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https://communities.sas.com/t5/SAS-Procedures/Proc-Expand/td-p/142278?nobounce | 1,531,871,942,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00173.warc.gz | 612,070,178 | 29,495 | New Contributor
Posts: 2
# Proc Expand
Hello all,
I'm relatively new here and this is my first post on SAS forum. Hope I can get the help I need!
I'm doing a research project on index returns. I have daily data for a period of over 40 years. However, I do not have returns for Saturday and Sunday or any other holidays when stock market would be closed. I'm trying to convert this daily data into monthly using PROC EXPAND procedure. Here is my analysis:
- I used PROC EXPAND and found sum of the returns for each month. Subsequently, I calculated the same using excel (using pivot table). I found the results to be different. Is there any reason why this would happen?
- My data starts from January of 1973 until May 2014. When I used PROC EXPAND with missing values for Saturdays and Sundays, the resultant output didn't have any return value for January of 1973. So I tried to set all missing values to zero and voila, sum of return for January 1973 appeared. Is this a right approach?
proc expand data=lib.dly_1 from=day to=month;
convert dly_ret\observed=total;
id date;
by yr;
run;
- And also, I need to find Standard deviation for each month once converted from daily return using PROC EXPAND. Can I do this by moving standard deviation? Is there a way to do this using PROC EXPAND or any other procedure or way?
Karthik
Super User
Posts: 23,700
## Re: Proc Expand
You should post some sample code and data that result in the different numbers in SAS vs Excel.
New Contributor
Posts: 2
## Re: Proc Expand
Reeza,
My apologies. Here are my code and results:
Code:
data usd.dly_5(drop= lag_DTWEXM count lag_count numday);
set usd.dly_2;
if(dly_ret=".") then dly_ret=0;
run;
proc expand data=usd.dly_5 from=day to=month out=usd.dly_6;
convert dly_ret/observed=total;
id date;
by yr;
run;
Result using PROC EXPAND:
yr date dly_ret
1973 JAN1973 -0.002940597
1973 FEB1973 -0.023633396
1973 MAR1973 0.0081171767
1973 APR1973 -0.001223039
1973 MAY1973 -0.006476482
1973 JUN1973 -0.00692127
1973 JUL1973 -0.00099385
1973 AUG1973 0.0200460595
1973 SEP1973 0.0001108501
1973 OCT1973 0.0011257383
1973 NOV1973 0.0123855573
1973 DEC1973 0.0024655859
Results using Excel:
mth/yr dly_ret
11973 -0.00369745 21973 -0.0227457 31973 0.00824313 41973 -0.000343374 51973 -0.010354 61973 -0.0102187 71973 0.000427683 81973 0.0195624 91973 -0.00165992 101973 0.00217638 111973 0.0147721 121973 0.00364333
I have attached the daily raw data for the year 1973. Please let me know if you need anything else. Appreciate your help.
Karthik
Super User
Posts: 23,700
## Re: Proc Expand
So I apparently can't help much because my license for ETS is messed up and it'll take a while for IT to fix it.
That being said, an import of your data and proc means on it resulted in the same value in Excel and SAS of -0.003697452 for Jan 1973.
You haven't shown the rest of your code but my *GUESS* is that you haven't formatted the date in the way SAS expects it.
Try running a proc means on your data to see if the output matches your Excel.
This assumes that the date is a SAS date.
proc means data=input2 n sum;
format date monyy7.;
class date;
var dly_ret;
run;
Super User
Posts: 23,700 | 987 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-30 | latest | en | 0.803728 |
https://www.airmilescalculator.com/distance/hsn-to-kwe/ | 1,627,056,083,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00019.warc.gz | 656,245,589 | 37,487 | # Distance between Zhoushan (HSN) and Guiyang (KWE)
Flight distance from Zhoushan to Guiyang (Zhoushan Putuoshan Airport – Guiyang Longdongbao International Airport) is 976 miles / 1572 kilometers / 849 nautical miles. Estimated flight time is 2 hours 20 minutes.
Driving distance from Zhoushan (HSN) to Guiyang (KWE) is 1145 miles / 1843 kilometers and travel time by car is about 19 hours 25 minutes.
## Map of flight path and driving directions from Zhoushan to Guiyang.
Shortest flight path between Zhoushan Putuoshan Airport (HSN) and Guiyang Longdongbao International Airport (KWE).
## How far is Guiyang from Zhoushan?
There are several ways to calculate distances between Zhoushan and Guiyang. Here are two common methods:
Vincenty's formula (applied above)
• 976.488 miles
• 1571.506 kilometers
• 848.545 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
• 974.959 miles
• 1569.044 kilometers
• 847.216 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 Zhoushan Putuoshan Airport
City: Zhoushan
Country: China
IATA Code: HSN
ICAO Code: ZSZS
Coordinates: 29°56′3″N, 122°21′43″E
B Guiyang Longdongbao International Airport
City: Guiyang
Country: China
IATA Code: KWE
ICAO Code: ZUGY
Coordinates: 26°32′18″N, 106°48′3″E
## Time difference and current local times
There is no time difference between Zhoushan and Guiyang.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 149 kg (329 pounds).
## Frequent Flyer Miles Calculator
Zhoushan (HSN) → Guiyang (KWE).
Distance:
976
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
976
Round trip? | 517 | 1,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-31 | latest | en | 0.834532 |
https://www.nntp.perl.org/group/perl.perl6.users/2021/03/msg9802.html | 1,696,369,677,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00339.warc.gz | 1,003,398,453 | 2,246 | Front page | perl.perl6.users | Postings from March 2021
## Re: 'CALL-ME' Math problem?
From:
Wenzel P. P. Peppmeyer
Date:
March 13, 2021 20:16
Subject:
Re: 'CALL-ME' Math problem?
Message ID:
1a9477e8-9324-1304-e8a0-1c4cd79d6315@gmx.de
```On 02/03/2021 09:12, ToddAndMargo via perl6-users wrote:
> Math problem:
> x = 60÷5(7−5)
>
> raku -e 'say 60÷5(7−5)'
> No such method 'CALL-ME' for invocant of type 'Int'
> in block <unit> at -e line 1
>
> Seems raku does not like the (). How do I fix this
> and maintain the flow and look of the equation?
>
> -T
>
> The correct answer is 24
The correct answer can be found with the following code.
class SuperInt {
has \$.left-factor is rw;
has \$.right-factor is rw;
method new(\l, \r) {
my \SELF = self.CREATE;
SELF.left-factor = l;
SELF.right-factor = r;
SELF
}
}
multi sub infix:<÷>(Numeric:D \l, SuperInt:D \r) {
l ÷ r.left-factor * r.right-factor
}
Int.^add_method('CALL-ME', my method (\SELF: \v) { SuperInt.new(SELF, v) });
Int.^compose;
say 60÷5(7−5);
Thought, I do have the hunch that this might break with a slightly more
complex examle.
Have -Ofun
gfldex
``` | 400 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.723974 |
https://www.theshiponline.com/deal-or-no-deal-win-money/ | 1,675,923,446,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00833.warc.gz | 1,055,781,939 | 19,472 | # Deal Or No Deal Win Money
Deal Or No Deal Win Money – For those readers who are not familiar with the US television series Deal or No Deal, it works like this. The game has one opponent and 26 cases. Each case has a specific amount, ranging from \$0.01 to \$1,000,000. Of course, the competitor does not know how much money is involved in this case. The opponent chooses a case at the beginning of the game, that case is different from the others. The contestant then completes the other cases one by one. At key moments during the game, the player can make a (“deal”), end the game, or not (“no deal”) to continue clearing the charges. If at any time there are only two cases left, the player must select one from the box and win the corresponding amount (although he does not get the first “hill” offered).
An interesting game resulted in the contestant being left with only two cases: the one he chose and the other case. One case had \$1,000,000; Another dollar. He was given \$416,000 for “business”. The man said there was no deal, picked the wrong box and won \$1.
## Deal Or No Deal Win Money
Descriptive versus inferential statistics. What is the difference? – Consider this data-driven investor. You are a restaurant owner and you want to know how people perceive your business. So give it a go… www.datadriveninvestor.com
## Whoopi Goldberg Blasts Meghan’s ‘deal Or No Deal’ Remarks
It is not surprising that the selection of this competition without permission sparked a huge debate on the internet, some saying the man is greedy and others saying the man is rational.
More interestingly, some argue that a person must necessarily change cases: by sticking to the case they chose at the beginning of the game, they say they only get a 1 in 26 chance of winning. A particular situation has been compared to the famous Monty Hall crisis. Since there was so much disagreement, I decided to go public. In the remainder of this post, I will first perform an expected value analysis to discuss whether the no-deal option was reasonable. After this, I will show whether the change of cases was reasonable or not, given our particular situation.
So you can take \$416,000 or a 50/50 shot to be sure of getting \$1 or \$1,000,000. What should you do? However, the expected cash cost (multiplying the odds by the cash rewards) of choosing “no deal” is 0.50 * \$1 + 0.50 * \$1,000,000 = \$500,000.50, which is more than the \$416,000 deal. This suggests that “no deal” makes more sense than “a deal”, but we must consider the value that money adds to our competitor. For the average US citizen, I would say that the impact of \$416,000 on their life is greater than the impact of an additional \$1,000,000 compared to \$416,000. Therefore, it is not counted as expected value. It can be different if the person is already a millionaire, so the logical choice depends on the person’s situation.
Remember our scenario: there are 26 scenarios with different monetary values. At first you choose one of them. After 24 rounds there are 2 left: yours and 1 more charge. One of them has \$1, the other has a million dollars. Does switching cases make more sense than sticking with your case?
#### Vintage 2005 Crown & Andrews Deal Or No Deal Action Tv Board Game 9310281015251 9310281015251
First, it is important to calculate the exact odds in the game here. Others argued that the probability that the selected case would have a million dollars was only 50% because there were two cases left and the cases were randomly distributed. However, the odds were also said to be 1 in 26 in the favored case and 25 in 26 in the other case. You can see the logic: the game started with 26 cases, resulting in a 1 in 26 chance that the selected case contains millions of dollars and that 25 of the 26 cases are in one of the 25 cases. After finishing 24 of those 25, the one left should have a 25 out of 26 chance of containing millions, right?
The idea here is to draw an analogue to the Monty Hall dilemma, where a contestant is given a choice between three doors. Behind two of these doors is a goat; Behind one of the three is a car. After the player chooses one, the game master opens one of the other doors, behind which is a goat. It is noteworthy that the manager
Choose the door with the goat behind it. The host then asks if the contestant wants to change their decision and choose a different department. As it turns out, in this scenario, the contestant has a 2 out of 3 chance of winning if they switch cars and a 1 out of 3 chance if they don’t. Why? Since it originally had a 2 in 3 chance of being wrong, switching in this case guarantees a win and a 1 in 3 chance of being wrong.
So if change makes sense in the Monty Hall problem, then it makes sense in a defined purchase situation with or without a contract, right? Well, no. The Monty Hall dilemma analogy fails: In the Monty Hall dilemma, the host chooses a door to take out.
#### Deal Or No Deal Board Game Nbc Game Show Pressman. 2 To 6 Players .
. The Monty Hall problem proceeds differently depending on the choice of the first part, but it proceeds in the same way regardless of whether it chooses a deal or a case of disapproval. In the scenario described, the contestant had a 50/50 chance of choosing the correct case.
To support this, we have made a deal or not to simulate a deal in Python, simulating 2 million games. Of those 2 million games, 6019 ended with a \$1 pick versus \$1,000,000. Of these, 2,994 had millions of dollars in initially designated cases, and 3,025 had millions of dollars in other cases. This results in a 49.74% probability compared to a 50.26% probability. This probability will probably be closer to 50/50 if you use large simulations.
The rational choice of agreement or no agreement of the opponent in the described situation depends on the financial situation of the person. Average US Citizen, “deal” would probably be better; For a very (rich) person, “no deal” would be better. However, it is clear that the probability of winning a million by switching cases where two cases are left is 50% and not one in 26. Is it a deal or not
A (functional) decision scientist sponsored by the Machine Intelligence Research Institute. A leading author on decision theory and game theory. | 1,459 | 6,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-06 | latest | en | 0.957658 |
https://revistalaika.org/lorn/application-of-kinetic-energy-in-daily-life.php | 1,657,157,470,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00723.warc.gz | 527,532,759 | 9,773 | # Application Of Kinetic Energy In Daily Life
Work-energy theorem no equations Physics Forums. How Do Kinetic Energy and Potential Energy Apply to Everyday Life? The pitcher winds up, then pitches. He demonstrates both potential energy in the windup, and kinetic, ... to kinetic energy to potential energy occurs when you walk up a stairs. Give two more everyday examples of energy everyday life where electrical energy is.
### Thermal Energy Examples Softschools.com
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Students are introduced to both potential energy and kinetic energy as forms of mechanical energy. A hands-on activity demonstrates how potential energy can change Application Elastic Energy/of Springs. Application of Elastic Energy/ Springs thus converting elastic potential energy into kinetic energy.
What are good examples of elastic potential energy and how is conversion of kinetic energy into potential energy and of elastic energy in everyday life? 6/12/2006В В· Wht is an in life example of each gravitational potential energy and chemical Life example like What does kinetic energy,light
Science of Everyday Things; ENERGY As with many concepts in physics, energy—along with the related ideas of work and power Real-life applications 6/12/2006 · Wht is an in life example of each gravitational potential energy and chemical Life example like What does kinetic energy,light
Kinetic energy is used by exploiting the motion of an object. How Is Kinetic Energy Used? A: Quick Answer. Kinetic energy is used by exploiting the motion of an Kinetic energy is the energy of mass in motion. The kinetic energy of an object is the energy it has because of its motion. Kinetic energy is the
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power the things we use in everyday life. • Energy Skate Park Basic retrieved on September 30 from • Use real world data to predict kinetic energy outputs. Kinetic energy is used by exploiting the motion of an object. How Is Kinetic Energy Used? A: Quick Answer. Kinetic energy is used by exploiting the motion of an
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### 22 Examples of Mechanical Energy We See All Around Us
Wht is an in life example of each gravitational potential. Energy - Real-life applications One of the best—and most frequently used—illustrations of potential and kinetic energy involves Everyday …, 7/11/2007 · i have to do a project in advanced science on kinetic energy this is like the only thing im stumped on How do i use kinetic energy in everyday life?.
### Nuclear Energy in Everyday Life Department of Energy
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• 10 Types of Energy and Examples ThoughtCo
• 20 Examples of Potential Energy in Everyday Life Life
• Difference between Kinetic Energy and Kinetic energy have many applications, many of them are from our daily life which depicts how Kinetic energy plays an Kinetic energy is energy that a body possess as a result of its motion. (Life-Threatening) updated daily and weekly.
Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles. Any object in motion is using kinetic energy: a person Start studying Examples of potential and kinetic energy and 1 other. Learn vocabulary, terms, and more with flashcards, games, and other study tools.
Elastic Potential Energy: or kinetic energy, Gravitational Potential Energy: Definition, Formula & Examples Related Study Materials. 42 Applications of Kinetic Theory. that the average kinetic energy in any degree of freedom of a molecule or other object is found interesting applications.
TAKS Objective 4 The student will The student is expected to: Illustrate examples of potential and kinetic energy in everyday life such as objects at rest, How Do Kinetic Energy and Potential Energy Apply to Everyday Life? By Michael Stratford; Updated April 25, 2018
Analysis of Situations in Which Mechanical Energy is Conserved; Application and Practice Questions; And so the kinetic energy at the bottom of the hill is 100 J Work and energy are closely related in physics. In this lesson, you'll learn what that relationship is as well as how we can apply it to various...
Start studying Examples of potential and kinetic energy and 1 other. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Almost everything in real life is an example of conservation of mass, The relevant conservation law is the conservation of mass/energy.
13/10/2011В В· This Site Might Help You. RE: What are 5 examples of potential and kinetic energy that you see in everyday life? What are 5 examples of potential and In physics, the kinetic energy of an object is the energy that it possesses due to its motion. (e.g. in everyday phenomena on Earth),
Kinetic energy is used by exploiting the motion of an object. How Is Kinetic Energy Used? A: Quick Answer. Kinetic energy is used by exploiting the motion of an ... to kinetic energy to potential energy occurs when you walk up a stairs. Give two more everyday examples of energy everyday life where electrical energy is
12/05/2008В В· I dont' quite understand where we would use this theorem in everyday life.. Work-energy theorem, no then it's kinetic energy won't change. As for daily How Do Kinetic Energy and Potential Energy Apply to Everyday Life? By Michael Stratford; Updated April 25, 2018
Examples of Potential Energy This object has the capability of producing energy as these conditions change - this energy could be kinetic energy, Name 10 Types of Energy. Search the Chemistry in Everyday Life such as the potential energy of an object placed on a shelf or the kinetic energy of the Moon
Students are introduced to both potential energy and kinetic energy as forms of mechanical energy. A hands-on activity demonstrates how potential energy can change Work and energy are closely related in physics. In this lesson, you'll learn what that relationship is as well as how we can apply it to various...
## Application of work kinetic energy theorem By
Law of Conservation of Energy Examples YourDictionary. Radiation & Daily Life. One of the most commonly recognised examples is medical applications such as the A radioactive material will release energy overtime, How Do Kinetic Energy and Potential Energy Apply to Everyday Life? The pitcher winds up, then pitches. He demonstrates both potential energy in the windup, and kinetic.
### Nuclear Energy in Everyday Life Department of Energy
Law of Conservation of mechanical Energy in everyday life. Wave Applications DCIs - High School; Force and Motion PEs The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy., Definition Of Conservation Of Energy If a particle or body is acted upon only by conservative forces energy is conserved. This means that the total kinetic and.
20 Examples of Potential Energy in Everyday Life of potential energy That we can find in the energy that is transformed into kinetic energy when the toy How does energy change form? What are some examples? Learn with flashcards, games, and more — for free.
29/04/2016 · This interactive animation describes about kinetic and potential energy,expression of the formula for kinetic energy. It also describes about the physical power the things we use in everyday life. • Energy Skate Park Basic retrieved on September 30 from • Use real world data to predict kinetic energy outputs.
Definition Of Conservation Of Energy If a particle or body is acted upon only by conservative forces energy is conserved. This means that the total kinetic and Kinetic energy is the energy of motion. It is also referred to as energy in motion, or as the work required to take an object from being at rest to action/movement.
Work - kinetic energy theorem is not an alternative to other techniques available for analyzing motion. What we want to mean here is that it provides a specific Nuclear Energy in Everyday Life depending on its application, dura- Risk is something we live with daily.
... The potential energy is converted into kinetic energy Law of Conservation of mechanical Energy in everyday Law of conservation of energy in everyday life. The energy possessed by a body due to its position is called potential energy Potential energy examples in everyday life. kinetic and potential energy,
Kinetic energy is the energy of motion.It is a scalar quantity, its unit is joule.Electrical energy, light energy, sound energy, mechanical energy, thermal energy Applications of electromagnetic induction. More accurately, eddy currents transform more useful forms of energy, such as kinetic energy, into heat,
Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles. Any object in motion is using kinetic energy: a person Science of Everyday Things; ENERGY As with many concepts in physics, energy—along with the related ideas of work and power Real-life applications
Radiation & Daily Life. One of the most commonly recognised examples is medical applications such as the A radioactive material will release energy overtime The kinetic energy she possessed because of her These law of conservation of energy examples show how commonplace this physics concept is in everyday life.
Kinetic energy is used by exploiting the motion of an object. How Is Kinetic Energy Used? A: Quick Answer. Kinetic energy is used by exploiting the motion of an Name 10 Types of Energy. Search the Chemistry in Everyday Life such as the potential energy of an object placed on a shelf or the kinetic energy of the Moon
Work and energy are closely related in physics. In this lesson, you'll learn what that relationship is as well as how we can apply it to various... How does energy change form? What are some examples? Learn with flashcards, games, and more — for free.
... to kinetic energy to potential energy occurs when you walk up a stairs. Give two more everyday examples of energy everyday life where electrical energy is This section focuses on differentiating among forms of energy, transformation of energy, and their real-world applications. The material presented is designed to help
Energy is a physical quantity that follows precise natural laws. Kinetic energy is energy due to the motion of an object. meanings in daily life, 13/10/2011В В· This Site Might Help You. RE: What are 5 examples of potential and kinetic energy that you see in everyday life? What are 5 examples of potential and
20/08/2012В В· Application of pressure in daily life. Posted on August 20, 2012 by chituraja under Uncategorized. Massage is one type of application of pressure. Analysis of Situations in Which Mechanical Energy is Conserved; Application and Practice Questions; And so the kinetic energy at the bottom of the hill is 100 J
A diode is a two-terminal electronic component that conducts The kinetic energy can translate in only "How Are Diodes Used in Our Everyday Lives Almost everything in real life is an example of conservation of mass, The relevant conservation law is the conservation of mass/energy.
Kinetic energy is the energy of mass in motion. The kinetic energy of an object is the energy it has because of its motion. Kinetic energy is the Kinetic energy is the energy of motion. It is also referred to as energy in motion, or as the work required to take an object from being at rest to action/movement.
APPLICATION OF KINETIC ENERGY STORAGE SYSTEMS TO POWER SYSTEMS OPERATION by Figure 2.1: Daily load profile Kinetic energy is the energy of motion.It is a scalar quantity, its unit is joule.Electrical energy, light energy, sound energy, mechanical energy, thermal energy
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### Kinetic and Potential Energy of Motion Lesson
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### Forms of Energy Everyday Examples to Help Students
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Science of Everyday Things; ENERGY As with many concepts in physics, energy—along with the related ideas of work and power Real-life applications We use energy for everything in the home and in the office and basically to perform daily tasks. Energy use can be divided many dfferent ways but the most common is
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Kinetic energy is used by exploiting the motion of an object. How Is Kinetic Energy Used? A: Quick Answer. Kinetic energy is used by exploiting the motion of an Wave Applications DCIs - High School; Force and Motion PEs The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.
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Work and energy are closely related in physics. In this lesson, you'll learn what that relationship is as well as how we can apply it to various... This section focuses on differentiating among forms of energy, transformation of energy, and their real-world applications. The material presented is designed to help
Students are introduced to both potential energy and kinetic energy as forms of mechanical energy. A hands-on activity demonstrates how potential energy can change Difference between Kinetic Energy and Kinetic energy have many applications, many of them are from our daily life which depicts how Kinetic energy plays an
Visit the physics classroom and focus on the parts dealing with Potential and Kinetic Energy conserve energy in your daily life is an application to Elastic Potential Energy: or kinetic energy, Gravitational Potential Energy: Definition, Formula & Examples Related Study Materials.
16/08/2013 · Forms of Energy: Everyday Examples to Help Students. August 16, 2013 at 5:00 pm Leave a comment. Help students understand energy and … Description together with definition of Examples Light Energy. so it is a kinetic type of energy. we see in our routine life carrying light energy like
The piezoelectric composition used for the piezoelectric layers was specially tuned for the application of kinetic energy in the daily life, including 6/12/2006В В· Wht is an in life example of each gravitational potential energy and chemical Life example like What does kinetic energy,light
Some applications of entropy and free energy free energy equilibrium applications . the strain energy is instead stored as thermal (kinetic) We use energy for everything in the home and in the office and basically to perform daily tasks. Energy use can be divided many dfferent ways but the most common is
Difference between Kinetic Energy and Kinetic energy have many applications, many of them are from our daily life which depicts how Kinetic energy plays an What are good examples of elastic potential energy and how is conversion of kinetic energy into potential energy and of elastic energy in everyday life?
Kinetic energy is the energy associated with the movement of objects. Although there are many forms of kinetic energy, this type of energy is often associated with Some applications of entropy and free energy free energy equilibrium applications . the strain energy is instead stored as thermal (kinetic) | 4,569 | 22,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.893571 |
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I understand that the Trojan points are located 60 degrees ahead and behind a planet in its orbit. However, since there are quite a number of Trojans in Jupiter's orbit, they cannot all be exactly at that point. Presumably they orbit around it.
So what sort of orbits do they have? Are they circular or anywhere near so, or are they elongated along the orbit of Jupiter? If the latter how far can they get from the 60 degree point?
Trojan asteroids are in roughly circular orbits around the Sun at roughly the same distance as Jupiter, that are in 1:1 resonance with Jupiter and stay very roughly 60 degrees away from it.
Scott Manley's video below shows two classes of asteroids in resonance with Jupiter. The first one shown is confusing because it is in a 3:2 resonance and in the rotating frame it looks like they are cycling between L3, L4 and L5. Skip ahead to 33 seconds and you can see what "normal Trojan asteroids" do. Most of them stay within +/-20 degrees of L4 or L5, only a few exotic stragglers go farther than that away from their Lagrange points. There is some out-of-plane motion as well, as there is for all asteroids.
@JamesK's answer showing a rather exotic asteroid in 1:1 resonance with Earth is an extreme case, but the GIF does help to give some illustration of the back-and-forthness, even though it's pretty exaggerated compared to what normally happens.
After watching, go back to the beginning and see the more confusing 3:2 resonance orbits.
update: There's this!
As an example, look at Earth's only confirmed Trojan
By Phoenix7777 - Own work Data source: HORIZONS System, JPL, NASA, CC BY-SA 4.0, source
Now, to understand what is happening here. The yellow dot is the sun. The blue dot is the Earth. Although the Earth is orbiting the sun, the "camera" is turning so that it appears that the Earth is roughly stationary (it wobbles slightly due to the eccentricty of the earth's orbit)
The pink dot is the asteroid 2010 TK7. It has an orbit that is eccentric, and so sometimes it is much closer to the sun than the Earth, and at other times it is further, but its orbit takes nearly exactly one year. However the exact shape of the orbit changes, sometimes it moves around the sun in slightly less than a year, so it starts to catch up with the Earth, but as it nears the Earth, the Earth's gravity tends to pull it forward, and out, causing it to slow down and move away from the earth. The whole cycle takes hundreds of years.
Note, the actual shape of the orbits are elliptical about the sun, the odd shapes is a result of the camera turning at one revolution per year.
This kind of orbit is said to librate about the L4 point. It is called a "tadpole orbit" Trojans don't have to remain exactly at the L4 point, they can stably orbit in a tadpole orbit around the L4 point.
## Trojan Asteroids
### Asteroids Beyond the Main Belt
The Trojan asteroids move in the same orbit as Jupiter, though they keep either well ahead of or well behind the Giant Planet and are in no danger of being engulfed. Mars has several Trojans, and Neptune one. No true Earth Trojans are known, though 3753 Cruithne has almost the same orbital period and describes a curious sort of ‘horseshoe’ path with respect to the Earth. There are also asteroids, such as 944 Hidalgo and 5335 Damocles, with very eccentric orbits, very like those of comets. For example, Damocles has a period of 40.9 years its orbit crosses those of Mars, Jupiter, Saturn, and Uranus, but is in no danger of collision as its orbital inclination is 61°. It is no more than 15 km in diameter.
The ‘Centaur’ asteroids remain well beyond the Main Belt the first to be found (in 1977) was 2060 Chiron, which moves mainly between the orbits of Saturn and Uranus, in a period of 50 years. It shows traces of a coma at times, but seems much too large to be classed as a comet, even though it has been given a cometary number.
## Trojan asteroid
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Trojan asteroid, also called Trojan planet, any one of a number of asteroids that occupy a stable Lagrangian point in a planet’s orbit around the Sun.
In 1772 the French mathematician and astronomer Joseph-Louis Lagrange predicted the existence and location of two groups of small bodies located near a pair of gravitationally stable points along Jupiter’s orbit. Those are positions (now called Lagrangian points and designated L4 and L5) where a small body can be held, by gravitational forces, at one vertex of an equilateral triangle whose other vertices are occupied by the massive bodies of Jupiter and the Sun. Those positions, which lead (L4) and trail (L5) Jupiter by 60° in the plane of its orbit, are two of the five theoretical Lagrangian points in the solution to the circular restricted three-body problem of celestial mechanics. The other three stable points are located along a line passing through the Sun and Jupiter. The presence of other planets, however—principally Saturn—perturbs the Sun-Jupiter-Trojan asteroid system enough to destabilize those points, and no asteroids have been found near them. In fact, because of that destabilization, most of Jupiter’s Trojan asteroids move in orbits inclined as much as 40° from Jupiter’s orbit and displaced as much as 70° from the leading and trailing positions of the true Lagrangian points.
In 1906 the first of the predicted objects, (588) Achilles, was discovered by German astronomer Max Wolf near L4. Within a year two more were found: (617) Patroclus, located near L5, and (624) Hektor, near L4. It was later decided to continue naming such asteroids after participants in the Trojan War as recounted in Homer’s epic work the Iliad and, furthermore, to name those near the leading point after Greek warriors and those near the trailing point after Trojan warriors. With the exception of the two “misplaced” names already bestowed (Hektor, the lone Trojan in the Greek camp, and Patroclus, the lone Greek in the Trojan camp), that tradition has been maintained.
As of 2020, of the more than 7,000 Jupiter Trojan asteroids discovered, about two-thirds are located near L4, and the remainder are near L5. Astronomers estimate that 1,800–2,200 of the total existing population of Jupiter’s Trojans have diameters greater than 15 km (10 miles).
Nearly all of Jupiter’s Trojans are dark, having albedos (percentage of visual light reflected) between 0.04 and 0.10. (However, one Trojan, [4709] Ennomos, has an albedo of 0.15, which is greater than that of the Moon [0.12].) The majority belong to two compositionally distinct groups that are similar to the most-common classes of outer main-belt asteroids.
Since the discovery of Jupiter’s orbital companions, astronomers have searched for Trojan objects of Earth, Mars, Saturn, Uranus, and Neptune as well as of the Earth-Moon system. It was long considered doubtful whether truly stable orbits could exist near the Lagrangian points of the smaller planets because of gravitational perturbations by the major planets. However, in 1990 an asteroid later named (5261) Eureka was discovered librating (oscillating) about the L5 point of Mars, and since then eight others have been found, one at L4 and seven at L5. Twenty-four Trojans of Neptune, all but three associated with L4, have been discovered since 2001. The first Earth Trojan asteroid, 2010 TK7, which librates around L4, was discovered in 2010, and the first Uranus Trojan, 2011 QF99, which librates around L4, was discovered the next year. Although Trojans of Saturn have yet to be found, objects librating about Lagrangian points of the systems formed by Saturn and its moon Tethys and Saturn and its moon Dione are known.
## Dec 13th: Small Asteroids in Earth-like Orbits & Jupiter Trojan Confusion
Description: Today’s 2 topics:
• Four days after it made its closest approach to Earth, I found a 10 foot diameter asteroid with the NASA funded University of Arizona 60 inch telescope on Mt. Lemmon.
• Humans know of approximately half million main belt asteroids orbiting the Sun between Mars and Jupiter. We have also found approximately 11,000 Earth approaching asteroids.
Bio: Dr. Al Grauer is currently an observing member of the Catalina Sky Survey Team at the University of Arizona. This group has discovered nearly half of the Earth approaching objects known to exist. He received a PhD in Physics in 1971 and has been an observational Astronomer for 43 years. He retired as a University Professor after 39 years of interacting with students. He has conducted research projects using telescopes in Arizona, Chile, Australia, Hawaii, Louisiana, and Georgia with funding from NSF and NASA.
He is noted as Co-discoverer of comet P/2010 TO20 Linear-Grauer, Discoverer of comet C/2009 U5 Grauer and has asteroid 18871 Grauer named for him.
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91E: Small Asteroids in Earth-like Orbits
Four days after it made its closest approach to Earth, I found a 10 foot diameter asteroid with the NASA funded University of Arizona 60 inch telescope on Mt. Lemmon. At that point it was 838,000 miles from us and was moving away at about 3 miles/second. Previously, It had been about half the distance to the Moon from planet Earth. This small asteroid orbits the Sun every 349 days on a path which crosses our orbit twice a year. Calculations by the NASA – JPL group indicate that it can come within 2 Earth diameters of us.
A week later my Catalina Sky Survey Survey team leader, Eric Christensen found a slightly larger small asteroid. It orbits the Sun in 388 days and can closer than 4 Earth diameters to us.
Either of these small asteroids would fit into the bed of a dump truck. Their speeds relative to Earth are well within our rockets ability to catch them.
These two asteroids do not pose a threat to humans. If they ever did enter the Earth’s atmosphere, they would likely burn up at high altitude producing a supersonic boom. They are interesting because they could be targets for NASA’s Asteroid Redirect Mission. The concept for this project is to redirect a small asteroid into orbit around the Moon where it would be visited by Astronauts using the Orion Spacecraft. This mission will provide us with scientific data about Earth approaching objects and develop the capability for humans to explore the planet Mars.
92E: Jupiter Trojan Confusion
Humans know of approximately half million main belt asteroids orbiting the Sun between Mars and Jupiter. We have also found approximately 11,000 Earth approaching asteroids. Members of these two groups can be recognized by briefly observing their motion since their paths about the Sun are very different. Jupiter Trojans, on the other hand, occupy a point either 60 degrees in front or behind Jupiter as it moves about the Sun. Their orbit around the Sun takes about 5 and a half years so short pieces of one of their orbits can mimic the path of an Earth approaching object. When Jupiter Trojans are opposite to the Sun, their little full moon faces are pointing towards Earth. When they are in this position in the sky we have to sort through a number of them so we can separate them from real near Earth objects.
The Jupiter Trojans oscillate about a stable point 60 degrees in front of or behind a planet as it orbits the Sun. This geometry was predicted by Joseph-Louis Lagrange in 1772. More than 100 years later Astronomers began to discover Jupiter Trojans. Today we know of almost 6,000 of the perhaps several hundred thousand of these asteroids which are larger than a kilometer in diameter. The largest one, Hektor. It is about 120 miles in diameter. Most of the Jupiter Trojans are much smaller than Hektor and may be fragments of larger ones which collided with each other.
Someday humans may visit these distant asteroids and find them to be a rich sources of raw materials.
For Travelers in the Night this is Dr. Al Grauer.
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## How were the Trojan asteroids discovered and named?
Illustration of the Lucy mission's seven targets: the binary asteroid Patroclus/Menoetius, Eurybates, Orus, Leucus, Polymele, and the main belt asteroid DonaldJohanson. Credit: NASA's Goddard Space Flight Center Conceptual Image Lab
On Feb. 22, 1906, German astrophotographer Max Wolf helped reshape our understanding of the solar system. Again.
Born in 1863, Wolf had a habit of dramatically altering the astronomy landscape. Something of a prodigy, he discovered his first comet at only 21 years old. Then in 1890, he boldly declared that he planned to use wide-field photography in his quest to discover new asteroids, which would make him the first to do so. Two years later, Wolf had found 18 new asteroids. He later became the first person to use the "stereo comparator," a View-Master-like device that showed two photographs of the sky at once so that moving asteroids appeared to pop out from the starry background.
It is perhaps unsurprising, then, that on Feb. 22, 1906, Wolf made another important discovery: an asteroid with a particularly unusual orbit. As Jupiter moved, this asteroid remained ahead of Jupiter, as though it was somehow trapped in Jupiter's orbit around the Sun. German astronomer Adolf Berberich observed that the asteroid was nearly 60 degrees in front of Jupiter. This specific position reminded Swedish astronomer Carl Charlier of a peculiar behavior predicted by the Italian-French mathematician Joseph-Louis Lagrange over 100 years earlier. Lagrange argued that if a small body (such as an asteroid) is placed at one of two stable points in a planet's orbit around the Sun (called the L4 and L5 Lagrange Points), the asteroid would remain stationary from the planet's perspective due to the combined gravitational forces of the planet and the Sun. Charlier realized that Wolf's asteroid was actually caught in Jupiter's L4 Lagrange point. Until Wolf's discovery, Lagrange's prediction had only been a mathematical exercise. Now, these astronomers had photographic proof that Lagrange was right.
Eight months later, one of Wolf's graduate students August Kopff discovered an asteroid in Jupiter's other stable Lagrange point L5, as well as another asteroid caught in L4 a few months afterward.
Once three of these Lagrange point-inhabiting asteroids had been discovered, astronomers began wondering what to call them. At this point, most asteroids were given the names of women from Roman or Greek mythology, unless their orbits were particularly strange. The asteroids in question certainly had bizarre orbits, so Austrian astronomer Johann Palisa suggested the names Achilles, Patroclus, and Hektor after characters from The Iliad. Achilles was a nigh-invulnerable Greek hero (except for his heel), and Patroclus was a friend of his. Hektor, prince of the Trojans, eventually killed Patroclus, and Achilles exacted revenge by killing Hektor. The recently discovered asteroids were then given Iliad-inspired names.
As astronomers continued discovering asteroids hiding in Jupiter's Lagrange points, they continued naming them after heroes of the Trojan War and began referring to them as "Trojan asteroids." ("Trojan asteroids" would eventually refer to asteroids inhabiting any planet's stable Lagrange points, though names from The Iliad are reserved for Jupiter's Trojans.) It later became convention to name Jupiter's L4 asteroids after Greek characters and Jupiter's L5 asteroids after Trojan characters, so L4 and L5 became the "Greek camp" and the "Trojan camp" respectively. Palisa apparently did not foresee this tradition, for his naming of first three asteroids led to a Greek "spy" residing in the Trojan camp (Patroclus) and a confused Trojan (Hektor) who probably wandered into the Greek camp hoping to order some of their famous custom-built wooden horses.
No spacecraft has ever been to this population of small bodies, called the Trojan asteroids. Now, a new NASA Discovery Program mission called Lucy will fly by seven Trojan asteroids, plus a main belt asteroid, to survey the diversity of this population in a single 12-year record-breaking mission. The Lucy spacecraft launch window opens Oct. 16, 2021.
Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points, are known as Trojan asteroids. Although Trojan asteroids have been discovered for Mars (4 to date, 1 at L4 and 3 at L5) and Neptune (8 Trojans, 6 at L4 and 2 at L5) and even Earth (1 Trojan at L4), the term ‘Trojan asteroid’ generally refers to the asteroids accompanying Jupiter.
There are currently over 4,800 known Trojan asteroids associated with Jupiter. About 65% of these belong to the leading group (L4) located 60 o in front of Jupiter in its orbit, while the other 35% cluster around the L5 Lagrangian point and trail 60 o behind Jupiter. Although their orbits are stabilised at the Lagrangian points by gravitational interactions with Jupiter and the Sun, their actual distribution is elongated along the orbit. Perturbations by the other planets (primarily Saturn) cause the Trojans to oscillate around L4 and L5 by ∼20° and at inclinations of up to 40° to the orbital plane. These oscillations generally take between 150 and 200 years to complete.
Such planetary perturbations may also be the reason why there have been so few Trojans found around other planets. In particular, we assume that the Trojans formed at their present positions at the same time as Jupiter emerged from the solar nebula. If this is correct, we would also expect Saturn to be accompanied by families of Trojan asteroids. That no Trojans are found at the Lagrangian points of Saturn is most likely the result of Jupiter removing them from these stable orbits through gravitational perturbations.
The term ‘Trojan asteroid’ was coined when it was decided to name all asteroids discovered at the L4 and L5 points of Jupiter after warriors in the Trojan war, Greek and Trojan respectively. The exceptions are Hector (a Trojan spy in the Greek camp) and Patroclus (a Greek spy in the Trojan camp), the first two Trojan asteroids discovered and named before the two camps were established.
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## Why are they called Trojan Asteroids?
On February 22, 1906, German astrophotographer Max Wolf helped reshape our understanding of the solar system. Again.
Born in 1863, Wolf had a habit of dramatically altering the astronomy landscape. Something of a prodigy, he discovered his first comet at only 21 years old. Then in 1890, he boldly declared that he planned to use wide-field photography in his quest to discover new asteroids, which would make him the first to do so. Two years later, Wolf had found 18 new asteroids. He later became the first person to use the “stereo comparator,” a View-Master-like device that showed two photographs of the sky at once so that moving asteroids appeared to pop out from the starry background.
It is perhaps unsurprising, then, that on February 22, 1906, Wolf made another important discovery: an asteroid with a particularly unusual orbit. As Jupiter moved, this asteroid remained ahead of Jupiter, as though it was somehow trapped in Jupiter’s orbit around the Sun. German astronomer Adolf Berberich observed that the asteroid was nearly 60° in front of Jupiter. This specific position reminded Swedish astronomer Carl Charlier of a peculiar behavior predicted by mathematician Joseph-Louis Lagrange over 100 years earlier. Lagrange argued that if a small body (such as an asteroid) is placed at one of two stable points in a planet’s orbit around the Sun (called the L4 and L5 “Lagrange Points”), the asteroid would remain stationary from the planet’s perspective due to the combined gravitational forces of the planet and the Sun. Charlier realized that Wolf’s asteroid was actually caught in Jupiter’s L4 Lagrange point. Until Wolf’s discovery, Lagrange’s prediction had only been a mathematical exercise. Now, these astronomers had photographic proof that Lagrange was right.
Eight months later, one of Wolf’s graduate students August Kopff discovered an asteroid in Jupiter’s other stable Lagrange point L5, as well as another asteroid caught in L4 a few months afterward.
Once three of these Lagrange point-inhabiting asteroids had been discovered, astronomers began wondering what to call them. At this point, most asteroids were given the names of women from Roman or Greek mythology, unless their orbits were particularly strange. The asteroids in question certainly had bizarre orbits, so Austrian astronomer Johann Palisa suggested the names Achilles, Patroclus, and Hektor after characters from The Iliad. Achilles was a nigh-invulnerable Greek hero (except for, you know…), and Patroclus was a friend of his. Hektor, prince of the Trojans, eventually killed Patroclus, and Achilles exacted revenge by killing Hektor.
As astronomers continued discovering asteroids hiding in Jupiter’s Lagrange points, they continued naming them after heroes of the Trojan War and began referring to them as “Trojan asteroids.” (“Trojan asteroids” would eventually refer to asteroids inhabiting any planet’s stable Lagrange points, though names from The Iliad are reserved for Jupiter’s Trojans.) It later became convention to name Jupiter’s L4 asteroids after Greek characters and Jupiter’s L5 asteroids after Trojan characters, so L4 and L5 became the “Greek camp” and the “Trojan camp.” Palisa apparently did not foresee this tradition, for his naming of first three asteroids led to a Greek “spy” residing in the Trojan camp (Patroclus) and a confused Trojan (Hektor) who probably wandered into the Greek camp hoping to order some of their famous custom-built wooden horses.
• Cornish, N. J. (2020, January 4). What is a Lagrange Point? Retrieved June 8, 2020, from https://solarsystem.nasa.gov/resources/754/what-is-a-lagrange-point/ Editors of Encyclopaedia Britannica. (2019, September 29). Max Wolf. Retrieved June 8, 2020, from https://www.britannica.com/biography/Max-Wolf
• Lowell Observatory. (2016, June 8). Naming Asteroids. Retrieved June 8, 2020, from https://lowell.edu/sga/naming-asteroids/
• Michel, P., DeMeo, F. E., & Bottke, W. F. (2015). Asteroids IV. Tucson: The University of Arizona Press.
• Murdin, P. (2016). Rock Legends: the Asteroids and Their Discoverers. Springer International Publishing AG. doi: 10.1007/978-3-319-31836-3
• Nicholson, S. B. (1961). The Trojan Asteroids. Astronomical Society of the Pacific Leaflets, 8(381), 239–246.
• Tenn J. S. (1994) Max Wolf: The twenty-fifth Bruce Medalist. Mercury, 23(3), 27-28.
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## Trojan Asteroids Are in a Class of Their Own
By: Christopher Crockett October 26, 2018 0
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Hordes of debris trapped by Jupiter and Neptune have distinct colors that mark them as possibly the last remnants of the material that built the giant planets.
Trojan asteroids lead and trail Jupiter along its orbit around the Sun in this illustration.
NASA / JPL-Caltech
There’s a color conundrum among some of the smaller members of our solar system.
Jupiter and Neptune drag pockets of rocky debris on their treks around the Sun. Known as Trojans, these asteroid-like objects were likely swept up by the gravity of these two worlds. Swept up from where and when, though, has long been a mystery that could reveal details about how the young solar system evolved.
Well, the mystery keeps getting deeper. New observations build on previous hints that the Trojans at Jupiter and Neptune — despite being separated by billions of kilometers — have remarkably similar colors that don’t match any other small bodies in the solar system.
“Nature is giving us a little punch in the eye,” said David Jewitt (UCLA) during an October 23rd press conference at a meeting of the American Astronomical Society’s Division for Planetary Sciences.
Trojan asteroids lead and trail a planet as it orbits the Sun. They congregate at special points known as Lagrange points, where gravitational and orbital forces balance. One leading idea (of several) for their origin is that Trojans are lost souls from the Kuiper belt, the field of frozen debris beyond Neptune.
To test this idea, Jewitt compared the colors of 13 Neptunian Trojans to the colors of bodies in the Kuiper belt. They didn’t match. Neptune’s Trojans do, however, have similar colors to the Trojans near Jupiter, implying a common origin for both.
Astronomers have known for a while that Jupiter’s Trojans don’t match the Kuiper belt. Some researchers suspected that warmer temperatures near Jupiter could modify the surfaces of any visitors from the frigid Kuiper belt, thus changing their colors. But Jewitt said that doesn’t explain the color mismatch at Neptune, where the temperature is nearly the same as the Kuiper belt.
(Story continues after animation.)
Jupiter's Trojans (green) follow the giant planet (orange) on its orbit around the Sun. This animation shows their movements (along with the inner planets) during the course of NASA's Lucy mission to the Trojans.
Petr Scheirich (Astronomical Institute of the Czech Academy of Sciences)
“The Trojans are probably the last remnants of what formed the planets,” says Scott Sheppard (Carnegie Institution for Science). The planets are thought to have formed from lots of little things smashing together to make big things. The stuff that didn’t make it into a planet continued to orbit the Sun, and some of that debris could have gotten trapped where the Trojans live today.
Sheppard published a paper in 2006, along with Chad Trujillo (Northern Arizona University), that presented colors of the first four known Neptunian Trojans. Even then, there were hints that the Trojans were a separate family from everything else in the solar system. In the years since, that case has continued to build. “This new result doubles the sample size,” says Sheppard. “It continues to show this trend of being a very different population.” (See Sheppard's article in our June 2016 issue.)
Sarah Sonnett (Planetary Science Institute) agrees: “This is a very interesting result that provides constraints on where Neptune Trojans might have formed,” she says. “It consequently could help us understand how giant planets may have migrated early in solar system history.”
Most planetary scientists agree that the giant planets formed closer together and then migrated to their current orbits more than 4 billion years ago. If the Trojans were swept up during that time, they might have much to say about how the planets jockeyed for position. Some Jupiter Trojans, for example, are on highly inclined orbits. “That suggests a very chaotic environment,” Sheppard says. Planets effectively bouncing off one another could have stirred up the surrounding debris to create the Trojan orbits seen today.
“The Trojans have always been an overlooked population,” says Sheppard. They’re hard to observe and their spectra don’t reveal much. That’s where NASA’s Lucy mission might help. Scheduled to launch in 2021, it will spend seven years visiting six Jupiter Trojans (plus an asteroid in the belt between Mars and Jupiter), becoming the first probe to investigate this enigmatic population.
## Asteroid Groups
Asteroids are grouped according to their orbit range, as follows:
#### Near-Earth Asteroids (NEAs)
Objects in near-earth orbit. There are about 9,000 objects known ranging from 1 meter to 32 meters in diameter. These asteroids can be further grouped as follows:
• Atira group: Inside Earth's orbit semi-major axis less than 1.0 AU
• Aten group: Cross Earth's orbit semi-major axis less than 1.0 AU
• Apollo group: Cross Earth's orbit semi-major axis greater than 1.0 AU
• Amor group: Outside Earth's orbit semi-major axis greater than 1.0 AU
#### Main-Belt Asteroids
The total mass of the asteroid belt is just 4% that of the moon. Millions of asteroids located between the orbits of Mars and Jupiter (1.8 to 4.0 AU) are in the "asteroid belt".
#### Trojan Asteroids
Mainly, two large groups of asteroids located along the orbital path of Jupiter, 60 degrees ahead and behind the planet at the trojan points. There are also eight Mars Trojans and one Earth Trojan known.
Dwarf planet Ceres
Asteroid Vesta
Asteroid Pallas
Asteroid Eros | 6,673 | 29,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.957507 |
https://engineering.stackexchange.com/questions/4270/how-to-create-a-logic-circuit-that-multiplies-two-4-bit-numbers-using-only-16-2 | 1,656,471,816,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00363.warc.gz | 276,176,952 | 65,267 | # How to create a logic circuit that multiplies two 4 bit numbers, using only 16 2-input AND gates and 3 4-bit adders?
"Sketch a combinational circuit which takes as input, two 4-bit binary numbers, A3A2A1A0 and B3B2B1B0, and which outputs the 8-bit product P7P6P5P4P3P2P1P0. Assume that you have available sixteen 2-input AND gates and three 4-bit adders. Please explain the reasoning behind your design."
After a bit of fiddling I got close to the final circuit but I found that three 4-bit adders wouldn't be enough. To save time (and my complete lack of being able to explain anything) I found this, which is almost exactly what I need except it used two 4-bit adders and one 6-bit adder. I want to know if it's even possible to construct a circuit that can multiply two 4-bit numbers using the gates specified above, and if so, how can you do it?
• there are quite a few 0 inputs on that you could optimize out. Sep 3, 2015 at 10:07
The first step is ANDing all the combinations of inputs which takes all the 16 AND gates available.
Lets call those results $S_{00}$ $S_{01}$ $S_{02}$ $S_{03}$ $S_{10}$ etc.
Is we arrange them in how we need to add them we get:
$$\begin{matrix} & & & S_{03}&S_{02}& S_{01} & S_{00} \\ & & S_{13}&S_{12}& S_{11} & S_{10} & \\ & S_{23}&S_{22}& S_{21} & S_{20} & & \\ S_{33}&S_{32}& S_{31} & S_{30} & & \\ \end{matrix}$$
$S_{00}$ is the low bit of the output.
Then you use 1 4-bit adder to add {0,$S_{03}$, $S_{02}$, $S_{01}$} and {$S_{13}$, $S_{12}$, $S_{11}$, $S_{10}$} Output is {$A_c$, $A_3$, $A_2$, $A_1$, $A_0$}
If we substitute that result in the original table we see that this collapses the top 2 rows:
$$\begin{matrix} & A_c & A_3 & A_2 & A_1 & A_0 & S_{00} \\ & S_{23}&S_{22}& S_{21} & S_{20} & & \\ S_{33}&S_{32}& S_{31} & S_{30} & & \\ \end{matrix}$$
$A_0$ is bit 1 of the output
Then you use the second 4-bit adder to add {$A_c$, $A_3$, $A_2$, $A_1$} and {$S_{23}$, $S_{22}$, $S_{21}$, $S_{20}$} output is {$B_c$, $B_3$, $B_2$, $B_1$, $B_0$}
$$\begin{matrix} B_c & B_3 & B_2 & B_1 & B_0 & A_0 & S_{00} \\ S_{33}&S_{32}& S_{31} & S_{30} & & \\ \end{matrix}$$
$B_0$ is bit 2 of the output
Then you use the third 4-bit adder to add {$B_c$, $B_3$, $B_2$, $B_1$} and {$S_{33}$, $S_{32}$, $S_{31}$, $S_{30}$} output is {$C_c$, $C_3$, $C_2$, $C_1$, $C_0$}
The carry input on all the 4-bit adders is 0.
The total output is then: {$C_c$, $C_3$, $C_2$, $C_1$, $C_0$, $B_0$, $A_0$, $S_{00}$}
• Ah, thank you so much, I was so fixated with using the third adder to add the results of the other two that I didn't even think of doing it like this. Makes perfect sense. Thanks again. :) Sep 3, 2015 at 10:34 | 994 | 2,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-27 | longest | en | 0.812 |
https://www.collegeperk.org/book/weighted-residual-methods/ | 1,675,399,665,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00538.warc.gz | 714,355,296 | 12,301 | • Author : Snehashish Chakraverty
• Release Date : 20 March 2019
• Publisher : John Wiley & Sons
• Genre : Mathematics
• Pages : 256
• ISBN 13 : 9781119423447
## Advanced Numerical and Semi-Analytical Methods for Differential Equations Book Summary
Examines numerical and semi-analytical methods for differential equations that can be used for solving practical ODEs and PDEs This student-friendly book deals with various approaches for solving differential equations numerically or semi-analytically depending on the type of equations and offers simple example problems to help readers along. Featuring both traditional and recent methods, Advanced Numerical and Semi Analytical Methods for Differential Equations begins with a review of basic numerical methods. It then looks at Laplace, Fourier, and weighted residual methods for solving differential equations. A new challenging method of Boundary Characteristics Orthogonal Polynomials (BCOPs) is introduced next. The book then discusses Finite Difference Method (FDM), Finite Element Method (FEM), Finite Volume Method (FVM), and Boundary Element Method (BEM). Following that, analytical/semi analytic methods like Akbari Ganji's Method (AGM) and Exp-function are used to solve nonlinear differential equations. Nonlinear differential equations using semi-analytical methods are also addressed, namely Adomian Decomposition Method (ADM), Homotopy Perturbation Method (HPM), Variational Iteration Method (VIM), and Homotopy Analysis Method (HAM). Other topics covered include: emerging areas of research related to the solution of differential equations based on differential quadrature and wavelet approach; combined and hybrid methods for solving differential equations; as well as an overview of fractal differential equations. Further, uncertainty in term of intervals and fuzzy numbers have also been included, along with the interval finite element method. This book: Discusses various methods for solving linear and nonlinear ODEs and PDEs Covers basic numerical techniques for solving differential equations along with various discretization methods Investigates nonlinear differential equations using semi-analytical methods Examines differential equations in an uncertain environment Includes a new scenario in which uncertainty (in term of intervals and fuzzy numbers) has been included in differential equations Contains solved example problems, as well as some unsolved problems for self-validation of the topics covered Advanced Numerical and Semi Analytical Methods for Differential Equations is an excellent text for graduate as well as post graduate students and researchers studying various methods for solving differential equations, numerically and semi-analytically.
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This classic book covers the solution of differential equations in science and engineering in such as way as to provide an introduction for novices before progressing toward increasingly more difficult problems. The Method of Weighted Residuals and Variational Principles describes variational principles, including how to find them and how to ... | 1,410 | 6,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-06 | latest | en | 0.885611 |
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# وقاية المولد والمحول
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• 1. ﺑﺴﻢ اﷲ اﻟﺮﺣﻤﻦ اﻟﺮﺣﻴﻢ هﺬﻩ ﻣﻘﺪﻣﺔ ﻟﻜﺘﺎﺑﻲ وﻗﺎﻳﺔ اﻟﻤﻮﻟﺪ واﻟﻤﺤﻮل ﺳﺎﺋﻼ اﻟﻤﻮﻟﻰ ﻋﺰ وﺟﻞ أن ﻳﻨﻔﻊ ﺑﻬﺎ اﻟﻤﺨﺘﺼﻴﻦ ﻓﻲ ﺷﺘﻰ اﻟﻤﺠﺎﻻت وﻻ ﺗﻨﺴﻮﻧﺎ ﻣﻦ ﺻﺎﻟﺢ اﻟﺪﻋﺎء ﻣﻬﻨﺪس ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ Email- zwuitina@yahoo.com أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 2. وﻗﺎﻳﺔ اﻟﻤﻮﻟﺪ واﻟﻤﺤﻮل )Generator differential (87G اﻟﻮﻗﺎﻳﺔ اﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﻠﻤﻮﻟﺪ وﻓﻴﻬﺎ ﻳﺘﻢ ﺗﻔﺎﺿﻞ أو ﻣﻘﺎرﻧﺔ اﻟﺘﻴﺎر ﻋﻠﻰ ﻃﺮﻓﻲ اﻟﻤﻮﻟﺪ ﺣﻴﺚ ﻓﻲ اﻟﻀﺮ وف اﻟﻌﺎدﻳﺔ ﻳﻜﻮن اﻟﺘﻴﺎر ﻋﻠﻰ ﻃﺮﻓﻲ اﻟﻤﻮﻟﺪ ﻣﺘﺴﺎوي وآﻼهﻤﺎ ﻓﻲ ﻋﻜﺲ اﺗﺠﺎﻩ اﻷﺧﺮ وﺣﻴﺚ إن آﻞ ﻗﻮﺗﻴﻦ ﻣﺘﺴﺎوﻳﺘﻴﻦ ﻓﻲ اﻟﻤﻘﺪار وﻣﺘﻀﺎدات ﻓﻲ اﻷﺗﺠﺎة ﺣﺎﺻﻞ ﺟﻤﻌﻦ ﻳﺴﺎوى ﺻﻔﺮ ﻓﺎن اﻟﺘﻴﺎر اﻟﻨﺎﺗﺞ ﻋﻨﻬﻤﺎ ﻳﺴﺎوى ﺻﻔﺮ . وﻓﻰ ﺣﺎﻟﺔ ﺣﺪوث ﻋﻄﻞ داﺧﻞ اﻟﻤﻮﻟﺪ ﻳﻜﻮن اﻟﺘﻴﺎر اﻟﻨﺎﺗﺞ ﻋﻠﻰ ﻃﺮﻓﻲ ﻣﻜﺎن اﻟﻌﻄﻞ ﺣﺎﺻﻞ ﺟﻤﻊ اﻟﺘﻴﺎرﻳﻦ وﺑﺎﻟﺘﺎﻟﻲ ﻳﺘﻢ ﺗﺸﻐﻴﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ وﻋﺰل اﻟﻤﻮﻟﺪ ﻣﻦ اﻟﺸﺒﻜﺔ واﻟﺠﺪﻳﺮ ﺑﺎﻟﺬآﺮ إن اﻟﻮﻗﺎﻳﺔ اﻟﺘﻔﺎﺿﻠﻴﺔ ﻓﻲ اﻟﻤﺮﺣﻼت اﻟﻘﺪﻳﻤﺔ ﻻ ﺗﻌﻤﻞ إﻻ ﻋﻨﺪ ﺣﺪوث ﻋﻄﻞ داﺧﻞ ﻣﻨﻄﻘﺔ اﻟﺤﻤﺎﻳﺔ اﻟﻮاﻗﻌﺔ ﺑﻴﻦ ﻣﺤﻮﻟﻲ اﻟﺘﻴﺎر وﻓﻰ ﻧﻈﺎم اﻟﺤﻤﺎﻳﺔ اﻟﺮﻗﻤﻲ 612 REGﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻮﻗﺎﻳﺔ اﻟﺘﻔﺎﺿﻠﻴﺔ ﻋﻠﻰ أﺳﺎس ﺛﻼﺛﺔ ﻣﺘﻐﻴﺮات وهﻰ اﻟﻤﺘﻐﻴﺮ gوهﻰ اﻗﻞ ﻗﻴﻤﺔ ﻟﺘﻴﺎر ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ وﺗﻜﻮن اﻟﻘﻴﻤﺔ اﻻﻓﺘﺮاﺿﻴﺔ ﻟﻠﻤﺘﻐﻴﺮ 0.1 INﺣﻴﺚ إذا آﺎن اﻟﻔﺮق ﺑﻴﻦ اﻟﺘﻴﺎرﻳﻦ اآﺒﺮ ﻣﻦ أو ﺗﺴﺎوى 1.0 ﻣﻦ اﻟﺘﻴﺎر اﻟﻤﻘﻨﻦ ﻟﻠﻤﻮﻟﺪ ﻳﺘﻢ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ اﻟﻤﺘﻐﻴﺮ bﻟﺘﺤﺪﻳﺪ ﻣﻨﻄﻘﺔ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ وﺗﻜﻮن اﻟﻘﻴﻤﺔ اﻻﻓﺘﺮاﺿﻴﺔ ﻟﻠﻤﺘﻐﻴﺮ 5.1 اﻟﻤﺘﻐﻴﺮ vﻟﺘﺤﺪﻳﺪ ﻣﻨﺤﻨﻰ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ ﻟﻸﻋﻄﺎل اﻟﺨﺎرﺟﻴﺔ وﺗﻜﻮن اﻟﻘﻴﻤﺔ اﻻﻓﺘﺮاﺿﻴﺔ ﻟﻠﻤﺘﻐﻴﺮ 52.0 أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 3. وﻓﻰ اﻟﺸﻜﻞ اﻋﻼﻩ ﻧﻼﺣﻆ ﻣﻨﺤﻨﻰ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ ﺣﻴﺚ ﻻﻳﺘﻢ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ ﺧﺎرج اﻟﻤﺘﻐﻴﺮ b واﻟﻤﺤﻨﻰ vإﻻ إذا آﺎﻧﺖ ﻗﻴﻤﺔ I1/INاآﺒﺮ ﻣﻦ bأو ﻗﻴﻤﺔ I2/INاآﺒﺮ ﻣﻦ b ﺣﻴﺚ 1 Iﻗﻴﻤﺔ ﺗﻴﺎر ﻣﺤﻮل اﻟﺘﻴﺎر اﻷول و 2 Iﻗﻴﻤﺔ ﺗﻴﺎر ﻣﺤﻮل اﻟﺘﻴﺎر اﻟﺜﺎﻧﻲ وﻓﻰ هﺬﻩ اﻟﺤﺎﻟﺔ ﻳﻜﻮن اﻟﻌﻄﻞ ﺧﺎرﺟﻲ اى ﺧﺎرج ﻣﻨﻄﻘﺔ اﻟﺤﻤﺎﻳﺔ اﻟﺘﻔﺎﺿﻠﻴﺔ وهﺬا ﻳﻌﺘﺒﺮ ﻣﻦ ﻣﻴﺰات اﻟﻤﻨﻈﻮﻣﺔ وﻳﻤﻜﻦ اﻟﺘﺤﻜﻢ ﻓﻲ ﺣﺴﺎﺳﻴﺔ اﻟﻤﺮﺣﻞ ﻟﻸﻋﻄﺎل اﻟﺨﺎرﺟﻴﺔ ﻋﻦ ﻃﺮﻳﻖ ﺗﻐﻴﻴﺮ ﻗﻴﻤﺔ اﻟﻤﺘﻐﻴﺮ v آﻤﺎ ﻧﻼﺣﻆ إن ﻣﻨﻄﻘﺔ اﻟﺤﻤﺎﻳﺔ ﻣﺤﺼﻮرة ﻣﺎﺑﻴﻦ اﻟﻤﺘﻐﻴﺮ gواﻟﻤﺘﻐﻴﺮ b اﻟﻮﻗﺎﻳﺔ اﻟﺘﻔﺎﺿﻠﻴﺔ ﻟﻠﻤﺤﻮل )Transformer differential (87 T وﻓﻴﻬﺎ ﻳﺘﻢ ﺗﻔﺎﺿﻞ أو ﻣﻘﺎرﻧﺔ اﻟﺘﻴﺎر ﻋﻠﻰ ﻃﺮﻓﻲ اﻟﻤﺤﻮل ﺣﻴﺚ ﻓﻲ اﻟﻀﺮ وف اﻟﻌﺎدﻳﺔ ﻳﻜﻮن اﻟﺘﻴﺎر ﻋﻠﻰ ﻃﺮﻓﻲ اﻟﻤﺤﻮل ﻣﺘﺴﺎوي وآﻼهﻤﺎ ﻓﻲ ﻋﻜﺲ اﺗﺠﺎﻩ اﻷﺧﺮ وﺣﻴﺚ إن آﻞ ﻗﻮﺗﻴﻦ ﻣﺘﺴﺎوﻳﺘﻴﻦ ﻓﻲ اﻟﻤﻘﺪار وﻣﺘﻀﺎدات ﻓﻲ اﻷﺗﺠﺎة ﺣﺎﺻﻞ ﺟﻤﻌﻬﻦ ﻳﺴﺎوى ﺻﻔﺮ ﻓﺎن اﻟﺘﻴﺎر اﻟﻨﺎﺗﺞ ﻋﻨﻬﻤﺎ ﻳﺴﺎوى ﺻﻔﺮ . وﻓﻰ ﺣﺎﻟﺔ ﺣﺪوث ﻋﻄﻞ داﺧﻞ اﻟﻤﺤﻮل ﻳﻜﻮن اﻟﺘﻴﺎر اﻟﻨﺎﺗﺞ ﻋﻠﻰ ﻃﺮﻓﻲ ﻣﻜﺎن اﻟﻌﻄﻞ ﺣﺎﺻﻞ ﺟﻤﻊ اﻟﺘﻴﺎرﻳﻦ وﺑﺎﻟﺘﺎﻟﻲ ﻳﺘﻢ ﺗﺸﻐﻴﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ وﻋﺰل اﻟﻤﺤﻮل ﻣﻦ اﻟﺸﺒﻜﺔ وﻳﺘﻢ ﺗﻌﺪﻳﻞ ﻗﻴﻢ ﺗﻴﺎر اﻟﺤﻤﺎﻳﺔ آﻤﺎ ﻓﻲ اﻟﻤﺮﺣﻞ اﻟﺴﺎﺑﻖ ﻟﺤﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﺨﻄﺎء اﻷرﺿﻲ ﻟﻤﻠﻔﺎت اﻟﻤﻮﻟﺪ اﻟﺜﺎﺑﺘﺔ %59 Stator Ground Fault أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 4. وﻓﻰ اﻟﺸﻜﻞ اﻋﻼﻩ ﻳﻮﺿﺢ اﻟﻔﻜﺮة اﻟﻌﺎﻣﺔ ﻟﻤﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ ﺣﻴﺚ ﻳﺘﻢ رﺑﻂ ﻣﺤﻮل ﺟﻬﺪ ﻋﻠﻰ ﻃﺮﻓﻲ ﻣﻘﺎوﻣﺔ اﻟﺘﺂرﻳﺾ ﻟﺘﻮﺻﻴﻠﺔ ﻧﺠﻤﺔ وﻓﻰ اﻟﺤﺎﻟﺔ اﻟﻌﺎدﻳﺔ ﻳﻜﻮن اﻟﺘﻴﺎر ﻣﺘﺰن ﻓﻲ ﺟﻤﻴﻊ ﻣﻠﻔﺎت اﻟﻤﻮﻟﺪ وﺑﺎﻟﺘﺎﻟﻲ ﻳﻜﻮن اﻟﺘﻴﺎر ﻓﻲ اﻟﻨﻘﻄﺔ اﻟﻨﺠﻤﻴﺔ ﻳﺴﺎوى ﺻﻔﺮ وﻳﻜﻮن اﻟﺠﻬﺪ ﻋﻠﻰ ﻃﺮﻓﻲ اﻟﻤﻘﺎوﻣﺔ ﻳﺴﺎوى ﺻﻔﺮ وﻋﻨﺪ ﺣﺪوث ﻋﻄﻞ داﺧﻞ اﻟﻤﻠﻔﺎت ﻳﻜﻮن اﻟﺘﻴﺎر ﻣﺨﺘﻠﻒ ﻓﻲ اﻟﻨﻘﻄﺔ اﻟﻨﺠﻤﻴﺔ ﻓﻴﻤﺮ ﺗﻴﺎر ﻣﻦ ﺧﻼل اﻟﻤﻘﺎوﻣﺔ وﺑﺎﻟﺘﺎﻟﻲ ﻳﺘﻢ ﺗﺤﻮﻳﻞ اﻟﺠﻬﺪ ﻋﻠﻰ ﻃﺮﻓﻲ اﻟﻤﻘﺎوﻣﺔ ﻋﻦ ﻃﺮﻳﻖ ﻣﺤﻮل اﻟﺠﻬﺪ ﻟﻴﺘﻢ ﺗﺸﻐﻴﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ وﻋﺰل اﻟﻤﻮﻟﺪ وﻋﻨﺪ ﺣﺪوث ﺧﻄﺎء أرﺿﻰ ﻓﻲ ﺑﺪاﻳﺔ ﻣﻠﻔﺎت اﻟﻤﻮﻟﺪ ﻳﻜﻮن اﻟﺘﻴﺎر اﻟﻨﺎﺗﺞ ﻋﻦ اﻟﻌﻄﻞ ﻓﻲ ﺣﺪود ﺗﻴﺎر ﺗﺸﻐﻴﻞ اﻟﻤﻮﻟﺪ ﻟﺬﻟﻚ ﻻﻳﺘﻢ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ إﻻ ﻓﻲ ﺣﺪود اآﺒﺮ ﻣﻦ %59 ﻣﻦ ﻣﻠﻔﺎت اﻟﺜﺎﺑﺘﺔ ﻟﻠﻤﻮﻟﺪ ﺣﻴﺚ ﺗﻜﻮن ﻓﻌﺎﻟﺔ آﻠﻤﺎ آﺎن اﻟﻌﻄﻞ ﺑﻌﻴﺪ ﻣﻦ اﻟﻨﻘﻄﺔ اﻟﻨﺠﻤﻴﺔ وﺗﻌﺘﺒﺮ هﺬﻩ اﻟﻤﺮﺣﻠﺔ ﻋﻴﺐ ﻣﻦ ﻋﻴﻮب اﻟﻤﺮﺣﻞ وﻓﻰ ﻧﻈﺎم اﻟﺤﻤﺎﻳﺔ اﻟﺮﻗﻤﻲ 612 REGﻳﺘﻢ ﺗﻌﺪﻳﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ ﺑﺘﺄﺧﻴﺮ زﻣﻨﻲ ﺣﻮاﻟﻲ0.5S اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﺨﻄﺎء اﻷرﺿﻲ ﻟﻤﻠﻔﺎت اﻟﻤﻮﻟﺪ اﻟﺜﺎﺑﺘﺔ %001 Stator Ground Fault وﻓﻰ اﻟﺸﻜﻞ أﻋﻼﻩ ﻳﻮﺿﺢ اﻟﻔﻜﺮة اﻟﻌﺎﻣﺔ ﻟﻤﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ ﺣﻴﺚ ﻳﺘﻢ ﺣﻘﻦ اﻟﻤﻮﻟﺪ ﺑﺠﻬﺪ ﻋﺎﻟﻲ اﻟﺘﺮدد 1khzوﻳﺘﻢ اﺳﺘﻘﺒﺎل اﻹﺷﺎرة ﻋﻦ ﻃﺮﻳﻖ اﻟﻤﺮﺣﻞ وﻋﻨﺪ ﺣﺪوث ﻋﻄﻞ ﻓﻲ ﻣﻠﻔﺎت اﻟﻤﻮﻟﺪ ﻳﺘﻢ ﻓﻘﺪ اﻹﺷﺎرة وﺗﺸﻐﻴﻞ اﻟﻤﺮﺣﻞ وﻋﺰل اﻟﻤﻮﻟﺪ وﺗﻜﻮن ﻣﻨﻄﻘﺔ اﻟﺤﻤﺎﻳﺔ ﺗﺸﻤﻞ %001 ﻣﻦ ﻣﻠﻔﺎت اﻟﻤﻮﻟﺪ وﻳﻤﺘﺎز هﺬا اﻟﻤﺮﺣﻞ ﺑﺪﻗﺔ ﻋﺎﻟﻴﺔ إﻻ اﻧﻪ ﺑﺎهﻆ اﻟﺜﻤﻦ أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 5. اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﺨﻄﺎء اﻷرﺿﻲ ﻟﻤﻠﻔﺎت اﻟﻤﻮﻟﺪ اﻟﻤﺘﺤﺮآﺔ Rotor Ground Fault وﻓﻰ اﻟﺸﻜﻞ أدﻧﺎﻩ ﻳﻮﺿﺢ اﻟﻔﻜﺮة اﻟﻌﺎﻣﺔ ﻟﻤﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ ﺣﻴﺚ ﻳﺘﻢ رﺑﻂ ﻗﻨﻄﺮة ﺗﺘﻜﻮن ﻣﻦ ﻣﻘﺎوﻣﺘﻴﻦ وﻣﻜﺜﻔﻴﻦ ﻣﻊ ﺟﺴﻢ اﻟﻌﻀﻮ اﻟﺪوار ﻟﻠﻤﻮﻟﺪ وﻋﻨﺪ ﺣﺪوث ﺧﻄﺎ أرﺿﻰ ﻟﻤﻠﻔﺎت اﻟﻌﻀﻮ اﻟﺪوار ﻟﻠﻤﻮﻟﺪ ﻳﺘﻢ ﻓﻘﺪ اﺗﺰان اﻟﻘﻨﻄﺮة ﻟﻴﻤﺮ ﺗﻴﺎر ﻋﺒﺮ اﻟﻤﻘﺎوﻣﺔ ﻓﻴﺘﻢ ﺗﺤﻮﻳﻞ اﻟﺠﻬﺪ ﻋﻦ ﻃﺮﻳﻖ ﻣﺤﻮل اﻟﺠﻬﺪ وﺗﺸﻐﻴﻞ اﻟﻤﺮﺣﻞ وﻋﺰل اﻟﻤﻮﻟﺪ وﻓﻰ ﻧﻈﺎم اﻟﺤﻤﺎﻳﺔ اﻟﺮﻗﻤﻲ 612 REGﻳﺘﻢ ﺗﻌﺪﻳﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ ﺑﺘﺄﺧﻴﺮ زﻣﻨﻲ ﺣﻮاﻟﻲ0.5S اﻟﻮﻗﺎﻳﺔ ﻣﻦ ﻓﻘﺪ اﻟﺘﺤﺮﻳﺾ ﻟﻠﻤﻮﻟﺪ Loss of filed وﻓﻰ اﻟﺸﻜﻞ أﻋﻼﻩ ﻳﻮﺿﺢ ﻣﻨﺤﻨﻰ اﻟﻘﺪرة ﻟﻠﻤﻮﻟﺪ وﻓﻴﺔ ﻳﺘﻢ ﻓﺼﻞ اﻟﻤﻮﻟﺪ ﻣﻦ اﻟﺸﺒﻜﺔ ﻋﻨﺪﻣﺎ ﺗﻜﻮن اﻟﻘﺪرة اﻟﻐﻴﺮ ﻓﻌﺎﻟﺔ اآﺒﺮ ﻣﻦ -0.45puﻣﻦ اﻟﻘﺪرة اﻟﻤﻘﻨﻨﺔ ﻟﻠﻤﻮﻟﺪ وﻋﻨﺪهﺎ ﻳﺘﺤﻮل اﻟﻤﻮﻟﺪ اﻟﻰ ﻣﺤﺮك ﺣﺜﻲ ﺑﺴﺒﺐ ﻓﻘﺪ اﻟﺘﺤﺮﻳﺾ واﺳﺘﻬﻼك اﻟﻤﻮﻟﺪ ﻟﻠﻘﺪرة ﻏﻴﺮ اﻟﻔﻌﺎﻟﺔ ﻣﻦ اﻟﺸﺒﻜﺔ ﻣﻤﺎ ﻳﺴﺒﺐ ﻓﻲ زﻳﺎدة ﺗﺤﻤﻴﻞ Over أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 6. loadﻋﻠﻰ اﻟﻤﻠﻔﺎت اﻟﺜﺎﺑﺘﺔ ﻟﻠﻤﻮﻟﺪ وﺣﺚ ﻓﻲ ﺟﺴﻢ اﻟﻌﻀﻮ اﻟﺪوار ﻟﻠﻤﻮﻟﺪ و ﺑﺎﻟﺘﺎﻟﻲ زﻳﺎدة اﻟﺘﻴﺎرات اﻟﺪواﻣﻴﺔ ﻓﻴﻪ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻧﻌﻜﺎس اﻟﻘﺪرة Reverse power وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ اﻧﻌﻜﺎس اﻟﻘﺪرة ﻓﻴﻪ وﺑﺎﻟﺘﺎﻟﻲ ﻋﻤﻞ اﻟﻤﻮﻟﺪ آﻤﺤﺮك ﺑﺴﺒﺐ ﻓﺼﻞ اﻟﺘﺮﺑﻴﻨﺔ ) ﺑﺨﺎرﻳﺔ أو ﻏﺎزﻳﺔ( وﻳﺘﻢ ذاﻟﻚ ﻋﻦ ﻃﺮﻳﻖ ﻣﺮاﻗﺒﺔ اﺗﺠﺎﻩ اﻟﺘﻴﺎر ﻟﻠﻤﻮﻟﺪ وﻋﺎدﺗﺎ ﻳﻜﻮن ﺗﻌﺪﻳﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ > 2% Pnﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻧﺨﻔﺎض اﻟﺘﺮدد Under Frequency وﻓﻴﻬﺎ ﻳﺘﻢ اﻟﺤﻤﺎﻳﺔ ﻣﻦ اﻧﺨﻔﺎض اﻟﺘﺮدد اﻟﺬي ﻳﻮﺗﺮ ﺑﺪورﻩ ﻋﻠﻰ اﻵﻟﺔ اﻟﻤﺤﺮآﺔ ﻟﻠﻤﻮﻟﺪ ﺧﺎﺻﺎ اﻟﺘﺮﺑﻴﻨﺔ اﻟﻐﺎزﻳﺔ ﺣﻴﺚ ﻳﺴﺒﺐ ﺣﺪوث ﺣﺎﻟﺔ اﻧﻌﻜﺎس اﻟﻀﻐﻂ ﻟﻠﻀﺎﻏﻂ back pressureواﻟﺘﻲ ﺗﻌﺮف ب sergeوﻳﺘﻜﻮن اﻟﻤﺮﺣﻞ ﻣﻦ ﻣﺮﺣﻠﺘﻴﻦ اﺣﺪوهﻤﺎ إﻧﺬار واﻷﺧﺮى ﻓﺼﻞ وﻓﻰ ﻧﻈﺎم اﻟﺤﻤﺎﻳﺔ اﻟﺮﻗﻤﻲ 612 REGﻳﺘﻜﻮن اﻟﻤﺮﺣﻞ ﻣﻦ ﺛﻼﺛﺔ ﻣﺮاﺣﻞ اﻷوﻟﻰ إﻧﺬار وﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻤﺮﺣﻞ ب 48.5 HZ اﻟﺜﺎﻧﻲ ﻓﺼﻞ ﻗﺎﻃﻊ اﻟﺠﻬﺪ اﻟﻌﺎﻟﻲ HVوﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻤﺮﺣﻞ ب 47.5 HZ اﻟﺜﺎﻟﺜﺔ ﻓﺼﻞ اﻟﻤﻮﻟﺪ واﻟﺘﺮﺑﻴﻨﺔ وﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻤﺮﺣﻞ ب 45 HZ اﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﻟﻠﻤﻮﻟﺪ Generator Over Current وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﺧﺎﺻﺎ ﻓﻲ اﻷﻋﻄﺎل اﻟﺨﺎرﺟﻴﺔ ﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ ﻟﻤﺮﺣﻞ اﻟﻮﻗﺎﻳﺔ ﺣﻴﺚ ﻻﻳﺘﻢ ﻓﺼﻞ ﻗﺎﻃﻊ اﻟﻤﻮﻟﺪ إﻻ ﺑﻌﺪ ﻣﺮور زﻣﻦ ﺗﺄﺧﻴﺮ اﻟﻤﺮﺣﻞ ﻣﻬﻤﺎ آﺎﻧﺖ ﻗﻴﻤﺔ ﺗﻴﺎر اﻟﻌﻄﻞ وﺗﻌﺘﺒﺮ هﺬﻩ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﻮﻗﺎﻳﺎت اﻟﺜﺎﻧﻮﻳﺔ أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 7. اﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﻟﻠﻤﺤﻮل Transformer Over Current وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﺤﻮل ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﺧﺎﺻﺎ ﻓﻲ اﻷﻋﻄﺎل اﻟﺨﺎرﺟﻴﺔ ﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ ﻟﻤﺮﺣﻞ اﻟﻮﻗﺎﻳﺔ ﺣﻴﺚ ﻻﻳﺘﻢ ﻓﺼﻞ ﻗﺎﻃﻊ اﻟﻤﺤﻮل إﻻ ﺑﻌﺪ ﻣﺮور زﻣﻦ ﺗﺄﺧﻴﺮ اﻟﻤﺮﺣﻞ ﻣﻬﻤﺎ آﺎﻧﺖ ﻗﻴﻤﺔ ﺗﻴﺎر اﻟﻌﻄﻞ وﺗﻌﺘﺒﺮ هﺬﻩ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﻮﻗﺎﻳﺎت اﻟﺜﺎﻧﻮﻳﺔ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﺨﻄﺄ اﻷرﺿﻲ ﻟﻠﻤﺤﻮل Transformer Ground Fault وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﺤﻮل ﻣﻦ اﻟﺨﻄﺎء اﻷرﺿﻲ داﺧﻞ أو ﺧﺎرج اﻟﻤﺤﻮل اى ﺑﺎﻟﻤﻠﻔﺎت أو ﻋﻮازل اﻟﻤﺤﻮل أو ﻋﻮازل أﺑﺮاج ﻧﻘﻞ اﻟﻘﺪرة ﺣﻴﺚ ﻳﺘﻢ وﺿﻊ ﻣﺤﻮل ﺗﻴﺎر ﻋﻠﻰ اﻟﻨﻘﻄﺔ اﻟﻨﺠﻤﻴﺔ اﻟﻤﺆرﺿﺔ ﻟﻠﻤﺤﻮل وﻋﻨﺪ ﺣﺪوث ﺧﻄﺎء أرﺿﻰ ﻳﻤﺮ ﺗﻴﺎر ﻓﻲ اﻟﻨﻘﻄﺔ اﻟﻨﺠﻤﻴﺔ ﺑﺴﺒﺐ ﻋﺪم ﺗﺴﺎوى اﻟﺘﻴﺎر ﻓﻲ اﻷﻃﻮار اﻟﺜﻼﺛﺔ ﻓﻴﺘﻢ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ وﻋﺰل اﻟﻤﻮﻟﺪ وﻻ ﻳﺘﻢ ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ إﻻ إذا آﺎن اﻟﺘﻴﺎر ﻣﺴﺎوي ﻟﻘﻴﻤﺔ ﺗﻌﺪﻳﻞ ﻣﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ ﻣﻊ اﻟﺘﺄﺧﻴﺮ اﻟﺰﻣﻨﻲ وﺗﻌﺘﺒﺮ هﺬﻩ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﻮﻗﺎﻳﺎت اﻟﺜﺎﻧﻮﻳﺔ اﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﻟﻠﻤﻮﻟﺪ Generator Over Current وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر وهﻰ أﺷﺒﻪ ﺑﺎﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺤﻤﻞ Over loadﺣﻴﺚ ﻳﺘﻢ ﺗﻌﺪﻳﻞ ﻣﺮﺣﻞ اﻟﻮﻗﺎﻳﺔ ﻋﻨﺪ 1.5 INﻓﻤﺎ ﻓﻮق ﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ ﻟﻠﻤﺮﺣﻞ اﻟﻮﻗﺎﻳﺔ وﺗﻌﺘﺒﺮ هﺬﻩ اﻟﻮﻗﺎﻳﺔ ﻣﻦ اﻟﻮﻗﺎﻳﺎت اﻟﺜﺎﻧﻮﻳﺔ اﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺠﻬﺪ ﻟﻠﻤﻮﻟﺪ Generator Over voltage وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ زﻳﺎدة اﻟﺠﻬﺪ ﺣﻴﺚ ﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻤﺮﺣﻞ ﻋﻨﺪ 1.1VNﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ ﻟﻠﻤﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ اﻟﻮﻗﺎﻳﺔ ﻣﻦ هﺒﻮط اﻟﺠﻬﺪ ﻟﻠﻤﻮﻟﺪ Generator under voltage وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ هﺒﻮط اﻟﺠﻬﺪ ﺣﻴﺚ ﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻤﺮﺣﻞ ﻋﻨﺪ -1.1VNﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ ﻟﻠﻤﺮﺣﻞ اﻟﺤﻤﺎﻳﺔ وﻋﺎدﺗﺎ ﻳﻜﻮن 70% VN اﻟﺤﻤﺎﻳﺔ ﻣﻦ ﻋﺪم اﺗﺰان اﻟﺤﻤﻞ Unbal. load وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ ﺣﺎﻟﺔ ﻋﺪم اﺗﺰان اﻟﺤﻤﻞ واﻟﺘﻲ ﺗﻌﺮف ﺑﺤﺎﻟﺔ اﻟﺘﺘﺎﺑﻊ اﻟﻮﺟﻬﻰ اﻟﺴﺎﻟﺐ negative phase sequenceواﻟﺘﻲ ﺗﺴﺒﺐ ارﺗﻔﺎع ﺳﺮﻳﻊ ﻓﻲ درﺟﺔ ﺣﺮارة ﻣﻠﻔﺎت ﻋﻤﻮد اﻟﻤﻮﻟﺪ وأﻳﻀﺎ ﺣﺚ ﻓﻲ ﺟﺴﻢ ﻋﻤﻮد اﻟﻤﻮﻟﺪ أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ
• 8. وﻓﻰ ﻧﻈﺎم اﻟﺤﻤﺎﻳﺔ اﻟﺮﻗﻤﻲ 612 REGﻳﺘﻜﻮن اﻟﻤﺮﺣﻞ ﻣﻦ ﻣﺮﺣﻠﺘﻴﻦ إﻧﺬار وﻓﺼﻞ آﻤﺎ ﻓﻲ اﻟﺸﻜﻞ أﻋﻼﻩ ﻣﻊ وﺟﻮد ﺗﺄﺧﻴﺮ زﻣﻨﻲ آﺒﻴﺮ ﻧﺴﺒﻴﺎ اﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﻟﻠﻤﻮﻟﺪ ﺑﺰﻣﻦ ﻋﻜﺴﻲ Inverse time Over Current وﻓﻴﻬﺎ ﻳﺘﻢ ﺣﻤﺎﻳﺔ اﻟﻤﻮﻟﺪ ﻣﻦ زﻳﺎدة اﻟﺘﻴﺎر ﻣﻊ ﺗﺄﺧﻴﺮ زﻣﻨﻲ ﻋﻜﺴﻲ ﺑﺤﻴﺚ آﻠﻤﺎ زادة ﻗﻴﻤﺔ ﺗﻴﺎر اﻟﻌﻄﻞ آﻠﻤﺎ آﺎن اﻟﺘﺄﺧﻴﺮ اﻟﺰﻣﻨﻲ اﻗﻞ آﻤﺎ ﻓﻲ اﻟﺸﻜﻞ أدﻧﺎﻩ وﻓﻰ ﻧﻈﺎم اﻟﺤﻤﺎﻳﺔ اﻟﺮﻗﻤﻲ 612 REGﻳﺘﻢ ﺗﻌﺪﻳﻞ اﻟﻤﺮﺣﻞ ﻋﻨﺪ اﻗﻞ ﻗﻴﻤﺔ ﻟﺘﻴﺎر اﻟﻌﻄﻞ ب 1.1 IN وهﻰ اﻗﻞ ﻗﻴﻤﺔ ﻟﺘﻴﺎر ﺗﺸﻐﻴﻞ اﻟﺤﻤﺎﻳﺔ وﻳﻜﻮن زﻣﻦ اﻟﻔﺼﻞ ﻟﺤﻈﻲ ﻋﻨﺪﻣﺎ ﺗﻜﻮن ﻗﻴﻤﺔ ﺗﻴﺎر اﻟﻌﻄﻞ اآﺒﺮ ﻣﻦ 3IN آﻤﺎ ﺗﺴﺘﺨﺪم ﻣﺜﻞ هﺬﻩ اﻟﻤﺮﺣﻼت ﻓﻲ اﻟﻮﻗﺎﻳﺔ ﻣﻦ زﻳﺎدة اﻟﺤﻤﻞ Over loadﻋﻠﻰ اﻟﻤﻮﻟﺪ أو اﻟﻤﺤﻮل أﻋﺪاد م/ ﺻﺎﻟﺢ ﺳﻌﻴﺪ ﺑﻮﺣﻠﻴﻘﺔ | 8,375 | 8,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2015-18 | latest | en | 0.22758 |
https://www.jiskha.com/display.cgi?id=1301932601 | 1,516,213,933,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886952.14/warc/CC-MAIN-20180117173312-20180117193312-00473.warc.gz | 930,258,774 | 4,226 | # Physics 203 ( College)
posted by .
A disk of radius 12 cm, initially at rest, begins rotating about its axis with a constant angular acceleration of 7.8 rad/s2. What are the following values at t = 4.0 s?
(a) the angular velocity of the disk
(b) the tangential acceleration at and the centripetal acceleration ac of a point at the edge of the disk
at =
ac =
• Physics 203 ( College) -
I don't know how to find ac, but I can help you with at and angular velocity of the disk.
(a) (7.8 rad/s^2)/4.0s = 31.2=angular velocity
at=7.8*.12=0.936
• Physics 203 ( College) -
Tangential acceleration is the instantaneous linear acceleration associated with the particle at distance (r) from the center of rotation.
thus
Centripetal Accel= (angular accel)(r)
• Physics 1301 -
= (31.2^2)*(0.12 m)
= 116.8
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A disk with mass m = 8.1 kg and radius R = 0.33 m begins at rest and accelerates uniformly for t = 18 s, to a final angular speed of ω = 34 rad/s. What is the angular acceleration of the disk?
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https://slave2.omega.jstor.org/citation/text/10.2307/j.ctt6wpz0p.17 | 1,632,774,887,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00471.warc.gz | 553,016,323 | 1,554 | @inbook{10.2307/j.ctt6wpz0p.17, ISBN = {9780691161853}, URL = {http://www.jstor.org/stable/j.ctt6wpz0p.17}, abstract = {Here, we present applications to polynomial matrix equations, algebraic Riccati equations, and linear quadratic regulators. Without attempting to develop indepth exposition of the topics (this would take us too far afield), we present these applications in basic forms. Maximal invariant semidefinite or neutral subspaces will play a key role.The approach to studying polynomial equations using companion matrices of Section 5.12 extends to polynomial matrix equations. Consider the matrix equation${Z^n} + {A_{n - 1}}{Z^{n - 1}} - \cdots + {A_1}Z + {A_0} - 0$, (14.1.1)where${A_0}, \ldots ,{A_{n - 1}} \in {H^{m \times m}}$are given and$Z \in {H^{m \times m}}$is the unknown matrix. LetCbe theblock companion matrixcorresponding to equation (14.1.1):$\left( {\begin{array}{*{20}{c}} 0 & {{I_m}} & 0 & 0 & \cdots & 0 \\ 0 & 0 & {{I_m}} & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & {{I_m}} \\ { - {A_0}} & { - {A_1}} & { - {A_2}} & { - {A_3}} & \cdots & { - {A_{n - 1}}} \\ \end{array}} \right) \in {H^{(mn) \times (mn)}}$Theorem 14.1.1. There exists a one-to-one}, bookauthor = {Leiba Rodman}, booktitle = {Topics in Quaternion Linear Algebra}, pages = {328--338}, publisher = {Princeton University Press}, title = {Matrix equations}, year = {2014} } | 495 | 1,425 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-39 | latest | en | 0.606676 |
https://termsdepot.com/superconductivity/ | 1,679,776,370,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00214.warc.gz | 636,656,042 | 10,175 | # Superconductivity
Superconductivity is a phenomenon of zero electrical resistance occurring in certain materials when cooled below a critical temperature. It was discovered by Heike Kamerlingh Onnes in 1911, who found that the resistance of a mercury wire disappeared entirely when the wire was cooled to 4.2 Kelvin.
##### What is principle of superconductivity?
The principle of superconductivity is the ability of certain materials to conduct electricity with zero resistance. This means that electrical current can flow freely through these materials without losing any energy to heat. This is a very useful property for many applications, such as electrical power transmission and magnetic levitation.
Superconductivity is a quantum mechanical phenomenon that occurs when a material is cooled to very low temperatures. At these low temperatures, the material's electrons are able to pair up and form a new type of quantum state known as a Cooper pair. In this state, the electrons are able to move freely without resistance.
The Cooper pairs are held together by a force known as the electron-phonon interaction. This is a force that exists between the electrons and the lattice of atoms that make up the material. The lattice vibrations (phonons) help to hold the Cooper pairs together.
When a material is cooled to its superconducting critical temperature, the electrons are able to form Cooper pairs and the material becomes a superconductor.
#### What is superconductivity give an example?
Superconductivity is a phenomenon of exactly zero electrical resistance and expulsion of magnetic flux fields occurring in certain materials when cooled below a characteristic critical temperature. It was discovered by Heike Kamerlingh Onnes in 1911. He was studying the resistance of solid mercury when he observed that it suddenly dropped to zero at 4.2 Kelvin.
One of the most famous examples of superconductivity is the Meissner effect. This is the ability of a superconducting material to completely expel an applied magnetic field. This effect can be demonstrated by levitating a magnet above a superconducting material.
### What is superconductivity used for?
Superconductivity is used in a variety of electronic applications. One common use is in magnetic resonance imaging (MRI) machines, where superconducting magnets are used to generate a strong magnetic field. This magnetic field is then used to produce images of the inside of the human body.
Another common use for superconductivity is in particle accelerators, such as the Large Hadron Collider (LHC). In these devices, superconducting magnets are used to accelerate particles to very high energies. These particles can then be used for a variety of purposes, such as studying the structure of matter or testing theories of physics.
Finally, superconductivity is also used in a variety of other electronic devices, such as SQUIDs (superconducting quantum interference devices) and Josephson junctions. These devices exploit the unique properties of superconductors to create highly sensitive sensors or to generate very high-frequency signals.
##### What causes superconductivity?
There are many different types of superconductivity, but all of them involve the formation of pairs of electrons (called Cooper pairs) that are bound together by an attractive force. These pairs are able to move freely through the material without resistance.
The exact mechanism that causes the formation of Cooper pairs is still not fully understood, but it is thought to involve the interaction of the electrons with the lattice of the material. This interaction causes the electrons to become "correlated" with each other, meaning that their behavior is linked. This correlation is what gives rise to the superconductivity. Is water a superconductor? No, water is not a superconductor. Superconductivity is a quantum mechanical phenomenon that occurs in certain materials when they are cooled below a certain critical temperature. Water does not exhibit this behavior. | 771 | 4,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-14 | longest | en | 0.93727 |
https://betterlesson.com/lesson/543930/using-formulas-to-solve-trig-equations?from=mtp_lesson | 1,632,308,870,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00203.warc.gz | 179,586,434 | 23,764 | Using Formulas to Solve Trig Equations
Objective
SWBAT solve trig equations that must be simplified using trig identities.
Big Idea
The trig equations are getting more complex, how can we simplify them before solving?
Launch and Explore
15 minutes
Yesterday we solved equations that involved trigonometry and we found that there were an infinite number of solutions because the trig functions were periodic. Today we are going to build on that concept by continuing to solve trig equations, but today we are going to have to simplify them using the trig identities. Like yesterday, the most important points when solving these equations are staying organized and thinking of every angle that is a solution, not just the one our calculator gives us.
I give my students this worksheet and have them work with their table for about 15 minutes to get them thinking about the problems. They may notice that these equations look much more complex than yesterday’s. That is a good observation, so I tell them to try to simplify them to make them easier. For equation #1, for example, students may not want to work with the equation csc(4x) = 5/3, so I ask them if they could rewrite it using a more familiar trig function. Or for the third equation, many students recognize that the quantity on the left side of the equation can be simplified using the sin(A + B) formula.
I am not looking for students to get all of the correct answers at this point, but to start thinking about these problems. If some students finish a problem entirely, I will suggest that they check their answer on a graphing calculator.
Share
20 minutes
When it is time to share, I will usually begin with equation #3 from the worksheet since it is the most complicated. I will choose a student who recognized the need to simplify using the sin(A + B) formula and have them share their thoughts. After the equation has been simplified to sin(2x + 10°) = sqrt(2)/2, then we can use inverses to find the two angles expressions that would work. Here is an example of how to keep the work organized so no solutions are missed.
One common mistake is that students will use the (2x + 10°) = 45° + 360°n equation and solve for x to get x = 17.5° + 180°n. Then, they will use 17.5° to find another angle (162.5°) that has the same y-coordinate, and think that it is other correct answer. If students plug it in to the original equation, they will find that it does not work. I stress that it is important to find the two angle expressions before we algebraically solve for x.
After our algebraic method of solving, I put up this graph and ask students how we could use it to solve the equation. Then I ask them what would happen if we did not set up the second angle expression 135° + 360°n = (2x + 10°). Many students will realize that we would be missing half of the solutions. They may also notice that we would only be getting the solution on the left side of each peak, not on the right side.
There may not be time to go over the other two equations completely, but I at least get them going on the right track by having students explain what their thinking was to simplify. For example, for the second equation, choose a student who substituted 2cos2x – 1 in for cos(2x) and then factored. This will be enough to get students on the right track if they are not already.
Summarize
15 minutes
To end this lesson, I ask students to reflect on the three equations and ask what was the important step to simplifying all of these equations so that it could be rewritten as something we knew how to work with. During the discussion, most students realize that we had to use one of trig identities to rewrite it in a simpler form. I stress that these trig identities are tools that can be helpful to us, and allow us to solve equations that we wouldn't be able to solve otherwise.
Here is an assignment for students to get some practice with solving trig equations. I discuss one of the questions in the video below.
Unable to display content. Adobe Flash is required. | 887 | 4,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-39 | latest | en | 0.966197 |
https://blogs.msdn.microsoft.com/visualizeparallel/2010/11/23/reducing-memory-usage-to-increase-concurrency/ | 1,508,417,357,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823284.50/warc/CC-MAIN-20171019122155-20171019142155-00021.warc.gz | 663,700,931 | 14,781 | # Reducing Memory Usage to Increase Concurrency
There are multiple ways for a computer to solve permutation problems. In an exercise to show the performance of different solutions, I created a couple of different ways to solve the problem of knowing if a certain sequence could be transformed to another sequence given a set of legal permutations. All of my little puzzle solving applications take a sequence of numbers 0 to 8, and see if following valid permutation rules, the given sequence can be ordered. The amount of possible permutations for a sequence of nine numbers is 9! (362880). Depending upon what the valid permutations are, it can be impossible to transform one sequence to another sequence. The way my applications would solve this is by calculating all possible legal permutations of the given sequence. What makes this problem eligible for a concurrent solution is that a sequence may be able to be transformed into multiple other sequences at any given step, and all of those sequences can be analyzed independent of each other (i.e. concurrently).
For some of the “solutions”, the bottlenecks were practically intentional. I would run the application under the Concurrency Visualizer, look at places where a thread was blocked, and sure enough, it was blocked on a lock that is in my application’s code. That was the purpose of the exercise though: to practice analyzing performance sessions to find the application’s bottle-necks. The Concurrency Visualizer did help me find a contention which I wasn’t aware of.
One of my solutions was practically lock free. The application did very little work inside of the critical sections and should have efficiently utilized the CPU. For the most part it did. But the CPU utilization view of the Concurrency Visualizer showed that the application was spiking instead of exhibiting a near constant usage, which I would expect of a mostly lock free application. Looking at the Visualizer’s Threads view there was a recognizable pattern of having all of the threads, but one, stuck in contention and all waiting on the single running thread. I wasn’t too surprised, but I checked the Current stack tab of the blocked threads to double check that the critical section was one in the applications code. It never was.
All of the blocked critical sections looked something like
clr.dll!CLREvent::WaitEx
clr.dll!WKS::gc_heap::wait_for_gc_done
clr.dll!WKS::gc_heap::try_allocate_more_space
clr.dll!WKS::GCHeap::Alloc
clr.dll!AllocateArrayEx
clr.dll!JIT_NewArr1
mscorlib.dll!System.Collections.Generic.List`1[System.__Canon]..ctor
numberpuzzle.dll!NumberPuzzle.NumberPuzzleBase.EvaluateState
Okay, so Garbage Collector has a lock. There’s no surprise there, the GC has to manage memory. The frequency in which the GC performs operations is based off of the number of objects allocated, so perhaps there might be an object being allocated that I’m not aware of which is causing an increase of GC work. To find out, I ran the Visual Studio performance wizard, but this time instead of selecting the option to “Visualize the behavior of a multithreaded application”, I chose the option to “Track managed memory allocation”. After a few runs I did discover something. The second most allocated object was the System.Thread.WaitCallback object. This was a discovery because the application code never created a WaitCallback object. But the profiler told me that on average, the application allocated 88373454 bytes with %12.8 of the bytes being WaitCallback objects. Thus, I couldn’t account for on eighth of the allocated objects.
All of those allocations were coming from one line:
What is happening at this point in the application is that it has discovered that there is one more sequence permutation to evaluate and it’s adding that work to the ThreadPool. The documentation for QueueUserWorkItem shows that it takes a WaitCallback object in addition to another object. But EvalWrapper is a method with a method signature of
protected void EvalWrapper(Object obj)
, not a WaitCallback. So what’s going on here? The compiler sees that I want to call the EvalWrapper method and since the signature for EvalWrapper matches the signature of WaitCallback, the compiler is instantiating a WaitCallback instance on the applications behalf and passing it along to QueueUserWorkItem. And it’s allocating this WaitCallback every time.
With this new information in hand, I modify the application to create a WaitCallback member instance for this class (which takes EvalWrapper as a parameter) and then change the call to QueueUserWorkItem to use the member instance. After running the application through the memory profiler again the application now allocates on average 78214686 bytes. Since there’s noticeably less memory allocated, there’s less work for the Garbage Collector to do, which leaves more CPU time for the application and increases concurrency.
Jared Van Leeuwen – Parallel Computing Platform | 1,001 | 4,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-43 | latest | en | 0.956013 |
https://www.my-mooc.com/zh/mooc/big-data-analytics-adelaidex-analyticsx/ | 1,597,440,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739370.8/warc/CC-MAIN-20200814190500-20200814220500-00044.warc.gz | 698,667,438 | 36,617 | list 10个序列
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Gain essential skills in today’s digital age to store, process and analyse data to inform business decisions.
In this course, part of the Big Data MicroMasters program, you will develop your knowledge of big data analytics and enhance your programming and mathematical skills. You will learn to use essential analytic tools such as Apache Spark and R.
Topics covered in this course include:
• cloud-based big data analysis;
• predictive analytics, including probabilistic and statistical models;
• application of large-scale data analysis;
• analysis of problem space and data needs.
By the end of this course, you will be able to approach large-scale data science problems with creativity and initiative.
• How to develop algorithms for the statistical analysis of big data;
• Knowledge of big data applications;
• How to use fundamental principles used in predictive analytics;
• Evaluate and apply appropriate principles, techniques and theories to large-scale data science problems.
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前提
Candidates pursuing the MicroMasters program are advised to completeProgramming for Data Science,Computational Thinking and Big Data&Big Data Fundamentalsbefore undertaking this course.
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课程大纲
Section 1: Simple linear regression
Fit a simple linear regression between two variables in R;Interpret output from R;Use models to predict a response variable;Validate the assumptions of the model.
Section 2: Modelling data
Adapt the simple linear regression model in R to deal with multiple variables;Incorporate continuous and categorical variables in their models;Select the best-fitting model by inspecting the R output.
Section 3: Many models
Manipulate nested dataframes in R;Use R to apply simultaneous linear models to large data frames by stratifying the data;Interpret the output of learner models.
Section 4: Classification
Adapt linear models to take into account when the response is a categorical variable;Implement Logistic regression (LR) in R;Implement Generalised linear models (GLMs) in R;Implement Linear discriminant analysis (LDA) in R.
Section 5: Prediction using models
Implement the principles of building a model to do prediction using classification;Split data into training and test sets, perform cross validation and model evaluation metrics;Use model selection for explaining data with models;Analyse the overfitting and bias-variance trade-off in prediction problems.
Section 6: Getting bigger
Set up and apply sparklyr;Use logical verbs in R by applying native sparklyr versions of the verbs.
Section 7: Supervised machine learning with sparklyr
Apply sparklyr to machine learning regression and classification models;Use machine learning models for prediction;Illustrate how distributed computing techniques can be used for “bigger” problems.
Section 8: Deep learning
Use massive amounts of data to train multi-layer networks for classification;Understand some of the guiding principles behind training deep networks, including the use of autoencoders, dropout, regularization, and early termination;Use sparklyr and H2O to train deep networks.
Section 9: Deep learning applications and scaling up
Understand some of the ways in which massive amounts of unlabelled data, and partially labelled data, is used to train neural network models;Leverage existing trained networks for targeting new applications;Implement architectures for object classification and object detection and assess their effectiveness.
Section 10: Bringing it all together
Consolidate your understanding of relationships between the methodologies presented in this course, theirrelative strengths, weaknesses and range of applicability of these methods.
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教师
Lewis Mitchell
Lecturer in Applied Mathematics
Simon Tuke
Lecturer in Statistics
David Suter
Professor of Computer Science
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EdX est une plateforme d'apprentissage en ligne (dite FLOT ou MOOC). Elle héberge et met gratuitement à disposition des cours en ligne de niveau universitaire à travers le monde entier. Elle mène également des recherches sur l'apprentissage en ligne et la façon dont les utilisateurs utilisent celle-ci. Elle est à but non lucratif et la plateforme utilise un logiciel open source.
EdX a été fondée par le Massachusetts Institute of Technology et par l'université Harvard en mai 2012. En 2014, environ 50 écoles, associations et organisations internationales offrent ou projettent d'offrir des cours sur EdX. En juillet 2014, elle avait plus de 2,5 millions d'utilisateurs suivant plus de 200 cours en ligne.
Les deux universités américaines qui financent la plateforme ont investi 60 millions USD dans son développement. La plateforme France Université Numérique utilise la technologie openedX, supportée par Google.
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0/5 | 1,055 | 4,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-34 | latest | en | 0.79653 |
https://www.quizover.com/course/section/conceptual-questions-4-4-centripetal-force-by-openstax | 1,545,077,182,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00447.warc.gz | 993,348,585 | 21,992 | # 0.3 4.4 centripetal force (Page 3/9)
Page 3 / 9
$N\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{\mathrm{mv}}^{2}}{r}\text{.}$
Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is $N\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ , and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,
$N\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\text{mg}\text{.}$
Now we can combine the last two equations to eliminate $N$ and get an expression for $\theta$ , as desired. Solving the second equation for $N=\text{mg}/\left(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)$ , and substituting this into the first yields
$\text{mg}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }{\text{cos}\phantom{\rule{0.25em}{0ex}}\theta }=\frac{{\text{mv}}^{2}}{r}$
$\begin{array}{rrr}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{tan}\left(\theta \right)& =& \frac{{\mathrm{mv}}^{2}}{r}\\ \text{tan}\phantom{\rule{0.25em}{0ex}}\theta & =& \frac{{v}^{2}}{\text{rg.}}\end{array}$
Taking the inverse tangent gives
$\theta ={\text{tan}}^{-1}\left(\frac{{v}^{2}}{\text{rg}}\right)\phantom{\rule{0.25em}{0ex}}\text{(ideally banked curve, no friction).}$
This expression can be understood by considering how $\theta$ depends on $v$ and $r$ . A large $\theta$ will be obtained for a large $v$ and a small $r$ . That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that $\theta$ does not depend on the mass of the vehicle.
## What is the ideal speed to take a steeply banked tight curve?
Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.
Strategy
We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.
Solution
Starting with
$\text{tan}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{v}^{2}}{\text{rg}}$
we get
$v=\left(\text{rg}\phantom{\rule{0.25em}{0ex}}\text{tan}\phantom{\rule{0.25em}{0ex}}\theta {\right)}^{1/2}\text{.}$
Noting that tan 65.0º = 2.14, we obtain
$\begin{array}{lll}v& =& {\left[\left(\text{100 m}\right)\left(9.80 m{\text{/s}}^{2}\right)\left(2\text{.}\text{14}\right)\right]}^{1/2}\\ & =& \text{45.8 m/s.}\end{array}$
Discussion
This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.
Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.
## Section summary
• Centripetal force ${\text{F}}_{\text{c}}$ is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity $v$ and has magnitude
$\phantom{\rule{0.25em}{0ex}}{F}_{\text{c}}={\text{ma}}_{\text{c}}\text{,}$
which can also be expressed as
$\left(\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ \begin{array}{}\text{or}\\ {F}_{\text{c}}=\text{mr}{\omega }^{2}\end{array}\end{array},}$
## Conceptual questions
If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.
Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?
If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain.
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how to synthesize TiO2 nanoparticles by chemical methods
Zubear
how did you get the value of 2000N.What calculations are needed to arrive at it
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2039 Words 9 Pages
Register to read the introduction… This will be called the interactive role. The computer takes the remaining roles. On the game display, the interactive role is displayed in colour, and the remaining roles are displayed in gray. Also, the information for the interactive roles is displayed but the information for other roles is hidden (with the exception of backorder at the supply chain member immediately upstream from the interactive supply chain member). In the example screen displayed above, the distributor is the interactive role. Also, by downstream, we mean the direction of the supply chain leading to the external demand, and by upstream, we mean in the direction of the factory. In addition, we refer to components of the supply chain as …show more content…
Recall (you can see this from the previous screen shot, on page 2) that the initial inventory was 4, and both Delay 1 and Delay 2 contained 4 units. This holds true for each of the supply chain facilities. Now, steps one and two have been completed. As you can see from the Order Entry dialog box, there is initially no backorder and no order from the Wholesaler. Since the starting inventory was 8 (the initial 4 plus 4 from Delay 1), 8 remain in inventory. Delay 2 is now empty. This is the first round, so there are no previous orders from the Distributor to the Factory, so the recent order box read 0. However, it this was a later round, and there was a order placed by the Distributor to the Factory in the previous round, it would appear in the recent order …show more content…
Note that the backorder box (indicated by an arrow in the figure above) on the right side of the screen indicates the current level of backorder, while this dialog box shows the level of backorder at the beginning of this round, before the player (in this case, the Distributor) attempted to fill downstream orders (in this case, from the Wholesaler). At this point, enter a demand amount. This can be zero, or any other integer. Remember that you are trying to balance inventory holding costs and shortage costs. Also, by looking at the backorder box at your supplier (in this example, the Factory), you can see how much backorder your upstream supplier already has to fill. That is, how many items you have ordered in prior rounds but have not yet received. Once an amount has been entered, the remaining upstream supply chain members play automatically, and the screen is updated. If you enter an order of 3 for the Distributor, the remainder of Week 1 play is carried out, and the Distributor portion of the screen looks
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# Documentation Center
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## Estimate Model Parameters and Initial States (Code)
This example shows how to estimate the initial state and parameters of a model.
This example requires Simscape®.
RC Circuit Model
The Simulink model, sdoRCCircuit, models a simple resistor-capacitor (RC) circuit.
```open_system('sdoRCCircuit');
```
Estimation Problem
You use the measured data to estimate the RC model parameter and state values.
Measured output data:
• Capacitor voltage, output of the PS-Simulink Converter block
Parameter:
• Capacitance, C1, used by the C1 block
State:
• Initial voltage of the capacitor, C1
Define the Estimation Experiment
Get the measured data.
```load sdoRCCircuit_ExperimentData
```
The variables time and data are loaded into the workspace, where data is the measured capacitor voltage for times time.
Create an experiment object to store the experimental voltage data.
```Exp = sdo.Experiment('sdoRCCircuit');
```
Create an object to store the measured capacitor voltage output.
```Voltage = Simulink.SimulationData.Signal;
Voltage.Name = 'Voltage';
Voltage.BlockPath = 'sdoRCCircuit/PS-Simulink Converter';
Voltage.PortType = 'outport';
Voltage.PortIndex = 1;
Voltage.Values = timeseries(data,time);
```
Add the measured capacitor data to the experiment as the expected output data.
```Exp.OutputData = Voltage;
```
Compare the Measured Output and the Initial Simulated Output
Create a simulation scenario using the experiment and obtain the simulated output.
```Simulator = createSimulator(Exp);
Simulator = sim(Simulator);
```
Search for the voltage signal in the logged simulation data.
```SimLog = find(Simulator.LoggedData,get_param('sdoRCCircuit','SignalLoggingName'));
Voltage = find(SimLog,'Voltage');
```
Plot the measured and simulated data.
The model response does not match the experimental output data.
```plot(time,data,'ro',Voltage.Values.Time,Voltage.Values.Data,'b')
title('Simulated and Measured Responses Before Estimation')
legend('Measured Voltage','Simulated Voltage')
```
Specify the Parameters to Estimate
Select the capacitance parameter from the model. Specify an initial guess for the capacitance value (460 uF) and a minimum bound (0 F).
```p = sdo.getParameterFromModel('sdoRCCircuit','C1');
p.Value = 460e-6;
p.Minimum = 0;
```
Define the Estimation Objective Function
Create an estimation objective function to evaluate how closely the simulation output, generated using the estimated parameter value, matches the measured data.
Use an anonymous function with one input argument that calls the sdoRCCircuit_Objective function. We pass the anonymous function to sdo.optimize, which evaluates the function at each optimization iteration.
```estFcn = @(v) sdoRCCircuit_Objective(v,Exp);
```
The sdoRCCircuit_Objective function:
• Has one input argument that specifies the estimated circuit capacitance value.
• Has one input argument that specifies the experiment object containing the measured data.
• Returns a vector of errors between simulated and experimental outputs.
The sdoRCCircuit_Objective function requires two inputs, but sdo.optimize requires a function with one input argument. To work around this, estFcn is an anonymous function with one input argument, v, but it calls sdoRCCircuit_Objective using two input arguments, v and Exp.
For more information regarding anonymous functions, see "Anonymous Functions""Anonymous Functions".
The optimization solver minimizes the residual errors. For more details on how to write an objective/constraint function to use with the sdo.optimize command, type help sdoExampleCostFunction at the MATLAB command prompt.
To examine the estimation object function in more detail, type edit sdoRCCircuit_Objective at the MATLAB command prompt.
```type sdoRCCircuit_Objective
```
```function vals = sdoRCCircuit_Objective(v,Exp)
%SDORCCIRCUIT_OBJECTIVE
%
% The sdoRCCircuit_Objective function is used to compare model
% outputs against experimental data.
%
% vals = sdoRCCircuit_Objective(v,Exp)
%
% The |v| input argument is a vector of estimated model parameter values
% and initial states.
%
% The |Exp| input argument contains the estimation experiment data.
%
% The |vals| return argument contains information about how well the
% model simulation results match the experimental data and is used by
% the |sdo.optimize| function to estimate the model parameters.
%
% See also sdo.optimize, sdoExampleCostFunction, sdoRCCircuit_cmddemo
%
% Copyright 2012 The MathWorks, Inc.
%%
% Define a signal tracking requirement to compute how well the model output
% matches the experiment data. Configure the tracking requirement so that
% it returns the tracking error residuals (rather than the
% sum-squared-error) and does not normalize the errors.
%
r = sdo.requirements.SignalTracking;
r.Type = '==';
r.Method = 'Residuals';
r.Normalize = 'off';
%%
% Update the experiments with the estimated parameter values.
%
Exp = setEstimatedValues(Exp,v);
%%
% Simulate the model and compare model outputs with measured experiment
% data.
%
Simulator = createSimulator(Exp);
Simulator = sim(Simulator);
SimLog = find(Simulator.LoggedData,get_param('sdoRCCircuit','SignalLoggingName'));
Voltage = find(SimLog,'Voltage');
VoltageError = evalRequirement(r,Voltage.Values,Exp.OutputData(1).Values);
%%
% Return the residual errors to the optimization solver.
%
vals.F = VoltageError(:);
end
```
Estimate the Parameters
Use the sdo.optimize function to estimate the capacitance value.
Specify the optimization options. The estimation function sdoRCCircuit_Objective returns the error residuals between simulated and experimental data and does not include any constraints, making this problem ideal for the 'lsqnonlin' solver.
```opt = sdo.OptimizeOptions;
opt.Method = 'lsqnonlin';
```
Estimate the parameters.
```pOpt = sdo.optimize(estFcn,p,opt)
```
``` Optimization started 18-Jan-2014 17:15:59
Step-size First-order
Iter F-count f(x) optimality
0 3 54.999 1
1 6 21.0094 0.2124 17.3
2 9 11.5162 0.1273 5.9
3 12 9.59875 0.06504 1.93
4 15 9.32857 0.02715 0.645
5 18 9.28738 0.01003 0.143
6 21 9.28455 0.002316 0.0219
Local minimum possible.
lsqnonlin stopped because the final change in the sum of squares relative to
its initial value is less than the selected value of the function tolerance.
pOpt =
Name: 'C1'
Value: 1.1128e-04
Minimum: 0
Maximum: Inf
Free: 1
Scale: 0.0020
Info: [1x1 struct]
1x1 param.Continuous
```
Compare the Measured Output and the Simulated Output
Update the experiment with the estimated capacitance value.
```Exp = setEstimatedValues(Exp,pOpt);
```
Create a simulation scenario using the experiment and obtain the simulated output.
```Simulator = createSimulator(Exp);
Simulator = sim(Simulator);
```
Search for the voltage signal in the logged simulation data.
```SimLog = find(Simulator.LoggedData,get_param('sdoRCCircuit','SignalLoggingName'));
Voltage = find(SimLog,'Voltage');
```
Plot the measured and simulated data.
The simulated and measured signals match well, except for near time zero. This mismatch is because the capacitor initial voltage defined in the model does not match the initial voltage from the experiment.
```plot(time,data,'ro',Voltage.Values.Time,Voltage.Values.Data,'b')
title('Simulated and Measured Responses After Estimation')
legend('Measured Voltage','Simulated Voltage')
```
Estimate the Initial State
Add the capacitor initial voltage for the C1 block to the experiment. Set its initial guess value to 1 V.
```Exp.InitialStates = sdo.getStateFromModel('sdoRCCircuit','C1');
Exp.InitialStates.Value = 1;
```
Recreate the estimation function to use the experiment with initial state estimation
```estFcn = @(v) sdoRCCircuit_Objective(v,Exp);
```
Get the initial state and capacitance value that is to be estimated from the experiment.
```v = getValuesToEstimate(Exp);
```
Estimate the parameters.
```vOpt = sdo.optimize(estFcn,v,opt)
```
``` Optimization started 18-Jan-2014 17:16:14
Step-size First-order
Iter F-count f(x) optimality
0 5 4.82677 1
1 10 2.19679 1.57 23.6
2 15 1.34958 0.1596 0.0883
3 20 1.34358 0.05594 0.135
4 25 1.34355 0.001457 0.00085
Local minimum found.
Optimization completed because the size of the gradient is less than
the selected value of the function tolerance.
vOpt(1,1) =
Name: 'sdoRCCircuit/C1:sdoRCCircuit.C1.vc'
Value: 2.3592
Minimum: -Inf
Maximum: Inf
Free: 1
Scale: 1
dxValue: 0
dxFree: 1
Info: [1x1 struct]
vOpt(2,1) =
Name: 'C1'
Value: 2.2663e-04
Minimum: 0
Maximum: Inf
Free: 1
Scale: 0.0020
Info: [1x1 struct]
2x1 param.Continuous
```
Compare the Measured Output and the Final Simulated Output
Update the experiment with the estimated capacitance and capacitor initial voltage values.
```Exp = setEstimatedValues(Exp,vOpt);
```
Simulate the model with the estimated initial-state and parameter values and compare the simulated output with the experiment data.
```Simulator = createSimulator(Exp);
Simulator = sim(Simulator);
SimLog = find(Simulator.LoggedData,get_param('sdoRCCircuit','SignalLoggingName'));
Voltage = find(SimLog,'Voltage');
plot(time,data,'ro',Voltage.Values.Time,Voltage.Values.Data,'b')
title('Simulated and Measured Responses After Initial State and Model Parameter Estimation')
legend('Measured Voltage','Simulated Voltage')
```
Update the Model Parameter Values
Update the model with the estimated capacitance value. Do not update the model capacitor initial voltage (first element of vOpt) as it is dependent on the experiment.
```sdo.setValueInModel('sdoRCCircuit',vOpt(2));
```
Close the model
```bdclose('sdoRCCircuit')
```
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### Transcript of Haunted house game averages
1. You are a successful private detective.You have been hiredto investigate an old housewhere there are strange noises.At each stage of your investigationyou will need to answer a question. To start press the buttonWhether you get the answer correctwill determine what happens to you.Beware, if you get too many questions wrongyour story will not end well!To begin your investigation press the button 2. 1 3. The strange noiseshave always been heard at nightso you decide this is the best timeto visit the house to investigate.You drive up to the gateand get out of the car.As you walk towards the houseyou notice that there is a full moonand hope that this is not a bad omen. Press the button 4. Question 1 aThe heights (in centimeters) of 8 plants were measured.The results are shown below.12.1, 18.4, 16.5, 11.5, 17.9, 18.0, 5.8, 23.4Calculate the mean height of the plants. 14.7 cm103.1 cm 15.5 cm17.2 cm 5. Correct answerPress the button 6. Wrong answerPress the button 7. 2 8. You think that it is probably saferto look for a side entrance.You begin to walk carefullyaround the side of the houselooking for another way in.Press the button 9. Question 2 aLucy recorded her scores (out of 10) in spelling tests at school.The scores are shown below.5, 6, 8, 9, 6, 7, 9, 8, 9What is Lucys median score?97 6 8 10. Correct answerPress the button 11. Wrong answerPress the button 12. You approach the front door.You turn the handleand the door creaks open.It is dark inside.As you step insidea trapdoor opensand you plunge downwards. Press the button 13. Question 2 bGordon records his times (in seconds) for running 100 m.They are given below. 14.1, 14.2, 13.9, 14.1, 13.7, 13.5, 13.4What is Gordons median time?13.8 seconds 14.0 seconds14.1 seconds 13.9 seconds 14. Correct answerPress the button 15. Wrong answerPress the button 16. 3 17. You find the old servants entranceat the side of the house.It is overgrownand looks like it hasnt been used for years.You try the handle and it turns,the door creaks openand you enter into the kitchen. Press the button 18. Question 3 aJosie did a survey of pocket money (in \$) amongst her friends.The results of her survey are given below.10, 5, 5, 10, 20, 4, 5, 15Find the median amount of pocket money.\$9.25 \$15\$7.50 \$10 19. Correct answerPress the button 20. Wrong answerPress the button 21. To the side of the houseis a cemetery.At this time of night it is spooky,but you must go on. Press the button 22. Question 3 bThe ages of a group of students are given below. 10, 15, 10, 11, 13, 15, 15, 15Find the median age. 1412 13 15 23. Correct answerPress the button 24. Wrong answerPress the button 25. It is pitch black,you cant see your handin front of your face.You take out your torchand begin to look around.In the corner you see a coffin.Press the button 26. Question 3 cThe scores of a group of students in an English exam are given below. 82, 69, 40, 90, 55, 65, 78, 79, 58, 64Find the median score. 68676065 27. Correct answerPress the button 28. Wrong answerPress the button 29. As you land you feel sharp painsall over.You have fallen into a pit of spikes! Press the button 30. 4 31. You walk through the kitchenand into the hallway.Suddenly there is a loud clatter above.Startled, you look upand a colony of bats fly past your face. Press the button 32. Question 4 aKieran asked his friends how many brothers and sisters they have.The results are shown below. 1, 2, 0, 0, 1, 3, 1, 2, 2, 4What is the mode number of brothers and sisters? 1 and 2 1.52 1 and 3 33. Correct answerPress the button 34. Wrong answerPress the button 35. You walk over to the dooron the other side of the kitchen.As you reach for the handle youhear a spine-tingling screamfrom the other side of the door. Press the button 36. Question 4 bThe ages of a group of students are given below. 10, 14, 11, 12, 14, 13, 10, 15What is the mode age? 1210 and 1413 12 and 14 37. Correct answerPress the button 38. Wrong answerPress the button 39. You hear a scraping sound nearby.You look around and see the earthon one of the graves move slightly.Then a hand pushes upout of the ground.Press the button 40. Question 4 cFrank did a survey of pocket money (in \$) among his friends.The results of his survey are given below.10, 5, 5, 10, 20, 4, 2, 15What is the mode of the data?\$10 and \$20 \$15\$7.50 \$5 and \$10 41. Correct answerPress the button 42. Wrong answerPress the button 43. There is an open grave.As you walk past ityou are pushed hard from behind.You fall into the graveand see a figure standing over youbegin to fill the grave in with you in it! Press the button 44. You walk over the coffinand see that the lid is offand it is empty.You hear faint footstepscoming towards the room.Press the button 45. Question 4 dThe temperature (in C) in a classroom was recorded daily.The results are shown below. 19, 20, 18, 20, 19, 21, 18, 19, 22, 20What is the mode of the temperatures? 19.5C19 and 20C19 and 21C20C 46. Correct answerPress the button 47. Wrong answerPress the button 48. You hear a faint noise behind.You turn aroundand see a man standing in front of you.His face seems too smooth and white.The sharp fangs convince you of what he is.Press the button 49. 5 50. You hear strange soundscoming from upstairs.You walk towards the stairsto investigate. Press the button 51. Question 5 aThe number of people in each house on a street is given below.Number of people Frequency 0 2 1 5 2 7 3 6 4 4 5 3 6 1Calculate the mean number of people in a house. 10.62.6 2.7 4 52. Correct answerPress the button 53. Wrong answerPress the button 54. In the center of one wallis a large fireplace.You walk over to itand examine the intricate carving.One of the panels is loose.As you press it a secret door opensin the back of the fireplace.Press the button 55. Question 5 bThe number of As that a class got on a report is given below.Number of As Number of children05162434425365Calculate the mean number of As. 2.72.9 11.3 4.1 56. Correct answerPress the button 57. Wrong answerPress the button 58. Against your better judgementyou open the door and walk into the hallway.You see a man crouched over.Before your eyes he changes into a wolf.Press the button 59. Question 5 cThe number of brothers and sisters that a class have is given below. Number ofFrequencybrothers and sisters 0 6 1 8 2 7 3 5 4 2 5 1Calculate the mean number of brothers and sisters.8.34.81.9 1.7 60. Correct answerPress the button 61. Wrong answerPress the button 62. Then you hear a piercing howl.The kitchen door burst open,knocking you backwards.There in front of you is a wolfas tall as you are.Before you can react, it attacks you!Press the button 63. A women pushes up through the soiland climbs out of the grave.You turn and run awayas quickly as your legs can carry you.Press the button 64. Question 5 dThe number of goals scored by a soccer team over a seasonis given below. Number of goals Number of matches0 61 42 53 14 25 1Calculate the mean number of goals scored. 3.2 5 1.61.9 65. Correct answerPress the button 66. Wrong answerPress the button 67. A women pushes up through the soiland climbs out of the grave.You are so shockedyou cannot move.She grabs at youPress the button 68. You look aroundand see a small chamberto your side.You quickly hide inside. Press the button 69. Question 5 eThe number of pets that a group of children have is given below. Number of pets Frequency0 61 72 53 54 15 26 1Calculate the mean number of pets.3.97.42.1 1.9 70. Correct answerPress the button 71. Wrong answerPress the button 72. You hear a faint noise behind.You turn aroundand see a man stood in front of you.His face seems too smooth and white.The sharp fangs convince you of what he is.Press the button 73. 6 74. You creep upstairstrying not to make any noise.You are halfway upwhen a ghost appears at the top of the stairs.Press the button 75. Question 6 aThe number of people in each house on a street is given belowNumber of people Frequency 0 2 1 5 2 7 3 6 4 4 5 3 6 1Find the median number of people in a house. 14.52.5 2 and 3 2 76. Correct answerPress the button 77. Wrong answerPress the button 78. You creep upstairstrying not to make any noise.On the landing at the top of the stairsyou see a painting of a young womanhanging on the wall.You are sure you see the eyesin the painting move. Press the button 79. Question 6 bThe number of people in each house on a street is given below.Number of people Frequency 0 2 1 5 2 7 3 6 4 4 5 3 6 1Find the median number of people in a house. 14.52.5 2 and 3 2 80. Correct answerPress the button 81. Wrong answerPress the button 82. You step into the room beyond.Inside the rooma group of women dressed like witchesstand around in a circle. Press the button 83. Question 6 cThe number of As that a class got on a report is given below.Number of As Number of children05162434425365Find the median number of As.3 1521 84. Correct answerPress the button 85. Wrong answerPress the button 86. You step forward to look inside.The floor beneath your feet crumblesand you fall forward.You land with a bump.You look around and can seethat you are trappedat the bottom of a deep pit!Press the button 87. You cant believe what you saw.This only happens in horror stories,not in real life,but there where a few minutes agothere was a man is now a huge wolf.Press the button 88. Question 6 dThe number of brothers and sisters that a class have is given below. Number ofFrequencybrothers and sisters 06 18 27 35 42 51Find the median number of brothers and sisters. 2 15 2.51 89. Correct answerPress the button 90. Wrong answerPress the button 91. The wolf lunges at youand sinks its teeth into your arm.You pull awayand run back through the door,closing it in the approaching wolfs face.You ru | 2,786 | 9,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | latest | en | 0.836873 |
https://blog.reallymagazine.com/2016/12/16/bicycle-tracks-still-being-covered/ | 1,627,913,997,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154321.31/warc/CC-MAIN-20210802141221-20210802171221-00268.warc.gz | 153,482,360 | 7,425 | # Bicycle tracks – still being covered
As Sherlock Holmes aficionados will know, in the 1903 story ‘The Adventure of the Priory School’, Holmes determined the direction in which a bicycle was travelling simply by observing the tyre tracks which it had made – asserting that the deeper of the two wheel marks must have come from the heavier rear wheel …
[Scroll on 93 years]
But in 1996 the essay Which Way Did the Bicycle Go?…and Other Intriguing Mathematical Mysteries. (Dolciani Mathematical Expositions Series of the Mathematical Association of America, No. 18.) authors Joseph D. E. Konhauser, Daniel J. Velleman, and Stan Wagon questioned the absolute validity of Sherlock’s methodology. They described what they saw as a better method of bicycle-track-direction-detection based purely on calculus – and independent of the depth of track.
To clarify, their diagram above shows a notional tyre track in which the heavier track was made by the front wheel.
[Scroll on 6 years]
But then, in 2002, a new twist was exposed. David L Finn, (Associate Professor of Mathematics at the Rose-Hulman Institute of Technology) determined (by the use of differential geometry) that it is theoretically possible to construct a unicycle track with a bicycle. See: Can a Bicycle Create a Unicycle Track? (College Mathematics Journal, 2002). Thus the tracks which Holmes observed could have come from two bicycles instead of one. Considerably complicating the chances of solving what was an already tricky-enough case.
Professor Finn’s diagram above shows a mathematically generated version of a single tire track that can be created by a bicycle.
NOTE : Professor Stan Wagon, one of the co-authors of ‘Which Way Did the Bicycle Go?’ has designed, constructed and ridden a bicycle with square wheels. Here’s video of a tricycle built on Wagon’s principle:
ALSO SEE: There are two new (2013) books about the science of Sherlock Holmes, reviewed here by Jonathon Keats at New Scientist.
UPDATE Sep. 2020
Prof. Wagon and Prof. Velleman have launched a book of mathematical puzzles – including a chapter on bicycle/unicycle tracks, entitled Bicycle or Unicycle?: A Collection of Intriguing Mathematical Puzzles. (MAA Press: An Imprint of the American Mathematical Society)
Please note that to avoid comment spam, no e-mail addresses or web links are allowed in the message! If you include one, the message will be auto-deleted | 525 | 2,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-31 | latest | en | 0.944678 |
https://qa.answers.com/health/What_is_the_Average_number_of_weeks_in_a_month | 1,726,864,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00314.warc.gz | 434,746,712 | 52,423 | 0
# What is the Average number of weeks in a month?
Updated: 9/7/2023
Wiki User
14y ago
52 weeks in a year and 12 months gives 4.3333 weeks in a month
Wiki User
14y ago
Doug Jones
Lvl 1
3y ago
Not quite. That's the incorrect answer when 364 days are used in a year, when the actual number is closer to 365.2422 days in a year according to NASA.
Doug Jones
Lvl 1
3y ago
Not quite. That's the incorrect answer when 364 days are used in a year, when the actual number is closer to 365.2422 days in a year according to NASA.
Doug Jones
Lvl 1
3y ago
365.2422÷7= 52.1775 weeks in a year 52.1775 4.3481
Doug Jones
Lvl 1
3y ago
so there's no delete or edit option here on answers . com. what a fail. the answer is 4.3481 weeks per month on average.
Doug Jones
Lvl 1
3y ago
52.1775÷12=4.3481
Doug Jones
Lvl 1
3y ago
"average weeks in a month" shouldn't be in the "health" section either. answers . com is fairly low quality and disorganized it appears.
Wiki User
11y ago
• January = 31 Days = 4 weeks
• February = 28 Days (leap years 29) = 4 weeks
• March = 31 Days = 5 weeks
• April = 30 Days = 4 weeks
• May = 31 Days = 5 weeks
• June = 30 Days = 4 weeks
• July = 31 Days = 5 weeks
• August = 31 Days = 4 weeks
• September = 30 Days = 4 weeks
• October = 31 Days = 5 weeks
• November = 30 Days = 4 weeks
• December = 31 Days = 5 weeks
They all have 4 weeks, but February is the only one to have exactly 4 weeks (except in Leap years when it has 4 weeks and 1 day)
April, June September and November each have 30 days = 4 weeks and 2 days
The rest of the months each have 31 days = 4 weeks and 3 days
forallar ali
Lvl 2
1y ago
Well, every month includes a minimum of 4 weeks, since no month has less than 28 days. But since there are 52 weeks in a year, you could say that, on average, there are 4 1/3 weeks per month. If you wanted to be more specific, you could assume that since a year normally lasts 365.25 days, the average could be 4,348 weeks per month. In 2000, the average tropical year was 365.2421897 days, which is still the same number of significant digits as in the previous answer.
Wiki User
14y ago
there is 4 and a half weeks in 1 month
Wiki User
14y ago
Four? Do you not own a calendar?
Wiki User
14y ago
4 and then a few more days.
Wiki User
14y ago
4-5
Wiki User
12y ago
Four | 734 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.953906 |
https://stats.stackexchange.com/questions/359921/what-is-a-positively-skewed-distribution-that-can-include-zero?noredirect=1 | 1,585,885,774,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00170.warc.gz | 704,911,124 | 32,301 | # What is a positively skewed distribution that can include zero?
I'm modelling data from a behavioural task. Participants do a few hundred trials. On each trial, they see a sequence of letters at a point on the screen and one of these letters appears surrounded by a white circle. Their task is to report the letter within the circle. Any response they make can be mapped onto a point in time on the trial relative to the white circle because there are no repeats in the sequences that they see. A response of the letter in the circle would have a value of 0; one letter before would have a score of -1; the letter after would have a score of 1 and so on. We've been modelling the distribution of these temporal errors with a mixture of a uniform distribution and some other, non-uniform distribution. Up until now the non-uniform distribution has been Gaussian, but certain theoretical considerations have led us to consider that we need a positively skewed component with a domain that is bounded at zero instead of the Gaussian. I considered using the lognormal distribution, but this is a bad choice because it is undefined at zero.
What positively skewed distributions can model values of zero and greater?
I'm using Matlab. Something that has a PDF written in that language would be great (I'm a scientist, not a statistician).
• Truncating nearly any distribution below at $0$ is going to increase its skewness. That gives you a huge array of possible solutions (and every solution can be expressed in such a form). To keep this thread from being overly (and uselessly) broad, could you please be more specific about what you're modeling and what you're trying to accomplish? – whuber Jul 31 '18 at 11:41
• Thanks @whuber. I've edited the post with more detail – ivan.k Aug 1 '18 at 1:14
You could take any distribution defined on $(0,\infty)$ and simply shift it to the left by a small $\epsilon$. Then you would get negative values in $(-\epsilon,0)$, which I guess is not what you are looking for.
Alternatively, your skewed component might derive from a two-stage data generating process that might either generate a zero or a nonzero value. If this sounds like something that might be present in your application, you might want to look into , i.e., s between a point mass at zero and a second distribution, which in turn could be supported on $(0,\infty)$ or $[0,\infty)$. | 521 | 2,391 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.953546 |
https://www.topperlearning.com/forums/home-work-help-19/kindly-refer-to-question-number-5-of-ex-24-polynomials-mathematics-polynomials-63501/reply | 1,511,143,033,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805894.15/warc/CC-MAIN-20171120013853-20171120033853-00031.warc.gz | 889,111,188 | 39,931 | Question
Mon February 25, 2013 By:
In the solution of the question it is suggested that if
# Kindly refer to question number 5 of Ex 2.4 (Polynomials).
Tue February 26, 2013
Consider ax + b = 0
This implies a = 0 as well as b = 0
Reason:
ax + b = 0
=> ax + b = 0(x) + 0
Now, compare the coefficient of x and the contant term on both sides. We get:
a = 0 and b = 0
Related Questions
Fri October 27, 2017
# ðŸ™
Mon October 02, 2017 | 152 | 434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-47 | latest | en | 0.980315 |
https://espanol.libretexts.org/Matematicas/Precalculo_y_Trigonometria/Prec%C3%A1lculo_(OpenStax)/02%3A_Funciones_lineales/2.01%3A_Funciones_lineales | 1,721,335,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00413.warc.gz | 200,506,983 | 44,322 | Saltar al contenido principal
# 2.1: Funciones lineales
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##### Objetivos de aprendizaje
• Representar una función lineal.
• Determine si una función lineal es creciente, decreciente o constante.
• Interpretar la pendiente como una tasa de cambio.
• Escribir e interpretar una ecuación para una función lineal.
• Gráfica funciones lineales.
• Determinar si las líneas son paralelas o perpendiculares.
• Escribe la ecuación de una línea paralela o perpendicular a una línea dada.
Al igual que con el crecimiento de una planta de bambú, hay muchas situaciones que implican un cambio constante a lo largo del tiempo. Consideremos, por ejemplo, el primer tren maglev comercial del mundo, el Tren MagLev de Shanghai (tren Maglev en el mundo, el Tren MagLev de Shanghái (Figura$$\PageIndex{1}$$). Lleva a los pasajeros cómodamente para un viaje de 30 kilómetros desde el aeropuerto hasta la estación del metro en solo ocho minutos.
Supongamos que un tren maglev iba a recorrer una larga distancia, y que el tren mantiene una velocidad constante de 83 metros por segundo durante un periodo de tiempo una vez que se encuentra a 250 metros de la estación. ¿Cómo podemos analizar la distancia del tren a la estación en función del tiempo? En esta sección, investigaremos un tipo de función que sea útil para este propósito, y la usaremos para investigar situaciones del mundo real como la distancia del tren a la estación en un punto determinado en el tiempo.tren maglev iban a recorrer una larga distancia, y que el tren mantenga una velocidad constante de 83 metros por segundo por un periodo de tiempo una vez que esté a 250 metros de la estación. ¿Cómo podemos analizar la distancia del tren a la estación en función del tiempo? En esta sección, investigaremos un tipo de función que sea útil para este propósito, y la usaremos para investigar situaciones del mundo real como la distancia del tren a la estación en un momento dado.
## Representando funciones lineales
La función que describe el movimiento del tren es una función lineal, que se define como una función con una tasa de cambio constante, es decir, un polinomio de grado 1. Hay varias formas de representar una función lineal, incluyendo la forma de la palabra, la notación de función, la forma tabular y la forma gráfica. Describiremos el movimiento del tren como una función utilizando cada método.
### Representación de una función lineal en forma de palabra
Comencemos describiendo la función lineal en palabras. Para el problema del tren que acabamos de considerar, se puede usar la siguiente frase de palabra para describir la relación de función.
• La distancia del tren a la estación es función del tiempo durante el cual el tren se mueve a una velocidad constante más su distancia original de la estación cuando comenzó a moverse a velocidad constante.
La velocidad es la tasa de cambio. Recordemos que una tasa de cambio es una medida de la rapidez con la que cambia la variable dependiente con respecto a la variable independiente. La tasa de cambio para este ejemplo es constante, lo que significa que es la misma para cada valor de entrada. A medida que el tiempo (entrada) aumenta en 1 segundo, la distancia correspondiente (salida) aumenta en 83 metros. El tren comenzó a moverse a esta velocidad constante a una distancia de 250 metros de la estación.
### Representación de una función lineal en notación de funciones
Otro enfoque para representar funciones lineales es mediante el uso de la notación de funciones. Un ejemplo de notación de función es una ecuación escrita en la forma conocida como la forma pendiente-intercepción de una línea, donde xes el valor de entrada,$$m$$ es la tasa de cambio, y$$b$$ es el valor inicial de la variable dependiente.
\begin{align*} &\text{Equation form } &y=mx+b \\[4pt] &\text{Equation notation } &f(x)=mx+b \end{align*}
En el ejemplo del tren, podríamos usar la notación$$D(t)$$ en la que la distancia total$$D$$ es una función del tiempo$$t$$. La tasa,$$m$$, es de 83 metros por segundo. El valor inicial de la variable dependiente$$b$$ es la distancia original de la estación, 250 metros. Podemos escribir una ecuación generalizada para representar el movimiento del tren.
$D(t)=83t+250$
### Representación de una función lineal en forma tabular
Un tercer método para representar una función lineal es mediante el uso de una tabla. La relación entre la distancia desde la estación y el tiempo se representa en la Figura$$\PageIndex{2}$$. De la tabla, podemos ver que la distancia cambia 83 metros por cada incremento de tiempo de 1 segundo.
##### Q/A? ¿Puede la entrada en el ejemplo anterior ser algún número real?
No. La entrada representa el tiempo, por lo que si bien los números racionales e irracionales no negativos son posibles, los números reales negativos no son posibles para este ejemplo. La entrada consiste en números reales no negativos.
### Representación de una función lineal en forma gráfica
Otra forma de representar funciones lineales es visualmente, usando una gráfica. Podemos utilizar la relación de función desde arriba,$$D(t)=83t+250$$, para dibujar una gráfica, representada en la Figura$$\PageIndex{3}$$. Observe que la gráfica es una línea. Cuando trazamos una función lineal, la gráfica es siempre una línea.
La tasa de cambio, que es constante, determina la inclinación o pendiente de la línea. El punto en el que el valor de entrada es cero es la intercepción vertical, o intercepción y, de la línea. Podemos ver en la gráfica de la Figura$$\PageIndex{3}$$ que la intercepción y en el ejemplo del tren que acabamos de ver es$$(0,250)$$ y representa la distancia del tren desde la estación cuando comenzó a moverse a una velocidad constante.
Observe que la gráfica del ejemplo de tren está restringida, pero no siempre es así. Considera la gráfica de la línea$$f(x)=2x+1$$. Pregúntate qué números se pueden ingresar a la función, es decir, ¿cuál es el dominio de la función? El dominio está compuesto por todos los números reales porque cualquier número puede duplicarse, y luego tener uno agregado al producto.
##### Definición: Función lineal
Una función lineal es una función cuya gráfica es una línea. Las funciones lineales se pueden escribir en forma de pendiente-intercepción de una línea
$f(x)=mx+b$
donde$$b$$ es el valor inicial o inicial de la función (cuando se introduce,$$x=0$$), y$$m$$ es la tasa constante de cambio, o pendiente de la función. La intercepción y está en$$(0,b)$$.
##### Ejemplo$$\PageIndex{1}$$: Using a Linear Function to Find the Pressure on a Diver
La presión,$$P$$, en libras por pulgada cuadrada (PSI) sobre el buzo en la Figura$$\PageIndex{4}$$ depende de su profundidad por debajo de la superficie del agua$$d$$,, en pies. Esta relación puede ser modelada por la ecuación,$$P(d)=0.434d+14.696$$. Reafirmar esta función en palabras.
Para reafirmar la función en palabras, necesitamos describir cada parte de la ecuación. La presión en función de la profundidad equivale a cuatrocientos treinta y cuatro milésimas veces la profundidad más catorce seiscientos noventa y seis milésimas.
Análisis
El valor inicial, 14.696, es la presión en PSI sobre el buceador a una profundidad de 0 pies, que es la superficie del agua. La tasa de cambio, o pendiente, es de 0.434 PSI por pie. Esto nos dice que la presión sobre el buceador aumenta 0.434 PSI por cada pie que aumenta su profundidad.
## Determinar si una función lineal aumenta, disminuye o es constante
Las funciones lineales que usamos en los dos ejemplos anteriores aumentaron con el tiempo, pero no todas las funciones lineales lo hacen. Una función lineal puede ser creciente, decreciente o constante. Para una función creciente, como en el ejemplo del tren, los valores de salida aumentan a medida que aumentan los valores de entrada. La gráfica de una función creciente tiene una pendiente positiva. Una línea con pendiente positiva se inclina hacia arriba de izquierda a derecha como en la Figura$$\PageIndex{5}$$ (a). Para una función decreciente, la pendiente es negativa. Los valores de salida disminuyen a medida que aumentan los valores de entrada. Una línea con pendiente negativa se inclina hacia abajo de izquierda a derecha como en la Figura$$\PageIndex{5}$$ (b). Si la función es constante, los valores de salida son los mismos para todos los valores de entrada por lo que la pendiente es cero. Una línea con una pendiente de cero es horizontal como en la Figura$$\PageIndex{5}$$ (c).
Funciones crecientes y decrecientes
La pendiente determina si la función es una función lineal creciente, una función lineal decreciente o una función constante.
• $$f(x)=mx+b$$es una función creciente si$$m>0$$.
• $$f(x)=mx+b$$es una función decreciente si$$m<0$$.
• $$f(x)=mx+b$$es una función constante si$$m=0$$.
##### Ejemplo$$\PageIndex{2}$$: Deciding whether a Function Is Increasing, Decreasing, or Constant
Algunos estudios recientes sugieren que un adolescente envía un promedio de 60 textos diarios. Para cada uno de los siguientes escenarios, busque la función lineal que describe la relación entre el valor de entrada y el valor de salida. Luego, determine si la gráfica de la función es creciente, decreciente o constante.
1. El número total de textos que envía un adolescente se considera una función del tiempo en días. La entrada es el número de días, y la salida es el número total de textos enviados.
2. Un adolescente tiene un límite de 500 textos mensuales en su plan de datos. La entrada es el número de días, y la salida es el número total de textos restantes para el mes.
3. Un adolescente tiene un número ilimitado de textos en su plan de datos por un costo de 50 mensuales. La entrada es el número de días, y la salida es el costo total de enviar mensajes de texto cada mes. Solución Analizar cada función. 1. La función se puede representar como$$f(x)=60x$$ donde$$x$$ está el número de días. La pendiente, 60, es positiva por lo que la función va en aumento. Esto tiene sentido porque el número total de textos aumenta con cada día. 2. La función se puede representar como$$f(x)=500−60x$$ donde$$x$$ está el número de días. En este caso, la pendiente es negativa por lo que la función es decreciente. Esto tiene sentido porque el número de textos restantes disminuye cada día y esta función representa el número de textos que quedan en el plan de datos después de$$x$$ días. 3. La función de costo se puede representar como$$f(x)=50$$ porque el número de días no afecta el costo total. La pendiente es 0 por lo que la función es constante. ## Cálculo e Interpretación de Talud En los ejemplos que hemos visto hasta ahora, se nos ha proporcionado la pendiente. Sin embargo, a menudo necesitamos calcular la pendiente dados los valores de entrada y salida. Dados dos valores para la entrada,$$x_1$$ y$$x_2$$, y dos valores correspondientes para la salida,$$y_1$$ y$$y_2$$ —que pueden ser representados por un conjunto de puntos,$$(x_1,y_1)$$ y$$(x_2,y_2)$$ —podemos calcular la pendiente$$m$$, de la siguiente manera \begin{align*} m &= \dfrac{\text{change in output (rise)}}{ \text{change in input (run)}} \\[4pt] &= \dfrac{{\Delta}y}{ {\Delta}x} = \dfrac{y_2−y_1}{x_2−x_1} \end{align*} donde$${\Delta}y$$ está el desplazamiento vertical y$${\Delta}x$$ es el desplazamiento horizontal. Nota en notación de función dos valores correspondientes para la salida$$y_1$$ y$$y_2$$ para la función$$f$$,$$y_1=f(x_1)$$ y$$y_2=f(x_2)$$, así podríamos escribir de manera equivalente $m=\dfrac{f(x_2)-f(x_1)}{x_2-x_1} \nonumber$ La figura$$\PageIndex{6}$$ indica cómo se calcula la pendiente de la línea entre los puntos$$(x_2,y_2)$$,$$(x_1,y_1)$$ y,. Recordemos que la pendiente mide la inclinación. Cuanto mayor sea el valor absoluto de la pendiente, más pronunciada es la línea. ##### Q & A ¿Las unidades para pendiente son siempre$$\frac{\text{units for the output}}{ \text{units for the input}}$$? Sí. Piense en las unidades como el cambio del valor de salida para cada unidad de cambio en el valor de entrada. Un ejemplo de pendiente podría ser millas por hora o dólares por día. Observe que las unidades aparecen como una relación de unidades para la salida por unidades para la entrada. ##### Calcular pendiente La pendiente, o tasa de cambio, de una función se$$m$$ puede calcular de acuerdo con lo siguiente: $m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{{\Delta}y}{{\Delta}x}=\dfrac{y_2-y_1}{x_2-x_1}$ donde$$x_1$$ y$$x_2$$ son valores de entrada,$$y_1$$ y$$y_2$$ son valores de salida. Dados dos puntos de una función lineal, calcular e interpretar la pendiente. 1. Determine las unidades para los valores de salida y entrada. 2. Calcular el cambio de los valores de salida y el cambio de los valores de entrada. 3. Interpretar la pendiente como el cambio en los valores de salida por unidad del valor de entrada. ##### Ejemplo$$\PageIndex{3}$$: Finding the Slope of a Linear Function Si$$f(x)$$ es una función lineal, y$$(3,−2)$$ y$$(8,1)$$ son puntos en la línea, encuentra la pendiente. ¿Esta función está aumentando o disminuyendo? Solución Los pares de coordenadas son$$(3,−2)$$ y$$(8,1)$$. Para encontrar la tasa de cambio, dividimos el cambio en la salida por el cambio en la entrada. $m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{1-(-2)}{8-3}=\dfrac{3}{5}$ También podríamos escribir la pendiente como$$m=0.6$$. La función va en aumento porque$$m>0$$. Análisis Como se señaló anteriormente, el orden en que escribimos los puntos no importa cuando calculamos la pendiente de la línea siempre y cuando el primer valor de salida, o coordenada y, utilizado corresponda con el primer valor de entrada, o coordenada x, usado. ##### ¡Pruébalo! $$\PageIndex{1}$$ Si$$f(x)$$ es una función lineal, y$$(2, 3)$$ y$$(0,4)$$ son puntos en la línea, encuentra la pendiente. ¿Esta función está aumentando o disminuyendo? Contestar $$m=\frac{4−3}{0−2} =\frac{1}{-2}=-\frac{1}{2}$$; decreciente porque$$m<0$$. ##### Ejemplo$$\PageIndex{4}$$: Finding the Population Change from a Linear Function La población de una ciudad aumentó de 23,400 a 27,800 entre 2008 y 2012. Encontrar el cambio de población por año si asumimos que el cambio fue constante de 2008 a 2012. La tasa de cambio relaciona el cambio en la población con el cambio en el tiempo. La población aumentó en$$27,800−23,400=4400$$ personas durante el intervalo de tiempo de cuatro años. Para encontrar la tasa de cambio, divida el cambio en el número de personas por el número de años. $\dfrac{4,400 \text{ people}}{4 \text{ years}} =1,100 \dfrac{\text{people}}{\text{year}}$ Por lo que la población aumentó en mil 100 personas al año. Análisis Debido a que nos dicen que la población aumentó, esperaríamos que la pendiente fuera positiva. Esta pendiente positiva que calculamos es, por lo tanto, razonable. ##### ¡Pruébalo! $$\PageIndex{2}$$ La población de un pequeño pueblo aumentó de 1,442 a 1,868 entre 2009 y 2012. Encontrar el cambio de población por año si asumimos que el cambio fue constante de 2009 a 2012. Contestar $$m=\frac{1,868−1,442}{2,012−2,009} = \frac{426}{3} =\text{ 142 people per year}$$ ## Escribir la forma de punto-pendiente de una ecuación lineal Hasta ahora, hemos estado usando la forma pendiente-intercepción de una ecuación lineal para describir funciones lineales. Aquí aprenderemos otra forma de escribir una función lineal, la forma punto-pendiente. $y-y_1=m(x-x_1)$ La forma punto-pendiente se deriva de la fórmula de pendiente. \begin{align*} &m=\dfrac{y-y_1}{x-x_1} &\text{assuming }x{\neq}x_1 \\ &m(x-x_1)=\dfrac{y-y_1}{x-x_1}(x-x_1) &\text{Multiply both sides by }(x-x_1). \\ &m(x-x_1)=y-y_1 &\text{Simplify} \\ &y-y_1=m(x-x_1) &\text{Rearrange} \end{align*} Tenga en cuenta que la forma pendiente-intercepción y la forma punto-pendiente se pueden utilizar para describir la misma función. Podemos pasar de una forma a otra usando álgebra básica. Por ejemplo, supongamos que se nos da una ecuación en forma de punto-pendiente,$$y−4=− \frac{1}{2}(x−6)$$. Podemos convertirlo a la forma pendiente-intercepción como se muestra. \begin{align*} y-4&=-\dfrac{1}{2}(x-6) \\ y-4&=-\dfrac{1}{2}x+3 &\text{Distribute the }-\dfrac{1}{2}. \\ y&=-\dfrac{1}{2}x+7 &\text{Add 4 to each side.}\end{align*} Por lo tanto, la misma línea se puede describir en forma de pendiente-intercepción como$$y=\dfrac{1}{2}x+7$$. ##### Forma Punto-Pendiente de una Ecuación Lineal La forma punto-pendiente de una ecuación lineal toma la forma $y-y_1=m(x−x_1)$ donde$$m$$ está la pendiente,$$x_1$$ y$$y_1$$ son las$$y$$ coordenadas$$x$$ y de un punto específico por el que pasa la línea. ### Escribir la ecuación de una línea usando un punto y el talud La forma punto-pendiente es particularmente útil si conocemos un punto y la pendiente de una línea. Supongamos, por ejemplo, se nos dice que una línea tiene una pendiente de 2 y pasa por el punto$$(4,1)$$. Sabemos eso$$m=2$$ y eso$$x_1=4$$ y$$y_1=1$$. Podemos sustituir estos valores en la ecuación general de punto-pendiente. \begin{align*} y−y_1&=m(x−x_1) \\ y−1&=2(x−4) \end{align*} Si quisiéramos entonces reescribir la ecuación en forma de pendiente-intercepción, aplicamos técnicas algebraicas. \begin{align*} y−1&=2(x−4) \\ y−1&=2x−8 &\text{Distribute the 2.} \\ y&=2x−7 &\text{Add 1 to each side.} \end{align*} Ambas ecuaciones,$$y−1=2(x−4)$$ y$$y=2x–7$$, describen la misma línea. Ver Figura$$\PageIndex{7}$$. ##### Ejemplo$$\PageIndex{5}$$: Writing Linear Equations Using a Point and the Slope Escribe la forma punto-pendiente de una ecuación de una línea con una pendiente de 3 que pasa por el punto$$(6,–1)$$. Después reescribirlo en la forma pendiente-intercepción. Solución Averiguemos lo que sabemos de la información dada. La pendiente es de 3, entonces$$m=3$$. También conocemos un punto, entonces sabemos$$x_1=6$$ y$$y_1 =−1$$. Ahora podemos sustituir estos valores en la ecuación general de punto-pendiente. \begin{align*} y-y_1&=m(x-x_1) \\ y−(−1)&=3(x−6) &\text{Substitute known values.} \\ y+1&=3(x−6) &\text{Distribute −1 to find point-slope form.} \end{align*} Entonces usamos álgebra para encontrar la forma pendiente-intercepción. \begin{align*} y+1&=3(x−6) \\ y+1&=3x−18 &\text{Distribute 3.} \\ y&=3x−19 &\text{Simplify to slope-intercept form.} \end{align*} ##### ¡Pruébalo! $$\PageIndex{3}$$ Escriba la forma punto-pendiente de una ecuación de una línea con una pendiente de —2 que pasa por el punto$$(–2, 2)$$. Después reescribirlo en la forma pendiente-intercepción. Contestar $$y−2=−2(x+2)$$;$$y=−2x−2$$ ### Escribir la ecuación de una línea usando dos puntos La forma punto-pendiente de una ecuación también es útil si conocemos dos puntos a través de los cuales pasa una línea. Supongamos, por ejemplo, sabemos que una línea pasa por los puntos$$(0, 1)$$ y$$(3, 2)$$. Podemos usar las coordenadas de los dos puntos para encontrar la pendiente. \begin{align*} m&=\dfrac{y_2-y_1}{x_2-x_1} \\ &=\dfrac{2-1}{3-0} \\ &=\dfrac{1}{3} \end{align*} Ahora podemos usar la pendiente que encontramos y las coordenadas de uno de los puntos para encontrar la ecuación para la línea. Déjanos usar$$(0,1)$$ para nuestro punto. \begin{align*} y-y_1&=m(x-x_1) \\ y-1&=\dfrac{1}{3}(x-0) \end{align*} Como antes, podemos usar álgebra para reescribir la ecuación en la forma pendiente-intercepción. \begin{align*} y-1&=\dfrac{1}{3}(x-0) \\ y-1&=\dfrac{1}{3}x &\text{Distribute the }\dfrac{1}{3}. \\ y&=\dfrac{1}{3}x+1 &\text{Add 1 to each side.} \end{align*} Ambas ecuaciones describen la línea mostrada en la Figura$$\PageIndex{8}$$. ##### Ejemplo$$\PageIndex{6}$$: Writing Linear Equations Using Two Points Escribe la forma punto-pendiente de una ecuación de una línea que pasa por los puntos$$(5,1)$$ y$$(8, 7)$$. Después reescribirlo en la forma pendiente-intercepción. Empecemos por encontrar la pendiente. \begin{align*} m&=\dfrac{y_2-y_1}{x_2-x_1} \\ &=\dfrac{7-1}{8-5} \\ &=\dfrac{6}{3} \\ &= 2 \end{align*} Entonces$$m=2$$. A continuación, sustituimos la pendiente y las coordenadas por uno de los puntos en la ecuación general de pendiente puntual. Podemos elegir cualquiera de los dos puntos, pero vamos a utilizar$$(5,1)$$. \begin{align*} y-y_1&=m(x-x_1) \\ y-1&=2(x-5) \end{align*} La ecuación de punto-pendiente de la línea es$$y_2–1=2(x_2–5)$$. Para reescribir la ecuación en forma pendiente-intercepción, utilizamos álgebra. \begin{align*} y-1&=2(x-5) \\ y-1&=2x-10 \\ y&=2x-9 \end{align*} La ecuación pendiente-intercepción de la línea es$$y=2x–9$$. $$\PageIndex{4}$$: Escribir la forma punto-pendiente de una ecuación de una línea que pasa por los puntos$$(–1,3)$$ y$$(0,0)$$. Después reescribirlo en la forma pendiente-intercepción. Solución $$y−0=−3(x−0)$$;$$y=−3x$$ ## Escribir e interpretar una ecuación para una función lineal Ahora que hemos escrito ecuaciones para funciones lineales tanto en la forma pendiente-intercepción como en la forma punto-pendiente, podemos elegir qué método usar en función de la información que se nos da. Esa información puede ser proporcionada en forma de gráfica, punto y pendiente, dos puntos, y así sucesivamente. Mira la gráfica de la función$$f$$ en la Figura$$\PageIndex{9}$$. No se nos da la pendiente de la línea, pero podemos elegir dos puntos cualesquiera en la línea para encontrar la pendiente. Vamos a elegir$$(0,7)$$ y$$(4, 4)$$. Podemos utilizar estos puntos para calcular la pendiente. \begin{align*} m&=\dfrac{y_2-y_1}{x_2-x_1} \\ &=\dfrac{4-7}{4-0} \\&=-\dfrac{3}{4}\end{align*} Ahora podemos sustituir la pendiente y las coordenadas de uno de los puntos en la forma punto-pendiente. \begin{align*} y-y_1&=m(x-x_1) \\ y-4&=-\dfrac{3}{4}(x-4) \end{align*} Si queremos reescribir la ecuación en la forma pendiente-intercepción, encontraríamos \begin{align*} y-4&=-\dfrac{3}{4}(x-4) \\ y-4 &=-\dfrac{3}{4}x+3 \\ y&=-\dfrac{3}{4}x+7\end{align*} Si quisiéramos encontrar la forma pendiente-intercepción sin escribir primero la forma de punto-pendiente, podríamos haber reconocido que la línea cruza el eje y cuando el valor de salida es 7. Por lo tanto,$$b=7$$. Ahora tenemos el valor inicial$$b$$ y la pendiente$$m$$ para que podamos sustituir$$m$$ y$$b$$ en la forma pendiente-intercepción de una línea. Entonces la función es$$f(x)=−\frac{3}{4}x+7$$, y la ecuación lineal sería$$y=−\frac{3}{4}x+7$$. Dada la gráfica de una función lineal, escribir una ecuación para representar la función. 1. Identificar dos puntos en la línea. 2. Usa los dos puntos para calcular la pendiente. 3. Determinar dónde la línea cruza el eje y para identificar la intercepción y mediante inspección visual. 4. Sustituya la pendiente y la intercepción y en la forma pendiente-intercepción de una ecuación de línea. ##### Ejemplo$$\PageIndex{7}$$: Writing an Equation for a Linear Function Escribe una ecuación para una función lineal dada una gráfica de la que$$f$$ se muestra en la Figura$$\PageIndex{11}$$. Solución Identificar dos puntos en la línea, como$$(0, 2)$$ y$$(−2,−4)$$. Usa los puntos para calcular la pendiente. \begin{align*} m&=\dfrac{y_2-y_1}{x_2-x_1} \\ &=\dfrac{4-2}{-2-0} \\ &=\dfrac{-6}{-2} \\ &=3 \end{align*} Sustituya el talud y las coordenadas de uno de los puntos en la forma punto-pendiente. \begin{align*} y-y_1&=m(x-x_1) \\ y-(-4)&=3(x-(-2)) \\ y+4 &= 3(x+2)\end{align*} Podemos usar álgebra para reescribir la ecuación en la forma pendiente-intercepción. \begin{align*} y+4&= 3(x+2) \\ y+4&= 3x+6 \\ y & = 3x + 2 \end{align*} Análisis Esto tiene sentido porque podemos ver en Figura$$\PageIndex{12}$$ que la línea cruza el eje y en el punto$$(0, 2)$$, que es la intercepción y, entonces$$b=2$$. ##### Ejemplo$$\PageIndex{8}$$: Writing an Equation for a Linear Cost Function Supongamos que Ben inicia una empresa en la que incurre en un costo fijo de1,250 mensuales por los gastos generales, que incluye la renta de su oficina. Sus costos de producción son de 37.50 dólares por artículo. Escribe una función lineal$$C$$ donde$$C(x)$$ está el costo de los$$x$$ artículos producidos en un mes determinado.
Solución
El costo fijo está presente cada mes, $1,250. Los costos que pueden variar incluyen el costo para producir cada artículo, que es$37.50 para Ben. El costo variable, denominado costo marginal, está representado por 37.5. El costo en el que incurre Ben es la suma de estos dos costos, representados por$$C(x)=1250+37.5x$$.
Análisis
Si Ben produce 100 artículos en un mes, su costo mensual está representado por
\begin{align*} C(100)&=1250+37.5(100) \\ &=5000 \end{align*}
Por lo que su costo mensual sería de 5.000 dólares.
##### Ejemplo$$\PageIndex{9}$$: Writing an Equation for a Linear Function Given Two Points
Si$$f$$ es una función lineal, con$$f(3)=−2$$, y$$f(8)=1$$, encontrar una ecuación para la función en forma pendiente-intercepción.
Solución
\begin{align*} f(3)&= -2{\rightarrow}(3,2) \\ f(8)&=1{\rightarrow}(8,1) \end{align*}
Entonces podemos usar los puntos para calcular la pendiente.
\begin{align*} m&=\dfrac{y_2-y_1}{x_2-x_1} \\ &=\dfrac{1-(-2)}{8-3} \\ &=\dfrac{3}{5} \end{align*}
Sustituya el talud y las coordenadas de uno de los puntos en la forma punto-pendiente.
\begin{align*} y-y_1&=m(x-x_1) \\ y-(-2)&=\dfrac{3}{5}(x-3) \end{align*}
Podemos usar álgebra para reescribir la ecuación en la forma pendiente-intercepción.
\begin{align*} y+2&=\dfrac{3}{5}(x-3) \\ y+2&=\dfrac{3}{5}x-\dfrac{9}{5} \\ y&=\dfrac{3}{5}x-\dfrac{19}{5} \end{align*}
##### ¡Tri It! $$\PageIndex{1}$$
Si$$f(x)$$ es una función lineal, con$$f(2)=–11$$, y$$f(4)=−25$$, encontrar una ecuación para la función en forma pendiente-intercepción.
Contestar
$$y=−7x+3$$
## Modelado de problemas del mundo real con funciones lineales
En el mundo real, los problemas no siempre se enuncian explícitamente en términos de una función o se representan con una gráfica. Afortunadamente, podemos analizar el problema representándolo primero como una función lineal y luego interpretando los componentes de la función. Mientras sepamos, o podamos averiguar, el valor inicial y la tasa de cambio de una función lineal, podemos resolver muchos tipos diferentes de problemas del mundo real.
Dada una función lineal$$f$$ y el valor inicial y la tasa de cambio, evaluar$$f(c)$$.
1. Determinar el valor inicial y la tasa de cambio (pendiente).
2. Sustituir los valores en$$f(x)=mx+b$$.
3. Evaluar la función en$$x=c$$.
##### Ejemplo$$\PageIndex{10}$$: Using a Linear Function to Determine the number of Songs in a Music Collection
Marcus actualmente tiene 200 canciones en su colección de música. Cada mes, agrega 15 canciones nuevas. Escribir una fórmula para el número de canciones,$$N$$, en su colección en función del tiempo,$$t$$, el número de meses. ¿Cuántas canciones tendrá en un año?
Solución
El valor inicial para esta función es 200 porque actualmente posee 200 canciones, entonces$$N(0)=200$$, lo que significa que$$b=200$$.
El número de canciones aumenta en 15 canciones al mes, por lo que la tasa de cambio es de 15 canciones al mes. Por lo tanto, eso lo sabemos$$m=15$$. Podemos sustituir el valor inicial y la tasa de cambio por la forma pendiente-intercepción de una línea.
Podemos escribir la fórmula$$N(t)=15t+200$$.
Con esta fórmula, podemos entonces predecir cuántas canciones tendrá Marcus en 1 año (12 meses). En otras palabras, podemos evaluar la función en$$t=12$$.
\begin{align*} N(12)&=15(12)+200 \\ &=180+200 \\ &= 380 \end{align*}
Marcus tendrá 380 canciones en 12 meses.
Análisis
Observe que$$N$$ es una función lineal creciente. A medida que aumenta la entrada (el número de meses), la salida (número de canciones) aumenta también.
##### Ejemplo$$\PageIndex{11}$$: Using a Linear Function to Calculate Salary Plus Commission
Al trabajar como vendedor de seguros, Ilya gana un salario base más una comisión por cada nueva póliza. Por lo tanto, el ingreso semanal de Ilya, yo, depende del número de pólizas nuevas$$n$$,, vende durante la semana. La semana pasada vendió 3 nuevas pólizas, y ganó 760 dólares por semana. La semana anterior, vendió 5 nuevas pólizas y ganó \$920. Encontrar una ecuación para$$I(n)$$, e interpretar el significado de los componentes de la ecuación.
Solución
La información dada nos da dos pares entrada-salida:$$(3,760)$$ y$$(5,920)$$. Comenzamos por encontrar la tasa de cambio.
\begin{align*} m&=\dfrac{920-760}{5-3} \\ &=\dfrac{160}{2 \text{ policies}} \\ &=80 \text{ per policy} \end{align*}
Hacer un seguimiento de las unidades puede ayudarnos a interpretar esta cantidad. Los ingresos aumentaron 160 dólares cuando el número de pólizas aumentó en 2, por lo que la tasa de cambio es de 80 dólares por póliza. Por lo tanto, Ilya gana una comisión de 80 dólares por cada póliza vendida durante la semana.
Entonces podemos resolver por el valor inicial.
\begin{align*} I(n)&=80n+b \\ 760&=80(3)+b \text{ when } n=3, I(3)=760 \\ 760-80(3)&=b \\ 520 & =b \end{align*}
El valor de$$b$$ es el valor inicial de la función y representa los ingresos de Ilya cuando$$n=0$$, o cuando no se venden nuevas pólizas. Podemos interpretar esto como el salario base de Ilya para la semana, que no depende del número de pólizas vendidas.
Ahora podemos escribir la ecuación final.
$I(n)=80n+520 \nonumber$
Nuestra interpretación final es que el salario base de Ilya es de 520 dólares semanales y gana una comisión adicional de 80 dólares por cada póliza vendida.
##### Ejemplo de$$\PageIndex{12}$$: Using Tabular Form to Write an ecuación para una función lineal
El cuadro$$\PageIndex{1}$$ relaciona el número de ratas en una población con el tiempo, en semanas. Usa la tabla para escribir una ecuación lineal.
w, número de semanas P (w), número de ratas 0 2 4 6 1000 1080 1160 1240
Solución
Podemos ver en la tabla que el valor inicial para el número de ratas es de 1000, entonces$$b=1000$$.
En lugar de resolver para$$m$$, podemos decir al mirar la tabla que la población aumenta en 80 por cada 2 semanas que pasan. Esto significa que la tasa de cambio es de 80 ratas por 2 semanas, lo que puede simplificarse a 40 ratas por semana.
$P(w)=40w+1000 \nonumber$
Si no nos percatamos de la tasa de cambio de la tabla aún podríamos resolver para la pendiente usando dos puntos cualesquiera de la tabla. Por ejemplo, usando$$(2,1080)$$ y$$(6,1240)$$
\begin{align*} m&=\dfrac{1240-1080}{6-2} \\ &=\dfrac{160}{4} \\ &= 40\end{align*}
¿El valor inicial siempre se proporciona en una tabla de valores como Table$$\PageIndex{1}$$?
No. A veces el valor inicial se proporciona en una tabla de valores, pero a veces no lo es. Si ves una entrada de 0, entonces el valor inicial sería la salida correspondiente. Si no se proporciona el valor inicial porque no hay ningún valor de entrada en la tabla igual a 0, busque la pendiente, sustituya un par de coordenadas y la pendiente en$$f(x)=mx+b$$, y resuelva para$$b$$.
$$\PageIndex{5}$$: Se introdujo un nuevo alimento vegetal a un árbol joven para probar su efecto en la altura del árbol. $$\PageIndex{2}$$La tabla muestra la altura del árbol, en pies,$$x$$ meses desde que comenzaron las mediciones. Escribir una función lineal,$$H(x)$$, donde$$x$$ está el número de meses desde el inicio del experimento.
$$x$$ $$H(x)$$ 0 2 4 8 12 12.5 13.5 14.5 16.5 18.5
Solución
$$H(x)=0.5x+12.5$$
## Ecuaciones Clave
• forma pendiente-intercepción de una línea:$$f(x)=mx+b$$
• pendiente:$$m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{{\Delta}y}{{\Delta}x}=\dfrac{y_2-y_1}{x_2-x_1}$$
• forma de punto-pendiente de una línea:$$y−y_1 =m(x-x_1)$$
## Conceptos clave
• Los pares ordenados dados por una función lineal representan puntos en una línea.
• Las funciones lineales se pueden representar en palabras, notación de funciones, forma tabular y forma gráfica.
• La tasa de cambio de una función lineal también se conoce como pendiente.
• Una ecuación en la forma pendiente-intercepción de una línea incluye la pendiente y el valor inicial de la función.
• El valor inicial, o intercepción y, es el valor de salida cuando la entrada de una función lineal es cero. Es el valor y del punto en el que la línea cruza el eje y.
• Una función lineal creciente da como resultado una gráfica que se incline hacia arriba de izquierda a derecha y tiene una pendiente positiva.
• Una función lineal decreciente da como resultado una gráfica que se incline hacia abajo de izquierda a derecha y tiene una pendiente negativa.
• Una función lineal constante da como resultado una gráfica que es una línea horizontal.
• Analizar la pendiente en el contexto de un problema indica si una función lineal es creciente, decreciente o constante.
• La pendiente de una función lineal se puede calcular dividiendo la diferencia entre los valores y por la diferencia en los valores x correspondientes de dos puntos cualesquiera de la línea.
• La pendiente y el valor inicial se pueden determinar dada una gráfica o dos puntos cualesquiera en la línea.
• Un tipo de notación de función es la forma pendiente-intercepción de una ecuación.
• La forma punto-pendiente es útil para encontrar una ecuación lineal cuando se le da la pendiente de una línea y un punto.
• La forma punto-pendiente también es conveniente para encontrar una ecuación lineal cuando se le dan dos puntos a través de los cuales pasa una línea.
• La ecuación para una función lineal se puede escribir si se$$b$$ conocen la pendiente$$m$$ y el valor inicial.
• Una función lineal se puede utilizar para resolver problemas del mundo real.
• Una función lineal se puede escribir a partir de forma tabular.
## Notas al pie
1 www.chinahighlights.com/shang... glev-train.htm
2 www.cbsnews.com/8301-501465_1... ay-study-says/
## Glosario
función lineal decreciente
una función con pendiente negativa: Si$$f(x)=mx+b$$, entonces$$m<0$$.
aumentar la función lineal
una función con una pendiente positiva: Si$$f(x)=mx+b$$, entonces$$m>0$$.
función lineal
una función con una tasa de cambio constante que es un polinomio de grado 1, y cuya gráfica es una línea recta
forma de punto-pendiente
la ecuación para una línea que representa una función lineal de la forma\ (y−y_1=m (x−x_1)
pendiente
la relación entre el cambio en los valores de salida y el cambio en los valores de entrada; una medida de la inclinación de una línea
forma pendiente-intercepción
la ecuación para una línea que representa una función lineal en la forma$$f(x)=mx+b$$
Intercepción en Y
el valor de una función cuando el valor de entrada es cero; también conocido como valor inicial
This page titled 2.1: Funciones lineales is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. | 12,252 | 39,002 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-30 | latest | en | 0.187954 |
mjgoff.wordpress.com | 1,540,020,045,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512592.60/warc/CC-MAIN-20181020055317-20181020080817-00182.warc.gz | 744,069,704 | 20,986 | Math Manipulatives Help All Students
14 Mar
Pure mathematics is, in its way, the poetry of logical ideas. ~Albert Einstein
I never did very well in math – I could never seem to persuade the teacher that I hadn’t meant my answers literally. ~Calvin Trillin
Introduction
Manipulatives are proving to be effective tools for teaching certain math concepts to all students (Berkas, 2007). Students learn best through active experiences rather than lectures or passive listening. Visual images stick with people longer and better than abstract or language based concepts. (Stover, ret. 2011) Manipulatives are objects that can be touched and moved by students to introduce or reinforce a math concept. Since 1940 to the present day, the National Council of Teachers of Mathematics (NCTM) has encouraged all grade levels to use manipulatives in daily math instruction. Many concepts in math are abstract. Manipulatives are useful in helping students move from the concrete to abstract level. (Hartshorn, 1990) Students are being encouraged to use manipulatives to demonstrate understanding in representational and abstract stages (Berkas, 2007). They are beneficial for students who have been diagnosed with a learning disability. Manipulatives provide additional reinforcement of basic math skills and enables them to complete math problems independently. (Schreiner, 2010)
Manipulative Toolbox
Once a child is familiar with a manipulative it should be made available to use at any time to help them think, reason and solve problems. After a while, they begin to develop a great toolbox of ideas for solving problems. (Rudnicki, 2010)
Positive Impacts
Manipulatives show positive impacts when combined with:
1. virtual manipulative software
2. reflective practices
3. cooperative learning
4. activities that are exploratory and deductive in their approach
(Berkas, 2007)
Types and Uses
• learn measurement using nonstandard units
• visualize concept of area and geometric shapes
• percent, probability and fractions
Two-Color counters
• teach equivalence
• number concepts/ strategies for addition and subtraction
• create visuals showing ratio, proportion, fractions
• understand odd/even numbers
• create patterns
Playing cards, dominoes and number cubes
• illustrate decimals or whole numbers
• mental math games (addition, subtraction and/or multiplication war)
• understand probability and logical reasoning
Pattern blocks and tangrams
• understanding of spatial sense and geometry
• create shapes with certain number of sides, angles, complex patterns
• solving word problems
Measuring cups, volume containers, rulers, scales, and clocks
• concept of measurement
(Stover, ret. 2011)
Critique
Research has shown that long-term use of manipulatives is more effective than short-term use (Hartshorn, 1990).
Interesting Facts
“Some state and local systems have mandated the implementation of manipulatives through policy, law, or curriculum documents. Some have also provided funding.” (Hartshorn, 1990)
Manipulatives help hold the attention of the inattentive learner (Stover, ret. 2011).
Multiple Intelligences
Logical Mathematical
Using manipulatives in math help students to understand and analyze concepts with a concrete purpose.
Bodily Kinesthetic
In terms of mental and physical activities, manipulatives allow students to use their mind and parts of their body to solve problems. To show students how much a gallon is, start with cups of water, then pour those cups to make pints, then quarts to the eventual gallon. This is both a physical and mental process that show students the meaning of a gallon.
Interpersonal
Often times in the fields of math and science we work with a group of people to solve a problem. Using manipulatives in groups allow students to work together and to listen to the ideas of their peers to come to a conclusion.
Universal Design for Learning
I. Provide Multiple Means of Representation
1. Provide options for perception
• Options that customize the display of information
• Options that provide alternatives for visual information
2. Provide options for language and symbols
• Options that illustrate key concepts non-linguistically
II. Provide Multiple Means of Action and Expression
5. Provide options for expressive skills and fluency
• Options in the tools for composition and problem solving
III. Provide Multiple Means of Engagement
7. Provide options for recruiting interest
• Options that enhance relevance, value and authenticity
8. Provide options for sustaining effort and persistence
• Options that foster collaboration and communication
“7 Musts for Using Manipulatives”
Resources
Berkas, N. & Pattison, C. (2007). “Manipulatives: more than a special education intervention.” National Council of Teacher of Mathematics.
Hartshorn, R. & Boren, S. (1990). “Experiential learning of mathematics: using manipulatives. ERIC Digest
Rudnicki, A. (2010). “Why use manipulatives to teach math.” http://www.ehow.com/print/about_664188_use-manipulatives-teach-math_.html.
Schreiner, E. (2010). “How to use manipulatives to teach math to ld students.” http://www.ehow.com/print/how_5991197_use-teach-math-ld-student.html.
Stover, E. (ret. 2011). “About the use of manipulatives in math.” http://www.ehow.com/print/about_5048097_use-manipulatives-math.html.
Stover, E. (ret. 2011). “How to use math manipulatives in the classroom.” http://www.ehow.com/print/how_5031920_use-math-manipulatives-classroom.html.
Posted by on March 14, 2011 in Uncategorized
2 responses to “Math Manipulatives Help All Students”
1. March 20, 2011 at 11:06 pm
Hello Michael! | 1,289 | 5,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-43 | latest | en | 0.892394 |
http://de.metamath.org/mpeuni/2cshwcom.html | 1,653,340,861,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00792.warc.gz | 14,366,811 | 5,584 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 2cshwcom Structured version Visualization version GIF version
Theorem 2cshwcom 13413
Description: Cyclically shifting a word two times is commutative. (Contributed by AV, 21-Apr-2018.) (Revised by AV, 5-Jun-2018.) (Revised by Mario Carneiro/AV, 1-Nov-2018.)
Assertion
Ref Expression
2cshwcom ((𝑊 ∈ Word 𝑉𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((𝑊 cyclShift 𝑁) cyclShift 𝑀) = ((𝑊 cyclShift 𝑀) cyclShift 𝑁))
Proof of Theorem 2cshwcom
StepHypRef Expression
1 zcn 11259 . . . . 5 (𝑀 ∈ ℤ → 𝑀 ∈ ℂ)
2 zcn 11259 . . . . 5 (𝑁 ∈ ℤ → 𝑁 ∈ ℂ)
3 addcom 10101 . . . . 5 ((𝑀 ∈ ℂ ∧ 𝑁 ∈ ℂ) → (𝑀 + 𝑁) = (𝑁 + 𝑀))
41, 2, 3syl2anr 494 . . . 4 ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑀 + 𝑁) = (𝑁 + 𝑀))
543adant1 1072 . . 3 ((𝑊 ∈ Word 𝑉𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑀 + 𝑁) = (𝑁 + 𝑀))
65oveq2d 6565 . 2 ((𝑊 ∈ Word 𝑉𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑊 cyclShift (𝑀 + 𝑁)) = (𝑊 cyclShift (𝑁 + 𝑀)))
7 2cshw 13410 . . 3 ((𝑊 ∈ Word 𝑉𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑊 cyclShift 𝑀) cyclShift 𝑁) = (𝑊 cyclShift (𝑀 + 𝑁)))
873com23 1263 . 2 ((𝑊 ∈ Word 𝑉𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((𝑊 cyclShift 𝑀) cyclShift 𝑁) = (𝑊 cyclShift (𝑀 + 𝑁)))
9 2cshw 13410 . 2 ((𝑊 ∈ Word 𝑉𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((𝑊 cyclShift 𝑁) cyclShift 𝑀) = (𝑊 cyclShift (𝑁 + 𝑀)))
106, 8, 93eqtr4rd 2655 1 ((𝑊 ∈ Word 𝑉𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((𝑊 cyclShift 𝑁) cyclShift 𝑀) = ((𝑊 cyclShift 𝑀) cyclShift 𝑁))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 (class class class)co 6549 ℂcc 9813 + caddc 9818 ℤcz 11254 Word cword 13146 cyclShift ccsh 13385 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 ax-pre-sup 9893 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-sup 8231 df-inf 8232 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-n0 11170 df-z 11255 df-uz 11564 df-rp 11709 df-fz 12198 df-fzo 12335 df-fl 12455 df-mod 12531 df-hash 12980 df-word 13154 df-concat 13156 df-substr 13158 df-csh 13386 This theorem is referenced by: 3cshw 13415
Copyright terms: Public domain W3C validator | 2,152 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-21 | longest | en | 0.123949 |
http://www.scientificamerican.com/article.cfm?id=influential-few-predict-behavior | 1,386,594,764,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163973624/warc/CC-MAIN-20131204133253-00075-ip-10-33-133-15.ec2.internal.warc.gz | 524,580,573 | 29,365 | # Influential Few Predict Behavior of the Many
A new technique is helping to untangle complexity in systems ranging from metabolism to social networks
Image: Flickr/David.Asch
To completely understand how a living organism works one would have to take it apart, the great physicist Niels Bohr once observed—but then the organism would certainly be dead. In general, systems of high complexity, including living things but ranging from the Internet to social networks, are often impossible to track in all their details.
But what if you didn't have to? Network-theory researchers now have come up with some clever mathematics that reveals complex systems by tracking a selected few of their components.
Say, for example, that you wanted to find a biological marker that identifies people with a certain disease. You can track down all the genes that are expressed differently in people with the disease and assemble a network that shows their interactions, but how do you then pick out those that are specific to the illness?
The new work may help researchers to identify the key nodes in a network that determine the state of every other node, greatly simplifying the search.
“This paper shows how you can reduce a network to the really important component parts that drive the system’s behavior,” says Joseph Loscalzo of Harvard Medical School in Boston, Massachusetts. “It begins to make the system more tractable,” adds Loscalzo, who would like to apply the technique to medicine.
To demonstrate their technique, Yang-Yu Liu of Northeastern University in Boston and his colleagues looked at the entire human metabolic network and found that concentrations of about 10 percent of the body's 2,763 metabolites could be used to determine the levels of all the rest.
But the method could also be used in social networks to identify the people whose opinions determine everyone else’s, helping to predict the outcome of, say, a presidential election. Or it could help ecologists to single out the particular species to track to follow changes in an entire ecosystem, to name just a few potential applications.
Needle-like nodes
To imagine how this works, start with a simple network in which a chemical A becomes chemical B. Because any changes in B are exclusively determined by A, monitoring B over time will also enable you to determine the state of A. The same would not be true if you monitored only A: Without knowing the initial level of B, changes in A aren’t enough to determine the level of B. The team pictured the situation above as a pair of nodes, A and B, with an arrow going from B to A to represent how information about B leads to information about A but not vice versa.
More complex reaction systems don’t yield such obvious results. Liu's team tackled the problem by examining clusters of strongly connected components in a network, again represented by nodes with arrows connecting them. For clusters that have no incoming arrows, as is the case of B above, the researchers freely picked a single node from each cluster.
They found that most of the time (and almost always in real-world networks), these selected nodes alone are sufficient to determine the state of every other node in the network. The team published its results in the Proceedings of the National Academy of Sciences.
Although theoretically possible, it is extremely difficult to reconstruct the entire network from these nodes. In many applications, however, that will not be necessary, because the needle-like nodes in the haystack of links should reveal the network's most important properties.
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1. 1. gmperkins 02:51 PM 2/21/13
This is very interesting work but I know that it cannot predict as broadly as this article is presenting. Like many mathematical methods, they have their uses and that is why I like this work but I don't like how the media portrays them as 'silver bullet' solutions for a host of models/predictions.
2. 2. mounthell 04:38 PM 2/21/13
@gmperkins (1.) Yes, quite right.
This article presents a picture that is far too optimistic about the theoretical extent of our capability in controlling truly complex systems:
“This paper shows how you can reduce a network to the really important component parts that drive the system’s behavior.”
These are physicists, remember, not biologists (and certainly not ecologists). The physicist view is that, if some phenomenon can't be measured, e.g., an emergent property of a living system, it therefore does not exist. Physicists' anti-soft-science inculcation renders them clueless as to the extent of the spatiotemporally varying complexity by which even the simplest of living systems functions.
They should stick to systems that can't be easily accessed, like subatomic and extraterrestrial systems and leave messy life to those of us who don't mind getting their hands dirty and making fools of themselves.
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##### Arithmetic Circuit Complexity of Division and Truncation
Authors: Pranjal Dutta, Gorav Jindal, Anurag Pandey, and Amit Sinhababu
Published in: LIPIcs, Volume 200, 36th Computational Complexity Conference (CCC 2021)
##### Abstract
Given polynomials f,g,h ∈ 𝔽[x₁,…,x_n] such that f = g/h, where both g and h are computable by arithmetic circuits of size s, we show that f can be computed by a circuit of size poly(s,deg(h)). This solves a special case of division elimination for high-degree circuits (Kaltofen'87 & WACT'16). The result is an exponential improvement over Strassen’s classic result (Strassen'73) when deg(h) is poly(s) and deg(f) is exp(s), since the latter gives an upper bound of poly(s, deg(f)). Further, we show that any univariate polynomial family (f_d)_d, defined by the initial segment of the power series expansion of rational function g_d(x)/h_d(x) up to degree d (i.e. f_d = g_d/h_d od x^{d+1}), where circuit size of g is s_d and degree of g_d is at most d, can be computed by a circuit of size poly(s_d,deg(h_d),log d). We also show a hardness result when the degrees of the rational functions are high (i.e. Ω (d)), assuming hardness of the integer factorization problem. Finally, we extend this conditional hardness to simple algebraic functions as well, and show that for every prime p, there is an integral algebraic power series with its minimal polynomial satisfying a degree p polynomial equation, such that its initial segment is hard to compute unless integer factoring is easy, or a multiple of n! is easy to compute. Both, integer factoring and computation of multiple of n!, are believed to be notoriously hard. In contrast, we show examples of transcendental power series whose initial segments are easy to compute.
##### Cite as
Pranjal Dutta, Gorav Jindal, Anurag Pandey, and Amit Sinhababu. Arithmetic Circuit Complexity of Division and Truncation. In 36th Computational Complexity Conference (CCC 2021). Leibniz International Proceedings in Informatics (LIPIcs), Volume 200, pp. 25:1-25:36, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2021)
```@InProceedings{dutta_et_al:LIPIcs.CCC.2021.25,
author = {Dutta, Pranjal and Jindal, Gorav and Pandey, Anurag and Sinhababu, Amit},
title = {{Arithmetic Circuit Complexity of Division and Truncation}},
booktitle = {36th Computational Complexity Conference (CCC 2021)},
pages = {25:1--25:36},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-193-1},
ISSN = {1868-8969},
year = {2021},
volume = {200},
editor = {Kabanets, Valentine},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.CCC.2021.25},
URN = {urn:nbn:de:0030-drops-142990},
doi = {10.4230/LIPIcs.CCC.2021.25},
annote = {Keywords: Arithmetic Circuits, Division, Truncation, Division elimination, Rational function, Algebraic power series, Transcendental power series, Integer factorization}
}```
Document
##### Factorization of Polynomials Given By Arithmetic Branching Programs
Authors: Amit Sinhababu and Thomas Thierauf
Published in: LIPIcs, Volume 169, 35th Computational Complexity Conference (CCC 2020)
##### Abstract
Given a multivariate polynomial computed by an arithmetic branching program (ABP) of size s, we show that all its factors can be computed by arithmetic branching programs of size poly(s). Kaltofen gave a similar result for polynomials computed by arithmetic circuits. The previously known best upper bound for ABP-factors was poly(s^(log s)).
##### Cite as
Amit Sinhababu and Thomas Thierauf. Factorization of Polynomials Given By Arithmetic Branching Programs. In 35th Computational Complexity Conference (CCC 2020). Leibniz International Proceedings in Informatics (LIPIcs), Volume 169, pp. 33:1-33:19, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2020)
```@InProceedings{sinhababu_et_al:LIPIcs.CCC.2020.33,
author = {Sinhababu, Amit and Thierauf, Thomas},
title = {{Factorization of Polynomials Given By Arithmetic Branching Programs}},
booktitle = {35th Computational Complexity Conference (CCC 2020)},
pages = {33:1--33:19},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-156-6},
ISSN = {1868-8969},
year = {2020},
volume = {169},
editor = {Saraf, Shubhangi},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.CCC.2020.33},
URN = {urn:nbn:de:0030-drops-125854},
doi = {10.4230/LIPIcs.CCC.2020.33},
annote = {Keywords: Arithmetic Branching Program, Multivariate Polynomial Factorization, Hensel Lifting, Newton Iteration, Hardness vs Randomness}
}```
Document
##### Algebraic Dependencies and PSPACE Algorithms in Approximative Complexity
Authors: Zeyu Guo, Nitin Saxena, and Amit Sinhababu
Published in: LIPIcs, Volume 102, 33rd Computational Complexity Conference (CCC 2018)
##### Abstract
Testing whether a set f of polynomials has an algebraic dependence is a basic problem with several applications. The polynomials are given as algebraic circuits. Algebraic independence testing question is wide open over finite fields (Dvir, Gabizon, Wigderson, FOCS'07). Previously, the best complexity known was NP^{#P} (Mittmann, Saxena, Scheiblechner, Trans.AMS'14). In this work we put the problem in AM cap coAM. In particular, dependence testing is unlikely to be NP-hard and joins the league of problems of "intermediate" complexity, eg. graph isomorphism & integer factoring. Our proof method is algebro-geometric- estimating the size of the image/preimage of the polynomial map f over the finite field. A gap in this size is utilized in the AM protocols. Next, we study the open question of testing whether every annihilator of f has zero constant term (Kayal, CCC'09). We give a geometric characterization using Zariski closure of the image of f; introducing a new problem called approximate polynomials satisfiability (APS). We show that APS is NP-hard and, using projective algebraic-geometry ideas, we put APS in PSPACE (prior best was EXPSPACE via Gröbner basis computation). As an unexpected application of this to approximative complexity theory we get- over any field, hitting-sets for overline{VP} can be verified in PSPACE. This solves an open problem posed in (Mulmuley, FOCS'12, J.AMS 2017); greatly mitigating the GCT Chasm (exponentially in terms of space complexity).
##### Cite as
Zeyu Guo, Nitin Saxena, and Amit Sinhababu. Algebraic Dependencies and PSPACE Algorithms in Approximative Complexity. In 33rd Computational Complexity Conference (CCC 2018). Leibniz International Proceedings in Informatics (LIPIcs), Volume 102, pp. 10:1-10:21, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2018)
```@InProceedings{guo_et_al:LIPIcs.CCC.2018.10,
author = {Guo, Zeyu and Saxena, Nitin and Sinhababu, Amit},
title = {{Algebraic Dependencies and PSPACE Algorithms in Approximative Complexity}},
booktitle = {33rd Computational Complexity Conference (CCC 2018)},
pages = {10:1--10:21},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-069-9},
ISSN = {1868-8969},
year = {2018},
volume = {102},
editor = {Servedio, Rocco A.},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.CCC.2018.10},
URN = {urn:nbn:de:0030-drops-88786},
doi = {10.4230/LIPIcs.CCC.2018.10},
annote = {Keywords: algebraic dependence, Jacobian, Arthur-Merlin, approximate polynomial, satisfiability, hitting-set, border VP, finite field, PSPACE, EXPSPACE, GCT Chasm}
}```
Document
##### Algebraic Independence over Positive Characteristic: New Criterion and Applications to Locally Low Algebraic Rank Circuits
Authors: Anurag Pandey, Nitin Saxena, and Amit Sinhababu
Published in: LIPIcs, Volume 58, 41st International Symposium on Mathematical Foundations of Computer Science (MFCS 2016)
##### Abstract
The motivation for this work comes from two problems--test algebraic independence of arithmetic circuits over a field of small characteristic, and generalize the structural property of algebraic dependence used by (Kumar, Saraf CCC'16) to arbitrary fields. It is known that in the case of zero, or large characteristic, using a classical criterion based on the Jacobian, we get a randomized poly-time algorithm to test algebraic independence. Over small characteristic, the Jacobian criterion fails and there is no subexponential time algorithm known. This problem could well be conjectured to be in RP, but the current best algorithm puts it in NP^#P (Mittmann, Saxena, Scheiblechner Trans.AMS'14). Currently, even the case of two bivariate circuits over F_2 is open. We come up with a natural generalization of Jacobian criterion, that works over all characteristic. The new criterion is efficient if the underlying inseparable degree is promised to be a constant. This is a modest step towards the open question of fast independence testing, over finite fields, posed in (Dvir, Gabizon, Wigderson FOCS'07). In a set of linearly dependent polynomials, any polynomial can be written as a linear combination of the polynomials forming a basis. The analogous property for algebraic dependence is false, but a property approximately in that spirit is named as ``functional dependence'' in (Kumar, Saraf CCC'16) and proved for zero or large characteristic. We show that functional dependence holds for arbitrary fields, thereby answering the open questions in (Kumar, Saraf CCC'16). Following them we use the functional dependence lemma to prove the first exponential lower bound for locally low algebraic rank circuits for arbitrary fields (a model that strongly generalizes homogeneous depth-4 circuits). We also recover their quasipoly-time hitting-set for such models, for fields of characteristic smaller than the ones known before. Our results show that approximate functional dependence is indeed a more fundamental concept than the Jacobian as it is field independent. We achieve the former by first picking a ``good'' transcendence basis, then translating the circuits by new variables, and finally approximating them by truncating higher degree monomials. We give a tight analysis of the ``degree'' of approximation needed in the criterion. To get the locally low algebraic rank circuit applications we follow the known shifted partial derivative based methods.
##### Cite as
Anurag Pandey, Nitin Saxena, and Amit Sinhababu. Algebraic Independence over Positive Characteristic: New Criterion and Applications to Locally Low Algebraic Rank Circuits. In 41st International Symposium on Mathematical Foundations of Computer Science (MFCS 2016). Leibniz International Proceedings in Informatics (LIPIcs), Volume 58, pp. 74:1-74:15, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2016)
```@InProceedings{pandey_et_al:LIPIcs.MFCS.2016.74,
author = {Pandey, Anurag and Saxena, Nitin and Sinhababu, Amit},
title = {{Algebraic Independence over Positive Characteristic: New Criterion and Applications to Locally Low Algebraic Rank Circuits}},
booktitle = {41st International Symposium on Mathematical Foundations of Computer Science (MFCS 2016)},
pages = {74:1--74:15},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-016-3},
ISSN = {1868-8969},
year = {2016},
volume = {58},
editor = {Faliszewski, Piotr and Muscholl, Anca and Niedermeier, Rolf},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.MFCS.2016.74},
URN = {urn:nbn:de:0030-drops-65057},
doi = {10.4230/LIPIcs.MFCS.2016.74},
annote = {Keywords: independence, transcendence, finite field, Hasse-Schmidt, Jacobian, differential, inseparable, circuit, identity testing, lower bound, depth-4, shifte}
}```
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Please try again later or send an E-mail | 3,227 | 12,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.870073 |
http://mathhelpforum.com/calculus/33302-volume-oil-tanker-print.html | 1,526,833,187,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863626.14/warc/CC-MAIN-20180520151124-20180520171124-00604.warc.gz | 173,347,614 | 3,279 | # volume of oil tanker
• Apr 5th 2008, 01:31 PM
galactus
volume of oil tanker
Here's a neat integration problem. Not particularly difficult, but cool. Give it a go if you have a mind to.
"The tank of an oil truck is 18 feet long and has elliptical cross-sections that are 6 feet wide and 4 feet high. Find the volume of the oil in the tank in terms of the depth h of the oil".
This one is a little different in that the cross sections are elliptic rather than circular. We have all seen these types of tankers on the road, it got me to thinking about such a problem.
• Apr 6th 2008, 09:17 AM
galactus
I thought someone would bite on this problem. Here's how I done it. It appears to check. I just made up those dimensions from looking at one of those types of tankers on the road. we could've derived a general case instead.
Since the ellipse is 6 feet wide and 4 feet high, the ellipse has the equation
$\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
Solve for x in terms of y:
$\displaystyle x=\frac{3}{2}\sqrt{4-y^{2}}$
$\displaystyle V=\int_{-2}^{-2+h}(2)(3/2)\sqrt{4-y^{2}}(18)dy=54\int_{-2}^{-2+h}\sqrt{4-y^{2}}dy$
Now, integrate and we get:
$\displaystyle V=27\left[4sin^{-1}(\frac{h-2}{2})+(h-2)\sqrt{4h-h^{2}}+2\pi\right]$
For the general case, from $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\displaystyle V=\frac{ahL\sqrt{2bh-h^{2}}}{b}+\frac{abL\pi}{2}-absin^{-1}(\frac{b-h}{b})L-aL\sqrt{2bh-h^{2}}$
Kind of long connected and probably simplifies, but that's all I'm doing.
• Apr 6th 2008, 09:23 AM
janvdl
Quote:
Originally Posted by galactus
Here's a neat integration problem. Not particularly difficult, but cool. Give it a go if you have a mind to.
Quote:
Originally Posted by galactus
I thought someone would bite on this problem...
I didn't notice it before... Not that I would have been able to do it yet anyway.
Patience Galactus, soon I will be answering all these questions... (This term actually :D We'll be starting with this kind of work.) | 616 | 1,999 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.911727 |
https://isabelle.in.tum.de/repos/isabelle/file/180f1b3508ed/src/HOL/Induct/Tree.thy | 1,579,526,872,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598726.39/warc/CC-MAIN-20200120110422-20200120134422-00355.warc.gz | 497,840,736 | 3,932 | src/HOL/Induct/Tree.thy
author blanchet Tue Sep 09 20:51:36 2014 +0200 (2014-09-09) changeset 58249 180f1b3508ed parent 46914 c2ca2c3d23a6 child 58310 91ea607a34d8 permissions -rw-r--r--
use 'datatype_new' (soon to be renamed 'datatype') in Isabelle's libraries
``` 1 (* Title: HOL/Induct/Tree.thy
```
``` 2 Author: Stefan Berghofer, TU Muenchen
```
``` 3 Author: Lawrence C Paulson, Cambridge University Computer Laboratory
```
``` 4 *)
```
``` 5
```
``` 6 header {* Infinitely branching trees *}
```
``` 7
```
``` 8 theory Tree
```
``` 9 imports Main
```
``` 10 begin
```
``` 11
```
``` 12 datatype_new 'a tree =
```
``` 13 Atom 'a
```
``` 14 | Branch "nat => 'a tree"
```
``` 15
```
``` 16 primrec map_tree :: "('a => 'b) => 'a tree => 'b tree"
```
``` 17 where
```
``` 18 "map_tree f (Atom a) = Atom (f a)"
```
``` 19 | "map_tree f (Branch ts) = Branch (\<lambda>x. map_tree f (ts x))"
```
``` 20
```
``` 21 lemma tree_map_compose: "map_tree g (map_tree f t) = map_tree (g \<circ> f) t"
```
``` 22 by (induct t) simp_all
```
``` 23
```
``` 24 primrec exists_tree :: "('a => bool) => 'a tree => bool"
```
``` 25 where
```
``` 26 "exists_tree P (Atom a) = P a"
```
``` 27 | "exists_tree P (Branch ts) = (\<exists>x. exists_tree P (ts x))"
```
``` 28
```
``` 29 lemma exists_map:
```
``` 30 "(!!x. P x ==> Q (f x)) ==>
```
``` 31 exists_tree P ts ==> exists_tree Q (map_tree f ts)"
```
``` 32 by (induct ts) auto
```
``` 33
```
``` 34
```
``` 35 subsection{*The Brouwer ordinals, as in ZF/Induct/Brouwer.thy.*}
```
``` 36
```
``` 37 datatype_new brouwer = Zero | Succ "brouwer" | Lim "nat => brouwer"
```
``` 38
```
``` 39 text{*Addition of ordinals*}
```
``` 40 primrec add :: "[brouwer,brouwer] => brouwer"
```
``` 41 where
```
``` 42 "add i Zero = i"
```
``` 43 | "add i (Succ j) = Succ (add i j)"
```
``` 44 | "add i (Lim f) = Lim (%n. add i (f n))"
```
``` 45
```
``` 46 lemma add_assoc: "add (add i j) k = add i (add j k)"
```
``` 47 by (induct k) auto
```
``` 48
```
``` 49 text{*Multiplication of ordinals*}
```
``` 50 primrec mult :: "[brouwer,brouwer] => brouwer"
```
``` 51 where
```
``` 52 "mult i Zero = Zero"
```
``` 53 | "mult i (Succ j) = add (mult i j) i"
```
``` 54 | "mult i (Lim f) = Lim (%n. mult i (f n))"
```
``` 55
```
``` 56 lemma add_mult_distrib: "mult i (add j k) = add (mult i j) (mult i k)"
```
``` 57 by (induct k) (auto simp add: add_assoc)
```
``` 58
```
``` 59 lemma mult_assoc: "mult (mult i j) k = mult i (mult j k)"
```
``` 60 by (induct k) (auto simp add: add_mult_distrib)
```
``` 61
```
``` 62 text{*We could probably instantiate some axiomatic type classes and use
```
``` 63 the standard infix operators.*}
```
``` 64
```
``` 65 subsection{*A WF Ordering for The Brouwer ordinals (Michael Compton)*}
```
``` 66
```
``` 67 text{*To use the function package we need an ordering on the Brouwer
```
``` 68 ordinals. Start with a predecessor relation and form its transitive
```
``` 69 closure. *}
```
``` 70
```
``` 71 definition brouwer_pred :: "(brouwer * brouwer) set"
```
``` 72 where "brouwer_pred = (\<Union>i. {(m,n). n = Succ m \<or> (EX f. n = Lim f & m = f i)})"
```
``` 73
```
``` 74 definition brouwer_order :: "(brouwer * brouwer) set"
```
``` 75 where "brouwer_order = brouwer_pred^+"
```
``` 76
```
``` 77 lemma wf_brouwer_pred: "wf brouwer_pred"
```
``` 78 by(unfold wf_def brouwer_pred_def, clarify, induct_tac x, blast+)
```
``` 79
```
``` 80 lemma wf_brouwer_order[simp]: "wf brouwer_order"
```
``` 81 by(unfold brouwer_order_def, rule wf_trancl[OF wf_brouwer_pred])
```
``` 82
```
``` 83 lemma [simp]: "(j, Succ j) : brouwer_order"
```
``` 84 by(auto simp add: brouwer_order_def brouwer_pred_def)
```
``` 85
```
``` 86 lemma [simp]: "(f n, Lim f) : brouwer_order"
```
``` 87 by(auto simp add: brouwer_order_def brouwer_pred_def)
```
``` 88
```
``` 89 text{*Example of a general function*}
```
``` 90
```
``` 91 function add2 :: "brouwer \<Rightarrow> brouwer \<Rightarrow> brouwer"
```
``` 92 where
```
``` 93 "add2 i Zero = i"
```
``` 94 | "add2 i (Succ j) = Succ (add2 i j)"
```
``` 95 | "add2 i (Lim f) = Lim (\<lambda>n. add2 i (f n))"
```
``` 96 by pat_completeness auto
```
``` 97 termination by (relation "inv_image brouwer_order snd") auto
```
``` 98
```
``` 99 lemma add2_assoc: "add2 (add2 i j) k = add2 i (add2 j k)"
```
``` 100 by (induct k) auto
```
``` 101
```
``` 102 end
``` | 1,759 | 4,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-05 | longest | en | 0.666006 |
https://febo.com/pipermail/time-nuts/2005-April/018087.html | 1,656,575,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00171.warc.gz | 307,400,316 | 3,196 | # [time-nuts] Re: Phase noise with a lock-in amplifier.
Javier javier at nebulosa.org
Sun Apr 17 14:30:14 EDT 2005
```Hello,
I think that there is a problem with the mixing of the oscillator with a
delayed version of itself. Suppose that the noise affecting the
oscillator were exclusively a square wave lets say at 1KHz rate, lets
say modulating the oscillator at +/- 1Hz. At any given instant, and
suppossing that the cable length is far less that to produce a 1ms delay
(1ms delay would be around 200Km of RG-58/U, if I remember welll), the
frequencies entering the mixer would be exactly the same, producing a DC
- you would only have glitches at a 1KHz rate, corresponding to the
instants when the two frequencies differ.
This can be translated to that for measuring the phase noise mixing the
signal with a delayed version of itself, you must have a delay at least
comparable to the lowest noise frequency you want to determine, in order
to have truly uncorrelated phase noise at the two mixer inputs.
Regards,
Javier, EA1CRB
David Kirkby wrote:
> Bill Hawkins wrote:
>
>> Dr. David Kirkby wrote,
>>
>> "I was not thinking of impedance matching at all. If the mixer is
>> 50Ohm input (as most are) and the oscillator has a 50 Ohm output, the
>> cable length would have no effect on this at all."
>>
>> The cable length has no effect on frequency (if the output device is
>> buffered with an amplifier).
>
>
> But what if a 10MHz (for example) oscillator is not perfect? The
> output is no longer a pure sine wave at 10MHz, but could be considered
> as an infinite number of oscillators, all of different frequencies,
> with amplitudes and phases that are essentially random. Those
> oscillators closer to 10MHz would on average have higher amplitudes,
> but all oscillators will have some amplitude.
>
> So if the two inputs are fed to a mixer have different path lengths,
> the inputs to the mixers will *not* be identical.
>
>> As I understand it, a phase detector that has both inputs fed by the
>> same frequency is only capable of measuring systematic errors, such as
>> may be caused by differences in the FET response times. Of course, it
>> will also measure differences in cable length.
>
>
> But the inputs will not be of the same frequency normally, due to the
> phase noise on the oscillator.
>
>
>> There was a time when obscure discoveries could be rediscovered when
>> the right opportunity arose. These days, with information at the speed
>> of the Internet, the process of natural selection works much faster.
>> The phase noise question has been around longer than the Internet, so
>> it seems to be true that the best solutions have been found.
>
>
> Well, it never hurts to consider new ideas. That might be silly,
> fatally flawed in some way.
>
>
> I'm 99.9% sure the mixer, with two different cable lengths from a
> 10MHz oscillator will produce two ranges of frequencies. One will be
> close to DC, but will extend up a few hundred kHz or so. The other
> will be close to 20MHz, but extend a few hundred kHz either side.
>
> I'm pretty sure if you low-pass filter the output of that mixer, you
> will see the oscillator phase noise shifted from 10MHz down to near DC.
>
> I think if you put an audio spectrum analyser on the output of the
> mixer, after the low-pass filter, you would see the phase noise of the
> oscillator.
>
> In fact, would you not hear the oscillator phase noise on a speaker if
> you amplified the low-pass filtered signal from the mixer? (I'm
> ignoring the fact the noise on the audio system might be lower than
> the oscillator phase noise).
>
> I'm less sure, but it it possible a lock-in might be able to see this
> phase noise, by setting the reference to the offset you want.
>
> Question, what would you see on the lock-in if you used its internal
> oscillator as as reference and added to this (op-amp configured as an
> adder) the output from the mixer? The op-amp would measure its own
> reference amplitude, but you would have introduced noise on it from
> the 10MHz reference, so the lock-in should see that noise too.
>
> Perhaps I'll sketch out what I mean and put a circuit diagram.
>
> The Standford dual-phase lock-in has functions to measure the noise on
> the signal.
>
``` | 1,035 | 4,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.940988 |
http://nrich.maths.org/public/leg.php?code=-68&cl=1&cldcmpid=2372 | 1,448,599,403,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447913.86/warc/CC-MAIN-20151124205407-00095-ip-10-71-132-137.ec2.internal.warc.gz | 178,868,816 | 9,026 | # Search by Topic
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### Dicey
##### Stage: 2 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### Midpoint Triangle
##### Stage: 2 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Change Around
##### Stage: 1 Challenge Level:
Move just three of the circles so that the triangle faces in the opposite direction.
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### Six to Four
##### Stage: 1 Challenge Level:
Move four sticks so there are exactly four triangles.
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### Put Yourself in a Box
##### Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### Right or Left?
##### Stage: 2 Challenge Level:
Which of these dice are right-handed and which are left-handed?
### Endless Noughts and Crosses
##### Stage: 2 Challenge Level:
An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5.
### Penta Play
##### Stage: 2 Challenge Level:
A shape and space game for 2,3 or 4 players. Be the last person to be able to place a pentomino piece on the playing board. Play with card, or on the computer.
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
##### Stage: 1 Challenge Level:
Here are shadows of some 3D shapes. What shapes could have made them?
### World of Tan 19 - Working Men
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### World of Tan 20 - Fractions
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### World of Tan 26 - Old Chestnut
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 28 - Concentrating on Coordinates
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### World of Tan 25 - Pentominoes
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### Skeleton Shapes
##### Stage: 1 Challenge Level:
How many balls of modelling clay and how many straws does it take to make these skeleton shapes?
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### More Building with Cubes
##### Stage: 2 Challenge Level:
Here are more buildings to picture in your mind's eye. Watch out - they become quite complicated!
### Making Maths: Rolypoly
##### Stage: 1 and 2 Challenge Level:
Paint a stripe on a cardboard roll. Can you predict what will happen when it is rolled across a sheet of paper?
### Go Moku
##### Stage: 2 Challenge Level:
A game for two players on a large squared space.
### Cube Drilling
##### Stage: 2 Challenge Level:
Imagine a 4 by 4 by 4 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will not have holes drilled through them?
### The Path of the Dice
##### Stage: 2 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line. | 1,897 | 8,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-48 | longest | en | 0.874383 |
https://www.evi.com/q/how_many_cm_is_192_mm | 1,429,951,731,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246648209.18/warc/CC-MAIN-20150417045728-00175-ip-10-235-10-82.ec2.internal.warc.gz | 930,185,598 | 5,169 | how many cm is 192 mm
• tk10publ tk10canl
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• whats 192mm in cm | 445 | 1,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-18 | longest | en | 0.902795 |
https://www.esaral.com/q/factorize-each-of-the-following-algebraic-expression-62031 | 1,726,077,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00419.warc.gz | 721,394,641 | 11,591 | # Factorize each of the following algebraic expression:
Question:
Factorize each of the following algebraic expression:
x2 + 14x + 45
Solution:
To factorise $\mathrm{x}^{2}+14 \mathrm{x}+45$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=14$ and $\mathrm{pq}=45$.
Now,
$9+5=14$
and
$9 \times 5=45$
Splitting the middle term $14 \mathrm{x}$ in the given quadratic as $9 \mathrm{x}+5 \mathrm{x}$, we get:
$\mathrm{x}^{2}+14 \mathrm{x}+45=\mathrm{x}^{2}+9 \mathrm{x}+5 \mathrm{x}+45$
$=\left(\mathrm{x}^{2}+9 \mathrm{x}\right)+(5 \mathrm{x}+45)$
$=\mathrm{x}(\mathrm{x}+9)+5(\mathrm{x}+9)$
$=(\mathrm{x}+5)(\mathrm{x}+9)$ | 249 | 673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.38786 |
https://jarviscodinghub.com/product/cse-101-introduction-to-computational-and-algorithmic-thinking-lab-assignment-6-solution/ | 1,679,822,772,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00495.warc.gz | 381,835,069 | 22,059 | # CSE 101: Introduction to Computational and Algorithmic Thinking Lab Assignment #6 solution
\$25.00
Category:
## Description
This lab assignment will give you practice with nested lists and nested loops.
Getting Started
Visit Piazza and download the “bare bones” file lab6.py onto your computer. Open lab6.py in PyCharm and
fill in the following information at the top:
1. your first and last name as they appear in Blackboard
2. your Net ID (e.g., jsmith)
3. your Stony Brook ID # (e.g., 111999999)
4. the course number (CSE 101)
5. the assignment name and number (Lab #6)
Submit your final lab6.py file to Blackboard by the due date and time. Late work will not be graded. Code that
crashes and cannot be graded will earn no credit.
In this part, you will write a function student alert(). The function simulates a simplified student grade
alert system, which will evaluate a list of students’ grades, and return their corresponding “alert level” (more on
this later).
The function takes only one argument, students, which is a list of lists. Each sub-list contains one particular
student’s grades in the form of letter grade strings. Each letter grade has a corresponding integer value as shown
below:
’A’ 95
’B’ 85
’C’ 75
’D’ 65
’F 55
Example list: [[’A’,’B’,’A’,’F’,’B’], [’A’,’F’], [’F’,’C’,’D’,’A’,’A’], [’F’,’F’]]
CSE 101 – Spring 2018 Lab #6 Page 1
Note that the above example list stores the grades for four students.
The function iterates through each student’s sub-list that contains all of his/her grades, then calculates the average
grade for that student. Depending on the average grade for that student, a particular character is appended to a list
that the function will return:
• If a student’s average grade is less than 70, it is considered a red alert, so the function appends ’R’ to the
result list.
• If a student’s average grade is at least 70 but less than 80, it is considered a yellow alert, so the function
appends ’Y’ to the result list.
• If a student’s average grade is at least 80, it is considered green alert (safe), so the function appends ’G’ to
the result list.
Eventually, the function returns a list of strings that represent the corresponding alert levels for all the students in
the students list. For the example list given earlier, the average grades for the four students are 83, 75, 77 and
55, respectively. Therefore, the list returned by the function would be [’G’,’Y’,’Y’,’R’].
Examples:
Function Call Return Value
student alert([[’D’], [’A’, ’A’, ’C’, ’D’], [’B’],
[’C’, ’F’, ’C’], [’C’, ’A’, ’A’, ’F’, ’F’, ’D’],
[’D’, ’D’, ’C’, ’F’, ’B’]])
[’R’, ’G’, ’G’, ’R’,
’Y’, ’R’]
student alert([[’D’], [’D’, ’B’], [’D’, ’C’, ’F’,
’C’, ’A’, ’D’]])
[’R’, ’Y’, ’Y’]
student alert([[’F’], [’D’, ’C’, ’A’, ’B’], [’C’,
’F’, ’C’], [’F’, ’F’], [’C’, ’B’, ’B’, ’A’, ’F’,
’F’], [’B’, ’C’, ’C’, ’C’], [’F’, ’C’, ’B’, ’B’],
[’D’, ’B’, ’D’, ’B’, ’D’, ’C’]])
[’R’, ’G’, ’R’, ’R’,
’Y’, ’Y’, ’Y’, ’Y’]
student alert([[’D’, ’F’, ’C’, ’B’], [’C’, ’A’,
’A’, ’F’, ’A’], [’A’, ’F’], [’A’, ’B’], [’A’, ’A’,
’F’, ’A’]])
[’Y’, ’G’, ’Y’, ’G’,
’G’]
student alert([[’D’, ’B’, ’D’, ’F’], [’D’, ’B’,
’D’], [’A’, ’D’], [’D’, ’B’]])
[’R’, ’Y’, ’G’, ’Y’]
student alert([[’A’], [’B’, ’C’, ’A’, ’D’, ’B’,
’A’], [’F’, ’B’, ’C’, ’A’, ’F’, ’B’], [’A’, ’B’,
’F’, ’A’], [’A’, ’C’, ’A’], [’C’, ’B’, ’B’, ’F’,
’F’], [’A’, ’F’], [’D’]])
[’G’, ’G’, ’Y’, ’G’,
’G’, ’Y’, ’Y’, ’R’]
student alert([[’F’, ’B’], [’B’, ’B’, ’D’, ’F’,
’D’], [’D’], [’D’, ’B’, ’C’, ’D’, ’A’], [’A’, ’A’,
’D’, ’F’, ’D’, ’D’], [’A’], [’D’, ’B’], [’C’, ’F’,
’A’, ’B’, ’A’], [’C’, ’A’, ’A’, ’B’, ’A’]])
[’Y’, ’Y’, ’R’, ’Y’,
’Y’, ’G’, ’Y’, ’G’,
’G’]
CSE 101 – Spring 2018 Lab #6 Page 2
student alert([[’A’], [’C’, ’A’], [’C’, ’B’, ’C’],
[’C’], [’F’], [’A’], [’B’, ’B’], [’F’, ’C’, ’C’,
’B’, ’F’, ’F’], [’B’, ’A’, ’F’, ’D’, ’D’], [’F’]])
[’G’, ’G’, ’Y’, ’Y’,
’R’, ’G’, ’G’, ’R’,
’Y’, ’R’]
student alert([[’A’, ’A’, ’B’], [’A’, ’C’, ’B’,
’D’, ’B’, ’D’], [’A’]])
[’G’, ’Y’, ’G’]
Part II: Employee Searcher (20 points)
Some Preliminaries
In Python, a tuple is a collection similar to a list in that it is an ordered collection of items. An important difference,
however, is that a tuple is immutable – once created, a tuple cannot be changed. Also, tuples are denoted using
parentheses instead of square brackets. As with lists, elements of a tuple are accessed using indexing notation.
For example, given the tuple subjects = (’physics’, ’chemistry’, ’biology’), we could access the elements of the tuple using subjects[0], subjects[1] and subjects[2].
The Function to Write
In this part of the assignment, you will write a function find employee(), which takes three arguments, in
this order:
• employees: A list of lists, in which each sub-list contains information about one particular employee.
Inside each sub-list we always find five tuples, wherein each tuple has two values: the first value is a field
name, and the second value is the value of that field.
The fields available are listed below. Note that although the five tuples will always exist, their order inside
the sub-list may differ from employee to employee.
◦ ’Name’: this string is the employee’s name
◦ ’Age’: this integer is the employee’s age
◦ ’Salary’: this integer is the employee’s salary
◦ ’Experience’: this integer is the years of experience the employee has
◦ ’Level’: this integer is the employee’s rank
An example employees list containing the information about two employees is given below:
employees = [
[(’Name’, ’Tom’), (’Age’, 21), (’Salary’, 1000), (’Experience’, 1), (’Level’, 1)],
[(’Experience’, 2), (’Age’, 22), (’Name’, ’Sam’), (’Salary’, 2500), (’Level’, 3)]
]
• search field: this is the particular field you are looking for
• search value: this is the desired value of a field you are looking for
The function searches the employees list and returns a list of employee names whose search field’s value
is equal to search value. For example, if search field = ’Level’ and search value = 2, the
function would return a list of employee names whose Level is exactly 2.
CSE 101 – Spring 2018 Lab #6 Page 3
In brief, the function executes the following algorithm:
create an empty list (e.g., result) to store the names of employees
we will return
for each employee in the list of employees:
1. using a for-loop, search the list of tuples of this employee to find
the employee’s name (hint: have an if-statement look at index 0
of each tuple to find the tuple that contains ’Name’ so that we
can record the employee’s name, which will be at index 1 of
that particular tuple)
2. using a separate for-loop, search the list of tuples of this employee
to find the tuple that the search_field argument at index 0
2a. when we find that tuple, check if the value at index 1 of the
tuple equals the search_value argument of the function
2b. if so, append the employee’s name to the result list
return the result list
Example:
Imagine that employees contained the following data and we had search field = ’Experience’ and
search value = 1:
[ [(’Level’, 1), (’Age’, 25), (’Salary’, 30008), (’Experience’, 5), (’Name’, ’Janice’)],
[(’Name’, ’Jose’), (’Level’, 1), (’Salary’, 30011), (’Experience’, 5), (’Age’, 23)],
[(’Experience’, 1), (’Age’, 22), (’Name’, ’Chance’), (’Level’, 2), (’Salary’, 30015)],
[(’Name’, ’Erminia’), (’Salary’, 30006), (’Experience’, 4), (’Level’, 5), (’Age’, 18)],
[(’Salary’, 30014), (’Level’, 4), (’Experience’, 2), (’Name’, ’Wilton’), (’Age’, 19)],
[(’Name’, ’Bibi’), (’Age’, 19), (’Salary’, 30008), (’Experience’, 2), (’Level’, 2)],
[(’Salary’, 30000), (’Age’, 23), (’Level’, 4), (’Name’, ’Henry’), (’Experience’, 3)],
[(’Age’, 18), (’Salary’, 30009), (’Experience’, 3), (’Level’, 1), (’Name’, ’Tyesha’)],
[(’Name’, ’Bethany’), (’Age’, 21), (’Salary’, 30002), (’Experience’, 5), (’Level’, 5)],
[(’Salary’, 30013), (’Experience’, 3), (’Name’, ’Sofia’), (’Age’, 20), (’Level’, 3)],
[(’Name’, ’Tiffany’), (’Level’, 3), (’Salary’, 30007), (’Experience’, 3), (’Age’, 23)],
[(’Level’, 2), (’Age’, 20), (’Name’, ’Darryl’), (’Experience’, 3), (’Salary’, 30008)],
[(’Level’, 1), (’Experience’, 5), (’Name’, ’Twila’), (’Age’, 23), (’Salary’, 30014)],
[(’Salary’, 30014), (’Name’, ’Aleshia’), (’Experience’, 1), (’Age’, 23), (’Level’, 2)],
[(’Experience’, 1), (’Name’, ’Armandina’), (’Level’, 4), (’Age’, 21), (’Salary’, 30010)] ]
In this case, the function would return the list [’Chance’, ’Aleshia’, ’Armandina’] because these
are the three employess whose ’Experience’ tuple contains the value 1. Note that you may return the list of
names in any order you wish. | 2,900 | 8,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-14 | latest | en | 0.854976 |
http://gmatclub.com/forum/when-presented-with-only-circumstantional-evidence-a-juror-62612.html#p456361 | 1,481,254,109,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00302-ip-10-31-129-80.ec2.internal.warc.gz | 113,609,596 | 53,740 | When presented with only circumstantional evidence, a juror : GMAT Sentence Correction (SC)
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# When presented with only circumstantional evidence, a juror
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When presented with only circumstantional evidence, a juror [#permalink]
### Show Tags
16 Apr 2008, 08:34
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When presented with only circumstantional evidence, a juror tends to decide a case according to his or her gut instinct,which are not formally sanctioned or prohibited means of reaching a verdict.
A. which are not formally
B which are not a formally
C which is not a formally
D which is formally not a
E which is not formally
OA to follow
If you have any questions
New!
Manager
Joined: 09 Apr 2008
Posts: 60
Concentration: Strategy, Operations
Schools: CBS '15 (A)
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### Show Tags
16 Apr 2008, 08:50
My opinion...
A. which are not formally [are is plural]
B which are not a formally [are is plural]
C which is not a formally [the "a" makes "means" singular, which keeps parallel with "instinct"]
D which is formally not a [splits "is...not"]
E which is not formally [the lack of "a" makes "means" plural, which does not match "instinct"]
[C]
OA?
_________________
"The price of anything is the amount of life you exchange for it." -Thoreau
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### Show Tags
17 Apr 2008, 11:15
OA is C
Re: SC Kaplan 800 [#permalink] 17 Apr 2008, 11:15
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20 When presented with only circumstantial evidence, a juror 31 29 Apr 2009, 18:53
Display posts from previous: Sort by | 797 | 2,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-50 | latest | en | 0.926368 |
https://datascience.stackexchange.com/questions/8026/fittting-histograms-with-combination-of-an-unknown-number-of-gaussians | 1,701,859,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00554.warc.gz | 231,818,701 | 42,876 | # Fittting histograms with combination of an unknown number of Gaussians
I have many data in the form of one dimensional histogram, to give an example consider the data at http://pastebin.com/embed_js.php?i=1mNRuEHZ
I expect that these data are obtained from a pdf composed of a sum of Gaussian distributions but with their actual number being unknown.
I want then to fit these data with a model given by a sum of Gaussians
$$f_N(x) = \sum_{k=1}^N c_k \exp \left[ - \frac{(x-a_k)^2}{2 b_k^2} \right]$$
where $a_k$, $b_k$, $c_k$ and $N$ are in principle parameters to be fitted.
The problem can be viewed as a problem of model selection between different $f_N$ for all the possible values of $N$.
My idea is then to make an usual fit by least squares at fixed $N$ and then compare the results for different $N$ via some statistical measure of the quality of the fit, as e.g. Akaike information criterion (AIC). Is this an acceptable test?
For the data of the previous example (that I generated as a test from 3 gaussians) I will obtain
N AIC
-------------
1 +568.1
2 +557.4
3 -446.6
4 -443.5
5 -442.7
As you see, the AIC decreases quickly till $N=3$, where it starts to increase again very slowly (due to overfitting). So it is clear that $N=3$ is the best choice here.
I mention that in some cases I have more data histograms extracted from the same pdf, so I could do some cross validation. Although this always a good test, the availability of my data changes from sample to sample, so I'd prefer to have a criterion for the single histogram (if the quality of the test is not too much different).
• Perhaps you could explain a bit more what your data looks like? How many dimensions, etc? Do you have (x,y) data and you are trying to fit a curve through the points, or is it a density you are trying to match? Sep 7, 2015 at 16:29
• You don't have histograms nor do you have random samples from a mixture of normal distributions. (If so, you certainly wouldn't be getting negative numbers for the counts or for a probably density.) It appears that you have samples from a fixed set of x's where the expected values for the vertical axis are from a linear combination of curves with a Gaussian shape and you've added random noise from a separate normal distribution. I don't mean to be so negative and I'll be more constructive in my next comment.
– JimB
Sep 11, 2015 at 1:29
• But N=3 does have the smallest AIC value. -446.6 < -443.5 < -442.7. I also wonder if you've restricted the values of c to be non-negative.
– JimB
Sep 11, 2015 at 5:22
• You may have a look at a similar question I answered at Mathematica Stack Exchange: bit.ly/1XW2Iob. Sep 12, 2015 at 7:54
• Just a small caution: when fitting with linear combinations of gaussian curves, the number of peaks can be less than the number of gaussian curves. There are a variety of automated methods to find peaks (although none are perfect in part because a peak depends on the scale that you define what a peak is). You might google for "bump hunting" and "peaks" in this forum and the related Cross Validation and Mathematica forums. An example with Mathematica is at mathematica.stackexchange.com/questions/23828/….
– JimB
Sep 13, 2015 at 18:09 | 840 | 3,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.934055 |
https://api-project-1022638073839.appspot.com/questions/what-is-the-average-speed-of-an-object-that-is-still-at-t-0-and-accelerates-at-a-34 | 1,586,123,909,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00178.warc.gz | 339,069,796 | 6,342 | # What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =5t+3 from t in [0, 2]?
##### 1 Answer
Apr 22, 2016
$\text{I found :} {v}_{a} = 8$
#### Explanation:
$\text{The average value of a function for an interval of (a,b) is given as;}$
${v}_{a} = \frac{1}{b - a} {\int}_{a}^{b} a \left(t\right) d t$
${v}_{a} = \frac{1}{2 - 0} {\int}_{0}^{2} \left(5 t + 3\right) d t$
${v}_{a} = \frac{1}{2} \left[| 5 {t}^{2} / 2 + 3 t {|}_{0}^{2}\right]$
${v}_{a} = \frac{1}{2} \left[\left(5 \cdot {2}^{2} / 2 + 3 \cdot 2\right) - \left(5 \cdot {0}^{2} / 2 + 3 \cdot 0\right)\right]$
${v}_{a} = \frac{1}{2} \left[\left(10 + 6\right) - 0\right]$
${v}_{a} = \frac{1}{2} \cdot 16$
${v}_{a} = 8$ | 337 | 725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | latest | en | 0.553293 |
https://community.tableau.com/thread/127634 | 1,555,778,255,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529898.48/warc/CC-MAIN-20190420160858-20190420182858-00535.warc.gz | 395,416,527 | 27,141 | 5 Replies Latest reply on Sep 16, 2013 1:34 AM by bart.groots
# Average
Hello,
I like to have a row with average instead of a row with Grand Total.
See picture & packaged workbook.
Thanks.
• ###### 1. Re: Average
Hi Bart,
A quick and easy way to do this would be to combine two views onto a single dashboard so they're next to each other.
There are other solutions but they're quite complex to implement -- there are some great ideas, courtesy of Mr Jonathan Drummey, here: Customizing Grand Totals – Part 1 | Drawing with Numbers
1 of 1 people found this helpful
• ###### 2. Re: Average
Robin Thanks,
Unfortunately, to hear that this is not easy to accomplish in Tableau but great idea to combine two views!
Regards,
Bart
• ###### 3. Re: Average
So I put Sales per Year in one dashboard. Next I made a custom calculation for average and placed it in a 2nd sheet.
(I used this calculation: sum([Sales])/COUNTD(YEAR([Order Date]) )
Then I made a dashboard and placed the 2 tabs in it. It is acceptable and quick:
The blue line will dissapear when not hovered over, but I have it there to show you how close you can get the sections using the tiled approach. This seems like the quickest way to get a Grand Average.
• ###### 4. Re: Average
Hi Bart,
Your Calculation1 looks good for getting the desired outcome for Aandeel. ketels warmte [Gj]. Place MONTH(Maand (copy)) on the level of detail shelf. Then, right click on Calculation1 on the measure values card and select Compute Using...>Maand (copy). Finally, right click on Calculation1 in the data box and select Default Properties>Number Format... and select Number (Custom) and change the number of decimal places to 3.
Create calculated fields similar for the other two calculations.
Hope this helps!
-Tracy
• ###### 5. Re: Average
Thanks Tracy,
I do understand the formule but i dont understand how i can combine the packaged workbook with the AVG in one worksheet?
Regards,
Bart | 474 | 1,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-18 | latest | en | 0.870552 |
https://www.myhelper.tk/rd-sharma-solution-class-12-mathematics-chapter-2-functions/ | 1,600,413,648,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00500.warc.gz | 1,340,083,349 | 78,146 | # RD Sharma Solution CLass 12 Mathematics Chapter 2 Functions
## Chapter 2 – Functions Exercise Ex. 2.1
Question 1
Give an example of function which is one-one but not onto.
Solution 1
Question 2
Give an example of a function which is not one-one but onto.
Solution 2
Question 3
Given an example of a function which is neither one-one nor onto.
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = sin2x + cos2x
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28
Question 29
If A = { 1, 2, 3}, show that a ono-one function f : A → A must be onto.
Solution 29
Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.
Hence, f has to be onto.
Question 30
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Solution 30
Suppose f is not one-one.
Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.
Also, the image of 3 under f can be only one element.
Therefore, the range set can have at most two elements of the co-domain {1, 2, 3}
i.e f is not an onto function, a contradiction.
Hence, f must be one-one.
Question 31
Solution 31
Question 32
Solution 32
Question 33
Solution 33
Question 34
Solution 34
Question 35
Solution 35
Question 36
Solution 36
Question 37
Solution 37
## Chapter 2 – Functions Exercise Ex. 2.2
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and
h(z) = sin z for all x, y, z ϵ N.
show tht ho (gof) = (hog) of.
Solution 15
Question 16
Given Examples of two functions f : N → N and g ; N → N such that gof is onto but f is not onto.
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
## Chapter 2 – Functions Exercise Ex. 2.3
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Let f be a real function given by
Find each of the following
fof
Solution 19
Question 20
Let f be a real function given by
Find each of the following
fofof
Solution 20
Question 21
Let f be a real function given by
Find each of the following
(f0f0f) (38)
Solution 21
Question 22
Let f be a real function given by
Find each of the following
f2Also, show that fof ≠ f2 .
Solution 22
Question 23
Solution 23
## Chapter 2 – Functions Exercise Ex. 2.5
Question 1
Solution 1
Thus, h is a bijection and is invertible.
Question 2
Solution 2
Question 3
Solution 3
Question 4
Consider and defined as .
Show that are invertible. Find and show that
Solution 4
Question 5
Solution 5
f-1 = {(3, 1), (5, 2), (7, 3), (9, 4)}
f-1 0g-1 = {(7, 1), (23, 2), (47, 3), (79, 4)} ……. (A)
Now (A) & (B) we have (g0f)-1 = f-10g-1
Question 6
Show that the function defined by is invertible. Also find .
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Consider f : R+ → [-5,] given by f(x) = 9×2 + 6x – 5. Show that f is invertible with
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Let A = R – {3} and B = R – {1}. Consider the function defined by . Show that f is one-one and onto and hence find .
Solution 14
Question 15
Consider the function f ; R+
[-9, ∞]
given by
f(x) = 5x2
+ 6x – 9.
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
## Chapter 2 – Functions Exercise Ex. 2VSAQ
Question 1
Figure (a)
Figure (b)
Solution 1
Question 2
Figure (a)
Figure (b)
Solution 2
Question 3
If A = {1, 2, 3} B = {a, b}, write total number of functions from A to B.
Solution 3
A = {1, 2, 3} B = {a, b}
The total number of functions is 8
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28
Question 29
Solution 29
Question 30
Solution 30
Question 31
Solution 31
Question 32
Solution 32
Question 33
Solution 33
Question 34
Solution 34
Question 35
Solution 35
Question 36
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B.
State the whether f is one – one or onto
Solution 36
It is given that
A = {1, 2, 3}, B = {4, 5, 6, 7} and f = {(1,4), (2,5), (3,6)}
The function f is one-one from A to B
## Chapter 2 – Functions Exercise MCQ
Question 1
Let A = {x ε R : – 1 ≤ x ≤ 1} = B and C = {x ε R : x ≥ 0} and let S = {(x, y) ε A × B : x2 + y2 = 1} and s0 = {(x, y) ε A × C : x2 + y2 = 1}. Then
(a) S defines a function from A to B
(b) S0 defines a function from A To C
(c) S0 defines a function from A to B
(d) S defines a function from A to C
Solution 1
Question 2
(a) injective
(b) surjective
(c) bijective
(d) none of these
Solution 2
Question 3
If f : A → B given by 3f(x) + 2-x = 4 is a bijection, then
(a) A = {x ε R : – 1 < x < ∞], B = {x ε R ; 2 < x < 4]
(b) A = {x ε R : – 3 < x < ∞}, B = {x ε R ; 0 < x < 4}
(c) A = {x ε R : – 2 < x ε R : 0 < x < 4}
(d) none of these
Solution 3
Question 4
The function f : R → R defined by f(x) = 2x + 2|x| is
(a) one-one and onto
(b) many-one and onto
(c) one-one and into
(d) many-one and into
Solution 4
Question 5
(a) f is one-one but not onto
(b) f is onto but one-one
(c) f is both one-one and into
(d) none of these
Solution 5
Question 6
The function f : A → B defined by f(x) = -x2 + 6x – 8 is a bijection, if
(a) A = ( -∞, 3] and B = ( -∞, 1]
(b) A = [-3, ∞) and B = (-∞, 1]
(c) A = ( -∞, 3] and B = [1, ∞)
(d) A = [3, ∞) and B = [1, ∞)
Solution 6
Question 7
Let A = {x ε R : – 1 ≤ x ≤ 1} = B. then, the mapping f : A → B given by f(x) = x|x| is
(a) injective but not surjective
(b) surjective but not injective
(c) bijective
(d) none of these
Solution 7
Question 8
Let f : R → R be given by f(x) = [x]2 + [x + 1] – 3, where [x] denotes the greatest integer less than or equal to x. Then, f(x) is
(a) many-one and onto
(b) Many-one and into
(c) one-one and into
(d) one-one and onto
Solution 8
Question 9
let m be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then the function f : M → R defined by f(A) = |A| for every A ε M, is
(a) one-one and onto
(b) neither one-one nor onto
(c) one-one but-not onto
(d) onto but not one-one
Solution 9
Question 10
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) neither one-one nor onto
Solution 10
Question 11
The range of the function f (x) = 7-xPx-3 is
(a) {1, 2, 3, 4, 5}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3, 4}
(d) {1, 2, 3}
Solution 11
Question 12
A function f from the set of natural numbers to integers defined by
is
(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto both
Solution 12
Question 13
Let f be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false.
f(x) = 1, f(y) ≠ 1, f(z) ≠ 2.
The value of f-1 (1) is
(a) x
(b) y
(c) z
(d) none of these
Solution 13
Question 14
Which of the following functions from Z to itself are bijections?
(a) f(x) = x3
(b) f(x) = x + 2
(c) f(x) = 2x + 1
(d) f(x) = x2 + x
Solution 14
Question 15
Which of the following from A = {x : -1 ≤ x ≤ 1} to itself are bijections?
Solution 15
Question 16
Let A = {x : -1 ≤ x ≤ 1} and f; A → A such that f(x) = x|x|, then f is
(a) a bijection
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective
Solution 16
Question 17
(a) R
(b) [0, 1]
(c) (0, 1]
(d) [0, 1)
Solution 17
Question 18
if a function f : [2, ∞) → B defined by f(x) = x2 – 4x + 5 is a bijection, then B =
(a) R
(b) [1, ∞)
(c) [4, ∞)
(d) [5, ∞)
Solution 18
Question 19
The function f : R → R defined by f(x) = (x – 1)(x – 2)(x – 3) is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto
Solution 19
Question 20
(a) bijection
(b) injection but not a surjection
(c) surjection but not an injection
(d) neither an injection nor a surjection
Solution 20
Question 21
(a) f is a bijection
(b) f is an injection only
(c) f is surjection on only
(d) f is neither an injection nor a surjection
Solution 21
Question 22
(a) f is one-one onto
(b) f is one-one into
(c) f is many one onto
(d) f is many one into
Solution 22
Question 23
(a) one-one but not onto
(b) one-one and onto
(c) onto but not one-one
(d) neither one-one nor onto
Solution 23
Question 24
(a) one-one but not onto
(b) many-one but onto
(c) one-one and onto
(d) neither one-one nor onto
Solution 24
Question 25
The function f ; R → R, f(x) = x2 is
(a) injective but not surjective
(b) surjective but not injective
(c) injective as well as surjective
(d) neither injective nor surjective
Solution 25
Question 26
(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto both
Solution 26
Question 27
Which of the following functions from A = {x ε R : – 1 ≤ x ≤ 1} to itself are bijections?
Solution 27
Question 28
(a) onto but not one-one
(b) one-one but not onto
(c) one-one and onto
(d) neither one-one nor onto
Solution 28
Question 29
The function f ; R → R defined by f (x) = 6x + 6|x| is
(a) one-one and onto
(b) many one and onto
(c) one-one and into
(d) many one and into
Solution 29
Question 30
Let f(x) = x2 and g(x) = 2x. Then the solution set of the equation fog (x) = gof (x) is
(a) R
(b) {0}
(c) {0, 2}
(d) none of these
Solution 30
Question 31
if f : R → R is given by f(x) = 3x – 5, then f-1(x)
Solution 31
Question 32
Solution 32
Question 33
Solution 33
Question 34
Let A = {x ε R : x ≥ 1}. The inverse of the function f ; A → A given by f(x) = 2x(x – 1), is
Solution 34
Question 35
Let A = {x ε R : x ≤ 1} and f : A → 1 A be defined as f (x) = x(2 – x). Then, f-1 (x) is
Solution 35
Question 36
Solution 36
Question 37
If the function f: R→R be such that f(x) = x – [x], where [x] denotes the greatest integer less than or equal to x, then f-1(x) is
Solution 37
Question 38
Solution 38
Question 39
(a) x
(b) 1
(c) f(x)
(d) g(x)
Solution 39
Question 40
Solution 40
Question 41
The distinct linear functions which map [-1, 1] onto [0, 2] are
(a) f(x) = x + 1, g(x) = -x + 1
(b) f(x) = x – 1, g(x) = x + 1
(c) f(x) = – x – 1, g(x) = x – 1
(d) none of these
Solution 41
Question 42
Let f: [2, ∞) → X be defined by f(x) = 4x – x2. Then, f is invertible, if X=
(a) [2, ∞)
(b) ( -∞, 2]
(c) (-∞, 4]
(d) [4, ∞)
Solution 42
Question 43
Solution 43
Question 44
Solution 44
Question 45
(a) 2x – 3
(b) 2x + 3
(c) 2x2 + 3x + 1
(d) 2x2 – 3x – 1
Solution 45
Question 46
Solution 46
Question 47
Solution 47
Question 48
Let f(x) = x3 be a function with domain {0, 1, 2, 3}. Then domain of f-1 is
(a) {3, 2, 1, 0}
(b) {0, -1, -2, -3}
(c) {0, 1, 8, 27}
(d) {0, -1, -8, -27}
Solution 48
Question 49
Let f : R → R be given by f(x) = x2 – 3. Then domain of f-1 is
Solution 49
Question 50
Solution 50
Question 51
Solution 51
Question 52
Let A = {1, 2, …, n} and B = {a, b}. Then the number of subjections from A into B is
Solution 52
Question 53
If the set A contains 5 elements and the set B 6 elements, then the number of one-one and onto mappings from A to B is
(a) 720
(b) 120
(c) 0
(d) none of these
Solution 53
Question 54
If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is
Solution 54
Question 55
Solution 55
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1. ### Finding the number of arrangements
Which problem are u addressing : 1. 3 Reds and 3 Blues 2. 6 red and 3 blues If the FIRST one then here is my question to u : How can two red balls touch each other for it is given that the blue balls must be apart from each other . If any red ball was to touch another red ball, the two of...
2. ### Finding the number of arrangements
*amazed* You are right. So how would we proceed in this case , do u have any leads ?
3. ### Finding the number of arrangements
Lets put our method to test, what if the number of red balls and blue balls were not equal . There are 5 red balls (distinguishable) and 3 blue balls (distinguishable) , find the number of possible permutations in which all blue balls are not next to each other . I would speculate : for the...
4. ### Finding the number of arrangements
Homework Statement In how many ways can 3 red and 3 blue balls be arranged if no two balls of the same colour are next to each other ? Homework Equations .... The Attempt at a Solution Here is what i did (Which is wrong) 6 x 3 x 2 x 2 x 1 x 1 I did this since there are 6...
5. ### Electrolysis of Iron(II)Sulphate
It does indeed but that is something different. In this particular example we have : Fe(II) ions , water molecules, extremely low number of Hydrogen ions and hydroxide ions, Sulphate ions . Since the concentration of Hydrogen ions is almost negligible (as is the case in any aqueous...
7. ### Electrolysis of Iron(II)Sulphate
Homework Statement I am given a 1 M solution of Iron(II)sulphate(aq) and it is being electrolyzed using inert electrodes . What will form on the cathode ? Homework Equations Reduction potentials of : Fe =-0.44V of water to from hydrogen : -0.83V The Attempt at a Solution Since...
8. ### Moment of Inertia - Swinging arms
No, it is τ = Iα where α is the angular acceleration
9. ### Simple Harmonic motion : Is energy conserved
It is a needle which is being moved by an automated sewing machine I really can't see why do you need that information to answer the question : Why isn't Energy conserved in this problem ? In ANY case, the total energy at each and every point (displacement) during the SHM must be the same
10. ### Simple Harmonic motion : Is energy conserved
I thought i had uploaded it, sorry for that. Anyways, it is under gravity and it is going SHM. (why wouldn't it)
11. ### Simple Harmonic motion : Is energy conserved
Homework Statement I doubt energy is conserved in SHM, or it might be possible that i be doing something wrong. The particle (red dot) in the attachment is at its equilibrium position and oscillates with Simple Harmonic Motion between the two yellow colored plates. Amplitude A = 1.5 m...
12. ### The concept of rotational equilibrium
I cant see the logic in your answer . I think that even if all three forces were passing through the Center of mass, there MIGHT still be a net torque ! Let me prove it : https://www.physicsforums.com/attachment.php?attachmentid=57701 Consider this diagram in which the three forces are...
13. ### The concept of rotational equilibrium
This book was recommended as a great book for introductory physics by ppl here on PF Physics for Scientists & Engineers by Serway. As mentioned in the first post, there was this quiz on page 349~350 which asked whether the given object is in equilibrium, i stumbled across it while reading...
Not at all
15. ### The concept of rotational equilibrium
Here is a text from my Physics Book : The net external torque on the object about any axis must be zero for it to be in rotational equilibrium. I divide the torques into two categories, anticlockwise and clockwise. (This approach works fine for 2-D objects but will it work for 3-D objects...
16. ### Kinetic Energy Formula definition explanation
Work done on an object and one of the equation of motion for constant acceleration : V^2 - U^2 = 2AD (A= acceleration vector)(D= Displacement vector)(the rest is self explanatory) W=F.D
17. ### Approximating a Simple Harmonic motion
Understandable, my presentation is confusing, that is why you misunderstood my approach. The variable "r" has a value of "a" in this situation. (both of them are just lengths and have dimensions of length) Thanks for the explaining about the mathematical model for the signs . At least now...
18. ### Approximating a Simple Harmonic motion
No there is no typing error, i just rechecked the solution archive, it is exactly as i wrote it ! I really can't say anything about its direction since i have not defined them yet. Lets do that ! For the tension Let the clockwise tangent be -ve and the tangent which points in the anti...
19. ### Approximating a Simple Harmonic motion
Homework Statement Homework Equations F=ma The Attempt at a Solution I did the first three parts . The last part of this question is quite hard, i tried using newton's 2nd law of motion but ... here is what happens : T is the tension, as stated in the question . so the equation of...
20. ### The positioning of the thermometer
I find that plausible, but would you care to share some experimental results to back this hypothesis ? If i understand your claim correctly: (correct me if i misunderstood) The temperature at the surface is the true boiling temperature of a solution while that deep inside the liquid is...
21. ### The positioning of the thermometer
Simple, since the air just above the surface will be most saturated with the vapors (water) which would exert a pressure along with the atmospheric pressure. Thus the total pressure would just be the sum of the two pressures . (first by vapors and the second by atmosphere). Let me convince...
22. ### The positioning of the thermometer
While the solution is boiling, the pressure at the surface of the liquid will be greater than the atmospheric pressure (14.7 PSI). The pressure at the bottom of the surface will be the sum of the pressure due to the hieight of the liquid and the vapour pressure above the liquid. Will that...
23. ### The positioning of the thermometer
A thermometer doesn't "measure" pressure so i am guessing the key point over here might be that the thermometer ( which is in fact a glass thermometer) might break ? But if that were true, why the need for partial immersion ?
24. ### The positioning of the thermometer
Hello, I am instructed by my chemistry professor to design an experiment to measure the temperature at which a saturated solution of Sodium Nitrate will begin to boil. If the reagents are pure this temperature will be constant . Can someone tell me why does the lab instructor suggests...
25. ### Finding the moment of inertia
Homework Statement Two uniform square laminas are combined into a single body. One lamina ABCD has mass 5m and the other lamina PQRS has mass m. The lamina PQRS has side 2a, and its vertices are at the mid-points of the sides of ABCD, with P on AB and S on AD. The line PS meets AC at K, and...
26. ### Equilibrium, Torque and Friction (hard)
Ah, i see, i misunderstood the question (yet again) . CB = 1.2sin(30) = 0.6
27. ### Equilibrium, Torque and Friction (hard)
I am not quite sure what u mean by that statement. Aren't these the distances from the line of action of the force (extrapolated) and the pivot : Shown in Green colour in the diagram .
28. ### Equilibrium, Torque and Friction (hard)
That's simple : The perpendicular distance from the line of action of the forces to the Pivot D (i am taking moments about D)
29. ### Equilibrium, Torque and Friction (hard)
I still can't seem to solve this question : Here is what i did but failed : Taking moments about D : since the reaction force acts through D thus it produces no moment about D. Weight produced an anti clockwise moment about D and Normal force at (which i need to find out) , F produced a...
30. ### Equilibrium, Torque and Friction (hard)
Nothing really ! | 1,828 | 7,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-04 | latest | en | 0.916014 |
https://kidsworksheetfun.com/fifth-grade-place-value-worksheets-grade-5-pdf/ | 1,701,704,532,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00479.warc.gz | 409,635,541 | 25,336 | Your students will love these fill in matching and other creative worksheets that effortlessly teach key place value concepts from reading and writing numbers in standard and expanded form to ordering and comparing. Use these freeplace value worksheets 5th.
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Math Worksheets Place Value 3rd Grade Place Value Worksheets Math Worksheets Free Math Worksheets | 1,020 | 5,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-50 | latest | en | 0.861119 |
https://byjus.com/question-answer/f-43-to-a-f-x-x-2-2-is-a-function-then-which-of-1/ | 1,638,483,130,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00445.warc.gz | 206,213,573 | 27,343 | Question
# f:[−4,3)→A,f(x)=x2+2 is a function. Then which of the following represent's set A [2,18][2,11)[2,18)[2,11]
Solution
## The correct option is A [2,18]f(x)=x2+2 in x∈[−4,3) ∵x2+2≥2 Now, x2+2=2 at x=0∈[−4,3) and f(−4)=18,f(−3)=11 ∴f(x)∈[2,18], ∀ x∈[−4,3)
Suggest corrections | 151 | 287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.556421 |
http://profstewartmath.com/trigweb/equ_P1/1_examples.htm | 1,696,365,598,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00749.warc.gz | 33,246,302 | 5,788 | SOLVING SIMPLE TRIGONOMETRIC EQUATIONS
Example 1:
Solve for x in the interval . Express the solutions in EXACT radians.
Step 1:
The interval in which we are supposed to find the solutions is . We will now isolate the trigonometric ratio by dividing both sides by 2 to get
We notice that solution angles must be in QI and QII.
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .
Step 3:
Given the solution in Step 2, we will find its reference angle. Since angle 45o is a QI angle, its reference angle equals 45o.
Step 4:
The value of the sine ratio in Step 1 is positive and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the sine ratio is positive. It is best to draw a picture!
We need to find solution angles located in QI and QII with a reference angle of 45o. We find the solutions for angle x in the interval to be as follows.
Solution in QI:
x1 = 45o
Solution in QII:
x2 = 180o 45o = 135o
Since we are supposed to express the solutions in terms of radians we get
x1 = /4
and x2 = 3/4
Example 2:
Solve for x in the interval . Express the solutions in EXACT radians.
Step 1:
The interval in which we are supposed to find the solutions is . We will now isolate the trigonometric ratio by dividing both sides by 2 to get
We notice that solution angles must be in QIII and QIV.
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .
Step 3:
Given the solution in Step 2, we will find its reference angle. We know that the reference angle of a negative angle is equal to that of its positive counterpart.
Since 45o is a QI angle, its reference angle equals 45o. Therefore, the reference angle of 45o is 45o as well.
Step 4:
The value of the sine ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the sine ratio is negative. It is best to draw a picture!
We need to find solution angles located in QIII and QIV with a reference angle of 45o. We find the solutions for angle x in the interval to be as follows.
Solution in QIII:
x1 = 180o + 45o = 225o
Solution in QIV:
x2 = 360o 45o = 315o
Since we are supposed to express the solutions in terms of radians we get
x1 = 5/4
and x2 = 7/4
Example 3:
Solve for x in the interval . Express the solutions in EXACT radians.
Step 1:
The interval in which we are supposed to find the solutions is . The trigonometric ratio is already isolated on one side of the equal sign.
We notice that solution angles must be in QI and QIII.
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .
Step 3:
Given the solution in Step 2, we will find its reference angle. Since angle 60o is a QI angle, its reference angle equals 60o.
Step 4:
The value of the tangent ratio in Step 1 is positive and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is positive. It is best to draw a picture!
We need to find solution angles located in QI and QIII with a reference angle of 60o. We find the solutions for angle x in the interval to be as follows.
Solution in QI:
x1 = 60o
Solution in QIII:
x2 = 180o + 60o = 240o
Since we are supposed to express the solutions in terms of radians we get
x1 = /3
and x2 = 4/3
Example 4:
We are going to solve again, but now we only want solutions for x in the interval . Express the solutions in EXACT radians.
Step 1:
The interval in which we are supposed to find the solutions is . The trigonometric ratio is already isolated on one side of the equal sign.
We notice that solution angles must be in QI and QIII.
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely [180o,180o).
Step 3:
Given the solution in Step 2, we will find its reference angle. Since angle 60o is a QI angle, its reference angle equals 60o.
Step 4:
The value of the tangent ratio in Step 1 is positive and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is positive. It is best to draw a picture!
We need to find solution angles located in QI and QIII with a reference angle of 60o.
Here Step 4 is a bit more complicated because we are asked to find solutions in the interval [180o,180o). Drawing a picture is always a good idea.
Since we determined that the solutions are in QI and QIII, let's draw the reference angles in these quadrants AND the solution interval [180o,180o).
We now find the solutions for angle x in the interval [180o, 180o] with the help of the picture above.
Solution in QI:
x1 = 60o
Solution in QIII:
x2 = 180o + 60o = 120o
Since we are supposed to express the solutions in terms of radians we get
x1 = /3
and x2 = 2/3
Example 5:
Solve cos x = 0.858 for x in the interval [265o, 90o] in degrees. Round all calculations to 2 decimal places.
Step 1:
The interval in which we are supposed to find the solutions is [265o, 90o]. The trigonometric ratio is already isolated on one side of the equal sign.
cos x = 0.858
We notice that solution angles must be in QII and QIII.
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
cos x = 0.858
x = cos-1 (0.858)
x 149.09o
Step 3:
Given the solution in Step 2, we will find its reference angle. Since 149.09o is a QII angle, its reference angle is 180o 149.09o = 30.91o.
Step 4:
The value of the cosine ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is negative. It is best to draw a picture!
We need to find solution angles located in QII and QIII with a reference angle of 30.91o.
Here Step 4 it a bit more complicated because we are asked to find solutions in the interval [265o, 90o]. Drawing a picture is always a good idea.
Since we determined that the solutions are in QII and QIII, let's draw the reference angles in these quadrants and the solution interval [265o, 90o].
We now find the solutions for angle x in the interval [265o, 90o] with the help of the picture above.
Solution in QII:
x1 = [ 30.91o + 180o] = 210.91o
Solution in QIII:
x2 = 180o + 30.91o = 149.09o
Example 6:
Solve for x in the interval . Express the solutions in EXACT radians.
Step 1:
The interval in which we are supposed to find the solutions is . We will isolate the trigonometric ratio by dividing both sides by 1 to get
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .
Step 3:
Given the solution in Step 2, we will find its reference angle.We know that the reference angle of a negative angle is equal to that of its positive counterpart.
Since 60o is a QI angle, its reference angle equals 60o. Therefore, the reference angle of 60o is 60o as well.
Step 4:
The value of the tangent ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is negative. It is best to draw a picture!
We need to find solution angles located in QII and QIV with a reference angle of 60o. We find the solutions for angle x in the interval to be as follows:
Solution in QII:
x1 = 180o 60o = 120o
Solution in QIV:
x2 = 360o 60o = 300o
Since we are supposed to express the solutions in terms of radians we get
x1 = 2/3
and x2 = 5/3
Example 7:
Solve sin x + 1 = 0 for x in the interval . Express the solutions in degrees.
Step 1:
The interval in which we are supposed to find the solutions is. We will isolate the trigonometric ratio by adding 1 to both sides to get
sin x = 1
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Step 3:
is a quadrantal angle, which has no reference angle!
Step 4:
Since we do not have a reference angle, we are going to show the following picture:
From the picture above, we find that sin x = 1 only at 270o on interval .
Therefore, we have one solution, namely x = 270o.
Please note that the angle found in Step 2 is a perfectly good solution to the given equation. However, it is not part of the solution set because it does not lie in the solution interval [0o, 360o)
Example 8:
Solve sin x 1 = 0 for x in the interval . Express the solutions in degrees.
Step 1:
The interval in which we are supposed to find the solutions is . We will isolate the trigonometric ratio by adding 1 to both sides to get
sin x = 1
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Step 3:
is a quadrantal angle, which has no reference angle!
Step 4:
Since we do not have a reference angle, we are going to show the following picture:
From the picture above, we find that sin x = 1 only at 90o on interval .
Therefore, we have one solution, namely .
However, we must find the solutions in the interval . We will simple add another 360o rotation to 90o to find a second solution, namely, .
Example 9:
Find ALL solutions for . Express the solutions in EXACT radians.
Step 1:
We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by 2 to get
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .
Step 3:
Given the solution in Step 2, we will find its reference angle. Since 150o is a QII angle, its reference angle is 180o 150o = 30o.
Step 4:
The value of the cosine ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is negative. It is best to draw a picture!
We need to find solution angles located in QII and QIII with a reference angle of 30o. We find the solutions for angle x in the interval to be as follows:
Solution in QII:
x1 = 180o 30o = 150o
Solution in QIII:
x2 = 180o + 30o = 210o
Since we are supposed to express the solutions in terms of radians we get
x1 = 5/6
and x2 = 7/6
We are asked to find ALL solutions expressed in radians. We see in Step 4 that the solutions are more than (or 180o) apart,
Therefore, we add 2k to these solutions, where k is any integer.
That is, x1 = 5/6 + 2k and x2 = 7/6 + 2k.
Example 10:
Find ALL solutions for 2cos x = 0. Express the solutions in degrees.
Step 1:
We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by 2 to get
cos x = 0
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
Step 3:
is a quadrantal angle, which has no reference angle!
Step 4:
Since we do not have a reference angle, we are going to show the following picture:
From the picture above, we find that cos x = 0 at 90o and 270oon interval .
Therefore, we have two solutions, namely x1 = 90o and x2 = 270o.
We are asked to find ALL solutions expressed in degrees. We see in Step 4 that the cosine ratio is 0 at intervals of 180o.
Therefore, to show all solutions, we only need to add 180ok to the smaller of the two solutions in the interval between 0o and 360o where k is defined as any integer.
That is, all solutions for angle x are 90o + 180ok.
Example 11:
Find ALL solutions for 3tan x = 0. Express the solutions in degrees.
Step 1:
We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by 3 to get
tan x = 0
Step 2:
We are going to use the concept of Inverse Trigonometric Functions to solve for the angle x using the calculator.
tan x = 0
x = tan-1 0
x = 0o
Step 3:
0o is a quadrantal angle, which has no reference angle!
Step 4:
Since we do not have a reference angle, we are going to show the following picture:
From the picture above, we find that tan x = 0 at 0o and 180o on interval . Please note that the solution of 360o is excluded from the solution interval!
Therefore, we have two solutions, namely x1 = 0o and x2 = 180o.
We are asked to find ALL solutions expressed in degrees. We see in Step 4 that the tangent ratio is 0 at intervals of 180o.
Therefore, to show all solutions, we only need to add 180ok to the smaller of the two solutions in the interval between 0o and 360o where k is defined as any integer.
That is, x = 0o + 180ok. | 3,599 | 13,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-40 | longest | en | 0.923473 |
https://www.tutorialspoint.com/cplusplus-program-to-swap-numbers-in-cyclic-order-using-call-by-reference | 1,713,353,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00832.warc.gz | 929,341,469 | 23,640 | # C++ Program to Swap Numbers in Cyclic Order Using Call by Reference
Three numbers can be swapped in cyclic order by passing them to a function cyclicSwapping() using call by reference. This function swaps the numbers in a cyclic fashion.
The program to swap numbers in cyclic order using call by reference is given as follows −
## Example
Live Demo
#include<iostream>
using namespace std;
void cyclicSwapping(int *x, int *y, int *z) {
int temp;
temp = *y;
*y = *x;
*x = *z;
*z = temp;
}
int main() {
int x, y, z;
cout << "Enter the values of 3 numbers: "<<endl;
cin >> x >> y >> z;
cout << "Number values before cyclic swapping..." << endl;
cout << "x = "<< x <<endl;
cout << "y = "<< y <<endl;
cout << "z = "<< z <<endl;
cyclicSwapping(&x, &y, &z);
cout << "Number values after cyclic swapping..." << endl;
cout << "x = "<< x <<endl;
cout << "y = "<< y <<endl;
cout << "z = "<< z <<endl;
return 0;
}
## Output
The output of the above program is as follows −
Enter the values of 3 numbers: 2 5 7
Number values before cyclic swapping...
x = 2
y = 5
z = 7
Number values after cyclic swapping...
x = 7
y = 2
z = 5
In the above program, the function cyclicSwapping() swaps the three numbers in cyclic order using call by reference. The function uses a variable temp to do so. The code snippet for this is as follows −
void cyclicSwapping(int *x, int *y, int *z) {
int temp;
temp = *y;
*y = *x;
*x = *z;
*z = temp;
}
In the function main(), the values of the 3 numbers are provided by the users. Then these values are displayed before swapping them. The function cyclicSwapping() is called to swap the numbers and then the values are displayed after swapping them. This is given below −
cout << "Enter the values of 3 numbers: "<<endl;
cin >> x >> y >> z;
cout << "Number values before cyclic swapping..." << endl;
cout << "x = "<< x <<endl;
cout << "y = "<< y <<endl;
cout << "z = "<< z <<endl;
cyclicSwapping(&x, &y, &z);
cout << "Number values after cyclic swapping..." << endl;
cout << "x = "<< x <<endl;
cout << "y = "<< y <<endl;
cout << "z = "<< z <<endl;
Updated on: 25-Jun-2020
903 Views | 596 | 2,116 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.733323 |
https://www.doubtnut.com/qa-hindi/109837226 | 1,642,820,018,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00702.warc.gz | 745,469,820 | 93,502 | # p और Q क्रमशः त्रिभुज ABC की भुजाओं AB और BC के मध्य बिंदु है तथा R रेखाखण्ड AP का मध्य बिंदु है|दर्शाइए की ar (PBQ) =ar (ARC)
उत्तर
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 7-3-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
चूँकि S,QC का मध्य बिंदु है <br> therefore " "ar (Delta ROS ) =ar (Delta RSC ) <br> rArr " "ar (Delta RQC) =2ar (Delta ROS) <br> " "=2((1)/(2) )(RS) (qM) <br> " " =(RS) (QM) <br> लेकिन " "RS =(1)/(2) (PQ+AC) <br> " "=(1)/(2) ((1)/(2) AC+AC) <br> " "=(3)/(4)AC <br> इसी प्रकार " "ar (Delta RQC) =((3)/(4) AC) (QM) <br> " "=((3)/(4) AC) (SN) " "[because QM=SN] <br> " "=(3)/(2) ar (Delta RAC) <br> " " (3)/(2) ar (Delta PBQ) <br> लेकिन " "ar (Delta ABC)=4ar (Delta BPQ) <br> अतः " "(Delta RQC) =(3)/(8) ar (Delta ABC) <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KCS_HIN_MAT_IX_C11_SLV_023_S01.png" width="80%"> | 435 | 965 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-05 | latest | en | 0.384729 |
https://morioh.com/p/ade073b84d77 | 1,643,046,253,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00169.warc.gz | 474,662,894 | 9,591 | 1637306902
# How List Comprehension Works in Python
## List Comprehension in Python
Lists are a helpful and frequently used feature in Python.
And list comprehension gives you a way to create lists while writing more elegant code that is easy to read.
In this beginner-friendly article, I'll give an overview of how list comprehension works in Python. I'll also show plenty of code examples along the way.
Let's get started!
## How to use a `for` loop to create a list in Python
One way to create a list in Python is by using a `for` loop.
For example, you can use the `range()` function to create a list of numbers ranging from 0 - 4.
``````#first create an empty list
my_list = []
#iterate over the numbers 0 - 4 using the range() function
#range(5) creates an iterable, starting from 0 up to (but not including) 5
#Use the .append() method to add the numbers 0 - 4 to my_list
for num in range(5):
my_list.append(num)
#print my_list
print(my_list)
#output
#[0, 1, 2, 3, 4]
``````
What if you already have a list of numbers, but want to create a new list with their squares?
You could again use a `for` loop, like so:
``````#initial list of numbers
numbers = [1,2,3,4,5,6]
#create a new,empty list to hold their squares
square_numbers = []
#iterate over initial list
#multiply each number by itself
#use .append() method, to add the square to the new list, square_numbers
for num in numbers:
square_numbers.append(num * num)
#print new list
print(square_numbers)
#output
#[1, 4, 9, 16, 25, 36]
``````
But there is a quicker and more succinct way to achieve the same results – by using list comprehension.
## What is list comprehension in Python? A syntax overview
When you're analyzing and working with lists in Python, you'll often have to manipulate, modify, or perform calculations on every single item in the list, all at once.
You may also need to create new lists from scratch, or create a new list based on the values of an already existing list.
List comprehension is a fast, short, and elegant way to create lists compared to other iterative methods, like `for` loops.
The general syntax for list comprehension looks like this:
``````new_list = [expression for variable in iterable]
``````
Let's break it down:
• List comprehensions start and end with opening and closing square brackets, `[]`.
• Then comes the `expression` or operation you'd like to perform and carry out on each value inside the current iterable. The results of these calculations enter the new list.
• The `expression` is followed by a `for` clause.
• `variable` is a temporary name you want to use for each item in the current list that is going through the iteration.
• The `in` keyword is used to loop over the iterable.
• `iterable` can be any Python object, such as a list, tuple, string and so on.
• From the iteration that was performed and the calculations that took place on each item during the iteration, new values were created which are saved to a variable, in this case `new_list`. The old list (or other object) will remain unchanged.
• There can be an optional `if` statement and additional `for` clause.
## How to use list comprehension in Python
Using the same example from earlier on, here is how you'd create a new list of numbers from 0 - 4 with the `range()` function in just one single line, using list comprehension:
``````new_list = [num for num in range(5)]
print(new_list)
#output
#[0, 1, 2, 3, 4]
``````
This has the same output as the `for` loop example, but with significantly less code!
Let's break it down:
• the iterable in this case is a sequence of numbers from 0 to 4, using `range(5)`. `range()` constructs a list of numbers.
• You use the `in` keyword to iterate over the numbers.
• The `num` following the `for` clause is a variable, a temporary name for each value in the iterable. So `num` would be equal to `0` in the first iteration, then `num` would be equal to `1` in the next iteration and so on, until it reached and equalled the number 4, where the iteration would stop.
• The `num` before the `for` clause is an expression for each item in the sequence.
• Finally, the new list (or other iterable) that is created gets stored in the variable `new_list`.
You can even perform mathematical operations on the items contained in the iterable and the result will be added to the new list:
``````new_list = [num * 2 for num in range(5)]
print(new_list)
#output
#[0, 2, 4, 6, 8]
``````
Here each number in `range(5)` will be multiplied by two and the new value will be stored in the variable `new_list`.
What if you had a pre-existing list where you wanted to manipulate and modify each item in it? This would be similar to the example from earlier on, where we created a list of squares.
Again, you can achieve that with just one line of code, using list comprehension:
``````#initial list
numbers = [1,2,3,4,5,6]
#new list
#num * num is the operation that takes place to create the squares
square_numbers = [num * num for num in numbers]
print(square_numbers)
#output
[1, 4, 9, 16, 25, 36]
``````
### How to use conditionals with list comprehension in Python
Optionally, you can use an `if` statement with a list comprehension.
The general syntax looks like this:
``````new_list = [expression for variable in iterable if condition == True]
``````
Conditionals act as a filter and add an extra check for additional precision and customisation when creating a new list.
This means that the value in the expression has to meet certain criteria and a certain condition you speficy, in order to go in the new list.
``````new_list = [num for num in range(50) if num % 2 == 0]
print(new_list)
#output
#[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48]
``````
In the example above, only the values where the condition `num % 2 == 0` is checked and evaluates to True will enter `new_list`.
The modulo operator is used on every single one of the numbers in the sequence of numbers starting from 0 and ending in 49.
If the remainder of the numbers when divided by 2 is 0, then and only then does it enter the list.
So in this case, it creates a list of only even numbers.
You can then make it as specific as you want.
For example, you could add more than one condition, like so:
``````new_list = [num for num in range(50) if num > 20 and num % 2 == 0]
print(new_list)
#output
#[22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48]
``````
In this example, there are two conditions `num > 20` and `num % 2 == 0`.
The `and` operator indicates that both have to be met in order for the value to be added to the new list.
The values that don't meet the conditions are excluded and are not added.
### How to use list comprehension on strings in Python
You can create a new list with the individual characters contained in a given string.
``````fave_language_chars = [letter for letter in "Python"]
print(fave_language_chars)
#output
#['P', 'y', 't', 'h', 'o', 'n']
``````
The new list that gets created is comprised of all the separate letters contained in the string "Python", which acts as an iterable.
Just like numbers, you can perform operations on the characters contained in a string and customize them depending on how you want them to be in the new list you create.
If you wanted all letters to be uppercase, you would do the following:
``````fave_language_chars_upper = [letter.upper() for letter in "Python"]
print(fave_language_chars_upper)
#output
#['P', 'Y', 'T', 'H', 'O', 'N']
``````
Here you use the `.upper()` method to convert every single letter in "Python" to uppercase and add them to the `fave_language_chars_upper` variable.
The same goes if you wanted all your letters to be lowercase - you'd instead use the `lower()` method.
## Conclusion
And there you have it! You now know the basics of list comprehension in Python.
It offers an elegant and concise syntax for creating new lists based on existing lists or other iterables.
Original article source at https://www.freecodecamp.org
#python
1619518440
## top 30 Python Tips and Tricks for Beginners
Welcome to my Blog , In this article, you are going to learn the top 10 python tips and tricks.
### 8) Check The Memory Usage Of An Object.
#python #python hacks tricks #python learning tips #python programming tricks #python tips #python tips and tricks #python tips and tricks advanced #python tips and tricks for beginners #python tips tricks and techniques #python tutorial #tips and tricks in python #tips to learn python #top 30 python tips and tricks for beginners
1619510796
## Lambda, Map, Filter functions in python
Welcome to my Blog, In this article, we will learn python lambda function, Map function, and filter function.
Lambda function in python: Lambda is a one line anonymous function and lambda takes any number of arguments but can only have one expression and python lambda syntax is
Syntax: x = lambda arguments : expression
Now i will show you some python lambda function examples:
#python #anonymous function python #filter function in python #lambda #lambda python 3 #map python #python filter #python filter lambda #python lambda #python lambda examples #python map
1624429860
## How to Convert Loops to List Comprehension in Python
### Do the same but faster
List comprehension is used for creating lists based on iterables. It can also be described as representing for and if loops with a simpler and more appealing syntax. List comprehensions are relatively faster than for loops.
The syntax of a list comprehension is actually easy to understand. However, when it comes to complex and nested operations, it might get a little tricky to figure out how to structure a list comprehension.
In such cases, writing the loop version first makes it easier to write the code for the list comprehension. We will go over several examples that demonstrate how to convert a loop-wise syntax to a list comprehension.
Basic structure of list comprehension (image by author)
Let’s start with a simple example. We have a list of 5 integers and want to create a list that contains the squares of each item. Following is the for loop that performs this operation.
``````lst_a = [1, 2, 3, 4, 5]
lst_b = []
for i in lst_a:
lst_b.append(i**2)
print(lst_b)
[1, 4, 9, 16, 25]
``````
#python #programming #how to convert loops to list comprehension in python #convert loops #list comprehension #how to convert loops to list comprehension
1626775355
## Why use Python for Software Development
No programming language is pretty much as diverse as Python. It enables building cutting edge applications effortlessly. Developers are as yet investigating the full capability of end-to-end Python development services in various areas.
By areas, we mean FinTech, HealthTech, InsureTech, Cybersecurity, and that's just the beginning. These are New Economy areas, and Python has the ability to serve every one of them. The vast majority of them require massive computational abilities. Python's code is dynamic and powerful - equipped for taking care of the heavy traffic and substantial algorithmic capacities.
Programming advancement is multidimensional today. Endeavor programming requires an intelligent application with AI and ML capacities. Shopper based applications require information examination to convey a superior client experience. Netflix, Trello, and Amazon are genuine instances of such applications. Python assists with building them effortlessly.
## 5 Reasons to Utilize Python for Programming Web Apps
Python can do such numerous things that developers can't discover enough reasons to admire it. Python application development isn't restricted to web and enterprise applications. It is exceptionally adaptable and superb for a wide range of uses.
Robust frameworks
Python is known for its tools and frameworks. There's a structure for everything. Django is helpful for building web applications, venture applications, logical applications, and mathematical processing. Flask is another web improvement framework with no conditions.
Web2Py, CherryPy, and Falcon offer incredible capabilities to customize Python development services. A large portion of them are open-source frameworks that allow quick turn of events.
Python has an improved sentence structure - one that is like the English language. New engineers for Python can undoubtedly understand where they stand in the development process. The simplicity of composing allows quick application building.
The motivation behind building Python, as said by its maker Guido Van Rossum, was to empower even beginner engineers to comprehend the programming language. The simple coding likewise permits developers to roll out speedy improvements without getting confused by pointless subtleties.
Utilized by the best
Alright - Python isn't simply one more programming language. It should have something, which is the reason the business giants use it. Furthermore, that too for different purposes. Developers at Google use Python to assemble framework organization systems, parallel information pusher, code audit, testing and QA, and substantially more. Netflix utilizes Python web development services for its recommendation algorithm and media player.
Massive community support
Python has a steadily developing community that offers enormous help. From amateurs to specialists, there's everybody. There are a lot of instructional exercises, documentation, and guides accessible for Python web development solutions.
Today, numerous universities start with Python, adding to the quantity of individuals in the community. Frequently, Python designers team up on various tasks and help each other with algorithmic, utilitarian, and application critical thinking.
Progressive applications
Python is the greatest supporter of data science, Machine Learning, and Artificial Intelligence at any enterprise software development company. Its utilization cases in cutting edge applications are the most compelling motivation for its prosperity. Python is the second most well known tool after R for data analytics.
The simplicity of getting sorted out, overseeing, and visualizing information through unique libraries makes it ideal for data based applications. TensorFlow for neural networks and OpenCV for computer vision are two of Python's most well known use cases for Machine learning applications.
### Summary
Thinking about the advances in programming and innovation, Python is a YES for an assorted scope of utilizations. Game development, web application development services, GUI advancement, ML and AI improvement, Enterprise and customer applications - every one of them uses Python to its full potential.
The disadvantages of Python web improvement arrangements are regularly disregarded by developers and organizations because of the advantages it gives. They focus on quality over speed and performance over blunders. That is the reason it's a good idea to utilize Python for building the applications of the future.
#python development services #python development company #python app development #python development #python in web development #python software development
1622279504
## List Comprehension
List comprehension is nothing but a shorter and crisper version of the code and also memory efficient. By using this we can either create a new list or perform some operation in an existing list.
The normal code for creating a list of 0–9 will be like
``````x=[]
for i in range (10):
x.append(i)
print(x)
[0,1,2,3,4,5,6,7,8,9]
``````
By using list comprehension
``````x=[i for i in range(10)]
print(x)
[0,1,2,3,4,5,6,7,8,9]
``````
As you can see the normal code is long but the code that we did using list comprehension does the job just in one line so list comprehension is preferred over the traditional method.
#list-comprehension #lists #python #python-list-comprehension | 3,639 | 15,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.809771 |
https://www.systutorials.com/docs/linux/man/l-zgerfs/ | 1,726,073,184,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00692.warc.gz | 945,453,352 | 4,141 | # zgerfs (l) - Linux Manuals
## NAME
ZGERFS - improves the computed solution to a system of linear equations and provides error bounds and backward error estimates for the solution
## SYNOPSIS
SUBROUTINE ZGERFS(
TRANS, N, NRHS, A, LDA, AF, LDAF, IPIV, B, LDB, X, LDX, FERR, BERR, WORK, RWORK, INFO )
CHARACTER TRANS
INTEGER INFO, LDA, LDAF, LDB, LDX, N, NRHS
INTEGER IPIV( * )
DOUBLE PRECISION BERR( * ), FERR( * ), RWORK( * )
COMPLEX*16 A( LDA, * ), AF( LDAF, * ), B( LDB, * ), WORK( * ), X( LDX, * )
## PURPOSE
ZGERFS improves the computed solution to a system of linear equations and provides error bounds and backward error estimates for the solution.
## ARGUMENTS
TRANS (input) CHARACTER*1
Specifies the form of the system of equations:
= aqNaq: A * X = B (No transpose)
= aqTaq: A**T * X = B (Transpose)
= aqCaq: A**H * X = B (Conjugate transpose)
N (input) INTEGER
The order of the matrix A. N >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of columns of the matrices B and X. NRHS >= 0.
A (input) COMPLEX*16 array, dimension (LDA,N)
The original N-by-N matrix A.
LDA (input) INTEGER
The leading dimension of the array A. LDA >= max(1,N).
AF (input) COMPLEX*16 array, dimension (LDAF,N)
The factors L and U from the factorization A = P*L*U as computed by ZGETRF.
LDAF (input) INTEGER
The leading dimension of the array AF. LDAF >= max(1,N).
IPIV (input) INTEGER array, dimension (N)
The pivot indices from ZGETRF; for 1<=i<=N, row i of the matrix was interchanged with row IPIV(i).
B (input) COMPLEX*16 array, dimension (LDB,NRHS)
The right hand side matrix B.
LDB (input) INTEGER
The leading dimension of the array B. LDB >= max(1,N).
X (input/output) COMPLEX*16 array, dimension (LDX,NRHS)
On entry, the solution matrix X, as computed by ZGETRS. On exit, the improved solution matrix X.
LDX (input) INTEGER
The leading dimension of the array X. LDX >= max(1,N).
FERR (output) DOUBLE PRECISION array, dimension (NRHS)
The estimated forward error bound for each solution vector X(j) (the j-th column of the solution matrix X). If XTRUE is the true solution corresponding to X(j), FERR(j) is an estimated upper bound for the magnitude of the largest element in (X(j) - XTRUE) divided by the magnitude of the largest element in X(j). The estimate is as reliable as the estimate for RCOND, and is almost always a slight overestimate of the true error.
BERR (output) DOUBLE PRECISION array, dimension (NRHS)
The componentwise relative backward error of each solution vector X(j) (i.e., the smallest relative change in any element of A or B that makes X(j) an exact solution).
WORK (workspace) COMPLEX*16 array, dimension (2*N)
RWORK (workspace) DOUBLE PRECISION array, dimension (N)
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
## PARAMETERS
ITMAX is the maximum number of steps of iterative refinement. | 845 | 2,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.659005 |
https://www.edaboard.com/showthread.php?380023-Difference-between-the-two-coding-styles-for-clock-gating | 1,550,712,093,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247497858.46/warc/CC-MAIN-20190221010932-20190221032932-00603.warc.gz | 823,151,523 | 19,121 | # Difference between the two coding styles for clock gating
1. ## Difference between the two coding styles for clock gating
Hi all,
I'm trying to clk gate a few registers for saving power & couldn't understand the following. Below is a snippet of what I've tried
Method 1 : where i explicitly coarse & fine gater
```Code Verilog - [expand]1
2
3
4
5
6
always @(posedge clk) begin
for (i =0; i < 255; i = i+1) begin
if (we[i >> 5] & coarse_en)
mem[i][0] <= mem_d [i][0]
end
end```
Method 2 : where it's implicit and gating
```Code Verilog - [expand]1
2
3
4
5
6
7
8
always @(posedge clk) begin
if (coarse_en) begin
for (i =0; i < 255; i = i+1) begin
if (we[i >> 5])
mem[i][0] <= mem_d [i][0]
end
end
end```
Method 2 shows power saving but method 1 doesn't. Why do I see the difference ?
~rahdirs
- - - Updated - - -
Originally Posted by rahdirs
Method 1 : where i explicitly coarse & fine gater
```Code Verilog - [expand]1
2
3
4
5
6
always @(posedge clk) begin
for (i =0; i < 255; i = i+1) begin
if (we[i >> 5] & coarse_en)
mem[i][0] <= mem_d [i][0]
end
end```
I think I've figured it out. It wasn't showing savings probably because I was using the bit-wise and (&) instead of the logical and (&&) ? Will try with this change
•
2. ## Re: Difference between the two coding styles for clock gating
Originally Posted by rahdirs
Hi all,
I'm trying to clk gate a few registers for saving power & couldn't understand the following. Below is a snippet of what I've tried
Method 1 : where i explicitly coarse & fine gater
```Code Verilog - [expand]1
2
3
4
5
6
always @(posedge clk) begin
for (i =0; i < 255; i = i+1) begin
if (we[i >> 5] & coarse_en)
mem[i][0] <= mem_d [i][0]
end
end```
Method 2 : where it's implicit and gating
```Code Verilog - [expand]1
2
3
4
5
6
7
8
always @(posedge clk) begin
if (coarse_en) begin
for (i =0; i < 255; i = i+1) begin
if (we[i >> 5])
mem[i][0] <= mem_d [i][0]
end
end
end```
Method 2 shows power saving but method 1 doesn't. Why do I see the difference ?
~rahdirs
- - - Updated - - -
I think I've figured it out. It wasn't showing savings probably because I was using the bit-wise and (&) instead of the logical and (&&) ? Will try with this change
I think you have stumbled upon the problem of shared enable detection. The coding style of method 2 might be easier for some tools to pick.
•
3. ## Re: Difference between the two coding styles for clock gating
Method 2 should use flip-flop enable, and thus spend less logic.
Method 1 probably will use more logic to implement the (we[i>>5] && coarse_en) part. In this case coarse_en should not be part of flip-flop enable, but part of LUT logic.
Also, maybe if you make your for loop increment by 16, instead increment by 1 and shifting, you might save some logic and increase speed.
--[[ ]]-- | 865 | 2,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-09 | latest | en | 0.853817 |
https://www.sfu.ca/~vjungic/SN-Guide/sec_SN02_Guide.html | 1,695,507,899,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00591.warc.gz | 1,108,652,581 | 9,814 | ## Section2.1Small Number and the Old Canoe - Classroom Guide
View the full story: Small Number and the Old Canoe
### Subsection2.1.1Mathematics
1. counting
2. patterns
3. shapes
4. mathematical thinking
5. inclusion-exclusion formula
### Subsection2.1.2Mathematical Vocabulary
• small, number, 5-year-old, down, circle, the farthest, surface, quickly, to go far, smooth, flat, oval shaped, wanders far along the shore, tall, falling headfirst, into, stands up, looks around, stand around, along its smooth shape, very old, very big, How many?, How many generations ago?, long time, run back, huge, at least a hundred years old, the fastest, two of his brothers, all (the sons of my grandfather), three old totem poles, in front of the longhouse, each of them, by one of (my uncles), just before, two, three, four, five or more.
### Subsection2.1.3Cultural Components
1. Indigenous:
2. Grandpa’s blanket: Maybe you’ve noticed that some Indigenous people wear blankets during various ceremonies. Learn more about traditional button blankets. You may start here: Button Blankets.
3. Canoe: Canoes have been one of the most important means of transportation along the Pacific Coast. Learn more about the traditional use of canoes and the ways they were built. You may wish to start with: the SFU Museum of Archeology (Waterways), the SFU Bill Read Centre (Northwest Coast Canoes), and the Canadian Encyclopedia (Birchbark Canoes).
5. Feast dish: A potlatch is a feast practiced by Indigenous peoples of the Pacific Northwest Coast. Traditionally, during a potlatch food was served in so-called “feast dishes”. Learn more about feast dishes. You may wish to start with the UBC Museum of Anthropology: MOA - Feast Dish.
2. General:
1. Have you ever tried skipping stones across the water? If yes, share your story. Do you think that the shape of the stone matters?
2. Have you ever been in a canoe or a kayak? If yes, share your story.
3. Have you ever made your grandparents worried by something that you did?
5. Have you ever encountered a mathematical question that had more than one possible answer?
### Subsection2.1.4Mathematical Observations (Video)
Opening scene: Notice the shape of the canoe and the shapes of the carvings on its side.
0:15 - Notice the shapes of Grandpa’s hat, the mountain peaks, and the trees. How long is Small Number’s stick?
0:24 - How many boys are there? Notice their reflections in the water.
0:38 - 0:41 - What is the shape of the stone? Notice the pattern made by the skipping stone. How many times did it skip? Notice the perspective. “For a stone to go far it needs to be smooth, flat, and oval shaped.”
1:00 - What is the shape of the canoe?
1:17 - “How many people do you think [the canoe] could hold?”
1:22 - “How many generations ago was [the canoe] built?”
1:43 - Notice the geometrical shapes, the angles, and the perspective. Which objects are parallel to each other? Perpendicular?
1:47 - “Grandpa is carving the surface of a huge wooden dish.” What is the shape of the dish? Notice the patterns and shapes in the carving.
2:05 - “I found an old canoe down on the beach! It must be at least a hundred years old!”
2:10 - 2:16 - “I know that canoe. It was once the fastest canoe in our village. It was built by my father and two of his brothers.” How many bothers were building the canoe? What is the shape of the tools the brothers were using?
2:21 - “All the sons of my grandfather were known as great wood carvers.”
2:23 - 2:29 “You know those three old totem poles in front of the longhouse? Each of them was built by one of my uncles.” Notice the shapes and heights of the totem poles.
2:37 - What is the shape of the log? The diameter of the log? The circumference of the log?
2:42 - Notice the shapes and the perspective.
### Subsection2.1.5Answer: Why did Small Number think that his great-grandpa might have two, three, four, five or more brothers?
Grandpa: “I know that canoe. It was once the fastest canoe in our village. It was built by my father and two of his brothers.”
Small Number thought:
“So, my great-grandpa had at least two brothers!” (See Figure 2.1.1)
Grandpa: “You know those three old totem poles in front of the longhouse? Each of them was built by one of my uncles.” (See Figure 2.1.2)
Small Number continued:
“Maybe the totem poles were built by the same two brothers that built the canoe together with my great-grandpa? Maybe my great-grandpa didn’t have any other brothers? So therefore, maybe my great-grandpa had only two brothers?” (See Figure 2.1.3)
“Maybe there was another brother? Maybe each totem pole was built by a different brother? It could be that my great-grandpa had three brothers.” (See Figure 2.1.4)
“Maybe my great-grandpa had four brothers? Maybe each totem pole was built by a different brother, but only one was built by one of Grandpa’s uncle who built the canoe?” (See Figure 2.1.5)
Small Number was getting sleepy:
“But maybe each totem pole was built by a different brother and none of those brothers worked on the canoe? It could be that my great-grandpa had five brothers. Maybe there were other brothers? I have to ask Grandpa tomorrow how many brothers his father had. Two, three, four, five or more…”
### Subsection2.1.6Discussion/Activities
Prepare six cards: on three cards draw the same image of a canoe. On each of the remaining cards draw a (different) totem pole.
Continue by asking: “Who would like to be Small Number’s great-grandpa?” (Hand over one of the “canoe” cards to the volunteer, Alice.)
Next: “Now I need two of you to be Great-Grandpa’s brothers who helped him build the canoe.” (Hand over the other two “canoe” cards to Bob and Carol.)
To the class: “Let us talk about the phrase at least two. For example, I watched each Small Number film at least two times. How do you understand that? How many times did I watch each Small Number film?”
[Moderate the conversation. Make sure that students understand that “at least two” means “two or more”.]
To the class: “Do you agree that Small Number’s great-grandpa had at least two brothers?”
To the class: “Do you remember when Grandpa said: ‘You know those three old totem poles in front of the longhouse? Each of them was built by one of my uncles.’
Show the three cards with the totem poles to the class: “If Small Number’s great-grandpa had only two brothers, what should I do with these three cards?”
[Moderate the conversation. If needed lead the class to the conclusion that one of the already established brothers (Bob and Carol, those with a “canoe” card) should get two “totem” cards. Hand over one card to Bob and two to Carol.]
Conclude: “It is possible that Small Number’s great-grandpa had exactly two brothers?” [The class should answer “Yes!”]
Take one totem card from Carol and give it to Dave. “Now, Bob is a brother who built a canoe and one of the totem poles, Carol is a brother who also built a canoe and one of the totem poles, and Dave is a brother who built the third totem pole. How many brothers are there now?” [The class should answer “Three!”]
To the class: “Is it possible that Great-Grandpa had four brothers?” [This should lead to the conclusion that Bob’s totem card should be given to Erin.]
To the class: “What about five brothers?” [This should lead to the conclusion that Carol’s totem card should be given to Frank.]
To the class: “What about more than five brothers?” [This should lead to the conclusion that maybe there were other brothers, those who didn’t build either the canoe or one of the totem poles.]
1. Start by reminding students how Venn diagrams are used to represent sets.
You may wish to discuss that in mathematics we use the word set as a synonym for a “collection of objects”. You may also wish to remind (or introduce to) the class some basic terminology related to sets.
For example, consider the following sets:
$A$ = {students who were born in June} = {Alice, Bob, Carol}
$B$ = {students who were born in July} = {Dave, Erin, Frank}
$C$ = {students with an older brother} = {Alice, Erin, Frank}
$D$ = {students with a younger sister} = {Dave, Frank}
$E$ = {students who play soccer} = {Alice, Erin, Frank}
These five sets can be represented by Venn diagrams. (See Figure 2.1.8)
Continue by observing:
1. Sets $A$ and $B$ do not have any elements in common. [If you wish you may introduce the terminology, “$A$ and $B$ are two disjoint sets”.]
2. Sets $A$ and $C$ have only one element in common: Alice. See Figure 2.1.9. [If you wish you may introduce the phrase “Alice is the only element in the intersection of $A$ and $C\text{.}$” You may introduce the notation $A$ $\cap$ $C\text{.}$]
3. Observe that sets $B$ and $C$ and sets $B$ and $D$ have exactly two elements in common. See Figure 2.1.10.
Observe the difference between these two cases: All elements of the set $D$ belong to the set $B\text{.}$ [If you wish you may introduce the phrase “$D$ is a subset of $B$”. Also, you may introduce the appropriate notation $D$ $\subset$ $B\text{.}$]
4. Observe that sets $C$ and $E$ have all of their elements in common. We say that “sets $C$ and $E$ are equal”.
2. Watch the movie.
Repeat the question from the end of the movie: “Why did Small Number think that his great-grandpa might have two, three, four, five or more brothers?” Make sure that everyone understands the question. Then invite students to share their thoughts.
Say: “What if we try to use Venn diagrams to help Small Number?”
Continue: “What are the sets of interest that may help in finding the answer?” [Allow students to suggest “sets of interest”. You should end up with the following three sets. Students will probably need your guidance.]:
$B$ = {all of Great-Grandpa’s brothers}
$C$ = {Great-Grandpa’s brothers who built the canoe}
$T$ = {Great-Grandpa’s brothers who built the totem poles}
Start by noticing that sets $C$ and $T$ are subsets of the set $B\text{.}$
Ask: “Do we know how many elements the set $C$ has?” [The answer should be “two”.]
Next, ask: “Do we know how many elements the set $T$ has?” [The answer should be: “We don’t know”.]
Remind students about Grandpa’s words: “You know those three old totem poles in front of the longhouse? Each of them was built by one of my uncles.” Observe that Grandpa used the word “uncles” and ask: “Do you agree that there was more than one uncle involved in building the totem poles?” [Students may agree with this or they may not. It does not matter. Go with the majority opinion.]
Ask the class: “How do you understand Grandpa’s words: ‘Each of them was built by one of my uncles’? Is it possible that one brother built two totem poles?” [Again, let students share their thinking. Lead students to the conclusion that there were at most three elements in set $T\text{.}$]
Your strategy should be to discuss the following five cases:
1. Sets $C$ and $B$ are equal. In this case the set $T$ also must be equal to the set $C\text{.}$ See Figure 2.1.11.
2. Set $C$ is a subset of the set $T\text{,}$ but they are not equal. See Figure 2.1.12.
3. The set $T$ has three elements and sets $C$ and $T$ have one element in common. Make sure that you warn students about avoiding double-counting. Figure 2.1.13.
4. The set $T$ has three elements and sets $C$ and $T$ have no elements in common. Figure 2.1.14.
5. The set $T$ has three elements and sets $C$ and $T$ have no elements in common, but there is another brother. Figure 2.1.15.
• You may wish to discuss the following:
Ask students to pay attention to mathematics (words, geometrical shapes, patterns, numbers). Ask students to pay attention to the cultural, particularly Indigenous, aspects of the story. Ask students to make their own game that includes double counting.
### Subsection2.1.7Challenge
Suppose there are 30 students, 10 who have a dog and 15 who have a cat. If there are 5 students that have both a dog and a cat, how many students have neither a dog nor a cat as a pet? | 3,031 | 11,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-40 | latest | en | 0.935396 |
http://www.reference.com/browse/callippic+cycle | 1,409,168,168,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829754.11/warc/CC-MAIN-20140820021349-00382-ip-10-180-136-8.ec2.internal.warc.gz | 577,411,172 | 23,764 | Definitions
Nearby Words
# Callippic cycle
[kuh-lip-ik]
In astronomy and calendar studies, the Callippic cycle (or Calippic) is a particular approximate common multiple of the year (specifically the tropical year) and the synodic month, that was proposed by Callippus in 330 BC. It is a period of 76 years, as an improvement on the 19-year Metonic cycle.
A century before Callippus, Meton invented the cycle of 19 years that counted 6,940 days, which exceeds 235 lunations by almost a third of a day, and 19 tropical years by four tenths of a day. It implicitly gave the solar year a length of 6940/19 = 365 + 1/4 + 1/76 days = 365 d 6 h 18 min 56 s. But Callippus knew that the length of the year was more closely 365 + 1/4 day (= 365d 6h 00m 00s), so he multiplied the 19-year cycle by 4 to reach an integer number of days, and then dropped 1 day from the last 19-year cycle. Thus he constructed a cycle of 76 years that contains 940 lunations and 27,759 days, and has been called the Callippic after him.
Although the cycle's error has been calculated as one full day in 553 years, or 4.95 parts per million., in actuality 27,759 days in 76 years has a mean year of exactly 365 + 1/4 days, which relative to the mean northward equinoctial year is about 11 minutes too long per year, in other words the cycle drifts another day late per 130 + 10/11 years, which is considerably worse than the drift of the unrounded Metonic cycle. If the Callippic cycle is considered as closer to its unrounded length of 27,758 + 3/4 days (based on 940 lunations) then its accuracy is essentially the same as the unrounded Metonic cycle (within a few seconds per year). If it is taken as 940 lunations less one day then the Callippic mean year will be shortened by 1/76 day (18 minutes 57 seconds), making it grossly too short, and it will also grossly drift ahead with respect to the mean lunar cycle at the rate of 1/940 of a day (1 minute 31 seconds) per lunar month. If the cycle length is truncated to 27,758 days then the mean year is 365 days 5 hours 41 minutes 3 seconds, or almost 8 minutes too short per year, and it will drift ahead of the mean lunar cycle by about (3/4)/940 day (1 minute 9 seconds) per lunar month. Altogether, the purported accuracy of this cycle is not impressive, but it is of historical interest.
The first year of the first Callippic cycle began at the summer solstice of 330 BC (28 June in the proleptic Julian calendar), and was subsequently used by later astronomers. In Ptolemy's Almagest, for example, he cites (Almagest VII 3, H25) observations by Timocharis in the 47th year of the first Callippic cycle (283 BC), when on the eighth of Anthesterion, the Pleiades were occulted by the Moon.
The Callippic calendar originally used the names of months from the Attic calendar, although later astronomers, such as Hipparchus, preferred other calendars, including the Egyptian calendar. Also Hipparchus invented his own Hipparchic calendar cycle as an improvement upon the Callippic cycle. Ptolemy's Almagest provided some conversions between the Callippic and Egyptian calendars, such as that Anthesterion 8, 47th year of the first Callippic period was equivalent to day 29 in the month of Athyr, in year 465 of Nabonassar. However, the original, complete form of the Callippic calendar is no longer known. | 869 | 3,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2014-35 | longest | en | 0.951349 |
https://blog.herbert.top/problemset/subarray-product-less-than-k/readme_en/ | 1,726,177,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00217.warc.gz | 119,439,543 | 4,127 | | English | 简体中文 |
# 713. Subarray Product Less Than K
## Description
Given an array of integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than k.
Example 1:
Input: nums = [10,5,2,6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are:
[10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6]
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.
Example 2:
Input: nums = [1,2,3], k = 0
Output: 0
Constraints:
• 1 <= nums.length <= 3 * 104
• 1 <= nums[i] <= 1000
• 0 <= k <= 106 | 227 | 648 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.708502 |
https://www.litscape.com/word_analysis/duopoly | 1,537,796,220,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160454.88/warc/CC-MAIN-20180924125835-20180924150235-00550.warc.gz | 769,283,767 | 8,478 | # duopoly in Scrabble®
The word duopoly is playable in Scrabble®, no blanks required.
DUOPOLY
(102 = 52 + 50)
duopoly
DUOPOLY
(102 = 52 + 50)
DUOPOLY
(101 = 51 + 50)
DUOPOLY
(98 = 48 + 50)
DUOPOLY
(98 = 48 + 50)
DUOPOLY
(95 = 45 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(89 = 39 + 50)
DUOPOLY
(86 = 36 + 50)
DUOPOLY
(84 = 34 + 50)
DUOPOLY
(84 = 34 + 50)
DUOPOLY
(82 = 32 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(73 = 23 + 50)
DUOPOLY
(70 = 20 + 50)
DUOPOLY
(70 = 20 + 50)
DUOPOLY
(69 = 19 + 50)
DUOPOLY
(69 = 19 + 50)
DUOPOLY
(68 = 18 + 50)
DUOPOLY
(67 = 17 + 50)
DUOPOLY
(67 = 17 + 50)
DUOPOLY
(67 = 17 + 50)
DUOPOLY
(66 = 16 + 50)
DUOPOLY
(66 = 16 + 50)
DUOPOLY
(65 = 15 + 50)
DUOPOLY
(65 = 15 + 50)
DUOPOLY
(102 = 52 + 50)
DUOPOLY
(101 = 51 + 50)
DUOPOLY
(98 = 48 + 50)
DUOPOLY
(98 = 48 + 50)
DUOPOLY
(95 = 45 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(92 = 42 + 50)
DUOPOLY
(89 = 39 + 50)
DUOPOLY
(86 = 36 + 50)
DUOPOLY
(84 = 34 + 50)
DUOPOLY
(84 = 34 + 50)
DUOPOLY
(82 = 32 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(80 = 30 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(78 = 28 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(76 = 26 + 50)
DUOPOLY
(73 = 23 + 50)
DUOPOLY
(70 = 20 + 50)
DUOPOLY
(70 = 20 + 50)
DUOPOLY
(69 = 19 + 50)
DUOPOLY
(69 = 19 + 50)
DUOPOLY
(68 = 18 + 50)
DUOPOLY
(67 = 17 + 50)
DUOPOLY
(67 = 17 + 50)
DUOPOLY
(67 = 17 + 50)
DUOPOLY
(66 = 16 + 50)
DUOPOLY
(66 = 16 + 50)
DUOPOLY
(65 = 15 + 50)
DUOPOLY
(65 = 15 + 50)
LOOPY
(42)
PLOY
(39)
LOOPY
(39)
DULY
(36)
LOOPY
(36)
PLOY
(36)
LOOPY
(33)
LOOPY
(33)
DULY
(30)
PLOD
(30)
LOOPY
(30)
LOOPY
(30)
LOOPY
(30)
LOOPY
(28)
LOOPY
(28)
PLOY
(27)
PLOY
(27)
POOL
(27)
POLO
(27)
PLOD
(27)
PLOY
(27)
LOOP
(27)
PLOY
(27)
PLOY
(26)
YUP
(24)
PLY
(24)
PLOY
(24)
LOOPY
(24)
DULY
(24)
DULY
(24)
PLY
(24)
DULY
(24)
PLY
(24)
YUP
(24)
DULY
(24)
DULY
(24)
YUP
(24)
LOOPY
(22)
LOOPY
(22)
PLOD
(21)
PLOD
(21)
YOD
(21)
PLOD
(21)
PLOD
(21)
POLO
(21)
LOUD
(21)
POOL
(21)
LOOP
(21)
YOD
(21)
YOD
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LOOPY
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LOOPY
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LOOPY
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DULY
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LOOPY
(20)
PLOD
(20)
LOOPY
(20)
LOOPY
(20)
LOOP
(18)
LOOP
(18)
POLO
(18)
LOUD
(18)
POD
(18)
POLO
(18)
POD
(18)
POD
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POOL
(18)
YOU
(18)
POLO
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POLO
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YOU
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POLO
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YOU
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PLOY
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LOOP
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LOOP
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PLOY
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PLOY
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LOOP
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POOL
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PLOD
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PLOY
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POOL
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POOL
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POOL
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PLOY
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YUP
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PLY
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YUP
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YUP
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DULY
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DULY
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DULY
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PLY
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PLY
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DULY
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YUP
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PLY
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LOOPY
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DULY
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LOP
(15)
LOUD
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YUP
(15)
LOP
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LOUD
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PLOY
(15)
LOUD
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PLY
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LOUD
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LOP
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LOOPY
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YOD
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LOOPY
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YOU
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LOOPY
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PLOY
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YOD
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PLOD
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POOL
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PLOD
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PLOD
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PLOD
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YUP
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LOUD
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PLY
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YOD
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LOOP
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YOD
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POLO
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LOOPY
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PLOY
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YOD
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PLOD
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DULY
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PLOY
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POOL
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OLD
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PLY
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POLO
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LOOPY
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OLD
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LOOP
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LOOP
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DUO
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LOOP
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DULY
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PLOY
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POLO
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POOL
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POOL
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YUP
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POLO
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UP
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UP
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LOOP
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LOOPY
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OLD
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LOUD
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# duopoly in Words With Friends™
The word duopoly is playable in Words With Friends™, no blanks required.
DUOPOLY
(116 = 81 + 35)
DUOPOLY
(116 = 81 + 35)
duopoly
DUOPOLY
(116 = 81 + 35)
DUOPOLY
(116 = 81 + 35)
DUOPOLY
(104 = 69 + 35)
DUOPOLY
(104 = 69 + 35)
DUOPOLY
(104 = 69 + 35)
DUOPOLY
(98 = 63 + 35)
DUOPOLY
(98 = 63 + 35)
DUOPOLY
(95 = 60 + 35)
DUOPOLY
(95 = 60 + 35)
DUOPOLY
(95 = 60 + 35)
DUOPOLY
(92 = 57 + 35)
DUOPOLY
(92 = 57 + 35)
DUOPOLY
(92 = 57 + 35)
DUOPOLY
(77 = 42 + 35)
DUOPOLY
(73 = 38 + 35)
DUOPOLY
(73 = 38 + 35)
DUOPOLY
(73 = 38 + 35)
DUOPOLY
(71 = 36 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(67 = 32 + 35)
DUOPOLY
(67 = 32 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(58 = 23 + 35)
DUOPOLY
(58 = 23 + 35)
DUOPOLY
(58 = 23 + 35)
DUOPOLY
(57 = 22 + 35)
DUOPOLY
(57 = 22 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(55 = 20 + 35)
DUOPOLY
(55 = 20 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(53 = 18 + 35)
DUOPOLY
(53 = 18 + 35)
DUOPOLY
(53 = 18 + 35)
DUOPOLY
(52 = 17 + 35)
DUOPOLY
(52 = 17 + 35)
DUOPOLY
(52 = 17 + 35)
DUOPOLY
(51 = 16 + 35)
DUOPOLY
(51 = 16 + 35)
DUOPOLY
(50 = 15 + 35)
DUOPOLY
(116 = 81 + 35)
DUOPOLY
(116 = 81 + 35)
DUOPOLY
(104 = 69 + 35)
DUOPOLY
(104 = 69 + 35)
DUOPOLY
(104 = 69 + 35)
DUOPOLY
(98 = 63 + 35)
DUOPOLY
(98 = 63 + 35)
DUOPOLY
(95 = 60 + 35)
DUOPOLY
(95 = 60 + 35)
DUOPOLY
(95 = 60 + 35)
DUOPOLY
(92 = 57 + 35)
DUOPOLY
(92 = 57 + 35)
DUOPOLY
(92 = 57 + 35)
DUOPOLY
(77 = 42 + 35)
DUOPOLY
(73 = 38 + 35)
DUOPOLY
(73 = 38 + 35)
DUOPOLY
(73 = 38 + 35)
DUOPOLY
(71 = 36 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(69 = 34 + 35)
DUOPOLY
(67 = 32 + 35)
DUOPOLY
(67 = 32 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(65 = 30 + 35)
DUOPOLY
(58 = 23 + 35)
DUOPOLY
(58 = 23 + 35)
DUOPOLY
(58 = 23 + 35)
DUOPOLY
(57 = 22 + 35)
LOOPY
(57)
DUOPOLY
(57 = 22 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(56 = 21 + 35)
DUOPOLY
(55 = 20 + 35)
DUOPOLY
(55 = 20 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(54 = 19 + 35)
DUOPOLY
(54 = 19 + 35)
PLOY
(54)
DUOPOLY
(53 = 18 + 35)
DUOPOLY
(53 = 18 + 35)
DUOPOLY
(53 = 18 + 35)
DUOPOLY
(52 = 17 + 35)
DUOPOLY
(52 = 17 + 35)
DUOPOLY
(52 = 17 + 35)
DUOPOLY
(51 = 16 + 35)
DUOPOLY
(51 = 16 + 35)
LOOPY
(51)
PLOD
(51)
DUOPOLY
(50 = 15 + 35)
PLOY
(48)
POOL
(48)
LOOP
(48)
POLO
(48)
LOOPY
(45)
DULY
(45)
LOOPY
(44)
PLOD
(39)
LOOPY
(39)
DULY
(39)
POOL
(36)
LOOP
(36)
LOOPY
(34)
LOOPY
(33)
LOOPY
(33)
LOOPY
(33)
LOUD
(33)
LOUD
(33)
POLO
(30)
LOOPY
(30)
PLOY
(30)
PLOY
(30)
PLOY
(30)
PLOY
(30)
LOOPY
(30)
PLOY
(28)
LOOPY
(28)
YUP
(27)
PLOD
(27)
DULY
(27)
PLOD
(27)
YUP
(27)
PLOD
(27)
YUP
(27)
DULY
(27)
DULY
(27)
PLY
(27)
PLY
(27)
PLY
(27)
DULY
(27)
PLOD
(27)
PLOY
(26)
LOOPY
(26)
PLOD
(26)
POLO
(24)
LOOP
(24)
LOOP
(24)
LOOP
(24)
LOOP
(24)
POLO
(24)
LOOP
(24)
LOOPY
(24)
POLO
(24)
POLO
(24)
POLO
(24)
POOL
(24)
POOL
(24)
POOL
(24)
POOL
(24)
DULY
(24)
POOL
(24)
YUP
(23)
PLY
(23)
PLOD
(22)
LOOPY
(22)
LOOPY
(22)
DULY
(22)
LOOPY
(22)
LOOPY
(22)
LOOPY
(22)
LOUD
(21)
LOP
(21)
LOP
(21)
LOUD
(21)
LOOPY
(21)
POD
(21)
LOUD
(21)
LOP
(21)
LOUD
(21)
POD
(21)
LOOPY
(21)
POD
(21)
PLOY
(20)
PLOY
(20)
POLO
(20)
POOL
(20)
PLOY
(20)
LOOP
(20)
PLOY
(20)
PLOY
(20)
PLOY
(20)
LOP
(19)
LOOPY
(19)
POD
(19)
DULY
(19)
PLOD
(19)
LOOPY
(19)
LOOPY
(19)
YOD
(18)
YUP
(18)
DULY
(18)
PLOD
(18)
YOU
(18)
PLOD
(18)
DULY
(18)
PLY
(18)
YUP
(18)
PLOD
(18)
YOU
(18)
PLY
(18)
DULY
(18)
LOUD
(18)
LOUD
(18)
YOU
(18)
DULY
(18)
PLOY
(18)
YUP
(18)
PLY
(18)
PLOD
(18)
LOOP
(18)
POLO
(18)
POOL
(18)
UP
(18)
LOOPY
(18)
YOD
(18)
UP
(18)
YOD
(18)
PLOD
(17)
PLY
(17)
LOOPY
(17)
YUP
(17)
LOOPY
(17)
PLOY
(17)
PLOD
(17)
DULY
(17)
POLO
(16)
POOL
(16)
LOOP
(16)
# Word Growth involving duopoly
duo
## Longer words containing duopoly
(No longer words found) | 4,651 | 7,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-39 | latest | en | 0.483992 |
http://openbooks.library.umass.edu/toggerson-131/chapter/acceleration/ | 1,721,490,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00510.warc.gz | 22,507,554 | 30,259 | # 25 Acceleration
OpenStax and Brokk Toggerson
In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive.
Key Takeaways – Acceleration
Acceleration is the rate at which velocity changes,
where is acceleration, is velocity, and is time. Note that both and have arrows over them which indicates that these quantities have both a magnitude and a direction.
Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are meters per second squared or meters per second per second, which literally means by how many meters per second the velocity changes every second.
Examples – Units of Acceleration
Problem:
The imperial unit of acceleration is which is said as either
• feet per second-squared
• feet per second per second
Interpret an acceleration of .
Solution:
An acceleration of means that the speed increases by each second. If I start at rest, this means that:
time [s] Speed [ft/s] 0 0 1 32 2 64 3 96
# Touchstone Values of Acceleration
It is often useful to have some values two which you can compare your accelerations. Does the answer you get make sense? In order to do this, you need to have some values of acceleration in your head. We expect you to have memorized these values.
Situation Acceleration [m/s2] Maximum acceleration of an average gasoline powered car. 4.5 Maximum acceleration of the Space-X Dragon capsule on launch. 44 Acceleration at which most untrained humans begin to experience dizziness and fainting. 49 The maximum acceleration a fit and trained human can endure for any duration without losing consciousness. 88 Maximum acceleration of a jumping flea 3200 Typical acceleration of a hit baseball. 29 460
# Acceleration as a Vector
Recall that velocity is a vector—it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.
Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion. When an object slows down, its acceleration is opposite to the direction of its motion.
Negative acceleration, does NOT always imply a reduction in speed: negative acceleration is acceleration in the negative direction in the chosen coordinate system. Negative acceleration may or may not be slowing down and slowing down may or may not be considered negative acceleration. In general, if the acceleration and the velocity have opposite signs, the object is slowing down.
Key Takeaways – Sign of Acceleration
The sign of acceleration is a bit tricky for many students. A negative acceleration does not mean slowing down. To determine if an object is speeding up or slowing down, you need to look at the directions (signs in the 1-D case) of both velocity and acceleration.
• If the two have the same sign: the object is speeding up.
• If the two have opposite signs: the object is slowing down.
Situation Sign of Sign of Moving in the positive direction and speeding up. + (Moving in the positive direction.) + (Must be the same as the sign of v as we are speeding up.) Moving in the positive direction and slowing down. + (Moving in the positive direction.) – (Must be opposite the sign of v as we are slowing down.) Moving in the negative direction and speeding up. – (Moving in the negative direction.) – (Must be the same as the sign of v as we are speeding up.) Moving in the negative direction and slowing down. – (Moving in the negative direction.) + (Must be in the opposite direction of v as we are slowing down.)
Example 1: A Car
Problem:
Consider the figure below:
For each sub-figure, is the car speeding up or slowing down?
Solution:
(a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system.
(b) This car is slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the left. The car is also slowing down: the direction of its acceleration is opposite to its direction of motion.
(c) This car is moving toward the left, but slowing down over time. Therefore, its acceleration is positive in our coordinate system because it is toward the right. However, the car is slowing down because its acceleration is opposite to its motion.
(d) This car is speeding up as it moves toward the left. It has negative acceleration because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is speeding up.
Example 2: Calculating Acceleration as a Racehorse Leaves the Gate
Problem:
A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?
Solution:
Strategy
First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.
We can solve this problem by identifying and from the given information and then calculating the average acceleration directly from the equation .
Solution
1. Identify the knowns. , (the negative sign indicates direction toward the west), .
2. Find the change in velocity. Since the horse is going from zero to , its change in velocity equals its final velocity: .
3. Plug in the known values ( and ) and solve for the unknown .
.
Discussion
The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of due west means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as . This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.
The next several examples consider the motion of the subway train shown below. In (a) the train moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.
Example 3: Calculating Displacement: A Subway Train
Problem:
What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of the subway figure above?
Strategy
A drawing with a coordinate system is already provided, so we don’t need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation . This is straightforward since the initial and final positions are given.
Solution
1. Identify the knowns. In the figure we see that and for part (a), and and for part (b).
2. Solve for displacement in part (a).
3. Solve for displacement in part (b).
Discussion
The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.
Example 4: Calculating Acceleration: A Subway Train Speeding Up
Problem:
Suppose the train in the figure above accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?
Solution:
It is worth it at this point to make a simple sketch:
This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.
1. Identify the knowns. (the trains starts at rest), , and .
2. Calculate . Since the train starts from rest, its change in velocity is , where the plus sign means velocity to the right.
3. Plug in known values and solve for the unknown, .
4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Chapter 1.2 Physical Quantities and Units for more guidance.)
Discussion
The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.
Example 5: Calculate Acceleration: A Subway Train Slowing Down
Problem:
Now suppose that at the end of its trip, the train in Figure 7(a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?
Strategy
In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.
Solution
1. Identify the knowns. , (the train is stopped, so its velocity is 0), and .
2. Solve for the change in velocity, .
3. Plug in the knowns, and , and solve for .
4. Convert the units to meters and seconds.
.
The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.
### PHET EXPLORATIONS: MOVING MAN SIMULATION
Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you.
# Section Summary
• Acceleration is the rate at which velocity changes. In symbols, \textbf{ m/s}^2 \textbf{a}\$ is the acceleration at a specific instant in time.
• Slowing down is when acceleration has a direction opposite to that of the velocity.
Homework problems
15. Touchstone values for acceleration.
16. Intercontinental ballistic missiles.
17. Slowing down at the end of a race.
18. Catching a ball.
19. Backing out of a garage.
20. Signs of position velocity and acceleration.
21. Units of position velocity and acceleration. | 2,311 | 10,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.938257 |
https://www.physicsforums.com/threads/integral-of-cos-and-ln.164720/ | 1,540,267,698,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00430.warc.gz | 1,047,715,653 | 14,412 | # Homework Help: Integral of cos and ln
1. Apr 8, 2007
### sagita
Please, how can I solve this?
∫ cos x ln x dx
I get this:
ln x sin x - ∫sin x/x dx
but how do I continue from here?
2. Apr 8, 2007
### Mystic998
The antiderivative of sin(x)/x isn't expressible in terms of elementary functions so perhaps it would be better to change the role of u and dv in your integration by parts.
Edit: At least, I think that's the case. Couldn't hurt to try anyway.
3. Apr 8, 2007
### Hurkyl
Staff Emeritus
That won't change anything -- one cannot be expressed in terms of elementary functions iff the other cannot be expressed as well.
4. Apr 8, 2007
### fizzzzzzzzzzzy
look at it as an equation, and you need to integrate by parts at least twice
5. Apr 9, 2007
### dfx
No, this won't help. Even Wolfram gives an answer with Si(x) in it - the integral of Sinx/x.
6. Apr 9, 2007
### Vagrant
So, can't sinx/x be integrated?
7. Apr 9, 2007
### HallsofIvy
Yes, of course it can- its integral is Si(x)! It cannot, however, be integrated in terms of elementary functions.
8. Apr 9, 2007
### sagita
It seems difficult to continue from ln x sin x - ∫sin x/x dx ...
Thanks to averyone who posted. I'll tell you if something different appears.
Thanks again.
9. Apr 9, 2007
### Data
It is impossible to continue without introducing "special" functions or series expansions (from which you won't be able to obtain closed forms). So, play with it for a while, but don't spend too much time on it .
10. Apr 10, 2007
### Gib Z
If you really don't want Si(x), heres your only alternative:
$$\frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!}$$.
Integrate that, and there you go.
11. Apr 10, 2007
### dimensionless
And don't forget this one:
$$\frac{sin(x)}{x} = sinc(x)$$ | 546 | 1,797 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-43 | latest | en | 0.938142 |
https://nl.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/1484588 | 1,603,990,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904834.82/warc/CC-MAIN-20201029154446-20201029184446-00068.warc.gz | 450,742,244 | 16,916 | Cody
# Problem 8. Add two numbers
Solution 1484588
Submitted on 9 Apr 2018 by yubari
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 188 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-45 | latest | en | 0.708671 |
https://fred.stlouisfed.org/graph/?id=COMPHAI | 1,490,893,754,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218195419.89/warc/CC-MAIN-20170322212955-00141-ip-10-233-31-227.ec2.internal.warc.gz | 793,871,331 | 21,847 | # Housing Affordability Index (Composite)© (COMPHAI) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
Jan 2017: 162.0
Updated: Mar 17, 2017
Units:
Index,
Frequency:
Monthly
1Y | 5Y | 10Y | Max
EDIT LINE 1
Copyright, 2016, National Association of Realtors. Reprinted with permission.
Measures the degree to which a typical family can afford the monthly mortgage payments on a typical home.
Value of 100 means that a family with the median income has exactly enough income to qualify for a mortgage on a median-priced home. An index above 100 signifies that family earning the median income has more than enough income to qualify for a mortgage loan on a median-priced home, assuming a 20 percent down payment. For example, a composite housing affordability index (COMPHAI) of 120.0 means a family earning the median family income has 120% of the income necessary to qualify for a conventional loan covering 80 percent of a median-priced existing single-family home. An increase in the COMPHAI then shows that this family is more able to afford the median priced home.
This index is calculated as composite of the fixed and adjustable rate mortgages.
Select a date that will equal 100 for your custom index:
to
#### Customize data:
Write a custom formula to transform one or more series or combine two or more series.
You can begin by adding a series to combine with your existing series.
Now create a custom formula to combine or transform the series.
Need help? []
Finally, you can change the units of your new series.
Select a date that will equal 100 for your custom index:
FORMAT GRAPH
Log scale:
NOTES
#### Notes:
Copyright, 2016, National Association of Realtors. Reprinted with permission.
Measures the degree to which a typical family can afford the monthly mortgage payments on a typical home.
Value of 100 means that a family with the median income has exactly enough income to qualify for a mortgage on a median-priced home. An index above 100 signifies that family earning the median income has more than enough income to qualify for a mortgage loan on a median-priced home, assuming a 20 percent down payment. For example, a composite housing affordability index (COMPHAI) of 120.0 means a family earning the median family income has 120% of the income necessary to qualify for a conventional loan covering 80 percent of a median-priced existing single-family home. An increase in the COMPHAI then shows that this family is more able to afford the median priced home.
This index is calculated as composite of the fixed and adjustable rate mortgages.
#### Suggested Citation:
National Association of Realtors, Housing Affordability Index (Composite)© [COMPHAI], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/COMPHAI, March 30, 2017.
RELATED CONTENT
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Retrieving data.
Updating graph. | 655 | 2,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-13 | longest | en | 0.925457 |
http://mathhelpforum.com/number-theory/139357-jacobi-symbol-properties.html | 1,481,410,677,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00064-ip-10-31-129-80.ec2.internal.warc.gz | 174,005,660 | 10,420 | # Thread: Jacobi Symbol Properties
1. ## Jacobi Symbol Properties
This is a theorem about Jacobi symbols in my textbook:
Let n and m be ODD and positive. Then (a/nm)=(a/n)(a/m) and (ab/n)=(a/n)(b/n)
Moreover,
(i) If gcd(a,n)=1, then ( $a^2/n$) = 1 = ( $a/n^2$)
(ii) If gcd(ab,nm)=1, then ( $ab^2/nm^2$)=(a/n)
=====================================
(i) is easy and follows from the definition, but how can we prove (ii)? My textbook stated the theorem without proof and just says the proofs are easy, but I have no idea why (ii) is true.
Any help is appreciated!
2. Originally Posted by kingwinner
This is a theorem about Jacobi symbols in my textbook:
Let n and m be ODD and positive. Then (a/nm)=(a/n)(a/m) and (ab/n)=(a/n)(b/n)
Moreover,
(i) If gcd(a,n)=1, then ( $a^2/n$) = 1 = ( $a/n^2$)
(ii) If gcd(ab,nm)=1, then ( $ab^2/nm^2$)=(a/n)
=====================================
(i) is easy and follows from the definition, but how can we prove (ii)? My textbook stated the theorem without proof and just says the proofs are easy, but I have no idea why (ii) is true.
Any help is appreciated!
Think it is because...
$\bigg{(}\frac{ab^2}{nm^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\bigg{(}\frac{b^2}{n m^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\cdot 1$
$= \bigg{(}\frac{a}{n}\bigg{)}\bigg{(}\frac{a}{m^2}\b igg{)} = \bigg{(}\frac{a}{n}\bigg{)}\cdot 1 = \bigg{(}\frac{a}{n}\bigg{)}$
Was it supposed to be $\bigg{(}\frac{(ab)^2}{(nm)^2}\bigg{)}$? That was a bit unclear
3. Originally Posted by Deadstar
Think it is because...
$\bigg{(}\frac{ab^2}{nm^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\bigg{(}\frac{b^2}{n m^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\cdot 1$
$= \bigg{(}\frac{a}{n}\bigg{)}\bigg{(}\frac{a}{m^2}\b igg{)} = \bigg{(}\frac{a}{n}\bigg{)}\cdot 1 = \bigg{(}\frac{a}{n}\bigg{)}$
Why (b^2/nm^2)=1? How do you know that gcd(b,nm^2)=1?
Why (a/m^2)=1? How do you know that gcd(a,m)=1?
Was it supposed to be $\bigg{(}\frac{(ab)^2}{(nm)^2}\bigg{)}$?
No.
Thanks for explaining!
4. Originally Posted by kingwinner
Why (b^2/nm^2)=1? How do you know that gcd(b,nm^2)=1?
Why (a/m^2)=1? How do you know that gcd(a,m)=1?
Since $\left(\frac{xy}{r}\right) = \left(\frac{x}{r}\right)\left(\frac{y}{r}\right)$, we have that $\left(\frac{b^2}{nm^2}\right) = \left(\frac{b}{nm^2}\right)\left(\frac{b}{nm^2}\ri ght) = \left(\frac{b}{nm^2}\right)^2 =1$
The other case follows similarly. | 927 | 2,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-50 | longest | en | 0.783504 |
https://www.intmath.com/functions-and-graphs/what-is-projection-vector-in-geometry.php | 1,723,182,491,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00040.warc.gz | 637,602,879 | 22,999 | Search IntMath
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# What is a Projection Vector in Geometry?
Have you ever taken a course in geometry and heard your teacher mention something called “projection vector”? If so, you may be wondering what they are talking about. In this blog post, we will explain what projection vectors are and how they are used in geometry.
Projection vectors are a type of vector that can be used to represent the direction and magnitude of a point on a line or plane. A vector is an object that has both magnitude (or length) and direction. It can be represented by an arrow pointing from its starting point to its end point.
When it comes to projection vectors, the length of the vector represents the distance between two points on a line or plane, while the direction of the vector indicates which point is closer to the origin. For example, if there are two points with coordinates (x1,y1) and (x2,y2), then the projection vector would have coordinates (x2-x1, y2-y1). The first coordinate indicates how far along the x-axis we have traveled; while the second coordinate indicates how far along the y-axis we have traveled. Projection vectors can also be used to represent angles between lines or planes. When two lines or planes intersect at an angle, their projection vectors will form an angle as well. This angle can then be measured using basic trigonometric functions such as sine and cosine. For example, if two projection vectors form an angle of 45 degrees when drawn together on a graph paper, then their dot product will equal 1/sqrt(2).
## Conclusion:
In conclusion, projection vectors are useful for representing distances and angles between points on lines or planes in geometry. These vectors can help us calculate lengths between points as well as measure angles formed when two lines or planes intersect each other at different angles. Understanding these concepts can help students excel in their geometry classes!
## FAQ
### What is projection in geometry?
Projection in geometry is a vector that can be used to represent the direction and magnitude of a point on a line or plane. It has both magnitude (or length) and direction, which are determined by the coordinates of the two points being compared. The first coordinate indicates how far along the x-axis we have traveled; while the second indicates how far along the y-axis we have traveled. These vectors can also be used to represent angles between lines or planes.
### What is the formula of projection vector?
The formula for projection vector is (x2-x1, y2-y1), where x1 and y1 represent the coordinates of one point, and x2 and y2 represent the coordinates of another point. The first coordinate indicates how far along the x-axis we have traveled; while the second coordinate indicates how far along the y-axis we have traveled. This formula can also be used to calculate the angle formed when two projection vectors intersect each other by using basic trigonometric functions such as sine and cosine.
### What is scalar and vector projection?
Scalar projection is when a vector is projected onto a single point, such as when measuring the distance between two points on a line or plane. Vector projection is when a vector is projected onto another vector, such as when calculating the angle formed when two lines or planes intersect each other at different angles. Both scalar and vector projections can be useful in geometry, as they can help us measure distances and angles between points.
### What is the projection of a vector onto a plane?
The projection of a vector onto a plane is the vector that results when the original vector is projected onto the plane. This can be calculated by taking the dot product of the two vectors and then dividing it by the magnitude of the plane's normal vector, which is perpendicular to the plane. The result will be a new vector whose direction and magnitude indicate the projection of the original vector onto the plane. This can be used to measure angles between planes or lines and calculate distances between points. | 798 | 4,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-33 | latest | en | 0.936246 |
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# utkarsh singhvi
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nästan 2 år ago | 0 | 460 | 1,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-16 | latest | en | 0.593546 |
https://www.forex.academy/understanding-the-trading-costs-involved-in-usd-try-exotic-currency-pair/ | 1,582,554,692,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00101.warc.gz | 715,487,773 | 23,777 | Home Beginners Forex Education Forex Assets Understanding The Trading Costs Involved In USD/TRY Exotic Currency Pair
# Understanding The Trading Costs Involved In USD/TRY Exotic Currency Pair
27
0
#### Introduction
USDTRY, an exotic currency pair, is the abbreviation for the US Dollar against the Turkish Lira. One can expect high volatility in these pairs. Here, the US Dollar is called the base currency and TRY the quote currency.
#### Understanding USD/TRY
The value of USDTRY depicts the value of TRY equivalent to one USD. It is quoted as 1 USD per X TRY. So, if the market value of this pair is 5.9878, then 5.9878 Liras are required to buy one US Dollar.
Spread is the difference between the bid price in the market and the ask price in the market. These prices are set by the brokers. Hence, the prices from each broker differ. Moreover, it varies from the type of execution as well.
ECN: 12 pips | STP: 14 pips
#### Fees
The commission that you pay to your broker for taking a position in a currency pair is a fee on the trade. This, too, depends on the type of execution model. There is typically no fee on STP accounts. And on ECN accounts, there are a few pips of fees.
#### Slippage
Slippage is the difference between the trader’s requested price and the broker’s executed price. It depends on two factors, namely, the broker’s execution speed and market volatility.
The trading range is the range of the pip movement in a currency pair on different timeframes. With it, traders can determine their minimum, average, and maximum risk on a trade in a specified time frame.
#### Procedure to assess Pip Ranges
2. Set the period to 1
3. Add a 200-period SMA to this indicator
4. Shrink the chart so you can determine a large time period
6. Measure the floor level and set this value as the min
7. Measure the level of the 200-period SMA and set this as the average
8. Measure the peak levels and set this as Max.
#### USD/TRY Cost as a Percent of the Trading Range
Apart from knowing how many pips the market moves in a given timeframe, it is also necessary to understand the total cost variation in a trade. And below are two tables (for ECN and STP) that will help determine the best time of the day to trade in the currency pair with reduced costs.
#### ECN Model Account
Total cost = Slippage + Spread + Trading Fee = 3 + 12 + 3 = 18
#### STP Model Account
Spread = 14 | Slippage = 3 | Trading fee = 0
Total cost = Slippage + Spread + Trading Fee = 3 + 14 + 0 = 17
#### The Ideal way to trade the USD/TRY
The costs on major currencies are pretty low, and the volatility is great. So it is ideal to enter any time in the market to trade these pairs. But, when it comes to exotic pairs, the volatility, as well as the costs, are quite high. Hence, one must be aware of when exactly they should trade these currencies.
The percentages in the above tables are directly proportional to the volatility of the market. Hence, we can conclude that costs are when the volatility is low and vice versa.
To determine the ideal times of the day to trade, you must glance at the volatility table and check if the current volatility if nearby the average values mentioned in the tables. If they are more or less in that range, you are good to trade that currency pair because this will assure a balance between both volatilities as well as costs.
Also, another simple way to reduce costs is by getting rid of the slippage on the trade. This can be done by executing orders using limit orders instead of market orders. In doing so, the total costs will reduce by a significant amount, and so will the cost of the trade.
I am a professional Price Action retail trader and Speculator with expertise in Risk Management, Trade Management, and Hedging. | 870 | 3,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-10 | longest | en | 0.916152 |
https://www.adrian.idv.hk/2008-12-22-papagiannaki-elephants/ | 1,628,057,933,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00361.warc.gz | 564,724,927 | 5,202 | Objective: Identify elephant flows from all the flows, given the elephants and mice phenomenon (or mass-count disparity in statistics term) happens in the Internet.
Elephants and mice are defined according to their average bandwidth rather than file size or duration. Because of the power law heavy tail property, a flow is charactierised as an elephant when it is located at the tail of the flow bandwidth distribution. A threshold is determined so that the bandwidth higher than the threshold is said to be elephant. The threshold can be determined initially in two ways:
1. AEST methodology, which assumes flow bandwidth is Pareto, $$P[X>x] \sim cx^{-\alpha}$$, and therefore set the threshold $$\hat x = \min_x(\dfrac{d\log F(x)}{d\log x} - \alpha)$$.
2. Constant load, set the threshold to $$\hat x$$ such that $$\int_{\hat x}^\infty xf(x)dx=R\%$$. The threshold is determined continuously and we run an auto-regressive filter on $$\hat x$$: $$v(n) = (1-\beta)v(n)+\beta\hat x(n)$$, and decided a flow of bandwidth $$x(n)$$ is elephant if $$x(n)>v(n)$$.
However, because flows are bursty, the decision can be improved by introducing latent heat, which defined as: $$L(n)=\sum_{t=n-k}^{n} x(t)-v(t)$$. The new decision rule is that $$L(n)>0$$ for elephant. This improvement is to avoid reclassification for short-term fluctuations and let the membership of elephants change less frequently.
To read: AEST approach to determine the Pareto distribution parameters. See Crovella and Taqqu, “Estimating the Heavy Tail Index from Scaling Properties” (1999)
## Bibliographic data
@techreport{
title = "On the Feasibility of Identifying Elephants in Internet Backbone Traffic",
author = "K. Papagiannaki and N. Taft and S. Bhattacharya and P. Thiran and K. Salamatian and C. Diot",
insitution = "Sprint Labs",
howpublished = "Sprint Tech Report",
number = "ATL Technical Report TR01-ATL-110918",
month = "November",
year = "2001",
} | 498 | 1,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-31 | longest | en | 0.874293 |
https://www.hackmath.net/en/math-problem/8230 | 1,632,253,239,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00054.warc.gz | 824,160,658 | 12,465 | On the menu are 12 kinds of meal. How many ways can we choose four different meals into the daily menu?
n = 495
### Step-by-step explanation:
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There are 7 women and 5 men in a department. a) how many ways can a committee of 3 people be selected? b) how many ways can a committee of 2 men and 1 woman be selected? c) how many ways can a committee of at least 2 woman be selected (3 people total)?
• First class
The shipment contains 40 items. 36 are first grade, 4 are defective. How many ways can select 5 items, so that it is no more than one defective? | 736 | 2,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-39 | longest | en | 0.952545 |
https://tutorgig.info/ed/Belt_(mechanical) | 1,606,624,641,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00200.warc.gz | 529,078,980 | 11,033 | Search: in Tutorials Encyclopedia Videos Books Software DVDs Belt (mechanical) Encyclopedia
Tutorials Encyclopedia Videos Books Software DVDs
## Belt (mechanical)
A pair of vee-belts flat belt Hagley Museum]] A belt is a loop of flexible material used to link two or more rotating shafts mechanically. Belts may be used as a source of motion, to transmit power efficiently, or to track relative movement. Belts are looped over pulleys. In a two pulley system, the belt can either drive the pulleys in the same direction, or the belt may be crossed, so that the direction of the shafts is opposite. As a source of motion, a conveyor belt is one application where the belt is adapted to continuously carry a load between two points.
## Contents
• Power transmission
• Pros and cons
• Flat belts
• Round belts
• Vee belts
• Multi-groove belts
• Ribbed belt
• Film belts
• Timing belts
• Specialty belts
• Flying rope
• Standards for use
• Selection criteria
• Belt friction
• Belt tension
• Belt wear
• Specifications
• References
## Power transmission
Belts are the cheapest utility for power transmission between shafts that may not be axially aligned. Power transmission is achieved by specially designed belts and pulleys. The demands on a belt drive transmission system are large and this has led to many variations on the theme. They run smoothly and with little noise, and cushion motor and bearings against load changes, albeit with less strength than gears or chains. However, improvements in belt engineering allow use of belts in systems that only formerly allowed chains or gears.
Power transmitted between a belt and a pulley is expressed as the product of difference of tension and belt velocity:[1]
P=(T_1-T_2)v
where, T1 and T2 are tensions in the tight side and slack side of the belt respectively. They are related as:
\frac{T_1}{T_2} = e^{\mu\alpha}
where, is the coefficient of friction, and is the angle subtended by contact surface at the centre of the pulley.
### Pros and cons
Belt drive, moreover, is simple, inexpensive, and does not require axially aligned shafts. It helps protect the machinery from overload and jam, and damps and isolates noise and vibration. Load fluctuations are shock-absorbed (cushioned). They need no lubrication and minimal maintenance. They have high efficiency (90-98%, usually 95%), high tolerance for misalignment, and are inexpensive if the shafts are far apart. Clutch action is activated by releasing belt tension. Different speeds can be obtained by step or tapered pulleys.
The angular-velocity ratio may not be constant or equal to that of the pulley diameters, due to slip and stretch. However, this problem has been largely solved by the use of toothed belts. Temperatures ranges from to . Adjustment of center distance or addition of an idler pulley is crucial to compensate for wear and stretch.
### Flat belts
The drive belt: used to transfer power from the engine's flywheel. Here shown driving a threshing machine. Flat belts were used early in line shafting to transmit power in factories.[2] They were also used in countless farming, mining, and logging applications, such as bucksaws, sawmills, threshers, silo blowers, conveyors for filling corn cribs or haylofts, balers, water pumps (for wells, mines, or swampy farm fields), and electrical generators. The flat belt is a simple system of power transmission that was well suited for its day. It delivered high power for high speeds (500 hp for 10,000 ft/min), in cases of wide belts and large pulleys. These drives are bulky, requiring high tension leading to high loads, so vee belts have mainly replaced the flat-belts except when high speed is needed over power. The Industrial Revolution soon demanded more from the system, and flat belt pulleys needed to be carefully aligned to prevent the belt from slipping off. Because flat belts tend to climb towards the higher side of the pulley, pulleys were made with a slightly convex or "crowned" surface (rather than flat) to keep the belts centered. Flat belts also tend to slip on the pulley face when heavy loads are applied and many proprietary dressings were available that could be applied to the belts to increase friction, and so power transmission. Grip was better if the belt was assembled with the hair (i.e. outer) side of the leather against the pulley although belts were also often given a half-twist before joining the ends (forming a M bius strip), so that wear was evenly distributed on both sides of the belt (DB). Belts were joined by lacing the ends together with leather thonging,[3][4] or later by steel comb fasteners.[5] A good modern use for a flat belt is with smaller pulleys and large central distances. They can connect inside and outside pulleys, and can come in both endless and jointed construction.
### Round belts
Round belts are a circular cross section belt designed to run in a pulley with a 60 degree V-groove. Round grooves are only suitable for idler pulleys that guide the belt, or when (soft) O-ring type belts are used. The V-groove transmits torque through a wedging action, thus increasing friction. Nevertheless, round belts are for use in relatively low torque situations only and may be purchased in various lengths or cut to length and joined, either by a staple, a metallic connector (in the case of hollow plastic), glueing or welding (in the case of polyurethane). Early sewing machines utilized a leather belt, joined either by a metal staple or glued, to great effect.
### Vee belts
Belts on a Yanmar 2GM20 marine diesel engine. A multiple-V-belt drive on an air compressor. Vee belts (also known as V-belt or wedge rope) solved the slippage and alignment problem. It is now the basic belt for power transmission. They provide the best combination of traction, speed of movement, load of the bearings, and long service life. The V-belt was developed in 1917 by John Gates of the Gates Rubber Company. They are generally endless, and their general cross-section shape is trapezoidal. The "V" shape of the belt tracks in a mating groove in the pulley (or sheave), with the result that the belt cannot slip off. The belt also tends to wedge into the groove as the load increases — the greater the load, the greater the wedging action — improving torque transmission and making the V-belt an effective solution, needing less width and tension than flat belts. V-belts trump flat belts with their small center distances and high reduction ratios. The preferred center distance is larger than the largest pulley diameter, but less than three times the sum of both pulleys. Optimal speed range is 1000 7000 ft/min. V-belts need larger pulleys for their larger thickness than flat belts. They can be supplied at various fixed lengths or as a segmented section, where the segments are linked (spliced) to form a belt of the required length. For high-power requirements, two or more vee belts can be joined side-by-side in an arrangement called a multi-V, running on matching multi-groove sheaves. The strength of these belts is obtained by reinforcements with fibers like steel, polyester or aramid (e.g. Twaron or Kevlar). This is known as a multiple-V-belt drive (or sometimes a "classical V-belt drive"). When an endless belt does not fit the need, jointed and link V-belts may be employed. However they are weaker and only usable at speeds up to 4000 ft/min. A link v-belt is a number of rubberized fabric links held together by metal fasteners. They are length adjustable by disassembling and removing links when needed.
### Multi-groove belts
A multi-groove or polygroove belt[6] is made up of usually 5 or 6 "V" shapes along side each other. This gives a thinner belt for the same drive surface, thus it is more flexible, although often wider. The added flexibility offers an improved efficiency, as less energy is wasted in the internal friction of continually bending the belt. In practice this gain of efficiency causes a reduced heating effect on the belt and a cooler-running belt lasts longer in service.
A further advantage of the polygroove belt, and the reason they have become so popular, stems from the ability for them to be run over pulleys on the ungrooved back of the belt. Although this is sometimes done with Vee belts with a single idler pulley for tensioning, a polygroove belt may be wrapped around a pulley on its back tightly enough to change its direction, or even to provide a light driving force.[7]
Any Vee belt's ability to drive pulleys depends on wrapping the belt around a sufficient angle of the pulley to provide grip. Where a single-Vee belt is limited to a simple convex shape, it can adequately wrap at most three or possibly four pulleys, so can drive at most three accessories. Where more must be driven, such as for modern cars with power steering and air conditioning, multiple belts are required. As the polygroove belt can be bent into concave paths by external idlers, it can wrap any number of driven pulleys, limited only by the power capacity of the belt.[7]
This ability to bend the belt at the designer's whim allows it to take a complex or "serpentine" path. This can assist the design of a compact engine layout, where the accessories are mounted more closely to the engine block and without the need to provide movable tensioning adjustments. The entire belt may be tensioned by a single idler pulley.
### Ribbed belt
A ribbed belt is a power transmission belt featuring lengthwise grooves. It operates from contact between the ribs of the belt and the grooves in the pulley. Its single-piece structure is reported to offer an even distribution of tension across the width of the pulley where the belt is in contact, a power range up to 600 kW, a high speed ratio, serpentine drives (possibility to drive off the back of the belt), long life, stability and homogeneity of the drive tension, and reduced vibration. The ribbed belt may be fitted on various applications : compressors, fitness bikes, agricultural machinery, food mixers, washing machines, lawn mowers, etc.
### Film belts
Though often grouped with flat belts, they are actually a different kind. They consist of a very thin belt (0.5-15 millimeters or 100-4000 micrometres) strip of plastic and occasionally rubber. They are generally intended for low-power (10 hp or 7 kW), high-speed uses, allowing high efficiency (up to 98%) and long life. These are seen in business machines, printers, tape recorders, and other light-duty operations.
### Timing belts
Timing belt Belt-drive cog on a belt-driven bicycle Timing belts, (also known as toothed, notch, cog, or synchronous belts) are a positive transfer belt and can track relative movement. These belts have teeth that fit into a matching toothed pulley. When correctly tensioned, they have no slippage, run at constant speed, and are often used to transfer direct motion for indexing or timing purposes (hence their name). They are often used in lieu of chains or gears, so there is less noise and a lubrication bath is not necessary. Camshafts of automobiles, miniature timing systems, and stepper motors often utilize these belts. Timing belts need the least tension of all belts, and are among the most efficient. They can bear up to 200 hp (150 kW) at speeds of 16,000 ft/min.
Timing belts with a helical offset tooth design are available. The helical offset tooth design forms a chevron pattern and causes the teeth to engage progressively. The chevron pattern design is self-aligning. The chevron pattern design does not make the noise that some timing belts make at certain speeds, and is more efficient at transferring power (up to 98%).
Disadvantages include a relatively high purchase cost, the need for specially fabricated toothed pulleys, less protection from overloading and jamming, and the lack of clutch action.
### Specialty belts
Belts normally transmit power on the tension side of the loop. However, designs for continuously variable transmissions exist that use belts that are a series of solid metal blocks, linked together as in a chain, transmitting power on the compression side of the loop.
Belts used for rolling roads for wind tunnels can be capable of 250 km/h.[8]
### Flying rope
For transmission of mechanical power over distance without electrical energy, a flying rope can be used.[9] A wire or manila rope can be used to transmit mechanical energy from a steam engine or water wheel to a factory or pump which is located a considerable distance (10 to 100s of meters or more) from the power source. A flying rope way could be supported on poles and pulleys similar to the cable on a chair lift or aerial tramway. Transmission efficiency is generally high.
## Standards for use
The open belt drive has parallel shafts rotating in the same direction, whereas the cross-belt drive also bears parallel shafts but rotate in opposite direction. The former is far more common, and the latter not appropriate for timing and standard V-belts, because the pulleys contact both the both inner and outer belt surfaces. Nonparallel shafts can be connected if the belt's center line is aligned with the center plane of the pulley. Industrial belts are usually reinforced rubber but sometimes leather types, non-leather non-reinforced belts, can only be used in light applications.
The pitch line is the line between the inner and outer surfaces that is neither subject to tension (like the outer surface) nor compression (like the inner). It is midway through the surfaces in film and flat belts and dependent on cross-sectional shape and size in timing and V-belts. Calculating pitch diameter is an engineering task and is beyond the scope of this article. The angular speed is inversely proportional to size, so the larger the one wheel, the less angular velocity, and vice versa. Actual pulley speeds tend to be 0.5 1% less than generally calculated because of belt slip and stretch. In timing belts, the inverse ratio teeth of the belt contributes to the exact measurement. The speed of the belt is:
Speed = Circumference based on pitch diameter angular speed in rpm
### Selection criteria
Belt drives are built under the following required conditions: speeds of and power transmitted between drive and driven unit; suitable distance between shafts; and appropriate operating conditions. The equation for power is:
power (kW) = (torque in newton-meters) (rpm) (2 radians)/(60 sec 1000 W)
Factors of power adjustment include speed ratio; shaft distance (long or short); type of drive unit (electric motor, internal combustion engine); service environment (oily, wet, dusty); driven unit loads (jerky, shock, reversed); and pulley-belt arrangement (open, crossed, turned). These are found in engineering handbooks and manufacturer's literature. When corrected, the horsepower is compared to rated horsepowers of the standard belt cross sections at particular belt speeds to find a number of arrays that will perform best. Now the pulley diameters are chosen. It is generally either large diameters or large cross section that are chosen, since, as stated earlier, larger belts transmit this same power at low belt speeds as smaller belts do at high speeds. To keep the driving part at its smallest, minimum-diameter pulleys are desired. Minimum pulley diameters are limited by the elongation of the belt's outer fibers as the belt wraps around the pulleys. Small pulleys increase this elongation, greatly reducing belt life. Minimum pulley diameters are often listed with each cross section and speed, or listed separately by belt cross section. After the cheapest diameters and belt section are chosen, the belt length is computed. If endless belts are used, the desired shaft spacing may need adjusting to accommodate standard length belts. It is often more economical to use two or more juxtaposed V-belts, rather than one larger belt.
In large speed ratios or small central distances, the angle of contact between the belt and pulley may be less than 180 . If this is the case, the drive power must be further increased, according to manufacturer's tables, and the selection process repeated. This is because power capacities are based on the standard of a 180 contact angle. Smaller contact angles mean less area for the belt to obtain traction, and thus the belt carries less power.
### Belt friction
Belt drives depend on friction to operate but, if the friction is excessive, there will be waste of energy and rapid wear of the belt. Factors which affect belt friction include belt tension, contact angle and the materials from which the belt and pulleys are made.
### Belt tension
Power transmission is a function of belt tension. However, also increasing with tension is stress (load) on the belt and bearings. The ideal belt is that of the lowest tension which does not slip in high loads. Belt tensions should also be adjusted to belt type, size, speed, and pulley diameters. Belt tension is determined by measuring the force to deflect the belt a given distance per inch of pulley. Timing belts need only adequate tension to keep the belt in contact with the pulley.
### Belt wear
Fatigue, more so than abrasion, is the culprit for most belt problems. This wear is caused by stress from rolling around the pulleys. High belt tension; excessive slippage; adverse environmental conditions; and belt overloads caused by shock, vibration, or belt slapping all contribute to belt fatigue.
### Specifications
To fully specify a belt, the material, length, and cross-section size and shape are required. Timing belts, in addition, require that the size of the teeth be given. The length of the belt is the sum of the central length of the system on both sides, half the circumference of both pulleys, and the square of the sum (if crossed) or the difference (if open) of the radii. Thus, when dividing by the central distance, it can be visualized as the central distance times the height that gives the same squared value of the radius difference on, of course, both sides. When adding to the length of either side, the length of the belt increases, in a similar manner to the Pythagorean theorem. One important concept to remember is that as D1 gets closer to D2 there is less of a distance (and therefore less addition of length) until its approaches zero.
On the other hand, in a crossed belt drive the sum rather than the difference of radii is the basis for computation for length. So the wider the small drive increases, the belt length is higher.
• Belt track
• Conveyor belt
• Gilmer belt
• Lariat chain - a science exhibit showing the effects when a belt is run 'too fast'
• Roller chain
• Timing belt
## References
Source: Wikipedia | The above article is available under the GNU FDL. | Edit this article
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Regular Visitor
## Column Total does not match
Hi,
I need your help. I have a Purchase Price Variance report where I needed to show in a table:
These items are sourced from the same source table 'me2l' with a date range of 1 Jan 2021 to present 21 July 2024
1. Date Range - 1 Jan 2024 to 21 July 2024
1. Material item - sourced from "me2l"
2. Last year's average unit price - New Measure = Net Price per Unit LY = CALCULATE(AVERAGE('me2l'[Net Price per Unit]),DATEADD('DATE TAB'[Date],-1,YEAR); sourced from "me2l"
3. This year's average unit price - New Measure = Net Price per Unit = [Net Price]/[Price unit], sourced from "me2l"
4. This year's total ordered quantity - sourced from "me2l"
5. Currency - sourced from "me2l"
6. Exchange Rate - sourced from "me2l"
7. PPV Total = difference between 2 & 3 x 4 x 6. Unfortunately, the column total is incorrect when I download it in Excel and compare it. See New Measure = PPV TOTAL = SUMX(values('me2l'[Material]),[PPV Unit Price 2])*round(average('me2l'[Exchange Rate]),4). Where, another New Measure is created PPV Unit Price 2 = CALCULATE(AVERAGE([Net Price per Unit])-[Net Price per Unit LY])*sum('me2l'[Order Qty-adj])
See screenshot:
Unfortunately, the column total in PPV TOTAL doesn't jibe or is incorrect when you download it in Excel and validate the total column.
Noel Cacnio
Email: noel.cacnio@jgspetrochem.ph
1 ACCEPTED SOLUTION
Regular Visitor
Hi all,
We were able to identify the issue. Instead of averaging the Unit Price, we instead get the Average of PO value / Ordered Qty to arrive at the Average unit price. We also used the Round (x).
Thank you to Yangliu for your inputs and suggestions.
4 REPLIES 4
Regular Visitor
Hi all,
We were able to identify the issue. Instead of averaging the Unit Price, we instead get the Average of PO value / Ordered Qty to arrive at the Average unit price. We also used the Round (x).
Thank you to Yangliu for your inputs and suggestions.
Community Support
Hi @NC_03 ,
Are you referring to the fact that PPV Total is different in power bi and excel after exporting data to excel?
You can check if there are any filters applied on the visual object, sometimes the data transformations or filters applied in Power BI may not be reflected in the exported data.
Due to rounding differences between Power BI and Excel, round() may cause differences, you can try to remove this function to see if the values are still different in power bi and excel
Check the official documentation below for the limitations related to export data:
Export data from a Power BI visualization - Power BI | Microsoft Learn
This is the related document, you can view this content:
Solved: Sum Column Total of exported data is different to ... - Microsoft Fabric Community
Solved: Wrong totals, different than export - Microsoft Fabric Community
Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Regular Visitor
Noted and thanks for taking time responding to my inquiry.
Regular Visitor
Hi all,
I am validating the one in Power BI and the one extracted to Excel. See sample item below:
The one in Power BI report has no value in PPV since the Average Unit Rate in Current Year is the same as the Average Unit Rate in Last Year. However, the one extracted to Excel somehow erroneously generated a difference of (36,119.29).
How is that possible? Could it be that it has something to do with the PPV TOTAL measure?
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The Unfavorable Phone numbers Worksheet is a great way to commence instructing your young ones the concept of bad figures. A negative number is any variety that is less than absolutely no. It could be extra or subtracted. The minus sign indicates the adverse variety. You can also compose adverse amounts in parentheses. Beneath is actually a worksheet to acquire started. This worksheet has a variety of adverse numbers from -10 to 10. Subtracting Negative Numbers Worksheet Pdf.
Unfavorable figures are quite a lot as their importance is below zero
A poor variety includes a importance lower than absolutely no. It can be indicated on the quantity collection in just two approaches: with the beneficial amount written because the initial digit, and with the negative quantity written as being the very last digit. A good quantity is published having a additionally signal ( ) prior to it, however it is optionally available to publish it like that. If the number is not written with a plus sign, it is assumed to be a positive number.
## They are symbolized from a minus indication
In old Greece, negative figures were actually not applied. They were dismissed, as their mathematics was based upon geometrical concepts. When European scholars began converting ancient Arabic text messages from North Africa, they came to acknowledge adverse numbers and adopted them. These days, bad amounts are represented by a minus signal. To understand more about the history and origins of bad figures, read this article. Then, attempt these good examples to see how bad phone numbers have advanced with time.
## They are often additional or subtracted
Positive numbers and negative numbers are easy to subtract and add because the sign of the numbers is the same, as you might already know. They are closer to than positive numbers are, though negative numbers, on the other hand, have a larger absolute value. These numbers have some special rules for arithmetic, but they can still be added and subtracted just like positive ones. You may also subtract and add negative figures employing a quantity collection and use exactly the same guidelines for addition and subtraction as you do for beneficial phone numbers.
## They may be symbolized by way of a quantity in parentheses
A poor amount is depicted from a variety encased in parentheses. The unfavorable sign is changed into its binary comparable, and also the two’s accentuate is stored in a similar devote storage. Sometimes a negative number is represented by a positive number, though the result is always negative. In such cases, the parentheses has to be provided. If you have any questions about the meaning of negative numbers, you should consult a book on math.
## They can be divided up from a beneficial variety
Adverse amounts can be multiplied and divided like optimistic figures. They may also be separated by other negative numbers. They are not equal to one another, however. The 1st time you increase a negative number with a optimistic number, you will definitely get absolutely no because of this. To help make the best solution, you have to determine which signal your solution must have. It is much easier to bear in mind a negative amount after it is developed in mounting brackets. | 635 | 3,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-30 | latest | en | 0.965225 |
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# n comparing the ratios a1/a2, b1/b2 and c1/c2 ,find out whether the following pair of linear equations are consistent, or inconsistent. 3x + 2y = 5 ; 2x – 3y = 7 - CBSE Class 10 - Mathematics
ConceptGraphical Method of Solution of a Pair of Linear Equations
#### Question
On comparing the ratios a_1/a_2, b_1/b_2 and c_1/c_2 ,find out whether the following pair of linear equations are consistent, or inconsistent.
3x + 2y = 5 ; 2x – 3y = 7
#### Solution
3x + 2y = 5
2x - 3y = 7
a_1/a_2 = 3/2, b_1/b_2 = (-2)/3, (c_1)/(c_2)= 5/7
a_1/a_2 != b_1/b_2
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
Is there an error in this question or solution?
#### APPEARS IN
NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 3: Pair of Linear Equations in Two Variables
Ex. 3.20 | Q: 3.1 | Page no. 49
Solution n comparing the ratios a1/a2, b1/b2 and c1/c2 ,find out whether the following pair of linear equations are consistent, or inconsistent. 3x + 2y = 5 ; 2x – 3y = 7 Concept: Graphical Method of Solution of a Pair of Linear Equations.
S | 394 | 1,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-51 | latest | en | 0.793284 |
https://www.r-bloggers.com/2013/02/the-wisdom-of-crowds-clustering-using-evidence-accumulation-clustering-eac/ | 1,718,931,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00139.warc.gz | 839,758,646 | 32,694 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Today’s blog post is about a problem known by most of the people using cluster algorithms on datasets without given true labels (unsupervised learning). The challenge here is the “freedom of choice” over a broad range of different cluster algorithms and how to determine the right parameter values. The difficulty is the following: Every clustering algorithm and even any set of parameters will produce a somewhat different solution. This makes it hard to decide, which of the results should be kept. Because there is no reference when using clustering in an unsupervised fashion, the analyst has to decide whether the results describe some causal or artificial patterns. I will present a method, which tackles the described problem and is also very simple to apply. I think that this makes it really interesting for a lot of practical problems and time-bounded projects. Like for most of the data analytics problems, the rule “There is No Free Lunch for the Data Miner” is still valid and hence also the limitations of the approach will be discussed. For illustration I used three datasets: The first two are artificial datasets and their purpose is to demonstrate the benefits and the limitations from the presented method. This will be done by contrasting the results with those from “classical methods”. For this we use the datasets named “Aggregation” (1) and “Spiral” (2). You can download them from http://cs.joensuu.fi/sipu/datasets/ together with some information about the true number of clusters. The third dataset is about clustering the McDonalds menu and has more of a “real world character”. The clusters will consist of products featuring the same nutrition profile. You can find a different clustering with the same dataset at this blog post from Edwin Chen, where also the dataset was originally from. At the end it would be interesting to compare our result with those presented by the mentioned blog.
Our first step is to load all three datasets and to do some data preprocessing (standardizing the features and omitting all tuples containing NA’s). The plots for dataset 1 and 2 show some typical shapes creating real challenges for most cluster algorithms.
require(ggplot2)
require(cluster)
require(reshape)
# Dataset 1 - Aggregation dataset (http://cs.joensuu.fi/sipu/datasets/)
dataAggregatione <- read.csv("DATA/Aggregation.txt", sep = "\t",
dataAggregationeScaled <- scale(dataAggregatione[, -3]) # normalize data
dataAggregatione <- data.frame(dataAggregationeScaled,
name = as.character(c(1:nrow(dataAggregationeScaled))))
rownames(dataAggregatione) <- dataAggregatione$name ggplot(dataAggregatione, aes(V1, V2)) + geom_point() # Dataset 2 - Spiral dataset (http://cs.joensuu.fi/sipu/datasets/) dataSpiral <- read.csv("DATA/spiral.txt", sep = "\t", header = FALSE) dataSpiralScaled <- scale(dataSpiral[, -3]) # normalize data dataSpiral <- data.frame(dataSpiralScaled, name = as.character(c(1:nrow(dataSpiralScaled)))) rownames(dataSpiral) <- dataSpiral$name
ggplot(dataSpiral, aes(V1, V2)) + geom_point()
# Dataset 3 - the mcdonalds menu
# (https://github.com/echen/dirichlet-process)
dataMC$total_fat <- as.numeric(dataMC$total_fat)
numericAttr <- c("total_fat", "cholesterol", "sodium", "dietary_fiber",
"sugars", "protein", "vitamin_a_dv",
"vitamin_c_dv", "calcium_dv",
"iron_dv", "calories_from_fat",
"saturated_fat", "trans_fat", "carbohydrates")
dataMC <- na.omit(dataMC) # drop NAs
dataMCScaled <- scale(dataMC[, -ncol(dataMC)]) # normalize data
dataMC <- data.frame(dataMCScaled, name = dataMC$name) Now, let’s start with the analytical part by describing a typical challenge occurring in clustering. Most of the algorithms require some parameter $$k$$ which is used to determine the number of clusters (I will also use the word partitions synonymously) produced by the clustering/partitioning algorithm (e.g. k-means, spectral clustering) and hence the final partitioning. But it is inherent in the nature of the problem, that k cannot be specified a priori. This is because in unsupervised clustering the true number of partitions is not known prior. An approach to coop with that problem is to use some relative validation criteria. The logic behind this is the following: The user computes a score over partitions resulting from different parameterizations where the score expresses the quality of a partition. At the end, the partition with the highest score should be kept as final result (It depends on the meaning of the score if it should be minimized or maximized. But for simplicity I assume that every score could be expressed as such that a high value is associated with a high clustering quality). The package “clusterCrit” contains most relevant cluster validity indices coming from a broad range of research literature (e.g. “On the Comparison of Relative Clustering Validity Criteria”, Vendramin et al.). In our first experiment I will use the “Dunn-Index”, the “Calinski-Harabasz-Index” and the popular Silhouette measure) together with the k-means algorithm on the two synthetic datasets. To account for the uncertainty about the right value of k, all possible values between 2 and 50 will be evaluated: # ------------------------------- # Application of k-means # ------------------------------- require(clusterCrit) # Aggregation dataset set.seed(1234) # matrix to hold the score value vals <- matrix(rep(NA, 49 * 3), ncol = 3, dimnames = list(c(), c("Dunn", "Calinski-Harabasz", "Silhouette"))) for (k in 2:50) { cl <- kmeans(dataAggregatione[, c(1, 2)], k) vals[(k - 1), 1] <- as.numeric(intCriteria( as.matrix(dataAggregatione[, c(1, 2)]), cl$cluster,
"Dunn"))
vals[(k - 1), 2] <- as.numeric(intCriteria(
as.matrix(dataAggregatione[, c(1, 2)]), cl$cluster, "Calinski_Harabasz")) vals[(k - 1), 3] <- as.numeric(intCriteria( as.matrix(dataAggregatione[, c(1, 2)]), cl$cluster,
"Silhouette"))
}
vals <- data.frame(K = c(2:50), vals)
choosen_k <- matrix(c(vals[bestCriterion(vals[, 2], "Dunn"), "K"],
vals[bestCriterion(vals[, 3], "Calinski_Harabasz"), "K"],
vals[bestCriterion(vals[, 4], "Silhouette"), "K"]),
ncol = 3,
dimnames = list(c("Aggregation"),
c("Dunn", "Calinski_Harabasz", "Silhouette")))
choosen_k
## Dunn Calinski_Harabasz Silhouette
## Aggregation 46 45 4
# Spiral dataset
set.seed(1234)
vals <- matrix(rep(NA, 49 * 3), ncol = 3, dimnames = list(c(),
c("Dunn", "Calinski-Harabasz", "Silhouette")))
for (k in 2:50) {
cl <- kmeans(dataSpiral[, c(1, 2)], k)
vals[(k - 1), 1] <- as.numeric(intCriteria(
as.matrix(dataSpiral[, c(1, 2)]), cl$cluster, "Dunn")) vals[(k - 1), 2] <- as.numeric(intCriteria( as.matrix(dataSpiral[, c(1, 2)]), cl$cluster,
"Calinski_Harabasz"))
vals[(k - 1), 3] <- as.numeric(intCriteria(
as.matrix(dataSpiral[, c(1, 2)]), cl$cluster, "Silhouette")) } vals <- data.frame(K = c(2:50), vals) choosen_k <- matrix(c(vals[bestCriterion(vals[, 2], "Dunn"), "K"], vals[bestCriterion(vals[, 3], "Calinski_Harabasz"), "K"], vals[bestCriterion(vals[, 4], "Silhouette"), "K"]), ncol = 3, dimnames = list(c("Spiral"), c("Dunn", "Calinski_Harabasz", "Silhouette"))) choosen_k ## Dunn Calinski_Harabasz Silhouette ## Spiral 34 50 37 Hence the first surprise is how distinct some of the validity scores are for both datasets. Even though the special structure of the data is challenging for the k-means algorithm, the results indicates that the choice of the right validation criteria is non-trivial, difficult and shows significant impact on the results. If we plot the resulting clustering for the best value obtained with the Silhouette score, the figures show that the approach failed for both datasets, even in picturing the coarse structure: # ------------------------------- # Plot results # ------------------------------- # Aggregation dataset kmeansResultsAggreation <- kmeans(x = dataAggregatione[, c(1, 2)], centers = 3)$cluster
dataAggregatione$clusterSimpleKmeans <- as.character(kmeansResultsAggreation) ggplot(dataAggregatione, aes(V1, V2)) + geom_point(aes(colour = clusterSimpleKmeans)) + opts(legend.position = "none") # Spiral dataset kmeansResultsSpiral <- kmeans(x = dataSpiral[, c(1, 2)], centers = 37)$cluster
dataSpiral$clusterSimpleKmeans <- as.character(kmeansResultsSpiral) ggplot(dataSpiral, aes(V1, V2)) + geom_point(aes(colour = clusterSimpleKmeans)) + opts(legend.position = "none") Now, let me introduce a method which overcomes the problem of choosing the right k value and which gives better result even when a simple clustering algorithm like k-means is used. Even that this blog post focus on the choice of the parameter k, the method is a form of a general approach to “directly estimate” the final clustering without relying too much on a single set of more or less arbitrarily estimated parameters. It is called “Evidence Accumulation Clustering” and you can find some more deep information here and here. The notion behind this method is to build partitions with different algorithms and parameterizations and to aggregate all solutions into one final partition using every single partition as a voting if instances should be placed together. Hence if two venues will be placed together in most solutions, it is reasonable to assign them to the same cluster in the final partition. In this context, this method could also be understood as a tool to enhance the validity of the resulting partition by reducing the impact of a single non-optimal clustering. The algorithmic part is simple: The first part is about the creation of different partitions and aggregating the “votes”. For this we do two steps in each iteration: First, we cluster all data points with k-means using a randomly sampled value from a given interval for $$k$$. Second, we note if two instances ($$i$$ and $$j$$) are placed together inside of one cluster. In this case we increment the corresponding entry ($$A(i,j)$$ and $$A(j,i)$$) in our so called association matrix $$A$$ by 1. At the end of the first part we divide all entries by R, the number of iterations. In the second part the “votes” are aggregated through hierarchical clustering and we obtain our final partition by selecting an appropriate cutting point in the resulting dendogram. For this we transform the association matrix $$A$$ from the first part into a distance matrix and feed it into a single linkage clustering. On the resulting dendogram we calculate the so called “maximal cluster livetime”. It is the longest “gap” between two successive merges. Using this as our cutting point we obtain a final partition. The code below is some technically simple draft of this algorithm (e.g. no parallelization): # ------------------------------- # Evidence Accumulation Clustering # ------------------------------- createCoAssocMatrix <- function(Iter, rangeK, dataSet) { nV <- dim(dataSet)[1] CoAssoc <- matrix(rep(0, nV * nV), nrow = nV) for (j in 1:Iter) { jK <- sample(c(rangeK[1]:rangeK[2]), 1, replace = FALSE) jSpecCl <- kmeans(x = dataSet, centers = jK)$cluster
CoAssoc_j <- matrix(rep(0, nV * nV), nrow = nV)
for (i in unique(jSpecCl)) {
indVenues <- which(jSpecCl == i)
CoAssoc_j[indVenues, indVenues]
<- CoAssoc_j[indVenues, indVenues] + (1/Iter)
}
CoAssoc <- CoAssoc + CoAssoc_j
}
return(CoAssoc)
}
eac <- function(Iter, rangeK, dataset, hcMethod = "single") {
CoAssocSim <- createCoAssocMatrix(Iter, rangeK, dataset)
# transform from similiarity into distance matrix
CoAssocDist <- 1 - CoAssocSim
hclustM <- hclust(as.dist(CoAssocDist), method = hcMethod)
# determine the cut
cutValue <- hclustM$height[which.max(diff(hclustM$height))]
return(cutree(hclustM, h = cutValue))
}
Now let’s try this method on both artificial datasets using exactly the same set of possible values for k like in the first case above:
# -------------------------------
# Application of EAC
# -------------------------------
# # Aggregation dataset
set.seed(1234)
EACResults_Aggregatione <- eac(Iter = 200, rangeK = c(2, 50),
dataset = dataAggregatione[, c(1, 2)], hcMethod = "single")
table(EACResults_Aggregatione)
## EACResults_Aggregatione
## 1 2 3 4 5
## 170 307 232 45 34
dataAggregatione$clusterEAC <- as.character(EACResults_Aggregatione) ggplot(dataAggregatione, aes(V1, V2)) + geom_point(aes(colour = clusterEAC)) + opts(legend.position = "none") # Spiral dataset set.seed(1234) EACResults_Spiral <- eac(Iter = 200, rangeK = c(2, 50), dataset = dataSpiral[, c(1, 2)], hcMethod = "single") table(EACResults_Spiral) ## EACResults_Spiral ## 1 2 3 ## 106 101 105 dataSpiral$clusterEAC <- as.character(EACResults_Spiral)
ggplot(dataSpiral, aes(V1, V2)) + geom_point(aes(colour = clusterEAC)) +
opts(legend.position = "none")
The new results clearly show a real progress even when the algorithm slightly underestimates the true number of clusters for the Aggregation dataset (five instead of seven). But as the discussion in the mentioned paper on EAC depicts, the algorithm has some problems with not so well separated clusters - nobody is perfect. The results using the Spiral dataset look flawless. All three clusters are identified even though we use a simple k-means algorithm with moderate overhead for building the EAC algorithm. Let’s close this post with an application of the described method on the mentioned McDonalds dataset. The preprocessing of the data and the plotting is taken from the code published by Edwin Chen.
# -------------------------------
# -------------------------------
set.seed(1234)
EACResults_MC <- eac(Iter = 1000, rangeK = c(2, 50),
dataset = dataMC[, numericAttr],
hcMethod = "single")
table(EACResults_MC)
## EACResults_MC
## 1 2 3 4 5 6 7 8 9 10 11 12 13 14
## 19 126 39 2 9 15 16 1 21 20 4 27 18 8
dataMC$cluster <- as.character(EACResults_MC) # the following code snippet is taken from # "http://blog.echen.me/2012/03/20/infinite-mixture-models-with-nonparametric # -bayes-and-the-dirichlet-process/" # ignore duplicate food items x = ddply(dataMC, .(name), function(df) head(df, 1)) # for each cluster, take at most 5 items # (to avoid the plot being dominated by large clusters) x = ddply(x, .(cluster), function(df) head(df, 5)) # Reorder names by cluster # (so we can get a plot where all points in a cluster are together) x$name = factor(x$name, levels = x$name[order(x$cluster)], ordered = T) # Turn this into a tall-thin matrix m = melt(x, id = c("name", "cluster")) m$name <- paste("(", m$cluster, ") ", m$name, sep = "")
qplot(variable, weight = value, data = m, color = cluster, fill = cluster,
geom = "bar", width = 1,
xlab = "nutritional variable", ylab = "z-scaled value",
main = "McDonald's food clusters") +
facet_wrap(~name, ncol = 5) + coord_flip() + scale_colour_hue("cluster") +
opts(axis.text.y = theme_text(size = 8),
axis.text.x = theme_text(size = 8)) +
scale_fill_hue("cluster")
Here EAC gives us 14 partitions, 3 more than Edwin Chen found. In this example, it is more difficult to verify the solution. Hence our validation is more a qualitative one. The final plot shows five instances at maximum per partition together with their nutrition profiles. A first look at our results indicates that most of the clusters exhibit a good separation (except cluster 2 which includes some more heterogeneous items). I will not discuss every cluster in detail but here are some specific remarks: Cluster 3, 10, 12 and 14 show somehow similar profiles, nevertheless:
• cluster 12 (non-fat variants of coffee with syrup) is rich on carbohydrates and sugar
• cluster 10 (non-fat variants of coffee pure) has more protein and calcium
• cluster 3 is similar to cluster 12 but with less proteins and it contains more items with “extreme” profiles like apple dippers and
• cluster 14 - consisting of fruit smoothies with a lot of dietary fiber and proteins
Maybe it would be more plausible to shift the apple dippers in cluster 8, which only contains the apple slices.
So what could finally be said about the presented method? Although it has some limitations it poses an interesting and simple method, which gives much better results than single runs with an algorithm like k-means. The shown algorithm could be extended as well. One option could be to substitute the base method with a more sophisticated method. Another idea is to use a form of weighted aggregation of the partitions - mentioned here.
Like always, any helpful comment or idea is welcome. | 4,212 | 16,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.9292 |
https://edustrings.com/mathematics/2511267.html | 1,721,766,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00694.warc.gz | 202,529,588 | 7,595 | 9 November, 14:41
Conversion. Write a reduced fraction notation for 0.56.
+2
1. 9 November, 15:54
0
Reduced fraction: 14/25
Step-by-step explanation:
To write 0.56 in terms of fractions we need to multiply and divide the number by 100:
(0.56 * 100) / 100 = 56/100. Then we simplify the fraction by dividing the numerator and denominator by the common factor 2 as many times as possible:
56/100 = 28/50 = 14/25
The fraction 14/25 cannot be more simplified
2. 9 November, 17:05
0
0.56 = 14/25
Step-by-step explanation:
It is given a decimal number 0.56
To convert decimal to fraction
Here we have 0.56 as decimal
0.56 = 56/100
To convert 56/100 to reduced form
We can reduce 56/100 as simple for,
56/100 = 28/50
28/50 = 14/25
reduced form 0f 0.56 is 14/25
Therefore 0.56 = 14/25 | 267 | 795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.877554 |
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ctbmv.f
Go to the documentation of this file.
1*> \brief \b CTBMV
2*
3* =========== DOCUMENTATION ===========
4*
5* Online html documentation available at
6* http://www.netlib.org/lapack/explore-html/
7*
8* Definition:
9* ===========
10*
11* SUBROUTINE CTBMV(UPLO,TRANS,DIAG,N,K,A,LDA,X,INCX)
12*
13* .. Scalar Arguments ..
14* INTEGER INCX,K,LDA,N
15* CHARACTER DIAG,TRANS,UPLO
16* ..
17* .. Array Arguments ..
18* COMPLEX A(LDA,*),X(*)
19* ..
20*
21*
22*> \par Purpose:
23* =============
24*>
25*> \verbatim
26*>
27*> CTBMV performs one of the matrix-vector operations
28*>
29*> x := A*x, or x := A**T*x, or x := A**H*x,
30*>
31*> where x is an n element vector and A is an n by n unit, or non-unit,
32*> upper or lower triangular band matrix, with ( k + 1 ) diagonals.
33*> \endverbatim
34*
35* Arguments:
36* ==========
37*
38*> \param[in] UPLO
39*> \verbatim
40*> UPLO is CHARACTER*1
41*> On entry, UPLO specifies whether the matrix is an upper or
42*> lower triangular matrix as follows:
43*>
44*> UPLO = 'U' or 'u' A is an upper triangular matrix.
45*>
46*> UPLO = 'L' or 'l' A is a lower triangular matrix.
47*> \endverbatim
48*>
49*> \param[in] TRANS
50*> \verbatim
51*> TRANS is CHARACTER*1
52*> On entry, TRANS specifies the operation to be performed as
53*> follows:
54*>
55*> TRANS = 'N' or 'n' x := A*x.
56*>
57*> TRANS = 'T' or 't' x := A**T*x.
58*>
59*> TRANS = 'C' or 'c' x := A**H*x.
60*> \endverbatim
61*>
62*> \param[in] DIAG
63*> \verbatim
64*> DIAG is CHARACTER*1
65*> On entry, DIAG specifies whether or not A is unit
66*> triangular as follows:
67*>
68*> DIAG = 'U' or 'u' A is assumed to be unit triangular.
69*>
70*> DIAG = 'N' or 'n' A is not assumed to be unit
71*> triangular.
72*> \endverbatim
73*>
74*> \param[in] N
75*> \verbatim
76*> N is INTEGER
77*> On entry, N specifies the order of the matrix A.
78*> N must be at least zero.
79*> \endverbatim
80*>
81*> \param[in] K
82*> \verbatim
83*> K is INTEGER
84*> On entry with UPLO = 'U' or 'u', K specifies the number of
85*> super-diagonals of the matrix A.
86*> On entry with UPLO = 'L' or 'l', K specifies the number of
87*> sub-diagonals of the matrix A.
88*> K must satisfy 0 .le. K.
89*> \endverbatim
90*>
91*> \param[in] A
92*> \verbatim
93*> A is COMPLEX array, dimension ( LDA, N ).
94*> Before entry with UPLO = 'U' or 'u', the leading ( k + 1 )
95*> by n part of the array A must contain the upper triangular
96*> band part of the matrix of coefficients, supplied column by
97*> column, with the leading diagonal of the matrix in row
98*> ( k + 1 ) of the array, the first super-diagonal starting at
99*> position 2 in row k, and so on. The top left k by k triangle
100*> of the array A is not referenced.
101*> The following program segment will transfer an upper
102*> triangular band matrix from conventional full matrix storage
103*> to band storage:
104*>
105*> DO 20, J = 1, N
106*> M = K + 1 - J
107*> DO 10, I = MAX( 1, J - K ), J
108*> A( M + I, J ) = matrix( I, J )
109*> 10 CONTINUE
110*> 20 CONTINUE
111*>
112*> Before entry with UPLO = 'L' or 'l', the leading ( k + 1 )
113*> by n part of the array A must contain the lower triangular
114*> band part of the matrix of coefficients, supplied column by
115*> column, with the leading diagonal of the matrix in row 1 of
116*> the array, the first sub-diagonal starting at position 1 in
117*> row 2, and so on. The bottom right k by k triangle of the
118*> array A is not referenced.
119*> The following program segment will transfer a lower
120*> triangular band matrix from conventional full matrix storage
121*> to band storage:
122*>
123*> DO 20, J = 1, N
124*> M = 1 - J
125*> DO 10, I = J, MIN( N, J + K )
126*> A( M + I, J ) = matrix( I, J )
127*> 10 CONTINUE
128*> 20 CONTINUE
129*>
130*> Note that when DIAG = 'U' or 'u' the elements of the array A
131*> corresponding to the diagonal elements of the matrix are not
132*> referenced, but are assumed to be unity.
133*> \endverbatim
134*>
135*> \param[in] LDA
136*> \verbatim
137*> LDA is INTEGER
138*> On entry, LDA specifies the first dimension of A as declared
139*> in the calling (sub) program. LDA must be at least
140*> ( k + 1 ).
141*> \endverbatim
142*>
143*> \param[in,out] X
144*> \verbatim
145*> X is COMPLEX array, dimension at least
146*> ( 1 + ( n - 1 )*abs( INCX ) ).
147*> Before entry, the incremented array X must contain the n
148*> element vector x. On exit, X is overwritten with the
149*> transformed vector x.
150*> \endverbatim
151*>
152*> \param[in] INCX
153*> \verbatim
154*> INCX is INTEGER
155*> On entry, INCX specifies the increment for the elements of
156*> X. INCX must not be zero.
157*> \endverbatim
158*
159* Authors:
160* ========
161*
162*> \author Univ. of Tennessee
163*> \author Univ. of California Berkeley
164*> \author Univ. of Colorado Denver
165*> \author NAG Ltd.
166*
167*> \ingroup tbmv
168*
169*> \par Further Details:
170* =====================
171*>
172*> \verbatim
173*>
174*> Level 2 Blas routine.
175*> The vector and matrix arguments are not referenced when N = 0, or M = 0
176*>
177*> -- Written on 22-October-1986.
178*> Jack Dongarra, Argonne National Lab.
179*> Jeremy Du Croz, Nag Central Office.
180*> Sven Hammarling, Nag Central Office.
181*> Richard Hanson, Sandia National Labs.
182*> \endverbatim
183*>
184* =====================================================================
185 SUBROUTINE ctbmv(UPLO,TRANS,DIAG,N,K,A,LDA,X,INCX)
186*
187* -- Reference BLAS level2 routine --
188* -- Reference BLAS is a software package provided by Univ. of Tennessee, --
189* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
190*
191* .. Scalar Arguments ..
192 INTEGER INCX,K,LDA,N
193 CHARACTER DIAG,TRANS,UPLO
194* ..
195* .. Array Arguments ..
196 COMPLEX A(LDA,*),X(*)
197* ..
198*
199* =====================================================================
200*
201* .. Parameters ..
202 COMPLEX ZERO
203 parameter(zero= (0.0e+0,0.0e+0))
204* ..
205* .. Local Scalars ..
206 COMPLEX TEMP
207 INTEGER I,INFO,IX,J,JX,KPLUS1,KX,L
208 LOGICAL NOCONJ,NOUNIT
209* ..
210* .. External Functions ..
211 LOGICAL LSAME
212 EXTERNAL lsame
213* ..
214* .. External Subroutines ..
215 EXTERNAL xerbla
216* ..
217* .. Intrinsic Functions ..
218 INTRINSIC conjg,max,min
219* ..
220*
221* Test the input parameters.
222*
223 info = 0
224 IF (.NOT.lsame(uplo,'U') .AND. .NOT.lsame(uplo,'L')) THEN
225 info = 1
226 ELSE IF (.NOT.lsame(trans,'N') .AND.
227 + .NOT.lsame(trans,'T') .AND.
228 + .NOT.lsame(trans,'C')) THEN
229 info = 2
230 ELSE IF (.NOT.lsame(diag,'U') .AND.
231 + .NOT.lsame(diag,'N')) THEN
232 info = 3
233 ELSE IF (n.LT.0) THEN
234 info = 4
235 ELSE IF (k.LT.0) THEN
236 info = 5
237 ELSE IF (lda.LT. (k+1)) THEN
238 info = 7
239 ELSE IF (incx.EQ.0) THEN
240 info = 9
241 END IF
242 IF (info.NE.0) THEN
243 CALL xerbla('CTBMV ',info)
244 RETURN
245 END IF
246*
247* Quick return if possible.
248*
249 IF (n.EQ.0) RETURN
250*
251 noconj = lsame(trans,'T')
252 nounit = lsame(diag,'N')
253*
254* Set up the start point in X if the increment is not unity. This
255* will be ( N - 1 )*INCX too small for descending loops.
256*
257 IF (incx.LE.0) THEN
258 kx = 1 - (n-1)*incx
259 ELSE IF (incx.NE.1) THEN
260 kx = 1
261 END IF
262*
263* Start the operations. In this version the elements of A are
264* accessed sequentially with one pass through A.
265*
266 IF (lsame(trans,'N')) THEN
267*
268* Form x := A*x.
269*
270 IF (lsame(uplo,'U')) THEN
271 kplus1 = k + 1
272 IF (incx.EQ.1) THEN
273 DO 20 j = 1,n
274 IF (x(j).NE.zero) THEN
275 temp = x(j)
276 l = kplus1 - j
277 DO 10 i = max(1,j-k),j - 1
278 x(i) = x(i) + temp*a(l+i,j)
279 10 CONTINUE
280 IF (nounit) x(j) = x(j)*a(kplus1,j)
281 END IF
282 20 CONTINUE
283 ELSE
284 jx = kx
285 DO 40 j = 1,n
286 IF (x(jx).NE.zero) THEN
287 temp = x(jx)
288 ix = kx
289 l = kplus1 - j
290 DO 30 i = max(1,j-k),j - 1
291 x(ix) = x(ix) + temp*a(l+i,j)
292 ix = ix + incx
293 30 CONTINUE
294 IF (nounit) x(jx) = x(jx)*a(kplus1,j)
295 END IF
296 jx = jx + incx
297 IF (j.GT.k) kx = kx + incx
298 40 CONTINUE
299 END IF
300 ELSE
301 IF (incx.EQ.1) THEN
302 DO 60 j = n,1,-1
303 IF (x(j).NE.zero) THEN
304 temp = x(j)
305 l = 1 - j
306 DO 50 i = min(n,j+k),j + 1,-1
307 x(i) = x(i) + temp*a(l+i,j)
308 50 CONTINUE
309 IF (nounit) x(j) = x(j)*a(1,j)
310 END IF
311 60 CONTINUE
312 ELSE
313 kx = kx + (n-1)*incx
314 jx = kx
315 DO 80 j = n,1,-1
316 IF (x(jx).NE.zero) THEN
317 temp = x(jx)
318 ix = kx
319 l = 1 - j
320 DO 70 i = min(n,j+k),j + 1,-1
321 x(ix) = x(ix) + temp*a(l+i,j)
322 ix = ix - incx
323 70 CONTINUE
324 IF (nounit) x(jx) = x(jx)*a(1,j)
325 END IF
326 jx = jx - incx
327 IF ((n-j).GE.k) kx = kx - incx
328 80 CONTINUE
329 END IF
330 END IF
331 ELSE
332*
333* Form x := A**T*x or x := A**H*x.
334*
335 IF (lsame(uplo,'U')) THEN
336 kplus1 = k + 1
337 IF (incx.EQ.1) THEN
338 DO 110 j = n,1,-1
339 temp = x(j)
340 l = kplus1 - j
341 IF (noconj) THEN
342 IF (nounit) temp = temp*a(kplus1,j)
343 DO 90 i = j - 1,max(1,j-k),-1
344 temp = temp + a(l+i,j)*x(i)
345 90 CONTINUE
346 ELSE
347 IF (nounit) temp = temp*conjg(a(kplus1,j))
348 DO 100 i = j - 1,max(1,j-k),-1
349 temp = temp + conjg(a(l+i,j))*x(i)
350 100 CONTINUE
351 END IF
352 x(j) = temp
353 110 CONTINUE
354 ELSE
355 kx = kx + (n-1)*incx
356 jx = kx
357 DO 140 j = n,1,-1
358 temp = x(jx)
359 kx = kx - incx
360 ix = kx
361 l = kplus1 - j
362 IF (noconj) THEN
363 IF (nounit) temp = temp*a(kplus1,j)
364 DO 120 i = j - 1,max(1,j-k),-1
365 temp = temp + a(l+i,j)*x(ix)
366 ix = ix - incx
367 120 CONTINUE
368 ELSE
369 IF (nounit) temp = temp*conjg(a(kplus1,j))
370 DO 130 i = j - 1,max(1,j-k),-1
371 temp = temp + conjg(a(l+i,j))*x(ix)
372 ix = ix - incx
373 130 CONTINUE
374 END IF
375 x(jx) = temp
376 jx = jx - incx
377 140 CONTINUE
378 END IF
379 ELSE
380 IF (incx.EQ.1) THEN
381 DO 170 j = 1,n
382 temp = x(j)
383 l = 1 - j
384 IF (noconj) THEN
385 IF (nounit) temp = temp*a(1,j)
386 DO 150 i = j + 1,min(n,j+k)
387 temp = temp + a(l+i,j)*x(i)
388 150 CONTINUE
389 ELSE
390 IF (nounit) temp = temp*conjg(a(1,j))
391 DO 160 i = j + 1,min(n,j+k)
392 temp = temp + conjg(a(l+i,j))*x(i)
393 160 CONTINUE
394 END IF
395 x(j) = temp
396 170 CONTINUE
397 ELSE
398 jx = kx
399 DO 200 j = 1,n
400 temp = x(jx)
401 kx = kx + incx
402 ix = kx
403 l = 1 - j
404 IF (noconj) THEN
405 IF (nounit) temp = temp*a(1,j)
406 DO 180 i = j + 1,min(n,j+k)
407 temp = temp + a(l+i,j)*x(ix)
408 ix = ix + incx
409 180 CONTINUE
410 ELSE
411 IF (nounit) temp = temp*conjg(a(1,j))
412 DO 190 i = j + 1,min(n,j+k)
413 temp = temp + conjg(a(l+i,j))*x(ix)
414 ix = ix + incx
415 190 CONTINUE
416 END IF
417 x(jx) = temp
418 jx = jx + incx
419 200 CONTINUE
420 END IF
421 END IF
422 END IF
423*
424 RETURN
425*
426* End of CTBMV
427*
428 END
subroutine xerbla(srname, info)
Definition cblat2.f:3285
subroutine ctbmv(uplo, trans, diag, n, k, a, lda, x, incx)
CTBMV
Definition ctbmv.f:186 | 3,899 | 10,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-10 | latest | en | 0.492111 |
https://oeis.org/A267322 | 1,571,839,939,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833766.94/warc/CC-MAIN-20191023122219-20191023145719-00504.warc.gz | 641,398,079 | 4,450 | This site is supported by donations to The OEIS Foundation.
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A267322 Expansion of (1 + x + x^2 + x^4 + 2*x^5)/(1 - x^3)^3. 1
1, 1, 1, 3, 4, 5, 6, 9, 12, 10, 16, 22, 15, 25, 35, 21, 36, 51, 28, 49, 70, 36, 64, 92, 45, 81, 117, 55, 100, 145, 66, 121, 176, 78, 144, 210, 91, 169, 247, 105, 196, 287, 120, 225, 330, 136, 256, 376, 153, 289, 425, 171, 324, 477, 190, 361, 532, 210, 400, 590, 231, 441, 651 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS Triangular numbers alternating with squares and pentagonal numbers. LINKS Ilya Gutkovskiy, Extended illustration of initial terms Eric Weisstein's World of Mathematics, Triangular Number Eric Weisstein's World of Mathematics, Square Number Eric Weisstein's World of Mathematics, Pentagonal Number Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1) FORMULA G.f.: (1 + x + x^2 + x^4 + 2*x^5)/(1 - x^3)^3. a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9). a(3k) = A000217(k+1), a(3k+1) = A000290(k+1), a(3k+2) = A000326(k+1). Sum_{n>=0} 1/a(n) = 2 - Pi/sqrt(3) + Pi^2/6 + 3*log(3) = 5.1269715686... a(n) = (floor(n/3) + 1)*((n+1)*floor(n/3) - 3*floor(n/3)^2 + 2)/2. - Bruno Berselli, Apr 08 2016 EXAMPLE Illustration of initial terms: ========================================================== n: 0 1 2 3 4 5 6 7 8 ---------------------------------------------------------- o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o ========================================================== 1 1 1 3 4 5 6 9 12 ---------------------------------------------------------- MATHEMATICA LinearRecurrence[{0, 0, 3, 0, 0, -3, 0, 0, 1}, {1, 1, 1, 3, 4, 5, 6, 9, 12}, 70] Table[(Floor[n/3] + 1) ((n + 1) Floor[n/3] - 3 Floor[n/3]^2 + 2)/2, {n, 0, 70}] (* Bruno Berselli, Apr 08 2016 *) PROG (PARI) x='x+O('x^99); Vec((1+x+x^2+x^4+2*x^5)/(1-x^3)^3) \\ Altug Alkan, Apr 07 2016 CROSSREFS Cf. A000217, A000290, A000326, A123596, A124093, A271391. Sequence in context: A047427 A228235 A228895 * A218929 A088875 A022884 Adjacent sequences: A267319 A267320 A267321 * A267323 A267324 A267325 KEYWORD nonn,easy AUTHOR Ilya Gutkovskiy, Apr 07 2016 STATUS approved
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Last modified October 23 10:07 EDT 2019. Contains 328345 sequences. (Running on oeis4.) | 1,085 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-43 | latest | en | 0.387012 |
https://www.neetprep.com/questions/55-Physics/691-Electric-Charges-Fields?courseId=8&testId=1339018-Padma-Shri-H-C-Verma-Objective-Exercises-Based-MCQs&questionId=303965-body-charged-rubbing-its-weightremains-precisely-constantincreases-slightlydecreases-slightlymayincrease-slightly-may-decrease-slightly | 1,719,027,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00488.warc.gz | 814,751,884 | 56,428 | Select Chapter Topics:
# If a body is charged by rubbing it, its weight: 1. remains precisely constant. 2. increases slightly. 3. decreases slightly. 4. may increase slightly or may decrease slightly.
Subtopic: Electric Charge |
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An electric dipole is placed in a uniform electric field. The net electric force on the dipole:
1 is always zero. 2 depends on the orientation of the dipole. 3 can never be zero. 4 depends on the strength of the dipole.
Subtopic: Electric Dipole |
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Given below are four statements:
(a) The total charge of the universe is constant. (b) The total positive charge of the universe is constant. (c) The total negative charge of the universe is constant. (d) The total number of charged particles in the universe is constant.
Choose the correct option:
1 (a) only 2 (b), (c) 3 (c), (d) 4 (a), (d)
Subtopic: Electric Charge |
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A point charge is brought in an electric field. The electric field at a nearby point,
(a) will increase if the charge is positive (b) will decrease if the charge is negative (c) may increase if the charge is positive (d) may decrease if the charge is negative
Choose the correct option:
1 (a) only 2 (b), (c) 3 (c), (d) 4 (a), (d)
Subtopic: Electric Field |
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An electric dipole is placed in an electric field generated by a point charge.
(a) The net electric force on the dipole must be zero. (b) The net electric force on the dipole may be zero. (c) The torque on the dipole due to the field must be zero. (d) The torque on the dipole due to the field may be zero.
Choose the correct option:
1 (a) only 2 (b), (c) 3 (c), (d) 4 (d) only
Subtopic: Electric Dipole |
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A proton and an electron are placed in a uniform electric field.
(a) The electric forces acting on them will be equal. (b) The magnitudes of the forces will be equal. (c) Their accelerations will be equal. (d) The magnitudes of their accelerations will be equal.
Choose the correct option.
1 (a), (b) 2 (b) only 3 (c), (d) 4 (a), (d)
Subtopic: Electric Field |
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A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/m–1. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become
1. zero
2. 5 V m–1
3. 10 V m–1
4 20 V m–1
Subtopic: Electric Field |
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A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be
1. towards the plate
2. away from the plate
3. parallel to the plate
4. zero.
Subtopic: Electric Charge |
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Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius $$10~\mathrm{cm}$$ surrounding the total charge is $$25$$ V-m. The flux over a concentric sphere of radius $$20$$ cm will be:
1. $$25$$ V-m
2. $$50$$ V-m
3. $$100$$ V-m
4. $$200$$ V-m
Subtopic: Gauss's Law |
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The figure(a) shows an imaginary cube of edge $$L/2$$. A uniformly charged rod of length $$L$$ moves towards left at a small but constant speed $$u$$. At $$t=0$$, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in figure(b) represents the flux of the electric field through the cube as the rod goes through it?
1. a
2. b
3. c
4. d
Subtopic: Gauss's Law |
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Hints | 1,400 | 5,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.847897 |
https://blog.csdn.net/weixin_39656513/article/details/116513942 | 1,670,021,915,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00778.warc.gz | 160,871,830 | 22,194 | # matlab数组判断零点,请教一个matlab 求函数零点的问题
function beta_3
c=3*10^8;
lambda=1.55;
rho(1)=1;nc=1;n1=3;l1=0.13;n2=1.5;l2=0.265;
omega=2*pi*c/lambda;
x=fsolve(@(x) myfun(x,omega,l1,l2,n1,n2,nc,c),[0.5,4.0227,12.1507,6.0599,-1.2508-i*0.0461,0.7453-0.1052*i,3.1416-0.6942*i]')
function F=myfun(x,omega,l1,l2,n1,n2,nc,c)
beta=x(1);
kc=x(2);
k1=x(3);
k2=x(4);
kte=x(5);
Ate=x(6);
Bte=x(7);
F(1)=-besselj(kc*1,1)/(besselj(kc*1,0))+i*kc/k1*(exp(i*kte)-Ate-Bte)/(exp(i*kte)-Ate+Bte);
F(2)=kc-sqrt(nc^2*(omega/c)^2-beta^2);
F(3)=k1-sqrt(n1^2*(omega/c)^2-beta^2);
F(4)=k2-sqrt(n2^2*(omega/c)^2-beta^2);
F(5)=Ate-exp(i*k1*l1)*(i*(k1^2+k2^2)/(2*k1*k2)*sin(k2*l2)+cos(k2*l2));
F(6)=Bte-i*exp(-i*k1*11)*(k1^2-k2^2)/(2*k1*k2)*sin(k2*l2);
F(7)=kte+i*log(real(Ate)-sqrt(real(Ate)^2-1));
kc=sqrt(nc^2*(omega/c)^2-beta^2);
k1=sqrt(n1^2*(omega/c)^2-beta^2);
k2=sqrt(n2^2*(omega/c)^2-beta^2);
Ate=exp(i*k1*l1)*(i*(k1^2+k2^2)/(2*k1*k2)*sin(k2*l2)+cos(k2*l2));
Bte=i*exp(-i*k1*11)*(k1^2-k2^2)/(2*k1*k2)*sin(k2*l2);
kte=-i*log(real(Ate)-sqrt(real(Ate)^2-1));
-besselj(kc*1,1)/(besselj(kc*1,0))+i*kc/k1*(exp(i*kte)-Ate-Bte)/(exp(i*kte)-Ate+Bte)=0；
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• 非常有帮助 | 695 | 1,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-49 | latest | en | 0.162784 |
https://www.jiskha.com/display.cgi?id=1195304245 | 1,503,374,092,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00527.warc.gz | 940,966,456 | 4,073 | # English Grammar
posted by .
e.g. The bar graph below shows the TV viewing habits of Koreans.
In this sentence, what is the part of speech of 'below'?
Can we write like this?: The bar graph which is below shows the TV viewing habits of Koreans.
## Similar Questions
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What part of speech is the word "below "used in this sentence. From the ledge can you see below?
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The bar graph below shows the numbers of various colors of cars owned by the employees of a certain business. Write five ratios (expressed as fractions in lowest terms) that compare the least popular color to each of the other colors. …
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7. ### Math Bar graphs
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I need to create a bar graph for a retail store I created that sells clothing for men, women, kids, etc. I already made a circle graph, frequency table, and line graph to show my stores budget and what customers favorite items are, …
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Nathan made a triple bar graph showing the numbers of bronze, silver, and gold medals won by each country in the last Olympics. How can he use the information from this graph to create a bar graph that shows the total number of medals …
More Similar Questions | 517 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-34 | latest | en | 0.902779 |
http://www.diracdelta.co.uk/science/source/h/o/hookes%20law/source.html | 1,492,948,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118552.28/warc/CC-MAIN-20170423031158-00489-ip-10-145-167-34.ec2.internal.warc.gz | 505,300,431 | 4,262 | # Hooke's Law
A Law stating that, in a linear system, the restoring force is proportional to the displacement of the body, acting in a direction as to restore equilibrium.
where
F = restoring force [N]
k = spring constant [Nm-1]
x = elongation of material [m]
A material such as copper, when stretched beyond it's elastic limit it retains it's shape.
A material such as rubber, this does not follow Hooke's law and remains elastic until it snaps.
A material such as glass, this follows Hooke's law until it snaps.
The constant of that proportionality is the Young modulus of elasticity for that substance.
where E = Young's modulus [Nm-2] F = applied load [N] A = cross-sectional area [m2] x = extension [m] L = original length [m] Enter values into 4 of the 5 fields. The others will be calculated.
Stress-Strain curve for a material such as mild steel.
Hooke's Law would apply within the linear region up to the elastic limit of the material.
Subjects: Mechanical Engineering | 231 | 989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-17 | latest | en | 0.93034 |
https://ell.stackexchange.com/questions/76708/how-to-read-1-2-1-and-1-2-1 | 1,713,906,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00856.warc.gz | 207,614,710 | 43,972 | # How to read "1 + (-2) = -1" and "1 - 2 = -1"
How to read this to others, so they can write it down just by listening?
1 + (-2) = -1
Also:
1 - 2 = -1
I want to know the differences.
How you explain this really depends on how much you want to convey, and how much you think will be assumed. For instance, the most specific wording of the first example would go something like this:
One plus open parenthesis negative two close parenthesis equals negative one.
The second example would be as follows:
One minus two equals negative one.
The important difference (pun intended) is how to say the '-' sign. I have found, contrary to the other answers so far, that a '-' when used on only one number (such as on the right side of those equations) is referred to as a 'negative' operator, whereas when used on two numbers it's a 'minus' sign. Saying 'two plus minus one', to me, implies 'plus or minus', which might not apply in most contexts but is definitely worth avoiding. On the other hand, 'two plus negative one' is immediately clear to me.
This even makes the parentheses in your first example unnecessary, and you can just say
One plus negative two equals negative one.
Unless you actually want everyone to write down the parentheses, in which case you should be explicit.
• You woudn't happen to be american, would you? In the rest of the world, we do tend to use "minus" to refer to a negative number. "Negative one" sounds incredibly obtuse to my ear. Dec 23, 2015 at 15:25
• (not to negate the answer, because it is very valuable to touch on the fact that 'negative' can only be used in the unary sense, but 'minus' can be used for both). Dec 23, 2015 at 15:28
• @fabspro Yes i am American. Using 'minus' for everything sounds really confusing to me, especially in text. Maybe with the right emphasis it can be clear what operator you're referring to, but i think using 'negative' avoids that problem altogether. Dec 23, 2015 at 15:39
• From my undergrad experience in math and CS at an american a school, negative one is vastly more common (I dont recall ever hearing "minus one". So it might be regional.) If I heard "plus minus one" I'd think of ±. Dec 23, 2015 at 23:44
• @DaaaahWhoosh it's turned out to be an interesting discussion point - minus versus negative for the unary operator. I think we've both had our eyes opened a little to the regional differences. Dec 27, 2015 at 1:30
As a maths teacher, I would accept any of the following as correct (listing my most preferred reading first):
## 1 + (−2) = −1
One plus negative two equals negative one
One plus minus two equals minus one
One add negative two equals negative one
(Other arrangements of "negative" versus "minus", and "add" versus "plus", are also possible, although I'd be a little bemused if someone said "negative two" and then "minus one"!)
Other readings are, of course, possible, although I would regard something like "the sum of one and minus two is minus one" as an interpretation, or maybe a paraphrase—describing the meaning, rather than reading it as written.
## 1 − 2 = −1
One take two equals negative one
One subtract two equals negative one
One take away two equals minus one
One minus two equals minus one
This last one illustrates the potentially confusing status of "minus" as both "something done to two numbers" and "a type of number" (in grammatical terms, both a verb and an adjective; in mathematical terms, both a binary and a unary operation).
There's no much difference on those examples.
For example the second one is read as one minus two equals minus one. The first one can be read similarly (one plus minus two equals minus one), but having something like 1 - (1 - 2) = 2, that's different, you need to say there's a bracket, and then that's one minus, open bracket, one minus two, close bracket, equals two. You can also use parenthesis.
• Wouldn't you say bracket only for [...]? Dec 23, 2015 at 19:35
• Brackets can cover (),{},[], & <> in my experience. The last are called "angle brackets" and generally appear in mark-up languages. Dec 23, 2015 at 20:46
In the first example, i would personally say
One plus [pause] [quickly] minus two [pause] equals minus one
Depending on my mood, i might say "negative" instead of "minus".
It's shorter than saying aloud "parenthesis/bracket", but still conveys the separation.
I've never had anyone complain about it being hard to understand, at least not with short expressions.
A slightly more complex example to explain:
1 + (2 * 3) = 7
One plus [pause] [quickly] two-times-three [pause] equals seven
But
(1 + 2) * 3 = 9
[quickly] One-plus-two [pause] times three equals nine
You could simply say :
One plus minus two is equal to minus one.
and
One minus two is equal to minus one.
Logically, there is no difference between the two expressions. I'm sure you're aware of that. Generally brackets () are used to avoid confusions with the arithmetic operators. If you want to know more about the mathematical aspects of the expression, you should ask the same questions here. The math geniuses there will certainly help you. In these expressions, you need not specify the brackets as they are mostly used only for written purposes and not used verbally. Additionally, you could swap 'is equal to' with 'equals'.
• I would say "negative two" instead of "minus two"... Dec 23, 2015 at 3:49
• That's a regional thing, I say negative as well. Dec 23, 2015 at 7:03
• FWIW, I'd go with minus (but accept it's regional). Dec 23, 2015 at 10:48
You can say, with emphasis on the bolded part,
One plus minus two inside bracket equals one
and
one minus two, equals minus one
• It's not at all clear what your "inside bracket" applies to. Dec 23, 2015 at 18:58
• well, when we say emphasize on bold one, we mean to stop after 'One plus', and then say the whole phrase together so that other person understand, "minus two inside bracket". And it can't be explained further. Dec 24, 2015 at 7:16
Well, over here (Nigeria) it is quite different as it reads:
1 + (-2) = -1
one plus open-bracket minus 2 close-bracket equal-to minus 1
• That is how you literally read each character aloud. But this how teachers teach in schools and you don't use terms like "open-bracket" when an expression is used in a non-curricular context. Dec 24, 2015 at 6:11
I know this doesn't sound grammatically right.
I remember my Math teacher saying something like
1 + (-2) = -1
One plus of minus two equals minus one
And
1 - 2 = -1
One minus two equals minus one
But then this is Math! It's a wholly different language by itself!
• "One plus of minus two" really does not sound right. Where was this? Dec 24, 2015 at 7:30
• This was an Anglo-Indian teacher who taught Math at my school in India. Dec 24, 2015 at 7:44
the first for me would be the sum of one and negative two is negative one
the second would be: the difference of two from one is negative one
• Please edit to include an explanation of why this is correct; answers without explanation do not teach the patterns of the language well. Dec 23, 2015 at 22:28 | 1,785 | 7,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-18 | latest | en | 0.947594 |
http://mathlab.cit.cornell.edu/local_maple/mvcr7/Lecture1/surfaces_1.html | 1,544,557,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00423.warc.gz | 173,654,683 | 2,093 | surfaces_1.mws
Graphs_ I
Version .75 for MapleVR6
> with(plots): with(plottools):
Warning, the name changecoords has been redefined
Setting Defaults
Set some defaults for all pictures.
A Paraboloid
> gr0 := plot3d(x^2 + 4*y^2,x=-3..3,y=-3..3,view=0..9,style=patchcontour):
> display3d(gr0);
Contours in R^2 for this same function.
> gr1 := contourplot3d(x^2+4*y^2,x=-3..3,y=-3..3):
> display3d(gr1);
Twin mountains
> gr2 := plot3d((x^2+3*y^2)*exp(1-x^2-y^2),x=-2..2,y=-2..2, view = 0..3, style = contour, orientation= [157,64]):
> display3d(gr2);
A Misleading Plot of Contours ?
Note that the contours should have rotational symmetry !
> gr3 := contourplot3d(sin(x^2+y^2),x=-10..10,y=-10..10):
> display3d(gr3);
Rotated Result Looking Down:
> gr4 := plot3d(x^3-3*x*y^2,x=-2..2,y=-2..2,view = -2..2,thickness=3):
> display3d(gr4);
A swallowtail.
> gr5 := plot3d(x^3 - x*y,x=-3..3,y=-3..3,view=-5..5,orientation= [125,76]):
> display3d(gr5);
Implicit Plots
A one-sheeted hyperboloid.
> gr6 := implicitplot3d(x^2+y^2-z^2=4,x=-5..5,y=-5..5,z=-5..5,thickness=1):
> display3d(gr6);
Hyperboloid with Plane Section.
> xz_plane := plot3d([t,0,s],t=-5..5,s=-5..5,grid=[5,5]):
> display3d([xz_plane,gr6]);
rotated version pasted below.
Two sheeted hyperboloid.
> gr8 := implicitplot3d(x^2+y^2-z^2=-4, x=-5..5, y=-5..5, z=-5..5,thickness=3):
> display(xz_plane,gr8);
Hyperbolic Cylinder
> gr9 := implicitplot3d(x^2-z^2 = -4,x=-5..5,z=-5..5,y=-5..5):
> display(gr9);
A Few Interesting Surfaces
> plot3d([sin(u),sin(2*u),v],u=-Pi..Pi,v=-10..10,axes=boxed,style=patchnogrid);
The Whitney umbrella. | 629 | 1,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-51 | latest | en | 0.45471 |
https://www.mariakhalusova.com/posts/2019-04-11-ml-model-evaluation-metrics-p1/ | 1,718,551,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00001.warc.gz | 797,365,901 | 11,035 | ### Machine Learning Model Evaluation Metrics part 1: Classification
If you’re in the beginning of your machine learning journey, you may be taking online courses, reading books on the topic, dabbling with competitions and maybe even starting your own pet projects. If you do, you will inevitably start to notice that there’s more than one way to evaluate a trained model. This is especially apparent in competitions - some companies post a classification challenge where results are evaluated with ROC/AUC score, others with F1 score, log loss, or some other metric. Regression problems, of course, have their own zoo of evaluation metrics. At some point, you gotta ask yourself - why are there so many? What do they mean? How are they different? And, if I start my own project, how do I approach choosing a metric for it? I know I did ask myself those questions, and ended up spending a significant amount of time figuring these things out. So to save others the time and frustration I compiled what I now know about evaluation metrics into a talk called “Machine Learning Model Evaluation Metrics” which I have recently delivered at AnacondaCON in Austin, Texas. If you prefer reading blog posts over watching talks, I intend to publish a couple of blog posts on the topic, and this is the first one.
As a bonus, this format allows me to add more useful links :)
In this first post, I’ll focus on evaluation metrics for classification problems. And to make things a little simpler, I’ll limit myself to binary classification problems. In the next post I’ll talk about how some of these metrics can be extended to a multi-class problem.
Here’s what’s in this post:
### Classification accuracy
Everybody who has ever build a classifier or read about building one has encountered classification accuracy. Classification accuracy (or simply, accuracy) is an evaluation metric that is a portion of correct predictions out of the total number of predictions produced by a ML model:
$$\text{Accuracy} = \frac{\text{Number of correct predictions}} {\text{Total number of predictions}}$$
Accuracy can range from 0 to 1 (or 0 to 100%), and it is easy to intuitively understand. In scikit-learn, all estimators have a score method that gives you a default evaluation metric for your model, and for classifiers it is accuracy.
Here’s an example where I train a basic LogisticRegression on a dataset for a binary problem. Once I did that, I can call the score method on my model, pass the test data and get the accuracy of my model on the test data.
I got almost 96%! Woohoo! This is amazing! Or is it? That’s a trick question of course, because at this point we know too little and it’s impossible to say whether 96% accuracy is “good” or not. In this case, it turns out to be not that great.
To illustrate what I mean, I built a DummyClassifier on the same data. A DummyClassifier in scikit-learn is a classifier that doesn’t learn anything from the data but simply follows a given strategy. It can generate predictions by respecting the training set’s class distribution, or uniformly at random, or, like in my case, it will always return the most frequent label in the training set.
You can use DummyClassifier in the same fashion as any other scikit-learn estimator, and like other classifiers it has a score method that will return its accuracy. So, drumroll pleaseā¦
The DummyClassifier’s accuracy is 94%! This means my LogisticRegression model was only tiny bit better than predicting the most frequent label in the training set which doesn’t seem so good anymore. Why is this happening?
The reason is in the data, and I haven’t shown you my data yet. For this example I used a synthetic dataset with 10 000 samples, where only 5% of them represent samples of the positive class, while 95% of samples are of negative class:
So even simply predicting negative class all the time we’ll get 95% accuracy. If you’re wondering why I got 94% and not 95% with the DummyClassifier, that would have to do with how the data got split into train and test sets.
This type of dataset is what’s called a class-imbalanced dataset, and unfortunately they are quite common. If you don’t know that you are working with a class-imbalanced data, classification accuracy can give you a false impression that your model is doing great when in reality it is not. So first things first - be sure to get to know your data!
Another thing to consider is that even if the data is balanced, if we’d like to improve the model, accuracy alone doesn’t give enough information to diagnose the errors the model is making. Luckily, there are other evaluation metrics and diagnostic tools available for classification models.
### Confusion matrix
One of such tools is called Confusion Matrix. It is a table/matrix that shows how many samples a model classified correctly for what they are and how many were misclassified as something they are not.
Confusion matrix is more of a diagnostic tool rather than a metric - it can help you gain insight into the type of errors are model is making. Plus, there’s a number of evaluation metrics that can be derived from it.
To illustrate Confusion Matrix, I took Titanic dataset from kaggle where the goal is to predict survival of the passengers. Skipping all the data cleanup and feature engineering, here’s a basic RandomForestClassifier I built on the training data.
It has 83.4% accuracy on the test data, which is not too bad according to the Leaderboard :) But I still would like to learn what can be improved, and what the errors look like.
Let’s see what a confusion matrix would look like in this case. First, I need to import confusion_matrix from sklearn.metrics. Now, when calling confusion_matrix it’s important to remember that by convention, you need to provide the actual values first, then the predictions. The same convention is used for other metrics from sklearn.metrics.
Once I call confusion_matrix, I get this beautiful array where the numbers indicate what has been correctly classified and what wasn’t. If you haven’t seen a confusion matrix before, it may be unclear what’s what here.
To help understand it, confusion matrices are typically drawn as tables. I’ll do exactly the same:
Here the rows represent actual values, and the columns stand for predicted. It’s important to note that this is a convention that is used in scikit-learn, and, for example, tensorflow, but there are tools that have it the other way around, so you may see tutorials and articles that have it the other way around (I’m looking at you, Wikipedia!). As if we needed to justify the word “confusion” in “confusion matrix” by constantly confusing what goes on what axis, sigh.
We can learn a lot from the confusion matrix. On the diagonal, you’ll find all the correct predictions:
• True Positives (TP): a sample was positive, and the model predicted a positive class for it.
• True Negatives (TN): a sample was negative, and the model predicted a negative class for it.
We also have two types of errors:
• False Positives (FP): a sample was negative, but the model predicted positive.
• False Negatives (FN): a sample was positive, but the model predicted negative.
If we sum up the correct predictions on the diagonal, and divide by the sum of all cells (all the predictions), we’ll get classification accuracy. But, of course, we can get some other interesting metrics from the confusion matrix.
### Precision, Recall, F1 Score
Suppose we’re building a spam filter, where “spam” is a positive class, or a recommendation system where item being relevant is a positive class. In both cases, to make user experience better we really want to minimize False Positives. False Positive in a case of a spam filter would look like an important email classified as a spam, and that would make users annoyed. False Positive in a recommendation system would mean recommending something irrelevant, which again, can be annoying, and will contribute to negative user experience. How can we evaluate which model will be “better” in this case? One metric that can help is called Precision. Here’s the formula:
$$\text{Precision} = \frac {\text{TP}} {\text{TP} + \text{FP}}$$
We take the number of true positives, and divide it by the sum of true positives and false positives. If we happen to be so lucky that we don’t have any false positives, we’ll get a Precision of 1. And if we had some true positives and some false positives, and managed to reduce the number of false positives, this metric will get a little closer to 1 compared to what it was. If you care about minimizing the number of false positives, keep an eye on Precision.
Now let’s say we’re working on predicting cancer diagnosis. In this case the cost of False Negatives is high. If the model predicts a patient doesn’t have cancer (when in fact they do) and the patient is sent home, it can have much more serious implications compared to sending a healthy person for a few more tests. Or, if, say, we’re classifying fraudulent transactions, a false negative here can be quite costly. In cases like these, if you want to minimize the number of false negatives, the evaluation metric that can help is Recall. The formula is very similar to Precision, but now we take into account the number of false negatives instead of false positives:
$$\text{Recall} = \frac {\text{TP}} {\text{TP} + \text{FN}}$$
The choice of Precision or Recall as an evaluation metric depends largely on the business problem your model is intended to solve.
There’s also an evaluation metric that takes into account both Precision and Recall, and presents another way of summarising a confusion matrix in one number. It’s called F1 score, and it is the harmonic mean of Precision and Recall:
$$\text{F1 score} = \frac {2*\text{Precision}*\text{Recall}} {\text{Precision} + \text{Recall}} = \frac {2*\text{TP}} {2*\text{TP} + \text{FP} + \text{FN}}$$
To get precision, recall and f1 score for a model, you can import these metrics from sklearn.metrics and call them in the same manner as confusion_matrix (actual values first, then predictions):
Tip: Evaluation metrics differ - some you want to maximize, others - to minimize. Scikit-learn has a helpful naming convention. If the name of a metric has the word “score” in it, it’s a metric you want to maximize. If it has words “loss” or “error”, that’s a metric to minimize.
While it may be difficult to compare confusion matrices of different models, you can easily use Precision, Recall and F1 score to choose the best one. For instance, if you want to use GridSearchCV to find hyperparameters that will give you a model with the highest Recall, you can pass recall as a scoring parameter:
### Matthews Correlation Coefficient
Accuracy and F1 Score are not the only ways to sum up a confusion matrix in a single number. There’s also, for example, Matthews Correlation Coefficient:
$$\text{MCC} = \frac {\text{TP}\text{TN} - \text{FP}\text{FN}}{\sqrt{(\text{TP}+\text{FP})(\text{TP}+\text{FN})(\text{TN}+\text{FP})(\text{TN}+\text{FN})}}$$
One important thing to notice about this formula is that MCC takes into account all 4 cells of Confusion Matrix, unlike F1 Score. Indeed, F1 score is based on Precision and Recall, neither of which adds True Negatives to the equation. This has some interesting implications. Let’s see how MCC is different from F1 Score with a couple of examples.
### MCC vs F1 score
First, let’s take the following dataset and model:
• Data: 100 samples, where 95 of them are positive examples and only 5 are negative
• Model: DummyClassifier that always predicts a positive class
Let’s see what a confusion matrix looks like in this case:
Naturally, the accuracy of such classifier will be 95%. Interestingly F1 score is even better - 97.4%:
$$\text{F1 score} = \frac {2*\text{TP}} {2*\text{TP} + \text{FP} + \text{FN}} = \frac {2*95} {2*95 + 5} = \frac {190}{195} = 0.974$$
$$\text{MCC} = \frac {\text{TP}\text{TN} - \text{FP}\text{FN}}{\sqrt{(\text{TP}+\text{FP})(\text{TP}+\text{FN})(\text{TN}+\text{FP})(\text{TN}+\text{FN})}} = \frac {950 - 50} {\sqrt{100955*0}} = \text{undefined}$$
In this example scikit-learn will return 0 with a big red warning that it’s actually undefined, and this is because division by 0 is happening here.
From this example, we can already see that from all the three metrics - accuracy, F1 score and MCC, Matthews Correlation Coefficient was the only one that gave me a red flag that something’s fishy is about this model, which is indeed the case since the model is a DummyClassifier.
Ok, let’s take a less extreme example. This time the data will remain the same, but the model will actually predict something.
$$\text{F1 score} = \frac {2*\text{TP}} {2*\text{TP} + \text{FP} + \text{FN}} = \frac {2*90} {2*90 + 4 + 5} = \frac {180}{189} = 0.952$$
$$\text{MCC} = \frac {\text{TP}\text{TN} - \text{FP}\text{FN}}{\sqrt{(\text{TP}+\text{FP})(\text{TP}+\text{FN})(\text{TN}+\text{FP})(\text{TN}+\text{FN})}} = \frac {901 - 45} {\sqrt{94955*6}} = 0.135$$
Ok, but what if we switch what we call a positive class and a negative class, leaving everything else the same? In this case the confusion matrix will look a bit differently:
The data is the same, the model is the same, the only thing that has changed is what we call “positive” and what we call “negative”.
$$\text{F1 score} = \frac {2*\text{TP}} {2*\text{TP} + \text{FP} + \text{FN}} = \frac {2*1} {2*1 + 4 + 5} = \frac {2}{11} = 0.182$$
$$\text{MCC} = \frac {\text{TP}\text{TN} - \text{FP}\text{FN}}{\sqrt{(\text{TP}+\text{FP})(\text{TP}+\text{FN})(\text{TN}+\text{FP})(\text{TN}+\text{FN})}} = \frac {190 - 54} {\sqrt{6595*94}} = 0.135$$
We get the same MCC but F1 score is suddenly different. Why is that? This happens because F1 score takes into account true positives but not true negatives. This makes it sensitive to what we decide to call a “positive” class and what we refer to as “negative class”. MCC, on the other hand, is not sensitive to it.
So all things considered, I find that MCC is a better way to summarise a confusion matrix in a single number when you have a binary problem. Unfortunately, one downside of MCC is that it’s hard to extend it to multi-class problems.
Up until this point all the metrics I mentioned only took into account whether a model’s prediction was correct or not. But many classification models output a probability of a sample being positive or negative, and there are metrics and diagnostic tools that leverage that.
### ROC Curve/AUC score
One popular way to measure performance of a classifier that outputs probabilities is a ROC curve. ROC stands for Receiver Operating Characteristic which probably doesn’t help much in explaining how it works or what it means.
ROC curve is a plot. On the Y-axis it has the True Positive Rate of a model: out of all samples that were positive, how many the model classified as positive?
On the X-axis it has the False Positive Rate: out of all the samples that were actually negative, how many the model confused for positive?
If you’re asking yourself, why is it a curve and not a dot? That’s a good question! This is where the probabilities as output of a model come to play. When we have a model that outputs a probability of example belonging to a class, there’s always a probability threshold, by default 50%, that serves as a decision boundary for the prediction. If the probability of belonging to class 1 is higher than 50%, then we’ll predict class 1. But we can change this threshold! We can decide to predict that a sample belongs to a positive class only if the probability of that is higher than, say, 80%. But if we do that, the number of TP, FP, TN and FN will change. Why? If we had a sample of positive class that was previously predicted to be positive with only 65% probability, it will now become a false negative.
So what ROC curve represents is all the combinations of True Positive Rate and False Positive Rate as we change the probability threshold. Each dot on the plot corresponds to these rates at a certain probability threshold.
The next reasonable question would be - ok, what would be a “good” curve? To understand this, I find helpful to think of what would ROC curve look like in a perfect case where there’s a threshold that allows the model to split the classes without making mistakes. In this unrealistic scenario, at that threshold, the True Positive Rate will be equal to 1 (as we won’t have any false negatives), while the false positive rate will be 0 (no false positives). As we move the threshold one way or the other, one of those rates will remain the same, but the other will change:
The resulting ROC curve will be “glued” to the upper left corner of the plot. So, in practice, we want the ROC curve to be as close to that corner of the plot as possible.
Now, we could compare different models by plotting their ROC curves, but it still would be more convenient in many cases to have a single number metric for comparison. The evaluation metric that helps in this case is AUC score.
AUC stands for Area Under Curve and it is simply the percentage of the plot box that lies under the ROC curve. The closer it is to 1 (100%), the better.
### Precision/Recall Curve
Another curve that you may see being used is called Precision/Recall curve. The idea is exactly the same - we’ll move the the probability threshold and plot combinations of two metrics, in this case, however, we’ll have Precision on the Y-axis, and Recall on the X-axis. In exactly the same way we can calculate AUC for this curve.
So what’s the difference? I’ll illustrate with an example. Again, I’ll create a class-imbalanced dataset, and train a simple Logistic Regression on.
This time I’ll take the output probabilities, and plot both curves side by side.
I think by now you already got the idea that in case of a class-imbalanced dataset we’ll see two very different pictures from these curves ;)
### Log Loss
Logarithmic Loss aka Log Loss, is another evaluation metric that takes into account the output probabilities. It is commonly used as a loss function. (If you want to learn more about log loss as a loss function, check out this article )
Log Loss quantifies the performance of a classifier by penalising false classifications and taking into account the uncertainty of the predictions which gives you a more nuanced view into the performance of our model compared to accuracy.
In case of a binary problem log loss it is often presented as follows (the equation for multi-class looks a little different):
$$-\frac{1}{n} \sum_{i=1}^{n} (y_i\log p_i + (1 - y_i) \log (1 - p_i) )$$
Here:
• N is the number of observations
• $y_i$ in case of a binary problem ends up being the same as the true label (0 or 1) of the $i_th$ sample
• $p_i$ is the model’s predicted probability that sample i is 1.
Tip: The negative sign in the equation is a matter of convenience. It is easier to compare performance of models using a positive metrics, but the probabilities in the equation are always <1, and the log of numbers < 1 returns negative values. So without the negative sign we would have to deal with a negative metric, which can be confusing to work with.
This is what log loss looks like when a true label is 1.
If the model is confident in predicting 1 (probability of 1 is close to 1), log loss will be quite low. As the predicted probability decreases, however, the log loss increases rapidly. If the predictions are confident and wrong, the penalties will be high.
Minimizing log loss will also get you better classification accuracy, but if you want to make sure that the predictions of your model are not only correct but also confident, then log loss is the metric to go with.
If you made it this far - thank you! These are, of course, not the only existing evaluation metrics for classification problems, there’s more! But I find that these are essential to understand. | 4,644 | 19,970 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.933678 |
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If the marginal cost of production is \$3.25, what is the profit-maximising price? Now suppose that after you have determined the profit-maximising price, your manager tells you that dropping... | 601 | 2,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-47 | latest | en | 0.903228 |
http://aboutcivil.org/flow-characteristics-triangular-notch.html | 1,495,466,520,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463605188.47/warc/CC-MAIN-20170522151715-20170522171715-00017.warc.gz | 8,596,601 | 5,980 | hydraulic jump experiment, hydraulic jump lab experiment, procedure of laboratory experiment on hydraulic jump
# An Experiment on Hydraulic Jump
## Objectives of the experiments:
1. To create the hydraulic jump.
2. To verify the questions of fluid flow.
3. To determine the slatrility & characteristics of the hydraulic jump obtained in the lab using Impulse momentum & specific energy equations.
4. To compare measured flow depths with theoretical results.
## Theoretical background:
Hydraulic jumps are very efficient in dissipating the energy of the flow to make it more controllable & les erosive. In engineering practice, the hydraulic jump frequently appears downstream from overflow structures (spillways), or under flow structures (slvice gates), where velocities are height.
A hydraulic jump is formed when liquid at high velocity discharges into a zone of lower velocity only if the 3 independent velocities (y1, y2, fr1) of the hydraulic jump equation conform to the following equation:
Y2 = y1/2 [-1+√1+8Fr2 ]
Fr2 = 92/9y3
## Apparatus:
• Glass walled flume with sluice gates & a spillway arrangement
• Point gauges
• Manometer & scales
• Pump
## Procedure for Hydraulic Jump Experiment:
1. I started the pump to supply water to the flume.
2. Then I closed the tail gate to allow water to accumulate and to develop hydraulic jump.
3. I adjusted the position of the hydraulic jump by adjusting the amount of closure of slvice gate.
4. I then measured the depth of the bed of flume by using a point gauge.
5. In the next step , I measured water surface level before it had crossed the spillway.
6. Then I measured height of spillway & the depth of water over the spillway.
7. Using the point gauges I then determined the water surface levels downstream of the jump.
8. Then I measured y1 & y2.
9. I repeated the measurement steps again for a different flowchart.
## Results:
S.No Hm(m) Y1(mm) Y2(mm) Lj(m) H(mn) H1(mn) H2 1 0.8 342 46 2 6.5 24 106 0.45
## Sources of errors:
### Human errors:
1. Errors occurred during measurements i.e. by taking erroneous reading of depths or in micrometer.
2. Errors occurred in operation of slvice gates.
### Instrumentation error:
• Leakage from the flume
• Assumptions of ideal conditions did not prevail:
• Ideal conditions which prevailed in the theoretical equations were not there and frictional forces also had some effect on the experiment.
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Help with Module Inverses
0
97
1
What is $$3^{-1}+3^{-2}$$ (mod 25)? Express your answer as an integer from 0-24 inclusive.
Feb 19, 2021
#1
+25658
+2
What is $$3^{-1}+3^{-2} \pmod {25}$$?
My attempt:
$$\begin{array}{|rcll|} \hline && 3^{-1}+3^{-2} \pmod {25} \\ &\equiv& 3^{-1}+3^{-1}3^{-1} \pmod {25} \\ &\mathbf{\equiv}& \mathbf{3^{-1}\left( 1+3^{-1}\right) \pmod {25}} \\ \hline \end{array}$$
$$3^{-1} \pmod {25} = \ ?$$
$$\begin{array}{|rcll|} \hline 3^{-1} \pmod {25} &\equiv& \dfrac{1}{3} \pmod {25} \\ &\equiv& 3^{\phi(25)-1} \pmod {25} \quad | \quad \phi(25)=20 \\ &\equiv& 3^{20-1} \pmod {25} \\ &\equiv& 3^{19} \pmod {25} \\ &\equiv& 3^{3*6+1} \pmod {25} \\ &\equiv& 3^{3*6}*3 \pmod {25} \\ &\equiv& \left(3^3\right)^6*3 \pmod {25} \quad | \quad 3^3 =27 \equiv 2 \pmod{25} \\ &\equiv& 2^6*3 \pmod {25} \\ &\equiv& 192 \pmod {25} \\ \mathbf{3^{-1} \pmod {25}} &\equiv& \mathbf{17 \pmod {25}} \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline 3^{-1}+3^{-2} \pmod {25} &\mathbf{\equiv}& \mathbf{3^{-1}\left( 1+3^{-1}\right) \pmod {25}} \quad | \quad \mathbf{3^{-1} \pmod {25}} \equiv \mathbf{17 \pmod {25}} \\ &\equiv& 17 ( 1+17 ) \pmod {25} \\ &\equiv& 17*18 \pmod {25} \\ &\equiv& 306 \pmod {25} \\ \mathbf{3^{-1}+3^{-2} \pmod {25}} &\equiv& \mathbf{6 \pmod {25}} \\ \hline \end{array}$$
Feb 19, 2021 | 665 | 1,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-17 | latest | en | 0.171116 |
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Printable View
• May 19th 2007, 05:20 PM
red55
functions
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2
1+3
calculate
i) f(2)
ii)g(-3)
iii) fog(2)
write down an expression giving
iv) g o f (x)
v)f-1(x)
again thanks for any help
• May 19th 2007, 10:46 PM
Glaysher
Quote:
Originally Posted by red55
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2/1+3
Sure you've typed g correctly?
• May 25th 2007, 10:42 PM
ginafara
Quote:
Originally Posted by red55
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2
1+3
calculate
i) f(2)
ii)g(-3)
iii) fog(2)
write down an expression giving
iv) g o f (x)
v)f-1(x)
again thanks for any help
i) f(2) where 2 is your x value... this is the function f(x) evaluated at 2 so you plug in 2 where ever you have an x in your function
f(2) = 10-3(2) = 4
ii) Unforutnately your g function is not reasonable, I think there is an error...
but g(-3) would be done the same way, substitute x with -3 because you are evaluating the function g(x) at -3
iii) the composite is not as scary as it seems...(well depending on your functions anyway...)
f(g(x)) means that the functions f(x) is evaluated at g(x) so substitute x in f(x) with g(x)...
you will have
f(g(x)) = 10 - 3(g(x))
iv) g compositite f is the same as iii), but instead you use f(x) as your x-value and plug into all x's in g(x)
v) f-1 is f inverse.
Take your function,
f(x) = 10 - 3x Solve for x
f(x)= 10 - 3x ---> move the 3x to the left side of the = sign and the f(x) to the right of the = sign
10 - f(x) = 3x ----> get x by itself by dividing both sides by 3
(10 - f(x))/3 = x -----> no that you have this just swap x with f(x)
(10 - x)/3 = f(x) ------> this is your inverse function
I hope that helps
• May 26th 2007, 12:16 AM
earboth
Quote:
Originally Posted by red55
am having some problems with these equations
given tha f and g are functions where the target set and source sets are the set of real numbers and
f(x) = 10-3x
g(x) = 2
1+3
calculate
i) f(2)
ii)g(-3)
iii) fog(2)
write down an expression giving
iv) g o f (x)
v)f-1(x)
again thanks for any help
Hello,
I assume that the function g reads:
$\displaystyle g(x) = \frac{2}{x+3}$
If so you'll get at:
ii) $\displaystyle g(-3) \notin \mathbb{R}$ because you have to divide by zero.
iii) $\displaystyle f \circ g(2)=10 - 3 \cdot \left( \frac{2}{2+3} \right) = 10 -\frac{6}{5} = \frac{44}{5}$
iv) $\displaystyle g \circ f(x) = g(f(x))=\frac{2}{10 - 3x +3} = \frac{2}{13-3x}$ | 948 | 2,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-17 | latest | en | 0.903205 |
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West February 25, 2011 22:34
Newbie Problems
Hello guys, my first time here, so I'm sorry if my questions are to stupid..
I have to run a simulation to find the drag coefficient of various types of tennis balls considering the fluffs and the lines they have.. something like this:
I tried to make the geometry on ANSYS Design Modeler, but all i could make was a sphere. What i have to do to make those lines on the sphere? And can make the fluff on this program, or i have to use another one. I hope someone can help me. Thanks.
PSYMN February 26, 2011 07:51
Surface roughness...
Making the lines will be an interesting challenge for any CAD tool... Sorry, no idea.
But for the fluff, I think you want to use a surface roughness boundary condition (rather than actually modeling the fluff). Separate the surface of your smooth sphere into "fluf" and "lines" boundary conditions and set different surface roughness so the solver handles the boundary conditions differently.
Keep us posted, I am interested to hear your final solution to this tricky CFD problem.
West February 26, 2011 15:47
Thanks for the attention in first place.
I understood, but to set the roughness like you said i have to delimitate the lines on the sphere, don't I? And I don't know how to do so..
And which command I use to set the roughness of the surface?
Sorry if it's too stupid.
Thanks.
PSYMN February 26, 2011 18:54
Right, you would first need to draw your boundaries on the sphere so you can break it up into separate boundary conditions... Sorry, I don't know how to do that.
Once you have them as separate parts, you can mesh and export to the solver. Then in the solver you set those as viscous walls and adjust the setting for surface roughness (under boundary conditions => Edit).
West April 26, 2011 20:03
I made the geometry using Solidworks:
http://img96.imageshack.us/i/semttulowyq.jpg/
but now i have another problem, i can't import it to the ANSYS Design Modeler. I get an error that says:
Attach Failed"
Can somebody help me?
http://img96.imageshack.us/i/semttulowyq.jpg/
PSYMN April 27, 2011 06:04
But I can tell you that CAD interfaces require royalties to the various CAD vendors and are therefore usually sold separately. It appears to be saying that you do not have a Solidworks CAD interface. (or perhaps your license is just used up or not running).
If you agree that you don't have a Solidworks CAD interface, you could look into getting one or output in a different format (such as ACIS (SAT)).
Simon
West June 18, 2011 13:55
Hello again..
Got help from a friend and I did simulate it like I wanted. But, now I have to calculate the drag coefficient of the ball, and I'm having some problems doing that. Can anybody help me? Thanks..
PSYMN June 18, 2011 20:05
That depends on your solver, some have "Drag" listed... In others you can just sum the force in a particular direction (such as X) over the entire surface...
West August 18, 2011 19:42
Quote:
Originally Posted by PSYMN (Post 312592) That depends on your solver, some have "Drag" listed... In others you can just sum the force in a particular direction (such as X) over the entire surface...
Humm, I don't know if I'm doing it right, cause i'm getting the ball's drag coefficient 0.21, and that's half the expected..
this is my expression on ANSYS, is it right?
Cd = -2*(force_z()@Ball + force_z()@Lines)/(1.225[kg/m^3]*(1[m/s])^2*(0.003201928904[m^2]))
PSYMN August 18, 2011 23:40
It is late and I don't want to dig out a book to really confirm the expression, but it looks ok (assuming that "Ball" and "Lines" are the two surface components of the tennis ball and you are just summing up the forces)...
However, It could be a number of things.
Perhaps your ball really does have less drag due to the fuzzy surface you are modeling, or maybe the ridges break up the flow?
Perhaps you don't have these surface treatments right? Or perhaps other bocos or properties are wrong?
Or perhaps your units are not right?
Have you tried modeling it as a smooth sphere (just change the wall condition), does that give you what you expect? Have you checked the expression in a book or asked about it on the solver forum?
West August 19, 2011 00:24
Thankyou for the fast reply at first..
I did chech the expression and its correct, as the units..
I'll try to do what you said to the smooth ball, but what i'm trying right now is to raise the flow velocity, which is 1m/s right now.. i guess the reynolds number is too low to apply that formula, and maybe raising the velocity it will help.. can u please detail what u mean by this:
"Perhaps you don't have these surface treatments right? Or perhaps other bocos or properties are wrong?"
the only thing i did to the surface was apply the condition "Rough Wall" and set the Roughness Height (or something like that) to 0.65mm (which i estimated as the height of the fluff on the tennis ball) i don't know if there was any other thing to do... so..
thank u for your help, again
PSYMN August 20, 2011 23:51
New territory
Right, I just meant that maybe with that surface roughness, you get that Cd...
Perhaps it is not the right way to simulate the fuzz accurately. Perhaps it is not really intended (or calibrated) for the fuzzy nature of a tennis ball and would be better suited to a more granular solid surface roughness? I am not sure, but I just mean that a Cd being off doesn't mean your expression is wrong, there could be many factors. Maybe your Cd expectation is wrong? It is hard to tell.
Trying a smooth ball (more predictable and backed up with a lot of experimental and simulation data) will give you some confidence in your base level and then you can move forward. you could also check if anyone else has done any work with fuzz and see if they say what sort of boco they used to simulate it... It would be great if someone has already done some physical testing to calibrate the CFD... Or maybe that is what you will end up doing... (Adjust the surface roughness until it meets experiment over a range of velocities and conditions and then write a paper on that...)
West August 20, 2011 23:54
Thanks man, I'l try what you said.
PSYMN August 21, 2011 00:03
One other thought...
Any time you are dealing with a ball in a flow, it is likely an unsteady solution...
If you are doing a steady state solve, you can pretty much bet that you will get a different answer every time you run the solver... This also makes it very difficult on convergence, etc.
Make sure that you are not just trying to solve this with a steady state solution (as you might for an airfoil).
Simon | 1,692 | 6,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2016-40 | latest | en | 0.940503 |
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## Recommended Posts
Does anyone know what the limits of the turn rate in the turn coordinator in the C172 are? In theory books I found that it's 20deg/sec which is rather large (ie 1200deg/min). If this is linear (and since the needle is on a spring this is reasonable) my miniature plane rotates about 130 degrees max, which is a lot. The limit on the default gauge is about 40 degrees in comparison, so the miniature plane wings don't go over the balance ball cylinder, but (assuming a linear relationship) this corresponds to only 360 deg/min. Anyone knows how this works in real life?Cheers, Christian
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Should only be 180 degrees a minute. That is why putting the wings onto the dash on either side results in a "Two-minute turn"
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Sorry, I mustn't been very clear. I know the turn rate at the lines is 180 deg/min. I was wondering how much the maximum turn rate is that indicator will show (ie how much will the miniature plane move past the 180 deg/min lines)?Cheers, Christian
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If I understand you correctly Christian, you want to know the maximum angle of bank that the little airplane silhouette in the turn coordinator will display. This can vary by instrument (and keep in mind that I'm not a certificated instrument repairman); will be limited by internal stops in the instrument's dashpot. These stops are necessary to avoid stretching and possibly damaging the calibration spring.40 degrees is a very reasonable limit for the indicator, as I'm sure you're aware the turn coordinator is only meant to display rate information; rate of roll, and rate of turn once the roll rate is zero and the aircraft is established in the bank. I've never heard about the 20 degree per second limit (or is that the maximum rate?) for turn coordinators. That
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Thanks, Douglas.Yes, that was my question. I guess, you mean 40 deg tilt of the miniature plane. The limit I found in a book was max turn rate of 20 deg/sec which translates to 130 deg tilt of the indicator plane (which obviously is a lot more than 40 deg). The question remains, what does a 40 deg on the indicator translate to. Since a rate of 180 deg/min = 20 deg indicated, 40 deg indicated would be a rate of 360 deg/min, if the relationship is linear (given it's a spring, that's plausable). This doesn't fit with the 20deg/sec (which is 1200deg/min). The 20deg/sec limit stops the gimbal, so it's possible that there is a second limit on the spring?Cheers, Christian
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>>>>The question remains, what does a 40 deg on the indicator translate to.<<<< I'm going with the answer you've already worked out Christian (360), but the real answer is that it doesn't matter because the instrument is only certified to indicate a standard rate turn. The 20-degrees/sec limit you found is probably a theoretical limit on the gyro, you won't be making a procedure turn at that rate in a Skyhawk! And as I said before, internal stops will prevent such a rate from being displayed. To avoid confusion, and for those who are interested in the operation of the turn coordinator - note that the turn coordinator does not indicate the angle of bank with any precision. In a 172 performing a steady level turn at the standard rate (2 minutes) the wing of aircraft silhouette on the index mark will roughly correspond to a bank angle of 20 degrees. However, the instrument is only showing the rate of turn.The turn coordinator's gyro gimbal axis is tilted 30-35 degrees. This causes the instrument to precess when the aircraft rolls as well as when it yaws. This precession is responsible for driving the indicator in both the turn coordinator and the earlier turn and slip instrument. Unlike the earlier turn and slip indicator, which only senses yaw, the turn coordinator is not affected by adverse yaw at the beginning of the turn. Thus the turn coordinator will sense rate of roll as well as rate of turn.Try this experiment in MSFS, using the 172, establish the aircraft in a level turn at the standard rate. Now without changing the bank angle load a little G onto the aircraft with backpressure on the yoke to tighten the rate of turn. DO NOT increase the bank angle, check the bank angle before applying backpressure and maintain it as necessary. Now watch the aircraft silhouette indicate a higher rate of turn by banking further in the direction of turn.Try the same experiment with one of the newer aircraft that have instrument tooltips to indicate rate of turn, see how close you can come to 20 deg/sec.Douglas
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>I'm going with the answer you've already worked out Christian>(360), but the real answer is that it doesn't matter because>the instrument is only certified to indicate a standard rate>turn. ok. many thanks. I'll go with that.>The 20-degrees/sec limit you found is probably a>theoretical limit on the gyro, you won't be making a procedure>turn at that rate in a Skyhawk! I didn't say, I would :-lol. I was just interested in the theoretical possible max. I know that 20 deg/sec is quite a turn :).I'm not a real pilot, but do know a bit about how things work. I certainly wouldn't try to achieve such turn rates in reality. Having said that, I could imagine a situation of high turn (mainly roll) rates, and that's a spin (although I wouldn't rely on the TC in such a case). This could be tried with Rob Youngs C172 airfile, as the default C172 isn't capable of spinning. Anyway, I go with the 40 deg (360deg/min) angle. Anything higher is very unlikely anyway, I was just interested (I'm probably too much of a perfectionist).>Try this experiment in MSFS, using the 172, establish the>aircraft in a level turn at the standard rate. Now without>changing the bank angle load a little G onto the aircraft with>backpressure on the yoke to tighten the rate of turn. DO NOT>increase the bank angle, check the bank angle before applying>backpressure and maintain it as necessary. Now watch the>aircraft silhouette indicate a higher rate of turn by banking>further in the direction of turn.Another good way to show that the TC doesn't show bank angle is to fix a bank angle and vary the speed with the throttle. Note though that higher speed does not necessarily equate to higher turn rate. If my memory serves me well, just above stall speed for the selected bank angle will give best turn rate for that bank angle, and this makes sense if you think about it (faster car = bigger minimum turn radius). Ok, now I'm confusing things. Just above stall speed = tightest turn, but not highest turn rate. I think in most cases turn rate will increase with higher speed settings above stall speed, but then drop again at faster speeds, ie highest turn rate is when speed and turn radius are optimal taken together.Cheers, Christian
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>>>>I'm not a real pilot, but do know a bit about how things work. I certainly wouldn't try to achieve such turn rates in reality. Having said that, I could imagine a situation of high turn (mainly roll) rates, and that's a spin (although I wouldn't rely on the TC in such a case). This could be tried with Rob Youngs C172 airfile, as the default C172 isn't capable of spinning.<<<
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No, they go for about \$365.00, just checked yesterday. :D
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>No, they go for about \$365.00, just checked yesterday. :D>We're resurrecting a 2 1/2 year old thread here, but the \$365.00 units are the cheapo Falcon types. The model I have is \$879.00 & some run around a thousand. Mine has been replaced with an auto-pilot in the 3 1/8" hole, that includes an electronic turn coordinator and has a built in "ball". I need to sell it.L.Adamson
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While this is indeed an old thread, interesting read.Some misconceptions throughout it regarding the turn coordinator and some basic aerodynamic principles. ;)Rate of turn is based on:(PI()*velocity)/(RADIANS(180)*9.8*TAN(RADIANS(bank angle)))Radius of turn is based on:POWER(velocity,2)/(9.8*TAN(RADIANS(bank angle)))Rate of turn is proportional to velocity while radius is proportional to velocity squared.
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Master 2020/2021
# Formal Methods in Software Engineering
Type: Compulsory course (System and Software Engineering)
Area of studies: Software Engineering
When: 1 year, 1-4 module
Mode of studies: offline
Master’s programme: Software and Systems Engineering
Language: English
ECTS credits: 10
Contact hours: 128
### Course Syllabus
#### Abstract
In computer science and software engineering, formal methods are a particular kind of mathematically based techniques for the specification, development and verification of software and hardware systems. The use of formal methods for software and hardware design is motivated by the fact that, as in other engineering disciplines, performing appropriate mathematical analysis can contribute to the reliability and robustness of a design. Formal methods are best described as the application of a fairly broad variety of theoretical computer science fundamentals, in particular logic calculi, formal languages, automata theory, and program semantics, but also type systems and algebraic data types to problems in software and hardware specification and verification. Formal methods can: (a) be a foundation for describing complex systems; (b) be a foundation for reasoning about systems; (c) provide support for program development. In contrast to other system design approaches, formal methods use mathematical proof as a complement to system testing in order to ensure correct behavior. As systems become more complicated, and safety becomes a more important issue, the formal approach to system design offers another level of insurance. Formal methods differ from other design methods by the use of formal verification schemes, the basic principles of the system must be proven correct before they are accepted. Traditional system design uses extensive testing to verify behavior, but testing is capable of only finite conclusions. E. Dijkstra and others have demonstrated that tests can help to find fails and bugs, but cannot guarantee their absence. In contrast, once a theorem is proven true it remains true. It is very important to note that formal verification does not obviate the need for testing. Formal verification cannot fix bad assumptions in the design, but it can help identify errors in reasoning which would otherwise be left unverified. In several cases, engineers have reported finding flaws in systems once they reviewed their designs formally.
#### Learning Objectives
• To learn basic principles of using formal methods for the specification and analysis of software systems
• To learn how to specify the formal semantics of sequential and concurrent systems
• To learn how to use formalisms, such as process algebras and Petri nets, and corresponding methods for modeling and analyzing concurrent and distributed systems
• To learn how to use methods and algorithms for model checking of concurrent systems
• To master methods and tools of software specification, analysis and verification
#### Expected Learning Outcomes
• To model various classes of distributed systems within appropriate formalisms
• To model various classes of distributed systems within appropriate formalisms;
• To interpret and apply the formal languages of the formalisms for modeling distributed systems
• To apply specific techniques for the analysis and verification of distributed systems;
• To formulate and prove properties of distributed systems within studied formalisms
#### Course Contents
• Formal methods as a basis for software reliability
• Floyd method for verification of sequential programs. Hoare axiomatic semantics for sequential and parallel programs.
• Finite state machines (FSMs): basic definitions, operational semantics. Categories of FSMs. Extended FSMs. Modeling concurrent systems with communicating FSMs.
• Petri nets: basic notions, definitions and classification. Modeling distributed systems with Petri nets.
• Petri nets analysis. Checking structural and behavioral properties.
• High-level Petri nets. Colored Petri nets and CPNTools.
• Modeling distributed and concurrent system with process algebras. Structured operational semantics and its formalization (SOS). Algebra CCS: syntax, semantics, modeling technique.
• The notion and properties of bisimilarity relation.
• Verifying reactive concurrent systems with CCS. Hennesy-Milner logic and temporal properties. The notion of fixed point and Tarski’s fixed point theorem
• Transition systems and program graphs. Nondeterminism, parallelism and communication. Peterson algorithm
• Specifying distributed systems with Promela. Spin model checker.
• Temporal logics LTL and CTL for specification of behavioral properties of reactive systems.
• Automata-based approach for verification of LTL formulae.
• Model checking algorithm for verification of CTL formulae
#### Assessment Elements
• O_accum1
• O_exam1
• O_current2
• O_exam2
Письменный экзамен с помощью MS Teams / MS Forms - студенты авторизуются по студенческой почте в домене edu.hse.ru Технические требования: web-камера, микрофон, наушники / колонки
#### Interim Assessment
• Interim assessment (2 module)
0.5 * O_accum1 + 0.5 * O_exam1
• Interim assessment (4 module)
0.12 * Interim assessment (2 module) + 0.3 * O_current2 + 0.4 * O_exam2 + 0.18 * O_htask
#### Recommended Core Bibliography
• Baeten, J. C. M., Reniers, M. A., & Basten, T. G. H. (2010). Process Algebra: Equational Theories of Communicating Processes. Cambridge: Cambridge University Press. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=317653
• Baier, C., & Katoen, J.-P. (2008). Principles of Model Checking. Cambridge, Mass: The MIT Press. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=226091
• Fisher, M. (2011). An Introduction to Practical Formal Methods Using Temporal Logic. Chichester, West Sussex, U.K.: Wiley. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=509889
• Jensen, K., & Kristensen, L. M. (2009). Coloured Petri Nets : Modelling and Validation of Concurrent Systems. Dordrecht: Springer. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=285464 | 1,377 | 6,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.902428 |
https://www.edplace.com/worksheet_info/maths/keystage1/year1/topic/944/105/find-the-numbers-hidden-in-the-wordsearch-1-10 | 1,579,628,056,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00162.warc.gz | 850,685,045 | 29,714 | # Find the Numbers Hidden in the Wordsearch (1-10)
In this worksheet, students must find all the numbers hidden in five different wordsearches.
Key stage: KS 1
Curriculum topic: Number: Number and Place Value
Curriculum subtopic: Read and Write Number Words to 20
Difficulty level:
### QUESTION 1 of 10
In a word search, you have to try to find words hidden in a grid.
p h c t e n i a u e n b y o d o s i x c t h r e e v c w s r v d q a s e v e n h c v v q m f o u r o p f i v e q d a t z k z p b t e i g h t t y j l a n i n e o t w o b d d v c h f x h v d m o n e u v
In this word search you can see the number words:
10 ten 6 six 3 three 7 seven 4 four 5 five 8 eight 9 nine 2 two 1 one
Now it is your turn to find all the number words.
There are nine number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
f u a z n n i n e z j v s l f o u r w p z c y g s e v e n b u a h s i x a r r k j v z l o r e m x h w o e i g h t l t k t e n w v w n g i x i i y t h r e e z k f i v e v i u f v v a r w t w o f c e n
ten
eight
three
seven
four
one
five
nine
six
two
There are eight number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
p h c t e n i a u e n b y o d o s i x c t h r e e v c w s r v d q a s e v e n h c v v q m g o u r o p f i v e q d a t z k z p b t e i g h t t y j l a n i n l o t w o b d d v c h f x h v d m o n e u v
one
two
three
four
five
six
seven
eight
nine
ten
There are ten number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
z y i y o s e v e n t e n n n e x f j f u s g b z q l o n e r n i n e n g l k w c s i x h x q m x k e m e e i g h t d u q w t h r e e a s r k x r t w o g y j k o e i t h q f i v e j s c w h d f o u r
six
four
ten
eight
three
one
five
nine
two
seven
There are seven number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
w a q q x s i x r a h n v p i y t e n k o w t w o p q g g m x d f d h f o u r c t r e r s s e f e n l n e t l c z y j w o a j e i g h t n l s t h o e e y m j c f v s i f i v e s g c b n i n e g f b y
two
seven
ten
one
six
three
nine
eight
four
five
You're nearly there!
Search this final grid for the last numbers and tick the boxes of the ones you find.
u c v x k t w o m y r s r e i g h t g c a s j j r i h o n e l z h s h v f o u r d o r j f i v e t q d z t e n a c j y o f g p a m t h r e e k s e v e n i g e m l y o m a x h i x p w q h r d n i n e u
nine
five
seven
one
ten
three
six
two
eight
four
• Question 1
There are nine number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
f u a z n n i n e z j v s l f o u r w p z c y g s e v e n b u a h s i x a r r k j v z l o r e m x h w o e i g h t l t k t e n w v w n g i x i i y t h r e e z k f i v e v i u f v v a r w t w o f c e n
ten
eight
three
seven
four
five
nine
six
two
EDDIE SAYS
f u a z n n i n e z j v s l f o u r w p z c y g s e v e n b u a h s i x a r r k j v z l o r e m x h w o e i g h t l t k t e n w v w n g i x i i y t h r e e z k f i v e v i u f v v a r w t w o f c e n
• Question 2
There are eight number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
p h c t e n i a u e n b y o d o s i x c t h r e e v c w s r v d q a s e v e n h c v v q m g o u r o p f i v e q d a t z k z p b t e i g h t t y j l a n i n l o t w o b d d v c h f x h v d m o n e u v
one
two
three
five
six
seven
eight
ten
EDDIE SAYS
p h c t e n i a u e n b y o d o s i x c t h r e e v c w s r v d q a s e v e n h c v v q m g o u r o p f i v e q d a t z k z p b t e i g h t t y j l a n i n l o t w o b d d v c h f x h v d m o n e u v
• Question 3
There are ten number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
z y i y o s e v e n t e n n n e x f j f u s g b z q l o n e r n i n e n g l k w c s i x h x q m x k e m e e i g h t d u q w t h r e e a s r k x r t w o g y j k o e i t h q f i v e j s c w h d f o u r
six
four
ten
eight
three
one
five
nine
two
seven
EDDIE SAYS
z y i y o s e v e n t e n n n e x f j f u s g b z q l o n e r n i n e n g l k w c s i x h x q m x k e m e e i g h t d u q w t h r e e a s r k x r t w o g y j k o e i t h q f i v e j s c w h d f o u r
• Question 4
There are seven number words hidden in this wordsearch.
Can you find them?
When you have found them, tick the boxes below.
w a q q x s i x r a h n v p i y t e n k o w t w o p q g g m x d f d h f o u r c t r e r s s e f e n l n e t l c z y j w o a j e i g h t n l s t h o e e y m j c f v s i f i v e s g c b n i n e g f b y
two
ten
six
nine
eight
four
five
EDDIE SAYS
w a q q x s i x r a h n v p i y t e n k o w t w o p q g g m x d f d h f o u r c t r e r s s e f e n l n e t l c z y j w o a j e i g h t n l s t h o e e y m j c f v s i f i v e s g c b n i n e g f b y
• Question 5
You're nearly there!
Search this final grid for the last numbers and tick the boxes of the ones you find.
u c v x k t w o m y r s r e i g h t g c a s j j r i h o n e l z h s h v f o u r d o r j f i v e t q d z t e n a c j y o f g p a m t h r e e k s e v e n i g e m l y o m a x h i x p w q h r d n i n e u
nine
five
seven
one
ten
three
two
eight
four
EDDIE SAYS
u c v x k t w o m y r s r e i g h t g c a s j j r i h o n e l z h s h v f o u r d o r j f i v e t q d z t e n a c j y o f g p a m t h r e e k s e v e n i g e m l y o m a x h i x p w q h r d n i n e u
---- OR ----
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Get started | 2,372 | 6,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-05 | longest | en | 0.727202 |
https://wifusion.org/and-pdf/1232-elasticity-of-demand-and-supply-formula-pdf-471-735.php | 1,643,316,143,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00163.warc.gz | 655,407,989 | 9,167 | # Elasticity Of Demand And Supply Formula Pdf
File Name: elasticity of demand and supply formula .zip
Size: 27405Kb
Published: 12.12.2020
Actively scan device characteristics for identification. Use precise geolocation data.
Both the demand and supply curve show the relationship between price and the number of units demanded or supplied. Price elasticity is the ratio between the percentage change in the quantity demanded Qd or supplied Qs and the corresponding percent change in price. The price elasticity of demand is the percentage change in the quantity demanded of a good or service divided by the percentage change in the price.
## Elasticity of Demand and Supply (With Diagram)
Both the demand and supply curve show the relationship between price and the number of units demanded or supplied. Price elasticity is the ratio between the percentage change in the quantity demanded Qd or supplied Qs and the corresponding percent change in price. The price elasticity of demand is the percentage change in the quantity demanded of a good or service divided by the percentage change in the price. The price elasticity of supply is the percentage change in quantity supplied divided by the percentage change in price. We can usefully divide elasticities into three broad categories: elastic, inelastic, and unitary. Because price and quantity demanded move in opposite directions, price elasticity of demand is always a negative number. Therefore, price elasticity of demand is usually reported as its absolute value, without a negative sign.
## Elasticity of Demand and Supply (With Diagram)
A good's price elasticity of demand is a measure of how sensitive the quantity demanded of it is to its price. When the price rises, quantity demanded falls for almost any good, but it falls more for some than for others. The price elasticity gives the percentage change in quantity demanded when there is a one percent increase in price, holding everything else constant. If the elasticity is -2, that means a one percent price rise leads to a two percent decline in quantity demanded. Other elasticities measure how the quantity demanded changes with other variables e.
The midpoint formula is preferable when calculating the price elasticity of demand because it gives the same answer regardless of the direction of the change.
## Price elasticity of demand and price elasticity of supply
Read this article to learn about Elasticity of Demand and Supply: — 1. Subject Matter of Elasticity of Demand and Supply 2. Meaning of Price Elasticity of Demand 3. Different Kinds of Price Elasticities 4.
Both the demand and supply curve show the relationship between price and the number of units demanded or supplied. Price elasticity is the ratio between the percentage change in the quantity demanded Qd or supplied Qs and the corresponding percent change in price. The price elasticity of demand is the percentage change in the quantity demanded of a good or service divided by the percentage change in the price. The price elasticity of supply is the percentage change in quantity supplied divided by the percentage change in price.
Price elasticity measures the relationship between the supply and demand of a commodity and its price. Stay on top of your accounting and finances with Debitoor invoicing software. Sign up now and try Debitoor free for seven days.
### Examples of demand elasticity other than price elasticity of demand
If you're seeing this message, it means we're having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Login Sign up Search for courses, skills, and videos. Economics Microeconomics Elasticity Price elasticity of demand. Introduction to price elasticity of demand.
The price elasticity of supply is the measure of the responsiveness in quantity supplied to a change in price for a specific good. In economics, elasticity is a summary measure of how the supply or demand of a particular good is influenced by changes in price. Elasticity is defined as a proportionate change in one variable over the proportionate change in another variable:. There are numerous factors that directly impact the elasticity of supply for a good including stock, time period, availability of substitutes, and spare capacity.
How do I calculate price elasticity of demand? Price elasticity of demand = % change in price elasticity of supply is? demand formula which is.
#### Price Elasticity of Demand
ТРАНСТЕКСТ, подобно всем великим технологическим достижениям, появился на свет в силу необходимости. В 1980-е годы АНБ стало свидетелем революции в сфере телекоммуникаций, которой было суждено навсегда изменить весь мир разведывательной деятельности, - имеется в виду широкая доступность Интернета, а если говорить конкретнее - появление электронной почты. Преступники, террористы и шпионы, которым надоело прослушивание их телефонов, с радостью встретили это новое средство глобальной коммуникации. Электронная почта соединила безопасность обычной почты со скоростью телефонной связи. С тех пор как сообщения стали передаваться по подземным волоконно-оптическим линиям, а не с помощью радиоволн, они оказались полностью защищенными от перехвата - таков по крайней мере был замысел. В действительности перехват электронных писем, передвигаемых по Интернету, был детской забавой для технических гуру из АНБ. Интернет не был создан, как считали многие, в эру домашних персональных компьютеров.
Стратмор покачал головой: - Танкадо дал нам шанс. Это совершенно ясно. Тем не менее риск велик: если нас обнаружат, это, в сущности, будет означать, что он своим алгоритмом нас напугал. Нам придется публично признать не только то, что мы имеем ТРАНСТЕКСТ, но и то, что Цифровая крепость неприступна. - Каким временем мы располагаем.
Бринкерхофф ухмыльнулся. Деньги налогоплательщиков в действии. Когда он начал просматривать отчет и проверять ежедневную СЦР, в голове у него вдруг возник образ Кармен, обмазывающей себя медом и посыпающей сахарной пудрой.
У нас имеется пять уровней защиты, - объяснял Джабба. - Главный бастион, два набора пакетных фильтров для Протокола передачи файлов, Х-одиннадцать, туннельный блок и, наконец, окно авторизации справа от проекта Трюфель. Внешний щит, исчезающий на наших глазах, - открытый главный компьютер. Этот щит практически взломан.
- Блоки из четырех знаков, ну прямо ЭНИГМА. Директор понимающе кивнул. ЭНИГМА, это двенадцатитонное чудовище нацистов, была самой известной в истории шифровальной машиной. Там тоже были группы из четырех знаков. - Потрясающе, - страдальчески сказал директор.
Беккер шел, низко опустив голову. Халохот прокручивал в голове дальнейшие события. Все было очень просто: подойдя к жертве вплотную, нужно низко держать револьвер, чтобы никто не заметил, сделать два выстрела в спину, Беккер начнет падать, Халохот подхватит его и оттащит к скамье, как друга, которому вдруг стало плохо. Затем он быстро побежит в заднюю часть собора, словно бы за помощью, и в возникшей неразберихе исчезнет прежде, чем люди поймут, что произошло. Пять человек.
Беккер изумился. - Un punqui. - Si. Punqui. - Панк.
Росио покачала головой: - Это. Но вам ее не найти. Севилья - город большой и очень обманчивый.
Офицер был поражен этим открытием. - Кольцо? - Он вдруг забеспокоился. Вгляделся в полоску на пальце и пристыжено покраснел. - О Боже, - хмыкнул он, - значит, эта история подтверждается.
Получить ключ было необходимо, но Стратмор отлично понимал, что посылать глухого киллера в севильский морг было бы настоящим самоубийством. И тогда он стал искать иные возможности. Так начал обретать форму второй план. Стратмор вдруг увидел шанс выиграть на двух фронтах сразу, осуществить две мечты, а не одну.
Где ключ? - потребовала. Хейл с трудом пришел в .
## Sean H.
Arc elasticity of demand arc PED is the value of PED over a range of prices, and can be calculated using the standard formula:.
## Jason N.
The price elasticity of demand PED measures the change in demand for a good in response to a change in price.
## Gallmilreri
Figure shows the price elasticity of demand calculation. The price elasticity of demand is 2. By using the formula, the price elasticity of demand equals | 2,028 | 8,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-05 | latest | en | 0.917851 |
https://stats.stackexchange.com/questions/511734/calculate-the-probability-of-a-variable-on-a-bayesian-network | 1,719,109,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00133.warc.gz | 467,189,329 | 38,151 | Calculate the probability of a variable on a Bayesian Network
Let's assume we have a bayesian nework with discrete variables like shown below. For simplicity, assume all variables are binary.
Assume we know the states for V1 and V2 and we want to calculate the probabilities for V3 being in either of its states.
Is it sufficient to just observe the values in V3's CPT, given that we know the state of its parents or do we have to calculate the joint probabilities over all network variables?
It depends on what probability you want. If you want the marginal $$P(V_3=v_3)$$ then observing only realizations of $$V_3$$ is enough. However, your goal seems to be
given the states for $$V_1$$ and $$V_2$$, what is the probability for $$V_3$$ taking a specific value $$v_3$$.
In Bayesian Networks, one usually computes the kernels $$P(V_i\mid \mathrm{Pa}(V_i))$$ where $$\mathrm{Pa}(V_i)$$ are the parents of the node $$V_i$$. In this case, you need to observe the variable $$V_3$$ jointly with its parents $$\mathrm{Pa}(V_3) = \{V_1, V_2\}$$. This is because in a DAG the local Markov condition allows for the factorization: $$P(V_1, \dots, V_n) = \prod_{i=1}^nP(V_i\mid \mathrm{Pa}(V_i))$$ So it is enough to observe $$V_1, V_2, V_3$$ because of this factorization. You do not need to condition on the descendants or non-descendants of $$V_3$$. In your case the factorization becomes $$P(V_1,V_2,V_3, V_4, V_5) = P(V_1)P(V_2)P(V_3\mid V_1, V_2) P(V_4 \mid V_3) P(V_5\mid V_3)$$ | 444 | 1,478 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-26 | latest | en | 0.875758 |
https://vijayraghavendran.wordpress.com/2016/02/10/world-energy-needs-as-itaipu-dam-equivalents/ | 1,500,727,620,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424060.49/warc/CC-MAIN-20170722122816-20170722142816-00385.warc.gz | 732,003,918 | 46,319 | # World energy needs as Itaipu dam equivalents
Humans have an intuitive sense for numbers. We know that 100 is smaller than 1000 and that a 4GB memory card stores less data than a 16GB one. Bearing this in mind, it was decided to approach the energy needs of Brazil and the world. Nowadays, it is very common to see a pie chart or a bar graph depicting various forms of energy consumption, in newspapers, televisions and magazines. One often wonders if these numbers makes any sense to a layperson. Is it possible to build on their numeracy skills, to further understand, say, their country’s needs?
A survey was made with a set of questions, based on individual electricity consumption, using Google forms. By taking an average of the upper and lower guestimate for the country’s electricity consumption, the number of Itaipu dams could be calculated. This form could be used in any country to understand their electricity needs. Link to the survey here.
# Brazil’s electricity consumption
There has been a few text books (MacKay 2009; Milo and Phillips 2014) and blog post (Murphy) on the importance of back of the envelope calculations. When a layperson is asked how much electricity Brazil consumes per year, (s)he would perhaps struggle to guestimate. With the knowledge of their own electricity consumption, they could arrive at an estimate for the whole country.
A single person living in an apartment complex in Campinas would get an average electricity bill of around 50 Reais and consume around 100 kWh per month or 1200 KWh per year (in fact the per capita consumption of Brazil is 2437 kWh, and differs from the number just mentioned by a factor of two). We can then ask the next question: how much does Brazil consume every year?
Brazil has a population of 200 million and 97% of the population have access to electricity. When we guestimate, it is good to have an upper and lower limit as one can never guess the accurate value. The lower limit could be that of a single person that was just calculated, of 100 kWh per month and the upper limit could be 300 kWh per month (not so unreasonable for a household of 6 people). By taking an arithmetic average of 200 kWh per month (as most people know averages), we obtain a value of 480 TWh for total electricity consumption in Brazil. In fact, Brazil consumed around 580 TWh of electricity in 2014 (British Petroleum 2015). We find that the estimated value is not very far from the actual value and our guesses were fairly reasonable.
# Number of Itaipu dams Brazil would need
We then ask how many Itaipu dams Brazil needs to meet the yearly electricity consumption. Itaipu dam is considered as one of the seven modern wonders of the world. It can generate a maximum of 100 TWh per year and supplies 75% of electricity consumed in Paraguay and 17% of energy in Brazil. The catchment area of the Itaipu dam is 1.35 million sq.km. From our calculation above, we find that ~ 5 Itaipu dams are needed to power the whole of Brazil.
# Number of Itaipu dams the world would need
The total energy generated by all the hydroelectric power stations in the world was 3900 TWh, in 2014 which is equivalent to 39 Itaipu dams. World’s energy needs are met using petroleum, coal, natural gas, hydroelectric, nuclear, and renewables (32% , 30%, 23%, 7%, 4% and 2% respectively) at ~ 57,000 TWh or 205 EJ (British Petroleum 2015). The primary energy consumption of the world would then be equivalent to ~570 Itaipu dams.
# Conclusions
By using Itaipu as a basis for understanding the world energy needs, one is immediately able to understand the scale of energy consumption in the world. Each dam costs ~20 billion USD to construct. It also shows that it is impossible to build so many dams as we do not have the monetary resources or big rivers on which such dams can be built. Calculations such as these give us a quick estimate of the upper and lower limit that differs from the actual value by a factor of two. Implementing policies with the consent of an informed public must be the goal of the government officials and public participation must be strongly encouraged in the developing countries.
# References
British Petroleum. BP Statistical Review of World energy. 2015.
MacKay DJ. Sustainable Energy — without the hot air. UIT Cambridge Ltd. 2009.
Milo R, Phillips R. Cell Biology by the Numbers. 2014;(June).
Murphy T. Do the Math [Internet]. [cited 2016 Jan 19]. Available from: http://physics.ucsd.edu/do-the-math/useful-energy-relations/
Advertisements | 1,042 | 4,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-30 | longest | en | 0.969645 |
https://xarray.pydata.org/en/v0.9.0/generated/xarray.ufuncs.arctan.html | 1,558,724,449,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257731.70/warc/CC-MAIN-20190524184553-20190524210553-00360.warc.gz | 1,017,236,015 | 7,159 | # xarray.ufuncs.arctan¶
xarray.ufuncs.arctan = <xarray.ufuncs._UFuncDispatcher object>
xarray specific variant of numpy.arctan. Handles xarray.Dataset, xarray.DataArray, xarray.Variable, numpy.ndarray and dask.array.Array objects with automatic dispatching.
Documentation from numpy:
arctan(x[, out])
Trigonometric inverse tangent, element-wise.
The inverse of tan, so that if y = tan(x) then x = arctan(y).
Parameters: x : array_like Input values. arctan is applied to each element of x. out : ndarray Out has the same shape as x. Its real part is in [-pi/2, pi/2] (arctan(+/-inf) returns +/-pi/2). It is a scalar if x is a scalar.
arctan2
The “four quadrant” arctan of the angle formed by (x, y) and the positive x-axis.
angle
Argument of complex values.
Notes
arctan is a multi-valued function: for each x there are infinitely many numbers z such that tan(z) = x. The convention is to return the angle z whose real part lies in [-pi/2, pi/2].
For real-valued input data types, arctan always returns real output. For each value that cannot be expressed as a real number or infinity, it yields nan and sets the invalid floating point error flag.
For complex-valued input, arctan is a complex analytic function that has [1j, infj] and [-1j, -infj] as branch cuts, and is continuous from the left on the former and from the right on the latter.
The inverse tangent is also known as atan or tan^{-1}.
References
Abramowitz, M. and Stegun, I. A., Handbook of Mathematical Functions, 10th printing, New York: Dover, 1964, pp. 79. http://www.math.sfu.ca/~cbm/aands/
Examples
We expect the arctan of 0 to be 0, and of 1 to be pi/4:
>>> np.arctan([0, 1])
array([ 0. , 0.78539816])
>>> np.pi/4
0.78539816339744828
Plot arctan:
>>> import matplotlib.pyplot as plt
>>> x = np.linspace(-10, 10)
>>> plt.plot(x, np.arctan(x))
>>> plt.axis('tight')
>>> plt.show() | 540 | 1,880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-22 | latest | en | 0.759001 |
https://www.rapidtables.com/convert/length/mile-to-km.html | 1,656,803,286,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00242.warc.gz | 1,007,679,807 | 5,722 | # Miles to Kilometers Converter
mi
km
km
m
Kilometers to miles ►
## How to convert miles to kilometers
1 mile is equal to 1.609344 kilometers:
1mi = 1.609344km
The distance d in kilometers (km) is equal to the distance d in miles (mi) times 1.609344:
d(km) = d(mi) × 1.609344
#### Example
Convert 20 mi to kilometers:
d(km) = 20mi × 1.609344 = 32.18688km
### How many kilometers in a mile
One mile is equal to 1.609344 kilometers:
1mi = 1mi × 1.609344 = 1.609344km
### How many miles in a kilometer
One kilometer is equal to 1/1.609344 miles:
1km = 1km/1.609344 = 0.6213712mi
### How to convert 10mi to kilometers
Multiply 10 miles by 1.609344 to get kilometers:
10mi = 10mi × 1.609344 = 16.09344km
## Miles to kilometers conversion table
Miles (mi) Meters (km)
0.01 mi 0.01609344 km
0.1 mi 0.1609344 km
1 mi 1.609344 km
2 mi 3.218688 km
3 mi 4.828032 km
4 mi 6.437376 km
5 mi 8.046720 km
6 mi 9.656064 km
7 mi 11.265408 km
8 mi 12.874752 km
9 mi 14.484096 km
10 mi 16.093440 km
20 mi 32.186880 km
30 mi 48.280320 km
40 mi 64.373760 km
50 mi 80.467200 km
60 mi 96.560640 km
70 mi 112.654080 km
80 mi 128.747520 km
90 mi 144.840960 km
100 mi 160.934400 km
Km to miles ► | 471 | 1,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-27 | longest | en | 0.418466 |
https://au.mathworks.com/matlabcentral/cody/problems/43026-determinant-of-a-3x3-matrix/solutions/996620 | 1,580,278,168,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00368.warc.gz | 329,250,295 | 15,622 | Cody
Problem 43026. Determinant of a 3x3 Matrix
Solution 996620
Submitted on 5 Oct 2016 by Peng Liu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
filetext = fileread('dtrmnt_3x3.m'); assert(isempty(strfind(filetext, 'det')),'det() forbidden')
2 Pass
x = [3 0 2; 2 1 -2; 0 1 1]; y_correct = 13; assert(sum(sum(abs(dtrmnt_3x3(x)-y_correct)))<1e-3)
3 Pass
x = [1 40 2; 7 -9 10; 1 0 1]; y_correct = 129; assert(sum(sum(abs(dtrmnt_3x3(x)-y_correct)))<1e-3) | 215 | 578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-05 | latest | en | 0.558004 |
https://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv009.cgi?read=25206 | 1,638,669,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00106.warc.gz | 898,168,466 | 4,402 | The Museum of HP Calculators
HP Forum Archive 09
hp41 howto do simple linear regression?Message #1 Posted by David E on 17 Dec 2002, 7:17 p.m. HI. I have the stat pac for hp41, and neither of the linear regressions sets seems to do what I need. I need something that I can input my data (x and y) and from that get R, slope, and y (y=mx+b). Thanks for any guidance .
Re: hp41 howto do simple linear regression?Message #2 Posted by Dave Shaffer on 17 Dec 2002, 8:02 p.m.,in response to message #1 by David E Well, David, you can always do it the semi-hard way - with just the '41 itself. If you use the sigma+ button (top left on the keypad)(after clearing the summation registers using the CLSigma button - shift x<>y key, just below the sigma+ key), you can accumulate the sums of x, x^2, y, y^2, and xy in memory registers 11 through 15 respectively, while register 16 will get the value of n (the number of x,y pairs). To put in your numbers, enter y first, then "ENTER", and then enter your x value - then hit the sigma+ key. After you have entered all you pairs of values, you need to remember (or be told - or, left as an exercise for the student, derive!) the least-squares fitting expressions for slope, intercept, and correlation coefficient. In case you are rusty: the slope is (n*SUMxy - SUMx*Sumy)/(n*SUMx^2 - (Sumx)^2) the intercept has the numerator SUMy*SUMx^2 - SUMxy*SUMx (with the same denominator as for the slope) r is (n*SUMXY - SUMx*SUMy) / ( SQRT(n*SUMx^2 - (SUMx)^2) * SQRT(n*SUMy^2 - (SUMy)^2) ) I hope my notation is clear. SUMx means the summation of x, and (SUMx)^2 means to square this sum (whereas SUMx^2 means the summation of x-squared). Have fun.
Re: hp41 howto do simple linear regression?Message #3 Posted by David on 17 Dec 2002, 9:45 p.m.,in response to message #2 by Dave Shaffer Boy, that scares me just looking at it. lol. maybe i will try to write a prog for that. thanks
Re: hp41 howto do simple linear regression?Message #4 Posted by Mike (Stgt) on 18 Dec 2002, 10:07 a.m.,in response to message #3 by David Citation from Dave's append: "After you have entered all you pairs of values, you need to remember (or be told - or, left as an exercise for the student, derive!) the least-squares fitting expressions for slope, intercept, and correlation coefficient." To ease the exercise for the student just take the values of X as free of error. My little (basic) routine is based on this assumption. Next step in students exercise would be to take Y values for true and the fault is in X :-) And if you have some time and piles of paper left, just take X as faulty as Y ;-) Ciao.....Mike
Re: hp41 howto do simple linear regression?Message #5 Posted by Vieira, Luiz C. (Brazil) on 17 Dec 2002, 9:40 p.m.,in response to message #1 by David E Hi; Dave Shaffer's post contains ALL information you need, but you can do it in the "easy way" by using the STAT pack programs. Get your Stat Pack manual and open it on page 32, under "Curve Fitting". On page 33, top left, you'll see your Y = a + bX expression. Now go to page 34 and you'll see that you should [XEQ] [SIGMA]LIN program for a straight line sample (or any of the other three: LOG, EXP or POW), being [SIGMA] the summation symbol. You should enter each pair in the REVERSED orther (why do some applications are against Y and X registers?), say:` X [ENTER^] Y [A]` for each pair; to correct any mistakes:` X [ENTER^] Y [C]` After the last (x,y) pair, simply press [E] and you'll have:``` R2 = (corr. coef.) [R/S] a = (intercept) [R/S] b = (slope)``` Hope this is waht you want. BUT if you really want to know the mechanics, read D. Shaffer's post. It is complete reference. Success.
Re: hp42s how to operate it, do simple mathMessage #6 Posted by Kenneth on 17 Dec 2002, 11:10 p.m.,in response to message #5 by Vieira, Luiz C. (Brazil) hi i'm trying to operate my 42s rpn scietific and i can't do a simple problem with it, i don't have a manual or instructions on how to operate it can you help me?
Re: hp42s how to operate it, do simple mathMessage #7 Posted by Richard Garner on 18 Dec 2002, 1:32 p.m.,in response to message #6 by Kenneth Here is a quick primer 4+5=9 4 Enter 5 + (4+5)x10=90 4 Enter 5 + 10 x (4+5)/3=3 4 Enter 5 + 3 / I hope this helps. Depending on how the numbers are set in the stack, you can load the stack with you numbers and the perform the operations all at once. As in (4x5)+7=27 7 Enter 5 Enter 4 x + Have fun.
Re: hp42s how to operate it, do simple math - a light hearted responseMessage #8 Posted by Tom (UK) on 19 Dec 2002, 1:02 p.m.,in response to message #6 by Kenneth The HP42S is a tricky beast to master, and much harder without the manual. Best thing to do is pop it in the post to me and I'll return a much easier to use calculator in exchange... how about a Casio FX580? But seriously: The HP42S is a nice tool, once you have mastered it (doesn't take long) you will wonder how you got along without it. To start using it this site has a brief introduction on RPN. To get a manual buy the DVD/CD's from this web site, it will be well worth it in the long run.
Re: hp41 howto do simple linear regression?Message #9 Posted by Mike (Stgt) on 18 Dec 2002, 7:34 a.m.,in response to message #1 by David E I just did an 'edit/copy-edit/paste' from my emulator on the IBM-host, so the § stands for Sigma: ```01·LBL "§" 02 §REG 10 03 CL§ 04 MEAN 05 X<>Y 06 RCL 10 07 ST* Z 08 * 09 RCL 14 10 - 11 X<>Y 12 RCL 11 13 - 14 / 15 STO 04 16 MEAN 17 LASTX 18 * 19 - 20 STO 05 21 RCL 04 22 STOP 23 SDEV 24 RCL 04 25 * 26 X<>Y 27 / 28 END ``` That's the minimum code, to make life easier just use the Alpha capabilities of the HP-41 and add some literals and prompts. Ciao.....Mike
Re: hp41 howto do simple linear regression?Message #10 Posted by Vieira, Luiz C. (Brazil) on 18 Dec 2002, 9:02 a.m.,in response to message #9 by Mike (Stgt) Hi; I did not load the program and run it, but I am curious about one sequence: CL§ and MEAN. How can you compute MEAN if CL§ clears all statistic data? As you change the §REGS to R10 instead of original R16 and you execute CL§, I thought you created a buffer to compute partial data and return to §REG 16 before using MEAN. But right after that you clear statistic data and execute MEAN with all registers down to zero. Is that what you want? Did I miss something? If I am wrong, sorry. I just want to understand. Cheers.
Re: hp41 howto do simple linear regression?Message #11 Posted by Mike (Stgt) on 18 Dec 2002, 9:44 a.m.,in response to message #10 by Vieira, Luiz C. (Brazil) No, you did not miss something. I gave no clear explanation. This is my LinReg prgm that fits in every HP-41, independent what other BIG routines use up all memory. There are no literals and no prompts to ease the user input and result output. After line 03 you may add the two lines "Y enter X §+" PROMPT. Without it you just see DATA ERROR and must know that this is the point to enter data. Ciao.....Mike
Go back to the main exhibit hall | 2,024 | 6,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-49 | latest | en | 0.845121 |
http://mathematica.stackexchange.com/questions/8780/parameter-optimization?answertab=active | 1,461,973,263,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111518.82/warc/CC-MAIN-20160428161511-00086-ip-10-239-7-51.ec2.internal.warc.gz | 184,978,393 | 17,458 | # parameter optimization
Does anybody know of any tutorial material that explains how to use Mathematica to fit parameter values used in a set of ordinary differential equations to experimental values stored in a comma separated variable (CSV) file, and assumes very basic Mathematica programming experience?
The ordinary differential equations are required to provide a time course simulation of the variables that is fitted to experimental data, in order to use the parameter values to predict hypothesis on the behavior of the system.
Thanks for any assistance.
-
@Steve Welcome to the forum. Please search the forum before you post a question!! Its happens often that the problem you are facing is already addressed before. Please look at this question mathematica.stackexchange.com/questions/6751/…. I hope this helps you. – PlatoManiac Jul 27 '12 at 17:27
Also, it seems to me that you are posting two unrelated questions there. Please, if that is the case, separate them into two postings – Dr. belisarius Jul 27 '12 at 18:03
Another issue: the etcs in your functions (they are not equations, are they?) are not clear enough for me – Dr. belisarius Jul 27 '12 at 18:04
etc is "etcetera" to simplify what I was inserting. Sorry for posting a question that had been investigated in another post, if you are a moderator please feel free to delete this. I will investigate the forum post linked above and attempt to adapt it to my work. – Steve Jul 27 '12 at 18:44
What does ODE stand for? – FredrikD Jul 27 '12 at 19:02
I am not sure if I am following your problem. Perhaps this is a partial answer.
For solving ODEs like yours for a variable number of functions, you could do something like:
dims = 3;
k = RandomInteger[{-2, 2}, {dims, dims}];
DSolve[Table[
D[f[i, x], x] == Table[f[j, x], {j, dims}].k[[i]] &&
(D[f[i, x], x] /. x -> 0) == 1, {i, dims}],
Table[f[j, x], {j, dims}], x]
which of course is solvable only in some cases
Plot[Table[f[i, x] /. sol[[1]], {i, dims}], {x, 0, 1},Evaluated -> True]
Note that the system you are solving is something like:
$\left\{f^{(0,1)}(1,x)=f(1,x) p[[1,1]]+f(2,x) p[[1,2]]+f(3,x) p[[1,3]]\land f^{(0,1)}(1,0)=1,\\ f^{(0,1)}(2,x)=f(1,x) p[[2,1]]+f(2,x) p[[2,2]]+f(3,x) p[[2,3]]\land f^{(0,1)}(2,0)=1, \\f^{(0,1)}(3,x)=f(1,x) p[[3,1]]+f(2,x) p[[3,2]]+f(3,x) p[[3,3]]\land f^{(0,1)}(3,0)=1\right\}$
- | 727 | 2,359 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2016-18 | latest | en | 0.904998 |
flustercucked.blogspot.com | 1,469,535,221,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00121-ip-10-185-27-174.ec2.internal.warc.gz | 82,399,149 | 31,919 | ## Monday, July 5, 2010
### 40 Years of Lawyer Overproduction, a Data Table, and 2 Charts
In my last two posts I explored the rate of new lawyer production in terms of the inverse number of attorneys per capita that could be sustained by graduation rates (expressed in terms of one lawyer for every X-amount of people) using the assumption that a new lawyer would want to practice for 40 years. Let's call this number the Sustained Inverse Lawyers Per Capita, or SILPC.
I was curious about the historical trend, so I conducted a study and was surprised to discover that the law schools have been overproducing lawyers for almost 40 years! In other words, the rate of production in terms of SLPC has averaged one lawyer for every 171.9 people since 1973. I knew that lawyer overproduction had been a problem for decades, but I had never imagined that it was this bad! I had previously assumed that the SILPC had decreased steadily over time, but apparently this is not the case.
I calculated this data using U.S. Census Data for the U.S. population and ABA statistics for the number of JDs awarded each year since 1963.
Assuming that, on average, a lawyer would want to practice for 40 years, SILPC = Population / (JDs Awarded * 40)
Year JDs Awarded US Population Inverse Lawyers Per Capita (SILPC) 1963 9,638 189,242,000 490.9 1964 10,491 191,889,000 457.3 1965 11,507 194,303,000 422.1 1966 13,115 196,560,000 374.7 1967 14,738 198,712,000 337.1 1968 16,007 200,706,000 313.5 1969 16,733 202,677,000 302.8 1970 17,477 205,052,000 293.3 1971 17,006 207,661,000 305.3 1972 22,342 209,896,000 234.9 1973 27,756 211,909,000 190.9 1974 28,729 213,854,000 186.1 1975 29,961 215,973,000 180.2 1976 32,597 218,035,000 167.2 1977 33,640 220,239,000 163.7 1978 33,317 222,585,000 167.0 1979 34,590 225,055,000 162.7 1980 35,059 227,726,000 162.4 1981 35,604 229,966,000 161.5 1982 34,847 232,188,000 166.6 1983 36,390 234,307,000 161.0 1984 36,688 236,348,000 161.1 1985 36,830 238,466,000 161.9 1986 36,122 240,651,000 166.6 1987 35,479 242,804,000 171.1 1988 35,702 245,021,000 171.6 1989 35,521 247,342,000 174.1 1990 36,386 250,132,000 171.9 1991 38,801 253,493,000 163.3 1992 39,082 256,894,000 164.3 1993 39,915 260,255,000 163.0 1994 39,711 263,436,000 165.8 1995 39,355 266,557,000 169.3 1996 39,921 269,667,000 168.9 1997 41,115 272,912,000 165.9 1998 39,456 276,115,000 175.0 1999 39,072 279,295,000 178.7 2000 38,158 282,434,000 185.0 2001 37,910 285,545,000 188.3 2002 38,606 288,600,000 186.9 2003 38,875 291,221,000 187.3 2004 40,024 293,842,000 183.5 2005 42,672 296,463,000 173.7 2006 43,883 299,084,000 170.4 2007 43,518 301,705,000 173.3 2008 43,588 304,326,000 174.5 2009 44,000 306,947,000 174.4 2010 309,565,000
From 1963 to 1970 the SILPC decreased steadily until a huge jump occurred between 1971 and 1973. The worst year was 1983 when the rate bottomed out at 161.0. Perhaps the market was able to comfortably absorb this amount of lawyer overproduction in the Sixties and Seventies. Presumably, new attorneys have suffered difficulty finding career-building entry-level jobs and earning a living practicing law since the late Seventies or early Eighties, but the Internet was not available to chronicle it. It is also possible that lawyers were more easily able to obtain upwardly mobile white collar jobs in those decades at a time before hordes of people went to college. In other words, the Law School Scam took root 40 years ago.
If I can obtain the data, I would like to plot the number of attorneys who maintained licenses every year since 1963. A plot of the data shows that JD production has outpaced U.S. population growth by a tremendous margin since 1963:
The red data in the chart above is SILPC. It has remained fairly steady since 1973 with an average SILPC of 171.9.
This last chart expresses U.S. population growth and the growth in the amount of JDs awarded as a percentage since 1963. So, a data point of 50% population growth would mean that the population had increased by 50% since 1963. A data point of 300% JDs awarded means that the number of JDs awarded that year was four times the number awarded in 1963. (A number of 0% would mean that the number is the same as it was in 1963, and a number of 100% would mean that it was double the amount in 1963.)
------- EDITOR'S NOTE -------
March 11, 2011. I want to clarify that the 40 year average lawyer-to-population ratio that new JD production can sustain (which I eventually calculated to be 1 lawyer for every 171.9 people) is NOT the same thing as the actual lawyer-to-population ratio. The number I calculated for a given year of new JD production would only reflect the actual lawyer-to-population ratio if the U.S. population remained the same for the following 40 years. This is because while the U.S. population continues to increase, the number of JDs produced in a given prior year is static and cannot increase proportionally with population growth.
Consequently, Using ABA and BLS stats, the actual lawyer-to-population ratio is about 1 lawyer for every 215 people (only counting JDs minted over the past 40 years).
BL1Y said...
While lawyers per capita certainly is a useful stat, it's not the full picture of lawyer job market sustainability. Increases in government regulation mean more lawyers, both in the government, and in private practice either at firms or in house. Basically, they create more demand for lawyers.
What I find more interesting about this is that government regulations or not, most of what lawyers do does not build the economy. Doing work on contracts arguably helps build the economy (as much as the underlying deal does), but regulatory work does not, law suits do not, wills do not.
The increase in lawyers per capita shows that we have a lot of highly paid people who don't contribute to the economy. That's a recipe for economic disaster. I bet if you looked at the number of HR personnel and middle managers, you'd see a similar increase.
A Law School Victim said...
I hope you and Nando jump on this: http://www.bls.gov/k12/law02.htm, look at what it says: "How much does this job pay? Back to Top
Lawyers are some of the highest paid workers. Although some lawyers work for themselves, many other lawyers work for governments, law firms, and corporations. In May 2008, the average yearly wages for lawyers were \$124,750.
Lawyers held about 759,200 jobs in 2008. Most lawyers worked for themselves or in law firms. Some lawyers worked for other businesses or for government." I wanted to smack the screen when I say this.
Frank the Underemployed Professional said...
I'm sure that the average income statistic conveniently or perhaps naively only counts lawyers who work in the field and not unemployed and underemployed lawyers. Those types of career summary pages are normally worthless wastes of space written by laymen who know nothing about the field they're describing.
Nice find with the 759,200 stat.
"Lawyers held about 759,200 jobs in 2008."
It doesn't say what kinds of jobs or if they were legal jobs. If you take the data and add up the number of JDs produced in the 40 years from 1969 to 2008, then 1,412,328 lawyers should be in the job market, assuming that a lawyer would want to work for 40 years on average. According to the ABA licensure stats for 2008, there were 1,162,124 "resident and active attorneys".
I think I'll make a blog post to hash out some data from those numbers.
Frank the Underemployed Professional said...
This comment has been removed by the author.
C. Cryn Johannsen said...
Anonymous said...
I graduated from Law School at the University of Denver in 1983. The economy was terrible then,particularly in boom or bust states, but not as bad as now. When I was a first year, I bought running shoes from a guy who had just graduated from law school in Kansas and moved to Denver to live with his wife's family and look for work in the "big city." One of my friend's husbands went to Harvard Law but decided to move out west for the "quality of life." I recall being one of a few people who had job offers after my second year. I ended up being moved to New York City by Mobil Oil Company. I moved from there to an associate at a big law firm. Many of my friends struggled for years to get careers going. Many of them never made it, I was lucky.
The big difference was that my law school tuition, which I worked to pay for while in school, was about \$8,000 or so a year - less with scholarships.
My student loan payment was less than \$300 a month. (I went to a local state university and lived at home so I wouldn't incur debt because I knew I wanted to go to graduate school.)
Honestly, I was so financially conservative and worried about debt - because my parents, products of the depression, had no money to help me out- I don't think I would have gone to law school if the price tag meant I had to accumulate a staggering amount of debt.
Also, I grew up in a small town in the Midwest and never really thought about making staggering amounts of money. We had just been through a serious recession and oil crises. People where I lived didn't have access to, or real knowledge of, big law. I didn't even know about big firms until I was moved to New York.
I would never let my son go to law school unless we worked out a way to pay for it in cash and scholarships. No one should take on \$200,000 in debt they can't walk away from if things don't work out right.
I didn't realize how bad life was for many law students until I read some of these scam blogs.
Anonymous said...
I wanted to add that you should do a chart that shows how the cost of law school has gone up and the amount of debt needed to pay for it since 1983, or any other date you choose.
I also neglected to mention that in 1983, there were jobs to be had that you could support yourself on, even selling shoes. And people were hiring for jobs, though as I said, the economy was in a recession then as well.
A third point, when I went to law school it was the first time the class had been half women. When I graduated, companies were under government pressure to show they had women in highly paid, managerial positions. Hiring women lawyers was one of the fastest ways to improve their standing.
Finally, many companies then had the idea of building a great in-house law firm instead of using outside lawyers all the time.
JeffM said...
I tried to use the same reasoning as the author and found out I was wrong to use statistics that way. You can't compare new lawyers to overall population. You have to compare ALL lawyers to overall population. | 2,891 | 10,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-30 | longest | en | 0.918692 |
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### 12 Katesismo Hi Sch Public
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# 4.1 derivatives as rates linear motions
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### 4.1 derivatives as rates linear motions
1. 1. Derivatives as Rates and Linear Motions
2. 2. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.
3. 3. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..
4. 4. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..The output unit could be in meters (m) for distancemeasurement as in D = D(t) = distance of a movingobject from a fixed location. D(t)
5. 5. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..The output unit could be in meters (m) for distancemeasurement as in D = D(t) = distance of a movingobject from a fixed location.The output unit could be in centigrade (co) fortemperature measurement as inT = T(t) = temperature of an object at time t.
6. 6. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..The output unit could be in meters (m) for distancemeasurement as in D = D(t) = distance of a movingobject from a fixed location.The output unit could be in centigrade (co) fortemperature measurement as inT = T(t) = temperature of an object at time t.The output might be W kilograms (kg) for weightmeasurement as inW = W(t) = weight of an object at time t.
7. 7. Derivatives as Rates and Linear MotionsLinear Motions
8. 8. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.
9. 9. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt whereD = height in feet,t = time in secondsv = initial speed (ft /sec)
10. 10. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)
11. 11. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)A stone is thrown straight up at aspeed of 96 ft/sec, after t secondsit reaches the height D where D(t) =
12. 12. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)A stone is thrown straight up at aspeed of 96 ft/sec, after t secondsit reaches the height D where D(t) = –16t2 + 96t.
13. 13. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)A stone is thrown straight up at aspeed of 96 ft/sec, after t secondsit reaches the height D where D(t) = –16t2 + 96t.We make a few observations with some simple algebra.
14. 14. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.
15. 15. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0
16. 16. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.
17. 17. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.Hence it will stay in air for 6 seconds.It reaches the maximum height in half of that time or3 seconds.
18. 18. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.Hence it will stay in air for 6 seconds.It reaches the maximum height in half of that time or3 seconds. The corresponding maximum height itreaches is D(3) = –16(3)2 + 96(3) = 144 ft.
19. 19. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.Hence it will stay in air for 6 seconds.It reaches the maximum height in half of that time or3 seconds. The corresponding maximum height itreaches is D(3) = –16(3)2 + 96(3) = 144 ft.Note that the speed of the stone slows from96 ft/sec to 0 ft/sec as it reaches the height of 144 ft.
20. 20. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. D = -16t2 + 64t
21. 21. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t2 + 64tor simply “the rate” at time t.
22. 22. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t2 + 64tor simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.
23. 23. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).
24. 24. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).
25. 25. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).To see this, consider the “averagerate of change of a function f(t)from the time t = a to t = b”.
26. 26. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).To see this, consider the “averagerate of change of a function y = f(t) (b, f(b))from the time t = a to t = b” which is (a, f(a)) f(b)–f(a) = Δy b-a=Δxf(b) – f(a) = Δy = the slope of the b–a Δtcord as shown. a b
27. 27. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96tfrom t = 1 to t = 3 isD(3) – D(1) 3–1
28. 28. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96tfrom t = 1 to t = 3 isD(3) – D(1) = 144 – 80 3–1 2= 32ft/sec
29. 29. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/sec
30. 30. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from(1, 80) to ( 3, 144).
31. 31. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from(1, 80) to ( 3, 144). D D(t) = -16t2 + 96t t 1 3 6
32. 32. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from slope =32 (3, 144) = avg. rate(1, 80) to ( 3, 144). D D(t) = -16t2 + 96t (1, 80) t 1 3 6
33. 33. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from slope =32 (3, 144) = avg. rate(1, 80) to ( 3, 144). DThe average rate from t = 1 to t = 2 D(t) = -16t2 + 96tis D(2) – D(1)= 128 – 80 = 48 ft/sec (1, 80) 2–1 1 t 1 3 6
34. 34. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from slope =32 (3, 144) = avg. rate(1, 80) to ( 3, 144). DThe average rate from t = 1 to t = 2 D(t) = -16t2 + 96tis D(2) – D(1)= 128 – 80 = 48 ft/sec (1, 80) 2–1 1which is as expected because it’s ttraveling faster in the beginning. 1 3 6
35. 35. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.
36. 36. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96
37. 37. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96Hence D(1) = 64 ft/sec or that the upward speed ofthe stone is 64 ft/sec at t = 1.
38. 38. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96Hence D(1) = 64 ft/sec or that the upward speed ofthe stone is 64 ft/sec at t = 1.The rate of 64 ft/sec means that if the gravitydisappeared at the instant t = 1, then the stone wouldgo upward at a constant speed of 64 ft/sec.
39. 39. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96Hence D(1) = 64 ft/sec or that the upward speed ofthe stone is 64 ft/sec at t = 1.The rate of 64 ft/sec means that if the gravitydisappeared at the instant t = 1, then the stone wouldgo upward at a constant speed of 64 ft/sec.It also means that at the instant t = 1,per unit change in the input would give 64 units ofchange in output, i.e. 1 sec. in time t to 64 ft. in height.
40. 40. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For whatvalue(s) of t is y = 0?
41. 41. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For whatvalue(s) of t is y = 0?When t = 0 the position isy(0) = 12.
42. 42. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 y = –t2–t+12
43. 43. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0If the position y = 0, then–t2 – t + 12 = 0 or y = –t2–t+12
44. 44. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 If the position y = 0, then –t2 – t + 12 = 0 or t2 + t –12 = 0 y = –t2–t+12(t – 3) (t + 4) = 0or when t = 3 or –4 that y = 0.
45. 45. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what y yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 y= 0 at 0 If the position y = 0, then t= –4 –t2 – t + 12 = 0 or t2 + t –12 = 0 y = –t2–t+12(t – 3) (t + 4) = 0or when t = 3 or –4 that y = 0.
46. 46. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what y y yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 y= 0 at 0 0 y= 0 at If the position y = 0, then t= –4 t= 3 –t2 – t + 12 = 0 or t2 + t –12 = 0 y = –t2–t+12(t – 3) (t + 4) = 0or when t = 3 or –4 that y = 0.
47. 47. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?
48. 48. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.
49. 49. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7y(0) = –1y(3) = –7
50. 50. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7 yy(0) = –1y(3) = –7Respectively, the point is 0 y= 0 atgoing up when t = –4 as it t= –4passed y = 0. y = –t2–t+12
51. 51. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7 y yy(0) = –1 y= 12 aty(3) = –7 t=0Respectively, the point is 0 y= 0 at 0going up when t = –4 as it t= –4passed y = 0. It’s on theway down when t = 0 as itpasses y =12. y = –t2–t+12
52. 52. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7 y y yy(0) = –1 y= 12 aty(3) = –7 t=0Respectively, the point is 0 y= 0 at 0 0 y= 0 atgoing up when t = –4 as it t= –4 t= 3passed y = 0. It’s on theway down when t = 0 as itpasses y =12. When t = 3, y = –t2–t+12it passes y = 0 going down.
53. 53. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?
54. 54. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?When y reaches the maximum value, its rate y(t)must be 0 because the rate can’t be + or –(it’s not going up nor down at that instance).
55. 55. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?When y reaches the maximum value, its rate y(t)must be 0 because the rate can’t be + or –(it’s not going up nor down at that instance). So itreaches the max. when y(t) = -2t –1 = 0 or t = –½,
56. 56. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?When y reaches the maximum value, its rate y(t)must be 0 because the rate can’t be + or –(it’s not going up nor down at that instance). So itreaches the max. when y(t) = -2t –1 = 0 or t = –½,at y(–½ ) = –(–½)2 – (–½) + 12 = 12¼. y y y (–1/2,12.25) y y(–1/2)=0 y= 12 at (0,12) t=0 y(0)=–30 y= 0 at 0 0 y= 0 at t (–4, 0) (3,0) t= –4 t= 3 y(–4)=7 y(3)= –7 y = –t2–t+12 y = –t2–t+12
57. 57. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola.
58. 58. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 (0,12) t =–1/2 y(0)=–3 cast0 y= 0 at t t= –4 shadow (–4, 0) (3,0) y(–4)=7 y(3)= –7 y = –t2–t+12 y = –t2–t+12
59. 59. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 t t= –4 shadow (–4, 0) (3,0) y(–4)=7 y(3)= –7 y = –t2–t+12 y = –t2–t+12
60. 60. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 0 y=0 t t= –4 shadow (–4, 0) (3,0) at t=3 y(–4)=7 y(3)= –7 y = –t2–t+12
61. 61. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 0 y=0 t t= –4 shadow (–4, 0) (3,0) at t=3 y(–4)=7 y(3)= –7 y = –t2–t+12We define the velocity of the linear motion as the1st derivative y of the position function y.
62. 62. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 0 y=0 t t= –4 shadow (–4, 0) (3,0) at t=3 y(–4)=7 y(3)= –7 y = –t2–t+12We define the velocity of the linear motion as the1st derivative y of the position function y.The velocity y gives the rates of changes in thepositions y as in the phrase “the velocity is 60 mph”.
63. 63. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y).
64. 64. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities.
65. 65. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft)
66. 66. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3..
67. 67. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left?
68. 68. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left? When does it change direction?
69. 69. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left? When does it change direction? What are the right most and left most positions of the point on the x–axis?
70. 70. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left? When does it change direction? What are the right most and left most positions of the point on the x–axis? b. Find the acceleration x and when does x = 0? What’s the meaning of x = 0?
71. 71. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2.
72. 72. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative.
73. 73. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart:
74. 74. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0 t2(t – 1) = 0 t = 0, 1
75. 75. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0 t2(t – 1) = 0 t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1
76. 76. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0 t2(t – 1) = 0 t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1Hence the point is heading left when t < 1, i.e. from“t = –∞” up to t = 1,
77. 77. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0 t2(t – 1) = 0 t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1Hence the point is heading left when t < 1, i.e. from“t = –∞” up to t = 1, except at t = 0, it paused (why?).
78. 78. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0 t2(t – 1) = 0 t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1Hence the point is heading left when t < 1, i.e. from“t = –∞” up to t = 1, except at t = 0, it paused (why?).After t = 1, it changes direction and heads to theright. Here is the corresponding graphic projection.
79. 79. Derivatives as Rates and Linear MotionsThe graph of x(t) = t4/4 – t3/3 xx(t) = t4/4 – t3/3is shown to the right. t
80. 80. Derivatives as Rates and Linear MotionsThe graph of x(t) = t /4 – t /3 4 3 xx(t) = t4/4 – t3/3is shown to the right.Below is the projection of thisgraph (a compressed version) tonto the x–axis. x
81. 81. Derivatives as Rates and Linear MotionsThe graph of x(t) = t /4 – t /3 4 3 xx(t) = t4/4 – t3/3is shown to the right.Below is the projection of thisgraph (a compressed version) tonto the x–axis. The point is coming from the right,pauses (why?) at x = 0 when t = 0, x=0 x at t =0
82. 82. Derivatives as Rates and Linear Motions The graph of x(t) = t /4 – t /3 4 3 x x(t) = t4/4 – t3/3 is shown to the right. Below is the projection of this graph (a compressed version) t onto the x–axis. The point is coming from the right, pauses (why?) at x = 0 when t = 0, continues on to the left. At t = 1 it reaches the left most position x = –1/12,x = –1/12 x=0 xat t =1 at t =0
83. 83. Derivatives as Rates and Linear Motions The graph of x(t) = t /4 – t /3 4 3 x x(t) = t4/4 – t3/3 is shown to the right. Below is the projection of this graph (a compressed version) t onto the x–axis. The point is coming from the right, pauses (why?) at x = 0 when t = 0, continues on to the left. At t = 1 it reaches the left most position x = –1/12, then it changes direction and heads right after t = 1.x = –1/12 x=0 xat t =1 at t =0
84. 84. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?
85. 85. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.
86. 86. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0 t = 0, 2/3.
87. 87. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0 t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3
88. 88. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0 t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).
89. 89. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0 t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).The sign–chart tells us that the point “accelerates”from t = –∞ to t = 0, x=0 x at t =0
90. 90. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0 t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).The sign–chart tells us that the point “accelerates”from t = –∞ to t = 0, i.e. as it goes left it slows down. The point is “accelerating” because the 2nd derivative x(t) > 0, however the speed of the point is decreasing, or “decelerating” in daily usage of the word. It’s heading left and slowing down is a more apt answer. x=0 x at t =0
91. 91. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0 t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).The sign–chart tells us that the point “accelerates”from t = –∞ to t = 0, i.e. as it goes left it slows down.At t = 0, it stops at x = 0, and the acceleration is 0. x=0 x at t =0
92. 92. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
93. 93. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, Again, the point is “decelerating” because the 2nd derivative x(t) < 0, however the speed of the point is increasing, or “accelerating” in daily usage of the word. It’s heading left and speeding up is a more apt answer. + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
94. 94. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3. + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
95. 95. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
96. 96. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. At t = 1 the “acceleration” slows theleftward motion or the velocity to 0 with x-min. at -1/12. + – + x (t) = 3t2–2t 0 2/3x = –1/12 t=2/3 x=0 xat t =1 at t =0
97. 97. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. At t = 1 the “acceleration” slows theleftward motion or the velocity to 0 with x-min. at -1/12.Then the “acceleration” causes the point to changedirection and head right after t = 0. + – + x (t) = 3t2–2t 0 2/3x = –1/12 t=2/3 x=0 xat t =1 at t =0
98. 98. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. At t = 1 the “acceleration” slows theleftward motion or the velocity to 0 with x-min. at -1/12.Then the “acceleration” causes the point to changesdirection and head right, speeding up after t = 1.At t = 0, 2/3, + – + x (t) = 3t2–2tthe acceleration 0 2/3becomes decelerationand vice versa.x = –1/12 t=2/3 x=0 xat t =1 at t =0
99. 99. Derivatives as Rates and Linear MotionsHere is a summary about the acceleration x(t).1. Given that x(t) > 0 and a. its velocity is negative, i.e. x (t) < 0, it means its leftward motion is slowing down. b. its velocity is positive, i.e. x (t) > 0, it means its rightward motion is speeding up.2. Given that x(t) < 0 and a. its velocity is negative, i.e. x (t) < 0, it means its leftward motion is speeding up. b. its velocity is positive, i.e. x (t) > 0, it means its rightward motion is slowing down.3. Given that x(t) = 0. If it’s an inflection point, thenthe acceleration ceased and it begins to decelerate,or vice versa. If it’s not an inflection point, it’s a pause. | 11,881 | 34,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-39 | latest | en | 0.898981 |
http://gamepolitics.com/2007/07/06/xbox-360-failure-rate-do-the-math/ | 1,448,864,058,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461113.77/warc/CC-MAIN-20151124205421-00321-ip-10-71-132-137.ec2.internal.warc.gz | 97,102,524 | 31,251 | # Xbox 360 Failure Rate? Do the Math
Now that Microsoft has done the right thing about the Xbox 360’s rampant hardware problems, one key question remains:
What is the failure rate for the system? MS won’t say. The Seattle Times quotes MS games boss Robbie Bach (left):
Suffice it to say that with a billion-dollar charge and the focus we’re putting on this that it’s a meaningful number.
While normal console failures are apparently in the sub-5% range, some reports have estimated that 360’s throw the red rings at a rate as high as 33%.
Here are the numbers that are known. The Times (and other sites) are reporting that MS has shipped 11.6 million Xbox 360’s to date, but the failure rate remains a mystery. The addition of a three-year warranty for the rings of death failure is expected to cost MS a cool billion dollars, so:
(11,600,000 * X) * Y = 1,000,0000,000
…where 11.6m is the number of systems shipped; X is the failure rate and Y is the total cost to MS per replaced unit. This works out to:
Y=86/X
Now, math is not especially GP’s strong suit, but if each repair (Y) costs MS \$100, that translates to a staggering failure rate of 86%.
At \$200, the failure rate is 43%. At \$300 it’s 29%. At \$400, it’s 22%
At the high end, the \$400 repair cost on a 22% failure rate is very unlikely. Since that is the current system’s MSRP, why bother to repair? They could just send you a new 360 to replace your dead one.
On the other end, the 86% scenario seems pretty extreme (perhaps not if you’re one of the unfortunates who have lost multiple 360’s to the dreaded rings).
There is a lot of behind-the-scenes stuff we can’t know here. But, based on the numbers, a failure rate between 29% and 43% would not be unreasonable. So maybe those 33% reports aren’t so far off the mark, after all.
Categories Uncategorized
1. 0
travestibu says:
Don’t forget that the ESA represent the member companies and their policies, so if these are sticking point issues with you, travesti blame and fight the companies themselves. travesti i am.
2. 0
oto kirlama says:
Gallagher can araç kiralama say all he wants, but I strongly rent a car believe it’s due to his crappy leadership and E3 being a joke. ESA’s Board of Directors need to find a way to get out rent a car of this horrid contract with this Bush cronie before there’s no one left on the Board.
Btw, I think Atari and Midway will drop out too, but mostly travesti because these guys have done nothing ttnet vitamin or little and need to start saving costs.
3. 0
markofando says:
Want to start your private office arms race right now?
I just got my own USB rocket launcher Awsome thing.
Plug into your computer and you got a remote controlled office missile launcher with 360 degrees horizontal and 45 degree vertival rotation with a range of more than 6 meters – which gives you a coverage of 113 square meters round your workplace.
You can get the gadget here: http://tinyurl.com/2qul3c
Check out the video they have on the page.
Cheers
Marko Fando
4. 0
TIM says:
CONTACT SEC.GOV CONTACT INVESTORS PAGES RELEASE THE TRUTH! I NOW AWAIT MY FOURT RETURN TO GO BACK TO MC ALLEEN TEXAS WITH IS NOT MICROSOFT BUT A COMPANY CALLED JABIL JABIL.COM THEY RECENTLY PURCHASED A COMPANY IN INDIA? YOU KNOW THE PEOPLE YOU TALK TO WHEN YOU HAVE A PROBLEM!! IF ONE XBOX360 FAILS THEY ALL WILL? MICROSOFT CANT WAIT UNTILL YOUR THREE YEARS IS UP CAN YOU?
5. 0
Unsasysam says:
I’ve got an Amazon gift certificate burning holes in my pocket,
and I want to get the most bang for my buck.
Enter the Secret Amazon Web Pages:
This is where you’re going to find the “latest sales, rebates, and limited-time offers” from
Amazon, and you can score some pretty deep discounts if you’re a savvy shopper.
Next, there’s the special Sale link. This is open every Friday, and ONLY on Fridays.
You can find the same good discounts here as you would in hidden Deals, although some
Fridays you can really get lucky and make off like an Amazon bandit – I’ve seen discounts
there as low as 75% off sticker price.
6. 0
Axewielderx says:
I just recieved my 3rd xbox 360 less than3 weeks ago. It froze after being on for 15 mins.,while still being in the dashboard. I would love to see someone elxplain that.
Microsoft has no idea what the problem is and the warrenty is bogus. You are not honoring your warrenty if you keep sending defective products to your customers. I believe this is called Fraud.
7. 0
[…] James Robertson has brought up a topic of consoles again. While I totally agree that reported failure rates are absolutely unacceptable and point to a problem in either hardware design or shoddy assembly, the other important side of the story that often gets overlooked is how Microsoft deals with these failures when they do occur. To be honest, I was a little concerned about the perceived failures when I bought my own Xbox 360 for Christmas, and those concerns grew into uneasy anticipation when I started seeing intermittent error messages during bootup before network adapter died completely. I pretty much expected the repair process to cost me a couple hundred dollars and countless hours spent on the phone, but actually that couldn’t be further from the truth. Here’s how the process worked (in my case, other’s may differ, but the chatter in blogosphere indicates similar experiences), […]
8. 0
A.C.E. says:
@Brokenscope:
dont forget that SONY also probably follow a ROHS in Japan due to manufacturers being responsible for recycling of their products. if PS3 had the same issues it would be indicated at least on the Japan market by now whether Europe PS3s are RHOS or not. (its not really mandatory yet after all)
9. 0
Brokenscope ( User Karma: 0 ) says:
I’m beginning to wonder if this has something to do with ROHS directives in Europe?
The electronics industry is having a whole hell of trouble with it. For the type of soldered ball grid arrays they use…. without lead well, companies are making boards and the chips are falling off and they don’t know why sometimes.
My company currently has an xray machine set up in productions so we can see what where the failures are occurring, currently it isn’t pretty, we can’t find what is causing half the failures.
Alas,in computing it seems lead is just as important as silicon.
10. 0
Eli Mordino says:
Don’t forget that they’re reimbursing everyone who has paid for a repair over the past nearly-two years. That must account for a fair chunk of the billion dollars.
11. 0
G-Dog says:
That only took two years. Good thing I decided not to buy one, based on the fail rate, after X-Mas of 05. All I heard in my office for a week was how many of the damn things stopped working the first day out of the box.
Springs and paper mache people. The 360 is made from springs and paper mache.
12. 0
Marlowe says:
There may be a better chance for Xbox’s that previously broke to break again, that could explain the anecdotes of Jimmy who sent the thing back to Microsoft for the 12th time
13. 0
Dog_Welder ( User Karma: 0 ) says:
And it’s weird because I still haven’t had any issues with mine, which I purchased at launch. A friend of mine has sent his back twice now. He’ll be thrilled that he’s getting reimbursed for any costs for repairs.
14. 0
Robb ( User Karma: 0 ) says:
The suspicious side of me thinks that some of the sales-losses for each system may be rolled up in the retroactive costs to provide a bigger tax write-off. But I’m very cynical that way.
15. 0
ijed ( User Karma: 0 ) says:
Hm.
Well, Microsoft as done something practically unprecedented by saying they’ll repair broken ones for free (that’d be, treating the consumer with respect after they’ve given them money). Although they’re still looking at games revenue . . .
Which means the problem must be serious. It’s pretty loaded to think so, but that’s the logical conclusion.
I work in games and we’ve had five devkits die out of around fifteen. Another five need to be squeezed (as in pushing downwards when stood on end) every couple of hours to continue working.
It sounds like MS is going to lose a pile of money on this, but its f*ing excellent to see that they’ve approached the problem so well.
16. 0
laterose says:
When I sent my xbox in for the three red lights of doom, they did just send me a new one rather then repair it. I sent it in right after they changed the warrenty to a full one year, so it’s possible that they were a little swamped at the time. But my guess would be that the repairs are close to what it costs them to make a console.
17. 0
Ed Gardner ( User Karma: 0 ) says:
“At the high end, the \$400 repair cost on a 22% failure rate is very unlikely. Since that is the current system’s MSRP, why bother to repair? They could just send you a new 360 to replace your dead one.”
That’s what they did for me, so, maybe that’s not so far fetched. I might also factor in marketing communications around it, and legal fees, alot of things could go into that billion bucks.
I am not going to argue with an extended warranty for free though
18. 0
Dave ( User Karma: 0 ) says:
Seems my 360 has fallen into the “failure” category, as well.
This is just baffling. The original XBox didn’t have problems like this, so it’s not as if Microsoft is inexperienced.
19. 0
This reminds me of something. I feel like I’m getting de’ja vu. Oh, yeah. It was the original X-Box. Constant hardware failures, people not even wanting to touch them for months on end (I can remember the PX I went to in Germany just trying to get rid of their inventory on them), etc. Microsoft has made this kind of mistake before. I’m unimpressed with their ability to learn from their mistakes.
20. 0
~the1jeffy says:
You know, I still don’t think M\$ ‘did the right thing.’ This is a standard recall math situation: if X * Y * Sold Units > The Amount It Would Take To Keep It Quiet, you issue a recall (or in this case, warranty extension and retroactive reimbursement). It’s positive move, to be sure, but the RIGHT thing would have been to release the console when it was ready, and have your second console be as robust as the first.
And I find it odd that a week after GP complains, M\$ issues this move. Does GP have such awesome clout, or did GP get wind of this move early? Maybe my skepticism is a little too healthy …
21. 0
Brok3nHalo says:
Good article, just wanted to mention a correction to your math that might account for the vast difference in percentage assuming the \$100 per repair:
“MS has shipped 11.6 million Xbox 360’s to date”
They are obviously going to ship much more units over the next 3 years, and assuming it’s cheaper to repair the problem than actually fix, and that the rate of shipments per year is constant that number quickly moves from 11.6 to 29 million consoles, with this adjustment it comes out to roughly 34% of all 360s ships, also this doesn’t take into account that numbers of consoles produced would go up over time and shipping and handling costs (I think I read they’re handling that as well) among other factors. Those other factors may be negligible though, as the cost of the repair parts theoretically would go down causing the repair to be cheaper.
22. 0
Jabrwock ( User Karma: 0 ) says:
The math probably gets more complicated if you try to figure out how many users suffer multiple failures. I suspect that happens more often than not, because it happens in other industries too. Something goes, they patch it. But it wasn’t responsible for the overall bad state, so something else goes. Then another. They keep replacing the parts that are obviously broken, but not the underlying cause, so the system keeps failing.
Happened to my car. They took it back 3 times, because first they thought the battery, then the alternator, then then finally they traced it to a short in the circuitry.
23. 0
point09micron says:
Xbox 360 failure rate is 100%!
Anything with moving parts is going to break eventually.
24. 0
qAaRoN ( User Karma: 0 ) says:
“On the other end, the 86% scenario seems pretty extreme (perhaps not if you’re one of the unfortunates who have lost multiple 360’s to the dreaded rings).”
On my 3rd one and my 1yr warranty hasn’t even expired yet.
25. 0
Salen says:
Well, I had a problem with my first 360 with it locking up at random times. Not a Red Ring of Death but it was locking up. Instead of just fixing my original 360, they just mailed me a whole brand new system instead. My new system runs nicely now, and I’ve only had it give me issues 1 or 2 times at most over half a year now.
I’m going to guess that the cost per system is the fact that they’re having to replace all those dead systems completely, bringing the overall number of systems down. Another place that 1 billion might be going is to revamp the manufacturing process as well to make sure there are lot less issues later on, or a reserve of cash to pay for future replacements.
Either way, I love my 360, and I’m happy Microsoft decided to throw down the 3 year warranty extension.
26. 0
Slothy says:
There’s a problem with your math. They’re taking a one-time \$1.3b charge for extending the warranty to 3 years. So, the total cost is going to include consoles that die over the next 1.5 years as well – so the real failure rate is going to be tied up in their future sales projections and failure projections, which we don’t know.
27. 0
Brokenscope ( User Karma: 0 ) says:
I am a 360 fan, I will not own a sony product for the time until they change certain practices. I have no dislike of sony as a game company, I think the cell is an amazing piece of silicon and I would kill for the chance to program one of these things. I have been giddy about them since i first read about them from IBM.
The following is completely off the wall and hypothetical. Everything I say is based on limited information and minimal research. I am attempting to not give a point of view and maintain some neutrality. This is not a condemnation of any system. It is simply the musings of a concerned consumer and as someone with a significant interest in the continuing competition between the big 3 console makers. I am also concerned because I hope to graduate from college and make games. These are my theories and my genuine concerns.
[RAMBLING THOUGHTS AND CRAZY THEORIES]
Guys, this is not a part breaking. Looking back at the issue, Im wondering.
RoHS compliance means removing certain hazardous materials. Lead Cadmium and the like. Lead is a great material for securing a chip to a board. It doesn’t oxidize, it has a very low melting point and it is soft enough to deal with flexing and expansion due to heat. Europe stated RoHS would go into effect on July 22 2006. http://en.wikipedia.org/wiki/RoHS
In Europe RoHS has for the most part banned lead from being used in consumer electronics. This means computers, cells phones, and the like.
To replace lead they use silver. One issue, silver is brittle, very brittle. Really any stress (Expansion, vibration, anything) can pop a chip soldered to a board with silver. Its not that hard, silver just doesn’t work as well as lead. Oh yeah, silver also costs more.
I have heard many a time someone say “I have a box from launch and it is still chugging away.” Well I say the same thing, I got mine in early January 06, its still running. In fact all the people I know who got one within 5 month of launch none of them have had a 3 rings error. 2 Were sent back for repairs because ti stopped reading disc. Just a quick look at the games club at my college, this gives a 5% failure rate. I would like a better sample but this only occurred to me today while at work talking to the guys in production and my mom who works up in test engineering.
Many electronics manufactures rushed out a ton of stuff to beat the dead line. They wanted enough products in stock that they could sell under an exemption so they would have time to switch over to RoHS compliant manufacturing.
Looking at this I have realized something. When did we first really see the 360 supply shortage stop? Summer 06 if memory serves, when did the red ring of death stuff start happening in droves? Well if memory serves, fall 06. Just in time for RoHS compliant boxes to be rolled out and used for a time, enough time for them to start breaking.
The worst part is? The only solution is to find a replacement for silver that works like lead but isn’t banned. No amount of super cooling is going to solve the problem. Eventually the solder points are going to crack and break then the chip will fall off.
Now looking at the bigger picture I realized something. Big N could be avoiding the issue completely. They have a small quiet console that doesn’t generate a ton of heat, and thus by the time it breaks it won’t matter because the console will be rather old.
What about the PS3? Toshiba, Sony, and IBM, they in theory, could have the capacity to Produce RoHS compliant systems for Europe and non RoHS ones for the other major markets. What could this mean? Well PS3s could start failing droves in Europe too. However it RoHS like standards are being put together all over the world. The US is going to have a similar one soon. So this could be a problem for both MS and Sony. Who knows.
If I’m right, well console gamers get screwed. If I’m wrong well the world turns and the console war goes on.
[/RAMBLING THOUGHTS AND CRAZY THEORIES]
28. 0
Shan says:
My first-generation 360 is still running strong for a year and a half now while constantly remaining on.
I was also able to get my original XBox repaired for free two years after the warranty expired and had it back within two weeks. I guess it depends on the guy you get on the phone.
29. 0
Andrew says:
Think about this, also: there’s a cost involved in shipping to and from the repair center, there’s the general cost of repairs, which we can assume must be decently expensive considering the complexity of the procedure. They also have to pay for the hiring, training, and paychecks for all the additional workers they’ve hired to deal with things
(“I think we’ve got the infrastructure we got in place, but this is a very large policy change, and it’s going to require agent training and stuff of that nature…”)
30. 0
@Brokenscope
Wow. Very good point you made, I got my 360 back in about March 06 and it still works perfectly fine (other than some disc-reading problems). I think the reason you said might just be the root of the problem
31. 0
Akbar says:
There’s a problem with your math, I think. Now, I’m no accountant (and thank God for that), but I remember when Western Digital and IBM and the other big boys all reduced their warranties from either 5 to 3 or 3 to 1 years, their biggest claim was that it freed up a lot of money that they had to keep on the books. Looking at Wikipedia (not the greatest source, I know, but it’s all I found on such short notice), I found this:
“A manufacturer or distributor may be required to carry reserve funds on its financial balance sheet to cover potential services or refunds that may arise for any products still covered “under warranty”.” (http://en.wikipedia.org/wiki/Warranty#Extended_warranty)
So it’s possible that it’s not the amount of failures that Microsoft expects to have to pay, but perhaps they have to reserve a certain amount of money for absolutely every unit sold, just in case each one comes back. {That} can add up to a billion, pretty easily.
–Ak
32. 0
Jessica Mintz says:
Anyone want to chat about an Xbox 360 meltdown? I’m a reporter with the Associated Press, looking for some real-world red rings of death. jmintz@ap.org
33. 0
Pirra says:
[random bit] Doesn’t the formula for system failure remind you a bit of the one form the beginning of Fight Club where Edd Norton’s character is talking about how car companies decide weather to recall a car model or just go through some lawsuits? I was just wondering.[random bit]
Anyway… while the situation sucks I think M\$ has acted wisely by giving free repairs and extending warrantys.
34. 0
Randomd says:
The calculation above looks like a pretty good approximation to me. I think it takes into account all the factors that really matter.
As for the refunds of previous repairs, the calculations above Do address that issue. After all, each of those previous repairs were a portion of the of the 11.6 million units sold to date.
As for future systems. The Microsoft statement plainly states that Microsoft is changing the design to address the problems. Consider that it takes awhile for newly built machines to make it from the factory to the retail channel (at least a month if traveling by boat). I strongly suspect that Microsoft is currently building systems with a new design. I think they’re hoping and preying that the new systems won’t have nearly the return rate as the current design.
While the calculation above doesn’t take into account multiple failures of a specific unit, I don’t think that factor would impact the numbers very much. After all, many of the people to have suffered multiple malfunctions have received new units. It’s certainly possible that Microsoft doesn’t even try to repair badly over-heated units.
Overall, the calculations look shocking accurate. It was certainly the right thing for Microsoft to do. On the other hand, a class action suit was probably just about to happen. And while Microsoft can certainly afford this billion dollar charge, it is the admission of a massive failure by the company. It also makes clear that Microsoft has been boldly telling outright lies about 360 failure rates for some time.
35. 0
Brokenscope ( User Karma: 0 ) says:
RoHS Is causing trouble for anyone who makes integrated electronics and computer parts.
36. 0
Ace of Sevens ( User Karma: 0 ) says:
I suspect the \$1 billion includes the cost to redesign the system to prevent new units from having the same problem in the future.
37. 0
Randomd says:
You’re right, the charge probably does take redesign costs into account. But unless it’s a complete redesign, I wouldn’t imagine the redesign costs would suck up any more than a few percentage points of the 1.15 Billion dollar charge.
There are strong rumors in the industry that MS is about to drop the price on the standard unit by \$50. It wouldn’t surprise me if the new design were to coincide with this price drop.
(It also wouldn’t surprise me if MS were to discontinue the core unit at the same time. A \$50 difference in unit prices wouldn’t make much sense, and I haven’t heard any rumors about a price drop for the core price.)
38. 0
Salen says:
Uh… yeah. Nothing like free enterprise in the middle of a discussion on the 360 and Warranties. Something tells me having someone else fix the 360 will void said warranty.
39. 0
Randomd says:
Microsoft just held a conference call regarding the red-ring warranty problem. Here are the bits pertaining to the calculations above.
They refused to discussed defect rates other than to say it is a “meaningful” amount.
They mention that the Billion is to be split 50:50 between the existing 1 year warranty and the 3 year warranty. They say some of the existing warranty portion may be used to fix (re-manufacture?) currently warehoused units and that some of the money will be used to write off units which cannot be repaired cost effectively. Unless they have truly huge amounts of warehoused, defective 360s, this wouldn’t seem to have a large effect on the calculations above.
The Billion dollar charge does not seem to include any redesign fees. (there was no mention of this)
The Billion dollar charge does Not include warranty repairs made prior to today.
“We’ve been incurring some costs during the course of the year, both for the repairs that we’ve done to date and that; and this is over and above that (billion dollar charge).”
So remove those repairs from the stats above.
The Billion dollar charge does NOT include costs to fix newly made machines. Microsoft thinks the newly (redesigned?) Xboxes will not have the issue.
“And to be clear, clearly, with the Xboxes that we’re now manufacturing, even though the warranty is for 3 years, that’s as much a security blanket, if you like, rather than an expectation. We expect it to be a very good performance going forward, given the corrections we’ve made in the manufacturing process.”
Bottom Line:
Assuming that most of the billion will be spent on warranty repair, it would seem that Microsoft expects a truly Massive proportion of the Xbox 360s manufactured to date to fail within a 3 year time period!
http://www.joystiq.com/2007/07/06/microsofts-red-ring-warranty-conference-call-transcribed/
40. 0
logic says:
The 3 year warrenty does not just go for currently shipped consoles but also future consoles so the real equation is
s= total number of 360 that will be bought ever
f= failure rate
c= cost to fix defective units
(s*f)*c= 1×10^9
So the cost really depends on how many they sell.
But if they fix the problem then the [posted] math might be right
41. 0
Terminator44 says:
@Brokenscope
Yes, that theory may be one the right track, though I think more evidence is needed before we can start taking it too seriously. To add to the anecdote, many of my closest Live friends got their 360s at launch, and few have had to send it in (or won’t admit it) for El Circulo Rojo de la Muerte.
I for one would love to see the research RoHS presented in favor of banning lead from consumer electronics. If that is the cause of hardware unrealiability all around, they better have a good reason for it.
42. 0
Randalor ( User Karma: 0 ) says:
I’m on my third system now. The first one lasted… about half a year to a year before crapping out on me *Yay for Dead Rising killing the CPUs!*, sent that in, got a brand new system *the policy they had for Canadians at the time*, that one lasted… 3 months maybe? before crapping out, sent that one in, got a refurbished system that’s been going strong for well over a year now *Other than the disc drive making an odd sound whenever there’s a disk in it, it’s fine*. So yeah, that whole thing about the RoHS policy sounds reasonable to me.
43. 0
Terrible Tom ( User Karma: -1 ) says:
I guess it IS a moving number… it’s constantly going up.
44. 0
It’s plain and simple why the 360 has such a high failure rate: poor manufacturing. They skimp on materials to try and save every dollar, but as they just found out, it cost them a nice billion and some. I don’t blame ROHS, hell they’re helping the planet. PSP is built on all ROHS compliant solder, and have you seen the NAND flash pop because of it? No. It’s all because Microsoft saves a dollar a system, just to lose a dollar and change per system for that penny pinching.
45. 0
ZippyDSMlee ( User Karma: -1 ) says:
thats probably it,but the RoHS and removing lead from solder is a intersing angle to it.
I think the manufacturing spec is to high for china factories to do without a high fail rate,this is what they get for trusting china shops LOL
46. 0
[…] Xbox 360 failure rates July 8, 2007 Posted by Boris Popov in Life, Games. trackback James Robertson has brought up a topic of consoles again. While I totally agree that reportedfailure rates are absolutely unacceptable and point to a problem in either hardware design or shoddy assembly, the other important side of the story that often gets overlooked is how Microsoft deals with these failures when they do occur. To be honest, I was a little concerned about the perceived failures when I bought my own Xbox 360 for Christmas, and those concerns grew into uneasy anticipation when I started seeing intermittent error messages during bootup before network adapter died completely. I pretty much expected the repair process to cost me a couple hundred dollars and countless hours spent on the phone, but actually that couldn’t be further from the truth. Here’s how the process worked (in my case, other’s may differ, but the chatter in blogosphere indicates similar experiences), […]
47. 0
StiC says:
According to MS the failure rates went from industry norm (3 to 5% I think) to significant (currently estimated as high as 33%) over the course of a few months. What will the next few months have in store for X360 owners? 40%? 50%? And what about Forza bricking systems? MS says there is no software problem so perhaps it’s all the gaming marathons with this popular title. If so what impact will a title such as Halo 3 have on the failure rates?
48. 0
LotsofLuck says:
“(11,600,000 * X) * Y = 1,000,0000,000”
There’s one extra zero in there, right now it’s says \$11,600,000(X)(Y)=\$10 billion.
Anyways, that’s freaking crazy! 1 billion for repairs? I’m hoping that Y is small, because that means people will get their 360’s back quicker, but that means the failure rate is pretty high | 6,730 | 28,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2015-48 | longest | en | 0.918217 |
https://www.thestudentroom.co.uk/showthread.php?t=5702578 | 1,606,769,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00484.warc.gz | 865,876,055 | 36,828 | # OCR A physics help
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#1
A satellite of mass 3000 kg moves from a parking orbit of radius 6800 km to a geostationary
orbit of radius 42 000 km. The mass of the Earth is 6.0 × 1024 kg. What is the magnitude of the change in gravitational potential?
A 8.4 J kg-1
B 2.5×104Jkg-1
C 4.9×107Jkg-1
D 1.5 × 1011 J kg -1
0
1 year ago
#2
A satellite of mass 3000 kg moves from a parking orbit of radius 6800 km to a geostationary
orbit of radius 42 000 km. The mass of the Earth is 6.0 × 1024 kg. What is the magnitude of the change in gravitational potential?
A 8.4 J kg-1
B 2.5×104Jkg-1
C 4.9×107Jkg-1
D 1.5 × 1011 J kg -1
What is it that you don't know? It is applying the equation of gravitational potential .
0
#3
(Original post by Eimmanuel)
What is it that you don't know? It is applying the equation of gravitational potential .
I don’t know how to get the answer, it is C but I can’t get it. You are supposed to use square root GMm/r
0
1 year ago
#4
I don’t know how to get the answer, it is C but I can’t get it. You are supposed to use square root GMm/r
What is square root GMm/r?
0
#5
(Original post by Eimmanuel)
What is square root GMm/r?
That is the formula is think that you use
0
1 year ago
#6
That is the formula is think that you use
I am asking you what does it represent? And are you this is formula that you are supposed to apply. The formula that you should apply is gravitational potential is given to you in post #2.
https://www.s-cool.co.uk/a-level/phy...onal-potential
0
1 year ago
#7
A satellite of mass 3000 kg moves from a parking orbit of radius 6800 km to a geostationary
orbit of radius 42 000 km. The mass of the Earth is 6.0 × 1024 kg. What is the magnitude of the change in gravitational potential?
A 8.4 J kg-1
B 2.5×104Jkg-1
C 4.9×107Jkg-1
D 1.5 × 1011 J kg -1
I don’t know how to get the answer, it is C but I can’t get it. You are supposed to use square root GMm/r
You need to find the gravitational potential at radius 6800 km first. Then find the gravitational potential at radius 42 000 km. Lastly compute the difference.
0
#8
(Original post by Eimmanuel)
I am asking you what does it represent? And are you this is formula that you are supposed to apply. The formula that you should apply is gravitational potential is given to you in post #2.
https://www.s-cool.co.uk/a-level/phy...onal-potential
Thanks, was just using the wrong formula
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4.38% | 935 | 3,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-50 | latest | en | 0.918353 |
https://www.bellaonline.com/newsdtl.asp?name=jokes&date=11/13/2008%204:44:47%20PM | 1,717,009,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00269.warc.gz | 583,597,682 | 10,738 | ## November 13 2008 Jokes & Riddles Newsletter
Happy Thursday (almost Friday) Everyone!
1. The human body holds nine quarts of blood. Suppose you were alone in the desert and accidently cut a major artery. If you bleed at one cup a minute, how long will it take you to bleed to death?
2. Suppose you are on an elevator on the 16th floor of a building, when the cable breaks. As you plummet toward the ground, you recall that you once heard that by jumping up fast at the instant of impact, you can escape death. You also recall that the floors are twelve feet apart and that the acceleration of a falling object is 32 feet per second squared. When should you jump?
3. If a boy and a half could eat a hot dog and a half in a minute and a half, how many hot dogs could six boys eat in six minutes?
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Here's the latest article from the Jokes & Riddles site at BellaOnline.com.
Fry Thieves and Burger Bouncers
Do you like to people watch? Ever hope to catch someone doing something that's completely laugh worthy (excluding yourself)? Here's a true story about what happens if you ever think about stealing something...even french fries.
http://www.bellaonline.com/articles/art44882.asp
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Please visit jokes.bellaonline.com for even more great content about Jokes & Riddles.
To participate in free, fun online discussions, this site has a community forum all about Jokes & Riddles located here -
Please feel free to contact me either through the contact form or the Jokes Forum for suggestions on articles or links, or ideas about what you would like to see on the site. I would also be happy to review sites or products- humor books, CD's or DVD's and any other joke oriented materials.
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Answer to above common sense riddles:
1. Apply a tourniquet, and you won't.
2. You would impact between the 2nd and 3rd second after the cable broke, so you would want to jump 2 seconds after the break. However, considering the calculations involved, you would probably end up as a heap of screaming bloody mess at the bottom of the elevator shaft before you figured the answer out. Besides, there's no telling if jumping really works.
3. Who cares? I want to see how that half a boy can eat anything.
Have fun passing this message along to family and friends, because we all love humor!
Rose Thomason, Jokes & Riddles Editor
http://jokes.bellaonline.com
One of hundreds of sites at BellaOnline.com
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