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Orlov [11]
3 years ago
7

What is the center and radius of this circle

Mathematics
1 answer:
melamori03 [73]3 years ago
3 0
The center is the point (1,-3) while the radius is 3 units.
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Factor x^2 + 5x + 10 completley
k0ka [10]
This should be the answer.

5 0
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
Martin throws a ball straight up in the air. the equation h(t) = -16t^2 + 40t + 5 gives the height of the ball, in feet, t secon
Lisa [10]

Answer:

2.62 seconds

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> he height of the ball, in feet

we have

h(t)=-16t^2+40t+5

we know that

When the ball hits the ground, the height is equal to zero

so

-16t^2+40t+5=0

The formula to solve a quadratic equation of the form

at^{2} +bt+c=0

is equal to

t=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-16t^2+40t+5=0

so

a=-16\\b=40\\c=5

substitute in the formula

t=\frac{-40\pm\sqrt{40^{2}-4(-16)(5)}} {2(-16)}

t=\frac{-40\pm\sqrt{1,920}} {-32}

t=\frac{-40+\sqrt{1,920}} {-32}=-0.12\ sec

t=\frac{-40-\sqrt{1,920}} {-32}=2.62\ sec

therefore

The solution  is t=2.62 seconds

4 0
2 years ago
What is (6*2)+12*2/4
pishuonlain [190]

Answer:

(6)(2)+

(12)(2)

4

=18

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the inequality shown?<br> n<br> -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
Keith_Richards [23]

Answer:

0 answer is yes.........

5 0
2 years ago
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