The population, P (in 100's), of a town can be modeled y P = 2 /t -5/ = 6, where t = 0 represents 1990. During whuch two years does the town have a population of 12,000.?
To find the years when the town has a population of 12,000, we need to solve the equation P = 12,000.
The given equation is P = 2/t - 5/(t + 6), where t represents the year with t = 0 representing 1990.
1. Set P = 12,000 in the equation:
12,000 = 2/t - 5/(t + 6)
2. Get rid of any fractions:
Multiply every term by t(t + 6) to clear the denominators:
12,000t(t + 6) = 2(t + 6) - 5t
3. Simplify the equation:
12,000t^2 + 72,000t = 2t + 12 - 5t
4. Rearrange the terms to get a quadratic equation in standard form:
12,000t^2 + 72,000t - 2t - 5t - 12 = 0
12,000t^2 + 72,000t - 7t - 12 = 0
5. Combine the like terms:
12,000t^2 + 71,993t - 12 = 0
6. Solve the quadratic equation:
You can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation:
a = 12,000, b = 71,993, c = -12
Substituting these values into the quadratic formula:
t = (-71,993 ± √(71,993^2 - 4 * 12,000 * -12)) / 2 * 12,000
7. Calculate the values of t:
Using a calculator, compute the value of t. You will get two solutions for t, which represent the two possible years when the population is 12,000.
For example, the solutions may be t = 0.0128 and t = -5.9678.
8. Convert t to years:
Since t = 0 represents the year 1990, add the values of t to 1990 to find the corresponding years when the population is 12,000.
For the example above:
Year 1 = 1990 + 0.0128 = 1990.0128 (approx.)
Year 2 = 1990 - 5.9678 = 1984.0322 (approx.)
Therefore, the town has a population of 12,000 approximately in the years 1990 and 1984.