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2 years ago

PLEASE HHELP!!! Bring the fraction:

Mathematics

1 answer:

Anna35 [415]2 years ago

8 0

Answer:

$\frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}$

$\frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}$

Step-by-step explanation:

Given

$\frac{b}{7a^2c}$

Express the denominator as $35a^3c^3$

To do this, we divide $35a^3c^3$ by the denominator

$\frac{35a^3c^3}{7a^2c} = 5ac^2$

So, the required fraction is:

$\frac{b}{7a^2c} * \frac{5ac^2}{5ac^2}$

$\frac{5abc^2}{35a^3c^3}$

Hence:

$\frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}$

Given

$\frac{a}{a - 4}$

Express the denominator as $16 - a^2$

Multiply the fraction a+4/a+4

So, we have:

$\frac{a}{a - 4} * \frac{(a + 4)}{(a + 4)}$

Apply difference of two squares to the denominator

$\frac{a^2 + 4a}{a^2 - 16}$

Take the additive inverse of the numerator and denominator

$\frac{-(a^2 + 4a)}{-(a^2 - 16)}$

$\frac{-a^2 - 4a}{16 -a^2}$

Hence:

$\frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}$

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3 years ago

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

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

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

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

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

$z=t\implies z_t=1$

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

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

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.