The population can be modeled by P(t) = 82.5 − 67.5cos⎡ ⎣(π/6)t ⎤ ⎦, where t is time in months (t = 0 represents January 1) and P is population (in thousands). During a year, in what intervals is the population less than 20,000? Dur

The intervals in which the population is less than 20,000 include

(0 ≤ t < 0.74) and (11.26 < t ≤ 12)

Step-by-step explanation:

P(t) = 82.5 - 67.5 cos [(π/6)t]

where

P = population in thousands.

t = time in months.

During a year, in what intervals is the population less than 20,000?

That is, during (0 ≤ t ≤ 12), when is (P < 20)

82.5 - 67.5 cos [(π/6)t] < 20

- 67.5 cos [(π/6)t] < 20 - 82.5

-67.5 cos [(π/6)t] < -62.5

Dividing both sides by (-67.5) changes the inequality sign

cos [(π/6)t] > (62.5/67.5)

Cos [(π/6)t] > 0.9259

Note: cos 22.2° = 0.9259 = cos (0.1233π) or cos 337.8° = cos (1.8767π) = 0.9259

If cos (0.1233π) = 0.9259

Cos [(π/6)t] > cos (0.1233π)

Since (cos θ) is a decreasing function, as θ increases in the first quadrant

(π/6)t < 0.1233π

(t/6) < 0.1233

t < 6×0.1233

t < 0.74 months

If cos (1.8767π) = 0.9259

Cos [(π/6)t] > cos (1.8767π)

cos θ is an increasing function, as θ increases in the 4th quadrant,

[(π/6)t] > 1.8767π (as long as (π/6)t < 2π, that is t ≤ 12)

(t/6) > 1.8767

t > 6 × 1.8767

t > 11.26

Second interval is 11.26 < t ≤ 12.

Hope this Helps!!!

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

step-by-step explanation:

idk the answer i want to say 10 but then it would be equal so try 9 but then yeah so yeah we love edgen