The population of a town P(t)is modelled by the function P(t)=6t^2+110t+3000, where t is time in years. Note t=0 represents the years 2000. When will the population reach 6000?
11 years ago
11 years ago
yes thank u but i don't understand
simplified form please!!
11 years ago
just plug it in. 6t^2+110t+3000=6000, thus 6t^2+110t-3000=0 and we factor but first we divide by 2. Thus we get the equation 2(3t+100)(t-15)=0. FYI if you did not learn what just happened please study your algebra book. Anyway, x= -100/3 or 15, but it cannot be negative so t=15. Thus, 2000+15 = 2015
1 year ago
To find when the population will reach 6000, we need to solve the equation P(t) = 6000.
The given population function is P(t) = 6t^2 + 110t + 3000, where t represents the time in years.
So, we need to solve the equation:
6t^2 + 110t + 3000 = 6000
To solve this quadratic equation, we first rearrange it to be in the form of "ax^2 + bx + c = 0".
6t^2 + 110t + 3000 - 6000 = 0
6t^2 + 110t - 3000 = 0
Now we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the quadratic equation with the general form ax^2 + bx + c = 0, we have:
a = 6, b = 110, and c = -3000
Substituting these values into the quadratic formula:
t = (-(110) ± √((110)^2 - 4(6)(-3000))) / (2(6))
t = (-110 ± √(12100 + 72000)) / 12
t = (-110 ± √84100) / 12
t = (-110 ± 290) / 12
We have two possible solutions:
1. t = (-110 + 290) / 12
2. t = (-110 - 290) / 12
Simplifying each solution:
1. t = 180 / 12 = 15
2. t = -400 / 12 = -33.33
Since time cannot be negative in this case, we discard the second solution.
Therefore, the population will reach 6000 after approximately 15 years.
To convert this to the actual year, we add 15 years to the base year (2000):
Year = 2000 + 15
Year = 2015
So, the population will reach 6000 in the year 2015.