what would be the vertex based on the standard form


its supposed to be (1,16) tho according to my answer key


yes because you know how e divided by negative 1 in the beginning


thats the remove the negative to make complete the square much easier


but when we write it back in standard form we have to multiply the negative 1 back


so the equation would become -(x-1)^2+16


which would make the vertex (1,16)


does that make sesne


shouldn't the negative 1 multiply to the entire thing tho, not just the -16?


yeah sorry in the text the negative appears in the line up for some reason


does this make sense


yea I think so. but the don't you have to multiply the - to what's in the parentheses


no because muiplying it out of the parenthesis is the same thing as multiplying each term in the parenthesis by -1 so we keep the eat i’ve sign on the outside


not really


would it make it not in standard form anymore?


its still in standard form cause standard form is written as a(x-h)^2+k where a can be any constant and (h,k) is the vertex


in this case a is -1


so this is in standard form


so if e distribute the negative one, it wouldn't be in standard form anymore eis what ur saying


yes cause the parenthesis is being raised the the power so if you wanna distribute you have to do (x-1)^2 then distribute the negative 1 but if we do that it wouldn’t be in standard form anymore


could you do that and get it back to standard form or no


no cause if you put it back in standard form you’ll end up with whatever we started with originally. the x in the parenthesis must be positive for standard form as well so making it negative will not get you standard orm


got it


and what was wrong about what I was doing earlier? I was doing it how my teacher was doing it


its just the negative is causing problems to we have to distribute it back out


ill show you how


it doesn't look like the negative caused any problems. am I not seeing it right?


for me what i learned was o multiply the negative 1 later but based on what ur teacher is doing treat the -1 as a coefficient of x^2 to be factored out


its just in my opinion a lot ore confusing to have the negative 1 in there which is what i learned in school


so my teacher taught e to divide the negative 1 out then multiply it later


wait so then would you be able to teach me how to do it correctly my teachers way?


taking taking it out as a coefficient


yes so look at the bottom and ill show the steps


ohh ok thanks. idk why that was so hard for me to do


it’s fine everyone makes mistakes


that was all. thank you for helping me


no problem


have a good rest of your evening


bye


hey!


how are ya?


Good


How are you


pretty tired, but ready to tutor you in geometry :)


Okay let’s do it ¡


okay, what can i help you with?


I am ATTACHING THE PROBLEM


ohh


do you know what adjacent complementary angles are?


I’m sorry , just number 3 please


I just need to know a step by step with that problem


okay


so we know that angle(KJN) 180 degrees because its a straight line


and it already has a right angle at KJL


KJL+LJN=KJN because they are supplementary


KJL is 90 degree angle (because it is given that the angle is a box) and we know that KJN 180 because its a line


so 90+KJL=180


so KJL=90


So would KJL be my adjacent angle?


no


the things inside would be adjacent and complimentary


So like angles LM


oops!


i meant LJN-90


because it’s inside of the KJn


oh okay


So what’s my adjacent angle


so whats inside LJN is complementary


adjacent means the angles that are next to each other


so which angles are next to each other that make up LJN?


LJP?


nl


LJM and MJN


because if you add them together that equals LJN


I see it now


yes, thats great


Okay thanks


are there any other complementary adjecent angles?


That’s all I needed to know for know but that was really helpful


NJP is adjacent right ?


when youre saying adjacent, you have to say two angles


because adjacent by definition means "next to"


MJN IS ADJACENT TO NJP


Am I right ?


that is correct


but is it complementary? how would you check


Check if it equals 90 degree?


yes


ow


how*


Subtract 49from 90


does that equal MJN>


MJN?


No


I don’t think so


so then it is not adjacent complementary angles


meaning that there were only one pair


okay good job!


Thanks


have a good day :)


You too ,)


Hi! How can I help you?


Same thing as before!


so the firstone is 6


is this right?


wait why did you put 3 to the power of -2?


and also 1 squared is just 1


because 1x1=1


and 3 to the power of -2 would be 0.1111 going on forever


its (3)^-2


yea 3 to the power of -2 right?


because 3 squared is 6


but 3 to the power of -2 is not 6


oh right


yea


nah 3 to the power of -2 isn't -6


because when the exponent is negative it's different


ok now what do i do


okay so for the 1/2^2 did u mean (1/2)^2


mhm


no paranthesis


so 1/2 is the same as 0.5


and 0.5^2 is 0.25


and 2/2 is 1


which is not 0.25


actually


no i meant paranthesis


sorry about that


can you type it out for me in the chat?


yea thats 0.25


because 1/2 is 0.5


actually ill just send u the pdf


Alice


k


page 2 problem 23


I need a problem just like that except with diff numbers


I'm only allowed to see documents here on this platform on the document editor


You can access it as long as it doesn't show your personal info


Like it can't be alink to something where i can see your name and info


The pdf doesn't disclose that


Nah like I can't copy and paste the link you gave me


The program coordinator told me that


Or whatever she is she's the coordinator of something in Springfield


I had a meeting with her about it


the thing is I can


cannot copy and paste the link


She said coaches can access links as long as its safe and doesn't show your info


it doesn'twork


Here


okay so I see that you solved it


did you wanna check if its right?


No


I want a problem just like that but with different numbers


So I can solve the one you create


Oh for practice


yes


Yea um the picture quality on the image you sent me isn't great so I can't actually read the original question


i can only read the answer because it's in bigger numbers


ill write out what it says


k


okay let me think


can u write it on the whiteboard


sure


thanks


No problem!


ok so its


Not quite. This one is a bit tricky but the answer 12/16


no im asking about the first part


like how do u solve this


Okay let me walk you through it


so first we need to simplify the terms


4^-1 is the same thing as 1/4


ohh right


3^2 is the same thing as 9


mhm


and then its -8/3?


and (4/3)^-2 is the same thing as (3/4)^2 because we flip the fraction so we flip the sign. And (3/4)^2 is the same thing as just 9/16.


and so now we have 1/4 + 9 -9/16


wait


can u show this on the whiteboard please


yea


its gone


yea i wanted to make space for the new simplified equation


i have written down 1/4+9(-4/3)^2


oh its supposed to be what I've written on the board


because we had simplified (4/3)^2


sorry we had simplified (4/3)^-2


oh you flip it


k i see


Yup!


ok


now what


tims 1/4 by 4?


yea basically to make it so that its 4/16 because thats the same thing has 1/4


kk


so now


we need to make 9 a fraction


so its


with the denominator as 16


Yes!


so i do 4/16


not yet


we need to turn 9 into a fractoin


with a denominator of 16


oh right


so we now need to multiply 16 by 9


which would 144


now plug that in back into the equation


Yes! That is your answer!


Looks like your getting the hang of it now!


Alright is that all for today?


lets do 24


and 25


Oh shoot I have to go now. Got some classes to attend. If you need any help later, let me know!


ok, thanks! bye


hi Chicago


hello coach David


Welcome to your 2nd Sarah session!


are you uploading a problem?


yes


sorry for the wait here it is


I see it


okay


so if im right the lor is the midpoint?


betweent the 2 triangles


yeah, that's one way of saying it


oh


whats the other way?


a line is not really a midpoint


a midpoint is a point


but, if you find the midpoint between A and A', and the midpoint between C and C', and draw a line between those points, then you'll have the line of reflection


then divide it by 2?


divide what by 2?


like say the lor was 14 between A &A'


then you split it and youd get 7


or no?


you aren't describing this very clearly


sorry


find the difference in x-coordinate between A and A'.


Divide that in half. Move that many spaces from A to get the x-coordinate for the midpoint


Do the same for y


do i need to make a coordinate graph?


you need to draw on the image you uploaded


okay


from A, how many units to the right and up do you need to go?


no


remember to divide by 2 after finding the difference


so


i mean


which is 2


let's start with x


i dont get it


how far from A to A'


divide that by 2


yes


now do the same for y


so what did i do wrong


ohhh


did you tell me 3?


was it because i subtracted?


you told me how far from A to A'


I was asking how far from A to where you need to graph the midpoint


the midpoint is halfway, so you divide by 2


oh okayy


so now i do C and C'?


after you draw the midpoint for A and A' on the whiteboard where I can see it


like that?


draw a point where those lines intersect, then delete the lines


there


good


now do the same for C and C'


you don't need to draw intersecting lines


you can find the x and y differences, divide by 2, just count that many lines to the right and up


12 from C to C'


is it 12/2 which is 6


yes


yes, 6 for the midpoint


x


so


no


oh


your point has the same y as C'


it's supposed to be halfway between C and C'


im confused


what are you confused about? your new point is correct


ohh


the first one was wrong


now draw a line through those points and you'll have your answer


okay


like that?


looks right to me. does it look like the line of reflection to you?


it is correct


wait i have to put it


still working on it?


yes


keep saying its wrong


but i graphed it right


want to upload a picture of the answer you put it?


yes but it changed it but ik the midpoint for each one


what more can I do for you at this point? the answer you drew on the whiteboard here is correct


no like it put a different problem


but i know the midpoint to it


that lower blue dot


is its y the same as B or halfway between B and B'?


isnt it 9


I'm asking about y, not x


i dont know what your saying


up/down, not right/left


you picked a point whose y-coordinate is the same as point B


it is supposed to be halfway between B and B'


do you see?


so its not 9?


x-coordinates have difference of 18. 18/2 is 9.


I'm not talking about x


right


I'm talking about y


up/down


how far up from B to B' ?


divide by 2


you need to go up 3 lines from B


and 9 to the right, like you did


okay


you have to go halfway between the points for both x AND y


do you understand, now?


no


i do


but the graphing no


start at B


count 9 lines to the right, 3 lines up, pick that point


i got that


now the bottom point is right, the top one is wrong


y is OK, x is wrong


how far to the left from C to C' ?


yes


and how far to the right did you go from C?


count again


ohh 8


actually, 7


don't count that first dot


that one lines up with C


oh


only supposed to go 6, but you went 7


so go back one


yeah


keep y the same


ok tahnk you its rihg


my blue dot


right*


just remember to go halfway and halfway


and count carefully


so it has to be the same


same halfway point for both x- and y-coordinates


okay okay


thanks


yw


bye


hi


Hi! What can I help you with today?


What is 4*5


Help please


Unfortunately I cannot solve it for you, but I can help you rephrase the question so that it is a bit easier to understand


4*5 is the same as 4 multiplied by 5


Is it 20?


Yes!


Okay what about 56x+66=800


So for problems like these, the first step is to isolate the x variable


Can you write the question on the board please my writing is messy


Is this right?


You are on the right track! but when I divided 734 by 56 I got a slightly different answer to yours


Is this right now?


Yes, you got it!


Thanks


Do you have any other questions?


I have two more questions


Ok!


So same thing as the last question, the first step would be to isolate the x variable. I can help you write it on the board


Okay


You were on the right track, but I got a different answer than you when I divided 364 by 37


Should we move on to the next question?


Actually I have two more questions


Okay, no problem!


How do I do this


So this question is a bit different from the other one. We can take it one step at a time


you can add 78 + 856 first


get the new answer then add 345


Is this right?


You got it!


Last question


How do I do this?


With problems like these, I like to break it down so I would 500 + 745


654 + 987


345 + 689


and then add all three answers together


Yes, that's the same answer I got


Thank you so much


You're welcome


I appreciate your help


Have a great rest of your day


Thank you!


Bye


Hello


Hi


How are you?


I'm fine, how are you?


I'm doing well


Just a min


Ok so are you familiar with what domain  and range is


or should we go with a recap?


sure


So domain can be thought of as the input set (group of values) of a function. Basically, all those values which can be inputted into the function to get a valid (defined) output


So basically to check the domain, we would just look at the x-axis to see for what values of x is f(x) defined


Can you have a go at the example using this?https://mathhelper.com/question1y question 1 part A


yep


you've made a slight mistake in part A


what is it


at x=5 the function is not defined so you have rightly used an open interval but what about at x=-1?


oh


yeah


same with your range


perfecy


*t


you've switched up the answers


wait for x intercept do u not make y = 0


an x-intercept is the point where the function intersects the x-axis


yes


look at x=3


i see


no that's the same mistake again


ok look at x=3


so what axis do you see it intersecting


what do you do to find x-intercept


set y = 0?


and to find y intercept set x = 0


i understand it now


yeah, perfect


Any other questions?


yerp


ok yeah i can see it now


zeroes is the same as x intercept right coach?


yeah


find zeros make y = 0


yes


L math


do you just say x = 20/3 or


(20/3, 0)


both are correct, but I'd recommend the second one


alright what about y-intercept


one sec


so y intercept set x = 0


excellent


have a go at the second one


yup


last one


you forgot to divide the 2 when solving for x


um


whoops


the rest is corredct


so umm


one min


Alright so go ahead, I'll check


i dont rly know this ngl


ok Ill help you out


so if the function is symmetric about the x-axis, then for every value of x, I should have two values of y


y and -y


right?


so a parabola


not necessarily, look at the first graph itself


its symmetric, but isnt a parabola


hmm


so basically I can say that f(x) = y = -y


so what would i call dat


that would be your algebraic test


basically you can replace all the y's in the equation with -y's and the equation would remain the same


so the other one is f(x) = x = -x


no not exactly, just a min, we'll come to the second one just make a note in the first


you can't call this a function, so I would avoid writing f(x)


why cant u call it a function


isnt that just a prabola


the definition of a function is that for each value of x there can only be one value of y, if you are familiar with mapping and the concepts of sets, I could make it clearer but otherwise we can ignore that for now


just remember, if you see one x value giving you more than one y-value, the graph isnt a function


So a rather better way of phrasing the algebraic test is y and -y give same value of x, or that the rq remains the same


if you are familiar with the modulus function, you can identify graph 1 too(as an extra)


ok getting back to point, any questions in the algebraic test?


how do i confirm algebriaclly


replacing y with -y in the equation of the graph does not change the equation


Now we can move on to the second graph


You there?


Any questions?


yeah


just kinda lost


alright let's get back on track then


so we need to algebraically prove that a graph is symmetrical about the x-axis


so you will be provided with an equation and/or a graph of the equation


like in example 5


wait so which model is x-axis


the 1st one?


yes


model 1 is symmetry about the x-axis


ok so when it says confirm algebrically i have to put an equation right?


you have to use the equation given to you and perform the test that we mention there


or input the equation ig


the test will tell you if its symmetrical or not


do you get it


so in the case of symmetry about x


for each value of x you should have 2 y values


one as y and other as -y


so we substitute y and -y in the equation given to us, and we should get the same equation


if you still don't understand feel free to ask


i think an example would clear it up well though


ok


so


i think doing rexamples would prob help


yeah, let me explain the three models and then try the examples, it'll make it clear


so in the second case, if for each y-value we have two x values -> x,-x then we can substitute -x in the equation and we should get the same eq


now in the last case, for every (x,y) there is a (-x,-y) on the graph, so we can substitute -x and -y in the equation to get the same eq


Now let's get going on example 5 you should have a better understanding


erm


yeah you were correct


that was correct


but like


so xy = -6


if i made them both negaitve


what would the difference be


wait


so lets say x = 5


y = -6


xy = -30


make them both negative


it won't make a difference, that's the point


the equation remains the same


which means that the equation is symmetric about the origin


ok so how would it be wrong algebreically


say i have a function y=x+3


replacing y  with -y and x with -x


-y = -x+3


or y=x-3


which isn't the same function


hmm


so the function is not symmetrical about the origin


u know i learned this but thankfully


it wasnt on my quiz


so maybe it wont be on test :)


it's not too hard though, you're doing a good job


i mean horribly but close enough


don't worry about your test, you have a good understanding


these topics take a bit of time and practice to digest


so can you try the second one and we can wrap up for today


I have to leave a bit urgently


oh


same thing


so its symmetrical cause its the same


but different symmetry, it isnt symmetrical about the origin


look at the graph


i thought u replace x with -x


what axis is the function symmetric about


yeah that's correct


oh to the y axis


yeah I just meant here we won't replace y with -y because we're checking for symmetry about y axis


anyways yes, y = x^2+2 remains the same for x and -x


so it's symmetrical


about y axis


any questions?


nah


so you have to leave?


alright I'll end the session then, you did a gr8 job today, keep practiving


thank you


yeah sorry about that


have a good night


or day


same to you


hiii


Good evening Emily


can you help me for a few minutes with indepdnent and dependent quantities


like story problems


yes I can


tyyy


Do you have handout or document that we can work on


nope just make up problems i guess


alright


so would you please define idependent and dependent quantity in your own words?


indepent is what ur changing and like what doesnt depdnd on smth and its x and domain


dependent is like dpeending on x and its the y and its range


I understand you


ok


so we have 2 similar terms which are input and output. Which one is independent?


input is x


output is y


sorry


input is independent


output is dependent


correctttt


let's see how they relate to function


okay


base on the diagram of function, can you please tell me why output is dependent?


output is dependent because it relies on the x to go get the result


idk fi that makes sense


it is


but in reality


it depdns on x


so what is the role of function f(x)?


thats function notation


so nothing really


but


its like


the variable ig


idk how to explain it


hint: using the verb transform


Maya idk tbh


so we have the independent input x value. using the function f(x), we can transform that x value into dependent y value


for example


I get 2 bucks everyday from selling items. After 2 days, I gain 4 bucks.


days is independent


amount of money is dependent


bc how much u have


epends on how long u have spent getting it


correcttt


good reasoning


so number 2 in the last sentence is depedent or independent


where


independent


bc its days


and days are independent


correct


you are smartty


bc they are depended on by the money


so for formula


area = base times height


yes


which are independent and which is dependent?


so for the parallelogram


it would be


base is indepentent


and area is dependent


oh sorry


nd height is also indepentent


bc the area depends on them


that's truee


do you want a harder question?


ye


which subject do you like


idm


anythign tbh


alright


r u here still


ok


yes


hmm


for chemistry, each carbon atom has 6 valence electrons, each x atom have 1 valence electron. two x atoms and one oxygen atom in total make an water molecule. Which is independent and which is dependent?


there are 2 ways to solve


ok uhh


idk what that is to be honest


so i couldnt answer becasuse i dont know what those are


can u change it to be about something diffetent


you make 2 projects in a semester. Each of your friends makes 1 project in a semester. total amount of projects by you and your friends in this semester are 6. Which one is independent, which is dependent quantity?


there are many ways to describe


is independent


how many people


and dependent is projects made


corrects


okay


can you find another relationship?


not really tbh


think that your number of project can be change


eh dw


can we just do another situation


yes


for the first formula, which is independent?


the volume depends on.


oh


the minutes


correct


how about the second formula?


volume


is volume independent?


yes in the second part


truee


not in the volume problem


in the height proble


so using the first formula, but in another situation: I unpredictably have a meeting and I can't track the time. When I go back, I can mearsure the volume of water. Then I calculate the time that has flown. SO in this scenario, which one is dependent?


time right


bc time depends on the measure


idk


yesss


ok


so at the end of the day, the independent and dependent will depend on the situation


can we practice increasing and decreasing functions rq


yesssssss


like for example this confused me


what is rq?


really quick


like


fast


alright


would that be increasing or decreasing or


you mean increasing and decreasing slope?


liek increasing line decreasing line


do describe the graph


to*


u would say its increasing decreasing or both


I would say it is increasing


do you agree>


yes


because its not like


you got the visual reason


it is correct


ok


next


cuz that was like


idek how to describe it


a weird function


def not a parabola


increasing then decreasing


if you want to clarify if it is increasing or decreasing, just select the initial point and final point on the graph


wait huh


explain


OH


so for some functions that have both increasing and decreasing pattern


we need to check on which interval we are working


ohh


for example in this parabolla, from point A to B is increasing pattern while on interval from B to C is decreasing pattern


how is this a parabola if its not symmetrical like look


this is a parabola


but


the one we drew isnt


this is the one im drawing is


sorry no its not


yeah, it is quite distort


but it is fine because you know what a parabolla is


is this a good example


f(x) = x^2


that definitely a parabolla


ok


bc its quadratic


so in the interval from 1 to 2, is it deacrease or increase trend?


i would say increase


if ur talking about


this line


that's correct


okay lets try this again


which one?


actually


lets figure out if situations are discrete or not


do yk what discrete is


its this


yes


no line


do you think it is discrete or not?


this yes


it is


but i need a situation or smth


like can u make situations


and i guess if theyre discrete


for example


this is continuous


oh


I got you


so you count the number of drop of water when you make experiment. Then you draw a graph with time is input and number of drop is output. Is this discrete or continuous


discrete because


you cant have like


4.5 drops of water


can you double check


using graphic to visualize


ok


still discrete because


yes you can have 1.33222 minutes


but


you cannot have 3.5 drops


is that right


wait, I need to double check myself because I think it is continuous


okay


its because the nature of drops


because


oh yeah


can you have 2.5 drops of water


it is discrete


ye


I thought the graph would be like this


ohh


nah bc ur not measuring i dont think


well yes u are but


like


yeah the number of drop must be positive integer so it is discrete


lol


you are smart


do you want to try another problem?


yes


alright


you are driven on freeway and the speed of the car changes many times. You create the graph with time is input and the distance you have traveled is output. Is this concrete?


it is not discrete


its continuous


and it looks like this


lol did I chat concrete


my bad


:pensive:


nws


okay


is this the right graph


well


it could lowk be linear


exponential


quadratic


etc


bc ur speed isnt necessarily constan


constant


well not quadratic


bc ur speed wouldnt decrease


well wait


is distance or speed y


it depends


bc ur speed can change


yes


ur distance cant


just draw a random scenario


idk if this is speed or not tho


bc


if this is not speed


then i have to change the grapg


graph


to remind you, input is time and output is distance traveled


ohhhhh


okay then this is the graph most likely


yeah, it can look like that


no arrow to the negative quadrants because


ur distance wont go negative if ur on the freeway


well


if its from where u were


shouldnt it actually be an absolute value function


actually, the distance can't be negative at all time if you don't mention the direction of it


wait wdym


what are you trying to show me?


that that example


should be absolute value maybe


idk


think this way


without mention the direction, you walk forward 10 metres. Then, move backward 10 metres. What is the total distance you have traveled during the progress?


20 meters


meaning


correctttt


negatives dont matter here


but


linear piecewise actually


look


if I give you the direction. on a number line, the positive direction is to the right. If you move forward 10 numbers and then move back 10 number. How many number have you jumped from the start to final?


uhh


alright. It is quite hard to imagine.  Basically, it is physic lesson


you don't need to worry about it right now


okay


am i doing good


for sure ,yes


can we practice domain and range rq


like really fast


so that's all for your quiz?


domain and range also on my quiz


oh yeah, we can


kk


in inequalities btw


so you have basic knowledge about domain and range on equation, right?


ineaulity like < yes


inequality*


do you want me to give some question to test you like we have done or something else?


idm


i dont care either way


so what is domain and range for this one?


uhh


idk


you don't know how to find domain or you don't familiar with square root?


i do know domain and range


so you don't familiar with square root?


not square root no


but


so for this one


do you want me show how to do it?


yes


wait, this can't make a function


wdym


because for almost all input value x will provide 2 output value y


wait we arent learning that rn


only like this


which is like


oh I see


yk


you mean they will give the graph and you will find the range and domain


right?


ye


alright


I got it


ok


what is domain and range of this graph?


the domain is


-3<x<orequalto 1


or


you can type >=, I understand it


okk


correctttt


the range is


that's right


do you know about union symbol and intersect symbol?


nope


I have showed the union symbol


are you familiar with it?


no


alright


what is the domain and range of this graph? If you don't know, it is ok


but


i get the main idea


wait which graph


lemme see


the domain is going to be


-infinity <x<infinity for both


and the range


i cant tell because their maxximums and miniums are difficult to see


actually, the domain is every real number exceipt 1


why


and the range is every real number exceipt the interval (0.75, 3.75)


because the line wont ever touch the red line (which is 1)


I mean the function


and it won't ever go inside that Y interval


um


for example


can you use desmos?


ye


alright


type in this function: y = 1/x


done


it never touch the zero in both x axis and y axis


right?


yea


if you think it touches, zoom in


it always have distance from the axis line


yh


why is it? Because we can not divide any number by zero. Also, No number divide from 1 to get zero


do you know why we can not divide any number by zero?


because you will get undefined/divide by 0 error


can you prove it?


prove why it is undefine?


oh


if u have 2 people


and 0 cookies


they cannot get 0 cookies


because there just arent any


so in that case, no people get cooky


so yes they would get 0 cookies but mathematically


yes


I will prove it by math calculation


before, when we put a negative value under square root. It is undefined


However, thanks to mathematicians, they invented a definition of the imaginary number number


which is expressed by letter i


yea


so i^2 = -1


and square root of -1 is i


but we haven't invented a new definition for the value that make that equation true


so we still called it undefined


maybe in the future. Someone smart enough that can invent a new definition, we can change the way we call it


Hi


Sorry abt my lag


do you have any other concern


Am I good at this


yes, I think


you should select confident next time lol


Hmm?


the question John asks you before the session


Ohh


Ok


do you have any other concern?


or I will end the session here?


Sophia


hi


Hi! How's it going?


ok


What can I help you with?


For local maximums, you are trying to find all the points where the graph switches from increasing to decreasing and makes a ^ shape


for local minimums, you are trying to find all the points where the graph switches from decreasing to increasing and makes a V shape


there is a local maximum at the tip of each ^ and a local minimum at the bottom of each V


I will be right back


ok!


I'm back


Great!


Did that explanation make sense?


yes


Ok good!


Do you have any ideas where the local minimums of that graph might be?


Yep!


Now this question asks what the local minimum values are


this means it is asking for the y-value


and since these points both have the same y-value, the answer to the question should only be 1 number


Do you know what the y coordinate of these local minimums is?


Yes!


Great job!


Now the next question is asking where the graph has a value that is a local minimum


this means it is asking for the x coordinate of the local minimums


4 and 0


The 1st question asks what the local minimum values are, meaning it is asking how low the graph gets at its local minimums which means it is asking for the y coordinate since that is the coordinate that determines how high/low the point is


The 2nd question is asking where the graph has those y-values, which means it is asking for the x-values


Those types of questions can be confusing because it is hard to tell ig they are asking for the x or y coordinate


but you are correct the answer is 0 and 4!


Do you need help with anything else?


no


Ok! should I end the session then?


ok


Ok! Goodbye!


hi


Hii how can I help


OK. Um


can you tell me which are the local minimums of G?


Do you need help with understanding the concept?


yes


Would you like me to explain what a local minimum of function is.


yes please


OK, great. Um.


OK, let me try to


hm, clear it out for you. Um, if you want to find where the function has a local minimum, you have to look at the function, the lowest points in a certain um sector of the function.


For example


um, for G,


um


you have to look where the function changes from decreasing to increasing, like the valleys of the graph.


Um, for G, it's -2. You know why? Because from -4 to -1, the function is decreasing and the minimum points of the function is that


X equals -2.


So let me just write that down.


When


Did you understand what I explained or


would you like me to elaborate?


Great. Now, can you find the other local minimums of dysfunction.


based on this example


Actually, what you just found, uh, the point that you pointed out is the local maximum of the function because at this point of the curve, the curve has its maximum value, which is coming from -5 to -2.


Here it's the maximum value that minus 4 that you pointed. You need to find the point of the curve when it's the minimum, the lowest point.


For example, um, X equals 0, where Y equals 3, that's also a maximum point.


You should try finding the minimum point.


Yes, exactly. That's another


local minimum point of G.


Now, the local minimum value of G is another thing.


It wants you to put the local minimum value of X, the one you found is in X equals 3 and x equals minus 2.


Am I seeing that right? Is it -2 or minus, yeah, -2.


You put this value of X and you find the value of why in that specific point.


For example, for x equals minus 2. Let me type that down.


then y=-1


Now, based on this example, can you find the, the other local minimum value of G.


Would you like me to explain it once again?


yes


OK, let me try and be more specific. So you find, found the minimum values, the local minimum values of G, which were minus 2 and -3, right?


These are the values of X.


on the graph.


A point has two coordinates. Those are X and Y.


So, if a question asks all values at which G has local minimum, it wants the values of X.


But when the question asked all local minimum values of G, it wants the values of y.


Um


so


for this graph, for graph G,


uh the minimum value, one of the minimum values was minus 2 and the other one was 3.


So when you look at this point,


this point in -2, can you say what is its Y coordinate?


You should just look at the graph.


and at the scale of y.


and just say the corresponding coordinate.


can you tell me the coordinates of the red dot


do you know what coordinates mean?


yes


okay great


now find the coordinates of the red dot


yeah that's correct


that is what the question needs


Now you found the coordinates for the dots, and you remember that -2 was the uh value where G has a local minimum.


So you have your first answer for the first question.


The other coordinates, the other dots that we found was this one.


where X equals 3.


Can you find the Y value of this dot as well.


Yes, this is also correct. So now we've found both X's and Y coordinates of the local minimum points of the function, and that's all the question wanted.


The first question asks for the X values, and the second question asks for the Y values.


So you just have to type


minus 2 and 3 for the first question, and minus 3 and 0 for the second question. That was it.


Do you understand now?


okay then great job


Do you have any other questions?


no


alright


HI! How can I help you?


hiii


I need help with 5 problems including this one


sounds great! just for some background info, which theorems are you familiar with that you can use for this problem


Tangent, Sin, and Cos


can you use law of sines?


hmm ok


well


would you like to try using it? It's very simple!


sure


I said no cause SINE = opposite / hypotenuse


ok so basically - it is about how angles correlate with opposite sides


and 8 is adjacent and x = hypotenuse


wait actually you're right - we don't need law of sines?


could you tell me what m<JKL is?


58 deg


close! I meant the upper angle!


yk how all angles in a triangle add up 180


would it something like 90 - 58 or somthn like that to find the missing angle


almost!


Since there 180 degrees in a triangle


wait omg I'm so sry we don't need to do that lol


but just for reference the other angle would 180 - (90+58)!


but anyways


you know how


cos is adjacent over hypotenuse?


ok


so we can apply that to the side


do you know what cos 58 would be?


in terms of x and 8?


No problem if you don't - let me know!


let me solve


sounds good!


so it would be Cos (58) = 8/x


yesss


do you have a calculator?


because now that we do that


we have to isolate x


so since you have that equation - could you tell me what x is in terms of cos (58) and 8?


ok hold on


I got this


but also


the question is asking for 2 decimal places


so the number is right! but what would it be with just 2 decimal places?


Awesome!!!!


great job!!!


yayyy!!!


do you want to move on to the next problem?


yes


sounds great!


ok I got it!


so we can use the same method as before


ok


which trig operation could we use regarding 43 degrees?


either tan or sin?


yess


which one do u want to use?


either one is fine


idk


ok well we can use tan!


so that's opposite over adjacent


so what would tan 43 be equal to in terms of 36 and y?


Tan(43) =36/y


so what would y be equal to?


38.60 and rounded two decimal place is 38.6


Right


when u did ur calculation


it should be 38.605


so when u round the 5 up


the 0 becomes one


so it would be 38.61


but great great job u still did it!


ok


and now we can do the same for x


we can use m<QRP which is 90 degrees


wait what


or actually we can still do 43 degees


degrees


because we need to find x


so since it includes x


the hypotnuse


that means that we can use either sin or cos


does that make sense?


ok then would we do cos(43) = 38.61 /x?


yes! However, I would suggest doing sin 43=36/x since it is more precise since you already know those values are exact


because when rounded, the answer when it comes to dividing can be a bit off, so by using the exact numbers provided, you can ensure that the answer is more precise


yess - except remember that it wants 2 decimal places! so it would be 52.79!


great job!!


oh cause when I calculated it said 52.78


well when you calculate it fully, you have to look at the 3rd decimal since it can affect your answer


so It was 52.786


and since 6 means rounding up


then the 8 goes up by 1


so it would be 52.79


yes


so x = what again


ok


and y = 38.61


yess


I can do 1 more before my computer dies


sounds good!


what are the following questions?


oops


lem'me get that


dw lol


ok so first we can figure out m<B


do you know how to find it?


uuuuuuuhh nope


well like we did before, there 180 degrees in a triangle


since one angle is 90 and another is 75


m<B=180 - (75+90)


got it


great! do you want to move on the problem b?


m<B = 15 degrees


yes!!


wanna find bc now?


yea


sounds good


ok so we can give it a variable name


BC=x


now we can take the angle 75 degrees


ok


and use an operation that has both its opposite and adjacent


Tangent


yess


so tan 75 = x/23!


so what would x be?


x = 85.84


yes!!!!!


and since x=BC


now we can move on to AB


I'll make AB=y


so we can use 75 again


what operation uses adjacent and hypotnuse?


cos


so what would your equation be?


cos (75) =23/y?


yes!!!


yayyy!


so what would y be?


y = 88.87


YESSSS great job!!!!


and since y=AB


congratulations Pat did all of them!!


thank you soo much!!


no problem I'm glad I could help!!


make sure to charge your computer haha


I will!


byyee!!


bye!!


hey


what do u need help with


Hi! I need a bit help with this


ok


so first of all do u know how to rewrite a negative exponent


for example like 2^-1


mhm


so for a


its all diff variables


yea


we can work through it together in steps


oh so i do


im gonna write it below the screenshot


-4 times 2


6 times 2


3 times 2?


don't forget the coefficient though


do you know the next step after this


i solved it


give me one sec Emily writing on paper


ok


ok next one


b


so i do


did u get it


yes now #2


tell me if you have a qs


so i do 4 times 4


5 time 5


um actually its 4^-2


oh


so its actually 1/4^2


and same for the 5


ok nex one


do u mean 3 or are you still doing 2


so i would do -2 times -3


so -2^-3 is basically 1/2^3 which equals 1/8


k


ok #4


ok


so


would it be


umm not quite


lemme write smth on the board


you know that concept right


well you apply it to this problem


oh ok i see


wait sorry um I think I did something wrong


ok i finished all of it


um ok


this looks hard


its just a lot of numbers its not rly that hard


do u need help with the problems


i need help with a & b in #1


do for the first expression you distribute the value outside the parenthesis


actully ill work on this later im gonna do english instead


Hi there. How may I help you today?


Hi ! i need possible answers and help on this assignment, ill put it on screen.


Sure, no problem. By the way, this is my first time tutoring on Ava so bear with me please as I navigate the tools :)


Please let me know once you've posted the question. I don't see anything yet


Oh, Okay!


I see it now. Is it possible to make it a little bit bigger?


yeah!


ok. thank you


Are you familiar with the concept of domain and range and what they refer to?


not really


ok


So do you know what the horizontal axis is typically referred to by? Like what letter?


y


And also vertical axis - what letter?


x


close! It's the other way around.
Horizontal axis is X and vertical axis is Y


oh i thought you were saying do i know the axis


No worries :)


So for each of these graphs we have to say what the range is for X and what the range is for Y based on the shaped of the lines you're seeing


Let's start with an easier one maybe .. how about graph # 6


based on the shape that you see there on the horizontal axis what do you see the lowest and highest values being on the x (horizontal) axis?


let me see if I can draw on here


what are the values of those two points


as in the points?


yes


I'll help you


so the circle I have to the left is at zero, is that right?


1, 5


you're super close!


one less from each one :)


perfect!


So that means X is between 0 and 4, does that make sense ?


for that shape


mhm


let me draw another line to try and show you


see the thick line I drew? That's the X it goes from point 0 to 4 places to the right


which is your "domain"


so basically its the same for y


so


super close


1 sec


The thick vertical line is the Y (range)


so what's the lowest point? It's not -3 but you're close


it's two squares


Those two dots


what is the lower one? how many squares down below zero?


ah closer :)


2 is right for the top point


what about the bottom point


ohhh


does that make sense?


let me try to draw it bigger next to the graph


yeah its too little for me thats why i keep missing up


yep it's kind of small


but does what I drew make sense?


you have to count the squares and also give it the right sign based on whether is above the axis or below


yes


so now the domain (X) would be between 0 and 4


and the range (Y) would be between -2 and 2


do you know how to write that up on the sheet


I see little keyboard symbols next to domain and range, not sure what that means


i put that and it says it doesnt work


Do you have to click the keyboard symbol?


Do you guys use less than and great than symbols for domain and range <   and >


or do you use brackets and parentheses (x,y) or [x,y] ?


ohh ok lemme try again


sure


I'm still here if you need my help figuring out how to type it in.


Unfortunately I do need to end the session in about two minutes. I hope this has been helpful for you.


this has but okay


Let me type in two different ways you can write the domain and range for graph #6


I hope that helps guide you with the other ones


okay David


sorry if i wasted your time.


not a waste of time at all


I'm just now sure which way you're supposed to write them in


so I put both ways in there for you


okay ty!


no problem :)
Take care!


hi


Hi Maya


okay


i have this so far hold on let me writs


looks good to me, but let me double check


yes, you got it. but you can simplify it further


i can combine e^x and 3x right


that won't help. but you see there is a common factor between 3x and 6


does that work


yes! btw, you can drop that 1


thanks


you are very welcome.


any more question?


yeah, a chain rule question. Can you give me a quick second


sure


you want to give it a try first?


yeah


is this a combination of chain rule and product


or just product


it is just product rule


you got it!


i just have a problem understanding the formula


the chain rule formula?


like the middle part


I see


Let me think about how to explain it intuitively ok?


ok


let me try one explaination


we would like to compute the derivative


of h(x)


which is the instantaneous rate change in h(x)


I will write down the rate of change formula


this is the rate of change of h(x) when we increase x by a little bit


does this make sense to you?


Yes, I'm familiar


great


next, I will apply the definition of h(x)


I just replaced h() with f(g(())


next step is the key step


I just multiplied the rate of change by 1


right ok


we are almost there


I just swapped the two terms in the denominator


you see the second term is just instantaneous rate of change of g(x), so it is g'(x)


ok


if we set g(x)=u, then g(x+**) is like u+**


oh like u sub


yes


this is one way to think about


im thinking about how i would apply it here


oh I see, I thought you already got it


you got the g'(x) part


you got the f'(u)


right


almost. let me show you. It seems you got scared


actually you just have to follow the fomula


this is the formula you learned. right?


let's copy what you have written down


finally we have to replace u with terms of x


oh yeah you cant have u in an equation that started as x


thanks that makes a lot more sense


you are very welcome.


I think you are in very good shape.


thank you


bye.


bye


hello!


hi, I am having problems with my rationals test review


okay great! can you give me an example?


so you have to simplify?


The question for that equation is what is the EBA?


im sorry, can you define EBA?


EBA = end behavior asymptote


ahh i see


so that basically means the horizontal asymptote


Yes, that is the umbrella that the HA is under


okay greay


great*


so let me teach you a little trick


look at the highest degree of numerator and denominator


wait do you want to factor or find the HA first?


I was taught to factor then find HA


okay, lets factor first


so we would objectively want to cancel out the numerator and the denominator


are you familiar with the ac method?


Yes, and the A and M method


okay, i like the ac method so lets do that


so we are left with 3x+1


I just don't know where to go from there


so think about it on a graph


its just a regular line that goes up


as it goes to positive infinity, its going to go up up up


and goes down when going to negative infinity


Yes, how did you get the graph from your factored form?


well the graph i just did is 3x+1


because the (x-5) cancelled out on the numerator and denominator


Ahh, those are your zeros?


That you graphed?


no


its the same thing as f(x) but there is a hole at x=5


if you cancel out the factors, then youre left with the basic graph (3x+1)


okay, lets take a step back


did you understand how i did the ac method, right?


sort of


what part is a bit confusing?


how you got to -15


so the standard form of a quadratic is:


ok


a in the numerator is 3 and c is -5


so a times c (ac) is -15


ok, that makes a lot more sense. I may have forgotten how to do the ac method . that was s along time ago


yeah,haha me too


but practice just engraved it into my brain


what class is this for again?


pre-cal


okay cool


so lets go back to the problem


so the factors of -15 that make up -14 are 1 and -15


because you add the factors together to get b


yes, that is very obvious now


haha, okay so then i expanded the original polynomial into 3x^2-15x+x-5


i think its best to group the biggest number with Sarah because it always works


so thats why 3x^2 and -15x are togehter


what is the largest common factor between those two terms?


no, 5 is not a factor of 3x^2


yes!


so then you group it as i did on the board


so it looks like 3x(x-5)


and then you devid


not yet


because you need to group it


since we already have (x-5) of the other side, we can add what is multiplying (x-5) together to make it a term itself


so 3x(x-5) +1(x-5)


we can just add the outsides


ok


yes


and thenn you can divide


is just 3x+1


ohhh, then you set 3x+1 equal to 0?


you do not need to set anything equal to zero


well how did you get the point on the graph?


oh for the graph? i guess so


but the graph isnt too revelant, it was just to help you visualize


but just think about y=3x+1


and then thats the graph


ok, I understand


okay, so using the graph to help you visualize, what are the end behaviors?


down, up


could you explain


that would be the end behavior?


yes!


okay


do you wanna see an easier way?


yes!!!!!!


can you erase the part under the function


okay watch this


ok


these are the only terms you need to check for the end behavior


are you familiar with the term 'degree'?


okay great


so all you need to identity is the degree of the numerator and denominator


if the degree in the numerator is bigger than the degree in the denominator, then there is no horizontal asymptote


oh, yes now I am remembering this from class


and to determine the end behavior, just divide the degrees


great!


so if the quotient is an even, the end behaviors are the same


and if its odd, which is true in this case, theyre going to be opposite end behaviors


that is stupid easy, why did I not get taught this earlier?


well....


you did say you typically started with factoring


yes


but factoring is wayy too long


but make sure that the end behaviors change if the function is negative


like a is negative


ok


anyway, did you understand that?


yes!!!!!


any other questions?


thank you! nope I do not have any other questions.


okay amazing!


have a good day!


you too


Hello, how are you?


Other questions that I can help you with?


Any specific questions I can help you with?


Hello


Hi


I will put the question on the board for you


OK


I'm waiting


You can type the question on the board if that's easier.


or, or sharing your desktop


there


OK.


Are you gonna try it out and then I help you or how do you wanna go about it?


can we do it together


OK, you wanna take the first step.


Or do you want me to write down the first year?


The first step


Can you solve that


Hi there


Emily , do you see the board?


yes


ok go on


Did you see how we


separately multiplied the 4x.


and the -2 x 3.


Any questions so far


Hello? Can you hear me? Are you there?


yes I'm still here


I'm just writing down this


OK


Can you start the next step


no response


Hi Jamie


how can I help today?


Hi


I need help with understanding converse , inverse , contrapositive statements


ok


Let's look at an example.


Does this help?


Yes


The converse just switches the if and then.


The inverse keeps the original order, but makes them both negative.


So the then goes first in converse ?


we still write if, then but we switch the two ideas around.


see how "study" and "pass the exam" switched spots?


The contrapositive switches the order AND makes them both negative.  It's kind of a combination of converse and inverse, in a way.


Thank you now I understand it better 😆


okay, glad to help


have a great day, Jamie


Bye you too


hello


Hey Alex, was there a bug that wouldn't let you upload the question


im so sorry for what happened


I accidentally closed Tutor Provider out


No it's okay


and it took a while to reload for some reason


so heres the paper


I erased everything because I didn't understand it and I wanted to write it down as well


My teacher also grades the steps involved besides the answer


Okay so what do you need help specifically


factoring


mainly problems like these


We'll go problem by problem


Okay let's start with question 1


Just for the heads up, I won't be able to stay for long, so let's use our time wisely to solve as many problems as we can. Sorry for the late notice.


no worries


Okay so what do you know about factoring in general


we'd typically try to find two numbers that equal an equation like a quadratic equation


something like (x-1)(x-5)


Yes good job!


or 2x-5)(5x+2)


but these ones require a bit more steps


So that's what we'll be doing in this problem


Okay in this question, you notice that there is a common term amongst both expressions. Can you pinpoint what it is


like that?


Yes precisely


then how do we factor it then?


And now you're left with h^2 - k^2 = (h-k)(h+k)


those two only?


Also include h^2 - k^2 at the beginning of the factor


there


You did it somewhat correctly, but I'll show you


ok


In order to fully factor this you would need to know the difference of squares


oh alr


what if it was a cube


It would be the same process, but with cubes instead. For example h^3 - K^3 = (h-k)(h^2 + hk + k^2), which is the fully factor form if it was cube


But in this scenario we're working with squares, so the fully factor form will be substituting the h^2 in to a factor with k


ok bc I barely get cubes


so (h^2 + k)(h-k)(h+k)


problem 2


So what are you confused about in this question?


what do we do with the x


Okay we will use x later when we recognize the differences of squares


So first what do you notice in this expression


we have a quadratic equation


the factors for that is ((u-1)^2


Great job Alex!


So we notice that x - 2u + 1 is a perfect square, resulting in ((u-1)^2


So what have right now is (u - 1)^2 - v^2


So what do you notice now in this expression?


we have a perfect square and another term


That's true, but consider on what that truly means


This perfect square and another term tells us that there's a difference of squares


a difference of squares?


Yes, meaning (a^2 - b^2) = (a-b)(a+b)


And in this scenario, what do you think a and b is?


a is u-1 and b is v


Correct!


So all you have to do is fully factor it from this (u-1)^2 - v^2


like that?


Great job! That is perfect work


nice


problem 3


How do you feel so far on this topic?


factoring eh


I get some of it


but the bigger equations just mess me up


Oh okay with extra practice, I can guarantee that you'll do just fine on this upcoming test/quiz


we also have two other topics to go over as well but i'm more thorough with those


So in this problem, what do you see in common in these expressions


trying to factor it after the first step


like I typically dont know what variables to factor espiecially if its a cube


Don't worry we'll get to that problem soon, but let's focus on this one for now


So in this expression, what common factor do you see


Yes, you're on the right track


Yes good job Alex!


is that correct?


Yes you did it correctly, but be mindful of the operation that you used for y-1


Good job! Yes (y-1) should be (y+1)


wait im so confused


It's because you put (y-1) at first


ohh


would we combine terms?


I just wanted to guide you to fix that minor mistake


like y+3 and y+2


Nope, this will be your fully factored form/expression


oh


I got it right


thats something


ill try 4


and see what I get


After you finish, I'll guide you on what you did right or wrong


Also I'm very sorry that this may be my last question that I can help and support you on


its fine


Sorry for my inconveniences


is that it?


Not quite, but you got the first part down efficiently. Do you remember the cubes formula?


A^3-B^3= A+B and A^2-B^2


The cubes formula: x^3 + y^3 = (x+y)(x^2 -xy + y^2)


a = x and b = y


wait


do we use the coefficient?


Yes


a = 3a and b = 4c


Good job, you're learning very quick


What did you put for c? And also, be careful of "3a^2", you'll need to square it


You would need to multiply is by 4c which will give you 16c^2


huh


why


oh


we have to square the bases?


Yes precisely


Since it's in factor form, you'll have to multiply each one


is this something we only do for a cube equation?


Not really, there's other ways into solving cube equations, but I only demonstrated one of the few


so 3a and 4c become 9a and 16c


what about 3a+4c


Good job at executing each problem with effort


3a and 4c will remain the same as they only act as a factor


alright


Okay I hope I helped you effectively to understand and solve questions like these


wait


I cant think of a number multiplying to 16 and adding to 12


Are you trying to further simplify the fully factored expression?


do we have to square it again?


No you don't have to


oh wait


The final product should be: (3a + 4c)(9a^2 -12ac + 16c^2)


now?


No it wouldn't work


Because the last term of c will be 12c^2 and not 16c^2, which we are looking for


So your first answer that you wrote down was correct


(a-2c)^2 and (3a+4c)?


No, it won't be the same


what...


Because you will receive -8a^2c - 4ac^2


So it wouldn't align with 9a^2 -12ac + 16c^2


oh


Okay is there any more questions that you have before I end the session.


Also it was nice meeting you Alex


what are the two formulas?


The cube and square?


yes


Square formula: (a^2 - b^2) = (a-b)(a+b) and Cube formula: a^3 + b^3 = (a+b)(a^2 - ab + b^2)


Is there any other questions?


thats it


thank you for working with me


My pleasure. If you ever need help again, feel free to contact me since Tutor Provider is available 24/7


It was nice working and speaking to you Alex!


Alright then, see you another possible day


Bye then!


Of course, I'll look forward into working with you


Bye Alex!!!


bye!


Hi Liam! How is your day so far?


It's okay. thanks for asking


My pleasure, well I hope you're excited because by the end of this session, you'll be able to solve and understand the topic and questions that you need help with


Awesome. Thank you. I'm not really sure how this works. I just signed up


Oh it's okay, I'm just here for guidance and help on what you're confused about


Okay let's jump right into the question


Ok. I will write it on the board


Okay perfect.


After you write down the equations. Can you provide specific information on what the question is asking you to solve or do


But based on what you're writing, it seems like system of equations if I'm not mistaken


Yes it is. The question asks: Solve the following system of equations using any method. I remember doing this my freshman year but I don't know how to do it anymore


What methods do you know to solve system of equations?


Substitution is all


That I know


Oh okay, we can use that method then to solve the given equations


Do you know what to do first?


No


What do you remember from substitution?


That there has to be the same number to cancel each other out


Oh, you mixed elimination and substitution together where substitution requires you to solve for y through putting your equation into slope-intercept form


Then you substitute it into the other equation of y. For example y = 2x + 2, If I want to figure out whether it has a solution or not then I would plug in 6x - 2(2x + 2) = 10


Hi Alex!


How can I best help you today?


Hi. Well first I wanna make sure i'Northwood High algebra wasn't an option :(


I don'Northwood High algebra is an option...but depending on what it is I can probably help!


Ok thank you! Let me get the photo


Honestly I have been struggling to understand ANY of this unit


Okay no worries! I think I can help!


Let's start with the first one... any ideas on how to start? If not that's totally fine!


I have no idea


Okay that's all good! I'll do a mix of writing on the board and typing out explanations to hopefully make it clearer


okay let's start by multiplying both sides by x-5


what will that give us?


well I think on the first side, it cancels out the bottom..


Yep exactly right!


I'm not sure it would affect the 10 though


it would not affect the 10


so the other would just be x-5 right? Because 1x1 is still 1


Yep!


and then what would x be?


if I did my math right, I think 15?


Yeah, you got it!


Alright, ready for the next one?


yes


Great! So for this one since there are multiple fractions we will need a common denominator


Any idea how to do that?


nope


Okay no worries!


so let's start with the one on the right....if we pull a negative sign out front, it will look closer to the other denominators


does what I just did make sense so far?


kinda?


sorry, I'm just a little confused on why it would become negative if that makes sense


Okay no worries, that's a great question to ask. So what I did was basically multiplying by a -1 over a -1 which is just 1 right? So it doesn't actually change the fraction


but what it does do, is when the negative gets distributed to all the terms the x becomes positive and the three becomes negative


You also don't have to do this step, but it will make your other steps much easier.


oh ok


I'm hoping that makes a little more sense now?


Mostly, yeah


Okay great to hear! So now we have two different denominators: x+3 and x-3


to get a common denominator we'd have to multiply them together to get (x+3)(x-3)


So doing one fraction at a time for the (x-1) / x+3 what would we have to multiply by to get that common denominator?


oh jeez I'm sorry, I completly spaced that you asked me a question I'm so sorry-


It's all good!


the only thing I can think of is 1x3


Really close...I like to look at it by what is missing from each fraction if that helps. For example, in the first one, you have x+3 in the denominator so you'd be missing the x-3 and have to multiply the numerator and denominator by that if that makes sense.


So for the next one, which term are you missing and what would you then multiply by?


would it still be x-3?..


https://example.comis time you already have an x-3


oh so x+3


Yeah you got it!


Then for the last one, we already have the x+3 so it would be....?


Yes!!


Now what ideas do you have for possible next steps?


uh simplify?


Yeah! :)


And how would you go about doing that?


https://example.com not sure actually. normally, i'd just cancel out the number that's the same or opposite of it but there's both of those..


Okay so try thinking about it like this.... if you had something simple like 1/5 + 2/5 how would you simplify that?


If the denominators are the same we can just add right?


yeah


so then what happens is if you have one fraction equal to another fraction with the same denominator then the numerators must be equal right?


like if I had something simple like x/5 = y/5 then the x and y would be equal right?


So I know there's a lot more going on here, but we could apply the same idea


I'm so sorry, i'm still a little lost...:(


Okay I'll try breaking it down a bit simpler, where did you first start feeling a little lost and I'll go back to there


probably the message after my last response


Okay cool


okay let's start here, if I have x/5 and y/5 in what I just drew out we can agree that x and y will be equal right?


Great, so now if I take it a step further and add more stuff to the numerator the two expressions would still be equal right?


like the x+1 and the y-3


So they have the same denominator right? So if those expressions are equal what also must be equal?


the numerator?


Exactly!


So in your original question, if all of your terms have the same denominator and you simplified the one on the left by adding it....what must be the same?


the numerator...would it be x-3


Yep it would be the numerator! So you'd have the (x-1)(x-3) + (x-2)(x+3) = -(1-2x)(x-3) expression


does it make sense how we got there?


uh partially...


what can I clarify that doesn't quite make sense?


Sorry, I'm just trying to think of how x-3 would be the answer still


like the denominator


for the numerator part


Like why it is still included in the numerator? I'm not sure what you are asking sorry


wait wait wait is this the numerator answer?


Yes that's right!


since the denominators are the same we ONLY have to solve for the terms in the numerator! :)


so does it just stay like that or does it factor out into something from there?


So you would have to factor out the (x-1)(x-3) + (x-2)(x+3) = -(1-2x)(x-3)


and then solve for x by simplifying both sides


Like the top part I just circled in red is equal to the part you circled


Since I believe this problem is looking for the value of x not just an expression for it


oh ok


how would we factor it out


You can use the distributive property on each set of terms and then add the like ones if that makes sense?


huh?..


Have you talked about the distributive property? I can do a quick example if that would help?


I have but it didn't make a ton of sense


Okay I'll do a quick example down at the bottom


Did the example I just did make sennse?


I tried to use different colors, but I can type out an explanation if that helps too


yeah it makes sense


Yayy! I'm glad


Do you want to try the distributive property on your problem and then I can check your work?


yeah. is it ok if I do it on my paper irl and then say my answer before I type out what I did if something is wrong


Yeah that's totally fine!


You can also upload a picture of your work if you want me to check it or if you feel confident and don't need me to check it that's fine too


I'm here to help you so whatever is best for you!


I'll probably upload it


What you have there looks right! Now you just need to do the other side of the equation:(x-1)(x-3) + (x-2)(x+3)


wait what??


2x^2 -7x + 3 = (x-1)(x-3) + (x-2)(x+3)


I didn't see the simplification of the other side, but maybe I missed it?


I didn't know there was another thing I needing to do-


Oh okay I thought we clarified that earlier, my bad


Because if you have the 2x^2 -7x +3 it has to be equal to something to solve for x right?


Oh!


wait how do I start that...I wrote down the equation but don't know where to start


It would be the same as before, you start with distributive property for the terms on the other side. Or did you already do that?


uh nope I just have this written down- 2x^2 -7x + 3 = (x-1)(x-3) + (x-2)(x+3)


Okay so follow the same process you did before for the (x-1)(x-3) + (x-2)(x+3) stuff


where you use the distributive property and then simplify it out


ok I'll do that on paper


Sounds good!


should i simplify my answers down so they're in a shorter solution?


What do you mean by shorter solution?


Like when you are doing the distributive property?


Okay I see it!


The bottom circled ones


Now you have to move all the terms to one side of the equation and then solve for x using the quadratic formula!


We are almost there!


uh how do i do that


How to move the terms to one side or the quadratic formula part?


both


Okay for how to move the terms to one side, think about the first problem we did where 10 = x - 5. How did you solve for x there? What did you do to move the 5 term to the other side?


add 5


so it cancelled out on one side and added 5 to the 10


yep exactly so if you wanted to move the x^2 to the other side what would you do


And then what about the 5x and the 6?


minus the x^2, minus the 5x and minue the 6


minus


Correct!


woah sorry my internet cut for a moment


Let me do that on my paper real quick


wait


oh never mind I was looking at something wrong


ok so I moved that over so would I use those equations to find x?


Yep you'd use the quadratic formula to find x now unless you want to factor your solution and find x that way


Whichever way is easiest for you


uh how do I do that-


The formula


Did you talk about that formula in class?


nope


Okay then your teacher probably wants you to factor the solution and solve that way instead


Do you know how to do that?


not very well


okay give it your best shot and I'll help you fix it if its not right!


uhhh


what would your first step be or any ideas about how to solve it?


oh wait I do have the quadratic formula


I didn't realize


Okay you can use that then if you want since its just plugging numbers in


uh here's where Im confused though


This is my equation: x^2-12x-3=x^2-4x+3


wait what


I thought you simplified by bringing all of the terms to one side?


I only moved the one


do I need to move the others too?


Yep


oh shoot


Sorry, that's my mistake


It's all good


wait how do I do that...x^2-x^2 would be 0 wouldn't it


correct you'd have x^2 on one side since 2x^2 - x^2 = 1x^2 and 0 on the other side


I underlined it on the screenshot of your paper


oh but here's where I'm lost. I already did one which gave the equation I told you about so wouldn't it be zero or just not exist if I moved the other x^2 over


What do you mean you did one?


Like you already moved the x^2?


Because then yes it would be zero correct


Did you see the 1,2,3 on the one paper? I can explain what I mean if you saw them


Yeah I see that


ok so what I did is I took e.3 and put it into e.1 so I got the x^2-12x-3


wait that's wrong


where did equation 3 come from?


Oh wait nevermind I see


So I think you never added equation 2 and 3?


not together no


Okay start with that


I think that was the step you were missing that was throwing you off


makes sense


great that's right


now move those terms to the other side where equation 1 is


and solve for x


wait what-


I don't think that's right..


No that's right


oh


You have -8x - 6 = 0


which you can solve for x


So what would x be then?


Yes, let's go!!!


Great job persevering through that one!


Almost cried twice haha but would've been worth it


Proud of you!


any way we could go through at least #3?


Yep, I have a little time left - about 15 more minutes. Why don't you try to work through this one using the skills we worked on in the last one and I can help you when you get stuck?


oh uh-


So what was our very first step for the last problem?


common denominator


Great so any ideas for the common denominator here?


um


just give it your best guess


all of them are letters, I don't know-


pretend they are numbers, say 3, 7, 4


what would you do then to find a common denominator?


uh I don't remember how to find thst


that


okay that's find you'd just multiply the denominators together


so in the case with the numbers it would 3 x 7 x 4


or in your case with the letters it would be....?


rmj?


yes!!


Now go one by one and fill in the numerator for each fraction with the two letters that are missing


so for 1/r for example you are missing m and j right?


yes


great, now do the other two yourself!


would it be rj/rmj and rm/rmj


yes, exactly!


and all the terms again have the same denominator so what should we do next?


factor the numerators


you got it!


how would I do that with letters


Great question! I would recommend moving all of the terms with the letter you want to solve for (m) in this case to one side


so add rn to mj?


rm*


it would be subtract but right idea!


would it be r m^2 j?


not quite


let me write it out and see if that makes it clearer


what can I factor out of the left side?


m


Yes!


So now we have m times what on the left side?


uh j-r?


on the left


those are the correct letters but it will be plus or minus


so which one?


minus?


Yep!


Now we need to get rid of that (j-r) term so m is by itself


Any ideas? Hint: what do we do to cancel out multiplication?


Another hint -> try treating that term as just a number like 5


What would we do to get rid of it?


Hello!


Hello :) are you in need of any help?


yes one problem in my pre-calc work! i'll post it on the screen


Ok :) I should be able to help


thank you!


Ok I see it :)


So what are you looking for help with?


I want to solve a first, and then double check for me if that works because this is my first time learning the topic 1.4 on precalc


Ok no problem :) ill wait for you


wait oops lemme erase that


is this correct?


I would have a look at salute minimum again


Absalute*


I can’t spell


But, could you tell me what an absolute minimum is?


wait your right lemme fix it, its the lowest point.


Good! No problem :)


is it that?


Yes that looks a lot better :)


alr lemme place that in


Now you gotta identify the rest of the max and min values :)


alr imma try that out


is there a realitive maximum all the way from -3 to 0? and 3 to 5?


oh sorry minimum


can you tell me what a realitive minimum is?


a relative minimum where the graph switches from decreasing to increasing


good, that's good, however typically it isnt an interval, and ehre it isnt


a realitive minimum means that it just has to be the lowest point compared to what is next to it


so how many minimums does this graph have to you?


is there two?


good :)


and now it's asking at what X coordinate they are at


so where would they be located?


-3 and 3


good :) and that's what you would put there :)


and now we can look at relative maximums


how many maximums do you see?


One?


can you tell me what a maximum is?


highest point of a function


so theres two?


good, and then realative maximum is highest to all of the realitive points, correct?


meaning the ones around it'sself


right/


so its the one in blue i circled ?


close, you have 2, but i think i see 3


what else is highest compared to the points around it?


right there?


not quite, because the function continues above it


but what if we look at x=0?


was js abt to say


you see how it's the highest point compared to everything around it?


so -4, 0, and 5


good ! that is what i got :)


do you understand why that is then?


yep that makes so much sense!


good :)


thank you so much!


no problem!


alr have a great day!:)


you as well!


Hi Chloe


Hello! I will explain in one moment! I'm going to attach an image.


Ok.


I see what you uploaded.


Okay, so I need help solving this question. This is a question in ADV but that subject wasn't available on Tutor Provider, so I chose statistics. To give a tiny backstory, my classmates and I were given an overnight to solve this question, but I still don't have an answer. Today, she will put us in groups, to answer the question, which will give me a panic attack. So, I want to finish it before entering class! <3


What does "ADV" stand for?


ADV


Or aqr math class


Did you learn how to solve "simultaneous equations" ?


Umm, can you give an example?


For example: x+2y+z=0, 3x-5y-z=2, 3+6y+7z=-9


Uh, I recognize the equations but I don't think my teachers have labeled them as "simultaneous equations" just yet


I'll just say no


OK. Can you give me 10 minutes to see if I can solve? And then I'll explain.


Of course!


Look at the 4th deck. What do you see?


A polar bear equals three zebras


Correct. Now look at the third deck.


How many zebras are on the left?


And if 1 bear = 3 zebras, how many bears should you have on the right-hand side?


My bad. We need to put seals (not bears) on the 3rd deck.


No worries, I went through that too


OK. I'll keep solving...


Okay!


It's a trial-and-error problem. We try different things until it's solved. If you're not in a rush, you can leave the session open, and I'll keep trying. It's an interesting problem.


Yeah of course!


It is very intresting


It's a long problem, but I think I solved it.


We'll use a bit of algebra.


Are you still there?


SO sorry! My computer died for a second


OK. I can explain if you like.


im so sorry, it died again


Your battery is running out?


Before you do explain, can I ask a question? I think i may have a method as well


Umm, not exactly. it's on my charger, but the charger just kept falling out of the outlet


It should be good now!


OK. Ask your question.


So, I think the seals may be a fraction instead of a whole number itself. In the second row, two kangaroos and four seals somehow equals two polarpears. I thought if I divided 6/2 I might get somehwere, but I'm not sure


I guess that's not a question


Each animal has a different weight. Looking at the different decks, it would make sense that the elephant is the heaviest, correct?


Yes


So let's give the elephant a "weight", say 24 units.


24 is convenient, because this number can be divided many different ways. OK so far?


Okay!


Now look at deck 4. If one elements = 24, what is the weight of a bear?


... one *elephant


is this where i divide?


Good. So far, we have elephant weighs "24" and bear "12".


I meant deck 5, by the way. Now look at deck 4. 1 bear = 3 zebras, correct?


yes


So if the bear weighs "12", how much does a zebra weigh?


Perfect. I'm writing this on the board.


Now we look at deck 1.


Let's label the animals: E = elephant, B = bear, S = seal, Z = zebra, K = kangaroo. OK so far?


ok!


We'll write an equation for deck 1 that includes the weight of all animals on that deck.


How many elephants are on deck 1?


I'll write this as 1E. How many bears?


And one the left side, we also have 2S, correct?


Then on the right side, we have 6Z + 2K, right?


So we put 6Z+2K to the right of the equal sign. Now we plug in the known weights of the animals.


Now don't know S and K yet, so I write these in purple color.


but we dont know the kangroos weight yet


Yes. That's why we leave their weights as variables (S and K in purple).


ah ok


Now let's do the math. On the left, we have 1(24)+1(12) = ?


Good. And on the right, 6(4) = ?


To make things more tidy, we'll subtract 24 from both sides, then subtract 2S from both sides.


wait how did you get 6(4)? thats 6 zebras and 2 kangaroos


Yes. 1 zebra weighs 4 units, correct?


yes?


Look at deck 1. I'll write the weights above each animal.


On the right, we have 6 zebras. Each zebra weighs 4 units, so 6(4)=24 is the total weight of zebras on deck 1.


but then we'd be multiplying the zebra with itself


I'll add a picture to explain.


Do you see the picture I uploaded showing deck 1?


yes


Each elephant weighs 24. We have 1 elephant, so the total weight of elephants is 1(24), correct?


ohhh i think i get it now


Next, we add the weight of one bear = 1(12).


We don't know the seal's weight, so we write it as 2S.


OK so far?


yes!


After doing the math, we end up with 12 = 2K-2S as our last equation, right?


after subtracting 24 from boh sides?


Yes, After subracting 24 from both sides, and also subtracting 2S from both sides.


thats 36 + 2s = 24 + 2k
         -24          -24
        12 + 2s = 2k


where would 25 come in


It's not "25". It's 2*S or 2S.


ohhh


I'll write the S more clearly.


No it's fine! i wasn't looking at the board


There are just a couple of steps left to do.


so that's 36 + 2s = 24 + 2k
-24       -24
12 + 2s = 2k
     -2s       -2s
12 = k?
or 12 = 2k + s?


Also, i have to use the restroom. i will return in a couple of minutes


OK.


12 + 2S = 2K is correct. Subtract 2S from both sides, we get: 12+2S-2S = 2K-2S => 12 = 2K-2S.


Ah okay


and i'm back!


Lastly, we can factor out "2" from the last equation. So divide both sides by 2.


12 = 2k - 2s
/2               /2
6  = k-s


Yes. Let's call that Equation (1)


ok


Next, we do the same for deck 2. Do you see the uploaded picture?


yes


For example, there's 1 elephant on the left, so 1(24), etc.


Now let's do the math and simplify.


4(12) = ?


Let's subtract 24 from both sides.


2k + 4s = 24


Right. Now divide both sides by 2 to simplify.


wouldn't it be 12 = k + 4s?


ohhhh nevermind


i see


4/2 is 2 lol


OK. We now have our Equation 2.


We're going to solve Equations 1 and 2 to get the values of K and S.


To do that, we'll take Equation 1 and subtract Equation 2.


So on the left, we get 6-12 = ?


Right.


On the right, we get K-K, so the K cancels out.


Finally, we get -6 = -3S. Can you solve for S = ?


2 = 1?


i divided both sides by -3


Yes. -6 = -3S.


Divide both sides by -3, you get -6/(-3) = -3S/(-3) => 2 = S.


shouldn't s = 1?


6/3 = ?


ohhhhh


i forgot the 3 cancels out


Yes. So 1 seal weighs S = 2, correct?


yes


Now we can balance Deck 3 and find the number of seals. Do you see the picture?


There are 6 zebras on the right, each weighing "4". So 6(4).


There's 1 bear on the right = 1(12). Let "X" be the number of seals, each weighing 2, so X(2).


ok


We don't know "X" yet. That's what we want to find out.


Let's do the math.


We can subtract 12 from both sides to simplify.


Can you find X = ?


6 = x !!


Exactly. It takes 6 seals to balance deck 3. I'll show you the final picture.


Do you see it?


yes!


That completes the solution, and I'm running out of time. Any quick question before I go?


Thank you so much !!!


How do I explain to my teacher that i know the answer


She'll think i used ai


You can explain the steps that were used to solve the problem.


We started by assuming 1 Elephant weighs "24". Using that, we found Equations 1 and 2. And finally we solved for S = 2. And we need 6 seals to balance Deck 3.


Okayy. Thank you! Were you my tutor last time?


Nevermind! Thank you! Have a nice day!


Bye.


Hi


Hi there


I am working on a problem for calculus. I am stuck however on simplifing.


This is a quotient problem.


The question asks the derivative of f(x) at x=2 right?


I see how to get u'v-v'u/v^2


right


I don't see how to get from the first step (setting up the equation) to the the simplified form (x+1)-(x+1)


When I do it I get (1/x+1)-1/x+1)


ok i'll write down the steps, you do understand how they got the derivative right?


yes


ok we'll start there


now we just need to look at the numerator for now


ok


do you understand this step?


yes.


so now we're done with the simplification


the question asks the value of this at x=2


right.


so just substitute 2 in this


and that's your answer


any confusions?


I was making this way harder than it is.


i see, you understand it now right?


yes. thank you


ok great


hey Sophia! how can i help?


I need help in that


sure!


Yay!


ok so lets start with number 1


5+ root 19


what do u think is the answer?


wait sry before that, how much do u know abt adding rational and irrational numbers


Can I use my calculator


you dont need to for this


I’m not that good at it


Oh okay okay


its not asking u for a number, just whether its rational or irrational


did u read the explanation it gives u in number 1?


i think that'll help


Ohh okay


I’ll read it right now


lmk when ur ready to try number 1


ok!


Yes!


ok! so for 5+ root 19, is tell me whether 5 and root 19 are rational or irrational


Irrational?


5 is rational or irrational


Rational


good!


YAY


and root 19?


think about it. does 19 have a perfect square root?


No


ok, so....


Irrational


yes!!


whats wrong?


no I just like using that emoji


oh lol ok


u got it!


so when u add 5 and root 19, you get? an irrational or rational number?


remember 5 is rational, and root 19 is irrational


Uhm


look back at the description they gave u in number 1


not quite


yea!


YAY


why tho


Because when u add it up


The number


Goes on forever


technically yea but can u try explaining it the way the worksheet does?


cuz ur not always gonna have a calculator to tell u what the answer is


Ohh GESS GESS okay I got this


yes lock in :)


Cause the sum of a rational and irrational number is an irrational number


Perfect!!!


YAYAY


loving the enthusiasm!


alr next one!


Yes!


is 3/;7 rational or irrational


Uhh


How do I tell if it’s rational


Or


Irrational


a rational number is anything that can be expressed as a fraction


and an irrational number cannot be expressed as a fraction


Ohhhh


So does that make


and the numbers in the fraction must be integers


so...


Is uhm


Okay wait guys


I need to think


wait 5+root 19 is what?


uhuh, and ur line is pointing to what?


LOL whoopsies


lol


3/7? is it a fraction?


Yes


RIGTH


so is it rational?


YES! DONT SECOND GUESS URSELF


Rational?


YES!!!!


YAYAYAYAY


ur on fire!!


ok and 9/5?


is it a fraction?


Yes?


yes!


YAY


and if u add the two?


3/7 + 9/5?


rational or irrational


If we add two rational numbers that makes it a rational


YES MA'AM!


Sophia😎


ok next!!!


is root 6 rational or irrational?


Irrational


and 6? rational or irrational;


Rational


yes!!!


Yay


so if u multiply the two? rational or irrational?


awesome!!!!


algebra pro 🎉


ok try a-d on ur own now


Okay okay😎


I did it😎


YAY!!!!


YAYAY


AND EVERYTHING'S RIGHT!!


OMG OMG


RAHAHAHHA YAYAYAYA


girl u did not need me for this lol


Uhm heh…🥺🥺


ur a star 🤩


U helped me a lot😎


Aww thanks u😎


aww cmon all u


but ofcc im so happy i could help!


Gessss😎😎😎


alr! last 2!!


Yes yes YAY


do u think u can do those on ur own?


i'll be here if u need help


Owkyyy I’ll try


want me to erase the screen for u?


you'll have to reupload the worksheet tho


nvm u got it!


Yesss


ok almost done!!


I’m doneee


Is that right🥺


3a is correct!!


i dont see an answer under 3b tho\


But I did put it


maybe its a connectivity issue, just tell me what u said


oh wait i see it


yes! u might wanna explain it better tho


say root 9 is equal to 3, which is a rational number


Ohhhhh okaayay


and the product of 2 rational numbers is rational


but u got it!!!


im so proud of u <3


AWHWHWH YAYAYAY


There🥺🥺🥺


great job today Sophia! i hope you learned smth from me!


perfect!


Yes I learned a lot


Thank u so much🥰


you're very welcome! good night!


Nighty night😎


welcome Maya, how may I help you?


I have a problem on my homework, it's about 2 step equations with rational numbers.


ok.  I'll be glad to help, can you please show me what you are trying to do


Yes.


do u know what you are solving for?


Solving for X.


so how do u think you start first?


+1/2 on both sides


ok..why would u do that first?


You're supposed to put the variables and constants on different sides.


Oh wait no


-6/5x on both sides.


how can you group all the variables on one side and the constants on the other side in one step?


I'm not sure.


let's move the 2/5x to the other side would u do that?


You subtract 2/5x by 6/5x.


yup


so what answer would u get?


what is 6/5x-2/5x?


correct..


so what is the final answer?


can u write it on the board please?


Yes


awesome


you can simply further too


In decimal form?


So, moving the fractions


you want to solve for X, correct?


Yes.


how can we get x by itself on one side of the equation?


You divide 4/5 on both sides.


correct...u are doing great


go ahead and write on the board please


How do you divide by fractions?


do you know what you can multiply to 4/5 to get 1?


yes


so if you multiply the left side by 5/4..what do you get?


do you understand?


please show me on the board


You are missing something?


is it 1/2 or ?


correct


so now since you multiplied on the left by 5/4, you must do the same on the right side


so what is the final answer?


what does x=?


Great job....


did all of that make sense?


Im confused, what do I do with the 1x and the -5/8?


1x = x


Oh


the answer would be...


yup


do u know how to check to see if your answer is right?


No


so please right the original problem again


awesome


now everywhere you see an x, substitute it with -5/8


and solve the equation


what did u get?


let me delete that,  that ok?


Yes.


not much room to draw


Im sorry


no problem


can u please write the problem again at the top of the page please


yes


ok, solve the left side of the equation only for x=-5/8 please


ok


so what answer do u get for the left side of the equation?


-10/40 x


you have to remember that there is  no longer an x since you substituted a number


also you forgot to include the -1/2


oh oops


when you subtract fractions, do you go straight across?


you can only go straight across if they have the same common denominator


what is the common denominator?


why?


What happens when you don't have the same common denominator?


you have to find a number which is common for both?


You have to find a number which is common for both


what is the smallest number that is common to both numbers?


40 is the smallest number


is 40 divisible by 40?


yes


is 40 divisible by 2?


I think so.


yes it is


so if 40 is the common denominator, please solve the equation


don't solve it..write down all the steps to get the number for the numerator please


Can you show me how?


if you divide 40 by 40, what do u get?


correct


if you divide 40 by 2..what do you get?


does that make sense?


so the left side of the original equation is equal to -30/40


now we have to solve the right side of the original please solve


ok


i got the same answer.


So that means the answer you got for x is correct


Yes, thank you


do you remember the value for x?


yup


you will need to practice getting the common denominator though


i will


great...hopefully u learned something


i did thank you for teaching me


no problem.  anything else?


no it was just that question.


awesome..goodnight


goodnight


Another way to find the common denominator is to multiply the two numbers together.  If you remember from our session, we had 40 and 2 in the denominator.  We could have simply used 40*2=80 as the common denominator and then solved the fraction.  Then you would need to simplify the fraction.  Hopefully this makes sense.  Keep up the good work Maya!!!


Okay, how about now?


Yes!


Sweet !


Okay, so now maybe you post/draw somethign and see if it's working both ways?


Okay so basically in class we are doing review of algebra 1 but i have some gaps from that class and also it was 3 years ago so im struggling with simple concepts


Okey dokey.


Any particular topic then you want to review?


i started a problem just now but the tutor person was being unresponsive so i can show you what i have so far


Here’s the problem, im on #35


I see it !


Okay, so in general here, what they are asking you to do is to "solve each system" of equations by graphing.


What that means is that they are giving you the equation of two different lines and asking you "where do these lines cross"?


So what we'll do is to graph each line (as best we can) and then look to see where they intersect.


Okay so far?


Yes


In graphing the line, the easiest way to do it is to just plot two points and then connect the dots.


Or, you can convert each equation into a y = mx + b format like you've done here.


Then you can easily plot the y intercept at b, and then move up/down m units in slope to create your second point.  Then connect the dots.


Yes so i just went through the whole process of question #33 with another Tutor Provider tutor and i think i did fairly well but now here the y intercept is a fraction so is that incorrect


That's correct - the slope can be any number, including fractions.


So how did you find the point (-3, -4) in the previous problem?


I don't see any graphing...


Brooklyn !!!


I saw that.


So, if it were me, I'd not bother changing the format into the y = mx + b format.


Just get two points for each line and graph them.  Boom done.


Huh


So let's graph the first line - see what I'm writing?


And we only need 2 points to draw the line, right?


Right


So let's make it easy on ourselves and find the points by first setting x = 0 and solve for y


Then we set y = 0 and solve for x.


Okay?


Okay..


So if we set x = 0, we get y = 12/-8


or -1.5


And if we set y = 0, we get x = 12/-3 or -4


Okay so far?


Ive never seen anyone do this method


It's the easiest way to do these (by graphing).


When you solve them NOT by graphing, you'd solve each equation for y (or x, doesn't matter), then set them equal and solve for the other variable.


But that's now that we're being asked to do here..


Can you plot those two points on your graph paper (since your graphs are clean and neat and mine aren't) ???


Ok


So here when x = 0, y = -48/32 = -1.5


And when y = 0, we get x = -4


Hmm - notice anything?


They’re the same


Yup.


In general, as you got thru these type of problems, you'll get one of 3 different outcomes.


1. The lines will touch once, giving you one ordered pair (simple answer).


2. The lines are parallel but have different y intercepts.  Hence they NEVER touch (no answers).


3. The lines are parallel and line "on top of each other" which is the case here.


But backing up, does this "plot two points for each line and connect the dots" idea make sense?


No i am confused on how you came up with these coordinates


What I did (for each line) was to create two points.


In theory, you can plug any number in for x (like 27) and see what y value that gives you.


But, because I'm lazy and like to work with easy numbers, I chose x = 0 and solved for y.


But in general, you can plug in ANY number you want.  You're just finding one of the two points that you need.


Okay


Again, this is just a visual way to "solve the solution".


Wanna do another, or are you good with this technique?


Or, you can stay with your "solve them algebraically" method and graph once done...


So the answer for the question is no answer because theyre on top of eachother?


The answer (depends on the vocabulary of your text) would be that there are infinite solutions to the problem set.


There's not just one point where the lines intersect, ALL the points are intersecting.


Another thing I wanted to mention is that sometimes they won't give you the equations for the lines


They'll just give you a few points for each line and ask "where do these lines touch?"


That's why this "connect the dots" thingy is important.


Im sorry Mr. David but i like y = mx + b better because i can visualize everything and i feel like you just did some magic i dont understand your way


Yep, I get it.


So do you wanna do another one, or are you all set for now?


Let’s do the next one


It’s #39


My teacher assigns problems randomly


Chicago !


This is a vocabulary question, so to say.


What do those phrases mean?


Omg


I dont even know


Better ask Mr. Mr. Harrison.


It's those same 3 cases I mentioned a while ago methinks.


hi! how are you doing


good and you


great!


what do you need help with today


i need help with parallel and perpendicular lines


got it


do you have a specific question


that you want to work through?


yes i shared my screen so you're able to see the problem


yes


do you know where to start


i found the sqaure roots by doing a^2+b^2=c^2


and then i added them and got 6square root of five


i can show you my steps on the white board


which part are we talking about?


I finished my work and i got to that point and don't know what else to do


ok i see your work


you got those numbers from solving part a and b right


I got these numbers by drawing a graph and seeing the distance between each triangle


you got 6square root of 5 for the total distance from A-X-Y-B


the question wants to know if that is longer or shorter than just going from A-B


how would i do that


you know the coordinates of both A and B right


yes A is (8, 1) and B is (14, 9)


so you have to use the same formula you used earlier to find the distance between them right


what formula the a^2+b^2=c^2


yes


however:


are you sure that you have the correct points for X and Y


for x and y i have 10, 2


your coordinates for x are (10,2)?


yes is that correct


did you find those coordinates mathematically


like using a slope equation


I was with a person in Tutor Provider and they told me to set them equal to each other to find x and y


what was your equation for line U


I think i forgot to do that to find Point Y


my equation for line U is y=-2x+22


how did you get that


I found the slope which is 1/2 for H and S then i pluged it into y=mx+b with y and x being the coordinates for line u


ok thats good


(10,2) is right for X


have you solved for Y


yes now i need to find y


ok let me know what you get for Y


Y is (8, 6)


yes


so what do we do for part c


you need to redo your work for the a^2+b^2=c^2


using (10,2) for X and (8,6) for Y


what so i use the coordinates for my a and b


yes, (8,1) for A and (14,9) for B


no like for the a^2+b^2=c^2


no


you find the difference between X and Y values


so for A-X


you should draw out a graph


so you can visualize what you're finding


so the original square root i had is wrong


let me see


im going to check your work


okay


no your work is correct


now all you have to do is find the distance from A-B


so do the same thing you did for like A-X with the a^2+b^2=c^2 equation to find A-B


how do I know what plugs into A^2 and B^2


the same way you knew what plugged into A^2 and B^2 for the earlier equations


you subtract the y values and the x values


so if A is (8,1) and B is (14,9) the vertical distance is 8 and the horizontal distance is 6


which you'd plug into A^2 and B^2


to find C^2


wait im soo confused


so you find the right triangle, right?


so if i have coordinates how would i plug that into A^2 and B^2


because thats how A^2+b^2=c^2 works


the coordinates help you because you can make a right triangle out of them


to find the straight line distance


but like how do i know which coordinates to plug in if theres 4 points


there's only 2 points


because the second part of (c) is just asking for A to B


the first part was asking for the distance from A to X to Y to B


but the second part is asking what it would be to go straight from A to B


yea but you have your x and y value for the points a and b so how would i know to plug in the y or cx value into a and b


you don't plug in the points to a and b


you plug in the differences


how would i find the difference