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Dovator [93]
3 years ago
11

Evaluate 3/7r + 5/8s when r = 14 and s = 8.

Mathematics
1 answer:
Ahat [919]3 years ago
6 0
In this question, plug in 14 for r and 8 for s, which will look like 3/98 + 5/64. Solve, and you will get 0.109 = 0.11.
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A given line has the equation 10x + 2y = −2.
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Find the area of a square if its side length is: 3 inches
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zlopas [31]

9514 1404 393

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7 0
3 years ago
Find the flux of the vector field F = 〈e-z,4z,6xy) across the curved sides of the surface S = {(x,y,z): z= cos y, lys π, 0sxs4}
Len [333]

I'll go ahead and assume you meant to say that <em>S</em> is the surface given by

S = \left\{(x,y,z) \mid z = \cos(y)\text{ with } 0\le y\le \pi\text{ and }0\le x\le4\right\}

This immediately gives us a parameterization for the surface,

\vec r(x, y) = \left\langle x, y, \cos(y)\right \rangle

The upward-pointing normal vector to this surface is then

\vec n = \dfrac{\partial\vec r}{\partial x} \times \dfrac{\partial\vec r}{\partial y} = \left\langle0,\sin(y),1\right\rangle

Then the flux of \vec F(x,y,z) = \left\langle e^{-z}, 4z, 6xy\right\rangle across <em>S</em> is

\displaystyle \iint_S \vec F(x,y,z)\cdot\mathrm d\vec s = \int_0^4\int_0^\pi \vec F(x,y,\cos(y))\cdot\vec n\,\mathrm dy\,\mathrm dx \\\\ = \int_0^4\int_0^\pi \left\langle e^{-\cos(y)},4\cos(y),6xy\right\rangle \cdot \left\langle0,\sin(y),1\right\rangle \,\mathrm dy\,\mathrm dx \\\\ = \int_0^4\int_0^\pi (4\sin(y)\cos(y)+6xy)\,\mathrm dy\,\mathrm dx \\\\ = 2 \int_0^4\int_0^\pi (\sin(2y) + 3xy)\,\mathrm dy\,\mathrm dx = \boxed{24\pi^2}

8 0
3 years ago
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