<p>side length is what it asked</p>
<p>Yup, it said side length.</p>
<p>**** me it is 1/8 then -3 on math wow. hopefully i can pull off a 34 on the math</p>
<p>im -2 or -3 also so hoping for the same thing lol</p>
<p>well the n-1 thing is stil up for debate so -2 with two really ****ing dumb mistakes. im actually mad</p>
<p>thats what I meant, sorry</p>
<p>I am almost positive that n-1 is one. 99% sure.</p>
<p>what are you saying it is?</p>
<p>However, I will acknowledge that it is a terrible question.</p>
<p>Guys, it’s one. The structure was [(n-1)/n]x[n/(n+1)]. It does not take calculus to solve it, just choose a really, really big number. Like, n=1,000,000,000. </p>
<p>(999999999/1000000000)x(1000000000/1000000001)= 0.999999998 </p>
<p>It can obviously be seen that the result is approaching 1.</p>
<p>The funny thing is that at some point in this discussion everyone thought they reached a consensus that 0 was undoubtedly 100% the answer. but yes, Deziky is right.</p>
<p>I’m confused…is the answer 0 or 1? lol 
And for those of you that said 1, how did you account for the (3/4)*(4/5) etc in the beginning>?</p>
<p>I answered 1 for the test but the answer is actually 0 now that i think about it. (2/3)<em>(3/4)= (2/4). (2/4)</em>(4/5) is (2/5). (2/5)*(5/6) is (2/6). If you keep going you will eventually get something like (2/100000000) which approaches 0.</p>
<p>thats not how you handle it though! its 2/3x 3/4 is 1/2. then u go 3/4x 4/5 is 0.6. then u go 4/5 x 5/6 is 0.66. The numbers go higher and higher, but they will never reach one. they approach one. that’s why they give you those terms at the end. 0 is really not right!</p>
<p>Are you sure we weren’t supposed to do (2/3)<em>(3/4)</em>(4/5)<em>(5/6)…</em>(999999999/1000000000)*(1000000000/1000000001)? Because this number DOES come down to 0…</p>
<p>For those of you that say it is 1, you don’t ONLY do (999999999/1000000000)<em>(1000000000/1000000001). You have to account FOR ALL THE PREVIOUS multiplications starting from (2/3)</em>(3/4) (or whatever it was)</p>
<p>Someone correct me if I’m wrong.</p>
<p>I’m not sure if it’s been answered, but what was #59 (I think it’s that one) that was the paper being folded. I know it’s an easy one and I put 1/8 but I may have read the question incorrectly.</p>
<p>It was sides of 1; then 2 folds making sides of 1/2; then 2 additional folds making sides of 1/4; finally 2 folds making side of 1/8</p>
<p>Hopefully that was the question but my friend has me doubting myself.</p>
<p>Stop debating over the fraction question, it approaches zero.</p>
<p>No, you don’t need calculus. Nor do you need precalc. While those skills would help, just plug in your calculator to realize that the sequence does in fact approach zero.</p>
<p>If you still don’t understand it, there are plenty of explanations mentioned earlier in this thread. To argue that it approaches 1 because 2 random terms approach 1 is just poor reasoning.</p>
<p>I’m trying to come up with a conjecture/proof for it that can clear it up for those still unsure but having trouble. If I do end up coming up with one I’ll post it on here later.</p>
<p>The structure is not [(n-1)/n]x[n/(n+1)], those are only the last two terms of the sequence. </p>
<p>You are not looking for the product of those two terms, you are looking for the product of all of the terms. </p>
<p>A higher n value gives you a higher number of terms, which are each less than 1. The more times you multiply by numbers less than 1, the closer it gets to zero.</p>
<p>for the paper one, why is it decreased by 2 for every 2 folds? If i fold a paper in half, that would decrease it by 2, which means it is 1/64.</p>
<p>it’s 1. I really wish that you could get a score report from the ACT so that it can be proved. Maybe when it comes back and it gives you your breakdown you can pinpoint it if you did well enough. And its not just the last 2 terms. The entire question was written incorrectly on the forum.</p>