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bearhunter [10]

1 year ago

9

QUICK WILL MARK BRAINLIEST!!!

1 answer:

Arturiano [62]1 year ago

6 0

Answer:

The Second, Fourth, and Fifth points on the graph

Finn, Jeff, Kevin

Step-by-step explanation:

hopefully this helped, if not HMU and I will get a better 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

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

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

7 0

2 years ago

Need help please!!!!

r-ruslan [8.4K]

Answer:

ggyyyuuuuuuuuuujjiiii

7 0

1 year ago

using the formula v=l×w×h , find the volume of a right rectangular prism when the length of the prism is 45cm , the width is 12c

Natali [406]

The volume of the rectangular of prism is 5400

3 0

2 years ago

Read 2 more answers

6a^2-[-5a-(9a^2-3a)]-[8+(-19a-8)]

Mamont248 [21]

The answer is 15a^2+21a

5 0

2 years ago

Will give brainliest plz help

Dominik [7]

Answer:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Step-by-step explanation:

Given:

Center of the ellipse is, $(h,k)=(-4,-3)$

Minor axis length is, $2b=6$

A vertex of the ellipse is at (1, -3)

Now, distance between the center and the vertex is half of the length of the major axis.

Using distance formula for (-4, -3) and (1, -3), we get:

$a=\sqrt{(1+4)^2+(-3+3)^2}=5$

Therefore, the value of half of major axis is, $a=5$ . Also,

$2b=6\\b=\frac{6}{2}=3$

Now, equation of an ellipse with center $(h,k)$ is given as:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

Plug in $h=-4,k=-3,a=5,b=3$ and determine the equation.

$\frac{(x-(-4))^2}{5^2}+\frac{(y-(-3))^2}{3^2}=1\\\\\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Therefore, the equation of the ellipse is:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

6 0

2 years ago