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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Are you using radians for sin x? EDIT: Once y > 200, x and sin x will have no more solutions. Therefore, we're looking at the domain of x < 200 (since y = x, and y must be less than 200). In this domain, sin x completes 31.8...

10C4: four distinct numbers (pick 4 different numbers from 10) 3 x (10C3): 3 groups total: one group of two digits and two with one (pick 3 numbers from 10, pick one of the numbers to have two digits) 10C2: 2 groups of two digits (pick 2 numbers from the 10) 10P2: 2 groups: one with 3 digits...

My answer was wrong, but I fixed it. For the last case, I assumed there was order (which I was correcting rand for, lol).

I got 715, I think the 5950 accounts for the order of the number, which we don't need to worry about. No repeated numbers: 10C4 = 210 (pick any 4 from the 10) 2 unique numbers: 9C2 x 10 = 360 (pick one from the 10 to have 2 digits, then pick 2 from the remaining 9) 1 unique number: 10 x 9 = 90...

Re: HSC 2015 4U Marathon P(x) = x^3 - 3x + 1 $Substitute $ x = t + \frac{1}{t} \begin{align*} P(t + \frac{1}{t}) &= (t + \frac{1}{t})^3 - 3(t + \frac{1}{t}) + 1 \\ &= t^3 + 1 + \frac{1}{t^3} \end{align*} $Let $t = cis \theta $Therefore, $ \\ \begin{align*} t + \frac{1}{t} &= 2cos \theta...

Re: HSC 2015 3U Marathon Alternatively, multiply the top and bottom by [1 + x^(1/4)]. This makes the bottom [1 - x^(1/2)] and the top [(1 - x^(1/2))(1 + x^(1/4)). So, you're just integrating 1 + x^(1/4).

Re: HSC 2015 4U Marathon $Let the tension at $ C = T_1 $, the tension at $ B = T_2 $ and the total tension in the string $ = T$.$ $Note that the diagram will be two circles with centre A, with C circling with radius 2 and B circling with radius 3.$ $Horizontal component of the forces at C:$...

Re: HSC 2015 4U Marathon $If $z_1 - z_2 + z_3 - z_4 = 0$, then $z_1 - z_2 = z_4 - z_3$ and $z_1 - z_4 = z_2 - z_3$.$ $Therefore, the quadrilateral has two opposite sets of equal sides.$ $If $z_1 - iz_2 - z_3 + iz_4 = 0$, then $z_1 - z_3 = i(z_2 - z_4)$, i.e. the two diagonals of the...

MPC = (change in consumption)/(change in income) = 90/150 = 0.6 k = 1/(1-MPC) = 1/0.4 = 2.5 Change in equilibrium income = k x change in investment = 2.5 x 40bn = $100bn Therefore, the...

Re: 2015 permutation X2 marathon I had something similar to this, except I think that \binom{2(k-1)}{k-1} assumes that you can be broke or have negative dollars, and the game will continue. Instead, I had: [\binom{2(k-1)}{k-1} - \binom{2(k-2)}{k-2} + \binom{2(k-3)}{k-3} - ... ] + 1 for...

Think of the generator like a power pack for the external circuit: the current from a power pack enters the external circuit from the positive terminal then re-enters the power pack through the negative terminal. In this case, the current goes out of the generator through Terminal B (making it...

It says it's a generator, so we use the left hand push rule, meaning the current is flowing clockwise.

98.75 EDIT: 2015 HSC, and I like maths! :D

Re: 2015 permutation X2 marathon $Prove true for n=1: \begin{align*} 1 \: round &= 3 \: teams \\ &= 1 \: game \\ P &= 1 \\ &= 3^0 \\ &= 3^{1-1} \end{align*} $Therefore, this statement is true for $n=1$. $Assume true for $n=k$:$ $i.e. $ P(k\:rounds) = 3^{1-k} $Prove true for...

Re: 2015 permutation X2 marathon Oh, right! Yeah, something seemed off in that first round, I couldn't figure it out. Thanks, I'll try to fix that!

Re: 2015 permutation X2 marathon I hope this image is clear enough: http://imgur.com/dhLssUm EDIT: Latex added $Prove true for n=1: \begin{align*} 1 \: round &= 3 \: teams \\ &= 1 \: game \\ P &= 1 \\ &= 3^0 \\ &= 3^{1-1} \end{align*} $Therefore, this statement is true for...

No worries! Glad to help! (My teacher showed me how to do it, haha)

Since the circle touches both the x-axis and y-axis, and it has radius r, its centre will be (r,r) and its equation (x-r)^2 + (y-r)^2 = r^2 (this is a lot clearer with a diagram). Then we use the perpendicular distance formula from the centre to the tangent, letting the distance equal r. This...

Re: HSC 2015 4U Marathon I guess p, q and r could just be positive? ¯\_(ツ)_/¯

Re: HSC 2015 4U Marathon I'm not sure if there's much mathematical merit behind this, but if p, q and r are positive integers, wouldn't the only case for which (p+2)(q+2)(r+2) = 64 holds true be when p = q = r = 2 (pqr = 8)? p, q or r cannot equal 1, since 3 isn't a factor of 64. When...