SOLUTION: A woman bought 65 lb. of fertilizer to spread over her rectangular backyard. if she spread it at a rate of 20 lb. to 1000 ft. and the length of the yard exceeds the width by 10 ft.
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Question 936648: A woman bought 65 lb. of fertilizer to spread over her rectangular backyard. if she spread it at a rate of 20 lb. to 1000 ft. and the length of the yard exceeds the width by 10 ft., what were the dimensions of the yard?
This is a word problem in a certain chapter in our book which deals all about Quadratic Equations. I hope you could help me out. Thanks.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A woman bought 65 lb. of fertilizer to spread over her rectangular backyard. if she spread it at a rate of 20 lb. to 1000 ft. and the length of the yard exceeds the width by 10 ft., what were the dimensions of the yard?
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width = w ft
length = w+10 ft
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Area = w(w+10) = w^2 + 10w sq. ft.
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Assuming that the 65 lb of fertilizer covers her yard,
area of yard = 65/(20 lb/1000 ft) = (65/20)*1000 = 3250 sq yds
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Equation:
w^2 + 10w - 3250 = 0
w = 175.35 ft(width)
w+10 = 185.35 ft (length)
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Cheers,
Stan H.
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