SOLUTION: The length of a rectangle is 3 cm is less than twice its width. A second rectangle is such that each dimension is the reciprocal of the corresponding dimension of the first rectang
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Question 137249This question is from textbook algebra: structure and method: book1
: The length of a rectangle is 3 cm is less than twice its width. A second rectangle is such that each dimension is the reciprocal of the corresponding dimension of the first rectangle. The perimeter of the of the second rectangle is 1/5 the perimeter of the first. Find the perimeter of the first rectangle. PLEASE HELP AND IT IS MUCH APPRICIATED!!!!!!!!!!!!!!!! This question is from textbook algebra: structure and method: book1
You can put this solution on YOUR website! The length of a rectangle is 3 cm is less than twice its width.
Let width = x : then length = 2x-3
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A second rectangle is such that each dimension is the reciprocal of the corresponding dimension of the first rectangle.
Width = 1/x ; length = 1/(2x-3)
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The perimeter of the of the second rectangle is 1/5 the perimeter of the first. Find the perimeter of the first rectangle.
EQUATION:
2*(width + length) = (1/5)(width + length)
2*((1/x) + 1/(2x-3)) = (1/5)(x + 2x-3)
10[(2x-3+x)/(2x^2-3x)] = 3x-3
30x-10 = 3x(x-1)(2x-3)
30x-10 = 3x(2x^2-5x+3)
30x-10 = 6x^3 -15x^2+9x
6x^3-15x^2-21x+10 = 0
x = 3.387
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Width = 3.388 length = 2x-3 = 3.776
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Cheers,
Stan H.
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