i nd help Order the set of numbers from least to greatest: (4 points)

2.71 repeating 71, 2 and 3 over 4, square root 5, 5 over 2
Group of answer choices

2.71 repeating 71, 2 and 3 over 4, square root 5, 5 over 2.

square root 5, 5 over 2, 2.71 repeating 71, 2 3 over 4

5 over 2, square root 5, 2.71 repeating 71, 2 3 over 4

2 3 over 4, 2.71 repeating 71, 5 over 2, square root 5

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

∠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.

it has changed from the width and the height.

step-by-step explanation:

about 0.75 hours

step-by-step explanation: