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2 years ago

If y varies inversely with the square of x, and y = 26 when x = 4, find y when x = 2.

Mathematics

1 answer:

eduard2 years ago

3 0

Answer:

D. 104

Step-by-step explanation:

$y \: \alpha \: \frac{1}{ {x}^{2} } \\ \\ y = \frac{k}{ {x}^{2} }$

when y is 26, x is 4:

$26 = \frac{k}{ {(4)}^{2} } \\ k = 416$

when x is 2:

$y = \frac{416}{ {x}^{2} } \\ \\ y = \frac{416}{ {(2)}^{2} } \\ y = 104$

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$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

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

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

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nadya68 [22]

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Since, The side length of each side = 25

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