seymourangela
Answered

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an u please help me out​

An U Please Help Me Out class=

Sagot :

r3t40

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 :)

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