Answer :
Answer:
[tex](g\circ f)(x)=4x^2-32x+58[/tex]Explanation:
We were given that:
[tex]\begin{gathered} f(x)=2x-10 \\ g(x)=x^2+4x-2 \\ (g\circ f)(x)=? \end{gathered}[/tex]We will solve for this as shown below:
[tex]\begin{gathered} (g\circ f)(x)=g\mleft(2x−10\mright) \\ =(2x-10)^2+4(2x-10)-2 \\ \left(2x-10\right)\left(2x-10\right)+4\left(2x-10\right)-2 \\ 2x(2x-10\rparen-10\left(2x-10\right)+4\left(2x-10\right)-2 \\ \text{Expanding the bracket, we have:} \\ 4x^2-20x-20x+100+8x-40-2 \\ 4x^2-40x+8x+100-42 \\ 4x^2-32x+58 \\ \\ \therefore4x^2-32x+58 \end{gathered}[/tex]