jmc

Since $\sqrt{9}<\sqrt{10}<\sqrt{16}$, we know that $\sqrt{10}$ is a number between $3$ and $4$. Therefore, the smallest integer that is greater than or equal to $\sqrt{10}$ is $4$ and the greatest integer that is less than or equal to $\sqrt{10}$ is $3$. So, $\lceil \sqrt{10}\rceil +\lfloor \sqrt{10}\rfloor =4+3=\boxed{7}$.