a varies directly as the square of b and inversely as the cube of c. If a=158 when b=4 and c=8 , find a if b=6 and c=3 . Round your answer to two decimal places if necessary.

Answer 1

We know that a is associated with b² and 1/c³ with some constant being multiplied.

Together, this can be represented by the following equation: a = (k)(b²/c³), where k is a constant.

To solve for k, we can plug in a = 158, b = 4, and c = 8.

158 = (k)(4²/8³)

158 = (k)(16/512)

*multiply both sides by reciprocal*

k = 5056

Now we know that a = 5056(b²/c³).

We can plug in b = 6 and c = 3 to solve for a.

a = 5056(6²/3³) = 5056(36/27) ≈ 6741.333