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y ∝ 1/x y = K/x K = xy K = (1/2) * 4 ( when y = 4 and x = 1/2) K = 2 ∴y = 2/x 10 = 2/x (when y = 10) 10x = 2 x = 2/10 x = 1/5

-1 < 3 -2x and 3 - 2x \(\leq\) 5 2x < 3 + 1 and -2x \(\leq\) 5 - 3 2x < 4 and -2x \(\leq\) 2 x < 2 and x \(\geq\) 2/-2 = x \(\geq\) -1 If -1 \(\leq\) x < 2 it implies that x = -1, 0, 1

To find the sum of the first three terms of a geometric progression (G.P.) where the first term \( a \) and the fourth term \( ar^3 \) are given, we set up the following equations: Given: - \( a = 6 \) - \( ar^3 = 162 \) Find the common ratio \( r \) From the equation for the fourth term: \(ar^3\) = 162 Substituting \( a \): \(6r^3\) = 162 Dividing both sides by 6: \(r^3 = \frac{162}{6}\) = 27 Taking the cube root: r = \(\sqrt[3]{27}\) = 3 Find the first three terms The first three terms of the G.P. are: - First term: \( a = 6 \) - Second term: \( ar = 6 \cdot 3 = 18 \) - Third term: \( ar^2 = 6 \cdot 3^2 = 6 \cdot 9 = 54 \) Calculate the sum of the first three terms The sum \( S \) of the first three terms is: S = a + ar + ar\(^2\) = 6 + 18 + 54 Calculating this: S = 6 + 18 + 54 = 78 The sum of the first three terms of the progression is \( \boxed{78} \).

\(P\times Q =\sqrt{PQ}\) 4\(\times(8\times32) = 4\times\sqrt{8\times32}\) = 4\(\times\sqrt{256}\) =4\(\times16\) = \(\sqrt{4\times16}\) =\(\sqrt{64}\) = 8

x = -3 substitute x = -3 in 3x3 + 5x2 - 11x + 4 3(-3)3 + 5(-3)2 - 11(-3) + 4 -81 + 45 + 33 + 4 -81 + 82 = 1

Qn = 3 * 2n - 2 Um = 3 * 22m - 3 Q2 * U2 = (3 * 22 - 2) * (3 * 22(2) - 3) = 3 * 20 * 3 * 21 = 3 * 1 * 3 * 2 = 18

ac - 2bc - a\(^2\) + 4b\(^2\) ac - 2bc - (a\(^2\) - 4b\(^2\)) ac - 2bc - (a\(^2\) -(2b)\(^2\)) c(a - 2b) - {(a - 2b)(a + 2b)} (a - 2b)(c - (a + 2b)) (a - 2b)(c - a - 2b)

\(a=\frac{1}{2}\\r=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{6}\times\frac{2}{1}=\frac{1}{3}\\S_\infty = \frac{a}{1-r}\\S_\infty = \frac{\frac{1}{2}}{1-\frac{1}{3}}\\=\frac{\frac{1}{2}}{\frac{3}{2}}\\=\frac{1}{2}\times\frac{3}{2}\\=\frac{3}{4}\)

\(L \propto T^2\) \(L = kT^2\) \(k = \frac{L}{T^2}\) when T = 4, L = 64. \(k = \frac{64}{4^2} = 4\) \(L = 4T^2\) when T = 9, \(L = 4 \times 9^2 = 324cm\)

First teacher received \(\frac{2}{5}\) ∴Remainder \(= 1 - \frac{2}{5} = \frac{3}{5}\) 2nd teacher received; \(\frac{2}{15} of \frac{3}{5} = \frac{2}{15} \times \frac{3}{5} = \frac{2}{25}\) 3rd teacher will receive; \(1-\frac{2}{5}-\frac{2}{25}=\frac{25-10-2}{15}\\=\frac{13}{25}\)

6log\(_x2\) - 3log\(_x3\) = 3log\(_50.2\) = log\(_x2^6\) - 3log\(_x3^3\) = log\(_5(0.2)^3\) = logx(64/27) = log5(1/5)3 logx(64/27) = log5(1/125) let logx(64/27) = y ∴xy = 64/27 and log5(1/125) = y ∴ 5y = 1/125 5y = 125-1 5y = 5-3 ∴ y = -3 substitute y = -3 in xy = 64/27 implies x-3 = 64/27 1/x3 = 64/27 64x3 = 27 x3 = 27/64 x3 = 3\sqrt27/64 x = 3/4

4516 - P7 = 3056 P7 = 4516 - 3056 P7 = 1426 convert 1426 = 1 * 62 + 4 * 61 + 2 * 60 = 36 + 24 + 2 = 62 Convert 6210 to base 7 62/7 = 8 R 6 8/7 = 1 R 1 1/7 = 0 R 1 ∴P7 = 1167

3\sqrt42x = 16 this implies that (3\sqrt42x)3 = (16)3 42x = 42*3 42x = 46 ∴ 2x = 6 x = 3

Numerator: \(\frac{1}{10}\cdot\frac{2}{3} + \frac{1}{4} = \frac{1}{15} + \frac{1}{4} = \frac{4+15}{60} = \frac{19}{60}\) Denominator: \(\frac{\frac{1}{2}}{\frac{3}{5}} - \frac{1}{4} = \frac{1}{2}\cdot\frac{5}{3} - \frac{1}{4} = \frac{5}{6} - \frac{1}{4} = \frac{10-3}{12} = \frac{7}{12}\) Final: \(\frac{19}{60} \div \frac{7}{12}\) = \(\frac{19}{60} \times \frac{12}{7}\) = \(\frac{19}{35}\)

The shaded portion shows the elements in Q and R, but not in P, Q n R n P'