SOLUTION: I have an quadratic equation that has to do with graphing and its due tomorrow night (4/16/2010). P=7t^2+8000. 8000 represents the population (p0), t is the time meaning years. t=o
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-> SOLUTION: I have an quadratic equation that has to do with graphing and its due tomorrow night (4/16/2010). P=7t^2+8000. 8000 represents the population (p0), t is the time meaning years. t=o
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Question 292988: I have an quadratic equation that has to do with graphing and its due tomorrow night (4/16/2010). P=7t^2+8000. 8000 represents the population (p0), t is the time meaning years. t=o (2010), t=1 (2011), t=2 (2012), t=3 (2013), and t=6 (2016). This part is already figured out but the next part is not. It says:use your equation from the first part to approximate how many years it will take for the population to reach 12,000. Round to the nearset whole year when necessary. 1st answer then 2nd show the work. Can someone help me, please and thank you.
You can put this solution on YOUR website! ): I have an quadratic equation that has to do with graphing and its due tomorrow night (4/16/2010). P=7t^2+8000. 8000 represents the population (p0), t is the time meaning years. t=o (2010), t=1 (2011), t=2 (2012), t=3 (2013), and t=6 (2016). This part is already figured out but the next part is not. It says:use your equation from the first part to approximate how many years it will take for the population to reach 12,000. Round to the nearest whole year when necessary. 1st answer then 2nd show the work. Can someone help me, please and thank you.
Substitute 12000 for P and solve for t:
Subtract 8000 from both sides:
Divide both sides by 7
Take square roots of both sides:
So in 23 years it will be a little less than 12000 and in 24
years it will a little more than 12000.
To check, substitute t=23 and t=24 into
Substituting t=23
Substituting t=24
t=24 gives closest to 12000. so the answer is 24 years, or
the year 2034.
Edwin
