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A C E Answers | Investigation 5 Applications 1. Students may explain the result using an 7. a. Possible answers: The product of three equation such as (2x + 6) ÷ 2 – x, where x is the number picked. In order to simplify this equation, students will need to make sense of (2x + 6) ÷ 2. They need to rewrite 2x + 6 in two parts as x + x + 3 + 3, with the result that (2x + 6) ÷ 2 = x + 3. Thus (2x + 6) ÷ 2 – x = x + 3 – x = 3 for any number x a student picks. consecutive whole numbers will always yield an even number. The product of three consecutive whole numbers is always a multiple of 6. b. The product of three consecutive whole numbers is even. There are two cases for a set of 3 consecutive numbers. Case 1: Two evens and an odd. Using the facts from Exercise 5 and Problem 2.2, (even × odd) × even = even × even = even. Case 2: Two odds and an even, (odd × even) × odd = even × odd = even. The product of three consecutive whole numbers is a multiple of 6 since there is at least one even (multiple of 2) and at least one multiple of 3 (there are 3 numbers so one must be a multiple of 3), so the product must be a multiple of 6. 2. Students may explain the result using an equation such as (2(x + 4) – 6) ÷ 2 – x = 1. In order to simplify this equation, students will need to make sense of (2(x + 4) – 6) ÷ 2, which is equal to (2x + 2) ÷ 2. Since 2x + 2 = x + x + 1 + 1, dividing x + x + 1 + 1 in two parts results in x + 1, or [2(x + 4) – 6] ÷ 2 = (2x + 2) ÷ 2 = x + 1. Thus [(2(x + 4) – 6) ÷ 2] – x = x + 1 – x = 1 for any number x a student picks. 3. Answers will vary. 8. a. The product of four consecutive 4. Answers may vary, but they should make whole numbers will always be even. It is also divisible by 24. For any four consecutive numbers, one of them is divisible by 4, another is divisible by 2 (but not 4), and at least one of the four is divisible by 3. For example, (4 × 3 × 2) × 1 = 24 × 1. Therefore, 24 is a factor of the product of any four consecutive numbers. logical sense. Possible argument: Let n and m represent any integers. Then 2n + 1 and 2m + 1 are two odd numbers. But, (2n + 1) + (2m + 1) = 2n + 2m + 2 = 2(n + m + 1), and 2(n + m + 1) is an even number because it is a multiple of 2. So, the sum of two odd integers is even. b. In a set of four consecutive whole 5. Again, answers may vary; one possible numbers, there will be two odd whole numbers and two even whole numbers. This is symbolically shown with E = any even whole number and O = any odd whole number: O × E × O × E; since O × E = E, O×E×O×E=E×E=E argument: Let n and m represent any integers. Then 2n and 2m are two even numbers. But (2n)(2m) = 4nm = 2(2nm). 2(2nm) is an even integer. So, the product of two even numbers must be even. 9. a. rational 6. Answers may vary; one possible argument: Take two odd numbers represented by (2n + 1) and (2m + 1), where m and n are integers. Then 2n and 2m are two even numbers. But, (2n + 1)(2m + 1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1, and 2(2nm + n + m) + 1 is an odd number, since it is 1 more than a multiple of 2. So, the product of two odd integers is odd. Say It With Symbols b. irrational c. irrational d. rational e. rational f. rational 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Investigation 5 A C E Answers | Investigation 5 10. a. 10,034 is divisible by 2; 69,883 is not. Students may reason that they could tell because 4 is divisible by 2, but 3 is not. 11. Answers will vary, but students should note that the last two digits must be divisible by 4. For example, 4,516 is divisible by 4 whereas 4,519 is not, because 16 is divisible by 4 and 19 is not. This is not easy to see. Since all powers of 10 greater than 10 (100, 10,000, etc.) are divisible by 4, you only have to check the last two digits. b. Answers will vary. Some students may suggest that as long as the last digit is even, then the number is divisible by 2, using the reasoning of long division. One explanation is that any digit except the last is a multiple of 10, which is clearly divisible by 2, and thus the only important digit is the ones digit. 12. Answers will vary, but students should note that the last digit must be 0 or 5. This is true because the other digit is a multiple of 10, and thus is divisible by 5. The ones digit is the only important one to check. Connections 13. a. Answers will vary. Some possibilities are 14. a. Students may list all the combinations below. These buildings are composed of a central cube and 5 arms that contain 0 cubes, then 1 cube, then 2 cubes, then 3 cubes, and so on. This pattern is described in the student edition. The buildings are composed of a central tower that contains 1 cube, then 2 cubes, and so on. Each new building is the previous building with 5 cubes added. or make a table. For 2 participants, 2 games must be played; for 3 participants, 6 games must be played; for 4 participants, 12 games must be played; for 5 participants, 20 games must be played; and for 6 participants, 30 games must be played. b. The expression for the number of games played g in relation to the number of participants in the league n is n(n – 1) = g or n2 – n = g. b. There are many possible equivalent 15. a. n2, where n is the number of participants expressions for the number of cubes in the nth building. Some examples: 1 + 5(n – 1), n + 4(n – 1), or 5n – 4. in the league b. n; the diagonal is not filled in because participants do not play themselves. c. The fifth building contains 21 cubes, which can be found by substituting 5 into any of the above equations or other correct equations. c. Students should combine the two equations to create a new expression n2 – n, but some students may recognize that this is the same as n(n – 1) and use this as their expression. d. Answers will vary since they depend on what the students had for part (b). However, the expression 5n – 4 is the simplest form of the expression possible. This expression can be thought of as representing an addition of five blocks for every new tower starting with n = 1. d. Answers may vary since they depend on which expressions students used for parts (a) and (b). Students should note that, regardless, the expressions are equivalent. 16. 4 and –3; (x – 4)(x + 3) = 0, so x = 4 or x = –3. e. Students may either find a different pattern that relates to one of the previous expressions, or they may find a new expression altogether. Other possibilities include rewriting n + 4(n – 1) as n + 4n – 4. Say It With Symbols 17. 0 and –4; x2 + 4x = x(x + 4) = 0, so x = 0 or x + 4 = 0, and thus x = 0 or x = –4. 2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Investigation 5 A C E Answers | Investigation 5 18. –4 and –5; x2 + 9x + 20 = (x + 4)(x + 5) = 0, so (x + 4) = 0 or (x + 5) = 0, and thus x = –4 or x = –5. make sense since the number of dimes and quarters each should be greater than or equal to zero. So, x should be less than or equal to 1,000 and greater than or equal to zero, since you want to account for only a nonnegative number of coins. Thus, having the value of the coins add up to anything less than $100 is not possible. 19. –8 and 1; x2 + 7x – 8 = (x + 8)(x – 1) = 0, so (x + 8) = 0 or (x – 1) = 0, and thus x = –8 or x = 1. 20. 10 and 1; x2 – 11x + 10 = (x – 10)(x – 1) = 0, so (x – 10) = 0 or (x – 1) = 0, and thus x = 10 or x = 1. To check, substitute the values into the equation: 0.10x + 0.25(1,000 – x) 0.10(1600) + 0.25(1,000 – 1,600) 160 – 150 = 10, which is the amount of money in dollars in the bag. 21. 9 and –3; x2 – 6x – 27 = (x – 9)(x + 3) = 0, so (x – 9) = 0 or (x + 3) = 0, and thus x = 9 or x = –3. 22. 5 and –5; x2 – 25 = (x + 5)(x – 5) = 0, so x = –5 or x = 5. 28. a. P = 2(2w + 4) + 2w = 4w + 8 + 2w = 6w + 8 = 2(3w + 4) 23. 10 and –10; x2 – 100 = (x + 10)(x – 10) = 0, so x = –10 or x = 10. b. Find the dimension of the rectangle if the perimeter is 80. If the perimeter of a rectangle is 80, then you could substitute 80 into your simplest equation, solve for w, and calculate the length using the value of w. 80 = 6w + 8 80 – 8 = 6w + 8 – 8 72 = 6w 24. –0.5 and –1; 2x2 + 3x + 1 = (2x + 1)(x + 1) = 0, so x = –0.5 or x = –1. 25. – 4 and –2; 3x2 + 10x + 8 = 3 (3x + 4)(x + 2) = 0, so x = – 4 or x = –2. 3 26. a. 6r + 2(r – 30) = 8r – 60 = distance traveled 72 6 b. Possible question: How many miles are traveled if the rate is 70 miles per hour? Substitute 70 for r, the rate of speed, in the simplest expression and get 4(70) – 30 = 280 – 30 = 250 miles traveled. 29. a. 15x + 9(4,000 – x) = 15x + 36,000 – 9x = 6x + 36,000 = 6(x + 6,000) b. If the total amount of money collected at a concert was $60,000, find the number of reserved and unreserved seats. Set 60,000 equal to your simplest expression and solve for the number of tickets. 60,000 = 6x + 36,000 60,000 – 36,000 = 6x 24,000 = 6x b. Possible question: If the amount of money in the bag is $10, how many of each coin are in the bag? Set $10 equal to your expression and solve for the number of coins: 10 = 250 – 0.15x 10 – 250 = 250 – 250 – 0.15x –240 = –0.15x 24,000 6 = 6x 6 4,000 = x The number of reserved seats was 4,000 and the number of unreserved seats was 4,000 – 4,000, or 0. = –0.15 x –0.15 1,600 = x So, in theory, there are 1,600 dimes and –600 quarters. Note: Students may recognize that this answer does not Say It With Symbols 6 12 = w Therefore, the width is 12 feet and the length is 2(12) + 4 or 28 feet. 27. a. If x is the number of dimes, then 0.10x + 0.25(1,000 – x) = 0.10x + 250 – 0.25x = 250 – 0.15x = amount of money in dollars –240 –0.15 = 6w 3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Investigation 5 A C E Answers | Investigation 5 30. a. 32 feet. Students may substitute the value 2 into the equation for t and get 32. 33. The expression is linear. b. 36 feet. To find the maximum height, students would first find the x-intercepts (See part (c).). Then, the maximum height occurs between the 2 t= h= –16 3 2 2 + 48 3 2 –7 –5 –3 5 y = 4(x –7) + 6 –50 –42 –34 –2 34. The expression is linear. two intercepts, so it is at t = 3 . When 3 , 2 x = 36. x –7 –5 –3 5 y = –3– 7(x + 9) –17 –31 –45 –101 35. The expression is exponential. The values for y in the table are approximations. c. 3 seconds; students can use the equation h = –16t2 + 48t and the factored form h = –16t(t – 3), or h = 16t(–t + 3). They find the x-intercepts, which are 0 and 3. The intercepts are the points where the height is 0, so the ball is in the air from 0 to 3 seconds. –7 x y = 2(3)x –5 –3 5 0.001 0.008 0.074 486 36. The expression is quadratic. d. Answers will vary; students who use a graph may see the highest point on the graph as the maximum height and use the x-intercepts to find the total time the ball was in the air. Students who use a table will look for the greatest y-value in the table for part (b) and the x-values when y is 0 for part (c). x –7 –5 –3 5 y = 3x2 – x – 1 153 79 29 69 37. The expression is quadratic. x –7 –5 –3 5 y = 5(x – 2)(x + 3) 180 70 0 180 38. For Exercise 33, 4x – 28 + 6, or 4x – 22; for Exercise 34, –3 – 7x – 63, or –66 – 7x. 39. Table 1: linear, y = 9 – 3x Table 2: quadratic, y = x2 – 16 Table 3: none of these Table 4: exponential, y = 3(4x) Table 5: none of these. Note: This is 31–32. a. The two expressions in both Exercises 31 and 32 are not equivalent. For part (a), students should make a table or graph and find that the two equations have either the same table and graph, or that they are different, in which case they are not equivalent. inverse variation, y = 4 . x 40. a. Yes, the perimeter can be a rational b. They can simplify the expressions number if the three numbers form a using the Distributive Property and/ Pythagorean Triple. For example, a or Commutative Property to see 3–4–5 right triangle will have perimeter if the expressions are equivalent. 12 units, which is rational since all For Exercise 31, the first expression integers are rational numbers. simplifies to 4x – 5, and for Exercise 32, the first simplifies to 6x – 7 and the b. Yes, the perimeter can be irrational second to 6x – 3. To show that the if the three numbers do not form a expressions are not equivalent, Pythagorean Triple. For example, 1, 1, students can show that a value and 2 . for x when substituted into both expressions yields different values. Say It With Symbols 4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Investigation 5 A C E Answers | Investigation 5 41. a. No, the perimeter cannot be a rational number. If the length of the hypotenuse is an integer, then the length of the shortest leg will be rational. However, the length of the longer leg will be irrational because it is square root of three times the length of the shorter leg. The perimeter is the sum of two rational numbers plus an irrational number. For example, a 30–60–90 triangle with a hypotenuse of 10 inches will have a shorter leg measuring 5 inches, and a longer leg measuring 5 3 inches. The perimeter will be 15 + 5 3 inches. b. Using similar reasoning as in part (a), the perimeter will always be irrational. Extensions 42. a. 52 + 62 = 312 – 302 43. a. Answers may vary depending on the choice for n. b. Let a be the first number and a + 1 be the second number. Then a2 + (a + 1)2 = (a(a + 1) + 1)2 – (a(a + 1))2. b. George is correct. All it takes is one counterexample to disprove a conjecture. So, even if there are many examples that work, that is not sufficient proof. This conjecture fails when n is 41, for example. c. Begin with the right-hand side of the equation to show that it is equal to the left-hand side. a2 + (a + 1)2 = (a(a + 1) + 1)2 – (a(a + 1))2 = (a2 + a + 1)2 – (a2 + a)2 by the Distributive Property = (a2 + a + 1)(a2 + a + 1) – (a2 + a)(a2 + a) by Definition of Exponents = (a4 + 2a3 + 3a2 + 2a + 1) – (a4 + 2a3 + a2) by the Distributive Property = 2a2 + 2a + 1 by combining like terms = a2 + a2 + 2a + 1 by rewriting the first term = a2 + a2 + a + a + 1 by rewriting the third term = a2 + a(a + 1) + 1(a + 1) = a2 + (a + 1)2 by factoring the last two terms Since the left-hand side and the righthand side of the equation are equal, the conjecture is true. Say It With Symbols 44. a. b. 2 + 2 0.5 2 + 2 ; the sum 0.5 + c. 2 is an irrational number. 2 2 2 2 , which is irrational because it is the product of a rational number and an irrational number. 45. If the sum of the digits of a number is divisible by 3, then the number must also be divisible by 3. Also, if a number is divisible by 3, then the sum of the digits is divisible by 3. The following proofs are written symbolically. This is not something that students are expected to be able to do at this point. Students may talk informally about parts of the proof without the symbols. Suppose n is a three-digit number whose digits sum to a number that is divisible by 3. Then n = a(100) + b(10) + c, where a, b, and c are the digits of n. Rewriting, n = 99a + a + 9b + b + c. Using the Commutative Property, n = (99a + 9b) + (a + b + c). Using the Distributive Property, (99a + 9b) is divisible by 3, or 3(33a + 3b). In order for the entire number to be divisible by 3, a + b + c must also be divisible by 3. 5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Investigation 5 A C E Answers | Investigation 5 For example, n = 651 = 6 × 100 + 5 × 10 + 1 = (6 × 99) + (6 × 1) + (5 × 9) + (5 × 1) + 1 = (6 × 99) + (5 × 9) + (6 × 1) + (5 × 1) + 1 = (99 × 6 + 9 × 5) + (6 × 1 + 5 × 1 + 1) = 3(33 × 6 + 3 × 5) + (6 + 5 + 1) = 3(33 × 6 + 15) + (12) = 3(198 + 15 + 4) So, given that n is divisible by 3, 6 + 5 + 1 = 12 is divisible by 3. 46. If the sum of the digits of a number is divisible by 3 and it is even, then the number is divisible by 6. Suppose n is an even number whose digits sum to a number that is divisible by 3. Since it is even, it must be divisible by 2. It must also be divisible by 3, by Exercise 45. By the Fundamental Theorem of Arithmetic (the prime factorization of), n must be of the form n = 2 × 3 ×…If we rewrite, n = 6 ×…Thus, n is divisible by 6. Now let’s look at another case: If a number is divisible by 3, then the sum of its digits is divisible by 3. Say a number n is divisible by 3, then we can write n = 3k, where k is an integer. Thus, n = a(100) + b(10) + c = 3k. Again, n = 99a + 9b + a + b + c = 3k, so dividing each side by 3, k = 33a + 3b + 47. a. Answers will vary. abc . But 3 then a b c 3 b. Students may find it easiest to explain why this method works by forming an equation to represent the value of any number ending in five, such as 10x + 5, where x can be any whole number. Then a student taking the square of this value will get (10x + 5)(10x + 5) = 100x2 + 100x + 25 = 100(x2 + x) + 25. This equation represents Judy’s method of finding the square. since k is an integer, must be an integer, so a + b + c (the sum of the digits of n) must be divisible by 3. Say It With Symbols 6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Investigation 5