Question: Select all the squares in order of least to greatest
$\infty$
$\frac{6\pi}{2}$
$\sum_{i=2}^4 i$
$\int_2^8 x\,dx$
$\sqrt{10}$
$5!$
$\frac{11}{12}$
$e^4$
$\log_2(4)$
top-left: $\infty$
top-center: $\frac{6\pi}{2}$
top-right: $\sum_{i=2}^4 i$
center-left: $\int_2^8 x\,dx$
center: $\sqrt{10}$
center-right: $5!$
bottom-left: $\frac{11}{12}$
bottom-center: $e^4$
bottom-right: $\log_2(4)$
LEAST TO GREATEST
1. **State the problem:** We need to order the given mathematical expressions from least to greatest.
2. **Evaluate each expression:**
- $\infty$ is infinite, so it is the greatest.
- $\frac{6\pi}{2} = 3\pi \approx 3 \times 3.1416 = 9.4248$
- $\sum_{i=2}^4 i = 2 + 3 + 4 = 9$
- $\int_2^8 x\,dx = \left[ \frac{x^2}{2} \right]_2^8 = \frac{8^2}{2} - \frac{2^2}{2} = \frac{64}{2} - \frac{4}{2} = 32 - 2 = 30$
- $\sqrt{10} \approx 3.1623$
- $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
- $\frac{11}{12} \approx 0.9167$
- $e^4 \approx 54.5982$
- $\log_2(4) = 2$ because $2^2 = 4$
3. **Order from least to greatest:**
$\frac{11}{12} \approx 0.9167 < \log_2(4) = 2 < \sqrt{10} \approx 3.1623 < \sum_{i=2}^4 i = 9 < \frac{6\pi}{2} \approx 9.4248 < \int_2^8 x\,dx = 30 < e^4 \approx 54.5982 < 5! = 120 < \infty$
4. **Final answer:**
$$\frac{11}{12} < \log_2(4) < \sqrt{10} < \sum_{i=2}^4 i < \frac{6\pi}{2} < \int_2^8 x\,dx < e^4 < 5! < \infty$$