# p. 1

teacher s manual with answer key integrated algebra 1 ann xavier gantert amsco amsco school publications inc 315 hudson street new york n.y 10013

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# p. 2

contents answer keys for enrichment activities for suggested test items for sat preparation exercises for textbook exercises chapter 1 chapter 2 chapter 3 chapter 4 chapter 5 chapter 6 chapter 7 chapter 8 chapter 9 chapter 10 chapter 11 chapter 12 chapter 13 chapter 14 chapter 15 chapter 16 290 293 299 303 308 312 315 321 324 345 355 358 362 371 375 384 264 280 288

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# p. 3

answers for enrichment activity exercises enrichment activity 1-1 guessing a number by bisection a 8 b 6 c a maximum of 11 guesses would be needed to locate a number between 1 and 2,000 the first guess would be the average of 0 and 2,000 or 1,000 assuming that the number itself was not 1,000 this would divide the possible numbers into two equal sets numbers between 0 and 999 and numbers between 1,001 and 2,000 sets of this size require a maximum of 10 guesses to find the number use a simpler related problem to find a general solution range max number of numbers of guesses from 1 to 3 2 first guess would be 2 if 2 is not the number the next guess would be correct first guess would be 4 if 4 is not the number the number must be one of a set of three possible numbers and at most two more guesses are needed first guess would be 8 if 8 is not the number the number must be one of a set of seven possible numbers and at most three more guesses are needed powers of 2 22 23 24 25 26 27 28 4 8 16 32 64 128 256 to find a word in the dictionary choose a middle word alphabetically and ask if the chosen word comes before or after the word for which you are searching continue the process to find a name on a list choose a middle name and continue the process enrichment activity 1-2 repeating decimals 1 1 15 3 5 7 12 47 111 2 41 99 4 10 33 6 322 or 325 99 99 8 155 198 10 120 7 1 9 15 enrichment activity 1-3 a piece of pi 1 0 11 1 5 2 12 3 11 4 10 5 8 6 9 7 8 8 12 9 14 2 no the digit 9 appears most often while the digit 1 appears least often 3 0 12 1 11 2 12 3 8 4 12 5 12 6 7 7 4 8 13 9 9 4 answers will vary example the digits 0 1 4 5 and 8 appear more often than in the first group the digits 3 6 7 and 9 appear less often than in the first group and the digit 8 appears with same frequency as in the first group 5 0 23 1 16 2 24 3 19 4 22 5 20 6 16 7 12 8 25 9 23 6 answers will vary examples the digit 7 appears significantly less frequently than the other digits appearances of the digits 0 2 3 4 5 8 and 9 are evening out as more digits in the expansion are checked the digits may seem to appear the same number of times from 1 to 7 3 from 1 to 15 4 range max number of numbers of guesses from 1 to 3 from 1 to 7 from 1 to 15 from 1 to 31 from 1 to 63 from 1 to 127 from 1 to 255 2 3 4 5 6 7 8 enrichment activity 1-4 making 100 1 96 537 1,428 3 96 357 5 91 836 3,546 7 82 197 5,643 7,524 2,148 2 96 438 1,578 4 94 263 6 91 647 7,524 8 81 396 69,258 5,823 1,752 if 2n 1 the number of integers in the set from which the chosen number is selected n is the maximum number of guesses needed to identify the number 9 81 297 10 3 714 5,472 11 answers will vary examples 13 5 9 1,368 16 5 12 894 40 5 27 396 3,576 5,148 264

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# p. 4

enrichment activity 2-1 special ops 7 2 5 6.2 5 20 0.4 8 7 5 8 11 1 4 a a b a 2b b 4 13 answers will vary 1 4 7 10 12 3 6 9 23 5 8 5 8 3 z 0 2 4 6 8 0 0 0 0 0 0 2 0 4 8 2 6 4 0 8 6 4 2 6 0 2 4 6 8 8 0 6 2 8 4 enrichment activity 2-2a clock arithmetic a 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 2 3 4 5 6 7 8 3 4 5 6 7 8 4 5 6 7 8 5 6 7 8 6 7 8 7 8 8 9 10 11 12 2 3 4 5 6 7 8 9 9 10 11 12 1 2 3 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 6 7 8 4 a yes b yes c yes d 6 e number z inverse identity 2 z 8 6 4 z 6 6 6 z 4 6 8 z 2 6 5 yes conclusion this is a field all 11 properties hold 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 enrichment activity 2-5 paying the toll 17 111 13 17 111 23 17 111 33 17 2 3 3 3 3 1 3 2 3 6 7 1 3 9 13 10 8 4 5 1 2 111 43 17 111 53 17 111 111 17 27 11 12 23 13 15 33 23 17 43 23 16 17 18 19 20 13 14 111 211 27 83 211 311 93 47 211 37 17 22 33 24 13 1 7 27 28 17 33 29 30 25 26 23 13 21 9 10 11 12 1 2 3 4 5 6 9 10 11 12 1 2 3 4 5 9 10 11 12 1 2 3 4 9 10 11 12 1 10 11 12 1 11 12 1 12 1 b 1 2 3 4 5 2 2 3 9 10 9 10 11 9 10 11 12 yes yes yes 12 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 all purchases can be made since any whole number can be formed using multiples of the numbers above enrichment activity 2-8 graphing ordered pairs of numbers state street enrichment activity 2-2b digital addition digital multiplication 1 0 2 4 6 8 0 0 2 4 6 8 2 2 4 6 8 0 4 4 6 8 0 2 6 6 8 0 2 4 8 8 0 2 4 6 a hagds main street lcvompf 2 a yes b yes c yes d 0 e number inverse identity 0 0 0 2 8 0 4 6 0 6 4 0 8 2 0 b m is two blocks west and one block south of o 2 1 c the first number gives the number of blocks east or west of o the center of town if the number is positive the position is east of o if the number is negative the position is west of o the second number gives the number of blocks north or 265

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# p. 5

south of o if the number is positive the position is north of o if the number is negative the position is south of o 4 6 8 9 16x 13,600 5 x 150 x 300 x 450 16x 20,400 7 33x 34,000 33x 34,000 65,000 $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 and 4 are shown below col 1 1 2 3 4 5 6 7 8 9 15 3 38 3,800 3,600 360 400 20 15 col 4 n 2n 2n 8 200n 800 200n 600 20n 60 20n 100 n 5 n enrichment activity 4-3 consecutive integers 1 number of integers 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 first integer 1 4 7 x 1 4 7 x 1 4 7 x 1 4 7 x 1 4 7 x 7x 6x 28 49 70 21 5x 21 39 57 15 sum 6 15 24 3x 10 22 34 4x 15 30 45 10 6 3 even or odd even odd even even or odd even even even even odd even odd even or odd odd odd odd odd even odd even even or odd conclusion the result on line 9 will always equal the starting number on line 1 enrichment activity 3-2 variable codes 1 a paoik b zusuxxuc c sgznksgzoiy d kdzxguxjotgxe 2 answers will vary 3 i must study for an exam 4 a xivrk b glqqcv c rjjzxedvek d vogcfirkzfe 5 x 9 6 nzixs mlevnyvi the correspondence is symmetric if a line is drawn between m and n in the alphabet list the corresponding letters are equal distances from that line a corresponds to z b corresponds to y and so on 7 a mlbncfh b tuesday the result is the original word 7 x is the encoding and decoding expression 8 a ykbwtr b rdupmk no enrichment activity 3-7 formulas for health 1 4 7 10 13 22 31 24 114 bpm 19 bpm 2 5 8 11 14 24 26 34 19 bpm 84 3 6 9 12 17 22 bmi 5 703w 2 h 23 bpm enrichment activity 4-2 book value 1 let x represent the middle book because the values of the other books can be expressed in relation to it 2 16 books 3 x 100 x 200 x 300 2 let x be the first consecutive integer a odd 2x 1 is the sum of an even number and an odd number b even when x is odd 3x 3 is the sum of two odd numbers odd when x is even 3x 3 is the sum of an even number and an odd number c even 4x 6 is the sum of two even numbers d even when x is even 5x 10 is the sum of two even numbers odd when x is odd 5x 10 is the sum of an odd number and an even number e odd 6x 15 is the sum of an even number and an odd number f even when x is odd 7x 21 is the sum of two odd numbers odd when x is even 7x 21 is the sum of an even number and an odd number 266

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# p. 6

3 4 5 6 g even when the number of integers divided by 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 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 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 a no b the sum is 6 more than 4 times the first integer the sum is twice the sum of the second and third integers the sum is 6 less than 4 times the last integer a student tables b the sum of even integers is always even 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 2 x 4 3 x 3 4 all real numbers between 4 and 4 4 4 5 enrichment activity 4-4 law of the lever 1 2 3 4 5 2.4 ft 15 lb 25 lb heavier carton 9 ft lighter carton 12 ft kelly 49 lb laurie 70 lb a the side with the 32-lb weight b 3 ft 6 all real numbers between 1 and 1 1 1 enrichment activity 4-7 graphing an inequality 1 x 4 all real numbers greater than 1 1 since there is no real number that is the square root of a negative number the inequality is meaningless for x 0 267

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# p. 7

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 157 159 161 163 165 167 169 171 173 175 179 181 183 5 row number 1 2 3 4 5 6 7 sum of numbers in row 1 8 27 64 125 216 343 8 all positive real numbers 0 9 since x is always positive there is no real number that makes this inequality true 6 sum of the numbers in row n n3 7 a 1,728 b row 19 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 10 a row number 1 2 3 4 5 6 sum of all numbers 1 9 36 100 225 441 784 sum written as a square 12 32 62 102 152 212 282 enrichment activity 5-1 sums and squares 1 42 4 5 25 52 2 n n 1 n2 1 n n 2 n n 2 n 2n 2n 2n 2n 1 n 2 2n 1 2 1 1 5 n 1 5 n 1 n2 1 n n+1 400 n2 n2 n2 1 n 1 2n 1 12 b c nn 1 1 2 d 2 7 3 202 20 21 4 102 100 5 n 12 enrichment activity 5-4 products sums and cubes 1 34 5 4 60 4 64 43 45 6 5 120 5 125 53 56 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 20 21 441 268

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# p. 8

3 67 8 7 336 4 78 9 8 504 5 n 1 n n 1 7 8 n 343 73 512 83 nn 1 n 1 n n2 n n 1 n n3 n2 n2 n n n3 1 5 n 1 5 n 1 5 n 1 n3 enrichment activity 6-1b ratio estimates and comparisons columns 1 and 2 will vary according to student estimates column 3 actual ratio of size 1 australia brazil canada china india japan mexico russia spain united states bonus australia brazil canada china india japan mexico russia spain united states 0.80 0.88 1.04 1.00 0.34 0.04 0.20 1.77 0.05 1.00 actual ratio of size 3.90 4.32 5.06 4.87 1.67 0.19 1.00 8.66 0.26 4.88 source for teacher area sq mi 2,967,908 3,286,487 3,855,101 3,705,405 1,269,345 145,883 761,606 6,592,769 194,897 3,718,709 6 n n n n2 n 3 1 n 3n 3n n3 2 2 2 2n 3n2 n n n 3n 13 1 n 1 n 3n2 2 1 n 2n 1 3n 1 enrichment activity 6-1a fibonacci sequence and the golden ratio part 1 for clarity the first 26 terms of the fibonacci sequence are written below from smallest to largest in three-column format 1 1 2 3 5 8 13 21 34 part 2 13 8 34 21 89 55 233 144 610 377 1,597 987 4,181 2,584 10,946 6,765 28,657 17,711 75,025 46,368 55 89 144 233 377 610 987 1,597 2,584 4,181 6,765 10,946 17,711 28,657 46,368 75,025 121,393 5 1.625 5 1.619047 5 1.618 5 1.61805 1.618037135 1.618034448 1.618034056 1.618033999 1.61803399 1.618033989 21 13 55 34 144 89 377 233 987 610 2,584 1,597 6,765 4,181 17,711 10,946 46,368 28,657 121,393 75,025 5 1.615384 1.617647059 1.617977528 1.618025751 1.618032787 1.618033813 1.618033963 1.618033985 1.618033988 1.618033989 part 3 1 12 5 1.618033989 as the terms of the fibonacci sequence increase the ratio comparing the greater of two consecutive fibonacci numbers to the smaller approaches the value of the golden ratio in a and b student responses will vary a with mexico s area used as the base of comparison its ratio changes from 0.20 to 1.00 in turn the ratio for every nation becomes about 5 times as large as the ratio using the united states as the base of comparison b the ratios in both lists show the sizes of the countries in relation to one another japan has the smallest area followed by spain up to russia with the largest area 269

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# p. 9

enrichment activity 6-2 population density part 1 state land area population population sq mi density alaska california florida montana new jersey new york north carolina texas 648,818 35,484,453 17,019,068 917,621 8,638,396 19,190,115 8,407,248 22,118,509 571,951 155,959 53,927 145,552 7,414 47,214 48,711 261,797 1 228 316 6 1,165 406 173 84 9 a false b leads to the contradiction y a a1c c 10 b b 1 d d 0 enrichment activity 7-4 hero s formula 1 a 24 sq in b 24 sq in 2 a 431.6 m2 b 11,161.36 sq ft 3 a 388.8 cm2 b since the area and base are known substitute the values into the formula a 5 1bh b 18 cm 2 c right triangle enrichment activity 7-5 rectangle cover-up 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 70 35 2 a 43 steps b 5 steps 10 steps 25 c 43 25 d 43 25 evan is gaining on marisa 70 37.5 75 3 51 or 102 50 62.5 or 125 75 134 75 59 67 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 2 8 ways enrichment activity 6-4 ratios and inequalities 1 approach 1 proof abab b d ad ad approach 2 proof a d badb ad bd a b c abbd b bc bd ad bc 2 a ad bc b ad bc c d b d d bc 1 1 3 b b 1 1 must be true because 1b or b 1 b a 4 b 1 is true because a2 cdcd b d cb bc c d 3 rectangles of width 2 length of rectangle 0 1 2 3 4 number of ways to cover with dominoes 1 1 2 3 5 8 13 21 34 55 89 1 b1 5 6 7 8 9 10 5 6 7 8 a b a b is true because ab c ba c so ab ac ba bc and ac bc a false b leads to contradiction x 0 a true b consistent with x 0 y 0 a false b leads to contradiction y 0 b2 so a b and 4 each number is the sum of the two previous numbers this is the fibonacci sequence 270

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# p. 10

enrichment activity 7-6 area of polygons 1 a 60 cm2 b 1red 2 4 1nsr 2 2 a 480 cm2 b 4red b 96 3 cm2 e 800 mm2 3 a b 88 sq in 5 2 ryz 5 a 9 3 sq in d 28,500 cm2 c 8 m2 enrichment activity 8-1 pythagorean triples 1 a u 2 3 4 5 6 v 1 1 1 1 1 u2 3 8 15 24 35 v2 2uv 4 6 8 10 12 u2 5 10 17 26 37 v2 6 69 260 269 or 69 92 115 other answers are possible 7 a yes 6 8 10 b no if a number is odd its square is odd the sum of the squares of two odd numbers is even therefore the third number must be even c yes 3 4 5 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 then c must be even therefore it is not possible to have exactly one odd number in a pythagorean triple enrichment activity 8-3 polytans 1 4 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 2u [5 3 24 the sum of the three numbers is twice the product of u and the next consecutive integer [3 4 5 22 3 other relationships are possible 2 a u 3 4 5 6 7 v 2 3 4 5 6 u 2 2 4 1 2 2 1 3 2 3 2 32 2 32 2 32 42 6 4 22 4 22 4 22 v 5 7 9 11 13 2 2uv 12 24 40 60 84 u 2 v 13 25 41 61 85 2 4 22 4 22 4 22 4 22 b the difference between u2 v2 and 2uv is 1 the sum of u2 v2 and 2uv is the square of 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 the sum of u and v other relationships are possible 3 a u 4 5 6 7 8 v 2 3 4 5 6 u2 12 16 20 24 28 v2 2uv 16 30 48 70 96 u2 20 34 52 74 100 v2 4 4 2 6 4 1 2 2 2 1 4 2 4 42 4 42 4 42 4 42 4 42 enrichment activity 8-6 trigonometric identities 1 a 1 b 1 c 1 d 1 e 1 f 1 2 tan a tan 90° a 1 where 0° a a 3 tan a 5 b b tan b 5 a a tan a 3 tan b 5 b 3 b 5 1 a b 90° a tan a tan 90° a 1 4 a 1 b 1 c 1 d 1 e 1 f 1 5 sin a2 cos a2 1 2 6 sin a 5 a sin a 2 5 a a b 2 5 a2 c c c 2 cos a 5 b cos a 2 5 a b b 2 5 b2 c c c 2 2 2 2 2 90° b the difference between u2 v2 and 2uv is 4 the sum of u2 v2 and u2 v2 is twice u2 [20 12 242 the sum of u2 v2 and 2uv is the square of half the smallest number u2 v2 [20 16 12 22 other relationships are possible 4 10 and 7 5 33 56 65 or 42 56 70 other answers are possible sin a 2 1 cos a 2 5 a2 1 b2 5 a 1 b 5 c2 5 1 c c c2 c note the use of the pythagorean theorem 271

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# p. 11

sin 7 a cos 208 5 tan 208 208 sin c cos 788 5 tan 788 788 sin e cos 158 5 tan 158 158 sin b cos 428 5 tan 428 428 sin d cos 348 5 tan 348 348 sin f cos 78 5 tan 78 78 11 z sin 8 cos a 5 tan a where 0° a 90° a 9 sin a 5 a c cos a 5 b c a tan a 5 b sin aabaca cos a 5 c 4 c 5 c 3 b 5 b 5 tan a 0 0 4 0 4 0 2 0 0 1 1 1 y enrichment activity 9-3 graphing with three variables 14 2 5 7 3 2 6 z enrichment activity 9-7 iterating linear functions 1 1 5 4 1 1 1 y x y starting value 5 8 9.2 9.68 9.872 9.9488 9.97952 0.4x 6 starting value 15 12 10.8 10.32 10.128 10.0512 10.02048 x 4 2 3 5 octant 6 8 octant 8 10 6 octant 5 7 octant 4 9 answers will vary 0 y 0 z 0 0 4 1 1 1 0 3 0 y 6 0 0 x 2 the iterates are approaching 10 for the starting value 5 the iterates approach 10 from below for the starting value 15 the iterates approach 10 from above 3 all the y-values outputs are exactly 10 4 a 3 b 4 c 0 d 2 5 y x b 6 a x 5 1 2 m b undefined for m 1 since the denominator would be 0 functions of the form y x b b 0 cannot have a fixed point because the y-value can never be the same as the x-value if a nonzero number is being added to the x-value 7 a starting point 0 2 6 14 30 62 126 starting point 4 6 10 18 34 66 130 b no c here m 0 in question 1 0 m 1 272

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# p. 12

in 811 answers will vary 8 no fixed point starting point 0 3 6 9 12 15 18 each iterate is b more than the previous 9 fixed point 100 starting point 101 100.8 100.64 100.512 100.4096 100.32768 100.262144 starting point 99 99.2 99.36 99.488 99.5904 99.67232 99.737856 iterates approach the fixed point 10 fixed point 3.5 starting point 4.5 6.5 12.5 30.5 84.5 246.5 732.5 starting point 2.5 0.5 5.5 23.5 77.5 239.5 725.5 iterates move away from the fixed point 11 fixed point 0 starting point 1 6 36 216 1,296 7,776 46,656 starting point 1 6 36 216 1,296 7,776 46,656 iterates move away from the fixed point c y 40 36 32 28 24 20 16 12 8 4 0 cost of parking x enrichment activity 9-10 graphing step functions a hours 1 4 2 3 cost $4.00 $4.00 $4.00 $6.00 time out hours 2 21 4 6 121 2 cost $6.00 $8.00 $14.00 $28.00 time in hours 173 4 18 19 201 3 time out cost $38.00 $38.00 $40.00 $40.00 2 4 6 8 10 12 14 16 18 20 22 24 number of hours d from 1 to 19 hours the graph would be a straight line y 40 36 32 28 24 20 16 12 8 4 cost of parking 0 2 4 6 8 10 12 14 16 18 20 22 24 number of hours 1 11 2 b time in x cost $4.00 $4.00 $6.00 $8.00 cost $12.00 $26.00 enrichment activity 9-11 holes holes and more holes an exponential investigation part i task 1 of folds of holes 0 1 1 2 21 2 4 22 3 8 23 4 16 24 5 32 25 9:15 a.m 9:50 a.m 9:30 a.m 10:29 a.m 9:30 a.m 10:35 a.m 10:00 a.m 12:45 a.m 10:00 a.m 2:30 p.m 10:10 a.m 10:00 p.m 12:15 a.m 8:00 a.m $40.00 12:40 a.m 11:00 a.m $40.00 of holes expressed 20 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 of holes 0 2 1 4 22 2 8 23 3 16 24 4 32 25 5 64 26 of holes expressed 21 as a power of 2 273

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# p. 13

a the pattern is similar but begins with 21 rather than 20 b 21 2 20 22 2 21 23 2 22 24 2 23 25 2 24 26 2 25 c h 2 2n part ii a h 84 78 72 66 60 54 48 h 2n 42 36 30 24 18 12 6 n 5 the function is maximized at 6 2 the maximum profit of $18 is obtained when 6 bracelets and 2 necklaces are produced exercises 1 max 20 at 4 4 min 0 at 0 0 2 max 68 at 12 4 min 14 at 2 2 3 a s 30c 40t b 2c 4t 800 c t 300 c 0 t 0 c y 400 300 0 200 100 x y 300 200 100 2x 4y 800 0 1 2 3 4 5 number of folds part iii answers will vary example yes it is appropriate because as n the number of fold increases h the number of holes punched increases rapidly or exponentially 0 0 100 200 300 0 400 500 x d 0 0 0 200 200 100 300 0 e at 0 0 s 0 at 0 200 s 8,000 at 200 100 s 10,000 at 300 0 s 9,000 f the function is maximized at 200 100 the maximum sales of $10,000 is obtained when 200 chairs and 100 tables are produced enrichment activity 10-5 solving systems using matrices 1 2 3 4 5 x 2 y 1 x 1 y 6 x 4 y 1 x 3 y 5 a error message results b there is no solution to the system enrichment activity 11-1 finding primes the sieve of eratosthenes 1 17 2 composite numbers can be written as pairs of factors such that when the number divided by a factor is less than the factor all positive integral factors have been found in this case the greatest number 200 divided by 15 yields 131 therefore all composite numbers between 3 15 and 200 that have a factor greater than 15 must also have a factor less than 15 and have already been crossed out 3 19 the largest prime less than 20 4 a x 1 12 1 41 41 x 2 22 2 41 43 x 3 32 3 41 47 x 4 42 4 41 53 x 5 52 5 41 61 x 6 62 6 41 71 x 7 72 7 41 83 x 8 82 8 41 97 x 9 92 9 41 113 x 10 102 10 41 131 b x 41 412 41 41 1,681 41 41 enrichment activity 10-6 systems with three variables 1 3 5 2 3 2 1 2 5 10 3 7 2 1 2 2 4 4 3 0.5 6 6 4 2 enrichment activity 10-8 linear programming example 3 b 0 5 c 6 2 d 7 0 4 at b 0 5 2x 3y 15 at c 6 2 2x 3y 18 at d 7 0 2x 3y 14 274

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# p. 14

enrichment activity 11-5 differences of squares 1 a 92 82 b 72 52 c 302 292 2 2 2 2 d 16 14 e 36 35 f 262 242 g 512 502 h 732 722 i 492 472 2 2 2 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 so using the rule 2n 1 can be written as n2 n 12 to check n2 n 12 n2 n2 2n 1 2n 1 3 4n 4 n which is between n 1 and n 1 so using the rule 4n can be written as n 12 n 12 to check n 12 n 12 n2 2n 1 n2 2n 1 4n 2 square abcde equation relating sides 18 5 3 2 32 5 4 2 50 5 5 2 8 5 2 2 3 side square root of side decimal square area area approximation f 5 20 45 equation relating sides enrichment activity 11-7 factoring trinomials 1 1 4x2 10 2 8 and 5 3 4x2 8x 4 4xx 2 5 x 2 4x 2 a x 2 2x c x 5 3x e 2x 3 3x g x 5 8x i x 3 7x 40x2 5x 10 5x 2 5 5 1 2 3 1 g h 20 45 5 2.236067977 4.472135955 6.708203932 45 5 3 5 20 5 2 5 enrichment activity 12-7 operations with radicals b d f h x 3 4x 3 x 4 3x 2 3x 4 4x 1 x 5 5x 2 1 a 3 2 c 3 2 g 0 i 0 b 8 11 e 2 10 k 10 6 f 5 2 2 3 j 5 2 2 3 l 4 1 2 3 p 3 r 2 5 n 6 3 1 3 5 h 2 5 d 6 3 1 3 5 enrichment activity 12-2 square root divide and average 1 4 7 10 7.07 9.38 3.32 0.77 2 5 8 11 6.32 10.39 23 0.55 3 6 9 12 529 2.45 24.78 24.78 50.25 m 11 7 o 11 7 s 5 w 3 q 10 6 u 8 11 y 5 2 matches a c b u d n f j g i h r 3 a 100 b yes 13 529 is rational because 232 14 3.162 15 10.100 t 4 1 2 3 x 23 5 common answer 8 11 3 2 common answer 4 1 2 3 11 7 3 10 6 v 23 5 enrichment activity 12-4 equivalent radical expressions in 1 and 3 answers will vary depending on the number of decimal places displayed on the calculator 1 side square side decimal square area root of area approximation abcde 2 8 18 32 50 2 matches k q l t m o p w v x 2 8 6 3 1 3 5 5 2 2 3 2 5 0 1.414213562 2.828427125 4.242640687 5.6565854249 7.071067812 18 32 50 23 5 275

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# p. 15

enrichment activity 13-2 carpet squares 1 4 with fringe on two sides 8 with fringe on one side 4 with no fringe 2 dimensions fringe on fringe on no total of carpet two sides one side fringe squares 2 ft by 2 ft 3 ft by 3 ft 4 ft by 4 ft 5 ft by 5 ft 6 ft by 6 ft 7 ft by 7 ft 8 ft by 8 ft 9 ft by 9 ft 10 ft by 10 ft 4 4 4 4 4 4 4 4 4 0 4 8 12 16 20 24 28 32 0 1 4 9 16 25 36 49 64 4 9 16 25 36 49 64 81 100 5 the graphs for squares with fringe on two sides and one side are linear the graph for squares with no fringe is steepest 6 a 4 b 4x 2 c x 22 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 9 no for the number of squares to be equal 4x 2 x 22 simplifying gives x2 8x 12 0 which factors to x 6 x 2 0 the only solutions for x are 2 and 6 10 n 12 n2 2n 3 enrichment activity 13-4 quadratic inequalities 1 y 1 1 3 the number of squares with fringe on two sides is 4 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 with no fringe starts at 0 and is the sequence of square numbers the total number of squares is also the square numbers starting with 4 4 y o 1 x 2 3 4 5 6 0 0 is not in the region 0 3 is in the region see graph all points inside of the graph of y x2 6x 8 2 x 4 y 64 60 56 52 48 44 40 36 32 28 24 20 16 12 8 4 0 number of each type of square 1 1 1 o 1 x 1 2 3 4 5 6 7 8 9 10 length of side of square x 276

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