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Kennesaw - MATH - 4381

! " # $ " # $ % & # $ %& ' %& %& %& % & % & %& %& %& ' %& %& %& ( " # $ !)! %& ' %& %&

Kennesaw - MATH - 4381

!"#$ ! $ !"#$ ! $ !"#$ ! $ %& ! $ ' ( ' ( ) ! $ ! *$ + & & " , '( - - . . / ! $ ! *$ - , - ! $ . , &

Kennesaw - MATH - 4381

!" # $ !$ % $& '

Kennesaw - MATH - 4381

Kennesaw - MATH - 4381

! " " # $ % ! " # &# % ' (!" )* + ,! -) ! " ) ! " " ! % ! " % " % " (* % ! (.* % . . % . ! % " % . ! % . "

Kennesaw - MATH - 4381

! ! " # # # $ # # # ! # $# # # % # # # # # ! # #& # & ' & ! #

Kennesaw - MATH - 4381

! "# $ " # " # "%# & " # '()* "%# '* "# " # $ "# " # + "# ! ! , , "# ",# $ ",# " ,# "# ! ",

Kennesaw - MATH - 4381

!" " # # $ % & & !" " # # $ %

Kennesaw - MATH - 1111

Graphs for Exam 2 (Version 1), Problem 8Graph 16 5 4 3 2 1654321 1 2 3 4 5 6123456Graph 26 5 4 3 2 1654321 1 2 3 4 5 61234561Graph 36 5 4 3 2 1654321 1 2 3 4 5 6123456

Kennesaw - MATH - 1111

S. F. EllermeyerMATH 1111 Exam 1 (Version 1) Solutions June 11, 2008 NameInstructions. You may not use any books or notes on this exam. You may use a calculator. You must show all of your work in order to receive credit! Also, write in complete s

Kennesaw - MATH - 1111

S. F. EllermeyerMATH 1111 Exam 1 (Version 2) Solutions June 11, 2008 NameInstructions. You may not use any books or notes on this exam. You may use a calculator. You must show all of your work in order to receive credit! Also, write in complete s

Kennesaw - MATH - 1113

!" # $ % %" " & " ' () " () " () " () " () " ! * " " % + " , ( ) -,. & ( ) /-,. 0-. & / ' % %

Kennesaw - MATH - 1113

Exponential and Logarithmic FunctionsS. F. Ellermeyer October 8, 19981.What are Exponential Functions?Let be a positive real number. The base exponential function is the function with domain defined by . Example 1 The base 2 exponential

Kennesaw - MATH - 1113

The Concept of FunctionS. F. Ellermeyer December 28, 1999 Let D with D 2 A function, f, on D is an assignment of each number x 5 D to a single real number, y. If the function f assigns the number x to the number y, then we write fx = y which is r

Kennesaw - MATH - 1113

Kennesaw - MATH - 1113

MATH 1113/02 Exam 1 Results8% 0%32%36% 24%A (80-100) B (60-79) C (40-59) D (20-39) F (0-19)MATH 1113/08 Exam 1 Results8% 0% 27%22%43%A (80-100) B (60-79) C (40-59) D (20-39) F (0-19)MATH 1113/05 Exam 1 Results20%4%0%31%45%

Kennesaw - MATH - 1113

MATH 1113/02 Practice Exam (to be graded but grade doesnt count) September 4, 2002 S. F. Ellermeyer Name Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other w

Kennesaw - MATH - 1113

S. F. EllermeyerMATH 1113 Final Exam July 24, 2003 NameInstructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (mathema

Kennesaw - MATH - 1113

S. F. EllermeyerMATH 1113 Exam 1 (Version 1) June 20, 2005 NameInstructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (

Kennesaw - MATH - 1113

S. F. EllermeyerMATH 1113 Exam 2 (Version 2) July 7, 2005 NameInstructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (m

Kennesaw - MATH - 1113

MATH 1113/02 Exam 3 Results Average Score was 7024%0% 3% 33%40%A (80-100) B (60-79) C (40-59) D (20-39) F (0-19)MATH 1113/05 Exam 3 Results Average Score was 7010% 15%3% 33%39%A (80-100) B (60-79) C (40-59) D (20-39) F (0-19)MATH

Kennesaw - MATH - 1113

Solutions to Homework Problems from Section 7.3 of Stewart 1. Given that sin x 5/13 and that x is in quadrant I, we have 2 144 cos 2 x 1 sin 2 x 1 5 169 13 so 144 12 . 13 169 Since x is in quadrant I we know that cos x 0 so cos x 12/13. Fro

Kennesaw - MATH - 1113

Solutions to Homework Problems from Section 7.5 of Stewart The solutions of the basic equations cos x k, sin x k, and tan x k are as follows: 1. If k is a number such that 1 k 1, then the solutions of the equation cos x k are x arccosk 2n (o

Kennesaw - MATH - 1113

Kennesaw - MATH - 1509

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Kennesaw - MATH - 1509

! " " # $ % $ & & ' & () ) * & & & ( *" +, + , +, +, 6 4 2 -20 -10 0 0 -2 10( -50 40 30 20 10 5 10 15 20 25( -3 2.5 2 1.5 1 0.5 00 0.5 1 1.5 2

Kennesaw - MATH - 1509

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Kennesaw - MATH - 1509

Kennesaw - MATH - 1509

Here are solutions for the seven homework problems on the handout which was given in class.1. The seventh degree Taylor polynomial (centered at 0) of the functionfx 1 sin2x 1 x 4 2 is f 2 0 2 f 3 0 3 f 4 0 4 f 5 0 5 f 6 0 6 f 7 0

Kennesaw - MATH - 4361

Subgroups of GroupsS. F. Ellermeyer November 8, 2006Suppose that G is a group and that H G. If H is also a group (under the same operation as G), then we say that H is a subgroup of G. If G is a group with identity element e, then E = feg is a subg

Kennesaw - MATH - 4361

Solutions to Exercises from Section 1.1 Notes S. F. Ellermeyer 1. Use the Well-Ordering Principle to explain why the set S = f37 + 5x j x 2 Z and 37 + 5x > 0g must have a smallest member. Find this smallest member. Solution: If x is an integer, then

Kennesaw - MATH - 4361

Prime NumbersAugust 27, 20061The Fundamental Theorem of ArithmeticDenition 1 A prime number is an integer greater than 1 whose only positive divisors are itself and 1. Any positive integer greater than 1 that is not prime is said to be composi

Kennesaw - MATH - 4361

Solutions to Section 2.4 Homework Problems S. F. Ellermeyer 2. This is similar to the proof given for addition in the notes. 3. (a) In Z13 , 6 + 7 = 0. (b) In Z13 ; 6 7 = 3 (c) In Z10 , 4 + 8 = 2 (d) In Z10 , 4 8 = 2. 5. (a) In Z5 , we solve 3 + x =

Kennesaw - MATH - 4361

Rings of ResiduesS. F. Ellermeyer September 18, 2006If m is a positive integer, then we obtain the partition C = f[0]m ; [1]m ; : : : ; [m 1]m gof Z into m congruence classes. (This is discussed in detail in the course notes that accompany Sectio

Kennesaw - MATH - 4361

Answers and Solutions to Selected Homework Problems From Section 2.5 S. F. Ellermeyer 5. Since gcd (2; 4) 6= 1, then 2 is a zero divisor (and not a unit) in Z4 . In fact, we see that 2 2 = 0 in Z4 . Thus 2x = 2y in Z4 does not imply that x = y. In pa

Kennesaw - MATH - 4361

Rings, Integral Domains, and FieldsS. F. Ellermeyer September 26, 2006Suppose that A is a set of objects endowed with two binary operations called addition (and denoted by "+" and multiplication (denoted by " " ) ). Let R = fA; +; g. R is said to b

Kennesaw - MATH - 4361

Solutions to Section 3.4 Homework Problems S. F. Ellermeyer October 27, 2006 1. 3x3 + 2x + 5 is primitive and 6x4 + 3x + 9 is not primitive. 2. Let p (x) 2 Q [x]. (We may assume that the coe cients of p (x) are written in lowest terms.) Let m be the

Kennesaw - MATH - 4361

Answers and Solutions to Section 4.1 Homework Problems S. Ellermeyer November 5, 2006 1. (a) fZ; g is not a group because the associative property is not satised for the subtraction operation. For example 7 (3 5) = 7 ( 2) = 9 but (7 3) 5 = 4 5 = 1. (

Kennesaw - MATH - 4361

Answers and Solutions to Section 4.4 Homework Problems S. Ellermeyer November 17, 2006 1. The trivial subgroup of Q f0g is f1g and, of course, Q f0g is a subgroup of Q f0g. In addition, f 1; 1g is a subgroup of Q f0g. Another subgroup of Q f0g is f3n

Kennesaw - MATH - 4361

Isomorphisms of GroupsS. F. Ellermeyer November 14, 2006The Cayley tables for the Klein 4 Group, K = Z2 U8 = f1; 3; 5; 7g are shown below. (0; 0) (0; 1) (1; 0) (1; 1) (0; 0) (0; 0) (0; 1) (1; 0) (1; 1) (0; 1) (0; 1) (0; 0) (1; 1) (1; 0) (1; 0) (1;

Kennesaw - MATH - 4361

Answers to Section 4.7 Homework Problems S. F. Ellermeyer November 27, 2006 1. Here is the Cayley table for Z4 : + 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2Here is an intermediate table with the substitutions 0 1 2 3 1 2 3 4 1 1 2 3 4 ! ! ! !

Kennesaw - MATH - 4361

Partitions of GroupsS. F. Ellermeyer November 15, 2006Suppose that G is a group and suppose that H is a subgroup of G. For any given element a 2 G, we may consider the left coset aH = fah j h 2 Hg and the right coset Ha = fha j h 2 Hg. If G is an a

Kennesaw - MATH - 4361

It is problems of the following type that inspired the Chinese Remainder Theorem. These problems were found on the Web page http:/www.cut-the-knot.org/blue/chinese.shtml. 1. (posed by Sun Tsu Suan-Ching (4th century AD) There are certain things whose

Kennesaw - MATH - 4361

MATH 4361 Abstract Algebra Study Guide for Exam 1 (September 13, 2006) Be able to state and apply the Well Ordering Principle, General Well Ordering Principle, and Principle of Induction. Be able to dene the following terms: 1. what it means for an i

Kennesaw - MATH - 4361

MATH 4361 Exam 1 (Version 1) Solutions September 13, 2006 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. The harder that I have to work to try to interpret what you are trying to say, the less partial credit you

Kennesaw - MATH - 4361

MATH 4361 Abstract Algebra Study Guide for Exam 2 (October 9, 2006) Be able to dene the following terms: 1. What is an equivalence relation on a set? 2. What is a partition of a set? 3. Given a positive integer, m, and integers a and b, what does it

Kennesaw - MATH - 4361

MATH 4361 Abstract Algebra Study Guide for Exam 2 (October 9, 2006) Be able to dene the following terms: 1. What is an equivalence relation on a set? 2. What is a partition of a set? 3. Given a positive integer, m, and integers a and b, what does it

Kennesaw - MATH - 4361

MATH 4361 Exam 2 (Version 2) Solutions October 9, 2006 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. The harder that I have to work to try to interpret what you are trying to say, the less partial credit you wil

Kennesaw - MATH - 4361

MATH 4361 Abstract Algebra Study Guide for Exam 3 (November 1, 2006) Be able to de.ne the following terms: 1. the degree of a polynomial in R [x] where R is a ring. 2. what is meant by a root (also called a zero) of a polynomial f (x) 2 R [x] where R

Kennesaw - MATH - 4361

MATH 4361 Exam 3 (Version 1) Solutions November 1, 2006 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. The harder that I have to work to try to interpret what you are trying to say, the less partial credit you wi

Kennesaw - MATH - 4361

MATH 4361 Exam 3 (Version 2) Solutions November 1, 2006 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. The harder that I have to work to try to interpret what you are trying to say, the less partial credit you wi

Kennesaw - MATH - 4361

MATH 4361 Study Guide for Exam 4 December 4, 2006 Be able to give accurate denitions of the following terms. 1. group 2. abelian group (also called commutative group) 3. subgroup of a group 4. cyclic group 5. generator of a cyclic group 6. the order

Kennesaw - MATH - 4361

MATH 4361 Study Guide for Exam 4 July 26, 2007 Be able to give accurate denitions of the following terms. 1. group 2. abelian group (also called commutative group) 3. subgroup of a group 4. cyclic group 5. generator of a cyclic group 6. the order of

Kennesaw - MATH - 4361

MATH 4361 Exam 4 (Version 2) Solutions December 4, 2006 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. The harder that I have to work to try to interpret what you are trying to say, the less partial credit you wi

Kennesaw - MATH - 4361

MATH 4361 Final Exam (Version 2) Solutions December 11, 2006 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. The harder that I have to work to try to interpret what you are trying to say, the less partial credit y

Kennesaw - MATH - 4361

MATH 4361 Abstract Algebra Study Guide for Exam 3 (Summer 2007) Be able to de.ne the following terms: 1. the degree of a polynomial in R [x] where R is a ring. 2. what is meant by a root (also called a zero) of a polynomial f (x) 2 R [x] where R is a

Kennesaw - MATH - 4361

MATH 4361 Exam 3 (Version 2) Solutions July 10, 2007 S. F. Ellermeyer Name Instructions. Please be detailed in your solutions and proofs. Write in complete sentences! The harder that I have to work to try to interpret what you are trying to say, the

Kennesaw - MATH - 4361

Rings of PolynomialsOctober 23, 2006Let R be a ring and let x be an indeterminateor place holder Then . the set of all expressions of the form f (x) = an xn + an 1 xn1++ a1 x + a0where an , an 1 , , a1 and a0 are members of R, and addition a

Kennesaw - MATH - 1113

vti_encoding:SR|utf8-nl vti_title:SR|Syllabus for MATH 1190 - Fall Semester, 1998 vti_cachedlinkinfo:VX| vti_cachedhastheme:BR|false vti_cachedhasborder:BR|false vti_filesize:IX|94208 vti_cacheddtm:TX|27 Aug 1998 17:58:18 -0400 vti_cachedhasbots:BR|f

CSU Fullerton - BTS - 530

BTS530 Test1 F2008 Name Student IDPage 1 of 9This test is out of 80 and is worth 15% of your course mark Question 1 (20 marks) Orienteering is a sport in which people use their navigational skills along with a map and a compass to reach specifi

CSU Fullerton - BTS - 530

BTS530 Exercise 3 Question 1 What problem(s) is/are illustrated in the sequence diagram shown below? Use the Pure Fabrication and Indirection to solve the problem(s).1: CustServClerk: NewCustScreenc : Customer: SQLDatabaseenter(custName,