1. **State the problem:** We need to order the given expressions from least to greatest. 2. **Evaluate each expression:** - $\sqrt{14} \approx 3.7417$ - $\int_2^9 x \, dx = \left[ \frac{x^2}{2} \right]_2^9 = \frac{9^2}{2} - \frac{2^2}{2} = \frac{81}{2} - 2 = 40.5 - 2 = 38.5$ - $\log_3(17)$: Since $3^2=9$ and $3^3=27$, $\log_3(17)$ is between 2 and 3. Using change of base formula: $$\log_3(17) = \frac{\ln 17}{\ln 3} \approx \frac{2.833}{1.099} \approx 2.577$$ - $\frac{4\pi}{5} \approx \frac{4 \times 3.1416}{5} = \frac{12.5664}{5} = 2.5133$ - $\frac{8}{13} \approx 0.6154$ - $\infty$ is infinitely large, so it is the greatest. - $\sum_{i=5}^7 i = 5 + 6 + 7 = 18$ - $e^3 \approx 2.7183^3 = 20.0855$ - $3! = 3 \times 2 \times 1 = 6$ 3. **Order from least to greatest:** $$\frac{8}{13} (0.6154) < \frac{4\pi}{5} (2.5133) < \log_3(17) (2.577) < \sqrt{14} (3.7417) < 3! (6) < \sum_{i=5}^7 i (18) < e^3 (20.0855) < \int_2^9 x \, dx (38.5) < \infty$$ 4. **Final answer:** $$\frac{8}{13} < \frac{4\pi}{5} < \log_3(17) < \sqrt{14} < 3! < \sum_{i=5}^7 i < e^3 < \int_2^9 x \, dx < \infty$$