Question 1058987: After a long period of heavy rain, water seeps into an underground cave. After a day there are 5000 liters of water in the cave. The next day 4000 liters of water seeps into the cave. The day after, 3200 liters of water seeps into the cave, and so on.
A) How much water seeps into the cave on the 5th day?
B) How much water in total has seeped into the cave after 5 days?
C) How long before 20,000 liters of water in total have seeped into the cave?
D) if the water keeps seeping into the cave indefinitely, and none is lost, how much water will be in the cave eventually?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It is 5000+4000+3200, and each day 0.80 of the prior day seeps in.
The eventual amount is 5000/(1-0.80)=5000/0.2=25,000 liters ANSWER D
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A*r^4, which is r^(5-1) is 5000*(.8^4)=2048 liters.
Can also continue from above, where day 4 is 2560 and day 5 is 2048. ANSWER A
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Through day 5, it is 12200+2560+2048=16,808 l ANSWER B
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20000=A (1-r^n)/(1-r)
4*(1-r)=(1-r^n)
0.8=1-r^n
-0.2=-r^n
r^n=0.2
ln both sides and ln r=-0.223
n ln r=ln (0.2)
n=ln(0.2)/ln 0.8)=-1.609/-.223
n=7.21 days ANSWER C
Day 6 will have 1638.4 l.
Day 7 will have 1310.7 l
The sum through day 7 is 12200+2560+2048+1638.4+1130.7=19576.7 l, which checks.
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