That question one, Mr. Lee, uh, question 2.


Yeah


I


Is right here, uh, so,


topic, the topic is.


ratios and proportions, as I said.


So, yeah, if you can help me out with these two questions, I would be really grateful.


do you want to go ahead and give that a try?


Yeah, Mr. Lee.


I'm sorry if there's a sound in the back, but yeah, I'll go give it a try.


Don't a lot, um, I'm not sure.


I'm having difficulty solving it. I'm not sure how to do it.


Um, we can also do like


You can have like a breakdown of it like.


I'll tell you what I know first, which


I think the problem is not written quite correctly


I will tell you that and then like you can.


You think the problem is not OK, give me one sec. Let me.


I mean, I got it from Cha GBT, so.


Let me see what happened. Let me ask her to, to give me a better one.


i think it should be " a bag contains blue and red marbles in a ratio of 3:5"


All right, Mr. Lee, uh, this is what she gave me back. Let me write it on it.


Who is "she"?


Yeah


Uh, um.


who is that?


How to say.


You


your teacher?


Oh, it's, no, it's um


Yeah you wanna It's a, it's a um AI.


Yeah, um.


All in a way, yeah, it is my teacher, yeah, in a way.


Hey.


You have AI as a teacher?


Mr. Lee, it's always like helpful, you know, if I have a question.


That maybe someone else cannot answer.


It's just something that you go.


Or, you know,


Organize AI as a teacher, yes, I did say in a way, but it's


It's, it helps you with everything.


properly


Hi, uh, give me one sec, uh, I'm going to solve it.


It should be 24 blue marbles in the bag


Nice catch.


Yeah, um,


I understood that later.


so blue/red  =3/5, right?


Yeah.


And then to solve that, you would do a butterfly method. That's what I would do to do that, you just need to do a 3.


For 6


And do this.


I mean 3 eggs don't know why I'm saying 6, but.


yes, cross multiply


3 X.


Yes, cross multiply, um, yeah, so do that. Give me one sec. Let me just solve that.


Right.


On


3 eggs.


Cross out the.


3, 120.


About about 3.


Would be a 4, 12 minus that would be a 0, that would be a 0, but what we have to do is put a 40, always remember because


The 20s, you know, they have to, because 3 times 4 will be a 12 without this 0.


That's rigth


This one right here, but we can't do that. That's something that


You are, yeah, yeah, you have to add the 0, so now we can make sense that 3 times 40 equals, not 3 times 4, which is 12. So, uh, 40.


Yeah.


A 40, 40 red marbles.


24/40 =3/5


24 divided by 40.


Um,


I didn't understand that part.


they are the same ratio


Why we are dividing 24 by 40. Is that what we, is that what we get as our final what we got 35. Is that what you're trying to say?


Oh, OK.


Yeah. That, that's just when they're


You know


More simplified, that's just not more, not, not a simplified form.


But yeah, I see what you're.


I think so.


So


24 out of 40, same thing as 3/5.


Just not simplified.


Am I right?


was


I am just saying that 40 is the correct answer because 3/5 is the same fraction as 24/40


they represent the same ratio


You don't have to do all the 24 divided by 40. It's just showing that it goes to the final, um,


part of action.


Because 2440.


Like I said, they


They are the same, but yeah, I get what you're saying.


That's 40.


Yeah.


So our final answer would be 40.


Right


right


OK.


Um, yeah, so.


You wanna do.


01 more.


you can write a ratio like 3:5 or 35


Yeah, because ratios are like fractions.


yes


Yeah, um.


OK, so,


You wanna do one more?


If that's fine with you.


On a different topic.


ok, one more


Thank you, Mr. Lee. Uh, let me take it out.


All right, Mr. Lee, um, let me write it out.


question.


Well,


All right, so,


Uh, let me try first, right? And then, uh,


You assist if I, and then you help me out later.


So let me try first.


Sure


Not sure what you are doing


Give me one second, I'll.


Let me just


Brainstorm.


Hi, Mr. Lee, uh, Mr. Lee, sorry. Uh, you can.


Come help me because I'm just, my way of doing it is just putting in numbers.


To see which one.


Uh, helps.


Which one solves X, which will take ages.


So


I would like to see the way you do it.


This way, next time I have this type of problem.


I can.


Do it with your help.


Meaning you're


Way


Go ahead.


Yes, that might take a while, athough it is one way to do it


Um


I mean, it's fine with me.


the strategy is get all the x terms on one side and all  the numbers on the other side


I was thinking that, so 7 X.


Plus 3 X.


If I'm right


Is that


Get the Exelon?


The exters


one way to do that is to subtract 3x from both sides


Both sides, you mean like from like.


Like, decide.


I


Decide and like a side.


Hm, OK.


Uh, this would be


32X1 is uh 7 minus that would be 4 eggs.


Or eggs.


Or uh eggs.


That


And then 3 eggs.


3 x minus 3 x would be 0.


That's a 0. So this is just 4 eggs.


Right, Mr. Lee


yes, 4x


OK, so that would equal 28.


So


I don't understand. Uh, did we solve the eggs?


So does that mean X is 28?


Now you subtract 4 from both sides


OK. Uh, so you would do.


Um, this?


That's


For no more.


Is that the way to go out?


Keep going


OK, so


That's 24.


24. I don't understand what we just solved. So like X is 24.


so  4x =24


You have 4x on the left and 24 on the right


Yeah, 4 on the left and 24 on the right. Yes, yes, I do. Uh, forex.


Right here


And 24 right here?


Oh


Oh, OK, so it's.


Let's get the


OK, so it would be 4.


X equals 24.


Now, we will do this, which would give us for for X


Right?


Take that away


That would be for, you would divide that.


24 divided by 4.


Now 4 goes into 24.


6 times


Right, 6 times, so 6 equals 24.


Right? So, so 24, so.


Minus that would be 0. So X is uh 6


Equals 6, right? Mr. Lee?


right


Right.


All right, Mr. Lee, um.


alright Mr. Lee thanks for all ur help


Thank you for your help. Uh, you have a good one.


You are very welcome.


You really helped me deep in my topic. That's what I clicked when I came here.


Uh, you really


Did help me throughout that.


I


Glad I could help


Yeah.


Have a great day!


All right, man. Uh, you have a good one.


Bye.


bye


Hello!


hello i wanna know about arithmatc sequence


okay


does your teacher want you to use the formula to solve these questions?


yea


1. 335


okay so ill write the formula on the board


do you see what i wrote


2. val=54.8 val=67.9 val=81


3. val=9


which question are you solving?


oh wait nvm


do you want me to check your answers?


4. val=130 val=136 val=142


4 seems right


for number 3


how did you get 9?


wait


also number 2 seems correct


wouldn't 3 be
42-6(4-1)?


i mean 45-33(4-1)


imean 45-12(3)


33+12= 45


oh wait


should use -6 as d
not -12


oho oki


are the rest all correct?


yes 2-5 are correct


but what about number 1


how did you solve?


300+50(8= 650


yeah great job
that should be right


do you need help with anything else?


okay so you would use the same formula here


Yea


since you know how to use the formula these should be easy


i can check over your answers once you are done


for what question?


if it is for question X then make sure it is -332


yes thats correct for question X


thats wrong for question 3


show me the plugged in formula for 3


273 for the next


yes


both are right


thats right


correct


plz cheeck


how did u figure that out?


-86 is right


but calculate for this problem?


wait it was for another assingment


alright I will have to go now


bye!


great work and bye!


Hi, hi


Hii! How are you?


You are very puctual


Im doing good


you are very punctual too


thank you, I try, *bowing to audience*


I tried to get the topics but they were not clearly mentioned


so, I got few tings


like inverse trigonometry domain and range with graphs


ok


how to solve inverse trigonmetry question, complex ones


i know how to solve questions like this


lemme get a hard question


2cot^-1(3) is a constant


such type of question, when there are multiple things inside a bracket.


This is for monday right?


yes


ok, I will prep for then. This is precalc or algebra 2?


is that (pi/2 - 2cot^-1(3)) inside the parenthesis?


please give me 10 min


I am having connection issue


Its ok. We can wait until Monday.


Hi, thank you for waiting


Yes, we can do on monday.


ok. sounds good to me


enjoy to your weekend


I have little doubt in a quadratic question.


can I ask right now?


sure


do I show you how i try to solve it?


If you want


squaring both sides


ok


I put the x values into g(x) = 0, to check if these roots statisfy the equation


OK.


to be minus.


wait a min


Good


so, according this , it means g(x)=0 doesn't have real roots righht?


Wait, what?


Why'd you get 17?


Or you check two different values for


Uh, the one.


yeah


Why did you, why did you check two different values?


Because the


plus/minus sqqr root 1 = + / - 1


You're, um,


Your one route.


Wasn't out of a square.


He didn't take the square root of it.


So it should just be positive one. I don't know why you tested um negative one as well.


-1 wasn't a route.


Right? Cause out of the um out of your algebra, you got two roots.


Um, your first route, uh, the 1/81, that uh


can be rewritten as a complex number to get you


To get you a, um,


The actual route.


But I think, I think it's just 11 real route.


how 1/81 can be written as complex numbeer?


I don't know. I was just trying stuff, and I think I found a solution.


So if you take


What did I do? Right.


If you go back to your 81, well, I solved it differently than you. I mean, we got the same result, but I solved this slightly different.


I said,


Let


okay


Why be equal to square root of x.


Therefore, my G of X G Y.


becomes


Y^2 minus 8Y.


Which just becomes


9 Y + 1.


Time


Why minus 1.


Which is what you got.


Except, um, you had it just basically squared, right? You got that.


X minus


Yeah, close enough.


Can I get, yeah, 9, yeah, and then if you


Factor back in what X is.


Or what why is you get 9 root X.


Plus one


Times


Root X minus 1.


You said equal to 0.


And for your first expression, you got square root of X minus 1 equals 0, so you got a square root of X equals 1 square both sides, you just get X is equal to 1.


Right, just a single root out of that.


Then you go to your 2nd 1.


Which is 9 root X + 1 equals 0.


Then you subtract.


Equals -1.


Um, but I'm a rewrite -1.


To equal


I squared.


So then if I square both sides,


Take the whole thing.


And squared, then I get 81.


X is equal to 14th.


So then X is equal to 14/81, which Ida the 4th is still just one.


But it helps the algebra later.


So you know your X is equal to 1 is is works as a route. So we plug in.


To G of X, we plug in GO


I to the 4th over 81.


Yes, to equal to 9.


Square root and the 4th.


For anyone


you chose only x=1 becuz when we put x = -1, it doesn't statisfy the equation righht?


Uh, then you evaluate, you get 1/9 out of the 4th.


Where did you get X is equal to -1?


So


X can't and in this instance, X can't be equal to -1 because it came from a square root.


That would mean, that would mean here that X is equal to -1


I squared.


It'll work the same too, but


Actually be Ida the 4th, so it'd still be a


Um,


ohh, got it! I was doing it wrong


Like you only get your plus or minus when you, when you take the square root, but in this answer, we were squaring it.


So X has to be positive.


If we have the square root of X equals 1, X has to be positive, so we can't consider the negative one.


yeah, right.


got it!


Right? So think of, so go back. So if you, if you even try to plug it in as a route, uh, to your


Other expression, just the 9 square of x + 1 into the square root of X minus 1. If you try to plug in a negative 1, you'll have 9 I + 1 times I minus 1.


Right, and that wouldn't


That obviously wouldn't work as a route.


So it wouldn't, it wouldn't even hold.


Anyway, uh, get back to my fancy pants.


So, minus 8/9.


I 2 minus 1, and we, we, um, substitute back in what eyes are.


So we get uh 1/9, because out of the 4th there's just 1, I squad is -1, so you get 19 plus 8/9 minus 1 gives you 0.


At least that's, I think that's how you do it. Normally, a complex number is like A + BI.


yeah


But you can't really do that with the square root.


At least I've never seen a square root of a complex number.


He asked that.


kind of square root.


complex number.


Huh, square root of a complex numbers always returns.


A conjugate pair


Mm.


I understand very little.


how squar root of complex number gives conjugate pair ?


That's what I'm saying.


They even bolded it for me, they highlighted it.


Uh, let's see.


I guess it's


The conjugate pair.


Of the square root of the


Oh, this is normalized. B over B.


Absolute value.


And A is 0.


Then you just get the


Really get


Be over the square root of B over 2.


you really see.


to calculate the magnitude.


Mm


OK, that kind of makes sense, really ugly.


Oh, it's gonna be absolute value.


So that's probably why it came out to um


This square root of X is equal to 1. This X is probably absolute value.


So you're right, it could have been -1.


But we don't see absolute value anywhere else.


B or be absolute value.


no, i wasn't right. sqr of x will be only 1


Oh, oops.


I went to another website and I'm still only showing you.


To close this.


I feel like


-1 would work if that square root was an abs was an absolute value of X.


So if it was the square root of absolute value of X.


Then yeah, that would work. You'd square both sides, and then you would, um,


Take the inverse of the absolute value in consider both sides, plus or minus.


But that would give two roots.


And we have a polynomial of


It's a degree of one, but we have the square roots, so it almost bumps it up.


I guess it guarantees that it's gonna be a um one reel and one unreal, or one complex root.


Let's see what's it.


What's the degree of a pole a meal with.


square roots.


Huh


If the square root is applied to the variable itself, the expression is not a polynomial.


Because it's got to be


In your exponents only.


Non-negative integer powers of the guru.


For this


problem, for this problem, what do you think it is?


Or at least um give me your logic.


To narrow the um multiple choice.


so, according to my method, I am getting one real root that staisfy g(x)=0


since as you told, we can only consider x=1, because that's what we are getting after solve g(x)=0


Right.


Is that it? Can you at least give uh


I guess that's it, isn't mean, no other way to explain away the other answers.


Though I'm hesitant to, um,


I keep thinking about my answer, my


Ida the 4th or 81, that's not a complex number.


And I'm trying to figure out


How would I solve this? I guess I'd have to do


Um


What if I did the


Pythagorean theorem.


For why. Let's do that.


If I had


Negative. Let's say why is equal to.


8 plus or minus the square root.


64 minus 4 times.


CA is 9.


C is -1.


Over.


18, 2 times A, which is 18.


8 square root of 64 plus 36 100.


Oh, that comes out pretty clean.


It becomes -8 plus or minus.


Over 18.


Which is, um,


4 plus or minus 5.


09. I think that is


I think there's no real roots.


Hm


This is a head scratcher.


how?


Because we get a conjugate pair out of this, which is two solutions.


What we


I mean, it's, it's not, it's not a, I mean,


It turns out that this is not a polynomial, because


It has a square root term.


In a polynomial has to have uh an expression with


Non-negative.


Non-negative exponents in the square root is not, is not an injury.


So this is not a polynomial.


Oh, wait, this is just why.


I got to take the square root of this.


Right, as you got, all right.


X is equal to.


Um


Sorry, Y is equal to square root of x. So X is equal to Y^2.


So if I square this,


It'd be


How do you square complex number? Do you have to take the


Determinant or do you


Do you know how to square a complex number? Good little trivia for you.


Oh


ABI 2 ABI wouldn't it? Oh, you played the squire, I square. I see what you did there.


Clever


So it should be


Plus 5 nights, which is where I get 36?


I'm done. That should be 16.


Minus


Uh, 4 times 5, I guess be plus, yeah, minus.


Over 81 I.


Which is equal to 16 and 25 is 41, so 41.


0 81.


minus 20/81 I.


So that is


The other route.


And I do not want to plug that in. Oh.


Because what would be


The square root.


Would it be the conjugate pair or just one of, one of the pair?


I had the chests both cases.


how did apply this formula, you have take b of x


but here it is b of root x


Oh, it's minus 81.


Whoops.


Yeah, that's what you got, right?


Because the B square, I 2 is minus, so it's be minus square. So this is actually


minus


So 16 minus 25 is 9.


So this is actually 19th.


negative one night.


-20 over 81 I.


Now good question here would be how do you take the square root of a complex number.


I got a lot to practice for Monday.


Oh, that's what we're looking up earlier, square root of a complex number.


How did you apply this formula, you have to take B of X, but here it is B of root x


Say that again.


So I, oh, whoops, I forgot to my eye here.


Um, I said,


A was -4 nights.


And B was


5 nights.


So then I just square, I did a squad, which is 16/81.


Um, it's supposed to be minus i^2 because I 2 is -1, so minus 25 or 81.


And uh


16 minus 25 is -9.


That's why I got the negative 1/9.


The of.


Um,


Did I get, oh, I see what you did wrong. I see what I did wrong. You're saying that


B is supposed to be.


You're right. So it should be a positive for nice.


You're positive for nights.


I guess it doesn't really matter?


i am getting y=1 and -1/9


The calculations, cause I'm still squaring it.


You're getting win, you go to 1 and negative or nice.


From the Pythagorean identity.


yeah


Like a theorem.


Hm.


All right, let me just delete all that. I got no room left.


Start over.


noo


It's not good.


No


Sorry, I'm pretty decisive, unfortunately.


Uh


you put the sign wrong


We said B.


for b


Yeah, yeah, I was, I would try to fix B this time.


So it's 8 plus or minus the square root of 64 minus


Um, for A, which is 9C, which is -1.


All over.


Uh, 2 x 9.


So


Over 18 plus or minus.


Square root 9.


Over 18, which is 4/9 plus or minus.


Um, 10 or


I.


Wait a minute.


That is not I.


how are you getting i?


What am I doing? This is not a complex number then.


yeah


I threw eye in there because I thought it was.


It's supposed to be a complex number, but now,


I'm throwing my hands in the sky.


I can, um, formula.


Is it?


if you solve it further, you will get right values of y


If I saw it further, you will get right values of wine.


Oh, I see what you're saying. So, going from this,


If I have


4 + 5 is just 1.


And the other answer is 4 minus 5 is -19.


Did you get negative one night?


feel like you did. You got 1/81, which is much different than what


We got earlier.


All right. And we know that one works.


Let's check. So, y.


This, oh, this is for why. Whoops.


yup!


So, y


is equal to x.


You go to the square root of it, so we get the square root of one, which is one, and we get the square root of -1/9, which is just


I times the square root of 19, which is equal to


I over.


Does that make sense? Huh.


no


you have to squre -1/9


it will be 1/81


Why do I, oh, you're right.


At the square, not square root, I went the opposite route.


So it just becomes 1/81.


yeah'


Hm


She's like every instinct is telling me.


Something's wrong.


Or everything is right, but so we put we already tried 1/81.


The only way that it would work.


is if


If we take a square root of 1/81.


It's always gonna be positive unless


This had an eye on it, but we can't.


No, cause if you take the square root of -1/9


That won't give you anything.


It's like the square root or the square root.


yeah


I feel like I'd put


It's one real route.


Because math


And leave it at that.


Just throw my hands in the air.


what?


Well, the second answer we found.


Isn't a root, because it doesn't.


Cause our function to go to zero.


By definition,


A root causes of function to go to 0.


Let's see the polynomial root theorem.


They said a constraint, rational rational solutions or polynomial equation.


not always, right?


The rational zero theorem is used to determine the rational routes.


Huh? Not always.


a root doesn't necessarily always make equation 0


A root doesn't necessarily always make.


Equation 0


is that right?


I feel like that goes against the very definition.


Um, is you said determine the rational routes that you can't use a route to, you can't use a definition that


Define it.


All right. Or you can't use a term to define a term.


like circular reasoning.


OK, here we go. So here the values of x that satisfy the equation F X equals 0 are known as the roots.


Or zeros of the polynomial.


I, like, by definition, a root has to zero the function.


yeah,


So if


A solution we found.


Does not cause it to zero, then


It's not a root.


when c, we still have roots


but they don't satisfy the equation


And see greater than 0, we still have roots, but they don't satisfy the equation.


I see what you're saying.


But remember, we said,


I see, I see what you're saying. So, when you're basically going back to a definition that when A is greater than 0, or when A and C are the same sign.


There are no, um,


There are no roots.


yeah no real roots


On the graph.


Hmm.


I can't help but think, um.


I can't help but feel that's wrong.


Cause you're right, graphing it, that comes true.


But then,


If we say X^2 plus.


2 X


Plus one


Right, which is


X + 1 squad.


The route would be x equals -1 twice with a multiplicity of 2.


This has a


Um, it has a C value. Remember we said that C is not, um, strictly


A vertical.


What term shift.


Um


If we can


yes


denigrate this back to


Our, um,


I feel like P and Q are in the same place. So this might have


My example might have a intercept.


Or, or the, I'm sorry, the


What's the term


My God, she's, it's only been a day and my mind has just completely gone off the rails.


vertex?


Vertex, yes.


Where is, uh, so I'm trying to look through my notes where we did all these derivations.


We said that


So B is equal to -2AH.


So be would be


Well, let's apply it to our original expression.


Um, I guess I'll keep it Y. 9 z minus 8 Y minus 1.


So if we solve for age,


For the horizontal.


My goodness, my brain is horizontal ****, the horizontal shift of the vertex.


We get


Wow, I'm getting, I'm getting flagged for words like shift.


And


Asympto


Apparently, I'm pronouncing them too close to bad language. I'd be very mindful of that.


K + AH squared is equal to.


C minus A H 2.


Rather than me calculating, I'm just gonna throw it up on Desmos. This is.


My mind is not.


can we consider 1/81 as complex root?


Sharp is it was yesterday.


All right, so I'll do both forms. So 9 X squared.


Minus 8 X minus 1.


OK. So we see,


We have


Two intercepts.


We got one at one.


One at -1 over 9, so we're right on that.


Mm


So let's try 9 X minus.


E. Where's my square root?


X


There it is.


Because of the square root,


Um, the function doesn't exist.


Uh


When X is less than 0.


I knew something was different.


Even veers off differently, look at that behavior.


Wow.


OK.


As a as a smaller


Um, slope doesn't explode. It's not as narrow, it's wider.


we consider 1/81 as complex root.


I want to say yes.


I feel like part of me wants to say yes.


because it exist as a root but does't satisfy the equation


I feel like this left hand side.


Because it exists as x < 0 doesn't satisfy the equation. Yeah, I feel like the when X is less than 0.


Um


It only exists in the complex plane.


Because we would by definition include I.


Because we have that square root of X.


So when Xs say,


equal to -1.


You'd have


9 times -1 to be -9 minus 8 times I.


So you have like -10.


Minus 8 I.


Yeah, I think this does only have one real root, which is an integer.


yeah


It is B


Man, this is annoying. Why can I not understand this unless I grafted it.


A square root threw me off.


Oh, and the behavior is different because


It's a line. It's supposed to be a line.


Because 9 X.


It's supposed to be alive, but the square root behavior.


try value. If X is equal to 0.5.


Let's say 0.25.


Then you have


9 times 0.25, which is 1.5. No.


It


Minus


8 times 0.5, which is 4. OK, so that is negative.


Hm


That's so weird, that a square root can override a line.


So much


To dictate his behavior like this.


It's basically linear.


Yeah, with a climb of 15 to 80.


is it necessary when x < 0, it goes in complex plain? becuz there are so many instance when w and it satisfy the equation too


Yeah, it's pretty consistent climb.


Once we get past a certain point, it just becomes a line. I think once we get past one.


It just becomes alive. Is it this?


This is annoying. I can't hear when you


Type in the chat, it should notify me with like a noise, but I don't get it. I keep


Missing when you type in.


Do you mean when X is less than 0?


yeah


Not when X is greater than 0.


OK.


I mean


x < 0


Yeah, I mean that by like by definition, unless you had um the square root of the absolute value.


Then you, we would have, um,


Then we'll be able to define it when X is less than 0.


But as soon as we


But as soon as we try to go,


With a negative X, it immediately turns complex.


how does it immediatey turn complex at -x?


Because the function, uh, sorry, go ahead, ask your question.


is this x root?


Well,


Let's give an example. Try using X is equal to.


Negative


Um, 0.25.


Right? And solve that into your


Um, G of X.


I had to clear the board.


All your wonderful work unfortunately, has to.


pay the piper.


i can solve on my notebook


OK. So try solving, um, the 9 X minus 8 square root of x.


Um, minus 1 when X is equal to -0.25.


when i put x=-0.25 in g(x)


it gives -16


Did you solve it on your calculator?


Or did you solve it by hand?


calculator


That doesn't seem right.


Did, you're not doing square, you're not doing y squad, are you?


no


Cause you should get 5 minus 4 I.


oh wait.


no, my g(x) = 9x^x - 82x - 1


Right, that's not the right equation, your G of X should be 9 X.


Wait, X to the X?


To be next to a second. Anyway,


sorry 9x^2


Your GMX should be 9 x minus 8 square root x minus 1.


yeah,


But you have no square root X in your


Oh, I see what you said 82 X.


Your terms are way off. How do you have an X 2 in a square to x.


I am baffled. I have no idea where you got -16.


Can you do your work on the board. Let me, um,


Mm


yeah


Make some space unless you want to do it for me.


To my work in the years up there.


do i erase work from the board


Sure.


I'm already working on it.


ok then


Um


I made some room close to the bottom.


Where I started working.


yes, I see it


Oops.


Why are you doing all this algebra again?


Are we not just plugging in


X is equal to


To our original expression g(x) to see if what happens when X is less than 0.


I thought, let's make sqr root dissappear then, plug in x value.


I think i did it wrong


Yeah, cause if you get rid of the square root, then yeah, obviously you're, you're gonna be able to define um when is less than 0, but that's not original expression. Now you're manipulating it, um.


But that's not the original expression.


That's the same as saying,


If we look at the


If we look at X + 2.


Well let's go with X minus 2.


Over X^2 minus 4.


Right? And we know there are asymptotes that X cannot equal.


Negative.


Um


Actually, let me rewrite that. It should be.


The other way around.


If we had y minus 4 over X minus 2.


Right?


Now, in this expression,


We have an asymptote ver vertical as until X is equal to 2, right? Cause at that point,


It becomes undefined.


You with me?


yeah


right


OK, but what happens when we do geometry, when we do algebra and we factor out X + 2.


X minus 2 in the numerator.


Over X minus 2, and we cancel out, then that just becomes X + 2.


Now, the point is, is, is X is the graph of x + 2 the same.


As these two are they interchangeable.


And I would say, sorry, go ahead.


What do you think?


they are same


same graphs


They're the same? OK, good.


Let's


See if that's true.


Uh, I'll leave this up here. Let's go with simple one X + 2, right? Just mine.


Now, let's try. X to the 2nd.


Divided by X minus 2.


They are


The same.


Yeah, so they are the same. It's the same. So it's the, huh, it's defined at


I go too.


Interesting.


But then why are


So that makes sense, I guess. It's the same expression, but then,


Our original expression is, if we consider the two expressions, these have different


Values


9 X word. Oh, it's because of the


Um, let me try this. If I change my coefficients to 81.


And 64.


This set out to be plus one?


Let's try plus one. Yeah, no.


One.


Yeah, no, still not the same.


Where did you get? You ended up getting 81.


Thanks


Let me see 2 weeks.


I see.


81 X^2 minus 82 x + 1.


So it is plus one.


But you're saying


I mean, it has the same


Doesn't behave the same.


Um, continue, sorry, continue doing your


I know you did it earlier. Did I delete it?


Now you did up here.


You had 81 x minus 1 times X minus 1. Let's try that.


no, it's correct


81 X minus 1 times X minus 1.


Yeah.


Those aren't the same.


Oh, I'm sorry, I didn't show you the graph.


But we see we get um you can't tell, but they're the red graph, the original 81 a minus 82 + 1.


They're, they're both the same, these two expressions when it's factored. But you see, you still don't get the same behavior that you do with our G of X.


yeah


This is like this black or the equation you're finding is G of X^2.


So what if we took


The


Square root of this whole thing.


Uh.


Would you look at that.


it's different from g(x)\


Let me try.


It's diff you're saying it's different, so let's try the negative. If I have to include.


A go though, huh, there it is.


Just as drastic squares it.


It's different from GOX.


But you're taking your GFX.


Um


And at some point you're squaring the whole equation.


Oh, you're saying you specifically define G of X is equal to 0.


And then squared it. So only at x equals 0 with this.


Expression hold.


81 X minus 1 times X minus 1.


yeah


in the function, when the function is equal to 0.


Oh, basically the X intercepts.


Yeah


yes


Damn, you're outsmarting me.


Hm


Oops did not curse. I said dang.


I apologize.


I can use.


I thought you knew it.


therfore i didn't mention it


I think I got confused because I was graphing um


G of Y, which was


Basically where Y is equal, where X is equal to k.


So my coefficients were the exact same as G of X.


Yeah, still.


But I still believe


That


The function G oh, what just happened?


Sorry, Sam.


To be relaxed for a second.


Um,


I still think


I still think because of this square root of X.


X has to be positive. It can't be negative.


Even if, cause if you have to, if you have to manipulate it.


Well


Uh, I guess even if you manipulate it because the parabola is still a parabola, right? A function is still a function even if it's just a different form or you.


yeah


it won't change


Still, though,


Um, if we look at the answers,


They're specifying integers.


So we focus on that. How does that change?


What we've been doing.


looking for.


ohh, I didn't focus on that


I think 1/81 got dismissed becuz it isn't an integer


it is real but not integer


Yeah.


I think, I think because just like we're having this big debate.


On whether there's two real roots or not, or if there's just one in one complex. I think they wanted to mitigate that and just.


Just so there's no ambiguity.


In the test.


yeah, both are real roots but only of em is integer


OK. Well, that's that problem.


Do you have any more? I gotta, I gotta get going.


You wake up early.


I don't have any more problem


if you want, we can take tomorrow off


Yeah, so if you want to do that, we can wait till Monday evening.


I mean monday off, then start from tuesday.


Oh, OK.


is to tomorrow night.


OK, that's fine. We want to start again on Tuesday.


Take 2 days off.


it;s just one day off --> monday


we did class today


Right, and today is, oh, it's Sunday right now.


Right


So we can


Meat


2 Tuesday night, Tuesday evening.


yes


All right. Well then, have a good couple of days, and I'll see you.


Tuesday.


Sharp


yup!


Bye.


Get well soon! bye


Was that you? I'm ready now


Hello


How may I help you?


Hi Chris


I have doubt in question of quadratic equation


Can you show me the problem you are working on?


If f(x)=9x−8 sqr root x such that g(x)=f(x)−1, then which one of the following is correct?


(a) g(x)=0 has no real roots
(b) g(x)=0 has only one real root which is an integer
(c) g(x)=0 has two real roots which are integers
(d) g(x)=0 has only one real root which is not an integer


I am not quite sure of the equations. did I write them correctly?


it is right


You should start by writing the equation for g(x)


ok


g(x) = 9x - 8 sqr root (x) - 1


OK, we need to solve that for x


yes


I took 1 on right side, and square both sides


Usually in this kind of equation, you get the √x on one side by itself and then square both sides. That gets rid of the square roots


ok, got it!


so, do i keep "1" on LHS or take it RHS?


If you brought the 1 over to the RHS, then when you squared both sides, you would still have a square root term


now, when you square both sides, there will be no square root terms at all.  Make sense?


I got the first para, that taking "1" on RHS will have root x remain in equation


i didnot understand second one


Let's square both sides now


ohhh


you took 8 root x other side.


Got it!


Now you will end up with a quadratic equations with no square roots anywhere.


yes, that's right


But when you square things, you must be very careful to check your answers in the original equation, because you can introduce extraneous roots


can you solve that equation, do you think?


yes, I did it.


I am getting two values for x


x = 81/3 and 1


sorry, x = 1/81


You need to check them in the original equation


Do they both work?


x = 1/81 , doesnt make equation equal to 0


but x = 1 statisfies the equation


Right, 1/81 is an extraneous root.


so x =1 is the only root


what is extraneous root? root which doesn't satisfy the equation?


An extra root that doesn't satisfy the original equation. WHen you square equations, you can introduce roots that don't actually satisfy the original equation. Those are called extraneous roots


Ohh, ok, understood.


so, for this question, it will be option (b)


Yes, I believe so.


yes, it is right!


I have one more question


ok


can I erase the board? and write question there?


Yes, or you can also upload an image


I upload the picture, is it visible?


yes


I don't know what they mean by AP, GP, and HP


it is arithmeic progresion, Geometric Progression, harmonic progression


We can leave this question, then.


I have its second part which is only related to quadratic


My first impession is that Descartes's rules would be helpful here, but I will have to work it out.


i tried to apply it but I didn't understand how to solve it further


Yes, it uses Descartes's rules


The sum of the two roots equals -b/a for ax^2 +bx +c


but they tell us that alpha=beta


how did you know alpha=beta?


Because it says that the roots of the equation are equal


yes right! I didn't read it properly


understood!


I have one more question


ok


can you explain like how do i approach such questions? what should I be thinking first? I donot understand these type of questions


What kind of curve is x^2 +2x+11 ?


Not sure?


wait a min


is this right?


So what is the minimum value of the quadratic?


will we take lowest value of x here?


as minimum value?


what is the minimum value of the quadratic?


I donot know


Yes, you do. You have written it on your graph.


-1 is the x coordinate of the quadratic. But the minimum VALUE of the quadratic (that is, the y value) is 10


Any questions about that?


so, the lowest value of y coordinate is minimum value


WHen we talk about the "value" of a function, we are talking about y, not x


so the minimum value of the quadratic is 10


oh, so when we talk about x, how do we say it?


x is the x coordinate.


So now you need to calculate the log base 10 of 10


that will be 1


option (b) is right answer.


yes


I got one more question


How many more questions do you have?


One one last question, but that's is of intergral calculus


Where are these questions coming from?


“Intermediate Algebra” by Descartes


On this, all I can think of at the moment is for you to expand it out


ok


would you like me to do on screen or write the final answer?


You might want to work on this question by yourself for a while.


alright


Perhaps then you can request another session


Can i request another session with you or new teacher?


You won't have any difficulty getting someone to help you.


ok!


I have one last question


You always have one last question


last for sure


are you in summer school?


Yes, I’Maplewood Academy entrance exams.


can we think like, no. of roots is equal to power of equation.


like quadratic equation hae 2 roots


so if power is 4, then roots should be 4 too


Yes, but some of the roots might be complex, not real.


oh! in quadratic equation, real root = complex roots, like one real, one complex


so, if roots are 4, so will it be 2 real , and 2 complex?


No, complex roots always come in conjugate pairs


you are right, the equation is same for all the roots


Perhaps there is a simpler way, but I would do is expand it, and try simple roots using synthetic division and reduce it to a quadratic. then use the quadratic formula for the quadratic.


I will try it, then get back to you!


Ok.  good luck


bye


can you tell me one last thing, how can i request your session?


You request a session just likk you requested this one.


you will come?


It won't necessarily be me.


you are really good. and i have some familarity. that's i am asking


but there are many math tutors


i will have few question next time


ok


ok, thank you


you are very welcome


Hi, how can I help?


Hello


how far have you gotten into this problem?


for example, based on the behavior of the graph. what function is it?


tangent


right


Now, what shrinks or enlarges the tan or trig functions?


i got that the period is pi / 6 and that B = 6 but im not sure if that is correct


the B value


Im assuming you are talking about the Atan(Bx-C)+D


form


yes


the B value looks good to me


What do you think C is


ok


im not sure how to find c


Well, It's the phase shift


there is none


so C would be?


rightt


now D?


what does D represent?


midline ?


close


C is to x-axis what D is to y-axis


also to clarify, C is not phase shift, C/B is phase shift


I miswrote before


vertical shift


C is shift


correct


what is the vertical shift for this graph


wait vertical shift means shift means updown right


right


think y-intercept


for tan x the x int is at 0,0 so its the same here the same x ints so wouldnt it be the same


right, but we are asking about y-intercepts


oh its 0 for here then


example of a vertical shift


yes


so d = 0


last is to check A


how do we do that


what does it represent?


like how far apart the VA s are?


not quite


think of a parabola


what does the coefficient to the x^2 mean


eg. y=ax^2


what happens when a is big?


further from the y axis


doesn't affect intercepts


a stands for amplitude


It is what either shrinks or stretches the function vertically


okay


a big amplitude will stretch and a small amplitude will shrink


this is based on a normalized A=1


so
A>1 stretch
0<A<1 shrink


got it


sorry


|A|


so
|A|>1 stretch
0<|A|<1 shrink


A can be negative as well


can you see the graphs im sharing?


yes


If you can see the blue graph


So it gets stretched


I switched to A=10 to make it more obvious


yellow is the standard tanx graph


green is when 0<A<1


see how it is squeezed vertically


A<0 flips tan around y-axis


yes its like hugging the x axis more


'hug' was the exact word I was thinking but didn't want to confuse


you


purple is a negative amplitude


all that to say that amplitude is A


now back to our question


how to find A


easiest way is to find where pi/4 is


on x-axis


as tanpi/4 = 1


oh


now, what 'pi/4' is depends on the phase shift and what B is


'pi/4' is dependent on the period


you can think of it as period over 4


okay


we git that the period was pi/6 rigjt


right


so it would be what for this problem


the important x-value


hello?


i dont understand


what would (pi/6)/4 be?


ohh


pi over 24


so what is the y-value for pi/24


that's what I get


Now put it all together into the general form of tanx and what is the function?


y = A tan(B(x - C)) + D


yes


plug in the values


is A = 3?


for 'ABCD'


y = 3 tan(6x)


Another tip for finding the amplitude


You need at least one data point that is not on the x-axis


so I recommended using the period/4


You may not always need this as they may give you some (x,y)


Then once you find the shifts and phase


wdym by one data point not on x axis


assuming you don't have a vertical shift or 'D=0'


using a point of form (0,y)


Wont help you find A


as y=Atan(0) = 0


ohh


you want x=/=0


so that you can calculate A in form A=y/tan(bx-c)


now if you do have a vertical shift


A = (y-D)/tan(bx-c)


I can show you this on the boad


board*


yes please


does that makes sense?


we used the point (pi/24 , 3) for this problem


yeah


does that math make sense?


yeah it does


Now, this is what my first instinct is to try to get pi/4 inside tan


because I know tan(pi/4)=1


imagine using the point (pi/10,y)


you will need a calculator and then have to use an estimate as the decimal number will be long


okay


using x=period/4 gives me a cleaner solution


another way of thinking it is just to take the distance between the vertical asymptotes and divide by 4


will that always work?


the distance between vertical asymptotes? yes


as that is the period


the period/4 will work as well


this is a methodology that you will only get from doing it


oh right


I had to learn it intuitively


okay well i have to go now


but thank u so much for being patient and helping me with this problem!


No problem glad to help


was that all the help you needed?


for now, yes!


great!


Hello! How are you? I need help completing a homework assignment that's due soon!


Hi


Hi, are you there


hello


You can turn on your mic if you wanna speak, or else you can type in the chat box. You can also listen to me by increasing the volume.


Okay I hear you!


OK, yeah


Good


So you're working on this problem, right?


yes!


OK


So


how do you think should we approach this problem?


I already know the givens which are lebeled by since this is a very new topic to me I am confused


Mm OK so a paper license just.


OK, no problem. Uh let's, let's see this together, right?


we know that angle B is equal to angle D and BC is parallel to AD.


OK. Uh, so, wait.


hm


Do you know this concept that if we have two parallel lines


let's say these are our two parallel lines.


L and M, and if there is one transversal, then


this angle will be equal to this angle if this angle is X and this angle is Y, then X will be equal to y. Do you know this point?


This is the property of parallel lines, right?


Yes!


Yeah, so we call this alternate alternate interior angles, alternate interior angles are equal. So if we try to use the same concept here. So we know that


uh just give me the


Yeah, so we know that this is one line that we have


right? And this is another line that we have, and both of these lines are parallel


If we consider this our transversal.


So can we say that this angle will be equal to this angle.


Yes! So angle BCA


So angle BCAS.


angle BCA equals angle


Tell me the name of the other angle.


is congruent to angle ACD


CAD


It's congruentumangal ACD.


AC


not ACD


BCA is uh your


this angle, this angle is BCA


and this one is equal to CAD, yeah, it's correct now.


And see


I'm sorry


Let's see AD.


OK. So now we know this point.


Um, now, what do you think should be our next step? So we have these two angles which are equal uh in our both the triangles, what else do we need?


Hmm


Do you have any ideas or something.


Yeah, you can draw if you want on the whiteboard.


I think we also need do show


Hello


Hello. Hi


Hi.


Hi. Sorry, my, my.


dizzy


Uh, yes, you're audible.


So


Mm


I like started this topic, but um,


I


I only started it like on Friday and we were given a homework assignment for it, so I do know the property that you already explained. Um,


OK


I'm just like confused on like how to apply it into proofs.


Um, OK. So, uh, do you know how to, how do we prove two triangles to be congruent.


Um


OK, I, I'll teach you.


Yeah


so how did you learn like side, side, side and like side angle side.


Uh, yes, let me write that down for you. Side, side, side.


uh side angle, side. What else?


And we also did Angle side angle.


Angle siderounding


and RHS right? or HL?


Yeah. So, let's see what can we apply here?


So we know that this angle is equal to this angle, and this angle is equal to this angle.


So we know our two angles, right?


Mhm


Yeah.


So


that the properties were SSS.


ASA or


So we already know we side, side, side.


So side, side, side, uh, we can't apply this here because we haven't proved that all the three sides are equal. Do you know that the, we have this site which, which is equal in both the triangles, right?


Mhm. So it can't be side, side, side. And I don't think it can be side angle side either because we already have um two angles.


I do


Yeah. So if we think about ASA, you might think, OK, so we, we have proved that these two angles are equal and this side is equal. So, so we can say that this


Can we do reflexive property for ACA's congruent to AC.


AC is congruent to, you know, what is reflexive property? Can you explain that?


Um, like


it's pretty much just repeating that like one segment is equal to itself and because Triangle ABC has segment AC and triangle ADC also has segment AC. Um, we can use reflexive property to show that they're both equal.


Yeah, we can do that


OK, so now we have two angles equal, uh, what is this one and the other is angle this one. So, and we also have one side which is equal. So we might think that we can now prove uh that most of these triangles are congruent by ASA, but this is not true. For ASA, the site should be the included side, the side between those two angles. So if I'm talking about these two angles, then this one should be our side, right? So


Yeah


we can't use ASA on these three properties that we know. This is the first one. This is the 2nd 1, and this is the 3rd 1. So, now we know that this angle at this side so we will try to prove that this angle is equal to this angle.


Then we can use ASA for these, right? Because the site will be the included site, like the side between these two angles.


Yes


OK.


So how, uh, how can you, can you prove that let me change the color.


How can you prove that this angle?


is equal to this one


Um


Um, yeah.


I'm not sure what it's called exactly, but can we say like, because we already know that um


Engle BCA is congruent to Engle CAD because it's technically half.


um, like the other side would be equal.


So, we can't say that it's half. How do we know that it's half?


Because


We don't


oh OK


Yeah.


Yeah, we don't know that. So, again, we'll use the


um, what, what else can we use? OK. OK. He is out. We know that angle B equals angle D and let's say if this angle is, let's say X


and this angle is Y, so we know that this angle will also be X because both of these angles are equal, right?


Yes


Yes, and also, if this angle is by that, this angle will also be why because both of these angles are equal.


Uh, so, uh we also know this property that for a triangle.


Um, we can say


like


uh that some of all the angles. So this is B A B C Agula plus angle B + angle C 180 degree, right?


yes


This is the angle some property. So if I say that one of the angles is X, the other one is Y, and I need to find this angle. How can I do that? I'll just substitute these values, right? A is X, is B is Y, plus angle C is equals 180 degree uh minus X minus Y minus X minus Y.


uh and the C equals 18T minus X minus.


Is this clear


OK. So, if I use the same thing here, so can I say that this angle


which is angle BAC.


Ale B A C equals 180 degree minus x minusy.


because for


for our this triangle that we have here.


Mhm


some of all three angles should 180 degree, and if two angles are X and Y, so the third angle should 180 minus X minus Y.


Then only the sum of all the three angles will 180.


Yeah


Uh, OK, all right. Uh, is this clear or not?


Yeah, I understand it


Yeah, or else we can say that for this triangle,


triangle A, B, C


Um, I'm gonna be AC + that BSC is this one, right?


That's ACB


A C B ACB is your this


plus angle CBA.


C B A


equals 180 degree


right?


And we know that angle ACB.


ACB is Y.


So I'll write BAC + Y + CBA CBA is X


equals 180 degree.


and


from here


I'll say and the BAC equals 180 minus X minus Y.


Is this clear now


Yes


Has it subtracted uh X and Y from both sides.


Mm.


and then you would get angle


BAC's equal 180 minus, yeah, OK.


OK. Yeah


Mhm


Uh, in the similar way


if we look at this triangle, angle A, D, C, so I'm sorry, triangle ADC.


No triangle ADC.


Um,


we would do the same.


Yeah, we'll do the same for a triangle ADC


uh I see AD plus


right and we CAD plus


uh


and then ABC.


plus


CAD ADC and DCA.


Yeah


And the DCA and the sum should be 1ically because of, because it's about all angles in a triangle 180 degree.


Uh, now, I'll, I'll do the same thing. So, can you help me with the next steps?


Um


Oh


well, like we would do another um


statement after this one.


See, like, I understand what you did to show that they're like congruent.


but


um


wait, I'll add a picture to show you, like these are the statements that um I've like been using to show the reasoning that we just showed.


um, like right here.


OK


Uh, reflexive cogitive substitution, vertical angles are congruent all right under the communal final angles opposite of congruent sides.


are congruent


OK. And I'm will bisected divides and into two.


congruentambles.


I'm sorry.


Uh, but lines


The transversal


from congruent alternate interior angles.


but uh you can use this statement, of course.


To say that um to say the angle, BAC is congruent to PCD.


Oh


uh, can you please repeat? I'm sorry.


To say that CAD is congruent to ACD just like um we just proved.


Yes, yes.


Yeah, you can do that.


Um


it says that it like


it doesn't work because um they're either not alternate exterior angles or we haven't proven the applic applicable lines to be parallel.


Uh, no. Oh, these are, which option did you choose, the last or the second last one?


Um, I just tried both, and that's what they say.


How is that possible? So it says parallel lines cut by a transversal form congruent alternate interior angles, this should be true because


um


because we know that line BC is parallel to AD because this is given and we also know that AC acts as a transversal hair. So


so your angle BCA should be congruent to CAD AD.


because of this property.


are you sure


Um, this is like also what I have so far, maybe like I'm missing something.


And then like when I click try it like says that um


it doesn't work for that proof.


Mm


Uh, can you share your screen if that's possible?


Um, how


Mm.


Oh, OK. Um


I said like a picture of my screen, like so far.


um, on the website.


below where we did the math.


OK


then you of other BCA


is I went to CAD PCA.


E C A C A D, OK.


Uh panel lines set by transversal um form alternated interior angles. This is correct, and AC is AC is congruent to AC.


uh reflexive property and then


angle BAC and ACD.


BAC


B A C N A C D OK.


Uh pal and cut patterns were still form concrete alternate interior angle.


Yeah, this is, this is what I'm telling you.


Uh, for your


So far, like up until question number 3, it says that everything is right. Um, step number 3, and then when it comes to step number 4, it's, that's when it gets wrong.


Yeah, exactly. You, you can't say that this is like, I mean, over here, it's correct because say


uh this is a, this is, right? And you know that these signs are parallel, so this angle should be


just a second


This is your line.


We can say that this angle will be equal to this angle because we know that these two lines are parallel and this acts as a transversal, but how can you say that this angle will be equal to this angle. We don't know whether A, B, and CD are parallel or not.


So you can't use this seasoning over here. I mean, uh, the statement is correct, but the reasoning here is wrong.


So.


yeah.


Mm


Mhm.


um


Mm.


I'm


trying other possible reasonings, but


Oh, what are the other questions?


there's none of them seem to work.


Reflexive property won't work here.


Uh


that


Um, well, technically reflexive property. I don't think reflexive property would work because it has to be the same exact angle.


Yeah, about to saying the same thing like it won't work.


All right, I'm


Yeah, none of them work for that statement.


this help to work.


vertical angles are congruent, these are not vertical angles, in triangle angles opposite to confluent sides are congruent.


Um.


angles opposite to congruent sides.


So do we know that these are consument sides.


No, we don't know that, so.


this will also not work.


An angle bisector divides an angle into two congruent angles.


but we also don't know whether this is an angle bisector or not.


So that is also not a correct reasoning.


if I also eliminate this


and


this also won't work


uh exterior angles for exterior angles, what can we say?


Um, these are not exterior angles.


So this also won't work.


So the only two options we have is transitive


or substitution


No


Um, when I used


both of them, none of them work.


they don't, oh, they don't work. OK.


Uh, OK.


No, maybe we have to do like a statement before we get to the statement.


Um, so you are saying that we should change the statement.


yes


Um, OK


Mm


So we have


sorry my mice is glitching again


ASA, SSS.


SAS


Just, just give me a minute. Um, I am thinking about this


I figured it out


OK. That's great


The answer is AAS


with the final proof


AAS Yeah, AES is another criteria which we can use.


So AAS because this angle this angle and this site


Yeah


we can use that


So we can just say triangle ABC is congruent to Triangle


ABC is congruent to.


CDA


It it may


you are saying it's coming to CDA.


A B C C D A


A B C C D A. Yeah. Yeah, yeah, correct.


Well done


Basically, we have 5


Thank you!


uh congruency criteria.


Yeah, sure


Do uh do you want me to help you with any other question?


Yes Please!


OK


Yeah


Uh, can I just clear the whiteboard or do you wanna


write something down


we can clear is!


Just uh share the next question and we can work on it together.


So


uh, is the figure same, or, or, or the figure is not given, I guess.


That's good


So given angle


A B D


is congruent to anger DCA.


BC is perpendicular to AD and.


AD is perpendicular to CD through triangle BAD is compliment to CDA and


and it's OK, given is over here.


So the 5 criteria that we have


SSS


ASA


SAS


A A S


RHS


Triangle BAD.


Let, let me just highlight these triangles for you.


For this one Im taking notes so it may take longer. Just a bit tho!


OK, no problem


B A D C D A.


So one thing which is uh very clear.


is that both of them have uh these 90 degree angles, right?


Yes!


So we can very easily say that, uh, these, these angles will be congruent. So I can say that anger BAD.


is equal to angle


Ace ADC.


because both of them are 90 degree.


and this is sort of a given statement.


Uh, what else can we say?


We can say that this airline AD is common in both of those, so we can use that as a reflexive property. So


the equals


AD


reflexive property.


So now we have one angle and we have one side.


What else can we do


Mm


We also know that angle ABD, ABD this angle.


is equal to anger DCA this angle.


These angles are equal because it's given


angle


A B D equals angle A, C, D


because this is given


So now we have


two angles and one side, and the side, uh, this is the site that we are talking about. The site is not included. So by AAS.


we can say that our two triangles are congruent.


right


Is this clear


Yes!


OK. So by AAS


uh, now let's try naming those triangles.


or, or it's the name is already given, we can just


uh write that from here. Triangle BAD.


tangle BAD is congruent to triangle


CDA


Yeah, basically, that's it


This is it. So, if you have any questions, then ask.


For the first line you have


whats a reasoning


For the 1st 9 year


Uh, what's the reasoning?


Um


uh, can you just share what the, what type of reasonings do you have that, that list of reasonings?


I can tell you from that


OK


So these are not vertical angles.


So this one is wrong.


All right angles are congruent.


Yeah, maybe you can use this one.


because both of these are right angles, so they should be congruent.


Its still not working


It's not working. Oh my God, this is strange.


I tried that before


uh, what else could be the issue?


Mm


We know that these are right angles because it's sort of given.


you know, what a substitution? Can you explain substitution to me?


Like if we have a variable and we can substitute that for a number we may know


Like we have a variable and we can substitute that for uh


value that we know. OK.


Um, OK


So I don't think this is substitution.


If I also eliminate this in a triangle angle opposite of congruent sides are.


congruent, we also would use this because we don't know whether these sides are congruent or not.


An angle bisector, there is no angle bisected here.


parallel lines cut by a transversal


Do we know that these lines are parallel?


no


or congruit alternate interior, so, yeah, and also they're talking about alternative ear and there is no such rule of alternate interior angles here. So this also shouldn't be our answer. Parallel lines cut by or transversal form congruent alternate exterior angles


alternate exterior angles. We're not talking about them.


They shouldn't also be an answer, so ideally it should be


all right angles are congruent, but I don't know why is that not working?


Uh, maybe we can think of something else then.


So


hm


Uh, they have given us that BC is perpendicular to AD.


BC is


at least one of them to.


It shows this when I right that


is not proven or is not proven a right angle.


Let's let's see that


So BC is perpendicular to AD


which means that angle BAD should be 90 degree and AD is perpendicular to CD, which means that this angle should also a 90 degree.


um


B A D


Uh, what do you think should be the problem here?


Because according to me it's, it's correct.


It should be correct


But it's not working, so.


hmm


at least one of the two angels I mentioned is not proven right angle


How is it not a proven right angle. It's it's, it's given in the question.


OK


So what else can we do?


Uh, have you entered the other statements correctly?


I did this so far


A convent to AD reflexive.


and


PAD OK.


APT is


we just need one more angle


uh, everything seems correct to me. I don't know what the problem is.


we have angle and side so far


B A D N A D C B A D N A D C.


We have other, oh, so you, you need a third statement?


So you can use uh this, this should be your, this also should be one of your statements, right?


We also have this here


So this is our angle, this is our side. This is angle. So by AAS, we are proving this


can we use vertical angles y saying that BEC is congruent to AED


So we can use vertical angles


as saying that the ECH congruent.


BEC


to AED.


It worked!


Uh, I'm sorry, I'm not able to understand what you did.


I said that


BEC is congruent to AED


BEC is congruent to AED.


by saying they are vertical angles


BEC for going to AED right because they are, they are vertically opposite angles.


Yeah, correct


The vertical angles, right?


Yeah


and we can say the same for BEA and CED


And we can say the same for BEA B E A


and CED C E D.


OK


These numbers are also equal because they are vertical albums.


But how will that help you to, to prove that trianger BAT is congruent to CDA.


Um


Maybe something about like how they are both having something to do with E


so that these angles like.


or another reflecive perhaps


OK, OK. I, I got the point now.


OK, so, so


um


if doesn't this angle are equal,


then these sides should also be equal. There's a property like this, so by this we can say that AB will be equal to CD and that we have this angle which, which is equal so we have this angle here, the side here


uh


and we also have this side here


but we don't have any criteria called a


ASSR S S A


So, how is it working?


Mm


Well, you can turn your mic on if you wanna say.


My mic is not working since the mic itself is not in best condition do its not labeling me


abling


OK, OK. Oh, no problem.


No problem


hmm we can say that like


Yeah, maybe we can say like that.


But uh the, this is still strange that if we have a very clear tra solution like this, why is uh


it not accepting this as a solution. It's, it's still strange, don't you think so?


Because this solution is very clear


um, but it's fine. OK. Um, we can move to the next question if you want.


It wont let me move on until I complete it


Uh, so you told me that


I


you just used this property and it worked so, so isn't your uh question solved now?


I, I thought you saw that


OK, so we still need to think about this, so.


um OK


This is how far I am


OK, OK


and 6 is wrong


So you use reflexive property and then you used vertical angles congruency. So you can, you can say uh that is exactly what I'm saying.


that


uh, you can't use these properties here.


because they won't give you anything.


I mean, this is correct your statement is also correct. Your reasoning is also correct, but how will you prove that these triangles are congruent by these statements.


You're using AC congruent to AC.


OK. But uh what will you do from that? How will you prove that these two triangles are congruent.


Im not sur


sure


Yeah, exactly. So, yeah, these won't work. Yeah. We need to think of something else. OK. Let, let's just think again, right?


So


hm


up until 5 the app says its right. Not for fully solving but true


uh doctor Fully solving, but true.


so that's how far I got even tho some statements don't help


up until 5. Yeah, it's, it's true. I won't say that any of these statements are false, but these won't help you to uh get to your final


uh, congruency statement, right?


Because just by these, how can you say that your triangles are congruent by SSS.


How are you even applying that?


yes. and also I still can


You haven't proved that above, right?


change 6


as it is default


You can change 6, yeah. Oh


but


I can also make it an angle


maybe we, we first need to prove that angle BAD is 90 degrees.


even though it's given that BC is perpendicular to AD. Maybe we need to prove that the BAD is 90 degree because if we see this notice over here, it says that at least one of the two angles you mentioned is not a proven right angle. It doesn't say that you're those angles are not 90 degrees, says that they aren't proven right angle, so maybe we first need to prove that those angles are 90 degree.


What do you think


so perpendicular angles form right angles


Yeah, I know that


Uh, but we need to explain this to the


saying that BAD is a right angle


a pro the the solution, right?


ded that


Perpendicular lines form


a right angles.


Yes


Um


OK, then, uh, maybe try using this first add this as a statement that angle that angle be AD is 90 degrees and then add this as a statement that I will CDA is 90 degree, and then, and then add the statement.


and then say substitution as a reasoning for this


Maybe it will work this way.


I did the reasoning for both as perpendicular angles and it worked


As perpendicular angles and it worked?


OK


And now use uh this


this statement, uh, with the reasoning substitution.


substitution still does not work


Substitution doesn't work. Oh my God. Um.


uh, can you share the screenshot with me?


Do you substitution for congruency, there must be a third angle which has already been proved


to be congruent to both BAD and ADC.


BAD


P A D and ADC.


Oh


Uh, OK, so, uh, do one thing that remove this statement and also remove this one, also remove this one


Uh, these are not useful.


once it excepts it you cant remove. but even tho they don't help, it wont effect


OK, it won't help, it won't affect. OK, OK. Then it's fine.


Um


as long as the data base sees that I complely proved


OK


its on IXL math


I got you


Um, OK, so


uh, are you trying to explain something to me?


yes I am examining it now


You can write on the, OK, OK, you're examining. Oh, it's fine. OK.


You can also write something on the whiteboard if you want.


So


OK, so


we said that


well we proved angle ABD and angle DCA


those angles, angles are 90 degrees. So,


um


yeah, we proved ABD and DCA are equal.


are congruent, here, we have proved that, uh, but we are thinking about the reasoning we should give there.


because substitution is not working there, but other reasoning can be used?


But we need to prove how angle A and angle D work to be proved


the first thing I said is already prved


How Uncle A and Uncle D work to be proved.


was given


Um


something with cut by transversal


Yeah, we are doing the same sole A and angle D basically means angle BAD and angle ADC. They are one and the same thing.


So I think you get this message when you write that angle BAD equals angle ADC and new substitution as the reason, so it says that no, it's wrong. Uh, so we are thinking about the correct reason we should give the. Uh, what are the list of reasons do we have


over here?


For the statement that I BAD is equal to an ADC.


these


The these are the same thing. Uh, what was transitive property again. No, I'm sorry, I forgot that.


transitive property.


like if a isqual


if a is equal to e and e is equal to d then a is equal to d


OK, then, yeah, OK, if, yeah, transitive property is basically that


uh A equals E and E equals d, then A equals d. This is exactly what we are trying to say. We are saying that our angle A is equals 90 degree, and then we are saying that angle D is equals is also equal to 90 degrees, and from this we are implying that angle A should be equal to our anger D So yeah, use transitive property instead, as the reason.


Yeah, because when I was, when I was looking here for the reasonings, there was nothing like transitive property. So I was thinking what could be the possible reasoning for this.


Now try transitive reasoning for that.


still no


Still no? Oh my God.


Um


it should work now


its not working yet


Mm.


What notice does it show now?


it shows this


No, you tru use transitive property.


I told you that no, don't you substitution use transitive property.


I did


that's what it gives


to use substitution for congruent, but we are not using substitution. Why does it say substitution then?


This is strange


Uh, now try using this all right angles are congruent.


because now we have proved that both of those are right angles, so maybe it will accept this now.


yes it worked!


OK, yeah. Good.


So, OK, so let's think why the transitive property thing didn't work.


So


so it's basically A equals A E equals t, then we say that a should be equal to D We were saying that angle A equals 90 degrees. Then we said angle D is equals 90 degree. And then we were trying to imply that angle A should be equal to angle angle D


But instead our reasoning uh should have been like 90 degree equals angle.


So this equals this and this again equals this, then we can prove that anguanggu D uh, but, but somehow that should also have, should have worked, but it's fine if it doesn't. So, yeah, I mean, it's OK.


yes


So all right angles are con. So do you understand why it was not working previously because we just assumed that they automatically knew that if it's given that both of these, uh, sides are perpendicular, then the angle should be 90 degrees. So we were wrong with that assumption. We first had to tell them that these angles will be 90 degrees because of that


uh statement that perpendicular lines form 9 right angles, and then we had to say that all right hungles are congruented. So is the whole problem clear now?


Like, why, why was it not working previously?


yes but its still not finished, like I understand so far


Yeah, yeah, yeah


Uh, yeah, we, it's not finished, but we have written the three statements now. We now, we just now need to write that congruent triangle statement. So just uh write this statement and for the reason, what do you think should be the reason? We have 5 congruency criteria ESA


SSS


S uh SAS AAS.


A is


So we have this angle.


here. We have side and we have angle. So it should be AAS.


uh try to protect this as the last statement, and I think it should work now.


Yes it worked!


OK, yeah, great. Finally, finally.


So do you want me to summarize the whole problem again? I can do that.


yes please! so I can take notes!


Uh, OK. So just give me a minute.


We knew that


um, OK.


So if you wanna take notes


uh, OK, let me just clear this thing.


Uh, these are not important.


OK


So yeah, I'll, I'll write the whole


whole solution for you again.


So the first uh statement we used here was that angle BAD is equal to 90 degree.


Then we use that angle A, D, C is equal to 90-degree.


Uh, and the reasoning for both of these was that uh


perpendicular lines.


make


right angle.


This was our first statement


Then our second statement was that AD is


congruent to AD.


because of the reflexive property.


Then we said that


angle A, B, D equals


angle


ACD.


uh because this is given.


And then from these two statements, we said that


Angle BAD, AD equals angle ADC because


all right angles are


congruent


Uh, just ask me if my handwriting is not very clear.


It's, it's a bit busy, I know.


Uh just ask me if it, if anything is not clear. And then from this we said that by


AAS


triangle


B A D


is congruent to Trale CDA.


So yes, this was basically the whole thing and


thank you! I will write it down!


and this was the angle that we used here, this angle we used this angle we use this angle, and this was the site.


Yeah, you can write it down.


Uh, do you have any more questions, by the way?


I can help you


yes I do ! I have a few more like this one\


OK, OK, we can work on those together.


Uh, by the way, uh, was the solution clear to you? I know we can be confused the whole thing a lot.


uh, but is this clear at the end?


yes!


OK, perfect.


Uh, just write this down and then you are done, just let me know and then they'll move on to the next question.


okay Im ready!


Hi, OK


Uh, just, uh, share the next question, we can work on it together.


I'm sorry


Oh, OK


So AD is


congruent to AB, OK. And AB is


perpendicular to BC. So basically, we know that this angle is 90 degrees.


and AD is perpendicular to CD. So this angle is also 90 degrees.


Prove that triangle A, B, C is congruent to ADC.


OK, so that's again think of the five criterias of congruency that we have. SSS SAS


A A S


A S A


and RHS


So, uh, what do you think? Can we use SSS? We don't have 3 sides.


uh, which are equal. So let's first write down what do we know? We know that AB is equal to Ad because this is given, right?


And we also know that we first, we first write that anger A, B, C is equal to 90 degree directly write that angle ABC is equal to angle ADC as we did in the last question because that whole concept would be wrong, right? So we'll first write this and the reasoning for this will be that uh perpendicular lines,


make


right angles


And similarly, we'll say that angle ADC is also 90 degree.


and from this, then we will say that angle A, B, C equals angle A.


ADC


because right angles are congruent.


I'm sorry


OK, so, uh


now let's think of the next steps. So we know that one angle is equal at this


here we have our angle


This is our side. This is another side. We also know that this site is common in both so by reflexive property, we can say that


AC is congruent to


AC


by reflexive property.


So we have our, another side here.


OK


Um


so we can't use SSS here because we don't have 3 sites. We also can't use essays. We we might think that we can use SAS because we have two sides and we have one angle.


Uh, but this angle should be included angle, which means they should be between both of these sides. So if I'm taking these two sides, then this should have been our angle, but our angle is this, so we also can't use this one. If you use AAS


we can use AAS because we don't have two uncles, so this is also wrong


Again, we don't have two angles, so we can't use this. If you think about RHS.


or, or do you also call that HL?


Uh


I call it HL


hypotenuse leg


OK, OK. You call it HL then I'll, I'll also call it the same. I'm sorry.


A hypoten leg. OK.


Uh, so basically, I'm from another country, so I kind of don't know the term terminologies that you use in the UK, so I'm sorry for that. I'm sorry for the inconvenience, uh, yeah.


its okay!


By HL


So how, how are we using HL here HL is basically hypertonous and leg. So we have, so, uh, what does hypoten use? It's basically the site opposite to the right angle in a triangle. So if


if this is our chronicle


right


And this is a right angle to the side opposite to this is this side, right? And we know that this side is equal in both. We prove that by reflexive property. So we have the hyperterus equal about the neck


This is our leg, right? We have this equal in both the trigals, and we also have, uh, this, so by HL we can say that our triangle.


ABC


is congruent to triangle


ADC


Um, is this clear? Just, just try this and tell me if it's working or not.


I think it should work now.


everything works up until we got to HL


So HL is not working.


Um


why is that you not working?


I think it should work because we have already proved that


AB is equal to AD, AD is equal to AD because this is given in our question.


and C is equal to AC because it's reflexive and this is the hypertenus.


This is the leg, so I think it should work.


Uh, can you show me the error message that you're getting.


You cannot use HL theorem yet, because you have not proved that each of these triangles is a right triangle.


So you also need to prove that these triangles are right triangle.


Oh


uh, how can you do that?


Basically, uh, if you have one right angle in your triangle. So it is a, it is a right triangle, right?


So


uh, can you think of another statement which you can use, maybe something like


um triangle A, B, C


is a right triangle, a statement like that


and then the reason will be because because angle ABC is 90 degree.


Uh, just try something like that if it's possible.


Uh, just let me know if you are able to try or not, and then we can think of something else.


it says it has already been proved


It, it has already been proved.


Uh, but in the notice we get that these triangle are because we haven't proved yet that that each of these tiles is a right triangle.


Uh, what is the problem?


So for the thing that you are saying that it has already been proved, uh, can you share a screenshot of that.


of like what you entered and what was the notice that you got.


OK, so, no, don't use this as your reason.


No, don't use this. So what you are using, the reason you're using is that an angle congruent to a right angle is a right angle.


So you are basically saying that if we know that angle is 90 degrees, OK?


Uh


basically


basically at, at the


it's like if you're saying that angle A is congruent to angle B, and this angle B is 90 degrees. Then your angle A should also be rating you. This is what you are trying to uh save from this reason, but if you're uh if you're telling the reason for this statement that


no, you have to


your statement should be that


it should be triangle ABC.


is a right triangle.


This should be your statement and your reasoning should be that angle A.


angle ABC is a right angle, something like this.


Uh, this is something I want you to do.


Yeah, because we have already proved that angle ABC is a right angle by saying that then two perpendicular lines meet they form 90 degrees. So, yeah, you don't have to prove this thing again. Why are we doing this? Because when we uh tried using HL on those two triangles, the error message appeared that you haven't proved that you're two triangles are right triangles. So first we are trying to prove that our two triangles are right triangles


and for that we are going to use this uh statement and reasoning. I try this if it's possible.


nothing yet still same notice


Hi, I'm sorry. Uh


uh, still the same notice?


Uh, I'm sorry, but can you share the screenshot.


I tried all types


Uh, so you don't have to prove that anger to be a right angle, you who have to prove the triangle to be a right triangle.


Does this make sense


yes


Uh, now try this


Try


adding this as a statement that triangle ABC is a right triangle. You don't have to prove that it's a right angle. You have already proved that angalabi is a right angle by saying that when two perpendicular lines meet, they form a right angle. You now have to prove that your triangle ABC is a right triangle.


Is this clear


Oh okay!


Yeah


yes I see it!


Is it working now? Uh, what will you use for the reason? You should use that. OK, so how can you prove that a triangle is right triangle.


Just tell me


the trianglecontains a right angle


Yeah, exactly. The triangle contains the right angle, so, uh, use that as a reason for your statement.


Now, see if it works


yes it does!


Oh, great. OK


Uh, we can move on the next steps now. So, basically, now you have proof that an triangle ABC is a right triangle. Similarly, you need to prove that a triangle ADC is a right triangle. Uh, just quickly do that and


uh let me know when you, when you, when you are done.


I did it!


OK, great. So, basically, now you have


I guess the statements that you have right now are that AB is congruent to Ad, that is given, and then you have that AC is congruent to AC because of reflexive property, then you should have that angle ABC is 90 degree, then you should have angle ADC is 90 degree, then you should have


triangle ABC. Is a right triangle then.


triangle ADC is a right triangle. Am I right?


Yes!


OK, so we have all the material now to apply HL.


and say that triangle ABC is congruent to triangle ADC.


Now just try that and uh let me know.


Yes! I completed it! it worked, thanks!


OK, perfect. So, uh, I can summarize the whole thing for you again if you want.


No I think I understand this one better, tthanks!


OK. OK, perfect. So, well, OK, we can move on to the next question if you want. Or do you wanna note down, note down this uh solution.


I did as we were doing it!


OK


So, uh, the issues which we faced here in this question were uh


for applying HLU first need to prove that your triangle is a right triangle, and we assume that, uh, this assignment already understood that if we are saying that the anger is 90 degrees, and the triangle should be a right triangle. Uh, so we were wrong with that assumption. Uh, then, then we just wrote that another statement proving that a triangle is a right triangle if the angle is


90 degrees


So I think this was the, uh, this was the learning that we gained from this question. So, yeah, we can keep this in mind. Uh, let's move on to the next question. You can share that next question with me.


okay! So this one is a little quicker i belive


Uh, uh, can you move, uh, both of these images to the top of the page.


because I can't do that


OK, OK, this is fine. Thank you.


Thank you


Oh, OK.


your welcome!


So it's given that


FT is congruent to FD.


FD, uh, I think I should mark this


uh


FD


is congruent to FD.


and angle FDE FTE.


The angle is equal to other FCE this sent


and we need to prove that.


Unle DAB D sorry, triangle DAB. No angle ADAB, I'm sorry.


And DAV


should be equal to CBA.


Hm, OK


Uh, it's


it's given that


T is congruent to FD and angle FDE is converted under the FCE we need to enter the 2nd statement for the 3rd statement we have triangle ACF.


ACF.


I'll just part the struggle.


struggle ACF.


is


Conran Triangle BDF.


um BDF is this one.


So we need to prove these two triangles to be congruent and then we have something else. OK.


So our main motto here is to fill the 2nd statement


so that we can prove that these two triangles are congruent by ASA. This is our main motor over here.


uh, for ASA, we need one angle, one side, and one angle.


Uh, if we look at these two triangles, we know that FD side


we are missing an angle for ASA


congruent to FD side, because this is given


Yeah, exactly, you are correct. We have the site and for the angle


put the angle also we have that angle FCA.


FC


A is


congruent to triangle, uh, I'm sorry, angle.


uh, this angle which is FDB angle FDB because this is also given.


So we have another angle


Uh, now we need to have this 3rd angle and we can very easily know which angle it should be because


uh, the side is included. The side is between both of these angles. So, uh, if we see this


so we do, this one is the


side and


this is the angle, so another angle should be?


uh, tell me, should it be this or this?


This is one and this is 2.


What do you think


The side should be included between these two angles. It should be between the two angles. So should the angle be 1 or 2?


know how to


how 2?


no


No


it would be wrong because of side angle side


I think you're confused


Yeah, exactly. It would be wrong because of side.


No, we


and so it is on because angle side angle


uh, we are not using side angle side. We are using angle, side angle.


Uh, so it is on because


uh, it is one, yeah, it's one. I think you mean lung.


So yeah, it should be one, right? because uh the site should be between these struggle. If I take this angle, the site is not between these two angles, but if I take this angle, then the side is between these two angles.


Uh, so,


in this triangle, basically the green triangle that you see.


uh angle one is this. What do you think in the blue blue triangle should angle 1B. Should it be the same one, it's basically common in both the triangles, right?


Does this make sense


yes!


Yeah


So


we just need to name this angle in two different ways, basically, to say that uh it's equal.


Uh, I'll just


uh erase some stuff from here.


So yes


uh, in one way I can write this angle is angle CFA.


is equal to a DFB.


yes I understand the statement


What is, I will see if you understood it. And we'll see if a CFA is this angle. The angle DFB DFB again is the same and so we are saying that this angle is equal to this angle, so


basically, we are using the reflexive property.


Uh, don't you think this should be our statement over here.


this cant be reflexive since it would have to be the exact same letters


OK, so it needs the same letters.


So instead of using the full CFA name, you can simply write on the left, equals angle left.


Try this. We can also use this, right?


yes! I will try


Yeah.


So here are the main challenges like uh what will the, the system considered correct because we know the concept and everything, uh, but it's more like what will the


that algorithm consider connect


yes! it workd


thanks!


OK, perfect. Uh, do you have any doubt on this problem?


no I understand it


OK. Um, any other questions?


that all for now! Thanks!


OK, thank you. Uh, in fact, I also learned so many things today, um, the terminologies and stuff, and it was, it was nice teaching you. Thank you.


Have a good night


Thank you so much! Have a good night! bye!


Hi


hi


how can i help you?


So I'm doing angular & linear speed problems in pre calc and I don't get them. I have word problems and I can paste one here


sure


which one?


We can start with the first one. Can you also show me how to draw the diagram circle?


you have to draw a diagram?


My teacher doesn't require it but it helps me visualize.


oh


okay do you know the formula for Distance along the surface between two latitudes


s = theta * radius


yup


so the first one is Sarah


equator


but you have to convert the theta from degree to radians


so 19.43 * 180/pi?


basically your question says that this angle is given


and from this you have to calculate the rest


Got it


for equator the formula should be theta*pi


if you want to convert it to radian you to multiply pi/180 with it


what is the result?


then?


this is correct


but you have to convert it to radian


19.43* pi/180 =0.339


how?


i get it


but you missed to multiply it with the radius


ooh okay wait


so that times 3960 = 1342.91


yes


perfect


you got the first one


omg there was confetti


yay


can we go to the second?


the second one says we have to calculate the distance of the city from north pole


so look at the figure


we have to calculate the red arcdistance


can you guess how can we do it?


not sure


no worries


we that formula that we used for the first question


theta*R


how can we calculate the theta for red?


we know north to eqador is 90 degrees


to equator?


yup


north to equator


perfect


now if we want to determine the distance what should we do?


you already know the theta


yes


I love this confetti


i love it toooo


you got the second one too Maya


yayyyy


yay im writing this down


sure


i am here, take your time


thanks. So for South Pole, do I add 90 to 19.43


since it's different in the south


Exactly!


wow


im cooking


you are so smart Maya


lol thank you


CONFETTI YAY


yaayyyyyy


so 19.43*pi/180*3960=7,563.259


yup


perfect


yay I wrote it down


yay


do you need help with anything else, Maya?


can we do 3? the formulas are different there I'm pretty sure, and I haven't gotten them drilled into my head yet


sure sure


the whole thing with revolutions and v


whats wr?


linear velocity


oh okay


v= 2pi/1 revolution 360 degrees/1 revolution


so do you know any formula related to this?


v=rw


I forgot what w is


or w= theta/time


w is 2pi*N


what's N


600 rpm


N = rotational speed


oh I see


okay first we calculate w


can you try?


yes hmmmm


lowkey I'm not sure what to multiply because there are so many values


put the values


in this equation


here N = 600


yup


now by doing this


we have converted rpm to radians per min


next step is to follow v = rw formula


can you do it?


r is 9 ft , given in the question


v= 9*5280


oh wow I was just plugging in hoping it's right


but


why did you do it?


if the radius is 9, the w of angular speed can be plugged in radians per minute


but why?


not exactly sure


see the problem says you have to and it with miles per hour unit


but radius is given in feet


not in miles


so we're converting


yup


check one more time if you conversion is right


your*


we are converting ft to mile


9 ft * 1 mile/5280ft


then we are dividing it 5280?


right?


yes


now plug it in with w


v = rw


w= 600 revolution/min * 2pi/revolution= 1200pi/1 minute


1200pi/1 minute * 9ft * 1mile/5280 ft


whats the final value?


386 mile/hr because we also multiple by 60min/1 hour


and you forgot to convert one thing


yay


you got it


confetti!!!


yup


okay wait thank you so much


I get it better now


you got it Maya


I'll just keep doing problems and applying the knowledge


Wow!


but do you know why are you converting 600 rpm to radians per min?


well it asks for the final answer to be in mph


but why you covert that?


because


600 is the no of spin


not the distance


so we convert that spin to the distance


thats why


yea


thanks


I get it


you did really well


thank you that's so kind. it was thanks to you


best of luck Maya


Also do you happen to be Indian


you are most welcome


just curious because of your name


yup


oh my gosh same


you too?


I'm Pakistani


same


I had a classmate named Anaya when I was younger


just spelled differently


ikr


so I was curious


oh okay twinsss


okay thank youuu. good night!


it's a very common name except I am a girl


OHH ME TOO


okay that's so surprising because I didn't realize it could be used for girls as well


My names similar except it's more feminine since it has an a at the end


yes


that's so cool


okay thank you!


good night


nice to meet you


nice to meet you too :)


OK, I hope you can hear me


Hey, I would like to um deepen my understanding for in statistics for Chapter 5 on hypothesis testing with the means of the sample.


Right. And you said you had a practice problem, right?


Yes


OK. Are you able to show me the practice problem and we can try it.


Let me just


OK


1. Why is the standard deviation of the distribution of means generally
smaller than the standard deviation of the distribution of the population of
individuals?
2. For a population that has a standard deviation of 10, figure the standard
deviation of the distribution of means for samples of size (a) 2, (b) 3, (c) 4,
and (d) 9.
3. For a population that has a standard deviation of 20, figure the standard
deviation of the distribution of means for samples of size (a) 2, (b) 3, (c) 4,
and (d) 9.


Let


These are the questions I want to practice first.


All right


OK, we can start with the first question. Do you have any, any gas?


Is, um, so the first one is because um


the um


it's because it wears among the um among the population.


Yes, if you have um


if you have a population of, of individuals and you just take


the value of one


one person


that can vary a lot, but when you take the mean, you add up


data from many people like can be a large sample.


and then you divide by the number of data points and that's your sample mean. So it's very difficult for that to vary. It will vary much less so that when you calculate, for example, if the standard deviation


of the population.


Well, the answer is that it bears among the population, um.


right.


Well, it varies among the population, but the sample mean will vary much less


than the individuals in the population.


I'll, I'll give you an example. I'll, I'll write on the board, if you have, if you have a variable x, you see what I'm writing on the board?


So let's say the standard deviation of X


is sigma, then if you


look at the


sample mean, so that's x bar. Do you see that?


Yes, I see that


OK. So, this is, this is.


That's also called the um the variants, right?


uh no, the variance is something else. The variance is the square of the standard deviation.


So the sample mean


And the X bars the sample mean.


that's right


And when you calculate the standard deviation of the sample mean, that will be


sigma, which is the standard deviation divided by the square root of N. N is the sample size. So when you divide by


the square root of a large number, the standard deviation of the sample mean will always be much smaller


So there's much less variation


So what is the stigma again?


Sigma, I'm gonna circle it in red. That's the standard deviation. I circled it in yellow.


Um


and access the like the wall score, right?


Well, yes, that's correct


So the correct answer to this question is, it varies among the sample, but if the, it will vary with the sample me, but, but it will be based on the sample mean, right?


That's right. And the variation will be much smaller


So it's basically like it varies among the sample mean of the population.


Right


OK, I'm ready for number 2.


For number 2, we wrote the same formula I wrote on the board. So I'll write, I'll just start to write the data for number 2.


So it tells us


OK, let me read number 2. So a population has a standard deviation of 10


So we write


so


for X


the standard deviation is 10


and they're asking for


the standard deviation


of the distribution of means.


So the mean is x bar


and the formula for the standard deviation of the sample mean is sigma


over square root of n and for question A, they tell us that sigma.


is


So it will be 10 divided by the square root of the um temple size, right?


That's right


which gave me like


which gave me 7.07.


Correct


And sometimes we write it like this


We write like a sigma and a little X bar to show that it's the standard deviation of the mean.


So like this


and then you do do the same for question B.


The only difference is question B is a different value of n.


OK.


because the way I solve it, it is like I put 10 square, which is 100 divided by 2, which gives me the um um and then I do the square root of 50, um, which also gives me the same answer, 7.07.


Yes, you can do it that way


And the answer to the first question was that it varies among the population based on the sample of the mean, right?


Yes


OK, I'm ready to move on to number 3.


Oh


Oh, let me, let me, let us solve all parts of number one first, sorry.


OK


Mm.


For number 2, I got phone number um 4B. I got um 10 divided by the square root of 3, which equals to 5.77.


That's correct


And you know it is that as the value of N.


Or C, I have an answer of 5.


And then for D, I have an answer of um 3.33, 10 divided by the square root of 9.


That's right


Correct


Let's move on to number 3 now.


Right


So for A, I got um


Yes, that's correct.


For me, I did the


um, so I need my square, which 400 divided by 3, and then I get a square root of 13133.3 and then that equals to 11.5.


Yes, that works


And then for C, I did 2 divided by the square root of 4 which is 10.


Yup, that's good


And then for D, I got, I, I 400 divided by 9, which equals to 44.4, the square root of 44.4, which equals to 6.66.


You got it. That's correct


OK, so I'm gonna give you a practice problem that requires um award problem type of thing now. We're gonna move on to that.


All right


Mm imbu.


OK.


you cannot


OK


Hm.


what's one name?


19. A researcher is interested in whether people are able to identify emotions
correctly when they are extremely tired. It is known that, using a particular method
of measurement, the accuracy ratings of people in the general population (who are
not extremely tired) are normally distributed with a mean of 82 and a variance of 20.
In the present study, however, the researcher arranges to test 50 people who had no
sleep the previous night. The mean accuracy for these 50 individuals was 78. Using
the .05 level, what should the researcher conclude? (a) Carry out a Z test using the
five steps of hypothesis testing. (b) Make a drawing of the distributions involved.
(c) Explain your answer to someone who knows about hypothesis testing with a
sample of a single individual but who knows nothing about hypothesis testing with
a sample of more than one individual. (d) 4: Figure the 95%
confidence interval and explain your answer to someone who is familiar with the
general logic of hypothesis testing, the normal curve, Z scores, probability, and the
idea of a distribution of means, but who has not heard of confidence intervals.


Solve this pro um


solve this problem with um all the five steps of hypothesis testing.


With the hypothesis testing procedure.


OK, I'm reading the problem


So it says a researcher is interested in whether people are able to identify emotions correctly when they are extremely tired. It is known that using a particular method of measurement


the accuracy ratings of people in the general population who are not extremely


are normally distributed


OK, so we'll note those numbers. So we have a first group of people who are not tired, so the meme


is 82 and the variance


that's the square of the standard deviation is 20. In the present study, however, the researcher arranges to test.


50 people who had no sleep the previous night. So obviously they're tired.


The mean accuracy for these 50 individuals


was 78, so that is the, that's our X bar is 78.


using the


0, the 0.05 level what?


should the researcher conclude?


Hey, carry out the Z test using 5 steps of hypothesis testing.


Make a drawing of the distributions involved. Explain your answer


etc


OK


OK, so


we have two things we're gonna do. We have a um


so we have a group


that is not typed, so I'm gonna write that on the board.


So, first, sorry, sorry. So first I wanted to like the the way my instructor did it is first we had to figure out the two populations being compared before we stayed the north and the research hypothesis.


Right


OK, so the first population is the people


who are


not tired


So that would be the research hypothesis, ages of one.


Uh, we haven't


we haven't given the hypothesis yet. I'm just writing.


Yeah.


Oh no. Oh, I, oh, the, figuring out the populations.


right


Oh.


So we have


so we have new


This is the, the known meme.


is 82


and the


variance


is


when


Oh, so, wait, wait a minute. Hold up. So the 82 is the popular population of the means, right?


uh, it's the, it's the mean of the population of the people who are not tired. So that's the first group. I'm gonna write it here. So first group


we'll call it not tired.


And then the second group


the second group, we got


X bar


the group that is tired


Xar


equals