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Sagot :
Let's solve this problem step-by-step.
### Step 1: Find the composite function [tex]\( N(T(t)) \)[/tex]
Given the functions:
[tex]\[ N(T) = 26T^2 - 124T + 90 \][/tex]
[tex]\[ T(t) = 2t + 1.1 \][/tex]
To find the composite function [tex]\( N(T(t)) \)[/tex], we substitute [tex]\( T(t) = 2t + 1.1 \)[/tex] into [tex]\( N(T) \)[/tex]:
[tex]\[ N(T(t)) = N(2t + 1.1) \][/tex]
Substitute [tex]\( T = 2t + 1.1 \)[/tex] into the function [tex]\( N(T) \)[/tex]:
[tex]\[ N(T) = 26T^2 - 124T + 90 \][/tex]
So,
[tex]\[ N(2t + 1.1) = 26(2t + 1.1)^2 - 124(2t + 1.1) + 90 \][/tex]
Calculate [tex]\( (2t + 1.1)^2 \)[/tex]:
[tex]\[ (2t + 1.1)^2 = (2t)^2 + 2(2t)(1.1) + (1.1)^2 = 4t^2 + 4.4t + 1.21 \][/tex]
Now, substitute and simplify:
[tex]\[ N(2t + 1.1) = 26(4t^2 + 4.4t + 1.21) - 124(2t + 1.1) + 90 \][/tex]
[tex]\[ = 26 \cdot 4t^2 + 26 \cdot 4.4t + 26 \cdot 1.21 - 124 \cdot 2t - 124 \cdot 1.1 + 90 \][/tex]
[tex]\[ = 104t^2 + 114.4t + 31.46 - 248t - 136.4 + 90 \][/tex]
Combine like terms:
[tex]\[ = 104t^2 + 114.4t - 248t + 31.46 - 136.4 + 90 \][/tex]
[tex]\[ = 104t^2 - 133.6t - 14.94 \][/tex]
Thus, the composite function is:
[tex]\[ N(T(t)) = 104t^2 - 133.6t - 14.94 \][/tex]
### Step 2: Find the time when the bacteria count reaches 18481
We need to solve for [tex]\( t \)[/tex] when [tex]\( N(T(t)) = 18481 \)[/tex]:
[tex]\[ 104t^2 - 133.6t - 14.94 = 18481 \][/tex]
Set up the equation:
[tex]\[ 104t^2 - 133.6t - 18495.94 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex]. We can solve it using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 104 \)[/tex], [tex]\( b = -133.6 \)[/tex], and [tex]\( c = -18495.94 \)[/tex].
Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-133.6)^2 - 4(104)(-18495.94) \][/tex]
[tex]\[ = 17844.96 + 7696473.6 \][/tex]
[tex]\[ = 7714318.56 \][/tex]
Now find [tex]\( t \)[/tex]:
[tex]\[ t = \frac{133.6 \pm \sqrt{7714318.56}}{208} \][/tex]
[tex]\[ t \approx \frac{133.6 \pm 2777.697}{208} \][/tex]
Thus, the solutions are:
[tex]\[ t_1 = \frac{2911.297}{208} \approx 13.9936 \][/tex]
[tex]\[ t_2 = \frac{-2644.097}{208} \approx -12.7090 \][/tex]
### Conclusion
For the bacteria count to reach 18481, the valid time (assuming positive time) is:
[tex]\[ \text{Time Needed} = 13.9936 \text{ hours} \][/tex]
Thus, the final answers are:
[tex]\[ N(T(t)) = 104t^2 - 133.6t - 14.94 \][/tex]
[tex]\[ \text{Time Needed} = 13.9936 \][/tex]
### Step 1: Find the composite function [tex]\( N(T(t)) \)[/tex]
Given the functions:
[tex]\[ N(T) = 26T^2 - 124T + 90 \][/tex]
[tex]\[ T(t) = 2t + 1.1 \][/tex]
To find the composite function [tex]\( N(T(t)) \)[/tex], we substitute [tex]\( T(t) = 2t + 1.1 \)[/tex] into [tex]\( N(T) \)[/tex]:
[tex]\[ N(T(t)) = N(2t + 1.1) \][/tex]
Substitute [tex]\( T = 2t + 1.1 \)[/tex] into the function [tex]\( N(T) \)[/tex]:
[tex]\[ N(T) = 26T^2 - 124T + 90 \][/tex]
So,
[tex]\[ N(2t + 1.1) = 26(2t + 1.1)^2 - 124(2t + 1.1) + 90 \][/tex]
Calculate [tex]\( (2t + 1.1)^2 \)[/tex]:
[tex]\[ (2t + 1.1)^2 = (2t)^2 + 2(2t)(1.1) + (1.1)^2 = 4t^2 + 4.4t + 1.21 \][/tex]
Now, substitute and simplify:
[tex]\[ N(2t + 1.1) = 26(4t^2 + 4.4t + 1.21) - 124(2t + 1.1) + 90 \][/tex]
[tex]\[ = 26 \cdot 4t^2 + 26 \cdot 4.4t + 26 \cdot 1.21 - 124 \cdot 2t - 124 \cdot 1.1 + 90 \][/tex]
[tex]\[ = 104t^2 + 114.4t + 31.46 - 248t - 136.4 + 90 \][/tex]
Combine like terms:
[tex]\[ = 104t^2 + 114.4t - 248t + 31.46 - 136.4 + 90 \][/tex]
[tex]\[ = 104t^2 - 133.6t - 14.94 \][/tex]
Thus, the composite function is:
[tex]\[ N(T(t)) = 104t^2 - 133.6t - 14.94 \][/tex]
### Step 2: Find the time when the bacteria count reaches 18481
We need to solve for [tex]\( t \)[/tex] when [tex]\( N(T(t)) = 18481 \)[/tex]:
[tex]\[ 104t^2 - 133.6t - 14.94 = 18481 \][/tex]
Set up the equation:
[tex]\[ 104t^2 - 133.6t - 18495.94 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex]. We can solve it using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 104 \)[/tex], [tex]\( b = -133.6 \)[/tex], and [tex]\( c = -18495.94 \)[/tex].
Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-133.6)^2 - 4(104)(-18495.94) \][/tex]
[tex]\[ = 17844.96 + 7696473.6 \][/tex]
[tex]\[ = 7714318.56 \][/tex]
Now find [tex]\( t \)[/tex]:
[tex]\[ t = \frac{133.6 \pm \sqrt{7714318.56}}{208} \][/tex]
[tex]\[ t \approx \frac{133.6 \pm 2777.697}{208} \][/tex]
Thus, the solutions are:
[tex]\[ t_1 = \frac{2911.297}{208} \approx 13.9936 \][/tex]
[tex]\[ t_2 = \frac{-2644.097}{208} \approx -12.7090 \][/tex]
### Conclusion
For the bacteria count to reach 18481, the valid time (assuming positive time) is:
[tex]\[ \text{Time Needed} = 13.9936 \text{ hours} \][/tex]
Thus, the final answers are:
[tex]\[ N(T(t)) = 104t^2 - 133.6t - 14.94 \][/tex]
[tex]\[ \text{Time Needed} = 13.9936 \][/tex]
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