How do you simplify #(2sqrt4)/(8sqrt3)#?

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Answer 1 · 2016-09-04T03:26:48

#sqrt3/6#

#(2sqrt4)/(8sqrt3) = (2xx2)/(8sqrt3) = 1/(2sqrt3)#

#1/(2sqrt3)xxsqrt3/sqrt3 = sqrt3/(2xx3) =sqrt3/6#

Answer 2 · 2016-09-04T07:42:31

#(sqrt(3)) / (6)#

We have: #(2 sqrt(4)) / (8 sqrt(3))#

#= (2 cdot 2) / (8 sqrt(3))#

#= (4) / (8 sqrt(3))#

#= (1) / (2 sqrt(3))#

Now, let's rationalise the denominator by multiplying both the numerator and denominator by #sqrt(3)#:

#= (1) / (2 sqrt(3)) cdot (sqrt(3)) / (sqrt(3))#

#= (sqrt(3)) / (2 cdot 3)#

#= (sqrt(3)) / (6)#