To solve this question, we need to determine the difference between the number of components assembled by an experienced employee and a new employee per day.
Given the functions:
[tex]\[
N(t) = \frac{50 t}{t + 4} \quad \text{and} \quad E(t) = \frac{70 t}{t + 3}
\][/tex]
To find the difference, we calculate:
[tex]\[
D(t) = E(t) - N(t)
\][/tex]
This involves subtracting [tex]\( N(t) \)[/tex] from [tex]\( E(t) \)[/tex]:
[tex]\[
D(t) = \frac{70 t}{t + 3} - \frac{50 t}{t + 4}
\][/tex]
To perform this subtraction, we first need a common denominator. The common denominator of the fractions [tex]\( \frac{70 t}{t + 3} \)[/tex] and [tex]\( \frac{50 t}{t + 4} \)[/tex] is [tex]\( (t + 3)(t + 4) \)[/tex].
Rewriting each fraction with the common denominator, we get:
[tex]\[
\frac{70 t (t + 4)}{(t + 3)(t + 4)} - \frac{50 t (t + 3)}{(t + 3)(t + 4)}
\][/tex]
Now let's simplify the numerators:
[tex]\[
\frac{70 t (t + 4) - 50 t (t + 3)}{(t + 3)(t + 4)}
\][/tex]
[tex]\[
= \frac{70 t^2 + 280 t - 50 t^2 - 150 t}{(t + 3)(t + 4)}
\][/tex]
Combine like terms in the numerator:
[tex]\[
= \frac{(70 t^2 - 50 t^2) + (280 t - 150 t)}{(t + 3)(t + 4)}
\][/tex]
[tex]\[
= \frac{20 t^2 + 130 t}{(t + 3)(t + 4)}
\][/tex]
Factor out the common factor in the numerator:
[tex]\[
= \frac{10 t (2 t + 13)}{(t + 3)(t + 4)}
\][/tex]
Thus, the simplified form of the difference function is:
[tex]\[
D(t) = \frac{10 t (2 t + 13)}{(t + 3)(t + 4)}
\][/tex]
Therefore, the correct answer is:
C. [tex]\( D(t) = \frac{10 t (2 t + 13)}{(t + 3)(t + 4)} \)[/tex]
