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This is a standard trigonometric derivative Therefore, \(\frac{d}{dx}\)(2 sec x) = 2 sec x tan x.

The given central angle of \(120^\circ\) corresponds to the minor arc The central angle corresponding to the major arc is therefore \(360^\circ - 120^\circ = 240^\circ.\) The length of an arc of a circle of radius \(r\) that subtends an angle \(\theta\) (in degrees) at the centre is given by \(\frac{\theta}{360^\circ} \times 2\pi r.\) Substitute \(r = 14\) cm and \(\theta = 240^\circ\): \(\frac{240}{360} \times 2\pi \times 14 = \frac{2}{3} \times 28\pi = \frac{56\pi}{3}\ \text{cm}.\) = 58.67 cm

The interquartile range (IQR) is given by: IQR = Q\(_3\) − Q\(_1\)

Total number of pears = 100 Number of unripe pears = 20 Number of ripe pears = 100 - 20 = 80. Probability that a randomly picked pear is ripe \(P(\text{ripe}) = \frac{\text{Number of ripe pears}}{\text{Total number of pears}} = \frac{80}{100} = \frac{4}{5}\) = 0.8

\(\frac{1}{2} + \frac{1}{3} \times \frac{3}{4}\) Using BODMAS \(\frac{1}{2} + \frac{1}{3} \times \frac{3}{4}\) = \(\frac{1}{2} + \frac{1}{4}\) = \(\frac{2 + 1}{4}\) = \(\frac{3}{4}\)

If Log\(_k243\) = 5 Convert to index form. k\(^5\) = 243 k\(^5\) = 3\(^5\) Since the powers on both sides are equal, equate the bases. k = 3

0.000371 has three significant figures (3, 7, and 1). To write it with two significant figures, retain the first two digits and examine the following digit: First two significant digits: 3 and 7 Next digit: 1 (which is less than 5), so no rounding up occurs. Therefore: 0.000371 \(\approx\) 0.00037.

P varies directly as N and inversely as M P = \(\frac{\text{k N}}{\text{M}}\) 9 = \(\frac{ 8k}{12}\) 8k = 12 x 9 K = \(\frac{108}{8}\) = 13.5 P = \(\frac{13.5 N}{\text{M}}\) Find N, when M = 17, P = 27 27 = \(\frac{13.5 N}{17}\) 13.5 N = 27 x 17 = 459 N = \(\frac{459}{13.5\) = 34.

\(\begin{pmatrix} 2 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & 4 & 1 \end{pmatrix}\) = 2\(\begin{pmatrix} 2& 4 \\ 4 & 1 \end{pmatrix}\) - 0\(\begin{pmatrix} 3 & 4 \\ 1 & 1 \end{pmatrix}\) + 1\(\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\) = 2(2 - 16) - 0 + 1(12 - 2) = 2(-14) + 1(10) = - 28 + 10 = - 18.

The parts labelled I, II and III are: I: Collenchyma – Located in the hypodermis (just below the epidermis), providing flexible mechanical support to the young stem. II: Parenchyma – Comprises the main bulk of the cortex, consisting of thin-walled cells responsible for food storage and metabolic functions. III: Sclerenchyma – Forms the protective pericycle caps (bundle caps) over the vascular bundles, providing rigid mechanical strength. Option C is the correct answer

\(\frac{\text{x}}{2}\) - \(\frac{\text{y}}{5}\) = 1 - - - - - - 1 y - \(\frac{\text{x}}{3}\) = 8 - - - - - 2 We clear the fractions by multiplying through by the lcm of 2 and 5 = 10 for eqn 1 and 3 for eqn 2 5x - 2y = 10 - - - - - - - 3 3y - x = 24 - - - - - - - - - 4 from eqn 4, make x the subject of formula x = 3y - 24 (after multiply thru by -1, so x can be positive) Put x = 3y - 24 into eqn 3 5(3y - 24) - 2y = 10 15y - 120 - 2y = 10 15y - 2y = 10 + 120 13y = 130 y = \(\frac{130}{13}\) = 10 y = 10 put in eqn 4 3y - x = 24 - - - - - - - - - 4 3(10) - x = 24 30 - x = 24 - x = 24 - 30 = -6 x = 6 ( minus cancels out minus) Therefore, x = 6, y = 10.

The distance d from the center to the chord is found using the Pythagorean theorem, where half the chord is 15 cm (\(\frac{30}{2}\)). So d \(\sqrt{(r^2 - (\frac{chord}{2})^2}\) = \(\sqrt{(17^2 - 15^2)}\) = \(\sqrt{(289 - 225)}\) = \(\sqrt{64}\) = 8 cm.

∴ ∠VRS = 94º (angle on a straight line) ∠VTS = 116º (adjacent angle on a straight line) ∠RVT = 50^\circ (corresponding angles) ∴ 94º + 50^\circ + 116^\circ + x = 360^\circ x + 260^\circ = 360^\circ x = 100^\circ

a. x 0º 30º 60º 90º 120º 150º 180º cos2x 1 0.5 -0.5 -1.0 -0.5 0.5 1 2cos2x 2 1.0 -1.0 -2.0 -1.0 1.0 2 y = 2cos2x - 1 1.0 0 -2.0 -3.0 -2.0 0 1.0 b See graph above c. See table and graph above d. 2cos2x + \(\frac{1}{2}\) = 0 Substract 1\(\frac{1}{2}\) from both sides 2cos2x + \(\frac{1}{2}\) - 1\(\frac{1}{2}\) = 0 - 1\(\frac{1}{2}\) From the graph, look for the value of x corresponding to where y = + 1\(\frac{1}{2}\). This occurs at x = 51º and 129º

Divide \( 137 \) by \( 5 \): \(137 \div 5 = 27 \quad \text{(remainder } 2\text{)}\) Divide \( 27 \) by \( 5 \): \(27 \div 5 = 5 \quad \text{(remainder } 2\text{)}\) Divide \( 5 \) by \( 5 \): \(5 \div 5 = 1 \quad \text{(remainder } 0\text{)}\) Divide \( 1 \) by \( 5 \): \(1 \div 5 = 0 \quad \text{(remainder } 1\text{)}\) Now, we collect the remainders in reverse order: \( 1, 0, 2, 2 \). Thus, \( 137 \) in base \( 5 \) is represented as \( 1022_5 \).