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△PQR is right-angled at P. Using Pythagoras theorem: QR\(^2\) = PQ\(^2\) + PR\(^2\) = 144 + 25 = 169 QR = \(\sqrt{169}\) = 13 m

To find the limit \(\lim_{x \to 2} \frac{x^3 - 2x^2 + 6x - 12}{x - 2},\) we first substitute \( x = 2 \): \(\frac{2^3 - 2(2^2) + 6(2) - 12}{2 - 2} = \frac{0}{0}.\) This is an indeterminate form, so we factor the numerator. Using synthetic division by \( x - 2 \): \(\begin{array}{r|rrrr} 2 & 1 & -2 & 6 & -12 \\ & & 2 & 0 & 12 \\ \hline & 1 & 0 & 6 & 0 \\ \end{array}\) Thus, \(x^3 - 2x^2 + 6x - 12 = (x - 2)(x^2 + 6).\) We simplify the limit: \(\frac{x^3 - 2x^2 + 6x - 12}{x - 2} = x^2 + 6 \quad \text{for } x \neq 2.\) Now, we find the limit: \(\lim_{x \to 2} (x^2 + 6) = 2^2 + 6 = 10.\) Thus, the limit is 10

Let 4x = u y = sin u \(\frac{\text{dy}}{\text{du}}\) = cos u \(\frac{\text{du}}{\text{dx}}\) = 4 \(\frac{\text{dy}}{\text{dx}}\) = \(\frac{\text{dy}}{\text{du}}\) x \(\frac{\text{du}}{\text{dx}}\) = 4 x cos u but u = 4x Therefore, \(\frac{\text{dy}}{\text{dx}}\) = 4cos4x

Integrate y = 4x\(^3\) + 2x + cos x. ∫4x\(^3\) dx = x\(^4\) + ∫2x dx = x\(^2\) + ∫cos x dx = sin x The integral is x\(^4\) + x\(^2\) + sin x + C

∫ (3x² + 2x - 5) dx = ∫ 3x² dx + ∫ 2x dx + ∫ (-5) dx = 3 · (\(\frac{x^3}{3}\)) + 2 · (\(\frac{x^2}{2}\)) + (-5) · x + C = x\(^3\) + x\(^2\) - 5x + C

Total weight = 15 \times 1000 = 15000 kg. New average = 1000 - 5 = 995 kg. Total for 14 bars = 14 \times 995 = 13930 kg. Heaviest = 15000 - 13930 = 1070 kg.

The pie chart represents a total of 360^\circ. Total number of people = 16 (Doctors) + 8 (Engineers) + 24 (Lawyers) + 21 (Information Technologists) + 27 (Teachers) = 96. The portion for doctors is 16 out of 96. Angle for doctors = \(\frac{16}{96} \times 360^\circ = 60^\circ\).

Total number of students = x + 1 + 2x + 2 + 3 = 3x + 6 = 12 Solve for x: 3x + 6 = 12 3x = 6 x = 2 Now substitute x = 2: - 54 kg: 2 students - 56 kg: 1 student - 58 kg: 4 students (2x = 4) - 60 kg: 2 students - 62 kg: 3 students The frequencies are: 2, 1, 4, 2, 3. The highest frequency is 4 (at 58 kg). Thus, Modal weight: 58 kg

The modal class is the class interval with the highest frequency. From the table: Frequency of 1-10: 2 11 - 20: 7 21 - 30: 10 (highest) 31 - 40: 3 41 - 50: 1 Therefore, the modal class is 21-30. The upper-class boundary of a class interval is the upper limit, considering the continuous nature of grouped data (typically 0.5 above the stated upper limit if classes are assumed continuous). For the class 21 - 30, the upper-class boundary is 30.5.

Ordered: 2.35, 2.50, 2.64, 2.72, 2.91 Median = middle value = 2.64.

Letters: S, C, H, O, O, L (6 letters, O repeated 2 times). Permutations = \(\frac{6!}{2!}\) = \(\frac{720}{2}\) = 360.

The word HANDIER consists of 7 letters: H, A, N, D, I, E, R. All 7 letters are distinct (no repetitions). The number of distinct ways to arrange these 7 letters is the number of permutations of 7 different items, which is 7!. 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 Therefore, HANDIER can be arranged in 5040 ways.

The die has faces 1, 2, 3, 4, 5, 6. Even numbers are 2, 4, 6 (3 outcomes). Total outcomes are 6. Probability = \(\frac{3}{6}\) = \(\frac{1}{2}\).

1: First survives, second dies: \( p \cdot q \) 2: First dies, second survives: \( q \cdot p \) The total probability can be expressed as: \(\text{Total Probability} = pq + qp = 2pq\) Since \( 2pq \) is not an option, the closest form is \( pq \).

Sum \ge 10 ⇒ 6 favourable outcomes out of 36 Pr(sum at least 10) = \(\frac{6}{36}\) = \(\frac{1}{6}\)