1. The problem is to order the given expressions from least to greatest. 2. The expressions are: $\sqrt{15}$, $2!$, $e^4$, $\frac{5}{9}$, $\int_1^8 x \, dx$, $\log_3(26)$, $\infty$, $\frac{6\pi}{5}$, and $\sum_{i=5}^7 i$. 3. Calculate each value: - $\sqrt{15} \approx 3.873$ (since $15$ is between $9$ and $16$) - $2! = 2 \times 1 = 2$ - $e^4 \approx 54.598$ (using $e \approx 2.718$) - $\frac{5}{9} \approx 0.555$ - $\int_1^8 x \, dx = \left[ \frac{x^2}{2} \right]_1^8 = \frac{8^2}{2} - \frac{1^2}{2} = \frac{64}{2} - \frac{1}{2} = 32 - 0.5 = 31.5$ - $\log_3(26)$: since $3^3=27$, $\log_3(26) \approx 2.965$ - $\infty$ is infinitely large - $\frac{6\pi}{5} \approx \frac{6 \times 3.1416}{5} = 3.7699$ - $\sum_{i=5}^7 i = 5 + 6 + 7 = 18$ 4. Now order them from least to greatest: $$\frac{5}{9} \approx 0.555 < 2! = 2 < \frac{6\pi}{5} \approx 3.77 < \sqrt{15} \approx 3.873 < \log_3(26) \approx 2.965 \text{ (reorder: actually } 2.965 < 3.77, so correct order is } 2 < 2.965 < 3.77 < 3.873)$$ Correct order: $$\frac{5}{9} < 2! < \log_3(26) < \frac{6\pi}{5} < \sqrt{15} < \sum_{i=5}^7 i < \int_1^8 x \, dx < e^4 < \infty$$ Numerically: $$0.555 < 2 < 2.965 < 3.77 < 3.873 < 18 < 31.5 < 54.598 < \infty$$ 5. Explanation: We approximated each expression to a decimal to compare easily. Factorials, integrals, sums, and logarithms were evaluated using their definitions or known approximations. Final answer: $$\frac{5}{9} < 2! < \log_3(26) < \frac{6\pi}{5} < \sqrt{15} < \sum_{i=5}^7 i < \int_1^8 x \, dx < e^4 < \infty$$