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    Can somone answer June 2010 Question 7a for me ?


    I dont get the answers on the MS


    Cheers
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    (Original post by fpmaniac)
    Where did the 2n come from in the end equation. If u add r=0 shouldnt it be just +2
    As 2n is the sum of 2 to n numbers and as it is r^2 +2 then for the sum you need to add 2n.
    With the r=0 you just add whatever the series equals when 0 is put in. For my example I added 2
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    No I have not, I'm on my phone so dont have the link.

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    (Original post by economicss)
    Ah just found it, thanks so much for that I don't suppose you've done part b? Thanks
    No I haven't as Im on my phone now so don't have the link

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    (Original post by Billsonbubbles)
    Can somone answer June 2010 Question 7a for me ?


    I dont get the answers on the MS


    Cheers
    check attachment
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    (Original post by kingaaran)
    What have you tried?
    awww cute
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    (Original post by Billsonbubbles)
    Can somone answer June 2010 Question 7a for me ?


    I dont get the answers on the MS


    Cheers
    Hope this helps!
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    (Original post by techfan42)
    check attachment
    Cheers man,


    How did you factorise it in part a ... I mean I get how you use the indice law in line 1 ... but then you went to 6( 2^k ... etc?


    P,S there is a good exam on the physics and maths tutor website for the gold silver and bronze papers for FP1 i am sure you knew anyway ah
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    (Original post by Brailey)
    Hope this helps!

    Thanks, I dont get the factorisation on the second line of working.


    Also I have taught myself AS further maths, do you have any tips that your teacher has mentioned , or you think is nice to know.


    Like for me I noted that


    Area of the new shape = Determinant of the transformation matrix x area of the old shape



    And this works for quadrilaterals and triangles when you want to work out the area of shape .


    Enjoy and thanks !
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    (Original post by Billsonbubbles)
    Cheers man,


    How did you factorise it in part a ... I mean I get how you use the indice law in line 1 ... but then you went to 6( 2^k ... etc?


    P,S there is a good exam on the physics and maths tutor website for the gold silver and bronze papers for FP1 i am sure you knew anyway ah
    You don't actually factorise, all you have to do is try and get the expression of f(k) multiplied by a number, in this case 6. Once you do that, you'll see that 6(2^k + 6^k) is basically 6f(k). However, remember that you can't change the equation. This is still an expression for f(k+1) Therefore, once you realise that, you see that 6f(k) gives you 4(2^k) extra. Therefore, you have to minus that in order to retain the original equation, giving you the required expression

    p.s. yeah, thanks for the tip, but I've already finished them hahahaha
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    (Original post by TomWeller)
    question: for the argument of z

    i know you do tan^-1 (b/a) of a + bi

    but do you keep a and b exactly how they are, or are they the mod of a and b?

    example:
    z = 1 - i

    would it be tan-1(-1/1) or simply tan-1(1/1)... and why?
    you dont take the mod. It depends on where it is on a argand diagram, as the angle for the argument is taken from the real numbers axis/x axis anticlockwise.

    so for your example the angle would be minus tan-1(1/1) or 360/2pie - tan(1/1).

    I always draw a diagram like a tiny sketch and see where it lies and then go from there!

    Hope this helps
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    (Original post by Billsonbubbles)
    Cheers man,


    How did you factorise it in part a ... I mean I get how you use the indice law in line 1 ... but then you went to 6( 2^k ... etc?


    P,S there is a good exam on the physics and maths tutor website for the gold silver and bronze papers for FP1 i am sure you knew anyway ah
    post 983; all he did was use the law of indices 2(2^k) is basically 2^k+1 and vice versa

    (p.s. I've also had to teach myself most of further maths, just started in jan
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    (Original post by Billsonbubbles)
    Thanks, I dont get the factorisation on the second line of working.


    Also I have taught myself AS further maths, do you have any tips that your teacher has mentioned , or you think is nice to know.


    Like for me I noted that


    Area of the new shape = Determinant of the transformation matrix x area of the old shape



    And this works for quadrilaterals and triangles when you want to work out the area of shape .


    Enjoy and thanks !
    So for the second line of working the questions asks for 6f(k) you already have 6.6^k so by changing the 2.2^k to 6^k-4.2^k you still get 2.2K^2 and can factorise to get in the form that the question is asking, its a case of jigging it around to get what the question asks.,

    In terms of tips I havnt learnt the transformations i prove them this takes maybe a minute longer but you save time if you cant remember instead of a guess!

    Cheers!
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    (Original post by techfan42)
    post 983; all he did was use the law of indices 2(2^k) is basically 2^k+1 and vice versa

    (p.s. I've also had to teach myself most of further maths, just started in jan
    Oh nice, cheers by the way.


    I take it you have finished it then ?


    I still have half of S2 to do aha FML ah.


    Not to mention going over pretty much off all of A2 economics.

    And A2 physics ah

    get it I just need to go over it in my head again for the last two

    Cheers
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    (Original post by iMacJack)
    Yep - silly me.
    Cheers
    explain to me please!!!!!!! stuck on it for 20 mins (i did c2 2 years ago and got 100 yet i still cant remember how to do it loooool)
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    (Original post by Billsonbubbles)
    Oh nice, cheers by the way.


    I take it you have finished it then ?


    I still have half of S2 to do aha FML ah.


    Not to mention going over pretty much off all of A2 economics.

    And A2 physics ah

    get it I just need to go over it in my head again for the last two

    Cheers
    I'm actually doing it AS, so they're making us doing FP1, FP2 and D1. Which is messed up because FP2 is usually an A2 module and you need C3/C4 knowledge, but it's fine cause I went over the C3/C4 books too
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    https://gyazo.com/bb2a0cc0002c1fdf91f31d04facbb76a

    part b help with the geometric series bit please
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    (Original post by tazza ma razza)
    explain to me please!!!!!!! stuck on it for 20 mins (i did c2 2 years ago and got 100 yet i still cant remember how to do it loooool)
    Split it up into the sum of r = 1 to 12 of (9r^2-4r) + k lots of the sum of r = 1 to 12 of 2^r

    put r = 1 in to get your a, your n = 12, use the sum of a geometric series formula, youll get a value, add the bits together, equate it to the value given, rearrange for k
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    (Original post by Billsonbubbles)
    Thanks, I dont get the factorisation on the second line of working.


    Also I have taught myself AS further maths, do you have any tips that your teacher has mentioned , or you think is nice to know.


    Like for me I noted that


    Area of the new shape = Determinant of the transformation matrix x area of the old shape



    And this works for quadrilaterals and triangles when you want to work out the area of shape .


    Enjoy and thanks !
    bruh it says that in the textbook
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    (Original post by tazza ma razza)
    https://gyazo.com/bb2a0cc0002c1fdf91f31d04facbb76a

    part b help with the geometric series bit please
    sum of 2^r from r=0 to r=n is 2^(n+1) -1
 
 
 
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