Question 192217
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I presume you performed the calculation thusly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(t) = 18t^2 + 174t + 6241]


Where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(t) = 35000]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 18t^2 + 174t + 6241 = 35000]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 18t^2 + 174t -28759 = 0]


Using the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{-174 \pm sqrt{174^2 - 4(18)(-28579)}}{2(18)} ]


Which, after a bit of prime factoring work and calculator button punching becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{-29 \pm 7\sqrt{397}}{6} \approx 35.4]


Now, reading the problem carefully, *[tex \LARGE C(0) = 6241] is the number of new cars sold in 1972.  That means that 6241 new cars were sold by the <i><b>end of</b></i> 1972.  So, 35 years later, you would be at the end of 1972 plus 35 = 2007, and at that time you would not have quite reached the 35,000 number. *[tex \LARGE C(35) = 18(35)^2 + 174(35) + 6241 = 34381].  So you trip the 35,000 number right around 3:22 AM June 5, 2008, roughly speaking.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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