Chapter R REVIEW OF BASIC CONCEPTS Section R.: Sets. The elements of the set {,,,, 0} are all the natural numbers from to 0 inclusive. There are 9 elements in the set, {,,,,, 7,, 9, 0}.. Each element of the set {,,,, } after the first is found b multipling the preceding number b. There are elements in the set, { },,,,,.. To find the elements of the set {7,, 7, 7}, start with 7 and add to find the net number. There are 7 elements in the set, {7,, 7,, 7,, 7}. 7. When ou list all elements in the set {all natural numbers greater than 7 and less than }, ou obtain {, 9, 0,,,, }. 9. The set {,,,, } has a limited number of elements, so it is a finite set.. The set { },,,, has an unlimited number of elements, so it is an infinite set.. The set { is a natural number larger than }, which can also be written as {, 7,, 9, }, has an unlimited number of elements, so it is an infinite set.. There are an infinite number of fractions between 0 and, so { is a fraction between 0 and } is an infinite set. 7. is an element of the set {,,, }, so we write {,,, }. 9. is not an element of {,,, 0}, so we write {,,, 0}.. 0 is an element of {, 0,, }, so we write 0 {, 0,, }.. {} is a subset of {,,, }, not an element of {,,, }, so we write {} {,,, }.. {0} is a subset of {0,,, }, not an element of {0,,, }, so we write {0} {0,,, }. 7. 0 is not an element of, since the empt set contains no elements. Thus, 0. 9. {,,, } Since is not one of the elements in {,,, }, the statement is false.. {,,,, } Since is one of the elements of {,,,, }, the statement is true.. 9 {,,, } Since 9 is not one of the elements of {,,, }, the statement is true.. {,,, 9} {,, 9, } This statement is true because both sets contain eactl the same four elements. 7. {,, 9} {,, 9, 0} These two sets are not equal because {,, 9, 0} contains the element 0, which is not an element of {,, 9}. Therefore, the statement is false. 9. { is a natural number less than } {, } Since and are the onl natural numbers less than, this statement is true.. {, 7, 9, 9} {7, 9,, } {7, 9} Since 7 and 9 are the onl elements belonging to both sets, the statement is true.. {,, 7} {,, 9} {} {,, 7} {,, 9} {,,, 7, 9}, while {,, 7} {,, 9} {}. Therefore, the statement is false.. {,,, 9} {, 7,, 0} {} Since is the onl element belonging to both sets, the statement is true. 7. {,, 9, 0} {,, 9, 0} In order to belong to the intersection of two sets, an element must belong to both sets. Since the empt set contains no elements, {,, 9, 0}, so the statement is false. 9. {,, } {,, } {,, } Since the two sets are equal, their union contains the same elements, namel,, and. Thus, the statement is true.
Chapter R: Review of Basic Concepts. Since the empt set contains no elements, the statement is true. For Eercises, A {,,,, 0, }, B {,,, 0}, C {, 0, }, D {, 0}, and U {,,,, 0,, }.. A U This statement sas A is a subset of U. Since ever element of A is also an element of U, the statement is true.. D B Since both elements of D, and 0, are also elements of B, D is a subset of B. The statement is true. 7. A B Set A contains two elements, and, that are not elements of B. Thus, A is not a subset of B. The statement is false. 9. A The empt set is a subset of ever set, so the statement is true.. {,, 0} B Since,, and 0 are all elements of B, {,, 0} is a subset of B. The statement is true.. B D Since B contains two elements, and, that are not elements of D, B is not a subset of D. The statement is false.. Ever element of {,, } is also an element of {,,,, }, so {,, } is a subset of {,,,, }. We write {,, } {,,,, }. 7. Since 0 is an element of {0,, }, but is not an element of {,,,, }, {0,, } is not a subset of {,,,, }. We write {0,, } {,,,, }. 9. The empt set is a subset of ever set, so {,,, }. For Eercises 7 9, U {0,,,,,,, 7,, 9, 0,,, }, M {0,,,, }, N {,,, 7, 9,, }, Q {0,,,,, 0, }, and R {0,,,, }. 7. M R The onl elements belonging to both M and R are 0,, and, so M R {0,, }. 7. M N The union of two sets contains all elements that belong to either set or to both sets. M N {0,,,,,,, 7,, 9,, } 7. M N There are no elements which belong to both M and N, so M N. M and N are disjoint sets. 77. N R {0,,,,,, 7, 9,, } 79. N The set N is the complement of set N, which means the set of all elements in the universal set U that do not belong to N. N Q or {0,,,,, 0, }. M Q First form M, the complement of M. M contains all elements of U that are not elements of M. Thus, M {,,, 7, 9, 0,,, }. Now form the intersection of M and Q. Thus, we have M Q {0, }.. R Since the empt set contains no elements, there are no elements belonging to both and R. Thus, and R are disjoint sets, and R.. N Since contains no elements, the onl elements belonging to N or are the elements of N. Thus, and N are disjoint sets, and N N or {,,, 7, 9,, }. 7. (M N) R First form the intersection of M and N. Since M and N have no common elements (the are disjoint), M N. Thus, (M N) R R. Now, since contains no elements, the onl elements belonging to R or are the elements of R. Thus, and R are disjoint sets, and R R or {0,,,, }. 9. (Q M) R First form the intersection of Q and M. We have Q M {0,,,, } M. Now form the union of this set with R. We have (Q M) R M R {0,,,,,, }.
Section R.: Real Numbers and Their Properties 9. (M Q) R First, find M, the complement of M. We have M {,,, 7, 9, 0,,, }. Net, form the union of M and Q. We have M Q {0,,,,,,, 7,, 9, 0,,, } U. Thus, we have ( M Q) R U R R or {0,,,, }. 9. Q ( N U) First, find Q, the complement of Q. We have Q {,,, 7, 9,, } N. Now find N, the complement of N. We have N {0,,,,, 0, } Q. Net, form the intersection of N and U. We have N U Q U Q Finall, we have Q ( N U) Q Q intersection of Q and ( N U) and ( N U) are disjoint sets. Since the is, Q 9. M is the set of all students in this school who are not taking this course. 97. N P is the set of all students in this school who are taking both calculus and histor. 99. M P is the set of all students in this school who are taking this course or histor or both. Section R.: Real Numbers and Their Properties. (a) 0 is a whole number. Therefore, it is also an integer, a rational number, and a real number. 0 belongs to B, C, D, F. (b) is a natural number. Therefore, it is also a whole number, an integer, a rational number, and a real number. belongs to A, B, C, D, F. 9 (c) is a rational number and a real 9 number. belongs to D, F. (d) is a natural number. Therefore, it is also a whole number, an integer, a rational number, and a real number. belongs to A, B, C, D, F. (e) is an irrational number and a real number. belongs to E, F. (f). is a rational number and 00 a real number.. belongs to D, F.. False. Positive integers are whole numbers, but negative integers are not.. False. No irrational numbers are integers. 7. True. Ever natural number is a whole number. 9. True. Some rational numbers are whole numbers.. and are natural numbers.., (or ), 0,, and are integers.. 7. 9... + 0+ 0 +. 9 ( ) + ( 7) + ( 7) () + ( 7) + ( 7) + ( ) 7. 9. 0 ( )( + ) ( )( + ) ( )() 9 9 9 + 7 7 + + + + 7 7 7.. Let p, q, and r 0. p 7q+ r 7 + 0 7 + 00 7 + 00 + 00 + 00
Chapter R: Review of Basic Concepts. Let p, q, and r 0. q+ r + ( 0) q+ p + ( ) 7. Let p, q, and r 0. q r p 0 0 + + 0 0 0 9. Let p, q, and r 0. + + q ( ) ( 0) ( 0) ( 0) + 0 ( p ) r ( ) ( 0). A, C, Y 09, T, I Passing Rating C Y. +. A A T I +. 9.07 A A 09. +. +. 9.07. + 9. +.9. 9.9. A 0, C 7, Y 9, T, I Passing Rating C Y. +. A A T I +. 9.07 A A 7 9. +. 0 0 +. 9.07 0 0.7 + 0.0 +. 0..0. BAC..07 90.0.0 7. Eercise : BAC..07.0.0 Eercise : BAC.0.07 0.0 0.0 The increased weight results in lower BACs. 9. distributive. inverse. identit. No; in general a b b a. Eamples will var, i.e. if a and b 0, then a b 0, but b a 0. 7. p p ( ) p p z z z+ 9. 0 0 z z 0z. m+ + m+ + m+.. 0 + z 9 7 9 0 + z 9 7 9 + z 9 7 + z 9 7. The process in our head should be the following: 7 7 + 7 (7 + ) ( 7) (00) 7 700 9. The process in our head should be the following: (00) (00). 0 7. This statement is false since and. A corrected statement would be or 7. This statement is true since 0 and 0 0. 7. This statement is false. For eample if ou let a and b then and a b. A corrected statement is a b b a, if b > a > 0.
Section R.: Polnomials 77. 0 0 79. 7 7. Let and.. Let and. + ( ) + () +. Let and. 7. Let and. + () + ( ) + ( ) 0 0 () 9. Propert 9. Propert 9. Propert 9. Since and ( ), the number of strokes between their scores is. 97. Pd P 9 9 The P d value for a woman whose actual sstolic pressure is and whose normal value should be is 9. 99. The absolute value of the difference in windchill factors for wind at mph with a 0 F temperature and wind at 0 mph with a 0 F temperature is 9 ( ) 7 7 F. 0. The absolute value of the difference in windchill factors for wind at 0 mph with a 0 F temperature and wind at mph with a 0 F temperature is 7 ( ) F. 0. d( P, Q ) ( ) + or d( P, Q ) ( ) + 0. d( Q, R ) ( ) + 9 9 or d( Q, R ) 9 9 07. > 0 if and have the same sign. 09. < 0 if and have different signs.. Since has the same sign as, and have the same sign. Section R.: Polnomials mn m n. Incorrect:. Incorrect: k k k + 7. Incorrect: 7. Incorrect: cd c c 9. Correct:. 0 + 9 9 9 9. ( )( ) ( )( ). + 7 + + n n n n n 7. ( m )( m )( m ) ( )( ) ( mmm) 9. + + 7m 7m 0... 0 > if 0 0 ( mn) ( m) ( n) 0 m n 0 mn ( ) m m m r ( r ) r r s ( s ) s s
Chapter R: Review of Basic Concepts 7. 9. (a) m ( ) ( m ) ( ) m m t t t t 0 ; B (b) 0 ; C (c) ( ) 0 ; B (d) 0. Answers will var... 7. 9. + ; C is a polnomial. It is a monomial since it has one term. It has degree since is the highest eponent. pq+ pqis a polnomial. It is a binomial since it has two terms. It has degree because is the sum of the eponents in the term p q, and this term has a higher degree than the term pq. + is a polnomial. It is a binomial since it has two terms. It has degree since is the highest eponent. rs rs + rs is a polnomial. It is a trinomial since it has three terms. It has degree because the sum of the eponents in the term rs is, and this term has the highest degree.. + + is not a polnomial since p p p positive eponents in the denominator are equivalent to negative eponents in the numerator.. ( + 7) + ( + ) + + + + 7+ + + +. ( + ) ( + ) ( ) ( ) + ( ) ( ) + + + 7. ( m m + m) ( m + m + m) + ( m m) 9.. m m + m m m m+ m m m m + + m + m m m + 7 m + m m m 7m m (r )(7r+ ) r(7 r) + r() (7 r) () r + r 7r r + r + + + ( ) ( ) 0 + 9 0 + 9 7 7 9 9. ( + + ) + + + 0 +. ( z )( z + z ) ( z )( z ) + ( z )( z) ( z ) z( z ) ( z ) + z( z) ( z) ( z) ( ) z + z + z z z z + 7z z+ We ma also multipl verticall. z + z z z z + ( z + z ) z + z z z( z + z ) z + 7z z+
Section R.: Polnomials 7 7. ( m n+ k)( m+ n k) ( m n+ k)( m) + ( m n+ k)( n) ( m n+ k)( k) 9.... m mn+ km+ mn n + kn km + kn k m + mn n km + kn k We ma also multipl verticall. m n+ k m+ n k km + kn k mn n + kn m mn + km m + mn n km + kn k (m+ )(m ) ( m) m 9 ( )( + ) ( ) ( ) (m+ n) ( m) + ( m)( n) + ( n) m + mn+ n (r t ) ( r) ( r)( t ) + ( t ) r 0rt + 9t 7. ( p ) + q ( ) ( ) 9. 7. p + p q + q p p + + pq q+ q p p+ 9+ pq q+ q [(q+ ) p][(q+ ) + p] (q+ ) p [( q) + ( q)() + ] p 9q + 0q+ p [( a+ b) ] ( a+ b) ( a+ b)() + (9a + ab+ b ) ( a+ b) + 9a + ab+ b a b+ 7. ( + ) ( + ) ( + ) ( + + )( + ) + + + + + + + + 7. ( q ) ( q ) ( q ) ( q q )( q q ) + + q q + q q + q q+ q q+ q q + q q+ 77. ( p p + p) ( p + p+ 7) 79.... 7. p p + p p p 7 p 7p p 7 (7m+ n)(7m n) (7 m) ( n) 9m n (q q+ ) + ( q + q ) q + 9q q + q q + q p( p ) + ( p ) p p+ p p ( ) + ( ) + + + + 7 + 7 7 + 0 7 0 0 0 7 + 0 + 7 + 7
Chapter R: Review of Basic Concepts 9. 9. 9. 9. 97. + 0 0 + 0 + + m + m m+ m + 7m m+ m + m m m m + m m + m + + m + m + m 7m m m+ m+ + + 0 + 9 + + 9 + 7 + 9 7 + + + + or 7 +. + 99 0 (00 )(00 + ) 00 0, 000 9999 0 (00 + ) 00 + (00)() + 0,000 + 00 + 0, 0 99. (a) The area of the largest square is s ( + ). (b) The areas of the two squares are and. The area of each rectangle is. Therefore, the area of the largest square can be written as + +. (c) Answers will var. The total area must equal the sum of the four parts. (d) It reinforces the special product for squaring a binomial: ( + ) + +. 0. (a) The volume is V h ( a + ab+ b ) (00)( 7 7 + + ) 0,0,000ft (b) The shape becomes a rectangular bo with a square base. Its volume is given b length width height or bh. (c) If we let a b, then ( V h a + ab+ b ) becomes ( V h b + bb+ b ), which simplifies to V hb. Yes, the Egptian formula gives the same result. 0. 90.00000907( 90).09( 90) +.0( 90), 9.. The formula is. high. 0. 97.00000907( 97).09( 97) +.0( 97), 9.. The formula is eact. 07. 09.. 00. 00 00,000,000....
Section R.: Factoring Polnomials 9 Section R.: Factoring Polnomials. The greatest common factor is. m+ 0 m + m+. The greatest common factor is k. () k + k k k + k k k +. The greatest common factor is. 7. The greatest common factor is pq. pq pq ( pq)( p) + ( pq)( q) pq p+ q 9. The greatest common factor is k m. k m + k m k m k m + k m k k m m () k m ( z m) +. The greatest common factor is (a + b) ( a+ b) + m( a+ b) [ ( a+ b) ]() + [ ( a+ b) ]( m) ( a+ b)( + m). (r )(r + ) (r )(r + ) (r + )[(r ) (r )] (r + )[r r + ] (r + )(r ). ( m ) ( m ) + ( m ) ( m ) ( m ) ( m ) + ( m ) m ( m m ) + + + ( m )( m+ + m m+ ) ( m )(m 7m+ 7) 7. The completel factored form of. is 9. st + 9t 0s (st + 9 t) (0s + ) ( t s+ ) (s+ ) (s+ )(t ). + (m ) ( am a) m am a + + ( m + ) a( m + ) ( m + )( a). 0 + pq q p q p p pq q p + 0 ( ) + ( ) ( p )( q + ). The positive factors of could be and, or and. Since the middle term is negative, we know the factors of must both be negative. As factors of, we could have and, or and. Tr different combinations of these factors until the correct one is found. a a+ a a 7. The positive factors of are and. Since the middle term is positive, we know the factors of must both be positive. As factors of, we could have and, or and. Tr different combinations of these factors until the correct one is found. m + m+ m+ m+ 9. The positive factors of are and, or and. Since the middle term is positive, we know the factors of must both be positive. As factors of, we could have and, or and. Tring different combinations of these factors we find that p + p+ is prime.. Factor out the greatest common factor, a: a + 0a a a(a + a ). Now factor the trinomial b trial and error: a + a (a+ 7)(a ). Thus, a + 0a a a(a+ 7)(a ).. The positive factors of could be and, or and. As factors of, we could have and, and, and, or and. Tr different combinations of these factors until the correct one is found. k + kp p k + p k p. The positive factors of can onl be and. As factors of, we could have and, and, and, or and. Tr different combinations of these factors until the correct one is found. a 7ab b a+ b a b 7. The positive factors of could be and, and, or and. The factors of are and. Tr different combination of these factors until the correct one is found. ( + )( )
0 Chapter R: Review of Basic Concepts 9. Factor out the greatest common factor,... 7. a : a + 0a b a b a a + ab b Now factor the trinomial b trial and error: a + ab b a b a+ b Thus, a + 0a b a b a a + ab b a a b a+ b + + ( m) ( m)() + 9m m ( m) m (m ) + + (a ab 9 b ) [( a) ab ( b) ] a ab b + + + + [( a) + ( a)( b) + ( b) ] (a+ b) + + + + ( ) + ( )(7) + 7 9 ( ) 7 ( + 7) ( a b) ( a b) + 9 ( a b) ( a b) + + ( a b) ( a b) () [( a b) ] ( a b ) 9. (a) Since ( + ) + 0+, a matches B... (b) Since ( ) 0+, b matches C. (c) Since ( + )( ), c matches A. (d) Since ( + )( ), d matches D. 9a ( a) (a+ )(a ) +. s 9 t ( s ) ( t) (s + t)(s t) 7. 9.... 7. 9. ( a+ b) ( a+ b) [( a+ b) + ][( a+ b) ] ( a+ b+ )( a+ b ) p ( p ) ( p + )( p ) ( p + )( p ) ( p + )( p+ )( p ) Note that p + is a prime factor. a a ( a)( + a+ a ) ( a)( + a+ a ) 7 ( ) ( ) ( ) + + ( )( + + 9) 9 7 + z ( ) + ( z ) ( + z ) ( ) ( )( z ) ( z ) + ( + z )(9 z + z ) ( r + ) ( r + ) ( r ) ( r ) ( r )( ) + + + + + ( r ) r r ( r )( ) + + + + + + [ r+ ] r r r + + + + + rr ( + r+ 0) 7 ( m+ n) ( m n) ( m n) + ( m+ n) + ( m+ n) ( m n) () + + + + m+ n + m + mn+ n + + + + + ( m n)( 9 m n m mn n ) 7. The correct complete factorization of is choice B: ( + )( + )( ). Choice A is not a complete factorization, since can be factored as ( + )( ). The other choices are not correct factorizations of.
Section R.: Factoring Polnomials 7. ( ) ( + )( ) or ( )( + ) Use the patterns for the difference of cubes and sum of cubes to factor further. Since ( )( + + ) and + ( + )( + ), we obtain the factorization ( )( + ) ( )( + + )( + )( + ). b + bc+ c (b + bc+ c ) ( b) ( b)( c) c + + ( b+ c) [( b+ c) + ][( b+ c) ] (b+ c+ )( b+ c ) 7. + ( + ) ( + ) ( ) ( ) ( )( ) + + + 7. From Eercise 7, we have ( )( + + )( + )( + ). From Eercise 7, we have ( )( + )( + + ). Comparing these answers, we see that + + ( + )( + + ). 77. The answer in Eercise 7 and the final line in Eercise 7 are the same. 79. m m 0 Let m, so that ( m ) m. 0 ( )( + ). Replacing with m gives m m 0 ( m )( m + ).. Let k. This substitution gives 7(k ) + (k ) 7 + (7 )( + ) Replacing with k gives 7(k ) + (k ) [7(k ) ][(k ) + ] (k 7 )(k + ) (k 9)(k + ) (7k )()( k + ) 9(7k )( k + ).. Let a. This substitution gives 9( a ) + 0( a ) + 9 + 0+ + + + Replacing b a gives 9( a ) + 0( a ) + [( a ) + ] (a + ) (a 7) 9. 9. 9. 9. 97. 99. 0. 0. p ( m n) + q( m n) ( m n)( p + q) + + + + z z 9 () z ()(7) z 7 (z + 7) + 000 (0 ) + (7 ) (0+ 7 ) (0 ) (0 )(7 ) (7 ) + (0 + 7 )(00 70+ 9 ) m ( m ) (m ) ( m ) m () + + (m )(m + 0m + ) + ( + ) + ( + ) ( ) ( ) ( ) + + + + + ( + ) + + + 9 + + + + + + + ( )( 9 ) ( )( 9 ) ( )( ) 9( )( ) + + + + + + 9 + 7 7 z + The sum of squares cannot be factored. z + is prime. ( + ) ( ) [( + ) + ( )][( + ) ( )] ( + + )( + + ) 0. Answers will var.
Chapter R: Review of Basic Concepts 07. 09. z + bz+ ( z) + bz+ 9 will be a perfect trinomial if bz ± ( z)( 9) bz ± z b ±. If b, z + z+ ( z+ 9). If b, z z ( z 9) +. 00r 0r+ c (0 r) (0 r)() + c 0r will be a perfect trinomial if c 9. If c 9, 00r 0r + 9 ( 0r ). Section R.: Rational Epressions +. In the rational epression, the solution to the equation 0 is ecluded from the domain. 0 The domain is { }. + 7. In the rational epression, the (+ )( ) solution to the equation ( + )( ) 0 is ecluded from the domain. (+ )( ) 0 + 0 or 0 The domain is,.. In the rational epression, the + + solution to the equation + + 0 is ecluded from the domain. + + 0 ( + )( + ) 0 + 0 or + 0 The domain is {, }. 7., (a) (b) + + +. + + 9. If, (a)... 7. 9.. (b). k + ( k + ) 9k + 9( k + ) 9 ( t) ( t) t + t t+ t t + + ( + ) ( + ) ( + ) + m m+ ( m )( m ) m m + m ( m )( m+ ) m+ m + m 9 (m+ )(m ) m+ m 9 (m+ )(m ) m+ p p p 0 p 0 p 9p 0p 9p p p p 9 p 9 p. k + k + k + k + ( k + )() ()( k + ) 9. 7. 9. + ( + ) + ( + ) ( + ) ( + ) a+ a 9 a+ a a 0 a 0 a a 0 a 0 a 9 ( a+ ) a a+ ( a ) a+ a ( a+ ) a+ a a p p p 9 p+ 0 p p p p+ ( p )( p+ ) ( p )( p ) ( p )( p+ ) ( p )( p )
Section R.: Rational Epressions... m + m+ m + m+ m + m+ m + 0m+ m + m+ m + 0m+ m + m+ m + m+ ( m+ )( m+ ) ( m+ )( m+ ) m+ ( m+ )( m+ ) ( m+ )( m+ ) m+ z w + z w z + w + z + w z w ( z w) + ( z w) ( z+ w)( z w) ( z+ w) + ( z+ w) ( + )( ) ( z w)( + ) ( z+ w)( z w) ( z+ w)( + ) ( + )( ) + + + + ( + )( + ) ( + )( ) ( )( + + ) ( + )( + ) + + + 7. Epressions (B) and (C) are both equal to, since the numerator and denominator are additive inverses. ( + ) B. + + 9. C. ( ) 9 0 9 + + + k k k k k k k. 0 + + + + m m m m m m 0 0 7 + + 0m 0m 0m 0m b a b a b a b a a a a a a a a.. 7. 9. ( 7+ ) ( 0 ) 7 + 0 9+ 7 9+ 7 9+ 7 7 + + 0+ 9 + 7 7 + 9 + 7 9 ( + 7) 9 + 7 ( z) ( + z) + + + z z ( + z)( z) ( z)( + z) ( z) + ( + z) ( z)( + z) z+ + z ( + z)( z) ( + z)( z). Since a ( )( a) we have ( ) ( ) a a a ( a) a a a a We ma also use a as the common denominator. ( ) a a ( a )( ) a a a a The two results, and a a, are equivalent rational epressions.. Since ( )( ) + + ( ) ( )( ) + + ( ) + + + We ma also use as the common denominator. In this case, our result will be. The two results are equivalent rational epressions.
Chapter R: Review of Basic Concepts. + + + + + + + + + ( + ) ( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + ) + ( + ) + + + ( + )( + ) ( + )( + ) + ( 7)( + ) 7 7 + + + + + 7. 9.. ( ) ( ) + ( )( + ) ( )( + ) ( )( + )( ) ( )( + )( ) ( ) ( ) + 9 ( ) ( )( + )( ) ( )( + )( ) ( )( + )( ) ( ) + + + + + + + + + + ( + ) ( + ) ( + ) + + +. b b. ( b) () ( )( + ) + ( )( + ) + ( )( + ) + b b b b b b b + b + b + b b b b b b ( b)( + b) + ( + b) ( )( + ) ( )( + ) ( )( + ) ( b)( + b) ( b) ( + b)[( b) + ] ( + b)( b) ( b)( + b) or ( b)[( + b) ] ( b) b b( b) m ( m )( m ) m+ m+ m m + Multipl both numerator and denominator b the LCD of all the fractions, (m + )(m ). m ( m+ )( m ) m ( m+ )( m )( m) ( m+ )( m ) m ( m+ )( m ) ( m+ )( m ) m+ ( m+ )( m ) ( m+ )( m ) m+ ( m+ ) mm+ m mm ( ) m m m m m ( + h) ( h ) 7. + h + h ( + h)( h) ( + h) ( + h) + h ( ) ( + h) h h ( + h)( h) + h h + h h + h h + h + ( ) 9. ( ) + + ( )( + ) + + ( ) + ( ) + ( ) + + +
Section R.: Rational Eponents 7. Altitude of 00 feet,. (thousand) The distance from the origin is 7..0, 0.9(.) +.7 which represents about 0 miles. Section R.: Rational Eponents. (a) ; B ; D (b) (c) ; B ( ).. Because is a relativel large number, it is generall acceptable not to simplif it to be,. r r ( ) r r r r r. m n m n ( m n ) mn m n 7 7 7 n m n 7 m 7. + r r r r r r ; (d). ( ) ( ). ( ) 7. 9.... ( ) 9, also 9 9 ( ) ( ) a a D + ( ) 9. ( a ) ( a ) a a a 0 a a Because is a relativel large number, it is generall acceptable not to simplif to. 0 a. 0 0 p 0 0 p p p p p 0 p p p. + pq q pq pq pq pq pq p q p p q p q pq. ( a ) + a a aa a a a q a a a a a 7. 9. 7 7 0 r r 0 r r r 7. 9.. / 9, because 9. /, because. /, because. 7 7
Chapter R: Review of Basic Concepts. ( ) / is not a real number, because no real. (a) number, when squared, will ield a negative quantit. / / ; E 9 9 7.. / / 0 / ( ) ( ) / / / 0/ 7/ / 0/ (b) (c) / / / 9 9 9 7 ; G / / 9 9 7 ; F 7. / / / / / / m 9n m 9 n / n m n m / / m n / n m / / m n ( ) / m n ( /) ( /) ( / m n ) ( /) ( /) ( /) ( / m n ) / / m n / / m n (d) 7. / / / 9 9 9 7 ; F / / 9. / 7/0 / /+ 7/0+ / p p p p ( p ) / / p /0+ 7 /0+ /0 /0 p p p p ( /0) ( /0 p ) 0 /0 p p /0 /0 / / 00 00 0 000 9. / / 7... 7. 9.. / / / 7 7 7 / / /+ / / 9 / / / / / 7/ / 7/ + ( /) / / / / k k k / / + ( ) /+ ( / k k k k ) / k k k / k / ( /) / 7. (a) Let w, 9, 9, 9 t. / / w () ( ), 9, 9 0 sec () (b) To double the weight, replace w with w to get, 9, 9 (); t so. /. / ( w) w the holding time changes b a factor of.. / 7. 7. 77. / / / / / / / ( 0 ) 0 0 0 /+ / /+ / 7/ / 7/ / kk ( k ) kk + kk 0 / / k + k / / ( + )( ) has the form ( a b)( a+ b) a b. / / / ( + )( ) ( )
Section R.: Rational Eponents 7 / / 79. ( r r ) / / / / ( r ) ( r )( r ) + ( r ) + r r + r r r + r r + r r + r r + r / / 0. Factor k + k, using the common factor k : k + k k (k + ). Factor t + t, using the common factor t : t + t t ( t + ) / /. Factor 9z + z, using the common / factor z : / / / 9z + z z (9+ z) / 7/ 7. Factor p p, using the common 7/ factor p : / 7/ 7/ / p p p ( p ) 7/ p ( p ) / 7/ 9. Factor a + a, using the common 7/ factor a : / 7/ 7/ a + a a ( a+ ) / / / 9. Factor ( p+ ) + ( p+ ) + ( p+ ), / using the common factor ( p + ). / / / ( p+ ) + ( p+ ) + ( p+ ) / ( p+ ) [ + ( p+ ) + ( p+ ) ] / ( p+ ) ( + p+ + p + p+ ) / ( p+ ) ( p + 9p+ ) 9. Factor / / / (+ ) + (+ ) + (+ ) / using the common factor ( + ) : / / / ( + ) + ( + ) + ( + ) / ( + ) + ( + ) + ( + ) / ( + ) + ( + ) + 9 ( + + ) / ( + ) 7 + + + + + / + 7 + + / ( ) ( ) / ( ) ( ) + 9 + + + 9 + + 9. Factor (+ ) + (+ ) (+ ) 97. 99. 0. 0. /9 /9 /9 /9 using the common factor (+ ) : /9 /9 /9 (+ ) + (+ ) (+ ) /9 (+ ) ( ) ( ) + + + /9 (+ ) + ( + 9) ( + + 9) /9 (+ ) 9 + + /9 (+ ) 0 + 9+ a b a + b + + a b a b b a ( ab) ab ab b+ a ab ab b+ a ab ab b+ a + r q q r q r + q r q r q r r+ q r + q ( )( + ) ( r q) ( r q) rq + r q rq r + q q+ r r q r q q r r + q q r r q ( r q) 9 9 + 9 9 9 9 9 ( 9) or 9 9 ( + ) ( ) ()( + ) ( ) ( + ) ( )( + ) ( + ) ( + ) ( + ) ( + ) ( ) ( + ) ( )( )
Chapter R: Review of Basic Concepts 0. 07. 09. ( ) + ( ) ( ) ( ) + ( ) ( ) + ( ) + ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 7 7 a 0 ( a ) 0 a 7,000. Let length of side of cube. Then length of side of bigger cube (side tripled). is the volume of the cube, and.. ( ) 7 is the volume of the bigger cube. The volume will change b a factor of 7. / / / /. 0 (. 0) ( ) / / / 000 000 / / 000 000 0 00 Section R.7: Radical Epressions /. (a) F; ( ) / (b) H; ( ) / ( ) / (c) G; ( ) (d) C; ( ) /. / or m m m. / ( ) 7. 9.. A ( m+ p) ( m+ p) or m+ p / k k / / / p ( p ) p. It is true for all 0.. ( ) k m k m k m k m 7. 9... This is not a real number since no real number raised to the fourth power will ield a negative quantit.. 7. 7 7 9. pqr pqr pqr 7 7... 7. 9 9 m m m n n n 9.... ( ) () ( ) z z z z + cannot be simplified further.
Section R.7: Radical Epressions 9 7. 9.... 7. 9. 9 z z z z z ( g h) ( 9r ) gh gh h h gh 9r 9r r r 9r h g h r h g hr r r r r r h 9g hr r / / / / / / / / / / / / /. Cannot be simplified further. + 7 + +. 7. + + 7 + + 7 0 ( ) ( ) 9. This product has the pattern ( a+ b)( a b) a b, the difference of squares. + 9 7. 7. This product has the pattern ( a b)( a + ab+ b ) a b, the difference of cubes. ( ) + + 0. 7. This product has the pattern ( a+ b) a + ab+ b, the square of a binomial. ( + ) + ( ) + ( ) + + + + + 7. This product can be found b using the FOIL method. 77. 79.. ( + )( ) + + + mn m mnm m n n n m n m n n n n 9 9 + + + + + + + + + + + +
0 Chapter R: Review of Basic Concepts + + + 7.. 7. 9. + + + 9 9 + + ( ) ( ) 7 7 + 7 7 7 + 7 ( 7 )( 7 ) ( 7 + )( 7 ) 7 7 7 7 + ( 7) ( ) 7 7 + 7 7 + 7 + 7+ + 7 7+ + 7 p p p p + p + p p p p p p ( ) ( p ) ( ) 9. 9. 9. 97. 99. 0. 0. 0. ( ) ( ) + + m m m+ n + m+ n + m+ n m+ n m m n ( + ) ( m+ n) ( + ) m m n ( m+ n) m m n m n 9 9 ( + ) S. n. 9. The speed of the boat with an eight-man crew is appro. 9. ft/sec. Windchill temperature.7 +.T.7 V +.7TV Windchill temperature..7 +. 0.7 0... +.7 0 0. The table gives. 0 0 0 0 9 9 7.99 and π.9, so it gives si decimal places of accurac. Chapter R: Review Eercises. The elements of the set {,, 0,, 0} are the even numbers from to 0 inclusive. The elements in the set are {,, 0,,,,, 0}.. True. The set of negative integers {,,,, }, while the set of whole numbers {0,,,, }. The two sets do not intersect, and so the are disjoint.. True
Chapter R: Review Eercises 7. False. The two sets are not equal because the do not have the same elements. 9. True. True. A {,, 9,0}. B E 7. D 9. ( C D) B {,} B {,,,,,}.,, (or ), 0, and are integers.. 0 is a whole number, an integer, a rational number, and a real number.. π is an irrational number and a real number. 7. Answers will var. Sample answer: The reciprocal of a product is the product of the reciprocals. 9. Answers will var. Sample answer: A product raised to a power is the product of each factor raised to the power.. commutative. associative. identit 7. In a sample of 000 students, % + % + % or 7% are epected to be over 9, or 000(0.7) 70 students. 9... ( )( ) ( )( ) 0 9 9 9 7 9 ( ) ( ) ( ) 9( ) [ ( )] [ + ] ( ) 9( ) () ( 7) + 7 0 0 0. Let a, b, c. c(a b) [( ) ( )] 0 + 0 () 7. Let a, b, c. 9a+ b 9( ) + ( ) a+ b+ c + ( ) + 9 + ( ) + 9... (q 9q + ) + (q q+ ) q 9q + + q q+ 7q 9q q+ 9 ( 7)(+ 7) + 9 + 9 (k m) ( k) ( k)( m) + ( m) 9k 0km+ m. (a) million (b) Evaluate.. +.0 +.+. when 0.. 0. 0 +.0 0 +. 0 +.. 0. 0 +.0 0 +. 0 +. 0 0+ 0+ 0+.. or approimatel million (c) The approimation is million high. 7. (a) million (b) Evaluate.. +.0 +.+. when... +.0 +. +... +.0 +. +. 9. 7. + 7 + +.. or approimatel million (c) The are the same.
Chapter R: Review of Basic Concepts 9. 0m 9m + m+ m + m m + m+ 0m 9m + m+ 0m + m m + m m m m + m + 0 Thus, + + m m + 0m 9m m m +. 7. 7. (r ) + (r ) Let r. With this substitution, (r ) + (r ) becomes +. Factor the trinomial b trial and error. + ( 7)(+ ) Replacing with r gives [ (r ) 7 ][ (r ) + ] (9r 7)(r + ) (9r 0)(r+ ) (9r 0)(r+ ). ( ) + ( )()( )() ( )[( ) + ( )()()] ( )[ + 0] ( )(9 )... 7. b b + b 0 b + Insert the missing term in the divisor with a 0 coefficient. b b + 0b+ b b + b 0 b + 0b + b b + 0b 0 b + 0b Thus, b b + b 0 b + b + b +. z ( z ) + 9( z ) ( z ) [ + ( z )] ( ) ( + z ) ( z ) (z ) z zk k ( z k)( z+ k) 7 a a b 90a b a (a ab b ) a (a+ b)(a b) 7. 77. 79. k + k k( k + )() k k k ( k + )( k ) k k ( k ) k ( k ) + + + + + + + + + + + + ( + )( ) ( + )( + ) + ( + )( + ) ( + )( ) + p q p pq q p ( p q) p pq+ q p + q p q p pq q p + q ( p q) p pq+ q p ( p+ q)( p q) ( p q)( p+ q) ( p q) p pq+ q ( p+ q)( p pq+ q ) p ( p+ q)( p+ q) p 9. 9m 9 n (7 m ) ( n) (7m + n)(7m n). m m m( ) m + + m m ( m)( ) m m m m + m m m We ma also use m as the common denominator. In this case, the result will be m. The two results are equivalent rational m epressions.
Chapter R: Test + pq ( + p q p q). pq pq ( pq ) ( p) ( q) pq() pq ( pq ). 7. 9. 9. 9. 9. 97. 99. 0. pq + pq q+ p pq + ( z )( z ) 0z 0z 0 0 ( pwm ) Definition of a 7 p p 7 ( ) p p ( p+ q) ( p+ q) ( p+ q) ( p+ q) + ( p+ q) ( p+ q) / / / /+ /+ / 7 / (7 r )( r )( r ) r r / / + ( ) / + ( /) / / / / / / ( /) /+ / / / / / / /+ / 0z z /+ / / 0z z 7/ / z (z ) z ( z ) z () 0z z 0. 00 00 00 0 0. 07. 0 p p p p p p 0p p p / / // 09. / m m ( m ) m / m m. This product has the pattern ( a+ b)( a ab+ b ) a + b, the sum of two cubes... 7. 9.. + + + + + + m m m + m 9m m m m + m m m m m m + m m m m m m + m m 9m m + m m or m 9 m + m + + 9 7 ( + ) ( + ) ( + ) + + ( m ) m m a b a a b b. One possible answer is a b a b. a + b a + b a b a b. (t + ) t or t 7. ( ) ( )( ) or Chapter R: Test. False. B {,,,7,}. True. False. D. False. ( B C) D {} D {, }. True. (a), (or ), 0, and 9 (or 7) are integers.
Chapter R: Review of Basic Concepts (b), (or ), 0, 9,.9 (or ), and 0 9 (or 7) are rational numbers. (c) All numbers in the set are real numbers. 7. Let,, z. + z ( ) + ( )() ( + z) ( + ) + ( 0) () 9. (a) associative propert (b) commutative (c) distributive (d) inverse 9. A 9, C 7, Y 07, T, I 0 Rating C Y. +. A A T I +. 9.07 A A 7 07. +. 9 9 0 +. 9.07 9 9 7.9 Matt Hasselbeck s rating is 7.9. 0... ( + ) ( ) + (+ ) + + + + + (r ) ( r) ( r)() + r 0r+ ( t + )(t t + ) t t+ t + t t+ t t + t t + t + t +. + + 0+ 0 + 0 + + + Thus, + +.. Adjusted povert threshold.7 +.+ 79.7 +. + 79 9. +.9 + 79 0. approimatel $0. Adjusted povert threshold.7 +.+ 79.7 +. + 79.9 + 77. + 79 797. approimatel $797. 7.. 9. 7+ 7 ( 7)( ) ( ) ( + )( ) + ( ) ( + )( + )( ) m m m m(m 7m ) m(m+ )(m ) 9 + 7 ( 9 ) ( 7) ( )( ) ( 9) ( 9) ( )( 9) ( )( + + )( + )( )
Chapter R: Test 0. 7 9 + + 0 + (+ )( ) ( + + ) (+ ) ( )( + ) (+ )( ) ( + )( + ) (+ ) ( + )( )( + ) ( + ) ( + ) ( + ) ( + ). + + + + ( + )( + ) ( )( + ) The least common denominator is ( + )( + )( ). ( ) ( + ) + ( + )( + )( ) ( )( + )( + ) + + ( + )( + )( ) ( + )( + )( ) + + + ( + )( + )( ) ( + )( + )( ) (+ ) ( + )( + )( ) a+ b a b a+ b ( a b)( ). a a a ( a)( ) a+ b a b a + a a a If a is used as the common denominator, a the result will be. The rational a a a epressions and are a a equivalent.... / / / ( ) ( ) / / / 7 7 7 9 (9 )( ) 9 7. + + 9 +. ( )( + ) ( ) ( ) 9. + 7 7 7 + 7 7 7 7 7( + 7) ( + ) ( + ) 0. Let L.. L. t π π. The period of a pendulum. ft long is approimatel. seconds.. ( ) ( ) ( ) ( + )( ) +