The sum of three consecutive integers is three more than twice the middle integer. create an equation and solve to find the integers. can someone me out? and explain and show work!

Let's assume the three consecutive integers are x, x+1 and x+2.

So, sum of x, x+1 and x+2 can be written as x+x+1+x+2.

Twice the middle integer (x+1) will be 2*(x + 1).

Hence 3 more than 2(x+1) can be represent as 3 + 2(x+1).

Hence, we can set u the following equation as following:

x + x + 1 + x + 2 = 3 + 2(x + 1).

3x + 3 = 3 + 2x + 2

3x + 3 = 2x + 5 Combine the like terms.

3x + 3- 3 = 2x + 5 - 3 Subtract 3 from each sides.

3x = 2x + 2

3x - 2x = 2x + 2 - 2x Subtract 2x from each sides.

x = 2

So, the first integer is 2.

Second integer is 2+ 1 = 3 and

Third integer 2+2 = 4

So, the three consecutive integers are 2, 3 and 4.

"Hope this helps!!"

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2