Question

Suppose that z varies jointly with the cube of x and the square of y. Find the constant of proportionality k if z = 76.8 when y = 2 and x = 2. k= Using the k from above write the variation equation in terms of x and y. Z = Using the k from above find z given that y = 6 and x = 11. Z= If needed, round answer to 3 decimal places. Enter DNE for Does Not Exist, oo for Infinity

Answer 1

• k=2.4

,

• z=114998.4

Step-by-step explanation:

If z varies jointly with the cube of x and the square of y, then we have:

`\begin{gathered} z=kx^3y^2 \\ k=\text{constant of proportionality} \end{gathered}`

If z = 76.8 when y = 2 and x = 2:

`\begin{gathered} 76.8=k*2^3*2^2 \\ 76.8=k*8*4 \\ 32k=76.8 \\ k=(76.8)/(32) \\ k=2.4 \end{gathered}`

Next, we write an equation for z in terms of x and y:

`z=2.4x^3y^2`

Finally, when y=6 and x=11:

`\begin{gathered} z=2.4*11^3*6^2 \\ z=114998.4 \end{gathered}`