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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

Respuesta :

Answer:

• k=2.4

,

• z=114998.4

Explanation:

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

[tex]\begin{gathered} z=kx^3y^2 \\ k=\text{constant of proportionality} \end{gathered}[/tex]

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

[tex]\begin{gathered} 76.8=k\times2^3\times2^2 \\ 76.8=k\times8\times4 \\ 32k=76.8 \\ k=\frac{76.8}{32} \\ k=2.4 \end{gathered}[/tex]

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

[tex]z=2.4x^3y^2[/tex]

Finally, when y=6 and x=11:

[tex]\begin{gathered} z=2.4\times11^3\times6^2 \\ z=114998.4 \end{gathered}[/tex]

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