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I think you are thinking too deeply about the problem. sin3x/sinx + cos3x/cosx = (sin3xcosx + cos3xsinx)/ sinxcosx =2sin4x/sin2x =4sin2xcos2x/sin2x =4cos2x

q2) should give you a circle, I think. Check your working out.

You can find k a little easier by realising that when T=120, dT/dt = 0.

But the quality of education courses, is, in general, quite atrocious.

I agree with you about doing more applied maths. It seems to me that the teaching of applied at UNSW in the first 2 years of the degree is distinctly unimpresive and this turns a lot of people off doing it in third year - I know many people who have decided never to touch applied again after...

Well Slidey, the discrete logistic equation is in the proposed 3u syllabus, but the continuous version isn't. Why? because you can't solve it without using partial fractions, which are also not in the 3u syllabus. It is all very ad-hoc. I don't think it makes much sense.

They are only introducing DEs in a very wishy-washy way. And the introduction of difference equations, the way it is suggested at the moment, makes very little sense, either.

Maybe its because many law graduates don't end up becoming lawyers and so they need something else to fall back on?

Suppose that the triangle has sides x-d, x, x+d. We take x as our unit. Then, in our measuring system, the length of the sides are 1-d*, 1, 1+d*, where d* = d/x. That is, we are introducing a new parameter, d*, to make the algebra easier by combining two of our unknowns into one parameter...

I think I was editing my last post while you were posting, Kurt. Like I said, find the angles in terms of d, and then use a scale factor (shall we call it x?) and the fact that the area of the triangle is t to find out what the scale factor should be. Interesting question.

Well, if t=6, then the 3,4,5 right angled triangle fits the criteria mentioned. So if t isn't 6, you can just rescale this triangle until the area is 6. But of course, we can do something similar with a 3,5,7 triangle (or any of an infinite number of different triangles, as long as we pick...

I think I read somewhere (can't remember where) that it is about 26 or 27.

You can get chromosomal abnormalities occuring during meiosis. Sometimes when the pairs of homologous chromosomes split up one of the daughter cells ends up with both of the pair and the other daughter cell has none. The cell that ends up with none is usually not viable, but the other (if it...

Would you consider numerical integration to be pure maths, then? I think the distinction between applied and pure is rather arbitrary, and essentially, not a very interesting question, as I like both. (although I haven't had much opportunity to learn applied maths yet.)

For Q2, remember that if a matrix is invertible then it has a non-zero determinant. For Q3, det(AB) = det(A) det(B), det(A^(-1)) = 1/det(A). For Q4, remember that A=[A^(-1)]^(-1), and [AB]^(-1) = B^(-1)A^(-1)

If you want to be a secondary teacher, I would pick the uni based on how well they teach your specialisation, rather than on their education faculty. For instance, if you want to be a maths teacher, go to a uni that offers a good maths degree, such as USyd or UNSW. UOW is also expanding what...

If it is a private school you could talk your daddy into making a $20,000 donation to the school building fund. That might help.

They have a whole topic called Financial Mathematics in General Maths where they learn about annuities, compound interest, loan repayments and stuff like that. There is a formula sheet available in the exam with all the finanacial mathematics formulas on it, so it is not necessary to memorize...

The answer is indeed $22,012.76. Iyounamu, you have forgotten that it is only the last $700 that will earn compound interest for the full three years. You should be using the present value of an annuity formula, not the compound interest formula.

Egyptian fractions!!! OK, 1/a + 1/b = 1/8 Multiplying through by 8ab, we have 8b+8a = ab So ab - 8a -8b = 0. Now we can add 64 to both sides of the equation, then we have ab - 8a -8b + 64= 64, which we can factor as (a-8)(b-8) = 64. Now I am assuming that you are looking for a and b...