Question

If a varies directly as the square of b and inversely as the cube of c , how does the value of a change when the values of both b and c are halved?

a remains the same

a is halved

a doubles

a is 25% of its original value

Answer 1

a varies directly as the square of b and inversely as the cube of c
original y = k(b^2)/c^3

b is halved ---> b/2, c is halved ---> c/2
new y = k(b^2/4) / (c^3/8)
= k(b^2)/4 * 8/c^3 = 2 [k b^2/c^3] = 2(original y)

Answer 2

a = k b^2/c^3
first let b = c = 1
then a = k (1)/(1) = k

now let b = c = 1/2
then b^2 = 1/4
and c^3 = 1/8
so
a = k (1/4) /(1/8) = k (8/2) = 2 k

Answer 3

I'm not quite sure what to make of that. Does it mean I'm correct or that a doubles?

Answer 4

It was 1 k and now it is 2 k

Answer 5

Gotcha. Thank you very much :)

Answer 6

You are welcome.

Answer 7

Batman catches Ms.Quinn