Step-by-step explanation:
[tex] \sqrt{10} = 3.2 \\ then \: the \: sign \: there \: is \: = \\ \sqrt{10 } = 3.2 \\ \\ \: \sqrt{12} = 3.5 \\then \: the \: sign \: there \: is \: = \\ \: \sqrt{12} = 3.5 \\ \\ 6 \frac{1}{3} = 6.3 \: and \: \sqrt{40 } = 6.3 \\ therefore \: 6 \frac{1}{3} = \sqrt{40} \\ \\ 2 \frac{2}{5} = 2.4 \: and \: \sqrt{5.76} = 2.4 \\ then \: \: 2 \frac{2}{5} = \sqrt{5.76} \\ \\ 5 \frac{1}{6} = 5.16 \: and \: 516 \: \: therefore \: \\ 5 \frac{1}{6} = 5.16 \\ \\ \sqrt{6.2} = 2.5 \\ and \: 2.4 \: therefore \: \\ \sqrt{6.2} > 2.4[/tex]
