The population, P (in thousands), of a town can be modeled by P= 2|t-6|+4, where t=0 represents 1990. During which two years does the town have a population of 8000?
Can someone show me the steps to this problem?
For 2|t-6| + 4 = 8 (thousand),
you must have
2|t-6| = 4
which can be rewritten
|t-6| = 2
t can be either -4 or 8
Since t = 0 represents 1990, the years would be 1986 or 1998
To find the years during which the town has a population of 8000, we need to solve the equation P = 8000:
2|t - 6| + 4 = 8000
First, subtract 4 from both sides of the equation:
2|t - 6| = 7996
Next, divide both sides by 2:
|t - 6| = 3998
At this point, we have an absolute value equation. To solve it, we need to consider two cases:
Case 1: t - 6 is positive:
In this case, the equation becomes:
t - 6 = 3998
Solving for t:
t = 3998 + 6
t = 4004
So one solution is t = 4004.
Case 2: t - 6 is negative:
In this case, the equation becomes:
-(t - 6) = 3998
Simplifying the equation:
-t + 6 = 3998
Solving for t:
-t = 3998 - 6
-t = 3992
Multiplying both sides by -1 to get t by itself:
t = -3992
So another solution is t = -3992.
Now we need to convert these t values to years. Given that t = 0 represents 1990, we need to calculate the years for t = 4004 and t = -3992.
For t = 4004:
Year = 1990 + 4004
Year = 5994
For t = -3992:
Year = 1990 + (-3992)
Year = -2002
Therefore, the town has a population of 8000 in the years 5994 and -2002.