Given that x varies inversely as the cube of y and that x=24 when y=3,

Question

i) what is the relationship connecting x and y

ii) Find the value of x when y=2

iii) Find the value of y when x=10 1/8

Answer 1

x = k/y^3

24 = k/3^3 = k/27

k= 24(27) = 648

x =648/y^3

x = 648/2^3 = 648/8 = 81 when y = 3

10 1/8 = 81/8 = 648/y^3

y^3 = 648/(81/8) = 81(8/81) = 8

y = cube root of 8 = 2

y= 2 when x = 10 1/8

Actually there are 3 cube roots, one real, two imaginary

y^3 = 2

y^3-2 = 0

(y-2)(y^2 + 2y + 4) = 0

set each factor = 0 and solve for y

y=2

and

y^2 +2y + 4 = 0, use the quadratic formula

y =-2/2 + or - 1/2sqr(4-16) = -1 + or - isqr3

y = 2, -1+isqr3, or -1-isqr3

but all you probably want is y=2 when x = 10 1/8 = 10.125