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Sum of a series using partial sums by Edwin Gunnerstedge - Sun, 06 Apr 2014 21:18:10 EST ID:MZ/8O5Np No.13858 Ignore Report Reply Quick Reply
1396833490994.png -(4909 B, 550x151) Thumbnail displayed, click image for full size. 4909
Hello, this is the last question on my math homework and I cannot figure it out for the life of me. I think it's a divergent series, but apparently it is not. Any help is appreciated
>>
Angus Clondlefield - Mon, 07 Apr 2014 01:12:58 EST ID:sxI6onv0 No.13860 Ignore Report Quick Reply
Assuming sn = 2 - 6 * .9^n is the sum of the infinite series, just take the limit of that function as it goes to infinity, no summation involved.
>>
Beatrice Goodlock - Mon, 07 Apr 2014 03:16:38 EST ID:dtJRvm7a No.13861 Ignore Report Quick Reply
>>13860
This. It sums to 2. You can also use the provided information to learn that a_1 = -3.4, and for subsequent terms, a_n = 0.6(0.9)^(n-1).
>>
Beatrice Goodlock - Mon, 07 Apr 2014 03:38:16 EST ID:dtJRvm7a No.13862 Ignore Report Quick Reply
>>13861
>a_n = 0.6(0.9)^(n-1)
Which can be simplified to a_n = 2/3(0.9)^n.
>>
Ernest Duckman - Wed, 09 Apr 2014 17:40:30 EST ID:LqF3f+Rw No.13870 Ignore Report Quick Reply
i tried to determine the sum another way and got a different result, though i agree that we just have to take the limit of s_n as n->inf actually, but i cant find a mistake:
as mentioned a_1=-3,4 and a_n=2/3(0.9)^n
so
the sum over n from 1 to inf a_n=-3.4+the sum over n from 2 to inf 2/3(0,9)^n
the last term is a geometric series missing the first two terms
so the last term equals 2/3*(1/(1-0.9)-1-0.9) [-1-0.9 are the missing two terms]
so the final result is -3.4+2/3*(1/(1-0.9)-1-0.9)=-3.4+2.7=-0.7

can anyone spot a mistake?
>>
Hedda Nenkinford - Wed, 09 Apr 2014 18:27:13 EST ID:dtJRvm7a No.13871 Ignore Report Quick Reply
>>13870
Seems to be a simple arithmetic error.

-3.4 + 2/3*(1/(1 - 0.9) - 1 - 0.9) = -3.4 + 2/3*(10 - 1 - 0.9) = -3.4 + 2/3*8.1 = -3.4 + 5.4 = 2


u-v substitution "determinant but not" mnemonic by Hedda Smallhood - Fri, 28 Mar 2014 22:00:40 EST ID:4spVkkBQ No.13819 Ignore Report Reply Quick Reply
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the other day my prof showed us a cool trick for remembering uv substitution (note: he's a physics prof so this may seem dirty)

it looked like a determinant, where you do the criss cross stuff, but you don't do the entire criss-crossing diagonal process

This probably sounds vague and strange if you don't know what i'm talking about, but if you do can you show it? I forgot and it's turning out to be hard to google!
4 posts omitted. Click Reply to view.
>>
Beatrice Bindlesirk - Sun, 30 Mar 2014 09:09:59 EST ID:gxFYvDAi No.13826 Ignore Report Quick Reply
>>13825
holy crap I'm high
sorry!!! that's meaningless garbage
nb because stupid garbage language infecting an idiot's rotting brain
>>
Albert Sillykadging - Wed, 02 Apr 2014 15:01:01 EST ID:o5vhj6+B No.13848 Ignore Report Quick Reply
>>13823
I'm not 100% following you and could you elaborate on "uv substitution"? Look up "Jacobian matrix" That might be it.

Or is it just integrating by parts multiple times but just substituting the original integral back into the equation after you've integrated by parts? And then you keep integrating by parts until you can solve the integral normally and just substitute the answer into equation.
>>
Shitting Feddlestock - Sun, 06 Apr 2014 14:35:48 EST ID:4spVkkBQ No.13856 Ignore Report Quick Reply
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OP here. So... integration by parts.

Basically my prof had integral which had the form of integral(udv) or equivalently, integral(u*dv/dx*dx)

Step by step, he started out with what looked like a 2x2 determinant
(It's not a determinant, just a similar looking mnemonic device)

1) u goes in top left, dv/dx in bottom right. These were given since they make up in the integrand
2) Calculate du/dx, put it in the bottom left. Calculate v from dv/dx (integrate dv/dx dx) put it in top right
3) ...actually I'm not too sure about this step. I'm working backwards since keep forgetting to ask my prof, but see pic for how the "determinant-ish crisscrossing" works. I'm confused because I remember(??) the arrows going differently, more similar to an actual determinant so step 1/2 might not even be right.

I've only glanced at the tabular method... I don't think this is it. In fact, I should just check out the tabular method since it looks pretty nifty. Anyway, does anyone recognize something out of the ugliness I have tried to re-create?
>>
William Menderstore - Mon, 07 Apr 2014 19:48:54 EST ID:AMc02gy4 No.13864 Ignore Report Quick Reply
>>13856
Hmm ... It kind of reminds me of changing the basis when dealing with double integrals. No idea really though. It would be cool if you could find out from your prof and post it back here.
>>
William Shakeworth - Tue, 08 Apr 2014 00:18:46 EST ID:4spVkkBQ No.13865 Ignore Report Quick Reply
>>13864
Just asked him today and that's it.


eigenwat by Jarvis Morringstot - Tue, 01 Apr 2014 17:57:15 EST ID:GBVWm2Fa No.13835 Ignore Report Reply Quick Reply
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I just watched an MIT lecture on linear algebra where the prof wrote the fibonnci sequence as a matrix equation and found the positive eigenvalue was the golden ratio. I went back to relearn linear algebra because I thought I was missing something (I just followed some algorithms to solve differential equations without much thought). Now it seems I've underestimated how crucial this eigenshit is to math and applied math especially. I want to restart my lin alg knowledge from the base and then build up to applications of it. Where do I start? Any suggestions? Thanks.
>>
Fucking Fidgelodge - Wed, 02 Apr 2014 01:18:51 EST ID:qz3c7Bt+ No.13845 Ignore Report Quick Reply
Any proofs based linear algebra course should be fine. It's all vector spaces.

The fibonacci trick relies on generating functions for certain recurrence relations being representable as linear transforms. It's not linear algebra per se.
>>
Shit Bipperway - Wed, 02 Apr 2014 03:29:09 EST ID:GBVWm2Fa No.13847 Ignore Report Quick Reply
>>13845
hmm maybe I should revisit discrete as well then
>>
Lillian Blondlehodge - Wed, 02 Apr 2014 22:55:24 EST ID:+KPsBlYP No.13849 Ignore Report Quick Reply
>>13847
There's thousands of websites/videos on linear algebra. Find one you like if you want to start from the ground up. But first, take a look at the wiki entry for eigenvalue/vectors. The mechanics of actually multiplying matrices/vectors is tricky, but eigenwats are a fairly simple concept.

Then search for Difference Equations, or recurrent relations. It's not "differential". It's difference equations. FUUUUUUCK. I hate that feeling when you realize you've forgotten everything about a class you took. You sound like you've had it with linear algebra, and I just realized I've forgotten most of my difference equation class.

This looks like a good summary, though it moves very fast: https://math.uc.edu/~halpern/Combinatorics/Hocombinatorics/Differenceequations.pdf


Riddle Me This by Fiend !!1C9jE+w+ - Tue, 01 Apr 2014 16:00:39 EST ID:8prwY8OH No.13833 Ignore Report Reply Quick Reply
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Now you may have heard the of the 10 dot problem, where you are tasked with arranging 10 dots so that they are lined up in 5 rows, with 4 dots in each row. The answer is a 5 pointed star (top left of picture). A more daunting challenge is to arrange only 9 dots, into 10 rows of 3 (answer on bottom left).

This got me thinking about the ratio of dots to rows and the number of dots in the row. So I created a ratio that would compare the number of dots in our shape to the number of dots in the rectangle assumed after the words "5 rows of 4."
5 rows of 4 would be 20 dots, but you can do it using only 10, making the ratio 2.

Lemme give you an equation because my words are failing me:
> x = rk/d
r is the number of rows, k is the number of dots in each row, and d is the total number of dots. That means the ratio of 9 dots in 10 rows of 3 is x = (10r3)/9 = 3.333.
In the top right part of my image I have the general formula for when not all the rows contain the same number of dots.

At this point you've probably noticed that you can use this generalized formula on any group of intersected lines to find out the average number of lines at any given intersection in the shape. Well my question is this:
> Can you create a shape with x > 3.333?
> If not can you prove that it is impossible?
Now granted, as pictured bottom right, turning any dot formation into a fractal will allow x to go to infinity, but is there any other way to average more than 3.333 lines per intersection?
5 posts and 1 images omitted. Click Reply to view.
>>
Hugh Pungerdale - Tue, 01 Apr 2014 23:11:05 EST ID:dtJRvm7a No.13841 Ignore Report Quick Reply
>>13840
>Can you figure out how?
Probably not. "The system" you're using isn't very clear. So go ahead, I'm curious where you're going with this.
>>
Hugh Pungerdale - Wed, 02 Apr 2014 00:32:53 EST ID:dtJRvm7a No.13842 Ignore Report Quick Reply
>>13841
I mean, I got x = 3.75 by just arranging >>13838 into a triangle. But you'll probably tell me this is cheating. Not allowing the same shape inside of itself? Why not? It's just arbitrary bro. Also from >>13838, you're demonstrating there doesn't need to be a dot at every intersection of rows. I'd rather it be a rule that there must be a dot at every intersection instead of the same shape thing.
>>
Fiend !!1C9jE+w+ - Wed, 02 Apr 2014 00:33:46 EST ID:8prwY8OH No.13843 Ignore Report Quick Reply
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>>13841
I made a 5x5 version of the above image. I'm sorry the drawing is sloppy, but I hope the different colors make it clear.
This arrangement of 25 dots has 14 rows of 5 dots each (black), then 4 rows of 4 dots (blue), 10 rows of 3 dots (orange), and 2 rows of 2 dots (ugly ass maroon color). So to ditch the notation I was using, x = (14*5 + 4*4 + 10*3 + 2*2) / 25 which simplifies to 120/25 or x = 4.8.
I assume that a 7x7 would yield yet a higher x value, and 9x9 and so on, meaning x has no limit, as the more you expand the same pattern, there are more angles to create rows in, and therefore more lines at each intersection.
>>
Whitey Blezzleridge - Wed, 02 Apr 2014 01:14:35 EST ID:HvBSycws No.13844 Ignore Report Quick Reply
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>>13843
Looks like a fractal to me
>>
Fiend !!1C9jE+w+ - Wed, 02 Apr 2014 01:21:22 EST ID:8prwY8OH No.13846 Ignore Report Quick Reply
>>13844
Does that type of expansion fall under the literal definition of 'fractal'?


Derivatives, Help!! by Martin Fittingput - Sun, 30 Mar 2014 20:17:02 EST ID:g5vmHJ4m No.13829 Ignore Report Reply Quick Reply
1396225022113.jpg -(1932285 B, 3264x2448) Thumbnail displayed, click image for full size. 1932285
This is a pretty easy question but I don't really understand how to explain it.
Any help?
>>
Eugene Pillyhall - Sun, 30 Mar 2014 20:39:37 EST ID:ZxnBDs99 No.13830 Ignore Report Quick Reply
If you've done derivative problems with position, velocity, etc. think of it this way: g(x) = distance/time (velocity)
g'(x) = distance/time^2 (acceleration, rate of change of velocity)

g(x) = water depth with respect to time x
g'(x) = rate of change of depth with respect to time x
g'(25)= rate of change at 25mins after the pool has started to fill

I'm very high, this could probably be explained in a better fashion. Hope it helps nonetheless
>>
Martin Fittingput - Sun, 30 Mar 2014 21:39:22 EST ID:g5vmHJ4m No.13831 Ignore Report Quick Reply
>>13830
that's exactly what I had but thank you very much for clarifying and confirming!


Help me understand this type of algebra please, I need you! by Reuben Cobbermere - Mon, 24 Mar 2014 14:59:33 EST ID:bTMSKJUg No.13806 Ignore Report Reply Quick Reply
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So i'm not expecting you to help but i'm really hoping because I really need help with this until tomorrow. Yes it's for school. It's not homework, but a question on a test I couldn't answer.

But i'm not needing help with just this particular problem, but help on solving these types of problems. Make me understand how this works.

So, a guy has 5 dollar bills and 10 dollar bills, in total 215 dollar bills. How many 10 dollar bills does he need to have for it to be at least 1500 dollars in total?

Thanks!

I'm hoping to not just be able to answer math questions in school but to really comprehent mathematics one day, like really GET it.
>>
Hamilton Nenkinlock - Mon, 24 Mar 2014 16:01:10 EST ID:MTIV7/tU No.13807 Ignore Report Quick Reply
So this dude has x 10$ bills. He also has (215 - x) 5$ bills because in total he has 215.
So the money he owns is 10*x + 5*(215 - x) $ = 5x + 1075 $.
The smallest x such that 5x + 1075 is at least 1500 is 425/5 = 85.
You can also check that with 85 bills of 10$ and 130 of 5$, he indeed has 1500$, the rich fucker.
>>
Hamilton Nenkinlock - Mon, 24 Mar 2014 16:04:53 EST ID:MTIV7/tU No.13808 Ignore Report Quick Reply
>>13807
Just in case this is blurry, the third step is actually this kind of reflection:
If you want 5x + 1075 to be at least 1500, then 5x has to be at least 1500-1075=425. This gives x >= 85, because then 5x >= 5*85 = 425.
>>
Reuben Cobbermere - Mon, 24 Mar 2014 16:14:51 EST ID:bTMSKJUg No.13809 Ignore Report Quick Reply
>>13807
>>13808
Thanks! Thanks a whole bunch! You're awesome!


Just a quick question and i'll be on my way by Charles Fuckingshit - Thu, 20 Feb 2014 02:31:01 EST ID:QD7zvUn8 No.13708 Ignore Report Reply Quick Reply
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I know it's probably a stupid question to ask, but if somebody could at least give me a hint on the solving for y part, I could figure the rest out.

It seems simple but I must be having a brain fart.
2 posts omitted. Click Reply to view.
>>
Angus Peshson - Fri, 28 Feb 2014 19:36:02 EST ID:3E9PS/tX No.13726 Ignore Report Quick Reply
This problem looks easy enough that you don't have to use implicit differentiation. Just do like the problem says. solve for y. I have no idea what in the sam fuck the previous poster in this thread is talking about when he says "write the left member" do you mean the left hand side? THE ONLY place in mathematics we ever use the word "member" is when we're talking about elements in sets. OP. Here's what you do.

1.) Solve for y : It should look like : y = (x^3 - 2) / ( (1-3)(2x^2) )
2.) Differentiate. Done.
3.) Looks like you're gonna be using quotient rule. Since I didn't further simplify the Right Hand side, I may be wrong and you could end up getting away using simple power rule. I don't know. But it looks like quotient rule.
>>
Augustus Dangerbere - Sat, 01 Mar 2014 00:44:16 EST ID:dtJRvm7a No.13731 Ignore Report Quick Reply
>>13726
No, look at the original problem.

y = (x^3 - 2) / (2x^2 - 3)

so

dy/dx = ((3x^2)(2x^2 - 3) - (x^3 - 2)(4x)) / (2x^2 - 3)^2
= (2x^4 -9x^2 + 8x) / (4x^4 - 12x^2 + 9)

/thread
>>
Shit Bannerstone - Sun, 02 Mar 2014 10:54:59 EST ID:ELymY0SU No.13736 Ignore Report Quick Reply
>>13726
English is not my native language, and we use the term member referring to either side of the equation. I didn't know it could put you in so much confusion, sorry about that.

Also my answer was incorrect, as I saw an x too many, see Augustus's post for the correct answer.
>>
Ernest Dorryhit - Tue, 18 Mar 2014 14:19:27 EST ID:weDVNH8d No.13779 Ignore Report Quick Reply
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I don't understand that there will be an extra minus sign using the second substitution method. Thanks in advance.
>>
Priscilla Duckcocke - Mon, 24 Mar 2014 06:03:24 EST ID:weDVNH8d No.13799 Ignore Report Quick Reply
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>>13779
mfw i integrated and had to undo the substitution.


Help by Eugene Chushchud - Mon, 24 Mar 2014 00:09:16 EST ID:0Uju5WgZ No.13791 Ignore Report Reply Quick Reply
1395634156862.jpg -(2082803 B, 3264x2448) Thumbnail displayed, click image for full size. 2082803
I don't understand these problems, help please
1 posts omitted. Click Reply to view.
>>
Eugene Chushchud - Mon, 24 Mar 2014 00:30:42 EST ID:0Uju5WgZ No.13793 Ignore Report Quick Reply
Obviously, I'm on my phone though so I can't change that.
>>
Caroline Blackgold - Mon, 24 Mar 2014 02:10:49 EST ID:yGgK6aCs No.13796 Ignore Report Quick Reply
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>>13793
try turnin the paper upways so u can read it better. might help
>>
Eugene Chushchud - Mon, 24 Mar 2014 02:19:38 EST ID:0Uju5WgZ No.13797 Ignore Report Quick Reply
>>13796
It all makes sense now
>>
Simon Murdcocke - Mon, 24 Mar 2014 06:07:41 EST ID:uXKndL+s No.13800 Ignore Report Quick Reply
Not only is it sideways, but also sort of bent, leaving turning one's head out of the question.
>>
Thomas Ponkinbidge - Mon, 24 Mar 2014 14:57:01 EST ID:dtJRvm7a No.13805 Ignore Report Quick Reply
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Here you go OP.


ONE LOVE, ONE HEART by God's a chick, bro. - Fri, 21 Mar 2014 23:30:23 EST ID:fFWeOOT5 No.13786 Ignore Report Reply Quick Reply
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Let's get together, and... do some math. <3

http://en.wikipedia.org/wiki/Cardioid
http://en.wikipedia.org/wiki/Visible_light
https://www.youtube.com/watch?v=vdB-8eLEW8g
https://www.youtube.com/watch?v=che5bHJVzLQ
https://www.youtube.com/watch?v=WzV6mXIOVl4
http://www.youtube.com/watch?v=3OT5jQCaM8Y
https://www.youtube.com/watch?v=E-9O_GEiaEY

Give thanks and praise to the Lord, and she'll suck your dick.

Sky-pussy forever. ∞
>>
God's a chick, bro. - Fri, 21 Mar 2014 23:32:06 EST ID:fFWeOOT5 No.13787 Ignore Report Quick Reply
1395459126549.png -(475621 B, 739x648) Thumbnail displayed, click image for full size. 475621
>>13786

"...and all her paths are peace."

https://www.youtube.com/watch?v=rR8v88UrlH0


derpherpburp by Hugh Socklecocke - Fri, 14 Mar 2014 20:34:40 EST ID:zMY2Vm0I No.13772 Ignore Report Reply Quick Reply
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lol i dont understand anything on this board
>>
Emma Finnerket - Sat, 15 Mar 2014 16:17:12 EST ID:Uq/TU4om No.13774 Ignore Report Quick Reply
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I don't understand anything on this site.
hurp a derp
>>
Edwin Niblingbury - Sat, 15 Mar 2014 22:05:21 EST ID:NL4zf5yO No.13775 Ignore Report Quick Reply
>>13772
haha thats real funny man i was just thinking about postin the same thread here so i clicked on /math/ to do this and then this is the first one oh well
>>
Nigel Sonderfield - Wed, 19 Mar 2014 17:51:01 EST ID:4spVkkBQ No.13783 Ignore Report Quick Reply
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http://en.wikipedia.org/wiki/Mathematics
this should clear things up
>>
dameunmcchiken.sys - Wed, 19 Mar 2014 18:53:00 EST ID:Gb85/5TK No.13784 Ignore Report Quick Reply
>>13783
I fucking love this guy
Spardoe should hire him to go on every board.
>>
Jack Dartville - Fri, 21 Mar 2014 12:54:24 EST ID:IHm5SL/4 No.13785 Ignore Report Quick Reply
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i dont understand anything


Algebraic Topology Book by Lydia Hemmlewill - Mon, 17 Mar 2014 09:19:07 EST ID:VJTocZIm No.13776 Ignore Report Reply Quick Reply
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I'm currently using Hatcher's algebraic topology book and I was wondering if anyone's found something better.
>>
Beatrice Gassleshaw - Mon, 17 Mar 2014 19:57:03 EST ID:8PJ0nVdr No.13777 Ignore Report Quick Reply
Have you tried Munkres?
>>
Barnaby Crollerfick - Mon, 17 Mar 2014 22:31:11 EST ID:VJTocZIm No.13778 Ignore Report Quick Reply
>>13777
I have not. I heard it was a bit dry but that it also tends to be rigorous, which I like. Do you know anything about May's "concise" books?
>>
Edwin Fiddlenon - Wed, 19 Mar 2014 14:09:31 EST ID:1s2mLncZ No.13782 Ignore Report Quick Reply
I like Munkres. It can be "dry" but he does an excellent job of explaining the concepts. I would not have succeeded in learning the subject without this book.


Need help /math/ by Priscilla Drivingsod - Tue, 18 Mar 2014 21:57:34 EST ID:LaCkf6nr No.13780 Ignore Report Reply Quick Reply
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I am sorry to bother all of you, but I am an english major, and am absolutely HORRIBLE at math. I've been trying for 30 minutes to figure this out myself, but I absolutely need help from someone who understands math. Tonight is the add/drop deadline at my college and I am unsure as to whether or not I need to drop my class. The grading scale is:

Participation - 5% Speaking - 10%
Online HW - 15% Midterm- 10%
Chapter Exams - 15% Writing - 10%
Quizzes - 10% Final - 10%
There have only been two tests, one on which I scored an 92, and one I missed, so my average is not currently doing well for the exam section. There are 5 total. There are also 5 quizzes, three speaking assignments, and three writing assignments. Mathematically, as long as my grades for that are okay, and I get decent grades on the rest of the tests do I stand a chance of passing this class? Or at least not destroying my GPA?

Thanks for the time/help /math/.
If there's anything I can do in return, please let me know.
>>
Isabella Padgeman - Tue, 18 Mar 2014 23:20:59 EST ID:HvBSycws No.13781 Ignore Report Quick Reply
>>13780
Just divide the percent by how many there are of that thing to see how much one of that thing is worth

So, Chapter Exams - 15%, If there are 5 of them, each are worth 15%/3 = 3%, so if you miss one you can get a max of 100%-3% = 97% in the class
Quizzes - 10%, 10%/5 = 2% per quiz
ect ect ect
>>
Simon Murdcocke - Mon, 24 Mar 2014 06:42:22 EST ID:uXKndL+s No.13801 Ignore Report Quick Reply
Does the syllabus tell you how many exams you will have total? If you have already taken two, I would expect there to be at least two more, if not three (including a cumulative final, if there is one).

If you were to post each grade by category we could give you a direct answer.


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