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Length of Arc = \(\frac{\theta}{360}\) x 2\(\pi\)r Length of arc = 33cm, diameter = 28cm → r = 14cm 33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7}\times14\) \(\theta\) = \(\frac{33 \times 180}{22 \times 2}\) \(\theta\) = 135º

Length of a chord = 2rSin\(\frac{\theta}{2}\) \(\theta\) = 60º not 30º and radius = 9cm Therefore, Length of the chord = 2 x 9 Sin\(\frac{60º}{2}\) = 2 x 9 x Sin 30º = 2 x 9 x \(\frac{1}{2}\) = 9cm.

The area of a triangle given two sides and the included angle = \(\frac{1}{2}\) x a x b x Sin\(\theta\) a = b = 8cm \(\theta\) = 60º Area = \(\frac{1}{2}\) x 8 x 8 x Sin 60º = \(\frac{1}{2}\) x 64 x \(\frac{\sqrt{3}}{2}\) ( Sin 60º = \(\frac{\sqrt{3}}{2}\) ) = \(\frac{64\sqrt{3}}{4}\) = 16\(\sqrt{3}cm^2\)

The construction of angle 30º (If you take a 60º angle and bisect it, you would divide it into two angles of 30º each.)

10111.11\(_{two}\) convert the number to base 10. 10111.11\(_{two}\) = 1 x 2\(^4\) + 0 x 2\(^3\) + 1 x \(^2\) + 1 x 2\(^1\) + 1 x 2\(^0\) + 1 x 2\(^{-1}\) + 1 x 2\(^{-2}\) = 16 + 0 + 4 + 2 + 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) = 23 + \(\frac{3}{4}\) = 23\(\frac{3}{4}\)

Subtract 14256\(_{seven}\) from 20045\(_{seven}\) = 2456\(_{seven}\) 2 0 0 4 5 1 4 2 5 6 0 2 4 5 6

16\(^x\) = 0.25 16\(^x\) = \(\frac{25}{100} = \frac{1}{4}\) 4(\(^2)^x\) = 4\(^{-1}\) 2x = -1 x = \(\frac{-1}{2}\)

\(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) multiply the denominator and the numerator by the conjugate of 3 + \(\sqrt{7}\) → 3 - \(\sqrt{7}\) \(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) x \(\frac{3 - \sqrt{7}}{3 - \sqrt{7}}\) \(\frac{15 - 5\sqrt{7} + 3\sqrt{7} - 7}{3^2 - (\sqrt{7})^2}\) \(\frac{8 - 2\sqrt{7}}{9 - 7}\) = \(\frac{ 8 - 2\sqrt{7}}{2}\) = 4 - \(\sqrt{7}\)

65 = 30 -x + x + 45 - x 65 - 75 = 10 = x ( number of students offering both subjects) Pr( \(\text{students offering both subjects}\)) = \(\frac{10}{65}\) = \(\frac{2}{13}\)

P = {Primes numbers b/w 0 and 10} = { 2, 3, 5, 7} Q = { Odd numbers b/w 0 and 10} = {1, 3, 5, 7, 9} Let W = { Union of elements in P not in Q and elements in Q not in P} = {1, 2, 9}

For a set with n elements, the total number of subsets is given by: 2\(^n\) where n = the number of members of the set = 5 so, 2\(^5\) = 32 (every set is an improper subset of itself)

144 = 42 - x + 30 - x + 34 + 4x + 3x + 6 + x + 2x 144 = 112 + 8x 8x = 144 - 112 = 32 x = 4. The number of students that registered for physics and mathematics = 4x + 2x = 4 x 4 + 2 x 4 = 24.

M = \(\begin{bmatrix}3 & 4 & 6\\2 & 1 & -1\\-1 & 3 & 5\end{bmatrix}\) |M| = 3\(\begin{bmatrix}1 & -1\\3 & 5\end{bmatrix}\) - 4\(\begin{bmatrix}2 & -1\\-1 & 5\end{bmatrix}\) + 6\(\begin{bmatrix}2 & 1\\-1 & 3\end{bmatrix}\) 3(5 + 3) - 4(10 - 1) + 6(6 + 1) = 24 - 36 + 42 = 30

A = \(\begin{pmatrix}2 & -4 & 3\\5 & 1 & 0\end{pmatrix}\), B = \(\begin{pmatrix}1 & 4 & -2\\-3 & 3 & -1\end{pmatrix}\) 2B = 2 x \(\begin{pmatrix}1 & 4 & -2\\-3 & 3 & -1\end{pmatrix}\) = \(\begin{pmatrix}2 & 8 & -4\\-6 & 6 & -2\end{pmatrix}\) A + 2B = \(\begin{pmatrix}2 & -4 & 3\\5 & 1 & 0\end{pmatrix}\) + \(\begin{pmatrix}2 & 8 & -4\\-6 & 6 & -2\end{pmatrix}\) = \(\begin{pmatrix}4 & 4 & -1\\-1 & 7 & -2\end{pmatrix}\)

Given that A = \(\begin{pmatrix}3 & 2 & 1\\4 & 2 & -1\end{pmatrix}\) and B = \(\begin{pmatrix}1 & 4\\0 & 1\\3 & 2\end{pmatrix}\) A\(^T\) = \(\begin{pmatrix}3 & 4\\2 & 2\\1 & -1\end{pmatrix}\) Find A\(^T\) + B = \(\begin{pmatrix}3 & 4\\2 & 2\\1 & -1\end{pmatrix}\) + \(\begin{pmatrix}1 & 4\\0 & 1\\3 & 2\end{pmatrix}\) = \(\begin{pmatrix}4 & 8\\2 & 3\\4 & 1\end{pmatrix}\)