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Hi Scarlets. We miss you! You can use this padlet to review what we've already covered in class.  Please email Mr. Spaine or me if you have questions. Stay strong,  we'll get through this.

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Slope dude video (best movie of all time)

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Slope dude video (best movie of all time)

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Review for Unit 6 Test 3a and 3b(with slope and dist)

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writing equations of circles and factoring practice

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notes 2/21- using pythag to find distance

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notes 2/21- using pythag to find distance

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Plotting a point (ordered pair) (video) | Khan Academy

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Plot the ordered pair 6, comma negative 8 into the coordinate plane. So this is a coordinate plane right over here. The horizontal axis here, this is the x-axis. The vertical axis here is the y-axis. And the convention, when we get an ordered pair like this, is that the first coordinate is the x-coordinate, and the second coordinate is the y-coordinate.

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Video-finding the distance between two points 

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Video-finding the distance between two points 

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- We are asked what is the distance between the following points. Pause this video and see if you can figure it out. There's multiple ways to think about it. The way I think about it is really to try to draw a right triangle where these points, where the line that connects these points is the hypotenuse and then we can just use the Pythagorean Theorem. Let me show you what I am talking about. Let me draw a right triangle, here. That is the height of my right triangle and this is the width of my right triangle. Then the hypotenuse will connect these two points. I could use my little ruler tool here to connect that point and that point right over there. I'll color it in orange. There you have it. There you have it. I have a right triangle where the line that connects those two points is the hypotenuse of that right triangle. Why is that useful? From this, can you pause the video and figure out the length of that orange line, which is the distance between those two points? What is the length of this red line? You could see it on this grid, here. This is equal to two. It's exactly two spaces, and you could even think about it in terms of coordinates. The coordinate of this point up here is negative five comma eight. Negative five comma eight. The coordinate here is X is four, Y is six. Four comma six, and so the coordinate over here is going to have the same Y coordinate as this point. This is going to be comma six. It's going to have the same X coordinate as this point. This is going to be negative five comma six. Notice, you're only changing in the Y direction and you're changing by two. What's the length of this line? You could count it out, one, two, three, four, five, six, seven, eight, nine. It's nine, or you could even say hey look, we're only changing in the X value. We're going from negative five, X equals negative five, to X equals four. We're going to increase by nine. All of that just sets us up so that we can use the Pythagorean Theorem. If we call this C, we know that A squared plus B squared is equal to C squared, or we could say that two squared ... Let me do it over here. Use that same red color. Two squared plus nine squared, plus nine squared, is going to be equal to our hypotenuse square, which I'm just calling C, is going to be equal to C squared, which is really the distance. That's what we're trying to figure out. Two squared, that is four, plus nine squared is 81. That's going to be equal to C squared. We get C squared is equal to 85. C squared is equal to 85 or C is equal to the principal root of 85. Can I simplify that a little bit? Let's see. How many times does five go into 85? It goes, let's see, it goes 17 times. Neither of those are perfect squares. Yeah, that's 50 plus 35. Yeah, I think that's about as simple as I can write it. If you wanted to express it as a decimal, you could approximate it by putting this into a calculator and however precise you want your approximation to be. That over here, that's the length of this line, our hypotenuse and our right triangle, but more importantly for the question they're asking, the distance between those points.

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Finding distance with Pythagorean theorem (video) | Khan Academy

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We are asked what is the distance between the following points. Pause this video and see if you can figure it out. There's multiple ways to think about it. The way I think about it is really to try to draw a right triangle where these points, where the line that connects these points is the hypotenuse and then we can just use the Pythagorean Theorem.

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Circle Equations - MathBitsNotebook(Geo - CCSS Math)

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MathBitsNotebook Geometry CCSS Lessons and Practice is a free site for students (and teachers) studying high school level geometry under the Common Core State Standards.

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Video- how to find the center and radius of a circle

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Video- how to find the center and radius of a circle

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How to find the center and radius of a circle in standard form

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Distance between two points | Analytic geometry (practice) | Khan Academy

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Use the Pythagorean theorem to find the distance between two points on the coordinate plane.

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VIdeo-features of a circle from the expanded form

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VIdeo-features of a circle from the expanded form

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We're asked to graph the circle. And they give us this somewhat crazy looking equation. And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x, and complete the square in terms of y, to put it into a form that we can recognize. So first let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here, because I'm going to complete the square. And then I have my y terms. I'll circle those in-- well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And then we have a minus 17. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need to review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2, square it to get 4. And that comes straight out of the idea if you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x, plus 2 squared. Now, we can't just willy-nilly add a 4 here. We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a, b, essentially, the point that makes both of these equal to 0. And that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2, and the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0, and the y value that makes this 0. So the center is negative 2, 2. And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2, 2. So our center is negative 2, 2. So that's right over there. X is negative 2, y is positive 2. And the radius is 5. So let's see, this would be 1, 2, 3, 4, 5. So you have to go a little bit wider than this. My pen is having trouble. There you go. 1, 2, 3, 4, 5. Let's check our answer. We got it right.

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Features of a circle from its expanded equation | Analytic geometry (video) | Khan Academy

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We're asked to graph the circle. And they give us this somewhat crazy looking equation. And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is.

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Video- write an equation in standard form for a circle

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Video- write an equation in standard form for a circle

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Writing standard equation of a circle | Analytic geometry (video) | Khan Academy

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Parallel & perpendicular lines intro | Analytic geometry (video) | Khan Academy

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Parallel & perpendicular lines on the coordinate plane

Parallel & perpendicular lines on the coordinate plane

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Parallel & perpendicular lines from graph | Analytic geometry (practice) | Khan Academy

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Sal classifies quadrilaterals by their properties (such as parallel sides).

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Practice-write an equation of a circle given radius and center

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Practice-write an equation of a circle given radius and center

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Write standard equation of a circle | Analytic geometry (practice) | Khan Academy

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Practice- graph a circle from an expanded form equation

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Practice- graph a circle from an expanded form equation

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Graph a circle from its expanded equation

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Given the expanded equation of a circle, graph the circle. For example, graph the circle whose equation is x²+y²+4x+8y+16=0.

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Classify quadrilaterals (practice) | Shapes | Khan Academy

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Classify and compare rectangles, rhombuses, and squares.

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Factoring quadratics: leading coefficient = 1 (article) | Khan Academy

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in class 2/19 & 2/20

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factoring trinomials 2/18

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handed out 2/18

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Handed out Mon. 2/17/20