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    High grade boundaries.


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    (Original post by Dan205)
    Did you get -32 for the product of v?

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    32 e^j(pi/2)


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    (Original post by Genty Boy)
    32 e^j(pi/2)


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    Uh oh... I got e^j(pi) :/

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    (Original post by Dan205)
    Uh oh... I got e^j(pi) :/

    Posted from TSR Mobile
    For the first z value I got 2e^j(pi/10)...

    Raise that to 5 I think is how you work it out...


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    (Original post by Genty Boy)
    For the first z value I got 2e^j(pi/10)...

    Raise that to 5 I think is how you work it out...


    Posted from TSR Mobile
    Nah that makes sense, I don't know what I did haha!

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    (Original post by Dan205)
    Nah that makes sense, I don't know what I did haha!

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    I may be wrong.


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    (Original post by Dan205)
    Did you get -32 for the product of v?

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    Fairly sure it was +32j
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    (Original post by Genty Boy)
    32 e^j(pi/2)


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    It said simplified form so you had to leave it as just +32j
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    Overall, nice exam IMO
    I think it'll be around 62 for an A
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    (Original post by chizz1889)
    Fairly sure it was +32j
    Yep, that's the same as 32e^pi/2 j
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    Well that went surprisingly well. I didn't even expect to finish.

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    Last integral was x/2 * ln ((1+x)/(1-x)) + 1/2 * ln(1-x^2) + c
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    (Original post by sarcasmrules)
    Yep, that's the same as 32e^pi/2 j
    Yeah but it said 'simplified form' which is 32j
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    Wow, that was such a nice exam, feeling pretty confident about that but probably will be around 66-68 for an A*

    I got v^5 = 32j
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    (Original post by chizz1889)
    Last integral was x/2 * ln ((1+x)/(1-x)) + 1/2 * ln(1-x^2) + c
    I got that also.
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    M^4 = 43m^2 + 42m
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    For The PDP^-1 = M^4

    The diagonal components of D were 1;0;2401
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    For the lambda3 the vector was (4,1,3)
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    Coordinates (a,b)?
    Radius sqrt(a^2+b^2)?


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    The eigenvalues were -1, 0, 7.

    The eigenvector for 7 was (4,1,3) I think
 
 
 

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