# data[].attributes.body
Solve real-world problems involving probabilities related to a single or repeated experiments, including making predictions based on theoretical probability

# data[].attributes.attachment_link.title
Using Theoretical Probability to Make Predictions

# data[].attributes.body
Find the theoretical probability of an event related to a repeated experiment

# data[].attributes.attachment_link.readable_text
- [Voiceover] What I'm hoping to explore in this video is the notion of a sample space. And it's a pretty, hopefully you'll find, straightforward idea. If you're doing a trial, something that is probabilistic, a trial or an experiment, a sample space is just the set of the possible outcomes. So a very simple trial might be a coin flip. So if you're talking about a coin flip, well then the sample space is going to be the set of all the possible outcomes. So you could get a heads or you could get a tails. That right over here is the sample space for the coin flip. And it's very useful because, for example, if these are equally likely outcomes and you say well what's the probability of the event of a heads, you say okay that's one out of the two equally likely outcomes. Or you could even construct, once you know all of the possible outcomes even if they aren't equally likely, you could say well let's create a probability distribution. We at least know what the sample space is. We know what the possible outcomes are, now let's think about the probability of each of those outcomes. But a lot of times when people talk about sample spaces, they're often, they tend to be most useful, I would say, when you have equally likely outcomes like in the case of a fair coin flip. Because then from the sample space, it's fairly straight forward to think about the probability of various events. But this is a simple sample space right over here, but let's make things a little bit more interesting. Let's imagine a world, so let's just put this aside a little bit. Let's imagine a world where there's a bakery and at that bakery there are three types, three flavors of cupcakes, but there's also three different sizes of cupcakes. So now we're essentially looking at two different ways in which the thing that we're going to be sampling can vary. So what we're doing, let me write this down. So we have our flavors, flavors of cupcakes at this bakery. And let's say that you have chocolate, chocolate, you have, let's say there's strawberry, strawberry, and let's say that there is vanilla, there is vanilla. And they come in three different sizes. So sizes. Sizes could be small, I'll just write it out, small, medium, or large. So if you were, and let's say each of these flavors come in each of these sizes. Or you could view it the other way around, each of these sizes come in all three flavors. So now how do you construct the sample space? If you were to say, look I'm gonna go, you know, I'm gonna blindfold myself and walk into this bakery and randomly somehow pickup a cupcake and my fingers can't tell the flavor or the size of the cupcake, what are the possible, what are the possible outcomes for the cupcake I'll pick? And the outcome would be both the flavor and the size of the cupcake. Well there's a bunch of ways to think about this. One way is you could draw a tree. You could say, okay well I'm gonna pick three different flavors. I could either pick chocolate, chocolate, I'm gonna pick strawberry, strawberry, or I'm going to pick vanilla, vanilla. And then for each of those flavors, I'm only gonna pick a small, medium, or large. So you could say small, medium, large. Small, so this is a small chocolate, this is a medium chocolate, this is a large chocolate. This is a small strawberry, medium strawberry, large strawberry. This is a small vanilla, medium vanilla, large vanilla. And so you see there's nine possible outcomes. Once again, this is a medium chocolate. You picked a chocolate and it was a medium one. This is a large vanilla. You picked a vanilla, and it is a large one. And you could have done it the other way around. You could have said, well okay, I'm going to either pick a small, medium, or large and then for each of those I'm gonna pick either a chocolate, strawberry, or vanilla. And I'll just use the first letters. So I'm either gonna pick a chocolate, a strawberry, or a vanilla. When I write the S over here in this magenta color, I'm talking about the flavor and if I write the S in green, I'm talking about small. So here you can have, if you have a medium cupcake, it could be chocolate, it could be strawberry, or it could be vanilla. If you have a large cupcake it could be chocolate, strawberry, or vanilla. So for example, this was a medium, chocolate cupcake. Over here a medium, chocolate cupcake is this one. It's medium, chocolate. It would be that one over here. So you could use these kind of a tree diagram like this to think about the sample space, to think about the nine possible outcomes here. But you could also, you could also do a, I guess you could say grid, where you could write the flavors, so you could have chocolate, chocolate, actually, let me just write the, well no, let me write them out. Actually, let me just, let me just write the letters since this is gonna take a long time to do. So you could have the flavors chocolate, strawberry, vanilla. So that's along that axis. And then you have your sizes. You can have a small, a medium, or large. And you can set up a grid here. So this is another way to do it. And notice this grid has nine boxes. So let's look at it. So set up the grid. Set up the grid. And so what is this one going to be? This is a small chocolate. Small. Small chocolate. Small chocolate. What is this one? This is going to be a small, a small strawberry, A small strawberry. And you could just keep constructing like this where everything in this row, there are all about small. Oops. Do it in the... This is small. I'm having trouble changing colors. Alright (laughing). There is a small and then this is a small, this is a small vanilla. This color changing is really, it's a difficult thing. Small vanilla. And all of these, these are all, this would be a medium chocolate. Medium strawberry. Medium vanilla. Large chocolate, large strawberry, large vanilla. And once again you have nine outcomes. This is another way to think about all of the possible outcomes when you're looking at these two ways in which my cupcakes could vary. Another way, a third way that you could do it, is you could literally just construct a table. Well you could say, okay I could have a chocolate, actually I'm going to use the letters again. So let's say we make, this is the flavor column. And then this is the size column. Size column. And so you could say I could have a chocolate that is, so let's see, there's three types of chocolate that I could have. And they could be they could be small, medium, or large. You could say there's three types of, three types of strawberry. It could be small, medium, or large. So let me write that in. Small, medium, large. Or you could say oh, there's three types of vanilla. There's three types of, I'm color changing again. Three types of vanilla. Once again, it could be small, medium, or large. So you have these nine possibilities. Now, sample space, the sample space isn't telling you if they're equally likely or not. It's just telling you, hey, if you're gonna do an experiment, what are all the different possibilities, the possible outcomes for that experiment? Now in the case where they are equally likely, it can be very, very useful because you can say you could do something like, if you said okay, it's equally likely to pick any one of these nine outcomes, you could say well what's the probability of, what's the probability of getting something that is either small or chocolate? And so you could see well how many of those events out of the total actually meet that constraint? But we'll do more of that in future videos. That's just a little bit of a clue of why we even care about things like sample spaces. Especially sample spaces like this, where we're looking along two ways or multiple ways that something can vary. And these types of sample spaces in particular are called compound sample spaces. So these right over here, this is a compound sample space, because we're looking at two different ways that it can vary. Not just a heads or tails, it can vary by size or by flavor. And you can even have compound sample spaces that vary in more than two ways.

# data[].attributes.attachment_link.title
Sample spaces for compound events (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Roll 2 dice. List all possible outcome (sample space)

# data[].attributes.attachment_link.title
Determine the sample space of an experiment

# data[].attributes.body
Determine the sample space for a repeated experiment

# data[].attributes.attachment_link.title
Experiment, Outcomes, Sample Space and Events

# data[].attributes.body
See KHAN practice in section DP.1.2

# data[].attributes.body
Given a splatter plot with a linear association, informally fit a straight line.

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

There are two graphs. One shows the relationship between study time and test grades. Do you get better grades if you study more? The other graph shows the relationship between shoe size and grades. Do you get better grades if you wear larger shoes? Created by Sal Khan.

Video transcript

The graphs below show the test grades of the students in Dexter's class. The first graph shows the relationship between test grades and the amount of time the students spent studying. So this is study time on this axis and this is the test grade on this axis. And the second graph shows the relationship between test grades and shoes size. So shoe size on this axis and then test grade. Choose the best description of the relationship between the graphs. So first, before looking at the explanations, let's look at the actual graphs. So this one on the left right over here, it looks like there is a positive linear relationship right over here. I could almost fit a line that would go just like that. And it makes sense that there would be, that the more time that you spend studying, the better score that you would get. Now for a certain amount of time studying, some people might do better than others, but it does seem like there's this relationship. Here it doesn't seem like there's really much of a relationship. You see the shoe sizes, for a given shoe size, some people do not so well and some people do very well. Someone with a size 10 and 1/2, it looks like, someone it looks like they flunked the exam. Someone else, looks like they got A minus or a B plus on the exam. And it really would be hard to somehow fit a line here. No matter how you draw a line, these dots don't seem to form a trend. So let's see which of these choices apply. There's a negative linear relationship between study time and score. No, that's not true. It looks like there's a positive linear relationship. The more you study, the better your score would be. A negative linear relationship would trend downwards like that. There is a non-linear relationship between study time and score and a negative linear relationship between shoe size and score. Well that doesn't seem right either. A non-linear relationship, it would not be easy to fit a line to it. And this one seems like a line would be very reasonable. And it doesn't seem like there's any type of relationship between shoe size and score. So I wouldn't pick this one either. There's a positive linear relationship between study time and score. That's right. And no relationship between shoe size and score. Well, I'm going to go with that one. Both graphs show positive linear trends of approximately equal strength. No, not at all. This one doesn't show a linear relationship of really any strength.

# data[].attributes.attachment_link.title
Example of direction in scatterplots (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Scatter Plots & Lines of Fit | HSS.ID.B.6 🖤

# data[].attributes.body
Given a scatter plot within a real-world context, describe patters on association

# data[].attributes.attachment_link.title
Scatter Plots, Association and Correlation

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

There are two graphs. One shows the relationship between study time and test grades. Do you get better grades if you study more? The other graph shows the relationship between shoe size and grades. Do you get better grades if you wear larger shoes? Created by Sal Khan.

Video transcript

The graphs below show the test grades of the students in Dexter's class. The first graph shows the relationship between test grades and the amount of time the students spent studying. So this is study time on this axis and this is the test grade on this axis. And the second graph shows the relationship between test grades and shoes size. So shoe size on this axis and then test grade. Choose the best description of the relationship between the graphs. So first, before looking at the explanations, let's look at the actual graphs. So this one on the left right over here, it looks like there is a positive linear relationship right over here. I could almost fit a line that would go just like that. And it makes sense that there would be, that the more time that you spend studying, the better score that you would get. Now for a certain amount of time studying, some people might do better than others, but it does seem like there's this relationship. Here it doesn't seem like there's really much of a relationship. You see the shoe sizes, for a given shoe size, some people do not so well and some people do very well. Someone with a size 10 and 1/2, it looks like, someone it looks like they flunked the exam. Someone else, looks like they got A minus or a B plus on the exam. And it really would be hard to somehow fit a line here. No matter how you draw a line, these dots don't seem to form a trend. So let's see which of these choices apply. There's a negative linear relationship between study time and score. No, that's not true. It looks like there's a positive linear relationship. The more you study, the better your score would be. A negative linear relationship would trend downwards like that. There is a non-linear relationship between study time and score and a negative linear relationship between shoe size and score. Well that doesn't seem right either. A non-linear relationship, it would not be easy to fit a line to it. And this one seems like a line would be very reasonable. And it doesn't seem like there's any type of relationship between shoe size and score. So I wouldn't pick this one either. There's a positive linear relationship between study time and score. That's right. And no relationship between shoe size and score. Well, I'm going to go with that one. Both graphs show positive linear trends of approximately equal strength. No, not at all. This one doesn't show a linear relationship of really any strength.

# data[].attributes.attachment_link.title
Example of direction in scatterplots (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
Given a set of bivariate numerical data, construct a scatter plot or a line graph as appropriate for the context

# data[].attributes.attachment_link.title
Constructing a scatter plot (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Choose Appropriate Graphs and Charts to Display Data

# data[].attributes.body
Given a real-world numerical or categorical data set, choose and create an appropriate graphical representation

# data[].attributes.attachment_link.title
Choose the appropriate graphical representation

# data[].attributes.attachment_link.title
Circle Graphs with Fractions & Percentages (A)

# data[].attributes.body
Use proportional reasoning to construct, display and interpret data in circle graphs.

# data[].attributes.attachment_link.title
Reading pie graphs (circle graphs) | Applying mathematical reasoning | Pre-Algebra | Khan Academy

# data[].attributes.attachment_link.readable_text
In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then, we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles. So they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles, and they are going to be congruent. Or you could say that, if you continue this transversal, you would have a corresponding angle with CDE right up here and that this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles and two of the corresponding angles are the same. And that by itself is enough to establish similarity. We actually could show that this angle and this angle are also congruent by alternate interior angles, but we don't have to. So we already know that they are similar. And actually, we could just say it. Just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar, even before doing that. So we already know that triangle-- I'll color-code it so that we have the same corresponding vertices. And that's really important-- to know what angles and what sides correspond to what side so that you don't mess up your, I guess, your ratios or so that you do know what's corresponding to what. So we know triangle ABC is similar to triangle-- so this vertex A corresponds to vertex E over here. It's similar to vertex E. And then, vertex B right over here corresponds to vertex D. EDC. Now, what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4. So this is going to be 2 and 2/5. And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here. Once again, we could have stopped at two angles, but we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. And once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar-- not congruent-- it is similar to triangle CAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio between CB to CA-- so let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross-multiply. We have 5CE. 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about that, 6 and 2/5. Now, we're not done because they didn't ask for what CE is. They're asking for just this part right over here. They're asking for DE. So we know that this entire length-- CE right over here-- this is 6 and 2/5. And so DE right over here-- what we actually have to figure out-- it's going to be this entire length, 6 and 2/5, minus 4, minus CD right over here. So it's going to be 2 and 2/5. 6 and 2/5 minus 4 and 2/5 is 2 and 2/5. So we're done. DE is 2 and 2/5.

# data[].attributes.attachment_link.title
Solving similar triangles (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
Solve mathematical and real-world problems involving proportional relationships between similar triangles

# data[].attributes.attachment_link.title
Proving two triangles are similar two different ways

# data[].attributes.body
See GR.2.2 for more videos

# data[].attributes.attachment_link.readable_text
- [Instructor] What we're going to do in this video is look at all of the ways of describing how to translate a point and then to actually translate that point on our coordinate plane. So, for example, they say plot the image of point P under a translation by five units to the left and three units up. So let's just do that at first, and then we're gonna think about other ways of describing this. So we want to go five units to the left. So we start right over here. We're gonna go one, two, three, four, five units to the left, and then we're gonna go three units up. So that's going to be one, two, three. And so the image of point P, I guess, would show up right over here, after this translation described this way. Now, there are other ways that you could describe this translation. Here, we described it just in plain English, by five units to the left and three units up. But you could, and this will look fancy, but, as we'll see, it's hopefully a pretty intuitive way to describe a translation. You could say, look, I'm gonna take some point with the coordinates x comma y. And the x coordinate tells me what's my coordinate in the horizontal direction to the left or the right. And so I want that to be five less. So I would say x minus five comma y. And what do we do to the y coordinate? Well, we're going to increase it by three. We're going to translate three units up, so y plus three. So all this is saying is whatever x and y coordinates you have, this translation will make you take five from the x. That's what, meaning this is, this right over here, is five units to the left. And then this right over here, is saying three units up. Increase your y coordinate by three. Decrease your x coordinate by five. And so let's just test this out with this particular coordinate, with this particular point. So at this point right over here, P has the coordinates, its x coordinate is three, and its y coordinate is negative four. So let's see how that works. If I have three comma negative four, and I want to apply this translation, what happens? Well, let me just do my coordinates. And so I started off with three and negative four, and I'm going to subtract five from the three. So subtract five here, we see that right over there, and we're going to add three to the y. So notice, well, instead of an x, now I have a three. Instead of an x, now I have a three. Instead of a y, now I have a negative four. Instead of a y, now I have a negative four. And so another way of writing this, we're going from three comma negative four to three minus five is negative two, and negative four plus three is negative one. So what are the coordinates right over here? Well, the coordinate of this point is indeed negative two comma negative one. So notice how this, I guess you could say this formula, the algebraic formula that shows how we map our coordinates, how it's able to draw the connection between the coordinates. And so you'll see questions where they'll tell you, hey, plot the image, and they'll describe it like this. Translate x units to the left or the right or three units up or down. You'll sometimes see it like this, but just recognize this is just saying just take your x and subtract five from it, which means move five to the left. And this just means take your y coordinate and add three to it, which means move three up. And sometimes they'll ask you, hey, what's the new coordinate? Or sometimes they'll ask you to plot something like that, but just realize that it's all the same underlying idea.

# data[].attributes.attachment_link.title
Translating points (video) | Translations | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
Describe and apply the affect of a single transformation on two-dimensional figures using coordinates and the coordinate plane

# data[].attributes.attachment_link.readable_text
- [Instructor] We're asked to plot the image of point A under a dilation about point P with a scale factor of three. So what they're saying when they say under a dilation, they're saying stretching it or scaling it up or down around the point P. Now so what we're going to do is just think about, well how far is point A and then we want to dilate it with a scale factor of three. So however far A is from point P, it's going to be three times further under the dilation. Three times further in the same direction. So how do we think about that? Well, one way to think about it is to go from P to A you have to go one down and two to the left, so minus one and minus two. And so if you dilate it with a factor of three, then you're going to want to go three times as far down. So minus three, and three times as far to the left, so you'll go minus six. So one, let me do this, so negative one, negative two, negative three, negative four, negative five, negative six. So you will end up right over there. And you can even see it, that this is indeed three times as far from P in the same direction. So we could call the image of point A, maybe we call that A prime, and so there you have it. It has been dilated with a scale factor of three. And so you might be saying, wait, I'm used to dilating being stretching or scaling. How have I stretched or scaled something? Well imagine a bunch of points here that represents some type of picture and if you push them all three times further from point P, which you could do as your center of dilation, then you would expand the size of your picture by a scale factor of three. Let's do another example with a point. So, here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1/3 so first of all we don't even see the point A here, so it's actually below the fold. So let's see, there we go, that's our point A. We want it to be about the origin, so about the point zero zero. This is what we want to, the dilation about the origin with a scale factor of 1/3, scale is 1/3. Scale factor, I should say. So how do we do this? Well here, however far A is from the origin we now want to be in the same direction, but 1/3 as far. So one way to think about it, to go from the origin to A you have to go six down and three to the right. So 1/3 of that would be two down and one to the right. Two is 1/3 of six and one is 1/3 of three, so you will end up right over here. That would be our A prime. So notice, you are 1/3 away from the origin as we were before because once again, this is point A under a dilation about the origin with a scale factor of 1/3.

# data[].attributes.attachment_link.title
Dilating points (video) | Dilations | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Dilation in the Coordinate Plane: Examples (Geometry Concepts)

# data[].attributes.body
Given a preimage and image generated by a single dilation, identify the scale factor that determines the relationship

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

1,100 possible mastery points

Level up on the above skills and collect up to 560 Mastery points Start quiz

Level up on the above skills and collect up to 320 Mastery points Start quiz

Level up on all the skills in this unit and collect up to 1100 Mastery points! Start Unit test

# data[].attributes.attachment_link.title
Transformations & similarity | Grade 8 math (FL B.E.S.T.) | Khan Academy

# data[].attributes.attachment_link.description
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.attachment_link.title
Translations Reflections and Rotations - Geometric Transformations!

# data[].attributes.attachment_link.title
Reflections Over the X-Axis and Y-Axis Explained!

# data[].attributes.body
Given a preimage and image generated by a single transformation , identify the transformation that describes the relationship.

# data[].attributes.attachment_link.title
Transformations - Rotate 90 Degrees Around The Origin

# data[].attributes.attachment_link.title
What is the formula to find the measure of one interior angle

# data[].attributes.body
Develop and use formulas for interior angles of regular polygons by decomposing them into triangles.

# data[].attributes.attachment_link.title
How To Calculate The Interior Angles and Exterior Angles of a Regular Polygon

# data[].attributes.attachment_link.title
Solving for Missing Interior and Exterior Angles of Triangles

# data[].attributes.attachment_link.title
Find the Measure of the EXTERIOR ANGLE | Triangle Exterior Angle Theorem | Geometry

# data[].attributes.body
Solve problems involving the interior and exterior angles of a triangle

# data[].attributes.attachment_link.title
Exterior Angle Theorem For Triangles, Practice Problems - Geometry

# data[].attributes.body
Supplementary: straight line: all angles add up to 180

Complementary: Right angle: angles add up to 90

Vertical: angles across from another: they are equal to each other.

Adjacent: angles right next to each other.

# data[].attributes.attachment_link.title
Finding Missing Angles Vertical and Supplementary Angles

# data[].attributes.attachment_link.title
Vertical Angles and Adjacent Angles - Geometry

# data[].attributes.attachment_link.title
Solving Problems Involving Complementary and Supplementary Angles with Algebra ‎@MathTeacherGon 

# data[].attributes.attachment_link.title
Using Equations To Solve Problems - Supplementary Angles

# data[].attributes.body
Solve mathematical problems involving supplementary, complementary, vertical and adjacent angles.

# data[].attributes.attachment_link.title
Determine if the sides of a triangle produce a right triangle ex 1

# data[].attributes.body
Use the triangle inequality theorem to determine if a triangle can be formed from a given set of sides. Use the converse of the Pythagorean Theorem to determine if a right triangle can be formed from a given set of sides.

# data[].attributes.attachment_link.title
Do these side lengths form a right triangle?

# data[].attributes.attachment_link.readable_text
In this video, we're going to learn how to take the distance between any two points in our x, y coordinate plane, and we're going to see, it's really just an application of the Pythagorean theorem. So let's start with an example. Let's say I have the point, I'll do it in a darker color so we can see it on the graph paper. Let's say I have the point 3 comma negative 4. So if I were to graph it, I'd go 1, 2, 3, and then I'd go down 4. 1, 2, 3, 4, right there, is 3 comma negative 4. And let's say I also have the point 6 comma 0. So 1, 2, 3, 4, 5, 6, and then there's no movement in the y-direction. We're just sitting on the x-axis. The y-coordinate is 0, so that's 6 comma 0. And what I want to figure out is the distance between these two points. How far is this blue point away from this orange point? And at first, you're like, gee, Sal, I don't think I've ever seen anything about how to solve for a distance like this. And what are you even talking about the Pythagorean theorem? I don't see a triangle there! And if you don't see a triangle, let me draw it for you. Let me draw this triangle right there, just like that. Let me actually do several colors here, just to really hit the point home. So there is our triangle. And you might immediately recognize this is a right triangle. This is a right angle right there. The base goes straight left to right, the right side goes straight up and down, so we're dealing with a right triangle. So if we could just figure out what the base length is and what this height is, we could use the Pythagorean theorem to figure out this long side, the side that is opposite the right angle, the hypotenuse. This right here, the distance is the hypotenuse of this right triangle. Let me write that down. The distance is equal to the hypotenuse of this right triangle. So let me draw it a little bit bigger. So this is the hypotenuse right there. And then we have the side on the right, the side that goes straight up and down. And then we have our base. Now, how do we figure out-- let's call this d for distance. That's the length of our hypotenuse. How do we figure out the lengths of this up and down side and the base side right here? So let's look at the base first. What is this distance? You could even count it on this graph paper, but here, where x is equal to-- let me do it in the green. Here, we're at x is equal to 3 and here we're at x is equal to 6, right? We're just moving straight right. This is the same distance as that distance right there. So to figure out that distance, it's literally the end x point. And you could actually go either way, because you're going to square everything, so it doesn't matter if you get negative numbers, so the distance here is going to be 6 minus 3, right? 6 minus 3. That's this distance right here, which is equal to 3. So we figured out the base. And to just remind ourselves, that is equal to the change in x. That was equal to your finishing x minus your starting x. 6 minus 3. This is our delta x. Now, by the same exact line of reasoning, this height right here is going to be your change in y. Up here, you're at y is equal to 0. That's kind of where you finish. That's your higher y point. And over here, you're at y is equal to negative 4. So change in y is equal to 0 minus negative 4. I'm just taking the larger y-value minus the smaller y-value, the larger x-value minus the smaller x-value. But you're going to see we're going to square it in a second, so even if you did it the other way around, you'd get a negative number, but you'd still get the same answer, so this is equal to 4. So this side is equal to 4. You can even count it on the graph paper if you like. And this side is equal to 3. And now we can do the Pythagorean theorem. This distance is the distance squared. Be careful. The distance squared is going to be equal to this delta x squared, the change in x squared plus the change in y squared. This is nothing fancy. Sometimes people will call this the distance formula. It's just the Pythagorean theorem. This side squared plus that side squared is equal to hypotenuse squared, because this is a right triangle. So let's apply it with these numbers, the numbers that we have at hand. So the distance squared is going to be equal to delta x squared is 3 squared plus delta y squared plus 4 squared, which is equal to 9 plus 16, which is equal to 25. So the distance is equal to-- let me write that-- d squared is equal to 25. d, our distance, is equal to-- you don't want to take the negative square root, because you can't have a negative distance, So it's only the principal root, the positive square root of 25, which is equal to 5. So this distance right here is 5. Or if we look at this distance right here, that was the original problem. How far is this point from that point? It is 5 units away. So what you'll see here, they call it the distance formula, but it's just the Pythagorean theorem. And just so you're exposed to all of the ways that you'll see the distance formula, sometimes people will say, oh, if I have two points, if I have one point, let's call it x1 and y1, so that's just a particular point. And let's say I have another point that is x2 comma y2. Sometimes, you'll see this formula, that the distance-- you'll see it in different ways. But you'll see that the distance is equal to-- and it looks as though there's this really complicated formula, but I want you to see that this is really just the Pythagorean theorem. You see that the distance is equal to x2 minus x1 minus x1 squared plus y2 minus y1 squared. You'll see this written in a lot of textbooks as the distance formula. And it's a complete waste of your time to memorize it because it's really just the Pythagorean theorem. This is your change in x. And it really doesn't matter which x you pick to be first or second, because even if you get the negative of this value, when you square it, the negative disappears. This right here is your change in y. So it's just saying that the distance squared-- remember, if you square both sides of this equation, the radical will disappear and this will be the distance squared is equal to this expression squared, delta x squared, change in x-- delta means change in-- delta x squared plus delta y squared. I don't want to confuse you. Delta y just means change in y. I should have probably said that earlier in the video. But let's apply it to a couple more, and I'll just pick some points at random. Let's say I have the point, let's see, 1, 2, 3, 4, 5, 6. Negative 6 comma negative 4. And let's say I want to find the distance between that and 1 comma 1, 2, 3, 4, 5, 6, 7, and the point 1 comma 7, so I want to find this distance right here. So it's the exact same idea. We just use the Pythagorean theorem. You figure out this distance, which is our change in x, this distance, which is our change in y. This distance squared plus this distance squared is going to equal that distance squared. So let's do it. So our change in x, you just take-- you know, it doesn't matter. In general, you want to take the larger x-value minus the smaller x-value, but you could do it either way. So we could write the distance squared is equal to-- what's our change in x? So let's take the larger x minus the smaller x, 1 minus negative 6. 1 minus negative 6 squared plus the change in y. The larger y is here. It's 7. 7 minus negative 4. 7 minus negative 4 squared. And I just picked these numbers at random, so they're probably not going to come out too cleanly. So we get that the distance squared is equal to 1 minus negative 6. That is 7, 7 squared, and you'll even see it over here, if you count it. You go, 1, 2, 3, 4, 5, 6, 7. That's that number right here. That's what your change in x is. Plus 7 minus negative 4. That's 11. That's this distance right here, and you can count it on the blocks. We're going up 11. We're just taking 7 minus negative 4 to get a distance of 11. So plus 11 squared is equal to d squared. So let me just take the calculator out. So the distance if we take 7 squared plus 11 squared is equal to 170, that distance is going to be the square root of that, right? d squared is equal to 170. So let's take the square root of 170 and we get 13.0, roughly 13.04. So this distance right here that we tried to figure out is 13.04. Hopefully, you found that helpful.

# data[].attributes.attachment_link.title
Distance formula | Analytic geometry (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Finding Distance using the Pythagorean theorem

# data[].attributes.body
Apply the Pythagorean theorem to solve mathematical and real-world problems involving the distance between two points in a coordinate plane

# data[].attributes.attachment_link.title
How to Find Distance Between Two Points Using Pythagorean Theorem

# data[].attributes.body
Apply the Pythagorean theorem to solve mathematical and real-world problems involving unknown side lengths in right triangles.

# data[].attributes.attachment_link.title
An Introduction to the Pythagorean Theorem | Math with Mr. J

# data[].attributes.attachment_link.title
Volume of cylinders (practice) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
How to Find the Volume of a Cylinder | Math with Mr. J

# data[].attributes.body
Solve mathematical and real-world problems involving volume of right circular cylinders.

# data[].attributes.body
See above videos for extra help

# data[].attributes.body
Solve real world problems involving surface area of right circular cylinders

# data[].attributes.attachment_link.readable_text
Let's find the volume of a few more solid figures and then if we have time, we might be able to do some surface area problems. So let me draw a cylinder over here. So that is the top of my cylinder. And then this is the height of my cylinder. This is the bottom right over here. If this was transparent, maybe you could see the back side of the cylinder. So you could imagine this kind of looks like a soda can. And let's say that the height of my cylinder, h, is equal to 8. I'll give it some units. 8 centimeters, that is my height. And then let's say that the radius of one of the top of my cylinder of my soda can, let's say that this radius over here is equal to 4 centimeters. So what is the volume here? What is the volume going to be? And the idea here is really the exact same thing that we saw in some of the previous problems. If you can find the surface area of one side and then figure out how deep it goes, you'll be able to figure out the volume. So what we're going to do here is figure out the surface area of the top of this cylinder, or the top of the soda can. And then we're going to multiply it by its height. And that'll give us a volume. This will tell us essentially, how many square centimeters fit in this top. And then if we multiply that by how many centimeters we go down, then that'll give us the number of cubic centimeters in this cylinder or soda can. So how do we figure out this area up here? Well the area at the top, this is just finding the area of a circle. You could imagine drawing it like this. If we were to just look at it straight on, that's a circle with a radius of 4 centimeters. The area of a circle with a radius 4 centimeters, area is equal to pi r squared. So it's going to be pi times the radius squared, times 4 centimeters squared, which is equal to 4 squared, is 16, times pi. And our units now are going to be centimeters squared. Or another way to think of these are square centimeters. So that's the area. The volume is going to be this area times the height. So the volume is going to be equal to 16 pi centimeters squared, times the height, times 8 centimeters. And so, when you do multiplication, the associative property, you can kind of rearrange these things, and the commutative property. It doesn't matter what order you do it if it's all multiplication. So this is the same thing as 16 times 8. Let's see. 8 times 8 is 64. 16 times 8 is twice that. So it's going to be 128 pi. Now you have centimeter squared times centimeters. So that gives us centimeters cubed. Or 128 pi cubic centimeters. Remember, pi is just a number. We write it as pi, because it's kind of a crazy, irrational number, that if you were to write it, you could never completely write pi. 3.14159 keeps going on, never a repeat. So we just leave it as pi. But if you wanted to figure it out, you can get a calculator. And this would be 3.14 roughly, times 128. So it would be close to 400 cubic centimeters. Now, how would we find the surface area? How would we find the surface area of this figure over here? Well, part of the surface area of the two surfaces, the top and the bottom. So that would be part of the surface area. And then the bottom over here would also be part of the surface area. So if we're trying to find the surface area, it's definitely going to have both of these areas here. So it's going to have the 16 pi centimeters squared twice. This is 16 pi. This is 16 pi square centimeters. So it's going to have 2 times 16 pi centimeters squared. I'll keep the units still. So that covers the top and the bottom of our soda can. And now we have to figure out the surface area of this thing that goes around. And the way I imagine it is, imagine if you're trying to wrap this thing with wrapping paper. So let me just draw a little dotted line here. So imagine if you were to cut it just like that. Cut the side of the soda can. And if you were to unwind this thing that goes around it, what would you have. Well, you would have something. You would end up with a sheet of paper where this length right over here is the same thing as this length over here. And then it would be completely unwound. And then these two ends-- let me do it in magenta-- these two ends used to touch each other. And-- I'm going to do it in a color that I haven't used yet, I'll do it in pink-- these two ends used to touch each other when it was all rolled together. And they used to touch each other right over there. So the length of this side and that side is going to be the same thing as the height of my cylinder. So this is going to be 8 centimeters. And then this over here is also going to be 8 centimeters. And so the question we need to ask ourselves is, what is going to be this dimension right over here. And remember, that dimension is essentially, how far did we go around the cylinder. Well, if you think about it, that's going to be the exact same thing as the circumference of either the top or the bottom of the cylinder. So what is the circumference? The circumference of this circle right over here, which is the same thing as the circumference of that circle over there, it is 2 times the radius times pi. Or 2 pi times the radius. 2 pi times 4 centimeters, which is equal to 8 pi centimeters. So this distance right over here is the circumference of either the top or the bottom of the cylinder. It's going to be 8 pi centimeters. So if you want to find the surface area of just the wrapping, just the part that goes around the cylinder, not the top or the bottom, when you unwind it, it's going to look like this rectangle. And so its area, the area of just that part, is going to be equal to 8 centimeters times 8 pi centimeters. So let me do it this way. It's going to be 8 centimeters times 8 pi centimeters. And that's equal to 64 pi. 8 times 8 is 64. You have your pi centimeters squared. So when you want the surface area of the whole thing, you have the top, you have the bottom, we already threw those there. And then you want to find the area of the thing around. We just figured that out. So it's going to be plus 64 pi centimeters squared. And now we just have to calculate it. So this gives us 2 times 16 pi, is going to be equal to 32. That is 32 pi centimeters squared, plus 64 pi. Let me scroll over to the right a little bit. Plus 64 pi centimeters squared. And then 32 plus 64 is 96 pi centimeters squared. So it's equal to 96 pi square centimeters, which is going to be a little bit over 300 square centimeters. And notice, when we did surface area, we got our answer in terms of square centimeters. That makes sense, because surface area, it's a two-dimensional measurement. Think about how many square centimeters can we fit on the surface of the cylinder. When we did the volume, we got centimeters cubed, or cubic centimeters. And that's because we're trying to calculate how many one by one by one centimeter cubes can we fit inside of this structure. And so that's why it's cubic centimeters. Anyway, hopefully that clarifies things up a little bit.

# data[].attributes.attachment_link.title
Cylinder volume & surface area (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Calculating the Surface Area of a Right Cylinder

# data[].attributes.body
Given a mathematical or real-world context, find the surface area of a right circular cylinder using the figures net.

# data[].attributes.attachment_link.title
How to Find the Surface Area of a Cylinder | Math with Mr. J

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

900 possible mastery points

Level up on the above skills and collect up to 480 Mastery points Start quiz

Level up on the above skills and collect up to 240 Mastery points Start quiz

Level up on all the skills in this unit and collect up to 900 Mastery points! Start Unit test

# data[].attributes.attachment_link.title
Scale copies and scale drawings | Grade 7 math (FL B.E.S.T.) | Khan Academy

# data[].attributes.attachment_link.description
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.attachment_link.title
Relating Scale Drawings to Ratios and Rates (Simplifying Math)

# data[].attributes.body
Turn it into a fraction and cross multiply

# data[].attributes.attachment_link.title
Scale Drawings and Scale Factors 7th Grade Math

# data[].attributes.body
Solve mathematical and real-world problems involving dimensions and areas of geometric figures, including scale drawings and scale factor

# data[].attributes.attachment_link.title
Identifying scale factor in drawings | Geometry | 7th grade | Khan Academy

# data[].attributes.attachment_link.title
How to Find the Area of a Circle | Area of a Circle Step by Step

# data[].attributes.body
Explore and apply a formula to find the area of a circle to solve mathematical and real-world problems