an u please help me out
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Rewrite the root be,
[tex]\sqrt{39}=\sqrt{3\cdot13}=\sqrt{3}\sqrt{13}[/tex]
We know that [tex]\sqrt{3}\approx1.7,\sqrt{13}\approx3.6[/tex]
Write decimals as fractions and multiply them,
[tex]\frac{17}{10}\cdot\frac{36}{10}=\frac{612}{100}=6.12[/tex]
So it should be on somewhere around the next tick from 6.
Hope this helps :)