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=-1, x=3, x=4
Step-by-step explanation:
1. Substitute 0 for f(x)
0=(x+1)(x-3)(x-4)
2. Set each set of parenthesis equal to 0
0=x+1, 0=x-3, 0=x-4
3. Solve for x
x=-1, x=3, x=4
1. To find the degree of the polynomial find the variable with the highest exponent, in this case is y^4 which means the degree of the exponent is 4.
2. 1, 2 and 3 are polynomials because there are more than one terms in the expression.
3. It would be a quadratic trinomial.
4. It would be a cubic binomial.
Answer:\
y=10
Step-by-step explanation:
5 times 6 to 30
30=20+y
Subtract 20 from both sides.
30-20=y
Simplify 30-2o to 10.
10=y
y=10
The answer to this question is INTEGER
