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    The y coordinates will be when x=0 so you sub x=0 into the equation, factorise the equation you end up with.

    Thats what I did anyway <erm>
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    (Original post by ReviseLastMinute)
    i think the question was x squared + y squared + 6x - 4y=12
    What are the possible y coordinates?
    substitute 0 into the x's and you'll get y^2-4y=12
    =>y^2-4y-12=0
    =>factorise to get (y-6)(y+2)
    => y=6 y=-2
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    (Original post by the_googly)
    If I remember correctly shouldn't it be 5/3

    and -3x

    ?
    Oh yeah my mistake, typing on the ipad is a ***** at times! That is what i meant though
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    (Original post by Fillly)
    I got C to be (12, -7) by using simultaneous equations
    This is what I did: I eliminated the x terms and I got -7 for y, then I substituted y=-7 into the first equation and I got 3x = 36 which means that x is 12 which means that the co-ordinates are (12,-7). Is that what you did? Or did you substitute y=-7 into the 2nd equation? Or did you eliminate the y terms? I'm just curious, that's all.
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    (Original post by ThomSDK)
    1)
    a) prove k=-4 i substituted k as y and the x value for x into the formula and you should get it
    b)gradient of AB = -10/6 = -5/2
    c) equation of the line perpendicular to AB through A in the form px + py + q= 0
    I used gradient of 2/5 and y - y1 = m(x - x1) to get 5y + 3x + 21 = 0 (not sure about the 3 i cant remember that bit)
    d) find the point that 5y + 3x = 1 meets the line 3y + 5x = 4 or something of that nature i just remember getting the coordinate (7,-12)

    2)
    a) rate of change = -3/2ms-1
    b) the height of the bird is decreasing as dy/dx<0
    c) d2y/dx2 = 4 therefore minimum

    3)
    a) put y= x^2 - 6x + 11 in the form (x-a)^2 + b
    I got (x-3)^2 + 2
    b) prove <b> using your completed square </b>that it has no solutions.
    I attempted a solution using the completed square and got x = 3 +- root -2 and as root-2 has no real solutions, it cannot be solved
    c) vertex (3,2)
    d) sketch it with the vertex at (3,2) and the statement "crosses y axis at y=11"
    e) translation that maps y= x^2 -6x + 11 <b> to <\b> y = x^2
    So answer [-3]
    [-2] and im 100% sure about that
    4) i shall use a } for a root sign for ease of reading ok
    a) }18 = 3}2
    b) }18/}32 + }8 = 2/7 (dont think the question is right but that is the right answer
    c) 5 - }6 was the answer, i dont remember the question

    5)
    a) find the remainder when divided by x+1
    I got p(-1) = 16
    b) prove x-3 is a factor: p(3) = 0 with statement "therefore x-3 is a factor"
    c) three linear factors (x-3)(x-3)(x+1)
    d) sketch and state where it crosses coordinate axes
    simple sketch of a cubic curve that goes through the x axis at -1 and touches the x axis at 3 and crosses the y axis at 18 if i remember correctly

    Im missing the last 3 questions from my memory lol
    This is my attempt at an unnofficial mark scheme, if anyone can just let me know what they remember from the last 3 questions and correct me if im wrong on any of these, its hard typing and remembering these questions haha. Hope everyone did okay!
    -10/6 = -5/3 you divide both numerator and denominator by 2
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    Erm, i seem to remember multiplying both equations by 5, so that i got ?x +40y = ? and ?x +40y= ?. I then subtracted one from each other to get the x value of 12. Well, either way, the fact we got the same answer is a good sign. I used a ? as i forgot the values :P
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    Woa I'm seeing some really dodgy answers lol
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    (Original post by SamTheMan95)
    Hahaha Hamas. it's Samatar

    I found the exam pretty easy tbh so the grade boundaries may be high. Hoping for 100%.
    Cool. What exam board? At JCC we do AQA. I also had my Physics and Chemistry exam last week which are OCR.
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    (Original post by SamTheMan95)
    -10/6 = -5/3 you divide both numerator and denominator by 2
    aint 5c the last factor (x+2)
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    (Original post by SamTheMan95)
    substitute 0 into the x's and you'll get y^2-4y=12
    =>y^2-4y-12=0
    =>factorise to get (y-6)(y+2)
    => y=6 y=-2
    oh nice that's what i got cheers man
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    I got the factors from (x^2-x-6) Meaning (x+2)(x-3)
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    (Original post by Ferrari_1996)
    Cool. What exam board? At JCC we do AQA. I also had my Physics and Chemistry exam last week which are OCR.
    im on AQA for Chemistry and maths
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    (Original post by ReviseLastMinute)
    oh nice that's what i got cheers man
    Which College/Sixth Form do you go to, if you don't mind me asking?
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    (Original post by ReviseLastMinute)
    oh nice that's what i got cheers man
    No worries
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    (Original post by SamTheMan95)
    im on AQA for Chemistry and maths
    Cool. Anyway I hope both of us are happy in March when we get our results.
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    (Original post by SamTheMan95)
    -10/6 = -5/3 you divide both numerator and denominator by 2
    The 2/7 for the surds is correct.
    Answer for:

    Question 8, Part B (II)

    K<1/2 or K>9/2

    I think thats right.
    It was factorised into:

    (2k-1)(2k-9) from 4k^2-20k+9>0
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    (Original post by Ferrari_1996)
    Which College/Sixth Form do you go to, if you don't mind me asking?
    Arthur Terry
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    (Original post by Sitrix)
    The 2/7 for the surds is correct.
    Answer for:

    Question 8, Part B (II)

    K<1/2 or K>9/2

    I think thats right.
    It was factorised into:

    (2k-1)(2k-9) from 4k-20k+9>0
    Wasn't it 4k^2-20k+9>0?

    EDIT: Why has my post been given negative ratings?
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    (Original post by ReviseLastMinute)
    Arthur Terry
    I have no idea where that is lol.
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    (Original post by ReviseLastMinute)
    oh nice that's what i got cheers man
    I got same too Cheers
 
 
 
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