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Chapter R REVIEW OF BASIC CONCEPTS Section R.1 Sets The elements of the set of natural number are {1, 2, 3, 4, …} Set A is a subset of set B if every element of set A is also an element of set B The set of all elements of the universal set U that not belong to set A is the complement of set A The intersection of sets A and B is made up of all the elements belonging to both set A and set B The union of sets A and B is made up of all the elements belonging to set A or set B (or both) The set 1, 13 , 19 , 27 , is infinite Using set notation, the set {x|x is a natural number less than 6} is {1, 2, 3, 4, 5} A 4, 5 10 16,18, 21, 50 15,16,17,18 16,18 16,18, 21, 50 15,16,17,18 15,16,17,18, 21, 50 11 The set {4, 5, 6, …, 15} has a limited number of elements, so it is a finite set Yes, 10 is an element of the set 12 The set {1, 2, 3, 4, 5, …, 75} has a limited number of elements, so it is a finite set Yes, 10 is an element of the set 13 The set 1, , , , has an unlimited number of elements, so it is an infinite set No, 10 is not an element of the set 14 The set {4, 5, 6, …} has an unlimited number of elements, so it is an infinite set Yes, 10 is an element of the set 15 The set {x | x is a natural number larger than 11}, which can also be written as {11, 12, 13, 14, …}, has an unlimited number of elements, so it is an infinite set No, 10 is not an element of the set 16 The set {x | x is a natural number greater than or equal to 10}, which can also be written as {10, 11, 12, 13, …}, has an unlimited number of elements, so it is an infinite set Yes, 10 is an element of the set 17 There are infinitely many fractions between and 2, so {x | x is a fraction between and 2} is an infinite set No, 10 is not an element of the set 18 The set {x | x is an even natural number} has no largest element Because it has an unlimited number of elements, this is an infinite set Yes, 10 is an element of the set 19 The elements of the set {12, 13, 14, …, 20} are all the natural numbers from 12 to 20 inclusive There are elements in the set, {12, 13, 14, 15, 16, 17, 18, 19, 20} 20 The elements of the set {8, 9, 10, …, 17} are all the natural numbers from to 17 inclusive There are 10 elements in the set, {8, 9, 10, 11, 12, 13, 14, 15, 16, 17} 21 Each element of the set 1, 12 , , , 32 after the first is found by multiplying the preceding number by 12 There are elements in the set, 1, , , 81 , 161 , 32 22 Each element of the set {3, 9, 27, …, 729} after the first is found by multiplying the preceding number by There are elements in the set, {3, 9, 27, 81, 243, 729} 23 To find the elements of the set {17, 22, 27, …, 47}, start with 17 and add to find the next number There are elements in the set, {17, 22, 27, 32, 37, 42, 47} 24 To find the elements of the set {74, 68, 62, …, 38}, start with 74 and subtract (or add –6) to find the next number There are elements in the set, {74, 68, 62, 56, 50, 44, 38} 25 When you list all elements in the set {all natural numbers greater than and less than 15}, you obtain {9, 10, 11, 12, 13, 14} Copyright © 2017 Pearson Education, Inc Chapter R Review of Basic Concepts 26 When you list all elements in the set {all natural numbers not greater than 4} you obtain {1, 2, 3, 4} 47 False These two sets are not equal because {5, 8, 9, 0} contains the element 0, which is not an element of {5, 8, 9} 27 is an element of the set {3, 4, 5, 6}, so we write {3, 4, 5, 6} 48 False These two sets are not equal because {3, 7, 12, 14, 0} contains the element 0, which is not an element of {3, 7, 12, 14} 28 is an element of the set {2, 3, 5, 9, 8}, so we write {2, 3, 5, 9, 8} 29 is not an element of {4, 6, 8, 10}, so we write {4, 6, 8, 10} 30 13 is not an element of the set {3, 5, 12, 14}, so we write 13 {3, 5, 12, 14} 31 is an element of {0, 2, 3, 4}, so we write {0, 2, 3, 4} 32 is an element of the set {0, 5, 6, 7, 8, 10}, so we write {5, 6, 7, 8, 10} 49 True and are the only natural numbers less than 50 True Both sets describe the same elements For Exercises 51−62, A = {2, 4, 6, 8, 10, 12}, B = {2, 4, 8, 10}, C = {4, 10, 12}, D = {2, 10}, and U = {2, 4, 6, 8, 10, 12, 14} 51 True This statement says “A is a subset of U.” Because every element of A is also an element of U, the statement is true 33 {3} is a subset of {2, 3, 4, 5}, not an element of {2, 3, 4, 5}, so we write {3} {2, 3, 4, 5} 52 True This statement says “C is a subset of U.” Because every element of C is also an element of U, the statement is true 34 {5} is a subset of {3, 4, 5, 6, 7}, not an element of {3, 4, 5, 6, 7}, so we write {5} {3, 4, 5, 6, 7} 53 True Because both elements of D, and 10, are also elements of B, D is a subset of B 35 {0} is a subset of {0, 1, 2, 5}, not an element of {0, 1, 2, 5}, so we write {0} {0, 1, 2, 5} 36 {2} is a subset of {2, 4, 6, 8}, not an element of {2, 4, 6, 8}, so we write {2} {2, 4, 6, 8} 37 is not an element of , because the empty set contains no elements Thus, 54 True Because both elements of D, 2, and 10, are also elements of A, D is a subset of A 55 False Set A contains two elements, and 12, that are not elements of B Thus, A is not a subset of B 56 False Set B contains two elements, and 8, that are not elements of C Thus, B is not a subset of C 38 is a subset of , not an element of The empty set contains no elements Thus we write, 57 True The empty set is a subset of every set 39 False is not one of the elements in {2, 5, 6, 8} 59 True Because 4, 8, and 10 are all elements of B, {4, 8, 10} is a subset of B 40 False is not one of the elements of {2, 5, 8, 9} 60 False Because is not an element of D, {0, 2} is not a subset of D 41 True is one of the elements of {11, 5, 4, 3, 1} 61 False Because B contains two elements, and 8, that are not elements of D, B is not a subset of D 42 True 12 is one of the elements of {18, 17, 15, 13, 12} 43 True is not one of the elements of {8, 5, 2, 1} 44 True is not an element of {7, 6, 5, 4} 45 True Both sets contain exactly the same four elements 58 True The empty set is a subset of every set 62 False There are three elements of A (2, 6, and 8) that are not elements of C, so A is not a subset of C 63 Every element of {2, 4, 6} is also an element of {2, 3, 4, 5, 6}, so {2, 4, 6} is a subset of {2, 3, 4, 5, 6} We write {2, 4, 6} {2, 3, 4, 5, 6} 46 True Both sets contain exactly the same five elements Copyright © 2017 Pearson Education, Inc Section R.1 Sets 64 Every element of {1, 5} is also an element of {0, 2, 3, 5}, so {1, 5} is a subset of the set {0, 2, 3, 5} We write {1, 5} {0, 2, 3, 5} 65 Because is an element of {0, 1, 2}, but is not an element of {1, 2, 3, 4, 5}, {0, 1, 2} is not a subset of {1, 2, 3, 4, 5} We write {0, 1, 2} {1, 2, 3, 4, 5} For Exercises 81–110, U = {0, 1, , 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, M = {0, 2, 4, 6, 8}, N = {1, 3, 5, 7, 9, 11, 13}, Q = {0, 2, 4, 6, 8, 10, 12}, and R = {0, 1, 2, 3, 4} 81 M R The only elements belonging to both M and R are 0, 2, and 4, so M R = {0, 2, 4} 66 Because is an element of {5, 6, 7, 8}, but is not an element of {1, 2, 3, 4, 5, 6, 7}, {5, 6, 7, 8} is not a subset of {1, 2, 3, 4, 5, 6, 7} We write {5, 6, 7, 8} {1, 2, 3, 4, 5, 6, 7} 82 M U Because M U, the intersection of M and U will contain the same elements as M M U = M or {0, 2, 4, 6, 8} 67 The empty set is a subset of every set, so {1, 4, 6, 8} 83 M N The union of two sets contains all elements that belong to either set or to both sets M N = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13} 68 The empty set is a subset of every set, including itself, so 69 True and are the only elements belonging to both sets 70 True and 11 are the only elements belonging to both sets 71 False {1, 2, 7} {1, 5, 9} {1, 2, 5, 7, 9}, while {1, 2, 7} {1, 5, 9} {1} 72 False {6, 12, 14, 16} {6, 14, 19} {6, 12, 14, 16, 19}, while {6, 12, 14, 16} {6, 14, 19} {6, 14} 73 True is the only element belonging to both sets 74 False The sets {6, 8, 9} and {9, 8, 6} are equal because they contain exactly the same three elements Their union contains the same elements, namely 8, 9, and 75 {3, 5, 9, 10} {3, 5, 9, 10} In order to belong to the intersection of two sets, an element must belong to both sets Because the empty set contains no elements, {3, 5, 9, 10} , so the statement is false 76 True For any set A, A A 77 True Because the two sets are equal, their union contains the same elements, namely 1, 2, and 78 False {1, 2, 4} {1, 2, 4} {1, 2, 4} 84 M R The union M and R is made up of elements which belong to M or to R (or to both) M R = {0, 1, 2, 3, 4, 6, 8} 85 M N There are no elements which belong to both M and N, so M N M and N are disjoint sets 86 U N Because N U, the elements belonging to U and N are all the elements belonging to N, U N = N or {1, 3, 5, 7, 9, 11, 13} 87 N R = {0, 1, 2, 3, 4, 5, 7, 9, 11, 13} 88 M Q Because M Q, the elements belonging to M or Q are all the elements belonging to Q M Q = Q or {0, 2, 4, 6, 8, 10, 12} 89 N The set N is the complement of set N, which means the set of all elements in the universal set U that not belong to N N Q or {0, 2, 4, 6, 8, 10, 12} 90 Q The set Q is the complement of set Q, which means the set of all elements in the universal set U that not belong to Q Q N or {1, 3, 5, 7, 9, 11, 13} 79 True 80 True Copyright © 2017 Pearson Education, Inc Chapter R Review of Basic Concepts 91 M Q First form M , the complement of M M contains all elements of U that are not elements of M Thus, M {1, 3, 5, 7, 9, 10, 11, 12, 13} Now form the intersection of M and Q Thus, we have M Q {10, 12} 92 Q R First form R , the complement of R R contains all elements of U the are not elements of R Thus, R {5, 6, 7, 8, 9, 10, 11, 12, 13} Now form the intersection of Q and R Thus, we have Q R {6, 8, 10, 12} 93 R Because the empty set contains no elements, there are no elements belonging to both and R Thus, and R are disjoint sets, and R 94 Q Because the empty set contains no elements, there are no elements belonging to both and Q Thus, and Q are disjoint sets, and Q 95 N Because contains no elements, the only elements belonging to N or are the elements of N Thus, and N are disjoint sets, and N N or {1, 3, 5, 7, 9, 11, 13} 99 (Q M) R First form the intersection of Q and M We have Q M = {0, 2, 4, 6, 8} = M Now form the union of this set with R We have (Q M) R = M R = {0, 1, 2, 3, 4, 6, 8} 100 (R N) M First form the union of R and N We have R N = {0, 1, 2, 3, 4, 5, 7, 9, 11, 13} Now find the complement of M We have M = {1, 3, 5, 7, 9, 10, 11, 12, 13} Now, find the intersection of these two sets We have (R N) M N or {1, 3, 5, 7, 9, 11, 13} 101 ( M Q) R First, find M , the complement of M We have M {1, 3, 5, 7, 9, 10, 11, 12, 13} Next, form the union of M and Q We have M Q = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} U Thus, we have ( M Q) R U R R or {0, 1, 2, 3, 4} 102 Q (M N) First, form the union of M and N We have M N = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13} Now form the intersection of Q with this set We have Q (M N) M or {0, 2, 4, 6, 8} 103 Q N U First, find Q , the complement of Q We have Q {1, 3, 5, 7, 9, 11, 13} N Now find N , the complement of N We have N {0, 2, 4, 6, 8, 10, 12} Q Next, form the intersection of N and U We have N U Q U Q Finally, we have 96 R Because contains no elements, the only elements belonging to R or are the elements of R Thus, and R are disjoint sets, and R R or {0, 1, 2, 3, 4} 97 (M N) R First, form the intersection of M and N Because M and N have no common elements (they are disjoint), M N Thus, (M N) R R Now, because contains no elements, the only elements belonging to R or are the elements of R Thus, and R are disjoint sets, and R R or {0, 1, 2, 3, 4} 98 (N R) M First form the union of N and R We have N R = {0, 1, 2, 3, 4, 5, 7, 9, 11, 13} Now form the intersection of this set with M We have (N R) M = {0, 2, 4} Q N U Q Q Because the intersection of Q and N U is , Q and N U are disjoint sets 104 U R Because U , and U U U , we have U R U U R U R U or {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} 105 x | x U , x M This means all elements of U except those in set M This gives the set {1, 3, 5, 7, 9, 10, 11, 12, 13} Copyright © 2017 Pearson Education, Inc Section R.2 Real Numbers and Their Properties 106 x | x U , x R (c) 94 is a rational number and a real This means all elements of U except those in set R This gives the set {5, 6, 7, 8, 9, 10, 11, 12, 13} 107 x | x M number 94 belongs to D, F and x Q This means all elements that are members of both set M and set Q, or M Q Note that set M is a subset of set Q because all members of set M are included in set Q Thus, the answer is set M or {0, 2, 4, 6, 8} 108 x | x M (e) 13 is an irrational number and a real 2.16 216 100 54 25 is a rational number and a real number 2.16 belongs to D, F 103 1000 10 10 or x Q x | x Q or x R This means all elements that are either members of set Q or set R, or Q R Thus, the answer is {0, 1, 2, 3, 4, 6, 8, 10, 12} Section R.2 36 is a natural number Therefore, it is also a whole number, an integer, a rational number, and a real number 36 belongs to A, B, C, D, F (f) This means all elements that are either members of set M or set Q, or M Q Note that set M is a subset of set Q because all members of set M are included in set Q Thus, the answer is set Q or {0, 2, 4, 6, 8, 10, 12} 110 (d) number 13 belongs to E, F x | x Q and x R This means all elements that are members of both set Q and set R, or Q R Thus, the answer is {0, 2, 4} 109 Real Numbers and Their Properties {0, 1, 2, 3, …} describes the set of whole numbers {…, –3, –2, –1, 0, 1, 2, 3, …} describes the set of integers In the expression 63 , is the base and is the exponent If the real number a is to the left of the real number b on a number line, then a < (is less than) b The distance on a number line from a number to is the absolute value of that number (a) is a whole number Therefore, it is also an integer, a rational number, and a real number belongs to B, C, D, F (b) 34 is a natural number Therefore, it is also a whole number, an integer, a rational number, and a real number 34 belongs to A, B, C, D, F 4 10 7 x y 7 x 28 y For Exercises 11−16, A 6, 124 , 85 , 3, 0, , 1, 2 , 3, 12 11 and are natural numbers 12 0, 1, and are whole numbers 13 –6, 124 (or –3), 0, 1, and are integers 14 6, 124 (or 3), 85 , 0, , 1, and are rational numbers 15 3, 2 and 12 are irrational numbers 16 All are real numbers 17 24 2 16 18 35 3 3 243 19 24 2 2 2 2 16 20 26 222222 64 21 35 3 3 3 3 3 243 22 (2)5 (2)(2)( 2)( 2)( 2) 32 23 2 34 2 3 3 2 81 162 24 4 53 4 5 5 4 125 500 Copyright © 2017 Pearson Education, Inc Chapter R Review of Basic Concepts For Exercises 35−48, p = −4, q = 8, and r = –10 25 2 12 10 12 10 6 26 35 p q r 4 10 16 27 16 27 23 3 28 5 3 2 5 3 16 30 3 16 30 48 30 48 18 29 (4 23 )(2 25) (4 8)(2 5) (4)(3) 12 5 16 23 5 9 23 4 8 4 4 16 31 18 9 36 36 18 18 17 25 17 36 18 36 36 36 11 32 10 25 16 15 11 40 40 10 10 40 10 16 25 40 40 40 33 34 4 100 16 100 16 56 100 72 100 28 27 4 9 8 7 2 4 1 7 2 4 1 7 7 4 56 60 30 8 4 6 12 8 24 12 3 43 8 6 7 15 3 6 5 6 5 12 28 6 1 6 5 36 p 2q r 4 10 16 10 16 16 10 32 10 42 37 q r (10) 2 q p (4) 38 p r 4 (10) 14 4 pq 39 40 41 42 43 44 3q 24 12 r p 10 4 10 4 4 12 48 25 23 20 20 20 3r 3(10) 30 10 q r 8 10 15 75 71 20 20 20 5r 5(10) 50 p 3r 2(4) 3(10) 8 30 50 25 22 11 3q 38 24 p 2r 4 10 12 10 24 24 24 3 12 20 12 20 q r 10 10 22 4 11 p q 3(4) 3 2 8 q r 10 10 2 4 8 p q 2 2 2 2 2 2 2 2 Copyright © 2017 Pearson Education, Inc Section R.2 Real Numbers and Their Properties p 2 3r 4 2 10 45 2q 28 2 2 10 6 4 10 4 30 6 6 13 4 30 26 6 6 q 6 p 8 6 4 4 p 4 46 4 44 2 3 4 4 p 4 p r 8 10 12 0 12 6 2 2 1 4 5q 1 p 48 r 3 10 3 27 3 7 7 40 27 40 54 7 7 14 2 7 10 10 22 z 22 z 20 z 11 11 3 3 3 64 r 12 12 r 12 r 9r 4 4 4 65 m 5 m 5 6 m 11 66 a 7 a 8 a 15 a (or a 15) 67 47 63 16 32 40 z y 8 27 16 32 40 y z 27 16 32 y z 8 27 y+ z 68 20m y 32 z 1 1 20m 8 y 32 z 4 4 20 m y 32 z 5m+ 2 y 8 z = 5m y z 69 p 14 p (8 14) p 6 p 70 15 x 10 x 15 10 x x 49 distributive 50 distributive 71 4 z y 4 z 4 y 4 z y 51 inverse 52 inverse 72 3 m n 3m 3n 3m 3n 53 identity 54 identity 55 commutative 56 commutative 57 associative 58 associative 59 closure 60 closure 61 No; in general a b b a Examples will vary, i.e if a = 15 and b = 0, then a − b = 15 − = 15, but b − a = − 15 = −15 62 No; in general a b c a b c Examples will vary, i.e if a = 15, b = 0, and c = 3, then a b c 15 0 15 12, but a b c 15 0 3 15 3 18 73 The process in your head should be the following: 72 17 28 17 (72 28) 17 (100) 17 1700 74 The process in your head should be the following: 32 80 32 20 32 80 20 32 100 3200 75 The process in your head should be the following: 5 5 1 123 23 123 23 1 8 8 2 1 (100) 1 150 2 Copyright © 2017 Pearson Education, Inc Chapter R Review of Basic Concepts 76 The process in your head should be the following: 3 2 3 17 14 17 17 14 5 5 4 17 10 17.4 10 174 77 This statement is false because and 2 A For Exercises 91−98, x = −4 and y = 91 92 x y 4 2 10 18 18 93 corrected statement would be 94 or 3x y 3(4) 2(2) 12 16 16 x y 3(4) 4(2) 12 20 20 y x 5 2 4 10 4 14 14 78 This statement is false because 33 27 27 and 95 33 27 27 A corrected statement would be 3 3 or 3 3 79 This statement is true because 5 30 and 5 30 30 80 This statement is true because 96 14 14 7 2 97 14 7 and 81 This statement is false For example if you let a and b then 4 and a b 4 A corrected statement is a b b a , if b > a > 82 This statement is true by the algebraic definition of absolute value stated in the text 83 10 10 85 4 7 87 8 8 84 15 15 86 y 3 x 2 3 4 xy 2 7 2 88 12 12 89 Pd P 120 116 120 The Pd value for a woman whose actual systolic blood pressure is 116 and whose normal value should be 120 is 90 Consider the relation, Pd P 120 17 Because 103 and 137 both differ from 120 by 17, these are the two possible values for the patient’s systolic blood pressure 98 2 4 12 8 1 8 8 x 4 y 4 4 4 x 4 2 16 24 6 4 8y x 8(2) (4) 16 (4) x 4 4 20 20 5 4 4 x 2 y 2 2 2 x 4 4 4 99 True because 25 25 and 25 25 100 True because the absolute value of any number is always greater than or equal to 101 False 13 8 while 13 13 18 102 False 12 4 while 12 12 4 103 True because 11 6 11 66 and 66 66 104 True because 10 10 2 Copyright © 2017 Pearson Education, Inc 10 5 and 2 Section R.2 Real Numbers and Their Properties 105 d P, Q 4 or d P, Q 1 106 d P, R 4 12 12 or d P, R 12 12 107 d Q, R 1 or d Q, R 108 d Q, S 12 1 13 13 or d Q, S 12 13 13 109 xy if x and y have the same sign 110 x y if y is positive, because x is positive for any nonzero x 111 x if x and y have different signs y 112 y2 if x is negative, because y is positive for any nonzero y x 113 Because x3 has the same sign as x, 114 x3 if x and y have the same sign y x if x and y have different signs y 115 Because 18 (1) 19 19 and (18) 19 19, the number of strokes between their scores is 19 116 18, 355 3224 15 21, 564 yards No, it is not the same, because the sum of the absolute values is 18, 355 3224 15 18, 355 3224 15 21, 594 The fact that | –15 | = 15 changes the two answers 117 BAC 48 3.2 0.075 190 0.015 0.031 118 BAC 36 4.0 0.075 135 0.015 0.035 119 BAC 20 3.8 0.075 200 0.015 0.026 If the man’s weight was greater, the BAC would be lower For example, if the man weighed 250 pounds, his BAC would be 20 3.8 0.075 250 0.015 0.0084 120 BAC 14 0.075 150 0.015 0.054 If the woman consumed the same two glasses of wine over a longer period of time, her BAC would be lower For example, if she drank the wine over hours, her BAC would be 14 0.075 150 0.015 0.039 Copyright © 2017 Pearson Education, Inc 10 Chapter R Review of Basic Concepts C T Y I 250 1000 12.5 6.25 1250 A A A A , where For exercises 121−132, we use the formula R CA 0.775, TA 0.11875, YA 12.5, and AI 0.095 121 First, find the value of the numerator C T Y I 250 1000 12.5 6.25 1250 A A A A 304 34 3705 250 1000 12.5 6.25 1250 339.73 435 435 435 435 R 339.73 113.2 122 First, find the value of the numerator C T Y I 250 1000 12.5 6.25 1250 A A A A 341 38 4381 250 1000 12.5 6.25 1250 336.56 520 520 520 520 R 336.56 112.2 123 First, find the value of the numerator C T Y I 250 1000 12.5 6.25 1250 A A A A 408 32 4952 250 1000 12.5 6.25 1250 309.95 608 608 408 608 R 309.95 103.3 124 First, find the value of the numerator C T Y I 250 1000 12.5 6.25 1250 A A A A 395 39 4727 15 250 1000 12.5 6.25 1250 304.55 597 597 597 597 R 304.55 101.5 125 First, find the value of the numerator C T Y I 250 1000 12.5 6.25 1250 A A A A 373 33 4109 250 1000 12.5 6.25 1250 292.10 582 582 582 582 R 292.10 97.4 Copyright © 2017 Pearson Education, Inc 40 Chapter R Review of Basic Concepts 103 Factor x(2 x 3) 5 / x (2 x 3) / x3 (2 x 3)13 / 108 using the common factor x(2 x 3) 5 / : x 2 y 2 x y x 2 y 2 x y x(2 x 3) 5 / x (2 x 3) / x3 (2 x 3)13 / x(2 x 3) 5 / 3x(2 x 3) x (2 x 3) x(2 x 3) 5 / x x x x 12 x 5 / x(2 x 3) x x 16 x 48 x3 36 x x(2 x 3) 5 / 16 x 48 x3 30 x x 2 104 Factor y (4 y 1) 3 / y (4 y 1)4 / 16 y (4 y 1)11/ 109 y (4 y 1) 3 / y (4 y 1)4 / 16 y (4 y 1)11/ y (4 y 1) 3 / 3 y y (4 y 1) 8(4 y 1) y (4 y 1) 3 / 3 y 16 y y 16 y y y (4 y 1) 3 / 3 y 16 y y 128 y 64 y 8 y (4 y 1) 3 / 115 y 60 y 8 105 a 1 b 1 (ab) 1 106 p 1 q 1 ( pq )1 107 a b ab ba ab ab p pq q p pq pq 1b a b = q r 1 q 1 r q r 1 q 1 r q ab q p pq q p pq r r rq rq q q rq rq r q r q rq rq 2 x y x y 1 y 3 3 x y x y 9 y2 x x y y x2 y x2 y y y xy 9 y x 9y or 2 2 y x 9 x y 9 1 a 16b 1 a 16 b a 4b a 4b a a 1 1 b a 16 b a 162 b pq 2 x x y x y 1 p q p qr rq rq qr rq qr rq 1 1(r q ) x 9y b a ab ba ab 1 q p q y2 x y x y x2 y x2 y2 or y x x y x y 2 1 110 1 a ba x2 x y x y y x x y y x2 x y y x2 x y y x y x x y using the common factor y (4 y 1) 3 / : 12 x2 y 1 2 x y 1 x2 y2 111 112 b a 16b b2 a b ab 16b b ab 16 a 2b 16 a 2b 16 ( x 1) (2 x) x (4)( x 1)3 (2 x) ( x 1)8 (2 x)( x 1)3 ( x 1) x ( x 1)8 (2 x) ( x 1) x (2 x)(1 3x ) ( x 1)5 ( x 1)5 ( y 2)5 (3 y ) y (6)( y 2) (3 y ) ( y 2)7 (3 y )( y 2) ( y 2) y ( y 2)7 y( y y3 ) ( y 2)3 Copyright © 2017 Pearson Education, Inc 162 b Section R.7 Radical Expressions 113 114 4( x 1)3 x( x 1) 16( x 1)3 4( x 1)3 1 x( x 1) 16( x 1)3 x( x 1) x3 x 4 10(4 x 9) 25 x(4 x 9)3 15(4 x 9)6 5(4 x 9) x(4 x 9) 15(4 x 9)6 x(4 x 9) 20 x3 45 x 3(4 x 9) 3(4 x 9) 2(2 x 3)1 ( x 1)(2 x 3) 2 115 (2 x 3) (2 x 3)1 ( x 1)(2 x 3)1 (2 x 3) 2xx13 ( x 1)(2 x 3) 1 (2 x 3)1 (2 x 3)1 2(2 x 3) ( x 1) 3x 43 (2 x 3) (2 x 3) 120 0.13/ 903/ (0.1 90)3/ 93/ (91/ )3 33 27 121 22 / 2/3 2000 2000 2/3 1/ 1000 1 10 100 1000 122 203 / 20 53 / Section R.7 2/3 3/ 43 / (41/ )3 23 Radical Expressions 64 641 27 27 32 3 (a) F; (3x)1/ 3 x (3 x)1/ (b) H; (3 x) 1/ 7(3t 1)1 (t 1)(3t 1) 3 116 (3t 1)3 (3t 1)1 7 (t 1)(3t 1) 1 (3t 1)3 t 1 3t 1 12 7 (t 1)(3t 1) 12 (3t 1) (3t 1) 7(3t 1) (t 1) 20t 4(5t 2) 32 32 (3t 1) (3t 1) (3t 1)3 1 117 Let x = length of side of cube Then 3x = length of side of bigger cube (side tripled) x3 is the volume of the cube, and (3x)3 33 x3 27 x3 is the volume of the bigger cube The volume will change by a factor of 27 118 Let r = radius of circle Then 2r = radius of the bigger circle (radius doubled) r is the area of the circle, and (2r ) 22 r 4r r is the area of the bigger circle The area will change by a factor of 119 0.22/ 402/3 (0.2 40) 2/ (81/3 )2 22 (c) G; (3x)1/ 3x (d) C; (3x)1/ 3x (a) B; 3 x1/ 3 x 3 3 3 1/ x x (b) D; 3x 1/ (c) A; 3x 1/ (d) E; 3x1/ 1/ x 3 x x t t5 t 5 6 24 36 36 12 50 25 25 36 7 36 xy xy 5 xy Copyright © 2017 Pearson Education, Inc 3 x 41 Chapter R Review of Basic Concepts 42 10 2 2 11 125 1251 12 81 811 15 125 125 5 16 343 343 7 17 81 81 13 14 13 13 43 1 34 Expression (D) is not simplified, because the radicand is a fraction This expression may be 216 2161 simplified by using the rule n 256 2561 14 4 x4 x 36 x6 x 18 256 256 This expression is not a real number 19 32 321 20 128 1281 2 15 22 343 343 24 p5/ p5 or 42 m 27 28 k k z z 5k m 5k m 4 p3q p3q 3 p q x y 2 4x y 5 2m4 2m 81 27 27 3 250 125 125 3 3 3 4 4 2m p 5r 3t 14 pqr 14 pqr 42 pqr 46 xt xt 35 xt 47 x y x y 14 xy 48 x y x y 36 xy 49 9 25 25 30 m y m(2 y )1/ 21/ my 5/ 50 16 16 49 49 51 3 5 3 8 52 3 4 16 16 3 1/ p 31 A 32 3, 5, 7, … (odd positive integers greater than or equal to 3) 33 It is true for all x ≥ 45 5/ 29 3 p 3(5 p ) 2/5 1/ 44 243 81 3 3/ b 2 81 p12 q p a n 43 32 16 16 2 26 (5r 3t ) / (5r 3t ) or 41 25 (2m p )2 / (2m p) or 40 7 13 39 21 32 32 23 m / m or 38 5k m 25k m 37 14 n 3 35 This expression is not a real number a b 53 m n4 m 4 n Copyright © 2017 Pearson Education, Inc m n : Section R.7 Radical Expressions 54 r s6 6 r s r s 68 56 343 7 7 35 3 69 25 32 g h5 9r 25(3) (5)3 52 34 53 52 33 53 3 33 53 52 52 15 25 15 75 8x z x x z 70 62 63 64 2 3x 3x 3x 3x 3x 3p 3p 3p 3x 9x 53p 9p 6x 3x 15 p 3p x xy y z g h5 r g h5 r g h ghr r x x x 3 2 x x x2 x2 6n 9r h g 3h r 32 r r r gh ghr r h g hr 2 r4 r4 32 r 32 r g 3h 9r 2 h 32 g 3hr r 34 r 3r 32 x5 y5 16 x x y4 y 72 53 53 / 51/ 3 41/ (41/ )1/ 41/ (22 )1/ 22 / 21/ 74 25 251/ (251/ )1/ 251/ (52 )1/ 52 / 51/ 21/ (21/ )1/ 21/12 53 91/ (91/ )1/ 91/15 12 15 77 x x 72 x x x 36 x x 2 x x 12 x 78 18k 72k 50k 2k 36 2k 25 2k 3 2k 2k 2k 12 2k 2k 2k 11 2k x x 2x 2x y4 y 4 y x xy 2x 2x y y y3 y2 34 34 / 31/ 76 g h5 r h g 3h 73 h 9 r r 71 75 x y xy x y xy z z z g h r3 x4 3p 3p x5 y 3 27 a3 cannot be simplified further 66 x y cannot be simplified further x5 y z2 65 67 32 x5 y5 m n 6n 2m n 61 24m n m n n g h5 x4 z8 x x2 z x 60 36 p 36 p 2p2p p2 3 59 2p 2p 3 3 3 p2 36 p p3 p p 2 p3 p3 16(2) (2)8 24 (2) 28 24 24 28 215 215 58 8p 2p 3 3 3 55 3125 5 5 15 57 9 16 p 16 p 43 x2 x2 xx x2 Copyright © 2017 Pearson Education, Inc Chapter R Review of Basic Concepts 44 3 89 This product has the pattern (a b) a 2ab b , the square of a binomial 79 24 81 3 27 3 2 3 3 3 8 3 80 3 3 3 3 3 5 6 3 81x y 16 x10 y 10 5 2 3x x y3 x x y3 82 256 x y 625 x y 3 256 x y xy 625 x8 xy 84 11 13 cannot be simplified further 2 3 5 3 3 93 mn m n2 mnm m3 n n2 3 3 94 3 32m n3 3 3 95 3 x x 7 3 3 n 3 m3 8m n3 2m 11 10 72 12 10 20 10 11 10 14 3 5 3 11 11 2 4 4 88 This product has the pattern (a b)(a ab b ) a b3 , the sum of two cubes 5 2 52 11 1 11 5 2 5 87 This product has the pattern (a b)(a ab b ) a b3 , the difference of cubes 2 3 4 32 7 86 This product has the pattern (a b)(a b) a b , the difference of squares 10 10 92 This product can be found by using the FOIL method 85 This product has the pattern (a b)(a b) a b , the difference of squares 6 3 23 6 5 83 10 cannot be simplified further 3 2 3 2 3 xy xy x xy 91 This product can be found by using the FOIL method 50 10 15 25 15 15 10 81x x y 16 x8 x y 2 90 This product has the pattern (a b) a 2ab b , the square of a binomial 3 24 11 11 2 11 3 3 32 108 27 81 3 27 34 Copyright © 2017 Pearson Education, Inc n2 n2 16m n3 m n2 n 3 16m n3 32m n3 32m n3 3 1 4 3 2 2 x x3 x x x x3 x x x x x x3 x3 Section R.7 Radical Expressions 96 t 4 12 t4 12 4 t 100 t t t t2 t t t2 t3 t t2 t3 t 4 1 32 42 16 2 1 4 4 4 11 11 4 2 11 11 42 97 26 13 13 33 33 13 13 13 3 101 12 15 7 12 12 12 7 12 36 4 3 4 3 24 81 83 4 27 1 4 3 3 102 3 3 32 32 5 3 15 53 7 21 4 21 7 1 1 4 4 2 4 2 4 1 4 2 4 2 3 22 3 3 3 3 37 21 25 5 3 3 3 6 3 3 2 24 24 25 3 6 6 5 5 5 3 5 103 3 5 99 5 3 3 3 3 16 54 28 27 2 1 6 30 6 5 12 27 48 43 93 16 3 62 1 35 3 3 12 15 12 12 12 98 45 25 25 63 18 Copyright © 2017 Pearson Education, Inc 2 1 1 2 4 2 14 4 16 14 14 28 32 14 14 4 2 7 14 2 4 7 14 2 46 104 Chapter R Review of Basic Concepts 110 The denominator is in the form of the difference of cubes, so multiply the numerator and 1 1 3 52 52 3 52 3 52 1 3 3 3 2 denominator by 105 106 p 2 15 25 in order to rationalize the denominator 15 25 3 15 25 15 25 p 2 p 4 p 2 p4 p 4 p p 2 9 r 3 r 9r 3 ( m n) 3m m n 4mn r x x y 4x y 15 25 15 25 45 75 45 75 15 25 2 3 9 9 0.4275 10 30 0.16 12 35.74 0.6215 30 35.75 150.16 114 Windchill temperature 35.74 0.6215T 35.75V 0.16 0.4275TV 0.16 Windchill temperature x x y x x y x y x y 15 25 35.74 0.6215 10 35.75 300.16 113 Windchill temperature 35.74 0.6215T 35.75V 0.16 0.4275TV 0.16 Windchill temperature a a b 1 a b 1 a b 1 a a b 1 a a b 1 a b 1 a b 12 a a b 1 109 3m m n 108 112 S 15.18 n 15.18 19.1 The speed of the boat with an eight-man crew is approx 19.1 ft/sec 3m 3m 2 mn 107 2 mn 2 mn 2 mn 3m m n 22 m n 111 S 15.18 n 15.18 17.7 The speed of the boat with a four-man crew is approx 17.7 ft/sec 9r r r 9 r r 3 r 3 r 3 r 32 r 22 p 2 32 3 52 or 45 12 15 33 p4 p4 p 2 p 2 0.4275 30 15 115 116 4 16 54 320 10 19 54 27 5 320 32 10 0.1 40 0.1 40 118 119 120 3 117 0.16 15 3 27 6 6 64 Copyright © 2017 Pearson Education, Inc Chapter R Review Exercises 121 122 123 355 3.1415929 … and 3.1415926…, so 113 it gives six decimal places of accuracy 377 3.14166 and 3.14159… , so it first 120 differs in the fourth decimal place 3927 3.1416 and 3.1415926…, so it 1250 first differs in the fourth decimal place Chapter R Review Exercises The elements of the set {6, 8, 10, …, 20} are the even numbers from to 20 inclusive The elements in the set are {6, 8, 10, 12, 14, 16, 18, 20} The set {x x is a decimal between and 1} has an unlimited number of elements, so it is infinite True The set of negative integers = {…, −4, −3, −2, −1}, while the set of whole numbers = {0, 1, 2, 3, …} The two sets not intersect, and so they are disjoint False True False is an element in {1, 3, 5, 7} False The two sets are not equal because they not have the same elements False True 10 True 11 True 12 False All of the elements in E are also elements in A, so E is a subset of A 13 A {2, 6, 9,10} 14 B A {4,8} 47 23 –12, –6, (or –2), 0, and are integers 24 The rational numbers are –12, –6, –0.9, (or –2), 0, , and 25 26 4 is an irrational number and a real number is not defined, so none of the terms apply 27 is a whole number, an integer, a rational number, and a real number 28 36 6 is an integer, a rational number, and a real number 29 Answers will vary Sample answer: The reciprocal of a product is the product of the reciprocals 30 Answers will vary Sample answer: The multiple of a difference is the difference of the multiples 31 Answers will vary Sample answer: A product raised to a power is the product of the factors raised to the power 32 Answers will vary Sample answer: The difference of the squares of two terms is the product of the sum of the terms and the difference of the terms 33 Answers will vary Sample answer: A quotient raised to a power is the quotient of the numerator raised to the power and the denominator raised to the power 34 Answers will vary Sample answer: The absolute value of a product is the product of the absolute values of the multiplicands 15 B E 35 commutative 36 distributive 16 C E {1, 3, 5, 7} C 37 associative 38 inverse 17 D 39 identity 40 identity 18 B {2, 4, 6,8} B 41 The year 2012 corresponds to x = 10 0.0112 102 0.4663 10 1.513 7.296 7.296 million strudents took at least one online college course in 2012 19 (C D ) B {1, 3} B {1, 2, 3, 4, 6,8} 20 ( D U ) E {4, 5, 6, 7,8, 9,10} U E {4, 5, 6, 7,8, 9,10} E {3, 4, 5, 6, 7,8, 9,10} 21 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 or U 42 47 29 43 (4 1)(3 5) 23 (5)( 8) 40 32 22 True Copyright © 2017 Pearson Education, Inc 48 Chapter R Review of Basic Concepts 44 (6 9)(2 7) (4) (3)(9) ( 4) 27 27 4 45 2 6 3 9 22 15 37 11 18 18 18 23 46 4 2 4 2 8 35 4 32 15 20 20 47 47 10 20 20 10 37 47 10 20 20 20 47 6(4) 32 (2)3 6(4) 9(8) 5[2 (6)] 5[2 6] 6(4) 9(8) 5(4) 24 72 24 72 20 20 48 12 20 (7)(3) (23 )(5) (7)(3) ( 8)( 5) 48 (4 2)(1 6) (22 2)(1 6) 21 40 19 (6)(7) 42 49 Let a 1 , b 2 , c = c(2a 5b) 4[2(1) 5(2)] 4 2 10 4 2 10 4(8) 32 50 Let a 1 , b 2 , c = a 2 b c 1 2 2 3 3 2 2 6 20 26 4 5 5 52 Let a 1 , b 2 , c = b c 4 3 4 ac 1 4 16 10 4 53 (3q3 9q 6) (4q3 8q 3) 3q3 9q 4q 8q q q 8q 54 2(3 y y y ) (5 y y ) y 18 y y y y y 18 y y 55 8 y 7 2 y y 3 y y2 y y2 y 16 y 56 y 24 y 14 y 49 y 21 16 y 42 y 73 y 21 2 56 (2r 11s )(4r s ) 8r 18rs 44rs 99 s 8r 26rs 99 s 57 (3k 5m) (3k ) 2(3k )(5m) (5m) 9k 30km 25m 58 (4a 3b)2 (4a ) 2(4a)(3b) (3b) 16a 24ab 9b 59 30m3 9m 22m 5m 6m 3m 5m 30m3 9m 22m s 30m3 6m 15m 22m 15m 3m 25m 25m Thus, 30m3 9m 22m 6m 3m 5m 51 Let a 1 , b 2 , c = 9a 2b 9(1) 2(2) a b c 1 (2) 9 (4) 13 13 3 Copyright © 2017 Pearson Education, Inc Chapter R Review Exercises 60 66 r rp 42 p Find two numbers whose product is –42 and whose sum is They are and –6 Thus, r rp 42 p (r p)(r p) 72r 59r 12 8r 9r 8r 72r 59r 12 72r 27 r 32r 12 32r 12 Thus, 72r 59r 12 9r 8r 61 67 48a8 12a b 90a 6b 6a (8a 2ab 15b ) 6a (4a 5b)(2a 3b) 3b3 8b 12b 30 b2 Insert the missing term in the divisor with a coefficient 3b b 0b 3b3 8b 12b 30 3b3 0b 12b 8b 0b 30 8b 0b 32 Thus, 62 49 3b3 8b 12b 30 3b 2 b 4 b 4 5m3 7m 14 m2 Insert each missing term with a zero coefficient 5m m 0m 5m3 m 0m 14 5m3 0m 10m m 10m 14 7 m 0m 14 10m Thus, 5m3 m 14 10m 5m m 2 m 2 63 3( z 4) 9( z 4)3 3( z 4) [1 3( z 4)] 3( z 4) (1 3z 12) 3( z 4) (3 z 11) 64 z z z z (7 z z 1) 65 z zk 16k ( z 8k )( z 2k ) 68 6m 13m The positive factors of could be and or and As factors of –5, we could have –1 and or –5 and Try different combinations of these factors until the correct one is found 6m 13m (3m 1)(2m 5) 69 49m8 9n (7 m ) (3n) (7 m 3n)(7m 3n) 70 169 y (13 y ) 12 (13 y 1)(13 y 1) 71 6(3r 1) (3r 1) 35 Let x = 3r – With this substitution, 6(3r 1) (3r 1) 35 becomes x x 35 Factor the trinomial by trial and error x x 35 (3 x 7)(2 x 5) Replacing x with 3r – gives [ 3(3r 1) ][ 2(3r 1) ] (9r 7)(6r 5) (9r 10)(6r 3) 3(9r 10)(2r 1) 72 y 1000 z 8( y 125 z ) 8[ y (5 z )3 ] 8( y z )[ y y (5 z ) (5 z ) ] 8( y z )( y yz 25 z ) 73 xy x y xy x y 2 x y y 2 y 2 x 1 74 15mp 9mq 10np 6nq (15mp 9mq ) (10np 6nq ) 3m(5 p 3q) 2n(5 p 3q ) (5 p 3q )(3m 2n) 75 (3x 4) ( x 5)(2)(3 x 4)(3) (3 x 4)[(3 x 4) ( x 5)(2)(3)] (3 x 4)[3x x 30] (3 x 4)(9 x 34) Copyright © 2017 Pearson Education, Inc 50 Chapter R Review of Basic Concepts 76 (5 x)(3)(7 x 8) (7) (7 x 8)3 (2) (7 x 8) [(5 x )(3)(7) (7 x 8)(2)] (7 x 8)2 [(5 x )(21) (7 x 8)(2)] (7 x 8) [105 42 x 14 x 16] (7 x 8) (121 56 x) 77 78 79 80 81 82 k2 k k (k 1)(4) 3 8k k 8k (k 1)(k 1) 4k 8k (k 1) 2k (k 1) 83 84 3r 9r 8r 3r 9r r r 3 8r r 9 r2 3r (r 3) (r 3) (r 3)(r 3) 8r 8r x x x2 3x x2 5x x2 x x2 x x2 x x x x 3x ( x 2)( x 1) ( x 3)( x 1) x ( x 3)( x 2) ( x 4)( x 1) x 27m3 n3 9m 3mn n 3m n 9m n 3 27m n 9m n 3m n 9m 3mn n (3m)3 n3 (3m) n 3m n 9m 3mn n (3m n)[(3m)2 3mn n ] (3m n)(3m n) 3m n 9m 3mn n 2 (3m n)(9m 3mn n )(3m n)(3m n) (3m n)(9m 3mn n ) (3m n)(3m n) p 36q p pq 6q p 12 pq 36q p pq q ( p 6q )( p 6q) ( p 6q )( p q ) ( p 6q ) ( p q)2 p 6q pq 1 84 32 y y y 5 y 20 y 20 y 32 37 20 y 20 y 85 m m(1) 3m 3m m m (4 m)(1) m m 3m 2m m4 m4 m4 We may also use – m as the common denominator In this case, the result will be 2 m The two results are equivalent rational 4m expressions x 4x x ( x 3)( x 1) ( x 1)( x 1) 3( x 1) 2( x 3) ( x 3)( x 1)( x 1) ( x 1)( x 1)( x 3) 3( x 1) 2( x 3) ( x 3)( x 1)( x 1) 3x x x9 ( x 3)( x 1)( x 1) ( x 3)( x 1)( x 1) p 1 q 1 pq 1 1 pq 1 pq pq q p pq(1) pq pq p q pq pq p p q pq q pq 86 3 2m m2 m2 3 2m ( m 2)( m 2) m2 (m 2)(m 2) (m 2)(m 2) 2m ( m 2)( m 2) ( m 2) 3(m 2)(m 2) 2m 5(m 2) 3m 12 2m 3m 2m 12 5m 10 5m 10 3m 2m 12 m 2 5 87 4 2 16 4 5 25 88 31 41 1 12 12 12 89 (5 z )(2 z ) 10 z 10 z Copyright © 2017 Pearson Education, Inc Chapter R Review Exercises 90 (8 p q )(2 p5 q 4 ) 16 p q3 16 p 16 p q 1 q 25m3n5 100 2 m n 1/ 25m5 n 1 91 (6 p w4 m12 )0 Definition of a 8 y p 2 8 y ( 4) p ( 2) ( 3) 8 y11 p y 4 p 3 6 8 1/ 101 a (a ) a 94 2 11 a 2 (a11 ) a a 14 = a 14 a 23 23 a a ( p q ) 5 2 97 (7 r 1/ )(2r 3/ )( r 1/ 1/ / 1/ ) 14r 14r17 /12 98 (a / 4b / )(a / 8b 5 / ) a / a / b / b 5 / a / / b / ( 5 / 6) a / / b / ( 5 / 6) a11/ a11/ 8b 1/ 1/ b 99 y / y 2 y / ( 2) y / ( 6 / 3) y 1/ 5 / 5 / 5 / 5 / y y y y 1/ ( 5 / 6) 2 / / y y y / y1/ q 24 16 q 1/ 25m5 n 1/ n1/ (25)1/ m5 / n1/ 5m / 15 4 12 4 p q p 20 q 16 18 12 24 3 16 3 p q p q p 20 ( 18) q 16 ( 12) p 2 q 4 p q 102 m3 / (8m1/ 4m 3 / ) m3 / (8m1/ ) m3 / (4m 3 / ) 8m3 / 1/ 4m3 / ( 3 / 2) 8m3 / / 4m3 / ( 6 / 4) 8m5 / 4m 3 / or 8m5 / 6 p ( m n) [ p ( m n) ] 2 p 2 (m n) 5 p (m n) 5 p 4 ( 2) (m n) 6 ( 5) p 2 (m n) 1 p ( m n) 96 4 ( p q )5 p p 15 12 6 ( 8) ( p q) ( p q ) 3 95 ( p q ) ( 3) ( p q )6 1/ n 25m5 92 (6 x y 3 z ) 2 (6) 2 x 4 y z 4 y6 y6 62 x z 36 x z 93 25m3 ( 2) n5 51 105 3 3 16 2 1250 625 625 4 4 16 16 4 3 3 106 107 3 2 25 p 3 3 p2 25 p p2 p2 108 m 200 100 100 10 103 104 3/ 27 y m3 25 p 125 p3 27 y 50 p 5p 26 y 26 y m3 m2m m2 m 4 y 2 y m m m m m m y 2m y 2m y 2m mm m2 m m2 109 m m 1/ m1/12 Copyright © 2017 Pearson Education, Inc (m1/ )1/ m1/ 31/ 12 m Chapter R Review of Basic Concepts 52 110 p q5 p3q p5q 4 p q5 p3q p5 q 115 16 p5 q 16q p q 16q 2q 116 111 This product has the pattern (a b)(a ab b ) a b3 , the sum of two cubes 4 22 16 3 3 113 9m 2m 5m m or m 9 2m m 114 2 7 7 72 14 14 49 46 7 7 46 23 120 x y x y xy 121 122 123 ba a b 2 2 2 3 2 or 8 1 23 8 124 (5) (5)(5) 25 or 52 8 a 125 7 b 1 8b a 7b 7b 8b 7a 7b Chapter R a 2b m n m n mn r r r r r 9m 2m 3m 16 2m m m 3m 2m 3m 2m 5m m 3m 2m 12m 2m 5m m k k 3 k k 3 k 9 k 3 k 3 k k 3 119 (m )3 m 23 m6 6 45 80 95 16 5 12 24 63 12 4 36 18 12 12 12 36 18 46 12 12 23 23 6 5 23 23 65 30 18m 3m 32m m 118 32 (32 ) 9 3 117 x( x 5) x x x x3 x 4 64 66 112 6 3 3 3 3 3 3 92 1 1 7b 8b a Test False B {2, 4, 6, 7,8} True False ( B C ) D {1} D {1, 4} True 12 (or –3), 0, and integers (a) –13, 49 (or 7) are 12 59 ), and (or –3), 0, , 5.9 (or 10 49 (or 7) are rational numbers (b) –13, (c) All numbers in the set are real numbers Copyright © 2017 Pearson Education, Inc Chapter R Test Let x = –2, y = –4, z = x yz (2) 2(4)(5) (40) 3( x z ) 3(2 5) 3(3) 36 4 4 2 12 (b) commutative property (c) distributive property (d) inverse property ( x x 2) ( x x ) x(2 x 1) x 3x x x x 3x 11x x 10 (6r 5) (6r ) 2(6r )(5) 52 36r 60r 25 11 (t 2)(3t t 4) 3t t t 2 6t 2t 3t t 4t 3t 5t 2t 2 x3 11x 28 x5 x2 x x x3 11x x 28 x3 10 x x2 x x2 5x x 28 x 25 Thus, x3 11x 28 x2 x x5 x5 (a) associative property A = 650, C = 446, Y = 5162, T = 39, I = 12 First, find the value of the numerator C T Y 250 1000 12.5 A A A I 6.25 1250 A 446 39 5162 250 1000 12.5 650 650 650 12 6.25 1250 650 313.98 313.98 R 104.7 Drew Brees’ quarterback rating was about 104.7 53 13 For the year 2005, x = Adjusted poverty threshold 2.719 x 196.1x 8718 2.719 62 196.1 8718 97.884 1176.6 8718 9992.484 $9992 14 For the year 2012, x = 13 Adjusted poverty threshold 2.719 x 196.1x 8718 2.719 132 196.1 13 8718 459.511 2549.3 8718 11, 726.811 $11, 727 15 x 17 x (3 x 7)(2 x 1) 16 x 16 ( x ) 42 ( x 4)( x 4) ( x 4)( x 22 ) ( x 4)( x 2)( x 2) 17 24m3 14m 24m 2m(12m 7m 12) 2m(4m 3)(3m 4) 18 x3 y x3 y 72 ( x y x3 ) (8 y 72) x3 ( y 9) 8( y 9) ( x3 8)( y 9) ( x 23 )( y 32 ) ( x 2)( x x 4)( y 3)( y 3) 19 a b2 a b a b a b 2 20 27 x x 3x x Copyright © 2017 Pearson Education, Inc 54 21 Chapter R Review of Basic Concepts 5x2 x x 3x 30 x x x x7 x6 x x 2 x8 x x 30 x3 x x4 3x2 (5 x 1)( x 2) x ( x x 2) x (5 x 1) ( x 4)( x 1) (5 x 1)( x 2) x ( x 2)( x 1) x (5 x 1) ( x 2)( x 2)( x 1) x ( x 1) x ( x 1) 2 x ( x 1) 3( x 1) 2x x 22 x 3x 2 x x 2x x ( x 2)( x 1) (2 x 3)( x 1) The least common denominator is (x + 2)(x + 1)(2x – 3) x( x 2) x(2 x 3) ( x 2)( x 1)(2 x 3) (2 x 3)( x 1)( x 2) x 3x x2 x ( x 2)( x 1)(2 x 3) ( x 2)( x 1)(2 x 3) x 3x x x x2 x ( x 2)( x 1)(2 x 3) ( x 2)( x 1)(2 x 3) x(4 x 1) ( x 2)( x 1)(2 x 3) (a b)(1) ab ab ab 23 2a 3 2a 2a (3 2a )(1) 2a ab ab 2a 2a 2a If – 2a is used as the common denominator, 2a The rational the result will be 2a 2a 2a and expressions are 2a 3 2a equivalent 24 18 x5 y (9 x y8 )(2 x) 25 x y8 x 3x y x 32 x x 18 x 16 x x x 2x 2x 2x 2x 26 27 28 x y x y x y x y 14 14 11 11 11 11 14 11 14 11 11 7 11 x 2 y 1/ x 6 y 1 29 5 / 2 / 5 2 y x y x x 6 ( 5) y 1 ( 2) x 1 y 64 30 27 2 / 27 64 2/3 27 1/ 64 2 3 4 16 31 False For all real numbers x, x2 x 32 Let L 3.5 L 3.5 2 2.1 t 2 32 32 The period of a pendulum 3.5 ft long is approximately 2.1 seconds y2 y ( y 2) y y y y y y y 4 y y ( y 2) y ( y 2)( y 2) y Copyright © 2017 Pearson Education, Inc y x ... 1) 128 x x x x x by the additive inverse property (0 in the form x x ) ( x x 1) x by the associative property of addition ( x 1) x by factoring a perfect square trinomial... First form the intersection of Q and M We have Q M = {0, 2, 4, 6, 8} = M Now form the union of this set with R We have (Q M) R = M R = {0, 1, 2, 3, 4, 6, 8} 100 (R N) M First form... Basic Concepts 91 M Q First form M , the complement of M M contains all elements of U that are not elements of M Thus, M {1, 3, 5, 7, 9, 10, 11, 12, 13} Now form the intersection of M
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