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A company manufactures computers. Function [tex]\( N \)[/tex] represents the number of components that a new employee can assemble per day. Function [tex]\( E \)[/tex] represents the number of components that an experienced employee can assemble per day. In both functions, [tex]\( t \)[/tex] represents the number of hours worked in one day.

[tex]\[ N(t) = \frac{50t}{t+4} \][/tex]
[tex]\[ E(t) = \frac{70t}{t+3} \][/tex]

Which function describes the difference in the number of components assembled per day by the experienced and new employees?

A. [tex]\( D(t) = \frac{10t(2t-13)}{t+4} \)[/tex]

B. [tex]\( D(t) = \frac{10t(2t+13)}{(t+3)(t+4)} \)[/tex]

C. [tex]\( D(t) = \frac{10t(2t-13)}{(t+3)(t+4)} \)[/tex]

D. [tex]\( D(t) = \frac{10t(2t+13)}{t+3} \)[/tex]

Sagot :

GinnyAnswer
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]
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