A newspaper is launching a new advertising campaign in order to increase the number of daily subscribers. The newspaper currently (t0) has 26,000 daily subscribers and management expects that number, S(t), to grow at the rate of subscribers per day, where t is the number of days since the campaign began. How long (to the nearest day) should the campaign last if the newspaper wants the number of daily subscribers to grow to 49,000?

57 days

Step-by-step explanation:

Given that:

t(0) = 26000

Growth rate (S'(t)) = 80t^1/2

Targeted number of subscribers = 49000

Take the integral of the growth rate :

∫80t^1/2dt = [(80t^(1/2 + 1)) / 1/2 + 1] + C

[(80t^(1/2 + 1)) / 1/2 + 1] + C = (80t^3/2) / 3/2 + C

(80t^3/2) * 3/2 + C = (160/3)t^3/2 + C

(160/3)t^3/2 + C

Let C = t(0)

(160/3)t^3/2 + 26000 = 49000

(160/3)t^3/2 = 49000 - 26000

(160/3)t^3/2 = 23000

Multiply both sides by 3/160

t^3/2 = 23000* 3/160

t^3/2 = 69000/160

t^3/2 = 431.25

t^(3/2 * 2/3) = 431.25^2/3

t = 57.080277

t = 57 days (approx.)

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

2x^2+5x+17

your welcome. : )