Question Library

xo = 35 + 20 (Exterior angle = sum of two interior opposite angles) x = 55o (∠ at the center twice ∠ at circumference) y = 110o ∠PSO = ∠SPO (base ∠S of 1sc \Delta b/c PO = SO) ∴ ∠PSO = (180 - 110)/2 = 35o

\(V = \frac{2}{3} \pi r^{3}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(\frac{\mathrm d V}{\mathrm d r} (r = 2) = 2\pi (2)^{2}\) = \(8\pi\)

\(\int_{1}^{3}(x^2\) - 1)dx = \(\left[\frac{1}{3}x^3 - x\right] ^{3}_{1}\) = (9-3) - \(\frac{1}{3}-1)\) = 6 - \(\left(-\frac{2}{3}\right)\) = 6+\(\frac{2}{3}\) =6\(\frac{2}{3}\)

The number of ways of selecting the debtors = \(^5C_2\) ways \(^5C_2\) = \(\frac{5!}{(5 - 2)!2!}\) = \(\frac{5 . 4 . 3!}{(3!)2!}\) = \(\frac{5. 4}{2. 1}\) = 10ways (3!s cancels ou)

The number of ways to choose 1 member from the house of assembly is: \(\binom{3}{1} = 3\) The number of ways to choose 5 members from the state commissioners is: \(\binom{9}{5}\) = 126 Thus, the total number of ways to form the committee is: \(\text{Total ways} = \binom{3}{1} \times \binom{9}{5}\) = 3 \(\times\) 126 = 378 The total number of ways to form the committee, including one member from the house of assembly, is 378 Ways.

Range = 18 - 15 = 3

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

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

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)

\(L \propto T^2\) \(L = kT^2\) \(k = \frac{L}{T^2}\) 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