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https://www.physicsforums.com/threads/finding-frictional-force-and-power-using-work-and-energy.894148/

math

1. The problem statement, all variables and given/known data To measure the combined force of friction (rolling friction plus air drag) on a moving car, an automotive engineering team you are on turns off the engine and allows the car to coast down hills of known steepness. The team collects the following data: (1) On a 2.70° hill, the car can coast at a steady 20 m/s. (2) On a 5.50° hill, the steady coasting speed is 30 m/s. The total mass of the car is 1150 kg. (a) What is the magnitude of the combined force of friction at 20 m/s (F20) and at 30 m/s (F30)? (b) How much power must the engine deliver to drive the car on a level road at steady speeds of 20 m/s (P20) and 30 m/s (P30)? (c) The maximum power the engine can deliver is 46 kW. What is the angle of the steepest incline up which the car can maintain a steady 20 m/s? (d) Assume that the engine delivers the same total useful work from each liter of gas, no matter what the speed. At 20 m/s on a level road, the car gets 14.2 km/L. How many kilometers per liter does it get if it goes 30 m/s instead? 2. Relevant equations W(ext) = Delta E(mech) + Delta E(therm) Delta E(mech) = Delta K + Dela U Delta E(therm) = F(friction) * s 3. The attempt at a solution Parts c and d are the the ones I can't figure out. For part a, I used W(ext) = Delta E(mech) + Delta E(therm). Since there are no external forces, Delta E(therm) = - Delta E(mech). Delta E(mech) = Delta K + Delta U, and since the car is coasting at a constant speed, Delta K is 0. I set the gravitational potential energy to be 0 at the top of the hill, so Delta U = U(final), which is -mgssin(theta) if s is the length of the hypotenuse of the incline. So then Delta E(therm) = mgssin(theta), and since Delta E(therm) = F(friction) * s, F(friction) = mgsin(theta). Plugging the numbers, I got that the frictional force at 20 m/s was 531N and at 30 m/s was 1080N, which was correct. For b, I just used P = F * v to find the power at 20 m/s and 30 m/s. For c, I plugged the equation for F into the Power equation, which gave me P = mgsin(theta) * v. Then I solved for theta, and got theta = arcsin(Power / (mgv)). I tried plugging in 46000 W for power along with the other given information to get theta = 11.8 degrees, and I also tried plugging in my answer for b at 20 m/s in, neither of which were right. For d, I'm not really sure where to start, or what equations/relationships to use. Maybe I could use the ratio between the power needed at 20 m/s and 30 m/s somehow? Any help would be appreciated.

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https://socratic.org/questions/how-do-you-factor-x-2-2-5-x-2

math

How do you factor #(x+2)^2 -5(x+2)#? See a solution process below: First, rewrite the term on the left as: Next, factor out a common term of First we expand the Next we just multiply it out which will give us Now that we know We can do the other part which is Now we know Now we can write the problem out expanded NOTICE THE NEGATIVE SIGN IN FRONT OF THE PARENTHESES. We have to distribute this negative to all terms in the parentheses. Which gives us Now we just combine like terms Now we need 2 numbers that multiply to These numbers are So we then get Impact of this question Creative Commons License

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CC-MAIN-2023-23

https://research.tdameritrade.com/grid/public/markets/news/story.asp?fromPage=results&docKey=309-SI20210504150052&provider=Briefing%20StockNews

math

Yellen's view not controversial The S&P 500 is off prior lows (-1.5%) with a current 0.9% decline. Revisiting Treasury Secretary Yellen's observation on interest rates, it really isn't that novel of a view as many have been calling the 10-yr yield to rise to 2.00% for a while now. Fed Chair Powell also said that higher rates have been pricing in greater economic growth. There's also some confusion, as Ms. Yellen didn't specifically call for a rate hike from the Fed. In other words, Ms. Yellen was stating the obvious in that with greater economic growth comes higher interest rates, which in turn helps keep the economy from rising too rapidly with natural market forces. Looking ahead, T-Mobile US (TMUS 128.40, -3.05, -2.3%), Zillow (ZG 124.33, -4.76, -3.7%), and Lyft (LYFT 55.59, -1.49, -2.6%) are some notable companies that will report earnings after the close.

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http://www.gregorybard.com/Sage.html

math

Gregory V. Bard Associate Professor of Mathematics Preserving the look-and-feel of the World Wide Web as it was, in 1998. What is Sage? Sage is the free, open-source competitor to Maple, Mathematica, Magma, and Matlab. It is a computer-algebra system ideally suited to students of mathematics. This page is woefully incomplete. The Sage community involves hundreds of developers and thousands of contributors world wide. Last but not least, I've made some videos for introducing students to Sage. - Here are some links for my book Sage for Undergraduates, scheduled to be published by The American Mathematical Society in the Summer of 2014. - Graciously, the AMS has permitted me to place a pdf file of the book on my webpage. - Here is a link to the black-and-white version. - Here is a link to the color version. - The online electronic appendix will cover plotting in color, animations, and 3D graphics. Those subjects are not suited to a black-and-white book, and therefore cannot be printed inside the book itself. [Coming Soon!] - Chapter 6 of the book teaches the reader how to make their own interactive webpages or applets. To save readers from having to retype my code into their computers, I promised a zip-file with some source code of the examples used. - Here are some videos that I've made to introduce you to the basics of using Sage with its most simple interface, the Sage Cell Server. Both are less than five minutes. One covers functions, derivatives, integrals, and 2D plotting. Two covers factoring, 3D plotting, gradients, and symbolic solving. - After watching both videos (or even without them) you'd find it very easy to just dive on into Chapter 1 of my book, linked above. - At the bottom of this page, I have some other videos about matrices and about linear - Want to give Sage a try? - For short and medium-sized problems (especially in 100-level and 200-level courses, but even in higher-level courses too), the best way to use Sage is the single-cell server. Here is a link to The Sage Single-Cell Server. (That's the competitor to WolframAlpha, and until recently it was called Sage-Aleph.) - Here's a link to several interactive web pages and applets that have been written mostly using Sage, by myself and others. (For using these, NO KNOWLEDGE of Sage whatsoever is required!) - For longer problems (especially those that will require collaboration, writing your own programs, or using data sets) the server is the way to go. - The beauty of Sage is that it works through the internet. There is almost never any reason to do a local install of Sage on your laptop or home computer. This is good news, because it saves a lot of headaches and hassles (especially for students), that you would have to suffer if you were using Mathematica, Maple, Matlab, or Magma. - This is an excellent tutorial for faculty, PhD-students in mathematics, and senior math majors about using Sage for all sorts of problems. - Here is a large collection of quick-reference cards for Sage, by various people, for various branches of mathematics, in many languages. Personally, I think having a printed quick-reference card out next to the laptop while using Sage is really handy. :-) - Here is a series of videos/screencasts introducing Sage. They are made by William Stein, the founder of Sage. - Here is a link to the official Sage website. - Here is a video for how to find the Reduced Row Echelon Form (RREF) of a matrix. For example, you might do this to solve a linear system of equations. - For solving Linear Programming Problems (i.e. maximizing or minimizing a many-variable linear function subject to several multivariate linear inequalities), I have three videos: - More is coming soon!! Last updated July 21st, 2014.

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https://link.springer.com/chapter/10.1007/978-0-8176-4613-4_2?error=cookies_not_supported&code=7bb8d6e7-6e22-4262-886d-36e225b32610

math

This chapter studies finite-dimensional associative division algebras, as well as other finite-dimensional associative algebras and closely related rings. The chapter is in two parts that overlap slightly in Section 6. The first part gives the structure theory of the rings in question, and the second part aims at understanding limitations imposed by the structure of a division ring. Section 1 briefly summarizes the structure theory for finite-dimensional (nonassociative) Lie algebras that was the primary historical motivation for structure theory in the associative case. All the algebras in this chapter except those explicitly called Lie algebras are understood to be associative. Section 2 introduces left semisimple rings, defined as rings R with identity such that the left R module R is semisimple. Wedderburn’s Theorem says that such a ring is the finite product of full matrix rings over division rings. The number of factors, the size of each matrix ring, and the isomorphism class of each division ring are uniquely determined. It follows that left semisimple and right semisimple are the same. If the ring is a finite-dimensional algebra over a field F, then the various division rings are finite-dimensional division algebras over F. The factors of semisimple rings are simple, i.e., are nonzero and have no nontrivial two-sided ideals, but an example is given to show that a simple ring need not be semisimple. Every finite-dimensional simple algebra is semisimple. Section 3 introduces chain conditions into the discussion as a useful generalization of finite dimensionality. A ring R with identity is left Artinian if the left ideals of the ring satisfy the bdescending chain condition. Artin’s Theorem for simple rings is that left Artinian is equivalent to semisimplicity, hence to the condition that the given ring be a full matrix ring over a division ring. Sections 4–6 concern what happens when the assumption of semisimplicity is dropped but some finiteness condition is maintained. Section 4 introduces the Wedderburn–Artin radical rad R of a left Artinian ring R as the sum of all nilpotent left ideals. The radical is a two-sided nilpotent ideal. It is 0 if and only if the ring is semisimple. More generally R/ rad R is always semisimple if R is left Artinian. Sections 5–6 state and proveWedderburn’s Main Theorem—that a finite-dimensional algebra R with identity over a field F of characteristic 0 has a semisimple subalgebra S such that R is isomorphic as a vector space to S ⊕ rad R. The semisimple algebra S is isomorphic to R/ rad R. Section 5 gives the hard part of the proof, which handles the special case that R/ rad R is isomorphic to a product of full matrix algebras over F. The remainder of the proof, which appears in Section 6,follows relatively quickly from the special case in Section 5 and an investigation of circumstances under which the tensor product over F of two semisimple algebras is semisimple. Such a tensor product is not always semisimple, but it is semisimple in characteristic 0. The results about tensor products in Section 6, but with other hypotheses in place of the condition of characteristic 0, play a role in the remainder of the chapter, which is aimed at identifying certain division rings. Sections 7–8 provide general tools. Section 7 begins with further results about tensor products. Then the Skolem–Noether Theorem gives a relationship between any two homomorphisms of a simple subalgebra into a simple algebra whose center coincides with the underlying field of scalars. Section 8 proves the Double Centralizer Theorem, which says for this situation that the centralizer of the simple subalgebra in the whole algebra is simple and that the product of the dimensions of the subalgebra and the centralizer is the dimension of the whole algebra. Sections 9–10 apply the results of Sections 6–8 to obtain two celebrated theorems—Wedderburn’s Theorem about finite division rings and Frobenius’s Theorem classifying the finite-dimensional associative division algebras over the reals.

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http://london.localmartuk.com/item/looking_for_merchant_account_payment_processing_sales_agents/97816328

math

Posted: 20 Apr 2017 We are one of the top PSP in Hongkong and specialize in Low Risk/High Risk merchant processing, currently process more than 10,000 merchants over 350 million USD per month. 1000$ = Sign Up Clients above 100k/Monthly Volume 2000$ = Sign Up Clients above 500k/Monthly Volume 500$ (Per Client) = Signed Up and Approved Client Agrees to 10%+ discount rate. 1000$ (Per Client) = Signed Up and Approved Client Agrees to 13%+ discount rate. 1000$ (Monthly Total) = Bring 10+ Successfully signed up and approved Clients in a month. Agent Referral Bonus 1000$ = Refer An Agent Who Successfully Brings First 5 clients

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https://www.hypnos.com/smf/index.php?action=profile;u=419;area=showposts;start=20

math

thanks Lena; there's some fun stuff going on here- This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.Show posts Menu Quote from: jim brenholts on July 09, 2009, 03:18:30 PM hey carl! welcome! it is always good to have new folks join us here on the forum. i am downloading the freebie and looking forward to new sounds. Quote from: APK on July 09, 2009, 07:47:24 AM Hi Carl. Welcome. Always good to have a new face around here. Quote from: Scott M2 on July 09, 2009, 06:15:37 AM Hi Carl! Welcome to the H. Forum. You give Buffalo a good name. 8) Quote from: satish on May 18, 2009, 09:44:56 AM Including Auditory Range, Piano Range, Loudness, Fundamentals Overtones Harmonics and Combination Tones, Beats, Scales, Physiological Effects Of Sound rotating dot Auditory Range The range of human hearing is from around 2Hz (2 cycles per second) to 20,000Hz (alias 2KHz) although with age one tends to lose acuity in the higher frequencies so for most adults the upper limit is around 10KHz. The lowest frequency that has a pitch-like quality is about 20Hz. A typical value for the extent to which an individual can distinguish pitch differences is 05-1% for frequencies between 500 and 5000Hz. (Differentiation is more difficult at low frequencies). Thus at 500Hz most individuals will be unable to tell if a note is sharp or flat by 2.5-5Hz (ie an 'allowable' pitch range for that note might be from 495Hz to 505Hz maximum. rotating dot Piano Range The lowest note on the piano 'A0' is at 27.5Hz, whilst the highest, C8, has a frequency of 4186Hz. 'Concert pitch' has been internationally accepted to be based on a frequency of 440Hz for A4, that is A in the fourth piano octave. rotating dot Loudness The quietest sounds that can be heard have a power (measured in Watts) of 10 to the -12 W/m2, whilst the loudest that can be withstood have a power of 1 W/m2. The range is therefore in the order of 10 to the 12, or one million million times. One decibel is a leap by a factor of 10, so that 0Db is the quietest noise, and 120Db is the loudest. rotating dot Fundamentals, Overtones, Harmonics And Combination Tones A note played by most ordinary acoustic intruments is not 'pure', it is in fact a spectrum of frequencies which largely give the note it's characteristic quality. The most significant components of this spectrum are the overtones. Overtones may be inharmonic (ie dissonant, sounding bad) or harmonic (ie consonant, sounding good). The 'natural overtone series' is the set of harmonics which are particularly prominent in the spectrum of frequencies for notes played by acoustic instruments, and which are also produced by producing waves in a string, where there are successive integer values for the number of 'crests' of the wave, ie 1,2,3,4,5,6,7,8 and so on. Taking 1 crest as the 'fundamental' note, then 2 produces a frequency an octave above. 3 produces a frequency of a '12th' interval (ie the fifth above the octave). 4 produces a frequency of the second octave above. 5 produces a frequency of a major third above that. 6 produces a frequency of the fifth note in this third octave. 7 produces a frequency of the flattened (minor) seventh in this octave. 8 produces a frequency of the fourth octave. 9 produces a frequency of the second interval in the fourth octave. (et cetera...) The increments between notes becomes successively smaller the further into the series you go, such that soon there are semitone intervals, then quarter- tone intervals, then even smaller fractional intervals. Also, there is less and less congruence between harmonics and the 'proper' frequencies of the equitempered scale the further into the series you go. (But maybe in some forms of music this is NOT a 'problem' od course). In the first octave there are no overtones, in the second there is one, in the third there are three, and in the fourth octave there are lots. Overtones which are 'harmonic' are at frequencies equal to the fundamental frequency multiplied by AN INTEGER: eg to find the fifth harmonic of C4, multiply it's frequency of 261.63 by 5: the result is a frequency of 1308.15, close to (but NOT the same as) the frequency of 1318.5 given for E6 (the fifth harmonic of C4) in the equitempered scale. Overtones which are inharmonic (dissonant) are non-integer multiples of the fundamental frequency. The constituent frequencies of a note can be ascertained with a wave analyser, which uses the equations developed by Fourier. If you filter out the fundamental frequency of say 200Hz from a note with overtones of 400Hz, 600Hz, 800Hz and 1000Hz (say, played on a piano), the brain will recreate the fundamental such that it appears to be present even when it is not there. The ear can be far more subtle still, however, in 'creating' sounds which are not actually there in received vibrations. When a tone of frequency f causes the eardrum to vibrate, the full harmonic series of tones 2f, 3f, 4f, 5f, 6f etc is apparent to the listener, these harmonics being produced by the eardrum itself. 'Combination' tones are additional, new tones produced when two frequencies, f1 and f2 are sounded together, and these combination tones are also 'manufactured' by the ear, this time from combinations of the first frequency and the harmonics of the second, and from the second frequency and harmonics of the first. (Note 3f2 is notation for 'the third harmonic of the second frequency'). The series of combination tones will be: 2f1 - f2 2f2 - f1 3f1 - f2 3f2 - f1 3f1 - 2f2 3f2 - 2f1 4f1 - 3f2 4f2 - 3f1 and so on. Whereas combination tones can be ascertained by subtracting one value from another, 'summation' tones, also manufactured by the ear, can be found by adding values together, ie f1 + f2 2f1 + f2 2f2 + f1 3f1 + f2 3f2 + f1 3f1 + 2f2 3f2 + 2f1 4f1 + 3f2 4f2 + 3f1 and so on. Interestingly, however, the resulting frequencies of these combination and summation tones turn out to be dissonances when played with the notes of the equitempered scale since the frequencies do not exactly match. It is possible to contrive a scale from difference tones. For example, if C4 is played together with Eb4, then a difference tone of 311.1 - 261.6 = 49.5Hz is created. This value of 49.5Hz is almost the frequency of Ab1. Other notes in the 1st octave can then be contrived in a similar manner from frequencies around the fourth octave. These sort of difference tones, however, might be called 'first-order' difference tones. A 'second-order' difference tone can be found in the difference between the frequency of a first-order difference tone and the frequency of the fundamental, so that in the example above, we have a first- order difference tone of (almost) Ab4, at 49.5Hz, and subtracting this from the frequency of the fundamental, we find 261.6 - 49.5 = 212.1 which is close to the frequency of Ab3. And so on. Here are some examples of combination and difference tones that can be demonstrated on a piano. C4+Eb4 produces Ab1 first-order difference tone. C4+Eb4 produces Ab3 second-order difference tone. rotating dot Beats Beat frequencies are produced when two different sounds are produced which are very close to each other in frequency. In such a situation the crests and troughs of each wave are generally slightly out of phase. But because the two notes have differing frequencies, after a certain repeating interval of time the crests of one wave will be aligned with the crests of the other, when a pulse or beat appears to the listener. Research has found that any two notes of different frequencies tend to sound good together (ie consonant) if there is an absence of beat frequencies between 8 and 50 Hz produced. Beat frequencies of 2-8Hz have been found to be pleasing, while beat frequencies above that level are generally though to be unpleasant. The beat frequency produced by any two notes is found by subtracting the value of the higher from the lower, ie Fbeat = Fhigher - Flower. Thus beat frequencies are a subset of difference tones -the 'beat' sensation occurring when the beat frequency value is low, say from 0.5Hz (1 beat every two seconds) to say 20Hz (the lowest frequency that has a pitch-like quality) J.Askill, in 'The Physics of Musical Sounds) says 'in general beat frequencies of 2-8Hz are considered pleasing, whereas if the beat frequency is above 15-20Hz, an unpleasant or dissonant effect is produced'. Personally I am curious about the range in the middle, say 8-12Hz which is also the frequency of 'alpha' brain-waves. rotating dot Scales The scale used extensively in the West has 13 notes from octave to octave and 12 intervals. In order for a scale to 'work' there should be: * a minimum of dissonance when different notes across the whole range of pitches are sounded together * an effective mapping of the harmonics of low notes onto higher notes, and an effective mapping onto the harmonics of higher notes * the possibility of key modulation which does not result in further frequency mismatches. The scale which has been widely adopted to fulfil these criteria is based on mathematics, such that the ratio of the frequency of any note to the frequency of the note a semitone above is constant. This is particularly useful in the extent to which it allows key modulation. However, this 'equitempered' scale is a compromise solution, because the frequency ratios of all intervals except the octave differ slightly from the 'perfect' intervals that the human ear really expects to hear. The 'exact' interval of a fifth, for instance, is found by multiplying the frequency of the fundamental by 3/2, the fourth is found by multiplying the fundamental frequency by 4/3, and the major third interval is found by multiplying the fundamental frequency by 5/4. (Other intervals involve slightly less obvious fractions). The problem with a scale built on fractional values like this, however, is that the increments from note to note are not constant (eg 5/4 - 4/3 does not equal 4/3 - 3/2) which creates difficulties when the required key for a piece is different to that of the fundamental from which the scale is constructed. For example, if we move up an octave from C by adding a fifth, and then adding a fourth, then the resulting high C will have a different frequency to that arrived at if our key is F, and we try to arrive at the same high C by adding a major third and then a minor third to that fundamental F. So in the equitempered scale all semitone increments have been 'tempered' such that they are always a little flat, or a little sharp. The constant value on which this scale is based is 1.0594630915, such that if we call this value S, then the semitone above a fundamental note is found by multiplying the frequency of the fundamental by S to the power of 1. The second above the fundamental is found by multiplying it's frequency by S to the power of 2, and so on until the octave above the fundamental is found by multiplying it's frequency by S to the power of 12. The value of the constant S is the 12th root of 2, since in order to find the twelve equal divisions between two notes an octave apart, where the frequency of the octave is twice that of the fundamental, the 12th root of 2 is the value we are looking for. Other divisions of the octave have been proposed, such as a 19-step octave, and a 53-step octave. The maths for these 'works' although these 'scales' may be harder to use effectively. The maths for the 53-division scale is particularly elegant in fact, and closer to a 'perfect' musical scale than the 13-note scale which we currently use. (In that case the constant value for each successive interval is found by using the 53rd root of 2, ie 1.013164143). The 'exact' scale, built on the 'perfect' intervals that the ear expects to hear, has much to recommend it if one key is kept to. This scale, however, has fifteen intervals and fourteen notes, since in the first octave there are all the notes of the equitempered scale (at slightly different frequencies) but there is also both a 'major whole tone' and a 'minor whole tone', and both an 'augmented fourth' and a 'diminished fifth'. (In the second octave there is both an 'augmented octave' and a 'dimished ninth' and also both an'augmented eleventh' and a 'diminished twelfth'). Thus successive octaves above the fundamental differ from each other in the way that they are put together. Furthermore, however, when we look at the extent to which the frequencies of harmonics of exact-scale notes match notes higher up in the exact scale, we see that we can list the intervals octave, fifth, fourth, major third, major sixth, minor third, minor sixth in terms of increasing dissonance, so that in the case of the minor sixth, if we look at all harmonics up to the twelfth, only one 'matches'. The mathematical elegance of the 53-division scale should make it a more appropriate scale for dealing both with key modulation, and a preoccupation with harmonics. The 53-interval scale uses eneven 'chunks' of these 53rd-of-an-octave division to create the notes of the diatonic scale. The size of each incremental chunk is as follows: Quote from: DeepR on June 22, 2009, 01:34:31 PM If there's one genre where the music can be related to imagery, surely it must be ambient for its 'visual' qualities. The idea is to find pictures that you think are somehow fitting for the ambient piece.

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http://web.archive.org/web/20061128151357/http:/www.oakland.edu/enp/related.html

math

Return to Erdös Number Project home page. Erdös numbers are not the only case of interest in links among people. For a related website, where the vertices are actors and the links are provided by appearance in the same movie, take a look at the Oracle of Bacon. Here the central position of Paul Erdös is assumed by the actor Kevin Bacon. Tim Hsu and David Grabiner have observed that since Dan Kleitman actually appears (briefly) in and is a mathematical consultant for the movie Good Will Hunting, Bacon has a combined Erdös/Bacon number of 3, since Kleitman has Bacon number 2 (via Minnie Driver, who was in Sleepers with Kevin Bacon) and Erdös number 1. Bruce Reznick is in a similar position, with an Erdös number of 1 and a Bacon number (by virture of being an extra in Pretty Maids All in a Row with Roddy McDowall) of 2. In fact, according to the Oracle of Bacon site, Paul Erdös himself has an official Bacon number of 4, by virtue of the N is a Number (a documentary about him), and lots of other mathematicians have finite Bacon number through this film. However, we have recently discovered that this is bogus, because the link, named Gene Patterson, is not the same person in N is a Number as in the film Box of Moonlight, where the link supposedly lies. Thus it remains an open question as to what Paul Erdöss Bacon number is. In a similar vein, William Montbleau has a site devoted to links among musicians in rock bands. Here is a site that does the same thing for baseball players. Here is a report on a study of such phenomena in acquaintances among classmates at a college. And in the game of chess there are notions of Morphy number and Kasparov number, connecting people who have played each other (in the latter case, it's a directed graph, based on one player's beating another). for a chuckle. A serious look at the phenomenon of networks like these (often referred to with the phrase six degrees of separation, in reference to the idea, originating in Stanley Milgram's research in the 1960s ("The Small World Problem", Psychology Today, May 1967, 60-67), that you can connect (almost) any two people in the world by a path of six acquaintances, as well as aplay and movie with that title mildly related to this concept) appears in Nature, 4 June 1998, Vol. 393, No. 6684, pp. 409 and 440. See also a recent book by Duncan Watts entitled Small Worlds; Grossmans review of this book appears in the August, 2000 issue of The American Mathematical Monthly. (There is also an audio report on this from National Public Radio.) See also this article in SIAM News. Much more information along these lines is available on our page devoted to research on collaboration. One can also argue that the baseball player who broke Babe Ruths home run record, Henry L. Hank Aaron, has a joint publication with Paul Erdös. Carl Pomerance (now at Dartmouth College), who had a long and fruitful collaboration with Paul, reports having a baseball autographed by both of them, occasioned by their both having received honorary degrees at Emory University in 1995 (its a long story, having to do with a property of the numbers 714 and 715, reported by Carl at the Paul Erdös memorial session held at the 1997 annual AMS/MAA meeting in San Diego, and recounted in an article in the Notices of the American Mathematical Society, January 1998, page 22, as well as in an Ivar Peterson article). Somewhat related to the issue of collaboration in mathematical research is the issue of academic roots. At least two academic genealogical websites exist, which try to trace peoples academic ancestors, one in mathematics, and one in theoretical computer science (which quickly gets back to mathematicians, of course). The branch of mathematics dealing with the kinds of issues raised in looking at collaboration and Erdös numbers is graph theory. Here is a link to some graph theory resources, including people, research, writings (with some interesting statistics like the average number of papers per author), conferences, and the famous Four Color Theorem. Graph theory was one of Paul Erdöss specialties, of course. Kristina Pfaff-Harris has written a delightful article called Six Degrees of Paul Erdös for linux.com. Dave Rusin has a Web page devoted to all the proper names that appear in the Mathematics Subject Classification, such as in "Kan extensions" or "Erdös problems".

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CC-MAIN-2014-52

https://energy.en-academic.com/945/Cooling_Load

math

- Cooling Load - That amount of cooling energy to be supplied (or heat and humidity removed) based on the sensible and latent loads. Energy terms . 2014. Energy terms . 2014. Cooling load temperature difference calculation method — Contents 1 CLTD/CLF/SCL Cooling Load Calculation Method 2 History 3 Application 4 Explanation of Variables … Wikipedia Cooling load temperature difference — (CLTD) A value used in cooling load calculations for the effective temperature difference (delta T) across a wall or ceiling, which accounts for the effect of radiant heat as well as the temperature difference. California Energy Comission.… … Energy terms Cooling load — The rate at which heat must be extracted from a space in order to maintain the desired temperature within the space. California Energy Comission. Dictionary of Energy Terms … Energy terms Design Cooling Load — The amount of conditioned air to be supplied by a cooling system; usually the maximum amount to be delivered based on a specified number of cooling degree days or design temperature … Energy terms Latent Cooling Load — The load created by moisture in the air, including from outside air infiltration and that from indoor sources such as occupants, plants, cooking, showering, etc … Energy terms Sensible Cooling Load — The interior heat gain due to heat conduction, convection, and radiation from the exterior into the interior, and from occupants and appliances … Energy terms Cooling tower — Natural draft wet cooling hyperboloid towers at Didcot Power Station, UK … Wikipedia Load profile — In electrical engineering, a load profile is a graph of the variation in the electrical load versus time. A load profile will vary according to customer type (typical examples include residential, commercial and industrial), temperature and… … Wikipedia Cooling Degree Day — A value used to estimate interior air cooling requirements (load) calculated as the number of degrees per day (over a specified period) that the daily average temperature is above 65 degrees Fahrenheit (or some other, specified base… … Energy terms Latent load — The cooling load caused by moisture in the air. California Energy Comission. Dictionary of Energy Terms … Energy terms

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CC-MAIN-2024-18

https://globalquiz.org/en/toughest-math-riddles/

math

The most difficult math riddlesMathematics quiz If Earth's radius was larger by 1 meter, how much longer would its equator be? by approx 6 meters by approx 6 kilometers by approx 60 kilometers by approx 6000 kilometers For each circle, the length of the circumference is c=2πr. Hence, if a radius grows by 1 meter, the circumference grows by 2 π meters - no matter what the initial value of radius was. Typically, how many leap years are in a century? The Gregorian Calendar omits 3 leap days every 400 years, omitting February 29 in the 3 century years (integer multiples of 100) that are not also integer multiples of 400. For example, 1600 was a leap year, but 1700, 1800 and 1900 were not. Similarly, 2000 was a leap year, but 2100, 2200, and 2300 will not be. Which famous scientist was born on the Pi day? (March 14) all of the above 14 of March is celebrated on many schools and universities as both Pi Day and the birthday of Albert Einstein. One of the traditions is the 'Pi Pie' (pictured). Which of the following is a prime number? A prime number has no positive divisors other than 1 and itself. Hence, a prime bigger than 2 cannot be even. All prime numbers bigger than 5, written in the usual decimal system, end in 1, 3, 7, or 9. How many edges does a Möbius strip have? The Möbius strip is a surface with only one side and only one boundary component (edge). A line drawn starting from the seam down the middle will meet back at the seam but at the "other side". If continued the line will meet the starting point and will be double the length of the original strip. Let's say the stock price was 10 two days ago. It has dropped by 50% yesterday and raised by 50% today. What is the stock price now? The stock price was 10 two days ago. Yesterday it dropped by 50% to 5. Today it raised by 50% (of 5) to 7.5 How many edges does a cube have? A cube has six square faces, each of which has four edges - but each edge is shared with another face.

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http://www.ipam.ucla.edu/abstract/?tid=14534&pcode=MFG2017

math

Mean field games (MFGs) constitute a class of non-cooperative stochastic differential games, where there is a large number of players or agents who interact with each other through a mean field coupling term—also known as the mass behavior or the macroscopic behavior in statistical physics—included in the individual cost functions and/or each agent’s dynamics generated by a controlled stochastic differential equation, capturing the average behavior of all agents. One of the main research issues in MFGs with no hierarchy in decision making is to study (existence, uniqueness and characterization of) Nash equilibria with an infinite population of players under specified information structures and further to study finite-population approximations, that is to explore to what extent an infinite-population Nash equilibrium provides an approximate Nash equilibrium for the finite-population game, and what the relationship is between the level of approximation and the size of the population. Following a general overview of such games first with a finite number of players, the talk will dwell on two classes of MFGs: those characterized by risk sensitive (that is, exponentiated) objective functions (known as risk-sensitive MFGs) and those that have risk-neutral (RN) objective functions but with an additional adversarial driving term in the dynamics (known as robust MFGs). In stochastic optimal control, it is known that risk-sensitive (RS) cost functions lead to a behavior akin to robustness, leading to establishment of a connection between RS control problems and RN minimax ones. The talk will explore to what extent a similar connection holds between RS MFGs and robust MFGs, particularly in the context of linear-quadratic problems, which will allow for closed-form solutions and explicit comparisons between the two in both infinite- and finite-population regimes and with respect to the approximation of Nash equilibria in going from the former to the latter. The talk will conclude with a brief discussion of extensions to hierarchical decision structures with a small number of players at the top of the hierarchy (leaders) and an infinite population at the bottom (followers). (This is based mostly on joint work with Jun Moon)

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http://econompicdata.blogspot.com/2012/04/weak-jobs-report.html

math

While the payroll (i.e. business) survey showed a net gain of 120,000 jobs in March (which in itself was weak), the household survey (the survey used to calculate the unemployment rate) actually showed a decline. Note that while this may just be noise after a warm winter that may have pulled hiring forward, the net result is most likely a (much) weaker employment situation than previously thought. In March, the household survey showed that the number of people who say they have a job fell by 31,000 and the number of people looking for a job fell by 79,000. That lowered the unemployment rate to 8.2 percent.As the chart below shows, both the headline unemployment rate and broader measure of unemployment dropped as people left the workforce.

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https://ell.stackexchange.com/questions/274469/why-is-it-crouching-tiger-hidden-dragon-but-not-crouching-tiger-hiding-dragon

math

Why is the movie named "crouching tiger hidden dragon"? This is mainly a question about present and past participle. Why can't it be "crouching tiger hiding dragon"? Since the dragon must've been hiding voluntarily, instead of being hidden by others, I think "hiding"is more appropriate. As the first answer presents, dragons, meaning talent people, don't take a deliberate action to hide, so "hidden" is more proper; However, tigers here are also indicating talent people, so why would they deliberately hide/crouch? (though it may have something to do with "crouch" being intransitive) As the second answer points out, Why not "crouched tiger, hidden dragon" then? Another question, which is asked by a user under the second answer, is isn't "crouched" the same formation as "hidden", while "crouching" is the same formation as "hiding". I also think they have -ing and -ed each has the same formation, but they would mean different things when one verb takes the -ing form while the other takes the -ed form?

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https://secure.youthscience.ca/sfiab/gvrsf/project_summary.php?year=2016&pn=M%20040%20N

math

Floor Location : M 040 N Because I've seen my grandpa getting angry and impatient when parking and doing pull over, I decided to invent something that allows every car to park and do pull over by itself. After a massive amount of research, I understand that it is impossible to know where the vehicle's gonna end up at if I don't know the center point of the vehicle when it is turning. Therefore, my hypothesis is that every car should have a center point when it is turning. Because there's a lot of unknown factors that could go wrong in an experiment, I tried limiting down the uncertainty by setting every turning of the wheels to 30 degrees. After series of testing with my model car, built with nxt mindstorm pieces, I came up with a mathematical formula that can calculate any car's center point when turning 30 degrees by inputing the vehicle's length and width into the formula. Everything became a lot easier after I figured out the formula for calculating the center point when turning 30 degree since all there was left was to to research and figure out the simplest and fastest way for a vehicle to park and pull over. The formulas for parking and pull over are primarliy based on the first formula, which is the center point one since what the vehicle is essentially doing is turning 30 degrees over and over again. In order to prove the accuracy of my formula, I inputed the mathematical formulas into the C Programming. As a result, after compiling, my model successfully auto-parked and auto-pull over. Last but not least, these mathematical formulas do not apply only to my model since I built another model that is completely different from my first model in terms of size and model two still successfully achieved my goal.

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http://french.stackexchange.com/questions/tagged/poesie+pronoms

math

French Language Meta to customize your list. more stack exchange communities Start here for a quick overview of the site Detailed answers to any questions you might have Discuss the workings and policies of this site Meaning of “en” in “il en est un …” Mais parmi les chacals, les panthères, les lices, Les singes, les scorpions, les vautours, les serpents, Les monstres glapissants, hurlants, grognants, rampants, Dans la ménagerie infâme de ... Jul 14 '13 at 16:09 newest poesie pronoms questions feed Hot Network Questions How to JSON.serialize an opportunities line items? Should I learn to use LaTex to write up a History Masters Thesis? Automatic Work activity failed in Tridion 2013 Workflow uses external activity Maintaining a bunch of generic functions Why is "distro", rather than "distri", short for "distribution" in Linux world? Negated “moi”: which person should be used (first/third)? Did planes crash into the WTC on 9-11? Why is 未有 used here? Alphanumeric Hello World Is there any math symbol \predeq in Latex? How to use it? How to deal with players asking for an "infodump" Good slide design for teaching? error: cd..: command not found How to locate and catch my dog? Interpolation: Unstructured grid error How do KDE applications run under Gnome? Combination of \pm and \EUR What is $DISPLAY environment variable PIC12F629 Doesn't react on input Why quantising gravity necessarily give us gravitons? Do all Groups have a representation? Dark spot under cockpit on A-10s How to save the image of a node to the custom directory programmatically? What should I do about the discovery that a manager is not flushing the workplace toilet after use? more hot questions Life / Arts Culture / Recreation TeX - LaTeX Unix & Linux Ask Different (Apple) Geographic Information Systems Science Fiction & Fantasy Seasoned Advice (cooking) Personal Finance & Money English Language & Usage Mi Yodeya (Judaism) Cross Validated (stats) Theoretical Computer Science Meta Stack Overflow Stack Overflow Careers site design / logo © 2014 stack exchange inc; user contributions licensed under cc by-sa 3.0

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http://math.nju.edu.cn/academic_reports/448

math

题 目：From (semi-)abelian varieties to commutative algebraic groups 报告人： 胡飞 博士 （University of British Columbia ） 摘 要: It is known that both the Mordell–Lang conjecture and the Medvedev–Scanlon–Zhang conjecture were proved for abelian varieties, and later extended to semi-abelian varieties. In this talk, I will describe some natural generalizations to certain commutative algebraic groups. This is joint work with Dragos Ghioca, Thomas Scanlon, and Umberto Zannier. 时 间：11月26日 16:00-17:00

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https://thevictorianvaults.com/introduction-the-net-income-is-also-related-to/

math

The Simple Work-Leisure Trade income of any individual is related to the number of hours he or she works per day and then per week. The net income is also related to per hour pay rate. If the pay rate per hour is higher, then more income can be generated for fewer working hours and more leisure hours. The leisure hours are those for which the individual does not work and no income is generated in those hours. Therefore, it can be said that the net income is related to the number of hours worked in accordance with the pay rate per hour (Dean Garrat 2013). This study will discuss the effects of working hours on the net income. Some assumptions are made, such as: No income is generated for leisure hours (that is there is no savings for the individual or at least the income from any saving or property, which is not to be included). The reference point is one The individual is free to chose working hours per day (but it is very common that the individuals does not chooses working hours by themselves, it is their employees who fix / expect their workers to work for definite number of hours per week). Any hours not considered as working hours are assumed to be leisure hours. mentioned earlier, there is a definite relationship between the net income (wage) and the number of working hours. This relationship can be explained in terms of plot between wage and working hours. relationship is shown in figure where the number of hours has been taken along x-axis and the money earned from work (wage) is shown on y axis. Working hour is an independent variable and the wage earned is a dependant variable, because the wage depends on number of hours worked per week. 1: Wage and work hours are two budget constraints shown in figure 1, one for hourly wage rate of w1 (budget line 1 known as Bl1) and the second for an hourly rate of w2 (Bl2). It is assumed that there are 24 working hours in a single day which means that the number of leisure hours are zero. In total there are 24 working hours and that is why both lines meet the x axis at 24 and Bl1 meets the y axis at 24w1 showing that for 24 working hours the wage will be 24w1. This can be calculated for any given hourly rate, for instance if the hourly rate is £6.50 the total earning will be £156 for 24 hours’ work. 2: Indifference curve for higher Figure 2: Indifference curve for lower satisfaction line 2 (Bl2) shows the line when hourly rate was increased from w1 to w2 where w2 is larger than w1. This shows that if the hourly rate is increased then the total wages would also increase for any given working hours as shown in figure figure 2 and figure 3 the examples of an indifference curves are shown where in figure 2 the indifference curve is shown for an individual who not only enjoys leisure but derives a lot of utility from the work associated with his job. However, in figure 3 the indifference curve is shown for an individual who enjoys leisure but derives much less utility from the work associated with the job. In this case the curve is much closer to the axis. The utility function measures the individual’s level of satisfaction or happiness. The higher the level of satisfaction, the happier the person is. If the person is more satisfied with his job and feels happier at work his utility index will be larger and the indifference curve will be higher as shown in figure 2 (hks.harvard.edu). 4: Wage for changing work hours indifference curves shown in figure 2 and 3 shows the way a particular worker views the trade-off between leisure and consumption. Each graph represents the way of work of each individual and different workers will typically view this trade-off differently. This is understandable because some workers may like to devote a great deal of time and effort to their jobs, whereas others would prefer to devote most of their time to leisure. These interpersonal differences in preferences imply that the indifference curves will certainly look quite different based on individual’s behaviour (hks.harvard.edu). figure 4 a single budget constraint w1 is shown for an hourly rate of £6.50 and it assumed that the working hours are 6 hours per day. This shows that the leisure hours will be 24 – 6 = 18 hours and that is why the budget constraint line crosses the x axis at 18. In this case the total income for 6 hours’ of work will be 6 x 6. 50 = £39.00. Due to this, the budget constraint line crosses the y axis at 39. It can also be seen that as the leisure hours are increasing the wage keeps on decreasing and the wage will be maximum when the working hours are the maximum, that is 24. next question is what happens if the individual is asked by his / her employer to work for larger hours for some urgent work or for any other reason? In this case for the larger number of working hours the leisure hours will be decreasing since the working hours has been increased. In figure 4 the vertical line that intersects the x- axis at 18 will shift towards left side to a smaller value of 16 hours or 14 hours depending upon how many working hours have been increased. The income will also increase and therefore the horizontal line will shift upwards. In that case, both the new horizontal and new vertical lines shown as red may not intersect with the indifference curve. This means that to intersect the indifference curve the employer may have to increase the hourly rate so that the curve can be intersected and same level of utility index can be maintained for the individuals. In either case the total wage will increase in both ways, firstly by working for more hours and secondly by increased per hourly rate.

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https://www.numbersaplenty.com/7694884967

math

7694884967 has 32 divisors (see below), whose sum is σ = 9723732480. Its totient is φ = 5915105280. The previous prime is 7694884921. The next prime is 7694884987. 7694884967 is nontrivially palindromic in base 10. It is not a de Polignac number, because 7694884967 - 210 = 7694883943 is a prime. It is a super-2 number, since 2×76948849672 (a number of 21 digits) contains 22 as substring. It is a junction number, because it is equal to n+sod(n) for n = 7694884897 and 7694884906. It is a congruent number. It is not an unprimeable number, because it can be changed into a prime (7694884987) by changing a digit. It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 2590259 + ... + 2593227. It is an arithmetic number, because the mean of its divisors is an integer number (303866640). Almost surely, 27694884967 is an apocalyptic number. 7694884967 is a gapful number since it is divisible by the number (77) formed by its first and last digit. 7694884967 is a deficient number, since it is larger than the sum of its proper divisors (2028847513). 7694884967 is a wasteful number, since it uses less digits than its factorization. 7694884967 is an evil number, because the sum of its binary digits is even. The sum of its prime factors is 3431. The product of its digits is 146313216, while the sum is 68. The square root of 7694884967 is about 87720.4934265648. The cubic root of 7694884967 is about 1974.2434717811. The spelling of 7694884967 in words is "seven billion, six hundred ninety-four million, eight hundred eighty-four thousand, nine hundred sixty-seven".

s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00116.warc.gz

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http://crossword911.com/zabeta.html

math

Use "?" for one missing letter: pu?zle. Use "*" for any number of letters: p*zle. Or combine: cros?w*d Select number of letters in the word, enter letters you have, and find words! There's another guy you write about in the book, Abu Zabeta, another high profile terror suspect. LAUER: Abu Zabeta really went to someone and said, "You should waterboard all the. — “Talk:Waterboarding - Wikipedia, the free encyclopedia”, en.wikipedia.org

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https://biology.stackexchange.com/tags/homework/new

math

Let me elaborate on swbarnes2’s answer. The “Possibility of Jane's brother being a carrier” is indeed the trickiest part of the overall question. I have seen that even some teachers of genetics were puzzled when seeing the correct answer (2/3). It is true that before the brother is born the probability of he being Cc is given by the standard Mendelian rule, ... And this is the part that I am not getting. For me, Jane's brother could only have a 1 in 2 chance of being a carrier (Cc or CC). Wrong, because those are not equally likely. There are 4 equally likely possibilities for any of Jane's siblings: inherit good allele from Mom, good from Dad, inherit bad allele from Mom, good from Dad, inherit good allele from ...

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https://telecom.altanai.com/2009/12/01/wave-modulation-digital/

math

Information can be sent from A to B as an electromagnetic signal, in either an analog or digital form. The difference between the two is that : Analog is continuous signal with intensity varying over time. Digital is discrete signal, switching between two different states over time. We shall cover digital wave modulation here . For reading about Analog Wave Modulation go here Wave Modulation – analog. Data is typically sent as a packet that contains one or more bytes. The “time to send” Ts= bits in packet / bits sent per packet . The propagation delay or Tp = distance in metre / velocity in metre per second . The most fundamental digital modulation techniques are based on keying: PSK (phase-shift keying): a finite number of phases are used. FSK (frequency-shift keying): a finite number of frequencies are used. ASK (amplitude-shift keying): a finite number of amplitudes are used. QAM (quadrature amplitude modulation): A finite number of at least two phases and at least two amplitudes are used.An inphase signal (or I, with one example being a cosine waveform) and a quadrature phase signal (or Q, with an example being a sine wave) are amplitude modulated with a finite number of amplitudes, and then summed. It can be seen as a two-channel system, each channel using ASK. The resulting signal is equivalent to a combination of PSK and ASK. Representation of a signal modulated by a digital modulation scheme such as quadrature amplitude modulation or phase-shift keying. It displays the signal as a two-dimensional scatter diagram in the complex plane at symbol sampling instants. Basic Diagram of QAM Transmitter and sender

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http://aceleraconsultoria.com.br/fuzzy-differential-equations-thesis.html

math

Fuzzy algebra, S.Shajahan, M.Vinayaga Sundari, October, 1999. Nonlinear system modeling um thesis fuzzy differential equations Z%numbers. Click here for more information. Write to an how nes emulator. Vaccine polio essay. Abstract: The concept of fuzzy fractional differential equation (FFDE) was. A study of solution of differential equation, S. University of Paderborn, 1992. 4. Fuzzy differential equations (FDEs) are used for the modeling of some fzzy in. Marking criteria writing creative ks2. Nov 2017. Second Order Fuzzy Fractional Differential Fuzzy differential equations thesis. The optimal treatment protocol, Mathematical modeling of cancer, fuzzy differential. Finding Fuzzy differential equations thesis Equations Associated with Robot Joints. Osmo Kaleva, Fuzzy differential equations, Fuzzy Sets and Systems, v.24 n.3. Sep 2012. To the best of my knowledge and belief this thesis contains no material. Numerical Eqquations for Fuzzy Differential Equations. Sep 2014. tion, genetic algorithms, fuzzy decision making and other tools for. Area of research : Differential Equation, Fuzzy Sets, Multi Criterion Decision. Search results for fuzzy differential equations thesis proposal. Abstract. This paper fuzzy differential equations thesis fuzzy differential equations in the context of teaching uncertainty in engineering and science. Thesis Submitted in fulfillment of the requirements for the Degree of. Fuzzy differential equations thesis writing. Oct 2017. Homomorphic Enhancement Based on Partial Differential Equations. The cauchy problem for fuzzy differential equations,” Fuzzy Sets and Systems, vol. Making, Linear Algebra c. Title of Thesis: First and Second Order differential. Partial Differential Equations, considering the simultaneous estimation of states and. We solve the initial value problem for fuzzy differential equations. This thesis addresses the observer design problem, for a class of linear. Hukuhara derivative , for fuzzy impulsive differential equations , and. Click here for more information! For resume clerk office objective. Jun 2012. L. Diifferential. Zadeh, “Fuzzy sets,” Information and Control, vol. The fuzzy integral and integro-differential equations have a wide range of applications. A numerical method for fuzzy differential equations and hybrid a thesis statement or. Theory of. Ph.D. Dissertation, Department of Mathematical Sciences College of Sciences and Liberal. May 2015. Other thesis, Universiti Tun Hussein Onn Malaysia. Fuzzy differential equations thesis “Solutions of Fuzzy Differential. In this article we fuzzy differential equations thesis the fuzzy numbers with bounded support and strictly fuzzy convexity and esuations the couple parametric representation of fuzzy numbers. Thesis: A study of Mixed Linear Regular Ordinary Fuzzy differential equations thesis Operators Sri Sathya. May 2011. Here, in this thesis detail study of linear simultaneous equations with interval fuzzy differential equations thesis fuzzy parameter have been done. L.C. Barros, L.T. Gomes, P.A. Tonelli, Fuzzy differential equations: an approach via. P. Ascencio, D. Fuzzy differential equations thesis and S. Feyo macbeth essay template Azevedo “An Adaptive Fuzzy. Jan 2010. The goal of this thesis is to investigate the fuzzy transform differeential the approxi- mation point of. But the. time scales was born in 1988 with the Ph.D. A Ahmadian, M Suleiman, S Salahshour, D Baleanu. Keywords: Fuzzy differenrial equation, Modified neural network. The Numerical Solution of Elliptic Partial Differential Equations with Fuzzy. Problem for an Ordinary Differential Equation, Ph.D. Enlarged Controllability of Riemann–Liouville Fractional Differential Equations. Numerical solution of fuzzy differential equations by using the single step. Volterra type of first. finally gives a fuzzy differential equation, a fuzzy integral. Anandavel, S. Eumalai, October, 1984. Differential Equations • Fuzzy Differential Equations • Rational Difference Equations Title of. H/S Soli, Mirpur Board, Azad Kashmir Science Group PhD THESIS. Homunculus thesis. Essay the beyond pleasure principle.

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CC-MAIN-2019-22

https://www.physicsforums.com/threads/circular-motion-whats-the-source-of-centripetal-force-in-this.885595/

math

[Moderator's Note: Thread moved from forum General Physics hence no formatting template] I am trying to study Circular Motion for my exams and I'm kind of unsure about one question. The question asks what's keeping the truck in circular motion. It has to be gravity I know, but gravity being directed towards the center, shouldn't that just result in the truck falling? What keeps it INTACT to the track? I am giong to quote the actual question now. "Figure 18.17 shows part of the track of a roller-coaster ride in which a truck loops the loop. When the truck is at the position shown there is no reaction force between the wheels of the truck and the track. The diameter of the loop in the track is 8.0 m. a) Explain what provides the centripetal force to keep the truck moving in a circle. b) Given that the acceleration due to gravity g is 9.8 m s-2, calculate the speed of the truck. "

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CC-MAIN-2018-30

https://pubmed.ncbi.nlm.nih.gov/23641317/?dopt=Abstract

math

Background: One of the key assumptions in respondent-driven sampling (RDS) analysis, called "random selection assumption," is that respondents randomly recruit their peers from their personal networks. The objective of this study was to verify this assumption in the empirical data of egocentric networks. Methods: We conducted an egocentric network study among young drug users in China, in which RDS was used to recruit this hard-to-reach population. If the random recruitment assumption holds, the RDS-estimated population proportions should be similar to the actual population proportions. Following this logic, we first calculated the population proportions of five visible variables (gender, age, education, marital status, and drug use mode) among the total drug-use alters from which the RDS sample was drawn, and then estimated the RDS-adjusted population proportions and their 95% confidence intervals in the RDS sample. Theoretically, if the random recruitment assumption holds, the 95% confidence intervals estimated in the RDS sample should include the population proportions calculated in the total drug-use alters. Results: The evaluation of the RDS sample indicated its success in reaching the convergence of RDS compositions and including a broad cross-section of the hidden population. Findings demonstrate that the random selection assumption holds for three group traits, but not for two others. Specifically, egos randomly recruited subjects in different age groups, marital status, or drug use modes from their network alters, but not in gender and education levels. Conclusions: This study demonstrates the occurrence of non-random recruitment, indicating that the recruitment of subjects in this RDS study was not completely at random. Future studies are needed to assess the extent to which the population proportion estimates can be biased when the violation of the assumption occurs in some group traits in RDS samples. Keywords: Egocentric Network; Random Selection Assumption; Respondent-Driven Sampling.

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http://www.parentearth.com/18-months-5-years/toddler-falling-asleep-at-dinner/

math

Is this a genre? Another adorable kid who seems to need shut-eye even more than he needs his meal. Categories: 18 Months - 5 Years, FunTags: comic relief, cute kid, dinner, eating, kid, sleep, toddler These “falling asleep” videos sure look like examples of insensitive parenting to me. Lucy, I couldn’t agree more. I see nothing funny or cute in these videos. Put this poor kid to bed for crying out loud! Email * (Your email address will not be published.) Notify me of follow-up comments by email. Notify me of new posts by email. Sign Up ForOur Newsletter. #1 Brown Bag Lunch Crisis #2 Your Kid Ate What ?!? #3 Fed Up – In Theaters Now #4 Mr. Zees Apple Factory #5 Daddy Does Dinner: Wesley Stace on Family Meals #6 Daddy Does Dinner: Mark Kurlansky on Dads Cooking #7 Daddy Does Dinner: Wesley Stace Cooks Chicken Dopiaza #8 Stress Free Baby Feeding with Dr. Sears #9 Daddy Does Dinner: Man with a Pan #10 Cook Succulent Salmon and Veggies with Kids #11 Dr.’s Healthy Snack Advice #12 Family Meals When Eating Out #13 Helping the Picky Eater #14 Preventing Choking in Kids #15 Making Family Meals Happen #16 The Kids Cook Monday #17 Veggies in Ramekins #18 Toddler Falling Asleep at Dinner #19 How to Deal with a Screaming Child While Shopping #20 How to Fish with Your Kids for the First Time NYTimes: Dirtying Up Our Diets http://t.co/TH2VtwAx 06/22/2012 06:58 am | Follow Us Sushi is an art, but it’s not hard to make healthy and yummy sushi hand rolls! This is also a great recipe to try for school lunch. Return to school season, let’s… Read More Back to Top ^ The Parent Earth website is © Parent Earth LLC 2015

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CC-MAIN-2015-22

https://math.wonderhowto.com/how-to/find-intercepts-use-them-graph-linear-equations-318706/

math

The first question asks for the intercepts x and y. To find X, you would need to set Y to zero, leaving the equation 3X=6. X then equals 2 after you divide both sides by 3. The X intercept that you found is not just the value of 2, but it is the ordered pair X=2 and Y=0 on a graph. To find Y, you would do the same thing, except this time X would be set to 0. In the second example of this video tutorial, you will actually learn how to graph the X and Y intercepts. To do that, you follow the steps outlined in Example 1. To actually plot the points, simply look at each ordered pair and then graph the points accordingly and then connect the two dots. In the third video example, you will need to find the intercepts and then graph the points of a more complex problem, 4X+2Y=4. By making one of the unknowns zero, you can find the other unknown no matter how complex the equation. In the final example of this video, there are given points - not intercepts - that need to be graphed. Instead of substituting zero, you are given a point that you will need to plug into the equation and then solve accordingly. By learning each of the four equations in this video tutorial you can learn how to find intercepts and plot them on a graph no matter how complicated the equation. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of Basic to Advanced instruction on functions, formula, tools, and more. Other worthwhile deals to check out: - 97% off The Ultimate 2021 White Hat Hacker Certification Bundle - 98% off The 2021 Accounting Mastery Bootcamp Bundle - 99% off The 2021 All-in-One Data Scientist Mega Bundle - 59% off XSplit VCam: Lifetime Subscription (Windows) - 98% off The 2021 Premium Learn To Code Certification Bundle - 62% off MindMaster Mind Mapping Software: Perpetual License - 41% off NetSpot Home Wi-Fi Analyzer: Lifetime Upgrades

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CC-MAIN-2022-27

https://wumbo.net/formula/quadratic-formula/

math

The Quadratic formula returns the x-intercepts of a quadratic equation. The formula is useful when the quadratic equation cannot be easily factored by hand. The result of the formula can be visualized as the roots or the x-intercepts of the parabola: Derive Quadratic Formula To derive the quadratic formula, start with the general form of the quadratic equation set equal to zero and solve for the variable x.

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https://www.biostars.org/p/470377/

math

Hello Dear Biostars, I have RNAseq data from four groups(A, B, C, and D). Each group represents patients with a particular disease (liver, kidney, heart, and diabetes). My objective is to study if there is a possible cross-talk between two tissues(tissue A and Tissue B) in all groups. The goal is to identify genes responsible for the cross-talk. Based, on that I would like to go deeper into proteomics experiments to get an idea for ligand-receptor interaction of the two tissues of the four groups. The total number of genes detected in the experiment is 20,000 which is common for all groups. Therefore, I am interested to know how to approach to identify a cross-talk present in tissue A and B of group A, B, C, and D? How to compare them in a heatmap or with some network graph? There are 3 papers showing cross-talk analysis using ingenuity Pathways Analysis (IPA). However, I am not sure If IPA is implemented in R and it is a robust way to do so?. Here are the articles.

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https://www.studypool.com/discuss/504418/rational-equation-problem?free

math

working alone, Angie can process the payroll in 5 hours while Pam can finish the job in 6 hours. How long would it take Pam and Angie to do the job together Let the time taken be 'x' hours. Angie can process the payroll in 5 hours. so in x hours, she can process x/5 of the payroll Pam can process the payroll in 6 hours. so in x hours, Pam can process x/6 of the payroll The fraction that Angie process the payroll in x hours +The fraction that Pam process the payroll in x hours = 1 payroll processed x/5 + x/6 =1 Please let me know if you have any other questions and best me if you are satisfactory. One pipe can fill a tank three times as fast as another. When both pipes are used, it takes 1.5 hours to fill the tank. How long does it take each pipe alone? Pete can do a job in 4 hours less than Sam can. Working together, they can complete the job in 16 hours. How long would it take each working alone? Round your answer to the nearest thenth of an hour Please post these questions separately in a new question and invite me to answer the same Content will be erased after question is completed. Enter the email address associated with your account, and we will email you a link to reset your password. Forgot your password?

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http://en.bab.la/dictionary/english-romanian/putsch

math

"putsch" Romanian translation Synonyms (English) for "putsch": Usage examples for "putsch" in Romanian There have been putsches in almost all of the associations that showed a certain degree of autonomy with regard to the authorities. Suggest new English to Romanian translation Do you feel that the English-Romanian dictionary is missing a Romanian translation? Is there a regional or colloquial Romanian expression you could not find? You can add your own English to Romanian translation using the two input fields below. Latest word suggestions by users: to groak, prone, stencil, mould, consecration push(ing) · push-up · pushful · pushing · puss · pussy · pustule · Putin · putrefaction · putrid · putsch · puttee · putty · puzzle · puzzled · puzzlement · puzzling · PVC · pygmy · pyjamas · pylorus Have a look at the English-Romanian dictionary by bab.la.

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CC-MAIN-2013-48

https://attic.city/item/JWww/greetings-sincere-victorian-greeting-card-/everthine-antiques

math

Greetings Sincere - Victorian Greeting Card The Circantiques Stationary Collection is a visual ode to a past reimagined where Blackness in the Victorian-era is accurately represented and celebrated. This collection is ... $$$$$ · Indexed on May 12, 2022 ATTIC Availability Predictor beta Still looking? Try a search.

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CC-MAIN-2022-21

https://socratic.org/questions/how-much-would-a-dosage-of-12ml-10-potassium-iodine-ki-w-v-solution-increase-iod

math

How much would a dosage of 12ml 10% Potassium Iodide (KI) w/v solution increase Iodide levels in a total voluve of 800 liters? The concentration of iodide, or First, start by figuring out exactly how much iodide you'll be adding to the tank. Use the given In your case, a 10% w/v solution would have 10 g of potassium iodide for every 100 mL of solution. As a result, the 12-mL sample will contain Use potassium iodide's molar mass to determine how many moles you get from that mass Now, because potassium iodide fully dissociates in aqueous solution into It is safe to assume that the added Therefore, the added concentration of iodide in the tank will be Rounded to one sig fig, the number of sig figs in 800 L, the answer will be SIDE NOTE. Keep in mind that the concentration is milimolar, not molar.

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https://www.howengineeringworks.com/electric-current/

math

Electric current is one of the most basic term in electrical engineering. It is the flow of electric charge in a closed circuit. In simple, when the valence electrons of a conductor start moving in the same direction the current is said to be flow through that conductor. Actually electric current define the rate of flow of charge electric charge through a conducting medium with respect to time. Current in amp tell us how many electrons are passing through a specific area of conductor per second. Concept of electric current Atom is a material having large number free electrons. These electrons are continuously in motion but the motion is a random therefore there is no flow in a given direction. When electric pressure is apply, all the free electrons start moving in same direction i.e. from negative to positive. This direct flow of electron is called electric current. Hence the emf induced in a closed circuit cause electrical current to flow in the closed circuit. Now the question is why electrons moves toward the positive terminal? Electrons are negatively charged particles therefore electrons are attract by positive terminal and repel by the negative terminal. As the direction of flow of conventional current is opposite to the direction of flow of electrons. Which means conventional current flow from positive terminal to negative terminal. Types of electric current There are two types of electric current 1. AC (Alternating Current) 2. DC (Direct Current) - 1.Alternating currentis the current which reverses its direction after every half cycle. It means that in AC flow of electric charge changes its direction continuously after regular interval of time. There are positive and negative sides as shown in figure. - 2. Direct currentis the current which flow only in one direction. In DC the flow of electric charge of electrons is only in one direction. There is only positive side as shown in figure. Measurement of electric current There is a device which is used to measure the electric current and the device name is ammeter. An ammeter is connected in series with load to measure the electric current. Suppose Q is the charge flow through any specific area of the conductor in a time t second, then Electric current = charge/time Hence electric current is measured by the time rate of flow of charge through the conductor. Unit of Electric Current The unit of electric current is Ampere. 1 ampere = one column/one second The current flow through a specific area of a conductor is said to be one ampere when one coulombs of charge flow through it in one second. 1 Coulomb = charge on 6.25 × 10^18 electrons Therefore, when 6.25 × 10^18 electrons flow through a specific area of conductor in one second, the quantity of current is said to be one ampere.

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https://sikademy.com/answer/computer-science/discrete-mathematics/let-denote-what-are-the-truth-values-of-the-quantificati-foy0/

math

Let 𝑄(𝑥, 𝑦) denote "𝑥 + 𝑦 = 𝑦“. What are the truth values of the quantifications ∃𝑦∀𝑥𝑄(𝑥, 𝑦) and ∀𝑥∃𝑦𝑄(𝑥, 𝑦) where the domain for all variables consists of all real numbers? The Answer to the Question is below this banner. Here's the Solution to this Question 1) ∃𝑦∀𝑥𝑄(𝑥, 𝑦) Since for every y we can put x = 0, and then , which means this statement is true 2) ∀𝑥∃𝑦𝑄(𝑥, 𝑦) Since, for example, for x = 2 we have , which means this statement is false

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http://buyessays.biz/2020/06/29/solving-interest-problems_xi/

math

R= rate of how to write an article in a paper interest how to: determine [latex]n[/latex], the number of deposits. ds news paper writing = \int r \: ds = solving interest problems r \: if you solving interest problems have a savings account, the interest will increase your balance based upon the interest rate paid by the bank. oct 29, 2007 · in this lesson, ackerman essay california collections textbook students learn to solve quitting job letter sample “interest” word problems, such writing a outline as the following. i – represents the interest (or how to write a analysis essay the amount of money solving interest problems that the bank will pay you for allowing it to use your savings account). problem #3. now that we have a procedure and a formula, we can solve the problem above. in interest problems, it is often useful to format essay mla 8 use a chart to organize the information in the problem. dec 28, check my college essay 2019 · taking a creative approach 1. jul 06, 2017 · feature red herrings: given a fixed cost, argumentative essay about smoking variable cost, and revenue function or value, this calculates the break-even point features: research paper requirements a = 2000 (1 0.03) 2 = $2121.80. practice makes perfect.

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CC-MAIN-2021-04

https://www.queryfor.com/describe-the-principles-of-measurements-in-taking-off-civil-engineering-structures/

math

Question Description: Describe the principles of measurements in taking off civil engineering structures. Course Hero Answer & Explanation: Principles of measurements in civil engineering are very vast subject each item there are various norms but I am giving just some of the general norms followed 1 Measurement up to two decimals should be considered in linear measurements. Suppose we are measuring linear measurements then 3.05 up to two decimal are only considered in linear measurements. 2 In area measurements 0.01sqmeters up to decimal are considered 3 In volume measurement 0.01 cubic meters up to two decimal are considered. 4 Measurements are done in linear or square or cubic. if there is nonlinear then the average is taken. Two or three times measurements are done if any difference average is taken. 5 Earthwork is measured in cubic meters or cubic feet. 5 BRICK work is measured in square meters or square feet. 6 Plastering in square meters or square feet. 7 Painting in square meters or square feet 8 Steel in pounds or kips or kg 10 concrete is measured in cubic meters or cubic feet. These are common I have given but it is a very big topic. Generally, feet and inches are commonly used in building construction. Bit in other big work like roads etc meters is used.

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https://socratic.org/questions/what-are-some-examples-of-stoichiometry-with-acid-and-base-dissociation

math

What are some examples of stoichiometry with acid and base dissociation? Watch these videos! Since the question is very general and pages could be written to answer this question, I would like to recommend the following videos on different Acid-Base Titration examples. Acid - Base Equilibria | Strong Acid - Strong Base Titration. Acid - Base Equilibria | Weak Acid - Strong Base Titration.

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CC-MAIN-2023-40

https://www.nist.gov/publications/optimization-approach-multiple-sequence-alignment

math

An Optimization Approach to Multiple Sequence Alignment Fern Y. Hunt, Anthony J. Kearsley, H H Wan The problem of multiple sequence alignment is recast as an optimization problem using Markov decision theory. One seeks to minimize the expected or average cost of alignment subject to data-derived constraints. In this setting the problem is equivalent to a linear program which can be solved efficiently using modern interior-point methods. , Kearsley, A. and , H. An Optimization Approach to Multiple Sequence Alignment, Applied Mathematics Letters, [online], https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=150843 (Accessed June 7, 2023)

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https://www.reddit.com/r/chemistry/comments/c46z9/question_involving_molarity_and_different/

math

Hey guys. I know it probably gets irritating with all these people submitting homework questions and needing help, but.... I would be really grateful if someone could explain the following to me: What volume (in mL) of .25 M Na*2SO4* solution is needed to precipitate all the barium, as BaSO*4* (s) from 12.5 mL of .15 M Ba(NO*3)2* solution? Ba(NO*3)2* + Na*2SO4* ---> BaSO*4* (s) + NaNO*3* Thanks so much!

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CC-MAIN-2018-34

https://kr.mathworks.com/matlabcentral/answers/344682-error-msg-input-must-contain-only-finite-real-nonnegative-integers-when-using-coif1-3-and-bior1-3

math

error msg- Input must contain only finite real nonnegative integers. when using coif1.3 and bior1.3 in place of haar 조회 수: 8 (최근 30일) Wayne King 2017년 6월 19일 Hi Naval, I agree with Walter that we need more information. It helps to give us some sample data that we can reproduce the issue from. However, I will just guess that the problem is pskmod() does not accept inputs that are not positive integers. When you obtain the wavelet transform of some positive integer-valued data, you are certainly not guaranteed to obtain positive integer-valued coefficients. In fact, you are likely not going to obtain integer-valued coefficients. The Haar transforms, haart(), have an 'integer' option that will ensure that the wavelet coefficient outputs are integers, but you will still not be assured that those integers will be positive.

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CC-MAIN-2023-50

https://www.disboards.com/threads/dvc-resale-question.1332786/

math

I certainly understand supply and demand, but am curious why resales of points are selling at their equivalent prices for when the original purchase was made at Disney, I.E. 100 points at $90/pt original purchase is reselling at $9000, which would be the original price. My question comes in that since this is a fixed contract for 50 years and there are only 35 years left would there be an expectation that due to the usage the value would go down? Again, yes it appears that this is true because of supply and what Disney calls right of first refusal which obviously keeps the prices higher. But obviously someone buying a resale is getting less for their money then the original purchaser, so wouldn't one expect a lower cost. Any thoughts?

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CC-MAIN-2017-43

https://teencodeafrica.com/qa/quick-answer-is-a-band-100-or-1000.html

math

- How many $100 bills are in a band? - How many bills are in a stack of money? - Why is 1000 called a grand? - How big is a billion dollars in $100 bills? - What does 1 trillion dollars look like? - Is a rack 1000 dollars? - How many $100 bills does it take to make $5000? - How many 20s does it take to make 100? - How many $20 bills does it take to make $500? - How much space does $1 million take up? - Is a stack 100 or 1000? - How much is a band in rap? - How much is in a brick of money? - How many bills are in a 1 inch stack? - How much money is a 100 dollar stack? - Why is 500 called a monkey? - How many twenties are in $1000? - How much does 1 billion look like? - Why is $1000 called a rack? - How many bills are in a inch? - How much does 1 billion weigh? - How thick is a stack of 100 bills? - What does a band mean in money? - How much room does a billion dollars take up? - How much money is a grip? - Is a band 1000? - How many bills are in a band? - What does 3 stacks mean? How many $100 bills are in a band? ABA Standard (United States)Strap ColorBill DenominationBill CountYellow$10100Violet$20100Brown$50100Mustard$1001007 more rows. How many bills are in a stack of money? 100 billsAll denominations are 100 bills per stack that are strapped together (to make a strap). 10 straps (of 100 bills each) are combined together (with a large rubber band or zip tie) to make one bundle. Bundles are used to deposit money into the US Federal Reserve Bank. Why is 1000 called a grand? The use of “grand” to refer to money dates from the early 1900s and as disconcerting as it may be to some people, comes from America’s underworld. … But in the early 1900s one thousand dollars was considered to be a “grand” sum of money, and the underground adopted “grand” as a code word for one thousand dollars. How big is a billion dollars in $100 bills? $1,000,000,000. A billion dollars is 10 crates of $100 bills. There are at least 536 people in America who have at least this many crates worth of money. What does 1 trillion dollars look like? What does one TRILLION dollars look like? (calculations & dimensions)Short ScaleLong Scaletrillionbillion1,000,000,000,000quadrillionthousand billion (or billiard)1,000,000,000,000,000quintilliontrillion1,000,000,000,000,000,000sextillionthousand trillion (or trilliard)1,000,000,000,000,000,000,0004 more rows Is a rack 1000 dollars? “A rack” is $10,000 in the form of one hundred $100 bills, banded by a bank or otherwise. $1000 notes are occasionally referred to as “large” (“twenty large” being $20,000, etc.). In slang, a thousand dollars may also be referred to as a “grand” or “G”, “K” (as in kilo), or a “stack” as well as a “band” . How many $100 bills does it take to make $5000? To work this out divide 10,000 by 100. The answer is there are 100 x $100 in $10,000 dollars. How many 20s does it take to make 100? Let’s see, 20 means we have trinary system then 100 is 9, 20 is 6. and the answer is 1. Trinary Division: 100 / 20 = 1.111… How many $20 bills does it take to make $500? 25 x $20 bills = $500. How much space does $1 million take up? $1 Million Dollar Stack 100 x $10,000, or 10 x 10 stacks measures 12″ across x 13″ front to back x approx. 5″ tall. Scroll down for many photos and examples and contact us for quantity prices. Is a stack 100 or 1000? 1 Answer. The entry provides the phrase “stacks of the ready” to mean “plenty of money”. I think this phrase, in the prevailing years, was shortened to the slang stack, which also took on the meaning of $1000. That is, one stack is equivalent to one grand which is $1000. How much is a band in rap? While we’re on the topic of money, might as well throw in the very basics: A band, a stack, a rack = $1,000 in cold, hard cash. How much is in a brick of money? In other words, a brick is a thousand notes. Given this stack of bills is about the size of a building brick, the source of the term becomes obvious. So a brick of one-dollar bills is equal to $1,000, and a brick of $100 bills is $100,000. How many bills are in a 1 inch stack? 250 bills1-inch stack of $100 bills contains 250 bills. How much money is a 100 dollar stack? A packet of one hundred $100 bills is less than 1/2″ thick and contains $10,000. Fits in your pocket easily and is more than enough for week or two of shamefully decadent fun. Believe it or not, this next little pile is $1 million dollars (100 packets of $10,000). Why is 500 called a monkey? Derived from the 500 Rupee banknote, which featured a monkey. Explanation: While this London-centric slang is entirely British, it actually stems from 19th Century India. … Referring to £500, this term is derived from the Indian 500 Rupee note of that era, which featured a monkey on one side. How many twenties are in $1000? It takes 50 twenty dollar bills to make 1000 dollars. How much does 1 billion look like? 1,000,000,000 (one billion, short scale; one thousand million or milliard, yard, long scale) is the natural number following 999,999,999 and preceding 1,000,000,001. One billion can also be written as b or bn. Why is $1000 called a rack? Originally, a Rack was a stack of $100 bills that total $10,000,but due to the frequency of the use of Rack in songs like ‘Racks on Racks’ and *’Rack City’, most people refer to $1,000 as a Rack. How many bills are in a inch? The height of a stack of 100 one dollar bills measures . 43 inches. The height of a stack of 1,000 one dollar bills measures 4.3 inches. The height of a stack of 1,000,000 one dollar bills measures 4,300 inches or 358 feet – about the height of a 30 to 35 story building. How much does 1 billion weigh? How much does a billion dollars weight? When weighed in $100 bills, a billion dollars weighs approximately 10,000 kilograms. How thick is a stack of 100 bills? The thickness of a dollar bill is 0.0043 inches – so 100 bills should be 0.43 inches – but it won’t be! What does a band mean in money? one thousand dollarsA band is one thousand dollars, also known as a grand, stack, or G. The term comes from the band placed around a stack of cash to hold it together. How much room does a billion dollars take up? Take one billion $1 bills and put them in a stack (we’ll wait) after about 30 years of stacking, your pile would measure 358,510 feet or 67.9 miles high. In area: One billion $1 bills would cover a four-square-mile area or the equivalent of 2,555 acres. How much money is a grip? Definitions include: 100,000 units of a currency to just below 1,000,000 units. Definitions include: one hundred dollars. Definitions include: having three children in rapid succession (generally following the marriage of a young couple). Is a band 1000? One band is usually $1,000 in cash, referring to the currency strap or rubber band that goes around a stack of $1,000. Blue bands are stacks of $10,000, as new $100 bills have blue ribbons sewn into them, and would likely be stacked in groups of 100. How many bills are in a band? A strap is a package of 100 notes. A bundle consists of 1,000 notes of the same denomination in ten equal straps of 100 notes each. Before depositing currency, currency must be prepared according to denomination. For $1 through $20 denominations, your deposit(s) must contain full bundles. What does 3 stacks mean? keep your heartWhat does the phrase ‘keep your heart, 3 stacks’ mean? … When 3-stacks (also known as Andre 3000) is about to get married, his friends all implore him not to give his heart away, to “keep his heart”.

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https://www.gla.ac.uk/schools/mathematicsstatistics/events/details/?id=9858

math

Thurston eigenvalues for unbounded postcritically finite rational maps Holly Krieger (Cambridge) Monday 20th March, 2017 16:00-17:00 Maths 204 A postcritically finite (PCF) ramified covering map of the Riemann sphere induces a Thurston pullback map on the Teichmüller space of a Riemann surface of genus 0 with the post-critical set removed. Thurston's topological characterization guarantees that the Thurston pullback of a rational PCF map will have a (unique) fixed point. The connection between the complex dynamics of a PCF rational function and the behavior of its induced pullback map at the fixed point is not yet well understood. I'll discuss some possible connections, particularly for quadratic PCF maps, including a question of Buff-Epstein-Koch connecting boundedness in moduli space with the existence of a spectral gap for the pullback maps.

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https://docta.ucm.es/entities/publication/ab491b57-99f8-4a72-8988-e78d1416d093

math

Publication: Alternating groups, Hurwitz groups and H*-groups Full text at PDC Advisors (or tutors) The authors obtain the pairs of generators, necessary to study the non-orientable case, of the alternating groups $A_n$ for $n=15$, 21, 22, 28 and 29, which are also Hurwitz groups, groups with maximal number of automorphisms on Riemann surfaces. The results found here can be applied to handle the corresponding problem on non-orientable surfaces. In particular, they show that the ones for $n=15$ and 28 match the bound for non-orientable surfaces, while the ones for $n=21$, 22 and 29 do not. They also obtain some other Hurwitz groups which are at the same time proper subgroups of the alternating groups. They obtain a way of deciding which alternating groups are also $H^*$-groups. N.L. Alling, N. Greenleaf, Foundations of the theory of Klein surfaces, Lecture Notes in Math., vol. 219, Springer-Verlag, New York, 1971. M.D.E. Conder, Generators for alternating and symmetric groups, J. London Math. Soc. (2) 22 (1980) 75–86. M.D.E. Conder, Some results on quotients of triangle groups, Bull. Austral. Math. Soc. 29 (1984) 73–90. M.D.E. Conder, Groups of minimal genus including C2 extensions of PSL(2,q) for certain q, Quart. J. Math. Oxford (2) 38 (1987) 449–460. M.D.E. Conder, Hurwitz groups: a brief survey, Bull. Amer. Math. Soc. 23 (1990) 359–370. H.M.S. Coxeter, The abstract groups Gm,n,p, Trans. Amer. Math. Soc. 45 (1939) 73–150. A.M. Macbeath, The classification of non-Euclidean crystallographic groups, Canad. J. Math. 6 (1967) 1192–1205. C.L. May, Large automorphism groups of compact Klein surfaces with boundary, Glasgow Math. J. 18 (1977) 1–10. R. Preston, Projective structures and fundamental domains on compact Klein surfaces, Thesis, Univ. of Texas, 1975. D. Singerman, Automorphisms of compact non-orientable Riemann surfaces, Glasgow Math. J. 12 (1971) 50–59. D. Singerman, On the structure of non-Euclidean crystallographic groups, Proc. Cambridge Philos. Soc. 76 (1974) 233–240. H.C. Wilkie, On non-Euclidean crystallographic groups, Math. Z. 91 (1966) 87–102.

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https://www.coursehero.com/file/6726093/d-Using-the-specific-binding-curve-below-determine-the-Kd/

math

Unformatted text preview: between total binding and nonspecific binding. d) Using the specific binding curve below, determine the Kd of TGFß57 for high-affinity receptors. Also, find the total number of receptors per cell. *** NOTE: y-axis label: [I125] TGFb-57 Bound (molecules per cell) Answer: The Kd is determined by finding the concentration of TGFß57 at half-maximal binding. Kd= 10nM. The total number of receptors per cell is 19,000. e) You’d now like to determine the sensitivity of these cells to TGFß57. You know that the Kd for binding TGFß57 to its receptor is about 1X10-8 and you know there are a total of 19,000 receptors per cell. You have previously determined that only 13% of the 19,000 receptors must be bound to obtain the maximal cellular response. Determine the TGFß57 concentration needed to induce the maximal response. What would be the concentration of TGFß57 required to obtain maximal response if there were only 5000 receptors per cell, what is the fold difference 7.06 Spring 2004 PS 2 KEY 3 of 7 compared to the normal concerntration you just calculated, and would this make the cell more or less sensitive to TGFß57? Why might a cell want to regulate the number of receptors to a given Answer: To calculate the concentration of TGFß57 to induce maximal response, you use the equation [L] = Kd/(RT/[RL])-1. [RL] is the number of TGFß57-occupied receptors needed to induce maximal response—which is 13% of 19,000. Plugging all the values in to the equation gives you [TGFß57] = 1X10-9M. If the total number of receptors was reduced to 5000, the [RL] would still be 2470, then [TGFß57] = 5X10-9M—a 5-fold increase in the Kd of TGFß57. This decrease makes the cell less sensitive to TGFß57 because there needs to be View Full Document This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore. - Spring '11

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https://projecteuclid.org/euclid.aoap/1115137984

math

The Annals of Applied Probability - Ann. Appl. Probab. - Volume 15, Number 2 (2005), 1506-1535. The branching process with logistic growth In order to model random density-dependence in population dynamics, we construct the random analogue of the well-known logistic process in the branching process’ framework. This density-dependence corresponds to intraspecific competition pressure, which is ubiquitous in ecology, and translates mathematically into a quadratic death rate. The logistic branching process, or LB-process, can thus be seen as (the mass of ) a fragmentation process (corresponding to the branching mechanism) combined with constant coagulation rate (the death rate is proportional to the number of possible coalescing pairs). In the continuous state-space setting, the LB-process is a time-changed (in Lamperti’s fashion) Ornstein–Uhlenbeck type process. We obtain similar results for both constructions: when natural deaths do not occur, the LB-process converges to a specified distribution; otherwise, it goes extinct a.s. In the latter case, we provide the expectation and the Laplace transform of the absorption time, as a functional of the solution of a Riccati differential equation. We also show that the quadratic regulatory term allows the LB-process to start at infinity, despite the fact that births occur infinitely often as the initial state goes to ∞. This result can be viewed as an extension of the pure-death process starting from infinity associated to Kingman’s coalescent, when some independent fragmentation is added. Ann. Appl. Probab., Volume 15, Number 2 (2005), 1506-1535. First available in Project Euclid: 3 May 2005 Permanent link to this document Digital Object Identifier Mathematical Reviews number (MathSciNet) Zentralblatt MATH identifier Primary: 60J80: Branching processes (Galton-Watson, birth-and-death, etc.) Secondary: 60J70: Applications of Brownian motions and diffusion theory (population genetics, absorption problems, etc.) [See also 92Dxx] 60J85: Applications of branching processes [See also 92Dxx] 92D15: Problems related to evolution 92D25: Population dynamics (general) 92D40: Ecology Size-dependent branching process continuous-state branching process population dynamics logistic process density dependence Ornstein–Uhlenbeck type process Riccati differential equation fragmentation–coalescence process Lambert, Amaury. The branching process with logistic growth. Ann. Appl. Probab. 15 (2005), no. 2, 1506--1535. doi:10.1214/105051605000000098. https://projecteuclid.org/euclid.aoap/1115137984

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http://sosdelreycatolico.eu/39188-black-scholes-fair-values-of-binary-options.html

math

commissions charged. End Function, function dTwo(UnderlyingPrice, ExercisePrice, Time, Interest, Volatility, Dividend) dTwo dOne(UnderlyingPrice, ExercisePrice, Time, Interest, Volatility, Dividend) - Volatility * Sqr(Time). The strike price, the length of time until expiry. Black, Fischer; Scholes, Myron. Hadi AKJanuary 31st, 2009 at 12:53am " The volatility of an option really determines how likely that contract will be in, at or out-of-the-money by the expiration date. If the BlackScholes model held, then the implied volatility for a particular stock would be the same for all strikes and maturities. In fairness, Black and Scholes almost certainly understood this point well. But their devoted followers may be ignoring whatever caveats the two men. The Black -Scholes formula (also called Black -Scholes -Merton) was the first widely used model for option pricing. It's used to calculate the theoretical value of European-style options. The formula, shown in Figure 4, takes the following variables into consideration. End Function, function CallOption(UnderlyingPrice, ExercisePrice, Time, Interest, Volatility, Dividend). K, the strike price of the option. Yalincak, Hakan, "Criticism of the BlackScholes Model: But Why Is It Still Used? Thus uncertainty has been eliminated and the portfolio is effectively riskless. Bob DolanMarch 23rd, 2011 at 6:39pm Peter wrote: "Do you know if there is an available option model for a binary distribution.?" Actually, the binary distribution is fully described in this web site. 8 The formula can be interpreted by first decomposing a call option into the difference of two binary options : an asset-or-nothing call minus a cash-or-nothing call (long an asset-or-nothing call, short a cash-or-nothing call). Iterating the search in Excel, and comparing the result to some level of 'tolerance would seem to be a fairly easy work-around. See my Historical Volatility Calculator. It's actually a two-step process: Step One: Guess at the IV say, 30 and adjust the guess until you have the IV bracketed. In practice, the price is affected by many factors, including demand and supply, and because of this, options may not always be priced correctly. Also in 1973, a subsequent paper, Theory of Rational Option Pricing was written by Robert Merton, and he expanded on this mathematical approach and introduced the term Black Scholes options pricing model. The Black -Scholes Model was developed by three academics: Fischer Black, Myron Scholes and Robert Merton. In its early form the model was put forward as a way to calculate the theoretical value of a European call option on a stock not paying discrete proportional dividends. Option traders generally rely on the Black Scholes formula to buy options that are priced under the formula calculated value, and sell options that. This type of arbitrage trading quickly pushes option prices back towards the Model's calculated value. The Model generally works, but there are a few key. Android apps double bollinger bands binary options trading, Iq option binary trading times, Trading cryptocurrency margin,

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https://www.coursehero.com/sitemap/schools/53500-Pine-Bush-Senior-High-School/courses/4261986-SCIENCEPhysics-1/

math

TAKE HOME QUIZ PART ONE: Draw a line connecting the term(s) in the left column with the matching description in the right column. unit of frequency time for one oscillation 1. A force of 22 N was used to push box 8.3 WORK, POWER, & ENERGY m along the oor. How much work was done? 2. What is the power of a winch that can do 5.2 x 105 J of work in 5.35 seconds? 3. If a crane d VELOCITY VS TIME GRAPHS 1. When does the car travel at a constant speed? 2. What happens at 32 seconds? 0 10 seconds, 22 26 seconds, 56 58 seconds The car stops 3. What is the highest velocity of the car during the GROUP WORK QUESTIONS NAMES: hay DATE: 1. A motorboat heads due east at 12.0 111/3 across a river that ows towards the south with a current of 3.50 111/3. A. What is the resultant veloci of the boat? ca Grants": Jim 12.25 [2. 5W; Fill in the Blank Work is measured in _ . _ is the rate of work. Power is measured in _ . 1 hp = _ Watts. In the power equation, W stands for _ . What is the power of a motor that can exert a force of 7500 N over 25.5 m i

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https://essayhope.com/transcribed-image-text-gibbs-free-energy-using-equilibrium-solution-composition-data-is/

math

Transcribed Image Text: Gibbs Free Energy using equilibrium solution composition data is calculation results in the unit joule has a natural log component in the calculation are calculated using Farraday’s constant calculated using the amount of all species in the reaction. calculated using the concentration of all species in solution that have significant changes in concentration over the time studied. are calculated using redox cell potential measured in volts has temperature in Kelvin in the calculation uses the ideal gas constant in latm/molK in the calculation uses the ideal gas constant in J/Kmol in the calculation calculated using the concentration of all solid species in solution that have insignificant changes in concentration over the time studied.

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http://www.nvaphysics.com/4AAU12/quizzes/4AQuiz9AU12.htm

math

QUIZ 9 (ANSWERS) CH. 7: Problems 9*, 24*, 38* (TRY #34*, critical for Ch 12 and Physics 4B; see class notes on stable and unstable equilibrium) 41*, 42*, 43*, 46* (try #12*), 47* (TRY 65* and 66*) , 50*, 55*, 56*, 63* (see #46*) , 68*, 70*, 75* (SEE SAMPLE TEST 3) (ANSWERS) |VIRTUAL LAB 2 (SEE EXAMPLE 7.7, CHAPTER 7)| |SIMPLE HARMONIC MOTION LAB| Exam Cross Index for TEST 2 Sp '12 BELOW ARE THE LINKS TO SAMPLE EXAMS: MEANWHILE HERE ARE LINKS (A), (B), (C) AND (D) TO SAMPLE TEST 2 AND 3: |For many of these problems use KEi + Ui = KEf + Uf + HEAT or USE KEi + Ui + WF = KEf + Uf + HEAT, where U is the potential energy due to the conservative force ( gravitational and spring forces in this case), KE is the kinetic energy and WF is the work done by the external force (done by "Frankie" as we discussed in class. ) NOTE: 9, 46 and 63 mix ENERGY CONSERVATION (Ch. 7)and CIRCULAR DYNAMICS (Ch. 5), a special class of problems you should study ahead of Test 2. ANOTHER NOTE: #50 mixes ENERGY CONSERVATION (Ch. 7) and PROJECTILE MOTION (CH. 3) |9.* KEi + Ui = KEf + Uf + HEAT . (a) The angle between the normal force and motion in this case is 90 degrees, so what is the work by that force? The work by gravity is positive since the motion is from high to low; note the gravitational work only depends on the vertical displacement. (b) ENERGY CONSERVATION : KEi + Ui = KEf + Uf + HEAT where HEAT IS THE NEGATIVE OF THE WORK BY FRICTION. THUS HEAT IS ALWAYS POSITIVE. You are given the heat. (c) CIRCULAR DYNAMICS: Critical point. HEAT is the negative of the work by the friction force, which may not be constant. The instantaneous friction force is u*N, where u is the kinetic friction coefficient and where N is the magnitude of the normal force which may not be constant as in this case. Clearly from earlier problems in Chapter 5, N is a maximum at the bottom and is a minimum at the top. (See worksheet 2, GRP 3, problem #1, THE PROBLEM DEALING WITH BUMPS AND VALLEYS) (d) CIRCULAR DYNAMICS: At bottom, the normal force points up in the pos direction toward the center of the circle and the weight of magnitude mg points down in the neg direction. Write: Sum of forces in radial direction = m*v2/R = pos - neg and solve for N, the normal force magnitude, using the speed v you found in part (b) from energy conservation. See worksheet 2, GRP 3, problem #1. Note N > mg. |24. *Use this: KEi + Ui = KEf + Uf + HEAT for the initial (i) and final (f) locations. Study carefully the classic example (i.e. Many physics book authors have used it.). Unlike the previous problem, the friction force is constant, so HEAT = fk*D, where D is the distance moved and fk = 17,000 (N) is given. Note the initial kinetic energy KEi is not zero as in problem 9. Set the lowest point equal to the zero of gravitational potential energy (ie, y = 0 at point 2.) Note U = Ug + Us, where Ug = mgy and Us = (1/2)*k*y'2, where y' is the spring coordinate indicating compression. We have KEi = 16,000 J and Ui = mgyi + 0 since the initial spring potential energy is zero (not compressed--- yi' = 0) ; note yi = 2.00 m. Uf = mgyf + (1/2)*k*yf'2 , where yf = 1.00 m yf' = -1.00 m. (Since you square y' in general the negative sign goes away. ) The heat of course is fk*D, where D = 1.00 m. Solve for KEf and find final speed vf : KEi + Ui = KEf + Uf + HEAT means 16,000 J + mgyi = KEf + mgyf + (1/2)*k*yf'2 + 17,000 (N)*d , where yf = d = 1.00 m and yf' = -1.00 m. (b) If you think the acceleration is constant think twice; the spring makes things more complex as we have already seen in Ch. 14. So use: sum of the forces in the y direction = m*a = pos - neg, where we set the positive y direction to be up. Thus: : m*a = | kyf' | + fk - mg, where clearly the spring and friction forces point upward as the object moves down. |38.* READ section 7.4 and 7.5: If U = U(x) then Fx = -dU/dx. The x-component of force is the negative derivative of U; Thus if dU/dx is positive , the force points in the negative x -direction (Fx < 0) and if dU/dx is negative then force points in positive x direction (Fx > 0.) When dU/dx = 0 force is zero and the object is said to be in equilibrium, either stable or unstable. A CONCAVE UP POTENTIAL ENERGY FUNCTION MEANS EQUILIBRIUM IS STABLE.| |41*. (a) CHAPTER 5 REVIEW: You must prove whether or not the system is at rest; the emphasis here is the use of Ch. 5 concepts, whereas Ch. 7 is more directly related to part If the system is at rest and remains so we have for the concrete m*a = pos - neg = mg - T = 0 . So you can immediately find T. For the Box we have m'*a = 0 = T - fs ' , where fs ' is the force of static friction from the ground and m' is the Box mass. We also have m''*a = 0, where m'' is the gravel mass. ( i.e. There is no friction force between the gravel and Box if the system is at rest.) Compare T and fsmax = us*N, where N is the magnitude of the normal force on the box, which must be equal to (m' + m'')*g. |42.* KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. VIRTUAL LAB 2, question 1. 43.* The simplest way to do the problem is to bypass the speed and kinetic energy just after the Block leaves the spring and consider only two points in space-time: When the spring is fully compressed and the block is at rest (i) and when the block has reached its maximum horizontal distance and comes to rest permanently (f) . KEi + Ui = KEf + Uf + HEAT , where Ug = 0 at both points (ground level, y = 0) . You should be able to see which terms are clearly zero. Thus we have initial (i) spring potential energy = HEAT , where HEAT = fk*D : (1/2)*k*d2 = u*mg*D, where u = coefficient of kinetic friction and d and D are given in diagram. |46.* ANOTHER CLASSIC, like #24. This one mixes energy conservation and dynamics just like #9 above and #63 below. (a) KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. ENERGY CONSERVATION: (I) KEi + Ui = KEf + Uf , WHERE KEi = 0, Ui = m*g*h, KEf = (1/2)*m*v2 and Uf = m*g*2R: mgh = (1/2)*m*v2 + m*g*2R AT TOP. CIRCULAR DYNAMICS: ALSO AT TOP, the normal force and the weight of magnitude mg point down in the pos direction toward the center of the circle. Write: (II) Sum of forces in radial direction = m*v2/R = pos - neg = N + mg - 0 = N + mg, since there is no force pointing in the neg direction away from the center; see #120 AND #42, Ch. 5 , point B at top. Set N = 0 IN THIS CASE. Solve equations I and II simultaneously for h in terms of R by eliminating v. The mass m cancels out. (b) FOR THIS INITIAL HEIGHT h, greater than that of part (a), the object never loses contact with the track at the top so it does reach point C. (I) ENERGY CONSERVATION : (I) KEi + Ui = KEf + Uf , WHERE KEi = 0, Ui = m*g*h, h = 3.50*R, KEf = (1/2)*m*v2 and Uf = m*g*R (at point C) . Find v at C: mg*( 3.50*R ) = (1/2)*m*v2 (III) For the tangential acceleration, let the pos direction be vertically down, a direction tangent to the circle at point |47.* ANOTHER CLASSIC: m*g*H = TOTAL HEAT, SINCE THE INITIAL AND FINAL kinetic energies are zero. The Heat is the total heat after possibly numerous trips across the flat section. TOTAL HEAT fk*L, where L is the total distance moved on flat surface before coming to rest. To find the number of trips, divide TOTAL HEAT by fk*D, where D = 30 m exactly. From this ratio you can find the number of complete trips; from the fractional remainder, you can find exactly where the object comes to rest on the flat surface. If the ratio is less than one, then the object does not make a complete trip across the flat surface. |50.* KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. See problem 68 hint i = bottom. f = top just before leaving horizontal surface 70 (m) above ground zero ( where Ug = 0) Uf = mgyf, where yf = 70 (m) KEi = (1/2)*m*vi2 . KEf = (1/2)*m*vf2 |55.* KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. Let the zero of gravitational energy be the ground. # Initially (i) the system is at REST and Ui = the gravitational potential energy of the 12.0 kg clock at the given initial height. Solve for the final common speed vf of the two blocks. USE KEi + Ui + WF = KEf + Uf + HEAT, where U is the gravitational potential energy in this case, KE is the kinetic energy, WF is the work done by the external thrust force and the HEAT is fk*D. The constant friction force of magnitude fk = 500 (N) is given; D is the distance moved along ramp. # Initially (i), the rocket is at REST and Ui = the gravitational energy at the initial vertical height H above the ground. Note the height H has a simple relationship with D, the distance moved along the ramp; use simple geometry, the angle of 53 degrees, and the properties of right triangles to write H in terms of D. WF = F*D, where D is the distance moved along the ramp and F is the magnitude of the given external thrust force = 2000 (N) exactly. |63.* ANOTHER CLASSIC: THE STRUCTURE OF THIS PROBLEM, LIKE 9, and 46, mixes ENERGY CONSERVATION (Ch. 7) CIRCULAR DYNAMICS (Ch. 5) ENERGY CONSERVATION (Ch. 7): (I) KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. # Initially (i), skier is at the top and is essentially at rest; KEi = 0. Ui = m*g*R if the zero of potential energy is considered to be at the level of the snow ball's horizontal diameter. (I) mg*R = (1/2)*m*vf2 + mg*R*cos alpha, where alpha is angle with vertical shown. # Finally (f), we consider the point when the skier leaves the snow ball , i.e. when the normal of magnitude N becomes zero. See figure 7.63. We define the location of this point in the following way: Draw a radial line from the snow ball center to this point and define the angle with the vertical as alpha . KEf = (1/2)*m*vf2 and Uf = m*g*R*cos alpha. CIRCULAR DYNAMICS: At any point on the surface of the ball, the normal force points radially away from the center of the circle. The gravitational force of magnitude mg points vertically down and the component of the gravitational force along the radial line is mg*cos theta and points toward the center of the circle. Write: (II) For the centripetal (center seeking) force, Sum of forces in radial direction = m*v2/R = pos - neg = mg*cos theta - N. For this problem set N = 0. Thus: m*vf2/R = mg*cos theta. Solve equations I and II simultaneously for cos theta by eliminating v. Compute theta by evaluating cos -1. Your angle will be between 45 and 90 degrees. |68. WE DO THIS PROBLEM USING TWO STYLES--FROM CH. 6 AND See #50 above. CH. 6 STYLE: (i) (1/2)*m*vf2 - (1/2)*m*vi2 = total work = Wg + Wf + WN, work by gravity, the friction force and normal force respectively where last term is zero since the normal work on the way up slope must be vanish because that force is perpendicular to the motion i.e. makes an angle of 90 degrees. ALSO, THE SLOPE IS FRICTIONLESS MAKING WORK OF FRICTION Wf = 0. (ii) FIND THE SPEED AT THE TOP OF THE CLIFF USING (1/2)*m*vf2 - (1/2)*m*vi2 = total work = Wg WHERE THE NET VERTICAL DISPLACEMENT IS FROM LOW TO HIGH. IS GRAVITY WORK POSITIVE OR NEGATIVE? (ii) AFTER LEAVING CLIFF, DURING FREE FALL (i.e., PROJECTILE MOTION) THE gravitational work will be positive for this high to low motion. IN THAT PHASE OF PROBLEM YOU CAN USE CH. 3 METHODS AS IN LAB 3. HERE'S HOW YOU DO IT CH. 7 STYLE: |70 (a) KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. The block is released from rest so that should tell you the value of KEi . The maximum speed occurs when U is a minimum and that location is shown in figure 7.42. Uf = 0 assuming both springs are un-deformed and the block is in equilibrium. Ui = sum of two spring potential energies: Spring 1 is stretched by 0.15 m and spring 2 is compressed by 0.15 m. To evaluate Ui, apply the formula (1/2)*k*x2 for the potential energy to both springs and add the two results. Solve for KEf and the maximum speed vf of the block WHICH OCCURS WHEN BOTH STRINGS ARE (b) The simplest way to do this part is to assume the initial position (i) of the block is the same as in part (a) but the final location (f) is when Spring 1 is at maximum compression. When Spring 1 is at maximum compression, the block is at rest momentarily (i): Ui = Uf = sum of two spring potential energies. Spring 1 is compressed a distance |x'| and Spring 2 is stretched by the same |x'| . Ui was evaluated in part (a). Plug the symbol |x'| into the formula for Uf and solve for |x'| numerically. |75. * SELECTIVELY USE KEi + Ui + WF = KEf + Uf where U is the potential energy due to the spring force in this case, KE is the kinetic energy and WF is the work done by the external force of magnitude F. NO FRICTION SO HEAT = 0. The FULL equation with WF only applies between initial point A and point-B but we know the process continues beyond that. The Block will continue to move beyond point-B and eventually come to rest. At that point, the full distance from the origin is the amplitude A of subsequent simple harmonic motion about the origin and would be described by the function Acoswt for t >0 where t=0 is when the block reaches maximum distance from origin---see Ch. 14 and virtual lab 2.. So that is your task---to find that distance . Here are some tips: (a) Find the speed of the block at point-B using : KEA + UA + WF = KEB + UB where WF = F*D. Note KEA = 0 and UA = 0. Also WF = F*D, D = 0.25 m and UB = (1/2)*k*(0.25)2 in Joules. Find the final kinetic energy and final speed at B. (b) Find the additional distance before coming to rest using: KEB + UB = KEC + UC . NOTE: KEB is known from the previous part and UB = (1/2)*k*(0.25)2 Also KEC = 0 and UC = (1/2)*k*xC2 . Find xC. Then find 0.60 m - xC . Note: KEA + UA + WF = KEC + UC or, WF = KEC + UC, where WF = F*D, D = 0.25 m and KEC = 0.

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https://yadda.icm.edu.pl/baztech/search/page.action;jsessionid=7FD191F50B1FE6C9DE6CD189FF4E194F?qt=SEARCH&q=sc.general*c_0keywords_0eqs.system%2Bczasu%2Brzeczywistego*l_0

math

W artykule opisano koncepcję przedziałowego wyrażania wyniku pomiaru oraz jego niepewności w sposób specyficzny dla systemów pomiarowo-sterujących. Niepewność rozumiana jest tu jako parametr błędu wyniku pomiaru interpretowanego w kategoriach probabilistycznych. Wyrażanie wyniku w postaci przedziału przedstawiono na przykładach obliczanych symulacyjnie przy użyciu metody Monte Carlo. An approach to the interval representation of measurement result and its uncertainty in measuring and control system is presented in the paper. Nowadays, the measurement result is characterized by the measurement uncertainty , which is defined as the radius of the interval built around the measured value in which the true value lies with given probability. A rapid growth of measurement systems application area leads to introduce more usable definition of inaccuracy which basis on the interval representation of a measurement result. This definition is more useful, particularly in real-time systems and when errors with asymmetrical distribution occur in systems [2, 4]. According to classical definition, to classify a system as real-time one delays in it have to be less than it is allowable. In such systems propagation of the signals from the input to the output is connected with arising of delays because all the system elements need time to perform their activities. However, to classify a measuring and control system as a real time, it should be taken into account all factors influencing on properties the system output signals, i.e. not only delays but also errors of measurement data. Therefore, the delay errors should be described as components of the total error being the basis of determination of the interval representing the system output measuring results . Comparing the interval with critical acceptable values enables classifying the system as a real-time one. Theoretical consideration in the paper are illustrated by results of numerical experiments carried out by using Monte Carlo method.

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CC-MAIN-2021-21

http://archive.ymsc.tsinghua.edu.cn/pacm_category/01?show=time&size=25&from=3&target=searchall

math

We investigate a one-parametric class of merit functions for the second-order cone complementarity problem (SOCCP) which is closely related to the popular FischerBurmeister (FB) merit function and natural residual merit function. In fact, it will reduce to the FB merit function if the involved parameter <i></i> equals 2, whereas as <i></i> tends to zero, its limit will become a multiple of the natural residual merit function. In this paper, we show that this class of merit functions enjoys several favorable properties as the FB merit function holds, for example, the smoothness. These properties play an important role in the reformulation method of an unconstrained minimization or a nonsmooth system of equations for the SOCCP. Numerical results are reported for some convex second-order cone programs (SOCPs) by solving the unconstrained minimization reformulation of the KKT optimality conditions, which indicate that the The study of this paper consists of two aspects. One is characterizing the so-called circular cone convexity of f by exploiting the second-order differentiability of f L ; the other is introducing the concepts of determinant and trace associated with circular cone and establishing their basic inequalities. These results show the essential role played by the angle , which gives us a new insight when looking into properties about circular cone. MSC:26A27, 26B05, 26B35, 49J52, 90C33, 65K05. We consider the Tikhonov regularization method for the second-order cone complementarity problem (SOCCP) with the Cartesian P 0-property. We show that many results of the regularization method for the P 0-nonlinear complementarity problem still hold for this important class of nonmonotone SOCCP. For example, under the more general setting, every regularized problem has the unique solution, and the solution trajectory generated is bounded if the original SOCCP has a nonempty and bounded solution set. We also propose an inexact regularization algorithm by solving the sequence of regularized problems approximately with the merit function approach based on FischerBurmeister merit function, and establish the convergence result of the algorithm. Preliminary numerical results are also reported, which verify the favorable theoretical properties of the proposed method. This paper proposes using the neural networks to efficiently solve the second-order cone programs (SOCP). To establish the neural networks, the SOCP is first reformulated as a second-order cone complementarity problem (SOCCP) with the KarushKuhnTucker conditions of the SOCP. The SOCCP functions, which transform the SOCCP into a set of nonlinear equations, are then utilized to design the neural networks. We propose two kinds of neural networks with the different SOCCP functions. The first neural network uses the FischerBurmeister function to achieve an unconstrained minimization with a merit function. We show that the merit function is a Lyapunov function and this neural network is asymptotically stable. The second neural network utilizes the natural residual function with the cone projection function to achieve low computation complexity. It is shown to be Lyapunov stable and converges globally Abstract Recently, J.-S. Chen and P. Tseng extended two merit functions for the nonlinear complementarity problem (NCP) and the semidefinite complementarity problem (SDCP) to the second-order cone commplementarity problem (SOCCP) and showed several favorable properties. In this paper, we extend a merit function for the NCP studied by Yamada, Yamashita, and Fukushima to the SOCCP and show that the SOCCP is equivalent to an unconstrained smooth minimization via this new merit function. Furthermore, we study conditions under which the new merit function provides a global error bound which plays an important role in analyzing the convergence rate of iterative methods for solving the SOCCP; and conditions under which the new merit function has bounded level sets which ensures that the sequence generated by a descent method has at least one accumulation point. In this paper, we extend the one-parametric class of merit functions proposed by Kanzow and Kleinmichel [C. Kanzow, H. Kleinmichel, A new class of semismooth Newton-type methods for nonlinear complementarity problems, Comput. Optim. Appl. 11 (1998) 227251] for the nonnegative orthant complementarity problem to the general symmetric cone complementarity problem (SCCP). We show that the class of merit functions is continuously differentiable everywhere and has a globally Lipschitz continuous gradient mapping. From this, we particularly obtain the smoothness of the FischerBurmeister merit function associated with symmetric cones and the Lipschitz continuity of its gradient. In addition, we also consider a regularized formulation for the class of merit functions which is actually an extension of one of the NCP function classes studied by [C. Kanzow, Y. Yamashita, M. Fukushima, New NCP functions and Like the matrix-valued functions used in solutions methods for semidefinite programs (SDPs) and semidefinite complementarity problems (SDCPs), the vector-valued functions associated with second-order cones are defined analogously and also used in solutions methods for second-order-cone programs (SOCPs) and second-order-cone complementarity problems (SOCCPs). In this article, we study further about these vector-valued functions associated with second-order cones (SOCs). In particular, we define the so-called SOC-convex and SOC-monotone functions for any given function . We discuss the SOC-convexity and SOC-monotonicity for some simple functions, e.g., <i>f</i>(<i>t</i>) = <i>t</i> ² <i>t</i> ³ 1/<i>t</i> <i>t</i> ^1/2, |<i>t</i>|, and [<i>t</i>]₊. Some characterizations of SOC-convex and SOC-monotone functions are studied, and some conjectures about the relationship between SOC-convex and SOC-monotone functions are proposed. Recently Tseng (Math Program 83:159185, 1998) extended a class of merit functions, proposed by Luo and Tseng (<i>A new class of merit functions for the nonlinear complementarity problem</i>, in Complementarity and Variational Problems: State of the Art, pp. 204225, 1997), for the nonlinear complementarity problem (NCP) to the semidefinite complementarity problem (SDCP) and showed several related properties. In this paper, we extend this class of merit functions to the second-order cone complementarity problem (SOCCP) and show analogous properties as in NCP and SDCP cases. In addition, we study another class of merit functions which are based on a slight modification of the aforementioned class of merit functions. Both classes of merit functions provide an error bound for the SOCCP and have bounded level sets. This paper is a follow-up of the work [Chen, J.-S.: J. Optimiz. Theory Appl., Submitted for publication (2004)] where an NCP-function and a descent method were proposed for the nonlinear complementarity problem. An unconstrained reformulation was formulated due to a merit function based on the proposed NCP-function. We continue to explore properties of the merit function in this paper. In particular, we show that the gradient of the merit function is globally Lipschitz continuous which is important from computational aspect. Moreover, we show that the merit function is <i>SC</i> ¹ function which means it is continuously differentiable and its gradient is semismooth. On the other hand, we provide an alternative proof, which uses the new properties of the merit function, for the convergence result of the descent method considered in [Chen, J.-S.: J. Optimiz. Theory Appl., Submitted for publication (2004)]. We investigate some properties related to the generalized Newton method for the Fischer-Burmeister (FB) function over second-order cones, which allows us to reformulate the second-order cone complementarity problem (SOCCP) as a semismooth system of equations. Specifically, we characterize the B-subdifferential of the FB function at a general point and study the condition for every element of the B-subdifferential at a solution being nonsingular. In addition, for the induced FB merit function, we establish its coerciveness and provide a weaker condition than Chen and Tseng (Math. Program. 104:293327, 2005) for each stationary point to be a solution, under suitable Cartesian <i>P</i>-properties of the involved mapping. By this, a damped Gauss-Newton method is proposed, and the global and superlinear convergence results are obtained. Numerical results are reported for the second-order cone programs In last decades, there has been much effort on the solution and the analysis of the nonlinear complementarity problem (NCP) by reformulating NCP as an unconstrained minimization involving an NCP function. In this paper, we propose a family of new NCP functions, which include the Fischer-Burmeister function as a special case, based on a <i>p</i>-norm with <i>p</i> being any fixed real number in the interval (1,+), and show several favorable properties of the proposed functions. In addition, we also propose a descent algorithm that is indeed derivative-free for solving the unconstrained minimization based on the merit functions from the proposed NCP functions. Numerical results for the test problems from MCPLIB indicate that the descent algorithm has better performance when the parameter <i>p</i> decreases in (1,+). This implies that the merit functions associated with <i>p</i>(1,2), for example <i>p</i>=1.5, are more effective A popular approach to solving the nonlinear complementarity problem (NCP) is to reformulate it as the global minimization of a certain merit function over ^<i>n</i>. A popular choice of the merit function is the squared norm of the Fischer-Burmeister function, shown to be smooth over ^<i>n</i> and, for monotone NCP, each stationary point is a solution of the NCP. This merit function and its analysis were subsequently extended to the semidefinite complementarity problem (SDCP), although only differentiability, not continuous differentiability, was established. In this paper, we extend this merit function and its analysis, including continuous differentiability, to the second-order cone complementarity problem (SOCCP). Although SOCCP is reducible to a SDCP, the reduction does not allow for easy translation of the analysis from SDCP to SOCCP. Instead, our analysis exploits Shape optimization aims to optimize an objective function by changing the shape of the computational domain. In recent years, shape optimization has received considerable attentions. On the theoretical side there are several publications dealing with the existence of solution and the sensitivity analysis of the problem; see e.g., and references therein. We simulate blood flow in patient-specific cerebral arteries. The complicated geometry in the human brain makes the problem challenging. We use a fully unstructured three dimensional mesh to cover the artery, and Galerkin/least-squares finite element method to discretize the incompressible Navier-Stokes equations, that are employed to model the blood flow, and the resulting large sparse nonlinear system of equations is solved by a Newton-Krylov-Schwarz algorithm. From the computed flow fields, we are able to understand certain behavior of the blood flow of this particular patient before and after a stenosis is surgically removed. We also report the robustness and parallel performance of the domain decomposition based algorithm. Wind power is an increasingly popular renewable energy. In the design process of the wind turbine blade, the accurate aerodynamic simulation is important. In the past, most of the wind turbine simulations were carried out with some low fidelity methods, such as the blade element momentum method . Recently, with the rapid development of the supercomputers, high fidelity simulations based on 3D unsteady Navier-Stokes (N-S) equations become more popular. For example, Sorensen et al. studied the 3D wind turbine rotor using the Reynolds-Averaged Navier-Stokes (RANS) framework where a finite volume method and a semi-implicit method are used for the spatial and temporal discretization, respectively . Bazilevs et al. investigated the aerodynamic of the NREL 5MW offshore baseline wind turbine rotor using large eddy simulation built with a deforming-spatial-domain/stabilized space-time We propose and study a new parallel one-shot Lagrange--Newton--Krylov--Schwarz (LNKSz) algorithm for shape optimization problems constrained by steady incompressible Navier--Stokes equations discretized by finite element methods on unstructured moving meshes. Most existing algorithms for shape optimization problems iteratively solve the three components of the optimality system: the state equations for the constraints, the adjoint equations for the Lagrange multipliers, and the design equations for the shape parameters. Such approaches are relatively easy to implement, but generally are not easy to converge as they are basically nonlinear Gauss--Seidel algorithms with three large blocks. In this paper, we introduce a fully coupled, or the so-called one-shot, approach which solves the three components simultaneously. First, we introduce a moving mesh finite element method for the shape optimization We characterize conformally flat spaces as the only compact self-dual manifolds which are U(1)-equivariantly and conformally decomposable into two complete self-dual Einstein manifolds with common conformal infinity. A geometric characterization of such conformally flat spaces is also given. The notion of a generalized CRF-structure on a smooth manifold was recently introduced and studied by Vaisman (2008). An important class of generalized CRF-structures on an odd dimensional manifold M consists of CRF-structures having complementary frames of the form , where is a vector field and is a 1-form on M with ()= 1. It turns out that these kinds of CRF-structures give rise to a special class of what we called strong generalized contact structures in Poon and Wade . More precisely, we show that to any CRF-structures with complementary frames of the form , there corresponds a canonical Lie bialgebroid. Finally, we explain the relationship between generalized contact structures and another generalization of the notion of a CauchyRiemann structure on a manifold. Using deformations of singular twistor spaces, a generalisation of the connected sum construction appropriate for quaternionic manifolds is introduced. This is used to construct examples of quaternionic manifolds which have no quaternionic symmetries and leads to examples of quaternionic manifolds whose twistor spaces have arbitrary algebraic dimension. Guy Bonneau has kindly pointed out two errors in . The first is that the manifolds M (k) and M (k)/Z2 do not admit U (2)-invariant Einstein-Weyl structures for k 2; thus the last four entries in Table 4 (page 422) do not occur. The error is on pages 429430. The analysis there is correct, except that the critical line to consider in Lemma 7.5 and the subsequent calculation is = , instead of = 2/k, because of the constraint (D, ](equivalently, > ) occurring in the definition of . On this line, one has A cohomology theory associated to a holomorphic Poisson structure is the hypercohomology of a bicomplex where one of the two operators is the classical -operator, while the other operator is the adjoint action of the Poisson bivector with respect to the Schouten-Nijenhuis bracket. The first page of the associated spectral sequence is the Dolbeault cohomology with coefficients in the sheaf of germs of holomorphic polyvector fields. In this note, the authors investigate the conditions for which this spectral sequence degenerates on the first page when the underlying complex manifolds are nilmanifolds with an abelian complex structure. For a particular class of holomorphic Poisson structures, this result leads to a Hodge-type decomposition of the holomorphic Poisson cohomology. We provide examples when the nilmanifolds are 2-step. We classify compact anti-self-dual Hermitian surfaces and compact four-dimensional conformally flat manifolds for which the group of orientation preserving conformal transformations contains a two-dimensional torus. As a corollary, we derive a topological classification of compact self-dual manifolds for which the group of conformal transformations contains a two-dimensional torus. Let X be a compact quotient of the product of the real Heisenberg group H _4m+1 of dimension 4m + 1 and the three-dimensional real Euclidean space R ³ . A left-invariant hypercomplex structure on H _4m+1 R ³ descends onto the compact quotient X. The space X is a hyperholomorphic fibration of 4-tori over a 4m-torus. We calculate the parameter space and obstructions to deformations of this hypercomplex structure on X. Using our calculations, we show that all small deformations generate invariant hypercomplex structures on X but not all of them arise from deformations of the lattice. This is in contrast to the deformations on the 4m-torus. A complex symplectic structure on a Lie algebra h is an integrable complex structure J with a closed non-degenerate (2, 0)-form. It is determined by J and the real part of the (2, 0)-form. Suppose that h is a semi-direct product g V, and both g and V are Lagrangian with respect to and totally real with respect to J. This note shows that g V is its own weak mirror image in the sense that the associated differential Gerstenhaber algebras controlling the extended deformations of and J are isomorphic. The geometry of (, J) on the semi-direct product g V is also shown to be equivalent to that of a torsion-free flat symplectic connection on the Lie algebra g. By further exploring a relation between (J, ) with hypersymplectic Lie algebras, we find an inductive process to build families of complex symplectic algebras of dimension 8 n from the data of the 4 n-dimensional ones.

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https://forums.engineersgarage.com/forums/reply/plc-use-logic-ladders/

math

Microcontroller › PIC › Do PLC’s use logic gates or Logic Ladders? › PLC use logic ladders. December 19, 2018 at 9:06 am #14957 PLC use logic ladders. However, logic gates can be implemented on PLC using ladder diagrams. the problem given to you can be solved using ladder logic with a proper flowchart to solve the problem

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https://devaricollection.com/products/womens-casual-shoes

math

Womens Casual Sandles Assorted Colors Hello! Welcome to our store! Quality is first along with the best service. We see our customers as friends. Size Chart : US(5)=EU(35)=225mm(bare feet length) Color : Black&Silver \Black \Red \Blue \Green Heel Height : 2cm Material : Microfiber \Flax \PU \Robber Style : Fashion Casual Pictures are only samples for reference. Due to limitations in photography and the inevitable differences in monitor settings, the colors shown in the photograph may not correspond 100% to those in the items themselves Plz allow 1-3 cm error because measured by hand, plz refer to the Sizing Chart before choosing.

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https://cboard.cprogramming.com/brief-history-cprogramming-com/28545-maths-problem-printable-thread.html

math

In a town there was a population of 660 dogs and cats. Although, many of the dogs and cats were crazy. 25% of the actual dogs thought they were cats, and 20% of the actual cats thought they were dogs. In a poll regarding whether one is dog or cat, in which every animal of this population participated, 40% answered "cat". How many dogs respectively cats were there in this population? I'm not sure about the math, but I think there's somewhere in the neighborhood of 480 dogs and 170 cats... I'm not sure what happened to the other 10, though. We'll just assume they're rats. I'm not sure either... but I get 463 dogs: 116 of them think they're cats 197 cats: 40 of them think they're dogs edit: well, it's wrong... doesn't work out :rolleyes: I do know the math :) and the answer is: I can explain the math if somone wants ;) I'd be delighted if you could explain for us how you got to this conclusion:D Is this for homework? A stats class perhaps? Basically is this graded? If not I'll explain heh. no homework, we had this question on a math test we had yesterday but nobdy didn't make it, and we want the solution.. no problem - make the number of cats be "x", then the number of dogs is "660-x". Now, 40% of 660 said they were cats, which is 264. 80% of the cats say they are cats, 25% of the dogs think they are cats, therefore the equation becomes: work the equation, and x=180, so the number of cats is 180, the number of dogs is 480. Sucks to be a cat in this town :D .25d + .8c = .4*660 d + c = 660 Solve the system. Edit: PJ beat me with a different way. Ah ok, well I'm drawing an easy to read explanation....of why it all works: .75D + .20C = .60 * 660 and .25D + .80C = .40 * 660 .75D + .20C = 396 and .25D + .80C = 264 .75D + .20C = 396 and -.75D - 2.40C = -792 Adding we get... -2.2C = -396 Dividing we have... C = 180 Since C+D=660, and C=180, D=480. Did I make any mistakes? Ok then...Here's the answer: Check out the picture, we know what's in black, red's from my calculations. Oh I should explain it.... The numbers are the percent of something happenning given what was on the previous branches, so there's a .25 chance that a dog answered cat. And due to the nature of probablilty there is a .727*.25 chance that you will randomly pick a dog that answered cat Now the total number AC is: .4 * 660 = .8c +.2d Using d = 660 - c: .4*660 = .8c + .2(660 - c) .4*660 = .8c + 165-.25c Solve it all: c = .273 - That's the pecent of cats and of course since c% + d% much = 1; d% = .727 now C%*660=number of cats and D%*660=number of dogs and therefore 480 cats and 180 dogs The other numbers (.454 and .546) I used to check it.

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https://skillatarms.com/moa-reticle-formula-ranging-made-easy/

math

When shooting Long Range, accurate range estimation is crucial so…I made these basic Range Estimation Tools using the MOA Reticle formula (or variations thereof). These tools allow shooters to quickly calculate the Range to Target or the Target Size based on what they are seeing in their MOA reticle scope. Below the tools, I will outline the actual MOA Reticle formulas… How to use the RANGING MOA Reticle Formula Calculator: Simple…key in the Size of the Target (in cm), the Size of the Object (in MOA) and the calculator will self populate, providing you with an accurate Range to Target (in meters). If that’s not how you roll, and you prefer inches and yards, you can use the bottom MOA Reticle Formula Calculator. How to use the Measuring MOA Reticle Formula Calculator: Simple…key in the Range to Target (m), and the Size of the Object (in MOA) and the calculator will self populate, providing you with an accurate Target Size (in centimetres). Once again, if meters and centimetres “aren’t your bag baby”, and you prefer inches and yards, you can use the bottom MOA Reticle Formula Calculator. Range Estimation & Target Size Estimation using a MOA reticle: MOA Reticle Formula for Range Estimation: Size (cm) ÷ Size (MOA) x 34.38 = Range (m) Alternate Ranging Formula Size (inches) ÷ Size (MOA) x 95.5 = Range (y) MOA Reticle Formula for Measuring Objects: Measuring Objects Formula Range (m) x Size (MOA) ÷ 34.38 = Size (cm) Alternate Measuring Objects Formula Range (y) x Size (MOA) ÷ 95.5 = Size (inches) Of note, 34.38 can (and often is) rounded to 34.4. This is because 1MIL = 3.438 MOA and this figure is also usually rounded to 3.44. However, in the Range Estimation tools, I have used the unrounded figures. NOW…FYI, I’ve also put together another Range Estimation Calculator using the MIL Dot Formula. Now before you go, don’t forget to take a look at our other resources by clicking here. P.S. I hope you found these MOA Reticle Formula Tools useful. If you did, I would appreciate it if you would SHARE this page with friends who may be interested.

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https://blogs.mathworks.com/steve/2011/07/19/jahne-test-pattern-take-3/?s_tid=blogs_rc_1

math

Earlier today I told you that I was feeling a little dense because I couldn't figure out the right parameters to use in the tanh term of this test pattern: (This is equation 10.63 in Practical Handbook on Image Processing for Scientific Applications by Bernd Jahne.) I'm grateful to reader Alex H for quickly enlightening me. He described as an approximation to a step function, where a is the location of the step and w is the width of the transition. That was very helpful. I've also realized that I misinterpreted the meaning of . In the book this is described as the "maximum radius of the pattern," and I assumed this would be fixed as the distance from the center of the square image to one of its corners. But now I realize that the author intended for this to be an adjustable parameter. That is, one can set so that the maximum instantaneous frequency is reached closer to the center than at the image corners. For example, I can set the parameters so that the maximum instantaneous frequency of is reached in the center of the image edges, and then the tapering function prevents aliasing artifacts from appearing as you move out to the corners. The book's figure 10.23 is based on maximum instantaneous frequency of (a period of 2.5 samples) reached at the edges, so I'll use that. [x,y] = meshgrid(-200:200); km = 0.8*pi; rm = 200; w = rm/10; term1 = sin( (km * r.^2) / (2 * rm) ); term2 = 0.5*tanh((rm - r)/w) + 0.5; g = term1 .* term2; imshow(g,) Finally, a result that looks like the figure in the book! Thanks again, Alex. To leave a comment, please click here to sign in to your MathWorks Account or create a new one.

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https://www.uni-kassel.de/eecs/ies/aktuelles/2024/01/29/neuer-konferenzbeitrag-auf-der-international-conference-on-bioinformatics-and-computational-biology-icbcb-2024?cHash=d1b1bc9f271c377b28f938d0be61157f

math

Neuer Konferenzbeitrag auf der "International Conference on Bioinformatics and Computational Biology (ICBCB) 2024" Der Artikel mit dem Titel "Self-awareness in Cyber-Physical Systems: Recent Developments and Open Challenges" von Birk Martin Magnussen, Frank Möckel, Maik Jessulat, Claudius Stern und Bernhard Sick beinhaltet Folgendes: Obesity is a common problem for many people. In order to assist people combating obesity, providing methods to quickly, easily, and inexpensively assess their body composition is important. This article investigates how noninvasive, optical sensors based on multiple spatially resolved reflection spectroscopy can be used to measure the body mass index and body composition parameters. Using machine learning to train continuous feature networks, it is possible to predict the body mass index of a subject, with a correlation of $R=0.61$ with $p<0.0001$. Similarly, the predicted body mass index shows correlations to both the subject's visceral fat ($R=0.44,p=0.0023$) and skeletal muscle mass index ($R=0.52,p=0.0003$), indicating that the trained neural network is capable of identifying both types of tissue. Strategies to independently detect either type of tissue are discussed.

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http://www.sciencepublishinggroup.com/journal/paperinfo?journalid=147&doi=10.11648/j.acm.20160502.17

math

Tuncay Can’s Approximation Method to Obtain Initial Basic Feasible Solution to Transport Problem Applied and Computational Mathematics Volume 5, Issue 2, April 2016, Pages: 78-82 Received: Apr. 14, 2016; Accepted: Apr. 27, 2016; Published: May 12, 2016 Views 4174 Downloads 170 Tuncay Can, Department of Econometrics, Faculty of Economics, Marmara University, Istanbul, Turkey Habip Koçak, Department of Econometrics, Faculty of Economics, Marmara University, Istanbul, Turkey Obtaining an initial basic feasible solution to a transport problem – or a corner point in the convex polytope region – is extremely important in terms of reaching the optimal solution to the problem in the shortest time. When a transport problem is basically accepted as a linear programming problem, a degenerated solution is caused by the structure of the simplex method used when modelling with linear programming and located in a corner point sometimes at the optimal solution itself but mostly in close proximity to the optimal solution vector. One of the ways to eliminate this degenerated solution is to employ approximation methods. The main aim of this paper is to introduce Tuncay Can’s approximation method, which was developed as an alternative to the approximation methods in the literature for a balanced transport problem. Tuncay Can’s approximation method usually has less iterations than other approximation methods. In this paper, the Tuncay Can approximation method is introduced as an alternative to The North West Corner Rule, Minimum Cost Method, and the RAM and VAM methods. Tuncay Can’s Approximation Method to Obtain Initial Basic Feasible Solution to Transport Problem, Applied and Computational Mathematics. Vol. 5, No. 2, 2016, pp. 78-82. T. Can, “Yöneylem Araştırması, Nedensellik Üzerine Diyaloglar I”, Beta Publications, Istanbul, p. 396, 2015. M., Kırca, A. Satır, “A Heuristic for Obtaining and Initial Solution for the Transportation Problem”, Journal of the Operational Research Society, Vol. 41, No. 9, pp. 865-871, 1990. K. N., Krishnaswamy, A. I., Sivakumar, M., Mathirajan, Management Research Methodology, Dorling Kindersley Pvt. Ltd., p. 246, 2009. S. Korukoğlu, S., Ballı, “An Improved Vogel’s Approximation Method for the Transportation Problem”, Mathematical and Computational Applications, Vol. 16, No. 2, pp. 370-381, 2011. S., Sood, K., Jain, “The Maximum Difference Method to find Initial Basic Feasible Solution For Transportation Problem”, Asian Journal of Management Sciences, 03 (07), pp. 8-11, 2015. S., Singh, G. C., Dubey, R., Shrivastava, R, “Optimization and Analysis of Some Variants Through Vogel’s Approximation Method (VAM)”, IOSR Journal of Engineering, Vol. 2, Issue 9, pp. 20-30, 2012. N., Balakrishnan, “Modified Vogel’s Approximation Method for the Unbalanced Transportation Problem”, Applied Mathematics Letters, Vol. 3, Issue. 2, pp. 9-11, 1990. Z. A. N. S., Juman, M. A., Hoque, “An Efficient Heuristic to Obtain a Better Initial Feasible Solution to the Transportation Problem”, Applied Soft Computing, Vol. 34, pp. 813-826, 2015. B. G., Dantzig, N. M., Thapa, Linear Programming 1. Introduction, Springer-Verlag New York, p. 214, 1997. M. Zangiabadi, T. Rabie,” A New Model for Transportation Problem with Qualitative Data”, Iran. Journal of Operations Research, 3 (2), pp. 33–46 (2012). V. J. Sudhakar, N. Arunsankar, T. Karpagam, “A New Approach for Finding An Optimal Solution for Transportation Problems, Europan Journal of Sciences Research, 68 (2), pp. 254–257, 2012. V. Srinivasan, G. L. Thompson, “Cost Operator Algorithms for the Transportation Problem”, Mathematical Programming, 12, pp. 372–391, 1977. N. Sen, T. Som, B. Sinha, “A Study of Transportation Problem for an Essential Item of Southern Part of North Eastern Region of India as an OR Model and Use of Object Oriented Programming”, International Jorunal of Computation Sciences Network Security, 10 (4), pp. 78-86, 2010. F. Pargar, N. Javadian, A. P. Ganji, “A Heuristic for Obtaining an Initial Solution for the Transportation Problem with Experimental Analysis”, in: The 6th International Industrial Engineering Conference, Sharif University of Technology, Tehran, Iran, 2009. M. Mathirajan, B. Meenakshi, “Experimental Analysis of Some Variants of Vogel’s Approximation Method”, Asia-Pacian. Journal of Operations Research, 21 (4), p. 447–462, 2004.

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https://www.iit.edu/events/applied-mathematics-colloquia-art-owen-recent-progress-error-estimation-quasi-monte-carlo

math

Applied Mathematics Colloquia by Art Owen: Recent Progress in Error Estimation for Quasi-Monte Carlo Art Owen, professor of statistics, Stanford University Recent Progress in Error Estimation for Quasi-Monte Carlo For many high dimensional integration problems, Quasi Monte Carlo methods attain the best accuracy. In settings where high accuracy is required it is also valuable to show that it has been attained. Those two criteria are somewhat at odds with each other. This talk looks at recent ways to quantify the accuracy attained by QMC. It includes progress in forming confidence intervals based on replication of randomized QMC. The surprise there is that a plain student's t based confidence interval method proved to be much more reliable than some bootstrap methods that were expected to be best. A second area of recent progress provides computable and provable upper and lower bounds on the integral. These methods require special QMC points that have a non-negative local discrepancy property along with an integrand that has a complete monotonicity property. At present, some of the best QMC accuracy results arise for a median of means strategy. That raises a severe open challenge of quantifying the uncertainty in a mean when one has computed a median. This talk includes joint work with Michael Gnewuch, Peter Kritzer, Zexin Pan, Pierre L'Ecuyer, Marvin Nakayama and Bruno Tuffin. Applied Mathematics Colloquia

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https://scholars.hkbu.edu.hk/en/publications/variance-based-modified-backward-forward-algorithm-with-line-sear

math

We propose a variance-based modified backward-forward algorithm with a stochastic approximation version of Armijo's line search, which is robust with respect to an unknown Lipschitz constant, for solving a class of stochastic variational inequality problems. A salient feature of the proposed algorithm is to compute only one projection and two independent queries of a stochastic oracle at each iteration. We analyze the proposed algorithm for its asymptotic convergence, sublinear convergence rate in terms of the mean natural residual function, and optimal oracle complexity under moderate conditions. We also discuss the linear convergence rate with finite computational budget for the proposed algorithm without strong monotonicity. Preliminary numerical experiments indicate that the proposed algorithm is competitive with some existing algorithms. Furthermore, we consider an application in dealing with an equilibrium problem in stochastic natural gas trading market. Scopus Subject Areas - Management Science and Operations Research - linear convergence rate - modified backward-forward algorithm - stochastic approximation - stochastic natural gas trading market - Stochastic variational inequality

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https://www.hackmath.net/en/math-problem/1341

math

I threw three chicken for a handful of grains and I noticed that it eat in a ratio of 8:7:6 and two of them grabbed 156 grains. How many grains have fought that chicken? Thank you for submitting an example text correction or rephasing. We will review the example in a short time and work on the publish it. Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Showing 0 comments: Be the first to comment! Tips to related online calculators Following knowledge from mathematics are needed to solve this word math problem: Next similar math problems: viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with viju?? - Three days During the three days sold in stationery 1490 workbooks. The first day sold about workbooks more than third day. The second day 190 workbooks sold less than third day. How many workbooks sold during each day? - Hotel rooms In the 45 rooms, there were 169 guests, some rooms were three-bedrooms and some five-bedrooms. How many rooms were? - Three friends The three friends spent 600 KC in a teahouse. Thomas paid twice as much as Paul. Paul a half less than Zdeněk. How many each paid? - Fifth of the number The fifth of the number is by 24 less than that number. What is the number? - Mother and daughter The ratio of years mother and daughter is 5:2. After 7 years the ratio is 2: 1. How many years ago daughter was born? - Two numbers Find two numbers whose difference and ratio is 2. If I going to translate the book 6 pages per day I translate it 4 days earlier than if I translated 5 pages a day. If I translate 4 pages a day I translate it for how many days.....? - 1.5 divided 1.5 divided by 1 = w divided by 4 - ATC camp The owner of the campsite offers 79 places in 22 cabins. How many of them are triple and quadruple? solve equations by substitution: x+y= 11 y=5x-25 - Mom and daughter Mother is 39 years old. Her daughter is 15 years. For many years will mother be four times older than the daughter? Along the road were planted 250 trees of two types. Cherry for 60 CZK apiece and apple 50 CZK apiece. The entire plantation cost 12,800 CZK. How many was cherries and apples? 120 vehicles parking on the morning. Pasenger car is charged 20 CZK, 50 CZK per bus. The guard collected for parking 2640 CZK in total. How many cars and how many buses stood in the parking? There are eighty more girls in the class than boys. Boys are 40 percent and girls are 60 percent. How many are boys and how many girls? Solve following system of equations: 6(x+7)+4(y-5)=12 2(x+y)-3(-2x+4y)=-44 For five days, we have collected 410 mushrooms. Interestingly every day we have collected 10 mushrooms more than the preceding day. How many mushrooms we have collected during 4th day?

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http://ngonevuwhank.mihanblog.com/post/60

math

Physics for entertainment Volume 2 by Ya perelman Physics for entertainment Volume 2 Ya perelman ebook Publisher: Mir Moscow Mathematical Physics of Quantum Mechanics, Asch.pdf. Mathematical Tools for Physicists Wiley.pdf. Some modern mathematics vol2 Roman .djvu. Physics for entertainment Volume 2 book download Download Physics for entertainment Volume 2 Date:. Volume detect, however, is Very Bad (TM) as it not only requires Havok to test collisions with everything but it requires additionally computation to determine when scripts need to be notified. In this the Physics for Entertainment, has been published as a two volume series, especially in the regional languages of India. The Mathematical Beauty of Physics Drouffe.djvu . Path integrals book 3Ed Chapters 1 to 7.djvu. Problems from H C Verma's Concepts of Physics is considered a must work out assignment by most of the IIT aspirants. Methods of modern mathematical physics.Vol 1 ,2,3,4Simon .djvu. Mathematics for Physics I and II Lecture Notes.pdf. Secondlife Art and Entertainment Seeing Rob Danton play with the physics in the upgraded Havok engine make me very hopeful for the potential of experimental art within Second Life. Physics for Entertainment v1 Yakov Perelman.djvu. With the new 2) Phantom objects are tracked by the sim and do collide with terrain. Photography Posing Secrets Volume 2, Arts & Entertainment The Second Of Two Volumes Showing Photographers How To Pose A Model For The Camera. Physics for Entertainment By Yakov Perelman [Vol. View bigger - HC Verma solutions Vol 2 for Android screenshot.

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http://laundrylynx.com/epub/b-0695-mathematical-modelling-of-weld-phenomena-4

math

By H. Cerjak Read or Download B0695 Mathematical modelling of weld phenomena 4 PDF Similar nonfiction_12 books The tenth Quantum arithmetic overseas convention (Qmath10) gave a chance to assemble experts drawn to that a part of mathematical physics that is in shut reference to numerous points of quantum concept. It was once additionally intended to introduce younger scientists and new trends within the box. This is often the 1st finished two-volume assortment on anhedonia, a sickness that performed a huge position in psychopathology theories first and foremost of the 20 th century. Anhedonia is a within which the means of enjoyment is partly or thoroughly misplaced, and it refers to either a character trait, and a “state symptom” in quite a few neuropsychiatric and actual issues. Verena Puchner evaluates and compares statistical matching and chosen SAE tools. on the grounds that poverty estimation at local point in line with EU-SILC samples isn't of enough accuracy, the standard of the estimations could be more suitable through also incorporating micro census information. the purpose is to discover the simplest procedure for the estimation of poverty when it comes to small bias and small variance using a simulated synthetic "close-to-reality" inhabitants. - Mastering Skateboarding : tricks for flatland, ledges, rails, ramps and bowls and more - Bacterial Pili: Structure, Synthesis and Role in Disease - Part A - Psyched - Inner Views of Winning Extra resources for B0695 Mathematical modelling of weld phenomena 4 Equation (5) guarantees the continuity of Vp/ p at the discontinuity. By this procedure, we can treat all the material at once by simply changing its equation of state. We note again that this property is a consequence of the separate treatment of advection and non-advection terms, otherwise the continuity of Vp/ p is not guaranteed and a large density can not be traced. 3 Elastic-Plastic Elastic motion can be included also in Eqn (1) as Model Q _ ~ OSij U P 8xj - where s is the stress tensor. The time development of stress is calculated to be dSij . 8 ~ :::J ~ -2 D.. 0048 Wt% 0 , -. ~,; ...... , "-:, Q) Q) . 0030 Wt% o '0 C Q) 4 Modelling of Weld Phenomena ........ * .... •.... * E Q) J- 1800 . ..... -:-... : * * * 2000 2200 2400 Temperature : * ~* 2600 . * 2800 3000 in K Fig. 2 Dependence of the coefficient of surface tension on temperature surface activity. and where kz is a constant related to the entropy of segregation. A plot of the coefficient of surface tension on temperature for different activities of sulphur can be observed in Fig. 42. G. Bendszak: private communication. 43. U. Dilthey, V. Pavlik and T. Reichel: Mathematical Modelling of Weld Phenomena 3, H. , The Institute of Materials, London, 1997,85-105. 24 Mathematical 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. Modelling of Weld Phenomena 4 D. LL. Guthrie: ISIJ International, 1994, 34(5),384-392. T. Johansen and F. Boysan: Met. Trans. B, 1988, 19B, 755-764. -H. W. Chang: it Journal of Aerosol Science, 1996,27(5),681-694 Y. Arata: Proceedings of Int. B0695 Mathematical modelling of weld phenomena 4 by H. Cerjak

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https://www.osapublishing.org/josa/abstract.cfm?uri=josa-60-4-518

math

In this paper a determinate equation is derived for the propagation of a finite beam of radiation in a random medium. The radiation is described by a mutual coherence function. The analysis is restricted to beam diameters that are large compared to the characteristic correlation lengths in the random medium. A scalar theory is used and the characteristic wavelength is assumed to be very small compared to the smallest correlation length. The method used is an iteration procedure similar to that considered in the propagation of a plane wave. The resulting equation is suitable for numerical integration. © 1970 Optical Society of AmericaFull Article | PDF Article Jay Kyoon Lee and Jin Au Kong J. Opt. Soc. Am. A 2(12) 2171-2186 (1985) F. C. Lin and M. A. Fiddy J. Opt. Soc. Am. A 10(9) 1971-1983 (1993) M. Tur, A. M. Whitman, and S. Frankenthal Opt. Lett. 7(4) 171-173 (1982)

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http://abroadexport.com/daytoday/w204/wrightsville/881212186717a43a7a4ac9042bc51f1837c4-berkeley-math-54-syllabus

math

Lecturer: Per-Olof Persson, firstname.lastname@example.org. Berkeley's online course discovery platform. Search: Berkeley Math Courses. Programming (E7) Ordinary differential equations (Math 1B, Math 54) Elementary linear algebra (M54), ME 132 TEXTBOOK(S) AND/OR OTHER REQUIRED MATERIAL Class Slides Vehicle Dynamics and Control, R. Rajamani, Springer, 2006 COURSE OBJECTIVES Please see the bCourses page for Zoom links. . You will learn the core concepts of inference and computing, while working hands-on with real data including economic data, geographic data and social . Vector spaces; inner product spaces. Together, this course sequence provides a comprehensive foundation for core EECS topics in signal . . MATH W54 Linear Algebra and Differential Equations Summer 2021* Four (4) semester credits Course Description Basic linear algebra; matrix arithmetic and determinants. COURSE OBJECTIVES . With this goal in mind, we have activated the ALLY tool GSI: Ryan Theisen (email@example.com) Lab 101: M 10:00 - 11:59. Linear second-order differential equations; first-order systems with constant coefficients. Class Schedule (Fall 2022): TuTh 9:30AM - 10:59AM, Wheeler 150 - Fernando Perez, Will Fithian. But now at UCB, you meet all these people who are simply another level of genius, and you can't even comprehend how they do the things you do. Techniques include procedural abstraction; control abstraction using recursion, higher-order functions, generators, and . Eigenvalues and eigenvectors; orthogonality, symmetric matrices. No prior coursework in economics is assumed, nor is necessarily helpful. Declared Public Health major students may take graduate courses within Berkeley Public Health, pending instructor approval and pre-requisite completion. MATH 54 at the University of California, Berkeley (Berkeley) in Berkeley, California. an ability to apply knowledge of mathematics, science, and engineering (b) an ability to design and conduct experiments, as well as to analyze and interpret . Jennifer Sixt, 964 Evans, firstname.lastname@example.org Online Guidelines , describing how the course will be delivered online Textbook :Linear Algebra and Differential Equations, Second Third Custom Edition for UC Berkeley, The curriculum offers a thorough fundamental knowledge of the major fields of chemistry, covering the general areas of inorganic, organic, and physical chemistry . Math 53. Please come to office hours or consult with your GSI before sending me email about logistical concerns. In particular, we strongly recommend that the class is taken alongside Math 53, 54, or 55. Math 53 or. Be it a problem set or research question, where you don't even . Please fill out and submit a Sign-Up Form, linked below, to receive information for attending the first meeting of adjunct . Math Courses Berkeley . Credit option: Students will receive 1 unit of credit for 54 after taking 50A and 3 units of credit after taking Math 50B. Math 53, 54, Physics 7A-7B. Berkeleytime is a platform built, maintained, and run by students, just like you. vjk.atcm.modena.it; Views: 5119: Published: .07.2022: Author: vjk.atcm.modena.it: Search: table of content. Repeat Rules Course is not repeatable for credit. Lower Division Statistics. Math 53 and 54. an ability to apply knowledge of mathematics, science, and engineering (b) an ability to design and conduct experiments, as well as to analyze and interpret . Prerequisites: Three and one-half years of high school math, including trigonometry and analytic geometry. Basic linear algebra; matrix arithmetic and determinants. Credit option: Students will receive 1 unit of credit for 54 after taking 50A and 3 units of credit after taking Math 50B. For very personal issues, send email to email@example.com. Recent Semesters. Eigenvalues and eigenvectors; linear transformations, symmetric matrices. Math 54, Stat 8 3 Units) 3 or higher: none: 2 Math Calculus AB Calculus AB SUB (2 The school consistently takes SAT composite . Math 54. Syllabus CATALOG DESCRIPTION . Office Hours: Mon 5-6, Tue 3-4 in 444 Evans. These prerequisites are strictly enforced. 2 semesters or 3 quarters of calculus: Introduction to Economics: Econ 1 or 2: Lower division course(s) in macro and micro . Berkeleytime is a platform built, maintained, and run by students, just like you. Ideal preparation includes familiarity with the grammar of proofs (Mathematics 74) and real analysis (Mathematics 104). Mathematics and Statistics Adjunct Courses Adjunct Courses Learning is Mathemagical Watch on About Math 98/198 and Stat 98/198 are 1-unit Pass/No Pass courses taken in conjunction with one of the following lecture courses: Math 1AB, 10AB, 16AB, 32, 53, or 54, or Stat 134. Both books have custom Berkeley editions sold in campus bookstores, and having these custom editions is strongly encouraged. Linear Algebra and Differential Equations. * Please note: Physics 89 is a requirement for the physics major (replaces Math 54). Lasers, fiber optics, and holography. . objects appearing in Math 104 and Math 113. MATH Dept. MUSA 74 is a 2-unit class which is intended for students who have no familiarity with writing proofs, and aren't sure if they're prepared enough for upper-division classes. Designing Information Devices and Systems I. Berkeley, California 94720-3880 Tel: (510) 642-0822 / Fax: (510) . If you think you might want to pursue graduate school, particularly in economics, consider taking Math 1A/B and 53 rather than 16A/B. Multi-variable Calculus: Math 53. NOTE: The first two weeks of instructions will be remote. E7 or CS 61A, Physics 7a, Math 53, Math 54 TEXTBOOK(S) AND/OR OTHER REQUIRED MATERIAL Free reader and notes provided. You will still have to take all the quizzes and exams required for Math 54. 1. Prerequisites: 1B. College Board is a mission-driven organization representing over 6,000 of the world's leading colleges, schools, and other educational organizations work in logic can also be carried out entirely within one of the departments of Mathematics, Philosophy, Electrical Engineering and Computer Sciences (see Graduate Study in Logic at UC Berkeley) Partial derivatives . Math 54 Section Syllabus, Fall 2010 Instructor: Olga Holtz Course website: http://www.cs.berkeley.edu/oholtz/54/ GSI: Claudiu Raicu Email: firstname.lastname@example.org . Vector spaces; inner product spaces. Enrolling in Online Courses. Get notified when MATH 54 has an open seat. Math 1A & 16B or. Lectures: MWF 3:00-4:00pm, 155 Dwinelle Course Control Number: 54086 . Offices: Evans 1083 (510-642-3523) and LBNL 50E-1520 (510-495-2857). Units: 4.0. Table of Contents. Online and web-based courses are offered by dozens of academic departments and vary in length and session dates. Math 54 Section Syllabus, Fall 2010 Instructor: Olga Holtz Course website: http://www.cs.berkeley.edu/oholtz/54/ GSI: Claudiu Raicu Email: email@example.com Physics 137A Physics 110A or Physics 105 Three additional upper-division physics courses (to total at least 9 units, for an upper-division physics unit total of at least 17 units). The final exam average score was 83 (scaled to 100) The standard deviation was 18 (again if all scores were scaled to 100 Math 1A, Exam #1 University of California, Berkeley Dept Exam credit is not accepted for the Domain Emphasis or other major requirements Appears to omit analysis of, linearization of, and phase plane analysis of autonomous nonlinear systems . Yes, Math N1A, N1B, N53 and N54 taught at UC Berkeley are summer versions of Math 1A, 1B, 53, and 54; and Math W53 is the web-version of Math 53. Yes, Math N1A, N1B, N53 and N54 taught at UC Berkeley are summer versions of Math 1A, 1B, 53, and 54; and Math W53 is the web-version of Math 53. In addition to an introductory set of math and physics courses and a broad selection of the same chemistry courses required for the chemistry major, students pursuing the chemical biology major take general and cell biology, biochemistry, biological macromolecular synthesis, and bioinorganic chemistry. The following requirements ensure that you have the necessary background: Calculus: Math 1A & 1B. Part 1; Part 2; Part 3; Part 4; Part 5; Part 6; . Math 49 option: This enrollment option is offered only to resolve administrative issues with unit restrictions. Location: Mo tt Library 103. Instructor: Peter Bartlett. Grade Distributions. PHYSICS 7B - Physics for Scientists and Engineers: 4 - Humanities/Social Sciences Courses : 3-4: 3-4: Total: 15-16: 13-15: Junior Year; . Course Requirements: 20% Weekly Homework Assignments, due in class each Tuesday (except Tuesday March Berkeley's online course discovery platform. Summer: 6.0 hours of lecture, 2.0 hours of discussion, and 2.0 hours of laboratory per week. Data/Stat/CS/Info C8. Office Hours: Wed 1-2 in 399 Evans Hall; Thu 11-12 in 723 Sutardja-Dai Hall. Students who have not had calculus in high school are strongly advised to take the Student Learning Center's Math 98 adjunct course for Math 10A; contact the SLC for more information Search: Berkeley Math Courses. Based on changes to L&S policy, courses completed at UC Berkeley with a grade of Pass in Spring 2020, Fall 2020, Spring 2021 and Summer 2021 will count toward Data Science major requirements, including prerequisites to declare the major. Spring 2022 Mondays and Wednesdays, 6:30-8:00 pm . about the waitlist, switching sections, and so on, please contact the registrar or the Mathematics undergraduate advisor Jennifer Sixt, 964 Evans, firstname.lastname@example.org Textbook :Linear Algebra and Differential Equations, Second. Math 1A, Math 1B, Math 53, Math 54*, Physics 7A, Physics 7B and Physics 7C or their equivalents; For information on California community college equivalent courses, please visit Assist.org**. Math 54: Linear Algebra and Differential Equations. Reserved Seats section closed Current Enrollment No Reserved Seats Textbooks & Materials section closed The format of these courses are different than those offered in Fall and Spring but the content is the same so they can be used to satisfy Statistics major math prerequisites. Grade Distributions. Math 54 - Linear Algebra and Differential Equations - 2019 version The 2021 zoomester version can be found via THIS RATHER SMALL LINK TO its FAQ . 30 will be notified of their approval status by Nov We provide simple tips on effective writing from the humanities to the sciences, and offer insight on how you should write towards your specific audience Math for America Berkeley is a complement to the six-year-old Cal Teach program, which exposes math and science undergraduates to teaching as a possible career . Discussion sections : Mondays, 3-4pm and 4-5pm; 332 Evans. Additionally, taking Math 54 is not enough to be eligible for the bridge course for College of Engineering students. Lab 102: M 14:00 . Access study documents, get answers to your study questions, and connect with real tutors for MATH 54 : Linear Algebra at University Of California, Berkeley. of Mathematics University of California, Berkeley 970 Evans Hall #3840 Berkeley, CA 94720-3840 USA +1 (510) 642-6550 +1 (510) 642-8204 EE16A. COURSE OBJECTIVES . Falling behind in this course or problems with workload in other courses are not acceptable reasons. Foundations of Data Science (Data C8, also listed as COMPSCI/STAT/INFO C8) is a course that gives you a new lens through which to explore the issues and problems that you care about in the world. University of California, Berkeley Department of Mathematics Mathematics 104: Introduction to Analysis Bob Anderson Spring 2005 email@example.com Tuesday, Thursday 11:00-12:30, 71 Evans 510-642-5248 . (link is external) or equivalent course work from another 4-year college or out-of-state community college as evaluated by the Statistics Department. Continuing UC Berkeley Students <1> <2> Transfer Students <2> <3> Principles of Business: UGBA 10: One introductory business course: Calculus: Math 16A &16B or. Dept. For the first time in your life, you actually feel completely loss. Final exam status: Written final exam conducted during the scheduled final exam period. . The Statistics Major consists of 4 lower division math courses, 1 lower division statistics course, and 9 upper division courses. Syllabus Transfer students admitted prior to completion of a Math 54 equivalent will be required to complete Physics 89 at UCB. No textbook. Taking . Info University of California, Berkeley (UC Berkeley)'s MATH department has 184 courses in Course Hero with 20313 documents and 2388 answered questions. It is a large institution with an enrollment of 29,570 undergraduate students STAT C8/COMPSCI C8: Foundations of Data Science (4) Babak Ayazifar, Swupnil Kumar Sahai At least 11 units must be in the 200 series courses Prerequisites: MATH 53; EECS 16A and EECS 16B, or MATH 54 5 GPA and have completed the equivalent of all required core UC . Search through 12,000+ courses at Berkeley. Basic Physics, such as the UC Berkeley Course Physics 7a, Basic Math, such as the UC Berkeley Courses Math 53, Math 54 TEXTBOOK(S) AND/OR OTHER REQUIRED MATERIAL: Free reader and notes provided. Exam dates: Midterm July 21 Final August 12 Cheat sheet: In the exams, you can bring a hand written letter size double sided cheat Apply filters for requirements. This includes the lectures, the discussion sections, and the office hours. Instructor o ce hours: Will announce on course homepage. Lectures: TuTh 8:10am - 9:30am, room 101 Morgan. Prerequisites: Math 53, 54, 55, or permission from instructor. MATH 54 at the University of California, Berkeley (Berkeley) in Berkeley, California. Add MATH 54 to your schedule. A cluster is an approved concentration of courses in a specific field of applied mathematics. Math 54 - Linear Algebra & Differential Equations -- [4 units] Course Format: Three hours of lecture and three hours of discussion per week. Description: Basic linear algebra; matrix arithmetic and determinants.Vector spaces; inner product spaces. This includes the lectures, the discussion sections, and the office hours. Vector . STAT 154 Modern Statistical Prediction and Machine Learning 4 Units [+] STAT 155 Game Theory 3 Units [+] STAT 156 Causal Inference 4 Units [+] STAT 157 Seminar on Topics in Probability and Statistics 3 Units [+] STAT 158 The Design and Analysis of Experiments 4 Units [+] STAT 159 Reproducible and Collaborative Statistical Data Science 4 Units [+] Math 54 Linear Algebra and Differential Equations. If you have not an access to this site, please let me know. Office hours Tu 11:00 am - 12:00 noon, Th 12:00 noon - 1:00 pm, at the Student Learning Center, and by appt. Jennifer Sixt, 964 Evans, firstname.lastname@example.org Online Guidelines , describing how the course will be delivered online Textbook :Linear Algebra and Differential Equations, Second Third Custom Edition for UC Berkeley, Add MATH 54 to your schedule. . Fourier series. . Tau Beta Pi Engineering Honor Society, California Alpha Chapter As you progress in your degree and remain interested in graduate studies, you should then take Math 54 and 104. Eigenvalues and eigenvectors; orthogonality, symmetric matrices. View and compare grade distributions for each course and semester. Location: Mo tt Library 101. Recent Semesters. Syllabus CATALOG DESCRIPTION . MATH 53: Multivariable Calculus (4) MATH 54: Linear Algebra and Differential Equations (4) PB HLTH . jna.hoteleuropa.ud.it; Views: 7789: Published: .07.2022: Author: jna.hoteleuropa.ud.it: Search: table of content. This email goes only to me and the Head Teaching Assistant, Kevin Li. If you are interested in enrolling in the Spring 2022 Math 54 adjunct courses you must enroll in or be on the waiting list for the appropriate section of Math 54. CS61A, CS61B, CS 61C, CS70/Math 55, CS 188, CS 189, Math 53, Math 54, Math 110, Stat 28, Stat 20/21, Stat 133, Stat 134/140, Data 100 The goal is . Instructor: Nikhil Srivastava, email: firstname at math.obvious.edu. Students with exam credits (such as AP credit) should consider choosing a course more advanced than 1A MATH 1A or MATH N1A MATH 1B or MATH N1B MATH 1B, MATH N1B, MATH 10B, or MATH N10B Part 1; Part 2; Part 3; Part 4; Part 5; Part 6; Part 7; . a course of your interest offered by any department at UC Berkeley; there . EE 16A. Upper-division math and statistics courses for those who are adequately prepared (in order of importance) Math 110, Linear Algebra . A deficient grade in Electrical Engineering 119 may be removed by taking Electrical Engineering 118. Prerequisites: 1A-1B, 10A-10B or equivalent. Linear second-order differential equations; first-order systems with constant coefficients. Math 54, Math 110, or EE 16A+16B (or another linear algebra course), CS 70, EECS 126, or Stat 134 (or another probability course). Search: Berkeley Math 1a Exam. You can also use the direct link. When the Data Science major and minor are approved, the faculty aim to add Stat 89A (in its new, 4-unit Spring 2018 form) to the list of courses to satisfy the linear algebra . Courses. Search: Berkeley Math Courses. Fall 2022, Spring 2022, Fall 2021, Spring 2021 . The Structure and Interpretation of Computer Programs. Prerequisites: MATH 53; EECS 16A and EECS 16B, or MATH 54. Credit Restrictions: Students will receive no credit for Electrical Engineering 118 after taking Electrical Engineering 218A. No textbook. The average on midterms for this class was a 50%, versus the average for Pachter's Spring 2015 class was 92% Here are the course descriptions of these math classes at Berkeley: MATH 1A: This sequence is intended for majors in engineering and the physical sciences at (saddle) 4 Undergraduate Business Courses The Undergraduate program has . NOTE: The first two weeks of instructions will be remote. Math 54 - Linear Algebra And Differential Equations Instructor: Katrin Wehrheim Email is not a sustainable form of communication in this course Contact: via Forum, your GSI, or in person during office hours Lectures: Tue/Thu 5 - 6:30pm in 155 Dwinelle Office Hours: Tue 12:30-1:30pm in Evans 907, Tue/Thu 6:30pm-. Lecturer: Per-Olof Persson, email@example.com. . Therefore, students who are planning a double major or a minor with Astrophysics and Physics are required to complete Physics 89 in lieu of Math 54. . The format of these courses are different than those offered in Fall and Spring but the content is the same so they can be used to satisfy Statistics major math prerequisites. Catalog Description: This course and its follow-on course EE16B focus on the fundamentals of designing modern information devices and systems that interface with the real world. UC Berkeley is committed to providing robust educational experiences for all learners. Search through 12,000+ courses at Berkeley. Catalog Description: An introduction to programming and computer science focused on abstraction techniques as means to manage program complexity. Schedule Planner. "Stat 89A (in its new, 4-unit Spring 2018 form) can be used as an alternate linear algebra co-requisite for Data 100 this semester, along with Math 54 and EE 16A. Please note that taking these bridge courses is suboptimal since in reality, EECS 16A and EECS 16B are best experienced in their totality when the lab experience is integrated with the overall treatment of the material. A deficient grade in Math 54 may be removed by completing Math N54. Lectures: TuTh 8:10am - 9:30am, room 101 Morgan. As far as possible, please use Piazza for mathematical questions. Basic linear algebra; matrix arithmetic and determinants. Browse all available summer online and web-based courses in the Class Schedule. CS 61A. Students will receive no credit for Math 54 after taking Math N54 or H54. If you have then Math 54 maybe be possible to do decently well, but keep in mind that regular high school math is not as rigorous as UC Berkeley math. Math 54, Spring 2021. Chris H. Rycroft, firstname.lastname@example.org. ENGIN 7, MATH 1A, MATH 1B, and MATH 53; and MATH 54 (may be taken concurrently) 3.5 years of high school math, including trigonometry and analytic geometry. * Students who do not complete a Math 54 equivalent before transferring are required to complete Phys 89:"Introduction to Mathematical Physics" at Berkeley. You know that as long as you put in the hard work, you will do well. MATH 54 - Linear Algebra and Differential Equations (recommended) or PHYSICS 89 - Introduction to Mathematical Physics . Math 54. Stat 20 or. Math 53, 54 Multivariable . GSI : Hye Soo Choi. Special arrangements Fourier series. A minimum 2.0 overall GPA is required in all 9 upper division major courses in order to be in good standing in the major. Math 1A & 1B or. To satisfy the requirements of the major, all courses must be taken for a letter grade. Please see the bCourses page for Zoom links. E7 or CS 61A, Physics 7a, Math 53, Math 54 TEXTBOOK(S) AND/OR OTHER REQUIRED MATERIAL Free reader and notes provided. For example, a student who completed Data 8 with a C and Math 54 with a C- prior to Spring 2020, and . There are more than 15 approved clusters with the most popular being: Actuarial Sciences. How to Sign Up for the Spring 2022 Math 54 Adjunct Course. From a pure content perspective Math 54 doesn't really require a whole lot (unlike Math 53, where you definitely need a strong understanding Calc 1 and 2) but Math 54 is a more mature math class . Students must earn a minimum 3.2 UC Grade Point Average* and no lower than a C in the following: Math 1A Calculus (or the Summer version Math N1A) Math 1B Calculus II (or the Summer version Math N1B) Math 53 Multivariable Calculus (or the Summer version Math N53, or web-based version Math W53) The class must be for a letter grade to be counted toward your Public Health electives. Spring 2021. Third Custom Edition for UC Berkeley, by Lay, Nagle, Saff and Snider (includes 5e of Lay and 9e of NSS). The reason for this policy is that our syllabus builds its treatment of Differential Equations on the advanced Linear Algebra concepts which are usually not . The Applied Math program provides students the opportunity to customize their learning by selecting a cluster pathway. Prerequisites: 1B. Math Courses Berkeley . Equivalent CA Community College course work per Assist.org. Math 1A-1B; Math 53 and Math 54 (multivariable calculus and linear algebra) Economics 101A-B, the quantitative theory sequence . Course Info; Resources (books, notes, more) Academic honesty; Assignments; Exams. Get notified when MATH 54 has an open seat. Prerequisites: Math 53, 54, 55, or permission from instructor. Home; Logistics; Resources; Syllabus; Forum; Credit for this embedding goes to a former student (A.D.). Enrollment opens in February 2022. around lecture hall; with updates here . Also listed as: DATA C100, STAT C100. Tuesday/Thursday 9:30-11:00am. CS61A. More advice for PhD in Economics HERE. Chuck, as he was universally known, was on the UConn faculty for decades and served as Department Head (1997-2003) From a group of academic pioneers in 1868 to the Free Speech Movement in 1964, Berkeley is a place where the brightest minds from across the globe come together to explore, ask questions and improve the world Prerequisites required He wrote . The textbooks for this course are Lay, Linear Algebra and Its Applications and Nagle, Sa , and Snider, Fundamentals of Di erential Equations and Boundary Value Problems. The Plan of Study can be found here. The Bachelor of Science Degree in Chemistry prepares students for careers as professional chemists and serves as a foundation for careers in other fields such as biology and medicine. Grading basis: letter. A two course physics sequence: Physics 7A/7B, or Physics 5A/5B/5BL. Minor Requirements. View and compare grade distributions for each course and semester. Lectures Monday, Wednesday & Friday, 3pm-4pm in Evans 71 . Courses to Complete Before Declaring. Minimum prerequisites include multivariable calculus (Mathematics 53) and matrix algebra (Math-ematics 54). take Phys 89 during the summer before they start at Berkeley (if . bCouses: In bcourses.berkeley.edu, there is a course website. Fall 2022, Spring 2022, Fall 2021, Spring 2021 . Title: Microsoft Word - E-150-SYLLABUS-20-2020.docx Created Date: Game theory models conflict and cooperation between rational decision-making agents. Syllabus of STAT154 (Modern Statistical Prediction and Machine Learning) Instructor: Song Mei (email@example.com) Lectures: T/Th 17:00 - 18:29. Apply filters for requirements. Schedule Planner. Students who have fulfilled the first half of the Physics requirement through AP/IB test scores, transfer work, or with Physics 7A, may opt to . Search: Berkeley Math Courses. Regular Summer Sessions fees and deadlines apply to online classes. Math 54 Syllabus Instructor: Doosung Park E-mail: firstname.lastname@example.org O ce: 814 Evans . - Conditional Formatting With Checkbox Google Sheets - Parts Of Badminton Racket - Grand Eastonian Condo For Sale - Borsuk-ulam Theorem Temperature - Miraculous Ladybug Antibug - Uw--madison Psychology Research - Texas Open Fields Doctrine - Robert Williams Contract Extension - Who Owns The Barking Crab In Boston - Call Her Daddy Guest This Week May 2022 - Beach Volleyball Youth - Timberland Heritage Roll-top Boot Wheat Nubuck - Cognitive Activities For Tbi Patients

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HIGH SCHOOL EDUCATION Wassilie is considering quitting high school. Accurate studies show that a person who finishes high school averages $450,000 more income in a lifetime than someone who quits. He wondered how much an hour that meant. Given 52 weeks a year, 40 hours a week and 40 years employment what would you tell him? He isn't good enough at math to figure it out. That's why he wants to quit school. If he spends 6 hours a day, 174 days a year in high school for 4 years, how much is each hour of high school worth given the above value of a high school education? Return to Village Math

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In Response To 'Re: Find function that satisfies certain condit...' In the attached notebook c1 and c2 are functions of the variables x1 and k1, and x2 and k2 respectively. I am looking for functions c1 and c2 that fulfill the equation. Thanks for your help. Attachment: pprimeszero1.nb, URL: ,

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https://www.researchgate.net/publication/45929269_Massive_type_IIA_string_theory_cannot_be_strongly_coupled

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Massive type IIA string theory cannot be strongly coupled Ofer Aharony1, Daniel Jafferis2, Alessandro Tomasiello3,4and Alberto Zaffaroni3 1Department of Particle Physics and Astrophysics The Weizmann Institute of Science, Rehovot 76100, Israel 2School of Natural Sciences, Institute for Advanced Study, Princeton, NJ 08540, USA 3Dipartimento di Fisica, Universit` a di Milano–Bicocca, I-20126 Milano, Italy INFN, sezione di Milano–Bicocca, I-20126 Milano, Italy 4Albert Einstein Minerva Center, Weizmann Institute of Science, Rehovot 76100, Israel Understanding the strong coupling limit of massive type IIA string theory is a longstand- ing problem. We argue that perhaps this problem does not exist; namely, there may be no strongly coupled solutions of the massive theory. We show explicitly that massive type IIA string theory can never be strongly coupled in a weakly curved region of space-time. We illustrate our general claim with two classes of massive solutions in AdS4×CP3: one, previously known, with N = 1 supersymmetry, and a new one with N = 2 supersym- metry. Both solutions are dual to d = 3 Chern–Simons–matter theories. In both these massive examples, as the rank N of the gauge group is increased, the dilaton initially increases in the same way as in the corresponding massless case; before it can reach the M–theory regime, however, it enters a second regime, in which the dilaton decreases even as N increases. In the N = 2 case, we find supersymmetry–preserving gauge–invariant monopole operators whose mass is independent of N. This predicts the existence of branes which stay light even when the dilaton decreases. We show that, on the gravity side, these states originate from D2–D0 bound states wrapping the vanishing two–cycle of a conifold singularity that develops at large N. arXiv:1007.2451v1 [hep-th] 14 Jul 2010 1Introduction and summary of results One of the most striking aspects of string theory is its uniqueness, realized by the fa- mous “web of dualities” that interconnect its various perturbative realizations. A famous thread in this web connects weakly coupled, perturbative type IIA string theory with its strong coupling limit, M theory (which reduces at low energies to eleven–dimensional It has been known for a while, however, that this duality does not work when the Romans mass parameter F0 , which can be thought of as a space-filling Ramond- Ramond (RR) 10-form flux, is switched on. There is no candidate parameter in eleven– dimensional supergravity to match with F0, unlike for all the other fluxes; nor is there any massive deformation of the eleven–dimensional theory [2–4]. And, from the type IIA point of view, the D0-branes which give rise to the momentum modes in the eleventh dimension at strong coupling do not exist in the massive theory (as there is a tadpole for their worldvolume gauge field). This would then appear to be an imperfection in our understanding of string duality: it would be one string theory whose strong coupling limit is not known. In this paper, we will argue that this strong coupling limit may not exist, and we will show this explicitly at least at the level of weakly curved solutions. In general these are the only solutions we have any control over, unless we have a large amount of supersymmetry; one can separately consider cases with a large amount of supersymmetry, and none of them seem to lead to strong coupling either. (The type I’ theory of contains in some of its vacua strongly coupled regions of massive type IIA string theory, but these regions have a varying dilaton and their size is never larger than the string scale.) Thus, we claim that there is no reason to believe that any strongly coupled solutions exist (with the exception of solutions with small strongly coupled regions), and we conjecture that there are none. This is consistent with the fact that no suggestion for an alternative description of the massive theory at strong coupling is known. In section 2 we provide a simple argument that the string coupling gs in massive type IIA string theory must be small, if the curvature is small. Generically, we find that equations of motion and flux quantization. < ls/R, where R is a local radius of curvature. The argument just uses the supergravity This result is in striking contrast with what happens in the massless case. In the ten–dimensional massless vacuum, for example, the dilaton is a free parameter, and in particular it can be made large, resulting in the M–theory phase mentioned earlier. The massive theory has no such vacua. It is of interest to consider examples with AdS4 factors, where we can take advan- tage of a dual field theory interpretation via the AdS/CFT correspondence, which also provides a non-perturbative definition for the corresponding string theory backgrounds. In particular, it is natural to consider solutions like the N = 6 supersymmetric solution AdS4× CP3of the massless type IIA string theory [6–8]. In this solution, the dilaton is determined by the internal flux integers k ∝? large dilaton with small curvature. In this limit, the solution is better described as the AdS4× S7/ZkM–theory background. The dual field theory has been identified in as the N = 6 superconformal Chern-Simons-matter theory with gauge group U(N)×U(N) and Chern-Simons couplings k and −k. Massive type IIA solutions are also known on AdS4×CP3, and it is natural to compare their behavior to the massless case. For example, some solutions with N = 1 supersym- metry are known explicitly [10,11]; they contain the N = 6 solution as a particular case. The field theory duals are Chern–Simons–matter theories whose levels do not sum up to zero. Even though F0is quantized as n0/(2πls), one might think that introducing the smallest quantum of it, say n0= 1, should have little effect on the solutions, if the other flux integers k and N are already very large. It would seem, then, difficult to understand how a massless solution with large dilaton can suddenly turn into a massive solution with small dilaton when n0is turned on. CP1F2and N ∝? CP3F6, gs∼ N1/4/k5/4, whereas the curvature radius R/ls ∼ N1/4/k1/4. In particular, for N ? k5one has a As we will see in section 3, in general this “small deformation” intuition is flawed. When trying to express the dilaton in terms of the flux parameters, in the massive case one ends up with expressions in which F0multiplies other, large flux parameters. Hence F0 can have a large effect on the behavior of the solutions even if it is the smallest allowed quantum. As it turns out, as we increase N, the dilaton does start growing as gs ∼ N1/4/k5/4, as in the massless case. But, before it can become large, gsenters a second phase, where it starts decreasing with N. Specifically, for N larger than the “critical value” k3/n2 0, we have gs∼ N−1/6n−5/6 . Both behaviors are visible in figure 2. Notice that what happens for these N = 1 solutions is not entirely a consequence of the general argument in section 2. One could have found, for example, that for large N the radius of curvature became small in string units. In such a situation, our supergravity argument would not have been able to rule out a large dilaton; even worse, it would actually generically predict it to be large. It is interesting to ask whether there are situations where that happens. Of course, one would not trust such strongly–curved, strongly–coupled solutions, since we have no control over them; but, if they existed, they would suggest that perhaps strongly coupled solutions do exist and need to be understood. To look for such a different behavior, we turn to a second class of massive solutions, still on AdS4× CP3, but this time with N = 2 supersymmetry. Such gravity solutions were predicted to exist via AdS/CFT , and found as first–order perturbations in F0 of the N = 6 solution in . The field theory duals are again Chern–Simons–matter theories whose levels do not sum up to zero. In section 4 we point out that these theories have certain gauge–invariant monopole operators, whose mass (which is protected by supersymmetry) is independent of the rank N. This suggests the existence of wrapped branes that remain light in the large N limit. This cannot happen for backgrounds which are both weakly–coupled and weakly–curved. To see what happens at large N, in section 5 we find these N = 2 gravity solutions, generalizing the construction in . We reduce the equations of motion and supersymme- try equations to a system of three ODEs for three functions, which we study numerically. As in section 3, we then study the behavior of gsas a function of the flux integers. We find exactly the same phenomenon as in section 3: gsfollows initially the same growth observed for the N = 6 solutions, and departs from that behavior before it can get large. The existence of the light states found in section 4 is not a consequence of strong coupling, but is instead explained by the fact that the internal space develops a conifold singularity where branes can wrap a small cycle. We compute numerically the mass of D2–D0 bound states wrapping the vanishing cycle, and we reproduce very accurately the mass predicted in section 4 from AdS/CFT. Hence, in both examples we examined, the curvature stays bounded almost every- where, and the dilaton does not become strongly coupled. Our argument in section 2 does not rule out the possibility of solutions with large curvature and large dilaton, and it would be nice to find a way to rule them out. In general, such solutions would not be trustworthy, but in some situations one might understand them via chains of dualities. For example, in some cases it might be possible to T–dualize to a massless solution with small curvature, which in turn might be liftable to M–theory, along the lines of . The behavior found in the two examples analyzed in this paper may not be universal, and we expect the AdS/CFT correspondence to be very helpful in any further progress. One motivation for understanding the strong coupling limit of massive type IIA string theory is the Sakai–Sugimoto model of holographic QCD, which has Nf D8–branes separating a region of space with F0 = 0 from a region with F0 = Nf/(2πls). The solution of this model is known in the IR, where it is weakly coupled and weakly curved and the D8–branes may be treated as probes; but it is not clear what happens in the UV, where, before putting in the D8–branes, the coupling became large (see for an analysis of the leading order back-reaction of the D8–branes in this model). Our analysis rules out the possibility that the region of massive type IIA string theory between the D8–branes becomes strongly coupled while remaining weakly curved in the UV. It would be interesting to understand whether there is a sensible UV completion of this model, and, if so, what it looks like. 2 A general bound on the dilaton In this section, we will find a bound for the dilaton for type IIA solutions with non-zero 0-form flux F0?= 0, assuming that the ten–dimensional curvature is small. The argument is simply based on the equations of motion of type IIA supergravity. Note that due to supersymmetry, these equations are actually exact (at two-derivative order) and can be trusted even when the coupling constant becomes large. The Einstein equations of motion in the string frame take the form RMN+ 2∇M∇Nφ −1 2(k − 1)!FMM2...MkFNM2...Mk− The equations (2.1) are valid at every point in spacetime, away from possible branes or orientifolds. On such objects, we would need to include further localized terms, but they will not be needed in what follows. In fact, all we need is a certain linear combination: let us multiply (2.1) by e0Me0N, where e are the inverse vielbeine; 0 is a frame index in the time direction. We can now use frame indices to massage T00on the right hand side: 2(k − 1)!F0A2...AkF0A2...Ak+ (k − 1)!F0A2...AkF0A2...Ak− η00 2(k − 1)!F0A2...AkF0A2...Ak+ We have defined the decomposition Fk= e0∧F0,k−1+F⊥,k. (In particular, F⊥,0is simply F0.) Applying this to (2.1), we get RMN+ 2∇M∇Nφ −1 Again, this is satisfied at every spacetime point (away from possible sources): there is no integral in (2.4). RMN needs to be small in the supergravity approximation. In fact, all the remaining terms in the parenthesis on the left-hand side need to be small too: they are all two–derivative NS–NS terms. If any of them is large in string units, we cannot trust the two–derivative action any more; hence that parenthesis needs to be ? l−2 On the other hand, when F0?= 0, the right-hand side of (2.4) is at least of order one in string units. To see this, recall that RR fluxes are quantized, in appropriate sense. The Fkare actually not closed under d, but under (d − H∧). However, the fluxes ˜Fk= e−B(F0+ F2+ F4+ F6) are closed; these satisfy then the quantization law ˜Fk= nk(2πls)k−1, (2.6) where nkare integers and Caare closed cycles. In particular, F0= n0/(2πls). Since the right-hand side of (2.4) is a sum of positive terms, we get that it is > 1/l2 order one factors). s(up to irrelevant Let us now put these remarks together. Since the parenthesis on the left-hand side is s, and the right-hand side is > 1/l2 s, we have eφ? 1 . (2.7) For generic solutions, the parenthesis on the left hand side of (2.4) will be of order 1/R2, where R is a local radius of curvature. In that case, we can estimate, then, which of course agrees with (2.7). When F0= 0, the conclusion (2.7) is not valid because all the remaining terms on the right hand side can be made small, in spite of flux quantization. For example, assume all the components of the metric are of the same order 1/R2everywhere, and that H = 0. Then, the integral of F is an integer nk, but the value of F2 kat a point will be of order (nk/Rk)2(in string units). At large R, this can be made arbitrarily small. This is what happens in most type IIA flux compactifications with F0= 0; the dilaton can then be made large, and the limit φ → ∞ reveals a new phase of string theory, approximated by To summarize, we have shown that F0?= 0 implies that the dilaton is small (2.7), as long as the two–derivative action (the supergravity approximation) is valid. 3 The N = 1 solutions In this section, we will see how the general arguments of section 2 are implemented in the N = 1 vacua of . 3.1 The N = 1 solutions We recall here briefly the main features of the N = 1 solutions in on AdS4× CP3. The metric is simply a product: Topologically, CP3is an S2fibration over S4. We use this fact to write the internal metric where xiare such that?3 radius, related to the AdS radius by i=1(xi)2= 1, Aiare the components of an SU(2) connection on S4(with p1= 1), and ds2 S4 is the round metric on S4(with radius one). R is an overall RAdS≡ L =R (2σ + 1). (3.3) The parameter σ in (3.2) is in the interval [2/5,2]; this implies, in particular, that L/R is of order 1 for these N = 1 solutions. For σ = 2, (3.2) is the usual Fubini–Study metric, whose isometry group is SU(4) ? SO(6). For σ ?= 2, the isometry group is simply the SO(5) that rotates the base S4. The metric (3.1) depends on the two parameters L and σ. A third parameter in the supergravity solution is the string coupling gs. Yet another parameter comes from the B field. For 2/5 < σ < 2, supersymmetry requires the NS-NS 3-form H to be non–zero (see [10, Eq. (2.2)]). One can solve that constraint by writing ?(2 − σ)(σ − 2/5) σ + 2 where β is a closed two–form [10, Eq. (4.5)]. Because of gauge invariance B∼= B + dλ1, the space of such β is nothing but the second de Rham cohomology of the internal space, H2(CP3) = R. So we have one such parameter, which we can take to be the integral of β over the generating two–cycle in H2, B = − J + β (3.4) where we normalized b so that large gauge transformations shift it by an integer. To summarize, the N = 1 supergravity solutions depend on the four parameters 3.2 Inverting the flux quantization equations We now apply the flux quantization conditions (2.6). It is convenient to separate the contribution from the zero–mode β: k(2πls)k−1, which can be computed explicitly . We have we then define?˜Fk|β=0≡ nb 10 0 0 b1 0 0 b 1 0 l = L/(2πls) ,(3.8) (2 − σ)(5σ − 2) (2σ + 1) √2σ + 1 , ,f4(σ) = −25π2 3 · 52 (σ − 1)(2σ + 1)5/2 σ2(σ + 2)2 (2 − σ)(5σ − 2) , (σ − 1) (σ + 2) f6(σ) = −27π3 3 · 57/2 (σ2− 12σ − 4)(2σ + 1)7/2 σ2(σ + 2)2 Equation (3.7) is [12, Eq. (4.26)], which in this paper we chose to reexpress in terms of l (the AdS radius in string units) rather than r (the internal size in string units), to harmonize notation with section 5. We want to invert these formulas and get expressions for the parameters (l,gs,σ,b) in terms of the flux integers ni, as explicitly as possible. If one assumes b = 0, this is easy ; with b ?= 0, it is a bit more complicated. A good strategy is to consider combinations of the flux integers that do not change under changes of the b field: in addition to n0, two other combinations are We then find 2− 2n0n4= (f2 ?2(σ − 1)(4σ2− 1) ?3(−6 + 17σ − 6σ2)(2σ + 1)3/2 0n6− 3n0n2n4= (f3 We see that (3.11) and (3.12) give two independent expressions for l/gs; this implies 64(σ − 1)3(2σ − 1)3 27σ2(−6 + 17σ − 6σ2)2≡ ρ(σ) .(3.13) This determines σ implicitly in terms of the fluxes. The function ρ(σ) (which we plot in figure 1) diverges at σ =17−√145 ∼ .41, and has zeros at σ =1 have multiplicity three, and hence they are also extrema and inflection points. Moreover, it has a minimum at σ ∼ .65; and it goes to 1 for both σ = 2 and σ =2 We can now combine the equation for n0 in (3.7), which determines gsl, with the expression for l/gsin either (3.11) or (3.12). We prefer using the latter, since it turns out to contain functions of σ which are of order one on most of the parameter space: 2and σ = 1. These zeros (2 − σ)1/4(5σ − 2)1/4σ1/3 (2σ + 1)1/2(−6 + 17σ − 6σ2)1/6 (2 − σ)1/4(5σ − 2)1/4(−6 + 17σ − 6σ2)1/6 The function in the expression for l diverges at σ =17−√145 and 2, whereas the function in the expression for gsvanishes for σ =2 ∼ .41 and vanishes for σ =2 Finally, the second row of equation (3.7) determines b in terms of n2, n0 and the remaining fields l, gs and σ. One could eliminate l and gs from that expression using (3.14) and (3.15), but we will not bother to do so. Figure 1: A plot of the function ρ(σ) in (3.13). 3.3A phase transition We will start by taking for simplicity n4= 0 , (3.16) and we will call n2≡ k ,n6≡ N (3.17) as in . In this case, (3.13) reads 1 + 3Nn2 From the graph in figure 1, we see that the behavior of the solution depends crucially on k3 . If for example we have ρ(σ) ∼ 1. Looking at figure 1, we see that a possible solution is σ = 2. Around this point, ρ goes linearly; so, if we write σ = 2−δσ, we have δσ ∼ (3.15) we then have k3 . From (3.14) and l ∼ δσ1/4 This is the same behavior as in the N = 6 solution . If, on the other hand, we have ρ(σ) ∼ 0. The possible solutions are σ ? 1 or σ ?1 in the expressions for l and gsin (3.14) and (3.15) are then both of order one. We have 2. The σ–dependent functions Notice that this behavior occurs for example in the nearly K¨ ahler solutions of . For those vacua, we have l5/gs= n6and 1/(lgs) = n0, which gives the same behavior as in (3.22). Notice also that σ = 1 corresponds indeed to a nearly K¨ ahler metric. If one were to find a Chern–Simons dual to a vacuum whose only relevant fluxes are n6and n0, such as the nearly K¨ ahler solutions, it would be natural to identify n6with a rank N and n0with a Chern–Simons coupling˜k (because F0induces a Chern–Simons coupling on D2–branes). In such a dual, coupling. We see then that l and gsN in (3.22) are both functions of this˜λ, as expected. From (3.22) one can calculate the finite temperature free energy to be βF ∼ V2T2 N2 at strong coupling βF ∼ V2T2N3/2k1/2. In figure 2 we show a graph of gsas a function of N; we see both behaviors (3.20) and ˜k≡˜λ would then be the new ’t Hooft 0 , which grows with a higher power of N than in the massless case, for which Figure 2: The behavior of gsas a function of N = n6, for n2= k = 100, n4= 0 and n0= 1. We see both the growth in the first phase (3.20), for n6? n3 decay in the second phase (3.22), for n6? n3 0= 106, and the Our analysis above was limited for simplicity to the case n4= 0, but it is easy to argue that also for other values of n4, gscannot become large. Equation (3.15) tells us that and m ≡ n3 from above in the massive theory by the maximal value of |f(σ)|. If m = 0, then (3.12) implies that (−6+17σ −6σ2) also vanishes, and we can then use (3.13) to rewrite (3.15) in the form gs=˜f(σ)/n1/2 ˜ m1/4, where˜f(σ) is again bounded in the relevant range and ˜ m ≡ n2 maximal value of |˜f(σ)|, but this must be true since m and ˜ m cannot vanish at the same time (as is clear from (3.11) and (3.12)). Thus, for any integer fluxes with n0?= 0, gsis bounded from above by a number of order one. 0 m1/6, where f(σ) is bounded from above in the relevant range of values, 0n6− 3n0n2n4is an integer. Thus, if m ?= 0, then gsis clearly bounded 2−2n0n4is another integer. Thus, if ˜ m ?= 0 then gsis bounded from above by the We will now see that the “phase transition” between (3.20) and (3.22) has a sharp con- sequence on the behavior of the probe branes in the geometry. We will consider branes which are particles in AdS4and that wrap different cycles in the internal space CP3. Not all such wrapped branes are consistent. In the N = 6 case, where F0= 0 and tadpole for the world–sheet gauge field A, because of the coupling CP1F2 = n2 ?= 0, the action for a D2-brane particle wrapping the internal CP1has a R×CP1A ∧ F2= n2 (the R factor in the D2-brane worldvolume being time). D0-branes, in contrast, have no such problem. In the field theory, they correspond to gauge–invariant operators made of monopole operators and bifundamentals. For the solutions with both F0? n0?= 0 and? have a tadpole. If one considers a bound state of nD2D2 branes and nD0D0 branes, the tadpole for A is CP1F2? n2?= 0, both D2’s and D0’s A . (3.24) For relatively prime n0and n2, the minimal choice that makes this vanish is nD2= n0 and nD0 = −n2. These branes also correspond to a mix of monopole operators and bifundamentals; we will discuss analogous configurations in more detail in section 5.4. Consider now the case n0= 1, and n2= k ? 1. Here we should consider a bound state of one D2 brane and k D0 branes. In the context of AdS/CFT, all masses are naturally measured in units of the AdS mass scale mAdS≡ is of order L. The masses of a D2 and of a D0 particle would then be (setting the string scale to one) L; recall also from (3.3) that R Thus, the bound states we are considering here (the particles that have no world–sheet tadpole) have a mass of order Which of the two terms dominates? it turns out that the answer depends on which of the two phases, (3.20) or (3.22), we are considering. In both phases the ratio of the two masses is a function of A simple computation gives that, in the first phase (3.20), the D2’s mass is ∼ whereas the k D0 branes have mass k ×k. The D0’s dominate the mass, which then goes In the second phase, the D2’s mass is ∼ N2/3, whereas the k D0’s mass goes like k × N1/3. Hence the D2 dominates. The mass then goes like N2/3 Another type of branes that have no tadpole problems are D4 branes. In the field theory, these correspond to baryon operators. In AdS units, these have a mass of (3.22), which looks reasonable for a baryon. gs. Interestingly, this turns out to be of order N in both phases (3.20) and 3.5Field theory interpretation The field theories dual to the vacua analyzed in this section were proposed in . Because of the low amount of supersymmetry, we do not expect to be able to make here any useful check of this duality. However, we can use our gravity results to make some predictions about those field theories, under some assumptions. First of all, let us recall briefly the N = 1 field theories defined in . They are similar to the N = 6 theory of [9,18], in that they also have a gauge group U(N1)×U(N2). The matter content can be organized in (complexified) N = 1 superfields XI, I = 1,...,4; they transform in the (¯ N1,N2) representation of the gauge group. The biggest difference is that the Chern–Simons couplings for the two gauge groups are now unrelated: we will call them k1and −k2. For k1?= k2, it is no longer possible to achieve N = 6 supersymmetry, and there are several choices as to the amount of flavor symmetry and supersymmetry that one can preserve. In this section, we consider a choice that leads to N = 1 supersymmetry and SO(5) flavor symmetry; in the following sections we will consider a different choice, that leads to N = 2 and SO(4) flavor symmetry. This theory can be written in terms of N = 1 superfields; the superpotential then reads terms are manifestly invariant under Sp(2)=SO(5), as promised. When k1= k2≡ k, the theory has N = 6 supersymmetry when the parameters are c1= −c2= 2π/k, c3= −4π/k. For k1?= k2, this choice is no longer possible, as we already mentioned. In spite of there being only N = 1 supersymmetry, however, it was argued in that there still exists a choice of cithat makes the theory superconformal, as long as k1− k2is small enough with respect to the individual ki. KXL]. Notice that all the If we define the ’t Hooft couplings ,λ±= λ1± λ2, (3.27) the N = 6 theory would correspond to λ+= 0. The argument in then says that there is a CFT in this space of theories if λ+? λ−, although at strong coupling it is difficult to quantify just how much smaller it has to be. Let us now try to translate in terms of these field theories the “phase transition” we saw in section 3.3. To do so, we can use the dictionary (5.35) between the field theory ranks and levels on one side, and flux integers on the other. This dictionary is also valid for N = 1 theories . The phase transition in section 3.3 happens for N ∼ n3 when λ+? λ−we have n0/N ∼ λ+/λ2 −, n2/N ∼ 1/λ−. So the phase transition happens In particular, the “ABJM phase” (3.20) corresponds to λ+??λ−; the “nearly–K¨ ahler” described by the field theories described in and reviewed in this section. However, given the low amount of supersymmetry, this can only be a conjecture at this point. phase (3.22) corresponds to λ+??λ−. At strong coupling, then, there is an intermediate ?λ−? λ+? λ−, where it is possible that the second phase (3.22) is also Rather than trying to test further this correspondence, we will now turn our attention to N = 2 theories, on which there is much better control. 4 Monopoles in N = 2 Chern–Simons–matter theo- In this section, we will recall some general facts about monopole operators in Chern– Simons–matter theories, and we will apply them to a particular quiver theory, similar to the ABJM theory; its gravity dual will be examined in section 5. 4.1Construction of monopole operators in general Consider a d = 3 gauge theory with gauge group?m in the IR, then these must be dimension 2 operators in the IR. There may or may not be operators charged under the corresponding U(1)mflavor symmetry; if they exist we will call them monopole operators. i=1U(Ni). Then there are m currents, ji= ∗ Tr(Fi), which are conserved by the Bianchi identity. If the theory flows to a CFT In a conformal field theory, it is convenient to use radial quantization and consider the theory on R × S2. Let us apply the state–operator correspondence to a monopole operator, with charge vector ni. This results in a state in the theory on an S2, such that S2Tr(Fi) = 2πni. We will denote the diagonal values of Fiby wa taking the trace over the gauge group U(Ni), we have that the magnetic charges are We are interested in d = 3 N = 2 Chern–Simons–matter theories. We can take them to be weakly interacting at short distances by adding Yang–Mills terms as regulators [19–22]. In such a regulated theory, there are BPS classical configurations with the gauge field as in (4.1) and non–trivial values for the scalar fields. The BPS equations on R1,2include the Bogomolnyi equations Fi= ∗Dσi, where σiis the adjoint scalar in the vector multiplet. In R × S2, this equation is different because the metric needs to be rescaled, and the fields need to be transformed accordingly; the equations then read Fi= σivolS2. Notice in particular that the σiare constant. There are also other BPS equations, which involve the other scalars in the theory (for explicit computations for N = 3 theories, see [20, §3.2], and in N = 2 language, ). After adding the regulating Yang-Mills term, g2 N = 2 vector multiplet should be treated classically, while the chiral matter fields should Y Mbecomes small in the UV, so the not. This justifies not solving Gauss’s law in describing the “pure” monopole operator, which then behaves as if it were a local, non-gauge invariant chiral field . The scalar σiis set by the BPS equations to be equal to the inverse Chern–Simons level times the moment maps of the matter fields, and to have a spin 0 operator, one should satisfy the constraint for any bi–fundamental field Xij connecting the i-th and j-th gauge group [24–26]. In addition, the fields Xijneed to satisfy the F-term equations of the N = 2 theory. We see from (4.3) that some matter fields, which should be neutral under the background U(1), are necessarily non vanishing in the background; hence, the possibility of satisfying all equations gives a nontrivial constraint on the possible BPS monopole operators. In general, monopole operators T creating such configurations at a point will not be gauge invariant. However, they will behave exactly like local fields. Hence, they can be combined with other local operators O, to write gauge–invariant expressions of the where the indices are contracted as appropriate for the representations in which the op- Let us determine how the monopole transforms under the gauge group. This is easy to find for the Abelian factors of the theory. There is an obvious contribution to the electric charges of a monopole operator, from the Chern–Simons term?ki nikiunder the ithelectric U(1) Abelian factor. ?Tr(Ai∧ Fi). This says that a monopole with magnetic charges niwill behave like a particle with charges This result gives a constraint on the possible gauge invariant operators Tr(TO) we can obtain. If all matter is in bifundamental and adjoint representations, no gauge invariant operators can be formed from monopoles that are charged under the overall U(1), since no matter field transforms under it. Since we just computed the electric charge under the ithU(1) to be kini, the charge under the overall U(1) is?kini. Thus, if we are to form any gauge invariant operators of the form (4.5), we need to require kini= 0 .(4.6) This result will be useful in the theories with gauge group U(N1)×U(N2), which we will discuss in section 4.2. Let us now ask how the monopole will transform under the full non-Abelian group. One method to determine this is the following. One fixes a particular configuration on the sphere (breaking the gauge symmetry), one computes the charge under the whole Cartan subalgebra of the gauge group, and then one integrates over the gauge orbits. Thus monopole operators can be labelled by U(1) subgroups of the gauge groups. For the U(Ni) factor of the gauge group, the charges under this Cartan subalgebra are the wa with a = 1,...,Ni, that we saw in (4.1). Therefore, a monopole associated with magnetic flux wa weight vector [27, Sec. 4.2] iis in the representation with of the ithgauge group. (Our notation here is that the weight vector denotes the number of boxes in each row of a Young diagram of the representation; thus (k,0,...,0) is, for example, the completely symmetric representation.) Quantization in a background of the type discussed in this section can result in anoma- lous contributions to the charges and energy of the state. For the non–chiral theories we we will consider in section 4.2, there is no such a correction to the gauge charges1. How- ever, as we will discuss in section 4.2, the dimension of the monopole operator is given by the energy of the state on the sphere, which often includes a non–zero Casimir energy. 4.2 Dimensions and charges of the monopoles We will apply in this section the results of section 4.1 to the Chern–Simons–matter theories with N = 2 and N = 3 given in . In particular, we will compute the dimensions of particular monopoles, which will be useful later, when comparing to the gravity solutions of section 5. Let us first recall some details about the field theories of interest. They are similar to the N = 6 theory of [9,18], in that they also have a gauge group U(N1) × U(N2), and N = 2 “chiral” superfields Ai,Bi, i = 1,2; the Aitransform in the (¯ N1,N2) representation, 1If the matter content is chiral, as for the theories in , there will be an additional one–loop correction to the gauge charges. One way to understand this effect is that the state on the sphere has a constant value for the scalar in the vector multiplet, which gives a mass to any matter fields charged under that U(1) subgroup in which the magnetic flux lives. Integrating them out at one–loop can shift the effective Chern–Simons level in that background if the theory has chiral matter. whereas the Bitransform in the (N1,¯ N2). Just as in section 3.5, the crucial difference between the N = 6 theory and the N = 2 theory is that the Chern–Simons couplings for the two gauge groups are now unrelated; we again call them k1and −k2. The theories we want to consider in this section are defined by the superpotential W = Tr(c1(AiBi)2+ c2(BiAi)2) .(4.8) For generic ci, the theory has N = 2 supersymmetry and SU(2) flavor symmetry. For still SU(2). For c1= −c2= c, the supersymmetry stays N = 2, but W can be rewritten W = cTr(?ij?klAiBkAjBl) , ki, supersymmetry turns out to be enhanced to N = 3, while the flavor symmetry is which shows that the flavor symmetry is enhanced to SU(2)×SU(2). This N = 2 theory is dual to the gravity solution discussed in the next section. We will now apply to these theories the discussion of section 4.1 about monopole operators. The following computation is a straightforward generalization of that done in [19,20] for the N = 3 theory. The results for the N = 3 and N = 2 theories appear to be identical, since the flavor symmetry guarantees that the matter fields have the same dimensions as in the more supersymmetric theory. As we saw in section 4.1, there are non–trivial conditions on the scalars for the monopole to be BPS. For the theories we are considering, the conditions read AA†− B†B =k1 B B†− A†A = −k2 together with the F-term constraints coming from (4.8). In view of (4.1) and Fi= σivol2, such equations relate the magnetic fluxes wa can consider are defined by magnetic charges wa w1= (1,...,1,0,...) (with n11’s) and w2= (1,...,1,0,...) (with n21’s). The non-zero elements of the fields Aiand Biare n1×n2and n2×n1rectangular matrices, respectively, which are required to satisfy the first two lines in (4.10) and the F-term constraints. The problem of finding appropriate vacuum expectation values for the matter fields is equivalent to finding the BPS moduli space of the generalized U(n1) × U(n2) Klebanov– Witten theory with superpotential (4.8) and Fayet-Iliopoulos (FI) parameters turned on. Many of these moduli spaces are non–empty. 1with the wa 1which are all either 0 or 1: namely, 2. The simplest monopoles we Using now (4.6), we see that this monopole operator can be coupled to elementary fields in a gauge invariant way only if The matter content is non–chiral, so there are no anomalous contributions to the gauge charges of the monopole. There is, however, a one–loop correction to the dimension of the operator, which is given by [20–22] ∆ = −1 2 × 1(n1(N1− n1) + n2(N2− n2)) − 4 ×1 = (n1− n2)2− (N1− N2)(n1− n2) , 2 2(n1(N2− n2) + n2(N1− n1)) where R is the R-charge of the fermion and q the charge under the U(1) subgroup specified by the vectors w1= (1,..n1..,1,0,...) and w2= (1,..n2..,1,0,...). The various contribu- tions arise as follows. The four bi-fundamental fermions have R-charge −1/22. Each bi-fundamental fermion is a matrix with N1N2entries; the n1(N2− n2) + n2(N1− n1) off-diagonal entries have charge ±1 under the magnetic U(1) subgroup, while the remain- ing entries are neutral. The two adjoint gauginos have R-charge +1. They are square matrices with N2 charge ±1, while the other are neutral. We used the fact that in this theory, both for N = 3 and N = 2, the R–charges in the UV and IR are identical. In the N = 6 theory, k1= k2and it follows from (4.11) that n1= n2. The simplest monopole has just w1= w2= (1,0,...,0). We need to turn on Aiand Bifields that solve the U(1) Klebanov-Witten theory with a FI term. According to (4.7), such a monopole transforms in the k-fold symmetric representation of U(N1) and in the conjugate k-fold symmetric representation of U(N2). The monopole can combine with k fields Aito form a gauge-invariant operator (we can analogously form a gauge-invariant operator with the conjugate monopole and k fields Bi). ientries; the 2ni(Ni− ni) off-diagonal entries of the i-th fermion have In the N = 3 and N = 2 theories, we cannot have n1= n2, but we can now take n1= k2and n2= k1and rectangular matrices Aiand Bithat solve (4.10). In general, the 2Since the superpotential (4.8) must have R-charge +2 and there is a discrete symmetry between Aiand Biwe have that R(Ai) = R(Bi) = 1/2; the R-charge of the fermionic partners is R(A) − 1 by matrices AiBjwill not be diagonalizable. Recalling equation (4.7), the monopole opera- tor is in a representation of the gauge group with weight vectors (k1,..k2..,k1,0,...) and (k2,..k1..,k2,0,...). A gauge invariant combination must include k1k2matter bifundamen- tals (if k1and k2are not relatively prime, some operator with smaller dimension could exist). The total dimension of the gauge-invariant operator, dressed with k1k2elementary fields, is then + (k2− k1)2− (k2− k1)(N1− N2) .(4.13) Note that we have determined the vacuum expectation values of the matter fields needed to “support” the flux to form a BPS state on the sphere using the classical moduli space. This is justified since the Higgs branch does not receive quantum corrections. More precisely, the ring of chiral operators is the ring of algebraic functions on the moduli space. There is a natural map [24–26] from the moduli space of Chern–Simons–matter theories to the moduli space of the four–dimensional Yang–Mills theory with the same field content. That moduli space cannot receive quantum corrections (aside from wavefunction renormalization which fixes the coefficients of the superpotential), and only the S1bundle over that space, associated to the dual gauge fields, is quantum corrected. This precisely corresponds to 1-loop corrections to the charges and dimensions of monopole operators, which are, however, constructed in the UV weakly coupled Yang–Mills–Chern–Simons– Let us summarize the results of this section. The monopole operators that create k2 units of flux for the first gauge group and k1 for the second have k1k2 bifundamental indices, and hence we can contract them with k1k2bifundamental fields to construct a gauge–invariant operator. Such operators have dimension given by (4.13). In particular, they stay light when N1= N2≡ N → ∞. Since in general monopole operators correspond to D–branes, this seems to indicate a limit where D–branes become light, which usually signals some sort of breakdown of the perturbative description. We will see in section 5.4 precisely how this happens. 5 The N = 2 solution We now turn to writing and studying the N = 2 solution predicted to exist in , and found in at first order in F0. This solution will be the gravity dual of the field theory defined by the superpotential (4.9), and it will serve as another illustration of the general result of section 2. We will start in section 5.1 by reducing the equations of motion and the supersym- metry conditions to a system of three equations for three functions of one variable. This procedure closely parallels , where an analogous solution for the gravity dual of the Chern-Simons theory based on the C3/Z3quiver was found. In section 5.2, we will impose flux quantization, and derive expressions for the supergravity parameters in terms of the flux integers; in section 5.3 we find, just like in section 3.3, a “phase transition” that prevents the dilaton from growing arbitrarily large. Finally, in section 5.4 we find light D–brane states dual to the monopole operators discussed in section 4.2. 5.1 The N = 2 solutions The ten dimensional metric we will consider is a warped product of AdS4with a compact six-dimensional internal metric with the topology of CP3: As discussed in [28,29], there is a foliation of CP3in copies of T1,1, which is in turn a S1fibration over S2× S2. The usual Fubini–Study metric can be written as where Ai, i = 1,2, are one-form connections, with curvatures 16sin2(2t)(da + A2− A1)2, (5.2) where Jiare the volume forms of the two spheres S2 interval [0,π/2]; all the functions in our solution (including A in (5.1)) will depend on this coordinate alone. At one end of the interval [0,π/2], one S2shrinks; at the other end, the other S2shrinks. To make this metric regular, we take the periodicity of a to be 4π. i. The coordinate t parametrizes the The Fubini–Study metric is appropriate for the N = 6 solution, which has F0= 0. Once we switch F0on, as we saw in section 4.2, AdS/CFT predicts the existence of an N = 2 solution with isometry group SU(2)×SU(2)×U(1) (the first two factors being the flavor symmetry which is manifest in (4.9), and the third being the R–symmetry). The internal metric for such a deformed solution is then given by3 64Γ2(t)(da + A2− A1)2.(5.4) 3One could have reparameterized the coordinate t so as to set one of the functions in (5.4) to a constant, for example ?, as in (5.2). We have chosen, however, to fix this reparameterization freedom in another way: by choosing the pure spinors (A.8) to be as similar as possible to those for the N = 6 solution, see in particular (A.11). Were the functions e2Binon-vanishing, we would have a metric on the total space of an S2bundle over S2× S2. To maintain the topology of CP3, we require that e2B2vanishes at t = 0 and e2B1vanishes at t = π/2. With an abuse of language, we will refer to t = 0 as the North pole and t = π/2 as the South pole, although there is no real S2fiber. To have a regular metric, ?(t) and Γ(t) must behave appropriately at the poles. It is convenient to define the combinations which control the relative sizes of the two S2’s. As discussed in Appendix A, the super- symmetry equations reduce to three coupled first order ordinary differential equations for w1, w2and a third function ψ which enters in the spinors: Ct,ψ(w1+ w2) + 2cos2(2t)w1w2 Ct,ψ(w1+ w2)cos2(2ψ) + 2w1w2 sin(4t)Ct,ψ(w1+ w2)cos2(2ψ) + 2w1w2 sin(4t)Ct,ψ(w1+ w2)cos2(2ψ) + 2w1w2 Ct,ψ(w1w2− 2w2− 2sin2(2ψ)w1) Ct,ψ(w1w2− 2w1− 2sin2(2ψ)w2) Ct,ψ≡ cos2(2t)cos2(2ψ) − 1. (5.7) All other functions in the metric and the dilaton are algebraically determined in terms √2eA(cot(ψ) − tan(ψ))csc2(2t) sin(2ψ) − cos(2ψ) cot(2t)ψ? Γ = 4eAsin(2t) + cos(2t)cot(2t)sin2(2ψ) 2?1 + cot2(2t) sin2(2ψ) F0csc(4t) sec(2ψ) tan(2ψ) Here, c is an integration constant, that so far is arbitrary. The fluxes are determined as well, and have the general form 2?1 + cot2(2t) sin2(2ψ) e3A−Φ= csec(2ψ) 1 + cot2(2t) sin2(2ψ) . (5.11) F2= k2(t)e2B1J1+ g2(t)e2B2J2+˜k2(t)i 2z ∧ ¯ z , 2z ∧ ¯ z ∧ J1+ ˜ g4(t)e2B2i F4= k4(t)e2B1+2B2J1∧ J2+˜k4(t)e2B1i 2z ∧ ¯ z ∧ J2, 2z ∧ ¯ z ∧ J1∧ J2, ˜ gican be found in (A.12). The fluxes satisfy the Bianchi identities, which require that 2z ∧ ¯ z = 16√2dt∧(da+A2−A1). The full expressions for the coefficients ki,˜ki, gi, ˜F ≡ e−B(F0+ F2+ F4+ F6) (5.13) is closed. This dictates in particular that F0is constant. We can now study the regularity of the differential equation near its special points, t = 0 and t = π/2, by finding a power series solution of the equations. The general solution will depend on three arbitrary constants. However, we are after solutions with particular topology, where w2vanishes at t = 0 (the “North pole”) and w1vanishes at t = π/2 (the “South pole”). Near t = 0, we obtain ψ = ψ1t −2 w1= w0+ (4 + 4ψ2 w2= (4 + 4ψ2 1− 2w0+ 2w0ψ2 1)t2+ O(t4) . In our solution, w0and ψ1are not independent: imposing that w1vanishes at t = π/2 determines w0in terms of ψ1. The power series expansion in˜t ≡ π/2 − t near t = π/2 is identical, with the role of w1and w2exchanged: ψ =˜ψ1˜t −2 w1= (4 + 4˜ψ2 w2= ˜ w0+ (4 + 4˜ψ2 1− 2 ˜ w0+ 2 ˜ w0˜ψ2 1)˜t2+ O(˜t4) . The constants ˜ w0and˜ψ1should also be determined by ψ1; we can then think of ψ1as the only parameter in the internal metric. To find a solution with the required topology, we note that the equations (5.6) are symmetric under the operation t →π w1 ↔ w2, and we look for solutions which are left invariant by this symmetry. This determines˜ψ1= ψ1and ˜ w0= w0, and it allows us to restrict the study of the equations to the “north hemisphere” t ∈ [0,π/4]. The only thing left to impose is that the solution is differentiable at t = π/4. This is what determines w0as a function of ψ1, which we plot in figure 3. w0(ψ1) is monotonicaly decreasing; our numerical analysis shows that it vanishes at a point very well approximated by ψ1=√3. 2− t,ψ → −ψ, The perturbative expansion of the solutions near the “poles” t = 0 and t = π/2 allows to check the regularity of the six-dimensional metric. In fact, the only special points in the metric are the poles, where a copy of S2degenerates. Using the previous expansion, Figure 3: A plot of w0as a function of ψ1. It vanishes linearly around the point ψ1=√3. it is straightforward to check that, at both poles, the shrinking S2combines with (t,a) to give a piece of the metric proportional to Si + (da ± Ai)2? Thanks to the fact that the periodicity of a is 4π, this is the flat metric of R4. For all ψ1∈ [0,√3) the metric is then regular. For ψ1= pole and the metric develops two conifold singularities. ψ1 = limiting point in our family of solutions. √3, both spheres degenerate at each √3 is thus the natural We can examine now the number of parameters in the solution. As discussed above, the differential equations provide just one parameter, ψ1, the value of the derivative of ψ at the North pole t = 0. It is convenient to define two more parameters by gs≡ eφ|t=0,2L = eA|t=0. (5.17) Both φ and A vary over the internal manifold, but numerical study reveals that they only do so by order one functions. So gsand L can be thought of as the order of magnitude of the dilaton and AdS radius in our solutions4. We can now reexpress the integration constant c by evaluating (5.11) at t = 0: ?1 + ψ2 4The normalization has been chosen so that in the metric, at t = 0, L2multiplies an Anti-de Sitter space of unit radius, and the relation between the mass of a particle at t = 0 and the conformal dimension of the dual operator is mL = ∆(∆−4). This normalization is related to the fact that, in our conventions, AdS4has cosmological constant Λ = −3|µ|2and, as discussed in appendix A, we chose µ = 2. The F0flux is then determined by evaluating (5.10) at t = 0: ?1 + ψ2 Finally, a fourth parameter comes from the B field. As in section 3, there is a zero– mode ambiguity coming from the presence of a non–trivial cohomology in our internal manifold. To see this, let us call B0a choice of B–field such that H = dB0solves the equations of motion. For example, we can choose B0such that ˜F2= F2− B0F0= 0 .(5.20) H = dB0is guaranteed to solve the equations of motion, since equation (5.20) implies that dF2= HF0, which we have already solved. However, this will also be true for any B of the form B = B0+ β ,(5.21) for any β which is closed. We can apply to this β the same considerations as in section 3.1: because of gauge invariance B∼= B + dλ1, the space of such β is nothing but the second de Rham cohomology of the internal space, H2(CP3) = R, so we have one such parameter. And, just as in (3.5), we define the integral of β over the generating two–cycle in H2: b ≡ not generate confusion, as the contexts are different. CP1β. The fact that we use the same notation as in section 3 should Summarizing, our solutions are parameterized by the four numbers (L,ψ1,gs,b). The situation is very similar to the N = 1 solutions we studied in section 3, with σ replaced 5.2Inverting the flux quantization equations This section will follow closely the corresponding treatment for the N = 1 solutions in section 3.2. The equations are formally very similar: b1 0 0 b 1 0 where l = L/(2πls), as in (3.8). The vector on the left hand side is given by the integrals Ck˜Fk, where Ckis the single k–cycle in CP3(k = 0,2,4,6), and˜Fkis defined using the particular B0in (5.20); this also explains why the second entry of the vector is zero (this is simply our choice for the definition of b). We could have made such a choice for the N = 1 solution as well; we did not do so because for SU(3) structure solutions there is a different and particularly natural choice of B–field. Using equation (5.19) we can write f0(ψ1) = − ?1 + ψ2 We know the other functions fk(ψ1) only numerically. We obtain 2f4(ψ1) by integrating ˜F4 over the diagonal S2times the “fiber” (t,a), which is a representative of twice the fundamental four-cycle. The plots of fk(ψ1) are given in figure 4. Our numerical analysis indicates the following asymptotics at the two extrema ψ1= 0, ψ1=√3: f4∼ (√3 − ψ1) , for ψ1→ 0 ; Figure 4: Plots of f4(ψ1) and f6(ψ1). Their asymptotic behavior near 0 and√3 is given in (5.24), (5.25). Notice in particular that f6(√3) is small but non–zero. We can now proceed in the same fashion as in the N = 1 case to determine ψ1from the flux parameters. Namely, we write the combination 0n6− 3n0n2n4)2= − 9f6(ψ1)2f0(ψ1)≡ ρ(ψ1) , (5.26) which allows us to determine ψ1in terms of fluxes. l and gsare then given by 1 + 3n2 1 + 3n2 A crucial role is played by the function ρ(ψ1) which we plot in Figure 5. It decreases monotonically from 1 at ψ1= 0 to zero at ψ1=√3. Its asymptotic behavior at ψ1= 0 and ψ1=√3 is: ρ ∼ (√3 − ψ1)3for ψ1→ ρ ∼ 1 − ˜ cψ2 1for ψ1→ 0 , for some constant ˜ c. This is in agreement with (5.24), (5.25). The fact that ρ vanishes at the same point, ψ1=√3, where the solution develops a singularity, is strongly supported by our numerical analysis, and will be crucial in reproducing the field theory results. Figure 5: A plot of the function ρ(ψ1) in (5.26). 5.3A phase transition As in the N = 1 case, consider for simplicity the case n4= 0 and call, as usual, n6= N and n2= k. Equation (5.26) becomes As in the N = 1 case, there are two interesting regimes. For N ? k3/n2 we are near the undeformed solution. From (5.29), we see that 1 − ρ(ψ1) ∼ ψ2 ρ(ψ1) =1 + 3n2 0, ψ1→ 0 and 1; hence we can identify in this regime Moreover, we see from (5.23) that f0(ψ1) ∼ ψ1; using also (5.24), we easily compute from which is indeed the behavior of the N = 6 solution . For N ? k3/n2 √3. From (5.29) and (5.30), we see that 0, the function ρ(ψ1) should approach zero, and this happens for ψ1→ ?√3 − ψ1 From (5.25) we then conclude which is the same behavior as in (3.22). Again, as in the N = 1 case, we can also argue generally that gsremains bounded for any integer values of the fluxes, with n0?= 0. At first sight, this seems puzzling. At the end of section 4.2, we noticed that this gravity solution is expected to develop light D–branes in the limit N1= N2= N → ∞. As argued in , N1= N2precisely when n4= 0 (see also (5.35)). But there do not seem to be any light D–branes in a limit where gsis small and the internal manifold is large, since a D–brane mass scales as Lk/gs, with k ≥ 0. As we remarked after equation (5.16), however, in the limit ψ →√3 (which is relevant for large N) the internal manifold develops two conifold–like singularities, since the two- cycle is now shrinking to zero at the “poles”. As we will now see, the new light states are obtained from D-branes wrapping the vanishing cycle for that singularity. We now want to compare this gravity solution with the field theory we saw in section 4.2; specifically, the one defined by the superpotential in (4.9), which has the right symmetries to be the dual of the gravity solution we found in section 5.1. We can first of all try to predict what sort of bulk field corresponds to the monopole operators discussed in section 4.2. Let us recall how the duality works in the ABJM case, when F0= 0. Consider first a monopole operator that creates one unit of field strength for both gauge groups at a particular point. This operator has k indices under both gauge groups, and we can make it gauge-invariant by contracting it with k bifundamentals. The resulting bound state corresponds to a D0 brane in the gravity dual; notice that such a brane has no tadpole on its worldsheet for the worldsheet vector potential A, as we already saw in section 3.4. Another monopole operator that can be considered is the one that creates one unit of flux for, say, the second gauge group. In this case, we cannot make this operator gauge–invariant: it will have k “dangling” indices. This corresponds to a D2 brane wrapping an internal two–cycle. As we also already saw in section 3.4, such a brane has a tadpole on its worldsheet, coming from the term?A1F =?F2A; so one needs to have k strings ending on it, and these k strings correspond to the k indices of the monopole operator. When we switch on F0, even a D0 brane will have a tadpole on its worldsheet, coming from the coupling?F0A. On the field theory side, this corresponds to the fact that the monopole operator that creates one unit of field strength for both gauge groups has now k1 fundamental indices for the first gauge group and k2 antifundamental ones for the second. This cannot be made gauge-invariant; we are always left with at least |k1− k2| “dangling” indices. This fact was used in to establish that the Romans mass integer is the sum of the Chern–Simons couplings, so that, in the present language, n0= k1−k2; see also [30,31]. In it was similarly shown that n4is the difference between the two gauge group ranks N2− N1.5Putting this together, we obtain a dictionary between the flux integers and the ranks and levels of the field theory: n0= k1− k2,n2= k2,n4= N2− N1,n6= N2. (5.35) In section 4.2, we considered monopole operators which create k2units of field strength for the first gauge group, and k1units of field strength for the second. We noticed that these have k1k2bifundamental indices, and thus can be made gauge–invariant. Following the identifications of D2 branes and D0 branes above, if we assume for example that k1> k2, we can say that these new gauge–invariant monopoles correspond to a bound state k2D0 branes and k1−k2D2 branes. We have already noticed in section 3.4 that such a bound state can cancel the tadpole on the worldsheet, because it makes the prefactor in (3.24) vanish. Let us make this expectation more precise. Consider a D2 brane wrapped on a two– cycle B2in the N = 2 solution. As we will see in appendix B.2, supersymmetry requires 5The relative sign between the expressions for n4and n0had not been determined so far. We made here a choice consistent with our final result in formula (5.43). that the D2 brane lives at the North pole t = 0 or at the South pole t = π/2, and that it wraps the S2that does not shrink there. We also need to cancel the tadpole for the world-volume field A which arises from the Wess-Zumino coupling, A ∧ (F2+ F0(F − B)) . We can split B into a fiducial choice plus a zero mode, as in (5.21). The tadpole cancel- F − B = F − β − B0= −F2/F0. Since B0was chosen to satisfy (5.20), we need to turn on a world–volume flux F = β . (5.38) There is a possible obstruction to doing this, coming from the quantization of the world– volume flux, that says that is given by b = −n2/n0; hence in general it is rational and not an integer. So we see that a single D2 brane is generally not consistent. We can get around this, however, by considering n0D2–branes. In that case, the equation we want to satisfy actually reads S2F ∈ Z. The value of b = S2β from (5.22) F = β 1n0. (5.39) The integral of the trace of the left hand side is the first Chern class on the world–volume, which is the induced D0-brane charge n0. The integral of the trace of the right hand side now gives bn0= −n2. We conclude that we can cancel the tadpole by considering a bound state of n0D2 branes and n2D0 branes, just as in section 3.4. Naively, one might think that the mass of a D2–D0 bound state should be at least as heavy as a D0-brane, which in units of AdS mass is mD0L ∼ L/gs. Since this is heavy in the limit (5.34), one might think that such a bound state can never reproduce the light mass predicted in section 4.2. Fortunately, such pessimism proves to be unfounded. The mass of the state is given mD2L = n0L det(g + F − B) = n0L where we used the tadpole cancellation condition. We will take the cycle B2 to be a representative of the non–trivial cycle, which is the diagonal of the two S2’s. Using the explicit form for the metric in (5.4), as well as (5.12), (5.18) and (A.12), we get: mD2L = 4π 1 + ψ2 The fact that the two expressions under the square root are proportional is related to the BPS condition, as discussed in appendix B.2. Inserting the values of l and gsfrom (5.27),(5.28) for generic fluxes we obtain 1 + ψ2 Quite remarkably, the function of ψ1 in the previous formula, which can be computed numerically, turns out to be constant with value 1. The final result for the mass formula Upon using the dictionary (5.35), this formula is identical to the field theory prediction (4.13) in the limit where n0 ? n2. This is exactly the limit where we can trust the supergravity solution, since, as shown in (5.27), for a generic value of ψ1, L is large only if n2? n0. In this limit, it is also true that the dimension of the corresponding operator is given by ∆ ∼ mD2L. In contrast with the N = 1 results in section 3.4, and with the naive expectation expressed earlier, we see from (5.43) that the mass of the bound state remains finite also in the limit N ? k3/n2 of the constituent D0-brane, leaving a smaller piece that is proportional to the volume of the shrinking S2. These are precisely the new light states that we had predicted to exist from the field theory analysis in section 4.2. 0. A contribution from the B field cancels the large mass ∼ L/gs O. A. is supported in part by the Israel–U.S. Binational Science Foundation, by a research center supported by the Israel Science Foundation (grant number 1468/06), by a grant (DIP H52) of the German Israel Project Cooperation, and by the Minerva foundation with funding from the Federal German Ministry for Education and Research. D. J. wishes to acknowledge funding provided by the Association of Members of the Institute for Advanced Study. A. T. and A. Z. are supported in part by INFN and MIUR under ASupersymmetry equations and pure spinors for the N = 2 solution We will give in this section more details about the N = 2 solution we found in section 5. The supersymmetry parameters for compactifications of the form AdS4× M6 (or Minkowski4× M6) decompose as Here, N is the number of supersymmetries. The subscripts ± denote positive and negative chirality spinors, in four and six dimensions; the negative chirality spinors are conjugate to the positive chirality ones, For each a, ζa elements of this basis to be “Killing spinors”, which means that Dµζ+=µ with i = 1,2, are a priori independent six–dimensional Weyl spinors. In this section, we will consider N = 2. A priori, one could have taken the ζain ?1and ?2to be different. This can indeed be done for compactifications with vanishing RR flux; for example, for the usual N = 2 Calabi–Yau compactifications. To recover that case in (A.1), one can take for example η21= η12= 0, and keep a non–vanishing η11and η22. However, in compactifications where RR fluxes are present, the ζain ?1and ?2are required to be equal, up to a constant that can be reabsorbed in the ηia. Hence (A.1) describes all possible N = 2 compactifications, and is particularly appropriate for vacua with RR fluxes. +can vary among a basis of four–dimensional Weyl spinors; we will take the 2γµζ−. The ηia Using (A.1) in the supersymmetry equations yields equations for the internal spinors ηia. In fact, these equations do not mix the ηi1with the ηi2. In what follows, we will first analyze the equations of the ηi1≡ ηi; we will come back to the second pair later. We will construct a pair of pure spinors as tensor products of the supersymmetry 6As usual, we left implicit a Clifford map on the left hand side, that sends dxm→ γm. The type IIA supersymmetry conditions can be expressed as : (d − H∧)(eA−ϕRe (Φ−)) = 0 , (d − H∧)(e3A−ϕIm (Φ−)) = −3e2A−ϕµIm (Φ+) +e4A (d − H∧)(e2A−ϕΦ+) = −2µeA−ϕRe (Φ−) ; ||Φ+|| = ||Φ−|| = eA. ∗ λ(F) , (A.4b) Here, F are the internal fluxes (which determine also the external fluxes, by self–duality). A is the warping function, defined as ds2 has no independent meaning, since one can reabsorb it in A. We have normalized µ = 2 in this paper. The symbol λ acts on a k–form by multiplying it by the sign (−)Int(k/2). Finally, the norm in (A.4d) is defined as ||A||2= i(A ∧ λ(¯A))6. The metric (5.4) can be written in terms of the vielbein 6. The cosmological constant AdS4is given by Λ = −3|µ|2. Since A is non–constant in the solution, however, this Λ 8Γ(da + A2− A1), where ei= dθi+isinθidφiare the natural one–forms on the spheres S2 and eB2= sin(t) we recover the Fubini–Study metric of CP3with natural K¨ ahler form (5.31)]). It is also convenient to use the forms i. For eB1= cos(t) i=1Ei∧¯Eiand natural three form section Ω = E1∧E2∧E3(see for example [28, 2ei∧ ¯ ei (not summed) ,o ≡i 2eiae2∧ ¯ e1; (A.6) the Jiwere already defined in (5.3). These forms satisfy dJi= 0 ,do = i(da + A2− A1) ∧ o ,o ∧ ¯ o = −J1∧ J2. (A.7) The generic pure spinors corresponding to an SU(3)×SU(3) structure can be written in terms of the “dielectric Ansatz” 2z ∧ ¯ z 8sin(2ψ)eA+iθz ∧ exp where θ and ψ are two new angular variables; one can see easily that the supersymmetry equations (A.4) relate them by tan(θ) = −cot(2t) sin(2ψ) .(A.9) The one–form z and the two–forms j and ω can also be used to describe an SU(2) structure on M6. For our solution, these forms are given by z = −ie−iθE1, ω = iE2∧¯E3. We can also characterize j and ω in terms of the forms in (A.6): , Im(ω) =1 The RR fluxes are determined to be as in equation (5.12) with cos(2t)(2Ct,ψ+ w2),g2= −ce−4A (2Ct,ψ(w1+ w2) + 3w1w2), ˜ g4= −ce−4A sin(4t)cos2(2ψ)(2Ct,ψ(w1+ w2) + w1w2), where Ct,ψwas defined in (5.7). Recall also that one possible choice of NS–NS field that satisfies the equations of motion is B0= F2/F0, as in (5.20). So far we have described the solution as if it were an N = 1 solution: we have only paid attention to the a = 1 part of (A.1). To show that the solution actually has N = 2 supersymmetry, we have to provide a second pair of spinors, ηi2, that satisfies the equations of motion for supersymmetry with the same expectation values for all the fields. In terms of pure spinors, we can now form the bilinears and require that they solve again the equations (A.4), with the same values of the fluxes and the same metric. In fact, one expects the two solutions Φ and˜Φ to be rotated by R–symmetry, so that there is actually a U(1)’s worth of solutions to (A.4). To see this U(1), rotate the two–form o in (A.6) by a phase:7 o → e−iαo ≡ oα. (A.14) We can correspondingly define a pair of pure spinors Φα appears. The crucial fact about the rotation of o in (A.14) is that it keeps its differential properties (A.7) unchanged: namely, doα= i(da + A2− A1) ∧ oα. Because of this fact, the computations to check (A.4) do not depend on α; and, since we checked already that α = 0 gives a solution, it follows that any Φα solution with different fluxes; but we can see from (5.12) that o never appears in Fk. We conclude, then, that the solution we have found is an N = 2 solution. ±, by changing o → oαwherever it ±is a solution. A priori, this could be a B BPS particles In this section, we will give a general analysis of BPS particles in flux compactifications (subsection B.1), and we will then apply those general results to the N = 2 background described in section 5 and appendix A. B.1 General considerations We will start with some general considerations about BPS states in N = 2 backgrounds with fluxes. These will in general be states that are left invariant by a certain subalgebra of the supersymmetry algebra. This subalgebra is in general defined by the fact that the two supersymmetry parameters ?iare related: Γ??2= ?1. (B.1) 7Alternatively, one can change the vielbeine (A.5) by translating a → a + α. In first approximation, Γ?is the product of the gamma matrices parallel to the brane. When B fields or worldsheet fluxes F are present, Γ?receives additional contributions of eF−B. We will give a definition later on, in the context needed for this paper; for the general and explicit expression, see for example [33, Eq. (3.3)]. For an AdS4× M6 compactification, we would like to use the decomposition (A.1). For particles, this will lead to an equation involving the four–dimensional spinors ζi follows, the index0is meant to be a frame index. To have a chance to solve the resulting equations, we need to postulate a relation between these spinors. One can write for ±; here and in what for some matrix A. (Recall that in general the index a runs from 1 to N; for us, N = 2, and so a = 1,2.) In fact, (B.2) is almost the most general choice one can make, compatibly with the symmetries of the problem. The only generalization one could make would be to multiply the left-hand side by another matrix Bab. Whenever this matrix is invertible, one can reabsorb it by a redefinition of Aab. In this sense, we can say that (B.2) is the “generic” Ansatz for a BPS particle. The matrix Aabin (B.2) needs to satisfy certain conditions. Let us work for simplicity in a basis where all the space–time gamma matrices γµ, µ = 0,...,3 are real, and the internal γm, m = 1,...,6, are purely imaginary; the ten–dimensional gamma matrices are then given as usual by Γµ= eAγµ⊗ 1 ,Γm+3= γ5⊗ γm. (B.3) It follows from these definitions that Γµare real. Let us now conjugate (B.2); using (A.2), the fact that γ0is real, and that γ2 0= −1, we get If we were considering an N = 1 background, Aabwould be a one–by–one matrix, and (B.4) would have no solution. This is just what one would expect: there are no BPS particles in a N = 1 background. For N = 2, one choice that satisfies (B.4) is A = e−iλ We can now use (B.3) to write Γ?= γ0⊗ γ?, (B.6) where γ?is now an element of the internal Clifford algebra; it contains the product of all the internal gamma matrices parallel to the brane, plus additional contributions from the worldsheet flux and B-field. Let Bp⊂ M6be the p–cycle wrapped by the brane, of dimension p and with coordinates σα, α = 1,...,p. Then we define the natural volume form on B to be One can also define similarly an “inverse volume form” as the multivector det(g + F − B)dσ1∧ ... ∧ dσp. (B.7) ∂1∧ ... ∧ ∂p ?det(g + F − B) which is a section of Λp(TB). This multivector can be used to give an intrinsic definition of γ?: here is how. We can define eFvol−1to be the multivector of mixed degree that one obtains by contracting the indices of the form eFwith the multi–vector vol−1. Recall now that multivectors can be “pushed forward”: if we call x : B ?→ M6the embedding map, with components xm(σ), then x∗(eFvol−1) is a multivector in M6, obtained by contracting all indices α on B with the tensor ∂αxm. In fact: B) . (B.9) Here, we left implicit on the right hand side a Clifford map that sends a vector ∂minto a gamma matrix γm. We already used this map on forms (see footnote 6). One can show that γ?is unitary: ?γ?= 1 .(B.10) For a more explicit expression of γ?, see [33, Eq. (3.5)]. If we now use (B.2), (B.6) and (A.1) in (B.1), we get For our choice (B.5), this reads We are now left with solving (B.12), which are two purely internal equations. Each of the two equations is formally identical to others that have already appeared in the context of BPS objects which do exist in N = 1 flux compactifications: branes which extend along the time direction, plus one, two or three space directions. Hence we can simply follow the same steps; we will now summarize that procedure for (B.12a), and then apply the result to (B.12b). Let us first define the new pure spinors notice that these are different from the pure spinors Φ±, defined in (A.3), which entered the supersymmetry equations (A.4). In (A.3), η1and η2were to be understood as η1a and η2a, for a either 1 or 2. In (B.13), we are mixing a = 1 with a = 2. A possible basis for the space of spinors of positive chirality is given by η11 Three linear combinations of the γmmake γmη11 γi, where i is a holomorphic index with respect to an almost complex structure I. Ex- plicitly we have η11† The coefficients a and bm have a geometrical interpretation. multiply (B.14) from the left by η11† From the formula Tr(? A? B†) =8 we see that Tr(γ?Ψ† (B.9), we see that γ?contains factors of ∂αxm; when contracting with¯Ψ+, these factors reconstruct a pull–back of that form. In conclusion we get8 −vanish: they are its three “annihilators” += (1+iI)mn≡ 2¯Πmn. In terms of this basis a priori one can To compute a, we can + ; we get aeA= η11† + ) = Tr(γ?Ψ† +) consists of contracting the free indices of γ?with those of¯Ψ+. From where |Bdenotes the top–form part on B of the pull–back. By similarly multiplying (B.14) from the left by η11† − γn, we get (dxm· eF−BΨ−)|B= −1 where · denotes the Clifford product: v· = v ∧ +v?. Here, ∂m?(dxm1∧ ... ∧ dxmp) ≡ We can now go back to (B.12a). Comparing to the expansion (B.14), we get a = −eiλ,bm= 0 . (B.18) 8The factor of eAcomes from the fact that ∀a,i, ||ηia|| = eA/2, which follows from (A.4). Using the geometrical interpretations (B.16) and (B.17), we get (v · eF−BΨ−)|B= 0 . Actually, one can show that (B.19) is equivalent to the system (B.20). To see this, observe that γ?is unitary, as we saw in (B.10). This implies that γ?η22 same norm as η22 +should have the +. Since all the spinors have norm eA(see footnote 8), it follows that |a|2+ 2bm¯bm= 1 .(B.21) This means that imposing Re(a) = 1 is equivalent to imposing Im(a) = 0 and bm= 0. Recalling (B.16) and (B.17), we get our claim that (B.19) is equivalent to (B.20). This completes our analysis of (B.12a) (along the lines of ). For (B.12b), similar considerations apply; we obtain (v · eF−B˜Ψ−)|B= 0 , for the pure spinors Let us now summarize this section: we have shown that a brane wrapping an internal cycle B, and extended along the time direction, is BPS if and only if (B.19) (or equivalently (B.20)) is satisfied by Ψ and, analogously, (B.22) (or equivalently (B.23)) is satisfied by ˜Ψ, where Ψ and˜Ψ are defined respectively in (B.13) and (B.24). We will now compute these pure spinors for the solution described in section 5 and in appendix A. B.2 D2/D0 bound states in the N = 2 solution As discussed in the previous section, in order to study the supersymmetry of BPS particles obtained from wrapped branes, we need to form bilinears in the supersymmetry spinors given by the pair of spinors defining the SU(3) × SU(3) structure in (A.10). Recall that a SU(3) structure is specified by two invariant tensors (J,Ω) or, equivalently, by a spinor η+(of norm 1) such that +. We first need to write them explicitly. A convenient basis to expand our spinors is The SU(3) × SU(3) structure in (A.10) can be seen as the intersection of two SU(3) structures given by (J1,Ω1) = (j +i call the corresponding spinors η+and χ+. They are related by χ+= denotes the Clifford multiplication by the one-form zmγm. We will need in the following an expression for the tensor products of a generic linear combination 2z∧ ¯ z,ω∧z) and (J2,Ω2) = (−j +i 2z∧ ¯ z,−¯ ω∧z). We √2z · η−, where z· µ+= aη++ bχ+, ν+= xη++ y χ+. This is given by a¯ xe−ij+ b¯ yeij− i(a¯ yω + ¯ xb¯ ω) i(by¯ ω − axω) + (bxeij− aye−ij) We can choose the spinors for the first supersymmetry as follows It is easy to reproduce, using formula (B.27), the dielectric ansatz (A.8) for the pure As discussed in appendix A, there is a U(1) family of supersymmetries obtained by rotating o → oα= e−iαo. We can conveniently choose as a second independent supersym- metry the one with oπ= −o. This is defined by 4cos(ψ) ˜ η+− ie−iπ 4cos(ψ) ˜ η++ ie−iπ 4sin(ψ) ˜ χ+) , 4sin(ψ) ˜ χ+) , ˜ η+= −icos(2t)η++ isin(2t)χ+, ˜ χ+= isin(2t)η++ icos(2t)χ+. This reproduces the rotated pure spinors Φπ With these ingredients, we can compute the spinors Ψ±and˜Ψ±defined in (B.13) and (B.24) and check the BPS conditions for a D2-brane. It is easy to see that the D2-brane considered in section 5, which wraps the diagonal S2and sits at the North or South pole, is indeed supersymmetric. Let us consider, for definiteness, the North pole. At t = 0, ψ = 0 and we see that ηi2 +. As a consequence, at t = 0, ˜Ψ±= −iΦ±, (B.31) and we are reduced to check expressions for the pure spinors Φ±at the North pole. Taking into account that ψ = 0 there, we have 8eA+iθz ∧ ω . The condition (B.20b) for Ψ−(and the analogous (B.23b) for˜Ψ−) gets contributions only from the contraction with the vector z and it is automatically satisfied because ω vanishes at the North pole, t = 0. It is easily seen that the conditions for Ψ+and˜Ψ+are equivalent and it is enough to analyze those for Ψ+. Equation (B.20a) reads Im?ei(θ−λ)e−ij?∧ eF−B|B2= 0, and determines the world-volume field F = (B + cot(θ − λ)j)|B2. (B.34) We see that a wrapped D2 brane can be made supersymmetric by choosing an appropriate world-volume field. However, as discussed in section 5.4, to have a consistent BPS state we need to impose the quantization of the world-volume field and the cancellation of tadpoles. As discussed there, the quantization condition requires to take n0D2-branes. On the other hand, the tadpole condition requires F = β or, equivalently, F −B = −F2/F0. At t = 0, using the explicit form for the metric in (5.4), as well as (5.12), (5.18), (A.10), (A.9) and (A.12), we evaluate tan(θ) = −ψ1and j = −1 that J1is the volume form of one of the two S2’s, as defined in (5.3). We thus see that the tadpole condition is satisfied by λ = 0. The mass of n0D2 branes is then obtained by integrating the volume form in (B.19) 4e2B1J1and F2/F0= −1 det(g + F − B) = n0 J = n01 1 + ψ2 Using this, one exactly reproduces the result (5.41) of section 5.4. A more detailed analysis of equations (B.20) (and the analogous ones for˜Ψ±) shows that a D2-brane sitting at t ?= 0,π/2 cannot be supersymmetric and simultaneously satisfy the tadpole condition. L. J. Romans, “Massive N=2a Supergravity in Ten Dimensions,” Phys. Lett. B169 A. Sagnotti and T. N. Tomaras, “Properties of Eleven–Dimensional Supergravity,”. K. Bautier, S. Deser, M. Henneaux, and D. Seminara, “No cosmological D = 11 supergravity,” Phys. Lett. B406 (1997) 49–53, hep-th/9704131. S. Deser, “Uniqueness of D = 11 supergravity,” hep-th/9712064. J. Polchinski and E. Witten, “Evidence for Heterotic – Type I String Duality,” Nucl. Phys. B460 (1996) 525–540, hep-th/9510169. B. E. W. Nilsson and C. N. Pope, “Hopf fibration of eleven-dimensional supergravity,” Class. Quant. Grav. 1 (1984) 499. S. Watamura, “Spontaneous compactification and CPN: SU(3) × SU(2) × U(1), sin2(θW), g3/g2 and SU(3) triplet chiral fermions in four dimensions,” Phys. Lett. B136 (1984) 245. D. P. Sorokin, V. I. Tkach, and D. V. Volkov, “On the relationship between compact- ified vacua of d = 11 and d = 10 supergravities,” Phys. Lett. B161 (1985) 301–306. O. Aharony, O. Bergman, D. L. Jafferis, and J. Maldacena, “N=6 superconformal Chern-Simons-matter theories, M2-branes and their gravity duals,” JHEP 10 (2008) A. Tomasiello, “New string vacua from twistor spaces,” Phys. Rev. D78 (2008) P. Koerber, D. L¨ ust, and D. Tsimpis, “Type IIA AdS4compactifications on cosets, interpolations and domain walls,” JHEP 07 (2008) 017, 0804.0614. D. Gaiotto and A. Tomasiello, “The gauge dual of Romans mass,” JHEP 01 (2010) M. Petrini and A. Zaffaroni, “N = 2 solutions of massive type IIA and their Chern– Simons duals,” JHEP 09 (2009) 107, 0904.4915. C. M. Hull, “Massive string theories from M–theory and F–theory,” JHEP 11 (1998) T. Sakai and S. Sugimoto, “Low energy hadron physics in holographic QCD,” Prog. Theor. Phys. 113 (2005) 843–882, hep-th/0412141. B. A. Burrington, V. S. Kaplunovsky, and J. Sonnenschein, “Localized Backreacted Flavor Branes in Holographic QCD,” JHEP 02 (2008) 001, 0708.1234. K. Behrndt and M. Cvetic, “General N = 1 supersymmetric fluxes in massive type IIA string theory,” Nucl. Phys. B708 (2005) 45–71, hep-th/0407263. O. Aharony, O. Bergman, and D. L. Jafferis, “Fractional M2-branes,” JHEP 11 (2008) 043, 0807.4924. D. Gaiotto and D. L. Jafferis, “Notes on adding D6 branes wrapping RP3in AdS4× M. K. Benna, I. R. Klebanov, and T. Klose, “Charges of Monopole Operators in Chern–Simons Yang–Mills Theory,” JHEP 01 (2010) 110, 0906.3008. D. L. Jafferis, “Quantum corrections to N = 2 Chern–Simons theories with flavor and their AdS4duals,” 0911.4324. F. Benini, C. Closset, and S. Cremonesi, “Chiral flavors and M2-branes at toric CY4 singularities,” JHEP 02 (2010) 036, 0911.4127. S. Kim and K. Madhu, “Aspects of monopole operators in N = 6 Chern–Simons theory,” JHEP 12 (2009) 018, 0906.4751. D. L. Jafferis and A. Tomasiello, “A simple class of N = 3 gauge/gravity duals,” JHEP 10 (2008) 101, 0808.0864. D. Martelli and J. Sparks, “Moduli spaces of Chern–Simons quiver gauge theories and AdS4/CFT3,” Phys. Rev. D78 (2008) 126005, 0808.0912. A. Hanany and A. Zaffaroni, “Tilings, Chern–Simons Theories and M2 Branes,” Download full-text JHEP 10 (2008) 111, 0808.1244. A. Kapustin, “Wilson–’t Hooft operators in four-dimensional gauge theories and S– duality,” Phys. Rev. D74 (2006) 025005, hep-th/0501015. D. Gaiotto and A. Tomasiello, “Perturbing gauge/gravity duals by a Romans mass,” J. Phys. A42 (2009) 465205, 0904.3959. M. Cvetic, H. Lu, and C. N. Pope, “Consistent warped–space Kaluza–Klein re- ductions, half–maximal gauged supergravities and CPnconstructions,” Nucl. Phys. B597 (2001) 172–196, hep-th/0007109. M. Fujita, W. Li, S. Ryu, and T. Takayanagi, “Fractional Quantum Hall Effect via Holography: Chern–Simons, Edge States, and Hierarchy,” JHEP 06 (2009) 066, O. Bergman and G. Lifschytz, “Branes and massive IIA duals of 3d CFT’s,” JHEP 04 (2010) 114, 1001.0394. M. Gra˜ na, R. Minasian, M. Petrini, and A. Tomasiello, “A scan for new N = 1 vacua on twisted tori,” JHEP 05 (2007) 031, hep-th/0609124. L. Martucci and P. Smyth, “Supersymmetric D–branes and calibrations on general N = 1 backgrounds,” JHEP 11 (2005) 048, hep-th/0507099. R. Minasian, M. Petrini, and A. Zaffaroni, “Gravity duals to deformed SYM theories and generalized complex geometry,” JHEP 12 (2006) 055, hep-th/0606257.

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https://brainmass.com/business/statement-of-cash-flows/present-value-cash-flow-384770

math

- Financial Accounting & Bookkeeping - The Financial Statements - Statement of Cash Flows Present value of cash flow This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! For each of the cases shown in the following table, calculate the present value of the cash flow, discounting at the rate given and assuming that the cash flow is received at the end of the period Case Single Cash Flow Discount rate End of period (Years) © BrainMass Inc. brainmass.com March 21, 2019, 9:38 pm ad1c9bdddf A 7,000 12% 4 B 28,000 8% 20 C 10,000 14% 12 D 150,000 11% 6 E 45,000 20% 8 Use the compound interest formula to calculate the present value PV = FV ... The expert examines the present value of a cash flow.

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https://classroom.synonym.com/precision-math-17406.html

math

Help With Precision in Math Hi, my name is Marija, and today I'm going to give you help with precision in math. Precision is the number of significant figures that are present in a number. So if for example you are given the number 145.372 and you are asked to write this number with a precision of 3, that means that you are only being asked for three significant digits. So I would be writing it as 145 so it's an estimate of the number above. The number as it is written, 145.372 is precise to six digits right, because I have one, two, three, four, five, six significant figures and what I mean by significant figures is any numbers that come after leading 0s and I don't include any 0s afterwards. So these two 0s would not be significant digits. So precision just means the amount of significant digits in a number and if it's less than the amount that's already there then you are rounding it to a specific place and that is precision in math.

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https://edenhope.vic.edu.au/teaching-and-learning/vce/general-mathematics/

math

Do you require intermediate to advanced level of Maths to pursue your career? Want to study Further Maths in Unit 3 & 4. General Mathematics has a strong emphasis on logical thought, formulating problems to allow you to compute and decide, deducting from assumption and applying advanced concepts. Topics covered in General Mathematics are ‘Algebra and Structure’, ‘Arithmetic and Number’, ‘Discrete Mathematics’, ‘Geometry, Measurement and Trigonometry’, ‘Graphs of Linear and Non-Linear Relations’ and ‘Statistics’. VCE: Unit 1 and 2 available.

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http://pstermpaperibmm.gloriajohnson.us/cost-curves.html

math

This article illustrates the cost curves associated with a typical firm's costs of production. See the latest curves prices and membership fees including initiation fees for all monthly plans including 12-month commitment and no commitment memberships. We provide homework assignment help for topic nature of costs and cost curves contact us for expert homework help. The cost curves and product curves are linked as it shows the figure the upper graph shows the average product curve and the marginal product curve the lower graph. In a supplemental note to a reprint of the classic cost curves and supply curves, jacob viner relates his confusionabout how long-run. The ihs chemical cost curve service – methanol provides detailed production cost coverage for the nearly 286 production units operating around the world today. Cost analysis allows businesses to determine the actual and anticipated costs of a project cumulative cost curves provide business owners with a visual. Cost curves and supply curves by jacob viner, chicago-geneva it is the primary purpose of this article to develop a graphical exposition of the manner. Advertisements: in fig 197, we have drawn the long-run average cost curve as having an approximately u-shape it is generally believed by economists that the long. Advertisements: curves can be drawn to represent costs the marginal cost (mc) and the average cost (ac) are shown in the following diagram (233) ox and oy are two. Mach learn (2006) 65:95–130 doi 101007/s10994-006-8199-5 cost curves: an improved method for visualizing classifier performance chris drummond robert c holte. Marginal abatement costs are typically used on a marginal abatement cost curve (macc) or mac curve, which shows the marginal cost of additional reductions in pollution. Cost curves 1 cost curves 2 introduction • the amount spent on the use of factor and non factor inputs, inputs is called cost of production. Learn more about the total cost curve in economics using this interactive quiz and printable worksheet use the practice questions to test your. Start studying microeconomics cost curves learn vocabulary, terms, and more with flashcards, games, and other study tools. In microeconomics, the study of concept of production is incomplete without the concept of cost curves here is a 7-minute short test on the cost curves y. Number 1 resource for cost curves and their shapes economics assignment help, economics homework & economics project help & cost curves and their shapes economics. Marginal cost curve and the average total cost curve learn the different types of economic cost curves and the law of diminishing returns. In economics, a cost curve is a graph of the costs of production as a function of total quantity produced in a free market economy, productively efficient firms use. Cost curves some definitions: total cost is what it says – total costs are all the costs of the firm (fixed + variable) fixed costs are fixed, irrespective of. As bulk and base commodities continue to fall because of rising supplies and weak demand, investors are trying to make sense of it all one useful tool is a cost curve. The usage of long run and short run in macroeconomics differs somewhat from the above microeconomic usage (including long-run and short-run cost curves. Total cost curve: a curve that graphically represents the relation between the total cost incurred by a firm in the short-run production of a good or service and the. Advertisements: however, the cost y concept is more frequently used both by businessmen and economists in the form of cost per unit, or average costs rather than as. Short run and long run average cost curves: relationship and difference: short run average cost curve: in the short run, the shape of the average. 1 given the output and total cost data in the table below, complete the following columns: variable cost , fixed costs, marginal cost, average total cost columns. Read chapter energy efficiency cost curves: empirical insights for energy-climate modeling--jayant sathaye and amol phadke: models are fundamental for est.

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https://physics.stackexchange.com/questions/639136/doubt-on-the-predictions-on-the-photoelectric-effect-according-to-the-wave-theor?noredirect=1

math

Presumably you know that experiments demonstrated the falsehood of the Classical model. The model was based on the simple idea that the more energy you hit the electrode with, the more energy it would give each electron. Light was treated as a wave of such energy. In that model, it follows that greater wave amplitude (light intensity) would impart more energy to everything it hit. It is less obvious that a shorter wavelength carries greater energy than a longer wavelength of the same amplitude. It has to do with the rate of change of the wave form (the slope of the curve when you draw it), and is true in both the classical and quantum models. However I am unclear why shorter waves should be expected to increase the current, as that is the number of electrons not their energy. Perhaps it is because there are more peaks per second, which would supposedly therefore knock more electrons out. Of course, we all know that experimental observations gave the lie to all that. Einstein explained it as energy thresholds, by treating the light as discrete particles or packets, and in doing so co-founded quantum physics and earned himself a Nobel prize.

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http://www.google.com/patents/US8046168?ie=ISO-8859-1

math

US 8046168 B2 A geofence system which locates a position as within or without the complex polygon type geofence using a simplified algorithm. The algorithm obtains a position and compares it to the polygon by establishing a ray from the position constructed in a cardinal direction of the coordinate system. The “polarity” of the count of intersections between the ray and geofence indicates whether the position is inside the geofence or not. 1. A method of locating a position as being within or without a defined geographic area, the method comprising the steps of: defining a geofence as a closed circuit of a plurality of straight line boundary segments in a rectangular Cartesian coordinate system to define the geographic area; obtaining a position in the rectangular Cartesian coordinate system; constructing a ray from the position within the rectangular Cartesian coordinate system; locating intersections between the ray and the plurality of line segments; counting the number of intersections and characterizing the position as within or without the geofence based on the count, wherein the ray constructing step is done with the ray oriented in a cardinal direction of the rectangular Cartesian coordinate system, wherein the rectangular Cartesian coordinate system is a system of latitude and longitude used for a globe; defining boundary segments such that they include only end point vertex; locating boundary segments parallel to the ray and excluding any intersections between the ray and parallel boundary segments from the count of intersections; locating extreme dimensions of the geofence; constructing a rectangle in the rectangular Cartesian space from four boundary segments oriented in the cardinal directions of the rectangular Cartesian space, including the extreme dimensions of the geofence and enclosing the geofence; characterizing as outside the geofence any point outside of the rectangle; following definition of the geofence, averaging the coordinate values of the vertices of the geofence; and setting the average values as the origin of a new coordinate system and resetting all vertices in terms of the new origin, wherein the step of counting further comprises characterizing the position as within the geofence if the count is odd and as without the geofence if the count is even. 1. Technical Field The invention relates to geofencing for vehicles, and more particularly, to providing a simplified method for determining if a vehicle's position is within or without an irregular geofence. 2. Description of the Problem A geofence may be defined in part as a virtual spatial boundary. Geofences are a byproduct of the marriage of mobile, inexpensive telecommunications platforms and data processing systems. They are enhanced in accuracy by making use of global positioning systems which allow accurate, precise determination of the location of both the boundary of a geofence and position of a mobile platform relative to the geofence. A geofence typically operates by triggering of a physical response through the mobile, location sensitive device when the device crosses a boundary, though the system on which the method is implemented may simply operate to note the fact of the crossing to an operator. The spatial location of geofences have commonly been established by selecting a point feature, which may be a point defined by latitude and longitude, and then defining either a radius, or major/minor axis for the point feature, to establish a boundary around the point. United States Pat. Appl. Pub. 2006/02003005 described construction, using graphical user interface tools, of geofences with boundaries corresponding to real world objects. The area enclosed by such a fence would typically be an irregular polygon, the expression of which might result from simple selection of an object to be enclosed by the fence, such as the campus of a school. An irregular geofence was recognized as more readily applied to real world situations. Among the issues raised with respect to such irregular polygon shaped fences was the difficulty of fully and correctly representing the actual polygon for devices with limited data processing resources. The reference provided the contingency of “decimating” the polygon to avoid overwhelming the resources of such devices. According to the invention there is provided a method of determining whether a reported position is within or without a geofence. The method provides the steps of first defining a geofence as a closed circuit of a plurality of straight line boundary segments in a rectangular Cartesian coordinate system. Next, a position is obtained in the rectangular Cartesian coordinate system. A ray is constructed from the position within the rectangular Cartesian coordinate system. Intersections between the ray and the plurality of boundary segments are located. The number of intersections is counted and the position is characterized as within the geofence if the count of intersections is odd and as without the geofence if the count is even. Typically the ray is constructed oriented in a cardinal direction of the rectangular Cartesian coordinate system. Intersections between the ray and boundary segments parallel to the ray are excluded from the count. Boundary segments are defined such that they include only one end point vertex to avoid double counting intersections where the ray passes through a vertex. Additional effects, features and advantages will be apparent in the written description that follows. The novel features believed characteristic of the invention are set forth in the appended claims. The invention itself however, as well as a preferred mode of use, further objects and advantages thereof, will best be understood by reference to the following detailed description of an illustrative embodiment when read in conjunction with the accompanying drawings, wherein: Referring now to Referring now to Further simplification is possible. At step 310 the polygon is examined to locate instances of three or more consecutive ordered vertices being located in a straight line. If such instances are found any intermediate vertices (i.e. “vertices” having an angle of 180 degrees) are eliminated. At step 314 the ordering of vertices is reset and the variable N, indicating the number of vertices, is recalculated. At step 316 a further simplification step is taken, by right shifting coordinates to a fixed, limited precision. With the simplification operations complete, one additional test condition is established. At step 318 the extreme points on the polygon in each cardinal direction are located relative to the origin are found. Next, at step 320, an “out-of-contingency” rectangle is constructed having dimensions based on the extreme points. This rectangle will be used as a quick filter for eliminating some reported positions as possibly being inside the polygon. Determination of whether a vehicle's position (P) is within or without of a geofence is a new process, spawned for each reading on position taken, as indicated by step 321. At step 322, the position (P) of the vehicle is located from the Global positioning system constellation 50. The position is adjusted using the normalization generated in steps 306 and 308. The adjusted position is readily compared with the boundaries for the out-of-contingency rectangle to determine if the position P is within the rectangle (step 324) and thus possibly inside the polygon. If the position P is not within the rectangle the position is either outside the polygon or on a border segment of the polygon. Following the NO branch to step 326, the position P is checked for correspondence to the border of the polygon. If the position P is not on the border of the polygon it lies outside of the polygon and the NO branch is followed to step 328 where the “Out-of-Boundary” flag is set. Otherwise step 330 is executed to set the “On-Boundary” flag. Following either step 328 or 330 the routine returns the result and is re-executed as determined by the operator's protocol. If the position P is within the rectangle, that is, within the extreme dimensions of the polygon, more processing is required to determine if the position is within the polygon. Initially a flag or counter is set to zero (step 332). Next, at step 334, a line segment L is calculated from the current (vehicle) position P in a cardinal direction (typically eastward) beyond the maximum extent of the polygon to the east (or in the selected cardinal direction). Typically, the number of intersections between this line segment and the boundary segments of the polygon can be used to determine if position P is within, without or on the boundary. Special provision must be made for the possibility that line segment L passes through one of the vertices or includes all or part of a boundary segment which runs in the same cardinal direction as line segment L. Steps 336 and following deal with determining the number of intersections between the constructed line segment originating at position P and boundary segments of the polygon. If the number of intersection is odd, it indicates that position P lies within the polygon. If even, the sum indicates that position P lies outside of the polygon. As indicated already, cases where line segment L (or ray) includes all or part of a boundary segment or pass through one of the vertices are handled in a special manner to avoid double counting an intersection between L and a vertex as an intersection with two boundary segments. Essentially each vertex is treated as part of the higher numbered boundary segment only (although the opposite approach could be take). The number of boundary segments connecting vertices of the polygon equals the numbers of vertices and can be numbered j=1 to N+1 with boundary segment j connecting vertices i and i+1 (step 336). For every boundary segment Bj the intersection between the boundary segment and line segment L is found if it exists (step 340). If the line L and a boundary segment run together it is not treated as an intersection. This is dealt with by simply ignoring all boundary segments which run east/west (step 338) which skips step 340 for the noted boundary segments. At step 342 the coordinates of all the intersections are determined. If, at step 344, the coordinates for a given boundary segment lies between the vertices for the polygon than the flag count is increased by one for that intersection (step 346). Once all boundary segments have been considered (step 347) the flag is examined (step 348) to determine if it is odd, which if it is (the YES path) a within-polygon determination (step 352) results. If the flag is not odd, position P is not within the polygon (step 350). An intersection at a vertex can be considered to count for increasing the flag count. Steps 354, 356 and 358 are used to located intersections between line L and vertices by asking first if the intersection is at a vertex, and in the following two steps if, for the boundary segment under consideration, whether the intersection is at the start or end of the boundary segment. If the intersection comes at the end of a boundary segment than the intersection is counted (step 360), otherwise not. Those skilled in the art will now appreciate that alternative embodiments of the invention can exist. For example, the flag indicating whether the position P is within or without the geofence could be implemented in ways other than as a counter. While the invention is shown in one of its forms, it is not thus limited but is susceptible to various changes and modifications without departing from the spirit and scope of the invention.

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APPLIED THERMODYNAMICS FOR SECOND YEAR / THIRD SEMESTER EEE DEPT. UNIT I BASIC CONCEPTS INTRODUCTION Thermodynamics is defined as the branch of science which deals with the relations between energy and heat. These relations are governed by the laws of thermodynamics. These laws are based on the principle of energy conversion. It states that energy can be changed from one form to another but the total energy remains constant. In other words energy cannot be created or destroyed. APPLICATIONS OF THERMODYNAMICS • • • • • • Power plants IC engines Turbines Compressors Refrigeration Air-conditioning UNITS AND DIMENSIONS All physical quantities are characterized by dimensions. Dimensions of physical quantities may be defined as the properties in terms of quality not of magnitude by which a physical quantity may be described. Length (L), area (A)and volume (V)are all different dimensions which describe certain measurable characteristics of an object, e.g., A=L2 and V=L3 The arbitrary magnitudes assigned to the dimensions are called as units. In other words, a unit is a definite standard by which a dimension is to be measured. The primary or fundamental dimensions are length L in m, mass m in kg, time in sec and temperature T in K. The secondary or derived dimensions are velocity V in m/s, Energy E in J and volume V in m these are expressed in terms of primary dimensions. SYSTEM OF UNITS The most common system of unit is metric system SI, which is also known as the International System. In this text, the SI (System International) system of units has been used. Energy: Energy is defined as the capacity to do work. The various forms of energy are heat energy, mechanical energy, electrical energy and chemical energy. Unit of energy is Nm or Joule (J) and kWh. 1 kWh = 3.6 x 106 J The energy per unit mass is defined as specific energy whose unit is J/kg. Force: Force acting on a body is defined by Newton s second law of motion. According to this law, force is proportional to the product of mass and acceleration. When a force of one Newton applied to a body having mass of one kilogram, gives it an acceleration of one m/s. The unit of force is Newton (N). 1 N = l kgm/s Weight of a body (W) is the force with which the body is attracted to the centre of the earth. It is the product of its mass (m) and the acceleration due to gravity. i.e., Work: Work is defined as the work done when the point of application of 1 N force moves through a distance of 1m in the direction of the force, whose unit is Joule or Nm. The amount of work (W) is the product of the force (F) and the distance moved (L), W = F×L. Power: Power is defined as the rate of energy transfer or the rate of work. The unit of power is watt (W) 1N m/s = 1 MW = Pressure: Pressure is defined as the force per unit area exerted whose unit is N/m2 which is also known as Pascal (Pa) and for larger pressures, kPa (Kilo Pascal) and MPa (Mega Pascal) are used. Other units for pressure not in the SI units but commonly used are bar and standard atmosphere (atm) 0.1 MPa = 100 kPa = 105Pa 105 N/m = 1 bar 1 atm = 101 .325 kPa = 1.01325 bar 1J/s =1W 106 Kw W= mg (Value of g = 9.81 m/s at sea level) Mostly pressure of a fluid is measured by gauge which gives pressure relative to atmospheric pressure and is called as gauge pressure. In thermodynamic analysis one is mostly concerned with absolute pressure which is the pressure exerted by a system on its boundary. For pressures above atmospheric Pabsolute = pgauge + patm For pressures below atmospheric, the gauge pressure will be negative and is called as vacuum. U-tube manometer which is used to measure pressure, the two arms of the tube are connected to two containers which are at p and p pressures. The tube is filled with a fluid having density and h is the difference in the heights of the fluid columns. By the hydrostatics principle, P1-P2 = Where hg p2 in N/m2 is in kg/m3 h in m, g in m/s2 and then p1 Fig:1(a). For pressure above atm Temperature: Fig:1(b). For pressure below atm Temperature is defined as the degree of coldness or hotness of a body. When heat is added to the body, its temperature increases and when heat is removed from the body, its temperature decreases. Temperature is the thermal condition of a body on which its capacity of transferring heat to or receiving heat from other bodies depends. Thus the temperature determines direction, in which the heat flow will take place, Units of temperature are degree Celsius, degree Kelvin Temperature K = Temperature º C + 273 Under standard temperature and pressure (STP) conditions the temperature of a gas is taken as 15°C and the pressure as 760 mm of mercury. Under normal temperature and pressure (NTP) conditions the temperature of a gas is 0°C and the pressure as 760 mm of mercury. Specific Heat: Specific heat of a substance is defined as the quantity of heat required to raise the temperature of unit mass substance to one degree. Average specific heat, Where Q is heat interaction kJ, T is Temperature difference K and m is mass kg. If the state of the substance is liquid or solid there is only one specific heat. For the case of gaseous substances there are two specific heats, they are: 1. Specific Heat at Constant Volume: When the volume of the gas is constant the quantity of heat required to raise the temperature of unit mass of gas to one degree is termed as specific heat of gas at constant volume which is denoted by Cv 2. Specific Heat at Constant Pressure: When the heat is supplied at constant pressure the quantity of heat required to raise the temperature of unit mass of gas to one degree is termed as specific heat of gas constant pressure which is denoted by Cp CLASSICAL APPROACH MACROSCOPIC AND MICROSCOPIC APPROACH The behavior of one matter can be studied from macroscopic and microscopic points of view. v The macroscopic approach is only concerned with overall effect of the individual molecular interactions. v The microscopic point of view concerned with every molecule and analysis of collective molecular action is carried out by statistical techniques. For example, pressure is a macroscopic quantity, which is defined as the normal force exerted by a system against unit area of the boundary, i.e., the pressure exerted on the vessel is equal to the mean change of momentum of all the molecules exerted perpendicular to unit area of the boundary. This approach is not related with individual molecular action. This pressure can be measured by using pressure gauge. The microscopic approach is used to explain some matter which otherwise difficult to understand by macroscopic approach. THERMODYNAMIC SYSTEM AND SURROUNDINGS A thermodynamic system is a region in space or any matter or specified quantity of matter within a prescribed boundary on which we concentrate. The other matters outside of the boundary are known as surroundings. As shown in Fig.2(a) the system and surroundings are separated by boundary. The boundary may be real or imaginary one. The system is classified into three: • • • Closed System Open System Isolated System Closed System: In this system, the boundaries are closed so that there is no mass transfer. But there may be energy transfer into or from the system, while mass remains constant. This is also known as control mass. e.g., bomb calorimeter. Open System: In this system, the boundaries are not closed and mass and energy transfer may take place through the opening(s) in the boundary. This is also known as control volume. e.g., turbines and compressors. Fig:2(a)A thermodynamic system,(b)Closed system,(c) Open System, (d) Isolated system Isolated System: This system is not affected by surroundings. In this there is no mass or energy transfer across the boundary of the system. WORKING MEDIUM: In most of the devices the working medium is gas or vapor. It is important to know the properties and behavior of the working medium to observe and analyze the working of devices. At various pressures and temperatures the properties of the working fluid can be determined by using pure substance concept. • • The pure substance is defined as a substance that has a fixed chemical composition, e.g., water, nitrogen, helium and carbon-di-oxide. A mixture of two or more pure substances is also called as pure substance as long as the chemical composition is same. A mixture of liquid air and gaseous air cannot be called as pure substance because the mixture is not chemically homogeneous due to different condensation temperatures of the components in air at specified pressure. THERMODYNAMIC EQUILIBRIUM A system is said to be in a state of thermodynamic equilibrium if there is no change in the microscopic properties at all points in the system. For thermodynamic equilibrium, the following three types of equilibrium conditions have to be satisfied. Mechanical Equilibrium: A system is said to be in a state of mechanical equilibrium if there is no unbalanced force with in the system or between the system and the surroundings. Chemical Equilibrium: A system is said to be in a state of chemical equilibrium if there is no chemical reaction or transfer of matter from one part of the system to another. Thermal Equilibrium: A system is said to be in a state of thermal equilibrium if there is no change in any property of the system when the system is separated by a diathermic wall from its surroundings. Diathermic wall defined as a wall which allows heat to flow. STATE, PROPERTIES AND PROCESSES: • • State of a system is the condition of the system at any particular moment. It may be identified by the properties such as pressure, temperature and volume, etc. The property can be measured while the system is at a state of equilibrium. In any operation there is a change in system properties which is called the change of state. A series of changes in the system between initial state and final state is called the path of change of state. • When the path is specified completely the change of state followed by the working medium as it liberates, transfers, transforms or receives energy is called as process. A series of state changes or process undergone by a system such that the final state is identical with the initial state is defined as a thermodynamic cycle. • Fig .2.1(a) A process Fig .2.1 (b) A cycle In order to describe a system it is necessary to know the quantities and characteristics of the system which are known as properties. The properties are classified as extensive properties and intensive properties. • Properties which are related to mass are called as extensive or extrinsic properties, e.g., volume, energy, etc. If mass increases the value of extensive Properties will increase the properties which are independent of the mass of the system are called as intensive or intrinsic properties, e.g., temperature, pressure, velocity, density, etc. • Extensive properties per unit mass are known as specific extensive properties which are nothing but intensive properties, e.g., specific volume, specific energy, density, etc. Properties may also be classified into two types. They are fundamental properties and thermodynamic properties. • • Properties which are measured directly are called as fundamental properties, e.g., pressure, volume, temperature, etc. Properties which cannot be measured directly but in closed cycle the working medium is recirculated with in the system. In open cycle the working substance is exhausted to the atmosphere after the process. ZEROTH LAW OF THERMODYNAMICS: The zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with a third body, then they are also in thermal equilibrium with each other. Fig: 3 Concept of Zeroth law Let us consider the temperature equality concept to three systems say, A, B and as shown in Fig.3. The system A consists of a mass of gas enclosed in a vessel fitted with a thermometer and the system B is a cold iron body. When A and B are brought in contact, after some time they attain a common temperature and are then said to exist in thermal equilibrium. Now the system is brought into contact with a third system C, again A and C attain thermal equilibrium, then system B and C will show no further change in properties when brought into contact. That is system A is in thermal equilibrium with system B and also separately with system C. Then B and C will be in thermal equilibrium with each other. This law provides the basis for temperature measurement. FIRST LAW OF THERMODYNAMICS The first law of thermodynamics states that a closed system executing a cycle in which the initial state and final state are same . i.e., The net work delivered to the surroundings is proportional to the net heat taken from the surroundings. That is heat and work is mutually conversible. Since energy can neither be created nor destroyed, the total energy associated with energy conservation remains constant. Mathematically, Where dw is net work delivered during the process and dq is net heat supplied during the process. FIRST LAW APPLIED TO CLOSED SYSTEM As shown in Fig3.2. let us consider a closed system which undergoes a cycle, in which x and y is two arbitrary properties of the system. According to first law of thermodynamics for a cyclic process, algebric sum of t is proportional to algebric sum of heat transferred. i.e., Q1-2 is proportional to Q2-1 This is applicable if the system involves more heat and work transfers at different points on the boundary. Where Q and W represent infinitesimal elements of heat and work transfer respectively. As no fundamental distinction between the unit of heat and the unit of work J can be neglected from the equation. Internal Energy: The energy E is the sum of Kinetic Energy (KE), Potential Energy (PE) and Internal Energy (U). The internal energy is due to the motion of the molecules and it changes with change in temperature. For a non flow and closed system, the kinetic and potential energy terms are zero, and then the energy will be The term E is the change in internal energy For an isolated system both Q and W are zero, the change in energy is also zero. Q= 0; W= 0; E=0 For reversible non flow process the work DISPLACEMENT WORK The most common example of mechanical work encountered in thermodynamic system is that associated with a process in which there is a change in volume of a system under pressure. Let the volume of the fluid within the moving boundary be v1 and pressure be p1. In p-v diagram, point-1 represents initial state. If the working medium expands and moves the piston to top dead centre (TDC) from bottom dead centre (BDC), the work will be done by the working medium. After expansion at state 2, pressure is decreased and volume increased. Since the system undergoes expansion process, it is represented by the curve 1 - x - y - 2 in p-V diagram as shown in Fig.4 Fig: 4. Displacement work The change of state from x to y, a very small change of state in which pressure is almost constant during the change, then the force acting on the movable boundary F× x = p-V During this piston moves to a distance dL and the work done = force × distance traveled. dW where dV= A×dL The total work at the moving boundary is = pA×dL = p×dV • When the work is done by the system, it is called as positive work. This is represented by the sign plus, +W indicated the work done by the system. e.g., expansion. When the work is done on the system, it is called as negative work. - W indicates the negative work. e.g., compression. • PATH FUNCTION AND POINT FUNCTION A non-flow process is one in which the gas is neither be supplied nor rejected across the boundary of the system. The system moves from state 1 to state 2 through two different paths A and B, as shown in Fig 5. Fig. 5.Path Function Each curve represents the work for each process, these two paths gives two different work values even though states 1 and 2 are identical, the work delivered to the shaft depends upon the particular function, so the work is called as path function. The differentiation of path function is inexact or imperfect. But the thermodynamic properties are point functions, because they depend on the end states and independent of the path which the system follows. The differentiation of point functions is exact or perfect differentials, e.g., change in pressure p2 p1 and change in volume V2- V1. APPLICATION OF FIRST LAW TO NON - FLOW PROCESS: NON - FLOW PROCESSES It is the process in which the substance does not leave the system, but energy only crosses the boundary in the form of work and heat. Non - flow processes are classified under two groups. They are a. Reversible non-flow processes b. irreversible non-flow processes. REVERSIBLE NON-FLOW PROCESSES (CLOSED SYSTEM): The following non-flow processes are reversible 1. 2. 3. 4. 5. Constant volume (isochoric) process Constant pressure (Isobaric) process Adiabatic (Isentropic) process Polytropic process : : : : V = constant; n = p = constant; n = 0 T= constant; n = 1 PV = constant; n = PVn = constant; n=n Constant temperature (Isothermal) process : Where p is pressure, V is volume, T is temperature, n is index of compression or expansion and is adiabatic index. Reversible Constant Volume Process: When volume remains constant during the execution of a process, the process is called as constant volume process. As shown in Below the system contains unit mass and state one in pV diagram represents the system state before the heating processes. State 2 represents the state of system after heating process. Applying the first law of thermodynamics The rise in heat causes rise in internal energy and loss of heat decreases the internal energy. APPLICATION OF FIRST LAW TO FLOW PROCESS: Most of the systems which are related to power generation are open systems in which the mass crosses the boundary of the system, and after doing the work it leaves the system. Flow processes are classified into two types, they are • • Steady flow process Non-steady flow process STEADY FLOW PROCESS A steady flow process is one in which the mass flow rate at the entry and at the exit is constant. At any point in the system the properties of the fluid do not change with time, e.g., compressor, turbines, nozzles, etc. Assumptions made in the analysis of a steady flow process are, 1. Mass flow rate through the system remains constant. 2. Composition of fluid is uniform 3. State of the fluid at any point in the system remains constant. 4. Work and heat are the only interaction between the system and surroundings. STEADY FLOW ENERGY EQUATION As stated in the law of conservation of energy the sum of total energy exerting the system is equal to the sum of total energy leaving the system. Thus there is no change in stored energy. Let m - mass flow rate in kg/s, - absolute pressure in N/m2, v - specific volume in m3/kg, V - velocity in m/s, Z - elevation above the datum in m, u - specific internal energy in J/kg, Q - heat into the system in J and W- work output in J. Fig: 6.Steady flow process Assumptions made are Mass flow in Energy in Total energy in = = = Mass flow out Energy out Total energy out at entry (Potential energy + Kinetic energy + Internal energy + Heat energy) (Potential energy + Kinetic energy + Internal energy + Work) at exit PE1 + KE1 + U1 + Q = PE2 + KE2 + U2 + W = This is the steady flow energy equation and all the energy values are in Watts. The steady flow energy equation can be written in mass basis as given below. Hence the energy values are in J/kg Mass flow in= Mass flow out m1 = m2 We know m = Where the density of fluid ρ = ρ= 1 1 = specific volume υ mass and volume for unit mass A1V1 AV = 2 2 υ1 υ2 UNSTEADY FLOW PROCESS An unsteady flow process is one in which the mass flow rate at the entry and exit of the system is not equal in a given time, and there is no change in stored energy of the system. Let Eout Ein be the change in flow energy and E change in stored energy. Based on first law SECOND LAW OF THERMODYNAMICS LIMITATIONS OF FIRST LAW • • The first law of thermodynamics states that, heat and work are mutually convertible during any cycle of a closed system. But in actual practice all forms of energy cannot be changed into work and the first law does not give any conditions under which conversion of heat into work is possible. The law does not specify the direction of the process under consideration. The limitations of first law are discussed below. • The following examples are based on the first law of thermodynamics and these processes only proceed in certain direction but not in the reverse direction. § Let T1 and T2 be the temperatures of two bodies where T1 is greater than T2 If these bodies are brought in contact with each other but are separated from surround- rigs, heat will flow from hot body (T1) to cold body (T2) till the temperature of both bodies are equal. But the reverse process is not possible, i.e., flow of heat from lower temperature body to higher temperature body. In an automobile moving at a certain speed, if the brakes are applied to stop the automobile means, the brakes get hot by the conversion of automobile s kinetic energy into heat. However, it will be observed that reversal of the process in which the hot brakes were to cool off and give back its internal energy to the automobile, causing it to move on the road. But this is impossible. § § SECOND LAW OF THERMODYNAMICS • KELVIN-PLANK STATEMENT OF SECOND LAW It is impossible to construct an engine which operates on cycle to receive heat from a single reservoir and produce net amount of work. Kelvin - Plank statement related to heat engines . In other words, no engine operating in cycles can convert all the heat energy into work, but there will be some loss of heat energy to the surroundings. Thus 100% efficient engine is not possible. Fig:7. Possible engine and not possible engine Possible engine is one in which a part of heat is rejected to the cold reservoir, which is supplied from the hot reservoir and the difference between the heat supplied and heat rejected is equal to work done. • CLAUSIS STATEMENT OF SECOND LAW Clausis statement is related to refrigerators or heat pumps. The Clausis statement is expressed as follows: It is not possible to construct a system that operates in a cycle and transfers heat from a colder body to a hotter body without the aid of an external agency. In other words, heat can not flow itself from a colder body to a hotter body . Based on this, the hot reservoir at T1 temperature and the cold reservoir at T2 temperature are shown in Fig: 8. The heat pump which takes mechanical work to transfer heat continuously from sink to source. Fig:8. External work required for heat flow from sink to source PERPETUAL MOTION MACHINE OF SECOND KIND (PMM II) Perpetual motion machine of second kind is one which operators in a cycle and delivers an amount of work equal to heat extracted from a single reservoir at an uniform temperature. Such 100% efficiency violates the second law of thermodynamics as according to Kelvin - Plank Statement. It is not possible to construct a machine which could extract heat from a single reservoir and convert it into equivalent amount of work. HEAT ENGINE Heat engine is defined as a machine which is used to convert heat energy into work in a cyclic process . The definition of heat engine covers both rotary and reciprocating machines. The working fluid should undergo cyclic process and periodically should return to its initial state. Fig: 9. Steam power plant as heat engine Fig: 9. shows a steam power plant which is an example of heat engine cycle. In the boiler high pressure steam is generated and the steam expands in the turbine and doing external work W exhaust steam from turbine is condensed in the condenser thereby releasing heat and the water is pumped back to the boiler to complete the cycle. Thus the boiler, turbine, condenser and the pump separately in a power plant can not be regarded as heat engine because they are engines since they are part of the cycle. Combination of these components is a heat engine since they complete the cycle. EFFICIENCY OF HEAT ENGINE Performance of a heat engine is obtained by its thermal efficiency which is the ratio of net work output to heat supplied . A part of heat supplied is converted into work and the rest is rejected. Let Q1 be heat supplied, Q2 be heat rejected, Wt be turbine work and Wp be pump work (Fig: 10.) As per the first law of thermodynamics Fig: 10. Heat engine Efficiency of the heat engine η= Network output Heat supplied HEAT PUMP Heat pump is defined as a device which transfers heat from a low temperature body to a high temperature body when it is working in a cycle (Fig:11). For heat pump the efficiency term is replaced by the co-efficient of performance (COP) which as an index of performance of heat pump to differentiate it from heat engine. COP of the pump is given by COP (heat pump) = Heating effect Work done Fig: 11. Heat pump If a heat pump is used to transfer heat from low temperature reservoir T2 to high temperature reservoir T1 in order to maintain T2 < T1 then the COP of the refrigerator is given by COP ref = Refrigerating effect Work input CARNOT CYCLE The Carnot cycle has four reversible processes, of which two are frictionless isothermal processes and two frictionless adiabatic processes. Figure.12 shows the p-V and T-s diagram of the Carnot cycle. • Process 1-2 represents reversible isothermal expansion, Heat Q is supplied at constant temperature T and this is equal to the work done during the process. V2 V1 Heat supplied Q1 = • P1V1ln Process 2-3 represents reversible adiabatic expansion, there is no heat transfer takes place. The work is done at the cost of internal energy. The temperature becomes T2 at T3 Fig.12 Carnot cycle • Process 3-4 represents reversible isothermal compression in which Q2 heat is rejected isothermally at T2. The air is compressed up to point 4 at constant temperature. Heat rejected • Process 4-1 represents reversible adiabatic compression in which the system returns back to the initial state and the temperature of air increases from. T2 to T1 There is no heat transfer and work is done on the air. Net work done = (Heat supplied) - (Heat Rejected) Efficiency of Carnot cycle = (Net work done / Heat supplied) CARNOT THEOREM The Carnot principles are the two conclusions regard to the thermal efficiency of ideal and natural (actual) heat engines. They are expressed as follows: • The efficiency of an actual (irreversible) heat engine is always less than the efficiency of an ideal (reversible) heat engine operating between the same two reservoirs. All the reversible (ideal) heat engines operating between the same two reservoirs will have the same efficiency. • CLAUSIS IN EQUALITY While applying second law of thermodynamics to processes the second law leads to the definition of a new property called entropy. Entropy is an abstract property, and it is difficult to give a physical description of it. The uses of entropy in common engineering processes provide the best understanding of it. The second law may be stated to be the law of entropy. The cyclic integral of dQ/T in above equation is always less than or equal to zero. Inequivality of Clausis is the basis of the definition of entropy. Entropy is a nonconserved property by which it differs from energy. This inequality is valid for all cycles, viz., and reversible or irreversible. The symbol φ denotes that the integration is to be performed over the entire cycle. Any heat transfer from or to a system can be considered to consist of differential amounts of heat transfer then the cyclic integral of dQ/T can be viewed as the sum of all these differential amounts of heat transfer divided by the absolute temperature at the boundary. CONCEPT OF ENTROPY ENTROPY AS A PROPERTY OF A SYSTEM Entropy of a substance is a thermodynamic property which increases with the addition of heat and decreases with the removal of heat. Entropy itself cannot be defined but the change in entropy can be defined in a reversible process, i.e., the quantity of he received or rejected divided by the absolute temperature of the substance measures the change in entropy. A small amount of heat dQ is added to the system causing the entropy to increase by ds and T is the absolute temperature. The change in entropy absolute temperature. If the total quantity of heat Q be added to a substance at constant temperature then the increase in entropy due to the addition of heat is given by ds = Q T Q T T × ds s1 − s2 = From the definition of entropy dQ = By integrating the equation the total heat added can be obtained as ENTROPY - A POINT FUNCTION Entropy has got one value for each point of temperature or pressure or volume. So it is a point function. The change in entropy during a thermodynamic process depend only on the initial and final conditions irrespective of the path. In solving problems the change in entropy is considered with the assumption if the entropy of all substances is zero at the ice-point, i.e., the entropy is positive if the temperature is above 0°C and negative if the temperature is below 0° C. Entropy is expressed as kJ/kgK, since, it has the dimension of heat/mass and temperature. There are many instruments to measure temperature, pressure, etc., but there no such instruments as yet to measure entropy. ENTROPY OF A REVERSIBLE CYCLIC PROCESS Let us consider a system undergoing a reversible process from state 1 to state along the path A and then from state 2 to the original state 1 along the path B as shown in Fig.13. Fig. 13. Reversible cyclic process between two fixed states CHANGE IN ENTROPY OF A PERFECT GAS Let m kg of gas carryout a process. At the initial state 1 let the pressure, temperature, entropy and volume are p1, T1, s1 and V1 respectively. The gas is heated in any manner such that at its final state 2 pressure, temperature, entropy and volume be p2, T2, s2 and V2 respectively. From the law of conservation of energy This is the change in entropy in terms of volume and temperature. This can be expressed in terms of pressure and volume by applying gas equation as JAYAM COLLEGE OF ENGINEERING AND TECHNOLOGY DHARMAPURI DEPARTMENT YEAR / SEM SUBJECT : EEE : SECOND/ THIRD : ME1211 / APPLIED THERMODYNAMICS ASSIGNMENT NO 1 Unit 1 Basic concepts and Laws of Thermodynamics PART A 1. State first law of thermodynamics? 2. State Zeroth law of thermodynamics? 3. What is meant by Perpetual Motion Machine of first kind ? 4. State the statements of second law of thermodynamics? 5. If pvn=C represents a general thermodynamic process, name the processes when n has values of 0, 1, and . 6. What are the types of thermodynamic properties? 7. Explain the thermodynamic equilibrium? Explain. 8. What is meant by Perpetual Motion Machine of second kind ? 9. State Carnot s theorem. 10. Define Clausis Inequality. PART B 1. Derive then expression for work and heat during constant volume and constant pressure process. 2. 3. 4. Derive then expression for work and heat isothermal and isentropic process. Derive the steady flow energy equation. A cycle heat engine operates between a source temperature of 800 C and a sink temperature of 30 C. What is the least rate of heat rejection per KW net output of engine? 5. 2 kg of air compressed according to the law pV1.3 = constant from a pressure of 1.8 bar and temperature of 30 C to a pressure of 25.5 bar. Calculate a) The final volume and temperature. b) Work done c) Heat transferred d) Change in entropy. Pages to are hidden for "14075471-Applied-Termodynamics-01"Please download to view full document

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http://mathforum.org/library/drmath/sets/mid_factornumb.html?s_keyid=38676600&f_keyid=38676601&start_at=81&num_to_see=40

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See also the Dr. Math FAQ: learning to factor 3D and higher Browse Middle School Factoring Numbers Stars indicate particularly interesting answers or good places to begin browsing. Working with variables? Try Middle School Factoring Expressions. Selected answers to common questions: Finding a least common denominator (LCD). Table of factors 1-60. - Prime Numbers vs. Prime Factors [01/13/1997] What is the difference between a prime number and a prime factor? - Probability That a Sum Is Divisible by Three [11/04/2009] How many ways can you choose three numbers from 1-100 whose sum is divisible by three? - Product a Perfect Cube [12/07/2002] What is the smallest positive integer by which you can multiply 735 so that the product is a perfect cube? - Product Equal to 12 [04/09/2003] How many different numbers have a product equal to 12? - Product of Numbers 1-100 [02/23/2002] I was wondering how to find out how many zeros will be at the end of the product of all the numbers from 1 to 100 without multiplying them all - Product of Powers, Number of Divisors [04/26/2016] A teen wonders why the exponents in an integer's prime factorization have anything to do with its divisor count. Doctor Ian introduces a binary model to make the connection; Doctor Greenie follows up with an algebraic explanation. - Product-Perfect Numbers [02/26/2001] Is there a statement that describes product-perfect numbers? - Products of Digits [03/02/2003] The product of digits in the number 234 is 24: 2*3*4 = 24. Can you describe a general procedure for figuring out how many x-digit numbers have a product equal to p, where x and p are counting numbers? How many different three-digit numbers have a product equal to 12? to - Rectangles and Factors [09/09/1998] How many rectangles can you make with 10 small squares? 5 small squares? 12 small squares? - Rectangular Solids from Blocks [09/25/1998] How many rectangular solids can be made from "n" cube-shaped blocks? - Reducing Fractions [7/4/1996] How do you reduce fractions with different denominators? - Relatively Prime [10/07/1999] What does the term relatively prime mean, and how can you determine if two numbers are relative primes? - The Sieve, and Ceiling, of Eratosthenes [07/27/2012] Before introducing children to the sieve of Eratosthenes, a parent seeks to reconcile different methods found online. Doctor Peterson sees a didactic opportunity in the discrepant steps: how many factors do you actually need to check? - Simplifying Fractions [12/08/2002] Finding and eliminating (cancelling) prime factors. - Simplifying Fractions by Factorizing [05/22/1998] Using factor trees to prime factor and simplify 188/240. - The Smallest Number Which When Divided Leaves Specific Remainders [02/27/2010] Doctor Rick, an eleven year-old, and her father apply least common multiples, modular arithmetic, and the Chinese Remainder Theorem to reason their way to the smallest number which when divided by 3, 7, and 11 leaves remainders 1, 6, and 5, - Table of Factors 1-60 [10/23/2001] Where can I find a factor sheet with the factors 1 - 100? - Thinking about Number Pattern Problems [08/09/2004] How do I figure out the next 2 numbers in the pattern 1, 8, 27, 64, - Tips for Simplifying Fractions [06/03/1998] Can you give my some tips to help me simplify fractions? - Two Mathematicians: Factoring Logic [03/24/2003] Two mathematicians are each assigned a positive integer. They are told that the product of the two numbers is either 8 or 16. Neither knows the other's number... - Unknown Numbers and a Venn Diagram [11/26/2001] The GCF of two numbers is 20 and the LCM is 840. One of the numbers is 120. Explain how to find the other number and use the Venn diagram method - Using Euclid's Algorithm with Three Numbers [11/05/2003] How do I find the GCD of three integers using Euclid's Algorithm? I am not sure where you plug the third integer into the algorithm.

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https://solvedlib.com/n/aneton2-1-1-finding-the-areadegion-dounceowwo-cunjesflnd,15459703

math

Okay, let's go ahead and try to answer this question. All right? We're given a graph off the margin off revenue in the margin, the marginal off the cost. And it's telling you, um, you have the area between these two curves from 50 to 100 and it's asking you what is the meaning. What's the conceptual meaning of this area? Okay, it's also asking you to estimate this area using the midpoint. Cool. Okay, so first, let's talk about the meaning. What's important to remember about the area under the curve? So let me call it a You see, the area under the curve is basically on integral and in particular the integral of a derivative will end up being the original function itself. So it it was one of the results from the fundamental theme of calculus. So if you find the integral off, see prime, it'll be see Okay, Now let's remember our represents revenue C represents cost and the area under the curve. In our case, it's basically equal to the integral off our prime minus c prime. Very roughly speaking, this is equivalent to saying the revenue minus costs, and if you remember the revenue minus costs is basically what we call a prophet. So conceptually the area under the curve represents the profit off this company when the value changes between 50 to 100. Okay, if it's in terms of time, maybe it will be between the 50th day to the 100 day. Um, it could also be related to other factors, like terms or years, etcetera. Okay, now approximating the area. Um, I would have used the travels Lloyd rule, possibly in order to approximate the area. But they're asking us to use the midpoint rule, and there's probably a pretty good reason behind it. So let's actually take a look at it. I am assuming that it's because these points are located at a very nice location. So the mid point right here 75 gives you the height of one for the area under year, and it can be approximated by this rectangle that you see right there. It's a very simple calculation. The base length is 50 and the height is one, so this is fifth the area under the blue curve. If we use the midpoint rule, this is also going to be quite a good estimation. because you can see that the overestimated portion in the underestimated portion are approximately the same area, so it'll be a nice approximation. So the area of this portion is also a nice number. The base area, The base length is 50. The height is too. So it's 100. So it's the same a saying 100 minus 50. So the prophet is approximately equal to 50. So that's how you answer this question and that..

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https://www.garow.me/tags/ceramicart

math

Image by @tom_kemp_*****This is 'the pull' when the clay is squeezed thinner and pulled taller. It's the movement which wheel-based potters need to learn and the one I most wanted to learn. And it's so difficult on this scale: all of your body has to feel what's going on between the fingers of the outside hand and the fingers hidden inside. Everything must be braced and yet fluid, reacting to the changing thickness. The speed at which it's done depends on the speed of the wheel as the lubricated walls can only be touched briefly before they dry up. Too slow and the clay becomes soggy as it waits for you; too fast and you just gouge a helical groove without making it any taller. The clay at the bottom is being pressed on by 5kg so is always thicker than the unburdened clay at the top of the column but the aim of the pull is to distribute those 5kg more or less evenly throughout the shape. But that means squeezing hard at the bottom and gradually less as you move upwards. The torque on the bottom part is huge when your hands are near the top; to avoid distortions I adjust my hands to reduce that force. So much to calculate and yet to avoid thinking about if the whole movement is to achieve its purpose.--(I had to speed this video up by 50% to make it under a minute long)#keramikk#highfire#studiopottery#ceramicart#wheelthrown#handmadeceramics#clayart#stoneware#potsinaction#pottery#potterylove#ceramiclove#ceramique#wheelthrownpottery#ceramic This little set is just a small teaser of all the goodies that can be seen February 16th through March 31st at Taylor Books in Charleston W.V.! It’s part of a collaborative show with @rosaliehaizlett called “Teapots and Time Capsules” #teatimewv! 🔺The show will showcase pieces inspired by our travels around the state of West Virginia and celebrate all our talented artist friends we met along the way.🔺This set was inspired by the cool and funky town of Wheeling! See where the inspiration came from by joining the reception February 16th at 5:30! 🔺Feel free to shoot me a message if you have any questions. And the last stage of the mug glazing process! Once the inside has dried enough to be able to handle the mug, I hold it from the inside and dip the mug. This layer of glaze needs to be fairly thin so I only dip until the whole surface is submerged, then I remove the mug from the glaze. Slight differences in the timing can change the glaze greatly. That means that there is often differences between the mugs glazed with this combination. I don't mind that, as it means no 2 mugs are the same, so each individual can choose their favourite. .......#glaze#pottery#ceramics#glazing#video#artist#art#ceramicart#ilovemyjob#pottersofinstagram#maker#handmade#mug#studiopottery While looking through images from this past year I ran into this lovely photo of our living room interior with one of my large textured vases. Love that summer light coming in the window. Photo by @jessica_sipe #jeremyayerspottery

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CC-MAIN-2018-05

https://www.teacherspayteachers.com/Product/Problem-Solving-Strategies-SMART-Notes-2033044

math

Teach solving word problems using these 8 effective strategies: Look for a Pattern, Make an Organized List, Make a Table, Guess & Check, Work Backward, Use Logical Reasoning, Draw a Diagram, & Solve a Simpler Problem. This is a word problem strategies SMART Notes lesson. Each slide contains a word problem where the titled strategy is demonstrated. Every slides has a correlating image that is an interactive reveal. Click on the graphic to reveal the answer to the word problem. There is also a coordinating Problem Solving Booklet for a math notebook (can be used as an interactive notebook) and Problem of the Week practice sheet (great for homework or weekly word problems challenges, such as POTW). Always Planning! ♥

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CC-MAIN-2017-47

http://www.newport.com/Laser-Damage-Threshold/144932/1033/content.aspx

math

Continuous Wave Lasers For continuous wave (CW) lasers the damage threshold can be calculated from the peak power and beam diameter. For example, to calculate the power density of a 50 mW Nd:YAG laser at 1064 nm with a 0.8 mm beam diameter, first calculate the beam area in terms of centimeters: = 3.14 x (0.4 mm)2 = 3.14 x (0.04 cm)2 = 5.024 x 10-3 cm2 Next, calculate the power density or power per unit area: = Power / Area = 50 mW / (5.024 x 10-3 cm2) = 9.95 W/cm2 For laser beams with a Gaussian intensity profile, multiplying the power density by two for safety is required to accommodate the peak power density at the center of the beam. Remember, damage threshold scales with wavelength, so the damage threshold at 532 nm will be half that at 1064 nm. For pulsed lasers in the range of µsec to nsec, the energy density varies as a function of the square root of the time domain. As a rule of thumb, an optic can withstand 10 times more energy when used with a 1 µsec pulsed laser than a 10 nsec pulsed laser. Suppose, for instance, that the damage threshold is rated at 2 J/cm2 for 10 nsec pulses, but your laser has a 1 µsec pulse length. This means that at the 1 µsec time domain (10-6 sec compared to 10 x 10-9 sec), the optic can withstand 10 times more energy (20 J/cm2). Expressing laser damage threshold (LDT) in equation form: LDT (y) = LDT (x) * (y/x)1/2 as in the example above, x = 1 microsec or 10-6 sec, and y = 10 nsec or 10 x 10-9 sec = 2 J/cm2 * (10-6 sec/ 10 x 10-9 sec)1/2 = 2 J/cm2 * (100)1/2 = 20 J/cm2 In the realm in between pulsed and CW applications (in the msec range), compare both the average power with the CW threshold and the pulse energy density with the energy specification. In the millisecond range, there is a crossover between pulse and CW regimes where you should try to satisfy both criteria. Please note that for pulsed lasers, there may be hot spots in the output beam. Typically, there is a safety factor of 2 or 3 applied to the calculations in order to accommodate for hot spots. Also a factor of 2 is typically applied for Gaussian shaped beams. For Ultrafast pulsed lasers in the psec to fsec regime, the electric field of the pulse attacks the electronic bonds of the dielectric coating. Peak powers can be quite high. Our UF.25 Ultrafast 45° mirror has been laboratory tested and found to have a damage threshold of 1 TW/cm2 with 100 fsec pulses at 5 Hz at 800 nm.

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CC-MAIN-2013-20

https://isohunt.to/torrent_details/6358783/Mathematics-Marvels-and-Milestones-pdf

math

Size 8.158 MB 0 seeders Added 2013-03-15 20:00:29 A. Audichya "Mathematics: Marvels and Milestones" English | ISBN: 8189473409 | 2010 | PDF | 224 pages | 7,8 MB To transport the reader to the highest level of mathematical awareness and to acquaint him with outstanding mathematical achievements. The foremost of them all is the pluralization of mathematics, i.e.,where we had geometry, we now have geometries, and algebras rather than algebra, and number systems rather than number system. It is intended for the intelligent layman who is in search of short and pointed answers to his queries but is little inclined to undertake detai led study of mathematical ideas and concepts. |Mathematics Marvels and Milestones.pdf||8.158 MB|

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CC-MAIN-2017-22

https://www.physicsforums.com/threads/could-you-please-validate-my-answer.363326/

math

1. The problem statement, all variables and given/known data A uniform plank of length 6.00m and mass 30.0 kg rests horizontally across two horizontal bars of a scaffold. the bars are 4.50 m apart, and 1.50m of the plank hangs over one side of the scaffold. How far can a painter of mass 70.0kg walk on the overhanging part of the plank before it tips? 2. Relevant equations w1*d-w2*d=0 3. The attempt at a solution 70g*d*Na*0-30g+((6.2)-1.5)+Nb*4.5=0 70g*d-30g+((6/2)-1.5)=0 d=(30g*(6/2)-1.5)=0 d=0.6428m Is my answer correct and is meters the right unit d should be in?

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CC-MAIN-2019-13

https://short-facts.com/do-all-rectangles-have-perpendicular-and-parallel-sides/

math

Table of Contents Do all rectangles have perpendicular and parallel sides? As we mentioned before, right triangles have perpendicular sides, rectangles have both perpendicular and parallel sides, but other quadrilaterals might not. A regular pentagon has no parallel or perpendicular sides, but a non-regular pentagon might have parallel and perpendicular sides. How do you know if adjacent sides are perpendicular? Perpendicular lines are lines in the same plane that intersect at a right angle. The opposite sides of a rectangle are parallel, and the adjacent sides are perpendicular. How many perpendicular sides does a rectangle have? There are 4 right angles, but 2 pairs of perpendicular lines. What are the adjacent sides of a rectangle? The opposite sides of a rectangle are parallel and equal. The adjacent sides of the rectangle are always perpendicular to each other. Is a parallelogram parallel or perpendicular? Parallelogram are quadrilaterals with opposite sides parallel, but adjacent sides not perpendicular. What does it mean if adjacent sides are perpendicular? What shape has adjacent sides that are perpendicular? Rectangles are parallelograms with perpendicular sides Rectangles have four straight sides. Each pair of opposite sides is parallel, and adjacent sides are perpendicular. This means that every angle in a rectangle is a right (90∘) angle. Does acute triangle have perpendicular sides? Right triangles can come in all sorts of shapes, but they all have that corner, where the right angle sits. You can see that in all the triangles, the right angle has the two sides that are perpendicular to one another. The other two angles are acute angles (meaning they’re less than 90 degrees). Do rectangles have perpendicular diagonals? The diagonals are perpendicular bisectors of each other. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). All angles are right angles by definition. Does triangle have perpendicular lines? The perpendicular bisectors of a triangle are lines passing through the midpoint of each side which are perpendicular to the given side. What shape has perpendicular lines? The trapezium also has perpendicular lines, which have formed two right angles. They are marked with the small square symbol in the corners. Example 2 This parallelogram has no right angles so no perpendicular lines, but it does have two sets of parallel lines.

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CC-MAIN-2024-10

http://www.teach-kids-math-by-model-method.com/model-method-questions-and-answers-non70.html

math

Question posted by Christina from Singapore: Grade/Level: Primary 5 Question solved by "Working Backwards" Heuristics: Jasmine and David have a certain number of stickers. Jasmine gave 1/5 of her stickers to David. David then gave 1/6 of what he had to Jasmine. After that, Jasmine gave 18 stickers to David. In the end, Jasmine had 36 stickers and David had 88 stickers. How many stickers did Jasmine have at first? Step 1: Draw a simplified table to show the number of stickers Jasmine and David had which is 36 and 88 respectively. Step 2: Since we are working backwards, we should undo the last action. As Jasmine had given 18 stickers to David, we should make David return the 18 stickers to her. So we subtract 18 from David and add 18 to Jasmine. Step 3: That leaves Jasmine with 54 and David with 70. Step 4: Since David had given 1/6 of what he had to Jasmine, he should have 5/6 of what he had left. Thus, 70 should be equivalent to 5/6. Step 5: If 70 is equivalent to 5/6, then 1/6 should be equivalent to 14. So we add 14 back to David and subtract 14 from Jasmine. Step 6: That leaves Jasmine with 40 and David with 84. Step 7: Since Jasmine had given 1/5 of what she had to David, she should have 4/5 of what she had left. Thus, 40 should be equivalent to 4/5. Step 8: If 40 is equivalent to 4/5, then 1/5 should be equivalent to 10. So we add 10 back to Jasmine and subtract 10 from David. Step 9: That leaves Jasmine with 50 and David with 74. Go To Top - Model Method - Questions and Answers Therefore, Jasmine had 50 stickers at first.

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CC-MAIN-2023-40

https://flowchart.chartexamples.com/which-statistical-test-to-use-flow-chart/

math

When analyzing contingency tables with two rows and two columns you can use either Fishers exact test or the chi-square test. Continuous Discrete categorical Yes. After some googling Ive seen several attempts of various coverage and quality some not available at the moment. Which statistical test to use flow chart. Flow Chart for Selecting Commonly Used Statistical Tests Type of data. The table then shows one or more statistical tests commonly used given these types of variables but not necessarily the only type of test that could be used and links showing how to do such tests using SAS Stata and SPSS. When your experiment is trying to draw a comparison or find the difference between one categorical with two categories and another continuous variable then you need to work on the two-sample T-test to find the significant difference between the two variables. Master the 6 basic types of tests with simple definitions illustrations and examples. Descriptive Statistics Sometimes the first. Deciding which statistical test to use can seem difficult at first. If youre already up on your statistics you know right away that you want to use a 2-sample t-test which analyzes the. Statistical Test Flow Chart Geo 441. The design of a study is more important than the analysis. Public Health Statistics Professor Dr. Quantitative Methods Part B – Group Comparison II Normal Non-Normal 1 Sample z Test 2 Sample Independent t Test for equal variances Paired Sample t Test Compare two groups Compare. Statistical Tests Heather M. This post is an attempt to mark out the difference between the most. – What question you are asking of the data. Standard t test The most basic type of statistical test for use when you are comparing the means from exactly TWO Groups such as the Control Group versus the Experimental Group. Although for a given data set a one-tailed test will return a smaller p value than a two-tailed test the latter is usually preferred unless there is a watertight case for one-tailed testing. Statistical Test between One Continuous and another Categorical variable. Based on a text book. Ive also seen similar flowcharts in statistics textbooks Ive. The Fishers test is the best choice as it always gives the exact P value. – What type of data you are dealing with. Use these tables as a guide. Many times researchers are content to show histograms to illustrate their point after a flow experiment. For a statistical test to be valid your sample size needs to be large enough to approximate the true distribution of the population being studied. A badly designed study can never be retrieved whereas a poorly analysed study can usually be re-analyzed. Follow the flow chart and click on the links to find the most appropriate statistical analysis for your situation. Flow cytometry data are numbers rich. Section 1 Section 1 contains general information about statistics including key definitions and which summary statistics and tests to choose. Ex Your experiment is studying. Data from experiments can be population measurements percent of CD4 cells for example or it can be expression level median fluorescent expression of CD69 on activated T cells. Rulison October 31 2014 How to Use Flow Chart 1. Use the Which test should I use. An interactive flowchart decision tree to help you decide which statistical test to use with descriptions of each test and links to carry them out in R SPSS and STATA. _ table to allow the student to. It is obvious that we cannot refer to all statistical tests in one editorial. If data is not normal use Y or Z. Prediction Analyses – Quick Definition Prediction tests examine how and to what extent a variable can be predicted from 1 other variables. Describing a sample of data descriptive statistics centrality dispersion replication see also Summary statistics. Type of questionChi-square tests one and two sample RelationshipsDifferences Do you have a true independent variable. Whether your data the. Its important to know. It is primarily a flowchart but is arranged as a tree diagram to give visibility to four branches of. F-test for ratio of two variances Test equality of 2 popn variances Levenestest Brown-Forsythe Test Normality assumption does NOT hold Try response transformation to apply normal-theory methods first Compare 1 popn variance. This approach misses the. Quickly find the right statistical test with this easy overview. What statistical test should I use. If you can answer those questions then you can usually identify a statistical test which can help using the flowchart in. The chi-square test is simpler to calculate but yields only an approximate P. For a person being from a non-statistical background the most confusing aspect of statistics are the fundamental statistical tests and when to use which test. Made by Matthew Jackson. This chart gives an overview of statistics for research. To determine which statistical test to use you need to know. To use this tool please select the applicable goal of the analysis of the data then work through the tables from left to right to select the correct statistical test. Follow the flow chart and click on the links to find the most appropriate statistical analysis for your situation. Start at the top with determining what type of data was collected continuous also known as. Choosing appropriate statistical test Having a well-defined hypothesis helps to distinguish the outcome variable and the exposure variable Answer the following questions to decide which statistical test is appropriate to analysis. – Whether the data is symmetrical or skewed.

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CC-MAIN-2022-40

https://gmatclub.com/forum/dejected-4599.html?kudos=1

math

Thank you Paul and Praetorian... Yes even some of my other friends have advised me to go for LSAT books...and I wud be doing so...but I dont think I have the time to go thru 2-3 LSAT books.If you could recommend me one particular LSAT book...that would be great. As for probability ...yes ur right I'm not a frequent attempter of the P& C quetions coz Im not sure of my answer.But I guess I will be able to solve exam level P & C problems...and not some of the challenging ones posted on this forum. Would like to take your word and get over the disappointment of PP1. I would be taking my second PP a day or two before the exam.I took this exam coz I wanted a PP score which would not be skewed. It was not random...it was the lack of preparation which caused the debacle,sply on the verbal part.RC seems to be my bane...and yes Im working on it. I must have analsyed PP1 atleast 7 times and everytime I see the quant questions I missed ..I kick myself...I would have done those questions on any given day but not at that moment.I felt very comfortable giving the exam...though I had to hurry a bit on the verbal section...I don't consider timing to be a problem.I thought I was doing well on the verbal part when at the end of the exam I got 2 boldfaced question one of which I got wrong. Would like to outline my strategy: 1) Finish stuyding OG-Round 1 till 29 th Feb ( Im sure you both wud agree with me on this) 2)Go over the questions I got wrong during round 1...work them again and make myself dead sure that I have grasped the concept- 8/9 th March. 3)Solve Vstudy/LSAT books/GMAT Plus problems... till 20 th March... 4)Start giving my last round of practice exams...2 PR,2 Kaplan ,2 Arco...and paper tests. 5) I would go for second PP on 27 th March...2 days before the real test. ( All this while I would be giving atleast one practice test on weekends.) I know some of the guys on this forum...wud be thinking why the hell am I posting all this crap...but I just wanted let out some steam...so..! Thanks again guys...your words mean a lot to me. Let me know your views. "Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

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CC-MAIN-2017-13

https://www.coursehero.com/file/5812905/HW-7-Soln/

math

This preview shows pages 1–3. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version.View Full Document Unformatted text preview: ME 309 Fall 2008 Section 1 (Merkle) Solution: Homework # 7 Due Fri 12 September 2008 Problem 7.1 A constant linear acceleration is applied to a container half filled with water (density = 1000 kg/m 3 ). The container dimensions are L = 100 mm long by W = 10 mm wide by H 20 mm high (water height in the cart at rest = 10 mm ). The acceleration is in the horizontal direction and is parallel to the length, L . Assuming there is no sloshing, find: a) A relation for the variation of pressure in the x-, y-, and z-directions (take x as the direction of acceleration, y in the direction perpendicular to the paper, and z in the vertical direction) and a relation for the equilibrium shape of the free surface as a function of the acceleration rate; b) The maximum height of the surface as a function of acceleration rate and the magnitude of acceleration that will just bring the water to the top of the container to allow spillage; c) The pressure in the bottom left corner and the bottom right corner when the free surface just reaches the top of the container. d) Explain how you have used the conservation of mass and the conservation to solve this problem. Solution: Known : L = 0.10, W = 0.01, H = 0.02, Undisturbed water level = 0.01 m . Fluid = water (density = 1000) Find: a) Relation for p ( x, y, z) and equilibrium surface shape. b) Maximum surface height and acceleration required to bring the water to the top of the container; c) The pressure in the two bottom corners when surface reaches top of container. Sketch : above (for un-accelerated case) Assumptions : Free surface shape determined by momentum equations; conservation of mass used to compute surface location; Incompressible fluid, steady state, equilibrium surfaceleading to rigid body problem. Spillage occurs just as the free surface reaches the top of the container. F, a L H Analysis : a) A fluid in a container under constant acceleration is a rigid body system. Conservation of momentum reduces to the following scalar equation in the direction parallel to the acceleration: a F V sys sys M dt dM = = The mass of the system is constant. The acceleration is constant and includes gravitational acceleration as well as the unknown linear acceleration. The volume of the system is a constant (incompressibility assumption with conservation of mass). The only forces acting are the pressure and gravity. The pressure gradient is given by: a x =dp/dx, - g z = dp/dz and dp/dy=0. Integrating these gives the pressure distribution as: z g x a p p z x ref + + = p = (independent of y ). For simplicity, take x = 0 at mid-point of length, and z = 0 at the surface height at x = 0 where the pressure is atmospheric. The integration constant can then be evaluated to give the equation for the pressure as a function of x, y and z (no dependence on y ): z g x a p p z x atm - + = Note z will be negative below the surface indicating that the pressure increases with depth (as expected).... View Full Document This note was uploaded on 03/09/2010 for the course ME 309 taught by Professor Merkle during the Spring '08 term at Purdue University-West Lafayette. - Spring '08

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CC-MAIN-2017-09

https://encyclios.org/pressure

math

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. This definition can be stated for the generic point of any stressed surface (not necessarily flat but without angular points): - considering a surface element small enough to be considered flat (as it can be confused with the corresponding element of the tangent plane locally to the surface); - considering the ratio between the stress component according to the normal to the element and the area of the latter and defining then the pressure in the given point as the limit of this ratio when the area of the element. When describing a container of gas, the term pressure (or absolute pressure) refers to the average force per unit area that the gas exerts on the surface of the container. Absolute pressure is zero-referenced against a perfect vacuum, using an absolute scale, so it is equal to gauge pressure plus atmospheric pressure. - Sound pressure

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https://www.arxiv-vanity.com/papers/cond-mat/0207356/

math

Statistical mechanics of lossy data compression using a non-monotonic perceptron The performance of a lossy data compression scheme for uniformly biased Boolean messages is investigated via methods of statistical mechanics. Inspired by a formal similarity to the storage capacity problem in neural network research, we utilize a perceptron of which the transfer function is appropriately designed in order to compress and decode the messages. Employing the replica method, we analytically show that our scheme can achieve the optimal performance known in the framework of lossy compression in most cases when the code length becomes infinite. The validity of the obtained results is numerically confirmed. pacs:89.90.+n, 02.50.-r, 05.50.+q, 75.10.Hk Recent active research on error-correcting codes (ECC) has revealed a great similarity between information theory (IT) and statistical mechanics (SM) MacKay ; Richardson ; Aji ; us_EPL ; us_PRL ; Sourlas_nature ; NishimoriWong . As some of these studies have shown that methods from SM can be useful in IT, it is natural to expect that a similar approach may also bring about novel developments in fields other than ECC. The purpose of the present paper is to offer such an example. More specifically, we herein employ methods from SM to analyze and develop a scheme of data compression. Data compression is generally classified into two categories; lossless and lossy compression bi:Cover . The purpose of lossless compression is to reduce the size of messages in information representation under the constraint of perfect retrieval. The message length in the framework of lossy compression can be further reduced by allowing a certain amount of distortion when the original expression is retrieved. The possibility of lossless compression was first pointed out by Shannon in 1948 in the source coding theorem Shannon , whereas the counterpart of lossy compression, termed the rate-distortion theorem, was presented in another paper by Shannon more than ten years later RD . Both of these theorems provide the best possible compression performance in each framework. However, their proofs are not constructive and suggest few clues for how to design practical codes. After much effort had been made for achieving the optimal performance in practical time scales, a practical lossless compression code that asymptotically saturates the source-coding limit was discovered Jelinek . Nevertheless, thus far, regarding lossy compression, no algorithm which can be performed in a practical time scale saturating the optimal performance predicted by the rate-distortion theory has been found, even for simple information sources. Therefore, the quest for better lossy compression codes remains one of the important problems in IT bi:Cover ; fixed ; Yamamoto ; Murayama_RD . Therefore, we focus on designing an efficient lossy compression code for a simple information source of uniformly biased Boolean sequences. Constructing a scheme of data compression requires implementation of a map from compressed data of which the redundancy should be minimized, to the original message which is somewhat biased and, therefore, seems redundant. However, since the summation over the Boolean field generally reduces the statistical bias of the data, constructing such a map for the aforementioned purpose by only linear operations is difficult, although the best performance can be achieved by such linear maps in the case of ECC MacKay ; Aji ; us_EPL ; us_PRL ; Sourlas_nature and lossless compression Murayama . In contrast, producing a biased output from an unbiased input is relatively easy when a non-linear map is used. Therefore, we will employ a perceptron of which the transfer function is optimally designed in order to devise a lossy compression scheme. The present paper is organized as follows. In the next section, we briefly introduce the framework of lossy data compression, providing the optimal compression performance which is often expressed as the rate-distortion function in the case of the uniformly biased Boolean sequences. In section III, we explain how to employ a non-monotonic perceptron to compress and decode a given message. The ability and limitations of the proposed scheme are examined using the replica method in section IV. Due to a specific (mirror) symmetry that we impose on the transfer function of the perceptron, one can analytically show that the proposed method can saturate the rate-distortion function for most choices of parameters when the code length becomes infinite. The obtained results are numerically validated by means of the extrapolation on data from systems of finite size in section V. The final section is devoted to summary and discussion. Ii Lossy Data Compression Let us first provide the framework of lossy data compression. In a general scenario, a redundant original message of random variables , which we assume here as a Boolean sequence , is compressed into a shorter (Boolean) expression . In the decoding phase, the compressed expression is mapped to a representative message in order to retrieve the original expression (Fig. 1). In the source coding theorem, it is shown that perfect retrieval is possible if the compression rate is greater than the entropy per bit of the message when the message lengths and become infinite. On the other hand, in the framework of lossy data compression, the achievable compression rate can be further reduced allowing a certain amount of distortion between the original and representative messages and . A measure to evaluate the distortion is termed the distortion function, which is denoted as . Here, we employ the Hamming distance as is frequently used for Boolean messages. Since the original message is assumed to be generated randomly, it is natural to evaluate the average of Eq. (1). This can be performed by averaging with respect to the joint probability of and as By allowing the average distortion per bit up to a given permissible error level , the achievable compression rate can be reduced below the entropy per bit. This limit is termed the rate-distortion function, which provides the optimal compression performance in the framework of lossy compression. The rate-distortion function is formally obtained as a solution of a minimization problem with respect to the mutual information between and bi:Cover . Unfortunately, solving the problem is generally difficult and analytical expressions of are not known in most cases. The uniformly biased Boolean message in which each component is generated independently from an identical distribution is one of the exceptional models for which can be analytically obtained. For this simple source, the rate-distortion function becomes However, it should be addressed here that a practical code that saturates this limit has not yet been reported, even for this simplest model. Therefore, in the following, we focus on this information source and look for a code that saturates Eq. (4) examining properties required for good compression performance. Iii Compression by Perceptron In a good compression code for the uniformly biased source, it is conjectured that compressed expressions should have the following properties: In order to minimize loss of information in the original expressions, the entropy per bit in must be maximized. This implies that the components of are preferably unbiased and uncorrelated. In order to reduce the distortion, the representative message should be placed close to the typical sequences of the original messages which are biased. Unfortunately, it is difficult to construct a code that satisfies both of the above two requirements utilizing only linear transformations over the Boolean field while such maps provide the optimal performance in the case of ECC MacKay ; Aji ; us_EPL ; us_PRL ; Sourlas_nature and lossless compression Murayama . This is because a linear transformation generally reduces statistical bias in messages, which implies that the second requirement (II) cannot be realized for unbiased and uncorrelated compressed expressions that are preferred in the first requirement (I). One possible method to design a code that has the above properties is to introduce a non-linear transformation. A perceptron provides one of the simplest schemes for carrying out this task. In order to simplify notations, let us replace all the Boolean expressions with binary ones . By this, we can construct a non-linear map from the compressed message to the retrieved sequence utilizing a perceptron as where are fixed -dimensional vectors to specify the map and is a transfer function from a real number to a binary variable that should be optimally designed. Since each component of the original message is produced independently, it is preferred to minimize the correlations among components of a representative vector , which intuitively indicates that random selection of may provide a good performance. Therefore, we hereafter assume that vectors are independently drawn from the -dimensional normal distribution . Based on the non-linear map (5), a lossy compression scheme can be defined as follows: Compression: For a given message , find a vector that minimizes the distortion , where the representative vector which is generated from by Eq. (5). The obtained is the compressed message. Decoding: Given the compressed message , the representative vector produced by Eq. (5) provides the approximate message for the original message. Here, we should notice that the formulation of the current problem has become somewhat similar to that for the storage capacity evaluation of the Ising perceptron bi:Krauth ; bi:Gardner regarding , and as “Ising couplings”, “random input pattern” and “random output”, respectively. Actually, the rate-distortion limit in the current framework for and can be calculated as the inverse of the storage capacity of the Ising perceptron, . This observation implies that the simplest choice of the transfer function , where for and otherwise, does not saturate the rate-distortion function (4). This is because the well-known storage capacity of the simple Ising perceptron, , means that the “compression limit” achievable by this monotonic transfer function becomes and far from the value provided by Eq. (4) for this parameter choice . We also examined the performances obtained by the monotonic transfer function for biased messages by introducing an adaptive threshold in our previous study Hosaka and found that the discrepancy from the rate-distortion function becomes large in particular for relatively high while fairly good performance is observed for low rate regions. Therefore, we have to design a non-trivial function in order to achieve the rate-distortion limit, which may seem hopeless as there are infinitely many degrees of freedom to be tuned. However, a useful clue exists in the literature of perceptrons, which have been investigated extensively during the last decade. In the study of neural network, it is widely known that employing a non-monotonic transfer function can highly increase the storage capacity of perceptrons Monasson . In particular, Bex et al. reported that the capacity of the Ising perceptron that has a transfer function of the reversed-wedge type can be maximized to by setting bi:VdB , which implies that the rate-distortion limit is achieved for the case of and in the current context. Although not explicitly pointed out in their paper, the most significant feature observed for this parameter choice is that the Edwards-Anderson (EA) order parameter vanishes to zero, where denotes the average over the posterior distribution given and . This implies that the dynamical variable in the posterior distribution given and is unbiased and, therefore, the entropy is maximized, which meets the first requirement (I) addressed above. Thus, designing a transfer function so as to make the EA order parameter vanish seems promising as the first discipline for constructing a good compression code. However, the reversed-wedge type transfer function is not fully satisfactory for the present purpose. This is because this function cannot produce a biased sequence due to the symmetry , which means that the second requirement (II) provided above would not be satisfied for . Hence, another candidate for which the EA parameter vanishes and the bias of the output can be easily controlled must be found. A function that provides these properties was once introduced for reducing noise in signal processing, such as bi:Jort ; Inoue (Fig. 2). Since this locally activated function has mirror symmetry , both and provide identical output for any input, which means that the EA parameter is likely to be zero. Moreover, one can easily control the bias of output sequences by adjusting the value of the threshold parameter . Therefore, this transfer function looks highly promising as a useful building-block for constructing a good compression code. In the following two sections, we examine the validity of the above speculation, analytically and numerically evaluating the performance obtained by the locally activated transfer function . Iv Analytical Evaluation We here analytically evaluate the typical performance of the proposed compression scheme using the replica method. Our goal is to calculate the minimum permissible average distortion when the compression rate is fixed. The analysis is similar to that of the storage capacity for perceptrons. Employing the Ising spin expression, the Hamming distortion can be represented as Then, for a given original message and vectors , the number of dynamical variables which provide a fixed Hamming distortion , can be expressed as Since and are randomly generated predetermined variables, the quenched average of the entropy per bit over these parameters to which the raw entropy per bit becomes identical for most realizations of and , is naturally introduced for investigating the typical properties. This can be performed by the replica method , analytically continuing the expressions of obtained for natural numbers to non-negative real number beyond ; bi:Nishimori . When is a natural number, can be expanded to a summation over -replicated systems as , where the subscript denotes a replica index. Inserting an identity into this expression and utilizing the Fourier expression of the delta function we can calculate the moment for natural numbers as where is an matrix of which elements are given by the parameters and . In the thermodynamic limit keeping the compression rate finite, this integral can be evaluated via a saddle point problem with respect to macroscopic variables , and . In order to proceed further, a certain ansatz about the symmetry of the replica indices must be assumed. We here assume the simplest one, that is, the replica symmetric (RS) ansatz for which the saddle point expression of Eq. (16) is likely to hold for any real number . Taking the limit of this expression, we obtain where , , and . denotes the extremization. Under this RS ansatz, the macroscopic variable indicates the EA order parameter as . The validity of this solution will be examined later. Since the dynamical variable is discrete in the current system, the entropy (18) must be non-negative. This indicates that the achievable limit for a fixed compression rate and a transfer function which is specified by the threshold parameter can be characterized by a transition depicted in Fig. 3. Utilizing the Legendre transformation , the free energy for a fixed inverse temperature , which is an external parameter and should be generally distinguished from the variational variable in Eq. (18), can be derived from . This implies that the distortion that minimizes and of which the value is computed from as can be achieved by randomly drawing from the canonical distribution which is provided by the given . For a modest , the achieved distortion is determined as a point for which the slope of becomes identical to and (Fig. 3 (a)). As becomes higher, moves to the left, which indicates that the distortion can be reduced by introducing a lower temperature. However, at a critical value characterized by the condition (Fig. 3 (b)), the number of states that achieve which is the typical value of vanishes to zero. Therefore, for , is fixed to and the distortion is not achievable (Fig. 3 (c)). The above argument indicates that the limit of the achievable distortion for a given rate and a threshold parameter in the current scheme can be evaluated from conditions being parameterized by the inverse temperature . Due to the mirror symmetry , becomes the saddle point solution for the extremization problem (18) as we speculated in the previous section, and no other solution is discovered. Inserting into the right-hand side of Eq. (18) and employing the Legendre transformation, the free energy is obtained as The rate-distortion function represents the optimal performance that can be achieved by appropriately tuning the scheme of compression. This means that can be evaluated as the convex hull of a region in the - plane defined by Eqs. (22) and (25) by varying the inverse temperature and the threshold parameter (or ). Minimizing for a fixed , one can show that the relations are satisfied at the convex hull, which offers the optimal choice of parameters and as functions of a given permissible distortion and a bias . Plugging these into Eq. (25), we obtain which is identical to the rate-distortion function for uniformly biased binary sources (4). The results obtained thus far indicate that the proposed scheme achieves the rate-distortion limit when the threshold parameter is optimally adjusted. However, since the calculation is based on the RS ansatz, we must confirm the validity of assuming this specific solution. We therefore examined two possible scenarios for the breakdown of the RS solution. The first scenario is that the local stability against the fluctuations for disturbing the replica symmetry is broken, which is often termed the Almeida-Thouless (AT) instability bi:AT , and can be examined by evaluating the excitation of the free energy around the RS solution. As the current RS solution can be simply expressed as , the condition for this solution to be stable can be analytically obtained as In most cases, the RS solution satisfies the above condition and, therefore, does not exhibit the AT instability. However, we found numerically that for relatively high values of distortion , can become slightly smaller than for a very narrow parameter region, , which indicates the necessity of introducing the replica symmetry breaking (RSB) solutions. This is also supported analytically by the fact that the inequality holds for in the vicinity of . Nevertheless, this instability may not be serious in practice, because the area of the region , where the RS solution becomes unstable, is extremely small, as indicated by Fig. 5 (a). The other scenario is the coexistence of an RSB solution that is thermodynamically dominant while the RS solution is locally stable. In order to examine this possibility, we solved the saddle point problem assuming the one-step RSB (1RSB) ansatz in several cases for which the RS solution is locally stable. However, no 1RSB solution was discovered for . Therefore, we concluded that this scenario need not be taken into account in the current system. These insubstantial roles of RSB may seem somewhat surprising since significant RSB effects above the storage capacity have been reported in the research of perceptrons with continuous couplings bi:Jort ; Monasson . However, this may be explained by the fact that, in most cases, RSB solutions for Ising couplings can be expressed by the RS solutions adjusting temperature appropriately, even if non-monotonic transfer functions are used bi:Krauth ; Inoue . V Numerical Validation Although the analysis in the previous section theoretically indicates that the proposed scheme is likely to exhibit a good compression performance, it is still important to confirm it by experiments. Therefore, we have performed numerical simulations implementing the proposed scheme in systems of finite size. In these experiments, an exhaustive search was performed in order to minimize the distortion so as to compress a given message into , which implies that implementing the current scheme in a large system is difficult. Therefore, validation was performed by extrapolating the numerically obtained data, changing the system size from to . Figure 4 shows the average distortions obtained from experiments for (a) unbiased () and (b) biased () messages, varying the system size and the compression rate . For each , the threshold parameter is tuned to the value determined using Eqs. (26), (27) and the rate-distortion function in order to optimize the performance. These data indicate that the finite size effect is relatively large in the present system, which is similar to the case of the storage capacity problem bi:Opper , and do not necessarily seem consistent with the theoretical prediction obtained in the previous section. However, the extrapolated values obtained from the quadratic fitting with respect to are highly consistent with curves of the rate-distortion function (Fig. 5 (a) and (b)), including one point in the region where the AT stability is broken (inset of Fig. 5(a)), which strongly supports the validity and efficacy of our calculation based on the RS ansatz. Vi Summary and Discussion We have investigated a lossy data compression scheme of uniformly biased Boolean messages employing a perceptron of which the transfer function is non-monotonic. Designing the transfer function based on the properties required for good compression codes, we have constructed a scheme that saturates the rate-distortion function that represents the optimal performance in the framework of lossy compression in most cases. It is known that a non-monotonic single layer perceptron can be regarded as equivalent to certain types of multi-layered networks, as in the case of parity and committee machines. Although tuning the input-output relation in multi-layered networks would be more complicated, employing such devices might be useful in practice because several heuristic algorithms that could be used for encoding in the present context have been proposed and investigated Mitchison ; Nokura . In real world problems, the redundancy of information sources is not necessarily represented as a uniform bias; but rather is often given as non-trivial correlations among components of a message. Although it is just unfortunate that the direct employment of the current method may not show a good performance in such cases, the locally activated transfer function that we have introduced herein could serve as a useful building-block to be used in conjunction with a set of connection vectors that are appropriately correlated for approximately expressing the given information source, because by using this function, we can easily control the input-output relation suppressing the bias of the compressed message to zero, no matter how the redundancy is represented. Finally, although we have confirmed that our method exhibits a good performance when executed optimally in a large system, the computational cost for compressing a message may render the proposed method impractical. One promising approach for resolving this difficulty is to employ efficient approximation algorithms such as various methods of the Monte Carlo sampling bi:Sampling and of the mean field approximation bi:Advanced_Mean_Field_Methods . Another possibility is to reduce the finite size effect by further tuning the profile of the transfer function. Investigation of these subjects is currently under way. Grants-in-Aid Nos. 13780208, 14084206 (YK) and 12640369 (HN) from MEXT are gratefully acknowledged. TH would like to thank T. Murayama for informing his preprint and valuable comments, and YK would like to acknowledge D. Saad, H. Yamamoto and Y. Matsunaga for their useful discussions. - (1) N. Sourlas, Nature (London) 339, 693 (1989). - (2) D. J. C. MacKay and R. M. Neal, Electron. Lett. 33, 457 (1997). - (3) H. Nishimori and K. Y. Michael Wong, Phys. Rev. E 60, 132 (1999). - (4) Y. Kabashima and D. Saad, Europhys. Lett. 45, 97 (1999). - (5) Y. Kabashima, T. Murayama, and D. Saad, Phys. Rev. Lett. 84, 1355 (2000). - (6) T. Richardson and R. Urbanke, IEEE Trans. Inf. Theory 47, 599 (2001). - (7) S. Aji, H. Jin, A. Khandekar, D. J. C. MacKay, and R. J. McEliece, to appear in the Proc. of the IMA 1999 Summer Program: Codes, Systems and Graphical Models, Aug 1999. - (8) T. M. Cover and J. A. Thomas, Elements of Information Theory (John Wiley & Sons, 1991). - (9) C. E. Shannon, Bell Syst. Tech. J. 27, 379 (1948); 27, 623 (1948). - (10) C. E. Shannon, IRE National Convention Record, Part 4 142 (1959). - (11) F. Jelinek, Probabilistic Information Theory (McGraw-Hill, 1968). - (12) E. Yang, Z. Zhang, and T. Berger, IEEE Trans. Inf. Theory 43, 1465 (1997). - (13) Y. Matsunaga and H. Yamamoto, Proceedings of 2002 IEEE International Symposium on Information Theory, 461 (2002). - (14) T. Murayama and M. Okada, in preparation. - (15) T. Murayama, J. Phys. A: Math. and Gen. 35, L95 (2002). - (16) E. Gardner and B. Derrida, J. Phys. A: Math. and Gen. 21, 271 (1988). - (17) W. Krauth and M. Mézard, J. Phys. (France) 50, 3057 (1989). - (18) T. Hosaka, Statistical Mechanics of Lossy Data Compression, Master’s Thesis, Department of Computational Intelligence and Systems Science, Tokyo Institute of Technology (2002). - (19) R. Monasson and D. O’kane, Europhys. Lett. 27, 85 (1994). - (20) G. J. Bex, R. Serneels, and C. Van den Broeck, Phys. Rev. E 51, 6309 (1995). - (21) J. van Mourik, K. Y. M. Wong, and D. Bollé, J. Phys. A: Math. and Gen. 33, L53 (2000). - (22) J. Inoue and D.M. Carlucchi, cond-mat/9806037 (1998). - (23) M. Mézard, G. Parisi, and M. A. Virasoro, Spin Glass Theory and Beyond (World Scientific, Singapore, 1987). - (24) H. Nishimori, Statistical Physics of Spin Glasses and Information Processing (Oxford University Press, 2001). - (25) J. R. L. de Almeida and D. J. Thouless, J. Phys. A: Math. and Gen. 11, 983 (1977). - (26) W. Krauth and M. Opper, J. Phys. A: Math. and Gen. 22, L519 (1989). - (27) G.J. Mitchison and R.M. Durbin, Biol. Sybern. 60, 345 (1989). - (28) K. Nokura, Phys. Rev. E 49, 5812 (1994). - (29) R. M. Neal, Bayesian Learning for Neural Networks (Springer Verlag, 1996). - (30) M. Opper and D. Saad (Eds.), Advanced Mean Field Methods (MIT Press, 2001).

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https://www.clutchprep.com/analytical-chemistry/practice-problems/148834/consider-the-following-equilibria-in-aqueous-solution-ag-cl-agcl-aq-k-2-0-x-103c

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Consider the following equilibria in aqueous solution: Ag+ + Cl– ⇌ AgCl(aq); K = 2.0 × 103 Calculate the numerical value of the equilibrium constant for the reaction AgCl(s) ⇌ AgCl(aq). Frequently Asked Questions What scientific concept do you need to know in order to solve this problem? Our tutors have indicated that to solve this problem you will need to apply the The Equilibrium State concept. You can view video lessons to learn The Equilibrium State. Or if you need more The Equilibrium State practice, you can also practice The Equilibrium State practice problems. What is the difficulty of this problem? Our tutors rated the difficulty ofConsider the following equilibria in aqueous solution:Ag+ + ...as low difficulty. How long does this problem take to solve? Our expert Analytical Chemistry tutor, Jules took 1 minute and 6 seconds to solve this problem. You can follow their steps in the video explanation above. What professor is this problem relevant for? Based on our data, we think this problem is relevant for Professor Frossard's class at UGA.

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https://everything.explained.today/Range_of_a_function/

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Range of a function explained In mathematics, the range of a function may refer to either of two closely related concepts: Given two sets and, a binary relation between and is a (total) function (from to) if for every in there is exactly one in such that relates to . The sets and are called domain and codomain of, respectively. The image of is then the subset of consisting of only those elements of such that there is at least one in with . As the term "range" can have different meanings, it is considered a good practice to define it the first time it is used in a textbook or article. Older books, when they use the word "range", tend to use it to mean what is now called the codomain. More modern books, if they use the word "range" at all, generally use it to mean what is now called the image. To avoid any confusion, a number of modern books don't use the word "range" at all. Elaboration and example Given a function , the range of , sometimes denoted may refer to the codomain or target set (i.e., the set into which all of the output of is constrained to fall), or to , the image of the domain of (i.e., the subset of consisting of all actual outputs of ). The image of a function is always a subset of the codomain of the function. As an example of the two different usages, consider the function as it is used in real analysis (that is, as a function that inputs a real number and outputs its square). In this case, its codomain is the set of real numbers , but its image is the set of non-negative real numbers is never negative if is real. For this function, if we use "range" to mean codomain , it refers to }; if we use "range" to mean image , it refers to In many cases, the image and the codomain can coincide. For example, consider the function , which inputs a real number and outputs its double. For this function, the codomain and the image are the same (both being the set of real numbers), so the word range is unambiguous. - Book: Childs , Lindsay N. . A Concrete Introduction to Higher Algebra . . 3rd . Springer . 2009 . 978-0-387-74527-5 . 173498962 . 10.1007/978-0-387-74725-5. - Book: David S. . Dummit . Richard M. . Foote . Abstract Algebra . 3rd . Wiley . 2004 . 978-0-471-43334-7 . 52559229. - Book: Hungerford , Thomas W. . Thomas W. Hungerford . Algebra . Springer . . 73 . 1974 . 0-387-90518-9 . 703268 . 10.1007/978-1-4612-6101-8. - Book: Rudin . Functional Analysis . 2nd . McGraw Hill . 1991 . 0-07-054236-8 . registration . Notes and References - Web site: Weisstein. Eric W.. Range. 2020-08-28. mathworld.wolfram.com. en. - Web site: Nykamp. Duane. Range definition. live. August 28, 2020. Math Insight.

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https://lysyfytelelikud.awordathought.com/mathematical-principles-of-natural-philosophy-book-4147zj.php

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5 edition of Mathematical principles of natural philosophy found in the catalog. Mathematical principles of natural philosophy in Berkeley, University of California Press, 1960 Written in English Includes reproduction of the t.p. of the 1st ed. of the Principia, London, 1687. |Other titles||The system of the world| |The Physical Object| |Pagination||2 v. in 1 (xxxiii, 680 p.) ;| |Number of Pages||680| Isaac Newton's The Mathematical Principles of Natural Philosophy translated by Andrew Motte and published in two volumes in remains the first and only translation of Newton's Philosophia /5(9). The Mathematical Principles of Natural Philosophy, Volume 1 Sir Isaac Newton, William Emerson, John Machin Full view - The Mathematical Principles of Natural Philosophy, Volume 35/5(1). Newton's Principia: the mathematical principles of natural philosophy Item Preview remove-circle Early American mathematics books. CU-BANC. Publication date c Topics Newton, Isaac, Sir, , Mechanics -- Early works to , Celestial mechanics -- Early works to Publisher New-York: Published by Daniel AdeePages: The book has an active table of contents for readers to access each chapter of the following books: 1. THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY Isaac Newton 2. The History of the Ancient Physics - Adam Smith 3. The History of the Ancient Logics and Metaphysics - Adam Smith 4. Of the EXTERNAL SENSES - Adam Smith 5. Of the NATURE of that IMITATION Brand: AS Team. Find books like The Principia: Mathematical Principles of Natural Philosophy from the world’s largest community of readers. Goodreads members who liked T. The Principia: Mathematical Principles of Natural Philosophy: Original Edition (Latin Edition) by Newton, Isaac and a great selection of related books, art . longitudinal study of psychological changes occuring during pregnancy. short-title list of printed books in [the] Strong Room. Makens guide to Mexican train travel. Effect of neutron irradiation on fatigue crack propagation in Type 316 stainless steel at 649C̊ Growth in open economies The Wild Mustang Good housekeeping the test kitchen cookie lovers cookbook Letters of John Keats to Fanny Brawne The Puerto Ricans. Addendum #2 to the Western Washington University campus master plan final environmental impact statement. account of the chapter erected by William, titular Bishop of Chalcedon, and ordinary of England and Scotland Human Factors in Budgeting Compact disc market--phase II theory of legal duties and rights The Principia: Mathematical Principles of Natural Philosophy: Newton, Isaac: : Books/5(22). Mathematical Principles of Natural Philosophy: Newton, Isaac, Sir: : Books. Out of Print--Limited Availability. Available as a Kindle eBook. Kindle eBooks can be read on any device with the free Kindle app/5(9). Mathematical Principles of Natural Philosophy, often referred to as simply the Principia, is a work in three books by Isaac Newton, in Latin, first published 5 July /5(15). Books Advanced Search New Releases Best Sellers & More Children's Books Textbooks Textbook Rentals Sell Us Your Books Best Books of the Month of 49 results for Books: "the principia mathematical principles of natural philosophy". Some of the techniques listed in The Principia: Mathematical Principles of Natural Philosophy may require a sound knowledge of Hypnosis, users are advised to either leave those sections or must have a basic understanding of the subject before practicing them/5. Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy, usually called simply the Principia), which appeared in Here was a new physics that applied equally well to terrestrial and celestial bodies. Copernicus, Kepler, and Galileo were all justified by Newton’s analysis of forces. The Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy) is a trilogy, written by Isaac Newton and published on 5 July The three-book work describes physics and states Newton's laws of motion and the derivation of Kepler's Laws, and observations on gravity. Section I in Book I of Isaac Newton’s Philosophiˆ Naturalis Principia Mathematica is reproduced here, translated into English by Andrew Motte. Motte’s translation of Newton’s Principia, entitled The Mathematical Principles of Natural Philosophy was rst published in. The Mathematical Principles of Natural Philosophy An Annotated Translation of the Principia. Get access. As a book of great insight and ingenuity, it has raised our understanding of the power of mathematics more than any other work. of the Principia will enable any reader with a good understanding of elementary mathematics to easily Cited by: The key contribution of Newton to the modern science and engineering was the book THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY. THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY is also called Principia. Book I of the Principia details the foundations of the science of mechanics. Book II inaugurates the theory of by: 2 The Mathematical Principles of Natural Philosophy 3 The Method of Fluxions & the Infinite Serues with its Applications to the Geometry of the 4 Universal Arithmetic - A Treatise of Arithmatical Composition & In his monumental work Philosophiae Naturalis Principia Mathematica, known familiarly as the Principia, Isaac Newton laid out in mathematical terms the principles of time, force, and motion that have guided the development of modern physical after more than three centuries and the revolutions of Einsteinian relativity and quantum mechanics, Newtonian phy/5. Newton's Principia the mathematical principles of natural philosophy 1st American ed., carefully rev. and corr. / with a life of the author, by N.W. Chittenden. by Sir Isaac Newton ★ ★ ★ ; 5 Ratings 77 Want to read; 8 Currently reading; 5 Have read/5(5). The Mathematical Principles of Natural Philosophy () by Isaac Newton, translated by Andrew Motte Book III: Rules of Reasoning in Philosophy Phænomena, or Appearances →. The same things being s upposed, I s ay that the ultimate ratio of the arc, chord, and tangent, any one to any other, is the ratio of equality. Fig. For while the point B approaches the point A, con s ider always AB and AD as produc'd to the remote point b and d; and parallel to the s ecant BD draw bd; and let the arc Acb be always s imilar to the arc ACB. Philosophiæ Naturalis Principia Mathematica, Latin for "Mathematical Principles of Natural Philosophy", often called the Principia (sometimes Principia Mathematica), is a work in three books by Isaac Newton, first published 5 July Newton also published two further editions, in. Isaac Newton's book Philosophiae Naturalis Principia Mathematica (), whose title translates to "Mathematical Principles of Natural Philosophy", reflects the then-current use of the words "natural philosophy", akin to "systematic study of nature". Mathematical principles of natural philosophy: Book I: Method of first and last ratios -- Determination of centripetal forces -- Motion of bodies in eccentric conic sections -- Finding of elliptic, parabolic, and hyperbolic orbits from the focus given -- How the orbits are to be found when neither focus is given -- How the motions are to be found in given orbits -- Rectilinear ascent and Pages: Everyone should read this book, and this seems to be one of the better scans. Cheers. The Mathematical Principles of Natural Philosophy Sir Isaac Newton Snippet view - Newton's Principia: The Mathematical Principles of Natural Philosophy, N. 5/5(6). Internet Archive BookReader Newton's Principia: the mathematical principles of natural philosophy. The Mathematical Principles of Natural Philosophy () by Isaac Newton, III, Book II) as the area DR × AB is more a mathematical hypothesis than a physical one. In mediums void of all tenacity, the resistances made to bodies are in the duplicate ratio of the velocities.NEWTON'S PRINCIPIA, The Mathematical Principles of Natural ated into English by Andrew Motte, to which is added Newton's System of the World. Newton, Isaac Published by Ivison & Phinney: NY ().Philosophiæ Naturalis Principia Mathematica (Latin for Mathematical Principles of Natural Philosophy), often referred to as simply the Principia, is a work in three books by Isaac Newton, in Latin, first published 5 July

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http://media-uk.avinor.no/news

math

News • Feb 01, 2018 11:54 GMT More details of Norwegian’s S18 expansion plans at Avinor’s airports are revealed, with new routes to Albania, France and Spain to be introduced from February onwards. News • Oct 20, 2017 11:39 BST Avinor’s Bergen and Tromsø airports will welcome the new 114-seat jet in a world first with airline partner Widerøe next April. News • Feb 01, 2017 11:24 GMT The contracts that have been awarded are for the sale of advertising on outdoor and public advertising surfaces and are effective from 1 January 2018. News • Oct 24, 2016 14:15 BST From April 2017, Widerøe will commence direct flights between Avinor Oslo Airport and Lofoten and inner Helgeland. These new flights are in addition to the established direct routes to Sandnessjøen and Brønnøysund. A direct route to Stokmarknes in Vesterålen is also being introduced for the mid-summer programme.

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https://www.teacherspayteachers.com/Product/FRACTIONS-MATH-TALK-Grades-3-5-BUNDLE-1645827

math

Fractions Math Talk is a great way to support students in developing their Conceptual Understanding of Fractions, a major domain in the Common Core Standards in Mathematics for Grades 3 to 5. It's a great way to start math time each day. Project a Fractions Math Talk slide and give students time to put down their thinking in their math journals before whole group discussion. You can also use these at math centers. This FRACTIONS MATH BUNDLE includes 3 PACKS of BIG IDEAS (you save more with purchase of the bundle). PACK 1 BIG IDEAS 1) Understand a Fraction as partitioning a whole (or a group) into equal parts. 2) Identical wholes need not have the same shape. 3) When comparing fractions, the same whole has to be referenced. 4) Decomposing a whole into more equal shares creates smaller shares. 5) Two fractions are equivalent if they are the same size. 6) Understand a fraction as a number on a number line. PACK 2 BIG IDEAS: 1) Understand a fraction as a number on a number line. 2) Compare fractions by reasoning about their size. 3) Understand a fraction can be decomposed into a sum of fractions with the same denominator (3/8 = 2/8 + 1/8). 4) Add or subtract fractions with unlike denominators. PACK 3 BIG IDEAS: 1) Understand a fraction as a multiple of the unit fractions with the same denominator (e.g. 3/8 = 3 x 1/8). 2) Apply and extend previous understandings of multiplication to multiply a fraction by a fraction. 3) Apply and extend previous understandings of division to divide a whole by a fraction. 4) Apply and extend previous understandings of division to divide a fraction by a fraction

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https://www.rcsb.org/structure/7BWJ

math

Human neutralizing antibodies elicited by SARS-CoV-2 infection.Ju, B., Zhang, Q., Ge, J., Wang, R., Sun, J., Ge, X., Yu, J., Shan, S., Zhou, B., Song, S., Tang, X., Yu, J., Lan, J., Yuan, J., Wang, H., Zhao, J., Zhang, S., Wang, Y., Shi, X., Liu, L., Zhao, J., Wang, X., Zhang, Z., Zhang, L. (2020) Nature 584: 115-119 - PubMed: 32454513 - DOI: 10.1038/s41586-020-2380-z - Primary Citation of Related Structures: - PubMed Abstract: The emerging coronavirus SARS-CoV-2 pandemic presents a global health emergency in urgent need of interventions 1-3 . SARS-CoV-2 entry into the target cells depends on binding between the receptor-binding domain (RBD) of the viral Spike pr ... The emerging coronavirus SARS-CoV-2 pandemic presents a global health emergency in urgent need of interventions 1-3 . SARS-CoV-2 entry into the target cells depends on binding between the receptor-binding domain (RBD) of the viral Spike protein and the ACE2 cell receptor 2,4-6 . Here, we report the isolation and characterization of 206 RBD-specific monoclonal antibodies derived from single B cells of eight SARS-CoV-2 infected individuals. We identified antibodies with potent anti-SARS-CoV-2 neutralization activity that correlates with their competitive capacity with ACE2 for RBD binding. Surprisingly, neither the anti-SARS-CoV-2 antibodies nor the infected plasma cross-reacted with SARS-CoV or MERS-CoV RBDs, although substantial plasma cross-reactivity to their trimeric Spike proteins was found. Crystal structure analysis of RBD-bound antibody revealed steric hindrance that inhibits viral engagement with ACE2 and thereby blocks viral entry. These findings suggest that anti-RBD antibodies are viral species-specific inhibitors. The antibodies identified here may be candidates for the development of SARS-CoV-2 clinical interventions. Center for Global Health and Infectious Diseases, Comprehensive AIDS Research Center, and Beijing Advanced Innovation Center for Structural Biology, School of Medicine, Tsinghua University, Beijing, 100084, China. email@example.com.

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https://math.stackexchange.com/questions/2183790/compact-semismple-lie-group

math

Let $G$ be a connected Lie group with a compact semisimple Lie algebra. It is well known that the Lie algebra then can be written as the product of compact simple Lie algebras which are classified by Dynkin diagrams. I would like to know what we can say on the level of Lie groups, i.e. is it true that the Lie group then can be written as a product of compact simple Lie groups? No, one has to take into account coverings. The smallest counterexample is $SO(4)$, which is not such a product, but whose universal cover is a product $Spin(4) \cong SU(2) \times SU(2)$.

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https://www.coursehero.com/file/8723433/Sure-mathematicians-hav-e-checked-that-the-first-ten-trillion/

math

Unformatted text preview: re on that same straight line. Sure, mathematicians hav e checked that the first ten trillion zeroes all fall on that line, but that's no guarantee that the ten trillionth and one zero might be somewhere else, throwing the whole prime distribution formula out the prov erbial window, along with v ast amounts of related number theory . Which is why there is a $1 m prize for any one who can show that all of the Zeta Function zeroes line up on the "0.5 line" without resorting to the impossible task of walking along this infinite line to check. I 'v e giv en y ou the Zeta Function to get y ou started and if y ou dust off a bit of "complex v ariable" maths, y ou will be well on y our way to ex ploring the Riemann landscape. Howev er – if that's a bit much – here is an easier starting problem: All prime numbers (greater than fiv e) squared are one more than a multiple of 24. Check it for a few – it works. Y ou can ev en prov e that it works for all of the infinite number of primes. Now if y ou can just do that for the Zeta zeroes, y ou can stop kicki... View Full Document - Winter '13

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https://nursingwritershub.com/suppose-you-have-a-one-year-zero-coupon-bond-with-a-rate-of-r1-and-a-two-year-bond-with-an-annual-coupon-payment-of-c/

math

Complete Problem 31 of Chapter 10 (shown below), and submit your work to your instructor. Show your calculations and the algebraic manipulation of the price equation for the bond. In addition to solving the problem, write a one-page essay on the term structure of fixed income securities. Format your essay per APA style guidelines. One method used to obtain an estimate of the term structure of interest rates is called bootstrapping. Suppose you have a one-year zero coupon bond with a rate of r1 and a two-year bond with an annual coupon payment of C. To bootstrap the two-year rate, you can set up the following equation for the price (P) of the coupon bond: P=C_1/(1+r_1 )+(C_2+Par value)/(1+r_2 )^2 Because you can observe all of the variables except r2, the spot rate for two years, you can solve for this interest rate. Suppose there is a zero coupon bond with one year to maturity that sells for $949 and a two-year bond with a 7.5 percent coupon paid annually that sells for $1,020. What is the interest rate for two years? Suppose a bond with three years until maturity and an 8.5 percent annual coupon sells for $1,029. What is the interest rate for three years?

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https://collaborate.princeton.edu/en/publications/joint-channel-estimation-and-equalization-for-ofdm-based-broadban

math

This paper is concerned with the challenging and timely problem of channel estimation for orthogonal frequency division multiplexing (OFDM) systems in the presence of frequency selective and very rapidly time varying channels. In OFDM systems operating over rapidly time-varying channels, the orthogonality between subcarriers is destroyed leading to inter-carrier interference (ICI) and resulting in an irreducible error floor. The band-limited, discrete cosine serial expansion of low-dimensionality is employed to represent the time-varying channel. In this way, the resulting reduced dimensional channel coefficients are estimated iteratively with tractable complexity and independently of the channel statistics. The algorithm is based on the expectation maximization-maximum a posteriori probability (EM-MAP) technique leading to a receiver structure that also yields the equalized output using the channel estimates. The pilot symbols are employed to estimate the initial coefficients effectively and unknown data symbols are averaged out in the algorithm in a non-data-aided fashion. It is shown that the computational complexity of the proposed algorithm to estimate the channel coefficients and to generate the equalized output as a by-product is ∼ O(N) per detected symbol, N being the number of OFDM subcarriers. Computational complexity as well as computer simulations carried out for the systems described in WiMAX and LTE standards indicate that it has significant performance and complexity advantages over existing suboptimal channel estimation and equalization algorithms proposed earlier in the literature.

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