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Re: HSC 2015 4U Marathon Find alpha + beta + gamma All squares are of side lengths 1. There's a way using complex numbers

I got 549 but I'm not confident with this. nevermind, it's definitely wrong if 0 can be the first number I got 670 including 0 as first number

^ Note that also, horizontal points occur when f'(x) = 0 AND f"(x) = 0 Which means its the stationary point for the f'(x) and also when it cuts the x axis for the f'(x) curve

Re: HSC 2015 3U Marathon Most likely he did and thought you asked the question.

Re: HSC 2015 3U Marathon Just downloaded the app. He's referencing the style of questions they ask. They always ask to find the pronumerals and then give the result when we add them

Re: 2015 permutation X2 marathon I think when they said they play with everyone once, it's similar to a group of people all giving handshakes to everyone only once. So no-one is left out.

Re: HSC 2015 3U Marathon $ let $$ S $ = \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n} $ 2S $ = 1 +\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{n}{2^{n-1}} $ 2S - S $ = $ S $ = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ... + \frac{1}{2^{n-1}} - \frac{n}{2^{n}} $ S $ = 2(1 -...

Re: HSC 2015 3U Marathon Arrange the girls first = 5! Ways for boys to slip between the girls = 6P3 Arranging everyone, with the children as one item = 11! Total = 5! * 6P3* 11! = 5.7x10^11 ... a very large number

Re: 2015 permutation X2 marathon Is there a way to simplify the summation? Like starting off with a binomial expansion and then substituting some value in

Re: 2015 permutation X2 marathon Is my method valid for n= even?

Re: 2015 permutation X2 marathon For those cases I used the insertion method which would mean it's 6C2 for that example

Re: 2015 permutation X2 marathon The amount of heads (H) and the amount of tails (T) must fall under the condition H =< T for no consecutive heads to occur. Consider the case when n T's are thrown (n = even) = 1 way n-1 T's and 1 H is thrown = n!/(n-1)! = n ways n-2 T's and 2 H is thrown =...

re: HSC Chemistry Marathon Archive Oops you're right. Forgot and only added the barium hydroxide volumes.

re: HSC Chemistry Marathon Archive Ah yeah we should multiply by 2 before using the log. I got a lower answer (pH 13.28)

re: HSC Chemistry Marathon Archive $ Ba(OH)_{2} _{(aq)} $ + $ 2HNO_{3} _{(aq)} $ \rightarrow $ Ba(NO3)_{2} _{(aq)} $ + $ 2H_{2}O _{(l)} $ V_{HNO_{3}} = 0.02235 L [HNO_{3}] = 0.08 \frac{mol}{L} \therefore n_{HNO_{3}} = 0.08 * 0.02235 \therefore n_{Ba(OH)_{2}} = \frac{0.08 *...

3 marks? Was the marking criteria only like 'marks for correct solution.'?

Re: HSC 2015 3U Marathon 2. x = \frac{1}{u} dx = \frac{-1}{u^2 } du \therefore $ I $ = \int_{\frac{1}{2}}^{2} \frac{ln(\frac{1}{u})}{u^2 (1+ \frac{1}{u^2 }) } du = \int_{\frac{1}{2}}^{2} \frac{ln(\frac{1}{u})}{u^2 + 1} du = \int_{\frac{1}{2}}^{2} \frac{ln(\frac{1}{x})}{x^2 + 1} dx...

Re: HSC 2015 3U Marathon let u = x^(1/4) ==> x = u^4 dx = 4u^3 du I = integration of (1-u^2)/(1-u) * 4u^3 du I = integration of (1+u) * 4u^3 du expand and integrate

Re: HSC 2015 3U Marathon $ Let the first integer be x $ $ product of four consecutive integers $ = x(x+1)(x+2)(x+3) = (x^2 + x)(x^2 + 5x + 6) = x^4 + 6x^3 + 11x^2 + 6x = x^4 + 6x^3 + 11x^2 + 6x + 1 - 1 = (x^2 + 3x + 1)^2 - 1 = n^2 - 1 ( $where n is an integer $ )

Lol I use it for basic arithmetic. Can't trust myself with addition