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PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)

P1 (4, 3), P2 (x, y) y = 2x + 4 .....(1) y = 7 - x .....(2) Substitute (2) in (1) 7 - x = 2x + 4 7 - 4 = 2x + x 3 = 3x x = 1 Substitute in eqn (2) y = 7 - x y = 7 - 1 y = 6 P2 (1, 6) Distance between 2 points is given as D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\) D = \(\sqrt{(-3)^2 + (3)^2}\) D = \(\sqrt{9 + 9}\) D = \(\sqrt{18}\) D = \(\sqrt{9 \times 2}\) D = \(3\sqrt{2}\)

\(\theta\) = 135o Cos 135o = Cos(90 + 45)o = cos90ocos45o - sin90osin45o = 0cos45o - (1 x \(\frac{\sqrt{2}}{2}\)) = \(-\frac{\sqrt{2}}{2}\)

\(\int^{\frac{\pi}{4}}_0sec^2 \theta d \theta\) = \([\tan \theta]_{0} ^{\frac{\pi}{4}}\) = \(\tan \frac{\pi}{4} - \tan 0\) = \(1 - 0\) = 1.

\(\begin{array}{c|c} x & (x - \varkappa) & (x - \varkappa)^2 \\ \hline 2 & -5 & 25 \\ \hline 3 & -4 & 16 \\ \hline 8 & 1 & 1 \\ \hline 10 & 3 & 9 \\ \hline 12 & 5 & 25 \\ \hline & & 76 \end{array}\) S.D = \(\sqrt{\frac{(x - \varkappa)^2}{n}}\) S.D = \(\sqrt{\frac{76}{5}}\) S.D = 3.9

\(\frac{n + 1 (n - 2)}{(n + 1)!}\) \(\frac{(n + 1) + (n - 2)!(n - 2)!}{(n + 1)!}\) \(\frac{(n + 1)(n + 1 -1)(n+1-2)(n+1-3)!}{3!(n-2)!}\) \(\frac{(n + 1)(n)(n-1)(n-2)!}{3!(n-2)!}\) \(\frac{(n + 1)(n)(n-1)}{3!}\) Since n = 15 \(\frac{(15 + 1)(15)(15-1)}{3!}\) \(\frac{16 \times 15 \times 14}{3 \times 2 \times 1}\) = 560

< PQR = 90^\circ (angle in a semi-circle) < QRP = (90 - x)^\circ

\(\frac{720 \times 1000}{8 \times 60 \times 60}\) = \(\frac{20 \times 10}{8}\) = \(\frac{200}{8}\) = 25 m/s

Let 0 = angle of the minor sector angle of the major sector = 360 - \(\theta\)(angle at a point) 2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc) 2\(\pi r = 180^o\) r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\) Lenght of ninor arc = \(\frac{\theta}{360} \times 2 \pi r\) 54 = \(\frac{\theta}{360} \times 3 \pi r\) \(\theta = \frac{360 \times 54}{2 \pi r}\) but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r \(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\) \(\theta = 2 \times 54 = 108^o\) angle of the major sector = 360 - 108o = 252o

/YZ/\(^2\) = /XZ/\(^2\) + /XY/\(^2\) (Verify using Pythagoras' theorem)

cos(x + 40)o = 0.0872 x + 40 = cos-10.0872 x + 40o = 84.99o x = 84.99o - 40o x = 44.99 x = 45o

From the diagram above, Sin 54º = \(\frac{\text{OPP}}{\text{HYP}}\) Sin 54º = \(\frac{\text{x}}{50}\) x = 50 Sin 54º

\(\begin{array}{c|c} x & x - x & (x - \bar{x})^2\\ \hline 15 & -6 & 36\\21 & 0 & 0\\17 & -4 & 16\\ 26 & 5 & 25 \\ 18 & -3 &9 \\ 29 & 8 & 64 \end{array}\) \(E(x - \bar{x})^2\) = 150 N = 6 S.D = \(\sqrt{\frac{(x - x)^2}{N}}\) S.D = \(\sqrt{\frac{150}{6}}\) = 5

x + 0.4y = 3...(i) y = \(\frac{1}{2}\)x x = 2y x - 2y = 0....(ii) solve simultaneously; x + 0.4y = 3 - x - 2y = 0 2.4 = 3 y = \(\frac{3 \times 10}{2.4 \times 10} \) = \(\frac{30}{24} = \frac{5}{4}\) x - 2(\(\frac{5}{4}\)) = 0 x - \(\frac{5}{2}\) = 0 x = \(\frac{5}{2}\) x + y = \(\frac{5}{2} + \frac{5}{4}\) \(\frac{10 + 5}{4} = \frac{15}{4}\) = 3\(\frac{3}{4}\)

Sum = (n - 2)180 1800 = (n - 2)180 divide both sides by 180o \(\frac{1800}{180}\) = (n - 2)\(\frac{180}{180}\) 10 = n - 2 10 + 2 = n n = 12