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The Soc for People of 'GRDCT2008' Mk VI Watch

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    Cheers guys! :perv:
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    What's the offer? :teeth:
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    (Original post by Champagne Supernova)
    What's the offer? :teeth:
    Check the post above yours.

    There's a hidden url link somewhere...
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    Well done, Ghoshy!

    I'm proud of you :woo: :jumphug:
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    that offer :gasp:
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    (Original post by GHOSH-5)
    Check the post above yours.

    There's a hidden url link somewhere...
    Sneaky! :p:

    And very well done mate, very well deserved! Now, get the grades. :hmmm: :p:
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    Thanks very much!
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    (Original post by GHOSH-5)
    CAMBRIDGEEEE!!!! :woo:
    not shocked at all, well deserved bud! congraaaaats!!
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    Well done Ghosh, again! :p:

    ANNDDD
    Someone math me up

    How the hell do you get the range from a function like so:

    f:x = (2x+3)/(x-1), X = any real number and x > 1.

    Apparently it's x > 2, the solution bank says something like as it approaches infinity, but I don't understand. HELP ME PLZ.
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    woo, well done ghosh:woo: and :perv: :p: :teehee: =]
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    (Original post by ArchedEdge)
    Well done Ghosh, again! :p:

    ANNDDD
    Someone math me up

    How the hell do you get the range from a function like so:

    f:x = (2x+3)/(x-1), X = any real number and x > 1.

    Apparently it's x > 2, the solution bank says something like as it approaches infinity, but I don't understand. HELP ME PLZ.
    Write:

     \dfrac{2x+3}{x-1} = \dfrac{2(x-1) + 5}{x-1} = 2 + \dfrac{5}{x-1} > 2 for all x, as x > 1, so x-1 > 0, so 5/(x-1) > 0.

    Furthermore, as x approaches infinity, 5/(x-1) approaches zero, so f(x) approaches 2.
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    (Original post by John Locke)
    not shocked at all, well deserved bud! congraaaaats!!
    (Original post by Malsy)
    woo, well done ghosh:woo: and :perv: :p: :teehee: =]
    Cheers guys!
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    (Original post by GHOSH-5)
    Cheers guys!

    You must be on such a high! really amazing you cambridge mathmo! :hugs:
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    (Original post by Malsy)
    You must be on such a high! really amazing you cambridge mathmo! :hugs:
    Well, yeh on a high, but I know that, due to STEP, nothing is assured yet.
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    (Original post by GHOSH-5)
    Well, yeh on a high, but I know that, due to STEP, nothing is assured yet.
    you're an absolute STEP machine though! especially after you destroyed I :p:
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    (Original post by GHOSH-5)
    Write:

     \dfrac{2x+3}{x-1} = \dfrac{2(x-1) + 5}{x-1} = 2 + \dfrac{5}{x-1} > 2 for all x, as x > 1, so x-1 > 0, so 5/(x-1) > 0.

    Furthermore, as x approaches infinity, 5/(x-1) approaches zero, so f(x) approaches 2.
    THANKS DUDE

    Maths is so gay.

    Like

    Homoexpolosion

    JUST like that
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    (Original post by GHOSH-5)
    Well, yeh on a high, but I know that, due to STEP, nothing is assured yet.

    Well good luck with everything ghoshy :]
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    (Original post by John Locke)
    you're an absolute STEP machine though! especially after you destroyed I :p:
    Lol, cheers.
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    (Original post by GHOSH-5)
    Well, yeh on a high, but I know that, due to STEP, nothing is assured yet.
    Which college are you? I'll point you out some people you may find nearer the time pls so you can freak them out - I think there are three mathmos I know of going so that could be fun
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    (Original post by ArchedEdge)
    THANKS DUDE

    Maths is so gay.

    Like

    Homoexpolosion

    JUST like that
    Almost as gay as PPS. [The backdoor course of 2009] :awesome:
 
 
 
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