[tex]\bf g(x)=x-3\qquad h(x)=x^2+6
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(h\circ g)(1)\implies h[\ g(1)\ ]\impliedby \textit{now, let's find g(1) first}
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g(1)=(1)-3\implies g(1)=\boxed{-2}\qquad thus
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h[\ g(1)\ ]\implies h\left( \boxed{ -2}\right)\implies h(-2)=(-2)^2+6
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h(-2)=4+6\implies h(-2)=10\impliedby (h\circ g)(1)[/tex]
