Simplify: Negative two and one-third minus negative ten and one-sixth Negative twelve and one-half Twelve and one-half Seven and five-sixths Negative seven and five-sixths

∠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.

i got 10.28 but that rounds up to 10.3

3

step-by-step explanation:

answer: the center is the median and/or mean of the data. the spread is the range of the data. and, the shape describes the type of graph. the four ways to describe shape are whether it is symmetric, how many peaks it has, if it is skewed to the left or right, and whether it is uniform. good luck

step-by-step explanation: