Answer:
Explanation:[tex]\begin{gathered} g^{-1}(-1)=9 \\ \\ \text{This is the inerse} \end{gathered}[/tex][tex]h^{-1}(x)=\frac{x-10}{3}[/tex][tex]\begin{gathered} (h^{-1})(-5)=\frac{-5-10}{3}=-\frac{15}{3}=-5 \\ \\ (h\circ h^{-1})(-5)=3(-5)+10^{}_{} \\ =-15+10 \\ =-5 \end{gathered}[/tex]
