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3 years ago

Solve for x, m-3x=2x+p

Mathematics

1 answer:

just olya [345]3 years ago

5 0

<span>m-3x=2x+p
m-x=p (- 2x on both sides)
-x=p-m (- m on both sides)
answer: x=-p+m (divide -1 on both side)

</span>

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A fence completely surrounds a pool,that is 30 feet by 10 feet.what is the approximate length,in feet,of the fence

tamaranim1 [39]

Because the fence's dimensions is 30 feet by 10 feet, you can assume that one side of the pool is 30 feet while the other, non-congruent side, is 10 feet. The formula for perimeter, which you are trying to find, is 2L+2W=P. In this problem, 30 will be your L and 10 will be your W. Therefore, 2(30) +2(10)=80.

So 80 feet will be your answer.

4 0

3 years ago

What is 9 times 52 using mental math property

arsen [322]

Everyone works differently in mental meth.

To calculate 9*52, we note that 9=10-1, so mentally calculate
9(52)
=(10-1)(52)
=520-52 [ subtract 20 from each number]
=(520-20)-(52-20)
=500-32
=468.

Alternatively,
9*52=9*50+9*2=450+18=468

If you're good with multiplication and division,
9*52=3*3(52)=3*156=450+18=468

3 0

3 years ago

HELP I NEED HELP ASAP

tatiyna

Answer:

D) 8x²-172x+120

Step-by-step explanation:

Length=2(-2x+20)+6

Width=-2x+20

(-4x+40+6)(-2x+20)

(-4x+46)(-2x+20)

8x²-80x-92x+920

8x²-172x+920

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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

How do i write the opposite of one-third of a number is greater than 9

gtnhenbr [62]

1/3x<9. This is the....may be

7 0

3 years ago

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