The opposite of a negative number is positive.. The absolute value of a negative number is positive.. The quotient of a negative number and a positive num-ber is negative... ↑ 7 places S
Trang 1⫺1
0
⫺1
0
0
Chapter R
Review of Basic Algebra
Exercise Set R.1
RC2 The correct answer is (b) See page R-2 in the text
RC4 The correct answer is (f) See page R-3 in the text
RC6 The correct answer is (d) See page R-4 in the text
2 0, 1, 12,√
25
4 −6, 0, 1, −4, 12,√25, −123
6 √
3, 0.131331333133331
8 12
10 −3.43, −11, 12, 0, 1134, −137
12 All of them
14 {s, o, l, v, e}
16 {1, 3, 5, 7, 9, 11}
18 {−3, −2, −1}
20 {x|x is an integer greater than 3 and less than 11}
22 {x|x is a rational number or x is an irrational number}, or
{x|x is a real number}
24 {x|x is a real number and x ≤ 21}, or {x|x ≤ 21}
26 18 > 0
28 7 > −7
30 0 > −11
32 −6 < −3
34 −7 > −10
36 −1315< 11
250
38 −13.99 < −8.45
40 −1415 = −0.933 and −2753= −0.509433962 , so
−1415 < −2753
42 7 > x
44 t ≤ 102
3
46 True, since −7 = −7 is true
48 True, since −11 > −1312is true
50 x < −1
52 x ≥ −1
54 x < 0
56 x ≤ 0
58 3
60 16
62 127
64 13 8
66 16.4
68 465 70
0
−15
= |0| = 0
72 | − 5| ≥ | − 2|
74 | − 8| ≤ |8|, or | − 8| ≥ |8| since | − 8| = |8|
Exercise Set R.2
RC2 The opposite of a negative number is positive RC4 The absolute value of a negative number is positive RC6 The sum of 0 and a negative number is negative RC8 The quotient of a negative number and a positive num-ber is negative
2 −25
4 2
6 −12
Trang 28 −3
10 −10
12 −32
14 1.9
16 −11.19
18 −2
7
20 −54
22 −34+1
8= −68+1
8= −58
24 −56+−78= −2024+−2124= −4124
26 9
28 0
30 2
3
32 2x
34 −5
36 −20
38 46
40 0
42 5
44 44
46 −17.7
48 −34.8
50 −135
52 −78−−52= −78+ 20
8 =
13 8
54 −2
3−4
5= −10
15+
−12 15
= −22 15
56 −47−−59= −3663+35
63 = −631
58 −40
60 −45
62 35
64 −152
66 42.7
68 −5512
70 3
72 432
74 71.04
76 65 14
78 −12564
80 −8
82 −9
84 8
86 0.7
88 Not defined
90 0
92 Not defined
94 10 9
96 −6 5
98 −1 65
100 1 0.8 This can also be expressed as follows:
1 0.8 =
1 0.8·10
10 =
10
8 =
5
4, or 1.25
102 8x
104 3
5÷−67= 3
5·−76= −107
106 −125 ÷−103= −125 ·−103= 8
108 −3
110 110
112 −4000
114 5
8÷−12= 5
8·−21= −54
116 −59÷−56= −59·−65= 2
3
118 −35÷−58= −35·−85= 24
25
120 −5
8÷−5
6
= −5
8·−6 5
= 3 4
Trang 3
122 7
124 −1.9
20 =
−1.9
20 ·1010 = −20019 , or − 0.095
126 −17.8
3.2 =
−17.8 3.2 ·1010 = −17832 = −8916, or − 5.5625
128 Not defined
130 3
8, −83; −107, 10
7; 1, −1; 0, does not exist; 6.4, 1
−6.4, or
−6.41 ; −a,1a
132 26
134 √
3, π, 4.57557555755557
136 All of them
138 5 >3
8
140 123 > −10
142 Use the definition of subtraction The number that
can be added to 11.7 to obtain −73
4 is −73
4− 11.7, or
−734− 11107 = −19209, or −19.45
Exercise Set R.3
RC2 True; see page R-23 in the text
RC4 False; see page R-22 in the text
RC6 False; see page R-20 in the text
RC8 True; see page R-22 in the text
2 63
4 x4
6 t5
8 (3.8)5
10
−4
5
3
12 9 · 9 · 9 = 729
14 (−7) · (−7) = 49
16 (0.1)(0.1)(0.1)(0.1)(0.1)(0.1) = 0.000001
18 (−3)(−3)(−3)(−3) = 81
20 2
3·2
3·2
3·2
3=
16 81
22 √
6
24 5 2
26 1
28 1
30 1 5
−3
= 1
1 5
3 = 11 125
= 1 ·1251 = 125
32 5 2
−4
= 1
5 2
4 = 6251 16
= 1 ·62516 = 16
625
34 1
x6
36 y7
38 1 (−4)3 = −641
40 9−2
42 n−5
44 (−8)−6
46 [6 − 4(8 − 5 )] = [6 − 4 · 3]
= [6 − 12]
= −6
48 10[7 − 4(8 − 5)] = 10[7 − 4 · 3] = 10[7 − 12] = 10[−5 ] = −50
50 [9(7 − 4) + 19] − [25 − (7 + 3)] = [9 · 3 + 19] − [25 − 10] = [27 + 19] − 15=
46 − 15= 31
52 [48 ÷ (−3)] ÷−14= −16 ÷−14= 64
54 30 · 10 − 18 · 25= 300 − 450
= −150
56 (9 − 12)2= (−3)2= 9
92− 122= 81 − 144 = −63
58 7 · 8 − 32− 23= 7 · 8 − 9 − 8 = 5 6 − 9 − 8 = 39
60 43+ 20 · 10 + 72− 23 = 64 + 20 · 10 + 49 − 23 =
64 + 200 + 49 − 23 = 290
62 (9 · 8 + 3 · 3)2= (72 + 9)2= (81)2= 6561
64 5000 · (4 + 1.16)2= 5000 · (5.16)2=
5000 · (26.6256) = 133, 128
Trang 466 (43 · 6 − 14 · 7)3+ (33 · 34)2=
(258 − 98)3+ (1122)2= (160)3+ (1122)2=
4, 096, 000 + 1, 258, 884 = 5, 354, 884
68 18 − (2 · 3 − 9) = 18 − (6 − 9) = 18 − (−3) = 21
70 (18 − 2)(3 − 9) = 16(−6) = −96
72 [(−32) ÷ (−2)] ÷ (−2) = 16 ÷ (−2) = −8
74 30 · 20 − 15 · 24 = 600 − 360 = 240
76 16 ÷ (19 − 15)2− 7 = 16 ÷ (4)2− 7 =
16 ÷ 16 − 7 = 1 − 7 = −6
78 53+ 20 · 40 + 82− 29 = 125+ 20 · 40 + 64 − 29 =
125+ 800 + 64 − 29 = 960
80 4000 · (3 + 1.14)2= 4000 · (4.14)2=
4000 · (17.1396) = 66, 558.4
82 8(7 − 3)
4 =
8 · 4
4 =
32
4 = 8
84 43
8 =
64
8 = 8
86 53− 72= 125 − 49 = 76
88 10(−5 ) + 1(−1) = −50 − 1 = −51
90 14 − 2(−6) + 7 = 14 + 12 + 7 = 33
92 −32 − 8 ÷ 4 − (−2) = −32 − 2 − (−2) =
−32 − 2 + 2 = −32
94 (3 − 8)2= (−5)2= 25
96 28 − 103 = 28 − 1000
= −972
98 2 × 103− 5000 = 2 × 1000 − 5000 =
2000 − 5000 = −3000
100 6[9 − (3 − 4)] = 6[9 − (−1)] = 6[9 + 1] =
6[10] = 60
102 1000 ÷ (−100) ÷ 10 = −10 ÷ 10 = −1
104 20 − 62
92+ 32 = 20 − 36
81 + 9 =
−16
90 = −8 45
106 4|6 − 7| − 5 · 4
6 · 7 − 8|4 − 1|=
4| − 1| − 5 · 4
6 · 7 − 8|3| =
4 · 1 − 5 · 4
6 · 7 − 8 · 3=
4 − 20
42 − 24 =
−16
18 = −89
108 53− 32+ 12 · 5
−32 ÷ (−16) ÷ (−4)=
125 − 9 + 12 · 5
2 ÷ (−4) =
125 − 9 + 60
−12
= 176
−12
= −352
110 2.3
112 900
114 −79
116 −79
118 23 120
−2 3
−15 16
= 2 · 15
3 · 16
= 2 · 3 · 5
3 · 2 · 8
= 2/ · 3/ · 5 3/ · 2/ · 8
= 5 8
122 2(61· 6−1− 6−1· 60)
= 2
6 ·16−16· 1
= 2
1 −16
= 2 ·56= 10
6
= 5 3
124 12345679 · 9 = 111, 111, 111
12345679 · 18 = 222, 222, 222
12345679 · 27 = 333, 333, 333 According to this pattern, we have 12345679 · 36 =
444, 444, 444
126 (π)√2≈ 5.047497267 (√
2)π≈ 2.970686424 Thus, (π)√2 is larger than (√
2)π
Exercise Set R.4
RC2 (c) RC4 (e) RC6 (g) RC8 (f)
2 t + 11, or 11 + t
4 d − 0.203
6 18 + z, or z + 18
8 d + c, or c + d
10 c ÷ h, or hc
Trang 5Exercise Set R.5 5
12 t + s, or s + t
14 q − p
16 a + b, or b + a
18 2z
20 −6t
22 Let w represent the number of women attending Then we
have 48%w, or 0.48w
24 d − $19.95
26 65t
28 57y = 5 7(−8) = −456
30 x
y=
30
−6 = −5
32 5
p + q =
5
20 + 30 =
5
50 =
1 10
34 18m
n =
18 · 7
18 = 7
36 4x − y = 4 · 3 − (−2)
= 12 − (−2)
= 12 + 2
= 14
38 n3− 2 + p ÷ n2= 23− 2 + 12 ÷ 22
= 8 − 2 + 12 ÷ 4
= 8 − 2 + 3
= 6 + 3
= 9
40 a2− 3(a − b) = 102− 3(10 − (−8))
= 102− 3(10 + 8)
= 102− 3(18)
= 100 − 3(18)
= 100 − 54
= 46
42 I = $18, 000(0.046)(2) = $1656
44 A = (3.6 m)(1.9 m) = 6.84 m2
46 (−3)(−3)(−3)(−3)(−3) = −243
48 (−5.3)2= (−5.3)(−5.3) = 28.09
50 (4.5)0= 1 (For any nonzero number a, a0= 1.)
52 (3x)1= 3x (For any number a, a1= a.)
54
56 N = P + P (2.7%)(1), so we have N = P + 2.7%P , or
N = P + 0.027P , or N = 1.027P
58 2.56y 3.2x =
2.56(3) 3.2(4) =
7.68 12.8= 0.6
Exercise Set R.5
RC2 The statement 4 · 2 + 4 · 8 = 4(2 + 8) illustrates a distributive law
RC4 The multiplicative identity is the number 1
RC6 In the expression 7(5+ x), 5and x are terms 2
7x + 2x 5x 7x − 2x
x = −2 −18 −10 −10
x = 5 45 25 25
x = 0 0 0 0 5x and 7x − 2x are equivalent; 7x + 2x and 5x are not equivalent; 7x + 2x and 7x − 2x are not equivalent 4
5(x − 2) 5x − 2 5x − 10
x = −1 −15 −7 −15
x = 4.6 13 21 13
x = 0 −10 −2 −10 5(x − 2) and 5x − 10 are equivalent; 5(x − 2) and 5x − 2 are not equivalent; 5x − 2 and 5x − 10 are not equivalent
6 4
3=
4
3· 1 =43·aa = 4a
3a
8 3
10 =
3
10· 1 = 3
10·5y 5y =
15y 50y
10 36y 18y =
2 · 18y
1 · 18y =
2
1·18y18y = 2
12 −625t 15t =
−125 · 5t
3 · 5t =
−125
3 ·5t5t= −1253
14 5 + y
16 dc
18 pq + 14 = 14 + pq;
pq + 14 = 14 + pq = 14 + qp;
pq + 14 = qp + 14
20 s + qt = qt + s;
s + qt = qt + s = tq + s;
s + qt = s + tq
22 (5 · p) · q
24 7 + (p + q)
Trang 626 (4 + x) + y = 4 + (x + y) = (x + y) + 4;
(4 + x) + y = (x + 4) + y = x + (4 + y) =
(4 + y) + x;
(4 + x) + y = y + (4 + x) = y + (x + 4);
others are possible
28 (8 · m) · n = 8 · (m · n) = 8 · (n · m) = (8 · n) · m;
(8 · m) · n = 8 · (m · n) = (m · n) · 8 = (n · m) · 8;
(8 · m) · n = n · (8 · m) = n · (m · 8);
others are possible
30 3c + 3
32 7b − 7c
34 −6c − 10d
36 5xy − 5xz + 5xw
38 P + P rt
40 1
4πr +
1
4πrs
42 5x, −9y, 12
44 5a, −7b, −9c
46 9(a + b)
48 22(x − 1)
50 6(y − 6)
52 a(b + 1)
54 3(x + y − z)
56 4(a + 2b − 1)
58 4(2m + n − 6)
60 6(3a − 4b − 8)
62 x(y − z + w)
64 1
2h(a + b)
66 Let x and y represent the numbers; x2+ y2
68 n5
70 10 · 25 − 3 · 23= 10 · 25 − 3 · 8
= 25 0 − 24
= 226
72 Let a = −1 and b = 2 Then (a−b)(a+b) = (−1−2)(−1+
2) = −3 and a2− b2= (−1)2− 22= −3
Let a = 3 and b = −2 Then (a−b)(a+b) = (3−(−2))(3−
2) = 5and a2− b2= 32− (−2)2= 5
In fact, the expressions have the same value for all values
of x, so they are equivalent
74 Let x = 2 Then x
8
x4 = 2
8
24 = 256
16 = 16, but x
2= 22 = 4 The expressions are not equivalent
Exercise Set R.6
RC2 The terms 9y and 9c have different variables, so they are not similar terms The given statement is false RC4 −(5 − x) = −5 + x = x − 5; the given statement is true RC6
given statement is false
2 15a
4 −3c
6 14x
8 14x
10 −5x
12 −6x
14 14a − 9b
16 7a + 9b
18 9p + 12
20 −11.83a − 36.3b
22 −34x +3
4y − 34
24 P = 2(l + w) = 2(360 ft + 160 ft) =
2 · 520 ft = 1040 ft;
P = 2l + 2w = 2 · 360 ft + 2 · 160 ft =
720 ft + 320 ft = 1040 ft
26 5y
28 −a − 9
30 −x + 8, or 8 − x
32 −r + s, or s − r
34 −r − s − t
36 −9a + 7b − 24
38 4x − 8y + 5w − 9z
40 x − 2y +23z + 5 6.3w
42 6x + 9
44 5a − (4a − 3) = 5a − 4a + 3 = a + 3
46 6x − 7 − (9 − 3x) = 6x − 7 − 9 + 3x = 9x − 16
Trang 7
48 −9(y + 7) − 6(y − 3) = −9y − 63 − 6y + 18 =
−15y − 45
50 8y − 4(5y − 6) + 9 = 8y − 20y + 24 + 9 =
−12y + 33
52 −5t + (4t − 12) − 2(3t + 7)
= −5t + 4t − 12 − 6t − 14
= −7t − 26
54 −12(10t − w) +14(−28t + 4) + 1
= −5t +12w − 7t + 1 + 1
= −12t +12w + 2
56 14b − [7 − 3(9b − 4)] = 14b − [7 − 27b + 12]
= 14b − [19 − 27b]
= 14b − 19 + 27b
= 41b − 19
58 7{−7 + 8[5 − 3(4 + 6)]} = 7{−7 + 8[5 − 3(10)]} =
7{−7 + 8[5 − 30]} = 7{−7 + 8[−25]} =
7{−7 − 200} = 7{−207} = −1449
60 [9(x + 5 ) − 7] + [4(x − 12) + 9] =
[9x + 45 − 7] + [4x − 48 + 9] =
9x + 38 + 4x − 39 = 13x − 1
62 [6(x + 4) − 12] − [5(x − 8) + 11]
= [6x + 24 − 12] − [5x − 40 + 11]
= [6x + 12] − [5x − 29]
= 6x + 12 − 5x + 29
= x + 41
64 4{[8(x − 3) + 9] − [4(3x − 7) + 2]}
= 4{[8x − 24 + 9] − [12x − 28 + 2]}
= 4{[8x − 15] − [12x − 26]}
= 4{8x − 15 − 12x + 26}
= 4{−4x + 11}
= −16x + 44
66 3{[6(x − 4) + 52] − 2[5(x + 8) − 102]}
= 3{[6(x − 4) + 25 ] − 2[5(x + 8) − 100]}
= 3{[6x − 24 + 25 ] − 2[5x + 40 − 100]}
= 3{[6x + 1] − 2[5x − 60]}
= 3{6x + 1 − 10x + 120}
= 3{−4x + 121}
= −12x + 363
68 7b − {5[4(3b − 8) − (9b + 10)] + 14}
= 7b − {5[12b − 32 − 9b − 10] + 14}
= 7b − {5[3b − 42] + 14}
= 7b − {15b − 210 + 14}
= 7b − {15b − 196}
= 7b − 15b + 196
= −8b + 196
70 3d ÷ 2c = 3 · 5 ÷ 2 · 10
= 15 ÷ 2 · 10
= 7.5 · 10
= 75
72 x2÷ 3(y − z) = 62÷ 3(8 − 10)
= 62÷ 3(−2)
= 36 ÷ 3(−2)
= 12(−2)
= −24
74 16
76 −3
8÷9
4 = −3
8·4 9
= − 3/ · 4/ · 1
2 · 4/ · 3/ · 3
= −1 6
78 −16a + 24b − 32
80 16x − 8y + 10
82 8(3a − 2b)
84 5(3p + 9q − 2)
86 2 · (7 + 32· 5) = 104
88 (2 − 7) · 22+ 9 = −11
90 −3[9(x − 4) + 5x] − 8{3[5(3y + 4)] − 12}
= −3[9x − 36 + 5x] − 8{3[15y + 20] − 12}
= −3[14x − 36] − 8{45y + 60 − 12}
= −42x + 108 − 8{45y + 48}
= −42x + 108 − 360y − 384
= −42x − 360y − 276
92 {x + [f − (f + x)] + [x − f]} + 3x
= {x + [f − f − x] + [x − f]} + 3x
= {x − x + x − f} + 3x
= x − f + 3x
= 4x − f
Trang 8Exercise Set R.7
RC2 (b)
RC4 (e)
RC6 (d)
2 88
4 9−2= 1
92
6 9−7= 1
97
8 a
10 1
12 (9y)2· (2y)3= 81y2· 8y3= 648y5
14 −18x7y
16 −24x−12y = −24yx12
18 −36x−12n= −x3612n
20 15t−5a= 15
t5a
22 76
24 513
26 12−12= 1
1212
28 2−2= 1
22, or 1
4
30 y−11t= 1
y11t
32 m−2t= 1
m2t
34 y9
36 −3ab2
38 −7a−4b
2
4 = −7b
2
4a4
40 7a16b5
9
42 520
44 9−12= 1
912
46 740
48 32x15y20
50 −271a−6b15= − b
15
27a6
52 8−4x16y−20z−8= x
16
84y20z8
54 5−6
49 = 1
56· 49
56 −4x4y−2
5x−1y4
−4
= 5x−1y4
−4x4y−2
4
= 5y6
−4x5
4
=
54y24
44x20
58 −200x3y−5
8x5y−7
−4
= 8x
5y−7
−200x3y−5
4
= x
2
−25y2
4
=
x8
254y8, or x
8
58y8
60 9−2x−4y
3−3x−3y2
8
= 3−4x−4y
3−3x−3y2
8
= 1 3xy
8
= 1
38x8y8
62 [(−4a−4b−5)−3]4= [(−4)−3a12b15]4= 4−12a48b60=
a48b60
412
64 2x2y−2
3x8y7
9
= 2 3x6y9
9
= 2
9
39x54y81
66 11b +2−(3b−3)= 11b +2−3b+3= 11−2b+5
68 −3xa +1−(2−a)= −3xa +1−2+a= −3x2a−1
70 −4xb +5−(b−5)y4+c−(c−4)= −4xb +5−b+5y4+c−c+4=
−4x10y8
72 76pq
74 x3ab−3b
76 45cx15acy10bc
78 −5xa +b−(a−b)yb−a−(b+a)= −5xa +b−a+byb−a−b−a=
−5x2by−2a, or −5x
2b
y2a
80 2.6 × 1012
82 2.63 × 10−7
84 340, 000 = 3.4 × 105; 322, 000, 000 = 3.22 × 108
86 200 billionths = 200 × 0.000000001 = 0.0000002 0.0000002
↑
7 places Small number, so the exponent is negative
0.0000002 = 2 × 10−7
Trang 9
Chapter R Summary and Review:Concept Reinforcement 9
88 92,400,000
90 0.000000000000000000000000000911 g
92 3,240,000 tracks
94 33.8 × 10−5= 3.38 × 10−4
96 26.732 × 103= 2.6732 × 104
98 1.5 × 103
100 3 × 10−5
102 First we find the number of seconds in 31 days
31 days ×1 day24 hr×60 min1 hr ×60 sec1 min= 2, 678, 400 sec =
2.6784 × 106 sec
Also, we have 818 = 8.18 × 102
Now we find the number of hot dogs consumed
(8.18×102)×(2.6784×106) = (8.18×2.6784)×(102×106)
≈ 21.9 × 108
= (2.19 × 10) × 108
= 2.19 × 109 hot dogs
104 4.704 × 1013
5.88 × 1012 = 0.8 × 10 = 8 light years
106 2π(6.71 × 107) ≈ 42.16 × 107
= (4.216 × 10) × 107
= 4.216 × 108mi
108 1 hour = 60 minutes = 60(60 seconds) =
3600 seconds
The amount of water discharged in one hour is
(4, 200, 000 ft3/sec) × (3600 sec)
= (4.2 × 106ft3/sec) × (3.6 × 103 sec)
= (4.2 × 3.6)(106× 103) ft3
= 15.12 × 109ft3
= (1.512 × 101) × 109 ft3
= 1.512 × 1010ft3
1 year = 365days = 365(24 hours) = 365· 24 · 3600
sec-onds = 31,536,000 secsec-onds
The amount of water discharged in one year is
(4, 200, 000 ft3/sec) × (31, 536, 000 sec)
= (4.2 × 106ft3/sec) × (3.1536 × 107sec)
= (4.2 × 3.1536)(106× 107) ft3
= 13.24512 × 1013ft3
= (1.324512 × 101) × 1013ft3
= 1.324512 × 1014ft3
110 1
4· 70 × 1026= 17.5 × 1026= 1.75 × 10 × 1026=
1.75 × 1027oxygen atoms
112 −6t − (5t − 13) + 2(4 − 6t)
= −6t − 5t + 13 + 8 − 12t
= −23t + 21
114 54− 38 · 24 − (16 − 4 · 18)
= 625 − 38 · 24 − (16 − 4 · 18)
= 625 − 38 · 24 − (16 − 72)
= 625 − 38 · 24 − (−56)
= 625 − 38 · 24 + 5 6
= 625 − 912 + 5 6
= −287 + 56
= −231
116 20 − (5 · 4 − 8) = 20 − (20 − 8)
= 20 − 12
= 8
118 (−3x−2y5)−3
(2x4y−8)−2
2
=
(−3)−3x6y−15
2−2x−8y16
2
=
(−3)−3x14
2−2y31
2
= 3−6x
28
2−4y62
= 16x
28
729y62
120 (mx 2
−bxnx 2 +bx)(mbxn−bx) = mx 2
nx 2
122 (xa+bya+b)c= xac+bcyac+bc
Chapter R Vocabulary Reinforcement
1 The sentence x > −4 is an example of an inequality
2 In the notation 74, the number 7 is called the base
3 A variable is a letter that can represent various numbers
4 In the expression 6x, the multipliers 6 and x are factors
5 The number 5.93 × 107 is written in scientific notation
6 The product of reciprocals is 1
7 The sum of opposites is 0
8 The commutative law for addition states that a+b = b+a
Chapter R Concept Reinforcement
1 The statement is false For example, let a = 1 and b = 2 Then a − b = 1 − 2 = −1, but b − a = 2 − 1 = 1
2 The statement is true See page R-2 in the text
Trang 10⫺4 0
0 1
4 Zero is neither positive nor negative The given statement
is false
5 The statement is false because |0| = 0
6 The statement is true See page R-15in the text
7 The statement is true See page R-12 in the text
8 The statement is true See page R-50 in the text
Chapter R Review Exercises
1 The rational numbers can be named as quotients of
in-tegers with nonzero divisors Of the given numbers, the
rational numbers are 2, −23, 0.4545, and −23.788
2 We specify the conditions by which we know whether a
number is in the set
{x|x is a real number less than or equal to 46}
3 Since −3.9 is to the left of 2.9, we have −3.9 < 2.9
4 19 > x
The inequality x < 19 has the same meaning
5 −13 ≥ 5is false since neither −13 > 5nor −13 = 5is true
6 7.01 ≤ 7.01 is true since 7.01 = 7.01 is true
7 x > −4
We shade all numbers to the right of −4 and use a
paren-thesis at −4 to indicate it is not a solution
8 x ≤ 1
We shade all the numbers to the left of 1 and use a bracket
at 1 to indicate it is also a solution
9 The distance of −7.23 from 0 is 7.23, so | − 7.23| = 7.23
10 The distance of 9 − 9, or 0, from 0 is 0, so |9 − 9| = 0
11 6 + (−8)
We find the difference of the absolute values, 8 − 6 = 2
Since the negative number has the larger absolute value,
the answer is negative
6 + (−8) = −2
12 −3.8 + (−4.1)
The sum of two negative numbers is negative We add
the absolute values, 3.8 + 4.1 = 7.9, and make the answer
negative
−3.8 + (−4.1) = −7.9
We find the difference of the absolute values, 13
7 − 3
4=
52
28− 21
28=
31
28 Since the negative number has the larger absolute value, the answer is negative 3
4+
−137
= −3128
14 −8 − (−3) = −8 + 3 = −5
15 −17.3 − 9.4 = −17.3 + (−9.4) = −26.7
16 3
2−
−13 4
= 3
2+
13
4 =
6
4+
13
4 =
19 4
17 (−3.8)(−2.7) The product of two negative numbers is positive We mul-tiply the absolute values, 3.8(2.7) = 10.26, and make the answer positive
(−3.8)(−2.7) = 10.26
18 −23 9
14
The product of a negative number and a positive number
is negative We multiply the absolute values and make the answer negative
2
3· 9
14 =
2 · 9
3 · 14=
2 · 3 · 3
3 · 2 · 7=
2 · 3
2 · 3·
3 7 Thus, −23 9
14
= −37
19 −6(−7)(4) = 42(4) = 168
20 When a negative number is divided by a positive number, the answer is negative
−12 ÷ 3 =−123 = −4
21 When a negative number is divided by a negative number, the answer is positive
−84
−4 = 21
22 When a positive number is divided by a negative number, the answer is negative
49
−7 = −7
23 5
6÷
−10 7
=5
6·
− 7 10
= −5 · 7
6 · 10 = −
5 · 7
6 · 2 · 5=
− 7
6 · 2·
5
5= − 7
6 · 2 = −
7 12
24 −5
2÷
−15 16
= −5
2·
−16 15
= 5 · 16
2 · 15=
5 · 2 · 8
2 · 3 · 5=
2 · 5
2 · 5·
8
3=
8 3
25 25
0 Not defined: Division by 0.