7. Let the set $A=\left\{n \left\lvert\, \frac{n}{3} \in \mathbf{N}_{+}\right.\right\}, B=\left\{y \mid y=x+4+\sqrt{5-x^{2}}\right\}$, then the number of elements in the set $A \cap B$ is
A. 1
B. 2
C. 3
D. 4
Let $x=\sqrt{5} \sin \theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $y=\sqrt{5} \sin \theta+\sqrt{5} \cos \theta+4=\sqrt{10} \sin \left(\theta+\frac{\pi}{4}\right)+4$, since $\theta+\frac{\pi}{4} \in\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right] \Rightarrow \sin \left(\theta+\frac{\pi}{4}\right) \in\left[-\frac{\sqrt{2}}{2}, 1\right]$,
thus $y \in[4-\sqrt{5}, 4+\sqrt{10}] \Rightarrow A \cap B=\{3,6\}$. Therefore, the answer is $B$.