To solve this problem, we need to determine the distance Margarita covers in the given time. The distance [tex]\( d \)[/tex] is given by the product of speed and time:
[tex]\[ d = \text{speed} \times \text{time} \][/tex]
Given:
- Speed: [tex]\(\frac{s-4}{s^2 - 9s + 20}\)[/tex] miles per hour
- Time: [tex]\(\frac{s-5}{4}\)[/tex] hours
First, we multiply these two expressions:
[tex]\[ d = \left(\frac{s-4}{s^2 - 9s + 20}\right) \times \left(\frac{s-5}{4}\right) \][/tex]
We combine the fractions:
[tex]\[ d = \frac{(s-4)(s-5)}{4(s^2 - 9s + 20)} \][/tex]
Next, we need to factorize the quadratic expression in the denominator:
[tex]\[ s^2 - 9s + 20 \][/tex]
This factors as:
[tex]\[ s^2 - 9s + 20 = (s-4)(s-5) \][/tex]
So, we substitute this back into the expression for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{(s-4)(s-5)}{4(s-4)(s-5)} \][/tex]
Since [tex]\((s-4)(s-5)\)[/tex] appears in both the numerator and the denominator, they cancel each other out:
[tex]\[ d = \frac{1}{4} \][/tex]
Therefore, the distance Margarita covers is [tex]\( \boxed{\frac{1}{4}} \)[/tex] miles.
