Answer
$p(x)=(x+4)*\sqrt {10-x}$
with domain $x=10$
Work Step by Step
$h(x)=x+4$
$v(x)=\sqrt {10-x}$
$p(x)=h(x)*v(x)=(x+4)*\sqrt {10-x}$
The domain of $h(x)$ is $x\geq 10$
The domain of $v(x)$ is $[0,10]$
The only number that is both in the domain of $h$ and $v$ is $10$, so that will be the domain of $p$.
$1$ is not in the domain of $h$, so it is not in the domain of $p$. Therefore it is not defined in $a=1$
