You have five numbers [tex]\sqrt{6},\ 2\dfrac{1}{2},\ -\sqrt{1},\ \dfrac{7}{2}, \ 3.\overline{3}.[/tex]
Among them there is one negative number [tex]-\sqrt{1},[/tex] this number is smallest.
Note that
[tex](\sqrt{6} )^2=6 \text{ and } \left(2\dfrac{1}{2}\right)^2=(2.5)^2=6.25.[/tex]
Since 6.25>6, then
[tex]2\dfrac{1}{2}>\sqrt{6}[/tex]
and these numbers are less than 3
Note that
[tex]\dfrac{7}{2}=3.5 \text{ and } 3.\overline{3}=3.333333....,[/tex]
then [tex]3.3333333...<3.5[/tex] and these numbers are greater than 3.
Therefore, the correct order from the smallest to the greatest is:
[tex]-\sqrt{1}, \quad \sqrt{6},\quad 2\dfrac{1}{2},\quad 3.\overline{3}, \quad \dfrac{7}{2}.[/tex]
