SOLUTION: The number of Facebook members from the start of 2006 to mid-2008 can be approximated by S(t) = 15t2 − 76t + 101 million members (2 ≤ t ≤ 4.5) in year t (t = 0

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: The number of Facebook members from the start of 2006 to mid-2008 can be approximated by S(t) = 15t2 − 76t + 101 million members (2 ≤ t ≤ 4.5) in year t (t = 0      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 418542: The number of Facebook members from the start of 2006 to mid-2008 can be approximated by
S(t) = 15t2 − 76t + 101 million members (2 ≤ t ≤ 4.5)
in year t (t = 0 represents the start of 2004).
[Source: www.facebook.com/, http://insidehighered.com (Some data are interpolated.)]
(a) Compute S'(t).
S'(t) =
1
According to the model, how fast was Facebook membership changing at the start of 2008 (t = 4)? (Be careful to give correct units of measurement.)
S'(4) = 2 million members per year
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
S'(t) = 30t - 76 (million members per year), by the power rule.

When t = 4, S'(t) = 30(4) - 76 = 44 million members per year.