Correct Answer: A. 285°

Explanation

∠KPM = 360° - (30 + 45)° = 360° - 75° = 285°

Explanation

With the 6N force acting in the direction 090°, the diagram is redrawn as :

Resultant = \(\begin{pmatrix} 4 \cos 65° \\ 4 \sin 65° \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 8 \cos 145° \\ 8 \sin 145° \end{pmatrix} + \begin{pmatrix} 4 \cos 245° \\ 4 \sin 245° \end{pmatrix} + \begin{pmatrix} 5 \cos 295° \\ 5 \sin 295° \end{pmatrix}\)

= \(\begin{pmatrix} 4 \cos 65 \\ 4 \sin 65 \end{pmatrix} + \begin{pmatrix} 6 \cos 90 \\ 6 \sin 90 \end{pmatrix} + \begin{pmatrix} -8 \cos 35 \\ 8 \sin 35 \end{pmatrix} + \begin{pmatrix} -4 \cos 65 \\ -4 \sin 65 \end{pmatrix} + \begin{pmatrix} 5 \cos 65 \\ -5 \sin 65 \end{pmatrix}\)

= \(\begin{pmatrix} 4 \times 0.4226 \\ 4 \times 0.9063 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix} + \begin{pmatrix} -8 \times 0.8192 \\ 8 \times 0.5736 \end{pmatrix} + \begin{pmatrix} -4 \times 0.4226 \\ -4 \times 0.9063 \end{pmatrix} + \begin{pmatrix} 5 \times 0.4226 \\ -5 \times 0.9063 \end{pmatrix}\)

= \(\begin{pmatrix} 1.6904 \\ 3.6252 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix} + \begin{pmatrix} -6.5536 \\ 4.5888 \end{pmatrix} + \begin{pmatrix} -1.6904 \\ -3.6252 \end{pmatrix} + \begin{pmatrix} 2.113 \\ -4.5315 \end{pmatrix}\)

= \(\begin{pmatrix} -4.4406 \\ 6.0573 \end{pmatrix}\)

\(|Resultant| = \sqrt{(-4.4406)^{2} + (6.0573)^{2}} = \sqrt{56.41}\)

= 7.5107N.

The magnitude of the resultant = 7.5N.

(b) Let \(\theta\) be the direction of the resultant.

\(\tan \theta = \frac{6.0573}{-4.4406} = - 1.364\)

\(\theta = -53.75° \)

= \(180° - 53.75° = 126.25°\) with the positive x- axis.

Explanation

W = 8i + 3J

X = 6i - 5J

Y = 2i + 3J

w + x - 2y = 8i + 3j + 6i - 5j - 2(2i + 3j)

= 10i - 8j

Z = \(\frac{10i - 8j}{|Z|}\)

= \(\frac{10i - 8j}{4}\)

= \(\frac{10i}{4}\)

\(\frac{8j}{4}\)

= \(\frac{5}{2}\)i - 2j

Correct Answer: B. 14,784 cm\(^3\)

Explanation

Volume of the cylinder = \(\frac{θ}{360} \times \pi r^2h\)

θ = 360\(^o\) - 90\(^o\) = 270\(^o\)

∴ Volume of the cylinder = \(\frac{270}{360} \times \frac{22}{7} \times \frac{14^2}{1} \times \frac{32}{1} = \frac{37,255,680}{2520} = 14,784cm^3\)

Correct Answer: A. 3\(\sqrt{2}\) + 4

Correct Answer: B. 252°

Explanation

Let 0 = angle of the minor sector

angle of the major sector = 360 - \(\theta\)(angle at a point)

2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)

2\(\pi r = 180^o\)

r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)

Lenght of ninor arc

= \(\frac{\theta}{360} \times 2 \pi r\)

54 = \(\frac{\theta}{360} \times 3 \pi r\)

\(\theta = \frac{360 \times 54}{2 \pi r}\)

but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r

\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)

\(\theta = 2 \times 54 = 108^o\)

angle of the major sector = 360 - 108^o

= 252^o

Correct Answer: D. \(ab^2 - a^2b\)

Explanation

\(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\) = \(\frac{a^2 b^24(b^2 - a^2}{ab(a + b)}\)

= \(\frac{ab [(b - a) (b + a)]}{a + b}\)

= ab(b - a)

= \(ab^2 - a^2b\)

Correct Answer: C. 4

Explanation

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10

\(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10

\(\frac{6x - 4}{2}\) = 10

6x - 4 = 2 x 10

= 20

6x = 20 + 4

6x = 20

x = \(\frac{24}{6}\)

x = 4

Correct Answer: D. -2x + 3y - 6 =0

Explanation

Equation of a straight line : y = mx + b

where m = the slope of the line

b = y- intercept

Given the two points (-3, 0) and (0, 2).

\(m = \frac{2 - 0}{0 - (-3)} = \frac{2}{3}\)

\(y = \frac{2}{3}x + 2 \implies 3y = 2x + 6\)

\(-2x + 3y - 6 = 0\)

Explanation

(a)

Note that triangles AXC and BXD are similar.

\(\therefore \frac{6}{|XD|} = \frac{3}{4}\)

\(|XD| = \frac{6 \times 4}{3}\)

= \(8 cm\)

(b)(i)

From \(\Delta ZYX\),

\(\stackrel\frown{YZX} = 90° - 22° = 68°\)

\(\stackrel\frown{YXZ} = 312° - 270° = 42°\)

\(\stackrel\frown{ZYX} = 180° - (68° + 42°)\)

= \(180° - 110°\)

= \(70°\)

\(a^{2} = b^{2} + c^{2} - 2bc \cos A\)

\(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \cos 70\)

= \(2500 +1024 - 3200 \times 0.342\)

= \(3524 - 1094.46\)

\(|XZ|^{2} = 2429.54\)

\(|XZ| = \sqrt{2429.54} = 49.29 km\)

\(\approxeq 49.3 km\) (to 1 d.p)

(ii)

\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

\(\frac{50}{\sin \theta} = \frac{49.3}{\sin 70}\)

\(\sin \theta = \frac{50 \times \sin 70}{49.3}\)

\(\sin \theta = 0.9530\)

\(\theta = \sin^{-1} (0.9530)\)

= \(72.37°\)

\(\therefore\) The bearing of Z from X = \(312° - 72.37°\)

= \(239.63° \approxeq 239.6°\)