Is there some type of graph or something
Answer:
dh/dt = 2/3*π ft/min
Step-by-step explanation:
We have a right circular cylinder, and water is pouring at a constant rate, we must expect the height of the water will rise a constant rate, therefore it does not matter the height
V(c) = π*r²*h ⇒ V(c) = π*(3)²*h ⇒ V(c) =9*π*h
DV(c)/dh = 9*π
DV(c)/dh = DV(c)/dt * dt/ dh ⇒ 9*π = DV(c)/dt * dt/ dh
dh/dt = DV(c)/dt / 9*π
DV(c)/dt = 6 ft³/m ( from problem statement )
Then
dh/dt = 6/9*π ft/min
dh/dt = 2/3*π ft/min
Answer:
c = -4/9; c = 2
Step-by-step explanation:
Answer:
x = 4/3.
Step-by-step explanation:
x-2/3 + 1/6 =5/6
Add 2/3 and subtract 1/6 from both sides:
x - 2/3 + 2/3 + 1/6 - 1/6 = 5/6 + 2/3 - 1/6
x = 5/6 + 2/3 - 1/6
x = 5/6 + 4/6 - 1/6
x = 8/6 = 4/3.
Answer:
give me a second and i will answer it
Step-by-step explanation: