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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
    WOAH there

    no no ~!
    he's not taunting anyone

    being able to work out even harder questions will make the actual exam questions seem easier
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    heres something
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  1. File Type: pdf a_star1h_new2.pdf (613.0 KB, 83 views)
  2. File Type: pdf a_star2h_new2.pdf (659.9 KB, 54 views)
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    ty , i get it now.
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    (Original post by ihatehannah)
    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.
    no that's not how this works

     \left(2^2\right)^2= 2^4

    2^2=2^2





2^2 \not = \sqrt {2^2}
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    (Original post by ScrewTheExams)
    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.
    Not sure but they're worked solutions and yes you can find them for free too.
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    someone give me a rly hard question im bored
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    (Original post by okey)
    someone give me a rly hard question im bored
    Solve the simultaneous equations
     \displaystyle x^2+2y^2=1
     \displaystyle x+y=1 .
    Shade the area that satisfies the following inequalities
     x^2+y^2<1 and  x+y>1 .
    (Might be an easy question, I don't know).
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    (Original post by strikeh)
    heres something
    these look helpful, thank you!
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    (Original post by chrlhyms)
    thank you so much!!!
    This predicted paper is really easy


    Posted from TSR Mobile
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    (Original post by ihatehannah)
    lol ok well it's a starter, then in that case, try dis http://prntscr.com/b6fzjk, gonna replace my first link with this one actually.
    Initially it looked difficult but I put my head into it and it was pretty easy! Not sure if it is correct though but my final answer I suppose proves it.

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    (Original post by thefatone)
    no that's not how this works

     \left(2^2\right)^2= 2^4

    2^2=2^2





2^2 \not = \sqrt {2^2}
    could you say the the square root of 210 is the same as 210x1/2 which is 25
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    (Original post by BTAnonymous)
    Initially it looked difficult but I put my head into it and it was pretty easy! Not sure if it is correct though but my final answer I suppose proves it.

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    ikr idk why so many people are saying it's so hard
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    (Original post by there'snoneed)
    could you say the the square root of 210 is the same as 210x1/2 which is 25
    yes that works too
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    (Original post by okey)
    ikr idk why so many people are saying it's so hard
    I guess this is what separates A* candidates from A and B candidates. Everyone who's attempted the question knows the basic knowledge but doesn't know how to use it.
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    (Original post by ihatehannah)
    ahhh 1600, thought it said 1000, very hard to read , pic was blurry for me sry.
    how do you even do 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
    I still don't understand how you got from:

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

    to

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

    Like the other guy said, why don't you square root 10 but instead divide it by 2?
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    (Original post by BTAnonymous)
    I still don't understand how you got from:

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

    to

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

    Like the other guy said, why don't you square root 10 but instead divide it by 2?
     \sqrt{2^n}=2^{n/2} .
    More generally,  \displaystyle \sqrt[c]{a^b} = a^{\frac{b}{c}} .
    Just law of indices.
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    (Original post by BTAnonymous)
    I still don't understand how you got from:

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

    to

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

    Like the other guy said, why don't you square root 10 but instead divide it by 2?
    i'm not quite sure what you want to say? can you explain it to me in some sort of calculation?

    to get from step 1 to step 2 you square root both sides
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    Here's a question.
    Find the value of  x such that

     \displaystyle \left (\frac{2}{3} \right )^{x-1} = \left ( \frac{9}{4} \right )^{x+2} .
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    (Original post by thefatone)
    i'm not quite sure what you want to say? can you explain it to me in some sort of calculation?

    to get from step 1 to step 2 you square root both sides
    Ahh I see... just a misunderstanding.

    For some stupid reason I thought you rooted the power of 10 instead of the whole calculation.

    So 2^10 = 1024

    and square root 1024 to get 32 which is 2^5
 
 
 
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