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\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\) 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

The locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance. PV = PW \(\sqrt{(x-1)^{2} + (y-1)^{2}} = \sqrt{(x-3)^{2} + (y-5)^{2}}\). Squaring both sides of the equation, (x-1)2 + (y-1)2 = (x-3)2 + (y-5)2. x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25 Collecting like terms and solving, x + 2y = 8.

\(y = x(2-x) \Rightarrow y= 2x - x^{2}\); \(\int^{2}_{0}(2x-x^{2} = (x^{2}-\frac{x{3}}{3})^{2}\) solving further gives (4 - \(\frac{1}{3} * 8) - (0) = \frac{4}{3} sq\hspace{1 mm}unit\)

\(\int_{0}^{z} (\sin x - \cos x) \, dx\) 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: \(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = 1 - \sqrt{2}\)

\(y = 2x \\ V = \int\pi y^{2}dy \\ but\hspace{1mm}y = 2x \\ V = \int\pi4x^{2}dx\\ V = \frac{4(3)^{3}\pi}{3} - \frac{4(0)^{3}\pi}{3}\) V = \(\frac{4*27\pi}{3}\) = 36\(\pi \hspace{1mm}cubic\hspace{1mm}units\) (Note: the limit of integration x = 0 → x = 3)

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.

To evaluate the integral \(\int_{-2}^{1} (x - 1)^{2} \, dx,\) We first expand the integrand: \((x - 1)^{2} = x^{2} - 2x + 1\) Thus, the integral becomes: \(\int_{-2}^{1} (x^{2} - 2x + 1) \, dx = \int_{-2}^{1} x^{2} \, dx - 2\int_{-2}^{1} x \, dx + \int_{-2}^{1} 1 \, dx\) Now, we evaluate each integral: For \(\int_{-2}^{1} x^{2} \, dx\): \(\left[ \frac{x^{3}}{3} \right]_{-2}^{1} = \frac{1}{3} - \left(-\frac{8}{3}\right) = 3\) For \(-2\int_{-2}^{1} x \, dx\): \(-2\left[ \frac{x^{2}}{2} \right]_{-2}^{1} = -2\left(\frac{1}{2} - 2\right) = 3\) For \(\int_{-2}^{1} 1 \, dx\): \([x]_{-2}^{1} = 1 + 2 = 3\) Combining the results: 3 + 3 + 3 = 9.

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: \(x = \frac{1}{\sqrt{3}} \quad \text{or approximately} \quad x \approx 0.577\)

To find \( h \) for the function \(y = 1 + hx - 3x^2\) with a minimum value of 13: 1. The vertex \( x \) is given by \(x = \frac{h}{6}\) 2. Substitute into the function: \(y = 1 + \frac{h^2}{6} - 3\left(\frac{h}{6}\right)^2 = 1 + \frac{h^2}{6} - \frac{h^2}{12}\) 3. Combine terms: \(y = 1 + \frac{2h^2}{12} - \frac{h^2}{12} = 1 + \frac{h^2}{12}\) 4. Set equal to 13: \(1 + \frac{h^2}{12} = 13 \implies \frac{h^2}{12} = 12 \implies h^2 = 144 \implies h = 12 \text{ or } -12\) \( h = 12 \) or \( h = -12 \).

A Vernier caliper is an instrument used to measure both the outer and inner diameters of a tube.

Let the forces be F1, and F2. F1 = m1 a1 = 3 x 2 = 6N F2 = m2 a2 = 1 x 9 = 9N F1 : F2 = 6:9 = 2:3 Thus ratio of forces is 2: 3