Answers for Chapter 1 Masters Scaffolding Answers Scaffolding for Getting Started Activity (Master) p. 65 C. 1 1 15 1 18 4 4 4 6 6 6 1 1 1 5 1 1 15 Yes No Yes No No Yes Yes No 1 18 4 No Yes Yes Yes Yes Yes No No No No Yes No No Yes No No 4 4 No No No Yes No Yes No No 6 6 6 No No Yes No Yes Yes No No 1 1 1 No No No No No Yes No No 5 No No No No No Yes Yes No No No No No No Yes No Yes 1 1 1,, 6 6 6 D. For example, if each dimension of a smaller box is a factor of at least one dimension of the larger box, it can be used to fill the larger box. Scaffolding for Do You Remember? (Master) pp. 66 67 1. a) Factors of 1: 1,,, 4, 6, 1 Factors of 18: 1,,, 6, 9, 18 Common factors of 1 and 18: 1,,, and 6 b) Multiples of 1: 1, 4, 6, 48, 60, 7, 84, 96,108, 10, Multiples of 18: 18, 6, 54, 7, 90, 108, Three common multiples of 1 and 18: 6, 7, 108. a) composite; factors: 1,,, 4, 6, 8, 1, 16, 4, 48 b) prime c) composite; factors: 1,,, 4, 6, 9, 1, 18, 6 d) composite; factors: 1,,, 4, 5, 6, 8, 10, 1, 15, 0, 4, 0, 40, 60, 10 7. 64 = = 6 64 = 4 4 4 4 = 4 11. a) multiply 9; 1 + 9 = 1 + 7 = 9 So, 1 + 9 15 b) multiply ; add 1 + 9 inside both parentheses; = (1 + ) + 8 = (1 + 9) + 8 = (10) + 8 = 100 + 8 = 108 So, (1 + ) + 8 = 108 Scaffolding for Lesson 1.1 (Master) p. 68 7. You don t need to check 0, 04, 06, and 08 because they have as a factor. You don t need to check 05 because it has 5 as a factor. 01 has as a factor, so it is not prime. 0 has 7 as a factor, so it is not prime. 07 has 9 as a factor, so it is not prime. 09 has 11 as a factor, so it is not prime. Copyright 006 by Thomson Nelson Chapter 1 Answers 77
11. even; odd An even number cannot be prime because it s always divisible by. Any consecutive numbers will always have an odd and an even number after each other. Even numbers cannot be prime. 14. The number of students in a class is probably between 0 and 40. Numbers divisible by :, 4, 6, 8, 0,, 4, 6, 8 Numbers divisible by : 1, 4, 7, 0,, 6, 9 Numbers divisible by 5: 5, 0, 5 The only number between 0 and 40 divisible by,, and 5 is 0. Scaffolding for Lesson 1. (Master) p. 69 7. a) 100 d) 75 g) 51 5 4 15 5 4 18 5 5 100 5 b) 10 e) 4 17 10 17 5 5 5 75 5 41 06 10 41 10 4 8 4 4 51 9 c) 0 f) 055 h) 675 0 16 5 411 5 147 4 5 4 4 17 5 5 49 0 6 5 055 5 17 7 7 675 5 7 Scaffolding for Lesson 1. (Master) p. 70 6. a) 48 = ; 60 = 5 c) 4 = ; = Three common factors are,, 4. Three common factors are, 4, 8. Three common multiples are 40, 480, 70. Three common multiples are 96, 19, 88. 5 b) = ; 64 = d) 51 = ; Three common factors are, 4, 8. 648 = Three common multiples are 64, 18, 19. Three common factors are, 4, 8. Three common multiples are 41 47, 8 944, and 14 416. 78 Chapter 1: Number Relationships Copyright 006 by Thomson Nelson
Scaffolding for Lesson 1.5 (Master) p. 71 17. a) For example, 80 60 4 65 1 = 546 64 000 heartbeats in 1 years 546 64 000 + (115 00 45) = 551 808 000 heartbeats in 1 years 45 days 5.5 10 8 b) For example, 5 60 4 65 1 = 4 164 000 blinks in 1 years 4 164 000 + (700 45) = 4 488 000 blinks in 1 years 45 days.45 10 7 c) 6 000 60 4 65 1 = 45 980 800 000 45 980 800 000 + (51 840 000 45) = 48 1 600 000.48 10 11 Chapter Test (Master) pp. 7 7 Answers 1. a) prime b) composite: 1,,, 4, 5, 6, 8, 10, 1, 15, 0, 4, 0, 40, 60, 10. a) 5 5 b) 5 7. a) 6 cm by 4 cm, cm by cm b) For example, cm by 6 cm, 7 cm by 7 cm c) Each side of a photograph has to be a factor of the side of the bulletin board. 4. a) 51. cm b) The height of each poster is 0.8 of the previous poster. So the height of the fourth poster is 1 m 0.8 0.8 0.8 or 1 0.8 or 0.8. c) The height of the seventh poster will be 0.97 cm, about 1 cm. 5. 96 000 1000 = 96 000 000; 9.6 10 7 6. a).5 10 5 b) 4. 10 6 c) 6.8 10 7 d) 9.91 10 8 7. For example, because 565 is between 4900 and 6400, 565 must be between 4900 and 6400, or between 70 and 80. Because 565 ends in a 5, the square root must end in 5. The answer is 75. 8. For example, if the square root of a whole number is about 9.7, the whole number must be about 9.7 which is 94.09. The whole number is 94. 9. a) 10.5 b) 49.6 c) 48. d) 9.04 Chapter 1 Task (Master) pp. 74 75 A. prime factorization: 10 5 9 expanded form: 6 10 9 scientific notation: 6.0 10 9 B. For example, because 6 billion = 6 000 000 000. If you regroup the digit 6 for 100 million, you get 60 100 millions; if you regroup 60 100 millions as 10 millions, you get 600 10 millions, and if you regroup 600 10 millions as 1 million, you get 6000 million. C. For example, if the square has an area of 6 billion square units, the side lengths must be the square root of 6 billion. (6 000 00) 0 77 459.667 or about 77 460 units. D. For example, if the cube has a volume of 6 billion cubic units, I need to find three numbers that are multiplied together to get 6 billion or = 6 billion. I know 1 billion = 1000 1000 1000, so the length of the cube is about 1000. I tried 1500 1500 1500 or 1500 and got 75 000 000; I tried 1600 1600 1600 and got 4 096 000 000, which is getting closer; I tried 1700 1700 1700 and got 4 91 000 000; I tried 1800 1800 1800 = 5 8 000 000, which is close to 6 billion. So the length of the cube will be about 1800 units. Copyright 006 by Thomson Nelson Chapter 1 Answers 79
E. For example, we measured the distance of the 4 students in our class when we held our arms outstretched. It was 5.56 m, which means the average length for each student from fingertip to fingertip is 5.56 4 = 1.55 m. We figured that distance would be about average because lots of people would have shorter arms and others would have longer arms. So we used 1.5 m as our estimate. So 1.5 m 6 000 000 000 = 9 000 000 000 m or 9 000 000 km. We looked up the distance around the equator and it is about 40 000 km. 90 000 000 40 000 = 50 times. So Jay s claim makes sense. F. For example, It can be written as 07 1 95 15 and neither number ends in zero. It can be written as 5 999 999 999 + 1. I think writing it as 07 1 95 15 is most interesting because most of the time when you multiply numbers that don t have a lot of zeros in them, you get a product that doesn t have many zeros. Answers to Reflecting, Lesson (continued from p. 0). a) For example, Sheree could have used the divisibility rule that any number with 0 as its last digit is divisible by 10: 1050 10 = 105. She could then have used the divisibility rule that any number with 5 as its last digit is divisible by 5: 105 5 = 1. Finally, she could have used the divisibility rule that any number with digits that add up to is divisible by : 1 = 7. b) For example, Teo could have used the divisibility rule that any number with digits that add up to is divisible by : 0 = 10. Next, he could have used the divisibility rule that any number with 5 or 0 as its last digit is divisible by 5: 10 5 =. 4. Yes. For example, each number has only one prime factorization. It doesn t matter which method you use to get it. 90 15 6 90 5 45 9 5 Answers to Key Assessment of Learning Question, Lesson (continued from p. 0) c) 0 e) 41 g) 51 0 16 06 4 18 4 5 4 4 10 4 41 10 8 4 0 6 5 4 51 9 d) 75 f) 055 h) 675 15 5 5 411 5 147 5 5 5 75 5 17 055 5 17 5 5 49 7 7 675 5 7 80 Chapter 1: Number Relationships Copyright 006 by Thomson Nelson
Answers to Reflecting, Lesson (continued from p. 4). For example, each number in the overlap area, as well as the product of any two or more numbers, is a common factor of 6 and 48. The GCF is the product of all the numbers in the overlap.. For example, when you multiply two numbers, the product is a common multiple of both numbers. However, if the two numbers have common factors, the product of the two numbers will not be the LCM. The product of the numbers in all three sections of the Venn diagram is the LCM. Answers to Key Assessment of Learning Question, Lesson 5 (continued from p. ) b) For example, I have lived about 4. 10 8 s. Assuming I blink when I am asleep and I blink about once every 5 seconds on average, I have blinked about 4. 10 8 5 = 84 000 000 times or about 8.4 10 7 times in my lifetime. c) For example, I have lived about 4. 10 8 s. I assume my eyes don t process information while I am asleep, and I have been awake for about of my life. of 4. 108 s is about 0.67 4. 10 8, which is about.8 10 8 s or.8 10 8 600 = 77 777.78 h. If my eyes process about 6 000 bits of information every hour, they have processed about 6 000 77 777.78 = 800 000 000 or about.8 billion or.8 10 9 bits of information in my lifetime. Answers to Learn about the Math, Lesson 6 (continued from p. 41) D. For example, I know the side length of the square with 11 square units is 11, and the side length of the square with 144 square units is 1. The side length of the square with 15 square units must be between 11 and 1. Calculating 15 is the same as determining the side length of a square with an area of 15 square units. Since 11 = 11 and 144 = 1, then 15 must be between 11 and 1. E. 15 = 11.180989 or about 11.18. So 15 11.18. F. 0.8 15 = 0.8 11.18 4.5 or about 4 cm. The ice is 0 cm thick, so it will support their total mass of 15 kg. Answers to Explore the Math, Lesson 7 (continued from p. 45) D. For example, the area of a slanted square is half the area of an upright square if the diagonal length of the slanted square equals the side length of the upright square. So, to calculate the area of a slanted square, you square the diagonal length of the slanted square (or square the side length of the upright square), which gives you the area of the upright square. Then you divide this area by ; I predict the area of a slanted square with a diagonal length of 10 cm will be one half the area of an upright square whose side length is 10.0 cm; area = 10 = 50.0 cm. E. For example, the side length squared equals the area, or A = l. So, to calculate the side length, you calculate the square root of the area. F. For example, I take the area of each slanted square in steps A to E, then find the square root of that number to find the side length of each slanted square. Diagonal length (cm) Area (cm ) Side length (cm).0.0 1.4 4.0 8.0.8 6.0 18.0 4. 8.0.0 5.7 10.0 50.0 7.1 G. For example, on centimetre dot paper, I can draw a slanted square with a diagonal length of 18 cm. The area will have one half of the area of an upright square with a side length of 18 cm. The upright square has an area of 4 cm. So the area of the slanted square is 16 cm. Copyright 006 by Thomson Nelson Chapter 1 Answers 81
Answers to Checking, Lesson 8 (continued from p. 49) 5. a) For example,.5 is less than 9 and.5 + 16 = 19.5, 19.5 9 is just over 10; so, an answer of 1.5 is reasonable. b) 4. + 5 1 9.8: for example, 16 = 4, 4 9.8 is about 40, 4. 40 is about 160, so an answer of 164.64 is reasonable. c) 6 9.5 10.6 + 1.5.6: for example, 6 9.5 is about 60, 10.6 is about 5, 1.5.6 is about 5, and 60 5 + 5 = 60, so an answer of 57.1 is reasonable. d) (7. + 5 7 ) : 7 = 49, 15 + 49 is over 60, 60 is about 8, 7. = 7 7 7 or 49 7 or about 50 7 = 50, 50 8 = 4; 00 = 90 000 and 400 = 160 000, 4 should be between 00 and 400 so an answer of 145 17.954 is reasonable. Answers to Communicate about the Math, Lesson 9 (continued from p. 5) C. For example: My Solution The mass of electronic waste each year is 140 000 t. Because each tonne equals 1000 kg, I can multiply 140 000 1000 to get 1 400 000 kg. I used mental math to multiply by 1000 by multiplying the place value of each digit by 1000. So the place value of the front digit 1 is 100 000 1000 = 100 million. Because 1 year has 1 months, I used my calculator to divide 140 000 000 by 1 to get a monthly mass of about 11 666 667 kg. The answer is reasonable because 140 million is close to 144 million and 144 million divided by 1 is 1 million. I know the mass is about 11.7 million kg because 11 666 667 kg can be written as 11.666 667 million or about 11.7 million. To express 11.7 million in scientific notation, I could write 11.7 million as 11.7 10 6. But the decimal factor must be less than 10 so I wrote 11.7 10 6 as 1.17 10 7. I know this answer is reasonable because 1.17 10 7 is the same as 1.17 10 000 000, which is just over 10 million and close to 11.7 million. Answers to Self-Test (continued from p. 55) 11. For example, I need to determine two equal numbers that have a product of 9 984 670. 9 984 670 = 159.858. The square would have side lengths of about 160 km. 1. a) 0 4.47. For example, the square root of 0 should be between 4 and 5, so 4.5 is reasonable. b) 45 18.574. For example, the square root of 45 should be just less than 0, so 18.6 is reasonable. c) 775 7.89. For example, the square root of 775 should be between 0 and 0 and closer to 0, so 8 is reasonable. d) 05 = 45. For example, the square root of 05 should be between 40 and 50, so 45 is reasonable. 1. a) 8.8. 4 + (5.4 4.) = 4 4 + 1. = 16 + 1.44 = 17.44 b) 4.8 1 5 + 4.8 5 = 11. 1 + 4 = 4. c) 1 0 + 1 +.5 = 1 + 5 + 7.5 = 6 + 7.5 = 6 + 7.5 = 1.5 d) [1.5 ( +.6) ] = (156.5 5.6) = (156.5 1.6) = 14.89 = 15 597.511 e) [(10 ) ] = [(10 6) ] = (4 ) = 16 = 4096 f) 5 (11 7) (1 + ) = 5 4 (5) = 5 16 (5) = 5. = 1.8 8 Chapter 1: Number Relationships Copyright 006 by Thomson Nelson