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Solution manual for college algebra 12th edition by lial

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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 24  2  2  2  2  16 20 26  222222  64 21 35  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 43 8   6   7 15     3    6   5   6   5  12   28   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  pq 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 38 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  2q 28 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   44   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 33   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 ba ab ab p  pq q p pq pq 1b 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  rq rq r  q r  q rq  rq 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 qr rq rq   qr rq qr rq   1 1(r  q )  x  9y b  a ab  ba 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 ba  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)  2xx13  ( 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  43 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  2m4   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 53p 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  2p2p 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  xx 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  52   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  23   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 42 16     2 1     4 4    4 11 11   4 2 11 11   42 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 83 4   27  1  4 3  3 102 3 3 32 32 5 3 15   53      7 21  4 21  7 1 1 4   4 2 4 2 4 1   4 2 4 2 3 22  3  3 3 3   37  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 28  27    2   1    6 30    6  5      12 27 48 43 93 16     3 62 1 35     3 3 12 15    12 12 12 98 45 25 25  63 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 52   52 3 52 3 52 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 p4   p  4   p  p 2 9  r  3  r 9r   3       ( m  n)  3m  m  n  4mn 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 mn 107   2 mn 2 mn 2 mn 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 9r  r  r 9  r   r    3 r 3 r 3 r 32  r    22 p 2 32  3   52 or 45  12   15   33 p4 p4   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 35           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  pq 1 84 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    m4 m4 m4 We may also use – m as the common denominator In this case, the result will be 2 m The two results are equivalent rational 4m 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  x9   ( 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  m2   3 2m ( m  2)( m  2) m2  (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 31  41  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   62 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    mm m2 m m2 109 m   m 1/  m1/12  Copyright © 2017 Pearson Education, Inc  (m1/ )1/  m1/ 31/ 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    23 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 23  m6 6      45 80 95 16     5  12 24 63     12  4  36 18    12 12 12 36   18 46   12 12 23 23    6 5 23 23   65 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   92 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 x5 x2  x  x  x3  11x  x  28 x3  10 x  x2  x  x2  5x  x  28  x  25 Thus, x3  11x  28  x2  x   x5 x5 (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  b2  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) ab ab ab 23    2a  3  2a 2a  (3  2a )(1) 2a ab ab    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 y2 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

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