a 174 ft long fence is placed around the perimeter of a rectangular pool that has a 3 ft sidewalk around it. the actual pool w/o the sidewalk is twice as long as it is wide. what is the dimensions of the pool w/o the sidewalk?
what formulas do i need, p=2l+2w ? thanks for the help.
2006-09-13 09:33:11 · 6 answers · asked by shih rips 6 in Science & Mathematics ➔ Mathematics
unfortunately, all the answers are wrong.
2006-09-14 09:34:06 · update #1
since there are 3 feet betwenn pool and fence, the perimeter of pool would be
174-(3x4)=162
width=x, sidewalk=2x
(x+2x)*2=162
6x=162
x=27
sidewalk-- 27x2=54feet
2006-09-13 09:39:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The algebra is 174 = 2 (L+6) + 2 (W+6), and L = 2W. The solution is straightforward.
2006-09-13 09:39:46 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Don't forget the 3' sidewalk. The 3' serves to take 6' off the width and 6' off the length. You know that the length is twice the width so do the right substitutions and obtain the answer that someone else will give you.
2006-09-13 09:35:44 · answer #3 · answered by Anonymous · 1⤊ 0⤋
174-12=162 (perimiter of pool)
162/3=54 so pool is 54ft long and 27ft wide
I did not use any formulas
2006-09-13 09:46:11 · answer #4 · answered by Ha! Invisible! 3 · 0⤊ 0⤋
locate the area of the circle: A = ?D²/4 = one hundred? in.squarelocate the quantity of the paint: V = T x A = 0.25 in.cu the place T = Thickness of the paint replace and remedy for T: T = V/A = 0.25 in.cu/one hundred? in.sq = a million/4 hundred? in.
2016-11-07 06:26:19 · answer #5 · answered by belschner 4 · 0⤊ 0⤋
ummmm ummmmm ummmmmm idk
2006-09-13 09:35:05 · answer #6 · answered by shizzlechit 5 · 0⤊ 1⤋