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

2 Answers
Sep 4, 2016

Answer:

#sqrt3/6#

Explanation:

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

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

Sep 4, 2016

Answer:

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

Explanation:

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)#