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Everyday Mathematics Family Night September 22, 2010.

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Presentation on theme: "Everyday Mathematics Family Night September 22, 2010."— Presentation transcript:

1 Everyday Mathematics Family Night September 22, 2010

2 Background Developed by the University of Chicago School Mathematics Project Based on research about how students learn and develop mathematical power Provides the broad mathematical background needed in the 21 st century

3 You can expect to see… … a problem-solving approach based on everyday situations …an instructional approach that revisits concepts regularly …frequent practice of basic skills, often through games …lessons based on activities and discussion, not a textbook …mathematical content that goes beyond basic arithmetic

4 A Spiral Approach to Mathematics The program moves briskly and revisits key ideas and skills in slightly different contexts throughout the year. Multiple exposure to topics ensures solid comprehension. Strands are woven together-no strand is in danger of being left out.

5 More Spiraling… Mastery is developed over time. The Content by Strand Poster depicts the interwoven design. Homework problems will have familiar formats, but different levels of difficulty.

6 Everyday Mathematics Website Each student will receive login for home access. (available from your child’s teacher) Website contents: games and student reference book (SRB) http://www.everydaymathonline.com

7 Something to think about… “Even though it doesn’t look quite like what you did when you went to school, yes, this is really good, solid mathematics.”- 2001 Education Development Center Inc.

8 Focus Algorithms Algorithm slides created by Rina Iati, South Western School District, Hanover, PA

9 Partial Sums An Addition Algorithm

10 268+ 483 600 Add the hundreds ( 200 + 400) Add the tens (60 +80) 140 Add the ones (8 + 3) Add the partial sums (600 + 140 + 11) + 11 751

11 785+ 641 1300 Add the hundreds ( 700 + 600) Add the tens (80 +40) 120 Add the ones (5 + 1) Add the partial sums (1300 + 120 + 6) + 6 1426

12 329+ 989 1200 100 + 18 1318

13 An alternative subtraction algorithm

14 In order to subtract, the top number must be larger than the bottom number 9 3 2 - 3 5 6 To make the top number in the ones column larger than the bottom number, borrow 1 ten. The top number become 12 and the top number in the tens column becomes 2. 12 2 To make the top number in the tens column larger than the bottom number, borrow 1 hundred. The top number in the tens column becomes 12 and the top number in the hundreds column becomes 8. 12 8 Now subtract column by column in any order 5 6 7

15 Let’s try another one together 7 2 5 - 4 9 8 15 To make the top number in the ones column larger than the bottom number, borrow 1 ten. The top number become 15 and the top number in the tens column becomes 1. 15 1 To make the top number in the tens column larger than the bottom number, borrow 1 hundred. The top number in the tens column becomes 11 and the top number in the hundreds column becomes 6. 11 6 Now subtract column by column in any order 2 7 2

16 Now, do this one on your own. 9 4 2 - 2 8 7 12 3 13 8 6 5 5

17 Last one! This one is tricky! 7 0 3 - 4 6 9 13 9 6 2 4 3 10

18 Partial Products Algorithm for Multiplication

19 Calculate 50 X 60 67 X 53 Calculate 50 X 7 3,000 350 180 21 Calculate 3 X 60 Calculate 3 X 7 + Add the results 3,551 To find 67 x 53, think of 67 as 60 + 7 and 53 as 50 + 3. Then multiply each part of one sum by each part of the other, and add the results

20 Calculate 10 X 20 14 X 23 Calculate 20 X 4 200 80 30 12 Calculate 3 X 10 Calculate 3 X 4 + Add the results 322 Let’s try another one.

21 Calculate 30 X 70 38 X 79 Calculate 70 X 8 2, 100 560 270 72 Calculate 9 X 30 Calculate 9 X 8 + Add the results Do this one on your own. 3002 Let’s see if you’re right.

22 Partial Quotients A Division Algorithm

23 The Partial Quotients Algorithm uses a series of “at least, but less than” estimates of how many b’s in a. You might begin with multiples of 10 – they’re easiest. 12 158 There are at least ten 12’s in 158 (10 x 12=120), but fewer than twenty. (20 x 12 = 240) 10 – 1st guess - 120 38 Subtract There are more than three (3 x 12 = 36), but fewer than four (4 x 12 = 48). Record 3 as the next guess 3 – 2 nd guess - 36 2 13 Sum of guesses Subtract Since 2 is less than 12, you can stop estimating. The final result is the sum of the guesses (10 + 3 = 13) plus what is left over (remainder of 2 )

24 Let’s try another one 36 7,891 100 – 1st guess - 3,600 4,291 Subtract 100 – 2 nd guess - 3,600 7 219 R7 Sum of guesses Subtract 691 10 – 3 rd guess - 360 331 9 – 4th guess - 324

25 Now do this one on your own. 43 8,572 100 – 1st guess - 4,300 4272 Subtract 90 – 2 nd guess -3870 15 199 R 15 Sum of guesses Subtract 402 7 – 3 rd guess - 301 101 2 – 4th guess - 86


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