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Mathematics 16 Online
OpenStudy (anonymous):

The perimeter of a rectangle is 124 m. The length is 5 m more than twice the width. Find the dimensions

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

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?

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>

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 :)

OpenStudy (stormfire1):

yea, I just can't read :P

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