Exponential growth of the form:
F=Ar^t
F=100(1.22)^t
F(5)=100(1.22)^5
F(5)=270.27
F(5)=270 to the nearest frog...
....
1.22=r^12
r=1.22^(1/12)
F=100(1.22^(1/12)^t)
F=100(1.22^(t/12)) now it will produce monthly populations...
Say we did the same as before but instead of 5 years you have 60 months...
F(60)=100(1.22^(60/12))
F(60)=270 to the nearest frog... :P
6=.75x
x= 8 hours for the whole race
The only 'small number' that I know that would affect the answer is an exponent that looks like ³ so the problem would be x³+x
remember that x³=x times x times x
if x=3 then
3³+3=27+3=30
Step-by-step explanation:
=(y-y1)=Y2-Y1/X2-X1(X-X1)
