see http://www.mathhelpforum.com/math-he...ariations.html to see what "direct variation" and "inverse variation means"

let d be the distance the object falls

let t be the time it takes to fall

then we have,

d = k(t^2)

when t = 6, d = 1296

=> 1296 = k(6^2)

=> k = 1296/36 = 36

so our relationship is:

d = 36t^2

when d = 2304, we have:

2304 = 36t^2

=> t^2 = 2304/36 = 64

=> t = sqrt(64)

=> t = 8

so it takes 8 seconds for the object to fall 2304 feet

we have:2. x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled?

x = k(s^2) and x = k/t

if s is doubled:

x = k((2s)^2) = 4k(s^2)

so we see when s is doubled, x increases by a factor of 4

when t is doubled:

x = k/(2t) = (1/2)(k/t)

when t is doubled, x is halved. so if both s and t are doubled, x will increase by a factor of 4 and decrease by a factor of 2 at the same time. that means it will change by a factor of 4(1/2) = 2, so x will double