Respuesta :

in short, (h o g)(a) is just h(    g(a)    ).

so what we can do is simply get g(a) first and then plug that in h(x).

[tex]\bf \begin{cases} g(x)&=2x\\ h(x)&=x^2+4\\ (h\circ g)(a)&=h(~~g(a)~~) \end{cases} \\\\\\ g(a)=2(a)\implies g(a)=2a \\\\\\ h(~~g(a)~~)\implies h(~~2a~~)=(2a)^2+4 \\\\\\ h(~~2a~~)=(2^2a^2)+4\implies h(~~2a~~)=4a^2+4[/tex]

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