To find which function describes the difference in the number of components assembled per day by experienced and new employees, we need to subtract [tex]\( N(t) \)[/tex] from [tex]\( E(t) \)[/tex]:
Given:
[tex]\[ N(t) = \frac{50t}{t+4} \][/tex]
[tex]\[ E(t) = \frac{70t}{t+3} \][/tex]
We need to find [tex]\( D(t) = E(t) - N(t) \)[/tex].
Let’s compute this step-by-step.
First, express [tex]\( E(t) \)[/tex] and [tex]\( N(t) \)[/tex] with a common denominator:
[tex]\[ E(t) = \frac{70t}{t+3} \][/tex]
[tex]\[ N(t) = \frac{50t}{t+4} \][/tex]
The common denominator for the two fractions is [tex]\((t + 3)(t + 4)\)[/tex].
So, we rewrite [tex]\( E(t) \)[/tex] and [tex]\( N(t) \)[/tex] with the common denominator:
[tex]\[ E(t) = \frac{70t \cdot (t+4)}{(t+3)(t+4)} = \frac{70t^2 + 280t}{(t+3)(t+4)} \][/tex]
[tex]\[ N(t) = \frac{50t \cdot (t+3)}{(t+3)(t+4)} = \frac{50t^2 + 150t}{(t+3)(t+4)} \][/tex]
Now, subtract [tex]\( N(t) \)[/tex] from [tex]\( E(t) \)[/tex]:
[tex]\[ D(t) = \frac{70t^2 + 280t - (50t^2 + 150t)}{(t+3)(t+4)} \][/tex]
[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]
[tex]\[ D(t) = \frac{10t(2t + 13)}{(t+3)(t+4)} \][/tex]
Therefore, the correct 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]
So, the correct option is:
A. [tex]\(\quad D(t)=\frac{10 t(2 t+13)}{(t+3)(t+4)}\)[/tex]
