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luda_lava [24]
3 years ago
8

Which shows the graph of the solution set of 6x + 4y < 12? On a coordinate plane, a dashed straight line has a negative slope

and goes through (0, 3) and (3, 1). Everything below and to the left of the line is shaded. On a coordinate plane, a dashed straight line has a negative slope and goes through (0, 3) and (3, 1). Everything above and to the right of the line is shaded. On a coordinate plane, a dashed straight line has a negative slope and goes through (0, 3) and (2, 0). Everything to the left of the line is shaded. On a coordinate plane, a dashed straight line has a negative slope and goes through (0, 3) and (2, 0). Everything to the right of the line is shaded.
Mathematics
2 answers:
natta225 [31]3 years ago
7 0

Answer:

C

Step-by-step explanation:

irga5000 [103]3 years ago
3 0

Answer:

for edg 2020 its b / the second answer

Step-by-step explanation:

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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
DiKsa [7]

Answer:

We require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

Step-by-step explanation:

Let n₁ be the population of A required and n₂ be the population of B required.

Now we require 2 units of the first nutrient for species A and one unit of the first nutrient for species B. The total nutrients required by species A is 2n₁ and that by species B is 1n₂ = n₂. So, the total nutrients required by both species A and B is 2n₁ + n₂. Since this equals the quantity of the first nutrient which is 10,560, then  2n₁ + n₂ = 10,560 (1)

Now we require 5 units of the second nutrient for species A and 6 units of the second nutrient for species B. The total nutrients required by species A is 5n₁ and that by species B is 6n₂. So, the total nutrients required by both species A and B is 5n₁ + 6n₂. Since this equals the quantity of the first nutrient which is 31,510, then  5n₁ + 6n₂ = 31,510 (2).

So, we have two simultaneous equations which we would solve to find the populations of A and B which satisfy both equations.

2n₁ + n₂ = 10,560  (1)

5n₁ + 6n₂ = 31,510 (2)

From (1) n₂ = 10,560 - 2n₁ (3)

Substituting equation (3) into (2), we have

5n₁ + 6(10,560 - 2n₁) = 31,510

expanding the brackets, we have

5n₁ + 63,360 - 12n₁ = 31,510

collecting like terms, we have

5n₁ - 12n₁ = 31,510 - 63,360

simplifying, we have

- 7n₁ = -31,850

dividing both sides by -7, we have

n₁ = -31,850/-7

n₁ = 4,550

Substituting n₁ = 4,550 into (3), we have

n₂ = 10,560 - 2(4,550)

n₂ = 10,560 - 9,100

n₂ = 1,460

So, we require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

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Elena-2011 [213]

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