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# A closed cylindrical tank contains 36pi cubic feet of water

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Math Expert
Joined: 02 Sep 2009
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A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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14 Jun 2012, 02:54
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Difficulty:

55% (hard)

Question Stats:

67% (02:13) correct 33% (02:31) wrong based on 2271 sessions

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A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

Notice that some editions of OG have a typo saying that the height of the water in the tank is 2 feet, it should read "the height of the water in the tank is 4 feet".

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Question: 5
Page: 20
Difficulty: 650

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Posts: 59236
Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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14 Jun 2012, 02:54
23
23
SOLUTION

Notice that some editions of OG have a typo saying that the height of the water in the tank is 2 feet, it should read "the height of the water in the tank is 4 feet".

A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

Look at the diagram below:

Since the tank is half full when placed upright then naturally it'll also be half full when placed on its side, so the level of the water (when placed that way) will be half of the diameter, so $$r$$.

Now, given that $$V_{water}=\pi{*r^2}*H_{water}$$ --> $$36\pi=\pi{r^2}*4$$ --> $$r=3$$.

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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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14 Jun 2012, 11:11
2
Volume of water inside cylinder = 36pi = pi$$r^2$$h

Here water is filled up to a height of 2 feet, so h=2

$$r^2$$ = 18

r=3 $$sqrt2$$

There might be a mistake in the given problem.
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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16 Jun 2012, 03:56
2
1
gnan wrote:
Volume of water inside cylinder = 36pi = pi$$r^2$$h

Here water is filled up to a height of 2 feet, so h=2

$$r^2$$ = 18

r=3 $$sqrt2$$

There might be a mistake in the given problem.

I agree to your comment partially, in my opinion this question has some ambiguity when it states that the tank contains 36pi cubic feet of water and is filled to half its capacity, so we may assume that 36pi is the volume when half capacity. So it would be better to state that the tank, when full, can be filled with 36pi or the full capacity of the tank is 36pi, or something alike. But i think this is one of the small tricks of GMAT. But anyway if you solved this way and did not come up with answer you should see what else GMAT could think by saying 36pi, then you see that only possible answer is 3 (B)
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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22 Oct 2013, 09:34
1
I wanted to clarify one point. Is there any rule that we refer when we claim that regardless of position of cylinder the water in it will occupy its half?
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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22 Oct 2013, 09:46
1
Ergenekon wrote:
I wanted to clarify one point. Is there any rule that we refer when we claim that regardless of position of cylinder the water in it will occupy its half?

How else? If water occupies half of the capacity of the tank when it is placed upright, can it occupy other fraction when it is placed on its side???
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A closed cylindrical tank contains 36pi cubic feet of water and  [#permalink]

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10 Feb 2014, 17:03
how do you know the height is 4 feet when placed on its side?
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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10 Feb 2014, 21:29
6
2
Mackam1234 wrote:
how do you know the height is 4 feet when placed on its side?

There are a few different relevant heights in the question:

A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. when the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet.

The height of the water in the tank is given to be 4 feet. Since the tank is filled to half its capacity, the height of the cylinder is 8 ft (though we don't need this).

Volume of the water $$= 36*\pi = \pi*r^2*4$$
r = 3

This is all we need to find the required height. Why? We know that the tank is filled to half its capacity. Imgine you have placed it on its side. If it were full of water, the height of the water in this position would be the diameter of the circular base (which is now the height). Since it is half full, the height of the water will come up to the mid point of the circle i.e. its radius. Since radius is 3, the height of water now will be 3 feet.
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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11 Jan 2017, 07:51
2
Bunuel wrote:
A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

We are given that a closed cylindrical tank that is half full contains 36π cubic feet of water, and the height of the water is 4 feet. We can thus say that the full tank would have 72π cubic feet of water at a height of 8 feet. Using the volume formula, we can now determine the radius of the circular base:

volume = π(r^2)h

72π = π(r^2)(8)

9 = r^2

r = 3 feet

We see that the radius is 3 feet.

We need to determine the height of the water when the tank is placed on its side on level ground. When the cylinder is turned on its side, the diameter now represents the new height, and since the tank is half full, the new height of the water is equivalent to the radius, so the new height of the water is 3 feet.

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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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29 Jun 2018, 03:25
1
Volume of cylinder = pi*r^2*h = pi * r^2 * 4 ft = 36 pi
r^2= 9 pi
r=3ft
A cyl which is half full, placed on its side will also be filled up to half its height which is r.

Therefore, r = 3ft.
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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16 Sep 2019, 23:56
When the cylinder is placed on its side how is the base circular? Maybe a diagram would have helped.
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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17 Sep 2019, 00:20
1/2 volume = 36 pi ft^ 3 when the height of water = 4

Thus max height = 8 and max volume = 72pi ft ^3
Notice if you flip the right cylinder on its side and the cylinder is filled 1/2 way then the water will actually share the height of the radius, so we are looking for the radius.

72pi = pi*r^2*8
r^2=9
r=3
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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16 Oct 2019, 15:58
Bunuel wrote:
SOLUTION

Notice that some editions of OG have a typo saying that the height of the water in the tank is 2 feet, it should read "the height of the water in the tank is 4 feet".

A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

Look at the diagram below:

Since the tank is half full when placed upright then naturally it'll also be half full when placed on its side, so the level of the water (when placed that way) will be half of the diameter, so $$r$$.
Now, given that $$V_{water}=\pi{*r^2}*H_{water}$$ --> $$36\pi=\pi{r^2}*4$$ --> $$r=3$$.

Bunuel

Two questions on the highlight

Q1) Is this the case for EVERY figure --- what about if it were a cube or an empty trapezium for example ?

Q2) Does your assumption not depend on the length and radius or does these factors not count at all -- this rule can be memorized for every geometrical figure
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Re: A closed cylindrical tank contains 36pi cubic feet of water  [#permalink]

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16 Oct 2019, 20:41
jabhatta@umail.iu.edu wrote:
Bunuel wrote:
SOLUTION

Notice that some editions of OG have a typo saying that the height of the water in the tank is 2 feet, it should read "the height of the water in the tank is 4 feet".

A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

Look at the diagram below:

Since the tank is half full when placed upright then naturally it'll also be half full when placed on its side, so the level of the water (when placed that way) will be half of the diameter, so $$r$$.
Now, given that $$V_{water}=\pi{*r^2}*H_{water}$$ --> $$36\pi=\pi{r^2}*4$$ --> $$r=3$$.

Bunuel

Two questions on the highlight

Q1) Is this the case for EVERY figure --- what about if it were a cube or an empty trapezium for example ?

Q2) Does your assumption not depend on the length and radius or does these factors not count at all -- this rule can be memorized for every geometrical figure

If something is filled to half its capacity, then it's filled to half its capacity no matter how the vessel is placed. How else?
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Re: A closed cylindrical tank contains 36pi cubic feet of water   [#permalink] 16 Oct 2019, 20:41
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# Problem using environment variables in RosJava for Android

Hey could someone please explain me how can I use environment variables on rosjava in android?

I'm trying to run the android_core pubsub example, but like some people, I'm having trouble using it at the emulator. At a real phone it works flawlessly, but at the emulator, I can see the nodes and the topics being created, but nothing comes at the topic after that and no error message is shown.

My problem is exactly the same as the one in this question (https://answers.ros.org/question/2236...), and I saw many answers (as @gdvhoorn in the link above) that say that the many times these kind of problems with the emulator are caused because of incorrect network configuration, so I have two questions:

1 - How can I set the values of the environment variables in an Android Emulator? I searched, but couldn't find a setter function, and when I go to the function definition it just returns me an EnvironmentVariables.class file that have the variables set like this:

public interface EnvironmentVariables {
String ROS_MASTER_URI = "ROS_MASTER_URI";
String ROS_IP = "ROS_IP";
String ROS_HOSTNAME = "ROS_HOSTNAME";
String ROS_NAMESPACE = "ROS_NAMESPACE";
String ROS_ROOT = "ROS_ROOT";
String ROS_PACKAGE_PATH = "ROS_PACKAGE_PATH";
}

And when I print these variables in the runtime, the only one that has changed is the Master URI, because of the nodeConfiguration.setMasterUri. So how can I change them to make it work in the emulator?

2 - How should I config my Environment Variables to make it possible for my computer to communicate with the emulator? The emulator runs on port localhost:5554, but when it runs, it creates a kind of brigde and sets the emulator own IP to 10.0.2.15 or 16 depending on the interface used, and host computer (where the roscore is running) to 10.0.2.2. But from the host computer I can't ping on 10.0.2.15 or 10.0.2.16 because they don't really exist. When the android app connnects, it shows the message: Registrar: MasterXmlRpcEndpoint URI: http://10.0.2.2:11311/ and that's right, because that's the address the emulator finds my computer, but the Publisher is registered at http://10.0.2.16:40719/ so when I'm able to change , as can be seen in the following message:

Registrar: Registering publisher: Publisher<PublisherDefinition<PublisherIdentifier<NodeIdentifier</android_gingerbread/ros_text_view, http://10.0.2.16:40719/>, TopicIdentifier</rosout>>, Topic<TopicIdentifier</rosout>, TopicDescription<rosgraph_msgs/Log, acffd30cd6b6de30f120938c17c593fb>>>>

But when my roscore try to find it, it won't be able to connect to it, because for my computer there's nothing with the IP 10.0.2.16 that my host computer can see. So, how should I config my Environment Variables to make it possible for my computer to communicate with the emulator?

I really need to use the emulator, because the project I'm working on will need 6 phones working together in ROS, so it makes really better if I can use the emulator. I have already some experience with ...

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# Smtp Mailer That Supports Html And Attachments.

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Hi guys when I tried to use the code I had error

$ObjEmail.Send$ObjEmail.Send^ERROR

Error: The requested action with this object has failed.

PS.Sorry to rise up all threads.

You didn't use the Com error handler portion of the posted code..

That should tell you more about why the send fails....

Edited by JdeB

SciTE4AutoIt3 Full installer Download page   - Beta files

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; ;################################## ; Include ;################################## #Include<file.au3> ;################################## ; Variables ;#

Agree, but first needs to be properly tested to see if everything keeps working and I haven't done that yet.  So ...  input/feedback is welcome before doing so!   My preference is to do this wi

$Body = "one" & @CRLF & "two" & @CRLF & "three" ..... .....$Body = StringReplace($Body,@CRLF,'<br>') ;~ And now you can send your body message.... #### Posted Images I'm having problems with this, I have this in the code:$rc = _INetSmtpMailCom("smtp.mail.yahoo.com", "name", "email1@yahoo.ca", "email2@yahoo.ca", "subject", "message", "", "","", "email1username", "email1password")

I get error :

Error code:2 Description:The transport failed to connect to the server

Any help is appreciated, thanks

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I'm having problems with this, I have this in the code:

$rc = _INetSmtpMailCom("smtp.mail.yahoo.com", "name", "email1@yahoo.ca", "email2@yahoo.ca", "subject", "message", "", "","", "email1username", "email1password") I get error : Error code:2 Description:The transport failed to connect to the server Any help is appreciated, thanks Does it work when you use the same Servername/Userid/Password in outlook express ? SciTE4AutoIt3 Full installer Download page - Beta files Live for the present, Dream of the future, Learn from the past. ##### Link to post ##### Share on other sites • 4 weeks later... Great script, I do have one minor question. When I beta run the script I keep getting an error "C:\Program Files\AutoIt3\beta\Autoemail.au3(34,65) : ERROR: _PathFull(): undefined function.$S_Files2Attach[$x] = _PathFull ($S_Files2Attach[$x])" The script will complete and run fine even including the attachment when there is one, just looking for some illumination as to what I might have incorrect. Thanks in Advance, Badger ##### Link to post ##### Share on other sites • Developers Great script, I do have one minor question. When I beta run the script I keep getting an error "C:\Program Files\AutoIt3\beta\Autoemail.au3(34,65) : ERROR: _PathFull(): undefined function.$S_Files2Attach[$x] = _PathFull ($S_Files2Attach[$x])" The script will complete and run fine even including the attachment when there is one, just looking for some illumination as to what I might have incorrect. Thanks in Advance, Badger You probably get an error when running from SciTE ? Done any updates lately for SciTE? SciTE4AutoIt3 Full installer Download page - Beta files Live for the present, Dream of the future, Learn from the past. ##### Link to post ##### Share on other sites You probably get an error when running from SciTE ? Done any updates lately for SciTE? Thank you, that was the problem. Badger ##### Link to post ##### Share on other sites • 1 month later... #Include <smtp.au3>$s_SmtpServer = "smtp.gmail.com"
$s_FromName = "Azkay"$s_FromAddress = "Azkay@gmail.com"
$s_ToAddress = "azkay-s@hotmail.com"$s_Subject = "Okay"
$as_Body = "Test."$s_AttachFiles = ""
$s_CcAddress = ""$s_BccAddress = ""
$s_Username = "azkay.s@gmail.com"$s_Password = "mypass"
_INetSmtpMailCom($s_SmtpServer,$s_FromName, $s_FromAddress,$s_ToAddress, $s_Subject,$as_Body, $s_AttachFiles,$s_CcAddress, $s_BccAddress,$s_Username, $s_Password) Exit I get error 1. Why? EDIT:: ### COM Error ! Number: 80020009 ScriptLine: 51 Description:The transport failed to connect to the server. Edited by AzKay # MY LOVE FOR YOU... IS LIKE A TRUCK- # ##### Link to post ##### Share on other sites • Developers I get error 1. Why? EDIT:: ### COM Error ! Number: 80020009 ScriptLine: 51 Description:The transport failed to connect to the server. Google uses port 465 and SSL try it with this version of _InetSmtpMailCom and add the last parameters with 465 and 1. Moved updated script to first post... Edited by JdeB SciTE4AutoIt3 Full installer Download page - Beta files Live for the present, Dream of the future, Learn from the past. ##### Link to post ##### Share on other sites Gmail uses SSL security, might be the issue... Don't take my pic to serious...~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~You Looked, but you did not see!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##### Link to post ##### Share on other sites Wow, Very nice! Works. ^^, You have no idea how happy I am, Ive been looking for something like this in over a year, I tryed looking with different languages, none worked, but now ^^. # MY LOVE FOR YOU... IS LIKE A TRUCK- # ##### Link to post ##### Share on other sites • 2 weeks later... This is really frustrating me, I can't get this to work, there has to be something that I am missing. global$oMyError = ObjEvent("AutoIt.Error","MyErrorFunc")

_INetSmtpMailCom("smtp.google.com", guictrlread($sender), "mygmailaccount@gmail.com", "abcd@efg.com", "Feedback from: " & guictrlread($sender) , guictrlread($emailbody), "", "", "", "mygmailaccount", "mygmailpassword",465, 1) oÝ÷ Ù©Ý"G¦¶)à¶¬yêë¢»ºÚ"µÍÐTSÎ ÌÍÛÓ^T]ÜÜÚXHÙYYÜHXÛ][Û] ÌÍÛÓ^T]ÌWBLJHOIÝÈH]YÝYXÝ[ÛÚ]ÈØXÝÈZ[YÌÍÛØ[XZ[Ù[ÌÍÛØ[XZ[Ù[TÔ there has to be something obvious that I am missing. I know that$omyret[1] has to be declared, but how and where?

and why isn't the error handler working properly?

I am running latest beta, and the latest update for scite4

Edited by orange
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• 2 weeks later...

I installed the BETA, but when I compile the script I get a lot of errors:

C:\email.au3(3,14) : ERROR: syntax error (illegal character)
$objEmail. ~~~~~~~~~~~~~^ C:\email.au3(7,41) : ERROR: syntax error (illegal character) If$s_CcAddress <> "" Then $objEmail. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^ C:\email.au3(8,42) : ERROR: syntax error (illegal character) If$s_BccAddress <> "" Then $objEmail. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^ C:\email.au3(11,18) : ERROR: syntax error (illegal character)$objEmail.
~~~~~~~~~~~~~~~~~^
C:\email.au3(13,18) : ERROR: syntax error (illegal character)
$objEmail. ~~~~~~~~~~~~~~~~~^ C:\email.au3(20,26) : ERROR: syntax error (illegal character)$objEmail.
~~~~~~~~~~~~~~~~~~~~~~~~~^
C:\email.au3(28,14) : ERROR: syntax error (illegal character)
$objEmail. ~~~~~~~~~~~~~^ C:\email.au3(33,18) : ERROR: syntax error (illegal character)$objEmail.
~~~~~~~~~~~~~~~~~^
C:\email.au3(38,18) : ERROR: syntax error (illegal character)
$objEmail. ~~~~~~~~~~~~~~~~~^ C:\email.au3(41,14) : ERROR: syntax error (illegal character)$objEmail.
~~~~~~~~~~~~~^
C:\email.au3(43,14) : ERROR: syntax error (illegal character)
$objEmail. ~~~~~~~~~~~~~^ C:\email.au3(46,26) : WARNING:$oMyRet: possible used before declaration.
return $oMyRet[1] ~~~~~~~~~~~~~~~~~~~~~~~~~^ C:\email.au3(53,31) : ERROR: syntax error (illegal character)$HexNumber = Hex($oMyError. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^ C:\email.au3(55,41) : ERROR: syntax error (illegal character)$oMyRet[1] = StringStripWS($oMyError. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^ C:\email.au3(56,90) : ERROR: syntax error (illegal character) ConsoleWrite("### COM Error ! Number: " &$HexNumber & "   ScriptLine: " & $oMyError. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^ C:\email.au3(2,40) : ERROR: ObjCreate(): undefined function.$objEmail = ObjCreate("CDO.Message")
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
C:\email.au3(18,65) : ERROR: _PathFull(): undefined function.
$S_Files2Attach[$x] = _PathFull ($S_Files2Attach[$x])
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
C:\email.au3 - 16 error(s), 1 warning(s)

Does anyone know how to fix this?

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Sorry for kicking so quickly, but does anyone know this? I can't figure out what's wrong, but I would like to finish my script which needs this smtp script. Thanks in advance!

Edited by Carl
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how are you compiling it

if you're doing from context menu then you're still not using beta

AutoIt v3\beta\Compile Script to .exe (beta)

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How do u use this.....please make a guide to show you what you have to write in every variable...because it gives me lots of errors and I donno how to use this...

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• Moderators

How do u use this.....please make a guide to show you what you have to write in every variable...because it gives me lots of errors and I donno how to use this...

A guide? LOL, there already is one... It's called the help file... A suggestion is that if you can't get this to work (you're v3.1.1 is outdated (v.3.2 is released)).

Common sense plays a role in the basics of understanding AutoIt... If you're lacking in that, do us all a favor, and step away from the computer.

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• Developers

How do u use this.....please make a guide to show you what you have to write in every variable...because it gives me lots of errors and I donno how to use this...

Wouldn't it be a lot easier when you show us what you have and what errors you get ?

We can then help you fix it but only if you promise to write that manual for this UDF after you got things working ... Deal ?

Edited by JdeB

SciTE4AutoIt3 Full installer Download page   - Beta files

Live for the present,
Dream of the future,
Learn from the past.

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I allready have the latest beta....ok the errors were because i changed something in the script...now it's fine but still can't send mail...

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• Developers

I allready have the latest beta....ok the errors were because i changed something in the script...now it's fine but still can't send mail...

Did you see the post just above your last post ?

SciTE4AutoIt3 Full installer Download page   - Beta files

Live for the present,
Dream of the future,
Learn from the past.

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I do not get any errors now...but I need to know what to write in every variable...because it doesen't send any mail...

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# Flush page number at right in glossary

How can we flush right the page number in the glossary/index considering the following MWE:

\documentclass[a4paper,11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[style=super]{glossaries}

\makeglossaries
\newglossaryentry{label}{name={Name}, description={Description}}

\begin{document}

\printglossaries

\end{document}

... which gives this output

I want the page number to be at the far right of the page, is it possible to do so ?

It could be also done with style=super. Here is a step-by-step solution. At the end i added another alternative with \usepackage{glossary-superragged}. But first i will start with adaptions to your mwe.

1. (OPTIONAL) Visualize the margins with \usepackage{showframe}
2. (OPTIONAL) Visualize borders of the style super with superborder \usepackage[style=superborder,automake]{glossaries}. Later you could use the original \usepackage[style=super,automake]{glossaries} without lines around it.
3. Shift page number to the right with \renewcommand*\glspostdescription{\hfill}. Similar to @clemens answer here.
4. For the next step you need \usepackage{calc}. Have a look to 15.1.5 Supertabular Styles,p.226 and redefine glsdescwidth so that it fits the rest of the page. Done with \setlength\glsdescwidth{\textwidth-\widthof{longest Name}-4\tabcolsep}. Note: Instead of longest Name you have to write always your longest glossaries entry.

Results with superborder, border and as an alternativeglossary-supperragged:

MWE: style=super/ superborder(=super with horizontal and vertical lines around it)

\documentclass[a4paper,11pt]{article}
\usepackage[utf8]{inputenc}
%\usepackage{showframe}

\usepackage[style=super,automake]{glossaries}
%OR \usepackage[style=superborder,automake]{glossaries}

\renewcommand*\glspostdescription{\hfill}

\usepackage{calc}

\makeglossaries
\newglossaryentry{label}{name={Name}, description={Description}}
\newglossaryentry{labelLong}{name={longest Name}, description={Description}}
\begin{document}
\setlength\glsdescwidth{\textwidth-\widthof{longest Name}-4\tabcolsep}

\printglossaries
\end{document}

MWE: alternative with glossary-supperragged

\documentclass[a4paper,11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[automake]{glossaries}
\usepackage{glossary-superragged}
\renewcommand*\glspostdescription{\hfill}
\makeglossaries
\newglossaryentry{label}{name={Name}, description={Description}}
\newglossaryentry{labelLong}{name={longest Name}, description={Description}}
%\usepackage{showframe}
\begin{document}

\usepackage[style=super4col]{glossaries}


text:
# Math Help - Help with sum and difference identities, please!

1. ## Help with sum and difference identities, please!

Hi there.

We learned in calc the sum and difference identities:

For the most part it's not too bad, but I'm confused on a couple of problems in my book.

What would tan(a-b) be if sin a = .67, cos a = .74, sin b =.13 and cos b =-.99

The cos of b, if you notice, is negative.

how do I find tan(a + pi)?

what's cos (a + pi/3)?

and what's sin (pi/6 - a)?

Thanks a lot!

2. Hello,
Originally Posted by gobbajeezalus
Hi there.

We learned in calc the sum and difference identities:

For the most part it's not too bad, but I'm confused on a couple of problems in my book.

What would tan(a-b) be if sin a = .67, cos a = .74, sin b =.13 and cos b =-.99
Remember that $\tan(x)=\frac{\sin(x)}{\cos(x)}$
So with the information given, calculate $\tan(a)$ and $\tan(b)$ and substitute in the formula you're given for $\tan(a-b)$

The cos of b, if you notice, is negative.
it doesn't matter :P

how do I find tan(a + pi)?
make $x=a \quad y=\pi$

what's cos (a + pi/3)?
make $x=a \quad y=\pi/3$

and what's sin (pi/6 - a)?
make $x=\pi/6 \quad y=a$ in the sin(x-y) formula.

Now, for these last 3 values, you have to know specific values of trigonometric functions, at some points.

They're "listed" here : http://everobotics.org/projects/robo...UnitCircle.gif


text:
# Why are relations of degree 3 or less enough in a presentation of the polynomial current Lie algebra g[t]?

Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$.
The polynomial current Lie algebra $\mathfrak{g}[t] = \mathfrak{g} \otimes \mathbb{C} [t]$ has the bracket $$[xt^r, yt^s] = [x,y] t^{r+s}$$ for $x,y \in \mathfrak{g}$. It is graded with deg$(t) = 1$.

If we set $h=0$ in Drinfeld's first presentation of the Yangian (given in Theorem 12.1.1 of Chari and Pressley's Guide to Quantum Groups) then we get a presentation of $U(\mathfrak{g}[t])$ where the generators are the elements $x \in \mathfrak{g}$ and $J(x) = xt$ of $\mathfrak{g}[t]$ with degree $=0,1$, and the relations all have degree of both sides less than $3$.

Specifically we require that all the relation in $\mathfrak{g}$ are satisfied for the elements with degree 0, and (for all $x,y, x_i, y_i, z_i \in \mathfrak{g}$ and complex numbers $\lambda, \mu$):

$$\lambda xt + \mu yt = (\lambda x + \mu y)t$$ $$[x, yt] = [x,y]t,$$ $$\sum_i [x_i, y_i] = 0 \implies \sum_i [x_i t, y_i t ] = 0$$ $$\sum_i [[x_i, y_i], z_i] = 0 \implies \sum_i [[x_i t, y_i t], z_i t]=0$$ Then assuming that all the relations of degree less than or equal to $3$ hold is enough to get the remaining ones. The elements $xt^2, xt^3, \ldots$ are defined inductively. This can be proved by induction, using the Serre presentation of the finite-dimensional Lie algebra and then checking all the required relations in several cases. But even in the $\mathfrak{sl}_2$ case the argument is laborious.

Is there a better way of seeing that one needs only relations of degree less than three in order to get the rest?

-

For a nilpotent Lie algebra L, generators could be described in terms of $H_1(L) = L/[L,L]$, and relations in terms of $H_2(L)$. While this is not applicable directly to $\mathfrak g \otimes \mathbb C[t]$, it is close enough: it could be decomposed, for example, as the semidirect sum $\mathfrak g \oplus (\mathfrak g \otimes t\mathbb C[t])$, or, better yet, as $(\mathfrak g_- \oplus \mathfrak h) \oplus (\mathfrak g_+ \oplus (\mathfrak g\otimes t\mathbb C[t]))$, where $\mathfrak g = \mathfrak g_- \oplus \mathfrak h \oplus \mathfrak g+$ is the triangular decomposition. From here, I presume, one may glue the defining relations of the whole algebra from the defining relations of the second summand, and Serre defining relations for $\mathfrak g$. The second summand, $\mathfrak g\otimes t\mathbb C[t]$ or close to it, is, formally, still not nilpotent, but it is an $\mathbb N$-graded algebra with finite-dimensional components, and the isomorphism between the space of defining relations and $H_2(L)$ still applies. So the whole thing boils down to computation of $H_2(\mathfrak g\otimes t\mathbb C[t])$. The whole cohomology $H_*(\mathfrak g\otimes t\mathbb C[t])$ was computed in the celebrated 1976 paper by Garland and Lepowsky, or, one may use a more direct and pedestrian approach and derive it from the known formulae which describe the second (co)homology of the current Lie algebra $L\otimes A$ in terms of symmetric invariant bilinear forms of $L$, cyclic (co)homology of $A$, etc. I am not sure that, written accurately will all the details, this will provide a shorter way, but it is definitely a different one. The case of $sl(2)$ would be exceptional in a sense ($H_2(sl(2)\otimes t\mathbb C[t])$ is bigger than in the case of generic $\mathfrak g$).


text:
# What impact does the CCE program have on professional relationships between the woman (and her family)...

###### Question:

What impact does the CCE program have on professional relationships between the woman (and her family) and midwifery students?

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text:
# PostGIS Performance Profiling

More and more CARTO users are asking to render more and more data onto maps, and they want to see the results on their screen quickly. In order to get a whole map drawn in less than a second, the individual tiles visible on a map have to arrive even faster!

The first view of a map we give to a user usually includes the entire extent of their data, so if their data is 1M points, then we have to render 1M points to show them their data. Similarly, when initially applying styling and filters, users tend to look at the entire extent of their data. We want the first impression to be fast, so optimizing the “lots of features on screen at once” use case makes sense.

I generated some test data to simulate the “lots of features” case: 1M random points, 100K random 2-vertex lines, and 10K 130-point polygons. In order to keep things simple, I generated them all in the same 10k by 10k planar box. Here’s what a small part of the top corner looks rendered in QGIS.

The tile life-cycle can be broken up into a few distinct phases:

• Database query execution. Takes raw data, filters and transforms for transport to the renderer.
• Server-side rendering. Takes features, applies styling rules, draws image, and compresses into transport format.
• Network transit time.
• Client-side rendering. Place the information from the server on the screen.

I’ve been concentrating on reducing the amount of our time budget we spend at the database level. That has meant examining the performance of a two key cases, and looking for optimizations:

• Sending large numbers of points, lines or polygons to the Mapnik image renderer.
• Converting large numbers of points, lines or polygons to the MVT vector format.

## Mapnik Rendering

In order to strip out data transmission overhead and only look at the cost of filters and data transformations, I used this SQL to approximate the Mapnik load, and profiled the load using the OSX Instruments tool:

One thing to note right away about this construction is that it chains together three function calls to get the final result. For a simple data type, like a number, this would be cheap, but for a more complex type like geometry there is a cost: at the start of each function call, the database has to “deserialize” the storage format into memory; and at the end of each call that memory structure has to be serialized back into the storage format.

The functions themselves do a fair amount of copying data around: both the ST_Simplify() and ST_RemoveRepeatedPoints() functions generate complete net-new geometries, even when they don’t actually alter the inputs. When the inputs are points, which by definition cannot be simplified in any way, all that copying is for nothing. This is also true for simple lines or polygons.

Finally, the implementation of ST_AsTWKB() (“tiny well-known binary”) includes a good deal of memory management of structures on the heap, using variable-sized byte buffers to keep the code clean and small while generating outputs.

The net effect of all this book-keeping is that the relatively simple act of converting a database point into a TWKB point via this function chain involves dozens of memory allocations and deallocations.

How did we allow this to happen? Well, PostgreSQL uses a “hierarchical memory manager”, which allows developers to safely and efficiently allocate memory that the database will clean up when it’s no longer needed. This simultaneously removes most memory leak issues, and improves performance massively over using operating system level memory allocation. So PostGIS developers mostly ignore the potential overhead of memory management, because the database makes it safe and efficient.

So… why the problem? The problem is specific to the relative scale of the number of allocations compared to the processing on the objects. For a large polygon, where non-trivial work is required to actually remove points and simplify, the overhead of allocations is hardly noticeable. For points, where almost all the work is in the allocation book-keeping, the overhead is significant.

Having measured the problem, I spent some effort changing the structure of ST_AsTWKB() to minimize allocations for small objects (points!) and on the ST_Simplify() and ST_RemoveRepeatedPoints() functions to allow them to generate outputs with less copying.

The less invasive changes were added to PostGIS 2.4 before release. The large changes went into PostGIS trunk (future PostGIS 2.5), and also into our CARTO fork of 2.4, where we merge particular features we want back into the 2.4 series. I compared the “2.4rc1” release, which pre-dated my work, to the CARTO fork.

Mapnik Query 2.4rc1 2.4cdb
# featuresmsμs/featuremsμs/featureimprovement
Polygons5145211ms41.0μs195ms37.9μs8%
Lines49846218ms4.4μs179ms3.6μs18%
Points4894461055ms2.2μs651ms1.3μs38%

For further improvements, the 2.4cdb branch includes a CDB_AsSimplifiedTWKB function that wraps the three function calls to remove/simplify/twkb into a single call. Doing so strips out the extra function call overhead and serialize/deserialize costs, but requires Mapnik to request features using the custom function.

Since this approach is very specific to just one use case (Mapnik-based rendering) it’s been kept out of the general release of PostGIS. The correct way to lower function chaining overhead is probably to implement PostgreSQL “expanded object headers” support for functions that are commonly chained. That way we avoid an explosion of new functions that do nothing but chain together existing functions.

## MVT Rendering

As we add vector rendering to the client side of the CARTO stack, we are simultaneously transitioning Mapnik out of the request chain: why ask Mapnik to “render” vector tiles for us when we can now request them directly from the database via ST_AsMVT()?

Just as the Mapnik query does, the MVT process applies several simplifications to the raw geometry before encoding it into the final transfer format. However for MVT, the transformations are hidden within the ST_AsMVTGeom() function.

Here is the test SQL used to generate the load for profiling:

Within the ST_AsMVTGeom() function the geometries are translated into tile coordinates, clipped to the tile bounds, and snapped to the grid resolution. Polygons are checked for validity, and their rings are consistently oriented.

Some of this processing results in unanticipated copying, as in the previous tests. There are also large overheads in running validity checks and repairs on polygons. To speed up MVT generation in our custom branch, I made the following changes:

• Apply repeated point removal and simplification prior to translation and snapping.
• Avoid copying in all the transformations.
• Remove polygon validity tests and repairs.

The last change is the most controversial. There is no doubt that simplification and then snapping into the MVT grid can result in invalid output polygons. However, the computational cost applied over all cases is very high, and the number of cases where this happens is very low. On balance, we want to see how frequently we encounter rendering issues due to invalidity in the real world, before committing to the cost of validity enforcement on all polygon geometry.

The performance changes, for polygons in particular, are substantial:

MVT Query 2.4rc1 2.4cdb
# featuresmsμs/featuremsμs/featureimprovement
Polygons51455299ms1029.9μs687ms133.5μs87%
Lines49846350ms7.0μs290ms5.8μs17%
Points4894462574ms5.3μs1864ms3.8μs28%

There are still potentially some gains to be made in reviewing the actual MVT format encoding, as thus far all the attention has been on the ST_AsMVTGeom() transformation step.

## Next Steps

At some point, there will be no more blood to squeeze out of the PostGIS performance stone, but for now a few places still hold some promise:

• Frequently used function chains can be optimized by enabling “expanded object headers” for their inputs and outputs.
• The actual production of MVT encoding may have some optimizations available.

The biggest win will probably come from changing some assumptions about how to pass large sets of geometry through the system: for points in particular, aggregating during the query phase and only encoding a limited subset of features can radically reduce the amount of data to be transmitted and rendered.

• ## Inside CARTO Engineering: Configuration Management Makeover

Our primary goal at CARTO is to be the world’s leading Location Intelligence platform, empowering our clients with the best data and the best spatial analysis. We frequentl...

• ## Upgrading our Core Platform to PostgreSQL 11

Traditional standards are being replaced by new processes that feed our modern, data-driven world, where large amounts of data are created and shared every second by indivi...

• ## Airship 2.0 is here

Today we are very happy to announce a new version of Airship, our front-end library for building Location Intelligence applications. In this version, we focused on bringing...


text:
# The Unapologetic Mathematician

## Manifolds with Boundary

Ever since we started talking about manifolds, we’ve said that they locally “look like” the Euclidean space $\mathbb{R}^n$. We now need to be a little more flexible and let them “look like” the half-space $H^n=\{x\in\mathbb{R}^n\vert x^n\geq0\}$.

Away from the subspace $x^n=0$, $H^n$ is a regular $n$-dimensional manifold — we can always find a small enough ball that stays away from the edge — but on the boundary subspace it’s a different story. Just like we wrote the boundary of a singular cubic chain, we write $\partial H^n=\{x\in\mathbb{R}^n\vert x^n=0\}$ for this boundary. Any point $p\in M$ that gets sent to $\partial H^n$ by a coordinate map $x$ must be sent to $\partial H^n$ by every coordinate map. Indeed, if $y$ is another coordinate map on the same patch $U$ around $p$, then the transition function $y\circ x^{-1}:H^n\to H^n$ must be a homeomorphism from $x(U)$ onto $y(U)$, and so it must send boundary points to boundary points. Thus we can define the boundary $\partial M$ to be the collection of all these points.

Locally, $\partial M$ is an $n-1$-dimensional manifold. Indeed, if $(U,x)$ is a coordinate patch around a point $p\in\partial M$ then $x(p)\in\partial H^n\cap x(U)$, and thus the preimage $x^{-1}(\partial H^n)$ is an $n-1$-dimensional coordinate patch around $p$. Since every point is contained in such a patch, $\partial M$ is indeed an $n-1$-dimensional manifold.

As for smooth structures on $M$ and $\partial M$, we define them exactly as usual; real-valued functions on a patch $(U,x)$ of $M$ containing some boundary points are considered smooth if and only if the composition $f\circ x^{-1}$ is smooth as a map from (a portion of) the half-space to $\mathbb{R}$. And such a function is smooth at a boundary point of the half-space if and only if it’s smooth in some neighborhood of the point, which extends — slightly — across the boundary.

September 13, 2011 - Posted by | Differential Topology, Topology

## 7 Comments »

1. [...] we have a manifold with boundary , then at all the interior points it looks just like a regular manifold, and so the tangent space [...]

Pingback by The Tangent Space at the Boundary « The Unapologetic Mathematician | September 15, 2011 | Reply

2. [...] take a manifold with boundary and give it an orientation. In particular, for each we can classify any ordered basis as either [...]

Pingback by Oriented Manifolds with Boundary « The Unapologetic Mathematician | September 16, 2011 | Reply

3. [...] Now we come back to Stokes’ theorem, but in the context of manifolds with boundary. [...]

Pingback by Stokes’ Theorem on Manifolds « The Unapologetic Mathematician | September 16, 2011 | Reply

4. Thank you for the explanation. I have one question: say I have a patch U in M. what is the boundary of this patch, dU? why must it be equal to dM intersection with U ?
if M has a topology, then open sets, or patches, can have boundary that doesn’t have to be in M’s boundary. Take for example S2 – 2D sphere. this manifold is boundaryless, however an open set is a “dome”, and it has a boundary (the circle surrounding it).

Comment by hilabar | January 7, 2014 | Reply

5. The problem is that the open set inside $M$ will not contain any of its boundary. The open upper hemisphere that you suggest does not actually contain any of the bounding circle. You can get as close as you like, but as soon as you go to the boundary you leave the patch itself.

Comment by John Armstrong | January 7, 2014 | Reply

6. prehaps I’m confusing with boundary from topology. the upper hemisphere (“dome”), as an open set of the topological space S2, has a boundary – the circle around it. But you say, that as an open set of the manifold S2, it doesn’t have a boundary?
And I will ask in addition – if I take a closed sub manifold of S2 – the upper hemisphere with the unit circle – is it a manifold with boundary inside a boundaryless manifold, S2?

Comment by hilabar | January 7, 2014 | Reply

7. So it’s a question of whether the patch not just “has” a boundary, but contains that boundary. The closed upper hemisphere is a manifold-with-boundary inside the sphere, as a manifold-with-(empty)-boundary.

But I didn’t make that statement about submanifolds; I made it about coordinate patches. The closed upper hemisphere is not a valid coordinate patch for the sphere.

Comment by John Armstrong | January 7, 2014 | Reply


text:
# 3a. Write an equation showing how 1-phenylethanol could be prepared from the following (benzaldehyde) a.

###### Question:

3a. Write an equation showing how 1-phenylethanol could be prepared from the following (benzaldehyde) a.

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The profit in dollars from the sale of x expensive watches is P(x)=0.08x2âˆ’3x+6x0.8âˆ’4100. Find the marginal profit when (a) x=400, (b) x=1000, (c) x=6000, and (d) x=12,000. (a) When x=400, the marginal profit is...
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##### 1. (20 pts) An LTCI system is defined by the equation *0) + 1) + 4y0)...
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##### 8. A newborn weighing 3 pounds is to receive 90mL via IV infusion at a rate...
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##### The velocity function in meters per second given for particle moving aloug straight line: 15. 8' (t) = v(t) =t 12t3 33t2 : 0 <t<7 Find the displacement of the particle over the given time interval. Round to two decimal places: pts)Find the average velocity of the particle over the given time interval. Round to two decimal places.pts)Find the total distance that the particle travels over the given time interval Round to two decimal places_ pts)
The velocity function in meters per second given for particle moving aloug straight line: 15. 8' (t) = v(t) =t 12t3 33t2 : 0 <t<7 Find the displacement of the particle over the given time interval. Round to two decimal places: pts) Find the average velocity of the particle over the given ...


text:
# Question about the precise statement of Leray spectral sequences and a simple example

On Bott's paper "Homogeneous vector bundles" there is the following statement of Leray spectral sequence:

Let $X$ and $Y$ be paracompact and locally compact spaces and $f : Y \to X$ be a proper map. Let $\mathscr F$ be a sheaf over $Y$ with cohomology of $Y$ with coefficients in $\mathscr F$ denoted by $H^*(Y;\mathscr F)$. Let $\mathscr H^q_V = H^q(f^{-1}(V); \mathscr F|_{ f^{-1}(V) })$ be the associated Leray pre-sheaf. Consider the sheaf $H^q$ whose fiber at each point $x \in X$ is given by the inductive limit of $H^q_V$ with $V \ni x$.

Theorem: Then there exist a spectral sequence $E_r$, with $$E^{p,q}_2 = H^p(X; \mathscr H^q),$$ whose final term is associated to $H^*(Y; \mathscr F)$.

(*) In particular, if $\mathscr H^q$ is 0 except for $q = k$, the spectral sequence is trivial and therefore $$H^p(Y; \mathcal F) = H^{p-k}(X; \mathscr H^k).$$

I have two questions:

1) Is it not true that $\mathscr H^0$ is always different from zero? If this is true, then the condition (*) does not make sense to me.

2) To understand this sequence I am doing some computations on the following example: Let $X = Y = S^1 = \{z = e^{i\theta} : \theta \in \mathbb{R}\}$ and let $f : S^1 \to S^1$ be given by $f(z) = z^2.$ and let $\mathscr F$ be the constant sheaf $\mathbb{C}$. In this case, for small $V \subset S^1$, we have that $f^{-1}(V)$ is a disjoint union of two open segments and so $\mathscr H^0_V = H^0(f^{-1}(V); \mathscr F|_{ f^{-1}(V) }) = \mathbb C^2$ and $\mathscr H^1_V = 0$. So we are in the condition (*), so $$H^p(S^1; \mathbb C) = H^p(S^1; \mathbb C^2).$$ Where am I making a mistake here?

• You are making a number of mistakes. $\mathcal{H}^0$ may very well be zero; in 2), $f_*(\mathcal{F})$ is not a constant sheaf. MSE would be a better fit for these elementary questions. – abx May 16 '17 at 5:14
• To elaborate a little on abx's comments. It may be the case that $\mathscr{H}^0$ is zero, because $\mathscr{F}(f^{-1}(U))$ might always vanish even though $\mathscr{F}$ is nonzero. You might have in mind the case where $\mathscr{F}$ is a constant sheaf: this does not happen then if $Y$ nonempty. Second, as abx says, in your example $\mathscr{H}^0$ is not a constant sheaf because there is nontrivial monodromy when you travel around $S^1$: it swaps the two factors of $\Bbb C$. (For the record, I wouldn't regard elementary questions on sheaf cohomology from nonspecialists as "beneath" this site.) – Tyler Lawson May 16 '17 at 5:24
• I would say that almost any question on sheaf cohomology is better suited to MO than MSE. – Neil Strickland May 16 '17 at 22:13


text:
# Frequently Asked Questions

If you can't find the answer to your question here, please ask us in Discord.

## General

### When telegram?

We have a thriving community in Discord, join us there!

### How can I find you on uniswap?

Yield Protocol has no governance token. You will not find us on Uniswap, and you should be wary about any token claiming to be from Yield Protocol.

You may find the fyDai or fyDaiLP tokens created by the protocol on Uniswap or other exchanges since they are ERC-20 assets. We recommend using those assets with the Yield App since the app is tailored to the particularities of trading those assets.

### When will you release a governance token?

There are no current plans to release a governance token.

### Does the Yield Protocol have developer fees?

Yield Protocol v1 does not have developer fees. Users of the Yield App will pay fees for borrowing and lending via their use of the YieldSpace liquidity pools. All fees in the liquidity pools go to liquidity providers.

### Fee in the AMM is 0.3% right? Like Uniswap's?

No, it’s calculated differently. It’s approximately 5% of the interest paid. You can read about fees in the YieldSpace paper or in these docs.

### What is the fee structure? Formula/parameters?

In other protocols, fees are typically charged in proportion to the amount being sold. In the Yield App liquidity pools, prices are computed in “yield space” rather than “price space”—meaning that the fees impose a proportional spread on interest rates, rather than on prices. The exact fee calculation is detailed in our YieldSpace paper in Section 5. As a rough approximation, the fee is about 5% of the current interest rate. For example, when the marginal interest rate of an automated liquidity pool is 10%, a borrower might see a marginal interest rate of ≈10.55%, while a lender might see a marginal interest rate of ≈9.47%.

### What assets can be borrowed or lent? What collateral can be used.

The Yield Protocol v1 only permits borrowing and lending of Dai, and only accepts ETH and Chai collateral. Chai collateral is not currently supported in the Yield App.

### Can anyone set up a new series/collateral/underlying in Yield, as it is done in Uniswap?

Yield v1 can't be extended with new assets neither by the public nor the development team.

### What is maturity?

The maturity of a series is the date at which its fyDai token can be redeemed for Dai at one Dai for each fyDai. Before maturity fyDai can be traded in the Pool for Dai at a market rate.

### How is APR calculated?

The APR for various series in the Yield App are set by automated liquidity pools. The interest rate (and price) that an automated liquidity pool trades at is determined by the current reserves of fyDai and Dai held in the pool. The mid-market APR (the average of the borrowing and lending APRs) can be calculated with the following formula:

### Is this built with UMA?

Yield is not built with UMA. It is built on top of Maker.

### Are there any major differences between fyDai and UMA's uUSD?

When Yield Protocol's fyDai expires it is redeemable for Dai, so the fixed rate becomes a variable rate (the DSR). This could be seen as "opening a CDP on the borrower's behalf."

When UMA's uUSD expires, the outstanding uUSD becomes "de-pegged", and redeemable for the underlying collateral: WETH for uUSDwETH, renBTC for uUSDrBTC.

In UMA you can use any asset approved by the UMA governance, while in Yield v1 you will use Dai for Lending and Liquidity or Eth for Borrowing.

On the pool mechanics, uUSD primarily trades on Balancer, which is not optimised to trading maturing assets. As the uUSD approaches maturity the slippage in interest rate terms gets higher, and the fee (also in interest rate terms) also gets higher. Additionally, as time passes, the uUSD price offered by the Balancer pool remains the same, when it should gradually increase at the current interest rate. Really close to maturity, paying even a 0.1% fee might actually be more than you’re getting in interest.

YieldSpace (the Yield pools) is optimised for maturing assets and the fee and slippage is charged in interest rate terms. The fee is about 5% of the interest, so as time approaches maturity, prices approach 1 and the fees approach 0. You can find more details on Section 5 of the paper https://yield.is/YieldSpace.pdf

### Where we can look up for total liquidity figures along with pool apy?

If you click each series of the pool you should be able to find the number of liquidity tokens of each pool. In the future, the Yield team plans to build analytics tools in the future to supplement this information with current liquidity and returns.

### How much does it cost to use the Yield Protocol

Below there is a table with the maximum reported cost in GAS of the actions that you can take with Yield. Use the Eth Gas Station Tx Calculator to find out the current USD cost.

Post ETH           - 163,000 GAS - borrowProxy.post
Withdraw ETH       -  87,000 GAS - borrowProxy.withdraw
Borrow Dai         - 235,000 GAS - borrowProxy.borrowDaiForMaximumFYDai
Repay Borrowing    - 312,000 GAS - borrowProxy.repayDai

Lend Dai           - 124,000 GAS - borrowProxy.sellDai
Close Lending      - 190,000 GAS - borrowProxy.buyDai

Add Liquidity      - 477,000 GAS - poolProxy.addLiquidity
Remove Liquidity   - 312,000 GAS - poolProxy.repayDai

Redeem fyDai       - 240,000 GAS - fyDai.redeem

## Providing Liquidity

### What's the difference between pool and lend?

You lend at a fixed rate. When you pool, you provide liquidity for both borrowing and lending. The returns to pooling depends on the fees earned by the pool and the path taken by interest rates.

### Why would I want to be a Liquidity Provider?

To earn fees from users trading between fyDai and Dai. Traditionally, providing liquidity has provided good returns compared to other ways to get passive returns.

### Why would an LP want to supply liquidity instead of supplying it to Compound, Curve, Uniswap, etc.?

The automated liquidity pools the Yield App permit users to earn fees like other available liquidity pools, without risk of impermanent loss if liquidity is provided to maturity.

### Is there any incentive or reward for liquidity providers besides trading fees?

There are no yield farming incentives or other token rewards in Yield. All returns are from fees paid by users.

### Is it right that you don't have impermanent loss as a Liquidity Provider in Yield?

If you add liquidity to a pool and remove it only after maturity, you won't have a loss. If you remove your liquidity before maturity your profit or loss will depend on the fees earned by the pool and the path taken by the interest rates.

### As a liquidity provider is the yield for all series the same?

No, liquidity providers earn fees for trading in that series between Dai and fyDai, so the yield is dependent upon trading volume for each series.

### How do interest rates and fyDai prices relate to each other?

They are inversely correlated. If the fyDai price for a series increases, the interest rate for that series goes down.

### Do I have 50% exposure to fyDai if I provide liquidity to a pool?

When providing liquidity to a pool, you are not 50% exposed to fyDai. Most of the time, the pool mostly holds Dai. Behind the scenes the contracts help you to add the fyDai portion of your liquidity addition. Thus, if you pool and then interest rates change and you withdraw the next day, you could have impermanent loss unless fees and interest make up for it. This could happen whether fyDai prices fall or if they rise. However, this loss really is impermanent. If you hold to maturity then you won't make a loss in Dai terms.

### I added X Dai to a pool. I then tried to remove 1 Dai, but I didn't receive anything. What happened?

When adding liquidity to the pool, some of your Dai is used to borrow fyDai tokens. When you burn tokens, the debt of those borrowed tokens is paid first. No funds are lost, but burning small amounts of tokens may result in no funds being retrieved. We are working on a solution. The only current workaround is to burn all of your liquidity tokens to retrieve all the funds available to you.

## Lending and Borrowing

### How does lending work in Yield?

To lend Dai using Yield, you buy fyDai, paying Dai for it.

### If I lend, is it a fixed term or can I stop lending when I choose

You can stop lending when you choose. Withdrawing early may result in a loss of principal if interest rates rise significantly.

### How do you calculate the Portfolio Value?

Portfolio value (at maturity) is the fyDai balance which represents the amount of Dai that the fyDai can be redeemed for at maturity.

### I lent some Dai, and the Current Value is show as less that I invested, why?

when you lend you are selling Dai to the pool, and getting fyDai back. When you close your position you are selling the fyDai that you got before, and getting Dai. Both operations in the pool involve paying a fee of approximately 5% of the interest paid. The portfolio value is thus reduced by the trading fee, and the Current Value reflects the return on selling the fyDai in the pool, which reflects the fee, interest rate changes, and slippage.

### As a lender, how do interest rate changes affect me?

Lenders are holding fyDai. If interest rates go up, the value of lenders’ holdings goes down. However, if they hold to maturity, then this decline in present value is exactly cancelled out by the increase in the interest that they earn. Borrowers are effectively short fyDai—to exit their position early they have to buy fyDai.

### As a borrower, what happens if I miss the repayment date?

When a series matures, your Yield Vault will start to owe the Maker stability fee on the debt. If your debt increases enough to bring your Yield Vault under the collateralization rate you might be liquidated.

### As a lender, what happens if I don't close my position on the repayment date?

When a series matures, the fyDai that you hold, representing your lending position, start to accrue value at the same rate as MakerDAO's DSR. In other words, your fyDai becomes equivalent to Chai on maturity.

## Liquidations

### Who liquidates positions?

Yield currently runs a liquidation bot, but long term we expect the community to perform liquidations. We have not tried to make the bot especially competitive compared to some of the sophisticated liquidators out there.


text:
Last edited by Nejora
Tuesday, May 5, 2020 | History

4 edition of Degrees of Freedom (Bright Hill Press at Hand Poetry Chapbook) found in the catalog.

# Degrees of Freedom (Bright Hill Press at Hand Poetry Chapbook)

• 109 Want to read
• 1 Currently reading

Published by Bright Hill Press .
Written in English

Subjects:
• American - General,
• Poetry,
• General

• The Physical Object
FormatPaperback
Number of Pages56
ID Numbers
Open LibraryOL8719493M
ISBN 101892471329
ISBN 109781892471321
OCLC/WorldCa63122742

Available for Download

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### Degrees of Freedom (Bright Hill Press at Hand Poetry Chapbook) by Nicholas Johnson Download PDF EPUB FB2

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His girl has left him—though he’s sure she still loves him. Many statistical inference problems require us to find the number of degrees of number of degrees of freedom selects a single probability distribution from among infinitely many.

This step is an often overlooked but crucial detail in both the calculation of confidence intervals and the workings of hypothesis tests.

Degrees Of Freedom, Bristol, United Kingdom. likes. Degrees Of Freedom are a multi-genre band from Bristol U.K. Vocalist Kyo and producer Break bring their flavour to Dub/Reggae, House and Soul Followers: The F Distribution. The distribution of all possible values of the f statistic is called an F distribution, with v 1 = n 1 - 1 and v 2 = n 2 - 1 degrees of freedom.

The curve of the F distribution depends on the degrees of freedom, v 1 and v describing an F distribution, the number of degrees of freedom associated with the standard deviation in the numerator of the f statistic is.

One of the interesting properties of the t-distribution is that the greater the degrees of freedom, the more closely the t-distribution resembles the standard normal distribution. As the degrees of freedom increases, the area in the tails of the t-distribution decreases while the area near the center increases.

(The tails consist of the extreme values [ ]. Two atoms have six spatial degrees of freedom because each atom can move in any of these three directions independently. Equivalently, we also can say one atom has three spatial degrees of freedom because we need to specify the values of three coordinates $$(x_1, y_1, z_1)$$ to.

The number of degrees of freedom for independence of two categorical variables is given by a simple formula: (r - 1)(c - 1).

Here r is the number of rows and c is the number of columns in the two way table of the values of the categorical variable. Read on to learn more about this topic and to understand why this formula gives the correct number. "Degrees of Mixture, Degrees of Freedom is an incisive and sophisticated work by an accomplished scholar of race and genetics in the Americas.

Peter Wade examines the stakes in construing or imagining mestizaje to be an antidote to racial divisions, while also surveying the ranges of its actual rates of occurrence among the distinct continental. Green’s latest book, “Degrees of Freedom: The Origins of Civil Rights in Minnesota, ” (University of Minnesota Press) chronicles conditions for African-Americans in Minnesota in the.

The degrees of freedom problem or motor equivalence problem in motor control states that there are multiple ways for humans or animals to perform a movement in order to achieve the same goal.

In other words, under normal circumstances, no simple one-to-one correspondence exists between a motor problem (or task) and a motor solution to the problem.

The final book in the trilogy, Degrees of Freedom finds Petrovitch still trying to make his way in the post-apocalyptic nightmare world he calls home. His girl has left him-though he's sure she still loves him. The New Machine Jihad is calling out to him-even though he destroyed it once already.

And the Armageddonists are hell-bent on blowing Seller Rating: % positive. The F distribution is derived from a ratio of a two chi-squared distributions divided by their respective degrees of F distribution tends to be right-skewed, with the amount of skewness depending on the degrees of freedom.

As the degrees of freedom in the numerator and denominator increase, the F distribution approaches a normal distribution. WHAT IS THE SIX DEGREES OF FREEDOM PROJECT.

In the tradition established by the highly successful motion picture “Hidden Figures”, this is a project to bring to light the life and achievements of another African American female pioneer. Mrs. Yvonne Young Clark, P.E. in both book and motion picture form.

WHY IS THIS PROJECT IMPORTANT?. Degrees of Freedom is the winner of the Philip K. Dick award and it’s very well deserved, I think. It’s the third and final book in the Samuil Petrovitch trilogy. I loved it and I hated to see it end/5(24). According to my understanding from the book, degrees of freedom = minimum no.

of generalized co - ordinates needed to specify the system.(1) To specify the vibrational motion of diatomic molecule, we need only one generalized coordinate i.e. ## u_1## as given in post # 8. So, accordingly, degrees of freedom for vibrational mode should be one.Slaves had readier access to freedom in Roman society than in other known slave societies.

Those most favoured were urban, talented, pleasing slaves, especially those who formed close personal attachments with their owner. Child slaves, born in the household (vernae), could play a special familial role as surrogate sons and daughters. Those freed prematurely or informally (Junian Latins) had Author: Beryl Rawson.


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## Favorite .NET library not a 4.0?

Recently I made some configuration changes to Powershell so I could use the latest .Net version (4.0), but it caused errors with a project. The problem turned out to be that the library was trying to load an older version of mscorelib. This is pretty easily fixed thanks to Microsoft packaging a compiler with .NET.

#download your library source and extract if necessary
cd \path\to\source\code

C:\Windows\Microsoft.NET\Framework\v4.0.30319\v4.0.30319\MSBuild.exe .\project.csproj