Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Prove If x

OpenStudy (anonymous):

well (x+y)/2 is the average of x and y, so if x<y it's bound to be larger than x and less than y algebraically you can start with x<y add x to both sides x+x < y+x 2x <y+x then divide both sides by two x<(y+x)/2

OpenStudy (anonymous):

you can do something very similar to show that (x+y)/2 <y

OpenStudy (anonymous):

x<y hence x/2<y/2 or x/2+x/2<y/2+x/2 or x<(x+y)/2

OpenStudy (anonymous):

again x<y hence x/2<y/2 x/2+y/2<y/2+y/2 or (x+y)/2 <y

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!