Split (1)

train · 116k rows

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
volunteer	case	15,082	1,024	[]
volunteer	for absolute value	15,082	1,025	[]
volunteer	that satisfies	15,082	1,026	[]
volunteer	this expression	15,082	1,027	[]
volunteer	So, you know, sign fee minus 2 has to be absolute value for it to be taken by log. However, 2 minus sign fee.	15,082	1,028	[]
volunteer	doesn't need absolute value because it's always positive.	15,082	1,029	[]
student	to keep it positive	15,082	1,030	[]
volunteer	Right, cause the range would be from -1 to 3.	15,082	1,031	[]
volunteer	I'm sorry, sorry. The range would be from 1 to 3. That was wrong.	15,082	1,032	[]
volunteer	The, the range of the expression inside the log. There we go.	15,082	1,033	[]
student	yeah, right	15,082	1,034	[]
volunteer	All right, well, that was fun	15,082	1,035	[]
volunteer	JK	15,082	1,036	[]
volunteer	Um.	15,082	1,037	[]
volunteer	but that's interesting	15,082	1,038	[]
volunteer	if you consider this problem	15,082	1,039	[]
volunteer	that whenever you have	15,082	1,040	[]
volunteer	the	15,082	1,041	[]
volunteer	denominator as a polynomial expansion.	15,082	1,042	[]
volunteer	right?	15,082	1,043	[]
volunteer	where it's either the squared term here.	15,082	1,044	[]
volunteer	you will have a solution of the form	15,082	1,045	[]
volunteer	naturallog	15,082	1,046	[]
volunteer	and one over the expression.	15,082	1,047	[]
volunteer	right? These two separate	15,082	1,048	[]
volunteer	um	15,082	1,049	[]
volunteer	forms	15,082	1,050	[]
volunteer	It'll be log of whatever the expression is, and then negative 1 over	15,082	1,051	[]
volunteer	whatever the expression is	15,082	1,052	[]
volunteer	That makes sense	15,082	1,053	[]
student	yes, makes sense	15,082	1,054	[]
volunteer	You know, I didn't think to do this earlier.	15,082	1,055	[]
volunteer	but if you go to the board, let's look at our table.	15,082	1,056	[]
volunteer	specifically the, the blue one.	15,082	1,057	[]
volunteer	um	15,082	1,058	[]
volunteer	if we look at the first form.	15,082	1,059	[]
volunteer	right? We got PQ over X divided by X minus A, X minus B.	15,082	1,060	[]
volunteer	right	15,082	1,061	[]
volunteer	The solution comes, the rewritten expression.	15,082	1,062	[]
volunteer	is a	15,082	1,063	[]
volunteer	improper refraction. I'm sorry, it's a proper fraction, which is a constant divided by each term, right? Each expression.	15,082	1,064	[]
volunteer	So our solution	15,082	1,065	[]
volunteer	um would actually be of the form.	15,082	1,066	[]
volunteer	of two logs	15,082	1,067	[]
volunteer	two separate logs because they cannot be	15,082	1,068	[]
volunteer	Actually, no, I guess they can be um	15,082	1,069	[]
volunteer	divided by. So you can probably combine, you can probably combine the two logs.	15,082	1,070	[]
volunteer	because as long as they have the same base,	15,082	1,071	[]
volunteer	Does that make sense	15,082	1,072	[]
volunteer	I don't feel like I'll lose you	15,082	1,073	[]
volunteer	I don't want, yeah, um, cause then you end up with some constant times log of X minus A.	15,082	1,074	[]
volunteer	plus.	15,082	1,075	[]
volunteer	B times log of x minus B.	15,082	1,076	[]
volunteer	and then depending on, you know, kind of the sign will determine if you combine when you combine the logs, which one's going to be on the	15,082	1,077	[]
volunteer	numerator and denominator and so on and so forth, or if they're both in numerator or both in denominator.	15,082	1,078	[]
volunteer	or if you can just leave them separated as logs.	15,082	1,079	[]
volunteer	I, I don't know why they didn't, um	15,082	1,080	[]
volunteer	where is it	15,082	1,081	[]
volunteer	Uh	15,082	1,082	[]
volunteer	yeah	15,082	1,083	[]
volunteer	So for example, if you look back at example 11,	15,082	1,084	[]
volunteer	they take	15,082	1,085	[]
volunteer	this	15,082	1,086	[]
volunteer	right, and combine it	15,082	1,087	[]
volunteer	Are you there? Hello	15,082	1,088	[]
student	yes, Im here. chat working slow	15,082	1,089	[]
volunteer	Oh, OK. Um, so I was trying to figure out why they	15,082	1,090	[]
volunteer	you know, they combine the log terms here.	15,082	1,091	[]
volunteer	however, they keep them separate here.	15,082	1,092	[]
volunteer	right?	15,082	1,093	[]
volunteer	And I think it's because of the coefficients.	15,082	1,094	[]
volunteer	because the	15,082	1,095	[]
volunteer	add them together is multiplication.	15,082	1,096	[]
volunteer	but you have separate coefficients	15,082	1,097	[]
volunteer	So this would actually to combine them would be log or natural log.	15,082	1,098	[]
volunteer	X minus 25th.	15,082	1,099	[]
volunteer	times	15,082	1,100	[]
volunteer	X minus 310 inside the same log.	15,082	1,101	[]
student	we can't combine log if their cofficient don't match	15,082	1,102	[]
volunteer	I mean, is that true? Uh	15,082	1,103	[]
volunteer	because the coefficient is just the exponent in on the inside.	15,082	1,104	[]
volunteer	For example, um,	15,082	1,105	[]
volunteer	let's go to the board.	15,082	1,106	[]
volunteer	Make some room, cause I could say	15,082	1,107	[]
volunteer	uh,	15,082	1,108	[]
volunteer	2 natural log of X.	15,082	1,109	[]
volunteer	minus	15,082	1,110	[]
volunteer	3 natural log of X.	15,082	1,111	[]
volunteer	Um, let me see, let me get another expression, X minus 2.	15,082	1,112	[]
volunteer	right	15,082	1,113	[]
volunteer	Technically, I could combine these.	15,082	1,114	[]
volunteer	right, cause this would just be natural log of X^2 minus natural log of x minus 2 cubed.	15,082	1,115	[]
volunteer	right, which is then becomes natural log of	15,082	1,116	[]
volunteer	z over x minus 2	15,082	1,117	[]
volunteer	cubs	15,082	1,118	[]
volunteer	right?	15,082	1,119	[]
volunteer	I just think it might be easier to leave at this expression.	15,082	1,120	[]
volunteer	rather than try to	15,082	1,121	[]
volunteer	convolute it this way	15,082	1,122	[]
volunteer	cause they're both equal the same thing.	15,082	1,123	[]

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