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)
The standard form: Ax + By = C
3x - y = -6 YES - A = 3, B = -1, C = -6
3x + y = 6 YES - A = 3, B = 1, C = 6
x + 6y = 9 YES - A = 1, B = 6, C = 9
x - 6y = -9 YES - A = 1, B = -6, C = -9
your answer should be 58 correct me if I'm wrong.
Answer:
6 Years
Step-by-step explanation:
Orlando invests $1000 at 6% annual interest compounded daily.
Orlando's investment = 
Bernadette invests $1000 at 7% simple interest.
Bernadette's investment = A = 1000(1+0.07×t)
By trail and error method we will use t = 5
Bernadette's investment will be after 5 years
1000(1 + 0.07 × 5)
= 1000(1 + 0.35)
= 1000 × 1.35
= $1350
Orlando's investment after 5 years
= 
= 
= 1000(1.349826)
= 1349.825527 ≈ $1349.83
After 5 years Orlando's investment will not be more than Bernadette's.
Therefore, when we use t = 6
After 6 years Orlando's investment will be = $1433.29
and Bernadette's investment will be = $1420
So, after 6 whole years Orlando's investment will be worth more than Bernadette's investment.
