Respuesta :
[tex]\bf h(p)=p+7~\hspace{5em}\stackrel{\textit{when p = q+8}}{h(q+8)}~\hspace{7em}h(q+8)=[q+8]+7 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill h(q+8)=q+15~\hfill[/tex]
h(p) = p + 7
substitute p = q + 8
h(q + 8) = (q + 8) + 7 = q + 8 + 7 = q + 15