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    (Original post by Abdullah IGCSE)
    ndk123 im in the same position as you paper 2 was hella hard but its not possible for the A* to be 188 this time believe me i reckon they will put an A to 140-145 and an A* 160+.Depending on the difficulty of tomorrows paper,i have an aim of getting at least 100/130 or 95 at min.The only tips i know are read the question carefully even once finished and re check.For now id say skim through mark schemes and past papers you've done to see where you went wrong so those mistakes dont occur in the exam.I hope to have scored a minimum of 50 in paper 2 and 100 in paper 4 so fingers crossed ndk.
    Same... I really hope the grade boundaries are lower. Yep i'm just going to review all the papers. I hope i don't do any silly mistakes. Thanks! Good Luck for tomorrow.
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Size:  7.7 KBcould someone please help me with this question. Its from the winter 2015 qp.43. The answer is 36 but i have no idea how to get there. Thanks
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    (Original post by ndk123)
    could someone please help me with this question. Its from the winter 2015 qp.43. The answer is 36 but i have no idea how to get there. Thanks
    its a really big method and u need to divide the shape into triangles and then use concepts of parallel lines which are to separated by a traversal. Then place the angles accordingly and and all of them up to 360 .
    I know its confusing to understand here but just use basic concepts and that five marks would be yours.
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    (Original post by ndk123)
    Name:  1.PNG
Views: 75
Size:  7.7 KBcould someone please help me with this question. Its from the winter 2015 qp.43. The answer is 36 but i have no idea how to get there. Thanks
    co-interior angles: RQP = 180 - 2m
    angles in quadrilateral add up to 180: ROP = 360 - m - 2m - (180 - 2m) = 180 - m
    angle at the circumference is half the angle at the centre: RMP (see diagram) = (180 - m)/2
    opposite angles in a cyclic quadrilateral add up to 180: RQP + RMP = (180-m)/2 + 180-2m = 270 - (5m/2) = 180
    (5m/2) = 90
    5m = 180
    m = 36
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    (Original post by JickDee)
    co-interior angles: RQP = 180 - 2m
    angles in quadrilateral add up to 180: ROP = 360 - m - 2m - (180 - 2m) = 180 - m
    angle at the circumference is half the angle at the centre: RMP (see diagram) = (180 - m)/2
    opposite angles in a cyclic quadrilateral add up to 180: RQP + RMP = (180-m)/2 + 180-2m = 270 - (5m/2) = 180
    (5m/2) = 90
    5m = 180
    m = 36
    Thanks had difficulty with it too
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    How did paper 42 0580 go for you guys today? hard? ez?
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    (Original post by In Kieu)
    How did paper 42 0580 go for you guys today? hard? ez?
    really good compared to paper 2 i feel like i scored 110+
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    (Original post by Abdullah IGCSE)
    really good compared to paper 2 i feel like i scored 110+
    Yea it was reallyyyyyyyyy easy but I just hope that the grade boundaries are not very high
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    I hope the physics alternative to practical is as easy as today's math paper.
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    (Original post by Abdullah IGCSE)
    I hope the physics alternative to practical is as easy as today's math paper.
    Yea i wouldn't mind if its more easier than the math paper LOL
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    the answer to the interior angle polygon question is 45 sides
 
 
 
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