Question 1083573

If {{{x}}} varies indirectly as {{{y}}}, than

 {{{x=k/y}}}
 
 and if {{{y}}} varies directly as {{{z}}}, than 

{{{y=k*z}}}->{{{z=y/k}}}

so, the relationship between {{{x}}} and {{{z}}} is:

 {{{x/z=(k/y)/(y/k)}}}

 {{{x/z=k^2/y^2}}}

 {{{x=zk^2/y^2}}}->{{{x}}} varies directly as the  {{{z}}} and inversely as the square of {{{y}}}