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
Reflexive property of Equality is your answer!
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212= 2 X 2 X 53
212= 4X 53
Answer:
7m + 29
Step-by-step explanation:
You add 2m and 5m to get 7m
Then add -16 and 45 to get 29
Answer:
It would be much easier to answer of there was some diagram or something but if there is only just Team C and D then, it would be Team D
