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### 1150 day 1

1. 1. Math 1150Summer 2012
2. 2. Problem Solving andNumber Patterns
3. 3. How do you solve aproblem?
4. 4. George Polya(1887-1985)
5. 5. Polya’s 4-Step problem-solving process (pg. 4)1) Understand the problem2) Devise a plan • Look for a pattern • Examine related problems • Look at a simpler case • Make a table or list • Draw a picture • Write an equation • Guess and Check • Work backward • Use direct or indirect reasoning
6. 6. 3) Carry out the plan4) Look back • Does your answer make sense? • Did you answer the question that was asked? • Is there another way to find the solution?
7. 7. How doyousolve amathproblem?
8. 8. 4th grade - 1959
9. 9. 4th grade - 2008
10. 10. 1. A pen at the zoo holds giraffes and ostriches.Altogether, here are 30 eyes and 44 feet on the animals.How many of each type of animal are there? How would you solve this problem? Let x = the number of giraffes Let y = the number of ostriches 2x + 2y = 30  4x + 4y = 60 4x + 2y = 44 –4x – 2y = –44 2y = 16 y=8 x=7
11. 11. 1. A pen at the zoo holds giraffes and ostriches.Altogether, here are 30 eyes and 44 feet on the animals.How many of each type of animal are there? How could a 2nd grader do this problem?
12. 12. 1. A pen at the zoo holds giraffes and ostriches.Altogether, here are 30 eyes and 44 feet on the animals.How many of each type of animal are there? Each animal has two eyes 30 2 15 Each animal has at least two feet We still need 14 more feet (44 – 30) 7 giraffes 8 ostriches
13. 13. 1. A pen at the zoo holds giraffes and ostriches.Altogether, here are 30 eyes and 44 feet on the animals.How many of each type of animal are there? How did a 7th grader do this problem?
14. 14. 1. A pen at the zoo holds giraffes and ostriches.Altogether, here are 30 eyes and 44 feet on the animals.How many of each type of animal are there?
15. 15. 2. Find the sum: 1 + 2 + 3 + … + 48 + 49 + 50
16. 16. Gauss’ Method 2. Find the sum: 1 + 2 + 3 + … + 48 + 49 + 50 51 51Sum of each pair = 51Number of pairs = Number of numbers / 2 = 25Sum of numbers = (Number of pairs)(Sum of each pair) = (25 )(51) = 1275
17. 17. 3. If ten people are in a room, how many handshakes can theyexchange?People Handshakes 1 0 +1 2 1 +2 3 3 +3 4 6 +4 5 10 6 15 +5 7 21 +6 8 28 +7 9 36 +8 45 handshakes 10 45 +9
18. 18. 4. Mark and Bob began reading the same novel on the sameday. Mark reads 6 pages a day, and Bob reads 5 pages a day.If Mark is on page 78, what page is Bob on? Mark: 78 / 6 = 13 days of reading Bob: 5 (13) = page 65
19. 19. 4. Mark and Bob began reading the same novel on thesame day. Mark reads 6 pages a day, and Bob reads 5pages a day. If Mark is on page 78, what page is Bobon?
20. 20. 5. If a bag of potato chips and a Snickers bar togethercost \$3.00, and the chips cost \$1.50 more than thecandy bar, how much does each item cost separately?
21. 21. 5. If a bag of potato chips and a Snickers bar together cost \$3.00,and the chips cost \$1.50 more than the candy bar, how muchdoes each item cost separately? Guess-and-check must show at least one incorrect guess and the correct guess Chips Candy \$1.50 \$1.50 Chips and candy cost the same \$2.50 \$ .50 Chips cost \$2.00 more \$2.25 \$ .75 Chips cost \$1.50 more Chips: \$2.25 Candy: \$0.75
22. 22. 6. Billy spent 2/3 of his money on baseball cards and 1/6 of hismoney on a candy bar, and after that he still had 50 cents left inhis pocket. How much money did he start with? Billy’s whole box of money 50₵ 50₵ 50₵ 50₵ 50₵ 50₵ Baseball cards Candy bar 6 x 50₵ = \$3.00
23. 23. 6. Billy spent 2/3 of his money on baseball cards and1/6 of his money on a candy bar, and after that he stillhad 50 cents left in his pocket. How much money did hestart with?
24. 24. NumberPatterns
25. 25. From MODESE Model Curriculum Unit (6th grade)The student council plans to build a flower box withseveral sections in front of the school cafeteria. Eachsection will consist of squares made with railroad ties ofequal length. Due to money constraints, the box will bebuilt in several phases. Below is the plan for the firstthree phases. (Each side of the square represents onetie.) Phase 1 Phase 2 Phase 3Create a table to show how many ties will be needed for each ofthe first six phases. Explain how you know whether the patternrepresents a linear or a nonlinear function.
26. 26. Phase 1 Phase 2 Phase 3 Phase Ties 1 4 Linear +3 2 7 because the 3 10 +3 phase number +3 increases at a 4 13 constant rate of 5 16 +3 1, and the ties 6 19 +3 increase at a constant rate of 3
27. 27. A sequence is a pattern of numbers or symbols.A term is a number in a sequence. Phase Ties In an arithmetic 1 4 sequence, we add a 2 7 constant number (called 3 10 the common difference) 4 13 to find the next term in 5 16 the sequence. 6 19 For this sequence, d = 3
28. 28. A sequence is a pattern of numbers or symbols.A term is a number in a sequence. Term Ties 1 4 2 7 3 10 4 13 5 16 6 19
29. 29. A sequence is a pattern of numbers or symbols.A term is a number in a sequence. Term Ties a1 4 a2 7 a3 10 a4 13 a5 16 a6 19
30. 30. A sequence is a pattern of numbers or symbols.A term is a number in a sequence. Term Value a1 4 a2 7 a3 10 a4 13 a5 16 a6 19
31. 31. A sequence is a pattern of numbers or symbols.A term is a number in a sequence. Term Value Pattern a1 4 4 +3 a2 7 4 + 1(3) +3 a3 10 4 + 2(3) a4 13 +3 4 + 3(3) a5 16 +3 4 + 4(3) +3 a6 19 4 + 5(3) an 4 + (n-1)(3)
32. 32. What is the 100th term in the sequence?4 + (100 – 1)(3) = 4 + 297 = 301 Term Value Pattern a1 4 4 +3 a2 7 4 + 1(3) +3 a3 10 4 + 2(3) a4 13 +3 4 + 3(3) a5 16 +3 4 + 4(3) +3 a6 19 4 + 5(3) an 4 + (n-1)(3)
33. 33. nth term formula for an arithmetic sequence: an = a1 + (n – 1)d Term Pattern a1 4 a2 4 + 1(3) a3 4 + 2(3) a1 = 4 a4 4 + 3(3) a5 4 + 4(3) d = 3 a6 4 + 5(3) an 4 + (n-1)(3)
34. 34. 5, 9, 13, 17, … d=4Find the next three terms 21, 25, 29Find an nth term formula for the sequence an = a1 + (n – 1)d a1 = 5 d=4 an = 5 + (n – 1)4 an = 5 + 4n – 4 an = 4n + 1Find the 100th term of the sequence a100 = 4(100) + 1 = 401
35. 35. 5, 9, 13, 17, …What position in the sequence is the number609? an = 4n + 1 609 = 4n + 1 -1 -1 608 = 4n 4 4 609 is the 152nd term 152 = n
36. 36. DESE sample MAP item, 3rd grade
37. 37. 4, 12, 36, 108, … In a geometric sequence, we multiply by a constant (calledX3 X3 X3 the common ratio) to find the next term in the sequence. For this sequence, r = 3Find the common ratio for the sequence 128, 64, 32, 16, … 16 1 = 32 2
38. 38. 4, 12, 36, 108, … Term Value Pattern a1 4 4 x3 a2 12 4 x (3) x3 a3 36 4 x (3)2 a4 108 x3 4 x (3)3 an 4 x (3)n - 1 What is the 100th term of the sequence? a100 = 4 x 3100 – 1 = 4 x 399
39. 39. Term Pattern a1 4 a2 4 x (3) a = 4 2 1 a3 4 x (3) r=3 a4 4 x (3)3 an 4 x (3)n - 1nth term formula for a geometric sequence an = a1 · rn - 1
40. 40. What are the next three terms in this sequence: 1, 3, 4, 7, 11, 18, 29, 47In a Fibonacci sequence, we add twoconsecutive terms to find the next term.
41. 41. The first two terms of a sequence are 1 and 5.Find the next two terms if the sequence isa) Arithmetic 1, 5, 9, 13 +4 +4 +4b) Geometric 1, 5, 25, 125 x5 x5 x5c) Fibonacci 1, 5, 6, 11
42. 42. A finite sequence has a limited number of terms.One way to count the number of terms in asequence is to find a one-to-one correspondencewith the counting numbers. How many terms are in the sequence? A) 4, 8, 12, …, 40 divide each term by 4 1, 2, 3, …, 10 The sequence has 10 terms
43. 43. How many terms are in the sequence?B) 4, 9, 14, …, 59 add 1 to each term 5, 10, 15, …, 60 divide each term by 5 1, 2, 3, …, 12 The sequence has 12 terms
44. 44. How many terms are in the sequence?C) 62 , 63 , 64, …, 620 subtract 1 from each exponent 61 , 62 , 63, …, 619 look at exponents 1, 2, 3, …, 19 The sequence has 19 terms