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2 years ago

Question 3-17 i really stink at math :( it just doesnt make sense to me. alsi im failing. plz help

Mathematics

1 answer:

Trava [24]2 years ago

6 0

Answer:

Step-by-step explanation:

graph is a sad face, therefore a is negative. so 1 and 3 are out. the c is the y intercept, and the y intercept is at tye positve side. so 4 is the correct one.

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2 years ago

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Step-by-step explanation:

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3 years ago

Help this is due tomorrow but she didn’t teach us how.

mario62 [17]

Answer:

see below for drawings and description

Step-by-step explanation:

For geometry problems involving translation, rotation, and reflection—transformations that change location, but not size ("rigid" transformations)—it might be helpful for you to trace the image onto tracing paper or clear plastic so that you can manipulate it in the desired way. Eventually, you'll be able to do this mentally, without the aid of a physical object to play with.

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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:

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

4 0

3 years ago

Please help ASAP with this question

jeka57 [31]

Answer:

B. 7.9 in^2

Step-by-step explanation:

Part I.

Finding the fraction of the circle that is the sector

To start, let’s find the fraction of the circle that the sector covers. Since the measure of the central angle is 36, we get 36/360 or 1/10 of the circle.

Part II.

Finding the area of the Circle

The area of a circle is pi*r^2. Since r=5, then the area is 25pi.

Part III.

Finding the area of the sector.

Since the sector is 1/10 of the circle, its area is 1/10 of the circle’s area. So the area of the sector is 25/10*pi or (5/2)pi. This is approximately 7.9 in^2.

6 0

3 years ago