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 = ?
Step-by-step explanation:
z ∝ √x / y³
z = k √x / y³
when x = 36 and y = 9 then z = 61
![z=\text{ }\frac{k\text{ }√(x)}{y^3}](https://img.qammunity.org/2023/formulas/mathematics/college/fm67801dgka8uws5ke4ud1ka2tec0ydxsa.png)
![\begin{gathered} 61=\frac{k\text{ * }√(36)}{9^3} \\ 61=(k*6)/(729) \\ k=(61*729)/(6)=(61*243)/(2)=(14823)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dqpfkbcuddg84di67065ad9ws6499yndvi.png)
when x = 64 and y = 6
![z=\text{ }(14823)/(2)*(√(64))/(6^3)=(14823*8)/(2*216)=(4941*4)/(72)=(4941)/(18)=274.5](https://img.qammunity.org/2023/formulas/mathematics/college/yhnizjmakndtkddgqdj37bmd4vx1zem38n.png)
the value of z = 274.5
final answer:
z = 274.5 ≈ 300 when rounded off to the nearest hundredth