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3 years ago

Which equation represents grants path? A. y=2-4xB. y=4-x/2C. y=6-x/4D. y=8-2x

Mathematics

2 answers:

andre [41]3 years ago

8 0

Answer:

Step-by-step explanation:

EleoNora [17]3 years ago

6 0

Answer:

Step-by-step explanation:

recall that the point-slope form of a linear equation is given by

y = mx + c

Rearranging so that it is in the same form as the answer choices, we get

y = c + mx

where m is the slope and c is the y-intercept

we note the following observations

1) The line goes from top left to bottom right, this means that the slope is negative, hence m has to be negative. (in our case, all the choices have negative x)

2) The line crosses the y-axis at y = 4, this means the y-intercept is 4 and hence c = 4.

If we look at the choices, only choice B satisfies both observations (that m is negative and c = 4)

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Mr. Britt is pouring a 36' x 58' concrete pad for a metal building. If concrete costs

Helga [31]

Answer:

Step-by-step explanation:

8 in × (1 ft)/(12 in) = ⅔ ft

36 ft × 58 ft × ⅔ ft = 1392 ft³

1392 ft³ × (1 yd³)/(27 ft³) ≅ 52 yd³

52 yd³ × $123/yd³ = $6396

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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:

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

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3 years ago

Select the expressions that are equivalent to 10 + 2x + 5y + 3x.

Ierofanga [76]

:D :D :D :D :D :D :D :D

3 0

3 years ago

I need help with This question no links or I will report you

brilliants [131]

Answer:

Below.

Step-by-step explanation:

So let's do this Divide both sides.

equals to,

y+2>2

Cancel out equal terms.

y>0.

5 0

2 years ago

Someone help me . what’s the infinite solution or just what it says help please make sure right .

Gala2k [10]

Answer:

yes it is right

Step-by-step explanation:

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3 years ago

Which equation represents grants path? A. y=2-4xB. y=4-x/2C. y=6-x/4D. y=8-2x​

Which equation represents grants path? A. y=2-4xB. y=4-x/2C. y=6-x/4D. y=8-2x