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Edexcel Core 3 - 21st June 2016 AM Watch

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    Lilly1234567890 It's from Jan 14 (IAL)
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    (Original post by edothero)
    It's essentially asking for the greatest value the equation could output.
    You should know that as t tends to infinity, e^-t tends to 0

    As proof, type e^-10000000 into your calculator, should give you 0
    thanks
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    Guys I don't understand how I would get the root 2 part, I got everything else correct but when I find the 6th root of 8 in my calculator, I don't get root2 as its form but I get the decimal form (the decimal = root2)
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    the c3 bits on the jan 2016 paper are so easy ho hopefully the paper wil be like that too
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    (Original post by Supermanxxxxxx)
    the c3 bits on the jan 2016 paper are so easy ho hopefully the paper wil be like that too
    So please, help a brother out.

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    (Original post by KINGYusuf)
    Guys I don't understand how I would get the root 2 part, I got everything else correct but when I find the 6th root of 8 in my calculator, I don't get root2 as its form but I get the decimal form (the decimal = root2)
    The decimal you get is root(2). The calc doesn't give a high root number like that in surd form. The decimal is exactly the same as looking at the decimal form of root(2)
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    Just a quick last-minute question, how do you know whether to use degrees or radians for iteration questions?
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    http://www.madasmaths.com/archive/iy...apers/c3_v.pdf

    Question 2, anyone?
    If you could just give me a hint, that would be great
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    (Original post by KINGYusuf)
    Guys I don't understand how I would get the root 2 part, I got everything else correct but when I find the 6th root of 8 in my calculator, I don't get root2 as its form but I get the decimal form (the decimal = root2)
    Yeah, that's a weird one. If you type in √2 the value is 1.4142... If you type in 6√8 and click on the 'change' button (on the sharp calculator) so the value is a decimal, it should read '1.4142... They are the same numbers so I have no idea why it doesn't come through as √2.
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    (Original post by jammyjosh)
    Yeah, that's a weird one. If you type in √2 the value is 1.4142... If you type in 6√8 and click on the 'change' button (on the sharp calculator) so the value is a decimal, it should read '1.4142... They are the same numbers so I have no idea why it doesn't come through as √2.
    Wtf? are we supposed to memorise roots and their corresponding decimals or something
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    (Original post by Pyslocke)
    The decimal you get is root(2). The calc doesn't give a high root number like that in surd form. The decimal is exactly the same as looking at the decimal form of root(2)
    But how would I know it's root2?

    Am I supposed to trial and error or something
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    June 2013 was a joke wtf
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    (Original post by zombaldia)
    I got Y>2 but not sure if that's right as i'm prone to getting these wrong. Basicalls as x tends to infinity I got y tends to 2 and as x tends to 2.0000000001 y tends to infinity. Hence y>2

    Is this what it said in mark scheme?
    https://b3755649dbd1afe3db91a899c3b9...%20Edexcel.pdf
    This is the markscheme I don't really get it and you got the right answer but how I don't understand?
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    (Original post by KINGYusuf)
    Guys I don't understand how I would get the root 2 part, I got everything else correct but when I find the 6th root of 8 in my calculator, I don't get root2 as its form but I get the decimal form (the decimal = root2)
    Does anyone know why this method doesn't work?

    ln2x + ln(2x^2) + ln(2x^3) = 6
    ln2x + 2ln2x + 3ln2x = 6
    6ln2x = 6
    ln2x = 1
    e^ln2x = e^1
    2x = e^1
    x = (e^1)/2
    ?
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    (Original post by pineneedles)
    http://www.madasmaths.com/archive/iy...apers/c3_v.pdf

    Question 2, anyone?
    If you could just give me a hint, that would be great
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    Might be looking at it too easily, but x = 112.62 because sin-1(12/13) = 67.38 and 180 - 67.38 for the obtuse angle. y is 28.07. Just put this into the sin equation it gives you and you will come out with the fraction as the answer. Can't understand the reasoning to give this 5 marks though.
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    (Original post by ryanroks1)
    Please could someone help with this question?
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    which paper is this from?
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    (Original post by KINGYusuf)
    But how would I know it's root2?

    Am I supposed to trial and error or something
    Write 8 as a power of 2 and the root as an indice...

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    (Original post by pauux)
    June 2013 was a joke wtf
    You found it easy?
    I'm aiming for a B overall and just got 56% on that paper, completely demoralized now and doubting whether I'll be getting that B .
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    (Original post by KINGYusuf)
    Wtf? are we supposed to memorise roots and their corresponding decimals or something
    Nah see if you have 6th root of 8, that is the same as saying 8(1/6).

    We know that 8 = 23 ---> 8(1/6) = (23)(1/6).

    Using the rule (am)n = amn,

    23)(1/6) = 2(3/6) = 2(1/2) = root(2).

    So, 8(1/6) = root(2).
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    (Original post by Economistician)
    Does anyone know why this method doesn't work?

    ln2x + ln(2x^2) + ln(2x^3) = 6
    ln2x + 2ln2x + 3ln2x = 6
    6ln2x = 6
    ln2x = 1
    e^ln2x = e^1
    2x = e^1
    x = (e^1)/2
    ?
    I did this question earlier. You need to do the following...
    ln(2x . 2x^2 . 2x^3) = 6
    ln(8x^6) = 6
    8x^6 = e^6
    x^6 = 1/8 e^6
    x = 1/root2 e
 
 
 
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