Find three consecutive even integers such that the sum of the squares of the first two is four less than twice the square of the third.

the three consecutive even integers are -2, 0, 2

Step-by-step explanation:

Let the first even integer = n

Let the second even integer = n + 2

Let the third even integer = n + 4

From the given statement, we form the equation below

(n²) + (n + 2)²  + 4 = 2(n + 4)²

Expand this equation;

n² + (n + 2)(n + 2) +  4 = 2(n + 4)(n + 4)

n² + n² + 2n + 2n +  4 +  4 = 2(n² + 4n + 4n + 16)

2n² + 4n + 8 = 2(n² + 8n + 16)

2n² + 4n  + 8 = 2n² + 16n + 32

collect like terms together;

2n² - 2n² + 4n - 16n = 32 - 8

- 12n = 24

-n = 24/12

-n = 2

n = -2

Second integer = n + 2

                          = -2 + 2 = 0

Third integer = n + 4

                      = -2 + 4

                      = 2

Therefore, the three consecutive even integers are -2, 0, 2

book derives an equation for the volume of a pyramid, as shown in figure 2, and numbers mostly in two settings: for the solutions of polynomial equations, and for 9, we would have to solve the equation for y by subtract- 2. the point (1,. 2) is the center of the circle whose equation is x2 y2. 2x. 4y p.w.c. davies.

step-by-step explanation:

answer: it's upside down and i can't read it

step-by-step explanation: