ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 2) - By Lumos Learning
00:0-1 | Here's part two of all grown in math in 60 | |
00:03 | minutes or less . So hopefully you've watched Part one | |
00:05 | . The algebra unit , um , got about 25 | |
00:08 | minutes left to get through the rest of grade nine | |
00:10 | . Um , this section will be shorter than the | |
00:12 | algebra section in the last section . Geometry over the | |
00:14 | shortest section . Um , so this is part two | |
00:16 | linear relations . Once again , I'm going to go | |
00:18 | quickly through the concepts in just a few examples of | |
00:21 | things . If you want more in depth explanations , | |
00:23 | go to Jensen math dot c A . There's videos | |
00:25 | for each section there . Okay , so here we | |
00:27 | go linear relations . So first thing you would have | |
00:30 | looked at probably a scatter plot . Scatter plots is | |
00:33 | a type of graph that shows the correlation between two | |
00:36 | quantitative variables . So , for example , we have | |
00:38 | two variables here year and movie attendance and millions . | |
00:42 | And this is this is Canadian data . We have | |
00:44 | two variables . It's important to understand difference between the | |
00:47 | independent , dependent variable . If we have a table | |
00:49 | of data and horizontal rows like this X , our | |
00:52 | independent variable is usually the first row , and why | |
00:55 | are dependent ? Variable is usually the bottom row . | |
00:58 | It's important to know the difference between the two variables | |
01:00 | , because that depends where that determines where we put | |
01:03 | the variables on our graph . The independent variable goes | |
01:06 | on the horizontal axis dependent on the vertical axis , | |
01:10 | Um , and also , um , so independent variable | |
01:13 | is year in this one dependent variable is movie attendance | |
01:17 | and another way to tell which variable is , which | |
01:21 | is that . Attendance depends on the year , so | |
01:24 | that makes our dependent variable . The year causes a | |
01:27 | change in the attendance is another way to think about | |
01:29 | it , so it makes it our independent variable . | |
01:31 | If you're given data in vertical columns , remember that | |
01:35 | X is usually given to us on the left , | |
01:37 | so independent on the left Y on the right . | |
01:39 | And when we graph it , make sure you remember | |
01:41 | that we are going to put our independent variable X | |
01:45 | on the horizontal axis . So here's our year , | |
01:47 | and our dependent variable movie attendance is going to go | |
01:53 | on a vertical axis . Dependent always goes on the | |
01:55 | Y axis . The vertical axis independent always goes on | |
01:58 | the x axis , the horizontal axis , and make | |
02:00 | sure when you're choosing your scales , make sure you | |
02:03 | choose a scale that goes up by an even amount | |
02:05 | each time . So attendance notice that I should have | |
02:07 | a break in here because I skipped some values . | |
02:11 | But after that skipped values between zero and 80 it | |
02:13 | goes up by a constant amount . Each time it | |
02:15 | goes up by five . Each time , it's important | |
02:18 | that your scale goes up by an even amount . | |
02:20 | Each time and years , I started 1994 went up | |
02:22 | by an even number of years , each time by | |
02:24 | one year . Um , for each unit , Uh | |
02:28 | , and why did I choose 80 to start out | |
02:29 | for attendance while I needed my lowest value is 83.8 | |
02:33 | . My highest value is 1 25.6 . So I | |
02:36 | needed a scale that goes at least between those ranges | |
02:39 | . So I started at 80 and then I went | |
02:42 | all the way up to 1 . 45 because I'm | |
02:44 | going to be making some estimations later on this graph | |
02:47 | and years , I started 94 I went up to | |
02:50 | 2000 and six because later we're going to making an | |
02:53 | estimation for a year outside of our range . So | |
02:56 | if you plot all of your points on this graph | |
02:59 | . So this is going to represent a point on | |
03:00 | the graph . This is going to represent a point | |
03:02 | on the graph and so on . So in 1994 | |
03:04 | movie attendance was 83.8 million . So in 1994 83.8 | |
03:09 | would be roughly right about here . I put a | |
03:11 | dot at that point right there , 1995 movie tens | |
03:15 | was 87.3 million . So put a dot at about | |
03:19 | 87.3 , which is right about there and so on | |
03:22 | . If we put a point for all of our | |
03:24 | data values and then draw a line that shows the | |
03:26 | trend in the data , you would have something that | |
03:28 | looks like this have my break in my graph day | |
03:33 | , and it would fall roughly a linear trend . | |
03:35 | When you draw a trendline , make sure you follow | |
03:37 | two rules . Make sure you're lying goes through as | |
03:39 | many points as possible , and the points that it | |
03:42 | doesn't go through like here , here , here , | |
03:45 | here , here , here are evenly distributed above and | |
03:51 | below the line . So there's three above the line | |
03:53 | three below the line . But make sure dryer line | |
03:55 | that goes through as many as possible . The ones | |
03:57 | that doesn't go through evenly distribute above and below , | |
03:59 | and you'll have a good trend line . The trend | |
04:01 | line does not have to start at the origin . | |
04:03 | It can start anywhere along the Y axis anywhere along | |
04:05 | the X axis , just as long as it follows | |
04:07 | the general trend in the data . Now we can | |
04:10 | use that trend line to make estimations . But first | |
04:12 | , let's describe the correlation between year and attendance . | |
04:16 | I want to describe it in two ways . First | |
04:19 | of all , since our line is going up to | |
04:22 | the right , we say it's a positive correlation . | |
04:24 | And since all the dots are fairly close to the | |
04:27 | line , we say it's a strong correlation . So | |
04:29 | it's a strong , positive correlation we have here . | |
04:31 | So up to the right is positive and the dots | |
04:34 | are close to line . So strong . So it's | |
04:35 | a strong , positive linear correlation . And what that | |
04:38 | tells us is that as the year increases , the | |
04:41 | movie attendance is also increasing and now notice from our | |
04:45 | data table there's no data for 2000 and one , | |
04:47 | so we want to make an estimation . I want | |
04:51 | to make an estimation for the movie tenants using our | |
04:53 | line of best fit . So in 2000 and one | |
04:56 | , we go up to see where just 2000 and | |
04:57 | 1 m line of best fit it meets it right | |
05:00 | here . We look across on ry access . That's | |
05:03 | about right here . That's about 121 million would be | |
05:06 | our estimation for the year 2000 and one . So | |
05:10 | you make sure you're using your line of best fit | |
05:12 | properly here . So 121 million And then we also | |
05:17 | have to know , was this estimation something we call | |
05:20 | an interpolation , or was it an extrapolation ? So | |
05:23 | what's the difference between the two ? If we look | |
05:25 | at the data were originally given , were given data | |
05:28 | between 94 2000 and three ? That's the range of | |
05:31 | our data . It goes all the way from 1994 | |
05:34 | to 2000 and three . If we make an estimation | |
05:37 | for a year between our lowest and our highest year | |
05:40 | , we call that an interpolation . If we make | |
05:42 | an estimation that's outside of our range to either less | |
05:45 | than 94 or bigger than 2000 and three , we | |
05:47 | call that an extrapolation So we made an estimation for | |
05:50 | 2000 and one that's between 94 2000 and three . | |
05:53 | So we call that an interpretation . Let's make an | |
05:59 | estimation for 2000 and five . So if we look | |
06:02 | at our line of best fit so 2000 and five | |
06:04 | this year see where our line of best fit says | |
06:07 | the attendance would be for 2000 and five . We | |
06:09 | put a point on the line of Best fit at | |
06:11 | 2000 and five and then look across to the y | |
06:13 | axis . It's about right here about 1 43 or | |
06:16 | 1 44 . I'm gonna say my estimation is about | |
06:20 | 143 . Remember , we're in millions here . Did | |
06:24 | we use interpolation or extrapolation ? While 2000 and five | |
06:27 | is outside of the range of data were originally given | |
06:29 | , it's bigger than our highest value 2000 and three | |
06:31 | . So we say that that is in fact , | |
06:33 | an extrapolation . Yeah , okay , next thing you | |
06:37 | would have done and the distance time graphs . How | |
06:39 | do we analyze the distance ? Time graphs describe motion | |
06:41 | Well , here , have a distance . Time graphs | |
06:43 | . If we read here , it says described the | |
06:45 | following graph that represents a person's distance from home over | |
06:47 | a period of time . So we have a period | |
06:50 | of time seven minutes and the graph tells us the | |
06:53 | distance the person is in meters from their house , | |
06:56 | Um , during that period of time . So at | |
06:58 | the very beginning of the graph , when zero minutes | |
07:01 | have elapsed there 0 m from the house , which | |
07:04 | means they're at their house . But then two minutes | |
07:06 | later will notice they are 100 m from their house | |
07:09 | . So this rising line up to the right tells | |
07:13 | us the person is moving away from their house , | |
07:16 | right . If they were 0 m away from their | |
07:18 | house and now they're 100 m away from their house | |
07:19 | , they must be moving away from their house . | |
07:21 | So this rising line up to the right that tells | |
07:24 | us movement away from the center or , in this | |
07:27 | case , away from the person's house . So any | |
07:29 | line moving up to the right is movement away . | |
07:31 | And since it's a straight line with no curves , | |
07:33 | we say they're moving away at a constant rate , | |
07:36 | and I'm going to say this is a slow movement | |
07:38 | compared to the next line segment because It takes two | |
07:42 | minutes to travel 100 m in this line segment , | |
07:45 | but this line segment here from B to C two | |
07:48 | minutes later , they've traveled 1 to 300 m . | |
07:52 | They traveled 300 m in the same time it took | |
07:55 | them to travel 100 m . So B C is | |
07:57 | a faster movement . You can tell there between fast | |
07:59 | and slow , because faster movement as a steeper line | |
08:02 | , slower movement is a less steep line . So | |
08:05 | let's describe the motion here of line segment A . | |
08:08 | Be so up to the right , which means away | |
08:10 | from the House . And we're going to say at | |
08:12 | a slow , steady pace . Steady pace means that | |
08:16 | the speed is constant . They're not changing . Speeds | |
08:18 | are not accelerating or decelerating . NBC . The line | |
08:21 | is still going up to the right , but it's | |
08:23 | steeper . They're traveling a further distance in the same | |
08:25 | amount of times they're moving faster , so we'll say | |
08:28 | they're moving at a fast , steady pace away from | |
08:31 | their house , so away from home at a fast | |
08:33 | , steady pace . How about C D line segment | |
08:36 | CD at the beginning , So at four minutes , | |
08:40 | there are 400 m away from their house and then | |
08:42 | two minutes later , at six minutes , there's still | |
08:44 | 400 m away from their house . And why didn't | |
08:46 | go up or down at all between those two , | |
08:48 | So the person must not have moved at all . | |
08:51 | So for two minutes they didn't move . So any | |
08:53 | horizontal line represents no movement . So in this case | |
08:58 | , sorry . In this case , C d . | |
09:00 | No movement D E is the only line segment going | |
09:03 | down to the right . So they were 400 m | |
09:06 | away from their house . But now they're only at | |
09:07 | the end of the line segment there , only 200 | |
09:09 | m away from their house , they must moving towards | |
09:11 | their house . Any line segment going down to the | |
09:13 | right represents movement towards the sensor , in this case | |
09:16 | towards the person's house . And it's a steep line | |
09:19 | . So they're moving . They're moving fast towards their | |
09:21 | house . Um , and there's no curves , so | |
09:24 | it's at a steady rate , a constant speed . | |
09:27 | So line segment D . E . We would say | |
09:30 | the person is moving at a fast , steady pace | |
09:33 | toward home and in distance time graphs . Be careful | |
09:37 | to look for curves . If you have a curved | |
09:39 | line , you have either an acceleration or deceleration . | |
09:42 | You have an acceleration if the line starts off not | |
09:44 | very steep than gets steeper and steeper and steeper and | |
09:47 | steeper and steeper . That means their rate of movement | |
09:48 | is increasing , so that be an acceleration . If | |
09:51 | we have a line that starts off very steep and | |
09:54 | then gets less deep and less deep and less deep | |
09:56 | and less deep as we move from left to right | |
09:58 | , that represents a deceleration . So be careful of | |
10:02 | that when we're doing distance time graphs . Okay , | |
10:05 | the main section for linear relations would have looked at | |
10:07 | Is the y equals MX plus B section . Basically | |
10:09 | , if we have two variables that form a perfect | |
10:12 | linear relationship , that happens when , um , as | |
10:15 | one variable increases , Um , we have another variable | |
10:19 | that increases at a constant rate , so one variable | |
10:22 | is a constant multiple of the other variable . So | |
10:26 | this equation , you would have learned all these variables | |
10:28 | mean specifically would have learned the M represents the slope | |
10:31 | or rate of change actually have that written right below | |
10:35 | rate of change or called the constant variation . B | |
10:39 | represents the Y intercept , graphically speaking or the initial | |
10:41 | value , and we have Y and X right because | |
10:44 | we have our two variables are have a relationship between | |
10:47 | the two . Why being are dependent ? You know | |
10:49 | that X being are independent . So if we have | |
10:51 | two variables to form a linear relationship with one variable | |
10:53 | to constant multiple , the other variable Um , we | |
10:57 | say M is the rate of change between those two | |
10:59 | variables . And if we have a table of data | |
11:01 | , we can calculate them the rate of change by | |
11:03 | figuring out what is why changing by divide that by | |
11:06 | what X is changing by . If we have a | |
11:07 | graph , we can figure out the rate of change | |
11:09 | by counting our rise and dividing it by our run | |
11:12 | and in an equation , you'll refer to them as | |
11:15 | the slope a rate of change of constant variation . | |
11:18 | So what we're gonna do in this section is be | |
11:21 | able to work between the table graph and equation forms | |
11:24 | of linear relationship . Before we do that , let | |
11:27 | me just quickly show you what I'm talking about here | |
11:29 | . So if we have a relationship here between distance | |
11:34 | and time , so the distance traveled by bus varies | |
11:37 | directly with time . So we have distance and times | |
11:39 | are two variables distance depends on time . So distances | |
11:43 | . Why time of exit , says the bus travels | |
11:45 | 240 kilometers in three hours or distance and our time | |
11:48 | . Once again it travels 243 hours . But if | |
11:52 | we want to figure out the rate of change or | |
11:54 | the slope or constant variation where we want to call | |
11:56 | it at the rate of change , we want to | |
11:57 | figure out what is the distance changing by for every | |
12:00 | one hour . So to calculate that we call it | |
12:02 | our M . Our rate of change is m . | |
12:03 | We do our change in Y , which is 240 | |
12:07 | in this case , kilometers and we divided by our | |
12:10 | change in X , our independent variable which is times | |
12:12 | of three hours . If we don't do this division | |
12:16 | , it'll tell us how many kilometers the bus travels | |
12:19 | for every one hour . And that's a rate of | |
12:20 | change . We need to know what is why change | |
12:22 | changing by everyone . Unit increase in x 024 divided | |
12:25 | by three that gives us 80 kilometers per hour . | |
12:29 | So the buses going 80 kilometers per hour . That's | |
12:31 | RM . That's our rate of change . Another thing | |
12:34 | you would have looked at before . I get into | |
12:37 | more specific things about linear relations , as you would | |
12:39 | have looked at between direct variation and partial variation relationships | |
12:43 | . So there's two types of linear relationships . It | |
12:45 | could be a direct variation relationship , where the initial | |
12:47 | value is 00 for the relationship between the variables , | |
12:53 | meaning graphically . The line that represents the relationship between | |
12:57 | two variables would pass through the origin in the equation | |
13:00 | . You don't see anything added after the variable X | |
13:03 | . There's nothing added off at the end Here . | |
13:05 | You can write a plus zero if you want to | |
13:07 | , but that's a direct variation relationship . Initial value | |
13:10 | zero passes through the origin . The equation has no | |
13:13 | nothing added . At the end , you don't see | |
13:14 | a plus B because the B value is zero . | |
13:16 | The Y intercept zero partial variation relationship has an initial | |
13:20 | value that is not zero when x zero . Why | |
13:23 | is something else in this case , I gave an | |
13:25 | example of why being one when x zero , that | |
13:27 | means graphically speaking , the line is not going to | |
13:29 | pass through the origin . In this case , it's | |
13:30 | going to pass through one in the equation . You're | |
13:33 | going to see that Y intercept That initial value added | |
13:35 | after the X and we call that a partial variation | |
13:38 | relationship . So what we're gonna do now , we're | |
13:42 | going to be able to given a linear relationship between | |
13:44 | two variables . We're gonna be able to represent that | |
13:46 | relationship between the variables in a table of values . | |
13:49 | Um , we're going to be able to represent the | |
13:51 | relationship using an equation , and we're going to represent | |
13:54 | the relationship between the two variables using a graph . | |
13:57 | So let me quickly show you those three things , | |
13:59 | and then we'll work at moving between those three things | |
14:01 | . So costs to electoral work varies directly with time | |
14:03 | . So it tells us it's a direct variation , | |
14:06 | and it's the relationship between cost and time . In | |
14:09 | this case , cost depends on times that cost is | |
14:11 | dependent . Time is independent . They charge $25 per | |
14:15 | hour . It tells us how much it costs for | |
14:16 | every one hour , so we know our rate of | |
14:18 | change . We know our M value is 25 in | |
14:22 | this case through electric work . If we want to | |
14:24 | show this relationship in a table in a table , | |
14:27 | we put X on the left . Why on the | |
14:28 | right and remember access time so they could do work | |
14:32 | for 0123 And we can make this table as big | |
14:35 | as we want to stop there . Those are ours | |
14:37 | now . How much would it cost for each amount | |
14:39 | of these hours For them to do work Well , | |
14:40 | if they don't work , it's not gonna cost anything | |
14:43 | . If they work for one hour . Since they | |
14:45 | charge $25 per hour , it's gonna cost $25 . | |
14:48 | They work for two hours . It's going to cost | |
14:51 | $50 right ? It's 25 bucks per hour . So | |
14:53 | two times 25 is 50 . If they work for | |
14:55 | three hours three times 25 75 it would cost $75 | |
14:58 | . And so on . This shows the relationship between | |
15:02 | cost and time for this electrical company . We could | |
15:06 | represent all these numbers who want equation . We can | |
15:08 | make an equation that shows the relationship between X and | |
15:11 | Y and indirect variation relationship . The equation is always | |
15:15 | y equals MX . Any linear relationship is why equals | |
15:18 | MX plus B , but you'll see why we don't | |
15:19 | need to plus B for this equation , because the | |
15:21 | initial value for this equation the cost from X zero | |
15:24 | is zero . So we don't need the plus B | |
15:26 | because B is zero in this case , the initial | |
15:28 | value . The Y intercept is zero . So the | |
15:31 | relationship between cost remember wise cost , excess time cost | |
15:36 | is equal to $25 . That's a rate of change | |
15:39 | times the amount of time that you work for . | |
15:41 | And this equation shows us the relation between Y and | |
15:43 | X for any value of X , right . If | |
15:45 | we wanted to figure out how much it costs from | |
15:47 | you three hours of work , you plug in three | |
15:49 | into our equation for X £25.3 or 75 will get | |
15:52 | cost equals $75 . We don't know how much it | |
15:55 | costs to do 10 hours of work . You would | |
15:57 | plug in 10 forex and we'd get cost equals $250 | |
16:01 | and so on . Now , graphically speaking , if | |
16:03 | we were to plot these points on our graph , | |
16:05 | let's see what it would look like . Zero hours | |
16:08 | of work cost $0.1 hour of work cost $25.2 hours | |
16:12 | Costs $50 . 3 hours costs $75 . Notice how | |
16:15 | this is going up by a constant amount each time | |
16:19 | . That's what makes this a linear relationship , because | |
16:21 | the rate of change is constant . And then we | |
16:23 | would connect these points with a line to show the | |
16:26 | relationship . And there's the graph of the relationship between | |
16:30 | cost and time for these two variables , so we | |
16:34 | could represent the relationship between costs and time in three | |
16:36 | different ways . Here , in a graph using an | |
16:39 | equation or in a table of values . All three | |
16:42 | of those ways show the relationship between cost and time | |
16:44 | for this electric company and notice . This is a | |
16:48 | direct creation because it starts at 00 and goes up | |
16:51 | by a constant amount each time . That makes it | |
16:53 | a constant for sure . That makes it a direct | |
16:55 | variation . Also , we have a plus zero here | |
16:57 | that also shows us it's a direct variation , and | |
16:59 | from a table we know it's a direct variation because | |
17:01 | the initial value is zero Okay , let's look at | |
17:06 | what if I gave you just a table ? We | |
17:08 | don't know what X and y are here . We | |
17:10 | know exes are independent wise , are different . We | |
17:12 | don't know exactly what they are . We have a | |
17:13 | table of values here What if I wanted to come | |
17:15 | up in the equation to represent the relationship between X | |
17:18 | and Y here ? What I'm going to need for | |
17:20 | any linear relationship equation . It's always going to be | |
17:24 | the form y equals MX plus B . I'm always | |
17:26 | going to need to determine what is the value of | |
17:29 | and what is the value of B to be able | |
17:31 | to write the equation of the relationship between X and | |
17:33 | Y . So what I'm going to do first is | |
17:36 | I'm going to figure out the value of B . | |
17:38 | That's the easiest thing to figure out the value of | |
17:40 | B . That is our initial value or you'll hear | |
17:45 | it referred to graphically as the Y intercept the Y | |
17:47 | intercept . So that means , um , where does | |
17:52 | it cross the Y axis ? So when x zero | |
17:55 | , What is why ? So I'm just gonna write | |
17:57 | that here When X equals zero . Why equals what | |
18:04 | ? And we can get that from our table when | |
18:06 | X zero y is four . So are y intercept | |
18:09 | our B value . Our initial value B is equal | |
18:14 | to four . So you just look in your table | |
18:16 | when x zero . What's why that's your initial value | |
18:19 | . That's your y intercept . If we were to | |
18:20 | plot this 00.4 that would be right on the Y | |
18:22 | Axis , which makes our line across the UAE except | |
18:25 | four . So that's our B value . Mm . | |
18:27 | How do we figure out m from the table of | |
18:30 | values ? Remember , I showed you equations before your | |
18:33 | rate of change , you're going to have to figure | |
18:34 | out what is why changing by and divide that by | |
18:36 | what X is changing by and that will tell you | |
18:39 | for everyone . Unit increase in X . What is | |
18:41 | ? Why changing by So to figure out the change | |
18:44 | in y an organized way to do it is to | |
18:46 | do your second Y value minus your first y value | |
18:49 | . And divide that by two . Figure your change | |
18:51 | in X to your second X value minus your first | |
18:54 | X value . So y tu minus y one over | |
18:56 | x two minus x one That will tell your change | |
18:58 | and y divided by your change in X . Now | |
19:00 | you have to take two points to be your 1st | |
19:02 | and 2nd points . I'm just gonna choose the first | |
19:04 | two points for it to be simple here . So | |
19:06 | this is my 1st 20.4 So That's my first X | |
19:09 | value . My first y value . I'll choose . | |
19:11 | This is my second point . This is my second | |
19:13 | X value , my second y value . If I | |
19:15 | plug those values into my equation , I have eight | |
19:18 | minus 4/2 minus zero , and I get 4/2 , | |
19:23 | which is two . So my M values to my | |
19:26 | B values for I can write my equation by plugging | |
19:30 | in m and B into ecos and exports . B | |
19:33 | y equals two X plus four . This equation shows | |
19:37 | the relationship between X and Y and notice this equation | |
19:41 | works , right ? If I plugged in eight for | |
19:44 | X into this equation , what should I get my | |
19:46 | answer for ? Why to be It should be 22 | |
19:49 | times 8 to 16 plus four is 20 . It | |
19:51 | shows the relationship between the two variables . Now , | |
19:53 | I just want to point out you didn't have to | |
19:55 | choose . This is your first point . This is | |
19:57 | your second point when you plugged into y two months | |
19:59 | . Why ? Whenever x two , minus x one | |
20:02 | , let's say we could have chosen any two points | |
20:04 | . I could have chosen this one to my second | |
20:07 | point x two y two . If I plugged into | |
20:09 | my equation here , I would get 16 minus 4/6 | |
20:13 | , minus zero . And that would give me 12/6 | |
20:15 | , which is still the same answer of two . | |
20:19 | So that's how to go from a table to the | |
20:21 | equation . What if I give you an equation ? | |
20:23 | I want you to draw the graph . Well , | |
20:28 | to be able to do that , we know two | |
20:30 | things from our equation here . Em . We call | |
20:34 | that graphically speaking . We call that our slope M | |
20:38 | Equal Slope , which in this case is three and | |
20:42 | any whole number is over one . And we're talking | |
20:44 | about Slope . Remember , The top is our change | |
20:46 | and why I'm going to call that rise . And | |
20:50 | the bottom is our change in X . We call | |
20:52 | that run , but where do I start from ? | |
20:56 | Don't forget this year , that is our Y intercept | |
20:59 | . That's our initial values . We know our graph | |
21:01 | starts right here on the Y axis at negative two | |
21:04 | . And then from there , I need to use | |
21:06 | my slope of 3/1 plot more points . So that | |
21:09 | means rise three . Run one . Remember , A | |
21:11 | positive rise means up a positive run means right so | |
21:15 | up . Three right one and then continue up . | |
21:17 | Three . Right , one up , three , right | |
21:20 | , one and two plot points on the other side | |
21:22 | of the line . Do the opposite of up three | |
21:24 | . Right home , which is down three . Left | |
21:25 | one and we get more points on the same line | |
21:29 | . Connect all the points should be on the line | |
21:32 | if you've done it properly . And that's how to | |
21:34 | go from the equations of the graph . Plot your | |
21:36 | Y intercept to use your B value to plot an | |
21:38 | intercept and use your slope . Use the rise and | |
21:41 | the run . The numerator is the rise . And | |
21:42 | then there's the run for rise . Counting up is | |
21:45 | positive . Counting to the right is positive for run | |
21:48 | . If we had a negative rise , we would | |
21:50 | count down . If we had a negative run , | |
21:52 | we would count left and so on . What if | |
21:55 | we wanted to go from a graph to the equation | |
21:57 | ? Well , there's a couple ways we can do | |
21:59 | this . We know our equation is why cause MX | |
22:02 | plus B and we know we're going to need to | |
22:03 | figure out the values of M and B , so | |
22:06 | B is easier one to find , right ? The | |
22:08 | is just the Y intercept . It tells us when | |
22:10 | . X zero . What's why ? So where's across | |
22:12 | the Y axis that crosses the Y axis right there | |
22:13 | at two . Might be value is , too . | |
22:16 | But to calculate the rate of change , we have | |
22:17 | a couple options . We could choose to lattice points | |
22:20 | , so points in the corners of the boxes and | |
22:22 | we could count are rising , are run right M | |
22:25 | equals rise over . Run Graphically speaking . So count | |
22:30 | your rise . To get from one point to the | |
22:31 | other , we would have to count up 1234 Up | |
22:36 | is positive . So positive forces arise . And then | |
22:39 | we'd have to go to the right one to going | |
22:41 | to the right as a positive run . So four | |
22:43 | or two , which is two . That's fine . | |
22:46 | But what if you wanted to go from instead of | |
22:51 | going from here to here ? What if we wanted | |
22:52 | to go from here to here ? We would have | |
22:54 | to count down 1234 So that would be negative for | |
22:58 | as our rise . And then we would have to | |
23:00 | go left to which may get negative . Two is | |
23:02 | our run notice . Same answer negatively by negative is | |
23:05 | positive . It doesn't matter whether at which point you | |
23:08 | go from , you'll get the same answer either way | |
23:11 | . Also , you could use the method we used | |
23:13 | before using our change in y over change in X | |
23:19 | , which is y tu minus y one over x | |
23:22 | two minus X one that works perfectly fine as well | |
23:26 | , changing wise the same as rise change . Next | |
23:28 | is the same as Run . We could just choose | |
23:29 | our two points here . Let's say this is our | |
23:31 | first point . So that's our X one y one | |
23:33 | . Here's our second point x two y two . | |
23:36 | Plug into our equation . We get six minus 2/2 | |
23:40 | minus zero . Sorry , six minus 2/2 minus cereal | |
23:46 | . That gives us 4/2 , which is two . | |
23:49 | So either way you get the answer of two for | |
23:51 | M . Plug into your equation . Y equals MX | |
23:53 | plus B , and we get y equals two x | |
23:55 | plus to this equation shows the relation between X and | |
23:59 | Y . Um , for this graph , uh , | |
24:04 | we're going to do some equations of lines algebraic li | |
24:07 | quickly , and then we'll be done this section . | |
24:09 | So before we do that , you have to remember | |
24:10 | parallel lines lines in the same plane that never meet | |
24:13 | . Parallel lines have equivalent slopes , so you have | |
24:16 | to remember that parallel lines have equivalent slopes . Perpendicular | |
24:22 | lines , lines that meet at 90 degrees have slopes | |
24:24 | that are negative , reciprocal of each other . So | |
24:27 | you have to remember that negative reciprocal roles of each | |
24:31 | other , and I'll explain that more in a second | |
24:34 | . So determine the equation of a line algebraic given | |
24:37 | different information I could give you Slope member Slope is | |
24:41 | M and one point it passes through . So I | |
24:44 | didn't give you the Y intercept . Find the equation | |
24:46 | of a line you always need to things you always | |
24:49 | need . Slope y intercept . I gave you slope | |
24:52 | . I didn't give you y intercept you can use | |
24:56 | at this point . Every point has an X and | |
24:59 | A Y in the equation of a line will satisfy | |
25:01 | X and Y for any point that's on the line | |
25:03 | . So what we can do is use this information | |
25:06 | m x and Y plug that in for M X | |
25:08 | and Y in the equation and then solve for B | |
25:10 | . So if we do that , if I plug | |
25:12 | in five for Y negative number two for M negative | |
25:15 | four for X and Y equals MX plus B . | |
25:18 | So I'll have something . Looks like this . Negative | |
25:20 | 1/2 negative four plus b solve for B . I | |
25:24 | will then know the B value . I can write | |
25:26 | my equation by plugging in for M and B , | |
25:28 | so I'm just gonna solve this quickly . Negative . | |
25:30 | One times negative four is four . So we have | |
25:32 | 4/2 . I know that four divided by two is | |
25:35 | two , so I can just write that as two | |
25:38 | plus be moved . The tour becomes a minus 25 | |
25:41 | minus two is three . So I have three equals | |
25:42 | . B semi final equation . I plug in for | |
25:44 | M and R B . I have negative 1/2 X | |
25:48 | plus three . How about here ? What if I | |
25:51 | give you a point and I tell you it's parallel | |
25:53 | to this line . We know parallel lines of equivalent | |
25:55 | slopes and the slope of this line . The number | |
25:57 | in front of the X is a one . So | |
26:00 | I have a parallel slope , is one . Those | |
26:03 | two lines are parallel . That's a similar parallel . | |
26:06 | The parallel slope is one , and I have a | |
26:08 | point on the line X and y I can use | |
26:11 | the X y and M plug into Why goes MX | |
26:14 | plus B and solve for B Remember So like MX | |
26:19 | plus B negative four is why one is m Xs | |
26:24 | four . Solve for B . It's a negative . | |
26:27 | Four equals four plus B . Move this for over | |
26:30 | negative four . Minus four is negative . Eight . | |
26:33 | So my final answer Y equals one x minus eight | |
26:37 | . So we know parallel lines have equivalent slopes . | |
26:39 | They don't have the same y intercept , so don't | |
26:41 | use that at all . All we know para lines | |
26:44 | have equivalent slope so I can use the slope from | |
26:46 | that line along with the point we know is on | |
26:48 | our parallel line . Plug it all in sulfur be | |
26:51 | . There's the equation of our line . How about | |
26:53 | perpendicular lines ? So we know our line goes through | |
26:57 | this point and it's perpendicular to this line here . | |
27:01 | So perpendicular lines of slopes , their negative reciprocal . | |
27:04 | So this slope is negative . 2/5 , a perpendicular | |
27:07 | slope . There's a symbol for perpendicular perpendicular slope . | |
27:10 | We flip the slope , so instead of two or | |
27:13 | five , we make it five or two and we | |
27:15 | change the sign instead of it being negative , we | |
27:17 | make it positive So we have a perpendicular slope . | |
27:20 | We have a point . We don't use this y | |
27:21 | intercept at all . All we know are perpendicular slope | |
27:24 | . Sorry . Perpendicular lines have slopes or negative . | |
27:26 | Reciprocal . So we use that perpendicular slope . At | |
27:29 | that point , plug into y equals MX plus B | |
27:33 | and solve for B . So why is zero And | |
27:36 | we know it's 5/2 and X is five and we | |
27:42 | don't know B If we solve this equation 25/2 plus | |
27:47 | B and sorry , last step here and moved out | |
27:51 | 25 or two to the other side . We get | |
27:52 | negative 25/2 equals beast and my equation of my perpendicular | |
27:57 | line . I use my perpendicular slope 5/2 X minus | |
28:02 | 25/2 . That's the equation . Alignments perpendicular to this | |
28:06 | line and goes through that point . The only other | |
28:09 | way I could ask you to find the equation of | |
28:11 | a line would be if I don't give you the | |
28:14 | slope or the Y intercept right . We need both | |
28:16 | mean m N b . But if I don't give | |
28:18 | you either , but I do give you two points | |
28:21 | , the line goes through . There's only one line | |
28:22 | that goes through these two points and we'd have to | |
28:25 | start by finding em by figuring out our change in | |
28:28 | y So y tu minus y one over our change | |
28:31 | in x x two minus x one . So if | |
28:34 | this is our first point , that's our X one | |
28:36 | y one . This is our second point . That's | |
28:38 | our X two y two plug into our equation . | |
28:41 | Be careful with your science with through four minus negative | |
28:44 | too . Over six minus four . And if we | |
28:50 | do this , we get 6/2 , which is three | |
28:53 | . My slope is three . So I know em | |
28:56 | . I'm going to now next step into y equals | |
28:58 | MX plus B . I can plug in my M | |
29:01 | and I can choose either of these points to be | |
29:03 | my X and Y . Let's just choose the first | |
29:05 | one here . This is my ex and my wife | |
29:07 | . So why is negative ? Two m s three | |
29:10 | and my ex is four and then we can solve | |
29:13 | for B negative two equals 12 plus B and we | |
29:18 | would get him in . The 12 were unable to | |
29:20 | minus 12 is negative . 14 is be the equation | |
29:23 | of my line y equals three x minus 14 , | |
29:27 | and I plug in M and B that I solved | |
29:29 | for it . And you didn't have to choose this | |
29:31 | point . You could have chosen this point to be | |
29:33 | your X and y here . You would have gotten | |
29:35 | the exact same answer . Um , the only other | |
29:40 | thing from this section B horizontal vertical lines . Let | |
29:42 | me just very , very quickly review this . If | |
29:44 | you have an equation , looks like this . Why | |
29:46 | equals the number ? You don't see an X at | |
29:48 | all what that is . That's a horizontal line that | |
29:50 | crosses through two on the Y axis . So you | |
29:53 | would just draw a line perfectly horizontal that crosses through | |
29:56 | two on the Y axis . That's the line y | |
29:59 | equals to the slope of all horizontal lines is zero | |
30:02 | . And the Y intercept in this case is to | |
30:05 | Why is the equation y equals two ? So just | |
30:08 | very briefly notice that every single point on this line | |
30:12 | has a Y coordinate of two . It never goes | |
30:14 | up or down . Right ? This is 20.62 This | |
30:18 | is point negative . Eight to every single Y . | |
30:21 | Coordinate of every point on this line is , too | |
30:23 | . That's why we say the equation line is y | |
30:25 | equals to have about an equation that says X equals | |
30:28 | the number you don't see a y . Well , | |
30:29 | that's actually a vertical line that crosses through three on | |
30:32 | the X axis . So we would just draw a | |
30:35 | vertical line that crosses through three on the X axis | |
30:38 | . And that is the line X equals three . | |
30:41 | The slope of all vertical lines is undefined . And | |
30:45 | the Y intercept while it never crosses the Y axis | |
30:48 | , so we would say none . Why is the | |
30:51 | equation X equals three for this will notice every single | |
30:53 | point on this line . Right . This is 30.0.36 | |
30:57 | This is point negative six . Sorry . Uh , | |
31:02 | yeah , this is 60.3 . Negative six . Sorry | |
31:05 | . This is 60.3 negative six . This is point | |
31:09 | three . Negative two . Notice that the X coordinate | |
31:14 | for every point on this line is three . So | |
31:16 | we call the line X equals three X equals numbers | |
31:19 | of vertical line y equals the number is a horizontal | |
31:21 | line . Slope of horizontal lines . Zero slope of | |
31:23 | vertical lines is undefined . All right , so that's | |
31:27 | it for linear relations . Make sure you watch part | |
31:30 | three geometry and then you will have done all of | |
31:32 | grade nine math |
DESCRIPTION:
Here is a great exam review video reviewing all of the main concepts you would have learned in the MPM1D grade 9 academic math course. The video is divided in to 3 parts. This is part 2: Linear Relations. In this video you will review everything there is to know about y=mx+b, scatterplots, and distance time graphs.
OVERVIEW:
ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 2) is a free educational video by Lumos Learning.
This page not only allows students and teachers view ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 2) videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.