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H will represent \(\frac{108}{630}\) x \(\frac{360^o}{1}\) ≈ 62^\circ

The direction of the force is determined by Fleming's left-hand rule: Field (Forefinger): N to S (right to left). Current (Middle finger): P to Q (downwards). Force (Thumb): When aligning fingers as described, the thumb points upwards. This force acts on the segment PQ, causing the initial upward movement.

To find n if 34n = 100112, convert both sides to base 10 = 3n + 4 = (1 x 24) + (0 x23) + (0 x 22) + (1 x 21) + 1 x 2o = 3n + 4 = 16 + 0 + 0 + 2 + 1 3n + 4 = 19 3n = 15 n = 5

% error in Area = \(\frac{\pi(5.1)^2 - \pi(5)^2 \times 100%}{\pi(5)^2}\) = \(\frac{\pi 26.01 - 25 \times 100%}{\pi(25)}\) = \(\frac{1.01}{25}\) x 100% = 4.04%

Let \(\log_{b} a^{n} = x\) \(\therefore a^{n} = b^{x}\) \(a^{n} = (a^{\frac{1}{n}})^{x}\) \(a^{n} = a^{\frac{x}{n}} \implies n = \frac{x}{n}\) \(x = n^{2}\)

\(\frac{4^{2x}}{4^{3x}}\) = 2 42x - 3x = 2 4-x = 2 (22)-x = 21 Equating coefficients: -2x = 1 x = -\(\frac{1}{2}\)

\(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}\) = \(\frac{1.25 \times 2}{6.25}\) x 10\(^{-4 - 1 - 5}\) \(\frac{2.50}{6.25}\) x 10\(^{-10}\) = \(\frac{250}{625}\) x 10\(^{-10}\) 0.4 x 10\(^{-10}\) = 4.0 x 10\(^{-11}\)

5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\) = 5(3\(\sqrt{2}\)) - 3(6\(\sqrt{2}\)) + 4(5\(\sqrt{2}\)) 15\(\sqrt{2}\) - 18\(\sqrt{2}\) + 20\(\sqrt{2}\) = 35\(\sqrt{2}\) - 18\(\sqrt{2}\) = 17\(\sqrt{2}\)

x = 3 - \(\sqrt{3}\) x2 = (3 - \(\sqrt{3}\))2 = 9 + 3 - 6\(\sqrt{34}\) = 12 - 6\(\sqrt{3}\) = 6(2 - \(\sqrt{3}\)) ∴ x2 + \(\frac{36}{x^2}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{36}{6(2 - \sqrt{3})}\) 6(2 - \(\sqrt{3}\)) + \(\frac{6}{2 - \sqrt{3}}\) = 6(- \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{4 - 3}\) 6(2 - \(\sqrt{3}\)) + 6(2 + \(\sqrt{3}\)) = 12 + 12 = 24

x = (all prime factors of 44) and y = (all prime factors of 60) ∴ x = (2, 11), y = (2, 3, 5) X ∪ Y = (2, 3, 5, 11), X ∩ Y = (2)

U = (1, 2, 3, 6, 7, 8, 9, 10) E = (10, 4, 6, 8, 10) F = (x : x\(^2\) = 2\(^6\), x is odd) ∴ F = \(\phi\) Since x\(^2\) = 2\(^6\) = 64 x = \(\pm 8\) which is even ∴ E ∩ F = \(\phi\) Since there are no common elements Since E ∩ F = \(\phi\), then ( E ∩ F)' = U = (1, 2, 3, 6, 7, 8, 9, 10)

\(9p^{2} - q^{2} + 6qr - 9r^{2}\) = \(9p^{2} - (q^{2} - 6qr + 9r^{2})\) = \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\) = \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\) = \(9p^{2} - (q - 3r)^{2}\) = \((3p + (q - 3r))(3p - (q - 3r))\) = \((3p + q - 3r)(3p - q + 3r)\)

\(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\) Squaring both sides, \(y^{2} + 48y + 576 = 121y\) \(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\) \(y^{2} - 64y - 9y + 576 = 0\) \(y(y - 64) - 9(y - 64) = 0\) \((y - 9)(y - 64) = 0\) \(\therefore \text{y = 64 or y = 9}\)

Given S = ut + \(\frac{1}{2} at^2\) S = ut + \(\frac{1}{2} at^2\) ∴ 2S = 2ut + at2 = at2 + 2ut - 2s = 0 t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\) = -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\) = \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))

Let the amounts invested at 4% and 5% respectively be x and y. \(\therefore x + y = 280 ... (i)\) Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\) Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\) \(\therefore 0.04x + 0.05y = 12.80\) \(\implies 4x + 5y = 1280 ... (ii)\) From (i), \(x = 280 - y\). Put into (ii), \(4(280 - y) + 5y = 1280\) \(1120 - 4y + 5y = 1280\) \(1120 + y = 1280 \implies y = 1280 - 1120 = N160\) \(\therefore\) N160 was invested at the rate of 5% per annum.