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mmend98
 one year ago
Diana lost 35 points on a project for school. She can make changes and receive threetenths of the missing points back. She can do that 10 times. Create the formula for the sum of this geometric series and explain your steps in solving for the maximum number of points Diana can earn back.
mmend98
 one year ago
Diana lost 35 points on a project for school. She can make changes and receive threetenths of the missing points back. She can do that 10 times. Create the formula for the sum of this geometric series and explain your steps in solving for the maximum number of points Diana can earn back.

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.3dw:1439420130504:dw

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2is there any way you could type that haha its very hard to interpret this. but thank you @triciaal

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1i'll type it :=) \[\huge\rm S_n =a+ar^{n1}\] \[\large \rm (\frac{ 3 }{ 10 })^9 =a(1+r^{n1})\] \[\large \rm 35 =\frac{ a(1+(\frac{ 3 }{ 10 })^9) }{ 1 \frac{3}{10} }\] (formula: \[S_n = \frac{ a(1r^n) }{ 1r }\])

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2So how would I plug this in?

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2Help ! haha please good person @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2triciaal seems to be having the right method  i think you just need to add up the first 10 terms

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2remember the "partial sum formula" for geometric series ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large S_n=a_1*\dfrac{1r^n}{1r}\] does it look something like above in ur notes ?

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2yes. but there is the sigma thing in the front

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \sum\limits_{i=1}^na_i=a_1*\dfrac{1r^n}{1r}\] like that ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2No wait, there is a mistake

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2scratch that, il type it again fixing the mistake

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Okay, good. Lets see whats going on in the given word problem. She can make changes and receive threetenths of the missing points back, so this becomes the common ratio : \(r=\dfrac{3}{10}\) To start with, the missing points were 35, so the first term must be : \(a_1 = 35*\dfrac{3}{10} = 10.5\) Plug them in the partial sum formula and evaluate

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \sum\limits_{i=1}^{10}a_i=10.5*\dfrac{1\left(\frac{3}{10}\right)^{10}}{1\frac{3}{10}}\] simplify..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2let me know what you get and please medal triciaal, not me

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Wow! 149 ? She cannot receive 149 points back! double check ur work

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2oh sorry put it in way wrong haha

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2wait yea i got 149 again lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lets check with wolfram

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=10.5*%281%283%2F10%29%5E10%29%2F%2813%2F10%29

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2wolf says\(14.9999\), which approximates to \(15\) looks you have just missed one decimal point somewhere..

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2but doesn't the sigma change it to 149

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what do you know about sigma

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2like when u account for sigma in the problem.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what do you mean by that, could you elaborate

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2yea you have to take in to account the numbers above the sigma

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2idk maybe I'm crazy haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\sum\) is just a short cut for "sum of terms" don't let that confuse you here

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2check this quick http://www.mathsisfun.com/algebra/sigmanotation.html

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so how many points is she gonna get back after 10 attempts ? whats the maximum possible points that she can get back if she is allowed to attempt infinitely many times ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2careful here, \(14.999\) is not exactly same as \(15\) In fact, she can never get back \(15\) points even if she is allowed to attempt 100 or 1000 or any other big finite number of times

mmend98
 one year ago
Best ResponseYou've already chosen the best response.2ah. so it would just be 14.999

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2She gets back 14.999 after 10 attempts The maximum points that she can get back is 15 if she is allowed to try "infinite" times (which we know is practically impossible here)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Notice that with each attempt, the sum is by increasing a tiny bit and getting closer to 15, but it never quite reaches 15. We say the sum "converges" to 15.
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