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A histogram is described by the following; Rectangular bars of equal width II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval. III. Rectangular bars have common sides with no gaps in between.

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: \(\frac{19}{60} \div \frac{7}{12}\) = \(\frac{19}{60} \times \frac{12}{7}\) = \(\frac{19}{35}\)