Suppose the population of a town is 567 in 2001. the population decreases at a rate of 1.5% every year. what will be the population of the town in 2010?

Exponential decay can be expressed as:

f=ir^t, f=final value, i=initial value, r=common ratio (or rate) and t=times rate is applied (or time).

In this case i=567 and r=(100-1.5)/100=0.985, so the equation is:

p(t)=567(0.985^t), in this case t=year-2001=y-2001 so we can say:

p(y)=567(0.985)^(y-2001)  so in the year 2010

p(2010)=567(0.985)^(2010-2001)

p(2010)=567(0.985^9)

p(2010)≈494.89  (to nearest hundredth)

Since we are dealing with people, population needs to be an integer amount

p(2010)≈495  (to nearest whole person :P)

2a - 5b = 1
3a + 5b = 14

Let's solve this system using elimination. Add the equations together to cancel the b-terms, then solve for a.

    2a - 5b = 1
   3a + 5b = 14
+
    5a - 0 = 15
         5a = 15
           a = 3

Next, plug 3 for a into either of the equations and solve for b.

  2a - 5b = 1
2(3) - 5b = 1
    6 - 5b = 1
       - 5b = -5
           b = 1

Lastly, check all work by plugging both values for their respective variables.

     2a - 5b = 1
2(3) - 5(1) = 1
         6 - 5 = 1  --  This is true

     3a + 5b = 14
3(3) + 5(1) = 14
         9 + 5 = 14  --  This is true

The solution to the system of equations is (3, 1).