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Stells [14]
2 years ago
11

Anyone know how to solve?

Mathematics
1 answer:
sergejj [24]2 years ago
3 0

Answer:

was there more

Step-by-step explanation:

(i will edit answer)

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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
(x+40)(x+40) show steps
pishuonlain [190]

(x + 40) * (x + 40) =

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6 0
3 years ago
Sebastian and his two brothers read and equal amount each week very for a combined total of 135 minutes each week, very the same
ipn [44]

Answer:

270 minutes per boy.

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3 years ago
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If x to the 12th power is divided by x to the 9th power:

Since they share same bases, and are being divided, it would be the same as x^{12 - 9), or x^{3}

Now, simply find the cube root of 125, which is 5.

Therefore, the number (x) must be equal to 5.

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Find the value of x. Then tell whether the side lengths from a Pythagorean triple sides 20,16 find x
Tema [17]

Answer:

x = 12

Step-by-step explanation:

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