Mathematics: Ascientist 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

185-7=178So 178÷185=0.96216210.96×100=96%OR APPROXIMATELY 96 percent...

Ascientist is growing bacteria in a lab for study

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cheerkennedym2594

04.01.2022

ANSWER

t=15 minutes.

EXPLANATION

The equation that models the growth rate of the first type of bacteria is,

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

The growth rate of a second type of bacteria is modeled by:

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

Since y is the number of bacteria after t minutes in both equations, we equate both equations to find the time when there is an equal number of both types of bacteria.

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

We rewrite as standard quadratic equation,

$3 {t}^{2} - 2 {t}^{2} + t - 3t + 300 - 500 = 0$

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

We use the quadratic formula to obtain,

$t = \frac{ - - 2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 200) } }{2(1)}$

$t = \frac{ -2 \pm \sqrt{ 804} }{2}$

$t = \frac{ -2 \pm 2\sqrt{ 201} }{2}$

$t = 1 - \sqrt{201} \: or \: t = 1 + \sqrt{201}$

t=-13.177 or t=15.177

We discard the negative value. Hence there is an equal number of both types of bacteria after approximately 15 minutes.

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Ascientist 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

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