\( 1<\quad \) Suppose \( f(x)=\sqrt{x}+7 \) and \( g(x)=x^{2}+\pi \). If \( f(g(6))=8 \), what is the value of \( g(n) \) ? \( g(n)=\square \) (Simplify your answer. Type an integer or a fraction )

Let's first break down the given functions. We need to calculate \( g(6) \) first. Using the function \( g(x) = x^2 + \pi \): \[ g(6) = 6^2 + \pi = 36 + \pi \] Now, we plug \( g(6) \) into \( f(x) = \sqrt{x} + 7 \): \[ f(g(6)) = f(36 + \pi) = \sqrt{36 + \pi} + 7 \] According to the problem, we know that \( f(g(6)) = 8 \): \[ \sqrt{36 + \pi} + 7 = 8 \] Subtracting 7 from both sides gives: \[ \sqrt{36 + \pi} = 1 \] Squaring both sides results in: \[ 36 + \pi = 1 \] This leads to: \[ \pi = 1 - 36 = -35 \] However, this is incorrect since \( \pi \) is positive. Therefore, let's rewrite \( g(n) \) rather than making an assumption about \( n \) directly. We need to find \( g(n) \) in terms of \( n \): \[ g(n) = n^2 + \pi \] This gives us \( g(n) \) as \( n^2 + \pi\). To finalize, \( g(n) = \boxed{n^2 + \pi} \).