In the air, it had an average speed of 16 \text{m/s}m/start text, m, slash, s, end text. In the water, it had an average speed of 333 \text{m/s}m/start text, m, slash, s, end text before hitting the seabed. The total distance from the top of the cliff to the seabed is 127 meters, and the stone's entire fall took 12 seconds.

Incomplete question

This is the complete question

A stone falls from the top of a cliff into the ocean.

In the air, it had an average speed of 16 m/s. In the water, it had an average speed of 3 m/s before hitting the seabed. The total distance from the top of the cliff to the seabed is 127 meters, and the stone's entire fall took 12 seconds.

How long did the stone fall in air and how long did it fall in the water?

Step-by-step explanation:

The stone speed in air Sa =16m/s

The stone speed in water Sw=3m/s

Total distance travelled is 127m

Total time taken is 12 sec

Let time in air be ta

Tune in water be tw

Therefore

Total time =ta +tw

12=ta+tw. ,Equation 1

ta=12-tw. ,Equation 2

Also

Speed is given as

Speed = distance /time

Distance =speed × time

So, total distance = distance traveled in air + distance travelled in water

127= Sa×ta + Sw×tw

127=16ta+3tw. , Equation 3

Substitute equation 2 into. 3

127=16(12-tw)+3tw

127=192-16tw+3tw

127-192=-13tw

-65=-13tw

tw=-65/-13

tw=5 seconds

Also from equation 2

ta=12-tw

ta=12-5

ta=7 seconds

The stone spent 7 seconds in air and spent 5 seconds in water before getting to the seabeds

y = 0.77(x)

step-by-step explanation:

the jacket retains this value after 23% cut or retains 77% of original value. plug original 87 jacket price for x and the remaining value should be y.

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