Sure, let's solve this step-by-step together.
### Initial Temperature
To find the initial temperature of the soda when it was first placed in the cooler, we need to evaluate the function [tex]\( T(x) \)[/tex] at [tex]\( x = 0 \)[/tex].
The temperature function is given by:
[tex]\[ T(x) = -4 + 22 e^{-0.045 x} \][/tex]
Let’s substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ T(0) = -4 + 22 e^{-0.045 \cdot 0} \][/tex]
[tex]\[ T(0) = -4 + 22 e^0 \][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[ T(0) = -4 + 22 \cdot 1 \][/tex]
[tex]\[ T(0) = -4 + 22 \][/tex]
[tex]\[ T(0) = 18 \][/tex]
The initial temperature of the soda is:
[tex]\[ \boxed{18} \, ^\circ{C} \][/tex]
### Temperature After 18 Minutes
Next, we will find the temperature of the soda after 18 minutes, i.e., we need to evaluate [tex]\( T(x) \)[/tex] at [tex]\( x = 18 \)[/tex].
The temperature function is given by:
[tex]\[ T(x) = -4 + 22 e^{-0.045 x} \][/tex]
Let’s substitute [tex]\( x = 18 \)[/tex] into the function:
[tex]\[ T(18) = -4 + 22 e^{-0.045 \cdot 18} \][/tex]
First, compute the exponent:
[tex]\[ -0.045 \cdot 18 = -0.81 \][/tex]
Now, evaluate the expression:
[tex]\[ e^{-0.81} \][/tex]
Multiplying 22 by the result of [tex]\( e^{-0.81} \)[/tex]:
[tex]\[ T(18) = -4 + 22 \cdot e^{-0.81} \][/tex]
Combining these:
[tex]\[ T(18) = -4 + 22 \cdot (some value less than 1) \][/tex]
After performing the calculation more precisely,
[tex]\[ T(18) = -4 + 6.2375 \][/tex]
Now rounding this to the nearest degree, we get:
[tex]\[ T(18) \approx 6 \][/tex]
The temperature after 18 minutes is:
[tex]\[ \boxed{6} \, ^\circ{C} \][/tex]
So, to summarize:
- The initial temperature of the soda is [tex]\( \boxed{18} \, ^\circ{C} \)[/tex].
- The temperature after 18 minutes is [tex]\( \boxed{6} \, ^\circ{C} \)[/tex].
