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    :goodluck: Tomorrow everyone!
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    (Original post by Ahsen2015)
    has anyone got a good way of explaining over-estimate and under-estimate?
    I assume you mean after using to trapezium rule to find the area under a curve?
    Well if the curve is has a positive gradient then the trapezium rule answer will be an overestimate and if it's a negative gradient then it's an underestimate.
    I apologise for doing these in paint I know they're crap but they're an exaggerated version of what each trapezium will be like. It all about it the top line of the trapezium is above or below the curve
    don't click the link click the thumbnail

    Attachment 402629402631 Attachment 402629402631
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    (Original post by Anon-)
    Haven't seen a thread for this yet, thought I should make one.

    Specification: http://qualifications.pearson.com/co...hs_Issue_3.pdf

    Here's a graph and table I made of the previous C2 grade boundaries, make of it what you will: http://i.imgur.com/g9IBQDF.png

    If there's anything you want me to add to this post, let me know.

    Good luck revising.
    lol this is such a **** graph
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    (Original post by physicsmaths)
    That is c2 may 20th 2015 paper leaked. gimme 300000pound and I can give u the real c2 paper tmmrw?


    Posted from TSR Mobile
    Would you please add appropriate notation to clarify what physical quantity you mean to receive? My overworked physics brain is confusing the said amount with a mass of 1.36x10^5 kg - which would seem a bit much for just 1 paper. Sure, for 2 I might consider it, but for 1 1:30 mins long paper I am not going to hire an aircarft transporter to get a full Boeing767-200/-200ER (max take off weight) to your designated place.

    (I have too much time on my hands)
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    Can someone please help me with this logs question, a bit stuck:

    The curves y = (1/3)^x and y = 2(3^x) intersect at the point P.

    Find the coordinates of P to 2 decimal places.
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    (Original post by mursalxxo)
    Ohh swear down? Okay sure I'll arrange the money for ya xxx
    Swear down fam


    Posted from TSR Mobile
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    could anyone demonstrate how to use a CAST diagram for:

    Solve, for 0<x<180

    sin(2x-30)+1 = 0.4

    I know how to do the question, sin-1(-0.6) = -36.9

    then 180 --36.9 = 216.9

    and 360+-36.9 = 323.1

    216.9 + 30 / 2 = 123.45

    323.1 + 30/2 = 176.55

    those are the answers and I get 123.45 but I'm not sure why you add -36.9 to 360
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    (Original post by .JC.)
    Can someone please help me with this logs question, a bit stuck:

    The curves y = (1/3)^x and y = 2(3^x) intersect at the point P.

    Find the coordinates of P to 2 decimal places.
    Make them equal each other, then log both sides ao you'll have log(1/3)^x = log(2)^3x

    then use the power law to put the x in front
    so: xlog(1/3) = (3x)log2.

    3xlog2 = 3log2 + xlog2
    xlog(1/3) - xlog2 = 3log2
    x[log(1/3) - log2] = 3log2

    and then divide

    Edit: I am using thw log to base 10 for those that may be confused about it.
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    So ****ing nervous
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    (Original post by .JC.)
    Can someone please help me with this logs question, a bit stuck:

    The curves y = (1/3)^x and y = 2(3^x) intersect at the point P.

    Find the coordinates of P to 2 decimal places.
    Not sure if i'm right, but

    (1/3)^x = 2(3^x)
    x ln(1/3) = 2x ln3
    x ln(1/3) - 2x ln3 = 0
    x(ln(1/3) - 2 ln3) = 0
    x( (ln(1/3)/(ln3^2) ) =0
    nope screw that, doesn't work xD
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    (Original post by AJC1997)
    could anyone demonstrate how to use a CAST diagram for:

    Solve, for 0<x<180

    sin(2x-30)+1 = 0.4

    I know how to do the question, sin-1(-0.6) = -36.9

    then 180 --36.9 = 216.9

    and 360+-36.9 = 323.1

    216.9 + 30 / 2 = 123.45

    323.1 + 30/2 = 176.55

    those are the answers and I get 123.45 but I'm not sure why you add -36.9 to 360
    Going round anticlockwise into quadrant 4, meeting the line. The total angle is 360 - the 36.9.
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    (Original post by .JC.)
    Can someone please help me with this logs question, a bit stuck:

    The curves y = (1/3)^x and y = 2(3^x) intersect at the point P.

    Find the coordinates of P to 2 decimal places.
    1/3^x=2(3^x)
    so xlog1/3= log2 + xlog3
    xlog1/3-xlog3=log2
    x(log1/3-log3)=log2
    x=log2/ (log1/3-log3)
    Why you taken a selfie in a public toilet? that's gay af
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    1st question i soo confusing, can anyone help me to solve it thru?

    the 2nd page, what does it mean by using " TRIGNOMETRY" solve the equation?
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    Any help on how to solve this, not too sure how to do it?
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    (Original post by frozo123)
    1/3^x=2(3^x)
    so xlog1/3= log2 + xlog3
    xlog1/3-xlog3=log2
    x(log1/3-log3)=log2
    x=log2/ (log1/3-log3)
    Why you taken a selfie in a public toilet? that's gay af
    Cheers, seems so simple now written out.

    It's in the toilet at my gym bro, it's allowed
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    (Original post by jshep000)
    Any help on how to solve this, not too sure how to do it?
    is the answer 1/6 pi and -5/6 pi?
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    Why do I have a good feeling about this exam
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    (Original post by .JC.)
    Can someone please help me with this logs question, a bit stuck:

    The curves y = (1/3)^x and y = 2(3^x) intersect at the point P.

    Find the coordinates of P to 2 decimal places.
    I haven't got a mark scheme to check it, but this seems to work:

    (1/3)^x = 2(3^x)
    log(1/3)^x = log2(3^x)
    xlog(1/3) = log2 + log3^x
    xlog(1/3) = log2 + xlog3
    xlog(1/3) - xlog3 = log2
    x[log(1/3) - log3] = log2
    xlog[(1/3)/3] = log2
    xlog(1/9) = log2
    x = log2/log(1/9)
    x = -0.315...

    y = (1/3)^x
    y = (1/3)^-0.315
    y = 1.413...

    P: (-0.32, 1.41)
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    Is it just me doesn't understand CAST diagrams and draws out the sine/cos/tan graphs instead?
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    Also could someone post the formulas for volumes we may need to know?
 
 
 
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