Answers for Chapter 4 Masters

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1 Answers for Chapter 4 Masters Scaffolding Answers Scaffolding for Getting Started Activity A. For example, The sum of the numbers in one diagonal is 22. The sum of the number in the other diagonal is 22. I notice that the sums of the diagonals are equal. B. For example, Box 2: The sum of the diagonals is 32. Box 3: The sum of the diagonals is 52. Box 4: The sum of the diagonals is 16. I notice that the sum of the diagonals for any 2-by-2 box are equal. C. a) Box 1: The sum of the numbers in the columns are 21 & 23. Box 2: The sum of the numbers in the columns are 31 & 33. Box 3: The sum of the numbers in the columns are 51 & 53. Box 4: The sum of the numbers in the columns are 15 & 17. b) The pattern I found when I add the number in the columns is the second column is 2 more than the first column. D. Box 1: The sum of the numbers in the rows are 15 & 29. Box 2: The sum of the numbers in the columns are 25 & 39. Box 3: The sum of the numbers in the columns are 45 & 59. Box 4: The sum of the numbers in the columns are 9 & 23. b) The pattern I found when I add the number in the rows is the second row is 14 more than the first column. E. The patterns in steps A, B, and C work because the numbers on the calendar form a pattern. Each row has 7 consecutive numbers and the numbers in each column is 7 more than the number in the row directly above. Scaffolding for Do You Remember? Question 6 6. a) Figure 1 Figure 2 b) Figure number Number of Cubes To get from the number of cubes in Figure 1 to the number of cubes in Figure 2, I add 4 cubes. b) I start with 5 cubes and continue to add 4 more cubes to the ends of the cross to get each new figure. c) For Figure 6, I will need 25 cubes. I know this because I add 4 more cubes to Figure 5. So, = 25. Copyright 26 by Thomson Nelson Chapter 4 Answers 65

2 Scaffolding for Lesson 4.2, Questions :blue; 3:green; 5:yellow; 7:orange; 9:red The sum of is 4. The sum of the first 2 odd number is shown by the 2-by-2 square made up of the blue square and the green squares. The sum of the first 2 odd number is 2 squared = 4. The sum of = 9. The sum of the first 3 odd numbers is shown by the 3-by-3 square made up of the blue square, the green squares, and the yellow squares. The sum of the first 3 odd number is 3 squared = 9. The sum of = 16. The sum of the first 5 odd number is 25. The sum of the first 5 odd numbers is shown by the 5-by-5 square made up of the blue square, the green squares, the yellow squares, the orange squares, and the red squares. The sum of the first 5 odd numbers is 5 squared = 25. The sum of the first x odd numbers is x squared. Scaffolding for Lesson 4.4, Question 1 1. Day Number Cranes made on this day Total number of cranes made Pattern rule: begin at 12 and add 3 each day I added the number of crane s Heather made on that day and then added the total number of cranes she made from the previous day. Yes, she actually made 5 more than her goal. Scaffolding for Lesson 4.5, Question 8 Sections Number of posts Number of rails a) For a fence that is 8 sections long, Mohammed will need 41 posts and rails. b) For a fence that is 14 sections long, Mohammed will need 71 posts and rails. There is one post at the start and 1 post added for each section. So, the number of posts is 1 more than the number of sections. There are 4 rails for each section. So, the number of rails is 4 times the number of sections. From the table of values, the coordinates are (1, 4) (2, 11), (3, 16), (4, 21), (5, 26), (6, 31), (7, 36), (8, 41), (9, 46), (1, 51), (11, 56), (12, 61), (13, 66), and (14, 71). Number of rails and posts Number of Rails and Posts vs. Number of Sections Number of sections 66 Chapter 4: Patterns and Relationships Copyright 26 by Thomson Nelson

3 Chapter Test (Master) p a) 21, 25, 29; The pattern is start at 1 and add 4. b) 36, 216, 15 12; The pattern is start at 3 and multiply by consecutive numbers, i.e., 18 4 = 72; 72 5 = 36; 36 6 = 216; = c) 364, 193, 328; The pattern is start at 1, multiply by 3 and add a) Use the diagram to complete the table of values for the first six terms of the pattern. Term number Number of squares b) The pattern rule is add 1 to the term number and multiply by a) Term number Term value b) Number Sequence Term number 1 67 Term value c) I could join the points with a straight line and continue the line until it cross the line for the 16th term number. I then move horizontally from that point to the line for the term value and read the number there. The value for the 16th term is a) 168 tickets b) 12th day 5. a) Maggie s sister is right. On the 21st day, Maggie runs (21 1) or 55 m, which is 5.5 km. On the 19th day, Maggie runs (19 1) or 5 m, which is 5 km. b) 9th day c) 65 m or 6.5 km games; i.e., = 66, 66 3 = 198. Chapter 4 Task (Master) p. 59 MAKE A PLAN 1 Understand the Problem I know that Rana designed her necklace using the following: top row = x(green bead), second row = y(purple bead), and third row = x + 2(green beads) until the top row reaches 1. Then, the number of beads (x, y) will be decreased by 1. There are 7 white beads used for each section and is between each design made out of green and purple beads. I need to find out how many green and purple beads are used if Rana has 2 white beads. Once I determine the number of beads needed, I will need to add an extra white section at either one or both ends of the necklace. I will now design my own necklace using three different colours. I will follow the same rules that Rana used but I will change the white pattern so it increases as well as the pattern made by the other beads. Copyright 26 by Thomson Nelson Chapter 4 Answers 67

4 2 Make a Plan I will sketch out what the necklace will look like using the number of beads for each row. I will work it out until the top row is at 1 beads and then go back down until the top row has 1 bead and the 3rd row has 3 beads. This will help me see the pattern. Then, I will use the pattern to determine my table of values. 3 Carry out the Plan Using the Table of values, I can use a scatter plot to confirm the pattern First Row (x) Second Row (y) Third Row (x + 2) Total: 1 Total: 1 Total: 138 A. The total number of green beads is = 238. The total number of purple beads is 1. B. The number of white sections is as follows: White / 1 section / W / 2 section / W / 3 section / W / 4 section / W / 5 section / W / 6 section / W / 7 section / W / 8 section / W / 9 section / W / section 1 / W / 9 section / W / 8 section / W /7 section / W / 6 section / W / 5 section / W / 4 section / W / 3 section / W / 2 section / W / 1 section / Total number of white beads = 19 7 = 133. Since the necklace began with a white section and ended with a green and purple section, I will need to add one more white section to have a white section at each end of the necklace. Thus, the total number of white beads is = Look Back I can use a scatter plot graph to make sure that I have used the right number of beads. Row One (x) Row Three (x + 2) The numbers for the colours follow this pattern, so I think my solution is correct. Perimeter of beads Beaded Necklace Row 1 Row Section number 68 Chapter 4: Patterns and Relationships Copyright 26 by Thomson Nelson

5 Answers to Getting Started Activity (continued from p. 11) F. For example, When you add the numbers in the columns, the sum of each column is 3 more than the column to its left. When you add the numbers in the rows, the sum of each row is 21 more than the row above. When you add the numbers in the diagonals, the sums are equal. This works because the numbers in any 3-by-3 square could be: If you add the columns, the sums would be n + (n + 7) + (n + 14) = 3n + 21, (n + 1) + (n + 8) + (n + 15) = 3n + 24, and (n + 2) + (n + 9) + (n + 16) = 3n The difference of the sums would be (3n + 24) (3n + 21) = 3, (3n + 27) (3n + 24) + 3. The differences are equal. If you add the rows, the sums would be n + (n + 1) + (n + 2) = 3n + 3, (n + 7) + (n + 8) + (n + 9) = 3n + 24, and (n + 14) + (n + 15) + (n + 16) = 3n The difference of the sums would be (3n + 24) (3n + 3) = 21, (3n + 45) (3n + 24) = 21. The differences are equal. The sums of the diagonals would be n + (n + 8) + (n + 16) = 3n + 24 and (n + 2) + (n + 8) + (n + 14) = 3n The sums are equal. Answers to Lesson 4.2, Learn About Math (continued from p. 19) B. C. D. Any new term in the sequence is the sum of the last two terms. E. 144 pairs n n + 1 n + 2 n + 7 n + 8 n + 9 n + 14 n + 15 n + 16 Number of pairs of Month guinea pigs January 1 February 1 March 2 April 3 May 5 June 8 July 13 August 21 September 34 October 55 November 89 December 144 Answers to Lesson 4.3, Reflecting (continued from p. 25) 1. Each of Tynessa s tables has three columns. The first column in each table has the term number and the last column in each table has the term value. The contents of the middle column of each table are different. The middle column of the first table has pictures of the pattern, while the middle column of the second table has the rule for the pattern. 2. The strategy for the second table is look for a relationship between the term number in the first column and the corresponding term value in the last column. Copyright 26 by Thomson Nelson Chapter 4 Answers 69

6 3. To find the 2th term in the sequence it is better to use the rule that uses the position of a term in the sequence. This method uses the relationship between the term number and the term value. The other method uses the relationship between consecutive term values. Unless the value of the 19th term is given, the 2th term could not be calculated. Answers to Lesson 4.4, Reflecting (continued from p. 31) 4. a) She could make a model, use guess and test, or make an organized list. b) The numbers would be the same using each strategy and you would have to add the same numbers to each row of cases: guess and test might not be very accurate to start, making a model would involve manipulatives or drawings, and organized lists might omit some columns from the table of values. Answers to Lesson 4.5, Learn about the Math (continued from p. 36) D. Number of Perimeter of Number of Triangles path (units) border pieces E. 25 Omar s Path Perimeter of path (units) Number of triangles F Chapter 4: Patterns and Relationships Copyright 26 by Thomson Nelson