How do you do this question?

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

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

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step-by-step explanation:

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