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
student	Yea	13,140	88	[]
student	I’ll find that too now	13,140	89	[]
volunteer	You only need to find the area of one base because the pyramid is on the other one	13,140	90	[]
student	Now I got 5,760	13,140	91	[]
volunteer	Yes, thats right	13,140	92	[]
student	Okay	13,140	93	[]
volunteer	Now, lets find the SA of the pyramid.	13,140	94	[]
student	Don’t you do the base x height and divide that by two?	13,140	95	[]
volunteer	Yes	13,140	96	[]
student	Okay	13,140	97	[]
student	Is it 4,704?	13,140	98	[]
volunteer	How did you get that?	13,140	99	[]
student	I did 1200 then divided that by two which is 600 then I multiplied that by 4 which is 2400 then I added that to 48x48 which is 4,704	13,140	100	[]
volunteer	You actually don't need to add that to 48x48 because it is not on the "surface" so it is not part of the surface area.	13,140	101	[]
volunteer	It would just be 2400	13,140	102	[]
student	Ohh okay	13,140	103	[]
volunteer	5760+2400=?	13,140	104	[]
student	8160	13,140	105	[]
volunteer	Yes, so that is the surface area of the entire figure.	13,140	106	[]
student	Okay	13,140	107	[]
volunteer	Sorry but I only have time for one more problem. Which one would you like to do?	13,140	108	[]
student	Can we maybe do the rectangle one that is question 7?	13,140	109	[]
volunteer	Ok	13,140	110	[]
volunteer	We will need to use pythagorean theorem again for this problem	13,140	111	[]
student	Okay	13,140	112	[]
volunteer	Do you any ideas on where to start?	13,140	113	[]
student	not really	13,140	114	[]
volunteer	Lets try to find all the side lengths of this rectangle	13,140	115	[]
student	Okay	13,140	116	[]
volunteer	Can you use the grid and pythagorean theorem to solve for side AB	13,140	117	[]
student	Yea	13,140	118	[]
student	I got 20?	13,140	119	[]
student	I’m not sure if that’s right though for the side	13,140	120	[]
volunteer	Lets go through this together	13,140	121	[]
student	Okay	13,140	122	[]
volunteer	It looks like you found c^2	13,140	123	[]
volunteer	if c^2=20, then what is c	13,140	124	[]
student	Do you square root 20?	13,140	125	[]
volunteer	yes	13,140	126	[]
volunteer	Can you try finding CB?	13,140	127	[]
student	Is it 4.47	13,140	128	[]
student	Or do you just leave it like square root 20	13,140	129	[]
volunteer	We should keep that as square root of 20 because it is technically more precise.	13,140	130	[]
student	Okay	13,140	131	[]
student	I got the square root of 45	13,140	132	[]
volunteer	Yes, thats correct.	13,140	133	[]
student	Okay	13,140	134	[]
student	Do you multiply them now?	13,140	135	[]
volunteer	Yeah, now we have to multiply them.	13,140	136	[]
student	Okay	13,140	137	[]
volunteer	You can just do 45x20 and put a square root over the answer.	13,140	138	[]
student	It’s 30	13,140	139	[]
volunteer	Great, lets find the perimeter.	13,140	140	[]
student	Okay	13,140	141	[]
student	Do you add them now?	13,140	142	[]
student	Should I do 20 + 45 +20 + 45 then square root it?	13,140	143	[]
volunteer	When adding square roots, whatever is inside the square root needs to be the same.	13,140	144	[]
volunteer	For example square root of 7 + square root of 3 can not be added together.	13,140	145	[]
student	Oh okay	13,140	146	[]
student	So how do we do this?	13,140	147	[]
volunteer	we need to simplify the square roots first	13,140	148	[]
volunteer	Do you know how to do this?	13,140	149	[]
student	No	13,140	150	[]
volunteer	Since square root of 20 = square root of 5 x 2 x 2, we can take out the 2.	13,140	151	[]
student	Why do you take out the two?	13,140	152	[]
volunteer	This is how we simplify square roots. Hopefully we can simplify square root of 45 to have the same root.	13,140	153	[]
volunteer	do you want to try simplifying square root of 45?	13,140	154	[]
student	I’m a little confused	13,140	155	[]
volunteer	Yeah sorry this is a more confusing topic.	13,140	156	[]
volunteer	Let me explain it again.	13,140	157	[]
student	Okay	13,140	158	[]
volunteer	When we have a number inside of a square root such as 20, we can break it up into its factors and take out numbers that have a pair.	13,140	159	[]
volunteer	20 = 5 x 2 x 2	13,140	160	[]
student	Oh	13,140	161	[]
volunteer	Since 2 has a pair, we can take the 2 outside of the square root	13,140	162	[]
volunteer	If we were finding the square root of 4, 4 = 2 x 2, since 2 has a pair, we can take the 2 out of the square root.	13,140	163	[]
volunteer	We are basically just square rooting a certain part of the number and leaving the rest inside.	13,140	164	[]
student	Okay	13,140	165	[]
volunteer	Can you try doing this with the square root of 45	13,140	166	[]
student	I’m sorry I don’t know	13,140	167	[]
volunteer	Ok, lets do this together.	13,140	168	[]
volunteer	What are some factors of 45	13,140	169	[]
student	15x3	13,140	170	[]
volunteer	Can we break up 15 even more?	13,140	171	[]
student	5x3	13,140	172	[]
volunteer	Since 3 has a pair, we can take that out of the square root and leave the 5 inside.	13,140	173	[]
student	So it would be 5 then the square root of 3	13,140	174	[]
student	Oh wait	13,140	175	[]
student	The opposite	13,140	176	[]
volunteer	Yeah	13,140	177	[]
student	3 then the square root of 5	13,140	178	[]
volunteer	Now that we simplified the square roots, lets get back to the problem.	13,140	179	[]
student	Okay	13,140	180	[]
volunteer	Now that the square root is the same, we can just add 2+2+3+3	13,140	181	[]
student	Okay	13,140	182	[]
volunteer	If you put 10 root 5 into a calculator you should get 22.4	13,140	183	[]
volunteer	Ok, I need to go now	13,140	184	[]
student	Yes	13,140	185	[]
student	Okay	13,140	186	[]
volunteer	Great work today!	13,140	187	[]

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.