Suppose that $a^2$ varies inversely with $b^3$. If $a=7$ when $b=3$, find the value of $a^2$ when $b=6$.
+36915 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
+4609 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 !
+2094 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
