level 34: mathematics\nselect all the squares in order of least to greatest\n$int_{3}^{7} xdx$\n$\frac{2pi}{4…

Explanation:

Step1: Calculate $\int_{3}^{7}xdx$

Using the power - rule $\int xdx=\frac{1}{2}x^{2}+C$, we have $\int_{3}^{7}xdx=\left[\frac{1}{2}x^{2}\right]_{3}^{7}=\frac{1}{2}(7^{2}-3^{2})=\frac{1}{2}(49 - 9)=20$.

Step2: Simplify $\frac{2\pi}{4}$

$\frac{2\pi}{4}=\frac{\pi}{2}\approx1.57$.

Step3: Calculate $\sum_{i = 4}^{5}i$

$\sum_{i = 4}^{5}i=4 + 5=9$.

Step4: Evaluate $e^{2}$

$e^{2}\approx7.39$.

Step5: Evaluate $\log_{4}(24)$

Let $y = \log_{4}(24)$, then $4^{y}=24$, $y=\frac{\ln(24)}{\ln(4)}=\frac{\ln(3\times8)}{\ln(4)}=\frac{\ln(3)+3\ln(2)}{2\ln(2)}\approx\frac{1.099 + 3\times0.693}{2\times0.693}=\frac{1.099+2.079}{1.386}=\frac{3.178}{1.386}\approx2.29$.

Step6: Evaluate $\frac{3}{5}$

$\frac{3}{5}=0.6$.

Step7: Evaluate $3!$

$3!=3\times2\times1 = 6$.

Step8: Evaluate $\sqrt{9}$

$\sqrt{9}=3$.

Step9: Compare values

$\frac{3}{5}=0.6<\frac{2\pi}{4}\approx1.57<\log_{4}(24)\approx2.29<\sqrt{9}=3<3!<e^{2}\approx7.39<\sum_{i = 4}^{5}i = 9<\int_{3}^{7}xdx=20<\infty$.

Answer:

$\frac{3}{5},\frac{2\pi}{4},\log_{4}(24),\sqrt{9},3!,e^{2},\sum_{i = 4}^{5}i,\int_{3}^{7}xdx,\infty$