N+(2n+1)+(2n+3)+(2n+5)=56
in order of operations by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

2 Answers

????????????? yu want 4...4...4...4 konsequitiv odd numbers ??????????

if yu start at 1, 1+3+5+7=16=4^2

3+5+7+9=24

????????? but maebee N kan be even ??????????

11+13+15+17=56
by
the sum of n and three more consecutive odd integers is 56. what are the four numbers

Starting with the fact that n is odd, the next odd integer is n+2
The next odd integer is n+2+2, or n+4
Follow that with n+2+2+2, or n+6

We now have n, n+2, n+4 and n+6

n + (n+2) + (n+4) + (n+6) = 56

4n + 12 = 56

4n = 56 - 12

4n = 44

n = 11
n+2 = 13
n+4 = 15
n+6 = 17

11 + 13 + 15 + 17 = 56
by Level 11 User (78.4k points)

Related questions

1 answer
1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,542 questions
99,804 answers
2,417 comments
523,290 users