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Re: HSC 2013 4U Marathon Still testing the waters here, and I'm a bit tired at the moment. We know that each cross section is a segment of the semi-circle. So I decided that I would slice up everything and integrate, now I know it doesn't have a common cross section, but I decided to do...

Re: HSC 2013 4U Marathon $Find$ I=\int e^{x+e^x} \ dx J = \int \frac{1}{x^{\alpha} + x } \ dx

Re: HSC 2013 4U Marathon $let$ deg(p) = n deg(r) = o deg(d) = m deg(q) = l $It must be true that there exists non-negative integers such that$ n=ml + o The condition: o < l exists if the polynomials are all unique, since if \ \ o \geq l Then there exists...

Re: HSC 2013 4U Marathon Have you found a neat geometric way of doing it? I can do it quite easily with manually equating the equations and showing that it has no real roots. But I was wondering if you found a nicer way of doing this.

The Ottoman Caliphate was the proper Islamic system. Saudi Arabia is based off of British imperialism, it rose through rebellion to the Ottoman Empire, although it may have some elements that are in order with Islam, most of the corruption in Arabia is due to politics and elements outside of...

Maybe they don't want filthy eavesdroppers like you listening to what they want to say, which could be private. Proof? You have none

Re: HSC 2013 4U Marathon Ok I may have figured it out, first lets let the probability of winning be p for simplicity sake. We break the games into W and L, where W and L denotes win and loss respectively. We can create a string of W's and L's to represent the game, each game we can pair a W...

Re: HSC 2013 4U Marathon But for example, if we denote W as win, and L as Loss, he can go: WLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWL WWWWW WWWWW and then win. And this can go on infinitely many times, so infinite series are involved. I was experimenting with $2, whether he wins up to 4. When...

Re: HSC 2013 4U Marathon But he can lose say 9$ then win 19 to win, your probability is for when he wins 10 times in a row, it is merely one case.

re: HSC Physics Marathon Archive I can't read it well but if you showed it, you showed it, nice work.

Yes that's pretty much what the first method was but the logic behind it. But they are not interchangeable with respect to the theta that they have specified in the question. Parametrically speaking yes they are interchangeable but the thetas are different. If the co-ordinates were...

Notice that since the equation of the circle is x^2+y^2=1, then it has a radius of 1. This means that OT = 1 (the length). So if you draw out a right angle triangle, by drawing the perpendicular of T to the x-axis (lets call this point M), you have a right angle triangle with hypotenuse 1...

Re: HSC 2013 4U Marathon $A cylinder has a height of$ \ 2r \ \ $and radius$ \ \ r $Inside the cylinder a cone with height$ \ \ 2r \ \ $and radius$ \ \ r \ \ $is fit inside$ $Also a sphere of radius$ \ \ r \ \ $is fit inside, such that all 3 solids share a certain volume$ $Show that this...

re: HSC Physics Marathon Archive $By using the fact that work is the change in kinetic energy$ $Prove that for a projectile in projectile motion$ (v_y)^2 = (v_0 \sin \theta)^2 -2gy $Where$ \ \ v_y \ \ $is the vertical component of velocity$ \theta \ \ $is the angle of projection$ v_0 \...

This, though I like to be a little risky and take 15 min breaks every 2 hours I might regret it in the end though :/

Re: HSC 2013 4U Marathon Redone

From the previous question, remember our quadratic equation. y^2+y(1-2c)+c^2-r^2 = 0 The solutions to this equation are the y-coordinates of the intersection of the parabola and circle. But it is evident that at both intersections they have the same y-coordinate, and hence discriminant is...

Re: HSC 2013 4U Marathon $i) Explain why, for$ \ \ x \geq 3 \frac{1}{x(x+1)(x+2)} < \frac{1}{x^3} < \frac{1}{x(x-1)(x-2)} $ii) Hence prove that$ \frac{7}{6} < \sum_{n=1}^{\infty} \frac{1}{n^3} < \frac{11}{8}

Re: HSC 2013 4U Marathon I_n = t(1+t^2)^n |_0^x - 2n \int_0^x t^2(1+t)^{n-1} \ dt (1+t^2)^n-(1+t^2)^{n-1} = t^2(1+t^2)^{n-1} \therefore \ \ I_n = x(1+x^2)^n - 2n(I_n - I_{n-1}) (2n+1)I_n = (1+x^2)^n x +2nI_{n-1} \therefore \ \ I_n = \frac{1}{2n+1} (1+x^2)^n x + \frac{2n}{2n+1} I_{n-1}

Re: HSC 2013 3U Marathon Thread Let theta be an angle in the triangle. c^2=a^2+b^2-2ab\cos \theta \cos theta = \frac{a^2+b^2-c^2}{2ab} A=\frac{1}{2}ab \sin \theta 1-\cos^2 \theta = 1-\left(\frac{a^2+b^2-c^2}{2ab} \right)^2 \therefore \ \sin \theta =...