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Answers for Enrichment Activity Exercises Enrichment Activity 1-1: Guessing a To find a word in the dictionary, choose a middle Number by Bisection word alphabetically and ask if the chosen word a. 8 b. 6 comes before or after the word for which you are c. A maximum of 11 guesses would be needed to searching. Continue the process. To find a name locate a number between 1 and 2,000. The first on a list, choose a middle name and continue the guess would be the average of 0 and 2,000, or process. 1,000. Assuming that the number itself was not 1,000, this would divide the possible numbers Enrichment Activity 1-2: Repeating into two equal sets: numbers between 0 and 999 Decimals 1 and numbers between 1,001 and 2,000. Sets of 1. 15 2. 41 99 this size require a maximum of 10 guesses to 3. 7 4. 10 12 33 find the number. 47 5. 111 6. 322 or 325 99 99 Use a simpler related problem to find a general solution. 7. 1 8. 155 198 9. 15 10. 120 Range Max. Number of Numbers of Guesses Enrichment Activity 1-3: A Piece of Pi From 1 to 3 2 First guess would be 2. 1. 0 11 1 5 2 12 3 11 4 10 If 2 is not the number, 5 8 6 9 7 8 8 12 9 14 the next guess would 2. No. The digit 9 appears most often while the digit be correct. 1 appears least often. From 1 to 7 3 First guess would be 4. 3. 0 12 1 11 2 12 3 8 4 12 If 4 is not the number, 5 12 6 7 7 4 8 13 9 9 the number must be 4. Answers will vary. Example: The digits 0, 1, 4, 5, one of a set of three and 8 appear more often than in the first group, possible numbers, and the digits 3, 6, 7, and 9 appear less often than in at most two more the first group, and the digit 8 appears with same guesses are needed. frequency as in the first group. From 1 to 15 4 First guess would be 8. 5. 0 23 1 16 2 24 3 19 4 22 If 8 is not the number, 5 20 6 16 7 12 8 25 9 23 the number must be 6. Answers will vary. Examples: The digit 7 appears one of a set of seven significantly less frequently than the other digits. possible numbers, and Appearances of the digits 0, 2, 3, 4, 5, 8, and 9 are at most three more evening out. As more digits in the expansion are guesses are needed. checked, the digits may seem to appear the same Range Max. Number number of times. of Numbers of Guesses Powers of 2 From 1 to 3 2 22 4 Enrichment Activity 1-4: Making 100 2,148 1,752 From 1 to 7 3 23 8 1. 96 537 2. 96 438 From 1 to 15 4 24 16 1,428 1,578 3. 96 357 4. 94 263 From 1 to 31 5 25 32 7,524 5,823 From 1 to 63 6 26 64 5. 91 836 6. 91 647 3,546 7,524 From 1 to 127 7 27 128 7. 82 197 8. 81 396 From 1 to 255 8 28 256 5,643 9. 81 297 10. 3 714 69,258 If 2n 1 the number of integers in the set 5,472 11. Answers will vary. Examples: 13 5 9 1,368, from which the chosen number is selected, 3,576 5,148 n is the maximum number of guesses needed to 16 5 12 894 , 40 5 27 396 identify the number. 264 Enrichment Activity 2-1: Special Ops 3. z 0 2 4 6 8 5 0 0 0 0 0 0 1. 7 2. 5 3. 8 5 2 0 4 8 2 6 4. 6.2 5. 20 6. 8 7. 0.4 8. 75 9. 23 4 0 8 6 4 2 10. 8 11. 14 6 0 2 4 6 8 12. a. a b a 2b 8 0 6 2 8 4 b. 4 13. Answers will vary. 4. a. Yes b. Yes c. Yes d. 6 Enrichment Activity 2-2A: Clock Arithmetic e. Number z Inverse Identity 2 z 8 6 a. 1 2 3 4 5 6 7 8 9 10 11 12 4 z 6 6 1 2 3 4 5 6 7 8 9 10 11 12 1 6 z 4 6 2 3 4 5 6 7 8 9 10 11 12 1 2 8 z 2 6 5. Yes 3 4 5 6 7 8 9 10 11 12 1 2 3 Conclusion: This is a field; all 11 properties hold. 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 Enrichment Activity 2-5: Paying the Toll 6 7 8 9 10 11 12 1 2 3 4 5 6 1(7) 2( 3) 1 1(11) 11 1(11) 1(7) 1(3) 21 1(11) 3( 3) 2 4(3) 12 2(11) 22 7 8 9 10 11 12 1 2 3 4 5 6 7 1(3) 3 1(7) 2(3) 13 2(7) 3(3) 23 8 9 10 11 12 1 2 3 4 5 6 7 8 1(7) 1( 3) 4 1(11) 1(3) 14 8(3) 24 9 10 11 12 1 2 3 4 5 6 7 8 9 1(11) 2( 3) 5 5(3) 15 2(11) 1(3) 25 10 11 12 1 2 3 4 5 6 7 8 9 10 2(3) 6 1(7) 3(3) 16 3(11) 1( 7) 26 1(7) 7 1(11) 2(3) 17 9(3) 27 11 12 1 2 3 4 5 6 7 8 9 10 11 1(11) 1( 3) 8 1(11) 1(7) 18 4(7) 28 12 1 2 3 4 5 6 7 8 9 10 11 12 3(3) 9 1(7) 4(3) 19 2(11) 1(7) 29 b. 1. Yes 1(7) 1(3) 10 2(7) 2(3) 20 3(7) 3(3) 30 2. Yes All purchases can be made since any whole 3. Yes number can be formed using multiples of the 4. 12 numbers above. 5. Number 1 2 3 4 5 6 7 8 9 10 11 12 Inverse 11 10 9 8 7 6 5 4 3 2 1 12 Enrichment Activity 2-8: Graphing Ordered Pairs of Numbers Enrichment Activity 2-2B: Digital Addition; a. State Street Digital Multiplication 1. { 0 2 4 6 8 H S 0 0 2 4 6 8 2 2 4 6 8 0 A L V C Main Street 4 4 6 8 0 2 O G M P F 6 6 8 0 2 4 8 8 0 2 4 6 D 2. a. Yes b. Yes c. Yes d. 0 b. M is two blocks west and one block south of e. Number { Inverse Identity O( 2, 1). 0 { 0 0 c. The first number gives the number of blocks east 2 { 8 0 or west of O, the center of town. If the number is 4 { 6 0 positive, the position is east of O; if the number 6 { 4 0 is negative, the position is west of O. The second 8 { 2 0 number gives the number of blocks north or 265 south of O. If the number is positive, the position 4. 16x 13,600 5. x 150; x 300; x 450 is north of O; if the number is negative, the 6. 16x 20,400 7. 33x 34,000 position is south of O. 8. 33x 34,000 65,000 9. $3,000 10. $3,500 11. $45,200 Enrichment Activity 3-1: Number Play Columns 2 and 3 will vary. The answers for columns 1 Enrichment Activity 4-3: Consecutive and 4 are shown below. Integers Col. 1 Col. 4 1. Number First Even of Integers Integer Sum or Odd 1. 15 n 2. 3 2n 3 1 6 even 3. 38 2n 8 3 4 15 odd 4. 3,800 200n 800 3 7 24 even 5. 3,600 200n 600 6. 360 20n 60 3 x 3x 3 even or odd 7. 400 20n 100 4 1 10 even 8. 20 n 5 4 4 22 even 9. 15 n 4 7 34 even Conclusion: The result on line 9 will always equal the 4 x 4x 6 even starting number on line 1. 5 1 15 odd Enrichment Activity 3-2: Variable Codes 5 4 30 even 1. a. PAOIK b. ZUSUXXUC 5 7 45 odd c. SGZNKSGZOIY d. KDZXGUXJOTGXE 5 x 5x 10 even or odd 2. Answers will vary. 3. I MUST STUDY FOR AN EXAM 6 1 21 odd 4. a. XIVRK b. GLQQCV 6 4 39 odd c. RJJZXEDVEK d. VOGCFIRKZFE 6 7 57 odd 5. x 9 6 x 6x 15 odd 6. NZIXS, MLEVNYVI; the correspondence is symmetric. If a line is drawn between M and N in 7 1 28 even the alphabet list, the corresponding letters are 7 4 49 odd equal distances from that line. A corresponds to 7 7 70 even Z, B corresponds to Y, and so on. 7 x 7x 21 even or odd 7. a. MLBNCFH b. TUESDAY; the result is the original word. 2. Let x be the first consecutive integer. 7 x is the encoding and decoding expression. a. Odd; 2x 1 is the sum of an even number and 8. a. YKBWTR b. RDUPMK; no an odd number. b. Even when x is odd; 3x 3 is the sum of two Enrichment Activity 3-7: Formulas for odd numbers. Health Odd when x is even; 3x 3 is the sum of an 1. 22 2. 24 3. 17 even number and an odd number. 4. 31 5. 26 6. 22 c. Even; 4x 6 is the sum of two even numbers. d. Even when x is even; 5x 10 is the sum of 7. 24 8. 34 9. BMI 5 703W 2 H two even numbers. 10. 114 bpm 11. 19 bpm 12. 23 bpm Odd when x is odd; 5x 10 is the sum of an 13. 19 bpm 14. 84% odd number and an even number. e. Odd; 6x 15 is the sum of an even number Enrichment Activity 4-2: Book Value and an odd number. 1. Let x represent the middle book because the f. Even when x is odd; 7x 21 is the sum of two values of the other books can be expressed in odd numbers. relation to it. Odd when x is even; 7x 21 is the sum of an 2. 16 books 3. x 100; x 200; x 300 even number and an odd number. 266 g. Even when the number of integers divided by 2. x 4 2 is even. Odd when the number of integers divided by 2 is odd. h. Even when the first integer is even and the number of integers after the first, divided by 2, is even. Odd when the first integer is odd and the number of integers after the first, divided by 2, is even. Even when the first integer is odd and the 3. x 3 number of integers after the first, divided by 2, is odd. Odd when the first integer is even and the number of integers after the first, divided by 2, is odd. 3. a. 3 b. 5 c. 7 d. The sum always has the number of consecutive integers as a factor. The sum is always the number of integers times the middle number. 4. a. No b. The sum is 6 more than 4 times the first integer. 4. All real numbers The sum is twice the sum of the second and between 4 and third integers. 4; [ 4, 4] The sum is 6 less than 4 times the last integer. 5. a. Student tables b. The sum of even integers is always even. 6. a. Student tables b. The sum of odd integers is even if there is an even number of integers and odd if there is an odd number of integers. 5. All real numbers between 1 and Enrichment Activity 4-4: Law of the Lever 1; ( 1, 1) 1. 2.4 ft 2. 15 lb, 25 lb 3. Heavier carton, 9 ft; lighter carton, 12 ft 4. Kelly, 49 lb; Laurie, 70 lb 5. a. The side with the 32-lb weight b. 3 ft 6. All real numbers Enrichment Activity 4-7: Graphing an greater than 1; Inequality (1, ). Since there 1. x 4 is no real number that is the square root of a nega- tive number, the inequality is mean- ingless for x 0. 267 7. All real numbers; Enrichment Activity 5-2: An Odd Triangle ( , ). 1. Row 6: 31, 33, 35, 37, 39, 41 Row 7: 43, 45, 47, 49, 51, 53, 55 2. 1, 9, 25, 49; the middle number in odd-numbered Row n is n2. 3. 4, 16, 36; the average of the middle numbers in even-numbered Row n is n2. 4. a. 169 b. 6 c. 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 8. All positive real 179, 181, 183 numbers; (0, ). 5. Row Number Sum of Numbers in Row 1 1 2 8 3 27 4 64 5 125 6 216 9. Since x is always positive, there is 7 343 no real number 6. Sum of the numbers in Row n n3 that makes this 7. a. 1,728 b. Row 19 inequality true; . 8. a. 3 22 1, 7 32 2, 13 42 3, 21 52 4, 31 62 5, 43 72 6 b. n2 (n 1) n2 n 1 9. a. 5 22 1, 11 32 2, 19 42 3, 29 52 4, 41 62 5, 55 72 6 b. n2 (n 1) n2 n 1 Enrichment Activity 5-1: Sums and Squares 10. a. Row Sum of All Sum Written 1. 42 4 5 25 52 Number Numbers as a Square 1 1 12 2 9 32 3 36 62 4 100 102 2. 5 225 152 n + 1 6 441 212 2 2 n n (n 1) n 2n 1 7 784 282 n n2 n C D 2 ? 2 (n 2n 1) 5 (n 1) n(n 1 1) 2 ? b. 2 n2 2n 1 5 (n 1)(n 1) + n2 2n 1 n2 2n 1 1 n+1 n2 2n 1 (n 1)2 Enrichment Activity 5-4: Products, Sums, and Cubes 3. 202 20 21 400 20 21 441 1. 3(4)(5) 4 60 4 64 43 4. 102 100 4(5)(6) 5 120 5 125 53 5. (n 1)2 5(6)(7) 6 210 6 216 63 2. The product of three consecutive integers plus the middle integer is equal to the cube of the middle integer. 268 3. 6(7)(8) 7 336 7 343 73 Enrichment Activity 6-1B: Ratio: Estimates 4. 7(8)(9) 8 504 8 512 83 and Comparisons 5. (n 1)(n)(n 1) n n(n 1)(n 1) n Columns 1 and 2 will vary according to student (n2 n)(n 1) n estimates. n3 n2 n2 n n Column 3 n3 ? Actual Ratio Source for Teacher: 6. (n)(n 1)(n 2) (n 1) 5 (n 1)3 of Size Area (sq mi) ? (n)(n2 3n 2) (n 1) 5 (n 1)(n 1)(n 1) Australia 0.80 2,967,908 ? 3 2 2 (n 3n 2n) (n 1) 5 (n 1)(n 2n 1) Brazil 0.88 3,286,487 n3 3n2 3n 1 n3 3n2 3n 1 Canada 1.04 3,855,101 Enrichment Activity 6-1A: Fibonacci China 1.00 3,705,405 Sequence and the Golden Ratio India 0.34 1,269,345 Part 1. For clarity, the first 26 terms of the Fibonacci Japan 0.04 145,883 sequence are written below from smallest to Mexico 0.20 761,606 largest in three-column format. Russia 1.77 6,592,769 1 55 4,181 Spain 0.05 194,897 1 89 6,765 United States 1.00 3,718,709 2 144 10,946 3 233 17,711 Bonus: Actual Ratio 5 377 28,657 of Size 8 610 46,368 Australia 3.90 13 987 75,025 Brazil 4.32 21 1,597 121,393 Canada 5.06 34 2,584 China 4.87 13 21 India 1.67 Part 2. 8 5 1.625 13 5 1.615384 34 55 Japan 0.19 21 5 1.619047 34 < 1.617647059 89 144 Mexico 1.00 55 5 1.618 89 < 1.617977528 233 377 Russia 8.66 144 5 1.61805 233 < 1.618025751 610 987 Spain 0.26 377 < 1.618037135 610 < 1.618032787 1,597 2,584 United States 4.88 987 < 1.618034448 1,597 < 1.618033813 4,181 6,765 2,584 < 1.618034056 4,181 < 1.618033963 In a and b, student responses will vary. 10,946 17,711 a. With Mexico’s area used as the base of compari- 6,765 < 1.618033999 10,946 < 1.618033985 son, its ratio changes from 0.20 to 1.00. In turn, 28,657 46,368 17,711 < 1.61803399 28,657 < 1.618033988 the ratio for every nation becomes about 5 times 75,025 121,393 as large as the ratio using the United States as 46,368 < 1.618033989 75,025 < 1.618033989 the base of comparison. Part 3. 1 12"5 < 1.618033989 b. The ratios in both lists show the sizes of the countries in relation to one another: Japan has As the terms of the Fibonacci sequence the smallest area, followed by Spain, up to Russia increase, the ratio comparing the greater with the largest area. of two consecutive Fibonacci numbers to the smaller approaches the value of the golden ratio. 269 Enrichment Activity 6-2: Population Density 9. a. False b. Leads to the contradiction (y 0) a a1c c Part 1. 10. b , b 1 d , d State Land Area Population Population (sq mi) Density Enrichment Activity 7-4: Hero’s Formula Alaska 648,818 571,951 1 1. a. 24 sq in. b. 24 sq in. 2. a. 431.6 m2 b. 11,161.36 sq ft California 35,484,453 155,959 228 3. a. 388.8 cm2 Florida 17,019,068 53,927 316 b. Since the area and base are known, substitute Montana 917,621 145,552 6 the values into the formula A 5 1bh; b 18 cm 2 New Jersey 8,638,396 7,414 1,165 c. Right triangle New York 19,190,115 47,214 406 Enrichment Activity 7-5: Rectangle North Carolina 8,407,248 48,711 173 Cover-Up Texas 22,118,509 261,797 84 1. 5 ways Part 2. a. Student estimates will vary. b. Canada, 9; China, 361; India, 943; Japan, 880; Mexico, 141; Russia, 22 Enrichment Activity 6-3: Catching Up 1. a. 35 steps b. 5 steps c. 12.5 or 25 2. 8 ways 35 70 2. a. 43 steps b. 5 steps; 10 steps 25 c. 43 d. 43 . 25 ; Evan is gaining on Marisa. 25 70 37.5 75 3. 51 or 102, 50, 62.5 or 125, 75 59 67 134 75 4. Evan continues to gain on Marisa until he catches her after he has taken a total of 30 steps and she has taken a total of 75 steps. Enrichment Activity 6-4: Ratios and Inequalities 1. Approach 1 proof: a c b , d a c b (b)(d) , d (b)(d) 3. Rectangles of Width 2 ad , cb Number of Ways to ad , bc a c Length of Rectangle Cover with Dominoes Approach 2 proof: , , AbBd bAdB b d a d c 0 1 b ad bc 1 1 , bd bd 2 2 ad , bc 2. a. ad bc b. ad bc 3 3 c. d b d. d bc 4 5 1 1 3. b . b 1 1 must be true because 1(b 1) b(1) 5 8 or b 1 b. 6 13 a 4. b . 1 is true because a2 b2, so "a . "b and 7 21 a b. 8 34 5. a b is true because a(b c) b(a c), so ab ac ba bc and ac bc. 9 55 6. a. False b. Leads to contradiction (x 0) 10 89 7. a. True b. Consistent with x 0, y 0 4. Each number is the sum of the two previous 8. a. False b. Leads to contradiction (y 0) numbers. This is the Fibonacci sequence. 270 Enrichment Activity 7-6: Area of Polygons 6. 69, 260, 269 or 69, 92, 115 (Other answers are possible.) 1. a. 60 cm2 2. a. 480 cm2 3. a. 88 sq in. 7. a. Yes; 6, 8, 10 b. 1r(ED) 5 b. No. If a number is odd, its square is odd. The 2 b. 4r(ED) b. 2 r(YZ) sum of the squares of two odd numbers is even. 4. 1nsr 2 Therefore, the third number must be even. 5. a. 9"3 sq in. b. 96"3 cm2 c. 8 m2 c. Yes; 3, 4, 5 d. 28,500 cm2 e. 800 mm2 d. No. If either a or b is odd and the other even, then c must be odd. If a and b are both even, Enrichment Activity 8-1: Pythagorean Triples then c must be even. Therefore, it is not 1. a. U V U2 V2 2UV U2 V2 possible to have exactly one odd number in a Pythagorean triple. 2 1 3 4 5 3 1 8 6 10 Enrichment Activity 8-3: Polytans 4 1 15 8 17 1. 5 1 24 10 26 6 1 35 12 37 4 + √2 2 + 3√2 2 + 3√2 2 + 3√2 2. 4 1 "2, 2 1 3"2 b. The difference between U 2 V 2 and U 2 V 2 is 2. The sum of U 2 V 2 and U 2 V 2 is U times 3. 2U [5 3 2(4)]. The sum of the three num- bers is twice the product of U and the next con- secutive integer [3 4 5 2(2 3)]. Other relationships are possible. 4√2 6 4 + 2√2 4 + 2√2 4 + 2√2 2 2 2 2 2. a. U V U V 2UV U V 3 2 5 12 13 4 3 7 24 25 4 + 2√2 4 + 2√2 4 + 2√2 4 + 2√2 5 4 9 40 41 6 5 11 60 61 7 6 13 84 85 b. The difference between U2 V2 and 2UV is 1. The sum of U2 V2 and 2UV is the square of 4 + 4√2 4 + 4√2 4 + 4√2 4 + 4√2 4 + 4√2 4. 4"2, 6, 4 1 2"2, 2 1 4"2 U2 V2. The sum of the three numbers is the product of U2 V2 and U2 V2 1 [5 12 13 5 6]. The smallest number, U2 V2, is Enrichment Activity 8-6: Trigonometric the sum of U and V. Other relationships are Identities possible. 1. a. 1 b. 1 c. 1 3. a. U V U2 V2 2UV U2 V2 d. 1 e. 1 f. 1 4 2 12 16 20 2. tan A tan (90° A) 1, where 0° A 90° a 5 3 16 30 34 3. tan A 5 b b 6 4 20 48 52 tan B 5 a a tan A 3 tan B 5 b 3 b 5 1 a 7 5 24 70 74 B 90° A 8 6 28 96 100 tan A tan (90° A) 1 b. The difference between U2 V2 and 2UV is 4. 4. a. 1 b. 1 c. 1 The sum of U2 V2 and U2 V2 is twice U2 d. 1 e. 1 f. 1 [20 12 2(42)]. The sum of U2 V2 and 5. (sin A)2 (cos A)2 1 6. sin A 5 a, (sin A) 2 5 A a B 2 5 a2 2 2UV is the square of half the smallest number, c c c cos A 5 b, (cos A) 2 5 A b B 2 5 b2 U2 V2 [20 16 (12 2)2]. Other relation- 2 ships are possible. c c c 2 2 2 2 2 4. 10 and 7 (sin A) 2 1 (cos A) 2 5 a2 1 b2 5 a 1 b 5 c2 5 1 c c c2 c 5. 33, 56, 65 or 42, 56, 70 (Other answers are possible.) (Note the use of the Pythagorean Theorem.) 271 sin 7. a. cos 208 5 tan 208 208 sin b. cos 428 5 tan 428 428 11. z sin c. cos 788 5 tan 788 788 sin d. cos 348 5 tan 348 348 sin e. cos 158 5 tan 158 158 sin f. cos 78 5 tan 78 78 (0, 0, 4) sin 8. cos A 5 tan A, where 0° A 90° A 9. sin A 5 ac cos A 5 bc (0, –4, 0) 1 a tan A 5 b y sin A a b a c a 1 cos A 5 c 4 c 5 c 3 b 5 b 5 tan A (2, 0, 0) 1 Enrichment Activity 9-3: Graphing with Three Variables 1–4. z x (2, –5, 7) (3, 2, 6) Enrichment Activity 9-7: Iterating Linear Functions 1 1. y 0.4x 6 y Starting Value 5 Starting Value 15 1 8 12 1 (5, –4, 1) 9.2 10.8 9.68 10.32 9.872 10.128 (4, 2, –3) x 9.9488 10.0512 9.97952 10.02048 2. The iterates are approaching 10; for the starting 5. Octant 6 6. Octant 5 7. Octant 4 value 5, the iterates approach 10 from below; 8. Octant 8 9. Answers will vary; (0, y, 0) for the starting value 15, the iterates approach 10. z 10 from above. 3. All the y-values (outputs) are exactly 10. 4. a. 3 b. 4 c. 0 d. 2 (0, 0, 4) 5. y x b 6. a. x 5 1 2 m b. Undefined for m 1 since the denominator 1 would be 0. Functions of the form y x b y (b 0) cannot have a fixed point because the 1 (0, 3, 0) y-value can never be the same as the x-value 1 if a nonzero number is being added to the x-value. 7. a. Starting point 0: 2, 6, 14, 30, 62, 126 (6, 0, 0) Starting point 4: 6, 10, 18, 34, 66, 130 b. No x c. Here, m 0; in question 1, 0 m 1. 272 In 8–11, answers will vary. c. y 40 8. No fixed point; Starting point 0: 3, 6, 9, 12, 15, 18 36 Each iterate is b more than the previous. 32 Cost of parking 9. Fixed point 100; 28 Starting point 101: 100.8, 100.64, 100.512, 24 100.4096, 100.32768, 100.262144 20 Starting point 99: 99.2, 99.36, 99.488, 99.5904, 16 99.67232, 99.737856 Iterates approach the fixed point. 12 10. Fixed point 3.5; 8 Starting point 4.5: 6.5, 12.5, 30.5, 84.5, 246.5, 4 732.5 x Starting point 2.5: 0.5, 5.5, 23.5, 77.5, 2 4 6 8 10 12 14 16 18 20 22 24 0 239.5, 725.5 Number of hours Iterates move away from the fixed point d. From 1 to 19 hours, the graph would be a 11. Fixed point 0; straight line. Starting point 1: 6, 36, 216, 1,296, 7,776, 46,656 y Starting point 1: 6; 36; 216; 1,296; 40 7,776; 46,656 36 Iterates move away from the fixed point. 32 Cost of parking 28 Enrichment Activity 9-10: Graphing Step Functions 24 a. 20 16 Hours Cost Hours Cost Hours Cost 12 1 4 $4.00 2 $6.00 173 4 $38.00 8 2 3 $4.00 21 4 $8.00 18 $38.00 4 x 1 $4.00 6 $14.00 19 $40.00 0 2 4 6 8 10 12 14 16 18 20 22 24 11 2 $6.00 121 2 $28.00 201 3 $40.00 Number of hours b. Time Time Time Time Enrichment Activity 9-11: Holes, Holes, and In Out Cost In Out Cost More Holes: An Exponential Investigation Part I 9:15 A.M. 9:50 A.M. $4.00 10:00 A.M. 2:30 P.M. $12.00 Task 1 9:30 A.M. 10:29 A.M. $4.00 10:10 A.M. 10:00 P.M. $26.00 # of folds 0 1 2 3 4 5 9:30 A.M. 10:35 A.M. $6.00 12:15 A.M. 8:00 A.M. $40.00 # of holes 1 2 4 8 16 32 10:00 A.M. 12:45 A.M. $8.00 12:40 A.M. 11:00 A.M. $40.00 # of holes expressed 20 21 22 23 24 25 as a power of 2 a. The total number of holes doubles with each fold, or the total number of holes is a power of 2, the power being the number of folds. b. 2n c. H 2n Task 2 # of folds 0 1 2 3 4 5 # of holes 2 4 8 16 32 64 # of holes expressed 21 22 23 24 25 26 as a power of 2 273 a. The pattern is similar but begins with 21 rather 5. The function is maximized at (6, 2). The than 20. maximum profit of $18 is obtained when 6 b. 21 2 20, 22 2 21, 23 2 22, 24 2 23, bracelets and 2 necklaces are produced. 25 2 24, 26 2 25 Exercises c. H 2 2n 1. Max 20 at (4, 4), min 0 at (0, 0) Part II 2. Max 68 at (12, 4), min 14 at (2, 2) a. H 3. a. S 30c 40t 84 b. 2c 4t 800, c t 300, c 0, t 0 78 c. y 72 66 400 60 54 48 300 H = 2n 42 36 x + y = 300 (0, 200) 30 24 (200, 100) 18 100 2x + 4y = 800 12 6 n 0 1 2 3 4 5 (0, 0) 100 200 (300, 0) 400 500 x d. (0, 0), (0, 200), (200, 100), (300, 0) Number of folds e. At (0, 0), S 0; at (0, 200), S 8,000; at (200, 100), S 10,000; at (300, 0), S 9,000 Part III f. The function is maximized at (200, 100). The Answers will vary. Example: Yes. It is appropriate maximum sales of $10,000 is obtained when because as n, the number of fold, increases, H, 200 chairs and 100 tables are produced. the number of holes punched, increases rapidly (or exponentially). Enrichment Activity 11-1: Finding Primes—The Sieve of Eratosthenes Enrichment Activity 10-5: Solving Systems 1. 17 Using Matrices 2. Composite numbers can be written as pairs 1. x 2, y 1 of factors, such that when the number divided 2. x 1, y 6 by a factor is less than the factor, all positive 3. x 4, y 1 integral factors have been found. In this case, 4. x 3, y 5 the greatest number, 200, divided by 15 yields 5. a. Error message results. 131 . Therefore, all composite numbers between 3 b. There is no solution to the system. 15 and 200 that have a factor greater than 15 must also have a factor less than 15 and have Enrichment Activity 10-6: Systems with already been crossed out. Three Variables 3. 19, the largest prime less than 20 1. (3, 5, 2) 2. (1, 2, 2) 4. a. x 1, 12 1 41 41 3. (2, 1, 2) 4. (4, 3, 0.5) x 2, 22 2 41 43 5. (10, 3, 7) 6. ( 6, 4, 2) x 3, 32 3 41 47 x 4, 42 4 41 53 Enrichment Activity 10-8: Linear x 5, 52 5 41 61 Programming x 6, 62 6 41 71 Example x 7, 72 7 41 83 3. B (0, 5), C (6, 2), D (7, 0) x 8, 82 8 41 97 4. At B (0, 5), 2x 3y 15 x 9, 92 9 41 113 At C (6, 2), 2x 3y 18 x 10, 102 10 41 131 At D (7, 0), 2x 3y 14 b. x 41, 412 41 41 1,681 41 41 274 Enrichment Activity 11-5: Differences of 2. Square Equation Relating Sides Squares a — 1. a. 92 82 b. 72 52 c. 302 292 d. 16 2 14 2 e. 36 2 35 2 f. 262 242 b "8 5 2"2 g. 512 502 h. 732 722 2 2 2 2 i. 492 472 c "18 5 3"2 j. 152 151 k. 97 95 l. 5032 5022 2n 2 1 1 2. 2 5 n 2 2 , which is between n 1 and n, d "32 5 4"2 so using the rule, 2n 1 can be written as e "50 5 5"2 n2 (n 1)2. To check, n2 (n 1)2 3. n2 (n2 2n 1) 2n 1. 3. 4n 4 n, which is between n 1 and n 1, Side (square Equation so using the rule, 4n can be written as root of Side (decimal Relating (n 1)2 (n 1)2. To check, (n 1)2 (n 1)2 Square Area area) approximation) Sides n2 2n 1 (n2 2n 1) 4n. f 5 "5 2.236067977 — "20 "20 5 2"5 Enrichment Activity 11-7: Factoring g 20 4.472135955 Trinomials 1. (1) 4x2( 10) 40x2 h 45 "45 6.708203932 "45 5 3"5 (2) 8 and 5 (3) 4x2 8x 5x 10 Enrichment Activity 12-7: Operations with (4) 4x(x 2) 5(x 2) Radicals (5) (x 2)(4x 5) 1. a. 3"2 b. 8"11 c. 3"2 d. 6"3 1 3"5 2. a. (x 2)(2x 5) b. (x 3)(4x 3) e. 2"10 f. 5 2 2"3 c. (x 5)(3x 1) d. (x 4)(3x 2) e. (2x 3)(3x 2) f. (3x 4)(4x 1) g. (x 5)(8x 3) h. (x 5)(5x 2) g. 0 h. 2"5 i. (x 3)(7x 1) i. 0 j. 5 2 2"3 k. 10"6 l. 4 1 2"3 m. 11"7 n. 6"3 1 3"5 Enrichment Activity 12-2: Square Root: Divide and Average 1. 7.07 2. 6.32 3. 2.45 o. 11"7 p. 3 4. 9.38 5. 10.39 6. 24.78 q. 10"6 r. 2"5 7. 3.32 8. 23 9. 24.78 s. "5 t. 4 1 2"3 u. 8"11 v. 23"5 10. 0.77 11. 0.55 12. 50.25 13. "529 is rational because 232 x. 23"5 529. w. 3 y. 5"2 14. 3.162 15. 10.100 Enrichment Activity 12-4: Equivalent 2. Common Common Radical Expressions Matches Answer Matches Answer In 1 and 3, answers will vary depending on the number a, c 3"2 k, q 10"6 of decimal places displayed on the calculator. b, u 8"11 l, t 4 1 2"3 1. Side (square Side (decimal Square Area root of area) approximation) d, n 6"3 1 3"5 m, o 11"7 a 2 "2 1.414213562 f, j 5 2 2"3 p, w 3 b 8 "8 2.828427125 g, i 0 v, x 23"5 c 18 "18 4.242640687 h, r 2"5 d 32 "32 5.6565854249 3. a. 100 e 50 "50 7.071067812 b. Yes 275 Enrichment Activity 13-2: Carpet Squares 5. The graphs for squares with fringe on two sides 1. 4 with fringe on two sides, 8 with fringe on one and one side are linear. The graph for squares side, 4 with no fringe. with no fringe is steepest. 6. a 4, b 4(x 2), c (x 2)2 7. 4 with fringe on two sides, 192 with fringe on one side, 2,304 with no fringe. 8. The formulas are graphed in Exercise 4 for the domain of 2 x 10, where x is an integer. In general, the formulas hold true for the domain x 2. The graph of c is quadratic. 2. 9. No. For the number of squares to be equal, 4(x 2) (x 2)2. Simplifying gives x2 8x Dimensions Fringe on Fringe on No Total 12 0, which factors to (x 6)(x 2) 0. The of Carpet Two Sides One Side Fringe Squares only solutions for x are 2 and 6. 2 ft by 2 ft 4 0 0 4 10. (n 1)2 n2 2n 3 3 ft by 3 ft 4 4 1 9 4 ft by 4 ft 4 8 4 16 Enrichment Activity 13-4: Quadratic Inequalities 5 ft by 5 ft 4 12 9 25 1. y 6 ft by 6 ft 4 16 16 36 7 ft by 7 ft 4 20 25 49 8 ft by 8 ft 4 24 36 64 9 ft by 9 ft 4 28 49 81 10 ft by 10 ft 4 32 64 100 1 3. The number of squares with fringe on two sides is 4 O x –1 1 for any size because all squares have 4 corners. The number of squares with fringe on one side starts at 0 and increases by 4 each time the length of the side increases by 1 foot. The number of squares 2. (0, 0) is not in the region, (0, 3) is in the region with no fringe starts at 0 and is the sequence of 3. See graph square numbers. The total number of squares is 4. All points inside of the graph of y x2 6x 8 also the square numbers starting with 4. 5. 2 x 4 4. y 6. y 64 60 Number of each type of square 1 56 O x 52 –1 –1 1 48 44 40 36 32 28 24 20 16 12 8 4 * * * * * * * * * x 0 1 2 3 4 5 6 7 8 9 10 Length of side of square 276 7. All points outside the graph of y x2 x 12 2. Student-generated examples. a1c 8. x 3 or x 4 3. No. Cite any one example to show that b 1 d is a c not the average of b and d . Enrichment Activity 14-2: Is It Magic or Is 4. a. 3 , 3 1 9 , 9, 3 , 12 , 9, 0.75 , 1.5 , 2.25 4 414 4 4 8 4 It Math? 6 b. 1 , 1 1 5 , 5, 1 , 12 , 5, 0.16 , 0.5 , 0.83 6 616 6 6 6 1. a. 3 S 5 S 13 S 13 S 21 S 34 ; yes 5 8 8 21 34 55 c. 2 , 9 1 10 , 10, 9 , 19 , 10, 4.5 , 4.75 , 5 9 212 2 2 4 2 13 b. 13 S 15 S 15 S 28 S 43 S 114 ; yes 2 28 43 71 71 5. a. If b , b, then b , a 2b c , b (a 0, b 0, c 0) a c a 1 c 2. Student-generated results. Each result should a c b. Yes. The average of b and b is equal to equal 0.6 when rounded to the nearest tenth. a1c a1c x y x1y x 1 2y 2x 1 3y 3x 1 5y b 4 2 5 2b . 3. y S x 1 y S x 1 2y S 2x 1 3y S 3x 1 5y S 5x 1 8y 3x 3x 1 5y 5y 3 3x 1 5y 5 Enrichment Activity 14-8: Estimating 4. 5x , 5x 1 8y , 8y is equivalent to 5 , 5x 1 8y , 8 3x 1 5y Solutions to Quadratic Equations or 0.6 , 5x 1 8y , 0.625 1. a. x 1.32 or 1.33 b. x 1.316 or 1.317 3x 1 5y Therefore, 5x 1 8y 5 0.6 to the nearest tenth. 2. y 2; x 5.32 or 5.33 (to the nearest tenth), 5. a. 0.62 x 5.316 or 5.317 (to the nearest hundredth) b. The seventh term in the sequence beginning 3. Calculator check. 4. x 1.45, x 3.45 x 5x 1 8y 5. x 6.54, x 0.46 6. x 1.09, x 10.09 with y (x . 0, y . 0) is 8x 1 13y . Apply this 7. x 2.27, x 5.73 algebraic fraction to the rule illustrated 8. The product of x and x 4 equals 7, not each in Exercise 4, placing terms in a correct factor. If each factor were to equal 7, then their 8y 8y 1 5x 5x product would be 49. order, to discover: 13y , 13y 1 8x , 8x , 8 8y 1 5x 5 Enrichment Activity 15-3: Probability and which is equivalent to 13 , 13y 1 8x , 8 or 8y 1 5x a Digital Clock 0.615384615 13y 1 8x 0.625. Therefore, 1. Answers will vary; student guesses 8y 1 5x 450 5 13y 1 8x 0.62 to the nearest hundredth. 2. a. P(0) 1,440 5 16 .3125 720 b. P(1) 1,440 5 12 .5 Enrichment Activity 14-5: Operations with 540 Fractions c. P(2) 1,440 53 8 .375 450 5 A B A B A B A B d. P(3) 1,440 5 16 .3125 x1w x2w xw x 450 5 1. y y w e. P(4) 1,440 5 16 .3125 y2 450 5 x 1 wy x 2 wy xw x f. P(5) 1,440 5 16 .3125 2. y y y wy 252 7 g. P(6) 1,440 5 40 .175 3. (2) 4. (4) 5. (4) 252 7 6. (2) 7. 1 h. P(7) 1,440 5 40 .175 8. If the sum and the difference in Exercise 1 are 252 7 i. P(8) 1,440 5 40 .175 equal, then w 0. This would contradict the 252 7 j. P(9) 1,440 5 40 .175 given statement, w 0. 3. The digit appearing most often on a digital clock 9. No. If x 1 and y 1, then x 1 w 5 xw becomes 2 y y is 1, and the digit appearing the second most 1 w w (an impossible statement), often is 2. The digits 0, 3, 4, and 5 appear the same or 1 0 (also impossible). number of times over a 24-hour period. The digits 10. w 2 11. x 5 5 4 12. y 5 2 3 6, 7, 8, and 9 also appear the same number of times over a 24-hour period, and these digits Enrichment Activity 14-7: Fractions appear less than all the others. Between Fractions 4. The answers will be exactly the same for any 1. a. 2 , 2 1 4 , 4, 2 , 6 , 4, 0.6 , 0.75 , 0.8 3 315 5 3 8 5 12-hour period as for the 24-hour period discussed earlier because, in a 24-hour period, a 12-hour 9 b. 14 , 9 1 14 , 14, 9 , 23 , 14, 2.25 , 2.5 , 2.8 415 5 4 9 55 cycle is repeated twice. Doubling (or halving) the c. 7 , 7 1 19 , 19, 7 , 26 , 19, 8 8 1 20 20 8 28 20 numerator and the denominator of the probability 0.875 0.9285714… 0.95 fraction does not change the value of the fraction. 277 Enrichment Activity 15-6: Probability and Enrichment Activity 15-7: Probability on Area: And, Or, Not a Dartboard 7 1. a. P(D) 25 .28 1. a. 18 b. 3 (from 4 ) b. P(not D) 25 .72 7 2. a. P(D) 25 .28 c. 5 (from 9 4 ) 18 d. 7 (from 16 9 ) b. P(not D) 25 .72 7 e. 9 (from 25 16 ) 3. a. P(D) 25 .28 1 3 18 2. a. 25 .04 b. 25 .12 b. P(not D) 25 .72 5 7 c. 25 5 1 .2 5 d. 25 .28 4. Answers will vary. Example: The areas of the 9 three shaded regions in Exercises 1–3 are equal; e. 25 .36 10 each is 7 square units. The probability that a 3. a. 25 5 2 5 .4 14 b. 25 .56 point on the 5-by-5 grid lies in the shaded region 15 c. 25 5 3 .6 d. 24 .96 7 5 25 is 25 for each region. 9 e. 25 .36 5. a. A and B intersect in two squares. Example: 4 9 4. a. 25 .16 b. 25 .36 A c. 1 625 .0016 d. 576 625 .9216 B e. 81 .1296 256 f. 625 .4096 625 49 81 g. 625 .0784 h. 625 .1296 Enrichment Activity 15-11: Expectation 1. a. Responses will vary. b. P(A or B) 10 5 2 .4 25 5 b. Most should say “left” or “negative.” 2. E(game) 5 A 1 B 3 (13) 1 A 5 B 3 (21) 6. a. A and B intersect in three squares. Example: 6 6 3 25 5 61 6 22 21 A 5 6 5 3 3. The expectation of 21 indicates that, on average, 3 the player expects to move 1 place in a negative B direction for every 3 turns taken. 9 4. a. E(game) 5 A 1 B 3 (15) 1 A 5 B 3 (21) 5 0 6 6 b. P(A or B) 25 .36 b. The expectation of 0 indicates that, on 7. a. B is a subset of A. A and B intersect in four average, a player does not expect to advance squares. Example: or fall behind over a long period of play. A After many turns, the player should still be in the START box, although the marker may have moved to the left and to the right during different turns. 5. Answers will vary. To win on this board in an average of 24 turns, a player must advance 8 spaces to the right in that time. Therefore, B E(game) should equal 18 , or 11 . Here is one 8 24 3 b. P(A or B) 25 .32 possible set of rules to obtain E(game) 1: 3 8. P(A or B) is found by either: (1) If a player guesses correctly, move 7 (1) counting the number of square units in the spaces in the positive direction (to the 5-by-5 grid that are shaded by A, by B, or right). by both, and then dividing this area by 25, (2) If a player guesses incorrectly, move 1 which is the area of the grid; or space in the negative direction (to the (2) using the formula P(A or B) P(A) P(B) left). P(A and B) Thus: E(game) A 1 B 3 (17) 1 A 5 B 3 (21) 5 2 5 1 9. No. P(A and B) cannot exceed P(A), and since 4 6 6 6 3 P(A) 25 .16, the statement P(A) .16 is false. 278 Enrichment Activity 15-12: A “Pick Six” Enrichment Activity 16-7: The Lottery Median-Median Line C ? C 1 ? 1 1 1. Yes 1. a. 6 6 C48 0 5 25,827,165 5 25,827,165 54 6 b. .0000000387 100 6C5 ? 48C1 6 ? 48 288 2. a. 5 5 95 54C6 25,827,165 25,827,165 b. .0000112 90 C ? C 15 ? 1,128 16,920 3. a. 6 4 C48 2 5 25,827,165 5 25,827,165 85 54 6 Posttest b. .000655 80 C ? C ? 47C2 20 ? 1 ? 1,081 21,620 75 4. a. 6 3 1 C 1 5 25,827,165 5 25,827,165 54 6 b. .000837 70 5. a. Add the solutions to Exercises 1–4 part a: 65 38,829 60 25,827,165 b. Add solutions to Exercises 1–4 part b: .0015 55 6. Yes, the claim is essentially correct: 50 1 333 5 .003 By choosing two sets of 6 numbers, the probabil- 0 50 55 60 65 70 75 80 85 90 95 100 ity of winning a prize is doubled, so we multiply Pretest the probability found in Exercise 5 by two: 2. See graph 3. (58, 71); see graph 2 .0015 .003, which is close to the given 4. (69, 84); see graph 5. (85, 91); see graph probability. 6. 100 Enrichment Activity 16-1:Taking a Survey: 95 Designing a Statistical Study + 90 In 1–4, student responses will vary. 85 + Posttest Enrichment Activity 16-4: Locating the 80 Median Value 75 a. 4th 70 + b. 5th 65 c. 8th d. 20th 60 e. 41st 55 f. 119th 50 N11 g. 2 h. 4th and 5th 0 50 55 60 65 70 75 80 85 90 95 100 i. 5th and 6th Pretest j. 9th and 10th In 7 and 8, answers will vary but should be close to k. 21st and 22nd those given. l. 44th and 45th 7. Using the points (55, 70) and (75, 85), m. 115th and 116th y 0.75x 28.75 N N n. 2 and 2 1 1 8. a. 0.75 b. 28.75 General rule: Arrange the data values in order. Let the 9. a. 82 b. 74 c. 92 or 93 number of data values be N. 10. The regression equation (values rounded to the If N is odd, the median is the value that is N 2 1 from 1 nearest hundredth) is y 0.75x 30.24, which is either end. close to the median-median line. If N is even, the median is the average of the values 11. Student results will vary. The median-median line that are N and N 1 1 from either end. 2 2 should be close to the regression line. 279