1. ## application 2

The number N of bacteria in a refrigerated food is given by:

N(T)=25T^2-50T+300, 2(less then or equal to)T(less then or equal to)20

Where T is the temperature of the foodin degrees celsius. When the food is removed from refrigeration, the temperature of the food is given by:

T(t)=2t+1, 0(less then or equal to)t(less then or equal to)9

Where t is the time in hours. Find the time when the bacteria count reaches 750.

Ok so im thinking you use equation 2 and make it equal to 750 and then workout t. is it that simple or u need to defferintiate equation 1 and make it equal to equation 2s answear????? so confused, i know its probably easy its just that i havent done application questions in ages.

2. ## Re: application 2

Originally Posted by iFuuZe
The number N of bacteria in a refrigerated food is given by:

N(T)=25T^2-50T+300, 2(less then or equal to)T(less then or equal to)20

Where T is the temperature of the foodin degrees celsius. When the food is removed from refrigeration, the temperature of the food is given by:

T(t)=2t+1, 0(less then or equal to)t(less then or equal to)9

Where t is the time in hours. Find the time when the bacteria count reaches 750.

Ok so im thinking you use equation 2 and make it equal to 750 and then workout t. is it that simple or u need to defferintiate equation 1 and make it equal to equation 2s answear????? so confused, i know its probably easy its just that i havent done application questions in ages.

$\displaystyle N(t)=25(2t+1)^2-50(2t+1)+300 \Rightarrow 750=25(2t+1)^2-50(2t+1)+300$
$\displaystyle t \approx 2.18$