Doug because his sequence follows a rule of multiplying by -1/2
Hope I got it right for your sake.
Let 1993 = time 0 = 0.
Let 1999 = time 6 = 6
Let 2012 = time 19 = 19
So, a = 171 (million). First solve for k.
176 = 171 e^k6
176/171 = e^(k*6)
ln (176/171) = 6k
k = 1/6 ln (176/171)
So, in 2012 we have: P(19) = 171 e^(19k), where k = 1/6 ln (176/171)
Hope this helped!
A: $456,000
B; 4,560 bumpercarts
X(x+2)-(1/3)x=150
x²+2x-(1/3)x-150=0
x²+(5/3)x-150=0
3x²+5x-450=0
this does not produce an integer, because b²-2ac=5²+4*3*450=5425, and √5425=73.6546 is not an integer.
Please double check your question.