\begin{tabular}{l} Find the domain for $$ (g \circ f)(x) $$. \ $$ \qquad f(x)=\sqrt{x-10} $$ and $$ g(x)=x^{2}+4 $$ \ \hline\end{tabular}

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To find the domain of the composite function $$(g \circ f)(x)$$, let's analyze each function step by step.

**Given:**
- $$ f(x) = \sqrt{x - 10} $$
- $$ g(x) = x^2 + 4 $$

**Step 1: Determine the Domain of $$ f(x) $$**
The function $$ f(x) = \sqrt{x - 10} $$ is defined when the expression inside the square root is non-negative:
$$
x - 10 \geq 0 \implies x \geq 10
$$
So, the domain of $$ f(x) $$ is:
$$
[10, \infty)
$$

**Step 2: Determine the Domain of $$ g(x) $$**
The function $$ g(x) = x^2 + 4 $$ is a polynomial and is defined for all real numbers:
$$
\text{Domain of } g(x) = (-\infty, \infty)
$$

**Step 3: Determine the Domain of $$ (g \circ f)(x) $$**
The composite function $$ (g \circ f)(x) = g(f(x)) $$ will be defined as long as $$ f(x) $$ is within the domain of $$ g(x) $$. Since $$ g(x) $$ accepts all real numbers and $$ f(x) $$ is defined for $$ x \geq 10 $$, the domain of $$ (g \circ f)(x) $$ is the same as the domain of $$ f(x) $$.

**Final Answer:**
$$
\boxed{\, [10,\ \infty) \,}
$$
The domain of $$(g \circ f)(x)$$ is all real numbers $$ x $$ such that $$ x \geq 10 $$.

Find the domain for .

and

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