To solve the given problem, we need to determine the difference in the number of components assembled per day by an experienced employee compared to a new employee. This difference can be expressed as a function, which we'll denote as [tex]\(D(t)\)[/tex], where [tex]\(t\)[/tex] represents the number of hours worked in a day.
Given the functions representing the number of components assembled:
[tex]\[ N(t) = \frac{50t}{t+4} \][/tex]
[tex]\[ E(t) = \frac{70t}{t+3} \][/tex]
To find the difference function [tex]\(D(t)\)[/tex], we subtract [tex]\(N(t)\)[/tex] from [tex]\(E(t)\)[/tex]:
[tex]\[ D(t) = E(t) - N(t) \][/tex]
Substituting the given expressions for [tex]\(E(t)\)[/tex] and [tex]\(N(t)\)[/tex]:
[tex]\[ D(t) = \frac{70t}{t+3} - \frac{50t}{t+4} \][/tex]
To combine 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].
We'll rewrite each fraction 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]
So, we can write:
[tex]\[ D(t) = \frac{70t(t+4)}{(t+3)(t+4)} - \frac{50t(t+3)}{(t+3)(t+4)} \][/tex]
Now, combine the fractions:
[tex]\[ D(t) = \frac{70t(t+4) - 50t(t+3)}{(t+3)(t+4)} \][/tex]
Simplify the numerator:
[tex]\[ 70t(t+4) - 50t(t+3) = 70t^2 + 280t - 50t^2 - 150t \][/tex]
[tex]\[ = (70t^2 - 50t^2) + (280t - 150t) \][/tex]
[tex]\[ = 20t^2 + 130t \][/tex]
Thus, we have:
[tex]\[ D(t) = \frac{20t^2 + 130t}{(t+3)(t+4)} \][/tex]
Factoring the numerator:
[tex]\[ 20t^2 + 130t = 10t(2t + 13) \][/tex]
Which gives us:
[tex]\[ D(t) = \frac{10t(2t + 13)}{(t+3)(t+4)} \][/tex]
So, the function that describes the difference in the number of components assembled per day by the experienced and new employees is:
[tex]\[ D(t)=\frac{10t(2t + 13)}{(t+3)(t+4)} \][/tex]
Therefore, the correct answer is:
B. [tex]\( D(t)=\frac{10t(2t+13)}{(t+3)(t+4)} \)[/tex]
