The answer is (x-y)^2
hope it helps!!!!!!
the formula is
(a-b)^2
and when you expand it, it becomes
a^2-2ab+b^2
Step-by-step explanation:
(a) dP/dt = kP (1 − P/L)
L is the carrying capacity (20 billion = 20,000 million).
Since P₀ is small compared to L, we can approximate the initial rate as:
(dP/dt)₀ ≈ kP₀
Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.
20 = k (6,100)
k = 1/305
dP/dt = 1/305 P (1 − (P/20,000))
(b) P(t) = 20,000 / (1 + Ce^(-t/305))
6,100 = 20,000 / (1 + C)
C = 2.279
P(t) = 20,000 / (1 + 2.279e^(-t/305))
P(10) = 20,000 / (1 + 2.279e^(-10/305))
P(10) = 6240 million
P(10) = 6.24 billion
This is less than the actual population of 6.9 billion.
(c) P(100) = 20,000 / (1 + 2.279e^(-100/305))
P(100) = 7570 million = 7.57 billion
P(600) = 20,000 / (1 + 2.279e^(-600/305))
P(600) = 15170 million = 15.17 billion
Answer:
group 69
Step-by-step explanation:
John Krasinski is Mr. Fantastic in Multiverse of Madness
Answer:
i think -0.02
Step-by-step explanation:
Answer:
1/ab en (c/be^-x+c)
Step-by-step explanation:
Sure is a harsh question! Here's my Explanation
b+ce^x = t
ce^x an = dt
e^xan = dt/c
an = dt/ce^x = dt/c(t-b/c) = at/(t-b)
en = t-b/c
A/b+ce^x dx = a/t dt/t-b
a ∫1/t (t-b) dt = 1/a∫ (1/(t-b) - 1/t) dt
= 1/ab [∫1/(t-b) dt + ∫-1/t dt]
= 1/ab [en (t-b) - en(t)]
= 1/ab en ((t-b)/t)
t = b + ce^x
= 1/ab en (b+ce^x -b/b+ce^x)
=1/ab en (ce^x/b+ce^x)
= 1/ab en (c/be^-x+c)