1 2
01.49.15
Two-fifths of one less than a number is less than three-fifths of one more than that number. What numbers are in the
solution set of this problem?
X<-5
O x>-5
O x>-1
O XC-1

9514 1404 393

  x > -5

Step-by-step explanation:

  (2/5)(x -1) < (3/5)(x +1) . . . . . given

  2(x -1) < 3(x +1) . . . . . . . . . . .multiply by 5

  2x -2 < 3x +3 . . . . . . . . . . .  eliminate parentheses

  -5 < x . . . . . . . . . . . . . . . . . . add -2x-3 to both sides

This matches the second answer choice: x > -5.

∠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.