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zlopas [31]
3 years ago
15

Can anyone show me how to figure this out.

Mathematics
1 answer:
galina1969 [7]3 years ago
5 0

Answer:

Option A (5, 0)

Step-by-step explanation:

<u>Step 1:  Determine what is a solution</u>

To find a solution of two lines, it has to be in the area where both of the colors intersect.  As you can see the red line goes up and the blue line goes down, but there is a shaded region where there is just white, just blue, just red, and the red and blue.  The shaded region where both red and blue is, that is where the solutions are.  

(5, 0) -> This means that the x-value is 5 and the y-value is 0 meaning that it is in the area where there is both red and blue.

(1, -3) -> This means that the x-value is 1 and the y-value is -3 meaning that it is in the area where there is only red.

(3, 3) -> This means that the x-value is 3 and the y-value is 3 meaning that it is in the area where there is only blue.

(2, 1) -> This means that the x-value is 2 and the y-value is 1 meaning that it is in the area where there is only blue.

Answer:  Option A (5, 0)

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Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

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In calculus terms, we would first find an expression for the surface element, which is given by

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\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
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So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
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=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
3 years ago
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Answer:

b. the association is consistent when results are repeated in studies in different settings using different methods

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Step-by-step explanation:

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3 years ago
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tankabanditka [31]
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5 0
4 years ago
In ΔEFG, the measure of ∠G=90°, the measure of ∠E=6°, and GE = 71 feet. Find the length of EF to the nearest tenth of a foot.
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3 years ago
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