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3 years ago

11

What is the answer to this?

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

C

Step-by-step explanation:

Neither linear nor exponentional make sense.

Hope this helps!!! XD

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The volume of a sphere can be described by the following formula, where V = volume and r = radius of the sphere.

Anika [276]

Option C: $r = \sqrt[3]{\frac{3V}{4\pi} }$ is the right answer

Step-by-step explanation:

Given formula for volume of a sphere is:

$V = \frac{4}{3}\pi r^3$

We have to make r the subject of the formula

Multiplying equation by 3/4

$\frac{3}{4}V = \frac{4}{3} * \frac{3}{4} \pi r^3\\\frac{3}{4}V = \pi r^3$

Dividing both sides by pi

$\frac{3}{4\pi}V =\frac{ \pi r^3}{\pi}\\\frac{3V}{4\pi} = r^3$

Taking cube root on both sides

$\sqrt[3]{\frac{3V}{4\pi} } = \sqrt[3]{r^3} \\ r = \sqrt[3]{\frac{3V}{4\pi} }$

Hence,

Option C: $r = \sqrt[3]{\frac{3V}{4\pi} }$ is the right answer

Keywords: Volume, Sphere

Learn more about volume at:

#LearnwithBrainly

5 0

3 years ago

Which fraction is equivalent to 1/2 • 4/3 divided by 5/6 ?

Katen [24]

I believe the answer you are looking for is 4/5.

Solution:

1/2 • 4/3=2/3

2/3÷5/6=2/3 • 6/5

4/5

5 0

2 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

3 years ago

How do you do ratio charts

Alla [95]

Write several instances of the ratio,then plot the ratios as points on the graph

8 0

3 years ago

Read 2 more answers

7-C=4<br><br><br> Hint: equation and rational coefficient

nalin [4]

Answer:

you stupid and dumb person

5 0

3 years ago