Math, asked by zecedesa, 1 year ago

An elevator descends into a mine shaft at the rate of 8m/min. If the descend starts from 10m above the ground level, how long will it take to reach the shaft 390 m below the ground level?

Answers

Answered by anandparkash
20
speed=390
time=8min
distance=s×t
390×8=3120
above 10 m ground level=
390-10 =380m
Answered by Vanshu27july
8

speed=390

time=8min

distance=s×t

390×8=3120

above 10 m ground level=

390-10 =380m

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