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Also, it might be useful to know these: \sqrt{x} produces \sqrt{x} and \sqrt[n]{x} produces \sqrt[n]{x} _______________________________________________ For inequalities, you can use special commands: \neq \neq (not equal to) \geq \geq (greater than or equal to) \leq \leq (less...

Generally with these types of questions you should write out the primitive function including any unknown constants, then use the points that are given to you to figure out what the constants are.

You don't need tutoring to do well. If you plan on doing engineering at uni, take 4U. Bumping up your ATAR by .05 by doing 2U sounds good, but your ATAR doesn't matter as long as it meets the cutoff. Also in the long term, not doing 4U puts you behind everybody else in uni physics. If you...

Re: HSC 2018 MX2 Marathon w = \frac{3x}{4} + \frac{5y}{4}i \begin{cases} u = \frac{3x}{4} \\v = \frac{5y}{4}\end{cases} Squaring, we have \begin{cases} u^2 = \frac{9}{16}x^2 \\v^2 = \frac{25}{16}y^2\end{cases} x^2 = \frac{16}{9}u^2, \, \text{and}\, y^2 = \frac{16}{25}v^2 |z| = 2...

Maybe you could use three sine waves with differing periods?

On line 2 I think they are just taking the conjugate of the denominator, which doesn't matter because the absolute value function ignores the sign of the imaginary part. As stated in the justification, |z| = |\bar{z}|, so let z = \alpha - \beta and we have |\alpha-\beta| =...

When differentiating trigonometric functions it's less confusing to write them out fully. e.g. rewrite \tan^3 (5x-4) as (\tan (5x-4))^3. It's easy to see that you can just differentiate this with the chain rule. There are three nested functions in that particular example so you will need...

Why is it i and not -i ?

When we use take the square root using the radical sign √ in the reals we take only the positive root. So \sqrt{64} = 8 and not -8 . But what does the radical sign √ actually mean when used on complex numbers? What does \sqrt{-1} evaluate to? i or -i ?

Ask around. Check if your career options make use of 2U Math/calculus. If your uni courses involve lots of math you'll end up behind when you reach them. Why is 2U Maths too hard for you?

My teacher sets us Cambridge, and occasionally Fitzpatrick (the old one, I think he said the new one was worse) and Coroneos. I think Cambridge is pretty solid overall. Coroneos is very old and can be a bit confusing but the questions are still challenging. I don't think the old Fitzpatrick is...

b) \begin{aligned} A_\text{sector}&= \frac{1}{2} r^2 \theta \\ &= \frac{\theta}{2} \end{aligned} \begin{aligned} A_\text{triangle}&= \frac{1}{2} ab \sin C \\ &= \frac{1}{2}\sin{\theta} \end{aligned} \begin{aligned} A_\text{sector} - A_\text{triangle} = 1 = \frac{\theta}{2} - \frac{1}{2}...

The wording hints that you're expected to provide an answer without solving the equation. Solving the equation seems rather hard. I can't think of any ways to express \cos x in terms of x without losing the x on the RHS. Wolfram Alpha also has trouble providing exact answers to it. Although...

b) use formulae for area of sector and area of triangle. c) repeat b) for the new area. By trying to solve for \theta you should arrive at that equation naturally.

Use the identity: \cos x - \cos y = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)

I just realised I made a mistake with that diagram. The radius should be 4cm, not 2cm. \angle OAB = 2 \cos^{-1}(0.5) = \frac{2\pi}{3} Area of sector OAB = \frac{1}{2} (4)^2 \left(\frac{2\pi}{3}\right) = \frac{16\pi}{3} Area of triangle OAB = \frac{1}{2} (4) (4) \sin...

Gonna need a bit more information.... what question?

I'm not familiar with that specific terminology, but if "dilation of factor 1/2 from the y-axis" means compressing the function horizontally, then you replace x with 2x and to shift it right you would replace x with x - \pi/4 , in that order. y = \sin x Dilate: y = \sin 2x...

22) Find the area of the sector OAB and subtract it from the area of the triangle OAB . _________________________________________________________________________________________ 23) a) The arc AB is equal to the circumference of the cone's base. With that information you can...

I've never heard those terms used specifically, but it appears to be related to the neutron-proton ratio in the nucleus that determines stability. So a neutron-rich isotope has too many neutrons, and a neutron-deficient isotope would have too many protons, with respect to the ideal...