The number of concurrent users of a social networking site has increased dramatically since . By , this social networking site could connect concurrently 70 million users online. The function P(t)(), where t is the number of years after , models this increase in millions of users. Estimate the number of users of this site that could be online concurrently in , in , and in . Round to the nearest million users.

Complete question:

The number of concurrent users of a social networking site has increased dramatically since 2005. By 2014, this social networking site could connect concurrently 70 million users online. The function P(t) = 2.459(1.475)^t, where t is the numbe of years after 2005, models this increase in millions of users. Estimate the number of users of this site that could be onlin concurrently in 2006, in 2010, and in 2013. Round to the nearest million users

Kindly check explanation

Step-by-step explanation:

Given the function :

The function P(t) = 2.459(1.475)^t, where t is the numbe of years after 2005

Estimate the number of users of this site that could be online concurrently in:

2006:

t = (2006 - 2005) = 1

P(t) = 2.459(1.475)^t

P(1) = 2.459(1.475)^1

P = 2.459(1.475)

= 3.627025 = 4 million users (nearest million)

2010 :

t = (2010 - 2005) = 5

P(t) = 2.459(1.475)^t

P(5) = 2.459(1.475)^5

P = 2.459(6.981682607421875)

= 17.1679575 = 17 million users (nearest million)

2013:

t = (2013 - 2005) = 8

P(t) = 2.459(1.475)^t

P(8) = 2.459(1.475)^8

P = 2.459(22.404546753)

= 55.09278 = 55 million users (nearest million)

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: