Direct variation common core algebra 2 homework answers

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Hello and welcome to another common core algebra two lesson by E math instruction. My name is Kirk weiler and today we're going to be doing unit three lesson number one on direct variation. So unit three is our unit on linear functions in linear modeling. And this starts by looking at a very common topic that you've seen for quite a few years known as direct variation. So let's jump right into that and eventually we'll see how it will form the basis of all linear functions. All right, direct variation is a pretty simple idea. Two variables have a direct or proportional relationship either way. If for every ordered pair XY, we have Y divided by X is equal to K or rearranged Y equals K times X so at the end of the day, what direct variation just means is that one of the two variables is always just a multiple of the other. So Y is twice X or Y is one third X or Y is 5 halves times X Y is always some multiple of X or when you divide Y and X, you always get a constant. So let's kind of see how this plays out in a variety of different problems. Exercise one, this is the most basic thing you can do with direct variation. It says, in each of the following, Y and X are directly related, solve for the missing value. So Y is 15 when X is 5, what is Y when X is 9? Now, one way to do this is to simply set up a proportion, right? Which is that Y divided by X always must be a constant. In which case 15, if I take that Y and I divide it by this X is defied as my unknown Y is to 9. Of course we can then multiply both sides by 9. All right? We could even divide this into this and get three. And we could get Y equals 27. Now, some of you might automatically have said, well, Y is equal to 27, because it's clear that here, Y is simply three times X, right? The relationship here is Y equals three times X and therefore, when X is 9, Y would be 27. Now that's pretty clear, X is 5 Y is 15, not too tough. Here it's a little bit trickier. When X is four, Y is negative 6. That's not nearly as easy of a relationship. Y equals what? Times X, right? I mean, the question mark there, you know? So this is where that proportionality works quite well if we put negative 6 is to four as Y is to negative ten. That's not that bad. Some of you may have the urge to just cross multiply. Of course, that's okay, but there's no real need to. Eventually we'll get Y equals 60 divided by four, so Y is 15. All right. Let's do one more. And again, this one's a little bit weird, you know what I mean? What would we have to multiply 16 by to get 12? Well, there's definitely something out there. It would be three fourths. But for right now, let's not worry about it. Again, the idea is if I took that Y and I divided it by its X 12 divided by 16, I'd have to get this Y divided by the X 24. Again, I can multiply by 24 on both sides. Okay, here maybe do a little bit of simplifying 8 goes into that three times. It goes into that two times. So Y will be 36 divided three times 12 is 36 divided by two. Y is 18. All right. So directly proportional means that when you divide Y and X, you always get the same number. We can use this to sort of produce missing Y values. So pause the video now and write down anything you need to. Okay, here we go. Let's get into some applied problems. Exercise two, the distance a person can travel varies directly with the time they've been traveling if going at a constant speed. So if I'm driving at a constant speed, then how far I go in terms of miles is directly proportional or very directly with how long I've been driving. If Phoenix travels 78 miles in 1.5 hours going at a constant speed, how far will he travel in two hours at the same speed? Okay, so this is a situation where we know one Y value, we know one X value, and then we know a second X value. So it's kind of like Y one, X one, and then we have X two. And we're looking for Y two. See if you can set up a proportion to figure out how far Phoenix will go in two hours. All right. Well, let's go through it. We can take the approach that we did in the last problem. We'll put 78 divided by 1.5. And that's going to be equal to that Y value I don't know after two. Again, I can multiply both sides by two. All right, and ultimately speaking now, I just have to work this out on my calculator. That's all I have to do. Now, if you wanted to do two times 78, kind of as an intermediate step you could. That would be one 56 divided by 1.5. And that's equal to a 104. Miles. All right, Phoenix drove a 104 miles in two hours. All right. Let's clear this out. Pause the video now if you need to. Okay, here we go. Exercise three, Jenna works a job where her pay varies directly with the number of hours she works. She has worked. In one week, she worked 35 hours and made 274 75. How many hours would she need to work in order to earn three 37 55? Again, direct proportion problem. Why don't you go ahead and play around with it, okay? Now in my mind, in a certain extent here, you're given X one and you're given Y one. And then you're given Y two. It doesn't really matter what you call the variables, though. You can easily do 274 75. Divided by 35 equals 337 55. Divided by that input that we don't know. How many hours did she have to work? Here, I think cross multiplying is actually kind of helpful. Because we'll get two right, so a little cross multiplying. We'll get two 74 75 times X is equal to 35. Times 337 .5. All right. Now what we can do is we can then divide both sides by the 274 75. And we can simply figure out what this calculation is equal to. That comes out to be exactly 43. Hours. All right, so that to be careful with our calculator work. All right, so we got these ratios, right? Now what I'd like to do is I'd like to take it out of the applied realm and go back to just the pure algebraic realm where we start to kind of look at what these relationships look like when graft. All right? So pause the video now and write down anything you need to and then we'll move on to that. Okay, here we go. All clear. Exercise four. Two variables, X and Y, very directly. When X is four, sorry, when X is 6, Y is four. The point is shown plotted below. So here's that point, right? The .6 comma four. It says find the Y values for each of the following X values and plot each point and connect. So why don't you go ahead and do that, all right? So in other words, set up some ratios like we did before, maybe kind of divide it off. Figure out the Y values that go with those X values. And then we'll plot them, okay? Pause the video now and take a little bit of time. All right. Well, here we go. Why is four when X is 6? So when X is three, what's Y? Well, you may have already kind of figured out a nice way to do this. But again, if I just kind of move on and do that, I'll get Y is 12 6. So Y is two. So we have the .3 comma two. That's cool. Let's do negative 6. Again, we've got four is to 6 as Y is to negative 6. In which case, we'll multiply by negative 6 on both sides. And we'll get Y equals negative four. Hey, look at that. Negative 6 comma negative four. Right there. Plot each point and connect, I think, I'll connect with a nice. Nice, straight line. All right, there it is. And then you put some arrows there. And look at that. We have a nice linear relationship. It says, what is the constant of variation in this problem? What does it represent in the line? Well, remember the constant of variation is what we get when we divide Y and X so we can choose any two Y and X values we want. I'll just start with the ones that we were given. Four divided by 6, that's two thirds. Now what is the two thirds represent about this line? Think about that for a minute. Take you figure it out. It represents its slope. That's all that constant of variation is. Watch, let me do it in blue. Let me pick this point in this point. Do you see it? If I go to the right three and up to the right three and up to, so that constant of variation is simply the slope of the line. Notice it says write the equation of the line. That's real easy watch. We'll just do Y divided by X is equal to two thirds. It always is. And then we'll just and we'll just multiply both sides apparently by the wrong variable. And the equation of our line is Y equals two thirds times X that equation describes every point that lies on that line. All right. Well, pause the video now and then we're going to clear this up and keep working. Okay, here it goes. All right, back to a little application. Exercise three. The miles driven by a car, D, varies directly with the number of gallons, G of gasoline used. Okay, so we got gallons of gasoline and the number of miles that we've driven. Abigail is able to drive 336 miles on 8 gallons of gasoline in her hybrid vehicle. Good for her. Calculate the constant of variation for a relationship include proper units in your answer. So do this for a minute. Pause the video. Calculate that constant of variation. All right. Well, we could take any pair that we had, but we only have one pair. 336 miles, 8 gallons, and when we do that division, we get 42 miles per gallon. Right? 42 miles per gallon. Let her be give a linear equation that represents the relationship between D and G, express your answer as an equation solved for D all right, well this is similar to the last problem, D divided by G equals 42. Therefore, D must be 42 times G in other words, if I told you I used three gallons of gasoline, how far did I drive? You could say, well, 42 times three and get, I guess it would be a 126 miles. And there you go. Let her see how far can Abigail drive on 6 gallons of gas? Well, that's the point of an equation is to use it to predict stuff. So I'll just take that 6 and substitute it in and I'll find that Abigail can drive 252 miles. Letter D, how many gallons of gas will Abigail need in order to drive 483 miles? Well, in that case, we're given the value of D D equals 483. So I can substitute that into my equation 483 equals 42 times G so G will be 483 divided by 42. And she's going to need 11.5 gallons. Don't forget those units. All right. So this is a neat situation where we can look at something that we should feel kind of comfortable with a proportional relationship. I love that proportionality constant. 42 miles per gallon. That's your car's fuel efficiency. A lot of times the units on these proportionality constants can give us a nice physical feel for what's going on miles per gallon, right? All right, I'm going to clear this out. So pause the video now if you need to. All right. And let's wrap it up. So proportional relationships or direct variation, right? A very important topic. It's used a lot in the real world, where one variable is always proportional or a multiple of another variable. Distance and time or a great example when you're moving at a constant speed. I like that last exercise where we had the number of miles we could drive being a directly proportional relationship with the gallons of gas we use. So there's a lot of situations and what's important is that they form lines. Lines that go through the origin, but they form lines. In the next lesson, we'll start looking at lines that don't just go through the origin. All right. For now, I want to thank you for joining me for another common core algebra two lesson by E math instruction. I'm Kirk weiler. And until next time, keep thinking. I keep solving problems.

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direct variation common core algebra homework answers