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The acceleration is not constant, i.e. its direction changes although the magnitude is constant (III is wrong), another thing that makes III wrong is the fact that the acceleration is not directed away from the centre but towards the centre. - No work is done on a body undergoing circular motion - Note that when a body is undergoing circular motion ], its velocity is not constant, i.e. the direction changes, although the magnitude is constant. - Since the acceleration is directed towards the centre, the centripetal force is directed towards the centre.

Efficiency = \(\frac{\text{M.A}}{\text{V.R}}\) 75% = \(\frac{\text{M.A}}{\text{V.R}}\) \(\frac{75}{100} = \frac{\text{M.A}}{5}\) M.A = \(\frac{75(5)}{100}\) = 3.75 But M.A = \(\frac{\text{Load}}{\text{Effort}}\) 3.75 = \(\frac{150}{\text{Effort}}\) Effort = \(\frac{150}{3.75}\) Effort = 40N

a * b = a\(^b\) a * 2 = a\(^2\) = 2 - a a\(^2\) + a - 2 = 0 a\(^2\) + 2a - a - 2 = 0 a(a+2) - 1(a+2) = 0 (a-1)(a+2) = 0 a = 1, -2

n = 3, U\(_3\) = 4x - 2y, U\(_n\) = a + (n-1)d, 4x - 2y = a + 2d... (1) U\(_9\) = 10x - 8y, 10x - 8y = a + 8d... (2) Solving (1) and (2), subtract eqn i from ii a-a + 8d - 2d = 10x -4x - 8y- (-2y) 6d = 6x - 8y+ 2y. 6d= 6x - 6y 6d = 6( x-y) d = x - y

If P\(^{-1}\) is the inverse of P and O is the identity, Then P\(^{-1}\) * P = P * P\(^{-1}\) = 0 i.e. P\(^{-1}\) + P - P\(^{-1}\).P = 0 P\(^{-1}\) - P\(^{-1}\).P = -P P\(^{-1}\)(1 - P) = -P P\(^{-1}\) =\(\frac{-P}{(P-1)}\) = \(\frac{P}{(P-1)}\)

S\(_{\infty}\) = \(\frac{\text{a}}{(1-r)}\), where a = \(\frac{1}{2}\), r = -\(\frac{1}{2}\). S\(_{\infty}\) = \(\frac{\frac{1}{2}}{1 - \frac{-1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{3}{2}}\) = \(\frac{1}{2}\) x \(\frac{2}{3}\) \(\frac{1}{3}\) - 1 = -\(\frac{2}{3}\)

This is a polynomial of the 3rd order, thus x should have three answers. Use the factors given to get values of x as 1, -1 and -2. Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2 Equate each factor to zero to get the values of x x-1 = 0, x = 1 x +1 = 0, x = -1 x - 2 = 0, x = 2 substitute these value of x in the given equation \(ax^3 + bx^2\) + cx - 1 when x = 1 \(ax^3 + bx^2\) + cx - 1 = a + b + c - 1= 0 ------------------------(1) when x = -1 \(ax^3 + bx^2\) + cx - 1 = -a + b - c - 1 = 0-------------------------(2) when x = 2 \(ax^3 + bx^2\) + cx - 1 = 8a + 4b + 2c - 1 = 0 ----------------------(3) add equations 1 and 2 a + b + c - 1= 0 + -a + b - c - 1 = 0 b = 1 put b = 1 into equation (1) a + b + c - 1= 0 = a + 1 + c -1 = 0 a = - c put a = - c into equation(3) 8a + 4b + 2c - 1 = 0, -8c + 4 + 2c - 1 = 0 ( b = 1 remember) -6c = -3 c = \(\frac{-3}{-6}\) = \(\frac{1}{2}\) put c = \(\frac{1}{2}\) and b = 1 into equation (1) a + b + c - 1= 0 a + 1 + \(\frac{1}{2}\) - 1 = 0 a + \(\frac{1}{2}\) = 0 then a = \(\frac{-1}{2}\)

y = 10x - x\(^2\) dy/dx = 10 - 2x As dy/dx = 0, 10 - 2x = 0 2x = 10 x = 5

2 - x > x\(^2\) 2 - x - x\(^2\) > 0 (x+2)(x-1) < 0 Either (x + 2) > 0 and (x - 1) < 0. x + 2 > 0 \(\implies\) x > -2. -2 < x < 1

1/\alpha + 1/\beta = \beta+\alpha/\alpha\beta 3x\(^2\) + 5x - 2 = 0 x\(^2\) + 5x/3 - 2/3 = 0 \alpha\beta = -2/3 \beta+\alpha = -5/3 Thus; \beta+\alpha/\alpha\beta = -5/3 \div -2/3 = 5/2

The lateral view of the frustrum \(\tan \theta = \frac{6}{1.5} = 4\) \(\theta = \tan^{-1} 4 = 75.96^\circ\) Extending to the full height of the pyramid, we have The height of the pyramid which formed the frustrum = (6 + x)m, \(\tan 75.96 = \frac{6 + x}{2.5}\) \(6 + x = 2.5 \times 4 = 10.0\) \(\therefore \text{The height of the pyramid} = 10.0m\)

See Figure above. Since ST is parallel to UV => angle< U = 50^\circ(alternate to Angle UST = 180^\circ - 50^\circ = 130^\circ (angle on a straight line)

Since the inscribed triangle is equilateral, therefore the angles at all the points = 60^\circ Using the formula for inscribed circle, 2R = \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) where R = radius of the circle; a, b and c are the sides of the triangle. ⇒ 2R = \(\frac{\sqrt{3}}{\sin 60}\) 2R = \(\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}\) 2R = 2 R = 1cm

Grad of 3y = 4x - 1 y = \(\frac{4x}{3}\) - \(\frac{1}{3}\) Grad = \(\frac{4}{3}\) Grad of Ky = x + 3 y = \(\frac{\text{x}}{\text{k}}\) + \(\frac{3}{4}\) Grad = \(\frac{1}{\text{k}}\) Since the two lines are perpendicular, \(\frac{1}{\text{k}}\) = -\(\frac{3}{4}\) -3k = 4 k = -\(\frac{4}{3}\)

A perpendicular bisector always divides a line segment through its midpoint. The term bisect itself means dividing equally or uniformly.