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anonymous
 one year ago
Please help, geometric series..
Finn bought a yoyo from a company that claims that, with each retraction, the string rolls up by 80% of the original length. He sets up a tape measure and throws the yoyo 3 times. His data are charted below.
Throw Length of string (feet)
1 3
2 2.4
3 1.92
Finn wants to find the sum of the length of string after 10 throws. What is the sum of the lengths, rounded to the nearest hundredth?
13.39 feet
15.00 feet
4.46 feet
0.40 foot
anonymous
 one year ago
Please help, geometric series.. Finn bought a yoyo from a company that claims that, with each retraction, the string rolls up by 80% of the original length. He sets up a tape measure and throws the yoyo 3 times. His data are charted below. Throw Length of string (feet) 1 3 2 2.4 3 1.92 Finn wants to find the sum of the length of string after 10 throws. What is the sum of the lengths, rounded to the nearest hundredth? 13.39 feet 15.00 feet 4.46 feet 0.40 foot

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's either A or B, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, and any explanation needed? @breezy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay! Give me a second to type all of this and I'll explain c:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The formula for a the sum of a geometric sequence is \[\frac{ t _{1(r^n1)} }{ r1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0r equals common ratio you get it by dividing term2 by term1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Makes sense now... Thank you!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n is the term you want to calculate to.  the r in the equation would be 4/5 3/2.4= 4/5 n is equal to 10.  find 4/5 to the 10th power then subtract one and multiply t1 to it. the bottom is 4/51 and divide the top by bottom, and there's the answer!
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