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{:[g(x)=(8x-1)/(x+4)],[h(x)=3x+10]:} Write (g@h)(x) as an expression in terms of x. (g@h)(x)=

g(x)=8x−1x+4h(x)=3x+10 \begin{array}{l} g(x)=\frac{8 x-1}{x+4} \\ h(x)=3 x+10 \end{array} \newlineWrite (g∘h)(x) (g \circ h)(x) as an expression in terms of x x .\newline(g∘h)(x)= (g \circ h)(x)=

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Q. g(x)=8x−1x+4h(x)=3x+10 \begin{array}{l} g(x)=\frac{8 x-1}{x+4} \\ h(x)=3 x+10 \end{array} \newlineWrite (g∘h)(x) (g \circ h)(x) as an expression in terms of x x .\newline(g∘h)(x)= (g \circ h)(x)=
  1. Understanding (g@h)(x)(g@h)(x): First, we need to understand what (g@h)(x)(g@h)(x) means. The notation (g@h)(x)(g@h)(x) represents the composition of the functions gg and hh, which means we need to substitute the output of h(x)h(x) into the function gg. So, we will first find h(x)h(x) and then substitute it into g(x)g(x).
  2. Finding h(x)h(x): Let's find h(x)h(x). The function h(x)h(x) is given by:\newlineh(x)=3x+10h(x) = 3x + 10
  3. Substituting h(x)h(x) into g(x)g(x): Now we will substitute h(x)h(x) into g(x)g(x) to find (g@h)(x)(g@h)(x). The function g(x)g(x) is given by:\newlineg(x)=8x−1x+4g(x) = \frac{8x - 1}{x + 4}\newlineSo, (g@h)(x)=g(h(x))=g(3x+10)(g@h)(x) = g(h(x)) = g(3x + 10)
  4. Performing the substitution: We will now perform the substitution: \newline(g@h)(x)=g(3x+10)=8(3x+10)−1(3x+10)+4(g@h)(x) = g(3x + 10) = \frac{8(3x + 10) - 1}{(3x + 10) + 4}
  5. Simplifying the expression: Next, we simplify the expression:\newline(g@h)(x)=24x+80−13x+14(g@h)(x) = \frac{24x + 80 - 1}{3x + 14}\newline(g@h)(x)=24x+793x+14(g@h)(x) = \frac{24x + 79}{3x + 14}

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