select all the squares in order of least to greatest\n∞\n$e^{3}$\n$\\int_{1}^{4} x d x$\n$3!$\n$\\frac{6…

select all the squares in order of least to greatest\n∞\n$e^{3}$\n$\\int_{1}^{4} x d x$\n$3!$\n$\\frac{6 \\pi}{2}$\n$\\sqrt{10}$\n$\\sum_{i = 1}^{3} i$\n$\\frac{2}{8}$\n$\\log_{3}(12)$
Answer
Explanation:
Step1: Calculate each value
- $\frac{2}{8}=\frac{1}{4} = 0.25$
- $\log_{3}(12)=\frac{\ln(12)}{\ln(3)}\approx2.26$
- $\sqrt{10}\approx3.16$
- $\sum_{i = 1}^{3}i=1 + 2+3 = 6$
- $3!=3\times2\times1 = 6$
- $\int_{1}^{4}xdx=\frac{1}{2}x^{2}\big|_{1}^{4}=\frac{1}{2}(4^{2}-1^{2})=\frac{1}{2}(16 - 1)=\frac{15}{2}=7.5$
- $\frac{6\pi}{2}=3\pi\approx9.42$
- $e^{3}\approx20.09$
- $\infty$ is the largest.
Step2: Order the values
Ordering from least to greatest: $\frac{2}{8},\log_{3}(12),\sqrt{10},\sum_{i = 1}^{3}i,3!,\int_{1}^{4}xdx,\frac{6\pi}{2},e^{3},\infty$
Answer:
- $\frac{2}{8}$
- $\log_{3}(12)$
- $\sqrt{10}$
- $\sum_{i = 1}^{3}i$
- $3!$
- $\int_{1}^{4}xdx$
- $\frac{6\pi}{2}$
- $e^{3}$
- $\infty$