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# Maybe A Little bit of help

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The values of  and  are always positive, and  and  vary inversely. If  is 10 when  is 2, then find  when  is 4000.

Jul 15, 2021

#1
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The values of $x$ and $y$ are always positive, and $x^2$ and $y$ vary inversely. If $y$ is 10 when $x$ is 2, then find $x$ when $y$ is 4000.

$y = \frac{k}{x^2}, k \in \mathbb{R}$

$k = x^2 \cdot y = 2^2 \cdot 10 = 40$

$y = \frac{40}{x^2}$

$4000 = \frac{40}{x^2}$

$100 = \frac{1}{x^2}$

$100x^2 = 1$

$x^2 = \frac{1}{100}$

$x = \frac{1}{10}$

Jul 15, 2021
#2
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Thank you MathProblemSolver101

MismatchedCubing  Jul 15, 2021
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Anytime!

MathProblemSolver101  Jul 15, 2021
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Good work , MPS    !!!!!

CPhill  Jul 15, 2021
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Since  x^2 and  y are inversely proportional, their product is constant. So,

2^2 times 10=x^2 times 4000,

x=1/10

Jul 15, 2021
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I guess that works, too!

MathProblemSolver101  Jul 15, 2021