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I think it's possible for anyone. You just gotta start early(like year 9/10) or possess a really smart brain

You could do it the long way, let f(x) = (x-3)/(x+3) Better way is ln(x-3)-ln(x+3) d/dx= 1/(x-3) - 1/(x+3)

In the end of this argument, two things will happen: -Shadowdude will realise what he's writing on BoS. (Trust me man, you don't sound cool right now) -Chromatic will realise he's spent too much time on Bos arguing a pointless argument without any evidence(i.e. making up shit that seems right)...

Why did you censor fuck? So in one simple sentence, you're saying that you are intelligent (Reading your first few words). Lucky bastard @ thread. All i've read from past threads was: -Bos takes the ranks and internal marks -they get the external marks -they do some comparing stuff. Rank...

You gotta admit but...resources is resources

That is one big fucking diet! 13kg in 7 weeks, that's a lot man, nice. Though considering your reps per set,I doubt a lot of it was muscle...but that's one hell of a start

Oh ok thanks. Just found that part in cambridge 4u, unlike terry lee... @Trebla what books did you use when you were taught maths; or did you get everything you needed from your teacher?

Complete the square, then square root it and then isolate x

gamma = z z-4=(9-z)(1-e^2) Found e to be [(2z-13)/(z-9)]^0.5 Then solved the inequality 0 < e < 1 Got the answer 4< z <6.5

I found an expression for the eccentricity and then went: 0 < e < 1

Determine the real values of gamma for which the equation defines an ellipse (terry lee 5.1 q7) \[\frac{x^2}{9-\gamma } +\frac{y^2}{\gamma -4}=1\] I got 4 < gamma < 6.5 Worked solution: 9-#>0 #<9 #-4>0 #>4 Therefore 4<#<9 The way the book did it doesn't seem right...

y=f(x) is impossible to sketch ... y=1/f(x) x intercepts become vertical asymptotes the vertical asymptotes become 0 in the reciprocal function It should be easy from here. If it's low and positive, it'll be high and positive y=f[|x|] Pretty much two f(x) for x>0, with the y axis being...

More solutions? You mean ways to solve a question...because there are fixed number of answers for a question Ohhh..i should read everything first

that is elite man...

wow @ thread

THX MAN haha, I can tell it's mostly if not all from Sam Davis

Yea

I got the same answer... and everything you did looks right . Need another person to determine if the book is wrong again

I doubt most students in year 9 would've learnt this term. You must have a good teacher. Your second statement will be offensive to a lot of people. So please, get off your high horse

Almost everything in 5.3 maths is revisited in the first half-ish of yr 11