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Bacteria of species A and species B are kept in a single environment, where they are fed two nutrients. Each day the environment is supplied with 19,660 units of the first nutrient and 31,890 units of the second nutrient. Each bacterium of species A requires 4 units of the first nutrient and 5 units of the second, and each bacterium of species B requires 1 unit of the first nutrient and 6 units of the second. What populations of each species can coexist in the environment so that all the nutrients are consumed each day?

Respuesta :

Answer:

4570 species of the type A and 1540 species of the type B.

Step-by-step explanation:

A:

1st: 4 units

2nd: 5 units

B:

1st: 1 unit

2nd: 6 units

A=x, B=y

4x + 1y = 19820  units of the first nutrient

5x + 6y = 32090  units of the second nutrient

5x + 6*(19820-4x) = 32090.

       32090 - 6(19820-4x)

x= ------------------------------------- = 4570

                     5-24

y = 19820 - 4*4570 = 19820 - 18280 = 1540.

ANSWER:  4570 species of the type A and 1540 species of the type B.

Hopes this solves it.

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