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    (Original post by TheWolf)
    For one of the q you have to find when [modulus(x-1) -2] intersects 5x

    it would only intersect with -(x-1) -2

    so its -x + 1 -2 = 5x

    6x = -1
    x = -1/6

    i cant remember the q totally correctly, so this might be wrong
    gots the same
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    hey all

    i found that exam pretty hard

    i got a and b = -1 and -2 (dont know which is which)

    i got R=2root13 or root52

    i got the same alpha as above

    i got x=-1/6 , i eliminated something like x=1/4

    i got 14 years, although maybe it should have been rounded down to 13???, thats what one of my friends did

    for one of the trig questions i got x=60, 300, 131.8, 228.2

    for another trig question (i think the complicated one inclolving R and alpha etc) i got 0, pi, and something else but eliminated pi because it was out of range (0<=x<pi)

    graphs looks like v through yaxis and W

    thats all i remember for now

    please give me your feedback
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    i think i got al tha same bar a few...cant memba which 1 ... but u did aite me thinkz
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    For the trig where involves R, i think you put √52sin(x+0.588) = 3 rather than =0, which most of you seem to have done? =o Hmm does anyone remember the exact q?
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    It's over
    Now to revise for Biology tomorrow... =(
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    that -8/(x-1)^3 was as easy few marks (think it was 6 )
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    was the answer 13 or 14 years? do we round up or down or give a decimal???

    i rounded up to 14, is that right?
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    Yea that should be right. Rounding down to 13 just doesn't make sense. =o
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    model solutions for this paper should be ready by tomorrow evening :tsr:
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    (Original post by Hash)
    2.12 radians for the long trig question (as well as 0 and pi)
    pi was not included in the range, so that isn't right... (0<= x < pi)

    The answers were: 0 and 2.12 (radians)
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    (Original post by ArVi)
    pi was not included in the range, so that isn't right... (0<= x < pi)

    The answers were: 0 and 2.12 (radians)
    Did you put the R equation = 0. At the time I was pretty sure to put it = 3?
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    (Original post by TheWolf)
    For one of the q you have to find when [modulus(x-1) -2] intersects 5x

    it would only intersect with -(x-1) -2

    so its -x + 1 -2 = 5x

    6x = -1
    x = -1/6

    i cant remember the q totally correctly, so this might be wrong
    Same answer and method! hope we're right!
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    (Original post by r2they)
    was the answer 13 or 14 years? do we round up or down or give a decimal???

    i rounded up to 14, is that right?
    I rounded up to 14 too cuz it was like 13.93 or something.. When checking, i substituted back both 13 and 14 - 13 gave an answer too low and 14 was a little above, so I'm guessing 14 is right.
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    It should be 14 since there isn't going to be the right amount at the beginning of year 13. But there will be at year 14.
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    (Original post by TheWolf)
    Did you put the R equation = 0. At the time I was pretty sure to put it = 3?
    nope, i got the final equation =0

    it was something like:

    sin (x) [ √52 sin (x+0.588) - 3 ] = 0

    So Sin(x) = 0 or √52 sin (x+0.588) - 3 = 0

    So sin (x+0.588) = 3 ÷ √52
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    Yes I got sin (x+0.588) = 3 ÷ √52 as well. Might have calculated the final answers wrong.
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    i didnt manage to do that one.....
    C3 was quite hard, harder than C2.....
    i think the last q was the hardest of the paper...
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    (Original post by mackin boi)
    did u get root 52 for R and .588 for alpha??
    yup although I put it as 2sqr13
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    (Original post by ArVi)
    nope, i got the final equation =0

    it was something like:

    sin (x) [ √52 sin (x+0.588) - 3 ] = 0

    So Sin(x) = 0 or √52 sin (x+0.588) - 3 = 0

    So sin (x+0.588) = 3 ÷ √52
    That's right
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    (Original post by TheWolf)
    For the trig where involves R, i think you put √52sin(x+0.588) = 3 rather than =0, which most of you seem to have done? =o Hmm does anyone remember the exact q?
    I got the same thing although I don't recall the making it equal to 3 at all for that bit as that actual question gives you (what is at the time) a new formula and thus isn't equal to anything else...or I could be talkin aload of b*llox
 
 
 
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