Which graph has a slope of ? A coordinate plane with a straight line. The line starts at (negative 5, negative 4) and passes through points at (0, 1) and (4, 5). A coordinate plane with straight line. The line starts at (negative 5, negative 2) and passes through points at (negative 1, 0) and (4, 5). A coordinate plane with straight line. The line starts at (negative 5, negative 1) and passes through points at (negative 2, 0) and (3, 2). A coordinate plane with straight line. The line starts at (negative 4, negative 5) and passes through points at (0, 0) and (4, 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.

-2

7 - 9 = -2

reversed:

-2 + 9 = 7

35 is an outlier

step-by-step explanation:

-1   1/6.   you dont subtract the denominators

step-by-step explanation: