To solve this problem, we need to find the difference between the number of components that an experienced employee and a new employee can assemble per day, represented by [tex]\( E(t) - N(t) \)[/tex].
Given:
[tex]\[ N(t) = \frac{50t}{t + 4} \][/tex]
[tex]\[ E(t) = \frac{70t}{t + 3} \][/tex]
First, we'll find [tex]\( D(t) = E(t) - N(t) \)[/tex]:
[tex]\[ D(t) = \frac{70t}{t + 3} - \frac{50t}{t + 4} \][/tex]
To subtract these fractions, we need a common denominator. The common denominator for [tex]\( \frac{70t}{t + 3} \)[/tex] and [tex]\( \frac{50t}{t + 4} \)[/tex] is [tex]\( (t + 3)(t + 4) \)[/tex].
Next, we'll rewrite both fractions with this common denominator:
[tex]\[ E(t) = \frac{70t (t + 4)}{(t + 3)(t + 4)} \][/tex]
[tex]\[ N(t) = \frac{50t (t + 3)}{(t + 3)(t + 4)} \][/tex]
Now, we can subtract the numerators:
[tex]\[ D(t) = \frac{70t (t + 4) - 50t (t + 3)}{(t + 3)(t + 4)} \][/tex]
Simplify the numerator:
[tex]\[ D(t) = \frac{70t^2 + 280t - 50t^2 - 150t}{(t + 3)(t + 4)} \][/tex]
[tex]\[ D(t) = \frac{70t^2 - 50t^2 + 280t - 150t}{(t + 3)(t + 4)} \][/tex]
[tex]\[ D(t) = \frac{20t^2 + 130t}{(t + 3)(t + 4)} \][/tex]
Factor the numerator:
[tex]\[ D(t) = \frac{10t(2t + 13)}{(t + 3)(t + 4)} \][/tex]
Thus, the correct function that describes the difference, [tex]\( D(t) \)[/tex], is:
[tex]\[ D(t) = \frac{10t(2t + 13)}{(t + 3)(t + 4)} \][/tex]
Therefore, the correct answer is:
C. [tex]\( D(t) = \frac{10t(2t + 13)}{(t + 3)(t + 4)} \)[/tex]
