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3 years ago

MZQRS = [?] HELP PLEASE !!

Mathematics

1 answer:

motikmotik3 years ago

6 0

Answer:

m∠QRS = 49°

Step-by-step explanation:

Complementary angles are angles that add up to 90°. So m∠ABC + m∠QRS = 90°. When you simplify you get 41° + m∠QRS = 90°. So m∠QRS = 49°.

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Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

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I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

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Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

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To capture all the partial derivatives of $w$ , compute its gradient:

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2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.

If that's the case, then

$z_s=z_xx_s+z_yy_s$

$z_t=z_xx_t+z_yy_t$

as the hint suggests. We have

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$y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}$

Putting everything together, we get

$z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)$

$z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)$

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Answer:

Step-by-step explanation:

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