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https://math.answers.com/Q/What_is_the_mixed_number_for_2_and_3_over_7_minus_8_over_14 | 1,701,257,113,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100081.47/warc/CC-MAIN-20231129105306-20231129135306-00759.warc.gz | 453,612,165 | 44,212 | 0
# What is the mixed number for 2 and 3 over 7 minus 8 over 14?
Updated: 11/1/2022
Wiki User
10y ago
2 and 3/7 minus 8/14 = 13/7 or 1 and 6/7
Wiki User
10y ago | 73 | 167 | {"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} | 3.15625 | 3 | CC-MAIN-2023-50 | latest | en | 0.820461 |
http://mathhelpforum.com/geometry/129699-construction-hw-proble.html | 1,529,420,586,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863043.35/warc/CC-MAIN-20180619134548-20180619154548-00002.warc.gz | 195,505,496 | 11,795 | 1. construction hw proble
We are doing basic geometric constructions in class, and I have a problem on my hw that I dont know how to do. Please see the attached sheet if you can help. Thanks
2. where is the sheet?
3. I forgot to upload, i will post again
4. Hello, igottaquestion!
Mark is standing at G.
There is a chair H in the next room
and a mirror CD on the wall.
By looking in the mirror, can Mark see the chair?
Construct the line of sight from G to the mirror to H.
Code:
C D
* - - - - * - - * - - - *
| |
| |
| | |
| | o G |
| | |
* - - o - - * - - - - - *
H
Draw HP perpendicular to the far wall.
Extend HP to Q so that PQ = HP.
Draw line GQ, intersecting the far wall at X.
Code:
Qo
:\
: \
: \
: \
P: \ X
* - - o - - o - - - - - *
| : \ |
| : \ |
| : | \ |
| : | oG |
| : | |
* - - o - - * - - - - - *
H
If point $\displaystyle X$ occurs on the mirror $\displaystyle (CD)$,
. . Mark can see the chair.
Justification: Draw line segment $\displaystyle HX.$
Code:
Qo
:\
: \
: \
: \
P: θ\ X
* - - o - - o - - - - - *
| : α/ \β |
| : / \ |
| : / \ |
| : / oG |
| :/ |
* - - o - - * - - - - - *
H
The line of sight is: .$\displaystyle GX + XH$
Right triangles $\displaystyle QPX$ and $\displaystyle HPX$ are congruent.
. . Hence: .$\displaystyle \angle \alpha = \angle\theta$
$\displaystyle \angle\theta$ and $\displaystyle \angle\beta$ are vertical angles.
. . Hence: .$\displaystyle \angle\theta = \angle\beta$
Therefore: .$\displaystyle \angle\alpha \,=\,\angle\beta$
The angle of incidence equals the angle of reflection.
5. awesome, thanks so much man
6. I answered this for him elsewhere, but gave the hint, the same method you used here, rather than simply do his homework for him. You spoil everyone else's efforts by this practice, and incidentally spoil the efforts of conscientious students who do their own work! Perhaps you saw my hint? What good is it for the rest of us to try to guide him if you butt in, simply fluffing your own feathers?
You might think that you are helping, but you are not. You won't be there at exam time. If he wants to learn to play baseball, he has to pick up the bat, not just watch you hit home runs. You do this in every single forum that you attend, and it's extremely annoying for us to see our own efforts sabotaged like this.
I'm not going to change my ways
anymore than you'll change yours.
8. Originally Posted by Diagonal
I answered this for him elsewhere, but gave the hint, the same method you used here, rather than simply do his homework for him. You spoil everyone else's efforts by this practice, and incidentally spoil the efforts of conscientious students who do their own work! Perhaps you saw my hint? What good is it for the rest of us to try to guide him if you butt in, simply fluffing your own feathers?
You might think that you are helping, but you are not. You won't be there at exam time. If he wants to learn to play baseball, he has to pick up the bat, not just watch you hit home runs. You do this in every single forum that you attend, and it's extremely annoying for us to see our own efforts sabotaged like this.
It is MHF policy that members do not provide further help or solutions when someone is already helping a member and is waiting for the OP to give feedback. Infractions are given for violating this policy. Having said this, I give the benefit of the doubt to Soroban that your efforts (which I greatly appreciate, by the way) elsewhere was not seen by him in this instance. I sincerely hope I am correct in doing so .....
MHF strongly encourages that hints rather than complete solutions be given. Having said this, there are some members who feel that providing complete solutions is more beneficial. This style of help is not condoned by the majority - and I personally do not agree with it either - but it is not currently illegal. It is something that gets debated from time to time, and will continue to be debated. | 1,080 | 4,246 | {"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} | 3.859375 | 4 | CC-MAIN-2018-26 | latest | en | 0.825121 |
https://byjus.com/question-answer/let-f-x-satisfy-the-requirements-of-lagrange-s-mean-value-theorem-in-02-if/ | 1,723,596,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00196.warc.gz | 113,464,025 | 41,139 | 1
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Question
# Let f(x) satisfy the requirements of Lagrange's mean value theorem in [0,2]. If f(0)=0 and |f′(x)|≤12 for all x∈[0,2], then
A
f(x)2
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B
|f(x)|2x
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C
|f(x)|1
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D
f(x)=3, for at least one x[0,2]
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Solution
## The correct option is C |f(x)|≤1By Lagrange's mean value theorem in [0,2], there exists c∈(0,2) such thatf′(c)=f(2)−f(0)2⇒2f′(c)=f(2)−f(0)⇒2f′(c)=f(2) , since f(0)=0Now, since |f′(x)|≤12 for all x∈[0,2],⇒|f′(c)|≤12⇒2|f′(c)|≤1⇒|f(2)|≤1Also, |f′(k)|=∣∣∣f(k)−f(0)k∣∣∣≤12∴|f(k)|≤k2 for k∈[0,2]∴|f(x)|≤1
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Join BYJU'S Learning Program | 435 | 1,009 | {"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} | 3.421875 | 3 | CC-MAIN-2024-33 | latest | en | 0.786675 |
https://www.eevblog.com/forum/beginners/lm338-powering-a-stepper-motor/ | 1,590,832,388,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00077.warc.gz | 705,923,788 | 13,356 | ### Author Topic: LM338 powering a stepper motor? (Read 4123 times)
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#### aboodi
• Contributor
• Posts: 18
##### LM338 powering a stepper motor?
« on: January 10, 2014, 01:31:08 pm »
Hello guys/girls,
In this project I'm working with two stepper motors (3V, 2A each) from a 6V battery. I used a voltage regulator (LM338) to get 3V. I built the typical circuit provided in the LM338 datasheet. Then I used the output to power the two motor drivers.
The problem is that the measured current going out of the battery doesn't exceed 400mA. I tried to directly power the motor driver, and it was approximately 2.5A.
My guess is that the input impedance has something to do with the problem, and to come over it a MOSFET is needed. That is my guess, and i'm waiting to hear your interesting explanation/solutions.
thanks,
Abdullah
#### GeoffS
• Supporter
• Posts: 1267
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #1 on: January 10, 2014, 01:36:58 pm »
A circuit diagram might help.
#### aboodi
• Contributor
• Posts: 18
##### Re: LM338 powering a stepper motor?
« Reply #2 on: January 10, 2014, 02:18:33 pm »
A simple diagram is attached.
#### Zbig
• Frequent Contributor
• Posts: 871
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #3 on: January 10, 2014, 03:39:16 pm »
I'd rather look for some motors with voltage rating that matches your supply voltage or use a dedicated, current-limited stepper motor driver. It's very innefficient to use linear regulator for powering motors.
#### idpromnut
• Supporter
• Posts: 613
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #4 on: January 10, 2014, 03:39:48 pm »
~2A out of a 6V battery? If I understand correctly, the power that will drive the stepper is coming from the battery as well right? I think you should tell us what type of battery you are using/link the specs. I would suspect that the battery does not have the capability to source 2A (unless when you said you "directly powered the motor driver" you were directly powering it from the battery?).
#### aboodi
• Contributor
• Posts: 18
##### Re: LM338 powering a stepper motor?
« Reply #5 on: January 10, 2014, 03:58:25 pm »
The battery is a lead-acid battery. And yes, It is capable of sourcing 4A.
Datasheet of battery: http://www.yuasabatteries.com/pdfs/NP_4_6_DataSheet.pdf
#### Kremmen
• Super Contributor
• Posts: 1283
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #6 on: January 10, 2014, 04:10:34 pm »
You really need to check some L298 appnotes from the web because as it is now, your circuit will not work.
Nothing sings like a kilovolt.
Dr W. Bishop
#### aboodi
• Contributor
• Posts: 18
##### Re: LM338 powering a stepper motor?
« Reply #7 on: January 10, 2014, 04:11:13 pm »
The problem is supply voltage. The minimum supply voltage for the L298N has to be >Vih+2.5. Where Vih is the input high voltage, i.e. 5V. So the minimum supply voltage has to be 7.5V!!!!!!!!
Is there a solution guys??
#### SeanB
• Super Contributor
• Posts: 15379
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #8 on: January 10, 2014, 04:15:01 pm »
You need to use the L298's built in current limiting to provide the current limiting function. The appnote has this when used with the L297 controller. the IC does need at least 2.5V over the logic voltage to operate, so it really needs a 7V5 supply to operate properly with a 5V logic supply. Not going to operate well with a 6V SLA battery.
#### Kremmen
• Super Contributor
• Posts: 1283
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #9 on: January 10, 2014, 05:28:02 pm »
There are a number of issues here. Firstly the voltages:
You have a microcontroller (MCU) that obnviously is supposed to control the steppers, right. That needs a supply and you have marked it as 5V. That is OK but what is not OK is where it is connected to. More about that.
The L298 bridge/driver also needs a supply voltage Vss for its internal logic. Normally you would want to make that the same as the MCU supply to ensure both have compatible signal levels. --> First modification you need to do is to create a +5V regulator and supply the MCU and L298 Vss from that.
Then there is the L298 Vs supply. If you check the L298 datasheet it specifies an absolute maximum rating of 50V for that input. This should be a clue. In fact it is the positive power rail for the motors. So you need a hefty supply capable of producing the amperage the motor connected to this bridge needs.
Another non-obvious fact is that the voltage rating indicated in the stepper specs is _not_ an indication of the actual voltage you should provide to the Vs rail. Typically the working voltage is much, much higher than the nominal voltage of the motor. It could easily be like 20-30 volts or more for your motors, depending on exactly how the drive coils are connected and what the resulting coil inductance will turn out to be. Some high-inductance steppers need closer to 100 V before they are able to produce their nominal torque at any appreciable speed. So please _really_ do read some L298 application notes freely available in the net.
Finally, your drawing provides no clue how you have planned to provide the step sequencing necessary to make the stepper actually move. Your safest bet is to do what SeanB suggests and hook up an L297 sequencer chip that will handle the phasing signals for you. Additionally, it will provide the non-trivial logic needed for pwm current control without which you either burn your motors or not get them to move at all. Again, _do_ find a L297 app note and get familiar with it.
I don't want to discourage you but there is a wide gap between your drawing and a functional stepper control and drive circuit. The bright side of the coin is that all info needed to close that gap is easily available behind a couple of googles.
Nothing sings like a kilovolt.
Dr W. Bishop
#### aboodi
• Contributor
• Posts: 18
##### Re: LM338 powering a stepper motor?
« Reply #10 on: January 10, 2014, 06:17:01 pm »
Thanks for the explanation.
For the logic supply, I use an Arduino and it has a built-in 5V regulator, so I use that to supply the motor drive.
The Arduino Mega controls the motor driver and supplies the logic voltage (5V), as well.
So, the only problem I see is the motor supply voltage. And you explained that the motor voltage needed is not the rated voltage, not even close for high inductance motors. Thanks.
You asked me to google app notes for the L298. Isn't that the datasheet itself?
Abdullah
#### woodchips
• Frequent Contributor
• Posts: 527
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #11 on: January 10, 2014, 06:26:21 pm »
I agree with Kremmen, the voltage rating of a stepper is irrelevant.
The problem is that the motor is inductive, so slowing the rate of rise of the current. The way this is overcome is to run the motor from a much higher voltage with a resistor in series. Time constant of an LR circuit is L/R. The resistor actually greatly increases the achievable step rate of the motor.
Obviously the current needs to be limited, but only after the initial current spike.
#### Kremmen
• Super Contributor
• Posts: 1283
• Country:
##### Re: LM338 powering a stepper motor?
« Reply #12 on: January 10, 2014, 11:48:34 pm »
Thanks for the explanation.
For the logic supply, I use an Arduino and it has a built-in 5V regulator, so I use that to supply the motor drive.
No, you will not. You can try but it will only crash your Arduino. What you can do is to power the _logic_ supply i.e. Vss of the L298, but definitely not the power rail Vs. Please make a clear difference between the _drive_ part which is power and the logic part.
Quote
The Arduino Mega controls the motor driver and supplies the logic voltage (5V), as well.
Yes, that will work.
Quote
So, the only problem I see is the motor supply voltage. And you explained that the motor voltage needed is not the rated voltage, not even close for high inductance motors. Thanks.
You asked me to google app notes for the L298. Isn't that the datasheet itself?
Abdullah
Well, the datasheet contains all you really need including how to apply the L297 chip to create the sequencing and current limitation. If you understand what is written there then the info will be enough to correctly implement a stepper drive. But other than the datasheet there are numerous application notes scattered around the net to illuminate various points of the design task.
Nothing sings like a kilovolt.
Dr W. Bishop
#### aboodi
• Contributor
• Posts: 18
##### Re: LM338 powering a stepper motor?
« Reply #13 on: January 11, 2014, 06:36:34 pm »
yeah yeah, I meant the logic supply. Sorry, I wasn't careful when I was posting the reply.
Of course, the Arduino wouldn't be able to supply over 40 mA/pin.
And I will definitely check the "L298 app note"
Thank you all really much, especially Kremmen. ^_^
Smf | 2,307 | 9,007 | {"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-2020-24 | latest | en | 0.90178 |
https://metanumbers.com/255199800000000 | 1,623,725,576,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00242.warc.gz | 377,903,254 | 8,253 | ## 255199800000000
255,199,800,000,000 (two hundred fifty-five trillion one hundred ninety-nine billion eight hundred million) is an even fifteen-digits composite number following 255199799999999 and preceding 255199800000001. In scientific notation, it is written as 2.551998 × 1014. The sum of its digits is 39. It has a total of 19 prime factors and 360 positive divisors. There are 68,053,120,000,000 positive integers (up to 255199800000000) that are relatively prime to 255199800000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 15
• Sum of Digits 39
• Digital Root 3
## Name
Short name 255 trillion 199 billion 800 million two hundred fifty-five trillion one hundred ninety-nine billion eight hundred million
## Notation
Scientific notation 2.551998 × 1014 255.1998 × 1012
## Prime Factorization of 255199800000000
Prime Factorization 29 × 3 × 58 × 425333
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 19 Total number of prime factors rad(n) 12759990 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,199,800,000,000 is 29 × 3 × 58 × 425333. Since it has a total of 19 prime factors, 255,199,800,000,000 is a composite number.
## Divisors of 255199800000000
360 divisors
Even divisors 324 36 18 18
Total Divisors Sum of Divisors Aliquot Sum τ(n) 360 Total number of the positive divisors of n σ(n) 8.49837e+14 Sum of all the positive divisors of n s(n) 5.94637e+14 Sum of the proper positive divisors of n A(n) 2.36066e+12 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.5975e+07 Returns the nth root of the product of n divisors H(n) 108.105 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,199,800,000,000 can be divided by 360 positive divisors (out of which 324 are even, and 36 are odd). The sum of these divisors (counting 255,199,800,000,000) is 849,836,834,414,568, the average is 236,065,787,337,3.8.
## Other Arithmetic Functions (n = 255199800000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 68053120000000 Total number of positive integers not greater than n that are coprime to n λ(n) 2126660000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7938485835904 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 68,053,120,000,000 positive integers (less than 255,199,800,000,000) that are coprime with 255,199,800,000,000. And there are approximately 7,938,485,835,904 prime numbers less than or equal to 255,199,800,000,000.
## Divisibility of 255199800000000
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 2 0 3
The number 255,199,800,000,000 is divisible by 2, 3, 4, 5, 6 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (255199800000000)
Base System Value
2 Binary 111010000001101001010101011011001111111000000000
3 Ternary 1020110120212021211212002012110
4 Quaternary 322001221111123033320000
5 Quinary 231422143142300000000
6 Senary 2302433100350455320
8 Octal 7201512533177000
10 Decimal 255199800000000
12 Duodecimal 24757520804540
20 Vigesimal 14i8egha0000
36 Base36 2igl63utxc
## Basic calculations (n = 255199800000000)
### Multiplication
n×i
n×2 510399600000000 765599400000000 1020799200000000 1275999000000000
### Division
ni
n⁄2 1.276e+14 8.50666e+13 6.38e+13 5.104e+13
### Exponentiation
ni
n2 65126937920040000000000000000 16620381531806623992000000000000000000000000 4241518042840744081433601600000000000000000000000000000000 1082434556229349321433038841599680000000000000000000000000000000000000000
### Nth Root
i√n
2√n 1.5975e+07 63429.8 3996.87 760.985
## 255199800000000 as geometric shapes
### Circle
Diameter 5.104e+14 1.60347e+15 2.04602e+29
### Sphere
Volume 6.96193e+43 8.18409e+29 1.60347e+15
### Square
Length = n
Perimeter 1.0208e+15 6.51269e+28 3.60907e+14
### Cube
Length = n
Surface area 3.90762e+29 1.66204e+43 4.42019e+14
### Equilateral Triangle
Length = n
Perimeter 7.65599e+14 2.82008e+28 2.2101e+14
### Triangular Pyramid
Length = n
Surface area 1.12803e+29 1.95873e+42 2.0837e+14
## Cryptographic Hash Functions
md5 7b2339ce4eca5eae1b18e82de38269f7 f87b14a118fbe173917e47f8962fdc6632d034c4 2a316c482af3b9340ef369162f993b76242a57aff293112b01fb5b6eb7f7246f e988024be5b2247276dd56dc03ca4a7efd694c56ffbb9f85ba01cfa428149938417364c8bf0ad8d25ffd7e0e7225dc9cfc9b07b88045a776fdcd06c8cd06e11f 4d0386c46283dcf66f2528bd0ea4eb9eb74eb1d9 | 1,770 | 4,928 | {"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} | 3.328125 | 3 | CC-MAIN-2021-25 | latest | en | 0.767906 |
https://nuprl.org/wip/Standard/arithmetic/modulus-is-rem.html | 1,674,957,436,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00802.warc.gz | 449,535,824 | 3,124 | ### Nuprl Lemma : modulus-is-rem
`∀[a:ℕ]. ∀[n:ℤ-o]. (a mod n ~ a rem n)`
Proof
Definitions occuring in Statement : modulus: `a mod n` int_nzero: `ℤ-o` nat: `ℕ` uall: `∀[x:A]. B[x]` remainder: `n rem m` sqequal: `s ~ t`
Definitions unfolded in proof : uall: `∀[x:A]. B[x]` member: `t ∈ T` uimplies: `b supposing a` nat: `ℕ` so_lambda: `λ2x.t[x]` so_apply: `x[s]` modulus: `a mod n` has-value: `(a)↓` int_nzero: `ℤ-o` nequal: `a ≠ b ∈ T ` not: `¬A` implies: `P `` Q` false: `False` prop: `ℙ` all: `∀x:A. B[x]` bool: `𝔹` unit: `Unit` it: `⋅` btrue: `tt` uiff: `uiff(P;Q)` and: `P ∧ Q` less_than: `a < b` less_than': `less_than'(a;b)` top: `Top` true: `True` squash: `↓T` decidable: `Dec(P)` or: `P ∨ Q` nat_plus: `ℕ+` le: `A ≤ B` guard: `{T}` int_lower: `{...i}` iff: `P `⇐⇒` Q` rev_implies: `P `` Q` subtract: `n - m` subtype_rel: `A ⊆r B` bfalse: `ff` exists: `∃x:A. B[x]` sq_type: `SQType(T)` bnot: `¬bb` ifthenelse: `if b then t else f fi ` assert: `↑b` gt: `i > j`
Lemmas referenced : subtype_base_sq nat_wf set_subtype_base le_wf int_subtype_base value-type-has-value int-value-type equal_wf lt_int_wf bool_wf eqtt_to_assert assert_of_lt_int top_wf less_than_wf decidable__lt rem_bounds_1 less_than_transitivity1 less_than_irreflexivity rem_bounds_4 decidable__le false_wf not-le-2 not-equal-2 condition-implies-le minus-zero add-zero minus-add minus-minus add-swap add-commutes add-associates zero-add add_functionality_wrt_le le-add-cancel not-lt-2 eqff_to_assert bool_cases_sqequal bool_subtype_base iff_transitivity assert_wf bnot_wf not_wf iff_weakening_uiff assert_of_bnot not-gt-2 int_nzero_wf
Rules used in proof : sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut thin instantiate extract_by_obid sqequalHypSubstitution isectElimination cumulativity hypothesis independent_isectElimination sqequalRule intEquality lambdaEquality natural_numberEquality hypothesisEquality callbyvalueReduce remainderEquality setElimination rename because_Cache lambdaFormation independent_functionElimination voidElimination unionElimination equalityElimination equalityTransitivity equalitySymmetry productElimination lessCases sqequalAxiom isect_memberEquality independent_pairFormation voidEquality imageMemberEquality baseClosed imageElimination dependent_functionElimination dependent_set_memberEquality minusEquality addEquality applyEquality dependent_pairFormation promote_hyp impliesFunctionality
Latex:
\mforall{}[a:\mBbbN{}]. \mforall{}[n:\mBbbZ{}\msupminus{}\msupzero{}]. (a mod n \msim{} a rem n)
Date html generated: 2017_04_14-AM-07_19_04
Last ObjectModification: 2017_02_27-PM-02_53_16
Theory : arithmetic
Home Index | 904 | 2,712 | {"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} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.244765 |
https://www.doubtnut.com/qna/642715947 | 1,718,214,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00650.warc.gz | 683,386,901 | 38,021 | A space - ship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000 kg. Mass of the sun =2×1030kg. Mass of the Mars =6.4×1023 kg. Radius of Mars = 3395 Radius of the orbit of Mars = =2.28×1011m,G=6.67×10−11Nm2kg−2.
Text Solution
Verified by Experts
Let ,R be the radius of orbit of Mars and R. be the radius of the Mars. M be the mass of the Sun and M. be the mass of Mars. M be the mass of the Sun and M. be the mass of Mars. If m is the mass of the space - ship , then Potential energy of space - ship due to gravitational attraction of the Sun = -GMm/R Potential energy of space -ship to gravitational attraction of Mass - GM.m/R. Since the K.E. of space - ship is zero. therefore , total energy of space - ship =−GMmR−GM.mR.=−Gm(MR+M.R.) ∴ Energy required to rocket out the spaceship from the solar system = - (total energy of spaceship) =−[−Gm(MR+M.R.)]=Gm(MR+M.R.) =−6.67×10−11×1000×[2×10302.28×1011+6.4×10233395×103] =6.67×10−8[202.28+6.433.95]×1018J=5.98×1011J
|
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https://homework.cpm.org/category/MN/textbook/cc1mn/chapter/8/lesson/8.1.1/problem/8-14 | 1,620,580,891,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00613.warc.gz | 323,496,883 | 14,704 | ### Home > CC1MN > Chapter 8 > Lesson 8.1.1 > Problem8-14
8-14.
If $6$ rabbits can eat $24$ daisies, how many daisies would $4$ rabbits eat?
Try finding how many daisies one rabbit can eat. Division is helpful here!
$16$ daisies | 78 | 232 | {"found_math": true, "script_math_tex": 4, "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} | 2.953125 | 3 | CC-MAIN-2021-21 | longest | en | 0.745178 |
https://advice.thisoldhouse.com/showthread.php?119900-Deck-weight-limit&mode=hybrid | 1,484,997,606,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281069.89/warc/CC-MAIN-20170116095121-00318-ip-10-171-10-70.ec2.internal.warc.gz | 804,204,753 | 21,096 | #### Hybrid View
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## Deck weight limit
My driveway ends with a 4' drop onto my side lawn. I was thinking of extending the driveway by one car length by building a pressure treated deck. I'd like it to be strong enough to hold a car but the only purpose is to park my motorcycle trailer on it. With both bikes in it the trailer it could weigh as much as 3k pounds. I believe a normal deck is built for a 40 psf live load. What would one strong enough for a car be?
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## Re: Deck weight limit
Let's do some calculations. Let's say your trailer has four tires, each inflated to 30 psi. 3000 pounds divided by 30 psi means a tire contact area of 100 square inches (25 square inches per tire). A square foot is 144 square inches. So each of your tires weighs 750 pounds in an area a sixth of a square foot (about 5x5 - further calculation puts this at about 4300 psf); your deck needs to be able to support that much weight in that spot.
That's not to say the the entire structure must be able to support 4300 pounds loaded on every square foot simultaneously, but your average backyard deck isn't built to handle that much point load. Were the 4300 pounds evenly distributed across the entire deck, it might be OK. A standard substructure may be able to handle the load, but the decking itself must be able to transfer the weight to a larger part of the structure than standard decking would.
P.S. -- at 200 lbs, a man with a size 10 shoe exerts probably around 600 psf point load. And don't even ask about the lady with stilettos.
Last edited by Fencepost; 02-05-2012 at 11:26 PM.
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## Re: Deck weight limit
Thanks, Fencepost. That explains a lot. I would like to have an architect draw up the plans to take to the building inspector. I was interested, before i do that, in finding out what kind of structure would be necessary. IF it was pretty much 8 x 12 deck with extra support it would be feasible. If it would turn out to be an elaborate structure with steel beams and massive footings the cost wouldn't warrant a parking place for a utility trailer.
I was thinking something like 4 short 6 x6 posts on typical concrete columns below frost line. Each directly under where the parked tie would rest. 2 triple 2x10 beams running the same direction as the travel of the wheels onto the deck at 7' apart with 2 x 8 joists and 2x4 decking. The posts closest to the driveway would be 16" high and farthest from the driveway would be about 36" high.
A little over built for 8 x 12 deck but cost wise would make sense for the convenience I’m looking for. If more would be required then it’s probably not worth it.
4. dj1
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## Re: Deck weight limit
A deck like you are planning needs to be permitted and to get approved you need to show your plan checker a plan signed off by an engineer.
Since there will be costs involved, you have to ask yourself if it's all worth it for you.
A car parked on the deck is like having the entire New York Giants Defense squad on it...without collapsing.
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## Re: Deck weight limit
Wouldn't some type of fill, perhaps surrounded by and contained by, some appropriate landscaping blocks, etc. be better?
The problem with a deck is that, although you will all you can to prevent it, somebody will drive on it. It may be somebody turning around, backing up, the UPS or FEDEX guy accidentally getting a wheel on its edge, etc.
Good luck.
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## Re: Deck weight limit
Good point about a retaining wall. Look at the dry stacked block like Pavestone's Anchor retaining wall. Lots of companies make similar products.
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## Re: Deck weight limit
A retaining wall is a good idea, though more involved. Plans, permit and inspections required.
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## Re: Deck weight limit
With only a 4' drop, a vertical retaining wall made of concrete then pour a concrete pad. It might not cost as much as you think.
Otherwise, time to look for a lighter folding open trailer. Sounds like you have an enclosed trailer... or at least soemthing pretty substantial.
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## Re: Deck weight limit
@ Cougars1996,
We did consider a retaining wall but it will bury the Basement windows and also prevent my neighbor from getting his lawnmower from the back of his house to the front. His driveway was just like mine and he filled his in. Now he has to use my yard to get his lawnmower up front. It’s much higher on his property.
I do plan on having the plans drawn and approved by building inspector. I was just wondering if my plan was feasible, can you build a deck to hold a car (just in case as cougars1996 stated). Just wondered if any one knew what the specs would have to be for that type of load. To get any idea of what was involved before going to the Building inspector.
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## Re: Deck weight limit
To get your plans approved by the building department, you'll need them signed off by an engineer (that's true in my city, and I'm sure in all other cities). Why don't you check with your bldg dept first, then get estimates from two engineers. | 1,374 | 5,563 | {"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-2017-04 | latest | en | 0.965796 |
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Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
# Bases
See all sections
Sections
Acids Introduction
Bases Introduction
Binary Acids
Oxyacids
Bases
Amphoteric Species
Arrhenius Acids and Bases
Bronsted-Lowry Acids and Bases
Lewis Acids and Bases
The pH Scale
Auto-Ionization
Ka and Kb
pH of Strong Acids and Bases
Ionic Salts
pH of Weak Acids
pH of Weak Bases
Diprotic Acids and Bases
Diprotic Acids and Bases Calculations
Triprotic Acids and Bases
Triprotic Acids and Bases Calculations
Conjugate Acids and Bases
Bases represent compounds that neutralize acids when mixed
###### Strong & Weak Bases
Concept #1: A strong base can be formed when a Group 1A and some Group 2A metals combine with certain basic anions.
Example #1: Which of the following represents a strong base?
a) Be(OH)2 b) NaO2 c) LiNH2 d) Mg(OH)2 e) KIO
Concept #2: Strong Bases represent strong electrolytes and weak bases represent weak electrolytes.
Example #2: Which of the following bases will partially dissolve when placed in water?
a) LiOH b) NaNH2 c) Al(OH)3 d) Cs2O e) KOH
Concept #3: Amines represent a class of covalent compounds that can be either acidic or basic based on their charge.
Example #3: Identify the amine that will weakly accept a proton (H+) when in the presence of an acid.
a) H2NNH3+ b) (CH3)2NH c) CH3SH d) CH3NH3+ e) CH3CH2CH3
Practice: Which of the following bases would more greatly favor the product side of a chemical reaction?
a) BeH2 b) H2Se c) SrH2 d) Pb(OH)4 e) HF
Practice: Which of the following compounds would be found as mostly molecules when placed into water?
I. Be(OH)2 II. HNO3 III. LiOH IV. (CHÂ3)2NH V. CaO
a) I, IV, V
b) I only
c) I and IV
d) IV only
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Mar 14, 2018
$35 - 12 \sqrt{3}$
Explanation:
To multiply out this you do the same thing as what you would do with a polynomial. Therefore you need to multiply each term in the bracket by each term in the other bracket:
$8 \times 4 = 32$
$8 \times \left(- \sqrt{3}\right) = - 8 \sqrt{3}$
$\left(- \sqrt{3}\right) \times 4 = - 4 \sqrt{3}$
$\left(- \sqrt{3}\right) \times \left(- \sqrt{3}\right) = 3$
Then you can collect like terms and simplify:
$32 + 3 - 8 \sqrt{3} - 4 \sqrt{3} = 35 - 12 \sqrt{3}$ | 199 | 543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 4.625 | 5 | CC-MAIN-2020-16 | longest | en | 0.69838 |
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# platinum conversion
## Amount: cubic foot (cu ft - ft3) of platinum volume Equals: 0.037 cubic yards (cu yd - yd3) in platinum volume
Calculate cubic yards of platinum per cubic foot unit. The platinum converter.
TOGGLE : from cubic yards into cubic feet in the other way around.
### Enter a New cubic foot Amount of platinum to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## platinum from cubic foot to cubic yard Conversion Results :
Amount : cubic foot (cu ft - ft3) of platinum
Equals: 0.037 cubic yards (cu yd - yd3) in platinum
Fractions: 37/1000 cubic yards (cu yd - yd3) in platinum
CONVERT : between other platinum measuring units - complete list.
## Platinum Amounts (solid platinum)
Here the calculator is for platinum amounts (solid platinum volume; dense, precious, gray to white metal rare in abundance on the planet earth. Its annual production is only a very few hundred tons. It is a very highly valuable metal. Platinum performs real well in resisting corrosion. Not only beautiful jewellery is made out of platinum, this metal enjoys quite a wide variety of uses. For instance in electronics, chemical industries and also in chemotherapy applications against certain cancers. Traders invest money in platinum on commodity markets, in commodity future trading as this material is also one of the major precious commodity metals. Thinking of going into investing in stocks? It would be a wise idea to start learning at least basics at a commodity trading school first, to get used to the markets, then start with small investments. Only after sell and buy platinum.)
Is it possible to manage numerous units calculations, in relation to how heavy other volumes of platinum are, all on one page? The all in one Pt multiunit calculation tool makes it possible to manage just that.
Convert platinum measuring units between cubic foot (cu ft - ft3) and cubic yards (cu yd - yd3) of platinum but in the other direction from cubic yards into cubic feet.
conversion result for platinum: From Symbol Equals Result To Symbol 1 cubic foot cu ft - ft3 = 0.037 cubic yards cu yd - yd3
# Precious metals: platinum conversion
This online platinum from cu ft - ft3 into cu yd - yd3 (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this platinum calculator are ...
With the above mentioned units calculating service it provides, this platinum converter proved to be useful also as a teaching tool:
1. in practicing cubic feet and cubic yards ( cu ft - ft3 vs. cu yd - yd3 ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with platinum's density values including other physical properties this metal has.
International unit symbols for these two platinum measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for cubic foot is: cu ft - ft3
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for cubic yard is: cu yd - yd3
### One cubic foot of platinum converted to cubic yard equals to 0.037 cu yd - yd3
How many cubic yards of platinum are in 1 cubic foot? The answer is: The change of 1 cu ft - ft3 ( cubic foot ) unit of a platinum amount equals = to 0.037 cu yd - yd3 ( cubic yard ) as the equivalent measure for the same platinum type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of platinum can have a crucial/pivotal role in investments. If there is an exact known measure in cu ft - ft3 - cubic feet for platinum amount, the rule is that the cubic foot number gets converted into cu yd - yd3 - cubic yards or any other unit of platinum absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many cubic yards ( cu yd - yd3 ) of platinum are contained in a cubic foot ( 1 cu ft - ft3 ). Or, how much in cubic yards of platinum is in 1 cubic foot? To link to this platinum - cubic foot to cubic yards online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: platinum from cubic foot (cu ft - ft3) to cubic yards (cu yd - yd3) metal conversion.
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• Grade Range: 3–12
• Subject: ELL Support
The ELPS Instructional Tool outlines how to plan focused, targeted, and systematic instruction to meet the linguistic needs of ELLs identified at the beginning and intermediate proficiency levels in grade 3 or higher.
11. ELPS Linguistic Instructional Alignment Guide (LIAG)
• Resource ID: ELPS-LIAG01
• Grade Range: K–12
• Subject: ELL Support
The ELPS Linguistic Instructional Alignment Guide (LIAG) is designed to help teachers gather information needed to ensure that classroom instruction meets the individual academic and linguistic needs of English language learners (ELLs).
12. Sheltered Instruction Training Series
• Resource ID: SI001
• Grade Range: K–12
• Subject: ELL Support
Sheltered instruction is an instructional approach that uses various strategies to ensure that grade-level instruction provided in English addresses both content and language objectives. This resource introduces the Sheltered Instruction Training Series (20 CPE hours).
13. ESTAR/MSTAR
• Resource ID: E_MSTAR001
• Grade Range: 2–8
• Subject: Math
The ESTAR/MSTAR Universal Screeners and Diagnostic Assessments are a formative assessment system administered to students in grades 2–8 to support instructional decisions in mathematics. | 778 | 3,308 | {"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-2019-13 | latest | en | 0.760698 |
http://sciencesbookreview.com/four-bar-linkage-torque-calculator-electric-motor.php | 1,632,355,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00357.warc.gz | 50,401,169 | 7,545 | # Four bar linkage torque calculator electric motor
As most motors (electrical or internal combustion) provide a rotating drive shaft, some way is needed to convert the rotary engine motion into reciprocating pump motion. A pumpjack is a drive mechanism to achieve this, consisting of a four-bar linkage as shown below. The heavy rotating counterweight is arranged so that it is falling while the. Fuzzy control of a dc motor driven four-bar mechanism. Author links open overlay panel Ö to regulate the crank angular speed fluctuations introduced by the inertia of the rotor and the rotating bars in a four-bar linkage when an electric motor is used to drive the mechanism. T m in the equation represents the magnetic motor torque, and Cited by: The torque required to drive a 4-bar linkage over a complete cycle can be approximated by the following function T_L = - 50 * cos(0). A flywheel of mass 10kg is to be used with an average size motor that produces a constant torque of N.m.
# Four bar linkage torque calculator electric motor
## Watch Now Four Bar Linkage Torque Calculator Electric Motor
Simulation and motion study of crank-rocker mechanism, time: 13:28
Tags: Paint the town red tpb , , Now behold the lamb lyrics townsend , , Lil wayne moment sharebeast . Dec 20, · How to Calculate Torque of DC Motor? In: Electric Current, Electric Motors. Tagged: DC Machine Torque Calculation, dc motor, dc motor speed control, Torque is estimated as a cross product of estimated stator flux linkage vector and measured motor current vector. The estimated flux magnitude and torque are then compared with their reference Author: MS Chaudhry. 4 Bar Linkage Kinematics; Kinematics of an Off-centered Circular Cam; Statics Examples. Force and Spring Equilibrium; Crank Slider Torque; Forces in a Simple Structure; Inverse Dynamics Examples. Geneva Mechanism; Quick Return Mechanism; Torque in a Four-bar Linkage; Dynamics Examples. Trebuchet; Wheeled Trebuchet; Modeling Planetary Motion. TORQUE CALCULATORS STANDARD , non-convex four-bar linkage and turn it into a convex four-bar linkage by flexing the linkage Gen-set is needed to start a 3 phase electric motor Direct on Line (DOL) start Calculator.
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Should you tell it — error. | 666 | 2,869 | {"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} | 2.953125 | 3 | CC-MAIN-2021-39 | latest | en | 0.833811 |
https://www.jiskha.com/display.cgi?id=1205190830 | 1,516,547,105,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00266.warc.gz | 906,450,914 | 3,727 | # math
posted by .
a punch recipe calls for 1 .5 liters of juice .how many millilitres is this
• math -
1 liter = 1,000 milliliters. Multiply 1.5 * 1000 to find the answer.
## Similar Questions
1. ### Math
If a recipe calls for 6 cups of juice for each of the eight punch bowls, how many gallons of juice will be needed?
2. ### Math
A recipe for fruit punch calls for 2 1/2 c of orange juice, 1 1/3 c of pineapple juice, and 1 3/4 c of soda water. If June makes 4 times the recipe, how many cups of fruit punch will she have?
3. ### math
To make the punch, the staff adds 250 milliliters of fruit juice for every 5 liters of punch. How many liters of fruit juice would be added to 20 liters of punch?
4. ### math
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5. ### ratio and proportions
a fruit punch recipe calls for 3 parts of apple juice to 4 parts of cranberry juice . how many liters would be added to 4.5 L of apple juice?
6. ### math
A punch recipe calls for half gallon (64 oz) of punch requires one pint (16 oz) of grape juice. How many quarts (1 qt = 32 oz) of grape juice are required for 2 1/2 gallons of punch?
7. ### math
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8. ### Math
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9. ### math
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10. ### Math
Dave uses 500 milliliters of juice for a punch recipe. He mixes it with two liters of ginger ale. How many milliliters of punch does he make?
More Similar Questions | 574 | 2,140 | {"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-2018-05 | latest | en | 0.919402 |
https://number.academy/7285 | 1,656,477,064,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00752.warc.gz | 474,252,840 | 14,293 | # Number 7285
Number 7,285 spell 🔊, write in words: seven thousand, two hundred and eighty-five . Ordinal number 7285th is said 🔊 and write: seven thousand, two hundred and eighty-fifth. The meaning of number 7285 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 7285. What is 7285 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 7285.
## What is 7,285 in other units
The decimal (Arabic) number 7285 converted to a Roman number is (V)MMCCLXXXV. Roman and decimal number conversions.
The number 7285 converted to a Mayan number is Decimal and Mayan number conversions.
#### Weight conversion
7285 kilograms (kg) = 16060.5 pounds (lbs)
7285 pounds (lbs) = 3304.5 kilograms (kg)
#### Length conversion
7285 kilometers (km) equals to 4527 miles (mi).
7285 miles (mi) equals to 11725 kilometers (km).
7285 meters (m) equals to 23901 feet (ft).
7285 feet (ft) equals 2221 meters (m).
7285 centimeters (cm) equals to 2868.1 inches (in).
7285 inches (in) equals to 18503.9 centimeters (cm).
#### Temperature conversion
7285° Fahrenheit (°F) equals to 4029.4° Celsius (°C)
7285° Celsius (°C) equals to 13145° Fahrenheit (°F)
#### Power conversion
7285 Horsepower (hp) equals to 5357.38 kilowatts (kW)
7285 kilowatts (kW) equals to 9906.19 horsepower (hp)
#### Time conversion
(hours, minutes, seconds, days, weeks)
7285 seconds equals to 2 hours, 1 minute, 25 seconds
7285 minutes equals to 5 days, 1 hour, 25 minutes
### Codes and images of the number 7285
Number 7285 morse code: --... ..--- ---.. .....
Sign language for number 7285:
Number 7285 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 7285
### Multiplications
#### Multiplication table of 7285
7285 multiplied by two equals 14570 (7285 x 2 = 14570).
7285 multiplied by three equals 21855 (7285 x 3 = 21855).
7285 multiplied by four equals 29140 (7285 x 4 = 29140).
7285 multiplied by five equals 36425 (7285 x 5 = 36425).
7285 multiplied by six equals 43710 (7285 x 6 = 43710).
7285 multiplied by seven equals 50995 (7285 x 7 = 50995).
7285 multiplied by eight equals 58280 (7285 x 8 = 58280).
7285 multiplied by nine equals 65565 (7285 x 9 = 65565).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 7285
Half of 7285 is 3642,5 (7285 / 2 = 3642,5 = 3642 1/2).
One third of 7285 is 2428,3333 (7285 / 3 = 2428,3333 = 2428 1/3).
One quarter of 7285 is 1821,25 (7285 / 4 = 1821,25 = 1821 1/4).
One fifth of 7285 is 1457 (7285 / 5 = 1457).
One sixth of 7285 is 1214,1667 (7285 / 6 = 1214,1667 = 1214 1/6).
One seventh of 7285 is 1040,7143 (7285 / 7 = 1040,7143 = 1040 5/7).
One eighth of 7285 is 910,625 (7285 / 8 = 910,625 = 910 5/8).
One ninth of 7285 is 809,4444 (7285 / 9 = 809,4444 = 809 4/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
7285
#### Is Prime?
The number 7285 is not a prime number. The closest prime numbers are 7283, 7297.
7285th prime number in order is 73783.
#### Factorization and factors (dividers)
The prime factors of 7285 are 5 * 31 * 47
The factors of 7285 are 1 , 5 , 31 , 47 , 155 , 235 , 1457 , 7285
Total factors 8.
Sum of factors 9216 (1931).
#### Powers
The second power of 72852 is 53.071.225.
The third power of 72853 is 386.623.874.125.
#### Roots
The square root √7285 is 85,352211.
The cube root of 37285 is 19,385478.
#### Logarithms
The natural logarithm of No. ln 7285 = loge 7285 = 8,893573.
The logarithm to base 10 of No. log10 7285 = 3,86243.
The Napierian logarithm of No. log1/e 7285 = -8,893573.
### Trigonometric functions
The cosine of 7285 is -0,938214.
The sine of 7285 is 0,346056.
The tangent of 7285 is -0,368845.
### Properties of the number 7285
More math properties ...
## Number 7285 in Computer Science
Code typeCode value
PIN 7285 It's recommendable to use 7285 as a password or PIN.
7285 Number of bytes7.1KB
Unix timeUnix time 7285 is equal to Thursday Jan. 1, 1970, 2:01:25 a.m. GMT
IPv4, IPv6Number 7285 internet address in dotted format v4 0.0.28.117, v6 ::1c75
7285 Decimal = 1110001110101 Binary
7285 Decimal = 100222211 Ternary
7285 Decimal = 16165 Octal
7285 Decimal = 1C75 Hexadecimal (0x1c75 hex)
7285 BASE64NzI4NQ==
7285 SHA149e1a9d5b2d953ea65bda6aba32035a40edc76bc
7285 SHA2241ee509e01f34193b2d258efa549a0b0cfbc9d4029315ff55dcec369a
7285 SHA256ffb56b8d39faa60b53c15957c255c08e977b6fbe96c4ce9dcf1214bf48b91bc2
7285 SHA38465201d6fa9a179775860fc9d3dc292393b80d76d580484a3d644135df9a7aa75ec2033f69d439983aeb9080f7a472297
More SHA codes related to the number 7285 ...
If you know something interesting about the 7285 number that you did not find on this page, do not hesitate to write us here.
## Numerology 7285
### The meaning of the number 8 (eight), numerology 8
Character frequency 8: 1
The number eight (8) is the sign of organization, perseverance and control of energy to produce material and spiritual achievements. It represents the power of realization, abundance in the spiritual and material world. Sometimes it denotes a tendency to sacrifice but also to be unscrupulous.
More about the meaning of the number 8 (eight), numerology 8 ...
### The meaning of the number 7 (seven), numerology 7
Character frequency 7: 1
The number 7 (seven) is the sign of the intellect, thought, psychic analysis, idealism and wisdom. This number first needs to gain self-confidence and to open his/her life and heart to experience trust and openness in the world. And then you can develop or balance the aspects of reflection, meditation, seeking knowledge and knowing.
More about the meaning of the number 7 (seven), numerology 7 ...
### The meaning of the number 5 (five), numerology 5
Character frequency 5: 1
The number five (5) came to this world to achieve freedom. You need to apply discipline to find your inner freedom and open-mindedness. It is about a restless spirit in constant search of the truth that surrounds us. You need to accumulate as much information as possible to know what is happening in depth. Number 5 person is intelligent, selfish, curious and with great artistic ability. It is a symbol of freedom, independence, change, adaptation, movement, the search for new experiences, the traveling and adventurous spirit, but also of inconsistency and abuse of the senses.
More about the meaning of the number 5 (five), numerology 5 ...
### The meaning of the number 2 (two), numerology 2
Character frequency 2: 1
The number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, super-sensitivity towards the needs of others.
The number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2 person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.
More about the meaning of the number 2 (two), numerology 2 ...
## Interesting facts about the number 7285
### Asteroids
• (7285) Seggewiss is asteroid number 7285. It was discovered by E. W. Elst from La Silla Observatory on 3/2/1990.
### Distances between cities
• There is a 4,527 miles (7,285 km) direct distance between Adelaide (Australia) and Yangon (Burma / Myanmar).
• There is a 4,527 miles (7,285 km) direct distance between Al Başrat al Qadīmah (Iraq) and Bandarlampung (Indonesia).
• There is a 7,285 miles (11,723 km) direct distance between Alīgarh (India) and Toronto (Alberta).
• There is a 7,285 miles (11,724 km) direct distance between Antananarivo (Madagascar) and Manaus (Brazil).
• More distances between cities ...
• There is a 7,285 miles (11,723 km) direct distance between Antananarivo (Madagascar) and Sapporo (Japan).
• There is a 7,285 miles (11,723 km) direct distance between Bandarlampung (Indonesia) and Valencia (Spain).
• There is a 4,527 miles (7,285 km) direct distance between Beirut (Lebanon) and Guankou (China).
• There is a 4,527 miles (7,285 km) direct distance between Bishkek (Kyrgyzstan) and Zaria (Nigeria).
• There is a 7,285 miles (11,723 km) direct distance between Brazzaville (Congo) and Kaohsiung (Taiwan).
• There is a 4,527 miles (7,285 km) direct distance between Budta (Philippines) and Mashhad (Iran).
• There is a 7,285 miles (11,723 km) direct distance between Çankaya (Turkey) and Ecatepec (Mexico).
• There is a 7,285 miles (11,723 km) direct distance between Daejeon (South Korea) and Guadalajara (Mexico).
• There is a 7,285 miles (11,723 km) direct distance between Dammam (Saudi Arabia) and Goiânia (Brazil).
• There is a 7,285 miles (11,724 km) direct distance between Delhi (India) and The Bronx (USA).
• There is a 7,285 miles (11,724 km) direct distance between Depok (Indonesia) and London (United Kingdom).
• There is a 7,285 miles (11,723 km) direct distance between Douala (Cameroon) and Suzhou (China).
• There is a 7,285 miles (11,723 km) direct distance between Gorakhpur (India) and Philadelphia (USA).
• There is a 7,285 miles (11,723 km) direct distance between Guadalupe (Mexico) and Tangshan (China).
• There is a 4,527 miles (7,285 km) direct distance between Hyderabad (India) and Johannesburg (South Africa).
• There is a 7,285 miles (11,723 km) direct distance between Kaohsiung (Taiwan) and Kinshasa (Zaire).
• There is a 4,527 miles (7,285 km) direct distance between Karachi (Pakistan) and Lomé (Togo).
• There is a 4,527 miles (7,285 km) direct distance between Kitakyūshū (Japan) and Tbilisi (Georgia).
• There is a 4,527 miles (7,285 km) direct distance between Lubumbashi (Zaire) and Minsk (Belarus).
• There is a 7,285 miles (11,724 km) direct distance between Maracaibo (Venezuela) and Mecca (Saudi Arabia).
• There is a 7,285 miles (11,724 km) direct distance between Mexico City (Mexico) and Novosibirsk (Russia).
• There is a 4,527 miles (7,285 km) direct distance between Mogadishu (Somalia) and Semarang (Indonesia).
• There is a 4,527 miles (7,285 km) direct distance between Montréal (Alberta) and Nizhniy Novgorod (Russia).
• There is a 4,527 miles (7,285 km) direct distance between Multān (Pakistan) and Niamey (Niger).
• There is a 7,285 miles (11,724 km) direct distance between Niamey (Niger) and Tangerang (Indonesia).
• There is a 4,527 miles (7,285 km) direct distance between Qom (Iran) and Quezon City (Philippines).
• There is a 7,285 miles (11,724 km) direct distance between Riyadh (Saudi Arabia) and Vancouver (Alberta).
• There is a 7,285 miles (11,724 km) direct distance between San Antonio (USA) and Zhu Cheng City (China).
• There is a 4,527 miles (7,285 km) direct distance between Yaoundé (Cameroon) and Yekaterinburg (Russia).
### Mathematics
• 7285 has a 7th power that contains the same digits as 54410.
## Number 7,285 in other languages
How to say or write the number seven thousand, two hundred and eighty-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 7.285) siete mil doscientos ochenta y cinco German: 🔊 (Anzahl 7.285) siebentausendzweihundertfünfundachtzig French: 🔊 (nombre 7 285) sept mille deux cent quatre-vingt-cinq Portuguese: 🔊 (número 7 285) sete mil, duzentos e oitenta e cinco Chinese: 🔊 (数 7 285) 七千二百八十五 Arabian: 🔊 (عدد 7,285) سبعة آلاف و مئتانخمسة و ثمانون Czech: 🔊 (číslo 7 285) sedm tisíc dvěstě osmdesát pět Korean: 🔊 (번호 7,285) 칠천이백팔십오 Danish: 🔊 (nummer 7 285) syvtusinde og tohundrede og femogfirs Hebrew: (מספר 7,285) שבע אלף מאתיים שמנים וחמש Dutch: 🔊 (nummer 7 285) zevenduizendtweehonderdvijfentachtig Japanese: 🔊 (数 7,285) 七千二百八十五 Indonesian: 🔊 (jumlah 7.285) tujuh ribu dua ratus delapan puluh lima Italian: 🔊 (numero 7 285) settemiladuecentottantacinque Norwegian: 🔊 (nummer 7 285) syv tusen, to hundre og åtti-fem Polish: 🔊 (liczba 7 285) siedem tysięcy dwieście osiemdziesiąt pięć Russian: 🔊 (номер 7 285) семь тысяч двести восемьдесят пять Turkish: 🔊 (numara 7,285) yedibinikiyüzseksenbeş Thai: 🔊 (จำนวน 7 285) เจ็ดพันสองร้อยแปดสิบห้า Ukrainian: 🔊 (номер 7 285) сiм тисяч двiстi вiсiмдесят п'ять Vietnamese: 🔊 (con số 7.285) bảy nghìn hai trăm tám mươi lăm Other languages ...
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# Python NumPy Tutorial for Beginners
In this article, we learn about the Python NumPy Tutorial for Beginners. So you’ve learned the basics of Python and you’re looking for a more powerful way to analyse data? NumPy is what you need.NumPy is a module for Python that allows you to work with multidimensional arrays and matrices.
It’s perfect for scientific or mathematical calculations because it’s fast and efficient. In addition, NumPy includes support for signal processing and linear algebra operations. So if you need to do any mathematical operations on your data, NumPy is probably the library for you.
In this tutorial, we’ll show you how to use NumPy to its full potential. You’ll learn more about arrays as well as operate on them using mathematical functions.
NumPy, which stands for Numerical Python, is a library consisting of multidimensional array objects and a collection of routines for processing those arrays. Using NumPy, mathematical and logical operations on arrays can be performed. In this Python Numpy Tutorial, we will be learning about NumPy in Python, What is NumPy in Python, Data Types in NumPy, and more.
## What is NumPy in Python?
NumPy in Python is a library that is used to work with arrays and was created in 2005 by Travis Oliphant. NumPy library in Python has functions for working in domain of Fourier transform, linear algebra, and matrices. Python NumPy is an open-source project that can be used freely. NumPy stands for Numerical Python.
## How to install NumPy Python?
Installing the NumPy library is a straightforward process. You can use pip to install the library.Go to the command line and type the following:
pip install numpy If you are using Anaconda distribution, then you can use conda to install NumPy. conda install numpy Once the installation is complete, you can verify it by importing the NumPy library in the python interpreter. One can use the numpy library by importing it as shown below. import numpy If the import is successful, then you will see the following output. >>> import numpy >>> numpy.__version__ '1.17.2'
NumPy is a library for the Python programming language, and it’s specifically designed to help you work with data.
With NumPy, you can easily create arrays, which is a data structure that allows you to store multiple values in a single variable.
In particular, NumPy arrays provide an efficient way of storing and manipulating data.NumPy also includes a number of functions that make it easy to perform mathematical operations on arrays. This can be really useful for scientific or engineering applications. And if you’re working with data from a Python script, using NumPy can make your life a lot easier.
Let us take a look at how to create NumPy arrays, copy and view arrays, reshape arrays, and iterate over arrays.
## NumPy Creating Arrays
Arrays are different from Python lists in several ways. First, NumPy arrays are multi-dimensional, while Python lists are one-dimensional. Second, NumPy arrays are homogeneous, while Python lists are heterogeneous. This means that all the elements of a NumPy array must be of the same type. Third, NumPy arrays are more efficient than Python lists.NumPy arrays can be created in several ways. One way is to create an array from a Python list. Once you have created a NumPy array, you can manipulate it in various ways. For example, you can change the shape of an array, or you can index into an array to access its elements. You can also perform mathematical operations on NumPy arrays, such as addition, multiplication, and division.
One has to import the library in the program to use it. The module NumPy has an array function in it which creates an array.
Creating an Array:
``````import numpy as np
arr = np.array([1, 2, 3, 4, 5])
print(arr)
``````
``````Output:
[1 2 3 4 5]
``````
We can also pass a tuple in the array function to create an array. 2
``````import numpy as np
arr = np.array((1, 2, 3, 4, 5))
print(arr)
``````
The output would be similar to the above case.
Dimensions- Arrays:
0-D Arrays:
The following code will create a zero-dimensional array with a value 36.
``````import numpy as np
arr = np.array(36)
print(arr)
``````
``````Output:
36
``````
1-Dimensional Array:
The array that has Zero Dimensional arrays as its elements is a uni-dimensional or 1-D array.
The code below creates a 1-D array,
``````import numpy as np
arr = np.array([1, 2, 3, 4, 5])
print(arr)
``````
``````Output:
[1 2 3 4 5]
``````
Two Dimensional Arrays:
2-D Arrays are the ones that have 1-D arrays as its element. The following code will create a 2-D array with 1,2,3 and 4,5,6 as its values.
``````import numpy as np
3
arr1 = np.array([[1, 2, 3], [4, 5, 6]])
print(arr1)
``````
``````Output:
[[1 2 3]
[4 5 6]] ``````
Three Dimensional Arrays:
Let us see an example of creating a 3-D array with two 2-D arrays:
``````import numpy as np
arr1 = np.array([[[1, 2, 3], [4, 5, 6]], [[1, 2, 3], [4, 5, 6]]]) print(arr1)
``````
``````Output:
[[[1 2 3]
[4 5 6]]
[[1 2 3]
[4 5 6]]] ``````
To identify the dimensions of the array, we can use ndim as shown below:
``````import numpy as np
a = np.array(36)
d = np.array([[[1, 2, 3], [4, 5, 6]], [[1, 2, 3], [4, 5, 6]]])
print(a.ndim)
print(d.ndim)
``````
``````Output:
0
3
``````
## Operations using NumPy
Using NumPy, a developer can perform the following operations −
• Mathematical and logical operations on arrays.
• Fourier transforms and routines for shape manipulation.
• Operations related to linear algebra. NumPy has in-built functions for linear algebra and random number generation.
NumPy – A Replacement for MatLab
NumPy is often used along with packages like SciPy (Scientific Python) and Matplotlib (plotting library). This combination is widely used as a replacement for MatLab, a popular platform for technical computing. However, Python alternative to MatLab is now seen as a more modern and complete programming language.
The most important object defined in NumPy is an N-dimensional array type called ndarray. It describes the collection of items of the same type. Items in the collection can be accessed using a zero-based index.
Every item in a ndarray takes the same size as the block in the memory. Each element in ndarray is an object of the data-type object (called dtype).
Any item extracted from ndarray object (by slicing) is represented by a Python object of one of array scalar types. The following diagram shows a relationship between ndarray, data-type object (dtype) and array scalar type −
An instance of ndarray class can be constructed by different array creation routines described later in the tutorial. The basic ndarray is created using an array function in NumPy as follows-
numpy.array
It creates a ndarray from any object exposing an array interface, or from any method that returns an array.
numpy.array(object, dtype = None, copy = True, order = None, subok = False, ndmin = 0)
The ndarray object consists of a contiguous one-dimensional segment of computer memory, combined with an indexing scheme that maps each item to a location in the memory block. The memory block holds the elements in row-major order (C style) or a column-major order (FORTRAN or MatLab style).
The above constructor takes the following parameters −
Take a look at the following examples to understand better.
### Example 1
Live Demo
``````import numpy as np
a = np.array([1,2,3])
print a``````
The output is as follows –
[1, 2, 3]
### Example 2
Live Demo
``````# more than one dimensions
import numpy as np
a = np.array([[1, 2], [3, 4]])
print a``````
The output is as follows −
[[1, 2]
[3, 4]]
### Example 3
Live Demo
``````# minimum dimensions
import numpy as np
a = np.array([1, 2, 3,4,5], ndmin = 2)
print a``````
The output is as follows −
[[1, 2, 3, 4, 5]]
### Example 4
Live Demo
``````# dtype parameter
import numpy as np
a = np.array([1, 2, 3], dtype = complex)
print a``````
The output is as follows −
[ 1.+0.j, 2.+0.j, 3.+0.j]
The ndarray object consists of a contiguous one-dimensional segment of computer memory, combined with an indexing scheme that maps each item to a location in the memory block. The memory block holds the elements in row-major order (C style) or a column-major order (FORTRAN or MatLab style).
## NumPy – Data Types
Here is a list of the different Data Types in NumPy:
1. bool_
2. int_
3. intc
4. intp
5. int8
6. int16
7. float_
8. float64
9. complex_
10. complex64
11. complex128
bool_
Boolean (True or False) stored as a byte
int_
Default integer type (same as C long; normally either int64 or int32)
intc
Identical to C int (normally int32 or int64)
intp
An integer used for indexing (same as C ssize_t; normally either int32 or int64)
int8
Byte (-128 to 127)
int16
Integer (-32768 to 32767)
float_
Shorthand for float64
float64
Double precision float: sign bit, 11 bits exponent, 52 bits mantissa
complex_
Shorthand for complex128
complex64
Complex number, represented by two 32-bit floats (real and imaginary components)
complex128
Complex number, represented by two 64-bit floats (real and imaginary components)
NumPy numerical types are instances of dtype (data-type) objects, each having unique characteristics. The dtypes are available as np.bool_, np.float32, etc.
## Data Type Objects (dtype)
A data type object describes the interpretation of a fixed block of memory corresponding to an array, depending on the following aspects −
• Type of data (integer, float or Python object)
• Size of data
• Byte order (little-endian or big-endian)
• In case of structured type, the names of fields, data type of each field and part of the memory block taken by each field.
• If the data type is a subarray, its shape and data type
The byte order is decided by prefixing ‘<‘ or ‘>’ to the data type. ‘<‘ means that encoding is little-endian (least significant is stored in smallest address). ‘>’ means that encoding is big-endian (a most significant byte is stored in smallest address).
A dtype object is constructed using the following syntax −
numpy.dtype(object, align, copy)
The parameters are −
• Object − To be converted to data type object
• Align − If true, adds padding to the field to make it similar to C-struct
• Copy − Makes a new copy of dtype object. If false, the result is a reference to builtin data type object
### Example 1
Live Demo
``````# using array-scalar type
import numpy as np
dt = np.dtype(np.int32)
print dt``````
The output is as follows −
int32
### Example 2
Live Demo
``````#int8, int16, int32, int64 can be replaced by equivalent string 'i1', 'i2','i4', etc.
import numpy as np
dt = np.dtype('i4')
print dt ``````
The output is as follows −
int32
### Example 3
Live Demo
``````# using endian notation
import numpy as np
dt = np.dtype('>i4')
print dt``````
The output is as follows −
>i4
The following examples show the use of a structured data type. Here, the field name and the corresponding scalar data type is to be declared.
### Example 4
Live Demo
``````# first create structured data type
import numpy as np
dt = np.dtype([('age',np.int8)])
print dt ``````
The output is as follows – [(‘age’, ‘i1’)]
### Example 5
Live Demo
``````# now apply it to ndarray object
import numpy as np
dt = np.dtype([('age',np.int8)])
a = np.array([(10,),(20,),(30,)], dtype = dt)
print a``````
The output is as follows –
[(10,) (20,) (30,)]
Each built-in data type has a character code that uniquely identifies it.
• ‘b’ − boolean
• ‘i’ − (signed) integer
• ‘u’ − unsigned integer
• ‘f’ − floating-point
• ‘c’ − complex-floating point
• ‘m’ − timedelta
• ‘M’ − datetime
• ‘O’ − (Python) objects
• ‘S’, ‘a’ − (byte-)string
• ‘U’ − Unicode
• ‘V’ − raw data (void)
We will also discuss the various array attributes of NumPy.
## ndarray.shape
This array attribute returns a tuple consisting of array dimensions. It can also be used to resize the array.
### Example 1
Live Demo
``````import numpy as np
a = np.array([[1,2,3],[4,5,6]])
print a.shape``````
The output is as follows − (2, 3)
### Example 2
Live Demo
``````# this resizes the ndarray
import numpy as np
a = np.array([[1,2,3],[4,5,6]])
a.shape = (3,2)
print a ``````
The output is as follows -[[1, 2][3, 4] [5, 6]]
## ndarray.ndim
This array attribute returns the number of array dimensions.
### Example 1
Live Demo
``````# an array of evenly spaced numbers
import numpy as np
a = np.arange(24)
print a``````
The output is as follows −
[0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23]
### Example 2
Live Demo
``````# this is one dimensional array
import numpy as np
a = np.arange(24)
a.ndim
# now reshape it
b = a.reshape(2,4,3)
print b
# b is having three dimensions``````
The output is as follows −
[[[ 0, 1, 2]
[ 3, 4, 5]
[ 6, 7, 8]
[ 9, 10, 11]]
[[12, 13, 14]
[15, 16, 17]
[18, 19, 20]
[21, 22, 23]]]
## numpy.itemsize
This array attribute returns the length of each element of array in bytes.
### Example 1
Live Demo
``````# dtype of array is int8 (1 byte)
import numpy as np
x = np.array([1,2,3,4,5], dtype = np.int8)
print x.itemsize``````
The output is as follows −
1
### Example 2
Live Demo
``````# dtype of array is now float32 (4 bytes)
import numpy as np
x = np.array([1,2,3,4,5], dtype = np.float32)
print x.itemsize``````
The output is as follows −
4
## numpy.flags
The ndarray object has the following attributes. Its current values are returned by this function.
### Example
The following example shows the current values of flags.
Live Demo
``````import numpy as np
x = np.array([1,2,3,4,5])
print x.flags``````
The output is as follows −
C_CONTIGUOUS : True
F_CONTIGUOUS : True
OWNDATA : True
WRITEABLE : True
ALIGNED : True
UPDATEIFCOPY : False
## NumPy – Array Creation Routines
A new ndarray object can be constructed by any of the following array creation routines or using a low-level ndarray constructor.
numpy.empty
It creates an uninitialized array of specified shape and dtype. It uses the following constructor −
numpy.empty(shape, dtype = float, order = ‘C’)
The constructor takes the following parameters.
### Example
The following code shows an example of an empty array.
Live Demo
``````import numpy as np
x = np.empty([3,2], dtype = int)
print x``````
The output is as follows −[[22649312 1701344351]
[1818321759 1885959276] [16779776 156368896]]
## numpy.zeros
Returns a new array of specified size, filled with zeros.
numpy.zeros(shape, dtype = float, order = ‘C’)
The constructor takes the following parameters.
### Example 1
Live Demo
``````# array of five ones. Default dtype is float
import numpy as np
x = np.ones(5)
print x``````
The output is as follows −
[ 1. 1. 1. 1. 1.]
## NumPy – Indexing & Slicing
Contents of ndarray object can be accessed and modified by indexing or slicing, just like Python’s in-built container objects.
As mentioned earlier, items in ndarray object follows zero-based index. Three types of indexing methods are available − field access, basic slicing and advanced indexing.
Basic slicing is an extension of Python’s basic concept of slicing to n dimensions. A Python slice object is constructed by giving start, stop, and step parameters to the built-in slice function. This slice object is passed to the array to extract a part of array.
### Example 1
Live Demo
``````import numpy as np
a = np.arange(10)
s = slice(2,7,2)
print a[s]``````
Its output is as follows −
[2 4 6]
n the above example, an ndarray object is prepared by arange() function. Then a slice object is defined with start, stop, and step values 2, 7, and 2 respectively. When this slice object is passed to the ndarray, a part of it starting with index 2 up to 7 with a step of 2 is sliced.
The same result can also be obtained by giving the slicing parameters separated by a colon : (start:stop:step) directly to the ndarray object.
### Example 2
Live Demo
``````import numpy as np
a = np.arange(10)
b = a[2:7:2]
print b``````
Here, we will get the same output − [2 4 6]
If only one parameter is put, a single item corresponding to the index will be returned. If a: is inserted in front of it, all items from that index onwards will be extracted. If two parameters (with: between them) is used, items between the two indexes (not including the stop index) with default step one are sliced.
### Example 3
Live Demo
``````# slice single item
import numpy as np
a = np.arange(10)
b = a[5]
print b``````
Its output is as follows −
5
### Example 4
Live Demo
``````# slice items starting from index
import NumPy as np
a = np.arange(10)
print a[2:]``````
Now, the output would be −
[2 3 4 5 6 7 8 9]
### Example 5
Live Demo
``````# slice items between indexes
import numpy as np
a = np.arange(10)
print a[2:5]``````
Here, the output would be −
[2 3 4]
The above description applies to multi-dimensional ndarray too.
It is possible to make a selection from ndarray that is a non-tuple sequence, ndarray object of integer or Boolean data type, or a tuple with at least one item being a sequence object. Advanced indexing always returns a copy of the data. As against this, the slicing only presents a view.
There are two types of advanced indexing − Integer and Boolean.
## Integer Indexing
This mechanism helps in selecting any arbitrary item in an array based on its N-dimensional index. Each integer array represents the number of indexes into that dimension. When the index consists of as many integer arrays as the dimensions of the target ndarray, it becomes straightforward.
In the following example, one element of the specified column from each row of ndarray object is selected. Hence, the row index contains all row numbers, and the column index specifies the element to be selected.
### Example 1
``````import numpy as np
x = np.array([[1, 2], [3, 4], [5, 6]])
y = x[[0,1,2], [0,1,0]]
print y``````
Its output would be as follows −
[1 4 5]
The selection includes elements at (0,0), (1,1) and (2,0) from the first array.
In the following example, elements placed at corners of a 4X3 array are selected. The row indices of selection are [0, 0] and [3,3] whereas the column indices are [0,2] and [0,2].
Advanced and basic indexing can be combined by using one slice (:) or ellipsis (…) with an index array. The following example uses a slice for the advanced index for column. The result is the same when a slice is used for both. But advanced index results in copy and may have different memory layout.
## Boolean Array Indexing
This type of advanced indexing is used when the resultant object is meant to be the result of Boolean operations, such as comparison operators.
### Example 1
In this example, items greater than 5 are returned as a result of Boolean indexing.
Live Demo
``````import numpy as np
x = np.array([[ 0, 1, 2],[ 3, 4, 5],[ 6, 7, 8],[ 9, 10, 11]])
print 'Our array is:'
print x
print '\n'
# Now we will print the items greater than 5
print 'The items greater than 5 are:'
print x[x > 5]``````
The output of this program would be −
Our array is:
[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]
[ 9 10 11]]
The items greater than 5 are:
[ 6 7 8 9 10 11]
The term broadcasting refers to the ability of NumPy to treat arrays of different shapes during arithmetic operations. Arithmetic operations on arrays are usually done on corresponding elements. If two arrays are of exactly the same shape, then these operations are smoothly performed.
### Example 1
``````import numpy as np
a = np.array([1,2,3,4])
b = np.array([10,20,30,40])
c = a * b
print c``````
Its output is as follows −[10 40 90 160]
If the dimensions of the two arrays are dissimilar, element-to-element operations are not possible. However, operations on arrays of non-similar shapes is still possible in NumPy, because of the broadcasting capability. The smaller array is broadcast to the size of the larger array so that they have compatible shapes.
Broadcasting is possible if the following rules are satisfied −
• Array with smaller ndim than the other is prepended with ‘1’ in its shape.
• Size in each dimension of the output shape is maximum of the input sizes in that dimension.
• An input can be used in calculation if its size in a particular dimension matches the output size or its value is exactly 1.
• If an input has a dimension size of 1, the first data entry in that dimension is used for all calculations along that dimension.
A set of arrays is said to be broadcastable if the above rules produce a valid result and one of the following is true −
• Arrays have exactly the same shape.
• Arrays have the same number of dimensions and the length of each dimension is either a common length or 1.
• Array having too few dimensions can have its shape prepended with a dimension of length 1, so that the above stated property is true.
The following figure demonstrates how array b is broadcast to become compatible with a.
## NumPy – Iterating Over Array
NumPy package contains an iterator object numpy.nditer. It is an efficient multidimensional iterator object using which it is possible to iterate over an array. Each element of an array is visited using Python’s standard Iterator interface.
Let us create a 3X4 array using arrange() function and iterate over it using nditer.
## NumPy – Array Manipulation
Several routines are available in NumPy package for manipulation of elements in ndarray object. They can be classified into the following types −
Changing Shape
Transpose Operations
Changing Dimensions
Joining Arrays
Splitting Arrays
NumPy – Binary Operators
Following are the functions for bitwise operations available in NumPy package.
## NumPy – Mathematical Functions
Quite understandably, NumPy contains a large number of various mathematical operations. NumPy provides standard trigonometric functions, functions for arithmetic operations, handling complex numbers, etc.
### Trigonometric Functions
NumPy has standard trigonometric functions which return trigonometric ratios for a given angle in radians.
Example
Live Demo
``````import numpy as np
a = np.array([0,30,45,60,90])
print 'Sine of different angles:'
# Convert to radians by multiplying with pi/180
print np.sin(a*np.pi/180)
print '\n'
print 'Cosine values for angles in array:'
print np.cos(a*np.pi/180)
print '\n'
print 'Tangent values for given angles:'
print np.tan(a*np.pi/180) ``````
Here is its output −
Sine of different angles:
[ 0. 0.5 0.70710678 0.8660254 1. ]
Cosine values for angles in array:
[ 1.00000000e+00 8.66025404e-01 7.07106781e-01 5.00000000e-01
6.12323400e-17]
Tangent values for given angles:
[ 0.00000000e+00 5.77350269e-01 1.00000000e+00 1.73205081e+00
1.63312394e+16]
arcsin, arcos, and arctan functions return the trigonometric inverse of sin, cos, and tan of the given angle. The result of these functions can be verified by numpy.degrees() function by converting radians to degrees.
### Functions for Rounding
numpy.around()
This is a function that returns the value rounded to the desired precision. The function takes the following parameters.
numpy.around(a,decimals)
Where,
### NumPy – Statistical Functions
NumPy has quite a few useful statistical functions for finding minimum, maximum, percentile standard deviation and variance, etc. from the given elements in the array. The functions are explained as follows −
numpy.amin() and numpy.amax()numpy.amin() and numpy.amax()
These functions return the minimum and the maximum from the elements in the given array along the specified axis.
### Example
Live Demo
``````import numpy as np
a = np.array([[3,7,5],[8,4,3],[2,4,9]])
print 'Our array is:'
print a
print '\n'
print 'Applying amin() function:'
print np.amin(a,1)
print '\n'
print 'Applying amin() function again:'
print np.amin(a,0)
print '\n'
print 'Applying amax() function:'
print np.amax(a)
print '\n’
print 'Applying amax() function again:'
print np.amax(a, axis = 0)``````
It will produce the following output −
Our array is:
[[3 7 5]
[8 4 3]
[2 4 9]]
Applying amin() function:
[3 3 2]
Applying amin() function again:
[2 4 3]
Applying amax() function:
9
Applying amax() function again:
[8 7 9]
numpy.ptp()
The numpy.ptp() function returns the range (maximum-minimum) of values along an axis.
Live Demo
``````import numpy as np
a = np.array([[3,7,5],[8,4,3],[2,4,9]])
print 'Our array is:'
print a
print '\n'
print 'Applying ptp() function:'
print np.ptp(a)
print '\n'
print 'Applying ptp() function along axis 1:'
print np.ptp(a, axis = 1)
print '\n'
print 'Applying ptp() function along axis 0:'
print np.ptp(a, axis = 0)
numpy.percentile()``````
Percentile (or a centile) is a measure used in statistics indicating the value below which a given percentage of observations in a group of observations fall. The function numpy.percentile() takes the following arguments.
Where,
A variety of sorting related functions are available in NumPy. These sorting functions implement different sorting algorithms, each of them characterized by the speed of execution, worst-case performance, the workspace required and the stability of algorithms. Following table shows the comparison of three sorting algorithms.
numpy.sort()
The sort() function returns a sorted copy of the input array. It has the following parameters −
numpy.sort(a, axis, kind, order)
Where,
## NumPy – Byte Swapping
We have seen that the data stored in the memory of a computer depends on which architecture the CPU uses. It may be little-endian (least significant is stored in the smallest address) or big-endian (most significant byte in the smallest address).
numpy.ndarray.byteswap()
The numpy.ndarray.byteswap() function toggles between the two representations: bigendian and little-endian.
## NumPy – Copies & Views
While executing the functions, some of them return a copy of the input array, while some return the view. When the contents are physically stored in another location, it is called Copy. If on the other hand, a different view of the same memory content is provided, we call it as View.
### No Copy
Simple assignments do not make the copy of array object. Instead, it uses the same id() of the original array to access it. The id() returns a universal identifier of Python object, similar to the pointer in C.
Furthermore, any changes in either gets reflected in the other. For example, the changing shape of one will change the shape of the other too.
### View or Shallow Copy
NumPy has ndarray.view() method which is a new array object that looks at the same data of the original array. Unlike the earlier case, change in dimensions of the new array doesn’t change dimensions of the original.
NumPy – Matrix Library
NumPy package contains a Matrix library numpy.matlib. This module has functions that return matrices instead of ndarray objects.
matlib.empty()
The matlib.empty() function returns a new matrix without initializing the entries. The function takes the following parameters.
numpy.matlib.empty(shape, dtype, order)
Where,
### Example
Live Demo
``````import numpy.matlib
import numpy as np
print np.matlib.empty((2,2))
# filled with random data``````
It will produce the following output −
[[ 2.12199579e-314, 4.24399158e-314]
[ 4.24399158e-314, 2.12199579e-314]]
## numpy.matlib.eye()
This function returns a matrix with 1 along the diagonal elements and the zeros elsewhere. The function takes the following parameters.
numpy.matlib.eye(n, M,k, dtype)
Where,
### Example
Live Demo
``````import numpy.matlib
import numpy as np
print np.matlib.eye(n = 3, M = 4, k = 0, dtype = float)``````
It will produce the following output −
[[ 1. 0. 0. 0.]
[ 0. 1. 0. 0.]
[ 0. 0. 1. 0.]]
## NumPy – Matplotlib
Matplotlib is a plotting library for Python. It is used along with NumPy to provide an environment that is an effective open-source alternative for MatLab. It can also be used with graphics toolkits like PyQt and wxPython.
Matplotlib module was first written by John D. Hunter. Since 2012, Michael Droettboom is the principal developer. Currently, Matplotlib ver. 1.5.1 is the stable version available. The package is available in binary distribution as well as in the source code form on www.matplotlib.org.
Conventionally, the package is imported into the Python script by adding the following statement −
from matplotlib import pyplot as plt
Here pyplot() is the most important function in matplotlib library, which is used to plot 2D data. The following script plots the equation y = 2x + 5
Example:
``````import numpy as np
from matplotlib import pyplot as plt
x = np.arange(1,11)
y = 2 * x + 5
plt.title("Matplotlib demo")
plt.xlabel("x axis caption")
plt.ylabel("y axis caption")
plt.plot(x,y)
plt.show()
``````
An ndarray object x is created from np.arange() function as the values on the x axis. The corresponding values on the y axis are stored in another ndarray object y. These values are plotted using plot() function of pyplot submodule of matplotlib package.
The graphical representation is displayed by show() function.
The above code should produce the following output −
Instead of the linear graph, the values can be displayed discretely by adding a format string to the plot() function. Following formatting characters can be used.
## NumPy – Using Matplotlib
NumPy has a numpy.histogram() function that is a graphical representation of the frequency distribution of data. Rectangles of equal horizontal size corresponding to class interval called bin and variable height corresponding to frequency.
numpy.histogram()
The numpy.histogram() function takes the input array and bins as two parameters. The successive elements in bin array act as the boundary of each bin.
``````import numpy as np
a = np.array([22,87,5,43,56,73,55,54,11,20,51,5,79,31,27])
np.histogram(a,bins = [0,20,40,60,80,100])
hist,bins = np.histogram(a,bins = [0,20,40,60,80,100])
print hist
print bins ``````
It will produce the following output −
[3 4 5 2 1]
[0 20 40 60 80 100]
plt()
Matplotlib can convert this numeric representation of histogram into a graph. The plt() function of pyplot submodule takes the array containing the data and bin array as parameters and converts into a histogram.
``````from matplotlib import pyplot as plt
import numpy as np
a = np.array([22,87,5,43,56,73,55,54,11,20,51,5,79,31,27])
plt.hist(a, bins = [0,20,40,60,80,100])
plt.title("histogram")
plt.show()``````
It should produce the following output –
I/O with NumPy
The ndarray objects can be saved to and loaded from the disk files. The IO functions available are −
• load() and save() functions handle /numPy binary files (with npy extension)
• loadtxt() and savetxt() functions handle normal text files
NumPy introduces a simple file format for ndarray objects. This .npy file stores data, shape, dtype and other information required to reconstruct the ndarray in a disk file such that the array is correctly retrieved even if the file is on another machine with different architecture.
## numpy.save()
The numpy.save() file stores the input array in a disk file with npy extension.
``````import numpy as np
a = np.array([1,2,3,4,5])
np.save('outfile',a)
``````
To reconstruct array from outfile.npy, use load() function.
``````import numpy as np
print b ``````
It will produce the following output −
array([1, 2, 3, 4, 5])
The save() and load() functions accept an additional Boolean parameter allow_pickles. A pickle in Python is used to serialize and de-serialize objects before saving to or reading from a disk file.
savetxt()
The storage and retrieval of array data in simple text file format is done with savetxt() and loadtxt() functions.
### Example
``````import numpy as np
a = np.array([1,2,3,4,5])
np.savetxt('out.txt',a)
print b ``````
It will produce the following output −
[ 1. 2. 3. 4. 5.]
We’d also recommend you to visit Great Learning Academy, where you will find a free NumPy course and 1000+ other courses. You will also receive a certificate after the completion of these courses. We hope that this Python NumPy Tutorial was beneficial and you are now better equipped.
NumPy Copy vs View
The difference between copy and view of an array in NumPy is that the view is merely a view of the original array whereas copy is a new array. The copy will not affect the original array and the chances are restricted to the new array created and many modifications made to the original array will not be reflected in the copy array too. But in view, the changes made to the view will be reflected in the original array and vice versa.
Let us understand with code snippets:
Example of Copy:
``````import numpy as np
arr1 = np.array([1, 2, 3, 4, 5])
y = arr1.copy()
arr1[0] = 36
print(arr1)
print(y)
``````
``````Output :
[42 2 3 4 5]
[1 2 3 4 5]
``````
Example of view:
Notice the output of the below code; the changes made to the original array are also reflected in the view.
``````import numpy as np
arr1 = np.array([1, 2, 3, 4, 5])
y= arr1.view()
arr1[0] = 36
print(arr1)
print(y)
``````
``````Output:
[36 2 3 4 5]
[36 2 3 4 5]
``````
NumPy Array Shape
The shape of an array is nothing but the number of elements in each dimension. To get the shape of an array, we can use a .shape attribute that returns a tuple indicating the number of elements.
``````import numpy as np
array1 = np.array([[2, 3, 4,5], [ 6, 7, 8,9]])
print(array1.shape)
``````
``Output: (2,4) ``
NumPy Array Reshape
1-D to 2-D:
Array reshape is nothing but changing the shape of the array, through which one can add or remove a number of elements in each dimension. The following code will convert a 1-D array into 2-D array. The resulting will have 3 arrays having 4 elements
``````import numpy as np
array_1 = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
newarr1 = array_1.reshape(3, 4)
print(newarr1)
``````
``````Output:
[[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]]
``````
1-D to 3-D:
The outer dimension will contain two arrays that have three arrays with two elements each.
``````import numpy as np
array_1= np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
newarr1 = array_1.reshape(2, 3, 2)
print(newarr1)
``````
``````Output:
[[[ 1 2]
[ 3 4]
[ 5 6]]
[[ 7 8]
[ 9 10]
[11 12]]]
``````
Flattening arrays:
Converting higher dimensions arrays into one-dimensional arrays is called flattening of arrays.
``````import numpy as np
arr1= np.array([[4,5,6], [7, 8, 9]])
newarr1 = arr1.reshape(-1)
print(newarr1)
``````
``````Output :
[1 2 3 4 5 6]
``````
NumPy Array Iterating
Iteration through the arrays is possible using for loop.
Example 1:
``````import numpy as np
arr1 = np.array([1, 2, 3])
for i in arr1:
print(i)
``````
``````Output: 1
2
3
``````
Example 2:
``````import numpy as np
arr = np.array([[4, 5, 6], [1, 2, 3]])
for x in arr:
print(x)
Output: [4, 5, 6]
[1, 2, 3]
``````
Example3:
``````import numpy as np
array1 = np.array([[1, 2, 3], [4, 5, 6]])
for x in array1:
for y in x:
print(y)
``````
## NumPy Array Join
Joining is an operation of combining one or two arrays into a single array. In Numpy, the arrays are joined by axes. The concatenate() function is used for this operation, it takes a sequence of arrays that are to be joined, and if the axis is not specified, it will be taken as 0.
``````import numpy as np
arr1 = np.array([1, 2, 3])
arr2 = np.array([4, 5, 6])
finalarr = np.concatenate((arr1, arr2))
print(finalarr)
``````
``````Output: [1 2 3 4 5 6]
``````
The following code joins the specified arrays along the rows
``````import numpy as np
arr1 = np.array([[1, 2], [3, 4]])
arr2 = np.array([[5, 6], [7, 8]])
finalarr = np.concatenate((arr1, arr2), axis=1)
print(finalarr)
Output:
[[1 2 5 6]
[3 4 7 8]]
``````
## NumPy Array Split
As we know, split does the opposite of join operation. Split breaks a single array as specified. The function array_split() is used for this operation and one has to pass the number of splits along with the array.
``````import numpy as np
arr1 = np.array([7, 8, 3, 4, 1, 2])
finalarr = np.array_split(arr1, 3)
print(finalarr)
Output: [array([7, 8]), array([3, 4]), array([1, 2])]
``````
Look at an exceptional case where the no of elements is less than required and observe the output
Example :
``````import numpy as np
array_1 = np.array([4, 5, 6,1,2,3])
finalarr = np.array_split(array_1, 4)
print(finalarr)
Output : [array([4, 5]), array([6, 1]), array([2]), array([3])]
``````
## Split into Arrays
The array_split() will return an array containing an array as a split, we can access the elements just as we do in a normal array.
``````import numpy as np
array1 = np.array([4, 5, 6,7,8,9])
finalarr = np.array_split(array1, 3)
print(finalarr[0])
print(finalarr[1])
Output :
[4 5]
[6 7]
``````
Splitting of 2-D arrays is also similar, send the 2-d array in the array_split()
``````import numpy as np
arr1 = np.array([[1, 2], [3, 4], [5, 6], [7, 8], [9, 10], [11, 12]])
finalarr = np.array_split(arr1, 3)
print(finalarr)
Output:
[array([[1, 2],
[3, 4]]), array([[5, 6],
[7, 8]]), array([[ 9, 10],
[11, 12]])]
``````
NumPy Array Search
The where() method is used to search an array. It returns the index of the value specified in the where method.
The below code will return a tuple indicating that element 4 is at 3,5 and 6
``````import numpy as np
arr1 = np.array([1, 2, 3, 4, 5, 4, 4])
y = np.where(arr1 == 4)
print(y)
Output : (array([3, 5, 6]),)
``````
## Frequently Asked Questions on NumPy in Python
#### 1. What is NumPy and why is it used in Python?
Numpy- Also known as numerical Python, is a library used for working with arrays. It is also a general-purpose array-processing package that provides comprehensive mathematical functions, linear algebra routines, Fourier transforms, and more.
NumPy aims to provide less memory to store the data compared to python list and also helps in creating n-dimensional arrays. This is the reason why NumPy is used in Python.
#### 2. How do you define a NumPy in Python?
NumPy in python is defined as a fundamental package for scientific computing that helps in facilitating advanced mathematical and other types of operations on large numbers of data.
#### 3. Where is NumPy used?
NumPy is a python library mainly used for working with arrays and to perform a wide variety of mathematical operations on arrays.NumPy guarantees efficient calculations with arrays and matrices on high-level mathematical functions that operate on these arrays and matrices.
#### 4. Should I use NumPy or pandas?
Go through the below points and decide whether to use NumPy or Pandas, here we go:
• NumPy and Pandas are the most used libraries in Data Science, ML and AI.
• NumPy and Pandas are used to save n number of lines of Codes.
• NumPy and Pandas are open source libraries.
• NumPy is used for fast scientific computing and Pandas is used for data manipulation, analysis and cleaning.
#### 6. What is a NumPy array?
Numpy array is formed by all the computations performed by the NumPy library. This is a powerful N-dimensional array object with a central data structure and is a collection of elements that have the same data types.
#### 7. What is NumPy written in?
NumPy is a Python library that is partially written in Python and most of the parts are written in C or C++. And it also supports extensions in other languages, commonly C++ and Fortran.
#### 8. Is NumPy easy to learn?
NumPy is an open-source Python library that is mainly used for data manipulation and processing in the form of arrays.NumPy is easy to learn as it works fast, works well with other libraries, has lots of built-in functions, and lets you do matrix operations.
NumPy is a fundamental Python library that gives you access to powerful mathematical functions. If you’re looking to dive deep into scientific computing and data analysis, then NumPy is definitely the way to go.
On the other hand, pandas is a data analysis library that makes it easy to work with tabular data. If your focus is on business intelligence and data wrangling, then pandas are the library for you.
In the end, it’s up to you which one you want to learn first. Just be sure to focus on one at a time, and you’ll be mastering NumPy in no time!
Original article source at: https://www.mygreatlearning.com
#numpy #python
1624291780
## Learn Python - Full Course for Beginners [Tutorial]
This course will give you a full introduction into all of the core concepts in python. Follow along with the videos and you’ll be a python programmer in no time!
⭐️ Contents ⭐
⌨️ (0:00) Introduction
⌨️ (1:45) Installing Python & PyCharm
⌨️ (6:40) Setup & Hello World
⌨️ (10:23) Drawing a Shape
⌨️ (15:06) Variables & Data Types
⌨️ (27:03) Working With Strings
⌨️ (38:18) Working With Numbers
⌨️ (48:26) Getting Input From Users
⌨️ (52:37) Building a Basic Calculator
⌨️ (1:03:10) Lists
⌨️ (1:10:44) List Functions
⌨️ (1:18:57) Tuples
⌨️ (1:24:15) Functions
⌨️ (1:34:11) Return Statement
⌨️ (1:40:06) If Statements
⌨️ (1:54:07) If Statements & Comparisons
⌨️ (2:00:37) Building a better Calculator
⌨️ (2:07:17) Dictionaries
⌨️ (2:14:13) While Loop
⌨️ (2:20:21) Building a Guessing Game
⌨️ (2:32:44) For Loops
⌨️ (2:41:20) Exponent Function
⌨️ (2:47:13) 2D Lists & Nested Loops
⌨️ (2:52:41) Building a Translator
⌨️ (3:04:17) Try / Except
⌨️ (3:21:26) Writing to Files
⌨️ (3:28:13) Modules & Pip
⌨️ (3:43:56) Classes & Objects
⌨️ (3:57:37) Building a Multiple Choice Quiz
⌨️ (4:08:28) Object Functions
⌨️ (4:12:37) Inheritance
⌨️ (4:20:43) Python Interpreter
📺 The video in this post was made by freeCodeCamp.org
The origin of the article: https://www.youtube.com/watch?v=rfscVS0vtbw&list=PLWKjhJtqVAblfum5WiQblKPwIbqYXkDoC&index=3
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Thanks for visiting and watching! Please don’t forget to leave a like, comment and share!
#python #learn python #learn python for beginners #learn python - full course for beginners [tutorial] #python programmer #concepts in python
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
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
1602968400
## Python Tricks Every Developer Should Know
Python is awesome, it’s one of the easiest languages with simple and intuitive syntax but wait, have you ever thought that there might ways to write your python code simpler?
In this tutorial, you’re going to learn a variety of Python tricks that you can use to write your Python code in a more readable and efficient way like a pro.
### Let’s get started
Swapping value in Python
Instead of creating a temporary variable to hold the value of the one while swapping, you can do this instead
``````>>> FirstName = "kalebu"
>>> LastName = "Jordan"
>>> FirstName, LastName = LastName, FirstName
>>> print(FirstName, LastName)
('Jordan', 'kalebu')
``````
#python #python-programming #python3 #python-tutorials #learn-python #python-tips #python-skills #python-development
1574339477
## Python Programming Tutorials For Beginners
Description
Hello and welcome to brand new series of wiredwiki. In this series i will teach you guys all you need to know about python. This series is designed for beginners but that doesn’t means that i will not talk about the advanced stuff as well.
As you may all know by now that my approach of teaching is very simple and straightforward.In this series i will be talking about the all the things you need to know to jump start you python programming skills. This series is designed for noobs who are totally new to programming, so if you don’t know any thing about
programming than this is the way to go guys Here is the links to all the videos that i will upload in this whole series.
In this video i will talk about all the basic introduction you need to know about python, which python version to choose, how to install python, how to get around with the interface, how to code your first program. Than we will talk about operators, expressions, numbers, strings, boo leans, lists, dictionaries, tuples and than inputs in python. With
Lots of exercises and more fun stuff, let’s get started.
Dropbox: https://bit.ly/2AW7FYF
Who is the target audience?
First time Python programmers
Students and Teachers
IT pros who want to learn to code
Aspiring data scientists who want to add Python to their tool arsenal
Basic knowledge
Students should be comfortable working in the PC or Mac operating system
What will you learn
know basic programming concept and skill
build 6 text-based application using python
be able to learn other programming languages
be able to build sophisticated system using python in the future
To know more:
#python #Python Programming #Python Programming Tutorials #Python Programming Tutorials For Beginners | 13,189 | 50,295 | {"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-2023-06 | longest | en | 0.872427 |
https://semaths.com/2-is-what-percent-of-4 | 1,670,047,104,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00442.warc.gz | 541,782,639 | 7,067 | SEMATHS.ORG
Updated: 22 September 2021 06:26:00 AM
# 2 is what percent of 4?
FractionPercent
1/425%
2/450%
3/475%
1/520%
## From these considerations, what number is 0.2 percent of 4?
Percentage Calculator: What is 0.2 percent of 4.? = 0.008.
## Furthermore, what is the percentage of 4?
Solution for 4 is what percent of .4: 4 is 100% since it is our output value.
## Besides, what would be a 2 out of 4?
2 Out of 4 is 50%.
#### What is 30 out of 100 as a percentage?
Therefore the fraction 30/100 as a percentage is 30%.
#### What number is 20% of 35?
Percentage Calculator: What is 20. percent of 35? = 7.
#### What is 5 out of 20 as a percentage?
Percentage Calculator: . 5 is what percent of 20? = 2.5.
#### What grade is a 56 out of 80?
Convert fraction (ratio) 56 / 80 Answer: 70%
#### How do I calculate 2/3 of a number?
To find 23 of a whole number, you need to multiply the number by the numerator 2 and divide that product by the denominator 3 .
#### What number is 40% of 75?
Percentage Calculator: What is 40 percent of 75.? = 30.
#### Is 16 out of 20 an A or B?
Convert fraction (ratio) 16 / 20 Answer: 80%
#### What is 2 out of 100 as a percentage?
Therefore the fraction 2/100 as a percentage is 2%.
#### What number is 40% of 18?
What is 40 percent (calculated percentage %) of number 18? Answer: 7.2.
#### What is 2 out of 3 as a percentage?
Fraction to percent conversion table
FractionPercent
2/366.67%
1/425%
2/450%
3/475%
#### What is 2/3 as a percentage?
Fraction to percent conversion table
FractionPercent
2/366.67%
1/425%
2/450%
3/475%
#### What percentage is 2.5 out of 4?
Percentage Calculator: 2.5 is what percent of 4? = 62.5.
460%
550%
640%
730%
#### What is 2/3 as a percent with 2 decimal places?
Some common decimals and fractions
FractionDecimalPercent
1/20.550%
1/30.333?33.333?%
2/30.666?66.666?%
1/40.2525%
#### What percent is 20 out of 16?
How much is 20 out of 16 written as a percent value? Convert fraction (ratio) 20 / 16 Answer: 125%
#### What is 4 As a percentage of 20?
Convert fraction (ratio) 4 / 20 Answer: 20%
#### What number is 0.2 of 80?
Percentage Calculator: What is 0.2 percent of 80? = 0.16.
#### Whats 16 out of 20 as a percentage?
Percentage Calculator: 16 is what percent of 20? = 80.
#### What number is 2% of 35?
What is 2 percent (calculated percentage %) of number 35? Answer: 0.7.
#### What is a 16 out of 20 grade?
Convert fraction (ratio) 16 / 20 Answer: 80%
#### How do you calculate 0.2 percent?
Percentage Calculator How to find 0.2% of a number? Take the number and multiple it by 0.2. Then multiply that by . 01.
#### What is 5 out of 7 as a percentage?
Fraction to percent conversion table
FractionPercent
3/742.857143%
4/757.142858%
5/771.428571%
6/785.714286%
#### What is 2.5 on a 4.0 scale?
A 2.5 GPA, or Grade Point Average, is equivalent to a C+ letter grade on a 4.0 GPA scale, and a percentage grade of 77–79.
#### What number is 50% of 15?
What is 50 percent (calculated percentage %) of number 15? Answer: 7.5.
#### What percent is 2 out of?
How much is 2 out of 1 written as a percent value? Convert fraction (ratio) 2 / 1 Answer: 200% | 1,011 | 3,177 | {"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} | 4.15625 | 4 | CC-MAIN-2022-49 | latest | en | 0.901084 |
https://mlmnewsglobal.com/qa/question-what-is-the-pip-20-metre-rule.html | 1,606,455,337,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00076.warc.gz | 406,522,520 | 7,719 | # Question: What Is The Pip 20 Metre Rule?
## How do I get higher rate PIP?
Check what the mobility scores mean If you get between 8 and 11 points in total, you’ll get the mobility component of PIP at the standard rate.
If you get at least 12 points in total, you’ll get the mobility component at the enhanced rate..
## Do I have to pay council tax if I get PIP?
If you get PIP you may be entitled to extra money on top of your existing benefits, a reduction in your council tax or road tax bills and discounts on travel. You’ll need your PIP award letter before you can apply for this extra help.
## How far can you walk to get PIP?
You can stand and then move between 1 and 20 metres without any help. You can stand and then move between 1 and 20 metres with a special aid. You can’t stand, even with a special aid. You can’t move more than 1 metre, even with a special aid.
## How long does it take to walk 20 Metres?
20 meters is 1/300th of distance taken in an hour, so measured time walking that distance will be 36/3 seconds when leaving the unnecessary zeroes out = 12 seconds assuming that there is no change of pace (or steepness) at all. A typical adult walks at about 3 miles per hour. 3 miles is 4.82803 kilometers.
## How many buses is 20 meters?
2 buses20 metres is the length of 2 buses.
## Can you get PIP for multiple sclerosis?
Many people with MS can claim Personal Independence Payment (PIP) – a benefit that can help cover the extra costs you may face if you need help doing everyday tasks or find it difficult to get around your home. PIP has two components, you can be awarded either or both of these components.
## How many Metres can you walk in 5 minutes?
400 metersBased on the average walking speed a five-minute walk is represented by a radius measuring ¼ of a mile or about 400 meters. This rule of thumb is used to calculate public transport catchment areas or to determine access to destinations within neighborhoods.
## What conditions automatically qualify you for PIP UK?
You can get PIP whether you’re working or not. You must be aged 16 or over and usually have not reached State Pension age to claim. You must also have a health condition or disability where you: have had difficulties with daily living or getting around (or both) for 3 months.
## How long does Pip decision take 2020?
On average, it takes the DWP 12 weeks from the date you started your claim to make a decision. Some claims take less time, some take more.
## Does MS qualify for a blue badge?
You need to first of all consider if you are eligible to receive a blue badge. You may qualify if you: are receiving the higher rate of the Mobility Component of the Disability Living Allowance. … have a permanent and substantial disability which causes an inability to walk, or very considerable difficulty in walking.
## Is MS considered a permanent disability?
If your multiple sclerosis is advanced, you have a good chance of getting approved for Social Security disability benefits. Multiple sclerosis (MS) is a chronic autoimmune disease that affects your central nervous system, including your brain, spinal cord, and optic nerves.
## Is MS considered a disability?
If you have Multiple Sclerosis, often known as MS, you may qualify for Social Security disability benefits if your condition has limited your ability to work. To qualify and be approved for disability benefits with MS, you will need to meet the SSA’s Blue Book listing 11.09.
## What are the payment levels for PIP?
PIP ratesComponentWeekly rateDaily living – standard rate£59.70Daily living – enhanced rate£89.15Mobility – standard rate£23.60Mobility – enhanced rate£62.25
## What conditions automatically qualify you for PIP?
But which specific conditions are entitled to PIP?preparing or eating food.washing, bathing and using the toilet.dressing and undressing.reading and communicating.managing your medicines or treatments.making decisions about money.engaging with other people.
## How much is 2020 PIP?
PIP ratesPIP rateWeekly rates 2020/2021PIP Daily Living Enhanced Rate£89.15PIP Daily Living Standard Rate£59.70PIP Mobility Enhanced Rate£62.25PIP Mobility Standard Rate£23.60 | 934 | 4,192 | {"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} | 2.9375 | 3 | CC-MAIN-2020-50 | latest | en | 0.938168 |
http://www.circuitdiagramworld.com/switch_circuit_diagram/Switchmode_Constant_Current_Source_9265.html | 1,513,074,153,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515313.13/warc/CC-MAIN-20171212095356-20171212115356-00727.warc.gz | 343,300,127 | 4,981 | Switchmode Constant Current Source_Circuit Diagram World
Position: Index > Switch Circuit >
Switchmode Constant Current Source
2015-09-09 06:11
Declaration:We aim to transmit more information by carrying articles . We will delete it soon, if we are involved in the problems of article content ,copyright or other problems.
Operating a stepper motor using a fixed (constant) voltage supply results in poor torque at high speeds. In fact, stepper motors tend to stall at fairly low speeds under such conditions. Several approaches can be used to overcome this problem, one of which is to use a constant current supply in place of the more conventional constant voltage supply. A disadvantage of many constant current supplies is that simple circuits are inefficient but that doesn't apply to switchmode supplies such as the circuit shown here.
Basically, this circuit is a conventional switchmode regulator adapted for constant current output and is specially designed for stepper motor drivers - although it could be used for other applications as well. The circuit works as follows: IC1 (LM2575T) and its associated components (D1, L1, C1, etc) operate as a switchmode power supply. Normally, for constant voltage operation, the output is connected - either directly or via a resistive divider - back to the feedback input (pin 4) of IC1.
Circuit diagram:Switch-Mode Constant Current Source Circuit Diagram
In this circuit, however, Q1 senses the current flowing through R1 and produces a corresponding voltage across R3. This voltage is then fed to pin 4 of IC1. As a result, the the circuit regulates the current into a load rather than the voltage across the load. Only one adjustment is needed: you have to adjust VR1 for optimum stepper motor performance over the desired speed range. The simplest way to do this is to measure the motor current at its rated voltage at zero stepping speed and then adjust VR1 for this current. The prototype worked well with a stepper motor rated at 80O per winding and a 12V nominal input voltage. Some components might have to be modified for motors having different characteristics. | 435 | 2,131 | {"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-2017-51 | latest | en | 0.915731 |
https://www.shikhajha.com/2020/09/number-analogy-questions.html | 1,603,787,974,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00081.warc.gz | 896,250,104 | 35,403 | # Number Analogy Questions
Question 1:
8 : 81 : : 64 : ?
625
Solution:
As,
(2)3 : (2+1)(3+1)
= (2)3 : (3)4
=8 : 81
Similarly,
(4)3 : (4+1)(3+1)
= (4)3 : (5)4
=64 : 625
Question 2:
9 : 8 : : 16 : ?
27
Solution:
As,
(3)2 : (3-1)(2+1)
= (3)2 : (2)3
=9 : 8
Similarly,
(4)2 : (4-1)(2+1)
= (4)2 : (3)3
=16 : 27
Question 3:
85:40::78:?
56
Solution:
As,
85=8*5=40
Similarly,
78=7*8=56
Question 4:
42:56::110:?
132
Solution:
As,
6*7=42 : 7*8=56
(6,7,8)=(a,b,c) so, a*b=b*c i.e. 42:56
Similarly,
10*11 = 110 : 11*12 = 132
(10,11,12)=(a,b,c) so, a*b=b*c i.e. 11:132
Question 5:
48:122::168:?
downpour
Solution:
As,
48+1=49=72=perfect square
122-1=121=112=perfect square
Here, 7+4=11
Similarly,
168+1=169=132=perfect square
So, 13+4=17
So, 172=289=perfect square
So, 289+1=290
To understand,
168+1=169=132=perfect square
290-1=289=172=perfect square
So, 13+4=17
Practice More Reasoning Questions: | 437 | 944 | {"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} | 3.8125 | 4 | CC-MAIN-2020-45 | latest | en | 0.447221 |
http://mathoverflow.net/questions/40463/is-a-rhombus-rigid-on-a-sphere-or-torus-and-generalizations?sort=oldest | 1,462,486,497,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861623301.66/warc/CC-MAIN-20160428164023-00087-ip-10-239-7-51.ec2.internal.warc.gz | 180,276,164 | 22,117 | # Is a rhombus rigid on a sphere or torus? And generalizations.
If a rectangle is formed from rigid bars for edges and joints at vertices, then it is flexible in the plane: it can flex to a parallelogram. On any smooth surface with a metric, one can define a linkage (e.g., a rectangle) whose edges are geodesics of fixed length, and whose vertices are joints, and again ask if it is rigid or flexible on the surface. This leads to my first, specific question:
Q1. Is a rhombus, or a rectangle, always flexible on a sphere?
It seems the answer should be Yes but I am a bit uncertain if there must be a restriction on the edge lengths. (In the above figure, the four arcs are each $49^\circ$ in length, comfortably short.)
Q2. The same question for other surfaces: Arbitrary convex surfaces? A torus?
I am especially interested to learn if there are situations where a linkage that is flexible in the plane is rendered rigid when embedded on some surface. It seems this should be possible...?
Q3. More generally, Laman's theorem provides a combinatorial characterization of the rigid linkages in the plane. The $n{=}4$ rectangle is not rigid because it has fewer than $2n-3 = 5$ bars: it needs a 5th diagonal bar to rigidify. Has Laman's theorem been extended to arbitary (closed, smooth) surfaces embedded in $\mathbb{R}^3$? Perhaps at least to spheres, or to all convex surfaces?
Thanks for any ideas or pointers to relevant literature!
Addendum. I found one paper related to my question: "Rigidity of Frameworks Supported on Surfaces" by A. Nixon, J.C. Owen, S.C. Power. arXiv:1009.3772v1 math.CO In it they prove an analog of Laman's theorem for the circular cylinder in $\mathbb{R}^3$. If one phrases Laman's theorem as requiring for rigidity that the number of edges $E \ge 2 V - 3$ in both the graph and in all its subgraphs, then their result (Thm. 5.3) is that, on the cylinder, rigidity requires $E \ge 2 V -2$ in the graph and in all its subgraphs. This is not the precise statement of their theorem. They must also insist that the graph be regular in a sense that depends on the rigidity matrix achieving maximal rank (Def. 3.3). They give as examples of irregular linkages on a sphere one that contains an edge with antipodal endpoints, or one that includes a triangle all three of whose vertices lie on a great circle. But modulo excluding irregular graphs and other minor technical details, they essentially replace the constant 3 in Laman's theorem for the plane with 2 for the cylinder.
Theirs is a very recent paper but contains few citations to related work on surfaces, suggesting that perhaps the area of linkages embedded on surfaces is not yet well explored. In light of this apparent paucity of information, it seems appropriate that I 'accept' one of the excellent answers received. Thanks!
Addendum [31Jan11]. I just learned of a 2010 paper by Justin Malestein and Louis Theran, "Generic combinatorial rigidity of periodic frameworks" arXiv:1008.1837v2 (math.CO), which pretty much completely solves the problem of linkages on a flat 2-torus, generalizing to flat orbifolds. They obtain a combinatorial characterization for generic minimal rigidity for "planar periodic frameworks," which encompass these surfaces.
-
For an arbitrary surface, it's not clear to me that one can even slide just a rigid bar around on the surface arbitrarily - for instance, a geodesic of sufficient length might run into itself. – j.c. Sep 29 '10 at 17:42
@jc: Interesting point! But in that case I would let the segment self-intersect. View the bars as rigid but they can interpenetrate. In the plane one imagines them sliding over one another at $\epsilon$-separated layers. – Joseph O'Rourke Sep 29 '10 at 17:58
Right, I was afraid you'd say that. What puzzles me is that obviously you can do that for an infinitesimal path with no thickness, but what happens if one had a real stick made out of wood embedded in such a surface? i.e. when is it possible to transport a small "tube" of geodesics? Even outside of the centerline of this tube intersecting itself, there are caustics that emerge as nearby geodesics run into each other... See the beautiful images here for what I'm hinting at math.harvard.edu/~knill/caustic/exhibits/torus/index.html – j.c. Sep 29 '10 at 18:04
I want to argue that nonzero Gaussian curvature causes this sort of thing to happen (I interpret the above as a bar of nonzero thickness being sheared or strained as you move it in a surface) vis a vis its effect on the Jacobi field. But isn't this another kind of rigidity? Perhaps not the one that you're after here. – j.c. Sep 29 '10 at 18:06
@Joe: Just for completeness, the generalizations from the torus to some other orbifolds can be found in a second preprint arxiv.org/abs/1108.2518 – Louis Feb 14 '12 at 8:57
Q3: Laman's theorem is the same on the sphere.
Indeed, a configuration with $n$ vertices and $m$ edges is defined by a system of $m$ equations in $2n-3$ variables (there are $2n$ coordinates of points, but we may assume that the first point is fixed and the direction of one of the edges from the first point is fixed too). The the left-hand sides are analytic functions of our variables and the right-hand sides are the squares of the lengths of the bars (on the sphere, cosines rather than squares).
Consider this system as a map $f:\mathbb R^{2n-3}\to\mathbb R^m$. The rigidity means that a generic point $x\in\mathbb R^{2n-3}$ cannot be moved within the pre-image of $f(x)$. This implies that $rank(df)=2n-3$ on an open dense set. Choose a configuration from this set and project it to the sphere of radius $R\to\infty$. The equations on the sphere converge to those in the plane, hence the rank of the linearization on the sphere will be maximal ($=2n-3$) for all large $R$. So we get an open set of configuration on the sphere where the linearization has the maximal rank (and this implies rigidity). Since all functions involved are analytic and the rank is maximal on an open set, it is maximal generically. So our linkage is generically rigid on the sphere.
Conversely, consider a flexible linkage on the plane. If $m<2n-3$, it will be flexible on the sphere by a dimension counting argument. Otherwise, by Laman's theorem, there is a subgraph with $r$ vertices and more than $2r-3$ edges. Consider such a subgraph for which $r$ is minimal. Then, by Laman's theorem, we can remove some edges so that this subgraph remains rigid. And, by the above argument, it is rigid on the sphere too. So the edges that we removed were redundant both in the plane and in the sphere. Let's forget about them and repeat the procedure. Eventually we will get a linkage with fewer than $2n-3$ edges.
-
Outstanding! As I searched a bit for this before posting without luck, it may be a new result. – Joseph O'Rourke Sep 29 '10 at 20:55
I wonder if it is possible on the plane that a graph is generically rigid but some exceptional instances of it are non-rigid. It can easily happen on the sphere since geodesic segments between opposite points are not unique. – Sergei Ivanov Sep 29 '10 at 21:26
An example of a graph in the plane that is generically rigid but has a non-rigid realization is figure 1.3c in Graver, Servatius and Servatius's book on Combinatorial Rigidity: books.google.com/books?id=0XwvY1GVNN4C&pg=PA3 – j.c. Sep 30 '10 at 0:41
This might be a cheat, but if you consider a linkage turning once around the meridian of a torus, it might be not flexible. Actually, a linkage homeomorphic to a circle that minimizes length in its homology class is not flexible.
-
@Pierre: Ah, nice observation about linkages that minimize length! – Joseph O'Rourke Sep 29 '10 at 13:15
It would be interesting to learn if this is the only type of obstruction, the only conversion (flexible in plane) $\rightarrow$ (rigid on surface). – Joseph O'Rourke Sep 29 '10 at 16:07
Any loop in the linkage that is not contractible on the torus essentially gives an extra bar (a fixed distance in the universal cover). So there are more examples that are rigid on the torus. For example, any non-contractible loop of two edges is rigid, no matter minimal or not. – Sergei Ivanov Sep 29 '10 at 18:53
@Sergei: Nice digon example---Thanks! – Joseph O'Rourke Sep 29 '10 at 19:01
Actually any reasonable analog on the plane will be rigid there too, so this example is not quite on topic. – Sergei Ivanov Sep 29 '10 at 21:29
No rhombus on the round sphere is rigid, if we define a rhombus $ABCD$ to be a union of four geodesic segments $\overline{AB}\cup \overline{BC}\cup \overline{CD}\cup \overline{DA}$, such that $A$, $B$, $C$, and $D$ are all distinct points, no three of which are collinear, and such that all four segments have the same length, with self-intersections and overlaps allowed. Suppose $ABCD$ is a rhombus. First note that the hypotheses imply that $B$ is neither equal to nor antipodal to $A$: they're not equal because the vertices are distinct; while if they were antipodal, then $C$ would have to be equal to $A$ because $A$ is the only point at the correct distance from $B$. These facts imply that the endpoints of the geodesic segments starting at $A$ and having length equal to $AB$ trace out a circle, not a point. Let $B'$ be any point on this circle, and let $C'$ be the point obtained by reflecting $A$ through the great circle containing $B'$ and $D$. Because reflection through a great circle is an isometry, it follows that $AB'C'D$ is also a rhombus with the same side lengths as $ABCD$.
I would guess that this argument could be made to work in any symmetric space (with a suitable interpretation of "antipodal"), but I have no idea what would happen with more general metrics.
-
@Jack: Very pretty proof! Nice that your specifications rule out $B$ being equal or antipodal to $A$. – Joseph O'Rourke Sep 29 '10 at 19:35
The first paragraph of Some notes on the equivalence of first-order rigidity in various geometries by Franco V. Saliola and Walter Whiteley states:
In this paper, we explore the connections among the theories of first-order rigidity of bar and joint frameworks (and associated structures) in various metric geometries extracted from the underlying projective space of dimension n, or $\mathbb{R}^{n+1}$. The standard examples include Euclidean space, elliptical (or spherical) space, hyperbolic space, and a metric on the exterior of hyperbolic space.
Section 4 of the notes proves that a framework is first-order rigid in the upper hemisphere $S^n_+$ iff a corresponding framework in $\mathbb{R}^n$ is first-order rigid.
Section 5 extends this to geometries on the space $X^n_{c,k}=\{x\in\mathbb{R}^{n+1}|\langle x,x\rangle_k=c,x_{n+1}>0\}$ where $\langle x,x\rangle_k=x_1y_1+\cdots+x_{n-k+1}y_{n-k+1}-x_{n-k+2}y_{n-k+2}-\cdots-x_{n+1}y_{n+1}$. Note that $k=1,c=-1$ is hyperbolic space $\mathbb{H}^n$, the case $k=0,c=1$ is $S^n_+$.
-
@jc: Thanks for finding this! Interesting that they restrict to a hemisphere... – Joseph O'Rourke Oct 11 '10 at 13:53
My impression is that the restriction to the hemisphere is not essential, but just technically convenient for their proof. – j.c. Oct 11 '10 at 14:50
There is an obvious metric on the sphere, which you are assuming applies. There is more than one way to put a nice metric on a torus - but I think you mean a torus embedded in three-space, which does not admit such a nice metric. You need to specify what metric you are assuming on the surfaces you are interested in.
-
@maproom: A valid point! I was assuming some geometric torus (a circle swept around a circle) and shortest geodesic paths on its surface. – Joseph O'Rourke Sep 29 '10 at 17:37 | 3,007 | 11,656 | {"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} | 3.421875 | 3 | CC-MAIN-2016-18 | longest | en | 0.91989 |
https://teacheradvisor.org/5/activities/chunk_5-1-14-1 | 1,611,750,966,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00522.warc.gz | 592,824,551 | 7,394 | Fluency Activity
# Multiply and Divide by Exponents
Lesson 14. Unit 1. Grade 5 EngageNY
EngageNY3 min(s)
This Fluency Activity is a part of the Lesson 14, Unit 1, Grade 5. In this lesson, students continue to divide decimals by whole numbers. This review fluency helps solidify student understanding of multiplying by 10, 100, and 1,000 in the decimal system. | 98 | 363 | {"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} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.859218 |
http://geometryofmolecules.com/pcl3-lewis-structure-bond-angles-and-hybridization/ | 1,553,164,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00553.warc.gz | 85,932,085 | 15,390 | Lewis Structure
# PCL3 Molecular Electron Geometry, Lewis Structure, Bond Angles and Hybridization
PCL3 – Phosphorus Trichloride is a chemical formula of phosphorus and chlorine. The shape of PCL3 is Trigonal Pyramidal. It is toxic and corrosive chemical. There are also some exposure limits for this chemical, which is set by the government of America. Thus, it is proved that PCL3 is risky and a dangerous chemical. In this article, I will show you the molecule equations of PCL3. So, if you love Geometry of Molecules and want to learn about this chemical by its roots, stay tuned and read further.
Contents
## PCL3 Molecular Geometry
Before starting any complicated explanations, let’s start with the basics. Here is the molecular geometry of PCL3. There are two dimensions of Phosphorus Trichloride. One aspect contains electron and the second dimension includes the bonds. Here one thing we all should keep in mind that even if we have a composition, that doesn’t mean that we can get the shape of it. To find out any shape, we have to take help of the VSEPR theory, which is also known as Valence Shell Electron Pair Repulsion theory. The outer atoms and the lone pair of electrons are not attracted to each other. This situation can help us to get a three-dimensional shape, which is helpful using the AX Notion method.
## PCL3 Lewis Structure
If you consider the AXN method, the A should be found as the central phosphorus atom, and the X is the number of particles attached to it. As there are three chlorines connected to the central phosphorus atom, it will be X3. The rest one is N. N is the total number of the lone pair of electrons, which is not at all bound by any other bonds. Considering the formula of PCL3, such lone pair of electrons is only one. So, after evaluating AXN formula of Phosphorus Trichloride, the outcome is A1X3N1 or AX3N.
## PCL3 Bond Angles and Shape
After getting the AX3N, we should look upon that table which can help us to know the molecular geometry of PCL3. As you know, there will be different geometry for each one in the table. We just need to look at them one by one and find out the AX3N. Initially, there will be AX2 to AX3, but we must go down where the AX3N is written. There we will find out that in front of AX3N, the given shape is ‘Trigonal Pyramidal.’ So, the actual shape of Phosphorus Trichloride is Trigonal Pyramidal.
The bond angle of PCL3 is 109 degrees. Many other formulas like Ammonia – NH3 also possess the same amount of bond angles, so this angle is quite common. Now, again look at the molecule. It shows that the phosphorus is in the center with the lone pair of electrons. The particle will look like as under;
So, we have now found out that the AXN method gives us the idea about the bond angles and also inform us about the shape of the molecule. Now, let’s move to the electron geometry of PCL3 and its polarity status.
PCL3 Electron Geometry
Now we all are clear that the Phosphorus has 5 valence electrons and the chlorine has 7 valence electrons. There are three chlorines, and so the seven must be multiplied with three, which gives the output of 21. Now, this 21 should be added in 5 – the valence electrons of Phosphorus. The final result is 5 + 3(7) = 26.
After getting the output of 26 valence electrons, now it’s time to subtract 26 from the highest multiple of 8. And naturally, as the final result is 26, the highest multiple of 8 must not cross 26. So, the number, which is a multiple of 8 and does not pass 26, is 24. The difference is 2, which means that there is one lone pair of 2 electrons on the central phosphorus atom.
When we examine the Lewis structure of PCL3, we can see that each chlorine atoms have 3 lone pairs and all of them must have 8 electrons around it. These chlorines want to satisfy their oxide requirement, and that is why the geometry for PCL3 is called Trigonal Pyramidal.
## Is PCL3 Polar or Nonpolar?
Many of my students have asked this question; Is this molecule polar or nonpolar? Let’s find out together. As you know, when there is any partial transfer of electron density between two different atoms and if the electronegativity is not equal, then the bonds become ionic. Those bonds which are not at all ionic are known as nonpolar bonds, and all the other bonds are polar.
The molecular geometry of PCL3 is trigonal pyramidal with the partial charge distribution on the phosphorus. It is the well-known fact that if there is a vast difference of the electronegativity, there are more chances of polarity. The phosphorus has an electronegativity value of 2.19 and chlorine comes with 3.16. So, the end difference is 0.97, which is quite significant. If the difference is in between 0 to 0.50, then it will be non-polar. But, as the difference is more than 0.5, PCL3 is polar.
## PCL3 Hybridization
As discussed, the phosphorus has one lone pair and 3 atoms. So, it has 4 groups in total. That is why the hybridization is S1P3. Just like Phosphorus, if we talk about the chlorine, there are 3 lone pairs attached to the center atom, which is P And, these 3 lone pairs of Chlorine is connected to the single lone pair, which makes them 4. Therefore, the hybridization remains same as it is – SP3.
So, that’s all for the Phosphorus Trichloride. I hope I make some sense and you are comfortable with all of these explanations given here. I will keep sharing Geometry of Molecules like this, but till then, stay tuned and keep learning!
### Janice Powell
I am interested in sharing articles related to Geometry of Molecules. It represents forms of Chemical equation with the help of structured atoms.
Close | 1,372 | 5,647 | {"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} | 2.71875 | 3 | CC-MAIN-2019-13 | latest | en | 0.94185 |
https://documen.tv/4-10-11-13-14-15-25-29-33-33-35-43-51-58-64-find-the-median-find-the-lower-quartile-find-the-upp-27148798-89/ | 1,679,742,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00545.warc.gz | 259,662,355 | 15,947 | Question
4,10,11,13,14,15,25,29,33,33,35,43,51,58,64
Find the median
Find the lower quartile
Find the upper quartile
1. Latifah
First Quartile Q1 = 13
Second Quartile Q2 = 29
Third Quartile Q3 = 43
Interquartile Range IQR = 30
Median = Q2 x˜ = 29
Minimum Min = 4
Maximum Max = 64
Range R = 60
2. vankhanh
Median (Q2): 29
Lower quartile (Q1) = 13
Upper quartile (Q3): 43
Step-by-step explanation:
There are a total of 15 numbers in the data set:
4, 10, 11, 13, 14, 15, 25, 29, 33, 33, 35, 43, 51, 58, 64
Minimum (Q0): 4
Lower quartile (Q1) = 13
Median (Q2): 29
Upper quartile (Q3): 43
Maximum (Q4): 64
Hope this helps! | 269 | 620 | {"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} | 3.6875 | 4 | CC-MAIN-2023-14 | latest | en | 0.71964 |
https://caseforbasicincome.com/tag/gravity-forms-google-sheets-plugin-settings/ | 1,623,724,735,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00535.warc.gz | 167,625,031 | 10,782 | # Gravity Forms Google Sheets Plugin Settings
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Published at Saturday, June 12th 2021, 20:09:54 PM. Calculator Spreadsheet By Caitlin Carr. When it comes to using Microsofts Excel Spreadsheet program there is no question that there are a variety of ways to accomplish many tasks. One of the things I like about Excel is the ability to format multiple columns at the same time giving them the same width. If you are unfamiliar with how this is done, read on and take a look at the steps I perform to achieve this task. You will see that it is very easy to do. The first thing you want to do, obviously, is open up Excel. Once Excel is displayed on your screen, go ahead and type in cell "A1" the month name "January". In the cell "B1" type in the month name, "February", in "C1" type in "March" and so on until you type in all twelve months each in their own cell on the spreadsheet.
## Box Culvert Calculator SpreadsheetBox Culvert Calculator Spreadsheet
Published at Saturday, June 12th 2021, 19:30:20 PM. Calculator Spreadsheet By Nellie Graham. From here, I start my measuring and counting, better known in construction as doing a "take off". I use a measuring wheel. I never use a tape measure any more. Tapes are too slow and usually only measure up to thirty-five feet. The measuring wheel can measure to one thousand feet and it costs the same as a big tape measure. Measuring wheels are usually made by the same companies that make tape measure, like Lufkin or Stanley, and they dont break as often as tape measures. Once you use a measuring wheel, you probably wont go back to a tape measure. I used to use the infra-red measuring device but I find them way too inaccurate. Usually, I measure before I start counting things like windows. Once I begin to count windows and molding, I make note of anything that will add or take away time, which means adding to the cost or lessening the cost. When I count windows, I make three columns, one for windows that are located below eight feet, one column for windows below fifteen feet, and one for windows twenty feet and above. This also applies to molding or anything else heights above eight feet, like dormers, ceiling medallions or whatever.
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Published at Saturday, June 12th 2021, 18:39:26 PM. Calculator Spreadsheet By Rene Paul. Is your job dealing with more and more paper work all the time? If you are an office clerk or reports assistant or even if you are a manager you will have to send reports on a monthly basis and more often they will have to do with comparing old and new files as well as updating them. To compare spreadsheet reports, you will have to understand the values being given and consolidate it with any changes made for a certain period of time. If you will be dealing with several files here to compare, you may find this a very daunting task. However, knowing that there is a tool you can use to make this task easier for you is a great relief. Working in a financial firm would entail a lot of reports to be submitted on a weekly, monthly, semi- monthly, quarterly and a yearly basis. All of these reports should be monitored, and most of the data where these reports come from are based from several spreadsheets that you will have to consolidate in a timely manner. To effectively have these reports updated, you can make use of a file comparison tool that can compare spreadsheet files and allow the user to make changes to it from one file to another.
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Published at Saturday, June 12th 2021, 17:56:45 PM. Calculator Spreadsheet By Lidia Chapman. If there is any change in the payroll calculations, there might be a change in the summaries as well. By using a query or an Office Data Connection you can link the source and retrieve data as required. The data in the Excel sheet will be refreshed when there are changes in the source. Of course, if there are problems such as corruption in the Access database, it might have an effect on the Excel worksheet as well. You might have to reconstruct the worksheet after carrying out an Access recovery. Similarly, you might want to move some Excel data permanently into an Access database in order to take advantage of the many features of Access such as multi-user capabilities, data management abilities and security. You can convert the data from an Excel worksheet to Access by converting an entire Excel range into an Access database. On the other hand, you might want to simply summarize or analyze the data from an Excel worksheet. You can create an Access report that will take the data from the range specified in the Excel spreadsheet.
##### Refrigerant Charge Calculator SpreadsheetRefrigerant Charge Calculator Spreadsheet
Published at Saturday, June 12th 2021, 16:47:10 PM. Calculator Spreadsheet By Laurel Sparks. After you have spent some time collecting figures and transforming them into a Microsoft Excel spreadsheet you may need to present your workbook to clients and colleagues. Thanks to the huge array of features offered by the programme, you can design your workbooks in many different forms. Some entrepreneurs prefer information to be presented in straightforward tables, while others benefit from the varied functions of the software and they create coloured pie charts to reflect the data. Whichever way you choose to present information, you are likely to find that MS Excel is the backbone of many workplace documents that require analysis and presentation of data. A great feature of the product is that is has been designed to help you produce workbooks that can be easily moved to other applications. For instance, if you have recently put-together a spreadsheet that shows which departments are underperforming, you may wish to transfer information to other MS products. | 1,406 | 6,881 | {"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-2021-25 | latest | en | 0.938832 |
http://mathhelpforum.com/calculus/3052-need-solutions-some-easy-calc-problems.html | 1,529,344,659,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00102.warc.gz | 209,694,442 | 11,485 | Thread: Need solutions for some easy calc problems
1. Need solutions for some easy calc problems
thanks
=============================
It makes it more difficult for someone else to benefit from the
solutions.
2. Originally Posted by JohnSena
2. Use logrithms to solve the quation 25(1.06)^x=100 for x
?
You have,
$\displaystyle 25(1.06)^x=100$
Divide sides by 25,
$\displaystyle (1.06)^x=4$
Take, logarithm
$\displaystyle \log (1.06)^x=\log 4$
Thus, the exponent "comes down"
$\displaystyle x\log (1.06)=\log 4$
Thus,
$\displaystyle x=\frac{\log 4}{\log 1.06}\approx 23.8$
3. Originally Posted by JohnSena
3. Solve for x using logs. a) 2e^4x=8e^6x b) 3^x+3=e^7x
a)You have,
$\displaystyle 2e^{4x}=8e^{6x}$
Divide by 2,
$\displaystyle e^{4x}=4e^{6x}$
Divide by $\displaystyle e^{4x}$ thus,
$\displaystyle 1=4e^{6x-4x}=4e^{2x}$
Divide by four,
$\displaystyle 1/4=e^{2x}$
Take natural logarithm of both sides,
$\displaystyle \ln .25=2x$
Thus,
$\displaystyle x=.5\ln .25\approx -.69$
4. damn that was quick thanks
still confused on how to figure out part b and the other problems though, if anybody could help with those that'd be awesome
5. Originally Posted by JohnSena
1. A town has 1000 people initially. Find the formula for the population of the town, P, in terms of the number of years, t. a) The town grows at an annual rate of 8% a year. b) The town grows by 70 people a year
a) the rate of growth of population is proportional to the existing population.
So the population satisfies this differential equation:
$\displaystyle \frac{dp}{dt}=\alpha p(t)$,
which has solution:
$\displaystyle p(t)=p(0) e^{\alpha t}$.
Now we are told that the population grows by $\displaystyle 8\%$ per year, so
after one year the population is $\displaystyle 1.08$ times its original value so:
$\displaystyle p(1)=1.08 p(0)=e^{\alpha}p(0)$,
so: $\displaystyle \alpha=\log(1.08)$ (thats a natural log by the way).
hence:
$\displaystyle p(t)=p(0) e^{\log(1.08) t}=1000 e^{\log{1.08}t}$.
6. Originally Posted by JohnSena
1. A town has 1000 people initially. Find the formula for the population of the town, P, in terms of the number of years, t. a) The town grows at an annual rate of 8% a year. b) The town grows by 70 people a year
b) The rate of population growth is a constant so the population satisfies
the differential equation:
$\displaystyle \frac{dp}{dt}=70$,
which has solution:
$\displaystyle p(t)=70t+p(0)$,
plugging in the given initial population gives:
$\displaystyle p(t)=70t+1000$.
RonL
7. Originally Posted by JohnSena
4. If the size of a bacteria colony doubles in 8 hours, how long will it take for the number of bacteria to be 5 times the original amount?
Missing assumption needed to solve this is that the rate of growth of the
population is proportional to the existing population. Hence the population
satisfies the diffrenetial equation:
$\displaystyle \frac{dp}{dt}=\alpha p$,
which has solution:
$\displaystyle p(t)=p(0) e^{\alpha t}$
Now we are told that the population doubles in 8 hours so:
$\displaystyle p(8)=2p(0)=p(0) e^{8\alpha}$,
so: $\displaystyle 2=e^{8\alpha}$, or: $\displaystyle \alpha=\frac{\log(2)}{8}$.
Therefore:
$\displaystyle p(t)=p(0)e^{\frac{\log(2)}{8}t}$.
Now at time $\displaystyle t_5$ the population is 5 times its original value so:
$\displaystyle p(t_5)=5p(0)=p(0)e^{\frac{\log(2)}{8}t_5}$,
so:
$\displaystyle 5=e^{\frac{\log(2)}{8}t_5}$
or laking logs:
$\displaystyle \log(5)=\frac{\log(2)}{8}t_5$,
or:
$\displaystyle t_5=\frac{8\log(5)}{\log(2)}$
RonL
8. Originally Posted by JohnSena
thanks | 1,149 | 3,579 | {"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-2018-26 | latest | en | 0.818242 |
https://schoollearningcommons.info/question/13-ir-a-triangle-abc-if-la-l-b-65-and-lb-lc-140-then-the-measure-of-la-is-a-to-b-25-c-115-d-6o-20858397-20/ | 1,632,584,080,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00142.warc.gz | 538,806,022 | 13,478 | ## 13) Ir a triangle ABC, if LA + L B = 65 and LB + LC= 140, then the measure of LA is a) to b) 25 c) 115 d) 6o.
Question
13) Ir a triangle ABC, if LA + L B = 65 and LB + LC=
140, then the measure of LA is
a) to b) 25 c) 115 d)
6o.
in progress 0
3 weeks 2021-09-05T05:37:35+00:00 2 Answers 0 views 0
1. ∠A + ∠B = 65°
∠B + ∠C = 140°
∠A + ∠B + ∠C = 180° (angle sum property)
∠C = 180 – (∠A + ∠B)
= 180 – 65
= 115°
∠B + ∠C = 140
∠B + 115 = 140
∠B = 140 – 115
= 25
∠A + ∠B = 65
∠A + 25 = 65
∠A = 65 – 25
= 40°
Therefore, the measure of ∠A is 40°
2. Step-by-step explanation:
plz inbox me I will explain you | 294 | 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} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.729507 |
https://ccssmathanswers.com/into-math-grade-3-module-9-lesson-1-answer-key/ | 1,685,318,281,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00316.warc.gz | 211,591,260 | 22,797 | # Into Math Grade 3 Module 9 Lesson 1 Answer Key Identify Number Patterns on the Addition Table
We included HMH Into Math Grade 3 Answer Key PDF Module 9 Lesson 1 Identify Number Patterns on the Addition Table to make students experts in learning maths.
## HMH Into Math Grade 3 Module 9 Lesson 1 Answer Key Identify Number Patterns on the Addition Table
I Can identify number patterns on an addition table. I can use the Identity and Commutative Properties of Addition to complete equations.
Jamie uses an addition table to find sums. As he records the sums in the table, he sees a pattern. What pattern could he describe?
Complete the addition table.
Turn and Talk Shade all squares that have a sum of 6 in the table. Explain a pattern you notice.
1. Jon uses patterns to find sums in an addition table.
A. Jon finds 0 + 3 and 5 + 0. Write other sums with 0 as one addend.
B. What pattern do you see when you find the sum of any number and o?
_____________
_____________
C. Jon finds 2 + 4 and 4 + 2. Choose a column and a row with the same addend. Write the sums. Record the equations.
_____________
_____________
Think: Look at the row for 5, write: 5 + 0 = 5, 5 + 1 = 6, 5 + 2 = 7, and so on.
Look at the column for 5, write: 0 + 5 = 5, 1 + 5 = 6, 2 + 5 = 7,and so on.
D. What pattern do you see in the equations you wrote?
Connect to Vocabulary
The Identity Property of Addition states that the sum of any number and zero is that number. The Commutative Property of Addition states that you can add two or more numbers in any order and get the same sum.
Turn and Talk Describe any other patterns you can use to find sums in the addition table.
Step It Out
2. Lea uses an addition table to explore even and odd numbers.
A. Choose a row with an even addend. Shade the sums that are even. What even or odd patterns do you see ¡n the addends and sums?
B. Choose a row with an odd addend. Shade the sums that are odd. 6
What even or odd patterns do 8 you see in the addends and sums?
Turn and Talk How can you decide if the sum of two addends will be even or odd?
Check Understanding Math Board
Question 1.
Use the Identity Property of Addition to find the sum. Then use the Commutative Property of Addition to write a related fact.
9 + 0 = ____ ____
Identity Property:
9 + 0 = 9
Commutative Property:
9 + 0 = 0 + 9 = 9
Explanation:
The addition of two numbers 9 and is 9
In Commutative Property the addition of two numbers 0+9 and 9+0 is 9.
Is the sum even or odd? Write even or odd.
Question 2.
8 + 4 = ___
8 + 4 = 12
Explanation:
The addition of 8 and 4 is 12
12 is an even number.
Question 3.
6 + 3 = ___
6 + 3 = 9
Explanation:
The addition of 6 and 3 is 9
9 is an odd number.
Question 4.
7 + 9 = ___
7 + 9 = 16
Explanation:
The addition of 7 and 3 is 9
16 is an even number.
Question 5.
Reason Mulan and Emily are playing a board game. They both start on the same space. Mulan moves 3 spaces and 8 spaces in her first two turns. Emily moves 8 spaces and 3 spaces in her first two turns. Compare the positions of Mulan and Emily after their first two turns. Explain.
Mulan moves in first turn =3
Mulan moves in second turn=8
Emily moves in first turn =8
Emily moves in second turn=3
The total moves of mulan in first and second turn is 3+8=11
The total moves of emily in first and second turn is 8+3=11
So they both are at the same positions.
Is the sum even or odd? Write even or odd.
Question 6.
5 + 4 _______
5 + 4 = 9
Explanation:
The addition of 5 and 4 is 9
9 is an odd number.
Question 7.
1 + 9 _______
1 + 9 = 10
Explanation:
The addition of 1 and 9 is 10
10 is an even number.
Question 8.
4 + 6 ____
4 + 6 = 10
Explanation:
The addition of 4 and 6 is 10
10 is an even number.
Use the Identity Property of Addition to complete the equation.
Question 9.
0 + 4 = ___
Identity Property:
0+a =a
0+4= 4
Explanation:
The addition of 0 and 4 is 4.
Question 10.
1 + 9 = ___
1 + 9 =10
Explanation:
The addition of 1 and 9 is 10.
Question 11.
3 = ___ + 10
3 = -7 + 10
Explanation:
The subraction of -7 from 10 is 3.
Find the sum. Use the Commutative Property of Addition to write the related addition fact.
Question 12.
5 + 7 = ___
Commutative Property:
a+b=b+a
5+7=7+5=12
Explanation:
The addition of both 5 +7 and 7+5 and is 12.
Question 13.
4 + 10 = ___
Commutative Property:
a+b=b+a
4+10= 10+4=14
Explanation:
The addition of both 4+10 and 10+4 and is 14.
Question 14.
6 + 8 = ___
Commutative Property:
a+b=b+a
6+8=8+6=14
Explanation:
The addition of both 6+8 and 8+6 and is 14.
Question 15.
Critique Reasoning Ryan says that the sum of two odd addends is an odd number. Is Ryan correct? Explain. | 1,386 | 4,667 | {"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} | 4.9375 | 5 | CC-MAIN-2023-23 | latest | en | 0.872172 |
https://mathoverflow.net/questions/278920/parallelotope-fundamental-domains-of-the-n-torus | 1,606,242,762,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176922.14/warc/CC-MAIN-20201124170142-20201124200142-00386.warc.gz | 389,803,476 | 30,830 | # Parallelotope fundamental domains of the n-torus
The group $\mathbb Z^n$ acts on the topological space $\mathbb R^n$ by translation: if $z = (z_1, \cdots, z_n) \in \mathbb Z^n$ and $x = (x_1, \cdots, x_n) \in \mathbb R^n$, then $z\cdot x := z+x$. The quotient space of this action is the $n$-dimensional torus $\mathbb R^n/\mathbb Z^n$. In this setting, a fundamental domain is a convex set $Z \subset \mathbb R^n$ such that $Z$ surjects onto $\mathbb R^n / \mathbb Z^n$ via the usual quotient map, and such that this map is injective on the interior of $Z$.
My question is: when is a parallelotope in $\mathbb R^n$ a fundamental domain of the $n$-dimensional torus?. By a parallelotope, I mean the $n$-dimensional analogue of a parallelogram. More precisely, a parallelotope is set of the form $\left\{ \sum_{i = 1}^n a_i v_i \mid 0 \leq a_i \leq 1 \right\}$ for some linearly independent set $\left\{ v_1, \cdots, v_n \right\}\subset \mathbb R^n$.
I'm fine with assuming that our parallelotopes are rational, meaning $v_i \in \mathbb Q^n$ for all $i$. I'm also inerested more generally in which parallelitopes in $\mathbb R^n$ surject onto $\mathbb R^n / \mathbb Z^n$ via the usual quotient map.
My initial guess was that any parallelotope with sufficiently large volume would at least surject onto the $n$-torus, but this dream was quickly crushed by the following example: if $Z$ is the rectangle $[0, 0.9999]\times [0,10000000]$ in $\mathbb R^2$, then $Z$ doesn't surject onto $\mathbb R^2 / \mathbb Z^2$, but a small rotation of $Z$ does surject.
On the other hand, it's easy to see that this property is preserved by the action of $SL_n(\mathbb Z)$ on $\mathbb R^n$. Thus we may assume that the matrix $[v_1 v_2 \cdots v_n]$ is in Hermite normal form (and in particular, upper-triangular). Using this trick, I found it's not too hard (though quite messy) to figure out the $n=2$ case by hand, but I'm not sure what to do in higher dimensions.
• Is the trigonal unipotent form given by Yoav Kallus what you already found/ what you were looking for, or do you have other expectations? – Luc Guyot Aug 19 '17 at 7:56
Instead of asking which unit-volume n-parallelotopes ($TC$, where $T\in SL_n(\mathbb{R})$, and $C=[0,1]^n$) tile $\mathbb{R}^n$ when translated by $\mathbb{Z}^n$, we can equivalently ask, which unit-determinant lattices $T^{-1}\mathbb{Z}^n$ tile space when applied to the cube $C$. This is the subject of Hajós's theorem, previously a conjecture of Minkowki, which says that such a lattice has an upper triangular basis with ones on the diagonal.
This is not an answer, but only an attempt to retrieve, and present, the early result mentioned by the OP when $n = 2$. Hopefully, this will trigger further input from MO readers.
The following is immediate.
Claim 1. The image of $[0, 1]^n$ by any matrix in $GL_n(\mathbb{Z})$ is a fundamental domain of the torus $\mathbb{R}^n/\mathbb{Z}^n$.
The converse is false, as shown for instance by the parallelotope $[0, 1](1, \frac{1}{2}) + [0, 1](0,1) \subset \mathbb{R}^2$.
But this is somehow, the only kind of exceptions we can get when $n = 2$.
Claim 2. A parallelotope in $\mathbb{R}^2$ is a fundamental domain of the torus $\mathbb{R}^2/\mathbb{Z}^2$, if and only if, it is the image of $[0, 1](1, x) + [0, 1](0, 1)$ by some matrix in $GL_n(\mathbb{Z})$ for some $x \in [0, 1]$. Equivalently, if and only if the matrix of the parallelotope can be reduced to $\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$ by means of elementary integral row operations for some $x \in [0, 1]$.
By integral elementary row operation I mean, adding or subtracting one row to the other, or negating a row.
Proof of Claim 2. If a parallelotope $P$ has the form mentioned in Claim 2, then it is easy to check that $P$ is a fundamental domain. We suppose now that $P$ is a fundamental domain. Let us write $P = [0, 1]v_1 + [0, 1]v_2$ where $v_1$ and $v_2$ are the rows of $V = \begin{pmatrix} v_{11} & v_{12} \\ v_{21} & v_{22}\end{pmatrix}$. By hypothesis, we have $P + \mathbb{Z^n} = \mathbb{R}^n$ and $\dot{P} \cap (P + z) = \emptyset$ for every non-zero $z$ in $\mathbb{Z}^n$. From this, we infer that $\vert \det(V) \vert = \mu(P) = 1$ where $\mu$ denotes the Lebesgue measure on $\mathbb{R}^2$. Without loss of generality, we can replace $V$ by $AV$ for any $A \in GL_2(\mathbb{Z})$. In particular, we can try to reduce $V$ by means of elementary row operations. If we don't manage to cancel one of the coefficients of the first column of $V$, then these coefficients can be made arbitrarily small, say less than $\frac{1}{2}$, and positive. This is impossible because $P$ would then be contained in a vertical strip of width smaller than $1$ and hence could not surject onto $\mathbb{R}^2/\mathbb{Z}^2$ via the natural map $\mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2$. Therefore we can assume, without loss of generality, that $v_{21} = 0$, $v_{11} > 0$, $v_{22} > 0$ and $0 \le v_{12} < v_{22}$. Necessarily, $v_{11} \ge 1$, since otherwise $P$ would be contained in a vertical strip of width smaller than $1$. If $v_{11} > 1$ and $v_{12} > 0$, then $v_1 - (1, 0)$ belongs to the interior of $P$, a contradiction. We are left with two cases.
Case 1. $v_{11} = 1$. As $\det(V) = 1$, this implies $v_{22} = 1$.
Case 2. $v_{12} = 0$. Then $P$ is a rectangle, and since $P$ surjects onto $\mathbb{R}^2/\mathbb{Z}^2$, we necessarily have $v_{11} = v_{22} = 1$.
In each case, the matrix $V$ has the desired form.
Edit. Elaborating on the above result, we can try to get a more general statement with
Claim $n$. A parallelotope in $\mathbb{R}^n$ is a fundamental domain of the torus $\mathbb{R}^n/\mathbb{Z}^n$ if and only if the matrix $V = V(P)$ whose rows are the generators of $P$ can be reduced by means of integral elementary row operations, including negating rows, to a matrix of the form $$\begin{pmatrix} 1 & a_{11} & \cdots & \cdots & a_{nn} \\ 0 & 1 & a_{23} & \cdots & a_{2n} \\ 0 & 0 & 1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 1 \end{pmatrix}$$ with $a_{ij} \in [0, 1]$ whenever $j > i$.
This may be exactly what the OP already mentioned in his question. This is certainly what Yoav Kallus refers to in his answer. (So I invite the OP to elaborate on his expectations regarding the description of fundamental domains, e.g., do we seek for uniqueness of the matrix? In which ways this description could be insufficient?).
Proof Proposal for Claim $n$. We can reduce $V$ to an upper triangular form via induction, arguing as in the case $n = 2$. Using induction on $n$ and the fact that we can suppose $\det(V) = 1$, we obtain the result by projecting $P$ on a direct factor $\mathbb{R}^{n -1}$ of $\mathbb{R}^n$.
As pointed out by Yoav Kallus, Claim $n$ is equivalent to Minkowski's conjecture on lattice cube-tiling of $\mathbb{R}^n$, so the reasoning sketched above is certainly flawed for $n > 3$. But, as Yoav Kallus already said, the conjecture holds true by Hajós' theorem. I found the lecture notes of P. Shor very helpful to understand the link between OP's question and Minkowski's cube-tiling conjecture. We learn in Section 3 that Minkowski established the result for $n = 3$ but failed to generalize it. Section 4 is dedicated to Hajós' theorem. | 2,331 | 7,295 | {"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} | 2.84375 | 3 | CC-MAIN-2020-50 | latest | en | 0.847266 |
https://www.ixambee.com/questions/quantitative-aptitude/probability/11787 | 1,718,618,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00403.warc.gz | 746,205,411 | 60,996 | # Two schools, A and B participate in a Quiz competition. The probability of A’s winning is 3/7 and the probability of B’s winning is 3/5. What is the probability that only one of them wins?
A 18/35
B 19/25
C 16/35
D 20/29
E None of these
×
× | 81 | 244 | {"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} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.963396 |
http://library.thinkquest.org/05aug/00666/thesite/quizz/quiz/quiz2.html | 1,386,691,824,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164022163/warc/CC-MAIN-20131204133342-00010-ip-10-33-133-15.ec2.internal.warc.gz | 109,118,525 | 4,050 | Chapter 2
1.) i ∙ i=-1.
2.) For z= (x, y) = x + i∙ yC,
x=Rez- imaginary part of z;
y=Rez- real part of z;
3.) z1=z2 if and only if Rez1 = Rez2 and Imz1 =Imz2
4.) zC*\R hence Rez=0, z is called a strictly real complex number.
5.) z1∙z2= (x1 + y1∙ i) ∙ (x2 + y2∙ i) = (x1 x2- y1 y1) + (x1 y2 + x1 y1) ∙ i
6.) zR if and only if Rez=0. | 182 | 361 | {"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-2013-48 | latest | en | 0.230311 |
https://askcryp.to/t/resource-topic-2024-1199-on-degrees-of-carry-and-scholzs-conjecture/22382 | 1,723,283,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00618.warc.gz | 86,352,038 | 5,102 | # [Resource Topic] 2024/1199: On degrees of carry and Scholz's conjecture
Welcome to the resource topic for 2024/1199
Title:
On degrees of carry and Scholz’s conjecture
Authors: Theophilus Agama
Abstract:
Exploiting the notion of carries, we obtain improved upper bounds for the length of the shortest addition chains \iota(2^n-1) producing 2^n-1. Most notably, we show that if 2^n-1 has carries of degree at most $$\kappa(2^n-1)=\frac{1}{2}(\iota(n)-\lfloor \frac{\log n}{\log 2}\rfloor+\sum \limits_{j=1}^{\lfloor \frac{\log n}{\log 2}\rfloor}{\frac{n}{2^j}})$$ then the inequality $$\iota(2^n-1)\leq n+1+\sum \limits_{j=1}^{\lfloor \frac{\log n}{\log 2}\rfloor}\bigg({\frac{n}{2^j}}-\xi(n,j)\bigg)+\iota(n)$$ holds for all n\in \mathbb{N} with n\geq 4, where \iota(\cdot) denotes the length of the shortest addition chain producing \cdot, \{\cdot\} denotes the fractional part of \cdot and where \xi(n,1):=\{\frac{n}{2}\} with \xi(n,2)=\{\frac{1}{2}\lfloor \frac{n}{2}\rfloor\} and so on.
Feel free to post resources that are related to this paper below.
Example resources include: implementations, explanation materials, talks, slides, links to previous discussions on other websites. | 394 | 1,195 | {"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} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.745319 |
https://www.numbersaplenty.com/7604 | 1,674,911,367,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00135.warc.gz | 908,909,777 | 3,258 | Search a number
7604 = 221901
BaseRepresentation
bin1110110110100
3101102122
41312310
5220404
655112
731112
oct16664
911378
107604
115793
124498
1335cc
142ab2
1523be
hex1db4
7604 has 6 divisors (see below), whose sum is σ = 13314. Its totient is φ = 3800.
The previous prime is 7603. The next prime is 7607. The reversal of 7604 is 4067.
It can be written as a sum of positive squares in only one way, i.e., 4900 + 2704 = 70^2 + 52^2 .
It is a plaindrome in base 9, base 13 and base 15.
It is a congruent number.
It is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 7604.
It is not an unprimeable number, because it can be changed into a prime (7603) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 947 + ... + 954.
It is an arithmetic number, because the mean of its divisors is an integer number (2219).
27604 is an apocalyptic number.
It is an amenable number.
7604 is a deficient number, since it is larger than the sum of its proper divisors (5710).
7604 is a wasteful number, since it uses less digits than its factorization.
7604 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1905 (or 1903 counting only the distinct ones).
The product of its (nonzero) digits is 168, while the sum is 17.
The square root of 7604 is about 87.2009174264. The cubic root of 7604 is about 19.6644001346.
The spelling of 7604 in words is "seven thousand, six hundred four".
Divisors: 1 2 4 1901 3802 7604 | 480 | 1,575 | {"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-2023-06 | latest | en | 0.907215 |
http://www.reddit.com/r/cheatatmathhomework/comments/19g3is/discrete_math_proof_by_contradiction_and_proof_by/c8nplsc | 1,416,625,934,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400376197.4/warc/CC-MAIN-20141119123256-00028-ip-10-235-23-156.ec2.internal.warc.gz | 857,434,920 | 11,383 | you are viewing a single comment's thread.
[–][S] 0 points1 point (3 children)
sorry, this has been archived and can no longer be voted on
No yes the logical quantifier is "there exists" ok awesome thank you that's good. I tried this attack: for the first one prove: that there exists irrational numbers R and S such that RS is irrational I turned it into a conditional and found the contrapositive:
If RS is rational then R and S are rational numbers for all numbers
my counterexample would be (((sqrt(2))sqrt(2))sqrt(2) (Sorry there are alot of square roots) well that jumble of square roots all equal 2 a rational number while R and S are irrational
the second one was done the same way
[–][deleted] (1 child)
sorry, this has been archived and can no longer be voted on
[deleted]
[–][S] 1 point2 points (0 children)
sorry, this has been archived and can no longer be voted on
That is a good point I might have to build cases then where sqrt(2)sqrt(2) is rational for one case and irrational for another
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
If RS is rational then R and S are rational numbers for all numbers
To prove this via contrapositive. Say that if r and s are irrational then so is rs. Well, an irrational number raised an irrational amount of times is not rational. But you may have to prove that, and I'm not sure if you have the tools.
Let me know if you want more advice. | 364 | 1,461 | {"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} | 3.25 | 3 | CC-MAIN-2014-49 | latest | en | 0.95787 |
https://minuteshours.com/138-4-minutes-in-hours-and-minutes | 1,656,365,736,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00093.warc.gz | 433,904,642 | 4,724 | # 138.4 minutes in hours and minutes
## Result
138.4 minutes equals 2 hours and 18.4 minutes
You can also convert 138.4 minutes to hours.
## Converter
One hundred thirty-eight point four minutes is equal to two hours and eighteen point four minutes. | 63 | 254 | {"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-2022-27 | latest | en | 0.838014 |
https://prosa.mpi-sws.org/branches/master/pretty/prosa.util.bigcat.html | 1,717,028,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00509.warc.gz | 406,681,116 | 12,894 | Library prosa.util.bigcat
Require Export mathcomp.zify.zify.
Require Export prosa.util.tactics prosa.util.notation.
From mathcomp Require Import ssreflect ssrbool eqtype ssrnat seq fintype bigop.
Require Export prosa.util.tactics prosa.util.list.
In this section, we introduce lemmas about the concatenation operation performed over arbitrary sequences.
Section BigCatNatLemmas.
Consider any type T supporting equality comparisons...
Variable T : eqType.
...and a function f that, given an index, yields a sequence.
Variable f : nat seq T.
In this section, we prove that the concatenation over sequences works as expected: no element is lost during the concatenation, and no new element is introduced.
Section BigCatNatElements.
First, we show that the concatenation comprises all the elements of each sequence; i.e. any element contained in one of the sequences will also be an element of the result of the concatenation.
Lemma mem_bigcat_nat :
x m n j,
m j < n
x \in f j
x \in \cat_(m i < n) (f i).
Proof.
intros x m n j LE IN; move: LE ⇒ /andP [LE LE0].
rewritebig_cat_nat with (n := j); simpl; [| by ins | by apply ltnW].
rewrite mem_cat; apply/orP; right.
destruct n; first by rewrite ltn0 in LE0.
rewrite big_nat_recl; last by ins.
by rewrite mem_cat; apply/orP; left.
Qed.
Conversely, we prove that any element belonging to a concatenation of sequences must come from one of the sequences.
Lemma mem_bigcat_nat_exists :
x m n,
x \in \cat_(m i < n) (f i)
i,
x \in f i m i < n.
Proof.
intros x m n IN.
elim: n IN ⇒ [|n IHn] IN; first by rewrite big_geq // in IN.
destruct (leqP m n); last by rewrite big_geq ?in_nil // ltnW in IN.
rewrite big_nat_recr // /= mem_cat in IN.
move: IN ⇒ /orP [HEAD | TAIL].
- move: (IHn HEAD) ⇒ [x0 [H /andP[H0 H1]]]; x0.
split; first by done.
by apply/andP; split; last by apply ltnW.
- n; split; first by done.
by apply/andP; split; last apply ltnSn.
Qed.
We also restate lemma mem_bigcat_nat in terms of ordinals.
Lemma mem_bigcat_ord :
(x : T) (n : nat) (j : 'I_n) (f : 'I_n seq T),
j < n
x \in (f j)
x \in \cat_(i < n) (f i).
Proof.
movex; elim⇒ [//|n IHn] j f' Hj Hx.
rewrite big_ord_recr /= mem_cat; apply /orP.
move: Hj; rewrite ltnS leq_eqVlt ⇒ /orP [/eqP Hj|Hj].
- by right; rewrite (_ : ord_max = j); [|apply ord_inj].
- left.
apply (IHn (Ordinal Hj)); [by []|].
by set j' := widen_ord _ _; have → : j' = j; [apply ord_inj|].
Qed.
End BigCatNatElements.
In this section, we show how we can preserve uniqueness of the elements (i.e. the absence of a duplicate) over a concatenation of sequences.
Assume that there are no duplicates in each of the possible sequences to concatenate...
Hypothesis H_uniq_seq : i, uniq (f i).
...and that there are no elements in common between the sequences.
Hypothesis H_no_elements_in_common :
x i1 i2, x \in f i1 x \in f i2 i1 = i2.
We prove that the concatenation will yield a sequence with unique elements.
Lemma bigcat_nat_uniq :
n1 n2,
uniq (\cat_(n1 i < n2) (f i)).
Proof.
intros n1 n2.
case (leqP n1 n2) ⇒ [LE | GT]; last by rewrite big_geq // ltnW.
rewrite -addnBA //; set delta := n2 - n1.
elim: delta ⇒ [|delta IHdelta]; first by rewrite addn0 big_geq.
rewrite cat_uniq; apply/andP; split; first by apply IHdelta.
apply /andP; split; last by apply H_uniq_seq.
rewrite -all_predC; apply/allP; intros x INx.
simpl; apply/negP; unfold not; intro BUG.
apply mem_bigcat_nat_exists in BUG.
move: BUG ⇒ [i [IN /andP [_ LTi]]].
apply H_no_elements_in_common with (i1 := i) in INx; last by done.
by rewrite INx ltnn in LTi.
Qed.
Conversely, if the concatenation of the sequences has no duplicates, any element can only belong to a single sequence.
Lemma bigcat_ord_uniq_reverse :
(n : nat) (f : 'I_n seq T),
uniq (\cat_(i < n) (f i))
( x i1 i2,
x \in (f i1) x \in (f i2) i1 = i2).
Proof.
case⇒ [|n]; [by movef' Huniq x; case|].
elim: n ⇒ [|n IHn] f' Huniq x i1 i2 Hi1 Hi2.
{ move: i1 i2 {Hi1 Hi2}; case; case⇒ [i1|//]; case; case⇒ [i2|//].
apply f_equal, eq_irrelevance. }
move: (leq_ord i1); rewrite leq_eqVlt ⇒ /orP [H'i1|H'i1].
all: move: (leq_ord i2); rewrite leq_eqVlt ⇒ /orP [H'i2|H'i2].
{ by apply ord_inj; move: H'i1 H'i2 ⇒ /eqP → /eqP →. }
{ exfalso.
move: Huniq; rewrite big_ord_recr cat_uniq ⇒ /andP [_ /andP [H _]].
move: H; apply /negP; rewrite Bool.negb_involutive.
apply /hasP; x ⇒ /=.
{ set o := ord_max; suff → : o = i1; [by []|].
by apply ord_inj; move: H'i1 ⇒ /eqP →. }
apply (mem_bigcat_ord _ _ (Ordinal H'i2)) ⇒ //.
by set o := widen_ord _ _; suff → : o = i2; [|apply ord_inj]. }
{ exfalso.
move: Huniq; rewrite big_ord_recr cat_uniq ⇒ /andP [_ /andP [H _]].
move: H; apply /negP; rewrite Bool.negb_involutive.
apply /hasP; x ⇒ /=.
{ set o := ord_max; suff → : o = i2; [by []|].
by apply ord_inj; move: H'i2 ⇒ /eqP →. }
apply (mem_bigcat_ord _ _ (Ordinal H'i1)) ⇒ //.
by set o := widen_ord _ _; suff → : o = i1; [|apply ord_inj]. }
move: Huniq; rewrite big_ord_recr cat_uniq ⇒ /andP [Huniq _].
apply ord_inj; rewrite -(inordK H'i1) -(inordK H'i2); apply f_equal.
apply (IHn _ Huniq x).
{ set i1' := widen_ord _ _; suff → : i1' = i1; [by []|].
by apply ord_inj; rewrite /= inordK. }
set i2' := widen_ord _ _; suff → : i2' = i2; [by []|].
by apply ord_inj; rewrite /= inordK.
Qed.
End BigCatNatDistinctElements.
We show that filtering a concatenated sequence by any predicate P is the same as concatenating the elements of the sequence that satisfy P.
Lemma bigcat_nat_filter_eq_filter_bigcat_nat :
{X : Type} (F : nat seq X) (P : X bool) (t1 t2 : nat),
[seq x <- \cat_(t1 t < t2) F t | P x] = \cat_(t1 t < t2)[seq x <- F t | P x].
Proof.
moveX F P t1 t2.
specialize (leq_total t1 t2) ⇒ /orP [T1_LT | T2_LT].
+ have EX: Δ, t2 = t1 + Δ by simpl; (t2 - t1); lia.
move: EX ⇒ [Δ EQ]; subst t2.
elim: Δ T1_LT ⇒ [|Δ IHΔ] T1_LT.
{ by rewrite addn0 !big_geq ⇒ //. }
rewrite filter_cat IHΔ ⇒ //.
by lia. }
+ by rewrite !big_geq ⇒ //.
Qed.
We show that the size of a concatenated sequence is the same as summation of sizes of each element of the sequence.
Lemma size_big_nat :
{X : Type} (F : nat seq X) (t1 t2 : nat),
\sum_(t1 t < t2) size (F t) =
size (\cat_(t1 t < t2) F t).
Proof.
moveX F t1 t2.
specialize (leq_total t1 t2) ⇒ /orP [T1_LT | T2_LT].
- have EX: Δ, t2 = t1 + Δ by simpl; (t2 - t1); lia.
move: EX ⇒ [Δ EQ]; subst t2.
elim: Δ T1_LT ⇒ [|Δ IHΔ] T1_LT.
{ by rewrite addn0 !big_geq ⇒ //. }
by rewrite size_cat IHΔ ⇒ //; lia. }
- by rewrite !big_geq ⇒ //.
Qed.
End BigCatNatLemmas.
In this section, we introduce a few lemmas about the concatenation operation performed over arbitrary sequences.
Section BigCatLemmas.
Consider any two types X and Y supporting equality comparisons...
Variable X Y : eqType.
...and a function f that, given an index X, yields a sequence of Y.
Variable f : X seq Y.
First, we show that the concatenation comprises all the elements of each sequence; i.e. any element contained in one of the sequences will also be an element of the result of the concatenation.
Lemma mem_bigcat :
x y s,
x \in s
y \in f x
y \in \cat_(x <- s) f x.
Proof.
movex y s INs INfx.
elim: s INs ⇒ [//|z s IHs] INs.
rewrite big_cons mem_cat.
move:INs; rewrite in_cons ⇒ /orP[/eqP HEAD | CONS].
- by rewrite -HEAD; apply /orP; left.
- by apply /orP; right; apply IHs.
Qed.
Conversely, we prove that any element belonging to a concatenation of sequences must come from one of the sequences.
Lemma mem_bigcat_exists :
{P} s y,
y \in \cat_(x <- s | P x) f x
x, x \in s y \in f x.
Proof.
moveP.
elim⇒ [|a s IHs] y; first by rewrite big_nil.
rewrite big_cons.
case: (P a) ⇒ [|IN];
last by move: (IHs y IN) ⇒ [x [INx INy]]; x; split ⇒ //; rewrite in_cons INx orbT.
rewrite mem_cat ⇒ /orP[HEAD | CONS].
- a.
by split ⇒ //; apply mem_head.
- move: (IHs _ CONS) ⇒ [x [INs INfx]].
x; split =>//.
by rewrite in_cons; apply /orP; right.
Qed.
Next, we show that a map and filter operation can be done either before or after a concatenation, leading to the same result.
Lemma bigcat_filter_eq_filter_bigcat :
xss P,
[seq x <- \cat_(xs <- xss) f xs | P x] =
\cat_(xs <- xss) [seq x <- f xs | P x] .
Proof.
movexss P.
elim: xss ⇒ [|a xss IHxss].
- by rewrite !big_nil.
- by rewrite !big_cons filter_cat IHxss.
Qed.
In this section, we show how we can preserve uniqueness of the elements (i.e. the absence of a duplicate) over a concatenation of sequences.
Section BigCatDistinctElements.
Consider a list xs, ...
Context {xs : seq X}.
... a filter predicate P, ...
Context {P : pred X}.
... assume that there are no duplicates in each of the possible sequences to concatenate...
Hypothesis H_uniq_f : x, P x uniq (f x).
...and that there are no elements in common between the sequences.
Hypothesis H_no_elements_in_common :
x y z,
x \in f y x \in f z y = z.
We prove that the concatenation will yield a sequence with unique elements.
Lemma bigcat_uniq :
uniq xs
uniq (\cat_(x <- xs | P x) (f x)).
Proof.
move: xs.
elim⇒ [|a xs' IHxs]; first by rewrite big_nil.
rewrite cons_uniq ⇒ /andP [NINxs UNIQ].
rewrite big_cons.
case Pa: (P a) ⇒ //.
rewrite cat_uniq.
apply /andP; split; first by apply H_uniq_f.
apply /andP; split; last by apply IHxs.
apply /hasPnx IN; apply /negPINfa.
move: (mem_bigcat_exists _ _ IN) ⇒ [a' [INxs INfa']].
move: (H_no_elements_in_common x a a' INfa INfa') ⇒ EQa.
by move: NINxs; rewrite EQa ⇒ /negP CONTRA.
Qed.
End BigCatDistinctElements.
In this section, we show some properties of the concatenation of sequences in combination with a function g that cancels f.
Consider a function g...
Variable g : Y X.
... and assume that g can cancel f starting from an element of the sequence f x.
Hypothesis H_g_cancels_f : x y, y \in f x g y = x.
First, we show that no element of a sequence f x1 can be fed into g and obtain an element x2 which differs from x1. Hence, filtering by this condition always leads to the empty sequence.
Lemma seq_different_elements_nil :
x1 x2,
x1 != x2
[seq x <- f x1 | g x == x2] = [::].
Proof.
movex1 x2 ⇒ /negP NEQ.
apply filter_in_pred0.
movey IN; apply/negP ⇒ /eqP EQ2.
apply: NEQ; apply/eqP.
move: (H_g_cancels_f _ _ IN) ⇒ EQ1.
by rewrite -EQ1 -EQ2.
Qed.
Finally, assume we are given an element y which is contained in a duplicate-free sequence xs. Then, f is applied to each element of xs, but only the elements for which g yields y are kept. In this scenario, concatenating the resulting sequences will always lead to f y.
Lemma bigcat_seq_uniqK :
y xs,
y \in xs
uniq xs
\cat_(x <- xs) [seq x' <- f x | g x' == y] = f y.
Proof.
movey xs IN UNI.
elim: xs IN UNI ⇒ [//|x' xs IHxs] IN UNI.
move: IN; rewrite in_cons ⇒ /orP [/eqP EQ| IN].
{ subst; rewrite !big_cons.
have → : [seq x <- f x' | g x == x'] = f x'.
{ apply/all_filterP/allP.
intros y IN; apply/eqP.
by apply H_g_cancels_f. }
have ->: \cat_(j<-xs)[seq x <- f j | g x == x'] = [::]; last by rewrite cats0.
rewrite big1_seq //; movexs2 /andP [_ IN].
have NEQ: xs2 != x'; last by rewrite seq_different_elements_nil.
apply/neqP; intros EQ; subst x'.
move: UNI; rewrite cons_uniq ⇒ /andP [NIN _].
by move: NIN ⇒ /negP NIN; apply: NIN. }
{ rewrite big_cons IHxs //; last by move:UNI; rewrite cons_uniq⇒ /andP[_ ?].
have NEQ: (x' != y); last by rewrite seq_different_elements_nil.
apply/neqP; intros EQ; subst x'.
move: UNI; rewrite cons_uniq ⇒ /andP [NIN _].
by move: NIN ⇒ /negP NIN; apply: NIN. }
Qed.
End BigCatWithCancelFunctions.
End BigCatLemmas.
In this section, we show that the number of elements of the result of a nested mapping and concatenation operation is independent from the order in which the concatenations are performed.
Consider any three types supporting equality comparisons...
Variable X Y Z : eqType.
... a function F that, given two indices, yields a sequence...
Variable F : X Y seq Z.
and a predicate P.
Variable P : pred Z.
Assume that, given two sequences xs and ys, their elements are fed to F in a pair-wise fashion. The resulting lists are then concatenated. The following lemma shows that, when the operation described above is performed, the number of elements respecting P in the resulting list is the same, regardless of the order in which xs and ys are combined.
Lemma bigcat_count_exchange :
xs ys,
count P (\cat_(x <- xs) \cat_(y <- ys) F x y) =
count P (\cat_(y <- ys) \cat_(x <- xs) F x y).
Proof.
elim⇒ [|x0 seqX IHxs]; elim⇒ [|y0 seqY IHys].
{ by rewrite !big_nil. }
{ by rewrite big_cons count_cat -IHys !big_nil. }
{ by rewrite big_cons count_cat IHxs !big_nil. }
{ rewrite !big_cons !count_cat. | 4,004 | 12,374 | {"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} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.706499 |
https://physics.stackexchange.com/questions/409968/number-of-microstates | 1,603,227,904,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874135.2/warc/CC-MAIN-20201020192039-20201020222039-00157.warc.gz | 476,888,773 | 33,030 | Number of microstates
I need some help with this problem:
Consider a system of $N$ distinguishable particles. Each of these particles can be in a state with energy $\epsilon$ or $-\epsilon$. Consider that the states with energy $\epsilon$ are M-fold degenerate, and those with energy $-\epsilon$ are P-fold degenerate. Using the micro canonical ensemble, determine the entropy $S(E,N)$
I haven't been able to correctly calculate the number of microstates $\Omega (E,N)$. I know that in the case where the states are not degenerate, it's just:
$$\Omega(E,N)= \frac{N!}{N_{+}!(N-N_+)!}$$
where $N_+$ is the number of particles with energy $\epsilon$, but in this case I do not know how to take into account the degeneracies.
Any help will be truly appreciated.
Every one of $N_+$ ($N_- = N - N_+$) particles with the energy $\epsilon$ ($-\epsilon$) is in one of $P$ ($M$) possible states. Hence there is the additional factor $P^{N_+}M^{N_-}$ and the answer is $$\Omega(E,N) = \frac{N!}{N_+! N_-!} P^{N_+} M^{N_-}$$ | 289 | 1,020 | {"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} | 3.28125 | 3 | CC-MAIN-2020-45 | latest | en | 0.843965 |
https://byjus.com/question-answer/radiation-of-wavelength-lambda-is-incident-on-a-photocell-the-fastest-emitted-electron-has-speed-8/ | 1,718,909,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00697.warc.gz | 132,390,476 | 34,327 | 1
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Question
Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3λ4, the speed of the fastest emitted electron will be :
A
<v(43)1/2
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B
=v(43)1/2
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C
=v(34)1/2
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D
>v(43)1/2
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Solution
The correct option is D >v(43)1/2According to the Einstein's photoelectric equation, E=ϕ+K.Emax For case (1), hcλ=ϕ+12mv2−−−(1) For case (2), hcλ′=ϕ+12m(v′)2 ⇒hc(3λ4)=ϕ+12m(v′)2−−−(2) [(1)×43]−(2) 4hc3λ−43hcλ=43ϕ+43(12mv2)−ϕ−12m(v′)2 ⇒43ϕ+43(12mv2)=ϕ+12m(v′)2 ⇒12m(v′)2=ϕ3+4312mv2 ⇒12m(v′)2>43(12mv2) ⇒v′>√43 v Hence, option (D) is correct.
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Join BYJU'S Learning Program | 425 | 1,091 | {"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} | 3.109375 | 3 | CC-MAIN-2024-26 | latest | en | 0.787021 |
http://www.jiskha.com/display.cgi?id=1192233279 | 1,496,150,160,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615105.83/warc/CC-MAIN-20170530124450-20170530144450-00499.warc.gz | 663,520,174 | 4,203 | # Science:Biology
posted by on .
Use the following genotypes to answer the question.
N= big nose
n= small nose
B= big nostril
b= small nostrl
H= hairy face
h= no hairy face
XY= man
XX= woman
Z= zits on the nose
z= no zits on the nose
If two parents who are heterozygous for all the traits were 'crossed', what are the chance of getting a child with this genotype: NnBBHhXXZz?
Can someone do the problem then explain to me how you would do this? I've tried but i don't know if my answer is right
• Science:Biology - ,
Lets do the sex gene first.
XX+ XY. The probabality of getting XX is 1/2
Nn+Nn = NN nn Nn nN so probability is 1/2
BB probability is 1/4 (BB, Bb, bB, bb)
Hh prob is 1/2
Zz prob is 1/2
Now multiply them all
(1/2)4 * (1/4)1
or .0156
check my thinking and math. | 261 | 787 | {"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-2017-22 | latest | en | 0.919323 |
https://dieghernan.github.io/202004_headtails/ | 1,686,347,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00476.warc.gz | 256,138,062 | 18,047 | Head/Tails breaks on the classInt package.
There are far more ordinary people (say, 80 percent) than extraordinary people (say, 20 percent); this is often characterized by the 80/20 principle, based on the observation made by the Italian economist Vilfredo Pareto in 1906 that 80% of land in Italy was owned by 20% of the population. A histogram of the data values for these phenomena would reveal a right-skewed or heavy-tailed distribution. How to map the data with the heavy-tailed distribution?
Jiang (2013)
Abstract
This vignette discusses the implementation of the “Head/tail breaks” style (Jiang (2013)) on the classIntervals function of the classInt package. A step-by-step example is presented in order to clarify the method. A case study using spData::afcon is also included, making use of other additional packages as sf.
Introduction
The Head/tail breaks, sometimes referred as ht-index (Jiang and Yin (2013)), is a classification scheme introduced by Jiang (2013) in order to find groupings or hierarchy for data with a heavy-tailed distribution.
Heavy-tailed distributions are heavily right skewed, with a minority of large values in the head and a majority of small values in the tail. This imbalance between the head and tail, or between many small values and a few large values, can be expressed as “far more small things than large things”.
Heavy tailed distributions are commonly characterized by a power law, a lognormal or an exponential function. Nature, society, finance (Vasicek (2002)) and our daily lives are full of rare and extreme events, which are termed “black swan events” (Taleb (2008)). This line of thinking provides a good reason to reverse our thinking by focusing on low-frequency events.
library(classInt)
# 1. Characterization of heavy-tail distributions----
set.seed(1234)
# Pareto distribution a=1 b=1.161 n=1000
sample_par <- 1 / (1 - runif(1000))^(1 / 1.161)
par(mar = c(2, 4, 3, 1), cex = 0.8)
plot(
sort(sample_par, decreasing = TRUE),
type = "l",
ylab = "F(x)",
xlab = "",
main = "80/20 principle"
)
abline(
h = quantile(sample_par, .8),
lty = 2,
col = "red3"
)
abline(
v = 0.2 * length(sample_par),
lty = 2,
col = "darkblue"
)
legend(
"topleft",
legend = c("F(x): p80", "x: Top 20%"),
col = c("red3", "darkblue"),
lty = 2,
cex = 0.8
)
hist(
sample_par,
n = 100,
xlab = "",
main = "Histogram",
col = "grey50",
border = NA,
probability = TRUE
)
par(opar)
Breaking method
The method itself consists on a four-step process performed recursively until a stopping condition is satisfied. Given a vector of values $$v = (a_1, a_2, ..., a_n)$$ the process can be described as follows:
1. On each iteration, compute $$\mu = \sum_{i=1}^{n} a_i \:\:\: \forall \: a_i \in v$$.
2. Break $$v$$ into the $$tail$$ and the $$head$$: $$tail = \{ a_x \in v | a_x \lt \mu \}$$ $$head = \{ a_x \in v | a_x \gt \mu \}$$.
3. Assess if the proportion of $$head$$ over $$v$$ is lower or equal than a given threshold: $$\frac{|head|}{|v|} \le thresold$$
4. If 3 is TRUE, repeat 1 to 3 until the condition is FALSE or no more partitions are possible (i.e. $$head$$ has less than two elements).
It is important to note that, at the beginning of a new iteration, $$v$$ is replaced by $$head$$. The underlying hypothesis is to create partitions until the head and the tail are balanced in terms of distribution.So the stopping criteria is satisfied when the last head and the last tail are evenly balanced.
In terms of threshold, Jiang, Liu, and Jia (2013) set 40% as a good approximation, meaning that if the $$head$$ contains more than 40% of the observations the distribution is not considered heavy-tailed.
The final breaks are the vector of consecutive $$\mu$$:
$breaks = (\mu_1, \mu_2, \mu_3, ..., \mu_n )$
Step by step example
We reproduce here the pseudo-code1 as per Jiang (2019):
Recursive function Head/tail Breaks:
Rank the input data from the largest to the smallest
Break the data into the head and the tail around the mean;
// the head for those above the mean
// the tail for those below the mean
End Function
A step-by-step example in R (for illustrative purposes) has been developed:
opar <- par(no.readonly = TRUE)
par(mar = c(2, 2, 3, 1), cex = 0.8)
var <- sample_par
thr <- .4
brks <- c(min(var), max(var)) # Initialise with min and max
sum_table <- data.frame(
iter = 0,
mu = NA,
prop = NA,
n_var = NA,
)
# Pars for chart
limchart <- brks
# Iteration
for (i in 1:10) {
mu <- mean(var)
brks <- sort(c(brks, mu))
stopit <- prop < thr & length(head) > 1
sum_table <- rbind(
sum_table,
)
hist(
var,
main = paste0("Iter ", i),
breaks = 50,
col = "grey50",
border = NA,
xlab = "",
xlim = limchart
)
abline(v = mu, col = "red3", lty = 2)
ylabel <- max(hist(var, breaks = 50, plot = FALSE)$counts) labelplot <- paste0("PropHead: ", round(prop * 100, 2), "%") text( x = mu, y = ylabel, labels = labelplot, cex = 0.8, pos = 4 ) legend( "right", legend = paste0("mu", i), col = c("red3"), lty = 2, cex = 0.8 ) if (isFALSE(stopit)) { break } var <- head } par(opar) As it can be seen, in each iteration the resulting head gradually loses the high-tail property, until the stopping condition is met. iter mu prop n_var n_head 1 5.6755 14.5% 1000 145 2 27.2369 21.38% 145 31 3 85.1766 19.35% 31 6 4 264.7126 50% 6 3 The resulting breaks are then defined as breaks = c(min(var), mu1, mu2, ..., mu_n, max(var)). Implementation on classInt package The implementation in the classIntervals function should replicate the results: ht_sample_par <- classIntervals(sample_par, style = "headtails") brks == ht_sample_par$brks
## [1] TRUE TRUE TRUE TRUE TRUE TRUE
As stated in Jiang (2013), the number of breaks is naturally determined, however the thr parameter could help to adjust the final number. A lower value on thr would provide less breaks while a larger thr would increase the number, if the underlying distribution follows the “far more small things than large things” principle.
opar <- par(no.readonly = TRUE)
par(mar = c(2, 2, 2, 1), cex = 0.8)
pal1 <- c("wheat1", "wheat2", "red3")
# Minimum: single break
print(paste("number of breaks", length(classIntervals(sample_par, style = "headtails", thr = 0)$brks - 1))) plot( classIntervals(sample_par, style = "headtails", thr = 0), pal = pal1, main = "thr = 0" ) # Two breaks print(paste("number of breaks", length(classIntervals(sample_par, style = "headtails", thr = 0.2)$brks - 1)))
plot(
classIntervals(sample_par, style = "headtails", thr = 0.2),
pal = pal1,
main = "thr = 0.2"
)
# Default breaks: 0.4
print(paste("number of breaks", length(classIntervals(sample_par, style = "headtails")$brks - 1))) plot(classIntervals(sample_par, style = "headtails"), pal = pal1, main = "thr = Default" ) # Maximum breaks print(paste("number of breaks", length(classIntervals(sample_par, style = "headtails", thr = 1)$brks - 1)))
plot(
classIntervals(sample_par, style = "headtails", thr = 1),
pal = pal1,
main = "thr = 1"
)
par(opar)
The method always returns at least one break, corresponding to mean(var).
Case study
Jiang (2013) states that "the new classification scheme is more natural than the natural breaks in finding the groupings or hierarchy for data with a heavy-tailed distribution." (p. 482), referring to Jenks’ natural breaks method. In this case study we would compare headtails vs. fisher, that is the alias for the Fisher-Jenks algorithm and it is always preferred to the jenks style (see ?classIntervals). For this example we will use the afcon dataset from spData package, plus some additional spatial information in order to create the data visualization.
library(spData)
data(afcon, package = "spData")
Let’s have a look to the Top 10 values and the distribution of the variable totcon (index of total conflict 1966-78):
# Top10
knitr::kable(head(afcon[order(afcon$totcon, decreasing = TRUE), c("name", "totcon")], 10)) opar <- par(no.readonly = TRUE) par(mar = c(4, 4, 3, 1), cex = 0.8) hist(afcon$totcon,
n = 20,
main = "Histogram",
xlab = "totcon",
col = "grey50",
border = NA,
)
plot(
density(afcon$totcon), main = "Distribution", xlab = "totcon", ) par(opar) The data shows that EG and SU data present a clear hierarchy over the rest of values. As per the histogram, we can confirm a heavy-tailed distribution and therefore the “far more small things than large things” principle. As a testing proof, on top of headtails and fisher we would use also quantile to have a broader view on the different breaking styles. As quantile is a position-based metric, it doesn’t account for the magnitude of F(x) (hierarchy), so the breaks are solely defined by the position of x on the distribution. Applying the three aforementioned methods to break the data: brks_ht <- classIntervals(afcon$totcon, style = "headtails")
print(brks_ht)
# Same number of classes for "fisher"
nclass <- length(brks_ht$brks) - 1 brks_fisher <- classIntervals(afcon$totcon,
style = "fisher",
n = nclass
)
print(brks_fisher)
brks_quantile <- classIntervals(afcontotcon, style = "quantile", n = nclass ) print(brks_quantile) pal1 <- c("wheat1", "wheat2", "red3") opar <- par(no.readonly = TRUE) par(mar = c(2, 2, 2, 1), cex = 0.8) plot(brks_ht, pal = pal1, main = "headtails") plot(brks_fisher, pal = pal1, main = "fisher") plot(brks_quantile, pal = pal1, main = "quantile") par(opar) It is observed that the top three classes of headtails enclose 5 observations, whereas fisher includes 13 observations. In terms of classification, headtails breaks focuses more on extreme values. The next plot compares a continuous distribution of totcon re-escalated to a range of [1,nclass] versus the distribution across breaks for each style. The continuous distribution has been offset by -0.5 in order to align the continuous and the discrete distributions. # Helper function to reescale values help_reescale <- function(x, min = 1, max = 10) { r <- (x - min(x)) / (max(x) - min(x)) r <- r * (max - min) + min return(r) } afconecdf_class <- help_reescale(afcon$totcon, min = 1 - 0.5, max = nclass - 0.5 ) afcon$ht_breaks <- cut(afcon$totcon, brks_ht$brks,
labels = FALSE,
include.lowest = TRUE
)
afcon$fisher_breaks <- cut(afcon$totcon,
brks_fisher$brks, labels = FALSE, include.lowest = TRUE ) afcon$quantile_break <- cut(afcon$totcon, brks_quantile$brks,
labels = FALSE,
include.lowest = TRUE
)
par(mar = c(4, 4, 1, 1), cex = 0.8)
plot(
density(afcon$ecdf_class), ylim = c(0, 0.8), lwd = 2, main = "", xlab = "class" ) lines(density(afcon$ht_breaks), col = "darkblue", lty = 2)
lines(density(afcon$fisher_breaks), col = "limegreen", lty = 2) lines(density(afcon$quantile_break),
col = "red3",
lty = 2
)
legend("topright",
legend = c(
"fisher", "quantile"
),
col = c("black", "darkblue", "limegreen", "red3"),
lwd = c(2, 1, 1, 1),
lty = c(1, 2, 2, 2),
cex = 0.8
)
par(opar)
It can be observed that the distribution of headtails breaks is also heavy-tailed, and closer to the original distribution. On the other extreme, “quantile” provides a quasi-uniform distribution, ignoring the totcon hierarchy
In terms of data visualization, we compare here the final map using the techniques mentioned above. On this plotting exercise, a choropleth map would be created.
Additionally, a high-granularity choropleth map is created with a greater number of classes, in order to compare and contrast the actual grouping options against a more granular approach.
library(sf)
library(giscoR)
library(cartography)
par(
mfrow = c(2, 2),
mar = c(1, 1, 1, 1),
bg = "white"
)
africa <- gisco_get_countries(resolution = 60, region = "Africa", epsg = 3857)
afcon.sf <- st_as_sf(afcon, crs = 4326, coords = c("x", "y"))
afcon.sf <- st_transform(afcon.sf, st_crs(africa))
# afcon.sf <- st_join(africa[, "admin"], afcon.sf)
afcon.sf <- afcon.sf[order(afcon.sf\$totcon), ]
# High granularity map
plot(st_geometry(africa), col = "grey80", border = NA)
propSymbolsLayer(
afcon.sf,
var = "totcon",
inches = 0.2,
col = adjustcolor("grey10", alpha.f = 0.5),
border = NA
)
title(main = "High granularity map")
# Quantile
pal <- hcl.colors(5, palette = "inferno", alpha = 0.6)
plot(st_geometry(africa), col = "grey80", border = NA)
propSymbolsTypoLayer(
afcon.sf,
var = "totcon",
inches = 0.2,
col = pal,
border = NA,
legend.var.pos = "n",
legend.var2.pos = "bottomleft",
var2 = "quantile_break"
)
title(main = "Quantile")
# Fisher
plot(st_geometry(africa), col = "grey80", border = NA)
propSymbolsTypoLayer(
afcon.sf,
var = "totcon",
inches = 0.2,
col = pal,
border = NA,
legend.var.pos = "n",
legend.var2.pos = "bottomleft",
var2 = "fisher_breaks"
)
title(main = "Fisher")
plot(st_geometry(africa), col = "grey80", border = NA)
propSymbolsTypoLayer(
afcon.sf,
var = "totcon",
inches = 0.2,
col = pal,
border = NA,
legend.var.pos = "n",
legend.var2.pos = "bottomleft",
var2 = "ht_breaks"
)
par(opar)
As per the results, headtails seems to provide a better understanding of the most extreme values when the result is compared against the high-granularity plot. The quantile style, as expected, just provides a clustering without taking into account the real hierarchy. The fisher plot is in-between of these two interpretations.
It is also important to note that headtails and fisher reveal different information that can be useful depending of the context. While headtails highlights the outliers, it fails on providing a good clustering on the tail, while fisher seems to reflect better these patterns. This can be observed on the values of Western Africa and the Niger River Basin, where headtails doesn’t highlight any special cluster of conflicts, fisher suggests a potential cluster, aligned with the high-granularity plot. This can be confirmed on the histogram generated previously, where a concentration of totcon around 1,000 is visible.
References
Jiang, Bin. 2013. "Head/Tail Breaks: A New Classification Scheme for Data with a Heavy-Tailed Distribution." The Professional Geographer 65 (3): 482–94. DOI.
———. 2019. "A Recursive Definition of Goodness of Space for Bridging the Concepts of Space and Place for Sustainability." Sustainability 11 (15): 4091. DOI.
Jiang, Bin, Xintao Liu, and Tao Jia. 2013. "Scaling of Geographic Space as a Universal Rule for Map Generalization." Annals of the Association of American Geographers 103 (4): 844–55. DOI.
Jiang, Bin, and Junjun Yin. 2013. "Ht-Index for Quantifying the Fractal or Scaling Structure of Geographic Features." Annals of the Association of American Geographers 104 (3): 530–40. DOI.
Taleb, Nassim Nicholas. 2008. The Black Swan: The Impact of the Highly Improbable. 1st ed. London: Random House.
Vasicek, Oldrich. 2002. "Loan Portfolio Value." Risk, December, 160–62.
1. The method implemented on classInt corresponds to head/tails 1.0 as named on this article.
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# Calling all Anderson(UCLA) FEMBA 2013 Applicants!
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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21 May 2013, 19:20
Got the admit recently as part of R4, been a silent observer in the group so far
Anybody traveling from Seattle in 2016 batch?
re posting for lack of replies.. anyone from Seattle area PING ..
Last edited by dairymilk on 01 Jun 2013, 21:43, edited 1 time in total.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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22 May 2013, 09:12
berkeleyboy wrote:
Advice Wanted:
I applied to the UCLA FEMBA in R5 and I'm trying to decide whether to apply to USC's FEMBA as a back-up. I have a 3.7/3.75 from UC Berkeley (BS/MS in Mechanical Engineering), 720 GMAT, but only 2.5 years of industry work experience. As for total work experience, I have been working since I was 14 years old, but I'm not sure that bears any merit. The 2.5 years is in a defense company where my role is typically technical, but I direct 10-12 technicians across various production shifts (though none are direct subordinates). My company also pays 100% of tuition. Anyone have any advice for whether I should burn another \$200 to apply for USC or whether my chances are good enough despite applying to UCLA in R5.
As someone who has interacted extensively with both UCLA Anderson and USC Marshall throughout the admissions process, I would highly encourage you to drop USC. UCLA is a no-brainer and USC is in a downward spiral where the admissions staff engages in shady unprofessional business practices. They aren't worth your money or time.
Best of luck with UCLA!
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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22 May 2013, 10:43
Congrats to everyone who got admitted to the 2016 class and all the very best to the people who are wait listed or just submitted their application.
I am a round 5 applicant and I couldn't get a chance to come down for the April 20th interview. I submitted my application on May 14th 2013. There has been no status change in my application and it still says "APPLICATION STATUS: SUBMITTED". I have also not heard back anything from the admissions team. Not that I am worried but I am just curious to understand the life cycle of application once it is submitted.
Any and every info from the group here will be appreciated.
Regards,
Sid.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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22 May 2013, 13:47
sidmba wrote:
Congrats to everyone who got admitted to the 2016 class and all the very best to the people who are wait listed or just submitted their application.
I am a round 5 applicant and I couldn't get a chance to come down for the April 20th interview. I submitted my application on May 14th 2013. There has been no status change in my application and it still says "APPLICATION STATUS: SUBMITTED". I have also not heard back anything from the admissions team. Not that I am worried but I am just curious to understand the life cycle of application once it is submitted.
Any and every info from the group here will be appreciated.
Regards,
Sid.
Sid,
I'd applied in R3 and it took about 2-2.5 weeks for my application status to change from Submitted to "Application is under review by Admissions committee". And it stayed on that for a few weeks until I got the admit decision call.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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22 May 2013, 13:57
Thanks aspirantmba007 and congrats for your admit. I guess I will keep my eyes open for a few weeks.
Did you attend one of their interview rounds? Now it seems that the interviews are by invitation only.
Regards,
Sid.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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22 May 2013, 14:28
sidmba wrote:
Thanks aspirantmba007 and congrats for your admit. I guess I will keep my eyes open for a few weeks.
Did you attend one of their interview rounds? Now it seems that the interviews are by invitation only.
Regards,
Sid.
R3 was a non-interview round like R5 but R3 applicants could register for an interview on R2's super Saturday. I did that.
Good luck to you!
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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30 May 2013, 13:56
Anyone here who applied in R5 heard back from the adcom?
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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30 May 2013, 15:57
R4 applicant, tried logging into the UCLA application site and says could not process login request. Wondering if this means I am out.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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30 May 2013, 18:31
I am having the same trouble..
Posted from my mobile device
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30 May 2013, 18:46
Email tech support. It happened to me a while ago even before I submitted my application. They'll send you a new username, but your other info will remain the same.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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30 May 2013, 22:37
I just emailed the tech support. Let's see.
On another topic, is there anyone in this group who got an admit call without an interview call? It has been 2 weeks since I submitted my application. The status has changed to COMPLETED, but nothing since then.
Regards,
S.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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31 May 2013, 07:05
I was having the same problem of not being able to log in yesterday but now I can log in, Application Status = Submitted. I guess I assumed that on 5/31 the status would automatically be updated to the admission decision. Maybe I'm supposed to wait for a call? Anyone other R4 applicant get a decision today?
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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31 May 2013, 08:52
I think this login problem cropped up very recently. I was having same problem for last couple of days. However, when I contacted support, they sent me new userid with the same original password that I had set up and now I am able to log in. Contact tech Support if you still see issues.
_________________
Keeping up the spirit. Target = 750
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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31 May 2013, 10:33
waitlisted. R4 decisions available online.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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31 May 2013, 11:31
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anandbha wrote:
waitlisted. R4 decisions available online.
Hold in there... I've been on the waitlist since mid December. I spoke to someone that's graduating from the FEMBA program this year. When she applied, she applied round 2 and she was not admitted until the last week of June.
Never give up, Never Surrender!
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31 May 2013, 11:33
racer50...did u update ur application after you got waitlisted ?
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31 May 2013, 12:05
I got waitlisted too. the waitlist form requires information to bolster my profile such as higher gmat scores, promotions, etc.
the problem is that the company I work for will not offer promotions or merit increases since there is little room for growth. the only thing I can think of at the moment is either finding another job or increasing my gmat score…but I don’t think I can take the gmat anytime soon since I need to prepare for it again. I’m still not going to give up though.
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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31 May 2013, 12:53
I'm sort of in the same boat as a few of you (by the way, hi everyone, first time poster, long time reader). I was just notified I have been placed on the wait list as well, and while there have been additional circumstances at work (promotion, new job title) that may prove to bolster my application, I have a feeling that they're looking for improvement on my GMAT score since I scored somewhat low on that to begin with.
However, in the event I decide to retake the GMAT before the next notification deadline (June 28th), I do not think the new score will be included in the re-review of my application as the scores will be transmitted after round 5 deadline of May 31 (today). Is anyone else's situation similar?
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31 May 2013, 12:56
What are the GMAT score of the people who got waitlisted?
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Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink]
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31 May 2013, 13:09
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anandbha wrote:
racer50...did u update ur application after you got waitlisted ?
I sent them my ticket for the CFA Level I Exam, which I take June 1. I'm going to email them after I take it, just to remind them of my progress and let them know that I'm still interested.
I also told them that I work for the federal government and there aren't going to be promotions or greater work responsibilities anytime soon, the only thing I'm getting is 5-7 days furlough.
I decided against retaking the GMAT because my score, while not great, certainly isn't bad. I really don't think I could do much better on the GMAT without a good amount of preparation. The weakest part of my application is my GPA which isn't going to change. I'm certain that my GPA is why I'm on the waitlist. I don't think an additional 20 points on the GMAT would greatly increase my chances of getting in. They're just holding out to see what the final applicant pool will be before making their final cut. If round 5 is weak, then maybe I get in. If round 5 is strong, then maybe I don't.
Re: Calling all Anderson(UCLA) FEMBA 2013 Applicants! [#permalink] 31 May 2013, 13:09
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# Calling all Anderson(UCLA) FEMBA 2013 Applicants!
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,031 | 14,180 | {"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-2017-04 | latest | en | 0.850646 |
https://tiss.tuwien.ac.at/course/courseDetails.xhtml?courseNr=104421&semester=2019S&locale=en | 1,601,484,019,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00214.warc.gz | 648,254,318 | 5,759 | # 104.421 Geometry II This course is in all assigned curricula part of the STEOP.\$(function(){PrimeFaces.cw("Tooltip","widget_j_id_21",{id:"j_id_21",showEffect:"fade",hideEffect:"fade",target:"isAllSteop"});});This course is in at least 1 assigned curriculum part of the STEOP.\$(function(){PrimeFaces.cw("Tooltip","widget_j_id_23",{id:"j_id_23",showEffect:"fade",hideEffect:"fade",target:"isAnySteop"});});
2019S, VO, 1.0h, 1.5EC
## Properties
• Semester hours: 1.0
• Credits: 1.5
• Type: VO Lecture
## Aim of course
Knowledge of analytic and constructive methods in geometry and how to apply it.
## Subject of course
basics of perspective drawing;
reconstruction of plane objects (collineations, homogeneous coordinates);
curves and surfaces;
special classes of surfaces (algebraic surfaces, surfaces of rotation, ruled surfaces);
freeform curves and surfaces.
## Course dates
DayTimeDateLocationDescription
Tue11:15 - 12:0005.03.2019 - 25.06.2019 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Geometry II - Single appointments
DayDateTimeLocationDescription
Tue05.03.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue12.03.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue19.03.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
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Tue09.04.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue30.04.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue07.05.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue14.05.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue21.05.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue28.05.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
Tue04.06.201911:15 - 12:00 Freihaus, "Grüner Bereich", 7.OG, Zeichensaal 3Vorlesungstermin
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## Examination modalities
Written and oral examination
## Exams
DayTimeDateRoomMode of examinationApplication timeApplication modeExam
Fri16:00 - 17:0009.10.2020Zeichensaal 3 written29.09.2020 08:00 - 05.10.2020 23:59TISSManhart, 16 Uhr Zeichensaal 3
Fri16:00 - 17:0027.11.2020Zeichensaal 3 written06.11.2020 08:00 - 23.11.2020 23:59TISSManhart, 16 Uhr Zeichensaal 3
Not necessary
## Curricula
Study CodeSemesterPrecon.Info
033 221 Geodesy and Geoinformation 2. Semester
## Literature
Das Skriptum aus dem Wintersemester wird weiter verwendet.
German | 1,089 | 2,923 | {"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-2020-40 | latest | en | 0.330795 |
https://physicsline.com/keplers-second-law/ | 1,680,013,589,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00088.warc.gz | 521,325,973 | 19,955 | Mechanics
# Kepler’s Second Law
Kepler’s second law , also known as the law of areas, states that the imaginary line connecting a planet to the Sun sweeps out equal areas in equal intervals of time. According to this law, the area covered by the radius vector that connects a planet to the Sun during a time interval is constant and is called the areolar velocity .
## Introduction to Kepler’s second law
Kepler ‘s second law follows directly from the principle of conservation of angular momentum . Angular momentum is the amount of motion related to rotating bodies , such as planets moving around the Sun. Angular momentum is a vector quantity , and its magnitude depends directly on the radius of the orbit and the speed with which the body moves. Thus, if the distance between the Sun and the planet increases, its speed must decrease and vice versa.
Kepler’s second law was able to show that, when we draw a line that goes from a planet to the Sun, the area swept by this line along the orbit will always be equal for equal intervals of time , regardless of what the initial position is. of the planet. This is because the area covered is shaped like an arc whose length is related to the speed at which the planet is moving, but its sides are determined by the initial and final distances from the Sun.
The rate at which an area is traversed by the radius vector that connects the planet to the Sun for a given time interval is known as areolar velocity , moreover, when a planet is in orbit around the Sun, this velocity is always constant , check:
Ω – areolar velocity (m²/s)
A – area (m²)
Δt – time interval (s)
It is important to note that the areolar velocity is different from the orbital velocity of the planet. The latter changes according to the distance between the planet and the Sun — near perihelion , the orbital speed increases, and at aphelion , it decreases, thanks to variations in the magnitude of gravitational attraction.
## Kepler’s second law summary
According to Kepler’s second law:
• The line connecting a planet to the Sun covers equal areas in equal intervals of time.
• The rate at which the area is traversed by the ray vector is known as the areolar velocity.
## Exercises solved on Kepler’s second law
Question 1) (UFRGS) The ellipse, in the figure below, represents the orbit of a planet around a star S. The points along the ellipse represent successive positions of the planet, separated by equal time intervals. The alternately colored regions represent the areas swept by the radius of the trajectory in these time intervals. In the figure, in which the dimensions of the stars and the size of the orbit are not to scale, the line segment SH represents the focal ray of the point H of length p.
Considering that the only force acting on the star-planet system is the gravitational force, the following statements are made:
I. The areas S 1 and S 2 , swept by the radius of the trajectory, are equal.
II. The orbit period is proportional to ap 3 .
III. The tangential velocities of the planet at points A and H, V A and V H are such that V A > V H .
Which ones are correct?
a) only I
b) Only I and II
c) Only I and III
d) Only II and III
e) I, II and III
Template: Letter C
Resolution:
Let’s analyze the alternatives:
I – TRUE
II – FALSE. The square of the orbit’s period is proportional to the cube of the mean radius, according to Kepler’s 3rd law.
III – TRUE
Question 2) (Acafe) Astronomers have found an exoplanet (a planet orbiting a star other than the Sun) with an eccentricity much greater than normal. The eccentricity reveals how elongated its orbit around its star is. In the case of Earth, the eccentricity is 0.017, much smaller than the value of 0.96 for that planet, which was called HD 20782.
In the following figures, the orbits of Earth and HD 20782 can be compared.
In this sense, mark the correct alternative:
a) Kepler’s laws do not apply to HD 20782 because its orbit is not circular like Earth’s.
b) Newton’s laws of gravitation do not apply to HD 20782 because its orbit is too eccentric.
c) The gravitational force between the planet HD 20782 and its star is maximum when it is passing aphelion.
d) Planet HD 20782 has an accelerated motion when moving from aphelion to perihelion.
Template : Letter D
Resolution:
At perihelion, the planet is at the shortest distance to the Sun, so it is subject to the greatest gravitational pull that its orbit allows, so the correct answer is the letter D.
Question 3) (UFSM) Advances in observational techniques have allowed astronomers to track an increasing number of celestial objects orbiting the Sun. The figure shows, on an arbitrary scale, the orbits of Earth and a comet (body sizes are not to scale). Based on the figure, analyze the statements:
I. Given the large difference between the masses of the Sun and the comet, the gravitational attraction exerted by the comet on the Sun is much smaller than the attraction exerted by the Sun on the comet.
II. The comet’s velocity is constant at all points in the orbit.
III. The comet’s translation period is longer than one Earth year.
Is(are) correct:
a) only I
b) only III
c) only I and II
d) only II and III
e) I, II and III
Template: letter B
Resolution :
Let’s analyze the alternatives:
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II – FALSE. In elliptical paths, the motion is accelerated when the star approaches the Sun.
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# Is x > 0, 1. |x + 3| = 4x - 3 2. |x - 3| = |2x -3|
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Is x > 0, 1. |x + 3| = 4x - 3 2. |x - 3| = |2x -3| [#permalink]
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18 Aug 2007, 15:13
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Is x > 0,
1. |x + 3| = 4x - 3
2. |x - 3| = |2x -3|
Guys need your help with absolute on both ends; they are tough to fathom
Last edited by asaf on 18 Aug 2007, 16:08, edited 1 time in total.
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18 Aug 2007, 17:53
A is suff
it hold sonly for x=2
I don't know for B
Any inequality masters?
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### Show Tags
18 Aug 2007, 17:56
I find the answer to be A..
Here is how I arrived at it..
Option 1
Substituting values, there is no such value for X , when X is negative that satisfies equation 1 , hence X has to be positive
Option 2
X can be negative or X can be positive to satisfy the equation
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18 Aug 2007, 17:59
for B x=0 or 2
for A x=2
No A cant be the answer...
form
1. |x + 3| = 4x - 3
x is 2 when we take x+3 = 4x -3
but its mod so other value would be 0 when u take x+3 =-(4x-3)
from 2. 2. |x - 3| = |2x -3
value are 0 and 2..
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### Show Tags
18 Aug 2007, 18:18
chiya wrote:
for B x=0 or 2
for A x=2
No A cant be the answer...
form
1. |x + 3| = 4x - 3
x is 2 when we take x+3 = 4x -3
but its mod so other value would be 0 when u take x+3 =-(4x-3)
from 2. 2. |x - 3| = |2x -3
value are 0 and 2..
when x=0
|x-3| = 3
4x-3 = -3
3 != -3
Thx ! I juss noticed.. A bit weak in inequatily. IS there any good reference which i can follow for it ?
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### Show Tags
18 Aug 2007, 18:26
chiya wrote:
chiya wrote:
for B x=0 or 2
for A x=2
No A cant be the answer...
form
1. |x + 3| = 4x - 3
x is 2 when we take x+3 = 4x -3
but its mod so other value would be 0 when u take x+3 =-(4x-3)
from 2. 2. |x - 3| = |2x -3
value are 0 and 2..
when x=0
|x-3| = 3
4x-3 = -3
3 != -3
Thx ! I juss noticed.. A bit weak in inequatily. IS there any good reference which i can follow for it ?
yah I am too lol. I graphed it with Graphmatica to check it out.
Ughmmmmm the way I do the problems is to take conditions, one when the absolute is negative, so in A's case, when x<3>-3
when you have 2 absolutes there are 4 cases, when one is neg, the other pos, both neg, both pos, etc......... inequalities are a biaaaaatch :D
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18 Aug 2007, 19:06
I am no master of absolute values. Giving it a shot anyways.
statement 1: |x+3| = 4x-3
So either, x+3 = 4x-3 i.e., x = 2
or -x-3 = 4x-3 i.e., x=0
Not sufficient.
statement 2: |x-3| = |2x-3|
Either, x-3 = 2x-3 i.e., x=0
or x-3 = -2x+3 i.e., x=2
or -x+3 = 2x-3 i.e., x=2
or -x+3 = -2x+3 i.e., x=0
Not sufficient.
Both combined also not sufficient. Answer E.
18 Aug 2007, 19:06
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Python: remove odd number from a list
I wrote a function to remove odd number from a list, like that:
``````def remove_odd(l):
for i in l:
if i % 2 != 0:
l.remove(i)
print l
return l
remove_odd([4,5,4])
remove_odd([4,5,4,7,9,11])
remove_odd([4,5,4,7,9,11,12,13])
``````
It returns:
``````[4, 4]
[4, 4, 9]
[4, 4, 9, 12]
``````
-> wrong
but when I change to remove even number:
``````def remove_even(l):
for i in l:
if i % 2 == 0:
l.remove(i)
print l
return l
remove_even([4,5,4])
remove_even([4,5,4,7,9,11])
remove_even([4,5,4,7,9,11,12,13])
``````
``````[5]
[5, 7, 9, 11]
[5, 7, 9, 11, 13]
``````
What is wrong with the remove_odd() func? I know people usually create the second list inside the func then append even number to that list, but can we solve this exercise with list.remove() ?
Thank you!
-
Don't remove items from a list while iterating over it. That's like changing the wheels of your car while driving – Andreas Jung Jan 11 '13 at 9:30
If you really have to, iterate over its indices from highest to lowest. But prefer to use built-in `filter` or similar. – Thorsten Kranz Jan 11 '13 at 9:39
Your function is working in another way than you would expect. The `for` loop takes first element, than second etc., so when you remove one element, others change their positions and can be skipped by it (and that happens in your case) when they are preceded by another odd number.
If you insist on using `.remove()` method, you must operate on a copy instead, like this:
``````def remove_odd(1):
for i in l[:]:
if i % 2 != 0:
l.remove(i)
return l
``````
(`l[:]` is a shallow copy of list `l`)
However, I think using list comprehension would be much clearer:
``````def remove_odd(l):
return [x for x in l if x % 2 == 0]
``````
-
Although this is obviously the preferred way to do the job in python, it looks to me that the poster is already aware of how to solve the problem by creating a new list. Their question is specifically how to achieve the same using `remove`. – georg Jan 11 '13 at 9:51
Yes, I have concentrated on what's wrong with my function question and missed the second part. Thank you, I'll extend my answer now. – Althorion Jan 11 '13 at 11:23
Now it's perfect! Take my +1. – georg Jan 11 '13 at 11:28
Thanks all you guys. Now I understand :) – neo0 Jan 12 '13 at 14:01
Hi, one quick question : Why did you do that ` for i in l[:]:` instead of only `for i in l:`? – Andy K Oct 14 '14 at 14:27
Python has a built-in method for this: `filter`
``````filtered_list = filter(lambda x: x%2==0, input_list)
``````
Be careful in Python 3, as here filter is only a generator, so you have to write:
``````filtered_list = list(filter(lambda x: x%2==0, input_list))
``````
-
What is wrong with the remove_odd() func?
You are iterating over a list while changing its size. This is causing it to skip one or more elements
Why don't you use list comprehension. Its more Pythonic, and readable
``````def remove_odd(l):
return [e for e in l if e % 2 == 0]
remove_odd([4,5,4,7,9,11])
[4, 4]
``````
Similarly you can write your remove_even routine
``````def remove_even(l):
return [e for e in l if e % 2]
remove_even([4,5,4,7,9,11])
[5, 7, 9, 11]
``````
-
You are trying to modify a list while you're iterating over it.
Try something like this:
``````In [28]: def remove_odd(l):
return [x for x in l if x%2 == 0]
....:
In [29]: remove_odd([4,5,4,7,9,11])
Out[29]: [4, 4]
In [30]: remove_odd([4,5,4,7,9,11,12,13])
Out[30]: [4, 4, 12]
``````
or to fix your your code only, you should iterate over `l[:]`.
`l[:]` returns a shallow copy of `l` which is equivalent to `list(l)`.
``````In [38]: def remove_odd(l):
for i in l[:]:
if i % 2 != 0:
l.remove(i)
return l
....:
In [39]: remove_odd([4,5,4,7,9,11,12,13])
Out[39]: [4, 4, 12]
In [40]: remove_odd([4,5,4,7,9,11])
Out[40]: [4, 4]
``````
-
Yeah. Great to know about shallow copy of a list. Thanks! :) – neo0 Jan 12 '13 at 14:03
the best way to modify entire list is using it's copy:
``````>>> range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> l=range(10)
>>> type(l)
<type 'list'>
>>> l[:]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> type(l[:])
<type 'list'>
>>>
``````
From off docs:
If you need to modify the sequence you are iterating over while inside the loop (for example to duplicate selected items), it is recommended that you first make a copy. Iterating over a sequence does not implicitly make a copy. The slice notation makes this especially convenient:
`````` >>>>>> for w in words[:]: # Loop over a slice copy of the entire list.
... if len(w) > 6:
... words.insert(0, w)
...
>>> words
['defenestrate', 'cat', 'window', 'defenestrate']
``````
http://docs.python.org/2/tutorial/controlflow.html
``````def remove_odd(l):
for i in l[:]:
if i % 2:
l.remove(i)
return l
``````
works just fine.
-
This answers the question exactly as asked. – georg Jan 11 '13 at 9:44
You can understand what's happening if you use `enumerate` in your example.
``````def remove_odd(l):
for n, i in enumerate(l):
print n, i
if i % 2 != 0:
l.remove(i)
print l
return l
remove_odd([4,5,4,7,9,11])
``````
It gives the result:
``````0 4
1 5
2 7
3 11
[4, 4, 9]
``````
So in the first and second case the for loop uses the right values 4 and 5. But you remove the 5 from `l`. Then in the third step you call 7 instead of the 4 on third position. Therefore it's best to copy `l`, as already suggested by other answers.
-
Woah thanks for your suggestion. enumerate() is great function. Check it! – neo0 Jan 12 '13 at 14:06
you can try like this ..
``````def remove_odd(l):
a=[]
for i in range(len(l)):
if(l[i]%2 ==0):
a.append(l[i])
return a
print remove_odd([1,2,2,6,4,1,3])
``````
- | 1,917 | 5,883 | {"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} | 3.125 | 3 | CC-MAIN-2016-30 | latest | en | 0.8699 |
https://astronomy.stackexchange.com/questions/36517/why-do-gas-giants-have-clearly-delineated-surfaces-whereas-the-earths-atmosphe/36518 | 1,618,167,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064898.14/warc/CC-MAIN-20210411174053-20210411204053-00382.warc.gz | 237,474,400 | 39,113 | # Why do gas giants have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?
I've just seen this Forbes article.
Why do gas giants appear to have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?
Is it just a matter of scale? Or is there some form of "surface-tension" for the hydrogen gas?
In an isothermal atmosphere, the exponential scale height of the atmosphere is $$h \sim \frac{k_\mathrm B T}{\mu g},$$ where $$g$$ is the gravitational field, $$\mu$$ is the mean mass of a particle and $$T$$ is the temperature (in kelvin).
i.e. The pressure/density of the atmosphere falls exponentially, with an e-folding height given by the above expression.
I suppose what matters when you look at a photo, is how this height compares with the radius of the planet. $$\frac{h}{R} \sim \frac{k_\mathrm B T}{\mu g R}$$
Jupiter is half the temperature, 11 times the radius and with 3 times the gravity of Earth. However $$\mu$$ is about ten times smaller (hydrogen vs nitrogen/oxygen). Overall that means $$h/R$$ for Jupiter is of order 5–10 times smaller than for Earth and so it will appear "sharper".
EDIT: If you put some reasonable numbers in for Jupiter ($$T \sim 130$$ K, $$\mu=2$$, $$R=7\times 10^7$$ m), then $$h/R \sim 3 \times 10^{-4}$$. This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high.
• I'm curious how much (relative) viewing distance plays into this as well, for example comparing Juno pictures of Jupiter with ISS pictures of Earth, both of which orbit at an altitude roughly 6% of the planets' radii. – Kai Jun 12 '20 at 20:33
• @Kai Dividing by $R$ effectively does take viewing distance into account. – PM 2Ring Jun 13 '20 at 4:04
• @PM2Ring It seems that the relevant scale to look at here is the angular size of the atmospheric "layer" from the viewpoint, which is not captured by comparing the atmospheric height to planetary radius – Kai Jun 13 '20 at 4:48
• @Kai Sure, but we're (approximately) scaling by the viewing distance anyway, when we make the planets have the same apparent radii in the photos. So if photos of Jupiter & the Earth have both planets with a diameter of 1000 pixels, the viewing distance to Jupiter is about 11 times the viewing distance to Earth. – PM 2Ring Jun 13 '20 at 5:09
• @Kai If you "fill" your picture with a planet, what matters is height of the atmosphere vs radius of the planet, which is what I have given. – ProfRob Jun 13 '20 at 7:04 | 653 | 2,513 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-17 | longest | en | 0.889456 |
http://www.chegg.com/homework-help/repeat-problem-60-o-internal-angleproblem-link-length-planar-chapter-2-problem-41p-solution-9780201350999-exc | 1,472,686,336,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982954852.80/warc/CC-MAIN-20160823200914-00244-ip-10-153-172-175.ec2.internal.warc.gz | 352,776,627 | 17,594 | View more editions
# Kinematics and Dynamics of Machinery (3rd Edition)Solutions for Chapter 2 Problem 41PProblem 41P: Repeat problem for a 60 o internal angle.Problem: The link l...
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Chapter: Problem:
Repeat problem for a 60 o internal angle.
Problem: The link length of a planar four-bar mechanism are r0=120, r1=60, r2=140, and r3=80. Find the orientation of links 2 and 3 when the internal angle between the crank and the fixed link is 30o and the linkage is in the open phase (i.e. the coupler does not cross the fixed link). Use the vector cross-product method.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 3
Select a coordinate system so that the fixed link lies in the x direction as shown in figure. Then given values of link lengths can be represented as:
• Chapter , Problem is solved.
Corresponding Textbook
Kinematics and Dynamics of Machinery | 3rd Edition
9780201350999ISBN-13: 0201350998ISBN: | 297 | 1,145 | {"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-2016-36 | latest | en | 0.771905 |
https://www.allmathwords.org/en/a/arc.html | 1,716,365,006,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00081.warc.gz | 561,582,805 | 5,392 | # Arc
Pronunciation: /ɑɹk/ Explain
Figure 1: Arcs on a Circle
An arc is any smooth curve joining two points. This article will focus on arcs of circles. An arc of a circle is sometimes called a circular arc.
Any two points on a circle define two arcs. If the two points are on a diameter of the circle, each arc is called a semicircle. If the two points are not on a diameter of a circle, they define two arcs. The larger arc is called the major arc, and the smaller arc is called the minor arc.
Adjacent arcs are two arcs that share an endpoint.
### Arc Measure
Arcs are measured based on what portion of a whole circle they occupy. A whole circle measures 360° or rad. 'rad' is an abbreviation for radians. An arc occupying 1/2 of a circle would then measure 360° / 2 = 180° or rad / 2 = π rad.
### Arc Length
Click on the blue points and drag them to change the figure. What is the arc length of a arc of a circle with circumference 4 that covers 90 degrees? Manipulative 1 - Arc Length Created with GeoGebra.
Arc length is defined as the linear length of an arc. For arcs of circles, the arc length is calculated as a portion of the total circumference of a circle. if the arc measures 72°, and the circumference is 22cm, then the arc length is 72°/360° · 22cm = 4.4cm.
### References
1. McAdams, David E.. All Math Words Dictionary, arc. 2nd Classroom edition 20150108-4799968. pg 18. Life is a Story Problem LLC. January 8, 2015. Buy the book
### More Information
• Arc length. www.khanacademy.org. Kahn Academy. 6/19/2018. https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-arc-length-deg/e/circles_and_arcs.
### Cite this article as:
McAdams, David E. Arc. 4/11/2019. All Math Words Encyclopedia. Life is a Story Problem LLC. https://www.allmathwords.org/en/a/arc.html.
### Revision History
4/11/2019: Changed equations and expressions to new format. (McAdams, David E.)
12/21/2018: Reviewed and corrected IPA pronunication. (McAdams, David E.)
6/14/2018: Removed broken links, changed Geogebra links to work with Geogebra 5, updated license, implemented new markup. (McAdams, David E.)
3/2/2010: Added sentence on adjacent arcs. (McAdams, David E.)
1/2/2010: Added "References". (McAdams, David E.)
11/12/2009: Added arc length. (McAdams, David E.)
6/11/2008: Added paragraph on the measure of an arc. (McAdams, David E.)
6/7/2008: Corrected spelling. (McAdams, David E.)
4/18/2008: Initial version. (McAdams, David E.)
All Math Words Encyclopedia is a service of Life is a Story Problem LLC.
Copyright © 2018 Life is a Story Problem LLC. All rights reserved.
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Created by: Abhishek Jain
2
97
Problem 2817. Remove the positive integers.
Created by: Matthew Caron
1
220
2
41
Problem 46639. Transform a Decimal number into a Septimal number
Created by: Syed Shahed
1
21
Problem 42757. Convert integer to base26 using letters
Created by: Alberto Comin
Tags numbers
3
34
Created by: HH
0
31
2
62
Problem 44696. Sum of unique multiples of 3 and 5
Created by: Srishti Saha
1
45
Problem 42381. Dice roll - lateral faces
Created by: goc3
Tags game, dice, numbers
1
35
Problem 42981. Angle difference between Hour Hand and Minute Hand of clock
Created by: Vishal
Tags numbers, clock
1
29
Problem 45255. Find the sequence
Created by: Asif Newaz
0
21
Problem 43045. Display negative numbers
Created by: Abhishek Jain
8
77
Problem 43123. Sum of cubes
Created by: Abhishek Jain
Tags numbers, sum, cube
6
81
Problem 52442. Draw a x-by-x matrix British flag (Euro 2020)
Created by: Trần Nhật Tân
0
18
Created by: HH
2
24
Problem 2269. Create the following sequence : 0 1 1 4 9 25 64 169 ...
Created by: Debopam
2
100
Problem 49962. When the sum of the squares is a cubic...
Created by: Doddy Kastanya
Tags numbers, fun
2
15
Created by: CXD
0
7
Created by: HH
3
23
0
8
Problem 49967. When the sum of the cubes is a quartic...
Created by: Doddy Kastanya
2
13
Problem 44287. sum all digits
Created by: Zakaria Makhlouki
Tags digits, numbers, sum
1
89
Problem 322. Number of Circles in a Number
Created by: @bmtran (Bryant Tran)
3
206
Problem 1952. Count decimal digits of a number
Created by: Dimitris Kaliakmanis
Tags numbers
1
201
Created by: cdeliu
Tags numbers, age
4
27
Problem 291. Triangle Numbers Below N
Created by: @bmtran (Bryant Tran)
1
201
Created by: HH
3
17
Problem 45960. Set the Euler-Mascheroni constant with an identity using available MATLAB functions
Created by: ChrisR
1
9
Problem 1737. The sum of individual numbers...
Created by: Chris E.
3
69
Created by: Alex
5
392
Problem 54265. n-th digit of write-down all numbers
Created by: CXD
Tags series, numbers
1
9
0
40
Created by: Omer
0
45
Problem 44859. Get the value 100
Created by: Jesús Zambrano
1
23
Problem 2987. Hardy-Ramanujan number
Created by: Ritwik Rudra
0
72
Problem 2627. Convert to Binary Coded Decimal
Created by: Pritesh Shah
4
98
Problem 272. Generalized N-Cards Problem
Created by: @bmtran (Bryant Tran)
4
25
1 – 50 of 95 | 1,157 | 3,698 | {"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} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.729636 |
http://www.jiskha.com/display.cgi?id=1360598010 | 1,495,785,372,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608648.25/warc/CC-MAIN-20170526071051-20170526091051-00335.warc.gz | 720,943,677 | 3,810 | # geometry
posted by on .
Let O be the center of the circle Γ, and P be a point outside of circle Γ. PA is tangential to Γ at A, and PO intersects Γ at D. If PD=14 and PA=42, what is the radius of Γ.
• geometry - ,
In this case the secant - tangent theorem says
PA^2 = PD x PC , where D in on the extension of PO as it hits the circle , making CD a diameter
Let the diameter be 2x , (radius = x)
(14)(2x+14) = 42^2
28x + 196 = 1764
28x = 1568
x = 56 | 150 | 454 | {"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-2017-22 | latest | en | 0.853428 |
http://e-booksdirectory.com/listing.php?category=470 | 1,716,948,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00007.warc.gz | 8,719,581 | 4,560 | # Multivariable Calculus
## e-books in Multivariable Calculus category
Calculus Volume 3
- OpenStax ,
The textbook guides students through the core concepts of calculus. Volume 3 covers parametric equations and polar coordinates, vectors, functions of several variables, multiple integration, and second-order differential equations.
(8571 views)
Multivariable Calculus
by - National University of Singapore ,
Contents: Vector Functions; Functions of several variables; Limits and Continuity; Partial Derivatives; Maximum and Minimum Values; Lagrange Multipliers; Multiple Integrals; Surface Area; Triple Integrals; Vector Fields; Line Integrals; etc.
(10946 views)
Multivariable Calculus: Applications and Theory
by - Brigham Young University ,
This book presents the necessary linear algebra and then uses it as a framework upon which to build multivariable calculus. This is the correct approach, leaving open the possibility that at least some students will understand the topics presented.
(8964 views)
Multivariable Calculus
by - Reed College ,
A text for a two-semester multivariable calculus course. The setting is n-dimensional Euclidean space, with the material on differentiation culminating in the Inverse Function Theorem, and the material on integration culminating in Stokes's Theorem.
(14054 views)
Multivariable and Vector Analysis
by - Macquarie University ,
Introduction to multivariable and vector analysis: functions of several variables, differentiation, implicit and inverse function theorems, higher order derivatives, double and triple integrals, vector fields, integrals over paths, etc.
(17220 views)
Multivariable Calculus
by ,
The text covers Euclidean three space, vectors, vector functions, derivatives, more dimensions, linear functions and matrices, continuity, the Taylor polynomial, sequences and series, Taylor series, integration, Gauss and Green, Stokes.
(17010 views)
Calculus III
by - Lamar University ,
These lecture notes should be accessible to anyone wanting to learn Calculus III or needing a refresher in some of the topics from the class. The notes assume a working knowledge of limits, derivatives, integration, parametric equations, vectors.
(22953 views)
Vector Calculus
by - Schoolcraft College ,
A textbok on elementary multivariable calculus, the covered topics: vector algebra, lines, planes, surfaces, vector-valued functions, functions of 2 or 3 variables, partial derivatives, optimization, multiple, line and surface integrals.
(34236 views)
The Calculus of Functions of Several Variables
by - Furman University ,
Many functions in the application of mathematics involve many variables simultaneously. This book introducses Rn, angles and the dot product, cross product, lines, planes, hyperplanes, linear and affine functions, operations with matrices, and more.
(19981 views) | 587 | 2,838 | {"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-2024-22 | latest | en | 0.818081 |
https://www.ssusa.org/articles/2018/5/24/effects-of-cant/ | 1,726,428,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00670.warc.gz | 923,437,345 | 17,943 | # Effects of Cant
by
posted on May 24, 2018
From the June 1955 issue of American Rifleman, a question and answer session reviewing the effects of cant.
While I have seen answers giving the effect of cant for certain cartridges at a given range, I would like to have an explanation which will be adequate for determining this effect with any cartridge at any range.—George A. Williams, Brooklyn, NY.
Answer by E.H. Harrison: The effect of canting the rifle in firing is to move the impact of the bullet on the target in the direction of the cant, and also slightly downward.
The above illustration shows in perspective the effect of cant when firing at a vertical target with its center at the point T. The gun is at G, and the line GT is therefore the range. Because of the curvature of the trajectory resulting from gravity, the bullet must be discharged at an angle above the line to the target, and in the figure the point toward which the bullet must be discharged is P. If, for example, the rifle is canted to the right, the result is to discharge the bullet in the direction of the point P′. This obviously will result in the bullet striking to the right of the center of the target and also slightly lower. The latter effect is trifling and can be ignored, and the important result is to move the strike of the bullet from T to T′. We see that the magnitude of the effect is dependent on both the angle of elevation (which governs the height PT) and the angle through which the rifle is canted.
We can determine the total effect on the target from the magnitudes of the above two factors. PT obviously depends on the range and the angle of elevation. The sportsman seldom knows the latter, but he can obtain PT from the approximate relation that it is four times the midrange trajectory height. Trajectory heights for various factory loads at ranges up to 300 yards are given in factory ballistic tables. For example, the midrange height of the factory .270 Winchester 150-grain load when shooting at 300 yards is 7.8 inches, hence the height PT under those conditions is about 31 inches.
For the effect of the angle of cant C, note that the distance TT′ (which is the displacement on the target we are looking for) is given by the relation TT′ = PT sine C. The values of the sines of angles 1° to 6° are:
If in this case we cant the rifle 5° to the right, the displacement of the hilt will be 31 times .087 or about 2½ inches to the right.
In the above way the effect of cant can be readily computed for any cartridge of known ballistics. We see that at long ranges or with cartridges giving quite curved trajectories, the lateral displacement on the target can be considerable. This is notably the case with the .22 long rifle.
## Latest
### Results: 2024 NRA National Precision Pistol Championships
Jon Shue wins back to back NRA National Pistol titles after securing victory at Camp Atterbury in July.
| 640 | 2,927 | {"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} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.942272 |
https://brainmass.com/statistics/normal-distribution/probability-normal-distribution-heights-children-193838 | 1,620,705,414,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991641.5/warc/CC-MAIN-20210511025739-20210511055739-00053.warc.gz | 173,017,490 | 75,129 | Explore BrainMass
# Probability using normal distribution: heights of children
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The heights of children 2 years old are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches. Pediatricians regularly measure the heights of toddlers to determine whether there is a problem. There may be a problem when a child is in the top or bottom 5% of heights. Determine the heights of 2 year old children that could be a problem. | 136 | 642 | {"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-2021-21 | latest | en | 0.906434 |
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# Five Attitudes toward Quality Nursing Care
identify a sport organization facing a problem and propose thoughtful solutions to address the issue. Paper details Working individually, your first task is to pick a topic/issue that interests you. Topics should be recent (e.g., NOT whether the Montreal Expos should or should not relocate to Washington, etc.). And topics should relate to the business/management side of sport and NOT issues having to do with player development/coaching decisions.identify a sport organization facing a problem and propose thoughtful solutions to address the issue
Cuyamaca College Chemistry Specific Heat of Aluminum on A Sunny Day Lab Report.
3-3: Specific Heat of Al
On a sunny day, the water in a swimming pool may warm up a degree or two while the concrete around
the pool may become too hot to walk on with bare feet. This may seem strange since the water and
concrete are being heated by the same source—the sun. This evidence suggests that it takes more heat to
raise the temperature of some substances than others, which is true. The amount of heat required to raise
the temperature of 1 g of a substance by 1 degree is called the specific heat capacity or specific heat of
that substance. Water, for instance, has a specific heat of 4.18 J/Kg. This value is high in comparison
with the specific heats for other materials, such as concrete or metals. In this experiment, you will use a
simple calorimeter and your knowledge of the specific heat of water to measure the specific heat of
aluminum (Al).
1. Start Virtual ChemLab, select Thermodynamics, and then select Specific Heat of Al from the list of
assignments. The lab will open in the Calorimetry laboratory.
2. Record the mass of Al on the balance. If it is too small to read click on the Balance area to zoom in,
record the mass of Al in the data table below, and return to the laboratory.
3. Pick up the Al sample from the balance pan and place the sample in the oven. Click the oven door to
close. The oven is set to heat to 200C.
4. The calorimeter has been filled with 100 mL water. The density of water at 25C is 0.998 g/mL. Use
the density of the water to determine the mass of water from the volume and record the volume and
mass in the data table.
Make certain the stirrer is On (you should be able to see the shaft rotating). In the thermometer
window, click Save to begin recording data. Allow 20-30 seconds to obtain a baseline temperature of
the water. You can observe the temperature in the calorimeter as a function of time using the graph
window.
5. Click on the Oven to open it. Drag the hot Al sample from the oven until it snaps into place above the
calorimeter and drop it in. Click the thermometer and graph windows to bring them to the front and
observe the change in temperature in the graph window until it reaches a constant value and then wait
an additional 20-30 seconds. Click Stop in the temperature window. (You can click on the clock on
the wall labeled Accelerate to accelerate the time in the laboratory.) A data link icon will appear in
the lab book. Click the data link icon and record the temperature before adding the Al and the highest
temperature after adding the Al in the data table. (Remember that the water will begin to cool down
after reaching the equilibrium temperature.)
Data Table
Al
mass of metal (g)
volume of water (mL)
mass of water (g)
initial temperature of water (C)
initial temperature of metal (C)
max temp of water + metal (C)
6. Calculate the change in temperature of the water (Twater).
7. Calculate the heat (q), in J, gained by the water using the following equation:
qwater
= mwater ΔTwater Cwater
, given Cwater= 4.184 J/(Kg)
8. Calculate the changes in temperature of the Al (TAl).
9. Remembering that the heat gained by the water is equal to the heat lost by the metal, calculate the
specific heat of aluminum in J/Kg.
qwater
= −qmetal = mAl TAl CAl
and
( )( ) metal metal
metal
Al
m T
q
C
=
10. Calculate the percent error in the specific heat value that you determined experimentally. The
accepted value for Al is 0.903 J/ Kg.
100
% Error
=
% Error =
Cuyamaca College Chemistry Specific Heat of Aluminum on A Sunny Day Lab Report
## Hospitality Industry PESTEL Analysis
The process of conducting research on the business environment within which the organization operates and on the organization itself, in order to formulate and implementation of strategy for future business operations can be mentioned as Strategic analysis of hospitality organization. To do the assessment can use number of tools to process of strategic analysis, including PEST (sometimes PESTLE) for analyze external environment and, SWOT analysis use for the internal environmental scanning, and Michael Porter’s five forces model use to assists to understand the competitive forces, the attractiveness and current position in the industry. An effective way to know the past, present and future potentials regarding the industry development is analyzing the industry background as the part of strategic analysis. The External analysis can assess the factors affecting the industry to be existed including political, economical, social, technological, legal and environmental which have great impacts to run the business and use PESTLE analysis. In any industry which it is domestic or international, whether relating to products or services, the rules of competition are personified in five competitive forces of entry of new competitors, threat of substitutes, bargaining power of buyers, bargaining power of suppliers, and rivalry among the existing competitors. According to Porter, one of the crucial determinants of firm profitability is industrial attractiveness. In this assignment, a strategic analysis of the Galadhari hotel which is in hospitality industry has been done through the combination of both theoretical and practical facts regarding this property including organizational background and industry background of this hospitality sector in Sri Lanka, an assessment of the forces affecting the external environment (the use of PESTLE analysis) and an assessment of the attractiveness of this industry (the use of Porter’s 5 forces) regarding the future strategic action to grab more hospitality market share. Background of the industry and about Organization By concerning about the Sri Lankan market of tourism there can be seen growth in the tourist market. All this euphoria gives rise to the doubt about whether Sri Lanka Tourism is well on the way to recovery and growth or not. For the last seven months that ended July this year (2010), arrivals are up almost 50% year-on-year (YOY) (341,991), with income also keeping pace at 69% growth (Quarter 2; US\$ 244.5 million). The hotel and travel Colombo Stock Exchange (CSE) index has been increased by almost 200% for 2009. Today tourism is running on everyone’s minds, and it is difficult to open a local newspaper without seeing at least one written article on tourism. The reason for the dramatic improvement in Sri Lanka’s tourism data is the victory of war in May 2009. This would be a remarkable result when compared with other regional tourism destinations. Another reason for this improvement was the leader of the terrorist has killed and there is reason for cautious optimism that the social situation in Sri Lanka can improve rapidly. We can hope the company can take advantage of the ‘peace dividend’ by increasing the number of destinations the airline serves. Hotel Galadhari is one of the leading five star luxury hotels in the Sri Lanka. The story of the Galadari Hotel, Colombo which opened its doors in 1984, is a splendid tale of continual improvement of product and highest standard of quality in hospitality over the past 25 years. The vision
## HIS 340 Southern New Hampshire Historiography of Lewis and Clarks Expedition HW
python assignment help HIS 340 Southern New Hampshire Historiography of Lewis and Clarks Expedition HW.
Brundage provides a case study of the historiography of Lewis and Clark’s Expedition. In this model, the secondary sources are analyzed chronologically. Moulton, in his essay, approaches the same topic thematically. Cayton provides a comparative book review as a means of assessing Lewis and Clark historiography.In preparing your initial post, focus on the different approaches taken in the three readings. What approach seems most effective in terms of developing a comprehensive account of historians’ perspectives on the history of Lewis and Clark? How might these readings guide the development of your own historiographical review?Cite in Chicago/Turabian formatAttached are sourceshere is the link for the Brundage source : https://mbsdirect.vitalsource.com/#/books/97811192…
HIS 340 Southern New Hampshire Historiography of Lewis and Clarks Expedition HW
## BMCC The Circular Flow Diagram for The Economy & Economic Activities Discussion
BMCC The Circular Flow Diagram for The Economy & Economic Activities Discussion.
I’m working on a business project and need a sample draft to help me understand better.
Devote approximately 600 words, or two to four paragraphs preparing your mini essay. This equates to two full pages typewritten. Submit the text of your work directly into Blackboard without any attachment files. Accordingly, you are expected to draw on all assigned materials and course content to support your answers. This is not a research paper. Also, while you may share ideas, each student must ultimately construct and write individual essays. **Do not co-write or copy another student’s work in any form.**Grading: Each mini essay is awarded a grade of 20 points maximum. Evaluation is based primarily on content. Nevertheless, writing is expected to be at college level. Consequently, excellent writing quality will be rewarded with extra points, while a poorly written essay will get points deducted. Content quality is judged based on demonstrated mastery of course materials. This includes command of key terms, use of technical concepts, recognition of issues, coherent understanding of theory, and ability to illustrate ideas with concrete examples.
BMCC The Circular Flow Diagram for The Economy & Economic Activities Discussion
## Please answer the following questions from Tools For Mindful Living Chapters 2, 3 & 4
Please answer the following questions from Tools For Mindful Living Chapters 2, 3 & 4.
1. Explain each of the 4 steps involved in the MAC Model and provide a personal example (Use that same experience for all four steps – see student example attached below) for each of the steps. Please use complete sentences2. What happens in the body physically and psychologically (emotions) when it is in sympathetic vs parasympathetic. Please be detailed and use complete sentences. Chapter 4 talks about 3 different types of breath work – what was your favorite and why? Please use complete sentences. I will give you the access of the e-book
Please answer the following questions from Tools For Mindful Living Chapters 2, 3 & 4
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http://mofem.eng.gla.ac.uk/mofem/html/tutorial_level_set.html | 1,709,347,524,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00812.warc.gz | 25,691,584 | 55,140 | v0.14.0
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Note
Prerequisites of this tutorial include SCL-11: Discountinous Galrekin for Poisson problem
Note
Intended learning outcome:
• general structure of a program developed using MoFEM
• integartion on skeleton
• Upwind Discountinous Galerkin
• Level-set
/**
* @file level_set.cpp
* @example level_set.cpp
* @brief Implementation DG upwind method for advection/level set problem
* @date 2022-12-15
*
*
*/
#include <MoFEM.hpp>
using namespace MoFEM;
static char help[] = "...\n\n";
//! [Define dimension]
// We can have 2D elements embedded in 3D space or 2D element embedded in 2D
constexpr int FE_DIM = EXECUTABLE_DIMENSION; //< dimension of the element
constexpr int SPACE_DIM = FE_DIM; //< dimension of the space
constexpr int DIM1 = 1;
constexpr int DIM2 = 1;
// \todo We do not have implemented embedding of the edge elements in 3D space,
// thus integration on skeleton will not work for
// case FE_DIM = 2 and SPACE_DIM = 3. Since FE_DIM = 2, skeleton is 1D, i.e.
// edge elements.
constexpr AssemblyType A = AssemblyType::PETSC; //< selected assembly type
constexpr IntegrationType G =
IntegrationType::GAUSS; //< selected integration type
using FaceSideEleOp = FaceSideEle::UserDataOperator;
constexpr size_t potential_velocity_field_dim = FE_DIM == 2 ? 1 : 3;
// #ifndef NDEBUG
constexpr bool debug = true;
// #else
// constexpr bool debug = true;
// #endif
constexpr int nb_levels = 3; //< number of refinement levels
constexpr int start_bit =
nb_levels + 1; //< first refinement level for computational mesh
constexpr int current_bit =
2 * start_bit + 1; ///< dofs bit used to do calculations
constexpr int skeleton_bit = 2 * start_bit + 2; ///< skeleton elements bit
constexpr int aggregate_bit =
2 * start_bit + 3; ///< all bits for advection problem
constexpr int projection_bit =
2 * start_bit + 4; //< bit from which data are projected
constexpr int aggregate_projection_bit =
2 * start_bit + 5; ///< all bits for projection problem
struct LevelSet {
LevelSet(MoFEM::Interface &m_field) : mField(m_field) {}
private:
using MatSideArray = std::array<MatrixDouble, 2>;
/**
* @brief data structure carrying information on skeleton on both sides.
*
*/
struct SideData {
// data for skeleton computation
std::array<EntityHandle, 2> feSideHandle;
std::array<VectorInt, 2>
indicesRowSideMap; ///< indices on rows for left hand-side
std::array<VectorInt, 2>
indicesColSideMap; ///< indices on columns for left hand-side
std::array<MatrixDouble, 2> rowBaseSideMap; // base functions on rows
std::array<MatrixDouble, 2> colBaseSideMap; // base function on columns
std::array<int, 2> senseMap; // orientation of local element edge/face in
// respect to global orientation of edge/face
MatSideArray lVec; //< Values of level set field
MatSideArray velMat; //< Values of velocity field
int currentFESide; ///< current side counter
};
/**
*
* \note in current implementation is assumed that advection field has zero
* normal component.
*
* \note function define a vector velocity potential field, curl of potential
* field gives velocity, thus velocity is divergence free.
*
* @tparam FE_DIM
* @param x
* @param y
* @param z
* @return auto
*/
template <int FE_DIM>
static double get_velocity_potential(double x, double y, double z);
/**
* @brief inital level set, i.e. advected filed
*
* @param x
* @param y
* @param z
* @return double
*/
static double get_level_set(const double x, const double y, const double z);
/**
*
* @return MoFEMErrorCode
*/
/**
* @brief create fields, and set approximation order
*
* @return MoFEMErrorCode
*/
/**
* @brief push operators to integrate operators on domain
*
* @return MoFEMErrorCode
*/
/**
* @brief evaluate error
*
* @return MoFEMErrorCode
*/
std::tuple<double, Tag> evaluateError();
/**
* @brief Get operator calculating velocity on coarse mesh
*
* @param vel_ptr
* @return DomainEleOp*
*/
getZeroLevelVelOp(boost::shared_ptr<MatrixDouble> vel_ptr);
/**
* @brief create side element to assemble data from sides
*
* @param side_data_ptr
* @return boost::shared_ptr<FaceSideEle>
*/
boost::shared_ptr<FaceSideEle>
getSideFE(boost::shared_ptr<SideData> side_data_ptr);
/**
* @brief push operator to integrate on skeleton
*
* @return MoFEMErrorCode
*/
/**
* @brief test integration side elements
*
* Check consistency between volume and skeleton integral.
*
* @return MoFEMErrorCode
*/
/**
* @brief test consistency between tangent matrix and the right hand side
* vectors
*
* @return MoFEMErrorCode
*/
/**
* @brief initialise field set
*
* @param level_fun
* @return MoFEMErrorCode
*/
boost::function<double(double, double, double)> level_fun =
/**
* @brief initialise potential velocity field
*
* @param vel_fun
* @return MoFEMErrorCode
*/
boost::function<double(double, double, double)> vel_fun =
get_velocity_potential<FE_DIM>);
/**
* @brief dg level set projection
*
* @param prj_bit
* @param mesh_bit
* @return MoFEMErrorCode
*/
/**
*
* @return * MoFEMErrorCode
*/
/**
* @brief Wrapper executing stages while mesh refinement
*/
struct WrapperClass {
WrapperClass() = default;
/**
* @brief Set bit ref level to problem
*/
virtual MoFEMErrorCode setBits(LevelSet &level_set, int l) = 0;
/**
* @brief Run calculations
*/
virtual MoFEMErrorCode runCalcs(LevelSet &level_set, int l) = 0;
/**
* @brief Add bit to current element, so it aggregate all previious current
* elements
*/
virtual MoFEMErrorCode setAggregateBit(LevelSet &level_set, int l) = 0;
virtual double getThreshold(const double max) = 0;
};
/**
* @brief Used to execute inital mesh approximation while mesh refinement
*
*/
struct WrapperClassInitalSolution : public WrapperClass {
WrapperClassInitalSolution(boost::shared_ptr<double> max_ptr)
: WrapperClass(), maxPtr(max_ptr) {}
MoFEMErrorCode setBits(LevelSet &level_set, int l) {
auto simple = level_set.mField.getInterface<Simple>();
simple->getBitRefLevel() =
simple->reSetUp(true);
};
MoFEMErrorCode runCalcs(LevelSet &level_set, int l) {
}
auto bit_mng = level_set.mField.getInterface<BitRefManager>();
auto set_bit = [](auto l) { return BitRefLevel().set(l); };
Range level;
CHKERR bit_mng->getEntitiesByRefLevel(set_bit(start_bit + l),
BitRefLevel().set(), level);
->synchroniseEntities(level);
CHKERR bit_mng->setNthBitRefLevel(current_bit, false);
CHKERR bit_mng->setNthBitRefLevel(level, current_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level, aggregate_bit, true);
}
double getThreshold(const double max) {
*maxPtr = std::max(*maxPtr, max);
return 0.05 * (*maxPtr);
}
private:
boost::shared_ptr<double> maxPtr;
};
/**
* @brief Use peculated errors on all levels while mesh projection
*
*/
struct WrapperClassErrorProjection : public WrapperClass {
WrapperClassErrorProjection(boost::shared_ptr<double> max_ptr)
: maxPtr(max_ptr) {}
MoFEMErrorCode setBits(LevelSet &level_set, int l) { return 0; };
MoFEMErrorCode runCalcs(LevelSet &level_set, int l) { return 0; }
auto bit_mng = level_set.mField.getInterface<BitRefManager>();
auto set_bit = [](auto l) { return BitRefLevel().set(l); };
Range level;
CHKERR bit_mng->getEntitiesByRefLevel(set_bit(start_bit + l),
BitRefLevel().set(), level);
->synchroniseEntities(level);
CHKERR bit_mng->setNthBitRefLevel(current_bit, false);
CHKERR bit_mng->setNthBitRefLevel(level, current_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level, aggregate_bit, true);
}
double getThreshold(const double max) { return 0.05 * (*maxPtr); }
private:
boost::shared_ptr<double> maxPtr;
};
MoFEMErrorCode refineMesh(WrapperClass &&wp);
struct OpRhsDomain; ///< integrate volume operators on rhs
struct OpLhsDomain; ///< integrate volume operator on lhs
struct OpRhsSkeleton; ///< integrate skeleton operators on rhs
struct OpLhsSkeleton; ///< integrate skeleton operators on khs
// Main interfaces
G>::OpSource<1, DIM1 * DIM2>;
G>::OpSource<potential_velocity_field_dim, potential_velocity_field_dim>;
G>::OpBaseTimesVector<1, DIM1 * DIM2, 1>;
enum ElementSide { LEFT_SIDE = 0, RIGHT_SIDE = 1 };
private:
boost::shared_ptr<double> maxPtr;
};
template <>
double LevelSet::get_velocity_potential<2>(double x, double y, double z) {
return (x * x - 0.25) * (y * y - 0.25);
}
double LevelSet::get_level_set(const double x, const double y, const double z) {
constexpr double xc = 0.1;
constexpr double yc = 0.;
constexpr double zc = 0.;
constexpr double r = 0.2;
return std::sqrt(pow(x - xc, 2) + pow(y - yc, 2) + pow(z - zc, 2)) - r;
}
if constexpr (debug) {
}
maxPtr = boost::make_shared<double>(0);
CHKERR refineMesh(WrapperClassInitalSolution(maxPtr));
simple->getBitRefLevel() =
simple->reSetUp(true);
}
OpRhsDomain(const std::string field_name,
boost::shared_ptr<MatrixDouble> l_ptr,
boost::shared_ptr<MatrixDouble> l_dot_ptr,
boost::shared_ptr<MatrixDouble> vel_ptr);
private:
boost::shared_ptr<MatrixDouble> lPtr;
boost::shared_ptr<MatrixDouble> lDotPtr;
boost::shared_ptr<MatrixDouble> velPtr;
};
OpLhsDomain(const std::string field_name,
boost::shared_ptr<MatrixDouble> vel_ptr);
MoFEMErrorCode iNtegrate(EntData &row_data, EntData &col_data);
private:
boost::shared_ptr<MatrixDouble> velPtr;
};
OpRhsSkeleton(boost::shared_ptr<SideData> side_data_ptr,
boost::shared_ptr<FaceSideEle> side_fe_ptr);
private:
boost::shared_ptr<SideData> sideDataPtr;
boost::shared_ptr<FaceSideEle>
sideFEPtr; ///< pointer to element to get data on edge/face sides
};
OpLhsSkeleton(boost::shared_ptr<SideData> side_data_ptr,
boost::shared_ptr<FaceSideEle> side_fe_ptr);
private:
boost::shared_ptr<SideData> sideDataPtr;
boost::shared_ptr<FaceSideEle>
sideFEPtr; ///< pointer to element to get data on edge/face sides
};
int main(int argc, char *argv[]) {
// Initialisation of MoFEM/PETSc and MOAB data structures
const char param_file[] = "param_file.petsc";
try {
// Create MoAB database
moab::Core moab_core;
moab::Interface &moab = moab_core;
// Create MoFEM database and link it to MoAB
MoFEM::Core mofem_core(moab);
MoFEM::Interface &m_field = mofem_core;
// Register DM Manager
DMType dm_name = "DMMOFEM";
// Add logging channel for example
auto core_log = logging::core::get();
LogManager::createSink(LogManager::getStrmWorld(), "LevelSet"));
LogManager::setLog("LevelSet");
MOFEM_LOG_TAG("LevelSet", "LevelSet");
LevelSet level_set(m_field);
CHKERR level_set.runProblem();
}
// finish work cleaning memory, getting statistics, etc.
return 0;
}
// get options from command line
CHKERR simple->getOptions();
// Only L2 field is set in this example. Two lines bellow forces simple
// interface to creat lower dimension (edge) elements, despite that fact that
// there is no field spanning on such elements. We need them for DG method.
simple->getBitRefLevel() = BitRefLevel();
// Initialise bit ref levels
auto set_problem_bit = [&]() {
auto bit0 = BitRefLevel().set(start_bit);
for (auto s = 0; s != start_bit; ++s)
auto bit_mng = mField.getInterface<BitRefManager>();
Range level0;
CHKERR bit_mng->getEntitiesByRefLevel(BitRefLevel().set(0),
BitRefLevel().set(), level0);
CHKERR bit_mng->setNthBitRefLevel(level0, current_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level0, aggregate_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level0, skeleton_bit, true);
// Set bits to build adjacencies between parents and children. That is used
// by simple interface.
simple->getBitRefLevel() = BitRefLevel().set(current_bit);
#ifndef NDEBUG
if constexpr (debug) {
auto proc_str = boost::lexical_cast<std::string>(mField.get_comm_rank());
CHKERR bit_mng->writeBitLevelByDim(
BitRefLevel().set(0), BitRefLevel().set(), FE_DIM,
(proc_str + "level_base.vtk").c_str(), "VTK", "");
}
#endif
};
CHKERR set_problem_bit();
}
// Scalar fields and vector field is tested. Add more fields, i.e. vector
// field if needed.
// set fields order, i.e. for most first cases order is sufficient.
CHKERR simple->setFieldOrder("L", 4);
CHKERR simple->setFieldOrder("V", 4);
// setup problem
CHKERR simple->setUp();
}
boost::shared_ptr<MatrixDouble> l_ptr,
boost::shared_ptr<MatrixDouble> l_dot_ptr,
boost::shared_ptr<MatrixDouble> vel_ptr)
lPtr(l_ptr), lDotPtr(l_dot_ptr), velPtr(vel_ptr) {}
boost::shared_ptr<MatrixDouble> vel_ptr)
AssemblyDomainEleOp::OPROWCOL),
velPtr(vel_ptr) {
this->sYmm = false;
}
boost::shared_ptr<SideData> side_data_ptr,
boost::shared_ptr<FaceSideEle> side_fe_ptr)
sideDataPtr(side_data_ptr), sideFEPtr(side_fe_ptr) {}
boost::shared_ptr<SideData> side_data_ptr,
boost::shared_ptr<FaceSideEle> side_fe_ptr)
sideDataPtr(side_data_ptr), sideFEPtr(side_fe_ptr) {}
const auto nb_int_points = getGaussPts().size2();
const auto nb_dofs = data.getIndices().size();
const auto nb_base_func = data.getN().size2();
auto t_l = getFTensor2FromMat<DIM1, DIM2>(*lPtr);
auto t_l_dot = getFTensor2FromMat<DIM1, DIM2>(*lDotPtr);
auto t_vel = getFTensor1FromMat<SPACE_DIM>(*velPtr);
auto t_base = data.getFTensor0N();
auto t_diff_base = data.getFTensor1DiffN<SPACE_DIM>();
auto t_w = getFTensor0IntegrationWeight();
for (auto gg = 0; gg != nb_int_points; ++gg) {
const auto alpha = t_w * getMeasure();
t_res0(I, J) = alpha * t_l_dot(I, J);
t_res1(i, I, J) = (alpha * t_l(I, J)) * t_vel(i);
++t_w;
++t_l;
++t_l_dot;
++t_vel;
auto &nf = this->locF;
auto t_nf = getFTensor2FromPtr<DIM1, DIM2>(&*nf.begin());
int rr = 0;
for (; rr != nb_dofs; ++rr) {
t_nf(I, J) += t_res0(I, J) * t_base - t_res1(i, I, J) * t_diff_base(i);
++t_base;
++t_diff_base;
++t_nf;
}
for (; rr < nb_base_func; ++rr) {
++t_base;
++t_diff_base;
}
}
}
EntData &col_data) {
const auto nb_int_points = getGaussPts().size2();
const auto nb_base_func = row_data.getN().size2();
const auto nb_row_dofs = row_data.getIndices().size();
const auto nb_col_dofs = col_data.getIndices().size();
auto t_vel = getFTensor1FromMat<SPACE_DIM>(*velPtr);
auto t_row_base = row_data.getFTensor0N();
auto t_row_diff_base = row_data.getFTensor1DiffN<SPACE_DIM>();
auto t_w = getFTensor0IntegrationWeight();
for (auto gg = 0; gg != nb_int_points; ++gg) {
const auto alpha = t_w * getMeasure();
const auto beta = alpha * getTSa();
++t_w;
auto &mat = this->locMat;
int rr = 0;
for (; rr != nb_row_dofs; ++rr) {
auto t_col_base = col_data.getFTensor0N(gg, 0);
auto t_mat = getFTensor2FromPtr<DIM1, DIM2>(&mat(rr * DIM1, 0));
for (int cc = 0; cc != nb_col_dofs; ++cc) {
t_mat(I, J) +=
(beta * t_row_base - alpha * (t_row_diff_base(i) * t_vel(i))) *
t_col_base;
++t_col_base;
++t_mat;
}
++t_row_base;
++t_row_diff_base;
}
for (; rr < nb_base_func; ++rr) {
++t_row_base;
++t_row_diff_base;
}
++t_vel;
}
}
// Collect data from side domain elements
CHKERR loopSideFaces("dFE", *sideFEPtr);
const auto in_the_loop =
sideFEPtr->nInTheLoop; // return number of elements on the side
auto not_side = [](auto s) {
return s == LEFT_SIDE ? RIGHT_SIDE : LEFT_SIDE;
};
auto get_ntensor = [](auto &base_mat) {
&*base_mat.data().begin());
};
if (in_the_loop > 0) {
// get normal of the face or edge
auto t_normal = getFTensor1Normal();
const auto nb_gauss_pts = getGaussPts().size2();
for (auto s0 : {LEFT_SIDE, RIGHT_SIDE}) {
// gent number of DOFs on the right side.
const auto nb_rows = sideDataPtr->indicesRowSideMap[s0].size();
if (nb_rows) {
resSkelton.resize(nb_rows, false);
resSkelton.clear();
// get orientation of the local element edge
const auto opposite_s0 = not_side(s0);
const auto sense_row = sideDataPtr->senseMap[s0];
#ifndef NDEBUG
const auto opposite_sense_row = sideDataPtr->senseMap[opposite_s0];
if (sense_row * opposite_sense_row > 0)
SETERRQ(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"Should be opposite sign");
#endif
// iterate the side cols
const auto nb_row_base_functions =
sideDataPtr->rowBaseSideMap[s0].size2();
auto t_w = getFTensor0IntegrationWeight();
auto arr_t_l = make_array(
getFTensor2FromMat<DIM1, DIM2>(sideDataPtr->lVec[LEFT_SIDE]),
getFTensor2FromMat<DIM1, DIM2>(sideDataPtr->lVec[RIGHT_SIDE]));
auto arr_t_vel = make_array(
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[LEFT_SIDE]),
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[RIGHT_SIDE]));
auto next = [&]() {
for (auto &t_l : arr_t_l)
++t_l;
for (auto &t_vel : arr_t_vel)
++t_vel;
};
#ifndef NDEBUG
if (nb_gauss_pts != sideDataPtr->rowBaseSideMap[s0].size1())
SETERRQ(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"Inconsistent number of DOFs");
#endif
auto t_row_base = get_ntensor(sideDataPtr->rowBaseSideMap[s0]);
for (int gg = 0; gg != nb_gauss_pts; ++gg) {
t_vel(i) = (arr_t_vel[LEFT_SIDE](i) + arr_t_vel[RIGHT_SIDE](i)) / 2.;
const auto dot = sense_row * (t_normal(i) * t_vel(i));
const auto l_upwind_side = (dot > 0) ? s0 : opposite_s0;
const auto l_upwind = arr_t_l[l_upwind_side];
t_res(I, J) = t_w * dot * l_upwind(I, J);
next();
++t_w;
auto t_res_skeleton =
getFTensor2FromPtr<DIM1, DIM2>(&*resSkelton.data().begin());
auto rr = 0;
for (; rr != nb_rows; ++rr) {
t_res_skeleton(I, J) += t_row_base * t_res(I, J);
++t_row_base;
++t_res_skeleton;
}
for (; rr < nb_row_base_functions; ++rr) {
++t_row_base;
}
}
// assemble local operator vector to global vector
CHKERR ::VecSetValues(getTSf(),
sideDataPtr->indicesRowSideMap[s0].size(),
&*sideDataPtr->indicesRowSideMap[s0].begin(),
}
}
}
}
// Collect data from side domain elements
CHKERR loopSideFaces("dFE", *sideFEPtr);
const auto in_the_loop =
sideFEPtr->nInTheLoop; // return number of elements on the side
auto not_side = [](auto s) {
return s == LEFT_SIDE ? RIGHT_SIDE : LEFT_SIDE;
};
auto get_ntensor = [](auto &base_mat) {
&*base_mat.data().begin());
};
if (in_the_loop > 0) {
// get normal of the face or edge
auto t_normal = getFTensor1Normal();
const auto nb_gauss_pts = getGaussPts().size2();
for (auto s0 : {LEFT_SIDE, RIGHT_SIDE}) {
// gent number of DOFs on the right side.
const auto nb_rows = sideDataPtr->indicesRowSideMap[s0].size();
if (nb_rows) {
// get orientation of the local element edge
const auto opposite_s0 = not_side(s0);
const auto sense_row = sideDataPtr->senseMap[s0];
// iterate the side cols
const auto nb_row_base_functions =
sideDataPtr->rowBaseSideMap[s0].size2();
for (auto s1 : {LEFT_SIDE, RIGHT_SIDE}) {
// gent number of DOFs on the right side.
const auto nb_cols = sideDataPtr->indicesColSideMap[s1].size();
const auto sense_col = sideDataPtr->senseMap[s1];
// resize local element matrix
matSkeleton.resize(nb_rows, nb_cols, false);
matSkeleton.clear();
auto t_w = getFTensor0IntegrationWeight();
auto arr_t_vel = make_array(
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[LEFT_SIDE]),
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[RIGHT_SIDE]));
auto next = [&]() {
for (auto &t_vel : arr_t_vel)
++t_vel;
};
auto t_row_base = get_ntensor(sideDataPtr->rowBaseSideMap[s0]);
for (int gg = 0; gg != nb_gauss_pts; ++gg) {
t_vel(i) =
(arr_t_vel[LEFT_SIDE](i) + arr_t_vel[RIGHT_SIDE](i)) / 2.;
const auto dot = sense_row * (t_normal(i) * t_vel(i));
const auto l_upwind_side = (dot > 0) ? s0 : opposite_s0;
const auto sense_upwind = sideDataPtr->senseMap[l_upwind_side];
t_res(I, J) = t_w * dot;
next();
++t_w;
auto rr = 0;
if (s1 == l_upwind_side) {
for (; rr != nb_rows; ++rr) {
auto get_ntensor = [](auto &base_mat, auto gg, auto bb) {
double *ptr = &base_mat(gg, bb);
};
auto t_col_base =
get_ntensor(sideDataPtr->colBaseSideMap[s1], gg, 0);
auto t_mat_skeleton =
getFTensor2FromPtr<DIM1, DIM2>(&matSkeleton(rr * DIM1, 0));
t_res_row(I, J) = t_res(I, J) * t_row_base;
++t_row_base;
// iterate columns
for (size_t cc = 0; cc != nb_cols; ++cc) {
t_mat_skeleton(I, J) += t_res_row(I, J) * t_col_base;
++t_col_base;
++t_mat_skeleton;
}
}
}
for (; rr < nb_row_base_functions; ++rr) {
++t_row_base;
}
}
// assemble system
CHKERR ::MatSetValues(getTSB(),
sideDataPtr->indicesRowSideMap[s0].size(),
&*sideDataPtr->indicesRowSideMap[s0].begin(),
sideDataPtr->indicesColSideMap[s1].size(),
&*sideDataPtr->indicesColSideMap[s1].begin(),
}
}
}
}
}
LevelSet::getZeroLevelVelOp(boost::shared_ptr<MatrixDouble> vel_ptr) {
auto get_parent_vel_this = [&]() {
auto parent_fe_ptr = boost::make_shared<DomianParentEle>(mField);
parent_fe_ptr->getOpPtrVector(), {potential_velocity_space});
parent_fe_ptr->getOpPtrVector().push_back(
"V", vel_ptr));
return parent_fe_ptr;
};
auto get_parents_vel_fe_ptr = [&](auto this_fe_ptr) {
std::vector<boost::shared_ptr<DomianParentEle>> parents_elems_ptr_vec;
for (int l = 0; l <= nb_levels; ++l)
parents_elems_ptr_vec.emplace_back(
boost::make_shared<DomianParentEle>(mField));
for (auto l = 1; l <= nb_levels; ++l) {
parents_elems_ptr_vec[l - 1]->getOpPtrVector().push_back(
new OpRunParent(parents_elems_ptr_vec[l], BitRefLevel().set(),
BitRefLevel().set(0).flip(), this_fe_ptr,
BitRefLevel().set(0), BitRefLevel().set()));
}
return parents_elems_ptr_vec[0];
};
auto this_fe_ptr = get_parent_vel_this();
auto parent_fe_ptr = get_parents_vel_fe_ptr(this_fe_ptr);
return new OpRunParent(parent_fe_ptr, BitRefLevel().set(),
BitRefLevel().set(0).flip(), this_fe_ptr,
BitRefLevel().set(0), BitRefLevel().set());
}
auto pip = mField.getInterface<PipelineManager>(); // get interface to
// pipeline manager
pip->getOpDomainLhsPipeline().clear();
pip->getOpDomainRhsPipeline().clear();
pip->setDomainLhsIntegrationRule([](int, int, int o) { return 3 * o; });
pip->setDomainRhsIntegrationRule([](int, int, int o) { return 3 * o; });
pip->getDomainLhsFE()->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
pip->getDomainRhsFE()->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
auto l_ptr = boost::make_shared<MatrixDouble>();
auto l_dot_ptr = boost::make_shared<MatrixDouble>();
auto vel_ptr = boost::make_shared<MatrixDouble>();
pip->getOpDomainRhsPipeline().push_back(getZeroLevelVelOp(vel_ptr));
{L2});
pip->getOpDomainRhsPipeline().push_back(
pip->getOpDomainRhsPipeline().push_back(
pip->getOpDomainRhsPipeline().push_back(
new OpRhsDomain("L", l_ptr, l_dot_ptr, vel_ptr));
pip->getOpDomainLhsPipeline().push_back(getZeroLevelVelOp(vel_ptr));
{L2});
pip->getOpDomainLhsPipeline().push_back(new OpLhsDomain("L", vel_ptr));
}
boost::shared_ptr<FaceSideEle>
LevelSet::getSideFE(boost::shared_ptr<SideData> side_data_ptr) {
auto l_ptr = boost::make_shared<MatrixDouble>();
auto vel_ptr = boost::make_shared<MatrixDouble>();
struct OpSideData : public FaceSideEleOp {
OpSideData(boost::shared_ptr<SideData> side_data_ptr)
: FaceSideEleOp("L", "L", FaceSideEleOp::OPROWCOL),
sideDataPtr(side_data_ptr) {
std::fill(&doEntities[MBVERTEX], &doEntities[MBMAXTYPE], false);
for (auto t = moab::CN::TypeDimensionMap[FE_DIM].first;
t <= moab::CN::TypeDimensionMap[FE_DIM].second; ++t)
doEntities[t] = true;
sYmm = false;
}
MoFEMErrorCode doWork(int row_side, int col_side, EntityType row_type,
EntityType col_type, EntData &row_data,
EntData &col_data) {
if ((CN::Dimension(row_type) == FE_DIM) &&
(CN::Dimension(col_type) == FE_DIM)) {
auto reset = [&](auto nb_in_loop) {
sideDataPtr->feSideHandle[nb_in_loop] = 0;
sideDataPtr->indicesRowSideMap[nb_in_loop].clear();
sideDataPtr->indicesColSideMap[nb_in_loop].clear();
sideDataPtr->rowBaseSideMap[nb_in_loop].clear();
sideDataPtr->colBaseSideMap[nb_in_loop].clear();
sideDataPtr->senseMap[nb_in_loop] = 0;
};
const auto nb_in_loop = getFEMethod()->nInTheLoop;
if (nb_in_loop == 0)
for (auto s : {0, 1})
reset(s);
sideDataPtr->currentFESide = nb_in_loop;
sideDataPtr->senseMap[nb_in_loop] = getSkeletonSense();
} else {
SETERRQ(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY, "Should not happen");
}
};
private:
boost::shared_ptr<SideData> sideDataPtr;
};
struct OpSideDataOnParent : public DomainEleOp {
OpSideDataOnParent(boost::shared_ptr<SideData> side_data_ptr,
boost::shared_ptr<MatrixDouble> l_ptr,
boost::shared_ptr<MatrixDouble> vel_ptr)
: DomainEleOp("L", "L", DomainEleOp::OPROWCOL),
sideDataPtr(side_data_ptr), lPtr(l_ptr), velPtr(vel_ptr) {
std::fill(&doEntities[MBVERTEX], &doEntities[MBMAXTYPE], false);
for (auto t = moab::CN::TypeDimensionMap[FE_DIM].first;
t <= moab::CN::TypeDimensionMap[FE_DIM].second; ++t)
doEntities[t] = true;
sYmm = false;
}
MoFEMErrorCode doWork(int row_side, int col_side, EntityType row_type,
EntityType col_type, EntData &row_data,
EntData &col_data) {
if ((CN::Dimension(row_type) == FE_DIM) &&
(CN::Dimension(col_type) == FE_DIM)) {
const auto nb_in_loop = sideDataPtr->currentFESide;
sideDataPtr->feSideHandle[nb_in_loop] = getFEEntityHandle();
sideDataPtr->indicesRowSideMap[nb_in_loop] = row_data.getIndices();
sideDataPtr->indicesColSideMap[nb_in_loop] = col_data.getIndices();
sideDataPtr->rowBaseSideMap[nb_in_loop] = row_data.getN();
sideDataPtr->colBaseSideMap[nb_in_loop] = col_data.getN();
(sideDataPtr->lVec)[nb_in_loop] = *lPtr;
(sideDataPtr->velMat)[nb_in_loop] = *velPtr;
#ifndef NDEBUG
if ((sideDataPtr->lVec)[nb_in_loop].size2() !=
(sideDataPtr->velMat)[nb_in_loop].size2())
SETERRQ2(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"Wrong number of integaration pts %d != %d",
(sideDataPtr->lVec)[nb_in_loop].size2(),
(sideDataPtr->velMat)[nb_in_loop].size2());
if ((sideDataPtr->velMat)[nb_in_loop].size1() != SPACE_DIM)
SETERRQ1(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"Wrong size of velocity vector size = %d",
(sideDataPtr->velMat)[nb_in_loop].size1());
#endif
if (!nb_in_loop) {
(sideDataPtr->lVec)[1] = sideDataPtr->lVec[0];
(sideDataPtr->velMat)[1] = (sideDataPtr->velMat)[0];
} else {
#ifndef NDEBUG
if (sideDataPtr->rowBaseSideMap[0].size1() !=
sideDataPtr->rowBaseSideMap[1].size1()) {
SETERRQ2(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"Wrong number of integration pt %d != %d",
sideDataPtr->rowBaseSideMap[0].size1(),
sideDataPtr->rowBaseSideMap[1].size1());
}
if (sideDataPtr->colBaseSideMap[0].size1() !=
sideDataPtr->colBaseSideMap[1].size1()) {
SETERRQ(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"Wrong number of integration pt");
}
#endif
}
} else {
SETERRQ(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY, "Should not happen");
}
};
private:
boost::shared_ptr<SideData> sideDataPtr;
boost::shared_ptr<MatrixDouble> lPtr;
boost::shared_ptr<MatrixDouble> velPtr;
};
// Calculate fields on param mesh bit element
auto get_parent_this = [&]() {
auto parent_fe_ptr = boost::make_shared<DomianParentEle>(mField);
parent_fe_ptr->getOpPtrVector(), {L2});
parent_fe_ptr->getOpPtrVector().push_back(
parent_fe_ptr->getOpPtrVector().push_back(
new OpSideDataOnParent(side_data_ptr, l_ptr, vel_ptr));
return parent_fe_ptr;
};
auto get_parents_fe_ptr = [&](auto this_fe_ptr) {
std::vector<boost::shared_ptr<DomianParentEle>> parents_elems_ptr_vec;
for (int l = 0; l <= nb_levels; ++l)
parents_elems_ptr_vec.emplace_back(
boost::make_shared<DomianParentEle>(mField));
for (auto l = 1; l <= nb_levels; ++l) {
parents_elems_ptr_vec[l - 1]->getOpPtrVector().push_back(
new OpRunParent(parents_elems_ptr_vec[l], BitRefLevel().set(),
BitRefLevel().set(current_bit).flip(), this_fe_ptr,
BitRefLevel().set(current_bit), BitRefLevel().set()));
}
return parents_elems_ptr_vec[0];
};
// Create aliased shared pointers, all elements are destroyed if side_fe_ptr
// is destroyed
auto get_side_fe_ptr = [&]() {
auto side_fe_ptr = boost::make_shared<FaceSideEle>(mField);
auto this_fe_ptr = get_parent_this();
auto parent_fe_ptr = get_parents_fe_ptr(this_fe_ptr);
side_fe_ptr->getOpPtrVector().push_back(new OpSideData(side_data_ptr));
side_fe_ptr->getOpPtrVector().push_back(getZeroLevelVelOp(vel_ptr));
side_fe_ptr->getOpPtrVector().push_back(
new OpRunParent(parent_fe_ptr, BitRefLevel().set(),
BitRefLevel().set(current_bit).flip(), this_fe_ptr,
BitRefLevel().set(current_bit), BitRefLevel().set()));
return side_fe_ptr;
};
return get_side_fe_ptr();
};
auto pip = mField.getInterface<PipelineManager>(); // get interface to
pip->getOpSkeletonLhsPipeline().clear();
pip->getOpSkeletonRhsPipeline().clear();
pip->setSkeletonLhsIntegrationRule([](int, int, int o) { return 18; });
pip->setSkeletonRhsIntegrationRule([](int, int, int o) { return 18; });
pip->getSkeletonLhsFE()->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
pip->getSkeletonRhsFE()->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
auto side_data_ptr = boost::make_shared<SideData>();
auto side_fe_ptr = getSideFE(side_data_ptr);
pip->getOpSkeletonRhsPipeline().push_back(
new OpRhsSkeleton(side_data_ptr, side_fe_ptr));
pip->getOpSkeletonLhsPipeline().push_back(
new OpLhsSkeleton(side_data_ptr, side_fe_ptr));
}
std::tuple<double, Tag> LevelSet::evaluateError() {
struct OpErrorSkel : BoundaryEleOp {
OpErrorSkel(boost::shared_ptr<FaceSideEle> side_fe_ptr,
boost::shared_ptr<SideData> side_data_ptr,
SmartPetscObj<Vec> error_sum_ptr, Tag th_error)
sideFEPtr(side_fe_ptr), sideDataPtr(side_data_ptr),
errorSumPtr(error_sum_ptr), thError(th_error) {}
MoFEMErrorCode doWork(int side, EntityType type, EntData &data) {
// Collect data from side domain elements
CHKERR loopSideFaces("dFE", *sideFEPtr);
const auto in_the_loop =
sideFEPtr->nInTheLoop; // return number of elements on the side
auto not_side = [](auto s) {
return s == LEFT_SIDE ? RIGHT_SIDE : LEFT_SIDE;
};
auto nb_gauss_pts = getGaussPts().size2();
for (auto s : {LEFT_SIDE, RIGHT_SIDE}) {
auto arr_t_l = make_array(
getFTensor2FromMat<DIM1, DIM2>(sideDataPtr->lVec[LEFT_SIDE]),
getFTensor2FromMat<DIM1, DIM2>(sideDataPtr->lVec[RIGHT_SIDE]));
auto arr_t_vel = make_array(
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[LEFT_SIDE]),
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[RIGHT_SIDE]));
auto next = [&]() {
for (auto &t_l : arr_t_l)
++t_l;
for (auto &t_vel : arr_t_vel)
++t_vel;
};
double e = 0;
auto t_w = getFTensor0IntegrationWeight();
for (int gg = 0; gg != nb_gauss_pts; ++gg) {
t_diff(I, J) = arr_t_l[LEFT_SIDE](I, J) - arr_t_l[RIGHT_SIDE](I, J);
e += t_w * getMeasure() * t_diff(I, J) * t_diff(I, J);
next();
++t_w;
}
e = std::sqrt(e);
moab::Interface &moab =
getNumeredEntFiniteElementPtr()->getBasicDataPtr()->moab;
const void *tags_ptr[2];
CHKERR moab.tag_get_by_ptr(thError, sideDataPtr->feSideHandle.data(), 2,
tags_ptr);
for (auto ff : {0, 1}) {
*((double *)tags_ptr[ff]) += e;
}
};
}
private:
boost::shared_ptr<FaceSideEle> sideFEPtr;
boost::shared_ptr<SideData> sideDataPtr;
SmartPetscObj<Vec> errorSumPtr;
Tag thError;
};
auto error_sum_ptr = createVectorMPI(mField.get_comm(), PETSC_DECIDE, 1);
Tag th_error;
double def_val = 0;
CHKERR mField.get_moab().tag_get_handle("Error", 1, MB_TYPE_DOUBLE, th_error,
MB_TAG_CREAT | MB_TAG_SPARSE,
&def_val);
auto clear_tags = [&]() {
Range fe_ents;
CHKERR mField.get_moab().get_entities_by_dimension(0, FE_DIM, fe_ents);
double zero;
CHKERR mField.get_moab().tag_clear_data(th_error, fe_ents, &zero);
};
auto evaluate_error = [&]() {
auto skel_fe = boost::make_shared<BoundaryEle>(mField);
skel_fe->getRuleHook = [](int, int, int o) { return 3 * o; };
auto side_data_ptr = boost::make_shared<SideData>();
auto side_fe_ptr = getSideFE(side_data_ptr);
skel_fe->getOpPtrVector().push_back(
new OpErrorSkel(side_fe_ptr, side_data_ptr, error_sum_ptr, th_error));
skel_fe->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
simple->getSkeletonFEName(), skel_fe);
};
auto assemble_and_sum = [](auto vec) {
CHK_THROW_MESSAGE(VecAssemblyBegin(vec), "assemble");
CHK_THROW_MESSAGE(VecAssemblyEnd(vec), "assemble");
double sum;
CHK_THROW_MESSAGE(VecSum(vec, &sum), "assemble");
return sum;
};
auto propagate_error_to_parents = [&]() {
auto &moab = mField.get_moab();
auto fe_ptr = boost::make_shared<FEMethod>();
fe_ptr->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
fe_ptr->preProcessHook = []() { return 0; };
fe_ptr->postProcessHook = []() { return 0; };
fe_ptr->operatorHook = [&]() {
auto fe_ent = fe_ptr->numeredEntFiniteElementPtr->getEnt();
auto parent = fe_ptr->numeredEntFiniteElementPtr->getParentEnt();
auto th_parent = fe_ptr->numeredEntFiniteElementPtr->getBasicDataPtr()
->th_RefParentHandle;
double error;
CHKERR moab.tag_get_data(th_error, &fe_ent, 1, &error);
[&](auto fe_ent, auto error) {
double *e_ptr;
CHKERR moab.tag_get_by_ptr(th_error, &fe_ent, 1,
(const void **)&e_ptr);
(*e_ptr) += error;
EntityHandle parent;
CHKERR moab.tag_get_data(th_parent, &fe_ent, 1, &parent);
if (parent != fe_ent && parent)
};
};
CHKERR DMoFEMLoopFiniteElements(simple->getDM(), simple->getDomainFEName(),
fe_ptr);
};
CHK_THROW_MESSAGE(clear_tags(), "clear error tags");
CHK_THROW_MESSAGE(evaluate_error(), "evaluate error");
CHK_THROW_MESSAGE(propagate_error_to_parents(), "propagate error");
return std::make_tuple(assemble_and_sum(error_sum_ptr), th_error);
}
/**
* @brief test side element
*
* Check consistency between volume and skeleton integral
*
* @return MoFEMErrorCode
*/
/**
* @brief calculate volume
*
*/
struct DivergenceVol : public DomainEleOp {
DivergenceVol(boost::shared_ptr<MatrixDouble> l_ptr,
boost::shared_ptr<MatrixDouble> vel_ptr,
: DomainEleOp("L", DomainEleOp::OPROW), lPtr(l_ptr), velPtr(vel_ptr),
divVec(div_vec) {}
MoFEMErrorCode doWork(int side, EntityType type,
const auto nb_dofs = data.getIndices().size();
if (nb_dofs) {
const auto nb_gauss_pts = getGaussPts().size2();
const auto t_w = getFTensor0IntegrationWeight();
auto t_diff = data.getFTensor1DiffN<SPACE_DIM>();
auto t_l = getFTensor2FromMat<DIM1, DIM2>(*lPtr);
auto t_vel = getFTensor1FromMat<SPACE_DIM>(*velPtr);
double div = 0;
for (auto gg = 0; gg != nb_gauss_pts; ++gg) {
for (int rr = 0; rr != nb_dofs; ++rr) {
div += getMeasure() * t_w * t_l(I, I) * (t_diff(i) * t_vel(i));
++t_diff;
}
++t_w;
++t_l;
++t_vel;
}
}
}
private:
boost::shared_ptr<MatrixDouble> lPtr;
boost::shared_ptr<MatrixDouble> velPtr;
};
/**
* @brief calculate skeleton integral
*
*/
struct DivergenceSkeleton : public BoundaryEleOp {
DivergenceSkeleton(boost::shared_ptr<SideData> side_data_ptr,
boost::shared_ptr<FaceSideEle> side_fe_ptr,
sideDataPtr(side_data_ptr), sideFEPtr(side_fe_ptr), divVec(div_vec) {}
MoFEMErrorCode doWork(int side, EntityType type,
auto get_ntensor = [](auto &base_mat) {
&*base_mat.data().begin());
};
auto not_side = [](auto s) {
return s == LEFT_SIDE ? RIGHT_SIDE : LEFT_SIDE;
};
// Collect data from side domain elements
CHKERR loopSideFaces("dFE", *sideFEPtr);
const auto in_the_loop =
sideFEPtr->nInTheLoop; // return number of elements on the side
auto t_normal = getFTensor1Normal();
const auto nb_gauss_pts = getGaussPts().size2();
for (auto s0 : {LEFT_SIDE, RIGHT_SIDE}) {
const auto nb_dofs = sideDataPtr->indicesRowSideMap[s0].size();
if (nb_dofs) {
auto t_base = get_ntensor(sideDataPtr->rowBaseSideMap[s0]);
auto nb_row_base_functions = sideDataPtr->rowBaseSideMap[s0].size2();
auto side_sense = sideDataPtr->senseMap[s0];
auto opposite_s0 = not_side(s0);
auto arr_t_l = make_array(
getFTensor2FromMat<DIM1, DIM2>(sideDataPtr->lVec[LEFT_SIDE]),
getFTensor2FromMat<DIM1, DIM2>(sideDataPtr->lVec[RIGHT_SIDE]));
auto arr_t_vel = make_array(
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[LEFT_SIDE]),
getFTensor1FromMat<SPACE_DIM>(sideDataPtr->velMat[RIGHT_SIDE]));
auto next = [&]() {
for (auto &t_l : arr_t_l)
++t_l;
for (auto &t_vel : arr_t_vel)
++t_vel;
};
double div = 0;
auto t_w = getFTensor0IntegrationWeight();
for (int gg = 0; gg != nb_gauss_pts; ++gg) {
t_vel(i) =
(arr_t_vel[LEFT_SIDE](i) + arr_t_vel[RIGHT_SIDE](i)) / 2.;
const auto dot = (t_normal(i) * t_vel(i)) * side_sense;
const auto l_upwind_side = (dot > 0) ? s0 : opposite_s0;
const auto l_upwind =
arr_t_l[l_upwind_side]; //< assume that field is continues,
// initialisation field has to be smooth
// and exactly approximated by approx
// base
auto res = t_w * l_upwind(I, I) * dot;
++t_w;
next();
int rr = 0;
for (; rr != nb_dofs; ++rr) {
div += t_base * res;
++t_base;
}
for (; rr < nb_row_base_functions; ++rr) {
++t_base;
}
}
}
if (!in_the_loop)
break;
}
}
private:
boost::shared_ptr<SideData> sideDataPtr;
boost::shared_ptr<FaceSideEle> sideFEPtr;
boost::shared_ptr<MatrixDouble> velPtr;
};
auto vol_fe = boost::make_shared<DomainEle>(mField);
auto skel_fe = boost::make_shared<BoundaryEle>(mField);
vol_fe->getRuleHook = [](int, int, int o) { return 3 * o; };
skel_fe->getRuleHook = [](int, int, int o) { return 3 * o; };
auto div_vol_vec = createVectorMPI(mField.get_comm(), PETSC_DECIDE, 1);
auto div_skel_vec = createVectorMPI(mField.get_comm(), PETSC_DECIDE, 1);
auto l_ptr = boost::make_shared<MatrixDouble>();
auto vel_ptr = boost::make_shared<MatrixDouble>();
auto side_data_ptr = boost::make_shared<SideData>();
auto side_fe_ptr = getSideFE(side_data_ptr);
{L2});
vol_fe->getOpPtrVector().push_back(
vol_fe->getOpPtrVector().push_back(getZeroLevelVelOp(vel_ptr));
vol_fe->getOpPtrVector().push_back(
new DivergenceVol(l_ptr, vel_ptr, div_vol_vec));
skel_fe->getOpPtrVector().push_back(
new DivergenceSkeleton(side_data_ptr, side_fe_ptr, div_skel_vec));
auto dm = simple->getDM();
/**
* Set up problem such that gradient of level set field is orthogonal to
* velocity field. Then volume and skeleton integral should yield the same
* value.
*/
[](double x, double y, double) { return x - y; });
[](double x, double y, double) { return x - y; });
vol_fe->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
skel_fe->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
CHKERR DMoFEMLoopFiniteElements(dm, simple->getDomainFEName(), vol_fe);
CHKERR DMoFEMLoopFiniteElements(dm, simple->getSkeletonFEName(), skel_fe);
CHKERR DMoFEMLoopFiniteElements(dm, simple->getBoundaryFEName(), skel_fe);
auto assemble_and_sum = [](auto vec) {
CHK_THROW_MESSAGE(VecAssemblyBegin(vec), "assemble");
CHK_THROW_MESSAGE(VecAssemblyEnd(vec), "assemble");
double sum;
CHK_THROW_MESSAGE(VecSum(vec, &sum), "assemble");
return sum;
};
auto div_vol = assemble_and_sum(div_vol_vec);
auto div_skel = assemble_and_sum(div_skel_vec);
auto eps = std::abs((div_vol - div_skel) / (div_vol + div_skel));
MOFEM_LOG("WORLD", Sev::inform) << "Testing divergence volume: " << div_vol;
MOFEM_LOG("WORLD", Sev::inform) << "Testing divergence skeleton: " << div_skel
<< " relative difference: " << eps;
constexpr double eps_err = 1e-6;
if (eps > eps_err)
SETERRQ(PETSC_COMM_SELF, MOFEM_DATA_INCONSISTENCY,
"No consistency between skeleton integral and volume integral");
};
// get operators tester
auto opt = mField.getInterface<OperatorsTester>(); // get interface to
// OperatorsTester
auto pip = mField.getInterface<PipelineManager>(); // get interface to
// pipeline manager
auto post_proc = [&](auto dm, auto f_res, auto out_name) {
auto post_proc_fe =
boost::make_shared<PostProcBrokenMeshInMoab<DomainEle>>(mField);
if constexpr (DIM1 == 1 && DIM2 == 1) {
auto l_vec = boost::make_shared<VectorDouble>();
post_proc_fe->getOpPtrVector().push_back(
new OpCalculateScalarFieldValues("L", l_vec, f_res));
post_proc_fe->getOpPtrVector().push_back(
new OpPPMap(
post_proc_fe->getPostProcMesh(), post_proc_fe->getMapGaussPts(),
{{"L", l_vec}},
{},
{}, {})
);
}
CHKERR DMoFEMLoopFiniteElements(dm, simple->getDomainFEName(),
post_proc_fe);
post_proc_fe->writeFile(out_name);
};
constexpr double eps = 1e-4;
auto x =
opt->setRandomFields(simple->getDM(), {{"L", {-1, 1}}, {"V", {-1, 1}}});
auto dot_x = opt->setRandomFields(simple->getDM(), {{"L", {-1, 1}}});
auto diff_x = opt->setRandomFields(simple->getDM(), {{"L", {-1, 1}}});
auto test_domain_ops = [&](auto fe_name, auto lhs_pipeline,
auto rhs_pipeline) {
auto diff_res = opt->checkCentralFiniteDifference(
simple->getDM(), fe_name, rhs_pipeline, lhs_pipeline, x, dot_x,
SmartPetscObj<Vec>(), diff_x, 0, 1, eps);
if constexpr (debug) {
// Example how to plot direction in direction diff_x. If instead
// directionalCentralFiniteDifference(...) diff_res is used, then error
// on directive is plotted.
CHKERR post_proc(simple->getDM(), diff_res, "tangent_op_error.h5m");
}
// Calculate norm of difference between directive calculated from finite
// difference, and tangent matrix.
double fnorm;
CHKERR VecNorm(diff_res, NORM_2, &fnorm);
MOFEM_LOG_C("LevelSet", Sev::inform,
"Test consistency of tangent matrix %3.4e", fnorm);
constexpr double err = 1e-9;
if (fnorm > err)
SETERRQ1(PETSC_COMM_WORLD, MOFEM_ATOM_TEST_INVALID,
"Norm of directional derivative too large err = %3.4e", fnorm);
};
CHKERR test_domain_ops(simple->getDomainFEName(), pip->getDomainLhsFE(),
pip->getDomainRhsFE());
CHKERR test_domain_ops(simple->getSkeletonFEName(), pip->getSkeletonLhsFE(),
pip->getSkeletonRhsFE());
};
boost::function<double(double, double, double)> level_fun) {
// get operators tester
auto pip = mField.getInterface<PipelineManager>(); // get interface to
// pipeline manager
auto prb_mng = mField.getInterface<ProblemsManager>();
boost::shared_ptr<FEMethod> lhs_fe = boost::make_shared<DomainEle>(mField);
boost::shared_ptr<FEMethod> rhs_fe = boost::make_shared<DomainEle>(mField);
auto swap_fe = [&]() {
lhs_fe.swap(pip->getDomainLhsFE());
rhs_fe.swap(pip->getDomainRhsFE());
};
swap_fe();
pip->setDomainLhsIntegrationRule([](int, int, int o) { return 3 * o; });
pip->setDomainRhsIntegrationRule([](int, int, int o) { return 3 * o; });
auto sub_dm = createDM(mField.get_comm(), "DMMOFEM");
CHKERR DMMoFEMCreateSubDM(sub_dm, simple->getDM(), "LEVELSET_POJECTION");
CHKERR DMMoFEMSetDestroyProblem(sub_dm, PETSC_TRUE);
CHKERR DMMoFEMSetSquareProblem(sub_dm, PETSC_TRUE);
CHKERR DMSetUp(sub_dm);
remove_mask.flip(); // DOFs which are not on bit_domain_ele should be removed
CHKERR prb_mng->removeDofsOnEntities("LEVELSET_POJECTION", "L",
auto test_bit_ele = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
pip->getDomainLhsFE()->exeTestHook = test_bit_ele;
pip->getDomainRhsFE()->exeTestHook = test_bit_ele;
{L2});
{L2});
pip->getOpDomainLhsPipeline().push_back(new OpMassLL("L", "L"));
pip->getOpDomainRhsPipeline().push_back(new OpSourceL("L", level_fun));
CHKERR mField.getInterface<FieldBlas>()->setField(0, "L");
auto ksp = pip->createKSP(sub_dm);
CHKERR KSPSetDM(ksp, sub_dm);
CHKERR KSPSetFromOptions(ksp);
CHKERR KSPSetUp(ksp);
auto L = createDMVector(sub_dm);
auto F = vectorDuplicate(L);
CHKERR KSPSolve(ksp, F, L);
CHKERR VecGhostUpdateBegin(L, INSERT_VALUES, SCATTER_FORWARD);
CHKERR VecGhostUpdateEnd(L, INSERT_VALUES, SCATTER_FORWARD);
CHKERR DMoFEMMeshToLocalVector(sub_dm, L, INSERT_VALUES, SCATTER_REVERSE);
auto [error, th_error] = evaluateError();
MOFEM_LOG("LevelSet", Sev::inform) << "Error indicator " << error;
#ifndef NDEBUG
auto fe_meshset =
std::vector<Tag> tags{th_error};
CHKERR mField.get_moab().write_file("error.h5m", "MOAB",
"PARALLEL=WRITE_PART", &fe_meshset, 1,
&*tags.begin(), tags.size());
#endif
auto post_proc = [&](auto dm, auto out_name, auto th_error) {
auto post_proc_fe =
boost::make_shared<PostProcBrokenMeshInMoab<DomainEle>>(mField);
post_proc_fe->setTagsToTransfer({th_error});
post_proc_fe->exeTestHook = test_bit_ele;
if constexpr (DIM1 == 1 && DIM2 == 1) {
auto l_vec = boost::make_shared<VectorDouble>();
post_proc_fe->getOpPtrVector(), {L2});
post_proc_fe->getOpPtrVector().push_back(
new OpCalculateScalarFieldValues("L", l_vec));
post_proc_fe->getOpPtrVector().push_back(
post_proc_fe->getOpPtrVector().push_back(
new OpPPMap(
post_proc_fe->getPostProcMesh(), post_proc_fe->getMapGaussPts(),
{{"L", l_vec}},
{}, {})
);
}
CHKERR DMoFEMLoopFiniteElements(dm, simple->getDomainFEName(),
post_proc_fe);
post_proc_fe->writeFile(out_name);
};
if constexpr (debug)
CHKERR post_proc(sub_dm, "initial_level_set.h5m", th_error);
swap_fe();
}
boost::function<double(double, double, double)> vel_fun) {
// get operators tester
auto pip = mField.getInterface<PipelineManager>(); // get interface to
// pipeline manager
auto prb_mng = mField.getInterface<ProblemsManager>();
boost::shared_ptr<FEMethod> lhs_fe = boost::make_shared<DomainEle>(mField);
boost::shared_ptr<FEMethod> rhs_fe = boost::make_shared<DomainEle>(mField);
auto swap_fe = [&]() {
lhs_fe.swap(pip->getDomainLhsFE());
rhs_fe.swap(pip->getDomainRhsFE());
};
swap_fe();
pip->setDomainLhsIntegrationRule([](int, int, int o) { return 3 * o; });
pip->setDomainRhsIntegrationRule([](int, int, int o) { return 3 * o; });
auto sub_dm = createDM(mField.get_comm(), "DMMOFEM");
CHKERR DMMoFEMCreateSubDM(sub_dm, simple->getDM(), "VELOCITY_PROJECTION");
CHKERR DMMoFEMSetDestroyProblem(sub_dm, PETSC_TRUE);
CHKERR DMMoFEMSetSquareProblem(sub_dm, PETSC_TRUE);
CHKERR DMSetUp(sub_dm);
// Velocities are calculated only on corse mesh
remove_mask.flip(); // DOFs which are not on bit_domain_ele should be removed
CHKERR prb_mng->removeDofsOnEntities("VELOCITY_PROJECTION", "V",
auto test_bit = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(0);
};
pip->getDomainLhsFE()->exeTestHook = test_bit;
pip->getDomainRhsFE()->exeTestHook = test_bit;
{potential_velocity_space});
{potential_velocity_space});
pip->getOpDomainLhsPipeline().push_back(new OpMassVV("V", "V"));
pip->getOpDomainRhsPipeline().push_back(new OpSourceV("V", vel_fun));
auto ksp = pip->createKSP(sub_dm);
CHKERR KSPSetDM(ksp, sub_dm);
CHKERR KSPSetFromOptions(ksp);
CHKERR KSPSetUp(ksp);
auto L = createDMVector(sub_dm);
auto F = vectorDuplicate(L);
CHKERR KSPSolve(ksp, F, L);
CHKERR VecGhostUpdateBegin(L, INSERT_VALUES, SCATTER_FORWARD);
CHKERR VecGhostUpdateEnd(L, INSERT_VALUES, SCATTER_FORWARD);
CHKERR DMoFEMMeshToLocalVector(sub_dm, L, INSERT_VALUES, SCATTER_REVERSE);
auto post_proc = [&](auto dm, auto out_name) {
auto post_proc_fe =
boost::make_shared<PostProcBrokenMeshInMoab<DomainEle>>(mField);
post_proc_fe->exeTestHook = test_bit;
if constexpr (FE_DIM == 2) {
post_proc_fe->getOpPtrVector(), {potential_velocity_space});
auto potential_vec = boost::make_shared<VectorDouble>();
auto velocity_mat = boost::make_shared<MatrixDouble>();
post_proc_fe->getOpPtrVector().push_back(
new OpCalculateScalarFieldValues("V", potential_vec));
post_proc_fe->getOpPtrVector().push_back(
SPACE_DIM>("V", velocity_mat));
post_proc_fe->getOpPtrVector().push_back(
new OpPPMap(
post_proc_fe->getPostProcMesh(), post_proc_fe->getMapGaussPts(),
{{"VelocityPotential", potential_vec}},
{{"Velocity", velocity_mat}},
{}, {})
);
} else {
SETERRQ(PETSC_COMM_SELF, MOFEM_NOT_IMPLEMENTED,
"3d case not implemented");
}
CHKERR DMoFEMLoopFiniteElements(dm, simple->getDomainFEName(),
post_proc_fe);
post_proc_fe->writeFile(out_name);
};
if constexpr (debug)
CHKERR post_proc(sub_dm, "initial_velocity_potential.h5m");
swap_fe();
}
// get operators tester
auto pip = mField.getInterface<PipelineManager>(); // get interface to
auto prb_mng = mField.getInterface<ProblemsManager>();
auto sub_dm = createDM(mField.get_comm(), "DMMOFEM");
CHKERR DMMoFEMSetDestroyProblem(sub_dm, PETSC_TRUE);
CHKERR DMMoFEMSetSquareProblem(sub_dm, PETSC_TRUE);
CHKERR DMSetUp(sub_dm);
remove_mask.flip(); // DOFs which are not on bit_domain_ele should be removed
auto post_proc_fe = boost::make_shared<PostProcEle>(mField);
Tag th_error;
double def_val = 0;
CHKERR mField.get_moab().tag_get_handle(
"Error", 1, MB_TYPE_DOUBLE, th_error, MB_TAG_CREAT | MB_TAG_SPARSE,
&def_val);
post_proc_fe->setTagsToTransfer({th_error});
post_proc_fe->exeTestHook = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
auto vel_ptr = boost::make_shared<MatrixDouble>();
post_proc_fe->getOpPtrVector(), {L2});
post_proc_fe->getOpPtrVector().push_back(getZeroLevelVelOp(vel_ptr));
if constexpr (DIM1 == 1 && DIM2 == 1) {
auto l_vec = boost::make_shared<VectorDouble>();
post_proc_fe->getOpPtrVector().push_back(
new OpCalculateScalarFieldValues("L", l_vec));
post_proc_fe->getOpPtrVector().push_back(
new OpPPMap(
post_proc_fe->getPostProcMesh(),
post_proc_fe->getMapGaussPts(),
{{"L", l_vec}},
{{"V", vel_ptr}},
{}, {})
);
}
return post_proc_fe;
};
auto set_time_monitor = [&](auto dm, auto ts) {
auto monitor_ptr = boost::make_shared<FEMethod>();
monitor_ptr->preProcessHook = []() { return 0; };
monitor_ptr->operatorHook = []() { return 0; };
monitor_ptr->postProcessHook = [&]() {
if (!post_proc_fe)
SETERRQ(PETSC_COMM_WORLD, MOFEM_DATA_INCONSISTENCY,
"Null pointer for post proc element");
CHKERR DMoFEMLoopFiniteElements(dm, simple->getDomainFEName(),
post_proc_fe);
CHKERR post_proc_fe->writeFile(
"level_set_" +
boost::lexical_cast<std::string>(monitor_ptr->ts_step) + ".h5m");
};
boost::shared_ptr<FEMethod> null;
DMMoFEMTSSetMonitor(sub_dm, ts, simple->getDomainFEName(), monitor_ptr,
null, null);
return monitor_ptr;
};
auto ts = pip->createTSIM(sub_dm);
auto set_solution = [&](auto ts) {
auto D = createDMVector(sub_dm);
CHKERR DMoFEMMeshToLocalVector(sub_dm, D, INSERT_VALUES, SCATTER_FORWARD);
CHKERR TSSetSolution(ts, D);
};
CHKERR set_solution(ts);
auto monitor_pt = set_time_monitor(sub_dm, ts);
CHKERR TSSetFromOptions(ts);
auto B = createDMMatrix(sub_dm);
CHKERR TSSetIJacobian(ts, B, B, TsSetIJacobian, nullptr);
CHKERR TSSetUp(ts);
auto ts_pre_step = [](TS ts) {
auto &m_field = level_set_raw_ptr->mField;
auto simple = m_field.getInterface<Simple>();
auto bit_mng = m_field.getInterface<BitRefManager>();
auto [error, th_error] = level_set_raw_ptr->evaluateError();
MOFEM_LOG("LevelSet", Sev::inform) << "Error indicator " << error;
auto get_norm = [&](auto x) {
double nrm;
CHKERR VecNorm(x, NORM_2, &nrm);
return nrm;
};
auto set_solution = [&](auto ts) {
DM dm;
CHKERR TSGetDM(ts, &dm);
auto prb_ptr = getProblemPtr(dm);
auto x = createDMVector(dm);
CHKERR DMoFEMMeshToLocalVector(dm, x, INSERT_VALUES, SCATTER_FORWARD);
CHKERR VecGhostUpdateBegin(x, INSERT_VALUES, SCATTER_FORWARD);
CHKERR VecGhostUpdateEnd(x, INSERT_VALUES, SCATTER_FORWARD);
MOFEM_LOG("LevelSet", Sev::inform)
<< "Problem " << prb_ptr->getName() << " solution vector norm "
<< get_norm(x);
CHKERR TSSetSolution(ts, x);
};
auto refine_and_project = [&](auto ts) {
WrapperClassErrorProjection(level_set_raw_ptr->maxPtr));
simple->getBitRefLevel() = BitRefLevel().set(skeleton_bit) |
simple->reSetUp(true);
DM dm;
CHKERR TSGetDM(ts, &dm);
.flip(); // DOFs which are not on bit_domain_ele should be removed
};
auto ts_reset_theta = [&](auto ts) {
DM dm;
CHKERR TSGetDM(ts, &dm);
// FIXME: Look into vec-5 how to transfer internal theta method variables
CHKERR TSReset(ts);
CHKERR TSSetUp(ts);
CHKERR set_solution(ts);
auto B = createDMMatrix(dm);
CHKERR TSSetIJacobian(ts, B, B, TsSetIJacobian, nullptr);
};
CHKERR refine_and_project(ts);
CHKERR ts_reset_theta(ts);
};
auto ts_post_step = [](TS ts) { return 0; };
CHKERR TSSetPreStep(ts, ts_pre_step);
CHKERR TSSetPostStep(ts, ts_post_step);
CHKERR TSSolve(ts, NULL);
}
auto bit_mng = mField.getInterface<BitRefManager>();
ParallelComm *pcomm =
ParallelComm::get_pcomm(&mField.get_moab(), MYPCOMM_INDEX);
auto proc_str = boost::lexical_cast<std::string>(mField.get_comm_rank());
auto set_bit = [](auto l) { return BitRefLevel().set(l); };
auto save_range = [&](const std::string name, const Range &r) {
auto meshset_ptr = get_temp_meshset_ptr(mField.get_moab());
CHKERR mField.get_moab().write_file(name.c_str(), "VTK", "",
meshset_ptr->get_ptr(), 1);
};
// select domain elements to refine by threshold
auto get_refined_elements_meshset = [&](auto bit, auto mask) {
Range fe_ents;
Tag th_error;
CHK_MOAB_THROW(mField.get_moab().tag_get_handle("Error", th_error),
"get error handle");
std::vector<double> errors(fe_ents.size());
mField.get_moab().tag_get_data(th_error, fe_ents, &*errors.begin()),
"get tag data");
auto it = std::max_element(errors.begin(), errors.end());
double max;
MPI_Allreduce(&*it, &max, 1, MPI_DOUBLE, MPI_MAX, mField.get_comm());
MOFEM_LOG("LevelSet", Sev::inform) << "Max error: " << max;
auto threshold = wp.getThreshold(max);
std::vector<EntityHandle> fe_to_refine;
fe_to_refine.reserve(fe_ents.size());
auto fe_it = fe_ents.begin();
auto error_it = errors.begin();
for (auto i = 0; i != fe_ents.size(); ++i) {
if (*error_it > threshold) {
fe_to_refine.push_back(*fe_it);
}
++fe_it;
++error_it;
}
Range ents;
ents.insert_list(fe_to_refine.begin(), fe_to_refine.end());
CHKERR mField.getInterface<CommInterface>()->synchroniseEntities(
ents, nullptr, NOISY);
auto get_neighbours_by_bridge_vertices = [&](auto &&ents) {
Range verts;
CHKERR mField.get_moab().get_connectivity(ents, verts, true);
moab::Interface::UNION);
CHKERR mField.getInterface<CommInterface>()->synchroniseEntities(ents);
return ents;
};
ents = get_neighbours_by_bridge_vertices(ents);
#ifndef NDEBUG
if (debug) {
auto meshset_ptr = get_temp_meshset_ptr(mField.get_moab());
CHKERR mField.get_moab().write_file(
(proc_str + "_fe_to_refine.vtk").c_str(), "VTK", "",
meshset_ptr->get_ptr(), 1);
}
#endif
return ents;
};
// refine elements, and set bit ref level
auto refine_mesh = [&](auto l, auto &&fe_to_refine) {
Skinner skin(&mField.get_moab());
// get entities in "l-1" level
Range level_ents;
CHKERR bit_mng->getEntitiesByDimAndRefLevel(
set_bit(start_bit + l - 1), BitRefLevel().set(), FE_DIM, level_ents);
// select entities to refine
fe_to_refine = intersect(level_ents, fe_to_refine);
// select entities not to refine
level_ents = subtract(level_ents, fe_to_refine);
// for entities to refine get children, i.e. redlined entities
Range fe_to_refine_children;
bit_mng->updateRangeByChildren(fe_to_refine, fe_to_refine_children);
// add entities to to level "l"
fe_to_refine_children = fe_to_refine_children.subset_by_dimension(FE_DIM);
level_ents.merge(fe_to_refine_children);
auto fix_neighbour_level = [&](auto ll) {
// filter entities on level ll
auto level_ll = level_ents;
CHKERR bit_mng->filterEntitiesByRefLevel(set_bit(ll), BitRefLevel().set(),
level_ll);
// find skin of ll level
Range skin_edges;
CHKERR skin.find_skin(0, level_ll, false, skin_edges);
// get parents of skin of level ll
Range skin_parents;
for (auto lll = 0; lll <= ll; ++lll) {
CHKERR bit_mng->updateRangeByParent(skin_edges, skin_parents);
}
// filter parents on level ll - 1
for (auto lll = 0; lll <= ll - 2; ++lll) {
}
}
};
CHKERR fix_neighbour_level(l);
CHKERR mField.getInterface<CommInterface>()->synchroniseEntities(
level_ents);
// get lower dimension entities for level "l"
for (auto d = 0; d != FE_DIM; ++d) {
if (d == 0) {
CHKERR mField.get_moab().get_connectivity(
level_ents.subset_by_dimension(FE_DIM), level_ents, true);
} else {
level_ents.subset_by_dimension(FE_DIM), d, false, level_ents,
moab::Interface::UNION);
}
}
CHKERR mField.getInterface<CommInterface>()->synchroniseEntities(
level_ents);
// set bit ref level to level entities
CHKERR bit_mng->setNthBitRefLevel(start_bit + l, false);
CHKERR bit_mng->setNthBitRefLevel(level_ents, start_bit + l, true);
#ifndef NDEBUG
auto proc_str = boost::lexical_cast<std::string>(mField.get_comm_rank());
CHKERR bit_mng->writeBitLevelByDim(
set_bit(start_bit + l), BitRefLevel().set(), FE_DIM,
(boost::lexical_cast<std::string>(l) + "_" + proc_str + "_ref_mesh.vtk")
.c_str(),
"VTK", "");
#endif
};
// set skeleton
auto set_skelton_bit = [&](auto l) {
// get entities of dim-1 on level "l"
Range level_edges;
CHKERR bit_mng->getEntitiesByDimAndRefLevel(
set_bit(start_bit + l), BitRefLevel().set(), FE_DIM - 1, level_edges);
// get parent of entities of level "l"
Range level_edges_parents;
CHKERR bit_mng->updateRangeByParent(level_edges, level_edges_parents);
level_edges_parents = level_edges_parents.subset_by_dimension(FE_DIM - 1);
CHKERR bit_mng->filterEntitiesByRefLevel(
set_bit(start_bit + l), BitRefLevel().set(), level_edges_parents);
// skeleton entities which do not have parents
auto parent_skeleton = intersect(level_edges, level_edges_parents);
auto skeleton = subtract(level_edges, level_edges_parents);
FE_DIM, false, skeleton,
moab::Interface::UNION);
// set levels
CHKERR mField.getInterface<CommInterface>()->synchroniseEntities(skeleton);
CHKERR bit_mng->setNthBitRefLevel(skeleton_bit, false);
CHKERR bit_mng->setNthBitRefLevel(skeleton, skeleton_bit, true);
#ifndef NDEBUG
CHKERR bit_mng->writeBitLevel(
set_bit(skeleton_bit), BitRefLevel().set(),
(boost::lexical_cast<std::string>(l) + "_" + proc_str + "_skeleton.vtk")
.c_str(),
"VTK", "");
#endif
};
// Reset bit sand set old current and aggregate bits as projection bits
Range level0_current;
CHKERR bit_mng->getEntitiesByRefLevel(BitRefLevel().set(current_bit),
BitRefLevel().set(), level0_current);
Range level0_aggregate;
CHKERR bit_mng->getEntitiesByRefLevel(BitRefLevel().set(aggregate_bit),
BitRefLevel().set(), level0_aggregate);
for (auto s = 0; s != start_bit; ++s)
CHKERR bit_mng->lambdaBitRefLevel(
[&](EntityHandle ent, BitRefLevel &bit) { bit &= start_mask; });
CHKERR bit_mng->setNthBitRefLevel(level0_current, projection_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level0_aggregate, aggregate_projection_bit,
true);
// Set zero bit ref level
Range level0;
CHKERR bit_mng->getEntitiesByRefLevel(set_bit(0), BitRefLevel().set(),
level0);
CHKERR bit_mng->setNthBitRefLevel(level0, start_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level0, current_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level0, aggregate_bit, true);
CHKERR bit_mng->setNthBitRefLevel(level0, skeleton_bit, true);
CHKERR wp.setBits(*this, 0);
CHKERR wp.runCalcs(*this, 0);
for (auto l = 0; l != nb_levels; ++l) {
MOFEM_LOG("WORLD", Sev::inform) << "Process level: " << l;
CHKERR refine_mesh(l + 1, get_refined_elements_meshset(
set_bit(start_bit + l), BitRefLevel().set()));
CHKERR set_skelton_bit(l + 1);
CHKERR wp.setAggregateBit(*this, l + 1);
CHKERR wp.setBits(*this, l + 1);
CHKERR wp.runCalcs(*this, l + 1);
}
}
// get operators tester
auto pip = mField.getInterface<PipelineManager>(); // get interface to
auto bit_mng = mField.getInterface<BitRefManager>();
auto prb_mng = mField.getInterface<ProblemsManager>();
auto lhs_fe = boost::make_shared<DomainEle>(mField);
auto rhs_fe_prj = boost::make_shared<DomainEle>(mField);
auto rhs_fe_current = boost::make_shared<DomainEle>(mField);
lhs_fe->getRuleHook = [](int, int, int o) { return 3 * o; };
rhs_fe_prj->getRuleHook = [](int, int, int o) { return 3 * o; };
rhs_fe_current->getRuleHook = [](int, int, int o) { return 3 * o; };
auto sub_dm = createDM(mField.get_comm(), "DMMOFEM");
CHKERR DMMoFEMCreateSubDM(sub_dm, simple->getDM(), "DG_PROJECTION");
CHKERR DMMoFEMSetDestroyProblem(sub_dm, PETSC_TRUE);
CHKERR DMMoFEMSetSquareProblem(sub_dm, PETSC_TRUE);
CHKERR DMSetUp(sub_dm);
Range current_ents; // ents used to do calculations
CHKERR bit_mng->getEntitiesByDimAndRefLevel(BitRefLevel().set(current_bit),
BitRefLevel().set(), FE_DIM,
current_ents);
Range prj_ents; // ents from which data are projected
CHKERR bit_mng->getEntitiesByDimAndRefLevel(
BitRefLevel().set(projection_bit), BitRefLevel().set(), FE_DIM, prj_ents);
for (auto l = 0; l != nb_levels; ++l) {
CHKERR bit_mng->updateRangeByParent(prj_ents, prj_ents);
}
current_ents = subtract(
current_ents, prj_ents); // only restric to entities needed projection
auto test_mesh_bit = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
auto test_prj_bit = [&](FEMethod *fe_ptr) {
return fe_ptr->numeredEntFiniteElementPtr->getBitRefLevel().test(
};
auto test_current_bit = [&](FEMethod *fe_ptr) {
return current_ents.find(fe_ptr->getFEEntityHandle()) != current_ents.end();
};
lhs_fe->exeTestHook =
test_mesh_bit; // that element only is run when current bit is set
rhs_fe_prj->exeTestHook =
test_prj_bit; // that element is run only when projection bit is set
rhs_fe_current->exeTestHook =
test_current_bit; // that element is only run when current bit is set
remove_mask.flip(); // DOFs which are not on bit_domain_ele should be removed
CHKERR prb_mng->removeDofsOnEntities(
"DG_PROJECTION", "L", BitRefLevel().set(), remove_mask, nullptr, 0,
true); // remove all DOFs which are not
// on current bit. This case works for L2 space
{L2});
lhs_fe->getOpPtrVector().push_back(
new OpMassLL("L", "L")); // Assemble projection matrix
// This assumes that projection mesh is finer, current mesh is coarsened.
auto set_prj_from_child = [&](auto rhs_fe_prj) {
rhs_fe_prj->getOpPtrVector(), {L2});
// Evaluate field value on projection mesh
auto l_vec = boost::make_shared<MatrixDouble>();
rhs_fe_prj->getOpPtrVector().push_back(
// This element is used to assemble
auto get_parent_this = [&]() {
auto fe_parent_this = boost::make_shared<DomianParentEle>(mField);
fe_parent_this->getOpPtrVector().push_back(
new OpScalarFieldL("L", l_vec));
return fe_parent_this;
};
// Create levels of parent elements, until current element is reached, and
// then assemble.
auto get_parents_fe_ptr = [&](auto this_fe_ptr) {
std::vector<boost::shared_ptr<DomianParentEle>> parents_elems_ptr_vec;
for (int l = 0; l <= nb_levels; ++l)
parents_elems_ptr_vec.emplace_back(
boost::make_shared<DomianParentEle>(mField));
for (auto l = 1; l <= nb_levels; ++l) {
parents_elems_ptr_vec[l - 1]->getOpPtrVector().push_back(
new OpRunParent(parents_elems_ptr_vec[l], BitRefLevel().set(),
BitRefLevel().set(current_bit).flip(), this_fe_ptr,
BitRefLevel().set()));
}
return parents_elems_ptr_vec[0];
};
auto this_fe_ptr = get_parent_this();
auto parent_fe_ptr = get_parents_fe_ptr(this_fe_ptr);
rhs_fe_prj->getOpPtrVector().push_back(
new OpRunParent(parent_fe_ptr, BitRefLevel().set(),
BitRefLevel().set(current_bit).flip(), this_fe_ptr,
BitRefLevel().set(current_bit), BitRefLevel().set()));
};
// This assumed that current mesh is refined, and projection mesh is coarser
auto set_prj_from_parent = [&](auto rhs_fe_current) {
// Evaluate field value on projection mesh
auto l_vec = boost::make_shared<MatrixDouble>();
// Evaluate field on coarser element
auto get_parent_this = [&]() {
auto fe_parent_this = boost::make_shared<DomianParentEle>(mField);
fe_parent_this->getOpPtrVector().push_back(
return fe_parent_this;
};
// Create stack of evaluation on parent elements
auto get_parents_fe_ptr = [&](auto this_fe_ptr) {
std::vector<boost::shared_ptr<DomianParentEle>> parents_elems_ptr_vec;
for (int l = 0; l <= nb_levels; ++l)
parents_elems_ptr_vec.emplace_back(
boost::make_shared<DomianParentEle>(mField));
for (auto l = 1; l <= nb_levels; ++l) {
parents_elems_ptr_vec[l - 1]->getOpPtrVector().push_back(
new OpRunParent(parents_elems_ptr_vec[l], BitRefLevel().set(),
BitRefLevel().set(projection_bit).flip(),
this_fe_ptr, BitRefLevel().set(projection_bit),
BitRefLevel().set()));
}
return parents_elems_ptr_vec[0];
};
auto this_fe_ptr = get_parent_this();
auto parent_fe_ptr = get_parents_fe_ptr(this_fe_ptr);
auto reset_op_ptr = new DomainEleOp(NOSPACE, DomainEleOp::OPSPACE);
reset_op_ptr->doWorkRhsHook = [&](DataOperator *op_ptr, int, EntityType,
EntData &) {
l_vec->resize(static_cast<DomainEleOp *>(op_ptr)->getGaussPts().size2(),
false);
l_vec->clear();
return 0;
};
rhs_fe_current->getOpPtrVector().push_back(reset_op_ptr);
rhs_fe_current->getOpPtrVector().push_back(
new OpRunParent(parent_fe_ptr, BitRefLevel().set(),
BitRefLevel().set(projection_bit).flip(), this_fe_ptr,
// At the end assemble of current finite element
rhs_fe_current->getOpPtrVector().push_back(new OpScalarFieldL("L", l_vec));
};
set_prj_from_child(rhs_fe_prj);
set_prj_from_parent(rhs_fe_current);
boost::shared_ptr<FEMethod> null_fe;
getDMKspCtx(sub_dm)->clearLoops();
CHKERR DMMoFEMKSPSetComputeOperators(sub_dm, simple->getDomainFEName(),
lhs_fe, null_fe, null_fe);
CHKERR DMMoFEMKSPSetComputeRHS(sub_dm, simple->getDomainFEName(), rhs_fe_prj,
null_fe, null_fe);
CHKERR DMMoFEMKSPSetComputeRHS(sub_dm, simple->getDomainFEName(),
rhs_fe_current, null_fe, null_fe);
CHKERR KSPSetDM(ksp, sub_dm);
CHKERR KSPSetDM(ksp, sub_dm);
CHKERR KSPSetFromOptions(ksp);
CHKERR KSPSetUp(ksp);
auto L = createDMVector(sub_dm);
auto F = vectorDuplicate(L);
CHKERR KSPSolve(ksp, F, L);
CHKERR VecGhostUpdateBegin(L, INSERT_VALUES, SCATTER_FORWARD);
CHKERR VecGhostUpdateEnd(L, INSERT_VALUES, SCATTER_FORWARD);
CHKERR DMoFEMMeshToLocalVector(sub_dm, L, INSERT_VALUES, SCATTER_REVERSE);
auto [error, th_error] = evaluateError();
MOFEM_LOG("LevelSet", Sev::inform) << "Error indicator " << error;
auto post_proc = [&](auto dm, auto out_name, auto th_error) {
auto post_proc_fe =
boost::make_shared<PostProcBrokenMeshInMoab<DomainEle>>(mField);
post_proc_fe->setTagsToTransfer({th_error});
post_proc_fe->exeTestHook = test_mesh_bit;
if constexpr (DIM1 == 1 && DIM2 == 1) {
auto l_vec = boost::make_shared<VectorDouble>();
post_proc_fe->getOpPtrVector(), {L2});
post_proc_fe->getOpPtrVector().push_back(
new OpCalculateScalarFieldValues("L", l_vec));
post_proc_fe->getOpPtrVector().push_back(
post_proc_fe->getOpPtrVector().push_back(
new OpPPMap(
post_proc_fe->getPostProcMesh(), post_proc_fe->getMapGaussPts(),
{{"L", l_vec}},
{}, {})
);
}
CHKERR DMoFEMLoopFiniteElements(dm, simple->getDomainFEName(),
post_proc_fe);
post_proc_fe->writeFile(out_name);
};
if constexpr (debug)
CHKERR post_proc(sub_dm, "dg_projection.h5m", th_error);
}
static Index< 'o', 3 > o
static Index< 'L', 3 > L
static Index< 'J', 3 > J
std::string param_file
#define MOFEM_LOG_C(channel, severity, format,...)
Definition: LogManager.hpp:311
void simple(double P1[], double P2[], double P3[], double c[], const int N)
Definition: acoustic.cpp:69
static char help[]
int main()
static const double eps
constexpr int SPACE_DIM
ElementsAndOps< SPACE_DIM >::DomainEle DomainEle
ElementsAndOps< SPACE_DIM >::BoundaryEle BoundaryEle
@ NOISY
Definition: definitions.h:211
#define CATCH_ERRORS
Catch errors.
Definition: definitions.h:372
#define MAX_DOFS_ON_ENTITY
Maximal number of DOFs on entity.
Definition: definitions.h:236
@ AINSWORTH_LEGENDRE_BASE
Ainsworth Cole (Legendre) approx. base .
Definition: definitions.h:60
#define CHK_THROW_MESSAGE(err, msg)
Check and throw MoFEM exception.
Definition: definitions.h:595
FieldSpace
approximation spaces
Definition: definitions.h:82
@ L2
field with C-1 continuity
Definition: definitions.h:88
@ H1
continuous field
Definition: definitions.h:85
@ NOSPACE
Definition: definitions.h:83
@ HCURL
field with continuous tangents
Definition: definitions.h:86
#define MYPCOMM_INDEX
default communicator number PCOMM
Definition: definitions.h:215
#define MoFEMFunctionBegin
First executable line of each MoFEM function, used for error handling. Final line of MoFEM functions ...
Definition: definitions.h:346
#define CHK_MOAB_THROW(err, msg)
Check error code of MoAB function and throw MoFEM exception.
Definition: definitions.h:576
@ MOFEM_ATOM_TEST_INVALID
Definition: definitions.h:40
@ MOFEM_DATA_INCONSISTENCY
Definition: definitions.h:31
@ MOFEM_NOT_IMPLEMENTED
Definition: definitions.h:32
#define MoFEMFunctionReturn(a)
Last executable line of each PETSc function used for error handling. Replaces return()
Definition: definitions.h:416
#define CHKERR
Inline error check.
Definition: definitions.h:535
static const bool debug
FormsIntegrators< DomainEleOp >::Assembly< PETSC >::BiLinearForm< GAUSS >::OpMass< 1, SPACE_DIM > OpMass
auto get_ntensor(T &base_mat)
ElementsAndOps< SPACE_DIM >::FaceSideEle FaceSideEle
@ F
PetscErrorCode DMMoFEMCreateSubDM(DM subdm, DM dm, const char problem_name[])
Must be called by user to set Sub DM MoFEM data structures.
Definition: DMMoFEM.cpp:219
Definition: DMMoFEM.cpp:483
PetscErrorCode DMMoFEMSetSquareProblem(DM dm, PetscBool square_problem)
set squared problem
Definition: DMMoFEM.cpp:442
PetscErrorCode DMoFEMMeshToLocalVector(DM dm, Vec l, InsertMode mode, ScatterMode scatter_mode)
set local (or ghosted) vector values on mesh for partition only
Definition: DMMoFEM.cpp:509
PetscErrorCode DMMoFEMAddSubFieldRow(DM dm, const char field_name[])
Definition: DMMoFEM.cpp:242
PetscErrorCode DMRegister_MoFEM(const char sname[])
Register MoFEM problem.
Definition: DMMoFEM.cpp:47
PetscErrorCode DMMoFEMKSPSetComputeRHS(DM dm, const char fe_name[], MoFEM::FEMethod *method, MoFEM::BasicMethod *pre_only, MoFEM::BasicMethod *post_only)
set KSP right hand side evaluation function
Definition: DMMoFEM.cpp:623
PetscErrorCode DMoFEMLoopFiniteElements(DM dm, const char fe_name[], MoFEM::FEMethod *method, CacheTupleWeakPtr cache_ptr=CacheTupleSharedPtr())
Executes FEMethod for finite elements in DM.
Definition: DMMoFEM.cpp:572
auto createDMVector(DM dm)
Get smart vector from DM.
Definition: DMMoFEM.hpp:1003
auto createDMMatrix(DM dm)
Get smart matrix from DM.
Definition: DMMoFEM.hpp:988
PetscErrorCode DMSubDMSetUp_MoFEM(DM subdm)
Definition: DMMoFEM.cpp:1314
PetscErrorCode DMMoFEMKSPSetComputeOperators(DM dm, const char fe_name[], MoFEM::FEMethod *method, MoFEM::BasicMethod *pre_only, MoFEM::BasicMethod *post_only)
Set KSP operators and push mofem finite element methods.
Definition: DMMoFEM.cpp:664
boost::ptr_deque< UserDataOperator > & getOpDomainLhsPipeline()
Get the Op Domain Lhs Pipeline object.
boost::ptr_deque< UserDataOperator > & getOpSkeletonLhsPipeline()
Get the Op Skeleton Lhs Pipeline object.
virtual EntityHandle get_finite_element_meshset(const std::string name) const =0
IntegrationType
Form integrator integration types.
AssemblyType
[Storage and set boundary conditions]
#define MOFEM_LOG(channel, severity)
Log.
Definition: LogManager.hpp:308
#define MOFEM_LOG_TAG(channel, tag)
Tag channel.
Definition: LogManager.hpp:339
auto bit
set bit
FTensor::Index< 'i', SPACE_DIM > i
double D
constexpr int FE_DIM
[Define dimension]
Definition: level_set.cpp:19
constexpr int skeleton_bit
skeleton elements bit
Definition: level_set.cpp:65
FTensor::Index< 'I', DIM1 > I
Definition: level_set.cpp:29
constexpr int current_bit
dofs bit used to do calculations
Definition: level_set.cpp:63
constexpr FieldSpace potential_velocity_space
Definition: level_set.cpp:49
constexpr int nb_levels
Definition: level_set.cpp:58
FTensor::Index< 'i', SPACE_DIM > i
Definition: level_set.cpp:579
constexpr int SPACE_DIM
Definition: level_set.cpp:20
constexpr int DIM2
Definition: level_set.cpp:22
constexpr bool debug
Definition: level_set.cpp:53
PipelineManager::ElementsAndOpsByDim< FE_DIM >::DomianParentEle DomianParentEle
Definition: level_set.cpp:39
FTensor::Index< 'J', DIM1 > J
Definition: level_set.cpp:30
constexpr int start_bit
Definition: level_set.cpp:60
FaceSideEle::UserDataOperator FaceSideEleOp
Definition: level_set.cpp:45
constexpr int aggregate_projection_bit
all bits for projection problem
Definition: level_set.cpp:70
DomainEle::UserDataOperator DomainEleOp
Definition: level_set.cpp:43
constexpr AssemblyType A
Definition: level_set.cpp:32
LevelSet * level_set_raw_ptr
Definition: level_set.cpp:1931
constexpr int aggregate_bit
Definition: level_set.cpp:66
constexpr int projection_bit
Definition: level_set.cpp:68
constexpr IntegrationType G
Definition: level_set.cpp:33
constexpr int DIM1
Definition: level_set.cpp:21
MoFEM::TsCtx * ts_ctx
Definition: level_set.cpp:1932
constexpr size_t potential_velocity_field_dim
Definition: level_set.cpp:50
FTensor::Index< 'l', 3 > l
FTensor::Index< 'j', 3 > j
FTensor::Index< 'k', 3 > k
PetscErrorCode MoFEMErrorCode
MoFEM/PETSc error code.
Definition: Exceptions.hpp:56
std::bitset< BITREFLEVEL_SIZE > BitRefLevel
Bit structure attached to each entity identifying to what mesh entity is attached.
Definition: Types.hpp:40
UBlasVector< double > VectorDouble
Definition: Types.hpp:68
implementation of Data Operators for Forces and Sources
Definition: Common.hpp:10
auto createKSP(MPI_Comm comm)
PetscErrorCode DMMoFEMTSSetMonitor(DM dm, TS ts, const std::string fe_name, boost::shared_ptr< MoFEM::FEMethod > method, boost::shared_ptr< MoFEM::BasicMethod > pre_only, boost::shared_ptr< MoFEM::BasicMethod > post_only)
Set Monitor To TS solver.
Definition: DMMoFEM.cpp:1042
PetscErrorCode DMMoFEMSetDestroyProblem(DM dm, PetscBool destroy_problem)
Definition: DMMoFEM.cpp:424
SmartPetscObj< Vec > vectorDuplicate(Vec vec)
Create duplicate vector of smart vector.
auto createVectorMPI(MPI_Comm comm, PetscInt n, PetscInt N)
Create MPI Vector.
auto getDMKspCtx(DM dm)
Get KSP context data structure used by DM.
Definition: DMMoFEM.hpp:1017
constexpr auto make_array(Arg &&...arg)
Create Array.
Definition: Templates.hpp:1961
auto get_temp_meshset_ptr(moab::Interface &moab)
Create smart pointer to temporary meshset.
Definition: Templates.hpp:1857
auto createDM(MPI_Comm comm, const std::string dm_type_name)
Creates smart DM object.
auto getProblemPtr(DM dm)
get problem pointer from DM
Definition: DMMoFEM.hpp:976
int yc
Definition: sdf_hertz.py:8
int xc
Definition: sdf_hertz.py:7
int zc
Definition: sdf_hertz.py:9
float d
Definition: sdf_hertz.py:5
int r
Definition: sdf.py:8
constexpr IntegrationType I
constexpr AssemblyType A
OpPostProcMapInMoab< SPACE_DIM, SPACE_DIM > OpPPMap
#define EXECUTABLE_DIMENSION
Definition: plastic.cpp:13
constexpr double t
plate stiffness
Definition: plate.cpp:59
constexpr auto field_name
OpBaseImpl< PETSC, EdgeEleOp > OpBase
OpLhsDomain(const std::string field_name, boost::shared_ptr< MatrixDouble > vel_ptr)
Definition: level_set.cpp:590
MoFEMErrorCode iNtegrate(EntData &row_data, EntData &col_data)
Definition: level_set.cpp:655
boost::shared_ptr< MatrixDouble > velPtr
Definition: level_set.cpp:432
boost::shared_ptr< FaceSideEle > sideFEPtr
pointer to element to get data on edge/face sides
Definition: level_set.cpp:459
boost::shared_ptr< SideData > sideDataPtr
Definition: level_set.cpp:457
MatrixDouble matSkeleton
Definition: level_set.cpp:461
MoFEMErrorCode doWork(int side, EntityType type, EntitiesFieldData::EntData &data)
Operator for linear form, usually to calculate values on right hand side.
Definition: level_set.cpp:809
OpLhsSkeleton(boost::shared_ptr< SideData > side_data_ptr, boost::shared_ptr< FaceSideEle > side_fe_ptr)
Definition: level_set.cpp:604
boost::shared_ptr< MatrixDouble > lPtr
Definition: level_set.cpp:421
MoFEMErrorCode iNtegrate(EntData &data)
Class dedicated to integrate operator.
Definition: level_set.cpp:610
boost::shared_ptr< MatrixDouble > velPtr
Definition: level_set.cpp:423
OpRhsDomain(const std::string field_name, boost::shared_ptr< MatrixDouble > l_ptr, boost::shared_ptr< MatrixDouble > l_dot_ptr, boost::shared_ptr< MatrixDouble > vel_ptr)
Definition: level_set.cpp:583
boost::shared_ptr< MatrixDouble > lDotPtr
Definition: level_set.cpp:422
MoFEMErrorCode doWork(int side, EntityType type, EntitiesFieldData::EntData &data)
Operator for linear form, usually to calculate values on right hand side.
Definition: level_set.cpp:703
VectorDouble resSkelton
Definition: level_set.cpp:447
boost::shared_ptr< SideData > sideDataPtr
Definition: level_set.cpp:443
OpRhsSkeleton(boost::shared_ptr< SideData > side_data_ptr, boost::shared_ptr< FaceSideEle > side_fe_ptr)
Definition: level_set.cpp:598
boost::shared_ptr< FaceSideEle > sideFEPtr
pointer to element to get data on edge/face sides
Definition: level_set.cpp:445
std::array< VectorInt, 2 > indicesColSideMap
indices on columns for left hand-side
Definition: level_set.cpp:92
MatSideArray velMat
Definition: level_set.cpp:99
std::array< EntityHandle, 2 > feSideHandle
Definition: level_set.cpp:88
int currentFESide
current side counter
Definition: level_set.cpp:101
std::array< MatrixDouble, 2 > colBaseSideMap
Definition: level_set.cpp:94
std::array< MatrixDouble, 2 > rowBaseSideMap
Definition: level_set.cpp:93
std::array< VectorInt, 2 > indicesRowSideMap
indices on rows for left hand-side
Definition: level_set.cpp:90
MatSideArray lVec
Definition: level_set.cpp:98
std::array< int, 2 > senseMap
Definition: level_set.cpp:95
WrapperClassErrorProjection(boost::shared_ptr< double > max_ptr)
Definition: level_set.cpp:316
MoFEMErrorCode runCalcs(LevelSet &level_set, int l)
Run calculations.
Definition: level_set.cpp:320
MoFEMErrorCode setBits(LevelSet &level_set, int l)
Set bit ref level to problem.
Definition: level_set.cpp:319
double getThreshold(const double max)
Definition: level_set.cpp:335
MoFEMErrorCode setAggregateBit(LevelSet &level_set, int l)
Add bit to current element, so it aggregate all previious current elements.
Definition: level_set.cpp:321
boost::shared_ptr< double > maxPtr
Definition: level_set.cpp:338
virtual double getThreshold(const double max)=0
virtual MoFEMErrorCode runCalcs(LevelSet &level_set, int l)=0
Run calculations.
virtual MoFEMErrorCode setAggregateBit(LevelSet &level_set, int l)=0
Add bit to current element, so it aggregate all previious current elements.
virtual MoFEMErrorCode setBits(LevelSet &level_set, int l)=0
Set bit ref level to problem.
double getThreshold(const double max)
Definition: level_set.cpp:302
MoFEMErrorCode setAggregateBit(LevelSet &level_set, int l)
Add bit to current element, so it aggregate all previious current elements.
Definition: level_set.cpp:287
boost::shared_ptr< double > maxPtr
Definition: level_set.cpp:308
MoFEMErrorCode setBits(LevelSet &level_set, int l)
Set bit ref level to problem.
Definition: level_set.cpp:271
WrapperClassInitalSolution(boost::shared_ptr< double > max_ptr)
Definition: level_set.cpp:268
MoFEMErrorCode runCalcs(LevelSet &level_set, int l)
Run calculations.
Definition: level_set.cpp:281
FormsIntegrators< DomainEleOp >::Assembly< A >::BiLinearForm< G >::OpMass< potential_velocity_field_dim, potential_velocity_field_dim > OpMassVV
Definition: level_set.cpp:358
MoFEM::Interface & mField
integrate skeleton operators on khs
Definition: level_set.cpp:349
Definition: level_set.cpp:1934
boost::shared_ptr< double > maxPtr
Definition: level_set.cpp:370
std::array< MatrixDouble, 2 > MatSideArray
Definition: level_set.cpp:80
Definition: level_set.cpp:502
MoFEMErrorCode setUpProblem()
create fields, and set approximation order
Definition: level_set.cpp:560
MoFEMErrorCode initialiseFieldLevelSet(boost::function< double(double, double, double)> level_fun=get_level_set)
initialise field set
Definition: level_set.cpp:1693
boost::shared_ptr< FaceSideEle > getSideFE(boost::shared_ptr< SideData > side_data_ptr)
create side element to assemble data from sides
Definition: level_set.cpp:997
MoFEMErrorCode pushOpDomain()
push operators to integrate operators on domain
Definition: level_set.cpp:954
MoFEMErrorCode runProblem()
Definition: level_set.cpp:386
FormsIntegrators< DomainEleOp >::Assembly< A >::LinearForm< G >::OpSource< 1, DIM1 *DIM2 > OpSourceL
Definition: level_set.cpp:356
static double get_velocity_potential(double x, double y, double z)
MoFEMErrorCode pushOpSkeleton()
push operator to integrate on skeleton
Definition: level_set.cpp:1173
MoFEMErrorCode testOp()
test consistency between tangent matrix and the right hand side vectors
Definition: level_set.cpp:1602
MoFEMErrorCode dgProjection(const int prj_bit=projection_bit)
dg level set projection
Definition: level_set.cpp:2401
FormsIntegrators< DomainEleOp >::Assembly< A >::BiLinearForm< G >::OpMass< 1, DIM1 *DIM2 > OpMassLL
Definition: level_set.cpp:354
static double get_level_set(const double x, const double y, const double z)
inital level set, i.e. advected filed
Definition: level_set.cpp:378
MoFEMErrorCode refineMesh(WrapperClass &&wp)
Definition: level_set.cpp:2147
FormsIntegrators< DomainEleOp >::Assembly< A >::LinearForm< G >::OpBaseTimesVector< 1, DIM1 *DIM2, 1 > OpScalarFieldL
Definition: level_set.cpp:362
ForcesAndSourcesCore::UserDataOperator * getZeroLevelVelOp(boost::shared_ptr< MatrixDouble > vel_ptr)
Get operator calculating velocity on coarse mesh.
Definition: level_set.cpp:922
std::tuple< double, Tag > evaluateError()
evaluate error
Definition: level_set.cpp:1203
MoFEMErrorCode testSideFE()
test integration side elements
Definition: level_set.cpp:1387
MoFEMErrorCode initialiseFieldVelocity(boost::function< double(double, double, double)> vel_fun=get_velocity_potential< FE_DIM >)
initialise potential velocity field
Definition: level_set.cpp:1814
FormsIntegrators< DomainEleOp >::Assembly< A >::LinearForm< G >::OpSource< potential_velocity_field_dim, potential_velocity_field_dim > OpSourceV
Definition: level_set.cpp:360
Add operators pushing bases from local to physical configuration.
Managing BitRefLevels.
Managing BitRefLevels.
virtual moab::Interface & get_moab()=0
virtual MPI_Comm & get_comm() const =0
virtual int get_comm_rank() const =0
Core (interface) class.
Definition: Core.hpp:82
static MoFEMErrorCode Initialize(int *argc, char ***args, const char file[], const char help[])
Initializes the MoFEM database PETSc, MOAB and MPI.
Definition: Core.cpp:72
static MoFEMErrorCode Finalize()
Checks for options to be called at the conclusion of the program.
Definition: Core.cpp:112
base operator to do operations at Gauss Pt. level
Deprecated interface functions.
Data on single entity (This is passed as argument to DataOperator::doWork)
FTensor::Tensor1< FTensor::PackPtr< double *, Tensor_Dim >, Tensor_Dim > getFTensor1DiffN(const FieldApproximationBase base)
Get derivatives of base functions.
FTensor::Tensor0< FTensor::PackPtr< double *, 1 > > getFTensor0N(const FieldApproximationBase base)
Get base function as Tensor0.
MatrixDouble & getN(const FieldApproximationBase base)
get base functions this return matrix (nb. of rows is equal to nb. of Gauss pts, nb....
const VectorInt & getIndices() const
Get global indices of dofs on entity.
Basic algebra on fields.
Definition: FieldBlas.hpp:21
@ OPSPACE
operator do Work is execute on space data
MatrixDouble & getGaussPts()
matrix of integration (Gauss) points for Volume Element
VectorDouble locF
local entity vector
Calculate curl of vector field.
Get field gradients at integration pts for scalar filed rank 0, i.e. vector field.
Get value at integration points for scalar field.
Get time direvarive values at integration pts for tensor filed rank 2, i.e. matrix field.
Get values at integration pts for tensor filed rank 2, i.e. matrix field.
Post post-proc data at points from hash maps.
Operator to execute finite element instance on parent element. This operator is typically used to pro...
Calculate directional derivative of the right hand side and compare it with tangent matrix derivative...
PipelineManager interface.
Problem manager is used to build and partition problems.
Simple interface for fast problem set-up.
Definition: Simple.hpp:27
BitRefLevel & getBitRefLevel()
Get the BitRefLevel.
Definition: Simple.hpp:327
intrusive_ptr for managing petsc objects
Interface for Time Stepping (TS) solver.
Definition: TsCtx.hpp:17
MoFEMErrorCode getInterface(IFACE *&iface) const
Get interface refernce to pointer of interface.
Operator the left hand side matrix.
auto save_range | 23,330 | 81,815 | {"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-2024-10 | longest | en | 0.551869 |
https://www.cnblogs.com/iwtwiioi/p/3893410.html | 1,604,014,770,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00571.warc.gz | 658,295,937 | 6,575 | # 【BZOJ】1041: [HAOI2008]圆上的整点（几何）
http://www.lydsy.com:808/JudgeOnline/problem.php?id=1041
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define print(a) printf("%d", a)
#define debug(a) printf("%lld\n", a)
inline int getnum() { int ret=0; char c; for(c=getchar(); c<'0' || c>'9'; c=getchar()); for(; c>='0' && c<='9'; c=getchar()) ret=ret*10+c-'0'; return ret; }
typedef long long ll;
ll gcd(ll a, ll b) { return b?gcd(b, a%b):a; }
inline bool check(ll A, ll B) {
if(((ll)sqrt(B)*(ll)sqrt(B))==B && A!=B)
if(gcd(A, B)==1) return true;
return false;
}
int main() {
int ans=0;
ll d, d2, r, r2;
scanf("%lld", &r);
r2=r<<1;
ll m=sqrt(r2);
ll a;
for(d=1; d<=m; ++d) {
if(!(r2%d)) {
d2=d<<1;
for(a=1; a<=(ll)sqrt(r2/d2); ++a)
if(check(a*a, r2/d-a*a)) ++ans;
if(d!=r2/d) {
for(a=1; a<=(ll)sqrt(d/2); ++a)
if(check(a*a, d-a*a)) ++ans;
}
}
}
printf("%lld\n", (ll)(ans*4+4));
return 0;
}
r
4
4
n<=2000 000 000
## Source
posted @ 2014-08-05 22:59 iwtwiioi 阅读(474) 评论(0编辑 收藏 | 581 | 1,399 | {"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} | 2.96875 | 3 | CC-MAIN-2020-45 | latest | en | 0.09762 |
http://www.conversion-website.com/energy/watt-second-to-erg.html | 1,709,363,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00084.warc.gz | 42,746,510 | 4,252 | # Watt seconds to ergs (Ws to erg)
## Convert watt seconds to ergs
Watt seconds to ergs converter on this page calculates how many ergs are in 'X' watt seconds (where 'X' is the number of watt seconds to convert to ergs). In order to convert a value from watt seconds to ergs (from Ws to erg) simply type the number of Ws to be converted to erg and then click on the 'convert' button.
## Watt seconds to ergs conversion factor
1 watt second is equal to 10000000 ergs
## Watt seconds to ergs conversion formula
Energy(erg) = Energy (Ws) × 10000000
Example: Pressume there is a value of energy equal to 387 watt seconds. How to convert them in ergs?
Energy(erg) = 387 ( Ws ) × 10000000 ( erg / Ws )
Energy(erg) = 3870000000 erg or
387 Ws = 3870000000 erg
387 watt seconds equals 3870000000 ergs
## Watt seconds to ergs conversion table
watt seconds (Ws)ergs (erg)
770000000
990000000
11110000000
13130000000
15150000000
17170000000
19190000000
21210000000
23230000000
25250000000
watt seconds (Ws)ergs (erg)
2502500000000
3003000000000
3503500000000
4004000000000
4504500000000
5005000000000
5505500000000
6006000000000
6506500000000
7007000000000
Versions of the watt seconds to ergs conversion table. To create a watt seconds to ergs conversion table for different values, click on the "Create a customized energy conversion table" button.
## Related energy conversions
Back to watt seconds to ergs conversion
TableFormulaFactorConverterTop | 425 | 1,457 | {"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-2024-10 | latest | en | 0.559908 |
http://www.mathacademy.com/pr/prime/browse.asp?LT=L&ANCHOR=cantorbendixsontheorem00000000&TBM=&TAL=&TAN=Y&TBI=&TCA=&TCS=&TDI=&TEC=&TFO=&TGE=&TGR=&THI=&TNT=&TPH=&TST=&TTO=Y&TTR=&TAD=&LEV=B | 1,369,349,405,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704054863/warc/CC-MAIN-20130516113414-00071-ip-10-60-113-184.ec2.internal.warc.gz | 581,399,423 | 7,098 | BROWSE ALPHABETICALLY LEVEL: Elementary Advanced Both INCLUDE TOPICS: Basic Math Algebra Analysis Biography Calculus Comp Sci Discrete Economics Foundations Geometry Graph Thry History Number Thry Phys Sci Statistics Topology Trigonometry Cantor-Bendixson Theorem – closed set system Cantor-Bendixson Theorem Every closed subset of the real numbers is the disjoint union of a perfect set and a countable set.Cf. derived set. Cantor-Schröder-Bernstein Theorem For any sets A and B, if there exists an injective (one-to-one) function from A into B and also an injective function from B into A, then there exists a surjective (onto) function from A onto B. Cantor set ARTICLE A subset of the unit interval (on the real number line) which is a perfect set, nowhere dense, uncountable, and homeomorphic to the entire unit interval. See the article for a complete exposition. Cantor’s Theorem ARTICLE Given any set, a new set may be constructed which is cardinally larger, i.e., whose size is represented by a larger cardinal than the size of the initial set. More specifically, for no set X is there a bijection of X onto its power set. This theorem establishes both that there are different sizes of infinity, and that there is no greatest such “size.” Cantor’s proof was the first use of a diagonalization argument to derive a contradiction. Related MiniText: Infinity -- You Can't Get There From Here... Carathéodory’s Theorem See: outer measure. cardinal A cardinal number represents the size of a set irrespective of the order or structure of its elements. If X is a set, its cardinality is generally indicated by |X|. Two sets A and B are said to have the same cardinality if there is a function from A into B that is bijective (i.e., one-to-one and onto).A cardinal is an initial ordinal, i.e., an ordinal for which there does not exist a bijection onto any lesser ordinal. Thus, all finite ordinals (i.e., the natural numbers) are cardinals, but most transfinite ordinals are not cardinals.If a is a cardinal, then we denote by a+ the least cardinal greater than a. A cardinal k is called a successor cardinal if k = a+ for some cardinal a. Otherwise k is called a limit cardinal. The heirarchy of transfinite cardinals is defined by recursion, and denoted using the Hebrew letter aleph:where w is the first infinite ordinal.Cf. cardinal arithmetic, inaccessible cardinal, regular. Related MiniText: Infinity -- You Can't Get There From Here... cardinal arithmetic Let k and l denote cardinals and A and B denote sets such that k = |A| and l = |B|. Denote by BA the set of all functions from B to A, and by A × B the Cartesian product of A and B. The operations of cardinal arithmetic are then defined byk + l = |A union B|;k • l = |A × B|;kl = |BA|For finite cardinals, this corresponds to the operations of ordinary arithmetic, but is quite different in the case of transfinite cardinals. For instance, if k is not finite, then k + k = k.Cf. ordinal arithmetic. cardinality Two sets X and Y have the same cardinality if there exists a bijection between them. Specifically, the cardinality of X may be understood as a canonical object meant to represent the proper class of sets to which X is bijective. If the axiom of choice is assumed, then the cardinality of X is the unique cardinal number having the same cardinality as X Carroll’s Paradox ARTICLE Paradox published by Lewis Carroll as “What the Tortoise Said to Achilles.” See the article for a full exposition.Cf. Charles Lutwidge Dodgson. Cartesian product For any collection {Ai}, i = 1, 2, 3, ..., n, of sets, the Cartesian productis the set of ordered n-tuples (a1,a2, ... ,an) with a1 an element of A1, a2 an element of A2, etc. The Cartesian product R2 of the set of real numbers is called the Cartesian plane, and in general n-dimensional real space is the Cartesian product Rn. The assertion that the Cartesian product of an infinite collection of non-empty sets is non-empty is equivalent to the axiom of choice. Cauchy sequence A sequence (x1, x2, x3, ... ) of elements of a metric space X with metric d(x, y) is Cauchy if for any e greater than zero there is some natural number N such thatIn other words, in a Cauchy sequence, the elements eventually become “arbitrarily close together.” If the metric space X is closed, this condition is equivalent to the sequence being convergent. CH See: continuum hypothesis. chain If X is a partially ordered set, then a subset Y of X is called a chain if it is totally ordered, that is, if for any two elements a, b of Y, either a b or b a.Cf. antichain. chain condition For a an infinite cardinal, a partial order P is said to have the a-chain condition if every antichain in P has cardinality not greater than a. If a = w, this is called the countable chain condition, or “c.c.c.” characteristic function Given a subset E of a space X, the characteristic function cE is defined by cE(x) = 1 if x is in E, and cE(x) = 0 otherwise. All properties of sets and set operations may be expressed by means of characteristic functions. choice, axiom of See: axiom of choice. circumference Geometry: The distance around a circle in the plane, or around a great circle of a sphere.Graph Theory: The circumference of a graph G is defined as the length of the longest cycle of G. The circumference is ususally denoted by c(G), and is undefined if G has no cycles. class See proper class. closed interval An interval of the real number line (or any other totally ordered set) which includes its endpoints. An interval containing only one of its endpoints is called half-open. Cf. open interval. closed set Topology: A subset E of a topological space X is closed if X - E (set difference) is open. In a metric space, E is closed if every convergent sequence in E converges in E; equivalently, if every accumulation point of E is in E.Set Theory: If a is a limit ordinal, then a set C contained in a is called closed if and only if for every limit ordinal b less than a, if C b is unbounded in b, then b C. C is called c.u.b. (“cub set” or “club set”) if and only if C is closed and unbounded in a.Cf. stationary set. closed set system If X is a set (or proper class) and F is a family of subsets of X, then F is called a closed set system providedX is a member of F, andF is closed under arbitrary intersections.Cf. filter. Cantor-Bendixson Theorem – closed set system
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http://ebooks.asmedigitalcollection.asme.org/content.aspx?bookid=2706§ionid=222553074 | 1,566,310,602,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315544.11/warc/CC-MAIN-20190820133527-20190820155527-00306.warc.gz | 59,892,546 | 25,590 | 0
Walking Gaits and Motion Control
## Excerpt
An omni-directional walking robot requires legs whose leg-tips can independently access 3-dimensional space. ‘Omni-directional’ means the ability to walk in all directions such that the walking robot is able to walk sideways left, sideways right, forwards and backwards, and turn left and right. Furthermore, possession of 3dof legs enables the robot to turn while translating, which is similar to ballroom dancers when they dance the Viennese waltz as already discussed in chapter 1. In fact other complex manoeuvres are possible that include maintaining the body in a level attitude as the legs negotiate uneven terrain. Such motion can be analysed as the body rotating about any Instantaneous Axis of Rotation (note, ‘axis’ as compared to ‘centre’) which is a three dimensional vector as compared to a two dimensional point. (This interesting problem serves as a mathematical challenge to students). The leg tip motion and the leg tip locus have been analysed in detail in chapter 4 so we now consider how leg tip motion can achieve walking gaits and motion control.
8.1Introduction
8.2Review of the leg tip locus
8.3The “5-on” gait
8.4Microcomputer real-time program for the 5-on gait
8.5Wave or ‘ripple’ action of leg tip motion during walking
8.6The “4-on” gait
8.7The “3-on” or “double tripod” gait
8.8Morphing of the radical waveforms during gait transition
8.9Locomotion speed in each gait
8.10The support polygon for static stability
8.11Walking on 4 legs
8.12Walking robot turning about an Instantaneous Centre of Rotation, IC of R
8.13Steps in computing the curvature of the leg tip locus
8.14Computation of, plangle, P, and radius of curvature, L
\$25.00
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Topic Collections | 451 | 1,905 | {"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-2019-35 | latest | en | 0.861344 |
https://keisan.casio.com/exec/system/1225094602 | 1,680,421,922,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00449.warc.gz | 387,266,456 | 8,298 | # Projection (angle, height and distance) Calculator
## Calculates the initial angle, maximum height and travel distance of the projection from the initial velocity and flight duration.
Initial velocity v m/skm/hfpsmph Flight duration t sec [ Gravity g m/s2 ] 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit Initial angle θ deree Maximum height h Travel distance l
Projection (angle, height and distance)
[1-2] /2 Disp-Num5103050100200
[1] 2017/12/15 20:18 20 years old level / An engineer / Useful /
Purpose of use
exploring
Comment/Request
hi, not sure if its a bug but for "Projection (angle, height and distance) Calculator"
"l"formula no matter how i actually put in those number by hand or transpose onto a computer do not line up. the distance is indeed correct but not too sure about the formula. if the formula is correct was wondering if you could think of any reason why my computer and hand may not be interpriting it correctly?
from Keisan
The "l" formula can be obtained as below.
i) l = v*cosθ*t
l2 = v2*cos2θ*t2
ii)
0 = v*sinθ - (g*t/2)
v2*sin2θ = (g*t/2)2
From i) and ii),
l = sqrt(v2 - (g*t/2)2) * t
[2] 2016/08/04 16:02 60 years old level or over / A retired people / Useful /
Purpose of use
space station position when visible
Sending completion
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Purpose of use? | 406 | 1,458 | {"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} | 3.1875 | 3 | CC-MAIN-2023-14 | latest | en | 0.740727 |
http://eduqna.com/Homework-Help/2275-1-Homework-Help-7.html | 1,500,582,644,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423486.26/warc/CC-MAIN-20170720201925-20170720221925-00110.warc.gz | 98,460,541 | 3,095 | Question:1. Renee must solve the equation 4x+12=6x. If she subtracts 4x from the left side of the equation, what should renee write in the right side of the equation?
2. mark's cats eat 72 ounces of food in one week. how many pounds of food do mark's cats eat in one week?
3. jordan has \$608 in his savings account. he withdraws 15% of the money to pay for school clothes. which is the best estimate for the amount of money jprdan withdraws? \$40, \$90, \$400, or \$510?
please give me an explaination! I have no clue how to do these!
1. She must also subtract 4x from the right side of the equation. If you do something to one side of an equation you must do the same to the other side.
2. 72 ounces = 4.5 pounds , there are 16 ounces in a pound, so 72 ounces divided by 16 is 4.5
3. \$608 multiplied by 15% = 91.2 so \$90 would be the best estimate. Or you could mulitiply 608 times .15 and that would also equal out to 91.2
1. renee should write -4x on the right side too. That means 12=2x so x=6
2. 4.5 (16 ounces in a pound = 72 divided by 16)
3. 10% of 608 is \$60.80 and divide that by two to find out 5 percent. That is 30.40. Add that together and that is 15%. That is \$91.20. \$90 is your answer.
1.4x+12=6x
subtract 4x from both sides
4x+12=6x
4x...4x
which equals
12=2x
divide 2 into 12
equals 6
x=6
1. 4x+12=6x
12= 6x - 4x
12 = 2x
6 = x
2. 1 pound = 16 ounces
72/16 = 4.5
Therefore 4.5 pounds.
3. 608 x .15 = 91.2
Therefore \$90 would be the best estimate.
1) first off .. if your gonna subtract from one side .. then subtract the same amount from the other ..
4x+12=6x
( subtract 4x from both sides )
4x+12 (-4x )= 6x (-4x)
since +4x cancells out the -4x on the first side
12=6x (-4x)
now 6x - 4x would be 2x
12= 2x
devide both sides by 2
12/2 = 2x/2
6 = x
2) should be easy .. how many ounces in a pound ??
divide 72 by how many ounces in a pound
3) 90 is easy
quick way to look at this is how many 100's in 608. ( 6 right ?? )
ok .. 15% of 100 is ... 15.
since ya nead 6 sets of 100 ... multiply by 6
15 x 6 = 90
sorry i can't help ya much on #2 .. not good with weights and measures
1. she must subtract 4x from the right side , what ever is added /subtracted/ multi/divided from one side must be done to the other, this is because the two sides are supposed to equal each other if something is added to one side but not the other, the equation would no longer be equal
2. First off 1 ounce = .0625 pounds, the cats eat 72 ounces in a week so you convert 72 ounces to pounds (72 x .0625 = 4.5 pounds)
3. when dealing with percent, to find the percent of something you multiply the cost by the percent (608 x .15 = 91.2) so the answer would be 90. In these types of problems pay close attention to the wording if it was asking for the amount left you would first find the amount taken out (91.2) and subtract that from the total amount in the savings
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https://www.justintools.com/unit-conversion/area.php?k1=square-miles&k2=planck-areas | 1,718,665,951,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00052.warc.gz | 731,153,802 | 27,374 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# AREA Units Conversionsquare-miles to planck-areas
1 Square Miles
= 9.9153482255656E+75 Planck Areas
Category: area
Conversion: Square Miles to Planck Areas
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Miles] symbol/abbrevation: (mi2, sq mi)
[Planck Areas] symbol/abbrevation: (Ap)
How to convert Square Miles to Planck Areas (mi2, sq mi to Ap)?
1 mi2, sq mi = 9.9153482255656E+75 Ap.
1 x 9.9153482255656E+75 Ap = 9.9153482255656E+75 Planck Areas.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Square Miles (mi2, sq mi) is equal to 2589988.11 square-meters, while 1 Planck Areas (Ap) = 2.6121E-70 square-meters.
1 Square Miles to common area units
1 mi2, sq mi = 2589988.11 square meters (m2, sq m)
1 mi2, sq mi = 25899881100 square centimeters (cm2, sq cm)
1 mi2, sq mi = 2.58998811 square kilometers (km2, sq km)
1 mi2, sq mi = 27878411.999612 square feet (ft2, sq ft)
1 mi2, sq mi = 4014489599.4792 square inches (in2, sq in)
1 mi2, sq mi = 3097599.9995981 square yards (yd2, sq yd)
1 mi2, sq mi = 1 square miles (mi2, sq mi)
1 mi2, sq mi = 4.0144895994792E+15 square mils (sq mil)
1 mi2, sq mi = 258.998811 hectares (ha)
1 mi2, sq mi = 639.99943412918 acres (ac)
Square Milesto Planck Areas (table conversion)
1 mi2, sq mi = 9.9153482255656E+75 Ap
2 mi2, sq mi = 1.9830696451131E+76 Ap
3 mi2, sq mi = 2.9746044676697E+76 Ap
4 mi2, sq mi = 3.9661392902263E+76 Ap
5 mi2, sq mi = 4.9576741127828E+76 Ap
6 mi2, sq mi = 5.9492089353394E+76 Ap
7 mi2, sq mi = 6.9407437578959E+76 Ap
8 mi2, sq mi = 7.9322785804525E+76 Ap
9 mi2, sq mi = 8.9238134030091E+76 Ap
10 mi2, sq mi = 9.9153482255656E+76 Ap
20 mi2, sq mi = 1.9830696451131E+77 Ap
30 mi2, sq mi = 2.9746044676697E+77 Ap
40 mi2, sq mi = 3.9661392902263E+77 Ap
50 mi2, sq mi = 4.9576741127828E+77 Ap
60 mi2, sq mi = 5.9492089353394E+77 Ap
70 mi2, sq mi = 6.9407437578959E+77 Ap
80 mi2, sq mi = 7.9322785804525E+77 Ap
90 mi2, sq mi = 8.9238134030091E+77 Ap
100 mi2, sq mi = 9.9153482255656E+77 Ap
200 mi2, sq mi = 1.9830696451131E+78 Ap
300 mi2, sq mi = 2.9746044676697E+78 Ap
400 mi2, sq mi = 3.9661392902263E+78 Ap
500 mi2, sq mi = 4.9576741127828E+78 Ap
600 mi2, sq mi = 5.9492089353394E+78 Ap
700 mi2, sq mi = 6.9407437578959E+78 Ap
800 mi2, sq mi = 7.9322785804525E+78 Ap
900 mi2, sq mi = 8.9238134030091E+78 Ap
1000 mi2, sq mi = 9.9153482255656E+78 Ap
2000 mi2, sq mi = 1.9830696451131E+79 Ap
4000 mi2, sq mi = 3.9661392902263E+79 Ap
5000 mi2, sq mi = 4.9576741127828E+79 Ap
7500 mi2, sq mi = 7.4365111691742E+79 Ap
10000 mi2, sq mi = 9.9153482255656E+79 Ap
25000 mi2, sq mi = 2.4788370563914E+80 Ap
50000 mi2, sq mi = 4.9576741127828E+80 Ap
100000 mi2, sq mi = 9.9153482255656E+80 Ap
1000000 mi2, sq mi = 9.9153482255656E+81 Ap
1000000000 mi2, sq mi = 9.9153482255656E+84 Ap
(Square Miles) to (Planck Areas) conversions | 1,335 | 3,241 | {"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-2024-26 | latest | en | 0.658642 |
https://questions.examside.com/past-years/jee/question/pa-steady-current-i-flows-along-an-infinitely-long-hollow-jee-advanced-physics-units-and-measurements-bcw0h1lfcnk5em2a | 1,723,700,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00746.warc.gz | 370,039,073 | 43,159 | 1
JEE Advanced 2013 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is(are)
A
In the region 0 < r < R, the magnetic field is non-zero.
B
In the region R < r < 2R, the magnetic field is along the common axis.
C
In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centred on the axis.
D
In the region r > 2R, the magnetic field is non-zero.
2
JEE Advanced 2013 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2
A particle of mass M and positive charge Q, moving with a constant velocity $${\overrightarrow u _1} = 4\widehat i$$ ms$$-$$1 enters a region of uniform static magnetic field, normal to the xy plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 ms with a velocity $${\overrightarrow u _2} = 2\left( {\sqrt 3 \widehat i + \widehat j} \right)$$ ms$$-$$1. The correct statement(s) is(are)
A
The direction of the magnetic field is $$-$$z direction.
B
The direction of the magnetic field is +z direction.
C
The magnitude of the magnetic field is $${{50\pi M} \over {3Q}}$$ units.
D
The magnitude of the magnetic field is $${{100\pi M} \over {3Q}}$$ units.
3
IIT-JEE 2012 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2
Consider the motion of a positive point charge in a region, there are simultaneous uniform electric and magnetic fields $$\overrightarrow E = {E_0}\widehat j$$ and $$\overrightarrow B = {B_0}\widehat j$$. At time t = 0, this charge has velocity $$\overrightarrow v$$ in the xy-plane, making an angle $$\theta$$ with the x-axis. Which of the following option(s) is(are) correct for time t > 0 ?
A
If $$\theta$$ = 0$$^\circ$$, the charge moves in a circular path in the xy-plane.
B
If $$\theta$$ = 0$$^\circ$$, the charge undergoes helical motion with constant pitch along the y-axis.
C
If $$\theta$$ = 10$$^\circ$$, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.
D
If $$\theta$$ = 90$$^\circ$$, the charge undergoes linear but accelerated motion along the y-axis.
4
IIT-JEE 2011 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?
A
They will never come out of the magnetic field region.
B
They will come out travelling along parallel paths.
C
They will come out at the same time.
D
They will come out at different times.
EXAM MAP
Medical
NEET | 821 | 2,947 | {"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} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.790929 |
http://www.assignmenthelp.net/assignment_help/number-lines | 1,432,304,971,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207925274.34/warc/CC-MAIN-20150521113205-00231-ip-10-180-206-219.ec2.internal.warc.gz | 307,037,331 | 8,172 | # Math Assignment Help With Number Lines
## Chapter 15 Number Lines
15.1 Introduction: A number line is a line in which real numbers can be placed, according to their value. Each point on a number line corresponds to a real number, and each real number has a unique point that corresponds to it. For example, the number 1.5 (1 1/2) corresponds with the point on a number line that is halfway between one and two.
Often the only integers are shown as specially-marked points evenly spaced on the line.
Although in this image only the integers from −9 to 9 are shown, the line includes all numbers. The arrows on both the sides of the line mean that the line continues till infinity i.e. it contains all the real numbers from minus infinite to plus infinite.
The points on a number line are called coordinates. The zero point is called the origin. The numbers to the right of the origin are positive numbers; the numbers to the left of the origin are negative numbers.
### 15.2 Drawing the number line:
The number line is always represented as a horizontal line. Positive numbers lie on the right side of zero, and negative numbers lie on the left side of zero. An arrowhead is put on either end of the line which is meant to suggest that the line continues indefinitely in the positive and negative directions, as already suggested above.
### Characteristics of a number line
1. It is a horizontal line.
2. The points on a number line are called coordinates.
3. The zero point is called the origin.
4. The numbers to the right of the origin are called positive numbers and the numbers to the left of the origin are called negative numbers.
5. It has two arrow heads on either end of the line. This represents that the line extends up to infinity.
15.3 Decimal number line: To represent decimals on number line, divide each segment into 10 parts.
Example: represent 6.5 in the number line.
Draw a number line, dividing the segment between 6 and 7 into 10 equal parts
### 5.4 Graphing Inequalities on a Number Line
A number line is a horizontal line having points which correspond to numbers. The points are spaced according to the value of the numbers and are equally spaced.
We can graph real numbers by representing them as points on the number line. For example, we can graph 3 ¼ on the number line:
We can also graph inequalities on the number line.
The following graph represents the inequality x≤ 3 ¼ .
The dark line represents all the numbers that satisfy x≤ 3 ¼ .
Take any number on the dark line and plug it in for x, the inequality will be true. | 563 | 2,577 | {"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} | 5 | 5 | CC-MAIN-2015-22 | longest | en | 0.921648 |
http://studylib.net/doc/10259660/solving-one-step-equations | 1,531,953,015,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00586.warc.gz | 343,388,721 | 12,327 | ```Solving one step equations
Properties of Equality
1.
Addition: If a = b then a + c = b + c.
2. Subtraction: If a = b then a – c = b– c.
3. Multiplication: If a = b then ac = bc.
4. Division: If a = b and c ≠ 0 then a/c = b/c
Rules to Remember
Quick Facts and
Definitions:
When solving an equation,
the goal is to get the
variable by itself on one
side of the equation.
are inverse operations.
(opposites)
Multiplication and
division are inverse
operations. (opposites)
Solving Equations Requires
BALANCE!
the fulcrum (balancing
point)!
Just remember, whatever
you do to one side must
be done to the other side
to maintain
balance……..
If the number to move is positive, subtract it from
both sides of the equation.
If the number to move is negative, add its positive
value to both sides of the equation.
Solve the following equation for x:
x – 4 =
9
+4
+4
x
=
13
We are able to get the x by itself
by ADDING 4 to each side!
(you WILL show this step in your homework!!)
Solving with Subtraction
Solve the following equation for p:
p + 7 =
21
-7
-7
p
=
14
We are able to get the p by itself
by SUBTRACTING 7 from each side!
(you WILL show this step in your homework!!)
Solving with Division
Solve the following equation for m:
3m =
18
Divide both sides by 3 and simplify -- your work should
look like this
:
3m =
3
m =
18
3
6
(you WILL show this step in your homework!!)
Solving with Multiplication
Problems involving fractions are solved using
multiplication, and the knowledge of multiplicative
inverses (reciprocals)
Example :
Solve the following equation for n:
2n
=
3
5
7
Multiply each side by the inverse of 25 :
5 • 2n
2
5
=
n
=
3 • 5
7
2
15
14
Solve the following equations
with one step (show your
work!!):
1.
r + 4 = 17
2.
g - 15 = -11
3.
13 + p = - 9
4.
-5b = 85
5.
- 7v = - 91
6.
w = -2
3
9
1.
2.
3.
4.
5.
6.
r = 13
g = 4
p = - 22
b = - 17
v = 13
w = - 2/3
Word Problems
with One Step Equations
You will need to know how to solve word
problems involving these equations, by using
the four-step problem solving plan:
1.
Explore the problem
2.
Plan the solution
3.
Solve the problem
4.
Examine the solution
1. Explore the problem
problem carefully,
then:
Identify what
information is
given
Identify what you
2. Plan the Solution
Choose a variable to represent the unknown that you
are trying to solve for
Create an algebraic equation for solving
3.
Solve the Problem
4.
Examine the Solution
use algebraic
methods
“reasonable” ?
Did you check your
work to be sure?
Let’s try it!
One mile is equal to approximately 1.6 kilometers. A
local running club is holding a 10 kilometer race.
What is the length of the race in miles?
Define aVariable Let m = the number of miles
1.6 • m = 10
Divide both sides by 1.6 (calculators ok)
m = 6.25
The length of the race is 6.25 miles.
One more !
Write an Equation and Solve the Problem:
some athletic shoes for \$112, he had \$96 left.
How much was his paycheck?
Let p = paycheck
p - 112
=
96
+ 112
+112
p
=
208
Noah’s paycheck was \$208.
Homework:
You MUST show your work to get
credit on this assignment!!!
``` | 1,043 | 3,115 | {"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-2018-30 | latest | en | 0.906962 |
https://www.teacherspayteachers.com/Product/Equivalent-Fractions-Card-Deck-TEK-43-1743783 | 1,485,006,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281084.84/warc/CC-MAIN-20170116095121-00013-ip-10-171-10-70.ec2.internal.warc.gz | 988,505,119 | 51,789 | # Equivalent Fractions Card Deck (TEK 4.3)
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
16.81 MB | 60 cards pages
### PRODUCT DESCRIPTION
"But Miss, fractions are HARD!!" I've heard that a FEW times this year! To make it a little better, I created this deck of fraction cards for your students to enjoy. There are 15 fraction sets with a picture, reduced fraction, and two equivalents. You can use this set to order and compare, to sort, or to match classmates. You can also spice it up and play Spoons or Go Fish! I hope you enjoy these cards. I used them to support the NEW 4.3 TEK on fractions.
Standards:
(3) Number and operations. The student applies mathematical process standards to represent and explain fractional units. The student is expected to:
(G) explain that two fractions are equivalent if and only if they are both represented by the same point on the number line or represent the same portion of a same size whole for an area model; and
(H) compare two fractions having the same numerator or denominator in problems by reasoning about their sizes and justifying the conclusion using symbols, words, objects, and pictorial models.
The student applies mathematical process standards to develop and use strategies and methods for whole number computations in order to solve problems with efficiency and
accuracy. The student is expected to:
4.3 (A) Represent a fraction a/b as a sum of fractions 1/b, where a and b are whole numbers and b is greater than zero, including when a is greater than b.
4.3 (B) Decompose a fraction in more than one way into a sum of fractions with the same denominator using concrete and pictorial models and recording results with symbolic representations
******4.3 (C)Determine if two given fractions are equivalent using a variety of methods******
4.3 (D) Compare two fractions with different numerators and different denominators and represent the comparison using the symbols <, >, =
Could support or provide intervention for 5th Grade TEK 5.3
Common Core:
CCSS.MATH.CONTENT.3.NF.A.3
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
CCSS.MATH.CONTENT.3.NF.A.3.A
Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
CCSS.MATH.CONTENT.3.NF.A.3.B
Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3. Explain why the fractions are equivalent, e.g., by using a visual fraction model.
CCSS.MATH.CONTENT.3.NF.A.3.C
Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 = 6; locate 4/4 and 1 at the same point of a number line diagram.
CCSS.MATH.CONTENT.3.NF.A.3.D
Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
CCSS.MATH.CONTENT.4.NF.A.1
Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
CCSS.MATH.CONTENT.4.NF.A.2
Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Total Pages
60 cards
N/A
Teaching Duration
N/A
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
7 ratings | 997 | 4,154 | {"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-2017-04 | longest | en | 0.926568 |
phebach.blogspot.com | 1,601,458,700,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00493.warc.gz | 549,188,320 | 46,884 | Showing posts with label Chemistry. Show all posts
Showing posts with label Chemistry. Show all posts
## Wednesday, July 1, 2020
### Scientific Notation and Significant Figures Worksheet 3
Scientific Notation/Significant FiguresWorksheet 3
1. Give the number of significant figures in each of the following.
a) 10.0005 g ______
b) 0.003423 mm ______
c) 2900 + 100 ft ______
d) 8.9 x 105 L ______
e) the number of minutes (60) that make up an hour ______
2. Determine the answer for each of the following. Be sure to use the correct number of significant figures.
a) 27.34 b) 2.8023
6.90 - 4.762
+ 13.124
c) 0.32 x 14.50 x 120 = d) 24.1 / 0.005 =
3. Round each of the following to 3 significant figures.
707.5 ____________ 2,300.2 ______________
0.0003350 _____________ 10.26730 _______________
18.95 x 1021 ______________
4. Convert each of the following into correct scientific notation.
1747 _________________________________
0.00000984 ___________________________
3200.0 x 102 ____________________________
0.002014 x 102 __________________________
25600000000000000 ______________________ (use 4 sig. fig. for the last one only)
5. Calculate the following using the correct number of significant figures.
a) 2.34 x 1047 b) 9132.0
+ 9.2 x 1046 - 1.6 x 103
6. Calculate the following using the correct number of significant figures.
a) (1.54 x 1058)(3.5 x 1060)
b) (7.9 x 1034) / (8.32 x 1023)
1. Give the number of significant figures in each of the following.
a) 10.0005 g 6
b) 0.003423 mm 4
c) 2900 + 100 ft 2
d) 8.9 x 105 L 2
e) the number of minutes (60) that make up an hour infinite
2. Determine the answer for each of the following. Be sure to use the correct number of significant figures.
a) 47.36 b) -1.960
c) 5.6 x 102 d) 5 x 103
3. Round each of the following to 3 significant figures.
708 2.30 x 103
0.000335 10.3
19.0 x 1021
4. Convert each of the following into correct scientific notation.
1.747 x 103
9.84 x 10-6
3.2000 x 105
2.014 x 10-1
2.560 x 1016
5. Calculate the following using the correct number of significant figures.
a) 3.26 x 1047 b) 7.5 x 103
6. Calculate the following using the correct number of significant figures.
a) 5.4 x 10118
b) 9.5 x 1010
### Scientific Notation and Significant Figures Worksheet 2
Scientific Notation/Significant FiguresWorksheet 2
1. Convert each of the following into scientific notation.
727 _________________________________
172000 _________________________________
0.000984 _________________________________
200.0 x 102 _________________________________
0.014 x 102 _________________________________
2560000000000000000000000000000000 __________________________
(use 4 sig. fig. for the last one)
2. Convert each into decimal form.
1.56 x 104 _________________________________
3.6 x 10-2 _________________________________
736.9 x 105 _________________________________
0.0059 x 105 _________________________________
0.00059 x 10-1 _________________________________
3. Calculate the following. Give the answer in correct scientific notation.
a) 2.34 x 1065 b) 313.0
+ 9.2 x 1066 - 1.2 x 103
4. Calculate the following. Give the answer in correct scientific notation.
a) 8.95 x 1076/ 1.25 x 1056 b) (4.5 x 1029)(2.45 x 10100)
5. Give the number of significant figures in each of the following.
a) 1.05 g ______
b) 0.0003040 mm ______
c) 29000 + 10 ft ______
d) 0.90 x 1045 L ______
e) the number of eggs (12) that make up a dozen ______
6. Determine the answer for each of the following. Be sure to use the correct number of significant figures.
a) 17.34 b) 9.80
4.900 - 4.762
+ 23.1
c) 3.9 x 6.05 x 420 = d) 14.1 / 5 =
7. Round each of the following to 3 significant figures.
77.0653 ____________ 6,300,178.2 ______________
0.00023350 _____________ 10.2030 _______________
2.895 x 1021 ______________
1. Convert each of the following into scientific notation.
7.27 x 102
1.72000 x 105
9.84 x 10-4
2.000 x 104
1.4
2.560 x 1033
2. Convert each into decimal form.
15600 + 100
0.036
73690000 + 10000
590 + 10
0.000059
3. Calculate the following. Give the answer in correct scientific notation.
a) 9.434 x 1066 b) - 8.87 x 102
4. Calculate the following. Give the answer in correct scientific notation.
a) 7.16 x 1020 b) 1.1025 x 10130
5. Give the number of significant figures in each of the following.
a) 1.05 g 3
b) 0.0003040 mm 4
c) 29000 + 10 ft 4
d) 0.90 x 1045 L 2
e) the number of eggs (12) that make up a dozen infinite
6. Determine the answer for each of the following. Be sure to use the correct number of significant figures.
a) 45.3 b) 5.04
c) 9.9 x 103 d) 3
7. Round each of the following to 3 significant figures.
77.1 6.30 x 106 0.000234 10.2 2.90 x 1021
### Scientific Notation and Significant Figures Worksheet 1 with Answer keys
Scientific Notation/Significant Figures
Worksheet 1
1. Convert each of the following into scientific notation.
a) 3427 b) 0.00456 c) 123,453 d) 172 e) 0.000984 f) 0.502 g) 3100.0 x 102 h) 0.0114 x 104 i) 107.2 j) 0.0000455 k) 2205.2 l) 30.0 x 10-2 m) 0.982 x 10-3 n) 0.0473 o) 650.502 p) 3.03 x 10-1 q) 20.4 x 105 r) 1.29 s) 0.00565 t) 1362205.2 u) 450.0 x 103 v) 1000 x 10-3
2. Determine the number of significant figures in each of the following:
a) 3427 b) 0.00456 c) 123,453 d) 172 e) 0.000984 f) 0.502 g) 3100.0 x 102 h) 0.0114 x 104 i) 107.2 j) 0.0000455 k) 2205.2 l) 30.0 x 10-2 m) 0.982 x 10-3 n) 0.0473 o) 650.502 p) 3.03 x 10-1 q) 20.4 x 105 r) 1.29 s) 0.00565 t) 1362205.2 u) 450.0 x 103 v) 1000 x 10-3 w) 546,000 + 10 x) 546,000 + 1000
3. Convert each into decimal form.
1.56 x 104 0.56 x 10-2 3.69 x 10-2 736.9 x 105 0.00259 x 105 0.000459 x 10-1 13.69 x 10-2 6.9 x 104 0.00259 x 103 0.0209 x 10-3
4. Calculate the following. Give the answer in correct scientific notation.
a) 4.53 x 105 b) 1913.0
+ 2.2 x 106 - 4.6 x 103
c) 2.34 x 1024 d) 2.130 x 103
+ 1.92 x 1023 - 6.6 x 102
e) 9.10 x 103 f) 1113.0
+ 2.2 x 106 - 14.6 x 102
g) 6.18 x 10-45 h) 4.25 x 10-3
+ 4.72 x 10-44 - 1.6 x 10-2
5. Calculate the following. Give the answer in correct scientific notation.
a) 3.95 x 102/1.5 x 106 b) (3.5 x 102)(6.45 x 1010)
c) 4.44 x 107 /2.25 x 105 d) (4.50 x 10-12)(3.67 x 10-12)
e) 1.05 x 10-26 / 4.2 x 1056 f) (2.5 x 109)(6.45 x 104)
g) 6.022 x 1023 / 3.011 x 10-56 h) (6.88 x 102)(3.45 x 10-10)
Left increase, right decrease
1. Convert each of the following into scientific notation.
a) 3.427 x 103 b) 4.56 x 10-3 c) 1.23453 x 105 d) 1.72 x 102 e) 9.84 x 10-4 f) 5.02 x 10-1 g) 3.1000 x 105 h) 1.14 x 102 i) 1.072 x 102 j) 4.55 x 10-5 k) 2.2052 x 103 l) 3.00 x 10-1 m) 9.82 x 10-4 n) 4.73 x 10-2 o) 6.50502 x 102 p) 3.03 x 10-1 q) 2.04 x 106 r) 1.29 x 100 = 1.29 s) 5.65 x 10-3 t) 1.3622052 x 106 u) 4.500 x 105 v) 1.000 x 100 = 1.000
2. Determine the number of significant figures in each of the following:
a) 4 b) 3 c) 6 d) 3 e) 3 f) 3 g) 5 h) 3 i) 4 j) 3 k) 5 l) 3 m) 3 n) 3 o) 6 p) 3 q) 3 r) 3 s) 3 t) 8 u) 4 v) 4 w) 5 x) 3
3. Convert each into decimal form.
15600 + 100 0.0056 0.0369 73690000 + 10000 259 4.59e-05 0.1369 69000 + 1000 2.59 2.09e-05
4. Calculate the following. Give the answer in correct scientific notation.
a) 2.7 x 106 b) -2.7 x 103 c) 2.53 x 1024 d) 1.47 x 103 e) 2.2 x 106 f) -3.5 x 102 g) 5.34 x 10-44 h) -1.2 x 10-2
5. Calculate the following. Give the answer in correct scientific notation.
a) 2.63 x 10-4 b) 2.3 x 1013 c) 1.97 x 102 d) 1.65 x 10-23 e) 2.5 x 10-83 f) 1.6 x 1014 g) 2.000 x 1079 h) 2.37 x 10-7
## Thursday, December 11, 2014
### THINGS THAT YOU NEED TO KNOW FOR THE FINAL - HS Chemistry Final
THINGS THAT YOU NEED TO KNOW FOR THE FINAL
- Students can use the periodic table, a 5-8 “cheat sheet”, a scientific calculator, and the ion sheet on the final exam.
- There will be about 150 multiple choice questions.
- Your final grade will be posted by Friday of the final week.
Chapter 1 – INTRODUCTION TO CHEMISTRY
a. What is chemistry? 5 branches of chemistry. Scientific law and the scientific method. Law of conservation of mass, etc.
b. Know all of terminologies for chapter one; Distinguish between pure and applied chemistry; and why study chemistry.
#### Suggested problems: #34-64 even pg 35-37and whole page 37
Chapter 2 – MATTER AND CHANGE
c. All terminologies, different states and properties of matter.
d. Distinguish between substance and mixture (figure 2.11)
e. Identify chemical reaction, chemical and physical changes
Suggested problems: page58-59 # 33-69 odds
Chapter 3: SCIENTIFIC MEASUREMENT
a. Learn all terminologies and SI units system.
b. Conversion factor and Dimensional Analysis
c. Converting between units both simple and multi-step problems
d. Know rules for significant figures and scientific notation, etc
e. Conversion between temperature and other units, calculate density.
Suggested problems: pages 87-96, #38 to 70 evens
Chapter 4: Atomic Structure
a. Know who discover protons, electrons, neutrons
b. How to write atomic structure and its calculations: Atomic #, Atomic mass, etc
c. Isotopes of Element and calculating the atomic mass of an element
Suggested problems: pages 122-123 # 34 to 56 even.
Chapter 5: Electrons in Atoms
a. Know the development of Atomic models and all key terms
b. Know to write electron configuration, Lewis dot structure and draw orbital diagrams.
c. Know the physics and the quantum mechanical model; electromagnetic radiation; and calculate the frequency, wavelength, and energy.
Suggested problems: pages 149-150 # 23 to 47 odds and # 57-61 page 150.
Chapter 6: The Periodic Table
Everything about it; trends, patterns, location, names of groups/family.
Suggested problems: the first 30 problems of each chapter.
Chapter 7 and 8: CHEMICAL BONDING
Everything you need to know about bonding from metallic to non-polar covalent bonding. Suggested problems: the first 30 problems of each chapter.
Chapter 9: CHEMCIAL NAMES AND FORMULAS
a. Know the periodic table well
b. Know how to write and name element, ions, compounds
c. Must know how to write chemical formulas properly
d. Know the laws of definite and multiple proportions
Suggested problems: practice concepts 1-20 and page 136-138 # 33 to 51 odds.
Chapter 10 – Chemical Quantity
### 1. Know how to calculate molar mass.
1. Find the % composition of an element in a compound.
2. Conversions between mass, moles, and molecules of the same compound.
Suggested problems – the first 30 questions in the chapter.
## Monday, September 10, 2012
### Mr. Bach's Chemistry Course Outline and Syllabus
CP Chemistry Prerequisite – IM 2 or higher Graduation – Fulfill one year of physical science requirement Duration - 2 semesters Credit – 5 units per semester with “D-” or better for high school. “C-” for UC/CSU. 10 credits – 9th - 12th grade A-G approved science course for UC 1 year laboratory science “d”
Chemistry Course Outline and Syllabus
Science Department - Mira Loma High School
Teacher: Phe Bach, Ed.D.
Email: pbach@sanjuan.edu; Room: H-207
MLHS Vision/Mission Statement:
Embracing diversity and valuing excellence, Mira Loma High School's mission is to inspire and educate each student toward academic achievement, critical thinking, intrinsic success, and responsible contributions to a peaceful international society by providing innovative, rigorous, student-focused instruction through a rich tradition of high-quality programs and dynamic activities in a safe, compassionate, and collaborative learning community.
Course Goals:
- Achieving the defined California State Standards for High School Chemistry.
- Fulfilling the requirements of IB Middle Years Programs.
- Preparing students for college and presenting chemistry material that applies to their daily life with a positive attitude.
Organization/Course Standards: This course is followed closely the Prentice Hall Chemistry textbook. It is organized into ten units according to California State and SJUSD’s Standards for chemistry. Students will understand the following topics:
Unit 1 – Introduction to the Methods of Chemistry:
Measurements and Calculations; Lab Equipment and Lab Safety
Unit 2 – Matter and Changes / Scientific Measurement:
Matter, elements and compounds, mixtures, chemical reactions. Measurements and their uncertainty, Basic SI systems, Density.
Unit 3 – Atomic Structure, Electrons and Periodic Behavior
Atoms – The building blocks of Matter; Structure of the atoms; arrangement of Electrons in Atoms, the Periodic Law, Periodic Trends
Unit 4 –Chemical Bonding and Molecular Structure:
Chemical bonding; Structure of Covalent Molecules, Salts, and Metals
Unit 5 – Chemical Names and Formula/Conservation of Matter:
Chemical Formulas and Compounds, Chemical Equations and Reactions.
Unit 6 – Stoichiometry
The arithmetic of equations, chemical calculations, limiting reagent and percent yield.
Unit 7 – Gas Laws / Gas Behavior:
Physical Characteristics of Gases, Gas Laws, Molecular Composition of Gases
Unit 8 – Solids, Liquids and Solutions / Acids and Bases:
Liquids and Solids, Properties of Solutions and Solution Concentration, Acids and Bases, Characteristics of Acids and Bases, Titration, calculating pH, neutralization of acids and bases.
Unit 9 – Electrochemistry, Thermochemistry and/or Nuclear Chemistry
Electrochemical cells, reduction-oxidation reactions; The flow of energy, heat in changes of states, nuclear radiation, nuclear transformations, Fission/Fusion of Atomic nuclei.
Unit 10 – Introduction to Organic and Biochemistry.
Subjects outside of the State Standards, covered after the Content Standards Test in early May.
a. Classwork/Homework, standards testing, presentations 20%
b. Portfolio, Quizzes, and Unit Tests 25%
c. Laboratory notebook, lab quizzes, project and lab reports 25%
d. Comprehensive final exam 15%
e. Participation and attendance 15%
100% - 90% = A || 90% - 80% = B || 80% - 70% = C || 70% - 60% = D || Below 59.9% = F
I’ll use the normal grade scale with “plus” and “minus”.
I’ll not curve my tests, but I DO give extra credits.
Extra Credits:
You made earn up to 1.0% per semester. Bonus points are awarded from time to time at the instructor’s discretion for a variety of little extra reasons such as bring materials for the lab, writing summary of science/chemistry articles, asking brilliant questions and seeking the answer, helping out the room/lab, etc… or conducting an original research or writing an exceptional research paper and this work is no “Mickey Mouse” project. Ask the instructor for more information.
Besides HONESTY and PERSONAL INTEGRITY, RESPECT is the key. Being courteous, mindful and respectful to fellow students, T.A., substitute, your instructor, and our classroom settings are expected at all times. Unnecessary acting out, put down, talking, trashing class/campus, or other immature behaviors will lead to lowering of your citizenship grade. Finally, the citizenship grading policy will comply with Mira Loma and San Juan School Districts’ policies. This includes adhering to all rules in our handbook. For example, after 3 tardies (<30mins) results in parent-called plus 1-hour detention and after each truancy (>30mins) will results in lower 1 citizenship grade.
The following criteria will be used to determine the citizenship grade:
A: Student is attentive, stays on task, complies with teacher requests, is
courteous to others, treats equipment well, contributes positively to the
class learning environment, follows class rules and is on time to class.
B: Student demonstrates occasional transgressions of the expectations listed for an A, but makes an effort not to repeat them.
C: Student demonstrates occasional transgressions of the expectations listed for the “B” student and may be inattentive, off task, unprepared or occasionally tardy and may require reminders to remain on task.
D: Student is disruptive to the class and learning environment and makes little effort to correct his/her behavior even after disciplinary actions are taken. He/she may exhibit unacceptable or rude interpersonal behavior and may accumulate many tardies during the grading period.
F: Student persistently disrupts the classroom learning environment, fails to respect individual rights and property or school responsibilities along with possibly being chronically tardy to class.
NOTE: A major infraction of these rules may result in a more dramatic drop in the citizenship grade after conferencing with the student, legal guardian and/or administrator. Issues with academic malpractice will also negatively affect the student’s citizenship grade.
Electronic Devices: MUST BE TURNED OFF AND STORED OUT OF SIGHT WHILE STUDENTS ARE INSIDE THE CLASSROOM.
Electronic devices may be used before school, after school, during passing periods, break, and lunch. The school is not responsible for any lost or stolen items. Students will receive progressive discipline for using their cell phone during class (1st warning, 2nd phone taken away/‘phone jail’, 3rd detention/referral). Phone sleeves will be used during quizzes and tests.
At any time during the school year the privilege to use electronic devices during school hours may be revoked by school administration (phone lockers in the VP office).
Passes: Please use the restroom at lunch, passing periods, or before school, however, if you need to use the restroom, just raise your hand without interrupting the class. If this becomes a frequent occurrence, I will be forced to have a (remarkably awkward) conversation with you and your parents about the obvious medical issue you have. Per school rules, students are not allowed to use the restroom the first and last ten minutes of class. A forgetful mind is not an emergency, make sure all of your possessions are with you and not in the quad, or in you car, or in your last classroom, or . . .
Each student is responsible for performing academic tasks in such a way that honesty is not in question. Plagiarism is a serious violation of academic honesty, and students are expected to maintain the following standards of integrity: All tests, term papers, oral or written assignments, projects and recitations are to be the work of the student presenting the material. Any use of wording, ideas, or findings of other persons, writers, or researchers requires the explicit citation of the source; use of the exact wording requires a “quotation” format and citation (MLA format). Any student deliberately assisting another student in academic dishonesty is also culpable. (i.e. Letting a friend copy your homework, sharing test questions or answers, providing a photo of tests or quizzes) Students found in violation of the academic integrity policy will be subject to school discipline which may include, loss of, or zero credit on the assignment, lowered citizenship grade in the course, parent conference, and Saturday School. Teacher, counselor, or student may request restorative practices.
MAKE UP POLICY:
When you are absent, it is your responsibility to turn in work that was due during your absence the following day you’re back. If you missed a lab/test, it must make up within ONE week or it will results in a zero. Late works will NOT be accepted unless it is an excused absence. Homework are turn in every Wednesday and Friday. Late work is no longer accepted. All grade discrepancy must take care of within one week of posted grade.
Contact: I am almost always available during lunch (except when participating in intramural sports) and you may catch me during break, before or after school. To ensure you catch either of us, a quick email to set up an appointment is useful. The best way for you and your parents to reach us is through E-mail: pbach@sanjuan.edu. Students can also use Google Classroom (which goes to our email anyway). If needed, we will message students using Google Classroom so please ensure you have set up a method to see those messages (historically, students have found the app is VERY helpful for this).
Code of Conduct: Classroom policies, policies on attendance, tardiness, dress code, and behavior, are addressed in the student handbook, will be implemented along with district policies (Please see below)
Mr. BACH’S FIVE STUDYING AND LIVING PRINCIPLES
Or CODE OF CONDUCT
Classroom policies, policies on attendance, tardiness, dress code, electronic devices and behavior are addressed in the student handbook. They will be implemented along with district policies. Students cannot use electronic devices at any given time in the classroom. Exceptions will be allowed during class activities such as Kahoot, Google classroom, researching, and so on.
FIVE STUDYING AND LIVING PRINCIPLES
Mr. Bach will also implement these Five Studying/Living Principles in our classroom. It is a mindful way of living and we will practice reflective and mindfulness, as it is the key to peace and harmony with oneself in the classroom, in the family and ultimately in the world.
1. Do no harm. (Non-harming) – Abstain from intentionally hurting or killing any animate life. Do not harass or bully anyone or other beings physically, mentally or emotionally. For example, if you see spiders or other insects in the classroom please take them outside or let me know and I will do it. In Mr. Bach’s classroom remember this wise saying: think no evil; speak no evil; act no evil, text no evil, post no evil and comment no evil.
2. Have integrity. (Honesty) – Don’t take anything that does not belong to you without permission, including intellectual property. This means that you should avoid plagiarism, copying, or taking materials such as chemicals from the lab. Only take what is offered. Practice self-discipline and show self-esteem.
3. Respect yourself and others. (Respect) – Respect all others, including different living organisms, fellow classmates, other teachers, substitute teachers, and so on. Show courtesy and politeness. Being respectful starts with the Golden Rules. Be considerate of other people's feelings. Show manners, etiquette and sincere intention as it is a way to keep society in harmony. Respect the classroom space and property. Being respectful also means don't discriminate and being open-minded.
4. Communicate responsibly. (Communication) – Let’s put ourselves in each other’s shoes. Be mindful of your communication, including what you think, say and do. Please be responsive rather than reactive- think of a win-win situation. Please listen attentively; use deep listening skills and loving speech.
5. Show kindness and compassion. (Compassion) Show empathy and sympathy. Be kind, warm-hearted and caring to yourself and others around you. However, before you can show kindness, love or compassion for others, you must first practice self-kindness, self-love and self-compassion for yourself and family. In order to make our surroundings more kind, loving and compassionate, we have to be an agent of change. Happiness, love, kindness, and compassion start from the within.
Resources:
1. Good website to learn: www.sciencegeek.net | 6,696 | 24,386 | {"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-2020-40 | longest | en | 0.414911 |
http://elf.43.junowebmaillogin.com/6-grade-fraction-worksheet/ | 1,606,571,533,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00156.warc.gz | 35,905,805 | 18,497 | # 6 Grade Fraction Worksheet
## Grade 6 Fractions Worksheets Worksheets Buddy
Grade 6 Maths Fractions Multiple Choice Questions Mcqs 1 Write The Fraction Representing The Shaded Region In The Below Left Figure A B C D 2 Write The Fraction Representing The Shaded Region In The Above Sided Right Figure A B C D 3 Fill In The Boxes With The Correct Symbol A B Read More Grade 6 Fractions Worksheets
### Grade 6 Addition And Subtraction Of Fractions Worksheets
Fractions Worksheet These Grade 6 Fractions Worksheets Focus On Adding And Subtracting Fractions And Mixed Numbers With Unlike Denominators All Worksheets Are Files And Answer Keys Follow The Questions On A Separate Page
#### Grade 6 Simplifying And Converting Fractions Worksheets
Equivalent Fractions Math Worksheets These Grade 6 Math Worksheets Provide Practice In Simplifying Fractions Recognizing Equivalent Fractions And Converting Fractions To And From Mixed Numbers Exercises With Various Levels Of Difficulty Are Provided Answer Sheets Follow The Exercises
##### 6th Grade Fractions Worksheets Lessons And Printables
Fraction Worksheets And Printables 6th Grade Fractions Worksheets Lessons And Printables Number Theory Divisibility Circle All Of The Numbers That Are Divisible By The Number Divisibility Complete The Table Prime Or Composite Use The Clue To Fill In The Missing Digit Circle All Of The Numbers That Are Multiples Of A Number
###### Grade 6 Math Worksheets And Problems Fractions Edugain
Printable Worksheets And Online Practice Tests On Fractions For Grade 6 Fractions Mixed Review Word Problems
Grade 6 Multiplication And Division Of Fractions
Fraction Multiplication And Division Math Worksheets These Grade 6 Math Worksheets Cover The Multiplication And Division Of Fractions And Mixed Numbers We Believe Pencil And Paper Practice Is Needed To Master These Computations
Grade 6 Fractions Vs Decimals Worksheets Free
Worksheets Math Grade 6 Fractions Vs Decimals Converting Fractions To From Decimals Worksheets These Math Worksheets Provide Practice In Converting Fractions And Mixed Numbers To Decimal Numbers And Vice Versais Is A Fundamental Skill Which All Students Need To Master
Mixed Numbers And Fractions Worksheets 6th Grade
Mixed Numbers And Fractions Worksheets 6th Grade Are Helpful To Enable Kid S Easy Understanding Of Value Of Fractions Our 6 Th Grade Mixed Numbers And Fractions Exercises With Answers Are However Made Up Of Stimulating Models Guiding Kids To Visually See And Perceive The Relationship Between The Numerator And The Denominator One Of Such Amazing Models Is Fractions On Number Line Here
Printable Fraction Worksheets For Practice Grade 3 6
Fractions Worksheets For Grades 1 6 K5 Learning
Worksheets Math Math By Topic Fractions Fraction Worksheets For Grade 1 Through Grade 6 Our Fraction Worksheets Start With The Introduction Of The Concepts Of Equal Parts Parts Of A Whole And Fractions Of A Group Or Set And Proceed To Reading And Writing Fractions Adding Subtracting Multiplying And Dividing Fractions And Mixed Numbers
6 Grade Fraction Worksheet. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
You can begin with the uppercase cursives and after that move forward with the lowercase cursives. Handwriting for kids will also be rather simple to develop in such a fashion. If you're an adult and wish to increase your handwriting, it can be accomplished. As a result, in the event that you really wish to enhance handwriting of your kid, hurry to explore the advantages of an intelligent learning tool now!
Consider how you wish to compose your private faith statement. Sometimes letters have to be adjusted to fit in a particular space. When a letter does not have any verticals like a capital A or V, the very first diagonal stroke is regarded as the stem. The connected and slanted letters will be quite simple to form once the many shapes re learnt well. Even something as easy as guessing the beginning letter of long words can assist your child improve his phonics abilities. 6 Grade Fraction Worksheet.
There isn't anything like a superb story, and nothing like being the person who started a renowned urban legend. Deciding upon the ideal approach route Cursive writing is basically joined-up handwriting. Practice reading by yourself as often as possible.
Research urban legends to obtain a concept of what's out there prior to making a new one. You are still not sure the radicals have the proper idea. Naturally, you won't use the majority of your ideas. If you've got an idea for a tool please inform us. That means you can begin right where you are no matter how little you might feel you've got to give. You are also quite suspicious of any revolutionary shift. In earlier times you've stated that the move of independence may be too early.
Each lesson in handwriting should start on a fresh new page, so the little one becomes enough room to practice. Every handwriting lesson should begin with the alphabets. Handwriting learning is just one of the most important learning needs of a kid. Learning how to read isn't just challenging, but fun too.
The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused.
6 Grade Fraction Worksheet. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice. | 1,515 | 7,603 | {"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} | 2.71875 | 3 | CC-MAIN-2020-50 | latest | en | 0.654693 |
https://gmatclub.com/forum/5-dings-and-only-1-admit-with-a-710-on-gmat-150058.html?sort_by_oldest=true | 1,490,662,151,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189583.91/warc/CC-MAIN-20170322212949-00423-ip-10-233-31-227.ec2.internal.warc.gz | 794,918,246 | 54,157 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# 5 Dings and only 1 Admit with a 710 on GMAT
Author Message
Manager
Status: Preparing Apps
Joined: 04 Mar 2009
Posts: 91
Concentration: Marketing, Strategy
GMAT 1: 650 Q48 V31
GMAT 2: 710 Q49 V38
WE: Information Technology (Consulting)
Followers: 2
Kudos [?]: 222 [0], given: 4
5 Dings and only 1 Admit with a 710 on GMAT [#permalink]
### Show Tags
29 Mar 2013, 01:25
Hi Everyone,
I’ve taken the GMAT thrice and have gone through the application process twice.
With 6yrs of IT experience, a highest GMAT score of 710, 11 dings and 1 admit (@ SMU – Cox w ) over the last two years, and an aim to build a career in Strategy Consulting, is accepting the SMU – Cox admit a sensible decision? Or I should re-visit the application I prepared?
Your inputs will be highly appreciated.
Thanks!!
Joined: 02 Jul 2012
Posts: 382
Followers: 2
Kudos [?]: 45 [0], given: 0
Re: 5 Dings and only 1 Admit with a 710 on GMAT [#permalink]
### Show Tags
04 Apr 2013, 12:06
Hi there,
Congrats on SMU! I wish I could tell you yes or no, but as you know, this is a very personal decision. It does seem like you've put a lot into this process and tried your hardest. It's difficult for me to know how much room for improvement there is in your application approach without actually seeing it and getting to know you fully. At the surface it seems accepting SMU makes sense, but if you'd be really unhappy there, maybe it doesn't. Maybe it makes more sense to not pursue an MBA at all? Or, like you say, to try again for something else. But I think that if you're dead set on an MBA and you've already put in this much effort/this many attempts into the application process, accepting SMU is a logical choice.
In the end, it has to be your decision, but I hope this helps!
Mili
Mili Mittal
Senior Consultant
http://www.mbamission.com
Read the mbaMission Insider's Guides (16 individual school titles)
Visit the mbaMission Blog
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Display posts from previous: Sort by | 830 | 2,998 | {"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} | 2.640625 | 3 | CC-MAIN-2017-13 | longest | en | 0.916359 |
https://eanswers.in/math/question3159188 | 1,606,558,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00062.warc.gz | 281,418,707 | 14,367 | The sum of two number is twice the difference between two number. if the one number is 10 find the second number.
, 17.10.2019 23:00, nitesh3786
The sum of two number is twice the difference between two number. if the one number is 10 find the second number.
Answers: 2
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The sum of two number is twice the difference between two number. if the one number is 10 find the s...
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Total solved problems on the site: 16641358 | 475 | 1,410 | {"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-2020-50 | latest | en | 0.794565 |
https://www.imlearningmath.com/mr-grumper-grumbles-about-bad-time-keeping-trains-like-everybody-else/ | 1,632,453,078,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057496.18/warc/CC-MAIN-20210924020020-20210924050020-00268.warc.gz | 843,564,737 | 23,712 | # Mr. Grumper grumbles about bad time-keeping trains like everybody else
Math Riddle: Mr. Grumper grumbles about bad time-keeping trains like everybody else. On one particular morning, he was justified, though. The train left on time for the one hour journey and it arrived 5 minutes late. However, Mr. Grumper’s watch showed it to be 3 minutes early, so he adjusted his watch by putting it forward 3 minutes. His watch kept time during the day, and on the return journey in the evening, the train started on time, according to his watch, and arrived on time, according to the station clock. If the train traveled 25 percent faster on the return journey than it did on the morning journey, was the station clock fast or slow, and by how much?
Answer: The correct answer is the station clock is 3 minutes fast.
The morning journey took 65 minutes, and the evening journey, therefore, took 52 minutes, and the train arrived 57 minutes after it should have left, that is, 3 minutes early. | 226 | 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} | 4.09375 | 4 | CC-MAIN-2021-39 | latest | en | 0.972497 |
https://fixpython.com/2022/10/fix-python-what-s-a-good-rate-limiting-algorithm-0u8akzk/ | 1,679,647,914,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00408.warc.gz | 302,381,090 | 15,451 | # Fix Python – What’s a good rate limiting algorithm?
## Question
Asked By – miniman
I could use some pseudo-code, or better, Python. I am trying to implement a rate-limiting queue for a Python IRC bot, and it partially works, but if someone triggers less messages than the limit (e.g., rate limit is 5 messages per 8 seconds, and the person triggers only 4), and the next trigger is over the 8 seconds (e.g., 16 seconds later), the bot sends the message, but the queue becomes full and the bot waits 8 seconds, even though it’s not needed since the 8 second period has lapsed.
Now we will see solution for issue: What’s a good rate limiting algorithm?
## Answer
Here the simplest algorithm, if you want just to drop messages when they arrive too quickly (instead of queuing them, which makes sense because the queue might get arbitrarily large):
``````rate = 5.0; // unit: messages
per = 8.0; // unit: seconds
allowance = rate; // unit: messages
last_check = now(); // floating-point, e.g. usec accuracy. Unit: seconds
when (message_received):
current = now();
time_passed = current - last_check;
last_check = current;
allowance += time_passed * (rate / per);
if (allowance > rate):
allowance = rate; // throttle
if (allowance < 1.0):
discard_message();
else:
forward_message();
allowance -= 1.0;
``````
There are no datastructures, timers etc. in this solution and it works cleanly 🙂 To see this, ‘allowance’ grows at speed 5/8 units per seconds at most, i.e. at most five units per eight seconds. Every message that is forwarded deducts one unit, so you can’t send more than five messages per every eight seconds.
Note that `rate` should be an integer, i.e. without non-zero decimal part, or the algorithm won’t work correctly (actual rate will not be `rate/per`). E.g. `rate=0.5; per=1.0;` does not work because `allowance` will never grow to 1.0. But `rate=1.0; per=2.0;` works fine.
This question is answered By – Antti Huima
This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 | 547 | 2,095 | {"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-2023-14 | latest | en | 0.844929 |
https://decodedscience.org/deep-blue-and-chess-a-landmark-in-artificial-intelligence/ | 1,618,562,437,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00460.warc.gz | 312,086,441 | 15,776 | # Deep Blue and Chess: A Landmark in Artificial Intelligence?
Home / Deep Blue and Chess: A Landmark in Artificial Intelligence?
## AI vs. Human: Playing Kasparov
Garry Kasparov. Deep Blue (a Chess computer) defeated him in May 1997. Photo: Mary Ellen Mark
Deep Blue I played six games against Russian chess Grandmaster Garry Kasparov in February 1996, and Kasparov won 4-2.
In May, 1997, however, Deep Blue II played Kasparov again and beat the Grandmaster 3.5-2.5. This was the first time that a computer had defeated a reigning world champion when playing under standard tournament time controls.
There was some ill feeling after the match, with Kasparov claiming that there had been human intervention during the games. The Deep Blue team denied this, saying they had only consulted with humans and made changes between the games, which was allowed under the rules.
Kasparov requested a rematch, but IBM rejected his request, and dismantled Deep Blue.
## Methodologies used by Deep Blue
Using Chess chips is obviously important to Deep Blue’s success, as it was able to analyze more positions in the time available, an important factor as Chess is played against the clock.
Deep Blue also drew upon, and further developed, existing techniques, such as quiescence search and transposition tables.
• Quiescence search explores interesting positions to a deeper level than uninteresting positions. This attempts to overcome the horizon effect, where a computer might think it has found a good move but it might be running into trouble just a few moves ahead.
• Deep Blue also used transposition tables. These seek to reduce the amount of searching that has to be done by recognizing that a given position could be arrived at in a number of ways. If Deep Blue arrives at a position that has already been seen, but reached by a different set of moves, we can save ourselves a lot of time by simply recalling the outcome of the earlier analysis, rather than repeating it. Transpositions tables are designed to store this information so that it can be efficiently recalled when needed.
The evaluation function is a critical element of many, if not all, game playing programs. It returns a numerical value which gives an evaluation of the position being considered. It is now possible to compare two positions and say which one is best. The design of an evaluation function is often more of an art than a science and, in the case of Deep Blue, it was constantly being tweaked, often under the guidance of Chess Grandmasters.
As the evaluation function is often computationally expensive, using a lot of resources, it has to be as efficient as possible. Deep Blue had a fast evaluation function, when just an approximation was required, and a slow evaluation function when an accurate evaluation was needed.
Categories Uncategorized | 583 | 2,841 | {"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} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.972676 |
https://www.phpdeveloper.org.uk/project-euler-problem-4/ | 1,656,604,146,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00443.warc.gz | 992,401,617 | 10,419 | # Project Euler: Problem 4
Problem: Find the longest palindrome made from the product of two three-digit numbers.
Q: How do we work out if a number is a palindrome?
A: Convert it to a string and compare it with the reverse of the string. PHP has a built-in function, `strrev`, which makes this trivial.
Solution: Given the search space involved, we can simply test all the products of two three-digit numbers using a nested loop.
```<?php
declare(strict_types=1);
error_reporting(E_ALL);
function is_palindrome(int \$number) : bool
{
// A number is a palindrome if its string representation
// is the same when reversed
\$str = strval(\$number);
return (\$str === strrev(\$str));
}
\$longest_palindrome = 0;
for (\$i = 100; \$i <= 999; \$i++)
{
for (\$j = 100; \$j <= 999; \$j++)
{
\$product = \$i * \$j;
if (is_palindrome(\$product) && \$product > \$longest_palindrome)
{
\$longest_palindrome = \$product;
}
}
}
print("\$longest_palindrome\n");``` | 256 | 958 | {"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-2022-27 | latest | en | 0.570368 |
https://cloud.tencent.com/developer/article/1784157 | 1,638,833,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00403.warc.gz | 235,974,632 | 93,560 | # [013] 7种常见数据结构的图画解读
“数据结构在我们做数据过程中会经常遇到的，算法的实现更是依赖。今天给大家分享的这篇文章，对我们工作中最为常用的数据结构进行了结构化解释，同时也有一些图画，更加容易理解，希望可以帮助到大家哦！
Data structures are fundamental constructs that are used to build programs. Each data structure has its own way of organizing data, which may work efficiently in particular use cases. With their own particular structures, data structures offer alternative solutions to data organization, management, storage, access, and modification tasks.
Regardless you are building a machine learning model or a mobile app; you must rely on data structures when working on a project. Therefore, your prospective employer will also want to make sure that you have a good grasp of data structures.
In this post, we will cover seven fundamental data structures that you must know for your next job interview.
# 1. Arrays
Arrays are the most simple and widely used data structures. They are constructed as a collection of elements in a specific order. Arrays usually have fixed-size, and they hold items of the same data type. You can create an array from the collection of strings, integers, objects, or even arrays. Since arrays are indexed, you can have random access to any array element.
Figure 2. An Example Visualization of Arrays (Figure by Author)
Types of Arrays: The two most common types of arrays you may come across to are
• One-dimensional arrays
• Multi-dimensional arrays
Basic Operations on Arrays: The most common basic operations you can conduct on arrays are `insert`, `get`, `delete`, and `size` operations.
• `Insert`: Inserting a new element at a given index
• `Get`: Returning an element at a given index
• `Delete`: Deleting an element at a given index
• `Size`: Returning the total number of elements in an array
You can also conduct traverse, search, and update operations on arrays.
Implementation of Arrays:
• Arrays are usually used as building blocks of more complex data structures such as `arrays`, `lists`, `heaps`, `matrices`.
• In data science, arrays are used for efficient data manipulation operations. We often provide `NumPy`’s array objects to machine learning models for training.
# 2. Stacks
Stacks are data structures that work on a LIFO (Last in First Out) basis. In a LIFO based data structure, the data that are placed the last is accessed first. A real-life example of a stack could be a pile of rocks placed in vertical order, as shown in Figure 2.
Figure 3. Photo by Samrat Khadka on Unsplash
Basic Operations on Stacks: The most common basic operations you can conduct on stacks are `push`, `top`, `pop`, `isEmpty`, and `isFull` operations.
• `Push`: Inserting a new element to the top;
• `Top`: Returning an element at the top without removing it from the stack;
• `Pop`: Returning an element at the top after removing it from the stack; and
• `isEmpty`: Checking if the stack has any element.
• `isFull`: Checking if the stack is full.
Here is a visualization of stack structure:
Figure 4. An Example Visualization of Stacks (Figure by Author)
Commonly asked interview questions regarding stacks
• Evaluating postfix expression using a stack (e.g., Reverse Polish notation (RPN), or an abstract syntax tree (AST));
• Sorting values in a stack; and
• Checking balanced parentheses in an expression.
# 3. Queues
Queues are structurally very similar to stacks. The only difference is that queues are based on the FIFO (First In First Out) method. In a queue, we can either add a new element to the end or remove an element from the front. A real-life example of queues could be a line of people waiting in a grocery, as shown in Figure 4.
Figure 5. Photo by Adrien Delforge on Unsplash
Basic Operations on Queue: The most common basic operations you can conduct on queues are `enqueue`, `dequeue`, `top`, and `isEmpty` operations.
• `Enqueue`: Inserting a new element to the end of the queue;
• `Dequeue`: Removing the element from the front of the queue;
• `Top`: Returning the first element of the queue; and
• `isEmpty`: Checking if the queue is empty.
Here is a visualization of queue structure:
Figure 6. An Example Visualization of Queue (Figure by Author)
Implementation of Queues and Common Interview Questions
• Implement a queue using stacks;
• Retrieve the first n elements of a given queue; and
• Generate binary numbers from 1 to n.
Linked lists are structured as a linear collection of elements (i.e., node). Each node contains two information:
• `Data`: The value of the element
• `Pointer`: The pointer to the next node in the linked list.
Since linked lists are sequential structures, accessing elements randomly is not possible.
Figure 7. Photo by JJ Ying on Unsplash
Types of Linked List: We come across two main types of linked lists,
• Singly Linked List: We can only access the next node using the node before it.
• Doubly Linked List: We can access data both in forward and backward directions.
Here is a visualization of a linked list structure:
Figure 8. An Example Visualization of Single Linked List (Figure by Author)
Basic operations of Linked List: The most common basic operations you can conduct on linked lists are `insertAtEnd`, `insertAtHead`, `DeleteAtHead`, `Delete`, and `isEmpty` operations.
• `InsertAtEnd`: Inserting a new element to the end of the linked list
• `InsertAtHead`: Inserting a new element to the head of the linked list
• `DeleteAtHead`: Removing the first element of the linked list
• `Delete`: Removing a given element from the linked list
• `isEmpty`: Checking if the linked list is empty.
• Find the middle element of a singly linked list in one pass;
• Reverse a singly linked list without recursion; and
• Remove duplicate nodes in an unsorted linked list.
# 5. Trees
Trees are hierarchical data structures that consist of nodes and edges. Nodes represent values while edges connect them. The relationship between nodes in a tree is a uni or bidirectional one, not a cyclical one. Trees are widely used to build decision trees and ensemble methods.
Figure 9. Photo by Jazmin Quaynor on Unsplash
While tree types like binary trees and binary search trees are the most commonly used variations, the total number of different tree implementations reaches hundreds, where they can be grouped as:
• `Basic Tree Structures`
• `Multi-Way Trees`
• `Space-Partitioning trees`
• `Application-Specific trees`
Figure 10. An Example Visualization of Single Linked List (Figure by Author)
• Check if two given binary trees are identical or not;
• Calculate the height of a binary tree;
• Construct a full binary tree from a preorder and postorder sequence.
# 6. Graphs
Graphs consist of a finite set of nodes (i.e., vertices) with a set of unordered or ordered pairs of these vertices that together form a network. These pairs are also referred to as edges (i.e., links, lines, or arrows). An edge may contain weight or cost information to represent how much it cost to move from a to b.
Figure 11. An Example Visualization of Graph (Figure by Author)
Types of Graphs: There are two main types of graphs: (i) directed graphs and (ii) undirected graphs.
• `Directed Graphs`: If all the edges of a graph have direction info like in Figure 8, then we talk about a directed graph;
• `Undirected Graphs`: If none of the edges have direction info, then we talk about an undirected graph
Figure 12. Directed Graph vs. Undirected Graph (Figure by Author)
Basic operations of Graphs: The most common basic operations you can conduct on graphs are `adjacent`, `neighbors`, `add_vertex`, `remove_vertex`, `add_edge`, `remove_edge`, and `get` operations.
• `adjacent`: checks if there is an edge from the vertex a to the vertex b;
• `neighbors`: lists all vertices b such that there is an edge from the vertex a to the vertex b;
• `add_vertex`: adds the vertex a;
• `remove_vertex`: removes the vertex a;
• `add_edge`: adds the edge from the vertex a to the vertex b;
• `remove_edge`: removes the edge from the vertex a to the vertex b,
• `get`: returns the value associated with the vertex a or edges.
• Find the path between given vertices in a directed graph;
• Check if the given Graph is strongly connected or not; and
• Check if an undirected graph contains a cycle or not.
# 7. Hash Tables
Hash tables are associative data structures that store data in an array format with a unique index value. In hash tables, accessing random elements are very efficient and fast as long as we know the index value. Besides, data insertion, deletion, and search are also very fast in hash tables regardless of size. Hash tables use a hashing function to generate efficient index values, which gives them a further advantage of efficiency.
## Hashing
Hashing is a method to convert key values into a range of indices of an array. The reason for hashing is to reduce the complexity of key pairs and create an efficient indices list.
Figure 13. An Example Visualization of Hash Table (Figure by Author)
The performance of hashing data structure depends on (i) `Hash Function`, (ii) `Size of the Hash Table`, and (iii) `Collision Handling Method`. One of the combined methods used for hashing and collision handling is combining the modulo operator with linear probing so that the complexity can be reduced to a greater extent while ensuring unique index values.
## Commonly asked Hashing interview questions:
• Find symmetric pairs in an array;
• Trace the complete path of a journey;
• Find if an array is a subset of another array; and
• Check if given arrays are disjoint.
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http://slideplayer.com/slide/4235465/ | 1,568,785,777,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573184.25/warc/CC-MAIN-20190918044831-20190918070831-00053.warc.gz | 189,669,684 | 17,783 | Happyphysics.com Physics Lecture Resources Prof. Mineesh Gulati Head-Physics Wing Happy Model Hr. Sec. School, Udhampur, J&K Website: happyphysics.com.
Presentation on theme: "Happyphysics.com Physics Lecture Resources Prof. Mineesh Gulati Head-Physics Wing Happy Model Hr. Sec. School, Udhampur, J&K Website: happyphysics.com."— Presentation transcript:
happyphysics.com Physics Lecture Resources Prof. Mineesh Gulati Head-Physics Wing Happy Model Hr. Sec. School, Udhampur, J&K Website: happyphysics.com
Ch 14 Fluid Mechanics © 2005 Pearson Education
14.1 Density definition of density © 2005 Pearson Education SI unit is kg/m 3
14.2 Pressure in a Fluid © 2005 Pearson Education
Pressure, Depth and Pascal ’ s Law © 2005 Pearson Education
Pascal’s law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel © 2005 Pearson Education
14.3 Buoyancy © 2005 Pearson Education Archimedes’s principle states: When a body is completely or partially immersed in a fluid, the fluid exerts and upward force on the body equal to the weight of the fluid displaced by the body
14.4 Fluid Flow © 2005 Pearson Education
Ideal fluid: fluid that is incompressible and has no internal friction Ideal fluid: fluid that is incompressible and has no internal friction Flow line: The path of an individual particle in a moving fluid Flow line: The path of an individual particle in a moving fluid Steady flow: If the overall flow pattern does not change with time Steady flow: If the overall flow pattern does not change with time Streamline: is a curve whose tangent at any point is in the direction of the fluid velocity at that point. Streamline: is a curve whose tangent at any point is in the direction of the fluid velocity at that point. © 2005 Pearson Education
continuity equation, incompressible fluid volume flow rate © 2005 Pearson Education
14.5 Bernoulli’s Equation Bernoulli’s equation © 2005 Pearson Education
14.6 Viscosity and Turbulence Viscosity is internal friction in a fluid Viscosity is internal friction in a fluid When the speed of a flowing fluid exceeds a certain critical value, the flow is no longer laminar. Instead, the flow pattern becomes extremely irregular and complex, and it changes continuously with time; there is no steady-state pattern. This is irregular, chaotic flow is called turbulence When the speed of a flowing fluid exceeds a certain critical value, the flow is no longer laminar. Instead, the flow pattern becomes extremely irregular and complex, and it changes continuously with time; there is no steady-state pattern. This is irregular, chaotic flow is called turbulence © 2005 Pearson Education
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34c
Problem
# Problem 34c
## Chapter 1: Functions and Models
Textbook ExpertVerified Tutor
13 Oct 2021
#### Given information
Note that the given function is to draw two graphs of the natural logarithm function and the natural exponential each.
Since Natural logarithm;
And common logarithm;
is called the common logarithm.
#### Step-by-step explanation
Step 1.
Let us assume that,
1. Exponential function,
2. Natural logarithm, | 109 | 453 | {"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} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.759067 |
https://uk.mathworks.com/matlabcentral/cody/problems/2834-convert-kilometers-to-miles/solutions/1964539 | 1,600,777,253,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00093.warc.gz | 667,636,582 | 16,585 | Cody
# Problem 2834. Convert Kilometers to Miles
Solution 1964539
Submitted on 7 Oct 2019
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 Fail
x = 1; y_correct = 0.62; assert(isequal( Mile_from_KM(x),y_correct))
Undefined function 'convlength' for input arguments of type 'double'. Error in Mile_from_KM (line 2) y = convlength( x ,'km','mi') Error in Test1 (line 3) assert(isequal( Mile_from_KM(x),y_correct))
2 Fail
x = 10; y_correct = 6.2; assert(isequal( Mile_from_KM(x),y_correct))
Undefined function 'convlength' for input arguments of type 'double'. Error in Mile_from_KM (line 2) y = convlength( x ,'km','mi') Error in Test2 (line 3) assert(isequal( Mile_from_KM(x),y_correct))
3 Fail
x = 5.2; y_correct = 3.2240; assert(isequal( Mile_from_KM(x),y_correct))
Undefined function 'convlength' for input arguments of type 'double'. Error in Mile_from_KM (line 2) y = convlength( x ,'km','mi') Error in Test3 (line 3) assert(isequal( Mile_from_KM(x),y_correct))
4 Pass
filetext = fileread('Mile_from_KM.m'); assert(isempty(strfind(filetext, '*')),'sign * forbidden') assert(isempty(strfind(filetext, '==')),'sign == forbidden') | 387 | 1,260 | {"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} | 2.984375 | 3 | CC-MAIN-2020-40 | latest | en | 0.276915 |
https://gmatclub.com/forum/confused-need-insight-to-help-process-next-steps-288053.html | 1,571,106,096,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00304.warc.gz | 478,297,447 | 55,034 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Confused - Need insight to help process next steps
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Intern
Joined: 01 Nov 2018
Posts: 5
Confused - Need insight to help process next steps [#permalink]
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05 Feb 2019, 15:09
I have been preparing for GMAT for- 8 weeks. I know its a shorter time than ideal.
Here's a brief background of my activities and progress -
Mid-December - I took GMAT official practice test # 1 to know where I stand - (This was without any preparation)
Score:
610 - Q44, V27
I took an actual GMAT in early January after two weeks of preparation (This was to get an experience of real setting, environment, etc)
Score:
650 - Q48,V33
I prepare for a month - official guides, blogs, Manhattan books, official GMAT practice questions. Then I took Official Practice test # 2 on Jan 28th
Score:
650 - Q47, V35
I realized that I was making too many silly mistakes in Quant and I was missing application of concepts in Verbal - SC. When I see the correct answer, I immediately know why that is correct and what my mistake was. I practice some more with increased focus. I also noticed I was making mistakes in the first 10 questions that I tried to spend more time on. Gave Official Practice test # 3 on Jan 31st.
Score:
690 - Q49, V35
I know that its not all about just accuracy; but still to gauge some progress:- My typical accuracy in CR and RC is 80% and SC is 50%. I was getting on an average 12-14 questions incorrect in verbal with 7-8 SC incorrect. Applied myself more, practiced SC. I have Official Practice test # 4 on Feb. 3rd
Score:
750 - Q49, V44
I had only 4 questions incorrect in Verbal. 3 SCs and 1 CR. 6 questions incorrect in Quant.
In the last 10 days, I could feel the difference. I know that I was able to apply my knowledge with much more confidence in SC. I did not feel like I was guessing. that is why the improvement from 650 - 750 in 10 days.
Today I took the actual GMAT at the center.
Score:
640 - Q44, V34
I feel like I am back at where I started. The bigger problem is that i thought my test went quite well. I was hoping somewhere between 700 and 750. Couple of observations - My quant section in the middle was tough. i was thrown P&C problems, advanced geometry problems that I had not encountered in any practice. At least that told me that i was doing well or so I thought. Q44 certainly does not prove that.
I plan to take the test again in two weeks. I am confident that 640 is not reflective of my true abilities. Any words of wisdom? What should I do in the next two weeks? I am not going to lose heart but i need some direction as I feel a bit lost on where to start and what went so wrong in today's test. I have requested the ESR to see..
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Re: Confused - Need insight to help process next steps [#permalink]
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05 Feb 2019, 16:56
1
Hi rshah13.
Your situation is interesting, as it sounds as if you have a decent grasp of what it takes to score high on the GMAT and yet didn't match your practice test scores when you took the actual GMAT.
I guess there are a few questions to answer here.
Why didn't you score close to where you have been scoring on practice tests?
To what degree does that score reflect your current skill levels?
How should you proceed going forward?
The answer to the first question is likely some combination of careless errors, your happening to see some questions that matched some skill gaps that you had, and your getting tricked by trap answers and new twists in verbal.
My best guess answer to the second question is that your skill levels are higher than that score indicates they are and that, at the same time, there are some gaps in your skillset.
Of course, without my seeing your ESR, it's a little tough for me to say exactly how to proceed. At the same time, I can say the following.
For quant, you could lock in a higher score by filling in skill gaps. You may get some information on where skill gaps lie from the ESR for this test. You may know where some lie. Clearly, permutations and geometry seem to be areas you could work on. Also, you can identify some by going over your previous practice questions and tests. Wherever they are, strengthen weaker areas, so that, if you see questions of those types on the test, you answer them within two minutes and get them correct. You don't have to become a master of every area, but the more areas you master, the more likely you will be to hit your score goal.
Also, if you think you may have made some careless errors in quant, focus on accuracy in your practice. Seek to get long streaks of correct answers in your practice. It's one thing to know basically how to answer quant questions and another to correctly answer them consistently.
For some tips on quant accuracy, you could check out this post: https://blog.targettestprep.com/be-accu ... mat-score/
Regarding verbal, what you have done so far sounds great. At the same time, somehow you chose trap answers when you took the actual test. So, you have to continue to train in verbal and become even better at defining exactly why wrong answers are wrong and right answers are right. GMAT verbal is essentially a test of the level of sophistication of the thinking that one brings to bear when answering the questions. So, seek to make your reasons for choosing answers even more sophisticated.
For some ideas on how to think about SC, consider the following choices and explanations.
In 1988, the debut at the Metropolitan Opera of operatic singer Carlos Feller was made as Don Alfonso, regarded as his signature role, and returned there in 1990 as Dr. Bartolo in Rossini's Il Barbiere di Siviglia.
(A) the debut at the Metropolitan Opera of operatic singer Carlos Feller was made as Don Alfonso, regarded as his signature role, and returned
(B) operatic singer Carlos Feller made his debut at the Metropolitan Opera, in what was regarded as his signature role, Don Alfonso, and he returned
Explanation:
(A) Incorrect. The passive voice wording “the debut … was made” is suboptimal, as it places the focus of the sentence on the debut rather than on Carlos Feller, while Carlos Feller seems to be meant to be the main topic of the sentence, given what the sentence says about the debut, the role, and his returning in 1990.
Also, the wording “the debut … was made as Don Alfonso” does not clearly convey that Carlos Feller was in the role of Don Alfonso, instead conveying the nonsensical meaning that the debut itself somehow “was made as Don Alfonso.”
Also, since the past participial phrase “regarded as his signature role” is separated from the clause that precedes it by a comma, the phrase’s logical target is the subject of the preceding clause, and thus, the sentence seems to convey that Carlos Feller’s debut was regarded as his signature role.
Also, “the debut … was made … and returned” conveys the nonsensical meaning that the debut returned in 1990.
Also, since the sentence opens with “in 1988,” the sentence is essentially “in 1988, the debut ... was made … and returned … in 1990,” conveying that, in 1988, the debut returned in 1990. This issue could easily be missed by a test-taker who does not remember to consider the non-underlined portion along with the answer choice.
(B) Correct. The active voice wording “Carlos Feller made his debut … and returned” effectively conveys a meaning that makes sense.
Also, using the pronoun “what,” rather than “who,” to refer to “his signature role, Don Alfonso” makes sense, as a role is a thing, not a person.
To make your thinking more sophisticated, carefully analyze verbal questions, even ones that you have seen previously, identifying the logic that makes the answers wrong or correct and clearly articulating that logic as if you have to explain what you have found to someone else.
Overall, it sounds as if you are on the right track. To confirm that you are and get more practice in handling the test itself, take some more practice tests as you go along.
Also, it would make sense to come back here to get some ideas on how to further refine what you are doing after you get your ESR and as you proceed.
_________________
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Re: Confused - Need insight to help process next steps [#permalink]
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05 Feb 2019, 22:25
Hi Ravi,
I'm sorry to hear that Test Day didn't go as well as planned. When these types of score drops occur, the two likely "causes" involve either something that was unrealistic during practice or something that was surprising (or not accounted for) on Test Day. Before we talk about those issues though, I have some questions about your studies so far. What you have described in this post varies significantly from the plan that you outlined several months ago here:
(https://gmatclub.com/forum/need-some-in ... l#p2177541)
1) Can you describe how your plans changed in November/December of 2018? What studying - if any - did you do during that time? Did you end up taking any of those practice CATs (meaning that you took them a second time in January/February)?
2) What are your current application plans/timeline?
GMAT assassins aren't born, they're made,
Rich
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Re: Confused - Need insight to help process next steps [#permalink]
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06 Feb 2019, 15:54
Thank you Rich and Marty for your feedback.
Rich - To your question about my previous plan and the change...I planned to study for GMAT starting last November and took a date for 1st week in Jan (as per my previous post). There were few things happening in my personal life because of which I could not start preparations. I started preparing in mid-December but kept the Jan 1st week test date to use it for experience. I took another exam date for Feb. 5th. I did not give any CATs prior to January test due to lack of time. I just studied the concepts for two weeks and took the test.
I am targeting several EMBA programs.
I got my ESR for the test yesterday and here are the takeaways as I see it -
Quant - I had all questions correct for the 1st 1/4th and 80% correct for the second quarter. I think that's when i ran into P&C and geometry questions where time ran away from me. The most telling thing is i got all questions wrong in the last quarter spending average of under a minute on each one of them. so looks like I was on a great trajectory till middle of the section and completely lost my way after trying to rush through the questions.
Verbal - Similar story. All good till 12-14 questions. Then I remember getting tough RCs. It took away the time and accuracy. I reversed my usual scoring. I scored 90 percentile in SC; but 40 in RC. I am typically over 90 percentile in RCs...
My takewaway:
- Need to practice higher difficulty quant questions with accuracy and within time limit
- Practice difficult RCs
- Continue SC and CR practice as ''maintain'' the skill level
Any thoughts?
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Confused - Need insight to help process next steps [#permalink]
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06 Feb 2019, 16:54
rshah13 wrote:
Thank you Rich and Marty for your feedback.
I got my ESR for the test yesterday and here are the takeaways as I see it -
Quant - I had all questions correct for the 1st 1/4th and 80% correct for the second quarter. I think that's when i ran into P&C and geometry questions where time ran away from me. The most telling thing is i got all questions wrong in the last quarter spending average of under a minute on each one of them. so looks like I was on a great trajectory till middle of the section and completely lost my way after trying to rush through the questions.
Since you scored Q44 after missing all of the questions in the fourth quarter and rushing through the third quarter, you were totally rocking quant before things went downhill.
Had you simply guessed on a couple of questions that you clearly were not ready to answer and spent more time on ones that you were ready to answer, you may have scored 4 - 6 points higher in quant
Quote:
Verbal - Similar story. All good till 12-14 questions. Then I remember getting tough RCs. It took away the time and accuracy. I reversed my usual scoring. I scored 90 percentile in SC; but 40 in RC. I am typically over 90 percentile in RCs...
Well, at least SC was strong. You can't let RC drag you down. You have to practice RC until for you there is no such thing as a "tough RC."
Quote:
My takewaway:
- Need to practice higher difficulty quant questions with accuracy and within time limit
- Practice difficult RCs
- Continue SC and CR practice as ''maintain'' the skill level
I tend to agree with your plan.
At the same time, I don't entirely agree with what you said about answering higher difficultly quant questions within the time limit. I think your best bet is not to shoot for any time limit, at least not at first. Rather, you should work topic by topic in the following way.
Fill in any knowledge gaps you have for a topic. Then do dozens of questions of that one type, and not with a time limit. Just seek to get them correct. Your job is to arrive at correct answers, in two minutes or ten minutes, or however long it takes. Just get correct answers, and this approach works for verbal too.
To speed up, develop skill. Get it? If you are super good at, for instance, permutations questions, the clock won't even be an issue. You will naturally complete them in under two minutes. The same goes for any type of quant question. Your best bet is to increase speed by developing skill.
Sure, you can shoot for two minutes per question at some point, but doing so is something that you do later, as a refinement, once you are a total master at answering the questions.
For more on how to increase your quant score, you could check out this blog post. How To Increase Your GMAT Quant Score
Similarly for verbal, you can speed up by becoming better at analyzing answer choices. The better you are at seeing the differences between trap answers and correct answers, the more verbal questions you will answer correctly, and the faster you will answer them.
Overall, it sounds as if you are on the right track. So, make sure that you are realistic about what you have to do, and that you do what you have to do. Over the next two weeks, be sure to strengthen some weaker areas to the point of mastery. Permutations and combinations, for instance, can become some of the easiest questions that you will see, IF you know what you are doing.
_________________
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Re: Confused - Need insight to help process next steps [#permalink]
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06 Feb 2019, 22:23
Hi Ravi,
While the ESR doesn't provide a lot of information, there are usually a few data points that we can use to define what went wrong (and what you should work on to score higher). Since you have purchased the ESR, then I'll be happy to analyze it for you. If you would rather not post your ESR publicly, then you can feel free to PM or email it directly to me.
Given everything that you have described (including your Score Goal), I do not think that you should 'rush' back in to retest. There are likely a number of potential issues in terms of how you prepared for the Exam - and we need to define and fix them before you retest.
1) What application deadlines are you currently facing?
2) Going forward, how many hours do you think you can consistently study each week?
GMAT assassins aren't born, they're made,
Rich
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Re: Confused - Need insight to help process next steps [#permalink] 06 Feb 2019, 22:23
Display posts from previous: Sort by | 4,202 | 17,785 | {"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} | 3.078125 | 3 | CC-MAIN-2019-43 | latest | en | 0.957092 |
https://www.physicsforums.com/threads/left-handed-probability.266842/ | 1,542,573,836,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744649.77/warc/CC-MAIN-20181118201101-20181118223101-00201.warc.gz | 948,296,961 | 13,191 | # Left handed probability.
1. Oct 25, 2008
### Pumblechook
4 out of 5 recent US Presidents have been left handers? Given that 1 in 10 of the population is left handed what is the probability of this? Say it was taking socks out of draw with the light off.. 1 White to 9 Black. What are chances of having 4 White and 1 Black.. I get 0.00045 ??
More accurately what is the probability that a continuous sequence of 5 presidents out of 43 has 4 left handers?
2. Oct 25, 2008
### arildno
There is no particular reason, or improbability against, for a stronly skewed tendency within such a tiny sample.
3. Oct 25, 2008
### HallsofIvy
The probability of 4 out of 5 people being left handed is 5C4(.14)(.9)= 5(.0004)(.9)= .00045 as you have. The "out of 43" is irrelevant.
4. Oct 25, 2008
### Pumblechook
I don't think it is irrelevant.
The more presidents there has been the more likley you will get a sequence of 4 out of 5.
If there had been thousands or millions then the chances are that there would be a lot of '4 out of 5's
You would have to look at all the possible conbinations of L and R you could have with 43 presidents and for those that have at least one '4 out 5' work out those combimations' probabilities and add them up. There are maybe short cuts or at least estimates.
5. Oct 25, 2008 | 370 | 1,314 | {"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-2018-47 | latest | en | 0.970798 |
https://www.airmilescalculator.com/distance/ksj-to-jsy/ | 1,604,042,625,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00251.warc.gz | 604,288,200 | 17,509 | # Distance between Kasos Island (KSJ) and Syros Island (JSY)
Flight distance from Kasos Island to Syros Island (Kasos Island Public Airport – Syros Island National Airport) is 176 miles / 283 kilometers / 153 nautical miles. Estimated flight time is 49 minutes.
Driving distance from Kasos Island (KSJ) to Syros Island (JSY) is 447 miles / 720 kilometers and travel time by car is about 29 hours 29 minutes.
## Map of flight path and driving directions from Kasos Island to Syros Island.
Shortest flight path between Kasos Island Public Airport (KSJ) and Syros Island National Airport (JSY).
## How far is Syros Island from Kasos Island?
There are several ways to calculate distances between Kasos Island and Syros Island. Here are two common methods:
Vincenty's formula (applied above)
• 175.951 miles
• 283.165 kilometers
• 152.897 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
• 176.018 miles
• 283.274 kilometers
• 152.956 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 Kasos Island Public Airport
City: Kasos Island
Country: Greece
IATA Code: KSJ
ICAO Code: LGKS
Coordinates: 35°25′17″N, 26°54′35″E
B Syros Island National Airport
City: Syros Island
Country: Greece
IATA Code: JSY
ICAO Code: LGSO
Coordinates: 37°25′22″N, 24°57′3″E
## Time difference and current local times
There is no time difference between Kasos Island and Syros Island.
EET
EET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 51 kg (112 pounds).
## Frequent Flyer Miles Calculator
Kasos Island (KSJ) → Syros Island (JSY).
Distance:
176
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
176
Round trip? | 503 | 1,956 | {"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} | 2.53125 | 3 | CC-MAIN-2020-45 | latest | en | 0.799182 |
https://forums.space.com/threads/earth-velocity-has-me-in-a-spin.18842/ | 1,723,285,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00897.warc.gz | 211,877,685 | 29,167 | # Earth Velocity has me in a spin?
Status
Not open for further replies.
F
#### Fax_Monkey
##### Guest
OK I've lost the plot somewhere. Can somebody put me out of my misery with the following chain of ( probably faulty) assumptions.
1 - The Earth ambles round the Sun at 60,000 mph
2 - The Solar system belts round the galaxy at some scary speed
3- The Galaxy is moving through space away from universal centre at even more speed
4 - the speed of light ( C ) is a constant throughout the universe
OK my confusion is this - the earth as part of our galaxy must be moving at quite some speed from universe centre.
That velocity would have a value to a stationary observer watching our galaxy wizz by in space.
Enough velocity to suffer time dilation in comparison to the observer.
So on Earth we are always under a certain amount of constant time dilation ( as part of our galaxy).
So how can we say that we know the true speed of light - when every mesurement of distance over time would only be correct 'locally' The time function of the equation would surely be different to the stationary observer?
sorry for my stupidity in advance - fax -monkey
R
#### ramparts
##### Guest
All good questions Let's look at your "( probably faulty)" assumptions. As it turns out, you only missed one!
Fax_Monkey":35iyzhhr said:
1 - The Earth ambles round the Sun at 60,000 mph
I don't know the number, but sure, something like that.
2 - The Solar system belts round the galaxy at some scary speed
Indeed!
3- The Galaxy is moving through space away from universal centre at even more speed
Not quite. Yes, the universe is expanding, and our galaxy has some speed, but it's not moving away from a "universal center." The common analogy (so common, in fact, I just used it in another thread!) is of blowing up a balloon. Imagine a 2-D universe on the surface of a balloon, with little galaxies painted on. If you blow the balloon up, this 2-D universe expands - every point starts to move away from every other point. Of course, there is no center to the surface of the balloon So neither is there a center to the universe (at least, not that we know of!).
This is important because it means that there's no stationary point (like a center) with which we can reference the Earth's velocity. And that brings us to....
4 - the speed of light ( C ) is a constant throughout the universe
Oh, right. That's also true. But the above point brings us to...
OK my confusion is this - the earth as part of our galaxy must be moving at quite some speed from universe centre.
That velocity would have a value to a stationary observer watching our galaxy wizz by in space.
Enough velocity to suffer time dilation in comparison to the observer.
So on Earth we are always under a certain amount of constant time dilation ( as part of our galaxy).
So how can we say that we know the true speed of light - when every mesurement of distance over time would only be correct 'locally' The time function of the equation would surely be different to the stationary observer?
We can't have a "stationary observer" So yes, the Earth is moving, and as it feels acceleration from various gravitational pulls, there is indeed some associated time dilation - but it's miniscule, really, a part in billions or more. But here's the important thing - that only happens because of our accelerating motion, like that around the Sun. Just going at a fast velocity doesn't mean anything because, as Einstein famously pointed out (well, more like Galileo, but who's keeping track), who's to say that you're not at rest and the "rest" observer isn't travelling quickly?
C
#### chebby
##### Guest
Wait what about that example of getting into a near light speed spaceship example and coming back from a star trip thousands of years after all your relatives are dead? Can't you also view in that example that it's the spaceships that is at rest and it's earth that is moving away or accelerating away? So what determines which will have slow time and which regular time? i'm confused
O
#### origin
##### Guest
OK my confusion is this - the earth as part of our galaxy must be moving at quite some speed from universe centre.
That velocity would have a value to a stationary observer watching our galaxy wizz by in space.
Enough velocity to suffer time dilation in comparison to the observer.
So on Earth we are always under a certain amount of constant time dilation ( as part of our galaxy).
So how can we say that we know the true speed of light - when every mesurement of distance over time would only be correct 'locally' The time function of the equation would surely be different to the stationary observer?
sorry for my stupidity in advance - fax -monkey
First - there is no center of the universe.
If there was an observer that was not moving with our galaxy watching us whizz by he would infact note a time dialation affect. But WE would also look at him and say he is experiencing a time dialation affect - relativity you know, that is he would seem to have a velocity relative to us - there is NO stationary position so there is only relative motion.
The reason that we know the speed of light is because it ALWAYS measure the same speed. If we are whizzing towards the sun and we measure the speed of the light coming to us it will measure at c. If we where whizzing away from the sun and we measured the speed of light coming from the sun it would measure at c. The speed of light always is the same independent of the motion of the source or the observer.
edited to add, it is not a stupid question at all, it is a good, fundemental question.
M
#### MeteorWayne
##### Guest
Fax_Monkey":1pkphuvr said:
OK I've lost the plot somewhere. Can somebody put me out of my misery with the following chain of ( probably faulty) assumptions.
1 - The Earth ambles round the Sun at 60,000 mph
2 - The Solar system belts round the galaxy at some scary speed
Scary Speed= ~ 492,000 mph
[/quote]
S
#### SpeedFreek
##### Guest
chebby":i47w8luq said:
Wait what about that example of getting into a near light speed spaceship example and coming back from a star trip thousands of years after all your relatives are dead? Can't you also view in that example that it's the spaceships that is at rest and it's earth that is moving away or accelerating away? So what determines which will have slow time and which regular time? i'm confused
Whilst the people on Earth remain in the same reference frame throughout, your spaceship has to constantly change reference frames as it accelerates from being at rest in relation to the Earth to a relativistic speed, when it decelerates or turns around.
R
#### ramparts
##### Guest
chebby":6fmhkuf8 said:
Wait what about that example of getting into a near light speed spaceship example and coming back from a star trip thousands of years after all your relatives are dead? Can't you also view in that example that it's the spaceships that is at rest and it's earth that is moving away or accelerating away? So what determines which will have slow time and which regular time? i'm confused
Indeed - but as SpeedFreak says, when someone begins to accelerate, the equivalence of frames is lost. So if I'm moving at a velocity relative to you, I can very well say I'm at rest and you're moving. But if I start accelerating relative to you, well, then I can't say that anymore! Clearly I'm accelerating. So in the twin paradox you mentioned, the key is that when the spaceship leaves and comes back, it has to accelerate/decelerate. That's why the twin in the spaceship ages objectively less.
C
#### chebby
##### Guest
Ah ok, that makes sense. I think my confusion stems from the fact that acceleration is usually defined as change of velocity over time, which is clearly the same for planet and spaceship here. But what sets spaceship apart is that is experiencing force (that accelerates it) while the planet does not.
R
#### ramparts
##### Guest
chebby":3etgqjdd said:
Ah ok, that makes sense. I think my confusion stems from the fact that acceleration is usually defined as change of velocity over time, which is clearly the same for planet and spaceship here. But what sets spaceship apart is that is experiencing force (that accelerates it) while the planet does not.
Acceleration is always defined that way! In any kind of relativity, two observers at constant velocity can't tell who's at rest and who's moving. If your velocity is changing, though, then someone gets picked out.
Einstein's "equivalence principle" has a lot to do with this, actually. On one important level, it points out that an observer can't tell the difference between accelerating and being in the presence of a gravitational field. This is why you get things like gravitational time dilation - since we're on the surface of the Earth, time is actually a little bit slower here due to Earth's gravity than it is out in space!
An interesting tidbit copied from Wikipedia (since I'm too lazy to paraphrase!):
"Gravitational time dilation has been experimentally measured using atomic clocks on airplanes. The clocks that traveled aboard the airplanes upon return were slightly fast with respect to clocks on the ground. The effect is significant enough that the Global Positioning System needs to correct for its effect on clocks aboard artificial satellites, providing a further experimental confirmation of the effect."
C
#### chebby
##### Guest
Yes, I read about "equivalence principle" in brief history of time (can't say I understood a lot of it.) This actually made me think, when scientists calculate how much fuel stars have, do they take into account that fusion reactions actually happen slowly there?
Ramparts, what does your avatar formula mean?
F
#### Fax_Monkey
##### Guest
Thanks for clarifying it for me guys that's a great help - i guess i always had
this idea of the big bang being like a bomb blasting matter in all directions in
an already existing void - not realising that it was actually expanding void and matter together
I can see my understanding of time is incorrect as well - it's a lot more complicated than i thought!
In fact the whole universe seems more organic than clockwork to me now.
I'll get a few new books and try to get a better understanding of current thought.
Thanks again for the help, much appreciated
R
#### ramparts
##### Guest
chebby":w8icli8k said:
Yes, I read about "equivalence principle" in brief history of time (can't say I understood a lot of it.) This actually made me think, when scientists calculate how much fuel stars have, do they take into account that fusion reactions actually happen slowly there?
Oh, they take into account most everything As I learned the somewhat-hard way this past semester... If you want a more in-depth answer, I'll take a stab at it, but I'm not entirely sure what you're asking (or how it's related to the the equivalence principle!).
chebby":w8icli8k said:
Ramparts, what does your avatar formula mean?
My avatar is the Einstein field equation; I put it up because it's probably the single most beautiful mathematical statement in all of physics. Unfortunately it's in a slightly condensed form, to be short enough to fit on the site, but it works. It's the main equation of general relativity. The left side - the G - is called the Einstein tensor, a mathematical quantity that measures the curvature of spacetime. The right side - the 8 pi T - is the energy-momentum tensor, which measures the energy and matter content at a given point, including pressure, density, etc. So it tells you how a certain distribution of mass curves spacetime, in turn creating what we see as gravity
It takes the place, essentially, of Newton's law of gravitation, F=GmM/r^2, which relates the gravitational force to mass and distance. The second important equation, called the geodesic equation, takes the place of F=ma. The first tells you what gravity is created, the second tells you how things move in response to the gravity.
M
#### MetalMario
##### Guest
Wow, the Earth moves fast. But why is it, that if I jump up, I don't fly off the face of the Earth! The sheer speed should have some affect on my body.
M
#### MeteorWayne
##### Guest
Because the strongest force you feel is the earth's gravity. All the atmosphere moves with the earth, so we all move together around the sun. If you jump in the direction of the moon, you'll jump an immeasurably small amount higher
R
#### ramparts
##### Guest
Exactly - and don't forget that though the Earth is moving fast, you're on its surface... so you're moving fast, too! Don't believe me? Try jumping off the floor of a fast train some time. See if you get flung back since the train is moving so fast. People may look at you funny, but you won't move back an inch
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5K | 2,881 | 12,962 | {"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-2024-33 | latest | en | 0.928495 |
http://www.ciaoshen.com/algorithm/leetcode/2019/02/04/leetcode-random-pick-index.html | 1,627,549,261,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00194.warc.gz | 57,201,986 | 5,937 | ### 题目
Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.
Note: The array size can be very large. Solution that uses too much extra space will not pass the judge.
Example:
int[] nums = new int[] {1,2,3,3,3};
Solution solution = new Solution(nums);
// pick(3) should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(3);
// pick(1) should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(1);
### 直观解法
[3,1,3,2,3]
1 -> 1
2 -> 3
3 -> 0,2,4
### 蓄水池抽样（Reservoir Sampling）
当第一顶帽子来的时候：我只有1顶帽子，我以1/1的概率选择戴这顶帽子。
...
#### 代码
class Solution {
private int[] nums;
private Random r;
public Solution(int[] nums) {
this.nums = nums;
r = new Random();
}
public int pick(int target) {
int res = -1;
int count = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == target) {
if (r.nextInt(++count) == count - 1) res = i;
}
}
return res;
}
} | 335 | 1,054 | {"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} | 3.171875 | 3 | CC-MAIN-2021-31 | latest | en | 0.505647 |
http://www.math.purdue.edu/~bell/Course/latex.html | 1,516,495,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00171.warc.gz | 477,885,804 | 3,058 | # latex
I'll give you a quick example of a latex file that will explain the basic ideas of latex. (If you already know amstex, latex will be easy.)
Create a file called dad.tex containing the following lines.
\documentclass{article}
\pagestyle{empty} %this line makes NO PAGE NUMBERS
\begin{document}
\bigskip
I am happy to report that I have become one hell of an
excellent teacher since starting graduate school at Purdue
in math, as evidenced by the following little quiz that I
gave my recitation section this morning.
\bigskip
\bigskip
\begin{center}
MATH 161 QUIZ NUMBER ONE
\end{center}
\begin{enumerate}
\item Compute $\int_0^1 e^{-x^2}\,dx$
\item Compute $\int_0^1 \sin x^2\,dx$
\item Compute $\sum_{n=1}^\infty \frac{1}{n\ln n}\,dx$
\item Prove that the set of numbers satisfying a polynomial
equation with integer coefficients is a countable dense subset
of the real line. Is $\sqrt{e\pi}$ in this set?
\end{enumerate}
\bigskip
\bigskip
After class, I heard one of my students exclaim
\begin{quote}
\emph{Voil\a le commencement de la fin.}
\end{quote}
\bigskip
Imagine that!
\bigskip
Sincerely,
\bigskip
sutdent number 999-31-0000
\end{document}
After you have created this file exactly as shown, you can type
latex dad
to "LaTeX" the file. This command will create a file called dad.dvi and you can take a look at it by typing
xdvi dad &
in an xterm. (You cannot do this from a telnet session. If you try it from a telnet window, you will get "Unable to open display" and you can push RETURN to get a prompt back.)
Here is a list of handy one keystroke commands that you will want to use in xdvi instead of trying to mess around with the scroll bars.
u Moves the page UP
d Moves the page DOWN
n Moves to the NEXT page
p Moves to the PREVIOUS page
q QUITS xdvi and makes the window go away
x (for EXPERT) makes the buttons on the right
side of the xdvi window go away and come back
8g Moves to page 8
1g Moves to page 1, etc.
If you like what you see, you can print it on the printer lpub6 by typing
dvips -Plpub6 dad
Note: Here are some printers and their locations:
Printers: lpub Computer room, Rm 839
lpub7 Room 741
lpub6 Room 635
If your xterm is on peano.math.purdue.edu, you can print to the printer in LAEB B-286 by typing this:
dvips -f dad | lpr -Plaebb286hp@franklin.cc
Here as an example of a MATH 262 Final Exam in latex. Take a look at it and then study the latex source file at,
(You can download this latex source file by starting Netscape in a math xterm and then going to this URL and SHIFT-left mouse clicking on the link above. Then you can modify the file to fit your own needs.)
Here is the latex source file for a paper of mine paper.tex in AMS-LaTeX.
Here as an example of a TA Office Hour table in latex. Here is the latex source file, office.tex. To print it sideways on the paper, use the command
dvips -t landscape office.dvi
`
Here as an example of a PhD Plan of Study form in latex that was created by the Web page at http://www.math.purdue.edu/~bell/Graduate/planostudy.html and here is the latex source file, plan.tex.
Next, go to Professor Harold Boas' latex website at Texas A & M
for a latex tuturial.
Another very useful site for graduate students is Using LaTeX for your Purdue Thesis, a site maintained by Mark Senn. You'll find a nice latex tutorial and information for Purdue graduate students on how to use latex to create a PhD thesis that conforms to the excrutiating demands of the official Purdue Thesis Style Guide, complete with the latest version of a latex PU-thesis style file puthesis.cls that you can download to your TeX file directory.
An older LaTeX Math Dept Thesis style file is available on our system in the directory
/pkgs/teTeX/local/tex/latex/thesis
An example of how to use it can be found in the directory:
/pkgs/teTeX/local/doc/latex/thesis
WARNING: The LaTeX Math Dept Thesis style file was written for an old version of LaTeX. Mark Senn's puthesis.cls style file was written for LaTeX 2e, which is the default version of LaTeX that runs on the Math Dept SUNs.
Similar files exist for the AMS-TeX package. See
/pkgs/teTeX/local/tex/amstex/thesis
and
/pkgs/teTeX/local/doc/amstex/thesis
If you want to use AMS-TeX, TeX, or LaTeX at home on your own PC, check out MikTeX to download a free version of TeX for Windows. (It even has a dvi previewer called yap that is quite nice.) | 1,188 | 4,457 | {"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} | 2.890625 | 3 | CC-MAIN-2018-05 | latest | en | 0.839259 |
https://www.talkstats.com/threads/a-question-about-whether-dropping-interaction-terms-as-a-whole-or-not.58528/ | 1,660,787,425,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00271.warc.gz | 871,169,415 | 10,108 | # A question about whether dropping interaction terms as a whole or not
#### freezingswallow
##### New Member
Hello!
I'm taking a regression course this semester and I have a question regarding when to drop the interaction terms.
Suppose we have a regression model like this:
Y = a + b1*X1 + b2*X2 + b3*X3 + b4*X1_X3 + b5*X2_X3
where Y is the response variable (say, the outcome of a disease),
X1 and X2 are both indicator variables for the same factor (for example, there are three BMI categories, and I choose the third category to be the baseline group)
X3 is another variable ( eg. age)
X1_X3 is the interaction between X1 and X3
X2_X3 is the interaction between X2 and X3
After running the regression model on a data set, if we found the beta coefficient for X1_X2 is non-significant (p-value > 0.5) but the beta coefficient for X1_X3 is significant ( P-value < 0.5), should we drop X1_X3 only or both X1_X3 and X2_X3?
My question here is when the two interaction terms are actually between the same two factors ( just different indicator variables for different levels of the same variable), should we treat these interaction terms as a whole or treat them separately?
As we know, for main effect, we cannot drop X1 only and leave X2 in the model. I'm not sure if this is still the case regarding interaction terms!
Thank you!
#### hlsmith
##### Less is more. Stay pure. Stay poor.
I did not completely follow your description, you could probably benefit from some returns/spacing.
Did you enter BMI into the model as a single categorical variable, then the program kicks out more than one beta coefficient?
I believe you keep them all in when they cover the same variable (e.g., BMI).
#### freezingswallow
##### New Member
Sorry for causing misunderstanding.
I mean I'm trying to build a model to predict coronary heart disease, for example.
And my current model is:
Y = a + b1*BMI_Group1 + b2*BMI_Group2 + b3*age + b4*BMI1_Age + b5*BMI2_Age
where a, b1~b5 are just coefficients,
BMI1_Age is the interaction between BMI_Group1 and age
BMI2_Age is the interaction between BMI_group2 and age
And I run the model on my data set and found the p-value for one of the interaction terms is greater than 0.05 and p-value for the other interaction term is less than 0.05.
I'm not sure whether I should keep both interaction terms in or just the significant one.
Will this be more clear?
#### hlsmith
##### Less is more. Stay pure. Stay poor.
I totally got that part, at least that is what I assumed you meant. So, my response would still be the same as the above #2 post.
I believe you keep them all in when they cover the same variable (e.g., BMI). | 664 | 2,658 | {"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} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.913721 |
http://de.metamath.org/mpeuni/erclwwlktr.html | 1,721,489,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00357.warc.gz | 6,887,902 | 10,267 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > erclwwlktr Structured version Visualization version GIF version
Theorem erclwwlktr 26343
Description: ∼ is a transitive relation over the set of closed walks (defined as words). (Contributed by Alexander van der Vekens, 10-Apr-2018.) (Revised by Alexander van der Vekens, 11-Jun-2018.)
Hypothesis
Ref Expression
erclwwlk.r = {⟨𝑢, 𝑤⟩ ∣ (𝑢 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑤 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑤))𝑢 = (𝑤 cyclShift 𝑛))}
Assertion
Ref Expression
erclwwlktr ((𝑥 𝑦𝑦 𝑧) → 𝑥 𝑧)
Distinct variable groups: 𝑛,𝐸,𝑢,𝑤 𝑛,𝑉,𝑢,𝑤 𝑥,𝑛,𝑢,𝑤,𝑦 𝑧,𝑛,𝑢,𝑤,𝑥
Allowed substitution hints: (𝑥,𝑦,𝑧,𝑤,𝑢,𝑛) 𝐸(𝑥,𝑦,𝑧) 𝑉(𝑥,𝑦,𝑧)
Proof of Theorem erclwwlktr
Dummy variables 𝑚 𝑘 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 vex 3176 . 2 𝑥 ∈ V
2 vex 3176 . 2 𝑦 ∈ V
3 vex 3176 . 2 𝑧 ∈ V
4 erclwwlk.r . . . . . 6 = {⟨𝑢, 𝑤⟩ ∣ (𝑢 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑤 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑤))𝑢 = (𝑤 cyclShift 𝑛))}
54erclwwlkeqlen 26340 . . . . 5 ((𝑥 ∈ V ∧ 𝑦 ∈ V) → (𝑥 𝑦 → (#‘𝑥) = (#‘𝑦)))
653adant3 1074 . . . 4 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑦 → (#‘𝑥) = (#‘𝑦)))
74erclwwlkeqlen 26340 . . . . . . 7 ((𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑦 𝑧 → (#‘𝑦) = (#‘𝑧)))
873adant1 1072 . . . . . 6 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑦 𝑧 → (#‘𝑦) = (#‘𝑧)))
94erclwwlkeq 26339 . . . . . . . 8 ((𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑦 𝑧 ↔ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))))
1093adant1 1072 . . . . . . 7 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑦 𝑧 ↔ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))))
114erclwwlkeq 26339 . . . . . . . . . 10 ((𝑥 ∈ V ∧ 𝑦 ∈ V) → (𝑥 𝑦 ↔ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))))
12113adant3 1074 . . . . . . . . 9 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑦 ↔ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))))
13 simpr1 1060 . . . . . . . . . . . . . . 15 (((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) ∧ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))) → 𝑥 ∈ (𝑉 ClWWalks 𝐸))
14 simplr2 1097 . . . . . . . . . . . . . . 15 (((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) ∧ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))) → 𝑧 ∈ (𝑉 ClWWalks 𝐸))
15 oveq2 6557 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (𝑛 = 𝑚 → (𝑦 cyclShift 𝑛) = (𝑦 cyclShift 𝑚))
1615eqeq2d 2620 . . . . . . . . . . . . . . . . . . . . . . . . 25 (𝑛 = 𝑚 → (𝑥 = (𝑦 cyclShift 𝑛) ↔ 𝑥 = (𝑦 cyclShift 𝑚)))
1716cbvrexv 3148 . . . . . . . . . . . . . . . . . . . . . . . 24 (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) ↔ ∃𝑚 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑚))
18 oveq2 6557 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (𝑛 = 𝑘 → (𝑧 cyclShift 𝑛) = (𝑧 cyclShift 𝑘))
1918eqeq2d 2620 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (𝑛 = 𝑘 → (𝑦 = (𝑧 cyclShift 𝑛) ↔ 𝑦 = (𝑧 cyclShift 𝑘)))
2019cbvrexv 3148 . . . . . . . . . . . . . . . . . . . . . . . . 25 (∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛) ↔ ∃𝑘 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑘))
21 clwwlkprop 26298 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 (𝑧 ∈ (𝑉 ClWWalks 𝐸) → (𝑉 ∈ V ∧ 𝐸 ∈ V ∧ 𝑧 ∈ Word 𝑉))
2221simp3d 1068 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 (𝑧 ∈ (𝑉 ClWWalks 𝐸) → 𝑧 ∈ Word 𝑉)
2322ad2antlr 759 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → 𝑧 ∈ Word 𝑉)
24 simpr 476 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)))
2523, 24cshwcsh2id 13425 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (((𝑚 ∈ (0...(#‘𝑦)) ∧ 𝑥 = (𝑦 cyclShift 𝑚)) ∧ (𝑘 ∈ (0...(#‘𝑧)) ∧ 𝑦 = (𝑧 cyclShift 𝑘))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))
2625expdcom 454 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 ((𝑚 ∈ (0...(#‘𝑦)) ∧ 𝑥 = (𝑦 cyclShift 𝑚)) → ((𝑘 ∈ (0...(#‘𝑧)) ∧ 𝑦 = (𝑧 cyclShift 𝑘)) → ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
2726ancoms 468 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 ((𝑥 = (𝑦 cyclShift 𝑚) ∧ 𝑚 ∈ (0...(#‘𝑦))) → ((𝑘 ∈ (0...(#‘𝑧)) ∧ 𝑦 = (𝑧 cyclShift 𝑘)) → ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
2827expdcom 454 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (𝑘 ∈ (0...(#‘𝑧)) → (𝑦 = (𝑧 cyclShift 𝑘) → ((𝑥 = (𝑦 cyclShift 𝑚) ∧ 𝑚 ∈ (0...(#‘𝑦))) → ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))))
2928com4t 91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 ((𝑥 = (𝑦 cyclShift 𝑚) ∧ 𝑚 ∈ (0...(#‘𝑦))) → ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (𝑘 ∈ (0...(#‘𝑧)) → (𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))))
3029ex 449 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (𝑥 = (𝑦 cyclShift 𝑚) → (𝑚 ∈ (0...(#‘𝑦)) → ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (𝑘 ∈ (0...(#‘𝑧)) → (𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))))
3130com13 86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (𝑚 ∈ (0...(#‘𝑦)) → (𝑥 = (𝑦 cyclShift 𝑚) → (𝑘 ∈ (0...(#‘𝑧)) → (𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))))
3231imp41 617 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 (((((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) ∧ 𝑚 ∈ (0...(#‘𝑦))) ∧ 𝑥 = (𝑦 cyclShift 𝑚)) ∧ 𝑘 ∈ (0...(#‘𝑧))) → (𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))
3332rexlimdva 3013 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) ∧ 𝑚 ∈ (0...(#‘𝑦))) ∧ 𝑥 = (𝑦 cyclShift 𝑚)) → (∃𝑘 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))
3433ex 449 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) ∧ 𝑚 ∈ (0...(#‘𝑦))) → (𝑥 = (𝑦 cyclShift 𝑚) → (∃𝑘 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
3534rexlimdva 3013 . . . . . . . . . . . . . . . . . . . . . . . . 25 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (∃𝑚 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑚) → (∃𝑘 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑘) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
3620, 35syl7bi 244 . . . . . . . . . . . . . . . . . . . . . . . 24 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (∃𝑚 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑚) → (∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
3717, 36syl5bi 231 . . . . . . . . . . . . . . . . . . . . . . 23 ((((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸)) ∧ ((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦))) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → (∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
3837exp31 628 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) → (𝑧 ∈ (𝑉 ClWWalks 𝐸) → (((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → (∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))))
3938com15 99 . . . . . . . . . . . . . . . . . . . . 21 (∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛) → (𝑧 ∈ (𝑉 ClWWalks 𝐸) → (((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))))
4039impcom 445 . . . . . . . . . . . . . . . . . . . 20 ((𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → (((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))))
41403adant1 1072 . . . . . . . . . . . . . . . . . . 19 ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → (((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))))
4241impcom 445 . . . . . . . . . . . . . . . . . 18 ((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
4342com13 86 . . . . . . . . . . . . . . . . 17 ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸)) → (∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛) → ((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
44433impia 1253 . . . . . . . . . . . . . . . 16 ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛)) → ((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))
4544impcom 445 . . . . . . . . . . . . . . 15 (((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) ∧ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))) → ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))
4613, 14, 453jca 1235 . . . . . . . . . . . . . 14 (((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) ∧ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))) → (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛)))
474erclwwlkeq 26339 . . . . . . . . . . . . . . 15 ((𝑥 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑧 ↔ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
48473adant2 1073 . . . . . . . . . . . . . 14 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑧 ↔ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑥 = (𝑧 cyclShift 𝑛))))
4946, 48syl5ibrcom 236 . . . . . . . . . . . . 13 (((((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) ∧ (𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛))) ∧ (𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛))) → ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → 𝑥 𝑧))
5049exp31 628 . . . . . . . . . . . 12 (((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) → ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛)) → ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → 𝑥 𝑧))))
5150com24 93 . . . . . . . . . . 11 (((#‘𝑦) = (#‘𝑧) ∧ (#‘𝑥) = (#‘𝑦)) → ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛)) → ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → 𝑥 𝑧))))
5251ex 449 . . . . . . . . . 10 ((#‘𝑦) = (#‘𝑧) → ((#‘𝑥) = (#‘𝑦) → ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛)) → ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → 𝑥 𝑧)))))
5352com4t 91 . . . . . . . . 9 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → ((𝑥 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑦))𝑥 = (𝑦 cyclShift 𝑛)) → ((#‘𝑦) = (#‘𝑧) → ((#‘𝑥) = (#‘𝑦) → ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → 𝑥 𝑧)))))
5412, 53sylbid 229 . . . . . . . 8 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑦 → ((#‘𝑦) = (#‘𝑧) → ((#‘𝑥) = (#‘𝑦) → ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → 𝑥 𝑧)))))
5554com25 97 . . . . . . 7 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → ((𝑦 ∈ (𝑉 ClWWalks 𝐸) ∧ 𝑧 ∈ (𝑉 ClWWalks 𝐸) ∧ ∃𝑛 ∈ (0...(#‘𝑧))𝑦 = (𝑧 cyclShift 𝑛)) → ((#‘𝑦) = (#‘𝑧) → ((#‘𝑥) = (#‘𝑦) → (𝑥 𝑦𝑥 𝑧)))))
5610, 55sylbid 229 . . . . . 6 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑦 𝑧 → ((#‘𝑦) = (#‘𝑧) → ((#‘𝑥) = (#‘𝑦) → (𝑥 𝑦𝑥 𝑧)))))
578, 56mpdd 42 . . . . 5 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑦 𝑧 → ((#‘𝑥) = (#‘𝑦) → (𝑥 𝑦𝑥 𝑧))))
5857com24 93 . . . 4 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑦 → ((#‘𝑥) = (#‘𝑦) → (𝑦 𝑧𝑥 𝑧))))
596, 58mpdd 42 . . 3 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → (𝑥 𝑦 → (𝑦 𝑧𝑥 𝑧)))
6059impd 446 . 2 ((𝑥 ∈ V ∧ 𝑦 ∈ V ∧ 𝑧 ∈ V) → ((𝑥 𝑦𝑦 𝑧) → 𝑥 𝑧))
611, 2, 3, 60mp3an 1416 1 ((𝑥 𝑦𝑦 𝑧) → 𝑥 𝑧)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ∃wrex 2897 Vcvv 3173 class class class wbr 4583 {copab 4642 ‘cfv 5804 (class class class)co 6549 0cc0 9815 ...cfz 12197 #chash 12979 Word cword 13146 cyclShift ccsh 13385 ClWWalks cclwwlk 26276 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-map 7746 df-pm 7747 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 df-clwwlk 26279 This theorem is referenced by: erclwwlk 26344
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## #1 2005-11-04 00:45:39
jetsetwilly
Novice
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### y = mx + c. Why m?
Why is the letter m used to denote the gradient of a line? Why not g?
## #2 2005-11-04 01:32:49
mathsyperson
Moderator
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### Re: y = mx + c. Why m?
I think it's just convention. I don't know how that convention started though.
Why did the vector cross the road?
It wanted to be normal.
## #3 2005-11-04 08:26:59
MathsIsFun
Offline
### Re: y = mx + c. Why m?
Found this:
The earliest known use of "m" for slope is from an 1844 British text by Matthew O'Brien entitled "A Treatise on Plane Co-Ordinate Geometry". Later in 1848 George Salmon (1819-1904) referred to O'Brien's 1844 article within his "A Treatise on Conic Sections" and used the slope-intercept formula "y = mx + b", where "b" is the ordinate (vertical component) of the point where the line intersects the y-axis. It is also known that the four authors Isaac Todhunter in 1855 (Treatise on Plane Co-Ordinate Geometry), George A. Osborne in 1891 (Differential and Integral Calculus), and Arthur M. Harding and George W. Mullins in 1924 (Analytic Geometry) each used "m" to refer to slope in their mathematical writings.
(Source: http://www.bookrags.com/sciences/mathematics/slope-wom.html)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #4 2005-11-04 08:34:11
Curunen
Member
Offline
### Re: y = mx + c. Why m?
wow. It might be good if a new alphabet was made, so that you didn't end up with multiple meanings for a symbol.
Techniques are many, Principles are few.
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# Evaluate limx→2f(x) (if it exists), where f(x) =⎧⎪⎨⎪⎩x−[x],x<24,x=23x−5,x>2
Solution
## limx→2−f(x)=limx→2−{x−[x]} =limx→2−x−limx→2−[x] =2−1=1[∵limx→k−[x]=k−1] limx→2+f(x)=limx→2+(3x−5) =3(2)-5 =6-5 =1 Thus,limx→2−f(x)=1=limx→2+f(x) ⇒limx→2f(x)=1
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A MONTH ago I demonstrated one of my favourite studies: Mitrofanov's
exercise in wish fulfilment in which a king and three pawns ultimately defeat a good part of a complete army. Today another favourite, but in a quite different genre.
White to play and win
As with the Mitrofanov study, I refreshed my memory in Endgame Magic by John Beasley and Timothy Whitworth (BT Batsford, pounds 9.99). In fact the problem appears on the opposite page, but only because they are both in the chapter entitled "The Grand Manner".
Beasley and Whitworth tell the interesting story of how the Dutch composer John Selman had published a similar effort with the same finale in 1940 and improved it in 1949. But communications just after the war were bad and Korolkov produced this even better effort independently (unless you invoke Jung's Collective Unconscious) in 1951.
The solution depends completely on the logic stemming from White's efforts to promote his passed pawn and Black's to prevent this. The first move is obvious:
1 f7.
Now 1... Rf6 fails to 2 Bb2, so Black first gives check:
1... Ra6+
Posing a difficult problem, since 2 Kb2 would take the square away from the bishop, allowing 2... Rf6 while if 2 Kb1? Bxf5+ is check. There remains only:
2 Ba3! Rxa3+ 3 Kb2.
Now, the rook can't get back but it can harass the white king, which must avoid the diagonals from f5 to b1 (... Bxf5+) and e6 to a2 (...Be6+)
3... Ra2+!
If 3... Rb3+? 4 Ka2, White wins immediately:
4 Kc1!
Perhaps the hardest move of the solution. It turns out that 4 Kc3 Rc2+ would lead to a draw, since the white king can escape the checks neither by crossing the d file - when the rook can get to d8, eg 5 Kd4 Rd2+ 6 Ke5 Rd8; nor via the seventh rank, when a rook check on that rank will win the pawn, eg 5 Kb4 Rb2+ 6 Kc5 Rc2+ 7 Kb6 Rb2+ 8 Kc7 Rb7+.
4... Ra1+ 5 Kd2 Ra2+ 6 Ke3 Ra3+ 7 Kf4 Ra4+ 8 Kg5 Rg4+! 9 Kh6! Rg8.
Or 9... Rg6+ 10 Kxg6 Bxf5 + 11 Kf6
10 Ne7 Be6
11 fxg8Q+
I happened to have a computer on at this moment - Hiarchs - and to my amusement it assessed the position as equal right up to the point when Black recaptured: 11... Bxg8, allowing 12 Ng6 mate.
All this with just seven pieces. A masterpiece!
jspeelman@compuserve.com | 668 | 2,293 | {"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-2016-50 | longest | en | 0.922616 |
https://wikivisually.com/wiki/Working_capital | 1,547,941,743,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00170.warc.gz | 721,245,364 | 30,513 | # Working capital
Working capital (abbreviated WC) is a financial metric which represents operating liquidity available to a business, organisation or other entity, including governmental entities. Along with fixed assets such as plant and equipment, working capital is considered a part of operating capital. Gross working capital is equal to current assets. Working capital is calculated as current assets minus current liabilities.[1] If current assets are less than current liabilities, an entity has a working capital deficiency, also called a working capital deficit.
A company can be endowed with assets and profitability but may fall short of liquidity if its assets cannot be readily converted into cash. Positive working capital is required to ensure that a firm is able to continue its operations and that it has sufficient funds to satisfy both maturing short-term debt and upcoming operational expenses. The management of working capital involves managing inventories, accounts receivable and payable, and cash.
## Calculation
Working capital is the difference between the current assets and the current liabilities.
The basic calculation of the working capital is done on the basis of the gross current assets of the firm.
${\displaystyle {\text{Working Capital}}={\text{CURRENT ASSETS}}-{\text{CURRENT LIABILITIES}}}$
### Inputs
Current assets and current liabilities include three accounts which are of special importance. These accounts represent the areas of the business where managers have the most direct impact:
The current portion of debt (payable within 12 months) is critical because it represents a short-term claim to current assets and is often secured by long-term assets. Common types of short-term debt are bank loans and lines of credit.
An increase in net working capital indicates that the business has either increased current assets (that it has increased its receivables or other current assets) or has decreased current liabilities—for example has paid off some short-term creditors, or a combination of both.
## Working capital cycle
### Definition
The working capital cycle (WCC) is the amount of time it takes to turn the net current assets and current liabilities into cash. The longer the cycle is, the longer a business is tying up capital in its working capital without earning a return on it. Therefore, companies strive to reduce their working capital cycle by collecting receivables quicker or sometimes stretching accounts payable.
### Meaning
A positive working capital cycle balances incoming and outgoing payments to minimize net working capital and maximize free cash flow. For example, a company that pays its suppliers in 30 days but takes 60 days to collect its receivables has a working capital cycle of 30 days. This 30-day cycle usually needs to be funded through a bank operating line, and the interest on this financing is a carrying cost that reduces the company's profitability. Growing businesses require cash, and being able to free up cash by shortening the working capital cycle is the most inexpensive way to grow. Sophisticated buyers review closely a target's working capital cycle because it provides them with an idea of the management's effectiveness at managing their balance sheet and generating free cash flows.
As an absolute rule of funders, each of them wants to see a positive working capital. Such situation gives them the possibility to think that your company has more than enough current assets to cover financial obligations. Though, the same can’t be said about the negative working capital.[2] A large number of funders believe that businesses can’t be sustainable with a negative working capital, which is a wrong way of thinking. In order to run a sustainable business with a negative working capital, it’s essential to understand some key components.
1. Approach your suppliers and persuade them to let you purchase the inventory on 1-2 month credit terms, but keep in mind that you must sell the purchased goods, to consumers, for money. 2. Effectively monitor your inventory management, make sure that it’s often refilled and with the help of your supplier, back up your warehouse.
Plus, big companies like McDonald’s, Amazon, Dell, General Electric and Wal-Mart are using negative working capital.[citation needed]
## Working capital management
Decisions relating to working capital and short-term financing are referred to as working capital management. These involve managing the relationship between a firm's short-term assets and its short-term liabilities. The goal of working capital management is to ensure that the firm is able to continue its operations and that it has sufficient cash flow to satisfy both maturing short-term debt and upcoming operational expenses.
A managerial accounting strategy focusing on maintaining efficient levels of both components of working capital, current assets, and current liabilities, in respect to each other. Working capital management ensures a company has sufficient cash flow in order to meet its short-term debt obligations and operating expenses.
### Decision criteria
By definition, working capital management entails short-term decisions—generally, relating to the next one-year period—which are "reversible". These decisions are therefore not taken on the same basis as capital-investment decisions (NPV or related, as above); rather, they will be based on cash flows, or profitability, or both.
• One measure of cash flow is provided by the cash conversion cycle—the net number of days from the outlay of cash for raw material to receiving payment from the customer. As a management tool, this metric makes explicit the inter-relatedness of decisions relating to inventories, accounts receivable and payable, and cash. Because this number effectively corresponds to the time that the firm's cash is tied up in operations and unavailable for other activities, management generally aims at a low net count.
• In this context, the most useful measure of profitability is return on capital (ROC). The result is shown as a percentage, determined by dividing relevant income for the 12 months by capital employed; return on equity (ROE) shows this result for the firm's shareholders. Firm value is enhanced when, and if, the return on capital, which results from working-capital management, exceeds the cost of capital, which results from capital investment decisions as above. ROC measures are therefore useful as a management tool, in that they link short-term policy with long-term decision making. See economic value added (EVA).
• Credit policy of the firm: Another factor affecting working capital management is credit policy of the firm. It includes buying of raw material and selling of finished goods either in cash or on credit. This affects the cash conversion cycle.
### Management of working capital
Guided by the above criteria, management will use a combination of policies and techniques for the management of working capital. The policies aim at managing the current assets (generally cash and cash equivalents, inventories and debtors) and the short-term financing, such that cash flows and returns are acceptable.
• Cash management. Identify the cash balance which allows for the business to meet day to day expenses, but reduces cash holding costs.
• Inventory management. Identify the level of inventory which allows for uninterrupted production but reduces the investment in raw materials—and minimizes reordering costs—and hence increases cash flow. Besides this, the lead times in production should be lowered to reduce Work in Process (WIP) and similarly, the Finished Goods should be kept on as low level as possible to avoid overproduction—see Supply chain management; Just In Time (JIT); Economic order quantity (EOQ); Economic quantity
• Debtors management. Identify the appropriate credit policy, i.e. credit terms which will attract customers, such that any impact on cash flows and the cash conversion cycle will be offset by increased revenue and hence Return on Capital (or vice versa); see Discounts and allowances.
• Short-term financing. Identify the appropriate source of financing, given the cash conversion cycle: the inventory is ideally financed by credit granted by the supplier; however, it may be necessary to utilize a bank loan (or overdraft), or to "convert debtors to cash" through "factoring". | 1,583 | 8,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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-2019-04 | latest | en | 0.951711 |
https://www.slideserve.com/kipling/ratio-transformation-for-stationary-time-series-with-special-application-in-consumer-price-index-in-qatar | 1,601,055,569,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00236.warc.gz | 1,051,972,058 | 16,525 | # Ratio Transformation for Stationary Time Series with Special Application in Consumer Price Index in Qatar - PowerPoint PPT Presentation
Ratio Transformation for Stationary Time Series with Special Application in Consumer Price Index in Qatar
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Ratio Transformation for Stationary Time Series with Special Application in Consumer Price Index in Qatar
## Ratio Transformation for Stationary Time Series with Special Application in Consumer Price Index in Qatar
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##### Presentation Transcript
1. Ratio Transformation for Stationary Time Series withSpecial Application in Consumer Price Index in Qatar By: AdilYousif, Hind Alrkeb, Doha Alhashmi Department of Mathematics and Statistics Qatar University Doha, Qatar
2. Abstract This article was intended to perform a comprehensive time series analysis for Consumer Price Index (CPI) in Qatar. The data was obtained from the Statistics Authority in Qatar. For the period between: (2002-2009) a quarterly data was analyzed using several time series techniques such as Exponential Trend method, Holts’ Trend method, and ARIMA. These methods were used to examine trends and built a forecast model. Ratio transformation technique was used to obtain a stationary time series and found to be efficient with small size of data. Despite the small size of the data the analysis indicated that ARIMA model is more adequate forecast one. Key words: Time series, index, CPI, ARIMA, Ratio Transformation
3. Introduction • Qatar witnessed a tremendous increase in the economy in the last decade. • The economic policy in Qatar is moving towards economic diversifications through investing the returns of oil and gas in profitable projects. • In addition to that, the state adopts new methodologies in economic and trade liberalization to attract foreign investments and strengthen the private sector.
4. The GDP has increased rapidly during the last period. It reached 268 billion Qatari Riyals (73.4 billion US Dollar) in 2009 as per the estimation of Statistics Authority. The growth rates of the GDP during last three years were 34%, 13%, and 15% in the years 2006, 2007 and 2008 respectively. As Qatar’s economy is in the booming phase, the prices level increased rapidly during last years as a result of expanding in the Building & Construction sector. This expansion caused inflation in rent rates, and hence affected the prices level in Qatar
5. Index Number • The index number is defined as a statistical indicator expresses the proportional change in value of specific phenomenon or group of phenomena compared to specific base. • When the change is over the time the previous period is called the base period, where as the measured period is called the comparison period. Several methods are used to calculate the index number and some of them are discussed below.
6. Laspere’s index number • It is assumed within this index that individuals will consume, in the new period and in the shadow of change in prices, the same original group of goods, weighted by base year quantities, as shown in the following equation:
7. Paasche’s index number • This index is based on an assumption differs from Laspere’s assumption, which is, individuals will consume in the new period and in the shadow of change in prices, a new group or quantities of goods for each year separately, provided that prices are weighted by comparison year quantities, as shown in the following equation:
8. Fisher’s optimum index number • It is the index which comprises the Laspere’s index and the Paasche’s index by finding the geometric mean of both numbers, as shown in the following equation:
9. Collecting Price Data • The Statistics Authority in Qatar is the only official body responsible of collecting data in the country. • Price data were collected through personal interview by the trained data collectors. • Data collection is carried out by registering prices from shelves of commercial stores( trade malls, etc). • Or through personal interview with establishment managers education centers, clinics, real estate bureaus, government establishments,…etc). • Each data collector has been provided with questionnaire for each group of prices and sources of each questionnaire. • The data were collected quarterly for the period of 2002-2009 with the base year considered to be 2001. • After price data sets are collected, it will be audited at the office and entered to the computer, and then the average price and consumer prices index number are calculated. Laspere’s formula was used to calculate the index number weighted by 2001 prices.
10. Time series plot of HOUSEHOLD dataFig[1]
11. Time Series Plot of HOUSEHOLD and GDPFig[2]
12. Methodology • The original data was used to fit the exponential trend and the Holts Winter model. • Since ARIMA requires a stationary set of data and the original data is violating this condition transformation was used. • The traditional difference transformation was used and the third difference yield in a stationary set. • Since the data is relatively small a new approach of transformation in which ratios was considered to avoid the loss in the sample size as a result of several differences. Results of the two transformation methods are almost identical see table
13. ARIMA • When using this transformation with first order autoregressive model • It becomes: • Which eliminate the constant terms and the random shock term will be a multiple of the response variable. The final model will be
14. Ratio for ARIMA • When the ratio transformation ( ) to the first order moving average model is used • Substituting the ratio form we get: • Which involve the response variable directly unlike the original model, however the constant term is again eliminated
15. Double Exponential Smoothing • Double exponential smoothing technique was introduced by Holt (and Brown as a special case) smoothes the data. • Double exponential smoothing provides short-term forecasts. • This procedure can work well when a trend is present but it can also serve as a general smoothing method.
16. Both exponential trend method and Holts’ Double Exponential Smoothing were used and the last outperformed the first when their MADs are compared. Figure [3] shows the fitted values versus actual values and one year ahead of forecast together with their 95% prediction interval for the Holt’s exponential model. It appears that there is an increasing trend in the Household variable with a MAD (1.21962). The one year a head forecasts of this model are displayed in table [1]
17. Holt’s Trend Exponential Smoothing for HouseholdFig[3]
18. Auto Regressive Integrated Moving Avareg (ARIMA) • The data shows an exponential trend in the time series plot, without seasonal variation , so we can apply non-seasonal ARIMA model. • The analysis of the time series of the CPI is conducted by using MINITAB version 16.0. After the examination of the behavior of the sample autocorrelation (SAC) function and partial autocorrelation (SPAC) function for the regular difference transformation data. • Referring to figures [4] and [5], it can be concluded that the data became stationary after the third difference. It can also be observed that both of the SAC and SPAC cut off after lag 1.
19. ARIMA • However, since the SPAC cut off more fairly quickly, we tentatively identify the non-seasonal autoregressive model of order 1. • The ARIMA for this data shows that the two parameters in the AR model of order 1 have p -value less than 0.05. This implies that we have very strong evidence that each term in this is important since can be rejected at the significance level α equals to 0.05.
20. Also from the table [1] it can be noticed that Ljung-Box test has a p –value associated with for lags K = 12 and K = 24 are both greater than 0.05. Hence, it can be concluded that model is adequate. Moreover since =0.4095 < 1 then the third difference data is stationary. Modified Box-Pierce (Ljung-Box) Chi-Square statistic
21. Ratio Transformation • For the ratio transformation the sample autocorrelation function dies cuts off after lag 1, whereas the partial autocorrelation dies in a dammed fashion and a suggested model is a moving average of order 1. • The p-values for the two parameters are 0.000 which implies they are significantly different from zero. From table [2] below the Ljumg-Box test has p-values at lag K =12 and K=24 equal to 0.000 and therefore we can confidently say the model MA of order 1 is adequately fit the ratio transformation data. The invertibility condition is also satisfied since
22. Modified Box-Pierce (Ljung-Box) Chi-Square statistic
23. Forecast • The forecasts for one year ahead, (2010) for the three models along with actual values (obtained from Qatar Statistical Authority) are then compared and listed in the table [3] below. • The ARIMA model for the ratio transformation data has the best estimates followed by the ARIMA model for the regular transformation data. • Although the Holt’s exponential model is less accurate than ARIMA still its error forecast is very small and its predicted values are not far from the actual ones.
24. One Year Ahead Forecast (2010)
25. Conclusion • From time series analysis it appeared Double Exponential Smoothing model is better than exponential trend model since the values of MAD and MSD in Double Exponential Smoothing are smaller. • On the other hand ARIMA was used for two different sets of data and both of them provided forecasts more accurate than the Double Exponential Smoothing model. • For the ARIMA in the first set the regular difference transformation was used and a third difference yield into a stationary time series, while in the second set the ratio difference was used for the purpose of preventing the decrease the sample size due to more than one differences. • The ARIMA model for the ratio transformation produced more accurate forecast.
26. References • [1] Aka, B. F. and Pieretti, P., Consumer Price Index Dynamics in a Small Open Economy: Structural Time Series Model for Luxembourg, International Journal of Applied Economics, 5910, p 1-13, 2008. • [2] Bascand G., Product Price Index: June Quarter Highlights, Statistics New Zealand, ISSN 1178-0622, 2009 • [3]Bulletin of Prices Index Numbers 2006 -5th Issue July 2007 • [4] Bower B., O’Conell R. and Koehler A., Forecasting Time Series and Regression: An Applied Approach • [5] Bowles, T. J. and Cris Lewis, W., A Time Series Analysis of the Medical Care Price Index: Implication for Appraising Economic Losses, Journal of Forensic Economics 13(3), pp 245-254, 2000. • [6] Box G., Jenkins G. and Reinsel G., Time Series Analysis: Forecasting and Control. Prentice Hall. • [7] US Bureau of Labor Statistics: Response Rate for the Consumer Price Index: Consumer Price index Program, Report 2004
27. THANKYOU | 2,360 | 10,944 | {"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-2020-40 | latest | en | 0.873037 |
https://math.stackexchange.com/questions/318169/determinant-of-matrix-exponential/318689 | 1,568,686,439,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573011.59/warc/CC-MAIN-20190917020816-20190917042816-00305.warc.gz | 583,749,625 | 33,760 | # Determinant of matrix exponential?
Suppose $A$ is a $n \times n$ constant matrix. How can I prove $\det(e^A) = e^{\displaystyle \sum_{\lambda_i\in\sigma(A)} \lambda_i}$,
where $\sigma(A)$ is the multiset of eigenvalues of $A$?
The following matlab code shows this is true in $n = 3$ case:
A = rand(3)
detA = exp(sum(eig(A)))
detmA = det(expm(A))
• prove it for diagonal matrices and use denseness of diagonalizable matrices. notice that the trace of $A$ is the sum of the eigenvalues – cats Mar 1 '13 at 20:34
## 4 Answers
If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.
Then $$\mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U.$$
Now observe that $e^U$ is upper triangular with $\{e^{\lambda_j}\}$ on the diagonal.
So $$\mbox{det} \;e^A=\mbox{det} \;e^U=e^{\lambda_1}\cdots e^{\lambda_n}=e^{\lambda_1+\ldots+\lambda_n}.$$
• Anyone trying to understand this solution: The main idea is that if the eigenvalues of $A$ are $\lambda_i$, then the eigenvalues of $e^A$ are $e^{\lambda_i}$, which you can see by using the series definition of $e^A$ and multiplying nonzero eigenvector $v_i$ as in $e^A v_i = \lambda_i v_i$. Distribute the $v_i$ through the series. Finally, the determinant of a matrix is the product of the eigenvalues, and the trace of a matrix is the sum of the eigenvalues. This explains the second to last equality in the proof above. – wkschwartz Oct 25 '15 at 23:28
You can reduce problem for diagonal matrix using the fact that every matrix can be approximated with any given precision and both functions: $e^X$ and $\det X$ are continuous. The problem with diagonal matrix is obvious.
P.S. It seems that lyj was faster than me.
For an analytic method, using differential equations:
Let $f(t)= \det(e^{tA})$. Then $f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$. But $A$ and $e^{tA}$ commute, and $^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n$. Therefore, $f'(t)=\text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{\text{tr}(A)t}$. For $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.
• What kind of commutator is $^t\text{com}(e^{tA})$? Never seen this notation. – Your Majesty Jan 29 '14 at 10:13
• $\mathrm{com}(A)$ is the matrix of cofactors associated to $A$. Possibly, it is a French notation, where $\mathrm{com}$ stands for comatrice. – Seirios Jan 29 '14 at 11:56
• Hello from 2016! Could you please give a reference to this proof? – Rubi Shnol May 17 '16 at 6:35
You just use the Jordan normal form, $A$ is your matrix, $D$ is the Jordan normal form of $A$ and $S$ is the transformation matrix, for the second equal just remember the definition of matrix exponential, and that $$(SAS^{-1})^n = S A S^{-1}S A S^{-1} S A \dots S A S^{-1}=S A^n S^{-1}$$ \begin{align*} \det(\exp(A))&=\det(\exp(S D S^{-1}))\\ &=\det(S \exp(D) S^{-1})\\ &=\det(S) \det(\exp(D)) \det (S^{-1})\\ &=\det(\exp (D))\\ &=\prod_{i=1}^n e^{d_{ii}}\\ &=e^{\sum_{i=1}^n{d_{ii}}}\\ &=e^{\text{tr}D} \end{align*} As the trace is invariant this works. | 1,074 | 3,158 | {"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} | 4.09375 | 4 | CC-MAIN-2019-39 | latest | en | 0.791222 |
https://flatearthsoc.com/discount-flat-earth-news-blog-online-nick-davies-book.html | 1,566,196,519,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00307.warc.gz | 459,322,706 | 10,841 | ```175) Professional photo-analysts have dissected several NASA images of the ball-Earth and found undeniable proof of computer editing. For example, images of the Earth allegedly taken from the Moon have proven to be copied and pasted in, as evidenced by rectangular cuts found in the black background around the “Earth” by adjusting brightness and contrast levels. If they were truly on the Moon and Earth was truly a ball, there would be no need to fake such pictures.
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Another false law of Newton is that gravity increases with the increase of the mass of the object. There is no such thing as mass (no one in the world can define it) - there is only density of the object (total density volume of the object, including it's electric field that surrounds it), and it is enough to understand how the laws work. Rubber ball pumped with helium goes up irrespective of the "gravity law" which supposed to bring everything down. Ball goes up because the density of the helium is smaller than the density of air above it. There is also no resistance of the environment above the ball.
```Considerably more than a million Earths would be required to make up a body like the Sun -the astronomers tell us: and more than 53,000 suns would be wanted to equal the cubic contents of the star Vega. And Vega is a "small star!" And there are countless millions of these stars! And it takes 30,000,000 years for the light of some of those stars to reach us at 12,000,000 miles in a minute! And, says Mr. Proctor, "I think a moderate estimate of the age of the Earth would be 500,000,000 years! "Its weight," says the same individual, "is 6,000,000,000,000,000,000,060 tons!" Now, since no human being is able to comprehend these things, the giving of them to the world is an insult - an outrage. And though they have all risen from the one assumption that Earth is a planet, instead of upholding the assumption, they drag it down by the weight of their own absurdity, and leave it lying in the dust - a proof that Earth is not a globe.
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# If the earth is flat we are held hostage to one of the biggest hoaxes of the last 500 hundred years of human existence. Every day I see evidence of the flat earth. Just look at how at time the clouds are not moving at all --completely still. If the earth was spinning at 1600 klms per hour --how could it be that at times the clouds are completely still not moving in the sky at all.(how could the clouds be moving at the same speed as the rotating earth?)When you look at a long distant horizon --from right to left --there is no curviture in the earth whatsoever. If you try to point this out to people they will think you have lost your mind.. we have been brainwashed to believe a lie which is very difficult to prove to the world populous that there is a flat earth. The deception is wrapped in scientific mumbo jumbo which is very easy to accept as it is backed up by a lot of scientific explanation--if you want to take the easy way out--you just accept that the earth is a globe spinning at 1600 klms per hour . But how does all the water of the seas stay stuck to the planet when the earth is spinning around at 1600 klms per hour? (gravity cannot hold water upside down--water will flow to the lowest point--how can that happen on a spinning globe?)
3) The natural physics of water is to find and maintain its level. If Earth were a giant sphere tilted, wobbling and hurdling through infinite space then truly flat, consistently level surfaces would not exist here. But since Earth is in fact an extended flat plane, this fundamental physical property of fluids finding and remaining level is consistent with experience and common sense.
In the past 60 years of space exploration, we’ve launched satellites, probes, and people into space. Some of them got back, some of them still float through the solar system (and almost beyond it), and many transmit amazing images to our receivers on Earth. In all of these photos, the Earth is (wait for it) spherical. The curvature of the Earth is also visible in the many, many, many, many photos snapped by astronauts aboard the International Space Station. You can see a recent example from ISS Commander Scott Kelly's Instagram right here:
Great material, THANK YOU SO MUCH!!! I am very happy to be able to arrive at this information. By the way, they lied not only about earth shape and universe, but about physics, chemistry and even in math, inventing nonexistent values and rules. Example: they tell you 2x0=0. This is nonsense, because 2 has to denote something real, the actual values. Therefore 2x0=2, not 0, as they lie to us. Prove? Take 2 airplanes and multiply them on 0. How many airplanes will you get,-zero?...)) No, you will still have 2 airplanes! But 2 as abstract (nonexistent) number representing nothing, can not give any result but nothing.
```In general, we at the Flat Earth Society do not lend much credibility to photographic evidence. It is too easily manipulated and altered. Many of the videos posted here to "prove a round earth" by showing curvature will show no curvature or even concave curvature at parts. The sources are so inaccurate it's difficult to build an argument on them in either case. Furthermore, barrel distortion and other quirks of modern cameras will cause a picture to distort in ways which may not be immediately obvious or apparent, especially without references within the picture. Photographs are also prone to distortion when taken through the bent glass of a pressurized cabin as well as atmospheric conditions on the outside. With this litany of problems, it's easy to see why photographic evidence is not to be trusted.
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56.) The Sun and Moon may often be seen high in the heavens at the same time – the Sun rising in the east and the Moon setting in the west – the Sun's light positively putting the Moon's light out by sheer contrast! If the Newtonian theory were correct, and the moon had her light from the Sun, she ought to be getting more of it when face to face with that luminary – if it were possible for a sphere to act as a reflector all over its face! But as the Moon's light pales before the rising Sun, it is a proof that the theory fails; and is gives us a proof that the Earth is not a globe.
107) Ring magnets of the kind found in loudspeakers have a central North pole with the opposite “South” pole actually being all points along the outer circumference. This perfectly demonstrates the magnetism of our flat Earth, whereas the alleged source of magnetism in the ball-Earth model is emitted from a hypothetical molten magnetic core in the center of the ball which they claim conveniently causes both poles to constantly move thus evading independent verification at their two “ceremonial poles.” In reality the deepest drilling operation in history, the Russian Kola Ultradeep, managed to get only 8 miles down, so the entire ball-Earth model taught in schools showing a crust, outer-mantle, inner-mantle, outer-core and inner-core layers are all purely speculation as we have never penetrated through beyond the crust.
From the second equation above, the index of refraction at one atmosphere of pressure and a temperature of 310 K (50 degrees F) is 1.000284, while the index of refraction at one atmosphere of pressure and a temperature of 320 K (68 degrees F) is 1.000275. These values yield a critical angle of 89.76 degrees. Hence, when air attempts to pass from 310 K to 320 K air at one atmosphere of pressure, the light will be totally internally reflected if the angle of incidence is greater than 89.76 degrees, or less than about a quarter of a degree from grazing incidence. If the temperature difference is greater, the critical angle will be less; hence the angle from grazing incidence will be greater.
Astronomers tell us that, in consequence of the Earth's "rotundity," the perpendicular walls of buildings are, nowhere, parallel, and that even the walls of houses on opposite sides of a street are not! But, since all observation fails to find any evidence of this want of parallelism which theory demands, the idea must be renounced as being absurd and in opposition to all well-known facts. This is a proof that the Earth is not a globe.
I really do love this idea of flat earth. It brought me sense of peace and safety living on a flat and stationary Earth. It also makes us feel closer as brothers and sisters. It also is so much compatible with my holigraphic view of the universe. So thank you. But still i can't properly conceive how day and night happens how sunset and sunrise can be explained usong this theory.
There are rivers which flow east, west, north, an south - that is, rivers are flowing in all directions over the Earth's surface, and at the same time. Now, if the Earth were a globe, some of these rivers would be flowing up-hill and others down, taking it for a fact that there really is an "up" and a "down" in nature, whatever form she assumes. But, since rivers do not flow up-hill, and the globular theory requires that they should, it is a proof that the Earth is not a globe.
Supertramp became one of the first acts to sign to the emerging UK branch of A&M Records, and by the summer of 1970 they had recorded their first album, simply called Supertramp. Hodgson performed the lion's share of the lead vocals on this first effort, but by the time of their second album Indelibly Stamped, Davies had stepped up as a singer, and he and Hodgson were sharing lead vocal duties equally.
125) Another proof the Sun is not millions of miles away is found by tracing the angle of sun-rays back to their source above the clouds. There are thousands of pictures showing how sunlight comes down through cloud-cover at a variance of converging angles. The area of convergence is of course the Sun, and is clearly NOT millions of miles away, but rather relatively close to Earth just above the clouds.
It is a fact not so well known as it ought to be that when a ship, in sailing away from us, has reached the point at which her hull is lost to our unaided vision, a good telescope will restore to our view this portion of the vessel. Now, since telescopes are not made to enable people to see through a "hill of water," it is clear that the hulls of ships are not behind a hill of water when they can be seen through a telescope though lost to our unaided vision. This is a proof that Earth is not a globe.
99) Viewed from a ball-Earth, Polaris, situated directly over the North Pole, should not be visible anywhere in the Southern hemisphere. For Polaris to be seen from the Southern hemisphere of a globular Earth, the observer would have to be somehow looking “through the globe,” and miles of land and sea would have to be transparent. Polaris can be seen, however, up to over 20 degrees South latitude.
53) At places of comparable latitude North and South, the Sun behaves very differently than it would on a spinning ball Earth but precisely how it should on a flat Earth. For example, the longest summer days North of the equator are much longer than those South of the equator, and the shortest winter days North of the equator are much shorter than the shortest South of the equator. This is inexplicable on a uniformly spinning, wobbling ball Earth but fits exactly on the flat model with the Sun traveling circles over and around the Earth from Tropic to Tropic.
14 And God said, Let there be lights in the firmament of the heaven to divide the day from the night; and let them be for signs, and for seasons, and for days, and years: 15 And let them be for lights in the firmament of the heaven to give light upon the earth: and it was so. 16 And God made two great lights; the greater light to rule the day, and the lesser light to rule the night: he made the stars also. 17 And God set them in the firmament of the heaven to give light upon the earth, 18 And to rule over the day and over the night, and to divide the light from the darkness: and God saw that it was good. 19 And the evening and the morning were the fourth day.
As the mariners' compass points north and south at one time, and as the North, to which it is attracted is that part of the Earth situated where the North Star is in the zenith, it follows that there is no south "point" or "pole" but that, while the centre is North, a vast circumference must be South in its whole extent. This is a proof that the Earth is not a globe.
13) In a 19th century French experiment by M. M. Biot and Arago a powerful lamp with good reflectors was placed on the summit of Desierto las Palmas in Spain and able to be seen all the way from Camprey on the Island of Iviza. Since the elevation of the two points were identical and the distance between covered nearly 100 miles, if Earth were a ball 25,000 miles in circumference, the light should have been more than 6600 feet, a mile and a quarter, below the line of sight!
Since this image is visible above where the object is, it is called a superior mirage. Because cooler air has no physical reason to rise, a temperature inversion is a stable situation, with little convection as with the condition that produces an inferior mirage. Therefore, superior mirages can be very steady, much steadier than inferior mirages. Furthermore, since the refraction acts almost continually rather than at one point, superior mirages normally are erect rather than inverted. If one gains a little altitude, one can get out of the inversion layer, and thus avoid seeing a superior mirage. In my earlier article, I pointed out that this is what Alfred Russell Wallace did when he repeated the Bedford level experiment. Russell did not see the distant object that was his target, which is consistent with a spherical earth. Russell correctly accounted for this effect, but Rowbotham did not.
"By actual observation," says Schoedler, in his " Book of Nature," we know that the other heavenly bodies are spherical, hence we unhesitatingly assert that the earth is so also." This is a fair sample of all astronomical reasoning. When a thing is classed amongst "other" things, the likeness between them must first be proven. It does not take a Schoedler to tell us that "heavenly bodies" are spherical, but " the greatest astronomer of the age" will not, now, dare to tell us that THE EARTH is - and attempt to prove it. Now, since no likeness has ever been proven to exist between the Earth and the heavenly bodies, the classification of the Earth with the heavenly bodies is premature - unscientific -false! This is a proof that Earth is not a globe.
The circumstances which attend bodies which are caused merely to fall from a great height prove nothing as to the motion or stability of the Earth, since the object, if it be on a thing that is in motion, will participate in that motion; but, if an object be thrown, upwards from a body at rest, and, again, from a body in motion, the circumstances attending its descent will be very different. In the former case, it will fall, if thrown vertically upwards, at the place from whence it was projected; in the latter case, it will fall behind the moving body from which it is thrown will leave it in the rear. Now, fix a gun, muzzle upwards, accurately, in the ground; fire off a projectile; and it will fall by the gun. If the Earth traveled eleven hundred miles a minute, the projectile would fall behind the gun, in the opposite direction to that of the supposed motion. Since, then, this is NOT the case, in fact, the Earth's fancied motion is negatived and we have a proof that the Earth is not a, globe.
Mr. Proctor says.- "The Sun is so far off that even moving from one side of the Earth to the other does not cause him to be seen in a different direction - at least the difference is too small to be measured." Now, since we know that north of the equator, say 45 degrees, we see the Sun at mid-day to the south, and that at the same distance south of the equator we see the Sun at mid-day to the north, our very shadows on the round cry aloud against the delusion of the day and give us a proof that Earth is not a globe.
2. Another related thing I don’t understand: if the sun and moon are always above the disk of the Earth, why can’t everyone in the world see them at all times? Surely they should always be visible, at least at a low angle. I can’t draw myself any diagram where they are not always visible, but we see that that doesn’t happen. I can’t see how night time happens. Help!
I just watched a time lapse of the night sky that shows all the stars of heaven moving at once. This proves to me that they are something other than planets and suns far away. But I also noticed that there were really fast moving objects in the sky. Planes? meteors? I have looked up in the night sky and seen shooting stars. If its possible there is no ´´space´´, then what could they be? What are we really seeing when we look into the night sky?
10.) That the mariners' compass points north and south at the same time is a fact as indisputable as that two and two makes four; but that this would be impossible if the thing, were placed on a globe with "north" and "south' at the centre of opposite hemispheres is a fact that does not figure in the school-books, though very easily seen: and it requires no lengthy train of reasoning to bring out of it a pointed proof that the Earth is not a globe.
6) If Earth were a ball 25,000 miles in circumference as NASA and modern astronomy claim, spherical trigonometry dictates the surface of all standing water must curve downward an easily measurable 8 inches per mile multiplied by the square of the distance. This means along a 6 mile channel of standing water, the Earth would dip 6 feet on either end from the central peak. Every time such experiments have been conducted, however, standing water has proven to be perfectly level.
#### If the Earth were a globe, rolling and dashing through "space" at the rate of "a hundred miles in five seconds of time," the waters of seas and oceans could not, by any known law, be kept on its surface - the assertion that they could be retained under these circumstances being an outrage upon human understanding and credulity! But as the Earth - that is, the habitable world of dry land - is found to be "standing out of the wafer and in the water" of the "mighty deep," whose circumferential boundary is ice, we may throw the statement back into the teeth of those who make it and flaunt before their faces the flag of reason and common sense, inscribed with a proof that the Earth is not a globe.
On November 12, 2016, I had the opportunity to conduct this experiment. I was near the water’s edge, just beyond the surf, at Virginia Beach from middle to late afternoon. When I began, the air temperature was 50 degrees F, and the temperature dropped a degree or two by the time that I was done, near sunset. The water temperature was 62–64 degrees F, so the air immediately above the water was at least ten degrees warmer than the air temperature a short distance above the water. I photographed two cargo ships as they made their way out to sea from the port at Hampton Roads. I mounted a digital SLR camera on a 3.5-inch Questar telescope, having a 1,200-mm focal length. The ISO setting on the camera was 100 for all photographs.
65) Also Quoting Dr. Rowbotham, “On the shore near Waterloo, a few miles to the north of Liverpool, a good telescope was fixed, at an elevation of 6 feet above the water. It was directed to a large steamer, just leaving the River Mersey, and sailing out to Dublin. Gradually the mast-head of the receding vessel came nearer to the horizon, until, at length, after more than four hours had elapsed, it disappeared. The ordinary rate of sailing of the Dublin steamers was fully eight miles an hour; so that the vessel would be, at least, thirty-two miles distant when the mast-head came to the horizon. The 6 feet of elevation of the telescope would require three miles to be deducted for convexity, which would leave twenty-nine miles, the square of which, multiplied by 8 inches, gives 560 feet; deducting 80 feet for the height of the main-mast, and we find that, according to the doctrine of rotundity, the mast-head of the outward bound steamer should have been 480 feet below the horizon. Many other experiments of this kind have been made upon sea-going steamers, and always with results entirely incompatible with the theory that the earth is a globe.”
33.) If the Earth were a globe, people – except those on the top – would, certainly, have to be "fastened" to its surface by some means or other, whether by the "attraction" of astronomers or by some other undiscovered and undiscoverable process! But, as we know that we simply walk on its surface without any other aid than that which is necessary for locomotion on a plane, it follows that we have, herein, a conclusive proof that Earth is not a globe. | 4,639 | 20,904 | {"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-2019-35 | latest | en | 0.959881 |
https://mazurekgravity.in/question/a-hexagonal-plate-i-s-acted-upon-by-the-force-p-and-the-couple-shown-determine-the-magnitude-and-the-direction-of-the-smallest-force-p-for-which-this-system-can-be-replaced-with-a-single-force-at-e/ | 1,680,188,253,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00205.warc.gz | 456,684,150 | 24,900 | # A hexagonal plate i.s acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.
Question-AnswerCategory: Engineering MechanicsA hexagonal plate i.s acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.
S cornwell asked 1 year ago
A hexagonal plate i.s acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.
Step: 1
Draw the schematic of the hexagonal plate.
Step: 2
Identify that the triangle is an equilateral triangle. Hence, the distance is equals to .
Step: 3
Determine the net resultant force.
…… (1)
Calculate the distance
Substitute for , and for .
Determine the magnitude of the net moment about point .
…… (2)
Step: 4
Represent the net resultant force, and the net moment at point on the figure.
Fig. 2
Now let us consider that the moment, and force, may be replaced by force making an angle of with the horizontal axis at point as shown below.
Write the vector notation of force
…… (3)
Calculate the distance
Determine the magnitude of the net moment about point (see fig. 3)
…… (4)
Step: 5
Equate the components of equations (1) and (3).
Therefore, the magnitudes of the angles and are equal.
Equate the equations (2) and (4).
Substitute for .
…… (5)
Step: 6
Observe the equation (5).
The magnitude of the force is minimum when the value of is maximum.
The maximum value of sine of an angle is 1.
Equate the to 1.
Substitute for in the equation (5).
Therefore, the smallest force, is making an angle of with the horizontal. | 482 | 1,848 | {"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} | 3.84375 | 4 | CC-MAIN-2023-14 | latest | en | 0.686236 |
https://www.jmp.com/en_in/customer-stories/shibaura-institute-of-technology.html | 1,601,252,768,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00202.warc.gz | 892,756,298 | 19,318 | # A new generation of statistics-literate educators
## Empowering future educators with statistics savvy
Challenge To find new and innovative ways to introduce the next generation of teachers to statistical analysis. Shibaura Institute of Technology (SIT) encouraged all students and faculty to expand their understanding of statistics using easy-to-learn JMP® Pro statistical discovery software. Students and faculty members across SIT adopted JMP Pro to augment their use of statistical analysis in everything from basic coursework to graduate research. With the help of JMP Pro, SIT made statistics education both accessible and enjoyable for future educators.
The Shibaura Institute of Technology (SIT) first installed JMP® Pro in 2014 with the signing of a special licensing agreement that would allow all faculty members and students unlimited access to the statistical discovery software. Although JMP Pro has a wide range of capabilities spanning diverse application areas, its use in teacher training is unique. Since adopting JMP Pro for its classrooms, SIT has organized intensive summer courses for future teachers who will one day teach statistics at junior and senior high schools across Japan. Approximately 80 students and faculty members have participated.
### SIT takes a hands-on approach to teaching statistics
Associate Professor Hideyo Makishita of SIT’s College of Engineering is making the rounds in his classroom. Taking a stack of origami paper from a large bag on his desk, Makishita distributes one sheet to each student. "You must fold the paper only twice to find a spot that divides the side equally by three," he says, referencing Haga’s theorems of geometry. Circulating again to dispense pens, Makishita suggests adding dots to demarcate the paper’s divisions. When isosceles triangles appear in the corners of the paper surrounded by the broken and folded lines of an Egyptian triangle (a right triangle wherein the ratio of the three sides is 3:4:5), the point generated by the two triangles is that which divides the side into three equal parts.
"I always tell my students that it is natural for teachers to be able to solve problems, but we need to nurture teachers who also have the ability to inspire students to pursue statistics and math further," says Makishita. Using origami is just one small example of a hands-on approach to statistics.
Makishita, a longtime junior high and senior high school mathematics teacher at the University of Tsukuba, currently focuses on pedagogical methods. Among his interests are teaching practice and guidance, practical training, education theory, probability theory and statistics. He currently teaches at the SIT College of Systems Engineering and Science and the College of Engineering and Design, where his students pursue teaching certificates.
For Makishita, problem solving is a crucial part of any teacher training course. And what better way to think logically about problem solving than with statistics? In fact, the government of Japan has recently made a move to promote statistics education starting as early as junior high school. A member of the editorial committee for high school mathematics textbooks authorized by Japan’s Ministry of Education, Makishita understands the need to provide the next generation of educators with the most current information.
"Students learn to adopt a statistical way of thinking in the first year of junior high school in a course called Utilization of Materials," he says. "Once they reach high school, students learn the basics of descriptive statistics and data analysis, including correlative coefficients and statistical estimation. Since future mathematics teachers will be teaching statistics content, I want students in teacher training courses to have the opportunity to learn statistics themselves." That’s where JMP Pro comes in.
### Accelerated learning with JMP® Pro
At SIT, JMP Pro has become an invaluable tool for teaching statistics to future teachers. The intuitive, visual nature of JMP Pro makes it not only easy to learn, but easy to master. For this reason, Makishita sought to make JMP Pro the universitywide standard for statistics and mathematics courses at SIT. To demonstrate how this software could be used on a wide scale, Makishita held a special lecture about how the unique capabilities of JMP Pro were especially well-suited for teaching methods classes.
This lecture, which stemmed from a collaboration between Makishita and local JMP engineers, was later offered at the Toyosu and Omiya campuses of SIT as an intensive summer course for students and faculty members alike. At SIT, students have access to JMP Pro throughout their time on campus – so they really get a feel for how this software assists in the learning process.
"It was my intention that this course provide an opportunity for students to learn what JMP Pro can do. And it’s something that they can offer to their students when they themselves become teachers in the future," says Makishita.
With SIT’s JMP Academic Suite, JMP Pro is available for both Windows and Macintosh and can be installed on students’ personal computers simply by accessing the university’s website. This kind of unlimited access enables students and faculty to use JMP Pro outside of the classroom, whether for advanced quantitative research or basic coursework exercises.
In his lectures, Makishita emphasizes the expressive power of the graphs drawn in JMP Pro, a feature that he himself has found useful in his research and scholarship on pedagogical theory and practice. Data visualization through features like the JMP Pro Graph Builder may be designed to augment exploratory data analysis, but Makishita has found that JMP Pro also serves another purpose: getting students excited about the world of statistics.
### Visual capabilities enable hands-on learning
The statistical training of new Gore Japan recruits involves a design optimization exercise. By the end of the program, they’ll be able to meet required specs with minimal variation. With the repeated process of prototype creation, experimentation and analysis, participants follow the same steps as actual business operations. In the process, they learn the importance of DOE, how to use statistics correctly and how to use JMP in operational processes.
Today’s more thorough training has produced a new work cycle. Engineers consult with the statistics team before an experiment, planning and implementing DOE based on the team’s advice. In the future, Gore Japan plans to enhance the business systems database with a similar approach.
And JMP will likely be used here, too, paving the way to still newer uses of data.
I am reminded of the age-old Japanese expression senmitsu. Statistics surround us, though we may not see it. Using JMP Pro, that veneer of imperceptibility melts away and statistics begins to feel more familiar.
##### Hideyo Makishita
Associate Professor | 1,315 | 6,968 | {"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} | 2.71875 | 3 | CC-MAIN-2020-40 | longest | en | 0.942984 |
http://philosophy.enacademic.com/3302/rules_of_inference | 1,506,378,707,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693459.95/warc/CC-MAIN-20170925220350-20170926000350-00278.warc.gz | 270,608,029 | 12,958 | rules of inference
Philosophy dictionary. . 2011.
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• Rule of inference — In logic, a rule of inference (also called a transformation rule) is a function from sets of formulae to formulae. The argument is called the premise set (or simply premises ) and the value the conclusion . They can also be viewed as relations… … Wikipedia
• rule of inference — Lewis Carroll raised the Zeno like problem of how a proof ever gets started. Suppose I have as premises (1) p and (2) p →q . Can I infer q ? Only, it seems, if I am sure of (3) (p & p →q ) →q . Can I then infer q ? Only, it seems, if I am sure… … Philosophy dictionary
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• JBoss Rules — Drools Drools Développeur JBoss Dernière version … Wikipédia en Français
• Ripple down rules — is a knowledge acquisition methodology. Knowledge acquisition is a method to transfer knowledge from human experts to knowledge based systems. Introductory material Ripple Down Rules (RDR) is an incremental knowledge acquisition methodology. RDR… … Wikipedia | 588 | 2,638 | {"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} | 2.5625 | 3 | CC-MAIN-2017-39 | latest | en | 0.874312 |
akozbohatnutnyag.firebaseapp.com | 1,685,736,312,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648850.88/warc/CC-MAIN-20230602172755-20230602202755-00028.warc.gz | 124,406,595 | 7,103 | # E ^ x + y = xy potom dy dx
If x3dy + xy dx = x2 dy + 2y dx; y(2) = e and x > 1, then y(4) is equal to : (1) 3/2 + √e (2) 3/2(√e) (3) 1/2 + √e (4) √e/2. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries.
Find E (X x d dx [y]+y x d d x [ y] + y Reform the equation by setting the left side equal to the right side. eyy' = xy'+ y e y y ′ = x y ′ + y Solve for y' y ′. How to solve: Find dy/dx by implicit differentiation. e^{xy} = x - y By signing up, you'll get thousands of step-by-step solutions to your homework Find dy/dx of e^(xy)-x^2+y^2=1 y = ln( C_0 e^(-e^x)+e^x-1) Making the substitution y = ln u we have the transformed differential equation (u'-e^(2x)+e^x u)/u = 0 or assuming u ne 0 u'+e^x u -e^(2x) = 0 This is a linear non homogeneous differential equation easily soluble giving u = C_0 e^(-e^x)+e^x-1 and finally y = ln( C_0 e^(-e^x)+e^x-1) Jan 03, 2020 · Ex 5.5, 15 Find 𝑑𝑦/𝑑𝑥 of the functions in, 𝑥𝑦= 𝑒^((𝑥 −𝑦)) Given 𝑥𝑦= 𝑒^((𝑥 −𝑦)) Taking log both sides log (𝑥𝑦) = log 𝑒^((𝑥 −𝑦)) log (𝑥𝑦) = (𝑥 −𝑦) log 𝑒 log 𝑥+log𝑦 = (𝑥 −𝑦) (1) log 𝑥+log𝑦 = (𝑥 −𝑦) (As 𝑙𝑜𝑔(𝑎^𝑏 )=𝑏 . 𝑙𝑜𝑔𝑎) ("As " 𝑙𝑜𝑔𝑒 In this tutorial we shall evaluate the simple differential equation of the form $$\frac{{dy}}{{dx}} = {e^{\left( {x - y} \right)}}$$ using the method of separating the variables. Homogeneous Differential Equation (y^2 + yx)dx - x^2dy = 0If you enjoyed this video please consider liking, sharing, and subscribing.You can also help suppor Simple and best practice solution for (x-y)dx+xdy=0 equation.
2(x - y)(1 - dy/dx) = 1 + dy/dx Derivative of y with respect to x simply means the rate of change in y for a very small = 2e2x. 7 cos(x + 7y). -. 1. 7 . 8. Find dy/dx if ln(xy) = ex+y.
## Solved: Find dy/dx by implicit differentiation. e^x/y=x-y - Slader.
Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. Answer to (x + xy^2)dx + e^x^2 ydy = 0 dy / dx = 2 sqrt y = 1 cos x, y (pie) = 0 sqrt ydx + (1 + x) dy = 0, y(0) = 1 1、y'+y=e^-x是常系数线性非齐次方程法一：求出齐次方程y'+y=0的通解为y=Ce^-x 再求y'+y=e^-x的一个特解，设解为y=Cxe^-x代入得C=1 The solution of xdy/dx - y = (x - 1)e^x is . y = e^x + Cx • We have , xdy/dx - y = (x - 1)e^x. Dividing the above equation by x, we get.
### Free implicit derivative calculator - implicit differentiation solver step-by-step
1. 7 . 8. Find dy/dx if ln(xy) = ex+y. Answer: Again I use implicit differentiation:.
Given eqn. xe^(xy) – y = sin^2x, Differentiate y w.r.t. x as: (1)e^(xy) + xe^(xy)*[1.y + xdy/dx] – d Jul 04, 2016 Usually when you see the y' derivative notation in this context it just means the derivate of y wrt x, i.e. \frac{dy}{dx}. So x^2y\prime\prime + xy\prime + (x^2 - v^2)v \equiv x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x… Jul 27, 2018 Sep 06, 2011 It’s a Bernoulli type equation.We assume y(x) = v(x)^(1/4) and obtain the equation : v’(x) + (2/x)v(x) = 4x.The integrating factor is exp(integral(2/x)) = x^2 so Nov 08, 2018 JEE Main 2019: The solution of the differential equation , (dy/dx) = (x-y)2 , when y(1) = 1, is :- (A) loge |(2-y/2-x)| = 2 (y-1) (B) loge |(2-x/2- Mar 06, 2011 A simpler solution would be v=y' and then it becomes v'+v=x^2 which has an integrating factor of e^x which makes it \left (ve^x\right )'=x^2e^x and integrating both sides ve^x=e^x(x^2-2x+2)+C_1 A simpler solution would be v = y ′ and then it becomes v ′ + v = x 2 which has an integrating factor of e x which makes it ( v e x ) ′ = x 2 X(x + 1) Dy/dx + Xy = 1, Y(e) = 1 Y(x) = Give The Largest Interval I Over Which The Solution Is Defined.
𝑙𝑜𝑔𝑎) ("As " 𝑙𝑜𝑔𝑒 Get an answer for 'If x + y = xy, then dy/dx = Please explain step by step.' and find homework help for other Math questions at eNotes Question: Dy/dx-y=e^x Y^2. This problem has been solved! See the answer. dy/dx-y=e^x y^2. Expert Answer 100% (6 ratings) Previous question Next question If x3dy + xy dx = x2 dy + 2y dx; y(2) = e and x > 1, then y(4) is equal to : (1) 3/2 + √e (2) 3/2(√e) (3) 1/2 + √e (4) √e/2. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries.
Separable Equations: A first order differential equation is said to be separable if it is of the form dy dx! g!x"h!y" 2. Method of Separation of Variables: Observe that a separable equation can be written as 1 Using the joint probability density function obtained in problem 6, determine marginal distribution with respect to Y ,f Y (y). f Y (y) = ∫ 0 ∞ f (x , y) dx = ∫ 0 ∞ xy e − x − y dx = y e − y ∫ 0 ∞ x e − x dx As we have shown in previous problems, ∫ 0 ∞ x e − x dx = 1 Consequently, f Y (y) = y e − y 10. Find E (X x d dx [y]+y x d d x [ y] + y Reform the equation by setting the left side equal to the right side. eyy' = xy'+ y e y y ′ = x y ′ + y Solve for y' y ′. How to solve: Find dy/dx by implicit differentiation.
Why the price of food and gas is creeping higher. 6 Dr. Seuss books won't be published for racist images 优质解答 xy=e^(x+y)求dy/dx 这是隐函数求导问题：正统方法是用：隐函数存在定理来做；另一方法是等式两边对x求导,再解出y'来：方法1：f(x,y)=xy-e^(x+y)=0 dy/dx=-f'x/f'y f'x=y-e^(x+y) f'y=x-e^(x+y) dy/dx=-[y-e^(x+y)]/[ Mar 19, 2015 · Free Online Scientific Notation Calculator. Solve advanced problems in Physics, Mathematics and Engineering. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. Free implicit derivative calculator - implicit differentiation solver step-by-step Find dy/dx y=xe^x. Differentiate both sides of the equation. The derivative of with respect to is .
Then the derivative of the function f, in Leibniz's notation for differentiation, can be written as (()).The Leibniz expression, also, at times, written dy/dx, is one of several notations used for derivatives and derived functions.A common alternative is Lagrange's notation Simple and best practice solution for (X^2-y^2)dy=xy*dx equation.
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### Mar 25, 2012
dy %3D y sin x dy 1+x dy 14. e*y dx 2x-7 dy 3x+2y 13. dx x d dx [y]+y x d d x [ y] + y Reform the equation by setting the left side equal to the right side. eyy' = xy'+ y e y y ′ = x y ′ + y Solve for y' y ′. How to solve: Find dy/dx by implicit differentiation. | 2,449 | 6,550 | {"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} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.790524 |
http://percentagecalculator.pro/_6_percent-of_thirteen_ | 1,524,736,108,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00557.warc.gz | 249,788,491 | 13,950 | # How much is 6 percent of thirteen
Using this tool you can find any percentage in three ways. So, we think you reached us looking for answers like: What is 6 (6%) percent (%) of thirteen (13)? Or may be: How much is 6 percent of thirteen. See the solutions to these problems just after the percentage calculator below.
See the solutions to these problems just after the percentage calculator below.
What is % of ?
### Percentage Calculator 2
is what percent of ?
is % of what?
## How to work out percentages - Step by Step
Here are the solutions to the questions stated above:
### 1) What is 6% of 13?
Always use this formula to find a percentage:
% / 100 = Part / Whole replace the given values:
6 / 100 = Part / 13
Cross multiply:
6 x 13 = 100 x Part, or
78 = 100 x Part
Now, divide by 100 and get the answer:
Part = 78 / 100 = 0.78
### 2) 6 is what percent of 13?
Use again the same percentage formula:
% / 100 = Part / Whole replace the given values:
% / 100 = 6 / 13
Cross multiply:
% x 13 = 6 x 100
Divide by 13 and get the percentage:
% = 6 x 100 / 13 = 46.153846153846% | 308 | 1,103 | {"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} | 4.25 | 4 | CC-MAIN-2018-17 | latest | en | 0.88777 |
http://mathhelpforum.com/differential-geometry/168253-supremum.html | 1,529,393,034,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00197.warc.gz | 209,111,678 | 9,921 | 1. ## supremum
let inf S=m and sup S=M.
if T={|x-y| :x,y belong to S},
then prove that sup T= M-m
2. Originally Posted by luckylawrance
let inf S=m and sup S=M.
if T={|x-y| :x,y belong to S},
then prove that sup T= M-m
What ideas do you have?
3. Intuitively the difference between to elements in S is the biggest when one is the biggest element while the other one is the smallest (or vice versa).
4. Respected, the idea i have is that difference between two numbers is greatest when one is largest and other is smallest in S. but i don't know how to write the proof in proper way.
5. Originally Posted by luckylawrance
Respected, the idea i have is that difference between two numbers is greatest when one is largest and other is smallest in S. but i don't know how to write the proof in proper way.
Use the characterization that if $\displaystyle A\subseteq\mathbb{R}$ is bounded then $\displaystyle \sup A=\alpha$ if and only if $\displaystyle \alpha$ is an upper bound of $\displaystyle A$ and for every $\displaystyle \varepsilon>0$ there exists $\displaystyle a\in A$ such that $\displaystyle \alpha-\varepsilon<\alpha$. | 299 | 1,131 | {"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} | 3.125 | 3 | CC-MAIN-2018-26 | latest | en | 0.893702 |
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