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\(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\) \(\sqrt{50} - \frac{2}{\sqrt{2}}\) = K\(\sqrt{8}\) = \(\sqrt{2} \times 25 - \frac{2}{\sqrt{2}}\) = K \(\sqrt{4 \times 2}\) \(\frac{5\sqrt{2}}{1} - \frac{2}{\sqrt{2}}\) = 2K\(\sqrt{2}\) \(\frac{5\sqrt{4} - 2}{\sqrt{2}} = 2K\sqrt{2}\) \(\frac{10 - 2}{\sqrt{2}} = 2K \sqrt{2}\) \(\frac{8}{\sqrt{2}} = \frac{2K\sqrt{2}}{1}\) = 2k\(\sqrt{2} \times \sqrt{2}\) = 8 2k \(\sqrt{4}\) = 8 2k x 2 = 8 4k = 8 k = \(\frac{8}{4}\) k = 2

% error = \(\frac{error}{\text{actual value}} \times 100\) error = N72 - N68 = 4 actual value = N72 %error = \(\frac{4}{72} \times 100\) = \(\frac{100}{18} = \frac{50}{9}\) = 5\(\frac{5}{9}\)%

4 oranges sell for Nx, 1 orange will sell for \(\frac{Nx}{4}\) 24 oranges will sell for: \(\frac{Nx}{4} \times 24\) = n6x 3 mangoes sell for Ny, 1 mango will sell for \(\frac{Ny}{3}\) 12 mangoes will sell for \(\frac{Ny}{3} \times 12\) = 4Ny total money pay N6x + N4y = N(6x + 4y)

\(\frac{x^2 - y^2}{(x + y)^2} \div \frac{(x - y)^2}{(3x + 3y)}\) \(\frac{(x + y)(x - y)}{(x + y)(x + y)}\div \frac{(x - y)(x - y)}{3(x + y)}\) = \(\frac{3}{x - y}\)

\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\) substitute x = 64 and y = 27 \(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\) = \(\frac{8 - 3}{27 - 16}\) = \(\frac{5}{11}\)

Volume of a cube with side a cm = a\(^3\) a\(^3\) = 64 \(\implies\) a = 4cm Base area of a cube = a\(^2\) = 4\(^2\) = 16 cm\(^2\) \(\implies\) Base area of the cuboid = 16 cm\(^2\) Volume of cuboid = Base area x height 80 = 16 x h h = \(\frac{80}{16}\) = 5 cm

Tan 50o = \(\frac{h}{6}\) h = 6 tan 50 = 6 x 1.1917 = 7.1505 = 7.15

2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2) \(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\) \(\log_x \frac{27}{8} = 3\) \(\frac{27}{8} = x^3\) \(x^3 = \frac{27}{8}\) x = \(\sqrt{\frac{27}{8}}\) = (\(\frac{27}{8}\))\(\frac{1}{3}\) = \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\) = \(\frac{3}{2}\)

p = (y : 2y \(\geq\) 6) 2y \(\geq\) 6 y \(\geq \frac{6}{2}\) y = \(\geq\) 3 and Q = (y : y -3 \(\leq\) 4) y - 3 \(\leq\) 4 y \(\leq\) 4 + 3 y \(\leq\) 7 therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....} P\(\cap\)Q = {3, 4, 5, 6, 7}

y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\) 6 = k\(\sqrt{3 + 1}\) 6 = k\(\sqrt{4}\) 6 = 2k k = \(\frac{6}{2}\) = 3 y = \(\sqrt{(x + 1)}\) 9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3) \(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\) 3 = \(\sqrt{x + 1}\)(square both sides) 9 = x + 1 x = 9 - 1 x = 8

y = x2 + 2x + k at point(2,0) x = 2, y = 0 0 = (2)2 + 2(20 + k) 0 = 4 + 4 + k 0 = 8 + k k = -8

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

\(\frac{2}{x + 3} - \frac{1}{x - 2}\) = \(\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)}\) = \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\) = \(\frac{x -7}{x^2 + x - 6}\) = \(\frac{x - 7}{x^2 + x - 6}\)

P = x2 + 4x - 2, Q = 2x - 1 Q - p = 2, (2x - 1) - (x2 + 4x - 2) = 2 2x - 1 - x2 - 4x + 2 = 2 -2x - x2 + 1 -x2 - 2x - 1 = 0 x2 + 2x + 1 = 0 x2 + x + x + 1 = 0 x(x + 1) + 1(x + 1) = 0 (x + 1)(x + 1) = 0 x + 1 = 0 or x + 1 = 0 x = -1 or x = -1 x = -1

Volume of pyramid = \(\frac{1}{3}\) x base area x height = \(\frac{1}{3} \times 124 \times 8 \times 14\) = 4 x 8 x 14 = 448m\(^3\)