role stringclasses 2
values | content stringlengths 0 2.1k | session_id int64 10 21.7k | sequence_id int64 0 2.38k | annotations listlengths 0 8 |
|---|---|---|---|---|
student | Got it | 17,208 | 13 | [] |
volunteer | Then add the 1 and -6 | 17,208 | 14 | [] |
volunteer | Then repeat the process | 17,208 | 15 | [] |
volunteer | So now we would do -6 x -5 | 17,208 | 16 | [] |
volunteer | Now the results are at the bottom | 17,208 | 17 | [] |
volunteer | The last term is always the remainder | 17,208 | 18 | [] |
student | oh okioki | 17,208 | 19 | [] |
volunteer | And the other terms are applied to increasing powers of x starting from 0 | 17,208 | 20 | [] |
student | so whts hte fa | 17,208 | 21 | [] |
volunteer | So for -5 it would be -5x^0 which is just -5 | 17,208 | 22 | [] |
volunteer | Then 1x^1 which is x | 17,208 | 23 | [] |
volunteer | If there's any remainder, you would write that over the existent divisor | 17,208 | 24 | [] |
volunteer | Since the remainder is 0 it makes no difference | 17,208 | 25 | [] |
student | Hmm | 17,208 | 26 | [] |
student | then the next numberr?? | 17,208 | 27 | [] |
volunteer | Which number? | 17,208 | 28 | [] |
student | x+2 so i can do 2 --2 | 17,208 | 29 | [] |
volunteer | Ok | 17,208 | 30 | [] |
volunteer | Do you understand this? | 17,208 | 31 | [] |
student | yup | 17,208 | 32 | [] |
volunteer | That's great! | 17,208 | 33 | [] |
volunteer | Do you still want my help for the other questions? | 17,208 | 34 | [] |
student | what why is 2 -2 when in x+2 is x+2 | 17,208 | 35 | [] |
student | you change signs? | 17,208 | 36 | [] |
volunteer | For the denominator (i.e. divisor), you're supposed to solve for the root of x | 17,208 | 37 | [] |
volunteer | So in this case you would do x+2=0, which would give x=-2 | 17,208 | 38 | [] |
student | ohhh I see | 17,208 | 39 | [] |
student | wait | 17,208 | 40 | [] |
student | plz check | 17,208 | 41 | [] |
student | 4-3-12-32 | 17,208 | 42 | [] |
student | if you have any corrections feel free to correct in the thing i did | 17,208 | 43 | [] |
volunteer | Ok | 17,208 | 44 | [] |
volunteer | What's this? | 17,208 | 45 | [] |
student | 4 as we bring the first one down(?) | 17,208 | 46 | [] |
volunteer | Yeah, that's correct, it was just a bit hard to read | 17,208 | 47 | [] |
student | then abt abt -3-12-32 | 17,208 | 48 | [] |
volunteer | The final answer is wrong | 17,208 | 49 | [] |
volunteer | Wait I'll write in red | 17,208 | 50 | [] |
student | Understood | 17,208 | 51 | [] |
student | whast the fa | 17,208 | 52 | [] |
volunteer | Perfect | 17,208 | 53 | [] |
student | plz check | 17,208 | 54 | [] |
volunteer | Sure | 17,208 | 55 | [] |
student | The solution is below the worksheet | 17,208 | 56 | [] |
volunteer | Ok | 17,208 | 57 | [] |
student | can you check | 17,208 | 58 | [] |
volunteer | Yeah | 17,208 | 59 | [] |
volunteer | -3x-2=6 | 17,208 | 60 | [] |
student | okay next | 17,208 | 61 | [] |
student | oki please checkk | 17,208 | 62 | [] |
volunteer | You should put the values in descending order of powers of x | 17,208 | 63 | [] |
volunteer | So 3 should be first not lasr | 17,208 | 64 | [] |
student | oh | 17,208 | 65 | [] |
student | plz check | 17,208 | 66 | [] |
student | can you write it with the variables? | 17,208 | 67 | [] |
volunteer | Yes that's correct | 17,208 | 68 | [] |
student | can you check | 17,208 | 69 | [] |
student | can you do (2x^3 - 9x^2 + 4x + 12) / (2x-3) | 17,208 | 70 | [] |
student | also check for (2x^3 - 9x^2 + 4x + 12) / (2x-3) i did the fa | 17,208 | 71 | [] |
volunteer | Yes it's correct | 17,208 | 72 | [] |
volunteer | 9+2=1 | 17,208 | 73 | [] |
volunteer | 9+2=11 | 17,208 | 74 | [] |
student | also can you do (2x^3 - 9x^2 + 4x + 12) / (2x-3) plz? | 17,208 | 75 | [] |
volunteer | It's the same concept, equate the denominator to 0, solve for x, then apply synthetic division | 17,208 | 76 | [] |
student | i dont get it plz helpp | 17,208 | 77 | [] |
volunteer | For the divisor, you need to solve for x in 2x-3=0 | 17,208 | 78 | [] |
volunteer | And then the rest of the steps are the same | 17,208 | 79 | [] |
student | huh i dont get it tot | 17,208 | 80 | [] |
student | can you solve (2x^3 - 9x^2 + 4x + 12) / (2x-3) plz | 17,208 | 81 | [] |
student | huh | 17,208 | 82 | [] |
volunteer | It's the same steps, you're just multiplying and adding fractions instead of whole numbers | 17,208 | 83 | [] |
student | plz can you do itt?? | 17,208 | 84 | [] |
student | dont get it?? | 17,208 | 85 | [] |
volunteer | I've written it in red | 17,208 | 86 | [] |
volunteer | I'll explain it to you step by step | 17,208 | 87 | [] |
volunteer | First for the divisor, we solve for x in 2x-3=0, which gives x=3/2 | 17,208 | 88 | [] |
student | lets do it together | 17,208 | 89 | [] |
volunteer | I'll solve it in the gray box | 17,208 | 90 | [] |
student | Okay | 17,208 | 91 | [] |
volunteer | Do you understand so far? | 17,208 | 92 | [] |
student | yup | 17,208 | 93 | [] |
volunteer | 3/2 x 3 = 9/2 | 17,208 | 94 | [] |
volunteer | -9+9/2=-9/2 | 17,208 | 95 | [] |
volunteer | 3/2 x -9/2 = -27/4 | 17,208 | 96 | [] |
student | oh okoki | 17,208 | 97 | [] |
volunteer | -4+(-27/4)=-43/4 | 17,208 | 98 | [] |
student | oh okiki | 17,208 | 99 | [] |
volunteer | 3/2 x -43/4 = -129/8 | 17,208 | 100 | [] |
volunteer | 12+(-129/8)=-33/8 | 17,208 | 101 | [] |
volunteer | That's it | 17,208 | 102 | [] |
student | oh okioki | 17,208 | 103 | [] |
student | is it correct?? | 17,208 | 104 | [] |
volunteer | The denominator should be 2x-3 | 17,208 | 105 | [] |
volunteer | For the last term | 17,208 | 106 | [] |
student | huh | 17,208 | 107 | [] |
volunteer | We always use the denominator of the original question | 17,208 | 108 | [] |
student | thank youu | 17,208 | 109 | [] |
volunteer | No problem! | 17,208 | 110 | [] |
volunteer | Happy to help | 17,208 | 111 | [] |
volunteer | hello | 17,194 | 0 | [] |
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