Evaluate $\left\lceil\sqrt{\frac{9}{4}}\right\rceil+\left\lceil\frac{9}{4}\right\rceil+\left\lceil\left(\frac{9}{4}\right)^2\right\rceil$.
I know that $\sqrt{\frac{9}{4}}=\frac{3}{2}$, because $\left(\frac{3}{2}\right)^2=\frac{9}{4}$. So, rounding up $\frac{3}{2}$, I get $\left\lceil\frac{3}{2}\right\rceil=2$. I also know that $\frac{9}{4}$ is already a fraction in lowest terms, and it is greater than 2, so rounding it up, I get $\left\lceil\frac{9}{4}\right\rceil=3$. Finally, I know that $\left(\frac{9}{4}\right)^2=\frac{81}{16}$, which is greater than 5, but less than 6, so rounding it up, I get $\left\lceil\frac{81}{16}\right\rceil=6$. Now, I just need to add up these three numbers: $2+3+6=11$. 