the lore, the background, the world.


because the Lord of the Rings itself is only covering


um


I forget what the name of the age is. I don't know if it's a dark age or the middle age.


because I know Valinor is


Anyway, but there's more background, there's the whole story of how Morgoth came to be when we originally this um


advisor to this corrupt king, not this evil king who was also creating an empire of evil, um, but then that king was defeated in Sauron created, like, how they created the orcs, which were defiled elves and


um elves were these like sort of immortal beings that were among the people and


the corrupt king was actually amassing the um


all the humans together became king of the humans to attack the elves, and there was this great standoff. Then you have the dwarves and there's dragons and


there's a lot going on


Dragon trolls and


Anyway, that's


that's geek, that's nerd talk.


yeah right


with me geeking out over.


Um


anyway, Sherlock is a good show to get back to originally, and then we can work on for example 12.


okay


Sorry. I noticed you've lost interest about 20 minutes ago. I didn't mean to go into that long.


tirade


Um, all right. So, example 12. Now we have um


an improper fraction


yes


i solve it?


Yes, you solve it.


I feel like that's the best way to do um


the example problems because we actually have a solution and have it all written out. I think it's best that you kind of practice it on your own.


And then we can compare


the methodology


okay, on it


Don't forget I have the table on the board.


So if you want to minimize the, if you want to minimize my screen to look at it.


yes, i m looking at it


Oh


done


All right, let's confirm.


Did you get


X minus 5 log.


X minus 2 + 10 logx minus 3 natural long, of course.


oh no, i didn't


i didn't change in proper frac


Oh, you just try to do


straight


X^2 + 1, try to integrate that straight.


What did you use as


your substitution


or your parameter


So that's what I was curious about


i directly did like last question. i did it wrong way


Hm


Uh, you tried to find A and B immediately.


which you would have to do.


A + B + C. We had to have a a C term


cause you'd have


you have 3 equations, right? You got


an X^2 term, an X term and a 1 term, which doesn't really work out. And this is um


well, when we started this morning, this is what, what I


I was doing problems from the exercises. And there was one of them that had


um an improper fraction like this, and I was getting stuck the same way you are.


It wasn't until that we were reading the


the theory that I realized why it has to be an improper fraction to use this method.


And that we can't really


figure out straight, like it gets more complicated, we try to just keep using substitution method to solve these kinds of


uh improper functions.


Anyway, um,


I mean, you can do it the other way, but I'd recommend using


the methodology they give cause that's what we're here to practice.


we can't use substitution here, bcuz how?


You could also use your method too to practice that. But right now they're just kind of they didn't want you to practice using this methodology.


because how we can't use substitution? Well, here's a question if we use substitution


what would we use


right? Because the numerator can't be


um


reduced any further


The denominator can be reduced to two expressions


and say you set T is equal to


one of the expressions


and you rewrote the numerator, you would still have an improper fraction.


or say you


yeah, i reduced in two expression


converted um


say you converted this expression and instead of using tea, you decide to use a trig function.


and then you'd have to


try solving that way and I'd still be complicated.


Um, because you'd have, say, say you said that um


cosine theta is equal to X minus 2.


right


And then the denominator, you would have cosine, cosine theta, times, cosine plus one, right? So you'd end up with like co-signed squared plus cosine in the denominator.


But in the numerator, you have to rewrite it.


and it would still get complicated, right? You'd have like cosine plus something, cosine squared plus something, and


it will be very complicatded, bcuz now we have polynomal andtrig


denominator cosines, cosine square plus cosine squared, and then you try to use the trigger identities and it's just too complicated.


Yeah.


Right. So those, cause that's just using the two methodologies that they've shown so far, right? Which would be substitution and trig identities.


Um


Now for the, the method they're giving here, well, I mean, I guess the other um methodology they give is just an antiderivation where you just look for


um


things that you know


are the derivatives of something


Um, but that isn't as straightforward here either.


So, um, and if we would go back, remember?


uh right here.


If P over Q is improper, right? And they define


proper fractions


such that the degree of P is less than the degree of Q.


right?


Then you have to do long division.


I know you try to do this problem without doing long division.


And that's why I, um


I don't know if you heard me when we were going through this, but I was kind of


trying to emphasize, you know, how do they define what a proper and improper is.


and, you know


the long division process.


yeah, I heard you. but i didn't took notice of it in question i forgot


Yeah, and that's one of those things that like, you don't realize is important until you start doing problems and you're like, oh, OK, I guess I do need to know why they phrase it this way or why we're using this methodology or in what conditions we use it in.


yes right


Yeah, so don't, don't, don't be upset. That's the same trap that I fell into.


And I was


forced


out of


kind of frustration to go back and read the theory.


to figure out the conditions, yeah.


All right


right,


i'll do it again


I believe in you


done


All right, now that you get


the expression they got.


the solution they got


i have cofficient switched for log


You mean you got


the 5, you got the 10 going with X minus 2.


and the 5 going with X minus 3.


yeah


OK, well, that's not right.


I think you


may have just I think you just may have switched your coefficients.


cause that solution um is not the same.


Your methodology is probably right, but I think you just


probably switched your solutions


So did you get something of this form?


A over X minus 2 and B over x minus 3.


oh understood, i did got same form. but when integrating, i did a/x-3


Right


So the, remember it has to be the opposite term


opposite expression


Your coefficient is multiplied by the opposite expression to find what your coefficient is.


Yeah, that's what I assumed you did. That's all you did was just swap the coefficients.


but just know that you would get the wrong answer.


yeah, understood


OK


And then we can do, how many examples are there? 1415 looks good fun.


This is kind of what we're talking about earlier.


rig inside our


um inner grains


Uh, then there's 16, so they have


looks like they have examples for each form in the table.


Well, close enough. They probably don't, I don't think they have one for problems too.


but they definitely have one for 3 and


Um, I think that's enough. Well, we should have done.


sample 15. We can do example 15.


and then move on to, I don't know if you have time to do.


any examples


We can do


I'm sorry, exercises


Well, I guess you can work on that on your own, but


i'll do it


it'll be good to handle something like


where did they get tangent? Oh, dissolutions.


The solution, sir. All right, so


for example 15.


Now we have


expressions that need to be rewritten.


right, let's try it


All right, let it rip


Now they give you, well, I guess you don't want to look at the solution. I said they give you what


the parameter is


They say why here.


By the way, this symbol


um, with the zero with the vertical slash, it's called phi. It's a Greek letter.


This becomes more prevalent when you start talking about um


radio coordinates in 3 dimensions.


They it's, I think it's actually spherical coordinates. Yeah, it's spherical coordinates.


Anyway


done


you there?


Oh, sorry


I was distracted. Yeah, um, let's see. Did you get


3 log 22 minus sign there, um, plus 4/2 minus sine theta.


When do I have


oh, they flipped it


to get rid of the absolute value.


Yeah, cause this is what I got.


i got sin fi - 2


sci 5 minus 2. Yeah, that's, that's basically this form. They just rewrote it.


because you have a negative term here. So they changed the Y minus 2 to 2 minusy, and they changed the absolute value here.


um because fee can never be greater than one.


They changed it to 2 minus 5.


just to kind of solve the absolute value.


no, in log


So you're right


wait in long, yeah, yeah, 2 minus.


sine 5 minus 2. So the absolute value of Si phi minus 2.


is only ever between


I think, 3 and -1.


So


to change the absolute value, they distributed the negative


So you have 2 minus sign data sign phi, right? Sign only fluctuates between -1 and 1.


So 2 minus -1 is 3 and 2 minus 1 is 1, so they keep it this form to get rid of absolute value.


But this form is still acceptable.


ohh, understood


Yeah.


Cause remember whenever we do the natural oulo that they include the absolute value because log, natural log can never be um 0 or less than 0.


So with the district restriction on that domain.


you can modify the expression.


cause we're considering absolute value.


Right? And in this instance, there's only one


case


for absolute value


that satisfies


this expression


So, you know, sign fee minus 2 has to be absolute value for it to be taken by log. However, 2 minus sign fee.


doesn't need absolute value because it's always positive.


to keep it positive


Right, cause the range would be from -1 to 3.


I'm sorry, sorry. The range would be from 1 to 3. That was wrong.


The, the range of the expression inside the log. There we go.


yeah, right


All right, well, that was fun


JK


Um.


but that's interesting


if you consider this problem


that whenever you have


the


denominator as a polynomial expansion.


right?


where it's either the squared term here.


you will have a solution of the form


naturallog


and one over the expression.


right? These two separate


um


forms


It'll be log of whatever the expression is, and then negative 1 over


whatever the expression is


That makes sense


yes, makes sense


You know, I didn't think to do this earlier.


but if you go to the board, let's look at our table.


specifically the, the blue one.


if we look at the first form.


right? We got PQ over X divided by X minus A, X minus B.


right


The solution comes, the rewritten expression.


is a


improper refraction. I'm sorry, it's a proper fraction, which is a constant divided by each term, right? Each expression.


So our solution


um would actually be of the form.


of two logs


two separate logs because they cannot be


Actually, no, I guess they can be um


divided by. So you can probably combine, you can probably combine the two logs.


because as long as they have the same base,


Does that make sense


I don't feel like I'll lose you


I don't want, yeah, um, cause then you end up with some constant times log of X minus A.


plus.


B times log of x minus B.


and then depending on, you know, kind of the sign will determine if you combine when you combine the logs, which one's going to be on the


numerator and denominator and so on and so forth, or if they're both in numerator or both in denominator.


or if you can just leave them separated as logs.


I, I don't know why they didn't, um


where is it


Uh


yeah


So for example, if you look back at example 11,


they take


this


right, and combine it


Are you there? Hello


yes, Im here. chat working slow


Oh, OK. Um, so I was trying to figure out why they


you know, they combine the log terms here.


however, they keep them separate here.


right?


And I think it's because of the coefficients.


because the


add them together is multiplication.


but you have separate coefficients


So this would actually to combine them would be log or natural log.


X minus 25th.


times


X minus 310 inside the same log.


we can't combine log if their cofficient don't match


I mean, is that true? Uh


because the coefficient is just the exponent in on the inside.


For example, um,


let's go to the board.


Make some room, cause I could say


uh,


2 natural log of X.


minus


3 natural log of X.


Um, let me see, let me get another expression, X minus 2.


right


Technically, I could combine these.


right, cause this would just be natural log of X^2 minus natural log of x minus 2 cubed.


right, which is then becomes natural log of


z over x minus 2


cubs


right?


I just think it might be easier to leave at this expression.


rather than try to


convolute it this way


cause they're both equal the same thing.


I mean, we can take that um


that example 11


right? That solution has -5 log.


means natural log


X minus 2


+ 10


natural log


X minus 3.


right, and rewrite them as


um natural log. I'll skip a couple steps.


X minus 3 to 10th.


over X minus 2.


right? And they're the same thing


It's just kind of what's easier to look at.


Just coefficients and expressions and natural log expressions or like a natural log of a


bunch of exponents


Anyway, that was just kind of my mental tangent. I was thinking about this earlier. Oh no, the chat, your network.


i get it what you were saying. we cando like that


I mean, it's


I mean, I mean the same thing. I think they just like to use this form.


Um


anyway, um, if we go back to, we're saying uh what I was expressing is that


what happened to the board


OK


That scared me


All right


So the


first form


will give us two natural logs.


The second form will give us a natural log


i get what you were saying. it is property of log


and a um


yeah.


So I was, I was just trying to figure out their mentality on why they


didn't combine them.


um, or if you could even combine them.


im sorry, can we wrap for today cuz my network seem to leave soon?


Reason being there isn't an exercise problem.


Oh, OK


That's fine. Um, Chris, have a good day.


Bye


have a good day


bye


ready


hi


Hi Emily! How are you?


I'm good and you


I'm good as well :) How can I help you today?


I need help with homework


Okay! Is the screenshot your first question?


yes


Okay! Do you have an idea where to start?


part A


Do you have an idea how to start solving part A?


we read the scenario


to see what graph it would be


Exactly! What information do we get from reading the scenario?


I need help


What is the ship doing during the first hour? And how close is it to shore?


stop


Exactly! Is the ship stopped at shore or away from shore?


away


excellent!


So, we're looking at graphs of distance to shore vs time.


During the first hour, the ship is not moving at all. Thus, its distance to shore is not changing. So what kind of line should we see on the graph over the first stretch of time?


Remember, the line represents the distance from shore during that time, which is not changing


constant line


Very good! So, we can eliminate any possible answers that don't start their graph with a constant line


yes


Then, the captain starts heading back to shore and docks at shore for a while


When the ship is at shore, what is the distance to shore?


decrease


That's right! The line should decrease. Should it decrease all the way to zero?


very good!


Which option starts with a constant line, decreases, and reaches 0?


the last option


yes :)


Any questions on part a?


no


Awesome! Let's look at part b next. Here we're graphing speed vs time.


So Emily begins her jogging workout by running faster and faster for about a minute. So for the first stretch of time on the graph, what should be happening to Emily's speed? What kind of line should we see?


If Sarah is getting faster and faster at first, is her speed increasing or decreasing?


increase


exactly! Then should the graph of her speed vs time increase or decrease at first?


yes!


Then, which options can we eliminate, knowing that?


So after Emily gets faster for a minute, she jogs a constant pace for several minutes


If Emily's pace is constant next, is her speed changing?


exactly! What kind of line would we expect to see on her graph then?


The graph shows how her speed changes over time and her speed is not changing


yes


On a speed vs time graph, what kind of line indicates that speed is staying the same over time?


constant


exactly! So we should see a constant line next


Can we rule out one more option now?


very good!


Any other problems you'd like to go over?


Or further questions on this one?


yes I have more


okay!


For part a, the scenario tells us that the pool starts half full of water and ends half full of water


What does this tell us about the amount of water in the pool at the beginning and end of the stretch of time we are looking at?


Should the amount of water be the same or different at the beginning and end?


decrease


If the pool is half full at the beginning and half full at the end, is the amount of water in the pool at the beginning the same as the amount of water in the pool at the end?


no


Why do you say that?


oh wait I mean yes


Yeah! Then, what options can we eliminate, knowing that the start and end amount should be the same?


the third option


second option


So we want to see the amount of water at time 0 be the same as the amount of water at the furthest x value


Which options have that quality?


last option


To me, the last option looks like at time 0, there's a lot of water and at the furthest time, there's a lot less water


We want to see that y value at the beginning and end of the x range be the same


In other words, if we drew a line between the beginning and end points, we should have a constant line


first option


The first option doesn't fit that description unfortunately. The first point is the point of the line on the y axis, at time 0. The last point is the end of the graph line drawn. Does that make sense?


yes


So we want to look for options where the y values of those two points are the same


That means the amount of water in the pool started and ended at the same value because y is amount of water in the pool


And the y values of those two points being the same means they are at the same level on the y axis :)


If this way of looking at the problem isn't clear, there's another way we can approach this one? Would you like to try a different approach?


So a month ago, the pool was half full. Then, three weeks ago, the pool was filled.


So did the amount of water in the pool increase or decrease?


increase


Very good!


Do any of the options show an increase in water level near the beginning of the month?


We would see the line on the graph go up then because the amount of water is increasing.


Do any of the graphs have an up line near the start?


Which ones/one?


the first, second, and last options


Where do you see upward heading lines on those graphs?


Could you circle those areas?


Those lines represent that as time increases, amount of water decreases.


We want to find a line that represents as time increases, water amount increases


This is because the pool was filled near the beginning of the month


Which option has an line that represents amount of water increasing as time increases?


So as the x value, time, gets bigger, the line goes farther away from 0 on the y axis :)


Do you understand the question? :)


yes


awesome :)


How are you feeling about this problem, Emily? I haven't heard from you in a bit.


ok


Okay! Let me know if you have any questions


can we work on something else


Of course!


What else would you like to work on?


Do you have an idea where to start calculating f(-5)?


no


So we're given the function definition for f(x)


the parameter x can take on the value of any real number


When we see f(-5), -5 is in the place of x. So we know this is the value of f(x) when x = -5


Any questions so far?


Awesome


So the definition of function f gives us two case: 4 if x is not -2 and 2 if x is -2


in this case, f(-5), x is -5. Which case are we in then? x is -2 or x is not -2?


x equals -5. Can x equal -5 and -2 at the same time?


Then x must not equal -2


Since x equals -5


Thus, we are in the case if x does not equal -2


The text I circled in green says, if x does not equal -2, f(x) equals 4


Can you use this to evaluate f(-5)?


Close! But that's not correct


f(x) can only be two values: 4 or 2


f(x) is 4 if x does not equal -2


f(x) is 2 if x equals -2


Since for f(-5), x equals -5, we know x does not equal -2


Since x does not equal -2, is f(-5) 4 or 2?


yes!


What do you think for f(-2)?


How about f(4)?


Why do you say that?


from the first part


The first part of what?


Remember, f(x) is defined above to only be 4 or 2


so -5 is not correct


f(x) is 4 when x does not equal -2 and 2 when x equals -2


When calculating f(4), x equals 4


Since x equals 4, does x equal -2 or not equal -2?


not


exactly!


So if x does not equal -2, what is f(x)?


f(x) is 2 if x equals -2


But x equals 4 not -2


But x equals 4 not 2, I mean :)


No x equals 4 not -2, I mean. Sorry!


If x does not equal -2, what is f(x)?


I have to leave at 2:15. Just a heads up :)


ok


Was there anything else you wanted to ask?


no


Okay! Good work today :) This is tricky stuff


I hope you have a great day


Hello How are you?


I'm fine, hru?


I am fine tnx how can I help?


so basically we've been working with functions, limits, IVT, and etc... Now I will have a midterm about all of these we've been learning about and I need help getting all of these things in my head because I feel like I have some concepts done but then confused on the rest


OK How can O help?


I help*


well, the teacher did give us 3 midterm samples and I guess we can work with that


OK that sounbds like a good idea


alr one min


OK


okay let's js go step by step Sarah, I wasn't able to bring every page of it yet, but I guess we can start with page 1 which this is


OK I will copy to 2nd screen Pls wait


OK should we start with (a)?


I see that u are typing


Are you here?


yea I'm here


what do u think about the intervals of continuity?


there


I will scroll down pls wait


I think these intervals should all be in terms of x values I think I see some y values instead


how? I thought i did x-values?


I will circle


are those really the x vlaues that u want?


ok


I made a mistake on 0, but isn't infinity right?


IMO no


don't we want the x-coord of the verticla dotted line?


so 5?


yes


but doesn't the continuity line go all the way up at 5? does it mean that it stays at 5?


Remeber the interval is *x* values


okay


y might go to inf yes but we need x values


okay next one ig


OK b)?


Good


thx


for this one are we gonna start by plug 3 to f(x)?


Hmm I think  not


I think u need to ask about the lim of f(x) as x -> 3+


is it 2?


I think so yes


now can u use that to answer this  part?


like this?


Yes


keep going


ok


what is the vlaue of that sin?


the value of that is 1


I disagree


I think sin(pi/2) = 1


that's what I said


but you do not say sin(1)


so do I just put 1?


well the answer is just the number 1


so the whole answer is just 1? we stop there?


sorry


No problem go ahead


Yes 1


Good


I agree


do we just need to plug when the form is number over 0?


plug for x*


well I think the lim is undefined


but I am 90% sure


how do I get to where it tells me if it's undefined or not, do I js plug it?


well nbr/0 is undefined


so I think u can jsut say that


in fact in this case, I think we have 1/0


wouldn't be 2/0?


since 1+1^2 is 2?


I don't think so


Liam the x^2 become 0^2 ?


well x is going to 1-


no


where did u get the 0 from?


the one 1- means we are approaching +1 from the left side


okay so it's js zero?


yes the x vlaue is jsut zeor


so 1 + x^2 = 1


OK


Now when I was in school (about 60 years ago) inf and undefined were the same but I am not so sure that is what they say now


but u need to find thtat out


I think this is right


OK It is Ok with me


I have time for c) Then I must go


wait for the last one does it mean there is a vertical asymptote at x=1?


do u mena for c) or b) iv ?


b) IV?


well maybe but all we are asked for is the limit but I suspect you are right


you see that from 0 to just beofre 1 we have a striaght line


you culd get the eq of that lien and use it and grpah the function


okay Liam for see, it's asking for the inverse right?


Yes


ow would u do that?


doesn't js mean that we switch x and y values?


exactly


like (-3,-2) becomes (-2,-3)


yes


I will tell you another way that I like


If you draw the graph on thin apaper that you can see thru


you then rotate it 90 deg counter clockwise


that makes the x acxis poitn up


unfortuatley the y axis now points to the left


so you must flip the paper "around the new y axis!


but using points the way u do is probably best


I think the left one curves the other way I will draw


I think that is it I may be wrong


on the origianl what is the y vlaue when x = -2 ?


Np, how abt the last line, I'm having trouble drawing that


OK let me think I htink my pink one is worng


but let's thiknk about the last one


I think there will now be a horiz asy at y = 5


like that?


sort of yes let  Liam


yes the open circle went from (3,2) to (2,3) yes


good


I have to go now


Does the graph look right to you?


yes I mean it could be a little samoother


that last curve should not dip down


but it is hard to draw here


okay before you go, can you tell me why the pink one looks more like yours than mine? can you also arrange the last curve the way u say it?


I now think the pink may be wrong


let me thin


No you're right, i check on the answer ket


you see in my head I first rotate the orifginal grpah 90 deg countter clockwise


then I iamgine flipping it around the new y axis


I think your blakc is better


No yours is better, I checked


You could try to estimate y when x = -2


then swap the coords for that point See what I mean?


Yea I see


do u agree that in the original (-2, -1.4) might be close?


yea


I will put a dot where I mean


sort of the left  edge of the pink dot


the dot is too big but I htink it would be (-2,-1.4) which woudl become (-1.4,-2)


now wait


for the left hand curve all the x and y in the original are negative do u agree?


yes


so when you swap x and y all the new pairs whoudl have only negative x and y vlaues see what I mena?


I will circle


so that curve shoudl be in quadrant III


Pls think about that


I msut go now


Hope to "see" you agian soon


alright bye


BYE BYE


have a good night


You too


thx


Hi


oh hi! how are you


were you not expecting me?


also did you sleep?


yes! I did.


I wasn't expecting you


How weird is it running into each other like this.


Hm


yeah! we also kinda set time anyways.


so, about my exam


Right


yeah, i think it was okay. There were questions I could do. I kind of mess up due to time bound


Oh no


Um, what do you mean by questions you could do?


Were there a lot of questions you couldn't do?


there were lot of questions, 100 questions


I attempted 30  , time was over in that


Oh man


100 questions


What, what were the top, I mean


hmm,


Was it just a bunch of questions you didn't understand or hadn't done yet, like integrals and


um probability


yeah, there were topics i didn't study


lemme show you the questions


Please


It's not going online. I might need some more time.


OK.


No rush


im casting it


can you see it?


yes


okay!


Oh, matrices.


I did #2


A substances.


out of this page


Yeah, I'm, I'm looking at these questions and I'm thoroughly confused.


What is AP and GP?


arithmetic progression


geometric progression


All right, geometric progression and arithmetic progression.


I just wasn't expecting that. Um.


What did you say for the answer


Uh, non-empty subsets of, of a set.


and AMBC represent their compliments.


Oops, um


Uh, so A and B are subsets of the same set.


So A, C, and BC represent their compliments.


seems like i didn't attempt this one too.


I was confuse bcuz i was getting both statement wrong


Huh. It's


so all we do know is that the compliments, um


A plus A C is equal to the whole set U, which is equal to B + BC.


So if you subtract


AC from both sides and BC from both sides, you get A minus BC is equal to B minus AC.


I got 2


Oh


Oh, I see what you doing


You just is a different color to show a B minus A.


yeah


Oh, I see what was wrong


So the B complement


Are you talking? Sorry, I don't have my microphone.


where my headphones on


Oh, I see what you're saying


I don't understand what a' - b'


Basically red minus green, is that what you're saying?


I think 1st statement is correct


green line is b compliment


Oh.


red line is a'


Also green minus red


So it says B complement minus A complement.


Yeah, I'd say one is the same.


I'd say one is right


Do you get


BC minus AC.


You said green minus red.


and then statement 2 says,


A minus B complement.


is equal to


a complement minus B.


So A


minus B complement.


You just get the sliver in the middle, won't you?


for statement 2?


You're just getting 0


No, I'm sorry, that's a lie


You get a minus B complement, which is everything around it.


yeah


including a


A minus B. And then


I think 2 is right.


Could be way off. You would just get basically


um


kind of the universal set minus A and B.


I should get the negative.


All right, that's


BC


B complement


in red


He said A minus B complement you would get rid of.


Right, but when you, oh, wouldn't you still have the overlap of A + B.


or kind of that


convex shape


It looks like you made a face, by the way.


I didn't really study this topic. I forgot it


I was asking is that the part is a-b'


Mm


I don't think that internal portion would be


A minus B complement, because B complement includes that


a portion


outside of the.


So it technically cancel and you'd be left with the subset, um.


kind of outside AMB


and the portion inside um that overlaps between A and B.


But if you said a complement plus B minus B.


then you would get


the outer


By the way, I never, I don't, I don't think I ever really studied this either. None of this is really


like refreshing any memories.


oh it's fine


I can't, I don't even know what


let's move on


course it'd be or what topic. Oh, sorry, yeah.


I'm sure you're, I'm sure you're done with this.


Um


OK. That was number 2.


Well, yeah.


100 questions and I would, these are just the, the ones out the gate.


Was there any more? How many terms of the


what is number 6 say


i didn't any of these questions here


OK


you can look at them if u want


But, and Y equal x factorial and Z equals 2 x factorial.


So you have


an odd, uneven factorial divided by a factorial factorial.


where


um I guess you'll just be left with the odd numbers.


he said


you know, to times 4 times 6 times 8, times 10.


divided by 1 x 2 times.


3 times 4 times 5.


then 2 and 4s cancel.


and you'd be left with 6 times 8 times 10 over.


oops, more than that.


I'm sorry. I don't


if we can move on if you'd like.


you can try if u want.


Yeah.


Oh, it's 2 to the


it's 2 to the 5th.


Oh, so Z over Y


is equal to 25 or 2 to the end then, whatever the factorial is.


because for every X, you can divide two X by X.


But a Z over N is equal CALC 201.


Um, long face, so N is equal to.


This doesn't seem right.


Where's my desmuss?


So roughly 7


And then what is the value of 3 of X?


which between 3 to 7 times.


um 7 factorial is equal to.


And I'm wrong.


Oh the joy


Hm


Is it me or does


Yeah, I guess so


So C is twice as D, B is twice a C, and A is twice as B.


So really just it comes down to what um


what could what basically what factorial you get the value for X.


Well, I never evaluated what X was.


That was dumb


dumb, da dum dum dum.


Oh wait, can't you just do


um Y times Ethan Is that the same? X-factorial times 2 x factorial.


to become a 3 X-factorial.


No, that would give you kind of an X square factorial.


How much time did they give you


I can see why you


2 hours 30 min


2.5 hours MATH 300 questions.


So that's


150 minutes


ALGEBRA 101 questions. So you basically got 90 seconds per


question


So any confusion


and you just gotta move on.


do i move on to next page or ?


sounds sounds.


um


120 is just 5 factorial.


Um, let me, let me write down the question. This is gonna bother me. I'm gonna be very upset about this entire exam.


even though I didn't take it, but living vicariously through you.


sure!


OK, you can move on.


And then what is 60? How many terms of this series of odd numbers amount to a sum equal to my goodness, look at that number.


I don't even know what to


what to do about that


Oh no, what happened to the screen


No more questions you wanted to know


Why did I say no, I didn't say no.


Just go over


We can do it some other time then.


Can you share the link for those questions?


You mean the little in the time I have.


free time I have


kind of pour over those questions because they were.


stumpers


it's not online. and not going on google drive either.


Oh man


kind of anti-tamperproof.


idk how to share them


what's that?


I'm sorry, I just tamperproof.


Just tamper proof


Anyway, it's hard to share it because of the way they


um have guarded it.


Um, but wow, um, I forgot that you


have to face those kinds of questions.


That's the kind of stuff that will keep me up at night.


set theory


I could have done more if i had studied more.


You are smarter than me


I don't necessarily need not to do PHYSICS 205 questions. bcuz it got like criteria for passing


Oh really


So you could technically have strengths.


like certain ones you can um


certain topics you can only do.


but still get enough to pass


The Morgan's law complement laws.


no, I don't think so. bcuz you won't know if topic will lead the race. It's dangerous


Oh, that is true


Your word of warning


old merit


I went to the Professor Davies for set theory, trying to figure out compliments. They only made me more confused.


Huh.


I think we need to attempt like 50 questions or so for okay marks. bcuz they also do negative marking


Do they really


Very stringent


Hm


I, I missed our infinity board. I don't know why they got rid of it.


Is there any way we can ask for it back again?


Or do you have to ask her? I don't know if I've ever been asked.


ask what? for infinity whiteboard?


Yeah, Infinity whiteboard.


You know, another thing, sorry, I'm, I'm, I'm, I'm still looking up.


the set theory


There's also, there was no


nothing saying that AMB held an value set common in value, well, I'm sorry, Honey


values


in the set that were common


So it could have just been two separate.


oh maybe have write email ?


um


sets


or subsets


Maybe they write email


Oh, you mean like the, the AI tutor, the AI thing.


I think if you asked him the AI chat.


let me see


to get infinity board back.


By the way, can you see what I type into the UP bot whatever.


typically click click on the whiteboard icon.


in your toolbar or menu options.


Toolbar


the whiteboard


icon


yes I can see


Mm


This, yeah, it was not very helpful.


TODC


O MTH that math, 44APTO. What a


Oh, and you can't share


PDFs too


my network kinda slowing down. i see in chat, u said honey. why did you say that?


Can you upload it?


I said, honey, did I really?


That's a typo


Oh jeez, yeah. It says I'm nothing saying that AMB.


wells


That's


I don't remember saying, honey, I'm, I apologize if that's


Yeah, I didn't, I didn't mean that.


Um


where the heck did I get that from


Sorry, I was in between words. I gotta stop muttering into the mic.


How did it pick? Why did, why did you even think that?


There, there's been quite a few times where


um I don't know where they got the words from.


or or how the sounds I made, they chose those words for.


This mic is dangerous.


yeah, okay. let's be careful


Noted


So what do you wanna, um,


I don't wanna take up too much of your time. What do you, what do you wanna do today? Um, what do you want to focus on


Do you need kind of a mental break?


after the exam, or do you wanna focus on some questions from you. It's only 18 pages. That it'd be longer, um


Do you wanna go back to doing intervals? Like, what do you wanna, where do you wanna go from here?


I don't mind either or


let's continue with integrals.


These questions will take a lot of time


That's also true.


OK, so,


we pick up


where we left off.


We were on


um integration by substitution.


We're going through the examples


Um, we had


I completed 7.2


we had just finished discussing why single trig, um.


Oh, you finished 7.2


Oh, you do the exercises


So you wanna move on then


How'd you find the question


some of them were hard


Yeah, I bet


Did you wanna go over some of them or just continue on?


Do I memorize these integrals of tanx, ...


Um,


I don't think you need this, you don't need the integrals of the single.


um


you just need to know like the standard derivatives.


That way you can do the easy um antiderivatives for the intervals, but the other ones, you may have to like either derive on the spot.


or just kind of like remember the tricks, cause those are the more uncommon ones.


like the inner role of um tangent, stuff like that, where it's like log


um, those are, those aren't very standard.


They may be useful at one point, but it'd be kind of a niche.


But then again, I don't like, considering what you're your exams are like, I don't think they're going to be that cut and dry.


They're gonna be some very like


abstract or niche thing.


Where am I? You said you finished 7.2, or there's 7.3.


they not gonna ask integral of tan x, for sure


Good. Yeah, I don't think so either.


I, I, I hadn't like to this point, I don't think I ever knew what the inner role of tangent was.


I mean, once I saw it, it made sense, but I didn't think I never really needed to know it.


okay, so I just need to know the process


Yeah, the like,


for the majority of this stuff.


it's the process


the methodology, the thinking.


right? You don't need to know


the knickknacks. You just need to know kind of what tools to use when.


um


and also like


doing these problems helps you get used to thinking in a certain way.


like quicker to pick up the right form of um like what tea should be or what you should be, um,


but that just comes with practice and experience.


just kind of seeing, you know, typically what


becomes easier to do.


just from doing problems that end up becoming really hard.


yeah, right


OK. So if you've done 7.2 or 7.3, I guess. So now we can do um trick identities.


Here's where you're gonna become more versed in your um


trait intervals and derivatives.


and just more practice with substitution.


And um I think they use double angle and triple angles, double and half angles here.


Imagine so


All right, so now, um, rather than substitution,


well, rather than that form of substitution, now we're going to just start solving using um or doing intervals just using trig identities.


okay


which is an application of the chain rule.


after you subs it up. So for the first example, did you want to try, um, we'll go through the first example.


and then if you wanted to try to on your own, and then we can kind of go through 3.


or you want to do 2 together, 1 and 2 together, and then go through 3.


I'm sorry


So these end up being pretty short.


yes, I can try them


Did you at least want to try the first one together or you, you feel confident enough.


You should.


You're pretty. You should feel confident. I'm sorry.


These half thoughts


I'll try it first. if i can't, then we can see where i mess up


All right


I believe in you


thanks! i do too


can you zoom in ?


yes! looks good


done!


Good. How'd you do


idk if they are correct. they were moderate


They were moderate, oh, and difficulty. Uh, we can check the first one.


Did you get X over 2 + 1/4 2 X.


s square


oh, no i forgot to divide by 2


Yeah, 2 co-signs for X minus 1.


Still good job though if they were only off by a factor of 5.


Sorry.


an a factor of 5


factor of 2 i think?


Oh, it is a factor of 2


I don't know why it keeps


saying 5 instead of half.


Huh


i got this one wrong.


Can you see the


the UP bot when I type into it.


It says that you should be able to.


your student can see our chat and ask me questions too.


It just solved the


inner growl of cosine squared x.


Smartypants


chatbot showing off.


You got the


I can't see


you got the first example problem wrong, the cosine squared.


Oh


yes!


you can't see the


chat, never mind then


Did you want to go through it together


Or was it just um


yeah!


All right


So


oh, I see it on the board. I didn't see it.


I'm sorry, I had the screen blocking the entire time.


So you got one plus cosine 2 X.


I think, I think you got the sign wrong. It should be 1 minus.


Is it 1 plus?


Yeah, I guess it is one plus


then the half gets drawn out.


I mean, you're off by 2, and then you skipped


you skipped 22 should have been


oh, I think he did them backwards


Yeah, the second one should have been signed the third one is sine cube, but we can work on that.


i did 3 then 2


So you have


3 minus


Oh, you actually wrote out the


identity. You have 3 and 2, OK.


Let's see


well they use the identity as well.


So yeah, 3/45 cosine x.


Yeah, that's right


And then question 2.


Sign to like cosine 3X.


times 2.


Oh, you did 15 times 2. OK.


Oh, I see what you did.


Yeah, you got it right. Uh.


You are very clever


I is notifications going on.


Great job.


Smarter than me.


thank you!😅


So it can be shown.


Yeah, I don't know how much of a compliment that is.


but to each their own. For me, it's a compliment. I hope it doesn't offend you.


Uh.


So a decent treatment there that both answers are.


always talking about


This was actually one of the um


example problems from the previous chapter, something similar.


which sine cubed, cosine squared.


i practiced these identity in trig a lot. so i kind of have them in mind


You've done triple angles and


other ones


Cause those are typically aren't as memorized.


aren't as well known, I'm sorry


I didn't memorize these. it's muscle memory lol


Oh, OK


Um, did you wanna do some actual exercises or kind of leave you on your own and we can just move on to um the next section of trig.


which you should just kind of like,


other, other integrals that you can kind of like just memorize.


let's do some hard ones


I don't know how relevant they'd be


You wanna do some hard ones


Hm


then move on to next section. yeah


That might be too easy.


to incubed


No, that's just the easy.


substitution


might pick, this might just be the integral of that.


Um, let's see


What do you think we can go with


Oh, and 21


So inverse


Is that OK with you? Problems 19 and 21?


yeah


can i try them first?


yes


we can do them separately and then compare answers


ohh, sounds good!


is it cos^3 in 19?


okay!


don't check answer right now


ok


can you give a little hint for 21?


tbh I'm still working through it


I haven't looked at the answer


oh, how did you start? did you let anything?


I tried rewriting the inverse


how did you write the values for side?


I'm not sure if it's actually gonna help though.


ohh okay! got it


So it could be useless or not relevant.


let's see


forgot square root


where?


ohh i see


right


I know how to do this, but i can't remember the right thing to let


that's the tricky part


That's why I said this stuff will come with time. Like it, like you know how to do it, but it's just like on the tip of your tongue.


Man, this is a hard problem.


It's like right there, but I just can't put my finger on it.


So we know about theta and X is that they're


yeah, i just can't remember that. i almost got it


phase shifted


aren't so syax is equal to cosinax. I'm sorry, sin the as you go to cosinemax.


and cosine data is equal to sin X.


I wonder if we can use uh some


and um


some properties


cause we know that


cosine X


is equal to, I'm sorry, we know that.


cosine


um let's just use X


is equal to sin.


X


90 I think.


or minus 90.


I think that's true


because that'd be sign X.


times cosine of 2.


minus


cosine X.


maybe.


I guess it's pie, I guess it's plus plus 90.


Does that help


I think it does cause then that would rewrite the arc sign


from cosine to sign.


of X + 90.


and then that will just become X + 90 or pie or 2 whatever.


DX


huh


Is it really that simple


I almost feel scared


1/2 X. oh, I'm sorry, it's it.


1/2 x 2 plus.


i over 2


X


C


that makes sense


Does that make sense what I just did


Can you see anything wrong? I can make it bigger.


It just kind of scribbles.


no, I can see.


gimme a min


i can't understand this


Oh yeah, I'm sorry. I was trying to do the um


the summoned property. So sign, it's supposed to be signed.


um


A + B.


where I said A is X and B is night is pirate 2, it should be pirate 299.


To many scribbles. Let me try that again.


Sign.


I did X minus, I'll just use Pi too instead.


I couldn't remember what the shift was.


So I tried um


theta plus


Sine, cosine, cosign, sign.


So we sign


X cosine pi over 2.


plus.


cosine X sine i over 2, I ran in a room.


Yeah


There we go


And then cosiney 2 is 0, so you're just left with.


cosine X. Sorry, I was just trying to prove, um,


the relationship between


sine and Cosina.


And that um cosina and sinex are just face face shifted from each other.


Sorry, phase


You can't see that. Oh, I'm sorry. I keep forgetting you can't see that.


do this


Is that better


Uh I really wrote small. I'm sorry.


oh okay, then you re write the equation?


Yeah, and then at the bottom, I just, I, once I once I proved that cosine x is equal to sin of x + i over 2, I keep saying 90 should be pi over 2.


Um, then I just substituted that back in for, you know, arc sign of sign, just canceled out.


😦 it was this easy?


Yeah, I'm almost


I almost feel doubtful. It's OK if I look at the answers now?


Huh, that is the answer.


Though I got a negative somewhere wrong.


Let me, um.


Can you see it?


yeah


So I got


I got the pie over 2 X, right? But I guess I got the negative


x or 2. Where did they get the negative?


Unless they did


I think they did um


sign of pie over 2 minus X.


cause then that would cancel out too.


But then I'd have to prove that cosine x is equal to sin of x, i over 2 minus X.


i think sin(x+90) should be -cosx?


Is that right


Would it


do anything though? I feel like the negative should be within the phase shift.


It would make the whole


it would make the entire um


thing be negative rather than just one of the values being negative.


oh no, that's wrong. you're right


Anyway, looking at your 19, did you get 10 squared? I don't remember if you got 10 squared.


You got seekin X


log tangent you get express log tangent. So you get the log tangent, right, but you're missing the one half


tangent square, which you got for sequent.


We got to double back to that.


Your work seems correct


Oh, sorry, let's get back to 21.


And let's prove 21.


Um


do you wanna work on 21 yourself? Just kind of prove where I went wrong.


Or do you want to keep going through it together?


Just to be honest, it's kind of helping my self-esteem after looking at your exam problems.


are you upset due to those problems?


A little bit


It may just be


too early in the morning


Not really, but


um


those are also on the tip of my tongue. I just


anyway, I didn't, I didn't, um, I kind of followed you on the previous problem. I had no, I was so confused


I've always struggled with the um


with rig, uh


with these kind of trigger identities


or struggle is a strong word, but


tougher than most


I didn't really find any errors in your


formulation for number 19.


21. Let me


it's fine. there were easier problem too. you 'd plenty of them


let me retry that property. So if we say sign of.


you don't have to try to soothe my soap, my, my ego here. I'm fine.


Um


so they say 9, so you said 90, say pie or two.


minus X Where does that come from?


This would give us negative cosine.


What about


sign of


X minus 52.


This would also give us negative cosine.


Does it though


of course I your device.


minus


no, really. my teacher said to us try to do easier problem, you won't able to finish all of them due to time.


Oh, for the exam


Yeah


That's basically what you do for the SATs as well. Anything that's timed really.


You get the low hanging fruit, then you can focus on the harder problems.


cosine is an even function.


So this becomes


one, so this just becomes cosine of X.


and then co-sign 5 to 0.


Huh


So that's the identity. So I got it wrong.


It's not


um X + y or 2 should be 2 minus X.


i think both identities are right, but they went for second one.


Oh yeah, you're right


They're both right


Well, there you go


And I'm, I was


also surprised on oak. Shoot, what was the page number again? It's like 32.


Oh, so close


Um, I was also surprised why there was only one


inverse


problem


It was kind of, it, it kind of made it a little suspicious. I was like, are they just that difficult or are they that supposed to be easy?


But it seemed it's more towards the


easy side


to me to scratch that. Let me not say that because this is a very, this is one problem out of, you know, however many difficult problems there are.


may not generalize


Let me not generalize.


there's another way to do it too. Gimme a min to think


Oh, really? Oh, I wanna, I wanna see your


Mm.


live in this one.


Oops sorry


im unable to remember it.


What do you think it was?


Or what did I have


to do with


i think it was letting such cos theta = x


sin theta d theta = dx


But then you'd be stuck with the extras sign the term.


sine theta term.


I saw you working earlier.


You were um


trying that approach, but it just kind of dead ends.


im not getting sin thetea in numerator


yeah, not seem to work


Yeah, that's where I was getting stuck to.


Um


because once you have


your, your parameter, your D parameter where you just say T equal to


something


DX


If you can't rewrite this something in terms of tea.


say GFT, then


you can't really use it


in your inner gra


Ingra


yeah, right. we need dt


right? And you need and you really need is um DX by itself.


yeah


because it becomes kind of useless.


if you


introduce this parameter term, but then you get stuck


with some


you know, original ex term inside your


uh inner role when you have it defined as your parameter. Your parametric integration.


And that's basically what the um


not the next section, but the section after that integration by parts. That's when we start getting into


um


you know, how do we introduce


you know, how do we deal with problems like that you can't just


directly do


one parameter.


They would if you have to do, you have, would have to use two parameters instead.


or more specifically, what if your um


if you're integral kind of takes the form like X


Y DY


where you can create some separate um


parameters to say U is equal to X.


um, and YDY is equal to DV.


Right? Cause you already know what this


kind of like did this why crime and stuff is a derivative of.


So you take the innergral of it


I'm sorry, that's not right. That's


we'll just wait till the section to get there.


That's where it is supposed to be the integral of uv.


That's what makes sense


Yeah.


where Y is some function of X.


So that when you do this


you get you equal to X.


or, or whatever X term it is could be X and could be cosine x and then you have your own separate DUDX versus your second term, which we define as DV equals DY.


Then you integrate that, you get V equal to Y of X.


and then that has its own um kind of reverse chain rule.


Um, it's proven, it'll, it'll show it that it proves that the, that the inner role


of sort of um


what is it? UDV.


is equal to


UV +.


the integral


of VDU


and this becomes like


very important going on.


this little property.


because you're more likely to get problems, I think that, that you need


um this sort of integration by parts


then you will just a simple substitution.


Anyway, um, hope I didn't lose you there.


Hopefully you're still with me


im here. chat not working. i can hear you


OK, good


Um


right, so we can move, um, are you OK with 7 section 7.3.


with the um


trigonometric for uh functions.


identities


You seem pretty strong in trig identities. This one was just more of a um


just kind of remembering certain aspects, like certain identities.


the phase shift and


yes! can we review #19, what was wrong with it?


and how to do your, how to do your um inverse trig triangle.


Oh yes, of course


So


let me clear this problem.


So, 19, 1 over sign


X cosine cut.


X


So, I see what you did. You divide, the first thing you did was to


use the identity, um


like the original tri I did a sin X square, sine square plus cosine squared equals 1 and you separated terms. That makes sense.


Then you use substitution to solve the um


kind of definite integral.


which makes sense. It should just be C and D


ohh no! i just drag whole thing


However


well, you just dragged it where? You lost it?


I still see it


You, you probably hit undo


Let me see if I can drag everything up.


Yeah, I moved everything up.


OK


no, I mean like i didn't see tanxsecx


Oh, I see what you're saying


kind of the square root of


the derivative of, um,


Siganex. I'm pretty sure Seconex is just


yeah, it's just Econex TNX.


Is there a sign difference


I mean, you proved the derivative of seinex tanix just by using your integration uh substitution, so that's fine.


There's no.


problem with that


I just don't know where they got


tan squared.


1/2 10 squared. So they probably use the property.


unless there's some tea.


that they probably look, actually it looks like a


what's the answer?


so you said sine makes you go to TE and sine X D X you go to negative BT.


So they said


oh sorry


I'm not showing it


Uh, let me show this to.


Can you see it


So I agree with you. I, I agree with them for the log tan. I don't know where they got the 1/10 squared.


right


To, to be honest, that looks like a um


uh, a trig identity.


It could be


kind of seek and squared plus 1 equals tangent squared, or it could be


um


kind of a half angle or double angle.


did you got the answer?


No, I didn't.


I'll be honest, I did not get it.


oh okay


That's why I was getting a little frustrated. Uh


I mean, you work


is good


one over.


i got it


I, I


cause one over


signed one of sign cosine XL but I probably would have used 1/2 sign 2 theta 1 sine 2 x is what I was trying.


I got stuck with cosine 2 theta.


or cosine too, right, yeah, co-sequent.


2 X


D X


which


which the others, there is


Yeah, hmm


is a log of Marcus get X minus cotangent X.


which wouldn't help us.


TQ


Oh, you're right


It'd be 1/3 then


or I'm sorry, -3


No, you're right. I'm sorry


Yeah, that's what it was


Good catch


I don't know why I didn't see it.


is that right now?


Yeah, that's it. 1/2.


Yeah, that's right.


Wait, wait


Yeah, cause they, the one


the one gets included in the C, so that's right.


All the constants just get thrown into the C term.


ohh


All I care about is all you care about are the um X terms.


right


cause the C is arbitrary


without


a specific point.


on a function. It's arbitrary.


or a range


domain.


yeah, right


OK


We got like


15 minutes left.


We can kind of ram through.


not ram through, it's not, not the right term.


We can address section 7.4 if you like, which just seems to be um


some more memorization


These examples you can kind of work through on your own, but look, they just, they just literally just give you a bunch of um


you know, formulas to memorize.


which I'm pretty, yeah, I'm pretty sure this is in that's inverse tangent, that's right.


And then there's the inverse sign.


do i have to memorize them?


You don't have to memorize them. They're typically like


either in a textbook or you can just look them up.


It's just good, it's just good to


um


be able to look at them and be like, hey, I think that's


you know, if you're dealing with some


ratio where you have x or some positive


exponent in the bottom.


with some um


I forget what the term is.


shift


I mean, it could just be like x minus 3, that's still A could still um work that way.


But it's just, it's just a way of knowing just like, hey, but this actually already has solutions.


without you having to go through the whole um


substitution, integration by substitution method.


That's pretty much it


which is a lot to say a little


okay, so i can solve without these formulas


And then, oh, they actually used the


innovation by part


Yes


All you care about is the results.


right?


That like the, the integral of one over x + a is just the inverse tangent.


That's all you care about, um, stuff like that.


Just like we used for the


integra integral of like tangent and sequent. You didn't need to do it yourself. All you need to do is just be able to go back and refer to it. Like, basically like


you gotta know when to use the tools.


These, these are all just tools.


You gotta know when to use it. That's the main thing, like the, the methodology, the process.


too much tools for small toolbox, they might drop out.


when and how. I guess it's the best way to say it.


This this is also true. This is why the textbooks are so thick.


I mean, look at it, you have this this class 12 textbook had to be split in half and still the second book is still 540 pages. This is easily like


700 pages of textbook.


as a whole


Um, I will say though, it is


useful to know how to derive these, right, integration by substitution.


just in case you um


end up potentially get using um using it wrong, like getting the wrong answer somehow.


Or more specifically, it'd be, it'd be say you're using it for data.


and your data comes back very confusing.


And sometimes you have to kind of go into the, get your hands dirty and do the integration yourself to kind of


get the right formulation.


which you may not be doing data analysis if you want to get in the in the


cybersecurity, but um you never know.


what skills you're gonna need until you need, until you're, until they ask you to use it.


Oh


control there


Do you have something you're trying to say or you just


says you're typing but it's been several minutes.


what? no, I didn't type anything


Oh, I said that you were typing. That's why I was kind of waiting for you to


anyway,


um we can get back to it.


So these are


like I said, tools you can use and then we can kind of go.


to the problem. I you're gonna need


some of these tools, like for example, this


is kind of important


Uh


to see what it is. I'll see it right the answer.


reduced to the form 1/1TT. 0, it just says to use this form.


um, by completing the square.


So that makes sense


but it's kind of ugly


Um


and then you have a square root


mutan using a standard formulae.


Oh, right, cause we have


we have the solution for one over the square root of a binomial.


which is something like this. Yeah, it's this one.


Did you want to do some problems or do you want me to just kinda keep going through um


each form of it


So we can


each


yeah, plus C.


we have like 8 min left. we can't cover any topic rn


That's also true. Um, hmm


Do you have any more comments? I know that exam was brutal.


Oh, how was your day going


How is your day going?


I am heading over to kanji little bit


Oh, we actually have that next or soon?


You wanna


kinda end it early then


You really have enough time to cover anything.


Um


just kind of enjoy your day or


hope the day gets better


You was up early too.


yeah, Let's wrap up.


All right, we're gonna wrap it up. Great job today.


Um.


you really had no


issues with the questions other than


the


uh inverse trig


which was just like, once you get the trick,


that's it. You you can kind of get it, you can get the rest of them.


just a skill


yeah, it slipped my mind so bad


No, it's not that bad. It's


you just had to prep for such an ordeal


It's OK if one or two things slip.


for something that's not being graded.


im gonna take it again


Are you? When


They should have more practice tests, more prac I, I'm not used to seeing those kind of questions


to help prep you for him


next year. I really want to pass it


Is it the exact same exam?


it will be in march or april may be


I'm sorry


yes!


March or April maybe. Um, is it not the exact same, I mean, um, same topics covered.


OK


So we can kind of learn from this.


So when you say March, that's like 6 months away.


So we have half a year.


or a semester


Are you


um


Yes! right.  I have to increase my speed solving questions


yeah, considering the type of problems you go through, like, I feel like the SATs will be like a breeze compared to it, compared to this.


cause SATs aren't as


idk you never know


thought provoking. There, there are, not to say that there's no problems that aren't thought-provoking.


This is also true. I, I guess I don't want to give you


um, kind of a false sense of security.


Stay vigilant.


All right, I'll let you go


Have a great day


yeah!


have a good day.


Good


by e


Sayonara


sayonara!


There you go.


ready


Hii! sorry i missed your text. i over slept. you still there?


ywa


okay, I need to wait 4 min before I start session.


ok


Hi Clara! How are you? :)


Good and you


Good to hear! I'm also good, thanks! What can I help with today?


2 3/5


6 3/5


2 3/5 ÷ 6 3/5 ?


Yes


Awesome! Let's work this problem out together.


Ok


So to make this easier we can convert these mixed numbers into fractions. Do you need me to walk you through doing that?


*improper fractions


Yes please


Sure! No problem.


Ok thanks


Ok so we have 2 wholes and 3/5 of another one for the first fraction


Do you remember the steps to change a mixed number into an improper fraction?


Keep change flip?


yes! We do use keep, change, flip for this problem. But first, we need to turn these back into improper fractions


Do you remember which one is the denominator and numerator? H


One is the top and one is the bottom of a fraction!


Numerator is top denominater is bottom


Perfect!


Thank you


Mhm! First we need to Multiply the whole number (2) by the denominator (5)


What would that be?


10 + 1= 11


Nope! 2 x 5?


Perfect! And then add the numerator (3). What would that be?


10 + numerator (3)


Perfect! And then what's the denominator again?


Good try! Remember the denominator is the number on the bottom though!


6 × 5= 30


For the denominator on the improper fraction, we don't need to multiply :) You just switch the denominator over


Like this!


Does that make sense?


I don't understand


Okay! One moment


Ok


:D


Okay! Do you see what I wrote there?


Yes


That's how we make whole numbers into improper fractions!


If you write that down in your notes it'll help you to make improper fractions so we're able to divide them :D


Okay! So now we have the first part but we still need to make 6 3/5 into an improper fraction before we can divide


So remember we need to multiple the whole number by the denominator (bottom number)


Can you do that for me? 6 x 5


Awesome!


And then what would that bottom number be?


The denominator :)


Yes, perfect!


Did that make sense?


Okay! So do you know the next step? You mentioned it earlier! :)


Multiply both numbers


Sort of, yes! But we need to keep, change, flip


But yes, when we do the change, it changes to multiplication


Ok


And then what do we do for the second fraction?


But we need to flip! :)


And don't forget about that 3 we added, so it's 33 !


Ok 5/30


Yes! 5/33


Awesome job!


Thank you


Now, we multiply! Do you know 13 x 5?


Orry


No worries! :)


Sorry


Do you need some help with 13 x 5?


Perfect! And then 5 x 33?


Amazing, yes!


Nice


Do you know the last step? :)


It's usually the final step when it comes to fractions! ;)


Add?


Close!


Starts with an S!


Subtract


Also super close! But it's simplify!


Ok


Do you know how we would simplify 65/165?


No


No worries!


I can walk you through it!


How do you simplify things


We need to find the greatest common factor (GCF) of 65 and 165


Do you know the factors of 65?


Yes! 5 would be the GCF of 65


What about the GCF of 165?


Close! It's a pretty small number


Im


Factors of 165: 1,3,5,11,15,33,55,165


Which one does it have in common with 65?


Oh


Sorry


Nope! 165 isn't a factor of 65, but close!


No worries! Out of these, which number do you see in both


Factors of 65: 1,5,13, 65
Factors of 165: 1,3,5,11,15,33,55,165


Close! But 65 isn't a factor of 165


Yes!!


So we need to divide 5 by both numerator and denominator


Ok


Can you do 65 divided by 5?


Aka how many times does 5 go into 65!


Close!


Try one more time :)


Yes! Now can you do the same with 165 divided by 5? :)


You did it! That's your final answer :)


Ok thank you


Of course! Is that all for right now?


Yes


Awesome! Have a great rest of your night, and feel free to request another session if you have any other questions! Great work!


Bye 👋  thank you


Anytime!!


Hello How are you?


hi doing good needed some help with problems


OK Do u know there was famous Greek named Euclid?


no


I will look for a link later.


How can I help now?


limits help


OK I will copy to my second screen Pls wait


OK Where should we start?


part a


OK do u know any rules for limits? Like a product of l;imits etc?


not really


OK wait I see


it metniones the queeze theorem


So I think u need to use that


What can u tell me about that theorem?


not sure what that is


OK I will look it up and u shoudl too BUT first


I will tell you what I *think* it means here


what is max vlaue of x)?


OK and the min value of x)?


OK so I *think we can say that out limit here is between the linit for -5x^2  and +5x^2 do u see why?


our*


yes


OK


now what are bot of those limtis (I mean lim x -> 0 for -5x^2 and for 5x^2)?


both*


do i just substitute


Yes for this oen yes


ok now wait


I think we do not the sine and here is  why I think that


we know that the vlaue is squeezed between -5x^2 and 5x^2 See what I mean?


and since the limits of those two without the sin are zero I *think* we can say that our limit is too.


Now we each need to look up the squeeze theorem.


It has soem other names too, but too maybe the pinch theorem


Here is a video maybe you can look at it later:


thanks


I have *not* yet looked at it but I trsut https://www.khanacademy.org/math/calculus/limits-and-continuity/squeeze-theorem/v/squeeze-theorem-intro


should we try b now?


One commetn first


you see we know that the lim in a is between -5x^2 and 5x^2 because the min vlaue for sin -1 and the max is 1 OK?


got it


OK I will ook at b now


IMO b is quite simialr


can u explain it?


is it between -3 and 3


OK


I am looking in a book pls wait


I will type on the board oK?


ok


see that so far?


not really


does that help?


yes


OK so for b, x -> 0


and can u tell what g(x) and h(x) are?


3x amd cps 2/x


Student


what does cps mean?


*cos


I somewhat disagree


I think g(x) = -3x and h(x) = 3x


AND


f(x) x*cox(2/x)


i mean f(x) =


oh ok


so we have -3x <= 3xcos(2/x) <= 3x


we know that because cos max is 1 and cos min is -1


ok


and what is the lim x-> 0 of -3x ?


am i substituting -3(0)cost (2/0)?


no u just sub 0 in 3x


ok good and what abot  lim x-> 0 of 3x ?


good


so limx->0 3x*cos(2/x) is smadwiched between 0 and 0 !!!!


Good Yes


I see what you wrote


I will cricle in green and blue


The limit we want is squeezed between the green and blue limits See whagt I mean?


so we just substite 0 for both expressions


yes the green and blue are quite simple functions and for *those* we can jsut sub


Student is not always true but it *is* true here


how to approach part c


OK I will read it pls wait


Interesting I am not sure but here is how I would start


sub x = 1 for g(x) = x^3 - 3x +4


and also in h(x) = -3x^2 +6x -1 OK?


OK that is good for the left (g(x))


ok so f(x) is squeezed (or sandwiche) between 2 and 2 !


and those 2 and 2 are the limits for g(x) and h(x) so what do u think?


makes sense


so what is lim x -> 1 for f(x)?


I agree


Now I am not sure why they give the interval you might chk with your tchr about that


ok


Is this helping?


yes


OK good please be aware that the squeeze theorem is sometimes called the pinch theorem and sometimes called the sandwich theorem


Do u have more?


OK first I will send another link


This one is good but be aware that Wikipedia oftenn gets highly technical after the first few paragrpahs


The do however, have some nice examples for this


especially "First example"


They say this too:


Also known as the pinching theorem, the sandwich rule, the police theorem


I never heard of the police theorem !!!


is part b referencing squeeze theorem or is special limits another thing


part b from the previous group that we did?


on the screenshot posted now it says Special limits using special limitzs


I do not yet see a new scrnshot


OK I had to scroll down


OK I think they want you to use the special limits given at the start of #2


Thos -- I htink -- are derived using the sq. theorem


But you need to use them as given


Can u do (a)?


I think what you wrote is undefined


I think you need to rewrite (a) using sin/cos


then we can proceed


can you show me on board what you mean


yes but pls wait while I enable stylus


pone more minute


See what I wrote in blue?


yes


Have u seen that before?


i think i have


OK u should rvw that I will go to next step now


OK?


got it


OK Now next step


I mult by 1/x  / 1/x OK?


ok


almost done


do u follow the algegra?


algebra


how do you get 1/cosx and sinx/x


I will dd soem red to previous step


does that help?


and I will circle in green I hope


OK


now we have a prpdcit


product


I will do next step


Therefore if both of those limits exist (and we know their vlaues) we are done!


I wihat is the first limit?


good


now for the seond one look at the limits given at start of question


we need the one I circled in black


ok


it is given in the question


yes !!


so what is the answer?


yes !!!


did that all make sense?


yes


ready for b?


can I erase so work t make room?


OK what do u think about b?


substitute 0?


well yes good start


but what happens if you do that?


undefined


I agree


Now does b resemble anything given in the question?


sin x / x =1


but not exactly the same


can u get it into a form like sinx/x ?


divide both sides by 3


very clsoe


why not replce 3 by 2*(3/2)?


to start


or (3/2)*2 which is the same


can you show on board


Yes


I will try to use blue


The liitle x menas multiply OK?


why did we multiply by 3/2


I will cricle and maybe u eill see why


oops I forgot the x


does that help?


wasn't the bottom 3x originally?


yes


and it still is but in a different form!


you see 3/2 times 2x = 3x  !!!


oh ok


now the green part has exactly the same form as sin(x)/x I mean we could let big X = 2x and we would have sim(X)/X see what I mean?


OK so what do u think the limit is for the green part?


kind of stuck still


OK two comment:


1. I will explain what I meant 2. Your tchr may have wanted u to use a trig identity here butI do not think thaa is needed


I will explain


we know limit as x -> 0 sin(x)/x  is what (it is given above)?


so what is its vlae as given?


for lim x -> 0 sin(x)/x ?


it is given in the ques


OK now the green has the same *form as that


I mean 2x can be if we cnahge it in bot num adn denom see wha I mean?


yes, would you continue to draw out step by step


OK


I changed the name to w OK?


ok


Now I will circle


now from the question we know hte limit of what is in black as x -> 0 what is it?


hint w is lust like x


yes


and what is 1/(3/2)?


Maybe 2/3 ?


and that is the answer !!!


I thnk I can show another way if you want


ok thanks for all the help, makes sense now


OK good


these take practice


yeah


I suspect you can find more examples on the Web


see you later


pls wait I want to look up Sandwich Theorem if u have time?


sure


Here it is:


OK I hope all this helped and I hope I did not go too fast


no it was good


Do u have a textbook for ur course?


yes i believe so


OK do u know hwo the author is?


who*


not sure


OK well ur book might have moe exmaples like thoe ones we did


ok


OK All set for now?


yes bye


OK BYE BYE


hiiii


how r u today


I'm good!! How are you Jamie?


doing good


I just need help with this problem its asking me to find AD


Sure! Do you have an idea where to start?


nope


That's okay! Let's start by looking at the tick marks on the sides of triangle ABC


Do you know what the tick marks mean?


share the same length?


Yes! Very good :) So triangle ABC is an equilateral triangle, a triangle where all sides are the same length.


What does that say about the angles of triangle ABC?


shares the same angle as the other triangle


Are you saying each angle in the triangle shares the same value as the other angles?


yea


Yes! That's correct :) What is the value of each angle in triangle ABC then?


the value?


yeah, how many degrees is it?


Close! If we add together all the angles in any triangle, we 180 degrees. Here we have 3 angles that are all equal in degree that add up 180 degrees. How can we solve for the degree of an angle in triangle ABC, x, if we know 3x = 180?


very good!


So we know each angle in triangle ABC is 60 degrees. Can you find then the degree of each angle in triangle ABD?


60deg , 90 deg and 45?


angle A would be 30 right


Yes that's correct!


So...


Triangle ABD is a 30-60-90 triangle because the degrees of its angles are 30, 60, and 90 degrees.


gotcha


Have you learned the ratios of a 30-60-90 triangles hypotenuse to its side lengths?


I'm not sure


We can review those ratios! They'll be important for solving this problem


Any questions so far?


nope


Do you feel like you have an idea how to apply the ratios to your problem?


sorta


Do you want to give it a shot?


yea


Awesome! :)


so would I just plug in sqrt(3) / 2 h


Yeah! Except h here is 19 m


ok


so sqrt(3) /2 (19 ) m?


Yes!


So the ratios of the hypotenuse to side lengths of a 30-60-90 triangle are good to memorize


16.45 meters


Same for 45-45-90 triangles. I'll put those ratios up on the board as well


Is there anything else you'd like help with? :)


is the answer I got correct


ok thank u for ur help today


No problem! :)


wait


should I round my answer


even tho it doesn't ask


I would ask your teacher how many decimal places they want you to round too. There's not a standard, so it depends what they're looking for. Good question :)


ok byyyyeee


bye


Hey Alex!


hi! how are you today?


Good! How are you!


I'm good as well :)


That's awesome!


How can I help you today?


could you help me with my algebra hw pls?


Of course!


okay give me one moment


Is it ok if I use mic?


yes :)


OK, let me know if you can hear me.


I can


OK, that's good. So basically for this equation,


we can see over here, um, I'm gonna annotate, OK, so we can see GFX is this, OK, we're gonna do different colors. GFX is pink.


HFX is gonna be blue. And then for green


I'm gonna do Ben, so we can see on the chart that we have F of X, OK? So whenever we have like a variable, like a variable is basically just the same thing as like a letter, right? So, um


let's use like the letter, like let's like just take explain this to you. So E of X


it's just like a random number, OK, just like help you out. EFX is just the same thing as saying why, OK? And why is what we're using to define our function.


OK, so why we're basically saying like everything that's equal to Y is the function, OK? So over here we have GFX, which is just another way of saying why, OK? So we're saying GFX is 1/3 of F of X. So can you use like a different color? Can you use the red and can you go to F of X on the graph and put like a little dot, so I can see that you're following along.


wait on what


on the F of X.


like in the in the table.


just like next to it or anything.


Yeah, exactly. OK, perfect. So you know where the FX is, right? And


over here, we see that X over here, we see the values for X are next to what we see 6, we see 0, and we see 7, OK?


So these value or negative 6, sorry 6 and 7. These are the values for X. Now for F of X, the values are 18


15, and 24.


So, what we're gonna do is first, we're going to define that G(x) is equal to 1/3 of fx, OK? The second step


1 2nd


I'm trying to figure out a way to explain this like easier for you. OK.


So


an equation of is H X. This means that


to find HFX we need to take the G of X value, make it negative, so multiply it by -1, and then we're going to add 4, OK? So we're saying, see how it's negative, we're gonna make it negative, and then we're gonna add 4, OK?


And then


so that's our second one. This is just like understanding the functions, OK? So we understand for GFX we're taking any value from F of X and we're dividing it by 1/3.


OK? And then we see over here, for H of X,


over here for H of X, we're seeing we're making the, the value for um GFX negative and then we're adding 4 to that value, OK? So our second step is going to be to calculate the GFX values. So we are given F of X values for certain x values. So we can see over here what I'm talking about. For F(x), we see these three values, right, right next to it we see -18, 15, and 24. Can you just let me know if you understand so far what I'm trying to say because I don't want you to be lost.


yes


OK, that's good. So we see these values for F of X, OK?


So the next thing that we're gonna do,


is


we are going to use the formula for GFX, which is 1/3 of FFX to calculate GFX for each X. That sounds like a lot, but I'm gonna walk you through it. So first, for X equals -6, because do you see how over here we have G and then instead of the X right here, instead of the X, we're replacing it with a -6.


OK, wait, sorry, one second. I just got a call from someone. I will call them back, but, so you see right here, we're gonna do G, and then we're gonna plug in the negative6, and that's going to be the same thing as doing 1/3 times F of -6. So, so to write it down underneath.


ok


Also, I'm gonna do the first one like with you, and then for the second one, I want you to try it on your own. So,


over here


for GFX


we would do G.


times 6


OK, hold on. Let me just erase all of this like Y equals stuff over here.


OK, wait, I don't know why it's not letting me


erase it. OK, let me just erase this right here. OK.


So here we would have GF 6 is equal.


1 2nd. GF 6


OK, I don't know why it's not letting me draw, OK, there we go. Perfect. So we have GF 6 is equal to 1/3, because you see how the 1/3 is in the in the regular formula up here, 1/3 of F of X, so 1/3


of F


of F of X


So 1/3 of F of -6.


because you see how the 6 is at the top right here.


OK, this is just gonna be the same thing.


as 1/3


times -18, which is just going to give us negative 6.


OK. So we're saying here that G of -6 is -6.


OK. So all we're doing is we're taking the value that we're given. I'm gonna do like a complete, I'm gonna do like this teal color. We're taking the value that we're giving.


that we're given, and we're going to put it where the X is, OK?


We're gonna put it where the X is.


and we know that this this um value is gonna be equal to 15.


So


G 0 is going to be equal to 1/3 times F of 0.


And then over here we see F of 0, right? F of 0 is going to be 15.


So I'm gonna do this. I'm gonna give you 5 minutes, um, I'll be right back, because I have to go get my supplies to help you out really quick, but I will be back in just about 5 minutes, and I want you to go ahead and try these next two problems by yourself.


ok


hi


OK hi. So I'm back and let me know if you have any questions or you want me to explain it


cause it's a little confusing. So if you need help, I'm here to help you.


I got 5 and 8


You got 5 and 8.


Yeah.


So for G(0), that is 5 great job and for GF 7 that is 8, so great job you did that right.


So using the same pattern, do you want to go ahead and try H of X? Or do you want me to walk you through it for the first one?


I'll try


OK, perfect. Let me know if you have any questions.


I think I need help


ok no problem!


what part do you need help on


the negative thing


okay give me on sec!


OK. So to approach this, the first thing that we're going to do is


we're gonna fill out the chart for what we already know. So


give me one second. I'm trying to make room


um, for this. OK. So we know the answers for the G values, right? So we know that the answers are gonna be -6


Oh, hold on one second.


We know that the values


I don't know why it's not letting me draw, but


OK, we know that the values are going to be -6.


um 5.


and 8, right? So we know the values for these.


Now, what we're gonna do to find the HVX for H X is that number one, we know the GFX for each value of X, so all we're going to do is we're gonna take the negative of the GFX and add 4. So, for example, if we're looking at age of 6, what we're going to do


is we're going to just take


-6, and we're going to multiply it by


a negative, so basically we're replacing the X over here in the equation. So we're doing negative G of X, right? Our GFX right here, the GFX is just the value we got up here.


So the value we got up here is -6, so we're replacing the GFX with -6, and on the outside of the GFX, do you see how we have a negative sign. So that's why I'm putting a negative sign right here. And then we're going to go ahead and add 4 to this value. So negative times the negative is always going to be a positive, so -1, because again, writing a negative sign is just the same as it meaning -1. So 1 times 6 is the same as 6.


OK?


Then we're gonna have more. So we're gonna have 6 + 4.


And this is gonna go ahead and give us 10.


OK, so that's how we got the number 10. Let me know if you understand. Let me know if you want me to walk you through the next problem. Um, or if you want to try it on your own.


ohhh yes I get it


or if you have like a specific question. OK, perfect. So literally all we're doing is we're replacing the GFX with the value that we got before. So it'd be 6 for GF 65 for G of 0, and um 8 for GF 7. So we're just gonna take those values and replace it of G(x) and then just literally just do it like normal. So yeah, I think you got it. I'm gonna put a 10 here just for reference, and then move these numbers back up, so -6.


5 and 8, and I want you to go ahead and try these last two for me and then I will check.


your


with mine. See if we got the same um values. Let me know when you're done.


is -(0)+4 right


it would be -(5)+5! :)


bc remeber were taking g(0) which is 5


I forgot what I was plugging in mb


no worries! im here to help


-1 and -4


Perfect those are correct!


So how do you feel about the topic now?


and is there anything else you need help with?


I feel good


could u check my answers for smth else


Yay! Thats good


Yep of course!


okay !


can you let me kow ur answer and ill check


I chose the second to last one


ok lets check!


OK, so this is inverses, um, so we know that GFX is just another way of saying why, and the H is like basically the X, right? So the first step is we're gonna flip the X and Y. So we're gonna get H


hold on


This pen thing always stops working.


OK, there we go


So


we have um


H is equal to 10


um, to the square square root of


Y, right? Because G(x) is just why.


of why


and then remember the minus 6 is on the outside.


So the first thing we're gonna do is we're gonna add the 6, OK?


So now we're gonna get, sorry, also, um, yeah, hold on. So we're gonna get the H+ 6, right? And the whole point of this is to isolate the Y. So the H+ 6 is gonna be equal to 10


to the square root of Y. So now what we're going to do is we're going to divide both sides by 10.


OK


So now we're gonna get H+ 6/10. And now, since we're done with the 10, we're gonna get H+ 6.


over 10.


is equal


to the square root


of Y, and how do we get rid of this square root, we're going to go ahead and square this whole side. So then this whole side would be squared.


And now, since we got rid of the square root, it's literally just gonna be


um, why is equal to


H + 6/10 squared, which is exactly what you put, so great job.


So, and also always remember like, when we're putting it in back into inverse form, we would do the same thing up here. So do you see how we have GFX. We would do the same thing except now we would do G of negative1 of H. sorry, not GFX GFH GF 1 of H, and then you also got that same input that I got. So you did it right.


yay


thank you


Yep


Of course. Let me know if you if you need me to check anything else or if you have any questions with anything else.


I have one more


OK, perfect. Um, here, let's just erase this, and then you can put that one up for me.


I can


OK, so I'm back. I'm not sure if you can hear me, but let me know, um, like your next


thing that you need help with, and we will get started on that.


I chose the second one


OK, so


OK, suppose that F of X is equal to x + 2, and that GFX is the result of shifting F(x) down 6 units and left one unit. Which of the following is the correct equation for GF X. OK, so automatically on the first step for this is we need to note out our word translation. So we can see over here,


um


our translations is going down by 6, OK, and then we're going left by one. So automatically we can cross out literally.


all of these, right? We can cross out every single one because none of these are going down by 6. And also when we're going left one, it's gonna be the opposite in the equation. Usually we would think, oh, if we're going to the left, we would do minus 1, right? Like we would do -1 inside of the parentheses, like right here.


But no, we're always gonna do plus one, and then when we're going right, we do minus. So automatically this is really easy because you can just cancel everything out.


to see


OK, wait, hold on, I actually forgot about the entire first part of the equation. So actually that's my mistake.


Since we have an x + 2, if we're going down 6, can you tell me what that would be?


If we're doing 2-6, what would we get? See, that was actually a trick, that was a trick.


So we can automatically actually cross that one out.


So can you let me know what is 2, if we have 2 and we're going down 6, what would that be?


Yeah, perfect. -4. OK. So now with the, now that we know that we're going to be -4.


We're going to go ahead and cross out.


the plus 4 and the +1 and the +2. Now we have these two options to choose from, OK? So you went ahead and chose the second one. I'm just gonna go ahead and um like on a race this really quick that's just actually erased the entire thing. The reason you chose the second one is because you saw the the minus 6 and you just put down the minus 6, but that's not what we're doing. We're looking at the original equation, right, the original equation


we're starting with the 2.


OK


So on the original equation, we're starting with the with the plus 2, a 2. So let's say the graph is like over here + 2, OK? And it's saying, oh, we're going down by 6 units. So if we're going down by 6 units, we're gonna end up at -4. OK, so that's why we're going -4.


because this is talking about the results of the function. So, -4, and then since we're going left by one unit again, same thing. When we're going left, we're going to do a plus one or like a plus whatever unit. So we're going left one, so that's going to be plus one. If we were going right one unit, then we would do minus 1. So do you see now how it's Y equals x + 12 minus 4. So that automatically cancels out this one and our correct answer would


be.


um GFX equals x + 12 minus 4.


So the third one, do you understand or do you have any questions with that?


yes I got it


It's OK if you don't understand or, OK, that's great, that's amazing. Let me know if you need help with anything else.


Today, this app is being really glitchy, so


I am trying to erase everything. OK, perfect.


I'm good :) thank you so much for helping me tonight


I hope you have a great night


Yup, of course. Um, I'm glad I could help you. And I hope you have a great night. And also always come back and let me know if you need help with anything.


Have a great night. Bye.


Hello :) are you in need of assistance?


Hello, can you hear me?


Can you please turn on your mic so that I could learn better.


yes i can hear you :)


Hello


Wait, are you speaking? Wait, hold on, let me turn on my volume.


OK, I can hear you


Yes, I'm speaking


OK. Um, so, um, I have a test tomorrow.


Mhm.


and I'm taking APPO.


Yes, I can help you with that


OK. So basically, I, I'm gonna share my screen um


what I'm showing you is like what is gonna be on a test. So I just need like help to like


yeah.


All right, no problem. Uh, allow me to, let me see if I can.


Can you see my screen


Yes, now I can


OK


So this, this the first.


first, um


mhm.


Do you want me to screenshot it and place it in a whiteboard or you just want me to clean.


No, that's fine. I can, I can see your screen now.


So what is it you are looking for help with?


Basically, like


the topic like the entire, like, like, like guide me like as I like um answer the questions cause


the type of question that is here is what's gonna be on the test.


All right.


Let's see.


So, do you know how to turn the chat off?


How to turn the chat off, I do not.


Mm


OK, so this is the first question.


For that to below its defined such that H of X is equal to F times g.


And then there's examples also that it shows.


Mhm


multiple


0 factor X minus 8.


Mhm.


like I'm really bad at factoring.


No problem


I'm just trying to figure it out myself right now cause it's been a while since I've done this.


All right. Can I see the problem?


All right. So, given the function H of X is defined such that H of x is equal to FX times GX


with the functions F and G given by that. What is the multiplicity of the zero x equals 6 in h x.


All right


So,


so we would multiply FX by G of X.


but


that's, but i


Yes, I think that's what it's saying to do here.


so like


OK


So there's 3


X plus 6


X minus 8


and X minus 36 2.


And then we'd multiply that by 4 X cubes and then all of that.


but we've multiply like terms first or?


One moment


and um


Hm.


it's my work. All right.


Hmm. So, I do not believe they want you to


actually do all of this multiplication. I don't believe that would


be necessary


What I'm seeing right now.


is there's an X minus 6 in F of X, that would make FOX 0.


So it's basically saying


how many of those are there.


But you know what a multiplicity is, right?


Uh, kind of forgot


No problem. A multiplicity is


how many times it hits a 0.


Mm.


on that X value. Like, if it was, like, if we're just looking at FOX right now, the multiplicity would be one, because it's X minus 6 to the one power. There's no other value there.


Mhm


But if it was X minus 6 squared, it would hit that x value twice at 6.


So the most specific multiplicity.


is one


I'm looking right now.


I


Oh, wait a minute, at 6 because um if you look at this value on the right.


if you put in 6 there, it would also equal 0.


that would


because it would be 36 minus


it's


36. Uh, like if you look at the function F of X and look at the right


You see that X minus 36.


um


That could be X.


basically, if you have a 6 there, it would also equal 0, right?


yeah


All right.


Pretty sure


So that would be


X minus 6, and then


X


All right. So what are you thinking here?


Uh.


So what I'm currently seeing is if you plug in X equals 6, you would have basically 3 things equaling 0.


on X equals 6


if you what


Oh, if you plug in 6


Oh wait, that's, yes.


what's the, what's the third one?


Um, you see how it's X minus 36 2.


Yeah, that's, yeah, that.


Yeah, so that would technically mean that I believe that there is 2 there.


because that would, it would bex minus X minus 36, and then x minus 36 again.


You understand how that works because there's two of them there.


yeah


All right. So, from looking at this currently,


I think there are 3


Like if, could you click show examples again?


I'm sorry. I also have to relearn this myself.


Could you scroll up


I just want to look at the first part.


So HFX is formed by multiplications, so exponents add


so you


X equals a factor X minus 8.


OK, yeah. And then you find that fact. So the factor in two functions. So it'd be 3 x minus 8 cubed and then X + H.


I don't get this far. Like how do they, like what do they do with the functioning?


Is this was the original function, right?


Um, scroll up


Yes, that would be the original F of X and then G of X.


So like what is going on here?


All right, so


00, GF X.


yes


But like, oh, the factor out this for 8.


M + 8


Mhm. So,


Oh, OK, yeah


if it looks like they would, you would add it together, like basically, you see how it's X minus 8 cubs.


and then


It's good, but I have a question. If the factor is X minus A, why are we also adding X + 8.


Mhm.


because like they factored it out to be x minus A So like, why are we adding X per se with it.


that's state X minus 14 and then X minus 8.


That is a good question.


3 X minus 8 cubed.


X +


cubed


and oh, OK, I see. So you see how


that's just saying what it's, that's what H of X is equal to, right?


Right.


But you see how we're not using that in the bottom expression.


It would be 3 + 3.


+ 1


because you see how there's a, you see how the 3 is highlighted in that area, and the one has highlighted in the diagram.


Yeah


So I think that would be the reason


And then there's a 3 again in F of X.


So you would add 3 from the F of X.


3 from the H of X and then the 1.


because, oh yeah, cause it's more I fear


Mhm.


when I think


one like here


Yeah, well, I just didn't understand where they got blood from.


No problem. I also have to figure it out because I don't I don't remember doing it.


I need a, maybe it's an error.


Let me see this


OK.


All right, let's see. So we can see the X +1 in H in F of X and we see the X + 1 in G of X.


And so if we're trying to find X is equal to -1.


Yeah, 1 + 1 equals 2. Yeah, so that one makes sense.


it makes sense. So this one would be like


I think so we are finding it for X equals 6, right?


X minus 6, so this one.


And this


So I think it would be X minus 6, that's 1, and then the y minus 36.


OK


that's


would be


2. So you would add those together, it would be 3.


So try 3 1st.


Yay


So that was the correct answer


Yeah


Good


All right. So we


Coke. Oh, OK


mhm, you see how it basically adds 3 in the


All right


So let's go on to the next one


OK.


What is the leading term of the polymomia. Oh, this is um 4.


Are you sure


No, I gave 4 posts the negative one.


Yup


So can you explain to me why it's -1?


Oh


So.


apparently it's not negative 1. OK, let's look at it again.


I thought it was -1


How is it not negative one. Whoops.


10 + 4 X.


minus X the 4th.


+ 7 x cubed


Is there


is that a more? Is there more?


No, that's it.


No, that's it, but let me see what it says.


Oh, it's a term, not the, oh, it's not actually


Yeah


Yeah.


for like


Oh, not coefficient term. OK, that was both of our bats, but I'll, yeah, OK. So now it says, what is the leading term? and so this would be X to the 5th, good.


eggs, yeah


and those


And then it's asking for the leading term again. And so this would be


good.


oi fish. OK, now it's I can pull you pushing.


Now it's coefficient. So what would this be? What would the highest


on the


exponent be


Yep


Then leading coefficient again, 2. Good.


Did they


So what is the highest, what is the degree of the polynomial F of X.


And so can you tell me what


hm?


it's for degree. Is it asking for eyes or like, it didn't say IS furlough, it just says.


Well, it doesn't have to be the highest. The degree of a polynomial is the highest ex bonum.


Yeah, oh, OK


So you would, so it would


so if it, if, yes, it would be the highest numbered exponent there.


Yeah


OK


So it'll be 3


Good


Is this studying or is this an assignment?


It's both, but like


but like if you get like kind of, if you get


a question wrong, you could explain it to you, then you could do it over with a different question.


All right


I like both


All right, so what is, mhm, yeah, I got you. So what is the Y intercept of the polynomial? Do you know what the Y intercept is?


uh


is the EpiPen?


Yep, so basically the Y intercept is basically where X would touch the y axis, right?


Mhm


And so at the Y axis is when X equals 0.


Yeah. Also, uh,


So you would plug in 0 to all of the X values, and that would be your Y intercept.


so F of 0, please.


What times 0 minus


What do you think in here?


So the 0.


That's what I got cause we would just plug in 0 for all of the X values.


and it's uh


So 0 minus 0 minus 0.


Good.


yeah


So remember that the Y intercept is just the


Was it always gonna be 0,000?


it will not always be zero. Basically the Y intercept would be when X is equal to 0.


So there are some like if there wasn't, like if there was just like a plus.


Oh


Oh yeah, yeah


Yeah, so if there was like a +3 at the end, then it would be 3.


Yeah, without, without the eggs.


Yeah. E.


Oh, yeah


Oh yeah