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OCR C1 Mark Scheme 16/05/12 (NOT MEI)

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    (Original post by As_Dust_Dances_)
    The total perimeter was 10y+3 and it had to be between 20 and 54 right?
    so 20<10y+3<54
    17<10y<51
    1.7<y<5.1
    no it was 10y +6 the perimeter!
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    (Original post by As_Dust_Dances_)
    The total perimeter was 10y+3 and it had to be between 20 and 54 right?
    so 20<10y+3<54
    17<10y<51
    1.7<y<5.1
    total perimeter was 10Y+6
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    (Original post by As_Dust_Dances_)
    The total perimeter was 10y+3 and it had to be between 20 and 54 right?
    so 20<10y+3<54
    17<10y<51
    1.7<y<5.1
    total perimeter was 10y+6
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    (Original post by As_Dust_Dances_)
    The total perimeter was 10y+3 and it had to be between 20 and 54 right?
    so 20<10y+3<54
    17<10y<51
    1.7<y<5.1
    10y +6

    check out http://www.thestudentroom.co.uk/show....php?t=2002389

    unofficial mark scheme
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    i got -7<x<4 and 1.7<x<5.1 ...
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    (Original post by I_am_god_123)
    total perimeter was 10Y+6
    20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!
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    (Original post by max25)
    how did you find the last question?
    it was ok, pretty sure i got it right

    check out

    http://www.thestudentroom.co.uk/show....php?t=2002389
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    (Original post by As_Dust_Dances_)
    For the perimeter question I got 17/10<x<51/10
    I thought it was in terms of y?

    I think I got something like this...

    34 <= 10y + 6 <= 54

    When simplified gets:

    28 <= 10y <= 48

    then:

    14/5 <= y <= 24/5
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    (Original post by I_am_god_123)
    total perimeter was 10Y+6
    eeek apologies, wrong question hehe!were both right
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    (Original post by I_am_god_123)
    nah it was -7<x<4
    but then x must be +ve
    for part i) (x+3)(4x)<112 right?

    so 4x^2 +12x-112<0

    x^2+3x-28

    (x+7)(x-4)

    You're talking about part 1 right, I was on about part 2..

    So yeah it is x<-7 and x>4
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    (Original post by As_Dust_Dances_)
    for part i) (x+3)(4x)<112 right?

    so 4x^2 +12x-112<0

    x^2+3x-28

    (x+7)(x-4)

    You're talking about part 1 right, I was on about part 2..

    So yeah it is x<-7 and x>4
    But x cannot be negative - you cannot have a negative length as one of the sides was 4x negative dimensions are impossible so it had to be 0>x>4
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    100 ****ing percentaaah!
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    (Original post by Ilyas)
    100 ****ing percentaaah!
    check this out...
    http://www.thestudentroom.co.uk/show...389&p=37573386
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    (Original post by max25)
    20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!
    yaeh that's correct
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    Question 6

    Equation of normal is

    4x - 6y -29 = 0
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    would i lose a mark for having it as -4x + 6y + 29?
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    (Original post by As_Dust_Dances_)
    for part i) (x+3)(4x)<112 right?

    so 4x^2 +12x-112<0

    x^2+3x-28

    (x+7)(x-4)

    You're talking about part 1 right, I was on about part 2..

    So yeah it is x<-7 and x>4
    this is right.
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    (Original post by PhysicsMan)
    Question 6

    Equation of normal is

    4x - 6y -29 = 0
    This rings a bell has the gradient was like 2/3 or something?
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    (Original post by As_Dust_Dances_)
    This rings a bell has the gradient was like 2/3 or something?
    Yep, the gradient of tangent was -3/2 therefore the gradient of normal is 2/3.
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    (Original post by max25)
    20< 10y + 6 >54!!!!!!!!!!!!!!!!!!!!!!!!!!! !!
    That can't be correct (I didnt sit the paper though).. You are saying 20 < value > 54... so 20 is bigger than 54? 20 >54: you needed two statements.

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