### Instruction: Evaluate $\left\lceil {-\!\sqrt{23}}\right\rceil$.### Response: I need to find the smallest integer that is greater than or equal to $-\!\sqrt{23}$.
I know that $\sqrt{23}$ is between 4 and 5, since $4^2 = 16$ and $5^2 = 25$.
So, $-\!\sqrt{23}$ is between -5 and -4, and the smallest integer that is greater than or equal to it is -4.
Therefore, $\left\lceil {-\!\sqrt{23}}\right\rceil = -4$.