In a different plan for area codes, the first digit could be any number from 2 through 5, the second digit was either 2, 3, 4, or 5, and the third digit could be any number except 4, 5, or 8. With this plan, how many different area codes are possible?

Step-by-step explanation:

First digit 2,3,4,5 = 4 different digits.

Second digit 2,3,4,5 = 4 different digits.

Third digit = 1,2,3,6,7,9 = 6 different digits.

Total number of possibilities = 4*4*6 = 96

∠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.