Evaluate $\lfloor\sqrt{63}\rfloor$.
To evaluate $\lfloor\sqrt{63}\rfloor$, I need to find the largest integer that is less than or equal to $\sqrt{63}$. I know that $\sqrt{64} = 8$, so $\sqrt{63}$ must be slightly less than 8. Therefore, $\lfloor\sqrt{63}\rfloor$ must be 7, since that is the largest integer that is less than 8.

