Question Library

Using the sine rule \(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\ \frac{x}{sin43^o}=\frac{5}{sin53^o}\\ x = \frac{5\times sin43^o}{sin53^o}\\ =\frac{5\times 0.6820}{0.7986}\\ =4.3\)

Sin 53 = \(\frac {h}{6} \) h = 6 sin 53 = 6 x 0.7986 = 4.792m

\(Sin\theta^{\circ} = \frac{opp}{hyp}\\ Sin X^{\circ}=\frac{5}{10}=0.5000\\SinX^{\circ}\\ X^{\circ} = sin^{-1}0.5000\\ X^{\circ}=30^{\circ}\)

Sum of interior angles in a quadrilateral is 360 (x + 20)o + (x+ 10)o + (2x - 45)o + (x - 25)o = 360o 5xo - 40o = 360o x = 400/5 = 80o

Sum of interior angle of the diagram equals 360o 180o - 5yo + 136o + 180o + 180o - 3yo = 360o -8yo + 136o = 0 -8yo = -136; y = 17

< QPS = < PRS = 65^\circ (angles in the same segment) < PSR + 40^\circ + 65^\circ = 180^\circ < PSR + 105^\circ = 180^\circ < PSR = 75^\circ < PSR = < PSQ + < QSR 75^\circ = < PSQ + 20^\circ \(\implies\) < PSQ = 75^\circ - 20^\circ = 55^\circ

\(p \propto q^2\) \(q\propto\frac{1}{\sqrt{r}\) \(p = kq^2\) \(q = \frac{c}{\sqrt{r}}\) where c and k are constants. \(q^2 = \frac{c^2}{r}\) \(p = \frac{kc^2}{r}\) If k and c are constants, then kc\(^2\) is also a constant, say z. \(p = \frac{z}{r}\) p varies inversely as r.

Prob (passing English) = x Prob (passing Maths) = Y Prob (failing English) = 1 - x Prob (failing Maths) = 1 - y Prob (failing both test) = Prob(failing English) and Prob(failing Maths) = (1 - x)(1 - y) =1 - y - x + xy =1 - (y + x) + xy

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option C.

Considering the triangle in the diagram \(tan\theta = \frac{opp}{hyp}\) tan 63\(^{\circ} = \frac{h}{4}\) h = 4 tan 63\(^{\circ}\) Area of trapezium = \(\frac{1}{2}h(a+b)\) \(\frac{1}{2} \times 4 tan 63^{\circ}[5+9]\right)\) = 28\(\times 1.963 = 54.96 = 55cm^2\)

15% loss = N595 \(\implies\) (100 - 15)% of CP = N595 85% of CP = N595 CP = \(\frac{595 \times 100}{85}\) = N700.00

Bearing of Q and P \(180^{\circ} – (360^{\circ}-x)\) 180\(^{\circ}-360^{\circ}\)+x -180\(^{\circ}\) + x x - 180\(^{\circ}\)

The graph should start at its lowest point(-1) at -\(\frac{\pi}{2}\), rise to cross the origin (0, 0), and continue upward to reach its peak (1) at \(\frac{\pi}{2}\). It then turns downward, crossing the x- axis(0) at \(\pi\) and ends back at its lowest point (-1) at \(\frac{3\pi}{2}\)

(a) \(\sqrt{50} - 3\sqrt{2}(2\sqrt{2} - 5) - 5\sqrt{32}\) \(\sqrt{25 \times 2} - (3 \times 2)\sqrt{2 \times 2} + (3 \times 5)\sqrt{2} - 5\sqrt{16 \times 2}\) = \(5\sqrt{2} - 6(2) + 15\sqrt{2} - 5(4\sqrt{2})\) = \(5\sqrt{2} - 12 + 15\sqrt{2} - 20\sqrt{2}\) = \((5 + 15 - 20)\sqrt{2} - 12\) = -12. (b) \(\frac{1}{2} \log_{10} (\frac{25}{4}) - 2 \log_{10} (\frac{4}{5}) + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{25}{4})^{\frac{1}{2}} - \log_{10} (\frac{4}{5})^{2} + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{5}{2}) - \log_{10} (\frac{16}{25}) + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{\frac{5}{2} \times \frac{320}{125}}{\frac{16}{25}}\) = \(\log_{10} (\frac{\frac{160}{25}}{\frac{16}{25}}\) = \(\log_{10} 10 = 1\)