Question 102455
Given:
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{{{P(t) = 120 + 4t + 0.05t^2}}}
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This function models the population (in thousands) of a town starting on January 1, 1995. 
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The clock starts running on January 1, 1995. On that date t is equal to zero. So if you go
to the given function and substitute zero for t, you get the population of the town on that 
date. Let's do it ...
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{{{P(0) = 120 + 4*(0) + 0.05*(0)^2 = 120 + 0 + 0 = 120}}}
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So on January 1, 1995 the population of the town is 120 thousand or 120,000.
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Five years later [you get that from P(5)] is January 1, 2000. We can find the population of 
the town on that date by substituting 5 into the given equation in place of t. When we do, the
equation becomes:
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{{{P(5) = 120 + 4*(5) + 0.05*(5^2) = 120 + 20 + 0.05*25 = 120 + 20 + 1.25}}}
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Adding up the three terms tells us that on January 1, 2000 the population of the town is:
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{{{120 + 20 + 1.25 = 141.25 }}} thousand or 141,250.
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This means that in the first five years, the town's population grows from 120,000 to
141,250 ... a gain of 21,250.  
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Hope this helps you to see your way through this problem.  
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