a is directly proportional to m and n^2, and inversely proportional to y^3. If a=9 when m=2, n=8 and y=4, solve for a when m=6, n=7 and y=3.
I know the formula for direct and inverse proportions (y=kx and y=k/x), but I'm not entirely sure what to do with this problem. When I try solving individually for a k with a and m, a and n, and a and y, the 'k' values are different, so I don't know how to preceed.
a is directly proportional to m and n^2, and inversely proportional to y^3.
means a = k mn^2 / y^3
or a y^3 / (mn^2) = k a constant
9 (4^3) / (2(8^2)) = k
k = 4.5
then a (3)^3 / ( 6(7^2)) = k = 4.5 solve for 'a'