Question

Three consecutive integers have a sum of 87 .. What are the integers ?!

Answer 1

The options are endless...Any answer choices?

Answer 2

No...

Answer 3

Let n-1 be an integer.
The next integer is n.
The next integer after that would be n+1.

so just solve the equation:
(n-1)+n+(n+1)=87

Answer 4

What he said xD
Any questions?

Answer 5

I get 2n^3=86 then what?

Answer 6

oh mine how did you get that

Answer 7

Did you tell her to divide 86 into 3 first THEN do this?

Answer 8

how many n's do you see here
n, n , n?

Answer 9

3..

Answer 10

n+n+n=3n

Answer 11

-1+1=0

Answer 12

\[(n-1)+n+(n+1)=87 \\(n+n+n)+(-1+1)=87 \\ 3n+0=87 \\ 3n=87\]

Answer 13

n+n+n=3n
just like
4+4+4=3(4) or 12

Answer 14

So answer is 29..30..31?

Answer 15

3n=87
gives n=29
n-1=?
n+1=?

Answer 16

28,29,30

Answer 17

put n to be 29
these questions I just asked you are equivalent to:
n-1=29-1=28
n+1=29+1=30
yes

28+29+30=87

Answer 18

yep we had to find the number before and after 29
since we used (n-1) and n and (n+1) to be are integer
where n we found to be 29

Answer 19

Thank you .. Can u stick around and help me with some other problems?

Answer 20

I can probably help with one more

Answer 21

Write 7/3 as a mixed number

Answer 22

How do I start?

Answer 23

how many times does 3 go into 7?
or how many 3's are in 7?

Answer 24

2.3

Answer 25

7=3+3+1
how many 3's do you see?

Answer 26

2

Answer 27

right
so far we have:
\[\frac{7}{3}=2 \frac{?}{3}\]

Answer 28

the question mark is the remainder

Answer 29

1

Answer 30

7=3+3+1
right that 1 right there was left over

Answer 31

so the question mark is 1

Answer 32

Thank you!

Answer 33

np