consecutive even integers whose product is 1,520 2n = 1st even integer 2n + 2 = 2nd even integer Product = 2n(2n + 2) = 4n^2 + 4n = 4(n^2 + n) 1520 = 4(n^2 + n) 1520/4 = 4(n^2 + n)/4 380 = n^2 + n 0 =
1,236 questions posted
Usually answered in minutes!
2,410 Questions
809 Questions
789 Questions
510 Questions
405 Questions
