Solution:
Let the following functions:
[tex]h(a)=a^2-5a[/tex]and
[tex]g(a)\text{ = 3a+1}[/tex]notice that the composition of the above function is :
[tex](h\circ g)(a)=h(g(a))=h(3a+1)=(3a+1)^2-5(3a+1)[/tex]this is equivalent to:
[tex](h\circ g)(a)=(3a+1)^2-5(3a+1)=(3a)^2+2(3a)+1\text{ -15a-5}[/tex]this is equivalent to:
[tex](h\circ g)(a)=9a^2+6a+1\text{ -15a-5}[/tex]putting together the similar terms, we get:
[tex](h\circ g)(a)=9a^2-9\text{a}-4[/tex]now, replacing a = -2 in the above equation, we get:
[tex](h\circ g)(-2)=9(4)^{}-9(-2)-4=\text{ 36+18-4 = 50}[/tex]then, we can conclude that the correct answer is:
[tex](h\circ g)(-2)\text{ = 50}[/tex]