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lmao, what school you at, so i know in the future not to send my kids there :P

basically i did..she only helped me with some tough questions...i pretty gave up in the end couldnt be bothered anymore

yeh i had that last yr...but my teacher was like ceebs you do it yoruself lol

go crazy with http://users.tpg.com.au/nanahcub/me1.html

i was in the same situation as you last yr i still got e4 for ex2..only depends on your hsc mark

well for the division induction questions subbing the assumption is the best way to do it

if you look at 4d of http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_ext1_05.pdf we have to prove: 4^{n} - 1 - 7n > 0 for n\geq 2 now step 1 is easy so i'll skip it and go to step 2. Step 2 Assume true for n=k 4^{k} - 1 - 7k > 0 Step 3 To prove...

i know where you are coming from lol...im just trying to point out both the + or - solution will give the same answer as they are both roots of the polynomial so you only need to sub in one of them to find the polynomial with the new desired roots

The HSC is generally nice when marking they wanna give you marks...i reckon just add one line in saying equating real and imaginary parts gives: then you should be fine, doing this extra line demonstrates to the marker you know what you are doing

we know that: \sum_{k=0}^{2r}(-1)^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C})= (-1)^r(_{r}^{n}\textrm{C}) from i) Now we break the above sum on the LHS to: \sum_{k=0}^{2r}(-1)^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C})=...

it doesnt matter which you use, because we have a = (x \pm \sqrt{x^2 - 4})/2, where a is our root of the new eqn, as a is our root P(a)= 0, so that means P(a = (x + \sqrt{x^2 - 4})/2)=P(a = (x - \sqrt{x^2 - 4})/2)=0 so you can use either cause both will give right answer

subbing either the + or - root doesn't matter..they both give right answer..that's why i didnt bother with the neg one

lol thankyou...i didnt realise the latex thing till you said it on the other post

is that better :P

so you should work out that you have to sub a = (1/2)[x + \sqrt{x^2 - 4}] into the original eqn... doing so gives: (1/8)[x + \sqrt{x^2 - 4}]^3 + (3/2)[x + \sqrt{x^2 - 4}] + 2 = 0 ...now i will times everything by 8 [x + \sqrt{x^2 - 4}]^3 + 12[x + \sqrt{x^2 - 4}] + 16 = 0 ...here...

i will let theta equal x to make it simply for me z = cosx + isinx, z + 1 = cosx + 1 + isinx, to eliminate the 1 use cosx = 2cos^2(x/2) - 1 and also sub sinx = 2sin(x/2)cos(x/2) z + 1 = 2cos^2(x/2) - 1 + 1 + 2isin(x/2)cos(x/2) ...now we can factor cos(x/2) out and the first part is done z...

so you should work out that you have to sub a = (1/2)[x + sqrt(x^2 - 4)] into the original eqn... doing so gives: (1/8)[x + sqrt(x^2 - 4)]^3 + (3/2)[x + sqrt(x^2 - 4)] + 2 = 0 ...now i will times everything by 8 [x + sqrt(x^2 - 4)]^3 + 12[x + sqrt(x^2 - 4)] + 16 = 0 ...here expanding this...

nah man thats a weak mark lol

i got 95 for 2u maths 48 for ext 1 maths, im droppin maths this yr and doin ext 2 so im not fussed about that mark

well mi whole class, (just me), have finished complex numbers and graphing and now we are done for the term