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q. A game is played by n people A1,A2,....An sitting at a table. Each person has a card with their own name on it and all these cards are put in a box. Each person in turn starting with A1 draws a card from random from the box. If the person draws their own card, that person wins and the game...

fuaark

you wait until yr 12

read it backwards?

lol the cylindrical shells slicing annulus ones are the imaginary ones but ones like this are 'real'

5 min reading time obv read all the questions and get a good look at the bigger questions like conics and q7 and 8 cause I believe in the bullshit that your subconscious works on it also its so that when you get there you know a bit about it already and dont waste time reading freshly on what...

1+z+....+z^{2n+1}\\ \\ =(z^2-2cos\frac{2\pi }{2n+1}+1)(z^2-2cos\frac{4\pi }{2n+1}+1)...(z^2-2cos\frac{2n\pi }{2n+1}+1)\\ \\ then\\ \\cos2\theta=1-2sin ^{2}\theta \\\\ so\\\\ -2cos2\theta =4sin^{2}\theta \\ \\ \therefore 1+z+....+z^{2n+1}\\ \\ =(z^2+4sin^{2}\frac{\pi...

for the question from your trial are the answers (i)0,1,2,...n (ii)x(x-1)(x-2)....(x-n), so leading term A = 1 (iii)P(n+1) = [(n+1)/(n+2)][1+n!] ?

but they make up for it with the insanely higher cost

nais now give us a question

now try the one before

try this one it might help with the other L =\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\\ Find\ L

(x-1)^2\geq 0 \leftarrow 1\\ \\ (x^2+x+1)>0\ since\ \Delta< 0\leftarrow 2\\ \\ 1\times 2 :\\ \\ (x-1)^2(x^2+x+1)\geq 0\\ (x-1)(x-1)(x^2+x+1)\geq 0\\ (x-1)(x^3-1)\geq 0\\ x^4-x-x^3+1\geq 0\\ x^4+1\geq x^3+x\\ \\ then\ \div x^2\ (x^2\neq 0)\\ \\ x^2+\frac{1}{x^2}\geq x+\frac{1}{x}

idk about hsc style but yes it does use stuff youve learnt in the hsc course

if\ L =\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}}^{...}}}\\ \\ and\ it\ is\ known\ that\ L \ lies\ between\ 0.3\ and\ 0.7, \\ show\ that\ L\ is\ approximately\ equal\ to\ 0.64 thats 0.5 to the power of 0.5 repeatedly

restart find solutions to tan3x = cot2x between 0 and pi

(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}\\ =\sum_{k=0}^{n}\frac{n(n-1)...(n-k+1)(n-k)!}{k!(n-k!)(n^{k})}\\ =\sum_{k=0}^{n}\frac{1}{k!}\times \frac{n}{n}\times \frac{n-1}{n}\times ...\times \frac{n-k+1}{n}\\ =\sum_{k=0}^{n}\frac{1}{k!}(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{1...

mate

lol sack that just go from d/dx(dy/dx) > 0 and you'll get it

lold