Se differentials to estimate the amount of metal in a closed cylindrical can that is 30 cm high and 8 cm in diameter if the metal in the top and the bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. (Round your answer to two decimal places.)

the amount of metal is approximately 19.47 cm³

Step-by-step explanation:

the volume enclosed by the cylinder is

V= π*R²*L

then using differentials

dV = ∂V/∂L *dL + ∂V/∂R *dR

dV = π*R²*dL +  2*π*R*L *dR

ΔV ≈ π*R²*ΔL +  2*π*R*L *ΔR

replacing values

ΔV ≈ π* (4 cm)²* (0.1 cm*2) +  2*π*R*30 cm *0.05 cm = 19.47 cm³

then the amount of metal ( in volume ) Vmetal is

V metal = V(R+dR, L+dL) - V(R, L) = ΔV ≈ 19.47 cm³

< 6 cm to > 26 cm i believe

step-by-step explanation:

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.