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Mathematics 16 Online
OpenStudy (anonymous):

If X and Y are positive and X > Y , which is true? (A.) X > X+Y\2 (B.) Y > X+Y\2 (C.) X= X+Y\2 (D.) X < X+Y\2

OpenStudy (anonymous):

Are all the right sides \(\dfrac{X+Y}{2}\) or \(X+\dfrac{Y}{2}\)?

OpenStudy (anonymous):

The first one

OpenStudy (anonymous):

Okay. There are a few ways you can reason through this. One involves noting that that expression is the average of \(X\) and \(Y\). The average of any two numbers *always* lies between the two numbers. So if \(Z\) were the average, then \(Y<Z<X\). This would eliminate every option except ...?

OpenStudy (anonymous):

i have no clue :( i suck at this stuff.I was blowing through all the other equations this one just stumped me.

OpenStudy (anonymous):

\(Y<Z<X\) is the same as \(Y<\dfrac{X+Y}{2}< X\). Where does that put \(X\) relative to only \(\dfrac{X+Y}{2}\)?

OpenStudy (anonymous):

D. ?...im sorry i feel so dumb right now :/

OpenStudy (anonymous):

\(Y<\color{red}{\dfrac{X+Y}{2}< X}\) What's the red part the same as?

OpenStudy (anonymous):

If \(a<b\), then \(b>a\).

OpenStudy (anonymous):

A. ...i was reading everything backwards i'm sorry.Dislexia sucks

OpenStudy (anonymous):

That is righ,right? if it isn't then you should just give up on me

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