Question

y varies directly as the square of x. When x = 1, y = 8. Find x when y = 128.

Answer 1

\[\large{y=kx^2}\]

Answer 2

when x=1 and y=8\[\large{8=k(1)^2}\]\[\large{k=8}\]

Answer 3

so when y=128
\[128=8x\]\[x=\frac{128}{8}\]