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Damm [24]
4 years ago
7

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 differen

t 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.
Mathematics
1 answer:
Leya [2.2K]4 years ago
8 0

ANSWER

Approximately after 15 minutes.

EXPLANATION

The growth rate of the first bacteria is

y = 2 {t}^{2}   + 3t + 500

The growth rate of the first bacteria :

y = 3 {t}^{2}  + t + 300

To find the time that, there will be an equal number of bacteria, we equate the two equation;

3 {t}^{2}  + t + 300 = 2 {t}^{2}   + 3t + 500

3 {t}^{2}  - 2 {t}^{2}   + t  - 3t +  300 - 500 =00

{t}^{2}  - 2t  - 200= 0

We solve for t to get,

t = 15.177

Or

t =  - 13.177

We discard the negative value.

This implies that,

t \approx15

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