Respuesta :

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]

RELAXING NOICE
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