Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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]
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
