How do I solve ©? Also, can someone post the derivative, just to check my answer?

just one bump

it should just be N'(t) written as a percent.

ok

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...

Don't use the derivative at all for part a.

I think I got it but I want to dbl-check. When do you need it?

i needed it earlier tonight, its all good though. just out of curiosity, what did you get for b,c,d?

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)

Thank you so much!!!

MrL is our resident maths b**** now. ask him simple arithmetical questions and he will obey

 

sell out

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.

burn some magnesium or whatever you chemists do.

 

bahahahahaha