Answer:
At a combined speed of 6 in/min, it takes us 24 mins to clean the wall
Step-by-step explanation:
Since the question did not provide the speed with which each student cleans, we can make assumptions. This is so that we can solve the question before us
Assuming student 1 cleans at a speed of 2 inches per minute, student 2 cleans at a speed of 2½ inches per minute & student 3 cleans at a speed of 1½ inches per minute.
Let's list the parameters we have:
Height of wall (h) = 12 ft, Speed (student 1) = 2 in/min, Speed (student 2) = 2½ in/min, Speed (student 3) = 1½ in/min
Speed of cleaning wall = Height of wall ÷ Time to clean wall
Time to clean wall (t) = Height of wall ÷ Speed of cleaning wall
since students 1, 2 and 3 are working together, we will add their speed together; v = (2 + 2½ + 1½) = 6 in/min
1 ft = 12 in
Time (t) = h ÷ v = (12 * 12) ÷ 6 = 144 ÷ 6
Time (t) = 24 mins
Answer:
-54
Step-by-step explanation:
Answer:
All of the above
Step-by-step explanation:
dy/dt = y/3 (18 − y)
0 = y/3 (18 − y)
y = 0 or 18
d²y/dt² = y/3 (-dy/dt) + (1/3 dy/dt) (18 − y)
d²y/dt² = dy/dt (-y/3 + 6 − y/3)
d²y/dt² = dy/dt (6 − 2y/3)
d²y/dt² = y/3 (18 − y) (6 − 2y/3)
0 = y/3 (18 − y) (6 − 2y/3)
y = 0, 9, 18
y" = 0 at y = 9 and changes signs from + to -, so y' is a maximum at y = 9.
y' and y" = 0 at y = 0 and y = 18, so those are both asymptotes / limiting values.
Answer:
25.56 is the surface area of the pool.
Hope this helps! :)
Answer: 32 first 12 second 16 last Yw!
Step-by-step explanation:
