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    (Original post by money-for-all)
    ooooooo , check you fancy pants , desmos and ****, you still clapped tho
    Naw bruv, not into the alcohols and that kinda stuff - yolo remember. I'm glad the desmos image helped you. Feel free to PM me if you have more questions.
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    (Original post by craigie435)
    you had to use pythagaros which got you route 64
    It was √65:

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    What did we all get on the last integration question?
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    (Original post by Frazzles10)
    That was easy it was 9
    I got 9 aswell, annoying thing was if u got the first part of the question wrong you would've mucked the rest up
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    (Original post by Heidirennie)
    What did we all get on the last integration question?
    The integral came out to be 81/4 or 20.25.

    The shaded area was 45/4 or 11.25

    This has been answered before in the forum.
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    Do you think i will lose mark just because i did not simplify the answer 243/12 into 81/4?
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    oh gosh that paper went down hill all because of that circle question and the integration one, i literally couldn't find a trapezium or a triangle but as soon as people told me about the triangle i knew i failed
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    (Original post by goofyygoober)
    Do you think i will lose mark just because i did not simplify the answer 243/12 into 81/4?
    i doubt it unless the question said "simplify" or "in its simplest form" otherwise u would be able to get the OE (or equivalent) mark
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    How many marks would I get on the intergration question if i did it all right but somehow i managed to get the area of the triangle as 6 because i did 24 x 2/4 /2 instead of 24 x 3/4 /2 (i think this is what you were supposed to do?)

    Also for the length of CT I added it up wrong and got root80 instead of root81 so would i get method marks or no marks at all?
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    (Original post by beanigger)
    i doubt it unless the question said "simplify" or "in its simplest form" otherwise u would be able to get the OE (or equivalent) mark
    thx!!
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    (Original post by Fambox)
    I got the same I hope we are right !!! Mine was in fraction form like 135/12
    Same for some reason I never bothered to convert them in decimals and thought I was wrong...thanks the mental peace lol


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    did anyone get -5/4 as the x co-ordinate?
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    (Original post by aliyaahh)
    did anyone get -5/4 as the x coordinate?
    Yes, -5/4 was the x-coordinate of Q
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    (Original post by ella-1997)
    Exam was easy overall, but i just couldn't work out question 5)b where you had to find the coordinates of b?? how did everyone do this?
    I worked out that C was the midpoint, because AB was the diameter, so the distance AC was half of AB. Then just double the change in coordinates.
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    (Original post by aliyaahh)
    did anyone get -5/4 as the x co-ordinate?

    Yes. The Co-ordinates of Q were (-0.8,0) but you only needed to state the x.
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    (Original post by jakeheart291)
    how many marks for Q8?????/
    It was 7/8 marks
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    (Original post by SamCoulthard)
    I worked out that C was the midpoint, because AB was the diameter, so the distance AC was half of AB. Then just double the change in coordinates.
    You could do this, but I think it is easier and quicker to use the formula for a midpoint and sub in numbers you knew and worked out unknowns from this.
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    what was the equation that we had to integrate?
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    Does anyone remember what the integration equation after integrating was? 3x4/4 was one of them, right?
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    (Original post by SamCoulthard)
    I worked out that C was the midpoint, because AB was the diameter, so the distance AC was half of AB. Then just double the change in coordinates.
    Yeah I did that as well, seemed too simple at the time after thinking about it for a while. I said straight away the diameter would be 2*radius so using midpoint formula seemed than doing the midpoint thing below. Plus helped on the next question as I had already considered the gradient as -7/4 which made it easier to get the equation of the tangent to the circle. It was something like y-1=4/7(X+2) which simplified to 7x-4y+18=0


    (Original post by BenAtkinson)
    You could do this, but I think it is easier and quicker to use the formula for a midpoint and sub in numbers you knew and worked out unknowns from this.
 
 
 
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