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