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Maybe some of the stuff you did in Years 7-10 science could help in Who wants to be a millionaire, or trivia night. Otherwise, it's close to useless.

She's referring to the OP who posted a thread about being confident in attaining 99.95.

About 25c each, very smooth and very reliable

+1. English allows for development of a level of sophistication in your speech that perhaps no other HSC course will do. But in some cases English just makes you draw more meaning from a text than even the composer had ever intended.

Doesn't this mean "a" can be any number between 0 and 2pi because we are just adding two areas under the curve? Is it because we realise values of tan(a/2) go from positive to negative as we go a<pi to a>pi Maybe this is way beyond my knowledge =/ Thanks anyway for your help [Can't seem to...

Wow Thanks. There's just one bit I don't understand Why do you get a->pi What's the significance of pi that you can do that? Otherwise, really helpful!

$Use the method of t-substitution to evaluate:$\int_{0}^{2\pi}\frac{d\theta}{5+4cos\theta} I was wondering if this was possible, having one of the bounds to be 2\pi

1) Let theta = arccos(x) Draw the triangle represented by this ie. angle theta, side adjacent to theta = x, hypotenuse = 1, solve for other side which is \sqrt{1-x^2} \therefore sin(arccos(x))=sin(\theta)=\frac{\sqrt{1-x^2}}{1} 2) Same strategy 3) should be able to graph y=arctan(x) yes?

The question doesn't seem to make sense unless 1) isn't a question, but a condition to the question. Maybe it should read: Given the equation 2x^3 - x^2 + kx + 4 = 0 if x=1/2 is a zero for the above polynomial, find the other two zeros

BIG 1/3 page diagram.

So far my list would be (easiest to hardest) Graphing, Polynomials, Integration, Conics, Complex Numbers

I too would like to know!

New element copernicium wins a symbol at last - physics-math - 05 March 2010 - New Scientist

2cos^2(3x)-1=cos(6x)..............($Cosine double angle$) \\=cos(\frac{6(2\pi)}{9}) \\=cos(\frac{4\pi}{3}) \\=-cos(\frac{\pi}{3})............($cosine negative in the 3rd quadrant$) \\=-0.5

$Draw a diagram for this one. \\Let distance from tower to you initially be y metres$ \\tan\alpha =\frac{h}{y}\Rightarrow y=hcot\alpha -----(1) \\tan\beta =\frac{h}{y-x} \\y-x=hcot\beta \\x=y-hcot\beta \\$sub(1)into equation$ \\x=hcot\alpha -hcot\beta \\x=h(cot\alpha -cot\beta )

+1 That's nice.

3x^3 - 2x^2 + 1 = 0 \\$let roots be \alpha, \beta, \gamma$ \\$New roots are 2\alpha ,2\beta ,2\gamma$ \\$Let coefficients of new equation be A, B, C$ \\\frac{-b}{a}=\frac{+2}{3}=\alpha +\beta +\gamma \\\frac{-B}{A}=\frac{4}{3}=2\alpha +2\beta +2\gamma \\\frac{c}{a}=\frac{1}{3}=\alpha \beta...

In our 4unit exam last term, there was a question asking for you to express something in mod-arg form and apparently cis form isn't mod-arg form. So those who wrote cis lost marks, though you can use it in your working.

Generally with a question like this, you are expected to recognise the graph and take the alternate path to attaining the answer. I don't think they would ask you to find the area through purely integration.

Yeah reckon this one's better. Mistake in adding reals in 4th from bottom.