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    (Original post by thefatone)
    \left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}

    find x
    X= -1
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    (Original post by thefatone)
    \left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}

    find x
    nvm ):
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    (Original post by moggygeorgieee)
    X= -1
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    (Original post by chrlhyms)
    thank you so much!!!
    No worries.

    One more thing, after you've done a past paper and you're not sure how to do a question or maybe you just want to watch for revision check this YouTube channel called AchieveMaths and he has videos going through every question and explaining how to do them, but check if he a video going through the exam paper you did/are doing.

    https://www.youtube.com/channel/UCg4...6voZfG75oFso6w

    There's also videos teaching certain topics. :-)
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    (Original post by ihatehannah)
    -2/12?
    (Original post by chrlhyms)
    is it -3 ?
    nope nope unfortunately
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    (Original post by thefatone)
    nope nope unfortunately
    loll i deleted my comment bc i realised i forgot to square the bracket
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    -1
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    (Original post by chrlhyms)
    loll i deleted my comment bc i realised i forgot to square the bracket
    deleting comments again? xD
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    What was the method for this Name:  image.png
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    (Original post by thefatone)
    Why is it x = -1 ? help pls
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    (Original post by ihatehannah)
    well the topic is the same except this is a much harder version so it is actually very good practice.
    My working so far for the sin rule proof question. Just want to know if it is on the right track. Thank you
    Spoiler:
    Show
    Step 1)
    Sin30/AB=Sinx/BP
    0.5/AB=Sinx/BP
    AB/0.5=BP/Sinx
    AB=0.5BP/Sinx

    Step 2)
    180-30=150 degrees
    Sin150/AC=Sinx/PC
    AC/Sin150=PC/Sinx
    AC=0.5PC/Sinx

    So I would then have to prove BP and PC to reach a full solution?
    I've already noticed in these two answers, they bear some similarity.
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    (Original post by ihatehannah)
    well I guess we won't actually know it's been made harder untill the exam, but I figured cuz we're the last year doing this specification, they will make it harder maybe idk, because the new spec is much harder and they want to stretch us as much as they can in the current spec.
    That's the last question of a FM paper. Why are you giving people this in this thread? To make yourself feel better and higher?
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    (Original post by Kiniun)
    Why is it x = -1 ? help pls
    ok
    so

    \left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}

     \dfrac{2^{-3x}}{2^{-2}}=2^5

     2^{-3x}=2^3

    \dfrac{1}{2^{3x}}=2^3

    \dfrac{1}{2^3}=2^{3x}

    \dfrac{1}{2^3}=\left(2^x\right)^  3

    let  2^x = y

    \dfrac{1}{8}=y^3

    \dfrac{1}{2}=y

    2^x=\dfrac{1}{2}

    x=-1
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    (Original post by Gunjo)
    That's the last question of a FM paper. Why are you giving people this in this thread? To make yourself feel better and higher?
    omg, it's the same topic as u get in normal gcse maths but a little harder, it will prepare u for normal gcse maths much more , as it requires you to think a little more, so shh.
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    Mrbartonmaths.com has papers full of badly answered questions from recent Edexcel GCSEs and also papers of A/A* questions, which might be useful. They also have answers!
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    (Original post by ihatehannah)
    omg, it's the same topic but a little harder, it will prepare u for normal gcse maths much more , as it requires you to think a little more, so shh.
    You purposely put probably one of the hardest FM questions in order to taunt people about their abilities on an online forum. Pretty pathetic lmao
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    (Original post by Gunjo)
    You purposely put probably one of the hardest FM questions in order to taunt people about their abilities on an online forum. Pretty pathetic lmao
    are you joking or what?
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    (Original post by Ano123)
    I think, with all due respect, it was more a lack of understanding on your part if you didn't know what to do, as you say you know what to do for these types of questions now.
    I think it was lol, but it wasn't just me, but the majority of people doing the exam.

    If you read the examiner's report for that question on edexcel website it says: "Many students in this part tried to solve the quadratic equation..." and "...only a small minority of students were able to recognise and make the link between probability and the algebraic demands of this question."
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    (Original post by thefatone)
    ok
    so

    \left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}

     \dfrac{2^{-3x}}{2^{-2}}=2^5

     2^{-3x}=2^3

    \dfrac{1}{2^{3x}}=2^3

    \dfrac{1}{2^3}=2^{3x}

    \dfrac{1}{2^3}=\left(2^x\right)^  3

    let  2^x = y

    \dfrac{1}{8}=y^3

    \dfrac{1}{2}=y

    2^x=\dfrac{1}{2}

    x=-1
    but I do not understand , u rooted the (2^10) to get 2^5, but would you not have to root the 2 as well.
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    (Original post by _Xenon_)
    For probability & tree diagrams: http://www.mathsgenie.co.uk/resource...e-diagrams.pdf

    These are A* questions for Edexcel GCSE but he charges for the solutions. :/
    1H http://bland.in/a_star1h_new2.pdf
    2H http://bland.in/a_star2h_new2.pdf

    http://saveyourexams.co.uk/ is a site I use but you'll have to register and the edexcel GCSE maths doesn't have the question papers yet, however it does have the edexcel IGCSE maths questions + solutions by topic.
    Lol why does he charge for solutions, all of those questions are past paper questions and you can find them on the internet :/
    There's also a lot of videos of people going through exam papers, including june 2009 and 2010, which some of the questions are from.
 
 
 
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