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    (Original post by edothero)
    Can you do ^{0}\sqrt{4} ? :laugh:
    I meant 0^even number
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    (Original post by Middriver)
    I meant 0^even number
    Oh right, well it wouldn't make much of a difference whether if it were even or odd.

    0^{x} = \ 0

    for all x
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    (Original post by edothero)
    Oh right, well it wouldn't make much of a difference whether if it were even or odd.

    0^{x} = \ 0

    for all x
    0^0 is undefined.
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    (Original post by Zacken)
    0^0 is undefined.
    Yeah my bad, except 0
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    (Original post by Middriver)
    I meant 0^even number
    I just realised what you were asking me :laugh::laugh: sorry
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    Hi,
    Can someone please explain why, when solving a differential equation, you don't put constants of integration on both sides? Thanks in advance

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    (Original post by SeanFM)
    The line l intersects the curve but they do not necessarily have the same gradients at that point (eg think of the intersection of y = x^2 and y = x+2. They don't have the same gradient where they intersect.
    Ohh I see, thank you!
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    (Original post by Clovers)
    Hi,
    Can someone please explain why, when solving a differential equation, you don't put constants of integration on both sides? Thanks in advance

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    say your solution is of the form y = f(x) + c for an arbitrary constant c
    Well you could write y + b = f(x) + c, where b and c are both constants. But then y = f(x) + (c-b). since c and b are both arbitrary (they can be any constant you can think of), c-b can be any constant you can think of. So putting constants on both sides doesn't alter the situation.
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    http://www.examsolutions.net/a-level...uary/paper.php can somebody please help me with part A
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    Hi guys any hard integrals which i should do in preparation?
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    how would I draw the graph of
    y=3/2x-1
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    (Original post by imran_)
    how would I draw the graph of
    y=3/2x-1
    Basically just like you'd draw the graph of 1/x but shift the asymptote accordingly
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    (Original post by 13 1 20 8 42)
    Basically just like you'd draw the graph of 1/x but shift the asymptote accordingly
    what I did first was draw 3/(x-1)

    when I multiply it by the 2 it becomes 1/2 multiplied by 1 right>?
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    (Original post by imran_)
    how would I draw the graph of
    y=3/2x-1
    Vertical asymptote at x=1/2

    Horizontal asymptote at y=0
    • Very Important Poster
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    (Original post by imran_)
    how would I draw the graph of
    y=3/2x-1
    If you are ever stuck with graphs, plug in 'nice' values as well as test the limits of them.

    Eg what happens when x is less than 1/2 but increases to 1/2? What happens when x is greater than 1/2 and decreases 1/2?

    Answers are that when x is just under 1/2 then you are dividing by a very large negative number (eg put in x = 0.499999999) so 3/small -ve number = very large -ve number and that tells y is negative to the left ox f = 1/2, and similarly if you put in x = 0.50000001 then you are dividing by a very lage +ve number so y is a very large +ve number. And then as x tends to +ve infinity y tends to 0, same with - infinity.
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    (Original post by imran_)
    what I did first was draw 3/(x-1)

    when I multiply it by the 2 it becomes 1/2 multiplied by 1 right>?
    I'm not sure what you mean..
    Probably better to think in smaller steps. How would you draw 1/x? Then how would you draw 1/2x? Then 1/(2x - 1) is just a horizontal shift of this right? And then you multiply by 3
    Although to be honest the scale factors (going from 1/x to 1/2x and from 1/(2x - 1) to 3/(2x - 1)) don't really matter if you're just drawing this graph by itself

    Edit: Ah yeah well the horizontal shift is by a 1/2 if that's what you're referring to
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    Does anyone have the IAL c34 question paper only (no model answers just want to do the questions first)
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    (Original post by metrize)
    Does anyone have the IAL c34 question paper only (no model answers just want to do the questions first)
    Should be on physicsandmathstutor

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    Can someone help on jan 13 c4 question 8?

    https://googledrive.com/host/0B1ZiqB...%20Edexcel.pdf

    This is my method:
    -ln(3-theta all over A)=t/125
    So 3-theta over A=e^(-t/125)
    So 3-theta = Ae^(-t/125)
    So theta = 3-Ae^(-t/125)


    But the answer is meant to be theta = 3+Ae^(-t/125)
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    Name:  ImageUploadedByStudent Room1464640913.197247.jpg
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    Spent so long on a still couldn't do it


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