Instead of the three dots, write a digit to make the fraction reducible. (find all possible cases.) /

  {0, 1, 2, 4, 5, 6, 8}

Step-by-step explanation:

120/230 = 12/23

121/231 = 11/21

122/232 = 61/116

123/233 -- not reducible

124/234 = 62/117

125/235 = 25/47

126/236 = 63/118

127/237 -- not reducible

128/238 = 64/119

129/239 -- not reducible

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All even digits will make the fraction reducible, as will the digit 5. Then the digits we're not sure of are 1, 3, 7, 9. Appending 1 to the given numbers makes them divisible by 11, so that digit is included in our list.

Ending digits 3, 7, 9 result in fractions that are not reducible. (Numbers 127, 233, and 239 are prime numbers.)

step-by-step explanation:

∠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.