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Find the volume of a triangular prism and cube | Geometry | Khan Academy

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Surface Area using a Net- Triangular Prism 

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Surface Area using a Net- Triangular Prism 

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Volume of a rectangular prism or box examples | Measurement | Pre-Algebra | Khan Academy

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K12 Math Hip-Hop Videos & Lessons | Flocabulary

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Comprehensive math lesson plans with teacher guides, printable hand-outs, quizzes. Drive better standardized scores with Flocabulary!

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Some distributions are symmetrical, with data evenly distributed about the mean. Other distributions are "skewed," with data tending to the left or right of the mean. We sometimes say that skewed distributions have "tails."

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A histogram is a graphical display of data using bars of different heights. In a histogram, each bar groups numbers into ranges. Taller bars show that more data falls in that range. A histogram displays the shape and spread of continuous sample data.

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In this video we're going to talk about another way of visualizing data called the histogram, which is a very fancy word for a not so fancy thing. I think it's probably fair to say that the histogram is the most used way of representing statistical data. Let me just show you how to figure out a histogram for some data, and I think you're going to get the point pretty easily. So I have some data here and I want to represent it with a histogram. What we're going to see is how frequent are each of these numbers. And in order to figure that out, let me just write the numbers down, let me just categorize them in their respective buckets. So I have a 1 here. I have a 4, so I want to leave space for the 2, the 3, and put a 4 there. I have a 2. I have a 1. Let me put that 1 on a stack right above that 1. I have a 0-- let me put the 0 to the left of the 1. I want to put them in order. I have a 2, another 2. Let me stack that above my first 2. I have another 1. Let me stack that above my other two 1's. I have another 0. Let me stack it there. I have another 1. Then I have another 2. Another 1. I have two more 0's. 0, 0. I have two more 2's. I have a 3. I have two more 1's. Another 3. And then I have a 6. So no 5's, and then I have a 6. And that space right there was unnecessary. But right here I've listed-- I've just rewritten those numbers and I've essentially categorized them. Now what I want to do is calculate how many of each of these numbers we have. So let me go down here. So I want to look at the frequency of each of these numbers. So I have one, two, three, four 0's. I have one, two, three, four, five, six, seven 1's. I have one, two, three, four, five 2's. I have one, two 3's. I have one 4, and one 6. So we could write it this way. We could write the number, and then we can have the frequency. So I have the numbers 0, 1, 2, 3, 4-- we could even throw 5 in there, although 5 has a frequency of 0. And we have a 6. So the 0 showed up four times in this data set. 1 showed up seven times in this data set. 2 showed up five times, 3 showed up to two times, 4 showed up one time, 5 didn't show up, and 6 showed up one time. All I did is I counted this data set, and I did this first. But you could say how many times do I see a 0? I see it one, two, three, four times. How many times do I see a 1? One, two, three, four, five, six, seven times. That's what we mean by frequency. Now a histogram is really just a plot, kind of a bar graph, plotting the frequency of each of these numbers. It's going to look a lot like this original thing that I drew. So let me draw some axes here. So the different buckets here are the numbers. And that worked out because we're dealing with very clean integers that tend to repeat. If you're dealing with things that the exact number doesn't repeat, oftentimes people will put the numbers into buckets or ranges. But here they fit into nice little buckets. You have the numbers 0, 1, 2, 3, 4, 5, and 6. This is the actual numbers. And then on the vertical axis we're going to plot the frequency. So one, two, three, four, five, six, seven. So that's 7, 6, 5, 4, 3, 2, 1. So 0 shows up four times. So we'll draw a little bar graph here. 0 shows up four times. Draw it just like that. 0 shows up four times. That is that information right there. 1 shows up seven times. So I'll do a little bar graph. 1 shows up seven times. Just like that. I want to make it a little bit straighter than that-- 1 shows up seven times. 2-- I'll do it in a different color-- 2 shows up five times. Do a bar graph, go all the way up to five. 2 shows up five times. 3 shows up two times. We have one 3, two 3's. 4 shows up one time here. 5 doesn't show up at all. So it doesn't even get any height there. And then finally, 6 shows up one time. So I'll do 6 showing up one time. What I just plotted here, this is a histogram. This right here is a histogram. Very fancy word, but I think you will agree it's a fairly simple idea. Figure out the frequency of each of these numbers and then plot the frequency of each of these numbers and you get yourself a histogram.

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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.

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Stem-and-leaf plots (video) | Khan Academy

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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.

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The range is the difference between the largest and smallest numbers. The midrange is the average of the largest and smallest number. Practice with this problem.

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Examples looking at different features of distributions, such as clusters, gaps, peaks, and outliers for distributions

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Current time:0:00Total duration:2:25

Video transcript

let's see if we can figure out the surface area of this cereal box and there's a couple of ways to tackle it the first way is well let's figure out the surface area of the sides that we can see and then think about what the surface area of the sides that we can't see are and how they might relate and then add them all together so let's do that so the front of the box is 20 centimeters tall and 10 centimeters wide it's a rectangle so to figure out its area we can just multiply 20 centimeters times 10 centimeters and that's going to give us 200 centimeters 200 centimeters or 200 square centimeters I should say 200 square centimeters that's the area of the front and let me write it over here as well 200 now we also know there's another side that has the exact same area as the front of the box and that's the back of the box and so let's write another 200 square centimeters for the back of the box now let's figure out the area of the top of the box the top of the box is we see it's 3 the box is 3 centimeters deep so this right over here is 3 centimeters 3 it's 3 centimeters deep and it's 10 centimeters wide we see that the box is 10 centimeters wide so the top of the box is going to be 3 centimeters times 10 centimeters which is 30 square centimeters of area so that's the top of the box 30 square centimeters well the bottom of the box is going to have the exact same area we just can't see it right now so that's going to be another 30 and we have 2 more sides because this box has 6 sides we have this side panel that is 20 centimeters tall we see that the height of the box is 20 centimeters and 3 centimeters deep so 3 times 3 times 20 3 times let me write that a little bit neater 3 times 20 that's 20 centimeters right there 3 centimeters times 20 centimeters is going to give us 60 square centimeters 60 square centimeters now that's this side panel but there's another side panel that has the exact same air that's on the other side of the box so 60 centimeters squared or squared centimeters for this side and then another 60 for the corresponding side opposite to it that we can't see and now we can just add up all of these together and so we get 0 let's see this is going to be carry the one or regroup the one it's a 100 and then we have 500 so we get 580 square centimeters is the surface area of this box

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Surface area of a box (cuboid) (video) | Khan Academy

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Surface area is the sum of the areas of all faces (or surfaces) on a 3D shape. A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces.

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- [Instructor] How many faces does the following shape have? Pause this video, and see if you can figure that out. All right, I'm assuming you paused, and I'll see if we can work through it together. And I'm gonna actually try to color the faces. So, we have this face over, whoops, let me do it in my, this other tool, so we have this face over here in the back. So that's one face. So that's one. Then you have this face right over here, also in the back. The only way we can see this is because they've drawn it so that it is transparent. So that is the second face. Now you have this triangular face on top. So let me color that in. So you have this triangular face on top. So that's going to be our third face, third face. And then you have this triangular face on the bottom. That's gonna be our fourth face. That's going to be our fourth face. And then the key question is are we done? Looks like I've colored all the ones that I can see, but there's one a little bit tricky here. There's the one that we are actually seeing through. There is the face, there's, let me pick a color, there's the face out front that we can see through so that we can see faces one, two, and four. So that's actually going to be our fifth face. The way they've drawn it, it's like it's made out of glass, so we can see faces one, two, and four. But that is our fifth face. And so this thing has five faces. All right, let's do another example, but instead of faces, we're gonna think about edges. So how many edges does the following shape have? Pause the video, and see if you can work through this. Okay, let's work through this together now. And I'm just gonna count these edges. So the edges are where two faces meet. So this is an edge right over there. So that's one edge. There's an edge back here, we can see 'cause it's transparent. That is our second edge. We have this one over here. That is our third edge. We have this one over there. That is our fourth edge. Then we have this one over here. That is our fifth edge. Now we have this here. This is our sixth edge. Let's see, all we have left is this one, which is edge number seven. And then last but not least, this edge of here, which is edge number eight. And I actually found this very valuable to color 'em in to make sure that I wasn't missing an edge or double-counting an edge. So this thing has eight, eight edges. Actually, if you're curious, just for extra practice, how many faces does this have? Well, we can count those as well. This has one face back there, another face back there. So that's two faces. And then you have a third face, which is the base, this rectangular face. So that's three faces. And then you have your two faces out front. You have this face that we're seeing through, and then we have that face that we're seeing through. So even though they're asking edges, just for practice, we figured out that this thing has five faces, one for the square base and then four triangular faces for these sides to make this square pyramid. So it has eight edges and five faces.

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Counting faces and edges of 3D shapes (video) | Khan Academy

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Instructor] How many faces does the following shape have? Pause this video, and see if you can figure that out. All right, I'm assuming you paused, and I'll see if we can work through it together. And I'm gonna actually try to color the faces.

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Now we mixed up things a little bit more: on the left side of the scale, not only do we have these identical unknown masses with mass X, these three blue things, we also have some of the 1kg masses over here, actually, we have two of them. Now, we are going to figure out what X is. But before we even do that, I want you to think about a mathematical equation that can represent what is going on; that equates what we have on the left hand, with what we have on the right side of the scale. I will give you a few seconds to think about it... So let's think about what we have on the left side: we have 3 masses with mass X, so you can say we have 3x and then we have 2 masses of 1 kilogram, so in total we have 2 kg. So + 2. So one way to think about the total mass on the left-hand side is 3x + 2. Three masses with mass X, plus two kilograms. That is what we have on the left-hand side. Now, let us think about what we have on the right-hand side. We can simply count them: [counts to 14] Fourteen blocks, each has a mass of 1 kg, so the total mass will be 14 kg. And we see that the scale is balanced, not tilting down or upwards. So this mass over here must be equal to this total mass. The scale is balanced, so we can write an 'equal'-sign. (let me do that in a white coulour, I do not like that brown) Now, what I want you to think about, and you can think about it either through the symbols or through the scales, is: how would you go about -- let us think about a few things: how would you first go about at least getting rid of these little 1kg blocks? I will give you a second to think about that... Well, the simplest thing is: you can take these 1kg blocks off of the left-hand side, but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, our scale is still balanced. And our equation: 3x is now equal to 12. Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. If we want to keep a third of what we had here originally, -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced. So you see that the mass of this thing must be the same as the mass of these 4 left-over blocks. x must be equal to 4 kilograms.

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Intro to two-step equations | Algebra (video) | Khan Academy

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Now we mixed up things a little bit more: on the left side of the scale, not only do we have these identical unknown masses with mass X, these three blue things, we also have some of the 1kg masses over here, actually, we have two of them.

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Let's ease into this, shall we? Here's an introduction to basic algebraic equations of the form ax=b. Remember that you can check to see if you have the right answer by substituting it for the variable!

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Remember that what you do to one side, you have to do to the other. Will you multiply or divide both sides to dump the fraction, x/a? Let's do it together.

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Video transcript

Now that we're comfortable with the "why" of why we do something to both sides of an equation, let's see if we can apply it to some equations to solve for an unknown variable. So let's say that you have x plus seven is equal to ten, and I want to solve for x. All its saying is something plus seven is equal to ten, and you might be able to figure that out in your head, but if you want to do it a little bit more systematically, you're like well just all I want on the left hand side is an x. Well if all I want on the left hand side is an x I'd want to get rid of the seven. I want to subtract seven from the left-hand side, but if I want to maintain an equality here, whatever I do to the left-hand side I also have to do to the right-hand side going back to our scales that's so that we can keep our scale balanced, so that we can say that the left is still equal to the right. And so what we're going to be left with is x and then the sevens cancel out is equal to ten minus seven is equal to three. So that unknown is three, and you can verify it, three plus seven is indeed equal to ten. Let's try one more. Let's say we have a minus five is equal to negative two. So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. We just want an a over here on the left-hand side so we have to get rid of this negative five somehow. Well the best way of getting rid of a negative 5 is to add five to it. So I'll do that. So I will add five to the left-hand side. But if I want the left-hand side to stay equal to the right-hand side, whatever I do to the left I have to do the right. So I'm going to have to add five on the right-hand side as well, and so on the left-hand side I'm left with a, and then the negative five and the positive five cancel out and on the right hand side, and they're going to stay equal because I did the same thing to both sides, we have negative two plus five which is equal to three. So a is equal to three. Once again you can verify it. Three minus five is indeed equal to negative two.

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One-step addition & subtraction equations (video) | Khan Academy

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I'm starting to take a little bit more care of my health, and I start counting my actual calories. And let's say C is equal to the number of calories I eat in a given day. And I want to lose some weight. So, in particular, I want to eat less than 1,500 calories in a day. So how can I express that as an inequality? Well, I want the number of calories in a day to be less than-- and remember, the less than symbol, I make it point to the smaller thing. So I want the calories to be less than 1,500. So this is one way of expressing it. I say, look, the number of calories that I consume in a day need to be less than 1,500. Now, one thing to keep in mind when I write that is obviously if I eat no calories in a day, or if I eat 100 calories, or if I eat 1,400 calories, or if I eat 1,499 calories for C, those are all legitimate. Those are all less than 1,500. But what about 1,500 calories? Is it true that 1,500 is less than 1,500? No. 1,500 is equal to 1,500. So this is not a true statement. But what if I want to eat up to and including 1,500 calories? I want to make sure that I get every calorie in there. How can I express that? How can express that I can eat less than or equal to 1,500 calories, so I can eat up to and including 1,500 calories? Right now, this is only up to but not including 1,500. How could I express that? Well, the way I would do that is to throw this little line under the less than sign. Now, this is not just less than. This is less than or equal to. So this symbol right over here, this is saying that C is less than or equal to 1,500 calories. So now 1,500 would be a completely legitimate C, a completely legitimate number of calories to have in a day. And if we wanted to visualize this on a number line, the way we would think about it, let's say that this right over here is our number line. I'm not going to count all the way from 0 to 1,500, but let's imagine that this right over here is 0. Let's say this over here is 1,500. How would we display less than or equal to 1,500 a number line? Well, we would say, look, we could be 1,500, so we'll put a little solid circle right over there. And then we can be less than it, so then we would color in everything less than 1,500, and say, look, anything less than or equal to 1,500 is legitimate. And you might say, hey, but what about the situation where it wasn't less than or equal? What about the situation if it was just less than? So let me draw that, too. So going back to where we started, if I were to say C is less than 1,500, the way we would depict that on a number line is-- let's say this is 0, this is 1,500, we want to make it very clear that we're not including the value 1,500. So we would put an open circle around it. Notice, if we're including 1,500, we fill in the circle. If we're not including 1,500, so we're only less than, we were very explicit that we don't color in the circle. But then we show that, look, we can do everything below that. Now, you're probably saying, OK, Sal, you did less than, you did less than or equal, what if you wanted to do it the other way around? What if you wanted to do greater than and greater than or equal? Well, let's think about that for a second. Let's say that I'm also trying to increase the amount of water I intake. And so let's define some variable. Let's say W is equal to the number of ounces of water I consume per day. And I've read that I should have at least-- let me throw out a number-- 64 ounces of water per day. There's one way I could think about, where I always want to drink more than 64 ounces, so that would be W is greater than 64. W here is the thing that I want to be bigger, so the opening is to the W. W is greater than 64 ounces. How would I depict that? Well, let me do my number line right over here. Let's say that this is 0. This is 64. If I wanted to make strictly greater than, so in this situation it's not cool if I just drink exactly 64. That 64 is not greater than 64. I have to drink 64.01 ounces or 0.00001 ounces. It has to be something that is greater than 64. So I'm not going to include 64, but anything greater than that is completely cool. Now, what if I want to loosen things a little bit? It's OK if I drink exactly 64 ounces or more. Well, then I could write W is greater than or equal to 64. And the way that I would be depict that on the number line-- and obviously, I'm not showing all the numbers in between-- let's say this is 0, and then we go all the way up to 64. Well, now it's OK if I drink exactly 64 ounces, so I'm going to fill in the circle now. Here I opened it because 64 was not a cool number. Now, 64 is completely OK. I can drink exactly 64 ounces of water in the day or more, and then I just go up the number line just like that.

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Plotting inequalities (video) | Khan Academy

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To plot an inequality, such as x>3, on a number line, first draw a circle over the number (e.g., 3). Then if the sign includes equal to (≥ or ≤), fill in the circle. If the sign does not include equal to (> or <), leave the circle unfilled in.

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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.

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We're mixing it up by placing both fractions and decimals on the same number line. Great practice because you need to move effortlessly between the two. Created by Sal Khan and Monterey Institute for Technology and Education.

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Decimals & fractions on the number line (video) | Khan Academy

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We're mixing it up by placing both fractions and decimals on the same number line. Great practice because you need to move effortlessly between the two. Created by Sal Khan and Monterey Institute for Technology and Education.

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Sal puts -25, -30, -7, -10, and -40 in order from least to greatest.

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Learn to sort units of measure into two categories: U.S. Customary and Metric units.

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Video transcript

Let's see if we can figure out what 30% of 6 is. So one way of thinking about 30%-- this literally means 30 per 100. So you could view this as 30/100 times 6 is the same thing as 30% of 6. Or you could view this as 30 hundredths times 6, so 0.30 times 6. Now we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30/100-- and you could view this times 6/1-- this is equal to 180/100. And let's see. We can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9/5, which is the same thing as 1 and 4/5. And then if we wanted to write this as a decimal, 4/5 is 0.8. And if you want to verify that, you could verify that 5 goes into 4-- and there's going to be a decimal. So let's throw some decimals in there. It goes into 4 zero times. So we don't have to worry about that. It goes into 40 eight times. 8 times 5 is 40. Subtract. You have no remainder, and you just have 0's left here. So 4/5 is 0.8. You've got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would've gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6-- let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying. I only have the 3 to the right of the decimal. So I'm only going to have one number to the right of the decimal here. So I just count one number. It's going to be 1.8. So either way you think about it or calculate it, 30% of 6 is 1.8.

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Percent of a whole number (video) | Khan Academy

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Converting percents to decimals & fractions example

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Sal writes 18% as a decimal and as a fraction.

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- [Voiceover] What I want to explore in this video is the notion of a "rate." So, let's look at some examples of rates that you've probably encountered in your everyday life. So, if you're driving in your car down the road, and you're looking at the speedometer, you might see that it says that you are going 35 M-P-H, where the M-P-H stands for 35 miles per, per hour. Well, what's that saying? That's saying, well, every hour, how many miles are you going if you were to stay at that current rate. So, it could be a measure of speed. How much distance are you covering per unit time? And, most typically, when people talk about rates, that's what they're talking about. They're talking about how much of something that is happening per unit time. And, it doesn't have to be even distance per unit time, you might have a, you might have your hourly rate for someone who is doing some type of a job. They might say that they're making, they're making $10, so they're making $10. And, actually, let me write the dollars out so the units become a little bit more obvious, 10 dollars, dollars per hour, dollars per, dollars per hour. And so, once again, this is how much money. It's not talking about distance anymore. How much money is being earned per unit time? And, so, even though rates are often associated with how much something is happening per some unit time, and it could be miles per hour, or it could be meters per second, or, in this case, it could be a wage, it could be dollars per hour. Rates don't have to be just in those terms. In fact, you might say, "All right, "I have a dessert that I really enjoy, "but I'm very conscientious "about, about the number of calories that I consume." And, you might, you might see something like, there are 200 calories, calories per serving, per serving. And, so, this is telling us the number of calories per a serving. And they'll tell us what a serving is. A serving might be a cup or eight ounces or whatever else. And, so, I could say, "Okay, look, if I have two servings, "then I'm gonna have 400 calories. "Same way, if I work two hours, I'm gonna have 20 dollars. "If I, or if I go two hours, "I'm gonna go 70 miles." So, rates give you a sense. It's like, how fast is something happening? Or how much of one thing is happening for every time something else happens? Now, I can write rates so they look an awful lot like a ratio. And, these words are, actually, very related, 'cause you see that even how they're written. R-A-T, R-A-T. Their roots are coming from the exact same idea. In fact, this rate over here, 35 miles per hour, it could come from, "Hey, I just, I just went 35 miles in one hour, "what's the ratio?" So, the ratio of miles to hours. And, then, you could say, "Well, I went 35, "the ratio miles to hours "was 35 to one." Or it could have been, maybe it was 70 to two or something like that. But, that could have been reduced to 35 to one. So, as a ratio, you would typically see it written like this... Or maybe see it written like, see it written like this... And, sometimes, you might even see it written like this, 35 miles to one hour. But, now it's starting to resemble more of the special case of a ratio, which we call a "rate." Because, this is the same thing as 35. Instead of writing it out "miles per hour," you'll often see it written like this, miles per, miles per hour. So, these are very, very related ideas. If you find the ratio between calories and servings, well, then, you're going to be able to write, you're going to be able to express it as a rate and vice versa. Now, why do we care about rates? Well, especially if we're thinking about things like speed, without rates, it would be hard to quantify how fast things are happening. Otherwise, we'd be in a world where we're saying, "Hey, I'm faster than you," or "She's faster than me." But we wouldn't be able to quantify exactly how fast they are. But with rates, we can say, "Hey, that person ran "a hundred meters in 10 seconds, "they run 10 meters per second." We can quantify exactly how fast that thing is happening, the rate at which it is happening. Here, instead of saying, "Hey, a cup of that "is gonna give you, is gonna give you more energy, or, maybe, contribute more to your weight than a cup of that, and making these relative comparisons, here, you can actually, you can actually quantify things. And when we study rate, we're gonna study rate a lot in mathematics. It's gonna be essential in algebra when we look at the rate of change of a line, how far it moves in the vertical direction relative to the horizontal direction. We're gonna call that "slope". And you can even imagine the slope of a hill as how fast is it climbing for as much as you move forward. But we're also gonna study rates in detail when we go to calculus. In fact, the whole basis of differential calculus, that you might see later in high school and early college, is all about measuring instantaneous rate. How fast is something going right now? So, rates are really, really interesting, really, really important. And, I would guess that, if you just look around your life, even over the next few hours, you're going to encounter many, many, many rates.

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Voiceover] What I want to explore in this video is the notion of a "rate." So, let's look at some examples of rates that you've probably encountered in your everyday life. So, if you're driving in your car down the road, and you're looking at the speedometer, you might see that it says that you are going 35 M-P-H, where the M-P-H stands for 35 miles per, per hour.

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We're displaying ratios in a table format here, and then asking: given a ratio, solve for equivalent ratios. Here are a few examples to practice on.

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Voiceover:We've got some apples here and we've got some oranges and what I want to think about is, what is the ratio, what is the ratio of apples ... Of apples, to oranges? To oranges. To clarify what we're even talking about, a ratio is giving us the relationship between quantities of 2 different things. So there's a couple of ways that we can specify this. We can literally count the number of apples. 1, 2, 3, 4, 5, 6. So we have 6 apples. And we can say the ratio is going to be 6 to, 6 to ... And then how many oranges do we have? 1, 2, 3, 4, 5, 6, 7, 8, 9. It is 6 to 9. The ratio of apples to oranges is 6 to 9. And you could use a different notation. You could also write it this way. 6 to ... You would still read the ratio as being 6 to 9. But we don't have to just satisfy us with this because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. So one way to think about it ... If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges which is 6 to 9 or 2 to 3 if we reduce them. And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples.

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Voiceover:We've got some apples here and we've got some oranges and what I want to think about is, what is the ratio, what is the ratio of apples ... Of apples, to oranges? To oranges. To clarify what we're even talking about, a ratio is giving us the relationship between quantities of 2 different things.

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Drawing a quadrilateral on the coordinate plane example

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Sal draws a quadrilateral on the coordinate plane using the coordinates of its vertices.

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So right here, we have a four-sided figure, or a quadrilateral, where two of the sides are parallel to each other. And so this, by definition, is a trapezoid. And what we want to do is, given the dimensions that they've given us, what is the area of this trapezoid. So let's just think through it. So what would we get if we multiplied this long base 6 times the height 3? So what do we get if we multiply 6 times 3? Well, that would be the area of a rectangle that is 6 units wide and 3 units high. So that would give us the area of a figure that looked like-- let me do it in this pink color. The area of a figure that looked like this would be 6 times 3. So it would give us this entire area right over there. Now, the trapezoid is clearly less than that, but let's just go with the thought experiment. Now, what would happen if we went with 2 times 3? Well, now we'd be finding the area of a rectangle that has a width of 2 and a height of 3. So you could imagine that being this rectangle right over here. So that is this rectangle right over here. So that's the 2 times 3 rectangle. Now, it looks like the area of the trapezoid should be in between these two numbers. Maybe it should be exactly halfway in between, because when you look at the area difference between the two rectangles-- and let me color that in. So this is the area difference on the left-hand side. And this is the area difference on the right-hand side. If we focus on the trapezoid, you see that if we start with the yellow, the smaller rectangle, it reclaims half of the area, half of the difference between the smaller rectangle and the larger one on the left-hand side. It gets exactly half of it on the left-hand side. And it gets half the difference between the smaller and the larger on the right-hand side. So it completely makes sense that the area of the trapezoid, this entire area right over here, should really just be the average. It should exactly be halfway between the areas of the smaller rectangle and the larger rectangle. So let's take the average of those two numbers. It's going to be 6 times 3 plus 2 times 3, all of that over 2. So when you think about an area of a trapezoid, you look at the two bases, the long base and the short base. Multiply each of those times the height, and then you could take the average of them. Or you could also think of it as this is the same thing as 6 plus 2. And I'm just factoring out a 3 here. 6 plus 2 times 3, and then all of that over 2, which is the same thing as-- and I'm just writing it in different ways. These are all different ways to think about it-- 6 plus 2 over 2, and then that times 3. So you could view it as the average of the smaller and larger rectangle. So you multiply each of the bases times the height and then take the average. You could view it as-- well, let's just add up the two base lengths, multiply that times the height, and then divide by 2. Or you could say, hey, let's take the average of the two base lengths and multiply that by 3. And that gives you another interesting way to think about it. If you take the average of these two lengths, 6 plus 2 over 2 is 4. So that would be a width that looks something like-- let me do this in orange. A width of 4 would look something like this. A width of 4 would look something like that, and you're multiplying that times the height. Well, that would be a rectangle like this that is exactly halfway in between the areas of the small and the large rectangle. So these are all equivalent statements. Now let's actually just calculate it. So we could do any of these. 6 times 3 is 18. This is 18 plus 6, over 2. That is 24/2, or 12. You could also do it this way. 6 plus 2 is 8, times 3 is 24, divided by 2 is 12. 6 plus 2 divided by 2 is 4, times 3 is 12. Either way, the area of this trapezoid is 12 square units.

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Area of trapezoids (video) | Khan Academy

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So right here, we have a four-sided figure, or a quadrilateral, where two of the sides are parallel to each other. And so this, by definition, is a trapezoid. And what we want to do is, given the dimensions that they've given us, what is the area of this trapezoid.

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- [Voiceover] We know that we can find the area of a rectangle by multiplying the base times the height. The area of a rectangle is equal to base times height. In another video, we saw that, if we're looking at the area of a parallelogram, and we also know the length of a base, and we know its height, then the area is still going to be base times height. Now, it's not as obvious when you look at the parallelogram, but in that video, we did a little manipulation of the area. We said, "Hey, let's take this "little section right over here." So we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that, as long as the base and the height is the same, as this rectangle here, I'm able to construct the same rectangle by moving that area over, and that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height. So hopefully that convinces you that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So let's look at some triangles here. So that is a triangle, and we're given the base and the height, and we're gonna try to think about what's the area of this triangle going to be, and you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So let me copy, and then let me paste it, and what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram. I have now constructed a parallelogram that has twice the area of our original triangle, 'cause I have two of our original triangles right over here, you saw me do it, I copied and pasted it, and then I flipped it over and I constructed the parallelogram. Now why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base times this height. You also have height written with the "h" upside down over here. It's going to be base times height. That's going to be for the parallelogram, for the entire-- let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So our original triangle is just going to have half the area. So this area right over here is going to be one half the area of the parallelogram. One half base-- let me do those same colors. One half base times height. One half base times height. And you might say, "OK, maybe it worked for this triangle, "but I wanna see it work for more triangles." And so, to help you there, I've added another triangle right over here, you could do this as an obtuse triangle, this angle right over here is greater than 90 degrees, but I'm gonna do the same trick. We have the base, and then we have the height. Here, you can think of, if you start at this point right over here, and if you drop a ball, the length that the ball goes, if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height, it's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, "Well, this is the height." How far off the ground is it? Well, what's the area of this going to be? Well, you can imagine, it's going to be one half base times height. How do we feel good about that? Well, let's do the same magic here. So let me copy and paste this, so I'm gonna copy and then paste it. Whoops, that didn't work. Let me copy, and then paste it. And so, I have two of these triangles now, but I'm gonna flip this one over, so that I can construct a parallelogram. So I'm gonna flip it over, and move it over here, I'm gonna have to rotate it a little bit more. So, I think you get the general idea. So now I have constructed a parallelogram that has twice the area of our original triangle. It has twice the area of our original triangle. And so, if I talked about the area of the entire parallelogram, it would be base times the height of the parallelogram. Base times the height of the parallelogram. But if we're only talking about the area of -- If we're only talking about this area right over here, which is our original triangle, it's going to be half the area of the parallelogram, so it's gonna be one half of that. So our area of our original triangle is one half base times height. So hopefully that makes you feel pretty good about this formula that you will see in geometry, that area of a triangle is one half base times height, while the area of a rectangle or a paralleogram is going to be base times height.

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Area of a triangle (video) | Geometry | Khan Academy

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Voiceover] We know that we can find the area of a rectangle by multiplying the base times the height. The area of a rectangle is equal to base times height. In another video, we saw that, if we're looking at the area of a parallelogram, and we also know the length of a base, and we know its height, then the area is still going to be base times height.

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Area of a parallelogram (video) | Geometry | Khan Academy

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Understand why the formula for the area of a parallelogram is base times height, just like the formula for the area of a rectangle.

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Distributive Property to Simplify Expressions

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Distributive Property to Simplify Expressions

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We've learned about order of operations and combining like terms. Let's layer the distributive property on top of this.

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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.

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Learn how to evaluate an expression with variables using a technique called substitution (or "plugging in"). Created by Sal Khan and Monterey Institute for Technology and Education.

Video transcript

A local hospital is holding a raffle as a fundraiser. The individual cost of participating in the raffle is given by the following expression-- 5t plus 3, or 5 times t plus 3, where t represents the number of tickets someone purchases. Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10. So let's first take the situation where t is equal to 1. Then this expression right over here becomes-- and I'll use that same color-- becomes 5 times 1. 5 times 1 plus 3. 5 times 1 plus 3, and we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3, and then this is clearly equal to 8. Now let's do it when t is equal to 8. So when t is equal to 8, this expression becomes-- and I'll do the same colors again-- 5 times 8 plus 3. Same color of green. And once again, 5 times 8 is 40, and then we have the plus 3, there so this is equal to 43. And so we have the last situation, with t is equal to 10. I'll do that in blue. So we have 5 times 10. So 5t is 5 times 10. Instead of a t, put a 10 there. 5 times 10 plus 3. That's a slightly different shade of green, but I think you get the idea. 5 times 10 is 50. We do 50, and then we're going to have to add 3 to that, and that is equal to 53. And we're done.

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Evaluating an expression with one variable (video) | Khan Academy

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Lesson 3: Substitution and evaluating expressions

We've done a few examples together where we were faced with 1 variable. Why not try one that has 2 variables? Created by Sal Khan.

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When the equation comes to "6 - 2 + 9" would the answer not be minus 5? should you not add 9 and 2 and then subtract because of the order of operations ?

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Video transcript

Now, let's think about expressions with more than one variable. So say I had the expression a plus-- I'll do a really simple one, a plus b. And I want to evaluate this expression when a is equal to 7 and b is equal to 2. And I encourage you to pause this and try this on your own. Well, wherever we see the a, we would just replace it with the 7. And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x. Or another way of saying it plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. This 3 times 3 is 9. So it simplifies to 6 minus 2, which is 4, plus 9, which is equal to 13. So in this case, it is equal to 13.

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Evaluating expressions with two variables (video) | Khan Academy

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To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Lesson 3: Substitution and evaluating expressions We've done a few examples together where we were faced with 1 variable. Why not try one that has 2 variables? Created by Sal Khan.

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- [Voiceover] Let's figure out what 2.211 divided by 6.7 is. So the first thing I like to do is, I don't like to divide by a decimal. So I'm going to multiply 6.7, I'm going to multiply 6.7 by 10, which is essentially just taking this decimal and moving it one space to the right. Well I can't just do that to the 6.7, I also have to do that to the 2.21. So let me move its decimal one space to the right. And so now I can rewrite the problem. I can rewrite it as 22, 22 point, 22.11 divided by, divided by, instead of writing it as 6.7 I can now write it as 67. 67, let me do it in that same color. I can now write it as 67. I could write a decimal right over here but that's not going to change the fact that this is just now the whole number 67. And so now I'm ready to do some long division now that I'm dividing by a whole number. You might say, "Wait, I still have a decimal over here," but that's okay and you're going to see that in a second. So let's try to do some long division. So we're going to take, we're going to take 22.11 or 22 and 11 hundredths, and we are going to divide, let me write that a little bit bigger just so we have some space. 22.11, and we are going to divide that by... We are going to divide that by 67. 67. Alright, let's do some long division now. And actually, before we even do that, I like to keep track of my decimal. So my decimal is over here, I'm going to write my decimal right over here in the answer. And when you're doing these long division problems it's really important to write things neatly and keep things in nice columns and keep track of your place value because if you don't write things in neat columns then frankly you'll probably lose track of your place value. But let's do this. So 67 goes into 2 zero times, so let's move on. 67 goes into 22 zero times. 67 goes into 221. So let's think about that a little bit. This is pretty close to 70, 70 times 3 would be 210. So this looks like maybe three times, three times seems about right. So let's try, let's put a 3 up here. 3 times 7 is 21, carry the 2, 3 times 6 is 18, plus 2 is 20, is 20. So we get 201 and the difference between 221 and 201 is going to be... Well you just get a 0 here, you get a 2 here, then you get a 0 here, I don't have to write it. It's 20. So 3 was right, if 3 times 67 were higher than 221 then we'd be in trouble. And then if 3 times 67 were lower but it was so low that when you subtracted you got a number higher than 67 then that means you could have thrown in more 67s in there, but this number's just right. It's lower but our remainder is less than, is less than 67, so let's keep going. So now we can bring down, we can bring down the 1 and we see 67 goes into 201, well we just figured that out. 67 goes into 201 three times, three times. 3 times 7 is one, carry the 2, gotta make sure we don't... Well, if we're doing the same problem as we did before but it's good to not have the previous 2 there, sometimes you might want to erase it. Then you have 3 times 6 is 18 plus 2 is 20. And now you subtract, and we are done because we have no remainder and there's nothing here left to bring down. And so 22.11 divided by 67 is... And we can throw a 0 here just for good measure. It's 0.33. 0.33, this is equal to 0.33. Or this is equal to 0.33. And we're done.

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Long division with decimals (video) | Khan Academy

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Voiceover] Let's figure out what 2.211 divided by 6.7 is. So the first thing I like to do is, I don't like to divide by a decimal. So I'm going to multiply 6.7, I'm going to multiply 6.7 by 10, which is essentially just taking this decimal and moving it one space to the right.

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We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 1 is 5, plus 1 is 6. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're multiplying. So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. So 32.12 times 0.5 is 16.060. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense.

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Multiplying decimals example (video) | Khan Academy

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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.

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- [Instructor] Do some more examples subtracting decimals. So let's say we wanna figure out what two minus 1.2 is. Pause this video and see if you can calculate this. So there is multiple ways to tackle this. One way is you could say, "Look, this is the same thing "as two minus one "minus 2/10." This is two minus one and 2/10. So we're subtracting one, and we're subtracting 2/10. Now two minus one is pretty straightforward to compute. Two minus one is going to be one, and then we need to subtract 2/10 from that. So one is the same thing as 10/10. We could say this is 10/10. 10, write it this way. 10/10, and we're gonna subtract 2/10. What is that gonna give us? Well, that's going to give us, if we have 10/10, and we take away 2/10, that's going to give us 8/10. Lemme write it down here. 8/10, which is the same thing as 0.8, or 8/10. One way to think about this is if you're subtracting 0.2 from one, if you view this one as a 1.0, instead of expressing this as one, one, and 0/10, we are thinking about this as zero ones and 10/10. And when you think about it as zero ones and 10/10, well, 10/10 minus 2/10, it's easier to then think about, well, 10 of something minus two of something is gonna be eight of something. It's gonna be 8/10. Now we can also visualize this on a number line, so for example, let me draw a number line here, and so zero, one, two, three, four, five, six, seven, eight, nine. That's one, one, two, three, four, five, six, seven, eight, nine. That is two, so we're starting at two, and we're subtracting one and 2/10. So one way to think about it is we first subtracted one. We first subtracted one, and then we subtracted 2/10, and that got us to the point 8/10 on the number line. Another way to think about this, and the whole point here is to see multiple strategies and for you to think about what strategy you like the most and to realize they all get you to the same place. If you're thinking about it in a reasonable way, is you could use, what's the difference between two and 1.2? So 1.2 sits here on the number line, so what's the difference between two and one and 2/10? Another way to think about it is how many tenths do you have to add to one and 2/10 to get to two? Well, if you've already got 2/10, you need to add another 8/10 to get to the next whole. So you have to add 8/10, or you have to add 0.8. So the difference between two and 1.2, the difference between two and 1.2 is equal to 0.8. Let's do a few more examples that get a little bit more involved. So let's say we wanna calculate what 3.8 minus 1.5 is. Pause the video and see if you can calculate this. Well, just like before, we could view this as three minus one, so we're subtracting the ones plus 8/10 minus 5/10. Notice we have three and 8/10 minus one and 5/10, minus one and minus 5/10. And so now we can figure out, okay, three minus one, that's just going to be equal to two, and then 8/10 minus 5/10, well, that is 3/10, and so this is going to be two and 3/10, which of course we could write as 2.3. That seemed pretty straightforward. Let's do one that's a little bit more involved. Let's say we wanna calculate four and 5/10 or 4.5, minus two and 8/10 or 2.8. Pause the video and see if you can calculate this. So you might wanna do the exact same thing. You might say, "Okay, well, this is the same thing. "Let's think about the ones." This is four minus two plus 5/10 minus 8/10. Four and 5/10 minus two and minus 8/10, minus two and 8/10. That's exactly what we have up here. Then you would say, "Alright, four minus two, "that is two." But then you get to 5/10 minus 8/10, and there's multiple ways to tackle this, but you might say, "Oh, how do I take away 8/10 "if I've only got 5/10 here?" And there's a bunch of strategies that you could think about. You could say, "Hey, what if I can get "some more tenths here?" So this is 5/10 minus 8/10. What if I could get some more tenths here? The best way I can think of it is like what if I were to break up these ones because one is 10/10? So I could view a two as one plus one, or I could view this as one plus 10/10. So if you view a two as one and 10/10, you can then just add those. You could then figure out what is 10/10 plus 5/10 minus 8/10? What is that going to be? Well, this is a little bit more straightforward. 10/10 plus 5/10 is going to be 15/10, and if you have 15/10 and you take away eight of them, you're going to be left with 7/10. So this gets us to one plus 7/10, so all of this when you compute it, that is 7/10, which I could write as 0.7, which is equal to, so this is equal to one and 7/10. Now there's other strategies that you could do here. One strategy that, and this is the one that I typically do in my head is I write this as 4.5 minus 2.5 minus 0.3, or four and 5/10 minus two and 5/10 minus 3/10. Now why did I write it this way? Because I find this pretty straightforward to compute, and then once I get that answer, I just have to take away 3/10. So for example, well, if I have 5/10 here, and I'm taking away 5/10, those are gonna knock each other out, and so I'm just gonna be left with four minus two, which is gonna be two, and then I have to take away the 3/10. This is a pretty straightforward way of doing it in your head. So what's 3/10 less than two? Well, you can visualize a number line in your head. Well, that's gonna be one and 7/10, one and 7/10, and if this feels strange how I got, oh, one, I wrote 7/100, one and 7/10. And if this seems strange how I got there that fast, just think about this. This is the same thing as one plus one minus 3/10, and this one right over here is the same thing as 10/10, and so 10/10 minus 3/10. That over there is going to be, that is going to be 7/10, so it's going to be one and 7/10, which is what we got before. So the whole point here is to appreciate there's multiple strategies for subtracting decimals, some that you can do a little bit more automatically, but it's really good to think about what's going on in your head. And some strategies are actually better in your head than on paper, or at least more easy in your head.

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Strategies for subtracting more complex decimals with tenths (video) | Khan Academy

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Instructor] Do some more examples subtracting decimals. So let's say we wanna figure out what two minus 1.2 is. Pause this video and see if you can calculate this. So there is multiple ways to tackle this. One way is you could say, "Look, this is the same thing "as two minus one "minus 2/10."

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Video transcript

divide simplify the answer and write as a mixed number and we have two and one-fourth divided by one and three-fourths so the first thing we want to do since both of these are mixed numbers is to convert them both into improper fractions so let's start with two and one-fourth so we're still going to have four in the denominator but instead of two and one-fourth remember two is the same thing as eight forts so we have eight forts and then we have another 1/4 that gives us nine fourths or another way to come up with this 9 you take 4 you take 4 times 2 which is 8 plus 1 that gives you 9 and then the 1 and 3/4 same process you're going to have 4 in the denominator and then the numerator is going to be 4 times 1 which is 4 plus 3 which is 7 so this is the exact same problem here 2 and 1/4 divided by 1 and 3/4 is the same thing as 9 fourths divided by 7/4 and we saw in several videos already that dividing by a fraction is the same thing as multiplying by its reciprocal so this is equivalent to this is equivalent to so these are all equivalent this is equivalent to 9/4 x times the reciprocal of this we're taking changing the division operation to a multiplication and we're taking the reciprocal of the 7/4 the reciprocal of 7/4 you swap the numerator denominator or the top number in the bottom number and you get 4 over 7 now we could just multiply these we could just say this is 9 times 4 which would be 36 over 4 times 7 which is 28 and then try to put it in lowest terms or we could do it right now because it'd be simpler we have a 4 in the numerator we have a 4 in the denominator that will eventually be in the denominator so let's divide our eventual numerators on our debt ventral denominators both by 4 so you divide this 4 by 4 you get 1 this 4 by 4 you get 1 so now when you multiply it you get 9 times 1 which is 9 over 1 times 7 which is 7 so we have our answer but it's writ right now it's an improper fraction they want us to write it as a mixed number and to figure out as a mixed number we can do it in our now I think we've seen this enough times we say how many times does 7 how many times does 7 go into 9 what goes into it exactly one time but when you take 7 into 9 one time what do you have left over we're going to have 2 left over right 7 times 1 is 7 and you're gonna have 2 left over you need 2 more to get to 9 so you're going to have 2 left over so this is one and two sevens and we're done

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Dividing mixed numbers (video) | Khan Academy

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Video transcript

let's think about what it means to take 8/3 and divide it by one-third so let me draw a number line here so there is my number line this is 0 this is 1 and this is 2 maybe this is 3 right over here and let me plot 8/3 so to do that I just need to break up my each each hole into thirds so let's see it's one-third two-thirds three-thirds 4/3 5/3 6 thirds 7/3 8/3 so right over here and then of course 9/3 would get us to 3 so this right over here is 8/3 8/3 now one way to think about 8/3 divided by 3 is what if we take this length and we say how many jumps would it take to get there if we're doing it in jumps of 1/3 or essentially we're breaking this up if we were to break up 8/3 into sections of 1/3 how many sections would I have or how many jumps what I have well let's think about that if we're trying to take jumps of 1/3 we're gonna have to go 1 2 3 4 5 6 7 8 jumps so we could view this as we could you let me do this in a different color in this orange so we took these 8 jumps right over here so we could view 8/3 divided by 1/3 as being equal to 8 now why is this actually make sense well when you're dividing things in 2/3 for every hole you're now going to have 3 jumps so whatever value you're trying to get to you're going to have that number times 3 jumps so another way of thinking about it is that 8/3 8/3 divided by divided by 1/3 is the same thing as 8/3 8/3 times 3 now we could either write it like this we could write x 3 like that or if we want to write 3 as a fraction we know that 3 is the same thing three over one and when we already know how to multiply fractions a multiply the numerators eight times three so you have eight let me do that that same color you have eight times three in the numerator now eight times three and then you have three times 1 in the denominator three times three times one in the denominator which would give you 24 thirds which is the same thing as 24 divided by three which once again is equal is equal to is equal to is equal to eight now let's see if this still makes sense instead of dividing by one-third if we were divided by two-thirds so let's think about what 8/3 divided by two-thirds is divided by two-thirds well once again this is like asking the question if we wanted to break up this section from zero to 8/3 into sections of 2/3 or jumps of two-thirds how many sections or how many jumps would I have to make well let's think about it one jump just in a different color we could make one jump that's the same color as my 8/3 we could do one jump my computer is doing something strange we could do one jump two jumps three jumps and four jumps so we see 8/3 divided by 2/3 is equal is equal to four now does this make sense in this world right over here well if we take 8/3 and we do the same thing saying hey look dividing by a fraction is the same thing as multiplying by a reciprocal well let's multiply by three halves let's multiply by the reciprocal of 2/3 so we swap the numerator and the denominator so we multiply it times three halves and then what do we get in the numerator once again we get eight times three which is 24 and in the denominator we get 3 times 2 which is 6 so now we get 24 divided by 6 is equal to four now does it make sense that we got half the answer if you think about the difference between what we did here and what we did here these are almost the same except here we really just didn't divide or you could say you divided by one while here you / - well does that make sense well sure because here you jumped twice as far you jumped twice as far so you had to take half the number of steps and so in the first example you saw why it makes sense to multiply by 3 when you divide by a fraction for every hole you're making 3 jumps so that's why when you divide by this fraction you multiply or when you divide whatever's in the denominator you multiply by it and now when the numerator is greater than 1 every jump you're going twice as far as you did in this first one right over here and so you would have to do half as many jumps hopefully that makes sense it's easy to think about just mechanically how to divide fractions dividing 8/3 divided by 1/3 is the same thing as 8/3 times 3 over 1 or 8/3 divided by 2/3 is the same thing as 8/3 times 3 over 2 but hopefully this this video gives you a little bit more of an intuition of why this is the case

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Understanding division of fractions (video) | Khan Academy

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Let's think about what it means to multiply 2 over 3, or 2/3, times 4/5. In a previous video, we've already seen how we can actually compute this. This is going to be equal to-- in the numerator, we just multiply the numerators. So it's going to be 2 times 4. And in the denominator, we just multiply the denominator. So it's going to be 3 times 5. And so the numerator is going to be 8, and the denominator is going to be 15. And this is about as simple as we can make it. 8 and 15 don't have any factors common to each other, than 1, so this is what it is. It's 8/15. But how, why does that actually makes sense? And to think about it, we'll think of two ways of visualizing it. So let's draw 2/3. I'll draw it relatively big. So I'm going to draw 2/3, and I'm going to take 4/5 of it. So 2/3, and I'm going to make it pretty big. Just like this. So this is 1/3. And then this would be 2/3. Which I could do a little bit better job making those equal, or at least closer to looking equal. So there you go. I have thirds. Let me do it one more time. So here I have drawn thirds. 2/3 represents 2 of them. It represents 2 of them. One way to think about this is 2/3 times 4/5 is 4/5 of this 2/3. So how do we divide this 2/3 into fifths? Well, what if we divided each of these sections into 5. So let's do that. So let's divide each into 5. 1, 2, 3, 4, 5. 1, 2, 3, 4, 5. And I could even divide this into 5 if I want. 1, 2, 3, 4, 5. And we want to take 4/5 of this section here. So how many fifths do we have here? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And we've got to be careful. These really aren't fifths. These are actually 15ths, because the whole is this thing over here. So I should really say how many 15ths do we have? And that's where we get this number from. But you see if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Where did that come from? I had 3, I had thirds. And then I took each of those thirds, and I split them into fifths. So then I have five times as many sections. 3 times 5 is 15. But now we want 4/5 of this right over here. This is 10/15 right over here. Notice it's the same thing as 2/3. Now if we want to take 4/5 of that, if you have 10 of something, that's going to be 8 of them. So we're going to take 8 of them. So 1, 2, 3, 4, 5, 6, 7, 8. We took 8 of the 15, so that is 8/15. You could have thought about it the other way around. You could have started with fifths. So let me draw it that way. So let me draw a whole. So this is a whole. Let me cut it into five equal pieces, or as close as I can draw five equal pieces. 1, 2, 3, 4, 5. 4/5, we're going to shade in 4 of them. 4 of the 5 equal pieces. 3, 4. And now we want to take 2/3 of that. Well, how can we do that? Well, let's split each of these 5 into 3 pieces. So now we have essentially 15ths again. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. We want to take 2/3 of this yellow area. We're not taking 2/3 of the whole section. We're taking 2/3 of the 4/5. So how many 15ths do we have here? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So if you have 12 of something, and you want to take 2/3 of that, you're going to be taking 8 of it. So you're going to be taking 1, 2, 3, 4, 5, 6, 7, 8 or 8 of the fifteenths now. So either way, you get to the same result. One way, you're thinking of taking 4/5 of 2/3. Another way you could think of it as you're taking 2/3 of 4/5.

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Intro to multiplying 2 fractions (video) | Khan Academy

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Let's think about what it means to multiply 2 over 3, or 2/3, times 4/5. In a previous video, we've already seen how we can actually compute this. This is going to be equal to-- in the numerator, we just multiply the numerators. So it's going to be 2 times 4.

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Let's try to evaluate 7 and 6/9 minus 3 and 2/5. So like always, I like to separate out the whole number parts from the fractional parts. This is the same thing as 7 plus 6/9 minus 3 minus 2/5. And the reason why I'm saying minus 3 minus 2/5 is this is the same thing as minus 3 plus 2/5. And so you distribute the negative sign. You're subtracting a 3, and then you're subtracting the 2/5. And so now we can worry about the whole number parts, 7 minus 3. Well, 7 minus 3 is going to give us 4. So that's going to give us 4. And then we're going to have 6/9 minus 2/5. So let me think about what 6/9 minus 2/5 are. 6/9 minus 2/5, well, we're going to have to find a common denominator. So this is going to be the same thing. And I think the least common multiple of 9 and 5 is going to be 45. Literally, you then multiply. They have no common factors. So it's going to be over 45. To go from 9 to 45, I have to multiply by 5. So I'm going to have to multiply the numerator by 5. So 6 times 5 is 30. Then I'm going to subtract. To go from 5 to 45, I had to multiply by 9. So I have to multiply the numerator by 9 if I don't want to change the value. So 2 times 9 is 18. And 30/45 minus 18/45 is going to be something over 45. 30 minus 18 is 12. If I subtract these two fractions right over here, I get 12/45. So it's 4 plus 12/45. Or if we wanted to write it as a mixed number, this is equal to 4 and 12/45. But we're not done yet. We can simplify this further. 12 and 45 have common factors. They're both divisible by 3. Actually, they're both divisible by-- well, I think we can divide more after that. So let's see. If we divide the numerator by 3 and the denominator by 3, we end up with 4. And 12 divided by 3 is 4. And 45 divided by 3 is 15. 4 and 4/15. And actually, we're done. These two can't be simplified anymore. 4 and 4/15.

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Subtracting mixed numbers: 7 6/9 - 3 2/5 (video) | Khan Academy

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Let's try to evaluate 7 and 6/9 minus 3 and 2/5. So like always, I like to separate out the whole number parts from the fractional parts. This is the same thing as 7 plus 6/9 minus 3 minus 2/5.

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Adding and subtracting mixed number with unlike denominators

Adding and subtracting mixed number with unlike denominators

Video transcript

Let's add 19 and 3/18 to 18 and 2/3. So I like to separate out the whole number parts from the fraction parts. So 19 and 3/18 is the same thing as 19 plus 3/18. And to that, we are going to add 18 and 2/3, which is the same thing as 18 plus 2/3. Now we can separately add the whole number parts. So we could add the 19 to the 18. So we could do 19 plus 18. And then we can add the fraction parts-- let me do this in green-- plus 3/18 plus 2/3. Now 19/18, pretty straightforward. That is what? Let's see. 19 plus 19 would be 38. So this is going to be 1 less than that. It's going to be 37. So that gives me 37. And then 3/18 plus 2/3, to add them, I need to have the same denominator. And the least common multiple of 18 and 3 is 18. So let's convert 2/3 to something over 18. So 2/3, if I want to write it as something over 18, well, I multiplied the denominator by 6, so I'd also have to multiply the numerator by 6. So it's the same thing as 12/18. So I can rewrite 2/3 as 12/18. And now I can add these two things together. That's going to be-- so I have 37 plus-- it's going to be something over 18-- plus something over 18. 3 plus 12 is 15, plus 15/18. And so expressing this as a mixed number, I get 37 and 15/18. And that's the right number. But we can simplify it even more. We can simplify the 15/18. Both the numerator and the denominator are divisible by 3. So let's divide them both by 3. And we're not changing the value because we're doing the same thing to the numerator and the denominator. And so this gives us, we still have our 37, but the numerator is now 5, and the denominator is now 6. So we get 37 and 5/6. And we're done.

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Adding mixed numbers: 19 3/18 + 18 2/3 (video) | Khan Academy

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- [Voiceover] Let's say that we have the fraction 9/10, and I want to add to that the fraction 1/6. What is this, what is this going to equal? So when you first look at this, you say, "Oh, I have different denominators here. It's not obvious how I add these." And you'd be right and the way to actually move forward is to find a common denominator, to convert both of these fractions into fractions that have a common denominator. So how do you think about a common denominator? Well, a common denominator's gonna have to be a common multiple of these two denominators of 10 and six. So what's a common multiple of 10 and six? And it's usually simplest to find the least common multiple, and a good way of doing that is start with the larger denominator here, 10, and say, okay is 10 divisible by six? No. Okay, now, is 20 divisible by six? No. Is 30 divisible by six? Yes. 30 is divisible by six. So I'm just going through the multiples of 10 and saying, "Well what is the smallest multiple of 10 that is divisible by six?" And that's going to be 30. So I could rewrite both of these fractions as something over 30. So nine over 10. How would I write that as something over 30? Well I multiply the denominator, I'm multiplying the denominator by three. So I've just multiplied the denominator by three. So if I don't want to change the value of the fraction, I have to do the same thing to the numerator. I have to multiply that by three as well because now I'm just multiplying the numerator by three and the denominator by three, and that doesn't change the value of the fraction. So nine times three is 27. So once again, 9/10 and 27/30 represent the same number. I've just written it now with a denominator of 30, and that's useful because I can also write 1/6 with a denominator of 30. Let's do that. So 1/6 is what over 30? I encourage you to pause the video and try to think about it. So what did we do go from six to 30? We had to multiply by five. So if we multiply the denominator by five, we have to multiply the numerator by five as well, so one times five, one times five is five. So 9/10 is the same thing as 27/30, and 1/6 is the same thing as 5/30. And now we can add, now we can add and it's fairly straightforward. We have a certain number of 30ths, added to another number of 30ths, so 27/30 + 5/30, well that's going to be 27, that's going to be 27 plus five, plus five, plus 5/30, plus 5/30, which of course going to be equal to 32/30. 32 over 30, and if we want, we could try to reduce this fraction. We have a common factor of 32 and 30, they're both divisible by two. So if we divide the numerator and the denominator by two, numerator divided by two is 16, denominator divided by two is 15. So, this is the same thing as 16/15, and if I wanted to write this as a mixed number, 15 goes into 16 one time with a remainder one. So this is the same thing as 1 1/15. Let's do another example. Let's say that we wanted to add, we wanted to add 1/2 to to 11/12, to 11 over 12. And I encourage you to pause the video and see if you could work this out. Well like we saw before, we wanna find a common denominator. If these had the same denominator, we could just add them immediately, but we wanna find a common denominator because right now they're not the same. Well what we wanna find is a multiple, a common multiple of two and 12, and ideally we'll find the lowest common multiple of two and 12, and just like we did before, let's start with the larger of the two numbers, 12. Now we could just say well 12 times one is 12, so that we could view that as the lowest multiple of 12. And is that divisible by two? Yeah, sure. 12 is divisible by two. So 12 is actually the least common multiple of two and 12, so we could write both of these fractions as something over 12. So 1/2 is what over 12? Well to go from two to 12, you multiply by six, so we'll also multiply the numerator by six. Now we see 1/2, and 6/12, these are the same thing. One is half of two, six is half of 12. And how would we write 11/12 as something over 12? Well it's already written as something over 12, 11/12 already has 12 in the denominator, so we don't have to change that. 11/12, and now we're ready to add. So this is going to be equal to six, this is going to be equal to six plus 11, six plus 11 over 12. Over 12. We have 6/12 plus 11/12, it's gonna be six plus 11 over 12, which is equal to, six plus 11 is 17/12. If we wanted to write it as a mixed number, that is what, 12 goes into 17 one time with a remainder of five, so 1 5/12. Let's do one more of these. This is strangely fun. Alright. Let's say that we wanted to add, We're gonna add 3/4 to, we're gonna add 3/4 to 1/5. To one over five. What is this going to be? And once again, pause the video and see if you could work it out. Well we have different denominators here, and we wanna find, we wanna rewrite these so they have the same denominators, so we have to find a common multiple, ideally the least common multiple. So what's the least common multiple of four and five? Well let's start with the larger number, and let's look at its multiples and keep increasing them until we get one that's divisible by four. So five is not divisible by four. 10 is not divisible by four, or perfectly divisible by four is what we care about. 15 is not perfectly divisible by four. 20 is divisible by four, in fact, that is five times four. That is 20. So what we could do is, we could write both of these fractions as having 20 in the denominator, or 20 as the denominator. So we could write 3/4 is something over 20. So to go from four to 20 in the denominator, we multiplied by five. So we also do that to the numerator. We multiply by three times five to get 15. All I did to go from four to 20, multiplied by five. So I have to do the same thing to the numerator, three times five is 15. 3/4 is the same thing as 15/20, and over here. 1/5. What is that over 20? Well to go from five to 20, you have to multiply by four. So we have to do the same thing to the numerator. I have to multiply this numerator times four to get 4/20. So now I've rewritten this instead of 3/4 plus 1/5, it's now written as 15/20 plus 4/20. And what is that going to be? Well that's going to be 15 plus four is 19/20. 19/20, and we're done.

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Adding fractions with unlike denominators (video) | Khan Academy

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Voiceover] Let's say that we have the fraction 9/10, and I want to add to that the fraction 1/6. What is this, what is this going to equal? So when you first look at this, you say, "Oh, I have different denominators here. It's not obvious how I add these."

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Intro to order of operations (video) | Khan Academy

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This example shows the steps and clarifies the purpose of order of operations: to have ONE way to interpret a mathematical statement.

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Explore the commutative, associative, and identity properties of addition.

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Least common multiple exercise | Factors and multiples | Pre-Algebra | Khan Academy

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Greatest common factor explained | Factors and multiples | Pre-Algebra | Khan Academy

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You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 × 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication - a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) - which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, "Hey, maybe this is 2 × 3, which would be 6." But remember, I just said this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 × ... (And I'll use a little dot to signify multiplication.) ... 2 × 2 × 2. Well, what's 2 × 2 × 2? Well that is equal to 8. (2 × 2 × 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 - (Let me do that in yellow.) Let's multiply 3 × 3. So this is going to be equal to 9. Let's do a few more examples. What is, say, 5 to the - let's say - 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 × 5 × 5 × 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. We are not adding them. This is not 5 × 4. This is not 20. This is 5 × 5 × 5 × 5. So what is this going to be? Well 5 × 5 is 25. (5 × 5 = 25.) 25 × 5 is 125. (25 × 5 = 125.) 125 × 5 is 625. (125 × 5 = 625.)

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Intro to exponents (video) | Exponents | Khan Academy

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You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 × 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's.

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In this video, I'll introduce you to a new way of computing division, especially for larger numbers. And then we'll think a little bit about why it works. So we're going to try to compute what 96 divided by 4 is. And I'm going to write it a little bit differently. I'm going to write 96 divided by-- so I'm going to write this strange-looking symbol right over here, this thing that covers the 96. But you could view this as 96 divided by 4. And I'll show you in a second why we write it this way. This is actually a very useful way of actually computing it. So the first thing we'll do is we'll say, well, how many times does 4 go into 9? Well, we know that 4 times 2 is equal to 8 and that 4 times 3 is equal to 12. So 3 would be too much. We would go above 9. So we want to be below 9 but not have too much left over. We want the largest number that gets us into 9 without going over 9. So we'll say it goes two times. 4 goes into 9 two times. And then we say, what's 2 times 4? Well, 2 times 4 is 8. 4 times 2 is 8. Or 2 times 4 is 8. And now, we subtract. We subtract the 8 from the 9. And we get 1. And now we bring down the next digit, which is the 6. And then we ask ourselves, well, how many times does 4 go into 16? Well, in this case, we know that 4 goes into 16 exactly four times. 4 times 4 is 16. So we say 4 goes into 16 four times. Then we multiply 4 times 4 is 16. We subtract. And 16 minus 16, we have absolutely nothing left over. And there we have our answer. I know it seems kind of magical at this point. But in a few seconds, we're going to think about why this actually worked. We got that 96 divided by 4 is equal to-- I want to do that in a different color-- 24. Now, what I want you to do right now is pause this video and think about why did this actually work. How did we magically get the right answer here? And you can verify this. Multiply 4 times 24, and you will get 96. Well, I'm assuming you gave a go at it. And the important thing always is to keep track of the place value. And it really tells you what's going on when we do this process. When we looked at this 9 right over here, this 9 is in the tens place. This is actually representing 90. It represents 9 tens. So we're saying, well, how many times does 4 go into 90 if we're thinking about multiples of 10? Well, it goes 20 times. 4 times 20 is 80. And so we said, well, 4 times 20 is 80. But we still have 16 left over. You do 96 minus 80. You have 16 left over to divide 4 into. And then 4 goes into 16 four times. So really, a lot of this is just saying, well, we first figured out that we could go 20 times. And then we said, well, that doesn't get us all the way to 96. We have to go another four times. Hopefully, that helps.

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Intro to long division (no remainders) (video) | Khan Academy

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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.

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We now have the general tools to really tackle any multiplication problems. So in this video I'm just going to do a ton of examples. So let's start off with-- and I'll start in yellow. Let's start off with 32 times 18. Say 8 times 2 is 16. Well, I'll do it in our head this time because you always don't have all this space to work with. So 8 times 2 is 16. Put the 1 up there. 8 times 3 is 24. 24 plus 1 is 25. So 8 times 32 was 256. Now we're going to have to multiply this 1, which is really a 10, times 32. I'll underline it with the orange. 1 times 2-- oh, we have to be very careful here. 1 times 2 is 2. So you might say hey, let me stick a 2 down there. Remember, this isn't a 1. This is a 10, so we have to stick a 0 there to remember that. So 10 times 2 is 20. Or you say 1 times 2 is 2, but you're putting it in the 2's place, so you still get 20. So 10 times 2 is 20. It works out. Then 1 times 3. And we have to be very careful. Let's get rid of what we had from before. 1 times 3 is 3. There's nothing to add here, so you just get a 3. And so you get 10 times 32 is 320. This 1 right here, that's a 10. 10 plus 8 is 18. So now we just add up the two numbers. You add them up. 6 plus 0 is 6. 5 plus 2 is 7. 2 plus 3 is 5. Let's keep going. Let's do 99 times 88. So a big number. 8 times 9 is 72. Stick the 7 up there. And then you have 8 times 9 again. 8 times 9 is 72, but now you have the 7 up here. So 72 plus 7 is 79. Fair enough. Now we're done with this. Let's just delete it just so that we don't get confused in our next step. In our next we're going to multiply this 8 now times 99. But this 8 is an 80. So let's stick a 0 down there. 8 times 9 is 72. Stick a 7 up there. Then 8 times 9 is 72. Plus 7 is 79. 2 plus 0 is 2. Let me switch colors. 9 plus 2 is 11. Carry the 1. 1 plus 7 is 8. 8 plus 9 is 17. Carry the 1. 1 plus 7 is 8. 8,712. Let's keep going. Can't do enough of these. All right, 53 times 78. I think you're getting the hang of it now. Let's multiply 8 times 53 first. So 8 times 3 is 24. Stick the 2 up there. 8 times 5 is 40. 40 plus 2 is 42. Now we're going to have to deal with that 7 right there, which is really a 70. So we got to remember to put the 0 there. 7 times 3, and let's get rid of this. Don't want to get confused. 7 times 3 is 21. Put the 1 there and put the 2 up here. 7 times 5 is 35. Plus 2 is 37. Now we're ready to add. 4 plus 0 is 4. 2 plus 1 is 3. 4 plus 7 is 11. Carry the 1. 1 plus 3 is 4. 4,134. Let's up the stakes a little bit. So let's say I had 796 times 58. Let's mix it up well. All right, so first we're just going to multiply 8 times 796. And notice, I've thrown in an extra digit up here. So 8 times 6 is 48. Put the 4 up there. 8 times 9 is 72. Plus 4 is 76. And then 8 times 7 is 56. 56 plus 7 is 63. I'm sure I'll make a careless mistake at some point in this video. And the goal for you is to identify if and when I do. All right, now we're ready, so we can get rid of these guys up here. Now we can multiply this 5, which is in the 10's place. It's really a 50. Times this up here. Because it's a 50 we stick a 0 down there. 5 times 6 is 30. Put the 0 there, put the 3 up there. 5 times 9 is 45. Plus the 3 is 48. 5 times 7 is 35. Plus 4 is 39. Now we're ready to add. 8 plus 0 is 8. 6 plus 0 is 6. 3 plus 8 is 11. 1 plus 6 is 7. 7 plus 9 is 16. And then 1 plus 3 is 4. So 796 times 58 is 46,168. And that sounds about right because 796-- it's almost 800. You know, which is almost 1,000. So if we multiplied 1,000 times 58 we'd get 58,000. But we're multiplying something a little bit smaller than 1,000 times 58, so we're getting something a little bit smaller than 58,000. So the number is in the correct ballpark. Now let's do one more here where I'm really going to step up the stakes. Let's do 523 times-- I'm going to do a three-digit number now. Times 798. That's a big three-digit number. But it's the same exact process. And once you kind of see the pattern you say, hey, this'll apply to any number of digits times any number of digits. It'll just start taking you a long time and your chances of making a careless mistake are going to go up, but it's the same idea. So we start with 8 times 523. 8 times 3 is 24. Stick the 2 up there. Now 8 times 2 is 16. 16 plus 2 is 18. Put the 1 up there. 8 times 5 is 40. Plus 1 is 41. So 8 times 523 is 4,184. We're not done. We have to multiply times the 90 and by the 700. So let's do the 90 right there. So it's a 90, so we'll stick a 0 there. It's not a 9. And let's get rid of these guys right there. 9 times 3 is 27. 9 times 2 is 18. 18 plus 2 is 20. And then we have 9 times 5 is 45. 45 plus 2 is 47. I don't want to write that thick. 47. Let me make sure I did that one right, and let's just review it a little bit. 9 times 3 was 27. We wrote the 7 down here and put the 2 up there. 9 times 2 is 18. We added 2 to that, so we wrote 20. Wrote the 0 down there and the 2 up there. The 9 times 5 was 45. Plus 2 is 47. You really have to make sure you don't make careless mistakes with these. Then finally, we have to multiply the 7, which is really a 700 times 523. When it was just an 8 we just started multiplying here. When it was a 90, when we were dealing with the 10's place, we put a 0 there. Now that we're dealing with something that's in the 100's, we're going to put two 0's there. And so you have 7-- and let's get rid of this stuff. That'll just mess us up. 7 times 3 is 21. Put the 1 there. Stick the 2 up there. 7 times 2 is 14. 14 plus our 2 is 16. Put the 1 up there. 7 times 5 is 35. Plus 1 is 36. And now we're ready to add. And hopefully we didn't make any careless mistakes. So 4 plus 0 plus 0. That's easy. That's 4. 8 plus 7 plus 0. That's 15. Carry the 1. 1 plus 1 plus 1 is 3. 4 plus 7 plus 6. That's what's? 4 plus 6 is 10. It's 17. And then we have 1 plus 4 is 5. 5 plus 6 is 11. Carry the 1. 1 plus 3 is 4. So 523 times 798 is 417,354. Now we can even check to make sure. And so this is the moment of truth. Let's see if we have-- let's see. 523 times 798. There you go. Moment of truth. I don't have to re-record this video. It's 417,354. But we did it without the calculator, which is the important point.

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Multiplying multi-digit numbers (video) | Khan Academy

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We now have the general tools to really tackle any multiplication problems. So in this video I'm just going to do a ton of examples. So let's start off with-- and I'll start in yellow. Let's start off with 32 times 18. Say 8 times 2 is 16.

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XtraMath - 10 minutes a day for math fact fluency

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XtraMath is a free program that helps students master addition, subtraction, multiplication, and division facts.

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Practice Math and ELA in your Freckle account and earn coins for the piggy store.

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Amazing pets, epic battles and math practice. Prodigy, the no-cost math game where kids can earn prizes, go on quests and play with friends all while learning math.