Let the number of bags of feed type I to be used be x and the number of bags of feed type II to be used be y, then:
We are to minimize:
C = 4x + 3y
subject to the following constraints:

From the graph of the 4 constraints above, the corner points are (0, 5), (1, 2), (4, 0).
Testing the objective function for the minimum corner point we have:
For (0, 5):
C = 4(0) + 3(5) = $15
For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10
For (4, 0):
C = 4(4) + 3(0) = $16.
Therefore, the combination that yields the minimum cost is 1 bag of type I feed and 2 bags of type II feed.
30% of 70 is simply 0.3 * 70, which equals to 21
Answer:
D
Step-by-step explanation:
(given) -3(4x - 8) = -36
(distribute) -12x + 24 = -36
(subtract 24 from both sides) -12x = -60
(divide both sides by -12) x = 5
Answer:
its poop
Step-by-step explanation:
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