F(x) = 3x + 7.....g(x) = 4x - 2
A. (f + g)(x)
3x + 7 + 4x - 2
7x + 5 <====
B. (f * g)(x)
(3x + 7)(4x - 2) =
3x(4x - 2) + 7(4x - 2) =
12x^2 - 6x + 28x - 14
12x^2 + 22x - 14 <===
C. f[g(x)]
3(4x - 2) + 7 =
12x - 6 + 7 =
12x + 1 <===
Answer:
604
Step-by-step explanation:
add them together using collum method
Answer:
4 ft downwards
Step-by-step explanation:
old height off the ground :
7^2+b^2=25^2
49+b^2=625
b^2=576
b=24
new height off the ground :
7+8=15
15^2+ b^2 =25^2
225+b^2 = 625
b^2 = 400
b=20
difference between heights :
24 - 20 = 4
Answer:
12000
Step-by-step explanation:
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
