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The ancient Greeks used a geometric method for completing the square in which they literally transformed a figure into a square. a. Write a binomial in $x$ that represents the combined area of the small square and the eight rectangular stripes that make up the incomplete square on the left. b. What is the area of the region in the bottom-right corner that literally completes the square? c. Write a trinomial in $x$ that represents the combined area of the small square, the eight rectangular stripes, and the bottom-right corner that make up the complete square on the right. d. Use the length of each side of the complete square on the right to express its area as the square of a binomial. (THE IMAGES CANNOT COPY)

In this problem, I'm writing the reactions. So just look at it carefully. OK. FECN six came food, FT CN six plus six. He S CN k s c N will react to form, get to the f e SCN six and now I'm going to write the second reaction. They just look at it carefully Or FECL three. That's pretty careful. F E CN six, then it will give 54 F e CN six plus two old KCL. So option age correct here.

Welcome. We're here today to determine the area of each small square in the corner of the second figure. So second figure references the figures in your textbook on this white board here today I have simply added on to the original figure I already drew for exercises 61 through 64. So essentially, we are trying to calculate the area of each of these small squares, and we know that each of these small squares is gonna have the same exact area. So let's just focus on the area that I've colored here in red today. So as we can see, let's track all the way from the left hand side. Let's track the with or that rather the length of, um, one of these small squares. We can see that one of these small squares has a length of one because it is just simply tracking over to the side off the square. So we know that East Square has a length of one. Similarly, if we track the with of each small square, you know that each small square has a width of one as well. So for calculating our area, our area is gonna be equal to with times length, and in our case, that is equal to one times one. So the area of each small square in the bottom right of the figure of each one of these small squares is equal to one square unit. I hope this helps and keep at the great work.

Welcome. We're here today to determine the area of each strip in this diagram. This diagram is also reproduced in your textbook for work exercises 61 through 66. So each strip I'm gonna color one of the strips in blue so we can focus on just one. So let's just say we're focusing on this one strip. We know, um, that we added these strips and each trip has a width of one. That's what I've labeled right up here. This is our with. So each strip is essentially a rectangle, and we know that the area of erecting with the area of a rectangle is gonna equal with times length, length, right? So we can also see that our length is just the same as the length of this of this box right here. So our length for each strip is actually just gonna be X because it is the same exact, um, same exact length. Ah, drawn lengthwise for X. So given that we know are with our wit is gonna be equal to one one, and our length is equal to X. Each strip has an area off X units X units square. Um, I hope this helps and keep at the great work

In this question, a Squire is shown in red is one unit on a side. A second Blue Squire is constructed inside the first Squire by connecting the middle point of the first one. A third. Green square is constructed by connecting the middle point of the sites of the second Squire and so on. In the first part we need to find this formula for the area in of the Annette inscribed Squire. And the part B. We need to consider the sequence SN where S. And is close to even plus eight plus so on A. N. And we need to calculate the numerical value of the first 10th terms of this sequence. And in part C we need to make a conjugate about the convergence of sN. So let's see how to solve this question. Consider that consider the side of the outermost esquire. Is A not as equals to one. Four. Yeah the first he described esquire. Yeah let mhm. E one is the area and even is the side of this inscribed Squire. Similarly consider that four. The second inscribed. Yeah esquire. Let A two is the area of this square and a two is the side of this Squire in this manner we can consider the area inside of other Squires. Now to find the value of a small even let us consider a small part of the outermost Squire as shown below. So this is the small part of the outermost Squire and this is side even And this side is a note by two and this side is a note by two. No apply the bicycle store. um apply the pythagoras to your um to calculate even therefore even will be called to even esquire will be called to He notice required by four plus He not squired by four hands. Even Squire will be close to, He not required by two. Yeah, we know that the area of these queries vehicles to the Squire of the side. Therefore we can write area even, it equals two even Squire. And since even classical school you're not escorted by two hands even will be calls to He is not required by two. Similarly for the second in described esquire, they will have a two Esquire. Is the calls to a not esquire by four and area It will be equals two. He not required by to esquire similarly four. The third in described esquire, we will have A three square. Is the calls to A not required by eight and area a three is equals two. A not Squire by 2 to the Power three. No, we can observe a pattern between the areas even A. Two and a three. Therefore the reference formula for the area of the Senate. Area of the and it in described esquire can be written as E. N equals two. Not to the power to upon totally power And so this is the final answer for the part A. Now let's move to part B. Part B. The numerical value of 1st 10 terms of the sequence can be calculated as Area even equals two. He not esquire by totally power one and this will be equals two. Once acquired by two to the power one is the calls to 0.5. Area A two is the calls to a not Squire by to Squire and it will be called to one square by two square as equals to 0.25. Area A three X equals two. A notice choir bye took you. And this will be calls to one square by To Cube is equals to 0.125. Area a four equals 2. He not squired by two to the power for and this will be close to one to the power to upon two to the power for And this will become 0.0625. No the area 85 will be called to. He not required by two to the power five and this will be cause to one squired by two to the power five. And the further calculations provide a five physicals to 0.03125. Now area a six physicals too. He not required by 2 to the power six and this will be close to One square by totally power six equals 2 0.015 6 to 5. Now area a seven is equals two. He not esquire by two to the power seven which will be called to one inspired by two to the power seven. And this will be close to 0.0078125. No, the area A eight will be called to a not esquire by two to the Power Aid. And this will be calls to One square by totally power it is equals to 0.00 390 6 to 5. Area 89 will be close to he is not required by two to the power nine. And the sylvie calls to One square by totally Power nine. The further calculations provide, Area a nine is close to 0.0019 53125. Now area eight and will be calls to a not squired by Two to the power 10. And this will be calls to one is acquired by totally power 10 And this will be equals to 0.0009765. So these are the 10 terms of the sequence. And this is the answer for part B. Now let's move to part c part see. Mhm Yeah, the conjecture. How about the convergence of the S And he recalls to He went plus a two plus A three and so on. Up to E N. Yeah. So let's see how to solve this. We can write limit and tends to infinity A.N. two limit and tends to infinity. He not required by 2 to the power end. Therefore limit and tends to infinity Hey and it equals two limit and tends to infinity bon upon 2 to the power and since in articles 2, 1 on for the solving we can do it limit. Okay. and tends to Infinity one upon limit and tends to infinity totally power and yeah, no substitute infinity on the place of end. So we get limit and tends to infinity A N a Z equals to burn upon infinity and this will be equals to zero. Therefore it can be concluded that mhm the sequence mhm converges 20. So this is a final answer for this problem. I hope you understand the solution. Thank you.


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