blike Posted March 17, 2003 Posted March 17, 2003 How do I solve ©? Also, can someone post the derivative, just to check my answer?
blike Posted March 18, 2003 Author Posted March 18, 2003 for part a, i don't understand if I should take the derivative, THEN plug in 10 and t=0, or if i should just use the equation given...
Roark Posted March 19, 2003 Posted March 19, 2003 I think I got it but I want to dbl-check. When do you need it?
blike Posted March 19, 2003 Author Posted March 19, 2003 i needed it earlier tonight, its all good though. just out of curiosity, what did you get for b,c,d?
JaKiri Posted March 21, 2003 Posted March 21, 2003 b. Whatever the value of 2000*199*e^-5 / (1 + 199*e^-5)^2 is. c. % rate of change = (rate of change/total number so far) * 100% = N'(t)/N(t) * 100% = [2000*Ce^-t/(1+*Ce^-t)^2]/[2000/(1+Ce^-t)] * 100% = Ce^-t/(1+Ce^-t) * 100% = 19900e^-t/(1+199e^-t)% d. 1+199e^-t = 19900e^-t 19701e^-t = 1 e^-t = 1/19701 t = ln 19701 = 10 days (0dp)
Radical Edward Posted March 21, 2003 Posted March 21, 2003 MrL is our resident maths b**** now. ask him simple arithmetical questions and he will obey sell out
JaKiri Posted March 21, 2003 Posted March 21, 2003 Originally posted by Radical Edward MrL is our resident maths b**** now. ask him simple arithmetical questions and he will obey sell out Yo bitch, burn some magnesium or whatever you chemists do.
blike Posted March 21, 2003 Author Posted March 21, 2003 burn some magnesium or whatever you chemists do. bahahahahaha
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