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It's really important to know your trig identities when working with questions like these. Using the identity \cot^2\theta + 1 =\, $cosec$^2\theta, k = 4(\cot^2\theta + 1) - \cot^2\theta k = 3\cot^2\theta + 4 Now the RHS is \frac{k-1}{k-4}. We're clearly not going to get a fraction like...

If the resistive forces are k\dot x^2 and k\dot y^2 then we could write \ddot x = -k\dot x^2 (as the only horizontal force acting on the projectile AFTER launch is the resistance) And differentiating the given equation x = 50 \log(4t+1) should provide an easy solution for k . I...

i) Using the binomial theorem you have (1+x)^{n+k} =\sum_{r=0}^{n+k} \binom{n+k}{r} x^r Equate the co-efficients of x^n on both sides using the above for k = 0...m. On the LHS we have \binom{n}{n} + \binom{n+1}{n} + ... + \binom{n+m}{n} and on the RHS we want the co-efficients of...

I'm not sure, but I noticed on your fifth line you use \ddot x = \frac{d}{dx}\left(\frac12v^2\right) which, unless I've read the question wrong, isn't necessarily true since the overall velocity of the projectile is different from its horizontal and vertical velocities. Also you've...

Starting with the acceleration equation: \ddot x = -n^2x It is given that the maximum acceleration is 6 , and from the equation we can see that the maximum acceleration occurs when x is lowest. i.e. \ddot x_{\mathrm{max}} = -n^2\left(x_{\mathrm{min}}\right) Substituting in, we...

1. Usually I will do Cambridge first then other textbooks, after my teacher has finished covering the content. I'll do as many as I need to until I feel that I understand and am comfortable with the content. 2. I usually start about a month before or whenever we've finished all the required...

Where did you get the initial velocity 200\sqrt2 ?

The key is just to know that the derivative of \sec x is \sec x \tan x and hence the integral of \sec x \tan x must be \sec x plus a constant (since differentiation and integration are roughly opposites of each other). You could naturally derive the answer using a substitition but I don't...

By the Factor Theorem, P(2) = 0 and P(-1) = 0. \begin{aligned} P(3) = 28 &\implies a(\phantom{-}3)^3-b(\phantom{-}3)^2+c(\phantom{-}3)-8 = 28 \\ P(2) = \phantom{-}0 &\implies a(\phantom{-}2)^3-b(\phantom{-}2)^2+c(\phantom{-}2)-8= 0 \\ P(-1) = \phantom{0}0 &\implies a(-1)^3-b(-1)^2+c(-1)-8 =...

From the reference sheet we have \ddot x = -n^2(x-b), where the period T is given by T = \frac{2\pi}{n}. Clearly n = \sqrt 6 , so T = \frac{2\pi}{\sqrt6} = \frac{\pi\sqrt6}{3}. Now, \begin{aligned} \ddot x &= -6x \\ \frac{d}{dx} \left(\frac12v^2\right) &= -6x \\ v^2 &=...

One thing that might help is to show your teacher that you have a positive attitude towards maths. So e.g. participate in class more and maybe ask questions. As for revision, perhaps start by grinding out some past papers starting a month or so before your exam block. And also, I find that it...

Find the angles subtended at the circumference of the circle and then, using that information, focus on the diamond shaped quadrilateral formed at the top of the diagram to obtain a relation between the three angles.

whoops... my bad.

The coefficients of the polynomial are suspiciously symmetrical. Consider: \begin{aligned} P\left(\frac1z\right) &= \frac{2}{z^4} + \frac{5}{z^3} + \frac{7}{z^2} + \frac{5}{z} + 2 \\ &= \left(\frac{1}{z^4}\right)P(z)\end{aligned} Therefore, if \alpha is a (nonzero) root of P(z) then so...

wouldn't this count?

C4.2: Areas and the definite integral (page 59).

We need your school rank. Otherwise nobody can give you an accurate estimate. For example, if you got these marks at a top school you'd probably be looking at 95+ ATAR.

First start with the equilibrium equation: \mathrm{CO_{2(g)} + H_2O_{(l)} \rightleftharpoons H_2CO_{3 (aq)}}\,\,\,(\Delta H < 0) Le Chatelier's principle states that a system at equilibrium will move to counter any imposed changes. Roughly speaking, if you push it one way it will push back the...

If a solid metal reacts with a metal in solution, that means the solid metal has displaced the metal in solution. Therefore the solid metal was more reactive than the metal in solution. Keeping this in mind, look at which metals reacted: (Let (s) denote a solid metal and (aq) denote a metal in...

use [ tex ] and [ /tex ] (remove the spaces inside the square brackets). For example, [ tex ] x [ /tex ] gives x The equation of the tangent at any point (2t,\,-t^2) on the parabola x^2 = -4y is given by: \begin{aligned} y + t^2 &= -t(x-2t) \\ y&= t^2-tx\end{aligned} Solving...