Question 362007: 1/2 divide (2/3 divide 3/4) - (4/5 x 5/6) =
i tried
1/2 divide (2/3 x 4/3) - (4/5 x 5/6)
1/2 divide 8/9 - 2/3
1/2 divide 6/6
so my answer was 1/2 but my answers don't match. my answers have to be:
A. 1/16
B. 17/27
C. 13/18
D. 7/9
E. 5/6
im stuck! i dont know what to do please help!
Found 2 solutions by rfer, solver91311:
Answer by rfer(16322)   (Show Source):
You can put this solution on YOUR website! Follow the order of operations
do the division first then subtract
When subtracting be sure to find a common denominator.
I think the answer is (A)
But I am getting -1/6
So I am not sure if your choices are correct.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Something is very wrong, either with the way you have represented the problem, or with your list of answers.
Interpreting "divide," as you apparently did, to mean "divided by," the problem renders thus:
\ -\ \left(\frac{4}{5}\ \times\ \frac{5}{6}\right))
And then perform the operations inside of the parentheses as you did:
\ -\ \left(\frac{4}{5}\ \times\ \frac{5}{6}\right))
Here is where you made your error: In the first place, you must perform the division operation before the subtraction operation. In the second place,  , not no-how, not no-way. (Remember the thing about finding a common denominator and then adding the numerators once the denominators were the same?)
So, first the division:
Now we need a common denominator. Since the only prime factors of 16 are 2s, there are no factors in common between the two denominators, hence the common denominator is simply one denominator multiplied by the other, 16 times 3 is 48.
Apply the common denominator:
\ -\ \frac{2}{3}\left(\frac{16}{16}\right)\ =\ \frac{27}{48}\ -\ \frac{32}{48}\ =\ -\frac{5}{48})
Which, although it is the absolutely correct answer to the problem I just worked, it does not match the any of the set of answers you presented. So there are two possibilities. 1. You represented the problem incorrectly, or 2. You presented the wrong list of answers. It is possible that neither of these reasons are your fault -- it may be that your instructor or your textbook is in error.
John
My calculator said it, I believe it, that settles it
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