If n and k are positive integers, is n divisible by 6 ? (1) n = k(k + 1)(k – 1) (2) k – 1 is a multiple of 3.

yes, it is

Step-by-step explanation:

A number is divisible by 6 if it is divisible by 2 and 3 simultaniously.

n = k(k+1)(k-1)

If k-1 is a multiple of 3, n is divisible by 3, so one of the requirements is ok.

Now, if k-1 is a multiple of 3, it can be even or odd.

if k-1 is even, then it is divisible by 2 and as it is divisible by 3 as well, n is divisible by 6

if k-1 is odd, then k and k+1 is even, hence, divisible by 2.

As n = k(k+1)(k-1), n is also divisible by 6.

∠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.

its 50 miles away from the starting point