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g(b)=5b-9 h(b)=(b-1)^(2) Evaluate. (h@g)(-6)=

g(b)=5b−9 g(b)=5 b-9 \newlineh(b)=(b−1)2 h(b)=(b-1)^{2} \newlineEvaluate.\newline(h∘g)(−6)= (h \circ g)(-6)=

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Full solution

Q. g(b)=5b−9 g(b)=5 b-9 \newlineh(b)=(b−1)2 h(b)=(b-1)^{2} \newlineEvaluate.\newline(h∘g)(−6)= (h \circ g)(-6)=
  1. Function Composition: Understand the composition of functions.\newlineThe composition of functions h∘g)(b) meanswefirstapply$gh\circ g)(b)\ means we first apply \$g to bb, and then apply hh to the result of g(b)g(b).
  2. Evaluate g(b)g(b): Evaluate g(b)g(b) when b=−6b = -6.
    g(b)=5b−9g(b) = 5b - 9
    g(−6)=5(−6)−9g(-6) = 5(-6) - 9
    g(−6)=−30−9g(-6) = -30 - 9
    g(−6)=−39g(-6) = -39
  3. Evaluate hh at g(−6)g(-6): Evaluate hh at the result of g(−6)g(-6).
    h(b)=(b−1)2h(b) = (b - 1)^2
    h(g(−6))=h(−39)h(g(-6)) = h(-39)
    h(−39)=(−39−1)2h(-39) = (-39 - 1)^2
    h(−39)=(−40)2h(-39) = (-40)^2
    h(−39)=1600h(-39) = 1600

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