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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=3\$.

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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 .

RektTheNoob  Dec 18, 2017

#1
+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

Guest Dec 18, 2017
edited by Guest  Dec 18, 2017
Sort:

#1
+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

Guest Dec 18, 2017
edited by Guest  Dec 18, 2017
#2
+241
+1

Thanks man

RektTheNoob  Dec 23, 2017

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