Question

differentiation anyone?

Answer 1

ask away, i can try

Answer 2

i have seen it before...

Answer 3

A population grows according to the equation P(t)=6000-5500e^(-.159) for t greater or equal to 0, t measured in years. This population will approach a limiting value as time goes on. During which year will the population reach half of this limiting value?

Answer 4

\[P(t0=6000-5500e^{-.159t}\] yes?

Answer 5

yes!

Answer 6

you need a t on the right-hand side somewhere

Answer 7

not a differentination problem

Answer 8

oh i see it now

Answer 9

sorry.. its what my packet says:(

Answer 10

at t goes to infinity,
\[e^{-.159t}\] goes to zero, so the limiting values is 6000

Answer 11

ok.. makes sense so far

Answer 12

half of that is evidently 3000 so your job is to set
\[3000=6000-5500e^{-.159t}\] and solve for t

Answer 13

!! thank you so much.. that should help a lot!!

Answer 14

which gives about t=3.81

Answer 15

not too bad. get
\[5500e^{-.159t}=3000\]
\[e^{-.195t}=\frac{6}{11}\]
\[-.159t=\ln(\frac{6}{11})\]
\[t=\frac{\ln(\frac{6}{11})}{-.0159}\] check my algebra!

Answer 16

should be good!! thanks so so much!.. my friend on the phone thanks you too :)

Answer 17

i got what chessguy got

Answer 18

invoice is in the mail

Answer 19

you have a small typo in your last LaTeX equation