# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

5 Pages

### assignment5_sol

Course: LINEAR ALG MATH135, Spring 2008
School: Waterloo
Rating:

Word Count: 1710

#### Document Preview

135 MATH Assignment 5 Solutions Hand-In Problems Exercise 3-2: Which of the following integers are congruent modulo 6? -147, -91, -22, -14, -2, 2, 4, 5, 21, 185 Solution: Look at the quotients and remainders on division by 6: x q r -147 -25 3 -91 -16 5 -22 -4 2 -14 -3 4 -2 -1 4 2 0 2 4 0 4 5 0 5 21 3 3 185 30 5 Fall 2005 Then the following sets of numbers give members of the same equivalence class under...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Canada >> Waterloo >> LINEAR ALG MATH135

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
135 MATH Assignment 5 Solutions Hand-In Problems Exercise 3-2: Which of the following integers are congruent modulo 6? -147, -91, -22, -14, -2, 2, 4, 5, 21, 185 Solution: Look at the quotients and remainders on division by 6: x q r -147 -25 3 -91 -16 5 -22 -4 2 -14 -3 4 -2 -1 4 2 0 2 4 0 4 5 0 5 21 3 3 185 30 5 Fall 2005 Then the following sets of numbers give members of the same equivalence class under congruence modulo 6: {2, -22}, {21, -147}, {4, -2, -14}, {185, 5, -91} since integers which give the same reminder when divided by 6 are congruent modulo 6. (We could also approach this problem by looking at the difference between each pair and checking if this difference is divisible by 6 or not. This would require more work, though.) Exercise 3-4: Let N = 3729 . What is the last digit in the decimal representation of N ? What are the last digits in the base 9 and base 8 representations of N ? Solution: Since the last digit of a number written in base b is the remainder when the number is divided by b, then we must calculate N modulo b for each of b = 10, b = 9, and b = 8. Now 32 -1 (mod 10) and 3729 = 3364(2)+1 = (32 )364 3 and so 3729 3 (32 )364 (mod 10) 3(-1)364 (mod 10) 3 (mod 10) so the last digit in the decimal expansion of N is 3. Similarly, the last digit in the base 9 representation is the remainder upon division by 9. However 32 0 (mod 9), so 3729 = 32 3727 0 3727 (mod 9) 0 (mod 9) and the last digit in the base 9 representation of N is 0. Also, the last digit in the base 8 representation is the remainder upon division by 8. Now 32 1 (mod 8), so 3729 = 3364(2)+1 3 (32 )364 (mod 8) 3(1)364 (mod 8) 3 (mod 8) and the last digit in the base 8 representation of N is 3. Exercise 3-6: Is 617 + 176 divisible by 3 or 7? Solution: We consider 617 + 176 modulo 3 and modulo 7. For divisibility by 3, we see that 6 0 (mod 3) and 17 -1 (mod 3). Thus 617 + 176 017 + (-1)6 1 (mod 3) (mod 3) Therefore 3 does not divide 617 + 176 , since 617 + 176 0 (mod 3). For divisibility by 7, we have 6 -1 (mod 7) and 17 3 (mod 7). Therefore 617 + 176 (-1)17 + 36 (-1) + 3 3 2 3 (mod 7) (mod 7) (-1) + 9 (mod 7) (-1) + 23 (mod 7) -1 + 8 (mod 7) 0 (mod 7). Hence 617 + 176 is divisible by 7. Exercise 3-8: For the following congruence, determine whether there exists a positive integer k so that the congruence is satisfied. If so, find the smallest such k. 2k 1 (mod 11) Solution: Let us run through the first few positive integers (ie. the first few possible values for k): k 2k 1 2 2 4 3 8 Modulo 11 4 5 6 5 10 9 7 7 8 3 9 6 10 1 We conclude that there is a solution and the smallest value of k is 10. Exercise 3-10: For the following congruence, determine whether there exists a positive integer k so that the congruence is satisfied. If so, find the smallest such k. 2k 1 (mod 14) Solution 1: The number 2k is always even for k 1. Any number congruent to 1 modulo 14 is of the form 14q + 1 for some integer q, so is odd. Therefore, 2k can never be congruent to 1 modulo 14, since it cannot be both even and odd. Solution 2: 21 22 23 24 2 4 8 2 (mod (mod (mod (mod 14) 14) 14) 14) Therefore we can multiply the first congruence below by powers of 2: 21 22 23 24 24 25 26 27 (mod (mod (mod (mod 14) 14) 14) 14) Continuing this, we obtain 2 21 4 22 8 23 24 25 26 27 28 29 23r+1 23r+2 23r (mod 14) (mod 14) (mod 14) for every positive integer r. Thus no positive power of 2 is congruent to 1 modulo 14. Exercise 3-16: Find tests for determining whether an integer in base 12 is divisible by 13. Solution: Since 12 -1 (mod 13), we have (at . . . a3 a2 a1 a0 )12 12t at + + 124 a4 + 123 a3 + 122 a2 + 12a1 + a0 (mod 13) (-1)t at + + a4 - a3 + a2 - a1 + a0 (mod 13). Therefore, to determine whether a number in base 12 is divisible by 13, we just need to check if the alternating sum of its digits is divisible by 13. (Note that this is the same basic idea as the test for divisibility by 11 in base 10.) Exercise 3-28: Construct addition and multiplication tables for the following set of integers modulo m and find, if possible, multiplicative inverses of each of the elements in the set. Z3 Solution: Addition in Z3 + [0] [1] [2] [0] [0] [1] [2] [1] [1] [2] [0] [2] [2] [0] [1] Multiplication in Z3 [0] [1] [2] [0] [0] [0] [0] [2] [1] [0] [1] [2] [0] [2] [1] The multiplicative inverse of [1] is [1] and the multiplicative inverse of [2] is [2]. [0] does not have an inverse. Problem 3-56: If p is a prime, prove that x2 y 2 (mod p) if and only if x y (mod p). Solution: `` =" Suppose x2 y 2 (mod p). Then x2 - y 2 0 (mod p) or (x + y)(x - y) 0 (mod p) or p | (x + y)(x - y). Since p is prime, by Theorem 2.53 either p | x + y or p | x - y. Therefore, x -y (mod p) or x y (mod p). Thus, x y (mod p). "=" Conversely suppose x y (mod p). Then, squaring, x2 (y)2 y 2 (mod p), as required. Problem 3-57: If p is an odd prime, show that x2 a (mod p) has a solution for exactly half the values of a between 1 and p - 1 inclusive. Furthermore, if 1 a p - 1 and x2 a (mod p) has a solution, show that it has exactly two congruence classes of solutions modulo p. Solution: p-1 p+1 Since p is odd, the p possibilities x for modulo p are 0, 1, 2, . . ., , , . . . , p - 1. 2 2 2 2 By Problem 3-56, we know that if x -y (mod p), then x y (mod p). Therefore, since 1 -(p - 1) (mod p), we have 12 (p - 1)2 (mod p). Also, 2 -(p - 2) (mod p), so 22 (p - 2)2 (mod p). In general, if 1 m < 1 p, then m -(p - m) (mod p) so m2 (p - m)2 (mod p). 2 1 1 (Note that if 1 m < 2 p, then 2 p < p - m p - 1, so m and p - m are different numbers from the list 0, 1, 2, . . . , p - 1.) 2 2 p-1 p+1 p-1 p+1 The last congruence of this form will be - (mod p) (because - (mod p)). 2 2 2 2 So we have paired the numbers from 1 to p - 1 into 1 (p - 1) pairs of numbers whose squares are the same modulo p. 2 Is it possible for numbers in different pairs to have squares which are congruent modulo p? It is enough to look at the smaller numbers in two different pairs. Since if x2 y 2 (mod p), then x y (mod p), then it is impossible for any two different numbers i and j each less than 1 p to have congruent squares modulo p, since i and j cannot be congurent modulo p and since if i -j (mod p), 2 then we would have p | i + j which is impossible if both i and j are less than 1 p. 2 So we have divided up the squares 12 , 22 , . . ., (p - 1)2 modulo p into pairs which are congruent modulo p. No two squares in different pairs are congruent by the above remark. Also, if x 0 (mod p), then x2 0 (mod p). (If x2 0 (mod p), then p | x2 , so p | x so x 0 (mod p).) This tells us that 0 is not the square of any pair of numbers. Lastly, 02 0 (mod p). 1 Therefore, since there are 2 (p - 1) pairs of numbers with each pair giving a different non-zero square modulo p, then exactly half of the numbers from 1 to p - 1 are squares modulo p. Also, if we know that x2 a (mod p) (with 1 a p - 1) has a solution, then it has exactly two solutions modulo p, since the solution is exactly a pair of numbers from above. Recommended Problems Exercise 3-3: What is the remainder when 824 is divided by 3? Solution: We compute 824 modulo 3: 8 -1 (mod 3) 8 (-1)24 (mod 3) 1 (mod 3) 24 Hence the remainder when 824 is divided by 3 is 1. Exercise 3-5: What is the remainder when 1045 is divided by 7? Solution: We consider 1045 modulo 7 and try to find a small power of 10 which is congruent to 1 or -1 modulo 7: 10 102 103 1045 3 (mod 7) 32 2 (mod 7) 10 102 3 2 6 -1 (103 )15 (mod 7) (-1)15 (mod 7) -1 (mod 7) 6 (mod 7) (mod 7) Hence the remainder when 1045 is divided by 7 is 6. Exercise 3-7: Show that an integer of the form 5n + 3, where n P, can never be a perfect square. Solution: Every integer is congruent to 0, 1, 2, 3, or 4 modulo 5. Their squares have the following form. x x2 This table summarizes these facts. (1) The square of any integer divisible by 5 is divisible by 5. (2) The square of any integer of the form 5q + 1 or 5q + 4 has remainder 1 when divided by 5. (3) The square of any integer of the form 5q + 2 or 5q + 3 has remainder 4 when divided by 5. Hence the square of any integer has remainder 0, 1, or 4 modulo 5. Thus no integer has a square with congruent 3 modulo 5. That is, 5n + 3 is never a perfect square. Modulo 5 0 1 2 3 0 1 4 4 4 1 Exercise 3-13: Find tests for determining whether an integer in base 10 is divisible by 12. Solution: By Proposition 3.64, 12 | x if and only if 3 | x and 4 | x. Combining the tests for divisibility by 3 and 4 in base 10 we get that a number is divisible by 12 in base 10 if and only if the sum of its digits is divisible by 3 and the number determined by its last two digits is divisible by 4. Problem 3-58: Does x3 a (mod p) always have a solution for every value of a, whenever p is prime? Solution: The congruence x3 a (mod p) does not always have a solution for every a and every prime p. To show this, we need a counterexample. For example, when p = 7 : x x3 1 1 Modulo 7 2 3 4 5 1 6 1 6 6 6 0 0 So the equation x3 2 (mod 7), for example, has no solution. Problem 3-59: Choose any integer larger than 10, subtract the sum of its digits from it, cross out one nonzero digit from the result, and let the sum of the remaining digits be s. From a knowledge of s alone, is it possible to find the digit that was crossed out? Solution: Let the integer be x > 10, so it has at least 2 digits and the described procedure is possible. Let x = (dr dr-1 d1 d0 )10 be x in terms of its digits. Then x = 10r dr + 10r-1 dr-1 + + 10d1 + d0 and the sum of its digits is dr + dr-1 + + d1 + d0 . Therefore, the difference between x and the sum of its digits is (10r - 1)dr + (10r-1 - 1)dr-1 + + 9d1 which is divisible by 9, since each coefficient is of the form 99 9. Suppose that x - (dr + d1 + d0 ) = (10r - 1)dr + (10r-1 - 1)dr-1 + + 9d1 = (yt yt-1 . . . y1 y0 )10 which we know is divisible by 9. Then the sum of its digits y1 + + yt 0 (mod 9). If we cross out some digit, say yi , then the sum of the remaining digits, s = y0 + y1 + + yt - yi -yi (mod 9). Hence yi -s (mod 9). Since yi is nonzero, 1 yi 9 and yi can be determined from s. For example, if s 7 (mod 9), then yi = 2 -7 (mod 9). (Note that 0 9 (mod 9), so if we crossed out a zero digit, we could not distinguish that from crossing out a 9, from the knowledge of s alone.)
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Waterloo - LINEAR ALG - MATH135
MATH 135 Assignment 6 SolutionsHand-In ProblemsExercise 3-34:Solve the congruence 5x 7 Solution: Here, gcd(5, 15) = 5, but 5 | 7, so there is no solution by Theorem 3.54. (mod 15).Fall 2005Exercise 3-44:Find the inverse of [23] in Z41 . Solu
Waterloo - LINEAR ALG - MATH135
MATH 135 Assignment 8 SolutionsHand-In ProblemsExercise 7-1:The following is known to be a simple substitution cipher. Break the code.PCTPG BTHHP TUDAA ANJHT VTHHT DLXCV GDURG CIIDW HXBEA NEIDV XHIGD TBTIW GPEWN DEHLT DS LPHYJ GTSXH AXJHR VJXHT P
Waterloo - LINEAR ALG - MATH135
MATH 135 Assignment 9 SolutionsHand-In ProblemsExercise 8-64:Express following complex numbers in standard form. the (1 - 3i)8 Solution: |1 - 3i| = 2 and the argument of (1 - 3i) is = 5 , so (1 - 3i) = 2cis 3 By De Moivre's Theorem, we have
Waterloo - LINEAR ALG - MATH135
MATH 135 Assignment 10 SolutionsProblem ListChapter 9: 5, 23, 40, 44, 45, 48, 58, 63, 100, 159Fall 2005Recommended ProblemsExercise 9-5:Find the sum, difference and product of each of the following pairs of polynomials with coefficients in th
Waterloo - LINEAR ALG - MATH135
Faculty of Mathematics University of Waterloo MATH 135Monday 14 November 2004MIDTERM EXAM #2 Fall 200519:00 20:15Solutions[8] 1. Solve the system of congruences x 40 (mod 131) x 7 (mod 18) Solution: Since x 40 (mod 131), then x = 40 + 131y
Waterloo - MATH - MATH137
UBC - GEO - 122
Final Examination: Geography 122 Geographical Change in the Modernising World 17th. April 2007 8.30-10.30 amREAD CAREFULLY. You have two hours to complete the examination. Answer a total of four questions, and follow the instructions with care. Spe
UBC - GEO - 122
G122 Lecture 15The Globalisation of Sport 1. The Beckham phenomenon Athlete; lifestyle; TV/media; ambassador; advertising sponsor .and posh Spice! Transnational celebrity. 2. From folk cultures to global spectacle 19C modernisation and commercialis
UBC - GEO - 122
mike davisPLANET OF SLUMSUrban Involution and the Informal Proletariatometime in the next year, a woman will give birth in the Lagos slum of Ajegunle, a young man will flee his village in west Java for the bright lights of Jakarta, or a farmer
UBC - GEO - 122
.rom .ree Oil to .reedom Oil: Terrorism, War and US Geopolitics in the Persian GulfP H I L I P P E L E B I L L O N and . O U A D E L K H AT I BPersian oil . is yours. We share the oil of Iraq and Kuwait. As for Saudi Arabian oil, its ours. Preside
Oklahoma Christian - BIBL - New testim
Dates586-destruction of Jerusalem 535- Jews returned 400-end of the Old Testament. 336- Palestine by Greeks 63- Palestine's by romans 37- Reign of herod the great 6-4 Jesus birth 27-30- Jesus diedVerse1 Lords supper- Math 26 2- whoever believes mar
Oklahoma Christian - BIBL - New testim
Romans Theme; how can God save a sinner God's problem A. Author Paul B. Written to Christians in Rome C. Written from Corinth D. When- Third missionary journey 55-56 Eight chapter outline 1. Gentiles in sin and under God's wrath 2. Jews in sin and un
Oklahoma Christian - EDU - intro to e
Kimberly Thompson School in American Culture Article Summary Lifelong Learners The articles that I chose for life time learners were very informative. The first one talks about Hong Kong, and the growing rate of distance learners they have. This of c
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementAlphabet body tangoAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource ExtensionsK5 Phonics 1. The students will be able to reco
Oklahoma Christian - EDU - educationa
Calling All ParentsPARENT CORNERI feel a deep love and concern for each of my students. This month we will be holding a parent get together in the kindergarten classroom. The meeting will be on October 2nd at 7 pm. We will discuss topics such as ho
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Factor 4 Time Learning objection Day 1The students will be able to identify rhyming words with 80% accuracy.Instructional design Table Instructional Assessment Activity1. I will explain to the students what rhyming words are, an
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Factor 4.The activities were sequenced in this way, because the students would be challenged with a harder activity each day. The lesson gradually became a little harder each day. This is a good way to ease them into learning the
Oklahoma Christian - EDU - intro to e
Kimberly Thompson School in American Culture Journal Summary Pre K Spending and funding The article that I read was about the funding and support of early childhood education. It covered the facts about how much the states pay each year for pre K edu
Oklahoma Christian - EDU - intro to e
Kimberly Thompson Journal ReflectionThe journal entries have truly helped me. They have given me insight into another teacher's world. I believe that to get another teachers view is very helpful. This journal reflection proves that I am capable of
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio movementSpelling guessing gameAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions2nd Spelling 1. The students will be able to
Oklahoma Christian - EDU - literacy 1
1. Phonics makes it easier to read and learn new words. It helps children develop a large reading vocab. Contributes to fluent reading. Helps children become independent readers. B. Depends on the Childs development as readers. C. Synthetic. Starts w
Oklahoma Christian - EDU - literacy 1
Kimberly Fletcher Literacy 1 Reflection of practicumA learning ExperienceFor the past few months, I have had the privilege of working with a child named Sam. Sam is unlike many children his age. He is a slow learner. This was a very hard thing to
Oklahoma Christian - EDU - movement i
Category; Visual ArtHistory of Valentine's DayKimberly FletcherAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterialsDirectionsSource Extensions5th grade History The students wi
Oklahoma Christian - EDU - literacy 1
The Vowel SongWhen two vowels go walking the first one does the talking When two vowels go walking the first one says its name When two vowels go walking the first one does the talking When two vowels go walking the first one says its nameU and E
Oklahoma Christian - COMPUTER S - power poin
Alternate assignment requirements If you choose to do the alternate assignment for the PowerPoint class there are two requirements: 1. Read the materials below. You do not have to read the comments (if the page contains them). While there are several
Oklahoma Christian - EDU - movement i
Kimberly Fletcher MediaDefinition- Media is any form of visual art. It also includes the tools we use to create visual art. It is not limited to the everyday tools in the classroom. Here are some examples. Different forms of media are drawing and r
Oklahoma Christian - EDU - intro to e
Kimberly Thompson Lesson Plan The Solar System 1. I hope to introduce the planets to the students. I want to open the door to further education about the planets. I expect the students to be able to recite the planets. 2. I will need PowerPoint, and
Oklahoma Christian - EDU - intro to e
Kimberly Thompson Mini Lesson Reflection I felt that my mini lesson went well. I am very confident in the material. I chose the solar system because I felt that it would challenge my students to think about complex issues. I want to challenge their y
Oklahoma Christian - EDU - LIA
Monday= 10; 00-10:20-Explain and give examples of rhyming words. Use the Hand clap rhyme game in instruction. 10:20-10:30- Rhyming word sit down game and rotation of computer activity. 10:30-10:40- Rhyming word book. Read to the children. 10:40-10:50
Oklahoma Christian - BIBL - New testim
1 Timothy. Theme: order in the church Teachings. A. Warnings of false teachers (Jewish) B Pray for our authority. C. Woman (Modest Dress) 1. Jewelry, braided hair, expensive clothes. D. Woman in worship 1. Learn in quietness 2. Not to take authority
Oklahoma Christian - EDU - literacy 1
MODEL LESSON PLAN FORMAT Aligned with Oklahoma's Minimum Criteria for Effective Teaching and Teacher Work Sample SUBJECT MATTER _ _ Instructional Time/Period of the Day _DATE(s) _ Essential Elements of Planning Elements Notes about theEstablish Con
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Early Childhood(LIA) Factor TwoOklahoma Pass Standard 1; Print Awareness - The student will understand the characteristics of written language. Pass skill; 1. Demonstrate correct book orientation by holding book correctly (right
Oklahoma Christian - EDU - movement i
Kimberly FletcherStudio DramaAge Level Curriculum Area Objectives Adaptations For Special NeedsAdaptation for different age levels MaterialsDirectionsSource Extensions4th grade Science The kids will be able to give facts about the process
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementFlower walkAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials Directions5th Social Studies 1. Students will be able to recognize the state flower
Oklahoma Christian - EDU - movement i
So, who was Saint Valentine and how did he become associated with this ancient rite? Today, the Catholic Church recognizes at least three different saints named Valentine or Valentinus, all of whom were martyred.
Oklahoma Christian - EDU - movement i
Category: Visual Arts Age Level Curriculum Area ObjectivesRhyming WordsKimberly FletcherAdaptations For Special Needs Adaptation for different age levelsMaterialsDirectionsSource ExtensionsKindergarten Language arts The students will be
Oklahoma Christian - EDU - movement i
Category: Visual Arts Age Level Curriculum Area Objectives Adaptations For Special Needs7's FamilyKimberly FletcherAdaptation for different age levelsMaterialsDirectionsSource Extensions3rd Grade Math Students will be able to write thei
Oklahoma Christian - EDU - movement i
Kimberly FletcherDrama StudioAge Level Curriculum Area Objectives Adaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions4th grade Science The students will be able to identify characteristics
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementJump the creekAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials2nd Math 1. The student will learn to measure distance. 2. The students will learn
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementInterpretive danceAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions4th History 1. The students will have more knowl
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio LiteraryScience and litAge Level Curriculum Area ObjectivesAdaptations For Special Needs4th Language arts and Science 1. Students will be able to give key about the scientific method 2. Students will gain creative wri
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementDigestive dance party.Age Level Curriculum Area Objectives4th Science 1. Students will better understand the digestive system. 2. Students will be able to move around and release energy. 3. Students will have a
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio Visual ArtBird WatchAge Level Curriculum Area ObjectivesAdaptations For Special Needs1st Science 1. Students will be able to recognize state birds. 2. Students will learn listening skills. 3. Students will learn to gi
Oklahoma Christian - EDU - movement i
Category: Visual Arts Age Level Curriculum Area Objectives Adaptations For Special Needs Adaptation for different age levelsCloudsKimberly FletcherMaterials DirectionsSource Extensions4th Grade Science The students will be able to identify
Oklahoma Christian - EDU - movement i
Category: Visual Art Age Level Curriculum Area Objectives Adaptations For Special Needs Adaptation for different age levelsPilgrimsKimberly FletcherMaterialsDirectionsSource Extensions3rd Grade History The students will be able to identif
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MusicDigeridooAge Level 3rd Curriculu culture m Area Objectives 1. Students will be able to understand the musical culture of Australia 2. Students will be able to be creative with their hands 3. Students will learn to fo
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementPlanet SignAge Level Curriculum Area Objectives2nd Science1. Students will have knowledge of the planets 2. The students will learn to sign their planets. 3. The students will learn listening skills. 4. Studen
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementMath jumping jacksAge Level Curriculum Area Objectives1st Math 1. Students will learn their addition tables 2. Students will learn listening skills. 3. Students will be able to move around and get the wiggles ou
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementMath raceAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterialsDirections4th Math 1. Students will be able to Correctly answer the division problems
Oklahoma Christian - EDU - intro to e
Kimberly Thompson Philosophy ReflectionI believe that my philosophy paper shows that I am prepared for the teaching profession. I have shown that I have the desire and love for teaching. I have also shown that I have had experience in the field. I
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MusicAlphabet scrambleAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials Directions2nd Language art 1. Students will be able to put things in alphabetical
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementRibbon DanceAge Level Curriculum Area ObjectivesK5 Space around Us and science1. Students will understand what the space around us is. 2. Students will be able to understand a small idea of gravity. 3. Student
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MusicSpelling movement storyAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions3rd Spelling 1. The students will be able to c
Oklahoma Christian - EDU - movement i
Kimberly FletcherLiterary Studio5thAge Level Curriculum Area Objectives Adaptations For Special Needs Adaptation for different age levelsMaterials DirectionsSource Extensionsgrade history The students will be able to recite their states. I
Oklahoma Christian - BIBL - New testim
Word Identification Strategies. PG 134-171 1. Letter Sound Bingo- Read words spelled with the same letter sound patterns children are learning. B. Interactive bulletin boards- In a small group sort words according to shared letter sound patterns and
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Factor 1 Literacy 1 practicum. Contextual Factor Number of Students;22 Female;11 Male;11 Ethnic and Cultural make-up; Sam is a white male.Socio economic status. and make up The school seemed to be average. The homes in the area se
Oklahoma Christian - EDU - LIA
Kimberly FletcherFactor 3 Post Assessment 2. Students will be able to recognize rhyming words from a list of given words.Draw a line to the correct rhyming word.1. 2. 3. 4. 5. 6. 7. 8. 9.Hall Stop Make Bite Man Sea Light Keep ToreDrop Can F
Oklahoma Christian - EDU - LIA
Kimberly FletcherFactor 3 Post Test Key1. Fall 2. Drop 3. Take 4. Fight 5. Can 6. Tea 7. Tight 8. Leap 9. Door
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Factor 3 Type of Assessment 1. Pre Assessment The assessment specifically addresses the correct way to hold and read from a book. The test is reliable and valid because it shows that the student has mastered the challenge of holding
Oklahoma Christian - EDU - LIA
Kimberly Thompson Factor 1 Reflection Factor on has helped me in many ways. It has taught me how to evaluate my classroom. I learned that it is crucial to be familiar with my class makeup in order to be an effective teacher. I believe that this paper
Oklahoma Christian - EDU - movement i
Rhyming game activity.Like to rhyme? Well, now's the time! This game is for 2 or more players. You'll need pencils or pens and paper. You can play one on one or you can split into two even teams. You have one minute to think of as many pairs of wor