SOLUTION: The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.
Ok. I think I start out by
2l + 2w = 122
w+25=1
This is co
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-> SOLUTION: The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.
Ok. I think I start out by
2l + 2w = 122
w+25=1
This is co
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Question 457513: The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.
Ok. I think I start out by
2l + 2w = 122
w+25=1
This is confusing, can someone help? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.
Ok. I think I start out by
2l + 2w = 122
w+25=1
-----------------------
Let the width be "x"
Then length = "x+25"
----
Perimeter = 2(width + length)
122 = 2(x + x+25)
61 = 2x+25
2x = 36
x = 18 inches (width)
x+25 = 43 inches (length)
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Cheers,
Stan H.
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