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bump. Looking for a couple more students. School's coming up, so here's your chance to brush up and clarify and doubts before jumping into new material.

Khan academy has a lot of resources that could be helpful though it isn't catered for the HSC course. They're pretty amazing! http://www.youtube.com/user/khanacademy?blend=1&ob=4#g/u

Thanks Mitch! Bump. Looking for new students for the new year!

I usually leave it up to the student - although generally 1 hour lessons for Physics and 2 for Maths.

I had her for a teacher in Years 11 and 12 for Physics and I must say she wasn't a very good teacher. She relied too heavily on HSC Dotpoint, turned up late in half the lessons and made us do trivial activities like group presentations/ posters and power points. In terms of tutoring, she could...

You just used part i, substitute k = n. Expand each of the combinations like so: nC0 + \frac{nC1}{n} + \frac{nC2}{n^{2}} +...\\=1 + \frac{n!}{(n-1)!1!n} + \frac{n!}{(n-2)!2!n^{2}} + ... \\ = 1 + \frac{1}{1!} + \frac{n(n-1)}{2!n^{2}} + ... Then take the limit of the whole thing, and all the...

For the first one, I think you just calculate probabilities of all X's and compare them to find the biggest ie. X = 1. P(X=1) = 0.25 P(X=2) = 3/14 P(X=3) = 5/28 P(X=4) = 1/7 P(X=5) = 3/28 P(X=6) = 1/14 P(X=7) = 1/28 For part 2, just add up X = 5,6,7 to get your answer. [Edit: I'll double check...

In all solutions, you always have 5 Easts and 4 Norths, so if we represent the person's path as: EEEEENNNN it's just a matter of finding how many ways you can arrange them. 9!/4!5!

made a small change. C =/= 1

For simplicity in typing it out, I'll just replace phi with P. P(x) = x³Q(x)-----(1a) P(x) - 1 = (x-1)³q(x) -----(1b) P(1) - 1 = 0 P(1) = 1 -----------(2) P(1) = 1³Q(1) = 1 ----subbing (2) into (1a) Q(1) = 1 ----------(2b) P(x) - 1 = (x-1)³q(x) = x³Q(x) - 1 sub x = 0 -q(0) = -1...

Ahh I think I can see where the ambiguity arises. "carlo randomly chooses four tiles, how many distinct ways can he arrange them?" It sounds like the choosing has already happened, so all you are doing is calculating how many ways he can arrange the four tiles that he has. so 4*3*2*1...

Unless you overload/take summer classes in which case more majors are possible.

Before you buy real shares, why not play in a sharemarket game? Buy shares with fake money, see how you go and learn along the way.

Goodness, what was I thinking. Yep k = 9 works, but I was under the impression that k = 0 gives you 3 roots, 2 of which are equal.

f(x) = kx x(x-3)² = kx (x-3)² = k, x = 0 is always a solution Therefore find the situation where (x-3)² = k has only 1 solution Graph y = (x-3)² That has two solutions for y > 0 and a double root for y = 0. So no values of k work.

Replace x with x - 3, and the following justifies it. g(x) = f(x+3) = 5 - 3x g(x-3) = f(x) = 5 - 3(x-3) = 5 - 3x + 9 = 14 - 3x

This. +1

I think it would be enough if you said: 4x² - 1 > 0 x² > 0.25 x > 0.5, x < -0.5 Curve is undefined for x between 0.5 and -0.5, therefore integral doesn't exist ie. pretty much what the answer says. @Spiralflex, you're making it more complicated than it should be.

Getting so excited for Heart of the Swarm. The trailer and artwork look freaking epic.