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This means that when t=2;\,v=0,\,x=0 . Your integral of velocity will be x =4t^2-16t+C Knowing that the particle is at the origin after two seconds, you can work out the constant.

Don't think negatively like that. Just keep practicing and you'll improve. If you found OR,\,OP, then use Pythagoras' theorem: OR^2+OP^2=PR^2 and of course, 2PQ = PR.

Yep. The Trapezoidal Rule is exact when integrating any linear function. Simpson's Rule is exact when integrating any quadratic or cubic. Generally it's also more accurate with integrating any curvy function. It shouldn't matter how many subintervals there are. Because you're just...

Given that QR = PQ ... Let the height of Sydney Tower OT be h . (since the height of a tower won't change its bearing, we can use h in our calculations but it will not appear in the final answer) 1. Find OQ,\, OP,\,OR in terms of h . 2. Noting that \triangle OPR is a...

a) Use the formula x = \frac{mx_2+nx_1}{m+n} y = \frac{my_2+ny_1}{m+n} Calculate the point M using your ratio k:-1 and your two points (be careful, order matters). Then compare your point to the one given and you should be able to find k . c) Using the formula \tan\theta =...

Assuming you have done a) and b), you would have P(t) = P(t_0)e^{t/4\,\log(0.9)} Then substitute t=10 and you will get the population P(10) after 10 years as a percentage of the original population P(t_0) .

The picture is not there. Are you given the initial population of the city at t=0?

2a) You calculated the gradient of the curve only at the given point, and assumed it (the gradient) was the same for all other points on the curve. The question asked you to find the function which has gradient function {dy}/{dx} = \cos 2x for all values of x (in the domain), and which...

Re: HSC 2018 MX2 Integration Marathon These are nice practice: \int_0^{\frac\pi 6}\frac{x \sec^2x}{\tan x + \sqrt 3}\,dx = \frac{\pi}{6}\log\left(\frac{2\sqrt3}{3}\right) \int_0^{1}\frac{\log(1+x)}{1+x^2}\,dx = \frac\pi8\log2

You drew the graph of y = 2 \sin x instead of y = 2 \cos x .

Well actually they're reciprocals... the inverse functions \sin^{-1} etc. are different.

http://educationstandards.nsw.edu.au/wps/portal/nesa/mini-footer/privacy/!ut/p/z0/fY_BTsMwEES_hYOPKzuAKq6pigQVEm0BqfWl2iSr1MXYbnfT0L_HSTlUHLis3kijmVlt9VrbgCfXorgY0Ge9sZPt_fOTMXfm9uX1YTEz5Xy6_JguV8XjW6Hn2v5vyAlufzjYUts6BqFv0evUb0cOokw6xj3VAj1VygRiVAar2Mkges7Ce7jQcIElHh39UdCQoPPKdKlBocEPO67hC2XHgM0...

Note that the area bounded by the sine wave between -\pi,\,\,-{5\pi}/{6} is the same as the area bounded between \pi,\,\,{7\pi}/{6}.

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I believe it means that the first load is deposited 20km from the quarry and the last load is deposited 30km from the quarry in the same direction. i.e. there is a 10km distance between the first and last loads.

Let the AP be a,\, a+d,\,a+2d... Then, a,\, a+d,\,a+2d,\, a+3d =4a+6d=16 and a + 2d + a + 7d = 2a + 9d = 4

For a), note that the term inside the logarithm is halved each time. So, rewrite it as \log_b 36 + \log_b \left(36\cdot \frac 1 2\right) + \log_b \left(36\cdot \frac 1 2 \cdot \frac 1 2\right) + ... And use the law \log ab = \log a + \log b. b) can be solved similarly.

Looking at the intended answer, there's not enough information. For example, the given answer doesn't satisfy A(0,0) and B(0,1), so A and B should have been fixed points.

What are A and B ? Are you given any coordinates?

Take this previous HSC question, for example: You could do one paragraph on time and one on space but you have nothing left for the third and you can't integrate the two together as well. In general, how do you decide what three ideas will make up your body paragraphs?