THIS WORTH 50 POINTS!

Step-by-step explanation:

Let x represent the number of cases of 16-oz cups produced.

Let y represent the number of cases of 20-oz cups produced.

The limitation imposed by available production time is ...

  x + y ≤ 15·8 = 120 . . . . maximum number of cases produced in a day

The limitation imposed by raw material is ...

  14x +18y ≤ 1800 . . . . . maximum amount of resin used in a day

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The point of intersection of the boundary lines for these inequalities can be found using substitution:

  14(120- y)+18y = 1800

  4y = 120 . . . . . subtract 1680, simplify

  y = 30

  x = 120 -30 = 90

This solution represents the point at which production will make maximal use of available resources. It is one boundary point of the "feasible region" of the solution space.

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The feasible region for the solution is the doubly-shaded area on the graph of these inequalities. It has vertices at ...

  (x, y) = (0, 100), (90, 30), (120, 0)

The profit for each of these mixes of product is ...

  (0, 100): 25·0 +20·100 = 2000

  (90, 30): 25·90 +20·30 = 2850 . . . . uses all available resources

  (120, 0): 25·120 +20·0 = 3000 . . . . maximum possible profit

The family can maximize their profit by producing only 16-oz cups at 120 cases per day.

90 16-oz cases and 30 20-oz cases will maximize resin and time use

120 16-oz cases will maximize profit

Step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.