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3 years ago

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Mathematics

1 answer:

Natalija [7]3 years ago

7 0

Answer:

329 Kids and 151 Adults

Step-by-step explanation:

330 x 1.75= 577.50

150 x 2.50= 375.00

577.50+375.00= 952.50

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Makovka662 [10]

Answer:

1. all real numbers

2. y > 0

3. approaches y = 0

3 0

2 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:

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

4 0

3 years ago

Howard chose a candy from a bowl with 5 chocolate candies, 4 gummy candies and 6 hard candies. What is Howard's dependent probab

Hatshy [7]

Answer:

8.9%

Step-by-step explanation:

Here, we are to calculate the probability of Howard choosing a chocolate candy followed by a gummy candy.

The probability of selecting a chocolate candy = number if chocolate candy/ total number of candy

Total number of candy = 5 + 4 + 6 = 15

Number of chocolate candy = 5

The probability of selecting a chocolate candy = 5/15 = 1/3

The probability of selecting a gummy candy = number of gummy candies/total number of candies

Number of gummy candy = 4

The probability of selecting a gummy candy = 4/15

The probability of selecting a chocolate candy before a gummy candy = 1/3 * 4/15 = 4/45 = 0.088888888889

Which is same as 8.89 percent which is 8.9% to the nearest tenth of a percent

7 0

3 years ago

(X^3+3x^2-x+8)/(x-1)

ki77a [65]

Here is all of the work:

8 0

3 years ago

HELP ME WITH THIS QUESTION PLEASE

lukranit [14]

Answer:

The 3rd one is the correct answer.

Step-by-step explanation:

Hope this helps.

3 0

3 years ago

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