Explanation

(a) \(\frac{h}{100} = \sin 24\)

\(h = 100 \sin 24\)

= \(100 \times 0.4067\)

= \(40.67m\)

(b) \(\sin \theta = \frac{4}{5} = 0.8\)

\(\theta = \sin^{-1} (0.8) = 53.13°\)

\(< BAC = 2 \theta = 2(53.13°)\)

= \(106.26°\)

\(\approxeq 106°\) (to the nearest degree).

(c)

In \(\Delta PSQ\),

\(\frac{h}{PQ} = \tan 71.6\)

\(PQ = \frac{h}{\tan 71.6}\)

In \(\Delta QSR\),

\(\frac{h}{QR} = \tan 45\)

\(QR = \frac{h}{\tan 45}\)

\(PR = PQ + QR\)

\(\frac{h}{3} + h = 70 \implies \frac{4}{3}h = 70\)

\(h = \frac{70 \times 3}{4} = 52.5m\)

Correct Answer: B. x > -7

Correct Answer: A. 2.4

Explanation

(a)

(b)

(c) (i) From the graph, the roots of the equation 2x\(^2\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\(^2\) - 7x - 9.

The roots of the equation 2x\(^2\) - 7x = 26 : -2.2 and 5.8.

(ii) The coordinates of the minimum point of y : 1.8, -15.

(iii) The range of values for which 2x\(^{2}\) - 7x < 9 : - 0.9 < x < 4.5.

Correct Answer: A. \(\frac{2}{5}\)

Explanation

Gradient(slope) m = \(\frac{ y_2 - y_1}{ x_2 - x_1}\)

the points are \((\frac{1}{2}, \frac{- 1}{3}) and ( 3 , \frac{2}{3})\)

m = \(\frac{\frac{2}{3} - (\frac{-1}{3})}{3 - \frac{1}{2}}\)

= \(\frac{\frac{2}{3} + \frac{1}{3}}{3 - \frac{1}{2}}\)

= \(1 \div\frac{5}{2}\) = \(1\times\frac{2}{5}\)

Therefore, m = \(\frac{2}{5}\)

Correct Answer: D. x\(^2\) - 6x + 7

Explanation

dy/dx = 2x - 6

y = ∫ 2x - 6

y = \(\frac{2x^2}{2} - 6 + c\)

y = x\(^2\) - 6x + c

passes through (1,2)

2 = 1\(^2\) - 6(1) + c

2 = 1 - 6 + c

c = 7

y = x\(^2\) - 6x + c

y = x\(^2\) - 6x + 7

Correct Answer: B. 5

Correct Answer: B. {0, -1}

Explanation

6x(x + 1) = 0

When 6x = 0 and

x + 1 = 0

x = 0 and x = -1

(0, -1)

Explanation

Observe that the common difference = 2, substitute to get 165 = \(\frac{n}{2}\)[2(-3) + (n - 1)2]

n\(^2\) - 4n - 165 = 0

n = 15 or n = -11

Therefore, the number of terms is 15 as n cannot be negative.

Correct Answer: A. 13%

Explanation

Let the original volume be V with radius r.

\(V = \frac{4}{3}\pi r^{3}\)

45% increased volume = 145%V.

Let the %age increase in radius = m%r

\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)

\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)

\(1.45V = V(\frac{m}{100})^{3}\)

\(\implies 1.45 \times 10^{6} = m^{3}\)

\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)

\(\therefore \text{%age increase =} 113.2 - 100 = 13.2%\)

\(\approxeq 13%\)