Correct Answer: D. 60°

Correct Answer: D. 10%

Explanation

A = 3,500, P = N2,500

A = P + I

But I = N3,500 - N2,500

I = N1,000

I = \(\frac{PRT}{100}\)

N1,000 = \(\frac{2.500 \times R \times 4}{100}\)

1000 = 100R

R = \(\frac{1000}{100}\)

= 10%

Correct Answer: D. 27m

Explanation

sin30 = x/54 x =54 x 1/2 = 27m

Correct Answer: D. 4.2

Explanation

x = \(\frac{2 + 3 + 3 + 4 + 5 + 5+ 5+ 7 + 7 + 9}{10}\)

=\(\frac{50}{10}\)

= 5

Variance = \(\frac{\sum{(x - x})^2}{N}\)

\(\frac{9 + 4+ 4+ 1 + 4 + 4 + 16}{10}\)

= \(\frac{42}{10}\)

= 4.2

Correct Answer: (c) 1

Correct Answer: A. \(2^{-n}\)

Explanation

\(8^{n} \times 2^{2n} \div 4^{3n} = 2^{3n} \times 2^{2n} \div 2^{6n}\)

\(2^{2n + 3n - 6n}\)

= \(2^{-n}\)

Explanation

(a)(i) \(f(x) = x^{3} + px^{2} - 10x + q\)

\(x^{2} + x - 6 = 0 \implies x^{2} - 2x + 3x - 6 = 0\)

\(x(x - 2) + 3(x - 2) = 0 \implies (x - 2)(x + 3) = 0\)

\(x = 2 ; x = -3\)

\(f(-3) = (-3)^{3} + p(-3)^{2} - 10(-3) + q = 0\)

\(-27 + 9p + 30 + q = 0 \implies 9p + q = -3 .... (1)\)

\(f(2) = (2)^{3} + p(2)^{2} - 10(2) + q = 0\)

\(8 + 4p - 20 + q = 0 \implies 4p + q = 12 .... (2)\)

(2) - (1) :

\((4p + q) - (9p + q) = (12 - (-3)) \implies -5p = 15\)

\(p = \frac{15}{-5} = -3\)

Put p = -3 in (2), we have:

\(4(-3) + q = 12 \implies -12 + q = 12 \)

\(q = 12 + 12= 24\)

\(f(x) = x^{3} - 3x^{2} - 10x + 24\)

(ii) Using the method of long division,

\(\frac{x^{3} - 3x^{2} - 10x + 24}{x^{2} + x - 6}\)

we get the third factor = \(x - 4\).

(b) \(\frac{\mathrm d v}{\mathrm d t} = 2\frac{1}{2} cm^{3} s^{-1}\)

\(v = x^{3}\)

\(\frac{\mathrm d v}{\mathrm d x} = 3x^{2}\)

\(\frac{\mathrm d v}{\mathrm d t} = \frac{\mathrm d v}{\mathrm d x} \cdot \frac{\mathrm d x}{\mathrm d t}\)

\(\frac{5}{2} = 3x^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\)

When x = 2cm, we have

\(\frac{5}{2} = 3(2)^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\)

\(\frac{\mathrm d x}{\mathrm d t} = \frac{5}{2} \div 12 = \frac{5}{24} cm s^{-1}\)

Correct Answer: C. 19

Explanation

a + 3d = 37

-20 + 3d = 37

3d = 37 + 20 = 57

d = \(\frac{57}{3}\)

= 19

Correct Answer: B. 4.3

Explanation

Using the sine rule

\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\

\frac{x}{sin43^o}=\frac{5}{sin53^o}\\

x = \frac{5\times sin43^o}{sin53^o}\\

=\frac{5\times 0.6820}{0.7986}\\

=4.3\)

Explanation

(a) \(p = (4i - 3j) ; q = (-i + 5j)\)

\(2p + 3q = \begin{pmatrix} 8i - 3i \\ -6j + 15j \end{pmatrix}\)

= \(\begin{pmatrix} 5i \\ 9j \end{pmatrix}\)

Let \(r = k \begin{pmatrix} 5i \\ 9j \end{pmatrix} = \begin{pmatrix} 5ki \\ 9kj \end{pmatrix}\)

\(|r| = \sqrt{(5k)^{2} + (9k)^{2}} = 15\)

\(\sqrt{25k^{2} + 81k^{2}} = 15 \implies 25k^{2} + 81k^{2} = 225\)

\(106k^{2} = 225 \implies k^{2} = \frac{225}{106} = 2.123\)

\(k = \sqrt{2.123} = 1.457\)

\(\therefore r = \begin{pmatrix} 5i \times 1.457 \\ 9j \times 1.457 \end{pmatrix}\)

= \(\begin{pmatrix} 7.285i \\ 13.113j \end{pmatrix}\)

(b)(i)

Resolve the forces along the x and y- axis.

Let \(F_{1} = \begin{pmatrix} 8 \cos 120° \\ 8 \sin 120° \end{pmatrix}\)

\(F_{2} = \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix}\)

\(F_{3} = \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)

\(R = F_{1} + F_{2} + F_{3}\)

= \(\begin{pmatrix} 8 \cos 120° \\ 8 \sin 120° \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)

= \(\begin{pmatrix} -8 \cos 60° \\ 8 \sin 60 \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)

= \(\begin{pmatrix} -8 \times 0.5 \\ 8 \times 0.866 \end{pmatrix} + \begin{pmatrix} 6 \times 0 \\ 6 \times 1 \end{pmatrix} + \begin{pmatrix} 4 \times 0.866 \\ 4 \times 0.5 \end{pmatrix}\)

= \(\begin{pmatrix} -4 + 0 + 3.464 \\ 6.928 + 6 + 2 \end{pmatrix}\)

= \(\begin{pmatrix} -0.536 \\ 14.928 \end{pmatrix}\)

\(|R| = \sqrt{(-0.536)^{2} + (14.928)^{2}} = \sqrt{0.2873 + 222.8452}\)

= \(\sqrt{223.1325} = 14.94N\)

(ii)Direction \(\theta\)

\(\tan \theta = \frac{14.928}{-0.536} = -27.851\)

\(\theta = -87.9° = 92.1°\)

The direction is N 02.1° W or a bearing of 267.9°.