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Q: If a rope 36 feet long is cut into two pieces in such a way that one piece is twice as long as the other piece how long must the long piece be If a rope 36 feet long is cut into two pieces in such a w?

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One of them is 9.5 feet long and the other is 7.5 feet long.Explanation:Let us take piece one as X and piece two as Y.1. X + Y = 172. X - Y = 22X = 19 => X = 9.5Take statement 2 from above. 9.5 - Y = 2 => Y = 7.

You will get 36 pieces.

Using algebra it works out that the lengths are 8.4 feet and 33.6 feet because 4*(8.4) = 33.6 and 8.4+33.6 = 42 feet

One piece is 80 feet - the other is 8 feet !

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36 = 3 X 12. Therefore, one piece must be 12 feet and the other 24 feet.

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I think 9 and 3... no, its 5 and 7.

11 feet! y=2x+3 x+y=15 2x+3+x=15 3x=12 x=4 15-4=11 the two pieces are 4 feet and 11 feet

Suppose the shorter piece is x feet long. Then the longer piece is x+1 feet long. The total length of the two pieces is x + (x+1) = 2x+1 feet. So 2x+1 = 9 so 2x = 8 so x = 4' and then the other piece is 5'

9 pieces: 6 yards = 18 feet, then divide by 2 feet per piece to get 9 pieces.

One of them is 9.5 feet long and the other is 7.5 feet long.Explanation:Let us take piece one as X and piece two as Y.1. X + Y = 172. X - Y = 22X = 19 => X = 9.5Take statement 2 from above. 9.5 - Y = 2 => Y = 7.

4 ÷ 2 = 2 so each piece measures two feet in length.

Depends what size each piece is !

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You will get 36 pieces.

12 foot in 3 peicies in which two are of same lenght say as x and x and the third one is twice of other i.e. 2x thenx+x+2x = 124x=12x=3so two are of 3 foot each and one is of 6 feet long3,3,6