Answer:
8.20in³
Step-by-step explanation:
Given V = πr²h
r is the radius = 1.5in
h is the height = 6in
thickness of wall of the cylinder dr = 0.04in
top and bottom thickness dh 0.07in+0.07in = 0.14in
To compute the volume, we will find the value of dV
dV = dV/dr • dr + dV/dh • dh
dV/dr = 2πrh
dV/dh = πr²
dV = 2πrh dr + πr² dh
Substituting the values into the formula
dV = 2π(1.5)(6)•(0.04) + π(1.5)²(6) • 0.14
dV = 2π (0.36)+π(1.89)
dV = 0.72π+1.89π
dV = 2.61π
dV = 2.61(3.14)
dV = 8.1954in³
Hence volume, in cubic inches, of metal in the walls and top and bottom of the can is 8.20in³ (to two dp)
12 × 2 = 24
anything to ask please pm me
Answer:
the second equatin
2x+(-6)=14 that give x=10 and the other 2x=14+6 gives x= 10 so this pair is equal
Sol_
( 3 - 2 ) + ( 4 - 1 ) ^ 2
= 1 + 3^2
= 1 + 9
= 10
Answer is " - " .
hope it helped !
Thanks!
Answer:
6xy²z^4
Step-by-step explanation:
![\sqrt[3]{216x^3y^6z^12}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E3y%5E6z%5E12%7D)
= ∛(6³)(x³)(y³)(y³)(z³)(z³)(z³)(z³) = 6x·y·y·z·z·z·z = 6xy²z^4
