select all the squares in order of least to greatest \n\\(\\frac{9}{10}\\) \n\\(\\infty\\)…

Explanation:

Step1: Calculate each value

• $\frac{9}{10} = 0.9$
• $\infty$ is infinitely large, so it's the greatest (but we'll handle finite values first)
• $\sum_{i = 1}^{6}i=\frac{6(6 + 1)}{2}=21$ (using sum of arithmetic series formula $\sum_{i = 1}^{n}i=\frac{n(n + 1)}{2}$)
• $\frac{6\pi}{5}\approx\frac{6\times3.14}{5}\approx3.768$
• $\log_{3}(13)$: Since $3^{2}=9$, $3^{3}=27$, so $2\lt\log_{3}(13)\lt3$ (more precisely, $\log_{3}(13)\approx2.335$)
• $7!=7\times6\times5\times4\times3\times2\times1 = 5040$
• $\sqrt{5}\approx2.236$
• $\int_{1}^{7}xdx=\left[\frac{x^{2}}{2}\right]_{1}^{7}=\frac{7^{2}}{2}-\frac{1^{2}}{2}=\frac{49 - 1}{2}=24$

Step2: Order finite values from least to greatest

• $\frac{9}{10}=0.9$ (least)
• $\sqrt{5}\approx2.236$
• $\log_{3}(13)\approx2.335$
• $\frac{6\pi}{5}\approx3.768$
• $\sum_{i = 1}^{6}i = 21$
• $\int_{1}^{7}xdx = 24$
• $7!=5040$
• $\infty$ (greatest, but it's not a finite number in the usual ordering of real numbers; however, among finite values, the order is as above)

Now, ordering the squares (excluding $\infty$ first, then $\infty$ at last):

1. $\frac{9}{10}$
2. $\sqrt{5}$
3. $\log_{3}(13)$
4. $\frac{6\pi}{5}$
5. $\sum_{i = 1}^{6}i$
6. $\int_{1}^{7}xdx$
7. $7!$
8. $\infty$ (but since $\infty$ is not a real number with a finite value, if we consider only real numbers, the last finite one is $7!$, and $\infty$ is beyond all finite numbers)

Answer:

Order from least to greatest (finite values first, then $\infty$): $\boldsymbol{\frac{9}{10}}$, $\boldsymbol{\sqrt{5}}$, $\boldsymbol{\log_{3}(13)}$, $\boldsymbol{\frac{6\pi}{5}}$, $\boldsymbol{\sum_{i = 1}^{6}i}$, $\boldsymbol{\int_{1}^{7}xdx}$, $\boldsymbol{7!}$, $\boldsymbol{\infty}$