Given: z directly varies as √x and inversely varies as y³
when x = 36 and y = 9 then z = 61
To find:
when x = 64 and y = 6 then z = ?
explanation:
z ∝ √x / y³
z = k √x / y³
when x = 36 and y = 9 then z = 61
[tex]z=\text{ }\frac{k\text{ }\sqrt{x}}{y^3}[/tex][tex]\begin{gathered} 61=\frac{k\text{ * }\sqrt{36}}{9^3} \\ 61=\frac{k*6}{729} \\ k=\frac{61*729}{6}=\frac{61*243}{2}=\frac{14823}{2} \end{gathered}[/tex]
when x = 64 and y = 6
[tex]z=\text{ }\frac{14823}{2}*\frac{\sqrt{64}}{6^3}=\frac{14823*8}{2*216}=\frac{4941*4}{72}=\frac{4941}{18}=274.5[/tex]
the value of z = 274.5
final answer:
z = 274.5 ≈ 300 when rounded off to the nearest hundredth
