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\(1 - (\frac{7}{3} \times \frac{5}{4}) + \frac{3}{5}\) = \(1 - \frac{35}{12} + \frac{3}{5}\) = \(\frac{60 - 175 + 36}{60}\) = \(\frac{-79}{60}\)

\(\frac{\sqrt{98} - \sqrt{50}}{\sqrt{32}}\) = \(\frac{\sqrt{49 \times 2} - \sqrt{25 \times 2}}{\sqrt{16 \times 2}}\) => \(\frac{7\sqrt{2} - 5\sqrt{2}}{4\sqrt{2}}\) = \(\frac{2\sqrt{2}}{4\sqrt{2}}\) = \(\frac{1}{2}\)

45/2% + 95/2% = 70% 100% - 70% = 30% (i.e women) => 30% = 84 1% = 84/30 95/2% = (84/30 x 95/2) = 133.

\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\) \(\implies 9^{2x - 1} = 27^{x + 1}\) \((3^{2})^{2x - 1} = (3^{3})^{x + 1}\) \(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\) \(4x - 3x = 3 + 2 \implies x = 5\)

Convert to base 10, divide, and then re-convert to base 5.

U = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30} P = {6,12,18,24,30} Q = {4,8,12,16,20,24,28} P∪Q = {4,6,8,12,16,18,20,24,28,30} (P∪Q)c = {2,10,14,22,26}

\(x\propto \frac{uv}{u+v}\) x = \(\frac{kuv}{u+v}\) K = \(\frac{x(u+v)}{uv}\) \(\frac{3(3+1)}{3\times1}\) \(\frac{3\times4}{3}\) k = 4 x =\(\frac{4uv}{u+v}\) x = \(\frac{4\times 3\times3}{3+3}\) x = \(\frac{36}{6}\) x = 6

3x - 5y + 5 = 0 → eqn1 4x - 7y + 8 = 0 → eqn2 eqn1 * 4; 12x - 20y + 20 = 0 → eqn3 eqn2 * 3; 12x - 21y + 24 = 0 → eqn4 eqn3 - eqn4 = y - 4 = 0 ∴ y = 4 From eqn1, 3x - 5y + 5 = 0 3x - 5(4) + 5 = 0 3x - 20 + 5 = 0 3x - 15 = 0 3x = 15 x = 5 x and y = 5, 4 respectively

To find the common ratio of the geometric progression with terms \( n-2 \), \( n \), and \( n+3 \), we use the property that the square of the middle term is equal to the product of the other two terms: \(n^2 = (n - 2)(n + 3)\) Expanding the right side gives: \(n^2 = n^2 + 3n - 2n - 6\) Simplifying leads to: \(0 = n - 6 \implies n = 6\) The terms of the progression are: - First term: \( n - 2 = 4 \) - Second term: \( n = 6 \) - Third term: \( n + 3 = 9 \) The common ratio \( r \) is calculated as follows: \(r = \frac{n}{n-2} = \frac{6}{4} = \frac{3}{2}\) We can verify it with the third term: \(r = \frac{n+3}{n} = \frac{9}{6} = \frac{3}{2}\) Thus, the common ratio of the geometric progression is: \(\boxed{\frac{3}{2}}\)

4abx - 2axy - 12b2x + 6bxy = (4abx - 2axy) - (12b2x - 6bxy) = 2ax(2b - y) -6bx(2b - y) = (2ax - 6bx)(2b - y) = 2x(a - 3b)(2b - y)

\(\int^{3} _{2}(x^2 - 2x)\)dx =\(\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{3}_{2}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{3}_{2}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\) =\(\frac{-8}{3}\) + 4 = \(\frac{4}{3}\)

Range = (Highest - Lowest) items = 27 - 3 = 24

\(^{4}C_2 \times ^{3}C_1\) = \(\left(\frac{4!}{(4-2)!2!}\right)\times\left(\frac{3!}{(3-1)!1!}\right)\) = \(\frac{4!}{2!2!}\times \frac{3!}{2!}\) = \(\frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{3 \times 2!}{2!} = 18\)

\(\frac{4}{a-3}-\frac{1}{a+3}\) = \(\frac{4(a+2)-1(a-3)}{(a-3)(a+2)}\) \(\frac{4a+8 -a +3}{(a-3)(a+2)}\) = \(\frac{3a+11}{(a-3)(a+2)}\)

Q' ∩ R Q' = U - Q Q' = {a, b, c, d, g, h, i} R = {c, e, h, g} Q' ∩ R = {c, h, g}