Evaluate $\left\lceil\left(-\frac{5}{3}\right)^2\right\rceil$.
I know that $\left\lceil x \right\rceil$ is the smallest integer that is greater than or equal to $x$, also called the ceiling function. So I need to find the smallest integer that is greater than or equal to $\left(-\frac{5}{3}\right)^2$. To do that, I first need to evaluate $\left(-\frac{5}{3}\right)^2$. I know that negative numbers squared become positive, so I can ignore the negative sign and just square the fraction. Squaring a fraction means squaring both the numerator and the denominator, so I get $\left(-\frac{5}{3}\right)^2 = \frac{25}{9}$. Now I need to find the smallest integer that is greater than or equal to $\frac{25}{9}$. One way to do that is to divide 25 by 9 and see what the quotient and remainder are. I know that 9 goes into 25 two times with a remainder of 7, so 25 divided by 9 is 2 with a remainder of 7. That means that $\frac{25}{9} = 2 + \frac{7}{9}$. The fraction $\frac{7}{9}$ is between 0 and 1, so the smallest integer that is greater than or equal to $\frac{25}{9}$ is 3. Therefore, the answer is $\left\lceil\left(-\frac{5}{3}\right)^2\right\rceil = 3$.

