Answer:
13.695 m
Step-by-step explanation:
The assumption made here is that the boat/water interface is essentially frictionless, so that the center of mass of the system remains in the same place as the occupant of the boat moves around.
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We can find the sum of the moments of boat and child about the pier end:
(46 kg)(7.6 m) + (80 kg)((7.6 +9.6/2) m) = 1341.6 kg·m
After the child moves, the center of mass of boat and child is presumed to remain in the same place. If x is the new distance from the pier to the child, the sum of moments is now ...
46x +80(x-4.8)* = 1341.6
126x -384 = 1341.6
x = (1341.6 +384)/126 = 13 73/105 ≈ 13.695 . . . meters
The child is about 13.695 meters from the pier when she reaches the far end of the boat.
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* The center of mass of the boat alone is half its length closer to the pier than is the child, so is located at x-4.8 meters.
A= B+C
As given, A+B+C = 50000 seats
A+(B+C) = 50000
2A = 50000
A= 25000 seats
Now section A sells seats at $30 each
So, 
Now, B+C = 25000
B = 25000-C
B sells at $24 and C sells at $18
So, 24B+18C = 538800
Now as section A incurs an amount of $750000 from booking, so B and C will incur 1288800-750000 = 538800 where $1288800 is the total booking amount.
24B+18C = 538800
6C = 61200
C = 10200
Now, B = 25000-C
B = 25000 - 10200 = 14800
Hence, section A has 25000 seats
Section B has 14800 seats
Section C has 10200 seats
Answer:
+2 ft
Step-by-step explanation:
He descends -12 feet (negative descend is an ascend) so he ascends 12 feet
-14 + 12 = -2 so he needs to travel 2ft
Answer: the answer is 21
Step-by-step explanation:
It would stay the same, since the graph hasn’t moved
