1. **State the problem:** We need to evaluate each expression and then organize the results from least to greatest. 2. **Evaluate each expression:** - $\sqrt{19} \approx 4.3589$ - $\infty$ (infinity, considered largest) - $4! = 4 \times 3 \times 2 \times 1 = 24$ - $\sum_{i=1}^2 i = 1 + 2 = 3$ - $\log_2(12)$: Since $2^3=8$ and $2^4=16$, $\log_2(12)$ is between 3 and 4. More precisely, $\log_2(12) = \frac{\ln(12)}{\ln(2)} \approx \frac{2.4849}{0.6931} \approx 3.585$ - $\frac{3}{5} = 0.6$ - $e^5$: Using $e \approx 2.7183$, $e^5 \approx 2.7183^5 \approx 148.413$ - $\frac{4\pi}{2} = 2\pi \approx 6.283$ - $\int_3^9 x \, dx$: Use the formula $\int_a^b x \, dx = \frac{b^2 - a^2}{2}$ $$\int_3^9 x \, dx = \frac{9^2 - 3^2}{2} = \frac{81 - 9}{2} = \frac{72}{2} = 36$$ 3. **List all values:** - $\frac{3}{5} = 0.6$ - $\sum_{i=1}^2 i = 3$ - $\log_2(12) \approx 3.585$ - $\sqrt{19} \approx 4.3589$ - $\frac{4\pi}{2} \approx 6.283$ - $4! = 24$ - $\int_3^9 x \, dx = 36$ - $e^5 \approx 148.413$ - $\infty$ (largest) 4. **Order from least to greatest:** $$\frac{3}{5} < \sum_{i=1}^2 i < \log_2(12) < \sqrt{19} < \frac{4\pi}{2} < 4! < \int_3^9 x \, dx < e^5 < \infty$$ **Final answer:** $$0.6 < 3 < 3.585 < 4.3589 < 6.283 < 24 < 36 < 148.413 < \infty$$