1. **State the problem:** We need to order the given expressions from least to greatest. 2. **List the expressions:** - $\frac{6\pi}{6}$ - $\log_3(21)$ - $\sum_{i=5}^{10} i$ - $e^3$ - $\sqrt{17}$ - $6!$ - $\frac{6}{11}$ - $\int_2^6 x \, dx$ - $\infty$ 3. **Evaluate each expression:** - $\frac{6\pi}{6} = \pi \approx 3.1416$ - $\log_3(21)$: Since $3^2=9$ and $3^3=27$, $\log_3(21)$ is between 2 and 3. More precisely, $\log_3(21) \approx 2.77$ - $\sum_{i=5}^{10} i = 5 + 6 + 7 + 8 + 9 + 10 = 45$ - $e^3$: Since $e \approx 2.718$, $e^3 \approx 2.718^3 \approx 20.085$ - $\sqrt{17} \approx 4.123$ - $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ - $\frac{6}{11} \approx 0.545$ - $\int_2^6 x \, dx$: Use the formula for definite integral of $x$: $$\int_a^b x \, dx = \left. \frac{x^2}{2} \right|_a^b = \frac{b^2}{2} - \frac{a^2}{2}$$ Calculate: $$\int_2^6 x \, dx = \frac{6^2}{2} - \frac{2^2}{2} = \frac{36}{2} - \frac{4}{2} = 18 - 2 = 16$$ - $\infty$ is infinitely large. 4. **Order the values from least to greatest:** $$\frac{6}{11} \approx 0.545 < \pi \approx 3.1416 < \log_3(21) \approx 2.77 < \sqrt{17} \approx 4.123 < \int_2^6 x \, dx = 16 < e^3 \approx 20.085 < \sum_{i=5}^{10} i = 45 < 6! = 720 < \infty$$ Note: Correct the order of $\log_3(21)$ and $\pi$ since $\log_3(21) \approx 2.77$ is less than $\pi \approx 3.1416$. So the correct order is: $$\frac{6}{11} \approx 0.545 < \log_3(21) \approx 2.77 < \pi \approx 3.1416 < \sqrt{17} \approx 4.123 < \int_2^6 x \, dx = 16 < e^3 \approx 20.085 < \sum_{i=5}^{10} i = 45 < 6! = 720 < \infty$$