Respuesta :

yuumyllo
Let's go... They are consecutive and even integers.
Let call the first number as y.
y is an even integer. So its consecutive even integers must be:
y + 2, y + 4, y + 6. 
First even integer is y.
Second even integer is y + 2.
Third even integer is y + 4.
Fourth even integer is y + 6.

So, the sum is y + (y+2) + (y+4) + (y+6). This is equal to 44.
So, we have the equation:

y + (y+2) + (y+4) + (y+6) = 44

Let's add the terms...
y + y + y + y + 2 + 4 + 6 = 44
4y + 12 = 44

Let's subtract 12 from both sides...
4y + 12 - 12 = 44 - 12
4y = 32

Now, let's divide by 4...
4y/4 = 32/4

So, y = 8.

First even integer is y = 8
Second even integer is y + 2 => 8 + 2 = 10
Third even integer is y + 4 => 8 + 4 = 12
Fourth even integer is y + 6 => 8 + 6 = 14

So, the four consecutive even integers are 8, 10, 12 and 14.

The sum of four consecutive even integers is 44

Four consecutive even numbers are 8,10,12,14

Given :

The sum of four consecutive even integers is 44.

Let the four consecutive even integers are

[tex]n, n+2, n+4, n+6[/tex]

The sum of the four consecutive even integers is 44

[tex]n +n+2+ n+4+n+6=44\\4n+2+4+6=44\\4n+12=44\\Subtract \; 12\\4n=32\\\frac{4n}{4}=\frac{32}{4}\\n=8[/tex]

so the first even number is 8

second even number =[tex]n+2=8+2=10[/tex]

Third even number =[tex]n+4=8+4=12[/tex]

Fourth even number =[tex]n+6=8+6=14[/tex]

Four consecutive even numbers are 8,10,12,14

Learn more :  brainly.com/question/15206586

Otras preguntas