Question 209551: I need to translate this problem into an equation.
Since the beginning of the month, a local reservoir has been losing water at a constant rate.
On the 13th of the month the reservoir held 454 million gallons of water, and on the 21st it held only 191 million gallons.
How much water was in the reservoir on the 10th of the month?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! translate this problem into an equation.
Since the beginning of the month, a local reservoir has been losing water at a constant rate.
On the 13th of the month the reservoir held 454 million gallons of water, and on the 21st it held only 191 million gallons.
How much water was in the reservoir on the 10th of the month?
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You have two points relating day and amount of water: (13,454)and(21,191)
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Equation form is amt = (slope)(day)+ b
slope = (191-454)/(21-13) = -263/8
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to get the intercept:
191 = (-263/8)21 + b
b = 499.375
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Equation:
Amt(day) = (-263/8)(day) + 499.375
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How much water was in the reservoir on the 10th of the month?
A(10) = (-263/8)(10) + 499.375
A(10) = 170.625 gallons
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Cheers,
Stan H.
Reply to stanbon@comcast.net
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