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daser333 [38]
3 years ago
11

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:

B)

Step-by-step explanation:

EleoNora [17]3 years ago
6 0

Answer:

C.

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

4 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
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:

3 0
3 years ago
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