SOLUTION: I NEED HELP SOLVING THIS WORD PROBLEM. I DID TRY SOMETHING BUT I DON'T THINK IT IS RIGHT. THE QUESTION IS "the length of a rectangular is 5 more than twice its width. it's perimete
Algebra ->
Customizable Word Problem Solvers
-> Geometry
-> SOLUTION: I NEED HELP SOLVING THIS WORD PROBLEM. I DID TRY SOMETHING BUT I DON'T THINK IT IS RIGHT. THE QUESTION IS "the length of a rectangular is 5 more than twice its width. it's perimete
Log On
Question 183669: I NEED HELP SOLVING THIS WORD PROBLEM. I DID TRY SOMETHING BUT I DON'T THINK IT IS RIGHT. THE QUESTION IS "the length of a rectangular is 5 more than twice its width. it's perimeter is 88 feet. find it's dimensions?" THIS IS WHAT I TRIED:
L=w+5
2L+2w=88
2(W+5)+2w=68
2W+10+2W=68
4W=68-10
4W=58
W=58/4
W=14.5
L=15.5+5=20.5 INCHES Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! "the length of a rectangular is 5 more than twice its width. it's perimeter is 88 feet. find it's dimensions
---------------------
Perimeter = 2(L + W)
88 = 2((2W+5) + W)
44 = (3W + 5)
3W = 39
W = 13 feet (Width)
-----------------
L = 2W + 5
L = 2*13 + 5
L = 31 feet (Length)
==============================
Cheers,
Stan H.
(