#### Answer

All real numbers.

#### Work Step by Step

The domain of $g(t)=\sqrt {t^2+9}$ is the set of all real numbers $t$ for which $\sqrt{t^2+9}$ is defined as a real number. We know the sqare root of a negative number is not defined, so the domain of $g(t)=\sqrt{t^2+9}$ is all real numbers $t$ for which $t^2+9 \geq 0$.
We know $t^2 \geq 0$ for all real numbers $t,$ and, therefore, $$t^2 + 9 \geq t^2 \geq 0$$ for all real nmbers $t$, which gives $$t^2 +9 \geq 0$$ for all real numbers $t$. Thus the domain of this function is all real numbers.
In set notation, the domain is $\{t:t\in R\}.$
In interval notation, the domain is $(-\infty , \infty).$