THE POWER OF MEMES!!!!!!!!!!!!!!
Answer:
Population of mosquitoes in the area at any time t is:

Step-by-step explanation:
assume population at any time t = P(t)
population increases at a rate proportional to the current population:
⇒dP/dt ∝ P
----(1)
where k is constant rate at which population is doubled
solving (1)

---- (2)
initial population = 400,000
population is doubled every week
⇒P(1)=2P(0)
Using (2)


In presence of predators amount is decreased by 50,000 per day
Then amount decreased per week = 350,000
In this case (1) becomes
---(3)
solving (3) by calculating integrating factor

Multiplying I.F with all terms of (3)

Integrating w.r.to t




at t=0



So, population of mosquitoes in the area at any time t is

For this case we have an equation of the form:
y = A * (b) ^ t
Where,
A: initial amount
b: growth rate
t: time
We then have the following equation:
y = 57 (1.31) ^ t
The initial number of bacteria is 57 and the growth rate is 31% per hour.
Answer:
B. The initial number of bacteria is 57 and the growth rate is 31% per hour
Answer:
<u>-1133</u>
Step-by-step explanation:
<u>Formula for nth term</u>
<u>Finding d</u>
- d = -45 - (-28)
- d = -45 + 28
- d = -17
<u>Solving for a₆₆</u>
- a₆₆ = -28 + 65(-17)
- a₆₆ = -28 - 1105
- a₆₆ = <u>-1133</u>
Answer:
C and E has All real numbers are solutions.
Step-by-step explanation:
hope this helps