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We're on problem 221. And let's see, they have-- is all of that in the denominator? Well, I guess it is. Let's see, they wrote 1 over-- and I'll write it the way they did it. 1 over-- and they have this-- you know, I think it's got to be 1 plus. I think it's 1 plus. The way they wrote it is weird. 1 plus 1 over 1 plus 1/3. This looks a lot like another problem that we did. But anyway, I'll do this one. So let's see. If we do 1 plus 1/3, that's 3/3, plus 1/3, so that's 4/3. So then that turns into 4/3, 1, 1 plus, 1 over that, and then you have 1 plus that. And so then this equals 1 plus 1 over 1 plus-- what's 1 over 4/3? Well, that's the same thing as-- 1 divided by 4/3 is the the same thing as 1 times 3/4, and that's equal to 3/4. And what's 1 plus 3/4? Well, that's 4/4 plus 3/4, right? 4/4 plus-- so that's 7/4. So then that becomes 1 plus 1 over 7/4, which is equal to 1 plus 4/7, and that equals 7/7 plus 4/7 is equal to 11/7. And that's choice D. The way they wrote it was weird. They wrote it like this, 1 plus down here. But I assume that this is what they wanted, what I originally had done. Problem 222. In the rectangular coordinate system above-- well, let me draw it. That's the y-axis, that's the x-axis. y, x. And they have a point out here called point p, and they say that's at 4. x equals 4, y equals 0. In the rectangular coordinate system above, if point r, not shown, lies on the positive y-axis, so it lies someplace over here, and the area of triangle orp , so this is point o, and r is going to be some place on this, is 12, what is the coordinate of the point r? So the coordinate of the point r, let's just call it-- well, the x-value is going to be 0, right? And this y-value, I don't know, let's call it x. No, let's call it y. That's a better variable name for the y-value. So they essentially want to know what the area of this triangle right here is. Or actually, they want to know what y is, given that the area of that triangle, they tell us, is 12. The area of this triangle is 12. So what's the area of this triangle? It's the base times the height times 1/2. The base is 4, right? This is x is equal to 4. So it's 4 times the height. Well, the height is going to be y, right? Because the coordinate of r is 0 comma y. So 4 times y times 1/2-- if you don't do the 1/2, you're getting the area of the whole rectangle-- is equal to the area of the triangle is equal to 12. So you get 2y is equal to 12. y is equal to 6. So the coordinate of r is 0 comma 6. And that is choice-- oh, they just ask what is the y-coordinate of point r. So that's 6, and that's choice B. Problem 223. Car A is 20 miles behind car B. So let me see if I can draw that. So this is A, this is B-- and it's 20 miles behind it; I don't know if that's relevant-- which is traveling in the same direction along the same route as car A. Car A is traveling at a constant speed of 58 miles per hour, so car A is at 58 miles per hour, and car B is traveling at a constant speed of 50 miles per hour. Fair enough. How many hours will it take for car A to overtake and drive 8 miles ahead of car B? So they don't want just car A to catch up. They want car A to catch up and go ahead 8 miles, right? And drive 8 miles ahead. So another way to think about it is how long will it take for car A to go 28 miles further than car B? And the easiest way to think about it is just what their differential is, right? Car A is going 8 miles an hour faster than car B. So the relative velocity of A relative to B is 8 miles an hour. And you could say that, well, I don't want to use fancy notation, but car A, if you're sitting in car B and you looked at car A, you'd say, oh, it's going in my same direction, but I'm stationary and it's going at 8 miles per hour. So the question essentially says if I'm going at 8 miles per hour, how long does it take to go 28 miles, right? Because I have to catch up to 20 miles with B, and then I have to go 8 more miles. You could just pretend like you are in car B and you think that you are stationary. You ignore the fact that the world is speeding behind you. And you just see car A. You say, oh, car A must be moving at 8 miles an hour in the direction in which my car is pointed. So, anyway, if you have to go 28 miles, and you're going at 8 miles per hour, distance divided by rate is equal to time. That's equal to 3 and 1/2 hours, which is choice E. Next question, 224. For the past n days, the average daily production at a company was 50 units, so for n days, 50 unit per day average, if today's production of 90 units, so we have 90 units, and that raises the average to 55 units per day, what was the value of n? OK, so the old average would have been n times 50-- and this is kind of obvious -- divided by n is equal to 50, right? Now, if I want to average in this 55, what happens is I get to n times 50, right? That's the sum of the n days before, the sum of the total production for the n days before. And what I want to do is I want to have the production for today, and that is 90, plus 90, and now I'm going to divide by n plus 1 days, right? And now the average is 55. So we just have to solve this for n. So let's see what we can do. I'll do it in a different color. We get 50n plus 90 is equal to-- multiply both sides by n plus 1-- 55n plus 55. So let's subtract 50n from both sides. You get 90 is equal to 5n plus 55. Subtract 55 from both sides, 55 to 35 is equal to 5n, n is equal to 7. And that's choice E. Problem 225. OK, They write x plus 1 over x minus 1 squared. And they say if x does not equal 0 and x does not equal 1-- and that's good, because if x equaled 1, we'd be dividing by 0 here-- and if x is replaced by 1/x everywhere in the expression above-- so x is replaced with 1/x everywhere in the expression above-- then the resulting expression is equivalent to-- I don't know, let's just do it. So if we get 1/x-- we're just replacing x with 1/x-- plus 1 over 1/x minus 1 quantity squared, and that equals -- let's see. This is the same thing as 1/x plus x/x, right? 1 is the same thing as x/x, And this 1 is the same thing as x/x, right? That's an x/x. So then this equals 1 plus x/x over 1 minus x/x, everything squared. And this inside of the brackets, that's the same thing as 1 plus x/x-- instead of being divided by this, we could multiply by the reciprocal-- times x over 1 minus x. I just reciprocated this and multiplied. And this is all still in the squared . Now these x's cancel out, and lo and behold, we get 1 plus x over 1 minus x squared, which is the same thing as x plus 1-- well, let's think about it. x plus 1 over 1 minus x squared. And see, that's not one of the choices, right? But let's think about it. 1 minus x, that's the same thing as-- just let me rewrite it. This is the same thing as x plus 1 divided by-- now this 1 minus x is the same thing as x minus 1 times negative 1, right? And we're going to square all of this. Well, this over here, this is the same thing as x plus 1 over x minus 1 squared times 1 over negative 1 squared, right? If you take the square of a product, you could-- this is the same thing as that times 1 over negative 1. So that turns into this, and this, obviously if I'm squaring another number, this just becomes 1. So I'm left with x plus 1 over x minus 1 squared. And that's exactly what we started off with, and that's choice A. And that's the trick, to realize that this is just a negative of minus 1. And you can take a negative 1 out, but we're going to square it anyway, so it's going to turn it back positive. Next question. In choice E, actually, in that one, they tempt you by taking the negative outside, but you can't take the negative outside the square. You're squaring it, so the negative will disappear. Anyway, problem 200. Oh, I'm all out of time. See you in the next video.