All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that two sacks of sugar together with three sacks of flour weigh no more than 40 pounds and that the weight of a sack of flour is no more than 5 pounds more than the weight of two sacks of sugar. What is the largest possible weight (in pounds) of a sack of flour?

The largest possible weight of flour is 11.25  pounds.

Step-by-step explanation:

To start with, we will assume that the weight of 1 sack of sugar = x pounds

We will also assume that the weight of 1 sack of flour = y pounds

So, the weight of 2 sacks of sugar = 2 * (x) = 2x

Same thing goes for the weight of 3 sacks of flour  = 3 * (y) = 3y

Supposing that the weight of (2 sacks of sugar + 3 sacks of flour) ≤ 40 pounds

= 2x + 3y ≤ 40 we'll call that equation 1.

Also, suppose that the weight of ( 1 sack of flour) ≤ 2 sacks of sugar + 5 pounds

= y ≤ 2x + 5 we'll call that equation 2

Therefore, we'll solve for the values of x and y in the two equations and we will get:

2x + 3y ≤ 40

y ≤  2x + 5

Now, substitute the value of y into equation 1

2x + 3y ≤ 40 ⇒ 2x + 3(2x +5) =40

⇒ 2x + 6x + 15= 40

⇒ 8x + 15 = 40

⇒ 8x = 25

⇒ x = 25/8

⇒ x = 3.12

x cannot be more than 3.12 pounds, so we solve for y

Putting the value of x into equation 2, we'll get

⇒ 2y + 5 = 2(3.12) + 5

⇒ y = 11.25 pounds.

So, n cannot be more than 11.25 pounds

it 2

step-by-step explanation:

8^1/3=2

∠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.