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1. the complex numbers z1 + z2 and z1 - z2 are the diagonals of the parallelogram that's formed when you add z1 and z2 if the moduli of these complex numbers are equal, then the diagonals of the parallelogram are equal, which makes it a rectangle if it's a rectangle, then the angle between z1...

i misread the question and thought it meant the rate of change of the surface area as well -.- thanks bleakarcher! :D

The volume of a sphere is increasing at a rate numerically equal to its surface area at that instant. Show that dr/dt = 1 I thought this would be quite straightforward, and it might, but I don't really have any idea as to where to start :/

i don't think b is right since just subtracting a from 1 would count situations where only 1 of them were selected i think b) would be 7C4 * 6C5 = 210 .'. probability = 210/1470 = 1/7

Re: 2012 HSC MX2 Marathon 1. solving simultaneously with the ellipse and the line, (x^2)/2 + (mx+3)^2 = 1 x^2 + 2(mx+3)^2 = 2 x^2 + 2m^2x^2 + 12mx + 18 = 2 (1+2m^2)x^2 + 12mx + 16 = 2 since the line and the ellipse intersect when x = x1 and when x=x2, then x1 and x2 are solutions to this...

i think firstly you find how many moles of HCl are used up in the reaction between the CaCO3 and HCl by finding the number of moles of CaCO3 then using mole ratios. after that, you find the number of moles of NaOH that were used to neutralise the excess HCl, then use mole ratios again to find...

Re: 2012 HSC MX2 Marathon 1.since the equation is a cubic, there must be at least 1 real root, so therefore the last root must be real since alpha and beta are complex roots also, since the coefficients of the polynomial are all real, the 2 complex roots alpha and beta must be conjugates of...

i got 13.50 as well

i usually do my work with no music, but when i do, i don't even notice that music is playing :/ might as well just do my work with no music all the time..

Re: HSC 2012 Marathon :) 2^x = e^[log(2^x)] = e^[xlog(2)] = (e^x)*(e^[x(log2 - 1)]) .'. integral of 2^x dx = integral (e^x)*(e^[x(log2 - 1)]) dx let u = e^x du = (e^x)*dx when x = 1, u = e when x= 0, u = 1 .'. integral 2^x dx = integral u^(log2 - 1) du = [(u^log2)/log2] from 1 to e =...

English (Advanced) Written Examination Paper 1: Area of Study 15/10/2012 13:50 English (Advanced) Written Examination Paper 2: Modules 17/10/2012 09:25 Economics Written Examination 18/10/2012 09:25 Mathematics Extension 2 Written Examination 22/10/2012 09:25 Mathematics Extension 1...

the bold part is correct i think, but then you have to convert it to kilojoules since the molar heat of combustion is in kilojoules

yeah, you have to use the second part where st=-2 try rearranging that expression or to get the t's in terms of s or vice versa, so you're left with only 1 parameter edit: ^that's for the locus. for finding the intersection of the normal, it's fine to have both s and t, since the coordinates...

yeah, i understand why II is correct, but i wasn't exactly sure about I because although P(x) suggests that it's a cubic, if a was 0, then it wouldn't be so anymore, and if a, b and c were 0 then the polynomial wouldn't actually cut the x-axis provided d isn't 0

it was a question in my half-yearly that was the information given, and it was a multiple choice question asking which of the statements were correct A. I only B. II only C. Both D. Neither

ah, alright. thanks for that! :D

ok, so the first statement is correct unless the polynomial is P(x)=d where d isn't 0, and the second statement is correct no matter what?

yeah, i thought both were correct too. but what if a,b and c were all 0? wouldn't that mean that the first statement wouldn't be correct?

a polynomial P(x) has equation P(x) = ax^3 + bx^2 + cx + d where a, b, c and d are real I. the polynomial must cut the x-axis at least once II. if P(alpha) = 0, then P(conj. alpha)=0 are both these statements correct?

differentiate both sides with respect to x, RHS is self-explanatory, with LHS, you use the chain rule --> (d/dy)*(dy/dx) = d/dx .'.d/dx(lny) = d/dy(lny)*(dy/dx) = (1/y)(dy/dx)