Answer:
10
Step-by-step explanation:
Let the three consecutive positive even integers be x-1, x and x+1
First = x-1
Second = x
Third = x+1
If the product of the second and third integers is twenty more than ten times the first integer, then;
x(x+1) = 20 + 10(x-1)
Open the bracket
x² + x = 20 + 10x - 10
x² + x = 10x + 10
x² + x - 10x - 10 = 0
x² - 9x - 10 = 0
x² -10x + x - 10 = 0
x(x-10)+ 1(x-10) = 0
(x+1)(x-10) = 0
x+1 = 0 and x-10 = 0
x = -1 and 10
Hence the 2nd of three consecutive positive even integers is 10
