SOLUTION: I need to estimate, and then evaluate the expression 3(1-sqrt3)\6 the answer should be rounded to the nearest thousandth. I have sqrt of 27 is between 5 and 6, so the above express

Algebra ->  Square-cubic-other-roots -> SOLUTION: I need to estimate, and then evaluate the expression 3(1-sqrt3)\6 the answer should be rounded to the nearest thousandth. I have sqrt of 27 is between 5 and 6, so the above express      Log On


   

Question 32153: I need to estimate, and then evaluate the expression 3(1-sqrt3)\6 the answer should be rounded to the nearest thousandth. I have sqrt of 27 is between 5 and 6, so the above expression would have to be between -.333 and -.5. Then I did the problem like this: 3-sqrt27/6 = 3-sqrt(9*3)/6 = 3-3sqrt3/6 = 3(1-sqrt3)/6 I am not sure if I can simplify this anymore or not??? Does this look right so far and if so, how do I proceed? Thank you, Steven
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You started with "3" in the numerator and "6" in the
denominator. You can reduce that to 1/2.
So you have (1-sqrt3)/2
1-sqrt3 is > 1-sqrt4=-2
Since the numerator is greater than -2
dividing by "2" gives you a number greater
than -1.
Maybe it is around -0.5
Actually it is around -0.36
Cheers,
Stan H.