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
You use the Pythagorean theorem which is a^2+b^2=c^2
So....
.8^2+.6^2=c^2
.64+.36=c^2
1=c^2
1=c becaaue the square root of one is one
(<span>4x2</span><span> – </span><span>4x</span><span> – </span>7)(x<span> + </span>3)<span>
</span> = (<span>4x2</span><span> – </span><span>4x</span><span> – </span>7)(x) + (<span>4x2</span><span> – </span><span>4x</span><span> – </span>7)(3)<span>
</span><span> = </span><span>4x2</span>(x<span>) – </span><span>4x</span>(x<span>) – </span>7(x<span>) + </span><span>4x2</span>(3<span>) – </span><span>4x</span>(3<span>) – </span>7(3)<span>
</span><span> = 4x3 – 4x2 – 7x + 12x2 – 12x – 21</span><span>
</span><span> = 4x3 – 4x2 + 12x2 – 7x – 12x – 21</span><span>
</span><span> = 4x3 + 8x2 – 19x – 21</span>
