Answer:
82.52%
Step-by-step explanation:
118+25=143

solve for x

Math assignment = 3/8 hours
History assignment = 1/4 hours
We can not compare this as both fractions do not have alike denominators.
We can fix this problem easily by turning 3/8 and 1/4 into alike denominator fractions.
We can turn 1/4 into 2/8 as it is equivalent fractions.
1 x 2 = 2
4 x 2 = 8
(Whatever you do to the denominator, you have to do the same to the numerator)
Once we get this, we can see that Shen spent 2/8 hours on his homework while he spent 3/8 hours on his math assignment.
3/8 - 2/8 = 1/8
1/8 = 0.125
Shen spent 1/8 of an hour (or 0.125 of an hour) more on his math assignment than his history assignment.
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.
B) 16 + (5 + 2) - 1
I don’t know what C is
Answer:
2 - 2y^2 - x/y
Step-by-step explanation:
2x^2/x^2 = 2
and
x^2/xy = x/y
therefore
2x^2/x^2 - y^2 - x^2/xy - y^2
= 2 - y^2 - x/y - y^2
= 2 - y^2 - y^2 - x/y
= 2 - 2y^2 - x/y