Answer:
V' = -0.11552 *V\\= -0.11552(1.8)\\ \\=-0.20794 million per year
Step-by-step explanation:
Given that oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 3 million barrels of oil in the well; six years later 1,500,000 barrels remain.
i.e. if V stands for volume of oil, then
To find A and k
V(0) = A = 3 million
Hence V = 
V(6) = 1.5
i.e. 
a) Using the above value of k , we have
million per year.
Six million, one hundred seventy-three thousand, two hundred fifty three
Answer:490pi
Step-by-step explanation:just trust me
Answer:
In 1981, the Australian humpback whale population was 350
Po = Initial population = 350
rate of increase = 14% annually
P(t) = Po*(1.14)^t
P(t) = 350*(1.14)^t
Where
t = number of years that have passed since 1981
Year 2000
2000 - 1981 = 19 years
P(19) = 350*(1.14)^19
P(19) = 350*12.055
P(19) = 4219.49
P(19) ≈ 4219
Year 2018
2018 - 1981 = 37 years
P(37) = 350*(1.14)^37
P(37) = 350*127.4909
P(37) = 44621.84
P(37) ≈ 44622
There would be about 44622 humpback whales in the year 2018
