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The area of a rectangle is 33ft^2. The length is 8 feet more than the width. What are the dimensions of the rectangle?

Sagot :

Answer:

[tex]length = (8 + x) \: feet\\ width = x \: feet \\ area = length \times width \\ 33 = (8 + x)x \\ 33 = 8x + {x}^{2} \\ {x}^{2} + 8x - 33 = 0 \\ (x - 3)(x + 11) = 0 \\ (x + 11) \: is \: neglected \: since \: it \: is \: negative \\ therefore : x = 3 \\ dimensions \\ length = (8 + 3) = 11 \: feet \\ width = 3 \: feet[/tex]

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