Find the rate of change, in bacteria per day, of N with respect to t when the following values are true. (Round your answers to the nearest tenth.)

Question

N=600[1-3/(t^2+2^2]

when t=0, what does the bacteria per day equal?

when t=1 what does the bacteria per day equal?

when t =2 what does the bacteria per day equal?

when t=3 what does the bacteria per day equal?

when t=4 what does the bacteria per day equal?

Answer 1

Hi Jada H

First of all thanks for writing back.

N=600[1 minus the fraction of 3/(t^2+2^2)]

N=600*[1 minus the fraction of 3/(t^2+2^2)]

2² is 4 so you have

N=600*[1 minus the fraction of 3/(t^2+4)]

Then all you need do is to plug in the values given

N= 600[1 - 3/(0 + 2²)] = 600*[1 - 3/4] = 600*[4/4 - 3/4]=600*1/4=150

N= 600[1 - 3/(1²+ 4)] = 600[1 - 3/(1+ 4)] = 600[1 - 3/5]

Can you do the rest? If you have questions just write back,