If you mean standard 8.5x11 paper then 3 papers wide would make it 25.5in or 2.13 feet.
A linear approximation to the error in volume can be written as
... ∆V = (∂V/∂d)·∆d + (∂V/∂h)·∆h
For V=(π/4)·d²·h, this is
... ∆V = 2·(π/4)·d·h·∆d + (π/4)·d²·∆h
Using ∆d = 0.05d and ∆h = 0.05h, this becomes
... ∆V = (π/4)·d²·h·(2·0.05 + 0.05) = 0.15·V
The nominal volume is
... V = (π/4)·d²·h = (π/4)·(2.2 m)²·(6.8 m) = 25.849 m³
Then the maximum error in volume is
... 0.15V = 0.15·25.849 m³ ≈ 3.877 m³
_____
Essentially, the error percentage is multiplied by the exponent of the associated variable. Then these products are added to get the maximum error percentage.
The answer is C, hope this helps!
X=13; do you need the work? Also for future reference symbolab.com is where I get all my answers.
Write the equation of the line through (5,-4); m = 1
