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Svetllana [295]
3 years ago
9

The population of a certain species of insect is given by a differentiable function P, where P(t) is the number of insects in th

e population, in millions, at time t, where t is measured in days. When the environmental conditions are right, the population increases with respect to time at a rate that is directly proportional to the population. Starting August 15, the conditions were favorable and the population began increasing. On August 20, five days later, there were an estimated 10 million insects and the population was increasing at a rate of 2 million insects per day. Which of the following is a differential equation that models this situation?
a. P=2(t−5)+10
b. dP/dt=2/5t
c. dP/dt=1/5P
d. dP/dt=5P
Mathematics
1 answer:
Alisiya [41]3 years ago
6 0

Answer:

c. dP/dt = (1/5)P

Step-by-step explanation:

Given that the rate of change of population with respect to time dP/dt is directly proportional to the population, P, we have

dP/dt ∝ P

dP/dt = kP

Given that dP/dt = 2 million insects per day and P = 10 million insects after 5 days, So,

2 = k × 10

k = 2/10

k = 1/5

So, dP/dt = kP

dP/dt = (1/5)P

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3 years ago
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Can someone please help me
Readme [11.4K]
Surface Area of the figure is 1208 square centimeters
a=20
b=13
c=12
d=5
e=8
Top face:
A1=a×e=20×8
ae=160
Bottom face:
A2(a+2d)×e=(20+2×5)×8=(20+10)×8=30×8
(a+2d)e=240
Front face:
Rectangle
a×c=20×12=240
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Total surface area
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A=400+2(404)
A=400+808=1208
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8 0
2 years ago
If the revenue for the week is $2000, and labor cost consist of two workers earning $8 per hour who works 40 hours each, whay is
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