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# Counting money worksheets information
» » Counting money worksheets information
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Counting Money Worksheets. Worksheets include pictures of 2 1 50p 20p 10p 5p 2p and 1p coins. You can also limit the money value to less than 1 pound the answers will be. From a coin value chart to testing their recognition these are a great first step. You can choose to include or not include the 1-p coin 2-p coin 5-p coin 10-p coin 20-p coin 50-p coin 1-pound coin 2-pound coin 5-pound note and the 10-pound note.
Counting Money Worksheets 1st Grade First Grade Math Worksheets Kindergarten Money Worksheets Money Math From pinterest.com
You can choose to include or not include the penny nickel dime quarter half-dollar one dollar loonie two-dollar toonie five-dollar bill and ten-dollar bill. Counting Money Worksheets This group of money worksheets is designed to get kids to count up the value of a group of coins and write the value in cents on the line provided. If you wish to have more control on the options such as number of problems or font size or spacing of problems or range of numbers just click on these links to use the worksheet generators yourself. Explore the Australian money worksheets and acquire immense hands-on experience in counting coins and notes with varied exercises to comprehend the value of money and develop life skills like spending and saving. Using the Counting Money Worksheets. Counting money worksheets for grade 2.
### Bills and coins is sure to impart an in-depth knowledge in managing money as kids practice counting coins like pennies nickels dimes quarters and 1 5 10 100 bills.
Used by over 10 million students. Counting Money United Kingdom - Pounds Practice counting UK money with these printable worksheets. Counting pennies nickels and dimes a free lesson. You can choose to include or not include the 1-p coin 2-p coin 5-p coin 10-p coin 20-p coin 50-p coin 1-pound coin 2-pound coin 5-pound note and the 10-pound note. Counting money worksheets for grade 2. Children will progress to complete bar models to represent amounts.
Source: pinterest.com
You can choose to include or not include the 1-p coin 2-p coin 5-p coin 10-p coin 20-p coin 50-p coin 1-pound coin 2-pound coin 5-pound note and the 10-pound note. Real Life Problems Christine Sparke Sheet 1 PDF 2 PDF. Used by over 10 million students. Differentiated worksheets requiring children to count money pounds and write the amount shown. Combine childrens knowledge of money with counting in 2s 5s and 10s to count money efficiently with our range of money KS1 worksheets and activities.
Source: pinterest.com
There are also worksheets with coins in order by denomination as well as in random order just like they come out of your pocket. The worksheets on this page start with a certain number of specific bills then progress to mixed bills and finally to combinations of bills with coins. Counting pennies nickels and dimes a free lesson. Measure Problems Stephen Norwood PDF. Get kids to.
Source: pinterest.com
Counting coin worksheets including PDF printables for counting and calculating sets of the same coin or of multiple types of coins. Counting pennies nickels and dimes a free lesson. Pence This generator makes maths worksheets for counting British coins and notes sterling. Used by over 10 million students. The most comprehensive learning site for R-13.
Source: pinterest.com
Counting pennies nickels and dimes a free lesson. Children will progress to complete bar models to represent amounts. Bills and coins is sure to impart an in-depth knowledge in managing money as kids practice counting coins like pennies nickels dimes quarters and 1 5 10 100 bills. Students encounter money early on and they must be able to manage it themselves in their everyday lives and into adulthood. Using the Counting Money Worksheets.
Source: pinterest.com
Buying Items GP Magnificent Maths Shop James Maloney Money. Pence This generator makes maths worksheets for counting British coins and notes sterling. Get kids to. Pennies Only Worksheet 1. These kindergarten worksheets will help your students get there.
Source: pinterest.com
Ad Parents worldwide trust IXL to help their kids reach their academic potential. There are also worksheets with coins in order by denomination as well as in random order just like they come out of your pocket. Counting Money United Kingdom - Pounds Practice counting UK money with these printable worksheets. Differentiated sheets some have only 1s and 2s others have 1s 2s 5s and 10s. Pence This generator makes maths worksheets for counting British coins and notes sterling.
Source: pinterest.com
Students encounter money early on and they must be able to manage it themselves in their everyday lives and into adulthood. Money Counting Worksheet - Four of Four. This compilation of money worksheets on counting US. Worksheets include pictures of 2 1 50p 20p 10p 5p 2p and 1p coins. You can also limit the money value to less than 1 pound the answers will be.
Source: pinterest.com
Measure Problems Stephen Norwood PDF. Students encounter money early on and they must be able to manage it themselves in their everyday lives and into adulthood. Pence This generator makes maths worksheets for counting British coins and notes sterling. Worksheets include pictures of 2 1 50p 20p 10p 5p 2p and 1p coins. Counting Money Worksheets This group of money worksheets is designed to get kids to count up the value of a group of coins and write the value in cents on the line provided.
Source: pinterest.com
Children will progress to complete bar models to represent amounts. These printable money worksheets feature realistic coins and bills in problems for identifying coins making change counting coins comparing amounts of money. Bills and coins is sure to impart an in-depth knowledge in managing money as kids practice counting coins like pennies nickels dimes quarters and 1 5 10 100 bills. We also have our international section featuring Canadian British and Australian currencies. From a coin value chart to testing their recognition these are a great first step.
Source: pinterest.com
22022018 doc 247 MB doc 221 MB Worksheets to help children count different monetry values using 1ps 2ps 5ps and 10ps. Pennies Only Worksheet 1. Click on any of the images below. Money Worksheets Use the worksheets below to help teach counting coins counting dollars and cents and making change. This compilation of money worksheets on counting US.
Source: pinterest.com
You can choose to include or not include the 1-p coin 2-p coin 5-p coin 10-p coin 20-p coin 50-p coin 1-pound coin 2-pound coin 5-pound note and the 10-pound note. Combine childrens knowledge of money with counting in 2s 5s and 10s to count money efficiently with our range of money KS1 worksheets and activities. Get kids to. Ad Parents worldwide trust IXL to help their kids reach their academic potential. The most comprehensive learning site for R-13.
Source: pinterest.com
There are also worksheets with coins in order by denomination as well as in random order just like they come out of your pocket. Bills and coins is sure to impart an in-depth knowledge in managing money as kids practice counting coins like pennies nickels dimes quarters and 1 5 10 100 bills. Counting money is the next step when students know basic monetary values and can use skills like skip counting to figure out how many bills add up to a particular dollar value. Before your students are able to add subtract multiply and skip count with coins they will need to know the denomination of each coin. Explore the Australian money worksheets and acquire immense hands-on experience in counting coins and notes with varied exercises to comprehend the value of money and develop life skills like spending and saving.
Source: pinterest.com
British UK money worksheets pound. Addition of Money Kevin Kerr PDF. Buying Items GP Magnificent Maths Shop James Maloney Money. Before your students are able to add subtract multiply and skip count with coins they will need to know the denomination of each coin. These kindergarten worksheets will help your students get there.
Source: pinterest.com
This compilation of money worksheets on counting US. This page includes Money worksheets for counting coins and for operations with Dollars Euros and Pounds. Students encounter money early on and they must be able to manage it themselves in their everyday lives and into adulthood. Counting money worksheets for grade 2. Differentiated worksheets requiring children to count money pounds and write the amount shown.
Source: pinterest.com
The most comprehensive learning site for R-13. Counting pennies nickels and dimes a free lesson. This page includes Money worksheets for counting coins and for operations with Dollars Euros and Pounds. Real Life Problems Christine Sparke Sheet 1 PDF 2 PDF. Explore the Australian money worksheets and acquire immense hands-on experience in counting coins and notes with varied exercises to comprehend the value of money and develop life skills like spending and saving.
Source: pinterest.com
The most comprehensive learning site for R-13. Buying Items GP Magnificent Maths Shop James Maloney Money. Differentiated sheets some have only 1s and 2s others have 1s 2s 5s and 10s. Money Counting Worksheet - Four of Four. Differentiated worksheets requiring children to count money pounds and write the amount shown.
Source: pinterest.com
Get kids to. Explore the Australian money worksheets and acquire immense hands-on experience in counting coins and notes with varied exercises to comprehend the value of money and develop life skills like spending and saving. From a coin value chart to testing their recognition these are a great first step. Worksheets include pictures of 2 1 50p 20p 10p 5p 2p and 1p coins. These kindergarten worksheets will help your students get there.
Source: pinterest.com
Get kids to. You can choose to include or not include the penny nickel dime quarter half-dollar one dollar loonie two-dollar toonie five-dollar bill and ten-dollar bill. Before your students are able to add subtract multiply and skip count with coins they will need to know the denomination of each coin. British UK money worksheets pound. If you wish to have more control on the options such as number of problems or font size or spacing of problems or range of numbers just click on these links to use the worksheet generators yourself.
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Author Archimedes Ancient Greek Hydrostatics
On Floating Bodies (Greek: Περὶ τῶν ἐπιπλεόντων σωμάτων) is a work, originally in two books, by Archimedes, one of the most important mathematicians, physicists, and engineers of antiquity. Thought to have been written towards the end of Archimedes' life, On Floating Bodies I-II survives only partly in Greek and in a medieval Latin translation from the Greek. It is the first known work on hydrostatics, of which Archimedes is recognized as the founder.[1]
The purpose of On Floating Bodies I-II was to determine the positions that various solids will assume when floating in a fluid, according to their form and the variation in their specific gravities. The work is known for containing the first statement of what is now known as Archimedes' principle.
## Overview
Archimedes lived in the Greek city-state of Syracuse, Sicily, where he was known as a mathematician and as a designer of machines, some of which might have helped keeping Roman armies at bay during the Second Punic War.[2] Archimedes' interests in the conditions of stability for solid bodies are found both here and in his studies of the lever and centre of gravity in On the Equilibrium of Planes I-II.
Book one of On Floating Bodies begins with a derivation of the Law of Buoyancy and ends with a proof that a floating segment of a homogeneous solid sphere is always in stable equilibrium when its base is parallel to the surface of a fluid. Book two extends Archimedes' study from the segment of a sphere to the case of a right paraboloid and contains many sophisticated results.
Although the work is extant in Latin translation, the only known copy of On Floating Bodies I-II in Greek comes from the Archimedes Palimpsest.[3]
## Contents
### First book
In the first part of book one, Archimedes establishes various general principles, such as that a solid denser than a fluid will, when immersed in that fluid, be lighter (the "missing" weight found in the fluid it displaces). Archimedes spells out the law of equilibrium of fluids, and proves that water will adopt a spherical form around a centre of gravity.[4] This may have been a reference to contemporary Greek theory that the Earth is round, which is also found in the works of others such as Eratosthenes. The fluids described by Archimedes are not self-gravitating, since he assumes the existence of a point towards which all things fall in order to derive the spherical shape. Most notably, On Floating Bodies I contains the concept which became known as Archimedes' principle:
Any body wholly or partially immersed in a fluid experiences an upward force (buoyancy) equal to the weight of the fluid displaced
In addition to the principle that bears his name, Archimedes discovered that a submerged object displaces a volume of water equal to the object's own volume (upon which the story of him shouting "Eureka" is based). This concept has come to be referred to by some as the principle of flotation.[4]
### Second book
Book two of On Floating Bodies is considered a mathematical achievement unmatched in antiquity and rarely equaled until after the late Renaissance.[1] Heath called it "a veritable tour de force which must be read in full to be appreciated."[5] The book contains a detailed investigation of the stable equilibrium positions of floating right paraboloids of various shapes and relative densities when floating in a fluid of greater specific gravity, according to geometric and hydrostatic variations. It is restricted to the case when the base of the paraboloid lies either entirely above or entirely below the fluid surface.
Archimedes' investigation of paraboloids was possibly an idealization of the shapes of ships' hulls. Some of the paraboloids float with the base under water and the summit above water, similar to the way that icebergs float. Of Archimedes' works that survive, the second book of On Floating Bodies is considered his most mature work.[6]
## References
1. ^ a b "Archimedes (Greek mathematician) - Britannica Online Encyclopedia". Britannica.com. Retrieved 2012-08-13.
2. ^ Hoyos, Dexter. 2011. A companion to the Punic Wars. Malden, MA: Wiley-Blackwell. page 328
3. ^ Morelle, Rebecca (2007-04-26). "Text Reveals More Ancient Secrets". BBC News. Archived from the original on 19 February 2009. Retrieved 2009-03-31.
4. ^ a b "The works of Archimedes". Cambridge, University Press. 1897. p. 257. Retrieved 11 March 2010. Any solid lighter than a fluid will, if placed in the fluid, be so far immersed that the weight of the solid will be equal to the weight of the fluid displaced.
5. ^ Ivor Thomas. Greek Mathematical Works: Aristarchus to Pappus. Loeb Classical Library.
6. ^ "On Floating Bodies (Book II)". Math.nyu.edu. Archived from the original on 2013-09-18. Retrieved 2012-08-13.
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Off Topic #1000
14 posts / 0 new
getco
Offline
Last seen: 3 years 4 months ago
Joined: 2008-08-03 05:31
Off Topic #1000
Don't worry - nothing serious - just wanted to bring to your attention that we now have 1000 off topics and this is the 1000th off topic.
Mir
Offline
Last seen: 12 years 4 days ago
Joined: 2007-12-03 16:07
did you know there is 197000
did you know there is 197000 CU feet of nipple tissue in the us? just something worthy of being offtopic (~_0)
jnw222
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Last seen: 12 years 2 months ago
Joined: 2009-05-30 11:10
nice
nice
getco
Offline
Last seen: 3 years 4 months ago
Joined: 2008-08-03 05:31
"Post about other things that
"Post about other things that would be of interest to the community."
zero_snaz
Offline
Last seen: 14 years 1 month ago
Joined: 2009-08-01 17:37
If this doesn't interest you. . .
0.999... or 0.(9), how ever you want to represent a repeating decimal of 9, is equal to 1.
But I assume you already knew that.
If you didn't, think about it.
Mir
Offline
Last seen: 12 years 4 days ago
Joined: 2007-12-03 16:07
its not equal!
its .000....1 short of 1.0! take a math class d00d.
zero_snaz
Offline
Last seen: 14 years 1 month ago
Joined: 2009-08-01 17:37
Actually, it's very simple. . .
1/3 = 0.333..., 2/3 = 0.666..., 3/3 = 0.999... = 1
Mir
Offline
Last seen: 12 years 4 days ago
Joined: 2007-12-03 16:07
WRONG3/3=1/1=13/3 doesnt
WRONG
3/3=1/1=1
3/3 doesnt equal .99999999
Try putting 3/3 in google.
As i said take a math class.
Edit: oh you \$(%&*er
Some students interpret "0.999…" (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".[
3/3 is still 1 not 0.(9)
0.(9) can be rounded up to 1 but because there is a infinite .99...9... that doesnt end it never reaches 1. it would be asking for a end to PI, now stop with these mathimatical paradoxes. You might end up destroying the world or the universe. leave these for the lecture halls of your Mathamatics department of your local university or college.
OliverK
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Last seen: 3 years 3 weeks ago
Joined: 2007-03-27 15:21
.33 1/3; .66 2/3
.33 1/3; .66 2/3
Too many lonely hearts in the real world
Too many bridges you can burn
Too many tables you can't turn
Don't wanna live my life in the real world
Mir
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Last seen: 12 years 4 days ago
Joined: 2007-12-03 16:07
these are approximations.
these are approximations.
Oni-Neoxes
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Last seen: 11 years 1 week ago
Joined: 2008-08-02 12:02
O_o
ewwww
WHY!?
Mir
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Last seen: 12 years 4 days ago
Joined: 2007-12-03 16:07
because people need to know
because people need to know this fact!
roamer
Offline
Last seen: 14 years 7 months ago
Joined: 2007-02-21 16:01
Well,
I for one think it's interesting. ;P
OliverK> you don't live on a cow
IRC: It brings out the best in all of us...Especially when tired.
Mir
Offline
Last seen: 12 years 4 days ago
Joined: 2007-12-03 16:07
I just recently found a fact
I just recently found a fact that is interesting but disturbing and thus is too distrubing to post here. Also the fact might violate some rule of some sort (;>.>)
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getfem-users
[Top][All Lists]
## Re: [Getfem-users] Transient analysis-dirichlet conditions
From: Yves Renard Subject: Re: [Getfem-users] Transient analysis-dirichlet conditions Date: Wed, 4 Jun 2008 16:01:15 +0200 User-agent: KMail/1.9.5
```On Wednesday 04 June 2008 10:30, Sébastien Janas wrote:
> Hi all,
>
> I want to solve the transient linear equation of diffusion.
>
> It seems that there is no function pre-build in getfem for the
> transient problems. So I assemble the mass matrix (MM) and stiffness
> matrix (SM) "manually", and then construct the linear system A*X=B
> using the backward euler
>
> A = (MM + delta t * SM)
> B = MM * X(t-1)
>
> Then I have to impose dirichlet conditions on the surface. I import my
> mesh from gmsh, where the surface are tagged with a number (in my case
> 200). How can I modify my linear system to take into accompt the
> dirichlet conditions? All the functions I found in the documentation
> are in the case where you use model bricks or standard assembly
> procedures...
>
> If I can have the list of node numbers tagged by 200 I would use the
> penalty method. But I am not able to obtain these list.
>
> I am using the python interface of getfem.
>
>
> Best regards,
>
> Janas Sébastien
This is true that there is no general strategy proposed for transient
problems, especially with the matlab or python interfaces. This is however
not easy to build such a general strategy since the situation can vary a lot
for differente time integration schemes.
However, You can have the list of nodes on a boundary with the command
gf_mesh_fem_get(MF, 'dof on region', [bnum]);
but you have first to tag the boundary with the command
gf_mesh_set(M, 'region', int bnum, imat CVFLST);
(see the Matlab examples).
Yves.
--
Yves Renard (address@hidden) tel : (33) 04.72.43.87.08
Pole de Mathematiques, INSA de Lyon fax : (33) 04.72.43.85.29
20, rue Albert Einstein
69621 Villeurbanne Cedex, FRANCE
http://math.univ-lyon1.fr/~renard
---------
```
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# 93968342
## 93,968,342 is an even composite number composed of four prime numbers multiplied together.
What does the number 93968342 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.
93968342 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 93968342:
### 2 × 13 × 103 × 35089
See below for interesting mathematical facts about the number 93968342 from the Numbermatics database.
### Names of 93968342
• Cardinal: 93968342 can be written as Ninety-three million, nine hundred sixty-eight thousand, three hundred forty-two.
### Scientific notation
• Scientific notation: 9.3968342 × 107
### Factors of 93968342
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 35207
### Divisors of 93968342
• Number of divisors d(n): 16
• Complete list of divisors:
• Sum of all divisors σ(n): 153273120
• Sum of proper divisors (its aliquot sum) s(n): 59304778
• 93968342 is a deficient number, because the sum of its proper divisors (59304778) is less than itself. Its deficiency is 34663564
### Bases of 93968342
• Binary: 1011001100111010111110101102
• Base-36: 1JY2FQ
### Scales and comparisons
How big is 93968342?
• 93,968,342 seconds is equal to 2 years, 51 weeks, 2 days, 14 hours, 19 minutes, 2 seconds.
• To count from 1 to 93,968,342 would take you about four years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 93968342 cubic inches would be around 37.9 feet tall.
### Recreational maths with 93968342
• 93968342 backwards is 24386939
• The number of decimal digits it has is: 8
• The sum of 93968342's digits is 44
• More coming soon!
#### Copy this link to share with anyone:
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Keywords: Divisors of 93968342, math, Factors of 93968342, curriculum, school, college, exams, university, Prime factorization of 93968342, STEM, science, technology, engineering, physics, economics, calculator, ninety-three million, nine hundred sixty-eight thousand, three hundred forty-two.
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+0
How can I solve the first question ?
0
431
19
+1828
How can I solve the first question ?
and in this question I want to know if my answer is correct or no
http://i62.tinypic.com/2lkslxw.jpg
xvxvxv Jul 31, 2014
#14
+80785
+15
Melody is correct....no solution exists.......here's a graph of both functions....notice that there are no intersection points.....
CPhill Jul 31, 2014
Sort:
#1
+80785
+10
We have
1/x ≤ x multiply both sides by x .....note that we're assuming that x ≠ 0
1 ≤ x2 subtract x2 from both sides
1- x2 ≤ 0 I like to turn the inequaliy into an equality and factor ....this gives us some "interval" points to work with!!
(1 - x) (1 + x) = 0 Then x = ± 1
Our answer comes from two of four intervals (-∞, -1], [-1, 0), (0, 1] or [1, ∞)
If you pick a point in each interval, you'll find that [-1, 0) and [1, ∞) are the intervals that "work' in the original inequality.......
CPhill Jul 31, 2014
#2
+1828
0
why you use the zero ..؟
xvxvxv Jul 31, 2014
#3
+80785
+10
We aren't including 0 !!! That's what the parentheses indicates....if we were including it, we would have had a bracket....the interval is everything from -1 up to, but not including, 0 .......
CPhill Jul 31, 2014
#4
+91409
+10
Maybe I didn't look at your answer hard enough Chris but I think it has a floor in it.
If you multiply an inequality by x and x is negative then you have to turn the sign around.
Therefore you end up with 2 equations to solve.
It is better to multiply by something that you KNOW is positive. Squared things are always positive!
Edit: you cannot divide by 0 so $${x\ne0}$$
$$\begin{array}{rll} \frac{1}{x}&\le&x\\\\ x^2\times\frac{1}{x}&\le&x^2 \times x\\\\ x&\le&x^3\\\\ -x^3+x&\le&0\\\\ -x(x^2-1)&\le&0\\\\ -x(x-1)(x+1)&\le&0\\\\ \end{array}$$
Now the easiest way to finish this is with a graph of
$$y=-x(x-1)(x+1)$$ and see where it intersects with $$y\le0$$
I am going to try and do a quick sketch - I know how to do a proper graph but I want to try and replicate what I would do by hand.
It is a cubic because the highest power of x is 3 and it will finish in the 2nd and 4th quads because the leading coefficient is -1
I can see from the graph that the answer is (this answer has been very slightly edited)
$$-1\lex\le x<0 \qquad or \qquad x\ge1\\ [-1,0)\;\cup \; [1,\infty)$$
I think it is right now Chris - Do we have the same answer?
Melody Jul 31, 2014
#5
+80785
+10
Melody is correct in suggesting that I might should have reversed the signs......however, I couldn't say at that point whether x was negative or not.......all I could say originally is that x ≠ 0....it might have been that x only took on positive values...!!!! Besides, in these types of problems, I like to generally ignore the inequality signs and set up an equality that will give me some intervals to work with - which is what I did. By picking some test points in these intervals, we can determine which values make our original inequality true, and I don't have to necessarily worry about the direction of the inequality sign throughout the problem. This may not be the fastest way to solve such things, but it's the method I learned......
Here's the graph of both functions......y = 1/x is in red y = x is in blue
Note that the intervals I have noted in my answer appear to be correct....
CPhill Jul 31, 2014
#6
+1828
+5
Melody
I don't understand some points of your explain but I take your information and solve the question by this way
http://im44.gulfup.com/n3q0s0.jpg
xvxvxv Jul 31, 2014
#7
+91409
+5
Yes that is fine 15threefold. It is almost the same.
I remember I did not comprehend the relevance of the graph for a long time either.
It is really good when you do get it but your method will always work well and it is almost as easy.
Melody Jul 31, 2014
#8
+1828
+5
thank you melody
what about the second question .؟
xvxvxv Jul 31, 2014
#9
+91409
+5
Hi 15x3
I didn't even notice that question!
$$|2x-9|&=&3-x\\\\ \begin{array}{rlllrl} 2x-9&=&3-x\qquad &or \qquad 2x-9&=&-(3-x)\\ 3x&=&12\qquad &or \qquad 2x-9&=&-3+x\\ x&=&4\qquad &or \qquad x&=&6\\ \end{array} \\BUT\\ 3-x\ge 0\\ -x\ge -3\\ x\le 3\\$$
Therefore there are no solutions.
-----------------------------------------------------------
I am just not getting the hang of this aligning properly, I wanted the x up against the equal sign, and its not there!
This is my code.
\end{array} \\
Can you try and explain please?
Melody Jul 31, 2014
#10
+1828
+5
sory melody how do you know that 3-x bigger than 0 !
xvxvxv Jul 31, 2014
#11
+91409
+10
Because it is equal to an absolute value so it can't be negative!
Do you see what I mean?
Melody Jul 31, 2014
#12
+1828
0
yes yes ... now I understand this question ..
very nice idea
thank you melody
but why you call me 15×3 !
xvxvxv Jul 31, 2014
#13
+91409
+5
Because you are xvxvxv
xv is 15 in roman numerals and there are 3 of them so you are 15x3
Melody Jul 31, 2014
#14
+80785
+15
Melody is correct....no solution exists.......here's a graph of both functions....notice that there are no intersection points.....
CPhill Jul 31, 2014
#15
+91409
+5
Hey, 15x3 if you are feeling really generous you could give CPhill and me some thumbs up.
If you don't like me calling you 15x3 you'd better say so because otherwise you are stuck with it.
Melody Jul 31, 2014
#16
+1828
0
I'm sorry my friend ...
I forgot it
and i'm never get angry ... and I am grateful for you and CPhill
Ican't express my felling because English is not my mother tongue
xvxvxv Jul 31, 2014
#17
+80785
+5
No problem.....xvxvxv......English is my "mother tongue" and I'm still strugglig with it !!!!
CPhill Jul 31, 2014
#18
+1828
0
HaHaHa ....
xvxvxv Jul 31, 2014
#19
+91409
0
Thanks 15x3.
Melody Jul 31, 2014
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## Theoretical Aspects of Lexical Analysis/Exercise 15
### From Wiki**3
##### < Theoretical Aspects of Lexical Analysis
Compute the non-deterministic finite automaton (NFA) by using Thompson's algorithm. Compute the minimal deterministic finite automaton (DFA).
The alphabet is Σ = { a, b, c }. Indicate the number of processing steps for the given input string.
• G = { a*|c, a|b*, bc* }, input string = aaabcacc
## NFA
The following is the result of applying Thompson's algorithm. State 8 recognizes the first expression (token T1); state 16 recognizes token T2; and state 21 recognizes token T3.
## DFA
Determination table for the above NFA:
In α∈Σ move(In, α) ε-closure(move(In, α)) In+1 = ε-closure(move(In, α))
- - 0 0, 1, 2, 3, 5, 6, 8, 9, 10, 12, 13, 15, 16, 17 0 (T1)
0 a 4, 11 3, 4, 5, 8, 11, 16 1 (T1)
0 b 14, 18 13, 14, 15, 16, 18, 19, 21 2 (T2)
0 c 7 7, 8 3 (T1)
1 a 4 3, 4, 5, 8 4 (T1)
1 b - - -
1 c - - -
2 a - - -
2 b 14 13, 14, 15, 16 5 (T2)
2 c 20 19, 20, 21 6 (T3)
3 a - - -
3 b - - -
3 c - - -
4 a 4 3, 4, 5, 8 4 (T1)
4 b - - -
4 c - - -
5 a - - -
5 b 14 13, 14, 15, 16 5 (T2)
5 c - - -
6 a - - -
6 b - - -
6 c 20 19, 20, 21 6 (T3)
Graphically, the DFA is represented as follows:
The minimization tree is as follows. Note that before considering transition behavior, states are split according to the token they recognize.
Given the minimization tree, the final minimal DFA is as follows.
## Input Analysis
In Input In+1 / Token
0 aaabcacc\$ 14
14 aabcacc\$ 14
14 abcacc\$ 14
14 bcacc\$ T1 (aaa)
0 bcacc\$ 2
2 cacc\$ 6
6 acc\$ T3 (bc)
0 acc\$ 14
14 cc\$ T1 (a)
0 cc\$ 3
3 c\$ T1 (c)
0 c\$ 3
3 \$ T1 (c)
The input string aaabcacc is, after 13 steps, split into five tokens: T1 (corresponding to lexeme aaa), T3 (bc), T1 (a), T1 (c), T1 (c).
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## Welcome To TutorStats
Data Analytics and Data Sciences are in increasing use as we move into a more connected world. Statistics and Probability form the basis of data analysis. Tutor Stats was made as a gateway to learn more about statistics and how data affects our lives. Join us in this journey to learn more about Statistics and grow your knowledge.
## What is Statistics?
Statistics is a field of mathematics that helps us gain information, insights and conclusions for a given set of data. We are bombarded with statistics on a daily basis. Some of the most common phrases we hear are as follows: Shark attacks more likely to happen during the summer. 37% percent of voters approve of the candidate. A certain disease is more likely to happen to men than women. 64% of audience liked a movie. A product received 4.5 stars on average and many more. Where does this information come from? Who Collects the data? Who decides what data is to be collected? We will learn the answers to these questions and more!
## Data Drives The World.
Statistics has a wide range of use across platforms and industries.
• The Stock Market
• Industrial Engineering – Six Sigma
• Quality Engineering
• BioMedical Sciences
• Pharmaceutical Research
• Political Research
• Government Census
• Data Analytics and Data Science
• Applied Mathematics
If you plan to become and expert in any one of these fields, then you will have to learn statistics.
## Why TutorStats?
Tutor Stats knows that Statistics can become a daunting task. We wish to help it become easier to understand.
### Basics Explained
Every Topic will have its basics and Methodology explained.
### Problem Solving
Mathematical Problems of each topic will be solved to help the student understand it.
### Graphical Representation.
Some Problems require graphs. They will be solved and graphs will get included.
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## Glossary of statistics: What’s a parameter?
Check out this short paper on “Variables and parameters” by Douglas Altman and J. Martin Bland, who identify at least three different meanings of the statistical term “parameter” and attempt to distinguish parameters from variables as follows (note: the introductory labels in bold are mine):
1. Parameters as statistical quantities: “… parameters [unlike variables] do not relate to actual measurements or attributes but to quantities defining a theoretical model,” such as the mean and standard deviation of the data pictured in the figure pictured below.
2. Parameters as slope-intercepts: “Another use of the word parameter relates to its original mathematical meaning as the value(s) defining one of a family of curves,” i.e. the slope and intercept of a line or curve.
3. Parameters as variables(!): “In some contexts parameters are values that can be altered to see what happens to the performance of some system.” (Wait, isn’t this what a “variable” is?)
So, which is it? In the alternative, is the definition of the term parameter provided by Wikipedia (see here) even less helpful?
## About F. E. Guerra-Pujol
When I’m not blogging, I am a business law professor at the University of Central Florida.
This entry was posted in Uncategorized. Bookmark the permalink.
### 1 Response to Glossary of statistics: What’s a parameter?
1. Craig says:
I always thought of parameters as something like coefficients, i.e., the model sense — but having more “meaning” than a simple coefficient — something along the line of dimensionless constants, for example.
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# Rules for Phase 10 Dice Game - How Do You Play It.
In our game every turn you get you have to score a minimum of 350 before you can take a score other wise the rules are the same except that, if you score 0 in any turn its called a zilch and if you get 3 zilches in a row you lose 1000 points. The dice scores are very close to farkle. I make sets of these in little mini chests as gifts. I have always wondered where this game came from. now I.
The official rules for Liar's Dice. If you've lost your original rule set, you've come to the right place.
## Uno Rules - The Original Uno Card Game Rules.
Farkle - Best Dice Game. 1.6M likes. Welcome to the official Farkle fan page! Farkle is a dice game that you can play by yourself or against an opponent. Just throw your six dice and choose the best.Basic MathDice Rules Equipment MathDice is played with two 12-sided dice, and three 6-sided dice. Optionally, an electronic timer or watch with a second hand can also be used. Playing the Game In basic MathDice, two players compete against each other. (The game can also be played by three or more players, or by teams of players, with a slight adjustment to the scoring rules.) The rules below.
Five Dice Farkle Five dice are used instead of six. (Obviously, certain scoring combinations, such as three pair, are impossible using this variant.) Team Farkle The game is played in teams. Teammates sit opposite each other at the table and combine their scores. The game is typically played to 20,000 points instead of 10,000.
In every game of Star Wars: Destiny, you’ll gather your small team of iconic characters and battle to defeat your foes, using your dice and the cards in your deck. The last player with characters left standing wins the game, but to successfully outmaneuver your opponent, you’ll need to carefully consider your options and enhance your deck with new dice and cards. If you ever wondered who.
Rules of Backgammon: Setup: Backgammon is a game for two players, played on a board consisting of twenty-four narrow triangles called points. The triangles alternate in color and are grouped into four quadrants of six triangles each. The quadrants are referred to as a player's home board and outer board, and the opponent's home board and outer board. The home and outer boards are separated.
## LCR Dice Game Rules - Dice Game Depot - Dice store.
Dice-1000 game involves mental math and probability skills. Begin by rolling five dice at once. Collect 100 points for each roll of one and 50 points for each roll of five. If you throw three matching dice, multiply that number by 100 to calculate the point value (a set of three fours would be worth 400 points). You can roll as many times as you like, but if subsequent rolls result in zero.
Be the first player to get all ten of your dice to show the same number. How to Play: (2-4 players.) Each player chooses a set of dice. Players hold all ten dice in their hands. Someone says “Go” and everyone rolls at the same time. Quickly look at your roll and decide which number you are going to go for. (For example, if you have more 3.
To begin, place the game board on the table. Each player selects a token. Then they place their token on the table near the space labelled Go, placing it on Go only when their first turn to move arrives. One player becomes the Banker, who distributes assets from the Bank to the players. Only the player in question can use their money, money can only be lent via the Banker or by the player.
Shut the Box is also known as Canoga. Being a traditional pub game without any national governing body, variations of equipment and rules abound. Where there is doubt, locally played rules should always apply. Shut the Box can be played by any number of players although it is most enjoyable with two, three or four. Some people even play the.
Similar Rules: 10000 Dice Game. If you already have a cup and dice, the printable 10,000 dice game rules, which includes a scoring summary, can help you teach new players in minutes. Free printable score sheets are available and can be used for 10,000 or the Farkle dice game, but a pen and paper will work just as well. More Farkle Fun.
## Farkle Dice Game Alternatives and Similar Games.
Dice Rolls You must roll your dice in the beginning of the game, and after a completed challenge. Should one or more of your dice land on top of another -- you must roll again. You do not have to look at your dice immediatly. Face Values Unless called by the first bidder - 1's are wild. That is, if the first bid is 4 1's -- 1's are no longer wild.
Shake is an easy dice game that all comes down to the luck of the roll. There is little strategy in which dice you keep and which ones you roll again but in the end, luck will have to be on your side. Specialize dice make this game different than just a standard set of dice and adds to the fun of the game.
Rules for Phase 10 Dice Game. Don’t have time to play a full game of Phase 10, why not try the Phase 10 Dice game instead? Phase 10 Dice Instructions. You can play the Phase 10 Dice game with one player, to as many as you can tolerate waiting on. Begin play by having each player roll a dice numbered 5 to 10. The player with the highest roll is first. On your turn, roll all 10 dice one time.
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# The motion of an oscillator is described by the equation x(t)=sin t+sin2t. What is x if cos t=-0.25 and t is in the interval (180, 270)?
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
First of all, before calculate sin t, we must establish to what quadrant belongs. Due to the facts from hypothesis, t is in the interval (pi, 3pi/2), so the angle t belongs to the third quadrant, where the value of the function sine is negative.
cos a = -.25 = -1/4
sin a = sqrt[1- (-1/4) (from the fundamental formula of trigonometry,where (sin a)^2 + (cosa)^2 = 1).
sin a = -sqrt(15)/4
To determine x, first we have to calculate sin 2t.
We'll apply the formula for the double angle:
sin 2a = sin (a+a)=sina*cosa + sina*cosa=2sina*cosa
We'll substitute 2a by 2t and we'll re-write the equation x(t).
x(t) = sin t + 2sint*cos t
e= -sqrt(15)/4 + 2*(1/4)*sqrt(15)/4
We'll calculate the LCD of the ratios:
LCD = 16
We'll factorize by sqrt(15)/4:
x(t) = [sqrt(15)/4](-1 + 2/4)
x(t) = [sqrt(15)/4](-1 + 1/2)
x(t) = [-sqrt(15)/8]
neela | High School Teacher | (Level 3) Valedictorian
Posted on
The given oscillation function is x(t) = sint+sin2t
To find x if cost = -0.25 if tis in (180 , 270).
When cost = -0.25, sint = - sqrt(1-cos^2t) = -sqrt(1-0.25^2)
sint = -0.968245836...(1).
sin2t = 2sint*cost.
sin2t = 2{-sqrt(1-.25)^2}{-.25}
sin2t = 0.484122918....(2).
We substititute the obtained value of sint and sin2t in thye given equation, x(t) = sint+sin2t and get:
x(t) = -0.968245836 + 0.484122918.
x(t) = -0.484122918.
Therefore , when cost = -0.25 fot in (180,270), the value x = sint+sin2t = -0.484122918.
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# Profile Evaluation
Author Message
Intern
Joined: 13 Nov 2012
Posts: 11
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
14 Nov 2012, 17:39
Hi Veritas, I'd be extremely appreciative if you could evaluate my chances at the following schools: Stern, Columbia, Wharton. Thanks in advance!
GRE: Quant 164 (90 percentile); Verbal: 165 (95%), Written 4.5 (73%)
School: Queen's University (Canada's Dartmouth), 3.4/4.3 GPA
Volunteering: 2005 - youth mentoring program (Cape Town, South Africa townships); 2009 - member of local charity campaign; 2012 - created voluntary cross-entity research group at work
Work: 5 years experience at major Canadian financial firm; no promotions (flat organizational structure), but steady salary raises; I've been afforded a lot of responsibility/privilege for someone my age; strong recommendations (however neither is from a director or managing director).
Essays: Should be strong - I have a unique and (very biased, but I believe) compelling story why I want to get my MBA.
Other: Well-rounded extracurriculars; I'm a white Canadian guy that works in finance - this doesn't help!
Affiliations: Veritas Prep
Joined: 25 Jul 2010
Posts: 1910
Followers: 43
Kudos [?]: 182 [0], given: 3
### Show Tags
20 Nov 2012, 18:51
Actually Canada is fairly under-represented in US schools because there are so many good places to get an MBA in Canada, most students stay there. I think you should have a shot at your target schools, especially if you are correct in being able to cast your story in a convincing fashion to the adcoms. A bit odd maybe that you have never been promoted in five years, unless that's just how it's done there or something. In any case, you may want to address this in your essays, since the US equivalent of you would have had opportunities to move up in five years at least once if not twice.
_________________
Bryant Michaels
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# From a team of 6 students, in how many ways can we choose a captain and vice-captain assuming one person can not hold more than one position?
Harshit Singh
3 years ago
From a team of 6 students, two students are to be chosen in such a way that one student will hold only one position.
Here, the no. of ways of choosing a captain and vice-captain is the permutation of 6 different things taken 2 at a time.
So,6P2= 6! / ( 6 -2 )! = 6! / 4! = 30
Thanks
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# Linear Regression Algorithm – Applications and Concepts of Linear Algebra Using the Linear Regression Algorithm
Linear algebra, a branch of mathematics dealing with vectors and the rules for their operations, has many applications in the real world. One such application is in the field of machine learning, particularly in linear regression, a statistical method used to model the relationship between a dependent variable and one or more independent variables.
In this blog post, we’ll dive into the basics of linear regression, highlight the linear algebra concepts involved, and demonstrate a Python implementation with sample data.
## 1. Concepts in Linear Regression
a. Vector: A list or column of numbers.
b. Matrix: A 2-dimensional array of numbers.
c. Dot Product: The sum of products of the corresponding entries of two sequences of numbers.
d. Transpose of a Matrix: Reflecting a matrix over its main diagonal.
e. Matrix Multiplication: Combining matrices to produce another matrix.
Linear regression essentially revolves around the equation:
$Y = X \beta + \epsilon$
Where:
• $Y$ is the dependent variable (what you’re trying to predict),
• $X$ is the matrix of independent variables,
• $\beta$ is the matrix of coefficients, and
• $\epsilon$ is the error term.
The goal is to find $\beta$ that minimizes the sum of squared residuals, which can be represented in matrix notation as:
$\hat{\beta} = (X^TX)^{-1}X^TY$
## 2. Python Implementation using NumPy
Let’s consider a simple example where we’ll try to predict the price of houses based on their size (in square feet).
### Sample Data
import numpy as np
# Sample Data
# Independent Variable (Size in sq.ft)
X = np.array([1000, 1500, 2000, 2500, 3000]).reshape(-1, 1)
## Conclusion
Linear algebra, although a fundamental mathematical discipline, finds expansive applications in the modern world, particularly in the domain of machine learning. Linear regression, a seemingly simple algorithm, leverages these concepts to make predictions based on patterns in data.
As demonstrated, Python, along with the NumPy library, provides an intuitive platform to understand and implement these principles.
Course Preview
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# 192.098 Introduction to the Coq proof assistant This course is in all assigned curricula part of the STEOP.\$(function(){PrimeFaces.cw("Tooltip","widget_j_id_20",{id:"j_id_20",showEffect:"fade",hideEffect:"fade",target:"isAllSteop"});});This course is in at least 1 assigned curriculum part of the STEOP.\$(function(){PrimeFaces.cw("Tooltip","widget_j_id_22",{id:"j_id_22",showEffect:"fade",hideEffect:"fade",target:"isAnySteop"});});
2020S, VU, 2.0h, 3.0EC, to be held in blocked form
## Properties
• Semester hours: 2.0
• Credits: 3.0
• Type: VU Lecture and Exercise
## Learning outcomes
After successful completion of the course, students are able to understand and use the Coq proof assistant; in particular to formalize propositional logic, quantifiers inductive types and inductive predicates
## This course gives an introduction to the Coq proof assistant:
Coq is a formal proof management system. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. Typical applications include the certification of properties of programming languages (e.g. the CompCert compiler certification project, the Verified Software Toolchain for verification of C programs, or the Iris framework for concurrent separation logic), the formalization of mathematics (e.g. the full formalization of the Feit-Thompson theorem, or homotopy type theory), and teaching.
### 4 lecture sessions
1. Introduction to Coq, Propositional Logic
2. Quantifiers, equality and all that
3. Inductive types
4. Inductive predicates
### 4 lab sessions
1. Your first proofs, propositional logic
2. First Order Logic, Higher Order Logic
3. Proofs by induction on nat
4. Proofs by induction on proofs
## Teaching methods
To follow the lessons students need:
- a computer
- a microphone and headphones (mandatory)
- headphones to avoid audio feedback/larsen effect
- camera is a plus
- but we will certainly not use it much to save bandwith
- the Discord app: discordapp.com
- free of charge, allows screen sharing
- works in a navigator (tested on Chrome)
- standalone apps for linux, windows & ios
- need to register a (free) account on discordapp.com
- the account of the teacher is: Dominique Larchey-Wendling#0283
- For lab. sessions, the students need to install
Coq and CoqIDE on their own computer
- anything above Coq 8.8 (3 years old) will work
- the Coq main site: coq.inria.fr
- for Linux, try in that order:
1/ install coq & coqide available for your distribution
2/ https://github.com/coq/coq/wiki/Installation-of-Coq-on-Linux
3/ or install it through opam 2 (for experts)
- installing Coq for Windows or Mac users:
- there are packages in here
- https://github.com/coq/coq/releases/tag/V8.11.0
## Mode of examination
Immanent
The course will be held blocked May-June
`Ects Breakdown 8.0 h attendance of lecture (4 days x 2h) 8.0 h self study at home (4 x 2h) 8.0 h attendance to lab sessions (4 days x 2h) 8.0 h homework (4 x 2h)32.0 h project preparation 1.0 h project presentation 9.0 h preparation for oral exam 1.0 h oral exam-----------------------------------------------75.0 h = 3 Ects`
To be decided
## Course registration
Begin End Deregistration end
18.03.2020 10:00 04.05.2020 10:00 30.05.2020 10:00
## Literature
No lecture notes are available.
## Previous knowledge
- (short): https://coq.inria.fr/a-short-introduction-to-coq
- (less short): the first 20 pages of C. Paulin's introduction course
https://www.lri.fr/~paulin/LASER/course-notes.pdf
- (reference): The Coq'Art book,
https://www.labri.fr/perso/casteran/CoqArt
- their are many Coq courses available online
- (see eg.) https://github.com/coq/coq/wiki/CoqInTheClassroom
- we will roughtly follow the course of P. Casteran:
- slides and lab sessions
- https://www.labri.fr/perso/casteran/FM/Logique
English
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Are you more of a visual learner? Check out our online video lectures and start your course “Calculation Methods: Exercises” now for free!
Image: “Population” by geralt. License: CC0 1.0
## Estimation of Proportions Using Interval
Taking several multiple-sized samples helps with computing the proportion of successes in each sample. In this case, there are several different answers to be recorded from these different proportions. Therefore, sampling distributions are calculated to find out which one is the correct answer. The correct answer would be none of them.
The estimate of proportions using intervals measures results for an entire population, using a range of reasonable values for the population proportion. It is assumed that the intervals capture the population parameter’s true value. This interval is known as the confidence interval, which helps measure a suitable and average value extracted from multiple samples of a whole population. Using intervals helps provide multiple values to estimate the population proportion.
Confidence intervals give more details than point estimates. A point estimate has limited usefulness since it does not reveal the uncertainty associated with the estimate, i.e., it is unclear how far this sample mean might be from the population mean.
Example: Suppose we want to measure the proportion of undergraduates who are permanently residing in southern Mexico. We take a random sample of 25 students and compute the sample proportion. Suppose it comes out that 9 out of 25 students chosen are undergraduates and belong to southern Mexico. The population proportion comes out as follows:
ƥ = 9/25
ƥ = 0.36
Due to the “luck of the draw” factor, we are sure that the true population proportion is not exactly 0.36. If we want to find out an interval of a whole population, which draws the sample proportion probability of at least 95% confidence, we need to calculate it using the standard deviation formula.
ƥ = √(ƥ(1-ƥ))/n = 0.95
## Creation of a Confidence Interval
There are four basic steps involved in constructing a confidence interval:
• The first step involves identifying sample statistics. The sample statistic (i.e. sample mean, sample proportion) will be used to estimate the population parameter.
• The second step involves selecting a confidence level. The confidence level helps identify the uncertainty level of a sampling method. Normally, a 90%, 95%, or 99% confidence interval is used to find accurate results.
• The third step is calculating the margin of error. The margin of error is calculated using the following equation:
Margin of error = Critical value * Standard deviation of a statistic
• The fourth and last step is specifying the confidence interval. The uncertainty is interlinked with the confidence level by defining the range of confidence interval by the following equation:
Confidence interval = Sample statistic + Margin of error
A confidence interval is calculated using sampling distribution data of sample proportion ƥ.
Example: Suppose we have selected 100 people who believe in evolution. We currently don’t know the population proportion of the people who believe in evolution. If 47 out of 100 people say that they believe in evolution, then know that our sample proportion is ƥ = 0.47. In order to find out the confidence interval for the sample proportion, we need to find the sampling distribution of ƥ.
Here, the problem is that the sampling distribution depends on ƥ, and it is missing. First of all, we have to find out the standard deviation of ƥ; we can do that with the same formula:
= √(ƥ(1-ƥ))/n
= √(0.47(1-0.47)/100
= 0.0499
## Constructing the Interval
In order to construct an interval, we need to follow the 4 steps specified above.
Example: Suppose we need to calculate the average weight of an adult male in Baltimore, Maryland. We draw a random sample of 1,000 men from a total population of 1,000,000 men. The weighted average of samples chosen comes out at 180 pounds. Here, the standard deviation of the whole population comes out at 30 pounds.
We need to find the confidence interval. We can calculate it using the four steps mentioned above.
• We have identified the mean of 180 pounds as the sample statistic.
• The confidence level selected for this case is 95%.
• The margin of error can be calculated using the standard error of the mean as follows:
Standard error (SE) = s / sqrt (n) = 30 / sqrt (1000)
= 0.95
The critical value is a factor which is used to compute the margin of error here. The t score (t*) is calculated as follows:
Alpha (α): α = 1 – (confidence level / 100) = 0.05
Critical probability (p*): p* = 1 – α/2 = 1 – 0.05/2 = 0.975
Degrees of freedom (df): df = n – 1 = 1000 – 1 = 999
The critical value is the t statistics, with 999 degrees of freedom and a 0.975 cumulative frequency. Using the t distribution calculator, the critical value comes out to 1.96.
The margin of error is computed using the above-given equation.
Margin of error = Critical value * Standard deviation of a statistic
= 1.96 * 0.95
= 1.86
The confidence interval is specified by the sample statistic + margin of error. Here, the uncertainty is denoted by the confidence level. We have already chosen the confidence interval of 95%, which is 180 + 1.86.
## Meaning of Confidence
The confidence level is the probability that the targeted interval actually contains the target quantity. When we refer to a 95% confidence interval, it does not mean there is a 95% chance that the interval contains ƥ. The population proportion is either inside an interval or outside it. The population proportion is a fixed quantity. If we show a confidence interval of 95%, it means that 95% of samples in the population produce the true proportion.
It can also be stated like this: “We are 95% confident that the true proportion lies in the interval from which samples have been selected”. Here, the uncertainty refers to the probability that the selected sample belongs to 95% of the samples of the chosen interval and to the 5% that does not relate to the confidence interval.
## Margin of Error
The margin of error is the range of values below and above the sample statistic in a confidence interval. It represents the amount of sampling error in a survey result. The margin of error and confidence intervals are directly proportional to each other. If the margin of error is higher, the confidence in the sampling results is higher, showing that it is close to the true figure.
The margin of error is half the width of the interval. It refers to the extent on each side of the sample proportion. For a higher level of confidence, a higher level margin of error is required. Confidence and precision are inversely proportional to each other. A higher level of confidence indicates less precision.
Example: The common example of the margin of error is a group of people divided into two classes. One of them prefers product A, and the second group prefers product B. Measuring a global margin of error refers to the percentages from the whole population of people who like product A or B.
In case a statistic is a percentage, the maximum margin of error can be calculated using the radius of the confidence interval for 50% of the reported percentage. Here, the margin of error is known as an absolute quantity, which is equal to the confidence interval radius for the statistics. If the true value is 50%, the confidence interval radius is 5% points, and the margin of error is 5% points.
## Critical Values
The critical value is a factor used to compute the margin of error. It is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis.
It is the value that splits the probability of availability or rejection region, which includes or excludes the targeted value in an interval. Under the normal standard model, the critical value depicts the Z term linked with the central region. The critical value is used in case the sampling distribution is normal or close to normal. If the absolute value of the test statistic is greater than the critical value, you can declare statistical significance and reject the null hypothesis.
The critical value of the Z-score is used in case the sampling distribution is close to normal. Change the confidence level means changing the number of standard errors in order to extend the interval away from the sample proportion. There are three main types of critical values, including the t score, Chi-square, and z-tests. This number of standard errors is referred to as the critical value. The critical value is calculated using the Z-table to find out z*.
The Z-table shows a value of 95% confidence interval z* = 1.96 and for 90% confidence interval = 1.645.
### Using the Normal distribution
The critical value can be calculated using the normal distribution. The critical value at the 95% confidence interval refers to two values that lie between 95%. It shows that out of the total 5%, 2.5% lies on each tail. In the Z-table, we look up the probability of 0.4750 as 0 ≤ ≤ ∗ ≥ ∗ = 0.025, 0 ≤ ≤ ∗) = 0.4750. The value of 0.4750 in Z- table is = 1.96. Here, the cut-offs of the critical values for a 95% confidence interval are 1.96 and -1.96 on both sides.
The graph of 90% of the confidence interval is shown below:
Image: “Illustrating a 90% confidence interval on a standard normal curve.” by KendallVarent. License: Public Domain
The cut-offs in the case of the confidence interval 90 % have been shown above * = 1.645. The cut-offs, in this case, have a given value of -1.645 and 1.645 on both sides.
## The trade-off between Confidence and Precision
There is a close relationship between confidence and precision, but it is important to note that these terms are not direct complements of each other. The safest way to enhance the confidence level on the chosen sample for the inclusion of the targeted value is to choose a larger sample size. It helps improve the precision level as well. High levels of confidence can be achieved with wider intervals, while narrower or more precise intervals permit less confidence; thus, at a constant state, there is a trade-off between precision and confidence.
Most importantly, any statement of precision without a corresponding confidence level is termed incomplete and impossible to interpret.
The margin of error is calculated using the same formula, i.e. √(ƥ(1-ƥ)) /n. For that purpose, we are required to make the margin of error large enough to ensure the confidence interval is accurate. If we take the value of sampling proportion ƥ = 0.50, the desired sample size is calculated this way:
n = ((z* +(0.5))/ME)^2
The sample size is needed to obtain a margin of error of 0.03 for a 95% confidence interval from the whole population of people who believe in evolution. If we assume no sample has been taken yet, the calculations are shown as follows:
Z* = 1.96
ME = 0.30
n = ((1.96 +(0.5))/((0.03)))^2
n = 1,067.11
It is impossible to choose people in a rounded figure, so it will be chosen for the nearest whole number, i.e., 1,068 samples from the whole population.
• Absolute precision is often applied when estimating quantities such as population proportions, which exist in the form of percentages.
The standard error has similar physical units as the estimator; hence, the absolute precision is the same physical units as the estimation target.
• Relative precision is always unit-free and expressed as a percentage.
## Caution
The following points should be considered carefully in hypothesis testing:
• The population proportion in a fixed quantity does not vary.
• Different sample results in each interval don’t match with each other. They have different values; hence, each sample shows different results.
• It is impossible to be certain about a parameter; one can only be confident to a specific extent.
• The complete process revolves around estimating the population proportion instead of the sample proportion.
• Do not state confidently that more about the results is known than the interval actually tells.
• All sample intervals chosen are treated equally. The values near the center of the interval are not more plausible than the values near the edges.
• Beware of a margin of error that is too large to be useful. The margin of error between 10%—90% is not overly useful.
• The possibility of biased sampling should be taken into account.
• Consider the trials to be independent.
Learn. Apply. Retain.
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# Does Krugman Understand Ricardian Equivalence? (Wonkish)
Suppose that the government wants to acquire the resources necessary to implement a new expenditure program G = {g1, g2, g3 … }, where gt denotes government purchases of goods and services at date t.
Let us take G as given. To begin their evaluation of G, macroeconomists ask the following two questions. First, what are the likely macroeconomic consequences of implementing program G? Second, does the answer to first question depend on how G is financed? (Financing is assumed to take the form of taxes, deficits, and money creation, or some combination thereof).
The Ricardian Equivalence Theorem (RET) is a proposition that helps us answer the second question above. In particular, the RET lays out a set of conditions that must hold for the following proposition to hold: It does not matter how the government finances G.
Whether the set of conditions holds in reality is a separate issue that need not concern us here. (You may be interested to read this article from the Economist on the subject page 1 and page 2). For now, let me emphasize what the RET does not say: The RET does not say that G does not matter (it says that the method of financing G does not matter).
The G is so unimportant in the RET, that it is useful to ignore it completely when teaching the theorem to students for the first time. That is, set G = {0, 0, 0 …} and then ask whether it matters how G is financed. One way to finance such a program would be to cut taxes today and raise them tomorrow. Since G is fixed (at zero, in this case), this implies running a deficit today, which is matched by a surplus tomorrow. The RET states the conditions under which a deficit-financed tax cut like this does not matter. A deficit today simply represents a higher future tax bill; and people really don’t care whether they are kicked in the a\$\$ today or tomorrow–it’s still an a\$\$-kicking.
What I have just described is the stuff of elementary macro textbooks. We should all understand now that the RET has nothing to do with G. In particular, we should all know enough never to write a column with the title: A Note on the Ricardian Equivalence Argument Against Stimulus.
The title of Krugman’s post shows that it is he who does not understand the Ricardian proposition. There is no “Ricardian Equivalence argument against stimulus.” Indeed—the proposition can be used to defend bond-financed stimulus. (In particular, if deficits do not matter, then why not bond finance?)
Now, perhaps you might want to entertain the idea that Paulo knows all this and only chose the title to mock that horrible Bob Lucas fellow. Could be. Except for the fact that Lucas makes no reference to the RET in his informal speech.
In fact–good lord, can it be true–it appears to me that Lucas is making distinctly non-Ricardian arguments in his assessment of fiscal policy. Take a closer look at the passage quoted by Krugman. First, Lucas asserts that a money-financed increase in G will be stimulative; but that the stimulus part comes from the manner in which the spending is financed. (Because money can be thought of as zero-interest debt, this is virtually the same thing as saying that a deficit-financed increase in G will be stimulative.) Second, Lucas goes on to argue why he thinks a tax-financed increase in G will not be stimulative. In other words, his argument could be interpreted to be mean the method of finance matters. Needless to say, this is not a Ricardian Equivalence argument against stimulus.
One may agree or disagree with what Lucas has to say on any given issue (certainly, I do at times). But to come out and publicly declare the man to be ignorant of high-school economics–repeatedly–and on the basis of an informal speech–from a fellow Nobel-prize winner–who is himself is guilty of the charge leveled against his own colleague in the profession—well—holy cow, I don’t know what else to say.
PS. A couple of related blog posts on this subject:
Responding to Krugman on Ricardian Equivalence (Andrew Lilico, The Telegraph)
Ricardian Equivalence Redux (Stephen Williamson)
And in my defense:
On what is and is not an argument about Ricardian equivalence (Andrew Lilico)
Ricardian equivalence heat (Steve Williamson)
In PK’s defense:
The great Ricardian equivalence throwdown! (Noahopinion)
Affiliation: Simon Fraser University and St. Louis Fed
David Andolfatto is a Vice President in the Research Division of the Federal Reserve Bank of St. Louis. He is also a professor of economics at Simon Fraser University.
Professor Andolfatto earned his Ph.D. in economics from the University of Western Ontario in 1994, M.A. and B.B.A. from Simon Fraser University. He was associate professor at the University of Waterloo before moving to Simon Fraser University in 2000.
His current research is focused on reconciling theories of money and banking. His past research has examined questions relating to the business cycle, contract design, bank-runs, unemployment insurance, monetary policy regimes, endogenous debt constraints, and technology diffusion.
#### 1 Comment on Does Krugman Understand Ricardian Equivalence? (Wonkish)
1. SH
‘A deficit today simply represents a higher future tax bill; and people really don’t care whether they are kicked in the a\$\$ today or tomorrow–it’s still an a\$\$-kicking.’
Oh, dear.
a) Not necessarily. If a deficit today leads to higher productive capacity (and/or fuller capacity utilisation) tomorrow, then the tax bill as a share of income in the future may even be lower.
b) ‘Tomorrow’ may be a long way into a highly uncertain future. Often, people with weak arguments use this kind of language to obscure the logical flaws in their reasoning.
c) There is no empirical evidence to back up your quasi-religious belief.
d) I don’t think you understand Lucas, so it is unsurprising that you don’t understand Krugman on Lucas.
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# Physics Ch. 4&5
### Ch. 4 and 5 study cards
QuestionAnswer
Newton's First Law of Motion An object stays at rest or in motion (at a constant velocity) unless another net force changes its state. Often called the Law of Inertia
Newton's Second Law of Motion For a given net force, the magnitude of the acceleration is inversely proportional to the mass.
Newton's Third Law of Motion Often called acceleration or action-reaction law; states whenever 1 object exerts a force on a 2nd object, the 2nd object exerts a opposite directed force of = magnitude on the 1st object. For every action/force, there is a =, but opposite, reaction
Force A force is a push or pull and is a vector quality. The SI Unit of Force is Newton or kg*m/s2
Mass Mass is a property of matter that determines how difficult it is to accelerate or decelerate an object. Mass is a scalar quantity. 2)The mass of a body is a quantitative measure of inertia adn is measured in an SI unit of kg.
Weight W=mg; The weight of an object is the gravitational force that the earth exerts on the object. SI unit: Newton
Free Fall a=-g. The apparent weight is 0 b/c when both the person and the scale fall freely, they cannot push against one another.
Weightlessness the apparent weight is 0; apparent weight=mg +ma
Inertia Inertia is the natural tendency of an object to remain at rest or in motion at a constant velocity. The SI unit of Inertia and Mass is kg.
Kinetic Friction Kinetic Friction is usually less than Static Friction. The force of kinetic friction btw 2 surfaces sliding against one another opposs the relative motion of the surfaces.
Static Friction The force of static friction btw 2 surfaces opposes any impending relative motion of the surfaces.
1 Newton (N) =1 kg*m/s2
1 dyne (dyn) = 1 g*cm/s2
1 slug (sl) =1 lb*s2/ft
1 pound (lb) =1 ft/s2
Three fundamental forces gravitational force, strong nuclear force, and electroweak force
Normal Force The normal force is one component of the force that a surface exerts on an onject w/ which it is in contact--nameley, the component that is perpendicular to the surface.
Friction The force component parallel to the surface is called friction.
Kepler's third law The period is proportional to the 3/2 power of the orbital radius.
Created by: maddie9334
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## Sunday, April 26, 2015
### Shirts
Day 1
Thought a good title might be "Shi?ts" and might make you "notice and wonder". Students walked into my class on a Monday morning to this pile of shirts on my desk.
I immediately asked them to write down what they noticed and what they wondered. I did this with two classes for the record.
Here are their noticings:
And here are their wonderings:
After we got through all the noticings and wonderings we focused in on this question.
How many shirts are in that pile?
Asked students to - Guess too low, guess too high, best guess.
Students were randomly put in groups of three. I let them know I would give them a sample of 10 shirts.
Here are some samples of groups work on vertical surfaces.
Group 1
Group figured out the volume of all the shirts and then the volume of their ten shirts as a sphere and as a cone. Got 771 shirts using sphere and 258 shirts using the cone.
Group 2
Group figured out the volume of all the shirts and then the volume of their ten shirts as a cone. Got 304 shirts.
Group 3
Group figured out the volume of all the shirts and then the volume of one folded shirt (rectangular box). Got 146 shirts.
Day 2
At the start of class I modelled a solution to find out the number of shirts. It looked like this. Two solutions one for each class.
And the reveal.
The rest of the second day was spent figuring out how many shirts each student could fold per second. Class decided to include time to get shirt from pile, lay it out, fold it and pile it on their chair.
Students kept a table of values of time and number of shirts folded. We did it for 0, 20, 40, 60, 80, 100 and 120 seconds. Students plotted data, did line of best fit, and calculated slope to find their rate.
Here are some photos of one students data, scatterplot, equation, and graph.
Here are the final rates for both classes:
NAME RATE in Shirts per Second A 0.042 C 0.080 M 0.080 H 0.077 E 0.051 C 0.090 W 0.007 E 0.040 D 0.090 Z 0.117 W 0.083 K 0.046 M 0.067 L 0.081 A 0.046 E 0.041 E 0.082 F 0.092 L 0.033 K 0.039 D 0.047
NAME RATE in Shirts per Second S 0.063 R 0.044 T 0.080 L 0.036 R 0.047 J 0.084 A 0.068 L 0.046 A 0.053 S 0.044 A 0.067 H 0.058 L 0.067 L 0.038 B 0.046 O 0.054 E 0.027 M 0.032 T 0.046
And a photo of the madness.
Day 3
Today's question to focus on."How long would it take the class to fold all 282 shirts?" Asked students to - guess too low, guess too high, give best guess. Students were asked to devise and write out a strategy individually on how they might figure it out. Students were then put into random groups and asked to share their strategies. Once they talked it over they were to execute the group strategy on vertical white boards. Many groups wanted the data from the day before - I provided that. Some groups asked how many students were in the class - I provided that (23 for one class and 24 for the other). Some groups asked about attendance - how many students were in class on average. I told them to make their best guess based on a usual day. Groups were given the entire period to work on this.
Here are some samples of group work on vertical surfaces:
Day 4
Today's question to focus on, "What area could we cover with all those shirts?" Asked students to - guess too low, guess too high, give a best guess. Students were asked to devise and write out a strategy individually on how they might figure it out. Students were then put into the same groups as yesterday and asked to share their strategies. Once they talked it over they were to execute the group strategy on vertical white boards. I gave groups about 30 minutes to get through this problem.
A few photos of students doing some measuring.
And some of the groups solutions on vertical surfaces.
Time to do answer "How much area could we cover?" we went out in the hallway and got to work.
After some measurements with both classes, here is what we got.
I hustled students back into the class with all the shirts. We tossed them back into the pile and set up the room for folding. I posted the predicted times from the group work on day 3.
And then it was time to fold.
Actual time to fold them all for each class.
Final Reflection:
This was great. Based on the number of selfies that were taken during the area layout of all the shirts I would say this was a success. I bet you all have questions?
## Friday, April 3, 2015
### WODB Part 3
Today when students walked in I already had the room set up in a huge circle (15 sets of 2 desks) with the samples of WODB A to O that the two classes created yesterday. You can see them in the Day 2 post. As students walked in I handed them 15 choice and reason slips. Students were instructed to start at a letter and work their way around the room picking, top left, bottom left, top right, bottom right for each WODB and writing down their reason for choosing this option. They then folded their ballot and put it in the envelope which was also on the desks. It looked like this.
Once students worked their way around the room I had them analyse the data from the sample they started with. Here are the results:
Once this was done students filled in a sheet summarizing the reasons that were given. We then went back and looked at the criteria for what makes a good WODB. Here it is:
Students were then instructed to rank the A to O samples individually from strongest 1 to weakest 15 by looking at the samples, the final data of the distribution and comparing this to the criteria.
The back of the room was set up and looked like this for the ranking to occur.
Here are the slips.
And the final averages of the students rankings versus my rankings.
IMAGE OVERWIJK RANK STUDENT AVERAGE RANK A 6 10 B 7 2 C 5 6 D 14 8 E 15 5 F 9 7 G 2 4 H 10 12 I 1 3 J 11 1 K 9 15 L 13 14 M 4 13 N 3 11 O 12 9
From what I can tell students valued the equal distribution heavily and I valued the distribution and the math equally ( now there is a surprise! ).
Lastly I really liked the structure of this 3 day lesson. I think you could do this in any course - including non-math courses.
Hope you enjoyed. I did.
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# College Algebra
posted by .
Suppose you are offered a job at \$1600 a month with a guaranteed increase of \$70 every 6 months for 5 years. What will your salary be at the end of this period of time?
Thank you!!
• College Algebra -
Add 10 x \$70 to the starting monthly salary.
The number 10 is the number of raises received in 5 years
## Similar Questions
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1) Janet is offered a job with a starting salary of \$15,000, and a guaranteed annual increase of \$600. If she takes the job, what will be her guaranteed salary for the 10th year?
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Mandelbrot Set In TensorFlow – Compute Quickly
Free TensorFlow course with real-time projects Start Now!!
Today, in this TensorFlow Tutorial, we will see TensorFlow Mandelbrot Set. We’ll be visualizing the famous Chaos Theory Mandelbrot Set in TensorFlow. Moreover, in this TensorFlow Mandelbrot set.
We will see the set up for Mandelbrot Set and session and variable initialization in TensorFlow. Along with this, we will also look at the what Mandelbrot set actually is and computation of Mandelbrot Set in TensorFlow with an example.
So, let’s begin TensorFlow Mandelbrot Set.
What is Mandelbrot Set?
According to Wikipedia, “The Mandelbrot set is a famous example of a fractal in mathematics. The Mandelbrot set is important for the chaos theory. The edging of the set shows a self-similarity, which is not perfect because it has deformations.
The Mandelbrot set can be explained with the equation zn+1 = zn2 + c. In that equation, c and z are complex numbers and n is zero or a positive integer (natural number).
Starting with z0=0, c is in the Mandelbrot set if the absolute value of zn never becomes larger than a certain number (that number depends on c), no matter how large n gets.”
TensorFlow Mandelbrot Set
Visualizing the set has nothing to do with machine learning. It can be thought of as another TensorFlow example for mathematics. So, let’s learn how can we compute Mandelbrot set in TensorFlow.
Setup For Mandelbrot Set in TensorFlow
For Mandelbrot Set in TensorFlow, you’ll need a few imports to get started.
```# Import libraries for simulation
import tensorflow as tf
import numpy as np
# Imports for visualization
import PIL.Image
from io import BytesIO
from IPython.display import Image, display
```
Technology is evolving rapidly!
Now, define a function to actually display the image once you have iteration counts.
```def DisplayFractal(a, fmt='jpeg'):
"""Display an array of iteration counts as a
colorful picture of a fractal."""
a_cyclic = (6.28*a/20.0).reshape(list(a.shape)+[1])
img = np.concatenate([10+20*np.cos(a_cyclic),
30+50*np.sin(a_cyclic),
155-80*np.cos(a_cyclic)], 2)
img[a==a.max()] = 0
a = img
a = np.uint8(np.clip(a, 0, 255))
f = BytesIO()
PIL.Image.fromarray(a).save(f, fmt)
display(Image(data=f.getvalue()))
```
Session & Variable Initialization in Mandelbrot Set
Here an interactive session is used, but a regular session would work as well.
`sess = tf.InteractiveSession()`
Introducing NumPy with Tensorflow.
```# Use NumPy to create a 2D array of complex numbers
Y, X = np.mgrid[-1.3:1.3:0.005, -2:1:0.005]
Z = X+1j*Y
```
Initializing TensorFlow tensors.
```xs = tf.constant(Z.astype(np.complex64))
zs = tf.Variable(xs)
ns = tf.Variable(tf.zeros_like(xs, tf.float32))
```
Now, as you already aware that TensorFlow requires you to explicitly declare variables before using them.
`tf.global_variables_initializer().run()`
Running Computation – TensorFlow Mandelbrot Set
Now, we specify more of the computation for Mandelbrot set in TensorFlow.
```# Compute the new values of z: z^2 + x
zs_ = zs*zs + xs
# Have we diverged with this new value?
not_diverged = tf.abs(zs_) < 4
# Operation to update the zs and the iteration count.
#
# Note: We keep computing zs after they diverge! This
# is very wasteful! There are better, if a little
# less simple, ways to do this.
#
step = tf.group(
zs.assign(zs_),
)
```
and we run it for a couple hundred iterations:
for i in range(200): step.run()
Then, displaying the result using:
DisplayFractal(ns.eval())
Mandelbrot Set in TensorFlow- Running the Computation
So, this was all about Mandelbrot Set in TensorFlow. Hope you like and understand our explanation of computing Mandelbrot Set using TensorFlow.
Conclusion: TensorFlow Mandelbrot Set
Hence, we saw the concept of Mandelbrot Set in TensorFlow, just like the Mandelbrot Set solution. TensorFlow proves to be a great tool for visualizing and understanding not only machine learning concepts but complex mathematical functions including random processes and graph theory.
At last, we discussed how to run the computation for TensorFlow Mandelbrot Set. Next up, is solving PDEs using TensorFlow. Furthermore, for any query regarding Mandelbrot Set in TensorFlow, feel free to ask in the comment section.
Did you like our efforts? If Yes, please give DataFlair 5 Stars on Google
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1. ## Rearanging formula
Can anyone please explain how to get from 1 --> 2
1. $Q + r* dQ/dr + q*r = 0$
2. $1/r * d/dr * r*Q = -q$
Thanks
Calypso
2. Originally Posted by calypso
Can anyone please explain how to get from 1 --> 2
1. $Q + r* dQ/dr + q*r = 0$
2. $1/r * d/dr * r*Q = -q$
Thanks
Calypso
$Q+r\frac{dQ}{dr}+qr=0$
$\frac{Q}{r}+\frac{dQ}{dr}=-q$
Now, you can continue, using
$\frac{dQ}{dr}=\frac{d}{dr}\left(\frac{Qr}{r}\right )$ and use the quotient rule of differentiation,
$\frac{d}{dr}\left(\frac{u}{v}\right)=\frac{v\frac{ du}{dr}-u\frac{dv}{dr}}{v^2}$
$u=Qr,\ v=r$ gives
$\frac{r\frac{d(Qr)}{dr}-Qr(1)}{r^2}=\frac{1}{r}\frac{d(Qr)}{dr}-\frac{Q}{r}$
Hence,
$\frac{Q}{r}+\frac{dQ}{dr}=\frac{1}{r}\frac{d(Qr)}{ dr}$
3. Great thanks very much
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# hw3sol - Math 131A Section 2 Spring 2010 Homework 3...
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Unformatted text preview: Math 131A - Section 2 Spring 2010 Homework 3 Problems graded: 8.6 (2 points), 8.8 (3 points), 8.10 (2 points), 9.3 (2 points), 9.6 (3 points), 9.11 (3 points) 8.2(a). Let > 0 and set N =- 1 . Then, n n 2 + 1- 0 = n n 2 + 1 ≤ n n 2 = 1 n < for n > N . Hence, lim n →∞ n n 2 +1 = 0. (c). Let > 0 and set N = max { 1 , 42 41- 1 } . Then, 4 n + 1 7 n- 5- 4 7 = 41 7(7 n- 5) , and for n > N ≥ 5, we have that 7 n- 5 > 6 n , and so 41 7(7 n- 5) < 41 42 n < . We conclude that lim n →∞ 4 n +3 7 n- 5 = 4 7 . (e). Let > 0 and set N =- 1 . Then, for n > N , we have that 1 n sin( n )- 0 = 1 n | sin( n ) | ≤ 1 n < since | sin( x ) | ≤ 1 for all x ∈ R . Therefore, lim n →∞ 1 n sin( n ) = 0. 8.3. Let > 0. Since lim n →∞ s n = 0, there is N such that | s n- | < 2 for all n > N . Then, for n > N , we have that | √ s n- | = √ s n < , so lim n →∞ √ s n = 0. 8.6.(a). Suppose lim n →∞ s n = 0, and let > 0. Choose N such that | s n- | < for n > N . Then, || s n | - | = | s n | = | s n- | < for n > N , so lim n →∞ | s n | = 0. Now suppose lim n →∞ | s n | = 0 and let > 0. Choose N such that || s n |- | < for n > N . Then, | s n- |...
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## This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.
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https://nl.mathworks.com/matlabcentral/answers/1773140-choose-the-limit-of-two-y-axes
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# choose the limit of two y axes
3 views (last 30 days)
SYML2nd on 3 Aug 2022
Commented: KSSV on 3 Aug 2022
I would like to do a graph with two y axes. I have found this. I have two questions that have no answer here:
1.how to have the two axes simply in black and not in blu and orange (very strange default choice).
2.How can I choose the limit of each axes? I usually use ylim([]), but now how can I choose the right or left axis?
Thank you
KSSV on 3 Aug 2022
Edited: KSSV on 3 Aug 2022
x = linspace(0,25);
y = sin(x/2);
yyaxis left ;
plot(x,y,'k');
ax(1) = gca ;
r = x.^2/2;
yyaxis right
plot(x,r);
ax(2) = gca ;
% change axis colors
ax(1).YColor = 'g' ;
ax(2).YColor = 'm' ;
ax(1).YLim % set your value here
ans = 1×2
0 350
ax(2).YLim % set your value here
ans = 1×2
0 350
KSSV on 3 Aug 2022
Try using ylim after the plot twice.
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# [Plugin]Board, Layout2Board [Behavior] Grid Move
0 favourites
• 564 posts
• Is there anyway to set a value at points (0,0) like at arrays. I tried it with board instance or with hex tx instance.
For x 0 to 9
y 0 to 9
then on button click It gives only sum (100)
------------------------------------
With hex tx, I created a chess object. Then I tried to move it with grid move(hexagon) Up arrow is down, right arrow pressed It moves two squares right one square up, if two squares are available.
It would be really great if somene could share hex tx example.
Thank you
• ## Try Construct 3
Develop games in your browser. Powerful, performant & highly capable.
Construct 3 users don't see these ads
• It looks like that you want to move twice since "Up arrow is down, right arrow pressed" will trigger both of
"+ right arrow pressed" and ""+Up arrow is down, right arrow pressed". So that it move "Up-right" and "Right" at the same tick.
To pick chess at logic (0,0) , you might use "condition: pick chess at logic XYZ". Then do something after picking.
• My bad about movement. Thank you. I added inverted up-down arrows to solve that, and works like a charm now.
Well actually I wanted to ask to set value to points, like "set value at" (arrays)
Beceause I made something like this with arrays.
And I pretty liked the result.
First I randomly set values to all array points. Then for each xy, I look for neighbor values. %50 chanse to convert current point to neighbors sum.
<img src="https://dl.dropbox.com/u/56268958/randomkomsu.jpg" border="0" />
There are 4 green, 2 blue and 2 red neighbors. So mid square has a %50 chanse to convert to green. So it becomes like , continents at the end(with a 2-3 repeats).
It's like this logic
It divides squares. And gives chanse for child squares to look like mother square. But mine looks for neighbors. And looks like to them.
I wanted to do smth like this with your beatiful board plugin.
• The "array" is stored at each chess (cell) on the board. In this case, you might save the value at private value of each chess (cell).
Just pick chess (cell) by condition/action: pick chess at LXYZ, then read the private value. (or you could present the value at frame index)
• Update:
• Now grid move behavior could move tile (z=0) since some users might ask me why grid move could not work. It might be easy to use in a small case which only have tiles on the board.
• Hi All,
I have a few questions regarding the awesome plugins created here.
I'm working on extending the moving options demo. This is the one where you click a black chess piece then it shows you where you can move it etc.
Is it possible to make the layer bigger, when using the board object plug in I can't get a 13x20 board to fit on the layer :-(
I'm wondering is if I'm approaching this correctly.
I want to have say 4 pieces for each player, so player one puts down 4 pieces (clicks on four empty squares) and then player two does the same.
Should I set instance variables on the chess piece to mark it as being P1 or P2 etc. Then then anotehr Instance Variable for setting the piece type?
Thanks for all your help the first problem is the one I'm struggling with the most.
Thanks
Chris
• Finished reading the whole thread and it looks like some amazing stuff going on.
2 questions.
What is the easiest way to switch between movable chess tiles?
At the moment I've saving the x,y to global variables then when moving I'm using the board to pick them and then moving
Second questions is I've used the movement object to define the movable path then I move the piece but when I use inst->clean it doesn't remove my path tiles?
Thanks
Chris
• Uh, I'm not sure what's you mean "make the layer bigger".
If you want to make a bigger board, just set the width and height at properties table of board object, or use "action:Set board width", "action:Set board height".
About question 2, it depends on your design, these plugins did not handle the owner of chess.
• Thanks Rex,
I've managed to just add instance variables so I know what type of piece I'm working with and what player they belong to :-)
• To swap two chess, you could save one UID of chess, pick the other one, then use "action:Swap" in grid move behavior and put the first UID of chess in parameter. It will swap the logic index and physical position.
If you want to move the chess by your self, you could use "action:Swap chess by UID" in board object, it only swap the logic index.
About question 2, I'm not sure what's you means the "inst->clean". Uh... I did not remember I had made this action.
The moving path is stored by UID of tiles. User just pick tile by these UID and do what they want.
• Amazing thanks.
Yeah I ended up using the move to grid option and storing the x,y position of the logical chess
Is it possible to move diagnostically with the movement object (moveable path)
Lastly is it possible to only highly the target destination of the moveable area, instead of the full path?
for example if I you have 4 squares to move it only highlights the 4th square not the squares in the path?
Thanks
Chris
• Humm... one work-arond solution is --
1. pick the move-able are with moving points = 4, stored the tiles UID in instance group named "rang4", for example
2. pick the move-able are with moving points = 3, stored the tiles UID in instance group named "rang3", for example
3. instance group object has "Action:A - B" -> "rang4" - "rang3", to pick tiles which in set "range4" but not in set "rang3"
• <img src="http://dl.dropbox.com/u/34375299/Construct%202/bugs%20and%20fixes/SLGBoard%20error.JPG" border="0" />
What this error means? It just start showing up without a reason when I want to use an action: Sprite: GridMove move to Tiles_prite
• Uh, I'm not sure, please download the slg movement again... and give me the capx file which has problem.
• It was working fine until I've changed position of my board tiles on the layout. Then this error starts to pop up. I tried to undo my changes and play a lot with different settings for width and high in LayoutToBoard and SquareTX but it didn't help.
In the file below I've removed everything. All unnecessary layouts, layers, object and events.
Here's the .capx
It only works - but without moving sprite to target tile, when you disable
Human_Detector: GridMove move to Tile in event 10.
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# Algebra Survival e-Workbook arrives TODAY!!
eBook version of the Algebra Survival Guide is HERE! Check it out on Amazon.com Price is 20% lower than that of the paperback version, and the e-version has several unique advantages over the paperback book.
# How to Decrease Algebraic Mistakes – Part 6
Factoring by grouping is one of the trickier skills in Algebra 1. But by using a special form of notation, the double-slash, students can do this skill with greater ease. This post explains how this notation can help students tackle this challenging form of factoring.
# How to Decrease Algebraic Mistakes – Part 5
Combining like terms confuses many students. But by using a special notation, the double-slash, students get a visual aid that helps them combine like terms with greater care. This post shows how to use the double-slash to combine like terms. And it offers practice problems and their solutions, too.
# How to decrease Algebraic mistakes – Part 4
The double-slash notation comes to the rescue in algebra once again. In this post we learn how this clever notation device helps students combine positive and negative numbers, the trickiest pre-algebra concept of all. Read to find out how you can use this notation to make this concept easier for your students or children.
# How to Decrease Mistakes in Algebra – Part 3
The double-slash helps students for a variety of reasons. This post explains some of the reasons why it works.
# How to Decrease Mistakes in Algebra – Part 2
Algebra is an area of math that leads many students to make mistakes. Using a simple “Double-Slash” mark helps students wall off parts of algebra expressions from other parts, so that they avoid making mistakes. This post and the next several post will help educators learn how to use the “Double-Slash.”
# How to Decrease Algebraic Mistakes – Part 1
Algebraic expressions are covered with mental “land mines.” Step the wrong way, and an expression will blow up on the page, making it impossible for students to move in the right direction. Fortunately there’s a simple thing students can do to succeed at simplifying algebraic expressions, and that simple something is using what I call […]
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Lesson Plan:
# A.M. or P.M.?
no ratings yet
Subject
August 4, 2015
## Learning Objectives
Students will be able to distinguish between a.m. and p.m. throughout the day.
## Lesson
### Introduction (10 minutes)
• Talk with your class about time. Explain to your students that being able to read time is important because it ensures that they will be on time to appointments or meetings later on in life. Tell students that reading time can also help them keep track of everything they do in a day, and is necessary to create a schedule.
• Ask students if they know what a digital clock is. Write down your students' answers on the whiteboard.
• Ask students if they know what an analog clock is. Write down your students' answers on the whiteboard.
• Talk about what a.m. and p.m. mean. Define a.m. as meaning the time before noon, so from midnight to 11:59 in the morning. Define p.m. as meaning after noon, from the middle of the day until the middle of the night.
• Write down thee definitions on the whiteboard.
### Explicit Instruction/Teacher Modeling (5 minutes)
• Hang up the pocket chart somewhere the entire class can see it.
• Create two columns labeled A.M. and P.M.
• Tell the class that they will be identifying whether an activity is done in the morning (A.M.) or in the afternoon and evening (P.M.). Tell students that they will be placing the activity in the correct category on the pocket chart.
• Identify one activity and place it in the correct category together.
### Guided Practice/Interactive Modeling (5 minutes)
• Call on volunteer students to choose and place the notecards with activities on them under the correct category.
### Independent Working Time (10 minutes)
• Pass out the A.m. or P.m.? worksheet.
• Walk around the room to check for understanding as students complete the worksheet.
## Extend
### Differentiation
• Enrichment: Challenge students to create their own a.m. or p.m. question.
• Support Help struggling students by reading and explaining the questions to them.
## Review
### Assessment (5 minutes)
• Teacher observations during work time and worksheet correctness can be used as forms of assessment.
### Review and Closing (5 minutes)
• Have students share what the first thing they do in the morning is and what the last thing they do before bed is with the class.
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https://www.shaalaa.com/question-bank-solutions/heights-distances-from-top-50-m-high-tower-angles-depression-top-bottom-pole-are-observed-be-45-60-respectively-find-height-pole_23133
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Share
From the Top of a 50 M High Tower, the Angles of Depression of the Top and Bottom of a Pole Are Observed to Be 45° and 60° Respectively. Find the Height of the Pole. - CBSE Class 10 - Mathematics
Question
From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. Find the height of the pole.
Solution
Let H be the height of the pole, makes an angle of depression from the top of the tower to top and bottom of\ poles are 45° and 60° respectively.
Let AB = H , CE = hAD = x and DE = 50m.
∠CBE = 45^@ and ∠DAE = 60^@
Here we have to find height of pole.
The corresponding figure is as follows
=> tan A = (DE)/(AD)
=> tan 60^@ = 50/x
=> x = 50/sqrt3
Again in ΔBCE
=> tan B = (CE)/(BC)
=> tan 45^@ = h/x
=> 1 = h/x
=> h = 50/sqrt3
=> h = 28.87
Therefore H = 50 - h
=> H = 50 - 28.87
=> H = 21.13
Hence height pole is 21.13 m
Is there an error in this question or solution?
Video TutorialsVIEW ALL [5]
Solution From the Top of a 50 M High Tower, the Angles of Depression of the Top and Bottom of a Pole Are Observed to Be 45° and 60° Respectively. Find the Height of the Pole. Concept: Heights and Distances.
S
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2015-01-08T19:36:20-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
4(12-n)<12
48-4n<12 (i expanded the brackets)
-4n<-36 (i subtracted by 48 to both sides)
n>9 ( i divided both sides by -4 and the sign changes when dividing by a negative in inequalities)
therefore n is greater than 9
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https://www.physicsforums.com/threads/solving-system-of-two-differential-equations.721241/
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# Solving System of Two Differential Equations
1. Nov 6, 2013
### tsslaporte
The problem statement, all variables and given/known data
Find General Solution of the Following System
(2D+5)x - (2D+3)y = t
(D-2)x + (D+2)y = 0
https://dl.dropboxusercontent.com/u/32294083/Emath/New%20Doc%203_1.jpg [Broken]
Using the Quadratic Formula I get nothing so I am not sure what the complementary solution is.
After this what do I do to find the General Solution?
Last edited by a moderator: May 6, 2017
2. Nov 6, 2013
### Ray Vickson
Start over: your "characteristic equation" is wrong, so far as I can make out. You really should type this stuff out; your writing is borderline unreadable.
Last edited by a moderator: May 6, 2017
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http://www.docstoc.com/docs/125208858/Introduction-to-Artificial-Intelligence-236501
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Introduction to Artificial Intelligence 236501 by ewghwehws
VIEWS: 3 PAGES: 29
• pg 1
``` Intro to AI
Uncertainty
Ruth Bergman
Fall 2002
Why Not Use Logic?
• Suppose I want to write down rules about medical diagnosis:
Diagnostic rules: A x has(x,sorethroat) has(x, cold)
Causal rules: A x has(x,cold) has(x, sorethroat)
• Clearly, this isn’t right:
Diagnostic case:
• we may not know exactly which collections of symptoms
or tests allow us to infer a diagnosis (qualification problem)
• even if we did we may not have that information
• even if we do, how do we know it is correct?
Causal rules:
• Symptoms don’t usually appear guaranteed; note logical
case would use contrapositive
• There are lots of causes for symptoms; if we miss one we
might get an incorrect inference
• How do we reason backwards?
Uncertainty
• The problem with pure FOL is that it deals with black
and write
• The world isn’t black and write because of uncertainty:
1. Uncertainty due to imprecision or noise
2. Uncertainty because we don’t know everything about the
domain
3. Uncertainty because in practice we often cannot acquire all
the information we’d like.
– As a result, we’d like to assign a degree of belief (or
plausibility or possibility) to any statement we make
– note this is different than a degree of truth!
Ways of Handling
Uncertainty
• MYCIN: operationalize uncertainty with the rules:
• a b with certainty 0.7
• we know a with certainty 1
– ergo, we know b with 0.7
– but, we if we also know
• a c with certainty 0.6
• b v c d with certainty 1
– do we know d with certainty .7, .6, .88, 1, ....?
• suppose a ~e and ~e ~d ....
– In a rule-based system, such non-local
dependencies are hard to catch
Probability
• Problems such as this have led people to invent lots
of calculi for uncertainty; probability still dominates
• Basic idea:
– I have some DoB (a prior probability) about some proposition
p
– I receive evidence about p; the evidence is related to p by a
conditional probability
– From these two quantities, I can compute an updated DoB
about p --- a posterior probability
Probability Review
• Basic probability is on propositions or propositional
statements:
– P(A) (A is a proposition)
• P(Accident), P(phonecall), P(Cold)
– P(X = v) (X is a random variable; v a value)
• P(card = JackofClubs), P(weather=sunny), ....
– P(A v B), P(A ^ B), P(~A) ...
– Referred to as the prior or unconditional probability
• The conditional probability of A given B
P(A | B) = P(A,B)/P(B)
– the product rule P(A,B) = P(A | B) * P(B)
• Conditional independence P(A | B) = P(A)
– A is conditionally independent of B
Probability Review
• The joint distribution of A and B
– P(A,B) = x ( equivalent to P(A ^ B) = x)
A=1 A=2 A=3 P(A=1,B) = .1
P(A=1) = .1 + .2 = .3
B = T .1 .1 .2 .4 P(A =1 | B) = .1/.4 = .25
B = F .2 .1 .3 .6
.3 .2 .5 1
Bayes Theorem
• P(A,B) = P(A | B) P(B) = P(B | A) P(A)
P(A|B) = P(B | A) P(A) / P(B)
• Example: what is the probability of meningitis when a patient
has a stiff neck?
P(S|M) = 0.5
P(M) = 1/50000
P(S) = 1/20
P(M|S) = P(S|M)P(M)/P(S) = 0.5 * 1/50000 / 1/20 = 0.0002
• More general
P(A | B , E) = P(B | A , E) P(A | E)/ P(B | E)
Alarm System Example
• A burglary alarm system is fairly reliable at detecting
burglary
• It may also respond to minor earthquakes
• Neighbors John and Mary will call when they hear the
alarm
• John always calls when he hears the alarm
• He sometimes confuses the telephone with the alarm
and calls
• Mary sometimes misses the alarm
• Given the evidence of who has or has not called, we
would like to estimate the probability of a burglary.
Alarm System Example
• P(Alarm|Burglary) A burglary alarm system is fairly reliable at
detecting burglary
• P(Alarm|Earthquake) It may also respond to minor earthquakes
• P(JohnCalls|Alarm), P(MaryCalls|Alarm) Neighbors John and
Mary will call when they hear the alarm
• John always calls when he hears the alarm
• P(JohnCalls|~Alarm) He sometimes confuses the telephone with
the alarm and calls
• Mary sometimes misses the alarm
• Given the evidence of who has or has not called, we would like
to estimate the probability of a burglary.
P(Burglary|JohnCalls,MaryCalls)
Influence Diagrams
• Another way to present this information is an
influence diagram
burglary earthquake
alarm
John calls Mary calls
Influence Diagrams
1. A set of random variables.
2. A set of directed arcs
An arc from X to Y means that X has influence on Y.
3. Each node has an associated conditional probability table.
4. The graph has no directed cycle.
burglary earthquake
alarm
John calls Mary calls
Conditional Probability
Tables
• Each row contains the conditional probability for a possible
combination of values of the parent nodes
• Each row must sum to 1
burglary earthquake B E P(Alarm|B, E)
T F
T T 0.95 0.05
alarm T F 0.94 0.06
F T 0.29 0.71
F F 0.001 0.999
John calls Mary calls
Belief Network for the
Alarm
P(B) P(E)
0.001
burglary earthquake 0.002
B E P(A)
T T 0.95
T F 0.94
alarm
F T 0.29
F F 0.001
A P(A
)
John calls Mary calls
A P(A
)
T 0.90 T 0.70
F 0.05 F 0.01
The Semanics of Belief
Networks
• The probability that the alarm sounded but neither a burglary
nor an earthquake has occurred and both John and Mary call
– P(J ^ M ^ A ^ ~B ^ ~E) =
P(J | A) P(M | A) P(A | ~B ^ ~E) P(~B) P(~E) =
0.9 * 0.7 * 0.001 * 0.999* 0.998 = 0.00062
• More generally, we can write this as
– P(x1, ... xn) = πi P(xi | Parents(Xi))
Constructing Belief
Networks
1. Choose the set of variables Xi that describe the
domain
2. Choose an ordering for the variables
1. Ideally, work backward from observables to root causes
3. While there are variables left:
1. Pick a variable Xi and add it to the network
2. Set Parents{Xi} to the minimal set of nodes such that
conditional independence holds
3. Define the conditional probability table for Xi
• Once you’re done, its likely you’ll realize you need to
fiddle a little bit!
Node Ordering
• The correct order to add nodes is
– Add the “root causes” first
– Then the variables they
influence
– And so on…
Mary calls John calls
• Alarm example: consider the
ordering
– MaryCalls, JohnCalls, Alarm,
Burglary, Earthquake
– MaryCalls, JohnCalls, earthquake
Earthquake, Burglary, Alarm
burglary alarm
Probabilistic Inference
• Diagnostic inference (from effets to causes)
– Given that JohnCalls, infer that P(B|J) = 0.016
• Causal inference (from causes to effects)
– Given Burglary, P(J|B) = 0.86 and P(M|B) = 0.67
• Intercausal inference (between causes of a common
effect)
– Given Alarm, P(B|A) = 0.376
– If Earthquake is also true, P(B|A^E) = 0.003
• Mixed inference (combining two or more of the
above)
– P(A|J ^ ~E) = 0.03
– P(B|J ^ ~E) = 0.017
Conditional Independence
D-separation
• if every undirected path from a set of nodes X to a set of nodes
Y is d-separated by E, then X and Y are conditionally
independent given E
• a set of nodes E d-separates two sets of nodes X and Y if every
undirected path from a node in X to a node in Y is blocked
given E
X E Y
Z
Z
Z
Conditional Independence
• An undirected path from X to Y is blocked given E if there is a
node Z s.t.
1. Z is in E and there is one arrow leading in and one arrow
2. Z is in E and Z has both arrows leading out
3. Neither Z nor any descendant of Z is in E and both path
X E Y
Z
Z
Z
An Inference Algorithm for
Belief Networks
• In order to develop an algorithm, we will assume our networks are
singly connected
– A network is singly connected if there is at most a single
undirected path between nodes in the network
• note this means that any two nodes can be d-separated by
removing a single node
– These are also known as polytrees.
• We will then consider a generic node X with parents U1...Um,
and children Y1 ... Yn.
– parents of Yi are Zi,j
– Evidence above X is Ex+; below is Ex-
Singly Connected Network
Ex+
U1 … Um
X
Ex-
Z1j Z1j
…
Y1 Y1
Inference in Belief
Networks
• P(X|Ex) = P(X | Ex+, Ex-) = k P(Ex- | X, Ex+) P(X | Ex+)
k P(Ex- | X) P(X | Ex+)
– the last follows by noting that X d-separates its parents and
children
• Now, we note that we can apply the product rule to the second
term i
P(X | Ex+) = Σu P(X | u, Ex+) P(u | Ex+)
= Σu P(X | u) πi P(ui | EU/X)
again, these last facts follow from conditional independence
• Note that we now have a recursive algorithm: the first term in the
sum is just a table lookup; the second is what we started with on
a smaller set of nodes.
Inference in Belief
Networks
• P(X|E) = k P(Ex- | X) P(X | Ex+)
• The evaluation for the first expression is similar, but
more involved, yielding
P(X | Ex+) = k2 πi Σy P(Ex-| yi) Σz P(yi | X, zi ) πj P(zij | EZij/Yi)
• P(Ex-| yi) is a recursive instance of P(Ex- | X)
• P(yi | X, zi ) is a conditional probability table entry for Yi
• P(zij | EZij/Yi) is a recursive instance of the P(X|E)
calculation
The Algorithm
Support-Except(X,V) return P(X| Ex/v)
if EVIDENCE(X) then return point dist for X
else
calculate P(E-x/v| X) = evidence-except(X,V)
U parents(X)
if U is empty
then return normalized P(E-x/v| X) P(X)
else
for each Ui in U
calculate and store P(Ui|Eui/X) = support-except(Ui,X)
return k P(Ex- | X) Σu P(X | u) πi P(ui | EU/X)
The Algorithm
Evidence-Except(X,V) return P(E-X\V| X )
Y children[X] – V
if Y is empty
then return a uniform distribution
else
for each Yi in Y do
calculate P(E-Yi|yi) = Evidence-Except(Yi, null)
Zi = PARENTS(Yi) – X
foreach Zij in Zi
calculate P(Zij | Ezij\Yi) = Support-Except(Zij,Yi)
return k2 πi Σy P(Ex-| yi) Σz P(yi | X, zi ) πj P(zij | EZij/Yi)
The Call
• For a node X, call Support-Except(X,null)
PathFinder
• Diagnostic system for lymph node disease
• Pathfinder IV a Bayesian model
– 8 hrs devising vocabulary
– 35 hrs defining topology
– 40 hrs to make 14000 probability assessments
experts who designed it!
Other Uncertainty
Calculi
• Dempster-Shafer Theory
– Ignorance: there are sets which have no probability
– In this case, the best you can do, in some cases, is bound
the probability
– D-S theory is one way of doing this
• Fuzzy Logic
– Suppose we introduce a fuzzy membership function (a
degree of membership
– Logical semantics are based on set membership
– Thus, we get a logic with degrees of truth
• e.g. John is a big man bigman(John) w. truth value 0.
```
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# Does any one know the answer for Puzzle 70: Bungee..
Guest asks: Added Nov 7th 2011, ID #227797
#### Question for May's Mysteries: The Secret of Dragonville
Does any one know the answer for Puzzle 70: Bungee jump?
##### Guest answered:
Added 3rd Dec 2011, ID #461043
The Answer Is 50 This Took Me A While To Work Out Haha!
Side A-B Is 130m
You Then Work Out 36km/h To 10 m/s
10 m/s multiplied by 12 (The Seconds It Takes Him To Run) = 120m (Side B-C)
Pythagorean Theorem -
130 Squared = 16900
120 Squared = 14400
16900 - 14400 = 2500
Square root that and your answer is 50 :P
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Factorial Notation
Let n be a positive integer. Factorial of n denoted by n! and is defined as
n! = n(n - 1)(n - 2)(n-3) ... 3.2.1.
Examples:
We define 0! = 1.
2! = 2.1 = 2
3! = 3.2.1 = 6
4! = 4.3.2.1 = 24
5! = 5.4.3.2.1 = 120
So on......
Permutations (Arrangements
The different arrangements of a given number or things by taking some or all at a time, are called permutation.
Examples:
1. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
2. All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba).
Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
nPr = n(n - 1)(n - 2) ... (n - r + 1)
Examples:
• 6P2 = (6 x 5) = 30.
• 7P3 = (7 x 6 x 5) = 210.
Cor. number of all permutations of n things, taken all at a time = n!.
Note: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind;p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.
Then, number of permutations of these n objects is = n! (p1!).(p2)!.....(pr!)
Combinations
Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Examples:
• Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note: AB and BA represent the same selection.
• All the combinations formed by a, b, c taking ab, bc, ca.
• The only combination that can be formed of three letters a, b, c taken all at a time is abc.
• Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.
Note: that ab ba are two different permutations but they represent the same combination.
Number of Combinations:
The number of all combinations of n things, taken r at a time is:
Note:
1. nCn = 1 and nC0 = 1.
2. nCr = nC(n - r)
Examples:
11C4 = (11 x 10 x 9 x 8) = 330. (4 x 3 x 2 x 1)
16C13 = 16C(16 - 13) = 16C3 = 16 x 15 x 14 = 16 x 15 x 14 = 560. 3! 3 x 2 x 1
Choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations.
16! = 16! = 16! = 560 3!(16-3)! 13!(16-13)! 3!×13!
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# Documentation
Mathlib.Topology.ContinuousFunction.Ordered
# Bundled continuous maps into orders, with order-compatible topology #
def ContinuousMap.abs {α : Type u_1} {β : Type u_2} [] [] [] (f : C(α, β)) :
C(α, β)
The pointwise absolute value of a continuous function as a continuous function.
Instances For
instance ContinuousMap.instAbsContinuousMap {α : Type u_1} {β : Type u_2} [] [] [] :
Abs C(α, β)
@[simp]
theorem ContinuousMap.abs_apply {α : Type u_1} {β : Type u_2} [] [] [] (f : C(α, β)) (x : α) :
|f| x = |f x|
We now set up the partial order and lattice structure (given by pointwise min and max) on continuous functions.
instance ContinuousMap.partialOrder {α : Type u_1} {β : Type u_2} [] [] [] :
theorem ContinuousMap.le_def {α : Type u_1} {β : Type u_2} [] [] [] {f : C(α, β)} {g : C(α, β)} :
f g ∀ (a : α), f a g a
theorem ContinuousMap.lt_def {α : Type u_1} {β : Type u_2} [] [] [] {f : C(α, β)} {g : C(α, β)} :
f < g (∀ (a : α), f a g a) a, f a < g a
instance ContinuousMap.sup {α : Type u_1} {β : Type u_2} [] [] [] :
Sup C(α, β)
@[simp]
theorem ContinuousMap.sup_coe {α : Type u_1} {β : Type u_2} [] [] [] (f : C(α, β)) (g : C(α, β)) :
↑(f g) = f g
@[simp]
theorem ContinuousMap.sup_apply {α : Type u_1} {β : Type u_2} [] [] [] (f : C(α, β)) (g : C(α, β)) (a : α) :
↑(f g) a = max (f a) (g a)
instance ContinuousMap.semilatticeSup {α : Type u_1} {β : Type u_2} [] [] [] :
instance ContinuousMap.inf {α : Type u_1} {β : Type u_2} [] [] [] :
Inf C(α, β)
@[simp]
theorem ContinuousMap.inf_coe {α : Type u_1} {β : Type u_2} [] [] [] (f : C(α, β)) (g : C(α, β)) :
↑(f g) = f g
@[simp]
theorem ContinuousMap.inf_apply {α : Type u_1} {β : Type u_2} [] [] [] (f : C(α, β)) (g : C(α, β)) (a : α) :
↑(f g) a = min (f a) (g a)
instance ContinuousMap.semilatticeInf {α : Type u_1} {β : Type u_2} [] [] [] :
instance ContinuousMap.instLatticeContinuousMap {α : Type u_1} {β : Type u_2} [] [] [] :
theorem ContinuousMap.sup'_apply {β : Type u_2} {γ : Type u_3} [] [] [] {ι : Type u_4} {s : } (H : ) (f : ιC(β, γ)) (b : β) :
↑(Finset.sup' s H f) b = Finset.sup' s H fun a => ↑(f a) b
@[simp]
theorem ContinuousMap.sup'_coe {β : Type u_2} {γ : Type u_3} [] [] [] {ι : Type u_4} {s : } (H : ) (f : ιC(β, γ)) :
↑(Finset.sup' s H f) = Finset.sup' s H fun a => ↑(f a)
theorem ContinuousMap.inf'_apply {β : Type u_2} {γ : Type u_3} [] [] [] {ι : Type u_4} {s : } (H : ) (f : ιC(β, γ)) (b : β) :
↑(Finset.inf' s H f) b = Finset.inf' s H fun a => ↑(f a) b
@[simp]
theorem ContinuousMap.inf'_coe {β : Type u_2} {γ : Type u_3} [] [] [] {ι : Type u_4} {s : } (H : ) (f : ιC(β, γ)) :
↑(Finset.inf' s H f) = Finset.inf' s H fun a => ↑(f a)
def ContinuousMap.IccExtend {α : Type u_1} {β : Type u_2} [] [] [] [] {a : α} {b : α} (h : a b) (f : C(↑(Set.Icc a b), β)) :
C(α, β)
Extend a continuous function f : C(Set.Icc a b, β) to a function f : C(α, β).
Instances For
@[simp]
theorem ContinuousMap.coe_IccExtend {α : Type u_1} {β : Type u_2} [] [] [] [] {a : α} {b : α} (h : a b) (f : C(↑(Set.Icc a b), β)) :
↑() =
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# In a binary system, why $m_1 a_1 = m_2 a_2$?
In the center of mass coordinate, $$m_1 r_1 = m_2 r_2$$, which is straightforward. Yet in this detailed deviation of radial velocity page 27, it says that the $$r_1$$, which is the magnitude of the vector pointing from the CM to the star, is simply the semi‐major axis of the star’s orbit around the mutual CM, $$a_1$$. With the same reasoning, the $$r_2$$, which is the magnitude of the vector pointing from the CM to the planet, is the semi‐major axis of the planet’s orbit around the mutual CM, $$a_2$$. Therefore $$m_1 a_1 = m_2 a_2$$.
However, I do not see why $$r_1 (r_2)$$ can be identical to $$a_1(a_2)$$ since the $$r's$$ are both changing (in a elliptical orbit for example) while the $$a's$$ are fixed. Or the other way to ask this question is I do not see why $$r_1/a_1 = r_2/a_2$$?
It’s just poorly written. He doesn’t really mean that $$r$$ is $$a$$. He means that $$r$$ is $$a$$ at one point on the orbit, so if $$m_1r_1=m_2r_2$$ for the whole orbit than $$m_1a_1=m_2a_2$$.
• Now I see that $m_1 a_1 = m_2 a_2$ is a special case of $m_1 r_1 = m_2 r_2$. But at the moment when $r_1 = a_1$, how can you know $r_2 = a_2$? Commented Sep 8, 2019 at 0:18
• I think a clearer argument is to say that $2a$ is $r_\text{max}+r_\text{min}$. The two bodies have their min/max $r$ at the same time. So $m_1r_{1,\text{min}}=m_2r_{2,\text{min}}$, etc. That leads to $m_1a_1=m_2a_2$. Commented Sep 9, 2019 at 0:09
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# Depreciation Calculator Online With Graph (Straight Line Method)
Last updated on by Editorial Staff
# Depreciation Calculator with Graph
## Guide to Use Depreciation Calculator
To use the Depreciation Calculator with Graph, follow these simple steps:
• Select Your Currency: Choose the currency in which you want to calculate depreciation (USD, Euro, etc.).
• Cost of the Asset: Enter the total purchase price of the asset.
• Salvage Value: Input the estimated resale value of the asset at the end of its useful life.
• Useful Life: Specify the number of years you expect the asset to be in use.
• Calculate: Click the “Calculate” button to view the depreciation schedule and graph.
## Formula
Depreciation per year = Asset cost – Selvage value / Useful life
## Understanding Depreciation
Depreciation is the process of allocating the cost of tangible assets over their useful lives. It reflects the decrease in the value of an asset over time.
For businesses, depreciation is crucial for tax and accounting purposes, helping to match the cost of an asset with the revenue it generates.
## Who Can Use This Calculator?
• Business Owners: To estimate the depreciation of assets for accounting and tax purposes.
• Financial Analysts: For preparing financial models that require depreciation calculations.
• Students: Learning about accounting principles related to asset management.
• Individuals: Planning to depreciate personal assets for tax reporting.
### Where Is It Useful?
• Accounting and Bookkeeping: For accurate financial reporting and tax filings.
• Budget Planning: Helps in forecasting expenses and planning for future asset purchases.
• Tax Preparation: Essential for calculating deductions related to asset depreciation.
## FAQs
### Can I use this calculator for any type of asset?
Yes, this calculator is versatile and can be used for various tangible assets like machinery, vehicles, and office equipment.
### Is salvage value considered in the calculation?
Yes, the calculator takes into account the salvage value to provide a more accurate depreciation schedule.
### Can I use this for tax purposes?
While this calculator provides a good estimate, always consult with a tax professional for official purposes.
### Does this calculator show yearly depreciation?
Yes, it shows the annual depreciation expense, accumulated depreciation, and the book value at the end of each year.
## Conclusion
Our Depreciation Calculator with Graph is an easy-to-use, versatile tool designed to help individuals and businesses accurately calculate the depreciation of their assets.
By understanding the depreciation process, you can make more informed decisions regarding asset management, budgeting, and tax reporting.
Whether you’re a business owner, financial analyst, student, or individual, this calculator provides valuable insights into how assets depreciate over time, aiding in better financial planning and management.
Remember, while this tool offers a practical way to understand and calculate depreciation, always consult with financial or tax professionals when making official financial decisions or filings.
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# Question: A large city currently provides free water service to residents
A large city currently provides free water service to residents. The marginal social cost of making a gallon of water available per month is estimated to be 5 cents no matter how much water is used. Currently, city residents consume 500,000 gallons of water per month. The costs of making the water available are financed by a local tax on city residents.
a. Draw a graph to show that the current monthly consumption of water is not efficient.
b. Show the net gains in well-being possible by applying a user charge of 5 cents per gallon to residential users.
Assume that monthly consumption declines to 400,000 gallons after the user charge is imposed. Calculate the tax revenues that can be freed for other uses each month (including a reduction in taxes to local residents) after the user charge is imposed.
View Solution:
Sales0
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https://dvillers.umons.ac.be/wiki/teaching:methcalchim:system_of_linear_equations?do=diff&rev2%5B0%5D=1538998066&rev2%5B1%5D=1539850241&difftype=sidebyside
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teaching:methcalchim:system_of_linear_equations
# Différences
Ci-dessous, les différences entre deux révisions de la page.
teaching:methcalchim:system_of_linear_equations [2018/10/08 13:27]villersd teaching:methcalchim:system_of_linear_equations [2018/10/18 10:10]villersd 2018/10/18 10:10 villersd 2018/10/09 09:20 villersd 2018/10/08 16:06 villersd 2018/10/08 13:27 villersd 2017/09/28 10:39 villersd 2017/09/28 10:34 villersd 2016/11/25 11:15 villersd créée 2018/10/18 10:10 villersd 2018/10/09 09:20 villersd 2018/10/08 16:06 villersd 2018/10/08 13:27 villersd 2017/09/28 10:39 villersd 2017/09/28 10:34 villersd 2016/11/25 11:15 villersd créée Ligne 13: Ligne 13: * 2.2 Gaussian Elimination with Backsubstitution * 2.2 Gaussian Elimination with Backsubstitution * 2.3 LU Decomposition and Its Application * 2.3 LU Decomposition and Its Application + * Python [[https://docs.scipy.org/doc/numpy/|NumPy]] library : [[https://docs.scipy.org/doc/numpy/reference/index.html|NumPy Reference]] + * [[https://docs.scipy.org/doc/numpy/reference/routines.linalg.html|Linear algebra (numpy.linalg)]] : [[https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.solve.html#numpy.linalg.solve|numpy.linalg.solve]] * Time complexity analysis * Time complexity analysis * Hint : in Python, use the timeit module * Hint : in Python, use the timeit module + + ===== Jupyter notebooks ===== + * Example file (to be continued) : [[https://notebooks.azure.com/linusable/libraries/samples-public/html/notebooks/calculation_methods_applied_to_chemistry/Gauss-Jordan-01.ipynb]] ===== Exercices and applications ===== ===== Exercices and applications ===== Ligne 27: Ligne 32: * Using [[wp>Tridiagonal_matrix_algorithm|tridiagonal Thomas algorithm]] allows to save computational time thanks to n complexity * Using [[wp>Tridiagonal_matrix_algorithm|tridiagonal Thomas algorithm]] allows to save computational time thanks to n complexity * ? Python library with Thomas algorithm * ? Python library with Thomas algorithm + + ===== What you must have learned in this chapter ===== + * Except ill-conditionned, linear systems can be solved "exactly" using linear algebra algorithms in a finite and known number of arithmetic operations. + * The accuracy is determined by the number of numerical figures which are encoded in floating point description + * For a general system of n equations, diagonalisation requires of the order of n3 operations. Also for solving a system using these method. + * If the coefficient matrix is the same for different systems (only the independent coefficients are different), it is possible to solve systems with the order of n2 operations, if the matrix of coeeficients is decomposed in the product of two triangular matrix (Lower-Upper decomposition). This n3 step is realised only once. ===== References : ===== ===== References : =====
• teaching/methcalchim/system_of_linear_equations.txt
• Dernière modification: 2018/10/18 10:10
• de villersd
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# How do you solve the linear system by substitution 3x + y =2 and -x -3y=6?
Apr 28, 2018
Like so:
#### Explanation:
$3 x + y = 2$
$- x - 3 y = 6 \implies - 6 - 3 y = x$
Substitute this equation into the other system wherever you see $x$.
$3 \left(- 6 - 3 y\right) + y = 6$
Distribute $3$ onto $\left(- 6 - 3 y\right)$
$- 18 - 9 y + y = 2$
Combine alike terms
$- 8 y = 20$
Divide to solve for $y$
$\frac{- 8 y}{-} 8 = \frac{20}{- 8} \implies y = - \frac{20}{8}$
Then solve for $x$ using the value of $y$.
$- x - \frac{60}{8} = 6$
$x = - \frac{108}{8}$
$y = - \frac{20}{8}$
Hope that helps!
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# Quiz – 385
### Quant Quiz
Q1. A and B can complete a job in 24 days working together. A alone can complete it in 32 days. Both of them worked together for 16 days and then A left. The number of days B will take to complete the remaining job is
(a) 16 (b) 32 (c) 64 (d) 128 (e) None of these
Q2. In a hostel, there are 120 students and food stock is for 45 days. If 60 new students join the hostel, in how many days will the complete stock be exhausted?
(a) 38 (b) 40 (c) 32 (d) 36 (e) None of these
Q3. A certain number of men can do a piece of work in 80 days. If there were 10 men less, It could be finished in 20 days more. How many men are there in second case?
(a) 45 (b) 50 (c) 40 (d) 60 (e) 55
Q4. P takes twice the time taken by Q and thrice the time taken by R to do a particular piece of work. Working together, they can complete the work in 2 days. Find the number of days taken by R to complete the work alone.
(a) 4 (b) 6 (c) 8 (d) 2 (e) 5
Q5. 2 men or 3 women or 5 children can paint a house in 10 days. The painting is given to a couple and their 5 sons. They finish the job in
(a) 5(19/66) days (b) 5(10/11)days (c) 5(6/11)days (d) 5(1/5) (e) None of these
Q6. In a factory, there are equal number of men and children. Men work for 6 h a day and children for 4 h a day. During festival time, the work load goes up by 100%. The government rule does not allow children to work for more than 6 h a day. If they are equally efficient and the extra work is done by men, then extra hours of work put in by men everyday are
(a) 5 (b) 3 (c) 4 (d) 9 (e) 8
Q7. A contract is to be completed in 92 days and 234 men were set to work, each working 8 h a day. After 66 days, 3/7 of the work is completed. How many additional men may be employed, so that the work may be completed in time, each man now working 9 h a day?
(a) 160 (b) 230 (c) 620 (d) 700 (e) None of the above
Q8. Rahul can do a piece of work in 16 days. Ravinder can do the same work in 12 4/5 days, while Girish can do it in 32 days. All of them started to work together but Rahul leaves after 4 days. Ravinder leaves the job 3 days before the completion of the work. How long would the work last?
(a) 9 days (b) 6 days (c) 8 days (d) 5 days (e) 4 days
Q9. Amit and Bhanu can do a piece of work in 9 days and 18 days, respectively. As they were ill, they could do 45% and 90% of their efficiency, respectively. How many days will they take to complete the work together?
(a) 7 (b) 8 (c) 5 (d) 10 (e) 12
Q10. Ajit, Bikram and Chandu can do a piece of work individually in 8, 12 and 15 days, respectively. Ajit and Bikramstart working but Ajit quits after working for 4 days. After this, Chandu joins Bikram till the completion of work. In how many days will the work be completed?
(a) 5(8/9)days (b)5(7/9)days (c) 5(5/9)days (d)5(3/4)days (e) None of these
1.b 2.e 3.c 4.a 5.e 6.e 7.e 8.a 9.d 10.e
### Reasoning Quiz
Directions (Q. 1–5): Study the following information carefully and answer the questions given below:
A, B, C, D, E, G and H are sitting in a straight line facing north but not necessarily in same order. B sits third to the right of E. H sits second to the right of B. Only one person sits between D and A. G and C are immediate neighbours of each other. G is not an immediate neighbour of E. D is not an immediate neighbour of B.
Q1. How many persons sit between A and C?
1) None 2) Three 3) One 4) Four 5) Two
Q2. Which of the following pairs sit at the extreme ends of the line?
1) C, H 2) C, E 3) D, E 4) S, H 5) H, B
Q3. What is the position of E with respect to G?
1) Second to the left 2) Fourth of the left 3) Immediate right 4) Immediate left 5) Third to the right
Q4. Four of the following five are alike in a certain way based on their seating arrangement and so form a group. Which is the one that does not belong to this group?
1) E G 2) B H 3) C B 4) A D 5) G H
Q5. Who is fifth to the left of H?
1) G 2) D 3) E 4) C 5) A
Directions (Q. 6–10): Study the following arrangement of series carefully and answer the questions given below:
C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S
Q6. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both?
1) None 2) Three 3) One 4) More than three 5) Two
Q7. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end?
1) 8 2) 2 3) A 4) Other than given options 5) O
Q8. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter?
1) None 2) Two 3) One 4) Three 5) More than three
Q9. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group?
1) 6 & * 2) 5 X O 3) F L 4 4) G O 8 5) 9 K %
Q10. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement?
1) I 2) Other than given options 3) 8 4) J 5) *
1. 3
2. 2; 8 @ G, 4 \$ F, 9 star W
3. 5; Twelfth from the left is O.
4. 3
5. 1
1. 4; Seventh to the right of the eighteenth from the right
end means (18 – 7) = 11th from the right end, i.e. J.
### English Quiz
Directions (Q1-5): Choose the word which his most nearly the SAME in meaning as the word given in bold as used in the passage.
Q1. Conservation
(a) Preservation (b) Generation (c) Irrigation (d) Prevention (e) Application
Q2. Spate
(a) Epidemic of (b) Status of (c) Increase in (d) Wave of (e) Arrival of
Q3. Improvement
(a) Magnification (b) Exaggeration (c) Progress (d) Improvisation (e) Perfection
Directions (Q4-5): Choose the word which is most OPPOSITE in the meaning of the work given in bold as used in the passage.
Q4. Remarkable
(a) Wonderful (b) Graceful (c) Miraculous (d) Inexplicable (e) Insignificant
Q5. Extensively
(a) Briefly (b) Widely (c) Miserly (d) Rarely (e) Economically
Directions (Q6-10): Rearrange the following six comments (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph; then answer questions given below them.
(A) A computer, to corroborate further, can do simulations for a civilian product, for rapid prototyping or the suitability of market conditions for a future product; it can also simulate the performance of fighter aircraft or weapon performance.
(B) But they are the basis for their market dominance as well.
(C) While India needs to pay more attention to economic areas and employment generation, both crucial to making her a developed country, attention should also be paid to the strategic sectors.
(D) Such ‘dual technologies’ are closely guarded by the developed countries under the premise of non-proliferation of nuclear weapons or missiles.
(E) The confluence of civilian and defence technologies is leading to a situation where most new technologies are basically ‘dual use’ in nature.
(F) A carbon-composite material, for an example to substantiate the point, can go into making a tennis racquet or a FOR-caliper device for polio-affected patients and also for a missile system.
Q6. Which of the following should be the FOURTH statement after rearrangement?
(a) A (b) B (c) C (d) D (e) E
Q7. Which of the following should be the FIFTH statement after rearrangement?
(a) A (b) B (c) C (d) D (e) E
Q8. Which of the following should be the SECOND statement after rearrangement?
(a) A (b) B (c) C (d) D (e) E
Q9. Which of the following should be the SIXTH (LAST) statement after rearrangement?
(a) A (b) B (c) C (d) D (e) E
Q10. Which of the following should be the FIRST statement after arrangement?
(a) A (b) B (c) C (d) D (e) E
1.a 2.c 3.c 4.e 5.a 6.a 7.b 8.d 9.c 10.e
### Computer Quiz
Q1. Instructions that are required to start a computer referred to as bootstrap. Where is the bootstrap loader located in the CPU?
(a) SRAM (b) DRAM (c) ROM (d) Hard disk (e) Cache memory
Q2. Which of the following is not one of the manufacturers of MOTHERBOARD?
(a) Intel (b) ASUS (c) Microsoft (d) MSI (e) Gigabyte
Q3. Which of the following is true regarding the characteristics of ports?
(a) Can be a programmatic docking point from where a information flows. (b) These are slots on Motherboard
(c) Can also be used to connect external devices. (d) All of the above (e) None of the above
Q4.Which of the following are type of ports?
(a) Serial port (b) Parallel port (c) PS/2 port (d) VGA Port (e) All of the above.
Q5. Which among the following are example(s) of application software?
(a) Payroll software (b) Railway reservation software (c) Microsoft word
(d) Student attendance software (e) All of the above
Q6. Convert (10101)2 into decimal system.
(a) 31 (b) 21 (c) 11 (d) 12 (e) None of the above
Q7. Convert (19FDE)16 into decimal system
(a) 106462 (b) 100056 (c) 786 (d) 789 (e) None of the above
Q8. ___________ Store data temporarily and pass it on as directed by the control unit.
(a) Address (b) Register (c) Number (d) Memory (e) None of the above
Q9. To access properties of an object, which technique should be used _______.
(a) Dragging (b) Dropping (c) Right clicking (d) Shift-clicking (e) Double clicking.
Q10. __________ is a collection of web – pages and__________ is the very first page that we see on visiting a website.
(e) Web portal, Website
Q11. What’s considered the ‘backbone’ of the World Wide Web?
(a) URL (b) HTML (c) HTTP (d) FTP (e) None of the above.
Q12.Which of the following regarding Internet is correct:
(a) Internet used standard Internet Protocol.
(b) A special computer DNS (Domain Name Server) is used to give name to the IP Address
(c) IP Address is a unique set of numbers (such as 110.22.33.114) which identifies a computer’s location.
(d) All are correct.
(e) None of the above.
Q13. Where was India’s first computer installed and when?
(a) IIT Delhi, 1971 (b) IIT Madras, 1971 (c) Indian Institute of Science Bangalore, 1971
(d) India Statistical Institute, 1955 (e) None of the above
Q14. Which operation is not performed by the computer?
(a) Inputting (b) Processing (c) Controlling (d) Understanding (e) None of the above
Q15. USB in data cables stands for?
(a) Unicode smart Bus (b) Universal structural Bus (c) Unicode Serial Bus (d) Universal serial Bus
(e) None of these
S1. Ans.(c)
Sol. Bootstrap loader. Alternatively referred to as bootstrapping, bootloader, or bootprogram, a bootstrap loader is a program that resides in the computer’s EPROM, ROM, or other non-volatile memory. It is automatically executed by the processor when turning on the computer
S2. Ans. (c)
Sol. Microsoft is a headquartered in Redmond, Washington. MNC that develops, manufactures, licenses, supports and sells computer software, consumer electronics and personal computers and services.
S3. Ans.(d)
Sol. Ports are used to connect external devices or peripherals with computer.
S4. Ans.(e)
Soln. : There are various types of Ports for connecting several external hardware.
S5. Ans. (e)
Sol. Application software can be divided into two general classes: systems software and applications software. Applications software (also called end-user programs) include such things as database programs, word processors, Web browsers and spreadsheets.
S6. Ans. (b)
Sol. ((1 x 24) + (0 x 23) + (1 x 22) + (0 x 21) + (1 x 20))10
=16+4+1=21
S7. Ans. (a)
Sol.((1 x 164) + (9 x 163) + (F x 162) + (D x 161) + (E x 160))10
((1 x 164) + (9 x 163) + (15 x 162) + (13 x 161) + (14 x 160))10
(65536+ 36864 + 3840 + 208 + 14)10
=106462
S8. Ans.(b)
Sol. A (CPU register) is one of a small set of data holding places that are part of the computer processor. A register may hold an instruction, a storage address, or any kind of data. Some instructions specify registers as part of the instruction.
S9. Ans.(c)
Sol. Right clicking provides additional capabilities to a mouse.
S10. Ans.(b)
Sol. A website is a collection of related web pages, including multimedia content, typically identified with a common domain name, and published on at least one web server.
S11. Ans.(c)
Sol. The Hypertext Transfer Protocol (HTTP) is an application protocol. HTTP is the foundation of data communication for the World Wide Web.
S12. Ans.(d)
Sol. All the statements are correct.
S13. Ans.(d)
Sol.HEC-2M was the first computer installed in Indian Statistical Institute, Calcutta 1955.
S14. Ans.(d)
Sol. Except for Artificial Intelligence which still don’t have universal presence, a computer lacks from IQ, Self understanding.
S15. Ans.(d)
Sol. Universal Serial Bus (USB) is an industry standard developed in the mid-1990s that defines the cables, connectors and communications protocols used in a bus for connection, communication, and power supply between computers and electronic devices
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## Monday, January 26, 2015
My personal life has been turned upside down. I enjoy blogging about our new curriculum. I look forward to being able to look back next year in order to help me when we go through a 2nd time. However, when I have to cut some things out because of a lack of time blogging is the first thing cut. The thing is...blogging just helps me to organize my thoughts and see where I have been and where I am going. I think I might could really be diagnosed with ADHD...I doubt that any of my colleagues would disagree. I am kind of joking but I really do think it is true. Forcing myself to focus and "verbalize" what I am doing with my classes helps me to reflect and think. So...here is a brief summary of what we have done so far in Cookies.
How Many of Each Kind? - The students brainstorm the number of dozens of iced and plain cookies that meet the 4 constraints that are mentioned.
A Simpler Cookie - This activity allows the students to only worry about prep time. They assume unlimited amounts of dough, icing and oven space. Then they are asked to try to find the most profit. We believed the max profit in this case was 150 plain and 0 iced.
Manipulating Inequalities - This activity allowed the students to "discover" the rules for solving inequalities. They also practice graphing inequalities in one variable. When going over #3 ask the students what would make the inequalities where you multiplied or divided by a negative "ok" again.
My Simplest Inequality - In Part I the students practice solving inequalities in one variable. We supplemented with an additional worksheet on solving and graphing one-variable inequalities. In Part II we investigated equivalent inequalities. Then in #3 we took inequalities and wrote them in simpler ways.
Simplifying Cookies - In this activity you take the Cookies Constraints and write them as simpler inequalities. We did not spend much time on this but I did write the simpler inequalities on chart paper.
Picturing Cookies - In this activity students take the constraints and use 2 colors to indicate ordered pairs that either work or do not work for the cookies constraints. They will hopefully recognize a "dividing line" that halves the graphs into correct solutions and incorrect solutions. I gave each group one constraint to test and had them put it on chart paper after they did it individually on graph paper.
Inequalities Stories - I gave this as a bonus opportunity...
A Hat of a Different Color - This POW was assigned when I had a sub. I think it would be a good idea to use hats or objects and act out the situation a few times in class before the students try to figure out the color of Carletta's hat. Her hat is blue. If Arturo couldn't tell then the 2 girls could NOT have both had red hats. One of them must have been blue. If Belicia had opened her eyes and seen a red hat on Carletta she would have known her hat was blue. However, since she passed Carletta must have been wearing a blue hat.
Healthy Animals - You create 3 inequalities - one for fat, one for protein and one for total amount. I did not have my class graph this activity.
**Picturing Cookies - Part II - We used this to "reteach" graphing using slope-intercept form. We used Desmos to show the graphs but I don't know if I should have waited on the technology... Anyway, it is extremely important to have each student to graph these correctly and identify the feasible region. They will need it to solve the unit problem. I did the graph on chart paper and left it hanging in the room.
What's My Inequality - I really liked this activity to help us tie the work in the unit to "traditional algebra problems." This is the first time the text mentions the differences in when to make the line solid and when to make the line dashed.
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Posted by Nick .
Student on team #240, Mach V, from Jefferson Monroe High School and Visteon.
Posted on 10/14/2000 9:38 AM MST
I like to think I’m good in math but I must be missing something. 8.5’ x 3 is 25.5, correct? Well think I.D. of the rack is 21’, correct? Also can the rack leave the surface of the playing field? Or the whole field altogether?
Posted by Joe Johnson. [PICTURE: SAME | NEW | HELP]
Engineer on team #47, Chief Delphi, from Pontiac Central High School and Delphi Automotive Systems.
Posted on 10/14/2000 10:30 AM MST
In Reply to: Addition posted by Nick on 10/14/2000 9:38 AM MST:
The inside dimesion of the square rack (the distance between the two parallel tubes) is 26 inches.
The lengths of the straight edges are shorter because the elboes have a radius to them and therefor shorten the effective length of the straight tubes.
Joe J.
P.S. No rule against lifting the rack or pushing it out of bounds. Thinking about what to do if it IS pushed out… more later. JJ
Posted by Raul. [PICTURE: SAME | NEW | HELP]
Engineer on team #111, Wildstang, from Rolling Meadows & Wheeling HS and Motorola.
Posted on 10/14/2000 12:23 PM MST
In Reply to: Inside Dim of Square posted by Joe Johnson on 10/14/2000 10:30 AM MST:
How about scoring of the rack if it is partially out of bounds but still on your side of the field?
Raul
Posted by Joe Johnson. [PICTURE: SAME | NEW | HELP]
Engineer on team #47, Chief Delphi, from Pontiac Central High School and Delphi Automotive Systems.
Posted on 10/14/2000 3:11 PM MST
In Reply to: Re: Inside Dim of Square posted by Raul on 10/14/2000 12:23 PM MST:
1. If the rack is touching the ground out of bounds, it does not count as a doubler.
2. If the rack goes out of bounds, it will not be returned to the field by the officials.
3. A robot may attempt to bring the square rack back in bounds but if the robot touches the ground out of bounds while it is attempting to bring the rack back in bounds, it will be declared out of bounds itself – very perilous work, reclaiming out of bounds square racks.
Official rulings will follow, but I would like comments on this proposal if any exists.
Joe J.
Posted by Jason Rukes.
Engineer on team #109, Arial Systems & Libertyville HS, from Libertyville High School and Arial Systems Corp & SEC Design.
Posted on 10/16/2000 6:19 AM MST
In Reply to: Square Ball Rack & Out of Bounds posted by Joe Johnson on 10/14/2000 3:11 PM MST:
1. Is it legal for a robot to lift the square rack completely off of the playing surface and hold it there?
2. Is it legal for a robot to forcefully latch onto the square rack like a pair of vise-grips, as long as no noticeable damage is done to the square rack?
Posted by Joe Johnson. [PICTURE: SAME | NEW | HELP]
Engineer on team #47, Chief Delphi, from Pontiac Central High School and Delphi Automotive Systems.
Posted on 10/16/2000 8:04 AM MST
In Reply to: Re: Square Ball Rack & Out of Bounds posted by Jason Rukes on 10/16/2000 6:19 AM MST:
: 1. Is it legal for a robot to lift the square rack completely off of the playing surface and hold it there?
Yes, this is perfectly legal
: 2. Is it legal for a robot to forcefully latch onto the square rack like a pair of vise-grips, as long as no noticeable damage is done to the square rack?
Yes, but try to be reasonable with regard to clamping force. We will only have a limited supply of the square racks. If teams routinely start to crush them or otherwise damage them, we will have to start limiting this type of activity. So… Clamping on is OK, but don’t break the thing.
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``````#?global_economy
#?aus_production
#?aus_livestock
#?hh_budget
tail(aus_retail)``````
### 5.1. Produce forecasts for the following series using whichever of NAIVE(y), SNAIVE(y) or RW(y ~ drift()) is more appropriate in each case:
Australian Population (global_economy)
``aus.pop<-global_economy%>%filter(Country=="Australia")``
``aus.pop%>%autoplot(Population/1000000)+labs(x="Years", y="Population in Mlns", title= "Australian Population in millions")``
From the above plot, we can see that the population has been increasing at a steady rate. Because of this, using the Naive and SNaive models would not be appropriate since the Naive model would just forecast based on the last known value and SNaive would incorporate seasonality which doesn’t really exist in the above. Using the Random walk model with a drift seems to be the best choice here.
``aus.pop.fit<-aus.pop%>%model(RW(Population ~ drift())) ``
``````aus.pop.fc<-aus.pop.fit%>%forecast(h=5)
aus.pop.fc%>%autoplot(aus.pop)+ labs(title="Random Walk with drift - Forecast of Australian Population", subtitle = "h=5 years forecast", xlab="Years", ylab="Population")``````
Bricks (aus_production)
``aus_production%>%autoplot(Bricks)``
``## Warning: Removed 20 row(s) containing missing values (geom_path).``
From the plot above, it looks like there is seasonality in the production of bricks in Australia. We can generate the ggseason plot to verify this further. From the plot below, it looks like there is en element of seasonality wherein Q2 and Q3 production is typically higher than Q1 and Q4. Q1 is typically the lowest amount and Q3 is typically the highest amount.
So it would make sense to use the Seasonal Naive model with a lag = year, in this case.
``aus_production%>%gg_season(Bricks)``
``## Warning: Removed 20 row(s) containing missing values (geom_path).``
``````bricks<-aus_production%>%drop_na()%>%select(Quarter, Bricks)
bricks.fit<-bricks%>%model(SNAIVE(Bricks~lag("year")))
bricks.fc<-bricks.fit%>%forecast(h=8)
bricks.fc%>%autoplot(bricks)+ labs(title="Forecast of Australian Brick production based on Seasonal Naive model", subtitle = "h=8 quarters forecast", xlab="Quarters", ylab="Bricks Production")``````
NSW Lambs (aus_livestock)
``````nsw.lambs<-aus_livestock%>%filter(Animal=="Lambs",State=='New South Wales')%>%drop_na()
nsw.lambs%>%autoplot(Count)``````
From the plot above, we can see that the slaughter count seems to be seasonal in nature. To verify, we an plot the seasonal plot.
``nsw.lambs%>%gg_season(Count)``
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You are on page 1of 29
# STRENGTH OF MATERIALS
## ENGR. JOHN LESTER L. MORILLO BS CHE / ADAMSON UNIVERSITY
SIMPLE STRESS
Simple stress is the force per unit area.
P A
Where: P = force A = cross-sectional area = stress
SIMPLE STRESS
Shearing stress ( or Tangential stress ) is a stress caused by forces acting along or parallel to the area resisting the forces. Bearing stress is one which is caused by forces acting perpendicular to the area resisting the forces. Normal stresses, like tensile stress and compressive stress are examples of bearing stress.
SIMPLE STRAIN
Simple strain is the ration of deformation or elongation to the original length.
L
Where: = strain
## = elongation L = original length
HOOKEs LAW
Hookes Law states that within elastic limit the stress is proportional to strain, thus:
P A
## where: E = modulus of elasticity of the material
or Youngs modulus,
## STRESS - STRAIN DIAGRAM
Elastic limit refers to the stress beyond which the material will not return to its original when the load is removed. The permanent deformation caused by excessive stress is called permanent set.
Yield point refers to the point where is an appreciable elongation or yielding of the material even without any corresponding increase in load.
Ultimate stress (or ultimate strength) refers to the highest ordinate in the stress-strain diagram. Rapture strength is sometimes known as the stress at failure.
Working stress is the actual stress of the material when loaded. Allowable stress is the maximum safe stress which the material can carry. Factor of safety is the ratio of the ultimate stress to allowable stress. Shearing strain is the angular change between two perpendicular faces of a differential element. Modulus of rigidity (G) refers to the modulus of elasticity in shear.
## STRESSES IN THIN WALLED CYLINDERS AND SPHERES
FOR CYLINDERS:
A. Tangential Stress:
t
pD 2t
B. Longitudinal Stress: pD l 4t
FOR SPHERES pD t 4t
## STRESSES IN THIN WALLED CYLINDERS AND SPHERES
Where: p = pressure D = inside diameter. t = thickness Another term for tangential stress is circumferential stress, or hoop stress or girth stress.
Note that the longitudunal stress is one-half the value of the tangential stress.
TORSION
Torsion refers to the twisting of solid or hollow circular shafts.
A. Shearing stress where: T = torque applied Tp = radial distance from the center of cross-section J J = polar moment of inertia of the cross-section
B. Maximum shearing stress:
Max.
Tr J
## where: r = radius of the cross-section
TORSION
C. Maximum shearing stress of: 1. Solid shaft 16T Max. d3
2. Hollow shaft
Max.
16TD D4 d 4
## D = outer diameter of the shaft
TORSION
D. Angular deformation, :
TL JG
where: T = torque applied L = length J = polar moment of inertia of cross-section G = modulus of rigidity
E. Transmitted power, P
2 fT
## where: P = power T = torque f = frequency or speed in revolutions per second
HELICAL SPRINGS
A. Maximum shearing stress:
16 PR d 1 d3 4R
16 PR 4m 1 or d 3 4m 4
0.615 m
B. Spring deformation:
64 PR 3 n Gd 4
HELICAL SPRINGS
Where: P = axial load R = mean radius of helical spring
d = diameter of rod/wire of spring m = ratio of the mean diameter of the spring to the mean diameter of the spring rod or wire
2R d
D d
n = number of turns
G = modulus of rigidity
THERMAL STRESS
Thermal stress is the stress on the material caused by the internal forces due to change in temperature. The temperature deformation may be calculated using
L( T )
## Flanged Bolt Coupling
P 1 G1 A1 R1
T
Where: P = load A = cross sectional area of the bolt R = bolt radius
n i
P2 G2 A2 R2
P3 ... G3 A3 R3
Pi Ri ni
n = number of bolts T = Torque capacity of the flanged bolt coupling G = modulus of rigidity
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# Download A Course on Number Theory [Lecture notes] by Peter J. Cameron PDF
By Peter J. Cameron
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Additional info for A Course on Number Theory [Lecture notes]
Example text
INFINITE CONTINUED FRACTIONS Note that is y is approximable to order n, then it is approximable to any smaller order, since if m < n then c/qn ≤ c/qm for positive integer q. We will see that algebraic numbers (roots of polynomials over the integers) are not approximable to arbitrarily high order. Then, by writing down a number which is approximable to order n for every n, we will have exhibited a transcendental number (one which is not a root of a polynomial over Z). 7 (a) Positive rational numbers are approximable to order 1 and no higher.
We say that a rational number p/q is a best approximation to y if |y − p/q| < |y − a/b| for any rational number a/b with b < q. We see that the convergents from c2 on are best approximations to an irrational number. The proof involves quite a bit of work, which we isolate in a preliminary lemma. 6 Let [a0 ; a1 , a2 , . ] be the continued fraction for the irrational number y, and let [a0 ; a1 , . . , an ] = cn = pn /qn be the nth convergent. If gcd(p, q) = 1 and q ≤ qn , then |qy − p| ≥ |qn−1 y − pn−1 |, with equality if and only if p/q = pn−1 /qn−1 .
For which the limit of the sequence of convergents of [a0 ; a1 , . ] is y. Proof We take a0 = y , so that 0 < y − a0 < 1. Then we put y1 = 1/(y − a0 ), so that y1 is an irrational nummber greater than 1, and continue the process: ai = yi , yi+1 = 1 . yi − ai 32 CHAPTER 4. INFINITE CONTINUED FRACTIONS Then a0 , a1 , a2 , . . are positive integers and y0 = y, y1 , y2 are irrational numbers greater than 1, so the process continues infinitely and produces an infinite continued fraction [a0 ; a1 , a2 ].
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# Boolean Algebra Problems
Discussion in 'Homework Help' started by molder22, Oct 7, 2006.
1. ### molder22 Thread Starter New Member
Oct 7, 2006
3
0
I am having trouble reducing the following probems.
F = ABCD + ABCD+ABCD+ABCD
F = ABC+ ABC+ ACD+ ACD
F = (A′ + BD′ + AD) (B + C′) (A+ B′C)
Any help would be appreciated.
2. ### cat New Member
Sep 25, 2006
8
0
You can't simplify that function further. If you've learned karnaugh maps, just draw a karnaugh map for the function and you'll see why.
cat
3. ### mik3ca Active Member
Feb 11, 2007
189
0
Here is my solution:
using: F = AB’CD’ + A’BCD’+AB’C’D+A’BC’D
F = C(AB'D + A'BD') + C'(AB'D + A'BD)
F = C(AB'D + A'BD') + C'D(AB' + A'B)
F = C(AB'D + A'BD') + C'D(A xor B)
Mar 15, 2007
4
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# Non Linear Model
Let us consider an equation $Y = 10 + 5{X^2}$
By putting the values of $X = 0,1,2,3,4$ in this equation, we find the values of $Y$, as given in the table below. The first and second differences are calculated in the table.
$X$ $Y$ First differences $\Delta Y$ Second differences ${\Delta ^2}Y$ $\begin{array}{*{20}{c}} 0 \\ 1 \\ 2 \\ 3 \\ 4 \end{array}$ $\begin{array}{*{20}{c}} {10} \\ {15} \\ {30} \\ {55} \\ {90} \end{array}$ $\begin{array}{*{20}{c}} {15 – 10 = 5} \\ {30 – 15 = 15} \\ {55 – 30 = 25} \\ {90 – 55 = 35} \end{array}$ $\begin{array}{*{20}{c}} {15 – 5 = 10} \\ {25 – 15 = 10} \\ {35 – 25 = 10} \end{array}$
The second differences are exactly constant. The general quadratic equation or non linear model is written as
$Y = a + bX + c{X^2}$ $\left( {c \ne 0} \right)$
This is also called the second degree parabola or second degree curve. The graph of the data is shown in the figure below:
This figure is not a straight line; it is a curve, or we say that the model $Y = 10 + 5{X^2}$ is non-linear. Remember that if in certain observed data the second differences are constant or almost constant, we find the second degree curve close to the observed data. We shall face this type of situation in time series.
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# What Is 26/50 as a Decimal + Solution With Free Steps
The fraction 26/50 as a decimal is equal to 0.52.
The L0ng division method is used to convert fractional numbers into decimal numbers. The numerator is the dividend and the denominator is the divisor in the long division process. The result is taken from the quotient with a remainder left depending on the number.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 26/50.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 26
Divisor = 50
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 26 $\div$ 50
This is when we go through the Long Division solution to our problem. Given is the Long Division process in Figure 1:
Figure 1
## 26/50 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 26 and 50, we can see how 26 is Smaller than 50, and to solve this division, we require that 26 be Bigger than 50.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 26, which after getting multiplied by 10 becomes 260.
We take this 260 and divide it by 50; this can be done as follows:
 260 $\div$ 50 $\approx$ 5
Where:
50 x 5 = 250
This will lead to the generation of a Remainder equal to 260– 250 = 10. Now this means we have to repeat the process by Converting the 10 into 100 and solving for that:
100 $\div$ 50 $\approx$ 2Â
Where:
50 x 2 = 100
This, therefore, produces another Remainder which is equal to 100 – 100 = 0.
Finally, we have a Quotient generated after combining the two pieces of it as 0.52, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
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# COMSOL Blog
## Modeling Electromagnetic Waves and Periodic Structures
##### Walter Frei | January 17, 2014
We often want to model an electromagnetic wave (light, microwaves) incident upon periodic structures, such as diffraction gratings, metamaterials, or frequency selective surfaces. This can be done using the RF or Wave Optics modules from the COMSOL product suite. Both modules provide Floquet periodic boundary conditions and periodic ports and compute the reflected and transmitted diffraction orders as a function of incident angles and wavelength. This blog post introduces the concepts behind this type of analysis and walks through the set-up of such problems.
### The Scenario
First, let’s consider a parallelepided volume of free space representing a periodically repeating unit cell with a plane wave passing through it at an angle, as shown below:
The incident wavevector, \bf{k}, has component magnitudes: k_x = k_0 \sin(\alpha_1) \cos(\alpha_2), k_y = k_0 \sin(\alpha_1) \sin(\alpha_2), and k_z = k_0 \cos(\alpha_1) in the global coordinate system. This problem can be modeled by using Periodic boundary conditions on the sides of the domain and Port boundary conditions at the top and bottom. The most complex part of the problem set-up is defining the direction and polarization of the incoming and outgoing wave.
### Defining the Wave Direction
Although the COMSOL software is flexible enough to allow any definition of base coordinate system, in this posting, we will pick one and use it throughout. The direction of the incident light is defined by two angles, \alpha_1 and \alpha_2; and two vectors, \bf{n}, the outward pointing normal of the modeling space and \bf{a_1}, a vector in the plane of incidence. The convention we choose here is to align \bf{a_1} to the global x-axis and align \bf{n} with the global z-axis. Thus, the angle between the wavevector of the incoming wave and the global z-axis is \alpha_1, the elevation angle of incidence, where -\pi/2 > \alpha_1 > \pi/2 with \alpha_1 = 0, meaning normal incidence. The angle between the incident wavevector and the global x-axis is the azimuthal angle of incidence, \alpha_2, which lies in the range, -\pi/2 > \alpha_2 \geq \pi/2. As a consequence of this definition, positive values of both \alpha_1 and \alpha_2 imply that the wave is traveling in the positive x- and y-direction.
To use the above definition of direction of incidence, we need to specify the \bf{a_1} vector. This is done by picking a Periodic Port Reference Point, which must be one of the corner points of the incident port. The software uses the in-plane edges coming out of this point to define two vectors, \bf{a_1} and \bf{a_2}, such that \bf{a_1 \times a_2 = n}. In the figure below, we can see the four cases of \bf{a_1} and \bf{a_2} that satisfy this condition. Thus, the Periodic Port Reference Point on the incoming side port should be the point at the bottom left of the x-y plane, when looking down the z-axis and the surface. By choosing this point, the \bf{a_1} vector becomes aligned with the global x-axis.
Now that \bf{a_1} and \bf{a_2} have been defined on the incident side due to the choice of the Periodic Port Reference Point, the port on the outgoing side of the modeling domain must also be defined. The normal vector, \bf{n}, points in the opposite direction, hence the choice of the Periodic Port Reference Point must be adjusted. None of the four corner points will give a set of \bf{a_1} and \bf{a_2} that align with the vectors on the incident side, so we must choose one of the four points and adjust our definitions of \alpha_1 and \alpha_2. By choosing a periodic port reference point on the output side that is diametrically opposite the point chosen on the input side and applying a \pi/2 rotation to \alpha_2, the direction of \bf{a_1} is rotated to \bf{a_1'}, which points in the opposite direction of \bf{a_1} on the incident side. As a consequence of this rotation, \alpha_1 and \alpha_2 are switched in sign on the output side of the modeling domain.
Next, consider a modeling domain representing a dielectric half-space with a refractive index contrast between the input and output port sides that causes the wave to change direction, as shown below. From Snell’s law, we know that the angle of refraction is \beta=\arcsin \left( n_A\sin(\alpha_1)/n_B \right). This lets us compute the direction of the wavevector at the output port. Also, note that this relationship holds even if there are additional layers of dielectric sandwiched between the two half-spaces.
In summary, to define the direction of a plane wave traveling through a unit cell, we first need to choose two points, the Periodic Port Reference Points, which are diametrically opposite on the input and output sides. These points define the vectors \bf{a_1} and \bf{a_2}. As a consequence, \alpha_1 and \alpha_2 on the input side can be defined with respect to the global coordinate system. On the output side, the direction angles become: \alpha_{1,out} = -\arcsin \left( n_A\sin(\alpha_1)/n_B \right) and \alpha_{2,out}=-\alpha_2 + \pi/2.
### Defining the Polarization
The incoming plane wave can be in one of two polarizations, with either the electric or the magnetic field parallel to the x-y plane. All other polarizations, such as circular or elliptical, can be constructed from a linear combination of these two. The figure below shows the case of \alpha_2 = 0, with the magnetic field parallel to the x-y plane. For the case of \alpha_2 = 0, the magnetic field amplitude at the input and output ports is (0,1,0) in the global coordinate system. As the beam is rotated such that \alpha_2 \ne 0, the magnetic field amplitude becomes (\sin(\alpha_2), \cos(\alpha_2),0). For the orthogonal polarization, the electric field magnitude at the input can be defined similarly. At the output port, the field components in the x-y plane can be defined in the same way.
So far, we’ve seen how to define the direction and polarization of a plane wave that is propagating through a unit cell around a dielectric interface. You can see an example model of this in the Model Gallery that demonstrates an agreement with the analytically derived Fresnel Equations.
### Defining the Diffraction Orders
Next, let’s examine what happens when we introduce a structure with periodicity into the modeling domain. Consider a plane wave with \alpha_1, \alpha_2 \ne 0 incident upon a periodic structure as shown below. If the wavelength is sufficiently short compared to the grating spacing, one or several diffraction orders can be present. To understand these diffraction orders, we must look at the plane defined by the \bf{n} and \bf{k} vectors as well as in the plane defined by the \bf{n} and \bf{k \times n} vectors.
First, looking normal to the plane defined by \bf{n} and \bf{k}, we see that there can be a transmitted 0th order mode with direction defined by Snell’s law as described above. There is also a 0th order reflected component. There also may be some absorption in the structure, but that is not pictured here. The figure below shows only the 0th order transmitted mode. The spacing, d, is the periodicity in the plane defined by the \bf{n} and \bf{k} vectors.
For short enough wavelengths, there can also be higher-order diffracted modes. These are shown in the figure below, for the m=\pm1 cases.
The condition for the existence of these modes is that:
m\lambda_0 = d(n_B \sin \beta_m - n_A \sin \alpha_1)
for: m=0,\pm 1, \pm 2,
For m=0 , this reduces to Snell’s law, as above. For \beta_{m\ne0}, if the difference in path lengths equals an integer number of wavelengths in vacuum, then there is constructive interference and a beam of order m is diffracted by angle \beta_{m}. Note that there need not be equal numbers of positive and negative m-orders.
Next, we look along the plane defined by the \bf{n} and \bf{k} vectors. That is, we rotate our viewpoint around the z-axis such that the incident wavevector appears to be coming in normally to the surface. The diffraction into this plane are indexed as the n-order beams. Note that the periodic spacing, w, will be different in this plane and that there will always be equal numbers of positive and negative n-orders.
COMSOL will automatically compute these m,n \ne 0 order modes during the set-up of a Periodic Port and define listener ports so that it is possible to evaluate how much energy gets diffracted into each mode.
Last, we must consider that the wave may experience a rotation of its polarization as it gets diffracted. Thus, each diffracted order consists of two orthogonal polarizations, the In-plane vector and Out-of-plane vector components. Looking at the plane defined by \bf{n} and the diffracted wavevector \bf{k_D}, the diffracted field can have two components. The Out-of-plane vector component is the diffracted beam that is polarized out of the plane of diffraction (the plane defined by \bf{n} and \bf{k}), while the In-plane vector component has the orthogonal polarization. Thus, if the In-plane vector component is non-zero for a particular diffraction order, this means that the incoming wave experiences a rotation of polarization as it is diffracted. Similar definitions hold for the n \ne 0 order modes.
Consider a periodic structure on a dielectric substrate. As the incident beam comes in at \alpha_1, \alpha_2 \ne 0 and there are higher diffracted orders, the visualization of all of the diffracted orders can become quite involved. In the figure below, the incoming plane wave direction is shown as a yellow vector. The n=0 diffracted orders are shown as blue arrows for diffraction in the positive z-direction and cyan arrows for diffraction into the negative z-direction. Diffraction into the n \ne 0 order modes are shown as red and magenta for the positive and negative directions. There can be diffraction into each of these directions and the diffracted wave can be polarized either in or out of the plane of diffraction. The plane of diffraction itself is visualized as a circular arc. Note that the plane of diffraction for the n \ne 0 modes is different in the positive and negative z-direction.
All of the ports are automatically set up when defining a periodic structure in 3D. They capture these various diffracted orders and can compute the fields and relative phase in each order. Understanding the meaning and interpretation of these ports is helpful when modeling periodic structures.
#### Post Tags
Application Notes | Technical Content
1. Bryan Adomanis March 21, 2014 at 4:55 pm
Great explanation, Walter! You should lobby to have it put in the documentation!
2. Aneesh M Joseph March 31, 2014 at 7:59 am
subject: problem in assigning diffraction orders for 2d nano pillar grating structure in comsol 3d modeling
Can you help me in this regard?
I would like to find the electric filed intensity of reflection(or reflectance) from gold nano pillars from gold substrate, with the consideration of diffraction orders in case of reflection((1,1 ), (1,0 )reflection grating orders), for the different wave lengths starting from 200 nm to 1000 nm, for a fixed angle of incidence (normal incidence, or 30 degree angle of incidence)
I had made a 2D grating structures in 3D modeling Wizard of Comsol 4.4 version, While I am putting Diffraction orders( m=0, n=0 ; m=-1, n=0 ; m=+1, n=0) with “port reference points” as diagonally opposite end of Unit square cell with periodic boundary conditions on all four sides( by doing copay face). Unfortunately it gives me zero electric filed while i am running., If I run the same model without diffraction order it gives the result but only from one order diffraction (normal reflection) . Since I need to calculate reflection from at least two orders (0 th order and 1 st order) I have to include those diffraction orders to my model
Can you help me in this regard?
can you tell me how to put diffraction orders (with proper port reference points)in 2d grating on 3D model?
Also How can i implement PML layer over the port ( or it is really needed ), since if i add PMLl layer on the top of port, selecting port reference point is declined by default. I follow the pyramidal absorption model for making the infinite array of my 2d Grating nano structures( there a PML layer added over the port)
the plasmonic wire grating model available in Comsol library is just a 2D model , so we can’t able to follow the same thing in here in our model.
I used this blog for making port reference points and putting periodic ports
http://www.comsol.co.in/blogs/modeling-electromagnetic-waves-periodic-structures/
Thank you
Aneesh M Joseph
3. Illia Fedorin November 5, 2014 at 2:13 am
Great work, Walter!!
If you have some example models it would be great to see (.mph as an example).
Illia Fedorin.
4. Walter Frei November 7, 2014 at 2:41 pm
Dear Illia,
Right now the plasmonic wire grating example is the best place to get started: http://www.comsol.com/model/10032
And we are working on some other models as well.
Best,
Walter
5. Jalpa Soni March 31, 2015 at 10:30 am
Hello everyone …
Can anyone help me How to use Floquet boundary condition in case of 3d structure. I want to simulate a model which calculate scattering through array of nanodisc which lie on XY plane .. say i am supposed to incident an electric field through z- axis polarization along either x or y. So in that case what should be my floquet boundary condition??
6. Walter Frei March 31, 2015 at 10:39 am
We do recommend that you study the examples in our model library which demonstrate Floquet periodicity. The following examples are helpful:
http://www.comsol.com/model/10032 (RF Module)
http://www.comsol.com/model/14705 (Wave Optics Module, same example)
This demonstrates the diffraction orders (in 2D)
In 3D:
http://www.comsol.com/model/frequency-selective-surface-periodic-complementary-split-ring-resonators-15711
http://www.comsol.com/model/modeling-of-pyramidal-absorbers-for-an-anechoic-chamber-12129
http://www.comsol.com/model/fresnel-equations-12407
These demonstrate Floquet periodicity, but without the computation of diffraction orders.
7. Kalpana Singh April 1, 2015 at 4:21 am
Hello Walter
From last 6 months, I have been trying to learn COMSOL Multiphysics. I couldn’t find anyone to discuss my doubts while doing the tutorials for Plasmonic wire grating and FSS , split ring resonator. I finally found this blog where periodic structure is discussed as my work is also on periodic structures. Writing my questions here hoping that would be answered by you or anyone who will read this.
While going through the FSS/Split ring resonator I have following doubts:
1) the (1) emw.k0*sin(theta)*cos(phi) (2) emw.k0*sin(theta)*sin(phi) (3) emw.k0*cos(phi) are put in variables. Can we write them in parameters? And I think these components determine the direction of incident of light.
2)Difference between domain backed and PEC slit type. What actually their functions are and how would they affect the simulation?
3)Port orientation is been taken as reverse for both ports 1 and 2. Why? And the directions of the arrow of the port is same whether I click it for forward or reverse. What does it make difference?
4) While defining the properties of port, input quantity is given as magnetic field. Why it is magnetic field, can’t we put electric field as a input quantity ? And also exp(-i*k_x*x)*exp(-i*k_y*y) j A/m is given for the mode, it means wave is traveling in XY plane K has both the component kx and ky and the direction of magnetic field is along y direction. but magnetic field should be perpendicular to direction of propagation.
4) again if wave is propagating in XY plane than how the propagation constant beta = abs(k_z)?
5) Since both plasmonic wire and FSS are periodic structure then why port periodicity is considered in former one not in later one?
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# Wife started working on 3Q - how many exemptions can we claim?
I originally filled in my W-4 with "1" for items A, B and C of the "Personal Allowances Worksheet" since my wife wasn't working and I have only one job and we're filing jointly.
A Enter “1” for yourself if no one else can claim you as a dependent
B Enter “1” if:
• You are single and have only one job; or
• You are married, have only one job, and your spouse does not work; or
• Your wages from a second job or your spouse’s wages (or the total of both) are \$1,500 or less.
C Enter “1” for your spouse. But, you may choose to enter “-0-” if you are married and have either a working spouse or more than one job. (Entering “-0-” may help you avoid having too little tax withheld.)
However, my wife started working around July and she has very variable income. I am trying to work out our estimated taxes due to her new jobs using form 1040 ES. On the "2016 Estimated Tax Worksheet", item 4 (exemptions) it says:
4 Exemptions. Multiply \$4,050 by the number of personal exemptions. Caution: See Worksheet 2-6 in Pub. 505 to figure the amount to enter if line 1 is over: \$155,650
So originally this would be `3*4,050` but now that she has a job I would mark 0 on items B and C on the W-4 and this amount will be only `4,050` (our joint salaries is below \$155,650).
My questions:
• When I file my income tax for 2016, should I consider one exemption or three exemptions?
• If I consider one exemption, and considering my wife made LESS than the exemption amount we're not being granted due to the fact that she worked in 2016 (i.e. items A and B on the W-4: `2*4,050=8,100`), does that mean we have to pay more taxes than what she earned, and in the end she might as well have stayed at home?
• Form 1040-ES is normally only used if you are sending in estimated quarterly tax payments directly to the IRS. Are you sending in quarterly payments? Sep 22, 2016 at 4:49
• Are you or your wife self-employed? Sep 22, 2016 at 4:52
• @Ben no we are not self employed Sep 22, 2016 at 4:54
• And yes I was looking at the form for quarterly payments but likely will have to use the annualized income method instead. My biggest concern however is if I will have to pay a lot more taxes than I thought Sep 22, 2016 at 4:56
• Why would you have to send in quarterly payments if you are receiving a paycheck from your employer with taxes withheld? Sep 22, 2016 at 4:57
I think you are confused about a few things. (Either that, or I am confused by your question. :) Let's try to clear up the confusion first.
Form 1040-ES is used only if you are sending quarterly estimated tax payments directly to the IRS. This is normally done if you are self-employed. Since you are not self-employed and taxes are withheld from your paycheck, you most likely do not need to send in quarterly estimated tax payments, and you won't need Form 1040-ES.
Form W-4 is your control for how much tax is withheld from your paycheck. Generally, you can claim any number of exemptions on this form that you like, and you can change it at anytime. Fewer exemptions will result in more tax withheld; more exemptions will result in less tax withheld.
When you fill out your 2016 tax return next year (Form 1040), you claim the correct number of exemptions for your situation, which does not have to match the number you claimed on your W-4. If you had too much withheld from your paycheck, you get a refund; too little withheld, and you need to make a payment.
On your tax return, you will claim 2 exemptions: one for you, and one for your wife (assuming that you have no kids), and you'll get to deduct \$4050 for each exemption, for a total of \$8100. This would be true whether or not your wife was working, and is true no matter what you put on your W-4.
And, except for a few rare circumstances, you come out ahead when you earn more money. The tax brackets are set up in such a way that when you earn additional income, your tax burden won't go up by more than your income went up.
Now, to address the amount of tax withheld, so you don't end up with a high tax bill at the end of the year:
If you decide that you will have too little withheld from your paychecks and you want more withheld, you can submit a new W-4 to your employer. You said you originally claimed a "1" for A, B, and C on the W-4 Personal Allowances Worksheet, which I take to mean that you claimed 3 exemptions on your W-4. You can reduce this number to have more withheld. If you change it to 2, 1, or even 0, you'll have more tax withheld from your future paychecks.
If you want a finer control than that, you can keep the exemptions the same, but use box 6 on Form W-4. This box allows you to specify an additional amount of tax you want withheld from each check. For example, let's say that you estimate you will be \$2,000 short on your taxes at the end of the year, and you have 6 paychecks left for this year. You could submit a new W-4 to your employer and put \$333 in box 6, which will end up withholding an additional \$2000 by the end of the year. Then at the start of the next year, you may want to submit a new W-4 again, because your situation will have changed.
The IRS has an online withholding calculator designed to help you figure out how to adjust your W-4. Alternatively, you can take a look at Form 1040 and fill it out with estimates for what you think your income and withholdings will be at the end of the year, so you know how much you'll be short. (Note: The Form 1040 that is currently available is for 2015, but it will be close enough for estimating purposes.)
• Those were precisely my concerns :) so due to her being at work I must claim two exemptions (rather than three) and therefore must prepare myself to pay an extra \$4050 when I file my taxes. Sep 22, 2016 at 13:37
• @Phil, I think you are still confused. You can only claim 2 exemptions on your tax return (1040), and that doesn't change whether or not your wife has a job. The number (3) that you claimed on your W-4 only affects how much tax is withheld from your paycheck. Your wife having a job does not negatively affect your situation. Sure, you'll pay more tax, but your wife will make more money than your taxes will go up. As I discussed in the last paragraph, you should estimate your income and your tax for the year using one of the two methods I mentioned, and then adjust your W-4 accordingly. Sep 22, 2016 at 13:45
• @Phil Also, you are misunderstanding what a personal exemption is. You get to deduct \$4050 of household income for each person in your household. For two people, that is \$8100. But that is not \$8100 off of your taxes, it is \$8100 off of your income. Tax savings on that would depend on your tax bracket: in the 25% tax bracket, 2 exemptions results in a tax savings of \$2025. Sep 22, 2016 at 13:52
• @Phil I suggest you read my answer again and the answers on this question, and if there is something in here you don't understand, ask about it. Many of the statements you are making are simply incorrect. Sep 22, 2016 at 13:56
• Thanks @Ben. I simulated our situation using Form 1040 and now things are a lot clearer. I was confused about personal exemptions, and also the difference between declaring the exemptions on 1040 and the numbers I claimed on the W-4. Sep 22, 2016 at 14:33
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Home Metamath Proof ExplorerTheorem List (p. 329 of 394) < Previous Next > Browser slow? Try the Unicode version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
Color key: Metamath Proof Explorer (1-26406) Hilbert Space Explorer (26407-27929) Users' Mathboxes (27930-39313)
Theorem List for Metamath Proof Explorer - 32801-32900 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theorem2llnmat 32801 Two intersecting lines intersect at an atom. (Contributed by NM, 30-Apr-2012.)
Theorem2at0mat0 32802 Special case of 2atmat0 32803 where one atom could be zero. (Contributed by NM, 30-May-2013.)
Theorem2atmat0 32803 The meet of two unequal lines (expressed as joins of atoms) is an atom or zero. (Contributed by NM, 2-Dec-2012.)
Theorem2atm 32804 An atom majorized by two different atom joins (which could be atoms or lines) is equal to their intersection. (Contributed by NM, 30-Jun-2013.)
Theoremps-2c 32805 Variation of projective geometry axiom ps-2 32755. (Contributed by NM, 3-Jul-2012.)
Theoremlplnset 32806* The set of lattice planes in a Hilbert lattice. (Contributed by NM, 16-Jun-2012.)
Theoremislpln 32807* The predicate "is a lattice plane". (Contributed by NM, 16-Jun-2012.)
Theoremislpln4 32808* The predicate "is a lattice plane". (Contributed by NM, 17-Jun-2012.)
Theoremlplni 32809 Condition implying a lattice plane. (Contributed by NM, 20-Jun-2012.)
Theoremislpln3 32810* The predicate "is a lattice plane". (Contributed by NM, 17-Jun-2012.)
Theoremlplnbase 32811 A lattice plane is a lattice element. (Contributed by NM, 17-Jun-2012.)
Theoremislpln5 32812* The predicate "is a lattice plane" in terms of atoms. (Contributed by NM, 24-Jun-2012.)
Theoremislpln2 32813* The predicate "is a lattice plane" in terms of atoms. (Contributed by NM, 25-Jun-2012.)
Theoremlplni2 32814 The join of 3 different atoms is a lattice plane. (Contributed by NM, 4-Jul-2012.)
Theoremlvolex3N 32815* There is an atom outside of a lattice plane i.e. a 3-dimensional lattice volume exists. (Contributed by NM, 28-Jul-2012.) (New usage is discouraged.)
TheoremllnmlplnN 32816 The intersection of a line with a plane not containing it is an atom. (Contributed by NM, 29-Jun-2012.) (New usage is discouraged.)
Theoremlplnle 32817* Any element greater than 0 and not an atom and not a lattice line majorizes a lattice plane. (Contributed by NM, 28-Jun-2012.)
Theoremlplnnle2at 32818 A lattice line (or atom) cannot majorize a lattice plane. (Contributed by NM, 8-Jul-2012.)
Theoremlplnnleat 32819 A lattice plane cannot majorize an atom. (Contributed by NM, 14-Jul-2012.)
Theoremlplnnlelln 32820 A lattice plane is not less than or equal to a lattice line. (Contributed by NM, 14-Jul-2012.)
Theorem2atnelpln 32821 The join of two atoms is not a lattice plane. (Contributed by NM, 16-Jul-2012.)
Theoremlplnneat 32822 No lattice plane is an atom. (Contributed by NM, 15-Jul-2012.)
Theoremlplnnelln 32823 No lattice plane is a lattice line. (Contributed by NM, 19-Jun-2012.)
Theoremlplnn0N 32824 A lattice plane is non-zero. (Contributed by NM, 15-Jul-2012.) (New usage is discouraged.)
Theoremislpln2a 32825 The predicate "is a lattice plane" for join of atoms. (Contributed by NM, 16-Jul-2012.)
Theoremislpln2ah 32826 The predicate "is a lattice plane" for join of atoms. Version of islpln2a 32825 expressed with an abbreviation hypothesis. (Contributed by NM, 30-Jul-2012.)
TheoremlplnriaN 32827 Property of a lattice plane expressed as the join of 3 atoms. (Contributed by NM, 30-Jul-2012.) (New usage is discouraged.)
TheoremlplnribN 32828 Property of a lattice plane expressed as the join of 3 atoms. (Contributed by NM, 30-Jul-2012.) (New usage is discouraged.)
Theoremlplnric 32829 Property of a lattice plane expressed as the join of 3 atoms. (Contributed by NM, 30-Jul-2012.)
Theoremlplnri1 32830 Property of a lattice plane expressed as the join of 3 atoms. (Contributed by NM, 30-Jul-2012.)
Theoremlplnri2N 32831 Property of a lattice plane expressed as the join of 3 atoms. (Contributed by NM, 30-Jul-2012.) (New usage is discouraged.)
Theoremlplnri3N 32832 Property of a lattice plane expressed as the join of 3 atoms. (Contributed by NM, 30-Jul-2012.) (New usage is discouraged.)
TheoremlplnllnneN 32833 Two lattice lines defined by atoms defining a lattice plane are not equal. (Contributed by NM, 9-Oct-2012.) (New usage is discouraged.)
Theoremllncvrlpln2 32834 A lattice line under a lattice plane is covered by it. (Contributed by NM, 24-Jun-2012.)
Theoremllncvrlpln 32835 An element covering a lattice line is a lattice plane and vice-versa. (Contributed by NM, 26-Jun-2012.)
Theorem2lplnmN 32836 If the join of two lattice planes covers one of them, their meet is a lattice line. (Contributed by NM, 30-Jun-2012.) (New usage is discouraged.)
Theorem2llnmj 32837 The meet of two lattice lines is an atom iff their join is a lattice plane. (Contributed by NM, 27-Jun-2012.)
Theorem2atmat 32838 The meet of two intersecting lines (expressed as joins of atoms) is an atom. (Contributed by NM, 21-Nov-2012.)
Theoremlplncmp 32839 If two lattice planes are comparable, they are equal. (Contributed by NM, 24-Jun-2012.)
TheoremlplnexatN 32840* Given a lattice line on a lattice plane, there is an atom whose join with the line equals the plane. (Contributed by NM, 29-Jun-2012.) (New usage is discouraged.)
TheoremlplnexllnN 32841* Given an atom on a lattice plane, there is a lattice line whose join with the atom equals the plane. (Contributed by NM, 26-Jun-2012.) (New usage is discouraged.)
Theoremlplnnlt 32842 Two lattice planes cannot satisfy the less than relation. (Contributed by NM, 7-Jul-2012.)
Theorem2llnjaN 32843 The join of two different lattice lines in a lattice plane equals the plane (version of 2llnjN 32844 in terms of atoms). (Contributed by NM, 5-Jul-2012.) (New usage is discouraged.)
Theorem2llnjN 32844 The join of two different lattice lines in a lattice plane equals the plane. (Contributed by NM, 4-Jul-2012.) (New usage is discouraged.)
Theorem2llnm2N 32845 The meet of two different lattice lines in a lattice plane is an atom. (Contributed by NM, 5-Jul-2012.) (New usage is discouraged.)
Theorem2llnm3N 32846 Two lattice lines in a lattice plane always meet. (Contributed by NM, 5-Jul-2012.) (New usage is discouraged.)
Theorem2llnm4 32847 Two lattice lines that majorize the same atom always meet. (Contributed by NM, 20-Jul-2012.)
Theorem2llnmeqat 32848 An atom equals the intersection of two majorizing lines. (Contributed by NM, 3-Apr-2013.)
Theoremlvolset 32849* The set of 3-dim lattice volumes in a Hilbert lattice. (Contributed by NM, 1-Jul-2012.)
Theoremislvol 32850* The predicate "is a 3-dim lattice volume". (Contributed by NM, 1-Jul-2012.)
Theoremislvol4 32851* The predicate "is a 3-dim lattice volume". (Contributed by NM, 1-Jul-2012.)
Theoremlvoli 32852 Condition implying a 3-dim lattice volume. (Contributed by NM, 1-Jul-2012.)
Theoremislvol3 32853* The predicate "is a 3-dim lattice volume". (Contributed by NM, 1-Jul-2012.)
Theoremlvoli3 32854 Condition implying a 3-dim lattice volume. (Contributed by NM, 2-Aug-2012.)
Theoremlvolbase 32855 A 3-dim lattice volume is a lattice element. (Contributed by NM, 1-Jul-2012.)
Theoremislvol5 32856* The predicate "is a 3-dim lattice volume" in terms of atoms. (Contributed by NM, 1-Jul-2012.)
Theoremislvol2 32857* The predicate "is a 3-dim lattice volume" in terms of atoms. (Contributed by NM, 1-Jul-2012.)
Theoremlvoli2 32858 The join of 4 different atoms is a lattice volume. (Contributed by NM, 8-Jul-2012.)
Theoremlvolnle3at 32859 A lattice plane (or lattice line or atom) cannot majorize a lattice volume. (Contributed by NM, 8-Jul-2012.)
Theoremlvolnleat 32860 An atom cannot majorize a lattice volume. (Contributed by NM, 14-Jul-2012.)
Theoremlvolnlelln 32861 A lattice line cannot majorize a lattice volume. (Contributed by NM, 14-Jul-2012.)
Theoremlvolnlelpln 32862 A lattice plane cannot majorize a lattice volume. (Contributed by NM, 14-Jul-2012.)
Theorem3atnelvolN 32863 The join of 3 atoms is not a lattice volume. (Contributed by NM, 17-Jul-2012.) (New usage is discouraged.)
Theorem2atnelvolN 32864 The join of two atoms is not a lattice volume. (Contributed by NM, 17-Jul-2012.) (New usage is discouraged.)
TheoremlvolneatN 32865 No lattice volume is an atom. (Contributed by NM, 15-Jul-2012.) (New usage is discouraged.)
Theoremlvolnelln 32866 No lattice volume is a lattice line. (Contributed by NM, 15-Jul-2012.)
Theoremlvolnelpln 32867 No lattice volume is a lattice plane. (Contributed by NM, 19-Jun-2012.)
Theoremlvoln0N 32868 A lattice volume is non-zero. (Contributed by NM, 17-Jul-2012.) (New usage is discouraged.)
Theoremislvol2aN 32869 The predicate "is a lattice volume". (Contributed by NM, 16-Jul-2012.) (New usage is discouraged.)
Theorem4atlem0a 32870 Lemma for 4at 32890. (Contributed by NM, 10-Jul-2012.)
Theorem4atlem0ae 32871 Lemma for 4at 32890. (Contributed by NM, 10-Jul-2012.)
Theorem4atlem0be 32872 Lemma for 4at 32890. (Contributed by NM, 10-Jul-2012.)
Theorem4atlem3 32873 Lemma for 4at 32890. Break inequality into 4 cases. (Contributed by NM, 8-Jul-2012.)
Theorem4atlem3a 32874 Lemma for 4at 32890. Break inequality into 3 cases. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem3b 32875 Lemma for 4at 32890. Break inequality into 2 cases. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem4a 32876 Lemma for 4at 32890. Frequently used associative law. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem4b 32877 Lemma for 4at 32890. Frequently used associative law. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem4c 32878 Lemma for 4at 32890. Frequently used associative law. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem4d 32879 Lemma for 4at 32890. Frequently used associative law. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem9 32880 Lemma for 4at 32890. Substitute for . (Contributed by NM, 9-Jul-2012.)
Theorem4atlem10a 32881 Lemma for 4at 32890. Substitute for . (Contributed by NM, 9-Jul-2012.)
Theorem4atlem10b 32882 Lemma for 4at 32890. Substitute for (cont.). (Contributed by NM, 10-Jul-2012.)
Theorem4atlem10 32883 Lemma for 4at 32890. Combine both possible cases. (Contributed by NM, 9-Jul-2012.)
Theorem4atlem11a 32884 Lemma for 4at 32890. Substitute for . (Contributed by NM, 9-Jul-2012.)
Theorem4atlem11b 32885 Lemma for 4at 32890. Substitute for (cont.). (Contributed by NM, 10-Jul-2012.)
Theorem4atlem11 32886 Lemma for 4at 32890. Combine all three possible cases. (Contributed by NM, 10-Jul-2012.)
Theorem4atlem12a 32887 Lemma for 4at 32890. Substitute for . (Contributed by NM, 9-Jul-2012.)
Theorem4atlem12b 32888 Lemma for 4at 32890. Substitute for (cont.). (Contributed by NM, 11-Jul-2012.)
Theorem4atlem12 32889 Lemma for 4at 32890. Combine all four possible cases. (Contributed by NM, 11-Jul-2012.)
Theorem4at 32890 Four atoms determine a lattice volume uniquely. Three-dimensional analog of ps-1 32754 and 3at 32767. (Contributed by NM, 11-Jul-2012.)
Theorem4at2 32891 Four atoms determine a lattice volume uniquely. (Contributed by NM, 11-Jul-2012.)
Theoremlplncvrlvol2 32892 A lattice line under a lattice plane is covered by it. (Contributed by NM, 12-Jul-2012.)
Theoremlplncvrlvol 32893 An element covering a lattice plane is a lattice volume and vice-versa. (Contributed by NM, 15-Jul-2012.)
Theoremlvolcmp 32894 If two lattice planes are comparable, they are equal. (Contributed by NM, 12-Jul-2012.)
TheoremlvolnltN 32895 Two lattice volumes cannot satisfy the less than relation. (Contributed by NM, 12-Jul-2012.) (New usage is discouraged.)
Theorem2lplnja 32896 The join of two different lattice planes in a lattice volume equals the volume (version of 2lplnj 32897 in terms of atoms). (Contributed by NM, 12-Jul-2012.)
Theorem2lplnj 32897 The join of two different lattice planes in a (3-dimensional) lattice volume equals the volume. (Contributed by NM, 12-Jul-2012.)
Theorem2lplnm2N 32898 The meet of two different lattice planes in a lattice volume is a lattice line. (Contributed by NM, 12-Jul-2012.) (New usage is discouraged.)
Theorem2lplnmj 32899 The meet of two lattice planes is a lattice line iff their join is a lattice volume. (Contributed by NM, 13-Jul-2012.)
Theoremdalemkehl 32900 Lemma for dath 33013. Frequently-used utility lemma. (Contributed by NM, 13-Aug-2012.)
Page List
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https://thietkewebsitegiare.vn/how-can-i-implement-a-binary-tree-traversal-to-print-nodes-in-a-specific-order/
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#### How Can I Implement a Binary Tree Traversal to Print Nodes in a Specific Order?
I’m working on a project where I need to implement different types of tree traversals (in-order, pre-order, post-order) for a binary tree. I understand the basic structure of a binary tree, but I’m struggling with how to write the recursive functions for each traversal method.
Here’s the basic structure of my binary tree:
``````class Node:
def __init__(self, key):
self.left = None
self.right = None
self.val = key
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
``````
I’ve looked at some online resources, but I’m still confused about how the recursive calls work and how they ensure the correct order.
New contributor
kavindya ranasinghe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I suggest you to search the web
You can find solution in websites like geeksforgeeks
Here is an example
https://www.geeksforgeeks.org/tree-traversals-inorder-preorder-and-postorder/
New contributor
mu amp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I highly recommend her DEEPTI TALESRA `https://youtu.be/RJhh3Jcc9zw?si=Xw4tfPxIIdz8S921`. She explains so well and goes through examples step by step
Theme wordpress giá rẻ Theme wordpress giá rẻ Thiết kế website
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https://www.mail-archive.com/everything-list@googlegroups.com/msg05041.html
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# RE: Is the universe computable
At 1/21/04, David Barrett-Lennard wrote:
```This allows us to say the probability that an integer is even is 0.5, or
the probability that an integer is a perfect square is 0.
```
But can't you use this same logic to show that the cardinality of the even integers is half that of the cardinality of the total set of integers? Or to show that there are twice as many odd integers as there are integers evenly divisible by four? In other words, how can we talk about probability without implicitly talking about the cardinality of a subset relative to the cardinality of one of its supersets?
I'm not denying that your procedure "works", in the sense of actually generating some number that a sequence of probabilities converges to. The question is, what does this number actually mean? I'm suspicious of the idea that the resulting number actually represents the probability we're looking for. Indeed, what possible sense can it make to say that the probability that an integer is a perfect square is *zero*?
-- Kory
```
```
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https://www.physicsforums.com/threads/average-net-force-and-high-jumpers.754766/page-2
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# Average Net force and high jumpers
Instead of arguing and asking me all these questions show me a equation that I can use and that will work and show me how it works, otherwise if I try to do the calculations using that equation and get something wrong and my answer turns out wrong then what am I supposed to do and you asking me questions ain't helping.
The purpose of the questions is to try to lead you to discover the solution on your own. You are more likely to understand and remember the solution when you get it yourself than when someone just hands it to you.
It really would help if you wrote out the equations. That way you should see more clearly what your mistake is. Just start with
$$E_{initial} = E_{final} + \Delta E$$
and then substitute expressions for the initial and final energy, being careful not to reuse the same variable for different values.
SO I managed to ask my physics teacher and he told me this:
The high jumper’s energy at the top of his path is purely potential (as you have determined).
When he hits the mat he is doing work on the mat to compress it. To do this work he must give up energy. His final energy when he finally stops is also purely potential. Be careful when working out his final potential energy as you have to use both the height of the mat and how much it was compressed to find his final height.
Once you know his final energy, the energy he “lost” has gone into the mat. And since change in energy = work = force x displacement, you can determine the force by using: energy lost by high jumper = force applied to mat x compression of mat. Rearrange for force and Bob’s ya Uncle.
that is all I needed really. I was waiting for his reply and thought I would try out this website too, seems like teachers know what their doing. I only needed someone to explain to me simply what is going on and what equation to use. Thanks to him I got the answer
Last edited:
haruspex
Homework Helper
Gold Member
2020 Award
To do this work he must give up energy. His final energy when he finally stops is also purely potential. Be careful when working out his final potential energy as you have to use both the height of the mat and how much it was compressed to find his final height.
Precisely what I pointed out to you in post #18. You were using the full 2.13m for the lost height, not subtracting the remaining thickness of the mat.
Also, don't forget that the question is essentially wrong - the 'average force' cannot really be calculated this way. I'd be interested in your teacher's answer to that.
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# error using Solve with inequality
1 view (last 30 days)
Shan Chu on 7 Jun 2019
Commented: Shan Chu on 10 Jun 2019
Hi all,
I am facing a strange problem with using Solve with inequality.
Thanks
Torsten on 7 Jun 2019
Why strange ?
Isn't
4^2-4*4+3 > 0 and 0^2-4*0+3 > 0
as well as
e^2-4*e+3 < 0 and 2^2-4*2+3 < 0
?
Shan Chu on 10 Jun 2019
Hi,
I thought that the solve function would give me the intervals where the inequality is valid.
Instead, it gave me 2 specific values inside those intervals. (of course they are correct).
I saw the document about solve in matlab website. Why shoudn't it give me the intervals like in the website. I included the pics below
Thanks
John D'Errico on 10 Jun 2019
Edited: John D'Errico on 10 Jun 2019
You did not follow the example carefully.
syms x
>> S = solve(x^2 - 4*x + 3 < 0,'returnconditions',true)
S =
struct with fields:
x: [2×1 sym]
parameters: [1×2 sym]
conditions: [2×1 sym]
>> S.conditions
ans =
1 < x & x < 3
in(y, 'real')
Note the use of returnconditions. Without that, it gives a solution, but not the interval that you desire. So if I drop that property, we get a solution. It is valid, as you say. But just a solution.
S = solve(x^2 - 4*x + 3 < 0)
S =
exp(1)
2
#### 1 Comment
Shan Chu on 10 Jun 2019
Oh I see it now. Thank John.
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You can use a wide range of wild, cultivated or supermarket greens in this recipe. Consider nettles, beet tops, turnip tops, spinach, or watercress in place of chard. The combination is also up to you so choose the ones you like most.
— Y. Ottolenghi. Plenty More
In this post I’ll describe how I come up with geometry proposals for olympiad-style contests. In particular, I’ll go into detail about how I created the following two problems, which were the first olympiad problems which I got onto a contest. Note that I don’t claim this is the only way to write such problems, it just happens to be the approach I use, and has consistently gotten me reasonably good results.
[USA December TST for 56th IMO] Let ${ABC}$ be a triangle with incenter ${I}$ whose incircle is tangent to ${\overline{BC}}$, ${\overline{CA}}$, ${\overline{AB}}$ at ${D}$, ${E}$, ${F}$, respectively. Denote by ${M}$ the midpoint of ${\overline{BC}}$ and let ${P}$ be a point in the interior of ${\triangle ABC}$ so that ${MD = MP}$ and ${\angle PAB = \angle PAC}$. Let ${Q}$ be a point on the incircle such that ${\angle AQD = 90^{\circ}}$. Prove that either ${\angle PQE = 90^{\circ}}$ or ${\angle PQF = 90^{\circ}}$.
[Taiwan TST Quiz for 56th IMO] In scalene triangle ${ABC}$ with incenter ${I}$, the incircle is tangent to sides ${CA}$ and ${AB}$ at points ${E}$ and ${F}$. The tangents to the circumcircle of ${\triangle AEF}$ at ${E}$ and ${F}$ meet at ${S}$. Lines ${EF}$ and ${BC}$ intersect at ${T}$. Prove that the circle with diameter ${ST}$ is orthogonal to the nine-point circle of ${\triangle BIC}$.
## 1. General Procedure
Here are the main ingredients you’ll need.
• The ability to consistently solve medium to hard olympiad geometry problems. The intuition you have from being a contestant proves valuable when you go about looking for things.
• In particular, a good eye: in an accurate diagram, you should be able to notice if three points look collinear or if four points are concyclic, and so on. Fortunately, this is something you’ll hopefully have just from having done enough olympiad problems.
• Geogebra, or some other software that will let you quickly draw and edit diagrams.
With that in mind, here’s the gist of what you do.
1. Start with a configuration of your choice; something that has a bit of nontrivial structure in it, and add something more to it. For example, you might draw a triangle with its incircle and then add in the excircle tangency point, and the circle centered at ${BC}$ passing through both points (taking advantage of the fact that the two tangency points are equidistant from ${B}$ and ${C}$).
2. Start playing around, adding in points and so on to see if anything interesting happens. You might be guided by some actual geometry constructions: for example, if you know that the starting configuration has a harmonic bundle in it, you might project this bundle to obtain the new points to play with.
3. Keep going with this until you find something unexpected: three points are collinear, four points are cyclic, or so on. Perturb the diagram to make sure your conjecture looks like it’s true in all cases.
4. Figure out why this coincidence happened. This will probably add more points to you figure, since you often need to construct more auxiliary points to prove the conjecture that you have found.
5. Repeat the previous two steps to your satisfaction.
6. Once you are happy with what you have, you have a nontrivial statement and probably several things that are equivalent to it. Pick the one that is most elegant (or hardest), and erase auxiliary points you added that are not needed for the problem statement.
7. Look for other ways to reduce the number of points even further, by finding other equivalent formulations that have fewer points.
Or shorter yet: build up, then tear down.
None of this makes sense written this abstractly, so now let me walk you through the two problems I wrote.
## 2. The December TST Problem
In this narrative, the point names might be a little strange at first, because (to make the story follow-able) I used the point names that ended up in the final problem, rather than ones I initially gave. Please bear with me!
I began by drawing a triangle ${ABC}$ (always a good start\dots) and its incircle, tangent to side ${BC}$ at ${D}$. Then, I added in the excircle touch point ${T}$, and drew in the circle with diameter ${DT}$, which was centered at the midpoint ${M}$. This was a coy way of using the fact that ${MD = MT}$; I wanted to see whether it would give me anything interesting.
So, I now had the following picture.
Now I had two circles intersecting at a single point ${D}$, so I added in ${Q}$, the second intersection. But really, this point ${Q}$ can be thought of another way. If we let ${DS}$ be the diameter of the incircle, then as ${DT}$ is the other diameter, ${Q}$ is actually just the foot of the altitude from ${D}$ to line ${ST}$.
But recall that ${A}$, ${S}$, ${T}$ are collinear! (Again, this is why it’s helpful to be familiar with “standard” contest configurations; you see these kind of things immediately.) So ${Q}$ in fact lies on line ${AT}$.
This was pretty cool, though not yet interesting enough to be a contest problem. So I looked for most things that might be true.
I don’t remember what I tried next; it didn’t do anything interesting. But I do remember the thing I tried after that: I drew in the angle bisector, line ${AI}$. And then, I noticed a big coincidence: the first intersection of ${AI}$ with the circle with diameter ${DT}$ seemed to lie on line ${DE}$! I was initially confused by this; it didn’t seem like it could possibly be true due to symmetry reasons. But in my diagram, it was indeed correct. A moment later, I realized the reason why this was plausible: in fact, the second intersection of line ${AI}$ with the circle was on line ${DF}$.
Now, I could not see quickly at all why this was true. So I started trying to prove it, but initially failed: however, I managed to show (via angle chasing) that
$\displaystyle D, P, E \text{ collinear} \iff \angle PQE = 90^\circ.$
So, at least I had an interesting equivalent statement.
After another half hour of trying to prove my conjecture, I finally realized what was happening. The point ${P}$ was the one attached to a particular lemma: the ${A}$-bisector, ${B}$-midline, and ${C}$ touch-chord are concurrent, and from this ${MD = MP}$ just follows by some similar triangles. So, drawing in the point ${N}$ (the midpoint of ${AB}$), I had the full configuration which gave the answer to my conjecture.
Finally, I had to clean up the mess that I had made. How could I do this? Well, the points ${N}$, ${S}$ could be eliminated easily enough. And we could re-define ${Q}$ to be a point on the incircle such that ${\angle AQD = 90^\circ}$. This actually eliminated the green circle and point ${T}$ altogether, provided we defined ${P}$ by just saying that it was on the angle bisector, and that ${MD = MP}$. (So while the circle was still implicit in the condition ${MD = MP}$, it was no longer explicitly part of the problem.)
Finally, we could even remove the line through ${D}$, ${P}$ and ${E}$; we ask the contestant to prove ${\angle PQE = 90^\circ}$.
And that was it!
## 3. The Taiwan TST Problem
In fact, the starting point of this problem was the same lemma which provided the key to the previous solution: the circle with diameter ${BC}$ intersects the ${B}$ and ${C}$ bisectors on the ${A}$ touch chord. Thus, we had the following diagram.
The main idea I had was to look at the points ${D}$, ${X}$, ${Y}$ in conjunction with each other. Specifically, this was the orthic triangle of ${\triangle BIC}$, a situation which I had remembered from working on Iran TST 2009, Problem 9. So, I decided to see what would happen if I drew in the nine-point circle of ${\triangle BIC}$. Naturally, this induces the midpoint ${M}$ of ${BC}$.
At this point, notice (or recall!) that line ${AM}$ is concurrent with lines ${DI}$ and ${EF}$.
So the nine-point circle of the problem is very tied down to the triangle ${BIC}$. Now, since I was in the mood for something projective, I constructed the point ${T}$, the intersection of lines ${EF}$ and ${BC}$. In fact, what I was trying to do was take perspectivity through ${I}$. From this we actually deduce that ${(T,K;X,Y)}$ is a harmonic bundle.
Now, what could I do with this picture? I played around looking for some coincidences, but none immediately presented themselves. But I was enticed by the point ${T}$, which was somehow related to the cyclic complete quadrilateral ${XYMD}$. So, I went ahead and constructed the pole of ${T}$ to the nine-point circle, letting it hit line ${BC}$ at ${L}$. This was aimed at “completing” the picture of a cyclic quadrilateral and the pole of an intersection of two sides. In particular, ${(T,L;D,M)}$ was harmonic too.
I spent a long time thinking about how I could make this into a problem. I unfortunately don’t remember exactly what things I tried, other than the fact that I was taking a lot of perspectivity. In particular, the “busiest” point in the picture is ${K}$, so it makes sense to try and take perspectives through it. Especially enticing was the harmonic bundle
$\displaystyle \left( \overline{KT}, \overline{KL}; \overline{KD}, \overline{KM} \right) = -1.$
How could I use this to get a nice result?
Finally about half an hour I got the right idea. We could take this bundle and intersect it with the ray ${AI}$! Now, letting ${N}$ be the midpoint ${EF}$, we find that three of the points in the harmonic bundle we obtain are ${A}$, ${I}$, and ${N}$; let ${S}$ be the fourth point, which is the intersection of line ${KL}$ with ${AI}$. Then by hypothesis, we ought to have ${(A,I;N,S) = -1}$. But from this we know exactly what the point ${S}$. Just look at the circumcircle of triangle ${AEF}$: as this has diameter ${AI}$, we see that ${S}$ is the intersection of the tangents at ${E}$ and ${F}$.
Consequently, we know that the point ${S}$, defined very naturally in terms of the original picture, lies on the polar of ${T}$ to the nine-point circle. By simply asking the contestant to prove this, we thus eliminate all the points ${K}$, ${M}$, ${D}$, ${N}$, ${I}$, ${X}$, and ${Y}$ completely from the picture, leaving only the nine-point circle. Finally, instead of directly asking the contestant to show that ${T}$ lies on the polar of ${S}$, one can rephrase the problem as saying “the circle with diameter ${ST}$ is orthogonal to the nine-point circle of ${\triangle BIC}$”, concealing all the work that went into the creation of the problem.
Fantastic.
## One thought on “Writing Olympiad Geometry Problems”
1. Mauricio Rodriguez
Great Post!
You could have also seen the 90 degree angle on the first problem by applying spiral similarity on the tangent case.
Like
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# Flow Rate Technical Group ups the Ante for BP
( – promoted by buhdydharma )
Hot off the Presses …
Scoop: Gulf well may be gushing 25,000 to 30,000 barrels
By Joel Achenbach, washingtonpost.com — June 10, 2010; 4:00 PM ET
Just filed to the web some new figures we obtained this afternoon from the federal government that suggest the Deepwater Horizon well is gushing more oil than previously estimated.
The Flow Rate Technical Group has about five subgroups, and the one that is looking at the video of the leak, the so-called plume team,
has come up with an estimate of 20,000 to 40,000 barrels a day,
with the most likely range being 25,000 to 30,000 barrels.
The team had earlier estimated the flow at 12,000 to 25,000 barrels.
Study: Well most likely spewing more than 1M gallons of oil a day
By Joel Achenbach, Juliet Eilperin and David Fahrenthold
Washington Post Staff Writer
Thursday, June 10, 2010; 4:32 PM
If the team’s estimate is correct, and the flow has been more or less consistent,
approximately 1.3 million to 1.5 million barrels,
or 53.6 million to 64.3 million gallons,
of oil have emerged from the well since the April 20 blowout.
That is roughly five to six times the amount spilled in Alaskan waters in 1989 by the Exxon Valdez.
I knew it! So did a lot of us.
The Flow Rate Technical Group has
come up with an estimate of 20,000 to 40,000 barrels a day,
with the most likely range being 25,000 to 30,000 barrels.
What is that “in Gallons per day”?
since there are 42 Gallons per Barrel
the new Estimate, from Low to High estimates:
20,000 x 42 = 840,000 Gallons per day
40,000 x 42 = 1,680,000 Gallons per day
with a most likely range of:
25,000 x 42 = 1,050,000 Gallons per day
30,000 x 42 = 1,260,000 Gallons per day
How about an everyday benchmark for those Million Gallons?
How many Gallons of gas does a tanker truck hold?
SO How many 9,000-Gallon, “gasoline tanker trucks” are spouting from BP’s Gusher each day?
since there are 42 Gallons per Barrel
the new Estimate, from Low to High estimates:
840,000 / 9,000 = 93 Gas Tanker Trucks per day
1,680,000 / 9,000 = 187 Gas Tanker Trucks per day
with a most likely range of:
1,050,000 / 9,000 = 117 Gas Tanker Trucks per day
1,260,000 / 9,000 = 140 Gas Tanker Trucks per day
Yeah BUT What about the Oil that’s being recovered each day, now?
Coast Guard Toughens Oversight of BP’s Effort
By HENRY FOUNTAIN and CLIFFORD KRAUSS, NYTimes — June 9, 2010
BP said it captured about 15,000 barrels of oil on Tuesday, which it has said is about the daily limit that its collection ship, the Discoverer Enterprise, can process.
So, assuming the reported Recovery Rate 15,000 barrels per day is accurate,
15,000 x 42 = 630,000 Gallons per day
convert to Tankers
630,000 / 9,000 = 70 Gas Tanker Trucks per day being Recovered
SO updating those “Gas Tanker Truck” Benchmarks
the new Estimate, from Low to High estimates:
93 – 70 = 23 Gas Tanker Trucks per day (after Recovery)
187 – 70 = 117 Gas Tanker Trucks per day (after Recovery)
with a most likely range of:
117 – 70 = 47 Gas Tanker Trucks per day (after Recovery)
140 – 70 = 70 Gas Tanker Trucks per day (after Recovery)
[BTW, that last line is actually a 50% Recovery rateWoo Hoo!.]
That’s much better than a few weeks ago — BUT It’s STILL a LOT of Tanker Trucks, spilling their fuel, each and every day.
I sincerely hope they figure out how to capture that remaining flow (47-70 Trucks per day), very soon.
Whatever the ACTUAL Amount of Damage, finally turns out to be, when all is said and done.
Afterall “5 to 6 Exxon Valdez’s” are really WAY TOO much, already.
For more fun with Math and Benchmarks, see my recent posts:
How BIG is 12 Thousand Barrels, anyways?
by jamess — Tue Jun 01, 2010
UPDATED: Some of the Oil is being Captured, and SOME, of it is NOT
by jamess — Sat Jun 05, 2010
the floor is yours …
• jamess on June 10, 2010 at 23:47
Author
those scientists are fighting back.
1. this section:
SO updating those “Gas Tanker Truck” Benchmarks
the new Estimate, from Low to High estimates:
93 – 70 = 23 Gas Tanker Trucks per day (after Recovery)
187 – 70 = 117 Gas Tanker Trucks per day (after Recovery)
with a most likely range of:
117 – 70 = 47 Gas Tanker Trucks per day (after Recovery)
140 – 70 = 70 Gas Tanker Trucks per day (after Recovery)
2. by an ‘estimated’ 20%–maybe much much more.
And, note that the recovered oil they’re now getting from the top hat, is an oil / water mix–they report it as all oil, but it’s not by a long shot.
3. …. more oil sucking, gathering, and processing equipment and ships, they are going to be sucking, gathering, and processing more oil and gas than they previously admitted was coming forth from the well.
This is known as an “oops.”
They can’t brag about sucking 30,000 barrels of oil a day if the thing only officially leaks 15,000.
Prediction: The Flow Rate Team is going to declare they need a bigger Maxi Tanker soon, one with wings.
I believe Translator snarked this up as magickal on BP’s part- making oil out of seawater.
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$\DeclareMathOperator{\arccosh}{arccosh} \DeclareMathOperator*{\argmin}{arg\,min} \DeclareMathOperator{\Exp}{Exp} \newcommand{\geo}[2]{\gamma_{\overset{\frown}{#1,#2}}} \newcommand{\geoS}{\gamma} \newcommand{\geoD}[2]{\gamma_} \newcommand{\geoL}[2]{\gamma(#2; #1)} \newcommand{\gradM}{\nabla_{\M}} \newcommand{\gradMComp}[1]{\nabla_{\M,#1}} \newcommand{\Grid}{\mathcal G} \DeclareMathOperator{\Log}{Log} \newcommand{\M}{\mathcal M} \newcommand{\N}{\mathcal N} \newcommand{\mat}[1]{\mathbf{#1}} \DeclareMathOperator{\prox}{prox} \newcommand{\PT}[3]{\mathrm{PT}_{#1\to#2}#3} \newcommand{\R}{\mathbb R} \newcommand{\SPD}[1]{\mathcal{P}(#1)} \DeclareMathOperator{\Tr}{Tr} \newcommand{\tT}{\mathrm{T}} \newcommand{\vect}[1]{\mathbf{#1}}$
# The proximal map of an absolute difference squared
Proximal map $\prox_{\lambda f}$ for the function $f\colon\M^2\to\R,\quad f(x,y) = d^2_{\M}(x,y)$. Following [1], let $\geo{x}{y}$ denote the geodesic starting in $\geo{x}{y}(0)=x$ and reaching $\geo{x}{y}(1)=y$ at time $1$. Then
If a data item $x$ or $y$ contains a NaN it is set to the minimizer of the distance, i.e. initialized to the other argument of $f$.
### Matlab Documentation
1 2 3 4 5 6 7 8 9 10 11 12 13 % proxAbsoluteDifference(M,f1,f2,lambda) prox of d^2_M(f1,f2) in both args % with parameter lambda on an arbitrary manifold. Values containing NaN % are initialized to the other argument, which is the minimizer. % % INPUT % M : A manifold % f1,f2 : data columns % lambda : proxParameter % % OUTPUT % x1,x2 : resulting columns of the proximal map % --- % Manifold-valued Image Restoration Toolbox 1.0 ~ R. Bergmann, 2014-10-19
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# PCSGU250_ PcLab2000LT_DC Mean?
(PcLab2000LT.exe ( PCSGU250) - View -Waveforms parameters - DC mean)
Now I have Voltage value = fullscale *Databuf(n)-GND_level )/ 255, but it different.
Here is an example:
The full scale on 1V/Div range is 8V.
Here is the beginning of the data file:
[code]TIME STEP:
125 = 1ms
VOLTAGE STEP:
CH1: 32 = 1V
CH2: 32 = 1V
GND 75 86
N CH1 CH2
0 153 87
1 153 87
2 153 87
3 153 87
4 154 87
5 154 87
6 153 87
7 153 87
8 153 87[/code]
To calculate the DC mean for CH1 you can use following formula:
DC mean = fullscale * (Databuf(n)-GND_level )/ 255
Example:
Databuf(0) = 153
GND_level = 75
fullscale = 8V
DC mean = 8V * (153 - 75) / 255 = 2.447V
If you have a varying voltage, then you have to calculate first an average of the Databuf(n) values.
Hello, thank you for fast answer.
But: ( for DC)
Why voltage value from voltmeter (FLUKE 87) is not equals Voltage value = fullscale *Databuf(n)-GND_level )/ 255.
But DC Mean (from PcLab2000LT) is equals values from voltmeter (FLUKE 87) (it’s ok)
I need formula that equals values from voltmeter ( for DC).
May be need add setup ?
Thank you
Can you please post the values you get with the Fluke and the DC value from the “Waveform Parameters” window.
Also please post the beginning of the stored data file.
If your voltage is clean DC, then only about ten data values is enough to post.
Hello,
Source: Fluke 741 ( calibrator) NEW: 0.8 V
Measure: Fluke 287 NEW: 0.8001 v
Measure: PGSU250 DC MEAN: 0.84
Date from PGSU250:
CH2
Sample rate [Hz] 125000 125000
Full scale [mV] 8000 8000
GND level [counts] 127 127
Data (0) 127 155
Data (1) 127 155
Data (2) 128 155
Data (3) 128 154
Data (4) 127 155
Data (5) 127 154
Data (6) 128 154
Data (7) 128 154
Data (8) 127 154
Data (9) 128 155
Data (10) 127 155
DC Mean ( calc) = = 8 ((1556+154*5)/11-127 )/ 255 = 0.864 V
There seems to be strange difference.
The reason to this may be that the PCSGU250 calculates the DC mean over all the data.
Have you tried if the calibration helps: select Options -> Calibrate. ?
Yes, I did calibrated every time before. I have other device (new PCSGU250), I will try with new PCSGU250… I will let you know…
Hello,
I tried more times, but same results…
Can I get formula DC Mean from PcLab2000LT or get part of programme code from PcLab2000LT.exe.
Because formula for DC Mean (average of the Databuf(n) values) not equal DCMean from PcLab2000LT.
I think this is the main difference between your calculation and the oscilloscope’s calculation:
The oscilloscope uses values between 0 and 255 in calculation.
The sum of 4096 samples is divided by 4096 to get the average value.
Both the dividend and divider are integers.
`` average := sum div 4096;``
The result is an integer between 0 and 255.
I think, you are using floating point value to calculate the average of the 4096 samples.
This gives more accurate result.
This is the cause to the difference between the displayed value and the calculated value.
The calculated value is more accurate.
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# Proof: An Equivalence Relation Defines A Partition
So I was reading CS172 textbook chapter 0, and came across the equivalence relations. Equivalence relations, my favorite topic! But I almost forgot most of it since I reviewed my discrete math book briefly over the summer, and started to wonder why is that the equivalence relations defines a partition. So I started looking over my discrete math book. Well, there was no proof about why equivalence relations defines a partition, so here’s my (attempted) proof.
## Proof #
Let A be any finite set (I would let you figure out for infinite set), R be an equivalence relation defined on A; hence R is reflective, symmetric, and transitive.
The statement is trivially true if A is empty because any relation defined on A defines the trivial empty partition of A. Thus, we assume that A is not empty.
We construct non-empty subsets ${A}_{1},{A}_{2},...{A}_{n}$ in the following way:
• Basis: Pick an element $x\in A$, and let ${A}_{1}$ be a subset consisting of all elements $y\in A$ such that $yRx$.
• k-th step: Pick an element z in A such that $z\notin {A}_{1},...z\notin {A}_{k-1}$, and construct a new set ${A}_{k}=r\in A\mid rRz$. If there is no such element, we stop.
### Prove that ${A}_{1},{A}_{2},...{A}_{n}$ is a partition of A #
Now we show that $\mathrm{\forall }i\ne j,{A}_{i}\cap {A}_{j}=\mathrm{\varnothing }$ and ${A}_{1}\cup {A}_{2}\cup ,...\cup {A}_{n}=A$. Since above steps guarantee that every element in A is in at least one of ${A}_{1},...{A}_{n}$, the second property is trivially true. We're left to show that each subset is disjoint.
Let $x\in {A}_{k}$ for some k. (Since we assumed A is not empty, there must be at least one subset). Suppose that $x\in {A}_{j}$ for some j different from k. (This leads to a contradiction). Since $\mathrm{\forall }y\in {A}_{k},yRx$ and $\mathrm{\forall }z\in {A}_{j},zRx$ by the definition of subsets, every element in ${A}_{j}$ is related to every element in ${A}_{k}$ by R for its reflexivity and transitivity. Now if $j>k$, there must be at least one element in ${A}_{j}$ that is not present in sets ${A}_{1},{A}_{2},...{A}_{k}...{A}_{j-1}$ because j-th step picks an element in A that has not been picked in the previous steps. But this is a contradiction. A similar argument holds when $j. Thus, $x\notin {A}_{j}$ for all j different from k. Hence ${A}_{1},...{A}_{n}$ are disjoint subsets of A.
We have shown that given an equivalent relation R defined on A, we can construct disjoint subsets ${A}_{1},{A}_{2},...{A}_{n}$ whose union is A. Q.E.D.
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Chemistry
# Microscope Definition and Parts
A microscope is an instrument that is used to see very small things that cannot be seen with the naked eye. When we look at something through a microscope, it appears larger. The microscopes we use in our schools are light microscopes. These microscopes use light to show the images.
Nowadays scientists use electron microscopes to see very small objects inside the cell. An electron microscope can magnify the image up to 500,000 times. It shows clear images on a television screen. This microscope uses a beam of electrons instead of light.
A light microscope has a base, an arm, a tube, a stage, and two adjustment screws. Two lenses are, lifted on the two ends of the tube. The lens at the end of the tube through which we observe an object is called an Eyepiece.
The lens near the object to be seen is called an objective lens. The light is passed through the object from below, using a mirror. The object to be seen is placed on a glass slide and then on the stage. To focus the object clearly in the microscope, two adjustment screws are used.
We can view an object up to 1500 times bigger than its original size. Most of the cells are too small to be seen without a microscope. The slide is a rectangular piece of glass. The object is placed on it to observe under the microscope.
## Difference between Simple Microscope and Compound Microscope:
### Simple Microscope
A magnifying glass is a convex lens that is used to produce magnified images of small objects. The object is placed nearer the lens than the principal focus such that an upright, virtual, and magnified image is seen clearly at 25cm from the normal eye.
#### Magnifying power
Let θ be the angle subtended at the eye by a small object when it is placed at the near point of the eye. If the object is now moved nearer to the eye, the angle on the eye will increase and becomes θ’, but the eye will not be able to see it clearly.
In order to see the object clearly, we put a convex lens between the object and the eye, so that the lens makes a large virtual image of the object at the near point of the eye. In this way, the object appears magnified. The magnifying power in this case will be:
M =
It can be shown that the magnifying power is given by the relation:
M = = 1 +
Where f is the focal length of the object and d is the near point of the eye. It is clear from this relation that a lens of shorter focal length will have greater magnifying power.
#### Resolving power
The resolving power of an instrument is its ability to distinguish between two closely placed objects or point sources. In order to see objects that are close together, we use an instrument of high resolving power. For example, we use a high resolving power to see tiny organisms and telescope to view distant stars.
### Compound Microscope
A compound microscope has two converging lenses, the object, and the eyepiece, and is used to investigate the structure of small objects. Some features of the compound microscope are:
• It gives greater magnification than a single lens.
• The objective lens has a shorter focal length, f0 ˂ 1 cm.
• The eyepiece has a larger focal length, f0 of a few cms.
### Uses of the Compound Microscope
A compound microscope is used to study bacteria and other micro-objects. It is also used for research in several fields of sciences like Microbiology, Botany, Geology, and Genetics.
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Problem 8.
As part a of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. The block 1 has a weight of 400 N, and the block 2 has a weight of 600 N. The rope and the pulleys are massless and there is no friction.
(a) What is the acceleration of the lighter block?
(b) Suppose that the heavier block is removed, and a downward force of 600 N is provided by someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the remaining block.
Solution
Problem 9.
Part a of the drawing shows a block suspended from the pulley; the tension in the rope is 80 N. Part b shows the same block being pulled up at a constant velocity. What is the tension in the rope in part b ?
Solution
Problem 31.
A circus clown weighs 900 N. The coefficient of static friction between the clown's feet and the ground is 0.4. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?
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The average (arithmetic mean) of 3a + 4 and another number : PS Archive
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The average (arithmetic mean) of 3a + 4 and another number
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The average (arithmetic mean) of 3a + 4 and another number [#permalink]
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The average (arithmetic mean) of 3a + 4 and another number is 2a. What is the average of the other number and a?
2a
a - 4
a - 2
a + 2
a + 4
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shubhangi
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23 Dec 2003, 19:43
A)
I think this problem is from Kaplan
Let X be the other number
[(3a+4)+X]/2 = 2a
(3a+4)+X = 4a
X = a-4
Therefore:
[X+a]/2 = [(a-4)+a]/2 = [2a-4]/2 = a-2
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23 Dec 2003, 19:48
yes its from kaplan
its a-2 right ?? then how is it choice A? it should be C.
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shubhangi
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23 Dec 2003, 19:49
oops, I really meant C)
23 Dec 2003, 19:49
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# Ratio and Financial Statement Analysis Essay
Note: Sample below may appear distorted but all corresponding word document files contain proper formatting
Excerpt from Essay:
The data must be absolutely correct.
3. Effects of Price Level Changes: Price levels changes often make the comparison of figures difficult over a period of time. Changes in price affect the cost of production, sales and also the value of the assets. Therefore, it is necessary to make proper adjustment for price-level changes before any comparison.
4. Quality factors are ignored: Ratio analysis is a technique of quantitative analysis and thus, ignores qualitative factors, which may be important in decision-making. For example, an average collection period may be equal to standard credit period, but some debtors may be in the list of doubtful debts, which is not disclosed by ratio analysis.
5. Effect of window dressing: In order to cover up their bad financial position some companies resort to window dressing. They may record the accounting data according to the convenience to show the financial position of the company in a better way.
6. Costly Techniques: Ratio analysis is a costly technique and can be used by big business houses. Small business units are not able to afford it.
7. Misleading Results: In the absence of absolute data, the result may be misleading. For example, the gross profit of two firms is 25% whereas, the profit earned by one is just U.S.\$5,000 and sales are U.S.\$20,000 and profit earned by the other one is U.S.\$10,000 and sales are U.S.\$40,000. Even the profitability of the two firms is same but the magnitude of their business is quite different.
8. Absence of Standard University accepted terminology: There are no standard ratios, which are universally accepted for comparison purposes. As such, the significance of ratio analysis technique is reduced.
Financial Ratio Analysis and Formulas
Universal Teachers Publications (2010) outlines and explains the meaning, objective and method of calculation of the financial ratios as follows:-
(a.) Gross Profit Ratio:
This ratio shows the relationship between Gross Profit of the concern and its Net Sales; and it can be calculated as follows:-
Gross Profit Ratio = Gross Profit/Net Sales*100 whereby; Gross Profit = Net Sales
Cost of Goods Sold; and the Net Sale = Total Sales -- Sales Return.
Cost of Goods Sold = Opening stock + Net purchases + Direct Expenses -- Closing
Stock.
According to Universal Teachers Publications (2010), the Gross Profit Ratio provides guidelines to the concern whether it is earning sufficient profit to cover administration and marketing expenses; and is able to cover its fixed expenses. Universal Teachers Publications claims that the gross profit ratio of current year is compared to the previous year's ratios of the other concerns; hence, the minor change in the ratio knows about any departure from the standard mark-up and would indicate losses on account of theft, damage, bad stock system, bad sales policies and other reasons. However, Universal Teachers Publications advises that it is desirable that this ratio must be high and steady because any fall in it would put the management in difficulty in the realization of fixed expenses of the business.
(b.) Net Profit Ratio:
According to Universal Teachers Publications (2010), the Net Profit Ratio shows the relationship between Net Profit of the concern and its Net Sales. Universal Teachers Publications decrees that Net Profit Ratio can be calculated in the following manner:-
Net Profit Ratio = Net Profit / Net Sales *100 whereby
Net Profit -- Selling and Distribution Expenses -- Office and Administration Expenses
- Financial Expenses -- Non-Operating Expenses + Non-Operating Income; and Net Sales = Total Sales -- Sales Return
In order to workout overall efficiency of the concern, Universal Teachers Publications (2010) claims that Net Profit Ratio is calculated. This ratio is helpful to determine the operational ability of the concern. When comparing the ratio to the previous years' ratios, the increment shows the efficiency of the concern.
(c.) Operating Profit Ratio
Operating profit means earned profit by the concern from its business operation and not from the other sources. While calculating the net profit of the concern all incomes either they are not part of the business operation like Rent from tenants, interest in Investment etc. are added and all non-operating expenses are deducted. Therefore, while calculating operating profits these all are ignored and the concern comes to know about its business income from its business operations. Meaning, Operating Profit Ratio shows the relationship between Operating Profit and Net Sales; and it can be calculated as follows:
Operating Profit Ration = Operating Profit/Net Sales*100 whereby
Operating Profit = Gross Profit + Non-Operating Expenses -- Non-Operating Incomes
Net Sales = Total Sales -- Sales Returns
Therefore, Operating Profit Ratio indicates the earning capacity of the concern on the basis of its business operations and not from earnings from the order sources; hence it shows whether the businesses is able to stand in the market or not.
(d.) Operating Ratio
Operating Ratio matches the operating cost to the net sales of the business. Operating Cost means cost of goods sold plus Operating Expenses.
Operating Cost = Operating Cost / Net Sales*100 whereby
Operating Cost = Cost of Goods Sold + Operating Expenses
Cost of Goods Sold = Opening Stock + Net Purchases + Direct Expenses -- Closing
Stock
Operating Expenses = Selling and Distribution Expenses, Office and Administration
Expenses, Repair and Maintenance
Operating Ratio is calculated in order to calculate the operating efficiency of the concern. As this ratio indicates about the percentage of operating cost to the net sales, so it is better for a concern to have this ratio in less percentage. The less percent of operating cost to the net sales, so it is better for a concern to have this ratio in less percentage; therefore, less percentage of cost means higher margins to earn profit.
(e.) Return on Investment or Return on Capital Employed
This ration shows the relationship between the profits earned before interest and tax and the capital employed to earn such profit.
Return on Capital Employed = Net Profit before Interest, Tax, and Dividend / Capital Employed*100 whereby Capital Employed (Equity + Preference) + Reserves and Surplus + Long-term Loans -- Fictitious Assets
Or
Capital Employed = Fixed Assets -- Current Liabilities
Returns on capital employed measures the profit, which a firm earns on investing a unit capital. The profit being net result of all operations, the return on capital expresses all efficiencies and inefficiencies of a business. This ratio has a great importance to the shareholders and investors and also management. To shareholders it indicates how much their capital is earning and to the management as to how efficiently it has been working. This ratio influences the market price of the shares. The higher the ratio, the better it is.
(f.)
Return of Equity
Return on Equity is also known as shareholder's investment; and the ratio establishes relationship between profits available to equity shareholders with equity shareholder's funds.
Return on Equity = Net Profit after Interest, Tax and Preference Dividend/Equity Shareholders' Funds * 100 whereby;
Equity Shareholder's Funds = Equity Share Capital + Reserve and Surplus -- Fictitious Assets
Therefore, Return on Equity judges the profitability from the point-of-view of equity shareholders. This ratio has great interest to equity shareholders. The return on equity measures the profitability of Equity funds invested in the firm. The investors favor the company with higher ROE.
(g.) Earning Per Share
Earnings per share are calculated by dividing the net profit (after interest, tax and preference dividend) by the number of equity shares. Therefore,
Earnings per share = Net Profit after Interest, Tax and Preference Dividend/No. Of Equity Shares.
Otherwise, Earnings per share help in determining the market price of the Equity share of the company. I t also helps to know whether the company is able to use it equity share capital effectively which compare to other companies. It also tells about the capacity of the company to pay dividends to its equity shareholders.
CAMEL
According to Credit and Finance Risk Analysis (2010), the CAMELS approach was developed by bank regulators in the United States as a means of measurement of the financial condition of a financial institution. (Uniform Financial Institutions Rating System established by the Federal Financial Institutions Examination Council). Hence, Credit and Finance Risk Analysis (2010) declares that the acronym CAMELS stands for Capital Adequacy; Asset Quality; Management; Earnings (Profitability); Liquidity and Funding Sensitivity to Market Risk (losses arising from changes in market prices). Further, the Credit and Finance Risk Analysis claims that CAMELS analysis requires financial statements (the last three years and interim statements for the most recent 12-month period); cash flow projections; portfolio aging schedules; funding sources; information about the board of directors; operations/staffing; and macroeconomic information.
Recommendations and Conclusion
Personally, I recommend that both investors, financial institutions and department should analyze their financial statements in four stages i.e. (1) preliminary data adjustments; (2) ratio analysis; (3) assessment of accounting quality; and (4) valuation as recommend by De Mello-e-Souza and…[continue]
## Some Sources Used in Document:
"Request-Rejected"
"Financial-Ratio-Analysis-and-Formulas"
"Ratio-Analysis"
## Cite This Essay:
"Ratio And Financial Statement Analysis" (2010, April 04) Retrieved October 22, 2016, from http://www.paperdue.com/essay/ratio-and-financial-statement-analysis-1323
"Ratio And Financial Statement Analysis" 04 April 2010. Web.22 October. 2016. <http://www.paperdue.com/essay/ratio-and-financial-statement-analysis-1323>
"Ratio And Financial Statement Analysis", 04 April 2010, Accessed.22 October. 2016, http://www.paperdue.com/essay/ratio-and-financial-statement-analysis-1323
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# Permutations of Strings in C – Hacker Rank Solution | HackerRank Programming Solutions | HackerRank C Solutions
Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank of Programming Language C . At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight you profile to the recruiters.
In this post, you will find the solution for Permutations of Strings in C-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.
`Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.`
C is one of the most widely used Programming Languages. it is basically used to build Operating System. C was developed by Dennis Ritchie in 1972. Below are some examples of C Programming which might you understanding the basics of C Programming.
#### Objective
Strings are usually ordered in lexicographical order. That means they are ordered by comparing their leftmost different characters. For example abc < abd, because c < d. Also z > yyy because z > y. If one string is an exact prefix of the other it is lexicographically smaller, e.g., gh < ghij.
Given an array of strings sorted in lexicographical order, print all of its permutations in strict lexicographical order. If two permutations look the same, only print one of them. See the ‘note’ below for an example.
Complete the function next_permutation which generates the permutations in the described order.
For example, s = [ab, bc, cd]. The six permutations in correct order are:
```ab bc cd
ab cd bc
bc ab cd
bc cd ab
cd ab bc
cd bc ab
```
Note: There may be two or more of the same string as elements of s.
For example, s = [ab, ab, bc]. Only one instance of a permutation where all elements match should be printed. In other words, if s[0]==s[1] , then print either s[0] s[1] or s[1] s[0] but not both.
A three element array having three discrete elements has six permutations as shown above. In this case, there are three matching pairs of permutations where s[0] = ab and a[1] = ab are switched. We only print the three visibly unique permutations:
```ab ab bc
ab bc ab
bc ab ab
```
### Input Format :
The first line of each test file contains a single integer n, the length of the string array s.
Each of the next n lines contains a string s[i].
### Constraints :
• 2<=n<=9
• a<=s[i]<=10
• s[i] contains only lowercase English letters.
### Output Format :
Print each permutation as a list of space-separated strings on a single line.
```2
ab
cd
```
```ab cd
cd ab
```
```3
a
bc
bc
```
#### Sample Output 1 :
```a bc bc
bc a bc
bc bc a
```
#### Explanation 1
This is similar to the note above. Only three of the six permutations are printed to avoid redundancy in output.
`Permutations of Strings in C – Hacker Rank Solution`
```#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void swap(char **s, int i, int j)
{
char * tmp = s[i];
s[i] = s[j];
s[j] = tmp;
}
void reverse(char **s, int start, int end)
{
while(start<end)
{
swap(s,start++,end--);
}
}
int next_permutation(int n, char **s)
{
for(int i=n-2;i>-1;i--)
{
if(strcmp(s[i+1],s[i])>0)
{
//get min max
for(int j=n-1;j>i;j--)
{
if(strcmp(s[j],s[i])>0)
{
//do swap
swap(s,i,j);
// do reverse
reverse(s,i+1,n-1);
return 1;
}
}
}
}
return 0;
}
int main()
{
char **s;
int n;
scanf("%d", &n);
s = calloc(n, sizeof(char*));
for (int i = 0; i < n; i++)
{
s[i] = calloc(11, sizeof(char));
scanf("%s", s[i]);
}
do
{
for (int i = 0; i < n; i++)
printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
} while (next_permutation(n, s));
for (int i = 0; i < n; i++)
free(s[i]);
free(s);
return 0;
}
```
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# Fenchel duality, entropy, and the log partition function
[Update: As Max points out in the comments, this is really a specialized version of the Donsker-Varadhan formula, also mentioned by Mokshay in a comment here. I think the difficulty with concepts like these is that they are true for deeper reasons than the ones given when you learn them — this is a special case that requires undergraduate probability and calculus, basically.]
One of my collaborators said to me recently that it’s well known that the “negative entropy is the Fenchel dual of the log-partition function.” Now I know what these words meant, but it somehow was not a fact that I had learned elsewhere, and furthermore, a sentence like that is frustratingly terse. If you already know what it means, then it’s a nice shorthand, but for those trying to figure it out, it’s impenetrable jargon. I tried running it past a few people here who are generally knowledgeable but are not graphical model experts, and they too were unfamiliar with it. While this is just a simple thing about conjugate duality, I think it doesn’t really show up in information theory classes unless the instructor talks a bit more about exponential family distributions, maximum entropy distributions, and other related concepts. Bert Huang has a post on Jensen’s inequality as a justification.
We have a distribution in the exponential family:
$p(x; \theta) = \exp( \langle \theta, \phi(x) \rangle - A(\theta) )$
As a side note, I often find that the exponential family is not often covered in systems EE courses. Given how important it is in statistics, I think it should be a bit more of a central concept — I’m definitely going to try and work it in to the detection and estimation course.
For the purposes of this post I’m going to assume $x$ takes values in a discrete alphabet $\mathcal{X}$ (say, n-bit strings). The function $\phi(x)$ is a vector of statistics calculated from $x$, and $\theta$ is a vector of parameters. the function $A(\theta)$ is the log partition function:
$A(\theta) = \log\left( \sum_{x} \exp( \langle \theta, \phi(x) \rangle ) \right)$
Where the partition function is
$Z(\theta) = \sum_{x} \exp( \langle \theta, \phi(x) \rangle )$
The entropy of the distribution is easy to calculate:
$H(p) = \mathbb{E}[ - \log p(x; \theta) ] = A(\theta) - \langle \theta, \mathbb{E}[\phi(x)] \rangle$
The Fenchel dual of a function $f(\theta)$ is the function
$g(\psi) = \sup_{\theta} \left\{ \langle \psi, \theta \rangle - f(\theta) \right\}$.
So what’s the Fenchel dual of the log partition function? We have to take the gradient:
$\nabla_{\theta} \left( \langle \psi, \theta \rangle - A(\theta) \right) = \psi - \frac{1}{Z(\theta)} \sum_{x} \exp( \langle \theta, \phi(x) \rangle ) \phi(x) = \psi - \mathbb{E}[ \phi(x) ]$
So now setting this equal to zero, we see that at the optimum $\theta^*$:
$\langle \psi, \theta^* \rangle = \langle \mathbb{E}[ \phi(x) ], \theta^* \rangle$
And the dual function is:
$g(\psi) = \langle \mathbb{E}[ \phi(x) ], \theta^* \rangle - A(\theta^*) = - H(p)$
The standard approach seems to go the other direction by computing the dual of the negative entropy, but that seems more confusing to me (perhaps inspiring Bert’s post above). Since the log partition function and negative entropy are both convex, it seems easier to exploit the duality to prove it in one direction only.
# Data: what is it good for? (Absolutely Something): the first few weeks
So Waheed Bajwa and I have been teaching this Byrne Seminar on “data science.” At Allerton some people asked me how it was going and what we were covering in the class. These seminars are meant to be more discussion-based. This is a bit tough for us in particular:
• engineering classes are generally NOT discussion-based, neither in the US nor in Pakistan
• it’s been more than a decade since we were undergraduates, let alone 18
• the students in our class are fresh out of high school and also haven’t had discussion-based classes
My one experience in leading discussion was covering for a theater class approximately 10 years ago, but that was junior-level elective as I recall, and the dynamics were quite a bit different. So getting a discussion going and getting all of the students to participate is, on top of being tough in general, particularly challenging for us. What has helped is that a number of the students in the class are pretty engaged with the ideas and material, and we do in the end get to collectively think about the technologies around us and the role that data plays a bit differently.
What I wanted to talk about in this post was what we’ve covered in the first few weeks. If we offer this class again it would be good to revisit some of the decisions we’ve made along the way, as this is as much a learning process for us as it is for them. A Byrne Seminar meets for 10 times during the semester, so that it will end well before finals. We had some overflow from one topic to the next, but roughly speaking the class went in the following order:
• Introduction: what is data?
• Potentials and perils of data science
• The importance of modeling
• Statistical considerations
• Machine learning and algorithms
• Data and society: ethics and privacy
• Data visualizaion
• Project Presentations
I’ll talk a bit more on the blog about this class, what we covered, what readings/videos we ended up choosing, and how it went. I think it would be fun to offer this course again, assuming our evaluations pass muster. But in the meantime, the class is still on, so it’s a bit hard to pass retrospective judgement.
# Allerton 2014: hypercontracting interference channels while noisily feeding back what you detected on the graph
Before the expiration window passes, here are few more short takes from Allerton… for some talks I couldn’t take notes because I didn’t get a seat or I missed half the talk shuttling between sessions.
The Gaussian Channel with Noisy Feedback: Near-Capacity Performance Via Simple Interaction
Assaf Ben-Yishai, Ofer Shayevitz
This was a really nice talk by Ofer on trying to get practical codes for AWGN channels with noisy feedback by using the intuition given by the Schalkwijk-Kailath scheme plus some tricks from using the mod operation. This is reminiscent of lattices (which may be an interesting future direction). The SK scheme has a problem with noise accumulation, which they deal with using these mode operations, and can get to errors around 10^(-6) with around 19 rounds, or blocklength 19 at reasonable SNRs. Blocklength is misleading here since there is feedback every symbol. The other catch is that the feedback link must have much higher SNR than the forward link, but this is true in applications such as sensing, where the receiver may be plugged into the wall, but the transmitter may be on a swallowable medical monitoring device.
Point-To-Point Codes for Interference Channels: A Journey Toward High Performance at Low Complexity
Young-Han Kim
Continuing with my UCSD bias, I also wanted to mention Young-Han’s talk, which was on using COTS (commercial, off-the-shelf) coding schemes on the interference channel (in particular, the 2 user IC). He talked about rate splitting approaches and block Markov schemes. Much of this work is with Lele Wang, who may be graduating soon…
Signal Detection on Graphs
Venkatesh Saligrama
This was a hypothesis testing problem where the observations come from nodes on graph. Under the null, they are Gaussian noise, and under the other hypothesis, there is a connected subgraph with an elevated mean. How should we do detection in this scenario? This is a compound hypothesis testing problem because there are (too) many possible connected subgraphs to consider. He gets around this by looking at a convex program parameterized by a measure of the size/shape of the connected component. This is where my notes get messy though, so you might want to look at the paper if it sounds interesting to you…
Hypercontractivity in Hamming Space
Yury Polyanskiy
I’ve hypercontractivity before, and Yury talked about his paper on ArXiV, which is about functions on the binary hypercube. This talk felt more like a tour of results on hypercontractivity and less like a “here is my new result” talk, which I actually appreciated because I felt it tied together ideas well and made me realize how strange the hypercontractivity parameter of an operator is.
# Detection and Estimation: book recommendations?
It’s confirmed that I will be teaching Detection and Estimation next semester so I figured I would use the blog to conjure up some book recommendations (or even debate, if I can be so hopeful). Some of the contenders:
• Steven M. Kay, Fundamentals of Statistical Signal Processing – Estimation Theory (Vol. 1), Prentice Hall, 1993.
• H. Vincent Poor, An Introduction to Signal Detection and Estimation, 2nd Edition, Springer, 1998.
• Harry L. Van Trees, Detection, Estimation, and Modulation Theory (in 4 parts), Wiley, 2001 (a reprint).
• M.D. Srinath, P.K. Rajasekaran, P. K. and R. Viswanathan, Introduction to Statistical Signal Processing with Applications, Prentice Hall, 1996.
Detection and estimation is a fundamental class for the ECE graduate curriculum, but these “standard” textbooks are around 20 years old, and I can’t help but think there might be more “modern” take on the subject (no I’m not volunteering). Venu Veeravalli‘s class doesn’t use a book, but just has notes. However, I think the students at Rutgers (majority MS students) would benefit from a textbook, at least as a grounding.
Srinath et al. is what my colleague Narayan Mandyam uses. Kay is what I was leaning to before (because it seems to be the most widely used), but Poor’s book is the one I read. I think I am putting up the Van Trees as a joke, mostly. I mean, it’s a great book but I think a bit much for a textbook. So what do the rest of you use? Also, if you are teaching this course next semester, perhaps we can share some ideas. I think the curriculum might be ripe for some shaking up. If not in core material, at least in the kinds of examples we use. For example, I’m certainly going to cover differential privacy as a connection to hypothesis testing.
# MAP and ML in practice on the New Jersey Turnpike
Since I might be teaching detection and estimation next semester, I’ve been thinking a little bit about decision rules during my commute down the New Jersey Turnpike. The following question came to mind:
Suppose you see a car on the Turnpike who is clearly driving dangerously (weaving between cars, going 90+ MPH, tailgating an ambulance, and the like). You have to decide whether the car has New Jersey or New York plates [*]?
This is a hypothesis testing problem. I will assume for simplicity that New York drivers have cars with New York plates and New Jersey drivers have New Jersey plates [**]:
$H_0$: New Jersey driver
$H_1$: New York driver
Let $Y$ be a binary variable indicating whether or not I observe dangerous driving behavior. Based on my entirely subjective experience, I would say the in terms of likelihoods,
$\mathbb{P}(Y = 1 | H_1) > \mathbb{P}(Y = 1 | H_0)$
so the maximum likelihood (ML) rule would suggest that the driver is from New York.
However, if I take into account my (also entirely subjective) priors on the fraction of drivers $P(H_0), P_H(1)$ from New Jersey and New York, respectively, I would have to say
$\mathbb{P}(Y = 1 | H_1) P(H_1) < \mathbb{P}(Y = 1 | H_0) P(H_0)$
so the maximum a-posteriori probability (MAP) rule would suggest that the driver is from New Jersey.
Which is better?
[*] I am assuming North Jersey here, so Pennsylvania plates are negligible.
[**] This may be a questionable modeling assumption given suburban demographics.
# Postdoctoral position at University of Michigan
A postdoctoral position is available at the University of Michigan Electrical Engineering and Computer Science Department for a project related to anomaly detection in networked cyber-physical systems. The successful applicant will have knowledge in one or more of the following topics: convex optimization and relaxations, compressed sensing, distributed optimization, submodularity, control and dynamical systems or system identification. The project will cover both theory and algorithm development and some practical applications in fault and attack detection in transportation and energy networks. The position can start anytime in 2014 or early 2015. This is a one year position, renewable for a second year. Interested candidates should contact Necmiye Ozay at necmiye@umich.edu with a CV and some pointers to representative publications.
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# The property of the material which opposes the creation of magnetic flux in it is called
This question was previously asked in
KVS TGT WET (Work Experience Teacher) 8 Jan 2017 Official Paper
View all KVS TGT Papers >
1. reluctance.
2. conductance.
3. resistivity.
4. permeance.
Option 1 : reluctance.
Free
KVS TGT Mathematics Mini Mock Test
9.2 K Users
70 Questions 70 Marks 70 Mins
## Detailed Solution
The property of the material which opposes the creation of magnetic flux in it is called reluctance.
Key Points
• Magnetic reluctance, or magnetic resistance, is a concept used in the analysis of magnetic circuits.
• It is defined as the ratio of magnetomotive force (MMF) to magnetic flux.
• It represents the opposition to magnetic flux and depends on the geometry and composition of an object.
• Magnetic reluctance is a scalar extensive quantity, akin to electrical resistance. The unit for magnetic reluctance is inverse henry, H−1.
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# Distance between Ust-Kamenogorsk (UKK) and Aktau (SCO)
Flight distance from Ust-Kamenogorsk to Aktau (Oskemen Airport – Aktau Airport) is 1533 miles / 2467 kilometers / 1332 nautical miles. Estimated flight time is 3 hours 24 minutes.
Driving distance from Ust-Kamenogorsk (UKK) to Aktau (SCO) is 2540 miles / 4088 kilometers and travel time by car is about 49 hours 48 minutes.
## Map of flight path and driving directions from Ust-Kamenogorsk to Aktau.
Shortest flight path between Oskemen Airport (UKK) and Aktau Airport (SCO).
## How far is Aktau from Ust-Kamenogorsk?
There are several ways to calculate distances between Ust-Kamenogorsk and Aktau. Here are two common methods:
Vincenty's formula (applied above)
• 1532.785 miles
• 2466.779 kilometers
• 1331.954 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1528.711 miles
• 2460.222 kilometers
• 1328.414 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Oskemen Airport
City: Ust-Kamenogorsk
Country: Kazakhstan
IATA Code: UKK
Coordinates: 50°2′11″N, 82°29′39″E
B Aktau Airport
City: Aktau
Country: Kazakhstan
IATA Code: SCO
ICAO Code: UATE
Coordinates: 43°51′36″N, 51°5′31″E
## Time difference and current local times
The time difference between Ust-Kamenogorsk and Aktau is 1 hour. Aktau is 1 hour behind Ust-Kamenogorsk.
+06
+05
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 182 kg (400 pounds).
## Frequent Flyer Miles Calculator
Ust-Kamenogorsk (UKK) → Aktau (SCO).
Distance:
1533
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1533
Round trip?
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Chalice of the Void 虛空聖杯 常有規則上的問題
Chalice of the Void enters the battlefield with X charge counters on it.
Whenever a player casts a spell with converted mana cost equal to the number of charge counters on Chalice of the Void, counter that spell.
A mana cost of {X}{X} means that you pay twice X. If you want X to be 3, you pay {6} to cast Chalice of the Void.
(2018-03-16)
The number of counters on Chalice of the Void matters only at the time the spell is cast. Changing the number of charge counters on Chalice of the Void after a spell has been cast won’t change whether the ability counters the spell.
(2018-03-16)
If there are zero charge counters on Chalice of the Void, it counters each spell with a converted mana cost of 0. This includes face-down creature spells cast with morph’s alternative cost.
(2018-03-16)
Chalice of the Void has to be on the battlefield at the end of casting a spell for the ability to trigger. If you sacrifice Chalice of the Void as a cost to cast a spell, its ability can’t trigger. However, if it leaves the battlefield once its ability has triggered, that ability will still counter the spell.
(2018-03-16)
http://www.cranialinsertion.com/article/261
Published on 01/04/2010
Q: I control a Chalice of the Void with no counters on it. My opponent casts Engineered Explosives for to destroy my Chalice. Can he really do that?
A: Yes, he can, as long he has a source of actual colorless mana. He chooses X=1 and pays the cost of one generic mana with one colorless mana. Your Chalice won't counter his Explosives because it looks at the chosen value of X and sees a spell with a converted mana cost of 1. The Explosives' sunburst ability counts how many colors of mana were spent to cast it, and there are zero colors in , so the Explosives enter the battlefield with zero counters.
http://www.cranialinsertion.com/article/263
Published on 01/18/2010
Q: So let's say Player A has a Chalice of the Void on 0, and Player B has a Hive Mind. Now player B casts Pact of the Titan with nothing else on the stack on his turn. What exactly happens?
A: When the Pact is cast, both the Chalice's and Hive Mind's abilities trigger. The triggers go on the stack in Active Player, Non-Active Player order, so first the Hive Mind trigger will go on the stack, then Player A's Chalice trigger will go on the stack. Chalice of the Void will counter it, and since the game needs to know something about the Pact and it's not there anymore, it'll use last-known information to create the copy. Player A will need to find some way to make by his next upkeep!
http://www.cranialinsertion.com/article/286
Published on 06/28/2010
Q: My opponent controls a Chalice of the Void with two counters and I cast a kicked Breath of Darigaaz. Will the Chalice counter my spell?
A: Sadly, yes. A card's converted mana cost depends entirely on the mana cost that's printed in the top right corner of the card. Regardless of whether you pay the kicker, the converted mana cost of Breath of Darigaaz is 2, so you unfortunately wasted a card and on this play.
http://www.cranialinsertion.com/article/362
Published on 01/23/2012
Q: If there's a Chalice of the Void on 1, and I cast Flusterstorm, do I still get copies? Are the copies countered?
A: You will get copies, and they won't get countered! When you cast Flusterstorm, both Chalice of the Void and Flusterstorm will trigger. Chalice will counter the Flusterstorm, but no matter whose turn it is, Flusterstorm will still create copies even if it's countered before its storm trigger resolves. Chalice of the Void doesn't say you can't cast spells with converted mana cost equal to the number of counters on them, it just says that they're countered when they are cast. And the copies aren't cast, so they won't trigger Chalice again.
http://www.cranialinsertion.com/article/936
Published on 06/24/2013
Q: My opponent has locked me out of my cheap spells with a Chalice of the Void at 1. If there's a Trinisphere on the board, can I Brainstorm for 3 and not get it countered?
A: Sadly, no. Chalice of the Void looks at Brainstorm's converted mana cost, which is derived only by the mana cost in the top right corner. Trinisphere adds two mana to the total cost to cast Brainstorm, but Brainstorm's converted mana cost is still 1.
http://www.cranialinsertion.com/article/1236
Published on 03/17/2014
Q: Can Wear // Tear destroy a Chalice of the Void?
A: It depends on how many counters are on the Chalice — if there's 1 counter, then yes. If there are 2 counters, then no. And if there are 3 counters, then maybe!
On the stack, a split card only has the characteristics of the part of it that actually got cast. So if you cast Wear by itself, you're casting a spell with a converted mana cost of 2; it will successfully destroy a Chalice with either 1 or 3 counters, but be countered by a Chalice with 2 counters. If you fuse, casting both Wear and Tear, then you're casting a spell with converted mana cost 3; Chalice will only counter it if the Chalice has 3 counters (though keep in mind you'll need an enchantment to target in this case, as well as an artifact).
http://www.cranialinsertion.com/article/1306
Published on 06/02/2014
Q: My opponent has a Chalice of the Void with one counter; is there any way to destroy it with Shattering Spree?
A: There is! All you have to do is use replicate — although the original Shattering Spree was cast, and will be countered by the Chalice, copies created by replicate aren't cast (they just get created directly on the stack) and so the Chalice won't counter them.
http://www.cranialinsertion.com/article/1348
Published on 07/21/2014
Q: With Chief Engineer, can convoke pay for Chalice of the Void?
A: Yes, it can. When casting a spell with X in its cost, you first decide what value you want X to have, and then pay whatever cost you'd need for an X of that value. So if you wanted to play Chalice with one counter, X would be 1, which makes the mana cost 2, which you can pay by tapping two creatures.
http://www.cranialinsertion.com/article/1868
Published on 02/22/2016
Q: If my opponent controls a Chalice of the Void with one counter on it, is it legal for me to cast a one-mana spell and see if they miss the trigger, or do I have to point it out?
A: In tournament play, only the controller of a triggered ability is responsible for remembering it and ensuring its effect happens. So if your opponents miss their own Chalice triggers, you're within your rights to let them do so, and have no obligation to remind them about it when you cast spells the Chalice would counter.
Q: 如果我對手操控有一個指示物的虛空聖杯，我是否可以合法是放一個一費咒語，然後看我對手是否遺漏這個觸發？還是我必須要指出來？
A: 在競技比賽中，只有觸發的操控者有責任要記住它，並確保它發生。所以如果你的對手自己遺漏了聖杯的觸發，你有權利讓他們如此做，但是你沒有義務提醒他們你釋放你一個會被聖杯反擊的咒語。
http://www.cranialinsertion.com/article/1890
Published on 03/14/2016
Q: My opponent controls a Chalice of the Void with one counter on it, and I have an Isochron Scepter with a Lightning Bolt imprinted on it. If I cast the Bolt off of the Scepter, does my opponent's Chalice counter it?
A: Yes. Isochron Scepter tells you to copy the imprinted card and then to cast the copy. Even though the copy is not represented by a card, you're casting a spell, and that spell's converted mana cost is 1, so the Chalice's ability triggers and counters your Lightning Bolt.
http://www.cranialinsertion.com/article/2057
Published on 09/05/2016
Q: I know in a tournament if my opponent has a Chalice of the Void, I can try to sneak a spell through and hope they forget about the trigger; that works with Sanctum Prelate, too, right?
A: No. Chalice of the Void has a triggered ability which counters spells; Sanctum Prelate's ability isn't triggered, and just flat-out forbids casting spells of the chosen CMC. So there's no way to make someone "miss" it by casting a spell of that CMC, since casting the spell is forbidden
Q: 我知道在一場比賽中，如果對手施放了虛空聖杯，我可以試圖施放一個咒語，並盼望對手忘卻了觸發；這對聖所教長是否同樣適用？
A: 不行。虛空聖杯有一個反擊咒語的觸發式異能；聖所教長的異能不是觸發式的，它就是放在那阻止對手施放CMC為所選數字的咒語。所以沒有方法能讓某人"遺漏"它而施放該CMC的咒語，因為施放這些咒語是被禁止的。
http://www.cranialinsertion.com/article/2118
Published on 11/07/2016
Q: I have a Chalice of the Void with one counter and a Cavern of Souls in play, naming Merfolk. I know I'm responsible for remembering my own triggers, especially detrimental ones, but if I cast Cursecatcher through the Cavern, do I really need to indicate the Chalice's trigger when it doesn't do anything?
A: Not at all. Triggered abilities need to be acknowledged the first time they actually matter in some way, whether that's because you need to make a choice or because it has an impact on the visible game state. But a Chalice of the Void triggering and trying to counter an uncounterable spell never matters because it can't accomplish anything, so there's no need to demonstrate awareness of it at all, no matter whether it's beneficial or not.
Q: 我有帶一個指示物的虛空聖杯，還有靈魂洞窟在戰場，選擇了人魚。我知道我有責任記住我自己的觸發，尤其是有害的，但是如果我通過洞窟施放捕咒師，儘管聖杯的觸發不做任何事，但我是否依然要指出它？
A: 不用。觸發式異能只有在會發生影響時，要在第一時間生命，不論是它需要你做出選擇還是對可見遊戲狀態產生影響。但是虛空聖杯觸發並試圖釩基你不能被反擊的咒語對遊戲沒有影響，因為它不能完成任何事，所以你不用證明你意識到它，不論它是否對你有利。
http://www.cranialinsertion.com/article/2793
Published on 09/17/2018
Q: Andy controls a Chalice of the Void with one charge counter. Nancy is at one life. Suppose Andy casts a Gutshot targeting Nancy and saying nothing, fully intending to acknowledge Chalice's trigger once it would go to resolve but first wanting to see how Nancy would respond. Is this cheating under current policy? What if Nancy concedes?
A: Andy is allowed to do this.
He doesn't have to acknowledge the Chalice trigger until it would resolve, but he's not allowed to forget the trigger.
If Nancy concedes in response, that's her decision. Andy is in no way misrepresenting the game state or otherwise cheating and Nancy would be just fine if she just passed priority and forced Andy to counter his own Gut Shot.
This would be fine at any REL, though I would probably talk to Andy about sporting behaviour at Regular if Nancy complained. At Competitive REL, there's nothing to be said since Andy was playing by the rules and policies governing tournament Magic.
Q: Andy操控一個具有一個充電指示物的虛空聖杯 。Nancy現在只有一點生命。假如Andy對Nancy施放打擊內臟，並且什麼都沒說，並且打算看Nancy是否回應之後再在虛空聖杯的效果要結算時宣告。在當前的政策下，這是否算作弊？如果Nancy投降了呢？
A: Andy可以這麼做。
http://www.cranialinsertion.com/article/3019
Published on 05/27/2019
Q: I'm in a Regular REL tournament, and I control a Chalice of the Void with one counter on it. I cast Goryo's Vengeance targeting a legendary creature card in my graveyard, and my opponent pays 2 life to cast Surgical Extraction targeting that card. While we're in the middle of resolving Surgical Extraction, I realize that my Chalice should have countered the spell. What happens now? Did my opponent cheat?
A: Well, there's a lot to unpack here. First off, if something like this happens to you in a tournament, you should call a judge, and then the judge will decide what needs to be done. The document that guides the judge's ruling in this case is Judging at Regular REL, and this situation falls under "A player forgets a triggered ability." Since you didn't acknowledge the trigger at the time it had a visible in-game effect, you missed it. If the trigger were optional, the judge would simply assume you chose not to perform it, but Chalice's trigger is mandatory. The only possible fix is to put the trigger on the stack now unless doing so would be too disruptive to the game, but it doesn't really matter here since Surgical Extraction is already resolving. The Chalice trigger would resolve after Surgical Extraction has finished, so putting the trigger on the stack now is pointless. Backing up the game to before Surgical Extraction started to resolve is not an option, so Surgical Extraction will finish resolving no matter what the judge decides.
This leaves the question whether your opponent was cheating, and the answer to that is no. While what your opponent did was not very sporting, it wasn't illegal, so it can't be cheating. Your opponent is not responsible for your triggered abilities, and they are under no obligation to point out your missed triggers. I understand it feels bad that your opponent was able to sneak a spell past your Chalice, but if it's any consolation, players who rely on their opponent's mistakes to win tend not to make it very far in tournaments.
Q：我在一場一般級別的REL比賽中，我操控一個有一個指示物的虛空聖杯。我施放怨靈復仇指定了我墳場裡的一個傳奇生物，我對手支付2點生命施放了手術摘除指定了那張牌。在我們結算手術摘除的時候，我意識到我的聖杯應該反擊這個咒語。之後發生什麼？我的對手是否作弊了？
A：嗯，這裡有很多門道。首先，如果任何這樣的事情在你參加的比賽中發生了，你應該呼叫裁判，然後裁判會判斷需要做什麼。指導裁判判罰的檔叫一般級別REL判罰指南，這個情況是“一位元牌手忘記了觸發”。因為你在觸發產生對遊戲可視的影響時沒有聲明它，所以你遺漏了觸發。如果這個觸發是可選的，裁判會簡單的假設你選擇不使用它，但是聖杯的觸發是強制的。唯一可能修正就是，除非這個觸發會擾亂遊戲，否則將它立刻放入堆疊，因為手術摘除已經結算了，所以它無關緊要。聖杯的觸發會在手術摘除結算完後結算，所以將觸發放入堆疊無關緊要。將遊戲倒回到手術摘除開始結算前並不可以，所以手術摘除無論裁判如何決定都會順利結算。
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https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-1-introduction-to-algebraic-expressions-1-7-multiplication-and-division-of-real-numbers-1-7-exercise-set-page-59/123
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$-\dfrac{14}{15}$
Using the $LCD= 15$, the given expression, $\dfrac{-4}{15}+\dfrac{2}{-3}$, simplifies to \begin{array}{l}\require{cancel} \dfrac{1(-4)+(-5)(2)}{15} \\\\= \dfrac{-4-10}{15} \\\\= \dfrac{-14}{15} \\\\= -\dfrac{14}{15} .\end{array}
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https://www.physicsforums.com/threads/proving-two-sets-are-equivalent.767257/
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# Proving Two Sets Are Equivalent
1. Aug 23, 2014
### Bashyboy
To prove that two sets are in fact the same, do I actually have to prove that the two are subsets of each other; or could I prove that they are equivalent by some other means, such as invoking the definitions of the sets?
For instance, the I am trying to show that the binary set operator $+$ is commutative; that is,
$A + B = B + A$,
where
$A + B = (A-B) \cup (B-A)$.
By using the commutative property of the $\cup$ operator, I was able to prove the equality:
$A + B = (A-B) \cup (B-A) = (B-A) \cup (A-B) = B+A$
Is this a valid proof?
2. Aug 23, 2014
### micromass
Staff Emeritus
Yes, as long as you've already proven that union is commutative.
3. Aug 23, 2014
### Bashyboy
°Yes, I did prove that the union is commutative.
I have another question; it comes from the same problem. I am asked to determine whether $A + \emptyset = A$ is a true statement. Here is my proof:
$A + \emptyset = (A - \emptyset) \cup (\emptyset - A)$
Before we proceed, let us determine the nature of $(A - \emptyset)$ and $(\emptyset - A)$.
$(A - \emptyset) = \{ x~| x \in A \wedge x \notin \emptyset \}$. The statement $x \notin x \emptyset$ is always true by definition. Therefore,
$(A- \emptyset) = \{x~| x \in A \wedge T \} = \{x~| x \in A\} = A$.
Here is the portion of my proof that I am unsure of:
$\emptyset - A = \{x~| \underbrace{x \in \emptyset} \wedge x \notin A \}$. The underlined portion is always a false statement, as the empty set never houses any elements. As such,
$\emptyset - A = \{x~| F \wedge \underbrace{x \notin A}\}$. The truth value of the underlined portion is irrelevant to the truth value of the entire statement. Thus,
$\emptyset - A = \{x~| F \}$...
I would interpret this as being the empty set, but I do not have any basis for such an inference. How might I justly proceed from this last step in my proof?
EDIT:
What if I wrote $\emptyset - A = \{x~|\forall xp(x) = F\}$, where $p(x)$ is the condition that must be satisfied in order for the element $x$ to be a member. What $p(x)=F$ states is, that every $x$ makes the statement false, meaning that it can contain no elements.
Would this be sufficient reasoning?
Last edited: Aug 23, 2014
4. Aug 24, 2014
### HallsofIvy
Staff Emeritus
B- A is, by definition, "All members of B that are not in A". If B is empty, B- A is the empty set no matter what A is.
5. Aug 24, 2014
### vela
Staff Emeritus
I think you're fine simply saying that $\emptyset-A = \emptyset$, but if you're still worried, you can show that $\emptyset \subset \emptyset-A$ and $\emptyset-A \subset \emptyset$.
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## HyperCard & Tessellations
#### Suzanne Alejandre
HyperCard Tips || Tessellations with Rotations || ...with Reflections || Where's the Math?
#### Step One
Using the rectangle tool, draw a black square. If the shift key is
held down as the "rectangle" is drawn, it will make a square.
#### Step Two
Using the lasso, start at the top corner, wiggle around as you are going down, and exit the lower corner of the square. A section of the square will be selected.
#### Step Three
Hold the shift key and slide the chunk over to the right. Position the chunk on the right side, leaving no overlap and no gaps. [Also possible would be to start on the right side, select a chunk, and slide it to the left.]
#### Step Four
Then repeat the process but this time go from top to bottom
or bottom to top.
#### Step Five
Use the pencil or paint brush (with the white pattern selected) to decorate the figure. Do not add anything to the outer edges of the silhouette.
#### Step Six
Use the lasso and position the figure up higher on the card. Hold option, slide the copied figure to the side. In HyperCard this is a faster method to copy/paste if work is being done on only one card.
#### Step Seven
With one of the black figures still selected, go to "paint" in the menu and select "trace edges". The result is a white version of the black figure.
#### Step Eight
Using the lasso, hold "option" down and "copy" by sliding one black figure to the side. Lasso the white figure, hold "option" down, slide a white figure off and fit it into position up against the black figure. Hint: If it does not fit perfectly, review Steps 1 - 8 and try again. Continue this procedure until you have created a tessellation.
#### HyperCard Tessellation
Here are some inspirational pictures from the World of Escher.
Here is a Tessellation Contest your students can enter.
## Tessellation Survey
Do you like to tessellate?
How do you like to tessellate?
To submit your choices, press this button:
To reset the form, press this button:
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https://flatearthsoc.com/200-proof-earth-is-flat-30stm-edge-of-the-earth-lyrics.html
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```1.) The aeronaut can see for himself that Earth is a Plane. The appearance presented to him, even at the highest elevation he has ever attained, is that of a concave surface – this being exactly what is to be expected of a surface that is truly level, since it is the nature of level surfaces to appear to rise to a level with the eye of the observer. This is ocular demonstration and proof that Earth is not a globe.
```
Regarding the "traveling at the same speed" idea. When in an aircraft, two people can toss a ball to each other. From the POV of those Inside, the ball is moving at normal ball tossing speed. From the POV of those outside, the ball is being gently tossed to over 500kph and the catcher catchers it without problem. If this is the "everything on the planet moving at the same speed" idea in action. Which does make it possible to travel in either direction.
Besides the above difficulties or incompatibilities, many cases are on record of the sun and moon being eclipsed when both were above the horizon. The sun, the earth, and the moon, not in a straight line, but the earth below the sun and moon–out of the reach or direction of both–and yet a lunar eclipse has occurred! Is it possible that a “shadow” of the earth could be thrown upon the moon, when sun, earth, and moon, were not in the same line? The difficulty has been met by assuming the influence of refraction, as in the following quotations:–
The first photograph (Figure 4) is of a cargo ship bearing the name of the company on its hull. The company is the NYK line, a major Japanese shipping company. Notice that the bottoms of the letters are not visible. The letters on the hulls of cargo ships do not extend to the water line, even when fully loaded, so clearly the bottom of the hull is not visible. This is consistent with what we would expect on a spherical earth, but not on a flat earth. Notice the white bridge castle to the left. The shipping containers are multicolored, and they are stacked at least seven high above the hull directly in front of the bridge castle. Below the visible tiers of the multi-colored containers there is a level of what appears to be gray containers. It is not clear why the containers in this layer are the same color. Finally, notice that the image is a bit blurry. This is because of turbulence in the air between the ship and shore. With increasing distance, the turbulence will get worse, and the images will get blurrier.
17) “Olber’s Paradox” states that if there were billions of stars which are suns the night sky would be filled completely with light. As Edgar Allen Poe said, “Were the succession of stars endless, then the background of the sky would present us a uniform luminosity, since there could exist absolutely no point, in all that background, at which would not exist a star.” In fact Olber’s “Paradox” is no more a paradox than George Airy’s experiment was a “failure.” Both are actually excellent refutations of the heliocentric spinning ball model.
In the Libyan desert there is a mysterious glass, the purest natural silica glass ever found on earth. No other substance like it exists. Since natural glass, such as Libyan Desert Glass, can be formed only by lightning strike, volcanic activity, or meteorites striking the earth, it is a mystery to scientists how the glass got there. But none of these events make any sense since the glass dates thousands of years old and is spread out all over the desert, plus there is no sign of a crater and no volcanos.
The surveyor's plans in relation to the laying of the first Atlantic Telegraph cable, show that in 1665 miles - from Valentia, Ireland, to St . John's, Newfoundland - the surface of the Atlantic Ocean is a LEVEL surface - not the astronomers' "level," either! The authoritative drawings, published at the time, are a standing evidence of the fact, and form a practical proof that Earth is not a globe.
```87.) The theory of a rotating and revolving earth demands at theory to keep the water on its surface; but, as the. theory which is given for this purpose is as much opposed to all human experience as the one which it is intended to uphold, it is an illustration of the miserable makeshifts to which astronomers are compelled to resort, and affords, a proof that the Earth is not a globe.
```
## where P and T are the pressure and temperature of air, and P0 and T0 are the standard values of one atmosphere of pressure and 300K. Since with mirages there is no appreciable height difference, pressure differences are negligible, and so the temperature difference dominates differences in the index of refraction. When light travels from one medium to another, the path of the light is refracted, or bent. This is what causes the “bent stick” appearance of a long object partially inserted in water, such as a pole placed into the water of a pool. This behavior is described by Snell’s law:
98) NASA and modern astronomy say Polaris, the North Pole star, is somewhere between 323-434 light years, or about 2 quadrillion miles, away from us! Firstly, note that is between 1,938,000,000,000,000 - 2,604,000,000,000,000 miles making a difference of 666,000,000,000,000 (over six hundred trillion) miles! If modern astronomy cannot even agree on the distance to stars within hundreds of trillions of miles, perhaps their “science” is flawed and their theory needs re-examining. However, even granting them their obscurely distant stars, it is impossible for heliocentrists to explain how Polaris manages to always remain perfectly aligned straight above the North Pole throughout Earth’s various alleged tilting, wobbling, rotating and revolving motions.
### It is often said that the predictions of eclipses prove astronomers to be right in their theories. But it is not seen that this proves too much. It is well known that Ptolemy predicted eclipses for six-hundred years, on the basis of a plane Earth, with as much accuracy as they are predicted by modern observers. If, then, the predictions prove the truth of the particular theories current at the time, they just as well prove one side of the question as the other, and enable us to lay claim to a proof that the Earth is not a globe.
Great material, THANK YOU SO MUCH!!! I am very happy to be able to arrive at this information. By the way, they lied not only about earth shape and universe, but about physics, chemistry and even in math, inventing nonexistent values and rules. Example: they tell you 2x0=0. This is nonsense, because 2 has to denote something real, the actual values. Therefore 2x0=2, not 0, as they lie to us. Prove? Take 2 airplanes and multiply them on 0. How many airplanes will you get,-zero?...)) No, you will still have 2 airplanes! But 2 as abstract (nonexistent) number representing nothing, can not give any result but nothing.
When the sun rises in the morning its light is just coming into view. The sun's light then follows it as it journeys away from you, appearing to descend below the horizon. In reality it is not "going down" but moving away from you and going beyond the line of convergence and your eyesight. It takes it's light with it. You can clearly see this in time lapse videos of the sun moving away, causing a sunset.
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https://mathematica.stackexchange.com/questions/286217/find-roman-numerals-up-to-100-that-do-not-contain-i
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# Find Roman numerals up to 100 that do not contain “I"
The question is from Elementary Introduction to the Wolfram Language, section 28 Tests and Conditionals: Find Roman numerals up to 100 that do not contain “I”.
What am I doing wrong that the following code is not returning the desired output?
Select[RomanNumeral[Range[100]], MemberQ[#, "I"]&]
I can't quite figure out where I am going wrong.
Thanks!
• MemberQ is for lists. So, you could split the string into characters first (and use Not). Or you could just use StringFreeQ on the roman numeral strings. Commented Jun 7, 2023 at 16:40
The Roman numerals are strings, not lists (or another expression). So MemberQ will return False.
Select[RomanNumeral[Range[100]],
! MemberQ[Characters@#, "I"] &]
Or:
Select[RomanNumeral[Range[100]], StringFreeQ["I"]]
Also:
RomanNumeral[Range[5, 100, 5]]
• Roman zero N gets no respect. 😉 Commented Jun 9, 2023 at 16:04
• @chux-ReinstateMonica Mma respects it: RomanNumeral[Range[0, 100, 5]]. It's unclear whether nulla was considered a number, or even whether N was used by the Romans since, according to the Wiki article, "the earliest attested instances are medieval." Commented Jun 9, 2023 at 16:52
• Much of the rules about Roman numbers achieved their greater uniformity (as in this good answer) long after Roman times. Else XXXX, (and the like) should be on the list too. Commented Jun 9, 2023 at 17:51
I include some variations.
rn100 = RomanNumeral@Range@100;
Cases[rn100, _?(StringFreeQ["I"])]
DeleteCases[rn100, _?(StringContainsQ["I"])]
DeleteCases[rn100, _?(StringMatchQ[___ ~~ "I" ~~ ___])]
Complement[rn100, Cases[rn100, _?(StringMatchQ[___ ~~ "I" ~~ ___])]] //
SortBy[#, FromRomanNumeral] &
Pick[rn100, StringMatchQ["*I*"] /@ rn100, False]
Pick[rn100, StringMatchQ[Except["I"] ..]@rn100]
Select[rn100, # == StringDelete[#, "I"] &]
Select[rn100, StringPosition[#, "I"] == {} &]
Result:
{"V", "X", "XV", "XX", "XXV", "XXX", "XXXV", "XL", "XLV", "L", "LV",
"LX", "LXV", "LXX", "LXXV", "LXXX", "LXXXV", "XC", "XCV", "C"}
• Now that's just flexing. Commented Jun 9, 2023 at 9:16
Another way using GroupBy and Extract:
rn100 = RomanNumeral@Range@100;
Extract[GroupBy[rn100, FreeQ[Characters[#], "I"] &], Key[True]]
Or using GroupBy and Lookup:
Lookup[GroupBy[rn100, FreeQ[Characters[#], "I"] &], True]
Or using ReplaceAll:
rn100 /. x_?StringQ /; ! StringFreeQ[x, "I"] :> Nothing
Edit Thanks to the comment by @eldo, there's a shorter version using GroupBy:
GroupBy[rn100, FreeQ[Characters[#], "I"] &][True]
• +1 - GroupBy[rn100, FreeQ[Characters[#], "I"] &][True] would suffice
– eldo
Commented Apr 3 at 0:07
• Cheers, @eldo! I update the answer :) Commented Apr 3 at 0:10
Or use regex.
"Now you have two problems"
Range[100] // RomanNumeral // Select[StringMatchQ[RegularExpression["[^I]*"]]]
https://xkcd.com/1171/ Perl Problems
list = RomanNumeral[Range @ 10]
{"I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX", "X"}
Pick[list, StringFreeQ["I"] /@ list]
GroupBy[list, StringFreeQ @ "I"][True]
Replace[list, _?(StringContainsQ["I"]) :> Nothing, {1}]
`
All return
{"V", "X"}
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https://documen.tv/question/johnny-can-build-3-1-2-lego-planes-in-60-minutes-how-many-can-he-build-in-40-minutes-23238416-11/
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## Johnny can build 3 1/2 lego planes in 60 minutes. How many can he build in 40 minutes?
Question
Johnny can build 3 1/2 lego planes in 60 minutes. How many can he build in 40 minutes?
the answer needs to be a mixed fraction
in progress 0
3 years 2021-09-04T16:07:48+00:00 1 Answers 0 views 0
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https://www.bartleby.com/questions-and-answers/the-viscous-force-on-an-oil-drop-is-measured-to-be-equal-to-3.0-x-10-13-n-when-the-drop-is-falling-t/7112d9df-6cda-4937-9b63-59b6a090cd4d
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The viscous force on an oil drop is measured to be equal to 3.0 x 10-13 N when the drop is falling through air with a speed of 4.5 x 10-4 m/s. If the radius of the drop is 2.5 x 10-6 m, what is the viscosity of air?
Question
16 views
The viscous force on an oil drop is measured to be equal to 3.0 x 10-13 N when the drop is falling through air with a speed of 4.5 x 10-4 m/s. If the radius of the drop is 2.5 x 10-6 m, what is the viscosity of air?
check_circle
Step 1
Given,
Step 2
Viscosity of air can be calculate using Stoke’s law which can be mathematically expressed as below,
Step 3
By plugging in the g...
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See Solution
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Tagged in
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http://econsguide.blogspot.com/2011/01/tips-to-score-well-for-mcq-unit-3.html
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Thursday, January 20, 2011
Tips to Score Well for MCQ Unit 3: Business Economics and Economics Efficiency (Edexcel)
General technique
(1m) for correct option, (1m) for relevant definition and (2m) for explaining why the correct option is chosen
Specific techniques
I find that students are generally struggling especially to secure the remaining (2m) from explanation. Well, worry no more. I can assure you that if the techniques below are carefully applied, I can help you to increase your chance of scoring by 50%
• If you can support your answer with a diagram, please provide one. This is very applicable to questions related to topics like revenue and costs, perfect competition, monopoly and monopolistic
• If you are given a diagram, try to annotate or add curves. It may carry (1m) or (2m). Again this is applicable to topics outlined above
• If you are given a table with total costs or/ and total revenue, try to perform calculation such as marginal cost, average cost, marginal revenue and average revenue
• If you think that it is hard for you to explain a scenario say economies of scale, why AFC is falling as output increases etc try to provide simple numerical explanation. It will definitely help you
• If you have chosen the wrong option, under the new Edexcel Unit 3 specification, you can still secure up to (3m) from (4m). As such there is no need to panic if you are really not sure about the option
• If you have chosen the correct option but do not know how to explain it correctly, you can gain a maximum of (2m) by explaining why the other options are incorrect. Having said so, you ought to be careful in choosing which option to be eliminated. Make sure that you have sufficient knowledge. Secondly, you are reminded that not all options can be corrected as some of the answers are totally irrelevant that there is no way it can be corrected
What are the possible questions in MCQ?
Perfect competition
Watch out for topics like productive and allocative efficient. Also candidates must exhibit knowledge like why perfect market firms are only making normal profits in the long run. There is possibility that candidates may have to choose one out of five options, which resembles perfect market. They may provide options like coal mining, pharmaceutical, hairdressing, newspaper firms and wheat. Remember to choose answers that are agriculture related. Remember to draw diagram for questions on productive and allocative efficient as well as issues like supernormal or normal profit
Monopoly
Watch out for price discrimination. Remember the definition, assumptions, diagram and it what way price discrimination has been practiced. Productive and allocative efficient also likely to come out. Definitions of these two are essential and remember to draw diagram to support your answer. In the past, exam questions are extremely popular on change in objectives say from profit max to revenue max etc and change in costs (MC and AC) or change in revenue (MR and AR). However I have strong feeling that these topics are unlikely to be tested under new specification
Oligopoly
Make sure you know the following facts well. Oligopoly firms are few in the industry, they are highly interdependent, have high barriers to entry, price collusion/ price fixing/ tacit collusion is common, prefer non price competition, not contestable and has high sunk costs, high concentration ratio and often draw the attention of OFT, Competition Commission and European Commission. Watch out for few types of questions like benefits of mergers, game theory, concept of branding or advertisement and how is it related to high barriers of entry and in what way their act attracts the attention of regulators. It could be price fixing, agreements with distributors etc. Kinked demand curve is no longer in specification and therefore will not be tested
Monopolistic
Again, watch out for topics like productive and allocative efficient. Candidates have to remember that all firms that have price making ability are generally not efficient in both productive and allocative. Also you are reminded that only normal profits are made in the long run due to low barriers of entry and exit (perfect market has no barriers to entry and exit). Also they may ask you to choose which of the possible five options that represent monopolistic firm. It is easy to differentiate. Look for the business that is likely to have the lowest possible costs among the five
Regulators
Make sure you are familiar with roles of regulators like maintaining healthy competition and to safeguard consumers’ interest. Also expect possible questions on price capping like RPI-X and RPI+K. There are many ways to confuse students with these two terminologies. Under the new specification, candidates are required to have knowledge on price capping as well as profit capping and be able to identify the pros and cons between these two. Somehow the possibility of profit capping to reappear in this round of Unit 3 is remote since nothing much can be asked and considering that it is just recently tested (January 2010)
Hope this helps!! I will upload the next posting on Section B in a couple of hours. Stay tune.
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# IBPS Clerk Prelims 2018 – Reasoning Ability Questions Day-50
Dear Readers, Bank Exam Race for the Year 2018 is already started, To enrich your preparation here we have providing new series of Practice Questions on Reasoning Ability- Section. Candidates those who are preparing for IBPS Clerk Prelims 2018 Exams can practice these questions daily and make your preparation effective.
[WpProQuiz 4572]
Direction (1-5): Read the following information carefully and answer the questions given below.
       Eight persons are sitting in a square table in such a way that four of them are sitting in corners of the table and remaining of them are sitting in middle of the four sides. The persons who are sitting in the corners facing towards centre of the table and the persons who are sitting in the middle of the sides are facing away from the centre.
       Only two persons are sitting between R and P, who does not sit in the middle of the side.V sits second to the right of U, who is an immediate neighbour of R. As many persons sitting between R and W is same as the persons sitting between V and Q. W is an immediate neighbour of Q, who does not sit opposite to R. T sits second to the left of S.
1) Which of the following statements is true?
a) Q is an immediate neighbour of P
b) Only one person sits between T and Q
c) S sits one of the corners
d) T sits opposite to W
e) None is true
2) How many persons are sitting between Q and the one who sits opposite to R, when counted from right of Q?
a) None
b) One
c) Two
d) Three
e) More than three
3) Who among the following sits opposite to the person sits second to the left of W?
a) P
b) R
c) S
d) T
e) U
4) Four of the following five are alike in a certain way and hence form a group. Which one of the following that does not belong to the group?
a) T
b) W
c) Q
d) S
e) R
5) If W is related to P and U is related to S in a certain way. Then, V is related to which of the following?
a) W
b) U
c) R
d) Q
e) T
Direction (6-10): Read all the statements and then decides which of the given conclusions logically follows from the given statements disregarding commonly known facts.
a) If only Conclusions I follows
b) If only Conclusions II follows
c) If either Conclusions I or Conclusion II follows
d) If neither Conclusion I nor Conclusion II follows
e) If both Conclusion I and II follows
6) Statements:
All plastic are glass
No glass is sponge
All sponge are liquid
Some glass are liquid
Conclusions:
I. All plastics can be liquid
II. All sponge can never be plastics
7) Statements:
No lawyer is teacher
Some lawyer are engineer
Some engineer are teacher
Some teacher are artist
Conclusions:
I. Some artists are engineer is a possibility
II. All engineers are teacher is a possibility
8) Statements:
Some dog are horse
All dog are animal
No animal is lion
All horseare animal
Conclusions:
I. Some animals can be horse
II. Some lions are not dog
9) Statements:
Some red are blue
Some blue are orange
No blue is yellow
Conclusions:
I. All orange are yellow is a possibility.
II. Some yellow are red.
10) Statements:
No key is a chain
No chain is a lock
All locks are ring
Conclusions:
I. Some rings are chain is a possibility
II. Some key are not ring
Direction (1-5):
• Only two persons are sitting between R and P, who does not sit in the middle of the side.
• V sits second to the right of U, who is an immediate neighbour of R.
• As many persons sitting between R and W is same as the persons sitting between V and Q. W is an immediate neighbour of Q, who does not sit opposite to R.
• So, Case-1(b) will be dropped.
• T sits second to the left of S. So, Case-2 will be dropped.
Directions (6-10):
### Daily Practice Test Schedule | Good Luck
Topic Daily Publishing Time Daily News Papers & Editorials 8.00 AM Current Affairs Quiz 9.00 AM Current Affairs Quiz (Hindi) 9.30 AM IBPS Clerk Prelims – Reasoning 10.00 AM IBPS Clerk Prelims – Reasoning (Hindi) 10.30 AM IBPS Clerk Prelims – Quantitative Aptitude 11.00 AM IBPS Clerk Prelims – Quantitative Aptitude (Hindi) 11.30 AM Vocabulary (Based on The Hindu) 12.00 PM IBPS Clerk  Prelims – English Language 1.00 PM SSC Practice Questions (Reasoning/Quantitative aptitude) 2.00 PM IBPS Clerk – GK Questions 3.00 PM SSC Practice Questions (English/General Knowledge) 4.00 PM Daily Current Affairs Updates 5.00 PM Canara Bank PO Mains – Reasoning 6.00 PM Canara Bank PO Mains – Quantitative Aptitude 7.00 PM Canara Bank PO Mains – English Language 8.00 PM
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Constructing objects
POV-Ray gives you several simple objects that you can place in your scene. If you want more complex objects there are several ways of getting them. One of these ways is using constructive solid geometry, or CSG. With CSG you take simple objects and combine them into more complex objects. You can merge objects, unite them, cut them, and intersect them.
Once you’ve created a CSG object, CSG objects are themselves able to be used in merges, unions, differences, and intersections.
Let’s add a ring to our sphere. There is no such thing as a flat ring in POV-Ray (it does have a torus, but that’s a kind of donut shape). But it does have a cylinder. Below the sphere, add a cylinder object.
//ring around the sphere
cylinder {
<0, -.01, 0>, <0, .01, 0>, 3.2
pigment {
color Green
}
}
Cylinders are defined by their starting location, their ending location, and their radius. This cylinder starts at -.01 meters below the origin, ends at .01 meters above the origin, and is 3.2 meters in radius. It’s really more of a disc than a cylinder.
You should end up with a blue sphere with a green puddle around it on the plane.
We want this to be a ring, like a ring around a planet, so we don’t want the ring to go right up to the surface of the sphere. Let’s add another, smaller cylinder where we want the empty space to be.
cylinder {
<0, -.01, 0>, <0, .01, 0>, 2.8
}
This looks strange. There’s green all the way to the center, but little black semicircles on the edge. It might even look different on yours than it does in this picture.
This is happening because we have two coincident surfaces. We have a black plane and a green plane whose surfaces are exactly the same once you get inside the radius of 2.8. Both of those cylinders start at y -.01 and end at y .01. POV-Ray has no idea which surface to use as the “real” surface at those points.
This is similar to the problem of looking at a plane straight-on. Because these numbers are exact numbers, we can tell POV-Ray to put things or look at things at exactly the same locations.
In this case, the solution is to make the inner cylinder start at slightly lower and higher points than the outer cylinder does.
cylinder {
<0, -.02, 0>, <0, .02, 0>, 2.8
}
That’s better. This gives our green cylinder the basic shape we want, but we still have that black cylinder inside. We really want there to be nothing there. We want the cylinder to be, not a cylinder, but a ring.
This is what CSG is for. In CSG, we have a difference object that takes the difference between two other objects. We want the difference between our green cylinder and our inner cylinder. Surround the two cylinders with difference:
difference {
cylinder {
<0, -.01, 0>, <0, .01, 0>, 3.2
pigment {
color Green
}
}
cylinder {
<0, -.02, 0>, <0, .02, 0>, 2.8
}
}
Okay, this is what we want. The ring circles the sphere with space between the sphere and the inner edge of the ring. Let’s add a little more “action” to the scene. We can rotate the ring so that it is on an angle.
First, remove the plane from the scene so that we’ll be able to see the whole ring when we angle it. It was a nice plane, but we won’t be needing it any more.
Then, add a “rotate” line to the difference section, just above the final curly bracket:
rotate <0, 0, 30>
The complete difference section should be:
//ring around the sphere
difference {
cylinder {
<0, -.01, 0>, <0, .01, 0>, 3.2
pigment {
color Green
}
}
cylinder {
<0, -.02, 0>, <0, .02, 0>, 2.8
}
rotate <0, 0, 30>
}
The “rotate” line has a set of three numbers that look a lot like our other sets of three numbers. In this case, we’re rotating a number of degrees around the “pole” that we’ve specified. The numbers are still x, y, and z. We’ve told it to rotate 30 degrees around z. You can imagine z as a pole situated on zero x and zero y. It moves from front to back. Here’s a diagram (you can see how this diagram was made at The Persistence of Text).
Rotating the ring <0, 0, 30> is like resting it on the green pole and rotating it 30 degrees, with the right side moving up. We could also rotate it around x and y. In this case rotating the ring solely around y wouldn’t change anything in the image. No matter how much you rotate it around y, it is still a green ring circling the sphere, flat on the other two poles. Now, having rotated it around z first, we could rotate it around y and that would move the lower end either towards or away from us. Go ahead and add another rotate after the current rotate, and play around with rotating it around x and y.
When you’re done, remove that second rotate.
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http://java.worldbestlearningcenter.com/2013/04/with-maxfinder-program-you-can-find.html
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## Saturday, April 6, 2013
### Max Finder program
With the MaxFinder program, you can find a maximum number in a list of your numbers.
You can have a large list of numbers as you want. In this program, you will learn to use an array to store a collection of numbers. So the array in this program is the list of numbers.
MaxFinder source code:
1 import java.util.Scanner;
2 class MaxFinder{
3 public static void main(String[] args){
4 Scanner sc=new Scanner(System.in);
5 int n;
6 int i;
7 int max;
System.out.print("How many numbers in your list? ");
8 n=sc.nextInt();
9 Integer[] ls=new Integer[n];
10 for(i=0;i<n;i++)
{
System.out.format("Enter number %d:",i+1);
ls[i]=sc.nextInt();
}
11 max=ls[0];
12 for(i=1;i<n;i++){
if(max<ls[i]) max=ls[i];
}
13 System.out.format("The maximum number in the list is %d.\n", max);
14 }
15}
Program Output
Code Explanation:
1 Include Scanner class in the program so that you can use its nextInt method to receive input from keyboard.
2 Open MaxFinder class block.
3 Open the main method block.
4 Create Scanner object sc.
5 Declare local variable n to store the number of list items.
6 Declare local variable i to be used in for loop iteration.
7 Declare local variable max to store the maximum number in the list.
8 Receive the number of list items input form keyboards.
9 Create an array called ls to store n items.
10 Use a for loop to receive assign a value input from keyboard to every list item.
11 Let the variable max take the first item of the list.
12 The max (the first item) is compared to the next item of the list. If it is less than the next item, update the max by letting it take the next item. The comparison process repeats until every item of the list is examined.
13 Display the maximum number on the screen.
14 Close the main method block.
15 Close the MaxFinder class block.
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https://www.teachoo.com/9209/2896/Ex-10.2--1-(a)/category/Finding-area-by-counting-squares/
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1. Chapter 10 Class 6 Mensuration
2. Concept wise
3. Finding area by counting squares
Transcript
Ex 10.2, 1 (a) Find the areas of the following figures by counting square: Covered Area Number Area 1 Fully filled squares 9 1 × 9 = 9 2 Half filled squares 0 0 × 1/2 = 0 3 More than Half filled squares 0 0 × 1 = 0 4 Less than half filled squares 0 0 × 0 = 0 We have, 9 full squares So, Area = 9(1) = 9 square units
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https://hackaday.com/2021/05/13/somethings-up-in-switzerland-explaining-the-b-meson-news-from-the-large-hadron-collider/
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# Something’s Up In Switzerland: Explaining The B Meson News From The Large Hadron Collider
Particle physics is a field of extremes. Scales always have 10really big number associated. Some results from the Large Hadron Collider Beauty (LHCb) experiment have recently been reported that are statistically significant, and they may have profound implications for the Standard Model, but it might also just be a numbers anomaly, and we won’t get to find out for a while. Let’s dive into the basics of quantum particles, in case your elementary school education is a little rusty.
It all starts when one particle loves another particle very much and they are attracted to each other, but then things move too fast, and all of a sudden they’re going in circles in opposite directions, and then they break up catastrophically…
## The Standard Model
In the 1970s physicists started coalescing around a thing called the Standard Model, which is similar to the Periodic Table of Elements, but at a much smaller scale. It describes the particles that make up the protons, neutrons, and electrons (which in turn make up atoms), and the forces that act on them. The Standard Model has held up to most experiments so far, but the one earlier this year may throw a small wrench in that.
The Standard Model reuses a lot of pre-existing words in confusing ways, so let’s break it down a little bit from (relatively) large to small.
• A molecule is made of atoms.
• An atom is made of protons and neutrons surrounded by electrons.
• Protons and neutrons are called composite particles because they’re made of smaller elementary particles.
• Protons and neutrons are made of combinations of quarks. There are other composite particles, and in general these particles made up of combinations of quarks are called hadrons.
• There are 6 different types of quarks, named: up/down, charm/strange, top/bottom. Combinations of up and down quarks make up protons and neutrons.
• In addition to quarks, there’s another class of particle called the lepton. Leptons can have a charge, like an electron, or they can not have a charge, like a neutrino.
Besides the classification of these tiny particles, the Standard Model also describes how the fundamental forces interact. There are 4; electromagnetism, strong nuclear, and weak nuclear. Gravity is the fourth, but the Standard Model doesn’t like to talk about this black sheep of a force, and intentionally leaves it as a blank space, an exercise for the reader (or some janitor to Good Will Hunting it). Some day either the force of gravity will be incorporated into the Standard Model, or a new model will emerge that explains the universe better than the Standard Model AND incorporates gravity, but until that happens, some hand-waving will occur. Besides, gravity works on such a larger scale than the other three, that its effect over the scale of subatomic particles is considered negligible. *shrug* I’m just repeating what the physicists are saying.
## Keep Up with Bosons and Mesons
The way forces work is with force-carrying particles, a category called bosons. There are a few sub-categories here, because different bosons are responsible for the different forces. Photons and gluons are among them, and the recently discovered Higgs boson, which was theorized a long time ago as being a requirement for the Standard Model to work, so the physicists breathed a collective sigh of relief when it was finally found. There’s also a theorized graviton that would be the force-carrier for gravity, but it hasn’t been discovered… yet (looking at you, Matt Damon).
There’s one more kind of funny-word-on to know about, and that’s the meson. It’s a particle composed of quarks and antiquarks (which makes mesons a subset of hadrons), and which ones and how many lead to a large number of different variants. Mesons are very unstable and last less than a microsecond, decaying into various combinations of other types of particles.
Clear?
## The Experiment
Particles this small are impossible to measure directly, which is why we have the Large Hadron Collider. If one were to try to reverse engineer a cake, one could take pictures of it, do spectral analysis, maybe even taste it. These aren’t options at the subatomic level. Instead, the LHC makes particles move very fast, then bashes them against each other. The bits that result from these collisions have a lot of energy to dissipate, which results in them interacting with electromagnetic fields in tiny but measurable ways. From these collisions, we can work backwards to figure out the secrets of the universe, in much the same way that throwing a grenade at a cake and analyzing the spray pattern might allow you to determine whether the frosting was buttercream or fondant.
In the experiment announced recently (pdf of paper), there was a discrepancy in how a particular variant, the B meson (which isn’t a single variant but a whole class of variants), decayed. The Standard Model says that leptons all behave the same and are identical in every way except for their mass. So when the LHC bashed a whole bunch of particles together (specifically protons vs. protons) and measured the B meson decays, they theorized the subatomic soup that resulted would contain equal parts of electrons and muons. Instead they found 15% fewer muons than they expected.
They couldn’t exactly look at the nutrition label, so they’re left wondering if what they found was a fluke, or if the recipe was wrong. If it’s the former, then it will have beaten the odds by 3 sigma (which corresponds to a 1 in 740 chance of a fluke). In particle physics, this is merely eyebrow-raisingly interesting, as it’s not a large enough exponent to satisfy them. If it’s the latter, it means big things for the Standard Model. That could mean either revisions to the Standard Model, or possibly understanding if there’s a difference between the leptons (other than their mass alone), ending what was formerly called lepton universality.
Unfortunately the testing machine, much like a McDonald’s ice cream machine, is down for servicing, so we’ll have to wait until 2022 before the upgraded LHC can deliver some frosty mesons and give us an answer about the Standard Model.
## Fool Me Once, Well, Keep Fooling Me
Calling this eyebrow-raisingly interesting as opposed to earth-shatteringly amazing is an appropriate response, as we’re no strangers to the teasings of the physicists. There have been a number of anomalies over the last decade in particle physics that have all been overturned eventually as more data came in. There’s a vast difference between a 3 sigma anomaly and a 5 sigma (1:3.5 million) discovery, and we’ve been seduced by this before. Maybe you remember in 2011 the superluminal neutrinos that turned out to be an improperly attached fiber optic cable. In 2015 there was the 750 GeV bump with a significance of 3.9 that ended up being a statistical fluctuation the next year when data was collected again. In 2016 there was an anomaly with the B meson that seems to have faded, too.
Each of these anomalies leads to hundreds of papers and theorizations and new types of physics until the next set of data sends them into the shredders to make pulp for the next round of papers, and the machine grinds on, with journalists siphoning headlines from these papers and drawing wild conclusions about warp speed and time travel and new particles like the leadingmeon.
And yet, forward progress keeps happening, slowly and scientific methodically, as we move towards understanding the workings of the universe. Maybe in the decades to come they’ll laugh at our quaint Standard Model like we look at the four elements of Earth, Water, Air, and Fire, but we will have gotten to that point by moving through where we are today, so I’ll continue to read about it and nod like I pretend to understand what they’re talking about, and respect the complexities of the process of measuring things so small. After all, my day job as an electrical engineer relies on making electrons move in ways that were inconceivable a century ago; maybe the discovery of lepton non-universality will eventually lead to the downfall of Amazon and Uber.
## 50 thoughts on “Something’s Up In Switzerland: Explaining The B Meson News From The Large Hadron Collider”
1. William Wesley says:
unfortunately the publishing is the point, not the understanding, as in boxing or chess a final resolution is not actually desired, all the money is made through conflict in print.
1. Alex Rossie says:
Aptly the McDonalds ice cream machine if down for servicing because the money is made by selling the machine to the franchise and then repairing it constantly.
The money isn’t actually in delivering ice cream to customers…
2. Jimmy Wiggy says:
🤣 best physics review I have read in a long time. Hilarious!
1. Goody Rather says:
Agreed. Quite a few chuckles in here. Keep it comin’!
3. Gravis says:
I’m just glad to know that people are still probing the depths of reality. I am curious about the feasibility of a spaceborne collider, especially now that the lift capacity per launch has been drastically increased and the launch cost decreased. It’s not a very accessible for us earthbound humans but space seems like an ideal location for many of these experiments.
1. Chewy says:
What do you imagine would be advantageous about having a “spaceborne collider”? Perhaps I am missing something but I don’t see any new questions being answered. The collider instrumentation by design is meant to contain within fields so if I understand correctly the physical location should be moot, also assuming location did have some type of effect upon the experimental process wouldn’t that be a bad thing?
2. Eric Weatherby says:
If we can be reasonably certain that the colliders aren’t going to be causing any huge explosions or releasing dangerous amounts of radiation or so on, then the logistics advantages of building them on earth make that the logical choice.
I’d be interested in hearing your ideas on why spaces would be good. I do have to admit that building an equatorial collider around the moon would certainly be *cool*.
Hi there, accelerator physicist and operations specialist here. This idea was floated years ago because you don’t have to worry about radiation affecting surrounding areas, and you have no space constraints so your magnetic fields don’t have to be as high (i.e. larger circumference permitted). However, from an operational perspective, I can say that we’re constantly performing maintenance and responding to failures on our accelerators; putting our machines up in space makes accessing them even harder! Since colliders/accelerators are often pushing the boundaries of what the equipment can do (power supplies, computation, magnetic and RF fields), uptime is not 100%, so human intervention is a fact of life for us.
1. Anonymous says:
“accelerator physicist and operations specialist”
+1 That sounds impressive.
Nah just lots of syllables. I rely on real engineers and technicians to do the actual work, trust me.
Hi Adam. Could you please send me an email ? I would like to interview you. I have some documentation I am hoping to have validated as well. Thank you much.
Florida
2. No Nonsense..... says:
Greetings from a long time “electricity hacker”…. I’ve been able to figure out most of this “mess”, did get an intro to much of it decades ago. Had to do a lot of recalling! My forte’ is in electrical areas, but a lot of changes have happened over quite some time. Our lives have been influenced by the semiconductor industry, with transistors, tunnel diodes, light emitting diodes, The standard idea of sine-wave theory has been replaced with digital bits, thus revolutionizing everything. I’ve kept up with it enough to realize that “all that is needed” are a few Albert Einsteins again! Simply stated, today’s “ordinary” household computers are the result of the need for a human/machine interface to allow for interaction. Now, the idea of a research collider on the moon is within grasp, with problems of gravity, sun particles, et all, to be solved. Thanks to everyone for the insightful inputs! About math – I recall a very astute computer programmer who could dream up lots of ideas to accomplish quite vexing problems, yet she had no idea of multiplication tables and following algebra formulas! Everyone has some good ideas for solving problems!!
1. Jay Carlisle says:
Do consider that the language of precision is not a subject intended for the general public to hold any save a not nil understanding of suitable for a functional labor force while winnowing out natural talents for potential future development should circumstance warrant by those who oversee the social indoctrination programs tagged as education at the secondary level as if a passive act offering teaching could be mandated and upon leading the Houyhnhnms to water it follows that They can be made to think If the public were intended to hold fluency in communication of specifics it wouldn’t be taught as a stand alone abstract subject that dictates one route method as the single correct means of solving a given set of problems like anointing the cut/paste function as properly done only with the pulldown menu method with none of that fast and loose business which will result in the loss of a letter grade for each instance appearing in a students capture log
Were the intent ment to maximize lil Johnny’s knowledge meeting his full aptitude the mathematics would be incorporated into the subject mater as it is required in function and all valid approaches would be available with the specifics a subjective and contextual determination in details but shaking out in large part early on with the key differ being who’s allowed into the Lyceum and who’s in the number not By the time limit process might be conceptually abhorrent enough to spark preference for infinitesimal analysis its only gonna surprise at a mile off depending on observation point over/under
The focus is more on stand in line, raise Your hand, sign Your name in the upper right hand corner while quite a few so called subjects are not as purported When economics fails to make any practical predictions boom or bust and also fails in popularizing its findings such that its uncommonly rare to find a layman that can give a simple definition of dollar suggesting the subject instead is more accomplished in fostering disinformation in the public facilitating fractional reserve fiat dollars in a unsustainable grey goo perpetual growth model system taking on the properties of a sigel in public perception it makes some amount of practical sense how such disingenuous manipulation might be considered advantageous enough to implement
It was pretty chilling when the common core kid said “We’re both right” and I found order of operation was being argumentative trying to be right just to win
God help Us all should that kid end up designing a bridge some dark day
Anyway I told Y’all that to ask this
Aren’t the returns seriously diminished at really large circumferences?
And anyway the equatorial collider would suffer from no Corvallis effect. That’s how You find an equatorial point by finding where water just sloughs through a draining container without rotation
Catastrophic effect on spin
Please try and be more careful in the future
1. Jace says:
4. Dan says:
I believe Starlink is rapidly moving towards deploying a spaceborne collider.
1. Onetruegod says:
Yeah. I think its called Skynet…
2. Viz says:
It’s already done penning trap in space was already done and FTL micro quantum communications satellites deployed using the tech that hawkings developed back when his theoretical ideas went into practice. Primordial black hole drive engines were born and penning trap gwave drives just went online lhc is and engine. Non college grad thoughts from the autistic spectrum don’t really give two Fs
5. abjq says:
Don’t we get a better vacuum in LHC than we get in space (near earth)?
1. Pat says:
Depends on what you mean by “near earth.” Low earth orbit? Yeah, definitely. Geo? Nope. Surface of the moon? Pretty close.
Pressure at LHC is about 0.1 nanoTorr, or a few million “things” every cubic centimeter. Obviously space is thinner than that (standard rule of thumb is the ISM is an atom/cubic centimeter), but it takes a while to drop ~12 orders of magnitude in pressure.
Moon is about 50 times better vacuum than LHC (at night). An accelerator literally orbiting in space is a little silly – surface of the moon would be more reasonable. However as Adam pointed out above, I always find the people who think we can just happily build an unattended top-end scientific experiment *hilarious*.
1. RetepV says:
Only reason to have one in space, I think, would be to be in a place where gravity of the Earth interferes less with the experiments. I’m by far not into this stuff. But it occurs to me that we might have problems fitting gravity into the model because we are in a gravity well. It can’t be just a coïncidence that exactly *gravity* is what we are having trouble with. It’s all around us, we experience it all the time… But maybe *that* is actually the problem.
1. Pat says:
Meh, there are reasons like “you can’t practically build an accelerator that large on Earth because… because… politics, I guess?”
Going into space doesn’t actually “get rid” of gravity. You don’t experience it because you’re constantly in free fall (the equivalence principle). But even that’s an approximation for a macroscopic object.
“It can’t be just a coïncidence that exactly *gravity* is what we are having trouble with.”
It’s really not surprising that it’s difficult to fit gravity into any particle physics theory: gravity is like, 30+ orders of magnitude weaker than all the other forces.
6. Viz says:
It’s already a reality and in use for G wave penning trap engines to propel use to deep space utilizing quantum navigation quantum communications waypoints FTL
4. Mark Hambleton says:
Nice work. Thanks. Really good piece for the ‘totally out of their depth, but perversely interested nonetheless’.
5. Biomed says:
At this time it still sounds like a replay of Laurel and Hardy’s “Who’s on first?”.
It will continue to become more clear over a few more decades as we simply refuse to accept not knowing.
1. Piecutter says:
Perhaps you mean Abbott and Costello’s “Who’s on First?”?
1. Piecutter says:
And if you need it made clear, Lou would have been a great theoretical physicist. His mathematical skills could prove the impossible using multiple equations to come to the same answer:
1. Mike says:
Isn’t that the basis of the new math that they are teaching now?
6. David Misel says:
It’s good to mix humor with serious science stuff.
Makes me feel as if I understand what’s happening in the big world of tiny particles.
7. Anonymous says:
Earth, wind, air, and fire, huh?
Ha! Whoops. Fixed. Thanks for pointing it out.
2. J.J. A. says:
Hey I wonder what would happen if you mix them all together.
8. dukwon says:
“In 2016 there was an anomaly with the B meson that seems to have faded, too.”
Which one do you mean? The P5′, R(K*), R(ϕ), R(D) and R(D*) anomalies are alive and well.
By the way, the LHCb experiment is wholly located in France, not Switzerland, albeit right up against the border.
1. singlemale says:
are you sure. Some farmer may have moved the border like they did in Belgium
1. Patrick says:
That’s true. But the accelerator is across France and Switzerland.
9. Patrick says:
“maybe the discovery of lepton non-universality will eventually lead to the downfall of Amazon and Uber.” That’s a surprising angle 🤣. Someone who actually worked on these measurements.
10. Alice Lalita Heald says:
A video would worth 1000^2 words here.
11. Dan says:
I must admit, every time I read the word “meson”, I hear it in jar-jar binks’ voice.
12. J Titus says:
I suspect as physicists tools offer more resolution they will find smaller and smaller particles with shorter and shorter lives. It could go on forever…
1. Pat says:
Particle physicists call pretty much everything “a particle” but they’re really just bound states of quarks in new and different configurations. There’s no real thought or evidence that the fundamental particles (leptons, quarks) have *any* substructure whatsoever.
In other words, these particles aren’t getting “smaller”. They’re just new configurations of quarks. It’s basically like investigating the spectroscopy of atoms: the difference is that since the structure of the theory is *much* more complicated, the number and structure of bound states configurations is *much* more rich.
For instance, the delta-+ particle is really just an excited proton, much like if you excite hydrogen by heating it up and it flows at specific frequencies. But because it’s particle physics, it gets called a new particle instead of “excited proton.”
13. Mikey says:
Rather than the “stamp collecting” (taxonomic) approach to physics, wouldn’t it be nice to write a GPU shader in pycuda or numpy to plot SU(3)xSU(2)xSU(1) ?
The idea being, a Dirac Spinor (electron) is plotted as a soliton, a standing (Compton) wave. A nucleus would be illustrated as a shell of such tetragonal soliton lattice, exchanging boson-waves. It would be neat to see a dumbell shaped lattice fission as a moderated neutron-soliton, propagating at the appropriate velocity causes the shell-droplet to pop.
Need some clever ways to render bi-quaternions, that is an 8 dimensional wave function in three dimensions.
Something like the plots at https://github.com/portsmouth/fibre
I have trouble even getting my Vive sensors to work in Linux with Khronos drivers. I’ve used numpy to plot matrices, but the math is over my head, and I probably won’t ever find the time – https://www-fourier.ujf-grenoble.fr/~panchish/ETE%20LAMA%202018-AP/lecturesZETAS2018/Special%20unitary%20group%20-%20Wikipedia.pdf
1. Pat says:
Physics is *way more* than the gauge groups: pretty much all of the interesting stuff comes in how the symmetries are *broken*. The gauge structure tells you why, for instance the particles in the Eightfold Way all have *similar* energies and decays but the actual *physics* is in how it *differs* from exact.
The gauge groups also don’t tell you about the mass structure, which sectors the groups act on, and the alignment of the eigenstates between groups, or what currents the interactions couple to, for instance.
1. Pat says:
That should say *group* structure, not *gauge* for the Eightfold Way. And yes, that SU(3) (of up/down/strange) isn’t the QCD SU(3) (of color) but that’s partly the point: the mass structure and number of quarks add a *huge* amount to the richness of the theory and they’re not in the gauge group structure at all.
1. Mikey says:
After watching some youtube videos from Siggraph2019, Game2020 and Cohl Furey, I gathered that the 3-2-1 group structure signified a sort of recursive triple Maxwell equation, with divinely arbitrary coupling constants empirically derived. I’ve actually coded Maxwell equations and DSP.
Since some here are familiar with the electronics concepts, those might elucidate the physics for those unfamiliar with the parallel universe of physics conventions and nomenclature, since different professions like to use different terms for similar things. To me (now), the quantum vacuum is an 8-fold zero-bandgap semiconductor for soliton modes. Which is probably as far as I can go without years of math. I thought plots of configurations of the fields and momentum/spin of different particles would be more insightful than a table.
1. Pat says:
Yeah, science communication is really poor on this stuff. All of the descriptions I’ve seen are pretty universally bad. You’ve either got math-heavy theorists who are obsessed with naming math terms (… weird) or experimentalists who similarly ignore the underlying symmetry.
The reason for the table is actually the *broken* symmetry. Look at the table – the “SU(3)” part is a tiny label in the corner of the quarks, saying “oh yeah, there are really 3 of these guys but they all act identically”.
And the fact that there are 3 generations is some (presumably) broken higher symmetry, as is the fact that “up/down” and “electron/neutrino” are different particles. If the symmetries were unbroken, you’d just have a single “quark” with a little “x3 color/x3 generation/x2 weak isospin/arbitrary phase” label, and a single “lepton” with “x3 generation/x2 weak isospin/arbitrary phase”.
I’ve never really understood the point of calling things out as the full symmetry product space, because the only one that really matters for “taxonomy” are the *broken* symmetries. The unbroken symmetries just result in weird structure constants.
14. Well, the new theory is actually an old theory: the Aether Theory. Ooohhh. Tesla used this theory to come up with all of his inventions and it was the reigning theory of the late 1800’s, based on Newtonian physics.
It’s a very simple theory: the entire universe is filled with an electromagnetic fluid.
This theory explains gravity and dark matter (heavy bodies spinning fast enough to attract other bodies, outer edges of galaxies spinning as fast as the center) and anything electromagnetic (EM).
The theory also has this side hustle, where faster than gamma oscillations have vibrations so fast that particles form, based similarly to cymatics, where harmonics creates form at various overtones; these structures creating or absorbing radiation (a pressure modality in this Aether).
It’s where the word Aether or Ether comes from.
But in the late 1800, two scientists tried to prove this theory, the Michelson/Morley experiment (MMX).
They thought that if the Earth is traveling through this Aether, they should be able to measure this “Aether wind” as a pressure hitting the Earth.
So, they used an interferometer (laser split into two beams and reconverged at the same point), which could detect this Aether pressure.
What happend was they got a null signal, nothing!
It ruined science. No one knew what to think, Aether theorists were appalled and the scientific community scrambled to figure something out.
Einstein, who was also an Aether theorist, used math to break it down. He knew the Aether math was still valid and useful (aside weird anomalies like Mercury’s orbit weirdly out of sync with the common barycenter of the solar system and others), so he split the Aether theory into two: gravatational waves and E=MC2 (speed of light (C) = speed of Aether).
Thus the current foundation for our scientific models: standard model, quantum mechanics, string theory, etc.
There are a few problems with this though.
We use interferometers in space LIGO/LISA, etc, and those get signal from gravitational waves, also Aether theorists primarily believed in a God or intelligent design. Michelson was a huge proponent for the church; his invalidation of the Aether theory was primarily the basis for the scientific community to push atheism.
One of the reasons God was such an easy explanation: most believed in a God, but also, you can’t get something (like the big bang) from nothing. Enter Hawking, he saved the day! He came up with the idea of 4th dimensional universe bubbles, rubbing against each other, creating energy that would make particles.
This convinced the scientific community there was a possibility of this big bang from universe bubbles, but recent equations fall short, due to false vacuum decay, as well as the fact we would see more indication of pull or force from an outside universe; which there is none so far; so this is still being studied, but it’s rather impossible to study forces outside our universe.
But! Here’s the fun part!
The MMX is horribly invalid.
See, we have this fun protective EM field, which keeps us alive from the sun’s harmful rays. We see this in the Aurora borealis, we’ve since studied this field and know very well it’s there.
It also blocks all of any Aether wind. In fact, if you turn a spinning interferometer on its side, you actually get a signal (measuring the pressure of this EM field), as seen here: https://youtu.be/7T0d7o8X2-E
There are other obvious issues with the standard model, including the measurement of the Higgs boson, which puzzled scientists, which then came up with the standard model 2. Other theories like Higgs field and QM have stated that all these subparticles each have their own field, but it’s widely known that everything is made up from fields, due to when freezing particles, you get more of a wave function.
So, science is super close; everything is a field, but why not just one field, like the Aether theory?
Well, it makes a lot of sense.
See, after Tesla’s (and the German Nazi scientists, who were all Aether theorists) work was taken underground; all this Aether work vanished.
Aether not only explains light as an EM pressure in this medium, but it explains: flying saucers (instant quadrillionaire, if you mine the asteroid belt, hello Elon), Tesla’s death ray (a similar microwave device used by the military to disperse crowds), endless energy, mind control (TMS based devices), etc.
It’s basically has access the most destructive science ever. It also proves how the military could already be flying our own saucers (reverse EM pressure against “gravitational waves” (i.e. fluidic Aether).
Another weird thing is that it makes sense why “science” would continue to study this; these EM forces are being detected with EM sensors, which means all sorts of things will look correct… not to mention, those who “fund” “science” control the main narrative.
Problem is, the jig is up! The scientific community is finding too many errors, without “one medium to rule them all”, so yeah, B measons and all the other inconsistencies of the standard model.
I’m not suggesting that we ditch scientific research; it’s all quite useful, but math doesn’t tell you what to research, it only can validate studies on that research. We need to at least look through the eye of Aether again, to re-study each experiment done with gravatational waves, EM forces and particles.
Think of this, Aether has really never been disproven, but standard model has been constantly disproven and is disassociated from gravatational waves.
Also, maybe weird anomalies, like Mercury’s orbit will make more sense, when we study new data from the Parker Solar Probe (et al new solar satellites) and see that the sun doesn’t have a perfectly centered axis, which would account for the odd barycenter and why you “sometimes” have to use Einstein over Newton… maybe you don’t and Newton is just fine for everything.
It’s all how we think stuff works, Aether proves matrix theory or intelligent design; why not bring back spirituality, now that we know the pitfalls of religion?
1. Loren Heyns says:
You lost me with the last sentence. What about Aether proves intelligent design? We would we need to be spiritual in trying to prove matrix theory? Are you saying matrix theory is unprovable?
15. RubyPanther says:
My understanding is that gravity is left blank because it isn’t really believed to be a fundamental force. However, because we can’t sense or measure the shape of space-time directly, we have to use a fake force to describe the effects of the topology of space-time. Basically, mass changes the shape of space-time; or you could equivalently say that deformations in space-time create mass. And the direct path that this mass travels goes over these deformations. But we can’t see or intuit the deformations directly; we expect a “straight line” to be where space-time would be if it wasn’t deformed. So pretending that gravity is a force allows us to maintain this useful geometric concept of a straight line even when we’re talking about motion in the real world.
So it is not actually waiting to be solved at all. There are simply no gravitons to discover.
1. Pat says:
“My understanding is that gravity is left blank because it isn’t really believed to be a fundamental force.”
So to be clear, gravity gets left out because it doesn’t appear to be a *gauge* force like the others – not one we know how to deal with. Some gravity theorists are *really* obsessed with the whole idea of “there’s no native background spacetime, it’s matter matter matter” but that’s a philosophical point more than anything else.
“But we can’t see or intuit the deformations directly”
This is the odd thing about that idea – yes, we can. The Universe, for some reason, got *driven* to “flat background geometry” to *very* high precision. So the idea that “there’s no a priori background metric” is… well… silly. Maybe *fundamentally* there isn’t one, but *something* (canonically, inflation) shoved everything out in such a way as to *impose* a fundamental flat metric on the Universe.
So nominally you could say “well, yes, but we can *imagine* Weird Universe where inflation didn’t happen, and curvature can just be anything!” and – yeah, I mean, you *could*. So what? Still live in this universe, which means we can “group” inflation + gravity together, call the two of them “gravity” and hey look, now we’ve got a background metric.
“So it is not actually waiting to be solved at all. There are simply no gravitons to discover.”
Gravitons might end up being something weirdly emergent in the Universe, but there *will* be something we can call a graviton. It’s just a natural consequence of quantum mechanics. You get up to the Planck scale, and gravity is so strong that it’s just impossible to have certain configurations quantum mechanically, because the energy states don’t work.
The “solution” to “quantum mechanics + GR” can’t be “there’s no actual problem.” Get up to the Planck scale, and *neither* of the theories will work. And whatever the final solution is, there’s going to be *something* like a graviton.
That being said it’s important to understand that when we say “oh, there’s a particle,” it’s really just a way of naming things. An up quark is just an excitation of the “up” portion of the quark field (…it’s more complicated than that, even, but let’s just go with that), exactly like a fundamental mode of a drum. Just like you can’t create a standing wave on a drum other than in certain patterns, you can only create quarks in certain field configurations too. So a “graviton” in some sense ends up being the same way – it’ll just be the fundamental excitation of spacetime itself, which we *know* has to be limited in how it can be excited.
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# Computational Fluid Dynamics by T.J. Chung
By | 2016-02-26
Book Title : Computational Fluid Dynamics
Author(s) : T.J. Chung
Publisher : Cambridge University Press
Edition : First
Pages : 1022
PDF Size : 28 Mb
Book Description:
Computational fluid dynamics (CFD) techniques are used to study and solve fluid flow and heat transfer problems. Computational Fluid Dynamics by T.J. Chung book ranges from elementary concepts for the beginner to state of the art CFD for the practitioner. It discusses and illustrates the basic principles of finite difference, finite element and finite volume methods, with step-by-step hand calculations. Chapters go on to examine structured and unstructured grids, adaptive methods, computing techniques, and parallel processing. Finally, the author describes a variety of practical applications to problems in turbulence, reacting flows and combustion, acoustics, combined mode radiative heat transfer, multiphase flows, electromagnetic fields, and relativistic astrophysical flows. Students and practitioners particularly in mechanical, aerospace, chemical and civil engineering branch will be helpful for them to learn and apply numerical techniques to the solution of fluid dynamics problems.
Preface page
PART ONE: PRELIMINARIES
1. Introduction
2. Governing Equations
PART TWO: FINITE DIFFERENCE METHODS
3. Derivation of Finite Difference Equations
4. Solution Methods of Finite Difference Equations
5. Incompressible Viscous Flows via Finite Difference Methods
6. Compressible Flows via Finite Difference Methods
7. Finite Volume Methods via Finite Difference Methods
PART THREE: FINITE ELEMENT METHODS
8. Introduction to Finite Element Methods
9. Finite Element Interpolation Functions
10. Linear Problems
11. Nonlinear Problems/Convection-Dominated Flows
12. Incompressible Viscous Flows via Finite Element Methods
13. Compressible Flows via Finite Element Methods
14. Miscellaneous Weighted Residual Methods
15. Finite Volume Methods via Finite Element Methods
16. Relationships between Finite Differences and Finite Elements and Other Methods
PART FOUR: AUTOMATIC GRID GENERATION, ADAPTIVE METHODS AND COMPUTING TECHNIQUES
17. Structured Grid Generation
18. Unstructured Grid Generation
20. Computing Techniques
PART FIVE: APPLICATIONS
21. Applications to Turbulence
22. Applications to Chemically Reactive Flows and Combustion
23. Applications to Acoustics
24. Applications to Combined Mode Radiative Heat Transfer
25. Applications to Multiphase Flows
26. Applications to Electromagnetic Flows
27. Applications to Relativistic Astrophysical Flows
APPENDIXES
Appendix A: Three-Dimensional Flux Jacobians
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# Now What? A Math Tale
Author: Robbie H. Harris 2 5 candlewick.com 2019
Now What? A Math Tale, written by Robie H. Harris and illustrated by Chris Chatterton, is a short, hardback picture book. The book is published by Candlewick Press and is written for children between the ages of 2 and 5, much like Crash! Boom! A Math Tale (also by Harris and Chatterton). This story is about a tired puppy that wants to build a bed with blocks. While the puppy is building the bed, he notices a problem. There are not enough rectangular blocks to build a bed that fits him! Maybe he can use blocks with different shapes? Adjusting them in a way to fit all the blocks together should make a nice bed for a puppy his size.
The story’s narrative is a short, cute demonstration of how engineering thinking skills, such as spatial reasoning, mathematical skills, problem solving, analysis, and solution iteration, can be applied to simple problems. Spatial reasoning skills are expressed by how the puppy rearranges the blocks so that he can build a bed. The mathematical skills are the core of this book since, while enjoying the story, the child learns about basic geometry. For example, the puppy explains how a rectangle has 4 corners, 4 sides, and any way the block is turned it stays a rectangle. Then, the puppy solves problems throughout the whole story by coming up with different styles of block to make up a rectangle. For instance, instead of using a rectangle, he used two squares. After solving the problem, he analyzed his solution and realized that he can use other blocks to improve his bed further. He noticed one rectangle and two squares together were too skinny for him, and since he didn’t have any more squares or rectangles, he had to find another way to make the bed wider.
Now What? A Math Tale has a simple, easy-to-follow story and beautiful drawings that helps explain geometry concepts in a fun way, helping children develop engineering thinking and design skills.
Engineering thinking and design practices the gift encourages children to do or learn about:
Define a problem, Learn about the problem, Idea generation, Test the solution, Analyze a solution, Make improvements to the solution, Recognize patterns, Apply mathematical skills
Engineering text or context explicitly provided by the gift:
A problem to be solved by developing a new or improved object, tool, or process
Creative thinking, Problem Solving, Perseverance, Iterative design, Logical thinking
Overall ratings:
• Engineering & STEM Experts Reviews
• Rating: 4.0 out of 5.0
• Feedback:
• "It will entertain the child and at the same time it will introduce him to geometrical shapes."
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# Why Thomas Algorithm for tridiagonal systems?
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June 29, 1999, 08:23 Why Thomas Algorithm for tridiagonal systems? #1 Yogesh Talekar Guest Posts: n/a Sponsored Links Why onlt Thomas' alorithm is used for solving tri-diagonal system and Why not GAUSS elimination which is much simpler? IS there anything in literature.
June 29, 1999, 10:21 Re: Why Thomas Algorithm for tridiagonal systems? #2 Jim Park Guest Posts: n/a Am I forgetting something (entirely possible!)? Isn't the Thomas algorithm just a degenerate Gauss elimination operating on a tridiagonal coefficient matrix?
June 29, 1999, 10:52 Re: Why Thomas Algorithm for tridiagonal systems? #3 John C. Chien Guest Posts: n/a (1). There are a couple of books you can read. (2). "Numerical Methods fo Engineering Application" by Joel H. Ferzier, published by John Wiley & Sons 1981. ISBN-0-471-06336-3 (3). "Applied Iterative Methods" by Louis A. Hageman & David M. Young, published by Academic Press 1981,ISBN 0-12-313340-8. (4). Thomas algorithm is used to solve the tri-diagonal system of equations as part of line implicit method. There are also many point iterative methods available. Sometimes, I just use point S.O.R. method for easy programming. It is faster to use line implicit method when the total number of mesh points becomes large. Try to read the Ferzier's book first. (There are also other books by David Young).
June 29, 1999, 20:04 Re: Why Thomas Algorithm for tridiagonal systems? #4 Duane Baker Guest Posts: n/a Correction: Ferziger NOT Ferzier! good book!
June 30, 1999, 07:05 Re: Why Thomas Algorithm for tridiagonal systems? #5 John C. Chien Guest Posts: n/a (1). You are right. But it looks smoother and easier to write this way. (2). We will have to convince the author to change the name. Or we will give the author another name called "fuzzy" , "fuzzy joe". (3). Don't try to read between the keyboard characters.
June 30, 1999, 20:01 Re: Why Thomas Algorithm for tridiagonal systems? #6 T.J. Wanat Guest Posts: n/a The Thomas alogorithm is a specialized form of Gauss elimination that is faster and more accurate (you reduce the total number of operations, thus reducing the truncation and round-off error that accumulates for large systems). I recommend using it whenever possible.
July 15, 1999, 13:56 Re: Why Thomas Algorithm for tridiagonal systems? #7 Edward C. Chan Guest Posts: n/a Thomas' Algorithm is a special case of Gaussian elimination where the incident matrix consists of only the diagonal, and its immediate east and west neighbours (when the problem is 1D.) As with Gaussian elimination, it is an exact solver and it takes very little time to implement and execute. If you need the code, I can give it to you (in C++) Cheers, Ed.
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# Infinite Geometric Series and Limits
Infinite geometric series
Consider the series
nk=1 (2⋅½k-1) =2+1+½+¼+⅛+⋯
Consider also finding the partial sums for 10, 20 and 100 terms. The sums we are looking for are the partial sums of a geometric series. So,
As the number of terms increases, the partial sum appears to be approaching the number 4. This is no coincidence. In the language of limits,
This type of problem allows us to extend the usual concept of a ‘sum’ of a finite number of terms to make sense of sums in which an infinite number of terms is involved. Such series are called infinite series.
Limits
You may have noticed that in some geometric sequences, the later the term in the sequence, the closer the value is to 0. Another way to describe this is that as n increases, an approaches 0. The Value that the terms of a sequence approach, in this case 0, is called the limit of the sequence. Other types of infinite sequences may also have limits. If the terms of a sequence do not approach a unique value, we say that the limit of the sequence does not exist.
Example 1. Write the series
¼+⅛+1/16+1/32+⋯ for n terms
in sigma notation. Notice that the instructions specify that the lower limit of summation should be 1 and the variable k. So that would be a good thing to adhere to.
Solution:
Geometric, with a ratio of ½. Same deal as before; I need a formula for ak.
an=a1rn-1=¼⋅½k-1
Now, this one actually will simplify if I’d like it to, since ¼ is a power of ½. Check this out:
¼⋅½k-12⋅½k-12+k-1k+1.
So there are two different nice ways to write this.
nk=1 (¼⋅½k-1) =∑nk=1 ½k+1
One thing to be made clear about infinite series is that they are not true sums! The associative property of addition of real numbers allows us to extend the definition of the sum of two numbers, such as a+b, to three or four or n numbers, but not to an infinite number of numbers. For example, you can add any specific number of 5s together and get a real number, but if you add an infinite number of 5s together, you cannot get a real number! The remarkable thing about infinite series is that, in some cases, such as the example above, the sequence of partial sums (which are true sums) approach a finite limit L. The limit in our example is 4. This we write as
limn→∞ ∑nk=1 ak =limn→∞ (a1+a2+⋯+an)=L.
We say that the series converges to L, and it is convenient to define L as the sum of the infinite series. We use the notation
k=1 ak =limn→∞ ∑nk=1 ak =L.
We can, therefore, write the limit above as
k=1 (2⋅½k-1) =limn→∞ ∑nk=1 (2⋅½k-1) =4.
If the series does not have a limit, it diverges and does not have a sum. We are now ready to develop a general rule for infinite geometric series. As you know, the sum of the geometric series is given by
We will call this the sum of the infinite geometric series. In all other cases the series diverges. The proof is left as an exercise.
, as already shown.
Sum of an infinite geometric series
The sum, S, of an infinite geometric series with first term a1, such that the common ratio r satisfies the condition |r|<1 is given by:
Looking Ahead to Calculus: Infinite Series
As indicated above, each sequence {an} is associated with a sequence of partial sums {Sn}, where Sn=a1+a2+⋯+an. What happens to Sn as n gets larger and larger, that is, as we add more and more terms? We are considering an “infinite sum” written as a1+a2+a3+⋯, or in summation notation,
n=1 an
This is called an infinite series.
Since we cannot add an infinite set of numbers, we need instead the notion of a limit. In one sense, calculus is the study of limits. It is beyond the scope of this book to deal with infinite series in general, but for a geometric sequence {an}, we can at least get an intuitive feeling for what happens to Sn as n becomes large. For example, where
it is reasonable to assume that 1/2n gets close to 0 as n becomes large. In calculus notation limn→∞ 1/2n=0, from which limn→∞Sn=⅔.
We say that the infinite series,
converges to ⅔, and we write
In general, we associate each geometric sequence {a⋅rn-1} with an infinite geometric series
n=1 a⋅rn-1=a+a⋅r+a⋅r2+⋯+a⋅rn-1+⋯.
The only meaning we give to this infinite sum is the limit of the sequence of partial sums,
which depends on limn→∞rn. Looking at different values of r, we conclude that if r is any number between -1 and 1, then limn→∞rn=0, from which
Infinite geometric series
Associated with every geometric sequence { a⋅rn-1} is an infinite geometric series
n=1 a⋅rn-1=a+a⋅r+a⋅r2+⋯+a⋅rn-1+⋯.
If -1<r<1, then the series converges to a/(1-r), and we write
If |r|≥1, then the infinite series does not have a sum, and it diverges.
Example 2. Find the sum
Solution:
We substitute k=1 into the formula 13/100k and add successive terms until we reach k=4.
Equation-A. Sum of Geometric Sequence:
• The sum S of the first n terms of a geometric sequence ak=a⋅rk-1 for k≥1 is
We close this section with a peek into Calculus by considering infinite sums, called series. Consider the number 0.9. We can write this number as
0.9=0.9999…=0.9+0.09+0.009+0.0009+…
From Example 2, we know we can write the sum of the first n of these terms as
Using Equation-A, we have
It stands to reason that 0.9 is the same value of
Our knowledge of any exponential expressions tells us that
We have just argued that 0.9=1, which may cause some distress for some readers.7 Any non-terminating decimal can be thought of as an infinite sum whose denominators are the powers of 10, so the phenomenon of adding up infinitely many terms and arriving at a finite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series.
💎 Let you read a next post, how to express a recurring decimal as a fraction ?
Theorem-A. Geometric Series: Given the sequence ak=a⋅rk-1 for k≥1, where |r|<1,
If |r|≥1, the sum a+a⋅r+a⋅r2+⋯ is not defined.
The justification of the result in Theorem-A comes from taking the formula in Equation-A for the sum of the first n terms of a geometric sequence and examining the formula as n→∞,. Assuming |r|<1 means -1<r<1, so rn→0 as n→∞. Hence as n→∞,
As to what goes wrong when |r|≥1, we leave that to Calculus as well, but will explore some cases in the exercises.
_____
7 To make this more palatable, it is usually accepted that 0.3=⅓ so that 0.9=3(0.3)=3(⅓)=1. Feel better?
💎 A simpler thought than this page is the next page of Sigma Notation Examples about Infinite Geometric Series.
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# A body cools from 90°C to 80°C in 5 minutes. Under the same external conditions to cool from 80°C to 70°C the body will take
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1. 5 minutes
2. 4 minutes
3. 2.5 minutes
4. More than 5 minutes
Option 4 : More than 5 minutes
Free
ISRO Scientist ME 2020 Paper
1652
80 Questions 240 Marks 90 Mins
## Detailed Solution
Concept:
Heat transfer rate in transient heat transfer
$$\frac{{{T_{io}} - {T_\infty }}}{{T - {T_\infty }}} = {e^{\frac{{hA}}{{\rho V{C_p}}}t}}$$
Here h, A, ρ, V, and Cp are constant.
So, $$\frac{{{\rm{hA}}}}{{\rho V{c_p}}} = B\;$$
Now $$\frac{{{T_{io}} - {T_\infty }}}{{T - {T_\infty }}} = {e^{Bt}}$$
Here Tio is the initial temperature at t = 0 and T is temperature of body at any instant.
$$\begin{array}{l} \frac{{90 - {T_\infty }}}{{80 - {T_\infty }}} = {e^{B \times 300}}\\ \frac{{80 - {T_\infty }}}{{70 - {T_\infty }}} = {e^{Bt}} \end{array}$$
Let T = 0° C
$$\begin{array}{l} \frac{9}{8} = {e^{300\;B}}\\ \frac{{\ln \left( {\frac{9}{8}} \right)}}{{300}} = B\\ \frac{8}{7} = {e^{Bt}} \end{array}$$
B = 3.92 × 10-4
$$\begin{array}{l} \ln \left( {\frac{8}{7}} \right) = Bt\\ \Rightarrow \ln \left( {\frac{8}{7}} \right) = \frac{{\ln \left( {\frac{9}{8}} \right) \times t}}{{300}}\\ t = \frac{{\ln \left( {\frac{8}{7}} \right)}}{{\ln \left( {\frac{9}{8}} \right)}} \times 300 = 340\;sec > 5~minutes \end{array}$$
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http://mathoverflow.net/questions/45441/gelfand-tsetlin-bases-for-lie-groups-over-finite-fields
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# Gelfand-Tsetlin bases for Lie groups over finite fields
There is a method of constructing representations of classical Lie algebras via Gelfand-Tsetlin bases. It has also been applied to Symmetric groups by Vershik and Okounkov. Does anybody know of any application of the method to complex representations of $GL_n(\mathbb F_q)$? Or, at least, any results in this directions, like what is the centralizer of $GL_{n-1}$ in $\mathbb C[GL_n]$?
-
I don't have a precise answer, but keep in mind the Schur-Weyl duality between representations of general linear and symmetric groups. Whatever is done for the latter is likely to apply to the former, even over finite fields. In another direction, Dan Rockmore and colleagues have done quite a bit of practical work on fast Fourier transforms for finite linear groups, some with connections to Gelfand-Tsetlin basis methods. – Jim Humphreys Nov 9 '10 at 15:16
Maybe you should have a look at arXiv:0705.3605v1 [math.RT] (by Vershik and Kerov). This applies some approaches that worked successfully for the infinite symmetric group to the case of limits of finite linear groups. The constuction in the case of symmetric groups has much to do with GTs bases, as I remember. – Leonid Petrov Nov 9 '10 at 20:24
My earlier comment was not at all well-focused. After more thought, I'm inclined to be pessimistic about using a Gelfand-Tsetlin approach here (even if it has some success for symmetric groups). Though of course it would be interesting to be proven wrong.
As Matt Davis reminds me, my offhand reference to Schur-Weyl duality is not helpful here since the work of Benson, Doty, and others deals mainly with the representations of various groups over fields of prime characteristic. (See especially Doty's papers on arXiv.) Irreducible representations of finite general linear groups over $\mathbb{C}$ are very difficult to construct directly and have very little in common with the finite dimensional representations of general linear groups or their Lie algebras in characteristic 0. Instead, the theory imitates more closely the infinite dimensional Harish-Chandra approach to Lie group representations in which parabolic induction is exploited together with a study of "discrete series".
J.A. Green's 1955 TAMS paper followed somewhat this pattern in developing combinatorially the character theory of finite general linear groups. But there is little insight here into constructing the elusive discrete series characters; instead orthogonality relations and the like are exploited. The best approach to an actual construction of discrete series representations was given in Lusztig's 1974 Annals of Mathematics Studies No. 81. Soon after that, Deligne and Lusztig pioneered a more sophisticated method for constructing generalized characters of arbitrary finite groups of Lie type. This has become the dominant influence in the subject, since Lusztig's earlier techniques don't go far enough beyond the finite general linear case.
-
Thank you. I am now also pessimistic about the Gelfand-Tseitlin approach. What would be a good introduction to the Deligne-Lusztig theory? – Roman Fedorov Dec 21 '10 at 17:57
There are two detailed textbook treatments: Carter (Wiley Interscience, 1985), extensive but down-to-earth; Digne & Michel (Cambridge, 1991), more concise and sophisticated. Carter has also written useful surveys, including one translated into Russian: MR1170353 (93j:20034), Karter, R.U. [Carter, Roger William] (4-WARW), Representation theory of finite groups of Lie type over an algebraically closed field of characteristic zero. Translated from the English by N. A. Vavilov. Algebra, 9 (Russian), 5–143, 268, Itogi Nauki i Tekhniki, Vsesoyuz. – Jim Humphreys Dec 21 '10 at 18:42
Not an answer either, but in response to Jim - Schur-Weyl duality doesn't always apply over finite fields. See http://www.ams.org/mathscinet-getitem?mr=2563588 for one result and some discussion of the related issues.
-
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https://acm.timus.ru/forum/thread.aspx?id=22828&upd=633831622164363750
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ENG RUS Timus Online Judge
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## Discussion of Problem 1402. Cocktails
WA#13 WHY?
Posted by snowfly 14 Jul 2009 09:56
#include<iostream>
using namespace std;
int fac[100000],ans[100000],len,alen;
int main (void){
int n;cin>>n;
alen=1;
memset(ans,0,sizeof(ans));
for (int i=2;i<=n;++i){
memset(fac,0,sizeof(fac));
len=1;fac[0]=1;
for (int j=n;j>n-i;--j){
for (int k=0;k<len;++k)
fac[k]*=j;
for (int k=0;k<len;++k)
if (fac[k]>=10){
fac[k+1]+=fac[k]/10;
fac[k]%=10;
}
if (fac[len]!=0){
++len;
if (fac[len-1]>=10){
fac[len]=fac[len-1]/10;
fac[len]%=10;
++len;
}
}
}
for (int j=0;j<min(alen,len);++j)
ans[j]+=fac[j];
if (alen<len){
for (int j=alen;j<len;++j)
ans[j]=fac[j];
alen=len;
}
for (int j=0;j<alen;++j)
if (ans[j]>9){
ans[j+1]+=ans[j]/10;
ans[j]%=10;
}
while (ans[alen]!=0){
++alen;
ans[alen]+=ans[alen-1]/10;
ans[alen-1]%=10;
}
}
for (int i=alen-1;i>=0;--i)
cout<<ans[i];
cout<<endl;
return 0;
}
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https://educationexpert.net/engineering/2122532.html
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11 June, 13:42
# Determine the flow velocities at the inlet and exit sections of aninclined tapering pipe using fluid flow theory and given pressurereadings and flow rates.There is a sloping pipeline that has one end 1.35 m higher than theother. The pipe section tapers from 0.95 m diameter at the top end to0.44m diameter at the lower end. The difference in pressure betweenthe two sections is 12.35kPa, with pressure being greater at higherlevel.Your task is to determine the inlet and exit velocities and thevolumeflow rate of the inclined pipe.
+2
1. 11 June, 14:20
0
The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s
Explanation:
Using Bernoulli's equation
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²
P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²
ΔP + ρgΔh = 1/2ρ (v₂² - v₁²) (1)
where ΔP = pressure difference = 12.35 kPa = 12350 Pa
Δh = height difference = 1.35 m
From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.
Substituting v₁ into (1), we have
ΔP + ρgΔh = 1/2ρ (v₂² - (d₂²v₂/d₁²) ²)
ΔP + ρgΔh = 1/2ρ (v₂² - (d₂/d₁) ⁴v₂²)
v₂ = √[2 (ΔP + ρgΔh) / ρ (1 - (d₂/d₁) ⁴) }
substituting the values of the variables, we have
v₂ = √[2 (12350 Pa + 1000 kg/m³ * 9.8 m/s² * 1.35 m) / (1000 kg/m³ (1 - (0.44 m/0.95 m) ⁴)) }
= √[2 (12350 Pa + 13230 Pa) / (1000 kg/m³ * 0.954) ]
= √[2 (25580 Pa) / 954 kg/m³]
= √[51160 Pa/954 kg/m³]
= √53.627
= 7.32 m/s
v₁ = d₂²v₂/d₁²
= (0.44 m/0.95 m) ² * 7.32 m/s
= (0.954) ² * 7.32 m/s
= 6.66 m/s
The volume flow rate Q = A₁v₁
= πd₁²v₁/4
= π (0.95 m) ² * 6.66 m/s : 4
= 18.883 m³/s : 4
= 4.72 m³/s
So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s
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https://didactalia.net/comunidad/materialeducativo/recursos/tag/net
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## Net Children Go Mobile: los niños se inician en In...
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## CervanTest (juego interactivo)
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## Volume of Prisms Using Unit Cube? At grade
Use unit cubes to demonstrate the volume of a prism.
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Use a formula and given information to find the surface area of cones and truncated cones.
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You are Here: Home >< Maths
# Edexcel C3,C4 June 2013 Thread watch
1. (Original post by PoorLoser)
AB = OB - OA
so (0,4,1) - (9,-8,2)
Oh my.
can't tell between an i or a j
I am such a moron, hope I don't make mistakes like this tomorrow
Thanks !
2. What shapes do we need to know the area and volumes of?
3. (Original post by Jullith)
Hope this helps:
Attachment 227383
thanks! that made it a whole lot easier!
4. If the C4 paper was leaked, does anyone know where I could find it? Because I'd really like to attempt it as I'm pretty sure the replacement paper this Teusday isn't going to be anything like the past papers I've been doing...especially not after C3!
Posted from TSR Mobile
5. (Original post by sahjan)
If the C4 paper was leaked, does anyone know where I could find it? Because I'd really like to attempt it as I'm pretty sure the replacement paper this Teusday isn't going to be anything like the past papers I've been doing...especially not after C3!
Posted from TSR Mobile
i don't think the c4 paper has been leaked
6. (Original post by SuziieB)
...I think you're on your own there, mate...
Qui? this paper i supposed to be (hard i know) but they cant firetruck us up in this paper cuz they made c3 as hard as firtruck :3
7. (Original post by «WD»ddd)
Qui? this paper i supposed to be (hard i know) but they cant firetruck us up in this paper cuz they made c3 as hard as firtruck :3
we can only hope!
Edexcel can do what they want, I wouldn't be surprised if C4 is also hard tbh :/
8. -sintsin3t = cos2t-cos4t? hooow?
9. (Original post by hassassin04)
Those who want challenging vector questions- check AEA.
Those questions would just invoke more panic into the students, I'm just hoping a pleasant vectors question appears so it doesn't reduce my mark as I'm not very strong with my vectors
10. (Original post by Sunil_karan)
i recenly did a solomon intergration question where ln(something) became negative but they took it as positive because they used modulus, i heard in c4 exam conditions they wont do this?
I may be wrong, but I certainly thought they could ask us a question like that! Worth knowing just in case
11. (Original post by Zaphod77)
I may be wrong, but I certainly thought they could ask us a question like that! Worth knowing just in case
can you give an example of a question of this type?
12. (Original post by MathsNerd1)
Those questions would just invoke more panic into the students, I'm just hoping a pleasant vectors question appears so it doesn't reduce my mark as I'm not very strong with my vectors
I'm pretty sure I've missed something but at this point it's a little too late to worry . I'll probs peruse maths 247s videos this afternoon to see if anything's missing in my arsenal technique then I'll do specific hard Qs
13. (Original post by silentriver)
-sintsin3t = cos2t-cos4t? hooow?
Factor formulae from C3
14. (Original post by posthumus)
Factor formulae from C3
yeah i tried that, but im doing something wrong. need the steps!
15. (Original post by silentriver)
yeah i tried that, but im doing something wrong. need the steps!
-2sin(P+Q/2)sin(P-Q/2) = cos(P)-cos(Q)
P+Q/2=t
P+Q=2t
P-Q=6t
2P=8t
P=4t
Q=-2t
1/2[cos(4t) - cos(-2t)]
Is this what you got ?
16. (Original post by keromedic)
I'm pretty sure I've missed something but at this point it's a little too late to worry . I'll probs peruse maths 247s videos this afternoon to see if anything's missing in my arsenal technique then I'll do specific hard Qs
I've just been doing some lovely integration questions last night and hoping to god a nice vectors question comes up, otherwise I'm going to completely fail this!
17. Any predictions for c4
Posted from TSR Mobile
18. (Original post by posthumus)
-2sin(P+Q/2)sin(P-Q/2) = cos(P)-cos(Q)
P+Q/2=t
P+Q=2t
P-Q=6t
2P=8t
P=4t
Q=-2t
1/2[cos(4t) - cos(-2t)]
Is this what you got ?
Yes. thank you so much! looks like i switched the + and - signs. (As usual) -______-
19. Hey, On the C4 June 2012 paper Q)4) on differential equations, why cant you move the 3 and the y? Is there any rules or tips to avoid these mistakes?
20. http://www.edexcel.com/migrationdocu...s_20120816.pdf
quick question , on question 4)
if I separated the variables and got
integral y/3 dy = integral 1/cos^2 x dx
is this wrong? or will it lead to the same answer
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https://nursinghubwriters.com/a-plane-flies-horizontally-at-an-altitude-of-5-km-and-passes-directly-over-a-tracking-telescope-on-the-ground-when-the-angle-of-elevation-is-%CF%80-4-this-angle-is-decreasing-at-a-rate-of-%CF%80-3-r/
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# A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. when the angle of elevation is π/4, this angle is decreasing at a rate of π/3 radians p
A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. when the angle of elevation is π/4, this angle is decreasing at a rate of π/3 radians per minute. how fast is the plane traveling at that time?
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# foot, metric (long) to pace conversion
Conversion number between foot, metric (long) [lmf] and pace is 0.43744531933508. This means, that foot, metric (long) is smaller unit than pace.
### Contents [show][hide]
Switch to reverse conversion:
from pace to foot, metric (long) conversion
### Enter the number in foot, metric (long):
Decimal Fraction Exponential Expression
[lmf]
eg.: 10.12345 or 1.123e5
Result in pace
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 foot, metric (long) = (exactly) ((1/3)) / (0.762) = 0.43744531933508 pace
• 1 pace = (exactly) (0.762) / ((1/3)) = 2.286 foot, metric (long)
• ? foot, metric (long) × ((1/3) ("m"/"foot, metric (long)")) / (0.762 ("m"/"pace")) = ? pace
### High precision conversion
If conversion between foot, metric (long) to metre and metre to pace is exactly definied, high precision conversion from foot, metric (long) to pace is enabled.
Decimal places: (0-800)
foot, metric (long)
Result in pace:
?
### foot, metric (long) to pace conversion chart
Start value: [foot, metric (long)] Step size [foot, metric (long)] How many lines? (max 100)
visual:
foot, metric (long)pace
00
104.3744531933508
208.7489063867017
3013.123359580052
4017.497812773403
5021.872265966754
6026.246719160105
7030.621172353456
8034.995625546807
9039.370078740157
10043.744531933508
11048.118985126859
Copy to Excel
## Multiple conversion
Enter numbers in foot, metric (long) and click convert button.
One number per line.
Converted numbers in pace:
Click to select all
## Details about foot, metric (long) and pace units:
Convert Foot, metric (long) to other unit:
### foot, metric (long)
Definition of foot, metric (long) unit: ≡ 1⁄3 m .
Convert Pace to other unit:
### pace
Definition of pace unit: ≡ 2.5 ft.
← Back to Length units
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https://forum.allaboutcircuits.com/threads/pcm-spectral-efficiency.26589/
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# PCM spectral efficiency
Discussion in 'Wireless & RF Design' started by pregnant_joe, Aug 11, 2009.
1. ### pregnant_joe Thread Starter New Member
Aug 11, 2009
1
0
Hello all..
Im a newbie..I have question to ask for your favour..im studying abroad, i left my notes there and im in my home country..Please help me..
Below are the questions:
1. A baseband transmission channel with a raised cosine frequency response with a roll-off factor of 0.4. The channel has an absolute bandwidth of 1200 kHz. An analogue signal is converted into binary PCM with 64 level quantization before being transmitted over the channel. State the maximum limit on the bandwidth of the analogue signal?
________________________________________
2. 25 input signals, each band-limited to 3.3kHz are each sampled at an 8kHz rate and then time division multiplexed. Calculate the minimum bandwidth required to transmit this multiplexed signal in the presence of noise if the modulation used is:
i)PAM
ii)Quantised PPM with a required level resolution of 5%
iii)Binary PCM with a required level resolution of >0.5%
*Please orrect me if i'm wrong..what i understand from the question is
Bandwidth = 3.3kHz
fs = 8kHz
Signal channel = 25
Rs = 8 * 25 = 200kHz
i know that H = log2 M but i can't see how to find the answers.
im sure that i have to use n=(R*H)/B but i'm stuck.
________________________________________________________
3. A DSB-AM transmitter delivers a mean output power of 33W when a modulating signal comprising a signle tone is applied to it. If the power contained in one sideband is 4W, calculate
i)the carrier power
ii)The modulation index
iii)the peak voltage of the AM signal driving the transmitter antenna if the resistance of the lates is 50ohm
note: Pc=Pt-2Ps
______________________________________________________
thank you for you concern!!..
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https://www.thestudentroom.co.uk/showthread.php?t=2216873
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You are Here: Home
# Chemical reaction equation for a blend of Ethanol and Petrol watch
1. Hi,
I am struggling with a question that requires the chemical reaction of a combustion of a blend of Ethanol and Petrol. Below are the chemical equations of the cumbustion of Petrol and Ethanol on their own and I need to know the equation of the blend, containing %50 Ethanol and %50 Petrol.
Ethanol: C2H5OH + 3O2 -----> 2CO2 + 3H2O
Petrol: C8H18 + 12 1/2O2 -----> 8CO2 + 9H2O
2. (Original post by mech_eng)
Hi,
I am struggling with a question that requires the chemical reaction of a combustion of a blend of Ethanol and Petrol. Below are the chemical equations of the cumbustion of Petrol and Ethanol on their own and I need to know the equation of the blend, containing %50 Ethanol and %50 Petrol.
Ethanol: C2H5OH + 3O2 -----> 2CO2 + 3H2O
Petrol: C8H18 + 12 1/2O2 -----> 8CO2 + 9H2O
50% by mass or volume?
3. never seen a question like this.
I would solve like empirical formula problem
if 50% by molecular weight then
50/(24+5+16+1)= 1.08 mol
50/((8*12) +18) = 0.44mol 0.44/0.44=1
1.08/0.44= 2.45
double mols to get 2 mols of petrol and 5 mols of ethanol
make equation 5C8H18+ 2C2H5OH +O2 -------> CO2 + H2O
balance it 5C8H18+ 2C2H5OH + 67 1/2O2 -------> 44CO2 + 51H2O
i maybe very very wrong!!!!!!
you may just have to make into one equation.
4. (Original post by charco)
50% by mass or volume?
By the mass
5. (Original post by mech_eng)
By the mass
If it is by mass, then all you need to do is convert this into a molar ratio.
Then you can have two simultaneous basic combustion reactions (stoichiometrically adjusted according to the molar ratios). The actual mechanism may well not be so simplistic, but the end products will almost certainly be the individual combustion products.
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Graphs of Functions of Two Variables and Contour
52 – Reference – Graphs of eight basic types of functions
1. In the simplest case, a variable is drawn as a function of another, typically using rectangular axes; see plot (graphics) for details..
2. In the left of the links for the full Chapter and E-Book from the page you are on (if applicable), and links for notes, practice problems solutions to the practice problems and assignment problems.
3. A rectangular axes with two perpendicular lines, is known as coordinate axes and the point of intersection of these lines is known as the origin.
4. If the function input x is an ordered pair ( x 1, x 2 ) of real numbers, the graph is the collection of all ordered triples ( x 1, x 2, f ( x 1, x 2 )), and for a continuous function is a surface.
5. Once you have made your selection from this second menu, up to four connections (depending on whether or not the practice, and assignment problems are show is available for the page), to under the second menu, click to initiate the download.
6. The variable that we assign the value we call the independent variable and the other variable is the dependent variable because it is dependent on the value of the independent variable.
7. And remember if you ordered any problems with the drawing of the graph from the transformed pairs, just more points from the original graph to map to the new one.
8. to be included The links for the page you are on will be highlighted, so that you can find them easily.
If m is the slope is negative the functions value decreases with increasing x and the opposite, if we have a positive slope.
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https://www.sparknotes.com/math/algebra1/polynomials/section4/
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### Square of a Binomial
To square a binomial, multiply the binomial by itself:
(a + b)2 = (a + b)(a + b)
(a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 = a2 + ab + ab + b2 = a2 +2ab + b2
The square of a binomial is always the sum of:
1. The first term squared,
2. 2 times the product of the first and second terms, and
3. the second term squared.
When a binomial is squared, the resulting trinomial is called a perfect square trinomial.
Examples:
(x + 5)2 = x2 +2(x)(5) + 52 = x2 + 10x + 25
(100 - 1)2 = 1002 +2(100)(- 1) + (- 1)2 = 10000 - 200 + 1 = 9801
(2x - 3y)2 = (2x)2 +2(2x)(- 3y) + (- 3y)2 = 4x2 -12xy + 9y2
### Product of the Sum and Difference of Two Terms
When we multiply two polynomials that are the sum and difference of the same 2 terms -- (x + 5) and (x - 5) for example -- we get an interesting result:
(a + b)(a - b) = a(a) + a(- b) + ba + b(- b) = a2 - ab + ab - b2 = a2 - b2
The product of the sum and difference of the same two terms is always the difference of two squares; it is the first term squared minus the second term squared. Thus, this resulting binomial is called a difference of squares.
Examples:
(7 - 2)(7 + 2) = 72 -22 = 49 - 4 = 45
(x + 9)(x - 9) = x2 -92 = x2 - 81
(2x - y)(2x + y) = (2x)2 - y2 = 4x2 - y2
(3x2 -2)(3x2 +2) = (3x2)2 -22 = 9x4 - 4
(- y + 5x)(- y - 5x) = (- y)2 - (5x)2 = y2 -15x2
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https://math.stackexchange.com/questions/1005960/number-of-straight-line-through-point-forming-same-area-with-axes
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# Number of straight line through point forming same area with axes
I am given a point $P(2,3)$ thru which passing line forms triangle with axes of area $12$ , so how many lines will pass thru $P$ making same area with axes?
Writing intercept form of line
$$\frac{x}{a} + \frac{y}{b}=1$$ and then satisfying point in this equation I get $$2b+3a=ab$$ , now since area is $12$ hence $$1/2ab=12$$ we get $ab=24$ now substituting $ab$ and $b$ i get $$a^2 -8a +14=0$$ . Now we have 2 values for $a & b$ hence 2 lines should pass thru that points satisfying condition but correct answer is 3 , what I did wrong?
You focused exclusively on a quadrant 1 triangle (from the assumption that $ab=24$. But you get solutions in the other quadrants if you look for solutions where $ab=-24$).
By the way for the first quadrant: From your equations $ab=24$ and $\frac2a+\frac3b=1$, you get $b=\frac{24}{a}$, which leads to $a^2-8a+16=0$. This quadratic has only one (double) root of $a=4$.
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| 4.0625
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https://numbermatics.com/n/80178/
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# 80178
## 80,178 is an even composite number composed of five prime numbers multiplied together.
What does the number 80178 look like?
This visualization shows the relationship between its 5 prime factors (large circles) and 32 divisors.
80178 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of thirty-two divisors.
## Prime factorization of 80178:
### 2 × 3 × 7 × 23 × 83
See below for interesting mathematical facts about the number 80178 from the Numbermatics database.
### Names of 80178
• Cardinal: 80178 can be written as Eighty thousand, one hundred seventy-eight.
### Scientific notation
• Scientific notation: 8.0178 × 104
### Factors of 80178
• Number of distinct prime factors ω(n): 5
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 118
### Divisors of 80178
• Number of divisors d(n): 32
• Complete list of divisors:
• Sum of all divisors σ(n): 193536
• Sum of proper divisors (its aliquot sum) s(n): 113358
• 80178 is an abundant number, because the sum of its proper divisors (113358) is greater than itself. Its abundance is 33180
### Bases of 80178
• Binary: 100111001001100102
• Base-36: 1PV6
### Squares and roots of 80178
• 80178 squared (801782) is 6428511684
• 80178 cubed (801783) is 515425209799752
• The square root of 80178 is 283.1572001555
• The cube root of 80178 is 43.1206275761
### Scales and comparisons
How big is 80178?
• 80,178 seconds is equal to 22 hours, 16 minutes, 18 seconds.
• To count from 1 to 80,178 would take you about twenty-two hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 80178 cubic inches would be around 3.6 feet tall.
### Recreational maths with 80178
• 80178 backwards is 87108
• The number of decimal digits it has is: 5
• The sum of 80178's digits is 24
• More coming soon!
MLA style:
"Number 80178 - Facts about the integer". Numbermatics.com. 2022. Web. 5 July 2022.
APA style:
Numbermatics. (2022). Number 80178 - Facts about the integer. Retrieved 5 July 2022, from https://numbermatics.com/n/80178/
Chicago style:
Numbermatics. 2022. "Number 80178 - Facts about the integer". https://numbermatics.com/n/80178/
The information we have on file for 80178 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 80178, math, Factors of 80178, curriculum, school, college, exams, university, Prime factorization of 80178, STEM, science, technology, engineering, physics, economics, calculator, eighty thousand, one hundred seventy-eight.
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