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It's not too hard if you're reasonably competent at maths. If you have a solid foundation in calculus, are willing and able to grasp new concepts easily and you are good at algebra, then one should be able to keep up fairly easily. Although if you don't do the homework, or any kind of work to...

Not long at all. Just remember to wrap every line in the tags and you'll be fine. The script itself is pretty intuitive. If you want a fraction, you type in \frac{numerator}{denominator}. Square root? \sqrt{what's inside}. Integrate something? \int_{bottom limit}^{top limit} ...dx It has a...

Spoken like a true english advanced student (Y)

bleak have you done induction before? If you haven't, its pretty fun and you should self-learn it.

Ahhhhh I see where I went wrong. I foolishly assumed that sqrt(x) < x for all x, but this is only true when x>1. For 0 < x < 1, sqrt(x)>x. So my jumping from lnx < x to lnx < sqrt(x) is wrong. And I think you meant to say that instead of "just in case a>3", it should be "a<1", because if a>3...

Lol. I thought the mean value theorem would be easier because I he said that he could "cheat" with it. Meh. Anyways: Use\ mathematical\ induction\ to\ prove\ that\ 1^3+2^3+3^3+...+n^3>\frac{n^4}{4} There's a part 1 to this question but imo you don't need it. If you've worked with...

If this can occur: lnx>sqrtx, then why would they ask me to prove a false result? The question said prove lnx < sqrtx for ALL POSITIVE values of x. If lnx>sqrtx for some large value of x, then they wouldn't ask me to prove it true for all pos. values of x. Rather, they would ask me: "Is this...

Our sub for history was sleeping and a friend was forced into chucking increasing amounts of paper pellets, until she woke up. Shit ensued. Also, we nuggeted a bag, and the guy spent a minute trying find to the zipper, then realised that his bag was nuggeted. Our science teacher for year 10...

Mmmmkay. A few questions: Prove\ that\ lnx<x\ for\ all\ real\ positive\ x. But we were trying to prove that: Prove\ that\ lnx<\sqrt{x}\ for\ all\ real\ positive\ x. Was your reasoning that since x belongs in the all real number field, then \sqrt{x}<x, hence proving that lnx<x would answer...

Cheers shadow! I get it now (Y) Just wondering if there were any applications to it, besides solving inequalities like that one. And could you post up a solution to the inequality using the theorem? I'm interested in uni maths, and its use in 3/4U. It would be great if you could. Thanks!

Lol not my problem :D. You said show, not prove through the use of part a) or diagram 1.2 or whatever. The formula is quotable because it work in all cases with a point and a straight line. I applied it (Y) Although I wasn't surprised you'd say that. The question could have been done by anyone...

Isn't that the perpendicular distance formula? P(0,0)\ l:Ax+By+C \therefore\ x_1=0,\ y_1=0,\ a=A,\ b=B,\ c=C \begin{align*}d &=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\ \therefore\ d &=\frac{|A(0)+B(0)+C|}{\sqrt{A^2+B^2}}\\ &=\ \frac{|C|}{\sqrt{A^2+B^2}}\end{align*} For some stupid reason...

=_= If only I knew that. The I wouldn't have had to waste 5 min graphing that damn graph. I could've done another question... Probably failed the test >_> And shadow, could you perhaps rewrite what you were trying to write in tex? I don't understand what you wrote. And I'm gong to chuck...

Don't tutors make quite a lot of money? I mean, after all the advertising is done. Most parents would pay good money for their child to be tutored by someone who scored a top mark in HSC Maths, and most are oblivious that higher ATAR =/= good teaching skills. You could exploit that and flaunt...

Can that result be quoted in exams, or do you have to prove it first?

I did it with a graph. I don't think there's any other way to do it.

One problem: You didn't answer the question properly. You forgot to prove that the stationary point at θ=45 is a max T.P It's incredibly important that you prove that the stationary point is either a max or a min T.P. If you don't prove it, then you've just stated a value which could either give...

Lol it's 3U, not 4U projectile motion. I didn't mean 4U mechanics when I posted the question. My bad. You just apply the formulae for the x and y directions in creative ways then it basically turns into a max/min problem. I'll give you a hint bleak, as it's not fun spilling the beans on...

\int \frac{x^6-1}{x^4+x^3-x-1}dx =\int \frac{(x+1)(x^2-x+1)(x^3-1)}{x(x^3-1)+1(x^3-1)}dx =\int \frac{(x+1)(x^2-x+1)(x^3-1)}{(x+1)(x^3-1)}dx =\int (x^2-x+1)dx \therefore\ =\frac{x^3}{3} - \frac{x^2}{2} +x+C And a fundamental mechanics question: Prove that the angle of projection required...

Fair enough