Question 203246
Given {{{x[s] = Ks^2/t}}}  So you can see that x varies directly with s^2 and inversely with t.

How does x change when s is doubled? 
{{{x[s] = Ks^2/t}}}   doubling s means "the new value for s is now 2 times the original s ... so use 2s
{{{x[2s] = K(2s)^2/t}}} 
{{{x[2s] = 4ks^2/t}}}
compare {{{x[s]}}} to {{{x[2s]}}} and see that {{{x[2s]/x[s]}}} =
{{{( 4Ks^2/t) / (Ks^2/t)}}}
{{{4}}} thus doubling s, makes x 4 times larger (makes sense since 2^2 = 4)


When both s and t are doubled?

{{{x[2s2t] = K(2s)^2/(2t)}}}

{{{x[2s2t] = 4Ks^2/(2t)}}}
{{{x[2s2t] = 2Ks^2/t}}}
compare {{{x[s]}}} to {{{x[2s2t]}}} and see that {{{x[2s2t]/x[s]}}} =
{{{( 2Ks^2/t) / (Ks^2/t)}}}
{{{2}}} thus doubling both s and t, doubles x