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S50^\circW = 180^\circ + 50^\circ = 230^\circ

x 2 3 6 2 5 0 4 2 24 x - \(\bar{x}\) -1 0 3 -1 2 -3 1 -1 \((x - \bar{x})^2\) 1 0 9 1 4 9 1 1 26 Mean = \(\frac{24}{8} = 3\) Standard deviation = \(\sqrt{\frac{26}{8}}\) = \(\sqrt{3.25}\) = 1.802 \(\approx\) 1.8

P(Kodjo passing) = \(\frac{3}{4}\); P(Adoga passing) = \(\frac{3}{5}\) P(Kodjo failing) = \(\frac{1}{4}\); P(Adoga failing) = \(\frac{2}{5}\) P(both fail) = \(\frac{1}{4} \times \frac{2}{5}\) = \(\frac{1}{10}\)

< Q = < R (OQ = OR = radii) < QOR = 180^\circ - 2(40^\circ) = 100^\circ < QSR = < RPQ = \(\frac{1}{2}\) < QOR = \(\frac{100}{2} = 50^\circ\)

From the figure above In △BOC, 2x + 30º = 180º Therefore, x = \(\frac{180º - 30º}{2}\) = 75º Each interior angle = 2 times x = 2 x 75 = 150º OR To calculate the interior angle of a regular convex polygon where each side subtends an angle of \(30^\circ\) at the center, we follow these steps: Determine the number of sides (n) The angle subtended at the center by each side is given as \(30^\circ\). The total angle around a point is \(360^\circ\). \(n = \frac{360^\circ}{30^\circ}\) = 12 So, the polygon has 12 sides. Calculate the interior angle. The formula for the interior angle A of a regular polygon is: \(A = \frac{(n - 2) \times 180^\circ}{n}\) Substituting n = 12: \(A = \frac{(12 - 2) \times 180^\circ}{12} = \frac{10 \times 180^\circ}{12} = \frac{1800^\circ}{12} = 150^\circ\) Each interior angle of the regular convex polygon is \(150^\circ\).

l\(^2\) = 3 (k + m) Since they are consecutive positive numbers, we have l = k+1, m = k+2. \(\to\) (k+1)\(^2\) = 3(k + k + 2) k\(^2\) + 2k + 1 = 3(2k + 2) k\(^2\) + 2k + 1 = 6k + 6 k\(^2\) + 2k - 6k + 1 - 6 = 0 k\(^2\) - 4k - 5 = 0 k\(^2\) - 5k + k - 5 = 0 k(k - 5) + 1(k - 5) = 0 k = -1 or 5 Since k, l and m are positive, then k = 5. m = k + 2 = 5 + 2 = 7.

\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\) r = 4 units; centre (\(\alpha\), -2). \((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\) \((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\) \(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\) \(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\) \(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\) \(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\) \(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\) \((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)

Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30. x = 12.5m The height of the tree = 12.5 + 1.7 = 14.2 m

bisector theorem: \(\frac{|MN|}{|MO|}\) = \(\frac{|PO|}{|NP|}\) taking the bisected angle:x and y = |ON|=12 : x+y= 12 x = 12 - y |PO| = 12 - y \(\frac{6}{4}\)= \(\frac{12-y}{y}\) 6y = 4 (12-y) 6y = 48 - 4y = 4.8 Recall that x+y= 12 12 - 4.8 =7.2

Let \(\phi\) be the angle between the two lines. tan \(\phi\) = \(\frac{m_1 - m_2}{1 + m_1 m_2}\) where m\(_1\) = slope of line 1; m\(_2\) = slope of line 2. Line 1: 2x + y = 3 \(\implies\) y = 3 - 2x. Line 2: 3x - 2y = 5 \(\implies\) -2y = 5 - 3x. y = \(\frac{3}{2}\)x - \(\frac{5}{2}\). m\(_1\) = -2, m\(_2\) = \(\frac{3}{2}\). tan \(\phi\) = \(\frac{-2 - \frac{3}{2}}{1 + (-2 \times \frac{3}{2})}\) = \(\frac{\frac{-7}{2}}{-2}\) \(\therefore\) Tan \(\phi\) = \(\frac{7}{4}\).

\(\frac{150}{PR}\) = tan45º PR = \(\frac{150}{tan45º}\) = 150m

\(\int_{0}^{z} (\sin x - \cos x) \, dx\) where \( z = \frac{\pi}{4} \), we have: \(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = \left[-\cos x - \sin x\right]_{0}^{\frac{\pi}{4}}\) Evaluating at the bounds gives: \(-\sqrt{2} - (-1) = -\sqrt{2} + 1\) Thus, the value of the integral is: \(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = 1 - \sqrt{2}\)

t2 sin (3t - 5) = 2t sin ( 3t - 5) + t2 x 3 cos (3t - 5) = 2t sin (3t - 5) + 3t2 cos (3t - 5). Detailed answer below was provided by Ifechuks, a female prospective student of Okopoly.

We are given the function: \(y = x^3 - x\) To find the critical points, we set the derivative equal to zero: \(3x^2 - 1 = 0\) Solving this gives: \(x = \pm \frac{1}{\sqrt{3}}\) Using the second derivative test, we find that: \(\frac{d^2y}{dx^2} = 6x\) - For \( x = \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} > 0 \) (minimum). - For \( x = -\frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} < 0 \) (maximum). Thus, the function \( y = x^3 - x \) has a minimum value at: \(x = \frac{1}{\sqrt{3}} \quad \text{or approximately} \quad x \approx 0.577\)

We are to form two-digit numbers using the digits 0, 1, 2, 3, with repetition allowed, but the number cannot begin with 0. - Tens place (first digit): Cannot be 0 → Possible digits = 1, 2, 3 → 3 choices - Units place (second digit): Can be any of the 4 digits → 0, 1, 2, 3 → 4 choices Total numbers = 3 \times 4 = 12