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Lerok [7]
3 years ago
5

Science help plz! Giving brainlist:D

Biology
2 answers:
natka813 [3]3 years ago
7 0
18-the first one
19-the second one ???????????????????
Setler79 [48]3 years ago
7 0
#19 elastic potential energy i believe
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365 days

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Xylem _____.
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We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day
Nastasia [14]

Answer:

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Explanation:

From the given information:

For  food source A; we have:

3P₁ + P₂ + 2P₃ = 58    units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70   units of food B  ---- (2)

For food source C; we have:

P₁ + P₂  = 20   units of food C    ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

 3P₁ + P₂ + 2P₃ = 58

-

 2P₁ + 4P₂ + 2P₃ = 70

<u>                                       </u>

<u> P₁  -   3P₂   + 0    = - 12    </u>

P₁ = -12 + 3P₂   ---- (4)

Replace, the value of P₁  in (4) into equation (3)

P₁ + P₂  = 20

-12 + 3P₂ + P₂  = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂  = 20

P₁ + 8 = 20

P₁  = 20 - 8

P₁  = 12

To find P₃;  replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ =  7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

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