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.
Answer:
3.2
Step-by-step explanation:
Add up the first group then add up the second group then divide the group Y by group X
round to the nearest integer. The rule for rounding is simple: look at the digits in the tenth’s place (the first digit to the right of the decimal point).
Hope This Helps bud ^^
Answer:
B,5.4 units.
Step-by-step explanation:
Parallelogram RSTU
R(1,1),S(7,0),T(10,4),U(4,5)
slope of RU m=(5-1)/(4-1)=4/3
slope of RS=(0-1)/(7-1)=-1/6=-1/6
slope of ST=(4-0)/(10-7)=4/3
slope of TU=(5-4)/(4-10)=1/(-6)=-1/6
RU||ST
eq. of line RU is
y-y1=m(x-x1)
y-1=4/3(x-1)
3y-3=4x-4
4x-3y-4+3=0
4x-3y-1=0
perpendicular distance of a point (x1,y1) from line ax+by+c=0 is
perpendicular distance of (7,0) from 4x-3y-1=0
The perpendicular of that would be y=3x+12.
