sqrt{3}divided sqrt{18}
sqrt(3)/sqrt(18) (sqrt(3)*sqrt(18))/(sqrt(18)*sqrt(18)) (sqrt(3*18))/(sqrt(18*18)) (sqrt(54))/(sqrt(324)) (sqrt(9*6))/18 (sqrt(9)*sqrt(6))/18 (3*sqrt(6))/18 sqrt(6)/6 So sqrt(3)/sqrt(18) = sqrt(6)/6
sqrt{ab} = sqrt{a}sqrt{b} and since 18 = 6 x 3 sqrt{18} = sqrt{6}sqrt{3} so sqrt{3}/sqrt{18} = sqrt{3}/(sqrt{3}sqrt{6}) =1/sqrt{6}
got there before me xD
Both answers that I posted and eigenschmeigen posted are correct (and equivalent), but my answer has a rationalized denominator (which may be what your book is leaning towards). Of course, it may want a rationalized numerator, but my experience has seen it wants a rationalized denominator. As always, it's best to ask your teacher if you're not sure.
well put. rationalised denominators are the way forward.
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