Complete question:
The growth of a city is described by the population function p(t) = P0e^kt where P0 is the initial population of the city, t is the time in years, and k is a constant. If the population of the city atis 19,000 and the population of the city atis 23,000, which is the nearest approximation to the population of the city at
Answer:
27,800
Step-by-step explanation:
We need to obtain the initial population(P0) and constant value (k)
Population function : p(t) = P0e^kt
At t = 0, population = 19,000
19,000 = P0e^(k*0)
19,000 = P0 * e^0
19000 = P0 * 1
19000 = P0
Hence, initial population = 19,000
At t = 3; population = 23,000
23,000 = 19000e^(k*3)
23000 = 19000 * e^3k
e^3k = 23000/ 19000
e^3k = 1.2105263
Take the ln
3k = ln(1.2105263)
k = 0.1910552 / 3
k = 0.0636850
At t = 6
p(t) = P0e^kt
p(6) = 19000 * e^(0.0636850 * 6)
P(6) = 19000 * e^0.3821104
P(6) = 19000 * 1.4653739
P(6) = 27842.104
27,800 ( nearest whole number)
Answer:
True
Step-by-step explanation:
Both sides (right side and left side) of the equation are indeed equal.
I hope this helps!
Using the Quadratic formula
your answer would be A and C
The answer is -13. You figure it out by multiplying 2.5×-5.2=13
The best way to do this is to set up a proportion
2 5
--- = ----
5 x
then cross multiply and divide
5 times 5=25/2=12.5
he would move 12.5 feet in 5 hours
I hope I've helped!
