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Suppose that $a^2$ varies inversely with $b^3$. If $a=7$ when $b=3$, find the value of $a^2$ when $b=6$.

 Mar 20, 2020
 #1
avatar+23553 
+2

Varies inversly  means   a2 ~ 1/b3     or more correctly    a2 = k/b3    where k is a 'proportionality constant'

 

Sub in the values given to find the value of 'k'

  (7)^2 = k/(3^3)

49 = k/27

k = 1323

 

So    a^2 = 1323/b^3     now sub in the question numbers to find a^2

         a^2 = 1323/(6^3)    = 6.125

 Mar 20, 2020
 #2
avatar+4569 
+2

According to the information given in the problem....we have a^2*b^3=y, y is a random constant.

 

Thus, we have 7^2*3^3=1323.

 

 

a^2*6^3=1323, 

 

a=6.125

 

 

agree with ElectricPavlov !

 Mar 20, 2020
 #3
avatar+1955 
0

Okay. Here is another way to put it.

 

a^2=k/b^3.

 

So that means that 49=k/27.

 

Therefore, k would be 1323.

 

So, a^2=1323/(6^3).

       a^2=1323/216.

 

Therefore, a^2 would be equal to 6.125

 Mar 20, 2020

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