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4 years ago

A box containing 20 books costs $297.20. Which equation can be used to find c, the cost per book?

2 answers:

6 0

We need to set up an equation.

297.20 is our total, so it goes after the equal sign.
c is the amount per book.
20 is the amount of books.

297.20 = 20c

Now we need to isolate the variable to solve for c.

Do the inverse operation.

20c = 297.20
20c/20 = 297.20/20
c= 297.20/20
c= 14.86

Our final answer: Each book costs 14.86 dollars

Fynjy0 [20]4 years ago

3 0

C=297.20/20

Each book costs $14.86, assuming that each book costs the same

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Ashley is packing her bags for her vacation. She has 8 unique Fabergé eggs, but only 5 fit in her bag. How many different groups

irakobra [83]

Answer:

56 groups of 5 Fabergé eggs can be taken.

Step-by-step explanation:

It is given that Ashley is packing her bags for her vacation. She has 8 unique Fabergé eggs, but only 5 fit in her bag. In order to find how many different groups of 5 Faberge' eggs can she take, we apply the combination formula:

$8C_{5}=\frac{8!}{5!(8-5)!}$

$8C_{5}=\frac{8!}{5!3!}$

$8C_{5}=\frac{8{\times}7{\times}6{\times}5!}{5!{\times}3{\times}2}$

$8C_{5}=56$

Thus, 56 groups of 5 Fabergé eggs can be taken.

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4 years ago

Find the distance between the two numbers on a number line. Write your answer as a decimal.

dangina [55]

Answer:

-0.574

Step-by-step explanation:

Subtract the two numbers to get the solution

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3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

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4 years ago

Multiply 6/5 by the reciprocal of (-3/8)

weeeeeb [17]

Answer:

-48/15, simplified to -16/5

Step-by-step explanation:

The reciprocal of -3/8 is -8/3. So the equation will be 6/5 x -8/3. Multiply numerators then multiply denominators to get -48/15. This can simplify to -16/5 by dividing both by 3.

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