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blagie [28]
4 years ago
6

A gold mine has two​ elevators, one for equipment and another for the miners. The equipment elevator descends 66 feet per second

. The elevator for the miners descends 1616 feet per second. One​ day, the equipment elevator begins to descend. After 2323 ​seconds, the elevator for the miners begins to descend. What is the position of each elevator relative to the surface after another 1616 ​seconds? At that​ time, which elevator is​ deeper? The position of the equipment elevator relative to the surface is negative 234−234 feet. The position of the elevator for the miners relative to the surface is nothing feet.
Mathematics
1 answer:
OverLord2011 [107]4 years ago
7 0
Given that the equipment elevator descends at 6 feet per second.

The total number of seconds travelled by the equipment elevator is 23 + 16 = 39 seconds.

Thus the <span>position of the equipment elevator relative to the surface is given by -(6 x 39) = -234 feet.

Given that </span>the <span>miners elevator descends at 16 feet per second.

The total number of </span>seconds travelled by the miners elevator is 16 seconds.

Thus, the <span>position of the miners elevator relative to the surface is given by -(16 x 16) = -256 feet.

At that time the miners' elevator is deeper.
</span>
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