Since p^2=p*p,
(p*p)^5 = p*p * p*p * p*p * p*p * p*p = p^10
No -- other factors will influence whether your happy or not. Saying that is true is inductive reasoning, which is prone to error
Answer:
1) P(0) = 5000
2) P(t --> ∞) = 25000
Step-by-step explanation:
P(t) = (25t² + 125t + 200)/(t² + 5t + 40)
1) Population at the moment corresponds to population at t = 0
P(0) = (25(0²) + 125(0) + 200)/(0² + 5(0) + 40) = (0 + 0 + 200)/(0 + 0 + 40) = 5 thousand = 5000 (P was stated to be in thousands)
2) Population in the long term corresponds to the population as t --> ∞
P(t) = (25t² + 125t + 200)/(t² + 5t + 40)
Divide through the numerator and denominator by t²
P(t) = (25 + (125/t) + (200/t²))/(1 + 5/t + (40/t²))
P(t --> ∞) = (25 + 0 + 0)/(1 + 0 + 0) (Since, (1/∞) = 0)
P(t --> ∞) = 25 thousand = 25000
Answer:
Yes.
Step-by-step explanation:
Just like normal algebra, you factor our the common factor, in this case, 5.
Thus,
