A ball is thrown from a height of 248 feet. Every time it hits the ground it reboubs to 4/5 its original height. How far has the ball travled by the the time it comes to rest?
Total distance
= 248 + 2(248)(4/5) + 2(248)(4/5)^2 + 2(248)(4/5)^3 + ...
= -248 + 2(248) + 2(248)(4/5) + 2(248)(4/5)^2 + 2(248)(4/5)^3 + ...
= -248 + 2(248)/(1-(4/5))
= 2*248/(1/5) - 248
= 10*248 - 248
= 9 * 248
= 2232
thats a sigma
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Answers & Comments
Verified answer
Total distance
= 248 + 2(248)(4/5) + 2(248)(4/5)^2 + 2(248)(4/5)^3 + ...
= -248 + 2(248) + 2(248)(4/5) + 2(248)(4/5)^2 + 2(248)(4/5)^3 + ...
= -248 + 2(248)/(1-(4/5))
= 2*248/(1/5) - 248
= 10*248 - 248
= 9 * 248
= 2232
thats a sigma