Answer
$(f\circ h)(1)=-3$
Work Step by Step
$f(x)=x^{2}-6x+2;$ $h(x)=\sqrt{x}$
$(f\circ h)(1)$
First, find $h(1)$ by substituting $x$ by $1$ in $h(x)$:
$h(1)=\sqrt{1}=1$
Find $(f\circ h)(1)$ by substituting $x$ by $1$ in $f(x)$:
$(f\circ h)(1)=f(1)=1^{2}-6(1)+2=1-6+2=-3$