Question 1116932
If {{{e}}} varies directly as {{{r}}} and inversely as {{{d^2}}} (the square of d),
the relation is
{{{e=k*r*(1/d^2)}}} or {{{e=kd/r^2}}} , where {{{k}}} is a constant.
You know that when {{{r=8}}} and {{{d=5}}} , {{{e=32/25}}} .
Substituting those values into
{{{e=kd/r^2}}} ,
you get
{{{32/25=k*8/5^2}}} ,
which you can solve for {{{k}}} .
{{{32/25=k*8/5^2}}}
{{{32/25=k*8/25}}}
{{{32=k*8}}} 
{{{32/8=k}}}
{{{highlight(k=4)}}}
 
Now you know that the relation between {{{e}}} , {{{r}}} and {{{d^2}}} ,
or between {{{e}}} , {{{r}}} and {{{d}}} is
{{{e=4d/r^2}}} .
 
To calculate {{{r}}} when {{{e=16}}} and {{{d=3}}} ,
you substitute the known values into {{{e=4d/r^2}}} ,
and solve for {{{r}}} .
{{{16=4*3/r^2}}}
{{{r^2=4*3/16}}}
{{{r^2=3/4}}} , and as you know {{{r}}} is not a negative number,
{{{r=sqrt(3/4)}}}
{{{highlight(r=sqrt(3)/2)}}} .
 
When {{{r=5}}} and {{{e=5/6}}} , substituting into {{{e=4d/r^2}}} gives you
{{{5/6=4d/5^2}}}
{{{5/6=4d/25}}}
{{{25*5/(6*4)=d}}}
{{{highlight(d=125/24=5&5/24=approximately 5.21)}}}