Mathematics
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OpenStudy (anonymous):
The perimeter of a rectangle is 124 m. The length is 5 m more than twice the width. Find the dimensions
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OpenStudy (anonymous):
124 = 2m + 2.5m
OpenStudy (anonymous):
124= 12m m=10,3
OpenStudy (stormfire1):
P = 2L + 2W
124m = 2L + 2W
62m = L + W
L = W + 5
62m = (W + 5) + W
57m = 2W
W = 28.5
so L must be 33.5m
OpenStudy (anonymous):
dimensions: m 10,3 M= 51,6
OpenStudy (anonymous):
M = 5.m as he say
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OpenStudy (stormfire1):
Wow...some of those answers make no sense :)
OpenStudy (anonymous):
I need to know the length and the width..??
OpenStudy (stormfire1):
I wrote it.
OpenStudy (anonymous):
lenght = 51,6 width = 10,3
OpenStudy (anonymous):
62 length and 28.5 width?
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OpenStudy (anonymous):
stormfire1 is right
OpenStudy (stormfire1):
Think about the question...none of those answers can be right. As I wrote, the width is 28.5m and the length is 33.5m
OpenStudy (stormfire1):
wait a tick...
OpenStudy (stormfire1):
disregard...I didn't see the "5m more than twice the width"...my bad
OpenStudy (stormfire1):
<sheepish>
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OpenStudy (anonymous):
lol
OpenStudy (stormfire1):
P = 2L + 2W
124m = 2L + 2W
62m = L + W
L = 2W + 5
62m = (2W+5) + W
57m = 3W
W = 19
L = 43
OpenStudy (stormfire1):
Maybe I'll read it a bit more carefully next time :)
OpenStudy (stormfire1):
for some reason my mind skipped right past the word "twice"
OpenStudy (anonymous):
its okay.. at least you know how to do algebra :)
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OpenStudy (stormfire1):
yea, I just can't read :P