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Artyom0805 [142]
3 years ago
10

A scuba diver dove from the surface of the ocean to an elevation of −49 9/10 feet at a rate of −9.5 feet per minute. After spend

ing 10 minutes at that elevation, the diver ascended to an elevation of −28 9/10 feet. The total time for the dive so far was 17 1/8 minutes. What was the rate of change in the diver's elevation during the ascent? Round your answer to the nearest hundredth.
Mathematics
1 answer:
laila [671]3 years ago
7 0

Answer:

Rate of change  =  4,06 f/min

Step-by-step explanation:

The ascendent length  28 9/10   =  289/ 10

The time for that ascendent maneuver is total time 17 1/8 minutes  - spended time in the elevation 10 minutes, then

17 1/8  =  137/8

137/8  - 10   =  ( 137 - 80 ) /8   =  57/8 minutes

The rate of change in the diver´s elevation is

289 / 10 / 57/8

Rate of change  =  289*8/ 57*10

Rate of change  =  2312 / 570 ft/min

Rate of change  =  4,06 f/min

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