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!!!

the correct answer is 5 6 and 7

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi