1. Slide 1- 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Whole Numbers 1.1Standard Notation 1.2Addition and Subtraction 1.3Multiplication and Division; Rounding and Estimating 1.4Solving Equations 1.5Applications and Problem Solving 1.6Exponential Notation and Order of Operations 1.7Factorizations 1.8Divisibility 1.9Least Common Multiples 1

3. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley STANDARD NOTATION Give the meaning of digits in standard notation. Convert from standard notation to expanded notation. Convert between standard notation and word names. Use for to write a true sentence in a situation like 6 10. 1.1a b c d

4. Slide 1- 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Give the meaning of digits in standard notation.a

5. Slide 1- 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Place Value A digit is a number 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 that names a place-value location. For large numbers, digits are separated by commas into groups of three, called periods. Each period has a name: ones, thousands, millions, billions, trillions, and so on. OnesThousandsMillionsBillionsTrillions PLACE-VALUE CHART Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones

6. Slide 1- 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 845182048542 OnesThousandsMillionsBillionsTrillions PLACE-VALUE CHART Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones 245 billions, 840 millions, 281 thousands, 548 ones

7. Slide 1- 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A What does the digit 4 mean in each number? 1. 234,598 2. 456,901 3. 24,355,567,222 4 thousands 4 hundred thousands 4 billions Solution

8. Slide 1- 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Convert from standard notation to expanded notation.b

9. Slide 1- 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Whole Numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, … Natural Numbers: 1, 2, 3, 4, 5,… Standard Notation: 34,123 Expanded Notation: 34,123 = 3 ten thousands + 4 thousands + 1 hundred + 2 tens + 3 ones

10. Slide 1- 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Write expanded notation for 5280 feet, the number of feet in a mile. Solution 5280 = 5 thousands + 2 hundreds + 8 tens + 0 ones

11. Slide 1- 11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Write standard notation for 8 ten thousands + 4 thousands + 5 hundreds + 2 tens + 9 ones. Solution Standard notation is 84,529.

12. Slide 1- 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Convert between standard notation and word names.c

13. Slide 1- 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Write a word name for 123. Solution 123 = one hundred twenty-three

14. Slide 1- 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Write a word name for 123,456,789. One hundred twenty-three million, four hundred fifty-six thousand, seven hundred eighty-nine Solution 123, 456, 789

15. Slide 1- 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Write standard notation. Three hundred four million, two hundred thirty-five thousand, eight hundred eleven Standard notation is 304,235,811 Solution

16. Slide 1- 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use for to write a true sentence in a situation like 6 10.d

17. Slide 1- 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Order of Whole Numbers For any whole numbers a and b: 1. a < b (read “a is less than b”) is true when a is to the left of b on a number line. 2. a > b (“read a is greater than b”) is true when a is to the right of b on a number line. We call inequality symbols.

18. Slide 1- 18 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Use for to write a true sentence: 84 94. 84949888 Since 84 is to the left of 94 on a number line, 84 < 94. Solution

19. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley ADDITION and SUBTRACTION Add whole numbers. Use addition in finding perimeter. Convert between addition sentences and subtraction sentences. Subtract whole numbers. 1.2a b c d

20. Slide 1- 20 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Add whole numbers.a

21. Slide 1- 21 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Addition of whole numbers corresponds to combining or putting things together. The addition that corresponds to the figure above is 3 + 5 = 8 A set of 3 CD players A set of 5 CD players A set of 8 CD players We combine two sets. This is the resulting set. Addend Sum

22. Slide 1- 22 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Add: 8456 + 2484. 1 1 Add ones. We get 10 ones. Write the 0 in the ones column and 1 above the tens. This is called carrying, or regrouping. Add tens. We get 14 tens. Write 4 in the tens column and 1 above the hundreds. Add hundreds. We get 9 hundreds. Write 9 in the hundreds column. Add thousands. We get 10 thousands 8 4 5 6 + 2 4 8 4 Solution 04 9 1 0

23. Slide 1- 23 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Add: 3401 + 4387 + 9765 + 1356 Add ones. We get 19. Write 9 in the ones column and 1 above the tens. Add tens. We get 20. Write 0 in the tens column and 2 above the hundreds. Add hundreds. We get 19. Write 9 in the hundreds column and 1 above the thousands. Add thousands. We get 18. 1 21 3 4 0 1 4 3 8 7 9 7 6 5 + 1 3 5 6 Solution 90 9 1 8

24. Slide 1- 24 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use addition in finding perimeter.b

25. Slide 1- 25 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Perimeter The distance around an object is its perimeter.

26. Slide 1- 26 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Find the perimeter of the object shown. 5 m 8 m 9 m 8 m 5 m The letter m denotes meters (a meter is slightly more than 3 ft). Perimeter = 5 m + 8 m + 9 m + 8 m + 5 m = 35 m Solution

27. Slide 1- 27 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Find the perimeter of the figure shown. Solution Perimeter = 18 yd + 24 yd + 8 yd + 10 yd + 10 yd + 14 yd = 84 yd 18 yd 24 yd 10 yd 8 yd 10 yd 14 yd

28. Slide 1- 28 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Convert between addition sentences and subtraction sentences.c

29. Slide 1- 29 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtraction of whole numbers applies to two kinds of situations. The first is called “take away.” A restaurant starts with 8 pies and sells 5 of them. 88 – 5 = 3

30. Slide 1- 30 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The minuend is the number from which another number is being subtracted. The subtrahend is the number being subtracted. The difference is the result of subtracting the subtrahend from the minuend. 8  5 = 3 Minuend Subtrahend Difference

31. Slide 1- 31 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtraction The difference a  b is that unique number c for which a = c + b.

32. Slide 1- 32 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Write a related addition sentence: 9 – 4 = 5. Solution 9 – 5 = 4 This number gets added 9 = 4 + 5

33. Slide 1- 33 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Write two related subtraction sentences: 6 + 8 = 14. Solution 6 + 8 = 14. This addend gets subtracted from the sum. 6 = 14  8 8 = 14  6 6 + 8 = 14. This addend gets subtracted from the sum. The related subtraction sentences are 6 = 14 – 8 and 8 = 14 – 6.

34. Slide 1- 34 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The second kind of situation to which subtraction can apply is called “how many more.”. You have 2 notebooks, but you need 7. You can think of this as “how many do I need to add to 2 to get 7?” What must be added to 2 to get 7? The answer is 5. Have 2 notebooks 5 notebooks Need 7 notebooks

35. Slide 1- 35 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objectived Subtract whole numbers.

36. Slide 1- 36 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Subtract: 8453  2311. 241 6 Subtract ones. Solution 8 4 5 3  2 3 1 1 Subtract tens. Subtract hundreds. Subtract thousands.

37. Slide 1- 37 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Sometimes we need to borrow. Example H Solution 5 0 2 4  1 9 5 6 8603 1 14 9 4 11 Subtract: 5024  1956

38. Slide 1- 38 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I Subtract: 7000 – 2754. Solution 7 0 0 0 – 2 7 5 4 6424 699 10

39. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MULTIPLICATION AND DIVISION ; ROUNDING AND ESTIMATING Multiply whole numbers. Use multiplication in finding area. Convert between division sentences and multiplication sentences. Divide whole numbers. Round to the nearest ten, hundred, or thousand. Estimate sums, differences and products by rounding. 1.3a c b d e f

40. Slide 1- 40 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Multiply whole numbers.a

41. Slide 1- 41 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The multiplication 4  5 corresponds to this repeated addition: We combine 4 sets of 5 desks each. 5 desks + + + 4  5 desks = 20 desks Factor Product

42. Slide 1- 42 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Multiply: 5  652. Solution 6 5 2  5 1 0 Multiply the 2 ones by 5: 5  2 = 10 2 5 0 Multiply the 5 tens by 5: 5  50 = 250 3 0 0 0 Multiply the 6 hundreds by 5: 5  600 = 3000 3 2 6 0 Add.

43. Slide 1- 43 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Instead of writing each product on a separate line, we can use a shorter form. Solution 6 5 2  5 1 0632 2 Multiply the ones by 5: 5  (2 ones) = 10 ones = 1 ten + 0 ones. Write 0 in the ones column and 1 above the tens. Multiply the 5 tens by 5: 5  (5 tens) = 25 tens, 25 tens + 1 ten = 26 tens = 2 hundreds + 6 tens. Write 6 in the tens column and 2 above the hundreds. Multiply the 6 hundreds by 5 and add 2 hundred: 5  (6 hundreds) = 30 hundreds + 2 hundreds = 32 hundreds.

44. Slide 1- 44 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Multiply: 53  47. Solution 5 3  4 7 3 7 1 Multiplying by 7 2 1 2 0 Multiplying by 40. (We write a 0 and multiply 53 by 4) 2 4 9 1 Adding 2 1

45. Slide 1- 45 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Solution 3 2 5  6 7 4 1 3 0 0 Multiplying 325 by 4 2 2 7 5 0 Multiplying 325 by 70 1 9 5 0 0 0 Multiplying 325 by 600 2 1 9, 0 5 0 Adding Multiply: 325  674.

46. Slide 1- 46 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Multiply: 450  326. Solution 3 2 6  4 5 0 1 6 3 0 0 1 3 0 4 0 0 1 4 6,7 0 0 Multiplying by 5 tens. (We write 0 and then multiply 326 by 5. Multiplying by 4 hundreds. (We write 00 and then multiply 326 by 4. Adding

47. Slide 1- 47 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use multiplication in finding area.b

48. Slide 1- 48 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The area of a rectangular region can be considered to be the number of square units needed to fill it. Here is a rectangle 4 cm (centimeters) long and 3 cm wide. It takes 12 square centimeters (sq cm) to fill it. In this case, we have a rectangular array of 3 rows, each of which contains 4 squares. That is, A = l  w = 3 cm  4 cm = 12 sq cm. 1 cm 3 cm 4 cm

49. Slide 1- 49 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E An adult soccer goal is 24 feet wide by 8 feet high. Find the shooting area. Solution A = l  w = 24 ft  8 ft = 192 sq ft. 24 ft 8 ft

50. Slide 1- 50 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Convert between division sentences and multiplication sentences.c

51. Slide 1- 51 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Division of whole numbers applies to two kinds of situations. The first is repeated subtraction. Suppose we have 20 notebooks in a pile, and we want to find out how many sets of 5 there are. One way to do this is to repeatedly subtract sets of 5. How many sets of 5 notebooks each? 4 3 2 1 20  5 = 4 Dividend Divisor Quotient

52. Slide 1- 52 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We can also think of division in terms of rectangular arrays. Consider again the pile of 20 notebooks and division by 5. We can arrange the notebooks in a rectangular array with 5 rows and ask, “How many are in each row?” We can also consider a rectangular array with 5 notebooks in each column and ask, “How many columns are there?” The answer is still 4. In each case, we are asking, “What do we multiply 5 by in order to get 20?” 5  ?? = 20 20  5 = ??

53. Slide 1- 53 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Division The quotient a  b, where b  0, is that unique number c for which a = b  c.

54. Slide 1- 54 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Write two related division sentences: 9  8 = 72. This factor becomes a divisor. 9  8 = 72 9 = 72  8 This factor becomes a divisor. 8 = 72  9 9  8 = 72 Solution

55. Slide 1- 55 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Divide whole numbers.d

56. Slide 1- 56 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividing by 1 Any number divided by 1 is that same number: Dividing a Number by Itself Any nonzero number divided by itself is 1:

57. Slide 1- 57 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividends of 0 Zero divided by any nonzero number is 0: Excluding Division by 0 Division by 0 is not defined. (we agree not to divide by 0.)

58. Slide 1- 58 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Divide by repeated subtraction: 24  7. Solution 24  7 17  7 10  7 3 Subtracting 3 times Remainder

59. Slide 1- 59 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Divide 4369  6. Solution The answer is 728 R 1.

60. Slide 1- 60 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To do division of whole numbers: a) Estimate. b) Multiply. c) Subtract.

61. Slide 1- 61 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I Divide 9858  62. Solution

62. Slide 1- 62 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example J Divide 1223  4. Solution The answer is 305 R 3.

63. Slide 1- 63 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Round to the nearest ten, hundred or thousand. e

64. Slide 1- 64 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example K Round 57 to the nearest ten. Solution Since 57 is closer to 60, we round up to 60. 50605557

65. Slide 1- 65 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example L Round 52 to the nearest ten. Solution Since 52 is closer to 50, we round down to 50. 506055 52

66. Slide 1- 66 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example M Round 55 to the nearest ten. Solution We agree to round up to 60. 506055 When a number is halfway between rounding numbers, round up.

67. Slide 1- 67 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rounding Whole Numbers To round to a certain place: a) Locate the digit in that place. b) Consider the next digit to the right. c) If the digit to the right is 5 or higher, round up. If the digit to the right is 4 or lower, round down. d) Change all digits to the right of the rounding location to zeros.

68. Slide 1- 68 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example N Round 7564 to the nearest hundred. Solution a) Locate the digit in the hundreds place, 5. 7 5 6 4 b) Consider the next digit to the right, 6. 7 5 6 4 c) Since that digit is 5 or higher, round 5 hundreds up to 6 hundreds. d) Change all digits to the right of the hundreds digit to zeros. 7 6 0 0

69. Slide 1- 69 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example O Round 88,696 to the nearest ten. Solution a) Locate the digit in the tens place, 9. 8 8, 6 9 6 b) Consider the next digit to the right, 6. 8 8, 6 9 6 c) Since that digit is 5 or higher, round 9 tens to 10 tens and carry the 1 over to the hundreds. d) Change the digit to the right of the tens digit to zeros. 8 8, 7 0 0

70. Slide 1- 70 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Estimate sums and differences by rounding.f

71. Slide 1- 71 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example P Mario and Greta are considering buying a new computer. There are two models, and each has options beyond the basic price, as shown below. Mario and Greta have a budget of $1100. Make a quick estimate to determine if the XS with a monitor, memory upgrade to 80 gig and a printer is within their budget. XS ModelLT Model Basic price: $595Basic price: $825 Monitor: $220Monitor: $275 Memory upgrade: 40 gig: $75 80 gig: $90 Memory upgrade: 80 gig: $110 Printer: $120Printer: included

72. Slide 1- 72 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution First, we list the base price and then the cost of each option. We then round each number to the nearest hundred and add. XS $595 $600 Monitor $220 $200 Memory $90 $100 Printer+ $120 + $100 1000 The price of the computer is within their budget. XS Model Basic price: $595 Monitor: $220 Memory upgrade: 40 gig: $75 80 gig: $90 Printer: $120

73. Slide 1- 73 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Q Estimate the difference by first rounding to the nearest thousand: 8426  3840. 8 0 0 0  4 0 0 0 4 0 0 0 Solution 8 4 2 6  3 8 4 0

74. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING EQUATIONS Solve simple equations by trial. Solve equations like x + 28 = 54, 28  x = 168, and 98 ∙ 2 = y. 1.4a b

75. Slide 1- 75 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve simple equations by trial.a

76. Slide 1- 76 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solutions of an Equation A solution is a replacement for the variable that makes the equation true. When we find all the solutions, we say that we have solved the equation.

77. Slide 1- 77 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Solve x + 15 = 39 by trial. Solution We replace x with several numbers. If we replace x with 22, we get a false equation: 22 + 15 = 39. If we replace x with 23, we get a false equation: 23 + 15 = 39. If we replace x with 24, we get a true equation: 24 + 15 = 39. No other replacement makes the equation true, so the solution is 24.

78. Slide 1- 78 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Solve: 1. 8 + n = 35 2. 54  9 = y Solution 1. 8 + n = 35 (8 plus what number is 35?) The solution is 27. 2. 54  9 = y (54 times 9 is what?) The solution is 486

79. Slide 1- 79 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve equations like x + 28 = 54, 28  x = 168, and 98 ∙ 2 = y.b

80. Slide 1- 80 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving x + a = b To solve x + a = b, subtract a on both sides.

81. Slide 1- 81 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Solve: t + 37 = 83. Solution t + 37 = 83 t + 37  37 = 83 – 37 t + 0 = 46 t = 46. Subtracting 37 from both sides. Check: t + 37 = 83 46 + 37 83 83 true

82. Slide 1- 82 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Solve: 8365 + x = 9301. Solution 8365 + x = 9301 8365 + x  8365 = 9301 – 8365 x = 936 Check: 8365 + x = 9301 8365 + 936 = 9301 True The solution is 936. Subtracting 8365 from both sides.

83. Slide 1- 83 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving a  x = b To solve a  x = b, divide by a on both sides.

84. Slide 1- 84 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Solve: 18  y = 1476 Solution 18  y = 1476 18 18 y = 82 Check: 18  y = 1476 18  82 = 1476 1476 = 1476 True Dividing both sides by 18

85. Slide 1- 85 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Solve: 4032 = 56  y Solution 4032 = 56  y 56 56 y = 72 Dividing both sides by 56

86. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley APPLICATIONS AND PROBLEM SOLVING Solve applied problems involving addition, subtraction, multiplication, or division of whole numbers. 1.5a

87. Slide 1- 87 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems involving addition, subtraction, multiplication, or division of whole numbers.a

88. Slide 1- 88 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Five Steps for Problem Solving 1. Familiarize yourself with the situation. a) Carefully read and reread until you understand what you are being asked to find. b) Draw a diagram or see if there is a formula that applies to the situation. c) Assign a letter, or variable, to the unknown. 2. Translate the problem to an equation using the letter or variable. 3. Solve the equation. 4. Check the answer in the original wording of the problem. 5. State the answer to the problem clearly with appropriate units.

89. Slide 1- 89 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A The balance in Megan’s checking account is $712. She uses her debit card to buy the power saw shown in the ad. Find the new balance in her checking account. 1. Familarize. Visualize the situation. We let M = the new balance in her account. $712 New balance Now $112.00 Take away $112 Solution

90. Slide 1- 90 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued example A 2. Translate. 3. Solve. This sentence tells us what to do. Subtract. 712 – 112 = M 600 = M

91. Slide 1- 91 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued example A 4. Check. To check we can repeat our calculation. We can also estimate: 712 – 112  700 – 100 = 600. 5. State. Megan has a new balance of $600 in her checking account.

92. Slide 1- 92 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B The captain of a blimp is traveling between two cities that are 745 miles apart. The first day he traveled 375 miles. How much further does he need to travel? 1. Familarize. Visualize the situation. We let x = the remaining distance. 2. Translate. started distance traveled 375 distance to go x

93. Slide 1- 93 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley example B continued 3. Solve. To solve the equation we subtract 375 from both sides. 4. Check. Estimate: 750 – 380 = 370 5. State. The captain needs to travel 370 miles to the next city.

94. Slide 1- 94 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C What is the total cost of 6 calculators if each one costs $36? 1. Familiarize. Make a drawing or visualize the situation. Let T = total cost 2. Translate. $36

95. Slide 1- 95 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued example C 3. Solve. The sentence tells us what to do. Multiply. 4. Check. We can repeat the calculation or estimate. 40  6 = 240 5. State. The total cost of 6 calculators is $216.

96. Slide 1- 96 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Key Words, Phrases, and Concepts Addition (+)Subtraction (–) addsubtract added tosubtracted from sumdifference totalminus plusless than more thandecreased by increased bytake away how much more missing addend

97. Slide 1- 97 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Key Words, Phrases, and Concepts Multiplication (  )Division (  ) multiplydivide multiplied bydivided by productquotient timesrepeated subtraction ofmissing factor repeated additionfinding equal quantities rectangular arrays

98. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXPONENTIAL NOTATION AND ORDER OF OPERATIONS Write exponential notation for products such as 4  4  4. Evaluate exponential notation. Simplify expressions using the rules for order of operations. Remove parentheses within parentheses. 1.6a b c d

99. Slide 1- 99 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Write exponential notation for products such as 4  4  4.a

100. Slide 1- 100 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Notation The 5 is called an exponent. The 4 is the base.

101. Slide 1- 101 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Write exponential notation for 7  7  7  7  7  7. Solution Exponential notation is 7 6 7 is the base. 6 is the exponent.

102. Slide 1- 102 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Evaluate exponential notation.b

103. Slide 1- 103 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Evaluate: 8 4 and 10 4 Solution 8 4 = 8  8  8  8 = 4096 10 4 = 10  10  10  10 = 10,000

104. Slide 1- 104 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Simplify expressions using the rules for order of operations.c

105. Slide 1- 105 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rules for Order of Operations 1. Do all calculations within parentheses ( ), brackets [ ], or braces { } before operations outside. 2. Evaluate all exponential expressions. 3. Do all multiplications and divisions in order from left to right. 4. Do all additions and subtractions in order from left to right.

106. Slide 1- 106 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Simplify: 24  3  4 Solution There are no parentheses or exponents, so we start with the third step. 24  3  4 = 8  4 = 32 Doing all multiplications and divisions in order from left to right

107. Slide 1- 107 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Simplify: 20  4  3  (10 – 7) Solution 20  4  3  (10 – 7) = 20  4  3  3 = 5  3  3 = 15  3 = 5

108. Slide 1- 108 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Simplify: Solution

109. Slide 1- 109 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Average The average of a set of numbers is the sum of the numbers divided by the number of addends.

110. Slide 1- 110 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Mason has 4 math tests with scores of 76, 85, 92, and 91. Find the average of all four tests. Solution The average on Mason’s four tests is 86.

111. Slide 1- 111 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Remove parentheses within parentheses.d

112. Slide 1- 112 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When parentheses occur within parentheses, we can make them different shapes, such as [ ] (also called “brackets”) and { } (also called “braces”). All of these have the same meaning. When parentheses occur within parentheses, computations in the innermost ones are to be done first.

113. Slide 1- 113 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Simplify: Solution

114. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FACTORIZATIONS 1.7 Determine whether one number is a factor of another, and find the factors of a number. Find some multiples of a number, and determine whether a number is divisible by another. Given a number from 1 to 100, tell whether it is prime, composite, or neither. Find the prime factorization of a composite number.a b c d

115. Slide 1- 115 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether one number is a factor of another, and find the factors of a number.a

116. Slide 1- 116 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factor In the product a  b, a and b are factors. If we divide Q by d and get a remainder of 0, then the divisor d is a factor of the dividend Q.

117. Slide 1- 117 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Determine by long division whether 12 is a factor of 3915. Solution  Not 0 The remainder is not 0, so 12 is not a factor of 3915.

118. Slide 1- 118 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Find all the factors of 72. Solution Check sequentially the numbers 1, 2, 3, and so on, to see if we can form any factorizations. 1  72 2  36 3  24 4  18 6  12 8  9

119. Slide 1- 119 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find some multiples of a number, and determine whether a number is divisible by another.b

120. Slide 1- 120 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiples A multiple of a natural number is a product of that number and some natural number. We find multiples of 2 by counting by twos: 2, 4, 6, 8, and so on. We can find multiples of 3 by counting by threes: 3, 6, 9, 12, and so on.

121. Slide 1- 121 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Multiply by 1, 2, 3,… and so on, to find 6 multiples of seven. Solution 1  7 = 7 2  7 = 14 3  7 = 21 4  7 = 28 5  7 = 35 6  7 = 42

122. Slide 1- 122 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Divisibility The number a is divisible by another number b if there exists a number c such that a = b  c. The statements “a is divisible by b,” “a is a multiple of b,” and “b is a factor of a” all have the same meaning. Thus, 15 is divisible by 5 because 15 is a multiple of 5 (15 = 3  5) 40 is divisible by 4 because 40 is a multiple of 4 (40 = 10  4)

123. Slide 1- 123 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Determine whether 102 is divisible by 4. Solution Not 0 Since the remainder is not 0 we know that 102 is not divisible by 4.

124. Slide 1- 124 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Given a number from 1 to 100, tell whether it is prime, composite, or neither.c

125. Slide 1- 125 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Prime and Composite Numbers A natural number that has exactly two different factors only itself and one is called a prime number.  The number 1 is not prime  A natural number, other than 1, that is not prime is composite.

126. Slide 1- 126 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Determine whether the numbers listed below are prime, composite, or neither. 813 2433 8597 Has factors of 1, 2, 4 and 8, composite Has factors 1, 2, 3, 4, 6, 8, 12, 24, composite Has 5 as a factor, composite Has only two factors 1 and itself, prime Has factors 1, 3, 11, 33, composite Has only two factors 1 and itself, prime

127. Slide 1- 127 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A Table of Primes from 2 to 157 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157

128. Slide 1- 128 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the prime factorization of a composite number.d

129. Slide 1- 129 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Find the prime factorization of 50. Solution a) Since 50 is even, it must have 2 as a factor. b) Since 25 ends in 5, we know 5 is a factor. Because 5 is prime, we can factor no further. The prime factorization can be written as 2  5  5 or 2  5 2

130. Slide 1- 130 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Every number has just one (unique) prime factorization.

131. Slide 1- 131 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Find the prime factorization of 48 using a factor tree. Solution Had we begun with different factors (2 ∙ 24, or 4 ∙ 12), the same prime factorization would result. 2 2 2 2 3 6 8 2 3 4 48 2 · · · · = 48

132. Slide 1- 132 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Find the prime factorization of 220. Solution

133. Slide 1- 133 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I Find the prime factorization of 1424. Solution We use a string of successive divisions. 1424 = 2  2  2  2  89

134. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DIVISIBILITY Determine whether a number is divisible by 2, 3, 4, 5, 6, 8, 9, or 10. 1.8a

135. Slide 1- 135 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether a number is divisible by 2, 3, 4, 5, 6, 8, 9, or 10.a

136. Slide 1- 136 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 2 A number is divisible by 2 (is even) if it has a ones digit of 0, 2, 4, 6, or 8 (that is, it has an even ones digit).

137. Slide 1- 137 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Determine whether each of the following numbers is divisible by 2. 1. 4572. 3488 3. 32004. 7893 Solution 1. 457 2. 3488 2. 3200 4. 7893 is not divisible by 2; 7 is not even. is divisible by 2; 8 is even. is divisible by 2; 0 is even. is not divisible by 2; 3 is not even.

138. Slide 1- 138 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 3 A number is divisible by 3 if the sum of its digits is divisible by 3.

139. Slide 1- 139 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Determine whether the number is divisible by 3. 1. 122. 963. 3034. 374 Solution 1. 12 2. 96 3. 303 4. 374 1 + 2 = 3 3 + 7 + 4 = 14 9 + 6 = 15 3 + 0 + 3 = 6 Each is divisible by 3 because the sum of its digits is divisible by 3. The sum of the digits, 14, is not divisible by 3, so 374 is not divisible by 3.

140. Slide 1- 140 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 6 A number is divisible by 6 if its ones digit is 0, 2, 4, 6, or 8 (is even) and the sum of its digits is divisible by3.

141. Slide 1- 141 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Determine whether the number is divisible by 6. 1. 8402. 903. 83 Solution 1. 840 2. 90 3. 83 is even, divisible by 2. Also 8 + 4 + 0 = 12, so 840 is divisible by 3, 840 is divisible by 6. is even, divisible by 2. Also 9 + 0 = 9, so 90 is divisible by 3, 93 is divisible by 6. 83 is not divisible by 6 because it is not even.

142. Slide 1- 142 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 9 A number is divisible by 9, if the sum of its digits is divisible by 9.

143. Slide 1- 143 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Determine whether the number is divisible by 9. 1. 48242. 524 Solution 1. 4824 2. 524 4 + 8 + 2 + 4 = 18 and 18 is divisible by 9, so 4824 is divisible by 9. 5 + 2 + 4 = 11 and 11 is not divisible by 9, 524 is not divisible by 9.

144. Slide 1- 144 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By10 A number is divisible by 10, if its ones digit is 0.

145. Slide 1- 145 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Determine whether the number is divisible by 10. 1. 48102. 1524 Solution 1. 4810 2. 1524 is divisible by 10 because the ones digit is 0. is not divisible by 10 because the ones digit is not 0.

146. Slide 1- 146 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 5 A number is divisible by 5 if its ones digit is 0 or 5.

147. Slide 1- 147 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Determine whether each of the following numbers is divisible by 5. 1. 3402. 8853. 6721 Solution 1. 340 2. 885 3. 6721 is divisible by 5; because its one digit is 0. is divisible by 5; because its one digit is 5. is not divisible by 5; because its one digit is neither 0 nor 5.

148. Slide 1- 148 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 4 A number is divisible by 4 if the number named by its last two digits is divisible by 4.

149. Slide 1- 149 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Determine whether each of the following numbers is divisible by 4. 1. 77322. 8453 Solution 1. 7732 2. 8453 is divisible by 4 because 32 is divisible by 4. is not divisible by 4 because 53 is not divisible by 4.

150. Slide 1- 150 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley By 8 A number is divisible by 8 if the number named by its last three digits is divisible by 8.

151. Slide 1- 151 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Determine whether each of the following numbers is divisible by 8. 1. 12642. 43,911 Solution 1. 1264 2. 43,911 is divisible by 8 because 264 is divisible by 8. is not divisible by 8 because 911 is not divisible by 8.

152. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Least Common Multiples Find the least common multiple, LCM, of two or more numbers. 1.9 1.9a

153. Slide 1- 153 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the least common multiple, LCM, of two or more numbers.a

154. Slide 1- 154 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Least Common Multiple, LCM The least common multiple, or LCM, of two natural numbers is the smallest number that is a multiple of both.

155. Slide 1- 155 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Find the LCM of 40 and 60. Solution a) First list some multiples of 40 by multiplying 40 by 1, 2, 3, and so on: 40, 80, 120, 160, 200, 240, 280, … b) Then list some multiples of 60 by multiplying 60 by 1, 2, 3, and so on: 60, 120, 180, 240, … c) Now list the numbers common to both lists, the common multiples: 120, 240, … d) These are the common multiples of 40 and 60. Which is the smallest? 120

156. Slide 1- 156 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Method 1: To find the LCM of a set of numbers using a list of multiples: a) Determine whether the larger number is a multiple of the others. If it is, it is the LCM. That is, if the largest number has the others as factors, the LCM is that number. b) If it is not, check multiples of the largest until you get one that is a multiple of each of the others.

157. Slide 1- 157 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Find the LCM of 6 and 28. Solution 1.28 is the larger number, but it not a multiple of 6. 2. Check multiples of 28: 2  28 = 56 3  28 = 84A multiple of 6 The LCM = 84.

158. Slide 1- 158 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Find the LCM of 12 and 36. Solution 1. 36 is the larger number and 36 is a multiple of 12, so it is the LCM. The LCM is 36.

159. Slide 1- 159 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Method 2: To find the LCM of a set of numbers using prime factorizations a) Find the prime factorization of each number. b) Create a product of factors, using each factor the greatest number of times that it occurs in any one factorization.

160. Slide 1- 160 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Find the LCM of 35 and 90. Solution 1. We write the prime factorization of each number. 35 = 5  790 = 2  3  3  5 2. a) We note that 2  3  3  5 does not contain the other factorization 5  7. b) To find the LCM of 35 and 90, we multiply 2  3  3  5 by the factor of 35 that it lacks, 7: LCM = 2  3  3  5  7 35 is a factor. 90 is a factor.

161. Slide 1- 161 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Find the LCM of 36 and 48. Solution 1. Write the prime factorization of each number: 2. We multiply the factorization of 36, by any prime factors of 48 that are lacking. In this case, 2  2. LCM = 48 is a factor. 36 is a factor.

162. Slide 1- 162 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Find the LCM of 15, 30 and 25. Solution 1. Write the prime factorization of each number. 15 = 3  5 30 = 2  3  5 25 = 5  5 2. Consider 15 and 30, the factorization of 30 contains 15 2  3  5 Multiply by the factors of 25 that are still missing. 2  3  5  5 = 150