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\(1 + \sqrt [3]{x-3} = 4\) = \(\sqrt[3]{x - 3} = 4 - 1\) \(\sqrt[3]{x - 3} = 3 \) take the cube of both sides = x - 3 = 27 x = 27 + 3 ∴ x = 30

Let Mr Manu be x years old and Adu be y years old. But Mr Manu is four times as old as Adu then, x = 4y. 7 years ago, the sum of their ages was 76. ( x - 7) + ( y - 7) = 76 x + y - 14 = 76 x + y = 76 + 14 = 90 But x = 4y Therefore, 4y + y = 90 5y = 90 y = \(\frac{90}{5}\) y = 18 Therefore Adu is 18 years old.

Let the two numbers be x and y \(\frac{1}{3}( x + y) = 12\) then x + y = 36 ..........i 2( x - y) = 12 x - y = 6 ............ii add equations i and ii 2x = 42 x = 21, put x = 21 into equation i x + y = 36 21 + y = 36 y = 36 - 21 = 15 therefore the numbers are 21 and 15

2x - 3y = -11 --- (i) 3x + 2y = 3 --- (ii) Multiply equation (i) by 3 and equation (ii) by 2 6x - 9y = -33 --- (iii) 6x + 4y = 6 --- (iv) Subtract equation (iii) from (iv) 13y = 39 y = \(\frac{39}{13}\) = 3 substitute (3) for y in equation (ii) 3x + 2(3) = 3 3x + 6 = 3 3x = 3 - 6 3x = -3 x = \(\frac{-3}{3}\) = - 1 Now, \((y - x)^2 = (3 - (-1))^2\) = \((3 + 1)^2\) = \(4^2\) = 16

let x = number of tickets sold to children and y = number of tickets sold to adults. Given: x + y = 400 ...................................................i 3x + 5y = 1700 ..............................................ii from Eqn(i) x = (400 - y) .........................(iii) put (400 - y) into Eqn ii 3(400 - y ) + 5y = 1700 1200 - 3y + 5y = 1700 1200 + 2y = 1700 2y = 1700 - 1200 2y = 500 y = \(\frac{500}{2}\) y = 250 Therefore, the number of tickets sold to adults is 250 b. Total number of tickets Mr. Johnson sold = 250 Number of tickets he sold to adults = 175 Then, the number of tickets he sold to children = 250 - 175 = 75 The price of a ticket for an adult is D5.00 and for a child is D3.00 Total sales = 175(5) + 75(3) = 875 + 225 = D1,100.00

a. Let the numbers be p, r and q respectively Given: p + r + q = 81 --- (i) r = 2p --- (ii) q = r + 6 --- (iii) Substitute (2p) for r in equation (iii) q = 2p + 6 --- (iv) Substitute (2p) for r and (2p + 6) for q in equation (i) ⇒ p + (2p) + (2p + 6) = 81 ⇒ 5p + 6 = 81 ⇒ 5p = 81 - 6 ⇒ 5p = 75 p = \(\frac{75}{5}\) = 15. From equation (ii) ⇒ r = 2(15) = 30 From equation (iv) ⇒ q = 2(15) + 6 ⇒ q = 30 + 6 = 36 ∴ The numbers are 15, 30 and 36 b. \(m_1 = \frac{ y_2 - y_1}{ x_2 - x_1}\) Given points are P( 3, 5) and Q( -5, 7) = P(\( x_1 , y_1)\) , Q(\( x_2 , y_2)\) the slope of line PQ = \((\frac{ 7 - 5}{ -5 - 3}) = \frac{2}{-8} = -\frac{1}{4}\) Given that R( x, y) is the midpoint of the line formed by the points P( 3, 5) and Q( -5, 7) R(x, y) = \(\frac{3 + (-5)}{2}, \frac{5 + 7}{2} = \frac{ 3 - 5}{2} , \frac{ 5+7}{2}\) = -\(\frac{2}{2} , \frac{12}{2}\) = (-1, 6) but the line passes through the point R( -1, 6) and is perpendicular to LINE PQ \(m_2 = \frac{1}{m_1} = -\frac{1}{\frac{-1}{4}}\) = 4 Equation of the line can be obtained using \(y - y_1 = m_2( x - x_1)\) y - 6 = 4( x - (-1)) = y - 6 = 4( x + 1 ) y - 6 = 4x + 4 y = 4x + 4 + 6 Therefore, y = 4x + 10

\(y = \frac{ax^3 - b}{3z}\) cross multiply \(ax^3 - b\) = 3yz \(ax^3\) = 3yz + b divide both sides by a \(x^3 = \frac{3yz + b}{a}\) take cube root of both sides therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)

Find the roots of the equations: \(3m^2 - 2m - 65 = 0\) = \( m^2 - 15m + 13m - 65 = 0\) = 3m(m - 5) + 13( m - 5) = 0 ( m - 5)(3m + 13) = 0 m-5 = 0 or 3m + 13 = 0 therefore, m = 5 or \(\frac{-13}{3}\) therefore the roots of the quadratic equation = ( 5, \(\frac{-13}{3})\)

\(x^2 - (y + z)^2\) Using difference of two squares (\(a^2 - b^2 = (a + b)(a - b)|) ⇒ (x + (y + z))(x - (y + z)) ⇒ (x + y + z)(x - y - z)

Let p = \(\frac{2}{3}\) and q = \(\frac{- 3}{4}\) using (y - p)(y - q) = 0 = ( y - \(\frac{2}{3})\)( y - (\(\frac{- 3}{4})) = 0\) = (\( y - \frac{2}{3})( y + \frac{3}{4})\) = 0 \( y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{6}{12} = 0 \) \( y^2 + \frac{1}{12}y - \frac{1}{2}\) = 0 = multiply through by the l. c. m of 3 and 4 = 12 ∴ the quadratic equation is \(12y^2 + y - 6 = 0\)

a. x -3 -2 -1 0 1 2 3 \(2x^2\) 18 8 2 0 2 8 -18 - x 3 2 1 0 - 1 - 2 - 3 - 4 - 4 - 4 - 4 -4 -4 -4 -4 y 17 6 -1 -4 -3 2 11 b. the graph above, solution to question 8b. ci. The roots of the equation from the graph are -1.2 and 1.7. ii. Let y = \(2x^2 - x - 4\) \(\frac{dy}{dx}\) = 4x - 1 At critical point \(\frac{dy}{dx}\) = 0 so that 4x - 1 = 0; x = \(\frac{1}{4}\) Using test points: 1. we use a test point less than \(\frac{1}{4}\), e.g x = 0 \(\frac{dy}{dx}\)(0) = 4(0) - 1 = -1 at this point, y = \(2x^2 - x - 4\) is decreasing 2. we use a test point greater than \(\frac{1}{4}\) e.g x = 1 \(\frac{dy}{dx}\)(1) = 4(1) - 1 = 4 - 1 = 3. at this point, y = \(2x^2 - x - 4\) is increasing. so, the function y = \(2x^2 - x - 4\) increases as x increases when x is greater than \(\frac{1}{4}\) Therefore, the value of x for which y increases are x is greater than \(\frac{1}{4}\) iii. The minimum point of y = -4.1.

The fraction \(\frac{ x^2 + 2 }{ 10x^2 - 13x - 3}\) is undefined when the denominator is equal to zero \(then 10x^2 - 13x - 3 = 0\) by factorisation, \(10x^2 - 13x - 3\) = 0 becomes \( 10x^2 - 15x +2x -3\) = 0 \(5x(2x - 3) + 1(2x - 3) = 0\) \((5x + 1)(2x - 3) = 0\) \(then, x = \frac{-1}{5}\) or \(\frac{3}{2}\)

\(\theta = 115^\circ , radius = 3.4cm^2\) Area of a sector = \(\frac{\theta}{360} \times \pi r^2\) = \(\frac{115}{360} \times \frac{22}{7} \times 3.4\times 3.4\) = \(\frac{29246.4}{2520}\) = \(11.6cm^2\)

using Pythagoras theorem H = \(\sqrt{ 2^2 - 1^2}\) = \(\sqrt{ 4 - 1}\) = \(\sqrt3\)cm

Let each side be l, then area = \(l^2\) Using Pythagoras theorem \(l^2 + l^2 = 12^2\) \(2l^2 = 144\) divide both sides by 2 \(l^2 = 72\) Therefore, the Area of the square is \(72 cm^2\)