- #1

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if a length measured is x=2 and the error m

_{x}=+- 0.005

then x=2+-0.005

if i have x+y where y=3, m

_{y}=+-0.02

m

_{x+y}=[tex]\sqrt{m

_{y}

^{2}+m

_{x}

^{2}}[/tex]

m

_{x*y}=[tex]\sqrt{(y*m

_{x})

^{2}+(x*m

_{y})

^{2}}[/tex]

but if i have x^2 does this work the same

for example if the area of a rectangle is x*2x can i say 2x

^{2}

m

_{2x2}=4x*m

_{x}