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The geometric method (see your diagram) is the easiest to understand and the shortest.

So what is your answer for the range?

T Thanks; typo when in a hurry!

\therefore \frac {tan \frac {\pi}{4} + tan\theta}{1-tan\frac {\pi}{4} tan\theta} = 3 \frac{ tan \frac{\pi}{4} - tan\theta}{1+ tan{\frac {\pi}{4}}tan\theta} \\ \\ \therefore \frac {1+tan\theta}{1-tan\theta} = 3\times \frac {1-tan\theta}{1+ tan\theta} \\ \\ \therefore (1+tan\theta)^2 =...

I got |b| = 2.618 . . . instead of 4.441 . . .. With this correction,(multiplying my answers by 4.441/2.618) should end up with roughly the same answer.

Wonder if answers approx: 0.926 + 2.451 i and 1.797 - 1.906 i

A memorial service was held for Olivia Newton-John today. How many of you are aware that her maternal grandfather was the Nobel Prize winning physicist, Max Born?

So sorry kkk579. Naughty of me

your new?

By the way: it's "Dijkstra" (Edsger Dijkstra) rather than "djikstra". You are obviously a clever boy to be even aware of Dijkstra's algorithm. There are lots of dykes in Holland. I suspect the "dijk"(pronounced 'dike') refers to these, as is the name of American actor, Dick van Dyke.

Yes. Why the heck did you not mention the actual textbook instead of mentioning the publisher. If you mentioned Cambridge, we all have a fair idea of which text you are referring to. Pearson??

Most students here would not have heard of H Lamb. Sir Horace Lamb, FRS was at the University of Adelaide as its first Elder Professor of Mathematics from 1876 to 1885. His made major contribution to the field of Hydrodynamics. He was a Second Wrangler (what's that?), while the current Prime...

Hard to generalise. Depends on who the private maths tutor is. As they say: "There are maths tutors and there are maths tutors".

You can say: the answer is the intersection of the 2 solution sets.

Your x must satisfy these 2 inequalities simultaneously: \frac {2x}{x+1} > -1 $ and $ \frac {2x}{x+1} < 1 For the 1st, you need: x < -1 $ or $ x > -\frac {1}{3} \\ \\ $and for the 2nd: \\ \\ $ -1 < x < 1 \\ \\ $Combining, you need: $\\ \\ -\frac {1}{3} < x < 1 if I'm not mistaken. Note...

You can still make it, if you have a reasonable aptitude for Maths. Since you were doing 5.2, it's just possible that you do not have an aptitude for maths. But if you have an reasonable aptitude for maths, you can be coached to improve yourself. But you must do it urgently. Your normal coaching...

Didn't realise that. I never use my mobile to access Bored.

No wonder nobody comes to me. Didn't you notice my signature? Looks like no one notices it.

Talk to your parents. They may be able to advise you. Drag them along if necessary. Just ask your coordinator for an appointment to discuss your interest in doing MX1. What have you got to lose? At most your request may be turned down; you're no worse off. If you don't ask, you get nothing.

If you really want to do MX1, go for it. Yr 11 just started. Go see your Maths Coordinator to explain to him/her that you really want to do Ext 1, and ask to be allowed to do it. If necessary, acknowledge that your maths results last year was pretty crummy, but you are determined to put in more...