5 2 3 7 3 5 3 4 4 3 3 4 3 5 6 4 3 5 6 4 2 3 5 7 4 2 2 2 2 6 6 4 6 4 3 2 5 4 3 6 2 5 4 4 6 5 5 4 5 6 3 3 5 5 2 6 2 1 5 7 2 3 3 4 2 2 5 3 3 2 3 2 4 5 5 3 2 5 4 5 3 7 6 6 4 5 5 5 3 6 3.00 3.75 4.00 3.25 3.75 4.00 3.25 3.75 4.50 5.75 3.75 4.00 4.75 5.00 3.00 3.00 4.25 3.00 3.25 3.00 4.25 3.50 3.75 3.50 3.00 2.25 4.00 4.50 3.75 4.75 3.75 4.25 4.50 4.50 4.25 4.75 3.75 4.25 4.50 4.00 0.667 0.917 0.667 4.250 6.917 2.667 0.917 7.583 1.667 4.250 5.583 1.333 2.917 2.000 1.333 2.000 2.250 4.667 1.583 4.000 2.250 3.667 2.917 1.667 2.000 0.250 2.000 1.667 0.250 1.583 2.250 4.917 1.667 7.000 1.583 0.250 2.250 0.250 1.667 6.000 0.816 0.957 0.816 2.062 2.630 1.633 0.957 2.754 1.291 2.062 2.363 1.155 1.708 1.414 1.155 1.414 1.500 2.160 1.258 2.000 1.500 1.915 1.708 1.291 1.414 0.500 1.414 1.291 0.500 1.258 1.500 2.217 1.291 2.646 1.258 0.500 1.500 0.500 1.291 2.449 5 5 5 4 5 3 8 5 3 7 7 3 6 6 2 6 2 2 8 4 3 4 4 4 4 6 1 5 4 4 4 5 4 4 2 3 4 5 5 4 2 5 5 6 4 4 6 3 8 3 5 4 5 5 5 4 2 5 5 3 5 2 5 3 2 3 5 5 4 6 4 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0 5 4 1 5 6 4 2 3 5 6 6 2 5 3 4 2 2 3 4 1 2 7 2 5 3 6 5 3 4 3 4 4 3 5 2 3 7 3 4 4 3 5 3 1 4 5 5 4 4 6 4 2 2 6 2 5 4 4 3 4 5 4 4 2 4 5 4 4 2 2 3 4 3 6 3 4 7 2 6 3 3 3 3 3 6 4 3 4 5.00 4.25 5.00 4.00 3.00 4.75 3.50 3.25 3.25 4.25 3.75 3.50 3.50 5.00 4.25 3.50 3.75 4.75 2.75 3.50 3.25 4.50 3.50 3.00 3.25 4.50 3.75 4.25 6.25 2.75 4.00 2.50 3.75 3.75 2.50 2.75 5.25 4.50 3.75 4.00 2.000 4.250 1.333 6.000 2.000 1.583 1.667 2.250 0.917 6.917 2.917 3.000 1.667 2.000 2.917 3.000 0.250 0.917 4.917 1.667 0.917 5.667 0.333 1.333 0.250 1.667 2.917 1.583 2.250 0.917 2.000 3.000 0.917 0.917 1.000 2.917 0.917 0.333 2.250 0.667 1.414 2.062 1.155 2.449 1.414 1.258 1.291 1.500 0.957 2.630 1.708 1.732 1.291 1.414 1.708 1.732 0.500 0.957 2.217 1.291 0.957 2.380 0.577 1.155 0.500 1.291 1.708 1.258 1.500 0.957 1.414 1.732 0.957 0.957 1.000 1.708 0.957 0.577 1.500 0.816 3 4 3 3 7 3 5 2 4 6 4 5 3 6 1 4 2 4 3 7 1 4 5 5 6 4 5 4 3 1 4 1 5 5 5 2 6 4 6 5 2 5 6 4 4 4 6 3 6 5 4 5 4 7 7 3 3 6 5 3 4 3 3 5 6 7 5 4 2 4 4 5 5 5 3 4 4 4 4 4 6 3 1 3 9 5 6 5 4 2 3 6 5 5 5 5 5 2 0 4 4 6 5 4 7 3 2 5 4 3 2 2 5 4 4 5 2 1 4 3 7 3 6 6 4 5 8 3 6 3 6 6 4 3 6 5 4 3 7 2 4 3 6 6 3 6 5 4 6 4 4 6 4 4 6 5 4 4 4 5 4.50 3.75 4.00 4.00 6.00 4.25 6.25 3.25 5.00 4.00 4.25 5.50 4.00 5.25 4.75 4.25 3.50 3.75 3.75 4.00 3.25 4.00 4.75 5.00 5.50 5.00 4.25 4.25 3.75 3.00 3.50 3.50 4.75 4.50 4.50 4.00 4.00 3.25 4.50 4.25 5.667 0.917 6.000 2.000 6.000 0.917 1.583 1.583 1.333 3.333 1.583 0.333 0.667 2.917 6.917 0.917 1.667 2.917 8.917 4.667 2.250 2.000 1.583 0.667 3.000 3.333 2.250 0.250 2.917 2.000 1.000 5.667 0.250 0.333 1.667 2.000 2.667 2.250 1.000 0.917 2.380 0.957 2.449 1.414 2.449 0.957 1.258 1.258 1.155 1.826 1.258 0.577 0.816 1.708 2.630 0.957 1.291 1.708 2.986 2.160 1.500 1.414 1.258 0.816 1.732 1.826 1.500 0.500 1.708 1.414 1.000 2.380 0.500 0.577 1.291 1.414 1.633 1.500 1.000 0.957 5 5 5 4 3 5 4 3 2 2 3 2 2 5 6 4 5 4 3 4 5 3 9 2 5 3 4 3 3 2 6 5 4 5 2 4 8 3 5 4 3 4 4 5 4 2 4 4 9 3 2 7 5 3 5 4 2 2 1 3 3 6 2 4 3 3 6 1 5 4 3 2 3 3 3 4 4 1 5 6 7 4 7 3 5 5 2 6 2 1 3 2 6 3 5 4 4 4 7 4 3 7 3 6 2 3 6 4 6 3 2 5 4 4 4 4 5 4 4 3 5 5 2 4 5 5 4 3 5 2 3 5 4 3 4 2 4 4 4 1 6 3 3 1 3 4 8 2 4 3 5 3 3 2 3 5 4 5 4 4 5.00 4.50 4.50 4.00 4.25 4.25 3.50 4.00 4.50 2.00 2.75 4.00 4.25 3.50 5.00 3.50 3.75 3.50 3.75 3.00 4.25 4.75 4.25 3.25 3.25 3.25 6.00 2.50 4.50 3.00 4.00 3.75 3.50 3.50 3.00 4.25 5.25 3.25 4.50 4.25 2.667 0.333 4.333 0.667 0.917 2.250 1.000 2.000 11.000 0.667 0.250 6.000 2.917 1.000 0.667 1.000 1.583 1.000 6.250 2.000 2.250 4.250 10.250 4.917 1.583 0.250 2.667 1.667 1.667 0.667 3.333 2.250 0.333 1.667 0.667 0.250 3.583 2.917 0.333 1.583 1.633 0.577 2.082 0.816 0.957 1.500 1.000 1.414 3.317 0.816 0.500 2.449 1.708 1.000 0.816 1.000 1.258 1.000 2.500 1.414 1.500 2.062 3.202 2.217 1.258 0.500 1.633 1.291 1.291 0.816 1.826 1.500 0.577 1.291 0.816 0.500 1.893 1.708 0.577 1.258 5 5 1 6 4 5 3 1 3 2 6 8 4 5 6 4 4 5 1 4 4 4 4 7 4 2 4 4 4 5 2 2 3 3 5 2 6 3 2 6 3 3 4 1 2 2 6 4 2 4 6 6 5 3 7 4 5 4 4 3 4 3 4 7 5 4 1 1 7 4 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6 4 5 3 5 5 6 5 4 2 2 5 5 6 1 5 5 4 4 4 5 9 5 4 4 3 4 6 4 4 6 4 3 2 3 4 4 3 4 3 4 2 2 1 5 6 6 4 3 6 5 3 2 4 4 1 8 3 5 4 3 4 6 6 1 3 6 6 5 2 3 5 3 3 3 4 5 6 4 2 6 4 5 4 5 4 5 4 4 4 5 3.75 3.00 3.75 3.25 3.75 4.25 3.25 4.00 3.75 4.00 5.00 5.25 5.00 2.75 3.25 4.25 4.75 4.25 4.00 5.00 4.75 3.25 2.50 3.25 4.00 4.00 4.50 3.50 2.75 5.00 3.25 4.50 2.75 5.00 4.75 4.75 5.25 4.00 4.00 4.00 4.917 2.000 2.917 2.917 3.583 8.917 0.917 0.667 0.250 1.333 7.333 0.917 3.333 2.250 1.583 1.583 2.250 2.917 2.667 2.000 0.917 0.250 0.333 0.250 2.667 0.667 1.667 1.000 0.917 1.333 0.917 3.000 1.583 0.000 0.917 3.583 3.583 0.667 2.667 2.000 2.217 1.414 1.708 1.708 1.893 2.986 0.957 0.816 0.500 1.155 2.708 0.957 1.826 1.500 1.258 1.258 1.500 1.708 1.633 1.414 0.957 0.500 0.577 0.500 1.633 0.816 1.291 1.000 0.957 1.155 0.957 1.732 1.258 0.000 0.957 1.893 1.893 0.816 1.633 1.414 2 4 3 3 4 3 7 6 4 5 1 4 3 3 4 3 5 3 4 4 5 5 3 2 5 4 5 5 4 3 4 1 2 5 5 2 3 5 4 2 5 3 5 3 5 4 7 6 4 4 3 6 7 4 2 5 5 2 1 5 5 2 3 5 3 4 3 5 4 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3 2 3 2 5 5 4 3 6 4 2 4 5 4 9 5 6 4 3 4 2 5 3 1 5 4 4 7 3 2 2 6 7 5 6 5 5 4 6 4 4 6 3 6 5 5 6 3 1 4 6 2 6 6 3 5 6 4 6 4 6 8 4 5 7 2 4 2 4 1 5 5 2 5 2 5 3 2 4 6 5 6 3 4 3 5 2 3 6 6 3 4 4 7 4 3 4.75 3.25 5.00 3.75 4.00 5.25 4.50 4.00 4.75 3.75 4.50 4.50 4.25 3.25 4.00 5.25 4.25 3.75 4.75 3.75 3.00 2.50 4.25 4.75 4.25 4.50 2.50 4.00 4.00 5.00 4.25 3.25 4.75 5.00 4.00 3.50 5.50 5.00 5.25 4.00 6.250 2.250 2.000 1.583 3.333 4.250 5.667 6.667 4.250 1.583 0.333 4.333 0.917 4.250 2.000 0.917 4.250 2.250 3.583 2.250 2.000 1.667 1.583 0.917 0.917 3.667 1.000 2.667 1.333 0.000 2.917 0.250 6.250 1.333 3.333 1.000 1.667 3.333 6.917 1.333 2.500 1.500 1.414 1.258 1.826 2.062 2.380 2.582 2.062 1.258 0.577 2.082 0.957 2.062 1.414 0.957 2.062 1.500 1.893 1.500 1.414 1.291 1.258 0.957 0.957 1.915 1.000 1.633 1.155 0.000 1.708 0.500 2.500 1.155 1.826 1.000 1.291 1.826 2.630 1.155 4 4 5 7 3 4 3 5 2 6 4 4 4 1 6 4 2 4 3 3 5 2 6 3 3 2 3 3 4 4 2 5 3 1 2 6 2 4 2 5 4 2 4 7 5 4 7 7 6 1 2 5 4 5 6 4 2 3 3 5 5 1 1 3 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4 4 4 4 4 3 5 5 4 4 4 6 6 2 3 8 7 4 4 5 7 2 1 6 4 4 3 6 6 6 6 3 7 3 2 7 3 5 2 3 6 3 5 3 6 5 5 3 5 1 2 4 5 3 5 4 7 2 3 5 5 4 3 5 3 3 4 3 5 2 3 1 0 3 2 4 3 7 5 5 3 5 6 5 6 6 5 6 5 3 4 4 2 5 2 6 6 3 2 7 4 3.00 1.75 4.50 4.00 4.50 3.25 4.00 5.00 4.50 3.75 2.50 4.50 2.75 3.25 4.75 3.50 5.75 4.25 4.00 4.00 4.25 4.50 4.25 4.75 4.50 4.00 4.00 4.00 2.75 3.25 4.25 3.75 3.25 3.75 5.50 6.00 3.00 3.00 5.25 4.75 0.667 2.250 3.000 0.667 1.667 0.250 2.000 0.667 5.667 2.917 1.000 9.667 0.250 3.583 2.250 1.000 0.917 2.250 1.333 2.000 0.917 1.667 0.917 2.250 1.667 4.000 3.333 2.000 1.583 0.917 1.583 4.250 1.583 2.250 3.667 2.000 0.667 0.667 1.583 2.917 0.816 1.500 1.732 0.816 1.291 0.500 1.414 0.816 2.380 1.708 1.000 3.109 0.500 1.893 1.500 1.000 0.957 1.500 1.155 1.414 0.957 1.291 0.957 1.500 1.291 2.000 1.826 1.414 1.258 0.957 1.258 2.062 1.258 1.500 1.915 1.414 0.816 0.816 1.258 1.708 5 4 2 4 6 4 7 3 3 4 3 3 3 3 3 5 1 2 4 6 5 5 3 7 3 5 6 2 6 2 4 5 4 1 2 6 3 4 8 2 3 5 3 3 5 2 3 7 4 5 5 4 4 2 2 4 7 1 3 1 3 3 4 3 4 7 3 5 5 3 6 2 5 6 6 4 6 5 5 5 2 4 4 4 5 5 2 3 3 3 3 4 1 3 3 2 3 4 4 3 3 7 3 5 4 3 3 5 8 7 0 1 6 5 3 4 4 3 4 4 3 4 5 2 5 4 2 5 4 2 5 10 5 6 2 4 7 1 4 3 4 5 5 2 6 1 1 3 6 4 6 1 2 4 5 3 3 5 4 4 3.25 4.25 3.50 3.25 5.25 3.75 3.50 4.50 3.50 3.50 4.00 5.25 3.25 3.50 2.50 3.75 4.50 2.00 3.75 3.25 3.75 5.00 3.75 4.25 4.25 4.00 3.25 3.75 6.25 4.00 4.00 2.25 4.25 4.00 4.00 4.25 4.00 4.25 5.25 3.75 1.583 0.250 1.667 0.917 0.250 1.583 5.667 3.667 0.333 1.667 1.333 10.250 2.917 3.000 0.333 1.583 9.000 2.000 0.250 4.250 0.917 2.667 0.917 4.917 1.583 6.667 4.250 2.250 1.583 4.667 8.000 3.583 2.917 4.667 3.333 1.583 2.000 0.917 3.583 1.583 1.258 0.500 1.291 0.957 0.500 1.258 2.380 1.915 0.577 1.291 1.155 3.202 1.708 1.732 0.577 1.258 3.000 1.414 0.500 2.062 0.957 1.633 0.957 2.217 1.258 2.582 2.062 1.500 1.258 2.160 2.828 1.893 1.708 2.160 1.826 1.258 1.414 0.957 1.893 1.258 4 2 4 3 4 7 6 2 4 5 2 4 4 3 2 5 2 1 3 5 3 5 5 5 4 5 3 5 3 5 6 4 4 3 3 1 5 6 7 4 4 3 5 5 5 4 2 5 4 5 4 1 3 5 6 4 3 6 5 4 4.75 2.50 3.50 3.25 4.25 3.75 4.25 4.00 5.25 4.75 3.75 3.25 5.00 4.00 4.00 0.917 1.000 3.000 0.250 0.917 6.250 4.917 3.333 1.583 0.917 1.583 0.250 0.667 1.333 2.000 0.957 1.000 1.732 0.500 0.957 2.500 2.217 1.826 1.258 0.957 1.258 0.500 0.816 1.155 1.414 Variance for Standard Deviation for Sample of 20,000 Sample of 20,000 m's m's 0.599 times 4 2.396 0.774 times 2 1.548 ... 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Share # Prove that Tan {Pi/4 + 1by2 Cos^(-1) Abyb} + Tan {Piby4 - 1by2 Cos^(-1) Abyb} = (2b)Bya - CBSE (Commerce) Class 12 - Mathematics ConceptProperties of Inverse Trigonometric Functions #### Question Prove that tan {pi/4 + 1/2 cos^(-1)  a/b} + tan {pi/4 - 1/2 cos^(-1)  a/b} = (2b)/a #### Solution Let cos^(-1) (a/b) = 0 Then cos theta = a/b L.H.S: tan {pi/4 + 1/4 cos^(-1)  a/b} + tan (pi/4 - 1/2cos^(-1)  a/b) = tan (pi/4 + theta/2) + tan (pi/4 - theta/2) = (1 + tan  theta/2)/(1 - tan  theta/2) + (1 - tan  theta/2)/(1 + tan  theta/2) ((1 + tan  theta/2)^2 + (1 - tan  theta/2)^2)/(1 - tan^2   theta/2) = 2((1 + tan^2  theta/2)/(1 -  tan^2  theta/2)) = 2/(cos theta)   [∵ cos 2 theta = (1 - tan^2 theta)/(1 + tan^2 theta)] = (2b)/a =RHS LHS RHS Hence Proved Is there an error in this question or solution? #### APPEARS IN Solution Prove that Tan {Pi/4 + 1by2 Cos^(-1) Abyb} + Tan {Piby4 - 1by2 Cos^(-1) Abyb} = (2b)Bya Concept: Properties of Inverse Trigonometric Functions. S

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