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Let x and y be integers from 0 to 9 (inclusive) And x is the first digit and y is the second Let z be the number we are looking for, according to your property z=10x+ y z=x+y^2 10x+y=x+y^2 9x+y=y^2 y=\frac{1\pm \sqrt{1+36x}}{2} Now if put x in and it comes out that y is an integer...

So by limiting sum I am assuming you are allowed to use the formula: \frac{a}{1-\alpha} = a(1 + \alpha + \alpha^2 + \dots + \alpha^n + \dots) \ \ \ \ |\alpha| < 1 We can rewrite the decimal in f: $f)$ \ \ 0.2 \dot{3} = 0.2 + 0.0\dot{3} = 0.2 + (0.033333\dots) =0.2 + (0.03 + 0.003...

Re: HSC 2013 3U Marathon Thread Great question I came across: $The three sides in a triangle are in ratio$ \ \ 4 \ : 5 \ : \ 6 $Prove that one angle is twice the other$

Re: HSC 2013 4U Marathon Hmm not quite

Re: HSC 2013 4U Marathon $A polynomial$ \ \ S(y)=\sum_{k=0}^{2n}s_k y^k \ \ $has roots$ \ \ \alpha_1, \alpha_2, \dots , \alpha_{2n} $Find a polynomial with roots$ \ \ \pm \alpha_1, \pm \alpha_2, \dots, \pm \alpha_{2n}

Re: HSC 2013 4U Marathon $The points$ P(x_1,y_1) P'(x_1, -y_1) Q(-x_1,y_1 Q'(-x_1, -y_1) $lie on the ellipse$ \ \ \ \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 $Tangents are constructed at each of these points, the tangents at:$ $P and Q intersect at R$ $P and P' intersect at P''$ $Q and...

The points aren't deep at all, you are retelling the story half the time, and you don't seem to be going anywhere with the arguments 9-10/15

Re: HSC 2013 4U Marathon $2 points, P and Q lie on a circle with radius$ \ \ r \ \ $centred at the origin, O$ $The tangents at P and Q intersect at the point R$ \ \ \textbf{they intersect at a right angle} $The tangent at P is called p, and the tangent at Q is called q$ $The angle OP...

I made a mistake in my solution, I originally put (I edited my post) i(a-x) = b when it should be i(a-x)=b-x

Re: HSC 2013 4U Marathon Find points of intersection: \left (\pm \frac{ab}{\sqrt{a^2+b^2}}, \ \pm \frac{ab}{\sqrt{a^2+b^2}} \right ) k=\frac{ab}{\sqrt{a^2+b^2}} A=4 \cdot \left (b\int_{0}^k \sqrt{1-\frac{x^2}{a^2}} \ dx + a\int_k^b \sqrt{1-\frac{x^2}{b^2}} \ dx \right )

EDIT: I made a crucial mistake, the working has been changed Establish the properties of the square (that we will use) - Diagonals bisect at right angles - Diagonals bisect each other and are same length. Let A = a B = b C = c D = d X = x It must follow that: i(a-x) =...

Re: Help, Trig Well we know the formula: \cos 2\theta = 1-2\sin^2 \theta Substiute in theta = x/2 \cos x = 1-2\sin^2 \frac{x}{2} Then put this into the original expression: \cos x - 1 = 1-2\sin^2 \frac{x}{2} - 1 = -2\sin^2 \frac{x}{2}

For some topic patrickJMT doesn't really go into much detail in general, I recommend khanacademy

Re: HSC 2013 4U Marathon $Find the area enclosed by the ellipses$ \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 b\neq a

Re: HSC 2013 3U Marathon Thread $There exists a sequence of circular quadrants$ \ \ C_0, \ C_1, \ C_2, \ \dots $And a sequence of lines$ \ \ l_1, \ l_2, \ l_3 \ \dots C_0 \ \ $has a radius of 1$ $Subsequent circular quadrants and lines all undergo the same operation as shown$...

This was pretty mind blowing for me as well. Consider the series: 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} + \dots This series infact diverges (although very slowly), to see why this could be. First sketch a graph of y=1/x Draw the points, (1,1/1) (2,1/2) (3,1/3)...

Hey, just because it shouldn't be happening doesn't mean it doesn't happen. And no, I reckon there could be a possibility where everyone in a school does 4U.

I apologise I am not familiar with selective schools.

To each is his own :)

That question takes about 3 lines and I wouldn't mind if that wasn't in latex.