Answer:
[tex]\begin{equation} \sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6 \end{equation}[/tex]Explanation:
Given the irrational numbers:
[tex]$3 \sqrt{2}, \sqrt{3}-1, \sqrt{19}+1,6$, $2 \sqrt{10} \div 5,\sqrt{14}$[/tex]In order to arrange the numbers from the least to the greatest, we convert each number into its decimal equivalent.
[tex]\begin{gathered} 3\sqrt{2}=3\times1.414\approx4.242 \\ \sqrt{3}-1\approx1.732-1=0.732 \\ \sqrt{19}+1\approx4.3589+1=5.3589 \\ 6=6 \\ 2\sqrt{10}\div5=2(3.1623)\div5=1.2649 \\ \sqrt{14}=3.7147 \end{gathered}[/tex]Finally, sort these numbers in ascending order..
[tex]\begin{gathered} \sqrt{3}-1\approx1.732-1=0.732 \\ 2\sqrt{10}\div5=2(3.1623)\div5=1.2649 \\ \sqrt{14}=3.7147 \\ 3\sqrt{2}=3\times1.414\approx4.242 \\ \sqrt{19}+1\approx4.3589+1=5.3589 \\ 6=6 \end{gathered}[/tex]The given numbers in ascending order is:
[tex]\begin{equation} \sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6 \end{equation}[/tex]Note: In your solution, you can make the conversion of each irrational begin on a new line.
