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)
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
7 answers
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
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
I'm not quite sure what to make of that. Does it mean I'm correct or that a doubles?
It was 1 k and now it is 2 k
Gotcha. Thank you very much :)
You are welcome.
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