+0  
 
+1
314
2
avatar+402 

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

Best Answer 

 #1
avatar
+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
 #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+402 
+3

Thanks man

RektTheNoob  Dec 23, 2017

19 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.