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    (Original post by Blutooth)
    I'm guessing the law of the excluded middle features quite a bit in this paradox.
    The law of excluded middle is p \lor \lnot p. The formula you quote, \lnot (p \land \lnot p), is tautological even in weak fragments of intuitionistic logic. It is just an internalised modus ponens: (p \land (p \to \bot)) \to \bot.
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    Alright fellow mathmos?!
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    GCSE stats coursework help please:
    okay so i am doing coursework on track and field events in the Olympics i am focusing on the 2008 ones but i don't know if you can create a cumulative frequency diagram with the data that i have i have decathlon and individual data both with the events of 100m sprint and high jump help please its in for next week and i have no idea what i am doing please help!?!
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    [*] During the weekends, Eli delivers milk in the complex plane.

    On Saturday, he begins at z and delivers milk to houses located at z^3, z^5, z^7, \cdots, z^{2013}, in that order.
    On Sunday, he begins at 1 and delivers milk to houses located at z^2, z^4, z^6, \cdots, z^{2012}, in that order.

    Eli always walks directly (in a straight line) between two houses.

    If the distance he must travel from his starting point to the last house is \sqrt{2012} on both days, find the real part of z^2.
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    I am doing some quantitative analysis on marginal costs and revenue, where I have to find the figure where marginal cost and marginal revenue are equal. I have
    revenue = 720x - 4x^2
    cost = 96-17x^2 + 560x

    Ive got
    marginal revenue = 720 - 8x
    marginal cost = 560 - 34x

    so when marginal cost = marginal revenue

    720 - 8x = 560 - 34x
    26x = -160
    x = -6.15....

    is it even possible to get a negative and what does this mean? at this point is it still maximum profit? which means you have to sell -6 units to make the maximum profit??

    If someone could help today please
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    Somebody who doesn't like milk has negged me. :rolleyes:
    This crappy hand pointing downwards should be removed!
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    (Original post by gff)
    Milk Delivery Problem
    Here's a solution (I hope):

    Spoiler:
    Show


    From the problem statement, we have

    |z^3-z|+|z^5-z^3|+|z^7-z^5|+\cdots + |z^{2013}-z^{2011}| = \sqrt{2012} \ \ \ \ \ (1)
    |z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}| = \sqrt{2012} \ \ \ \ \ (2)

    From (1),

    |z|(|z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}|) = \sqrt{2012}

    so from (2),

    |z|\sqrt{2012} = \sqrt{2012} \implies |z| = 1

    Then, from (2),

    |z^2-1|+|z|^2|z^2-1|+|z|^4|z^2-1|+\cdots + |z|^{2010}|z^2-1| = \sqrt{2012}
    1006|z^2-1| = \sqrt{2012}
    |z^2-1| = \frac{\sqrt{2}}{\sqrt{1006}} = \frac{1}{\sqrt{503}}

    Let z = a+bi, where a,b \in \mathbb{R}. Then |z| = 1 implies a^2+b^2 = 1 and |z^2-1| = \frac{1}{\sqrt{503}} implies |(a+bi)^2-1| = \frac{1}{\sqrt{503}}, so |(a^2-b^2-1)+2abi| = \frac{1}{\sqrt{503}} and (a^2-b^2-1)^2+4a^2b^2 = \frac{1}{503}, so a^4+b^4+2a^2b^2-2(a^2-b^2)+1 = (a^2+b^2)^2-2(a^2-b^2)+1 = 1-2(a^2-b^2)+1 = \frac{1}{503}, so a^2-b^2 = \frac{1005}{1006}.

    We have z^2=(a+bi)^2 = (a^2-b^2)+2abi, so \Re (z^2) = a^2-b^2 = \frac{1005}{1006}.

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    (Original post by nohomo)
    Here's a solution (I hope):
    It is a solution.
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    (Original post by nohomo)
    Here's a solution (I hope):

    Spoiler:
    Show


    From the problem statement, we have

    |z^3-z|+|z^5-z^3|+|z^7-z^5|+\cdots + |z^{2013}-z^{2011}| = \sqrt{2012} \ \ \ \ \ (1)
    |z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}| = \sqrt{2012} \ \ \ \ \ (2)

    From (1),

    |z|(|z^2-1|+|z^4-z^2|+|z^6-z^4|+\cdots + |z^{2012}-z^{2010}|) = \sqrt{2012}

    so from (2),

    |z|\sqrt{2012} = \sqrt{2012} \implies |z| = 1

    Then, from (2),

    |z^2-1|+|z|^2|z^2-1|+|z|^4|z^2-1|+\cdots + |z|^{2010}|z^2-1| = \sqrt{2012}
    1006|z^2-1| = \sqrt{2012}
    |z^2-1| = \frac{\sqrt{2}}{\sqrt{1006}} = \frac{1}{\sqrt{503}}
    [/latex].

    Slightly neater from this point to...

    Spoiler:
    Show
    Note that |z^2-1| = \sqrt{(z^2-1)(\bar{z}^2-1)} = \sqrt{(z\bar{z})^2 - (z^2 + \bar{z}^2) + 1} = \sqrt{2-(z^2+\bar{z}^2)} so 2 - (z^2 + \bar{z}^2) = \dfrac{1}{503} \implies \dfrac{1005}{1006} = \dfrac{1}{2}(z^2+\bar{z}^2) = \mathrm{Re}(z^2)
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    A fair die is thrown four times and the number of times it falls with a 6 on the top, Y, is noted. Write down all the possible values of y.


    Why is 0 the first answer?
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    (Original post by Farhan.Hanif93)
    Slightly neater from this point to...

    Spoiler:
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    Note that |z^2-1| = \sqrt{(z^2-1)(\bar{z}^2-1)} = \sqrt{(z\bar{z})^2 - (z^2 + \bar{z}^2) + 1} = \sqrt{2-(z^2+\bar{z}^2)} so 2 - (z^2 + \bar{z}^2) = \dfrac{1}{503} \implies \dfrac{1005}{1006} = \dfrac{1}{2}(z^2+\bar{z}^2) = \mathrm{Re}(z^2)
    That's a good improvement to the solution.

    Does anyone have any more problems for people to solve?
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    (Original post by user)
    A fair die is thrown four times and the number of times it falls with a 6 on the top, Y, is noted. Write down all the possible values of y.


    Why is 0 the first answer?
    What result do you get if you roll 5 4 times?
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    (Original post by nohomo)
    Does anyone have any more problems for people to solve?
    I have tons of fun problems to solve, but, unfortunately, not enough time. Help yourself.
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    I know this is old but I just read the solutions by nohomo (and then the bit farhan posted); thought they were both a bit convoluted (algebra-wise) nearing the end (for me anyway)

    Where nohomo says let z = a+bi

    It's a lot easier to let z^2 = a+bi

    and we have |z^2-1| = \frac{1}{\sqrt{503}}
    or (a-1)^2 + b^2 = 1/503

    and since a^2 + b^2 = 1 then 2a-1 = 1 - 1/503 and then finally a = 1 - 1/1006

    Sorry for the lack of latex; couldn't really be bothered
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    Is this just random questions?

    5x0 = 0
    Re-arrange.
    5 = 0/0

    seems legit
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    (Original post by HypErTwisT)
    Is this just random questions?

    5x0 = 0
    Re-arrange.
    5 = 0/0

    seems legit
    0/0 is an indeterminate. And the answer proves that 0/0 is infinity as it provides no solution, so there can be an arbitrary constant equal to this equation, as in this example:5.
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    Posted for Usycool:

    When perpendicular lines meet, the product of their gradient is -1, right?

    However, say if we have the lines y = 3 and x = 5.

    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

    So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
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    (Original post by GreenLantern1)
    Posted for Usycool:

    When perpendicular lines meet, the product of their gradient is -1, right?

    However, say if we have the lines y = 3 and x = 5.

    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

    So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
    The gradient of the line x=5 is not 1.
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    (Original post by Slumpy)
    The gradient of the line x=5 is not 1.
    I only posted this for my friend not myself (so it has just been copied from what he has written to me). Can you please tell us what all the gradients are and if the prodyct of the gradients doesn't equal -1, then why!!
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    (Original post by GreenLantern1)
    The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.
    :lolwut:

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Updated: April 23, 2014
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