Question

m varies jointly with x, the cube of y, and the square root of z. m=160 when x=2 y=2 and z=4. Find m when x=3 y=3 and z=49.

Answer 1

Solution:

Given that:

m varies jointly with x, the cube of y, and the square root of z, The statement means

$m\propto xy^3\sqrt{z}$

We convert the expression above into an equation by introducing a constant, k,

It becomes

$m=kxy^3\sqrt{z}$

Where

$m=160,\text{ when x}=2,y=2\text{ and z}=4$

Substitute the values of variables into the equation above to find the constant, k.

$\begin{gathered} m=kxy^3\sqrt{z} \\ 160=k(2)(2^3)(\sqrt{4}) \\ 160=k(2\times8\times2) \\ 160=32k \\ Divide\text{ both sides by 32} \\ \frac{32k}{32}=\frac{160}{32} \\ k=5 \end{gathered}$

Thus, the equation becomes

$m=5xy^3\sqrt{z}$

To find the value of m when

$x=3,y=3\text{ and z}=49$

Substitute the values of the variables into the equation above

$\begin{gathered} m=5xy^3\sqrt{z} \\ m=5(3)(3^3)(\sqrt{49}) \\ m=5(3)(27)(7) \\ m=2835 \end{gathered}$

Hence, the value of m is 2835