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3 years ago

What is the approximate circumference of a circle with a radius of 12 ft? Use 3.14 for pi?

Mathematics

2 answers:

Svetach [21]3 years ago

6 0

Answer:

c = 75.4 ft

Step-by-step explanation:

I did this based on my math! Hope this helps!

jeka57 [31]3 years ago

5 0

Answer:

75.36 ft

Step-by-step explanation:

circumference of a circle is 2πr

2(3.14)(12ft)= 75.36 ft

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Answer:

There are 20 booths and (38 - 20), or 18, tables

Step-by-step explanation:

Represent the number of tables with t and the number of booths with b.

We need to find the values of t and b.

(6 people/table)(t) + (4 people/booth)b = 188 (units are "people")

t + b = 38 (units are "seating units")

Solving the second equation for t, we get 38 - b = t.

Substitute 38 - b for t in the first equation:

(6 people/table)(38 - b) + (4 people/booth)b = 188

Then solve for b: 6(38) - 6b + 4b = 188, or:

228 - 2b = 188, or 2b = 228 - 188, or 2b = 40. Thus, b = 20 (booths)

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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

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

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

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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

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

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

$=\boxed{-6\pi}$

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

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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Step-by-step explanation:

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What is the approximate circumference of a circle with a radius of 12 ft? Use 3.14 for pi?​

What is the approximate circumference of a circle with a radius of 12 ft? Use 3.14 for pi?