So I have less than equal to 3.


I multiply by 4.


Yeah, negative that's less than usual at all.


But I don't like this negative sign with the its just negative ets just makes your inequality really weird and usually when you write it inequality simplified it out. You only have X.


Mhm


So what I would do is I multiply by every one, right?


And the reason why I put it, the reason why I put the sign.


this


because, OK


How about I


I'm gonna show this to you graphically.


right?


Let's say 12 is this point.


Oh wait, I'm sorry. Let's say 12 is this and then 12 is that.


OK


Now, when I plug in positive when I plug in 12, I get 12 less than or equal to 12.


And that's true, right?


Yeah


Mm.


But now, if I plug in, say, -13, I get negative, 13.


Less than equal to 12, which is stating that 13 less than equal to 12, and that's not true at all.


So


this is what the solution would look like.


This is what the solution reach would look like on the draft. My sorry, my shading's bad.


cause it's like online, but


It's OK


it just goes from here to there, some, something like that.


OK. And that's the same as saying X is greater than 12.


that are equal tone of 12.


And look what we started with. We started with negative at less than or equal to 12.


right


Thank you


And we start, we start with negative


OK.


we start with negative ausomyrus wall and look where we are. We have the opposite side. Now,


and we have now have a negative 12 so a positive 12.


Is that like filling in the gaps or no?


Fill in the gap


Does, does it make sense now?


Yeah


OK, perfect


I


All right. This


I'll give you a new practice role. So if I have negative 2X3.


total 20


be dual Math 20


All right, do that prom


and let you know your stuff


OK


Sorry, I should have eased into an easier one. That's my fault.


This


No, it's OK. I can try, I can try this one.


OK, you should get like decently weird answers to heads up.


Yeah, definitely.


OK. I think I'm done.


Alex, tell me the answer.


4.5 less than equal to x, less than or equal to 16.5.


Mm, you forgot the negative sign.


Oh.


But other than that, you're right


uh


So this is where your answer should look like.


So


but yeah.


Are you reading that now


I


Did you get this answer now?


Oh I see. OK.


Uh


No, I think I must have did something wrong.


No, you haven't done like anything. It was just a neg sign.


because this is the answer that you told me.


like,


this


right


That's the answer you got originally.


Yes


Yeah


Yeah. This would be true for negatives.


And what do we say about Ninja vets?


We've


We'd have to


flipped the inequality sign.


yeah.


and then put the negative signs on here.


So now this is what it should look like.


Right


Oh, I got it now. Yeah. Thanks.


All right. You wanna do one more practice or?


One more.


Sure.


So


What.


There you go


All right, mate, let me know once you're done with this one and then write out your answer in this form.


So right ear two value here.


So that work


me


This is the practice problem. Let me know once you're done with this, and then we'll go over it. Does that work? Right in here.


OK


Mm.


So that, that's what I got.


You have to remember the negative sign there.


So this answer works again for negatives.


right


Yeah


So,


I


feel like what you're doing is you're forgetting about this negative sign or you just don't want to deal with it. Am I right?


No, I got it, but since the air was pointing to the left. I thought I maybe I was wrong cause originally


OK, wait, what was, what, what was your original answer?


uh


Just say it


graded in or equal to x, graded and or equal to -16.


Oh


Yeah, that's right. OK. You didn't change my arrows, like, I wanted you to write it in


yeah, I wanted you to write it in this form.


But


that's the form I wanted you to write it in, cause usually put the smaller number on the left, bigger number on the right, but your way was also right


OK


So yeah, that's my bet, but yeah, I see you've uh I say you got fractions


based on boundary qualities and what to do with the negative sign.


Is that all you had


Yeah


Perfect. Um,


yeah, if you want, you can do like, uh there's extra rat there the link I sent you has some extra practice you want over it, and


in case you're worried about negatives, I would say just add a negative to the coefficient of the practice problems that I sent you earlier, and you should be good.


OK, thanks


Yeah, no problem. Um, that's


so just checking. Well, no more questions, right?


All right. See you


Mhm.


Hello


Hi!


How're you!


Good


That's great, what do you need help on?


Is this correct?


Let me see


Hang on just a moment please


ok


Yes, it should be correct, nice job!


Honestly, I've been working on this and this problem that I was doing I was doign with my last coach and he left but I feel a lot more confident with Conic Sections.


I'm sorry to leave after just one problem but I think I'm doing now with hyperbolsa and stuff


I'm sorry:(


That's alright


Have a nice night


and good luck!


Thank you


The questions shows a graph and it asks me to write the equation of the rational function in FACTORED FORM


Those questions are tough, let's do this!


I am trying to upload a picture for you to see


Alright


So, what do you think the first step should be when you're coming at this problem?


identify the 2 vertical asymptotes


Bingo!


x=-1 and x=2


Yep! So that means what about the rational function?


so the demoniator is (x+1)(x-2)


Right on!


this is where im stuck for the x-intercepts


Well, let's think back to our polynomials. Can you tell me how the factored form related to the roots?


im not sure


maybe you can help me step by step becuase I am really lost


Ok, so basically, each factor in the polynomial will indicate an x intercept.


For example: In the polynomial (x-a)(x-b), it will have x-intercepts at x=a and x=b


Is that making sense so far?


a little


is the graph helping?


yes


So, keeping that in mind, no matter how many factors you add like (x-a)(x-b)(x-c)..., the polynomial will always have roots at a, b, c, ...


One more concept we will have to draw from polynomials is the multiplicities of the roots.


In our first equation, both factors have a multiplicity of 1, and this is shown in their behavior as they cross the x axis.


Actually, before we move on to that, make a list of the roots we can see on the rational function graph.


is it -2


Yep, one of them is at x=-2


and 1?


Yep!


There's one more


Now, to go back to multiplicity, differing multiplicities of the factors affect how the function behaves as it crosses the x axis


Close! If I plug in x=-1 into that equation, it outputs 0. However, it should not be defined at x=-1


x-1 squared?


Not quite; you said earlier that the vertical asymptotes should be in the denominator, but that's not how that equation is currently written


f(x)= numerator: (x+2)(x−1)(x−3)squared denominator: (x+1)(x-2)


Now the denominator is fully correct! However, we still need to consider multiplicity in the numerator


it bounces right


Yep, it bounces at x=3


which is even multiplcity


Right


so you square the x-3


Yep! You got it!


and the x+2 and x-1 stay


Yes, that's right.


omg 4 hours later


thank you so much


Of course! There's still one more thing


though


what is that


We have f(x) = A((x+2)(x-1)(x-3)^2)/((x+1)((x-2)), but we still need to solve for that coefficient A


The method for doing this is to take a point given on the graph, plug in y as f(x) and plug in x for all the x's, then to solve for A algebraically.


and that is any point?


or one that is touching


Any point that the graph goes through


I like to use the y intercept because I find it the easiest to work with, but I don't see it on the image.


so 1,0


oh ok so 0,-6


Are you sure (0,-6) is on the graph? From what I see, it looks like the y intercept should be positive


Ok, makes sense


So plug that in and solve for A, then let me know what you get.


so f(x)=2/3 x ((x+2)(x-1)(x-3)^2)/((x+1)((x-2))


solving 6=A(2*-1*(-3)^2)/(1*-2) you get 6=A(-18)/(-2) so 6=9A and A = 6/9 or 2/3


All good!


thank you have a good night


You too!


Howdy !


Hello


So how's your math class going?


Its stressful


Yeah, right!


Got something you wanna work on?


Tangent lines and linear approximations


Do you have a sample problem that you can post?


yes


So you have the function f(x) given


So let's try to draw that a sec just to get an idea of what this question is asking for...


So sorry for all that graphing, but this is what that function looks like (starting with the parent function of sin(x))


Okay so far?


Yes


So these graphs show a whole period, from 0 to 2Pi.


We are interested in the interval [0, Pi]


So let's find the graph of that tangent line shown in black.


Then we plug in 1.5 for x and see what the y value would be.


And we can find the equation of that line by evaluating the first derivative at Pi/2 and then using the point.


And note that Pi/2 = 1.57 so evaluating the line at 1.50 won't be far off.


So first, let's find the point that our tangent line will pass thru.


So there's our point.


Now let's take the first derivative and evaluate that at Pi/2.


Okay with that derivative?


So now we have the slope.


So there's our line.


Okay with that?


yes give me one moment please


okay got it!


Okay, so now we're trying to evaluate that line at x = 1.5 (instead of Pi/2 which is pretty close).


So we insert 1.5 for x and solve for y.


So it's answer D


do you think you could go over the problem in steps? for example: first you solve for F(pi/2)


Yep.


First step is to find the point which I did at top of screen.


I'll highlight with green pencil.


So there, I just plugged in Pi/2 for x.


And got f(x) or y = 1


So our point is (Pi/2, 1)


Let me know when you're good with that, and then we can find the point (to plug into the point-slope format of a line).


I'm good with that!


Sorry - I meant to say let's find the slope next.


So slope of this curve is evaluated with the first derivative.


Which I found to = -2 sin(x)


And to find the excact slope of the tangent there, we just plug in Pi/2


So now we have point and slope, so we plug that into the point-slope format for a line.


Then I just did some simple algebra to simplify and get the final equation in the blue box.


got it!


And finally, we evaluate that line at x = 1.5


Just plug it in and solve for y and we're DUN.


It makes a lot more sense now! thank you so much for your time.


Great - best of luck to you !!!


thank you!!


Hi! I was wondering if you could help me and let me know if I’m solving these problems correctly. It’s a review for an upcoming test. There’s an answer key at the bottom—I won’t look at it lol, but it’ll help you check if I’m on the right track. I am currently on objective 22, #2a, I need a bit help solving that one and the others. Here’s the link to the PDF: https://www.example.com/math_problems.pdf


Hello how is it going


Sure!


Thanks!


Should we start with objective 22, #2a then?


I meant to say 2c


alright


Okay, let me write the function and question for 2c on the board


One sec


Alright


What is the first step, any ideas?


thats what i'm stuck on all i know is y is 0


Indeed it is


g(x) = 0


We just have to solve for x from there


k


Sorry I'm writing with text on the board b/c i dont have a mouse :(


-2|x-1| + 4 = 0


What's the first step for solving for x?


We want x by itself


oh i see


kk


We subtract 4 from both sides first


now what


We need to divide both sides by -2 now


wait


i did -2-4


thats hw i solved it


But, you already subtracted 4 from the left hand side


You don't do it twice


Because remember


s how can i divide it by 2 if its gone


-2|x-1| + 4 is its own seperate side


You divide by -2 to cancel out the coefficient of -2


When you divide it


k


can u show all of this


You get |x-1| = 2


cause im confused


I am on the board


Are you able to see it?


you typed it


i need to actually see it


written


I already said I couldn't do that


oh


I don't have a mouse


Yeah...


Sorry


Hello, how are you?


Good


How about you?


I'm doing well


Working on conic sections


Alright sounds good


nic


Nice


These questions are so long


Yes they are


Let me know if you need help with them :)


is this right?


let me check


aren't the foci su[pposed to ber inside


the


ellipse


yes


oh i see


i did this completly wrong


the vertices are supposed to be


on the y


and the co vertice is on the x


cause a^2 = 25


bcs its bigger


Sorry, I have to go now. Please ask another coach for assistance.


oh okay


Thank you


Hi


Hello


How can I help you today?


I need help with powers of numbers with rational exponents


Are you able to use voice? because I understand better with voice


sure


Um, I uploaded the image.


so


OK


Um, let me take a look at it first, OK?


so tell me how i can help


Explaining in a fun way


So I can understand and solve


Are you there


Yes


so lets get started


Can you open your mic


Hi


Yeah. So,


when you say um that you want me to explain it in a fun way, um.


What do you mean, um?


So like


so I can understand.


OK. Uh, let's see if we can do that.


So


we're working with um


power of the power, right?


Yeah.


OK. So


first, let's see if you know this, um.


let's see if we have 2.


x 2. How do you write it with uh as a power.


due to the power of 2.


right


OK. So, um,


let me see


So when you like multiply, um.


8 times, 8 times, A, that would be like 3, right?


OK.


So when you like now let's talk about this one, so.


when you have


let's say a square.


and then


we raise it


to um 3. The power is, um, 3.


on top, um.


what happens


you multiply


You multiply 2 times, right?


Yeah


OK


So then what happens


A to the power of 8.


Hm


Yeah, so what happens, we're gonna just add those two.


here


So 2 like 2 + 2 + 2 is 6, right?


Yeah, so that's how it works. Now, let's work with the other number like other expressions. Let's see go a little bit harder.


Now, why this happens is just look at this one. Here we have.


8 times 8 times A. We can say we have a power, one on top of them, right? And we don't really write it because it's just redundant.


so what we do is just when we multiply 3 numbers like multiply 3 A's. So what happens, we're gonna


we actually add one plus one plus one.


So that actually gives us 80 power, um, 3.


Mm.


So let's say we do the same thing. Let's so we have.


um


let's do 3 square times


square, but it's gonna be.


that


Um


3 square and 3 square.


34, right?


Yeah


OK, so


let's see how we can derive the formula now.


so we can use it


Um, let's do another example. We can, uh, do this one to deeper 3.


4. It's a little hard.


How would you do it


negative 12.


12 right


You're saying this


negative


Yeah, negative toll. How did


you get there


multiply


the


Must possible


So you multiplied 3 times -4, right?


yep


OK


So now let's talk about if we have


this


would it be


82 3


M, right


Yeah


OK.


Could you solve the examples?


Examples.


Yeah, the one on I sent the board.


Yeah, sure. We can do that.


Mm


Let's look at the


example one, they said like simplified and expressed the answer using positive powers only?


This one?


Yeah.


OK


So we have


just write it here


The first one is


3 square.


and over cube, right?


Did you hide it


I, I think this is the one.


So


so we're gonna multiply 2 times 3, right?


2 times 3, that would be 6.


for um


number B, we do the same thing, 6 x 2 equals 12.


Mhm


When you do number C, so the same thing to cube over 4, so we multiply 3 times 4 is 12-12.


Now we can write this like when we have some.


something like this, we can write it on over A.


OK


We have to do the power 12, so we can write it as one over.


to do the negative, uh, 2 to about 12.


Does it make sense


Yeah


OK, can you uh tell me how to do number D


The example?


Yeah.


OK, um


you do 4 times 3.


which is 12.


Good


Um


do you want to do the activities too?


Yeah, I'll do it.


OK. Now walk me through them. Um, ah.


you're doing great, actually, so let's see if you can do them too.


Well, you do 5 times 2.


Uh-huh


OK


4 times 2.


Mhm


which is -8. Free to the power of -8.


Yeah


and it says use positive, so we make it


one to the power


Yes


1 over free to the power of -8, 8, sorry.


Yeah, 8, yeah


Um


uh-huh.


uh, for C, it's 3 times -2, which is 6. 11 to the powers.


for it's 3 times 5, which is -15, 7 to the power of -15, it becomes one to the power of.


one of


7 to 15.


mhm


You got it


I don't get the


what's the second question? I don't know it.


OK. So


says Anna claims that, so if we do ulti power.


I is it 3 square, I guess, right?


Yeah


Yeah, so it's


um to to the 9, and that's just you can, you know.


calculate 512.


And Ben claims that


OK, so whose claim is correct.


What do you think?


Um


is it Anna


Yes, why is that?


Mhm.


Because free to the power of 2 is 9, so it's 2/9.


It equals 1 equals 512.


OK, uh, why is Ben uh wrong?


because um


to, uh, due to the power of free to the power of 2 is not.


uh, that


um


Yeah, so the thing is here is the parenthesis. You can see that, right?


yeah


So


when we write


this, this actually say is.


2 to 9, right? Because this is 3 square.


But if we ride it like this


that actually means we're going to multiply 3 times 2, that's 6.


OK


So these are two different things, so when you do them, um, obviously make sure you, uh, you know.


look at the parenthesis because if there's no parenthesis, then it's different.


Oh


when you have


this


that actually means


we multiply m times n when you have


this, it means M is M has the power of N.


Yeah.


Yeah


Is that all you need help with today?


OK.


Bye


You did great today


Thank you


Good luck


Hello


Hi


I actually need help with something the total opposite


okay, what is it


Its Geography/Meterology, but Idk if you're knowledgeable in that no coachs are available in that section and it deals with numbers so I kinda requested a math section


so I am not knowledgeable in Geography/Meterology, but if it is with numbers, I can take a look at it


Ok


Give me one sec


Well you have to read a map


So that's the thing that is kinda challenging


Bc you have to know how to read the map


uhh, I am not sure, I think I am not very good at that


Ok, well I will try requesting in Redwood City again then. Thanks in advance!


hi cxan I get some help


can


Of course. What are you working on


iready


do u want me to share my scrren


If you paste something on the whiteboard it works well


I cant do u want me to share my screen


Lets try it


Have you done any of these before? Or should I work through it on the first one?


yes I have


What would you do first then to start?


thanx


Sorry if I wrote it in a confusing way. We multiply the numbers, then find the correct place for the decimal.


ok


Do you want to do another


Sure


Good luck


hi


can u help me with this problem


Hello


Yes I can help you with the problem.


Give me a second to look into it.


thank you sm


it says "https://example.come used at the same time' where it cuts off


Ok cool, this is a work rate problem.


Are you familiar with these types of problems?


a little


i was stuck in the lcd


Ok, so lets start with the large pump and small pump rates. What would be the rate of the large pump?


Yes its rate will be 1 pool / 48 hours


How about the small pump?


Great the rate of that will be 1 pool / 68 hours


Now we need to find if both pumps are used at the same time, how long would it take to empy the pool. Any ideas on how to do this?


using the formula? 1/a + 1/b =1/t


but we wpuld label


1/48 + 1/68 = 1/t?


Great!. Lets focus on simplifying 1/48 + 1/68


So we can solve for t.


im stcuk here


Okay, so to add these


We can multiply 1/48 by (68/68)


And we can multiply 1/68 by (48/48)


To get a common denominator.


can you write it out on the board?


Do you know why we are doing this?


no


wait to get a common denominator?


yes exactly.


We need to get a common denominator to add fractions.


yay


would i have to simplify from here?


Does your answer specify that it wants the answer in a decimal?


its said it before for a previous one but not for this one


but idk if it would make sense for it to not be in decimal for since were talking about time?


Okay so now we need to solve for the time.


ok


How would we solve for the time?


idk


The time is 1 / combined rate


And we already solved for the combine rate


So


so i would divide that?


So when we take 1 over a fraction


All you have to do to simplfiy it is take the reciprocal.


So we simply want to solve 3264/116


oh


OMG


oh my god thank u sm


i understand now


my savior


Happy that I was able to help :)


yay


have a good night/day


You as well :)


Hello


Hey Alex how's it going?


It's going really bad, but thanks for asking.


How is it going


It's going pretty good for me


That's good


Sorry I can't talk


OK


So what can I help you wiht today


Well, please help me with geometry proofs, as we all know, this is the hardest topic ever, right?


So we already know that they gave, they already gave us the option, but they it was given blood interior trans out a property of congruence, and I let me mark them to see if I if I know them. So IJ


and then GH right? and then HG.


I


so


we have something here.


My ankle


Well, an angle, right? But like there's like the transversal line like going through it. I don't think that counts.


Yes I remember doing proofs


They were so hard


I don't like them at all.


I completely agree with that


Alright give me one second to see this and then i'll get back to you


OK, thank you


Ok so we have that HGI equals GHJ


Do you want to mark that on the picture


OK, so HGI?


Yes, please mark that.


Oh, wait, I'll mark it, OK. So, I don't know how to market it. Can you mark it, please?


Yes sure no problem


Do you understand that?


I don't know if you are talking but your mic is off


Oh yeah, I'm a bit wasn't working that


OK, yeah, my my microphone wasn't working.


OK, so basically, yes. So I see that GNH were kind of a congruent, right? So I'm not sure if I'm right, but like, let's see, we have an angle, I don't know, but we see that um there's like separate, so here, right, this is your own triangle, and then the other one, it's a little confusing cause you have the transversal it's like a really hard to like


I don't know, but I, I definitely know that GNH are congruent


and then the black line that I wrote, I'm gonna take it off. I'm not sure if that's right.


I understand what you are saying


OK, let me just erase it. So, so far, am I good? Well,


what do you think?


Right now we're trying to find the theorem, right?


Like the congruent thing or whatever.


We are trying to find that the two triangles are congruent.


Yeah.


The black thing you drew was one of the triangle


What is the other one?


cause the transversal was like, the transversal was in between them, so that's why I erased it. So basically it's JK and then JKJ.


No it says we are trying to prove HIJ and GJI


OK, yeah, we're trying to find that.


Mm, I don't know how to do this. This is really hard.


Hey it's ok we are going to do this okay


So what do we need to find to prove that the triangles are congruent


OK, so basically, we're just trying to like see how.


triangle HIJ is congruent to GJI will, I always like to see in the middle, so the second one, that always shows that I is gonna run to J.


Did your teacher tell you about the requirements for 2 triangles to eb congruent


Okay


SAA


Well, side, angle side.


Yes do you know all of them?


angled side, side.


angled side, ankle.


Those are some I know


Yes I see


Angle side side is tricky because that one does not actually prove that the triangles are congruent


Every single time you try to look for them, I wanna look for them while I'm.


That wasn't background noise, OK. OK, yeah, yeah side, side is really tricky.


Did you put in statements 4 - 8 or were those already in the problem?


No, those are already on the problem.


Ok then we have that information too


Let's mark that in the picture


Woudl you like to try or do you just want to follow along for the first time


Is your mic off?


Yeah, my microphone was off


Um.


I want to follow along because this is really hard.


Ok no problem let's do it together the first time.


So we have HGi equals GIJ angles


Now HJI equals GHJ


HGI sorry


We already marked that


One second let em mark it


Then HJI equals GHJ


We just did that


GHJ equals IJK


Side Ij equals Ij


Yeah


No


Okay so what do we have in the two triangles we want to find


Well we want to have in those two triangles that we find.


We obviously want to find out like what, they shared together so IJ.


Perfect


Yes


IJ is one side


Now we need to find a few more things they share


How can we do that


OK.


OK, yeah, IJ is one side, so what does this tell us and tell us that this is a side, right?


Yes IJ is a side but look


We want to prove that those two triangles are congruent


I think that's


So, um, what you did the peach color.


hm, so.


Side IJ is shared by both triangles right>


it's just a crisscross apple sauce that's making me a little confused, but


um, basically telling us that JK and and then HKI.


Um, I don't know.


Think of it like that


They are just both overlapping


Oh, OK, so they're overlapping. So yeah


H


K J I.


So basically they're overlapping the shapes


I see what you are saying


Now side IJ is part of both triangels right


Yeah, so IJ is basically part of the triangles, right? So this is how, you know, the uh, um, it looks like I is like I don't know, but like it's like an angle I'm not sure, but they're like overlapping with eye.


Yes it is part of both triangles


So we have one side is equal in both triangles correct?


So yes, we have one side, so this is telling us already.


we can make a stripe, right, and then puts.


Great


You get it


Yay! So I would like to think of it, um, G, right? So G would be a, wouldn't it?


So this is telling me it could be *** or ASA.


Oh no.


Wait don't get ahead of yourself


Don't be me, you chief.


We still have to find more angles or sides


We can't assume that theyre equal


How do we find that more angles and sides are equal


Well, how do we find more angles and sides that are equal. Well, basically you can look at.


oh my gosh, OK.


Ha ha ha ha ha, hey, hey, hey. What happened?


OK


Um,


you can basically look at the statements, right?


So GI um GI is a side.


right? HJ


it's like, um, I don't know, like a, like line between them, so I'm not sure.


Mommy's call


It


Also, I have to leave like right now so I could eat. So I'm sorry, but like I'm not going to leave right now, but like when we're done with the problem.


Hello.


Yes, can you hear me


Yes


OK. So we have to find that multiple sides and angles are equal, right?


Yes, so we have the IJZ, so we have one side. Now we need to find like a couple more angles that are equal. So how can we do that? We've already put the givens in the problem, so that's everything we have. So how do you think we can find more angles. If you don't get it, it's completely fine.


Oh, I see that G and then J and then I have some type of marking so that could be an angle, right?


Yeah, but.


So look, alternative interior angles. So HGI and then is congruent to GIJ.


HG.


Yes, that is correct. Uh, one second.


SH


So this would be your ankle.


one


So that would be an angle


Yeah, no, HGI does equal GIJ, but do you see how uh both those angles are not in the triangles we're trying to find. So look.


Yeah


Yeah, HGI is here and GIJ is here. This angle is in this triangle, but you see how we still don't have uh angle in this triangle.


Mhm


Yeah.


So basically HGI isn't in the triangle, so it wouldn't work. It would have to be like in the pink only.


exactly


Yeah, H is in it in the triangle.


Yeah. OK. So, GIJ is in the triangle, right?


Yeah, so GJI is in a triangle.


No, no, GIJ.


Yeah, G I J.


Yes, OK. Sir, according, like they've given us a lot of stuff, so we know that HGI equals GIJ, right?


And we know that uh GHJ equals IGIJH, right?


Mhm.


Yes, so we have


we basically have this angle equals this angle.


OK


and we have this angle equals this angle, right?


Yeah, so that's one like angle.


Yeah, yeah, it


because HHJI, right?


Yeah


because, because we have this angle equals this angle, and we have this angle equals this angle, and it's


So it's angle side angle.


it says wait oh wait, we don't have enough ankles for that yet. See, now we, we know that this angle equals this angle because of this, right?


Mhm


Yeah. So we be because you see how because this angle equals this angle, the single equals this angle, so basically from that we can see how all of these angles are equal, right?


Yeah. So, think about it. We have this side


and we have this angle.


We have this angle, do you get that


Yeah. So


Yes.


So


we have a lot of angles.


yeah, right here, we have this side, this angle.


If this side


That's side, yeah, that's a side. So, yeah


OK


So right now, we just need one more side, one more angle, OK? Something like that.


So we always have to have 3 ankles and 3 sides.


Yes


No, that, that's the thing. There's multiple combinations. There's ASA, SAA, SAS. Did your teacher tell you about that?


Yeah


Yeah, so we just have to like know all the combinations, then we just have to try to find how different angles they equal and


yeah.


OK, so another one would be HJI.


Yeah, H J I, yeah, yeah, that's the one that we have.


Yeah, we have HJI equals GIJ, right?


Yeah, yeah.


Yes, and we have that side. So we're like halfway there. We have 2 of the 3 we need. Now we just need one more thing, you get that?


Yeah. Wouldn't it be like K? No OK, no, no.


See oh, yeah, K is not part of the


both the triangles.


Yeah. This is tricky because you can only have the angles inside.


the trainingles or whatever, the pink and peach.


Yeah, it is really tricky. So now we just have to think a little to try and find uh


another anglers.


OG, wouldn't GJ be considered like a um aside, I'm not sure.


No, GJ is a sassy if it's not an angle, it's a size. OK?


So we could put that


Mhm


No, OK, so we know that these are sides and angles. We just have to find that they're equal, OK. GJ is a% aside. Like if you can find that GJ equals HI, then yeah, we can use that.


Well, isn't that like, isn't like it like the thing they share or something.


OK


or, or G. No, no, no.


OK, you getting just a little confused and that's OK. OK. So, do you understand the difference between ankles and sides?


Yeah


Yes, so this line as a side and these like things or angles, you get that?


Yes. So don't get confused on that. We just need to find that certain sides are equal, like we need to find that this equals this.


Can I just put them like looking at the old one, can I just put them?


What older


Oh this


like these


OK, I see


That


So we have, we need to find that uh


angle side angle


yeah, but right now we only have side and angle. We need to find that other angle. So how do we do that?


I don't know. I just know it's single side angle.


Well, do you have the answer somewhere


Oh no


And how do you know it's an angle side angle?


Because I see that this, right? This is the ankle and then we move on to side and then side and then ankle.


Oh, we have to find that the other angle is equal. We don't know that yet.


So it's not


angle side angle. That's what you're saying.


No


no, no, I'm saying it could be angle side angle. We just don't know yet.


We haven't found that the other angle is equal. See, we have angle side. We don't have that the other angle angle is equal yet.


So I don't know what to do.


Yeah, so,


wait, do you know the names for like the, you know, like vertical angles.


Yeah, vertical angle serum.


Yeah, so because of vertical angles. Let me change the color.


So GK so GK J and then H K I.


Yeah, yeah, yeah. So those are equal, but those are still not part of the triangle, right?


Mm


We have to find.


Let me think for one quick second.


this.


OK


Oh, jump through so hard.


All my friends got 5s on the test. I don't know how.


A 5 on a geometry test.


Yeah, almost everyone in my school got a 5.


5 out of what


5 out of 5


Oh


And you know that I told them that they could like tutor me, right? They were like, I don't know anything. I just prayed. That's good advice, right?


Um, well, when you don't, when it's hard, I guess.


Oh, H, O, OK, I see. OK.


So


yeah, OK. So we have the this angle. Well, let me change the card.


Wait, yeah, this is getting really confusing. Let me change them. Do you want me to erase a lot of this?


Yes, you know, I feel like you have a European accent.


Really?


Oh OK


Does that make it a little better


What do you mean


Like iris a lot of it, so there's less stuff.


Yeah, more organized.


Yeah, OK, so let's just go back, so, so we have this angle, this angle, this side, right?


I see the little three dots, so that's showing us, no


Yeah


those are the angle, so we have 3 angles, so we're missing like one more side, I guess.


No, no, no, no, you are mixing up the triangle. See, look, this angle


is part of this triangle, you get that? Only this thing was part of that time.


We also, yeah, yeah, I get it, I get it.


And then this triangle.


Mhm.


has this angle, you got that?


mhm


Yes. So we have one side of each triangle, and we had one angle of each triangle, you get that?


OK, yeah, yeah


Now


to find another angle, let's do that.


So we know because of like this stuff that's given, uh, uh, yeah


equals HDI, right?


Yeah, and we know that GHJ equals HJI right?


Yes. So that means this angle.


equals this angle, OK? So we know that uh angle I J are equal, right?


Yeah.


So


if


let me change the color. So if this angle equals this angle


and this angle equals this angle and this angle equals this angle. We know that they're all equal, right?


Yeah


Yes, so all of these angles.


are equal, OK?


Great.


Now, what else?


do we need to know?


Mm


about just


This is vertical angletherm, right?


the vertical


angles


would not really help you here.


Since that time you were talking about vertical angles, so that's why I was asking.


Oh, I was, I was just asking if you knew that cause that's also a key concept.


So do you know the answer for this?


No, I did for a second, but then I just forgot. Give me one more second.


You should use AI


I


Hey, look.


I forgot when I was talking to him.


Can you hear me eating


No, right


No, no, it's OK.


But can you hear me


Yeah, I can hear you talking, but I hear like a bunch of like


noises


Dogs.


Do you want chicken.


Oh, this is OK. Yeah, do you see how they've given us a


GI equals HJ.


It was just a kid


Yeah, yeah, uh-huh, yeah.


um


Oh, so I was, you just have to see, oh, I see.


Yeah


Yeah, see, so we've got this side, this side. So we've got


Mhm


uh, we've got angle here.


Sorry.


the side here


Yeah, so yeah


You got angle here. So we got side angle, angle. No, no no no, side.


You see, I


side angle side.


No, OK.


OK, and we got


slide angle side. You got that?


Side, angle, side


Yeah, so because the side angle, side, both triangles are congruent.


Then we want Para. Let's go.


Uh uh


You get it


Yeah, OK. Now, what I prefer my statement number 9. Let's go.


Woohoo.


OK, see, that's the thing, there's a lot more lines, like you haven't, you see all that angle stuff we were doing.


Yeah


You have to like put that into, like, each line.


OK.


to look like


yeah


So there's gonna be a lot more lines than just one.


Stop


What?


I


say it


Hello


Yeah, so there's gonna be a lot more lines than just one.


It's, it's gone


So this is the side angle side, right?


It is side angle, side, but you can't just put that in statement 9.


Why


because we don't, we, we, what we just did, you have to like show that in each statement. Like you have to say, uh,


So basically a statement we got to put the actual like equation or something, right?


OK, wait, um, OK, they gave us a site


gave us the angle


No, I want Cheetos. No, that's, let's go.


Be here


Street


oh.


OK, yeah, you can. OK, they gave us everything, so just in statement 9, you would just have to put that the triangles are congruent.


OK, so in semen I, I will put, wait.


OK, that's lovely.


Why is she? You know what? I just want to do, play Roblox. Come on.


OK


OK, so number 9 is side angle side. Can you hear me?


Yeah, I can hear you


Do you know where Sarah is?


Yes, but we're doing geometry right now.


Do you know who John is


Uh no


So, are you good or?


What is number 9


Like the statement that's, I'm sorry, but I don't care with the statements I filling them in, that's like the death of me.


So we're trying to find the triangles are equal, so the statements that the triangles are equal, and the reasons why.


So what would I feel?


Cause, cause like, I know how to do this. I felt like not blame the man, but like I know how to get that answer, but like when it comes to feeling and like the stamens, you see the letters letters like, uh, I can't like I literally forgot how when you how how SAS statement looks like.


You just gotta go step, it's just.


Wait, wait, wait, don't we put HIJ congruent to G J I S S A S.


Exactly, yeah, triangle.


Yes, you see, I got it. Don't play, OK.


Prince anglo


I'm using that big brain. Wait.


OK. OK, so I basically put, put, um Triangle HIJ come to GJI.


Yup


Let's see if you're right


Wow


yep


OK. Thank you very much. You see, I learned I'm gonna, I'm gonna eat my dinner, watch a movie, and then come in to tutoring back.


OK


Bye


Yeah.


Thank you


Hi Alex


Sorry i was doing something


How can I help you today?


Can you hold on


Im lagging


sure


Now i got it


ok


I dont think she is correct


Or


let me work this a little. Why don't you think she is correct


We have to use gcf


And she has ()


We just have to say yes or no


I think she is correct.  Did you try multiplying out what she did?


Idk what to do


Im in 6th grade


Wow, and you are taking Algegra?


*algebra


This is extra work


Good.


Back


On


Topic


So to multiply it out, you would take 2x^7y^2 * 6y^7  first


2 and 6 are out


x^7 * 1 = x^7


You could also write this as


1 and x^6 are gone


x^7 * x^0 since x^0 = 1


I mean x^7


Sorry you are right


lets just move on to the the y's


Ok


And is no correct


I see your point.


The x^7 should be there and it is not


So you are right, Sarah is incorrect


Yea


Wait


I take that back


The answer key says that Sarah is not right


when you multiply out the terms, the x^7 will show up


The () was not there in the first place


So therefore I believe that she is not correct


Do you understand what I just did?


Y


E


A


So


So the first term checks out, as does the second, right?


Yes


So Sarah is right


The answer key says that she did not use gcf so she is not correct


And she did not fully factor it


That may be.  Let's look at how else she could have factored it


It need to go in 10 minutes


OK


So can we end the session


She could have taken out another 2


Ok


So yes, she was incorrect.


Bye


Sorry this took so long. bye


Hi


I am just here to explore sorry about that


Hello! No worries :)


Feel free to ask me any questions!


Ok by


Thanks i have to go


I will ask you questions later


Alrighty! Welcome to the platform, Sarah!


Have a fabulous rest of your day/night!


Ok you too hope you have a great weekend


Thank you :)


Your welcome


Ok by have to go


Bye!


By


Hello


Hey, how can i help you today?


Hi


I see that you want to learn something new!


Me and my classmate are trying Tutor Provider


So we are testing it


I see


Well, i am a tutor and if you have a question, ask me!


So am trying to learn something new


If you want to learn something new, I could teach you


Okay so, i will teach you?


Okay


Have you done much practice with Pythagoras theorem?


No


Okay, so would you like to learn that?


Are you there?


Yes


Okay!


So basically, the first thing to know is that we can see Pythagoras theorem in right angled triangles


So you can see on the whiteboard, that this angle is 90 degrees


Okay my teacher told me to explore this app but i got to go now.


I hope you learn more about this in the future!


Bye thank you for your time!


Have a good day!


Hey I am so sorry for ending the session!


no problem


It was my little brother!


I need help with a upcoming quiz on Trigonometry!?


is there a specific thing you want to work on in trigonometry


like word problems etc


Yes there is


okay, can you upload it onto here?


so i can help you work on it


Ok


that's correct nice job


are you solving for theta?


yes


or are you just trying to write the trig functions?


okay


do you need some help?


um if you want to help me


okay so to find the angle


we don't have to write a bunch of trig functions


we just need one


no these are apart of the problem


oh okay


so sin theta = 4/5


cos theta =  3/5


and you want to find the angle using those yes?


there is one more theta


sorry I have to go bye


Hi Emily, how can I help you?


hi i need some help on a few problems that’ll help my study for tmrws test


Can you post the first problem?


yes it’s loading


What does the circled root mean?


that’s a zero right?


but i’m not sure about the ^ symbol


Yes,, but the notation is confusing.


yeah


How many roots are indicated?  Remember that complex roots always come in pairs.


I count 6 roots.  x=3 with multiplicity of 2


x=3i means that x = -3i is also a root


oh yeaaahhh


x = 4-i means that x = 4+i is also a root


right


degree is at least 6?


so least possible degree is 6?


yes


How about 16?


I think so.


okayy


now these two questions


What does an odd function mean?


an odd function is when it’s symmetric with respect to the origin so basically in (x,y) both would turn into the opposite and become (-x, -y)


Close.  for odd functions, f(-x) = -f(x)


f(-x)=-f(x)


ohh


okay


In both problems f(negative numbers) are given.  The question is sketch f(positive numbers)


so the points would be the same just opposite?


Yes!!


like (-4,2) would be (4, -2)


yayy okay then i got that


and for even functions if it had a point of (-2, 5) would it be (2,5) on the positive side?


Yes!!!!


yayyy


then i think im gooddd


Can you sketch either 8 or 9?


it’s hard to draw on this device so i’m not sure


I'll draw 8, maybe


okay suree


wait also i want to make sure my understanding on this type of problem where it asks to find the complex zeros of the polynomial and the polynomial is like f(x) = x^4 + 5x^2 -36


would i use the substitute method?


Yes


okayy


Can you factor x^2+5x-36?


the final zeros would be -3i, 3i, 2, and 2


Good. How about the original problem?


-3i, 3i, 2 (multiplicity 2)


wait


-2 and 2


x= 3, -3, 2i, -2i


but isn’t it a negative 9 and positive 4?


(x^2-9)=0 gives x^2 = 9, x=3,-3


ohhh


i thought it turned into x^2 = -9 and x^2 = 4


x^2+4=0 gives x^2 = -4


right


OK?


yes got it


that’s all


thank you so much


Good work, Emily


byee


Hello! I have a question about this triangle, can it be proven by the Side Angle Side theorem?


let's see it


I think it's more AAS


Ah, I see


Yeah


I marked the sides in red, who know wheter they are congruent


ooh


I see, thank you!


yes


no prob


HI


Hi Maya! I am David, how can I help you today?


im bout to display it on the board


Alright!


its there


i tried to subtract the numbers but it wasnt right


I see! Alright, so at the bottom it is showing the hundreds, tenths, and ones as separate, are you currently learning subtracting separately?


i think soo


Alright! So lets start with the ones


but its hard to add since i cant write it on the paper


Yes that I can imagine would be difficult. You can use the whiteboard space :)


can you do it


like write it down


Yes :)


so our first step is the ones. What would we do here when 6 is bigger than 3?


yes


So we would borrow from the tens, right?


Alright. So 13-6=_?


oh mb


You're alright! That is correct


now 6-2?


Or you could think of it as 60 - 20


which is 40


Yes! And now we have 5-3 500-300


Yes! total?


Yes!


Is there any other problem I can help you with? Or any questions you have about this problem?


Alright, I am ready when you are! :)


its on there


Alright so, let's start the problem ourselves and see if we get the same steps.


The first thing we want to do is break 436 to hundreds tens and ones. Can you do that?


400 +  30 + 6


Yes! So let's look at Sarah's work, does it match?


no


So that would be her first error!


She has hundreds in the tens spot. If we 400+130+6 what do we get?


Yes, so by seeing that we can see that she set up the problem wrong. :)


nah


Are there any other problems you would like help with?


thats not the right answer


That should be correct.


Did it mark it wrong?


yes


Oh! I apologize I didn't see the bottom question in the first picture.


We still have to find the correct difference!


So we would start with the ones


Yes and now the tens


Hi Maya, do you need a five minute break or a moment to try and solve the problem?


Alright Maya, if you would still like help, please make another session! I am glad I was able to help you today.


Hello, I’d like to redo the problems I got wrong on my test in the same order. I understand why they were incorrect and have already corrected them, but I want to practice similar problems with different numbers to make sure I fully understand them before my retake tomorrow. Here is the link to the PDF: https://example.com/document.pdf


Hello Ben! How are you doing today?


Unfortunately, we can not use other links. Could you please paste the problems?


I'm great, thanks for asking! ^^^


You can actually!


Here's the proof


Give me a second Ben.


Ok!


Are you able to access it?


I am trying yo


to


I am unable to access your link. Could you paste the problems please?


hi


how are you doing?


scale drawlings


good hbu


ah ok


good thx


i was trying to predict wht you were gon' say lol


did you get it right?


bc most coaches just ask what i need they barely ask how i am


no lol


I'm a real person, not AI :)


ok scale drawlings


ok


yea


bc you said


can you put the question on the board?


'how are you' im supposed to say goon pal


i said good*


and i said 'scale drawlings' lol


4wait brb


yes typing is not my strong suit either


do you have a question you're working on?


are you still there?


yea wait b lil


ok me back


i was grabbin tires for my dad


tires? for his car?


yes


he has to switch his other tires for his workcar


ah ok, well nice that you're helping him


you ready for your question now?


yea now i need help


yessir


ok can you put it on the board?


is a pic ok?


sure


wrong one srry


yea this is horriable quality lol


maybe use a brighter light?


or the light on your phone?


no it's the camera and i don't have a phone


oh ok,


that's it ignore the scribble lol


can you write in the numbers


on the triangles


wdym like re-write them?


is this one 12?


some of them I can read but barely


my eyes are not that great


ok


thats a p btw in the left top lol


ok thx


just imagine all those number's with m after them lol


ok m for meters right


yea


so we are finding PN/P'N' ?


yea i think


yessir


ok so they gave us the first part


i actually did that lol


ok great


so now we have to simplify 18/12


do you know how to do that?


how i do dat?


no


i think


maybe start off and it'll come back to me


ok we have to think of a number that can divide BOTH the 18 and the 12


ok it came back


can you think of one?


yes but the number needs to be able to divide the 12 also


so not the 9


we 6 might work


yes


yea


try dividing the top and the bottom by the 6


wdym?


top of? bottom of?


what is 18/6 ?


srry if that sounded rude tw


btw*


that's ok


now divide the 12 by 6


i had put 3 there lol


but i erased it


good job, you had it right


now what goes on the bottom?


yes! excellent


one i did before this was all 2/3 lol


so the ratio of PN/P'N' is 3/2


and it was similar


oh ok


now PO/p'o'


yes, you start


what is PO = ?


12 and 8 right?


yes


so write in 12/8


bc you can also do 2x4


and get 12 and 8 right?


slow down a bit


oh my bad


write in 12/8 first


great


now we simplify


12 and 8 can be divided by 4


wait i wrote all this wanna see what i ut and just see if i had gotton it right cause so far it'sthe same thing


let's just carry on and finish here


ok


we divide the top and bottom by 4


what is 12/4 ?


yes!


write it in


so they have the same ratio right?


yessir


great!


but now for NO and N'O'


let's see if they have the same ratio


what is NO N'O' ?


12 and 10


yes


thanks


so we can divide the 12 by 2


yes we get 6


can we divide 10 by 2 ?


yes 5


write those numbers in


2 and 5


6 and 5


so 6/5 is not the same as 3/2


they would be not similar


right?


right! the triangles are not similar


ALL 3 sides need to have the same ratio ro be similar


oooh


ok


good work!


thxs


good teaching


thanks


have a good day!


you too


bye


HI Jamie! What do you need help with? Do you have a specific problem that you would like me to help you with?


No but I am having trouble understanding how to multiply integers:(


Okay! Do you have an example problem or would you like me to come up with one?


I do not have one on the top of my head can you please make one up :)


yes! i will put some example problems on the white board and you can try to solve them.


Ok


Just let me know if you are stuck


Okay lets start with these


I don’t know the last one :(


Thats okay, do you know what 8 x 8 is?


…no


Its 64!


Oh


Thats okay,  that was a tricky one


But all the rest you got right! good job


Lets do some more


BRB got to go to the bathroom wnt take long


Okay


IM BACK :)


welcome back :-)


Is 8x4 32???


yes! good job


Thanks


Is the first one for (-3)(-7) does that say +29?


Yea


I messed up on the writing


Oh okay, that one should be 21


Oh…


But you did get the positive sign for the integer! So you do understand that part


Good job! The rest are right


Ok :-)


So you do understand that a negative times a negative is always positive, and a positive times a negative is always going to be negative?


I think so


Great! I think I would just suggest you to work on memorizing your multiplication tables! That would help you the most since you seem to understand the rules for multiplying integers! Its just the number that is where it seems to go wrong! :-)


Ok will do ;-)


Did you want me to help with anything else? like with multiplication tables? or did you want to just practice that on your own?


I think im good for now thx


Okay! No problem! I hope I could help


THX FOR DA HELP


Do you feel more confident in multiplying integers now?


Great i am happy to hear!! And No problem it was my pleasure! :D


BYE


Bye bye!


I need help understanding what to do in questions 18 and 19


Hi Jamie, let's explore this.


I dont get why they put the angle numbers


degree*


Do you want to start with #18 or 19?


18 please


The angle is described as cos = root 5/3 in quadrant II


In quadrant II, what are the signs of cos and sin?


neg cos pos sin


Yes!! and cos is defined as adjacent/hyp, yes?


yes


So cos is neg and cos - adj/hyp?


cos = adj/hyp


Question is sin =?


opposite/hyp


What do we know about the relationship between sin and cos?


they both have the hyp in the denominator


Yes, what about numerically?


i dont know


does it have to do with the unit circle


yes, How about sin^2 + cos^2?


Have you learned that?


no i havent


sin^2+cos^2 = 1?


For any angle


my teacher did it that way


i havent seen the way you did it


What do you mean by "did it that way"?


she hasnt taught us that method


I just saw what you put on the white board....


that was her key


That's fine,  it works as well.


We know to take the positive square root of 4 because we know that the sine is positive.


yes


Do you understand what your teacher did to find sin theta?


yes she used soh and cah


since there wasnt an opposite she used what was given to the pyth theorem and solved


Yes and in the second quadrant.


can I get help on 19 now please


Yes


We are given tan theta = 3 = sin/cos


Shall we sketch that?


In which quadrant is the angle?


quadrant ll


Angle is given as greater 180 degrees and less 270 degrees.


are both signs less than?


180 less than theta and theta less 270 degrees


That would be read "180 is less than theta is less 270 degrees


180 degrees is less than theta, then theta is greater 180 degrees


yeah, no?


I don't understant "yeah, no?"


understand


like I agree but Im confused


Do you see why the angle is as I've drawn in the diagram


oooh nevermind I get it now


Great!!


Can you label sin and cos?


it quadrant 3 180 degrees is less than the angle we want to find but its also less 270 degrees


Sorry, I can't draw well  with a mouse. lol


its okay i can read it


to get tan it would need to be sin/cos


so the final answer is three


would it be 3/1?


so, if tan = 3, sin =3 and cos = 1


so secant would just be 1?


What is sec?


sec = 1/?


right?


Is sec = 1/cos?


no but sec is the reciprocal of cos so since cos is one wouldnt the reciprocal be 1?


sec = 1/cos


cos = -1


In quadrant 3


sec = -1


so are we looking at the points (-1,0) and (0,-1)?


Yes, that was given


because I thought that the reciprocal was not the opposite


ooh okay


The first point (-1,0)


The second point is (0,-1)


I added a negative sign to our sketch, sorry I forgot that part initially.


oh thats okay


Yes, your coordinates (-1,0) are correct


So, what is sec theta?


it would be -1


right?


Yes!!!


yay


okay I understand this now thank you


have a nice evening


Does all this make more sense now?


haii :)


hi Alex, how can i help you?


can u give me a sec


yah for sure


oki ill be back


oki sorry im back


no problem


oki so im doing verbal algebraic expressions


ive done this a lot but i cant seem to get it I_I


no problem we both together figure it out


so should i explain it Alex


uh i got some questions to do together


oh that's awesome


oki


so lets dive into it


so lets take x as small number and y as greater number


oki sounds good


so i want you to read the question two times


it actually helps to understand the question better


so you read it?


i mean the question


yea


ok so basically what i wrote is assumed x variable as smaller number


and what variable do you want to name our greater number


maybe y?


yeah sure


so in the question they said  greater of two numbers is 7 less than 6 times the smaller number


so which means y is 6x-7


so let's take this example it will improve intuition


a person y is = 6times of person x -7


oki


now i want you to try the next line and break it into equation


oki i can try


how would i start it srry im confused on this part?


no problem its okay lets do it together


so the question said that their sum is 42


alright which means if we add both of our smaller number x and greater number y it will be equal to 42


should i give a example if you like to ?


sure


suppose you have 5 candies and i have 7 candies if we both combine them together then how many candies we both will have


together


we would have


fantastic


now try this on the question


i think we should decode it together if you want to


am i allowed to speak if thats ok with you


So that caution says there sum is 42.


oki


and by their it means X and Y, because we assume that x and y


are are numbers, OK? So like uh the questioner said X


questioner's sum is for you, which means x + y is equal to 42.


oh oki


So now we have two equations. All right? The first is, why is equal to 6 x minus 7.


So now we can substitute Y value in the RS second equation, I want to try it.


I


Oh, sorry, I want you to try it by yourself.


and, if you will not be able to solve, we are here together. That's why this session is.


is that right?


yes it' correct


great job


now i want you to try to solve this and find value of x


would i add the like terms first or solve whats in the parentheses first?


you should solve parentheses first


then go for like terms


so 6x is a variable and 7 is a number we cannot actually directly  subtract them


in situation like this


we add/subtract the similar type things for example


so now try our question equation again


firstly go through the example


oki


awesome


yes it is


correct


so our smaller number is 7


now i want you to take this value into first equation y=6x-7


So, yes, that's all right. Uh, you got it.


So, actually, our question is finished, but I want you to think something else. Do you see the pattern like if we put the x value


in a sac equation and y value in the second equation, we will get 42, which is given in a question, which means somehow our coercion actually gives us answer. I mean, it already tells us the answer, but we just, it just rounds it so uh which uh confuse us. When I was young, I was also used to struggle with um Algebraic I actually.


oh oki


so what's our next question''


So whenever you face a question like this, OK? The first thing you have to do is


well im done for today on my math :)


you have to read the question, OK? After reading the question, you have to break the caution in different parts, you have to like the first lens is a is 3 times less than why you have to think it, OK, so axis 3 times less than by, OK? And in the sacculent, you have to read it and like they will give you two equations. You have to go through the question like 2 or 3 times, OK? And


you have to break it


It will make your question easy.


wow awesome


i hope you like it and sorry for aking this much time of yours


taking*


oh its fine :)


yah nice to meet you bye and you can always request for another session


oki bye


hi! how can i help you?


Hi, I have a quiz tomorrow on chain rule and implicit differentiation. I sort of understand it but definitely need to practice more.


do you have any practice problems?


i can help you through them


ok


one sec


ok. give them a try. i'll correct you where i see mistakes, or let me know if you get stuck.


im not sure how to determine the inside and outside


i can only see y' = 6x


are you writing more? i think it's going off the screen to the right.


ok hang on


can you see that?


no


its y = 3x^2 +2x +1


and i know for chain rule you have to find the inner and outer parts of the function and multiply their derivatives but im not sure how to identify it


for this one, i don't think you need the chain rule


so its just 6x +2?


correct


we could try the next one on your list.


Differentiate implicitly: x² + y² = 25


ok perfect


i dont know implicit differentiation well, could you explain it


so you know that y = f(x), but you can't write y in terms of x, like here


can you see my writing?


yes


so we differentiate both sides, keeping in mind that y = f(x)


so subtract x^2?


we can, but we don't need to


ok


what's the derivative of 25 (a constant)?


it gets canceled


ohhh


so is it 2x +2y but then you have to divide by 2


2x + 2y * y' = 0


divide by 2


we then get x + y * y' = 0


wait why y’


then solve for y' and we get


oh cause thats dy/dx?


correct


y' = dy/dx


so when we solve for y' we get:

y' = -x/y


and that's the solution


why is it over y


we had x + y * y' = 0


so we then solve for y'


rightt


does that make sense?


a little more yeah


could we do this one


Find dy/dx: e^y = y² + x²


yep. you want to give it a try? write just below where i wrote on the board, and i'll be able to see it think


perfect, i can see it


im assuming theres a difference between e^y and e^x so does that mean the derivative wont be e^y


it will be, but remember that y = f(x), so this is where you apply the chain rule:

d/dx(e^f(x)) = e^f(x) * f'(x)


not entirely


so d/dx(e^y) = e^y * y'


like the chain rule application


like how are we finding y'


we'll solve for it after we take the derivative of both sides


cause isnt that just e^y * e^y


it is not


wait its still d /dx


y' = dy/dx


so its d/dx(e^y) = 2y + 2x


almost, but not quite:

d/dx(e^y) = 2y * y' + 2x


is that the same as d/dx(e^y) = (2y)d/dx + 2x


when we differentiate the left side we get

e^y * y' = 2y * y' + 2x


almost. it's the same thing as (i added a y) :

d/dx(e^y) = (2y) * dy/dx + 2x


what do we do from here


when you differentiate the left side we get:

e^y * y' = 2y * y' + 2x


now we solve for y'


so 2y + 2x / e^y


when we solve for y' we get:

y' = 2x / (e^y - 2y)


and that's the answer


because u subtract both differentiated values so theyre together


exactly


and divide by those


yep


im started to get it now


you pull out the y' that's getting multiplied by both terms


ah gotcha


thank you for your help


no problem! good luck on your test/quiz!


hi


hello, I have 4 homework questions I have to solve


ok I can help you


I really need a step by step explanation though because I have a hard time following along with math


the first question is: Find the sum of the geometric series. a_1 = 43a_1 = 2752r = 4


the possible answers 687, 2795, 688, 2796


is it possible you could write it on the board please?


cause its kinda hard to understand in on text


ok so I need a second to think about, and then I can help


okayy


wait how is that a geometrix sequence?


I have no clue, thats why I was confused


I'm sorry, but to be honest I have never encountered this type of problem


I think you should try and find somone else


oh wait


I think I might be able to do it


so first lets look at 43 a_4 = 2752


so the equation for the geometric is a_1 * r ^ n-1


I got the geometric


sequence


did you learn anything about finding the sum


wait I remember the equation


ok so the equation is


lemme write it on the board


I'm really lost um


wait one second I'll solve it then I'll help you understand it


okay


I'm sorry I don't think I can explain as I might have gaps in my knowledge, I think its better if you find another tutor that can help you. Sorry


okay that's alright, thank you anyways


bye


Hi, i think the previous session got disconnected...


Yea


So are you able to use your mic


yeah sure, but let me know if there's background noise...


Hello


I think I can hear you


You can


OK, yeah


So basically, I have a quiz on triangle chores here and I have like 9 questions if you want to go over them.


OK, that's fine


OK, let me pull up the first one.


here.


You know, I love you so.


Yeah, so this is the previous question, right?


Yes that one


OK, so um


one second


Yeah


So first, I had asked you to list down the information that it gives. So first, you see that


uh line segment EF is bisected segment EF bisects, deviate DB


So what that tells us is that uh DM is equal to bm here.


because it's been bisected and


Yeah.


So that's one piece of information that we can use.


Then as you've said before, yes, you're right, these two are vertically opposite angles and hence they're equal.


I also know that since um


they bisect each other.


the DB is


bisex


like, since EF by 6 DBMM is the midpoint.


M is the midpoint of DB right?


Is this a live quiz or a practice assignment btw?


Um, this is a


this is like a practice.


for my tests. That's what my teacher gave me.


Yeah, OK


Uh, yeah, so,


Mhm


Mhm.


uh we got DM is equal to bm because it's the M is a quite of DB and we know that EMB


and DMF are vertically opposite and hence equal.


Now, what other information can we find from the


uh, you know, diagram that they've given.


Um


Note that CD and AB are parallel, so, and the uh BD is a transversal going through them.


So


so this would be


ASA


what, what, what angles are equal?


for what


You said ASA, right? So uh we, we only have um one angle and one side. So what we need one more angle.


and that angle would be angle D and V.


This one, right?


because it alternate interior angle.


OK


OK. Now, another thing, so we, we've confirmed that ASA is right.


Now


look at, look at EF. That is also a transversal going between Ab and CD.


So similarly,


these two also become alternate interior angles.


BEM and DSM.


Mhm. OK


Yeah, so because of that


I could also


prove the same congruency through AAS by taking the angles DFM


BEM, and the vertically opposite angles and the side, select.


So like if I took the red, the red angle, the black angle, and the two equal sides.


then this could also be proved through AS.


and through ASA


depending on what angles you would choose.


Do you have any questions


Um


wait, so they're asking us for triangle EMB.


and let's see EMD and


FMB


Yeah.


These ones


but


Sorry


so


if we're talking about those two triangles. Why, why are we talking about the whole?


What we want to prove them congruent, right?


So they're asking what triangle congruence method will prove that.


So


with


did you remove the question?


I think the whiteboard lagged out. Can you raise the question again, please?


Right, um, yeah, it's like.


OK


I, I hold it up again.


Yeah, got it


Yeah, so we're proving basically these two triangles.


right?


Mhm


So we want to prove them congruent, so to prove two triangles congruent will generally check if their angles are equal somehow or the sides are equal right?


So.


So, um, what I'm gonna do, what, what we did is like, like you pointed out, the first, the vertically opposite angles.


I


I can say that these two are vertically opposite, and I'm very good.


So and then


Right, let me see


EM is equal to BM.


I know BEM


is equal to BEM is equal to angle dfn


because


of alternate interior angles.


alternate interior angle property, I can say.


These two are equal. And then finally,


solve by fatoring


ok


do you know how to do this?


kinda


I'm refreshing for a test


so we need to find two numbers that multiply to -18 and add to 3


can you think of two?


6 and -3


yes


so know it's factored, just solve each equation in the brackets for x


perfect.


this is also solve by factoring


so this is pretty much the same thing


just first you multiply a and c


so what is 3 x -5


so now find two numbers that multiply to -15 and add to 2


5 and -3


good


so now


rewrite the equation


but instead of putting 2x


put 3x^2 + 5x -3x -5


same thing, but change the 2x to those two numbers


and now factor these two by fulling out the greatest common factor (GCF)


solve by factoring


ok so this is a difference of squares


both numbers in the equation are perfect squares


because the 25 is negative the 5s and + and -


and then just solve for x


okay that's it thank you


bye


Hello!


Hi! How can I help you today?


id like some help studying for a quiz i have on product rule and quotient rule for AB calc


Happy to help! Would you like to go over the product rule and quotient rule in general or do you have a specific problem you want to work on?


yea i have a ton of questions for us to look over


Perfect! Let's get started them


then*


is that first one right


yes, do you know why that is or do you want me to provide an example?


no I think i understand why, just wanted to check


okay!


and then the next one is true


that is correct!


and then true again?


yes!


and then false


it's just the sum rule for c


right


and d is false i think


yes you're right that d is false


cause product rule is needed


correct! and product rule is (fg)′=f′g+fg′.


skipping e cause we dont need it for this


sounds good


and for f, if a function is differentiable its continuous and if a function is continuous it might not be differentiable


that is exactly correct!


sweet


wait this is just asking me to make up a counter example


exactly! show that (fg)' does NOT equal f'g'


and that (f/g)' does not equal f'/g'


so i should find the answer the incorrect way, and then show the correct way?


yeah! use the correct product rule and use the incorrect product rule and then just say that they don't equal each other


like that?


yes! and you should probably make a statement on the test, comparing the two and saying that they are not equal which means that (fg)' does not equal f'g'


as well as label which is the true derivative and which uses the incorrect expression


you did great!


thanks!


i think ill just skip the quotient rule one because its basically the same thing


of course! if you feel confident about it, we can move on!


wait how come its asking me to use 2 different methods for each for a


because you can rearrange the equation so that different methods can work


ohhh


right let me try that


wait im unfamiliar with the constant multiple rule


sure! let me put it up for you


you're likely just not familiar with the name


oh isnt that just power rule


subtract by one, bring it down


technically it uses the power rule but it basically just means seperating the constant from the function then applying the power rule


i see it


yup!


i think i is the best option here


cause i can already tell its way more work for the other 2


you did i correctly! and that is true however, I think that the question is asking you to use both rather than pick the best one


ohh


but if you feel confident about the rest we can most definitely move on. let me know


i think im good!


okay!


b part i is also correct!


b part ii is correct!


c part i is correct!


well done!


how does that look


i would start off with a general expression so  h(x)=f(x)g(x) then do h'(x) = ... but other than that it looks good!


does that make sense?


oh yea makes sense


great!


we dont have to do all of these


just the hard ones


any of these catch your attention as hard


i would say d


and likely b


yea i saw that too


alright ill try them first


okay!


have you been taught u-substitution at all?


what do you mean


it's a method of simplifying equations but it is not necessary to solve these. i was just curious if you were taught it or not


it looks something like this, if you haven't been taught it, you can definitely ignore it


hm, it's not letting me upload


ill write it out


ill continue solving


yes! please do


if u = g(x) then du/dx = g'(x) and du = g'(x)dx so that when you have a form like f(g(x))⋅g′(x) you can let u = g(x) and replace g(x) by u and replace g'(x)dx by du, it sounds complicated but it can simplify hard equations though you might not have gotten that far yet


yea we havent gotten that far yet


is what im doing right though?


i dont really know how to simplify this any more


give me one second


am i allowed to move it down like that


oh wait


i see my mistake


is that right?


yes that is correct


thats like the most simplified i can get reasonably right


thats always the hardest part for me


i think you can simplify it further, are you required to on the test?


yes we always simplify but the extent always varies,


how could i simplify this more?


so basically since sqrt x is also equal to 2x divided by 2sqrtx you can rewrite the numerator


hold on


so numerator would look like -(x+1) divided by 2sqrtx


and you can then move the 2sqrtx to the denominator


yeaa i see


make sense?


i wish i had a key so i could see how far the key goes


yeah, it's definitely hard when you're not sure how far to go


but i hope that was helpful


yea it definitely was


lets try d now


good im glad!


do you want me to correct you as you go or after you're done?


yea that would be great


okay!


should i start by using quotient rule


for d


or just go right into it


what do you mean by just go right into it


like do product rule right away


i would start by rewriting the second factor


then do product rule


i mean i can do this


and then multiply it out


maybe


that's exactly what i would do


nice


and now i have my other equation


did i do it right


[student's calculation] - x^-1/2


yes!


i got this so far


let me see


i didnt have enough space for the full thing


looks good to me!


now i have no idea what to do from here


okay


there's not really much to do


wait


oh apologies was looking at the wrong parantheses, it's all correct


we can simplify it further by by expanding each part


then combine like terms


yea but on a test id assume we wouldnt have time for that


yeah it would be absolutely painful to do


so i'd say you did fantastic! good job!


alright i think thats good in this area for now let me get another type of question


okay!


am i thinking about this right


so for the derivative of hat


that*


oh wait


power rule


woops


yes!


there


and now i find the 3 different values


yes


i think id need a calculator for this


only the 0 one i wont need it for


are you allowed calculators on the test?


no


wait


wouldnt it just be -2/3


anything greater then -2/3 would be positive


-2/3 would be 0


for the slope is 0 it would be x = -2/3


and anything below -2/3 would be negative


yea so thats what im saying


and you're correct on + and -


yes! you got it all right, good job!


nice!


i think i was blessed with understanding calculus conceptually but being horrible at algebra


honestly, that is quite a blessing


i have 2 more questions and then i think im good for today


concepts are always the hardest to teach, algebra just needs a ton and a ton of practice.


sounds good to me!


is quotient rule a good place to start


I would say so, just don't forget the x^2 later


product rule would also work if you wanted to use it


i would definitly need a calculator for this too


since id be plugging in 3 to e^x


but i can atleast simplify it


yup


i think thats good enough


let me check


correct


nice


you're doing really good!


i can do b cause e^0 is just 1 so it should be easy


okay!


should just be 2


nice!


oh you should know though


since they asked for an equation


you should type it out fully and not just find the slope


ohh yea woops


forgot about that


so use point slope form and plug in your slope that you found ( which is 2 in part b) and x1 and y1 which were both 0 and you'd get y = 2x


no worries! i missed that too


in my class we use Taylors form which is basically the same thing


ohh interesting! okay!


is that right?


one sec


I can't tell what you did, did you start by finding h'(x)?


yes


i did power rule on both sides


and 2g(x)


and then i added both together


oh woops


i put 2 by accident i meant 3


where did you get the 2 or 3 from


it says it on top


you get the normal and the derived


so it would be 5f(-2) and f(-2) = -2 so the term would be 5 times -2


yea


then you get rid of the 5 cause the derivative of 5 is just 0


and then you have 3


or wait


am i supposed to just


sorry i wrote that wrong


remove that is it 0*3)


hold on


do you see the blue on the board


yea u see it


i*


does that make sense?


yea i understand


the way i did it i mixed myself up


yes it is definitely a very confusing problem, i think it is best to write all the steps out because i also confused myself at first


im gonna try b and I think im good


okay!


is that right


yes!


nice!


alright i think im good for tonight then


thanks so much for all the help!


great! you did really well and best of luck on your quiz tomorrow, you'll do great!


of course, anytime!


thanks alot!!


hello


How may I help you? Can you show me your question?


hi  ok


How may I help you with this?


It looks like you have the correct x intercepts


and the correct axis of symmetry


im confused by the differences


what is the lower x intercept?


You have two x intercepts. WHich is the lower one?


Hello?


are you still there?


Hello, are you there?


I'm sorry but if I don't hear from you in a minute or two, I will have to close the session.  You can always repost your request when you have more time.


Sorry, bye


I need to check my work for one math question


Hi there! Whats your question?


I understand what 25% 132 is. I know it is 33 but I need to know how to label each number line between zero 132.


I see, thats tricky


From looking at the number line, I can tell that it has 10 points


from 0-132


How did you get the middle number to be 66?


I 132÷2=66 so I know that’s 50%


Great!


Half of 66 is 33


Thats completely right


Is it a requirement for the question to put all of the points on the number line?


Just curious


I don’t know


Did your teacher say you had to?


No


Hmm, I don't think it is important when it comes to your question then


Would you still like me to help you try and figure out how to point the numbers?


It will be tricky because it will include decimal numbers


No, I think it’s OK. I will just leave it like this. Thank you for your help.


Of course, please come back to Tutor Provider anytime you need more help


Have an amazing day/night


Hey!


hi


What do you need help with?


Emma


okay!


For the question its asking how many equal parts are there


How many do you see?


yeah because we got fun Friday coming up and I want to participate so I need to do my lessons for the week and plus my birthday is coming up soon I will be turning 10


Okay!


Happy Early Birthday!


thankyou


Okay so how many parts do you see in total?


Not exactly


including the non-colered ones


Okay so there are 8 parts total


ok


SO it is divided into 8 equal parts


yes


So thats the answer


that's the next question


You know this!


You just siad it earlier


3 out of the 8 parts are part of the playground


no we are comparing it to the whole


The whole is split into 8 parts


so it would be 3/8


3 parts


David


Hi what do you need help with


Hello


Hi


uhh


you can see my messages?


because mine is bugged


yes I can


ok


refreshed


anyway


Leo, I need help with geometry because I struggle with it


okay


what topics in geometry


any specific ones


one second


I need help with angle based geometric questions


okay, do you have any specific examples


or problems that you want to  do


I don't have any specific problems


okay let me get some and we will work on them


is it like trigonometry geometry or just normal geometry


normal geometry


your textbook


Yes I should


why don'you and try a couple sections from Unit 9


those are the ones that relate angles to geometry


can you refer me a link?


you can do some and then come back to review the ones you are confused on


mathaid.org/geometry-practice


thanks


there you go


you can stay on the call


Im actually gonna go because I have smth else to do


my fault for wasting some time here


okay good luck


your good, I am here to help you


byr


bye


Hello


Hi hat can i help with?


Do you know how I could create an equation for this function?


to create the fucntion, you must analyze the behavior of the graph. For example, you must check where hte graph has horizontal / vertical asymtotes, where it has holes etc.


so for this graph it has vertical asymtoes at x=2 and x=-2 i beleive and it has a horziantal asymtoe at somehwere like y=2


wait sorry Alex asymtote at y=5


and a hole at (12,6)


Are you sure about the vertical asymptotes


And the point of the hole


wait the vertical asymtotes are x=-5 and x=2


and the hole is at (12,6)


I see it as (6, 6)


And the other vertical asymptote i see it as x = 1


oh sorry i read the thing wrong


yeah


sorry i thought they labeled the asyntote as x=2


yeah so the vertical asymtotes are x=1, x=-5


and the hole is 6,6


How would I start to create my equation


if there is a vertical asymtote at a certain value of x, the denominator of the fucntion would equal 0 if you plug in that number. for example if f(x)=1/(x-a) there is a horizontla asymtote at x=a


so you can start by mmaking the denominator of your equaiton as (x+5)(x-1))


then we will continue


So could I start with 1/(x+5)(x-1)


yeah that would mean there are Vertical asymtotes at x=-5 and x=1


Also if you have a hole at 6,6, the numerator and denominator of the fucntion should be 0/0 at x=6 whihc means you can multiply the equaiton by (x-6)/(x-6)


so your equation will turn into (x-6)/(x-6)(x-1)(x+5)


we also know tjat the y value of the hole is 6, because the hole is at (6,6). therefore lim x-->6 f(x) has to equal 6


does that make sense


Yes


Is 0/0 DNE or und


0/0 means it is undefined but it can be simplified further to find a vlaue


so if you cancel out x-6 from te numerator and denominator


hwoever this is not the final form of our fucntion yet, becuse we have to still use the horizontal asymtote and we have to make sure the y value of the hole is 6


becase right now if we do lim x-->6 (x-6)/(x-1)(x+5)(x-6) we get lim x-->6 1/(x-1)(x+5) = 1/55


but we wwant it to equal 6


How would we do this


we must multiply the numerator of the equaiton by something so that the limit euqals 6 and we have to also make sure that this equaiton satifies the fact that the horizontla asymtpote of the fucntion is y=5


give me a minute


we also know that the graph intersects the x-axis at -2,0


this means the numerator of the fucntion should also have (x+2)


Could I add 5 to make the horizontal asymptote


yes


So would the final equation be (x-6)(x+2)/(x-6)(x+5)(x-1) +5?


sorry you wouln't add 5, you would mtiply 5 on the numeraaotr


because horiznotnla asymtote att y=a means lim x--> inf f(x) = a


wait actually one second


that also works


however, then the x-int wouldn't be -2,0


so you would have to multiply 5 by the numeraotr


Could you give me a minute


Do you think 5(x-6)(x+2)^2/(x-6)(x+5)(x-1) would be the final equation?


you are almost there, but remember that lim x-->6 f(x) has to equal 6


so instead of squaring x+2 we can mulptiply the numerator by another  thing to make lim x--> 6 f(x)=6


right now we have the equation f(x) = 5(x+2)(x-6) / (x-6)(x-1)(x+5), so limx-->6 is 40/55


Ok


so to get the limit to equal 6, we must multiply the numerator by 8.25 because 40/55 *8.25 = 6


so if we multiply by (x+2.25), and we evaluate the limit again, we would get 6


therefore our final equation would be f(x) = 5(x-6)(x+2)(x+2.25) / (x-6)(x-1)(x+5)


Thanks, I got it now!


great!


i will end the session now! thank you for staying engaged! good job!


Goodbye!


bye


hello


what problem you need help with


Hi. My son is having trouble with this math work. Will you be able to assist


I should be able to


I uploaded a picture on the graph


Please make it bigger


Is that better?


Yup


I’m sorry let me upload a clean one


There you go?


My apologies


aight


Well he got part a right, but the graph for b is incorrect


it's not linear


What should he do?


It should look something like this


Note that I'm using a scale of 5 because going by 1 would take longer


Something like this


Could you explain?


in the graph I drew, y depends inversely on x


And since you can have negative number of volunteers we stick with quadrant I


As x increases, y decreases, and vice versa


heuuuuuuuuuuu


Howdy !


Yeah haw


I'm from Texas


So you have a specific problem to solve?


I put it  on the dashboard!


Okey dokey.


Artichoke


Oh, they are so tricky !!!


Flipping the 4 and the 7 - tsk tsk tsk.


yeperdoodle


But when I need help I come to upcheive!


Awesome sauce.


cool beans


So the remaining answers have all the right numbers, it's just a question of minus signs.


ok!


Thanks!


So it doesn't matter where in the fraction the minus sign is.


Just that it's there or not there.


Make sense?


yesp!


Sweet !


All set for now then?


Ohio


I like you better than Sarah


Later tater !!!


BYEQQQQQQQQQQQQQQQQQQQQQQQ


Hi Anna! How are you?


sgood


Great! What would you like help on today?


nothing


Is there something specific on math you need help on because you asked for a session on 9th grade math?


yes


hello


Okay so the problem 1/6+1/3 right?


Okay so to add fractions together, both need to haev the same denominator and you find that using what 2 numbers you are adding together have in common in the denominator.


Does that make sense?


For example, if you are given 1/8+1/4 you need to look at the denominator whicch is 8 and 4 and figure out how to make the numbers the same. You can do this by figuring what times 4 equals 8 to make them the same.


Does that make sense so far?


just hurry up and tell me the anwer


plssssssssssssssssssssssss


not trying to be mean


I'm sorry I can't give you the answer but if you understand the problem which I'm trying to explain then you can do all the problems easily yourself!


can you turn your mic on


No, I'm sorry my mic does not work.


But I can explain the priblem using the whiteboard


Would you like me to use the whiteboard so we can figure the problem out together?


Hello How are you?


hi, I'm fine


how are you?


Except for the spam calls, I am fine. How can I help?


haha. I'm preparing for a quarterly exam for my math class rn I just need some help reviewing each topic


OK where should we start?


I wanna start with comple conjugates. Thi problem is asking me to divide (1+i) by (2-i)


OK


Not sure what they have in mind, but see whethe this helps: first mult top and bottom by 2 + i


do u see why?


yeah can you model it on the whiteboard


pls


ok pls wait while I enable stylus


kk


ok here goes


the pink ratio is equal to one


ad 2 + i is complex cong of 2 - i


can you do the mult.?


no can you model it pls


OK do u know FOIL?


yeah


OK I will first use that on (2-i)(2+i)


I will rewrite


is that OK so far?


note i^2 = -1


u might need to scroll down


yes


Do u see the red now?


yup


now 1/3 + i is  one way to write the answer


I'm confused about the denominator


and there is another


me too this does not match my calculaor


how did u get 2+1 in the denominator


OK I will sue black


what..


2^2 + -(i^2)


see the black


the other terms camcel


cancel


so that gives 4 +1 oops


my bad


its fine


I get it now


Sorry for the error


I think to fats when I work alone LOL


and this matches my calculaotr


lol its okay it matches mine too


good


can I write a new prob on the board for u to help me with


Yes


May I erase the bd first?


yeah


OK good ahead


we have room


okay this is it


OK


I would rewrite the sqroot using i Can u do that?


I disagree


If you use sqroot u still need the 4


Remember sqroot(4) = 2


yes good


Now keep going if u can


what about the cross terms?


so you know what I mean?


no


OK will write


ok


ohhh


does that help?


follow the colors


unless you are color blind


yeah


I'm not colorblind lol


ok does thismake sense now?


yeah thank u


do u have another simila one?


similar*


yeah I'll write it


wait


may I clear the bd


yeas


OK start at top pls


that's it


OK


I think there are several ways to do this


I suggest you start by using i


first where are the i's ?


second


sqroot(4) is 2


but is qsqroot(3) = 3?


good keep going


good


now combine the i's


ok but what is i*i = i^2


I will write


Do u understand what I did?


yes


now what is i^2?


yes so what is the answer?


Yes


thank u


we got some confetti for that


loll


I have a new one


ok


this one is extracting using square roots


OK


are you supposed to solve for x?


ok what do u think u should do first?


I agree


pls keep going


good


and next?


almost


I htink there are two x values


One of them *is* 7


I am confused


x = 7 works


does x = -1 work?


OK now I agree


I will cricle


that should be plus or minsu 3


Do ua gree?


agree*


yes


OK good


I have time for one more


OK


solve by factoring


pls continue


idk what to do


can u factor out a 2y?


but that is not exactly factoring


I will show you what   had in mind OK?


yeah


does that help?


I iwll explain A and B soon


yes pls continue


OK Now A*B = 0 so either A or B (or both) are zero


so if 2y = 0 what is y?


good


and id 2y - 3 = 0 what is y?


if*'


I agree


I htink those are the answers


IF you had divided by 2y


youwould have lost one of them


ohh okay


unless you wnet back and though t that you cannot div by zero etc


Ok I msut go now


Must make businesss phone call


okay thx for the help


bye bye


You did well BYE for now


hello


Hi, I need some help with some polynomials please


I need to find all solutions of the equations, I got two problems till I finish this sheet and I completely forgot how to do it


what method have you been using in class? do they let you graph it?


no, we were using this formula x^2-(sum)x+(Product)


wait that's for the first half my bad


this is a cubic, not a quadratic


I can graph it yes


are you familiar with rational roots theorem?


Basically you need to find one solution, and then you can you use synthetic division to reduce it to a quadratic


If you graph it, that will suggest a solution.


Now you are all set.  very good


now what do I do?


Now you have a quadratic, so you can either factor it or use the quadratic formula.


1/3 ?


you can check and see if that works?  Does 1/3 work?


yes


that's good.


now I have x^4 + 3x^3 - 30x^2 - 6x + 56 = 0


Again, graphing should suggest a rational root


because of the rational root theorem, if there is a rational root, it must be an integer


an integer that divides into 56


Try it


use synthetic division to check.


synthetic div is often faster and easier than plugging in and, if it works, then you have a polynomial of lower degree to work with.


good. Now try -7


Why the question mark?


It looks like you are all set with this  problem.


Was making sure I wrote the quadratic properly


Awesome!


You said you had two polynomials. So are you set now?


Yep, now I just gotta finish 8 more sheets before next period


Thanks for your help!


You are very welcome


Have a good afternoon


bye


You too!


hi


hello, i have two problems i am confused about


they are derivates


this?


yup


Do you remember how chain rule works?


i think so


chain rule again


i got confused because also we learned the rule that d/dx e^u(x) = e^u(x) times u'(x)


yeah, that's exactly what we did, didn't we?


but with both cos and 4 times x, i don't understand


but what the difference?


you do completely the same thing


how am i supposed to realize that we need the derivatives of both cos and 4 x


because 4x is inside of cos


and it is still the chain rule


so would the first part be


true


ok i think i get that first part


does it make sense?


is it the product rule?


yes


yes! can you please go over though again the first part since i still don't get how the e^cos(4x) follows the e^u(x) rule


is it both that rule and chain rule combined?


after you did this step, you forget about e^cos(4x), b/c you need to find the derivative of this cos(4x)


oh i see wow


there is another problem i need help with please


5 is not related to x, so we can just put it outside


ok


I forgot a little bit, give me a second


sure


how is it just cosx?


yeah, you are right


ln(x) is the integral of 1/x, nevermind


i don't know integrals yet, we only learned limits and derivatives


yes, that's okay


there is your answer


can you simplify it fully


what?


i mean is this fully simplified or is there any other steps to further it


no, you cannot do anything else with this


sounds good


i have one last problem


it is a word problem so i will type it here


i really appreciate your time and patience! i have a big test tomorrow and this is super helpful


you are welcome


an equation of motion is given, where s is measured in meters and t in seconds.

s = f(t) = 2t^3-9t^2

a) find the time(s) at which the acceleration is 0

b) find the displacement (position) and velocity at these times

c) find the time period(s) when the particle is speeding up and when it is slowing down


the equation of motion it is your speed


the acceleration is a derivative of speed


isn't a when velocity is 0?


"acceleration is zero"


because isn't a(t) = v'(t) = f''(t)


f(t) = v(t)


oh nevermind


I am not right


ok


s is in meters


oh! i see, let me quickly jot this down on my paper


eh, what?


parts a and b make sense!


okay, that's good


my teacher said we need to see when the a(t) and v(t) are >0, so we can graph when they have the same vs. different signs for speeding up and slowing down


can you please explain it this way with all of those parts


and show them on the line graph


but for t=0 and t=3 would it be > or <


can you explain for figuring out forward and backward?


v(t) doesn't really give you any useful information, because if it is positive, then it move to the right, if it is negative it moves to the left


okay, then how do we figure the forward and backward out


a(t) is more valuable because we can clearly see where is is slowing down and speeding up


Do you need to find when it is forward and backward or you just want to know?


we need that as well to get the problem correct even though its not directly asked


okay


how did you figure that out from t=0 and t=3


I factored 6t


but like which was forward vs. backward


I took the random number in range of 0 to 3


1 for example


and substitute it in the equation


it is important, because that's how you generally find where are negatives and where are positives


so we have 6*1-18*1  = -12 <0


v < 0 --> backward


oh! so then t>3 is the general formula for forward and 0<t<3 for backward, because time cant be negative


yeah, you are right


would this be the final answer:


slow down: 3/2<t<3 and speed up: 0<t<3/2, t>3


oh, I understood why you need to find the velocity either. To change the direction car need to slow down


and then speed up


oh that makes sense


this clarifies a lot, thank you so much!


have a nice day!


well, it still doesn't make sense


oh, i thought it did


what am i missing?


car slowed down before 3/2, then speed up after that, then somehow slowdown again before t=3 to change the direction and then speed up again to actually move


i thought it sped up, slowed down, then sped up again, because isn't it figured out by same vs different velocity and acceleration signs


there must be slow down before 3, but we know that after 3/2 it was speed up


yeah you are right actually


ok phew


is there anything else we should work on here are my test topics


You are doing great


trig derivatives, trig proofs (sinx, cosx, tanx, secx), chain rule, derivative of a^x, e^u(x), and a^u(x), implicit differentiation, speeding up/slowing down problems


I would recommend you practice some problems with chain rule. Also memorize very well how d/dx for all trig formulas


can we work on those please


for the memorization


do you know how it works right?


the proofs?


they doesn't teach this trick because it doesn't really have mathematical basis under it (or maybe does, I don't know) but if you look on how positive one and negative one placed for cos and sin, it will look the same as trig derivative transformations


im kind of confused


that's kinda weird but it did really helped me, b/c I forgot every time


well, you know rthat how trig derivative tranformation works


it definitely helps but can you please explain it in differnet words


i know the derivative of sinx is cosx and the derivative of cosx is -sinx


oh! i get it now


this does help a loct


cool, right?


yes!


I hope it will help you in the future, but we need to do some problems on chain rule


sounds good


this one was a bit easy!


can we do this example


good


is that right unsimplified?


please, write d/dx when you are doing chain rule b/c it is very easy to lost which must be derivated


ok!


where did i go wrong?


well, you did everything right


well yeah, you did everything right actually, but I just did not understand what you wrote, so I solved it by myself


ok i get this now


is there a way to simplify my answer


or is it already in simplest form


no. you cannot go further


there is just nothing to simplify


ok sounds good can we do this example next


derivative of sin^2x/cosx


(simply first)


I mean, you could write x^2 + 1 -2x^2 as 1-x^2


couldn't we do that


of course not


oops that makes sense


i was mistaken


can you erase the board so we can do this derivative of sin^2x/cosx


it says simplify first how do we do so


I know


you know what, if we comeback to previous one, you could actually simplify it like that either. So you was kinda right


But here it actually make sense, because make it easier to solve


can you run through ir


the last step doesn't necessary


How are you feeling?


i think pretty good


cool, good luck on your test tomorrow


bye


Hey Alex how can I help you?


Hi.


Let me upload my picture real quick.


I'm doing great. How are you


great thanks!


All right, can you see it?


yes I can see it!


OK, I have a couple of questions, so you might want to use your mic.


Hey, Alex


Hi Jamie


What can I help you


OK, so a question on my screen is what I need help with.


OK. Do you know the equation to find a slope.


Not really


It's OK. So I'm gonna write it


OK


on the screen. So it's.


All right. So M


is slope


and then it's


minus y1


over


O minus x1, sorry, it's like so hard to write on this whiteboard.


No, you're fine


The question on


The question I already gave us two points.


So this should be pretty straightforward. So let's find


which one we're gonna do x1 and y1 and x2 and y2.


So


and then it's just a simple plug-in.


plug-in problem


So let's start with this 1st 1, -2.


That would be


we can use this, this is X, this is Y, X and Y.


So negative 2.


which one would that one be


Um


one minute


Oh, OK


Yeah


Wait, which one would what be? I'm sorry.


No, it's OK


x1 and y1 and x2 and y2, that just means


like, OK. So let's look at the first point. -2 and 2.


So we know that 2 is on the x-axis, right?


Yeah.


And this is why


And then same thing goes for 3 and 4. This is X


and this is Y


So 2 and 2, let's say we're gonna use that as our first point. So -2 would be x1, and then 2 will be y1.