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Put simply, alpha decay occurs when there is too many protons and beta decay occurs when there is too many neutrons.

If you believe you can do it, take Advanced maths. In order to keep the associated anxiety levels slow, I reccommend you take it slow, and study consistently throughout the year. If you dedicate a set amount of time each day (even as little as 30 minutes), and do this throughout the year you...

Ah, I like these kind of patterns. Got it in under a minute :p Nice thread.

I'm willing to pay $10AUD via Paypal ea. to a few people if they want to read over my english essay and make improvements. It's due in 9am tomorrow, so sorry for posting this late. I'm not too fussed if I don't get any replies, I'm pretty tired of English so I'm happy to let any mark slide. If...

So wait, it's cos^2 (t.pi/12)? The way he's done it it looks like its cos^2 (pi/12)t, which would act as a constant.

I got the same as alakazimmy.

Mind typing up the question?

Uh, I don't know if anybody answered your updated question or not, but I'll do it if you still need help. It's the same principle. \int \frac{t}{\sqrt[]{1-t^2}} \cdot dt Let, u = 1-t^2 \frac{du}{dt} = -2t du = -2t \cdot dt \therefore \int \frac{t}{\sqrt[]{1-t^2}} \cdot dt = -\frac{1}{2}...

Just think about it logically. A piece of wire is 20cm long. If you cut it up into two portions, with one portion x, then the other must be 20-x. Both portions must add up to 20cm. x + (20 - x) = 20 = true. Now let's just handle the x portion. We know that the length is twice the size of the...

\int \frac{t}{1-t^2} dt Let u = 1-t^2 Then, du = -2t \therefore \int \frac{t}{1-t^2} dt = -\frac{1}{2}\int \frac{du}{u} dt = -\frac{1}{2} lnu + c = -\frac{1}{2} ln(1-t^2) + c

y = xlogx. Differentiating that gives y' = log x + 1. Equating that to 0 gives logx + 1 = 0, logx = -1. Using the rule: y = logax x = a ^ y gives: x = e ^ -1 = 1 / e. y'' = 1/x = pos. Therefore min point. So a min.point exists at 1/e. --- I read it again, I did it again, same...

Question 2 has to be wrong. The derivative of xlogx = logx + 1. Can't have a negative value for x. A stationary point exists at 1/e but not -1/e

I'll do the first two, and I'll do the other two tomorrow if nobody has done them. Probability of a number being divisible by 8 is the same as saying the probability of a number being a multiple of 8. Between 1-100, that would be 8, 16, 24, ..., 96. Twelve possibilities. Therefore we start by...

If you just think about q2 for a second you'll realise m can be anything. x^2 + 2mx - 6 = 0 a=1 b=2m c=-6 Discriminant: b^2 - 4ac = 4(m^2) + 24 In order for roots to occur (aka the original curve to equal 0) discriminant >= 0 4(m^2) + 24 >= 0 True for all values of m (any number, whether it be...

He posted above the correct answer. If you don't understand it though, he just examed T30, saw it was 1+2+3+...+30, which in turn is the sum of an arithmetic progression with starting value 1, final value 30, difference 1, n=30. Sn = (n/2)(a+l) S30 = (30/2)(1+30) = 15x31 = 475

(x-a)^2 + (y-b)^2 = r^2 With centre (a, b) Think that's it.

Have you done calc yet? Turning points exist when y' = 0. Find the values of x that satisfy that equation and then sub them into y''. If y'' > 0, its concave up with a min. turning pt. If y''<0, its concave down with a max turning point.

Equation of a parabola is (x-h)^2 = 4a(y-q) Vertex is (2,4) therefore we know the formula is: (x-2)^2 = 4a(y-4) All we need is the value for a and we have the equation? This was year 11 wasn't it? I forget how to find the value of a. I know I could do it using algebra and just manipulating...

If you get a question identical to this in the HSC, or just the future you could probably just sub the values in to the formula we find at the end. It's better to get an understanding of how to reach the final answer though. There may probably be a better way of doing this. I never sat down and...