Let's begin by listing the first few multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44. So, between 1 and 37 there are 9 such multiples: {4, 8, 12, 16, 20, 24, 28, 32, 36}. Note that 4 divided into 36 is 9.
Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:
<span>How many multiples of 4 are there in {n; 37< n <101}? We could list and then count them: {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval. Try subtracting 40 from 100; we get 60. Dividing 60 by 4, we get 15, which is 1 less than 16. So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:
1000 -40 ------- 960
Dividing this by 4, we get 240. Adding 1, we get 241.
Finally, subtract 9 from 241: We get 232.
There are 232 multiples of 4 between 37 and 1001.
Can you think of a more straightforward method of determining this number? </span>
First we look for the area of the triangle which is given by: A = (1/2) * (4) * (6) A = 12 feet ^ 2 Area of the rectangles: Rectangle 1: R1 = (4) * (8) R1 = 32 feet ^ 2 Rectangle 2: R2 = (6) * (8) R2 = 48 feet ^ 2 Rectangle 3: R3 = (root (6 ^ 2 + 4 ^ 2)) * (8)) R3 = 57,68882041 feet ^ 2 The total area will be: A = 2A + R1 + R2 + R3 Substituting values: A = 2 * (12) + 32 + 48 + 57,68882041 A = 161.6888204 feet ^ 2 Answer: The total area of the prism is: A = 161.6888204 feet ^ 2
The equations we get are l = 6 + 2w (We get this from "The length of the rectangle is 6 more than 2 times the width.") and 2l + 2w = 140 (We get this from the perimeter. Two times the length plus two times the width equals the perimeter of a quadrilateral.)
The first equation can be written as l - 2w = 6
So now we have 2l + 2w = 140 l - 2w = 6 ____________ Add the two equations. We get 3l = 146