MATH SOLVE

3 months ago

Q:
# A scientist is growing bacteria in a lab for study. One particular type of bacteria grows at a rate of y=2t^2+3t+500. A different bacteria grows at a rate of y=3t^2+t+300. In both of these eqiations y is the number of bacteria after t minutes. When is there an equal number of both types of bacteria.

Accepted Solution

A:

ANSWERApproximately after 15 minutes.EXPLANATIONThe growth rate of the first bacteria is [tex]y = 2 {t}^{2} + 3t + 500[/tex]The growth rate of the first bacteria :[tex]y = 3 {t}^{2} + t + 300[/tex]To find the time that, there will be an equal number of bacteria, we equate the two equation;[tex]3 {t}^{2} + t + 300 = 2 {t}^{2} + 3t + 500[/tex][tex]3 {t}^{2} - 2 {t}^{2} + t - 3t + 300 - 500 =00[/tex][tex] {t}^{2} - 2t - 200= 0[/tex]We solve for t to get,[tex]t = 15.177[/tex]Or[tex] t = - 13.177[/tex]We discard the negative value.This implies that,[tex]t \approx15[/tex]