if y varies directly as the cube of x, what is the value of y in these ordered pairs? (4, 16) and (2, y)

Given, y varies directly as the cube of x.

So, [tex] y=kx^3 [/tex]

By using the order pair (4, 16) we can get x=4 and y=16.

Now plug in these values in the above equation.

So, 16=k*4^3

16 = k * 64

16/64 = k (By dividing each sides by 64).

So, k=1/4

So, the equation for this variation is y = 1/4*x^3

Now we need to find the value of y for x=2.

So, plug in x=2 in the above equation. Hence,

y=1/4 * 2^3

=1/4*8

=8/4

=2

So, y=2