Correct Answer: C. 6

Explanation

\(^nC_2\) = 15

\(\frac{n!}{(n - 3)! 2!}\) = 15

\(\frac{n(n - 1)(n - 2)!}{(n - 2)!2}\) = 15

n\(^2\) - n = 30

n\(^2\) - n - 30 = 0

(n - 6)(n + 5) = 0

n = 6 or n = -5

Correct Answer: A. B

Explanation

Boron (B) is highly immobile because of its size. Therefore, choice A is correct.

Correct Answer: A. joint supply

Explanation

Joint supply is a type of supply in which the production of one article in itself leads to the production of other article as well. for example, leather and beef etc. It is also a process that can yield two or more outputs.

Explanation

(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4.

\(T_{n} = a + (n - 1)d\) (terms of an A.P)

\(T_{25} = 5 + (25 - 1) \times 4\)

= \(5 + 24 \times 4\)

= \( 5 + 96 = 101\)

(b) \(\frac{8}{9}, \frac{-4}{3}, 2, ...\) is a G.P with first term = \(\frac{8}{9}\)

\(r = \frac{T_{2}}{T_{1}} = -\frac{4}{3} \div \frac{8}{9} = \frac{-4}{3} \times \frac{9}{8} = -\frac{3}{2}\)

\(T_{n} = ar^{n - 1}\) (terms of a G.P)

\(T_{5} = (\frac{8}{9})(-\frac{3}{2})^{5 - 1}\)

= \(\frac{8}{9} \times \frac{81}{16} = \frac{9}{2}\)

(c) \(T_{n} = ar^{n - 1}\) (terms of a G.P)

\(T_{3} = ar^{2} = 48 ... (1)\)

\(T_{6} = ar^{5} = 14\frac{2}{9} ... (2)\)

\((2) \div (1) : r^{3} = \frac{128}{9} \div 48 = \frac{128}{9} \times \frac{1}{48}\)

\(r^{3} = \frac{8}{27}\)

\(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)

From (1), \(ar^{2} = 48 \implies a \times (\frac{2}{3})^{2} = 48\)

\(4a = 48 \times 9 \implies a = \frac{48 \times 9}{4} = 108\)

\(T_{2} = 108 \times \frac{2}{3} = 72\)

\(T_{4} = 48 \times \frac{2}{3} = 32\)

\(\therefore\) The first four terms of the sequence are \(108, 72, 48, 32\).

Correct Answer: D. unsavory

Correct Answer: A. live to witness the second coming of Christ

Explanation

R = U\(\sqrt{\frac{2h}{g}}\)

h = \(\frac{R^2g}{2u^2}\)

= \(\frac{25^2 \times 10}{2 \times 4^2}\)

= 195.3m

Correct Answer: C. Investigation

Correct Answer: B. 132cm²

Explanation

A prism has 3 rectangular faces and 2 triangular faces and 2 rectangular faces = 10(3 + 4 + 5) = 120

Area of triangular faces = \(\sqrt{s(s - a) (s - b) (s - c)}\)

where s = \(\frac{a + b + c}{2}\)

= \(\frac{3 + 4 + 5}{2}\)

= \(\frac{12}{2}\)

= 6

Area of \(\bigtriangleup\) = \(\sqrt{6(6 - 30(6 - 4)(6 - 5)}\)

= \(\sqrt{6 \times 3 \times 2 \times 1}\) = 6

Area of triangle faces = 2 x 6 = 12cm²

Total surface area = Area of rectangular face + Area of \(\bigtriangleup\) = 120 + 12

= 132cm²

Correct Answer: A. 10km, 40km