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No expanded working has been attached for this question yet. The verified answer is Option C.

old area = 20\(^2\) = 400 new area = 21\(^2\) = 441 % increase in area \(=\frac{41}{400}\times \frac{100}{1}\) =10.25%

No expanded working has been attached for this question yet. The verified answer is Option B.

\(1011_2 + X_7 = 25_{10} = 1011_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\) = 8 + 0 + 2 + 1 = \(11_{10}\) \(X_7 = 25_{10} - 11_{10}\) = \(14_{10}\) \(\begin{array}{c|c} 7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2 \end{array}\) X = \(20_7\)

[\(\frac{1}{0.03}\) + \(\frac{1}{0.024}\)] = [\(\frac{1}{0.03 \times 0.024}\)]-1 = [\(\frac{0.024}{0.003}\)]-1 = \(\frac{0.03}{0.024}\) = \(\frac{30}{24}\) = 1.25

\(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)b b = \(a^{-2}\) and \(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)b b = \(a^{-\frac{2}{3}}\) \(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\) x \(a^{-\frac{2}{3}}\) = \(a^{-\frac{1}{6}}\) take the cube of both sides c = (\(a^{-\frac{1}{6}})^3\) = \(a^{-\frac{1}{2}}\)

\(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\) \(\frac{k}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = k\(\sqrt{3 - 2}\) K(\(\sqrt{3}\) - K\(\sqrt{2}\) ) = K\(\sqrt{3}\) - 2 K\(\sqrt{3}\) - K\(\sqrt{3}\) - K\(\sqrt{2}\) = - 2 - K\(\sqrt{2}\) = - 2 K = \(\frac{- 2}{- \sqrt{2}}\) K = \(\frac{2}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\) K = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\)

V\(\pi\)r2h = \(\pi\)(3)2(10) = 90\(\pi\)cm3 V = \(\pi\)(5)2 x 18 = 450\(\pi\)cm3 No of volume = \(\frac{450\pi}{90\pi}\) = 5 selling price = 5 x N15 = N75 profit = N75 - N50 = N25.00

\(\frac{x + a}{x - a}\) = m x + a = mx - ma a + ma = mx - x a(m + 1) = x(m - 1) \(\frac{a}{x}\) = \(\frac{m - 1}{m + 1}\)

No expanded working has been attached for this question yet. The verified answer is Option A.

\(\frac{11x + 2}{6x^2 - x - 1}\) = \(\frac{11x + 2}{(3x + 1)(2x - 1)}\) = \(\frac{A}{3x + 1}\) + \(\frac{B}{2x - 1}\) 11x + 2 = A(2x - 1) + B(3x + 1) put x = \(\frac{1}{2}\) \(\frac{15}{2} = \frac{5}{2}B\) B = 3. put x = \(-\frac{1}{3}\) \(-\frac{5}{3} = \frac{-5}{3}\)A \(\implies\) A = 1 ∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)

\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\) = \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\) = 4x(2 + 3x) > 6x = 12x\(^2\) - 2x = 0 = 2x(6x - 1) > 0 = x(6x - 1) > 0 Case 1 (-, -) = x < 0, 6x -1 < 0 = x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution) Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\) Combining solutions in cases(1) and (2) = x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)

Let the first term of the geometric progression (GP) be \( a \) and the common ratio be \( r \). The sum of the first three terms is: a + ar + ar\(^2\) = a(1 + r + r\(^2\)) \(S_3 = a(1 + r + r^2)\) The sum to infinity is: \(S_{\infty} = \frac{a}{1 - r}\) But: \(S_3 = \frac{1}{2} S_{\infty}\) Substituting the expressions gives: \(a(1 + r + r^2) = \frac{1}{2} \cdot \frac{a}{1 - r}\) Dividing by \( a \): \(1 + r + r^2 = \frac{1}{2(1 - r)}\) Cross-multiplying: \(2(1 + r + r^2)(1 - r) = 1\) Expanding and simplifying leads to: \(2(1 - r^3) = 1\) Thus: \(r^3 = \frac{1}{2}\) Taking the cube root: \(r = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}\) Therefore, the positive common ratio of the progression is: \(\frac{1}{\sqrt[3]{2}}\).

The identity element (e) under an operation, say \(\otimes\), is the element such that for any given element under the operation, say a, \(a \otimes e = e \otimes a = a\) From the table, q is the identity element. \(p \otimes q = q \otimes p = p\) Same as all through.

x \(\ast\) y = xy - y - x, x \(\ast\) 3 = 3x - 3 - x = 2x - 3 2 \(\ast\) x = 2x - x - 2 = x - 2 ∴ 2x - 3 = x - 2 x = -2 + 3 = 1