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    (Original post by DFranklin)
    What yo have is (a/b)/c. This is the same as a/(bc), not ac/b.

    Note that division is not associative, so once you have more than one division you should use brackets to avoid ambiguity. (And so I'm not even going to try to follow your attempted justification which has more divisions than the current Labour party...)

    Edit: Note also that there's a reason I advised you to find (and simplify) \Delta x and \Delta y separately, as you would then largely avoid constructs like this that you evidently find confusing.
    1/2/3 (a/b/c) = 1/2 * 3/1 = 3/2(ac/b) and calculator says I'm correct.
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    (Original post by ckfeister)
    1/2/3 (a/b/c) = 1/2 * 3/1 = 3/2(ab/c) and calculator says I'm correct.
    Repeating myself (for the last time before I plonk you into the "posters who don't listen and aren't worth wasting time on" ), what you have is of the form (a/b)/c and this equals a/(bc), not ab/c.

    When a=1, b=2, c=3, the LHS is (1/2)/3 = 1/6

    And again repeating myself, a/b/c is ambiguous. Either write it as (a/b)/c or as a/(b/c), depending on which you mean.
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    (Original post by DFranklin)
    Repeating myself (for the last time before I plonk you into the "posters who don't listen and aren't worth wasting time on" ), what you have is of the form (a/b)/c and this equals a/(bc), not ab/c.

    When a=1, b=2, c=3, the LHS is (1/2)/3 = 1/6

    And again repeating myself, a/b/c is ambiguous. Either write it as (a/b)/c or as a/(b/c), depending on which you mean.
    I edited it to ac/b
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    (Original post by ckfeister)
    I edited it to ac/b
    And as I said, it does't equal that, either.
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    Too complicated for MEng engineering student
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    (Original post by DFranklin)
    And as I said, it does't equal that, either.
    Explain it than,

     \frac {a}{b} \div \frac {c}{d}

     \frac {a}{b} \times \frac{d}{c}

    :. a(d)/bc ohhhhhh now I get it...
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    (Original post by RDKGames)
    Yeah those two are not equal to one another. If you multiply the numerator of the main fraction by h then you must multiply the denominator of the main fraction by h, thus giving you h^2 on the main denominator which not what you want. Also it should be -h on your main denominator by the looks of it anyway.

    To carry on correctly, simplify -\frac{1}{2}+\frac{1}{h-2} first.
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    I got this but I haven't even started doing divide by h on the bottom of the fraction and its the answer, what do I do here?
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    (Original post by ckfeister)

    I got this but I haven't even started doing divide by h on the bottom of the fraction and its the answer, what do I do here?
    Why did you take the reciprocal of \frac{1}{h-2}?? Your first line is not equal to what I said at the end of my last post.
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    (Original post by RDKGames)
    Why did you take the reciprocal of \frac{1}{h-2}?? Your first line is not equal to what I said at the end of my last post.
    I'm hoping this is correct... this question making me go insane.
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    (Original post by ckfeister)
    I'm hoping this is correct... this question making me go insane.
    Okay the working is fine, but having done the question myself at this point I noticed that you've went wrong earlier - when I assumed you were right.

    Starting from the beginning:

    y(-2)=-\frac{1}{2}
    y(-2+h)=\frac{1}{-2+h}

    You chose to work with the following form: \displaystyle \frac{y(-2)-y(-2+h)}{(-2)-(-2+h)} which gives -h on the denominator as I said earlier, but the numerator is -\frac{1}{2}-\frac{1}{h-2} which is nicely doable by factoring out the negative first and it should lead you to the correct answer. This is different as in your one a few posts above you had +\frac{1}{h-2} rather than minus.

    Next time though, be careful with your working. Lay it out clearly and be mindful of the negatives otherwise you're going to get very confused as it happened here.
 
 
 
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