Since the square root of 1 is ±1,
sqrt(1)/6 = ±1/6 = ±0.166...
1 over 6
1 over (3 times the square root of 6)
No. If you square 1/6, you get 1/36, which is considerably smaller than 3. The square root of 3 is actually an irrational number - i.e. cannot be represented by a ratio of integers (fraction). The approximate value of the square root of three is 1.73205101, which lies between the values of 1 2/3 and 1 5/6
8.1667
6, this answer is equivalent to asking the square root of 36
6!/6/6+square root of 6 times square root of six over six dunno how 2type square roots
The idea is to get rid of the square root in the denominator. For this purpose, you must multiply numerator and denominator by the square root of 6 in this case.
x2 - 6x + 1 = 0 x = [-(-6) +&- square root of [(-6)2 - 4(1)(1)]]/2(1) x = (6 +&- square root of 32)/2 x = [6 +&- 4(square root of 2)]/2 x = 3 +&- 2(square root of 2) x = 3 + 2(square root of 2) or x = 3 - 2(square root of 2) Check:
6/7 or 0.857
The square root of 6 is... 2.449489742783178 !
± 4.690416*i (to 6 dp) where i is the imaginary square root of -1.
0.0278