User should identify the fixed cost as $15. User should identify the variable cost as $0.50 per text. For part (a)- User should calculate cost for 80 texts. Cost = 15 + (0.50 × 80). Cost = 15 + 40 = 55. Answer for part (a)- $55. For part (b)- User should write the function using C for total cost and t for number of texts. Function- C = 15 + 0.50t or C = 0.50t + 15.For part (a)User should identify brother\'s weight as 32 kg. Emma\'s weight = (2 × brother\'s weight) - 9. Emma\'s weight = (2 × 32) - 9. Emma\'s weight = 64 - 9 = 55. Answer for part (a) 55 kg. For part (b)User should express the relationship as a function. E = 2b - 9.For part (a)User should verify the given values. Seats from 11 round tables=11 × 8 = 88. Seats from 7 rectangular tables=7 × 6 = 42. Total seats- 88 + 42 = 130. This equals 130, so it is correct. For part (b)Total tables = 20. Function- r + t = 20. For part (c)- If r = 15, then t = 20 - 15 = 5. Seats would be 15(8) + 5(6) = 120 + 30 = 150. This is more than 130 seats, so r cannot be 15.For part (a)- User should provide two different real-world interpretations. Example 1- p could be total cost in dollars, q could be number of items bought. The 25 represents a fixed fee, and 4 represents cost per item. The gradient (4) shows the rate of $4 per item. Example 2- p could be total pages read, q could be number of days. The 25 represents pages already read, and 4 represents pages read per day. The gradient (4) shows the rate of 4 pages per day. For part (b)- User should substitute q = 12. p = 4(12) + 25 = 48 + 25 = 73. For part (c)- User should substitute p = 89 and solve for q. 89 = 4q + 25. 4q = 89 - 25 = 64. q = 64 ÷ 4 = 16.For part (a)- Let x = number of 2-seater sofas, y = number of 3-seater sofas. Each 2-seater provides 2 seats, each 3-seater provides 3 seats. Function- 2x + 3y = 45. For part (b)- User should substitute x = 12 and y = 7. 2(12) + 3(7) = 24 + 21 = 45. This equals 45, so the values are correct. For part (c)- If y = 10 (3-seater sofas). 2x + 3(10) = 45. 2x + 30 = 45. 2x = 15. x = 7.5. You cannot have 7.5 sofas (must be a whole number). Therefore, the number of 3-seater sofas cannot be 10.User should substitute each x value into y = 5x - 7. For x = -1- y = 5(-1) - 7 = -5 - 7 = -12. For x = 0- y = 5(0) - 7 = 0 - 7 = -7. For x = 1- y = 5(1) - 7 = 5 - 7 = -2. For x = 2- y = 5(2) - 7 = 10 - 7 = 3. For x = 3- y = 5(3) - 7 = 15 - 7 = 8. For x = 4- y = 5(4) - 7 = 20 - 7 = 13. Complete table with x and y values where x values are -1, 0, 1, 2, 3, 4 and y values are -12, -7, -2, 3, 8, 13 respectively.User should substitute each x value into y = x² + 4. For x = -3, y = (-3)² + 4 = 9 + 4 = 13. For x = -2, y = (-2)² + 4 = 4 + 4 = 8. For x = -1, y = (-1)² + 4 = 1 + 4 = 5. For x = 0, y = (0)² + 4 = 0 + 4 = 4. For x = 1, y = (1)² + 4 = 1 + 4 = 5. For x = 2, y = (2)² + 4 = 4 + 4 = 8. For x = 3, y = (3)² + 4 = 9 + 4 = 13. Complete table with x and y values where x values are -3, -2, -1, 0, 1, 2, 3 and y values are 13, 8, 5, 4, 5, 8, 13 respectively.For part (a)- User should complete the table. For x = -1, y = 4(-1) - 1 = -4 - 1 = -5. For x = 0, y = 4(0) - 1 = 0 - 1 = -1. For x = 1, y = 4(1) - 1 = 4 - 1 = 3. For x = 2, y = 4(2) - 1 = 8 - 1 = 7. For x = 3, y = 4(3) - 1 = 12 - 1 = 11. For part (b)User should draw a horozonyal and vertical linr passing through horizontal line. This point is (0,0). user should mark points on the graph x-axis from -2 to 4 and y-axis from -6 to 12.Based on this mark points on the graph.The points are (-2,-9),(-1,-5),(0,-1),(1,3),(2,7),(3,11). Join these points to form a line. For part (c)The gradient is the coefficient of x. Gradient = 4. For part (d)The y-intercept is the constant term. Y-intercept = -1 (or point (0, -1)). For part (e)Graph crosses x-axis when y = 0. 0 = 4x - 1, so 4x = 1, x = 0.25 or x = 1/4.For part (a)- User should rearrange to y = (30 - 5x)/2. For x = 0- 2y = 30 - 0 = 30, so y = 15. For x = 2- 2y = 30 - 10 = 20, so y = 10. For x = 4- 2y = 30 - 20 = 10, so y = 5. For x = 6- 2y = 30 - 30 = 0, so y = 0. User must draw the table with x values are 0, 2, 4, 6 and y values are 15, 10, 5, 0 respectively. (b) User must draw the graph with x-axis as horizontal line and y-axis as vertical line which meet at (0,0). User must mark points on the graph x-axis in such way that x has 0-6.User can draw more but must include these points.Based on this mark points on the graph.The points are (0,15),(2,10),(4,5),(6,0). Join these points to form a line. For part (c)(i) Graph crosses x-axis at 6 when y = 0. X-intercept- (6, 0). For part (c)(ii) Graph crosses y-axis at 15 when x = 0. Y-intercept- (0, 15). For part (d) substitute x = 4 and y = 5 in the equation 5x + 2y = 30. LHS-5(4) + 2(5) = 20 + 10 = 30. RHS = 30. LHS = RHS.For part (a)User should observe that all four lines are parallel. For part (b)All lines have the same gradient (slope). The gradient is 3 for all four equations. Lines with the same gradient are parallel. For part (c)Y-intercepts are the constant terms. y = 3x + 8 crosses at (0, 8). y = 3x + 2 crosses at (0, 2). y = 3x - 4 crosses at (0, -4). y = 3x - 10 crosses at (0, -10). For part (d)y = 3x + 50 has gradient 3. Since it has the same gradient as the other lines, it is parallel.User should create tables for both equations and plot them. For y = 2x + 5- When x = 0, y = 5; when x = 1, y = 7; when x = 2, y = 9. For 3x + y = 10- Rearrange to y = 10 - 3x. When x = 0, y = 10;when x=1 then y=10-3=7; when x = 2, y = 4; when x = 3, y = 1. User must draw the graph with x-axis as horizontal line and y-axis as vertical line which meet at (0,0). User must mark points on the graph x-axis in such way that x has 0-3.User can draw more but must include these points.Based on this mark points on the graph.The points are (0,5),(1,7), (2,9). Join these points to form a line. User must mark points on the graph x-axis in such way that x has 0-3.User can draw more but must include these points.Based on this mark points on the graph.The points are (0,10),(2,4), (3,1). Join these points to form a line. For part (a)User should find intersection from graph. Intersection point (1, 7). For part (b)If x = 5 for y = x + 7, then x=0 then y=7,x=1 then y=8,x=2 then y=9,x=3 then y=10,x=4 then y=11,x=5 then y=12. user must plot these points (0,7),(1,8),(2,9),(3,10),(4,11),(5,12) on graph and join these points to form a line. part (c) From the graph, the lines y=2x+5 and 3x+y=10 intersect at one point. The line y=x+7 is drawn on the same axes but does not pass through that intersection point.The three lines do not intersect at one single point. For C bit user can write anthing but has to conclude that the three lines do not intersect at one single point.xxFor part (a) Initial volume is 200 liters. Water drains at 12 liters per minute (volume decreases). Function- V = 200 - 12t. For part (b) User should calculate V for each t value. When t = 0, V = 200 - 12(0) = 200. When t = 5, V = 200 - 12(5) = 200 - 60 = 140. When t = 10, V = 200 - 12(10) = 200 - 120 = 80. When t = 15, V = 200 - 12(15) = 200 - 180 = 20. When t = 20, V = 200 - 12(20) = 200 - 240 = -40 (not physically possible). (c)User must draw the graph with x-axis as horizontal line and y-axis as vertical line which meet at (0,0). User must mark points on the graph x-axis in such way that x has 0-20. User can draw more but must include these points. Based on this mark points on the graph. The points are (0,200), (5,140), (10,80), (15,20). Join the points to form a straight line. For part (d)(i) When t = 8, V = 200 - 12(8) = 200 - 96 = 104 liters. For part (d)(ii) When V = 80, 80 = 200 - 12t, so 12t = 120, t = 10 minutes. For part (e) When t = 25, V = 200 - 12(25) = 200 - 300 = -100. Negative volume is impossible; the tank would be empty before 25 minutes. Tank empties when V = 0, 0 = 200 - 12t, so t = 200/12 ≈ 16.67 minutes.
