If some one could answer this it would be great.
A population of bacteria is quadrupling in a glass of milk on the counter every half hour.
A)Write a function for this growth P(t), where "t" is in hours
B)Write a function for this growth P(t), where "t" is in minutes
c)If the current bacteria population is 250000, how many bacteria were there 58 minutes ago?
Thanks!
A good start but P(0)= 4^0= 1 and you were not told that. Put in a constant to represent the initial number of bacteria.
No, the problem said "the current bacteria is 250000". That is not true for any time, t. There are two ways to do this:(c) 4^(t/30)=250000
1) Take t= 0 as "current", find the constant so that P(0)= 250000, and then find P(-58).
2) Take t= 58 as "current", find the constant so that P(58)= 250000, and then find P(0).
4^(t/30 -58/30)=250000(4^(-58/30))
=17137.89
So for C I solve For P(0) for the constant, then when i have the constant I can solve for P(-58)? But when solving for P(0)= 250000 I get P(0)=250000=A(240^0/30) It turns to the same answer of course so what am I doing wrong?
This is what I set up P(-58)=250000(4^-58/30)=17137.89 Seems as though it can be right seeing as it is less then 250000 lol.