Intermediate Algebra (7th Edition)
Intermediate Algebra (7th Edition)
7th Edition
ISBN: 9780134196176
Author: Elayn Martin-Gay
Publisher: PEARSON
Question
Book Icon
Chapter 1.3, Problem 1P
To determine

a.

To add:

The given numbers as indicated

Solution:

 -6+-2=-8

Explanation:

1) Concept:

(i) A real number  a, with an absolute value, written as  |a|   is always positive or zero.

(ii)  Rule for addition of real numbers:

A) Add absolute values and attach their common sign, to add two numbers with the same sign.

B) Subtract the smaller absolute value from the larger absolute value and attach the sign of the number with the larger absolute value, to add two numbers with different signs.

2) Given:

-6+-2

3) Calculation:

First we find the absolute values of the given numbers.

Hence using the rule of absolute values,

|-6 | = 6 and | -2 | = 2.

According to rule of addition of real numbers, add their absolute values.

So, 6 + 2=8

Next, attach their common sign.

Hence, -6+-2=-8

Final statement:

-6+-2=-8

b.

To add:

The given numbers as indicated

Solution:

5+-8=-3

Explanation:

1) Concept:

(i) A real number  a, with an absolute value, written as  |a|   is always positive or zero.

(ii)  Rule for addition of real numbers:

A) Add absolute values and attach their common sign, to add two numbers with the same sign.

B) Subtract the smaller absolute value from the larger absolute value and attach the sign of the number with the larger absolute value, to add two numbers with different signs.

2) Given:

5+-8

3) Calculation:

First we find the absolute values of the given numbers.

Hence using the rule of absolute values,

|5 |=5 and | -8 | = 8.

According to rule of addition of real numbers, subtract the smaller absolute value from the larger absolute value.

8- 5= 3

Next, attach the sign of the number with the larger absolute value, so attach the sign of 8, which is negative.

Hence, 5+-8=-3

Final statement:

5+-8=-3

c.

To add:

The given numbers as indicated

Solution:

-4+9=5

Explanation:

1) Concept:

(i) A real number  a, with an absolute value, written as  |a|   is always positive or zero.

(ii)  Rule for addition of real numbers:

A) Add absolute values and attach their common sign, to add two numbers with the same sign.

B) Subtract the smaller absolute value from the larger absolute value and attach the sign of the number with the larger absolute value, to add two numbers with different signs.

2) Given:

-4+9

3) Calculation:

First we find the absolute values of given numbers.

Hence, using the rule of absolute values,

|-4 |=4 and | 9 | = 9.

According to rule of addition of real numbers, subtract the smaller absolute value from the larger absolute value.

9 - 4 = 5

Next, attach the sign of the number with the larger absolute value, so attach the sign of 9, which is positive.

Hence, -4+9=5

Final statement:

-4+9=5

d.

To add:

The given numbers as indicated

Solution:

-3.2+-4.9=-8.1

Explanation:

1) Concept:

(i) A real number  a, with an absolute value, written as  |a|   is always positive or zero.

(ii)  Rule for addition of real numbers:

A) Add absolute values and attach their common sign, to add two numbers with the same sign.

B) Subtract the smaller absolute value from the larger absolute value and attach the sign of the number with the larger absolute value, to add two numbers with different signs.

2) Given:

-3.2+-4.9

3) Calculation:

First we find the absolute values of given numbers.

Hence, using the rule of absolute values,

 |-3.2 | = 3.2 and | -4.9 | = 4.9.

According to rule of addition of real numbers, add their absolute values.

So, 3.2 + 4.9=8.1

Next, attach their common sign.

Hence, -3.2+-4.9=-8.1

Final statement:

-3.2+-4.9=-8.1

e.

To add:

The given numbers as indicated

Solution:

-35+23=115

Explanation:

1) Concept:

(i) A real number  a, with an absolute value, written as  |a|   is always positive or zero.

(ii)  Rule for addition of real numbers:

A) Add absolute values and attach their common sign, to add two numbers with the same sign.

B) Subtract the smaller absolute value from the larger absolute value and attach the sign of the number with the larger absolute value, to add two numbers with different signs.

2) Given:

-35+23

3) Calculation:

First we find the absolute values of given numbers.

Hence, using the rule of absolute values,

-35 =35 and 23  =23.

According to rule of addition of real numbers, subtract the smaller absolute value from the larger absolute value.

23 -35

To find it, first find the least common multiple of 3 and 5.

LCD (3, 5) = 15

23 -35=23+-35=23*55+-35*33=1015+-915=10-915=115

Next, attach the sign of the number with the larger absolute value, so attach the sign of 23, which is positive.

Hence,23 -35=115

Final statement:

-35+23=115

f.

To add:

The given numbers as indicated

Solution:

-511+322=-722

Explanation:

1) Concept:

(i) A real number  a, with an absolute value, written as  |a|   is always positive or zero.

(ii)  Rule for addition of real numbers:

A) Add absolute values and attach their common sign, to add two numbers with the same sign.

B) Subtract the smaller absolute value from the larger absolute value and attach the sign of the number with the larger absolute value, to add two numbers with different signs.

2) Given:

-511+322

3) Calculation:

First we find the absolute values of given numbers.

Hence, using the rule of absolute values,

-511 =511 and 322  =322.

According to rule of addition of real numbers, subtract the smaller absolute value from the larger absolute value.

511 -322

To find it, first find least the common multiple of 3 and 5.

LCD (11, 22) = 22

511 -322=511+-322=511*22+-322*11=1022+-322=10-322=722

Next, attach the sign of the number with the larger absolute value, so attach the sign of 511, which is negative.

Hence,-511+322=-722

Final statement:

-511+322=-722

Blurred answer
Students have asked these similar questions
The 173 acellus.com StudentFunctions inter ooks 24-25/08 R Mastery Connect ac ?ClassiD-952638111# Introduction - Surface Area of Composite Figures 3 cm 3 cm 8 cm 8 cm Find the surface area of the composite figure. 2 SA = [?] cm² 7 cm REMEMBER! Exclude areas where complex shapes touch. 7 cm 12 cm 10 cm might ©2003-2025 International Academy of Science. All Rights Reserved. Enter
You are given a plane Π in R3 defined by two vectors, p1 and p2, and a subspace W in R3 spanned by twovectors, w1 and w2. Your task is to project the plane Π onto the subspace W.First, answer the question of what the projection matrix is that projects onto the subspace W and how toapply it to find the desired projection. Second, approach the task in a different way by using the Gram-Schmidtmethod to find an orthonormal basis for subspace W, before then using the resulting basis vectors for theprojection. Last, compare the results obtained from both methods
Plane II is spanned by the vectors: - (2) · P² - (4) P1=2 P21 3 Subspace W is spanned by the vectors: 2 W1 - (9) · 1 W2 1 = (³)

Chapter 1 Solutions

Intermediate Algebra (7th Edition)

Ch. 1.2 - Prob. 3VRVCCh. 1.2 - Prob. 4VRVCCh. 1.2 - Prob. 5VRVCCh. 1.2 - Prob. 6VRVCCh. 1.2 - Prob. 7VRVCCh. 1.2 - Prob. 8VRVCCh. 1.2 - Prob. 9VRVCCh. 1.2 - Prob. 10VRVCCh. 1.2 - Prob. 1ECh. 1.2 - Prob. 2ECh. 1.2 - Prob. 3ECh. 1.2 - Prob. 4ECh. 1.2 - Prob. 5ECh. 1.2 - Prob. 6ECh. 1.2 - Prob. 7ECh. 1.2 - Prob. 8ECh. 1.2 - Prob. 9ECh. 1.2 - Prob. 10ECh. 1.2 - Prob. 11ECh. 1.2 - Prob. 12ECh. 1.2 - Prob. 13ECh. 1.2 - Prob. 14ECh. 1.2 - Prob. 15ECh. 1.2 - Prob. 16ECh. 1.2 - Prob. 17ECh. 1.2 - Prob. 18ECh. 1.2 - Prob. 19ECh. 1.2 - Prob. 20ECh. 1.2 - Prob. 21ECh. 1.2 - Prob. 22ECh. 1.2 - Prob. 23ECh. 1.2 - Prob. 24ECh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 47ECh. 1.2 - Prob. 48ECh. 1.2 - Prob. 49ECh. 1.2 - Prob. 50ECh. 1.2 - Prob. 51ECh. 1.2 - Prob. 52ECh. 1.2 - Prob. 53ECh. 1.2 - Prob. 54ECh. 1.2 - Prob. 55ECh. 1.2 - Prob. 56ECh. 1.2 - Prob. 57ECh. 1.2 - Prob. 58ECh. 1.2 - Prob. 59ECh. 1.2 - Prob. 60ECh. 1.2 - Prob. 61ECh. 1.2 - Prob. 62ECh. 1.2 - Prob. 63ECh. 1.2 - Prob. 64ECh. 1.2 - Prob. 65ECh. 1.2 - Prob. 66ECh. 1.2 - Prob. 67ECh. 1.2 - Prob. 68ECh. 1.2 - Prob. 69ECh. 1.2 - Prob. 70ECh. 1.2 - Prob. 71ECh. 1.2 - Prob. 72ECh. 1.2 - Prob. 73ECh. 1.2 - Prob. 74ECh. 1.2 - Prob. 75ECh. 1.2 - Prob. 76ECh. 1.2 - Prob. 77ECh. 1.2 - Prob. 78ECh. 1.2 - Prob. 79ECh. 1.2 - Prob. 80ECh. 1.2 - Prob. 81ECh. 1.2 - Prob. 82ECh. 1.2 - Prob. 83ECh. 1.2 - Prob. 84ECh. 1.2 - Prob. 85ECh. 1.2 - Prob. 86ECh. 1.2 - Prob. 87ECh. 1.2 - Prob. 88ECh. 1.2 - Prob. 89ECh. 1.2 - Prob. 90ECh. 1.2 - Prob. 91ECh. 1.2 - Prob. 92ECh. 1.2 - Prob. 93ECh. 1.2 - Prob. 94ECh. 1.2 - Prob. 95ECh. 1.2 - Prob. 96ECh. 1.2 - Prob. 97ECh. 1.2 - Prob. 98ECh. 1.2 - Prob. 99ECh. 1.2 - Prob. 100ECh. 1.2 - Prob. 101ECh. 1.2 - Prob. 102ECh. 1.3 - Prob. 1PCh. 1.3 - Prob. 2PCh. 1.3 - Prob. 3PCh. 1.3 - Prob. 4PCh. 1.3 - Prob. 5PCh. 1.3 - Prob. 6PCh. 1.3 - Prob. 7PCh. 1.3 - Prob. 8PCh. 1.3 - Prob. 9PCh. 1.3 - Prob. 10PCh. 1.3 - Prob. 11PCh. 1.3 - Prob. 12PCh. 1.3 - Prob. 13PCh. 1.3 - Prob. 1VRVCCh. 1.3 - Prob. 2VRVCCh. 1.3 - Prob. 3VRVCCh. 1.3 - Prob. 4VRVCCh. 1.3 - Prob. 5VRVCCh. 1.3 - Prob. 6VRVCCh. 1.3 - Prob. 7VRVCCh. 1.3 - Prob. 8VRVCCh. 1.3 - Prob. 1ECh. 1.3 - Prob. 2ECh. 1.3 - Prob. 3ECh. 1.3 - Prob. 4ECh. 1.3 - Prob. 5ECh. 1.3 - Prob. 6ECh. 1.3 - Prob. 7ECh. 1.3 - Prob. 8ECh. 1.3 - Prob. 9ECh. 1.3 - Prob. 10ECh. 1.3 - Prob. 11ECh. 1.3 - Prob. 12ECh. 1.3 - Prob. 13ECh. 1.3 - Prob. 14ECh. 1.3 - Prob. 15ECh. 1.3 - Prob. 16ECh. 1.3 - Prob. 17ECh. 1.3 - Prob. 18ECh. 1.3 - Prob. 19ECh. 1.3 - Prob. 20ECh. 1.3 - Prob. 21ECh. 1.3 - Prob. 22ECh. 1.3 - Prob. 23ECh. 1.3 - Prob. 24ECh. 1.3 - Prob. 25ECh. 1.3 - Prob. 26ECh. 1.3 - Prob. 27ECh. 1.3 - Prob. 28ECh. 1.3 - Prob. 29ECh. 1.3 - Prob. 30ECh. 1.3 - Prob. 31ECh. 1.3 - Prob. 32ECh. 1.3 - Prob. 33ECh. 1.3 - Prob. 34ECh. 1.3 - Prob. 35ECh. 1.3 - Prob. 36ECh. 1.3 - Prob. 37ECh. 1.3 - Prob. 38ECh. 1.3 - Prob. 39ECh. 1.3 - Prob. 40ECh. 1.3 - Prob. 41ECh. 1.3 - Prob. 42ECh. 1.3 - Prob. 43ECh. 1.3 - Prob. 44ECh. 1.3 - Prob. 45ECh. 1.3 - Prob. 46ECh. 1.3 - Prob. 47ECh. 1.3 - Prob. 48ECh. 1.3 - Prob. 49ECh. 1.3 - Prob. 50ECh. 1.3 - Prob. 51ECh. 1.3 - Prob. 52ECh. 1.3 - Prob. 53ECh. 1.3 - Prob. 54ECh. 1.3 - Prob. 55ECh. 1.3 - Prob. 56ECh. 1.3 - Prob. 57ECh. 1.3 - Prob. 58ECh. 1.3 - Prob. 59ECh. 1.3 - Prob. 60ECh. 1.3 - Prob. 61ECh. 1.3 - Prob. 62ECh. 1.3 - Prob. 63ECh. 1.3 - Prob. 64ECh. 1.3 - Prob. 65ECh. 1.3 - Prob. 66ECh. 1.3 - Prob. 67ECh. 1.3 - Prob. 68ECh. 1.3 - Prob. 69ECh. 1.3 - Prob. 70ECh. 1.3 - Prob. 71ECh. 1.3 - Prob. 72ECh. 1.3 - Prob. 73ECh. 1.3 - Prob. 74ECh. 1.3 - Prob. 75ECh. 1.3 - Prob. 76ECh. 1.3 - Prob. 77ECh. 1.3 - Prob. 78ECh. 1.3 - Prob. 79ECh. 1.3 - Prob. 80ECh. 1.3 - Prob. 81ECh. 1.3 - Prob. 82ECh. 1.3 - Prob. 83ECh. 1.3 - Prob. 84ECh. 1.3 - Prob. 85ECh. 1.3 - Prob. 86ECh. 1.3 - Prob. 87ECh. 1.3 - Prob. 88ECh. 1.3 - Prob. 89ECh. 1.3 - Prob. 90ECh. 1.3 - Prob. 91ECh. 1.3 - Prob. 92ECh. 1.3 - Prob. 93ECh. 1.3 - Prob. 94ECh. 1.3 - Prob. 95ECh. 1.3 - Prob. 96ECh. 1.3 - Prob. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 101ECh. 1.3 - Prob. 102ECh. 1.3 - Prob. 103ECh. 1.3 - Prob. 104ECh. 1.3 - Prob. 105ECh. 1.3 - Prob. 106ECh. 1.3 - Prob. 107ECh. 1.3 - Prob. 108ECh. 1.3 - Prob. 109ECh. 1.3 - Prob. 110ECh. 1.3 - Prob. 111ECh. 1.3 - Prob. 112ECh. 1.3 - Prob. 113ECh. 1.3 - Prob. 114ECh. 1.3 - Prob. 115ECh. 1.3 - Prob. 116ECh. 1.3 - Prob. 117ECh. 1.3 - Prob. 118ECh. 1.3 - Prob. 119ECh. 1.3 - Prob. 120ECh. 1.3 - Prob. 121ECh. 1.3 - Prob. 122ECh. 1.3 - Prob. 123ECh. 1.3 - Prob. 124ECh. 1.3 - Prob. 125ECh. 1.3 - Prob. 126ECh. 1.IR - Prob. 1IRCh. 1.IR - Prob. 2IRCh. 1.IR - Prob. 3IRCh. 1.IR - Prob. 4IRCh. 1.IR - Prob. 5IRCh. 1.IR - Prob. 6IRCh. 1.IR - Prob. 7IRCh. 1.IR - Prob. 8IRCh. 1.IR - Prob. 9IRCh. 1.IR - Prob. 10IRCh. 1.IR - Prob. 11IRCh. 1.IR - Prob. 12IRCh. 1.IR - Prob. 13IRCh. 1.IR - Prob. 14IRCh. 1.IR - Prob. 15IRCh. 1.IR - Prob. 16IRCh. 1.4 - Prob. 1PCh. 1.4 - Prob. 2PCh. 1.4 - Prob. 3PCh. 1.4 - Prob. 4PCh. 1.4 - Prob. 5PCh. 1.4 - Prob. 6PCh. 1.4 - Prob. 7PCh. 1.4 - Prob. 8PCh. 1.4 - Prob. 9PCh. 1.4 - Prob. 10PCh. 1.4 - Prob. 11PCh. 1.4 - Prob. 12PCh. 1.4 - Prob. 13PCh. 1.4 - Prob. 14PCh. 1.4 - Prob. 15PCh. 1.4 - Prob. 1VRVCCh. 1.4 - Prob. 2VRVCCh. 1.4 - Prob. 3VRVCCh. 1.4 - Prob. 4VRVCCh. 1.4 - Prob. 5VRVCCh. 1.4 - Prob. 6VRVCCh. 1.4 - Prob. 7VRVCCh. 1.4 - Prob. 8VRVCCh. 1.4 - Prob. 9VRVCCh. 1.4 - Prob. 10VRVCCh. 1.4 - Prob. 11VRVCCh. 1.4 - Prob. 12VRVCCh. 1.4 - Prob. 1ECh. 1.4 - Prob. 2ECh. 1.4 - Prob. 3ECh. 1.4 - Prob. 4ECh. 1.4 - Prob. 5ECh. 1.4 - Prob. 6ECh. 1.4 - Prob. 7ECh. 1.4 - Prob. 8ECh. 1.4 - Prob. 9ECh. 1.4 - Prob. 10ECh. 1.4 - Prob. 11ECh. 1.4 - Prob. 12ECh. 1.4 - Prob. 13ECh. 1.4 - Prob. 14ECh. 1.4 - Prob. 15ECh. 1.4 - Prob. 16ECh. 1.4 - Prob. 17ECh. 1.4 - Prob. 18ECh. 1.4 - Prob. 19ECh. 1.4 - Prob. 20ECh. 1.4 - Prob. 21ECh. 1.4 - Prob. 22ECh. 1.4 - Prob. 23ECh. 1.4 - Prob. 24ECh. 1.4 - Prob. 25ECh. 1.4 - Prob. 26ECh. 1.4 - Prob. 27ECh. 1.4 - Prob. 28ECh. 1.4 - Prob. 29ECh. 1.4 - Prob. 30ECh. 1.4 - Prob. 31ECh. 1.4 - Prob. 32ECh. 1.4 - Prob. 33ECh. 1.4 - Prob. 34ECh. 1.4 - Prob. 35ECh. 1.4 - Prob. 36ECh. 1.4 - Prob. 37ECh. 1.4 - Prob. 38ECh. 1.4 - Prob. 39ECh. 1.4 - Prob. 40ECh. 1.4 - Prob. 41ECh. 1.4 - Prob. 42ECh. 1.4 - Prob. 43ECh. 1.4 - Prob. 44ECh. 1.4 - Prob. 45ECh. 1.4 - Prob. 46ECh. 1.4 - Prob. 47ECh. 1.4 - Prob. 48ECh. 1.4 - Prob. 49ECh. 1.4 - Prob. 50ECh. 1.4 - Prob. 51ECh. 1.4 - Prob. 52ECh. 1.4 - Prob. 53ECh. 1.4 - Prob. 54ECh. 1.4 - Prob. 55ECh. 1.4 - Prob. 56ECh. 1.4 - Prob. 57ECh. 1.4 - Prob. 58ECh. 1.4 - Prob. 59ECh. 1.4 - Prob. 60ECh. 1.4 - Prob. 61ECh. 1.4 - Prob. 62ECh. 1.4 - Prob. 63ECh. 1.4 - Prob. 64ECh. 1.4 - Prob. 65ECh. 1.4 - Prob. 66ECh. 1.4 - Prob. 67ECh. 1.4 - Prob. 68ECh. 1.4 - Prob. 69ECh. 1.4 - Prob. 70ECh. 1.4 - Prob. 71ECh. 1.4 - Prob. 72ECh. 1.4 - Prob. 73ECh. 1.4 - Prob. 74ECh. 1.4 - Prob. 75ECh. 1.4 - Prob. 76ECh. 1.4 - Prob. 77ECh. 1.4 - Prob. 78ECh. 1.4 - Prob. 79ECh. 1.4 - Prob. 80ECh. 1.4 - Prob. 81ECh. 1.4 - Prob. 82ECh. 1.4 - Prob. 83ECh. 1.4 - Prob. 84ECh. 1.4 - Prob. 85ECh. 1.4 - Prob. 86ECh. 1.4 - Prob. 87ECh. 1.4 - Prob. 88ECh. 1.4 - Prob. 89ECh. 1.4 - Prob. 90ECh. 1.4 - Prob. 91ECh. 1.4 - Prob. 92ECh. 1.4 - Prob. 93ECh. 1.4 - Prob. 94ECh. 1.4 - Prob. 95ECh. 1.4 - Prob. 96ECh. 1.4 - Prob. 97ECh. 1.4 - Prob. 98ECh. 1.4 - Prob. 99ECh. 1.4 - Prob. 100ECh. 1.4 - Prob. 101ECh. 1.4 - Prob. 102ECh. 1.4 - Prob. 103ECh. 1.4 - Prob. 104ECh. 1.4 - Prob. 105ECh. 1.4 - Prob. 106ECh. 1.4 - Prob. 107ECh. 1.4 - Prob. 108ECh. 1.4 - Prob. 109ECh. 1.4 - Prob. 110ECh. 1.4 - Prob. 111ECh. 1.4 - Prob. 112ECh. 1.VC - Prob. 1VCCh. 1.VC - Prob. 2VCCh. 1.VC - Prob. 3VCCh. 1.VC - Prob. 4VCCh. 1.VC - Prob. 5VCCh. 1.VC - Prob. 6VCCh. 1.VC - Prob. 7VCCh. 1.VC - Prob. 8VCCh. 1.VC - Prob. 9VCCh. 1.VC - Prob. 10VCCh. 1.VC - Prob. 11VCCh. 1.VC - Prob. 12VCCh. 1.CR - Prob. 1CRCh. 1.CR - Prob. 2CRCh. 1.CR - Prob. 3CRCh. 1.CR - Prob. 4CRCh. 1.CR - Prob. 5CRCh. 1.CR - Prob. 6CRCh. 1.CR - Prob. 7CRCh. 1.CR - Prob. 8CRCh. 1.CR - Prob. 9CRCh. 1.CR - Prob. 10CRCh. 1.CR - Prob. 11CRCh. 1.CR - Prob. 12CRCh. 1.CR - Prob. 13CRCh. 1.CR - Prob. 14CRCh. 1.CR - Prob. 15CRCh. 1.CR - Prob. 16CRCh. 1.CR - Prob. 17CRCh. 1.CR - Prob. 18CRCh. 1.CR - Prob. 19CRCh. 1.CR - Prob. 20CRCh. 1.CR - Prob. 21CRCh. 1.CR - Prob. 22CRCh. 1.CR - Prob. 23CRCh. 1.CR - Prob. 24CRCh. 1.CR - Prob. 25CRCh. 1.CR - Prob. 26CRCh. 1.CR - Prob. 27CRCh. 1.CR - Prob. 28CRCh. 1.CR - Prob. 29CRCh. 1.CR - Prob. 30CRCh. 1.CR - Prob. 31CRCh. 1.CR - Prob. 32CRCh. 1.CR - Prob. 33CRCh. 1.CR - Prob. 34CRCh. 1.CR - Prob. 35CRCh. 1.CR - Prob. 36CRCh. 1.CR - Prob. 37CRCh. 1.CR - Prob. 38CRCh. 1.CR - Prob. 39CRCh. 1.CR - Prob. 40CRCh. 1.CR - Prob. 41CRCh. 1.CR - Prob. 42CRCh. 1.CR - Prob. 43CRCh. 1.CR - Prob. 44CRCh. 1.CR - Prob. 45CRCh. 1.CR - Prob. 46CRCh. 1.CR - Prob. 47CRCh. 1.CR - Prob. 48CRCh. 1.CR - Prob. 49CRCh. 1.CR - Prob. 50CRCh. 1.CR - Prob. 51CRCh. 1.CR - Prob. 52CRCh. 1.CR - Prob. 53CRCh. 1.CR - Prob. 54CRCh. 1.CR - Prob. 55CRCh. 1.CR - Prob. 56CRCh. 1.CR - Prob. 57CRCh. 1.CR - Prob. 58CRCh. 1.CR - Prob. 59CRCh. 1.CR - Prob. 60CRCh. 1.CR - Prob. 61CRCh. 1.CR - Prob. 62CRCh. 1.CR - Prob. 63CRCh. 1.CR - Prob. 64CRCh. 1.CR - Prob. 65CRCh. 1.CR - Prob. 66CRCh. 1.CR - Prob. 67CRCh. 1.CR - Prob. 68CRCh. 1.CR - Prob. 69CRCh. 1.CR - Prob. 70CRCh. 1.CR - Prob. 71CRCh. 1.CR - Prob. 72CRCh. 1.CR - Prob. 73CRCh. 1.CR - Prob. 74CRCh. 1.CR - Prob. 75CRCh. 1.CR - Prob. 76CRCh. 1.CR - Prob. 77CRCh. 1.CR - Prob. 78CRCh. 1.CR - Prob. 79CRCh. 1.CR - Prob. 80CRCh. 1.CR - Prob. 81CRCh. 1.CR - Prob. 82CRCh. 1.CR - Prob. 83CRCh. 1.CR - Prob. 84CRCh. 1.CR - Prob. 85CRCh. 1.CR - Prob. 86CRCh. 1.CR - Prob. 87CRCh. 1.CR - Prob. 88CRCh. 1.CR - Prob. 89CRCh. 1.CR - Prob. 90CRCh. 1.CR - Prob. 91CRCh. 1.CR - Prob. 92CRCh. 1.CR - Prob. 93CRCh. 1.CR - Prob. 94CRCh. 1.CR - Prob. 95CRCh. 1.CR - Prob. 96CRCh. 1.CR - Prob. 97CRCh. 1.CR - Prob. 98CRCh. 1.CR - Prob. 99CRCh. 1.CR - Prob. 100CRCh. 1.CR - Prob. 101CRCh. 1.CR - Prob. 102CRCh. 1.CR - Prob. 103CRCh. 1.CR - Prob. 104CRCh. 1.CR - Prob. 105CRCh. 1.CR - Prob. 106CRCh. 1.CR - Prob. 107CRCh. 1.CR - Prob. 108CRCh. 1.CR - Prob. 109CRCh. 1.CR - Prob. 110CRCh. 1.CR - Prob. 111CRCh. 1.CR - Prob. 112CRCh. 1.CR - Prob. 113CRCh. 1.CR - Prob. 114CRCh. 1.CR - Prob. 115CRCh. 1.CR - Prob. 116CRCh. 1.CR - Prob. 117CRCh. 1.CR - Prob. 118CRCh. 1.CR - Prob. 119CRCh. 1.CR - Prob. 120CRCh. 1.CR - Prob. 121CRCh. 1.CR - Prob. 122CRCh. 1.CR - Prob. 123CRCh. 1.CR - Prob. 124CRCh. 1.CR - Prob. 125CRCh. 1.CR - Prob. 126CRCh. 1.CR - Prob. 127CRCh. 1.CR - Prob. 128CRCh. 1.CR - Prob. 129CRCh. 1.CR - Prob. 130CRCh. 1.CR - Prob. 131CRCh. 1.CR - Prob. 132CRCh. 1.GRT - Prob. 1GRTCh. 1.GRT - Prob. 2GRTCh. 1.GRT - Prob. 3GRTCh. 1.GRT - Prob. 4GRTCh. 1.GRT - Prob. 5GRTCh. 1.GRT - Prob. 6GRTCh. 1.GRT - Prob. 7GRTCh. 1.GRT - Prob. 8GRTCh. 1.GRT - Prob. 9GRTCh. 1.GRT - Prob. 10GRTCh. 1.GRT - Prob. 11GRTCh. 1.GRT - Prob. 12GRTCh. 1.GRT - Prob. 13GRTCh. 1.GRT - Prob. 14GRTCh. 1.GRT - Prob. 15GRTCh. 1.GRT - Prob. 16GRTCh. 1.CT - Prob. 1CTCh. 1.CT - Prob. 2CTCh. 1.CT - Prob. 3CTCh. 1.CT - Prob. 4CTCh. 1.CT - Prob. 5CTCh. 1.CT - Prob. 6CTCh. 1.CT - Prob. 7CTCh. 1.CT - Prob. 8CTCh. 1.CT - Prob. 9CTCh. 1.CT - Prob. 10CTCh. 1.CT - Prob. 11CTCh. 1.CT - Prob. 12CTCh. 1.CT - Prob. 13CTCh. 1.CT - Prob. 14CTCh. 1.CT - Prob. 15CTCh. 1.CT - Prob. 16CTCh. 1.CT - Prob. 17CTCh. 1.CT - Prob. 18CTCh. 1.CT - Prob. 19CTCh. 1.CT - Prob. 20CTCh. 1.CT - Prob. 21CTCh. 1.CT - Prob. 22CTCh. 1.CT - Prob. 23CTCh. 1.CT - Prob. 24CTCh. 1.CT - Prob. 25CTCh. 1.CT - Prob. 26CTCh. 1.CT - Prob. 27CTCh. 1.CT - Prob. 28CTCh. 1.CT - Prob. 29CTCh. 1.CT - Prob. 30CT
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:PEARSON
Text book image
Contemporary Abstract Algebra
Algebra
ISBN:9781305657960
Author:Joseph Gallian
Publisher:Cengage Learning
Text book image
Linear Algebra: A Modern Introduction
Algebra
ISBN:9781285463247
Author:David Poole
Publisher:Cengage Learning
Text book image
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:9780135163078
Author:Michael Sullivan
Publisher:PEARSON
Text book image
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:9780980232776
Author:Gilbert Strang
Publisher:Wellesley-Cambridge Press
Text book image
College Algebra (Collegiate Math)
Algebra
ISBN:9780077836344
Author:Julie Miller, Donna Gerken
Publisher:McGraw-Hill Education