The number of cameras produces since May at a plant can be modeled by the function N (m) = 50m+600 and the cost per camera can be modeled by C (m) = 4m^2-15m+45, where m is the number of months since May. According to this model, what is the total amount of revenue generated by the plant's camera production on September?

Question

Mathematics

Guillermo Green

6 April, 07:39

The number of cameras produces since May at a plant can be modeled by the function N (m) = 50m+600 and the cost per camera can be modeled by C (m) = 4m^2-15m+45, where m is the number of months since May. According to this model, what is the total amount of revenue generated by the plant's camera production on September?

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Answers (2)

Know the Answer?

Fisher · Answer 1

39,200

Step-by-step explanation:

September is 4 months after may, so m = 4

No. of cameras:

N (4) = 50 (4) + 600 = 800

Cost per camera:

C (4) = 4 (4²) - 15 (4) + 45 = 49

Total cost:

N * C

800 * 49

39200

Julia Frank · Answer 2

39200

Step-by-step explanation:

Revenue = number of cameras * cost

R (m) = N (m) * C (m)

= (50m+600) * (4m^2-15m+45)

The months since may is June July August September = 4 months

R (4) = (50*4+600) * (4*4^2-15*4+45)

(200+600) * (64-60+45)

800*49

39200