## Precalculus (6th Edition) Blitzer

a. $10$ million. $9.97$ million. $9.94$ million. $9.91$ million. b. decreasing.
a. With $A(t)=10e^{-0.003t}$, for 2006, $x=2006-2006=0$, we have $A(0)=10e^{0}=10$ million. For 2007, $x=2007-2006=1$, we have $A(1)=10e^{-0.003}\approx9.97$ million. For 2008, $x=2008-2006=2$, we have $A(2)=10e^{-0.003(2)}\approx9.94$ million. For 2009, $x=2009-2006=3$, we have $A(3)=10e^{-0.003(3)}\approx9.91$ million. b. Based on the above results, the population is decreasing.