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It is 29.99m. Post your working out if you are still unsure.

Re: 2012 HSC MX1 Marathon Graphical. Drew it out and explained it. By the way it was a proof by induction. The question was worded differently but i'm pretty sure they are the same thing.

Re: 2012 HSC MX1 Marathon The proof has a nice trick to it. Teacher showed it to us in class.

Re: 2012 HSC MX1 Marathon You should also realise that we're all in year 12 and there's the occasional university student posting in here.

Re: 2012 HSC MX1 Marathon

$Remember that: $ log_ba = x b^x = a $Q10. $ y = lnx x = e^y $I'm sure you can do the rest yourself.$ $Q11. use the same property above$ $You can do this method by splitting the LHS in to two fractions or combining the RHS$ $If you want to split the LHS, do it this way:$ \frac{3x +...

I typed "(3) - (1) x 2" because i was lazy lol. What it means is multiply (1) by 2 first. Then minus it from (3) and you get the result i had. I multiplied it by 8 for (5).

I do, but i'm quite sure SpiralFlex has a much more extensive collection. Perhaps ask him?

Aim to obtain two equations, both that have the same variables. If you solve for these two variables first, you'll be able to find the third very easily. 2x - y - z = 17 ..... (1) x + 3y + 4z = - 20 ..... (2) 5x - 2y + 3z = 19 ..... (3) (1) + (2) + (3): 8x + 6z = 16 ..... (4) (3) - (1)...

What is a procedure that you could perform in a laboratory to distinguish between a gaseous alkane & alkene? First thing that pops in to my mind is bromine water test however this is with gases.. Perhaps bubbling the gas through bromine water? Maybe adding chlorine gas to the gas and see if...

Re: 2012 HSC MX1 Marathon tanA = \frac{1}{3} A = tan^{-1}\frac{1}{3} tanB = \frac{1}{2} B = tan^{-1}\frac{1}{2} tanC = 1 C = tan^{-1} A + B = tan^{-1}\frac{1}{3} + tan^{-1}\frac{1}{2} = tan^{-1}\frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{6}} =...

Re: 2012 HSC MX2 Marathon $As all the coefficients of p(x) are real, by conjugate root theorem, x = 5 - 6i is also a root$ p(x) = (x - 5 + 6i)(x - 5 - 6i)(ax + b) = (x^2 - 10x + 61)(ax + b) \equiv 2x^3 - 19x^2 + 112x + d $Comparing the terms of x^3, $ a = 2 $Comparing the terms of x^2, $ b...

Re: 2012 HSC MX2 Marathon 2x^3 - x^2 + 5 = 0 $Related roots: y = x^3 x = ^3\sqrt{y} \therefore 2y - y^{\frac{2}{3}} + 5 = 0 (2y + 5)^3 = y^2 8y^3 + 60y^2 + 150y + 125 = y^2 8y^3 + 59y^2 + 150y + 125 = 0

It's not blocked at school last time i checked.

12x^2 - 4xy + 11y^2 = 64.....(1) 16x^2 - 9xy + 11y^2 = 78.....(2) A)\; $Subbing y = mx in to both equations$ (1): 12x^2 - 4mx^2 + 11m^2x^2 = 64.....(3) (2): 16x^2 - 9mx^2 + 11m^2x^2 = 78.....(4) $Dividing (3) by (4)$ \frac{(12 - 4m + 11m^2)x^2}{(16 - 9m + 11m^2)x^2} = 32/39 39(12 - 4m +...

x^2 + 3y^2 = 2a^2 $Subbing $ y = \frac{1}{\sqrt{3}}(x - 2a) x^2 + (x - 2a)^2 = 2a^2 2x^2 - 4ax + 2a^2 = 0 x^2 - 2ax + a^2 = 0 (x - a)^2 = 0 $One intercept, therefore touches PQ touches the ellipse$

lol Shadowdude. Sorry for posting before you.

Find P by letting y = 0 in the tangent equation, P(2a, 0) Find Q by letting x = 0 in the normal equation, Q(0, -\frac{2a\sqrt{3}}{3}) Find the equation of PQ. Sub it in the equation of ellipse. Discriminant = 0 (or factorise it in to a perfect square) therefore touches it. Unsure if it's...

I'd have thought there would be more for 2U mathematics.

An increase in volume essentially means creating more space in that system. By creating more space, you alleviate the pressure of the gases in the system (decreasing the pressure).