Step-by-step explanation:
ANGLE STR AND ANGLE VWT IS THE RIGHT ANSWER
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
Steps to solve:
8j - k + 14; j = 0.25 and k = 1
~Substitute
8(0.25) - (1) + 14
~Simplify
2 - 1 + 14
~Subtract
1 + 14
~Add
15
Best of Luck!
Hi there!
Since the scale is 1 in = 4 ft, we divide both the length and width by 4 to get the new length and width with the scale factor applied.
16 / 4 = 4
12 / 4 = 3
So, the
length = 4
width = 3
Hope this helps!
I hope this helps you
(-5×-8)+1
40+1
41