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Gazelle T.

Calculus 1 / AB

2 months, 1 week ago

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Calculus homework question:

So this is a related rates type problem. So you have, you squared minus C squared is equal to 49. So we want to know what happens at U equals uh 25 cents. Um We know that D. You over D. T. T. Time is equal to -5 cents. It was equal to -5. So we want to so we want to know what is D. C. Over D. T. And to answer this question uh First we need to know what C. Is at that time. So to calculate, see we use the equation so we can see that by just doing algebra, C squared is equal to U squared minus 49. So plugging in you get, C squared is equal to 25 squared minus 49. And then you can check on your calculator that this is that's T. is equal to 24. Okay, so now we have, we know you we now see we know D. U. D. T. So now we need to find D. C. T. T. And the way we do that is by differentiating the equation. So let me rewrite again. Mhm. So we have U squared, my A C squared is equal to 49. So let's differentiate both sides with respect to T. D. Over DT on both sides. Well, the right hand sign is just a constant, so DDT is just zero. And the left hand side nelson here we do implicit differentiation. So using powerful, you get to you bu Over DT -2 C. D. C. Over D. T. Is equal to zero. Now let's solve for D. C. D. G. This implies D. C over D. T. Is equal to you. D you over DT over see uh So now let's actually plug in what we have. So we know you is 25 cents. We know d. U. D. T. is -5 And we know c. 24. So if you put this into your calculator, you get minus 5.21 uh since per month. And so this is the answer.

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Lectures

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In mathematics, precalculus is the study of functions (as opposed to calculus, which is the study of change, and algebra, which is the study of operations and their application to solving equations). It is generally considered to be a part of mathematics that prepares students for calculus.

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In mathematics, a function (or map) f from a set X to a set Y is a rule which assigns to each element x of X a unique element y of Y, the value of f at x, such that the following conditions are met: 1) For every x in X there is exactly one y in Y, the value of f at x; 2) If x and y are in X, then f(x) = y; 3) If x and y are in X, then f(x) = f(y) implies x = y; 4) For every x in X, there exists a y in Y such that f(x) = y.

01:17

\begin{equation}\begin…

02:24

$$\begin{array}{l}{\text {…

00:49

$u(t-1)-u(t-4)$

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24-as shown.

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