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
Answer:
x = 3; y = 2
Step-by-step explanation:
x + y = 5
y = 2x - 4
<em>Isolate the y in the first equation</em>
y = 5 - x
y = 2x - 4
<em>Substitute the values</em>
5 - x = 2x - 4
<em>Isolate the Variable</em>
9 = 3x
<em>Simplify</em>
3 = x
<em>Plug that into the first equation</em>
3 + y = 5
<em>Subtract 3 from both sides</em>
y = 2
Answer:
Step-by-step explanation:
What i would do is multiply both sides by 2z + 1
so you get z-6=20z+10
-19z=16
z=-16/19
After moving the 1 1/2 to the other side you will be able to solve it and get x=2
Jenna should say 35. Hope it helps
