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refer to attached diagram: AP is a line parallel to x axis You should be able to find X from the equation of the normal. noting that PAY and OXY are similar triangles, use the ratios of sides to get: Noting OX= x coordinate of X AP= x coordinate of P \frac{PY-PX}{PY}=\frac{OX}{AP} \\...

Try doing it using a diagram and similar triangles... its much easier

Assume we are in a world where we either spend on consumption goods/services or save. ie. Total Income (Y) = Total Consumption (C) + Savings (S) Consider the consumption component (C). There is a fixed component and a variable component which is dependent on our income. - The fixed component...

Draw a diagram to find out which region you need to integrate. In this question you should integrate along x axis. So it should be clear that: Area=\int_0^2 e^{2x}-1 dx

stuff like school captain, playing weekend sports, volunteering in salvation army, organising big events etc etc where you can rant on about your strong leadership/ teamwork/ management capabilities.

- mid 80s in everything at least. - possible but very difficult given the competitiveness. Not realistic unless you have really awesome extracurricular activities. Most UNSW Coop students have > 95 atars - Yes - Depends on what you want to do in the future, seems reasonable to me but you ll...

some companies automatically reject applicants with WAM < 65.

(1) Let P be (2p, p^2) and Q be (2q, q^2). Find q in terms of p by using the intersection between the normal at P and parabola. (2) remember PS=distance to focus = distance to directrix (defn of a parabola), which can easily be found by drawing a diagram. Do the same for QS in terms of p. (3)...

new question: differentiate xln(x) hence integrate ln(x)

On the other end of the scale, if you are doing too much algebra for conics, you are not doing it the way the question is meant to be done (ie. there is a better way) If you master 3u parametrics, conic is not really hard.

Conventional current is going down (+ to -). Magnetic field is going left at the current position of the wire (as it goes from N to S, so field is always going away from N). Use FBI left hand rule to deduce the wire is going clockwise.

No unless you involve complex numbers (4 unit stuff)

use sum of 2 cubes rule: a^3+b^3=(a+b)(a^2-ab+b^2) Try it yourself using this rule

12a - 30b + 4a^2 - 25b^2 = 6(2a-5b)+(2a-5b)(2a+5b) difference of 2 squares =(2a-5b)(6+2a+5b)

genius2014 is Bored of Fail 1,2,3,...

1. Find solutions to 1<(4x-4)/(2x+3) 2. Find solutions to (4x-4)/(2x+3)<3 3. Find where the solutions from 1 and 2 overlap

Also note that since |p|< 2, the discrimminant < 0 and hence roots must be complex and hence conjugates

For the later: Re(\frac{x-1+iy}{x-iy})=0 \\ Re(\frac{(x-1+iy)(x+iy)}{x^2+y^2})=0 \\ Re(\frac{(x-1)x-y^2+i(2x-1)y}{x^2+y^2})=0 Hence: x(x-1)-y^2 = 0

For the former: Let z=x+iy Re(x+iy-\frac{1}{x-iy}) = 0 \\ Re(x+iy-\frac{x+iy}{x^2+y^2}) = 0 Taking real parts we get: x-\frac{x}{x^2+y^2}=0 \\ x(1-\frac{1}{x^2+y^2})=0 hence either x=0 or x^2+y^2=1

Do you mean Re(z-\frac{1}{\bar{z}})=0 or Re(\frac{z-1}{\bar{z}})=0