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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
avatar+805 
+4

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
avatar+17 
+3

Thank you MathProblemSolver101smiley

MismatchedCubing  Jul 15, 2021
 #3
avatar+805 
+4

Anytime!

MathProblemSolver101  Jul 15, 2021
 #4
avatar+121065 
+3

Good work , MPS    !!!!!

 

 

cool cool cool

CPhill  Jul 15, 2021
 #5
avatar+17 
+4

My solution was this

 

 

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
 #6
avatar+805 
+1

I guess that works, too!

MathProblemSolver101  Jul 15, 2021

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