To address this problem, we need to find the composite function [tex]\( N(T(t)) \)[/tex] and determine at what time [tex]\( t \)[/tex] the bacteria count [tex]\( N(T(t)) \)[/tex] reaches 24130.
### Step 1: Substitute [tex]\( T(t) \)[/tex] into [tex]\( N(T) \)[/tex]
We start with the given functions:
[tex]\[ N(T) = 26T^2 - 94T + 45 \][/tex]
[tex]\[ T(t) = 9t + 1.5 \][/tex]
We need to find the composite function [tex]\( N(T(t)) \)[/tex] by substituting [tex]\( T(t) \)[/tex] into [tex]\( N(T) \)[/tex]:
[tex]\[ N(T(t)) = N(9t + 1.5) \][/tex]
### Step 2: Compute [tex]\( N(9t + 1.5) \)[/tex]
Substitute [tex]\( T = 9t + 1.5 \)[/tex] into [tex]\( N(T) \)[/tex]:
[tex]\[ N(9t + 1.5) = 26(9t + 1.5)^2 - 94(9t + 1.5) + 45 \][/tex]
We need to expand and simplify this expression:
[tex]\[ (9t + 1.5)^2 = (9t)^2 + 2(9t)(1.5) + (1.5)^2 \][/tex]
[tex]\[ = 81t^2 + 27t + 2.25 \][/tex]
Now substitute this back into [tex]\( N(T) \)[/tex]:
[tex]\[ N(9t + 1.5) = 26(81t^2 + 27t + 2.25) - 94(9t + 1.5) + 45 \][/tex]
[tex]\[ = 26 \cdot 81t^2 + 26 \cdot 27t + 26 \cdot 2.25 - 94 \cdot 9t - 94 \cdot 1.5 + 45 \][/tex]
[tex]\[ = 2106t^2 + 702t + 58.5 - 846t - 141 + 45 \][/tex]
[tex]\[ = 2106t^2 + 702t - 846t + 58.5 - 141 + 45 \][/tex]
[tex]\[ = 2106t^2 - 144t - 37.5 \][/tex]
Thus, the composite function is:
[tex]\[ N(T(t)) = 2106t^2 - 144t - 37.5 \][/tex]
### Step 3: Find the time [tex]\( t \)[/tex] when the bacteria count reaches 24130
We need to solve for [tex]\( t \)[/tex] when [tex]\( N(T(t)) = 24130 \)[/tex]:
[tex]\[ 2106t^2 - 144t - 37.5 = 24130 \][/tex]
Rearrange the equation:
[tex]\[ 2106t^2 - 144t - 37.5 - 24130 = 0 \][/tex]
[tex]\[ 2106t^2 - 144t - 24167.5 = 0 \][/tex]
### Step 4: Solve the quadratic equation
We apply the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 2106 \)[/tex], [tex]\( b = -144 \)[/tex], and [tex]\( c = -24167.5 \)[/tex]:
[tex]\[ t = \frac{-(-144) \pm \sqrt{(-144)^2 - 4(2106)(-24167.5)}}{2(2106)} \][/tex]
[tex]\[ t = \frac{144 \pm \sqrt{20736 + 203330940}}{4212} \][/tex]
[tex]\[ t = \frac{144 \pm \sqrt{203351676}}{4212} \][/tex]
[tex]\[ t = \frac{144 \pm 14257.522}{4212} \][/tex]
We get two potential solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{144 + 14257.522}{4212} \approx 3.421918 \][/tex]
[tex]\[ t = \frac{144 - 14257.522}{4212} \approx -3.401 \][/tex]
Since the time must be positive, the valid solution is:
[tex]\[ t \approx 3.421918 \][/tex]
Therefore, the time needed for the bacteria count to reach 24130 is approximately:
[tex]\[ \boxed{3.421918} \][/tex] hours.
