What is the smallest multiple of 18 of the form 2A945B, where A and B are digits?

Step-by-step explanation:

It must end in (divisible by 2 rule) and the digits must sum to a multiple of 9(divisible by 9 rule).

The digit sum is .

We want to make as low as possible because it's a higher digit than .

If is as low as possible, we have that the sum is . Uh-oh. has to be even! So doesn't work.

If , we have that the sum is . Yay!

So, the answer is

∠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.