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CAHSEE practice: Problems 43-46

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Problem 43. Mia found the area of this shape by dividing it into rectangles as shown. Mia could use the same method to find the area for which of these shapes? What technique can we use to figure out the area by dividing it into rectangles? So let's see. What do we have here? Can we divide this into rectangles? We can divide into a bunch of triangles, but we can't divide this into a bunch of rectangles. Or we could have done it this way as well. There's multiple ways to think about it. You could have it like this, and you can draw a bunch of triangles like that. But you can't divide this into rectangles. This one here, you can't divide it into rectangles. You could divide it into triangles and rectangles, but not necessarily just rectangles. And this here, I really don't know. It looks like there's some type of typo in my test. Because obviously I could divide both of these things right here. The way I'm looking at it, I see squares here. So obviously I could divide that into rectangles. But my guess is, if I had to guess, is that this was a triangle, maybe, before. If it was a triangle, then I couldn't divide this into rectangles. But maybe this was something like this. I don't know. I'm only guessing. If it did look like this, then we could divide it into rectangles the same way that she did up here, and figure out its area. So assuming that there was some type of typo in this problem, I'm going to go with choice D. If choice D was supposed to look like a figure like that, then you could actually divide it into a bunch of rectangles. Problem 44. Oh, I see. You know what these probably were? These were probably to show us that these are right angles right there. So that that's a right angle, that's a right angle. And maybe-- well, I don't know what else they're trying to show us. Who knows? You see what I'm going at. There's clearly some type of typo on that problem. 44. Simplify. x squared minus 3x plus 1, minus x squared plus 2x plus 7. So let's just write it out. We have x squared minus 3x plus 1. Then we have to distribute this minus sign. So minus x squared minus 2x, right? Minus times plus. Minus 2x. Minus times plus 7 is minus 7. So you have an x squared minus an x squared. They cancel out. Right? x squared minus x squared is 0. Then you have minus 3x minus 2x. Is minus 5x. And then you have 1 minus 7, which is minus 6. So you're left with minus 5x minus 6, which is choice C. Problem 45. What are the coordinates of the x-intercept of the line 3x plus 4y is equal to 12? They want to know the x-intercept. Remember, the x-intercept, if this is a line-- and this is a line. Let me draw the axes. Say the line looks-- I don't know what it looks like. Say the line looks something like this. The x-intercept is the value when y is equal to 0. The x-intercept is going to be some value x and 0. So let's figure out what this x-intercept is. So we can even look here. This is definitely not an x-intercept. This is a y-intercept. This is a y-intercept. In order to be an x-intercept, in order to be on the x-axis, your y value has to be 0. So the choices are either B or D. And so if y is 0, let's just solve the equation. So 3x plus 4 times 0 is equal to 12. Or we get 3x is equal to 12. That's just 0 right there. So x is equal to 4. So the x-intercept is 4 comma 0. x is 5, y is 0. So this is our choice right there. 46. Which of the following statements describes parallel lines? And parallel lines are just two lines that never intersect, that have the same slope an they don't sit on top of each other. So those would be two parallel lines right there. Same y-intercepts but different slopes. No, I just said, they have to have the same slope and they never intersect. So they can't have the same y-intercepts. So it's not that one. Same slope but different y-intercepts. That looks right. They're parallel, but they're not on top of each other. Or they have the same slope. Opposite slopes. If they had opposite slopes, they would intersect. Can't be that. Opposite x-intercepts but same y-intercepts. No, that's not necessarily the case. So our answer is B.