Answer:
thanks for the points !!! ;)
Step-by-step explanation:
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.
X=12
36 + 12 = 48
48/4 = 12
C = 2(pi)r....so we are looking for the radius (r)
42 = 2 * 3.14 * r
42 = 6.28r
42/6.28 = r
6.69 = r <===
Answer:
As a mixed number, 8/3 is equal to 2-2/3.
Step-by-step explanation: