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532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 896,029,219,413,438,500 | redeye | 02/06/2022 19:17:40 | okay because first we find out how many ways we can arrange the distinct letters which is n! and when we incorporate S into these strings we have k places to plug in S.. which is n+1 C k
Since they are related and happens together , we've got n! (n+1) C k | 1 | 0 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:18:09 | Yes, also you can see it another way. First, we place the distinct letters:```
xAxBxCx
xAxCxBx
xBxAxCx
xBxCxAx
xCxAxBx
xCxBxAx
``` | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:18:13 | Let's say k = 2. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:18:40 | So, there will be 4C2 = 6 variations for each row there. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:19:20 | ```
xAxBxCx
SASBxCx
SAxBSCx
SAxBxCS
xASBSCx
xASBxCS
xAxBSCS
``` | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:19:30 | That row turned into 6 final rows. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:19:39 | And so do all the other 3! - 1 rows. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:20:24 | And so, since we have 3! things that turn into 4C2 final rows each, that's 3! × 4C2. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 896,029,219,413,438,500 | redeye | 02/06/2022 19:25:36 | I'm not sure i understand this. but i understood the previous way | 1 | 0 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:25:51 | Well, you know that you get an arrangement of the ABC, right? | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 896,029,219,413,438,500 | redeye | 02/06/2022 19:26:03 | yes | 1 | 0 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:26:04 | And then you place the 2 Ss. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:26:50 | There are 4C2 ways of placing the 2 Ss, so if your arrangement is ACB, you get SASBC, SABSC, SABCS, ASBSC, ASBCS, ABSCS. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:27:09 | So, your 1 ACB arrangement becomes 6 arrangements with the Ss filled in. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:27:12 | Does that make sense? | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 896,029,219,413,438,500 | redeye | 02/06/2022 19:27:58 | yes.. so i just have to multiply with other arrangement of ACB and get final result | 1 | 0 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:28:05 | Yes. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:29:01 | 3! arrangements before Ss are added in, and each one turns into 4C2 final ones after the Ss are filled in, so 3! × 4C2. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:29:11 | It's basically the idea of why you multiply. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:29:34 | Each of the 3! is worth 4C2 final arrangements. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 896,029,219,413,438,500 | redeye | 02/06/2022 19:29:39 | yeah. thank you so much! it makes sense now | 1 | 0 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 537,482,727,557,955,600 | Chai T. Rex | 02/06/2022 19:29:45 | No problem. | 0 | 1 | null | null | null | null | null | |||||
532 | help-10 | 75b1bffbb1124175815afc02aa6e7131 | 896,029,219,413,438,500 | redeye | 02/06/2022 19:32:05 | .close | 1 | 0 | null | null | null | null | null | |||||
534 | help-10 | 6ea5c7314df84b5193c3a297bcf1234d | 434,889,445,792,022,600 | skoptec | 02/06/2022 20:48:52 | IMG_4232.jpg | 1 | 0 | null | null | null | null | null | |||||
534 | help-10 | 6ea5c7314df84b5193c3a297bcf1234d | 434,889,445,792,022,600 | skoptec | 02/06/2022 20:49:11 | Can someone help me | 1 | 0 | null | null | null | null | null | |||||
534 | help-10 | 6ea5c7314df84b5193c3a297bcf1234d | 434,889,445,792,022,600 | skoptec | 02/06/2022 21:07:26 | :thonk: | 1 | 0 | null | null | null | null | null | |||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 478,518,656,523,960,300 | scroll忍 | 11/06/2021 09:45:02 | how do we solve for question 1 b? | unknown.png | 1 | 0 | null | null | null | null | null | ||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 478,518,656,523,960,300 | scroll忍 | 11/06/2021 09:55:31 | isnt the answer supposed to be 7.85m/s? | unknown.png | 1 | 0 | null | null | null | null | null | ||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 178,897,388,507,758,600 | ShatteredSunlight | 11/06/2021 10:02:11 | This question does not make sense at all. It implies Jumbo phases through Penggy since $v_J > v_P$ after the collision, which either means further pushing or Jumbo passes through Penggy. | 0 | 1 | null | null | null | null | null | |||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 510,789,298,321,096,700 | TeXit | 11/06/2021 10:02:14 | **ShatteredSunlight** | 178897388507758592.png | 0 | 1 | null | null | null | null | null | ||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 478,518,656,523,960,300 | scroll忍 | 11/06/2021 10:03:21 | did it say phases through penggy? | 1 | 0 | null | null | null | null | null | |||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 478,518,656,523,960,300 | scroll忍 | 11/06/2021 10:11:50 | nevermind isok i will try to figure it out | 1 | 0 | null | null | null | null | null | |||||
535 | help-10 | ade39c480961464bad2f15a312987657 | 478,518,656,523,960,300 | scroll忍 | 11/06/2021 10:12:02 | .close | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:14:52 | Hi! I'm trying to disprove the following by contradiction.
$$c\sqrt{n} \leq log_2^3(n), \forall n \geq n_0$$ | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 510,789,298,321,096,700 | TeXit | 02/06/2022 22:14:55 | **Phascolarctos** | 308736672433045504.png | 0 | 1 | null | null | null | null | null | ||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:16:33 | My professor has told me that I cannot use limits or derivatives, but is this even possible? I've been trying to wrack my brain around this for several hours 😭 | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:18:37 | I've tried using properties such as $$log(n)<log^2(n)<log^3(n)$$ | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 510,789,298,321,096,700 | TeXit | 02/06/2022 22:18:40 | **Phascolarctos** | 308736672433045504.png | 0 | 1 | null | null | null | null | null | ||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:18:52 | $$log(n)<n$$ | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 510,789,298,321,096,700 | TeXit | 02/06/2022 22:18:55 | **Phascolarctos** | 308736672433045504.png | 0 | 1 | 478,518,656,523,960,300 | scroll忍 | 4820 | scroll忍 | false | ||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 356,908,116,429,897,700 | Kaynex | 02/06/2022 22:20:20 | Well, the contradiction would be to assume that
c√n ≤ log2(n)³
Is true for all n ≥ n0, and prove something obviously false. | 0 | 1 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:21:25 | I've been doing that, but I can't quite seem to get it to line up properly. | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:21:53 | The log and that pesky cube are especially hard to manipulate | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 356,908,116,429,897,700 | Kaynex | 02/06/2022 22:24:16 | You can manipulate this into
2^(³√[c√n]) ≤ n
But this doesn't seem to provide anything obvious | 0 | 1 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 356,908,116,429,897,700 | Kaynex | 02/06/2022 22:25:05 | What class is this? What theorems are you working with? | 0 | 1 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:28:24 | I hadn't actually thought of doing that! I guess I just thought that cube-root would be difficult to manipulate, but maybe I'll try playing around with this. But off the bat, I would agree with you 😛
It is for a Data Structures & Algorithms course, in regards to proving whether a function f(n) is is O(g(n)), Omega(g(n... | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:28:52 | We don't really have many theorems to play with unfortunately. Just algebraic manipulation and proof techniques. | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:36:27 | the bummer is that if you just write it as $$\frac{log_2^3(n)}{\sqrt{n}}$$ and take the limit as n goes to infinity, you get $c<0$, but the conditions of the problem state that $c>0$, so boom! Contradiction. | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 510,789,298,321,096,700 | TeXit | 02/06/2022 22:36:30 | **Phascolarctos** | 308736672433045504.png | 0 | 1 | null | null | null | null | null | ||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:37:06 | but apparently we can't do that 😦 | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:42:42 | Anyways, thanks for your help @Kaynex ! | 1 | 0 | null | null | null | null | null | |||||
536 | help-10 | eb1c7c71db7c44b08b5cf388754e219f | 308,736,672,433,045,500 | Phascolarctos | 02/06/2022 22:42:44 | .close | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:46:23 | IMG_1303.png | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:46:43 | I would like to ask part b plz | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:46:57 | The p=k+1 of the induction | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:47:18 | IMG_1322.jpg | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:47:44 | I used the assumption already | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:48:27 | Now I want to conclude the rightmost term is less than or equals to RHS, with a proper reason | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:49:48 | I wonder if n$\sum_{i=1}^n x_i^{k+1}$ $\sum_{i=1}^n x_i^{k+2} $ is less than or equal to $n^2$$(\sum_{i=1}^n x_i^{k+2} )^2$ | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 510,789,298,321,096,700 | TeXit | 02/06/2022 22:50:26 | **Trenton** | 355287686959071242.png | 0 | 1 | null | null | null | null | null | ||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:50:37 | @Helpers | 1 | 0 | null | null | null | null | null | |||||
537 | help-10 | 010495be9a2146149d440cd55421332e | 355,287,686,959,071,200 | Trenton | 02/06/2022 22:55:11 | .close | 1 | 0 | 356,908,116,429,897,700 | Kaynex | 7520 | Kaynex | false | |||||
538 | help-10 | 3811b9b970ad445ebee9ebdbff97a7b7 | 455,308,025,037,586,400 | LBK | 02/06/2022 23:24:22 | Take a pack of playing cards and discard the picture cards, that is the jack, queen and king, of each suit. Shuffle the cards and deal out 10 of them. Calculate the mean score by adding together the 10 face values and dividing by 10. Put the 10 cards back in the pack and shuffle again. Deal out a further 10 cards and c... | 1 | 0 | null | null | null | null | null | |||||
538 | help-10 | 3811b9b970ad445ebee9ebdbff97a7b7 | 455,308,025,037,586,400 | LBK | 02/06/2022 23:46:35 | @Helpers | 1 | 0 | null | null | null | null | null | |||||
538 | help-10 | 3811b9b970ad445ebee9ebdbff97a7b7 | 118,536,946,527,109,120 | Migillope | 02/06/2022 23:48:57 | so... it actually wants you to run the experiment huh | 0 | 1 | null | null | null | null | null | |||||
538 | help-10 | 3811b9b970ad445ebee9ebdbff97a7b7 | 118,536,946,527,109,120 | Migillope | 02/06/2022 23:49:01 | did you do it? | 0 | 1 | null | null | null | null | null | |||||
539 | help-10 | b7e44726b5644a61a86f102dd97aec53 | 932,850,347,582,570,500 | prnv | 02/07/2022 00:52:02 | unknown.jpeg | 1 | 0 | null | null | null | null | null | |||||
539 | help-10 | b7e44726b5644a61a86f102dd97aec53 | 932,850,347,582,570,500 | prnv | 02/07/2022 00:54:10 | .close | 1 | 0 | null | null | null | null | null | |||||
540 | help-10 | 4dd2077d5af74797a31e98fda5f0cca4 | 649,439,530,104,913,900 | Luczow | 02/07/2022 01:07:50 | Would any one be able to help me out on question 5 please | 1 | 0 | null | null | null | null | null | |||||
540 | help-10 | 4dd2077d5af74797a31e98fda5f0cca4 | 649,439,530,104,913,900 | Luczow | 02/07/2022 01:18:21 | It is a question about polynomials | 1 | 0 | null | null | null | null | null | |||||
540 | help-10 | 4dd2077d5af74797a31e98fda5f0cca4 | 649,439,530,104,913,900 | Luczow | 02/07/2022 01:24:19 | @Helpers | 1 | 0 | null | null | null | null | null | |||||
540 | help-10 | 4dd2077d5af74797a31e98fda5f0cca4 | 118,536,946,527,109,120 | Migillope | 02/07/2022 01:26:31 | Are you deleting the ping and reporting it every few minutes? | 0 | 1 | null | null | null | null | null | |||||
540 | help-10 | 4dd2077d5af74797a31e98fda5f0cca4 | 118,536,946,527,109,120 | Migillope | 02/07/2022 01:26:37 | Don't do that, just be patient | 0 | 1 | null | null | null | null | null | |||||
540 | help-10 | 4dd2077d5af74797a31e98fda5f0cca4 | 649,439,530,104,913,900 | Luczow | 02/07/2022 01:27:12 | Sorry | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 01:49:50 | IMG_0510.png | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 01:50:20 | Shouldn’t it be 4 instead of the 2 | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 01:50:35 | This is my third time asking, please help, I have an examination tomorrow. | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 510,789,298,321,096,700 | TeXit | 02/07/2022 02:01:58 | **Ansh**
```latex
For (ii), $\alpha = 2e^{i \frac{2\pi}{3}}$ I suppose
``` | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:02:47 | where do you it should be 4? | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:04:38 | $\implies (\alpha + \alpha^2)(\alpha - \alpha^2) =\alpha^2 - 8\alpha$ | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 510,789,298,321,096,700 | TeXit | 02/07/2022 02:05:16 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:05:54 | @Ansh_ this is from finding the cube roots from part (a) | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:06:07 | Which has a mod 2 for each of the cube roots | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:06:27 | But then it squared itself so I believe it should be 4e^i(2pi/3) | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:06:36 | huh? | unknown.png | 0 | 1 | null | null | null | null | null | ||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:07:19 | $\alpha= 2e^{i \frac{2\pi}{3}} \implies \alpha^2 = 4e^{i \frac{4\pi}{3}}$ no? | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 510,789,298,321,096,700 | TeXit | 02/07/2022 02:07:21 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:12:15 | Yeah | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:12:34 | So this is wrong right? | 1 | 0 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:19:37 | @Ansh_ | 1 | 0 | 475,966,350,322,303,000 | Brightness | 4998 | Brightness | false | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:20:07 | Sorry, yep it's wrong | 0 | 1 | 510,789,298,321,096,700 | TeXit | 0796 | TeXit | true | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:21:34 | actually | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:21:57 | maybe the arg(alpha) in (0, pi/2) has sth to do here and maybe the alpha you got isn't correct? | 0 | 1 | 706,934,233,012,371,600 | Ansh_ | 3502 | Arya | false | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:24:32 | Aaah got what's going on here lmao | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:24:53 | $\alpha = 2e^{-\frac{2\pi i}{3}}$ here | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 510,789,298,321,096,700 | TeXit | 02/07/2022 02:25:13 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:28:48 | @Brightness | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 706,934,233,012,371,600 | Ansh_ | 02/07/2022 02:29:32 | SO your equation would be: \\ $x^2 - (4e^{-\frac{2\pi i}{3}}) + ( 4e^{-\frac{4\pi i}{3}} - e^{-\frac{2\pi i}{3}}) = 0$ | 0 | 1 | null | null | null | null | null | |||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 510,789,298,321,096,700 | TeXit | 02/07/2022 02:29:56 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
541 | help-10 | 3fc8100e8dc34bc5af32342fe2688908 | 475,966,350,322,303,000 | Brightness | 02/07/2022 02:30:47 | Okay thanks and there is another quite challenging problem | 1 | 0 | null | null | null | null | null |
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