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3 years ago

What is 4 7/8 + 6 1/4

Mathematics

1 answer:

Svetlanka [38]3 years ago

5 0

Answer:

Step-by-step explanation:69

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Explain how to use a graph of the function f(x) to find f(3).

Nastasia [14]

Sample answer: For f(3), the input is 3 and you are looking for the output. To determine the function's value when x = 3, go to the value of 3 on the x-axis and then locate the graph for that value of x. Determine the value of y from the y-axis at that location on the graph.
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5 0

4 years ago

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Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/

devlian [24]

$u(x,y,z)=x^2+yz$

$\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}$

At the point $(p,r,\theta)=(2,2,0)$ , we have

$\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}$

Denote by $f_x:=\dfrac{\partial f}{\partial x}$ the partial derivative of a function $f$ with respect to the variable $x$ . We have

$\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}$

The Jacobian is

$\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}$

By the chain rule,

$u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y$

$u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}$

$u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y$

$u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}$

$u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta$

$u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}$

7 0

3 years ago

X ÷ 0.25 = 0 um... Ya pls help...

11Alexandr11 [23.1K]

Answer:

X = 0

Step-by-step explanation:

5 0

3 years ago

CAN SOMEONE HELP ASAP PLEASE!! I'll give 50 points to the person that's correct!

Gelneren [198K]

Choice 4 is incorrect. The angles aren’t similar nor congruent.

7 0

4 years ago

How is the graph of y=-2x^2-5 different from the graph of y=-2x^2?

ki77a [65]

In the first equation graph will touch x axis because equation have x intercept but in the 2nd equation graph will not touch the x axis. BECAUSE it doesn't have x intercept.

3 0

3 years ago

Read 2 more answers