So what is the radius in this case then?
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So what is the radius in this case then?
Partial Gamification
Banned
what you found:
\sqrt(p(\theta)^2 + p'(\theta)^2) = r
Gives: 1/2 int(p(\theta)^2 + p'(\theta)^2 d\theta) {d\theta is the differential} [edit]
matching the given, considering the limits of integration.
Ohhh, okay...but the other part that was bothering me is that I get p(\theta)^2 + p'(\theta)^2, but it should be p(\theta)^2 - p'(\theta)^2. Is that just a typo in the problem?
cpp_is_king
Member
i just realized an issue. if you use that method to obtain r, then you would not be integrating over d theta but rather some other angle. not sure how to relate d theta to the x and y coordinates
Edit: Or now that I think about it, perhaps I'm wrong again and you're doing it correctly, but it's actually a typo.
Partial Gamification
Banned
No, I am not going to assume there is typo. Your teacher "doesn't like the notation," so I must assume there are no errors as she has taken the time to decide the notation sucks. The difference, as opposed to the sum, is likely going to come from the trigonometry between the two triangles. You don't want to use theta, but rather the angle that leads to the point where the line is tangent to the curve. I feel like all the pieces are here but cannot put the together properly. In all honesty, I didn't realize the integrands were different at first but even the closeness of the answer leads me to believe that there is only one step missing. I believe that step involves the other angle and some trig (I can imagine you are reading this like "no shit sherlock," but I don't have anything hidden up my sleeve that would part the clouds here).
Here is a general rule for your situation, rigor and I are not on amicable terms right now:
Establish the convexity of a given set X
The set is one of the recognizable (simple) convex sets such as
polyhedral, simplex, norm cone, etc
Prove the convexity by directly applying the definition
For every x, y 2 X and 2 (0, 1), show that x+(1 − )y is also in X
Show that the set is obtained from one of the simple convex sets through
an operation that preserves convexity
SniperHunter
Banned
what math class is that troll?
Differential Geometry.
And thank you guys for all the help by the way, it all helps with this subject.
so hey yo guys, so i'm in introductory algebra class and i have a question...
im pretty confident with solving linear equations or whatever, ending up with something like x = 3 at the end, yay.
but when i end up with 0 = 3, what is that? or 3 = 3?
i remember professor saying something about it in class, like one is "no solution" and one is "all real numbers" or something, but which is which? i can't find anything in my textbook about it.
im pretty confident with solving linear equations or whatever, ending up with something like x = 3 at the end, yay.
but when i end up with 0 = 3, what is that? or 3 = 3?
i remember professor saying something about it in class, like one is "no solution" and one is "all real numbers" or something, but which is which? i can't find anything in my textbook about it.
Partial Gamification
Banned
A system of Linear Equations will have: one, none, or "infinite" solutions.
Consider that each equation is a defined relationship and you are seeking the values that satisfy that relationship (inclusive of all equations considered). I reccommend setting up the a "coefficient matrix" and finding the "reduced echelon form." It might seem a little alien, if you don't know what I mean; but it saves time and keep s things clean (depending on the linear system).
[edit] maybe graphically is a better representation. Think of the graphs of each equation and you are seeking where they intersect. A "linear equation" is an equation for a line. Think how two lines can cross, be parallel, or exist "on/as the same line."
I think coefficient matrices and ref will just further confuse someone in introductory algebra
Go with the graphical representation. You can represent each side of the equation by a line in the x-y plane. That could help.
If you end up with 3=0, the answer is "no solution". It should be easier to see if you take it a step back, lemme show you what I mean:
Say you start with
3x 7 = 6x 6 3x
You simplify to
3x 7 = 3x 6
Now instead of taking this to
-7 = -6
which would be comprable to the "3=0", just look at the equation. Could there possibly be any number for x such that that equation would hold true? It should be obvious that the answer is no, you can't take the same number on each side (3x), subtract different numbers and end up with the same result.
If your final line is something like 3=3, the answer is all real numbers. Let's look at
4x + 4 x = 5 + 3x 1
Then simplify,
3x + 4 = 3x + 4
Once again we could take this one step further to get something like 4=4, but I find it's more helpful to look at the line above: we have the exact same figure on both sides of the equation and x is still in our equation. No matter what x we choose, performing the same operations on it on both sides will give us the same result. Thus, the solution is all real numbers.
Hope this helps.
Very helpful Ander, thanks man!!! I've got a test tomorrow, feel pretty confident about most of it. I haven't been in a math class in like 10 years, I never even passed a math class in highschool .. the fact that i'm like six weeks into this one and doing pretty well is cool.
Sincere congrats on that, I'm sure it'd be hard to drop back into a subject like that. I know I haven't taken a biology class in 5 years and I'm sure going back into one now would terrify me. Good luck on the exam.
Thank you.
I've since looked into the material offered in this thread and it has helped immensely.
Partial Gamification
Banned
Yes but it is such a powerful technique I don't think it hurts to toss it out. You did a great job breaking it down, and I will concede that time will not be best spent between now and the test setting up single variable coefficient matrices (good luck dude).I think coefficient matrices and ref will just further confuse someone in introductory algebra
On a side, there is a great untapped potential for educational games that teach fundemental concepts, and are fun. Algebra is one of the first stumbling blocks for students. Tablets are making their ways into classrooms and the apps are not there. Perhaps they are around the corner; still, the last major games in schools were "Oregon Trail" and "Where in the World is Carmen Sandiegio" (US). I claim the holy grail of educational gaming, in the near future, will be some game that engages a diversity of learning styles through challenging and rewarding play while at the same time driving a deep understanding of mathematics into the student.
Meh, no problem. You must work for a better World now or be damned to an eternity of confusing notation!
Partial Gamification
Banned
cpp_is_king
Member
Here's a problem that has me stumped:
Find all functions f : Z -> Z with the property that for any 3 integers a, b, c such that a+b+c = 0, the following equation holds:
f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(a)f(c) + 2f(b)f(c)
I've made some significant progress, getting so far as to find an explicit answer for f(x) for all x being a power of 2. And then some other random facts about f for certain types of inputs, such as an expression for f(a+b) in terms of f(a) and f(b). But nothing that puts everything together. I can show what I've done, but I don't want to lead anyone down the wrong path in case I'm approaching it wrong, so let's see if anyone else can come up with something first.
Edit: I think I finally got the answer. Or at the very least I got an answer which appears to work, and I'm 95% sure there's no other non-trivial function that also satisfies the requirement.
Find all functions f : Z -> Z with the property that for any 3 integers a, b, c such that a+b+c = 0, the following equation holds:
f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(a)f(c) + 2f(b)f(c)
I've made some significant progress, getting so far as to find an explicit answer for f(x) for all x being a power of 2. And then some other random facts about f for certain types of inputs, such as an expression for f(a+b) in terms of f(a) and f(b). But nothing that puts everything together. I can show what I've done, but I don't want to lead anyone down the wrong path in case I'm approaching it wrong, so let's see if anyone else can come up with something first.
Edit: I think I finally got the answer. Or at the very least I got an answer which appears to work, and I'm 95% sure there's no other non-trivial function that also satisfies the requirement.
Partial Gamification
Banned
I played around with this briefly, just rewriting the equations. Are (a,b,c): distinct or is there only the restrictions: a+b=-c or a+c = -b or b+c=-a ?
Is the given map defined, bijective... etc?
This is a good one to think about, I'm sure I will be distracted by it at the wrong time today.
I decided to google it, matholympiad.org discussed at link. The internet is a crutch.
[edit:] and grats coldvein
cpp_is_king
Member
boo, shame on you
fwiw, i got the first 3 functions, but missed the last (the mod 4 piecewise)
the second one listed was the real tough one though, imo
if you only read the answer and not the solution, try to think about it anyway, the solution itself is pretty elegant
edit: actually my solution was different than all the ones given there
Partial Gamification
Banned
Indeed.boo, shame on you
LightOfTruth
Member
Never ever thought I would post here...
I was in a math competition, and this is the one I couldn't get:
3x - x^3 = (x+2)^1/2 (Im supposed to solve in real numbers)
Would anyone know how to do this?
I was in a math competition, and this is the one I couldn't get:
3x - x^3 = (x+2)^1/2 (Im supposed to solve in real numbers)
Would anyone know how to do this?
LightOfTruth
Member
eh I was looking for a step by step solution but I suppose that for a problem like this I wont get it online so easily. Thanks though.
You can sign in with facebook and get three free step by step solutions a day if that helps!
LightOfTruth
Member
Just tried it and the step by step feature isn't offered for the problem.
I was just really bothered that I couldn't get this, it isn't even for a grade or anything so I'll have to let this one go I suppose.
cpp_is_king
Member
my first thought is to square both sides, giving you a 6th degree polynomial which is missing its 3rd and 5th degree terms. Higher order polynomials with missing terms frequently have clever factorizations. for example, take out the x^2 and x terms, then you have degree 6, 3, and 0 terms in one poly, and degree 2, 1, and 0 in another poly. factor each of these as quadratics and see if it leads anywhere.
im on mobile right now, but when i get to a sheet of paper ill try it
Partial Gamification
Banned
cpp_is_right_as_usual 
Please do not enter into a habit-of-mind such as "I am too good to ask for help." I don't know the depth of this competition but no one could derive all of mathematics within a lifetime. "We all stand on the shoulders of giants." Anyone that looks down on someone for asking for help is just a pompous asshole.
Please do not enter into a habit-of-mind such as "I am too good to ask for help." I don't know the depth of this competition but no one could derive all of mathematics within a lifetime. "We all stand on the shoulders of giants." Anyone that looks down on someone for asking for help is just a pompous asshole.
cpp_is_king
Member
cpp_is_right_as_usual
Please do not enter into a habit-of-mind such as "I am too good to ask for help." I don't know the depth of this competition but no one could derive all of mathematics within a lifetime. "We all stand on the shoulders of giants." Anyone that looks down on someone for asking for help is just a pompous asshole.
usually my first idea isnt right though
. did you factor that by hand? it looks a little tricky, but i havent tried yetPartial Gamification
Banned
No, I'm a lazy bastard. Maple.usually my first idea isnt right though
Partial Gamification
Banned
I feel like you are looking for something to ponder, cpp.
This is a fun problem (unsolved) that is deceivingly simple: The Moving Sofa Problem. There have been some improvements since the publication of this paper but the "largest sofa" has not been found.
This problem is what are the dimensions (2D) of a sofa of greatest area that can pass down an "L-shaped" hallway?
Image from the wiki page (The Hammersley sofa):
This is a fun problem (unsolved) that is deceivingly simple: The Moving Sofa Problem. There have been some improvements since the publication of this paper but the "largest sofa" has not been found.
This problem is what are the dimensions (2D) of a sofa of greatest area that can pass down an "L-shaped" hallway?
Image from the wiki page (The Hammersley sofa):
I'd square both sides and it works out pretty easily.
Forgot to hit refresh when I opened Safari on my iPhone. Beaten like no other.
Above problem: relative motion with a "complex" shape. Damn I don't even want to begin trying that.
cpp_is_king
Member
I'm still looking at it and I don't agree that it works out very easily after you square both sides. For starters, since you've squared both sides you have to consider the case that LHS = RHS as well as the case that LHS = -RHS. Even if that weren't true though, It's not obvious how to factor the 6th degree polynomial. And even if you do manage to factor the polynomial, you have to make sure you check each solution back in the original equation due to the signs. For example, x-2 is a factor of the 6th degree polynomial that you get when you start with LHS = RHS, but x=2 is not one of the roots of the original equation.
But, since you have that x-2 is a root of LHS = RHS, you can always do the long division and find out that
(x-2)(x^5+2 x^4-2 x^3-4 x^2+x+1) = 0
But then you're stuck again trying to factor this 5th degree polynomial that has no rational roots, so good luck.
So I think that perhaps the squaring both sides approach may not be right, and there may be some other analysis and/or substitution that you can do first to transform the problem.
I'll look at it more later.
BTW, what math competition was this?
Partial Gamification
Banned
Here is what I put together, seeing you wouldn't let it go.
Squaring both sides would just require more work than is necessary (but so is arguably writing this stuff out when CAS can break it apart in the blink of an eye), here is where I got:
[edit 3:] changes made... this is a lot more challenging than i initially gave it credit. I too would be interested in where this problem came from. They have some great test-writers.
I love this type of distraction but I have check back later.
Squaring both sides would just require more work than is necessary (but so is arguably writing this stuff out when CAS can break it apart in the blink of an eye), here is where I got:
[edit 3:] changes made... this is a lot more challenging than i initially gave it credit. I too would be interested in where this problem came from. They have some great test-writers.
I love this type of distraction but I have check back later.
LightOfTruth
Member
Sorry guys I was helping someone with some physics homework. He didnt do a single question so it took me a long time.
I know, its just that I never saw myself going to the internet for help. I should be more open, as you say.
Thanks for attempting the question, first of all. I have tried this question multiple times and I would agree that squaring both sides wont help much.
Also, this was a math competition held at my University, sponsored by Virginia Tech (dont know if has an actual name or anything). Only I dont think its popular as only three people including myself showed up (and someone forced me to go in the first place).
I'll try your method and see what I can come up with. However, your method isnt matching what wolfram said was the answer.
cpp_is_right_as_usual
Please do not enter into a habit-of-mind such as "I am too good to ask for help." I don't know the depth of this competition but no one could derive all of mathematics within a lifetime. "We all stand on the shoulders of giants." Anyone that looks down on someone for asking for help is just a pompous asshole.
I know, its just that I never saw myself going to the internet for help. I should be more open, as you say.
Thanks for attempting the question, first of all. I have tried this question multiple times and I would agree that squaring both sides wont help much.
Also, this was a math competition held at my University, sponsored by Virginia Tech (dont know if has an actual name or anything). Only I dont think its popular as only three people including myself showed up (and someone forced me to go in the first place).
Here is what I put together, seeing you wouldn't let it go.
Squaring both sides would just require more work than is necessary (but so is arguably writing this stuff out when CAS can break it apart in the blink of an eye), here is where I got:
[edit 3:] changes made... this is a lot more challenging than i initially gave it credit. I too would be interested in where this problem came from. They have some great test-writers.
I love this type of distraction but I have check back later.
I'll try your method and see what I can come up with. However, your method isnt matching what wolfram said was the answer.
Partial Gamification
Banned
I swear, if I only had a dollar for all the arithmetic I botched, I could afford to hollow out a mountain and install and mad science lab (maybe my hypothetical typo money would need to be tapped for that).
(x+2)((3x-x^3)^2 - (x+2)) = 0
(x+2)(x^6 - 6x^4 + 9x^2 - x - 2) = 0
I can't see this factoring cleanly, off hand.
(x+2)((3x-x^3)^2 - (x+2)) = 0
(x+2)(x^6 - 6x^4 + 9x^2 - x - 2) = 0
I can't see this factoring cleanly, off hand.
cpp_is_king
Member
Careful how you interpret wolfram. I'm going to guess that when you plugged it into Wolfram, you were under the impression there was only 1 real root, right?
I'm beginning to think that any solution involving factoring is going to be a dead end.
LightOfTruth
Member
Careful how you interpret wolfram. I'm going to guess that when you plugged it into Wolfram, you were under the impression there was only 1 real root, right?
I'm beginning to think that any solution involving factoring is going to be a dead end.
It should be 3, no?
And yeah, if only you've seen all the methods Ive done. I probably tried this problem 6 times...
cpp_is_king
Member
Yea it's 3, but the first time I punched it into wolfram alpha it showed 1 real root and 2 complex roots. But then if you ran simplify on the complex roots, the imaginary parts all disappeared. Now when I punch them in it shows them all as real.
I managed to prove that there are exactly 3 real roots, as well as finding the intervals that they lie in, but I'm not sure how to solve for them explicitly.
The fact that Wolfram's solution is in the form of expressions involving imaginary numbers leads me to believe that on the off chance that there actually is a trick involved, it must involve an inverse trig substitution of some kind.
LightOfTruth
Member
I haven't tried this.
I should have the solution (step by step) on Monday, so I'll post it here afterwards. That is if I still didn't figure it out by then of course, cause this shit will keep haunting me...
cpp_is_king
Member
While thinking about this problem, I just stumbled across a group of identities which I feel could be extremely useful for certain types of problems. Everyone knows that cos(arctan(x)) (and related identities) have elegant algebraic simplifications. But what happens if you have cos(2 arctan(x)), or cos(10 arctan(x))? All methods of simplifying break down. I think I just came up with something though. If k is any integer, then I believe:
cos(k arctan(x)) = [1 / (x^2+1)^(k/2)] * Sum[p=0; p=Floor(k/2)] (-1)^p (k;2p) x^(2p)
Example:
cos(7 arctan(x))
Sum[p=0; 3] (-1)^p (7;2p) x^(2p) / (x^2+1)^(k/2)
[-(7;6)x^6 + (7;4)x^4 - (7;2)x^2 + (7;0)x^0] / (x^2+1)^(7/2)
= [-7x^6 + 35x^4 - 21x^2 + 1] / (x^2+1)^(7/2)
I'm going to try to prove this later, but something like this could be pretty useful in the right scenario. I think you could also come up with formulas for other combinations of func1(k*inversefunc2(x)). I can already think of occasions in the past where I got an expression like this and had to give up
I googled and can't find any related information about such an identity. Wonder if it's known, lol
cos(k arctan(x)) = [1 / (x^2+1)^(k/2)] * Sum[p=0; p=Floor(k/2)] (-1)^p (k;2p) x^(2p)
Example:
cos(7 arctan(x))
Sum[p=0; 3] (-1)^p (7;2p) x^(2p) / (x^2+1)^(k/2)
[-(7;6)x^6 + (7;4)x^4 - (7;2)x^2 + (7;0)x^0] / (x^2+1)^(7/2)
= [-7x^6 + 35x^4 - 21x^2 + 1] / (x^2+1)^(7/2)
I'm going to try to prove this later, but something like this could be pretty useful in the right scenario. I think you could also come up with formulas for other combinations of func1(k*inversefunc2(x)). I can already think of occasions in the past where I got an expression like this and had to give up
I googled and can't find any related information about such an identity. Wonder if it's known, lol
Partial Gamification
Banned
Not sure, but most excellent!
Playing around, I notice that two of the roots (and probably the third) have interesting values:
[edit:] Actually (post-posting hindsight 20/20 [?]), reading it wrong. (-1)*phi != (-1+sqrt(5))/2 (but close...)
[edit:] sign changed (Leftmost root)
cpp_is_king
Member
Are you sure that the left-most root is -phi? When I calculate it numerically I get approximately -1.80194
Partial Gamification
Banned
Its is my lack of ability to do correct arthimetic in my head
(-1+sqrt(5))/2 :Leftmost
[edit:] Its the form of the root, maybe I'm projecting something into the solution that isn't there.
[edit
Or type FFS. (changed sign)[edit:] I must be drinking from the stupid well... it is the middle root.
cpp_is_king
Member
Edit: Nvm
cpp_is_king
Member
Currently my money is on one of the following:
a) The problem asked you to find all RATIONAL roots, not all real roots (I believe it should be possible to prove prove there are 0)
b) The problem asked to show how many real roots there are, but not compute them explicitly.
c) You reproduced the equation incorrectly on this forum (unlikely)
d) There was a typo on the handout.
The more I look at this, the more I just don't think this is possible through normal means. Even if you managed to find the first one (golden ratio conjugate), which seems highly unlikely as it is, the other 2 just aren't going to crack. I mean their closed form solutions indicate they are the roots of a 3rd degree polynomial with irrational coefficients, so unless they expect you to be able to factor multiple 5th and 6th degree polynomials which have 0 rational roots (spoiler alert: not happening), I just don't see it.
Guess we'll just have to wait till Monday.
Edit: I've made some progress, but not much. All the roots of the original equation are roots of the following:
[(x+1)(x-1)Sqrt[x+2] + 1] * [(x+1)(x-1)Sqrt[x+2] - 1]
So it suffices to solve each of the following expressions over the reals:
(x+1)(x-1)Sqrt[x+2] + 1 = 0
(x+1)(x-1)Sqrt[x+2] - 1 = 0
Surprisingly, this is not that much easier than the original problem, despite how nice and symmetric the formulas look.
Edit: I finally managed to solve the above 2 equations, but all it led to was the original factorization that was computed by wolfram alpha, (x^2+x-1) (x^3+x^2-2 x-1). I guess with that, I'm satisfied if nothing else that i was able to figure out how to factor that expression. You're still left with the task of:
a) checking both the golden ratio and its conjugate in the original expression to see which (if either) of them works. which is really error prone considering you're going to be manipulating a fractional powers
b) solving the cubic x^3+x^2-2 x-1 = 0, which has 3 more real roots for you to check, and luckily they're all pretty like roots of cubic equations usually are, so that's not difficult at all! lol
c) verifying sure that there's no real roots to the expression (3x - x^3)^2 = x+2
So in summary: problem is broken, but at least now I can quit working on it
a) The problem asked you to find all RATIONAL roots, not all real roots (I believe it should be possible to prove prove there are 0)
b) The problem asked to show how many real roots there are, but not compute them explicitly.
c) You reproduced the equation incorrectly on this forum (unlikely)
d) There was a typo on the handout.
The more I look at this, the more I just don't think this is possible through normal means. Even if you managed to find the first one (golden ratio conjugate), which seems highly unlikely as it is, the other 2 just aren't going to crack. I mean their closed form solutions indicate they are the roots of a 3rd degree polynomial with irrational coefficients, so unless they expect you to be able to factor multiple 5th and 6th degree polynomials which have 0 rational roots (spoiler alert: not happening), I just don't see it.
Guess we'll just have to wait till Monday.
Edit: I've made some progress, but not much. All the roots of the original equation are roots of the following:
[(x+1)(x-1)Sqrt[x+2] + 1] * [(x+1)(x-1)Sqrt[x+2] - 1]
So it suffices to solve each of the following expressions over the reals:
(x+1)(x-1)Sqrt[x+2] + 1 = 0
(x+1)(x-1)Sqrt[x+2] - 1 = 0
Surprisingly, this is not that much easier than the original problem, despite how nice and symmetric the formulas look.
Edit: I finally managed to solve the above 2 equations, but all it led to was the original factorization that was computed by wolfram alpha, (x^2+x-1) (x^3+x^2-2 x-1). I guess with that, I'm satisfied if nothing else that i was able to figure out how to factor that expression. You're still left with the task of:
a) checking both the golden ratio and its conjugate in the original expression to see which (if either) of them works. which is really error prone considering you're going to be manipulating a fractional powers
b) solving the cubic x^3+x^2-2 x-1 = 0, which has 3 more real roots for you to check, and luckily they're all pretty like roots of cubic equations usually are, so that's not difficult at all! lol
c) verifying sure that there's no real roots to the expression (3x - x^3)^2 = x+2
So in summary: problem is broken, but at least now I can quit working on it
Partial Gamification
Banned
So in summary: problem is broken, but at least now I can quit working on it
It could be but I'm not sure. I think it is just ridiculously difficult. I am assuming that you didn't have access to a graphing calculator on the test (non-CAS). Here is how I came to factor (after squaring both sides):
[edit:] The inclusion of this problem could have been for a number of things: a catch for test-takers that had access to problems/solutions prior to the test, or even as a "tie breaker" in that it is supposed this would not be completely solved and those with equal numbers of correct answers would be judged on their approaches and "near-completeness" of this problem.
cpp_is_king
Member
It could be but I'm not sure. I think it is just ridiculously difficult. I am assuming that you didn't have access to a graphing calculator on the test (non-CAS). Here is how I came to factor (after squaring both sides):
ill post how i factored it later when im not on mobile. theres actually no magic involved, and perhaps even reasonable for a competition.
the thing is, the factorization doesnt really seem to help answer the original question.
looking at wolfram's closed form representation of the 2 difficult roots is enough to convince me that the problem is messed up, but like you said, maybe it was known to be too difficult
Partial Gamification
Banned
Graphing the equation: y = LHS 3rd Line
The orignial roots are preserved but additional ones are introduced (by squaring). The original graphs (from the given) y= RHS and y=LHS and the above mentioned all intersect at the appropriate points. Wolfram's (and Maple's) CAS algorithmic approach might be where the issue lies. I do think that this was likely a means to catch potential cheaters. You are probably right, but now I'm debating calculating the roots by hand...
The orignial roots are preserved but additional ones are introduced (by squaring). The original graphs (from the given) y= RHS and y=LHS and the above mentioned all intersect at the appropriate points. Wolfram's (and Maple's) CAS algorithmic approach might be where the issue lies. I do think that this was likely a means to catch potential cheaters. You are probably right, but now I'm debating calculating the roots by hand...
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