Find three consecutive even integers such that the square of the first is 80 less than the square of the third. [Only an algebraic solution will be accepted.]​

First number = 19

Second consecutive = 20

Third consecutive = 21

Step-by-step explanation:

Assume;

First number = x

So,

Second consecutive = x + 1

Third consecutive = x + 1 + 1 = x + 2

So,

(x + 2)² - x² = 80

x² + 2(2)(x) + 2² - x² = 80

x² + 4x + 4 - x² = 80

4x + 4 = 80

4x = 80 - 4

4x = 76

x = 76 / 4

x = 19

First number = x

First number = 19

Second consecutive = x + 1

Second consecutive = 19 + 1

Second consecutive = 20

Third consecutive = x + 2

Third consecutive = 19 + 2

Third consecutive = 21

i just took the test, the correct answer is .. 31.5 in.

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer