When we have a number that is jointly proportional to two other numebrs, the formula is:
![a=kcb](https://img.qammunity.org/2023/formulas/mathematics/college/dchn7ygjqzziz5dd9fuarssxzpky0e7z5t.png)
This means "a is jointly proportional to c and b with a factor of k"
Then, we need to find the factor k.
In this case z is jointly proportional to x² and y³
This is:
![z=kx^2y^3](https://img.qammunity.org/2023/formulas/mathematics/college/ll2p06nc5bq2viby8yqovo2083zefebdiq.png)
Then, we know that z = 115 when x = 8 and y = 3. We can write:
![115=k\cdot8^2\cdot3^3](https://img.qammunity.org/2023/formulas/mathematics/college/gztxcvlshr9n2lfnvc7oqk2rvaxhy4ndxu.png)
And solve:
![\begin{gathered} 115=k\cdot64\cdot27 \\ 115=k\cdot1728 \\ k=(115)/(1728) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/klkpua9v4kbaas3qymlena37wpedpj0g6d.png)
NOw we can use k to find the value of z when x = 5 and y = 2
![z=(115)/(1728)\cdot5^2\cdot2^3=(115)/(1728)\cdot25\cdot8=(2875)/(216)\approx13.31](https://img.qammunity.org/2023/formulas/mathematics/college/1o3gqgrzjcyeeolmz0e2em6giuqtcz5rdh.png)
To the nearest hundreth, the value of z when x = 5 and y = 2 is 13.31