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pogonyaev
3 years ago
13

5 of 8

Mathematics
2 answers:
Marina86 [1]3 years ago
4 0

Answer:

51 because you find the area of both rectangles then add them

GuDViN [60]3 years ago
4 0

Answer:

51m^2

Step-by-step explanation:

First, separate the shape into two, the bottom half of the shape would have an area of 18 square meteres because the sides are 6 meters and 3 meters.  You would find the area by multiplying those two sides.

For the top half of the shape, you would multiply the two sides as well and add teh totals together. The one side is clearly 3 meters, but to find the otehr side you would add 5 + 3 + 3 which would give you 11.

11 x 3 = 33

6 x 3 = 18

18 + 33 = 51

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I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

so we first compute the curl:

\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

Then parameterize S by

\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

Then the line integral is equal in value to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}

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3 years ago
What is the greatest common factor of 180 and 240
maria [59]
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4 years ago
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soldier1979 [14.2K]

Answer:

10, 50, 150

Step-by-step explanation:

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A horse farm is developing a new indoor arena in a rectangular buliding. They want the width of the building to be 3 wide feet a
erastovalidia [21]

Answer:

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Step-by-step explanation:

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