Answer
$V_{rms}=V_{avg}$
Work Step by Step
We know that
$V^2_+=(5.0)^2=25V^2$
and $V^2_-=(-5.0)^2=25V^2$
Now $V^2_{avg}=\frac{1}{2}(V^2_+ +V^2_-)=\frac{1}{2}(25+25)=25V^2$
$V_{rms}=\sqrt{V^2_{avg}}=\sqrt{25V^2}=5.0V=V_{max}$
Hence, we proved that $V_{rms}=V_{avg}$
