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Thanks CM_Tutor for pointing out my typo. Some of you may have noticed I struggle one line at a time when using LaTeX, for a reason.

\frac{dy}{dx} =\frac {d}{dx}xu = u + x\frac{du}{dx}\\ \\ y = xu \implies u = \frac {y}{x} \\ \\ \frac {dy}{dx} = \frac{1}{2}((\frac{y}{x})^2 + 1) \\ \\ \therefore u + x\frac{du}{dx} = \frac{1}{2}(u^2 + 1) \\ \\ \therefore x\frac{du}{dx} = \frac{1}{2}(u^2 -2u + 1) = \frac{1}{2}(u-1)^2\\ \\...

Booktopia

Don't give up. Don't know what your problems are. But you need to have your situation looked at to see what is a good way out. Give me a call. I may, or may not, be of help. If you like, you can send me a shot of your most recent Maths exam so I can have a look at your answers. This will give...

Q5 (3x-y)^{10} = \sum _{i=0} ^{10} \binom {10} i (3x)^{10 - i}(-y)^i \\ \\ $For i=3, term in $x^7 = \binom {10} 3 (3x)^7 (-y)^3 = \frac {10 \times 9 \times 8}{1 \times 2 \times 3} \times 2187x^7 \times (-y^3) = -262440 x^7y^3

Get a couple of Maths Ext 2 textbooks, and read up. Do a bit of self-study. Doesn't matter if at first you understand only 17% of what you read. If you read again, this will improve to 23.75%.

This sequence is an Arithmetic Sequence: 2\pi +1, 4\pi + 1, 6\pi + 1 \cdots with 1st term a = 2\pi + 1 and common difference d = 2\pi Sum of 1st 20 terms :S_{20} = \frac {20}{2}(2\times(2\pi + 1) + (20-1)\times 2\pi) = 420\pi + 20 Sum of 1st 4 terms : S_4 = \frac {4}{2}(2 \times(2\pi +...

Unless you are told to use another specific method, what's wrong or illegitimate about using "Terry Lee's" method. It's neat. It's very short. It's beautifully efficient. In fact it's cool. Do you want to use the usual method, requiring 10 to 15 lines? But it's not necessarily easy to use the...

Even I, with a postgraduate degree and over 20 years of tutoring experience, find it hard to charge $80/hr. I feel some of the people who achieved an ATAR of 99.95 think they are perfect. I think they are rather swell-headed and arrogant. But I'm sure an exceptional few do make very good tutors.

I like the method in Terry Lee's book: turn 15 - 8i into a square. 15 - 8i = 16 - 8i -1 = 4^2 + 2\times 4 \times (-i) + (-i)^2 = (4 - i)^2 \\ \\ \therefore \sqrt{15-8i} = \pm(4-i)

So beautifully set out! Such clarity. Such nice handwriting.

Oh really? Thanks for pointing this out.

P(AnB) = P(A)P(B|A) = P(B)P(A|B) P(AnB) = P(A) x P(B) only if events A and B are independent. I would not expect Bayes Thm to be in 2U Advanced.

I think the 2 areas are: S_1 = \sum_{i=1}^{n-2} \frac{2}{n}\times f(\frac{2i}{n}) = \frac{2}{n} \sum _{i=1} ^{n-2}\sqrt{\frac {2i}{n}(2-\frac {2i}{n})} = \frac {4}{n^2} \sum _{i=1} ^{n-2}\sqrt{(i(n-i))}\\ \\ S_2 = \sum_{n=1}^n \frac{2}{n} \times f(\frac {2i}{n}) = \cdots = \frac{4}{n^2} \sum...

Thanks. I mixed up the 0.02 and 0.03.

This requires use of Bayes' Theorem I think. P(Mgr on Duty|Defective) = 0.39 x 0.03 / (0.39 x 0.03 + 0.61 x 0.02) = 0.4895 . . . You can draw a tree diagram: 1st branch for Manager(prob: 0.39) vs No Manager(prob: 0.61), and from each branch, consider Defective vs Non-Defective. Edit I mixed...

I got 18 cm for BG; not sure if correct.

No, you're not necessarily struggling with Geometry. This question itself is not straightforward. I bet half of 2U Maths Advanced student will find this question hard.

You're lucky. What year? Which Uni?

I still think it's a better idea to do Maths Advanced! You will struggle with Standard, even if the maths in Chemistry is not that challenging; yes "dirt simple" for those already good in maths. The Standard students I know of struggle even with basic algebra (Yr 8, Yr 9 & Yr 10 level).