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  1. F

    numerically greatest term in binomial theorem

    then the 1st method somewhat is doing a similar thing once you know what's the largest number of the expansion of the sum then the one for the difference is just the previous expansion alternating in sign, hence their absolute values are the same
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    numerically greatest term in binomial theorem

    so if you just follow the second method and solve |Tr+1|>|Tr| you should be able to get r<3 which means that |T0|< |T1|< |T2|< |T3| > |T4|> |T5|> |T6|> ... (the strict inequality in |T3| > |T4| is verified by substituting into Tr, but you could just say |T3| >= |T4|) hence |T3| is the...
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    numerically greatest term in binomial theorem

    both of these methods are essentially doing a similar thing but the second one is better (intuitively makes more sense and also gives you the correct answer, IF you're careful) it's up to how you interpret 'numerically greatest', I think it's just the one with largest absolute value for these...
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    Is 3 unit mathematics (ext 1 maths) that hard?

    There's no definitive answer. It depends on who you are If interested, just choose it and see how it goes. You can always drop it after term3 and there's no harm learning a bit of prelim 3u maths for your 2u course
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    Trignometry?

    that's right just use d/dx x^n = n x^(n-1) , with n=-1( which you've already shown)
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    MATH1004 Discrete Math

    check out http://sydney.edu.au/summer/ for subjects offered some time next month but I don't think it's worth it even if it's offered
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    MATH1004 Discrete Math

    the number of subjects offered in summer school is quite limited though from what I can remember, only the popular ones are offered , like math1001 , maths2061 etc
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    Semester 2 USYD Chatter Thread 2013

    Does any one know what's happening with the strike next Tuesday? I fail to see why the students have to suffer because of the problems between the staff and the managing team. I do appreciate that a university should not be run like a business, but there's nothing much we, as students, can...
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    3D Trig.

    also just follow my hints (in white at #4) and then use the trig result cos(180-x)=-cos(x) to find an equation (which simplifies to a quadratic in h), from which the answer quickly follows
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    3D Trig.

    you can work out where p is by the angle of elevation from A B and C i.e. inverse tan(larger value) = larger angle = closer to P so the points , from closest to furthest are B, C , A which means, while A B and C being sequential , P lies between B and C with B closer to P Hopefully that...
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    3D Trig.

    I guess the 'hard' bit is trying to sketch the situation A B and C are sequential points with AB=90 and BC=30 From the angles given, P should be 'between' B and C Then you can express AP, BP and CP in terms of h=PQ then use the cosine rule in triangles ABP and CBP you will get a quadratic in h...
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    Should I Drop Extension 2 Maths?

    I wonder how you predicted your marks to put into atar calculator It would NOT be just your percentage mark from school assessments
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    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon $20 oranges are to be distributed among 15 students. The number of oranges given to any student can be any integer from 0 to 20. Find the probability that at least 1 student will have at least 3 oranges $
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    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Let $ u^2=2x-1$ \int \frac{dx}{x\sqrt{2x-1}} =2\int \frac{du}{1+u^2}=2\tan^{-1}u+C=2\tan^{-1}\sqrt{2x-1}+C
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    Further Trigonometry help

    Note: cos(\frac{\pi}{2}-\theta)=sin\theta
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    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread
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    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread Could you check this? Let M bet the midpoint of a chord PQ in a circle Two chords AD and BC are drawn so that they intersect at M AB and CD intersect the chord PQ at X and Y respectively Show that M is the midpoint of XY
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    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Let G be the centroid of triangle ABC which is inscribed in a circle Produce AG,BG and CG so that they intersect the circle again at points X,Y and Z respectively Show that \frac{AG}{GX}+\frac{BG}{GY}+\frac{CG}{GZ}=3
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