a.
N ( T ) = 10 T ^ 2 - 20 T + 600
T ( t ) = 3 t + 2
N ( T ( t ) ) = 10 * ( 3 t + 2 ) ^ 2 - 20 * ( 3 t + 2 ) + 600 =
10 * [ ( 3 t ) ^ 2 + 2 * 3 t * 2 + 2 ^ 2 ] - 20 * 3 t - 20 * 2 + 600 =
10 * ( 9 t ^ 2 + 12 t + 4 ) - 60 t - 40 + 600 =
90 t ^ 2 + 120 t + 40 - 60 t - 40 + 600 =
90 t ^ 2 + 60 t + 600 =
30 * 3 t ^ 2 + 30 * 2 t + 30 * 20 =
30 ( 3 t ^ 2 + 2 t + 20 )
b.
t = 0.5 h
N = 30 ( 3 t ^ 2 + 2 t + 20 )
N ( T ( 0.5 h ) ) = 30 ( 3 * 0.5 ^ 2 + 2 * 0.5 + 20 ) =
30 ( 3 * 0.25 + 1 + 20 ) =
30 ( 0.75 + 1 + 20 ) =
30 * 21.75 = 652.5
approx. 653
c.
30 ( 3 t ^ 2 + 2 t + 20 ) = 1500 Divide both sides by 30
3 t ^ 2 + 2 t + 20 = 50 Subtract 50 to both sides
3 t ^ 2 + 2 t + 20 - 50 = 50 - 50
3 t ^ 2 + 2 t - 30 = 0
The solutions are :
t = ( sqrt 91 - 1 ) / 3 = 2.846464 h
and
t = ( - 1 - sqrt 91 ) / 3 = - 3.51313 h
Time can't be negative so ;
t = 2.846464 h
P.S.
If you don't know how to solve equation
3 t ^ 2 + 2 t - 30 = 0
In google type :
quadratic equation online
When you see list of results click on :
w w w . w e b g r a p h i n g . c o m / q u a d r a t i c e q u a t i o n _ q u a d r a t i
When page be open in rectangle type :
3 t ^ 2 + 2 t - 30 = 0
then click option :
solve it!
You will see solution step - by - step.
Hi can you help me find the solution for this one? I have the answers at the end but I want to study how did it end with that. Thanks :)
The number N of bacteria in a refrigerated food is given by
N(T) = 10T^2 - 20T + 600, 1 ≤ T ≤ 20
where T is the temperature of food in degrees Celsius. When the food is removed from refrigeration, the temperature of the food is given by
T(t) = 3t + 2, 0 ≤ t ≤ 6
where t is time in hours.
a. Find the composition N(T(t)).
b. Find the bacteria count after 0.5 hour.
c. Find the time when the bacteria count reaches 1500.
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Answers:
a. N(T(t))= 30(3t^2+2t+20)
b. About 653 bacteria
c. 2.846 hour
1 answer
1 answer