Monday

July 28, 2014

July 28, 2014

Posted by **Brittany snow** on Monday, January 9, 2012 at 1:03am.

A rectangle has an area of 120 cm square we are trying to find possible perimeters. It's length and width are whole numbers. What are all the possibilities for the 2 numbers?

Give next two terms

3,4,7,11,18,29....

1,2,6,24,120....

77,49,36,18....

- Math -
**Anonymous**, Monday, January 9, 2012 at 2:22amOdd numbers: 1 , 3 , 5...

200th od number = 399

Now:

1 + 399 = 400

3 + 397 = 400

5 + 395 = 400

....

199 + 201 = 400

100 * 400 = 40,000

A = L * W = 120

L = 120 / W

120 = 2 * 2 * 2 * 3 * 5

Ohe possibilities for the 2 numbers all numbers dividable with 2 3 and 5

W = 120 , L = 120 / 120 = 1

W = 60 , L = 2

W = 30 , L = 4

W = 15 , L = 8

W = 10 , L = 12

W = 8 , L = 15

W = 6 , L = 20

W = 4 , L = 30

W = 3 , L = 40

W = 2 , L = 60

W = 1 , L = 120

3,4,7,11,18,29....

The previous two numbers make up the next number:

3 + 4 = 7

7 + 4 = 11

11 + 7 = 18

11 + 18 = 29

Last two are:

18 + 29 = 47

and

29 + 47 = 76

3 ,4 , 7 , 11 , 18 , 29 , 47 , 76

1 , 2 , 6 , 24 , 120..

The multiplier increases by one each time:

1 x 2 = 2

2 x 3 = 6

6 x 4 = 24

24 x 5 = 120

120 x 6 = 720

720 x 7 = 5040

1 , 2 , 6 , 24 , 120 , 720 , 5040

77 , 49 , 36 , 18 ...

77

7 x 7 = 49

4 x 9 = 36)

3 x 6 = 18

1 x 8 = 8)

0 x 8 = 0

77 , 49 , 36 , 18 , 8 , 0

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