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Relationships. 1.1. m. (a , b). 1 A straight line with gradient –3 passes through the point (– 2, 4). Determine the equation of this straight line. y – b = m(x – a) y – 4 = -3(x + 2 ) y – 4 = -3x – 6 y = -3x – 2. 1.1. 1.1. 1.1.

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## Relationships

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**1.1**m (a , b) 1 A straight line with gradient –3 passes through the point (– 2, 4).Determine the equation of this straight line. y – b = m(x – a) y – 4 = -3(x + 2) y – 4 = -3x – 6 y = -3x – 2**1.1**1.1 1.1 2 Solve the inequation 4p – 18 < p – 3. 3p – 18 < -3 3p < 15 p < 5 3 Gertrude buys 3 small tins and 5 medium tins of paint. Total cost is £44.92 Write an equation to represent this information. 3s + 5m = 4492 (-3 + 18) # 2.1**1.1**1.1 1.1 5a + 2b = 24 X2 → 2a – 2b = 4 4 Solve the following system of equations algebraically: 5a + 2b = 24 a – b = 2 Add 7a = 28 a = 4 4 – b = 2 b = 2 a = 4 , b = 2**1.1**1.1 1.1 5T – P = H r 5 Change the subject of the formula to T 5T = H + P r = r(H + P) 5T = r(H + P) T 5 1.1 Out of 10**1.2**y 18 6 The diagram shows the parabola with equation y = kx2What is the value of k? y = kx2 → 12 = k X 22 12 = 4k k = 3 16 14 (2 , 12) x y 12 10 8 6 4 2 x -3 -2 -1 0 1 2 3**y**1.2 1.2 8 6 4 2 x -2 -1 0 1 2 -4 -3 7 The equation of the quadratic function whose graph is shown below is of the form y = (x + a)2 + b, where a and b are integers. Write down the values of a and b. y = (x – 1)2 + 3 a = -1 , b = 3 3 1**1.2**1.2 1.2 8 Sketch the graph y = (x + 6)(x – 2) on plain paper.Mark clearly where the graph crosses the axes and state the coordinates of the turning point. For roots, y = 0 → (x + 6)(x – 2) = 0 x + 6 = 0 or x – 2 = 0 x = -6 or x = 2 Y Intercept, x = 0 at x = 0 y = (0 + 6)(0 – 2) = 6 X (-2) = -12 x value of turning pt = -6 + 2 = -2 at x = -2 y = (-2 + 6)(-2 – 2) = 4 X (-4) = -16 2 2 -6 -12 (-2 , -16)**1.2**1.2 1.2 9 A parabola has equation y = (x + 2)2 – 3. (a) Write down the equation of its axis of symmetry.(b) Write down the coordinates of the turning point on the parabola and state whether it is a maximum or minimum. x = -2 1.2 Out of 9 2 3 Minimum (-2 , -3)**1.3**10 Solve the equation (x – 5)(x + 2) = 0. x – 5 = 0 or x + 2 = 0 x = 5 or x = -2**1.3**1.3 1.3 1.3 ax2 + bx + c = 0 a = 1 b = 5 c = -2 + 11 Solve the equation x2 + 5x – 2 = 0 using the quadratic formula. -b √ x = b2 – 4ac – 2a + -5 √ 52 – 4X1X(-2) – 2X1 + -5 √ 33 – 2 + -5 ( ) √ -5 – 33 ( ) √ 33 OR 2 2 or = -5.4 = 0.4**1.3**1.3 12 Determine the nature of the roots of the equation 2x2 – 5x + 3 = 0 using the discriminant. a = 2, b = -5, c = 3 b2 – 4ac = (-5)2 – 4 X 2 X 3 = 25 – 24 = 1 Two real roots as b2 – 4ac > 0 1.3 Out of 7 # 2.2**1.4**1.4 0.75m 1.25m 1m 13 In a health and safety inspection this scaffold must have a right angle to be passed as safe. Will it meet the health and safety requirements? If it is a RAT then 0.752 + 12 = 1.252 LHS 0.752 + 12 = 1.5625 RHS 1.252 = 1.5625 LHS = RHS, therefore it is a RAT, so will meet H + S requirements. # 2.2 2.2 Out of 2**1.4**1.4 1.4 14 The diagram below shows the design stages of a kite. Kite STUV and a circle with centre T are detailed in the diagram. SV is the tangent to the circle at S and UV is the tangent to the circle at U. Given that angle STU is 124°, calculate angle SVU. LSVU = 360 – (90 + 90 + 124) = 560 1240**1.4**1.4 15 The large watering can holds 10 litres. It is 50cm high. There is a similar small version which is 30cm high. How much will it hold? Linear Scale Factor = 30/50 = 0.6 Volume Scale factor = 0.63 Volume Small Can= 0.63 X 10 = 2.16 litres**1.4**16 Here is a regular pentagon Calculate the size of the shaded angle Angle at centre: 360 ÷ 5 = 720 Angles at feet: (180 – 72) ÷ 2 = 540 Shaded angle: 2 X 54 = 1080 # 2.1 720 540 540 1.4 Out of 8 2.1 Out of 2**1.5**1.5 2 17 Sketch the graph of y = 2cosx° for 900 1800 2700 3600 -2 .**1.5**18 Write down the period of the graph of the equation y = sin 3x°. period = 360 ÷ 3 = 1200**1.5**1.5 1.5 1.5 Out of 6 4sinx – 2 = 0 A Change to sinx= S 19 Solve the equation 4 sin x° – 2 = 0 i ii 4sinx = 2 a0 180 – a sinx = 2/4 sinx = 0.5 Find Acute Angle iv iii Sin-1(0.5) = 300 360 – a 180 + a C T Quadrants? Sin Positive→ i and ii 180 – 30 x = 300 or Solutions = 1500 = 300

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