The sum of three consecutive even integers is –78. What are the integers? Describe how you would use the strategy of writing an equation to solve the given problem.

-28, -26, -24

Step-by-step explanation:

First, let's call the smallest number: n

Then, the next two consecutive even numbers would be:

n+2 and n+4 (this is because it's only even numbers so we add 2)

We know the sum of these is −78 so we can write this equation and solve for n:

n + (n + 2) + (n + 4) = -78

n + n + 2 + n + 4 = -78

n + n + n + 2 + 4 = -78

n + n + n + 6 = -78

1n + 1n + 1n + 6 = -78

3n + 6 = -78        Subtract 6 from each side, which cancels the 6 on the left.

3n + 6 - 6 = -78 - 6

3n = -78 - 6

3n = -84       Divide each side by 3

3n/3 = -84/3

n = -84/3

n = -28

n + 2 = -28 + 2 = -26

n + 4 = -28 + 4 = -24

The 3 consecutive odd integers would be:

-28, -26, -24

Let's check our work:

-28 + -26 + -24 = -78    We are correct!

(n^3 - n)

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

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

homozygous. one is homozygous, while two is heterozygous.