If $g(x) = \sqrt{10 − x}$ What is the Domain for $(g\circ g)(x) = g(g(x))$? What is the domain for $10 − \sqrt{10 − x} ≥$ ?
The expression you need to compute the domain of is
$$
g(g(x)=\sqrt{10-g(x)}=\sqrt{10-\sqrt{10-x}}
$$
Thus you have two conditions to fulfill:
\begin{cases}
10-x\ge0\\[1ex]
10-\sqrt{10-x}\ge0
\end{cases}
The first condition is equivalent to $x\le 10$; the second one to
$$
\sqrt{10-x}\le 10
$$
that can be squared, because both sides are surely non negative:
$$
10-x\le 100
$$
So the answer is…
$\sqrt{10-x}\le10$ gives $x\ge-90$ but be carefull $x\le10$ must be hold.