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6 - x - x2 = 0 a = -1; b = -1; c = 6 Sum of roots = c + k = -\(\frac{-b}{a}\) = \(\frac{-(-1)}{-1}\) = -1

Each interior angle = 140 \(\frac{(n - 2) \times 180}{n} = 140\) (n - 2) x 180 = 140n 150 - 360 = 140n 180m - 140n = 360 40n - 360 n = \(\frac{360}{40}\) n = 9 Sum of all interior angles = (n - 2) x 180 = (9 - 2) x 180 = 7 x 180 = 1260

No. of buckets of water = \(\frac{\text{Capacity of reservoir}}{\text{Capacity of buckets}}\) = \(\frac{800 \times 700 \times 500}{10 \times 1000}\) = \(\frac{28000 0000}{10000}\) = 28000

Total no of balls = 30 Let x = no. of red balls Pr(red) = \(\frac{x}{30}\) Pr(black) = \(\frac{3}{10} = \frac{9}{30}\) Pr(white) = \(\frac{2}{5} = \frac{12}{30}\) No. of black balls = 9 No. of white balls = 12 9 = 12 + x = 30 x = 30 - 21 x = 9 No. of red balls = 9 OR Let w = no of white balls B = no of black balls R = no of red balls Pr(black balls) = \(\frac{3}{10}\) Pr(white balls) = \(\frac{2}{5}\) But Pr(white balls) = \(\frac{2}{5}\) = \(\frac{W}{30}\) cross multiply W = no of white balls = 12 Pr(black balls) = \(\frac{3}{10}\) = \(\frac{B}{30}\) cross multiply B = no of black balls = 9 Total number of balls in the basket = W + B + R = 30 ( given) = 12 + 9 + R = 30 R = 30 - 12 - 9 = 9 Therefore, the no of red balls = 9.

\(\frac{1}{2}\sqrt{32} - \sqrt{18} + \sqrt{2}\) = \(\frac{1}{2} (4\sqrt{2}) - 3\sqrt{2} + \sqrt{2}\) = 2\(\sqrt{2} - 3\sqrt{2} + \sqrt{2}\) = 3\(\sqrt{2} - 3\sqrt{2} - 3\sqrt{2}\) = 0

(x + 10)o + (2x - 40)o + (3x - 90)o = 180 6x - 120 = 180 6x = 180 + 120 6x = 300 x = \(\frac{300}{6}\) x = 50 x + 10o = 50o + 10o = 60o 2x - 40 = 2(50o) - 40 = 60o 3x - 90 = 3(50o) - 90o = 60o Hence, it is an equilateral triangle

(x - 3y)2 - (x + 3y)2 = [(x - 3y) - (x + 3y)] [(x + 3y + x + 3y)] = [-6y] [2x] = -12xy

No expanded working has been attached for this question yet. The verified answer is Option A.

when y = 0, value of x = roots of the equation x2 - 5x + 4 = 0 x = 1, 4 (see graph)

h = ? \(\frac{h}{5cm}\) = sin 45o h = 5cm x 0.6691 = 3.35cm

Area = \(\frac{1}{2}(a + b)\)h where a = QR = 6cm b = PA = 10cm Area = \(\frac{1}{2}(6 + 10)\) x 3.35cm2 = \(\frac{1}{2}(16)\) x 3.35cm2 = 26.8cm2 = 27cm2 (approx.)

Tan \(\theta = \frac{6}{6} = 1\) \(\theta\) = tab - 1(1) = 45o o = 90o

Now, t + 60o + 50o = 180o t = 180o - 110o t = 70o

30 + x = 100 x = 100 - 30 = 70o

\(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\) \(\frac{x}{4 + x} = \frac{7}{7 + 5}\) \(\frac{x}{4 + x} = \frac{7}{12}\) (2x = 7)(4 + x); 12x = 28 + 7x 12x - 7x = 28 5x = 28 x = \(\frac{28}{5}\) x = 5.6