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    What did you guys get? Let's make an unofficial mark scheme
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    For the 6 mark simultaneous equation

    X was either 1/3 or -2
    Y was either 6 or -1
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    (Original post by johnthebaptist27)
    What did you guys get? Let's make an unofficial mark scheme
    (x-2)(X+5)^2
    Y=3/5x+25
    4√3 for the last one
    Solutions for simultaneous was (1/3, 6), (-2,-1) I think
    K was 17
    2x^2+16x+13 was 2(x-4)^2 -19
    I think matrices was r1 c1 (4), R1 c2 (16), forgot R2 c1 but R2 c2 was 17
    The first one I think was 3x^2 + x^3 I don't remember properly
    -2 -7 was a and b respectively for the second one I think or it was for the coordinates of P when another line was parallel to the line tangent to the curve
    (X+2)^2+(y-5)^2=25 given (-2,5)
    48 one of them
    Question with P and Q: x was 35
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    20. I'm pretty sure that I got 4 root 3.
    I got k as 17 for the quadratic expansion question.
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    (Original post by Jakazen)
    (x-2)(X+5)^2
    Y=3/5x+25
    4√3 for the last one
    Yeah same here
    The first one was 3x^5 +x^3 (because it was (x^6)/2 +(x^4)/4) I think
    And I'm pretty sure that R2 C1 was -8
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    For the line draw one I drew y = 2-x for answers of 3.6 and 0.6.
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    (Original post by Jakazen)
    (x-2)(X+5)^2
    Y=3/5x+25
    4√3 for the last one
    same here
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    (Original post by johnthebaptist27)
    For the line draw one I drew y = 2-x for answers of 3.6 and 0.6.
    How did you work that one out? I got really stuck and put some random working down in the hope I might get a method mark (unlikely lol)
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    (Original post by johnthebaptist27)
    For the line draw one I drew y = 2-x for answers of 3.6 and 0.6.
    Omg I was so unsure on that and wasn't really sure what I was doing. Do you think that was right?

    I also got that btw
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    (Original post by johnthebaptist27)
    For the line draw one I drew y = 2-x for answers of 3.6 and 0.6.
    i got 3.4 and 0.6
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    (Original post by LottieLouLou)
    How did you work that one out? I got really stuck and put some random working down in the hope I might get a method mark (unlikely lol)
    Well you were given a graph of 3x-x^2 and asked to solve x^2-4x+2, so if you add the two together, you get x^2 - x^2 +3x-4x + 2 = 2-x. THen draw this line on the graph and circle where 3x-x^2 and 2-x intersect
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    (Original post by 13dsaivaines)
    Omg I was so unsure on that and wasn't really sure what I was doing. Do you think that was right?

    I also got that btw
    Yh i am pretty sure because I checked in the exam and if you get the graph of x^2 -4x + 2 and complete the square, you get (x-2)^2 -2 = 0, hence x = 2+-sqrt(2), and i took the root of 2 to be 1.4 (remeber it as this), which gives 0.6 and 3.4
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    (Original post by Jakazen)
    (x-2)(X+5)^2
    Y=3/5x+25
    4√3 for the last one
    Solutions for simultaneous was (-1/3, -6), (2,1) I think
    K was 17
    2x^2+16x+13 was 2(x-4)^2 -19
    I think matrices was r1 c1 (4), R1 c2 (16), forgot R2 c1 but R2 c2 was 17
    The first one I think was 3x^2 + x^3
    -2 -7 was a and b respectively for the second one I think or it was for the coordinates of P when another line was parallel to the line tangent to the curve
    (X+2)^2+(y-5)^2=25 given (-2,5)
    I think I got 3x^5+x^3 for the first one and for the simultaneous I got 1/3 and 6, & -1 and -2
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    (Original post by somebrick)
    i got 3.4 and 0.6
    yeah thats what I got i remembered incorrectly
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    (Original post by johnthebaptist27)
    Yh i am pretty sure because I checked in the exam and if you get the graph of x^2 -4x + 2 and complete the square, you get (x-2)^2 -2 = 0, hence x = 2+-sqrt(2), and i took the root of 2 to be 1.4 (remeber it as this), which gives 0.6 and 3.4


    Yeah I did that as well and solved it using the quadratic formula but thought the root of 2 was 1.8 lol
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    (Original post by Flyn_Higginson)
    Yeah same here
    What was the question for
    Y = 3/5x +25
    Posted on the TSR App. Download from Apple or Google Play
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    • The question where you had to work out the coordinates of point P on the curve with its tangent parallel to a given equation of a line:
    (-0.70, -10.12).
    • Last question (trig/surds): 4√3 (in the form a√b)
    • Sketch question: obviously I can't show a sketch here but the coordinates, Max (-3, 13) & Min (1, 2⅓) and you roughly crossed the y-axis at (0,4).
    • K = 17
    • I didn't do quite well in the circle theorem but ig the main important angle (imo) that I worked out was 2x for one of the angles in the cyclic quadrilateral. I knew you ultimately needed to show two corresponding or alternate angles were equal to prove that the two lines were parallel, but I couldn't work the values
    • equation for circle (tick box q): (x+1)²+(y-6)² = 25. I'm not quite sure about the actual coordinates of the centre given to us but it defo was a '+' in the 1st bracket, a '-' in the 2nd then 25 (5² = 25).
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    (Original post by somebrick)
    i got 3.4 and 0.6
    same complete guess aha. apart from this i found everything else alright
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    (Original post by Jakazen)
    (x-2)(X+5)^2
    Y=3/5x+25
    4√3 for the last one
    Solutions for simultaneous was (-1/3, -6), (2,1) I think
    K was 17
    2x^2+16x+13 was 2(x-4)^2 -19
    I think matrices was r1 c1 (4), R1 c2 (16), forgot R2 c1 but R2 c2 was 17
    The first one I think was 3x^2 + x^3
    -2 -7 was a and b respectively for the second one I think or it was for the coordinates of P when another line was parallel to the line tangent to the curve
    (X+2)^2+(y-5)^2=25 given (-2,5)
    48 one of them
    yea i got these
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    What was the answer for x for the first circle theorem question with x + 75 and 2x?
 
 
 
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