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Re: HSC 2013 4U Marathon Yes that is the first step, now try and find that complex number in terms of x and y, then try and find Cartesian equation

Re: HSC 2013 4U Marathon Yep

Re: HSC 2013 4U Marathon $Let$ \ t \ $be variable$ $i) Show that the complex number$ \ \ z=2at + iat^2 \ \ $for constant$ \ a \ $describes the locus of a parabola$ $ii) Rotate this parabola by$ \ \ \frac{\pi}{4} \ \ $radians clockwise, and find the Cartesian equation for this new parabola$

Re: HSC 2013 4U Marathon I'm going to continue 2014 on here for now - We simply start over again, if people want to ask questions make sure to stick to Complex Numbers and maybe Polynomials firstly and then move on as time goes on - non-2014ers are encouraged not to answer straight away...

yep

I don't do 3u

You can leave it in exact form

If your final answer is equivalent then you won't lose the mark If the answer is wrong then yes you will lose something

1) c 2) d 3) c 4) d 5) a 6) b 7) a 8) d 9) b 10) c

Question 11 $a)$ \ \ \frac{-7}{2} $b)$ \ \ \int \frac{dx}{\sqrt{49-4x^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{49/4 - x^2}} = \frac{1}{2} \sin{-1} \frac{2}{7}x + c $c)$ \ \ \binom{10}{7} (0.25)^7 (0.75)^3 $d) i)$ f(x) = \frac{x}{4-x^2} f'(x) = \frac{4-x^2 - x(-2x)}{(4-x)^2} =...

Question 12 $a) i)$ 2\cos(x + \alpha) = 2\cos x \cos \alpha - 2\sin x \sin \alpha 2\cos \alpha = \sqrt{3} 2\sin \alpha = 1 \alpha = \frac{\pi}{6} \therefore \ \sqrt{3} \cos x - \sin x = 2\cos(x+ \frac{\pi}{6}) $a) ii)$ \sqrt{3} \cos x - \sin x = 1 \Rightarrow \ \cos(x + \pi/6) =...

r is a variable though

Yep

Question 13 $a)$ $i)$ \frac{dr}{dt} = \frac{dV}{dt} \times \frac{dr}{dV} \frac{dV}{dr} = 4\pi r^2 \Rightarrow \ \frac{dr}{dV} = \frac{1}{4\pi r^2} = \frac{1}{A} \therefore \ \frac{dr}{dt} = -10^{-4} A \times \frac{1}{A} = -10^{-4} = $constant$ $ii)$ \frac{dr}{dt} = -10^{-4} r(t) =...

$a)$ $i)$ \ \ \frac{1}{(k+1)^2} - \frac{1}{k} + \frac{1}{k+1} = \frac{1}{(k+1)^2} - \frac{1}{k(k+1)} = \frac{1}{k+1} \left( \frac{1}{k+1} - \frac{1}{k} \right) = \frac{-1}{k(k+1)^2} < 0 \ \ $as$ \ \ k> 0 $ii)$ $Step 1$ \ \ n=2 \frac{1}{1^2} + \frac{1}{2^2} < 2 - \frac{1}{2}...

$a) Identify TWO methods used to treat drinking water to reduce the risk of infection$ \ \ \fbox{2} $b) Outline how the use of each of these methods reduces the risk of infection$ \ \ \fbox{2}

Good luck everyone

You mean that's not normal?

Simple Harmonic Projection

I agree with this, the 4U paper was like that as well (imo)