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Rationalize using the reciprocal of the denominator to multiply through (i.e. Multiply both numerator and denominator using \(4 + 3\sqrt{6}\) ) Watch your signs in the course of this.

\(\frac{5^x \times 25^{x - 1}}{125^{x + 1}}\) = \(\frac{5^x \times 5^{2x - 2}}{5^{3x + 3}}\) = \(\frac{5^{x + 2x - 2}}{5^{3x + 3}}\) = \(\frac{5^{3x - 2}}{5^{3x + 3}}\) = 5\(^{3x - 2 - 3x - 3}\) = 5\(^{-5}\)

Volume of steel ball = \(\frac{4\pi r^2}{3}\) = \(\frac{4}{3}\) \(\pi\) x 1 = \(\frac{4 \pi}{3}\)cm3 Vol. of cylinder = \(\pi\)r2h = \(\pi\) x 22 x 4 Vol. of water = 16\(\pi\) - \(\frac{4 \pi}{3}\) = \(\frac{48 - 4 \pi}{3}\) = \(\frac{44 \pi}{3}\)cm\(^3\)

\(\frac{x^2 + y^2 + xy}{x + y}\) - \(\frac{x^2 + y^2}{x -y}\) = \(\frac{-y^3 - y^3}{x^2 - y^2}\) = \(\frac{2y^3}{x^2 - y^2}\) = \(\frac{2y^3}{y^2 - x^2}\)

If tan\(\theta\) = 2, 4 \(\theta\) = tan-1(2.4) = 67o 25 By sin formula \(\frac{x}{sin 90^o}\) = \(\frac{50}{sin 22^o 37}\) = \(\frac{50}{sin(9o - \theta)}\) sin90o = 1 x = 50 cosec 22o 37 = 50 x 2.604x = 130cm but 100cm = 1m 130cm = 1.3m

2\(\frac{5}{12}\) - 1\(\frac{7}{8}\) x \(\frac{6}{5}\) = \(\frac{29}{12}\) - \(\frac{15}{8}\) x \(\frac{6}{5}\) = \(\frac{29}{12}\) - \(\frac{9}{4}\) = \(\frac{29 - 27}{12}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Two similar triangles are not also congruent

PSO = 90o(angle in a semicircle) SPO = 180o - (90o + 15o) = 75o o - 75o = 105o(opposite angles in cyclic quadrilateral are supplementary)

a(b + c) + \(\frac{5}{d}\) - 2 = 0 ab + ac + \(\frac{5}{d}\) - 2 = 0 abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d acd = 2d - 5 - abd c = \(\frac{2d - 5 - abd}{ad}\)

0 < \(\frac{x + 3}{x - 1}\) < 2 Put x = 0, -3 and 9 0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2 i.e. 0 < 1.5 \(\leq\) 2 (true) but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2 i.e. 0 < -3 \(\leq\) 2 (not true) -3 \(\leq\) 2 -3 is not greater than 0

Let x rep. the usual speed = \(\frac{Distance}{Time}\) x = d/2 d = 2x ............. 1 x + 1 = \(\frac{d}{2 - 1/3}\) x + 1 = \(\frac{3d}{5}\) 5x + 5 = 3d d = \(\frac{5x + 5}{3}\) ................... 2 2x = \(\frac{5x + 5}{3}\) 6x = 5x + 5 x = 5km/hr

2x + 5y = 11.......(i) 7x + 4y = 2.......(ii) (i) x 7 \(\to\) 14x + 35y = 77.........(iii) (ii) x 2 \(\to\) 14x + 8y = 4........(iv) (iii) - (iv) 27y = 73 y = \(\frac{73}{27}\)

Given the equation \[x^3 - 12x - 16 = 0\] with \( x = -2 \) as a solution, we can factor it as follows: Divide by \( x + 2 \) gives: \[x^3 - 12x - 16 = (x + 2)(x^2 - 2x - 8)\] Finding Roots of \( x^2 - 2x - 8 = 0 \) Using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{36}}{2}\] This gives:\[x = 4 \quad \text{and} \quad x = -2 \text{ (with multiplicity 2)}\]

12 men in 9 days, 1 day in \(12 \times 9\) men In 6 days we have \(\frac{12 \times 9}{6}\) men = \(\frac{108}{6}\) = 18 men

Mean = 5, 42 Median = 6 Mode = 7 The mean is less than the median