Respuesta :

[tex]\bf \begin{cases} g(x)=x-3\\ h(x)=\sqrt{x} \end{cases}\qquad \begin{array}{llll} (g\circ h)(x)\implies g(~~h(x)~~)\\\\ (g\circ h)(25)\implies g(~~h(25)~~) \end{array} \\\\\\ h(25)=\sqrt{25}\implies h(25)=5\qquad thus\implies g(~~h(x)~~)=h(x)-3 \\\\\\ g(~~h(25)~~)=h(25)-3\implies g(~~5~~)=5-3[/tex]
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