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    (Original post by -jordan-)
    You loose the same percentage in the same time period.
    Not of the original value. That would mean an element would only ever have 2 half lives..
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    Can anyone remember how many marks q3 was and q8?
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    (Original post by df97)
    partial fractions and binomial expansion, i think answers are as follows:
    A=-5/2 B=9/2
    -1+7x-22/3x^2
    not sure on limit, think some people have something like mod x < 4/3
    wasnt there a smaller range mod x< 1/2
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    (Original post by C0balt)
    What he said
    So what form was the answer to the last question?
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    (Original post by df97)
    partial fractions and binomial expansion, i think answers are as follows:
    A=-5/2 B=9/2
    -1+7x-22/3x^2
    not sure on limit, think some people have something like mod x < 4/3
    mod x <1/2 i believe , since the range of 1/2 falls within the range of 3/4
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    (Original post by ihsanulhaq101)
    13/35 for me also.
    SAME
    Can we please verify if 1/5 or 13/35?
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    How did everyone find it? I thought it was alright, couldnt do last vectors or last page though
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    (Original post by Pingu7)
    So what form was the answer to the last question?
    3sin3y - 3cos3y =1/6 tan^-1(3x/2)+pi
    Or sth
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    (Original post by 1/2(p^2-p))
    wasnt there a smaller range mod x< 1/2
    yes thats right
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    (Original post by tanyapotter)
    does anyone remember the two parametric equations that you had to find dy/dx for?
    x = (4-4e^2-6t)4 y=e^3t/33t
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    (Original post by m00ngrk)
    Can anyone remember how many marks q3 was and q8?
    last one was 10 marks
    3 was the 5 marker i think? Trig quadratic
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    Whoops, did anyone get lamda=3 and mu=2?:confused:
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    (Original post by mickel_w)
    Not of the original value. That would mean an element would only ever have 2 half lives..
    After 100 days you have k^100 lots of the original value left. After 200 days you have (k^100)^2 of the original value. So therefore you loose (0.8)^3 over 300 days, given that k^100 is approximately 0.8 because you had roughly 8=(k^100)*10
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    (Original post by hi-zen-berg)
    SAME
    Can we please verify if 1/5 or 13/35?
    I got 13/35
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    (Original post by jtebbbs)
    Yeah sounds right. Just made me realise I put |x|<1/4 cause I was looking at (3-4x) and not (1-4/3x), there goes a mark lol

    Edit: think the right answer would be |x|<3/4, as the |x| in |x|<1 is replaced by |4/3x|, so |4/3x|<1, so |x|<3/4
    I got -1/2<x<1/2
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    What did everyone do for p(5-5^0.5)?
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    (Original post by Jm098)
    x = (4-4e^2-6t)4 y=e^3t/33t
    what was the gradient when you use these to find dy/dx
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    (Original post by AQslAyy)
    I got -1/2<x<1/2
    Me too, as this was the narrowest of the two ranges
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    (Original post by tanyapotter)
    does anyone remember the two parametric equations that you had to find dy/dx for?
    Attachment 551854551856
    Attached Images
      
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    (Original post by AQslAyy)
    I got -1/2<x<1/2
    Yeah don't listen to me mate, you're right, didn't even consider the (1+2x) part until just now, oh well it's only one mark
 
 
 
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