Algebra...!???!?

Question

"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?"

My dad said to do it i guess something like this to prove it...
(^2 = that means it's squared)

x = s^2
x = 2^2
x = 4
x = 4 times 4
x = 16 (Doubled?)
-------------------------
x = 1/t
x = 1/2
x = 1/2 times 1/2
x = 1/4 (Doubled?)

i'm not sure if this is what he ment but could you tell me what you all think?

First Off i'm Home-Schooled. Second off... The book i have to work with, says nothing about this sort of thing.... I've looked.... 1000000 times...

Answer 1

x varies with both x and t at the same time. they mus both appear in the equation.

x=(s^2)/t -- x varies as the square of s and the inverse of t

when s is doubled

([2s]^2)/t = (2^2 * s^2)/t = (4 * s^2)/t = 4 [(s^2)/t]

then x is quadrupled

when t is doubled

(s^2)/[2t] = [1/2] * [(s^2)/t] --- (the 2 comes out of the denominator as 1/2)

then x is halved


when bothe s and t are doubled then x is both quadrupled and halved

([2s]^2)/[2t] = (2^2 * s^2)/[2t] = (4 * s^2)/[2t] = [4 * 1/2] [(s^2)/t]

=2 [(s^2)/t]

so x is doubled!

Good Luck

Answer 2

x = s^2
x = 2^2
x = 4
x = 4 times 4
x = 16 (Doubled?) yes it is doubled in this case !
-------------------------
x = 1/t
x = 1/2
x = 1/2 divided by 1/2 gives 1
x = 1 is double of 1/2 (1/2 multiplied by 2 gives 1)

Answer 3

Take better notes in class

Answer 4

check with...................
directions in your math text
your dad
another student
your teacher