help me with math plz!?

Question

three consecutive even numbers have a sum between 84 and 96.
A. Write an equality to find the three numbers. Let n represent the smallest even number.
B.Solve the inequality

Answer 1

For question A, we simply follow the wording.

LET N be the smallest even number

Three consecutive even numbers means

n, n+2, and n+4 (since the interval between the numbers is by 2)

...have a sum.. is simply
the sum of n, n+2 and n+4, which is 3n + 6

...between 84 and 96...is simply

84 < 3n + 6 < 96

This is the inequality now.

Solving it

We subtract six on all sides.

84 - 6 < 3n + 6 - 6 < 96 - 6
78 < 3n < 90

And then we divide by three altogether
26 < n < 30

So n can be 28 only (since it is the only even number that falls in the inequality).

following the notion that n is the smallest even number

hence

the three numbers are 28, 30, 32

I hope this helps. :D Good luck

Answer 2

ok, assuming the first number to be the smallest, let's call it n.

the 2nd number after that would have to be two more than n because it is the next even number we come to (numbers fall the pattern, even, odd, even, odd, etc...)

So the three consecutive numbers are :
n
n+2
n+4

Therefore the sum of those three would be 3n + 6

therefore

84 < 3n + 6 < 96

subtract 6 from all three sides

78 < 3n < 90 divide by three
26 < n < 30

Therefore, the smallest of those numbers could be 27, 28, or 29.

Answer 3

n+(n+2)+(n+4) is less than 96.
n+(n+2)+(n+4) is greater than 84.

n is less than 30.
n is greater than 30.
Thusly, 30 is the median of the three consecutive even numbers.

Because the three numbers are consecutive numbers, the three numbers equal m-2, m, and m+2

30 -2 = 28
30 = 30
30 + 2 = 32

The three consecutive even numbers are 28,30,32

Answer 4

n=smallest number
n+2= median number
n+4= largest number

84<3n+6<96

subtract 6 from both sides
78<3n<90
26
since n is even then n=28
n+2=30
n+4=32

answer 28,30,32