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Let's investigate a curve that has an infinite length yet having finite area: Consider a sequence of curves generated by starting with an equilateral triangle ...

Question

Let's investigate a curve that has an infinite length yet having finite area: Consider a sequence of curves generated by starting with an equilateral triangle whose side has a length of 1 (stage 1} Successive stages (stages 2,3,4,-progress by:Stepl: Divide each side into three equal partsStep2: Draw an equilateral triangle on each middle part:Step3: Divide each outer side into thirds_Step4: Draw an equilateral triangle on each middle part(a) Find the length Ln of the n-th stage(b) Show that

Let's investigate a curve that has an infinite length yet having finite area: Consider a sequence of curves generated by starting with an equilateral triangle whose side has a length of 1 (stage 1} Successive stages (stages 2,3,4,-progress by: Stepl: Divide each side into three equal parts Step2: Draw an equilateral triangle on each middle part: Step3: Divide each outer side into thirds_ Step4: Draw an equilateral triangle on each middle part (a) Find the length Ln of the n-th stage (b) Show that lim Lr (c) Find the area Aof the n-th stage Show that lim An 8/5) A1 where Az is the area of the figure in stage 1 Stage Stage 2 Stare 6x 5tage Stage



Answers

Helga von Koch's snowflake curve Helga von Koch's snowflake is a curve of infinite length that encloses a region of finite area. To see why this is so, suppose the curve is generated by
starting with an equilateral triangle whose sides have length 1.
\begin{equation}\begin{array}{l}{\text { a. Find the length } L_{n} \text { of the } n \text { th curve } C_{n} \text { and show that }} \\ {\quad \lim _{n \rightarrow \infty} L_{n}=\infty} \\ {\text { b. Find the area } A_{n} \text { of the region enclosed by } C_{n} \text { and show that }} \\ {\quad \lim _{n \rightarrow \infty} A_{n}=(8 / 5) A_{1}}\end{array}\end{equation}

All right, everybody. And welcome back to New Murad. This is Kevin Chirac. Let's consider a really interesting fractal pattern which occurs when you take unequal lateral triangle, and then you remove the inner thirds of it of each side, and then you replace that with an equal lateral triangle of similar length, and you begin to get something of ah, snowflake pattern is called coaches snowflake pattern. And so the next one would look something like that all the way on through tough, tough drawing to do. But it's going to be worth its. We can have something to reference as we're working through this problem. All right, so that should be pretty good right about there. So let's take a look at what's happening mathematically, Every time that we move this this pattern along, we're going to consider both of the perimeter as well as the areas we're going to start up with the perimeter. And let's just kind of keep track of this perimeter. Now knows that this primer is going to be an addition. So each of these iterations of the perimeter is actually gonna be almost like a partial some themselves, so you can think of these printers is being partial sums. So the first, the perimeter of the first picture, is going to be three times the length, which is going to be one. The pruner of the second picture is going to be well, how many sides are there? 12? Well, can about one Teoh 3456789 10 11 12 There, 12 sides Now, which makes sense because I've essentially replaced each of these sides with four other sides. Let's take a look. This side right here has been replaced with 1234 different sides. So now this is increasing by four all kind of make a little note to myself times for got me up to 12 sides. And then how bigger each of those sides will. Each of the sides are 1/3 so I multiplying by 1/3 to get Teoh that next value. So it's going to be 12 times 1/3. And since this is a partial some, we're gonna add that to the value that was above it. So all this together gives us a little bit of a pattern of what we're working with through here. So we can write the infinite summation from n equals one. What's my first term when my first time is going to be that three and then each time after them, we multiplying that by four to the end and that's gonna be the first part of this. The second part is going to be that initial value of one times a 1/3 to the end. We should make that as a and minus 1 1/3 of the N minus one minus one. There we go. And that's gonna continue on forever. We can simplify this up a little bit if we'd like and showed this is going to be three times 4/3 to the N minus one. That's gonna be our equation for the perimeter. At any value are this would be our total perimeter After an infinite number of generations, we wanted to make this a partial Some. Let's fix this up a little bit will make this and, um and this will be an M Good. Let's consider what happens. Where does this approach? As M goes off to infinity? So is m goes to infinity. What happens will notice. This is a geometric series and this would be my our value. But this Our value is quite large. It's bigger than one. The absolute value is bigger than one. And we know that when geometric serious have a our value a common ratio that is bigger than one. That means that this serious diverges. So this perimeter is actually going to continue to grow without bounds or diverge off to infinity. Okay, so the perimeter is gonna grow up to infinity. What's going to happen with the area? Let's take a moment. We'll go and clear up some space here. So what's gonna happen with the area? Well, let's look again at what the total area is. And this is going to be a partial summation. Any given point. So are partial summation. Our area for the 1st 1 is going to be 1/2 bs inside because I'm working with the triangle here won't have based times I This is an equal lateral triangle. So this height is going to correspond with a triangle, a right triangle whose side length is 1/2 because the entire triangles length was one, and that would make that That's high it up here is going to be route through over two with some quick trigonometry. Let's be Route 3/2. So the area of the first triangle is going to be 1/2 times the base, which is 1/2 times the height of route 3/2. What about this? Second area will notice that this second area includes all of the original area, but now it just has some additional area. A little bit of, ah drying error there. So it has all of the initial area with just some extra. So we have all of this. Red is present on the inside with some extra. So this is gonna be best representatives, a partial some. So we know that we're gonna have a one which is going to play a role that's gonna show back up. And then how much my adding back each time. Well, again, it's gonna be a triangle equation. So it's 1/2 times the base. Now, these bases are the little dotted lines we have inside a year, so the base is going to be 1/3. So the base of 1/3 time's a height, and the height is just gonna correspond with what occurs when I scale this when I scaled this high it down, it's gonna be a similar triangle s. So if the basis scaled, how do we change this? We change this by multiplying by 2/3. So this is similarly going. Teoh changed by a factor of 2/3 which is going to get us Route three over three. Okay, that should be enough for us to start developing our summation formula. So we're gonna start from n equals one. We're gonna go to infinity, and then we're going to again take those terms. Those initial terms. I'm starting off with about a value of 1/2 times 1/2 times Route 3/2. That's my initial value. And I'm changing that by a common ratio of 2/3 times 2/3 or four nights. So I'm changing that by 4/9. I'll put the proper notation here that it goes to the n minus one power. So let's simplify this up. We're going to get 1/2 times. 1/2 is 1/4. Tenzer 3/2 is going to be 3/8 times. This gives implied down to two. Open up. We cannot be. So let's change this never keep. That is 4/3 4/3 four nights tripping on my own words Here we got 4/9 to the N minus one good, and it is a geometric syriza's well, but let's consider that again. This is the common ratio, and this, in this case, the common ratio for ninth is less than one. So we know that this Siri's actually is going to converge. So if we were able to simply categorize off in our mind that this was just our initial our initial area right here, then we could say that this is all going to converge. VR geometric summation equation, which would be a 1/1 minus R and r are here is for nineths, which a little bit of simplification from here is going to get us nine times a one over nine minutes for is five. We have that are infinite. Summation is going to go to 9/5 of the original area

Nobody. Welcome back to New Murad. This is Kevin Chirac. If you enjoyed this video, I'm gonna give you a little bit heads of I to do a lot of prep work for it. So if you did enjoy this, it was absolutely gruelling to make and it would be much appreciated thumbs up at the bottom. Did you just kind of give me some feedback to let me know that you appreciated and that all of this hard work was worthwhile? So let's go ahead and talk about a fractal like patterns that is called coaches snowflake. And essentially, it's created by taking some triangle and you take out a middle third of the trying all notes that we took out a middle third of each of the sides. And then we added on equal lateral rectus unequal little triangle. Excuse me with that those dimensions and we did that each and every time you were going to get a very beautiful snowflake type pattern. And let's put let's let's imagine that we are looking for the perimeter of this will notice that each time that I do this iteration, I am gaining for times the number of sides and you can see that if you take this side right here, we'll make it green. I'm taking that and I'm splitting it into four sides now. So now there are four sides where there used to be once. I've got four times as many asides each time, so we can use that to kind of talk about how many sides we have each time. So here we've got when N is equal to one, actually, let's let's make this an N minus one. When n is equal to one, that means that I would have four to the end minus one. Ah, time like number of sides mawr that I am kind of adding on are multiplying on. So I should actually probably put a three right here to show that I'm starting with three. And then I was multiplying that by four to the end minus one, which would be times one. So it's have three. The next iteration I would have three times for, which would be 12. I would then have three times four times forward to be three times 16 or 48 total sides. Let's talk about the length of each side if I suppose that the original length was size s then That means that all of these new triangles have a length of size s divided by three. So I can write that as s times 1/3 to the end. Okay, so we can write that out and say that if, for example, that first side was equal to a length of one, that when it switch that we're gonna have many, many rectangles that are now 1/3. So let's check in real quick here and make sure that we did this correctly. If we have any is equal to one that would be one that revolution make this an N minus one as well. So when N is equal to one, I have sides that are length one. When it is equal to two. I have 1/3 toothy tu minus one to be 1/3 toothy one so that I would have one times 1/3. So I have each of these would not be 1/3. That looks right. And then I would do that again for the third picture, and that would get me side lengths of one night. That looks like that looks great. now to find the total side length. In other words, the total perimeter. We could just multiply on down, so it's gonna be a total of three, and then we're gonna have 12. Divided by three is going before and then 48 divided by nine is going to be 48 nineths, which we can simplify up a little bit. We can make this into 16/3, and we begin to recognize a little bit of the pattern here. We're gonna be multiplying by 4/3 each time we go over here. So times 4/3. That's because we are increasing the number of sides by three. But we're decreasing the length by 4/3 so we can write this out as saying that the perimeter on the 10th step is going to be equal to at that original side length. You could call it one or s and maybe put as just a little bit more generic. It'll be s times four divided by three to the n minus one. Now we want to consider what happens to this perimeter as we continue to grow this snowflake, we could look at the limit as n goes towards infinity that goes towards infinity of PM Well, we know that this s is going to be a constant. That's gonna be a constant right there. This 4/3 3 and minus one, though, is a common ratio that is bigger than one. So that means that if I raise it to a higher and higher power, I'm going to grow without bounds. So that means that our limit is actually gonna be infinity. In other words, we're going to continue to increase the perimeter and eventually have a snowflake with infinite perimeter. All right, now let's talk about the area again. I mention that the area is that the harder of the two problems. So if you don't mind taking just one second and putting out like in the bottom boxer on the button below this video, it be much appreciated because, as you can see, there's quite a lot of work that I had to do for this one here. But I'm happy to walk through it with you guys. Let's consider what happens with the area of this stuff, like I've got the triangle and it's going to be increasing as I had on mawr. Many triangles each time. And so for my first triangle, I decided to just simply say that that is, that is gonna be my n equals one. And how many new triangles do I have here? I have zero new triangles, cause I'm just starting that off. That will be my first value. How many new triangles do I have by the time that I get to my second picture? Well, I'm going to be introducing three new triangles. How outlined them in green green triangles. One right there, there and there. How many do I have in the third? Well, the third. I've got one new rectangle for each of the many, many sides that I've created. And if you take a look back at the work that we just did for a perimeter, I'm going to be having four sides that are going to be appearing for each one of by iterations here. Progress. It gives you a little bit of indicators to where this four to the end minus one comes from as well as this four on the bottom. That kind of offsets it until the third iteration when it really starts to appear. So how many new triangles do I have in my second picture? Well, I would have three. So three divided by four to the end, minus one. That would be four to the first power and 3/4 times forward. Just be three. What about into the third power? That would be four toothy three minus one. So four square to be 16 divided by four would be, uh, four. And then times three would be 12. So I would be introducing 12 new triangles. Let's talk about the area that goes along with these new triangles. Well, if I consider my initial area to have an area a zero right in here, I've got a zero. How large are my might? Is the combined area for my new triangles. We can see here that I have divided up the first triangle into corresponding equilateral triangles that go along with these little blue ultra two minutes. Little, the little additions that we have here and we can see that each one of these little new triangles that appears is going to be the same size as one of these triangles I've drawn in. And if we count how many there are there 123 for 5678 There are nine triangles, so each one of these little blue triangles that's being added in over here, including this one in this one, are 1/9 of the original. So we can see that That's where I'm getting this idea of being 1/9 of the original. That's the area of the new triangles. Each new triangles, 1/9 of the original. So if we look at the area for this first generation is going to be one night. So a zero one night to the N minus one would be one night. So a 0/9, you can't say. And then if we did this again, it would be a zero toothy one minus nine to the second power, said the A 0/81. I can see why did a little bit of a head work here for us? So what's the total added area? Will? The total increase in area is going to be a combination of each of these stages, so we can kind of just multiply it on through and look at that 3/4 that four of the n minus one and that one night to the n minus one and kind of combine it all in. So, for example, in this second picture, I have added in 1/3 of my original area, we can see that by taking a sub zero divided by nine, multiplying that by three and seeing that we have a 0/3, so 1/3 of my original area has been added in. And let's check, we have 123 triangles and that would correspond with Let's see 123 triangles over here, which looks like it's about 1/3 cause if I colored this in with green and this and with red, those would all be three triangles. So I've added 1/3 of my initial area in the first iteration. What about the second generation? Will the second iteration is going to be 12/81 which you can simplify a little bit to be four over. Would that be 27 for 27 So four a 0/27 and this process is going to continue on. But now I'm gonna take a moment and just going to clear out some space and show you how a congenital allies this was something equations that over there on the right. All right, so now let's take a look at this total added area that we were able to calculate also grueling Lee. Well, now we can talk about that. This the total added area. Let's look over the total area will be I'm just gonna call that capital A So Capital A is going to be our original area we've started with, plus all of these additions. There's gonna be a zero times 3/4 times for over nine to the N minus one. But this was an iterative process. So we actually need to interrupt this this notation just a little bit and talk about this being a repeating edition. So we're eating this addition over and over again, And this started from an equal to up until infinity and we had our a sub zero here. All right, looking good. How can we take this a little bit further? Let's say a sub zero when factor out are a subzero from this second term. So it's gonna be three a sub zero over four infinite summation and equals two to infinity of the four overnight to the n minus one. Now we can see this is resembling a geometric structure. But I need to change my indexing in order to make this something that I can apply our geometric formula for. All right, So let's bring this up here and let's now play around a little bit a nice trick that you can dio if you can actually just pull out some of this and do it already in order to lower the index. Let me show you what I'm talking about. If I were to pull out a for over nine, then that would be three a zero times four over four times nine. And if I factored out a four night from everything, it's the same as me dropping the power of the four name that's left on the inside. It's not 4/9 toothy and minus two. Why would I want to do that? Because now I can just change my index accordingly. So now I can write this out as a subzero plus over some opportunity for simplification here will take away those fours will take away that three. Reduce that down to three. So now I'm left with a subzero over three times the infinite summation and equals one to infinity of 4/9 to the N minus one. I just dropped my end down one. Since it's an arbitrary index, Well, that's just great, because now I can finally apply my infinite summation equation. So I'm gonna get a 0/3. And then all of that inside can become my summation equation where I have my first value, which is one over one minus. R. R is 4/9. That's going to be equal to a sub zero, plus a subzero over three, multiplying through by nine in the fraction and that will get us nine on the top and then nine minus four on the bottom will give us 9/5. Almost there. That's gonna be a sub zero running out some space. Let's make some space here. All right, So now we get a sub zero Plus what it would leave up. We had a subzero plus all of that. There is a zero plus. It's a zero times three over five, right, That looks good. Last step will do five a subzero over five, a subzero plus three, a subzero over five. Correct that There should be No way! Subzero in the denominator there. That would give us eight fifth of our original area. So this is going to converge to eight fits of our original area. Oh, my God. That was an exhausting question. Thank you for bearing with it through me. If you guys don't mind. Please do. Put that.

So and spoke of this crash. So when you use Catholic, the the air off ho bags sign nerves. You know, between Sion axe in the X axis and starting bomb you Hello, Our two. It's the first. Let's story Grab it. What is it? Science seven. Lively in years. Originally, we need to Kathleen the arrow make a speech. So although we calculate IDs So according to question first you're the right destination, So basically we can know the arrow. Oh, news people, I bone and, uh, we have to remember the numerous invasions. Say, hey hard And here c k is home Hi o r times There should be great in our labs is hi more now Community Catholic of you, you hear sign Hi or I years Hi. First we need to carefully this summation of capital is Oh, and we need to do is use listens it killing dysfunction So basically these role equals in the apparently here, uh, pi work who is h in essence, it giving these pressure So basically we'd rather live the summation. Oh, all saying aye or toe miles barbs and then the wind Fine Four terms. This is you. It's Sam. We need to carry her to the limit. My own approaches you really? So or all this term here became We can get the liminal this time first where we can go. So you lied to Cano is full term. Is was a hero, Myers Right in the these ones cause on euros while this is there, that term goes Want to basically wait in the we only have this part in Ah mo Canady Thea are far apart. It's a brand new baseball is go zero and the, uh the bottom are possible This of all this time leaning that we can't Catholic, huh? You the law But basically way gather They believed you off this one on his own family. This you know we can cackling this evening on the Ganges So we finally the limit you

Okay. What we want to do is step through the process, uh, finding the area under the curve that is represented by why is equal to sign of X on the close interval from zero two pi over two. So the first thing I'm going to use I'm gonna actually kind of graph what this looks like. Um, and so we're going from zero to pi over two. Um, and we know sign starts off it zero and is a one at pi over two. So that's kind of what the graph looks like. And so what we want to do is to, um, partition this interval from zero to pi over two into in seven devils. Okay, so we know that, um, Delta X it okay? And we want the sub intervals to be equal in with. And so delta X of K is beam honest A over in. So, for our purposes is gonna be pi over two minus zero over in which is pie over to in. And what we want to do is to calculate the upper some, and so we know that if we're gonna be doing the upper some, the upper sound would be um, these rectangles right here. And so we want the function evaluated at each of those points that hit of the rectangle that hit that represent the height of the rectangle that hit the curb. So ceases que is a plus for the upper is, um, times a plus que times Delta X. Okay, so in this case is going to be, um, pai que over to in okay. And so that up or some is represented by the some Aziz k equals one to n, uh, sign Ah que pi over two in times pi over two in which is going to be equal to pi over two. In times this some as k equals one Teoh. End of sign of que pi over two in. Okay, so that is the up or some. Now the issue is okay. Um, how do we calculate that up or some. And so we need to use, um we know we have this formula that says Sign of age plus sign of to age plus sign of, um three h and so on. Plus sign of M H is equal to this, um, co sign of h over to minus co sign of M plus 1/2 times h divided by, um, divided by to sign of H over two. Okay. And so for our purposes, um, for our purposes are purposes. Um h is actually equal Teoh pie over to in. So when we had that the upper some was equal to hi over to in times of some k equal one to end of sign of que pi over two in that actually is represented by this string here. So if I actually expanded out this some I would get a son that looks like this, but h would be pi over two in and therefore that some is actually equal to this'll. So my upper some is pi over two in times, um, co sign of pi over four in because when I put in a chit of pi over two in over here, minus co sign of And then, of course, in this case, AM is actually in in plus 1/2 times pi over two in divided by to sign of pi over foreign. And so let's go ahead and kind of simplify that a little bit. So the upper some is equal to pi over to in Times co sign for not for Excuse me. Um, pi over. Foreign minus. Now. We have co sign of pi over two. Plus, um, pi over four in divided by to sign of pi over foreign. Okay. And then you should recognize that this is actually going to be a co sign that has shifted pi over two. And so that is that's gonna be equal to negative sign of, um, pi over foreign. So my upper sime is pi over two in times, um, co sign of pie over four end, plus a sign of pi over four end, divided by two. Sign of pi over for in. And so there is my upper some. Okay, so now what we want to do is to, um, find the limit as in goes to infinity. Um, but what I'm going to do is because, um let's let's see this. So here. Now we want to know because the area is, um, represented by the limit, as in goes to infinity of, um, pi over two in, um, times the sun. No, it wasn't the some we've changed to some, um of co sign of pi over four in plus sign of pi over four in divided by to sign of pi over for in. Okay. And so another way I'm gonna write this is I'm actually going to go back to help us out. Um, that, um if in ghost infinity this h value or you can say Delta X is out, you know, Also, the Delta X goes to zero on dso to simplify all the writing. I'm gonna change this to the limit as h goes to zero and where h is equal to pi Over to in, um of H um Times CO sign of 1/2 age plus sign of 1/2 h over to time Sign of 1/2 h. And you notice that when I insert zero in for H, I try to evaluate it. I get that in, determine it form, I get 0/0. And so now we have to be creative of how we can evaluate this. And so what we need to do is to, um, um use the hospital's rule and do some simplification. And so the first thing I'm going to do is to help us out is I am going to do the limit as h goes to zero, uh, eight times, co sign of 1/2 age over to sign of 1/2 h, um, plus the limit as h goes to zero of. And then when I break this up into sign of 1/2 h and then I also have an H multiplied here over on to sign of 1/2 age. Um, then I notice what happens here is this term the sign of 1/2 h is cancel out. And when I now do a direct substitution of H equals zero, this term actually goes 20 And so really, what I have to worry about is, um, this first term here and once again, this first charm is 0/0, which is still that indeterminant form. And so I'm gonna have to use low hospitals rule for that. And so let's go ahead and do that. And so, um, if we know that this is going to be equal to and let me go ahead and write that down first so we can remember what we were working with and this was a to sign of 1/2 page. And so remember LA hospitals rule says that we can take the derivative of the first of the numerator and the derivative of the denominator separately. So when I do that, I get the limit as age goes to zero. Um uh co sign of 1/2 H, um, minus h times sign of one note. And this is gonna have to be a h 02 because of what's inside the sine function, cause I'm taking a derivative of the co sign. Um, and then this is going to be the denominator is just gonna be co sign of 1/2 eight. Okay, so now let's see what happens if I do direct substitution. So basically, I would get co sign of zero minus zero times sign of zero, which is just zero over co sign of zero, which is one so that is represented of the area. Um for why equal to sign of X on the close interval from zero to pi over two


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Given that a solid is formed by z 2 2 and x' +y +2' < 20 . Find the volume of the solid. marks)...
5 answers
How many mL of 0.6 M HCl do you need to add to a 100 mL aliquot of 333 mM butyrate buffer at pH 5.5 to decrease the pH by 1 pH unit? The pKa of butyric acid is 4.82How many mL of 0.2 M HCl do you need to add to a 100 mL aliquot of 200 mM formate buffer at pH 4.2 to decrease the pH by 1 pH unit? The pKa of formic acid is 3.74
How many mL of 0.6 M HCl do you need to add to a 100 mL aliquot of 333 mM butyrate buffer at pH 5.5 to decrease the pH by 1 pH unit? The pKa of butyric acid is 4.82How many mL of 0.2 M HCl do you need to add to a 100 mL aliquot of 200 mM formate buffer at pH 4.2 to decrease the pH by 1 pH unit? The ...
4 answers
A sample of a substance is weighed and found to have mass of 2.198 g. If the substance has & molar mass of 173.1 g/mol, how many moles of the substance are in the sample? Use at least 3 significant figures in your answer: Your Answer:Answerunits
A sample of a substance is weighed and found to have mass of 2.198 g. If the substance has & molar mass of 173.1 g/mol, how many moles of the substance are in the sample? Use at least 3 significant figures in your answer: Your Answer: Answer units...
5 answers
Solyet petne L DETAILS 1 1 1 LARCOLALG1O 5.4.035_1 Januai1
Solyet petne L DETAILS 1 1 1 LARCOLALG1O 5.4.035_ 1 Januai 1...
5 answers
5. In the reaction of hydrochloric acid and sodium hydroxide,100.00 ml of 0.1234M hydrochloric acid reacts with 50.00 ml of0.5432M sodium hydroxide. Will the final solution be more acidic ormore basic? How many mole of excess acid or base will remain? Whatis the final (H+) and what is the final pH?
5. In the reaction of hydrochloric acid and sodium hydroxide, 100.00 ml of 0.1234M hydrochloric acid reacts with 50.00 ml of 0.5432M sodium hydroxide. Will the final solution be more acidic or more basic? How many mole of excess acid or base will remain? What is the final (H+) and what is the final ...
5 answers
65. If 0 is the angle between vectors a and b, show thatprojab projb a (a b) cos? 0
65. If 0 is the angle between vectors a and b, show that projab projb a (a b) cos? 0...
5 answers
Gme an erplyakon 05 ko M# Kzachw_Or9gn (0OS) 15, Mlso expkkin how #ey afkak # Q Orospoks[ m F wken Hxels kd akp ayc Ecla 6 sA umdv 8^ U6 } Vet I between 38onm_"rdslona
Gme an erplyakon 05 ko M# Kzachw_Or9gn (0OS) 15, Mlso expkkin how #ey afkak # Q Orospoks[ m F wken Hxels kd akp ayc Ecla 6 sA umdv 8^ U6 } Vet I between 38onm_"rdslona...

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