If x varies directly with the square root of y and inversely as the cube of z, and if x=6 when y=9 and z=3, find the values of x when y=16 and z=4

Question

If x varies directly with the square root of y and inversely as the cube of z, and if x=6 when y=9 and z=3, find the values of x when y=16 and z=4

Answer 1

This problem can be solved in two steps as follows:

1: you need to get rid of proportionality by identifying the proportionality constant

x p Sqrt(y) (here p is the proportional sign)

x p (1/z3)

combining these two will give

x p sqrt(y)/z3

x = k sqrt(y)/z3 (here k is the proportionality sign)

k = x * z3/ sqrt(y)

k = 6 *3*3*3/sqrt(16)

Now: use this value of k for the second step

x = k * sqrt(16)/4*4*4

Use value of k obtained in the first step to get numerical value of x.

I hope this helps.