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    I don't know why but I feel like I wrote the wrong numbers down on the last question. Like I know I got something like 8*10^-3, but I just have a feeling I had a brain meltdown and wrote down the wrong number...

    I also know that I got the area of the triangle wrong and the A stuff, as well as the whole of the stupid method is series as well. In total I probably lost around 15 marks not including method marks. So I hope you guys are right and the grade boundaries are low
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    (Original post by Adam_AFC)
    Yes to both questions. They made use the cosine rule and 0.5ABsinC for only 2 marks.
    You could treat it as two right angle triangles.
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    (Original post by Adam_AFC)
    Yes to both questions. They made use the cosine rule and 0.5ABsinC for only 2 marks.
    You could split it into two right-angled triangles and use 0.5bh
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    (Original post by hoakenfull)
    For the Inverse of B did anyone else multiply the matrix A by (1/25)?

    And 168 units squared for the first question? Area of the image of the triangle
    Oh god what was the first triangle question? Was it to work out the area because I've just written down the co-ord after transformation not the area
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    (Original post by Adam_AFC)
    Yes to both questions. They made use the cosine rule and 0.5ABsinC for only 2 marks.
    You only had to do (base x height)/2 of the triangle. You could either split it up into two triangles (one above/below x-axis) or turn the paper so the hypotenuse was the base and use that and the height to the point /2
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    here are my answers (that I can remember)

    2)Determinant was 0 so matrix was singular and plots to invariant points

    7)horizontal asymptote y=root2 verticals at 1 and -1
    8)roots 1,5,4-i,4+i
    10) converges to 3 as n~infinity
    for the differences, 1st and last terms are the only ones that don't cancel which gives the formula
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    (Original post by fp1wassocalm)
    I agree, but annoying because an a will be around 63/64
    that's just your opinion innit
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    (Original post by chemari1)
    that's just your opinion innit
    He's just a troll, A will be around 55.
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    Did anyone else get something like this for the Argand?
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    (Original post by Jamesinatr)
    Did anyone else get something like this for the Argand?
    YES. Thats what I got. I think it is right.
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    (Original post by -Gifted-)
    YES. Thats what I got. I think it is right.
    Awesome, thanks. Loci is my worst topic so I wanted to check.
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    (Original post by Jamesinatr)
    Awesome, thanks. Loci is my worst topic so I wanted to check.
    I think you gotta mention that its pi/4 from the y axis .
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    what were questions 7 and 9?
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    Did anyone get 420 for question 4.20
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    (Original post by -Gifted-)
    I think you gotta mention that its pi/4 from the y axis .

    At least I remembered to label the angles in the exam
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    (Original post by fp1wassocalm)
    here are my answers (that I can remember)

    2)Determinant was 0 so matrix was singular and plots to invariant points

    7)horizontal asymptote y=root2 verticals at 1 and -1
    8)roots 1,5,4-i,4+i
    10) converges to 3 as n~infinity
    for the differences, 1st and last terms are the only ones that don't cancel which gives the formula
    Yeah I agree, for the 3x3 matrices question did you get the eigenvalues (1/8, 7) and that the solutions form a sheaf?
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    (Original post by fp1wassocalm)
    here are my answers (that I can remember)

    2)Determinant was 0 so matrix was singular and plots to invariant points

    7)horizontal asymptote y=root2 verticals at 1 and -1
    8)roots 1,5,4-i,4+i
    10) converges to 3 as n~infinity
    for the differences, 1st and last terms are the only ones that don't cancel which gives the formula
    I'm pretty sure all of those are incorrect, the determinant was 12, roots were 2 and -1/2. Converges to 2/3 and 6 terms don't cancel.
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    For question 7 did anyone else get A=6 & B=51
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    For the graph question:

    Cut the x-axis at +/- root3
    Cut the y-axis at 9/4
    Vertical asymptotes at -4 and 1
    Horizontal asymptote at y=3
    As x approaches positive infinity, y approaches 3 (the horizontal asymptote) from below.
    As x approaches negative infinity, y approaches 3 from above.

    x=1.1 y is -ve
    x=0.9 y is +ve
    x=-3.9 y is -ve
    x=-4.1 y is +ve
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    (Original post by Jamesinatr)
    Did anyone else get something like this for the Argand?
    Yes
 
 
 
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