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b) dx/dt = dx/d@ x d@/dt d@/dt = 1/3 x = (45-36cos@)^1/2 dx/d@ = 1/2 x 36sin@ (45-36cos@)^(-1/2) dx/dt = 6sin@/(sqrt(45-36cos@) At pi/3 3sqrt3/sqrt(45-18) = 3sqrt3/3sqrt3 = 1

a) x^2 = 6^2 +3^2 - 2 x 6 x 3 cos@ x^2 = 45 - 36 cos@

e) (3-2x)^6/(-12) = -(3-2x)^6 /12 f) (x+1/x) (x-1/x) = x^2 + x/x-x/x -1/x^2 =x^2 - x^(-2) =(1/3)x^3 + x^-1 = (x^3)/3 + 1/(3x)

d) = Int(x+2)(x-2)/(x-2) =Int(x+2) x does not equal 2 = 1/2x^2 + 2x +c (x does not equal 2)

1a) 4x^3 - 4x + c b)Int[ x^4 + x - 2x^(2.5)] = 1/5x^5 +1/2x^2 - 4/7x^(3.5) + c c)2u +c

Yeah, complex numbers did seem a bit easy (probably why we did it first). I'm looking foward to mechanics (I'm a physics nerd - already done HSC Physics accelerated).

Our teacher has decided to start polynomials for just two lessons this year. He assures us it is easier than complex numbers.

sqrt(ab) does not equal sqrt(a) x sqrt (b) when dealing with complex numbers (when a and b are negative).

And if the question uses "at least" it is usually easier to find the probability of it not happening and subtract from one.

The Cambridge textbook is good for some more difficult questions.

Referring to this year's paper? The MC was hard but the rest was just half maths half social sciences - not much study required. Cover all the dot points - you may not do all of them in class. Remember that even the most obscure dot points can be tested - this year we had to assess the...

I did Quanta - this year it was easy in the exams. No one does geophysics - have you seen the thread for it?

That's the same as if it was not replaced after the first draw.

The syllabus is extremely important. Go through the syllabus and write a paragraph on each dot point after doing each section in class.

The majority of muslims are peaceful. Christianity and islam share many of the fundamental beliefs. While many horrible things are done in the name of islam, so have many horrible things been done in the name of christianity, but they are not advocated by either the Quran or the Bible.

Sum of all numbers that ARE multiples of six = 6+12+18+24..... Arithmetic series with a=6 d=6 To find no. of terms below 100 100 > a + d(n-1) 100 > 6 + 6(n-1) n < 16.67 So there are 16 terms in the series S = 8[12+ 6x15] S = 816 Sum of all integers between 1 and 100 = 50 (1...

S = n/2[2a + d(n-1)] Sum (5 terms) = 35 35 = 5/2[2a + d(5-1)] 14 = 2a + 4d Sum (10 terms) = 35 + 160 = 195 195 = 5[2a + d(10-1)] 39 = 2a + 9d 1) - 2) 25 = 5d d = 5 a = -3

I got 12. something as the term, which rounds up to the 13th year. Year 1 - 2005 2 - 2006 12 - 2016 13 - 2017 so the loan will be paid off in 2017