peter said that the set of data in part (b) describes a function, but the set of data in part (a) does not.

i leave the calculator work to you.

7. diameter 4 miles, circumference 4π miles

8. radius 0.2 miles, circumference 0.4π miles

[this is a bad problem because the actual radius of a car on the track is less than 0.2 miles so the answer is less than 0.4π miles]

9. radius 6 feet, circumference 12π feet

10. diameter 14 cm, circumference 14π cm

11. diameter 4.8 feet, circumference 4.8π feet

∠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.

first one because it can't have a fraction in it

step-by-step explanation: