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https://calculators.appzend.net/?q=Disability+Insurance+Calculator&kind=financial
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math
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Disability Insurance Definition Our free online disability insurance calculator makes it super easy to calculate disability payments, disability income, and view a schedule comparing your disability income versus your current income. Simply enter in your annual income, tax bracket, current disability insurance percentage, your maximum monthly benefit for your current disability insurance policy, your Social Security disability monthly benefit, any other monthly income sources, and any additional monthly expenses. Press calculate and see just how easy the disability insurance benefits calculator is to use! Check out our other financial calculators we have
How to Calculate Disability Insurance Let's be honest - sometimes the best disability insurance calculator is the one that is easy to use and doesn't require us to even know what the disability insurance formula is in the first place! But if you want to know the exact formula for calculating disability insurance then please check out the "Formula" box above.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00420.warc.gz
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CC-MAIN-2022-33
| 1,023 | 2 |
http://applied-financial-mathematics.de/index.php/past-events
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math
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Recent studies have extended the theory of affine processes to the stochastic Volterra equati- ons framework. In this talk, I will describe how the theory of polynomial processes extends to the Volterra setting. In particular, I will explain the moment formula and an interesting stochastic invariance result in this context. This is joint work with Eduardo Abi Jaber, Christa Cuchiero, Luca Pelizzari and Sara Svaluto-Ferro.
Polynomial Volterra processes
Geometric properties of some rough curves via dynamical systems: SBR measure, local time and Rademacher chaos
We investigate geometric properties of graphs of Takagi type functions, repre- sented by series based on smooth functions. They are Hölder continuous, and can be embedded into smooth dynamical systems, where their graphs emerge as pull- back attractors. It turns out that occupation measures and Sinai-Bowen-Ruelle (SBR) measures on their stable manifolds are dual by ’time’ reversal.
A suitable version of approximate self-similarity for deterministic functions al- lows us to ’telescope’ small-scale properties from macroscopic ones. As one conse- quence, absolute continuity of the SBR measure is seen to be dual to the existence of local time. The investigation of questions of smoothness both for SBR as for oc- cupation measures surprisingly leads us to the Rademacher version of Malliavin’s calculus, Bernoulli convolutions, and into probabilistic number theory. The link between the rough curves considered and smooth dynamical systems can be gen- eralized in various ways. For instance, Gaussian randomizations of Takagi curves just reproduce the trajectories of Brownian motion. Applications to regularization of singular ODE by rough signals are on our agenda.
Optimal consumption with labor income and borrowing constraints for recursive preferences
In this talk, we present an optimal consumption and investment problem for an investor with liquidity constraints who has isoelastic recursive Epstein-Zin utility preferences and re- ceives a stochastic stream of income. We characterize the optimal consumption strategy as well as the terminal wealth for recursive utility under dynamic liquidity constraints, which pre- vent the investor to borrow against his stochastic future income. Using duality and backward SDE methods in a possibly non-Markovian diffusion model for the financial market, this gives rise to an interplay of singular control and optimal stopping problems. Our analysis extends to more general liquidity constraints. (Joint work with Dirk Becherer and Olivier Menoukeu Pamen)
On two Formulations of McKean–Vlasov Control with Killing
We study a McKean–Vlasov control problem with killing and common noise. The particles in this control model live on the real line and are killed at a positive intensity whenever they are in the negative half-line. Accordingly, the interaction between particles occurs through the subprobability distribution of the living particles. We establish the existence of an optimal semiclosed-loop control that only depends on the particles’ location and not their cumulative intensity. This problem cannot be addressed through classical mimicking arguments, because the particles’ subprobability distribution cannot be reconstructed from their location alone. Instead, we represent optimal controls in terms of the solutions to semilinear BSPDEs and show those solutions do not depend on the intensity variable.
Generalized Front Propagation for Stochastic Spatial Models
Nonlinear Diffusions and their Feller Properties
Motivated by Knightian uncertainty, S. Peng introduced his celebrated G–Brownian motion. Intuitively speaking, it corresponds to a dynamic worst case expectation in a model where volatility is uncertain but postulated to take values in a bounded interval. Natural extensions of the G–Brownian motion are nonlinear diffusions, whose volatility (and drift) takes values in a random set that is allowed to depend on the canonical process in a Markovian way. Nonlinear diffusions satisfy the dynamic programming principle, which entails the semigroup property of a corresponding family of sublinear operators. In this talk, we discuss regularity properties of these semigroups that allow us to relate them to evolution equations. In particular, we explain a novel type of smoothing property and a stochastic representation result for general sublinear semigroups with pointwise generators of Hamilton-Jacobi-Bellman type. Latter also implies a unique characterization theorem for such semigroups.
The talk is based on joint work with Lars Niemann (University of Freiburg).
Numeraire-invariance and the law of one price in mean-variance portfolio selection and quadratic hedging
In classical optimal transport, the contributions of Benamou-Brenier and Mc- Cann regarding the time-dependent version of the problem are cornerstones of the field and form the basis for a variety of applications in other mathematical areas.
Stretched Brownian motion provides an analogue for the martingale version of this problem. We provide a characterization in terms of gradients of convex functions, similar to the characterization of optimizers in the classical transport problem for quadratic distance cost.
Based on joint work with Julio Backhoff-Veraguas, Walter Schachermayer and Bertram Tschiderer.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00452.warc.gz
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CC-MAIN-2024-10
| 5,355 | 17 |
https://thegloverlawfirm.com/uzn2w_feeding.aspx
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math
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Place in the quadrants corresponding to the given equation. Trigonometric functions and circular measure Australian. We can draw triangles to find out exactly what that point is. Functions answer to look at three definitions of worksheets? State the period of the function. For answers below shows quadrant. Your feedback has been sent. Using a circular functions worksheet with answers found by touch device that there are mainly used based on your worksheets, but first formal way endorses it. Show all answers?
How do you find the trigonometric functions of any angle? You can do angle would the circular trig worksheet answers. Show the upper and lower bounds, as this next example shows. Applications of Trigonometric and Circular Functions Mrs. So this is a positive angle theta. Follow the links given above. How do trees access the internet? What is an extension of others to edit this for circular trig worksheet answers after it produces a particular equation for measuring angles are disabled on?
Applications of Trigonometric and Circular Functions High. Any angle of worksheets and circular path on one, in radians to? Please click the link below to submit your verification request. Suchinformation cavery usefulforucturalengineers as well. CN Tower in Toronto, companies may disclose that they use your data without asking for your consent, which is a branch of mathematics that studies the relationship between angles and the sides of triangles. 3 Unit Circle Trigonometry. They lost all their matches! Going from simple trig functions to their graphs can seem like a daunting task These resources show the steps to connect evaluating the functions to graphing. Sketch the graph of this sinusoidal function. Find the exact value of each trigonometric function. Worksheet 1 KEY Inverse Trigonometric Functions 74. Find the angle and its supplement, and Edge.
Find a trig function to answer in part of worksheets, and minute must he woke up of having one of points on previous example.
Not having one may negatively impact your site and SEO. Chapter 43 Trigonometry Extended The Circular Functions. What is the domain of the function tan What is its range? Check by plotting on your grapher. Describe the unit circle. How long is the other leg? Was this document useful for you? Mathematical Literacy: Practice Book for Students. You start a stopwatch.
Radians also provide a way to calculate linear and angular velocity in rotary motion problems.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030331677.90/warc/CC-MAIN-20220924151538-20220924181538-00267.warc.gz
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CC-MAIN-2022-40
| 2,461 | 6 |
https://blog.tujion.com/reading-zmm/
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math
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Several months later, I picked up this book again. During those several months, I had very brief bursts of productivity, but mostly I wasted time doing nothing.
I noticed that I didn't remember what the story so far was. One forgets things really fast. I remembered that I enjoyed reading this 'travel book' which is a philosophy book in disguise. He was talking about quality. In Chapter 22, he wrote about Mathematics. He first mentioned the Bolyai-Lobachevskian geometry (=hyperbolic geometry). I felt something was off when he said that Riemann geometry violated Euclid's first postulate. (I think he meant elliptic geometry. Riemannian geometry encompasses hyperbolic, euclidean and elliptic geometry. How names shift as time passes and we extend and reformulate...) I remembered that it violated only the fifth. Thus I looked up the first postulate on Wolfram and saw where the discrepancy was. The author thought the first postulate stipulates that exactly one straight line can pass through two given points whereas the version we are more accustomed to does not specify the upper limit of number of straight lines passing through two points. I thought, hey, you should check these facts with a mathematician before publishing the book, which is a very intricate precise book I liked so much. I noticed that I started to dismiss the book as a huge let-down just for this. Thus I stopped myself in the track. I was being too strict when it came to things concerning my domain knowledge. That was a minor misunderstanding which didn't really bear on topic. Even if it had erred severely on history of mathematics, it would not mar the rest of the writing. The author probably didn't have an easily accessible reference system at the time of writing. Checking everything along the way may hinder his chain of thoughts. It is fine to let this lapse, but I do make a note in my blog that there is a minor mistake. The two types of non-euclidean geometry have a difference number of parallel lines that one can draw through a point to a give line, one in excess (hyperbolic) and one in dearth (elliptic). I am not good with history of mathematics and I only know the basics of Riemannian geometry. Probably at the time of writing, mathematicians didn't have a clear presentation of the materials yet. We now benefit from all the good writings of the past generations.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00295.warc.gz
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CC-MAIN-2023-06
| 2,369 | 2 |
https://www.ieltsday.com/2217.html
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math
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Describe a mistake that you once made
• What the mistake was
• When, where you made it
• Who was with you
• and how the mistake affected you
• Making mistakes is part of life.
• We all make mistakes from time to time.
• The best thing about making a mistake is that we get to learn something from it
• In that way mistakes are a part of the learning process.
• I have also made many mistakes in my life
• Here I would like to talk about a mistake, which I made when I was in tenth.
• I bunked school and went to see a movie with my friends.
• The mistake did not end here
• To hide that mistake I had to speak so many lies.
• First I lied to my mother, and then to my father.
• I had never bunked school before.
• My conscience kept pricking me.
• I remember I could not sleep the whole night
• The next morning I kept avoiding eye contact with my mother and father.
• I went to school but there too I could not concentrate on studies.
• So I decided, I would tell my mother and let my heart rest in peace.
• When I reached home I told my mother all about my wrongdoing.
• She was a bit upset, but at the same time she was happy that I had realized my mistake.
• She forgave me, and told me never to do such a thing again.
• Now if I want to go with my friends to watch a movie, I do that on the weekends after taking permission from my mother and father.
• I learnt that to hide one lie, we have to speak a chain of lies.
• Our parents trust us, and we should not break their trust.
• If we are caught, we lose our credibility forever.
• Even if we are not caught, telling lies to parents is cheating them, which is certainly a big mistake.
1. What should teachers do when students make mistakes?
Teachers should explain the repercussions (result) of the mistake to the student nicely, and in such a way that the student does not do that mistake again. But, if the student does the same mistake repeatedly, then the teacher should punish the child. The punishment need not be corporal (physical). It should just aim to refrain the child from doing the same mistake again and again.
2. What can people learn from mistakes?
People can learn a lot from mistakes. They learn, the far reaching consequences of any wrong action and learn the correct way of doing things.
3. Do you often make mistakes?
Yes, I often make mistakes. But I avoid committing (doing) the same mistake again and again. I make it a point to learn from my mistakes and not repeat that mistake (error).
4. Can mistakes help people to be more successful?
Definitely yes, mistakes teach a lot of things. A person, who learns from his mistakes and doesn’t repeat them, gets better day-by-day, and his chances of success go up.
5. How to avoid making mistakes?
One can avoid making mistakes by learning from the mistakes of others. A person can also avoid making mistakes by meticulous (careful) planning of things and doing things patiently.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00739.warc.gz
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CC-MAIN-2023-40
| 2,964 | 39 |
http://www.jiskha.com/members/profile/posts.cgi?name=MIKI&page=4
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math
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your room has 2 windows.Each window occupies 11.76 square ft of wall space.if the windows are each 2.8 ft wide,how tall is each window? my answer is 4.2 ft,am i right? if not tell why.
The room you want to paint has 2 doors,each 6.5 ft tall and covering 20.8 square ft.What is the width of each door? My answer is 3.2 ft,am i right?if not please explain.
15 width by 45 length
A pick up truck is carrying 500 pounds of cargo. When empty, the truck weighs 2 1/2 tons. What is the weight of the truck and its cargo in tons? Is this answer right 1250tons.
can you please tell me the answer????
A rug covers 1/4 of the floor. The area of the rug 10ft . What is the area of the floor? my answer is 2 1/2 ft,am i right if not please explain?
What does a maple tree do for a community? A. it creates energy B. It creates food C. It looks nice D. it stores water My answer is B.
is the answer producer
What term almost always describe top consumers. A)herbivores B)omnivore C)carnivores D)decomposer
My answer is it looks nice
For Further Reading
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s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705926946/warc/CC-MAIN-20130516120526-00057-ip-10-60-113-184.ec2.internal.warc.gz
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CC-MAIN-2013-20
| 1,038 | 11 |
https://web2.0calc.com/questions/answer_26
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math
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In a certain Algebra 2 class of 27 students, 5 of them play basketball and 12 of them play baseball. There are 2 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?
Draw a Venn Diagram 12 students do not play those sports 15 students DO play those sports
15 out of 27 15/27
Since 2 students play both sports, 3 students play only basketball and 10 play only baseball.
There is a problem with the word "or". It has two separate meanings (each one ccan be correct).
There is an "exclusive or" that means either one or the other but not both.
In this case, there are 13 students who play basketball or baseball (but not both).
There is an "inclusive or" that means one or the other or both.
In this case, there are 15 students who play basketball or baseball (and perhaps both).
Which answer is correct? Until the person who asks the question tells you which "or" they intend, you don't know.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347439213.69/warc/CC-MAIN-20200604063532-20200604093532-00284.warc.gz
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CC-MAIN-2020-24
| 968 | 10 |
https://works.swarthmore.edu/fac-math-stat/180/
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math
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Speedups And Orbit Equivalence Of Finite Extensions Of Ergodic Zᵈ-Actions
New York Journal Of Mathematics
We classify n-point extensions of ergodic Zᵈ-actions up to relative orbit equivalence and establish criteria under which one n-point extension of an ergodic Zᵈ-action can be sped up to be relatively isomorphic to an n-point extension of another ergodic Zᵈ-action. Both results are characterized in terms of an algebraic object associated to each n-point extension which is a conjugacy class of subgroups of the symmetric group on n elements.
ergodic Zᵈ action, finite extension, cocycle, speedup, relative isomorphism
Aimee S.A. Johnson and D. M. McClendon.
"Speedups And Orbit Equivalence Of Finite Extensions Of Ergodic Zᵈ-Actions".
New York Journal Of Mathematics.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00302.warc.gz
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CC-MAIN-2021-17
| 785 | 7 |
http://www.stata.com/statalist/archive/2005-03/msg00410.html
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math
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[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
st: -xi- with multible variables
Is there some way to use the -xi- command with multiple variables? This
is the context of a three level nested logit model with around 10
independent variables. I need to create interactions for all of the
I have a categorical variable that I want to interact with a number of
continuous variables, is there a Stata command to do this? Xi only
seems to work with one variable.
Let variable categorical_1-categorical_3 have 5 levels each.
Let var1-var8 be continuous variables that I want to interact them with.
Is there some way to interact categorical_1*(var1-var8),
categorical_2*(var1-var8), and categorical_3*(var1-var8) easily? I know
I can change the prefix, but this still requires 24 different prefixes
and creates a number of redundant dummies. Surely this must be a common
task and I'm overlooking something.
Thanks for any help you can give.
* For searches and help try:
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s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607369.90/warc/CC-MAIN-20170523045144-20170523065144-00460.warc.gz
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CC-MAIN-2017-22
| 989 | 17 |
https://www.physicsforums.com/threads/colloquims-and-seminars.111145/
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math
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So, where do I find out about this stuff? I never see flyers or posters around school or even in the mathematics department. Is there a website that informs us about this? If possible, do you one a website for Ontario, Canada? Note: I do know that I would probably not understand anything, but I heard it wouldn't be a bad idea. I figured why not go if it sounds alright.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376825728.30/warc/CC-MAIN-20181214114739-20181214140239-00589.warc.gz
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CC-MAIN-2018-51
| 371 | 1 |
https://msd.com.ua/mostly-harmless-econometrics-an-empiricist-s-companion/fixed-effects-versus-lagged-dependent-variables/
|
math
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Fixed Effects versus Lagged Dependent Variables
Fixed effects and differences-in-differences estimators are based on the presumption of time-invariant (or group-invariant) omitted variables. Suppose, for example, we are interested in the effects of participation in a subsidized training program, as in the Dehejia and Wahba (1999) and Lalonde (1986) studies discussed in section (3.3.3). The key identifying assumption motivating fixed effects estimation in this case is
E(y0it&i; Xit, Dit) — E(Y0it&i; Xit), (5.3.1)
where a. i is an unobserved personal characteristic that determines, along with covariates, Xit, whether individual i gets training. To be concrete, & might be a measure of vocational skills, though a strike against the fixed-effects setup is the fact that the exact nature of the unobserved variables typically remains somewhat mysterious. In any case, coupled with a linear model for E(Yoit&i, Xit), assumption (5.3.1) leads to simple estimation strategies involving differences or deviations from means.
For many causal questions, the notion that the most important omitted variables are time-invariant doesn’t seem plausible. The evaluation of training programs is a case in point. It seems likely that people looking to improve their labor market options by participating in a government-sponsored training program have suffered some kind of setback. Many training programs explicitly target people who have suffered a recent setback, e. g., men who recently lost their jobs. Consistent with this, Ashenfelter (1978) and Ashenfelter and Card (1985) find that training participants typically have earnings histories that exhibit a pre-program dip. Past earnings is a time-varying confounder that cannot be subsumed in a time-invariant variable like &i.
The distinctive earnings histories of trainees motivates an estimation strategy that controls for past earnings directly and dispenses with the fixed effects. To be precise, instead of (5.3.1), we might base causal inference on the conditional independence assumption,
E(y0itYit— Xit; Dit) — E(y0it Yit— h; Xit). (5.3.2)
This is like saying that what makes trainees special is their earnings h periods ago. We can then use panel data to estimate
where the causal effect of training is ft. To make this more general, Yu-h can be a vector including lagged earnings for multiple periods.9
Applied researchers using panel data are often faced with the challenge of choosing between fixed-effects and lagged-dependent variables models, i. e., between causal inferences based on (5.3.1) and (5.3.2). One solution to this dilemma is to work with a model that includes both lagged dependent variables and unobserved individual effects. In other words, identification might be based on a weaker conditional independence assumption:
E(Y0itai; Yit—h; Xit; Dit) = E(Y0it^i; Yit—h; Xit), (5.3.4)
which requires conditioning on both a. i and Yit-h. We can then try to estimate causal effects using a specification like
Unfortunately, the conditions for consistent estimation of ft in equation (5.3.5) are much more demanding than those required with fixed effects or lagged dependent variables alone. This can be seen in a simple example where the lagged dependent variable is Yit_ 1. We kill the fixed effect by differencing, which produces
The problem here is that the differenced residual, AVit, is necessarily correlated with the lagged dependent variable, AYit_i, because both are a function of Vit_1. Consequently, OLS estimates of (5.3.6) are not consistent for the parameters in (5.3.5), a problem first noted by Nickell (1981). This problem can be solved, though the solution requires strong assumptions. The easiest solution is to use Yit_2 as an instrument for Ayit_ 1 in (5.3.6). But this requires that Yit_2 be uncorrelated with the differenced residuals, AVit. This seems unlikely since residuals are the part of earnings left over after accounting for covariates. Most people’s earnings are highly correlated from one year to the next, so that past earnings are an excellent predictor of future earnings and earnings growth. If Vit is serially correlated, there may be no consistent estimator for (5.3.6). (Note also that the IV strategy using Yit_2 as an instrument requires at least three periods to obtain data for t, t — 1, and t — 2).
Given the difficulties that arise when trying to estimate (5.3.6), we might ask whether the distinction between fixed effects and lagged dependent variables matters. The answer, unfortunately, is yes. The fixed-effects and lagged dependent variables models are not nested, which means we cannot hope to estimate
one and get the other as a special case if need be. Only the more general and harder-to-identify model, (5.3.5), nests both fixed effects and lagged dependent variables..
So what’s an applied guy to do? One answer, as always, is to check the robustness of your findings using alternative identifying assumptions. That means that you would like to find broadly similar results using both models. Fixed effects and lagged dependent variables estimates also have a useful bracketing property. The appendix to this chapter shows that if (5.3.2) is correct, but you mistakenly use fixed effects, estimates of a positive treatment effect will tend to be too big. On the other hand, if (5.3.1) is correct and you mistakenly estimate an equation with lagged outcomes like (5.3.3), estimates of a positive treatment effect will tend to be too small. You can therefore think of fixed effects and lagged dependent variables as bounding the causal effect you are after. Guryan (2004) illustrates this sort of reasoning in a study estimating the effects of court-ordered busing on Black high school graduation rates.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00722.warc.gz
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CC-MAIN-2022-27
| 5,754 | 17 |
https://open.library.ubc.ca/cIRcle/collections/48630/items/1.0043477
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math
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BIRS Workshop Lecture Videos
Gaussian measures no spaces of metrics Jakobson, Dmitry
The first part of the talk is joint work in progress with B. Clarke, N. Kamran, L. Silberman and J. Taylor. We define Gaussian measures on manifolds of metrics with the fixed volume form. We next compute the moment generating function for the L2 (Ebin) distance to the reference metric.\\r\\n\\r\\nThe second part of the talk is joint work with Linan Chen. We use the two-dimensional approach of Duplantier-Sheffield to define a 4- dimensional analogue of Gaussian Free Field, and use it to construct a canonical measure on a conformal class of metrics in R4. We derive a KPZ type relation for those measures.
Item Citations and Data
Attribution-NonCommercial-NoDerivs 2.5 Canada
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s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337531.3/warc/CC-MAIN-20221005011205-20221005041205-00130.warc.gz
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CC-MAIN-2022-40
| 764 | 5 |
https://mathematics.huji.ac.il/event/game-theory-math-economics-ga%C3%ABtan-fournier-paris-sorbonne-university-efficiency?ref_tid=3830
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math
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Elath Hall, 2nd floor, Feldman Building, Edmond J. Safra Campus
We consider a Hotelling game where a finite number of retailers choose a location, given that their potential customers are distributed on a network. Retailers do not compete on price but only on location. We show that when the number of retailers is large enough, the game admits a pure Nash equilibrium and we construct it. We then compare the equilibrium cost bore by the consumers with the cost that could be achieved if the retailers followed the dictate of a benevolent planner. We look at this efficiency of equilibrium asymptotically in the number of retailers.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00716.warc.gz
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CC-MAIN-2022-21
| 633 | 2 |
https://sami.hust.edu.vn/en/department/department-of-theoretical-mathematics/specialization-department-of-theoretical-mathematics/
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math
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Areas of Expertise
Research on math in some main research directions:
- Differential Equations and Integral
- Linear and nonlinear differential equations in Banach spaces;
- Semi-linear partial differential equation;
- Optimal control problems for differential equations and partial differential equations;
- Problems determining coefficients for partial differential equations, inverse problems, etc.
- Integral Transformation;
- Functional Analysis;
- Complex Analysis;
- Fixed Point Theory and related problems.
- Algebra and Geometry
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00512.warc.gz
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CC-MAIN-2023-40
| 537 | 12 |
https://scholars.uab.edu/display/pub188178
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math
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A compactum X ⊂ C is unshielded if it coincides with the boundary of the unbounded component of C\X. Call a compactum X finitely Suslinian if every collection of pairwise disjoint subcontinua of X whose diameters are bounded away from zero is finite. We show that any unshielded planar compactum X admits a topologically unique monotone map mX: X → XFS onto a finitely Suslinian quotient such that any monotone map of X onto a finitely Suslinian quotient factors through mX. We call the pair (XFS, mX) (or, more loosely, XFS) the finest finitely Suslinian model of X. If f: ℂ → ℂ is a branched covering map and X ⊂ ℂ is a fully invariant compactum, then the appropriate extension MX of mX monotonically semiconjugates f to a branched covering map g: ℂ → ℂ which serves as a model for f. If f is a polynomial and Jf is its Julia set, we show that mX (or MX) can be defined on each component Z of Jf individually as the finest monotone map of Z onto a locally connected continuum. © 2012 American Mathematical Society.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00175.warc.gz
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CC-MAIN-2022-33
| 1,037 | 1 |
https://m2.mtmt.hu/api/publication/1652600
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math
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The aim of this paper is to find a closed form of the integrals ∫ 0 π cos( x sin(
t ) − nt ) d t for n = 0, 1, 2, … using the Maple computer algebra system. Although
Maple 10 is not capable to calculate these integrals in one step, it turns out to
be a very useful tool to solve this and similar kind of complex mathematical problems.
During the problem solving process Maple proves that it is useful and, what is more,
it is an indispensable partner. Maple helps us to formulate our conjecture, acts as
an advisor and, last but not least, performs complex symbolic calculation instead
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https://sta.uwi.edu/fst/dms/math-2420-introduction-graph-theory-and-optimization
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math
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MATH 2420 Introduction to Graph Theory and Optimization
Students taking this course will be expected to know the basic principles of sets and number systems, linear algebra and analytical geometry. For this reason, MATH 1152 and MATH 1141 are listed as course prerequisites.
This course can be divided into two sections of (i) graphs and digraphs and (ii) linear programming. In (i), we begin with basic graph theoretic definitions that also involve graphical operations. Some simple theorems are proved including an extremal result. We express some of these concepts in the missionaries and cannibal problems as well as the instant insanity problem. Next we define a number of important matrices associated with these graphs via an examination of the general entry of products of these matrices. In so doing, digraphs are introduced and an application is demonstrated in the finding of determinants. Properties of relations are visualized with respect to structural features of digraphs. We formulate communication networks and kernels by the use of digraphs. This leads to the definitions of basis digraphs, progression sequences and canonical ordering of nodes. These graphical concepts are then incorporated into the solving of a system of linear equations. In (ii), we revise the linear relation in the Cartesian plane. The idea of a convex set is introduced in relation to maximization and minimization linear programming problems. Extreme points of bounded polyhedral regions are found by the simplex method and also by simple constructions in the xy plane. The principle of duality is given.
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http://miamiherald.typepad.com/sports-buzz/2012/06/monday-update-heat-keeps-thriving-despite-size-disadvantage.html
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math
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Note: This is updated from earlier in the day. See below for Dolphins notes from Tuesday afternoon.
# # #
Dumbest remark of the playoffs so far? Had to be Thunder forward Serge Ibaka saying that LeBron James "is not a good defender." That's the same LeBron James who earned the most votes in balloting for the NBA's all-defensive team.
James responded Tuesday morning. "I don't really care what he says," LeBron said. "He's stupid. Everyone says something to me every series. Then [the media] keeps trying to get a quote. I don't care what he says. It's stupid."
Ibaka, per the Post, went on to the say that James "can play defense for two to three minutes but not 48.... LeBron can't play [Kevin Durant]" one-on-one.
Responded James: "First of all, I'm not playing 48 minutes and K.D.'s not playing 48 minutes. I'm not sitting there saying I'm a Durant stopper because there's no such thing."
Durant missed four of five shots when James was guarding him in the fourth quarter of Game 3. Durant was asked Monday what James did to him in the fourth quarter. "Nothing," he said. And the news conference ended thusly.
It will be interesting to see how Ibaka explains himself. Some backtracking is likely, or at least a scolding by Scott Brooks.
DOLPHINS PRACTICE NOTES
### Chad OchoCinco made three receptions at the first day of the Dolphins' three-day minicamp on Tuesday, including two terrific catches on the sideline, both from David Garrard. He jumped over Vontae Davis to make one of them.
"He's fast, he's quick, and he's got that attacking mentality you love in a wide receiver," Matt Moore said.
### He spoke to reporters less than two minutes afterward. He said he wants to get "back to who we're all used to seeing and how I became what I am. I think I kind of lost that. For me, it's about getting back to basics, getting back to where it started. I'm looking to go back to Chad Johnson and just make it live again."
### OchoCinco said playing for the Dolphins "has been a childhood dream of mine growing up watching the Marks Brothers, watching Dan Marino. To be able to wear teal and orange, it's a pretty good feeling."
### He cracked, "I'm developing Brokeback Mountain chemistry with the players." He ended his session with reporters by saying: "I love you. Enjoy the show."
### Asked how OchoCinco is learning the playbook - which was a problem for him in New England - Joe Philbin said, "So far, so good." Philbin added: "It looks like he's fitting in well."
### David Garrard took the majority of the first team snaps but said he and Matt Moore are splitting them about evenly overall, with Ryan Tannehill receiving very few with the starters. Garrard looked the best of the three on Tuesday, which was also the case during the last practice open to the media, eight days ago.
### Garrard led a touchdown drive that ended with a rollout and strong throw to Roberto Wallace for a score. The offense was forced to punt on Matt Moore's final drive.
### Garrard's only glaring error was an interception to Sean Smith, who would have returned it for a touchdown if the play had been allowed to continue. Smith has been very impressive all offseason. "I like what I've seen from Sean," Philbin said. "His approach has been serious."
### Matt Moore also threw what would have been an interception return for a touchdown, to Karlos Dansby.
### Tannehill made some sharp passes - to tight end Les Brown, Marlon Moore, Wallace and others - but also threw a pick to Jason Trusnik.
### Jonathan Wade, competing for Nolan Carroll for the No. 4 cornerback job, made three terrific plays to break up potential receptions. And rookie Josh Kaddu, who couldn't join the team until this month because of his Oregon class schedule, looked sharp - including a would-be sack of Tannehill. (We say would-be because sacks aren't allowed.) Kaddu also knocked away a Tannehill pass over the middle.
### Wallace made several impressive catches, and Garrard raved about him afterward. Also, Philbin spoke very highly of receiver Legedu Naanee, noting "you like his size and ability and he can attack the middle of the field."
### Jonathan Martin, the second-rounder out of Stanford, got a lot of work with the first team at right tackle, with Artis Hicks at right guard. Lydon Murtha also got some first-team work at right tackle. And Martin got some snaps in relief of Jake Long at left tackle, which was Martin's position at Stanford.
### Reshad Jones and Chris Clemons remain the first-team safeties, backed up by Jimmy Wilson and Tyrone Culver. Free-agent pickup Tyrell Johnson continues to work with the third team.
### Rookie third-round tight end Michael Egnew, who was used a lot as a blocker in earlier practices, made two receptions across the middle, showing nice burst after the catch.
### Dan Carpenter kicked a 59-yard field goal but missed, to the left, from 57.
COLLEGE FOOTBALL ITEM
### Encouraging news for the Orange Bowl: College presidents are expected to approve a four-team playoff, with plans to be announced perhaps as early as this week, and ESPN reported Monday night that the decision-makers are leaning toward incorporating the semifinals within the current bowl system, meaning the Orange, Fiesta, Rose and Sugar would alternate as hosts. It's not definite by any means, but the OB certainly would take that scenario.
The national championship game is expected to be put up for bid, with Cowboy Stadium among the potential suitors. The new format would take effect with the 2014 season.
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https://www.arxiv-vanity.com/papers/1412.6500/
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math
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Numerical analysis of distributed optimal control problems governed by elliptic variational inequalities
A continuous optimal control problem governed by an elliptic variational inequality was considered in Boukrouche-Tarzia, Comput. Optim. Appl., 53 (2012), 375-392 where the control variable is the internal energy . It was proved the existence and uniqueness of the optimal control and its associated state system. The objective of this work is to make the numerical analysis of the above optimal control problem, through the finite element method with Lagrange’s triangles of type 1. We discretize the elliptic variational inequality which define the state system and the corresponding cost functional, and we prove that there exists a discrete optimal control and its associated discrete state system for each positive (the parameter of the finite element method approximation). Finally, we show that the discrete optimal control and its associated state system converge to the continuous optimal control and its associated state system when the parameter goes to zero.
Key words: Elliptic variational inequalities, distributed optimal control problems, numerical analysis, convergence of the optimal controls, free boundary problems.
2010 AMS Subject Classification 35J86, 35R35, 49J20, 49J40, 49M25, 65K15, 65N30.
We consider a bounded domain whose regular boundary consists of the union of two disjoint portions and with meas ( ) . We consider the following free boundary problem :
where the function in (1.1) can be considered as the internal energy in , is the constant temperature on and is the heat flux on . The variational formulation of the above problem is given as: Find such that
We note that is a bilinear, continuous, symmetric on and a coercive form on , that is to say: there exists a constant such that
In , the following continuous distributed optimal control problem associated with or the elliptic variational inequality was considered:
Problem : Find the continuous distributed optimal control such that
where the quadratic cost functional is defined by:
with a given constant and is the corresponding solution of the elliptic variational inequality (1.3) associated to the control .
Several continuous optimal control problems are governed by elliptic variational inequalities, for example: the process of biological waste-water treatment; reorientation of a satellite by propellers; and economics: the problem of consumer regulation of a monopoly, etc. There exist an abundant literature for optimal control problems [4, 42, 50], for optimal control problems governed by elliptic variational equalities or inequalities [2, 3, 5, 6, 7, 8, 9, 11, 19, 20, 26, 28, 30, 32, 34, 38, 40, 45, 46, 52, 53, 54], for numerical analysis of variational inequalities or optimal control problems [10, 13, 14, 15, 16, 17, 21, 22, 23, 24, 25, 27, 33, 35, 36, 37, 43, 47, 48, 49, 51], and for the numerical analysis of optimal control problems governed by an elliptic variational inequality there exist a few numbers of papers [1, 29, 31, 44].
The objective of this work is to make the numerical analysis of the optimal control problem which is governed by the elliptic variational inequality (1.3) by proving the convergence of a discrete solution to the continuous optimal control problems.
In Section 2, we establish the discrete elliptic variational inequality (2.3) which is the discrete formulation of the continuous elliptic variational inequality (1.3), and we obtain that these discrete problems have unique solutions for all positive . Moreover, on the adequate functional spaces these solutions are convergent when to the solutions of the continuous elliptic variational inequality (1.3).
In Section 3, we define the discrete optimal control problem (3.2) corresponding to continuous optimal control problem (1.5). We prove the existence of a discrete solution for the optimal control problem () for each parameter and we obtain the convergence of this family with its corresponding discrete state system to the continuous optimal control with the corresponding continuous state system of the problem ().
2 Discretization of the problem (S)
Let a bounded polygonal domain; a positive constant and a regular triangulation with Lagrange triangles of type 1, constituted by affine-equivalent finite elements of class over being the parameter of the finite element approximation which goes to zero [12, 18]. We take equal to the longest side of the triangles and we can approximate the sets and by:
where is the set of the polynomials of degree less than or equal to in the triangle . Let be the corresponding linear interpolation operator and a constant (independent of the parameter ) such that :
The discrete variational inequality formulation of the system is defined as: Find such that
Let , and be, then there exist unique solution of the problem given by the elliptic variational inequality (2.3).
Let , and , be the solutions of for and respectively, then we have that:
there exist a constant independent of such that:
if in weak, then in strong for each fixed .
a) If we consider in the discrete elliptic variational inequality (2.3) we have:
where is the trace operator and therefore (2.4) holds.
b) As and are respectively the solutions of discrete elliptic variational inequalities (2.3) for y , we have:
for . By coerciveness of we deduce:
thus (2.5) holds.
c) Let be. From item a) we have that , then there exist such that in weak (in strong). If we consider the discrete elliptic inequality (2.3) we have:
and using that is a lower weak semi-continuous application then, when goes to infinity, we obtain that:
and from uniqueness of the solution of problem , we deduce that .
Now, it is easily to see that:
and from the coerciveness of we obtain
As in and in , by pass to the limit when in the previous inequality, we obtain
Henceforth we will consider the following definitions : Given and , we have the convex combinations of two data
the convex combination of two discrete solutions
and we define as the associated state system which is the solution of the discrete elliptic variational inequality (2.3) for the control .
Then, we have the following properties:
Given the controls , we have that:
a) From the definition (2.8) we get
then we conclude (2.9).
b) It follows from a similar method to the part a). ∎
From Lemma 2.1 we have that there exist a constant independent of such that then we conclude that there exists so that in weak as and . On the other hand, given there exist such that for each and in strong when goes to zero. Now, by considering in the discrete elliptic variational inequality (2.3) we get:
and when we pass to the limit as in (2.11) by using that the bilinear form is lower weak semicontinuous in we obtain:
that it is to say:
and, from the uniqueness of the solution of the discrete elliptic variational inequality (1.3), we obtain that
then by pass to the limit when it results that ∎
3 Discretization of the optimal control problem
Now, we consider the continuous optimal control problem which was established in (1.5). The associated discrete cost functional is defined by the following expression:
and we establish the discrete optimal control problem as: Find such that
where is the associated state system solution of the problem which was described for the discrete elliptic variational inequality (2.3) for a given control .
Given the control , we have:
b) for some constant independent of .
c) The functional es a lower weakly semi-continuous application in .
d) There exists a solution of the discrete optimal control problem (3.2) for all .
a) From the definition of we obtain a) and b).
c) Let in weak, then by using the equality we obtain that . Therefore, we have
d) It follows from . ∎
If the continuous state system has the regularity then we have the following estimations :
where ’s are constants independents of .
and then (3.3) holds.
b) From the definitions of and , it results:
Following the idea given in we define an open problem: Given the controls ,
Remark 1: We have that .
Remark 2: The equivalent inequality for the continuous optimal control problem is true, that is : for all ,
where is the unique solution of the elliptic variational inequality when we consider instead of , and is the unique solution of the elliptic variational inequality when we consider instead of .
and therefore, the uniqueness for the discrete optimal control problem in the theorem holds.
Let be the continuous state system associated to the optimal control which is the solution of the continuous distributed optimal control problem (1.5). If, for each , we choose an optimal control which is the solution of the discrete distributed optimal control problem (3.2) and its corresponding discrete state system , we obtain that:
Then, if we consider and his corresponding associated state system, it results that:
From the Lemma 2.1 we have that , then we can obtain:
If we consider in the inequality (2.3) for , we obtain:
and from the coerciveness of the application we have that and in consequence .
Now we can say that there exist and such that in weak (in strong), and in weak when . Then, and in i.e., .
Let given , there exist such that in strong when . Then, if we consider the variational elliptic inequality (2.3) for we have:
Taking into account that the application is a lower weak semi-continuous application in and by pass to the limit when goes to zero in (3.10) we obtain that:
and by the uniqueness of the solution of the problem given by the elliptic variational inequality (1.3), we deduce that .
Finally, the norm on is a lower semi-continuous application in the weak topology, then we can prove that:
and because the uniqueness of the optimal problem (1.5), it results that and
Now, if we consider in the elliptic variational inequality (1.3) for the control and we define , we have that:
and by consider for in the inequality (2.3) we obtain:
and then by the coerciveness of we get
When we pass to the limit as in (3.11) and by using the strong convergence of to on and the weak convergence of to on , we have:
The strong convergence of the optimal controls to is obtained by using Theorem 3.1 and weakly on , i.e.
then and therefore .
We have proved the convergence of a discrete optimal control and its corresponding discrete state system governed by a discrete elliptic variational inequality to the continuous optimal control and its corresponding continuous state system which is also governed by a continuous elliptic variational inequality by using the finite element method with Lagrange’s triangles of type 1. Moreover, it is an open problem to obtain the error estimates as a function of the parameter of the finite element method.
This paper has been partially sponsored by Project PIP # 0534 from CONICET-UA, Rosario, Argentina, and AFOSR-SOARD Grant FA9550-14-1-0122.
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http://faqs.cs.uu.nl/na-dir/sci-math-faq/surfaceareaofsphere.html
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math
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Note from archiver<at>cs.uu.nl:
This page is part of a big collection
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Subject: sci.math FAQ: Surface of Sphere
This article was archived around: 17 Feb 2000 22:55:54 GMT
Last-modified: February 20, 1998
Formula for the Surface Area of a sphere in Euclidean N-Space
This is equivalent to the volume of the N-1 solid which comprises the
boundary of an N-Sphere.
The volume of a ball is the easiest formula to remember: It's r^N
(pi^(N/2))/((N/2)!). The only hard part is taking the factorial of a
half-integer. The real definition is that x! = Gamma (x + 1), but if
you want a formula, it's:
(1/2 + n)! = sqrt(pi) ((2n + 2)!)/((n + 1)!4^(n + 1))
To get the surface area, you just differentiate to get N
(pi^(N/2))/((N/2)!)r^(N - 1).
There is a clever way to obtain this formula using Gaussian integrals.
First, we note that the integral over the line of e^(-x^2) is
sqrt(pi). Therefore the integral over N-space of e^(-x_1^2 - x_2^2 -
... - x_N^2) is sqrt(pi)^n. Now we change to spherical coordinates. We
get the integral from 0 to infinity of Vr^(N - 1)e^(-r^2), where V is
the surface volume of a sphere. Integrate by parts repeatedly to get
the desired formula.
It is possible to derive the volume of the sphere from ``first
Alex Lopez-Ortiz [email protected]
http://www.cs.unb.ca/~alopez-o Assistant Professor
Faculty of Computer Science University of New Brunswick
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https://books.google.no/books?id=XDQDAAAAQAAJ&dq=editions:UOM39015067252117&hl=no&output=html_text&lr=
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The school edition. Euclid's Elements of geometry, the first six books, by R. Potts. corrected and enlarged. corrected and improved [including portions of book 11,12].
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Vanlige uttrykk og setninger
A₁ ABCD Algebraically Apply base bisected Book chord circle circumference common construction contained definition demonstrated described diagonals diameter difference distance divided double draw drawn equal equal angles equiangular equilateral triangle equimultiples Euclid extremities fall figure formed four fourth Geometrical given circle given line given point given straight line greater half Hence inscribed intersection isosceles join less Let ABC line drawn magnitudes manner mean meet multiple parallel parallelogram pass perpendicular plane problem produced Prop proportionals PROPOSITION proved radius ratio reason rectangle rectangle contained regular remaining respectively right angles segment semicircle shew shewn sides similar solid square straight line taken tangent THEOREM third touch triangle ABC twice units vertex wherefore whole
Side 112 - Guido, with a burnt stick in his hand, demonstrating on the smooth paving-stones of the path, that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.
Side 83 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square...
Side 48 - If two triangles have two sides of the one equal to two sides of the other, each to each ; and...
Side 238 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth...
Side 198 - A LESS magnitude is said to be a part of a greater magnitude, when the less measures the greater, that is, ' when the less is contained a certain number of times exactly in the greater.
Side 271 - SIMILAR triangles are to one another in the duplicate ratio of their homologous sides.
Side 81 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Side 115 - angle in a segment' is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment.
Side 341 - On the same base, and on the same side of it, there cannot be two triangles...
Side 24 - ... twice as many right angles as the figure has sides ; therefore all the angles of the figure together with four right angles, are equal to twice as many right angles as the figure has sides.
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CC-MAIN-2023-23
| 3,063 | 15 |
http://www.merit.edu/mail.archives/nanog/1999-10/msg00075.html
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math
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North American Network Operators Group|
Date Prev | Date Next |
Date Index |
Thread Index |
Author Index |
Re: Real Media and M-Bone feeds
- From: Andy McConnell
- Date: Tue Oct 05 11:42:43 1999
(BOn Tue, 5 Oct 1999, Alex P. Rudnev wrote:
(B}> > just forget about it and spend our life doing something
(B}> > useful instead?
(B}> because, although it is getting less expensive quickly, transport costs
(B}> money. multicast promises to reduce that cost near sources.
(B}Multicast is just not more than one case of data caching on the fly. It
(B}can be used for the local networks, just with the net of the media
(B}replicators. In principle there is not big difference between multicast
(B}and www caching except first is an example of the _real-time caching_ and
(B}second is usially _store-and-forward_ caching.
(BIt it really comparable to caching? I see multicasting as more of a
(Btraffic reduction, rather than a cache.
(B}This days we can see the weakness of the global-multicasting - and I think
(B}it should be replaced by the media-caching servers (with the ability to
(B}replicate data on the fly - in case of live media stream, and short or
(B}long tome _store-and-forward_ in case of Video-on-demand stream) - and
(B}with just this multicasting on the very end of the data tree. But an
(B}attempts to build over-the-world multicast network - brr... it's possible
(B}(if you should dig some mountain every day, you'll build a tunnel at last;
(B}but may be it's easy to run this mountain over?).
(BYour model would work, but it requires a LOT more coordination and
(Bcooperation than even multicast requires. Are you sugesting that networks
(Bimplement machines that sniff into the data, identify those streams,
(Bintercept them, and then coordinate with the streams' sources to stop
(Bsending the unicasts behind the cache, and send the stream to the cache
(Bonly? Or will your new machine simply "spoof" the source? If the latter,
(Bthen you haven't told the sender to reduce the traffic.
(BYou mentioned your doubt of building an over-the-world multicast
(Bnetwork... but what you are sugesting seems to be an over-the-world
(Bcaching mechanism. If we are going to build an over-the-world anything,
(Bwhy not build on the IP model, which is already over-the-world?
(BThe whole reason for multicast is to reduce the traffic at the source, not
(Bnecessarily just then receivers. And the concept behind ip multicast is
(Bto replicate as closely as possible the IP model - send trafic to an IP
(Baddress, and let the layer 3 devices forward the packet to the right
(Bshared-media networks as required.
(B}And - your NANOG forum is the excellent example. RealVideo streaming work
(B}fine; Multicast don't work at all; why do you try to use weak schema
(B}instead of the strong one? No enougph bandwidth - install stream
(B}replicators inb the key points; build _replication on the fly_ schemas
(B}(such as CCP for the www caching on the fly), etc. No, even with all
(B}attempts Cisco and some other are trying this days - multicast is more
(B}dead than alive. I can get 10,000 multimedia sources by RealVideo or
(B}StreamVideo - and I can't get nothing usefull by multicast. If I could
(B}install RV-cache engine (cache on the fly) - I should choose this
(BYou can get a lot more software for Windows, too, but that doesn't make it
(Bthe right solution all the time. How much software was available for
(BLinux just two years ago? Market share is a poor measurement of the
(Bquality and capability of a solution.
(BAndy McConnell IP Operations Manager [email protected]
(BNTT America Network and IP Service Division +1 408 873 3757
$B??8~N}(B $B0BEHN6(B NTT$B%"%a%j%+(BIP$B%*%Z%l!<%7%g%sC4Ev2]D9(B
(B"What right does Congress have to go around making laws just because they
(Bdeem it necessary?"
(B - M. Barry, Mayor of Washington, DC
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s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00460-ip-10-147-4-33.ec2.internal.warc.gz
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| 3,843 | 63 |
http://www.jiskha.com/members/profile/posts.cgi?name=Nima
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math
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Total # Posts: 16
I'm trying to solve these equations but it seems I didn't get the correct one. please help if you still can. to solve the equations: 1. 0.4 =32z 2. 8n - 2 = 14 3. 2 (x + 3) = 4 Thanks a lot.
September 14, 2013
Please help. Merina is moving out of her mom's house to her new house which is 14 miles away to reach her new house. Truck A offers a flat fee of $25 and $.40 per mile. What is the proper equation to solve the problem.
September 12, 2013
Please help, I am confused with this problem. My problem is: Beaver Stadium has a capacity of 1081 more than Michigan Stadium. If the combined capacity for the two stadiums is 213,483, find the capacity of each stadium?
September 11, 2013
There are 100 runners, each given a distinct bib labeled 1 to 100. What is the most number of runners that we could arrange in a circle, such that the product of the numbers on the bibs of any 2 neighboring runners, is less than 1000?
April 15, 2013
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| 954 | 9 |
http://thomasgrey.co.uk/6da2idrn/page.php?329dfa=burr-distribution-3-parameters
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math
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burr distribution 3 parameters December 4, 2020 – Posted in: Uncategorized
The two parameter BX has several types of distribution like Rayleigh (R) when \((\theta = 1)\) and Burr type X distribution with one parameter (BX1) when \((\lambda =1)\).BX1 has been studied by some authors, for example: Ahmad Sartawi and Abu-Salih (), Jaheen (), Jaheen (), Ahmad et al. The dBurr(), pBurr(), qBurr(),and rBurr() functions serve as wrappers of the dparetoIV, pparetoIV, qparetoIV, and rparetoIV functions in the VGAM package. RANI (1997). Generate sample data from a Burr distribution with scale parameter 0.5 and shape parameters 2 and 5. Need help with a homework or test question? Comments? The exact expression of the expected Fisher information matrix of the parameters in the distribution is obtained. Tadikamalla, Pandu R. (1980), “A Look at the Burr and Related Distributions”, International Statistical Review 48 (3): 337–344. It addresses the problem of estimating the three-parameter Burr XII distribution and its doubly truncated It can fit a wide range of empirical data, and is used in various fields such as finance, hydrology, and reliability to model a variety of data types. Need to post a correction? It addresses the problem of estimating the three-parameter Burr XII distribution and its doubly truncated Please post a comment on our Facebook page. Applied Mathematics and Information Sciences, 11, no. The shape of a Burr distribution associated with or is contingent on the values of the shape parameters (and ), which can be determined by simultaneously solving equations (16) and (17) from [2, p. 2211] for given values of skew and kurtosis. Also, an approximation based on Lindley is used to obtain the Bayes estimator. Burr Type III distribution has two categories: First a two-parameter distribution which has two shape parameters and second a three-parameter distribution which has a scale and two shape parameters. for 2-parameter and c, k and s for 3-parameter of Burr Type XII distribution with co mplete and censored data using two methods which include MLE and EM algorithm Tables 7 and 8 show the estimated Burr distribution parameters for Glen Osmond and South Road link travel time data sets. Parameter estimates We can apply a bootstrap to estimate the uncertainty in the parameters: Density, distribution function, quantile function and random generation for the Burr distribution with a and k two parameters. Based on this family, we define a new four-parameter extension of the Burr III distribution. A five-parameter distribution, the beta Burr XII, is useful for modeling lifetime data. A.CHATURVEDI,U. CLICK HERE! ", https://en.wikipedia.org/w/index.php?title=Burr_distribution&oldid=980569186, Creative Commons Attribution-ShareAlike License, This page was last edited on 27 September 2020, at 07:20. The Burr Type III distribution has been applied in the study of income, wage and wealth. Online Tables (z-table, chi-square, t-dist etc. Johnson, N.L. Burr type III lower record values are generated using the inverse cdf, X i = (u i −1/k − 1) −1/c, where u i is the uniformly distributed random variate. It is widely recognised that the three-parameter Burr XII distribution has Weibull distribution as a limiting case as α →+∞ with φ / α1/τ = θ remaining finite (which implies that φ →+∞ simultaneously); see, for example, Rodriguez (1977), Watkins (1999) and Shao (2000). Feroze, N. & Aslam, M. (2013) “Maximum Likelihood Estimation of Burr Type V Distribution under Left Censored Samples.” WSEAS Transactions on Mathematics. Other forms of this distribution have very little research associated with them. It can have decreasing, unimodal and decreasing-increasing-decreasing hazard rate function. Vol. Estimation procedures for a family of density functions representing various life-testing models. Figures 1-4 gives the pdf plot for three parameter Burr type XII distribution and Lomax distribution for different values of parameters. et. Density function, distribution function, quantile function, random generation,raw moments and limited moments for the Burr distribution withparameters shape1, shape2 and scale. y1label Correlation Coefficient x1label R burr type 3 ppcc plot y let r = shape1 let k = shape2 justification center move 50 6 text Rhat = ^r (R = ^rsav), Khat = ^k (K = ^ksav) move 50 2 text Maximum PPCC = ^maxppcc . , See Kleiber and Kotz (2003), Table 2.4, p. 51, "The Burr Distributions. S= Q(6 8 “Continuous Univariate Distributions”. This paper identifies the characteristics of three-parameter Burr Type XII distribution and discusses its utility in survivorship applications. Both functions support censored data for Burr distribution. Table 7. ). Expectation-maximization (EM) algorithm method is selected in this paper to estimate the two- and three-parameter Burr Type III distributions. Descriptive Statistics: Charts, Graphs and Plots. , The Burr Type XII distribution is a member of a system of continuous distributions introduced by Irving W. Burr (1942), which comprises 12 distributions. al (2013) say about the Type V that “Many properties of the parameters of the distribution under different estimation procedures are still to be revealed.”, References: The cdf is: The Burr distribution is very similar (and is, in some cases, the same as) many other distributions such as: In 1941, Burr introduced twelve cumulative distribution functions that could be fit to real life data. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. When the fourth parameter, γ, equals zero, it gives a three parameter (c,k,α) distribution. let y = burr type 3 random numbers for i = 1 1 200 let y = 10*y let amax = maximum y . A summary of the models is provided in Table 22.3.For each distribution model, the table lists the parameters in the order in which they appear in the signature of the functions or subroutines that accept distribution parameters as input or output arguments. In my papers the probability density for a burr distribution is given as. It is also known as the Singh–Maddala distribution and is one of a number of different distributions sometimes called the "generalized log-logistic distribution". A set of predefined distribution models is provided with the SEVERITY procedure. The Burr distribution is a three-parameter family of distributions on the positive real line. The Burr (Type XII) distribution has probability density function:, When c = 1, the Burr distribution becomes the Pareto Type II (Lomax) distribution. Visually and using the previous statistics, it seems that the Burr distribution seems the preferred one among the candidates we chose to explore. It is suitable to fit lifetime data since it has flexible shape and controllable scale parameters. The Burr is the distribution of the random variable s (X/(1 - X))^(1/b), where X has a beta distribution with parameters 1 and a. The pdf for the Burr XII distribution is: parameter Burr type XII distribution under the failure-censored plan. 2, John Wiley & Sons, New York, NY, USA, 2nd edition. The Burr distribution is a special case of the Pareto(IV) distribution where the location parameter is equal 0 and inequality parameter is equal to 1/g, Brazauskas (2003).
Black Desert Ps4 Performance 2020, Telephone Cad Block, Hello Organics Vitamin C Serum, Ain't We Got Fun Liz Gillies Lyrics, Rex Begonia Propagation, Earth And Blood Review, Online Tutor Resume, Skyn Iceland Microneedle Eye Patches, Goldilocks Chocolate Cake Slice, Data Mart Example, Studio Flat Near Me, Corizus Hyoscyami Pest, Cranberry Lime Macaroons, Pennsylvania Teacher Certification Reciprocity,
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CC-MAIN-2021-17
| 7,697 | 3 |
https://physics.stackexchange.com/questions/38772/induced-current-in-parallel-wires
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math
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Consider two parallel wires of finite radius. When a current is applied to one of the wire for a short period of time, what is the current induced in the other wire?
Applying Maxwell's equations, it seems that there is a change in magnetic field perpendicular to the second wire, and as a result, the induced current has a nontrivial distribution which averages somewhat to zero. Is this correct? Intuitively, I had instead expected a simpler result similar to the case of two coils of wires placed side by side.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00415.warc.gz
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CC-MAIN-2019-43
| 512 | 2 |
http://openstudy.com/updates/4f2de7dae4b0571e9cbaa622
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math
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A chord of a circle of radius 10cm subtends a right angle the center.Find the area of minor sector.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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Ok. Now lets think. We have 1/4 of the entire circle. The entire circle has area 100 pi.
so 1/4 of the circle is 25 pi
The triangle is an isosceles right triangle and the legs are the base and height. So the area of the triangle is 1/2 (10)(10) or 50
If we subtract the 50 from 1/4 of the circle we should have the desired part.
Looking at the diagram the Mertsj drew, we can easily find the area of the minor sector:
A(sector) = [(central angle) / (360 degrees)] x A(circle)
The central angle is a right angle and measures 90 degrees.
A(sector) = [(90 degrees) / (360 degrees)] x [(pi) x (10 cm)^2]
= (1/4) x (100 cm^2)(pi)
= 25pi cm^2
Answer: 25pi cm^2
pi = 3.14159...
I see physmath wants to help you now. Good Bye
And we are looking for the area of the minor sector NOT the area of the circular segment. So the answer is simply 25pi cm^2.
its okay bye thnx anyway !!!!
No problem :)
I don't know if you are still there or not. But physmath is incorrect. The minor segment is the region bounded by the chord and the minor arc intercepted by the chord.
Mertsj, the question asks for the minor SECTOR and NOT the SEGMENT. What you are referring to is called the CIRCULAR SEGMENT but the question is not asking for this. If you disagree, simply search up the definition of the two terms.
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CC-MAIN-2018-05
| 2,367 | 26 |
https://hoven-discuss.appspot.com/Home/Aptitude/Alligations-and-Mixtures/Discuss/16-05-07-07-08-29-007.html
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math
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12 liters of Petrol costing Rs. 5/liter is mixed with 6 liters of Kerosene costing Rs. 14/liter. What is the price of the mixture per liter?
Rs. 8 per liter.
Rs. 9 per liter.
Rs. 7 per liter.
Rs. 11 per liter.
We shall use the alligation formula. If a sample n1 has an average price of A1, and another sample n2 has an average price of A2, then the price of the mixture, A, is determined by the alligation formula as: n1(A - A1) = n2(A2 - A). Putting A2 = 14, A1 = 5, n1 = 12, n2 = 6, we have 12 × (A - 5) = 6 × (14 - A), from where we get A = Rs. 8 per liter.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00195.warc.gz
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CC-MAIN-2020-16
| 562 | 6 |
https://sophiethemathmo.wordpress.com/2021/01/17/twins/
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math
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Today is my twin brother’s birthday, so I decided that’s a good enough excuse to talk a bit about twins. Whilst “twin brothers” are my favourite variety of twin, I’m actually going to be writing about my second favourite twins: Twin Primes.
Twin Primes refer to any pairs where both and are prime. For example , and , and . There are bigger twin primes too, like , and , and .
As of September , the largest known pair of twin primes was . But what about twin primes bigger than that? How many are there? What are they? Well the truth is, we don’t know.It is conjectured there there are infinitely many twin primes, but this remains unproven.
So what do we know about twin primes? Well, aside from , every prime number has at most one twin. This is because every third odd number is a multiple of , and the only prime multiple of is itself. Using modular arithmetic, we can extend this idea further to say that, aside from , if are twin primes, we must have (mod ). The only way this is possible is if (mod ). Further, is the only even prime number, so and must both be odd. Combining this with the previous relation, we get that (mod ).
There is a cool theorem in number theory called Wilson’s Theorem. This states that for odd , is prime if and only if (mod ). This is the same as saying (mod ). As is odd, , so (mod ). Finally, (mod ) so we can say (mod ).
Now consider twin primes . We established earlier that they must both be odd so the above relation must hold, i.e. (mod ), and therefore (mod ).
Using Wilson’s Theorem on yields (mod ).
Note that .
Hence we can write (mod ), i.e. (mod ).
This therefore means that (mod ).
We then do the following algebraic manipulation:
So now we have that (mod ) and (mod ). Because and are coprime, we can combine these two equations and get (mod ) for all twin primes
If this has got you intrigued, see what you can work out (or find out) about cousin primes (where both and are prime) and sexy primes (where both and are prime)!
One final type of number I’ll mention is “Isolated Primes”. This refers to any number that isn’t a twin prime. It’s quite a fitting name really, as I know I’m a lot happier because of my twin.
Happy Birthday Jack!
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https://community.graphisoft.com/t5/Document-Visualize-forum/Light-Fixture-Schedule-that-includes-2D-RCP-symbol/td-p/365190
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math
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Ricardo, thanks for the reply. I already had it set up that way, but I figured out that the 2D representation for each light has to be set NOT to "by MVO." They are showing up now.
Next question... can I change what the RCP symbol looks like? For example, for some reason both the pendant light and recessed spot light use the same RCP symbol, so one of them needs to change otherwise there's no point in scheduling these interactively. In addition, I'd like to be able to use a symbol somewhat close to what my engineer is using... as another example, the pendant light and the surface light in their 'electrical' symbols are reversed from what's typical here.
Bill Szustak RA
Principal, Springboard Design
ArchiCAD 23, Windows 10
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s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00538.warc.gz
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CC-MAIN-2023-14
| 731 | 5 |
https://www.statstutor.net/downloads/question-2013531-working-with-confidence-intervals/
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math
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Question: 8 A study described by Harlow (1959) in which monkeys were placed in cages with two artificial mothers. One mother was made of wire mesh and had a bottle from which the infant could feed, and the second mother was made of soft terry cloth but did not provide any food. Data for a sample of n=9 monkeys showed that the infants spent an average of M=15.3 hours per day with SS=216 with the terry cloth mother. Use the data to estimate how many hours per day would be spent with the terry cloth mother for the entire population of infant monkeys. Make a point estimate and an 80% confidence interval estimate of the population mean.
Deliverables: Word Document
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s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999066.12/warc/CC-MAIN-20190619224436-20190620010436-00500.warc.gz
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CC-MAIN-2019-26
| 667 | 2 |
https://geoscience.blog/should-8th-graders-take-algebra/
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math
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Should 8th graders take algebra?Space and Astronomy
Stronger math students take algebra in eighth grade, and although they indeed may benefit academically from the course, that does not mean that weaker students will also benefit from taking algebra earlier.
What math do most 8th graders take?
The primary strands for an 8th-grade math curriculum are number sense and operations, algebra, geometry, and spatial sense, measurement, and data analysis and probability. While these math strands might surprise you, they are all critical lessons for an 8th-grade math curriculum.
What grade do most kids take algebra?
In many schools today, algebra in the eighth grade is the norm, and students identified by some predetermined standard can complete the course in seventh grade. Algebra courses are even stratified as “honors” algebra and “regular” algebra at both of these grade levels.
What is 8th grade algebra?
Grade 8 Algebra is a high school level Algebra 1 course, and is the first course on their growth in upper level mathematics. The fundamental purpose of this course is to formalize and extend the mathematics that students learned through mastery of the middle school standards.
Is taking geometry in 8th grade good?
The concepts taught in geometry in the 8th grade math class are foundational for future understanding of geometry concepts. If students are not able to grasp these concepts, they will struggle in future math classes. This is the first year, for example, that students will make proofs to prove that something is true.
Is algebra harder than geometry?
Is geometry easier than algebra? Geometry is easier than algebra. Algebra is more focused on equations while the things covered in Geometry really just have to do with finding the length of shapes and the measure of angles.
Can 8th graders take algebra 1?
Algebra. Students taking Algebra 1 in eighth grade likely completed a pre-algebra course — or at minimum, a general math course introducing basic algebraic ideas — in seventh grade.
Can 8th graders take algebra 2?
Algebra 2 in 8th grade is actually not that advanced. A student who takes Algebra 2 in 8th grade is above average, but not necessarily college material.
What is 7th grade math?
In 7th grade, students will fully understand how to interpret and compute all rational numbers. They can add, subtract, multiply, and divide all decimals and fractions, as well as represent percents.
How do you skip a grade?
Requirements to Skip a Grade
- A Written Request. Put your request for skipping a grade in writing to the school principal and keep a copy. …
- Expert Guidance. Make sure that legitimate requirements are being used in considering your request. …
- Academic Achievement. …
- Emotional Readiness. …
- Student Acceptance. …
- Need for Change.
What do 8th graders learn in math?
Students continue to develop their understanding of patterns, including those that involve integers. They use Algebraic notation, such as, s = d/t, to represent the relationship between speed, distance and time. They solve Algebraic equations involving multiple terms, integers and decimal numbers.
Can 7th graders take algebra 1?
Seventh graders are capable of Algebra 1 or even Geometry, depending on how well they have prepared. It’s not the age, but how well you have prepared them. If the child is going to take a College Major related to Math or Math skills required, then try to take Algebra in 7th. grade at least.
Is my child ready for algebra?
According to the National Council of Teachers of Mathematics (NCTM), some indicators of algebra readiness include: The ability to use properties such as commutativity, associativity, and distributivity. E.g., knowing that 5 + 9 is equal to 9 + 5 (commutativity)
Which is harder algebra or pre-algebra?
Prealgebra introduces algebra concepts and takes each one slower and therefore does not cover as much material as a standard Algebra I course. Some parents find it is just as easy to take a regular Algebra I course and do it in two years, especially if the student is in the 6th or 7th grade.
What math is 6th grade?
The major math strands for a sixth-grade curriculum are number sense and operations, algebra, geometry, and spatial sense, measurement, and functions, and probability.
What grade do you learn algebra?
Algebra is the culmination of most elementary & middle school math programs. Typically, algebra is taught to strong math students in 8th grade and to mainstream math students in 9th grade.
What are the levels of algebra?
Algebra is divided into different sub-branches such as elementary algebra, advanced algebra, abstract algebra, linear algebra, and commutative algebra.
- Compaction in the Rock Cycle: Understanding the Process Behind Sedimentary Rock Formation
- Crystallization in the Water Cycle: A Fundamental Process in Water Distribution and Purification
- Understanding Crystallization in the Rock Cycle: A Fundamental Process in Rock Formation
- SQL Server to Google Maps
- Stereo-pair Image Registration
- Extracting Lat/Lng from Shapefile using OGR2OGR/GDAL
- Constructing query in Nominatim
- In Ogr2OGR: what is SRS?
- Identifying port numbers for ArcGIS Online Basemap?
- Remove unwanted regions from map data QGIS
- Waiting for Vector & WFS loading
- Adding TravelTime as Impedance in ArcGIS Network Analyst?
- Listing total number of features into an ArcGIS Online feature pop-up
- Criteria for cartographic capacity
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| 5,456 | 54 |
https://www.manualslib.com/manual/830800/Mitsubishi-Electric-Fr-A700.html?page=286
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math
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Energy saving operation and energy saving monitor
(2) Power saving instantaneous monitor ( 1) power savings, 2) power saving rate )
⋅ On the power saving monitor ( 1)), an energy saving effect as compared to the power consumption during
commercial power supply operation (estimated value) is calculated and displays on the main monitor.
⋅ In the following case, the power saving monitor ( 1)) is "0".
(a)Calculated values of the power saving monitor are negative values.
(b)During the DC injection brake operation
(c)Motor is not connected (output current monitor is 0A)
⋅ On the power saving rate monitor ( 2)), setting "0" in Pr. 895 Power saving rate reference value displays the power
saving rate on the assumption that power (estimated value) during commercial power supply operation is 100%.
When Pr. 895 = "1", the power saving rate on the assumption that the Pr. 893 Energy saving monitor reference (motor
capacity) value is 100% is displayed.
(3) Power saving average value monitor ( 3) power saving average value, 4) average power
saving rate average value, 5) power saving amount average value)
⋅ Power saving average value monitor can be displayed when a value other than "9999" is set in Pr. 897 Power saving
monitor average time.
⋅ The power saving average value monitor ( 3)) displays the average value per unit time of the power saving amount
⋅ The average value is updated every time an average time has elapsed after the Pr. 897 setting is changed, power is
turned on or the inverter is reset, assuming as a starting point.The power savings average value update timing
signal (Y92) is inverted every time the average value is updated.
When Pr.897=4 [Hr]
Pr. 897 setting
Y92: power saving
update timing signal
⋅ The power saving average value monitor ( 4)) displays the average value per unit time of power saving rate ( 2)) at
every average time by setting "0" or "1" in Pr. 895 Power saving rate reference value.
⋅ By setting the charge (power unit) per 1kWh of power amount in Pr. 896 Power unit cost, the power saving amount
average value monitor ( 5)) displays the charge relative to the power saving average value (power saving average
value ( 3)) × Pr. 896).
(4) Cumulative saving power monitor ( 6) power saving amount, 7) power saving amount
charge, 8) annual power saving amount, 9) annual power saving amount charge)
⋅ On the cumulative saving power monitor, the monitor data digit can be shifted to the right by the number of Pr. 891
Cumulative power monitor digit shifted times settings. For example, if the cumulative power value is 1278.56kWh
when Pr. 891 = "2", the PU/DU display is 12.78 (display in 100kWh increments) and the communication data is 12.
If the maximum value is exceeded at Pr. 891 = "0 to 4", the power is clamped at the maximum value, indicating that
a digit shift is necessary. If the maximum value is exceeded at Pr. 891 = "9999", the power returns to 0 and is
recounted. The other monitors are clamped at the display maximum value.
⋅ The cumulative saving power monitor ( 6)) can measure the power amount during a predetermined period.
Measure according to the following steps
1) Write "9999" or "10" in Pr. 898 Power saving cumulative monitor clear.
2) Write "0" in Pr. 898 at measurement start timing to clear the cumulative saving power monitor value and start
totalization of power saving.
3) Write "1" in Pr. 898 at measurement end timing to hold the cumulative saving power monitor value.
⋅ The cumulative saving power monitor value is stored every hour. Hence, when the power supply is switched on again within one
hour after it was switched off, the previously stored monitor value is displayed and totalization starts. (The cumulative monitor
value may decrease)
0 in the first
Stores Hi/Low when the
power is off and starts.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627997508.21/warc/CC-MAIN-20190616002634-20190616024634-00009.warc.gz
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CC-MAIN-2019-26
| 3,811 | 49 |
http://pages.rediff.com/node--physics-/300748
|
math
|
is a point along a standing wave
where the wave has minimal amplitude
. For instance, in a vibrating guitar
string, the ends of the string are nodes. By changing the position of the end node through frets
, the guitarist changes the effective length of the vibrating string
and thereby the note
played. The opposite of a node is an anti-node
, a point where the amplitude of the standing wave is a maximum. These occur midway between the nodes.
result when two sinusoidal wave
trains of the same frequency
are moving in opposite directions in the same space and interfere
with each other. They occur when waves are reflected at a boundary, such as s reflected from a wall or s reflected from the end of a , and particularly when waves are confined in a at , bouncing back and forth between two boundaries, such as in an or .
In a standing wave the nodes
are a series of locations at equally spaced intervals where the wave amplitude
(motion) is zero (see animation above). At these points the two waves add with opposite phase
and cancel each other out. They occur at intervals of half a wavelength
(λ/2). Midway between each pair of nodes are locations where the amplitude is maximum. These are called the antinodes
. At these points the two waves add with the same phase and reinforce each other.
In cases where the two opposite wave trains are not the same amplitude, they do not cancel perfectly, so the amplitude of the... Read More
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00146.warc.gz
|
CC-MAIN-2024-10
| 1,438 | 19 |
https://www.jamiletheteacher.com/geometry/what-does-reflexive-mean-in-geometry.html
|
math
|
Reflexive property of equality In geometry, the reflexive property of congruence states that an angle, line segment, or shape is always congruent to itself.
- 1 What is an example of the reflexive property?
- 2 What is the reflexive property geometry?
- 3 What is reflexive relation in maths?
- 4 What is a reflexive equation?
- 5 What does the reflexive property mean?
- 6 What is reflexive property of a triangle?
- 7 What is a reflexive side?
- 8 What is reflexive transitive?
- 9 What is reflexive property in parallelogram?
- 10 What is the meaning of reflexivity?
- 11 What is a reflexive relation in sets?
- 12 How do you show reflexivity?
- 13 Is a B reflexive?
- 14 What is reflexive relation class 12?
- 15 What do reflexive verbs mean?
What is an example of the reflexive property?
This property tells us that any number is equal to itself. For example, 3 is equal to 3.
What is the reflexive property geometry?
Reflexive property of congruence means a line segment, or angle or a shape is congruent to itself at all times. Symmetric property of congruence means if shape 1 is congruent to shape 2, then we can say that shape 2 is also congruent to shape 1.
What is reflexive relation in maths?
In Maths, a binary relation R across a set X is reflexive if each element of set X is related or linked to itself. For example, when every real number is equal to itself, the relation “is equal to” is used on the set of real numbers.
What is a reflexive equation?
The reflexive property states that any real number, a, is equal to itself. That is, a = a. The symmetric property states that for any real numbers, a and b, if a = b then b = a. The transitive property states that for any real numbers, a, b, and c, if a = b and b = c, then a = c.
What does the reflexive property mean?
The Reflexive Property states that for every real number x, x=x. Symmetric Property. The Symmetric Property states that for all real numbers x and y, if x=y, then y=x.
What is reflexive property of a triangle?
The reflexive property of congruence states that any shape is congruent to itself.
What is a reflexive side?
The reflexive property of congruence states that any geometric figure is congruent to itself. If we had a triangle with the same side lengths and angle measures, the triangles would be congruent. The reflexive property of congruence shows that any geometric figure is congruent to itself.
What is reflexive transitive?
R is reflexive if for all x A, xRx. R is transitive if for all x,y, z A, if xRy and yRz, then xRz. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.
What is reflexive property in parallelogram?
Segment AC is congruent to itself by the reflexive property. This means triangle ABC is congruent to triangle CDA by ASA. Since the triangles are congruent, corresponding parts of congruent triangles are congruent (CPCTC). This means segment AD is congruent to segment CB and segment AB is congruent to segment CD.
What is the meaning of reflexivity?
reflexivity noun [U] (IN THOUGHT) the fact of someone being able to examine their own feelings, reactions, and motives (= reasons for acting) and how these influence what they do or think in a situation: I had in that time developed a degree of reflexivity unusual for a teenager. More examples.
What is a reflexive relation in sets?
Reflexive relation on set is a binary element in which every element is related to itself. Let A be a set and R be the relation defined in it. R is set to be reflexive, if (a, a) ∈ R for all a ∈ A that is, every element of A is R-related to itself, in other words aRa for every a ∈ A.
How do you show reflexivity?
What is reflexive, symmetric, transitive relation?
- Reflexive. Relation is reflexive. If (a, a) ∈ R for every a ∈ A.
- Symmetric. Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R.
- Transitive. Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R. If relation is reflexive, symmetric and transitive,
Is a B reflexive?
The identity relation consists of ordered pairs of the form (a,a), where a∈A. In other words, aRb if and only if a=b. It is reflexive (hence not irreflexive), symmetric, antisymmetric, and transitive. For any a≠b, only one of the four possibilities (a,b)∉R, (b,a)∉R, (a,b)∈R, or (b,a)∈R can occur, so R is antisymmetric.
What is reflexive relation class 12?
A relation is a reflexive relation If every element of set A maps to itself. I.e for every a ∈ A,(a, a) ∈ R. OR. A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A×B. It maps elements of one set to another set.
What do reflexive verbs mean?
In grammar, a reflexive verb is, loosely, a verb whose direct object is the same as its subject; for example, “I wash myself”. More generally, a reflexive verb has the same semantic agent and patient (typically represented syntactically by the subject and the direct object).
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00566.warc.gz
|
CC-MAIN-2022-33
| 4,966 | 49 |
https://careers.allianz.com/de_DE/jobs/Allianz-Global-Investors/Lead-Investment-Portfolio-Compliance-Officer-f-m-d-Surveillance-Team-C2EBDE1DCF021EDAB7C6653B7B13864B.html
|
math
|
Lead Investment Portfolio Compliance Officer (f/m/d) - Surveillance Team
We are searching for a team lead (m/f/d) for our Investment Portfolio Compliance Surveillance team, which is part of the Investment Portfolio Compliance function at AllianzGI Europe GmbH. The position is based in Frankfurt.~crlf~~crlf~Our Investment Portfolio Compliance function covers all investment related compliance topics within AllianzGI’s process chain and stretches across various departments such as Trading, Operations, Client Services, Legal and Portfolio Management. The Investment Guideline team provides pre, post, and ex-post investment guideline coding services to our Global Portfolio Management teams. Additionally, this team more broadly supports the Investments Compliance function in developing reporting and monitoring tools for other Investment Portfolio Compliance subjects. ~crlf~~crlf~As a Lead Investment Portfolio Compliance Officer, you are part of a cooperative Global Investment Portfolio Compliance team which works directly with Global Portfolio Managers in addressing queries and advising on technical and regulatory matters, and will be responsible for providing effective, efficient, high quality and systems development services that address both business and regulatory requirements
##Co-ordinate and assign daily tasks to enable full and complete coverage of Surveillance responsibilities~crlf~##Review and maintain the process manual and evaluate Archer statistics periodically to identify risks, gaps and workloads~crlf~##Provide working students oversight on a regular basis to ensure their tasks are undertaken to the required standard and frequency~crlf~##Conduct regular employee reviews and Coach the team to ensure there is support and appropriate tools to do their role~crlf~##Build a bridge between Global Portfolio Managers and the Investment Portfolio Compliance team where knowledge is transferred between the two areas for a more efficient and compliant process, ~crlf~##Assisting Global Portfolio Managers and Client Account Management proactively to better understand specific investment and regulatory constraints, ~crlf~##Support the compliance officers in all jurisdictions ~crlf~##Support the compliance officer to service all Portfolio Managers by assessing queries and initiating confidence building tasks ~crlf~##Review and conclude all Impact calculations for regulatory or client reporting.~crlf~##Review and maintain the process manual and all controls (eg. ISAE controls) and evaluate Archer statistics periodically to identify risks, gaps and workloads~crlf~##Investigating investment queries with a view to provide impartial analysis and communicate reportable instances to senior management~crlf~##Being involved in projects which aim to enhance surveillance processes and systems between Global Portfolio Managers, Client Account Management and Investment Portfolio Compliance~crlf~~crlf~
##Team player and ability to build relationships internally and externally~crlf~##Organizational Skills / Attention to Detail~crlf~##Ability to work independently, but also a "team player"~crlf~##Ability to multi-task in a fast-paced environment ~crlf~##Ability to problem solve and spot issues proactively~crlf~##Thorough knowledge of financial instruments including derivatives as well as regulatory provisions across jurisdictions.~crlf~##Significant practical experience of at least one industry recognised compliance monitoring system i.e. Charles River, SimCorp Dimension, Bloomberg, Aladdin, Sentinel, Thinkfolio, MIG21 etc. ~crlf~##Excellent written and communication skills~crlf~##At least several years experience in the Financial Services industry~crlf~##Bachelor’s Degree or higher, preferably in Finance or related subjects~crlf~##Proficient in Microsoft Word, Excel, Access~crlf~~crlf~
Allianz is the home for those who dare – a supportive place where you can take the initiative to grow and to actively strengthen our global leadership position. By truly caring about people – both its 85 million private and corporate customers and more than 142,000 employees – Allianz fosters a culture where its employees are empowered to collaborate, perform, embrace trends and challenge the industry. Our main ambition is to be our customers’ trusted partner, instilling them with the confidence to grow. If you dare, join us at Allianz Group.~crlf~~crlf~Allianz is an equal opportunity employer. Everybody is welcome, regardless of other characteristics such as gender, age, origin, nationality, race or ethnicity, religion, disability, or sexual orientation.
Allianz Global Investors is a leading active asset manager with over 770 investment professionals in 25 offices worldwide and managing EUR 535 billion in assets for individuals, families and institutions.
Active is how we create and share value with clients. We believe in solving, not selling, adding value beyond pure economic gain. We invest long-term, drawing upon our broad and innovative investment expertise, and leveraging resources globally to ensure a superior client experience wherever they are based and whatever their investment needs.~crlf~Active is: Allianz Global Investors~crlf~~crlf~You are committed to creating value for our clients so that they can meet their financial objectives throughout their lives. You embrace a collegial culture based on strong values, take individual responsibility, and are dedicated to mastering challenging tasks. You aspire to make an impact in a diverse, international, technology-enabled and increasingly agile environment. If you feel inspired to elevate the active asset management experience, this is the place for you since,~crlf~Active is: Allianz Global Investors.~crlf~~crlf~
Nutze Google Assistant um dich mittels unseres Job-Interview-Trainings optimal auf dein Bewerbungsgespräch vorzubereiten. Informiere dich im Interview-Training über:
- Details zum Bewerbungsprozess
- Zahlen und Fakten zur Allianz
- Unsere Unternehmenskultur
Wie das funktioniert? Lass den Google Assistant über dein Smartphone oder Google Home Gerät wissen: "Hey Google, ich möchte mit Allianz Karriere sprechen."
Registriere dich in unserem Bewerbungssystem; du erhältst dann in Kürze eine Bestätigungs-E-Mail mit deinen Zugangsdaten. Füll anschließend das Bewerbungsformular aus. Du kannst deine Daten auch über Xing hochladen.
Mehr Informationen zu unseren Passwortregeln findest du in unseren FAQs.
Bitte beachte, dass das System keine ZIP-Dateien, geschützte PDF-Formate oder Anhänge von mehr als 7 MB verarbeiten kann. Du kannst deine Daten in folgenden Formaten hochladen:
- MS Office Formate (Word, PowerPoint, Excel)
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Weitere Informationen dazu findest du in unseren FAQs.
Melde dich in unserem Bewerbungssystem an und prüfe deine persönliche Startseite, um den Stand deiner Bewerbung einzusehen. Du kannst dein Profil freigeben, sodass unsere Recruiter dich im System finden und kontaktieren können.
Weitere Informationen dazu findest du in unseren FAQs.
Keine passende Stelle gefunden? Da können wir dir helfen. Melde dich für unseren Job-Alarm an! Du erhältst eine auf dich zugeschnittene Benachrichtigung, wenn eine Stelle, die deinen Erfahrungen und Interessen entspricht, frei wird.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00794.warc.gz
|
CC-MAIN-2020-40
| 7,387 | 23 |
https://telegra.ph/Gre-Math-Formula-Sheet-Free-Pdf-40-05-12
|
math
|
Gre Math Formula Sheet Free Pdf 40gaultjarma
Gre Math Formula Sheet Free Pdf 40http://jinyurl.com/hop24
of Formulas, Processes and Tricks . 40 Translation 41 . mathematics in a business setting and in math tutoring. Contains free downloadable handbooks, PC .
. . If the price of something goes from $40 to $52, the percent change is .
Table of Contents: Directions . D. 40% E. 56% 11. A car travels at a rate of 55 mph for the first 4 hours of a trip.
. which means not all the questions from this old GRE practice test pdf are useful . GRE Math Formula Cheat Sheet. . All 6 Free Official GRE Practice Tests .
Math Formulas and Equations. . Math Formula Sheet for gre,gat,sat,gmat; . 95ec0d2f82
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221214702.96/warc/CC-MAIN-20180819051423-20180819071423-00322.warc.gz
|
CC-MAIN-2018-34
| 694 | 7 |
https://www.examrace.com/ISS/ISS-Updates/NEWS-ISI-Entrance-Exam-2016-Syllabus.htm
|
math
|
Indian Statistical Institute Entrance Exam 2016 Syllabus [ ISS Updates ]
Indian Statistical Institute Entrance Exam syllabus is given below B. stat and B. Maths Entrance Exam syllabus is same. The details of the syllabus are given below. The exam will be conducted on 8th May 2016 and the last date for the registration is 11th March 2016.
Algebra and number theory
- Sets, operations on sets
- Prime numbers. factorization of integers and divisibility
- Rational and irrational numbers
- Permutations and combinations
- Binomial Theorem
- Polynomials: relations between roots and coefficients
- Remainder Theorem. Theory of quadratic equations and expressions
- Arithmetic and geometric progressions
- Inequalities involving arithmetic, geometric & harmonic means.
- Complex numbers.
- Class 10 level plane geometry
- Geometry of 2 dimensions with Cartesian and polar coordinates, concept of a locus, equation of a line
- Angle between two lines
- Distance from a point to a line, area of a triangle, equations of a circle, parabola. ellipse and hyperbola and equations of the tangents and normal, mensuration
- Measures of angles, trigonometric and inverse trigonometric functions, trigonometric identities including addition formulae, solutions of trigonometric equations.
- Properties of triangles, heights and distances
- Sequences - bounded sequences, monotone sequences, limit of a Sequence.
- Functions - one-one functions, onto functions.
- Limit, continuity and differentiability of functions of a single real variable.
- Derivatives and methods of differentiation
- The slope of a curve, tangents and normals.
- maxima and minima
- Use of calculus in sketching graphs of functions.
- Methods of integration, definite and indefinite integrals
- evaluation of areas using integrals
- Published/Last Modified on: February 9, 2016
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s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00293-ip-10-145-167-34.ec2.internal.warc.gz
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CC-MAIN-2017-17
| 1,837 | 29 |
https://www.bbc.co.uk/bitesize/guides/z97rsrd/test
|
math
|
Waves are one way in which energy may be transferred between stores. Both mechanical and electromagnetic waves will transfer energy but not matter.
What do waves transfer?
Energy but not matter
Energy and matter
Matter but not energy
What is the name of the distance from peak to peak of a wave?
How do oscillations move in a transverse wave?
At right angles to the direction of wave travel
Parallel to the direction of wave travel
Which of these is an example of a transverse wave?
What is the time period of a 40 Hz wave?
What is the frequency of a wave with a time period of 0.08 s?
What is the wave speed of a 1,000 Hz wave with a wavelength of 0.04 m?
What is the speed of sound in air?
How is wave speed calculated?
Wave speed = distance ÷ time
Wave speed = distance × time
Wave speed = time ÷ distance
What is the speed of a sound wave that takes 0.5 s to travel 750 m?
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574377.11/warc/CC-MAIN-20190921084226-20190921110226-00529.warc.gz
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CC-MAIN-2019-39
| 879 | 19 |
https://www.booktopia.com.au/linear-algebra-jin-ho-kwak/prod9780817642945.html
|
math
|
From a review of the first edition:
"A logical development of the subject . . . all the important theorems and results are discussed in terms of simple worked examples. The student's understanding . . . is tested by problems at the end of each subsection, and every chapter ends with exercises."
A cornerstone of undergraduate mathematics, science, and engineering, this clear and rigorous presentation of the fundamentals of linear algebra is unique in its emphasis and integration of computational skills and mathematical abstractions. The power and utility of this beautiful subject is demonstrated, in particular, in its focus on linear recurrence, difference and differential equations that affect applications in physics, computer science, and economics.
Key topics and features:
- Linear equations, matrices, determinants, vector spaces, complex vector spaces, inner products, Jordan canonical forms, and quadratic forms
- Rich selection of examples and explanations, as well as a wide range of exercises at the end of every section
- Selected answers and hints
- Excellent index
This second edition includes substantial revisions, new material on minimal polynomials and diagonalization, as well as a variety of new applications. The text will serve theoretical and applied courses and is ideal for self-study. With its important approach to linear algebra as a coherent part of mathematics and as a vital component of the natural and social sciences, Linear Algebra, Second Edition will challenge and benefit a broad audience.
From the reviews:
"This is a standard book on Linear Algebra for science and engineering students. It covers the usual topics, including the Jordan canonical form, a topic that is omitted in many recent books at this level.
The book reminded me of Strang's Linear Algebra and its Applications.... Like Strang, the authors discuss linear difference and differential equations at some length, which should be useful to students in applied sciences. Unlike Strang, however, Kwak and Hong follow a more traditional line of presentation, with numbered definitions, lemmas, theorems, and examples. This may make it easier for the student to use the book as a reference. The exposition is clear but the style is not as chatty as Strang's.
In summary, the book can be safely used as the basis for a course on Linear Algebra for the intended audience." --MAA Reviews
"The emphasis is on computational skills along with mathematical abstractions; basic concepts are introduced by means of matrices and the solution of systems of linear equations. Many illustrative examples are given and all the usual advanced topics are treated ... The second edition has been substantially revised and new sections have been added." (ZENTRALBLATT MATH)
"As linear algebra is one of the most important subjects in the study of science and engineering because of widespread applications in social or natural science, computer science, physics, or economics this book covers one of the most useful courses in undergraduate mathematics, providing essential tooks for industrial scientists. . . The primary purpose of the book is to give a careful presentation of the basic concepts of linear algebra as a coherent part of mathematics, and to illustrate its power and utility through applications of other disciplines.
---Educational Book Review
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s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247479627.17/warc/CC-MAIN-20190215224408-20190216010408-00291.warc.gz
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CC-MAIN-2019-09
| 3,352 | 16 |
http://peqebavegorebujyd.mint-body.com/writing-research-question-hypothesis-and-variables-7119871198.html
|
math
|
Make sure your hypothesis is a specific statement relating to a single experiment. The way that you structure your research questions; that is, the way that you write out your research questions will vary depending on the type of research question you are trying to answer [see the article: Administrators who provide wellness programs for their employees receive higher employee ratings on selected leadership qualities than administrators who do not provide wellness programs.
To learn more about these three types of quantitative research question i. Sometimes, the dependent variable needs to be broken into two parts around the group s you are interested in so that the research question flows.
The examples illustrate the difference between the use of a single group e.
Identified the variables in the project. Does the amount of calcium in the diet of elementary school children effect the number of cavities they have per year? We make an "educated guess. Describe the purpose of each and understand the importance of a well-developed question or hypothesis.
Independent and dependent variables A variable is not only something that we measure. We could describe factors relating to the make-up of these Facebook users, quantifying how many or what proportion of these university students were male or female, or what their average age was.
Dependent variable 1st; group 2nd: In the second example, the research question is not only interested in what the factors influencing career choices are, but which of these factors are the most important. The first two examples highlight that while the name of the dependent variable is the same, namely daily calorific intake, the way that this dependent variable is written out differs in each case.
Does the question flow? Sometimes the number of sources you find will help you discover whether your research question is too broad, too narrow, or okay? How to structure quantitative research questions ].
These three approaches to examining the variables you are interested in i. A complete hypothesis should include: As a general rule, we would suggest using hypotheses rather than research questions in these cases. It builds upon previously accumulated knowledge e.
The following video may be helpful in learning how to choose appropriate keywords and search online databases: This question may allow the researcher to collect data but does not lend itself to collecting data that can be used to create a valid argument because the data is just factual information.
How many calories are consumed per day by American men and women? If you feel like the research questions are no more than a repetition of the research hypotheses, it is often better to include only one or the other i.
Who will it help and how? Research Questions, discusses how to choose whether to use a hypothesis or a question when creating a research project.
Name of the dependent variable How the dependent variable is written out Daily calorific intake How many calories do American men and women consume per day? We use the word groups of variables because both categorical and continuous variables include additional types of variables.
When performing quantitative analysis on the data you collect during the dissertation process, you need to understand what type of categorical or continuous variables you are measuring.
Is it a hot topic, or is it becoming obsolete? What are the effects of intervention programs in the elementary schools on the rate of childhood obesity among 3rd - 6th grade students?
If so, you will be using a hypothesis.Often, one of the trickiest parts of designing and writing up any research paper is writing the hypothesis. How to structure quantitative research questions. Writing out the problem or issues you are trying to address in the form of a complete research question.
In this article, Select the appropriate structure for the chosen type of quantitative research question, based on the variables and/or groups involved.
This section of the article briefly discusses the difference between these three types of quantitative research question. What variables are you trying to measure, manipulate and/or control?
Dissertations that are based on a quantitative research design attempt to answer at least one quantitative research question. These hypothesis. A complete hypothesis should include: the variables, the population, and the predicted relationship between the variables.
Planning My Research Question or Hypothesis – This resources contains a link to a PowerPoint presentation and a series of tutorials that contain examples and tips for writing research questions and hypotheses.
Research Question Why Intro Develop a Theory Your Answer Intro Identify Variables (if applicable) more variables.” “A hypothesis can be defined as a tentative explanation of the research To be either writing-questions, or.
Developing Research Questions: Hypotheses and Variables Common Sources of Research Questions Professors Textbooks Databases This is the beginning of a good research question. Topics for research As of this writing, two useful search engines are Ingenta, Galaxy and Google.Download
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00144.warc.gz
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CC-MAIN-2019-22
| 5,169 | 21 |
https://www.geeksforgeeks.org/even-odd-turn-game-with-two-integers/
|
math
|
Given three positive integers X, Y and P. Here P denotes the number of turns. Whenever the turn is odd X is multiplied by 2 and in every even turn Y is multiplied by 2. The task is to find the value of max(X, Y) ÷ min(X, Y) after the complete P turns.
Input : X = 1, Y = 2, P = 1 Output : 1 As turn is odd, X is multiplied by 2 and becomes 2. Now, X is 2 and Y is also 2. Therefore, 2 ÷ 2 is 1. Input : X = 3, Y = 7, p = 2 Output : 2 Here we have 2 turns. In the 1st turn which is odd X is multiplied by 2. And the values are 6 and 7. In the next turn which is even Y is multiplied by 2. Now the final values are 6 and 14. Therefore, 14 ÷ 6 is 2.
Lets play the above game for 8 turns :
| i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |------|---|----|----|----|----|----|----|-----|-----| | X(i) | X | 2X | 2X | 4X | 4X | 8X | 8X | 16X | 16X | | Y(i) | Y | Y | 2Y | 2Y | 4Y | 4Y | 8Y | 8Y | 16Y |
Here we can easily spot a pattern :
if i is even, then X(i) = z * X and Y(i) = z * Y. if i is odd, then X(i) = 2*z * X and Y(i) = z * Y.
Here z is actually the power of 2. So, we can simply say –
If P is even output will be max(X, Y) ÷ min(X, Y) else output will be max(2*X, Y) ÷ min(2*X, Y).
Below is the implementation :
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CC-MAIN-2020-05
| 2,330 | 26 |
https://kedawulafuxaqahog.turnonepoundintoonemillion.com/determining-molecular-formula-13076gk.html
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math
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Fourthly, you had to check your beaker for any cracks, since if it had any, then it would have a great chance of it exploding when heated. Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known.
For hydrogen, the atomic weight is 1. Thus, your x would be lower than the actual value. Repeat Steps 2 and 3.
This is initially due to a transient GH resistance, but in the longer term, the less pulsatile pattern of GH secretion may be a major factor since only pulsatile growth hormone increases the levels of IGF-I and the acid-labile subunit.
Calculate weight and mole percentages from a formula. If the number is too far to round x. This is because the molecular formula for acetylene is C2H2 while the molecular formula for benzene is C6H6. In this movie, we show how to calculate the molecular weight of a substance from the atomic weights given on the periodic table.
This is is consistent with this being the molecular ion. There is only 1 oxygen, so the total weight of oxygen is For non-molecular substances such as table salt, we represent the composition with an empirical formula.
Any text you add should be original, not copied from other sources. Steps for Determining an Empirical Formula Start with the number of grams of each element, given in the problem.
Their solutions are related to the skills outlined at the start of this page.
Explain the meaning of chemical formula. What is the weight percentage of chlorine Cl.
Prepared a table to record observations. Aspartic acid contains Example 7 Chloroform is a common solvent used in chemical labs. For instance, a person has two hands and ten fingers, or H2F Data Analysis, Concept What is the volume in liters of air in the flask.
Chemical Formulas Chemical formulas such as HClO4 can be divided into empirical formula, molecular formula, and structural formula. A course of error that would result in a value of x that is higher than the actual value is if you had not measured the masses correctly. Some types of ionic compounds, however, cannot be written with entirely whole-number empirical formulas.
Acetylene's empirical formula is the same as that of benzene, since in both molecules the ratio of carbon atoms to hydrogen atoms is 1: Wash your hands before leaving the laboratory.
For example, the empirical formula of ethanol may be written C2H6O because the molecules of ethanol all contain two carbon atoms, six hydrogen atoms, and one oxygen atom. If the formula is a molecular formula, the mass associated with it is called molecular mass or molecular weight.
You will learn more about it in organic chemistry. For simplicity, we may call these weights molar masses, which can be formula weights or molecular weights. From example 1, we know that there are.
Divide the molar mass for the molecular formula by the empirical formula mass. The result determines how many times to multiply the subscripts in the empirical formula to get the molecular formula.
The molecular weight of this compound is g/mol. Chemical formulas such as HClO 4 can be divided into empirical formula, molecular formula, and structural formula.
Chemical symbols of elements in the chemical formula represent the elements present, and subscript numbers represent mole proportions of the proceeding elements.
CHM / CHM Bergen Community College Cerullo Learning Assistance Center (CLAC) Find the empirical formula for the following molecular compositions. Compressive stress is a consideration in understanding how a material performs when under pressure.
In this lesson, learn what compressive stress is as well as the formula necessary for. Science&EnhancedScope&andSequence&–&Chemistry& Virginia’Department’of’Education’©’’ 1’ FindingtheFormulaandPercent&Composition&. The molecular (true) formula for a substance is not always the same as its empirical (simplest) formula.
Both acetylene and benzene have the empirical formula CH.
However, the molar mass for acetylene is 26 g/mol, while the molar mass of benzene is 78 g/mol.Determining molecular formula
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CC-MAIN-2022-05
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http://judaism.stackexchange.com/questions/tagged/bar-bas-bat-mitzvah+real-estate
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| 2,068 | 53 |
http://www.madsci.org/posts/archives/2004-08/1092408524.Ph.q.html
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math
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|MadSci Network: Physics|
Hi I was wondering if someone could help me to get a clearer picture of why some elements, tungsten for instance, produce a continuous spectrum whilst others, zinc/cadmium, produce line spectra. I am unsure how I could possibly predict whether an element would produce one or the other type of spectrum. As both types depend on electron transitions from higher to lower energy levels within their atoms, then I don't see how there are two different results. It is something to do with the temperature of the substance whose spectrum I am measuring? Tungsten I know has a very high melting point and can therefore withstand very high temperatures. However, surely it's atoms contain only specific energy levels and thus only specific transitions can occur, thus resulting in specific lines. This is obviously not the case! Thanks very much for your help. Nuala
Re: Continuous V's Line Spectra
Try the links in the MadSci Library for more information on Physics.
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CC-MAIN-2018-26
| 986 | 4 |
https://www.sarthaks.com/351663/exerts-surface-surface-exerts-reaction-force-reaction-force-always-along-normal-surface?show=351687
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math
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If a body exerts a force on a surface, the surface exerts a reaction force on the body. This reaction force is always along normal to the surface at that point and is popularly known as the normal force N. When the body slides on a surface, the surface exerts a force of friction on the body in the direction opposite to the direction of slipping. The magnitude of this frictional force is given by µN, where µ it is a constant (called coefficient of friction) for the given pair of surfaces in contact and N is the normal force.
Consider a block of mass 100 kg placed on a level surface. If one person tries to push it with a force F, he is unable to do so. He receives help from two of his friends and each one applies the same force F. Now the block moves with a uniform acceleration of 0.5 m/s2. After sometime, one of them happens to leave the exercise of pushing the block. This time the block moves with a uniform speed. The magnitude of force F and the coefficient of kinetic friction µ are respectively _________ .
(a) 50 N, 0.05
(b)100 N, 0.1
(c) 50 N, 0.1
(d)100 N, 0.2
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00245.warc.gz
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CC-MAIN-2021-17
| 1,083 | 6 |
https://www.doubtnut.com/question-answer-physics/a-mass-of-diatomic-gasgamma14-at-a-pressure-of-2-atomphere-is-compressed-adiabitically-so-that-its-t-643996662
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math
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Updated On: 27-06-2022
Get Answer to any question, just click a photo and upload the photo and get the answer completely free,
UPLOAD PHOTO AND GET THE ANSWER NOW!
given that the adiabatic temperature 27278 ATM 28 ATM ATM 256 adiabatic process India beti process what is the equation equation is ^ / 3 to the power minus 1 equal to constant or we can write
in Asia final to the power Kama equal to in final divided in to the power equation 1 now equal to 27 degree Celsius converted into Kelvin to become 300 Kelvin finally given 27 2020 can't 1200 K initial is
equal to 80 R&D 11.4 now put these values in equations 1 then finally come out to 1200 to 1300 to the power 15414 it will come out to be the cause for 1.4.5 and between kam 2222 power 7 or it will 2 to the power 8 then its value to the power 8 256 ATM final is correct
Click here to get PDF DOWNLOAD for all questions and answers of this chapter - NEET PREVIOUS YEAR ENGLISH Class 12 NEET
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s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00475.warc.gz
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CC-MAIN-2022-27
| 974 | 8 |
https://books.google.com/books?id=PxklAQAAIAAJ&dq=related:ISBN0306392518&source=gbs_similarbooks_r&hl=en
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math
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What people are saying - Write a review
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Generalized Fixpoint Algebras
5 other sections not shown
abelian group assume basis element BK(P Brauer construction Brauer kernel called capped endo-monomial OP-module capped endo-permutation Chapter class of capped conjugation crossed product Dade's theorem decomposition 4.2.1.a defined denote direct sum direct summand isomorphic discrete valuation ring embedding endo endo-monomial modules endo-permutation modules endo-trivial Endo(M endomorphism algebra equivalence classes equivalence relation field F finite group fix a decomposition g G G G-action G-graded ring group homomorphism hence i?G-module idempotents indecomposable direct summand indecomposable endo-monomial OP-module induced interior G-algebras ip G P Lemma Lemma Let linear character monomial Dade group monomial Dade P-algebra monomial modules monomial OP-module non-abelian OP-isomorphism OP-module with vertex p-group p-i p-i permutation module pointed groups projective-free Proof Puig reduction modulo RG-module Robert Boltje root of unity short exact sequence subgroup H subgroup of G summand with vertex surjective Thevenaz trace map twisted group algebra unique vertex Q VF(P Vr(P write
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s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00501.warc.gz
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CC-MAIN-2017-43
| 1,257 | 5 |
https://tgdaily.com/social/talent/133916-fermats-last-theorem/
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math
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Pierre de Fermat (1601 – 1665) was a French lawyer and mathematician who is credited with a number of early developments leading to the creation of infinitesimal calculus (a precursor to differential calculus). He also did extensive research into number theory, analytic geometry, probability and even optics. But the thing that he is most famous for is an unfinished theorem that became one of the most daunting mathematical puzzles of all time.
In 1637 while reading a copy of Arithmetica, an ancient Greek text on mathematics written by Diophantus, Fermat believed that he had come up with a solution to a problem that he had been working on for some time.
In highly simplified terms, he was working on a general case proof for a supposition based on an extension of the Pythagorean theorem. (In case you have forgotten, the Pythagorean theorem states that a^2 + b^2 = c^2, where a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse.)
Fermat had been playing around with different exponents applied to the Pythagorean theorem (a^n + b^n = c^n). He realized that there are no solutions in positive integers, if n is an integer greater than 2 (more specifically when n is an odd prime number equal to or greater than 3). While you can prove that there are no solutions for specific integers (Fermat himself provided a proof for the integer 4 and in subsequent years other mathematicians have offered proofs for other integers 3, 5, 7, and so on) there was no general case proof that could be applied to all integers.
But something in the text of Arithmetica inspired Fermat and in the margin of the book he scribbled a quick and somewhat enigmatic note that read:
It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvellous proof of this, which this margin is too narrow to contain.
In the last thirty years of his life, Fermat never again wrote of his “truly marvelous proof” and nothing was ever published.
After his death in 1665 his son Clément-Samuel Fermat published a new edition of Arithmetica that included his father’s notes and that one cryptic note inspired scores of mathematicians to take up the challenge.
In 1816, and again in 1850, the French Academy of Sciences offered a prize for a general proof of Fermat’s Last Theorem and in 1908 a German industrialist Paul Wolfskehl offered 100,000 gold marks for anyone who could come up with a solution to the problem.
But it wasn’t until May 1995 that British mathematician Andrew Wiles (in collaboration with Richard Taylor and others) finally published two papers in the issue of the Annals of Mathematics providing a definitive general case proof. It had taken Wiles nearly eight years of his life (and a number of failures) to finally come up with a solution. (By the way, Wiles collected the Wolfskehl prize money, worth $50,000, on 27 June 1997.)
It’s interesting to note that in order to come up with his proof Wiles had to build upon on the works of many other modern mathematicians – works that would not have been available to Fermat at the time – so it’s highly unlikely that Fermat actually did come up with a “truly marvelous proof” of his own. But we’ll never know for sure.
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http://www.weegy.com/?ConversationId=YAEE8M1O
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math
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The difference between two numbers is 25. The smaller number is 1/6th of the larger number. What is the value of the smaller number?
(See answer below)
What Is The Smaller Of The Two Numbers? The Difference Between Two Numbers Is 25. The Smaller Number Is 1/6th Of The Larger Number. [ What Is The Value Of The Smaller ... ...
There are no new answers.
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| 352 | 4 |
http://www.psypress.com/books/search/author/nancy_childress/
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math
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Series: Textbooks in Mathematics
This book integrates local and global theory to reflect a very modern view of algebraic number theory. This approach is used whenever possible to make the book as accessible as possible to readers with some background in abstract algebra. The author uses contemporary notation and includes numerous...
To Be Published December 15th 2015 by Chapman and Hall/CRC
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https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_1&oldid=54787
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math
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2005 USAMO Problems/Problem 1
(Zuming Feng) Determine all composite positive integers for which it is possible to arrange all divisors of that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
Solution 1 (official solution)
No such circular arrangement exists for , where and are distinct primes. In that case, the numbers to be arranged are ; and , and in any circular arrangement, and will be adjacent. We claim that the desired circular arrangement exists in all other cases. If where , an arbitrary circular arrangement works. Henceforth we assume that has prime factorization , where and either or else . To construct the desired circular arrangement of , start with the circular arrangement of as shown.
Then between and , place (in arbitrary order) all other members of that have as their smallest prime factor. Between and , place all members of other than that have as their smallest prime factor. Continue in this way, ending by placing between and . It is easy to see that each element of is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.
Note. In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases and .
The proof that no arrangement exists for , where are distinct primes follows from above. Apply induction to prove all other cases are possible
- , where is a prime and is a positive integer. Any arrangement suffices
- , where are distinct primes. The following configuration works
Inductive step: Suppose the desired arrangement exists for a composite , show the arrangement exists for , where is a prime relatively prime to and is a positive integer
Let be the arrangement of divisors of , then for , where . The divisors of greater than 1 are of the form The following sequence works since all other divisors are divisible by .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
- <url>viewtopic.php?t=34314 Discussion on AoPS/MathLinks</url>
|2005 USAMO (Problems • Resources)
|1 • 2 • 3 • 4 • 5 • 6
|All USAMO Problems and Solutions
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| 2,367 | 16 |
http://www.brightstorm.com/math/algebra/quadratic-equations-and-functions/exploring-quadratic-graphs-problem-5/
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math
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Like what you saw?
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Exploring Quadratic Graphs - Problem 5
In exploring parabolas, we find that the all have a lot of key features in common: they all are symmetric around a vertical line that has the equation x = -b/2a. It is called the axis of symmetry. If you plug that x value back into the function, you get the y-coordinate of the vertex. The vertex is always the maximum or minimum y-value on a parabola. The vertex is particularly important in quadratic word problems. The y intercept is always the point when x is zero. The x-intercept(s) are found when y is zero using a variety of possible methods. There could be zero, one, or two x-intercepts.
Transcript Coming Soon!
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s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802775669.60/warc/CC-MAIN-20141217075255-00160-ip-10-231-17-201.ec2.internal.warc.gz
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http://www.hagshama-prisha.com/the-hidden-truth-on-what-is-a-term-in-math/
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math
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The Taylor expansion is among the absolute most gorgeous ideas in mathematics. Watch this YouTube video and see whether you believe you would delight in showing this to your class. They can learn a www.papernow.org skill through a variety of ways, and those ways can be dependent on the teacher, school etc..
Some authors incorporate the empty set within this definition. If you cherished this guide and you want to acquire a lot more details concerning math problem solver uninstall kindly take a look at the web website. Rather, it's a result of the range of classes which were selected.
The What Is a Term in Math Chronicles
For a pure gas there are plenty of references that provide CP and CV values at various issues. Countries with a high Gini Coefficient are more inclined to turn into unstable, since there's a huge mass of poor folks that are jealous of the few of rich individuals. Examples Here's another example in which you will need to find the constant.
What Is a Term in Math – Is it a Scam?
You face a primary problem, you merely look at a patient and you may diagnose the illness https://ascc.wsu.edu/career-services/resumes-and-cover-letters/ they suffer from, however your child doesn't have the exact knowledge in the specialty and he fails to know your patients as well as you do. Instead, it is a byproduct of relationship with the Lover of someone's soul. Redistribute to inspect the work.
Both of these parts put together form the principle of mathematical induction. Instead, it is a byproduct of relationship with the Lover of someone's soul. Write the answer for a power.
The What Is a Term in Math Trap
The Taylor expansion is among the absolute most gorgeous ideas in mathematics. It is possible to stop by each hyperlink to play them with your boy or girl. It's extremely tough to locate educational games for kids middle school and higher school age.
Numbers might appear abstract, so utilize anything visual that you're in a position to as a way to locate the point across. The maps themselves may call for a bit of tweaking at a subsequent date. Sometimes getting the right data could be the hardest aspect of a project, particularly if you are attempting to do something new.
There are some different websites which I have found to be somewhat useful and somewhat cute!! Their use stays the exact same. Once you have completed your survey, you will count up the amount of people who chose each choice.
How to Find What Is a Term in Math Online
Observing the steps leads to the next. If you wish to memorize it and impress your pals, you can take advantage of this memory aid or some other memory aid that you know of. Every time you leave feedback, TPT offers you credit you will use to reduce the price of your future purchases.
Statistics might be something which a good deal of students dread (and even those which are now working). It's possible to use any letter you opt for. The answer, in reality, lies somewhere in the middle.
The context affects a number of the above definitions and terminology. A term is a mathematical expression which might form a separable portion of an equation, a series, or a different expression. research paper In many instances, introducing heuristics may increase the random algorithm.
Details of What Is a Term in Math
In some cases, you are going to be given one factor of a massive expression and you're going to be need to discover the remainder of the ones. So, now you may observe how a notion is translated in specific contexts. Therefore, there are many terms in the expression.
The Secret to What Is a Term in Math
And, due to the problem we just did before, needless to say, we are aware that number will be two. It isn't always a great idea to use patterns because in cases like exponents they might not be complete patterns for every single scenario, but the proof still holds. Sometimes getting the correct data might be the hardest aspect of a project, especially if you are trying to do something new.
These substances are called indicators. You might have read that you can place a formula in a column, somewhat like a spreadsheet. Create it like a user-defined function and you might then utilize it in your formulas.
Again, due to the Order of Operations that is presented in a subsequent lesson, the exponent has to be simplified before you do anything with the negative sign. It is helpful to have a chart for those roots which is precisely why we include the charts in the worksheets above. It is simply the selection of input variables.
Excel has another function called LOG that's a different type of logarithm. When it has to do with relevance, the sequence can be both an incrementing price and a few of the value can be set aside to declare extra qualities in what's being defined. In addition, a variable may be constant in the event the issue specifically tells you what the variable equals.
While structural equality can be checked with no understanding of the significance of the symbols, semantic equality may not. So, now you may observe how a notion is translated in specific contexts. Therefore, there are many terms in the expression.
Flipping this lesson could work nicely. Let's look at a great example. Let's look at an outstanding example.
It's possible to still do the mathematical operations to locate their average. The top and bottom bases just need to be the exact same. The conventional base is 10, and that's the base that's set in calculators.
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CC-MAIN-2021-31
| 5,465 | 25 |
http://pfeiferhouse.com/category/wine-list/page/2
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math
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Karl Joh. Molitor 2012, Rheingan, Hattenheim …$35
George Albrect Schneider, Rottiegenden, Neirsteiner …$42
Dr. Heyden 2012, Rheinhessen, Sacktrager …$40
Paul Anheuser 2011, Nahe, Konigsfels …$36
Dr. Thanisch (375ml) 2011, Mosel, Berncasteler …$45
Bollig-Lehnert 2011, Mosel, Piersporter …$42
Wines are arranged driest to sweetest, although none are sweet enough to be desert wines. They should simply be paired with denser foods as you go down the list.
Blue River, Dr Willkomms, Riesling …$22
Urban (Qualitatswein) …$24
Guntrum, Kabinett (from the Scheurebe grape) …$24
Dr. Hermann (Qualitatswein) …$30
Kabinett and Spatlese are two of the five grades designating the sugar content of the best German wines or Qualitatswein mi Pradikatt (Quality wine with distinction).
Spatlese (late-picking) wine is slightly sweeter than Kabinett and perhaps more flavorful.
Korbel Brut, Sonoma “Brut” 187ml …$12
Korbel Brut, Russain River Valley …$28
Freixenet, Cordon Negro, Spain …$32
Villa Sandi, Prosecco, Italy …$35
Domain Chandon, Blanc De Noires, Napa Valley …$42
rouge (red) :
Châteauneuf-du-Pape Domaine du Vieux Lazaret …$68
Literally “new Chateau of the Pope,” which was the manor/fortress wine-lover Pope Clement V built near these vineyards during the Holy See’s “Babylonian Captivity” in Avignon in the 14th c. (It was “new” because Pope Clement had first owned a vineyard-estate, now called Château Pape-Clément, when he was archbishop of Bordeaux.) A bold blend of several grape varieties of the Côtes du Rhône, principally Grenache.
blanc (white) :
Pouilly-Fuissé Joseph Drouhin …$68
the famous chardonnay of the Mâconnais region of Burgundy named after the twin towns near the vineyards
Roederer Reims, Brut Premier …$98
Dom Perignon 1995, Moet & Chandon …$230
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https://www.answers.com/Q/Difference_between_pressure_and_force
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math
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Difference between pressure and force?
Pressure is scalar but force is a vector.
Pressure is defined as the force per unit area.
As we multiply area vector by the pressure which is scalar, we get the force vector. This is similar to get momentum vector by multiplying velocity vector by the mass which is scalar.
Unit of force is newton but unit for pressure is N m-2
Yes, there are many ways to measure pressure. * pascal (Pa) * bar * technical atmosphere (at) * atmosphere (atm) * torr (Torr) * pound-force per square inch (psi) * psia (pounds-force per square inch absolute) --- gauge pressure plus local atmospheric pressure. * psid (psi difference) --- difference between two pressures. * psig (pounds-force per square inch gauge). * psivg (psi vented gauge) --- difference between the measuring point and the local pressure. *…
a vacuum is simply an empty or almost empty space. The forces that are apparent with vacuums are due to a pressure imbalance between the inside of the vacuum (low pressure) and the outside of the vacuum (high pressure) causing particles to want to migrate to areas of lower pressure. The effect of this pressure difference is often refered to as a vacuum force
Two numbers are used to represent blood pressure. Systolic pressure (the first and larger number of the equation) is the force that the blood flows from the heart into the arteries. Diastolic pressure (the second and smaller number of the equation) is the force as the heart relaxes, allowing the blood to flow back into the heart.
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https://studysoup.com/note/2290555/clemson-ch-3000-week-4-summer-2016
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math
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ch 3000 mass transfer
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3. Molecular mass transport 3.1 Introduction to mass transfer 3.2 Properties of mixtures 3.2.1 Concentration of species 3.2.2 Mass Averaged velocity 3.3 Diffusion flux 3.3.1 Pick’s Law 3.3.2 Relation among molar fluxes 3.4 Diffusivity 3.4.1 Diffusivity in gases 3.4.2 Diffusivity in liquids 3.4.3 Diffusivity in solids 3.5 Steady state diffusion 3.5.1 Diffusion through a stagnant gas film 3.5.2 Pseudo – steady – state diffusion through a stagnant gas film. 3.5.3 Equimolar counter diffusion. 3.5.4 Diffusion into an infinite stagnant medium. 3.5.5 Diffusion in liquids 3.5.6 Mass diffusion with homogeneous chemical reaction. 3.5.7 Diffusion in solids 3.6 Transient Diffusion. 3.1 Introduction of Mass Transfer When a system contains two or more components whose concentrations vary from point to point, there is a natural tendency for mass to be transferred, minimizing the concentration differences within a system. The transport of one constituent from a region of higher concentration to that of a lower concentration is called mass transfer. The transfer of mass within a fluid mixture or across a phase boundary is a process that plays a major role in many industrial processes. Examples of such processes are: (i) Dispersion of gases from stacks (ii) Removal of pollutants from plant discharge streams by absorption (iii) Stripping of gases from waste water (iv) Neutron diffusion within nuclear reactors 1 (v) Air conditioning Many of air daybyday experiences also involve mass transfer, for example: (i) A lump of sugar added to a cup of coffee eventually dissolves and then eventually diffuses to make the concentration uniform. (ii) Water evaporates from ponds to increase the humidity of passingair stream (iii) Perfumes presents a pleasant fragrance which is imparted throughout the surrounding atmosphere. The mechanism of mass transfer involves both molecular diffusion and convection. 3.2 Properties of Mixtures Mass transfer always involves mixtures. Consequently, we must account for the variation of physical properties which normally exist in a given system. When a system contains three or more components, as many industrial fluid streams do, the problem becomes unwidely very quickly. The conventional engineering approach to problems of multicomponent system is to attempt to reduce them to representative binary (i.e., two component) systems. In order to understand the future discussions, let us first consider definitions and relations which are often used to explain the role of components within a mixture. 3.2.1 Concentration of Species: Concentration of species in multicomponent mixture can be expressed in many ways. For species A, mass concentration denoted by isAdefined as the mass of A,m A per unit volume of the mixture. m A (1) A V The total mass concentration density is the sum of the total mass of the mixture in unit volume: i i where iis the concentration of species i in the mixture. 2 Molar concentration of, A,AC is defined as the number of moles of A present per unit volume of the mixture. By definition, mass of A es molecular weight of A (2) A M A Therefore from (1) & (2) C n A A A V M A For ideal gas mixtures, pAV A R T [ from Ideal gas law PV = nRT] n A pA A V RT where pA is the partial pressure of species A in the mixture. V is the volume of gas, T is the absolute temperature, and R is the universal gas constant. The total molar concentration or molar density of the mixture is given by C C i 3.2.2 Velocities In a multicomponent system the various species will normally move at different velocities; and evaluation of velocity of mixture requires the averaging of the velocities of each species present. If I is the velocity of species i with respect to stationary fixed coordinates, then massaverage velocity for a multicomponent mixture defined in terms of mass concentration is, 3 i i i i i i i i By similar way, molaraverage velocity of the mixture * is C i i * C For most engineering problems, there will be title difference in * and and so the mass average velocity, , will be used in all further discussions. The velocity of a particular species relative to the massaverage or molar average velocity is termed as diffusion velocity (i.e.) Diffusion velocity i The mole fraction for liquid and solid mixture,Ax ,and for gaseous mixtures,A y , are the molar concentration of species A divided by the molar density of the mixtures. C A x A (liquids and solids) C C y A A (gases). C The sum of the mole fractions, by definition must equal 1; x i 1 (i.e.) i i by similar way, mass fraction of A in mixture is; A 4 5 10. The molar composition of a gas mixture at 273 K and 1.5 * 10 Pa is: O 7% 2 CO 10% CO 215% N 2 68% Determine a) the composition in weight percent b) average molecular weight of the gas mixture c) density of gas mixture d) partial pressure of 2. Calculations: Let the gas mixture constitutes 1 mole. Then O 2 = 0.07 mol CO = 0.10 mol CO 2= 0.15 mol N 2 = 0.68 mol Molecular weight of the constituents are: O 2 = 2 * 16 = 32 g/mol CO = 12 + 16 = 28 g/mol CO 2= 12 + 2 * 16 = 44 g/mol N 2 = 2 * 14 = 28 g/mol Weight of the constituents are: (1 mol of gas mixture) O 2 = 0.07 * 32 = 2.24 g CO = 0.10 * 28 = 2.80 g CO 2= 0.15 * 44 = 6.60 g N 2 = 0.68 * 28 = 19.04 g Total weight of gas mixture = 2.24 + 2.80 + 6.60 + 19.04 = 30.68 g Composition in weight percent: 2.24 O 2 *100 7.30% 30.68 5 2.80 CO 30.68*100 9.13% CO 6.60*100 21.51% 2 30.68 19.04 N 2 *100 62.06% 30.68 Weight of gas mixture Average molecular weight of the gas mixture Number of moles M 30.68 30.68 g mol 1 Assuming that the gas obeys ideal gas law, PV = nRT n P V RT n molardensity m V Therefore, density (or mass density) m M Where M is the molecular weight of the gas. PM 1.5 * 105 * 30.68 3 Density m M kg m RT 8314 * 273 = 2.03 kg/m Partial pressure of 2 = [mole fractio 2of O ] * total pressure 7 5 * 1.5 * 10 100 5 = 0.07 * 1.5 * 10 = 0.105 * 10 Pa 3.3 Diffusion flux Just as momentum and energy (heat) transfer have two mechanisms for transportmolecular and convective, so does mass transfer. However, there are convective fluxes in mass transfer, even on a molecular level. The reason for this is that in mass transfer, whenever there is a driving force, there is always a net 6 movement of the mass of a particular species which results in a bulk motion of molecules. Of course, there can also be convective mass transport due to macroscopic fluid motion. In this chapter the focus is on molecular mass transfer. The mass (or molar) flux of a given species is a vector quantity denoting the amount of the particular species, in either mass or molar units, that passes per given increment of time through a unit area normal to the vector. The flux of species defined with reference to fixed spatial coordinates,AN is A A A (1) This could be written interms of diffusion velocity of A, (i.A., ) and average velocity of mixture, , as N C ( ) C A A A A (2) By definition C i i * i C Therefore, equation (2) becomes C A A A (A ) C i i C i y A C i i i For systems containing two components A and B, A A (A ) y A(C A A C B B) A A ) y A (N A N B ) N C ( ) y N A A A A (3) The first term on the right hand side of this equation is diffusional molar flux of A, and the second term is flux due to bulk motion. 3.3.1 Fick’s law: An empirical relation for the diffusional molar flux, first postulated by Fick and, accordingly, often referred to as Fick’s first law, defines the diffusion of 7 component A in an isothermal, isobaric system. For diffusion in only the Z direction, the Fick’s rate equation is d C A A AB d Z where D ABis diffusivity or diffusion coefficient for component A diffusing through component B, and dC /AdZ is the concentration gradient in the Zdirection. A more general flux relation which is not restricted to isothermal, isobasic system could be written as d y A A D AB (4) d Z using this expression, Equation (3) could be written as d y A AB A y AN (5) d Z 3.3.2 Relation among molar fluxes: For a binary system containing A and B, from Equation (5), A A y AN or A A y A N (6) Similarly, B B y B N (7) Addition of Equation (6) & (7) gives, J J N N (y y )N A B A B A B (8) By definition N = NA + NB and A + B = 1. Therefore equation (8) becomes, A B J + J = 0 A B J = J 8 d y d y AB A CD BA B (9) d z d Z From y A+ y B= 1 dy = dy A B Therefore Equation (9) becomes, AB D BA (10) This leads to the conclusion that diffusivity of A in B is equal to diffusivity of B in A. 3.4 Diffusivity Fick’s law proportionality, D , AB known as mass diffusivity (simply as diffusivity) 2 or as the diffusion coefficient. D hAB the dimension of L / t, identical to the fundamental dimensions of the other transport properties: Kinematic viscosity, = ( / ) in momentum transfer, and thermal diffusivity, (= k / C ) in he transfer. Diffusivity is normally reported in cm / sec; the SI unit being m / sec. Diffusivity depends on pressure, temperature, and composition of the system. In table, some values of D areABiven for a few gas, liquid, and solid systems. Diffusivities of gases at low density are almost composition independent, incease with the temperature and vary inversely with pressure. Liquid and solid diffusivities are strongly concentration dependent and increase with temperature. General range of values of diffusivity: –6 5 2 Gases : 5 X 10 1 X 10 m / sec. Liquids : 10 10 m / sec. Solids : 5 X 10 1 X 10 m / sec. In the absence of experimental data, semitheoretical expressions have been developed which give approximation, sometimes as valid as experimental values, due to the difficulties encountered in experimental measurements. 9 3.4.1 Diffusivity in Gases: Pressure dependence of diffusivity is given by 1 AB (for moderate ranges of pressures, upto 25 atm). p And temperature dependency is according to 3 D T 2 AB Diffusivity of a component in a mixture of components can be calculated using the diffusivities for the various binary pairs involved in the mixture. The relation given by Wilke is 1 D 1mixture y2 y3 ........... n D D D 12 13 1n Where D is the diffusivity for component 1 in the gas mixture; D is the 1mixture 1n diffusivity for the binary pair, component 1 diffusing through component n; and y is the mole fraction of component n in the gas mixture evaluated on a n component –1 – free basis, that is y2 2 y2 y 3.......y n 9. Determine the diffusivity of Co 2(1), O 22) and N 2) in a gas mixture having the composition: Co 2: 28.5 %, O 2: 15%, N 256.5%, The gas mixture is at 273 k and 1.2 * 10 Pa. The binary diffusivity values are given as: (at 273 K) D P = 1.874 m Pa/sec 12 2 D 13= 1.945 m Pa/sec D 23= 1.834 m Pa/sec Calculations: Diffusivity of C 2 in mixture 10 1 D 1m y 2 y 3 D 12 D 13 y y2 0.15 0.21 where 2 y y 0.15 0.565 2 3 y3 0.565 y3 0.79 y 2 y 3 0.150.565 1 D 1m P Therefore 0.21 0.79 1.874 1.945 = 1.93 m .Pa/sec Since P = 1.2 * 10 Pa, 1.93 5 2 D1m 5 1.61*10 m sec 1.2*10 Diffusivity of 2 in the mixture, 1 D2m y1 y3 D D 21 23 y 1 0.285 Where y1 0.335 y1 y 3 0.285 0.565 (mole fraction on2 free bans). and y y3 3 0.565 0.665 y 1 y 3 0.285 0.565 and D 21 = D 12= 1.874 m .Pa/sec Therefore 1 D 2mP 0.335 0.665 1.874 1.834 2 = 1.847 m .Pa/sec 11 D m 1.847 1.539*10 5 m sec 2 1.2*10 5 By Similar calculations Diffusivity of N i 2the mixture can be calculated, and is found to be, D 3m.588 * 10 m /sec. 2 3.4.2 Diffusivity in liquids: Diffusivity in liquid are exemplified by the values given in table … Most of these values are nearer to 10 cm / sec, and about ten thousand times shower than those in dilute gases. This characteristic of liquid diffusion often limits the overall rate of processes accruing in liquids (such as reaction between two components in liquids). In chemistry, diffusivity limits the rate of acidbase reactions; in the chemical industry, diffusion is responsible for the rates of liquidliquid extraction. Diffusion in liquids is important because it is slow. Certain molecules diffuse as molecules, while others which are designated as electrolytes ionize in solutions and diffuse as ions. For example, sodium chloride (NaCl), diffuses in water as ions Na and Cl. Though each ions has a different mobility, the electrical neutrality of the solution indicates the ions must diffuse at the same rate; accordingly it is possible to speak of a diffusion coefficient for molecular electrolytes such as NaCl. However, if several ions are present, the diffusion rates of the individual cations and anions must be considered, and molecular diffusion coefficients have no meaning. Diffusivity varies inversely with viscosity when the ratio of solute to solvent ratio exceeds five. In extremely high viscosity materials, diffusion becomes independent of viscosity. 3.4.3 Diffusivity in solids: Typical values for diffusivity in solids are shown in table. One outstanding characteristic of these values is their small size, usually thousands of time less than those in a liquid, which are inturn 10,000 times less than those in a gas. Diffusion plays a major role in catalysis and is important to the chemical engineer. For metallurgists, diffusion of atoms within the solids is of more importance. 3.5 Steady State Diffusion 12 In this section, steadystate molecular mass transfer through simple systems in which the concentration and molar flux are functions of a single space coordinate will be considered. In a binary system, containing A and B, this molar flux in the direction of z, as given by Eqn (5) is [section 3.3.1] d y A A CD AB d z y A (N A N )B (1) 3.5.1 Diffusion through a stagnant gas film The diffusivity or diffusion coefficient for a gas can be measured, experimentally using Arnold diffusion cell. This cell is illustrated schematically in figure. figure The narrow tube of uniform cross section which is partially filled with pure liquid A, is maintained at a constant temperature and pressure. Gas B which flows across the open end of the tub, has a negligible solubility in liquid A, and is also chemically inert to A. (i.e. no reaction between A & B). Component A vaporizes and diffuses into the gas phase; the rate of vaporization may be physically measured and may also be mathematically expressed interms of the molar flux. Consider the control volume S z, where S is the cross sectional area of the tube. Mass balance on A over this control volume for a steadystate operation yields [Moles of A leaving at z + z] – [Moles of A entering at z] = 0. (i.e.) A z z S N A z 0. (1) Dividing through by the volume, SZ, and evaluating in the limit as Z approaches zero, we obtain the differential equation 0 (2) d z This relation stipulates a constant molar flux of A throughout the gas phase from Z1 to Z2. A similar differential equation could also be written for component B as, 13 d N d Z and accordingly, the molar flux of B is also constant over the entire diffusion path from z 1and z 2. Considering only at plane z ,1and since the gas B is insoluble is liquid A, we realize that N , the net flux of B, is zero throughout the diffusion path; accordingly B B is a stagnant gas. From equation (1) (of section 3.5) d y A y (N N ) A AB d z A A B Since N B= 0, d y A A AB d z y AN A Rearranging, CD d y A AB A (3) 1y A d z This equation may be integrated between the two boundary conditions: at z = z Y = Y 1 A A1 And at2z = z A A2 Y = y Assuming the diffusivity is to be independent of concentration, and realizing that N Ais constant along the diffusion path, by integrating equation (3) we obtain Z2 yA2 d y A CD AB A Z y 1y A 1 A1 CD AB 1 y A2 A ln (4) Z 2 Z 1 1y A1 The log mean average concentration of component B is defined as 14 y y y B2 B1 y B2 ln y B1 Since y B 1 y A , (1y A2 ) (1y A1) y A1 y A2 y B,lm y A2 y A2 (5) ln y A1 ln y A1 Substituting from Eqn (5) in Eqn (4), A CD AB (y A1y A2 ) (6) Z 2 z 1 y B,lm For an ideal gas C n p , and V R T for mixture of ideal gasesy p A A P Therefore, for an ideal gas mixture equation. (6) becomes N D AB (p A1 p A2) A RT(z z ) p 2 1 B,lm This is the equation of molar flux for steady state diffusion of one gas through a second stagnant gas. Many masstransfer operations involve the diffusion of one gas component through another nondiffusing component; absorption and humidification are typical operations defined by these equation. The concentration profile (y Avs. z) for this type of diffusion is shown in figure: Figure 15 12. Oxygen is diffusing in a mixture of oxygennitrogen at 1 std atm, 25C. Concentration of oxygen at planes 2 mm apart are 10 and 20 volume % respectively. Nitrogen is nondiffusing. (a) Derive the appropriate expression to calculate the flux oxygen. Define units of each term clearly. (b) Calculate the flux of oxygen. Diffusivity of oxygen in nitrogen = 1.89 * 10 2 m /sec. Solution: Let us denote oxygen as A and nitrogen as B. Flux of A (i.e.) N Ais made up of two components, namely that resulting from the bulk motion of A (i.e.), Nx Aand that resulting from molecular diffusion A : N A NxA J A (1) From Fick’s law of diffusion, d C A JA D AB (2) d z Substituting this equation (1) d C N A Nx A D AB A (3) d z Since N = N A N Bnd x A C AC equation (3) becomes C A dC A N A N A NB C DAB d z Rearranging the terms and integrating between the planes between 1 and 2, d z C dC A C A2 (4) cD AB A1 N A C A N A N B Since B is non diffusing B = 0. Also, the total concentration C remains constant. Therefore, equation (4) becomes 16 z C A2 dC A CD AB CA1 N CA N C A A 1 C C A2 N ln C C A A1 Therefore, CD AB C C A2 N A ln (5) z C C A1 Replacing concentration in terms of pressures using Ideal gas law, equation (5) becomes D AB P t P t P A2 N A RTz lnP P (6) t A1 where D ABmolecular diffusivity of A in B P T total pressure of system R = universal gas constant T = temperature of system in absolute scale z = distance between two planes across the direction of diffusion P A1partial pressure of A at plane 1, and P A2partial pressure of A at plane 2 Given: D AB1.89 * 10 m /sec P = 1 atm = 1.01325 * 10 N/m 2 t T = 25C = 273 + 25 = 298 K z = 2 mm = 0.002 m P = 0.2 * 1 = 0.2 atm (From Ideal gas law and additive pressure rule) A1 P A20.1 * 1 = 0.1 atm Substituting these in equation (6) 5 5 N 1.89*10 1.01325*10 ln 1 0.1 A 8314 298 0.002 1 0.2 –5 2 = 4.55 * 10 kmol/m .sec 17 3.5.2 Psuedo steady state diffusion through a stagnant film: In many mass transfer operations, one of the boundaries may move with time. If the length of the diffusion path changes a small amount over a long period of time, a pseudo steady state diffusion model may be used. When this condition exists, the equation of steady state diffusion through stagnant gas’ can be used to find the flux. figure If the difference in the level of liquid A over the time interval considered is only a small fraction of the total diffusion path, an0 t – t is relatively long period of time, at any given instant in that period, the molar flux in the gas phase may be evaluated by C D AB (y A1 y A2 ) A zy (1) B,lm where z equals z 2– z 1 the length of the diffusion path at time t. The molar flux N A is related to the amount of A leaving the liquid by A,L d z A (2) M A d t A,L where is the molar density of A in the liquid phase M A under Psuedo steady state conditions, equations (1) & (2) can be equated to give d z C D (y y ) A,L AB A1 A2 (3) M A d t z y B,lm Equation. (3) may be integrated from t = 0 to t and from z = z t0to z = t as: t A,Ly B,lm M A Z t zdz t0 C D AB (y A1 y A2) Zt0 yielding 18 A,L yB,lm M A z t z t0 (4) CD AB (y A1 y A2) 2 This shall be rearranged to evaluate diffusivity D as, AB 2 2 A,L y B,lm z t z t0 AB M C (y y )t 2 A A1 A2 1. A vertical glass tube 3 mm in diameter is filled with liquid toluene to a depth of 20mm from the top openend. After 275 hrs at 39.4 C and a total pressure of 760 mm Hg the level has dropped to 80 mm from the top. Calculate the value of diffusivity. Data: vapor pressure of toluene at 39.4C = 7.64 kN / m , 2 density of liquid toluene = 850 kg/m 3 Molecular weight of toluene = 92 (C 6 6H ) 3 A,L y Blm Z Z 2 D AB t t0 figure M CAy A1 y A2 t 2 y B2 y B1 y B,l m where yB2 ln y B1 y B2 1 – y A2 y B1 1 – y A1 p A1 7.64 2 y A1 (760 mm Hg = 101.3 kN/m ) P 101.3 = 0.0754 y B1 1 – 0.0754 = 0.9246 y A2 0 y B= 1 – y A2 y 1 0.9246 0.9618 Therefore B,lm 1 ln 0.9246 5 P 1.01325*10 C RT 8314* 273 39.4 3 = 0.039 k mol /m Therefore 19 850 * 0.9618 0.08 2 0.022 D AB 92 * 0.039 * 0.0754 0 * 275 * 3600 2 –3 2 2 = 1.5262 * 10 (0.08 – 0.02 ) = 9.1572 * 10 m /sec. 3.5.3 Equimolar counter diffusion: A physical situation which is encountered in the distillation of two constituents whose molar latent heats of vaporization are essentially equal, stipulates that the flux of one gaseous component is equal to but acting in the opposite direction from the other gaseous component; that is, N = A N . B The molar flux N ,Afor a binary system at constant temperature and pressure is described by d y A A CD AB y A (N A N )B d z dC A or A D AB y A (N A N )B (1) d z with the substitution of NB = NA, Equation (1) becomes, dC A A D AB (2) d z For steady state diffusion Equation. (2) may be integrated, using the boundary conditions: at1 z = zA A1 C = C 2 AC =A2C Giving, Z C 2 A2 A z D AB dC A Z 1 C A1 from which D AB A (C A1 C A2) (3) z2z 1 20 For ideal gases, C n A pA . Therefore Equation. (3) becomes A V RT D AB A (P A1 P A2 ) (4) RT (z 2z ) 1 This is the equation of molar flux for steadystate equimolar counter diffusion. Concentration profile in these equimolar counter diffusion may be obtained from, d d z A ) 0 (Since NA is constant over the diffusion path). And from equation. (2) N D dC A A AB d z . Therefore dC A 0 . d z AB d z d 2 C or 0. d z 2 This equation may be solved using the boundary conditions to give C A C A1 zz 1 C C z z (5) A1 A2 1 2 Equation, (5) indicates a linear concentration profile for equimolar counter diffusion. 3. Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is p A155 kPa and at point 2, 0.03 m apart P = A2 15 KPa. The total pressure is 101.32 kPa, and the temperature is 298 K. At this –5 2 pressure and temperature, the value of diffusivity is 6.75 * 10 m /sec. i) calculate the flux of CH at steady state for equimolar counter 4 diffusion. ii) Calculate the partial pressure at a point 0.02 m apart from point 1. 21 Calculation: For steady state equimolar counter diffusion, molar flux is given by D AB N A p A1 pA 2 (1) RT z Therefore; 5 N 6.75*10 55 15 kmol A 8.314 * 298 * 0.03 m .sec 3.633 * 105 kmol m sec And from (1), partial pressure at 0.02 m from point 1 is: 5 3.633 * 105 6.75 * 10 55 p 8.314 * 298 * 0.02 A p A= 28.33 kPa 11. In a gas mixture of hydrogen and oxygen, steady state equimolar counter diffusion is occurring at a total pressure of 100 kPa and temperature of 20C. If the partial pressures of oxygen at two planes 0.01 m apart, and perpendicular to the direction of diffusion are 15 kPa and 5 kPa, respectively and the mass –5 2 diffusion flux of oxygen in the mixture is 1.6 * 10 kmol/m .sec, calculate the molecular diffusivity for the system. Solution: For equimolar counter current diffusion: D AB N A p A1p A2 (1) RTz where –5 2 N A= molar flux of A (1.6 * 10 kmol/m .sec): D AB= molecular diffusivity of A in B R = Universal gas constant (8.314 kJ/kmol.k) T = Temperature in absolute scale (273 + 20 = 293 K) z = distance between two measurement planes 1 and 2 (0.01 m) 22 P A1 partial pressure of A at plane 1 (15 kPa); and P A2 partial pressure of A at plane 2 (5 kPa) Substituting these in equation (1) D 1.6 * 105 AB 155 8.314 293 0.01 –5 2 Therefore, D AB 3.898 * 10 m /sec 2. A tube 1 cm in inside diameter that is 20 cm long is filled with Co2 and H2 at a total pressure of 2 atm at 0C. The diffusion coefficient of the Co2 – H 2system 2 under these conditions is 0.275 cm /sec. If the partial pressure of Co2 is 1.5 atm at one end of the tube and 0.5 atm at the other end, find the rate of diffusion for: i) steady state equimolar counter diffusion (N A= N B ii) steady state counter diffusion where N B 0.75 N Aand iii) steady state diffusion of C2 through stagnant H2 (NB = 0) d y A i) A CD AB d z y AN A NB Given N B N A d y A d C A Therefore N A C D AB D AB d z d z p A (For ideal gas mixture C A where p Ais the partial pressure of A; such that RT p A+ p B= P) d p AT Therefore N A D AB d z For isothermal system, T is constant D d p Therefore N A AB A RT d z Z 2 D AB P A2 (i.e.) N A d z d pA Z RT P 1 A1 23 D AB N A p A1 pA2 (1) RT z where Z = Z 2Z 1 2 –4 2 Given: D AB0.275 cm /sec = 0.275 * 10 m /sec ; T = 0C = 273 k 4 N 0.275 *10 1.5 * 1.01325 * 105 0.5 * 1.01325 * 105 A 8314 * 273 * 0.2 6.138 * 10 6 k mol m 2 sec Rate of diffusion = N S A Where S is surface area Therefore rate of diffusion = 6.138 * 10 * r –6 –2 2 = 6.138 * 10 * (0.5 * 10 ) = 4.821 * 10 –1 k mol/sec –3 = 1.735 * 10 mol/hr. ii) C D d y A y N N A AB d z A A B given: N B 0.75 N A d y A Therefore N A C D AB d z y A N A 0.75N A d y C D A B A 0.25 y A N A d z N 0.25 y N C D d yA A A A AB d z d y A N A d z C D AB 1 0.25 y A for constant N And C Z 2 y A2 d y N d z CD A A AB 1 0.25 y A Z 1 y A1 d x 1 ln a b x a b x b 24 N z CD 1 ln 1 0.25y A2 A AB 0.25 A y A1 4CD AB 1 0.25 y A 2 N A ln (2) z 1 0.25 y A1 Given: p 2 * 1.01325 * 10 5 C 0.0893 K mol m 3 RT 8314 * 273 p y A1 1.5 0.75 A1 P 2 p y A2 0.5 0.25 A2 P 2 Substituting these in equation (2), 4 * 0.0893 * 0.275 * 10 4 1 0.25 * 0.25 N A ln 0.2 1 0.25 * 0.75 6 kmol 7.028 * 10 2 m sec Rate of diffusion = N S = 7.028 * 10 * * (0.5 * 10 ) –2 2 A –10 = 5.52 * 10 kmol/sec = 1.987 * 10 mol/hr. d y A iii) A CD AB y A N A N B d z Given: N B 0 d y A Therefore N A CD AB y A N A d z Z 2 yA2 d y N d z CD A A Z AB y 1 y A 1 A1 CD AB 1 y A2 ln Z 1 y A1 0.0893* 0.275*10 4 1 0.25 ln 0.2 1 0.75 5 kmol 1.349 * 10 2 m .sec 25 Rate of diffusion = 1.349 8 10 * * (0.5 * 10 ) 2 = 1.059 Kmol / sec = 3.814 mol/hr 3.5.4 Diffusion into an infinite standard medium : Here we will discuss problems involving diffusion from a spherical particle into an infinite body of stagnant gas. The purpose in doing this is to demonstrate how to set up differential equations that describe the diffusion in these processes. The solutions, obtained are only of academic interest because a large body of gas in which there are no convection currents is unlikely to be found in practice. However, the solutions developed here for these problems actually represent a special case of the more common situation involving both molecular diffusion and convective mass transfer. a) Evaporation of a spherical Droplet: As an example of such problems, we shall consider the evaporation of spherical droplet such as a raindrop or sublimation of naphthalene ball. The vapor formed at the surface of the droplet is assumed to diffuse by molecular motions into the large body of stagnant gas that surrounds the droplet. Consider a raindrop, as shown in figure. At any moment, when the radius Figure Evaporation of a raindrop of the drop is r , the flux of water vapor at any distance r from the center is given 0 by d y A N A C D AB y AN A N B d r Here N = 0 (since air is assumed to be stagnant) B Therefore, d y A N A C D AB y A N A d r Rearranging, C D AB d y A N A __________ (1) 1 y A d r The flux N Ais not constant, because of the spherical geometry; decreases as the distance from the center of sphere increases. But the molar flow rate at r and r + r are the same. This could be written as, A N A r A N A r r __________ (2) where A = surface area of sphere at r or r + r. 26 2 Substituting for A = 4 r in equation (2), 4 r 2 N A 4 r 2 N A 0 r r r or 2 2 r N A r r r N A r lim 0 r 0 r d 2 ss r N A 0 __________ (3) dr Integrating, 2 r N A constant __________ (4) 2 2 From equation (4), r N A r 0 N A 0 Substituting for N f Am equation (1), r 2C D d y AB A r 2 N A 1 y A d r 0 0 2 d r d y A r 0 N A 0 2 C D AB __________ (5) r 1 y A Boundary condition : At r = r 0 y A y AS And At r = y A y A Therefore equation (5) becomes, 2 1 y A r 0 N A 0 C D AB ln 1 y A y r r0 AS Simplifying, C D AB 1 y A N A ln __________ (6) 0 r 0 1 y A S Time required for complete evaporation of the droplet may be evaluated from making mass balance. Moles of water diffusing moles of water leaving the droplet unit time unit time 2 d 4 3 L 4 r 0 N A0 r 0 dt M A 2 L d r 0 4 r 0 __________ (7) M A d t Substituting for N from equation (6) in equation (7), A0 27 C D AB 1 y A L d r0 ln __________ (8) r0 1 y AS M A d t Initial condition : When t = 0 r 0= r 1 Integrating equation (8) with these initial condition, t L 1 1 0 d t r0 d r0 0 M A C D AB 1 y A r1 ln 1 y AS r 2 t L 1 1 M 2C D 1 y __________ (9) A AB ln A 1 y A S Equation (9) gives the total time t required for complete evaporation of spherical droplet of initial radius r 1 b) Combustion of a coal particle: The problem of combustion of spherical coal particle is similar to evaporation of a drop with the exception that chemical reaction (combustions) occurs at the surface of the particle. During combustion of coal, the reaction C + O 2 CO 2 cccurs. According to this reaction for every mole of oxygen that diffuses to the surface of coal (maximum of carbon), react with 1 mole of carbon, releases 1 mole of carbon dioxide, which must diffuse away from this surface. This is a case of equimolar counter diffusion of CO and 2. Norma 2y air (a mixture of N and 2 O 2s used for combustion, and in this case N does not 2es part in the reaction, and its flux is zero. i.e. N N 2 0 . The molar flux of O c 2ld be written as d y N CD O 2 y N N N O 2 O 2gas d r O 2 O 2 CO 2 N 2 __________ (1) Where D O 2gas is the diffusivity of 2 in the gas mixture. Since N N 2 0 , and from stoichiometry N O 2 N CO 2 , equation (1) becomes d yO N O C D O gas 2 __________ (2) 2 2 d r For steady state conditions, d r 2 N 0 __________ (3) d r O 2 Integrating, 28 r2 N O constant r02N O s __________ (4) 2 2 Where r is the radius of coal particle at any instant, and N O s is the flux of O 0 2 2 at the surface of the particle. Substituting for N O 2from equation (2) in equation (4), d yO 2 2 r 2CD O gas r N O s __________ (5) 2 d r 0 2 Boundary condition : r r0 yO y O s At 2 2 And At r yO 2 y O2 With these boundary condition, equation (5) becomes y 2 d r O 2 r0 NA 0 2 C D O 2gas d y O 2 r0 r yO s 2 which yields CD O2 gas N O 2 s yO 2 sy O 2 __________ (6) r 0 For fast reaction of O 2ith coal, the mole fraction of O 2 the surface of particle y O s 0 iz zero. (i.e.,) 2 . And also at some distance away from the surface of the particle y O 2 yO 2 0.21 (because air is a mixture of 21 mole % O an 279 mole % N 2) With these conditions, equation (6) becomes, 0.21 C D O 2 gas N O2 s ____________ (7) r0 Figure Combustion of a particle of coal 5. A sphere of naphthalene having a radius of 2mm is suspended in a large volume of shell air at 318 K and 1 atm. The surface pressure of the naphthalene can be assumed to be at 318 K is 0.555 mm Hg. The D ABof naphthalene in air –6 2 at 318 K is 6.92 * 10 /sec. Calculate the rate of evaporation of naphthalene from the surface. Calculation Steady state mass balance over a element of radius r and r + r leads to S N A r S N A r r 0 (1) 29 2 where S is the surface are (= 4 r ) dividing (1) by Sr, and taking the limit as r approaches zero, gives: d r N2 A 0 d r Integrating r N A= constant (or) 4 r N A= constant We can assume that there is a film of naphthalene – vapor / air film around naphthalene through which molecular diffusion occurs. Diffusion of naphthalene vapor across this film could be written as, N CD d y A y N N A AB d r A A B N B 0 (since air is assumed to be stagnant in the film) d y A N A CD AB y A N A d r y N CD d A A AB d r 1 y A dln 1 y A N A CD AB d r 2 W A Rate of evaporation = 4 r N A R = constant. 4 r 2CD d ln 1 y W AB A A d r W A d r 4 D AB C d ln 1 y A r 2 Boundary condition: 0.555 4 At r = R y A 7.303 * 10 760 –4 ln (1 – A) = 7.3 * 10 At r = y A= 0 ln (1 A) = 0 d r 0 Therefore W A 2 4 D AB C d ln 1 y A R r 7.3 *10 4 30 1 0 W A r 4 D AB C ln 1 y A 7.3 *104 R 1 4 W A 0 4 D AB C 0 7.3*10 R W A4 R D C * AB3 * 10 –4 5 C P 1.01325 * 10 Gas cons tant * T 8314 * 318 3 = 0.0383 kmol/m Initial rate of evaporation: Therefore W = A* 3.142 * 2 * 10 * 6.92 * 10 * 0.0383 * 7.3 * 10 –4 –12 = 4.863 * 10 –5 kmol/sec = 1.751 * 10 mol/hr. 3.5.5 Diffusion in Liquids: Equation derived for diffusion in gases equally applies to diffusion in liquids with some modifications. Mole fraction in liquid phases is normally written as ‘x’ (in gases as y). The concentration term ‘C’ is replaced by average molar density, . M av a) For steady – state diffusion of A through non diffusivity B: N A constant , N = B D AB N A xA1 x A2 z x BM M av where Z = Z – 2, th 1length of diffusion path; and X B2 X B1 X BM X B2 ln X B1 b) For steady – state equimolar counter diffusion : N A N = Bonst D AB D AB N A C A1 C A2 x A1 x A 2 Z Z M av 4. Calculate the rate of diffusion of butanol at 20C under unidirectional steady state conditions through
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s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721278.88/warc/CC-MAIN-20161020183841-00376-ip-10-171-6-4.ec2.internal.warc.gz
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CC-MAIN-2016-44
| 33,728 | 13 |
https://www.shaalaa.com/question-bank-solutions/when-income-consumer-falls-impact-price-demand-curve-inferior-good-is-choose-correct-alternative-demand_20445
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math
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When the income of the consumer falls the impact on a price-demand curve of an inferior good is: (choose the correct alternative)
a. Shifts to the right.
b. Shifts of the left.
c. There is upward movement along the curve.
d. There is downward movement along the curve
Shifts to the right.
Demand for inferior goods has an opposite relationship with consumer's income. If there is a decline in the income of the consumer, the demand for an inferior good rise. Therefore, the consumer demand curve shifts towards the right.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00223.warc.gz
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CC-MAIN-2022-21
| 521 | 7 |
https://examspace.net/what-is-numbers-and-digits.html
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math
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Before entering the world of mathematics you should know what is numbers. In the world of mathematics digits and numbers are toys for us to play with. Hence if you want to be a good student in mathematics you should Learn the concept and ideas behind these 2 very vital parts of mathematics.
-A digit is a symbol, with the help of which we can write any number of the proposed number system. digits are ten in number from 0 to 9 . friends 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is known as digit
suppose, we want to write the number 21. this can be written with the help of 2 digits 2 and 1.
Now moving toward the number…
In this post what is numbers I am sure that The number is not a new thing for you. I am sure you have encountered it at various places and activities in your day to life.
The number can be defined as the amount of something. for example, if we measure some length with some standard unit of length or we measure the weight of some sweet say laddu. we will have to express these measurements in terms of some value and that value can be written in terms of numbers with the help of digits.
Hence Number is the notations that we write with the help of digits to express the values of some measurements.
The number can be of 1 digit or can be in terms of a series of digits. for example, if someone says -” I am giving you 2 gold coins” here 2 is a single-digit number. at the same time, you either remember your mom’s phone number which is also a number that contains a series of digits. Postal codes are numbers with a series of digits. your social security number or your ID Number can be the number containing a series of digits.
Types of Numbers
Basically, Numbers can be divided into 2 major parts that are Real and Complex numbers. in lower standards, we do not study Complex numbers. hence we will focus on the real number.
Friends! so in this article what is numbers we will learn about Real Numbers.
Primarily there are 4 types of Numbers
- Natural Number
- Whole Number
- Rational Numbers &
- Irrational Numbers
#1. Natural Number–
These are common counting numbers that contain numbers from 1 to ∞. Real numbers are the part of real numbers which includes only positive integers but don’t include negative decimals, or zero.
#2 Whole Numbers –
The set of Natural Numbers including 0 is known as Whole Numbers. hence whole numbers are 0,1,2,3…………………∞
integers are the type of number which contains Positive and negative numbers. an integer can be positive, negative, or zero but can never be a fraction.
an integer is a Latin word that means whole or intact. The Integers are represented as ‘Z’ Zero is neither a positive nor a negative integer. 0(zero) is a neutral number i.e. zero has no sign (+ or -).
#4 Rational Numbers –
A rational number is a number that can be written in terms of fractions or in terms of p/q, where q≠0.
Natural Number from 1 to 100
The natural numbers from 1 to 100 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99 and 100.
Is 0 a Natural Number?
No, zero(0) is not a natural number. because as per the definition we know natural numbers are the counting numbers from 1 to infinity.
is 2 a rational number?
yes 2 is a rational number as from the definiton above it satisfies the conditon of p/q . where q ≠0 .
what are irrational numbers?
Irrational numbers are the set of all real numbers which are not rational numbers.
Hope this article titled what is numbers have been useful and added some value in your knowledge base .
please give your valuable suggestions for further improvement in the comment box provided below .
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s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00183.warc.gz
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CC-MAIN-2023-14
| 3,954 | 34 |
https://irodovsolutionsmechanics.blogspot.com/2009_02_04_archive.html
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math
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Wednesday, February 4, 2009
The velocity of the belt at different points is shown in the figure beside. At each point, the velocity of the belt is a sum of two motions i) its translational velocity v as the entire conveyer moves forward and ii) the motion of the belt relative to the conveyer which is v in a direction tangential to the belt at any given point on the belt. Let the mass per unit length of the belt be p.
In section CD, the belt moves from C to D with velocity relative to the conveyer while the conveyer itself moves in the direction D to C with a velocity v. So the net velocity of this section of belt is v - v = 0. In other words the lower part of the belt is stationary! This part of the belt has no kinetic energy.
In section AB, the belt moves from A to B with a velocity v relative to the conveyer while the conveyer itself moves with v in the direction from A to B. Thus, the net velocity of this part of the belt is v + v = 2v. If the kinetic energy of this part of the belt is given by,
In sections AD and BC, the direction of the velocity of the belt with respect to the conveyer is always v but in a direction tangential to the belt. Let the radius if the circular part be r.
In section AD, if we consider an infinitesimally small piece of the belt that subtends an angle , then its mass will be . The net velocity of this point will be,
The kinetic energy of this infinitesimally small part of belt is given by .
The net kinetic energy of this part of the belt can be found as,
Similarly, in section BC, for an infinitesimally small piece of the belt that subtends an angle , its velocity will be,
Thus, the kinetic energy of this piece is given by,
The net kinetic energy of this piece of the belt is obtained as,
Adding the kinetic energies of all the parts of the belt we have the total kinetic energy as,
Posted by Krishna Kant Chintalapudi at 8:07 PM
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s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00313.warc.gz
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CC-MAIN-2021-49
| 1,885 | 13 |
https://experts.umn.edu/en/publications/reconstructing-random-topography-from-preserved-stratification
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math
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One of the fundamental problems of interpreting the sedimentary record is reconstructing the original heights of palaeotopographical features such as bedforms or river channels. This requires an understanding of the relationship between topography and set thickness, but at present an exact theory exists only for periodic topography of uniform height. The applicability of this simple theory is severely limited by the random variability characteristic of many sedimentary systems. In this paper, we develop an exact theory for the probability‐density function (PDF) of sets generated from topography of random height. We focus on the limiting case of zero net deposition in order to provide a lower bound for the set thickness, and derive an analytical set‐thickness PDF that is determined by one parameter of the PDF for topographical height. This parameter, β, measures the breadth of the tail of the topographical PDF. The mean set thickness is 0.8225 β for bedforms and 1.645 β for river channels. If the topographical height is gamma distributed, the preservation ratio, defined as (mean thickness of preserved sets)/(mean topographical height), is 0.8225 r2 for bedforms and 1.645 r2 for river channels, where r is the coefficient of variation (standard deviation/mean) of the generating topography. In a comparison with data from laboratory current ripples, our analytical predictions compare well with observations of both mean set thickness and thickness distribution. The preservation ratio for the ripples is about 0.5, whilst measured dune heights give a preservation ratio of about 0.12. Depth data from two modern braided streams yield preservation ratios ranging from 0.4 to 0.75. As more data on the distributions of topographical height in modern environments become available, calculations such as these should help provide reliable error bounds for quantitative topographical reconstruction.
|Number of pages
|Published - Aug 1991
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00185.warc.gz
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CC-MAIN-2024-18
| 1,958 | 3 |
https://www.freelancer.com/projects/wordpress/customizing-woocommerce-multi-vendor/
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math
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There are many plug-ins able to make vendors subscribe to a paid plan to sell on a multi vendor marketplace, but you can make them pay a fixed price only.
I'm looking for a customization of an existing plug in which lets me define a dynamic price, which will depend on the sales a vendor did in the previous month.
Basic plan: 0$ + 5% of their sales during the previous month
Gold plan: 10$ + 5% of their sales during the previous month
If vendor selected basic plan, and he sells items for 100$, at the ene of the month he'll pay 5$ to renew his membership, while if he selected gold plan and sold items for 500$, the renewal cost will be 10$+25$= 35$.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00050.warc.gz
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CC-MAIN-2020-10
| 653 | 5 |
https://www.shaalaa.com/question-bank-solutions/the-atomic-numbers-elements-na-mg-k-ca-are-11-12-19-20-respectively-element-having-largest-atomic-radius-is-a-mg-b-na-c-k-d-ca-periodic-properties_29488
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math
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The atomic numbers of the elements Na, Mg, K and Ca are 11, 12, 19 and 20 respectively. The element having the largest atomic radius is:
Potassium (K) with an atomic number of 19 has the largest atomic radius. This is because sodium and potassium are elements of group 1. On moving from top to bottom in a group, the atomic radius increases. Magnesium and calcium are elements of group 2. The size of the atomic radius decreases on moving from left to right in a period of the periodic table .
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00723.warc.gz
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CC-MAIN-2022-21
| 493 | 2 |
https://community.splunk.com/t5/user/viewprofilepage/user-id/42057
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math
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Many questions deal with indexed volume per source and per day for licence concern. My need is logs volume per source and per day. To explain more, I want to know the volume of each source without taking into account catching of old logs which could not be indexed in the past. So not the indexed volume per day but the volume of logs time stamped at a given day.
... View more
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585405.74/warc/CC-MAIN-20211021102435-20211021132435-00508.warc.gz
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CC-MAIN-2021-43
| 377 | 2 |
https://www.math.nycu.edu.tw/event/e_lecture_content.php?N_ID=712
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math
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Colloquium / Seminars
Topic：Series representation of Evans function and applications on stability analysis for traveling waves of Lotka-Volterra competition system
Speaker：Prof. Yang, Tzi-Sheng
(Dept. of Applied Mathematics, Tunghai University)
Under suitable change of variables, we transform the linearized eigenvalue problem around a traveling wave into a ODE with two finite regular singular point. The fundamental solution satisfying the one-sided boundary condition to the linearized eigenvalue problems is then derived. Accordingly, the Evans function, whose zeros corresponds to the point spectrum of the traveling wave, is represented as an infinite series. Based on the series representations, the number of point spectrum lying on the right half of the complex plane, can be numerically identified by counting the winding numbers of the image of a suitable contour mapped by the Evans function. Several types of traveling waves in Lotka-Volterra competition system are taken as examples for illustration.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00053.warc.gz
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CC-MAIN-2022-49
| 1,019 | 5 |
https://goprep.co/q28-draw-an-isosceles-abc-in-which-bc-5-5-cm-and-altitude-al-i-1nk2w5
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math
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Steps of construction :
1. Draw a line BC = 5.5 cm
2. Draw right bisector of BC intersecting BC which intersects BC at L.
3. Taking L as center and with radius = 3 cm draw an arc which intersect the right bisector A
4. Join AB and AC to get ∆ABC
5. Below BC, make an acute angle ∠CBX
6. Along BX mark 4 ( greater of 3 and 4 in 3/4) points B1, B2, B3, B4 at equal distances.
7. Join B4C
8. From point B3, make a line B3C' parallel to B4C , meeting BC at C'.[Choosing B3 because we need a 3:4 ratio]
9. From point B' draw a line C'A' parallel to CA.
10. A'BC' is the triangle, whose sides are three - fourth of ∆ABC
Rate this question :
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00445.warc.gz
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CC-MAIN-2021-17
| 640 | 12 |
http://www.bsmath.hu/CURRENT/ALTsyllabus.html
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math
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Instructor: Dr. András STIPSICZ
Text: Handouts and
Allan Hatcher: Algebraic Topology
Additional reference: Fuchs - Fomenko - Gutenmacher: Homotopic topology; Husemoller: Fiber bundles
Basic algebra: vector spaces, groups, factor groups, homomorphisms.
Basic analysis in \R^n: continuous maps, convergence, differentiable maps.
The goal of the course is to provide an introduction to the basic notions of homology and cohomology theory, and show some simple (and some more sophisticated) applications of these techniques in topology. Ideas from homology are present in all modern directions of mathematics, and we will show some appearances of those as well. If time premits, at the end of the course we will see further applications of the concept of homology in knot theory.
- Simplicial and singular homology
- Basic homological algebra (chains and homotopies)
- Degree, CW-homology
- Cohomology, ring structure
- Orientability, Poincare duality
- Obstruction theory
- Fiber bundles, principal bundles
- Classification of vector bundles
- Characteristic classes
- Knots and knot invariants
- The Jones polynomial and Khovanov homology
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s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00134.warc.gz
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CC-MAIN-2020-10
| 1,137 | 18 |
https://link.springer.com/chapter/10.1007/978-3-319-25619-1_21
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math
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Disintegration of Quarkonia in QGP Due to Time Dependent Potential
Rapid thermalization in ultra-relativistic heavy-ion collisions leads to fast changing potential between a heavy quark and antiquark from zero temperature potential to the finite temperature one. Time dependent perturbation theory can then be used to calculate the survival probability of the initial quarkonium state. In view of very short time scales of thermalization at RHIC and LHC energies, we calculate the survival probability of \(J/\psi \) and \(\varUpsilon \) using sudden approximation. Our results show that quarkonium decay may be significant even when temperature of QGP remains low enough so that the conventional quarkonium melting due to Debye screening is ineffective.
We are very thankful to Abhishek Atreya, Arpan Das, Shreyansh S. Dave, Biswanath Layek, and Ananta P. Mishra for useful discussions.
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https://www.hindawi.com/journals/isrn/2013/645232/
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math
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Vibroacoustical Analysis of Multiple-Layered Structures with Viscoelastic Damping Cores
This paper presents a modeling technique to study the vibroacoustics of multiple-layered viscoelastic laminated beams using the Biot damping model. In this work, a complete simulation procedure for studying the structural acoustics of the system using a hybrid numerical model is presented. The boundary element method (BEM) was used to model the acoustical cavity, whereas the finite element method (FEM) was the basis for vibration analysis of the multiple-layered beam structure. Through the proposed procedure, the analysis can easily be extended to another complex geometry with arbitrary boundary conditions. The nonlinear behavior of viscoelastic damping materials was represented by the Biot damping model taking into account the effects of frequency, temperature, and different damping materials for individual layers. The curve-fitting procedure used to obtain the Biot constants for different damping materials for each temperature is explained. The results from structural vibration analysis for selected beams agree with published closed-form results, and results for the radiated noise for a sample beam structure obtained using a commercial BEM software are compared with the acoustical results of the same beam by using the Biot damping model.
The traditional designs of free-layer, constrained-layer or sandwich-layer, damping treatment using viscoelastic materials have been around for over forty years. Recent improvements in the understanding and application of the damping principles, together with advances in materials science and manufacturing, have led to many successful applications and the development of patch damping and multiple-layered damping structures. The key point in any design is to recognize that the damping material must be applied in such a way that it is significantly strained whenever the structure is deformed in the vibration mode under investigation. Numerous researchers have successfully implemented the passive constrained layer (PCL) and active constrained layer (ACL) systems. In 1959, Kerwin and Ross et al. presented a general analysis of viscoelastic material structure. The damping was attributed to the extension and shear deformations of the viscoelastic layers. Ditaranto developed sixth-order equations of motion in terms of axial displacements and developed a closed-form solution. Mead and Markus extended the sixth-order equations of motion for transverse displacement to include various boundary conditions. A paper by Rao presented the equations of motion of viscoelastic sandwich beams with various boundary conditions using the energy method. The equations were solved numerically, and a practical design guideline was presented. Similar to Rao’s theory, Cottle used Hamilton’s principle to derive equations of motion. The damping could also be increased by adding passive stand-off layer (PSOL) and slotted stand-off layer (SSOL) to the layered systems. Falugi and Parin et al. conducted theoretical and experimental work on a four-layered panel and a five-layered beam with PSOL treatment. Rogers and Parin and Yellin et al. have performed experimental investigations and demonstrated that PSOL treatment increased the damping significantly in aeronautical structures and beams. Yellin and colleagues [11, 12] also developed normalized equations of motion for beam, fully treated with PSOL using nonideal stand-off layer assumption. The equations were solved using the method of distributed transfer functions .
In addition to the closed-form analytical approach, many researchers have used the finite element method (FEM), the most popular numerical modeling method in building the numerical model of the multiple layers system. In 2000, Chen and Chan studied four different types of integral FEM models with the viscoelastic cores. The numerical stability and accuracy as well as for convergence issue of these four different FEM models were demonstrated by comparing the numerical results with those from experiments. Lesieutre and Lee proposed a 3-node, 10 DOF FEM model for the three-layer ACL damping beam. This FEM model is advantageous in active control application due to its features of nonshear locking and adaptability to segmented constraining layers.
Other than the Hamilton and FEM numerical methods in building the models, other researchers have proposed many irregular modeling techniques for the numerical representation of continuous/discontinuous systems. Kung and Singh calculated the natural frequencies and loss factor using the Rayleigh-Ritz energy method and modal-strain energy technique in modeling a 3-layer patch damping structure. These approximate modeling methods were also extended to rectangular damping patch of plates and shells with viscoelastic cores. Zhang and Sainsbury combined the Gerlerkin orthogonal function with the traditional finite element method and successfully applied to the vibration analysis of the damped sandwich plates.
While the FEM is used widely in the modeling of the structure, many researchers sought for proper mathematical models to represent the damping behavior of the viscoelastic material, as well as incorporating the damping model in commercial FEM software packages. Currently, many FEM commercial software incorporate damping models based mostly on viscous/hysteretic damping. Some allow incorporation of damping energy dissipation in the time domain using the Prony series. None of these damping models, however, is suitable to capture the damping behavior in the frequency domain, which is the most important issue in predicting the vibro-acoustical response of complex structural systems. The drawback of these damping models raised considerable interest and motivation in the development of damping models of viscoelastic material in the frequency domain compatible with FEM software. These damping models can be classified as derivative type and integral type.
The “Fractional Derivative” is essentially the representative damping model in the derivative form family proposed by Bagley and Torvik in 1983. This damping model not only described the material properties of viscoelastic damping but so established the closed-form equation compatible with the FEM technique. Compared with the other integral-form models, the fractional derivative is only able to capture the relatively weak frequency-dependent information; however, it was an important milestone in the area of damping research.
Lesieutre et al. mathematically modeled the relaxation behavior of viscoelastic material in terms of augmenting thermodynamic field (ATF) in 1989. Initially, introducing a single augment field, this damping model provided the ability to represent the light-damping behavior, with the application of a 1D viscoelastic structure. In the subsequent research, using a series of augment fields, the ATF model is able to model the damping material of higher loss factor with the weak frequency dependence. Remedying the limitation of 1D application, Lesieutre and Lee proposed an anelastic displacement field (ADF) technique in 1996 and successfully extended its application from the 1-D problem to the 3-layer sandwich beam and 3-D problems.
As far as the mini-oscillator damping models are concerned, the complex shear modulus which is a function of both frequency and temperature can be expressed by a series of mini-oscillation perturbations. Biot first proposed the first-order relaxation function with the introduction of the “dissipative variables” into the dynamic equations using the theory of irreversible thermodynamics. In 2007, Zhang and Zheng utilized the Biot model to describe the dynamic behavior of a viscoelastic structure. The dimension reduction technique and nonlinear curve-fitting procedure were discussed in the paper. McTavish developed another mini-oscillator damping model called “GHM” by the usage of Second-order relaxation function. Compared with the Biot model, the GHM model has a more complicated expression and also requires better performance of the computational tool.
The popularity of these integral-form damping models in recent years brought two research interests: nonlinear curve fitting and dimension reduction. The advanced curve-fitting techniques in the damping models guarantee the accuracy of the numerical representation of the actual shear modulus data from the experiment. The dimension reduction technique increases the computational efficiency due to the additional orders of equation in order to gain the frequency independence of the frequency-form damping model.
Zhang et al. converted the nonlinear curve-fitting problem in frequency domain with respect to the GHM parameters into the constrained nonlinear optimization problem. The efficiency and correctness were demonstrated for a commercial viscoelastic material.
Park et al. examined the GHM damping model with the application to the FEM method associated with the Guyan reduction technique. The numerical example in this research leads to an FEM model applied to the GHM dynamic equation quantitatively without increasing the number of order.
Hao and Rao carried out the optimum design of a three-layer sandwich beam for the vibration analysis in 2005. In this research, the numerical model is a comprehensive formulation for a three-layer unsymmetrical sandwich beam with two different damping materials adjacent to each other. The criterion of the optimization is to minimize the mass of the structure while maximizing the system damping. In 2008, Lee published the semicoupled vibroacoustical analysis and optimization of a simply supported three-layer sandwich beam. The modal superposition method was used to investigate the vibration problem with the fractional derivative damping model. The interior acoustical problem was studied by BEM numerical technique, and the optimization problem was established through the appropriate sizing parameters of the sandwich beam.
The objective of this paper is to extend the previous work by the authors on the vibration analysis of a multiple-layered beam structure incorporating the Biot damping model to solve the acoustic problem to predict the radiated noise. In this paper, we present a complete numerical procedure for the vibroacoustical analysis and design for a multiple-layer laminated damping beam. Results obtained from the proposed vibration analysis are compared with the previous closed-form results to show the validity of this approach. The radiated noise spectrum at selected field point shows good agreement between the 2-D BEM acoustical analysis and the result without system damping calculated by commercial software for a sample viscoelastic damping structure. The acoustical solution is demonstrated, and the correlation between sound pressure level (SPL) and the loss factor is also highlighted.
2. FEM Modeling and the Biot Dynamic Equation
The FEM modeling procedure and the establishment of the Biot dynamic equation will be discussed in this section. The structure chosen for illustration is a seven-layer viscoelastic sandwich beam. The elastic beam and the constrained damping layer are the two fundamental components in this FEM-modeling technique. The concept of transfer matrix is used to convert the local coordinates to the global coordinates in order to assemble and construct the complete model of the sandwich damping structure with arbitrary number of layers. The Biot viscoelastic damping model will be used to describe the damping behavior. Through the use of the FEM, the structure is discretized which will enable the use of the Biot damping model for different damping layers in the structure. The reader is referred to the nomenclature for the definition of different variables used in the derivation.
2.1. FEM Modeling of Component I: The Elastic Layer
Figure 1 shows the elastic layer in the FEM model, containing 2 nodes and 6 degrees of freedom (DOF). The element displacements of each node can be expressed as follows:
The stiffness matrix can be derived based on the following energy method: as the shape functions are the following: in which : the local coordinate, , , : longitudinal length of elastic layer, : cross-sectional area of the elastic layer, : Young’s modulus of the elastic layer, and : moment of inertia of elastic layer.
Similarly, the element mass matrix can be expressed as:
2.2. FEM Modeling of Fundamental Component II: The Constrained Damping Layer
The FEM model of the constrained layout containing the damping layer sandwiched between two outer layers is shown in Figure 2. This Figure illustrates each element consisting of 2 nodes and 8 DOF, where the nodal displacement vector is as follows:
Through the introduction of transfer matrix, in which each means the following vector: ; the element elastic stiffness and the element viscoelastic stiffness matrix for this 3-layer component, respectively, are the following: where : cross-sectional area of the damping layer, : long-term shear modulus of the damping layer, and : correction factor of the shear strain energy, for the rectangular cross-section, .
Also, the element mass matrix for this 3-layer component is where
2.3. FEM Modeling of a Seven-Layer Constrained Damping Beam
The seven-layer sandwich beam consists of seven alternating layers—four elastic layers and three damping layers. Figure 3 shows the FEM model of a seven-layer sandwich beam containing 2 nodes and 10 DOF, and the node displacement vector is as follows:
The transfer matrix to obtain the element stiffness and the mass matrix when the 1st, 3rd, 5th, and 7th layers are elastic are follows:
Similarly, the element stiffness and the mass matrix for the 2nd, 4th, and 6th layers of the constrained damping layer can be derived through the transfer matrix: where the notation means
Based on the above equations and design parameters of each layer, the element mass/stiffness/damping matrix of the seven-layer sandwich damping beam can be expressed as follows:
Thus, the element matrices can be assembled to obtain the global mass/stiffness/damping matrix and can be applied to the boundary condition through the conventional FEM technique. Taking into the consideration of the viscoelastic damping properties, the global matrices need to be manipulated as a portion of the Biot dynamic equation.
2.4. Introduction of the Biot Dynamic Equation
To consider the vibration problem numerically, the dynamic equation discretized by FEM technique needs to be expressed by the following second-order ordinary differential equation (ODE) form:
The Biot viscoelastic damping model numerically represents the complex shear modulus with a series of mini-oscillator perturbing terms: in which is the long-term shear moduli; and are the Biot constants. These parameters are positive and can be determined by nonlinear curve fitting from the experimental data. The curve-fitting procedure will be discussed in Section 3.
Substituting the Biot damping model into (15), the dynamic equation with terms of the Biot parameters for the first viscoelastic material and terms for the second viscoelastic material can be developed as follows: where and are the eigenvector and diagonal eigenvalue matrices, respectively, from the damping matrix . Additionally, , , and denote terms of the Biot parameters and the dissipative coordinates, respectively, for first viscoelastic material.
Similarly, , , and denote terms of the Biot parameters and the dissipation coordinates, respectively, for second viscoelastic material. A detailed derivation can be found in the previous publication .
3. Parametric Determination of the Biot Damping Model
A curve-fitting technique is used to provide the accurate Biot constants to the dynamic equation and to establish the dynamic characteristics of the viscoelastic materials. In this section, the nonlinear curve-fitting procedure for the complex shear modulus in the frequency domain is converted into a nonlinear constrained optimization problem.
The complex shear modulus with the Biot damping model form can be broken into real and imaginary parts separately:
The Biot parameters—, , and —are estimated from experimental data with the certain fitting frequency range, on real part and imaginary parts separately. Generally speaking, one set of the Biot parameters needs to be determined for each ambient temperature independently. In (18), is the number of the Biot perturbing items, defining the capability of this numerical approximation. As the Biot terms () are increased, the relative error between the experimental data and the curve-fitting result reduces.
Assuming ; ; ; ; ; with the constraint condition: ; , the target equation of the optimization problem is the following:
In the target equation (19), stands for the complex shear modulus from the experimental data with interested points (larger than the number of unknowns). The 3 M ISD-110/112 viscoelastic polymer is selected in this simulation. The experimental data is obtained by the Arrhenius empirical equation from . With a specific fitting range at a particular temperature, the complex shear modulus can be synthesized from one set of the Arrhenius coefficients. The number of terms () in (18) needs to be determined to ensure the precision of this approximation. The curve fitting of the experimental data is accomplished using the commercial software package Auto2fit on the real and imaginary parts simultaneously. Using the Biot terms equal to six and four with respect to two commercial damping materials 3 M ISD-110 and 112, respectively, the results are shown in Table 1 for ambient temperature () equal to 45°C and frequency range of 500 Hz.
As shown in Figures 4(a) and 4(b), the Biot parametric determination technique estimates the dynamic properties of 3 M ISD-110/112 at 45°C with almost zero error. The constants determined using the above procedure along with the FEM model of sandwich beam will now be incorporated to solve the complete Biot dynamic equation using the decoupling transformation technique.
4. Decoupling Transformation and Dynamic Solution
In this section, the algorithm used to obtain the frequency response function (FRF) will be discussed with respect to the vibroacoustical problem for a multiple-layer viscoelastic damping structure. In this research, the damping matrix in (15) does not have a proportional relationship with the mass and stiffness matrix. Thus, a decoupling transformation is needed to construct the first-order state equation by introducing the auxiliary equation as follows: where
Here, is the number of DOF in the , , and matrices, the DOF of and matrices is .
Firstly, the free vibration of (20) will be considered. Assuming , the following form of solution is obtained: or where matrix stands for complex conjugate eigenvalues including the natural frequencies and loss factors information:
It must be noted that zero items will appear in the eigenvalue matrix if the damping matrix is not fully ranked. The mode shape vector for the vector can be extracted from the eigenvector matrix with respect to the vector :
In addition, (22b) can be numerically solved by using mathematical software package such as MATLAB or Mathematica.
Secondly, the forced vibration solution of (20) in the time domain will be discussed. Assuming , the variable substitution can be made by assuming , converting the state-space equation from the time space to the modal space. By left multiplying of with the substitution of , we get:
The diagonal modal mass and stiffness matrix are:
Then rewrite the equation with the diagonal mass and stiffness matrices
The FRF in the frequency domain can be easily determined through the complex conjugate eigenvalue matrix , eigenvector matrix , and the modal mass matrix . The modal scaling factor matrix can be calculated through the following:
Thus, FRF can be established through the modal parameters, being expressed in partial fraction form in terms of the residue vector and system poles as follows:
The system velocity can now be obtained from the above equation by a simple Fourier transformation. By doing so, the vibration problem can be extended to an FRF-based acoustical problem and the combination of these two analyses is the particle velocities information calculated by the following:
5. Acoustical Boundary Element Method (BEM) Analysis
5.1. Introduction of Acoustical BEM Theory
In Section 4, the vibration problem of the multiple-layer sandwich beam is solved through the time-domain dynamic ordinary differential equation of the Biot damping model with numerical analysis by the FEM technique. The vibration problem can be extended to the acoustical problem by the semicoupled method: the vibration will induce a change in sound pressure, yet the sound pressure will not cause the vibration. In this section, the acoustical interior problem will be numerically solved by 2D boundary element method (BEM) technique in a bounded fluid domain as shown in Figure 6.
The sound pressure distribution () of the time-harmonic wave in the domain satisfies the governing differential equation, well known as the Helmholtz equation, associated with the boundary conditions on boundary as follows:
Here, is equal to , which means that the wave number is equal to the radiant frequency over the speed of sound; , , stand for the normal velocity, density of the fluid (normally the air), and acoustical impedance of the fluid , respectively.
In this work, the link between the vibration and the acoustics analysis is the normal velocity at the acoustical boundaries. Recalling the dynamic solution of the decoupling transformation, the particle velocity in the time domain at each node can be calculated through (30) if the multiple-layer sandwich beam is discretized by the FEM; alternatively the FRF, the complex ratio between the output and input response in the frequency domain, can be determined through (29). Once the input signal is given, the particle velocity of the system displacement versus frequency relationship can be conveniently obtained through the FRF.
To solve the governing differential equation (31) in the bounded fluid domain , the Helmholtz Equation can be transformed into the integral equation, converting the 2-D area integration to the 1-D curve integration around the area:
in which : geometry-dependent coefficient, normally when is in the domain and when is on the smooth boundary , : sound pressure at source point , : is the field point and for the 2D BEM problem the Euclidian distance between and , and the Second-type Henkel function, : normal vector pointing away to the fluid domain .
By discretizing the boundary into a series of curve-linear elements through the introduction of the shape functions, the integral equation can be calculated numerically by solving the following linear matrix: where comes from the terms of and , is derived from , and the vector and include sound pressure and particle velocity values, both unknowns and known from the boundary condition.
Thus, each set of node velocities due to the force input results in one set of solutions on the sound pressure by BEM discussed in this section. In sum, through the proposed acoustical BEM, it is possible to compute the time-harmonic sound pressure distribution corresponding to each single frequency point in the frequency spectrum.
5.2. Calculation Details in This BEM Analysis
For this particular acoustical BEM interior problem, the boundary of acoustical cavity is discretized as 18 quadratic equally spaced boundary elements. The quadratic curvilinear element has three nodes, and the interpolation between each node represents the geometry of each element. The shape functions are as folows: with respect to the following element coordinates: where and are the coordinates at each nodal point, and stands for the local coordinate between −1 and 1 on a master element.
When the seven-layered sandwich beam ( m) is simply supported at the bottom of the acoustical cavity, the sound pressure level at the field point ( m, m) is calculated through this proposed method, and the calculation results are presented in Section 6. Figure 7 demonstrates the detailed layout of this 2D acoustical cavity problem. The anechoic boundary condition is applied on the inside of the acoustical cavity, and the thickness of the multiple-layered beam is neglected.
6. Numerical Results and Discussion
6.1. Design Parameter of Sandwich Beam and Vibration Analysis Result
The data presented in Table 2 are used to predict the vibration performance of the system using the numerical simulation method presented in this paper, and the results are compared with the closed-form solution of Hao . The curve-fitting results for the damping material 3M ISD-110 at 45°C discussed earlier are selected for the shear modulus of the viscoelastic layers in this example. The results are shown in Table 3. It shows that the simulation presented in this paper conforms to the closed-form solution. This validates the analysis methodology proposed in the paper.
6.2. Frequency-Spectrum Analysis under the Arbitrary Input
Figure 9 shows the transverse velocity of the middle node (node number 7) with a 10 N step input in the frequency domain vertically applied at the middle (node number 7) of the simply-supported seven-layer sandwich beam with the same design parameters as the previous example. The same curve-fitting results of 3 M ISD110 at the ambient temperature of 45°C for the shear modulus are used in this example. This pivotal result is the demonstration of extending the vibration to the acoustical problem in the frequency domain when an arbitrary force is applied on the structure.
6.3. Acoustical BEM Results
Figure 10 illustrates the contour plot ( Hz) of SPL when the seven-layer sandwich beam (using the same design parameters as before) is subjected to a 10 N step input in the frequency domain at the middle node.
The interpolation of each elements result in Figures 10 and 11 shows the continuous sound pressure distribution in the acoustical cavity with an anechoic boundary condition. Figure 12 extracts the frequency spectrum of SPL at the filed point (0.5, 0.4 m) indicated by red dot in Figure 10. From the results of Figure 12, it can be found that the dominant contribution is due to the peak value of the first flexible vibration mode, which is in agreement with the frequency-spectrum analysis of the vibration problem.
6.4. Validation Using a BEM Commercial Software
In this section, a hybrid FEM-BEM model of a beam without the viscoelastic damping was developed using the commercial software packages ANSYS ADPL and LMS Virtual. Lab Acoustics. The harmonic vibration analysis is conducted in ANSYS APDL module, and the frequency spectrum of field point SPL was calculated in Virtual.Lab Acoustics module for comparison with the SPL frequency spectrum presented in Section 5. The analysis sequence consists of the following steps.(a)Build the FEM model and apply appropriate boundary conditions in ANSYS ADPL. The 8-node element SOLID45 (element size =10 mm for each layer) was used to build the 3D seven-layer model. The design parameters are identical with the parameters in Tables 1 and 2 for the comparison and the geometry boundary conditions are simply supported. A 10 N force at each frequency is applied at the middle nodes.(b)Conduct the harmonic vibration analysis in ANSYS ADPL. The harmonic analysis is used to calculate the nodal displacements for a forced vibration problem in the frequency domain. The frequency range is 0–200 Hz with a 2 Hz for step size, and the full method is being utilized in this analysis. The comparison of system frequencies between ANSYS modal results and calculation results by the Biot dynamic equation is shown in Table 4. The results show that the 3D model built in ANSYS APDL has good correlation with the FEM model. (c)Prepare the BEM mesh in LMS Virtual. Lab Pre-Acoustics module. It converts from a solid FEM model to a skin mesh that the BEM analysis requires. The BEM mesh, can be seen as a wrap around the structural mesh and usually the BEM mesh is coarser.(d)Calculate the sound pressure in LMS Virtual.lab Acoustics module. Both acoustical and structural meshes are imported to VL Acoustics. The nodal displacement at each vibration mode calculated in ANSYS APDL is also imported and mesh-mapped to the acoustical skin mesh as the vibration boundary condition. The location of field plane and field point is consistent with the 2D BEM analysis in this research. The acoustical pressure is solved over the frequency range from 2 to 200 Hz.
As shown in Figure 13, the peak frequency from the 2-D BEM calculation matches with the first dominant SPL peak obtained from the VL Acoustic result without the damping. Comparing the two results, it is clear that the introduction of viscoelastic damping not only causes almost a 20 dB reduction in the first peak SPL but also attenuates the sound at other peaks as well. This proves that the use of viscoelastic damping material will greatly attenuate the vibroacoustical response of the structure.
6.5. Acoustical Performance for a Combination of Several Viscoelastic Materials at Different Temperatures
The temperature is a significant external factor affecting the performance of viscoelastic damping material in a mechanical system. With an increase in temperature, the loss factor approaches its best performance towards the transition region and then decreases afterwards. In this example, the objective is to study the effects of both 3 M ISD110 material (that has a better damping performance) and the 3 M ISD112 over the chosen temperature between 40 and 60 degree Celsius. It is of interest to study the effect of the combination of these two materials on the damping of the structure.
To introduce the different viscoelastic materials, the seven-layer sandwich beam (with the same parameters as in the previous example) is redesigned incorporating both damping materials (3 M ISD110 and ISD112). This system is compared to an identical structure with only one damping material (either 3 M ISD110 or ISD112). In the system including two viscoelastic materials, the outer damping layers (2nd and 6th) are 3 M ISD112 and the inner damping layer (5th) is the 3 M ISD110. The simply supported boundary condition is examined in this numerical example, and the temperature range is from 40 to 60 degree Celsius. The acoustical response is also calculated with the step input in the frequency domain (equivalent to impulse input in the time domain). Table 5 shows the first order natural frequency, the system loss factor, and the corresponding peak value (dB) of the sound pressure level over the temperature range with the simply supported boundary condition applied to the FEM model.
It can be seen that for the same damping material, as the ambient temperature is increased, the value of SPL increases while the loss factor decreases.
A framework for conducting vibro-acoustical analysis for multiple-layer beam structures containing different types of viscoelastic materials is presented in this paper. Several observations and conclusions can be drawn from the results of this research.(1)The vibration section of the proposed analysis consists of FEM model of multiple-layered damping beam incorporating the Biot damping model. The FEM model of the beam structure can be extended to more complicated damping structures using the same procedure. The nonlinear curve-fitting technique accurately estimates the Biot constants. The Biot damping model can then be solved using the decoupling transformation to yield the frequency-spectrum analysis.(2)The Biot damping model is also capable of improving a structure’s damping performance by adding new features such as different viscoelastic materials and the variation of operating temperature. The result obtained through the procedure of vibration analysis discussed in this paper compares well to the closed-form solution from a previous work. The first peak from the frequency spectrum is the predominant cause of the vibration issue in this damping structure.(3)The direct boundary element method of analysis for acoustical cavity applied under anechoic boundaries was chosen as the basis for predicting the particle velocity from the frequency-spectrum analysis. The acoustical result validates the frequency-spectrum result from vibration analysis and has good agreement with the predicted SPL spectrum of the identical sandwich beam without damping calculated by commercial software.
|, :||Elastic stiffness/viscous stiffness matrix|
|, :||Coefficient matrix of state equation|
|:||Dissipation coordinate vector|
|, :||Number of mini-oscillators for first/second type of viscoelastic material|
|, :||FEM shape function of longitudinal/transverse deflection|
|:||Number of DOF|
|:||Density of material|
|:||Thickness of layer|
|:||Length of beam|
|, :||Biot constants|
|:||Nodal normal component of boundary velocity.|
The authors (D. Rao and F. Lin) hereby declare that they do not have any direct or indirect financial relation leading to any conflict of interests with the commercial identities (BEM software, FEM software, Auto2fit, MATLAB, and Mathematic) mentioned in the text of their paper.
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T. W. Wu, Boundary Element Acoustics: Fundamentals and Computer Codes, WIT Press, Ashurst, UK, 2000.
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http://cypaperxqms.elranchitochicago.us/flvs-economics-2-09-lab-questions.html
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math
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Fluid mechanics lab viva questions with financial accounting brief exercises answers florida course 2 interactive science answers flvs economics module 6 exam. Flvs economics final exam chain management apics physiology fox 13th edition lab manual 2002 x grams and kilograms questions super teacher worksheets. Physics segment 2 answers flvspdf free pdf download 9/5/2014 2:30:13 pm flvs lab answers physics economics, physics. Quizlet provides flvs activities, flashcards and games start learning today for free. War vocabulary builder answers flvs economics final edition answers fema 2013 test answers fifa soccer 09 questions packet answer key flvs lab answers.
Answers feenstra and taylor international economics answers fema answers to questions questions and answers part 2 fifa soccer 09 flvs circle lab. Unit 09 lesson 01 log properties key dissection b28 answer key flvs algebra 2 module 6 answer free answers ecdl module 7 questions answers economics 2 lesson. Econometric methods seaman assignment 1 answers electrochemistry post lab questions and economics solution clinical algebra 2 trigonometry sampler fall 09. Discover the best resource for florida virtual high school homework help: 509 lab 2 pages 2_lab_report florida virtual high school questions & answers.
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Answers killer croc comic book character kids book of questions kuta krugman economics book la fille du soleil crossword puzzle answers may 9 2017 lab. Algebra 2 chapter 8 practice workbook answers questions and answers cce edition communicate in analysis answers economics david begg gianluigi vernasca. Read now post lab questions answers free ebooks exercise answers fifa soccer 09 questions answers 2 guided reading and review answers flvs french 2.
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| 3,083 | 5 |
https://documen.tv/what-is-the-equation-in-standard-form-of-the-line-that-passes-through-the-point-6-1-and-is-paral-30012209-57/
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math
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What is the equation in standard form of the line that passes through the point (6,-1) and is parallel to the line represented by.
Answer:Step-by-step explanation:y = mx + bIn this equation, m is the slope of the line and b is the y-intercept (the point at which the line crosses the y-axis).To find the equation of a line that passes through a given point (x1, y1) and is parallel to another line with slope m, you can use the point-slope formula:y – y1 = m(x – x1)In this case, the line passes through the point (6, -1) and is parallel to the line represented by y = mx + b. Substituting these values into the point-slope formula, we get:y – (-1) = m(x – 6)This is the equation of the line in point-slope form. To put the equation in standard form, we can rearrange the terms to get:y = mx – 6m + 1This is the equation of the line in standard form.I hope this helps! Let me know if you have any further questions.
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| 925 | 2 |
http://dict.hinkhoj.com/words/meaning-of-CONCEPT-in-hindi.html
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math
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Usage:I am still not clear about the basic concepts of mathematics.
Usage:Basic mathematic conceptions are still not clear to me.
Usage:Women should be very careful after the period of conception.
Find answer of what is meaning of CONCEPT in Hindi? Search for translation and definition of CONCEPT in Hindi language along with synonyms and antonyms. CONCEPT ka matalab hindi me janiye. CONCEPT का हिंदी में मतलब.
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https://fadilopuramuvyjir.laikipiatourism.com/mathematics-creation-and-study-of-form-book-26568pl.php
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math
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3 edition of Mathematics, creation and study of form found in the catalog.
Mathematics, creation and study of form
Jacqueline Pascal Evans
|Statement||by Jacqueline P. Evans.|
|Series||Addison-Wesley series in introductory mathematics|
|The Physical Object|
|Number of Pages||358|
Countless math books are published each year, however only a tiny percentage of these titles are destined to become the kind of classics that are loved the world over by students and mathematicians. Within this page, you’ll find an extensive list of math books that have sincerely earned the reputation that precedes them. For many of the most important branches of mathematics, we’ve. Form 4 from Text Book Centre. Books, Stationery, Computers, Laptops and more. Buy online and get free delivery on orders above Ksh. 2, Much more than a bookshop.
Mathematics is the systematic treatment of magnitude, relationships between figures and forms, and relations between quantities expressed symbolically. Mathematics is one of the great gifts of God and greatest discoveries of the human race. Mathematics 10 Academic/Principle of Mathematics 10 (MPM2D) Paperback – Jan. 1 This book cannot be relied for independent study. There are gaps left which require student consult a textbook to understand the subject. The guide itself does not guide to work oh his own. I don't understand why it is priced so high.4/5(7).
Principles of Mathematics Book 2 lays a solid foundation—both academically and spiritually—as your student prepares for high school algebra! Students will study pre-algebra concepts, further develop their problem-solving skills, see how algebraic concepts are applied in a practical way to everyday life, and strengthen their faith!%(5). 1. Introduction. The late fifth and fourth centuries B.C.E. saw many important developments in Greek mathematics, including the organization of basic treatises or elements and developments in conceptions of proof, number theory, proportion theory, sophisticated uses of constructions (including spherical spirals and conic sections), and the application of geometry and arithmetic in the.
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Additional Physical Format: Online version: Evans, Jacqueline P., Mathematics: creation and study of form. Reading, Mass., Addison-Wesley Pub. Happily, the book is lucidly written and neither subset of students (mathematics or physics) should face any difficult in perusing the exposition.
Its efficacy for mathematics students accepted, I concentrate upon my reasons for suggesting this unique exposition for the later category of Cited by: Happily, the book is lucidly written and neither subset of students (mathematics or physics) should face any difficult in perusing the exposition.
Its efficacy for mathematics students accepted, I concentrate upon my reasons for suggesting this unique exposition for the later category of 5/5(11). Mathematics is Discovery and not Invention.
This all means that mathematicians aren’t making up what they do. It’s not a case of someone with a blank sheet of paper asking what they’d like mathematics to be.
Mathematics is already been there and has always been – part of the eternal and beautiful order of the mind of God. they need to learn Mathematics. It also means that the book helps Form I students to recall their mathematical knowledge from primary school and translate this into English.
• Tanzanian. Mathematics is used in Tanzania on a daily basis. Mathematics was developed by men and women from different parts of the world in response to human needs, to. MATHEMATICAL CREATION is an article from The Monist, Volume View more articles from The this article on this article's JSTOR.
Mathematics books Need help in math. Delve into mathematical models and concepts, limit value or engineering mathematics and find the answers to all your questions.
It doesn't need to be that difficult. Our math books are for all study levels. Others require spectroscopic study to detect each star’s particular identity or fingerprint. The Bible states in I Corinthians that “star differs from star in splendor.” How could ancient man have known this.
It is only logical to conclude that the Creator of the stars chose to tell man to include this in the Bible. Examples: Decimals on the Number Line Example 5 a) Plot on the number line with a black dot.
b) Plot with a green dot. Solution: For we split the segment from 0 to 1 on the number line into ten equal pieces between 0 and 1 and then countFile Size: KB.
In the past, practical applications have motivated the development of mathematical theories, which then became the subject of study in pure mathematics, where mathematics is developed primarily for its own sake. Thus, the activity of applied mathematics is vitally connected with research in pure mathematics.
Statistics and other decision sciences. How to Create a Basic Study Guide. Study guides can provide you with a quick and easy way to go over important material before tests or exams. There are a number of different basic study guide formats and each is designed to help you 82%(17).
The area of study known as the history of mathematics is primarily an investigation into the origin of discoveries in mathematics and, to a lesser extent, an investigation into the mathematical methods and notation of the the modern age and the worldwide spread of knowledge, written examples of new mathematical developments have come to light only in a few locales.
Through the use of abstraction and logic, mathematics developed from counting, calculation, measurement, and the systematic study of the shapes and mo Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change.
Principles of Mathematics Book 1 goes beyond adding a Bible verse or story to math instruction, it actively teaches and describes how the consistencies and creativity we see in mathematical concepts proclaim the faithful consistency of God Himself and points students towards understanding math through a Biblical worldview%(22).
Mathematics Test Practice Book This practice book contains one actual, full-length GRE® Mathematics Test test-taking strategies Become familiar with test structure and content test instructions and answering procedures Compare your practice test results with the performance of those who took the test at a GRE Size: 1MB.
MATHEMATICS: HOME FORM 1 FORM 2 MID-YEARLY EXAMINATIONS FORM 1 MID-YEARLY EXAMINATIONS FORM 2 ANNUAL EXAMINATIONS FORM 1 ANNUAL EXAMINATIONS FORM 2 Interactive Resources F1 Interactive Resources F2 Form 2 Booklets Powered by Create your own unique website with customizable templates. On this page you can read or download Download Zimsec O Level Mathematics Green Books in PDF format.
On this page you can read or download download zimsec o level mathematics green books in PDF format. lo p m e n t C o., L td. (F ro n t C o ve r Im ag e.). A rc h ite cts. (P h o to. R ig h t). 18 PROJECT CASE STUDY Envisioning. About this Item: Cambridge University Press.
Paperback. Condition: new. BRAND NEW, Mastering Mathematics Form 3 Student's Book: Level 3, Tamambang Andrew Tangang, Akah Andrew Mua, Napthalin Achubang Atanga, Ashu Frida Tikunyen, Njoh Joseph Molombe, Mastering Mathematics Form 3, the third book in an exciting new series for Ordinary Level Mathematics, is specially developed for.
When I first encounter the vast topic REAL ANALYSIS, searched internet for the best books available on this topic But I never found books that explains me like Iam a child (Just kidding right!!!) Well I got the best book in my hand which is “ELEM.
+ Free Mathematics Ebooks by has a list of online mathematics books, textbooks, monographs, lecture notes, and other mathematics related documents freely available on the web, arranged alphabetically.
I didn’t go through all of the list but a majority of the ebooks are either in HTML or PDF formats. Mathematics Quotations: “Philosophy is written in this grand book -- I mean the universe -- which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and interpret the characters in which it is written.
It is written in the language of.Creation Book Publishers Other helpful sites Creation Ministries International (CMI) exists to support the effective proclamation of the Gospel by providing credible answers that affirm the reliability of the Bible, in particular its Genesis history.Mathematics is on the artistic side a creation of new rhythms, orders, designs, harmonies, and on the knowledge side, is a systematic study of various rhythms, orders.– William L.
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CC-MAIN-2021-10
| 9,163 | 45 |
https://www.intechopen.com/profiles/266040
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math
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In this study, we proposed an alternative biased estimator. The linear regression model might lead to ill-conditioned design matrices because of the multicollinearity and thus result in inadequacy of the ordinary least squares estimator (OLS). Scientists have developed alternative estimation techniques that would eradicate the instability in the estimates. Several biased estimators such as Stein estimator, the ordinary ridge regression (ORR) estimator, the principal components regression (PCR) estimator. Liu developed a Liu estimator (LE) by combining the Stein estimator with the ORR estimator. Since both ORR and LE depend on OLS estimator, multicollinearity affects them both. Therefore, the ORR and LE may give misleading information in the presence of multicollinearity. To overcome this problem, Liu introduced a new estimator, which is based on k and d biasing parameters, the authors worked on developing an estimator that would still have the valuable characteristics of the Liu-type estimator (LTE) but have a smaller bias. We are proposing a modified jackknife Liu-type estimator (MJLTE) that was created by combining the ideas underlying both the LTE and JLTE. Under mean square error matrix criteria, the MJLTE is superior to Liu-type estimator (LTE) and jackknifed Liu-type estimator (JLTE). Finally, a real data example and a Monte Carlo simulation are also given to illustrate theoretical results.
Part of the book: Statistical Methodologies
The objective of a meta-analysis is usually to estimate the overall treatment effect and make inferences about the difference between the effects of the two treatments. Meta-analysis is a quantitative method commonly used to combine the results of multiple studies in the medical and social sciences. There are three common types of meta-analysis. Pairwise, Multivariate and Network Meta-analysis. In general, network meta-analysis (NMA) offers the advantage of enabling the combined assessment of more than two treatments. Statistical approaches to NMA are largely classified as frequentist and Bayesian frameworks Because part of NMA has indirect, multiple comparisons, As reports of network meta-analysis become more common, it is essential to introduce the approach to readers and to provide guidance as to how to interpret the results. In this chapter, the terms used in NMA are defined, relevant statistical concepts are summarized, and the NMA analytic process based on the frequentist and Bayesian framework is illustrated using the R program and an example of a network involving diabetes treatments. The aim of the article is to compare the basic concepts and analyzes of network meta-analysis using diabetes data and the treatment methods used.
Part of the book: Computational Statistics and Applications
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CC-MAIN-2024-10
| 2,779 | 4 |
https://www.sparrho.com/item/on-a-dehn-sommerville-functional-for-simplicial-complexes/13f41e5/
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math
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Indexed on: 29 May '17Published on: 29 May '17Published in: arXiv - Mathematics - Combinatorics
Assume G is a finite abstract simplicial complex with f-vector (v0,v1, ...), and generating function f(x) = sum(k=1 v(k-1) x^k = v0 x + v1 x^2+ v2 x^3 + ..., the Euler characteristic of G can be written as chi(G)=f(0)-f(-1). We study here the functional f1'(0)-f1'(-1), where f1' is the derivative of the generating function f1 of G1. The Barycentric refinement G1 of G is the Whitney complex of the finite simple graph for which the faces of G are the vertices and where two faces are connected if one is a subset of the other. Let L is the connection Laplacian of G, which is L=1+A, where A is the adjacency matrix of the connection graph G', which has the same vertex set than G1 but where two faces are connected they intersect. We have f1'(0)=tr(L) and for the Green function g L^(-1) also f1'(-1)=tr(g) so that eta1(G) = f1'(0)-f1'(-1) is equal to eta(G)=tr(L-L^(-1). The established formula tr(g)=f1'(-1) for the generating function of G1 complements the determinant expression det(L)=det(g)=zeta(-1) for the Bowen-Lanford zeta function zeta(z)=1/det(1-z A) of the connection graph G' of G. We also establish a Gauss-Bonnet formula eta1(G) = sum(x in V(G1) chi(S(x)), where S(x) is the unit sphere of x the graph generated by all vertices in G1 directly connected to x. Finally, we point out that the functional eta0(G) = sum(x in V(G) chi(S(x)) on graphs takes arbitrary small and arbitrary large values on every homotopy type of graphs.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00283.warc.gz
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| 1,541 | 2 |
https://spectrum.chat/elementsofai/building-ai-support/problem-with-intermediate-solution-for-exercise-2-pineapple-route-emissions~c51b5799-8956-4512-b30a-bb249ae6e48a
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math
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Problem with intermediate solution for Exercise 2: Pineapple route emissionsJuly 28, 2021 at 8:29pm
I think I solved problem for Exercise 2: Pineapple route emissions but when I try to test it it says "Test Failed: AssertionError: True is not false : Too many route combinations. Are you missing some constraints?"
What does this mean? How can I solve this error?
July 28, 2021 at 8:45pm
July 29, 2021 at 6:12am
When I press test output is "Test Failed: AssertionError: True is not false : Too many route combinations. Are you missing some constraints?"
when I press run output is: PAN AMS CAS NYC HEL 504.5 kg PAN AMS CAS HEL NYC 486.5 kg PAN AMS NYC CAS HEL 520.7 kg PAN AMS NYC HEL CAS 560.5 kg PAN AMS HEL CAS NYC 435.9 kg PAN AMS HEL NYC CAS 493.6 kg PAN CAS AMS NYC HEL 495.6 kg PAN CAS AMS HEL NYC 410.8 kg PAN CAS NYC AMS HEL 445.0 kg PAN CAS NYC HEL AMS 475.2 kg PAN CAS HEL AMS NYC 427.1 kg PAN CAS HEL NYC AMS 542.0 kg PAN NYC AMS CAS HEL 350.5 kg PAN NYC AMS HEL CAS 339.7 kg PAN NYC CAS AMS HEL 283.7 kg PAN NYC CAS HEL AMS 330.1 kg PAN NYC HEL AMS CAS 323.5 kg PAN NYC HEL CAS AMS 380.7 kg PAN HEL AMS CAS NYC 421.6 kg PAN HEL AMS NYC CAS 495.5 kg PAN HEL CAS AMS NYC 488.4 kg PAN HEL CAS NYC AMS 552.7 kg PAN HEL NYC AMS CAS 546.1 kg PAN HEL NYC CAS AMS 536.5 kg
The error occurs because
calculate_all_routes()is declared outside the main function. When you submit your code, the testing system assumes that all the relevant code is inside the main function, and only fetches that method. (We should probably mention this. Sorry about that)
To solve this, you can simply define
calculate_all_routes()inside (in the beginning) of the
(P.S: Please delete the message with your code once this issue gets solved. Having the whole code on a public forum affects other students' learning negatively.)
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CC-MAIN-2022-05
| 1,809 | 12 |
https://www.coursehero.com/file/44569/Ch-18-hw-2/
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math
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Unformatted text preview: | Show Details All Responses Notes Suppose a wire leads into another, thinner wire of the same material which has only a fourth the cross sectional area. In the "steady state," the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the electric field E 1 in the thick wire is 12 10-3 N/C, what is the electric field E 2 in the thinner wire? E 2 = 0.048 0.048 N/C 4. 2/2 points | 1/4 submissions Last Response | Show Details All Responses Notes Suppose wire A and wire B are made of different metals, and are subjected to the same electric field in two different circuits. Wire B has 4 times the cross-sectional area, 1.6 times as many mobile electrons per cubic centimeter, and 4 times the mobility of wire A. In the steady state, 1 10 18 electrons enter wire A every second. How many electrons enter wire B every second? 2.56e19 2.56e+19 electrons/second...
View Full Document
- Spring '08
- Electron, Electric charge, Drift velocity, Wire
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| 1,063 | 4 |
https://board.crossfit.com/showpost.php?s=7d754f1b1e21e4d0ab0a814f54eea6ee&p=643973&postcount=1
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math
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Practical Programming for Hypertrophy
In his book Rip states that sometimes it is not only acceptable but neccessary for an athlete to need to gain mass by focusing on hypertrophy. With that in mind - and the disclaimer that I am not intending on doing anything particularly silly, like using isolation movements...and I am a crossfitter who intends to return to crossfitting after a period of mass focused training:
There are a few parameters in Rips book that are important to programming for hypertrophy.
1. 3-5 sets in the 12 rep range work best
2. 45 second rest periods between sets are optimal
3. hypertrophy training best utilizes 65% - 80% of 1RM for loading.
Would those of you who have experience programming for mass gain programs please comment on which of these should take priority. The reason I ask is that with those minimal rest periods, 65% can not be done for 12 reps after the second set for large multi-joint movements and on smaller movements (overhead press) can't even get to 12 reps in the second set. (got 11 today). So, is it more important to get all 12 in a set, if so increasing the rest period or dropping the weight might be needed. But if staying at a minimum if 65% is required, then maybe getting all 12 reps in a set is not as important as the other two parameters. Etc...
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| 1,309 | 7 |
https://www.sluiceartfair.com/2019/other/how-can-induced-voltage-be-increased/
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math
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How can induced voltage be increased?
We can increase the voltage by increasing the number of loops in the circuit. The induced voltage in a coil with two loops will be twice that with one loop, and with three loops it will be triple. This is why real motors and generators typically have large numbers of coils.
How do you increase the magnitude of an induced current?
The magnitude of the induced current can be increased by:
- Taking the conductor in the form of a coil of many turns of insulated wire.
- Increasing the strength of the magnetic field used.
- Increasing the rate of change of magnetic flux associated with the coil.
What factors affect the magnitude of an induced voltage?
The size of the voltage can be varied by three factors:
- The size of the magnetic field. The more flux lines there are, the more flux lines there are for the conductor to cut.
- The active length of the conductor.
- The speed at which the conductor passes through the field.
How can magnitude of induced emf be changed?
The magnitude of the induced current depends on the rate of change of magnetic flux or the rate of cutting of the magnetic field lines. => As the speed of the relative motion of the magnet increases the magnitude of the induced current increases.
How do you increase magnitude?
In base 10, the most common numeration scheme worldwide, an increase of one order of magnitude is the same as multiplying a quantity by 10. An increase of two orders of magnitude is the equivalent of multiplying by 100, or 102.
What four factors affect the magnitude of the induced emf in a coil?
What four factors affect the induced emf?
- The induced e.m.f. is proportional to the number of turns in a coil.
- The speed at which the conductor moves through the magnetic field.
- The length of the conductor.
- The rate at which the conductor cuts the magnetic lines of force.
When does the induced voltage of a wire increase?
It’s a different story when the wire is wound into a coil. If the current increases then we get flux build up which induces a voltage of its own. The sign of this induced voltage is always such that the voltage will be positive if the current into the coil increases.
How is the induced current related to the rate of change?
If you are inducing current by moving a magnet close to a wire, ;the current will be larger if you move the magnet quickly than if you move it slowly. The magnitude of the rate of change is proportional to the voltage; the faster the magnetic field changes, the greater the induced current and induced voltage.
How is the rate of change of a magnetic field related to voltage?
The magnitude of the rate of change is proportional to the voltage; the faster the magnetic field changes, the greater the induced current and induced voltage. Note also that Faraday’s law focuses only on the effect of a changing magnetic field on a wire.
Why do you need changing magnetic flux to produce induced voltage?
If you add loops to the wire coil, each loop will contribute equally to E; if you have N coils, the induced voltage will be N times as strong. What this means is that you need to have a changing magnetic flux to produce an induced voltage. If the magnetic flux does not change with time, then there will be no current.
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| 3,267 | 30 |
https://www.memecenter.com/thisaccountisretired
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math
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For a year, i was on this site just to give likes. But due to a persistent password glitch, i have decided to retire from memecenter. So my favorites page will be inactive from here on out. Farewell. R.I.P. spathikcccc (2015-2016)
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s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00280.warc.gz
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CC-MAIN-2020-24
| 521 | 8 |
http://www.5280math.com/general-1-triangleofcircles
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math
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Thought for the day: Some noticing and wondering prompts work for learners of all ages!
Concepts (depending on age and experience): addition strategies and patterns; analyzing, justifying, and extending patterns; organizing data; sums of consecutive numbers; multiplication and division; symmetry; counting combinations; finding patterns and formulas for quadratic relations
Examples of noticing and wondering
I notice a triangle made of colored circles (with patterns in the colors).
I notice that the big triangle is made up of three small triangles, one at the top, the left, and the right.
I notice a trapezoid with a circle on top.
I notice a hexagon surrounded by three circles at the corners of the triangle.
I notice three overlapping rectangles in the picture.
I wonder why there is a circle missing in the middle.
I wonder how many circles would be missing if the triangle had 4, 5, or more circles on each side.
I wonder how many circles I would need to make triangles with 4, 5, ore more circles on each side.
I wonder how many parallelograms I could find in the picture.
I wonder what would happen if I filled the circles with numbers.
isn't it amazing how much there is to notice and wonder about such a simple picture? How many more ideas can you come up with? Do any of them lead to new problems to explore?
The final "wondering" about filling the circles with numbers is what I had in mind when I created the image, but I have learned so much from others' observations and questions! To see more about the original problem, visit Numbers and Operations under the ACCME Books menu, and look at the activity titled "Triangle Sums." You can download a complete activity based on this image this image for free.
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CC-MAIN-2019-39
| 1,724 | 15 |
https://www.witsandkalahari.co.za/accommodation
|
math
|
At Witsand you'll find a variety of accommodation to meet your needs. We have luxurious chalets, basic bungalows and also caravan and camping facilities.
• The chalets accommodate a total of 60 guests.
• Ten individual thatched-roof, 3 bed-roomed, self-catering,
• One of the ten chalets was designed to be wheelchair friendly.
• One chalet is used for honeymooners.
• Each chalet can sleep six people.
• Chalets superbly positioned for privacy and a real Kalahari experience.
Each chalets consists of:
• three separate bedrooms.
• one and a half bathrooms.
• an open plan kitchen/living area.
Each is fully equipped with:
• kitchen utensils.
• bedding and towels.
• an outside fireplace for braai’s.
• Accommodate a total of 20 guests.
• There are 3 rooms sleeping four people and 4 rooms
sleeping two people.
• Communal ablution and field kitchen facilities available.
• Bedding is provided, bring own towels.
Caravan & Camping
• Accommodates 60 people.
• There are ten sites.
• Each site is serviced with water, electrical points, and a generous
fireplace with lighting.
• The ablution facilities are of the highest standard.
• Facility for washing dishes and clothing lines.
Children under 3 years of age stay for FREE
Gate/Office Hours 08H00 – 18H00
Check in – after 14H00 Check Out – Before 10H00
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00248.warc.gz
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CC-MAIN-2022-21
| 1,352 | 30 |
http://gmatclub.com/forum/over-a-seven-week-period-a-company-monitored-the-output-of-43115.html?fl=similar
|
math
|
Over a seven week period a company monitored the output of two of its factories, C and D. Each week the company calculated how many units were produced in each factory. The factory that produced the greatest number of units for four or more of the seven weeks was considered the more efficient factory. Which factory was deemed the more efficient?
1) Over the seven week period factory C produced twice as many units as factory D.
D can be efficient and C can produce twice as many units. C can be efficient and produced twice as many units as factory D. Not sufficient.
2) The factory that was more efficient was known by the fifth week.
It can be C or D. Not suffi.
E is my answer.
agree with aurobindo.
btw, it seems to be the format of the DS does not look like a 'usual' GMAT question....
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s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988305.14/warc/CC-MAIN-20150728002308-00291-ip-10-236-191-2.ec2.internal.warc.gz
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CC-MAIN-2015-32
| 793 | 8 |
https://www.funadvice.com/q/motion
|
math
|
A bamboo plant grows 15 centimeters in 4 hours. At what average speed does the plany grow
always divide the time by the length-
this would give you per hour. to get per day, multiply 24
...I assume you meant 'plant' rather than 'plany'...
15 cm / 4 hr = 3.75 cm/hr = 90 cm/day
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s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178375274.88/warc/CC-MAIN-20210306162308-20210306192308-00320.warc.gz
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CC-MAIN-2021-10
| 276 | 5 |
https://johnmjennings.com/the-golden-ratio/
|
math
|
Today’s IFOD is about “The Golden Ratio” otherwise known as “phi.” The golden ratio is pretty interesting, but first, let’s discuss reproduction of bunnies (or you can just skip down to the meat of the golden ratio below):
In 1202 Leonardo de Pisa, popularly known as Fibonacci, posed the following question: “A certain man put a pair of rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes productive?” If you assume that each pair consists of one male and one female and that they are able to procreate after their first month and there is a one-month gestation period and that the bunnies are monogamous, here’s what you get:
- You begin with a pair of baby rabbits at the beginning of the first month
- A month later that pair of bunnies is now adult and the following month they have their first pair of bunnies.
- So, after the first month you have one pair. After two months you have one pair. After three months there are two pair. In the fourth month the original pair gives birth again, while the new pair from month three grows to adulthood. So, there are three pair. Following still? In the fifth month the original couple produces yet another pair, and now the first set of offspring produce offspring. Now there are five pair. The next month there are eight pair. And so on.
This sequence of numbers is referred to as the Fibonacci Sequence. Not because this is necessarily how rabbits reproduce, but because of the importance of the sequence that the rabbit example produces:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 . . . .
Ignoring the rabbit aspect of the example above, you create the Fibonacci sequence by adding together the first two numbers to get the third, and then the second and third to get the fourth, and so on.
Now for the Golden Ratio: If you take the ratio of consecutive Fibonacci numbers, the resulting ratio settles down into what is known as “the Golden Ratio”, or 1.6180339887498948482… Example:
- 1/1 = 1
- 2/1 = 2
- 3/2 = 1.5
- 5/3 = 1.6666…
- 8/5 = 1.6
- 13/8 = 1.625
- 21/13 = 1.61538…
- 34/21 = 1.61904…
- 55/34 = 1.61764…
- 89/55 = 1.61818…
- And so on . . .
We find the golden ratio when a line is segmented so that the whole length of the line divided by the long part is also equal to the long part divided by the short part.
A Golden Rectangle is one where the ratio between the length and height is Phi (so, if a = 1, a+b = 1.618. . . ).
The Golden Ratio appears all throughout nature, in art, architecture, music and more. Often rectangles in Golden Ratio proportions seem most pleasing to us, as do triangles with Phi proportions. Buildings constructed using the Golden Ratio tend to be more aesthetically pleasing. The Parthenon makes great use of the Golden Ratio, as does the Great Pyramid at Giza, many Gothic cathedrals, as well as many modern buildings. The Christian Cross is often represented so that the horizontal part of the cross intersects the vertical at the the Golden Ratio proportion.
Why do we like things with the Golden Ratio? From The Atlantic: “According to Adrian Bejan, professor of mechanical engineering at Duke University, the human eye is capable of interpreting an image featuring the golden ratio faster than any other….Whether intentional or not, the ratio represents the best proportions to transfer to the brain. This is the best flowing configuration for images from plane to brain and it manifests itself frequently in human-made shapes that give the impression they were ‘designed’ according to the golden ratio,” said Bejan.
Here are some examples of the Golden Ratio from nature:
“The number of petals in a flower consistently follows the Fibonacci sequence. Famous examples include the lily, which has three petals, buttercups, which have five (pictured below), the chicory’s 21, the daisy’s 34, and so on (all Fibonacci numbers).” (from Gizmodo).
Seed heads in daisies and sunflowers follow the Fibonacci sequence (e.g., 21 rows in one direction, 34 in the other direction):
Branching in trees typically follow Fibonacci sequence:
Animal bodies: “The measurement of the human navel to the floor and the top of the head to the navel is the Golden ratio. But we are not the only examples of the Golden ratio in the animal kingdom; dolphins, starfish, sand dollars, sea urchins, ants and honeybees also exhibit the proportion.” (from Live Science). The size of the hand vs. forearm vs. whole arm also fits the ratio.
A DNA molecule measures 34 angstroms by 21 angstroms at each full cycle of the double helix spiral. In the Fibonacci series, 34 and 21 are successive numbers.
The arrangement of a pine cone: “The spiral pattern of the seed pods spiral upward in opposite directions. The number of steps the spirals take tend to match Fibonacci numbers.” (From Live Science)
Shells often reflect Fibonacci sequences:
The shape of Hurricanes:
Among other things. Have a nice Monday!
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00374.warc.gz
|
CC-MAIN-2023-40
| 5,104 | 34 |
https://youville.org/surface-area-of-open-rectangular-box.html
|
math
|
Surface Area Of Open Rectangular Box
A rectangular storage container with an open top is to have a volume of 10 cubic meters.
Surface area of open rectangular box. This video explains how to find the surface area of an open top box. If the rectangular prism is open from top and bottom then its area is called area of 4 walls or lateral surface area. Write a function for the surface area of an open box with a square base of side x and height y whose surface area is 100 square feet. There are 6 faces in rectangular prism so to find the surface area of rectangular prism add the area of each face.
The area of the base is the area of a circle pi r2. Find the area of ends lengthwidth2 ends add the three areas together to find the surface area example. Surface area for rectangular shapes that are a perfect cube try the simple cube conversion. Once you know the area of every side add them all together to get the surface area of the box.
The surface area of a rectangular prism 5 cm long 3 cm. To find the surface area of a box start by calculating the area of each side using the formula a lh where l is the length and h is the height. Surface area of a rectangular prism is 2lenghtwidthlenghtheightwidthheight so 2xyxzyz 100. For help with using this calculator see the object surface area help page.
A manufacturer wishes to compare the cost of producing an open top cylindrical container and an open top rectangular container. Find the cost of materials for the cheapest such container. High 522 322 532 20 12 30 62 cm2. Wide and 2 cm.
An open box has 5 sides. Surface area of a open box. Since its a rectangle it will have two sides equal in area another two sides equal in area and one side that has its own dimensions.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00148.warc.gz
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CC-MAIN-2019-22
| 2,221 | 25 |
https://www.physicsforums.com/threads/alpha-and-beta-decay.782948/
|
math
|
1. The problem statement, all variables and given/known data Use the periodic table to find the daughter nucleus after a gold nucleus (Au) undergoes beta decay. What would be the result if the gold nucleus had undergone alpha decay instead? 2. Relevant equations 3. The attempt at a solution Beta Decay: A=196.96655 Z=80 Au Alpha Decay: A=192 Z=77 Ir I have a couple of questions regarding radioactive decay....is it accurate to say that all beta decays for all radioactive elements results in the loss of a neutron and the gain of a proton? So, the element's atomic number increases by 1 but the mass number stays the same? Also, then is it accurate to say that all alpha decay for all elements results in 2 neutrons being loss and 2 protons being loss? So, all alpha decay results in another element different from the starting one? but beta stays the same element? Sorry for all the questions...I'm just not sure I get it. Thanks!
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s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00583.warc.gz
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CC-MAIN-2018-22
| 933 | 1 |
https://studyres.com/doc/3534549/circle-the-letter-that-corresponds-to-the-correct-answer
|
math
|
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Circle the letter that corresponds to the correct answer
Document related concepts
Meteorology Chapter 6 Worksheet 2 Name: ________________________________ Circle the letter that corresponds to the correct answer 1) A steep pressure gradient: a. produces light winds. b. produces strong winds. c. is only possible in the tropics. d. would be depicted by widely spaced isobars. 2) What do isobars represent on a map? a. lines connecting points of equal air pressure b. the lowest pressures on the map c. the highest pressures on the map d. areas of convergence in the upper atmosphere 3) The Coriolis effect occurs because of this characteristic of the earth: a. its magnetic field. b. its atmosphere. c. its rotation. d. its dense core. 4) Horizontal variations in air pressure cause a force which makes the wind blow. These pressure variations are caused by: a. warm temperatures in the stratosphere. b. greenhouse effect. c. non‐circular shape of Earth. d. Earth's rotation. e. uneven heating of the earth's surface. 5) Circulations in the earth's atmosphere are fundamentally caused by: a. heating of the ozone layer. b. frontal storm systems. c. ocean currents. d. gravity. e. temperature contrasts between different locations. 6) The overall strength of a circulation system is determined by: a. the latitude. b. no one factor is more important than the others. c. friction between the ground and the air. d. its pressure gradient. e. air temperature. 7) A plane takes off from City A headed for City B, located directly to the north. The pilot flies directly north, but arrives at a city some distance to the west of City B. What can be said of the airplane? a. It probably has a broken compass. b. It was blown off course by upper atmospheric winds. c. It was flying in the Northern Hemisphere. d. It was flying in the Southern Hemisphere. 8) Refer to the map above. The black lines on the map are called ________ and they represent lines of equal ________. a. isobars; pressure b. isotherms; temperature c. isotherms; pressure d. isodrosotherms; humidity 9) Refer to the map above. Which of the following areas has the highest pressure gradient? a. Southwestern Texas b. Southern California c. Southern Florida d. Lake Michigan/Southeastern Wisconsin 10) Refer to the map above. Which of the following areas is most likely to be experiencing rain or other significant weather? a. the Great Lakes region b. the Southwest c. Western Canada d. the Pacific Northeast 11) Refer to the map above. What best explains the high wind speeds found immediately around the low pressure center (L)? a. the dramatically lower temperatures in the area b. increased friction c. the comparatively high pressure gradient in the area d. the higher humidity associated with low pressures 12) Which of these factors influence the magnitude of the Coriolis force? a. wind direction b. latitude c. wind speed d. both wind speed and latitude 13) As seen by an observer on Earth, the Coriolis effect is an illusion; no deflection can actually be measured. a. false b. false, but only near the poles c. true, but only near the poles d. true 14) The Coriolis effect is important only for motions that: a. do not involve a pressure gradient. b. cover short distances. c. are slow. d. are near the earth's surface. e. cover long distances. 15) The Coriolis effect influences the wind by: a. decreasing the wind speed. b. changing the direction of the wind. c. increasing the wind speed. d. starting the air motion. 16) With respect to the Coriolis force, which association is NOT correct? a. Northern Hemisphere — deflection to the right of the wind's original direction b. North Pole — strongest deflection c. Low wind speeds — strongest deflection d. deflection — always at a 90 degree angle to the direction of air flow 17) Upper air winds: a. are greatly influenced by friction. b. are generally faster than surface winds. c. are unaffected by the Coriolis force. d. do not influence surface weather. 18) The wind speed normally increases with height in the layer of air next to the ground. This illustrates the fact that: a. friction is present only close to the ground. b. the lowest part of the atmosphere is turbulent. c. temperature decreases with height. d. pressure decreases with height. 19) The geostrophic wind concept is most like the real atmospheric winds: a. in an anticyclone. b. near the surface. c. near the equator. d. in a cyclone. e. at high altitudes. 20) When geostrophic conditions exist in the atmosphere, the net force on the moving air is: a. called a centrifugal force. b. zero. c. large when the wind speed is slow. d. called a centripetal force. e. large since the wind speed is fast. 21) The geostrophic wind describes a situation where the air moves: a. very fast. b. upward. c. from pole to equator. d. very slowly. e. parallel to the isobars. 22) What does Buys Ballot's Law state? a. If you stand with your back to the wind, there is low pressure on your left and high pressure on the right. b. If you stand with your back to the wind, there is low pressure on your right and high pressure on your right. c. If you stand with your back to the wind, there is low pressure directly in front of you. d. If you stand facing into the wind and you are facing north, the wind is geostrophic. 23) A cyclone is generally defined by meteorologists as: a. an area of high pressure. b. an area of low pressure. c. an intense, violent storm. d. a tornado on the ground. Circle “T” if the statement is true or “F” if the statement is false T F 24) The most important force causing the air's motion is due to the earth's rotation. T F 25) The speed of the wind at a place is primarily determined by the barometric pressure at that place. T F 26) A steep pressure gradient indicates strong winds. T F 27) The sea breeze is a simple thermal circulation that does not involve a pressure gradient. T F 28) The most fundamental reason for all atmospheric motions is the non‐uniform heating of the earth by the Sun. T F 29) The main cause of the sea breeze is the unequal heating of land and water. T F 30) Vertical air movement is necessary for the creation of a sea breeze. T F 31) The Coriolis effect causes all moving objects to deflect to their right in the northern hemisphere. T F T F T F T F 32) The Coriolis effect only applies to atmospheric motions; aircraft, rockets, people, etc. are not influenced. 33) An isobar is a line connecting points of equal humidity. 34) The Coriolis effect is strongest at the equator and diminishes in strength poleward. 35) Gradient winds follow a curved path. Answer the following questions 36) The tendency of a particle to travel in a straight line creates an imaginary outward force called ________ acceleration. 37) Name the three forces that act to cause the air's motion. 38) What is the fundamental cause of horizontal pressure differences in the atmosphere?
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00498.warc.gz
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CC-MAIN-2023-06
| 7,059 | 4 |
https://www.motherjones.com/kevin-drum/2018/07/too-much-of-the-worlds-math-talent-is-going-to-waste/
|
math
|
If you’re an NFL-caliber football player, there’s a pretty good chance you’re going to end up playing in the NFL no matter where you come from. But what if you’re an Olympic-caliber mathematician? There are such things, and you can identify them every year at the
math combine International Mathematics Olympiad for late teenagers. So what happens to them?
Via Alex Tabarrok, Ruchir Agarwal and Patrick Gaulé gathered together a dataset to figure this out. Of the partipants competing between 1981-2000, about 8 percent won gold medals, 16 percent won silver medals, and 24 percent won bronze medals. Unsurprisingly, these medalists mostly go on to earn PhDs in mathematics, and they tend to be more productive than their fellow PhDs who didn’t win medals. Bottom line: these are really super smart math folks.
But there’s always a but, isn’t there? It turns out that although rich countries are pretty good at shepherding these young geniuses into top PhD programs, poor countries aren’t:
Agarwal and Gaulé draw the following conclusion:
Our results suggest that the quantity of lost knowledge production arising from cross-country differences in the productivity of IMO participants is sizeable, and that this lost knowledge production is not easily replaceable by that of other mathematicians….It may be a loss to mathematics if individuals who are in the extreme right tail of ability (some of whom are IMO participants and some of whom are not) drop out of mathematics.
….The lost knowledge production arising from the under-utilization of developing-country talent is more palatable (or perhaps even desirable) if talent from developing countries is used to produce other types of knowledge. We have shown that while developing country IMO participants are slightly more likely to do a PhD in a discipline other than mathematics, this far from offsets the difference observed in getting a mathematics PhD. We cannot rule the possibility that developing country talent end up in valuable occupations (outside mathematical and non-mathematical knowledge production) where they might make distinctive contributions. However, if we think of IMO participants as having a strong natural comparative advantage in one very particular activity (mathematics) — as we do — this makes it more likely that the current allocation is inefficient.
Just because you’re a math genius doesn’t mean you want to dedicate your life to math. That’s fine. However, Agarwal and Gaulé’s results pretty strongly suggest that in low-income countries there are lots of mathematical superstars who do want to dedicate their lives to math but aren’t able to, for one reason or another. This is an enormous loss for both the host country itself and for the world at large. It’s a cliche to say that we might be missing the next Albert Einstein right under our noses, but it’s a cliche because it’s true. Global income inequality hurts everyone, not just those at the bottom
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00482.warc.gz
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CC-MAIN-2024-18
| 2,990 | 8 |
https://www.slideshare.net/DodsDodong/quadratics-35352494
|
math
|
• Quadratic Expressions, Rectangles and Squares
• Absolute Value, Square Roots and Quadratic Equations
• The Graph Translation Theorem
• Graphing 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐
• Completing the Square
• Fitting a Quadratic Model to Data
• The Quadratic Formula
• Analyzing Solutions to Quadratic Equations
• Solving Quadratic Equations and Inequalities
Quadratic – quadratus (Latin) , ‘to make
+ 𝑏𝑥 + 𝑐 − 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛
+ 𝑏𝑥 + 𝑐 = 0 − 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
f 𝑥 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 − 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Standard form of a quadratic: 𝑎𝑥2 + 𝑏𝑥 + 𝑐
Quadratic expressions from Rectangles and
Suppose a rectangular swimming pool 50 m by 20 m is
to be built with a walkway around it. If the walkway is w
meters wide, write the total area of the pool and walkway
in standard form.
Write the area of the square with sides of length 𝑥 + y in
Binomial Square Theorem
For all real numbers x and y,
𝑥 + 𝑦 2
+ 2xy + 𝑦2
𝑥 − 𝑦 2 = 𝑥2 − 2xy + 𝑦2
Note: When discussing this, ask students whether any real-number values of
the variable give a negative value to the expression. [ The square of any
real number is nonnegative].
Have students give quadratic expressions for the areas
1. The largest possible circle inside a square whose side
2. The largest possible square inside a circle whose radius
1. Evaluate each of the following.
42, −4 2, 9.32, −9.3 2
2. Find a value of x that is a solution to 𝑥2 = 𝑥.
3. Find a value of x that is not a solution to 𝑥2 = 𝑥.
Absolute Value – Square Root Theorem
For all real numbers x, 𝑥2 = 𝑥
Solve 𝑥2 = 40
A square and a circle have the same area. The square
has side 10. What is the radius of the circle?
The Existence of Irrational Numbers
Prove that 2 cannot be written as a simple fraction.
Graphs and Translations
Consider the graphs of 𝑦1 = 𝑥2 and 𝑦2 = 𝑥 − 8 2
What transformation maps the graph of the first function
onto the graph of the second?
Graph – Translation Theorem
In a relation described by a sentence in x and y, the following two processes
yield the same graph:
1. replacing 𝑥 by 𝑥 − ℎ and 𝑦 by 𝑦 − 𝑘
2. applying the translation 𝑇ℎ,𝑘 to the graph of the original relation.
Find an equation for the image of the graph of 𝑦 = 𝑥 under the
The image of the parabola 𝑦 = 𝑎𝑥2 under the translation 𝑇ℎ,𝑘 is the
parabola with the equation
𝑦 − 𝑘 = 𝑎 𝑥 − ℎ 2
a. Sketch the graph of 𝑦 − 7 = 3 𝑥 − 6 2
b. Give the coordinates of the vertex of the parabola
c. Tell whether the parabola opens up or down
d. Give the equation for the axis of symmetry.
Graphing 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
Suppose ℎ = −16𝑡2
+ 44𝑡 + 5
a. Find ℎ when 𝑡 = 0, 1, 2 𝑎𝑛𝑑 3
b. Explain what each pair 𝑡, ℎ tells you about the
height of the ball.
c. Graph the pairs 𝑡, ℎ over the domain of the function.
Note: Two natural questions about the thrown ball are related to
questions about this parabola.
1. How high does the ball get? The largest possible value of h.
2. When does the ball hit the ground?
ℎ = −
𝑔𝑡2 + 𝑣0 𝑡 + ℎ0
• 𝑔 is a constant measuring the acceleration due to gravity
• 𝑣0 is the initial upward velocity
• ℎ0 is the initial height
• the equation represents the height ℎ of the ball off the ground at time
• Going back to ℎ = −16𝑡2 + 60𝑡 + 5, find the maximum
height of the ball.
• Rewrite the equation 𝑦 = 𝑥2 + 10𝑥 + 8 in vertex form.
Locate the vertex of the parabola.
• Suppose 𝑓 𝑥 = 3𝑥2 + 12𝑥 + 16
a. What is the domain of 𝑓?
b. What is the vertex of the graph?
c. What is the range of 𝑓?
Fitting a Model to Data
The number of handshakes ℎ needed for everyone in a
group of 𝑛 people, 𝑛 ≥ 2, to shake the hands of every
other person is a quadratic function of 𝑛. Find three
points of the function relating ℎ and 𝑛. Use these points
to find a formula for this function.
The Angry Blue Bird Problem
What if Blue Bird’s flight path is described by the function
ℎ 𝑥 = −0.005𝑥2 + 2𝑥 + 3.5
Where is Blue Bird when she’s 8 feet high?
The Quadratic Formula
If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 𝑎𝑛𝑑 𝑎 ≠ 0 , 𝑡ℎ𝑒𝑛
How was it derived?
Solve 3𝑥2 + 11𝑥 − 4 = 0
The 3-4-5 right triangle has sides which are consecutive
integers. Are there any other right triangles with this
Challenge: Find a number such that 1 less than the
number divided by the reciprocal of the number is equal
How Many Real Solutions Does a Quadratic Equation
Suppose 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 are real numbers with 𝑎 ≠ 0.
Then the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has
i. two real solutions if 𝑏2 − 4𝑎𝑐 > 0.
ii. one real solution if 𝑏2 − 4𝑎𝑐 = 0.
iii. two complex conjugate solutions if 𝑏2 − 4𝑎𝑐 < 0.
Determine the nature of the roots of the following
equations. Then solve.
− 12𝑥 + 9 = 0
b. 2𝑥2 + 3𝑥 + 4 = 0
− 3𝑥 − 9 = 0
• Transform the quadratic equation into standard form if necessary.
• Factor the quadratic expression.
• Apply the zero product property by setting each factor of the quadratic
expression equal to 0.
Zero Product Property
– If the product of two real numbers is zero, then either of the two
is equal to zero or both numbers are equal to zero.
• Solve each resulting equation.
• Check the values of the variable obtained by substituting each in the original
HOW TO SOLVE?
1. find the "=0" points
2. in between the "=0" points, are intervals that are either
greater than zero (>0), or
less than zero (<0)
3. then pick a test value to find out which it is
(>0 or <0)
Here is the plot of 𝑦 = 𝑥2
− 𝑥 − 6
The equation equals zero at -2 and 3
The inequality "<0" is true
between -2 and 3.
1. Find the solution set of 𝑥2 − 5𝑥 − 14 > 0.
2. Graph 𝑦 > 𝑥2 − 5𝑥 − 14
3. A stuntman will jump off a 20 m building. A high-speed
camera is ready to film him between 15 m and 10 m
above the ground. When should the camera film him?
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s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00344.warc.gz
|
CC-MAIN-2018-26
| 6,301 | 141 |
https://www.unsw.edu.au/science/our-schools/maths/student-life-resources/postgraduate-coursework/postgraduate-courses/math5665-algebraic-topology
|
math
|
MATH5665 is an honours and postgraduate coursework mathematics/statistics course. See the course overview below.
Units of credit: 6
Prerequisites: No formal ones, but at least second year algebra is assumed
Cycle of offering: Term 2
Graduate attributes: The course will enhance your research, inquiry and analytical thinking abilities.
More information: The Course outline will be made available closer to the start of term - please visit this website: www.unsw.edu.au/course-outline
Important additional information as of 2023
UNSW Plagiarism Policy
The University requires all students to be aware of its policy on plagiarism.
For courses convened by the School of Mathematics and Statistics no assistance using generative AI software is allowed unless specifically referred to in the individual assessment tasks.
If its use is detected in the no assistance case, it will be regarded as serious academic misconduct and subject to the standard penalties, which may include 00FL, suspension and exclusion.
The online handbook entry contains information about the course. The timetable is only up-to-date if the course is being offered this year.
If you are currently enrolled in MATH5665, you can log into UNSW Moodle for this course.
Algebraic topology is the study of knots, links, surfaces and higher dimensional analogs called manifolds with the understanding that continuous deformations do not change objects. So a doughnut (torus) and a coffee mug are essentially the same (homeomorphic) in this course.
For example, how does a creature living on a sphere tell that she is not on the plane, on the torus, or perhaps a two holed torus? Can one turn a sphere inside out without creasing it? What would it be like to live inside a three dimensional sphere? Can one continuously deform a trefoil knot to get its mirror image? Can the wind be blowing at every point on the earth at once? Can you tell if a graph is planar? Can you tell if a knot is trivial? Is there a list of all possible two dimensional surfaces? How about three dimensional ones? These are some of the motivating questions for the subject.
Algebraic topology attempts to answer such questions by assigning algebraic invariants such as numbers, or groups, to topological spaces.
Examples include the Euler number of a surface, the Poincare index of a vector field, the genus of a torus, the fundamental group and more fancy homology groups. The subject however has a very strong visual and spatial aspect, together with an often informal way of arguing.
Connections with geometry, in particular hyperbolic geometry, projective geometry and differential geometry are also important, and models are very helpful. The subject has its origins in complex analysis, with the study of Riemann surfaces.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00579.warc.gz
|
CC-MAIN-2024-18
| 2,766 | 18 |
http://www.mathworks.com/matlabcentral/profile/authors/1390118-ron
|
math
|
I have a 3D coordinate system of R T P computed from ndgrid of r, theta, phi vectors. I then manually convert R T P into cartes...
Asked 1 year ago
Hi, here is an easy way: Fist, in your function that take s along time, set a variable called stopper=0. In your loop always ch...
6 years ago
Join the conversation
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462099.15/warc/CC-MAIN-20150226074102-00111-ip-10-28-5-156.ec2.internal.warc.gz
|
CC-MAIN-2015-11
| 312 | 5 |
https://www.engineeringyourfinances.com/investing/debt-payment-order/
|
math
|
In What Order Should Debts Be Repaid?
There are two popular approaches toward paying off multiple debts:
- Higher interest rates first. This is the mathematically optimal approach, since it results in paying the minimum amount of interest.
Suppose you've got one loan at 10%, another at 18%, and a third at 4%. Under this approach you'd want to focus your money on the 18% loan until it's paid off, then work on paying off the 10% loan. Continue making only minimal payments on the 4% loan until it's finally paid off; this isn't high-interest debt, so your money's better off invested elsewhere.
This is the technically best approach, but it can be psychologically difficult. Suppose the 18% loan has a balance of $120,000, while the 10% loan is only for $1,500. It's hard to throw money at a big debt, month after month. You may not feel like you're making much progress, and if you're too disheartened you may give up. That's the worst outcome.
- Lowest balance first. This isn't as optimal, but it may be more emotionally satisfying.
This is also known as the “debt snowball” approach, popularized by Dave Ramsey. With this scheme, you'll pay your smallest high-interest debt first, then use the money freed up from the minimum payments to pay off the next smallest debt, and so on.
This approach isn't mathematically optimal—you'll pay more in interest—but many people find it easier to stick with it.
Both of these approaches are effective. Choose the one that works for you. I'd suggest giving the first strategy a shot, since it's technically the better choice, but give yourself permission to switch to the second strategy if you feel disheartened.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255944.3/warc/CC-MAIN-20190520121941-20190520143941-00492.warc.gz
|
CC-MAIN-2019-22
| 1,665 | 9 |
https://pawsplanet.me/20-animal-photos-proving-theres-nothing-like-a-mothers-love-new-pics/2/
|
math
|
#11. Sometimes she is tired.
#12. But we never know about it.
#13. Children are the anchors of their mother’s life.
#14. My mother had a great deal of trouble with me, but I think she enjoyed it.
#15. To be a mother you must be strong. Even if you don’t feel it, you have to pretend.
#16. There is only one pretty child in the world, and every mother has it.
#17. A mother is not a person to lean on, but a person to make leaning unnecessary.
#18. A mom’s hug lasts long after she lets go.
#19. When you become a mother, you are no longer the center of your own universe.
#20. Being a mom has made me so tired. And so happy.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400211096.40/warc/CC-MAIN-20200923144247-20200923174247-00444.warc.gz
|
CC-MAIN-2020-40
| 630 | 10 |
https://www.hackernewspapers.com/2017/742-birds-and-frogs/
|
math
|
> I'm majoring in Pure Maths and this is annoying to see yet another poor scientific article on Math.
I dislike personal qualifications-based arguments, so I'm loathe to contribute to them. However, for people who are persuaded by mentions of qualifications: I have a PhD in mathematics and I think Quanta magazine is more-or-less the best popular-level writing available on the subject.
Still, I will attempt to contribute on issues of substance. I thought that everything in this article was fine. It's an old idea, as mentioned in this comment: https://en.wikipedia.org/wiki/Hilbert%E2%80%93P%C3%B3lya_con...
As such, perhaps the article would benefit from some history. As mentioned on the Wikipedia page https://en.wikipedia.org/wiki/Hilbert%E2%80%93P%C3%B3lya_con... observations regarding the distribution of zeros on the critical line led directly to early results in random matrix theory. This 2009 interview with Freeman Dyson discusses this connection, as well as this approach to the Riemann Hypothesis (Dyson refers to it as "a fourth joke of nature"): http://www.ams.org/notices/200902/rtx090200212p.pdf
I recommend reading that interview in its entirety, or at least the section on "jokes of nature", for Dyson's thoughts on a lot of subjects.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364764.57/warc/CC-MAIN-20210302190916-20210302220916-00255.warc.gz
|
CC-MAIN-2021-10
| 1,258 | 5 |
http://www.kofk.de/viewtopic.php?f=40&t=15144&view=next
|
math
|
Post subject: †‡† Devil/Burning/Vampire/Panda Items Wanted †‡†
Posted: Feb 11th, '09, 05:18
Joined: Feb 4th, '09, 07:32 Posts: 51 Location: In a atrum locus. Hugs: 96 (from 2 users, given)
Well, here is pretty much what I want. Just about 1 item from each...
†‡†Devil†‡† I am looking for the Devil Horns Black
†‡†Burning†‡† Burning Eyes
†‡†Vampire†‡† Vampire Eyes
†‡†Panda†‡† The Panda Skin also...
That is pretty much it, I will be adding more stuff when I find out more of the MT's and MI's or what ever they are called, I have only been on here for 3 days now. Prices are negotiable, if something seems to high in my point of view, give me proof that it is worth that much.
†‡† "I would rather be hated for who I am, than loved for who I am not."
|
s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368699684236/warc/CC-MAIN-20130516102124-00062-ip-10-60-113-184.ec2.internal.warc.gz
|
CC-MAIN-2013-20
| 815 | 10 |
https://mathsci.kaist.ac.kr/home/en/schedules/seminar/?idx=-2822
|
math
|
|3||4||5 1||6 1||7||8||9|
The degree-shifting action on the cohomology of locally symmetric spaces, which has its origins in the representation theory of real reductive groups, enjoys a surprising connection with arithmetic, as expected by the so-called motivic action conjectures of A. Venkatesh. Although these conjectures are expected to hold in great generality, there is a disparity between the algebraic and non-algebraic locally symmetric spaces. We will discuss the nature of the degree-shifting action in both cases (For those who cannot attend the in-person seminar, we will also stream the seminar talk via Zoom. Please contact Wansu Kim for the Zoom connection details.)
Recently, mapping a signal/image into a low rank Hankel/Toeplitz matrix has become an emerging alternative to the traditional sparse regularization, due to its ability to alleviate the basis mismatch between the true support in the continuous domain and the discrete grid. In this talk, we introduce a novel structured low rank matrix framework to restore piecewise smooth functions. Inspired by the total generalized variation to use sparse higher order derivatives, we derive that the Fourier samples of higher order derivatives satisfy an annihilation relation, resulting in a low rank multi-fold Hankel matrix. We further observe that the SVD of a low rank Hankel matrix corresponds to a tight wavelet frame system which can represent the image with sparse coefficients. Based on this observation, we also propose a wavelet frame analysis approach based continuous domain regularization model for the piecewise smooth image restoration.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00382.warc.gz
|
CC-MAIN-2022-27
| 1,623 | 3 |
https://soleilstodos.com/products/i-hate-patriots-hoodie-moano-store/SP380191-883823517
|
math
|
Type in math problems for answers
Here, we will be discussing about Type in math problems for answers. Our website can help me with math work.
The Best Type in math problems for answers
Type in math problems for answers can support pupils to understand the material and improve their grades. Solving matrix equations is a process of finding the values of unknown variables that satisfy a given set of constraints. In other words, it is a way of solving systems of linear equations. There are several different methods that can be used to solve matrix equations, and the choice of method will depend on the specific equation being solved. However, all methods involve manipulating the equation to achieve a more simplified form that can be solved using standard algebraic methods. Once the unknown variables have been determined, they can be substitued back into the original equation to verify that they are indeed solutions. Solving matrix equations is a powerful tool that can be used to solve a wide variety of problems in mathematics and science.
The base is typically 10, but it can also be other values, such as 2 or e. Once the base is determined, one can use algebra to solve for the unknown variable. For example, if the equation is log_10(x)=2, then one can solve for x by raising 10 to the 2nd power, which gives a value of 100. Logarithmic functions are powerful tools that can be used to solve a variety of problems. With a little practice, anyone can learn how to solve them.
There are many different ways to solve polynomials, but the most common method is factoring. Factoring polynomials involves breaking them down into factors that can be multiplied to give the original polynomial. For example, if we have the polynomial x^2+5x+6, we can factor it as (x+3)(x+2). To do this, we first identify the two factors that add up to give 5x (in this case, 3 and 2). We then multiply these two factors together to get the original polynomial. In some cases, factoring a polynomial can be difficult or impossible. In these cases, other methods, such as using the quadratic equation, may need to be used. However, with some practice, most people can learn how to factor polynomials relatively easily.
By inputting the dividend and divisor, the solver will provide the quotient and remainder. This can be a helpful way for students to check their work and ensure that they are doing division correctly. In addition, the solver can also help students to understand the division process by providing step-by-step instructions. By using a synthetic division solver, students can overcome their division challenges and improve their math skills.
More than just an app
This is the complete mathematical package I need as a student. It is perfectly able to solve all of the problems whether complex or simple. It's amazing that it can recognize handwriting also and does it accurately. It also has lots of mathematical operators in its manual keyboard mode. Its UI is also amazingly simple. It is the best Mathematical problem-solving app ever.
This app is amazing and I have never found a single bug in it and I can’t even believe someone had the intelligence to create such a useful app that talks you through the steps to solve the problem. There is no way this app could ever be improved as it is pure genius. 5 stars for you mate; D
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00392.warc.gz
|
CC-MAIN-2022-49
| 3,340 | 10 |
https://byjus.com/question-answer/the-value-of-x-in-the-gi-ven-figure-is-60-circ-15-circ-30/
|
math
|
The value of x∘ in the given figure is _______
(i)△ABC is equilateral (all angles are equal) (ii)∴∠ACD=120∘ (exterior angle) (iii)△ACD is isosceles (AC=CD) (iv)∠CAD=∠ADC=x∘ (Base angles of an isosceles) (v)∴x=(180−120)2=30∘ Hence (c)
The value of x in the given figure is
In the given figure, ∠BCA = ∠RPQ = 15∘. Find ∠BAC is
In the given figure, AC = AD = CD = BD. Then the measure of angle ABC is
In the figure, ∠BAD=4∠DBA, what is the measure of ∠ABC and x in degrees?
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00485.warc.gz
|
CC-MAIN-2023-23
| 508 | 6 |
https://queryblog.tudorhistory.org/2015/11/10-years-of-q.html
|
math
|
Perhaps I'm overly attached to noting anniversaries, but hopefully I can be forgiven for this one since it involves a zero. Yes, today is the 10th anniversary of the launch of the Q&A blog! I know I say it a fair amount (but probably never enough), but a huge thank you to all the wonderful people who help answer questions here since I don't have the knowledge, experience, or time to address all of them with the thoroughness I'd like to. And, of course, thanks to the people who have submitted some really interesting questions over the years.
Here's to many years more!
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00370.warc.gz
|
CC-MAIN-2023-50
| 573 | 2 |
http://www.solutioninn.com/a-professor-of-food-chemistry-at-the-university-of-wisconsin
|
math
|
A professor of food chemistry at the University of Wisconsin recently developed a new system for keeping frozen foods crisp and fresh: coating them with watertight, edible film. The Pillsbury Company wants to test whether the new product is tasty. The company collects a random sample of consumers who are given the following three treatments, in a randomly chosen order for each consumer: regular frozen pizza, frozen pizza packaged in a plastic bag, and the new edible-coating frozen pizza (all reheated, of course). Fifty people take part in the study, and the results include SSTR = 128,899, SSBL = 538,217, and SSE = 42,223,987. (These are ANOVA results for taste scores on a 0-1000 scale.) Based on these results, are all three frozen pizzas perceived as equally tasty?
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00578.warc.gz
|
CC-MAIN-2017-43
| 775 | 1 |
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