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SVETLANKA909090 [29]

3 years ago

9

Which equation represents a line that has a slope of -1/2 and passes through point (4, –5)?

2 answers:

Wewaii [24]3 years ago

8 0

Y=-1/2x-3

You can rewrite the equation “y=mx+b” and fill in what you know. So your equation would look like y=-1/2x+b when filling in what you know. Now you can take the (x,y) coordinates that you were given and plug them into the equation and solve for b. So now it would be -5=-1/2(4)+b

Dominik [7]3 years ago

4 0

Y=-1/2x-3 is the best answer

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Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

Alik [6]

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

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3 years ago

Can someone help me

frozen [14]

Answer: A/D

Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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2 years ago

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Irina18 [472]

Answer:

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Step-by-step explaination:

=> ( -2 )^3*( -2 )^4

=> -8 * 16

=> -128...ans

<h2>HOPE IT HELPS U!!!! </h2>

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Iss D

Step-by-step explanation:

i did the test

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3 years ago

Read 2 more answers