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Timske, you found the equation of the tangent at that point of the curve

\frac{dy}{dx}= 34x Now we need the primitive function. y = 17x^2 +c Sub in the given point -1 = 17 + c c = -18 \therefore y = 17x^2 - 18

Re: HSC 2012 Chemistry Marathon Alpha particles can only travel a few cm in the air and can be stopped by a sheet of paper. Beta particles can travel even further, up to 2m, and can be stopped by a sheet of aluminium. Gamma rays are highly penetrating, able to travel long distances which...

Diagram first. First we want to find the points of intercepts. So solving both of the equations, x^2 = (x + 2)^2 x^2 - (x + 2)^2 = 0 (2x + 2)(-2) = 0\;\;\;$I just did difference of two squares$ x = -1 Now splitting the region in to two so we can work with them seperately. First...

Re: 2012 HSC MX1 Marathon Use the method of bisection to solve to 2 decimal places: log_e(x + 2) = e^x - 2

The HSC Chemistry Marathon is an open chain of questions between students. It works by answering a question then posting another question and allowing the cycle to repeat itself. Rules: - After answering a question, always provide a new one - this is what keeps the thread alive. - Allocate a...

a) Mg_{(s)} + H_2SO_{4(aq)} \rightarrow MgSO_{4(aq)} + H_{2(g)} $Decrease in hydrogen ions in solution, produced salt is neutral, if reaction goes to completion will have a pH of 7$ b) 2Na_{(s)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq)} + H_2{(g)} NaOH_{(aq)} \rightarrow Na^+_{(aq)} +...

For those who don't want to turn their heads.

It's not entirely all that simple. You have to know log integration to do it. \int \frac{x^2 + 1}{x}dx = \int x + \frac{1}{x} dx = \frac{x^2}{2} + lnx + c

Re: 2012 HSC MX2 Marathon If you want to use Latex in your post, use the opening and closing brackets [./tex] (without the dot). \int^4_1 xf''(x) dx = \left[ xf'(x) \right ]^4_1 - \int^4_1 f'(x) dx = 4f'(4) - 4f'(1) - \left [f(x) \right ]^4 _1 16 - 8 - f(4) + f(1) = 8 - 5 + 2 [tex]= 5...

Re: 2012 HSC MX1 Marathon 1. You need a calculator. You end up with 2x + 15 = ??? x = \frac{??? - 15}{2} Find the solutions between 0 to 360 2. Rearrange and solve. cos2y = -\frac{1}{2} 2y = ???...*$insert all of the solutions$* 3. Same approach for 1.

In year 11, you try both ways and see which one works best for you.

Re: 2012 HSC MX2 Marathon \int sin(lnx)dx $Let u = lnx$ e^u = x du = \frac{1}{x}dx \therefore \int sin(lnx) dx = \int e^usinudu $Applying I.B.P.$ = e^usinu - \int e^ucosudu = e^usinu - (e^ucosu + \int e^usinudu) \int e^u sinu du= e^usinu - e^ucosu - \int e^usinudu ..... (1) $Applying...

How embarrassing.. The original working out was up for less than 10 seconds. Thought i could get away with it.

I have no clue what you're talking about but i'm curious myself. :P

6tan^2x - 4sin^2x = 1 6\frac{sin^2x}{cos^2x} - 4sin^2x - 1 = 0 6sin^2x - 4sin^2xcos^2x - cos^2x = 0 6sin^2x - 4sin^2x(1 - sin^2x) - (1 - sin^2x) = 0 6sin^2x - 4sin^2x + 4sin^4x - 1 + sin^2x = 0 4sin^4x + 3 sin^2x - 1 = 0 4sin^4x + 4sin^2x - sin^2x -1 = 0 (4sin^2x - 1)(sin^2x + 1) = 0...

lol made a mistake wait

Re: 2012 HSC MX1 Marathon Yeah my mistake lol. Examine forgot to state the condition the question requires..

Re: 2012 HSC MX1 Marathon You had the discriminant, but you let it = 0 and solved for values of k. When the discriminant = 0, you find a quadratic equation with 1 real root. If discriminant is > 0, you look for values of k such that it has real roots. So you are solving the equation: (k +...

Double angle formula: sin2x = 2sinxcosx \therefore 2sinxcosx = cosx cosx(2sinx - 1) = 0 You can do the rest yourself.