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You're only in Yr 9? Wish I was as smart as you when I was your age.

Using ExtremeBoredUser's diagram above, this is a vector method: CD = AD - AC BD = AD - AB AB = -AC .: CD.BD = (AD - AC).(AD - AB) = (AD - AC).(AD + AC) = AD.AD + AD.AC - AC.AD - AC.AC = AD.AD - AC.AC = r^2 - r^2 = 0 (r being the radius) .:, since the dot product CD.BD = 0, CD and BD...

P(X<50) = 0 P(X<= 60) = P(50 <= X <= 60) P(X <= 75) = P(50 <= X <= 75) P(X >= 65) = 1 - P(50 <= X < 65) P(X >= 99) = 1 - P(50 <= X < 99) For a <= 50, P(X <= a) = 0 For a > 50: P(X <= a) = P(50 <= X < a) and: P(X \leq a) = \int _{50} ^a f(x) dx = \int _{50} ^a (15000x^{-3} - 750000x^{-4})dx \\...

As an afterthought, students may wish to note, in regard to divisibility problems in Proof by Induction: x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x^2 + x + 1) $ for n $ \ge 1\\ \\ 3^{2n^2} - 1 = 9^{n^2} - 1 = (9-1)(9^{n^2 -1} + 9^{n^2-2} + \cdots + 9^2 + 9 + 1) Therefore immediately...

$Let $ A = 3^{2n^2} - 1 = 9^{n^2} - 1\\ \\ $Let $B = 9^{2n+1} - 1\\ \\ Now by the usual procedure, you have shown that: 9^{(k+1)^2} - 1 = 9^{k^2 +2k+1} - 1 = 9^{2k+1}(9^{k^2} - 1 + 1) - 1\\ \\ = 9^{2k+1}(9^{k^2} -1) + 9^{2k+1} - 1\\ \\ = 9^{2k+1} \times 8M + B We now show that B is...

/_ BAC = /_ DBE = @ say (corresp angles, AC//BE) /_ EBC = /_ BCA = # say (alt. angles, AC//BE) Now /_ BAC = /_ BCA (base angles of isosceles triangle ABC) .: @ = # i.e. /_ DBE = /_ EBC .: BE bisects angle CBD. QED

Interesting how one error (a typo in this case) creates another: "Actuarial" to "acturial". Like "Hokkian"(my transliteration) is spelt "Hokkien"(to me, an inaccurate transliteration) by 99% of users, simply because someone long ago transliterated it that way. The followers don't know better or...

If I recall correctly: \frac {1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum ^{ \infty} _ {i =0 } x^i $ for $ |x| < 1\\ \\ \frac {1}{1+x} = \frac {1}{1 - (-x)} = 1 - x + x^2 - x^3 + \cdots = \sum ^{\infty} _ {i=0} (-x)^i $ for $|x| < 1

You made a very simple but fatal error in the 3rd line of your solution. Remember -t(1-t^2) = -t + t^3, not -t - t^3!!!

You have not shown your attempt: how can we tell where you are going wrong! One way: LHS = cos\theta(\frac{sin\theta}{cos\theta} - \frac {sin\frac{\theta}{2}}{cos \frac{\theta}{2}}) = sin\theta -(2cos^2 \frac{\theta}{2} - 1)\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}\\ \\...

Here's a derivation; not necessarily the best approach: cos2\theta = cos^2\theta - sin^2\theta = \frac {cos^2\theta - sin^2\theta}{cos^2\theta + sin^2\theta} = \frac{\frac{cos^2\theta}{cos^2\theta} - \frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta}{cos^2\theta} + \frac...

Instead of wasting your time worrying about your possible ATAR, why don't you just use the time to work harder for your forthcoming exam: you may add 1 or 2 units to your ATAR. And don't forget to take regular study breaks to avoid stressing yourself.

Why are you interested in Maths Ext 1, when you are only doing Standard 2? Edit: Sorry I didn't read the other posts.

It helps.

Easier with a diagram of course. The Normal curve is symmetrical about z = 0. P(-N <= z <= N) = P(-N <=z <= 0) + P(0 <= z <= N) = 0.1 + 0.1. P(z <= N) = 0.6 = P(z <= 0 ) + P(0 <= z <= N) = 0.5 + P(0 <= z <= N) Therefore P(0 <= z <= N) = 0.1 also = P(-N <= z <= 0) by symmetry.

C) 0.2

Strong Induction: Hypothesis step: Assume true for n <= k

= \left( x^2 -\frac{1}{x^2}\right)^{40}\\ \\ \therefore $ General term $ = \binom {40} i (x^2)^{40-i} (-x^2)^{-i} = \binom {40} i (-1)^i x^{80-2i -2i}\\ \\ $For term independent of x, power of x is 0 $ \\ \\ \therefore 80-4i =0 \implies i = 20\\ \\ \therefore $ constant term is $ \binom...

y = e^{kx} \implies y' = ke^{kx} \implies y'' = k^2e^{kx}\\ \\ \therefore k^2e^{kx} + 7ke^{kx} + 12e^{kx} = 0 \\ \\ \therefore e^{kx}(k^2 +7k +12) = 0 \implies k^2 + 7k + 12 = (k+4)(k + 3) = 0 $ since : $ e^{kx} \neq 0 \\ \\ \therefore k = -4 $ or $ -3

For 15b: h = \sqrt{pq} $ is equivalent to $ h^2 = pq $ or equivalently $ \frac{h}{p} = \frac {q}{h} It can be shown that triangles ADC and CDB are SIMILAR and therefore their corresponding sides are proportional. In particular: \frac{CD}{AD} = \frac{DB}{DC}\\ \\ $i.e. $ \frac{h}{p} =...