Cambridge Secondary 8 > Gradient, Intercept, and Interpreting Graphs
Gradient, Intercept, and Interpreting Graphs
Question 1 of 101 / 10
For part a: User should identify the equation is in the form y = mx + c. Gradient m = 9 and y-intercept c = -5. For part b: User should identify gradient m = -6 and y-intercept c = 11. For part c: User should rewrite the equation as y = -3x + 8. Gradient m = -3 and y-intercept c = 8. For part d: User should identify gradient m = 2/5 (or 0.4) and y-intercept c = 7.For part a: User should subtract 6x from both sides: 3y = 24 - 6x. User should divide all terms by 3: y = 8 - 2x. User can also write this as y = -2x + 8. For part b: From the equation y = -2x + 8, user should identify gradient m = -2. For part c: From the equation y = -2x + 8, user should identify y-intercept c = 8.For part a: User should use the form y = mx + c with m = 4 and c = -3. Equation: y = 4x - 3. For part b: The line passes through (0, 6), so y-intercept c = 6. Gradient m = -1/2. User should write y = -0.5x + 6 or y = -(1/2)x + 6. For part c: User should rearrange 2x - y = 10 to make y the subject. Subtract 2x from both sides: -y = 10 - 2x. Multiply by -1: y = 2x - 10. Gradient m = 2.For equation (a) 4x + y = 28: User should subtract 4x from both sides. y = 28 - 4x or y = -4x + 28. Gradient = -4, y-intercept = 28. For equation (b) 6x + 3y = 36: User should subtract 6x: 3y = 36 - 6x. User should divide by 3: y = 12 - 2x or y = -2x + 12. Gradient = -2, y-intercept = 12. For equation (c) 5x + 2y = 20: User should subtract 5x: 2y = 20 - 5x. User should divide by 2: y = 10 - 2.5x or y = -2.5x + 10. Gradient = -2.5 (or -5/2), y-intercept = 10.For part a: User should calculate the rate by dividing volume by time. Rate = 2700 ÷ 45 = 60 litres per minute. For part b: User should recognize that initially W = 0 when t = 0. The equation is W = rate × time. Equation: W = 60t. For part c: User should convert 2 hours to minutes: 2 hours = 120 minutes. W = 60 × 120 = 7200 litres. For part d: User should substitute W = 7200 into the equation: 7200 = 60t. Solve for t: t = 7200 ÷ 60 = 120 minutes or 2 hours.For part a: User should find the difference in costs: £19.50 - £12 = £7.50. User should find the difference in distances: 10km - 5km = 5km. Cost per kilometre = £7.50 ÷ 5 = £1.50 per km. For part b: User should calculate the cost for the distance at £1.50/km. For 5 km: 5 × £1.50 = £7.50. Total cost is £12, so booking fee = £12 - £7.50 = £4.50. For part c: User should write C = booking fee + (cost per km × distance). Equation: C = 4.50 + 1.50d. For part d: User should substitute d = 15 into the equation. C = 4.50 + 1.50(15) = 4.50 + 22.50 = £27. For part e: User should solve 31 = 4.50 + 1.50d. 1.50d = 31 - 4.50 = 26.50. d = 26.50 ÷ 1.50 ≈ 17.67 km (or 17.7 km rounded).For part a: User should calculate temperature drop: 22 - 18.5 = 3.5°C. Altitude increase = 500 metres. Rate = 3.5 ÷ 500 = 0.007°C per metre (or 7/1000). For part b: User should start with initial temperature of 22°C at ground level. T = 22 - (rate × altitude). Equation: T = 22 - 0.007h. For part c: User should substitute h = 1200. T = 22 - 0.007(1200) = 22 - 8.4 = 13.6°C. For part d: User should solve 10 = 22 - 0.007h. 0.007h = 22 - 10 = 12. h = 12 ÷ 0.007 ≈ 1714.29 metres (or 1714 m rounded). Final answer: (a) 0.007°C/m, (b) T = 22 - 0.007h, (c) 13.6°C, (d) 1714 metres.For part a: User should identify Zara starts with £80 and adds £15 per week. Equation: A = 80 + 15w. For part b: User should substitute w = 12 into Zara's equation. A = 80 + 15(12) = 80 + 180 = £260. For part c: User should solve 500 = 80 + 15w. 15w = 500 - 80 = 420. w = 420 ÷ 15 = 28 weeks. For part d: User should write Marcus's equation: A = 50 + 20w. User should set the two equations equal: 80 + 15w = 50 + 20w. User should solve: 80 - 50 = 20w - 15w. 30 = 5w. w = 30 ÷ 5 = 6 weeks. User should calculate the amount at 6 weeks using either equation. For Zara: A = 80 + 15(6) = 80 + 90 = £170. not compulsary but User can verify with Marcus: A = 50 + 20(6) = 50 + 120 = £170 For part (a): User should have drawn axes with data usage (GB) on horizontal axis and cost ($) on vertical axis.Have to mark points in such a way that the points to be marked must fit. Graph should show a straight line passing through origin (0, 0). Line should pass through point (8, 24). Line should be straight showing constant rate/proportional relationship. Overall figure check: Straight line from origin through (8, 24), properly labeled axes. For part (b): User should read from graph or calculate that 5 GB costs $15. Calculation method: (24 ÷ 8) × 5 = 3 × 5 = 15. Answer for part (b): 5 GB costs $15. For part (c): User should calculate gradient using change in y over change in x. Gradient = (24 - 0) ÷ (8 - 0) = 24 ÷ 8 = 3. Answer for part (c): Gradient is 3. For part (d): User should recognize gradient equals cost per GB. Cost per gigabyte = 24 ÷ 8 = 3. Answer for part (d): Cost per gigabyte is $3. For part (e): User should write formula y = 3x. Or equivalently: y = 3 × x or Cost = 3 × Data. For part (f): User should substitute x = 15 into formula. Calculation: y = 3 × 15 = 45. Answer for part (f): 15 GB costs $45. For part (g): User should solve 60 = 3x or 60 ÷ 3 = x. Calculation: x = 60 ÷ 3 = 20. Answer for part (g): You can buy 20 GB of data for $60.For part (a): User should draw axes with time (hours) on horizontal axis and distance (km) on vertical axis. Graph should show a straight line starting at point (0, 50) on the vertical axis. Line should pass through point (2, 230). Line should be straight showing constant speed. Overall figure check: Straight line from (0, 50) through (2, 230), properly labeled axes. For part (b): User should read from graph or calculate distance at 1.5 hours. First find speed: (230 - 50) ÷ 2 = 180 ÷ 2 = 90 km/h. Distance at 1.5 hours: 50 + (90 × 1.5) = 50 + 135 = 185. Answer for part (b): After 1.5 hours, total distance is 185 km. For part (c): User should calculate rate using change in distance over change in time. Rate = (230 - 50) ÷ (2 - 0) = 180 ÷ 2 = 90. Answer for part (c): The car is traveling at 90 km/hour. For part (d): User should recognize the equation form d = initial distance + rate × time. Equation: d = 50 + 90t. Or equivalently: d = 90t + 50. For part (e): User should substitute t = 4 into the equation. Calculation: d = 50 + 90(4) = 50 + 360 = 410. Answer for part (e): After 4 hours, total distance is 410 km. For part (f): User should solve 500 = 50 + 90t. Rearrange: 90t = 500 - 50 = 450. Solve: t = 450 ÷ 90 = 5. Answer for part (f): It will take 5 hours to travel 500 km in total. For part (g): This is a collaborative instruction, no specific calculation required.
