Answer:
P = 6200 / (1 + 5.2e^(0.0013t))
increases the fastest
Step-by-step explanation:
dP/dt = 0.0013 P (1 − P/6200)
Separate the variables.
dP / [P (1 − P/6200)] = 0.0013 dt
Multiply the left side by 6200 / 6200.
6200 dP / [P (6200 − P)] = 0.0013 dt
Factor P from the denominator.
6200 dP / [P² (6200/P − 1)] = 0.0013 dt
(6200/P²) dP / (6200/P − 1) = 0.0013 dt
Integrate.
ln│6200/P − 1│= 0.0013t + C
Solve for P.
6200/P − 1 = Ce^(0.0013t)
6200/P = 1 + Ce^(0.0013t)
P = 6200 / (1 + Ce^(0.0013t))
At t = 0, P = 1000.
1000 = 6200 / (1 + C)
1 + C = 6.2
C = 5.2
P = 6200 / (1 + 5.2e^(0.0013t))
You need to change the exponent from negative to positive.
The inflection points are where the population increases the fastest.
Answer:
$8,000
Step-by-step explanation:
Let the store earned $x in December.
Therefore,
Money spent to buy new inventory 
Remaining money 
Money used to pay bills 
Money still left over = $3,000
Total money earned in December 
Thus, total money earned in December is $8,000.
Answer:
false i think? and true
Step-by-step explanation:
Answer:
D. (4,3)
Step-by-step explanation:
I calculated it logically
