Question 1195509
<pre>

A 'gas tank' type problem...

The bowl has the shape {{{ y = -sqrt(100-x^2) }}}  and we can add 10 to get
the bottom of the bowl to sit exactly at the origin {{{ y = 10-sqrt(100-x^2) }}}.

So for water with radius 7:  {{{ y = -sqrt(100-7^2) + 10 = -sqrt(51)+10 }}}
which is approx 2.85857.

The height, h (green line), to one decimal place is therefore {{{ highlight(2.9cm) }}}

{{{ graph(400,400,-12,12,-12,12, y=-sqrt(100-x^2)+10, y=2.9 ) }}}