Question

The cube root of x varies inversely as the square of y. If x=27 when y =4, find the valye of x when y =6


18
64/27
27/4
4/27

Answer 1

\[\sqrt[3]{x}=\frac{ k }{ y^2 }\]
when x=27,y=4
\[\sqrt[3]{27}=\frac{ k }{ 4^2 },k=16*\left( 27 \right)^{\frac{ 1 }{ 3 }}=16*\left( 3^3 \right)^{\frac{ 1 }{ 3 }}=48\]
can you complete ?