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{:[g(y)=(y-15)/(y)],[h(y)=-3y]:} Evaluate. h(g(5))=

g(y)=y−15yh(y)=−3y \begin{array}{l} g(y)=\frac{y-15}{y} \\ h(y)=-3 y \end{array} \newlineEvaluate.\newlineh(g(5))= h(g(5))=

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Q. g(y)=y−15yh(y)=−3y \begin{array}{l} g(y)=\frac{y-15}{y} \\ h(y)=-3 y \end{array} \newlineEvaluate.\newlineh(g(5))= h(g(5))=
  1. Evaluate g(5)g(5): First, we need to evaluate g(5)g(5). The function g(y)g(y) is defined as g(y)=y−15yg(y) = \frac{y - 15}{y}. Let's substitute yy with 55.\newlineg(5)=5−155g(5) = \frac{5 - 15}{5}\newlineg(5)=−105g(5) = \frac{-10}{5}\newlineg(5)=−2g(5) = -2
  2. Evaluate h(g(5))h(g(5)): Now that we have g(5)=−2g(5) = -2, we need to evaluate h(g(5))h(g(5)). Since we found that g(5)=−2g(5) = -2, we can substitute −2-2 into the function h(y)h(y), which is defined as h(y)=−3yh(y) = -3y.
    h(g(5))=h(−2)h(g(5)) = h(-2)
    h(−2)=−3×(−2)h(-2) = -3 \times (-2)
    h(−2)=6h(-2) = 6

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