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3 years ago

For what value of x is line a parallel to line b? Enter your answer in the box. x =

1 answer:

3 0

Where are the values of B?
In order for X to be parallel to B you must have the values for B first. Then they will have the same slope.

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What expressions are equivalent to (p-q)(x)

Talja [164]

Answer:

px-qx

Step-by-step explanation:

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3 years ago

SOLVE THIS QUESTION

Serga [27]

Answer:

Step-by-step explanation:

dfs

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3 years ago

2x + 5y - 28 -2x - 5y = -18 Elimination

lidiya [134]

Answer:

no solution I believe

Step-by-step explanation:

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3 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

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3 years ago

Which pair of ratios forms a proportion?

bogdanovich [222]

Answer:

Step-by-step explanation:

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3 years ago

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