Question Library

using the trial method by inserting each option in the equation. Inserting 21º: sin([5 x 21] - 28) = cos([3 x 21] - 50) sin(105 - 28) = cos (63 - 50) sin 77º = cos 13º where: sin 77º = 0.9744 cos 13º = 0.9744

The sum of angles in a triangle = 180º (180 - 108)º = 72º Isosceles triangle has both two equal sides and two equal angles. ∠OSN = ∠SON = \(\frac{72}{2}\) = 36º

The sum of angles in a triangle = 180º (180 - 108)º = 72º Isosceles triangle has both two equal sides and two equal angles. ∠OSN = \(\frac{72}{2}\) = 36º

The exterior angle theorem states that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles. ∠M = ∠R + ∠N → 141 = 41 + ∠N ∠N = 141 - 41 = 100º ∠QNR = 180 - 100 → 80º

where x=-4 and y = 16, 16 = -4m - 4 4m = -16 - 4 4m = -20 M = -5

\((101_{two})^3\) = \((5_{ten})^3\) 5\(^3\) = 125 And 125 = \(1111101_{two}\)

Using the expansion method on (10110.011)\(_2\) (1 * 2\(^4\)) + (0 * 2\(^3\)) + (1 * 2\(^2\)) + (1 * 2\(^1\)) (0 * 2\(^0\)) + (0 * 2\(^{-1}\)) + (1 * 2\(^{-2}\)) + (1 * 2\(^{-3}\)) (1*16) + 0 (1*4) + (1*2) + 0 + 0 + (1*\(\frac{1}{4}\)) + (1*\(\frac{1}{8}\)) 16 + 4 + 2 + (0.25 + 0.125) 22 + 0.375 (10110.011)\(_2\) = 22.375\(_{10}\)

Convert from base 5 to base 10 432\(_{five}\) = (4 x 5\(^2\)) + (3 x 5\(^1\)) + (2 x 5\(^0\)) = (4 x 25) + (3 x 5) + (2 x 1) = 100 + 15 + 2 = 117\(_{ten}\) Then convert from base 10 to base 3 3 117 3 39 r 0 3 13 r 0 3 4 r 1 3 1 r 1 0 r 1 Selecting the remainders from bottom to top: 117\(_{ten}\) = 11100\(_{three}\) Hence; 432\(_{five}\) = 11100\(_{three}\)

\((8^{-1})^{k+2}\) = \((8^{0})\) base 8 cancel out on both sides -1(k+2) = 0 -k -2 = 0 : k = -2

4\(^{3x}\) = 4\(^{2(x+1)}\) 3x = 2x + 2 3x - 2x = 2 x = 2

where Log\(_2\) 8\sqrt2 → Log\(_2\) \sqrt128 → Log\(_2\) 128\(^\frac{1}{2}\) = \(\frac{1}{2}\) * (Log\(_2\) 128) → \(\frac{1}{2}\) * (Log\(_2\) 2\(^7\)) = 7 * \(\frac{1}{2}\) * (Log\(_2\) 2) where (Log\(_2\) 2) = 1 → 7 * \(\frac{1}{2}\) * 1 = \(\frac{7}{2}\) or 3.5

Recall that: log\(_3\) 27 → log\(_3\)3\(^3\) 3log\(_3\)3 → 3 * 1 = 3 Then log\(_3\) 27 = 2x + 1 → 3 = 2x + 1 3 - 1 = 2x 2 = 2x 1 = x

Rationalization with conjugate \({3+\sqrt2}\) \(\frac{1}{3-\sqrt2}\) → \(\frac{1 * [3+\sqrt2]}{[3-\sqrt2][3+\sqrt2]}\) = \(\frac{3+\sqrt2}{9 -3\sqrt2 + 3\sqrt2 - \sqrt4}\) = \(\frac{3+\sqrt2}{9 - 2}\) → \(\frac{3+\sqrt2}{7}\) = \(\frac{3}{7}\) + \(\frac{\sqrt2}{7}\)

\sqrt24 + \sqrt96 - \sqrt600 = y\sqrt6 convert all to basic surds \sqrt24 = 2\sqrt6 \sqrt96 = 4\sqrt6 - \sqrt600 = -10\sqrt6 2\sqrt6 + 4\sqrt6 - 10\sqrt6 = -4\sqrt6 value of y = -4

(x\(^2\) - 4x + p) Use the coefficient of the middle variable(-4x) = (\(\frac{-4}{2}\))\(^2\) = (-2)\(^2\) = 4