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3 years ago

Find the area of the shaded region. Show your work please

Mathematics

1 answer:

vitfil [10]3 years ago

4 0

So if the formula is A=(pi)r^2, find the area of the entire circle:
A=(pi)(13^2)=530.66cm
then find the area of the little circle:
A=(pi)(3^2)= 28.26cm
then subtract the little circle from the big circle:
530.66cm-28.26cm= 502.4cm
and there you go. Hope this helps!

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Sharon wants to buy a shirt that costs $50. the sales tax is 5%. how much is the sales tax? what is her total cost for the shirt

maria [59]

5% of $50 = 5/100 x 50 = $2.50
Total = 50 + 2.50 = $52.50

The saes tax is $2.50; her total cost of the shirt is $52.50

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3 years ago

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On Monday, the temperature at 10 a.m. at Sam’s house was –6° Fahrenheit. The

Alika [10]

Answer:

Step-by-step explanation:

to find the answer, subtract the second amount (2) from the first amount (-6), -6 - 2 = -8, the temperat decreased by 8

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3 years ago

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Find the measure of angle indicated

Rus_ich [418]

Answer:

there is no question

Step-by-step explanation:

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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

4 years ago

Select TWO values of x that are roots

Sauron [17]

$x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\y=x^2+3x+5\\\\a=1;\ b=3;\ c=5\\\\b^2-4ac=3^2-4(1)(5)=9-20=-11\\\\x_1=\dfrac{-3-\sqrt{-11}}{2\cdot1}=\dfrac{-3-\sqrt{-11}}{2}\\\\x_2=\dfrac{-3+\sqrt{-11}}{2}$

Answer: A and B.

3 0

3 years ago

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