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Pavlova-9 [17]
3 years ago
6

PLEASE HELP FAST

Mathematics
1 answer:
bazaltina [42]3 years ago
6 0
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ 1) You have to find the equation of the line first:

y = mx + c

First find the slope

slope = difference in y / difference in x

slope = 6/3

slope = 2

Substitute this into the equation, as well as one of the coordinates

2 = 2(2) + c

Subtract 4 from both sides

c = -2

The equation of the line is y = 2x - 2

I will attach the drawn graph (it won't be in blue but just note that the first graph is the first one)

2) Substitute the values into the equation:

5 = -3(10) + c

c = 35

The equation is y = -3x + 35

I will attach the graph (it will be the second one)

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Step-by-step explanation:

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Solve for x.<br> 7 inches<br> Х<br> 11 inches
jonny [76]

Answer:

8.49 inches

Step-by-step explanation:

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4 years ago
Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/
devlian [24]

u(x,y,z)=x^2+yz

\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}

At the point (p,r,\theta)=(2,2,0), we have

\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}

Denote by f_x:=\dfrac{\partial f}{\partial x} the partial derivative of a function f with respect to the variable x. We have

\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}

The Jacobian is

\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}

By the chain rule,

u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y

u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}

u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y

u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}

u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta

u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}

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Please help me smile this problem ASAP
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Answer:

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Photomath

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