To solve the given problem, let's analyze the functions that describe the number of components assembled by new and experienced employees, and find the difference between these two functions.
The function [tex]\( N(t) \)[/tex] describes the number of components a new employee can assemble per day:
[tex]\[ N(t) = \frac{50t}{t + 4} \][/tex]
The function [tex]\( E(t) \)[/tex] describes the number of components an experienced employee can assemble per day:
[tex]\[ E(t) = \frac{70t}{t + 3} \][/tex]
We need to find the difference between the number of components assembled by experienced and new employees:
[tex]\[ D(t) = E(t) - N(t) \][/tex]
Substitute the functions [tex]\( N(t) \)[/tex] and [tex]\( E(t) \)[/tex] into the difference equation:
[tex]\[ D(t) = \frac{70t}{t + 3} - \frac{50t}{t + 4} \][/tex]
To perform the subtraction, find a common denominator, which is [tex]\((t + 3)(t + 4)\)[/tex]. First, rewrite each term with the common denominator:
[tex]\[ \frac{70t}{t + 3} = \frac{70t(t + 4)}{(t + 3)(t + 4)} \][/tex]
[tex]\[ \frac{50t}{t + 4} = \frac{50t(t + 3)}{(t + 3)(t + 4)} \][/tex]
Now, express the difference with the common denominator:
[tex]\[ D(t) = \frac{70t(t + 4)}{(t + 3)(t + 4)} - \frac{50t(t + 3)}{(t + 3)(t + 4)} \][/tex]
Combine the numerators over the common denominator:
[tex]\[ D(t) = \frac{70t(t + 4) - 50t(t + 3)}{(t + 3)(t + 4)} \][/tex]
Expand and simplify the numerator:
[tex]\[ 70t(t + 4) = 70t^2 + 280t \][/tex]
[tex]\[ 50t(t + 3) = 50t^2 + 150t \][/tex]
Subtract these expressions:
[tex]\[ 70t^2 + 280t - (50t^2 + 150t) = 70t^2 + 280t - 50t^2 - 150t \][/tex]
[tex]\[ = 20t^2 + 130t \][/tex]
Thus, the difference function [tex]\( D(t) \)[/tex] simplifies to:
[tex]\[ D(t) = \frac{20t^2 + 130t}{(t + 3)(t + 4)} \][/tex]
Factor out the common term [tex]\( 10t \)[/tex] in the numerator:
[tex]\[ D(t) = \frac{10t(2t + 13)}{(t + 3)(t + 4)} \][/tex]
Thus, the function that describes the difference in the number of components assembled per day by experienced and new employees is:
[tex]\[ D(t) = \frac{10t(2t + 13)}{(t + 3)(t + 4)} \][/tex]
Hence, the correct answer is:
[tex]\[ B. \, D(t) = \frac{10t(2t + 13)}{(t + 3)(t + 4)} \][/tex]
