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i was gonna rep drongski's solution, but this "You must spread some Reputation around before giving it to Drongoski again." showed up anyways, you can try to do it on your own using my method now btw, I think it's more like a 2 unit question

let the amout of money each person has to pay (after the 2 ppl dogged them) be x dollars then you can express the number of ppl before they left in terms of x , since no. of ppl = total amout/individual amount using the same principle, the number of ppl after they left can also be...

srsly?

+1 also some one was saying that they into account of the relative gap between ranks as well

http://www.officeworks.com.au/retail/products/Office-Supplies/Document-Filing-Storage-and-Presentation/Document-Presentation/Manilla-Folders/AV81502 these ones don't bother if you don't have them, just staple them together and lable the front page or something NEVER use sheet protectors

Damn you, now I'm gonna fail SOR

P(-x) = -P(x) only when c=0 , which means it cannot have a constant term to be an odd function

if there is some non-zero constant , say c let P(x)= ax^3 + bx + c then P(-x)= a(-x)^3 +b(-x) + c = -ax^3 - bx + c =/= -P(x)

P(x) is odd , so there wouldn't be any constant or x^(even number) terms then set the coefficients of the odd powers (only x and x^3)to be some pronumerals (say a and b) then use factor and remainder theorems to find out these values

where a=22.5°

sin(2a) = 2 sin(a)cos(a)?

the question is explicitly asking you to "consider" the coefficients , I'd write them down to be apparent otherwise , you could state the values of a and b by inspection or use sum of roots since you only need the roots to graph it and then solve the inequality

so what was the "small difference" in your solution?

okay so let the other factor be ax+b then P(x)= 2x^3-x^2-13x-6 = (x-3)(2x+1)(ax+b) equating coefficient of x^3 , 2=2a, a=1 ... of x^0 , -6 = -3b , b=2 P(x)= 2x^3-x^2-13x-6 = (x-3)(2x+1)(x+2) , which has roots x=3, x=-1/2 and x=-2 after drawing a simple...

which textbook is it ? maybe it's a typo or something

are you sure it's divisible by 2x+1? cos P(-1/2) = -1/8 -2-9/2+18=91/8 =/= 0

also both answers are correct, with constants since ln |2x+sqrt(4x^2-1)| + c = ln 2*|x+sqrt(x^2-1/4)| + c =ln |x+sqrt(x^2-1/4)| + ln2 + c =ln |x+sqrt(x^2-1/4)|+ k where c and k are arbitary constants , k = c+ln2

give it a go , and even if you struggle with some of the concepts, the scaling would make it up scaling-wise, it gives you more better-scaling units since mx1 only counts as 1 unit if you do 3u maths also from your current results , you shouldn't find it too hard

after all, it's a preliminary assessment of YOUR school, so you should find past papers given by your teacher quite relevant

also you should ask for the past papers of your school from your teacher if he/she doesn't give them out