Question

z varies directly with x^2 and inversely with y. When x=2 and y=4, z=3.
what is the value of z when x=4 and y=9?

Answer 1

z varies directly with x^2 and inversely with y

means we have this equation

\[\Large z = \frac{kx^2}{y}\]

where k is some fixed number

Answer 2

Plug in x=2,y=4, z=3 and solve for k

\[\Large z = \frac{kx^2}{y}\]
\[\Large 3 = \frac{k(2)^2}{4}\]
\[\Large 3 = \frac{4k}{4}\]
\[\Large 3 = k\]
\[\Large k = 3\]

With me so far?