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P=$150,000, I=0.075/12=1/160 (monthly interest), n=120 monthly repayments i) A_1=P+I-M =150000+150000(1/160)-M =150000(1+1/160)-M A_2=A1+A1(I)-M =A1(1+1/160)-M =[150000(1+1/160)-M](1+1/160)-M =150000(1+1/160)^2-M(1+1/160)-M =150000(1+1/160)^2-M[1+(1+1/160)]...

I might come down to Melb soon and check out st hilda's actually, i've heard good things. Btw, how did you go with O-week? Did they let you live on the college during O-week? because i heard that you can only be admitted to college 3 days before uni commencement (march 1st). Do they let you go...

lol now that's just silly khorne

correct That's not correct, try counting the in two terms succession and you might see a pattern, the Q is really simple if you see it

New question: Find the values of r and n if in the expansion (1+x)^n in ascending powers of x, the (r+1)-th coefficient is twice the r-th and (r+10)-th coefficient is twice the (r+11)-th. What is the greatest coefficient in this expansion?

New Question: Find the sum to n terms of the series a) 1+5+9+... b) 3+7+11+... Hence or otherwise, find the sum of the series 1-3+5-7+9-11+.... to i)100 terms ii)101 terms

m(AP)=2m(BP) [The m just stands for gradient] Using m=(y2-y1)/(x2-x1) m(AP)=y/(x-a) m(GP)=y/x+a y/(x-a)=2y/(x+a) [Since m(AP)=2m(BP)] y(x+a)=2y(x-a) yx+ya=2yx-2ya yx=3ya x=3a, As required

That's really awesome dude, i really really love the university, cannot wait to commence. I think my first preference will be University College. I've only been to Queens, Ormond, Trinity and University Collegem and just like the atmosphere of it more than the others. Which College are you at?

Or maybe 2Re(z)+3Im(z)=4, converted it to 2U form, thus i retract my previous statement haha

It's easy but it's still 4U, this is a 2U thread if i'm not wrong

The question is when you have the gradient of the tangent to the curve and a point, hence to find an expression for f(x) you just integrate and find C. Should be right

equal roots means a double root so discriminant is=0 ∆=b^2-4ac=0 (k^2-2k+1)+4(2k+1)=0 k^2+6k+5=0 Factors: 5,1 (k+5)(k+1)=0 Hence k=-5 or k=-1

d/dx[(1+sinx)^5]=5(1+sinx)^4 x(cosx) =5cosx(1+sinx)^4 Question(Harder 2u): Using the fact that: (a-b)^2>=0 where a>b show that a^2+b^2+c^2>= ab+ac+cb

f'(x)=3x^2-2 f(x)=x^3-2x+C Sub in point (1,3) 3=1-2+C therefore C=4 f(x)=x^3-2x+4 Question: Factorise x^4-1

Papers 2000 and 2003 are probably the hardest in my opinion. Specifically the 10b from 2000, and the last two questions from 2003 are much harder than sourrounding years. 2004, 05, 06 and 07 have interesting Q 10's but they aren't relatively hard, just require intuitive thought.

Aww right, so moved down there for uni? what course and uni? and ahh right, not that far from albury at all. You liking it in the city? i'm going to uni of melb next year for BEngineering. Staying on college first year at least so that should be awesome!

Got mine first time in a manual The most fails i've ever heard of was a friend of mine who told us he failed 4 times, then one night we were talking to his mum whilst he wasn't there and she let it slip that he actually failed 9 times!!!!!

Concave up- f''(x)>0 Concave down- f''(x)<0

All values are in degrees: tan80=tan(60+20) =(tan60+tan20)/(1-tan60tan20) =[√3+(2tan10/1-tan^2(10)]/(1-√3tan20) Simplify gives: [√3tan^2(10)-2tan(10)-√3]/[tan^2(10)+2√3tan(10)-1] Since tan80=1/tan10 therefore tan10=[tan^2(10)+2√3tan(10)-1]/[√3tan^2(10)-2tan(10)-√3] Alternatively...

Haha yer i used to just unconsciously spin my pen in class, people would just be looking at me and i didn't realise why then later i would haha I can't help it now, whenever i got a pen in my hand and thinking or not using it i spin it lol I can do all the basics perfectly, around the thumb...