Find the value of n that satisfies 2(n+1)! + 6n! = 3(n+1), where $n! = n\cdot (n-1)\cdot (n-2) \cdots 2\cdot 1$.

aloha~! my name is zalgo and i am here to you out on this amazing day. to answer your question, yes, nonzero numbers can have a greatest common factor and that would be 1. still don't get it? let me considering every number is divisible by 1 and hence every number has 1 as a factor of theirs. so, if 2 nonzero numbers didn't share any gcf (greatest common factor), they would still have a factor in common and that would be 1.

i hope that this ! : p

"stay and stay proud! " - zalgo

(by the way, do you mind marking me as brainliest? i'd greatly appreciate it! mahalo! x3)

d. corresponding side lengths in similar triangles are proportional in length.

step-by-step explanation:

y = -1

x = 8

hope this !