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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 = \(3\sqrt{2}\)

\(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{y - 1}{x - -2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\) = \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\) y = 2x + 5

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

y = x2 - \(\frac{1}{x}\) y = x2 - x-1 \(\frac{\delta y}{\delta x}\) = 2x + x-2 \(\frac{\delta y}{\delta x}\) = 2x + \(\frac{1}{x^2}\)

\(\int^2_1(x^2 - 4x)dx\) \((\frac{x^3}{3} - \frac{4x^2}{2})\) substituting integrate values \([\frac{8}{3} - \frac{4 \times 4}{2}] - [\frac{1}{2} - 2]\) = \(-\frac{11}{3}\)

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

\(\frac{x}{7} = \frac{96}{1}\) ==> \(\frac{672 + x}{8} = 112\) Therefore x = 224

Arrange all the values in ascending order, 2,2,3,3,3,4,4,4,4,5,5,5,7,8,9,9

Range = Highest Number - Lowest Number Range = 11 - 2 = 9

\(\begin{array}{c|c} x & (x - \varkappa) & (x - \varkappa)^2 \\ \hline 2 & -5 & 25 \\ \hline 3 & -4 & 16 \\ \hline 8 & 1 & 1 \\ \hline 10 & 3 & 9 \\ \hline 12 & 5 & 25 \\ \hline & & 76 \end{array}\) 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

\(\frac{8!}{3!}\) \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2\times 1}\) ==> 8 x 7 x 6 x 5 x 4 = 6720

Pass(P) = \(\frac{2}{3}\), Fail(F) = \(\frac{1}{3}\) T = P.P.F ==> \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\) = \(\frac{4}{27}\)

Man lives = \(\frac{2}{3}\) not live = \(\frac{1}{3}\) Wife lives = \(\frac{3}{5}\) not live = \(\frac{2}{5}\) P(at least one lives to 80 years) = P(man lives to 80 not woman) + P(woman lives to 80 and not man) + P(both live to 80) \(P = (\frac{2}{3} \times \frac{2}{5}) + (\frac{2}{5} \times \frac{1}{3}) + (\frac{2}{3} \times \frac{3}{5})\) = \(\frac{4}{15} + \frac{3}{15} + \frac{6}{15}\) = \(\frac{13}{15}\)

Equation of the line \(\frac{y - 4}{x - 0} = \frac{0 - 4}{1 - 0}\) \(\frac{y - 4}{x} = \frac{-4}{1}\) \(\therefore -4x = y - 4\) \(y = -4x + 4\) \(\therefore \text{The shaded portion = } y \leq -4x + 4\)