The first one is < the second one is < and the last one is =
15,17,19
Suppose the three odd integers are #n#, #n+2# and #n+4#
Their sum is:
#n + (n+2) + (n+4) = 3n+6#
#13# more than twice the largest of the three is:
#2(n+4)+13 = 2n+21#
From what we are told these two are equal:
#3n+6 = 2n+21#
Subtract #2n+6# from both sides to get:
#n = 15#
So the three integers are:
#15, 17, 19#
Answer:
Step-by-step explanation:
