\begin{align*} This advice is for you! HSC XII BIOLOGY 2019 6TH March, 2019. trailer <]/Prev 271042>> startxref 0 %%EOF 367 0 obj <>stream Forum section is a place to get help with any of your subjects, clarify any misunderstandings. AUD$72.00, inc. GST. So we have named the trial papers “with solutions” to save your time. \end{align*}, Similarly, we can find the gradient of \(OT\): t &= -\frac{1}{k}ln\Big|\frac{g-1.1g}{g-1.6g}\Big|\\ y &= -4.9t^2 + 20\sin\Big(\frac{\pi}{3}\Big)t\\ \begin{align*} Hence the answer is \(D\). Thus, \(\frac{y}{w} > \frac{1 + \sqrt{1+4}}{2} \text{ or } \frac{y}{w} < \frac{1 – \sqrt{1+4}}{2}\) million.
\begin{align} \end{align*}, \begin{align*} & = \frac{1}{\sin 2x}\\ \nonumber\\ Since there are no other forces, resolving the forces gives us: Business Studies | &= \Big[\ln|x+1| + \frac{1}{x+1}\Big]_{0}^1\\ \frac{dy}{dx}(x+2y)&=-2x-y\\ \sin\angle BAC &= \sin\angle BDC\\ y – b\tan\theta &= \frac{b\sec\theta}{a\tan\theta} (x-a\sec\theta)\\
million, Page views: over 2I &= \int_{-1}^1\frac{e^{x} + e^{-x}}{e^x + e^{-x}} dx\\ 2005 In order for \(B\) to have a greater chance of winning, we need \(P(B~win) > \frac{1}{2}\). NESA is regularly updating its advice as the coronavirus outbreak unfolds. Other subjects. Since it has a double root, the roots must be \(1, 1, -3\) or \(1, -3, -3\). Hence, Object 1 hits the ground at \(y=0\), that is, \(-4.9t^2 + 10\sqrt{3}t=0\) or \(t(-4.9t + 10\sqrt{3})=0\). HSC XII ECONOMICS 9Th March 2019. Note that for \(0 < x < \frac{\pi}{4}\), \(0<\tan x<1.\) This means that \(0<\tan^2 x < \tan x<1.\) However, we also have that \(\tan x\) is concave up in this interval (that is, it lies under the straight line joining the origin and \((\frac{\pi}{4},1)\). Finally, we have that \(\frac{dx}{dt} = \frac{dx}{dy}\cdot \frac{dy}{dt} = \frac{13}{-11}\cdot 4 = \frac{-52}{11}\). ∴ r^3\cos^3\theta – \frac{3r^2}{4}(r\cos\theta)+q = 0\\ &= \cot x – \cot(2^kx) + \csc (2^{k+1} x)\\ &= \sum_{r=1}^{k} [\csc (2^r x)] + \csc (2^{k+1} x)\\ The equations for motion of Object 1 are \textrm{Let } P(z) = z^4 – 2kz^3 + 2k^2z^2 + mz + 1\\
\textrm{Since } \angle BAC &= \angle BDC\\ &=\frac{1}{2}(\alpha\beta + \alpha\bar{\beta} + \bar{\alpha}\beta + \bar{\alpha}\bar{\beta})\\ \end{align*} &= \int_{a}^{-a}\frac{f(u)}{f(-u) + f(u)} \cdot -du \text{ substituting } u = -x\\
&= 2 \pi x Ae^{-kx} HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019. &=\frac{b\sec\theta}{a\tan\theta} “HSC” is a registered trademark of the NSW Education Standards Authority (“NESA”). 0000023446 00000 n \end{align*}. Since it is projected two seconds after Object 1, its equations will be, \(x = v\cos\theta(t-2)\) and \(y = -4.9(t-2)^2 + v\sin\theta(t-2)\) at the point where \(x=\frac{100\sqrt{3}}{4.9}\), \(y=0\) and \(t=\frac{10\sqrt{3}}{4.9}\). See the exam paper, plus marking guidelines and feedback from markers, for the 2019 NSW Mathematics Extension 2 Higher School Certificate (HSC) exam. t &= -\frac{1}{k}ln\Big|\frac{-0.1g}{-0.6g}\Big|\\ 2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx}&=0 \\
V &= \int_{-1}^1 \frac{1-x^4}{2} dx\\ General Instructions. x^3+9x^2+15x-17&=0\\
\(LHS = \csc 2x\), \(RHS = \cot x – \cot 2x\). 0000008419 00000 n \begin{align*}
We would like to wish you good luck hope these resources help you prepare & ace your upcoming exams. D &= \frac{g}{k^2}\Big[0.5\frac{vk}{g} + ln\Big|g-1.6k\frac{g}{k}\Big| – ln\Big|g-1.1k\frac{g}{k}\Big| \Big]\\ Thus, by subtracting \((3)\) from \((4)\), we get that ∴ \frac{a}{\sin\angle CBD} &= \frac{d}{\sin\angle ACB}\\ &= \pi – E + \angle ACB \textrm{ (opposite angles on cyclic quad)}\\ \begin{align*} This means that a possible value of \(z\) is \(C\). (Re(\alpha))^2 + (Re(\beta))^2 &= (Re(\alpha) + Re(\beta))^2 – 2Re(\alpha)Re(\beta)\\
Hence \(f^{-1}(x)\) will also approach a line and thus \(g'(x)\) will approach a non-zero constant. 2015 Multiply both sides of \((2)\) by \(\tan\theta\): 2I &= 2 \\ 0000002265 00000 n \end{align*}. bx_0\sec\phi\tan\theta- bx_0\sec\theta\tan\phi &= ab\tan\theta – ab\tan\phi \\ 0000007133 00000 n All Rights Reserved.
\frac{dx}{dv} &= \frac{v}{g-kv}\\
D &= -\frac{1}{k}\int_{1.6w}^{1.1w} 1 -\frac{g}{g-kv} dv\\ &= 2\pi A \int_{10}^{40} x e^{-kx} dx\\
To prove that \(O\), \(M\) and \(T\) are collinear, we will show that \(m_{OM} = m_{OT}\). \frac{\sec\theta – \sec\phi}{\tan\theta – \tan\phi} &= \frac{\tan\theta + \tan\phi}{\sec\theta + \sec\phi} \Big(\frac{y}{w}\Big)^2 – \frac{y}{w} – 1 > 0