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if i recall correctly, the vertical asymptote would become a horizontal asymptote. e.g if the original function as x = 1 as the asymptote, then the inverse function would have an asymptote at y = 1

one does not simply do all the homework for you

Re: MX2 Integration Marathon $using by parts with$ \ \ u = tan^{-1}x \ \ dv = \frac{1}{1+x} \\ \\ \Rightarrow ln(2)\frac{\pi}{4} - \int_{0}^{1}\frac{ln(1+x)}{1+x^2}\ dx \\ \\ $Let I$ \ = \int_{0}^{1}\frac{ln(1+x)}{1+x^2}\ dx \ \ \ \ \ \ u = tan^{-1}x \\ \\ \Rightarrow...

most likely in the primary school section then, since there's no tutor in the high school section who "only" got an 82 ATAR

Which branch? I tried to apply for the one in homebush and got no response :(

heh my wam increased by 1/3. Solid effort :D

you sure it's not too early for this thread?

not checking your integration by differentiating your answer = -1/2

How did you go lol

YEAHHH 88 IN MATH1241 although i could have done better, I'm happy with my first HD :D

I guess it won't come out tonight :( Oh well, looks like i'll need one more sleep to get it

THEY ARE NOT OUT YET the wait is killing me lol

dem feels dem adrenaline

yeah i'd probably just do that haha

I'm assuming you mean \int x (2x + 7) \ dx your step is correct so far,regarding the bolded part it doesn't have to be. u = 2x+7 \Rightarrow \ x = \frac{u-7}{2} \\ \\ \therefore \ \int x (2x + 7) \ dx = \int (\frac{1}{2})(\frac{u-7}{2})(u) \ du

(1+2w+3w^3)(1+2w^2+3w) = (1+2w+3)(1+w+w^2+2w+w^2) = (2)(w+2)(2w+w^2) = (2w)(w+2)^2

4. expand the whole thing and simply with w^3 = 1 and 1+w+w^2 = 0 5.just sub them in ;) 6.x^2-(4+w+4+w^2)x+(4+w)(4+w^2)

Re: HSC 2014 4U Marathon 11C2 could include 2 Rs being together lol

Re: HSC 2014 4U Marathon 10(\frac{10!}{2!2!2!}) i'm not sure hehehe

it means there's an infinite number of solutions rather than 1 unique solution. you just need to find the conditions of the solution. Anyways the row echelon form is 1 4 1 0 1 1 0 0 -4 Let x_3 = k \\ \\ \therefore x_2+x_3 = x_2 + k = 0 \Rightarrow \ \ x_2 = -k \\ \\ x_1+4x_2+x_3 = 0...