Question

The 1 to 1 functions G & H are defined as follows:

Answer 1

Answer:

Step-by-step explanation:
`\begin{gathered} g^(-1)(-1)=9 \\ \\ \text{This is the inerse} \end{gathered}`
`h^(-1)(x)=(x-10)/(3)`
`\begin{gathered} (h^(-1))(-5)=(-5-10)/(3)=-(15)/(3)=-5 \\ \\ (h\circ h^(-1))(-5)=3(-5)+10^{}_{} \\ =-15+10 \\ =-5 \end{gathered}`