1. The problem is to order the given numbers from least to greatest: $\frac{5}{10}$, $\sqrt{2}$, $2!$, $\log_4(20)$, $\sum_{i=1}^2 i$, $\frac{6\pi}{4}$, $\int_2^5 x\, dx$, $e^5$, and $\infty$. 2. Let's evaluate each expression: - $\frac{5}{10} = 0.5$ - $\sqrt{2} \approx 1.414$ - $2! = 2 \times 1 = 2$ - $\log_4(20)$ means the logarithm base 4 of 20. Since $4^2=16$ and $4^3=64$, $\log_4(20)$ is between 2 and 3. More precisely, $\log_4(20) = \frac{\ln(20)}{\ln(4)} \approx \frac{2.9957}{1.3863} \approx 2.16$ - $\sum_{i=1}^2 i = 1 + 2 = 3$ - $\frac{6\pi}{4} = \frac{3\pi}{2} \approx 4.712$ - $\int_2^5 x\, dx$ is the definite integral of $x$ from 2 to 5. Using the formula $\int_a^b x\, dx = \frac{b^2 - a^2}{2}$: $$\int_2^5 x\, dx = \frac{5^2 - 2^2}{2} = \frac{25 - 4}{2} = \frac{21}{2} = 10.5$$ - $e^5 \approx 148.413$ - $\infty$ is infinity, which is greater than all finite numbers. 3. Now, order the numbers from least to greatest: $$0.5 < 1.414 < 2 < 2.16 < 3 < 4.712 < 10.5 < 148.413 < \infty$$ 4. Writing the original expressions in order: $$\frac{5}{10} < \sqrt{2} < 2! < \log_4(20) < \sum_{i=1}^2 i < \frac{6\pi}{4} < \int_2^5 x\, dx < e^5 < \infty$$ Final answer: The squares in order from least to greatest are: $\frac{5}{10}$, $\sqrt{2}$, $2!$, $\log_4(20)$, $\sum_{i=1}^2 i$, $\frac{6\pi}{4}$, $\int_2^5 x\, dx$, $e^5$, $\infty$.