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    (Original post by ChuckNorriss)
    guys has anyone come across a past paper question similar to example 13 from chapter 7?
    Nah.

    I think that example is there to help sketch polar curves. That is to say, remember the conditions it proves to assist you in drawing the curves.
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    11 days left.

    Get revising!
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    (Original post by edothero)
    11 days left.

    Get revising!
    AAAH! I still feel like it's months away... it's scary!
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    (Original post by oinkk)
    AAAH! I still feel like it's months away... it's scary!
    FP3 is still a month away :laugh:
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    (Original post by edothero)
    FP3 is still a month away :laugh:
    That is true.

    But FP2 was a month away almost a month ago, and look where we are now :cry2:
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    Does anyone have a Casio fx-CG20? I'm having a problem with the graph sketcher as when I input mod(any big quadratic) it doesn't show it crossing the x-axis when there are supposed to be two roots. I tried 3x^2-19x+20 which has two roots but on the calculator it shows it translates upwards with no roots. Does anyone know how to fix this?
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    Please could anyone explain how to do question 3c in here? Thanks Name:  image.jpg
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    (Original post by economicss)
    Please could anyone explain how to do question 3c in here? Thanks Name:  image.jpg
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    Hey,

    Your solutions are plotted in the Argand diagram from part b). You can then find the magnitude of the lengths between each of the three points on your diagram to find the perimeter (think C1 distance formula).

    Jack
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    (Original post by oinkk)
    Hey,

    Your solutions are plotted in the Argand diagram from part b). You can then find the magnitude of the lengths between each of the three points on your diagram to find the perimeter (think C1 distance formula).

    Jack
    Thanks for your help, I just gave it another go and these are the distances I got but they didn't give the right length, please could you have a look and see where I've gone wrong? Thanks Name:  image.jpg
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    (Original post by economicss)
    Thanks for your help, I just gave it another go and these are the distances I got but they didn't give the right length, please could you have a look and see where I've gone wrong? Thanks Name:  image.jpg
Views: 97
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    Allow me a few moments to do the question... I've had a bottle of prosecco so I hope my mathematical judgement is not harmed as a result
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    (Original post by oinkk)
    Allow me a few moments to do the question... I've had a bottle of prosecco so I hope my mathematical judgement is not harmed as a result
    Haha ofc, thank you wouldn't mind some prosecoo rn!
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    (Original post by economicss)
    Haha ofc, thank you wouldn't mind some prosecoo rn!
    Best way to revise

    I've just done the question from scratch and I agree with your solutions to the equation. But I am simply not getting the required perimeter either.

    Can you show your working for each of the lengths?
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    (Original post by economicss)
    Haha ofc, thank you wouldn't mind some prosecoo rn!
    (Original post by oinkk)
    Best way to revise

    I've just done the question from scratch and I agree with your solutions to the equation. But I am simply not getting the required perimeter either.

    Can you show your working for each of the lengths?
    You get:

    \displaystyle z = 2e^{i\pi/12}, 2e^{i3\pi/4}, 2^{-i7\pi/12}

    Which give you cartesian coordinates (2\cos \pi/12, 2\sin \pi/12), \quad (2\cos 3\pi/4, 2\sin 3\pi/4) and (2\cos (-7\pi/12), 2\sin (-7\pi/12)).

    The lengths between each one are \sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.

    Multiply by 3 to get the required perimeter. Not quite sure what the problem is here?
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    (Original post by Zacken)
    You get:

    \displaystyle z = 2e^{i\pi/12}, 2e^{i3\pi/4}, 2^{-i7\pi/12}

    Which give you cartesian coordinates (2\cos \pi/12, 2\sin \pi/12), \quad (2\cos 3\pi/4, 2\sin 3\pi/4) and (2\cos (-7\pi/12), 2\sin (-7\pi/12)).

    The lengths between each one are \sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.

    Multiply by 3 to get the required perimeter. Not quite sure what the problem is here?
    I totally had a sign issue... thanks for saving the day :-)
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    (Original post by oinkk)
    I totally had a sign issue... thanks for saving the day :-)
    Cheers, LaTeXing that was not pretty.
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    (Original post by Zacken)
    Cheers, LaTeXing that was not pretty.
    Haha, I can imagine!
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    (Original post by Zacken)
    You get:

    \displaystyle z = 2e^{i\pi/12}, 2e^{i3\pi/4}, 2^{-i7\pi/12}

    Which give you cartesian coordinates (2\cos \pi/12, 2\sin \pi/12), \quad (2\cos 3\pi/4, 2\sin 3\pi/4) and (2\cos (-7\pi/12), 2\sin (-7\pi/12)).

    The lengths between each one are \sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.

    Multiply by 3 to get the required perimeter. Not quite sure what the problem is here?
    Ah finally got it, thanks so much!
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    for this question (i have attached the mark scheme) can someone explain where the very first line of working came from? thanks
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    (Original post by Zacken)
    The lengths between each one are \sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.
    What's going on here? can't quite make out what you did
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    (Original post by edothero)
    What's going on here? can't quite make out what you did
    sqrt((x1 - x2)^2 + (y1-y2)^2) give the distance between (x1, y1) and (x2, y2).
 
 
 
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