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Suppose that a varies inversely with b^2. If a=9 when b=2, find the value of a when b=2 .

 Dec 18, 2017

Best Answer 

 #1
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+3

What is the second value of b? Is it 2 or 3 ?? Look at the title and the body of the question !

 

At any rate, I believe it goes like this:

a = k x 1/b^2, where k = constant

9 = k x 1/2^2 

k= 36 

 

a = 36 x 1/4 =9 when b=2, or

a = 36 x 1/9 =4 when b=3 

 Dec 18, 2017
edited by Guest  Dec 18, 2017
 #1
avatar
+3
Best Answer

What is the second value of b? Is it 2 or 3 ?? Look at the title and the body of the question !

 

At any rate, I believe it goes like this:

a = k x 1/b^2, where k = constant

9 = k x 1/2^2 

k= 36 

 

a = 36 x 1/4 =9 when b=2, or

a = 36 x 1/9 =4 when b=3 

Guest Dec 18, 2017
edited by Guest  Dec 18, 2017
 #2
avatar+474 
+3

Thanks man

RektTheNoob  Dec 23, 2017

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