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luda_lava [24]
3 years ago
13

Please help ASAP!!! I’m struggling

Mathematics
2 answers:
Arada [10]3 years ago
8 0

Answer:

eat well don't forget to eat your dinner okay

kykrilka [37]3 years ago
4 0

Answer:

10

Step-by-step explanation:

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Solve for x.<br> –4.5(x – 8.9) = 12.6<br> OVEC<br> Enter your answer, as a decimal, in the box.
Gnoma [55]

Step-by-step explanation:

just take -4.5 the right side

3 0
3 years ago
How to tell if an equation is linear or nonlinear?
gogolik [260]
A linear equation will contain  a term in x  has the highest degree.  For example  y = 2x - 5 . A non linear will contain terms with exponents 2 or higher 
eg x^2 - 2x - 5  is nonlinear.
6 0
3 years ago
Sam raked 3 bags of leaves in 16 minutes. If he continues to work at the same rate, about how long will it take him to rake 5 ba
kicyunya [14]

Answer:

26 minutes and 40 seconds

Step-by-step explanation:

5 0
3 years ago
8. A square has side length 6 inches. A parallelogram has a base of 6 inches,
wlad13 [49]

Answer:

The height is 6 inches.

Step-by-step explanation:

Area of the square = length of a side squared.

This square has area 6^2 = 36 in^2.

Area of the parallelogram =  base * height

= 6h.

As the areas are equal:

6h = 36

h = 36/6 = 6 inches.

6 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
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