What is #(-1/2) / ( [-sqrt3] / 2)#?

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Answer 1 · 2015-06-03T05:59:34

Answer: #(-1/2)/(-sqrt(3)/2)=sqrt 3/3# .

Problem: Solve #(-1/2)/(-sqrt(3)/2)# .

Multiply the numerator times the inverse of the denominator.

#-1/cancel2xxcancel2/-sqrt3# =

#(-1)/-sqrt 3# =

#1/sqrt 3#

Rationalize the denominator.

#1/sqrt 3xxsqrt3/sqrt3# =

#sqrt 3/3#