Three different bacterial species are cultured in one dish and feed on three nutrients. Each individual of species I consumes 1 unit of each of the first and second nutrients and 2 units of the third nutrient. Each individual of species Il consumes 4 units of the first nutrient and 4 of the third nutrient. Each individual of species III consumes 2 units of the first nutrient, 3 units of the second nutrient, and 5 units of the third nutrient. If the culture is given 5400 units of the first nutrient, 6600 units of the second nutrient, and 12,000 units of the third nutrient, how many of each species can be supported so all the nutrients are consumed? Set up the system of equations. Let x represent the number of individual bacteria of species I, let y = represent the number of individual bacteria species II, and let z represent the number individual bacteria species III. Select all that apply. O A. 2x + 4y + 5z = 12,000 O B. x+4y + 2z = 5400 O C. x+3z = 6600 O D. x+4y + 2z = 12,000 O E. x+3z = 5400 OF. 2x + 4y + 5z = 6600 How many of each species can be supported so all the nutrients are consumed? Select the correct choice and fill in the answer box(es) to complete your choice. O A. There is one solution. There will be of species I, of species II, and of species II. (Type integers fractions.) O B. There are infinitely many solutions. Let z be a free variable so that there will be of species I, of species II, and between and of species III. (Type expressions using z as the variable. Use ascending order. Type integers or fractions.)
Three different bacterial species are cultured in one dish and feed on three nutrients. Each individual of species I consumes 1 unit of each of the first and second nutrients and 2 units of the third nutrient. Each individual of species Il consumes 4 units of the first nutrient and 4 of the third nutrient. Each individual of species III consumes 2 units of the first nutrient, 3 units of the second nutrient, and 5 units of the third nutrient. If the culture is given 5400 units of the first nutrient, 6600 units of the second nutrient, and 12,000 units of the third nutrient, how many of each species can be supported so all the nutrients are consumed? Set up the system of equations. Let x represent the number of individual bacteria of species I, let y = represent the number of individual bacteria species II, and let z represent the number individual bacteria species III. Select all that apply. O A. 2x + 4y + 5z = 12,000 O B. x+4y + 2z = 5400 O C. x+3z = 6600 O D. x+4y + 2z = 12,000 O E. x+3z = 5400 OF. 2x + 4y + 5z = 6600 How many of each species can be supported so all the nutrients are consumed? Select the correct choice and fill in the answer box(es) to complete your choice. O A. There is one solution. There will be of species I, of species II, and of species II. (Type integers fractions.) O B. There are infinitely many solutions. Let z be a free variable so that there will be of species I, of species II, and between and of species III. (Type expressions using z as the variable. Use ascending order. Type integers or fractions.)
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter11: Systems Of Equations
Section11.CT: Test
Problem 24CT
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