Order Expressions Ab06Bb

1. **State the problem:** We need to order the following expressions from least to greatest: $\frac{5\pi}{6}$, $\log_{4}(12)$, $\sqrt{19}$, $4!$, $e^{3}$, $\frac{7}{13}$, $\infty$, $\sum_{i=5}^{6} i$, $\int_{3}^{10} x \, dx$ 2. **Calculate each value:** - $\frac{5\pi}{6} \approx \frac{5 \times 3.1416}{6} = 2.618$ - $\log_{4}(12)$: Change of base formula: $$\log_{4}(12) = \frac{\log_{10}(12)}{\log_{10}(4)} \approx \frac{1.07918}{0.60206} = 1.793$$ - $\sqrt{19} \approx 4.359$ - $4! = 4 \times 3 \times 2 \times 1 = 24$ - $e^{3} \approx 2.7183^{3} = 20.085$ - $\frac{7}{13} \approx 0.538$ - $\infty$ is infinitely large, so it will be the greatest. - $\sum_{i=5}^{6} i = 5 + 6 = 11$ - $\int_{3}^{10} x \, dx$: 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: $$\int_{3}^{10} x \, dx = \frac{10^{2}}{2} - \frac{3^{2}}{2} = \frac{100}{2} - \frac{9}{2} = 50 - 4.5 = 45.5$$ 3. **List all values:** - $\frac{7}{13} \approx 0.538$ - $\log_{4}(12) \approx 1.793$ - $\frac{5\pi}{6} \approx 2.618$ - $e^{3} \approx 20.085$ - $\sqrt{19} \approx 4.359$ - $\sum_{i=5}^{6} i = 11$ - $\int_{3}^{10} x \, dx = 45.5$ - $4! = 24$ - $\infty$ 4. **Order from least to greatest:** $\frac{7}{13} < \log_{4}(12) < \frac{5\pi}{6} < \sqrt{19} < \sum_{i=5}^{6} i < 4! < e^{3} < \int_{3}^{10} x \, dx < \infty$ Note: We must correct the order of $e^{3}$ and $4!$ and $\int_{3}^{10} x \, dx$ based on their values: - $4! = 24$ - $e^{3} = 20.085$ - $\int_{3}^{10} x \, dx = 45.5$ So the correct order is: $\frac{7}{13} < \log_{4}(12) < \frac{5\pi}{6} < \sqrt{19} < \sum_{i=5}^{6} i < e^{3} < 4! < \int_{3}^{10} x \, dx < \infty$