#### Answer

(a) 5 is not in the domain of $h$, but is in the domain of $f$ and $g$.
(b) $f(5)=10$, $g(5)=0$

#### Work Step by Step

We test if $5$ is in the domain of the functions by plugging it in:
$f(x)=x^2-3x$
$f(5)=5^{2}-3(5)=25-15=10$
$g(x)=\frac{x-5}{x}$
$g(5)=\frac{5-5}{5}=\frac{0}{5}=0$
$h(x)=\sqrt{x-10}$
$h(5)=\sqrt{5-10}=\sqrt{-5}$
Since we don't get a real number for $h$, 5 is not in the domain of $h$ (but is in the domain of $f$ and $g$).
(b) We found the values in (a).