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# Real World Problem Project-Perimeter

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## k17kekann Kanno

on 17 October 2012

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#### Transcript of Real World Problem Project-Perimeter

Real World Problem Project The length of one side of a rectangle is 56 yards more than 7 times the width. Its perimeter is 92 yards more than the length of a football field (without end zones). What's the width of the rectangle? Problem-Perimeter 2l+2w=p

2(7w+56)+2w=192 Equation w=width of the rectangle

l=length of the rectangle

p=perimeter of the rectangle Definition of Variables 2w + 2l = p
2(7w+56)+2w = 192
(2•7w)+(2•56)+2w = 192
14w+112+2w = 192
16w+112-112 = 192-112
16w = 80
16w÷16 = 80÷16
w = 5 How I Solved It The width of the rectangle is 5 yards. Solution
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