Add 11 + 16. Plug in 11 into b-2 since b equals eleven. You should now have 11+16+11-2. Do the same thing and plug in 16 for c and 11 for b for c-b. Your solution will now be 11+16+11-2+16-11.

4 0

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

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

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

7 0

3 years ago

How can I find the volume of this prism? Helpp

Alex

Answer:

152.971

Step-by-step explanation:

i think thats the answer but sorry if its not.

5 0

2 years ago

use the graph of g(x)=3 √x to help you explain why there is only one solution for every cube root equation of the form y=3 √x+a

Leni [432]

Answer:

G(x)=2x32+c

Step-by-step explanation:

4 0

2 years ago