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
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.
Sure, I'd be happy to help! Let's go through each part.
a. To find the composition N(T(t)), we need to substitute T(t) into N(T).
N(T(t)) = 10(T(t))^2 - 20(T(t)) + 600
N(T(t)) = 10(3t + 2)^2 - 20(3t + 2) + 600
N(T(t)) = 10(9t^2 + 12t + 4) - 60t - 40 + 600
N(T(t)) = 90t^2 + 120t + 40 - 60t - 40 + 600
N(T(t)) = 90t^2 + 60t + 600
So the composition N(T(t)) is 90t^2 + 60t + 600.
b. To find the bacteria count after 0.5 hour, we need to substitute t = 0.5 into N(T(t)).
N(T(t)) = 90(0.5)^2 + 60(0.5) + 600
N(T(t)) = 90(0.25) + 30 + 600
N(T(t)) = 22.5 + 30 + 600
N(T(t)) = 652.5
So, the bacteria count after 0.5 hour is approximately 653.
c. To find the time when the bacteria count reaches 1500, we need to set N(T(t)) equal to 1500 and solve for t.
1500 = 90t^2 + 60t + 600
90t^2 + 60t + 600 - 1500 = 0
90t^2 + 60t - 900 = 0
3t^2 + 2t - 30 = 0
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-2 ± √(2^2 - 4(3)(-30))) / (2(3))
t = (-2 ± √(4 + 360)) / 6
t = (-2 ± √364) / 6
t ≈ (-2 ± 19.07) / 6
t ≈ -3.51 or t ≈ 2.84
Since time cannot be negative in this context, the time when the bacteria count reaches 1500 is approximately 2.84 hours.
And there you have it! I hope that helps you understand how the answers were reached.
Sure, I can help you with that! Let's go step-by-step.
a. To find the composition N(T(t)), we need to substitute the expression for T(t) into N(T).
N(T(t)) = 10(T(t))^2 - 20(T(t)) + 600
Now, substitute T(t) = 3t + 2:
N(T(t)) = 10(3t + 2)^2 - 20(3t + 2) + 600
Simplifying further:
N(T(t)) = 10(9t^2 + 12t + 4) - 60t - 40 + 600
N(T(t)) = 90t^2 + 120t + 40 - 60t - 40 + 600
N(T(t)) = 90t^2 + 60t + 600
Finally, factor out a common multiple of 30 to simplify the expression:
N(T(t)) = 30(3t^2 + 2t + 20)
So, the composition N(T(t)) is equal to 30(3t^2 + 2t + 20).
b. To find the bacteria count after 0.5 hours, we need to substitute t = 0.5 into the expression for N(T(t)).
N(T(0.5)) = 30(3(0.5)^2 + 2(0.5) + 20)
N(T(0.5)) = 30(3(0.25) + 1 + 20)
N(T(0.5)) = 30(0.75 + 1 + 20)
N(T(0.5)) = 30(21.75)
N(T(0.5)) = 652.5
So, the bacteria count after 0.5 hour is approximately 653 (rounding to the nearest whole number).
c. To find the time when the bacteria count reaches 1500, we need to set N(T(t)) equal to 1500 and solve for t.
30(3t^2 + 2t + 20) = 1500
Divide both sides of the equation by 30:
3t^2 + 2t + 20 = 50
Rearrange the equation:
3t^2 + 2t - 30 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 3, b = 2, and c = -30.
t = (-2 ± √(2^2 - 4 * 3 * -30)) / (2 * 3)
t = (-2 ± √(4 + 360)) / 6
t = (-2 ± √364) / 6
t ≈ (-2 ± 19.06) / 6
So, the possible solutions for t are:
t ≈ (-2 + 19.06) / 6 ≈ 2.843
t ≈ (-2 - 19.06) / 6 ≈ -3.843
Since time cannot be negative in this context, we can disregard the second solution.
Therefore, the time when the bacteria count reaches 1500 is approximately 2.843 hours.
I hope this helps! Let me know if you have any further questions.
Sure! Let's start with the first question.
a. Find the composition N(T(t)).
To find the composition N(T(t)), we need to substitute T(t) into the original equation N(T).
N(T(t)) = 10(T(t))^2 - 20(T(t)) + 600
Now, let's substitute T(t) with its given equation:
N(T(t)) = 10(3t + 2)^2 - 20(3t + 2) + 600
To simplify this expression, we need to expand the square of (3t + 2):
N(T(t)) = 10(9t^2 + 12t + 4) - 60t - 40 + 600
N(T(t)) = 90t^2 + 120t + 40 - 60t - 40 + 600
N(T(t)) = 90t^2 + 60t + 600
Simplifying further, we can factor out 30 from the equation:
N(T(t)) = 30(3t^2 + 2t + 20)
So the composition N(T(t)) is equal to 30 times the equation 3t^2 + 2t + 20.
b. Find the bacteria count after 0.5 hour.
To find the bacteria count after 0.5 hour, we need to substitute t = 0.5 into the equation N(T(t)) = 30(3t^2 + 2t + 20) we derived in part a.
N(T(0.5)) = 30(3(0.5)^2 + 2(0.5) + 20)
N(T(0.5)) = 30(3(0.25) + 1 + 20)
N(T(0.5)) = 30(0.75 + 1 + 20)
N(T(0.5)) = 30(21.75) ≈ 652.5
So the bacteria count after 0.5 hour is approximately 653.
c. Find the time when the bacteria count reaches 1500.
To find the time when the bacteria count reaches 1500, we need to find the value of t that makes N(T(t)) equal to 1500.
Let's set N(T(t)) equal to 1500 and solve for t:
30(3t^2 + 2t + 20) = 1500
Divide by 30:
3t^2 + 2t + 20 = 50
Subtract 50 from both sides:
3t^2 + 2t - 30 = 0
Now, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 2, and c = -30. Let's apply these values to the equation:
t = (-2 ± √(2^2 - 4(3)(-30))) / (2(3))
t = (-2 ± √(4 + 360)) / 6
t = (-2 ± √364) / 6
Approximating the value of √364 to be around 19.07, we can further simplify:
t ≈ (-2 ± 19.07) / 6
t ≈ (-2 + 19.07) / 6 ≈ 2.846
Therefore, the time when the bacteria count reaches 1500 is approximately 2.846 hours.