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https://www.numbersaplenty.com/1034324497 | 1,721,447,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00173.warc.gz | 784,813,357 | 3,194 | Search a number
1034324497 is a prime number
BaseRepresentation
bin111101101001101…
…000101000010001
32200002021010110201
4331221220220101
54104241340442
6250345055201
734426420213
oct7551505021
92602233421
101034324497
11490938338
1224a487501
1313639763b
149b5245b3
1560c117b7
hex3da68a11
1034324497 has 2 divisors, whose sum is σ = 1034324498. Its totient is φ = 1034324496.
The previous prime is 1034324461. The next prime is 1034324539. The reversal of 1034324497 is 7944234301.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 647651601 + 386672896 = 25449^2 + 19664^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1034324497 is a prime.
It is a super-3 number, since 3×10343244973 (a number of 28 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (1034324407) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 517162248 + 517162249.
It is an arithmetic number, because the mean of its divisors is an integer number (517162249).
Almost surely, 21034324497 is an apocalyptic number.
It is an amenable number.
1034324497 is a deficient number, since it is larger than the sum of its proper divisors (1).
1034324497 is an equidigital number, since it uses as much as digits as its factorization.
1034324497 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 72576, while the sum is 37.
The square root of 1034324497 is about 32160.9156741533. The cubic root of 1034324497 is about 1011.3130316825.
Adding to 1034324497 its reverse (7944234301), we get a palindrome (8978558798).
The spelling of 1034324497 in words is "one billion, thirty-four million, three hundred twenty-four thousand, four hundred ninety-seven". | 577 | 1,894 | {"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-2024-30 | latest | en | 0.847045 |
https://numberworld.info/6601212 | 1,660,512,354,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572077.62/warc/CC-MAIN-20220814204141-20220814234141-00134.warc.gz | 380,134,784 | 4,078 | # Number 6601212
### Properties of number 6601212
Cross Sum:
Factorization:
2 * 2 * 3 * 3 * 29 * 6323
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
64b9fc
Base 32:
69efs
sin(6601212)
-0.12656450882234
cos(6601212)
-0.99195837871685
tan(6601212)
0.12759054365371
ln(6601212)
15.702763826501
lg(6601212)
6.8196236804795
sqrt(6601212)
2569.2823900848
Square(6601212)
### Number Look Up
Look Up
6601212 which is pronounced (six million six hundred one thousand two hundred twelve) is a very unique number. The cross sum of 6601212 is 18. If you factorisate the number 6601212 you will get these result 2 * 2 * 3 * 3 * 29 * 6323. The number 6601212 has 36 divisors ( 1, 2, 3, 4, 6, 9, 12, 18, 29, 36, 58, 87, 116, 174, 261, 348, 522, 1044, 6323, 12646, 18969, 25292, 37938, 56907, 75876, 113814, 183367, 227628, 366734, 550101, 733468, 1100202, 1650303, 2200404, 3300606, 6601212 ) whith a sum of 17264520. The figure 6601212 is not a prime number. The number 6601212 is not a fibonacci number. The number 6601212 is not a Bell Number. 6601212 is not a Catalan Number. The convertion of 6601212 to base 2 (Binary) is 11001001011100111111100. The convertion of 6601212 to base 3 (Ternary) is 110102101011100. The convertion of 6601212 to base 4 (Quaternary) is 121023213330. The convertion of 6601212 to base 5 (Quintal) is 3142214322. The convertion of 6601212 to base 8 (Octal) is 31134774. The convertion of 6601212 to base 16 (Hexadecimal) is 64b9fc. The convertion of 6601212 to base 32 is 69efs. The sine of the number 6601212 is -0.12656450882234. The cosine of the figure 6601212 is -0.99195837871685. The tangent of 6601212 is 0.12759054365371. The root of 6601212 is 2569.2823900848.
If you square 6601212 you will get the following result 43575999868944. The natural logarithm of 6601212 is 15.702763826501 and the decimal logarithm is 6.8196236804795. that 6601212 is unique figure! | 790 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-33 | latest | en | 0.698688 |
https://blog.shirui.me/2020/10/steady-compliance-linear-viscoelasty.html | 1,713,809,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00848.warc.gz | 128,888,295 | 11,783 | Boltzmann superposition: assume that there is no stress when $t\le 0$, so the integral starts from $0$ and $\sigma(0^{-})=0$:
$$$$\gamma(t)=\int_0^t J(t-t^\prime)\dot{\sigma}(t^\prime)\mathrm{d}t^\prime$$$$
Performing Laplace Transform yields:
\begin{align} \hat{\gamma}(s)&=\hat{J}(s)\hat{\dot{\sigma}}(s)\\ &=\hat{J}(s)\left(s\hat{\sigma}(s)-\sigma(0^{-})\right) \\ &=\hat{J}(s)\left(s\hat{G}(s)\hat{\dot{\gamma}}(s)-\sigma(0^{-})\right)\\ &=\hat{J}(s)\left(s\hat{G}(s)(s\hat{\gamma}(s)-\gamma(0^-))-\sigma(0^{-})\right)\\ &=s^2\hat{J}(s)\hat{\gamma}(s)\hat{G}(s) \end{align}
Here we let $\hat{f}(s):=\mathcal{L}\lbrace f(t)\rbrace(s)$ be the Laplace transformation and since the stress starts at time $0$, $\gamma(0^-)$ and $\sigma(0^-)$ are simply $0$. The 2nd to 3rd step is derived from the convolution relation $\sigma(t)=\int_0^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$. Cancelling out the $\hat{\gamma}(s)$ and rearranging the last equation give
$$$$\frac{1}{s^2}=\hat{J}(s)\hat{G}(s)$$$$
The inverse Laplace of above equation gives the convolution relation
$$$$t=\int_0^{t}J(t-t^\prime)G(t^\prime)\mathrm{d}t^\prime$$$$
The last step is substituting $J(t)=J_e + t/\eta$ into above convolution relation and making $t\to\infty$:
\begin{aligned} \color{blue}{t} &=\int_0^{t}\left(J_e+\frac{t-t^\prime}{\eta}\right)G(t^\prime)\mathrm{d}t^\prime\\ &= J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime + {\color{blue}{\frac{t}{\eta}\int_0^\infty G(t^\prime)\mathrm{d}t^\prime}} - \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime \end{aligned}
Since $\eta:=\int_0^\infty G(t^\prime)\mathrm{d}t^\prime$, the blue terms cancelled out, which gives $$J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime = J_e\eta= \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime = \frac{1}{\eta} \int_0^\infty G(t^\prime)t^\prime\mathrm{d}t^\prime$$ Therefore we have
$$J_e = \frac{1}{\eta^2}\int_0^\infty tG(t)\mathrm{d}t$$
$Q.E.D.$ | 761 | 1,967 | {"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": 1, "equation": 5, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-18 | latest | en | 0.511691 |
https://www.physicsforums.com/threads/trig-problem-cos-arctan-5-12.120571/ | 1,553,486,668,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203548.81/warc/CC-MAIN-20190325031213-20190325053213-00163.warc.gz | 867,049,982 | 15,095 | # Trig problem cos (arctan 5/12)
#### jacy
I have to calculate this without using the calculator.
cos (arctan 5/12)
So far i draw a triangle and i have the opposite side to be 5, adjacent to be 12, and hypotenuse to be 13. Please suggest me some hint, thanks.
Last edited:
#### Integral
Staff Emeritus
Science Advisor
Gold Member
Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan $\frac 5 {12}$
then use the definition of the cos of that angle to get your answer.
#### jacy
Integral said:
Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan $\frac 5 {12}$
then use the definition of the cos of that angle to get your answer.
am getting 12/13, am i correct
#### Curious3141
Homework Helper
jacy said:
am getting 12/13, am i correct
Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).
#### jacy
Curious3141 said:
Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).
Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
#### Curious3141
Homework Helper
jacy said:
Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
No, trig ratios have no unit. You're dividing a length by a length, so they're dimensionless.
#### arunbg
Just for another take on this problem (I find the construction of a rt triangle a bit cumbersome) we could use basic trig. results.
For eg, if you have to do something like cos(arctan(x))
we can proceed by taking arctan(x)=y
so x=tan(y)
$$\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+tan^2y}} = cos(y) or y=arccos(\frac{1}{\sqrt{1+x^2}})$$
So cos(arctan(x)) = cos(y) = $\frac{1}{\sqrt{1+x^2}})$
which gives the answer.This approach works for all such problems.
No messy triangles.
Arun
#### jacy
Thanks everyone for helping me out.
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• Solo and co-op problem solving | 633 | 2,332 | {"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} | 4.25 | 4 | CC-MAIN-2019-13 | latest | en | 0.875531 |
https://www.atilim.edu.tr/en/ects/site-courses/238/18453/detail | 1,695,335,667,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00086.warc.gz | 740,386,971 | 20,830 | # Introduction to Electrical Engineering (EE234) Course Detail
Course Name Course Code Season Lecture Hours Application Hours Lab Hours Credit ECTS
Introduction to Electrical Engineering EE234 3 1 0 3 5
Pre-requisite Course(s)
PHYS 102 or equivalent
Course Language English N/A Bachelor’s Degree (First Cycle) . Assoc. Prof. Dr. Reşat Özgür DORUK Instructor Dr. Mehmet BULUT The students who succeeded in this course; Will be able to write the current-voltage relationship for basic circuit elements. Will apply circuit analysis techniques. Will apply Thevenin and Norton theorems. Will define time constants for RL and RC circuits. Calculate the individual and total impedance of a circuit against sinusoidal voltages or currents Apply node, mesh, Thevenin and Norton methods to the analysis of alternating current circuits with sinusoidal inputs. Evaluate the active and reactive powers of an alternating current circuit. Can evaluate and correct the power factor. Definition of current, voltage, resistance, power, Kirchoff laws and resistive DC circuits, Thevenin and Norton equivalents, AC circuits, phasors, filters, reactive power, three-phase circuits and power, overview of combinational and sequential digital circuits and examples, diodes and transistors.
### Weekly Subjects and Releated Preparation Studies
Week Subjects Preparation
1 Basic concepts in electrical engineering
2 Voltage-Current Relationships, Ohm’s Law, Power Review last weeks topics
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4 Parallel and Series Circuits and their resistive versions Review last weeks topics
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### Sources
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Percentage of Semester Work 100 100
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# Program Qualifications / Competencies Level of Contribution
1 2 3 4 5
1 Adequate knowledge of mathematics, physical sciences and the subjects specific to engineering disciplines; the ability to apply theoretical and practical knowledge of these areas in the solution of complex engineering problems. X
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6 The ability to work efficiently in inter-, intra-, and multi-disciplinary teams; the ability to work individually.
7 (a) Sözlü ve yazılı etkin iletişim kurma becerisi; etkin rapor yazma ve yazılı raporları anlama, tasarım ve üretim raporları hazırlayabilme, etkin sunum yapabilme, açık ve anlaşılır talimat verme ve alma becerisi. (b) En az bir yabancı dil bilgisi; bu yabancı dilde etkin rapor yazma ve yazılı raporları anlama, tasarım ve üretim raporları hazırlayabilme, etkin sunum yapabilme, açık ve anlaşılır talimat verme ve alma becerisi.
8 Recognition of the need for lifelong learning; the ability to access information, follow developments in science and technology, and adapt and excel oneself continuously.
9 Acting in conformity with the ethical principles; professional and ethical responsibility and knowledge of the standards employed in engineering applications.
10 Knowledge of business practices such as project management, risk management, and change management; awareness of entrepreneurship and innovation; knowledge of sustainable development.
11 Knowledge of the global and social effects of engineering practices on health, environment, and safety issues, and knowledge of the contemporary issues in engineering areas; awareness of the possible legal consequences of engineering practices.
12 (a) Knowledge of (i) fluid mechanics, (ii) heat transfer, (iii) manufacturing process, (iv) electronics and control, (v) vehicle components design, (vi) vehicle dynamics, (vii) vehicle propulsion/drive and power systems, (viii) technical laws and regulations in automotive engineering field, and (ix) vehicle verification tests. (b) The ability to merge and apply these knowledge in solving multi-disciplinary automotive problems.
13 The ability to make use of theoretical, experimental, and simulation methods, and computer aided design techniques in automotive engineering field.
14 The ability to work in the field of vehicle design and manufacturing.
Activities Number Duration (Hours) Total Workload
Course Hours (Including Exam Week: 16 x Total Hours) 16 3 48
Laboratory 5 2 10
Application
Special Course Internship
Field Work
Study Hours Out of Class 14 2 28
Presentation/Seminar Prepration
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Prepration of Midterm Exams/Midterm Jury 2 10 20
Prepration of Final Exams/Final Jury 1 20 20 | 1,341 | 6,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.776949 |
https://justaaa.com/physics/432618-when-the-air-temperature-is-400c-a-hammer-strikes | 1,720,810,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00156.warc.gz | 266,611,407 | 9,412 | Question
# . When the air temperature is −40.0°C, a hammer strikes one end of an aluminum rail....
. When the air temperature is −40.0°C, a hammer strikes one end of an aluminum rail. A microphone located at the opposite end of the rail detects two pulses of sound, one that travels through the rail and the other travels through the air. The two pulses are detected at 20.0 ms apart. The speed of sound in aluminum is 6.42×103 m/s.
(a) Determine the length of the rail.
(b) If the air temperature increases to +45.0°C, determine the separation in time between the arrivals of the two pulses.
Air temperature is -40 0 C .
To calculate the sound in air at any given temperature,
Cair = 33.5 + (0.6 * Tc) m/s , where Tc = specific temperature
Therefore, speed of air at -400C = 331.5 + ( 0.6* (-40 ) m/s
= 307.5 m/s
Speed of aluminium = 6420 m/s (i'm assuming it to be this you have given as the actual speed of aluminium is 6320 m/s)
pulses detection time difference = 20 ms = .02s= T
therefore, x/S1 -x/S2 = T ( S1 = speed of sound, S2= speed of aluminium )
=> x/307.5 - x / 6420 = .02
x = 6.459 m
(b) speed of sound at air temperature +45 0C :- 331.5 + (06 * 45)
= 601.5 m/s
T1 = x/v1 = 6.45 / 601.5 = 0.0107 s
T2 = x/v2 = 6.45/ 6420 = 0.001
=> 0.0170-0.001 = 0.0097 s = 9.7 * 10 -3 s
(Difference between the two is the separation between the arriva; of the two pulses.)
#### Earn Coins
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https://www.codesansar.com/python-programming/augmented-assignment-operators.htm | 1,701,161,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00646.warc.gz | 769,506,604 | 8,397 | # Augmented Assignment Operators in Python with Examples
Unlike normal assignment operator, Augmented Assignment Operators are used to replace those statements where binary operator takes two operands says `var1` and `var2` and then assigns a final result back to one of operands i.e. `var1` or `var2`.
For example: statement `var1 = var1 + 5` is same as writing `var1 += 5` in python and this is known as augmented assignment operator. Such type of operators are known as augmented because their functionality is extended or augmented to two operations at the same time i.e. we're adding as well as assigning.
## List Of Augmented Assignment Operators in Python
Python has following list of augmented assignment operators:
1. Addition & Assignment (`+=`): `x+=y` is equivalent to `x=x+y`
PROGRAM
``````
a = 23
b = 3
a += b
``````
OUTPUT
```Addition = 26
```
2. Subtraction & Assignment (`-=`): `x-=y` is equivalent to `x=x-y`
PROGRAM
``````
# Subtraction
a = 23
b = 3
a -= b
print('Subtraction = %d' %(a))
``````
OUTPUT
```Subtraction = 20
```
3. Multiplication & Assignment (`*=`): `x*=y` is equivalent to `x=x*y`
PROGRAM
``````
# Multiplication
a = 23
b = 3
a *= b
print('Multiplication = %d' %(a))
``````
OUTPUT
```Multiplication = 69
```
4. Division & Assignment (`/=`): `x/=y` is equivalent to `x=x/y`
PROGRAM
``````
# Division
a = 23
b = 3
a /= b
print('Division = %f' %(a))
``````
OUTPUT
```Division = 7.666667
```
5. Remainder (or Modulo) & Assignment (`%=`): `x%=y` is equivalent to `x=x%y`
PROGRAM
``````
# Remainder or Modulo
a = 23
b = 3
a %= b
print('Remainder or Modulo = %d' %(a))
``````
OUTPUT
```Remainder or Modulo = 2
```
6. Power & Assignment (`**=`): `x**=y` is equivalent to `x=x**y`
PROGRAM
``````
# Power
a = 23
b = 3
a **= b
print('Power = %d' %(a))
``````
OUTPUT
```Power = 12167
```
7. Integer Division & Assignment (`//=`): `x//=y` is equivalent to `x=x//y`
PROGRAM
``````
# Integer Division
a = 23
b = 3
a //= b
print('Integer Division = %d' %(a))
``````
OUTPUT
```Integer Division = 7
```
8. Bitwise Shift Right & Assignment (`>>=`): `x>>=y` is equivalent to `x=x>>y`
PROGRAM
``````
# Bitwise Shift Right
a = 23
b = 3
a >>= b
print('Bitwise Shift Right = %d' %(a))
``````
OUTPUT
```Bitwise Shift Right = 2
```
9. Bitwise Shift Left & Assignment (`>>=`): `x<<=y` is equivalent to `x=x<<y`
PROGRAM
``````
# Bitwise Shift Left
a = 23
b = 3
a <<= b
print('Bitwise Shift Left = %d' %(a))
``````
OUTPUT
```Bitwise Shift Left = 184
```
10. Bitwise AND & Assignment (`&=`): `x&=y` is equivalent to `x=x&y`
PROGRAM
``````
# Bitwise AND
a = 23
b = 3
a &= b
print('Bitwise AND = %d' %(a))
``````
OUTPUT
```Bitwise AND = 3
```
11. Bitwise OR & Assignment (`|=`): `x|=y` is equivalent to `x=x|y`
PROGRAM
``````
# Bitwise OR
a = 23
b = 3
a |= b
print('Bitwise OR = %d' %(a))
``````
OUTPUT
```Bitwise OR = 23
```
12. Bitwise Exclusive OR & Assignment (`^=`): `x^=y` is equivalent to `x=x^y`
PROGRAM
``````
# Bitwise EXCLUSIVE OR
a = 23
b = 3
a ^= b
print('Bitwise EXCLUSIVE OR = %d' %(a))
``````
OUTPUT
```Bitwise EXCLUSIVE OR = 20
``` | 1,061 | 3,104 | {"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-2023-50 | latest | en | 0.844627 |
http://physics.stackexchange.com/questions/65724/difference-between-delta-d-and-delta | 1,466,948,502,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395346.6/warc/CC-MAIN-20160624154955-00035-ip-10-164-35-72.ec2.internal.warc.gz | 237,669,725 | 20,002 | # Difference between $\Delta$, $d$ and $\delta$
I have read the thread regarding 'the difference between the operators between $\delta$ and $d$', but it does not answer my question.
I am confused about the notation for change in Physics. In Mathematics, $\delta$ and $\Delta$ essentially refer to the same thing, i.e., change. This means that $\Delta x = x_1 - x_2 = \delta x$. The difference between $\delta$ and $d$ is also clear and distinct in differential calculus. We know that $\frac{dy}{dx}$ is always an operator and not a fraction, whereas $\frac{\delta y}{\delta x}$ is an infinitesimal change.
In Physics, however, the distinction is not as clear. Can anyone offer a clearer picture?
-
Don't forget $\partial$ either! :-) – b_jonas May 24 '13 at 12:35
"We know that $\frac{dy}{dx}$ is always an operator and not a fraction, whereas $\frac{\delta y}{\delta x}$ is an infinitesimal change." The operator would be $d/dx$, not $dy/dx$. Also, it is actually valid to consider $dy/dx$ as the quotient of two infinitesimal numbers. That's how physicists, mathematicians, and engineers throught about it for hundreds of years after the invention of calculus, and Abraham Robinson proved ca. 1960 that it didn't lead to logical inconsistency. There is even a freshman calc book using this approach: math.wisc.edu/~keisler/calc.html – Ben Crowell May 24 '13 at 12:52
You should have mentioned your related question on math.SE. – Zev Chonoles May 24 '13 at 13:06
@b_jonas Don't forget $\eth$ either! :) – Mike May 24 '13 at 13:43
– Qmechanic May 24 '13 at 13:43
The symbol $\Delta$ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small.
The symbols $d,\delta$ refer to infinitesimal variations or numerators and denominators of derivatives.
The difference between $d$ and $\delta$ is that $dX$ is only used if $X$ without the $d$ is an actual quantity that may be measured (i.e. as a function of time) without any ambiguity about the "additive shift" (i.e. about the question which level is declared to be $X=0$). On the other hand, we sometimes talk about small contributions to laws that can't be extracted from a well-defined quantity that depends on time.
An example, the first law of thermodynamics. $$dU = \delta Q - \delta W$$ The left hand side has $dU$, the change of the total energy $U$ of the system that is actually a well-defined function of time. The law says that it is equal to the infinitesimal heat $\delta Q$ supplied to the system during the change minus the infinitesimal work $\delta W$ done by the system. All three terms are equally infinitesimal but there is nothing such as "overall heat" $Q$ or "overall work" $W$ that could be traced – we only determine the changes (flows, doing work) of these things.
Also, one must understand the symbol $\partial$ for partial derivatives – derivatives of functions of many variables for which the remaining variables are kept fixed, e.g. $\partial f(x,y)/\partial x$ and similarly $y$ in the denominator.
Independently of that, $\delta$ is sometimes used in the functional calculus for functionals – functions that depend on whole functions (i.e. infinitely many variables). In this context, $\delta$ generalizes $d$ and has a different meaning, closer to $d$, than $\delta$ in the example of $\delta W$ and $\delta Q$ above. Just like we have $dy=f'(x)dx$ for ordinary derivatives in the case of one variable, we may have $\delta S = \int_a^b dt\,C(t)\delta x(t)$ where the integral is there because $S$ depends on uncountably many variables $x(t)$, one variable for each value of $t$.
In physics, one must be ready that $d,\delta,\Delta$ may be used for many other things. For example, there is a $\delta$-function (a distribution that is only non-vanishing for $x=0$) and its infinite-dimensional, functional generalization is called $\Delta[f(x)]$. That's a functional that is only nonzero for $f(x)=0$ for every $x$ and the integral $\int {\mathcal D}f(x) \,\Delta[f(x)]=1$. Note that for functional integrals (over the infinite-dimensional spaces of functions), the integration measure is denoted ${\mathcal D}$ and not $d$.
-
I think that we could in principle consider the total heat supplied to/released from the system, as well as the overall work done, taken since some moment. The internal energy in this way is not much better, because we usually do forget about some parts of it (oscillational modes, rest mass, binding energies, etc). I am used to think that we dont write $dQ$ or $dW$ because this would imply that we deal with total differentials, that is, $Q$ and $W$ are 'functions of state'. They are not, however, and we write $\delta Q$ and $\delta W$ in order to stress this fact. – Peter Kravchuk May 24 '13 at 14:57
Right, $Q$ and $W$ are not functions of state. That's equivalent to saying that the "total heat supplied to/released from the system" and "overall work done" since some moment depend on which moment we choose. – Luboš Motl May 25 '13 at 11:01
Well, their sum also depends on this moment, but it nevertheless is a function of state -- it is $U$ plus some constant, which depends on the moment. – Peter Kravchuk May 25 '13 at 11:08
Or difference, depending on your definitions. – Peter Kravchuk May 25 '13 at 11:14
In many books, the difference between $d$ and $\delta$ is that, in the first case, we have the differential of a function and, in the second case, we have the variation of a functional.
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https://chem.libretexts.org/Courses/Lumen_Learning/Book%3A_Statistics_for_the_Social_Sciences_(Lumen)/10%3A_9-_Inference_for_Two_Proportions/10.19%3A_Introduction-_Estimate_the_Difference_Between_Population_Proportions | 1,696,161,117,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00051.warc.gz | 183,974,498 | 28,325 | # 10.19: Introduction- Estimate the Difference Between Population Proportions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
## What you’ll learn to do: Construct and interpret confidence intervals to compare two population/treatment group proportions.
### LEARNING OBJECTIVES
• Construct a confidence interval to estimate the difference between two population proportions (or the size of a treatment effect) when conditions are met. Interpret the confidence interval in context.
• Interpret the meaning of a confidence level associated with a confidence interval and describe how the confidence level affects the margin of error.
• Given the description of a statistical study, evaluate whether conclusions are reasonable.
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November 1, 2010, 11:44
Pressure not symmertic in domain
#1
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Martin H.
Join Date: May 2009
Posts: 31
Rep Power: 9
Hi everyone,
I model a surface deposition model in 2D. Because I have a moving box (see picture) in my domain I don't want to use a symmetry axis, even though the model is y-axis symmetric.
While my model converges nicely when I apply y-symmetry, it won't converge when I use the full model. I think that is due to reversed flow in the pressure in and outlet as it often occurs in the first iterations which in turn causes a wrong split of the inlet flow.
Any thoughts on how I could get equal pressures on both sides of the model? I need an equal flow split of the inlet but often the flow fully goes in one direction. (see picture 2 for illustration of my problem)
Thank you very much
Attached Images
bitmap.png (50.9 KB, 33 views) bitmap1.jpg (40.7 KB, 32 views)
Last edited by bobmalaria; November 1, 2010 at 12:16.
November 2, 2010, 03:36 #2 Super Moderator Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 3,149 Rep Power: 32 are you sure your model is symetric if your box is moving? Despite reversed flow warning, let iterate __________________ In memory of my friend Hervé: CFD engineer & freerider
November 2, 2010, 05:20 #3 Member Martin H. Join Date: May 2009 Posts: 31 Rep Power: 9 Hi max, in an initial state I left the box stationary, basically to check my model. I expected to see a 50/50 split of the flow at least in the stationary case. When the box starts moving of course I expect to see a change of the velaocity vectors. But with the initial soution beeing not correct I saw no point in proceeding any further in the moment (I should have said this in the first place, sorry). I checked the solution several times and after initialization it seems to be better, but with ongoing iteration I will see this tilt towards one direction. If I change the pressure inlets to wall, which turns the domain into a -20Pa low pressure region, I get the expected flow split. But I need this opening to atmosphere since this is how the real process works... This finding and the fact that it works when I use a y-symmetry are what made me to assume a problem with the pressure. Thanky for helping
November 2, 2010, 05:37
#4
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Maxime Perelli
Join Date: Mar 2009
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Quote:
Originally Posted by bobmalaria in an initial state I left the box stationary, basically to check my model. I expected to see a 50/50 split of the flow at least in the stationary case.
you mean you compute at first your model without Moving Mesh, in steady state?
And you don't get a converged symmetrical result?
__________________
In memory of my friend Hervé: CFD engineer & freerider
November 2, 2010, 07:47 #5 Member Martin H. Join Date: May 2009 Posts: 31 Rep Power: 9 hi, exactly. I thought at first I do a steady state simulation and later I initialize from this solution and change to the transient solver method. The model converges, but looking at the velocity vectors, I can still see the tilt towards one direction. Also the reversed flow never disappears and occurs always at somewhat between 5 and 21 faces on the pressure-outlets. The geometry was created and meshed in Gambit and during design I used the y-symmetry and one side of the model is just a copy of the other so dimensions must be exactly the same. I use 12 cells to cover the 3mm gap between box and the above coating head, do you think this could be a problem?
November 2, 2010, 08:46 #6 Super Moderator Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 3,149 Rep Power: 32 In Gambit as you copied the flipped domain, did you check that the 2 domains are connected (along the y-axis)? in gambit you can check it by trying to move one domain. If you can , then the 2 domains aren't connected (else it returns you you cannot since another entity is connected) __________________ In memory of my friend Hervé: CFD engineer & freerider
November 2, 2010, 11:38 #7 Member Martin H. Join Date: May 2009 Posts: 31 Rep Power: 9 hi, when i created the flipped bit in gambit i refelected just the vertexes and connected them with edges and made faces from thos edges. So those are conneted. However, because this box is supposed to move at some point I have an interface in the model. But I defined the grid interfaces and checked the mesh in fluent. The flow across the interface also seems not to be the problem. I am really puzzled why in a fully symmetric geometry the solution is not symmetric. Thanks again for helping, I appreciate any input
November 3, 2010, 01:59 #8 Super Moderator Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 3,149 Rep Power: 32 at the other hand, a flow in a symmetrical geometry isn't necessarly symmetric... (cf von Karmann at specific Re) Is your steady state solution fully converged? Try to monitor the massflow at your inlet, and check if you get a converged solution __________________ In memory of my friend Hervé: CFD engineer & freerider
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``` Tony and Sunil are participating in a jogathon
to raise money for charity. Tony will raise Rs. 20, plus Rs. 2 for
each lap he jogs. Sunil will raise Rs. 30, plus Rs.1.50 for each lap he jogs. The total amount of money each
will raise can be calculated using the follwing expressions where n represents the number of laps run :
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``` Let laps be n
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n = 20 laps
after 20 laps,they raise same money of rs. 60 each
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``` after 20 laps they will raise the same amount
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### Grade 4 - Mathematics5.10 Division - One Digit Divisors without Remainder (WIZ Math)
Method: Look at the first number of the dividend. If this number is lesser than divisor, then include the second digit to divide. If the number is greater or equal to the divisor, then each number is divided one by one. Example: 126/7 1 is lesser than 7, so 12 is divided by 7. This leaves a remainder 5 and 6 is brought down. 7 completely divides 56 into 8. Answer: 56R0 Directions: Divide the following. Also write at least 10 examples of your own.
Question 1: 92 / 2
Question 2: 52 / 1
Question 3: 186 / 2
Question 4: 84 / 2
Question 5: 432 / 9
Question 6: This question is available to subscribers only!
Question 7: This question is available to subscribers only!
Question 8: This question is available to subscribers only!
Question 9: This question is available to subscribers only!
Question 10: This question is available to subscribers only! | 297 | 1,119 | {"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} | 3.5 | 4 | CC-MAIN-2022-33 | latest | en | 0.900784 |
http://mathforum.org/mathimages/index.php?title=Perko_pair_knots&diff=26010&oldid=26009 | 1,508,772,590,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00763.warc.gz | 230,226,780 | 13,307 | # Perko pair knots
(Difference between revisions)
Revision as of 11:17, 19 July 2011 (edit)← Previous diff Revision as of 11:23, 19 July 2011 (edit) (undo)m Next diff → Line 356: Line 356: {{Hide|1= {{Hide|1= -
[[Image:Perko_Dowker1.gif|right|thumb|250px|[[Perko pair knots#direction|Image 1]]. The first Perko knot with direction assigned.]]
+
[[Image:Perko_Dowker1.gif|left|thumb|150px|[[Perko pair knots#direction|Image 1]]. The first Perko knot with direction assigned.]]
-
[[Image:Perko_Dowker.gif|right|thumb|250px|[[#numbers|Image 2]]. The knot with numbers assigned to the crossings. Orange numbers are assigned when going over a crossing, pink numbers are assigned when going under.]]
+
[[Image:Perko_Dowker.gif|right|thumb|300px|[[#numbers|Image 2]]. The knot with numbers assigned to the crossings. Orange numbers are assigned when going over a crossing, pink numbers are assigned when going under.]]
To determine a knot's Dowker representation, first we need to assign direction to the knot. This is shown in [[#direction|Image 1]]. To determine a knot's Dowker representation, first we need to assign direction to the knot. This is shown in [[#direction|Image 1]]. - Next, we pick any crossing, and assign the number 1 to it. We follow the understrand out of this crossing, and travel around the knot in the direction specified by our arrows. Every time we encounter a crossing, we assign the next number to it. If we're on an even number, and we're going ''under'' a crossing, we assign a negative number instead. In [[#numbers|Image 2]], we can see that the second number is -2 instead of 2, because it is assigned while going under. + + Next, we pick any crossing, and assign the number 1 to it. We follow the understrand out of this crossing, and travel around the knot in the direction specified by our arrows. Every time we encounter a crossing, we assign the next number to it. If we're on an even number, and we're going ''under'' a crossing, we assign a negative number instead. In [[#numbers|Image 2]], we can see that the second number is -2 instead of 2, because it was assigned while going under. + We continue all the way around the knot until every crossing has two numbers, as shown in [[#numbers|Image 2]]. We continue all the way around the knot until every crossing has two numbers, as shown in [[#numbers|Image 2]]. + +
}} }}
## Revision as of 11:23, 19 July 2011
Perko pair knots
This is a picture of the Perko pair knots. They were first thought to be separate knots, but in 1974 it was proved that they were actually the same knot.
# Basic Description
In 1899, C. N. Little published a table of 43 nonalternating knots of 10 crossings that listed the two knots shown above as being distinct. Seventy-five years later, Kenneth Perko, a lawyer and part-time mathematician, discovered that these were actually the same knot[1].
To say that two knots are the same is to say that one can be deformed into the other without breaking the knot or passing it through itself. To prove that two knots are the same, we can create one of them out of actual rope, and tug at it and move it around until it looks like the other. As the story goes, that's how Perko figured out that these knots are the same - by working with rope on his floor.
We can also prove that two knots are the same by working with their projections. A projection of a knot is a flat representation of it, essentially a 2D drawing of the knot. There are many ways to use projections to show that certain knots are distinct from each other, but the main way of using projections to demonstrate that two knots are the same is to use the Reidemeister moves, which are described below.
# A More Mathematical Explanation
## Reidemeister moves
As was stated above, knots are considered to be the same if one can be rearra [...]
## Reidemeister moves
As was stated above, knots are considered to be the same if one can be rearranged into the other without breaking the string or passing it through itself. This kind of transformation is called an ambient isotopy. But when we're writing a written proof, we have to work with the knots projection, instead of the knot itself. What manipulations can we make on a knot’s projection that correspond to ambient isotopies in three dimensions?
The first answer is a planar isotopy. A planar isotopy is the sort of transformation you could make if the projection of a knot was printed on very stretchy rubber. The image can be stretched in all directions, but none of the crossings are affected:
The original image.
These two images are planar isotopies of the original image.
This is not a planar isotopy of the original image.
The second answer is the Reidemeister moves, a set of three changes we can make to a knot’s projection that do affect the knot’s crossings but are still ambient isotopies. Every change to a knot's projection that corresponds to an ambient isotopy can be described as some combination of these three moves. In the images below, we imagine that the line segments continue and connect in some sort of unspecified knot, and only the section of the knot we're looking at changes:
Type I Reidemeister Move: The first Reidemeister move allows you to create a twist in a strand that goes in either direction. Type II Reidemeister move: The second Reidemeister move allows you to slide one strand on top of or behind another. Type III Reidemeister move: The third Reidemeister move allows you to slide a strand to the other side of a crossing.
## Proving that the Perko knots are equivalent
In his paper "On the Classification of Knots", Kenneth Perko provided an abridged proof that the knots now known as the Perko pair are the same[2]. This proof is shown below:
Perko's proof relies on the ability of the reader to manipulate the knots in their head and verify that each projection can be manipulated to look like the next. To create a full, rigorous proof, we need to use planar isotopies and the Reidmeister moves, as described above.
Below is a step-by-step Reidemeister moves proof that follows the outline of Perko's shorter proof. The arrows between each step are labeled to show how we get from one image to the other: p.i. means we use a planar isotopy, I means we use the first Reidemeister move, II means we use the second move, and III means we use the third move. Mousing over a step will highlight the part of the knot that's about to move in pink, and display a dotted green line showing where it will move to.
## Dowker notation
Dowker notation is a way of describing knots with numbers so that anyone else who knows the system can reconstruct the knot. The Dowker notation for the Perko knots will be determined below.
Image 1. The first Perko knot with direction assigned.
Image 2. The knot with numbers assigned to the crossings. Orange numbers are assigned when going over a crossing, pink numbers are assigned when going under.
To determine a knot's Dowker representation, first we need to assign direction to the knot. This is shown in Image 1.
Next, we pick any crossing, and assign the number 1 to it. We follow the understrand out of this crossing, and travel around the knot in the direction specified by our arrows. Every time we encounter a crossing, we assign the next number to it. If we're on an even number, and we're going under a crossing, we assign a negative number instead. In Image 2, we can see that the second number is -2 instead of 2, because it was assigned while going under.
We continue all the way around the knot until every crossing has two numbers, as shown in Image 2.
# Why It's Interesting
The story of the Perko knots really helps to show that mathematics, and knot theory in particular, is not limited to academics and professional mathematicians. Anyone who is interested in math and spends time studying it has the potential to discover something new and interesting, no matter what their day job is.
Knot theory itself is one of the most accessible areas of math. What could be more familiar than knots? We use them every time we put on a pair of sneakers or a tie, and they're an important part of many crafts, from knitting to embroidery. Knots are also a common decorative element in artwork from many cultures. It's interesting that something so commonplace can become the object of mathematical study, and that even though the questions that mathematicians ask about knots are often simple, such as "How can we tell if two knots are the same?", the answers can turn out to be surprisingly complex.
### Connections to chemistry
Interest in comparing and tabulating knots actually grew out of a mistaken theory in chemistry. In the 1880s, chemists believed that space was filled with something called ether, and Lord Kelvin proposed that different atoms were different types of knots in the ether[1].
Belief in ether soon faded, but interest in identifying and cataloging different types of knots stayed. Knot theory soon grew into its own mathematical discipline, which for a while seemed to have nothing to do with chemistry. Recently, however, scientists have begun to see that large, complicated molecules are often knotted, and it looks as thought knot theory might once again be important to chemistry.
The Perko knots are sometimes used to illustrate the point that knot theory is still a relatively young mathematical field, and that it is accessible to non-mathematicians.
# References
1. 1.0 1.1 Adams, C. (2004). The knot book: An elementary introduction to the mathematical theory of knots. Providence, RI: American Mathematical Society.
2. Perko, K. (1974). On the classification of knots. Proceedings of the American Mathematical Society, 45, 262-266. | 2,193 | 9,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-43 | longest | en | 0.884936 |
https://www.daml.org/listarchive/joint-committee/1316.html | 1,725,894,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00176.warc.gz | 699,340,410 | 4,130 | # Re: Logic Layering Issues for DAML Rules
From: Peter F. Patel-Schneider ([email protected])
Date: 03/16/03
• Next message: Mike Dean: "Joint Committee telecon today 18 March"
```From: Sandro Hawke <[email protected]>
Subject: Logic Layering Issues for DAML Rules
Date: Sat, 08 Mar 2003 07:51:16 -0500
>
> We've talked about the DAML Rules language, like RuleML, having at
> least one of its concrete syntaxes be RDF-Graph-based like OWL. This
> raises concerns over logic layering; we need to handle universally
> quantified variables and some forms of negation. (An aside: when
> people say "Horn rules" are they thinking only definite clauses, or
> full Horn clauses, with headless rules, giving us classical negation?
> My understanding is that's not as widely implemented, but since I'm
> tempermentally inclined towards a full FOL syntax, I view it as a step
> in the right direction.)
>
> think I heard Pat and Ian say they knew how to do it, and maybe even
> saw Peter nod, but I didn't hear how.
Maybe you saw my head jerk from astonishment, but certainly not a semantic
nod. :-)
> When I've suggested doing it by
> encoding the syntactic structures of the rules into RDF and then using
> a limitted truth predicate (as in KIF [2]) to indicate which such
> structures are intended to be asserted as rules, people start to look
> very concerned.
Truth predicates are an extremely powerful tool, but one that requires
great care or else serious problems will result.
The problems are not limited to paradoxes. Suppose you (by accident)
closed off the universe of non-empty predicates, say by
sh:Rule rdf:subClassof _:x .
_:x owl:allValuesFrom _:y .
_:y owl:onProperty sh:predicate .
_:y owl:allValuesFrom [ list of predicates ] .
What is then the meaning of a rule that is missing a predicate?
> In any case, do we have a solution at hand, or is this going to be a
> major obstacle?
Major obstacle, provided that you want semantic layering to work.
If you
1/ want to only use RDF syntax, and
2/ want the syntax to have its RDF meaning,
you are going to have problems. The problems may be surmountable, with
great effort, but do you want to expend this effort, particularly when
relaxing either constraint above makes the problems largely go away?
> -- sandro
Peter F. Patel-Schneider
Bell Labs Research
Lucent Technologies
```
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https://metanumbers.com/1505720 | 1,643,182,342,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00133.warc.gz | 434,061,951 | 7,427 | 1505720 (number)
1,505,720 (one million five hundred five thousand seven hundred twenty) is an even seven-digits composite number following 1505719 and preceding 1505721. In scientific notation, it is written as 1.50572 × 106. The sum of its digits is 20. It has a total of 5 prime factors and 16 positive divisors. There are 602,272 positive integers (up to 1505720) that are relatively prime to 1505720.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 20
• Digital Root 2
Name
Short name 1 million 505 thousand 720 one million five hundred five thousand seven hundred twenty
Notation
Scientific notation 1.50572 × 106 1.50572 × 106
Prime Factorization of 1505720
Prime Factorization 23 × 5 × 37643
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 376430 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 1,505,720 is 23 × 5 × 37643. Since it has a total of 5 prime factors, 1,505,720 is a composite number.
Divisors of 1505720
16 divisors
Even divisors 12 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 3.38796e+06 Sum of all the positive divisors of n s(n) 1.88224e+06 Sum of the proper positive divisors of n A(n) 211748 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1227.08 Returns the nth root of the product of n divisors H(n) 7.11092 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,505,720 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 1,505,720) is 3,387,960, the average is 21,174,7.5.
Other Arithmetic Functions (n = 1505720)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 602272 Total number of positive integers not greater than n that are coprime to n λ(n) 75284 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 114293 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 602,272 positive integers (less than 1,505,720) that are coprime with 1,505,720. And there are approximately 114,293 prime numbers less than or equal to 1,505,720.
Divisibility of 1505720
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 6 0 2
The number 1,505,720 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
Base conversion (1505720)
Base System Value
2 Binary 101101111100110111000
3 Ternary 2211111110102
4 Quaternary 11233212320
5 Quinary 341140340
6 Senary 52134532
8 Octal 5574670
10 Decimal 1505720
12 Duodecimal 607448
20 Vigesimal 98460
36 Base36 w9tk
Basic calculations (n = 1505720)
Multiplication
n×y
n×2 3011440 4517160 6022880 7528600
Division
n÷y
n÷2 752860 501907 376430 301144
Exponentiation
ny
n2 2267192718400 3413757419949248000 5140162822365981698560000 7739645964892905963155763200000
Nth Root
y√n
2√n 1227.08 114.617 35.0297 17.2008
1505720 as geometric shapes
Circle
Diameter 3.01144e+06 9.46072e+06 7.1226e+12
Sphere
Volume 1.42995e+19 2.84904e+13 9.46072e+06
Square
Length = n
Perimeter 6.02288e+06 2.26719e+12 2.12941e+06
Cube
Length = n
Surface area 1.36032e+13 3.41376e+18 2.60798e+06
Equilateral Triangle
Length = n
Perimeter 4.51716e+06 9.81723e+11 1.30399e+06
Triangular Pyramid
Length = n
Surface area 3.92689e+12 4.02315e+17 1.22942e+06
Cryptographic Hash Functions
md5 e8c85d227998e41770c28ab6c4f518a2 f5d5d0881e8804c857b4ddb1d4383d4d58077923 8b2b62e47306c9ac59f856d45187642d3a58fdba172eda8efbe23ed2111e7f55 e75aea4c4a9291d7db97ba5ebd8451e3e4cc85e1d710c790b5e12fffe18f911f14dbc1b4acb2b1b7eef171fb1641614ef65f3afdda6150cf99ded9c0442b458d b50e298b4ed4f4b8859775c624983fc33bfb977c | 1,511 | 4,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-05 | latest | en | 0.808051 |
https://tht.fangraphs.com/asymmetric-dominance-and-trade-proposals/ | 1,716,252,703,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00861.warc.gz | 497,559,637 | 30,668 | Asymmetric dominance and trade proposals | The Hardball Times
# Asymmetric dominance and trade proposals
Imagine that you own Kevin Youkilis. You’re a Yankees fan and you hate the Red Sox, so you want to get rid of him out of spite. So you toss him up on the trading block and list outfielders and starting pitchers as your needs. The next morning, one manager offers you one of the following players for Youkilis:
Which offer are you more inclined to accept?
It turns out that you’re not sure which you one you want, so you let the trade offers linger for a couple days until the other owner cancels them. A week later, you find that that manager has now offered you a choice of three players for Youk:
1) Matt Kemp
3) Alex Rios
Now which offer are you more inclined to accept? Did your answer change at all?
The point I am trying to illustrate is one that was raised by one of our readers, Ben, here in the comments section of my previous article. That idea is the asymmetrical dominance effect , also known as the decoy effect.
The asymmetrical dominance effect is the phenomenon in which people show a change in preference between two items when a third item, dominated by either of the first two items, is introduced as an option. There are multiple research studies in which this effect is manifested, and I’ll describe two of them, beginning with the one mentioned in the comments section of the article linked above.
The first study is one that was conducted by Dan Ariely, and in this experiment, students at the University of North Carolina and Duke University were presented with prospective dating partners and instructed to pick a single person to ask out on a date. Three different types of situations were presented:
1). Situation 1 consisted of dating options A and B, both of whom were attractive but had varying degrees of attractive characteristics.
2). Situation 2 consisted of dating options A, B and C(a), with C(a) being almost but not quite as appealing as A.
3). Situation 3 consisted of dating options A, B and C(b), with C(b) being almost but not quite as appealing as B.
So a graphical illustration might look something like this:
The results of this experiment helped support the asymmetrical dominance effect. Participants were more likely to select dating option A over dating option B when the third dating option, C(a), who was slightly less appealing then A, was present. And conversely, participants were more likely to choose dating option B over dating option A when the third dating option, C(b), who was slightly less appealing than B, was present.
The effect manifests in situations outside of partner selection as well. The second study I want to describe is another study by Ariely. This time, subjects were divided into two groups. In group one, subjects decided between microwaves A and B, with microwave A being expensive and of high quality, and with microwave B being less expensive and of medium quality. Forty percent of the subjects in group one preferred microwave A and 60 percent preferred microwave B.
Subjects in group two had three microwaves to choose from: microwaves A and B, and then a third microwave, C(a), that was very similar in dimensions and quality to microwave A, but was more expensive. So in this scenario, it was very clear that microwave C(a) was dominated by microwave A.
A graphical illustration of this paradigm presented to group 2 might look something like this:
A Hardball Times Update
Goodbye for now.
While the majority of subjects in group one preferred microwave B to microwave A, the subjects in group two showed a different preference. This time, 56 percent of subjects chose microwave A, 36 percent chose microwave B and the remaining 8 percent chose microwave C(a). Somehow, this introduction of the third microwave, C(a), completely reversed the ratio of preference between microwaves A and B:
``` Microwaves A B C(a)
Group 1 40% 60% --
Group 2 56% 36% 8%```
Despite differing situations (people/dates vs. microwaves), the underlying theme in both of these studies is the same. That is, that the presence of a third option, one that has an asymmetric dominance relation with one of the two other alternatives, affects a person’s preference of a given alternative over a second alternative.
Let’s go back to my scenario at the top of the page. The construct, according to the studies mentioned earlier, would be as follows:
1) Player A
2) Player B: similar to A but has different attributes
3) Player C(a): dominated by A, meaning he has similar attributes but is slightly less appealing
Using Chone projections for 2009, and using the average draft positions (ADP) as sort of an anchoring point for market value, we’ll have this:
``` ADP Runs HR RBI BA SB W K ERA WHIP
Matt Kemp 38 84 16 74 0.311 26
Roy Halladay 46 -- -- -- -- -- 13 152 3.56 1.22
Alex Rios 39 86 17 75 0.285 19```
In looking at these stat lines, I think it’s pretty clearly that Alex Rios serves as Matt Kemp’s decoy. They both have similar attributes (R, HR, RBI), but Rios is slightly less appealing (lower BA and less SB). In other words, Rios is dominated by Kemp and would most likely never be chosen in this scenario given these numbers.
As Ben asked a couple weeks ago, the question then, is can this actually be applied to fantasy baseball?
My answer is yes, I do think that the asymmetric dominance effect can manifest itself within the realms of fantasy baseball. If we look at Ariely’s study, each potential dating option had varying degrees of attractive characteristics; some were funnier than others, some more intelligent, others a little more honest, and so on. Each option had different levels of attractive attributes that made it more (or less) appealing than the other options. We do the same thing with baseball players but instead of judging them by cost and quality like we did with the microwaves, we look at runs, home runs, batting average, etc.
Considering that these items are essentially the same, presenting our trade proposals in a similar format as with the dating options and microwaves should have some sort of effect since, what really matters, is the contextual configuration of the options. This doesn’t mean these types of trade proposals will always work, as there are many variables to consider. And, after all, the opposing manager can still decline all three proposals whereas the subjects in these studies were forced to pick. However, it does seem as if designing trade proposals in this way would have an effect in that it may increase the likelihood that a trade will happen in the first place or that the third, lesser player will really serve as a decoy.
Inline Feedbacks
Ed Schwehm
15 years ago
You could use this idea when making trade offers to try to influence which player you want the other manager to prefer.
In you example, Rios makes Kemp look better, so you might offer Rios to try to influence the manager to take Kemp instead of Halladay. Of course, you could always not offer Halladay, but if you really wanted Youkilis, maybe you’d accept moving Halladay but prefer to move Kemp.
15 years ago
Great article, I second what Ed said.
Ben
15 years ago
Hi Marco,
Yeah, that’s exactly what I was talking about. I’m a librarian, not a behavioral economist; I read Ariely’s book, which was geared toward a lay audience. So I didn’t retain (read?) the specific terminology—my apologies on that front.
Anyway, yes, Ed—that’s exactly what I was wondering: could you use that effect to subtly direct your trade partner to choose one offer over another? I’ve tried it a few times with no luck, but sample sizes being what they are…
Thanks,
Ben
Ed Schwehm
15 years ago
I can’t remember a specific instance of it working, but I know I still try it, so it hasn’t backfired at least
Nathaniel_g
15 years ago
Could this also work the other way in a trade deal? Meaning you have a player you know another team wants. In return, you give them three options. 2 quality players, but the third is similar to but considered more valuable than the one you want or expect in return.
Damasio
15 years ago
Youk’s owner shouldn’t even think about it in the first place, fan or no fan.
Kevin
15 years ago
Can the theory of relativity also be applied to fantasy trades?
John Dent
15 years ago
An interesting piece, although it’s not clear that Asymmetric Dominance really has any role to play in trading in Fantasy Baseball. The reason that it exerts an influence in every day choice is because it is so hard to value different attributes. However, this is not the case with most trades (or players). We have rankings, projections, dollar values, SGPs… all of which should make people less prone to Asymmetric Dominance (or Compromise Effects, etc) than for other choices… now there’s an undergraduate dissertation for someone to consider.
It should be noted as well that most theorists believe that Asymmetric Dominance is a cognitive shortcut which helps us to simplify choices, whereas fantasy players may be less motivated to use simplifying strategies (unless there was time pressure, for example). After all, we do seem to quite like numbers!
Marco Fujimoto
15 years ago
Again, guys, thanks for reading and leaving comments. And again, I apologize for my delay in response—I have to admit that my level of productivity, in life, decreases significantly during March Madness, especially when my alma mater made the tournament for the first time since 1998!
I think Ed has the right idea, in that we are trying to *influence* judgment and decision-making. In my example, by including Rios in our set of trade offers, we are trying to persuade the other owner to accept Kemp instead of Halladay. If we were intent on giving up Halladay instead of a hitter, we might include someone like Cliff Lee alongside Kemp and Halladay.
And John, I think an issue with the asymmetric dominance effect and other biases (and this applies to any field, not just fantasy baseball) is that the degree of effect these biases have are sort of tied to the level of expertise and amount of knowledge.
In other words, I can see something like this working to a higher degree in average public or private leagues, where the level of knowledge might still be stuck on superficial stats. For example, just look at ESPN’s fantasy section. They just recently began referencing BABIP, whereas BABIP has been common knowledge for a while now here at THT.
But would something like this work in a league full of Derek Carty’s and Paul Singman’s? Probably not. Now this doesn’t mean that cognitive biases or effects don’t exist at ‘expert’ levels; everyone is prone to some form of bias. I think the likelihood of bias just decreases.
get out of the friend zone
14 years ago
When the asymmetrically dominated option is present,a higher percentage of consumers will prefer the dominating option than when the asymmetrically dominated option is absent..
get out of the friend zone | 2,417 | 11,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.97508 |
https://www.enotes.com/homework-help/normal-distribution-384387 | 1,505,959,846,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687592.20/warc/CC-MAIN-20170921011035-20170921031035-00352.warc.gz | 753,166,261 | 9,561 | # The Normal DistributionA box of cookies contains a mean mass of 180 g with a standard deviation of 2 grams. If the mean mass of these boxes is assumed to be normally distributed, what is the...
The Normal Distribution
A box of cookies contains a mean mass of 180 g with a standard deviation of 2 grams. If the mean mass of these boxes is assumed to be normally distributed, what is the probability a box of cookies will contain less than 179 g? Express your answer to the nearest tenth of one percent. | 113 | 508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-39 | latest | en | 0.827079 |
https://www.ammarksman.com/is-rate-of-diffusion-same-as-rate-of-effusion/ | 1,713,457,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00782.warc.gz | 587,518,100 | 10,131 | Is rate of diffusion same as rate of effusion?
Is rate of diffusion same as rate of effusion?
Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
How does effusion differ from diffusion?
Effusion refers to the ability of the gas to travel through a tiny opening. Diffusion refers to the ability of the gases to mix, generally when there is an absence of a barrier. During the process of effusion, the particles tend to move faster than diffusion since there is no collision occurring between the molecules.
How do you determine the rate of diffusion?
Graham’s Law Formula Graham’s law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. See this law in equation form below. In these equations, r = rate of diffusion or effusion and M = molar mass.
What is diffusion rate?
The rate of diffusion, dn/dt, is the change in the number of diffusing molecules inside the cell over time. Since the net movement of diffusing molecules depends on the concentration gradient, the rate of diffusion is directly proportional to the concentration gradient (dC/dx) across the membrane.
What is the difference between effusion and diffusion quizlet?
-Diffusion occurs because the gas molecules are in continuous random motion. The molecules will eventually reach anywhere they can. -The rate of effusion measures the speed at which the gas travels through the tiny hole into a vacuum.
What is the law of effusion?
Effusion refers to the movement of gas particles through a small hole. Graham’s Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.
What are the 4 factors that affect the rate of diffusion?
Several factors determine the rate of diffusion of a solute including the mass of solute, the temperature of the environment, the solvent density, concentration, and solubility.
How do you find the rate of diffusion with volume?
Calculate % diffusion = Volume diffused /total volume x 100.
How is the effusion rate of a gas related to diffusion?
Effusion refers to the movement of gas particles through a small hole. Graham’s Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles. diffusion: movement of particles from an area of high concentration to one of low concentration
Why is diffusion faster at a higher temperature?
Key Points. Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. Effusion refers to the movement of gas particles through a small hole. Graham’s Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.
How does Graham’s Law of diffusion and effusion work?
Graham’s law of diffusion and effusion states the rate of diffusion or effusion for a gas is inversely proportional to the square root of the molar mass of the gas. r ∝ 1/(M)½. or. r(M)½ = constant. where. r = rate of diffusion or effusion. M = molar mass.
How is the rate of diffusion related to the molar mass?
Graham’s Law Formula. Graham’s law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. See this law in equation form below. In these equations, r = rate of diffusion or effusion and M = molar mass. Generally, this law is used to compare the difference in diffusion | 755 | 3,553 | {"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-2024-18 | latest | en | 0.935723 |
http://crypto.stackexchange.com/questions/tagged/cryptanalysis?sort=active&pagesize=50 | 1,466,902,652,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783394414.43/warc/CC-MAIN-20160624154954-00154-ip-10-164-35-72.ec2.internal.warc.gz | 67,747,030 | 26,124 | # Tagged Questions
Analysis of individual security aspects of a cipher or algorithm, not the security of a cipher or algorithm in general (which would lean towards “algorithm-design”).
17 views
### A5/2 Ciphertext Only Attack
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### Any benefit to writing a message as a mix of 2 languages before encryption?
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119 views
### Using e-ID Number for Encryption and Digital Signing
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### Reversing a known encryption algorithm (4-byte repeating key XOR)
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### How can I determine key length for message encrypted by many-time-pad?
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59 views
### Plotting Running Times of SAT solver
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207 views
### Mathematical formula for switching the key for OTP?
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30 views
### Has this forum been compromised by the NSA BULLRUN programme? [migrated]
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38 views
### Countering Cryptographic Attacks
By reading Camellia Design and Analysis document following is my understanding about few cryptographic attacks from designer point of view. Differential and Linear Crypt-analysis. Number of Active ...
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### Concept of “guessed ID” in Authenticated Identity-Based Encryption
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79 views
### Does a key derivation function always generate an even distribution of bits in the resulting key?
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202 views
### Hypothetical encryption technique, is it secure?
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18 views
### How to calculate how many trials I need to guess the plain text if a simple P-Box is used to make the cipher text
If I have 64 bit as input for a straight box that use a transposition process to encipher the plain text and the given 64 bits contains 60 bits as (1s), then how many trial I should do?
106 views
### Ciphertext with uneven letter distribution
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111 views
### Calculation of the avalanche effect coefficient
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8k views
### Now that quantum computers have been out for a while, has RSA been cracked?
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57 views
### Are transposition ciphers insecure?
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47 views
### What kind of analysis can I perform on my implementation of Salsa20 and Rabbit ciphers?
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7k views
### Security strength of RSA in relation with the modulus size
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105 views
### K Friedman - two texts explanation
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33 views
### How to find the keyword of the Playfair text knowing/not knowing the plaintext and having decrypted text?
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62 views
### Factorization of RSA modulus using a qubic residue
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88 views
### Is this a better involution S-Box for the Anubis cipher?
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1k views
### How does cryptanalysis of the Playfair cipher work?
I have a set of Playfair-enciphered data that I'm trying to crack without the key. I know I need to analyse bigrams; I've currently worked out what decrypts to th, <...
60 views
### Why is it that most messages seem to be expanded?
Whenever I have seen a more secure version of a message, the message seems to be expanded by some factor. One example is RSA encryption, but whenever a message does not appear to be expanded it is ...
34 views
### Security Strength of Double Substitution Ciphers
I am wondering why people are using RSA keys when some types of double substitution ciphers seem to be just as secure if not better off. By some types of double substitution ciphers I mean one Which ...
328 views
### Perfectly secure shift cipher
Prove that if only one character is encrypted using a shift cipher, then the shift cipher is perfectly secure. I want to show that $P(P=p | C=c)=P(P=p)$. But I don't know how to relate. Can anyone ...
47 views
106 views
### Comparison Vigènere vs. Monoalphabetic cipher
I have a question for a class of secure communication, but I have no idea about how to proceed in order to answer it. I'll be very grateful to whom replies to me. Assuming that brute force attack ...
39 views
Good evening, I'm about to write my own quadratic sieve implementation in C using GMP library for large numbers. I'm facing an issue while attempting to do the last factorization step for the number: ...
48 views
### How to brute force sqlcipher encrypted file with MD5 hash of a 7 bit key?
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35 views
### Not-So-Simple Substitution Cipher?
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869 views
### Using Chi-Square for Vigenère Cipher
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38k views
Suppose Alice wants to send encryptions (under a one-time pad) of $m_1$ and $m_2$ to Bob over a public channel. Alice and Bob have a shared key $k$; however, both messages are the same length as the ...
222 views
### How are constructs with data-dependent swaps and rotations cryptanalyzed?
Linear and differential cryptanalysis seem well suited for constructs with a (relatively) simple fixed structure of boolean expressions. But some ciphers incorporate swaps of array elements where the ...
909 views
### What is the smallest plaintext/ciphertext size for an algorithm like?
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1k views
### Security of N bit HMAC
Lets say that I am using 128 bit HMAC. How many operations are needed to find a "non secure" message. Is a birthday attack possible?
479 views
### Is there an algorithm to find the number of intersections of two sets?
Suppose both I and my friend have a set of integer numbers. We want to know the number of common elements in our two sets but without knowing elements of the sets of each other. So I don't want my ...
64 views
### Is this approach able to protect against padding oracle attacks?
I know that a protection against the padding oracle attack is to encrypt-then-mac. This is to avoid trying to decrypt a modified ciphertext, and leak information about padding errors. Suppose that we ...
34 views
### RC4 using a key without nonce's or IV's [duplicate]
I'm researching RC4 and I am trying to recreate a condition where RC4 is crackable. However I can't quite find the article/resource I need to find the weakness for an implementation like this (python):...
### Derive $x$ when given $g,g^x$ and $g^{(1/x)}$?
If an adversary has access to the generator g of a group G and is given access to $g^{x}$ and $g^{(1/x)}$, will it make it any easier to derive the value of $x$ compared to when he had access to only \$... | 2,707 | 11,874 | {"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} | 2.890625 | 3 | CC-MAIN-2016-26 | latest | en | 0.935642 |
http://www.enotes.com/homework-help/prove-that-w0-b1-j-j-1-b0-j-j-1-w0-b1-b0-j-1-b1-b0-383371 | 1,462,301,696,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121737.31/warc/CC-MAIN-20160428161521-00037-ip-10-239-7-51.ec2.internal.warc.gz | 496,665,509 | 11,308 | # prove that W0+b1*j’*(j’+1)-b0*j”*(j”+1) = W0+b1+b0*j”+1+(b1-b0)*(j”+1)(j"+1) where J’= j”+1
Posted on
You need to substitute `j'` for `j'' + 1` to the left and right side, such that:
`w_0 + b_1*j'(j' + 1) - b_0*(j' - 1)*j' = w_0 + b_1 + b_0*(j' - 1) + 1 + (b_1 - b_0)*j'*j'`
You need to open the brackets and reduce duplicate terms, such that:
`b_1*(j')^2 + b_1*j' - b_0*j' + b_0 = b_1 + b_0*j' - b_0 + 1 + b_1*(j')^2 - b_0*(j')^2`
`b_1*j' - b_0*j' + b_0 = b_1 + b_0*j' - b_0 + 1 - b_0*(j')^2`
`b_1*j'+ 2b_0 = b_1 + 2b_0*j' + 1 - b_0*(j')^2`
`b_1(j' - 1) -2b_0(j' - 1) = 1 - b_0*(j')^2`
`(j' - 1)(b_1 - 2b_0) = 1 - b_0*(j')^2`
Hence, testing the identity, using the provided condition `j' = j'' + 1` , yields that the identity does not hold. | 406 | 754 | {"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} | 3.703125 | 4 | CC-MAIN-2016-18 | longest | en | 0.530216 |
https://www.teachoo.com/22346/4630/Question-35-Choice-2/category/CBSE-Class-12-Sample-Paper-for-2024-Boards/ | 1,721,623,753,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00446.warc.gz | 858,873,953 | 24,940 | CBSE Class 12 Sample Paper for 2024 Boards
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
## An aeroplane is flying along the line r =λ( i ˆ - j ˆ + k ˆ ); where ' λ ' is a scalar and another aeroplane is flying along the line r =i ˆ-j ˆ+μ(-2j ˆ+ k ˆ ); where ' μ ' is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest? Find the shortest possible distance between them.
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### Transcript
ο»ΏThe given lines are non-parallel lines. Let Shortest distance = |(ππ) β | Since (ππ) β is shortest distance, (ππ) β β₯ Line 1 (ππ) β β₯ Line 2 Point P Since point P lies on Line 1 Position vector of P = π(π Λβπ Λ+π Λ) = (π) π Μ+(βπ)π Μ+(π)π Μ Now, (π·πΈ) β = Position vector of Q β Position vector of P = [π Μ+(β1β2π)π Λ+(π)π Μ ]β[(π) π Μ+(βπ)π Μ+(π)π Μ] = ο»Ώ(1 β π)π€Μ + (β 1 β 2π + π)π₯Μ + (π β π)π Μ Now, (π·πΈ) β β₯ Line 1 (π β = π(π Λβπ Λ+π Λ) ) Thus, (ππ) β β₯ (π Μβπ Μ+π Μ ) And (π·πΈ) β . (π Μβπ Μ+π Μ )=π ο»Ώ(1 β π)π€Μ + (β 1 β 2π + π)π₯Μ + (π β π)π Μ. (π Μβπ Μ+π Μ )=π (1β π)1 + (β1 β2π + π)(β1) + (π β π)1 = 0 (1β π) + (1 + 2π - π) + (π β π) = 0 (1 + 1) + (β π β π β π) + (2π + π) = 0 2 β 3π + 3π = 0 3π β 3π = β 2 Similarly (π·πΈ) β β₯ Line 2 (π β = π Λβπ Λ+π(β2π Λ+π Λ)) Thus, (ππ) β β₯(β2π Λ+π Λ) And (π·πΈ) β .(β2π Λ+π Λ)=π (1 β π)π€Μ + (β 1 β 2π + π)π₯Μ + (π β π)π Μ.(β2π Λ+π Λ )=0 (1 β π)(0) + (β 1 β 2π + π)(β2) + (π β π)(1) = 0 0 + (2 +4π β 2π) + (π β π) = 0 (2) + (β2π β π) + (4π + π) = 0 2 β 3π + 5π = 0 5π β 3π = β2 Thus, our equations are 3π β 3π = β2 β¦(1) 5π β 3π = β2 β¦(2) Solving (1) and (2) We get π = 0, π = π/π Point P Position vector of P = (π) π Μ+(βπ)π Μ+(π)π Μ Putting π = π/π = π/π π Μβπ/π π Μ+ π/π π Μ Point Q Position vector of Q = π Μ+(β1β2π)π Λ+(π)π Μ Putting π = 0 =( π) Μ+(β1 β0) π Μ+(0)π Μ = π Μβπ Μ Now, (π·πΈ) β = Position vector of Q β Position vector of P =[ π Μβ(π]) Μβ[ π/π π Μβπ/π π Μ+ π/π π Μ] = π/π π Μβ(π )/π π Μβπ/π π Μ And, Shortest distance = |(ππ) β | = β((1/3)^2+(β1/3)^2+(2/3)^2 ) =β( 1/9+1/9 +4/9) = β(6/9) = β(π/π) units | 1,895 | 2,961 | {"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} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.532918 |
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1) RE: [ap-calculus] Double Derivatives Posted: Oct 24, 2012 11:38 PM, by: Doug Kuhlmann 100%
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Pages: 33 [ 1 2 3 4 5 6 7 8 9 10 ... 33 | Next ] | 688 | 1,782 | {"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-2017-51 | longest | en | 0.74971 |
https://www.physicsforums.com/threads/solving-an-inertial-mystery-angular-acceleration-and-mud.1046854/ | 1,696,066,947,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00249.warc.gz | 997,067,736 | 18,288 | # Solving an Inertial Mystery: Angular Acceleration and Mud
• PhysicsRock
I threw a rock at a moving car. It hit the car about a foot behind the driver's seat. The rock had been thrown at an angle of about ##45^{\circ}##. The rock had been moving at about 30 MPH when it hit. The rock had been moving at about twice the speed of the car.f
#### PhysicsRock
Homework Statement
A wheel of radius ##R## is spinning at angular velocity ##\omega##. What distance does a pedestrian walking behind the vehicle have to walk, to not be hit by a piece of mud falling off the wheel.
Relevant Equations
None really, it's all about the thought.
So, my idea would be that this happens at an angle ##\theta = \frac{\pi}{2}##, or quarter of a whole rotation. At this point, the wheel starts moving right again, after going to the left. Due to it's inertia, the piece of mud would want to keep it's current direction of motion and therefore fall off. The problem I'm facing now is that the acceleration always points to the center of the wheel, thus the mud would be accelerated towards the cars movement direction, rather than being thrown backward where it could hit the person.
Is my idea of the angle wrong or did I calculate the acceleration vector incorrectly? Help is very apprechiated.
Homework Statement:: A wheel of radius ##R## is spinning at angular velocity ##\omega##. What distance does a pedestrian walking behind the vehicle have to walk, to not be hit by a piece of mud falling off the wheel.
Relevant Equations:: None really, it's all about the thought.
So, my idea would be that this happens at an angle ##\theta = \frac{\pi}{2}##, or quarter of a whole rotation. At this point, the wheel starts moving right again, after going to the left. Due to it's inertia, the piece of mud would want to keep it's current direction of motion and therefore fall off. The problem I'm facing now is that the acceleration always points to the center of the wheel, thus the mud would be accelerated towards the cars movement direction, rather than being thrown backward where it could hit the person.
Is my idea of the angle wrong or did I calculate the acceleration vector incorrectly? Help is very apprechiated.
I would suggest that mud or rain flies off of a spinning wheel at every possible rotation angle. It takes a variable length of time for a drop of mud to form and then drop free from the tire. Certainly, the seat of my pants has stood in mute testament to the fact that it sometimes takes more than a quarter of a rotation for the rear wheel in my bicycle to fling road grime up and forward.
Your observation that the acceleration of the mud is always inward is a good one. The rotating wheel cannot ever fling any material rearward. No portion of the rotating wheel is ever moving rearward relative to the ground.
The pedestrian can be hit by splashing mud. Or by mud flung forward by a rotating wheel.
I would suggest that mud or rain flies off of a spinning wheel at every possible rotation angle. It takes a variable length of time for a drop of mud to form and then drop free from the tire. Certainly, the seat of my pants has stood in mute testament to the fact that it sometimes takes more than a quarter of a rotation for the rear wheel in my bicycle to fling road grime up and forward.
Your observation that the acceleration of the mud is always inward is a good one. The rotating wheel cannot ever fling any material rearward. No portion of the rotating wheel is ever moving rearward relative to the ground.
The pedestrian can be hit by splashing mud. Or by mud flung forward by a rotating wheel.
Thank you so much! The last paragraph just gave me the idea I needed. Your answer was super helpful. :)
The rotating wheel cannot ever fling any material rearward. No portion of the rotating wheel is ever moving rearward relative to the ground.
I think "ever" is too absolute. Mud is notoriously slippery. If the wheel slips as it translates forward so that ##V<\omega R##, there is a good chance that mud can be flung rearward from ground level.
jbriggs444
I think "ever" is too absolute. Mud is notoriously slippery. If the wheel slips as it translates forward so that ##V<\omega R##, there is a good chance that mud can be flung rearward from ground level.
Yeah, a slipping wheel can do the trick. I was assuming rolling without slipping.
Another complication is that the pedestrian is walking. How fast? Mud thrown in the first quadrant of rise will have less forward speed than the bicycle, so could hit a pedestrian moving almost as fast as the rider.
When I was in college, I bought a cheap bike that didn't have a mudguard (fender?) over the rear wheel. The first time I rode it in wet weather, I ended up with a one-inch wide streak of filthy water running down the middle of the back of my shirt. That's when I consciously realized that the water that was flung from the top of the wheel traveled, relative to the ground, at about twice the speed of the back of my shirt.
bob012345 and jbriggs444
Back when my daily driver was a car that I had bought for \$100.00, some friends challenged my statement that driving skill was more important than four wheel drive when driving off road. I agreed to show them, provided that they wash the car afterward. The left front fender was missing.
I observed that the amount of mud flying off the front tire varied with speed. At low speed, the mud stuck to the tire. At "high" speed, it came off right away and very little was flung up into the air. At a certain speed, a large amount was flung up into the air, where it came down and covered the car. I think that speed was near either 15 or 25 MPH.
They had to spend some money and time at the car wash, but they did not complain. | 1,284 | 5,754 | {"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} | 3.96875 | 4 | CC-MAIN-2023-40 | latest | en | 0.963188 |
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# GPA Question
## Recommended Posts
Hi all,
I apologize in advance if this has been asked before. I go to York and the grade system it uses is:
90%+ --> A+
80-89.9% --> A
As you can see, there is no A-. Our transcripts only show these letter grades (and not percentages).
My question: would both A and A+ count as a 4.0 for McGill? Or is there something I'm missing here?
Thank you!
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Hey Dr. Octavius,
GPA's are calculated by OMSAS and not by your undergrad school. Regardless of your schools grading system, 90+ is a 4.0, 85+ is a 3.9 and 80+ is a 3.7.
Hope this helps.
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I did some more digging, and came across this thread:
There are several people there saying A = A+ = 4.0.
Now I'm confused
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18 minutes ago, Dr. Octavius said:
I did some more digging, and came across this thread:
There are several people there saying A = A+ = 4.0.
Now I'm confused
Yes that's true. McGill only has As so anything above an A is considered 4.0. That's how I entered my A+ grades into the workbook
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If you click the OMSAS conversion table, you shouldn't have a problem converting your grades across institutions: https://www.ouac.on.ca/guide/omsas-conversion-table/
You will see that McGill is the only school that uses column "8" for example. Whereas, York uses column "9". The OMSAS value scale (out of 4.0) is on the far left.
• Only your A+ at York would equate to 4.0 on OMSAS; your A at York would only equate to a 3.8.
• At McGill, an A would equate to 4.0, whereas an A- would be a 3.7.
Hope this helps.
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I think you guys are saying two different things lol.
Okay, what about this: If someone from York had 9 A+ and 1 A, what would their GPA be as an OOP McGill applicant?
A. 3.98 (using OMSAS, an A is a 3.8 and A+ is 4.0)
or
B. 4.00 (the A would be the same as the A+ for McGill)
I think the answer to that will settle things!
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44 minutes ago, Dr. Octavius said:
I think you guys are saying two different things lol.
Okay, what about this: If someone from York had 9 A+ and 1 A, what would their GPA be as an OOP McGill applicant?
A. 3.98 (using OMSAS, an A is a 3.8 and A+ is 4.0)
or
B. 4.00 (the A would be the same as the A+ for McGill)
I think the answer to that will settle things!
B for sure
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Sorry for the confusion lol. I used the appendix table in the instructions. Always thought that corresponded to the OMSAS table but I see that there are some differences for York
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My apologies - disregard my post! @SunAndMoon is right!
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Thanks everyone! Great - sounds like the conversion will work in my favour!
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890 191
#### Published on 24 Mar 2014 | over 4 years ago
See the rules of Red Black Tree's and each violation case and their respective solution below and check out my other video about this: youtu.be/axa2g5oOzCE . Also, see below for descriptions of inserting and deleting nodes from a Red Black Tree.
Rules of Red Black Trees (not in any particular order):
1. Every node is either RED or BLACK
2. The root node is BLACK
3. Every leaf node (null) is BLACK
4. Every RED node has two BLACK child nodes
5. Every path from node x to a leaf node has the same number of BLACK nodes (BLACK height)
VIOLATIONS and SOLUTIONS:
Case 1:
1. Uncle is Red
Solution for Case 1:
1. Swap colors of Parent, Uncle, and Grandparent
2. Grandparent becomes the new node to check for violations
Case 2:
1. Uncle is Black
2a. Node is a Left Child of a Right Child
2b. Node is a Right Child of a Left Child
Solution for Case 2:
1a. Right rotate around Parent for 2a = Case 3
1b. Left rotate around Parent for 2b = Case 3
2. Parent becomes the new node to check for violations
Case 3:
1. Uncle is Black
2a. Node is a Right Child of a Right Child
2b. Node is a Left Child of a Left Child
Solution for Case 3:
1a. Left rotate around Grandparent for 2a
1b. Right rotate around Grandparent for 2b
2. Swap colors of Parent and Grandparent
3. Grandparent becomes the new node to check for violations
INSERTION
To insert a node, you begin at the root of the tree, and follow the rules of a Binary Search Tree, to insert the new node. The only difference is that you have to give the node a starting color (RED) and then after the insertion, check for violations (as described above).
DELETION
To delete a node, you follow the rules of deletion for Binary Search Trees. There are three scenarios:
Assuming we want to delete node x:
1. If node x has no children, then just delete node x.
2. If node x has one child, then replace node x with it's child, and delete node x.
3. If node x has two children, then replace node x with either its predecessor or successor, and delete node x.
For scenario 2 or 3 above, you would then check for violations at the node that replaced node x.
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##### 8. What is the molarity of a 5.00 x 102 ml solution containing 290 g of FeS2?
label Chemistry
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8. What is the molarity of a 5.00 x 102 ml solution containing 290 g of FeS2?
Oct 16th, 2017
The molar mass of FeS2 is 55.845 + 2 * 32.06 = 120 g/mol. So, the number of moles of FeS2 in the solution is 290 g / 120 g/mol = 2.42 mol. The molarity of the solution is 2.42 / 0.500 L = 4.83 M
Apr 7th, 2015
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Oct 16th, 2017
...
Oct 16th, 2017
Oct 17th, 2017
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# Buoyancy, Flotation, and Stability
Buoyant force is the resultant fluid force acting on a stationary body that is completely
submerged or a floating body that is only partially submerged.
Buoyant force on
submerged and
floating bodies.
## If the specific weight of the fluid is constant, then
where A is the horizontal area of the upper (or lower) surface of the parallelepiped
## Simplifying, we arrive at the desired expression for the buoyant force
Therefore, the buoyant force has a magnitude equal to the weight of the fluid displaced by
the body and is directed vertically upward.
## This result is commonly referred to as Archimedes’ principle.
The buoyant force passes through the centroid of the displaced volume as shown in Fig.
16c. The point through which the buoyant force acts is called the center of buoyancy.
EXAMPLE 6.
Solution:
Stability
A body is said to be in a stable equilibrium position if, when displaced,
it returns to its equilibrium position. Conversely, it is in an unstable
equilibrium position if, when displaced (even slightly), it moves to a
new equilibrium position.
## From Fig. 17 as long as
the center of gravity
falls below the center
of buoyancy, the body
is in a stable
equilibrium position Fig. 17 Stability of a Fig. 18 Stability of a
with respect to small completely immersed completely immersed
rotations. body—center of gravity body—center of gravity
below centroid. above centroid.
## From Fig. 18, acompletely submerged body with its center of
gravity above its center of buoyancy is in an unstable
equilibrium position.
Surface and body forces
acting on small fluid
element.
The resultant surface force acting on a small fluid element depends only on the pressure
gradient if there are no shearing stresses present.
## The resultant surface force in the y direction is
or
Similarly, for the x and z directions the resultant surface forces are
and
The resultant surface force acting on the element can be expressed in vector form as
or
where i ̂ ,j ̂ , and k ̂ are the unit vectors along the coordinate axes. The group of terms in
parentheses in the resultant surface represents in vector form the pressure gradient and
can be written as
The resultant surface force per unit volume can be expressed as
## Since the z axis is vertical, the weight of the element is
where the negative sign indicates that the force due to the weight is downward
Newton’s second law, applied to the fluid element, can be expressed as
It follows that
or
and, therefore,
This the general equation of motion for a fluid in which there are no shearing stresses.
Pressure Variation in a Fluid with Rigid-Body Motion
Even though a fluid may be in motion, if it moves as a rigid body there will be no shearing
stresses present.
## The general equation of motion (Eq. 1):
Eq.1 in component form, based on rectangular coordinates with the positive z axis being
vertically upward, can be expressed as
There is no shear stress in fluids that move with rigid-body motion or with rigid-body
rotation.
Linear acceleration of a liquid with a free surface.
We first consider an open container of a liquid that is translating along a straight path with
a constant acceleration a as illustrated in Fig. 19
The change in pressure between two closely spaced points located at y, z, and y+dy, z+dz
can be expressed as
or in terms of the results of the pressure gradient
Along a line of constant pressure, dp = 0 and therefore from eq. above, it follows that the
slope of this line is given by the relationship
slope
EXAMPLE
Solution:
(a) For a constant horizontal acceleration the fuel will move as a rigid body, so that the
slope of the fuel surface can be expressed as
or
Since there is no acceleration in the vertical, z, direction, the pressure along the wall varies
hydrostatically. Thus, the pressure at the transducer is given by the relationship
where h is the depth of fuel above the transducer, and therefore
Some of the important equations in this chapter are: | 895 | 4,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.907259 |
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# hw31 - Î and m are the bare coupling constant and the bare...
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Homework Set No. 1, Physics 880.08 Deadline – Wednesday, April 13, 2011 1. (15 pts) Show that the light-by-light scattering QED diagram pictured below does not contain a UV divergence (even thought naively counting powers of momenta it seems like it is logarithmically divergent). The ellipsis in the figure represent all possible permutations of the vertices along the loop. (Hint: you may find formulas (A.41) and (A.42) in Appendix A.4 of Peskin and Schroeder useful.) 2. Using dimensional regularization find the one-loop beta-function β ( λ ) = μ 2 d λ d μ 2 of the real scalar ϕ 3 theory with the Lagrangian density L = 1 2 μ ϕ 0 μ ϕ 0 - m 2 0 2 ϕ 2 0 - λ 0 3! ϕ 3 0 (1) in six (!) space-time dimensions. Here ϕ 0 is the bare field, while
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Unformatted text preview: λ and m are the bare coupling constant and the bare mass correspondingly. a. (20 pts) Similar to how we did it in class for φ 4 theory (see also HW 4 from last quarter), rewrite the Lagrangian (1) in terms of renormalized physical Felds ϕ , coupling λ , mass m and the counterterms. By calculating the one-loop propagator and vertex corrections Fnd the divergent ( ∼ 1 /ǫ with ǫ = 6-d ) parts of the coe±cients of the counterterms δ λ and δ Z . (Ignore the tadpole graph.) Note that we do not need the Fnite parts of the counterterms to Fnd β ( λ )! b. (10 pts) Using the results of part a Fnd β ( λ ). 1...
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### Question 1
If the sides of a triangle are 13 cm, 14 cm and 15 cm respectively then its area is ____cm2.
A) 80
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### Question 2
If the sides of a triangle are 150 cm, 120 cm and 200 cm, then its area is ____ cm2.
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If two sides of a triangle are 8 cm, 11 cm and the perimeter is 32 cm, then its area is ____ cm2.
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If two sides of triangle are 8 cm and 11 cm and its perimeter is 64 cm, then its area is ____cm2.
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### Question 5
If two sides of a triangle are and the perimeter is 42 cm, then its area is ____cm2.
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If the sides of a triangle are 25 cm, 17 cm and 12 cm respectively, then its area is ____cm2.
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If the sides of a triangle are 13 cm, 12 cm and 5 cm, then its area is ____cm2.
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The sides of a triangle are 12 cm, 13 cm and 15 cm respectively, then its area is ____ cm2.
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If the perimeter of an isosceles triangle is 30 cm and each of the equal sides is 12 cm, then its area is ____cm2.
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B)
C)
D) | 477 | 1,339 | {"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} | 3.625 | 4 | CC-MAIN-2014-15 | latest | en | 0.711714 |
https://www.coursehero.com/file/5961684/Lecture-6/ | 1,529,904,109,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00567.warc.gz | 787,910,448 | 279,530 | Lecture 6 - Harmonic oscillator Algebraic approach...
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Harmonic oscillator Algebraic approach
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Classical harmonic oscillator ( 29 2 2 ; ; cos ; d x k k F kx x x A t dt m m ϖ δ ϖ = - = - = + = A block attached to an elastic spring is a typical example of a harmonic oscillator. One can find position as a function of time using Hook’s law in combination with 2 nd Newton’s law Energy of harmonic oscillator consists of kinetic and potential energies given as 2 2 2 2 2 1 1 1 ; ; 2 2 2 2 2 mv mv U kx K E kx kA = = = + =
Generalization of harmonic oscillator 0 0 2 2 0 0 2 2 0 0 0 2 0 1 ( ) ( ) ( ) ( ) 2 x x x x x x x x dU d U k dx dx dU d U U x U x x x x x dx dx = = = = = = + - + - + 1 4 442 4 4 43 1 442 4 43 K In most cases harmonic oscillators appear as a result of an approximation of a potential energy with a minimum at some points by its series expansion about this point
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Classical harmonic oscillator (continue) K=0 turning point K=0 turning point K<0 classically forbidden region K<0 classically forbidden region K>0 classically allowed region Classical harmonic oscillator exhibits so called finite motion: particle’s position is limited by some finite region of space, while remaining of it remains forbidden.
Quantum harmonic oscillator First step: Replace classical momentum and coordinate with the respective operators in order to find quantum Hamiltonian 2 2 2 2 2 2 2 2 ˆ 1 1 ˆ ˆ 2 2 2 2 p d H m x m x m m dx ϖ ϖ = + = - + h Second step: Set up stationary Schrödinger equation ˆ ( ) ( ) H x E x ψ ψ = 2 2 2 2 2 1 2 2 d m x E m dx ψ ϖ ψ ψ - + = h The ultimate task is to find wave function of a HO at any time given its initial state. The first step in achieving is goal is finding stationary states and respective energy levels. When those are known we can present a general solution as a linear combination of the stationary states with respective time-dependent factors, fit it to an initial condition, and we are done. Third step: Solve it.
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Energy levels Energy levels 1 , 0,1,2 2 n E n n ϖ = + = K h Energy levels of quantum harmonic oscillator are discrete and equidistant
Stationary states 2 2 2 2 2 2 2 2 1/2 /2 0 1/2 /2 1/2 2 /2 1/2 3 /2 1 1 0 ( ) 2 3 1 1 2 2 2 5 1 2 2 4 2 8 7 1 3 12 8 2 48 x a x a x a x a n E x e a x n E e a a x n E e a a x x n E e a a a ϖ ψ π ϖ π ϖ π ϖ π - - - - = = = = = = = - = = - h h h h 1. Wave functions are either symmetric (even) or antisymmetric (odd) 2. Number of zeroes is again level number minus one 3. Wave functions are a Gaussian function multiplied by a polynomial a m ϖ = h Characteristic length or “size” of the wave function
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Expectation values coordinate 2 2 1 1 0; ; 2 2 x x n a x a n = = + = + momentum 2 2 2 1 1 0; ; 2 2 p p n p n a a = = + = + h h 1 1 2 2 p x n ∆ ∆ = + h h Uncertainty relation Ground state has minimum possible uncertainty Kinetic energy 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 n p p K K n n E m m ma
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http://www.docstoc.com/docs/143820514/Field-Of-View-Calculator | 1,412,139,246,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663359.6/warc/CC-MAIN-20140930004103-00108-ip-10-234-18-248.ec2.internal.warc.gz | 479,916,912 | 15,526 | # Field-Of-View Calculator
Document Sample
``` A B C D E F G H
1
2
3
Field-Of-View Calculator
4
5
6
7 clip width p hAngle
8
9 9.75 6.35 75
10
11
12 Instructions
13
14 1. Find out what size (in 1:1) is your view clip
15 a. create new layer at 1:1
16 b. create temp. perspective:
17 b1. use Set 3D View
18 b2. observe clip outline
19 c. go to 1:1 scale layer
20 d. make sure your Layer Options are set to Show Others
21 e. use 2d Polygon tool to trace outline of view clip
22 f. See Obj. Info palette for view clip size.
23 g. Enter clip width in chart above.
24
25 2. Select desired 35mm lens angle of view from chart below
26 a. note: desired lens angle corresponds to hAngle above
27
28 3. Go to chart at top, and enter the desired lens angle under hAngle
29 a. p-value is automatically displayed.
30
31 4. Now apply the info to your drawing
32 a. Go back to original drawing layer (NOT 1:1 scale layer)
33 b. Use Set 3D View to create view--choose Orthogonal.
34 c. Go to View/Perspective/Set Perspective
35 d. Enter p-value you've chosen above.
36
37 8. The resulting view should represent the image generated by the chosen 35mm lens.
38
39
40 Horizontal Angles of View for 35 mm film Lenses
41
42 Focal Length Angle of View
43 20 83
44 24 74
45 28 64
46 35 53
47 50 39
48 55 36.1
49 85 23.9
A B C D E F G H
50 100 20
51 105 19.5
52 135 15
53 180 11.5
54 200 10.3
55
56
57 Additional info on lens angles and formulas can be obtained at the following URLs:
58 http://www.vidpro.org/lenschart.htm
59 http://www.eci.com/StudioOne/page26.html (formulas)
60 http://mir.com.my/rb/photography/fototech/chart/index.htm
```
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views: 2 posted: 2/3/2013 language: Unknown pages: 2 | 646 | 2,840 | {"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-2014-41 | longest | en | 0.591219 |
https://www.emathzone.com/tutorials/geometry/volume-of-a-cone.html | 1,712,975,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00290.warc.gz | 685,658,965 | 14,012 | # Volume of a Cone
The volume of a right circular cone is one-third of the volume of a right circular cylinder of the same base and same height.
$\therefore$ If $h$ is the height of the cone and $r$ is the radius of the base, then
${\text{Volume }} = {\text{ }}\frac{1}{3}{\text{ }} \times {\text{ area of the base }} \times {\text{ altitude}}$
$\therefore$ $V = \frac{1}{3}\pi {r^2}h$ (as the area of the base $= \pi {r^2}$)
Rule: The volume of a cone equals the area of the base times one-third the altitude.
Example:
The circumference of the base of a $9$m high conical tent is $44$m. Find the volume of the air contained in it.
Solution:
Circumference of the base $= 2\pi r = 44$m
$\therefore$ $2 \times \frac{{22}}{7} \times r = 44$
$r = \frac{{44 \times 7}}{{44}} = 7$m
$\because$ height of the conical tent $= 9$m
$\therefore$ volume of air $= \frac{1}{3}\pi {r^2}h$
$= \frac{1}{3} \times \frac{{22}}{7} \times 7 \times 9 = 462$ cubic m
Example:
The vertical height of a conical tent is $42$ dm and the diameter of its base is $5.4$ m. Find the number of people it can accommodate if each person is to be allowed $2916$ cubic dm of space.
Solution:
Here height $h = 42$ dm
Diameter $= 5.4$ m $= 54$ dm
Radius $r = 27$ dm
Volume $= \frac{1}{3}\pi {r^2}h$
$= \frac{1}{3} \times \frac{{22}}{7} \times 27 \times 27 \times 42$
$= 32076$ Cubic dm
Space allowed for $1$ person $= 2916$ cubic dm
$\therefore$ the required number of people $= \frac{{32076}}{{2916}} = 11$ people | 524 | 1,531 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-18 | latest | en | 0.669494 |
https://sciencetrends.com/the-formula-for-work-physics-equation-with-examples/?replytocom=28961 | 1,621,294,137,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991870.70/warc/CC-MAIN-20210517211550-20210518001550-00373.warc.gz | 539,233,221 | 40,019 | # The Formula For Work: Physics Equation With Examples
In physics, we say that a force does work if the application of the force displaces an object in the direction of the force. In other words, work is equivalent to the application of a force over a distance. The amount of work a force does is directly proportional to how far that force moves an object. The general formula for work and for determining the amount of work that is done on an object is:
• WF × D × cos(Θ)
where is the amount of work, F is the vector of force, D is the magnitude of displacement, and Θ is the angle between the vector of force and the vector of displacement. The SI unit for work is the joule (J), and its dimensions are kg•m2/s2. Another way to understand it is that one joule is equivalent to the amount of energy transferred when one newton of force moves an object a distance of one meter.
## Formula For Work
Whenever a force moves an object, we say that work has been done. When a ball rolls down a hill due to the force of gravity, when you pick up your backpack off the ground, when your car’s internal engine applies a force to make your wheels move; all of these events involve a force moving an object over a distance and so involve some work. In cases where a force is applied to an object but does not move it, no work has been done. So the force from a person pushing the side of a skyscraper is not doing any work as the skyscraper does not move. Let’s consider some simple examples to illustrate the concept of work.
## Example problems
### (1)
A 100 Newton force is applied to a 15kg box in the horizontal direction and moves it 5 meters horizontally. How much work was done?
In this case, we know the force is 100 N and the distance is 5 meters. We also know that since the force is applied in the same direction as the displacement, Θ is equal to 0. So we plug these values into our equation
• WF × D × cos(Θ)
and get:
• W = 100(5)cos(0)= 500 J
So, the 100 N force did 500 joules work moving the block 5 meters.
### (2)
There is a 2kg book lying on a table. A 64 N force is applied to the book at a 120° angle from horizontal and moves the book 3 meters in the horizontal direction. How much work was done?
In this case, we know the force of 64 N and the distance of 3 m. We also know that there is a 120°angle between the angle of the direction of applied force and the direction of motion. So plugging these values into our handy equation yields:
• W = 64(3)cos(120)= 156.32 Joules
So the 64 N force at a 120° angle did 156.32 joules of work moving the book 3 meters.
### (3)
Linda takes a 300 N suitcase up 3 flights of stairs for a total vertical distance of 16 meters. She then pushes the suitcase with 100 N of force the remaining 8 meters to her hotel room. How much work was done over her whole trip?
This question requires 2 separate steps. There are two main portions of her trip, so we can calculate the work done over each portion individually, then, combine the two values to get the total amount of work done. For the first part of her trip, she exerts 300 N of force to move a suitcase 16 meters vertically so the amount of work done is:
• W1=300(16)cos(0)=4800 Joules
So the first part of the trip did 4800 joules. For the second part, we know that a 100N force moved the suitcase horizontally 8 meters, so the total amount of work done on the second portion of the trip is:
• W2=100(8)cos(0)=800 Joules
Combining the two values from each portion of the trip yields:
• WtotalW1+W2= 4800+800= 5600 Joules
So over the entire course of Linda’s trip, 5600 Joules of work was done.
## Work/Energy Relationship
Work and energy in physics share a close relationship. According to the work-energy principle, an increase in a rigid body’s kinetic energy is caused by an equal amount of work done on that body by a force applied to that body. In more mathematical terms, the relationship can be expressed as:
• W = KEfinal−KEinitial
where KE stands for kinetic energy. In other words, the change in kinetic energy of a body is equal to the amount of work done on that body. In general, the formula for the kinetic energy of an object is:
• KE = (1/2) kg*v2
where v stands for an object’s velocity. The unit for kinetic energy is the same unit for work, the joule. Let’s look at some problems to examine these mathematical relationships.
### (4)
Donkey and Diddy Kong are sitting in a 90-kilogram minecart that is initially traveling horizontally at a velocity of 5 m/s. Rambi the rhino pushes the minecart from behind and speeds it up so it is now traveling 11 m/s. How much work did Rambi do on the minecart?
In order to solve this problem we first need to figure out the minecart’s initial kinetic energy and its final kinetic energy. Once we know those values, we can determine the total amount of work. We know both the velocity and mass of the minecart so we can determine the total kinetic energy at the beginning and at the end. The initial kinetic energy of the minecart is:
• KEinitial=(1/2)(90)(5)21125 J
The final kinetic energy of the minecart is
• KEfinal=(1/2)(90)(11)2=5445 J
Therefore, the total amount of work done on the minecart is 5545−1125= 4420 J.
“Science is the knowledge of consequences, and dependence of one fact upon another.” – Thomas Hobbes
### (5)
A car weighing 1300 kg is moving with a velocity of 18 m/s. If 60000 joules of work is done on the car, what will its final velocity be?
The question will require a bit of algebra. First, we must determine the initial kinetic energy of the car. The initial kinetic energy of the car is:
• (1/2)(1300)(18)2=210600 J
Since we know the total amount of work done on the system (60000 J) we can figure out the car’s final kinetic energy:
• 60000=KEfinal−210600
• 270600=KEfinal
Now, since we know the final kinetic energy and the mass of the car, we can determine its final velocity like this
• KEfinal=(1/2)kg*v2
• 270600 = (1/2)(1300)v2
• 270600 = 650v2
• 416.3 = v2
• v20.4 m/s
The car’s final velocity will be 20.4 m/s.
So in summation, we say work is done whenever a force moves an object over a distance. The magnitude of work is equal to the magnitude of force multiplied by the distance traveled. Work and kinetic energy are tightly intertwined and can be used to determine each other.
1. Sasha says:
How much work is done by a 60kg man in climbing a stair of 12 steps, each steps being 6 inches higher than the previous one?
2. Anonymous says:
The problem with 120 degree angle of force applied does not equal the number shown (its negative)
1. anonymous says:
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Science Trends is a popular source of science news and education around the world. We cover everything from solar power cell technology to climate change to cancer research. We help hundreds of thousands of people every month learn about the world we live in and the latest scientific breakthroughs. Want to know more? | 2,018 | 8,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-21 | longest | en | 0.917259 |
https://www.aqua-calc.com/convert/acceleration/centimeter-per-minute-squared-to-parsec-per-second-squared | 1,606,893,451,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00089.warc.gz | 558,422,385 | 10,078 | # Centimeters per (minute squared) to parsecs per (second squared)
## cm/min² to pc/s² (cm:centimeter, min:minute, pc:parsec, s:second)
### Convert cm/min² to pc/s²
#### an acceleration conversion table
to 100 with
centi-
meter per minute squared
×10−21,
parsec per second squared
centi-
meter per minute squared
×10−21,
parsec per second squared
centi-
meter per minute squared
×10−21,
parsec per second squared
centi-
meter per minute squared
×10−21,
parsec per second squared
centi-
meter per minute squared
×10−21,
parsec per second squared
10.09211.89413.69615.49817.29
20.18221.98423.78625.58827.38
30.27232.07433.87635.67837.47
40.36242.16443.96645.76847.56
50.45252.25454.05655.85857.65
60.54262.34464.14665.94867.74
70.63272.43474.23676.03877.83
80.72282.52484.32686.12887.92
90.81292.61494.41696.21898.01
100.90302.70504.50706.30908.10
110.99312.79514.59716.39918.19
121.08322.88524.68726.48928.28
131.17332.97534.77736.57938.37
141.26343.06544.86746.66948.46
151.35353.15554.95756.75958.55
161.44363.24565.04766.84968.64
171.53373.33575.13776.93978.73
181.62383.42585.22787.02988.82
191.71393.51595.31797.11998.91
201.80403.60605.40807.201009.00
### centimeters per minute squared to parsecs per second squared conversion cards
• 1
through
20
centimeters per minute squared
• 1 cm/min² to pc/s² = 9.0 × 10-23 pc/s²
• 2 cm/min² to pc/s² = 1.8 × 10-22 pc/s²
• 3 cm/min² to pc/s² = 2.7 × 10-22 pc/s²
• 4 cm/min² to pc/s² = 3.6 × 10-22 pc/s²
• 5 cm/min² to pc/s² = 4.5 × 10-22 pc/s²
• 6 cm/min² to pc/s² = 5.4 × 10-22 pc/s²
• 7 cm/min² to pc/s² = 6.3 × 10-22 pc/s²
• 8 cm/min² to pc/s² = 7.2 × 10-22 pc/s²
• 9 cm/min² to pc/s² = 8.1 × 10-22 pc/s²
• 10 cm/min² to pc/s² = 9.0 × 10-22 pc/s²
• 11 cm/min² to pc/s² = 9.9 × 10-22 pc/s²
• 12 cm/min² to pc/s² = 1.08 × 10-21 pc/s²
• 13 cm/min² to pc/s² = 1.17 × 10-21 pc/s²
• 14 cm/min² to pc/s² = 1.26 × 10-21 pc/s²
• 15 cm/min² to pc/s² = 1.35 × 10-21 pc/s²
• 16 cm/min² to pc/s² = 1.44 × 10-21 pc/s²
• 17 cm/min² to pc/s² = 1.53 × 10-21 pc/s²
• 18 cm/min² to pc/s² = 1.62 × 10-21 pc/s²
• 19 cm/min² to pc/s² = 1.71 × 10-21 pc/s²
• 20 cm/min² to pc/s² = 1.8 × 10-21 pc/s²
• 21
through
40
centimeters per minute squared
• 21 cm/min² to pc/s² = 1.89 × 10-21 pc/s²
• 22 cm/min² to pc/s² = 1.98 × 10-21 pc/s²
• 23 cm/min² to pc/s² = 2.07 × 10-21 pc/s²
• 24 cm/min² to pc/s² = 2.16 × 10-21 pc/s²
• 25 cm/min² to pc/s² = 2.25 × 10-21 pc/s²
• 26 cm/min² to pc/s² = 2.34 × 10-21 pc/s²
• 27 cm/min² to pc/s² = 2.43 × 10-21 pc/s²
• 28 cm/min² to pc/s² = 2.52 × 10-21 pc/s²
• 29 cm/min² to pc/s² = 2.61 × 10-21 pc/s²
• 30 cm/min² to pc/s² = 2.7 × 10-21 pc/s²
• 31 cm/min² to pc/s² = 2.79 × 10-21 pc/s²
• 32 cm/min² to pc/s² = 2.88 × 10-21 pc/s²
• 33 cm/min² to pc/s² = 2.97 × 10-21 pc/s²
• 34 cm/min² to pc/s² = 3.06 × 10-21 pc/s²
• 35 cm/min² to pc/s² = 3.15 × 10-21 pc/s²
• 36 cm/min² to pc/s² = 3.24 × 10-21 pc/s²
• 37 cm/min² to pc/s² = 3.33 × 10-21 pc/s²
• 38 cm/min² to pc/s² = 3.42 × 10-21 pc/s²
• 39 cm/min² to pc/s² = 3.51 × 10-21 pc/s²
• 40 cm/min² to pc/s² = 3.6 × 10-21 pc/s²
• 41
through
60
centimeters per minute squared
• 41 cm/min² to pc/s² = 3.69 × 10-21 pc/s²
• 42 cm/min² to pc/s² = 3.78 × 10-21 pc/s²
• 43 cm/min² to pc/s² = 3.87 × 10-21 pc/s²
• 44 cm/min² to pc/s² = 3.96 × 10-21 pc/s²
• 45 cm/min² to pc/s² = 4.05 × 10-21 pc/s²
• 46 cm/min² to pc/s² = 4.14 × 10-21 pc/s²
• 47 cm/min² to pc/s² = 4.23 × 10-21 pc/s²
• 48 cm/min² to pc/s² = 4.32 × 10-21 pc/s²
• 49 cm/min² to pc/s² = 4.41 × 10-21 pc/s²
• 50 cm/min² to pc/s² = 4.5 × 10-21 pc/s²
• 51 cm/min² to pc/s² = 4.59 × 10-21 pc/s²
• 52 cm/min² to pc/s² = 4.68 × 10-21 pc/s²
• 53 cm/min² to pc/s² = 4.77 × 10-21 pc/s²
• 54 cm/min² to pc/s² = 4.86 × 10-21 pc/s²
• 55 cm/min² to pc/s² = 4.95 × 10-21 pc/s²
• 56 cm/min² to pc/s² = 5.04 × 10-21 pc/s²
• 57 cm/min² to pc/s² = 5.13 × 10-21 pc/s²
• 58 cm/min² to pc/s² = 5.22 × 10-21 pc/s²
• 59 cm/min² to pc/s² = 5.31 × 10-21 pc/s²
• 60 cm/min² to pc/s² = 5.4 × 10-21 pc/s²
• 61
through
80
centimeters per minute squared
• 61 cm/min² to pc/s² = 5.49 × 10-21 pc/s²
• 62 cm/min² to pc/s² = 5.58 × 10-21 pc/s²
• 63 cm/min² to pc/s² = 5.67 × 10-21 pc/s²
• 64 cm/min² to pc/s² = 5.76 × 10-21 pc/s²
• 65 cm/min² to pc/s² = 5.85 × 10-21 pc/s²
• 66 cm/min² to pc/s² = 5.94 × 10-21 pc/s²
• 67 cm/min² to pc/s² = 6.03 × 10-21 pc/s²
• 68 cm/min² to pc/s² = 6.12 × 10-21 pc/s²
• 69 cm/min² to pc/s² = 6.21 × 10-21 pc/s²
• 70 cm/min² to pc/s² = 6.3 × 10-21 pc/s²
• 71 cm/min² to pc/s² = 6.39 × 10-21 pc/s²
• 72 cm/min² to pc/s² = 6.48 × 10-21 pc/s²
• 73 cm/min² to pc/s² = 6.57 × 10-21 pc/s²
• 74 cm/min² to pc/s² = 6.66 × 10-21 pc/s²
• 75 cm/min² to pc/s² = 6.75 × 10-21 pc/s²
• 76 cm/min² to pc/s² = 6.84 × 10-21 pc/s²
• 77 cm/min² to pc/s² = 6.93 × 10-21 pc/s²
• 78 cm/min² to pc/s² = 7.02 × 10-21 pc/s²
• 79 cm/min² to pc/s² = 7.11 × 10-21 pc/s²
• 80 cm/min² to pc/s² = 7.2 × 10-21 pc/s²
• 81
through
100
centimeters per minute squared
• 81 cm/min² to pc/s² = 7.29 × 10-21 pc/s²
• 82 cm/min² to pc/s² = 7.38 × 10-21 pc/s²
• 83 cm/min² to pc/s² = 7.47 × 10-21 pc/s²
• 84 cm/min² to pc/s² = 7.56 × 10-21 pc/s²
• 85 cm/min² to pc/s² = 7.65 × 10-21 pc/s²
• 86 cm/min² to pc/s² = 7.74 × 10-21 pc/s²
• 87 cm/min² to pc/s² = 7.83 × 10-21 pc/s²
• 88 cm/min² to pc/s² = 7.92 × 10-21 pc/s²
• 89 cm/min² to pc/s² = 8.01 × 10-21 pc/s²
• 90 cm/min² to pc/s² = 8.1 × 10-21 pc/s²
• 91 cm/min² to pc/s² = 8.19 × 10-21 pc/s²
• 92 cm/min² to pc/s² = 8.28 × 10-21 pc/s²
• 93 cm/min² to pc/s² = 8.37 × 10-21 pc/s²
• 94 cm/min² to pc/s² = 8.46 × 10-21 pc/s²
• 95 cm/min² to pc/s² = 8.55 × 10-21 pc/s²
• 96 cm/min² to pc/s² = 8.64 × 10-21 pc/s²
• 97 cm/min² to pc/s² = 8.73 × 10-21 pc/s²
• 98 cm/min² to pc/s² = 8.82 × 10-21 pc/s²
• 99 cm/min² to pc/s² = 8.91 × 10-21 pc/s²
• 100 cm/min² to pc/s² = 9.0 × 10-21 pc/s²
#### Foods, Nutrients and Calories
100% UNSWEETENED PINEAPPLE JUICE FROM CONCENTRATE WITH ADDED INGREDIENTS, PINEAPPLE, UPC: 036800294264 contain(s) 54 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Sand weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Ferrocene [(C5H5)2Fe] weighs 1 490 kg/m³ (93.01766 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Biodiesel with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
A microgram per cubic millimeter (µg/mm³) is a derived metric SI (System International) measurement unit of density used to measure volume in cubic millimeters in order to estimate weight or mass in micrograms
The online number conversions include conversions between binary, octal, decimal and hexadecimal numbers.
lb/in³ to short tn/metric tbsp conversion table, lb/in³ to short tn/metric tbsp unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a hollow cylinder and its surface area | 3,536 | 7,426 | {"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} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.374925 |
https://www.physicsforums.com/threads/whats-the-vector-theta.819600/ | 1,508,254,279,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00729.warc.gz | 957,485,354 | 16,473 | # What's the vector theta?
1. Jun 18, 2015
### Bruno Tolentino
If the vector r is (x,y), so, what is the vector θ? BY THE WAY is (y,-x) ?
Last edited: Jun 18, 2015
2. Jun 18, 2015
### HallsofIvy
Staff Emeritus
I have no idea what you mean by "the vector $\theta$". Could you please explain that? Where did you see a reference to a "vector $\theta$"?
3. Jun 18, 2015
### Fredrik
Staff Emeritus
The question "is (y,-x)?" doesn't make sense either.
4. Jun 18, 2015
### HallsofIvy
Staff Emeritus
I didn't notice the "(y, -x)"! If a vector is given by r= (x, y) then its length is |r|= $\sqrt{x^2+ y^2}$ and the angle it makes with the x-axis, it that is what you mean by "$\theta$", is given by $arctan(y/x)$ as long as x is not 0, $\pi/2$ if x= 0 and y is positive, $3\pi/2$ if x= 0 and y is negative.
Given a vector (x, y), the vector (y, -x) is the result of rotating (x, y) through an angle of $pi/2$ radians.
5. Jun 18, 2015
### WWGD
The angle $\theta$ depends on your frame of reference : the positive x-axis does not have to represent the angle $0$ , it can represent anything as long as the choices are made consistently, i.e., the angle with the negative x-axis must be $\pi$ larger than the choice on the positive x-axis, as is done, e.g., with branches of the Complex logarithm.
6. Jun 18, 2015
### HallsofIvy
Staff Emeritus
Well, that depends on exactly what Bruno Tolentino means by "the vector $\theta$". I asked that earlier and he still hasn't answered.
7. Jun 18, 2015
### WWGD
Another possible interpretation is that $\theta(x,y)=(-y,x)$ is a _vector field_ , assigning to each point/tangent space at $(x,y)$, the
vector $(-y,x)$.
8. Jun 18, 2015
### Bruno Tolentino
The ideia of vector θ come from following: if the vector dr is the tangent vector to parametric curve and the o vector dn is the normal vector:
And if the UNIT vector r^ is normal to UNIT vector θ^:
So: the vector dn = dθ and therefore θ = (-y,x)!?
EDIT: but confront with the following: if θ = (-y,x), so θ = (- r sin(θ), r cos(θ)) = r (- sin(θ), cos(θ)) = r θ^
Is known that the vector r = r r^
But, is correct to affirm that: θ = r θ^?
The vector θ wouldn't: θ = θ θ^
Last edited: Jun 18, 2015
9. Jun 19, 2015
### Fredrik
Staff Emeritus
In this context (polar coordinates), there's no standard definition of the notation θ. It's definitely non-standard to use that notation for the vector $r\hat{\theta}$.
I think I see what you were thinking now: Since $\hat{\mathbf r}$ is a normalized version of $\mathbf r$, it makes sense to ask if there's a vector that you can normalize to get $\hat\theta$. There is, but it's not denoted by θ.
The vectors $\hat{\mathbf r}$ and $\hat\theta$ are defined as what you get when you normalize $\frac{\partial\mathbf r}{\partial r}$ and $\frac{\partial\mathbf r}{\partial\theta}$.
\begin{align*}
&\mathbf r=(x,y)=(r\cos\theta,r\sin\theta)\\
&\hat{\mathbf r}=\frac{\frac{\partial\mathbf r}{\partial r}}{\left|\frac{\partial\mathbf r}{\partial r}\right|} =\frac{(\cos\theta,\sin\theta)}{1} =\frac{\mathbf r}{r}=\frac{(x,y)}{\sqrt{x^2+y^2}}\\
&\hat{\theta} =\frac{\frac{\partial\mathbf r}{\partial\theta}}{\left|\frac{\partial\mathbf r}{\partial\theta}\right|} =\frac{(-r\sin\theta,r\cos\theta)}{r} =\frac{(-y,x)}{\sqrt{x^2+y^2}}
\end{align*}
Last edited: Jun 19, 2015 | 1,076 | 3,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 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-2017-43 | longest | en | 0.909531 |
https://questions.llc/questions/1141839 | 1,716,296,066,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00126.warc.gz | 419,180,309 | 5,536 | # ``If 18g of a radioactive substance are present initially and 5 yr later only 9 g remain, how much of the substance will be present after 11yr ?
after 11yr there will be ___g of a radioactive substance.
round the final answer to the nearest thousandth as needed. round all intermediate values to the nearest thousandth as needed.''
## half life is 5 years
amount remaining= 18*e^(-.692*11/5)
Put this in the google search window
18*e^(-.692*11/5)=
## To find out how much of the radioactive substance remains after 11 years, we can use the formula for exponential decay:
A = A0 * (1/2)^(t/h)
Where:
A = the final amount of the substance remaining after time t
A0 = the initial amount of the substance
t = time elapsed
h = half-life of the substance
In this case, the initial amount A0 is 18g, 5 years have passed, and we know that 9g remain. However, we need to determine the half-life of the substance to use in the formula.
To find the half-life, we can use the formula:
A = A0 * (1/2)^(t/h)
9 = 18 * (1/2)^(5/h)
Since we are rounding intermediate values to the nearest thousandth, we can use trial and error or a calculator to solve for h.
By trial and error, let's assume the half-life (h) is 10 years:
9 = 18 * (1/2)^(5/10)
9 = 18 * (1/2)^(1/2)
9 = 18 * (√(1/2))
9 = 18 * 0.70710678118
9 = 12.72831606764
Since 12.72831606764 is not equal to 9, the half-life is not 10 years. We can further refine our calculation by trying a lower half-life value.
Let's try a half-life of 6 years:
9 = 18 * (1/2)^(5/6)
9 = 18 * (0.82287565553)
9 = 14.81176279854
Since 14.81176279854 is not equal to 9, we can conclude that the half-life is greater than 6 years.
Let's continue narrowing down the half-life value:
Trial 1: h = 7 years
9 = 18 * (1/2)^(5/7)
9 = 18 * (0.84589170773)
9 = 15.22605173894
Trial 2: h = 8 years
9 = 18 * (1/2)^(5/8)
9 = 18 * (0.86602540378)
9 = 15.58845977804
By trial and error, we can conclude that the half-life is approximately 8 years (rounded to the nearest whole number).
Now that we know the half-life is 8 years, we can substitute the values into the formula to find the amount of the substance that will be present after 11 years (t = 11):
A = A0 * (1/2)^(t/h)
A = 18 * (1/2)^(11/8)
Calculating this value:
A = 18 * (0.31622776602)
A = 5.69289978835
Rounding the final answer to the nearest thousandth, we find that after 11 years, approximately 5.693 g of the radioactive substance will remain. | 770 | 2,448 | {"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} | 4.28125 | 4 | CC-MAIN-2024-22 | latest | en | 0.887255 |
http://www.expertsmind.com/library/what-is-papier-mache-corp-dollar-value-of-cost-52280349.aspx | 1,624,036,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00105.warc.gz | 60,215,953 | 11,765 | What is papier mache corp dollar value of cost
Assignment Help Accounting Basics
Reference no: EM132280349
Question - Papier Mache Corp. uses the periodic inventory system, and implements LIFO. On March 1, it had a beginning inventory of 700 reams of paper at a cost of \$10 per ream.
During the month, it made the following purchases of paper:
• March 3: 800 reams at \$12 per ream
• March 12: 600 reams at \$13 per ream
• March 16: 1,500 reams at \$15 per ream
During the month, all of its sales were made to the Stan Still Stationery Store (which was fully paid for), as follows:
• March 4: 950 reams at \$28 per ream
• March 14: 300 reams at \$31 per ream
• March 19: 1,425 reams at \$25 per ream
Questions:
1. For the month of March, what is Papier Mache Corp.'s dollar value of cost of goods sold and ending inventory it would report on its financial statements? Round your answers to the nearest dollar.
2. On April 1, the Stan Still Stationery Store (not Papier Mache Corp.) sells 625 reams of paper at \$42 per ream to its customer. Assuming it has no other inventory other than the paper purchased from Papier Mache Corp., and it uses the average cost method, what is its cost of goods sold and gross profit it would report for this sale? Round the average cost per ream to the nearest cent.
Compute the organization program services ratio
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Market price and related interest expense
Determine the market price and related interest expense of an \$800,000, ten-year, 10% (pays interest semiannually) bond issue sold to yield an effective rate of 12%.
State and destroyed so that a new football stadium
1) Bob's building was taken over by the state and destroyed so that a new football stadium could be built. Bob's basis in his building was \$400,000. The state paid him \$1,25
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If the standard deviation of demand is six per week, demand is 50 per week, and the desired service level is 95%, approximately what is the statistical safety stock?
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Outline the provisions of the Worker Adjustment and Retraining Notification Ace (WARN) of 1988 stating its rules, liabilities, and loopholes. Discuss the reasons why this ac | 686 | 3,010 | {"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-2021-25 | latest | en | 0.960233 |
https://www.teachoo.com/19017/4114/Question-17/category/CBSE-Class-10-Sample-Paper-for-2023-Boards---Maths-Standard/ | 1,679,430,612,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00322.warc.gz | 1,136,055,539 | 35,454 | CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard
Class 10
Solutions of Sample Papers for Class 10 Boards
## (a)1/6 (b) 7/36 (c) 11/36 (d) 13/36
This question is Similar to Question 27 CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
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### Transcript
Question 17 Two dice are rolled simultaneously. What is the probability that 6 will come up at least once? (a)1/6 (b) 7/36 (c) 11/36 (d) 13/36 First dice Total Number of outcomes = 36 Number of outcomes where 6 will come up atleast once = 11 Probability that 6 will come atleast once = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑤ℎ𝑒𝑟𝑒 6 𝑤𝑖𝑙𝑙 𝑐𝑜𝑚𝑒 𝑎𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑐𝑒)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠) = 𝟏𝟏/𝟑𝟔 So, correct answer is (c) | 402 | 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} | 3.78125 | 4 | CC-MAIN-2023-14 | longest | en | 0.797173 |
https://financialoccultist.com/tag/econometrics/ | 1,670,393,598,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711150.61/warc/CC-MAIN-20221207053157-20221207083157-00096.warc.gz | 286,270,633 | 33,519 | ## A Review: “Two Centuries of Trend Following”
The paper “Two Centuries of Trend Following” by Lemperiere, Derenble, Seager, et al of Capital Fund Management purports to show that trend following has been profitable, over a wide range of markets, consistently over 200 years. It deserves to be reviewed as it represents a case study of the statistical practices, and armchair explanations that are sometimes used to justify a system that in the most recent five year period has lost its mojo. Rocky has asked me to review it.
The amazing thing is that the authors seem to know how to compute hyperbolic tangent regressions, and compute the duration of a drawdown given a sharpe ratio, yet they seem completely unaware of the problem of multicollinearity, overlapping observations, and lack of independent observations.
In a nutshell, they compute hundreds of thousands of means, and they combine them and measure how far away from randomness they are. Recall that the average of two random observations is about 0.7 times as variable as one observations. The average of 100,000 observations is about 1/320 as variable as 1 observation. (more…)
• Preparation: If you put yourself in the best possible position and you lose money at least you spent that money wisely. Good things happen to those that are prepared because 90% of people do not know how to do it or are unwilling.
• Purpose: Acting with purpose. You prepared, you knew the risks, you executed the way you wanted to execute. In cold blooded evaluation you would do it the same with the information you had at the time.
• Protection: Losing the invisible money is how I have seen many people blow up. Invisible money is not locking in profits or losing more than your plan allowed. If you lose what you intended to risk you own the trade, if you lose more the trade owns you.
Your goal as a trader is to always reduce the time it takes to analyze, react, and recover. The best traders do this effortlessly after much thought, experiment, and practice. I lacked confidence because I thought about the wrong things or not at all and I was doing random things all of which made it too costly, emotionally and financially, to practice.
## Constructing Diversified Futures Trading Strategies
• Once you reach a few million under management, hiring a research staff to improve details is a good idea.
• Wait for momentum to build in one direction and get on the bandwagon. Expect to lose about two thirds of the time and so make sure your winners can pay for the losers and leave enough over to cover the rent.
• Using a single strategy on a single instrument is for people with either extreme skill or for those who simply have a death wish
• If we put the same notional dollar amount in each trade the portfolio would immediately be dominated by the volatile instruments and not much impact at all would come from the less volatile.
• Trend following: Buying high and selling higher
• Non professionals tend to spend an excess of time and energy on the buy and sell rules and neglect diversification and risk
## 5 Points for Discretionary Traders
1) A discipline of pre-market preparation: All emphasize the importance of process and preparation: sticking to what you do best and being prepared for fresh opportunity–and threat–each market day.
2) Selectivity: All have some methods for screening stocks and focusing on a core group that offer opportunity. Often, these screens focus on stocks that are trading actively, that show good movement, and that are setting up for directional price moves because of earnings reports, breakout patterns, etc.
3) Patience: This follows from the first two. The experienced traders emphasize risk management and waiting for high quality trades, rather than overtrading. All stress understanding the current market environment and adapting to it.
4) Diversification: These traders don’t focus on one or two opportunities, but look at a range of promising shares and setups and trade more than one thing at a time. All the proverbial eggs are not in one basket.
5) Simplicity: My sense is that the traders are focused on understanding what is happening now, not predicting what will happen in the future. If I had to guess, I’d say that they are talented in detecting the flow of activity in and out of shares and are riding moves as they are getting under way. They don’t appear to be researching deep value and holding for long periods to wait for that value to be realized.
Go to top | 966 | 4,521 | {"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-2022-49 | latest | en | 0.960304 |
https://www.kelleysbookkeeping.com/how-to-calculate-safety-stock-safety-stock-formula/ | 1,709,409,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00847.warc.gz | 810,491,834 | 10,794 | # How To Calculate Safety Stock? Safety Stock Formula And Calculation
Content
It prevents you from running out of stock, especially on particularly popular items. Read on to see why safety stock is necessary and how to get it right for your business. All your products, customers, orders and transactions synced and secure in the cloud. Cloud Fulfillment Platform Logiwa has built a fully integrated WMS and cloud order fulfillment software solution for B2C and DTC businesses. The money you save likely won’t compensate for the money you can lose from dissatisfied customers who stop frequenting your business. When it comes to calculating safety inventory, it doesn’t pay to go with your gut.
In addition, businesses should also know their desired service level. In most B2B relationships, this is formalized in service level agreements. For B2C companies, senior management determines this number by balancing the cost of stock-outs or unfulfilled promises against the cost of losing customers.In this case, the safety stock is calculated considering the demand and supply variability risks during this period plus the replenishment lead time. In this case, safety stock is calculated considering the risk of only the replenishment lead time.Calculating safety stock plays a crucial role in preventing stockouts, which causes lost revenue and unhappy customers. If demand for your product fluctuates but your lead time is relatively predictable, this can be a helpful method. As a business owner, one of your biggest nightmares is missing out on sales because of stockouts.
## Limits Of The Normal Distribution For Your Safety Stock :
Positive numbers are the number of days over the expected time and negative numbers mean that the delivery arrived earlier than the expected time. With this information, we can find the standard deviation in lead time. The simplest method for calculating safety stock only requires a four-step process to calculate these variables.
• The definition of standard deviation is a quantity calculated to indicate the extent of deviation for a group as a whole.
• Lead time is the delay between the time the reorder point (inventory level which initiates an order) is reached and renewed availability.
• To calculate the standard deviation, you need your expected lead time and the actual time the shipments took.
• These are the terms you’ll need to know to calculate safety stock.
• Safety stock is inventory that a business holds to mitigate the risk of shortages or stockouts.
• Too little safety stock can result in lost sales and, in the thus a higher rate of customer turnover.
At this point, you haven’t been in operation long enough to accurately calculate supply and demand. Overestimating how much you’ll need could cripple your finances. This formula is a little more complicated, but it’s very useful. The main issue is that only established businesses will have the statistics necessary to get accurate results. This formula also assumes that lead time varies, but demand doesn’t, which isn’t always true. The majority of products have a service level of 85%-90%, with crucial goods going to 95% and more. But it’s hard to measure how well items are going to sell, and impossible to predict when things won’t get shipped on time.
## How much safety stock should be built into the forecast?
Many manufacturers have a lead time of 60 days, and most would consider 30 days of safety stock an ideal level of inventory. Given these two inputs, a minimum of 70% forecast accuracy is needed to ensure the business doesn’t run out of inventory.There are pros and cons of selecting a statistical-based safety stock. Although the statistical method is based on accurate mathematics, predicting business is not always as accurate. The statistical method to calculate safety stock is based on the premise that is possible to mathematically calculate the level of safety stock to prevent a stock-out situation. Remember, no inventory analysis is perfect, and the real world is more complex than an equation. That said, incorporating the safety stock formula can give you a leg up on your competition when demand spikes and provide a much-needed insurance policy against stockouts.
## Buffer Inventory Meaning: Define Buffer Inventory
As in this case, variability can impact sales and vice versa, even more, safety stock is needed. If you consider that the request and the lead time are dependent, i. That the lead time causes uncertainty on the request and vice versa, you have the following formula which will sum the safety stock of Formula 1 and Formula 2, giving 377 quantities in our example. Safety stock is one of the easiest ways to avoid running out of stock and not fulfilling orders. Without it, businesses would lose sales and, ultimately, customers. Safety stock is a valuable tool to combat stockouts, but it can have some disadvantages.It’s important to remember that the sales don’t only have to cover the sold products’ own carrying costs, but the carrying costs of the safety stock as well. The optimal level depends on several factors, including inventory velocity, current and future demand, sales volume and supplier lead times. As a rule of thumb, the safety stock amount should be the amount of inventory used per day multiplied by the lead time in days. Should extreme cases have an impact on stock and sales, there’s a risk that decision makers may not trust the safety stock formulas at all and strive for high service levels. As we have seen, a service level of 100% would mean having infinite stock and is not a financially viable or safe option.
Not only does this cut into potential revenue, but it can also sever ties with loyal customers you’ve worked hard to build relationships with. That person is now either going to buy from a competitor or not buy at all. It’s not a replacement for regular inventory; it’s simply a backup. It gives you peace of mind and operational flexibility when production halts, accidents happen, or customers buy more out of nowhere.If you have deliveries arriving earlier or later than expected, a safety stock formula will help you to cover unexpected delays and demand fluctuation to maintain a consistent output. The “out of stock” situation is beneficial to companies when the company has no demand for certain items, and zero inventory means no warehouse costs. Cycle stock is the amount of inventory that is planned to be sold for a certain period. This could mean the time period between orders or how long a production cycle takes.
## Time Period Based
It also affects you inventory carrying cost, so it’s important to control. The problem often is that we will put a safety threshold of 1000 quantities, for example, and sales will change but not the safety stock. It is, therefore, necessary to have dynamic formulas that evolve. The most important thing already will be the quality of the data, which is even more important than the method chosen in terms of sales and lead times.
This is important even for businesses who participate in a consignment inventory model or other non-traditional partnerships. This is basically all there is to know about how to calculate safety stock. Your stock holding formula will help your business navigate safely through all the fluctuations in demand and lead time. We come to put this famous safety stock to come and “hide” our problems. As another example, a bicycle manufacturer was recently featured in a popular biking magazine and experienced a sharp uptick in orders. As the manufacturer fulfills orders, inventory is depleted faster than the average lead time, increasing the risk of a complete stockout. | 1,489 | 7,672 | {"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-2024-10 | longest | en | 0.9348 |
https://www.instructables.com/id/Sparkling-LED-Ganesha/ | 1,566,120,410,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00096.warc.gz | 868,302,503 | 21,746 | # Sparkling LED Ganesha
487
2
2
This is the season of festivals in India and Lord Ganesha is one of the most revered and adorable God, especially for children. He is the first God to be worshipped in all religious ceremonies.
What better way to enjoy the festive season than assembling an illuminated Ganesha? Hope it sparks the creativity of electronics enthusiasts.
In this project, we will join LEDs in parallel connection by soldering to make a wonderful display.
Soldering is a very crucial 21st-century skill which must be mastered by every individual. Parallel connection means the positive terminal of all the LEDs will be joined together and all the negative terminals will also be joined together with the help of soldering using connecting wire. If one of the LEDs is not working or not connected properly, this will not affect the whole circuit. Remaining LEDs will keep glowing. The circuit is powered by two pencil cells (3V).
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Material Needed
We need the following:
One Vinyl board or cardboard.
LEDs: 10-15
Cell holder and two pencil cells
Toolkit containing soldering iron (25 watts)
Solder wire
Wire stripper
Ribbon wire
Take half cardboard and draw Lord Ganesha's figure.
After drawing the figure, mark the dots on places where you want to place the LEDs.
## Step 2: Inserting LED in the Circuit
Make two holes at the position where an LED is to be placed with the help of tweezer.
Insert the LEDs into these holes.
## Step 3: Labelling the Circuit
Turn the board and bend the legs of LED.
Mark the longer leg as ‘+’ and shorter leg as ‘-‘.
## Step 4: Placing LEDs on the Board
Likewise, place all the LEDs on the board at the marked places.
## Step 5: Soldering
Turn on the iron.
Clean the iron bit and let it heat.
Measure the length of the ribbon wire needed to join positives of the two nearby LEDs.
## Step 6: Preparing the Wire
There are three steps involved in preparing the wire before soldering.
Cut
Peel
Twist
Measure the length of insulating wire to ensure that the legs of LEDs are connected comfortably.
They should not stretch.
## Step 7: Peeling Insulation
Peel 1 cm insulation from both ends of the wire using wire stripper.
## Step 8: Time to Twist
Twist the uninsulated ends of the wire with the help of your fingers, so that no strand is protruding out from the tip of the wire.
Your wire is ready to be soldered now.
## Step 9: Wrapping Around the Wire
Wrap the uninsulated part of wire on the positive lead of the LED.
You need two wires to be connected to the LED’s leg, so wrap both the wires simultaneously and fix it on the lead.
## Step 10: Time to Solder
Check the soldering iron using solder wire.
If the wire melts, the iron is ready.
Hold the iron as you hold a pen or pencil in your active hand and solder wire in the other and solder the connection.
## Step 11: Joining the Positives
Likewise, join all the positive leads of the LED together.
Start the same procedure and join all the negative leads of the LEDs.
Now, we need to connect all the LEDs to two pencil cells through cell holder.
Soldering on cell holder is very critical to the circuit and needs a lot of skill.
## Step 12: Solder the Cell Holder
Here are steps to solder cell holder.
A cell holder has two metallic edges at one end. Hold the cell holder such that metallic edges face upside.
Put some solder on these metallic ends.
Be careful that hot solder iron should not touch the plastic body of the cell holder.
## Step 13: Tinning
Take two connecting wires preferably of different colors.
Prepare them for soldering (cut, peel, wrap).
Apply some solder wire on the uninsulated part of the connecting wire using hot iron.
This process is called tinning.
## Step 14: Joining the Wires
Now keep this wire on one of the metallic terminal of the cell holder.
Using soldering iron, join the wire to the terminal.
Similarly, join the other wire to the other terminal of the cell holder.
## Step 15:
The terminal of the cell holder with the spring attached, is the negative terminal.
We have joined grey wire to the negative terminal.
So grey wire is negative and red wire is positive.
## Step 16: Empowering the Circuit
Connect the positive terminal wire (RED) of the cell holder to the positive leg of any of the LED.
Connect the negative terminal (spring side) wire of the cell holder to the negative lead of any of the LED.
## Step 17: Fixing the Wires
Fix all the wires on the LED board using insulating tape to prevent them from entanglement.
Insert cells in the cell holder (Flat side of the cell to the spring side of the holder).
## Step 18: Time to Glow and Shine
All the LED will start glowing.
Isn’t it cool !!!!
## Recommendations
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## 2 Discussions
That looks really pretty :) | 1,129 | 4,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-35 | latest | en | 0.915442 |
https://www.cubrid.org/manual/en/10.0/sql/function/condition_op.html | 1,512,985,010,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513330.14/warc/CC-MAIN-20171211090353-20171211110353-00314.warc.gz | 729,252,500 | 28,535 | # Comparison Expression¶
## Simple Comparison Expression¶
A comparison expression is an expression that is included in the WHERE clause of the SELECT, UPDATE and DELETE statements, and in the HAVING clause of the SELECT statement. There are simple comparison, ANY / SOME / ALL, BETWEEN, EXISTS, IN / NOT IN, LIKE and IS NULL comparison expressions, depending on the kinds of the operators combined.
A simple comparison expression compares two comparable data values. Expressions or subqueries are specified as operands, and the comparison expression always returns NULL if one of the operands is NULL. The following table shows operators that can be used in the simple comparison expressions. For details, see Comparison Operators.
Comparison Operators
Comparison Operator Description Comparison Expression Return Value
= A value of left operand is the same as that of right operand. 1=2 0
<> , != A value of left operand is not the same as that of right operand. 1<>2 1
> A value of left operand is greater than that of right operand. 1>2 0
< A value of left operand is less than that of right operand. 1<2 1
>= A value of left operand is equal to or greater than that of right operand. 1>=2 0
<= A value of left operand is equal to or less than that of right operand. 1<=2 1
## ANY/SOME/ALL quantifiers¶
A comparison expression that includes quantifiers such as ANY/SOME/ALL performs comparison operation on one data value and on some or all values included in the list. A comparison expression that includes ANY or SOME returns TRUE if the value of the data on the left satisfies simple comparison with at least one of the values in the list specified as an operand on the right. A comparison expression that includes ALL returns TRUE if the value of the data on the left satisfies simple comparison with all values in the list on the right.
When a comparison operation is performed on NULL in a comparison expression that includes ANY or SOME, UNKNOWN or TRUE is returned as a result; when a comparison operation is performed on NULL in a comparison expression that includes ALL, UNKNOWN or FALSE is returned.
```expression comp_op SOME expression
expression comp_op ANY expression
expression comp_op ALL expression
```
• comp_op : A comparison operator >, = or <= can be used.
• expression (left): A single-value column, path expression (ex.: tbl_name.col_name), constant value or arithmetic function that produces a single value can be used.
• expression (right): A column name, path expression, list (set) of constant values or subquery can be used. A list is a set represented within braces ({}). If a subquery is used, expression (left) and comparison operation on all results of the subquery execution is performed.
```--creating a table
CREATE TABLE condition_tbl (id int primary key, name char(10), dept_name VARCHAR, salary INT);
INSERT INTO condition_tbl VALUES(1, 'Kim', 'devel', 4000000);
INSERT INTO condition_tbl VALUES(2, 'Moy', 'sales', 3000000);
INSERT INTO condition_tbl VALUES(3, 'Jones', 'sales', 5400000);
INSERT INTO condition_tbl VALUES(4, 'Smith', 'devel', 5500000);
INSERT INTO condition_tbl VALUES(5, 'Kim', 'account', 3800000);
INSERT INTO condition_tbl VALUES(6, 'Smith', 'devel', 2400000);
INSERT INTO condition_tbl VALUES(7, 'Brown', 'account', NULL);
--selecting rows where department is sales or devel
SELECT * FROM condition_tbl WHERE dept_name = ANY{'devel','sales'};
```
``` id name dept_name salary
======================================================================
1 'Kim ' 'devel' 4000000
2 'Moy ' 'sales' 3000000
3 'Jones ' 'sales' 5400000
4 'Smith ' 'devel' 5500000
6 'Smith ' 'devel' 2400000
```
```--selecting rows comparing NULL value in the ALL group conditions
SELECT * FROM condition_tbl WHERE salary > ALL{3000000, 4000000, NULL};
```
```There are no results.
```
```--selecting rows comparing NULL value in the ANY group conditions
SELECT * FROM condition_tbl WHERE salary > ANY{3000000, 4000000, NULL};
```
``` id name dept_name salary
======================================================================
1 'Kim ' 'devel' 4000000
3 'Jones ' 'sales' 5400000
4 'Smith ' 'devel' 5500000
5 'Kim ' 'account' 3800000
```
```--selecting rows where salary*0.9 is less than those salary in devel department
SELECT * FROM condition_tbl WHERE (
(0.9 * salary) < ALL (SELECT salary FROM condition_tbl
WHERE dept_name = 'devel')
);
```
``` id name dept_name salary
======================================================================
6 'Smith ' 'devel' 2400000
```
## BETWEEN¶
The BETWEEN makes a comparison to determine whether the data value on the left exists between two data values specified on the right. It returns TRUE even when the data value on the left is the same as a boundary value of the comparison target range. If NOT comes before the BETWEEN keyword, the result of a NOT operation on the result of the BETWEEN operation is returned.
i BETWEEN g AND m and the compound condition i >= g AND i <= m have the same effect.
```expression [ NOT ] BETWEEN expression AND expression
```
• expression : A column name, path expression (ex.: tbl_name.col_name), constant value, arithmetic expression or aggregate function can be used. For a character string expression, the conditions are evaluated in alphabetical order. If NULL is specified for at least one of the expressions, the BETWEEN predicate returns UNKNOWN as a result.
```--selecting rows where 3000000 <= salary <= 4000000
SELECT * FROM condition_tbl WHERE salary BETWEEN 3000000 AND 4000000;
SELECT * FROM condition_tbl WHERE (salary >= 3000000) AND (salary <= 4000000);
```
``` id name dept_name salary
======================================================================
1 'Kim ' 'devel' 4000000
2 'Moy ' 'sales' 3000000
5 'Kim ' 'account' 3800000
```
```--selecting rows where salary < 3000000 or salary > 4000000
SELECT * FROM condition_tbl WHERE salary NOT BETWEEN 3000000 AND 4000000;
```
``` id name dept_name salary
======================================================================
3 'Jones ' 'sales' 5400000
4 'Smith ' 'devel' 5500000
6 'Smith ' 'devel' 2400000
```
```--selecting rows where name starts from A to E
SELECT * FROM condition_tbl WHERE name BETWEEN 'A' AND 'E';
```
``` id name dept_name salary
======================================================================
7 'Brown ' 'account' NULL
```
## EXISTS¶
The EXISTS returns TRUE if one or more results of the execution of the subquery specified on the right exist, and returns FALSE if the result of the operation is an empty set.
```EXISTS expression
```
• expression : Specifies a subquery and compares to determine whether the result of the subquery execution exists. If the subquery does not produce any result, the result of the conditional expression is FALSE.
```--selecting rows using EXISTS and subquery
SELECT 'raise' FROM db_root WHERE EXISTS(
SELECT * FROM condition_tbl WHERE salary < 2500000);
```
``` 'raise'
======================
'raise'
```
```--selecting rows using NOT EXISTS and subquery
SELECT 'raise' FROM db_root WHERE NOT EXISTS(
SELECT * FROM condition_tbl WHERE salary < 2500000);
```
```There are no results.
```
## IN¶
The IN compares to determine whether the single data value on the left is included in the list specified on the right. That is, the predicate returns TRUE if the single data value on the left is an element of the expression specified on the right. If NOT comes before the IN keyword, the result of a NOT operation on the result of the IN operation is returned.
```expression [ NOT ] IN expression
```
• expression (left): A single-value column, path expression (ex.: tbl_name.col_name), constant value or arithmetic function that produces a single value can be used.
• expression (right): A column name, path expression, list (set) of constant values or subquery can be used. A list is a set represented within parentheses (()) or braces ({}). If a subquery is used, comparison with expression(left) is performed for all results of the subquery execution.
```--selecting rows where department is sales or devel
SELECT * FROM condition_tbl WHERE dept_name IN {'devel','sales'};
SELECT * FROM condition_tbl WHERE dept_name = ANY{'devel','sales'};
```
``` id name dept_name salary
======================================================================
1 'Kim ' 'devel' 4000000
2 'Moy ' 'sales' 3000000
3 'Jones ' 'sales' 5400000
4 'Smith ' 'devel' 5500000
6 'Smith ' 'devel' 2400000
```
```--selecting rows where department is neither sales nor devel
SELECT * FROM condition_tbl WHERE dept_name NOT IN {'devel','sales'};
```
``` id name dept_name salary
======================================================================
5 'Kim ' 'account' 3800000
7 'Brown ' 'account' NULL
```
## IS NULL¶
The IS NULL compares to determine whether the expression specified on the left is NULL, and if it is NULL, returns TRUE and it can be used in the conditional expression. If NOT comes before the NULL keyword, the result of a NOT operation on the result of the IS NULL operation is returned.
expression IS [ NOT ] NULL
• expression : A single-value column, path expression (ex.: tbl_name.col_name), constant value or arithmetic function that produces a single value can be used.
```--selecting rows where salary is NULL
SELECT * FROM condition_tbl WHERE salary IS NULL;
```
``` id name dept_name salary
======================================================================
7 'Brown ' 'account' NULL
```
```--selecting rows where salary is NOT NULL
SELECT * FROM condition_tbl WHERE salary IS NOT NULL;
```
``` id name dept_name salary
======================================================================
1 'Kim ' 'devel' 4000000
2 'Moy ' 'sales' 3000000
3 'Jones ' 'sales' 5400000
4 'Smith ' 'devel' 5500000
5 'Kim ' 'account' 3800000
6 'Smith ' 'devel' 2400000
```
```--simple comparison operation returns NULL when operand is NULL
SELECT * FROM condition_tbl WHERE salary = NULL;
```
```There are no results.
```
## LIKE¶
The LIKE compares patterns between character string data, and returns TRUE if a character string whose pattern matches the search word is found. Pattern comparison target types are CHAR, VARCHAR and STRING. The LIKE search cannot be performed on an BIT type. If NOT comes before the LIKE keyword, the result of a NOT operation on the result of the LIKE operation is returned.
A wild card string corresponding to any character or character string can be included in the search word on the right of the LIKE operator. % (percent) and _ (underscore) can be used. .% corresponds to any character string whose length is 0 or greater, and _ corresponds to one character. An escape character is a character that is used to search for a wild card character itself, and can be specified by the user as another character (NULL, alphabet, or number whose length is 1. See below for an example of using a character string that includes wild card or escape characters.
```expression [ NOT ] LIKE pattern [ ESCAPE char ]
```
• expression: Specifies the data type column of the character string. Pattern comparison, which is case-sensitive, starts from the first character of the column.
• pattern: Enters the search word. A character string with a length of 0 or greater is required. Wild card characters (% or _) can be included as the pattern of the search word. The length of the character string is 0 or greater.
• ESCAPE char : NULL, alphabet, or number is allowed for char. If the string pattern of the search word includes "_" or "%" itself, an ESCAPE character must be specified. For example, if you want to search for the character string "10%" after specifying backslash (\) as the ESCAPE character, you must specify "10%" for pattern. If you want to search for the character string "C:\", you can specify "C:\" for pattern.
For details about character sets supported in CUBRID, see Character Strings.
Whether to detect the escape characters of the LIKE conditional expression is determined depending on the configuration of no_backslash_escapes and require_like_escape_character in the cubrid.conf file. For details, see Statement/Type-Related Parameters.
Note
• To execute string comparison operation for data entered in the multibyte charset environment such as UTF-8, the parameter setting (single_byte_compare = yes) which compares strings by 1 byte should be added to the cubrid.conf file for a successful search result.
• Versions after CUBRID 9.0 support Unicode charset, so the single_byte_compare parameter is no longer used.
```--selection rows where name contains lower case 's', not upper case
SELECT * FROM condition_tbl WHERE name LIKE '%s%';
```
``` id name dept_name salary
======================================================================
3 'Jones ' 'sales' 5400000
```
```--selection rows where second letter is 'O' or 'o'
SELECT * FROM condition_tbl WHERE UPPER(name) LIKE '_O%';
```
``` id name dept_name salary
======================================================================
2 'Moy ' 'sales' 3000000
3 'Jones ' 'sales' 5400000
```
```--selection rows where name is 3 characters
SELECT * FROM condition_tbl WHERE name LIKE '___';
```
``` id name dept_name salary
======================================================================
1 'Kim ' 'devel' 4000000
2 'Moy ' 'sales' 3000000
5 'Kim ' 'account' 3800000
```
## REGEXP, RLIKE¶
The REGEXP and RLIKE are used interchangeably; a regular expressions is a powerful way to specify a pattern for a complex search. CUBRID uses Henry Spencer's implementation of regular expressions, which conforms the POSIX 1003.2 standards. The details on regular expressions are not described in this page. For more information on regular expressions, see Henry Spencer's regex(7).
The following list describes basic characteristics of regular expressions.
• "." matches any single character(including new-line and carriage-return).
• "[...]" matches one of characters within square brackets. For example, "[abc]" matches "a", "b", or "c". To represent a range of characters, use a dash (-). "[a-z]" matches any alphabet letter whereas "[0-9]" matches any single number.
• "*" matches 0 or more instances of the thing proceeding it. For example, "xabc*" matches "xab", "xabc", "xabcc", and "xabcxabc" etc. "[0-9][0-9]*" matches any numbers, and ".*" matches every string.
• To match special characters such as "\n", "\t", "\r", and "\", some must be escaped with the backslash (\) by specifying the value of no_backslash_escapes (default: yes) to no. For details on no_backslash_escapes, see Escape Special Characters.
The difference between REGEXP and LIKE are as follows:
• The LIKE operator succeeds only if the pattern matches the entire value.
• The REGEXP operator succeeds if the pattern matches anywhere in the value. To match the entire value, you should use "^" at the beginning and "\$" at the end.
• The LIKE operator is case sensitive, but patterns of regular expressions in REGEXP is not case sensitive. To enable case sensitive, you should use REGEXP BINARY statement.
• REGEXP, REGEXP BINARY works as ASCII encoding without considering the collation of operands.
```SELECT ('a' collate utf8_en_ci REGEXP BINARY 'A' collate utf8_en_ci);
```
```0
```
```SELECT ('a' collate utf8_en_cs REGEXP BINARY 'A' collate utf8_en_cs);
```
```0
```
```SELECT ('a' COLLATE iso88591_bin REGEXP 'A' COLLATE iso88591_bin);
```
```1
```
```SELECT ('a' COLLATE iso88591_bin REGEXP BINARY 'A' COLLATE iso88591_bin);
```
```0
```
In the below syntax, if expression matches pattern, 1 is returned; otherwise, 0 is returned. If either expression or pattern is NULL, NULL is returned.
The second syntax has the same meaning as the third syntax, which both syntaxes are using NOT.
```expression REGEXP | RLIKE [BINARY] pattern
expression NOT REGEXP | RLIKE pattern
NOT (expression REGEXP | RLIKE pattern)
```
• expression : Column or input expression
• pattern : Pattern used in regular expressions; not case sensitive
```-- When REGEXP is used in SELECT list, enclosing this with parentheses is required.
-- But used in WHERE clause, no need parentheses.
-- case insensitive, except when used with BINARY.
SELECT name FROM athlete where name REGEXP '^[a-d]';
```
```name
======================
'Dziouba Irina'
'Dzieciol Iwona'
'Dzamalutdinov Kamil'
'Crucq Maurits'
'Crosta Daniele'
'Bukovec Brigita'
'Bukic Perica'
'Abdullayev Namik'
```
```-- \n : match a special character, when no_backslash_escapes=no
SELECT ('new\nline' REGEXP 'new
line');
```
```('new
line' regexp 'new
line')
=====================================
1
```
```-- ^ : match the beginning of a string
SELECT ('cubrid dbms' REGEXP '^cub');
```
```('cubrid dbms' regexp '^cub')
===============================
1
```
```-- \$ : match the end of a string
SELECT ('this is cubrid dbms' REGEXP 'dbms\$');
```
```('this is cubrid dbms' regexp 'dbms\$')
========================================
1
```
```--.: match any character
SELECT ('cubrid dbms' REGEXP '^c.*\$');
```
```('cubrid dbms' regexp '^c.*\$')
================================
1
```
```-- a+ : match any sequence of one or more a characters. case insensitive.
SELECT ('Aaaapricot' REGEXP '^A+pricot');
```
```('Aaaapricot' regexp '^A+pricot')
================================
1
```
```-- a? : match either zero or one a character.
SELECT ('Apricot' REGEXP '^Aa?pricot');
```
```('Apricot' regexp '^Aa?pricot')
==========================
1
```
```SELECT ('Aapricot' REGEXP '^Aa?pricot');
```
```('Aapricot' regexp '^Aa?pricot')
===========================
1
```
```SELECT ('Aaapricot' REGEXP '^Aa?pricot');
```
```('Aaapricot' regexp '^Aa?pricot')
============================
0
```
```-- (cub)* : match zero or more instances of the sequence abc.
SELECT ('cubcub' REGEXP '^(cub)*\$');
```
```('cubcub' regexp '^(cub)*\$')
==========================
1
```
```-- [a-dX], [^a-dX] : matches any character that is (or is not, if ^ is used) either a, b, c, d or X.
SELECT ('aXbc' REGEXP '^[a-dXYZ]+');
```
```('aXbc' regexp '^[a-dXYZ]+')
==============================
1
```
```SELECT ('strike' REGEXP '^[^a-dXYZ]+\$');
```
```('strike' regexp '^[^a-dXYZ]+\$')
================================
1
```
Note
The following shows RegEx-Specer's license, which is library used to implement the REGEXP conditional expression.
```Copyright 1992, 1993, 1994 Henry Spencer. All rights reserved.
This software is not subject to any license of the American Telephone
and Telegraph Company or of the Regents of the University of California.
Permission is granted to anyone to use this software for any purpose on
any computer system, and to alter it and redistribute it, subject
to the following restrictions:
1. The author is not responsible for the consequences of use of this
software, no matter how awful, even if they arise from flaws in it.
2. The origin of this software must not be misrepresented, either by
explicit claim or by omission. Since few users ever read sources,
credits must appear in the documentation.
3. Altered versions must be plainly marked as such, and must not be
misrepresented as being the original software. Since few users
ever read sources, credits must appear in the documentation.
4. This notice may not be removed or altered.
```
## CASE¶
The CASE expression uses the SQL statement to perform an IF ... THEN statement. When a result of comparison expression specified in a WHEN clause is true, a value specified in THEN clause is returned. A value specified in an ELSE clause is returned otherwise. If no ELSE clause exists, NULL is returned.
```CASE control_expression simple_when_list
[ else_clause ]
END
CASE searched_when_list
[ else_clause ]
END
simple_when :
WHEN expression THEN result
searched_when :
WHEN search_condition THEN result
else_clause :
ELSE result
result :
expression | NULL
```
The CASE expression must end with the END keyword. A control_expression argument and an expression argument in simple_when expression should be comparable data types. The data types of result specified in the THEN ... ELSE statement should all same, or they can be convertible to common data type.
The data type for a value returned by the CASE expression is determined based on the following rules.
• If data types for result specified in the THEN statement are all same, a value with the data type is returned.
• If data types can be convertible to common data type even though they are not all same, a value with the data type is returned.
• If any of values for result is a variable length string, a value data type is a variable length string. If values for result are all a fixed length string, the longest character string or bit string is returned.
• If any of values for result is an approximate numeric data type, a value with a numeric data type is returned. The number of digits after the decimal point is determined to display all significant figures.
```--creating a table
CREATE TABLE case_tbl( a INT);
INSERT INTO case_tbl VALUES (1);
INSERT INTO case_tbl VALUES (2);
INSERT INTO case_tbl VALUES (3);
INSERT INTO case_tbl VALUES (NULL);
--case operation with a search when clause
SELECT a,
CASE WHEN a=1 THEN 'one'
WHEN a=2 THEN 'two'
ELSE 'other'
END
FROM case_tbl;
```
``` a case when a=1 then 'one' when a=2 then 'two' else 'other' end
===================================
1 'one'
2 'two'
3 'other'
NULL 'other'
```
```--case operation with a simple when clause
SELECT a,
CASE a WHEN 1 THEN 'one'
WHEN 2 THEN 'two'
ELSE 'other'
END
FROM case_tbl;
```
``` a case a when 1 then 'one' when 2 then 'two' else 'other' end
===================================
1 'one'
2 'two'
3 'other'
NULL 'other'
```
```--result types are converted to a single type containing all of significant figures
SELECT a,
CASE WHEN a=1 THEN 1
WHEN a=2 THEN 1.2345
ELSE 1.234567890
END
FROM case_tbl;
```
``` a case when a=1 then 1 when a=2 then 1.2345 else 1.234567890 end
===================================
1 1.000000000
2 1.234500000
3 1.234567890
NULL 1.234567890
```
```--an error occurs when result types are not convertible
SELECT a,
CASE WHEN a=1 THEN 'one'
WHEN a=2 THEN 'two'
ELSE 1.2345
END
FROM case_tbl;
```
```ERROR: Cannot coerce 'one' to type double.
``` | 5,537 | 24,023 | {"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} | 2.921875 | 3 | CC-MAIN-2017-51 | latest | en | 0.884419 |
https://www.geeksforgeeks.org/preorder-successor-node-binary-tree/?ref=lbp | 1,660,928,293,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00387.warc.gz | 678,127,167 | 29,718 | # Preorder Successor of a Node in Binary Tree
• Difficulty Level : Easy
Given a binary tree and a node in the binary tree, find the preorder successor of the given node. It may be assumed that every node has a parent link.
Examples:
```Consider the following binary tree
20
/ \
10 26
/ \ / \
4 18 24 27
/ \
14 19
/ \
13 15
Input : 4
Output : 18
Preorder traversal of given tree is 20, 10, 4,
18, 14, 13, 15, 19, 26, 24, 27.
Input : 19
Output : 26```
A simple solution is to first store the Preorder traversal of the given tree in an array then linearly search the given node and print the node next to it.
Time Complexity: O(n), as we will traverse the tree for searching the node.
Auxiliary Space: O(n), as we need extra space for storing the elements of the tree.
An efficient solution is based on the below observations.
1. If the left child of a given node exists, then the left child is the preorder successor.
2. If the left child does not exist, however, the right child exists, then the preorder successor is the right child.
3. If the left child and the right child does not exist and given node is the left child of its parent, then its sibling is its preorder successor.
4. If none of the above conditions are satisfied (left child does not exist and given node is not left child of its parent), then we move up using parent pointers until one of the following happens.
• We reach the root. In this case, a preorder successor does not exist.
• The current node (one of the ancestors of the given node) is the left child of its parent, in this case, the preorder successor is a sibling of the current node.
## C++
`// CPP program to find preorder successor of``// a node in Binary Tree.``#include ``using` `namespace` `std;` `struct` `Node {`` ``struct` `Node *left, *right, *parent;`` ``int` `key;``};` `Node* newNode(``int` `key)``{`` ``Node* temp = ``new` `Node;`` ``temp->left = temp->right = temp->parent = NULL;`` ``temp->key = key;`` ``return` `temp;``}` `Node* preorderSuccessor(Node* root, Node* n)``{`` ``// If left child exists, then it is preorder`` ``// successor.`` ``if` `(n->left)`` ``return` `n->left;`` ` ` ``// If left child does not exist and right child`` ``// exists, then it is preorder successor.`` ``if` `(n->right)`` ``return` `n->right;`` ``// If left child does not exist, then`` ``// travel up (using parent pointers)`` ``// until we reach a node which is left`` ``// child of its parent.`` ``Node *curr = n, *parent = curr->parent;`` ``while` `(parent != NULL && parent->right == curr) {`` ``curr = curr->parent;`` ``parent = parent->parent;`` ``}` ` ``// If we reached root, then the given`` ``// node has no preorder successor`` ``if` `(parent == NULL)`` ``return` `NULL;` ` ``return` `parent->right;``}` `int` `main()``{`` ``Node* root = newNode(20);`` ``root->parent = NULL;`` ``root->left = newNode(10);`` ``root->left->parent = root;`` ``root->left->left = newNode(4);`` ``root->left->left->parent = root->left;`` ``root->left->right = newNode(18);`` ``root->left->right->parent = root->left;`` ``root->right = newNode(26);`` ``root->right->parent = root;`` ``root->right->left = newNode(24);`` ``root->right->left->parent = root->right;`` ``root->right->right = newNode(27);`` ``root->right->right->parent = root->right;`` ``root->left->right->left = newNode(14);`` ``root->left->right->left->parent = root->left->right;`` ``root->left->right->left->left = newNode(13);`` ``root->left->right->left->left->parent = root->left->right->left;`` ``root->left->right->left->right = newNode(15);`` ``root->left->right->left->right->parent = root->left->right->left;`` ``root->left->right->right = newNode(19);`` ``root->left->right->right->parent = root->left->right;` ` ``Node* res = preorderSuccessor(root, root->left->right->right);` ` ``if` `(res) {`` ``printf``(``"Preorder successor of %d is %d\n"``,`` ``root->left->right->right->key, res->key);`` ``}`` ``else` `{`` ``printf``(``"Preorder successor of %d is NULL\n"``,`` ``root->left->right->right->key);`` ``}` ` ``return` `0;``}`
## Java
`// Java program to find preorder successor of``// a node in Binary Tree.``class` `Solution``{` `static` `class` `Node``{`` ``Node left, right, parent;`` ``int` `key;``};` `static` `Node newNode(``int` `key)``{`` ``Node temp = ``new` `Node();`` ``temp.left = temp.right = temp.parent = ``null``;`` ``temp.key = key;`` ``return` `temp;``}` `static` `Node preorderSuccessor(Node root, Node n)``{`` ``// If left child exists, then it is preorder`` ``// successor.`` ``if` `(n.left != ``null``)`` ``return` `n.left;`` ` ` ``// If left child does not exist and right child`` ``// exists, then it is preorder successor.`` ``if` `(n.right != ``null``)`` ``return` `n.right;`` ` ` ``// If left child does not exist, then`` ``// travel up (using parent pointers)`` ``// until we reach a node which is left`` ``// child of its parent.`` ``Node curr = n, parent = curr.parent;`` ``while` `(parent != ``null` `&& parent.right == curr)`` ``{`` ``curr = curr.parent;`` ``parent = parent.parent;`` ``}` ` ``// If we reached root, then the given`` ``// node has no preorder successor`` ``if` `(parent == ``null``)`` ``return` `null``;` ` ``return` `parent.right;``}` `// Driver code``public` `static` `void` `main(String args[])``{`` ``Node root = newNode(``20``);`` ``root.parent = ``null``;`` ``root.left = newNode(``10``);`` ``root.left.parent = root;`` ``root.left.left = newNode(``4``);`` ``root.left.left.parent = root.left;`` ``root.left.right = newNode(``18``);`` ``root.left.right.parent = root.left;`` ``root.right = newNode(``26``);`` ``root.right.parent = root;`` ``root.right.left = newNode(``24``);`` ``root.right.left.parent = root.right;`` ``root.right.right = newNode(``27``);`` ``root.right.right.parent = root.right;`` ``root.left.right.left = newNode(``14``);`` ``root.left.right.left.parent = root.left.right;`` ``root.left.right.left.left = newNode(``13``);`` ``root.left.right.left.left.parent = root.left.right.left;`` ``root.left.right.left.right = newNode(``15``);`` ``root.left.right.left.right.parent = root.left.right.left;`` ``root.left.right.right = newNode(``19``);`` ``root.left.right.right.parent = root.left.right;` ` ``Node res = preorderSuccessor(root, root.left.right.right);` ` ``if` `(res != ``null``)`` ``{`` ``System.out.printf(``"Preorder successor of %d is %d\n"``,`` ``root.left.right.right.key, res.key);`` ``}`` ``else`` ``{`` ``System.out.printf(``"Preorder successor of %d is null\n"``,`` ``root.left.right.right.key);`` ``}``}``}` `// This code is contributed by Arnab Kundu`
## Python3
`""" Python3 program to find preorder``successor of a node in Binary Tree."""` `# A Binary Tree Node``# Utility function to create a new tree node` `class` `newNode:` ` ``# Constructor to create a new node`` ``def` `__init__(``self``, data):`` ``self``.key ``=` `data`` ``self``.left ``=` `None`` ``self``.right ``=` `None`` ``self``.parent ``=` `None` `def` `preorderSuccessor(root, n):` ` ``# If left child exists, then it is`` ``# preorder successor.`` ``if` `(n.left):`` ``return` `n.left`` ``# If left child does not exist and right child`` ``# exists, then it is preorder successor.`` ``if` `(n.right):`` ``return` `n.right`` ``# If left child does not exist, then`` ``# travel up (using parent pointers)`` ``# until we reach a node which is left`` ``# child of its parent.`` ``curr ``=` `n`` ``parent ``=` `curr.parent`` ``while` `(parent !``=` `None` `and`` ``parent.right ``=``=` `curr):`` ``curr ``=` `curr.parent`` ``parent ``=` `parent.parent` ` ``# If we reached root, then the given`` ``# node has no preorder successor`` ``if` `(parent ``=``=` `None``):`` ``return` `None` ` ``return` `parent.right` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``root ``=` `newNode(``20``)`` ``root.parent ``=` `None`` ``root.left ``=` `newNode(``10``)`` ``root.left.parent ``=` `root`` ``root.left.left ``=` `newNode(``4``)`` ``root.left.left.parent ``=` `root.left`` ``root.left.right ``=` `newNode(``18``)`` ``root.left.right.parent ``=` `root.left`` ``root.right ``=` `newNode(``26``)`` ``root.right.parent ``=` `root`` ``root.right.left ``=` `newNode(``24``)`` ``root.right.left.parent ``=` `root.right`` ``root.right.right ``=` `newNode(``27``)`` ``root.right.right.parent ``=` `root.right`` ``root.left.right.left ``=` `newNode(``14``)`` ``root.left.right.left.parent ``=` `root.left.right`` ``root.left.right.left.left ``=` `newNode(``13``)`` ``root.left.right.left.left.parent ``=` `root.left.right.left`` ``root.left.right.left.right ``=` `newNode(``15``)`` ``root.left.right.left.right.parent ``=` `root.left.right.left`` ``root.left.right.right ``=` `newNode(``19``)`` ``root.left.right.right.parent ``=` `root.left.right`` ``res ``=` `preorderSuccessor(root, root.left.right.right)` ` ``if` `(res):`` ``print``(``"Preorder successor of"``,`` ``root.left.right.right.key, ``"is"``, res.key)` ` ``else``:`` ``print``(``"Preorder successor of"``,`` ``root.left.right.right.key, ``"is None"``)` `# This code is contributed``# by SHUBHAMSINGH10`
## C#
`// C# program to find preorder successor of``// a node in Binary Tree.``using` `System;`` ` `class` `GFG``{` `public` `class` `Node``{`` ``public` `Node left, right, parent;`` ``public` `int` `key;``};` `static` `Node newNode(``int` `key)``{`` ``Node temp = ``new` `Node();`` ``temp.left = temp.right = temp.parent = ``null``;`` ``temp.key = key;`` ``return` `temp;``}` `static` `Node preorderSuccessor(Node root, Node n)``{`` ``// If left child exists, then it is preorder`` ``// successor.`` ``if` `(n.left != ``null``)`` ``return` `n.left;`` ` ` ``// If left child does not exist and right child`` ``// exists, then it is preorder successor.`` ``if` `(n.right != ``null``)`` ``return` `n.right;`` ` ` ``// If left child does not exist, then`` ``// travel up (using parent pointers)`` ``// until we reach a node which is left`` ``// child of its parent.`` ``Node curr = n, parent = curr.parent;`` ``while` `(parent != ``null` `&& parent.right == curr)`` ``{`` ``curr = curr.parent;`` ``parent = parent.parent;`` ``}` ` ``// If we reached root, then the given`` ``// node has no preorder successor`` ``if` `(parent == ``null``)`` ``return` `null``;` ` ``return` `parent.right;``}` `// Driver code``public` `static` `void` `Main(String []args)``{`` ``Node root = newNode(20);`` ``root.parent = ``null``;`` ``root.left = newNode(10);`` ``root.left.parent = root;`` ``root.left.left = newNode(4);`` ``root.left.left.parent = root.left;`` ``root.left.right = newNode(18);`` ``root.left.right.parent = root.left;`` ``root.right = newNode(26);`` ``root.right.parent = root;`` ``root.right.left = newNode(24);`` ``root.right.left.parent = root.right;`` ``root.right.right = newNode(27);`` ``root.right.right.parent = root.right;`` ``root.left.right.left = newNode(14);`` ``root.left.right.left.parent = root.left.right;`` ``root.left.right.left.left = newNode(13);`` ``root.left.right.left.left.parent = root.left.right.left;`` ``root.left.right.left.right = newNode(15);`` ``root.left.right.left.right.parent = root.left.right.left;`` ``root.left.right.right = newNode(19);`` ``root.left.right.right.parent = root.left.right;` ` ``Node res = preorderSuccessor(root, root.left.right.right);` ` ``if` `(res != ``null``)`` ``{`` ``Console.Write(``"Preorder successor of {0} is {1}\n"``,`` ``root.left.right.right.key, res.key);`` ``}`` ``else`` ``{`` ``Console.Write(``"Preorder successor of {0} is null\n"``,`` ``root.left.right.right.key);`` ``}``}``}` `// This code is contributed by 29AjayKumar`
## Javascript
``
Output:
`Preorder successor of 19 is 26`
Time Complexity: O(h) where h is the height of the given Binary Tree, as we are not traversing all nodes. We have checked the child of each node that is equivalent to traversing the height of the tree.
Auxiliary Space: O(1), as we are not using any extra space.
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | 659 | 2,195 | {"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-2020-50 | longest | en | 0.853865 |
https://goprep.co/q28-show-that-the-height-of-a-closed-right-circular-cylinder-i-1nlewn | 1,621,170,247,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00334.warc.gz | 305,098,379 | 28,038 | # Show that the hei
Given; Let S be the surface area, r be the radius, h be the height, and V be the volume of the closed right circular cylinder.
S = 2πr2 + 2πrh
V = πr2h
Differentiating w.r.t x.
For maxima or minima
S = 6πr2
h = 2r
V is maximum when h = 2r
i.e When height of a closed right circular cylinder is equal to the diameter of its base.
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Prove that the leMathematics - Board Papers | 271 | 1,043 | {"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-2021-21 | latest | en | 0.840225 |
https://www.enotes.com/homework-help/2x-3y-z-10-y-2z-13-z-5-357551 | 1,488,205,479,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172831.37/warc/CC-MAIN-20170219104612-00282-ip-10-171-10-108.ec2.internal.warc.gz | 804,392,313 | 12,059 | # Solve the system using the substitution method: 2x-3y+z=10 y+2z=13 z=5
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The following system of equations has to be solved:
2x-3y+z=10 ...(1)
y+2z=13 ...(2)
z=5 ...(3)
From (3), z = 5
Substituting z = 5 in (2) gives y + 10 = 13 or y = 3
Substituting z = 5 and y = 3 in (1) gives 2x - 9 + 5 = 10
=> 2x = 14
=> x = 7
The solution of the equation is x = 7, y = 3 and z = 5 | 183 | 456 | {"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} | 4.0625 | 4 | CC-MAIN-2017-09 | longest | en | 0.848241 |
https://www.dlubal.com/en/support-and-learning/support/knowledge-base/001462 | 1,566,332,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315618.73/warc/CC-MAIN-20190820200701-20190820222701-00018.warc.gz | 788,395,037 | 21,953 | # Downstand Beams, Ribs, T-Beams: Minimum Reinforcement for Partial Cross-Sections According to 7.3.2
### Technical Article
001462
18 July 2017
According to Section 7.3.2 (2), the standard EN 1992‑1‑1 [1] states: ‘In profiled cross‑sections like T‑beams and box girders, minimum reinforcement should be determined for the individual parts of the section (webs, flanges).’
In the case of a T‑beam with a T‑section, the minimum reinforcement should be determined for both the chord and the web if the corresponding partial cross‑sections are in the tension area. Figure 01 shows the cross‑section classification.
Depending on whether a web or a chord is concerned, the coefficient kc, which takes into account the influence of the stress distribution within the partial cross‑section, is determined as follows:
$$\mathrm{Formula}\;7.2\;\mathrm{for}\;\mathrm{web}:\;{\mathrm k}_\mathrm c\;=\;0.4\;\cdot\;\left[1\;-\;\frac{{\mathrm\sigma}_\mathrm c}{{\mathrm k}_1\;\cdot\;{\displaystyle\frac{\mathrm h}{\mathrm h^\ast}}}\;\cdot\;{\mathrm f}_{\mathrm{ct},\mathrm{eff}}\right]\;\leq\;1$$
and
$$\mathrm{Formula}\;7.3\;\mathrm{for}\;\mathrm{chord}:\;{\mathrm k}_\mathrm c\;=\;0.9\;\cdot\;\frac{{\mathrm F}_\mathrm{cr}}{{\mathrm A}_\mathrm{ct}\;\cdot\;{\mathrm f}_{\mathrm{ct},\mathrm{eff}}}\;\geq\;0.5$$
In the case of pure tensile stress, kc = 1 applies for the entire cross‑section as well as the individual partial cross‑sections.
The mean concrete stress σc affecting the examined part of the cross‑section is determined using the concrete stress distribution applying the crack moment due to fct,eff, at the shear point of the respective partial cross‑section (see Figure 01 σc).
#### Options in RF-CONCRETE Members
The partial cross‑sections are determined automatically, depending on the underlying cross‑section. The RF‑CONCRETE Members add‑on module provides the following control options, which have an effect on the minimum reinforcement of the partial cross‑sections.
By selecting the minimum reinforcement layout, you specify at the same time on which side of the cross‑section the concrete tensile stress fct,eff is applied or in which direction the crack moment acts.
The corresponding options for stress distribution within the section prior to cracking control the distribution of the mean concrete stress.
If you select kc = 1.0, fct,eff acts constantly over the cross‑section. If you select kc = 0.4 or kc = variable depending on the defined load, the mean concrete stress distribution is determined due to bending about y or due to bending about y and the axial force. The minimum reinforcement is arranged in the partial cross‑sections where the stress fct,eff is reached.
If fct,eff is not reached on any cross‑section fibre in some partial cross‑sections, but nevertheless a wedge is formed, then the tensile force of the partial cross‑section is assigned to the governing tension side.
#### Results in RF-CONCRETE Members
Window 4 shows the governing design and the detailed results of the minimum reinforcement for the individual partial cross‑sections. For each partial cross‑section, the coefficient kc for considering the influence of the stress distribution, the area of the tension zone Act, and the absolute value of the maximum allowable steel stress are displayed, among other results.
Under ‘Design’, you can see the resulting design with the highest criterion. This is confronted with the calculated minimum reinforcement in the partial cross‑section of the provided reinforcement (Figure 07).
#### Reference
[1] Eurocode 2: Design of concrete structures - Part 1‑1: General rules and rules for buildings; EN 1992‑1‑1:2004 + AC:2010 [2] Zilch, K. & Zehetmaier, G. (2010). Bemessung im konstruktiven Betonbau - Nach DIN 1045‑1 (Fassung 2008) und EN 1992‑1‑1 (Eurocode 2) (2nd ed.). Berlin: Springer. | 997 | 3,845 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-35 | latest | en | 0.662317 |
http://www.cs.unb.ca/~bremner/teaching/cs4613/tutorials/tests/ | 1,685,716,226,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00757.warc.gz | 60,510,596 | 2,641 | UNB/ CS/ David Bremner/ teaching/ cs4613/ tutorials/ Testing in racket
# Unit Tests in Racket
Prerequisites
recursion, modules
This part is based on an example from RackUnit QuickStart. The Beautiful Racket unit test explainer is a better reference, since we'll skip some of the fancier features of `rackunit` and move straight to having a test submodule.
• Save the following as "arith.rkt"
```#lang racket/base
(define (my-+ a b)
(if (zero? a)
b
(define (my-* a b)
(if (zero? a)
b
(my-* (sub1 a) (my-+ b b))))
(provide my-+
my-*)
```
• add a test submodule with the following tests
```(check-equal? (my-+ 1 1) 2 "Simple addition")
(check-equal? (my-* 1 2) 2 "Simple multiplication")```
• run your "arith.rkt" in `DrRacket`; observe that one of the two tests fails.
• Fix the recursive definition of my-* so that it at least works for non-negative integers. There are several different approaches which work, one simple way is based on the equations
`````` 0 * b = 0 (base case)
a * b = (a - 1) * b + b (recursion)
``````
• Observe there is no output from the tests now when you run the code in `DrRacket`. This means success.
# Test Coverage
• This activity continues with the same file from the previous activity.
• Under `Language -> Choose Language -> Show Details -> Dynamic Properties`, enable
``````⊙ Syntactic test suite coverage
``````
• run your code in `DrRacket` again
• most likely you will have some code highlighted with orange text and black foreground. This means that code is not covered by your test suite. Add another test to cover each piece of uncovered code.
In this course, for all `racket` assignments you will lose marks if you don't have complete test coverage. | 456 | 1,732 | {"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-2023-23 | longest | en | 0.867777 |
https://aakashsrv1.meritnation.com/ask-answer/question/how-to-understand-congruence-of-triangles-please-tell-me-tod/congruence-of-triangles/2808047 | 1,695,992,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00676.warc.gz | 95,370,234 | 10,180 | # how to understand congruence of triangles . please tell me today because i have to prepare for test
## Congruent Triangles
Triangles that have exactly the same size and shape are called congruent triangles. The symbol for congruent is ≅. Two triangles are congruent when the three sides and the three angles of one triangle have the same measurements as three sides and three angles of another triangle. The triangles in Figure 1 are congruent triangles.
Figure 1 Congruent triangles.
## Corresponding parts
Example 1: If Δ PQR ≅ Δ STU which parts must have equal measurements?
These parts are equal because corresponding parts of congruent triangles are congruent.
## Tests for congruence
To show that two triangles are congruent, it is not necessary to show that all six pairs of corresponding parts are equal. The following postulates and theorems are the most common methods for proving that triangles are congruent (or equal).
Figure 2 The corresponding sides (SSS) of the two triangles are all congruent.
Figure 3 Two sides and the included angle (SAS) of one triangle are congruent to the corresponding parts of the other triangle.
Figure 4 Two angles and their common side (ASA) in one triangle are congruent to the corresponding parts of the other triangle.
Figure 5 Two angles and the side opposite one of these angles (AAS) in one triangle are congruent to the corresponding parts of the other triangle.
Figure 6 The hypotenuse and one leg (HL) of the first right triangle are congruent to the corresponding parts of the second right triangle.
Figure 7 The hypotenuse and an acute angle (HA) of the first right triangle are congruent to the corresponding parts of the second right triangle.
Figure 8 The legs (LL) of the first right triangle are congruent to the corresponding parts of the second right triangle.
Figure 9 One leg and an acute angle (LA) of the first right triangle are congruent to the corresponding parts of the second right triangle.
• 3
What are you looking for? | 452 | 2,021 | {"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} | 4.625 | 5 | CC-MAIN-2023-40 | latest | en | 0.845474 |
https://foreach.id/EN/common/power/calorieth%7Csecond-to-microwatt.html | 1,618,277,292,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00124.warc.gz | 369,380,992 | 10,944 | # Convert calorie (th)/second to microwatt (cal (th)/s to µW)
Batch Convert
• microwatt [µW]
• calorie (th)/second [cal (th)/s]
Copy
_
Copy
• microwatt [µW]
• calorie (th)/second [cal (th)/s]
## Calorie (th)/second to Microwatt (cal (th)/s to µW)
### Calorie (th)/second (Symbol or Abbreviation: cal (th)/s)
Calorie (th)/second is one of power units. Calorie (th)/second abbreviated or symbolized by cal (th)/s. The value of 1 calorie (th)/second is equal to 4.184 watt. In its relation with microwatt, 1 calorie (th)/second is equal to 4184000 microwatt.
#### Relation with other units
1 calorie (th)/second equals to 4.184 watt
1 calorie (th)/second equals to 0.004184 kilowatt
1 calorie (th)/second equals to 0.000004184 megawatt
1 calorie (th)/second equals to 4,184,000,000,000 picowatt
1 calorie (th)/second equals to 4,184,000,000 nanowatt
1 calorie (th)/second equals to 4,184,000 microwatt
1 calorie (th)/second equals to 4,184 milliwatt
1 calorie (th)/second equals to 418.4 centiwatt
1 calorie (th)/second equals to 41.84 deciwatt
1 calorie (th)/second equals to 0.4184 dekawatt
1 calorie (th)/second equals to 0.04184 hectowatt
1 calorie (th)/second equals to 4.184e-9 gigawatt
1 calorie (th)/second equals to 0.0056108 horsepower
1 calorie (th)/second equals to 0.0056887 horsepower (metric)
1 calorie (th)/second equals to 0.00042653 horsepower (boiler)
1 calorie (th)/second equals to 0.0056086 horsepower (electric)
1 calorie (th)/second equals to 0.0056083 horsepower (water)
1 calorie (th)/second equals to 0.0056887 pferdestarke
1 calorie (th)/second equals to 14.276 Btu (IT)/hour
1 calorie (th)/second equals to 0.23794 Btu (IT)/minute
1 calorie (th)/second equals to 0.0039657 Btu (IT)/second
1 calorie (th)/second equals to 14.286 Btu (th)/hour
1 calorie (th)/second equals to 0.2381 Btu (th)/minute
1 calorie (th)/second equals to 0.0039683 Btu (th)/second
1 calorie (th)/second equals to 0.000014276 MBtu (IT)/hour
1 calorie (th)/second equals to 0.014276 MBH
1 calorie (th)/second equals to 0.0011897 ton (refrigeration)
1 calorie (th)/second equals to 3.5976 kilocalorie (IT)/hour
1 calorie (th)/second equals to 0.05996 kilocalorie (IT)/minute
1 calorie (th)/second equals to 0.00099933 kilocalorie (IT)/second
1 calorie (th)/second equals to 3.6 kilocalorie (th)/hour
1 calorie (th)/second equals to 0.06 kilocalorie (th)/minute
1 calorie (th)/second equals to 0.001 kilocalorie (th)/second
1 calorie (th)/second equals to 3,597.6 calorie (IT)/hour
1 calorie (th)/second equals to 59.96 calorie (IT)/minute
1 calorie (th)/second equals to 0.99933 calorie (IT)/second
1 calorie (th)/second equals to 3,600 calorie (th)/hour
1 calorie (th)/second equals to 60 calorie (th)/minute
1 calorie (th)/second equals to 11,109 foot pound-force/hour
1 calorie (th)/second equals to 185.16 foot pound-force/minute
1 calorie (th)/second equals to 3.086 foot pound-force/second
1 calorie (th)/second equals to 41,840,000 erg/second
1 calorie (th)/second equals to 0.004184 kilovolt ampere
1 calorie (th)/second equals to 4.184 volt ampere
1 calorie (th)/second equals to 4.184 newton meter/second
1 calorie (th)/second equals to 4.184 joule/second
1 calorie (th)/second equals to 4.184e-9 gigajoule/second
1 calorie (th)/second equals to 0.000004184 megajoule/second
1 calorie (th)/second equals to 0.004184 kilojoule/second
1 calorie (th)/second equals to 0.04184 hectojoule/second
1 calorie (th)/second equals to 0.4184 dekajoule/second
1 calorie (th)/second equals to 41.84 decijoule/second
1 calorie (th)/second equals to 418.4 centijoule/second
1 calorie (th)/second equals to 4,184 millijoule/second
1 calorie (th)/second equals to 4,184,000 microjoule/second
1 calorie (th)/second equals to 4,184,000,000 nanojoule/second
1 calorie (th)/second equals to 4,184,000,000,000 picojoule/second
1 calorie (th)/second equals to 15,062 joule/hour
1 calorie (th)/second equals to 251.04 joule/minute
1 calorie (th)/second equals to 15.062 kilojoule/hour
1 calorie (th)/second equals to 0.25104 kilojoule/minute
### Microwatt (Symbol or Abbreviation: µW)
Microwatt is one of power units. Microwatt abbreviated or symbolized by µW. The value of 1 microwatt is equal to 0.000001 watt. In its relation with calorie (th)/second, 1 microwatt is equal to 2.3901e-7 calorie (th)/second.
#### Relation with other units
1 microwatt equals to 0.000001 watt
1 microwatt equals to 1e-9 kilowatt
1 microwatt equals to 1e-12 megawatt
1 microwatt equals to 1,000,000 picowatt
1 microwatt equals to 1,000 nanowatt
1 microwatt equals to 0.001 milliwatt
1 microwatt equals to 0.0001 centiwatt
1 microwatt equals to 0.00001 deciwatt
1 microwatt equals to 1e-7 dekawatt
1 microwatt equals to 1e-8 hectowatt
1 microwatt equals to 1e-15 gigawatt
1 microwatt equals to 1.341e-9 horsepower
1 microwatt equals to 1.3596e-9 horsepower (metric)
1 microwatt equals to 1.0194e-10 horsepower (boiler)
1 microwatt equals to 1.3405e-9 horsepower (electric)
1 microwatt equals to 1.3404e-9 horsepower (water)
1 microwatt equals to 1.3596e-9 pferdestarke
1 microwatt equals to 0.0000034121 Btu (IT)/hour
1 microwatt equals to 5.6869e-8 Btu (IT)/minute
1 microwatt equals to 9.4782e-10 Btu (IT)/second
1 microwatt equals to 0.0000034144 Btu (th)/hour
1 microwatt equals to 5.6907e-8 Btu (th)/minute
1 microwatt equals to 9.4845e-10 Btu (th)/second
1 microwatt equals to 3.4121e-12 MBtu (IT)/hour
1 microwatt equals to 3.4121e-9 MBH
1 microwatt equals to 2.8435e-10 ton (refrigeration)
1 microwatt equals to 8.5985e-7 kilocalorie (IT)/hour
1 microwatt equals to 1.4331e-8 kilocalorie (IT)/minute
1 microwatt equals to 2.3885e-10 kilocalorie (IT)/second
1 microwatt equals to 8.6042e-7 kilocalorie (th)/hour
1 microwatt equals to 1.434e-8 kilocalorie (th)/minute
1 microwatt equals to 2.3901e-10 kilocalorie (th)/second
1 microwatt equals to 0.00085985 calorie (IT)/hour
1 microwatt equals to 0.000014331 calorie (IT)/minute
1 microwatt equals to 2.3885e-7 calorie (IT)/second
1 microwatt equals to 0.00086042 calorie (th)/hour
1 microwatt equals to 0.00001434 calorie (th)/minute
1 microwatt equals to 2.3901e-7 calorie (th)/second
1 microwatt equals to 0.0026552 foot pound-force/hour
1 microwatt equals to 0.000044254 foot pound-force/minute
1 microwatt equals to 7.3756e-7 foot pound-force/second
1 microwatt equals to 10 erg/second
1 microwatt equals to 1e-9 kilovolt ampere
1 microwatt equals to 0.000001 volt ampere
1 microwatt equals to 0.000001 newton meter/second
1 microwatt equals to 0.000001 joule/second
1 microwatt equals to 1e-15 gigajoule/second
1 microwatt equals to 1e-12 megajoule/second
1 microwatt equals to 1e-9 kilojoule/second
1 microwatt equals to 1e-8 hectojoule/second
1 microwatt equals to 1e-7 dekajoule/second
1 microwatt equals to 0.00001 decijoule/second
1 microwatt equals to 0.0001 centijoule/second
1 microwatt equals to 0.001 millijoule/second
1 microwatt equals to 1 microjoule/second
1 microwatt equals to 1,000 nanojoule/second
1 microwatt equals to 1,000,000 picojoule/second
1 microwatt equals to 0.0036 joule/hour
1 microwatt equals to 0.00006 joule/minute
1 microwatt equals to 0.0000036 kilojoule/hour
1 microwatt equals to 6e-8 kilojoule/minute
### How to convert Calorie (th)/second to Microwatt (cal (th)/s to µW):
#### Conversion Table for Calorie (th)/second to Microwatt (cal (th)/s to µW)
calorie (th)/second (cal (th)/s) microwatt (µW)
0.01 cal (th)/s 41,840 µW
0.1 cal (th)/s 418,400 µW
1 cal (th)/s 4,184,000 µW
2 cal (th)/s 8,368,000 µW
3 cal (th)/s 12,552,000 µW
4 cal (th)/s 16,736,000 µW
5 cal (th)/s 20,920,000 µW
6 cal (th)/s 25,104,000 µW
7 cal (th)/s 29,288,000 µW
8 cal (th)/s 33,472,000 µW
9 cal (th)/s 37,656,000 µW
10 cal (th)/s 41,840,000 µW
20 cal (th)/s 83,680,000 µW
25 cal (th)/s 104,600,000 µW
50 cal (th)/s 209,200,000 µW
75 cal (th)/s 313,800,000 µW
100 cal (th)/s 418,400,000 µW
250 cal (th)/s 1,046,000,000 µW
500 cal (th)/s 2,092,000,000 µW
750 cal (th)/s 3,138,000,000 µW
1,000 cal (th)/s 4,184,000,000 µW
100,000 cal (th)/s 418,400,000,000 µW
1,000,000,000 cal (th)/s 4,184,000,000,000,000 µW
1,000,000,000,000 cal (th)/s 4,184,000,000,000,000,000 µW
#### Conversion Table for Microwatt to Calorie (th)/second (µW to cal (th)/s)
microwatt (µW) calorie (th)/second (cal (th)/s)
0.01 µW 2.3901e-9 cal (th)/s
0.1 µW 2.3901e-8 cal (th)/s
1 µW 2.3901e-7 cal (th)/s
2 µW 4.7801e-7 cal (th)/s
3 µW 7.1702e-7 cal (th)/s
4 µW 9.5602e-7 cal (th)/s
5 µW 0.000001195 cal (th)/s
6 µW 0.000001434 cal (th)/s
7 µW 0.000001673 cal (th)/s
8 µW 0.000001912 cal (th)/s
9 µW 0.0000021511 cal (th)/s
10 µW 0.0000023901 cal (th)/s
20 µW 0.0000047801 cal (th)/s
25 µW 0.0000059751 cal (th)/s
50 µW 0.00001195 cal (th)/s
75 µW 0.000017925 cal (th)/s
100 µW 0.000023901 cal (th)/s
250 µW 0.000059751 cal (th)/s
500 µW 0.0001195 cal (th)/s
750 µW 0.00017925 cal (th)/s
1,000 µW 0.00023901 cal (th)/s
100,000 µW 0.023901 cal (th)/s
1,000,000,000 µW 239.01 cal (th)/s
1,000,000,000,000 µW 239,010 cal (th)/s
#### Steps to Convert Calorie (th)/second to Microwatt (cal (th)/s to µW)
1. Example: Convert 37 calorie (th)/second to microwatt (37 cal (th)/s to µW).
2. 1 calorie (th)/second is equivalent to 4184000 microwatt (1 cal (th)/s is equivalent to 4184000 µW).
3. 37 calorie (th)/second (cal (th)/s) is equivalent to 37 times 4184000 microwatt (µW).
4. Retrieved 37 calorie (th)/second is equivalent to 154810000 microwatt (37 cal (th)/s is equivalent to 154810000 µW).
▸▸ | 3,598 | 9,540 | {"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-2021-17 | latest | en | 0.785586 |
https://digifesto.com/tag/interpretivism/ | 1,685,788,312,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00013.warc.gz | 246,479,922 | 21,378 | ## Tag: interpretivism
### System 2 hegemony and its discontents
Recent conversations have brought me back to the third rail of different modalities of knowledge and their implications for academic disciplines. God help me. The chain leading up to this is: a reminder of how frustrating it was trying to work with social scientists who methodologically reject the explanatory power of statistics, an intellectual encounter with a 20th century “complex systems” theorist who also didn’t seem to understand statistics, and the slow realization that’s been bubbling up for me over the years that I probably need to write an article or book about the phenomenology of probability, because I can’t find anything satisfying about it.
The hypothesis I am now entertaining is that probabilistic or statistical reasoning is the intellectual crux, disciplinarily. What we now call “STEM” is all happy to embrace statistics as its main mode of empirical verification. This includes the use of mathematical proof for “exact” or a priori verification of methods. Sometimes the use of statistics is delayed or implicit; there is qualitative research that is totally consistent with statistical methods. But the key to this whole approach is that the fields, in combination, are striving for consistency.
But not everybody is on board with statistics! Why is that?
One reason may be because statistics is difficult to learn and execute. Doing probabilistic reasoning correctly is at times counter-intuitive. That means that quite literally it can make your head hurt to think about it.
There is a lot of very famous empirical cognitive psychology that has explored this topic in depth. The heuristics and biases research program of Kahneman and Tversky was critical for showing that human behavior rarely accords with decision-theoretic models of mathematical, probabilistic rationality. An intuitive, “fast”, prereflective form of thinking, (“System 1”) is capable of making snap judgments but is prone to biases such as the availability heuristic and the representativeness heuristic.
A couple general comments can be made about System 1. (These are taken from Tetlock’s review of this material in Superforecasting). First, a hallmark of System 1 is that it takes whatever evidence it is working with as given; it never second-guesses it or questions its validity. Second, System 1 is fantastic at provided verbal rationalizations and justifications of anything that it encounters, even when these can be shown to be disconnected from reality. Many colorful studies of split brain cases, but also many other lab experiments, show the willingness people have to make of stories to explain anything, and their unwillingness to say, “this could be due to one of a hundred different reasons, or a mix of them, and so I don’t know.”
The cognitive psychologists will also describe a System 2 cognitive process that is more deliberate and reflective. Presumably, this is the system that is sometimes capable of statistical or otherwise logical reasons. And a big part of statistical reasoning is questioning the source of your evidence. A robust application of System 2 reasoning is capable of overcoming System 1’s biases. At the level of institutional knowledge creation, the statistical sciences are comprised mainly of formalized, shared results of System 2 reasoning.
Tetlock’s work, from Expert Political Judgment and on, is remarkable for showing that deference to one or the other cognitive system is to some extent a robust personality trait. Famously, those of the “hedgehog” cognitive style, who apply System 1 and a simplistic theory of the world to interpret everything they experience, are especially bad at predicting the outcomes of political events (what are certainly the results of ‘complex systems’), whereas the “fox” cognitive style, which is more cautious about considering evidence and coming to judgments, outperforms them. It seems that Tetlock’s analysis weighs in favor of System 2 as a way of navigating complex systems.
I would argue that there are academic disciplines, especially those grounded in Heideggerian phenomenology, that see the “dominance” of institutions (such as academic disciplines) that are based around accumulations of System 2 knowledge as a problem or threat.
This reaction has several different guises:
• A simple rejection of cognitive psychology, which has exposed the System 1/System 2 distinction, as “behaviorism”. (This obscures the way cognitive psychology was a major break away from behaviorism in the 50’s.)
• A call for more “authentic experience”, couched in language suggesting ownership or the true subject of one’s experience, contrasting this with the more alienated forms of knowing that rely on scientific consensus.
• An appeal to originality: System 2 tends to converge; my System 1 methods can come up with an exciting new idea!
• The interpretivist methodological mandate for anthropological sensitivity to “emic”, or directly “lived experience”, of research subjects. This mandate sometimes blurs several individually valid motivations, such as: when emic experience is the subject matter in its own right, but (crucially) with the caveat that the results are not generalizable; when emic sensitivity is identified via the researcher’s reflexivity as a condition for research access; or when the purpose of the work is to surface or represent otherwise underrepresented views.
There are ways to qualify or limit these kinds of methodologies or commitments that makes them entirely above reproach. However, under these limits, their conclusions are always fragile. According to the hegemonic logic of System 2 institutions, a consensus of those thoroughly considering the statistical evidence can always supercede the “lived experience” of some group or individual. This is, at the methodological level, simply the idea that while we may make theory-laden observations, when those theories are disproved, those observations are invalidated as being influenced by erronenous theory. Indeed, mainstream scientific institutions take as their duty this kind of procedural objectivity. There is no such thing as science unless a lot of people are often being proven wrong.
This provokes a great deal of grievance. “Who made scientists, an unrepresentative class of people and machines disconnected from authentic experience, the arbiter of the real? Who are they to tell me I am wrong, or my experiences invalid?” And this is where we start to find trouble.
Perhaps most troubling is how this plays out at the level of psychodynamic politics. To have one’s lived experiences rejected, especially those lived experiences of trauma, and especially when those experiences are rejected wrongly, is deeply disturbing. One of the more mighty political tendencies of recent years has been the idea that whole classes of people are systematically subject to this treatment. This is one reason, among others, for influential calls for recalibrating the weight given to the experiences of otherwise marginalized people. This is what Furedi calls the therapeutic ethos of the Left. This is slightly different from, though often conflated with, the idea that recalibration is necessary to allow in more relevant data that was being otherwise excluded from consideration. This latter consideration comes up in a more managerialist discussion of creating technology that satisfies diverse stakeholders (…customers) through “participatory” design methods. The ambiguity of the term “bias”–does it mean a statistical error, or does it mean any tendency of an inferential system at all?–is sometimes leveraged to accomplish this conflation.
It is in practice very difficult to disentangle the different psychological motivations here. This is partly because they are deeply personal and mixed even at the level of the individual. (Highlighting this is why I have framed this in terms of the cognitive science literature). It is also partly because these issues are highly political as well. Being proven right, or wrong, has material consequences–sometimes. I’d argue: perhaps not as often as it should. But sometimes. And so there’s always a political interest, especially among those disinclined towards System 2 thinking, in maintaining a right to be wrong.
So it is hypothesized (perhaps going back to Lyotard) that at an institutional level there’s a persistent heterodox movement that rejects the ideal of communal intellectual integrity. Rather, it maintains that the field of authoritative knowledge must contain contradictions and disturbances of statistical scientific consensus. In Lyotard’s formulation, this heterodoxy seeks “legitimation by paralogy”, which suggests that its telos is at best a kind of creative intellectual emancipation from restrictive logics, generative of new ideas, but perhaps at worst a heterodoxy for its own sake.
This tendency has an uneasy relationship with the sociopolitical motive of a more integrated and representative society, which is often associated with the goal of social justice. If I understand these arguments directly, the idea is that, in practice, legitimized paralogy is a way of giving the underrepresented a platform. This has the benefits of increasing, visibly, representation. Here, paralogy is legitimized as a means of affirmative action, but not as a means improving system performance objectively.
This is a source of persistent difficulty and unease, as the paralogical tendency is never capable of truly emancipating itself, but rather, in its recuperated form, is always-already embedded in a hierarchy that it must deny to its initiates. Authenticity is subsumed, via agonism, to a procedural objectivity that proves it wrong.
### How to tell the story about why stories don’t matter
I’m thinking of taking this seminar because I’m running into the problem it addresses: how do you pick a theoretical lens for academic writing?
This is related to a conversation I’ve found myself in repeatedly over the past weeks. A friend who studied Rhetoric insists that the narrative and framing of history is more important than the events and facts. A philosopher friend minimizes the historical impact of increased volumes of “raw footage”, because ultimately it’s the framing that will matter.
Yesterday I had the privilege of attending Techraking III, a conference put on by the Center for Investigative Reporting with the generous support and presence of Google. It was a conference about data journalism. The popular sentiment within the conference was that data doesn’t matter unless it’s told with a story, a framing.
I find this troubling because while I pay attention to this world and the way it frames itself, I also read the tech biz press carefully, and it tells a very different narrative. Data is worth billions of dollars. Even data exhaust, the data fumes that come from your information processing factory, can be recycled into valuable insights. Data is there to be mined for value. And if you are particularly genius at it, you can build an expert system that acts on the data without needing interpretation. You build an information processing machine that acts according to mechanical principles that approximate statistical laws, and these machines are powerful.
As social scientists realize they need to be data scientists, and journalists realize they need to be data journalists, there seems to be in practice a tacit admission of the data-driven counter-narrative. This tacit approval is contradicted by the explicit rhetoric that glorifies interpretation and narrative over data.
This is an interesting kind of contradiction, as it takes place as much in the psyche of the data scientist as anywhere else. It’s like the mouth doesn’t know what the hand is doing. This is entirely possible since our minds aren’t actually that coherent to start with. But it does make the process of collaboratively interacting with others in the data science field super complicated.
All this comes to a head when the data we are talking about isn’t something simple like sensor data about the weather but rather is something like text, which is both data and narrative simulatenously. We intuitively see the potential of treating narrative as something to be treated mechanically, statistically. We certainly see the effects of this in our daily lives. This is what the most powerful organizations in the world do all the time.
The irony is that the interpretivists, who are so quick to deny technological determinism, are the ones who are most vulnerable to being blindsided by “what technology wants.” Humanities departments are being slowly phased out, their funding cut. Why? Do they have an explanation for this? If interpetation/framing were as efficacious as they claim, they would be philosopher kings. So their sociopolitical situation contradicts their own rhetoric and ideology. Meanwhile, journalists who would like to believe that it’s the story that matters are, for the sake of job security, being corralled into classes to learn CSS, the programming language that determines, mechanically, the logic of formatting and presentation.
Sadly, neither mechanists nor interpretivists have much of an interest in engaging this contradiction. This is because interpretivists chase funding by reinforcing the narrative that they are critically important, and the work of mechanists speaks for itself in corporate accounting (an uninterpretive field) without explanation. So this contradiction falls mainly into the laps of those coordinating interaction between tribes. Managers who need to communicate between engineering and marketing. University administrators who have to juggle the interests of humanities and sciences. The leadership of investigative reporting non-profits who need to justify themselves to savvy foundations and who are removed enough from particular skillsets to be flexible.
Mechnanized information processing is becoming the new epistemic center. (Forgive me:) the Google supercomputer approximating statistics has replaced Kantian trancendental reason as the grounds for bourgious understanding of the world. This is threatening, of course, to the plurality of perspectives that do not themselves internalize the logic of machine learning. Where machine intelligence has succeeded, then, it has been by juggling this multitude of perspectives (and frames) through automated, data-driven processes. Machine intelligence is not comprehensible to lay interpretivism. Interestingly, lay interpetivism isn’t comprehensible yet to machine intelligence–natural language processing has not yet advanced so far. It treats our communications like we treat ants in an ant farm: a blooming buzzing confusion of arbitrary quanta, fascinatingly complex for its patterns that we cannot see. And when it makes mistakes–and it does often–we feel its effects as a structural force beyond our control. A change in the user interface of Facebook that suddenly exposes drunken college photos to employers and abusive ex-lovers.
What theoretical frame is adequate to tell this story, the story that’s determining the shape of knowledge today? For Lyotard, the postmodern condition is one in which metanarratives about the organization of knowledge collapse and leave only politics, power, and language games. The postmodern condition has gotten us into our present condition: industrial machine intelligence presiding over interpretivists battling in paralogical language games. When the interpretivists strike back, it looks like hipsters or Weird Twitter–paralogy as a subculture of resistance that can’t even acknowledge its own role as resistance for fear of recuperation.
We need a new metanarrative to get out of this mess. But what kind of theory could possibly satisfy all these constituents?
### several words, all in a row, some about numbers
I am getting increasingly bewildered by number of different paradigms available in academic research. Naively, I had thought I had a pretty good handle on this sort of thing coming into it. After trying to tackle the subject head on this semester, I feel like my head will explode.
I’m going to try to break down the options.
• Nobody likes positivism, which went out of style when Wittgenstein refuted his own Tractatus.
• Postpositivists say, “Sure, there isn’t really observer-independent inquiry, but we can still approximate that through rigorous methods.” The goal is an accurate description of the subject matter. I suppose this fits into a vision of science being about prediction and control of the environment, so generalizability of results would be considered important. I’d argue that this is also consistent with American pragmatism. I think “postpositivist” is a terrible name and would rather talk/think about pragmatism.
• Interpretivism, which seems to be a more fashionable term than antipositivism, is associated with Weber and Frankfurt school thinkers, as well as a feminist critique. The goal is for one reader (or scholarly community?) to understand another. “Understanding” here is understood intersubjectively–“I get you”. Interpretivists are skeptical of prediction and control as provided by a causal understanding. At times, this skepticism is expressed as a belief that causal understanding (of people) is impossible; other times it is expressed as a belief that causal understanding is nefarious.
Both teams share a common intellectual ancestor in Immanuel Kant, who few people bother to read.
Habermas has room in his overarching theory for multiple kinds of inquiry–technical, intersubjective, and emancipatory/dramaturgical–but winds up getting mobilized by the interpretivists. I suspect this is the case because research aimed at prediction and control is better funded, because it is more instrumental to power. And if you’ve got funding there’s little incentive to look to Habermas for validation.
It’s worth noting that mathematicians still basically run their own game. You can’t beat pure reason at the research game. Much computer science research falls into this category. Pragmatists will take advantage of mathematical reasoning. I think interpretivists find mathematics a bit threatening because it seems like the only way to “interpet” mathematicians is by learning the math that they are talking about. When intersubjective understanding requires understanding verbatim, that suggests the subject matter is more objectively true than not.
The gradual expansion of computer science towards the social science through “big data” analysis can be seen as a gradual expansion of what can be considered under mathematical closure.
Physicists still want to mathematize their descriptions of the universe. Some psychologists want to mathematize their descriptions. Some political scientists, sociologists, etc. want to mathematize their descriptions. Anthropologists don’t want to mathematize their descriptions. Mathematization is at the heart of the quantitative/qualitative dispute.
It’s worth noting that there are non-mathematized predictive theories, as well as mathematized theories that pretty much fail to predict anything. | 3,712 | 19,187 | {"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-2023-23 | latest | en | 0.956976 |
https://www.gamedev.net/forums/topic/634894-help-me-understand-rendering-a-quad-down-the-z-axis/ | 1,516,682,070,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891706.88/warc/CC-MAIN-20180123032443-20180123052443-00447.warc.gz | 896,148,171 | 37,647 | # OpenGL help me understand rendering a quad down the z-axis
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I've used the code from nehe's lesson2 to build a simple app that demonstrates the problem I'm having.
The code is original, except I've removed the gluPerspective line, made it non-fullscreen and changed the drawGLScene() routine.
Here is the new drawGLScene() :
int drawGLScene(GLvoid) { glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); glLoadIdentity(); glTranslatef(0, 0.0f, -0.5f); glColor3f(1, 1, 1); glBegin(GL_QUADS); glVertex3f(-0.5f, 0.5f, 0.2f); glVertex3f(0.5f, 0.5f, 0.2f); glVertex3f(0.5f, -0.5f, 0.2f); glVertex3f(-0.5f, -0.5f, 0.2f); glEnd(); glColor3f(1, 0, 0); glBegin(GL_QUADS); glVertex3f(-0.5f, 0.5f, -0.2f); glVertex3f(0.5f, 0.5f, -0.2f); glVertex3f(0.5f, -0.5f, -0.2f); glVertex3f(-0.5f, -0.5f, -0.2f); glEnd(); if (GLWin.doubleBuffered) { glXSwapBuffers(GLWin.dpy, GLWin.win); } return True; }
As you can see, I'm drawing 2 rectangles at the same position, except along the z-axis.
The first is colored white, and has its vectices for z at 0.2f. The second is red, and z is at -0.2f.
For some reason, the red quad appears in front of the white one. Seeing how opengl treats the negative z axis as going into the screen, I'm trying to understand why the white quad (z at 0.2f) isn't drawn in front of the red one.
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Because you probably didn't enable depth test. In that case OpenGL draws objects in the order you issue commands (like in 2D drawing).
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Depth testing is enabled. Even if I switch the drawing order, the red quad is still up front.
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If you use lesson 2, you set glDepthFunc(GL_LEQUAL);, and therefore, lesser depth values will be drawn over higher depth values.
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Two points:
1. The near and far clip plane values, as specified by the glOrtho or glFrustum (or equivalent functions) are along the negative Z-axis. That means a clip plane value of 5 means that the clip plane is located at z=-5.
2. If you leave out the projection matrix, the default matrix is the identity matrix which is equivalent to glOrtho(-1, 1, -1, 1, 1, -1).
Pay attention to the default clip plane values; near=1 and far=-1. This means that the near clip plane is located at z=-1, and the far clip plane is located at z=1. This means that a quad at z=-0.2 is close to the near clip plane than a quad at z=0.2, and thus your red quad occludes your white quad. Edited by Brother Bob
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Ok. I get the idea. Thanks for the help.
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• ### Similar Content
• By EddieK
Hello. I'm trying to make an android game and I have come across a problem. I want to draw different map layers at different Z depths so that some of the tiles are drawn above the player while others are drawn under him. But there's an issue where the pixels with alpha drawn above the player. This is the code i'm using:
int setup(){ GLES20.glEnable(GLES20.GL_DEPTH_TEST); GLES20.glEnable(GL10.GL_ALPHA_TEST); GLES20.glEnable(GLES20.GL_TEXTURE_2D); } int render(){ GLES20.glClearColor(0, 0, 0, 0); GLES20.glClear(GLES20.GL_ALPHA_BITS); GLES20.glClear(GLES20.GL_COLOR_BUFFER_BIT); GLES20.glClear(GLES20.GL_DEPTH_BUFFER_BIT); GLES20.glBlendFunc(GLES20.GL_ONE, GL10.GL_ONE_MINUS_SRC_ALPHA); // do the binding of textures and drawing vertices } My vertex shader:
uniform mat4 MVPMatrix; // model-view-projection matrix uniform mat4 projectionMatrix; attribute vec4 position; attribute vec2 textureCoords; attribute vec4 color; attribute vec3 normal; varying vec4 outColor; varying vec2 outTexCoords; varying vec3 outNormal; void main() { outNormal = normal; outTexCoords = textureCoords; outColor = color; gl_Position = MVPMatrix * position; } My fragment shader:
precision highp float; uniform sampler2D texture; varying vec4 outColor; varying vec2 outTexCoords; varying vec3 outNormal; void main() { vec4 color = texture2D(texture, outTexCoords) * outColor; gl_FragColor = vec4(color.r,color.g,color.b,color.a);//color.a); } I have attached a picture of how it looks. You can see the black squares near the tree. These squares should be transparent as they are in the png image:
Its strange that in this picture instead of alpha or just black color it displays the grass texture beneath the player and the tree:
Any ideas on how to fix this?
• This article uses material originally posted on Diligent Graphics web site.
Introduction
Graphics APIs have come a long way from small set of basic commands allowing limited control of configurable stages of early 3D accelerators to very low-level programming interfaces exposing almost every aspect of the underlying graphics hardware. Next-generation APIs, Direct3D12 by Microsoft and Vulkan by Khronos are relatively new and have only started getting widespread adoption and support from hardware vendors, while Direct3D11 and OpenGL are still considered industry standard. New APIs can provide substantial performance and functional improvements, but may not be supported by older hardware. An application targeting wide range of platforms needs to support Direct3D11 and OpenGL. New APIs will not give any advantage when used with old paradigms. It is totally possible to add Direct3D12 support to an existing renderer by implementing Direct3D11 interface through Direct3D12, but this will give zero benefits. Instead, new approaches and rendering architectures that leverage flexibility provided by the next-generation APIs are expected to be developed.
There are at least four APIs (Direct3D11, Direct3D12, OpenGL/GLES, Vulkan, plus Apple's Metal for iOS and osX platforms) that a cross-platform 3D application may need to support. Writing separate code paths for all APIs is clearly not an option for any real-world application and the need for a cross-platform graphics abstraction layer is evident. The following is the list of requirements that I believe such layer needs to satisfy:
Lightweight abstractions: the API should be as close to the underlying native APIs as possible to allow an application leverage all available low-level functionality. In many cases this requirement is difficult to achieve because specific features exposed by different APIs may vary considerably. Low performance overhead: the abstraction layer needs to be efficient from performance point of view. If it introduces considerable amount of overhead, there is no point in using it. Convenience: the API needs to be convenient to use. It needs to assist developers in achieving their goals not limiting their control of the graphics hardware. Multithreading: ability to efficiently parallelize work is in the core of Direct3D12 and Vulkan and one of the main selling points of the new APIs. Support for multithreading in a cross-platform layer is a must. Extensibility: no matter how well the API is designed, it still introduces some level of abstraction. In some cases the most efficient way to implement certain functionality is to directly use native API. The abstraction layer needs to provide seamless interoperability with the underlying native APIs to provide a way for the app to add features that may be missing. Diligent Engine is designed to solve these problems. Its main goal is to take advantages of the next-generation APIs such as Direct3D12 and Vulkan, but at the same time provide support for older platforms via Direct3D11, OpenGL and OpenGLES. Diligent Engine exposes common C++ front-end for all supported platforms and provides interoperability with underlying native APIs. It also supports integration with Unity and is designed to be used as graphics subsystem in a standalone game engine, Unity native plugin or any other 3D application. Full source code is available for download at GitHub and is free to use.
Overview
Diligent Engine API takes some features from Direct3D11 and Direct3D12 as well as introduces new concepts to hide certain platform-specific details and make the system easy to use. It contains the following main components:
Render device (IRenderDevice interface) is responsible for creating all other objects (textures, buffers, shaders, pipeline states, etc.).
Device context (IDeviceContext interface) is the main interface for recording rendering commands. Similar to Direct3D11, there are immediate context and deferred contexts (which in Direct3D11 implementation map directly to the corresponding context types). Immediate context combines command queue and command list recording functionality. It records commands and submits the command list for execution when it contains sufficient number of commands. Deferred contexts are designed to only record command lists that can be submitted for execution through the immediate context.
An alternative way to design the API would be to expose command queue and command lists directly. This approach however does not map well to Direct3D11 and OpenGL. Besides, some functionality (such as dynamic descriptor allocation) can be much more efficiently implemented when it is known that a command list is recorded by a certain deferred context from some thread.
The approach taken in the engine does not limit scalability as the application is expected to create one deferred context per thread, and internally every deferred context records a command list in lock-free fashion. At the same time this approach maps well to older APIs.
In current implementation, only one immediate context that uses default graphics command queue is created. To support multiple GPUs or multiple command queue types (compute, copy, etc.), it is natural to have one immediate contexts per queue. Cross-context synchronization utilities will be necessary.
Swap Chain (ISwapChain interface). Swap chain interface represents a chain of back buffers and is responsible for showing the final rendered image on the screen.
Render device, device contexts and swap chain are created during the engine initialization.
Resources (ITexture and IBuffer interfaces). There are two types of resources - textures and buffers. There are many different texture types (2D textures, 3D textures, texture array, cubmepas, etc.) that can all be represented by ITexture interface.
Resources Views (ITextureView and IBufferView interfaces). While textures and buffers are mere data containers, texture views and buffer views describe how the data should be interpreted. For instance, a 2D texture can be used as a render target for rendering commands or as a shader resource.
Pipeline State (IPipelineState interface). GPU pipeline contains many configurable stages (depth-stencil, rasterizer and blend states, different shader stage, etc.). Direct3D11 uses coarse-grain objects to set all stage parameters at once (for instance, a rasterizer object encompasses all rasterizer attributes), while OpenGL contains myriad functions to fine-grain control every individual attribute of every stage. Both methods do not map very well to modern graphics hardware that combines all states into one monolithic state under the hood. Direct3D12 directly exposes pipeline state object in the API, and Diligent Engine uses the same approach.
Shader Resource Binding (IShaderResourceBinding interface). Shaders are programs that run on the GPU. Shaders may access various resources (textures and buffers), and setting correspondence between shader variables and actual resources is called resource binding. Resource binding implementation varies considerably between different API. Diligent Engine introduces a new object called shader resource binding that encompasses all resources needed by all shaders in a certain pipeline state.
API Basics
Creating Resources
Device resources are created by the render device. The two main resource types are buffers, which represent linear memory, and textures, which use memory layouts optimized for fast filtering. Graphics APIs usually have a native object that represents linear buffer. Diligent Engine uses IBuffer interface as an abstraction for a native buffer. To create a buffer, one needs to populate BufferDesc structure and call IRenderDevice::CreateBuffer() method as in the following example:
BufferDesc BuffDesc; BufferDesc.Name = "Uniform buffer"; BuffDesc.BindFlags = BIND_UNIFORM_BUFFER; BuffDesc.Usage = USAGE_DYNAMIC; BuffDesc.uiSizeInBytes = sizeof(ShaderConstants); BuffDesc.CPUAccessFlags = CPU_ACCESS_WRITE; m_pDevice->CreateBuffer( BuffDesc, BufferData(), &m_pConstantBuffer ); While there is usually just one buffer object, different APIs use very different approaches to represent textures. For instance, in Direct3D11, there are ID3D11Texture1D, ID3D11Texture2D, and ID3D11Texture3D objects. In OpenGL, there is individual object for every texture dimension (1D, 2D, 3D, Cube), which may be a texture array, which may also be multisampled (i.e. GL_TEXTURE_2D_MULTISAMPLE_ARRAY). As a result there are nine different GL texture types that Diligent Engine may create under the hood. In Direct3D12, there is only one resource interface. Diligent Engine hides all these details in ITexture interface. There is only one IRenderDevice::CreateTexture() method that is capable of creating all texture types. Dimension, format, array size and all other parameters are specified by the members of the TextureDesc structure:
TextureDesc TexDesc; TexDesc.Name = "My texture 2D"; TexDesc.Type = TEXTURE_TYPE_2D; TexDesc.Width = 1024; TexDesc.Height = 1024; TexDesc.Format = TEX_FORMAT_RGBA8_UNORM; TexDesc.Usage = USAGE_DEFAULT; TexDesc.BindFlags = BIND_SHADER_RESOURCE | BIND_RENDER_TARGET | BIND_UNORDERED_ACCESS; TexDesc.Name = "Sample 2D Texture"; m_pRenderDevice->CreateTexture( TexDesc, TextureData(), &m_pTestTex ); If native API supports multithreaded resource creation, textures and buffers can be created by multiple threads simultaneously.
Interoperability with native API provides access to the native buffer/texture objects and also allows creating Diligent Engine objects from native handles. It allows applications seamlessly integrate native API-specific code with Diligent Engine.
Next-generation APIs allow fine level-control over how resources are allocated. Diligent Engine does not currently expose this functionality, but it can be added by implementing IResourceAllocator interface that encapsulates specifics of resource allocation and providing this interface to CreateBuffer() or CreateTexture() methods. If null is provided, default allocator should be used.
Initializing the Pipeline State
As it was mentioned earlier, Diligent Engine follows next-gen APIs to configure the graphics/compute pipeline. One big Pipelines State Object (PSO) encompasses all required states (all shader stages, input layout description, depth stencil, rasterizer and blend state descriptions etc.). This approach maps directly to Direct3D12/Vulkan, but is also beneficial for older APIs as it eliminates pipeline misconfiguration errors. With many individual calls tweaking various GPU pipeline settings it is very easy to forget to set one of the states or assume the stage is already properly configured when in fact it is not. Using pipeline state object helps avoid these problems as all stages are configured at once.
While in earlier APIs shaders were bound separately, in the next-generation APIs as well as in Diligent Engine shaders are part of the pipeline state object. The biggest challenge when authoring shaders is that Direct3D and OpenGL/Vulkan use different shader languages (while Apple uses yet another language in their Metal API). Maintaining two versions of every shader is not an option for real applications and Diligent Engine implements shader source code converter that allows shaders authored in HLSL to be translated to GLSL. To create a shader, one needs to populate ShaderCreationAttribs structure. SourceLanguage member of this structure tells the system which language the shader is authored in:
When sampling a texture in a shader, the texture sampler was traditionally specified as separate object that was bound to the pipeline at run time or set as part of the texture object itself. However, in most cases it is known beforehand what kind of sampler will be used in the shader. Next-generation APIs expose new type of sampler called static sampler that can be initialized directly in the pipeline state. Diligent Engine exposes this functionality: when creating a shader, textures can be assigned static samplers. If static sampler is assigned, it will always be used instead of the one initialized in the texture shader resource view. To initialize static samplers, prepare an array of StaticSamplerDesc structures and initialize StaticSamplers and NumStaticSamplers members. Static samplers are more efficient and it is highly recommended to use them whenever possible. On older APIs, static samplers are emulated via generic sampler objects.
The following is an example of shader initialization:
Creating the Pipeline State Object
After all required shaders are created, the rest of the fields of the PipelineStateDesc structure provide depth-stencil, rasterizer, and blend state descriptions, the number and format of render targets, input layout format, etc. For instance, rasterizer state can be described as follows:
PipelineStateDesc PSODesc; RasterizerStateDesc &RasterizerDesc = PSODesc.GraphicsPipeline.RasterizerDesc; RasterizerDesc.FillMode = FILL_MODE_SOLID; RasterizerDesc.CullMode = CULL_MODE_NONE; RasterizerDesc.FrontCounterClockwise = True; RasterizerDesc.ScissorEnable = True; RasterizerDesc.AntialiasedLineEnable = False; Depth-stencil and blend states are defined in a similar fashion.
Another important thing that pipeline state object encompasses is the input layout description that defines how inputs to the vertex shader, which is the very first shader stage, should be read from the memory. Input layout may define several vertex streams that contain values of different formats and sizes:
// Define input layout InputLayoutDesc &Layout = PSODesc.GraphicsPipeline.InputLayout; LayoutElement TextLayoutElems[] = { LayoutElement( 0, 0, 3, VT_FLOAT32, False ), LayoutElement( 1, 0, 4, VT_UINT8, True ), LayoutElement( 2, 0, 2, VT_FLOAT32, False ), }; Layout.LayoutElements = TextLayoutElems; Layout.NumElements = _countof( TextLayoutElems ); Finally, pipeline state defines primitive topology type. When all required members are initialized, a pipeline state object can be created by IRenderDevice::CreatePipelineState() method:
// Define shader and primitive topology PSODesc.GraphicsPipeline.PrimitiveTopologyType = PRIMITIVE_TOPOLOGY_TYPE_TRIANGLE; PSODesc.GraphicsPipeline.pVS = pVertexShader; PSODesc.GraphicsPipeline.pPS = pPixelShader; PSODesc.Name = "My pipeline state"; m_pDev->CreatePipelineState(PSODesc, &m_pPSO); When PSO object is bound to the pipeline, the engine invokes all API-specific commands to set all states specified by the object. In case of Direct3D12 this maps directly to setting the D3D12 PSO object. In case of Direct3D11, this involves setting individual state objects (such as rasterizer and blend states), shaders, input layout etc. In case of OpenGL, this requires a number of fine-grain state tweaking calls. Diligent Engine keeps track of currently bound states and only calls functions to update these states that have actually changed.
Direct3D11 and OpenGL utilize fine-grain resource binding models, where an application binds individual buffers and textures to certain shader or program resource binding slots. Direct3D12 uses a very different approach, where resource descriptors are grouped into tables, and an application can bind all resources in the table at once by setting the table in the command list. Resource binding model in Diligent Engine is designed to leverage this new method. It introduces a new object called shader resource binding that encapsulates all resource bindings required for all shaders in a certain pipeline state. It also introduces the classification of shader variables based on the frequency of expected change that helps the engine group them into tables under the hood:
Static variables (SHADER_VARIABLE_TYPE_STATIC) are variables that are expected to be set only once. They may not be changed once a resource is bound to the variable. Such variables are intended to hold global constants such as camera attributes or global light attributes constant buffers. Mutable variables (SHADER_VARIABLE_TYPE_MUTABLE) define resources that are expected to change on a per-material frequency. Examples may include diffuse textures, normal maps etc. Dynamic variables (SHADER_VARIABLE_TYPE_DYNAMIC) are expected to change frequently and randomly. Shader variable type must be specified during shader creation by populating an array of ShaderVariableDesc structures and initializing ShaderCreationAttribs::Desc::VariableDesc and ShaderCreationAttribs::Desc::NumVariables members (see example of shader creation above).
Static variables cannot be changed once a resource is bound to the variable. They are bound directly to the shader object. For instance, a shadow map texture is not expected to change after it is created, so it can be bound directly to the shader:
m_pPSO->CreateShaderResourceBinding(&m_pSRB); Note that an SRB is only compatible with the pipeline state it was created from. SRB object inherits all static bindings from shaders in the pipeline, but is not allowed to change them.
Mutable resources can only be set once for every instance of a shader resource binding. Such resources are intended to define specific material properties. For instance, a diffuse texture for a specific material is not expected to change once the material is defined and can be set right after the SRB object has been created:
m_pSRB->GetVariable(SHADER_TYPE_PIXEL, "tex2DDiffuse")->Set(pDiffuseTexSRV); In some cases it is necessary to bind a new resource to a variable every time a draw command is invoked. Such variables should be labeled as dynamic, which will allow setting them multiple times through the same SRB object:
m_pSRB->GetVariable(SHADER_TYPE_VERTEX, "cbRandomAttribs")->Set(pRandomAttrsCB); Under the hood, the engine pre-allocates descriptor tables for static and mutable resources when an SRB objcet is created. Space for dynamic resources is dynamically allocated at run time. Static and mutable resources are thus more efficient and should be used whenever possible.
As you can see, Diligent Engine does not expose low-level details of how resources are bound to shader variables. One reason for this is that these details are very different for various APIs. The other reason is that using low-level binding methods is extremely error-prone: it is very easy to forget to bind some resource, or bind incorrect resource such as bind a buffer to the variable that is in fact a texture, especially during shader development when everything changes fast. Diligent Engine instead relies on shader reflection system to automatically query the list of all shader variables. Grouping variables based on three types mentioned above allows the engine to create optimized layout and take heavy lifting of matching resources to API-specific resource location, register or descriptor in the table.
This post gives more details about the resource binding model in Diligent Engine.
Setting the Pipeline State and Committing Shader Resources
Before any draw or compute command can be invoked, the pipeline state needs to be bound to the context:
m_pContext->SetPipelineState(m_pPSO); Under the hood, the engine sets the internal PSO object in the command list or calls all the required native API functions to properly configure all pipeline stages.
The next step is to bind all required shader resources to the GPU pipeline, which is accomplished by IDeviceContext::CommitShaderResources() method:
m_pContext->CommitShaderResources(m_pSRB, COMMIT_SHADER_RESOURCES_FLAG_TRANSITION_RESOURCES); The method takes a pointer to the shader resource binding object and makes all resources the object holds available for the shaders. In the case of D3D12, this only requires setting appropriate descriptor tables in the command list. For older APIs, this typically requires setting all resources individually.
Next-generation APIs require the application to track the state of every resource and explicitly inform the system about all state transitions. For instance, if a texture was used as render target before, while the next draw command is going to use it as shader resource, a transition barrier needs to be executed. Diligent Engine does the heavy lifting of state tracking. When CommitShaderResources() method is called with COMMIT_SHADER_RESOURCES_FLAG_TRANSITION_RESOURCES flag, the engine commits and transitions resources to correct states at the same time. Note that transitioning resources does introduce some overhead. The engine tracks state of every resource and it will not issue the barrier if the state is already correct. But checking resource state is an overhead that can sometimes be avoided. The engine provides IDeviceContext::TransitionShaderResources() method that only transitions resources:
m_pContext->TransitionShaderResources(m_pPSO, m_pSRB); In some scenarios it is more efficient to transition resources once and then only commit them.
Invoking Draw Command
The final step is to set states that are not part of the PSO, such as render targets, vertex and index buffers. Diligent Engine uses Direct3D11-syle API that is translated to other native API calls under the hood:
ITextureView *pRTVs[] = {m_pRTV}; m_pContext->SetRenderTargets(_countof( pRTVs ), pRTVs, m_pDSV); // Clear render target and depth buffer const float zero[4] = {0, 0, 0, 0}; m_pContext->ClearRenderTarget(nullptr, zero); m_pContext->ClearDepthStencil(nullptr, CLEAR_DEPTH_FLAG, 1.f); // Set vertex and index buffers IBuffer *buffer[] = {m_pVertexBuffer}; Uint32 offsets[] = {0}; Uint32 strides[] = {sizeof(MyVertex)}; m_pContext->SetVertexBuffers(0, 1, buffer, strides, offsets, SET_VERTEX_BUFFERS_FLAG_RESET); m_pContext->SetIndexBuffer(m_pIndexBuffer, 0); Different native APIs use various set of function to execute draw commands depending on command details (if the command is indexed, instanced or both, what offsets in the source buffers are used etc.). For instance, there are 5 draw commands in Direct3D11 and more than 9 commands in OpenGL with something like glDrawElementsInstancedBaseVertexBaseInstance not uncommon. Diligent Engine hides all details with single IDeviceContext::Draw() method that takes takes DrawAttribs structure as an argument. The structure members define all attributes required to perform the command (primitive topology, number of vertices or indices, if draw call is indexed or not, if draw call is instanced or not, if draw call is indirect or not, etc.). For example:
DrawAttribs attrs; attrs.IsIndexed = true; attrs.IndexType = VT_UINT16; attrs.NumIndices = 36; attrs.Topology = PRIMITIVE_TOPOLOGY_TRIANGLE_LIST; pContext->Draw(attrs); For compute commands, there is IDeviceContext::DispatchCompute() method that takes DispatchComputeAttribs structure that defines compute grid dimension.
Source Code
Full engine source code is available on GitHub and is free to use. The repository contains two samples, asteroids performance benchmark and example Unity project that uses Diligent Engine in native plugin.
AntTweakBar sample is Diligent Engine’s “Hello World” example.
Atmospheric scattering sample is a more advanced example. It demonstrates how Diligent Engine can be used to implement various rendering tasks: loading textures from files, using complex shaders, rendering to multiple render targets, using compute shaders and unordered access views, etc.
Asteroids performance benchmark is based on this demo developed by Intel. It renders 50,000 unique textured asteroids and allows comparing performance of Direct3D11 and Direct3D12 implementations. Every asteroid is a combination of one of 1000 unique meshes and one of 10 unique textures.
Finally, there is an example project that shows how Diligent Engine can be integrated with Unity.
Future Work
The engine is under active development. It currently supports Windows desktop, Universal Windows and Android platforms. Direct3D11, Direct3D12, OpenGL/GLES backends are now feature complete. Vulkan backend is coming next, and support for more platforms is planned.
• By reenigne
For those that don't know me. I am the individual who's two videos are listed here under setup for https://wiki.libsdl.org/Tutorials
I also run grhmedia.com where I host the projects and code for the tutorials I have online.
Recently, I received a notice from youtube they will be implementing their new policy in protecting video content as of which I won't be monetized till I meat there required number of viewers and views each month.
Frankly, I'm pretty sick of youtube. I put up a video and someone else learns from it and puts up another video and because of the way youtube does their placement they end up with more views.
Even guys that clearly post false information such as one individual who said GLEW 2.0 was broken because he didn't know how to compile it. He in short didn't know how to modify the script he used because he didn't understand make files and how the requirements of the compiler and library changes needed some different flags.
At the end of the month when they implement this I will take down the content and host on my own server purely and it will be a paid system and or patreon.
I get my videos may be a bit dry, I generally figure people are there to learn how to do something and I rather not waste their time.
I used to also help people for free even those coming from the other videos. That won't be the case any more. I used to just take anyone emails and work with them my email is posted on the site.
I don't expect to get the required number of subscribers in that time or increased views. Even if I did well it wouldn't take care of each reoccurring month.
I figure this is simpler and I don't plan on putting some sort of exorbitant fee for a monthly subscription or the like.
I was thinking on the lines of a few dollars 1,2, and 3 and the larger subscription gets you assistance with the content in the tutorials if needed that month.
Maybe another fee if it is related but not directly in the content.
The fees would serve to cut down on the number of people who ask for help and maybe encourage some of the people to actually pay attention to what is said rather than do their own thing. That actually turns out to be 90% of the issues. I spent 6 hours helping one individual last week I must have asked him 20 times did you do exactly like I said in the video even pointed directly to the section. When he finally sent me a copy of the what he entered I knew then and there he had not. I circled it and I pointed out that wasn't what I said to do in the video. I didn't tell him what was wrong and how I knew that way he would go back and actually follow what it said to do. He then reported it worked. Yea, no kidding following directions works. But hey isn't alone and well its part of the learning process.
So the point of this isn't to be a gripe session. I'm just looking for a bit of feed back. Do you think the fees are unreasonable?
Should I keep the youtube channel and do just the fees with patreon or do you think locking the content to my site and require a subscription is an idea.
I'm just looking at the fact it is unrealistic to think youtube/google will actually get stuff right or that youtube viewers will actually bother to start looking for more accurate videos.
• i got error 1282 in my code. | 6,817 | 31,781 | {"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} | 2.53125 | 3 | CC-MAIN-2018-05 | longest | en | 0.853372 |
https://math.stackexchange.com/questions/3690547/analytic-estimates-of-limit-cycle-parameters | 1,716,805,846,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00245.warc.gz | 322,920,505 | 35,529 | # Analytic estimates of limit cycle parameters
Suppose we have a two-dimensional system of differential equations, say, the well-known Van der Pol oscillator:
$$\dot{x}=y, \dot{y}=\mu (1-x^2)y-x$$
Everyone knows that the study of limit cycles is a very complex problem. Each of them is unique in its own way, and there is no universal set of parameters characterizing each of them. As I understand it, in most cases the limit cycles are studied by numerical and graphical methods.
Are there approximate analytical methods that allow at least an average estimation of the amplitude and frequency of the limit cycle (for complex limit cycles, these concepts are very vague)?
Let me explain what I mean by the amplitude and frequency of limit cycles. The limit cycle of the Van der Pol oscillator has a very characteristic shape, therefore, parameters such as amplitude and frequency are not applicable to it. On the other hand, the amplitude can be considered the radius of the circle beyond which the limit cycle does not extend, and the frequency is the number of complete passage along the path of the limit cycle per second.
• I will supplement the question. Period $T$ can also be a parameter of the limit cycle.
– dtn
May 25, 2020 at 7:56
• Are you interested in the perturbation analysis where $\mu$ is small, or in the slow-fast dynamic you get when $\mu$ is large? See my comments to this question and the links in them for some partial answers, and esp. math.stackexchange.com/q/1564464/115115 for an exhaustive answer (the accepted one) on the periods of the Van der Pol oscillator. May 25, 2020 at 10:18
• I will look at these links. I construct various differential equations, not only Van der Pol equations, but also others where a limit cycle exists (for example, the Duffing Oscillator). I am not interested in the answer to the question: "the existence and decay of the limit cycle, depending on the value of the parameters." I'm interested in ways of more or less universal estimation of the amplitude and frequency of the limit cycle, which is suitable for a wide class of differential equations.
– dtn
May 25, 2020 at 10:25
• I answered your question, or is there something you need to clarify?
– dtn
May 25, 2020 at 10:25
• This can be seen as easy or complicated, often it is both. Most "random" ODE systems do not have limit cycles. They may have solutions that look for some while like a limit cycle but in the end this again deteriorates (like in the Lorenz attractor, or even with less structure). To have a guaranteed limit cycle needs a strong structural reason. This reason then often also suggests a nearby system with a computable cycle so that the wanted system can be treated as perturbation of it, as was done is the cited link with separate approaches for small and large $\mu$. May 25, 2020 at 10:33 | 647 | 2,839 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.917827 |
https://zacksinvest.com/gw17a89b/ | 1,611,641,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704798089.76/warc/CC-MAIN-20210126042704-20210126072704-00262.warc.gz | 1,099,340,169 | 9,674 | Finding Percent Of A Number Math Worksheet
Ami Maylis December 1, 2020 Worksheet
Have you ever noticed how many K-12 math content websites are devoted to math worksheets? There seems to be an increase in websites that cater to desperate teachers and parents by offering fast, free ”worksheet generation”, ”10 free fractions worksheets,” etc. Now, as a former teacher I am not saying that one should never use math worksheets; however, I do believe that many teachers are using a very superficial method of instruction that relies too much on low-level math worksheets and hands-off instructional approaches. Worksheet lessons move from reading the directions aloud, to doing sample problems as a group, to completing the worksheet independently (or at home with parents), day in and day out.
Most of even beginning algebra depends on being able to do two things–one, doing multiplication quickly and accurately in your head, two, knowing how to add, subtract, multiply, and divide fractions. You might remember a concept in algebra called ”factoring.” Factoring means breaking up into parts that are multiplied together to give you the whole. You can factor numbers. For instance, 6 factors into 2 and 3–2×3 =6. In elementary algebra we learn to factor expressions such as x^2+4x+4. This particular expression is easily factorable into (x+2)^2. If this doesn’t make any sense to you, don’t worry about it. Just trust me, if you don’t know your multiplication tables, you can’t factor. If you can’t factor, you won’t do well at all in algebra, geometry, or trigonometry.
Another advantage of these math worksheets is that kids and parents will be able to keep them to serve as their references for review. Since worksheets are easy to correct, students will be able to identify the items and areas that they had mistakes so that they will be able to correct those deficiencies. Keeping record is really a good thing; As a parent, you will be able to go back through them and assess their strong and weak areas. Keeping track you will be able to track your child’s progress as empirical evidence.
As a parent, I’m very aware of what my own children are learning in school. For the most part, I’ve been happy with their progress, but as they rise in grade level, I’m starting to see more emphasis on a loose understanding of the concepts and less emphasis on skills–particularly skills with arithmetic of fractions. The main problem with what I see with my students and my own children is that kids are taught ”concepts” and are not taught skills–unless they’re lucky enough to have a teacher who knows better. Most particularly, children are not taught mastery of arithmetic with fractions. Unfortunately, virtually all of their future math education depends on being able to do fractional arithmetic.
Math worksheets rarely ask students to think critically or creatively. They usually present multiple examples of the same problem type with the hope of reinforcing a skill or procedure. They do not challenge students to use higher order thinking skills such as comparing, analyzing, deducing, and synthesizing. These skills are built through activities in which students discover concepts, explore ideas, test a hypothesis, solve a problem, and discuss their thinking with their peers. Exploring concepts and problems in many different ways builds interest and promotes critical thinking.
Of course, there is a place for math worksheets. After some instruction has occurred, math worksheets can provide extended practice and support development in fluency, provided the teacher is engaged with students as they work. Teachers who are effective at grouping students can use math worksheets as a springboard for discussions, discovery, and communication. So the next time you do a search for curriculum materials, skip the worksheets. Instead, consider resources that provide interactive experiences or consider sites that provide students with challenging problems. These sites will more likely engage students, foster discussion, and build a true understanding of the purpose and joy of learning math.
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Dec 25, 2020 | 1,055 | 4,767 | {"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-2021-04 | latest | en | 0.958842 |
http://www.bayhealth.org/home-care/BayhealthContentPage.aspx?nd=839&parm1=2127&parm2=1&doc=true | 1,427,784,032,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131300441.51/warc/CC-MAIN-20150323172140-00162-ip-10-168-14-71.ec2.internal.warc.gz | 365,967,168 | 13,796 | In today's society, the media focus on celebrities and whether they lose or gain a pound, so it's hard to grasp the idea of a healthy weight.
Obesity is determined by percentage of body fat and weight, according to the National Heart, Lung and Blood Institute (NHLBI). Having a large percentage of body fat, regardless of how much you weigh, is unhealthy. You could be of normal weight or underweight and still have an unhealthy amount of body fat. Being overweight means that you have a heavy weight, but not necessarily too much body fat. People who are muscular weigh more than those who are not. Their extra weight comes from muscle, not body fat.
According to NHLBI guidelines, an evaluation of whether you are overweight involves using 3 key measures:
• Body mass index (BMI)
• Waist circumference
• Risk factors for diseases and conditions associated with obesity. These include high blood pressure, coronary artery disease, high level of LDL ("bad") cholesterol, high triglycerides, high blood sugar, and smoking.
## Body mass index
The first measurement to find out whether you are overweight or obese involves determining your body mass index, or BMI. To calculate your BMI, multiply your weight in pounds by 703, then divide by your height in inches. Divide by your height in inches again.
BMI = ( Weight in Pounds ) X 703
(Height in inches) X (Height in inches)
For example, if you weigh 162 pounds and are 69 inches (5 feet 9 inches) tall, your BMI is 23.9, or (162 x 703) ÷ (69 x 69), and is normal.
• A BMI of 18.4 or below: Underweight
• A BMI of 18.5 to 24.9: Normal
• A BMI of 25 to 29.9: Overweight
• A BMI of 30 or greater: Obese
For people who are considered obese or those who are overweight and have 2 or more risk factors (like high blood pressure, diabetes, abnormal blood fats, smoking, or coronary artery disease), the NHLBI guidelines recommend losing weight.
Although BMI is a reliable indicator of total body fat, it does have some limits. It may overestimate body fat in athletes and others with a muscular build. It may underestimate body fat in older adults and in others who have lost muscle mass.
## Belly fat or waist circumference
Another way to determine whether you are obese or overweight is to measure your belly fat. This can predict the risk for diseases associated with obesity. Determine your waist circumference by placing a measuring tape snugly around your waist. Your waist circumference is a good indicator of your belly fat. Your risk for diseases related to obesity increases with a waist measurement of more than 40 inches in men and more than 35 inches in women.
You should use your belly fat measurement along with your BMI to determine your risk. People who are overweight (BMI of 25 to 29.9), but don't have a large waist measurement and have fewer than 2 risk factors may need only to prevent additional weight gain rather than to lose weight.
If you are overweight and have other risk factors for diseases related to obesity, you should lose weight. If you are obese, you should lose weight even if you have no other risk factors.
## Skin fold thickness
A third way to determine whether you are obese or overweight is to have your skin fold thickness measured. This measurement is taken around the triceps (muscles of the upper arm), on the shoulder blades, and on the hips. The results can determine whether your weight is from muscle or from fat.
The BMI method is generally preferred because BMI shows a more accurate representation of total body fat that includes both height and weight, the NHLBI says. For the best determination of whether you're at a healthy weight, talk to your health care provider. Your health care provider will consider both your personal health and family history.
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Bayhealth is Southern Delaware’s healthcare leader with hospitals in Dover and in Milford. Bayhealth provides a wide range of medical services, including cardiovascular, cancer, orthopaedics and rehabilitation, pediatrics, respiratory care, sleep care, surgical weight loss, women’s services and walk-in medical care. Search for nursing jobs, and health classes and events. Find doctors affiliated with Bayhealth Medical Center or a Delaware hospital near you. | 916 | 4,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-14 | latest | en | 0.895991 |
http://mathhelpforum.com/advanced-statistics/36834-probability-number-line.html | 1,526,896,076,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00262.warc.gz | 186,013,986 | 10,113 | # Thread: Probability on a number line.
1. ## Probability on a number line.
A marker is placed at the origin of a number line. A fair six-sided die is rolled seven times. On each toss, if a 1 or 2 is tossed, the marker is moved one unit to the right, otherwise, the marker is moved one unit to the left.
a) Determine the possible positions on the number line where the marker could end up after the seven tosses.
b)What is the probability that the marker will end up 5 or 6 units to the left of the origin?
c) What is the probability that the marker will end up 3 or fewer units from the origin?
d) Determine the expected value of this random experiment.
Please help!
2. Originally Posted by digitalis77
A marker is placed at the origin of a number line. A fair six-sided die is rolled seven times. On each toss, if a 1 or 2 is tossed, the marker is moved one unit to the right, otherwise, the marker is moved one unit to the left.
a) Determine the possible positions on the number line where the marker could end up after the seven tosses.
b)What is the probability that the marker will end up 5 or 6 units to the left of the origin?
c) What is the probability that the marker will end up 3 or fewer units from the origin?
d) Determine the expected value of this random experiment.
Please help!
Let X be the random variable position of marker after seven rolls.
Let Y be the random variable number of times the marker moves right. Note that Y ~ Binomial(n = 7, p = 1/3).
Thinking about the possible values of Y should show you that the possible values of X are -7, -5, -3, -1, 1, 3, 5, 7. It should also let you calculate the probability of each value of X. Then it should be straight forward to calculate:
b) Pr(X = -5) = ......
c) Pr(X = 3 or 1 or -1 or -3) = ......
d) E(X) = ......
3. Originally Posted by digitalis77
A marker is placed at the origin of a number line. A fair six-sided die is rolled seven times. On each toss, if a 1 or 2 is tossed, the marker is moved one unit to the right, otherwise, the marker is moved one unit to the left.
Determine the expected value of this experiment.
The expected value after one roll is $\displaystyle (P(1) + P(2))\cdot 1 + (P(3) + P(4) + P(5) + P(6))\cdot -1$, which is
$\displaystyle \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$. So the expected value of the experiment after 7 rolls is $\displaystyle -\frac{7}{3}$. | 637 | 2,387 | {"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} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.890472 |
http://mathhelpforum.com/math-topics/46499-sketching-graph-logarithmic-function.html | 1,505,824,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685129.23/warc/CC-MAIN-20170919112242-20170919132242-00359.warc.gz | 210,864,028 | 11,485 | # Thread: Sketching a graph of a logarithmic function
1. ## Sketching a graph of a logarithmic function
So I have to sketch a graph of the function:
y= -2 log₁₀ x
Now I can write the log₁₀ x in exponential form as 10^y=x
But what do I do about the -2? Do I then do a table with x and y etc.?
2. Hi synchro,
To sketch a graph of any function you need to consider;
1. The value of $y$ when $x~=~0$ In this case undefined.
2. The value of $x$ when $y~=~0$ In this case $1$
3. What happens to the value of $y$ as $x\rightarrow 0$ Note that the $\log$ of a fraction is negative, thus in this case $y\rightarrow \infty$
4. What happens to the value of $y$ as $x\rightarrow \infty$ In this case $y\rightarrow -\infty$
5. And the result for $x\rightarrow -\infty$ In this case undefined since the $\log$ of a negative number is undfined.
3. Hi!
I have to say I'm a bit confuzzled by your post. I only have to sketch graphs of functions of this type and really the only thing I know is that one part of it is doing a table like this: (though probs not these values)
Heh.
4. Constructing a table like that implies that your not actually sketching a graph, more drawing one.
The table should consist of 2/3 columns, namely $x$ , $\log_{10}x$ , $y$ of which the middle one can be erased according to your preference.
If you have the $x$ values then you can sub into the equation $y~=~-2\log_{10}x$ to find the $y$ values.
If you have the $y$ values then you can rearrange the equation accordingly to help you find the $x$ values;
We have $y~=~-2\log_{10}x$ Multiplying both sides by $-\frac{1}{2}$ gives $\log_{10}x~=~-\frac{1}{2} \cdot y~$ Finally raising both sides to a power of $10$ (and a bit of rearranging) leads us to obtain the following equation $x~=~\frac{1}{\sqrt{10^y}}~$ Now subbing the $y$ values into this equation will give you the corresponding $x$ values.
5. Ahh okay, I was mistaken. Yep the stipulation is to "sketch a graph of". Could you please explain the "key points" in your first post in this thread? As in, why are they to be considered? How do you sketch a graph after having gone through that 'checking'? I don't really know much about sketching functions, of this type or otherwise (though I only need to know how to sketch of this type, i.e. with a "log" in).
Many thanks!!
6. Originally Posted by synchro
Ahh okay, I was mistaken. Yep the stipulation is to "sketch a graph of". Could you please explain the "key points" in your first post in this thread? As in, why are they to be considered? How do you sketch a graph after having gone through that 'checking'? I don't really know much about sketching functions, of this type or otherwise (though I only need to know how to sketch of this type, i.e. with a "log" in).
Many thanks!!
to graph logs, the best way is to use transformations. that is, you should know what the general shape of a log graph is, and then you can shift, flip or stretch it to get the graph you want, provided the function is simple enough, which in this case, it is. see if this helps. you can search for similar threads, there are a lot | 850 | 3,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 29, "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} | 4.34375 | 4 | CC-MAIN-2017-39 | longest | en | 0.85372 |
https://www.shaalaa.com/question-bank-solutions/calculate-number-moles-12044-1025-atoms-phosphorus-atoms-building-blocks-of-matter_74221 | 1,653,712,107,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00588.warc.gz | 1,128,677,360 | 9,757 | # Calculate the number of moles in 12.044 × 1025 atoms of phosphorus. - Science
Short Note
Calculate the number of moles in 12.044 × 1025 atoms of phosphorus.
#### Solution
12.044 × 1025 atoms of phosphorus
6.022 × 1023 atoms of phosphorus = 1 mole
12.044 × 1025 atoms of phosphorus = 12.044 × 1025/6.022 × 1023 = 199.9 moles or 200 moles
Hence, 12.044 × 1025 atoms of phosphorus have 200 moles.
Concept: Atoms: Building Blocks of Matter
Is there an error in this question or solution?
#### APPEARS IN
Lakhmir Singh Class 9 Chemistry - Science Part 2
Chapter 3 Atoms and Molecules
Short Answers | Q 23 | Page 173 | 202 | 622 | {"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} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.688192 |
https://electronics.stackexchange.com/questions/384483/high-current-low-voltage-transformer-not-working | 1,653,607,449,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00561.warc.gz | 286,269,567 | 66,643 | # High current low voltage transformer not working
I have rewound a microwave oven transformer to produce a low voltage on the secondary. I'm not sure what the ampere rating of the transformer is but its about 4mm thick wire on the primary. I wound the transformer to be 93:3 turns, yielding about 3 volts on the secondary. If the input max is 20A, that gives me around 600A on the secondary. Still about 2200W. Trouble is, the secondary appears to have very little current running through it. Why is this?
Only reason I can think of is the inductance of the primary being high enough to limit the current to less than 20A or something. 93 turns, 80 mm loop diameter, 4mm wire diameter, and relative permeability of 4000 (transformer steel) yields 5.35 H, which is huge. I'll bet if I plugged that into the impedance formula of 2 x pi x f x H (60Hz) it would be a huge resistance. Yeah, its 2kOhm, which makes the current max on the primary 54 mA @ 110V I'm guessing? One thing I may have confused is the reactance for a air core coil vs an iron core coil.
Anyways question is, what's wrong with my transformer? Why isn't the secondary high current? What can I do to get high current? I realize I'm probably using the wrong wire size for 600A, yes.
• have you left the magnetic shunts in, or taken them out? Post a picture of it, that could help. How are you measuring the secondary current? 20A sounds very high, even for a MOT, where do you get that figure from? Jul 12, 2018 at 18:52
• Is that the bars of steel separating the coils? Theyre in. Jul 12, 2018 at 18:52
• I see how that would screw with the flux... Jul 12, 2018 at 18:53
• The magnetic shunts increase the leakage inductance, to tune out the series capacitor in the voltage doubler. They will limit the current severely, you need to remove them. Jul 12, 2018 at 18:54
• Russell's answer here is useful.
– jonk
Jul 12, 2018 at 19:00 | 508 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | longest | en | 0.95173 |
https://1library.net/article/wave-propagation-in-an-anisotropic-medium.oy87oo0z | 1,638,787,646,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00162.warc.gz | 140,983,459 | 18,646 | Wave propagation in an anisotropic medium
In document Computation of synthetic seismograms in layered earth models (Page 72-78)
64 general anisotropy, such analytical expressions are not available.
4.2 Wave propagation in an anisotropic medium
For a linearly elastic solid, the following generalized Hook's law relating stress and strain applies
T i j Cijkl \ l 4.1
where is the stress tensor, e ^ the Cauchy infinitesimal strain tensor and c. .. , the 4th order tensor of elastic moduli. Because of
elements of c _ ^ for the mos‘t general anisotropic medium are distinct. The number of distinct elements of c. .. n is reduced if the
ijkl
medium in question possesses symmetry plane(s). When the medium is isotropic, only two constants, the Lam& constants, are needed to describe the elastic medium. Away from the isotropy assumption, it can be shown that in general there are three body waves with distinct velocities that vary with the direction of propagation and with particle motions that are neither parallel nor normal to the wave/slowness vector (see, for example, Auld, 1973, and Crampin, 1981). However, the particle motions of these body waves are still orthogonal to one another. The difference now is that, in general there is coupling among the three body waves. When the direction of propagation is along the symmetry plane (and also certain non-symmetry directions), the propagation modes have particle motions either parallel or normal to the wave vector (Auld, 1973). This is always the case for waves propagating in an isotropic medium regardless of their direction of propagation. Even for point sources, the anisotropy of material deformation in response to an applied force gives rise to non-spherical wave surface. In general, the radius vector of the wave surface is not parallel to the wave vector. As the wave vector is normal to the wave surface, it is now necessary to distinguish between the group and phase velocities. A point on the wave surface travels along the ray with the group velocity whereas the plane of constant phase moves along its normal with the phase velocity. The shape of the wave surface can be extremely complicated, especially those of the shear-waves. The effect of anisotropy on wave propagation causes azimuthal and polarization anomalies in seismic observation. There is also the possibility of more than one direct shear-wave arrival
66
showing up in the seismogram. This can be problematic in seismic
interpretation. A recent review of the theory and physical behaviour
of wave motion in anisotropic and cracked elastic medium was presented by Crampin (1981).
The simplest form of anisotropy is the hexagonal system. If the
axis of symmetry is the vertical axis, the medium is said to possess
transverse isotropy. In such a medium, 5 elastic moduli are needed to
fully describe the elastic behaviour. The simplification here is that
the particle motion of the SH-mode is always normal to the wave vector
and it always decouples from the quasi-longitudinal (qP) and
quasi-shear (qSV) modes. Once the phase -velocity for a particular
propagation mode travelling in a certain direction is known, the
Cartesian components of the intersection between the wave surface and
the xz-plane can be calculated from the following parametric
representation (see, for example, Postma, 1955, and Krey and Helbig, 1956).
x = c sin(j) + (dc/dj) cos(j)
z = c cos(j) - (dc/dj) sin(j)
(4.2)
where c is the phase velocity and j the angle the wave vector makes
with the vertical axis. The direction of the ray deviates from that
of the wave vector by the angle
0
arctan( c 1. dc/dj) (4.3)
The magnitude of the group velocity, v , is thus v =c/cos(0). If
§ §
presented in Figure 4.1. This occurs for the qSV wave surface when the anisotropy is strong. No such complexity exists in the case of the qP- and SH-waves. The SH wave surface is always an ellipsoid for transversely isotropic media. Figure 4.2 illustrates the geometrical relationship between the ray and wavefront normal for SH-waves in a transversely isotropic medium with the horizontal velocity 10 per cent higher than the vertical velocity. The direction of the ray and wavefront normal are coincident when the wavefront is normal or parallel to the symmetry axis. Similar results apply for the qP- and qSV-waves.
The extension of the matrix method for seismogram synthesis in an isotropic medium to a general anisotropic medium encounters a number of complications and involves additional computational efforts. The first is the difficulty in the description of the source. The second is that an analytical description of the wave vector (and hence the wavefield) has not been derived. A numerical root searching routine needs to be applied to the characteristic equation of the Kevin-Christoffel equation or its equivalent obtained from the condition for the existence of non-trivial solutions for the wavefield. The third complication relates to wave propagation in a spherical symmetric Earth. The theory for wave propagation in a spherically layered and generally anisotropic medium has not been developed.
In the simple case of transverse isotropy, the above difficulties disappear. For motion symmetric about the vertical axis, the transverse isotropy of the medium also makes the slowness integration in the azimuthal direction unnecessary thus greatly reducing the amount of computation required to simulate a curved wavefront. In the
V E R T IC A L w a v e s u r f a c e w a v e f r o n t n o r ma l
Figure 4-2 The geometric relationship between ray and wavefront normal for SH-waves propagating in a transversely isotropic medium with the horizontal velocity 10 per cent higher than the vertical velocity.
70
following, we shall present general algorithms for computing complete seismograms in a stratified, transversely isotropic medium. These algorithms are derived from the solution procedure described in Chapter 2. Apart from the definitions of the wave vectors and interfacial coefficients, the structure of the algorithms for the transversely isotropic medium follows those for the isotropic medium.
velocities) required to define the of density p are
(4.4)
where c ^ , c ^ , cgg> c ^ and c ^ are the elastic moduli of (4.1 ) in abbreviated subscripts. The subscripts h and v denote horizontal and vertical direction respectively. To refer the parameters to layer m of the stratification, a subscript m will be attached to them. The density and thickness of layer m bounded by the horizontal planes z=z , and z=z are denoted by p and d respectively.
For a transversely isotropic medium, an indication of the degree of anisotropy is the contrast between the horizontal and vertical velocities. Of course, the contrast between the horizontal and vertical velocities for qP-waves need not be the same as the contrast for the qSV-waves. The per cent difference between the horizontal and vertical velocities are defined to be
100 ( - Uy ) / Uh , u = a, 3 The five parameters (or
transversely isotropic medium
OL =
In document Computation of synthetic seismograms in layered earth models (Page 72-78) | 1,584 | 7,159 | {"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} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.883407 |
https://homework.zookal.com/questions-and-answers/find-an-approximate-solution-to-the-pendulum-problem-such-that-131662016 | 1,620,636,123,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00146.warc.gz | 332,847,480 | 26,968 | 1. Engineering
2. Computer Science
3. find an approximate solution to the pendulum problem such that...
Question: find an approximate solution to the pendulum problem such that...
Question details
Find an approximate solution to the pendulum problem such that d2 theta /dt2 +g/l theta = 0. Use an approximate solver in matlab to find the solution to the exact equation d2 theta/dt2 +g/l * sin( theta) = 0. Compare the two solutions when the initial angle is 10, 30, and 90. Find a way to quantify the difference.
One approximate method for solving differential equations is Runge-Kutta, which in Matlab goes by the name ode45. I have made a template of how to solve a differential equation in matlab using ode45. Note that these are two separate files: Equations.m (the function that contains the differential equations) and Driving_Script2.m (the script that drives the solver).
Make 3 plots, with each plot showing the exact and approximate solution for each of the three angles. (The command "subplot" will make multiple plots on a single page. You will see that the exact and approximate solution are different. What I want you to do is to quantify how different the solutions are. Also, remember that matlab uses radians...(2*pi/360 * angle).
Template for solving a differential equation using ode45 in matlab:
% Driving_Script2.m (You can change the name)
clear all;
close all;
tspan = [0 10] ; Interval to be solved
y0 = [1.0 0] ; %initial condition here ...the first is angle, the second is velocity
% specify the equations to be solved using the approximate solver within the function Equations.m
[t,y] = ode45('Equations',tspan,y0)
figure(100)
h1=plot(t, y(:,1))
% Equations.m
function dy = Equations(t,y)
g= 9.81, l=2; % Constants here
%
dy = zeros(2,1);
dy(1) = y(2);
dy(2) = -g/l*sin(y(1));
% | 462 | 1,839 | {"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} | 3.453125 | 3 | CC-MAIN-2021-21 | latest | en | 0.857305 |
http://www.inwardquest.com/questions/12642/do-you-consider-action-and-reaction-to-be-opposite-equals | 1,490,905,044,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218203515.32/warc/CC-MAIN-20170322213003-00273-ip-10-233-31-227.ec2.internal.warc.gz | 530,481,872 | 12,131 | # Do you consider action, and reaction to be opposite equals?
Consider: And how can you prove this to be so? asked 15 Mar '11, 06:38 Inactive User ♦♦ 259●1●10●137
1 not so sure of it being opposite equals, rather a need of balancing the initiated action, little proof other than short range consequences for now answered 15 Mar '11, 11:44 fred 18.3k●6●6 Well said, thank you. (16 Mar '11, 03:32) Inactive User ♦♦
0 There is a law in physics that says that for every action, there is an equal and opposite reaction. However, we have free will, so our reaction will be equal, but we choose the direction. We can turn it back on them which causes war, or we can internalize it and become sick, or we can deflect it to a positive reaction. Like if someone call you a fool, you can either call them a fool back and keep it going, maybe getting worse, or you can believe them that you are a fool, internalize and let it take you down, or you can smile and say, Thank you! That is your external response, you may still internalize, so learning a stress relief technique like EFT would be good. answered 15 Mar '11, 12:50 Fairy Princess (suspended) EFT would be good, thank you! (16 Mar '11, 03:20) Inactive User ♦♦
0 I am a sixth degree black belt in Karate and I can tell you the law of action/reaction works because that has been perfected in karate to give tremendous force and power to blocks, techniques, punches and to a lesser extent, kicks. Just for one punch alone takes four hip movements this creates a momentum like cracking a whip and explodes in the punch being thrown. This of course must be coupled with the torque of the arm and wrist counter the motion of the shoulders all in perfect sequence. In other words all of this motion is converted into energy of the actual punch, so all of this pre-power generation is the initial action, the reaction is the power it generates behind the punch. There is also a law of diminishing return, so if I take two balls and hang them in a row. I then pull back the first ball and release it, it crashes into the second ball the second ball moves out and the first stops dead. The second moves out not exactly the same distance but just a little less before returning to hit the first ball again. At this point the second ball stops and the first ball flies out at an angle again, not as much as the second ball. So in this way eventually both will come to rest. The interesting thing is on the spirit level these rules seem opposed, so something I send out returns back more than I had sent out. answered 15 Mar '11, 18:02 Wade Casaldi 36.1k●2●14●86 I enjoy reading your answer, and my brother in law is a Karate Teacher, and I have watched him in action, so I can relate to what you are saying. Thank you. (16 Mar '11, 03:30) Inactive User ♦♦
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# Municipalities have begun demanding that private developers
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29 Jan 2005, 13:51
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Municipalities have begun demanding that private developers pay an increased share of the costs of expanding the current road networks, sewer systems, and other public services to accomodate new development projects.
(A) demanding that private developers pay an increased share of the costs of expanding
(B) demanding private developers to pay for an increased share of the costs of expanding
(C) demanding payment by private developers for an increased share of the costs of expanding
(D) to demand that private developers pay for an increased share of the costs to expand
(E) to demand that private developers should pay an increased share of the costs to expand
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Re: SC - Municipalities [#permalink]
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29 Jan 2005, 14:15
Municipalities have begun demanding that private developers pay an increased share of the costs of expanding the current road networks, sewer systems, and other public services to accomodate new development projects.
Looking for to do for parallel:
Only (B) used "to pay".
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29 Jan 2005, 14:44
In fact, try again.
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29 Jan 2005, 18:05
I will go for D. "Pay for" seems to be more appropriate here. The only difference here in A/D is .. pay for and Pay an. I think Pay for sounds better.
Go for D.
Saurabh Malpani
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29 Jan 2005, 18:31
whats wrong with E?
to demand is the right idiom, should pay seems logical here?
D is my other choice....
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Re: SC - Municipalities [#permalink]
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29 Jan 2005, 19:35
qhoc0010 wrote:
Municipalities have begun demanding that private developers pay an increased share of the costs of expanding the current road networks, sewer systems, and other public services to accomodate new development projects.
(A) demanding that private developers pay an increased share of the costs of expanding
(B) demanding private developers to pay for an increased share of the costs of expanding
(C) demanding payment by private developers for an increased share of the costs of expanding
(D) to demand that private developers pay for an increased share of the costs to expand
(E) to demand that private developers should pay an increased share of the costs to expand
I will pick A.
Demand someone do something, you don't need "to" after "demand".
In addition, Pay for something is the correct usage
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Re: SC - Municipalities [#permalink]
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29 Jan 2005, 22:57
Yes, I see why I was wrong.
The sentence is:
Municipalities have begun to demand that private developers pay for an increased share of the costs to expand the current road networks, sewer systems, and other public services to accomodate new development projects.
So it is (D). Guess it is another trap that I just fell in. :p
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30 Jan 2005, 02:59
OA is (A)
DLMD, you are right!
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30 Jan 2005, 09:20
30 Jan 2005, 09:20
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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®. | 1,524 | 5,777 | {"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} | 3.046875 | 3 | CC-MAIN-2017-09 | longest | en | 0.904915 |
https://www.physicsforums.com/threads/maxwell-faraday-law-stealth-magnets-exploring-emf-induction.1044969/ | 1,695,292,114,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00246.warc.gz | 1,021,732,960 | 25,495 | # Maxwell-Faraday Law & Stealth Magnets: Exploring EMF Induction
• I
• MS La Moreaux
In summary, the author argues that the emf in a closed path is not the result of a flux change, but is instead due to the changing magnetic flux linking the closed path.f
#### MS La Moreaux
Consider a closed path consisting of a loop of wire with a nonconducting gap that completes the closed path. The wire is threaded through a toroidal permanent magnet, magnetized around the toroid (what I call a stealth magnet). The magnetic flux is considered to be confined to the magnet. The flux links the closed path. Now, slip the magnet off one end of the wire loop so that it now is threaded by the nonconductive part of the closed path. Then pull the magnet out away from the loop so that it crosses the nonconductive part of the path. An emf should be induced in the closed path while the magnet is crossing the path because the magnetic flux linking the closed path changes with time. This is in accordance with the Maxwell-Faraday Law, which is one of Maxwell's equations.
There is a problem, however, because the nonconducting part of the closed path does not have to be the shortest route between the wire ends. It can bow out (or in). In other words, it is arbitrary. So when does the emf appear?
Consider a closed path consisting of a loop of wire with a nonconducting gap that completes the closed path. The wire is threaded through a toroidal permanent magnet, magnetized around the toroid (what I call a stealth magnet). The magnetic flux is considered to be confined to the magnet. The flux links the closed path. Now, slip the magnet off one end of the wire loop so that it now is threaded by the nonconductive part of the closed path. Then pull the magnet out away from the loop so that it crosses the nonconductive part of the path. An emf should be induced in the closed path while the magnet is crossing the path because the magnetic flux linking the closed path changes with time. This is in accordance with the Maxwell-Faraday Law, which is one of Maxwell's equations.
There is a problem, however, because the nonconducting part of the closed path does not have to be the shortest route between the wire ends. It can bow out (or in). In other words, it is arbitrary. So when does the emf appear?
Welcome (back) to PF.
Is this question related to your old threads that were closed for cause?
Let us shed a little light on the subject by considering a case simpler than a coil, namely the Faraday Paradox. This employs two disks, say of the same size. One is made of copper and the other is a magnet with its faces the poles. These disks are arranged face to face, close but not touching. Each is mounted on an axle like a wheel and the axles are colinear. If the copper disk is spun while the magnet is stationary, a non-electrostatic emf appears between the copper disk's center and rim. If the magnet is spun and the copper disk remains stationary, there is no emf in the copper disk.
No, this is a new topic. By the way, the addition to my title, which I did not write, is wrong. There is no motion of the wire segment.
If you can post a compelling enough case for this new question, it should be allowed. Also, what is your background in E&M and using calculus to solve the differential equations involved in such problems? Thanks.
The closed thread was about the fact that Faraday's Law falsely implies that a changing magnetic flux linking a circuit is responsible for the motional emf in some cases.
The new question is about an application of one of Maxwell's equations. The equation in question is related to Faraday's Law, being a special class of cases, but that is where the similarity ends.
I have a bachelor's degree in electrical engineering from the University of Michigan.
Dale
The closed thread was about the fact that Faraday's Law falsely implies that a changing magnetic flux linking a circuit is responsible for the motional emf in some cases.
Falsely?
Yes. Motional emf is never the result of a flux change.
I have a bachelor's degree in electrical engineering from the University of Michigan.
Which means you are in a position to overthrow all of physics? I'm not so sure I buy that.
I found your description impossible to fathom. It reminds me of the idea that the reason we haven't built a perpetual motion machine is that past attempts just weren't complicated enough.
There are three things you could be trying to do:
1. Showing Maxwell's Equations do not match experiment. In this case, you need a real experiment, not a thought experiment.
2. Showing Maxwell's Equations are inconsistent. Too late - they are known to be consistent.
3. Trying to understand something you don't understand. Then show the simplest possible setup, perhaps with a drawing, in the words of Art Fleming "cast your response in the form of a question" and restrict yourself to physical measurements - i.e. "what will this ammeter read?"
berkeman and russ_watters
Obviously, your level of understanding is not sufficient.
Ibix and weirdoguy
By the way, the addition to my title, which I did not write, is wrong. There is no motion of the wire segment.
I updated the title to make it clearer that the motion is relative.
I think you have a C-shaped piece of wire and you have a toroidal magnet with a toroidal magnetic field that is only non-zero within the material of the torus. You move the magnet through the gap in the C in a plane perpendicular to the line that would notionally close the C into an O. I don't understand what you want to know about this scenario.
I updated the title to make it clearer that the motion is relative.
The wire segment does not move. How can I be any clearer?
The wire segment does not move. How can I be any clearer?
Relative to what? You have a BSEE from the U of M, so you know that there is no such thing as absolute motion. If the test coil and the toriod are moving relative to each other, that is what matters, no?
Yes. Motional emf is never the result of a flux change.
Obviously, your level of understanding is not sufficient.
Well, please state your question more clearly with an appropriate diagram. It sounds like it has something to do with a wire segment moving in relative fashion near a toroidally magnetized ferrite toriod. Can you please clarify exactly what the situation entails, and what your question is about? The more math and diagrams you can post, the better we can help you with this question.
I think you have a C-shaped piece of wire and you have a toroidal magnet with a toroidal magnetic field that is only non-zero within the material of the torus. You move the magnet through the gap in the C in a plane perpendicular to the line that would notionally close the C into an O. I don't understand what you want to know about this scenario.
When the magnet is situated so that the nonconductive part of the closed path passes through the hole in the magnet, the magnetic flux of the magnet links the closed path. When the magnet moves so that the portion of it that was inside the closed path crosses the closed path and ends up outside the closed path, there is no longer any flux linking the closed path. This should result in an emf in the closed path. The nonconductive portion of the closed path, however, is arbitrary. One cannot expect that the emf will appear when the magnet crosses a line that one has only imagined!
When the magnet is situated so that the nonconductive part of the closed path passes through the hole in the magnet, the magnetic flux of the magnet links the closed path. When the magnet moves so that the portion of it that was inside the closed path crosses the closed path and ends up outside the closed path, there is no longer any flux linking the closed path. This should result in an emf in the closed path. The nonconductive portion of the closed path, however, is arbitrary. One cannot expect that the emf will appear when the magnet crosses a line that one has only imagined!
So, still without any diagram we are forced to try to guess and ask if we are guessing correctly. This is getting tiresome, and you are risking having yet another thread closed for cause.
So here's my guess -- your C-shaped wire segment with the open gap large enough to be slid over the toroid is moved from encircling the toriod body to a position away from the toroid and you are asking what the induced EMF is in that wire segment. Is that basically correct?
If so, you can model this wire segment as that conducting wire and a parasitic capacitor that is formed across the open "non-conducting" portion of that C-shaped wire segment. Do you understand how to calculate that parasitic capacitance and how to use that in your DE calculations of the voltage induced in the wire segment?
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The wire segment does not move. How can I be any clearer?
I am reasonably convinced that the setup is as follows. We have a small toroidal magnet (shown from the side as a red line). We have a piece of wire curved into a C shape (blue). We can always draw an infinite number of arbitrary imaginary lines that connect the ends of the wire and pass through the hole in the toroid (one such is shown as a fine green line).
The blue and green lines together form the perimeter of a surface ##\Sigma## through which one side of the magnet passes.
I think that the relevant analysis is this: we can always choose an arbitrary green line arbitrarily close to the edge of the magnet. Thus any motion of the magnet takes it outside this particular loop, meaning that ##\iint_\Sigma\frac\partial{\partial t}\vec B\cdot d\vec S\neq 0##. Thus by the integral form of Maxwell's third equation, ##\oint\vec E\cdot d\vec l## around the blue+green loop is also non zero for such a choice of green line, wherever the magnet is and whenever it is moved.
I also think that @MS La Moreaux thinks that ##\oint\vec E\cdot d\vec l## should be non zero for only one choice of green line, but doesn't know how to select the "correct one". If so, the resolution is to note that any choice of green line will do. If not, OP needs to provide a better description of the problem.
(It's telling that after 17 posts, 7 by the OP, in a thread about electromagnetism, I'm the first one using LaTeX.)
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(It's telling that after 17 posts, 7 by the OP, in a thread about electromagnetism, I'm the first one using LaTeX.)
And first with a picture
hutchphd, berkeman, Vanadium 50 and 1 other person
And first with a picture
Yeah, but that's already been pointed out several times. The lack of LaTeX was forcibly brought to my attention because I had to do the preview-then-refresh trick to get it to render.
OP - you are going to find it very difficult to communicate about physics if you refuse to use maths and diagrams.
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Relative to what? You have a BSEE from the U of M, so you know that there is no such thing as absolute motion. If the test coil and the toriod are moving relative to each other, that is what matters, no?
The laboratory frame. No, relative motion is not what matters. Change of
Relative to what? You have a BSEE from the U of M, so you know that there is no such thing as absolute motion. If the test coil and the toriod are moving relative to each other, that is what matters, no?
Relative to the laboratory frame. Relative motion is not what counts. The Maxwell-Faraday Law does not address motion of the closed path.
I am reasonably convinced that the setup is as follows. We have a small toroidal magnet (shown from the side as a red line). We have a piece of wire curved into a C shape (blue). We can always draw an infinite number of arbitrary imaginary lines that connect the ends of the wire and pass through the hole in the toroid (one such is shown as a fine green line).
View attachment 313236
The blue and green lines together form the perimeter of a surface ##\Sigma## through which one side of the magnet passes.
I think that the relevant analysis is this: we can always choose an arbitrary green line arbitrarily close to the edge of the magnet. Thus any motion of the magnet takes it outside this particular loop, meaning that ##\iint_\Sigma\frac\partial{\partial t}\vec B\cdot d\vec S\neq 0##. Thus by the integral form of Maxwell's third equation, ##\oint\vec E\cdot d\vec l## around the blue+green loop is also non zero for such a choice of green line, wherever the magnet is and whenever it is moved.
I also think that @MS La Moreaux thinks that ##\oint\vec E\cdot d\vec l## should be non zero for only one choice of green line, but doesn't know how to select the "correct one". If so, the resolution is to note that any choice of green line will do. If not, OP needs to provide a better description of the problem.
(It's telling that after 17 posts, 7 by the OP, in a thread about electromagnetism, I'm the first one using LaTeX.)
Thanks for the great diagram. You are thinking correctly. Upon further reflection, I was able to answer my question. There is an infinite number of simultaneous nonconductive portions of the closed path which includes the wire. As the magnet moves, it crosses one after another. Sometimes the emf will be in one direction and sometimes in the opposite direction. I suspect that they all cancel each other out and that there is no net emf in the wire.
berkeman
Sometimes the emf will be in one direction and sometimes in the opposite direction. I suspect that they all cancel each other out and that there is no net emf in the wire.
Did you understand my comments about the parasitic capacitance? It is not some imaginary thing that varies with the path chosen. It is a real thing that is integrated over all of the paths between the two ends of your C-shaped wire...
Relative to the laboratory frame. Relative motion is not what counts.
Seriously?
I suspect
What we seem to have is a calculation that you didn't do but are guessing at the answer doesn't agree with your intuition of what the answer should be.
vanhees71 and berkeman
So, still without any diagram we are forced to try to guess and ask if we are guessing correctly. This is getting tiresome, and you are risking having yet another thread closed for cause.
So here's my guess -- your C-shaped wire segment with the open gap large enough to be slid over the toroid is moved from encircling the toriod body to a position away from the toroid and you are asking what the induced EMF is in that wire segment. Is that basically correct?
If so, you can model this wire segment as that conducting wire and a parasitic capacitor that is formed across the open "non-conducting" portion of that C-shaped wire segment. Do you understand how to calculate that parasitic capacitance and how to use that in your DE calculations of the voltage induced in the wire segment?
I was not asking what the induced emf was.
I was not asking what the induced emf was.
Yes. Motional emf is never the result of a flux change.
Sometimes the emf will be in one direction and sometimes in the opposite direction. I suspect that they all cancel each other out and that there is no net emf in the wire.
I'm getting dizzy...
vanhees71
The wire is threaded through a toroidal permanent magnet, magnetized around the toroid (what I call a stealth magnet).
BTW, since you are an EE, you do understand how a ferrous core transformer works, right? Is a permanently magnetized toroidal magnet any more "stealthy" than a toroidal core that has a primary coil with a DC current flowing through it?
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vanhees71
"Stealth toroid" is an anagram for "It's tool hatred".
berkeman | 3,476 | 15,606 | {"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-2023-40 | longest | en | 0.961364 |
http://teachers.henrico.k12.va.us/math/ito_08/07BasicTrig/7-3TrigFunc.html | 1,511,094,445,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805578.23/warc/CC-MAIN-20171119115102-20171119135102-00483.warc.gz | 287,183,614 | 3,029 | Lesson Materials
Lesson 7-3: Defining the Trig Functions
Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)
Objective:
The student will be able to evaluate the sine, cosine, tangent, secant, cosecant and cotangent of angles given a point on the terminal side of the angle in standard position.
Directions 1 Directions 2 Animation 2
1. In the applet to the left, the measurements on the circle have been labeled as SIN, COS, TAN and COT.
2. Drag the orange slider. This will increase and decrease the size of the triangle, keeping the same angle measures.
3. Look at the trig ratios. Even though the sides of the triangle change measures, the ratios are always equal!
4. Will this be true for all triangles?
1. Explore basic trig functions. C.a.R. Activity
2. Explore Learning
GNU: This site has an outstanding introduction to the concept of sine and cosine called "Discovering Trig." This would be a great investigation for students to go through independently or in pairs. The tutorial starts with some basic questions about trigonometry as a subject, as the students finish reading and interacting with each page they should click to go to the next page. This overview touches on the definitions of sine and cosine as well as the shapes of their graphs. GNU
Zona Land
No Puzzle yet | 316 | 1,407 | {"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-2017-47 | latest | en | 0.901801 |
https://www.askwillonline.com/2011/12/how-to-calculate-resolution-of-image.html | 1,506,022,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00253.warc.gz | 774,079,263 | 19,156 | [How to] Calculate Resolution Of An Image
In Physics AS, it is important to know how to work out the resolution of a given image. The resolution is…
The size, in real life, of one pixel or real life length / no. of pixels
For this reason, it is usually measured in metres per pixel.
Let’s have an example and calculate the resolution of the following picture of the River Thames:
You will have to estimate the size of the image and therefore how many metres of ‘real life’ it is showing unless told. For the case of the River Thames, I have estimated the width of the given part of the river is 250 metres. On the picture, this is represented by 20 pixels (for that, I just added them up). Now, remembering the equation for working out the resolution of a image:
We have everything to work out the resolution of the above image of the Thames. 250 / 20 given us a resolution of 12.5 metres per pixel.
It is important to remeber that anything smaller than 12.5 metres you will not be able to see.
If you are revising Physics Topics, please have a look at other articles I have done on Physics you may be interested in. | 269 | 1,129 | {"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} | 3.578125 | 4 | CC-MAIN-2017-39 | latest | en | 0.924999 |
https://www.riddledude.com/word-riddles/cap-flap/ | 1,675,158,984,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00227.warc.gz | 966,687,899 | 21,478 | # Cap Flap
Three men, Mr. White, Mr. Brown and Mr. Green, were in the habit of meeting in a local doughnut shop every morning for coffee and doughnuts.
One morning as they were sitting at their usual table, Mr. White remarked, “Hey, will you look at that. We’re each wearing a colored baseball cap today.” One white cap, one brown cap, and one green cap. But interestingly, no one was wearing a cap of the color that matches his name.
At this, the guy wearing the green cap says, “Oh you stup’, that doesn’t mean anything, it’s just coincidence. Shut up and eat your doughnut.”
The question is what color cap is each man wearing?
### 2 guesses to Cap Flap
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• Roolstar
Mr. White: Brown
Mr. Brown: Green
Mr. Green: White
All this because the guy in the green hat told Mr. White to shut up.
I just hope Mr. White is not schizophrenic nor crazy :)
• Dude
Correct, Rool.
Here’s why: If Mr. White says nobody is wearing a cap of his own color that obviously includes him, so he can be wearing either a green cap or a brown cap. But not a white one.
And he can’t be wearing the green cap because the guy with the green cap spoke up. So Mr. White must be wearing what? The brown cap!
And Mr. Green? He’s got to be wearing the white cap, and Mr. Brown has got to be wearing? The green cap!
Congrats! You’re today’s winner. | 394 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | longest | en | 0.934267 |
http://en.kakprosto.ru/how-26863-how-to-find-profitability | 1,487,825,224,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171078.90/warc/CC-MAIN-20170219104611-00066-ip-10-171-10-108.ec2.internal.warc.gz | 84,207,424 | 14,387 | You will need
• Calculator, analyzed company Balance sheet (Form №1), profit and loss statement (Form №2).
Instruction
1
On the basis of the statement of profit and loss statement (Form No. 2) to calculate the return on sales at the beginning and end of the reporting period. Profitability of sales is calculated as a ratio of profit from sales of products to revenue:
RP=PP (row 050) / (row 010)*100%
The figure indicates either rising prices or increased production costs.
2
On the basis of the statement of profit and loss statement (Form No. 2) to calculate product profitability at the beginning and end of the reporting period. Profitability of production is calculated as a ratio of profit from sales of products to the total cost of this production:
RP=PP (row 050) / SP (line 020)*100%
The figure indicates a decline of cost per unit or 1 RUB products, the growth of production, growth of prices for products while improving its quality.
3
On the basis of the statement of profit and loss statement (Form No. 2) to calculate a return on ordinary activities at the beginning and end of the reporting period. The profitability of ordinary activities is calculated by the ratio of net profit to revenue:Rd=FD (line 190) / V (line 010)*100%
The figure shows net profit.
4
Based on the data of balance sheet (Form №1) and profit and loss statement (Form No. 2) to calculate the economic rate of return at the beginning and end of the reporting period. Economic profitability is calculated by dividing net income by the average value of current assets:
ROA=PCH (line 190) / AOC (line 300)*100%
The ratio of economic profitability shows the efficiency of using the assets of the enterprise. The figure shows the growth of sales, increase the value of the property.
5
Based on the data of balance sheet (Form №1) and profit and loss statement (Form No. 2) to calculate the return on equity at the beginning and end of the reporting period. Profitability of own capital is calculated by dividing net income by average equity capital:RSK=PCH (line 190) / SKS (line 490)*100%
This ratio shows the efficiency of use of capital. Its meaning is that it shows how much profit falls on unit of own capital of the enterprise. | 506 | 2,217 | {"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-2017-09 | latest | en | 0.931693 |
http://www.johndcook.com/blog/2012/02/02/how-to-compute-jincx/ | 1,371,636,224,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708664942/warc/CC-MAIN-20130516125104-00088-ip-10-60-113-184.ec2.internal.warc.gz | 525,816,599 | 9,645 | # How to compute jinc(x)
The function jinc(x) that I wrote about yesterday is almost trivial to implement, but not quite. I’ll explain why it’s not quite as easy as it looks and how one might implement it in C and Python.
The function jinc(x) is defined as J1(x) / x, so if you have code to compute J1 then it ought to be a no-brainer. For example, why not use the following C code?
#include <math.h>
double jinc(double x) {
return j1(x) / x;
}
The problem is that if you pass in 0, the code will divide by 0 and return a NaN. The function jinc(x) is defined to be 1/2 at x = 0 because that’s the limit of J1(x)(x) / x as x goes to 0. So we try again:
#include <math.h>
double jinc(double x) {
return (x == 0.0) ? 0.5 : j1(x) / x;
}
Does that work? Technically, it could still fail — we’ll come back to that at the end — but we’ll assume for now that it’s OK.
We could write the analogous Python code, and it would be adequate as long as we’re only calling the function with scalars and not NumPy arrays.
from scipy.special import j1
def jinc(x):
if x == 0.0:
return 0.5
return j1(x) / x
Now suppose you want to plot this function. You create an array of points, say
x = np.linspace(-1, 1, 25)
and plot jinc(x). You’ll get a warning: “ValueError: The truth value of an array with one element is ambiguous. Use a.any() or a.all().” Incidentally, if we called linspace with an even integer in the last argument, our array of points would avoid zero and the naive implementation of jinc would work.
When Python tries to apply jinc to an array, it doesn’t know how to interpret the test x == 0. The warning suggests “Do you mean if any component of x is 0? Or if all components of x are 0?” Neither option is what we want. We want to apply jinc as written to each element of x. We could do this by calling the vectorize function.
jinc = np.vectorize(jinc)
This replaces our original jinc function with one that handles NumPy arrays correctly.
There is an extremely unlikely scenario in which the code above could fail. The value of J1(x) is approximately x/2 for small values of x. If the floating point value x is so small that 0.5*x returns 0, our function will return 0, even though it should return 0.5. The C code above works for values of x as small as DBL_MIN and even values much smaller. (DBL_MIN is not the smallest value of a double, only the smallest normalized double.) But if you set
x = DBL_MIN / pow(2.0, 52);
then jinc(x) will return 0. If you want to be absolutely safe, you could change the implementation to
#include <math.h>
double jinc(double x) {
return (fabs(x) < 1e-8) ? 0.5 : j1(x) / x;
}
Why test for whether the absolute value is less than 10-8 rather than a much smaller number? For small x, the error in approximating jinc(x) with 1/2 is on the order of x2/16. So for x as large as 10-8, the approximation error is below the resolution of a double. As a bonus, the function jinc(x) will be more efficient for |x| < 10-8 since it avoids a call to j1.
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Posted in Python
###### 8 comments on “How to compute jinc(x)”
1. Juanlu001 says:
I don’t understand the second part: jinx(x) is meant to be 1/2 for small x and not x/2, so I cannot see where’s the problem you are trying yo address with the last snippet.
On the other hand, the vectorize() thing is very useful for piecewise defined functions, functions with singular values such as sinc(x) if one had to define it, etc.
2. John says:
Jauanlu: Thanks for pointing that out. I had said “jinc” when I meant to say J1. For small x, J1(x) is approximately x/2. I corrected the post.
3. Juanlu001 says:
Ah right, I understand now, thanks.
4. Canageek says:
For those of us not familiar with higher level math could you explain what you would use a function like this for? I’m finishing a chemistry degree, so when I think odd function I think wavefunction of an electron, or of trying to model some physical process. I’m sure I’ve heard of sinc before (probably in my quantum mechanics class) but would have no idea what you would use either of those functions for.
5. John says:
Bessel functions often come up in physics problems stated in cylindrical coordinates, so often that they’re called “cylindrical functions.” They also come up in many other contexts.
I’m not as familiar with the jinc function in particular. One place it comes up is in the Fourier transform of a function that has value 1 inside a disk and 0 outside. In that sense it’s analogous to the sinc function, the Fourier transform of a function that’s 1 inside an interval and 0 elsewhere.
6. Alessandro says:
Why do you use 1e-8 in the last code snippet?
Because (1e-8)^2 (the size of the error in approximating jinc with 1/2) will be almost equal to std::numeric_limits<double>::epsilon() ?
7. John says:
Alessandro: The Taylor series for jinc(x) is x/2 + x^2/16 + O(x^4). If x = 1e-8, then x^2/16 = 6.25e-18. Adding the quadratic term would make no difference because 0.5 + 6.25e-18 = 0.5 in IEEE 754 double precision arithmetic.
8. Emery Carr says:
In Real Computing Made Real, Forman Acton gives an implementation of the sinc function which uses a truncated Taylor series expansion directly for small values of x, i.e. (fabs(x) < .1) ? (1.0 – x^2 * (1.0 – x^2/20.0)/6.0) ? sin(x)/x . The idea is to preserve more significant figures for small values. Perhaps something similar could be useful for jinc(x)?
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https://handwiki.org/wiki/Physics:Atomic_mass | 1,701,839,044,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.13/warc/CC-MAIN-20231206031946-20231206061946-00468.warc.gz | 338,143,251 | 19,200 | # Physics:Atomic mass
Short description: Rest mass of an atom in its ground state
Stylized lithium-7 atom: 3 protons, 4 neutrons, and 3 electrons (total electrons are ~14300th of the mass of the nucleus). It has a mass of 7.016 Da. Rare lithium-6 (mass of 6.015 Da) has only 3 neutrons, reducing the atomic weight (average) of lithium to 6.941.
The atomic mass (ma or m) is the mass of an atom. Although the SI unit of mass is the kilogram (symbol: kg), atomic mass is often expressed in the non-SI unit dalton (symbol: Da) – equivalently, unified atomic mass unit (u). 1 Da is defined as 112 of the mass of a free carbon-12 atom at rest in its ground state.[1] The protons and neutrons of the nucleus account for nearly all of the total mass of atoms, with the electrons and nuclear binding energy making minor contributions.[2] Thus, the numeric value of the atomic mass when expressed in daltons has nearly the same value as the mass number. Conversion between mass in kilograms and mass in daltons can be done using the atomic mass constant $\displaystyle{ m_{\rm{u}}= {{m({\rm{^{12}C}})} \over {12}} = 1\ \rm {Da} }$.
The formula used for conversion is:[3][4]
$\displaystyle{ 1\ {\rm{Da}}= m_{\rm{u}}={M_{\rm{u}} \over {N_{\rm A}}}={M(^{12}C) \over {12\ N_{\rm A}}} = 1.660\ 539\ 066\ 60(50)\times 10^{-27}\ \mathrm{kg} , }$
where $\displaystyle{ M_{\rm u} }$ is the molar mass constant, $\displaystyle{ N_{\rm A} }$ is the Avogadro constant,[5] and $\displaystyle{ M(^{12}\mathrm{C}) }$ is the experimentally determined molar mass of carbon-12.[6]
The relative isotopic mass (see section below) can be obtained by dividing the atomic mass ma of an isotope by the atomic mass constant mu yielding a dimensionless value. Thus, the atomic mass of a carbon-12 atom is 12 Da by definition, but the relative isotopic mass of a carbon-12 atom is simply 12. The sum of relative isotopic masses of all atoms in a molecule is the relative molecular mass.
The atomic mass of an isotope and the relative isotopic mass refers to a certain specific isotope of an element. Because substances are usually not isotopically pure, it is convenient to use the elemental atomic mass which is the average (mean) atomic mass of an element, weighted by the abundance of the isotopes. The dimensionless (standard) atomic weight is the weighted mean relative isotopic mass of a (typical naturally occurring) mixture of isotopes.
The atomic mass of atoms, ions, or atomic nuclei is slightly less than the sum of the masses of their constituent protons, neutrons, and electrons, due to binding energy mass loss (per E = mc2).
## Relative isotopic mass
Relative isotopic mass (a property of a single atom) is not to be confused with the averaged quantity atomic weight (see above), that is an average of values for many atoms in a given sample of a chemical element.
While atomic mass is an absolute mass, relative isotopic mass is a dimensionless number with no units. This loss of units results from the use of a scaling ratio with respect to a carbon-12 standard, and the word "relative" in the term "relative isotopic mass" refers to this scaling relative to carbon-12.
The relative isotopic mass, then, is the mass of a given isotope (specifically, any single nuclide), when this value is scaled by the mass of carbon-12, where the latter has to be determined experimentally. Equivalently, the relative isotopic mass of an isotope or nuclide is the mass of the isotope relative to 1/12 of the mass of a carbon-12 atom.
For example, the relative isotopic mass of a carbon-12 atom is exactly 12. For comparison, the atomic mass of a carbon-12 atom is exactly 12 daltons. Alternately, the atomic mass of a carbon-12 atom may be expressed in any other mass units: for example, the atomic mass of a carbon-12 atom is 1.99264687992(60)×10−26 kg.
As is the case for the related atomic mass when expressed in daltons, the relative isotopic mass numbers of nuclides other than carbon-12 are not whole numbers, but are always close to whole numbers. This is discussed fully below.
## Similar terms for different quantities
The atomic mass or relative isotopic mass are sometimes confused, or incorrectly used, as synonyms of relative atomic mass (also known as atomic weight) or the standard atomic weight (a particular variety of atomic weight, in the sense that it is standardized). However, as noted in the introduction, atomic mass is an absolute mass while all other terms are dimensionless. Relative atomic mass and standard atomic weight represent terms for (abundance-weighted) averages of relative atomic masses in elemental samples, not for single nuclides. As such, relative atomic mass and standard atomic weight often differ numerically from the relative isotopic mass.
The atomic mass (relative isotopic mass) is defined as the mass of a single atom, which can only be one isotope (nuclide) at a time, and is not an abundance-weighted average, as in the case of relative atomic mass/atomic weight. The atomic mass or relative isotopic mass of each isotope and nuclide of a chemical element is, therefore, a number that can in principle be measured to high precision, since every specimen of such a nuclide is expected to be exactly identical to every other specimen, as all atoms of a given type in the same energy state, and every specimen of a particular nuclide, are expected to be exactly identical in mass to every other specimen of that nuclide. For example, every atom of oxygen-16 is expected to have exactly the same atomic mass (relative isotopic mass) as every other atom of oxygen-16.
In the case of many elements that have one naturally occurring isotope (mononuclidic elements) or one dominant isotope, the difference between the atomic mass of the most common isotope, and the (standard) relative atomic mass or (standard) atomic weight can be small or even nil, and does not affect most bulk calculations. However, such an error can exist and even be important when considering individual atoms for elements that are not mononuclidic.
For non-mononuclidic elements that have more than one common isotope, the numerical difference in relative atomic mass (atomic weight) from even the most common relative isotopic mass, can be half a mass unit or more (e.g. see the case of chlorine where atomic weight and standard atomic weight are about 35.45). The atomic mass (relative isotopic mass) of an uncommon isotope can differ from the relative atomic mass, atomic weight, or standard atomic weight, by several mass units.
Relative isotopic masses are always close to whole-number values, but never (except in the case of carbon-12) exactly a whole number, for two reasons:
• protons and neutrons have different masses,[7][8] and different nuclides have different ratios of protons and neutrons.
• atomic masses are reduced, to different extents, by their binding energies.
The ratio of atomic mass to mass number (number of nucleons) varies from 0.9988381346(51) for 56Fe to 1.007825031898(14) for 1H.
Any mass defect due to nuclear binding energy is experimentally a small fraction (less than 1%) of the mass of an equal number of free nucleons. When compared to the average mass per nucleon in carbon-12, which is moderately strongly-bound compared with other atoms, the mass defect of binding for most atoms is an even smaller fraction of a dalton (unified atomic mass unit, based on carbon-12). Since free protons and neutrons differ from each other in mass by a small fraction of a dalton (1.38844933(49)×10−3 Da),[9] rounding the relative isotopic mass, or the atomic mass of any given nuclide given in daltons to the nearest whole number, always gives the nucleon count, or mass number. Additionally, the neutron count (neutron number) may then be derived by subtracting the number of protons (atomic number) from the mass number (nucleon count).
## Mass defects in atomic masses
Binding energy per nucleon of common isotopes. A graph of the ratio of mass number to atomic mass would be similar.
The amount that the ratio of atomic masses to mass number deviates from 1 is as follows: the deviation starts positive at hydrogen-1, then decreases until it reaches a local minimum at helium-4. Isotopes of lithium, beryllium, and boron are less strongly bound than helium, as shown by their increasing mass-to-mass number ratios.
At carbon, the ratio of mass (in daltons) to mass number is defined as 1, and after carbon it becomes less than one until a minimum is reached at iron-56 (with only slightly higher values for iron-58 and nickel-62), then increases to positive values in the heavy isotopes, with increasing atomic number. This corresponds to the fact that nuclear fission in an element heavier than zirconium produces energy, and fission in any element lighter than niobium requires energy. On the other hand, nuclear fusion of two atoms of an element lighter than scandium (except for helium) produces energy, whereas fusion in elements heavier than calcium requires energy. The fusion of two atoms of 4He yielding beryllium-8 would require energy, and the beryllium would quickly fall apart again. 4He can fuse with tritium (3H) or with 3He; these processes occurred during Big Bang nucleosynthesis. The formation of elements with more than seven nucleons requires the fusion of three atoms of 4He in the triple alpha process, skipping over lithium, beryllium, and boron to produce carbon-12.
Here are some values of the ratio of atomic mass to mass number:[10]
Nuclide Ratio of atomic mass to mass number
1H 1.007825031898(14)
2H 1.0070508889220(75)
3H 1.005349760440(27)
3He 1.005343107322(20)
4He 1.000650813533(40)
6Li 1.00252048124(26)
12C 1
14N 1.000219571732(17)
16O 0.999682163704(20)
56Fe 0.9988381346(51)
210Po 0.9999184461(59)
232Th 1.0001640242(66)
238U 1.0002133905(67)
## Measurement of atomic masses
Direct comparison and measurement of the masses of atoms is achieved with mass spectrometry.
## Relationship between atomic and molecular masses
Similar definitions apply to molecules. One can calculate the molecular mass of a compound by adding the atomic masses (not the standard atomic weights) of its constituent atoms. Conversely, the molar mass is usually computed from the standard atomic weights (not the atomic or nuclide masses). Thus, molecular mass and molar mass differ slightly in numerical value and represent different concepts. Molecular mass is the mass of a molecule, which is the sum of its constituent atomic masses. Molar mass is an average of the masses of the constituent molecules in a chemically pure but isotopically heterogeneous ensemble. In both cases, the multiplicity of the atoms (the number of times it occurs) must be taken into account, usually by multiplication of each unique mass by its multiplicity.
Molar mass of CH4
Standard atomic weight Number Total molar mass (g/mol)
or molecular weight (Da or g/mol)
C 12.011 1 12.011
H 1.008 4 4.032
CH4 16.043
Molecular mass of 12C1H4
Nuclide mass Number Total molecular mass (Da or u)
12C 12.00 1 12.00
1H 1.007825 4 4.0313
CH4 16.0313
## History
Main pages: Chemistry:History of chemistry and Physics:Unified atomic mass unit
The first scientists to determine relative atomic masses were John Dalton and Thomas Thomson between 1803 and 1805 and Jöns Jakob Berzelius between 1808 and 1826. Relative atomic mass (Atomic weight) was originally defined relative to that of the lightest element, hydrogen, which was taken as 1.00, and in the 1820s, Prout's hypothesis stated that atomic masses of all elements would prove to be exact multiples of that of hydrogen. Berzelius, however, soon proved that this was not even approximately true, and for some elements, such as chlorine, relative atomic mass, at about 35.5, falls almost exactly halfway between two integral multiples of that of hydrogen. Still later, this was shown to be largely due to a mix of isotopes, and that the atomic masses of pure isotopes, or nuclides, are multiples of the hydrogen mass, to within about 1%.
In the 1860s, Stanislao Cannizzaro refined relative atomic masses by applying Avogadro's law (notably at the Karlsruhe Congress of 1860). He formulated a law to determine relative atomic masses of elements: the different quantities of the same element contained in different molecules are all whole multiples of the atomic weight and determined relative atomic masses and molecular masses by comparing the vapor density of a collection of gases with molecules containing one or more of the chemical element in question.[11]
In the 20th century, until the 1960s, chemists and physicists used two different atomic-mass scales. The chemists used an "atomic mass unit" (amu) scale such that the natural mixture of oxygen isotopes had an atomic mass 16, while the physicists assigned the same number 16 to only the atomic mass of the most common oxygen isotope (16O, containing eight protons and eight neutrons). However, because oxygen-17 and oxygen-18 are also present in natural oxygen this led to two different tables of atomic mass. The unified scale based on carbon-12, 12C, met the physicists' need to base the scale on a pure isotope, while being numerically close to the chemists' scale. This was adopted as the 'unified atomic mass unit'. The current International System of Units (SI) primary recommendation for the name of this unit is the dalton and symbol 'Da'. The name 'unified atomic mass unit' and symbol 'u' are recognized names and symbols for the same unit.[12]
The term atomic weight is being phased out slowly and being replaced by relative atomic mass, in most current usage. This shift in nomenclature reaches back to the 1960s and has been the source of much debate in the scientific community, which was triggered by the adoption of the unified atomic mass unit and the realization that weight was in some ways an inappropriate term. The argument for keeping the term "atomic weight" was primarily that it was a well understood term to those in the field, that the term "atomic mass" was already in use (as it is currently defined) and that the term "relative atomic mass" might be easily confused with relative isotopic mass (the mass of a single atom of a given nuclide, expressed dimensionlessly relative to 1/12 of the mass of carbon-12; see section above).
In 1979, as a compromise, the term "relative atomic mass" was introduced as a secondary synonym for atomic weight. Twenty years later the primacy of these synonyms was reversed, and the term "relative atomic mass" is now the preferred term.
However, the term "standard atomic weights" (referring to the standardized expectation atomic weights of differing samples) has not been changed,[13] because simple replacement of "atomic weight" with "relative atomic mass" would have resulted in the term "standard relative atomic mass." | 3,521 | 14,849 | {"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} | 3.546875 | 4 | CC-MAIN-2023-50 | latest | en | 0.774529 |
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Polynomial # 1 (Posted on 2006-05-26)
Let f(x) be a nonconstant polynomial in x with integer coefficients and suppose that for five distinct integers a1, a2, a3, a4, a5, one has f(a1)= f(a2)= f(a3)= f(a4)= f(a5)= 2.
Find all integers b such that f(b)= 9.
No Solution Yet Submitted by Ravi Raja Rating: 5.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Solution (spoiler) | Comment 6 of 7 |
(In reply to re: Solution (spoiler) by Richard)
Actually, that h(x) has integer coefficients, in this case, is simple, and I didn't need Gauss' Lema.
First, (x-a1)(x-a2)(x-a3)(x-a4)(x-a5) is a polynomial with integer coeefficients, since a1 through a5 are integers. Let's write this product as r(x)= x^5+r4.x^4+...+r1.x+r0.
Suppose we have f(x)=fn.x^n+...+f2.x^2+f1.x+f0 with f0 through fn all integers.
Finally write h(x)=hm.x^m+...+h2.x^2+h1.x+h0. (In this case, m=n-5, but that doesn't matter.)
When we multiply r(x) and h(x) to get f(x), the highest order term (which must be fn) is hm... so hm is an integer. The next highest order term (f(n-1)) is r4.hm+h(m-1), so h(m-1) is also an integer. Keep working this way, and all h's turn out to be integers.
Edited on May 27, 2006, 8:11 am
Posted by e.g. on 2006-05-27 08:09:37
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Exponents, Roots (such as square roots, cube roots etc) and Logarithms are all related!
Let's use the simple example of 3x3=9:
3 Squared = = 3 × 3 = 9
Using Exponents we write it as:
32 = 9
When any of those values are missing, we have a question. And (unfortunately!) a different notation:
32 = ?
This is an exponent question, "What is 3 squared?": 32 = 9
?2 = 9
This is a root question, "What is the square root of 9?": √9 = 3
3? = 9
This is a logarithm question, "What is log base 3 of 9?": log3(9) = 2
So when you are stuck trying to solve questions with logs, roots or exponents just remember this.
One more example:
103 = 1000
103 = ?
"What is 10 cubed?": 103 = 1000
?3 = 1000
"What is the cube root of 1000?": ∛1000 = 10
10? = 1000
"What is log base 10 of 1000?": log10(1000) = 3
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# If a, b, c, and d are positive integers, is (a/b) (c/d) >
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If a, b, c, and d are positive integers, is (a/b) (c/d) > [#permalink]
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09 Nov 2008, 19:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If a, b, c, and d are positive integers, is (a/b) (c/d) > c/b?
(1) c > b
(2) a > d
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Re: a b c d --28 [#permalink]
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09 Nov 2008, 20:28
Jcpenny wrote:
If a, b, c, and d are positive integers, is (a/b) (c/d) > c/b?
(1) c > b
(2) a > d
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
(a/b) (c/d) > c/b ---> ac/db > c/b ---> a > d
B
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Re: a b c d --28 [#permalink]
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09 Nov 2008, 22:28
$$(a/b)(c/d) > c/b$$
or, $$c/b(a/d-1) > 0$$
and since a,b,c,d are positive integers,
$$a/d>1$$
or, $$a>d$$
Hence, B is sufficient.
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Re: a b c d --28 [#permalink]
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09 Nov 2008, 23:07
abc - bdc > 0
bc (a-d) > 0
b & c being positive , ensuring that a-d>0 is required
B tells us that precisely.
Hence B
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Re: a b c d --28 [#permalink]
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10 Nov 2008, 10:54
[quote="Jcpenny"]If a, b, c, and d are positive integers, is (a/b) (c/d) > c/b?
(1) c > b
(2) a > d
ac/bd > c/b
ac/bd - dc/ bd> 0
is c(a-d)/bd> 0 fo this to be +ve a has to be > d
from one
insuff
from 2 suff......B
Re: a b c d --28 [#permalink] 10 Nov 2008, 10:54
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So if you've just gotten SmileBASIC you might be wondering "How do i create a program?" Here is a step-by-step guide:Step 1:Create a project by using the "Manage Projects/Files button" and tapping "Add Project Folder" (Optional) Step 2:Go back to the main menu and choose "Create Programs with SmileBASIC" Step 3:Tap the "Edit" button with a 0 on it
### Book 1, Chapter 1.1
Now that you've gone to edit let's add some BASIC commands PRINT "Text String" - use to print a text string. change the text string if you want. PRINT NumValue - Print a number value PRINT StringValue\$ - Print a stored string ACLS - You should always put this at the start of your program it clears the following things: Sprites BG images Things cleared with GCLS SPDEFinition numbers Fonts However, it does NOT clear variables and BGM-Related things LOCATE X,Y,Z (Z is optional) - use to locate text COLOR Color - The included colors are #TBLACK #TBLUE #TCYAN #TGRAY #TGREEN #TLIME #TMAGENTA #TMAROON #TNAVY #TOLIVE #TPURPLE #TRED #TTEAL #TWHITE #TYELLOW RGB(R,G,B) R, G and B and be numbers or a number value (see equals) ==, =, !=, >, >=, < & <= - = is a version of LET from conventional BASIC, the rest are for IF, which will be explained later on == is for testing if a number value or string is a certain value or string. >= is for the same as equals except it's equal to or greater than, > is just greater than and nothing else. <= is equal to or less than. < is less than. != is NOT equal to GCLS - clear the graphic screen CLS - clear the console screen END - Stops the game prematurely (before reaching the last line) or ENDs a DEF (DEFs will be in the advanced tutorial (the third one) ) FADE Color - Use THESE colors for this: #BLACK #BLUE #CYAN #GRAY #GREEN #LIME #MAGENTA #MAROON #NAVY #OLIVE #PURPLE #RED #TEAL #WHITE #YELLOW GOTO and GOSUB @labelname - use in combination with @label. GOTO - goes to the label. RETURN doesn't work. GOSUB - Same as GOTO, but RETURN works @label - the label can be anything that doesn't include spaces. explained in GOTO and GOSUB. RETURN - Returns to the line after GOSUB was used after using GOSUB. Not using GOSUB and this being used leads to a nasty error. to be finished...maybe.
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• BASIC
• SmileBASIC
• Tutorial | 711 | 2,580 | {"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-2018-39 | longest | en | 0.816022 |
https://math.stackexchange.com/questions/3012864/how-to-prove-that-if-a-figure-in-a-number-changes-then-the-rest-also-changes | 1,566,428,518,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00193.warc.gz | 542,392,646 | 30,397 | How to prove that if a figure in a number changes, then the rest also changes.
We have a 8 figure number, lets call it $$N_1=a_1a_2a_3a_4a_5a_6a_7a_8$$. Prove that if we interchange two numbers (for example $$N_2=a_1a_2a_4a_3a_5a_6a_7a_8$$ and $$N_1 \neq N_2$$) then the rest we get by dividing $$N_1$$ and $$N_2$$ with 23 isn't the same. Then prove that if we change one number (for example: $$N_3=a_1a_2ba_4a_5a_6a_7a_8$$ and $$b\neq a_3$$) the rest we get by dividing $$N_1$$ and $$N_3$$ by 23 is not the same.
Then prove that this is not true if we divide the numbers by 24. I mean, that we can find $$N_1, N_2$$ and $$N_3$$ defined as I said before that after dividing by 24 the rest is the same.
I do not even know how to start. Thanks in advance.
• Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_i\neq a_j$, then $N_1-N_2=(a_i-a_j)\times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there. – lulu Nov 25 '18 at 14:04
• Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6\cdots a_0=\sum a_i10^i$. – lulu Nov 25 '18 at 14:12
• @lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure). – Andarrkor Nov 25 '18 at 14:49
• Sure. As I said, I figured you meant to add that condition. – lulu Nov 25 '18 at 15:02
• @lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers. – Andarrkor Nov 25 '18 at 15:09 | 604 | 1,726 | {"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} | 4.375 | 4 | CC-MAIN-2019-35 | latest | en | 0.882916 |
http://en.academic.ru/dic.nsf/cide/165478/Spherical | 1,498,492,883,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320823.40/warc/CC-MAIN-20170626152050-20170626172050-00660.warc.gz | 129,370,728 | 12,928 | Spherical trigonometry
Spherical Spher"ic*al, Spheric Spher"ic, a. [L. sphaericus, Gr. ???: cf. F. sph['e]rique.] 1. Having the form of a sphere; like a sphere; globular; orbicular; as, a spherical body. [1913 Webster]
2. Of or pertaining to a sphere. [1913 Webster]
3. Of or pertaining to the heavenly orbs, or to the sphere or spheres in which, according to ancient astronomy and astrology, they were set. [1913 Webster]
Knaves, thieves, and treachers by spherical predominance. --Shak. [1913 Webster]
Though the stars were suns, and overburned Their spheric limitations. --Mrs. Browning. [1913 Webster]
{Spherical angle}, {Spherical co["o]rdinate}, {Spherical excess}, etc. See under {Angle}, {Coordinate}, etc.
{Spherical geometry}, that branch of geometry which treats of spherical magnitudes; the doctrine of the sphere, especially of the circles described on its surface.
{Spherical harmonic analysis}. See under {Harmonic}, a.
{Spherical lune},portion of the surface of a sphere included between two great semicircles having a common diameter.
{Spherical opening}, the magnitude of a solid angle. It is measured by the portion within the solid angle of the surface of any sphere whose center is the angular point.
{Spherical polygon},portion of the surface of a sphere bounded by the arcs of three or more great circles.
{Spherical projection}, the projection of the circles of the sphere upon a plane. See {Projection}.
{Spherical sector}. See under {Sector}.
{Spherical segment}, the segment of a sphere. See under {Segment}.
{Spherical triangle},re on the surface of a sphere, bounded by the arcs of three great circles which intersect each other.
{Spherical trigonometry}. See {Trigonometry}. [1913 Webster] -- {Spher"ic*al*ly}, adv. -- {Spher"ic*al*ness}, n. [1913 Webster]
The Collaborative International Dictionary of English. 2000.
### Look at other dictionaries:
• Spherical trigonometry — Spherical triangle Spherical trigonometry is a branch of spherical geometry which deals with polygons (especially triangles) on the sphere and the relationships between the sides and the angles. This is of great importance for calculations in… … Wikipedia
• Spherical trigonometry — Trigonometry Trig o*nom e*try, n.; pl. { tries}. [Gr. ? a triangle + metry: cf. F. trigonom[ e]trie. See {Trigon}.] 1. That branch of mathematics which treats of the relations of the sides and angles of triangles, which the methods of deducing… … The Collaborative International Dictionary of English
• spherical trigonometry — n. the application of trigonometry to spherical triangles … English World dictionary
• spherical trigonometry — noun (mathematics) the trigonometry of spherical triangles • Topics: ↑mathematics, ↑math, ↑maths • Hypernyms: ↑trigonometry, ↑trig … Useful english dictionary
• spherical trigonometry — noun Date: circa 1728 trigonometry applied to spherical triangles and polygons … New Collegiate Dictionary
• spherical trigonometry — the branch of trigonometry that deals with spherical triangles. [1720 30] * * * … Universalium
• spherical trigonometry — spher′ical trigonom′etry n. math. the branch of trigonometry that deals with spherical triangles • Etymology: 1720–30 … From formal English to slang
• Spherical geometry — is the geometry of the two dimensional surface of a sphere. It is an example of a non Euclidean geometry. Two practical applications of the principles of spherical geometry are navigation and astronomy.In plane geometry the basic concepts are… … Wikipedia
• Spherical — Spher ic*al, Spheric Spher ic, a. [L. sphaericus, Gr. ???: cf. F. sph[ e]rique.] 1. Having the form of a sphere; like a sphere; globular; orbicular; as, a spherical body. [1913 Webster] 2. Of or pertaining to a sphere. [1913 Webster] 3. Of or… … The Collaborative International Dictionary of English
• Spherical angle — Spherical Spher ic*al, Spheric Spher ic, a. [L. sphaericus, Gr. ???: cf. F. sph[ e]rique.] 1. Having the form of a sphere; like a sphere; globular; orbicular; as, a spherical body. [1913 Webster] 2. Of or pertaining to a sphere. [1913 Webster] 3 … The Collaborative International Dictionary of English | 1,060 | 4,182 | {"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} | 3.296875 | 3 | CC-MAIN-2017-26 | longest | en | 0.869202 |
https://www.beatthegmat.com/test-13-25-t2391.html | 1,521,692,662,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647768.45/warc/CC-MAIN-20180322034041-20180322054041-00791.warc.gz | 753,993,550 | 33,339 | • Free Practice Test & Review
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## test 13 #25
This topic has 1 member reply
dunkin77 Master | Next Rank: 500 Posts
Joined
01 Apr 2007
Posted:
269 messages
#### test 13 #25
Mon Apr 23, 2007 2:43 pm
25. What is the greatest common divisor of positive integers m and n ?
(1) m is a prime number.
(2) m and n are consecutive integers.
Hi,
I thought the answer would be C), 2,3 would be consecutive prime numbers.... but the answer turned out to be B) -- Can anyone please explain why only 2) is sufficient?
Prasanna Master | Next Rank: 500 Posts
Joined
26 Feb 2007
Posted:
418 messages
24
Mon Apr 23, 2007 3:55 pm
dunkin77 wrote:
25. What is the greatest common divisor of positive integers m and n ?
(1) m is a prime number.
(2) m and n are consecutive integers.
Hi,
I thought the answer would be C), 2,3 would be consecutive prime numbers.... but the answer turned out to be B) -- Can anyone please explain why only 2) is sufficient?
-------------------------------------------------------------------------
The answer would be B. (1) tells us that m is a prime number. We do not know anything about n.
(2) tells us that m and n are consecutive integers. The greatest common divisor can be only the value of 1.
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https://www.teachoo.com/2582/1588/Misc-11---A-GP-consists-of-even-number-of-terms.-If-sum/category/Geometric-Progression(GP)--Formulae-based/ | 1,723,550,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00329.warc.gz | 766,993,578 | 24,860 | Geometric Progression(GP): Formulae based
Chapter 8 Class 11 Sequences and Series
Concept wise
### Transcript
Misc 5 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. Introduction Let G.P. 10,100,1000,10000,100000,1000000 Here, number of terms = 6 which is even Common ratio = 100/10 = 10 Sum of all terms = 10 + 100 + 1000 + 10000 + 100000 + 1000000 Sum of terms occupying odd places = 10 + 1000 + 100000 This is also a GP with common ratio = 1000/10 = 100 = 102 Misc 5 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. We know that Sn = (a(𝑟^𝑛− 1))/(𝑟 − 1) Since GP has even number of terms, We take number of terms as 2n (so that it is always even) Let G.P. be a, ar, ar2, ar3, … to 2n terms Now, finding sum of all terms & sum of terms occupying odd places G.P. be a, ar, ar2, ar3, … to 2n terms & Sn = (a(𝑟^𝑛− 1))/(𝑟 − 1) It is given that Sum of all the terms is 5 times the sum of terms occupying odd places i.e. S = 5S1 Putting values from (1) & (2) (a(𝑟^2𝑛 −1))/(𝑟 − 1) = 5 (a(𝑟^2𝑛 −1))/(𝑟2 − 1) using a2 – b2 = (a – b)(a + b) (a(𝑟^2𝑛 −1))/(𝑟 − 1) = 5(a(𝑟^2𝑛 −1))/((𝑟 − 1)(𝑟 + 1)) (a(𝑟^2𝑛 −1))/(𝑟 − 1) (𝑟−1) = 5(a(𝑟^2𝑛 −1))/((𝑟 + 1)) a(r2n – 1) = 5(a(𝑟^2𝑛 −1))/(𝑟 + 1) (a(𝑟^2𝑛 −1))/(a(𝑟^2𝑛 −1)) = 5/(𝑟 + 1) 1 = (5 )/(𝑟 + 1) 1 = (5 )/(𝑟 + 1) (r + 1) × 1 = 5 r + 1 = 5 r = 5 – 1 r = 4 Hence the common ratio is 4 | 685 | 1,521 | {"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} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.881783 |
http://search.cpan.org/~mamawe/Algorithm-CheckDigits-v1.2.1/lib/Algorithm/CheckDigits/M11_006.pm | 1,405,256,216,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776438008.40/warc/CC-MAIN-20140707234038-00029-ip-10-180-212-248.ec2.internal.warc.gz | 104,726,870 | 4,440 | Mathias Weidner > Algorithm-CheckDigits-v1.2.1 > Algorithm::CheckDigits::M11_006
Algorithm-CheckDigits-v1.2.1.tar.gz
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# CPAN RT
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Module Version: v1.1.2 Source Latest Release: Algorithm-CheckDigits-v1.3.0
# NAME
CheckDigits::M11_006 - compute check digits for Código de Cuenta Corriente (ES)
# SYNOPSIS
``` use Algorithm::CheckDigits;
\$ccc = CheckDigits('ccc_es');
if (\$ccc->is_valid('2420-0730-27-0050103552')) {
# do something
}
\$cn = \$ccc->complete('2420-0730- -0050103552');
# \$cn = '2420-0730-27-0050103552'
\$cd = \$ccc->checkdigit('2420-0730-27-0050103552');
# \$cd = '27'
\$bn = \$ccc->basenumber('2420-0730-27-0050103552');
# \$bn = '2420-0730- -0050103552';```
# DESCRIPTION
## ALGORITHM
1. Beginning right all digits are weighted 6,3,7,9,10,5,8,4,2,1.
2. The weighted digits are added.
3. The sum of step 2 is taken modulo 11.
4. The checkdigit is 11 minus the sum from step 3.
If the difference is 10, the checkdigit is 1.
If the difference is 11, the checkdigit is 0.
## METHODS
is_valid(\$number)
Returns true only if `\$number` consists solely of numbers and hyphens and the two digits in the middle are valid check digits according to the algorithm given above.
Returns false otherwise,
complete(\$number)
The check digit for `\$number` is computed and inserted into the middle of `\$number`.
Returns the complete number with check digit or '' if `\$number` does not consist solely of digits, hyphens and spaces.
basenumber(\$number)
Returns the basenumber of `\$number` if `\$number` has a valid check digit.
Return '' otherwise.
checkdigit(\$number)
Returns the check digits of `\$number` if `\$number` has valid check digits.
Return '' otherwise.
None by default.
# AUTHOR
Mathias Weidner, `<mamawe@cpan.org>` | 554 | 1,828 | {"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-2014-23 | longest | en | 0.43381 |
https://stat.ethz.ch/pipermail/r-sig-finance/2009q3/004479.html | 1,669,899,147,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00440.warc.gz | 564,846,903 | 2,760 | # [R-SIG-Finance] Help on constrained regression
spencerg spencer.graves at prodsyse.com
Fri Jul 3 02:19:19 CEST 2009
```
1. Have you made normal probability plots of the data
(function "qqnorm")? This could reveal outliers or a need for a
transformation.
2. Have you considered using "lme" with "corAR1", allowing
"a" to be negative? The examples in the help file for "corAR1" show how
to estimate this model with a different "b" for each series but with one
"a" shared between all series. Then an analysis of residuals should
3. Have you considered dropping one observation in each
series and using "lme" to estimate a distinct "a" and "b" for each
series, allowing "a" to be negative? If the grand mean of the
estimates for "a" are positive, this will reduce the number of negative
"a" values by shrinking them all towards the grand mean. Then you can
look more carefully at the series that still have negative "a" values.
They will either contain outliers, or you will likely identify some
other series to include in the model so it becomes something more like
the following:
y[t] = a*y[t-1] + b + c*x[t] + e
As your comment indicated, the seriously negative estimates for
"a" suggest some substantive violations of the assumptions you used to
conclude that "a" should be positive.
Hope this helps.
Spencer
R_help Help wrote:
> I did. The problem was the underlying process that's negative AR(1).
> So I just have to find other way to model it. Thank you.
>
> On Thu, Jul 2, 2009 at 7:31 PM, spencerg<spencer.graves at prodsyse.com> wrote:
>
>> Have you considered writing the model in terms of log(a) = g, say:
>>
>> y[t] = exp(g)*y[t-1]+b+epsilon?
>>
>> With this, you could estimate "g" and "b" using "nls". With multiple
>> series, you could use the "nlme" function in the "nmle" package. For the
>> "nlme" package, an excellent reference in Pinheiro and Bates (2000)
>> Mixed-Effects Models in S and S-PLUS (Springer).
>>
>> Hope this helps.
>> Spencer Graves
>>
>> R_help Help wrote:
>>
>>> Hi,
>>>
>>> I have an AR(1) model
>>>
>>> y[t] = ay[t-1]+b+epsilon
>>>
>>> I'm trying to force a to be positive. So I did the constrained
>>> regression with constraints 0 < a < 1. I used pcls in package mgcv.
>>> However, I found that the solution is not so stable. Most of my lag 1
>>> autocorrelation is negative. Forcing a to positive value makes the
>>> optimizer to stick a to the boundary value. All it does is varying b.
>>> I there anyway to solve this problem? I think the problem might be due
>>> to my initial value is not a smart choice.
>>>
>>> Thank you.
>>>
>>>
>>> _______________________________________________
>>> R-SIG-Finance at stat.math.ethz.ch mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance
>>> -- Subscriber-posting only.
>>> -- If you want to post, subscribe first.
>>>
>>>
>>>
>>
>
>
``` | 786 | 2,883 | {"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} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.926655 |
https://spreadsheetcenter.com/excel-functions/areas/ | 1,726,594,321,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00642.warc.gz | 488,252,591 | 9,394 | # AREAS
The AREAS function in Excel is used to count the number of areas in a reference. It is commonly used in conjunction with other functions and formulas to calculate the total number of non-contiguous ranges in a selected reference or range.
## Syntax 🔗
=AREAS(`reference`)
When you need to determine the total number of distinct areas within a specified reference or range in Excel, the AREAS function comes to the rescue. It facilitates the swift and accurate computation of non-contiguous ranges, offering a convenient solution for handling complex data organization scenarios or when working with multiple disjointed sections within a spreadsheet layout. By incorporating the AREAS function into your formulas, you gain the ability to efficiently manage and analyze diverse sets of data across various locations within your worksheet. This functionality proves particularly useful for summarizing and comprehending the composition of non-contiguous data segments, contributing to improved insights and decision-making in data analysis and reporting tasks.
## Examples 🔗
Suppose you have two non-contiguous ranges in your worksheet: A1:A5 and C1:C5. Using the AREAS function, you can calculate the total number of non-contiguous ranges in these references. The formula would be:
=AREAS(A1:A5,C1:C5)
This will return the count of areas, which should be 2 in this case.
## Notes 🔗
The AREAS function counts the total number of separate areas or non-contiguous ranges within the specified reference. It is important to ensure that the reference provided includes all the non-contiguous ranges that need to be counted. Also, the AREAS function does not work on its own; it is typically used as part of a larger formula or function to support specific analytical or computational requirements.
## Questions 🔗
Can the AREAS function be used to count non-contiguous ranges across different worksheets in a workbook?
Yes, the AREAS function can be used to count non-contiguous ranges across different worksheets within the same workbook. Simply provide the references for the non-contiguous ranges from different worksheets as arguments in the function to obtain the total count of separate areas.
Is the result of the AREAS function affected by the number of empty cells within the non-contiguous ranges?
No, the result of the AREAS function is not affected by the presence of empty cells within the non-contiguous ranges. It solely counts the distinct non-contiguous ranges within the specified reference, irrespective of the content or presence of empty cells. | 510 | 2,590 | {"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} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.853524 |
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# Shift Hours Calculation
## I have enclosed a sheet with the dilemma i currently face. Ive tried multiple variations on a solution none of which have been 100% accurate. Basically the work day is split into 3 shifts : Days ( 06:00 - 14:00 ) Afters (14:00 - 22:00 ) Nights ( 22:00 - 06:00 ) I have a report which tells me the total time the colleague will be getting paid for and there clock in and out times. I need to determine which shift bracket there hours fall into based on the time bands. Ie : David worked 8 hours , started at 10:00 finished at 18:10 , so thats 4 hours recorded in days and 4 in afters since he worked across both shifts. the 10 minutes is not being paid so it doesn't need to be recorded. the sheet should explain things better.
Related Forum Messages:
Calculate Pay For Shift Work With Different Rates Based On Shift Hours
a person works for certain hours and get paid according to the hours worked either by day or by night or a mix of both. Day payment is \$8 when worked between 08:00 and 19:59 , night payment is \$12 when worked between 20:00 and 07:59. The excel cell are formatted as datetime with yyyy-mm-dd hh:mm , the function works fine in getting the time information and checking whether the whole work is all day or all night , yet the if-then-else statements for calculation seems to be wrong!!
examples:
start = 2008-01-01 09:15 , end = 2008-01-01 11:40 , all day as it is between 08:00 and 20:00 and cost = 8/hr = 19.333
start = 2008-01-03 21:05 , end = 2008-01-04 02:05 , all night as it is between 20:00 and 08:00 and cost = 12/hr = 60.000
start = 2008-02-02 19:00 , end = 2008-02-02 20:05 , cost = 9.000 as 1 hour day = 8.000 plus 5minutes night = 1.000
Function prod(st As Date, en As Date) As Double
Dim shour As Integer
Dim smin As Integer
Dim ehour As Integer
Dim emin As Integer
Dim stod As String
Dim etod As String
pday = 8
pnight = 12
shour = Hour(st)
smin = Minute(st) + shour * 60
If (shour >= 8 & shour < 20) Then
stod = "day"
Else
stod = "night"
End If
ehour = Hour(en)
emin = Minute(en) + ehour * 60
If (ehour >= 8 & ehour < 20) Then.................
Calculating Night Shift Hours
I'm trying to calculate the hours worked for both my day shift and my night shift.
Day shift (thanks to search ) I have managed to figure out and worked quite well.
=ROUND((E7-D7)*96,0)/4
It totals adds up the time and converts it into a decimal of hours worked.
For example Joes starts at 1100 and finishes at 1330 it returns a total of 2.5 hours worked.
However I strike a problem with nightshift.
They start in the late afternnon and work thoguh into the am.
I have used the same formula but it doesn't seem to work:
=ROUND((K7-L7)*96,0)/4
I assume because once the clock strikes 12 it's a new day and it can't work out the maths.
Lets use the example form about but make it pm.
Joe starts at 2300 and finishes at 0130 it should give me a total of 2.5 hours instead it gives me 21.5 hours
Calculate The Number Of Hours In A Shift After Midnight
how would I calculate the number of hours in a shift after midnight. What I want to say is that the number of hours after midnight on a friday shift to be taken off from friday total hours and added to a saturday shift.
Operational Model - Shift Start Time Plus Amount Of Hours Worked In One Cell.
Is it possible that a cell contains both numeric and alphanumeric data and to do calculations on that?
For example: if a cell conatain the value "10a" or "8.5b" etc. Would it be possible to have a column that gives me the hours worked (the numeric value in the cell) and a line that gives me the amount of people that are working on shift "a" (the alphanumeric value in the cell).
Is this at all possible? Or does that require VBA/Macros and stuff (in which case this is posted in the wrong part of the forum )
Flying-Hours Calculation
One of my administrative duties is to keep a record of all of the flying-hours completed by a group of twenty pilots. I've constructed a spreadsheet and entered all of their flying records into it.
At the head of each column I have the date, aircraft type, registration number, pilot name, co-pilot name, other crew name, day flying, night flying, solo, dual, total captain hours, etc.
Whilst that I've completed the easy part of this project and that I can transfer each individual pilots flying-hours into his own seperate logbook (by filtering and copy/pasting into another worksheet), there are three other reports that I'm required to provide:
1. To be able to list the number of flying-hours completed during the previous 7 days (for each individual pilot).
2. To be able to list the number of flying-hours completed during the previous 30 days (for each individual pilot).
3. To be able to list the number of flying-hours completed during the previous 90 days (for each individual pilot).
Hours Worked Calculation ....
I am trying to create a spreadsheet that auto calculates my emp. time.
However I do not want to use military time. I can get it to work by =a2-a1 but only if it is 8.5 and 17.5. Any ideas how I can do clock in 8:30 clock out 4:30 = 8 hours?
I am using the following formula to calculate business hours.
=(NETWORKDAYS(R9,T9)-1)*("17:00"-"08:00")+IF(NETWORKDAYS(R9,T9),MEDIAN(MOD(T9,1),"17:00","08:00"),"17:00")-MEDIAN(NETWORKDAYS(R9,T9)*MOD(R9,1),"17:00","08:00")
The business hours considered here is 8AM - 5PM, Start time in R9 and End time in T9. Now the problem is its calculating the correct value when the days are same, for e.g.,
Condition 1
When I am giving "31 March 2009 15:00:00" as start time (R9) and "31 March 2009 23:00:00" in end time (T9), I am getting the correct value. i.e, "2:00:00"
Condition 2
While giving "31 March 2009 16:00:00" as start time and "01 April 2009 09:00:00" as end time I am getting a value of "1:00:00", actually the value should be "3:00:00".
Total Hours Calculation
I need to calculate below hours
07:50
07:50
07:50
07:50
07:50
The answer suppors to be 37:30 Hours but its showing total diffrent value. i used sum(E1:E5) Excel formula, but its not working.
Removing Non Work Hours From Calculation
I am trying to work out the minutes elapsed for a call monitoring
system. The hours monitored are between 05:30 and 19:00 - so if a call
gets logged outside of these hours then the minutes calculated will be
calculated from 05:30 the same day if logged on or after midnight or
05:30 the next day if logged before midnight (ie the next 05:30).
CALCULATION TAKING BUSINESS HOURS INTO CONSIDERATION
I need to do an hour calculation on two cells which have dates and times in both. the first cell is a call that we get from a customer and the second is the date and time in which that call is closed by us...meaning that call is complete.
I need to calculate how much time in hours did it take us to complete that call for the customer. I need this calculation to respect our business hours of Monday to Friday 8am-5pm and closed on Saturdays and Sundays.
here are some examples.
from - 2/12/2004 13:00 (thursday)
to - 2/13/2004 9:00 (friday)
from - 2/13/2004 14:00 (friday)
to - 2/16/2004 10:00 (monday)
Date/Time Calculation For WeekEnds/After Business Hours
What calculation would I enter in a results cell if I wanted to find the delta between 2 times in date format that repersent just the business hours of 8am-5pm, therefore excluding after hours and weekends.
eg.
Date 1 Date 2 Result Time
23/07/07 8:00 24/07/07 14:55 15:55
So far I can't get the caluclation that will compensate for the after hours and week ends.
Time Calculation: Calculate How Many Hours Have Elapsed Between To Entries
I am trying to build a spreadsheet to calculate how many hours have elapsed between to entries; start time (H10) e.g. 9:15 AM and end time (I10) e.g. 12:15 PM. The formula that I am using in the calculation cell field (J10) is (I10-H10+(I10<H10))*24. This formula works great till I wish to include in an IF statement. What I would like is if the total hours calculated with the formula (I10-H10+(I10<H10))*24 is less than 4, return 4 (hours) otherwise the value. As well if there is no start time nor end time entered then return zero.
Lookup Wage Calculations (calculate Pay Per Shift Dependant On The Type Of Shift)
I have the basics set up, but need to work out how to make it calculate my pay per shift dependant on the type of shift i have worked.
I have attached a screen shot of the current page,
In it i have currently used validation drop boxes for the location and worked columns with tables just to one side of the sheet.
The shift pay is the column i am having trouble with.
I would like it to change dependant on what is selected in the 'worked' column.
For most things it should just display basic plus holiday, however if supervisor is selcted in the work column, it should display basic plus holiday plus supervisor.
Stop Vacation Hours Calculation On Vacation Day
i didn't realize is that my current funcation that calculates vacation hours... will increase after a new year. i'd rather have it not increase until they are "reset" for the new year. how do i stop the function? =VLOOKUP(DATEDIF(A8,TODAY(),"y"),\$S\$8:\$T\$10,2)
basically goes to a lookup table with the caculations. PROBLEM: if a user's anniversery date passes, they may go from a 1-2 year status to 3 year bump... this will auto adjust the amount of vacation hours they have. if the reset button is ran to calculate the vacation hours, it might over calculate giving the user 40+ extra carryover hours. anyway i can make this vlookup stop when the current date is or has passed the anniversery date, yet has not been reset? maybe a count down timer, not sure. http://www.ozgrid.com/News/excel-eva....htm#ExcelTips
Start Shift And An End Shift
I have a Start Shift and an End Shift time,
Start Shift = 2009/11/10 09:27:06 (GMT-6:0)
End Shift Time= 2009/11/10 15:13:03 (GMT-6:0)
eg. Total Time = 5.3 hrs
I would like to take if from this format, and calculate the total time difference. Sometimes the GMT codes may be -5:0 if that means anything. For the cell "Total Time" I only need it to have a decimal format.
Solve Between 24 And 48 Hours OR Less Than 24 Hours OR Greater Than 48 Hours
In column A I have a date AND time entered. By the way, this is not via cell format, I have manually entered, say today's date and the current time. In column B I have a future date and time.
Basically, column A is the date and time a problem was given to me. Column B would be the date and time I resolved the problem. Now for the formula....Column C needs to spit out whether the problem was solved between 24 and 48 hours OR less than 24 hours OR greater than 48 hours.
Difference Between Dates And Times In Days , Hours Mins ( Working Hours )
The below formulae allows me to see the difference between two dates and only returns the difference in working hours ie :
Difference between
02/02/2010 08:00 & 03/02/2010 08:00 is 16 Hours 0 Minutes
=(INT(A3)-INT(C6))+MAX(MOD(A3,1)-MAX(MOD(C6,1)))
The following displays it in the Hrs and Mins format
=TEXT(B15,"[h]")&" Hour"&IF(OR(TEXT(B15,"[h]")+0=0,TEXT(B15,"[h]")+0>1),"s "," ")&MINUTE(B15)&" Minute"&IF(MINUTE(B15)1,"s ",""))
Format Time Cell For Greater Than 24 Hours: Hours & Minutes Only
Format Time Cell For Greater Than 24 Hours: Hours & Minutes Only .....
Time Scheduling: Take Out A 30 Min Break If The Hours Worked Is Over 6 Hours
I am making a schedule and I would like it to take out a 30 min break if the hours worked is over 6 hours.
I have so far
A B
1 11:00 7:30
=24*(B1-A1)
Gives me 8 hours, I would like it to subtract the 30 minutes only ifthe sum is over 6 and not alter the sum if it is under 6.
Format Total Hours To Days, Hours & Minutes
1) The output of an excel duration is : 22.00:8.00:25.00 ( day:hour:minutes ) - excel cannot average and work with this number format
2) resolution - =(LEFT(L2,4))+MID(L2, FIND(":",L2)+1,4)/24+MID(L2, FIND(":",L2,7)+1,4)/1440 as an array and Custom Format the cell as [h]:mm - works perfectly.
Q: to be conistent, the initial reporting is dd:hh:mm and then I convert to hh:mm so that excel can process the data. How can I convert from hh:mm to dd:hh:mm so that the excel report can be consistent in presenting the data to senior management?
example attached.
Count Hours Between 2 Times Based On Hours In Another Cell
A1 is 10 (10 hrs worked) , A2 is 10:30am (in time), A3 is 9:00pm (out time), A4 needs to be the total hours and minutes between A2 and A3 based on the hours listed in A-1. What i need is a formula that will calculate the hours and minutes between the 2 times based on hours entered in A1 but that will also compensate for a manadatory 30 minute lunch that needs to be deducted from the total hours if hrs listed in A1 are more than 6.
example: worked 10HRS, 10:30am to 9:00pm, Total hrs is 10hrs 30min, which should be just 10 since the lunch is a none work time and must be subtracted.
If a person worked more than 6hrs, they must take a lunch. if they worked less, than 6 then they don't have to. I need a calcuation to recognize the greater than, less than factor into the equasion also.
Convert Hours To Fractions Of Hours
I am attempting to convert a spreadsheet of times (listed in the format 06:15:39.62, where 06 is the hour, 15 is the minutes, 39 is the seconds, and .62 is in truncated miliseconds) into fractions of hours (so, 6.25 [NOT 6:25!]). I've so far been doing it manually for each value, which is quite tedious (doing basic division of seconds and minutes into hours, to find the fraction) but I'd like a single formula which I can then apply to the whole spreadsheet.
Convert Second To Hours And Mins (over 24 Hours)
i need to convert second to Hours and mins and can do so using:
Displaying Sums Of More Than 24 Hours, As Hours.
I am trying to compute a running total of hours (from row 1) in row 2 Example................
As you can see, when the sum exceeds 24 (moving to the right across row 2) the answer resets, so to speak. Cells are formatted as time. This format *seems* incapable of recognizing quantities of hours over 24 except as days, as it were. This is obviously useful in most sorts of cases but not in this sort of case.
If I simply want the aggregate number of hours expressed as such am I doomed to failure whenever the total exceeds 24? In reply to a somewhat similar enquiry elsewhere in this forum, advice was given to format a cell as Elapsed Time. I dont see such a choice in my dropdown menu.
Time Conversion: Convert A Time From Hours/Minutes To Hours/Tenths
When I am converting a time from Hours/Minutes to Hours/Tenths, Excel is not converting it consitantely. EXAMPLE: 1:15 = 1.25. When I format the cell to present only one place past the decimal point, sometimes the cell will round up to 1.3, and other times it will round down to 1.2. What am I missing?
The sum of (Monday!A1:A4) multiplied by the sum of (Monday!B1:B4) plus (Tuesday!A1:A4) multiplied by the sum of (Tuesday!B1:B4) and so on.
Delete Cells, Shift Everything Else Below It Up
I have a problem in making a delete function/sub and i'm applying it to a listbox in userform ... i'm trying to delete a row of data in the listbox which refers to cells (A12:D12) ... and at the same row (row=12), there are other data cells (E12:H12) ... after deleting the cells all other data below the deleted cells will shift up taking over the deleted cells ...
Data Shift Down And Duplicate?
I have a spreadsheet that has 3 columns. Out of which only 2 are in use.
I have attached an xls for your convenience.
The Date column is in play and Test column are in play.
Basically I want to know if there is a way, where I can
shift a value down and duplicate (in the Test column)
eg:....
Shift Key Down Arrow
1:
Does a code exist to do Shift Key_Arrow Right, Arrow Down and then select that range to Copy?
I know the arrows codes but was told by someone no code can ever exist for a Shift Key.
I need to select a range after a "Search For" but the Range is not determined by the actual Cell ranges.
My Code for Arrow keys are:
'Selection.Offset(0, 1).Select '*Right*
'Selection.Offset(0, -1).Select '*Left*
'Selection.Offset(-1, 0).Select '*Up*
'Selection.Offset(1, 0).Select '*Down*
'Selection.Offset(3, 0).Select '*Down * 3 Cells
2:
What is the macro code to insert more than 1 row ?
The code I use to insert 1 Row is :
Selection.EntireRow.Insert
Sometimes I need to insert 25 rows.
Weekend Shift Patterns
I'm trying to divide the hours between 2 given times in blocks:
i.e.: monday 0600 - 1400 = 8 hrs
2400-0700 [mon - fri] = 1
0700-1800 [mon - fri] = 7
1800-2400 [mon - fri] = 0
0000-2400 [weekend]
I got the first 3 blocks working but got stuck with the 4th one.
It should count only those hours between saturday morning 0000 and sunday night 2400 if it concerns a weekend day. and actually these hours should not be calculated in the first 3 time blocks.
Complicated Shift Differential
Currently I have developed a Time sheet for employees however I am having a lot of trouble with the shift differentials right now.
First, I used an If function to say If(B7="E",B6,0) E would be the evening shift and of course this works fine if everyone worked an perfect Evening shift within the time limits however, they dont.
I'm wonderin if it is possible to have one, or multiple formulas that can do the following:
First shift differential is from 14:00-23:00
Second shift differential is from 22:00-07:00, However lets say you start at 14:00 I don't want it calculating the shift differential of the first one for 22:00-23:00.
Third shift differential is a weekend one which I have figured out it's rather simple to just have a IF function for that.
Start time is B3
End time is B4
Breaks is B5 (However, Breaks is subtracted from B6)
Hours worked for the day is B6
Insert A Value And Shift Cells Value Down
If I put a length in (G13) such as 4.44 & position from (H13) is 1. Then I want that length of (G13) to move its value (4.44) into (C15) the 1st cell to add to is (C15)
Then Move (C15) to (C16) 11.12 becomes C16
and move (C16) to (C17) and so on all the way down the C column.
When it hits the last joint (C374) position 360. I want a messageBox that says you can not insert anymore.
Basically shift the existing values down untill (C374)
I have only column C to shift the calue from (G13)
I do NOT want to delete rows!
Below is how it looks before macro:
GH12LENGTHPOSITION134.441
C14LENGTH1511.121612.221711.561811.861913.112012.872113.062211.992312.03
Below is how it will look after macro: The RED moves to (C15)
C14LENGTH154.441611.121712.221811.561911.862013.112112.872213.062311.992412.03GH12LENGTHPOSITION134.441..........................
Using Control + Shift + Enter
For Excel 2007, does this key function only work after typing in a formula into a cell? I tried pasting the formula and am unable to get the {}.
Is there some setting in preferences to allow this?
Shift Colors To The Left
I need to change to shift the colors to the left instead of the right.
Delete Cells And Shift Up
I have data in cells A1:FM1326 I want to delete all of the cells that say FALSE and shift the remaining cells up.
I need a bit of a help here. What I'm looking for is basically a formula to calculate shift loading differences. For ex.: Total hrs worked 86. As a part timer, my ordinary hrs are 76. Whatever is over 76 comes as double time ("F").
A B C D E F
87 76 42.5 15 8 11
Thats how it looks initially. "F" is the double time hrs paid (the difference between Ordinary hrs and total hrs worked). Which means that I need a formula to automatically calculate my 20% ("C"), 50% ("D") and 100% ("E") loadings. Correct would be like this: A=87, B=76, C=42.5, D12 and E=0. So, basically this 11 hrs straight double time has to come off the 100% ("E") first and then move on to 50% ("D") and 20% ("C") if necessary. So, need a formula to fields C, D and E.
Verifying Employee Shift Coverage
I would like to create a formula that would verify that specific work shifts have been covered each day. The spreadsheet has already been created by someone else, so I am hoping not to recreate the wheel.
The goal is to make sure that all desired shifts are covered with a result in the last cell of the column that would indicate "covered", "not covered" or even a true or false statement.
As an example, an 8am shift is needed. The choices are 8A or 8ALEAD and only one of these is needed for each day.
Can something be set up with conditional formatting from a master list of required shifts.
such as: 7OR, 730*(for just 730 and 730LEAD), 11A, etc.
This is just a snippet of what the schedule looks like: ...
Make Cells Shift Down Automatically
i need a button to copy the content of A8:C28 and paste in another section of the worksheet, i get the button to copy and paste the selected range, but when i copy and paste again it overides my current paste, i need it to paste my new selection underneath my previous paste.
Shift Cells Up If Cell Equal 0
In the range A1:Y66, for every cell whose formula returns a 0, I want it to delete that cell and shift the cells up.
Here is the code I tried, but nothing happened:
Private Sub Worksheet_Change(ByVal Target As Range)
If Target.Column > 25 Then Exit Sub
If Target.Rows > 66 Then Exit Sub
If Target.Cells.Count > 1 Then Exit Sub
If Target.Value = 0 Then Target.Delete Shift:=xlUp
End Sub
VBA To Shift Cells To The Left
how can i get this
Sheet1
ABCDEFGHIJKL7WTXMF31326154RIO TINTOAUDMLWSINBFSALE 258414.13 ML W/S INT BD FD 16-Mar-0917-Mar-09-251307.75N
Ctrl Shift Enter Not Working???
i am trying to modify and existing array formula
=VLOOKUP(R2,TRIM(Codes!\$C\$3:\$D\$283),2,FALSE)
but ctrl shift enter does not seem to work. has anybody else encountered this problem?
FYI auto calculate is on, lookup value and table array are all formatted the same. as i said, the formula works but i need it updated for one extra row.
CTRL+Shift+Enter Not Working
I recently upgraded my computer. I noticed I can no longer enter array formulas. When I tried to enter an array formula using CTRL+Shift+Enter, nothing happens. I don't get any error, just nothing happens.
If I do the same exact thing on my old computer, it works - formula is converted and I see the braces {} added as part of my formula. Am I missing a macro or add-in? I'm using Excel 2002 - same version on my old computer. Is there another way to generate an array formula besides using CTRL+Shift+Enter?
Insert-shift Cells Down Macro
I want to be able to insert-shift cells down based on criteria in another cell. For instance, I would like a macro that would look at column B for saturday and sunday and then insert-shift cells down on the corresponding cells in column D...
Return The Value Of The Shift Time On Other Sheet
I have a worksheet that contains a 5 week shift pattern for workers. It is briefly laid out as:
JanFebAB
111330-21300700-1400
221230-20300700-1400
Months Jan to Dec 09 are there and shifts are divided up to ABCDE. I need another sheet to return the value of the shift time i.e. 1330-2130, probably via a lookup?? In my second sheet I have the Day number, the month and the shift letter.
Executing A SHIFT+xlUP Syntax
What is the syntax for executing a 'SHIFT+xlUP' function in an Excel macro?
Shift Cells Down Depends On Condition Of Others
Whenever "online" appears in the "Desc" column I need to shift the matching row down (shaded area). for example, "online" appears in H2, hence, C2:E2 need to be shifted down to C3:E3 leaving C2:E2 blank. Next, H3 has "online" there, C3:E3 need to be shifted down one row so the result will look like the data on the "result" tab.
Control Shift Enter In VBA
If ur using a formula that uses Control Shift Enter to activate it in Excel,
How do u use that formula in VBA? ....
Shift Cells Up, Removing Blank
How do I shift all the cells up labelled data, so that there are no blank rows in between? I tried using the ones found on the forum via search but it is stuck in an infinite loop.
Delete Cells Shift Left
I'm trying to do is check each Row and move the last 3 Columns of data in each Row to the left so they fall under the headings NHA2, NHA1, and OEM PN (Columns U, V, and W). Columns A and B are temporary. Column A utilizes the formula "= COUNTA(B2:AG2)" to count the number of cells in the row that contain data. Column B utilizes the formula "=COUNTA(C:C)" to count the total rows of data in the active worksheet. Columns A and B will be deleted at the end of the Macro.
The GOAL (END RESULT) is for Columns A through W (and ONLY these Columns) to contain data (keeping in mind that Columns A and B will be deleted at the end). If some rows have Column W (and beyond) blank, then I want to MOVE data from Columns U & V over to V & W and then COPY data from Column T into the [currently] blank Column U.
I've been trying to get the Macro to start in the last row and, using CASE Statements, delete the proper range of cells and SHIFT LEFT as it counts backwards towards the first row. I'm not too sure this is the best approach and could really use some advice from the experts! I can provide a "test" file if necessary ... the test file I've been working with is approximately 6.5MB, but I can delete most of the 14,287 rows and still give a good representation of how the data varies.
Sub b_DeleteCellsShiftLeft()
' DeleteShiftLeft Macro
' Macro recorded 3/14/2007 by George Nicholaou
' Need to assign variable for current row (?)
' r=ROW()
' Columns A and B are temporary
' Column A utilizes the formula "=COUNTA(B2:AG2)" to count the number of cells in
' the row that contain data
' Column B utilizes the formula "=COUNTA(C:C)" to count the total rows of data
' in the active worksheet
' They will be deleted at the end of the Macro
' What I want the Macro to do from this point is:
' 1. Check each Row and move the last 3 Columns of data in each Row to the left so
' they fall under the headings NHA2, NHA1, and OEM PN (Columns U, V, and W)
' 2. The GOAL (END/RESULT) is for Columns A through W (and ONLY these Columns)
' to contain data (keep in mind, Columns A and B will be deleted at the end)
' 3. If some rows have Column W blank, then MOVE data from Columns U & V over to V & W............
Shift Cells Left Based On Text Value
I need assistance in shifting cells to left based on three (3) specific text values in column B. There are tons of data and extra lines and stuff but consistently these three (3) text values are consistently not aligned and I simply need to shift entire row 1 space to the left. I can do the long and tedious process of going through them all but is there a macro which I can play that does it instantly... | 7,121 | 26,797 | {"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} | 3.28125 | 3 | CC-MAIN-2013-48 | longest | en | 0.916414 |
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### standardized canonical discriminant function coefficients interpretation
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Enable JavaScript use, and try again. Thai / ภาษาไทย 2. 2. Although comments of D.J. Standardized canonical discriminant function coefficients function1 lotsize .7845512 income .8058419 We requested both the unstandardized and standardized coefficients. statisticians, and ordinarily not highly technical. Strangely, Hair et al. However, these coefficients do not tell us between which of the groups the respective functions discriminate. Check out using a credit card or bank account with. Greek / Ελληνικά Test of Function(s) Wilks' Lambda Chi-square df Sig. into sections: Statistical Practice, General, Teacher's Corner, Statistical Example 1.A large international air carrier has collected data on employees in three different jobclassifications: 1) customer service personnel, 2) mechanics and 3) dispatchers. This preview shows page 1 - 12 out of 12 pages. JSTOR is part of ITHAKA, a not-for-profit organization helping the academic community use digital technologies to preserve the scholarly record and to advance research and teaching in sustainable ways. Experimental part. It is important to note, however, that these two equations are generated to yield the largest possible correlation be- Read your article online and download the PDF from your email or your account. The unstandardized coefficients apply to unstandardized variables. The number of discriminant functions is 1 minus the number of groups. . Select the purchase English / English Standardized discriminant coefficients can also be used as beta weight in regression. Show me. For canonical variates, the discussion includes standardized coefficients, correlations between variables and the function, rotation, and redundancy analysis. The standardized discriminant function coefficients in the table serve the same ... SPSS output: summary of canonical discriminant functions 1.The structure matrix table shows the correlations of each variable with each discriminant function. Czech / Čeština The score is calculated in the same manner as a predicted value from alinear regression, using the standardized coefficients and the standardizedvariables. For canonical variates, the discussion includes standardized coefficients, correlations between variables and the function, rotation, and redundancy analysis. unique) contribution of the variable to the discriminant function score, it is like regression coefficient. Wilks' Lambda . From standardized C1 and C2 canonical discriminant function coefficients, a DNA heterogeneity index (2c-HI) is proposed to characterize diploid cells providing a descriptive and predictive discriminant factor for solid tumour behaviour. Discriminant Attribute based perceptual mapping chart1.xlsx - Standardized Canonical Discriminant Function Coefficients Function 1 Function 2 1 2 Price. Only the first root is relevant since root two and three are not significant. You may look at both coefficients and correlations, keeping in mind the above differences. The prior argument given in the lda() function call isn’t strictly necessary because by default the lda() function will assign equal probabilities among the groups. Search in IBM Knowledge Center. each discriminant (canonical) function, and the larger the standardized coefficient, the greater is the contribution of the respective variable to the discrimination between groups. Each "ß" below: DF1 = ß1Xz1 + ß2Xz2 + ß3Xz3. The most important thing to remember is that the discriminant function coefficients denote the unique (partial) contribution of each variable to the discriminant function (s), while the structure coefficients denote the simple correlations between the variables and the function (s). Finnish / Suomi The score is calculated in the same manner as a predicted value from a linear regression, using the standardized coefficients and the standardized … Norwegian / Norsk in regression, are used to asses the relative classifying. These equations then yield the two synthetic variables illustrated in Figure 1. . . Portuguese/Brazil/Brazil / Português/Brasil These correlations generally alter the interpretation of the canonical functions. where a is a constant and b 1 through b m are regression coefficients. However I included this argument call to illustrate how to change the prior if you wanted. Italian / Italiano Standardized Discriminant Weights Standardized discriminant weights or coefficients are used when all the variables are in standard score from. Raw Canonical Coefficients for the Test Scores $$\bf{V}_1$$ scores1 scores2 scores3; create: 0.0697481411-0.192391323: 0.2465565859 : mech: 0.0307382997: 0.201574382-0.141895279: abs: 0.0895641768-0.495763258-0.280224053: math: 0.0628299739: 0.0683160677: 0.0113325936: Using the coefficient values in the first column, the first canonical variable for test scores is determined … The groups with the largest linear discriminant function, or regression coefficients, contribute most to the classification of observations. IBM Knowledge Center uses JavaScript. The standardized coefficients apply to variables standardized using the pooled within-group covariance. The interpretation of the results of a two-group problem is straightforward and closely follows the logic of multiple regression: Those variables with the largest (standardized) regression coefficients are the ones that contribute most to the prediction of group membership. etc. . Arabic / عربية Swedish / Svenska discriminant function coefficients are the discriminant function coefficients that are used as the multipliers when the variables have been standardised to a mean of 0 and a variance of 1. The CANDISC procedure performs a canonical discriminant analysis, computes squared Mahalanobis distances between class means, and performs both univariate and multivariate one-way analyses of variance. Canonical Discriminant Function Coefficients” table in SPSS). All Rights Reserved. The expert explains what the Box's M, the canonical correlation, the Chi-square, standardized canonical discriminant function coefficients, structure matrix, and the classification statistics mean. Danish / Dansk Linear discriminant analysis (LDA), normal discriminant analysis (NDA), or discriminant function analysis is a generalization of Fisher's linear discriminant, a method used in statistics, pattern recognition, and other fields, to find a linear combination of features that characterizes or separates two or more classes of objects or events. Pages 12. Nordlund and R. Nagel are welcomed, their arguments are not sufficient to accept the recommendation of using total variance estimates to standardize canonical discriminant function coefficients. Portuguese/Portugal / Português/Portugal Dutch / Nederlands Bulgarian / Български Building on two centuries' experience, Taylor & Francis has grown rapidlyover the last two decades to become a leading international academic publisher.The Group publishes over 800 journals and over 1,800 new books each year, coveringa wide variety of subject areas and incorporating the journal imprints of Routledge,Carfax, Spon Press, Psychology Press, Martin Dunitz, and Taylor & Francis.Taylor & Francis is fully committed to the publication and dissemination of scholarly information of the highest quality, and today this remains the primary goal. Correlation shows omnibus (i.e. Spanish / Español Request Permissions. There is Fisher’s (1936) classic example o… If standardized coefficients are used to help interpret a discriminant analysis, pooled within-group variance estimates should be used. 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Each ß '' below: DF1 = ß1Xz1 + ß2Xz2 + ß3Xz3 variables standardized using the pooled covariance! As a predicted value from alinear regression, are used when all the variables and the function rotation... The variables created by standardizing our discriminating variables job classifications appeal to different.... | 3,769 | 18,644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-25 | latest | en | 0.857793 |
https://apprize.best/python/unlocked/5.html | 1,725,999,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00041.warc.gz | 82,907,918 | 11,821 | Data Structures and Algorithms - Python Unlocked (2015)
# Python Unlocked (2015)
### Chapter 4. Data Structures and Algorithms
Data structures are the building blocks to solve programming problems. They provide organization for the data, and algorithms provide the logic to carve the perfect solution. Python provides many efficient built-in data structures that can be used effectively. There are other good data-structure implementations in the standard library as well as third-party libraries. Often, the more pressing question is when to use what, or what data-structure is good for the present problem description. To resolve this, we will cover the following topics:
· Python data structures
· Python library data structures
· Third-party data structures
· Algorithms on scale
Python built-in data structures
Key 1: Understanding Python's in-built data structure.
Before going in on how to use different data structures, we should take a look at the attributes of the object that are important for built-in data structures. For the default sorting to work, the object should have one of the __lt__, and __gt__ methods defined. Otherwise, we can pass a key function to the sorting method to use in getting the intermediate keys that are used to compare it, as shown in the following code:
def less_than(self, other):
return self.data <= other.data
class MyDs(object):
def __init__(self, data):
self.data = data
def __str__(self,):
return str(self.data)
__repr__ = __str__
if __name__ == '__main__':
ml = [MyDs(i) for i in range(10, 1, -1)]
try:
ml.sort()
except TypeError:
print("unable to sort by default")
for att in '__lt__', '__le__', '__gt__', '__ge__':
setattr(MyDs, att, less_than)
ml = [MyDs(i) for i in list(range(5, 1, -1)) + list(range(1, 5,))]
try:
ml.sort()
print(ml)
except TypeError:
print("cannot sort")
delattr(MyDs, att)
ml = [MyDs(i) for i in range(10, 1, -1)]
print("sorted", sorted(ml, key=lambda x: x.data))
ml.sort(key=lambda x: x.data)
print("sort", ml)
The output of the preceding code is as follows:
[1, 2, 2, 3, 3, 4, 4, 5]
cannot sort
[5, 4, 4, 3, 3, 2, 2, 1]
cannot sort
sorted [2, 3, 4, 5, 6, 7, 8, 9, 10]
sort [2, 3, 4, 5, 6, 7, 8, 9, 10]
Whether two objects are equal in value is defined by the output of the __eq__ method. Collections have the same value if they have the same length and the same value of all items, as shown in the following code:
def equals(self, other):
return self.data == other.data
class MyDs(object):
def __init__(self, data):
self.data = data
def __str__(self,):
return str(self.data)
__repr__ = __str__
if __name__ == '__main__':
m1 = MyDs(1)
m2 = MyDs(2)
m3 = MyDs(1)
print(m1 == m2)
print(m1 == m3)
setattr(MyDs, '__eq__', equals)
print(m1 == m2)
print(m1 == m3)
delattr(MyDs, '__eq__')
print("collection")
l1 = [1, "arun", MyDs(3)]
l2 = [1, "arun", MyDs(3)]
print(l1 == l2)
setattr(MyDs, '__eq__', equals)
print(l1 == l2)
l2.append(45)
print(l1 == l2)
delattr(MyDs, '__eq__')
print("immutable collection")
t1 = (1, "arun", MyDs(3), [1, 2])
t2 = (1, "arun", MyDs(3), [1, 2])
print(t1 == t2)
setattr(MyDs, '__eq__', equals)
print(t1 == t2)
t1[3].append(7)
print(t1 == t2)
The output of the preceding code is as follows:
False
False
False
True
collection
False
True
False
immutable collection
False
True
False
Hash function maps a larger value set to the smaller hash set. Hence, two different objects can have the same hash, but objects with a different hash must be different. In other words, equal value objects should have the same hash, and objects with a different hash must have different values for hash to be meaningful. When we define __eq__ in a class, we must define a hash function as well. By default, for user class instances, hash uses the ID of the object, as shown in the following code:
class MyDs(object):
def __init__(self, data):
self.data = data
def __str__(self,):
return "%s:%s" % (id(self) % 100000, self.data)
def __eq__(self, other):
print("collision")
return self.data == other.data
def __hash__(self):
return hash(self.data)
__repr__ = __str__
if __name__ == '__main__':
dd = {MyDs(i): i for i in (1, 2, 1)}
print(dd)
print("all collisions")
setattr(MyDs, '__hash__', lambda x: 1)
dd = {MyDs(i): i for i in (1, 2, 1)}
print(dd)
print("all collisions,all values same")
setattr(MyDs, '__eq__', lambda x, y: True)
dd = {MyDs(i): i for i in (1, 2, 1)}
print(dd)
The output of the preceding code is as follows:
collision
{92304:1: 1, 92360:2: 2}
all collisions
collision
collision
{51448:1: 1, 51560:2: 2}
all collisions,all values same
{92304:1: 1}
It can be seen that mutable objects do not have hash defined. Although this is not advised, we can, however, do so in our user defined classes:
· Tuples: These are immutable lists, slice operations are O(n), retrieval is O(n), and they have small memory requirements. They are normally used to group objects of different types in a single structure, such as C language structures, where the position is fixed for particular types of information, shown as follows:
· >>> sys.getsizeof(())
· 48
· >>> sys.getsizeof(tuple(range(100)))
848
Named tuples that are available from the collections module can be used to access values with object notation, as follows:
>>> from collections import namedtuple
>>> student = namedtuple('student','name,marks')
>>> s1 = student('arun',133)
>>> s1.name
'arun'
>>> s1.marks
133
>>> type(s1)
<class '__main__.student'>
· Lists : These are mutable data structures that are similar to tuples. They are good to collect objects of similar types. When analyzing their time-complexity, we see that insert, delete, slice, and copy operations require O(n), Retrieval require len O(1), and sort requires O(nlogn). Lists are implemented as dynamic arrays. It must resize to double of its previous on increase in size greater than current capacity. Insert and delete at the front of the list takes more time as it must move all references to other elements one by one:
· >>> sys.getsizeof([])
· 64
· >>> sys.getsizeof(list(range(100)))
1008
· Dictionary: These are mutable mappings. A key can be any hashable object. Getting a value for key, setting a value for a key, and deleting a key are all O(1), and copying is O(n):
· >>> d = dict()
· >>> getsizeof(d)
· 288
· >>> getsizeof({i:None for i in range(100)})
6240
· Sets: These can be thought as of groups of items where hash is used to retrieve items. Sets have methods to check union, and intersection, which is useful rather than checking the same with lists. Let's take an example of groups of animals, as follows:
· >>> air = ("sparrow", "crow")
· >>> land = ("sparrow","lizard","frog")
· >>> water = ("frog","fish")
· >>> # find animal able to live on land and water
· ...
· >>> [animal for animal in water if animal in land]
· ['frog']
· >>>
· >>> air = set(air)
· >>> land = set(land)
· >>> water = set(water)
· >>> land | water #animal living either land or water
· {'frog', 'fish', 'sparrow', 'lizard'}
· >>> land & water #animal living both land and water
· {'frog'}
· >>> land ^ water #animal living on only one land or water
{'fish', 'sparrow', 'lizard'}
Their implementation and time-complexity is very similar to dictionary, shown as follows:
>>> s = set()
>>> sys.getsizeof(s)
224
>>> s = set(range(100))
>>> sys.getsizeof(s)
8416
Python library data structures
Key 2: Using Python's standard library data structures.
· collections.deque: The collections module have a deque implementation. Deque is useful for the scenarios where item insertion and deletion occurs at both ends of structure as it has efficient inserts at the start of structure as well. Time-complexity is similar to copy O(n), insert—O(1), and delete—O(n). The following graph shows an insert at 0 position operation comparison between list and deque:
· >>> d = deque()
· >>> getsizeof(d)
· 632
· >>> d = deque(range(100))
· >>> getsizeof(d)
1160
The following image is the graphical representation of the preceding code:
· PriorityQueue: A standard library queue module has implementations for multiproducer, and multiconsumer queues. We can simplify and reuse its PriorityQueue for simpler cases using the heapq module, as follows:
· from heapq import heappush, heappop
· from itertools import count
·
· class PriorityQueue(object):
· def __init__(self,):
· self.queue = []
· self.counter = count()
·
· def __len__(self):
· return len(self.queue)
·
· def pop(self,):
· item = heappop(self.queue)
· print(item)
· return item[2],item[0]
·
· def push(self,item,priority):
· cnt = next(self.counter)
heappush(self.queue, (priority, cnt, item))
Other than these, queue modules have threadsafe, LifoQueue, PriorityQueue, queue, deque implementations. Also, lists can be used as stacks or queues. Collections also have orderedDict, which remembers the sequence of elements.
Third party data structures
Key 3: Using third-party data structures.
Python has a good bunch of data structures in the core language/library. But sometimes, an application has very specific requirements. We can always use third-party data-structure packages. Most of such modules are Python wrapper over C, C++ implementations:
· The blist module provides a drop-in replacement for list, sortedList, and sortedset. It is discussed in greater detail in later chapters.
· The bintrees module provides binary, AVL tree, and Red-Black trees.
· The banyan module provides Red-Black trees, splay tree, and sorted lists.
· The Sortedcontainers module provides SortedList, SortedDict, and SortedSet. So, one can get almost every data structure for Python easily. More stress should be given on why one data structure is better than another for a use case.
Arrays/List
For numeric calculations involving math, NumPy arrays should be considered. They are fast, memory-efficient, and provide many vector and matrix operations.
Binary tree
Trees have better worst-case insertion/removal, O(log(n)), min/max, and look-ups than dictionaries. There are several implementations that are available.
One module is bintrees, which have C implementations that are available for Red-Black trees, AVL tree, and Binary trees. For example, in Red-Black trees, it is easy to find max, and min, ranges as shown in the following example:
tr = bintrees.FastRBTree()
tr.insert("a",40)
tr.insert("b",5)
tr.insert("a",9)
print(list(tr.keys()),list(tr.items()))
print(tr.min_item())
print(tr.pop_max())
print(tr.pop_max())
tr = bintrees.FastRBTree([(i,i+1) for i in range(10)])
print(tr[5:9])
The output of the preceding code is as follows:
['a', 'b'] [('a', 9), ('b', 5)]
('a', 9)
('b', 5)
('a', 9)
FastRBTree({5: 6, 6: 7, 7: 8, 8: 9})
Sorted containers
These are pure-Python modules having SortedList, SortedSet, and SortedDict Data structures, which can keep the keys/items sorted. The SortedContainers module claims to have speed comparable to C extensions modules, shown as follows:
import sortedcontainers as sc
import sys
l = sc.SortedList()
l.update([0,4,2,1,4,2])
print(l)
print(l.index(2),l.index(4))
print(l[-1])
l = sc.SortedList(range(10))
print(l)
print(list(l.irange(2,6)))
seta = sc.SortedSet(range(1,4))
setb = sc.SortedSet(range(3,7))
print(seta - setb)
print(seta | setb )
print(seta & setb)
print([i for i in seta])
The output of the preceding code is as follows:
SortedList([0, 1, 2, 2, 4, 4], load=1000)
2 4
6
SortedList([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], load=1000)
[2, 3, 4, 5, 6]
SortedSet([1, 2, 3, 4, 5, 6], key=None, load=1000)
[1, 2, 3]
Trie
This is an ordered-tree data-structure, where the position in the tree defines the key. The keys are normally strings. In comparison to dictionaries, it has faster worst-case data retrieval O(m). Hash functions are not needed. If we are using strings only to be stored in the keys, it can take a lot less space then dictionary.
In Python, we have the marisa-trie package that provides this functionality as static Data structures. It is a Cython wrapper over the C++ library. We can associate values with the keys as well. It also provides memory mapped I/O, which is useful to decrease memory usage on cost of speed. The datrie is another package that provides read-write tries, The following are some basic usage of these libraries:
>>> import string
>>> import marisa_trie as mtr
>>> import datrie as dtr
>>>
>>>
... tr = mtr.Trie([u'192.168.124.1',u'192.168.124.2',u'10.10.1.1',u'10.10.1.2'])
>>> #get all keys
... print(tr.keys())
['10.10.1.1', '10.10.1.2', '192.168.124.1', '192.168.124.2']
>>> #check if key exists
... print(tr.has_keys_with_prefix('192'))
True
>>> # get id of key
... print(tr.get('192.168.124.1'))
2
>>> # get all items
... print(tr.items())
[('10.10.1.1', 0), ('10.10.1.2', 1), ('192.168.124.1', 2), ('192.168.124.2', 3)]
>>>
>>> # storing data along with keys
... btr = mtr.BytesTrie([('192.168.124.1',b'redmine.meeku.com'),
... ('192.168.124.2',b'jenkins.meeku.com'),
... ('10.5.5.1',b'gerrit.chiku.com'),
... ('10.5.5.2',b'gitlab.chiku.com'),
... ])
>>> print(list(btr.items()))
[('10.5.5.1', b'gerrit.chiku.com'), ('10.5.5.2', b'gitlab.chiku.com'), ('192.168.124.1', b'redmine.meeku.com'), ('192.168.124.2', b'jenkins.meeku.com')]
>>> print(btr.get('10.5.5.1'))
[b'gerrit.chiku.com']
>>>
>>> with open("/tmp/marisa","w") as f:
... btr.write(f)
...
>>>
... # using memory mapped io to decrease memory usage
... dbtr = mtr.BytesTrie().mmap("/tmp/marisa")
>>> print(dbtr.get("192.168.124.1"))
[b'redmine.meeku.com']
>>>
>>>
>>> trie = dtr.Trie('0123456789.') #define allowed character range
>>> trie['192.168.124.1']= 'redmine.meeku.com'
>>> trie['192.168.124.2'] = 'jenkins.meeku.com'
>>> trie['10.5.5.1'] = 'gerrit.chiku.com'
>>> trie['10.5.5.2'] = 'gitlab.chiku.com'
>>> print(trie.prefixes('192.168.245'))
[]
>>> print(trie.values())
['gerrit.chiku.com', 'gitlab.chiku.com', 'redmine.meeku.com', 'jenkins.meeku.com']
>>> print(trie.suffixes())
['10.5.5.1', '10.5.5.2', '192.168.124.1', '192.168.124.2']
>>>
>>> trie.save("/tmp/1.datrie")
>>> print(ntr.values())
['gerrit.chiku.com', 'gitlab.chiku.com', 'redmine.meeku.com', 'jenkins.meeku.com']
>>> print(ntr.suffixes())
['10.5.5.1', '10.5.5.2', '192.168.124.1', '192.168.124.2']
Algorithms on scale
Key 4: Thinking out-of-the-box for the algorithms.
An algorithm is how we solve a problem. The most common issue of not being able to solve the problem is to not being able to define it properly. Normally, we look to apply an algorithm only at a small level, such as in a small functionality, or sorting in a function. We, however, do not think about algorithms when the scale increases, mostly the stress is on how fast it is. Let's take a simple requirement of sorting a file and sending it to a user. If the file is, let's say 10-20 KB or so, it will be best to simply use the Python sorted function to sort the entries. In the following code, the file is of format where columns are ID, name, due, and due-date. We want to sort it based on dues, as follows:
10000000022,shy-flower-ac5,-473,16/03/25
10000000096,red-water-e85,-424,16/02/12
10000000097,dry-star-c85,-417,16/07/19
10000000070,damp-night-c76,-364,16/03/12
def dosort(filename,result):
with open(filename) as ifile:
with open(result,"w") as ofile:
for line in sorted(
),key=lambda x:int(x.split(',')[2])
):
ofile.write(line)
ofile.write('\n')
This works great, but as the file increases in size, the memory requirement increases. We cannot load all contents in the memory at the same time. Hence, we can use external merge-sort to divide the file into small parts, sort them, and then merge the sorted results together. In the following code, we used heapq.merge to merge iterators:
import tempfile
import heapq
while True:
if not ilines:
break
for line in ilines:
yield int(line.split(',')[2]),line
def dosort(filename, result):
partition = 5000
with open(filename,"r") as ifile:
with open(result,"w") as ofile:
tempfiles = []
while True:
if len(ilines) == 0 :
break
tfile = tempfile.TemporaryFile(mode="w+")
tfile.writelines(
sorted(
ilines,
key=lambda x:int(x.split(',')[2])
))
tfile.seek(0)
tempfiles.append(tfile)
lentempfiles = len(tempfiles)
res = []
res.append(line[1])
if len(res) > 100:
ofile.writelines(res)
res.clear()
if res:
ofile.writelines(res)
ofile.close()
Once we use up memory of a single computer, or have files distributed over multiple computers in a network, the file-based algorithm will not work. We will need to sort incoming streams from upstream servers, and send the sorted stream to the downstream. If we look at the following code carefully, we will see that we have not changed the underlying mechanism. We are still using heapq.merge to merge elements, but now, we are getting elements from the network instead. The following client code is simple, it just starts sending sorted lines by lines on receive of the next command from a downstream server:
import socket
import sys
from sort2 import dosort2
HOST = '127.0.0.1'
PORT = 9002
NCLIENTS = 2
class Client(object):
def __init__(self,HOST,PORT,filename):
self.skt = socket.socket(socket.AF_INET,socket.SOCK_STREAM)
self.skt.connect((HOST,PORT))
self.filename = filename
self.skt.setblocking(True)
def run(self):
for line in dosort2(self.filename):
print("for",line)
data = self.skt.recv(1024)
print("data cmd",data)
if data == b'next\r\n':
data = None
self.skt.send(line[1].encode())
else:
print("got from server",data)
print("closing socket")
self.skt.close()
c = Client(HOST,PORT,sys.argv[1])
c.run()
In the server code, the ClientConn class abstracts away network operations and provides an iterator interface to heapq.merge. We can greatly enhance the code using buffering. Here, the get_next method gets new line from the client. Simple abstraction solved a great problem:
import socket
import heapq
from collections import deque
HOST = '127.0.0.1'
PORT = 9002
NCLIENTS = 2
class Empty(Exception):
pass
class ClientConn(object):
self.conn = conn
self.buffer = deque()
self.finished = False
self.get_next()
def __str__(self, ):
def get_next(self):
self.conn.send(b"next\r\n")
try:
ndata = self.conn.recv(1024)
except Exception as e:
print(e)
self.finished = True
ndata = None
if ndata:
ndata = ndata.decode()
print("got from client", ndata)
self.push((int(ndata.split(',')[2]), ndata))
else:
self.finished = True
def pop(self):
if self.finished:
raise Empty()
else:
elem = self.buffer.popleft()
self.get_next()
return elem
def push(self, value):
self.buffer.append(value)
def __iter__(self, ):
return self
def __next__(self, ):
try:
return self.pop()
except Empty:
print("iter empty")
raise StopIteration
class Server(object):
def __init__(self, HOST, PORT, NCLIENTS):
self.nclients = NCLIENTS
self.skt = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
self.skt.setblocking(True)
self.skt.bind((HOST, PORT))
self.skt.listen(1)
def run(self):
self.conns = [] # list of all clients connected
while len(self.conns) < self.nclients: # accept client till we have all
self.conns.append(cli)
print('Connected by', cli)
with open("result", "w") as ofile:
for line in heapq.merge(*self.conns):
print("output", line)
ofile.write(line[1])
s = Server(HOST, PORT, NCLIENTS)
s.run()
Summary
In this chapter, we learned about the data structures that are available in the Python standard library and some third-party libraries, which are extremely useful for everyday programming. Knowledge of data-structure usage is very much important in choosing right tool for the job. Choosing of an algorithm is highly application-specific, and we should always try to find out a solution that is simpler to read.
In the next chapter, we will cover design patterns that provide great help in writing elegant solutions to the problems.
| 5,478 | 20,024 | {"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-2024-38 | latest | en | 0.793022 |
http://openstudy.com/updates/51adb293e4b06ee3ee25e594 | 1,516,733,441,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00132.warc.gz | 255,761,015 | 8,490 | • anonymous
Students Cost of land Cost of equipment Cost of mining and reclamation Time taken to mine the area 1 $5$2 $2.00 per minute 5 minutes 2$4 ? $2.00 per minute 6 minutes If the sum of the total mining costs for Student 1 and Student 2 is$40, what is the cost of equipment for Student 2? $4.00$6.00 $7.00$9.00
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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Looking for something else?
Not the answer you are looking for? Search for more explanations. | 343 | 1,246 | {"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": 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-2018-05 | latest | en | 0.364327 |
http://quizlet.com/8731996/exponents-quiz-flash-cards/ | 1,386,810,137,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164128316/warc/CC-MAIN-20131204133528-00075-ip-10-33-133-15.ec2.internal.warc.gz | 147,740,681 | 17,622 | 2
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1,024
9
27
81
16
64
256
25
125
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36
216
49
343
64
512
81
729
(ab)³ = a³b³
a³a² = a³⁺²
(a²)³ = a²*³
a³/a² = a³⁻²
(a/b)³ = a³/b³
#⁰ = 1
a⁻³ = 1/a³
1/a⁻³ = a³ | 133 | 217 | {"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} | 3.03125 | 3 | CC-MAIN-2013-48 | latest | en | 0.126062 |
https://www.distancesto.com/fuel-cost/sy/inkhil-to-quneitra-governorate/history/164004.html | 1,716,207,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00244.warc.gz | 643,693,981 | 14,617 | # INR1.12 Total Cost of Fuel from Inkhil to Quneitra Governorate
Your trip to Quneitra Governorate will consume a total of 0.45 gallons of fuel.
Trip start from Inkhil, SY and ends at Quneitra Governorate, SY.
Trip (17.9 mi) Inkhil » Quneitra Governorate
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Inkhil, Syria and end at Quneitra Governorate, Syria.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Inkhil to Quneitra Governorate plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Inkhil to Quneitra Governorate.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Inkhil to Quneitra Governorate.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Inkhil to Quneitra Governorate.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Quneitra Governorate are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Inkhil to Quneitra Governorate.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Inkhil to Quneitra Governorate.
Speaking of travel time, a flight to Quneitra Governorate takes up a lot less. How much less? Flight time from Inkhil to Quneitra Governorate.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Inkhil to Quneitra Governorate.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
Recent Fuel Calculations for Inkhil SY:
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Fuel Cost from Inkhil to Tasil | 679 | 2,875 | {"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} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.959635 |
https://www.glassdoor.co.in/Interview/Scenario-If-you-were-running-late-for-a-job-interview-and-you-had-no-lights-on-in-your-apartment-Let-s-say-you-went-into-QTN_64417.htm | 1,606,315,559,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182794.28/warc/CC-MAIN-20201125125427-20201125155427-00261.warc.gz | 690,119,891 | 53,463 | Eze Software Interview Question: Scenario: If you were running... | Glassdoor.co.in
## Interview Question
Q & A Analyst Training Program Interview Boston, MA (US)
# Scenario: If you were running late for a job interview and
you had no lights on in your apartment. Let's say you went into your dresser drawer and had to pull out one pair of socks. All you had was black or brown socks. And there were 5 socks total, what is the least amount of pulls you could do?
Tags:
logic
1
basically you have to keep in mind the probability of pulling out each sock. the shortest answer would be 2 pulls.
Interview Candidate on 26-Apr-2010
1
You have to pull out three socks to be sure that you had at least two of one color.
tjg on 02-May-2010
0
it would have to be three socks because if you pulled just two, you could get one black and one brown.
Anonymous on 02-May-2010
0
it's 2. You could be lucky and get 2 socks of the same color.
Anonymous on 16-Sep-2015 | 244 | 963 | {"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-2020-50 | latest | en | 0.976927 |
https://physics-network.org/what-is-terminal-velocity-of-a-falling-object/ | 1,709,577,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00598.warc.gz | 444,062,054 | 60,355 | # What is terminal velocity of a falling object?
The steady speed at which an object free falls is known as the terminal velocity. As an object falls, its speed increases up to a point where the gravitational pull and drag force are equal. At this point, the velocity of the object becomes the terminal velocity, and the acceleration becomes zero.
## What does Terminal mean in physics?
Terminal Velocity is a 1941 Willys monster truck owned by Jamey Garner of Over Bored Motorsports out of Fortville, Indiana. It is the fifth name in the Over Bored Motorsports stable and is driven by Jon Zimmer Jr.
## What speed is terminal velocity?
The speed achieved by a human body in free fall is conditioned of two factors, body weight and body orientation. In a stable, belly to earth position, terminal velocity of the human body is about 200 km/h (about 120 mph).
## How do you find terminal velocity?
In plain English, the terminal velocity of the object is equal to the square root of the quotient of twice the object’s weight over the product of the object’s frontal area, its drag coefficient, and the gas density of the medium through which the object is falling.
## Why do we have terminal velocity?
The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.
## What does terminal velocity depend on?
Terminal velocity is the point at which the drag force equals the force of gravity. Terminal velocity will depend on the mass, cross sectional area, and drag coefficient of the object as well as the density of the fluid through which the object is falling and gravitational accelleration.
## Do heavier objects fall faster?
Given two objects of the same size but of different materials, the heavier (denser) object will fall faster because the drag and buoyancy forces will be the same for both, but the gravitational force will be greater for the heavier object.
## Who drives terminal velocity?
In general, a person falling through the air on Earth reaches terminal velocity after about 12 seconds, which covers about 450 meters or 1500 feet. A skydiver in the belly-to-earth position reaches a terminal velocity of about 195 km/hr (54 m/s or 121 mph).
## What is terminal velocity in free fall?
Terminal velocity is the fastest speed that an object will reach as it falls through the air. As a skydiver jumps, gravity pulls them towards the earth, accelerating their fall.
## What height is terminal velocity?
“Especially me being the first and only female to drive Grave Digger, I feel like I have extra eyes on me.” The Grave Digger monster truck looms large behind Krysten Anderson, but it’s not as big as her father’s legacy. Feld Entertainment, Inc.
## Is there a terminal velocity in space?
Terminal Velocity is the velocity at which a the gravitational acceleration and the drag of the air cancel each other out to zero change in velocity. While one might think that space as a vacuum, it really isn’t. The interstellar medium has densities between 10^-4 and 10^6 molecules per cube centimetre.
## Why does rain not hurt?
Momentum is the product of the velocity and the mass of the body. Now we are lucky that these droplets are small and negligible in mass. Therefore a rain drop will not hurt us neither will it make holes on the ground because due to its very small size, it will have a very low terminal velocity.
## Can humans survive terminal velocity?
A human would need to be 8.5 m tall and 2.0 m wide in order to sufficiently slow the terminal velocity so they can land in 0.1 seconds. Unfor- tunately, this size human would most certainly not be able to survive, as they wouldn’t have enough muscle density to support their body.
## How high can a human fall without death?
A retrospective analysis of 101 patients who survived vertical deceleration injuries revealed an average fall height of 23 feet and 7 inches (7.2 meters), confirming the notion that survivable injuries occur below the critical threshold of a falling height around 20-25 feet [1].
## Can a rat survive terminal velocity?
A rat can fall as far as 50 feet and land unharmed – in theory! This is not a result found by live experiments, but by calculating the terminal velocity of an average rat at sea-level on Earth. Terminal velocity for animals is approximately 90 d0.
## Why can squirrels survive high falls?
The reason for this is because a squirrel has a large area/mass ratio. This means that gravity does not pull on it with too much force but relatively large aerodynamic resistance will be generated.
## Can a parachute reach terminal velocity?
At approximately 120mph, skydivers reach terminal velocity and ride air molecules that feel as stable as laying on a bed.
## Do heavier objects have a faster terminal velocity?
heavy objects will have a higher terminal velocity than light objects. (Why? It takes a larger air resistance force to equal the weight of a heavier object. A larger air resistance force requires more speed.)
## Can you go faster than terminal velocity?
Yes. The object will slow down to its terminal velocity if its speed starts higher than its terminal speed.
## What is the terminal velocity of a ping pong ball?
The measured terminal velocity is 9.5 m/s, 98% of which is attained after falling 12.5 m.
## Is terminal velocity initial or final?
g is the acceleration due to gravity, h is the height of object. Suppose an object is falling from a height h with an initial velocity of zero. It is known that the final velocity is termed as terminal velocity.
## What is terminal velocity in a vacuum?
Terminal velocity specifically refers to the velocity at which some accelerating force and some velocity-dependent drag force reach equilibrium. In vacuum with no drag force, terminal velocity is not a concept that makes sense. Of course space is not really a vacuum, but that’s not the point of the question.
## What reaches terminal velocity first?
Since the weight of the water-filled ball would be more, therefore it will have to attain more velocity to attain terminal velocity. Therefore, the ball filled with air reaches first to its terminal velocity.
## Does mass increase terminal velocity?
We see from this relation that the terminal velocity of an object is proportional to the object’s mass! The more massive an object, the faster it falls through a fluid. | 1,367 | 6,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-10 | latest | en | 0.924084 |
http://www.mashpedia.com/List_of_unusual_units_of_measurement | 1,369,142,961,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700014987/warc/CC-MAIN-20130516102654-00011-ip-10-60-113-184.ec2.internal.warc.gz | 585,906,558 | 54,234 | For units of measure primarily used in countries where English is not the main language, see the article specific to that country, a list of which can be found in the systems of measurement article.
For unusual units of measure invented primarily for amusement, see List of humorous units of measurement
An unusual unit of measurement is a unit of measurement that does not form part of a coherent system of measurement; especially in that its exact quantity may not be well known or that it may be an inconvenient multiple or fraction of base units in such systems. This definition is deliberately not exact since it might seem to encompass units such as the week or the light-year which are quite "usual" in the sense they are often used; if they are used out of context, they may be "unusual," as demonstrated by the FFF system of units.
Unusual units of measure are sometimes used by scientists, especially physicists and mathematicians, and other technicians, engineers and computer programmers, either humorously or for real convenience when dealing with objects in the real world whose quantities in more conventional units may be awkward to use. Scientists studying nuclear reactions use the barn quite frequently; however, its use is extremely rare for other scientific disciplines. Unusual units are also employed by journalists and science instructors in an attempt to help people who may not be able to grasp extremely large or small numbers or unfamiliar scientific units. "An area the size of Denmark" and the banana equivalent dose are classic examples. Whether such comparisons actually are meaningful is open to debate, since the comparative quantity itself may not be well known with much accuracy. (e.g., what size and cultivar of the banana is the unit based on?)
## Systems of measurement
### MTS units
During the twentieth century, the Soviets and French briefly used a variant of the metric system where the base unit of mass was the tonne.
### SI-imperial hybrids
In the US, mongrel units are sometimes formed by a combination of traditional units, which are widely used, and metric units. Thus, "grams per fluid ounce" and "grams per pound of body weight" are common units used in sports nutrition, for example, to express the concentration of carbohydrate in a beverage.
A hybrid standard quantity used in mining is the assay ton (AT), which is as many milligrams as there are troy ounces in a ton: 29.17 grams if the ton used is the short ton, and 32.67 g if the ton used is the long ton. So to find how many ounces of gold are in a ton of rock, one measures the number of milligrams of gold in an assay ton of rock.
There are also reports of engineers using base-ten SI prefixes in combination with Imperial or US customary units, for example the kiloyard (914.4 m). The kip or kilopound is regularly used in structural engineering. Similarly, the kilofoot is quite common in US telecommunication engineering, as significant distances in cable route planning are usually given in thousands of feet. Instruments like optical time-domain reflectometers usually have an option to display results in kilofeet. Perhaps most common is the use of the thou or mil, defined as 1/1,000 of an inch (25.4 µm), frequently used in the manufacturing industry and to measure the thickness of very thin materials like film and plastic sheeting. A related unit is the circular mil, used for measuring the cross-sectional area of wire.
In the UK, it is still (2007) not uncommon to find the "metric foot" in use in the domestic refurbishment market. A metric foot is 30 cm and usually it is used with lumber (timber in the UK) that is only available in metric yards or 90 cm multiples (see metric inch below). Its square form is also common in some fields: for example, carpet and other flooring materials, when supplied in "tiles" (squares), are often supplied in this size.
In commercial office fitout in the UK, most suspended ceilings and raised floors are based on a 60 cm module, or 2 "metric feet".
A useful unit when working with optical pathway lengths in the lab is the one foot per nanosecond approximation for the speed of light. In electronic circuitry, where the velocity of propagation is somewhat slower than the speed of light in a vacuum, similar approximations can be used for "signal races" in the circuitry. (See light-nanosecond below.)
### Planck units
Planck units, sometimes called "natural units", were proposed in 1899 by the physicist Max Planck, and are defined by setting certain physical constants to equal 1. In this system, one metre equals about 6.187×1034 Planck length units, and one second equals 1.854×1043 Planck time units. The speed of light is, by definition, exactly 1 Planck length/Planck time, and many other quantities take on simple forms as well. Because of this, these units are favoured in some branches of physics.
## Length
### Attoparsec
Parsecs are used in astronomy to measure enormous interstellar distances. A parsec is approximately 3.26 light-years or about 3.085×1016 m (1.917×1013 mi). Combining it with the "atto-" prefix yields attoparsec (apc), a conveniently human-scaled unit of about 3.085 centimetres (1.215 in) that has no obvious practical use.[1]
### Light-nanosecond
The light-nanosecond was popularized as a unit of distance by Grace Hopper as the distance which a photon could travel in one billionth of a second (roughly 30 cm or one foot): "The speed of light is one foot per nanosecond."[2] In her speaking engagements, she was well known for passing out light-nanoseconds of wire to the audience, and contrasting it with light-microseconds (a coil of wire 1,000 times as long) and light-picoseconds (the size of ground black pepper). Over the course of her life, she found many uses for this visual aid, including demonstrating the waste of sub-optimal programming, illustrating advances in computer speed, and simply giving young scientists and policy makers the ability to conceptualize the magnitude of very large and small numbers.[3]
### Metric foot
A metric foot (defined as 300 mm) has been used occasionally in the UK[4] but has never been an official unit.
### Horse
Similarly, horses are used to measure distances in horse racing - a horse length (shortened to merely a length when the context makes it obvious) equals roughly 8 feet or 2.4 metres. Shorter distances are measured in fractions of a horse length; also common are measurements of a full or fraction of a head, a neck, or a nose. "Neck and neck" is an example of a Siamese twin which refers to the neck as a unit of measurement in horse racing.
### Car
In vehicle driving tests in the UK and elsewhere, braking distances can be given in metres or in car lengths (with a car for this purpose being typically around 4 m long). A "car length" or "length" is also a common unit for estimating the distance between two cars in auto racing.[citation needed]
### Bus
In Britain, newspapers and other media will frequently refer to lengths in comparison to the length (8.4 metres or 28 feet) or height (4.4 metres or 14 feet) of a London Routemaster double-decker bus.
The UK Department of Transport has changed the definition of the maximum acceptable length dimensions about 40 times since 1900, as suggested by the UK documentary "100 Years of British Buses" (1996).[citation needed]
Similarly, in the United States the yellow school bus is a common point of comparison, with an average length of around 35 feet (11 m) .
Popular topics for such comparison include the blue whale (30 m), the IMAX screen (22 × 16.1 m), and the diplodocus (35 m).[5]
### Football field (length)
The length of an American football field is 100 yd (91 m). If the end zones are included, it is 120 yd (110 m), but 100 yards is used informally as a unit, to allow easier conversion from formal measurement in feet or yards.
The Canadian football field is 65 yards (59 m) wide and 110 yards (100 m) long with end zones 20 yards (18 m) deep.
Media in the UK also use the football pitch as a unit of length, although an area of the association football pitch is not fixed, but varies within a limit of 90–120 m (98–130 yd) in length and 45–90 m (50–100 yd) in width. The usual size of a football pitch is 105 m × 68 m (115 yd × 75 yd), the area used for matches in the UEFA Champions League.
### Tall buildings
The Empire State Building is one Empire State Building tall
In the US, buildings such as the Empire State Building (449 m or 1,473 ft), Willis Tower (519 m or 1,703 ft), and Seattle Space Needle (184 m or 604 ft) are used as comparative measurements of height, as are the Statue of Liberty and Washington Monument. The Empire State Building (but not the others mentioned) is also used in Britain for describing particularly large/tall objects. Before the September 11 attacks, the World Trade Center towers were frequently used in the same fashion.
In France, the Eiffel Tower (324 m or 1,063 ft), and the UK, Nelson's Column (169 ft or 52 m), Blackpool Tower (158 m or 518 ft),[6] the Tower of London (96.3 m or 316 ft), and St Paul's Cathedral (108 m or 354 ft) are commonly used by British newspapers or reference books to give the comparative heights of buildings or, occasionally, mountains.
In Canada (and occasionally elsewhere), the Toronto CN Tower (553 m or 1,814 ft) is used as a unit of length.[7][8]
### Block
A city block (in most US cities) is between 1/16 and 1/8 mi (100 and 200 m). In Manhattan, the measurement "block" usually refers to a north-south block, which is 1/20 mi (80 m). Sometimes people living in cities with a regularly-spaced street grid will speak of long blocks and short blocks. Within a typical large North American city, it is often only possible to travel along east-west and north-south streets, so travel distance between two points is often given in the number of blocks east-west plus the number north-south (known to mathematicians as the Manhattan Metric).[citation needed]
### The Earth's circumference
The circumference of a great circle of the Earth (about 40,000 km/25,000 mi/21,600 nmi/199,000 furlongs) is often compared to large distances. For example, one might say that a large number of objects laid end-to-end at the equator "would circle the Earth four and a half times".[9] According to WGS-84, the circumference of a circle through the poles (twice the length of a meridian) is 40,007,862.917 metres (131,259,392.772 ft) and the length of the equator is 40,075,016.686 m (131,479,713.537 ft). Despite the fact that the difference (0.17%) between the two is insignificant at the low precision that these quantities are typically given to, it is nevertheless often specified as being at the equator.[citation needed]
The definitions of both the nautical mile and the kilometre were originally derived from the Earth's circumference as measured through the poles. The nautical mile was defined as a minute of arc of latitude measured along any meridian. A circle has 360 degrees, and each degree is 60 minutes, so the nautical mile was defined as 1/21,600 of the Earth's circumference, or about 1,852.22 metres. However, by international agreement, it is now defined to be exactly 1,852 metres (approximately 6,076 feet).
The kilometre was originally defined as 1/10,000 of the distance from a pole to the equator, or as 1/40,000 of the Earth's circumference as measured through the poles. This definition would make the kilometre's length approximately 1,000.197 metres, whereas it is exactly 1,000 metres (approximately 3,281 feet), of course.
### Earth-to-Moon distance
The distance between the Earth's and the Moon's surfaces is, on average, approximately 380,000 km, or 240,000 miles. This distance is sometimes used in the same manner as the circumference of the Earth; that is, one might say that a large number of objects laid end-to-end "would reach all the way to the Moon and back two-and-a-half times."
The abbreviation for the Earth-to-Moon distance is "LD" which stands for "Lunar Distance", used in astronomy to express close approaches of Earth by minor planets.
### Siriometer
The siriometer is a rarely used astronomical measure equal to one million astronomical units, i.e., one million times the average distance between the Sun and Earth. This distance is equal to about 15.8 light-years, 149.6 Pm or 4.8 parsecs, and is about twice the distance from Earth to the star Sirius.
## Area
### Barn
One barn is 10−28 square metres, about the cross-sectional area of a uranium nucleus. The name probably derives from early neutron-deflection experiments, in which the uranium nucleus was described, comparatively, as being "big as a barn". Additional units include the microbarn (or "outhouse")[10] and the yoctobarn (or "shed").[11][12]
### Nanoacre
The nanoacre is a unit of real estate on a VLSI chip equal to 0.00627264 sq in (4.0468564224 mm2) or the area of a square of side length 0.0792 in (2.01168 mm). "The term gets its humor from the fact that VLSI nanoacres have costs in the same range as real acres in Silicon Valley once one figures in design and fabrication-setup costs."[13]
### Square
The square is an Imperial unit of area that is used in the construction industry in North America,[14] and was historically used in Australia by real estate agents. One square is equal to 100 square feet. A roof's area may be calculated in square feet, then converted to squares.
### Football field (area)
American football field
An association football pitch, or field, can be used as a man-in-the-street unit of area.[15][16] The standard FIFA football pitch is 105 m long by 68 m wide (7,140 m2, 0.71 ha, 1.76 acres); FIFA allows for a variance of up to 5 m in length and 4 m in width in either direction (and even larger discretions if the pitch is not used for international competition), which generally results in the association football pitch generally only being used for order of magnitude comparisons.[17]
An American football field, including both end zones, is 360 ft by 160 ft (ca. 110 m × 49 m), or 57,600 square-feet (5,350 m², 0.54 ha, 1.32 acres). A Canadian football field is 65 yards wide and 110 yards long (ca. 101 m × 59 m) with end zones adding a combined 40 yards to the length, making it 87,750 square-feet (8,156 m2, 0.82 ha, 2.01 acres)
An Australian rules football field may be approximately 150 metres (or more) long goal to goal and 135 metres (or more) wide, although the field's elliptical nature reduces its area to a certain extent. A 150 by 135 metre football field has an area of approximately 1.6 ha (3.93 acres, 15,900 m2), twice the area of a Canadian football field and three times that of an American football field.
### Various countries, regions, and cities
Wales (red) in the UK (pink)
The area of a familiar country, state or city is often used as a unit of measure, especially in journalism.
#### Wales
Equal to 20,779 km2 (8,023 sq mi), the country of Wales is used in phrases such as "an area the size of Wales" or "twice the area of Wales".[18][19] England is 6.275 times the size of Wales, and Scotland is roughly four times the size of Wales. Ireland is four times larger than Wales, and France is about twenty-five times larger.
The British comedy show The Eleven O'Clock Show parodied the use of this measurement, by introducing a news article about an earthquake in Wales, stating that an area the size of Wales was affected. The Radio 4 programme More or Less introduced the idea of "kiloWales" - an area 1000 times the size of Wales. The Register introduced the nanoWales (20.78 m2).
#### The United States
In the United States the areas of the smallest state, Rhode Island (1,545 sq mi or 4,000 km2), the largest contiguous state, Texas (268,601 sq mi or 695,670 km2), and, less commonly, Alaska (656,425 sq mi or 1,700,130 km2) are used in a similar fashion. Antarctica's Larsen B ice shelf was approximately the size of Rhode Island until it broke up in 2002. In the 1979 movie The China Syndrome, radiation is expected to contaminate "an area the size of Pennsylvania." Any state may be used in this fashion to describe the area of another country.
The US Central Intelligence Agency uses Washington, D.C. (61.4 sq mi or 159 km2) as a comparison for city-sized objects.[citation needed]
#### Other countries
In Canada, the standard unit of comparison is often Prince Edward Island (5,683 km2 or 2,194 sq mi),[20] the smallest Canadian province.
In Russia, France (551,695 km2 or 213,011 sq mi) is often used as a comparison for regions of Siberia.[21] This was so popular in the Soviet era that the phrase "как две Франции" (twice the size of France) became a stock phrase to denote any large area.
The country of Belgium (30,528 km2 or 11,787 sq mi) has also often been used when comparing areas, to the point where it has been regarded as a meme[22] and where there is a website dedicated to notable areas which have been compared to that of Belgium.
The Isle of Wight (380 km2 or 147 sq mi), an island off the south coast of mainland England, is commonly used to define smaller areas.
## Volume
### Pony
A pony is 0.75 ounces of liquor, or 34 of a shot.[23]
### Metric ounce
A metric ounce is an approximation of the imperial ounce, US dry ounce, or US fluid ounce. These three customary units vary. However, the metric ounce is usually taken as 25 or 30 ml when volume is being measured, or grams when mass is being measured.
The US Food and Drug Administration (FDA) defines the "food labeling ounce" as 30 ml, slightly larger than the 29.6 ml fluid ounce.[24]
The shot glass is a liquid volume measure that varies from country to country and state to state depending on legislation. It is routinely used for measuring strong liquor or spirits when the amount served and consumed is smaller than the more common measures of alcoholic "drink" and "pint". There is a legally defined maximum size of a serving in some jurisdictions. Typical sizes may range from a 25 to 30 ml (0.88 to 1.1 imp fl oz; 0.85 to 1.0 US fl oz) pony shot to a double shot of roughly 90 ml (3.2 imp fl oz; 3.0 US fl oz).[citation needed] In the UK, spirits are sold in shots of 25 ml, approximating the old fluid ounce.
### Board foot or super foot
A board foot is a United States and Canadian unit of volume, used for lumber. It is equivalent to 1 inch × 1 foot × 1 foot (144 cu in or 2,360 cm3). It is also found in the unit of density pounds per board foot. In Australia and New Zealand the terms super foot or superficial foot were formerly used for this unit.
### Hoppus foot
A system of measure for timber in the round (standing or felled), now largely superseded by the metric system except in measuring hardwoods in certain countries. Following the so-called "quarter-girth formula" (the square of one quarter of the circumference in inches multiplied by 1144 of the length in feet), the notional log is four feet in circumference, one inch of which yields the hoppus board foot, 1 foot yields the hoppus foot, and 50 feet yields a hoppus ton. The hoppus board foot, when milled, yields a board foot. The volume yielded by the quarter-girth formula is 78.5% of cubic measure.[25] Mathematically, this works out to about 36.05 dm³.
### Cord and rick
The cord is a unit of measure of dry volume used in Canada and the United States to measure firewood and pulpwood. A cord is the amount of wood that, when "ranked and well stowed" (arranged so pieces are aligned, parallel, touching and compact), occupies a volume of 128 cubic feet (3.62 m3).[26] This corresponds to a well stacked woodpile 4 feet deep by 4 feet high by 8 feet wide (122 cm × 122 cm × 244 cm), or any other arrangement of linear measurements that yields the same volume. An even more unusual measurement for firewood is the "rick". It is stacked 16 inches (40.6 cm) deep with the other measurements kept the same as a cord, making it 13 of a cord; however, regional variations mean that its precise definition is nonstandardized.[27]
### Twenty-foot equivalent unit
The twenty-foot equivalent unit (sometimes called a boatload) is the volume of the average shipping container or railroad boxcar, it is also called a carload, or shipload. It is equivalent to 1,360 ft3 (39 m3).
### Acre-foot
An acre-foot is a unit of volume commonly used in the United States in reference to large-scale water resources, such as reservoirs, aqueducts, canals, sewer flow capacity, and river flows. It is defined by the volume of one acre of surface area to a depth of one foot, which is ca. 1233.5 m3.
### Olympic-size swimming pool
Olympic swimming pool
For larger volumes of liquid, one measure commonly used in the media in many countries is the Olympic-size swimming pool.[28] An Olympic swimming pool with dimensions 50 m × 25 m × 2 m (approx 164 ft × 82 ft × 6.6 ft) holds 2,500 m3 (about 2 acre-feet, 550,000 imp gal or 660,000 US gal).
### Royal Albert Hall
The Royal Albert Hall, a large concert hall, is sometimes used as unit of volume in the UK, particularly when referring to volumes of rubbish placed in landfill.[29] The volume of the auditorium is between 3 and 3.5 million cubic feet (between 85,000 and 99,000 cubic metres).[30]
### Sydney Harbour
A unit of volume used in Australia for water. One Sydney Harbour, also called a Sydharb (or sydarb), is the amount of water in Sydney Harbour: approximately 500 gigalitres (400 thousand acre-feet).[31]
## Mass
### Grave
In 1793, the French term "grave" (from "gravity") was suggested as the base unit of mass for the metric system. In 1795, however, due in no small part to the French Revolution, the name "kilogram" was adopted instead.[32]
### Bag of cement and bag mix
The mass of an old bag of cement was one hundredweight ~ 112 lb, approximately 50 kg. The amount of material that, say, an aircraft could carry into the air is often visualised as the number of bags of cement that it could lift. In the concrete and petroleum industry, however, a bag of cement is defined as 94 pounds (~ 42.6 kg), because it has an apparent volume close to 1 cubic foot.[33] When ready-mix concrete is specified, a "bag mix" unit is used as if the batching company mixes 5 literal bags of cement per cubic yard (or cubic metre) when a "5 bag mix" is ordered.
The maximum takeoff weight of most earlier models of the Boeing 747 is 800,000 pounds, or 360 tonnes. In the media, multiples of this mass can be used to describe very heavy objects, especially freight trains, e.g. "It weighs as much as five fully loaded 747s."[citation needed]
### Jupiter
When reporting on the masses of extrasolar planets, astronomers often discuss them in terms of multiples of Jupiter's mass (1.9 ×1027 kg).[34] For example, "Astronomers recently discovered a planet outside our Solar System with a mass of approximately 3 Jupiters." Furthermore, the mass of Jupiter is nearly equal to one thousandth of the mass of the Sun.
### Sun
Solar mass (M = 2.0×1030 kg) is also often used in astronomy when talking about masses of stars or galaxies, for example the Milky Way has a mass of approximately 6 ×1011 M.
Solar mass also has a special use when estimating orbital periods and distances of 2 bodies using Kepler's laws: a3 = MtotalT2, where a is length of semi-major axis in AU, T is orbital period in years and Mtotal is the combined mass of objects in M. In case of planet orbiting a star, Mtotal can be approximated to mean the mass of the central object. More specifically in the case of Sun and Earth the numbers reduce to Mtotal ~ 1, a ~ 1 and T ~ 1.
## Time
### Light-distance
Isaac Asimov proposed in one of his essays that it would be useful to measure time by the linear interval needed by light to travel it.[citation needed] Thus, a "light-foot" would be the time (approximately 1 nanosecond) for light to travel 1 foot (30.48 cm) in a vacuum. Similarly, "light-mile", "light-centimetre". The construction is the inverse of "light-year", which names a time unit in order to measure distance.
### Shake
In nuclear engineering and astrophysics contexts, the shake is sometimes used as a conveniently short period of time. 1 shake is defined as 10 nanoseconds.[35]
### Jiffy
In computing, the jiffy is the duration of one tick of the system timer interrupt. Typically, this time is 0.01 seconds, though in some earlier systems (such as the Commodore 8-bit machines) the jiffy was defined as 1/60 of a second, roughly equal to the vertical refresh period (i.e. the field rate) on NTSC video hardware (and the period of AC electric power in North America).
### Microfortnight
One unit derived from the FFF system of units is the microfortnight, one millionth of the fundamental time unit of FFF, which equals 1.2096 seconds. This is a fairly representative example of "hacker humor",[36] and is occasionally used in operating systems; for example, the OpenVMS TIMEPROMPTWAIT parameter is measured in microfortnights.[37]
### Nanocentury
A unit sometimes used in computing. The term is believed to have been coined by IBM in 1969 from the design objective "never to let the user wait more than a few nanocenturies for a response".[38] A nanocentury is approximately 3.155 seconds although Tom Duff is frequently cited as saying that, to within half a percent, a nanocentury is pi seconds.[39]
### Moment
A moment is a medieval unit of time equal to 90 seconds[40] and a Hebrew calendar measurement equal to 5/114 of a second, or just under 0.044 seconds.[41]
### Decimal time systems
The measurement of time is unique in SI in that while the second is the base unit, and measurements of time smaller than a second use prefixed units smaller than a second (e.g. microsecond, nanosecond, etc.), measurements larger than a second instead use traditional divisions, including the sexagesimal-based minute and hour as well as the less regular day and year units. SI allows for the use of larger prefixed units based on the second, a system known as metric time, but this is unusual.
There have been numerous proposals and usage of decimal time, most of which were based on the day as the base unit. For instance, in dynastic China, the was a unit that represented 1/100 of a day. (It has since been refined to 1/96 of a day, or 15 minutes.) In France, a decimal time system in place from 1793 to 1805 divided the day into 10 hours, each divided into 100 minutes, in turn each divided into 100 seconds; the French Republican Calendar further extended this by assembling days into ten-day "weeks." Ordinal dates and Julian dates allow for the expression of a decimal portion of the day.[42] In the mid-1960s, to defeat the advantage of the recently introduced computers for the then popular rally racing in the Midwest, competition lag times a few events were given in centids (1/100 day, 864 seconds, 14.4 minutes), millids (1/1000 day, 86.4 seconds) and centims (1/100 minute, 0.6 seconds) the latter two looking and sounding a bit like the related units of minutes and seconds. Decimal time proposals are frequently used in fiction, often in futuristic works.
In addition to decimal time, there also exist binary clocks and hexadecimal time.
### Dog year
A unit of measurement equal to one seventh of a year, or approximately 52 days, is primarily used to approximate the equivalent age of dogs and other animals with similar life spans. It is based upon a popular myth regarding the aging of dogs that states that a dog ages seven years in the time it takes a human to age one year. (In fact, the aging of a dog varies by breed; dogs also develop faster and have longer adulthoods relative to their total life span than humans. Most dogs are sexually mature by 1 year old, which corresponds to perhaps 13 years old in humans.) While the typical convention is to identify the "dog year" as the shorter length of time than the standard "human year" (thus, a 12-year-old dog is said to be 84 "in dog years"),[43] it is also not unheard of to have the 52-day unit be identified as a "human year" and the full 365-day year as a "dog year". When the unit is used, measurements in both "dog years" and "human years" are generally included together, to more clearly indicate which name is used for each unit.[43]
### Galactic year
The most common large-scale time scale is millions of years (megaannum or "Ma"). However, for long-term measurements, this still requires rather large numbers. Using as a measure the time it takes for the solar system to revolve once around the galactic core (GY - not to be confused with Gyr for gigayear), approximately 250 Ma, yields some easily memorizable numbers. In this scale, oceans appeared on Earth after 4 GY, life began at 5 GY, and multicellular organisms first appeared at 15 GY. Dinosaurs went extinct about 1/4 GY ago, and the true age of mammals began about 0.2 GY ago. The age of the Earth is estimated at about 20 GY.[44]
## Angular measure
### Furman
The Furman is a unit of angular measure equal to 165,536 of a circle, or just under 20 arcseconds. It is named for Alan T. Furman, the American mathematician who adapted the CORDIC algorithm for 16-bit fixed-point arithmetic sometime around 1980.[45] 16 bits give a resolution of 216 = 65536 distinct angles. A related unit of angular measure equal to 1256 of a circle has found some use in controllers for high-speed machinery where fine precision is not required, most notably crankshaft and camshaft position sensing in internal combustion engine controllers. It has been called the 8-Bit Furman, the Small Furman and more recently, the miFurman, (milli-binary-Furman) but there is not yet a consensus as to its name.
A related unit of angular measure equal to 1256 of a circle, represented by 8 bits, has found some use in machinery control where fine precision is not required, most notably crankshaft and camshaft position in internal combustion engine controllers, but there is no consensus as to its name. This unit is also used in video game programming. These units are convenient because they form cycles: for the 8-bit unit, the value overflows from 255 to 0 when a full circle has been traversed. Measures are often made using a Gray code, which is trivially converted into more conventional notation.
### Angular mil
Estimating angular mils by hand
The angular mil is used by many military organisations to measure plane angle and so to triangulate distances, given an object's apparent and actual size. It is approximately the angle which has a tangent of 1/1000; in NATO standard, this is rounded to 1⁄6400 of a circle, although other definitions are in use. Its name derives from Latin: millesimus ("thousandth") and so the fact it is used mostly by the military is coincidental to its name.[46]
## Energy
### Gallon gasoline equivalent
In 2011 the United States Environmental Protection Agency introduced the gallon gasoline equivalent as a unit of energy because their research showed most US citizens do not understand the standard units. The gallon gasoline equivalent is defined as 33.7 kWh.[47]
### Tons of TNT
The explosive energy of various amounts of the explosive TNT (kiloton, megaton, gigaton) is often used as a unit of explosion energy, and sometimes of asteroid impacts and violent explosive volcanic eruptions. One ton of TNT is 4.184 × 109 joules, or 109 thermochemical calories (≈3.964 × 106 BTU). This definition is not based on the actual physical properties of TNT.
### Hiroshima bomb and Halifax explosion
Hiroshima bomb explosion
The energy released by the Hiroshima bomb explosion (about 15 kT TNT equivalent, or 6 × 1013 J) is often used by geologists as a unit when describing the energy of earthquakes, volcanic eruptions and asteroid impacts.
Prior to the detonation of the Hiroshima bomb, the size of the Halifax Explosion (about 3 kT TNT equivalent, or 1.26 × 1013 J), was the standard for this type of relative measurement. Each explosion had been the largest known man-made detonation to date.[48]
### Foe
A foe is a unit of energy equal to 1044 joules (≈9.478 × 1040 BTU) that was coined by physicist Gerry Brown of Stony Brook University. To measure the staggeringly immense amount of energy produced by a supernova, specialists occasionally use the "foe", an acronym derived from the phrase [ten to the power of] fifty-one ergs, or 1051 ergs. This unit of measure is convenient because a supernova typically releases about one foe of observable energy in a very short period of time (which can be measured in seconds).
## Other metric-compatible scales
### Flow: Amazon River
The volume of discharge of the Amazon River sometimes used to describe large volumes of water flow such as ocean currents. The unit is equivalent to 216,000 m3/s.[49]
### Energy
It is common in particle physics, where mass and energy are often interchanged, to use eV/c2, where c is the speed of light in a vacuum (from E = mc2). Even more common is to use a system of natural units with c set to 1, and simply use eV as a unit of mass.
1 amu = 931.46 MeV/c2
### Energy intensity
The langley (symbol Ly) is used to measure solar radiation or insolation. It is equal to one thermochemical calorie per square centimetre (4.184×104 J/m2 or ≈3.684 BTU/sq ft) and was named after Samuel Pierpont Langley.
### Kinematic viscosity
One of the few CGS units to see wider use, one stokes (symbol S or St) is a unit of kinematic viscosity, defined as 1 cm2/s, i.e., 10−4 m2/s (≈1.08×10−3 sq ft/s).
### Electromagnetic flux
In radio astronomy, the unit of electromagnetic flux is the jansky (symbol Jy), equivalent to 10−26 watts per square metre per hertz (= 10−26 kg/s2 in base units, about 8.8×10−31 BTU/ft2). It is named after the pioneering radio astronomer Karl Jansky. The brightest natural radio sources have flux densities of the order of one to one hundred jansky.
### Metre of water equivalence
A material-dependent unit used in nuclear and particle physics and engineering to measure the thickness of shielding, for example around a nuclear reactor, particle accelerator, or radiation or particle detector. 1 mwe of a material is the thickness of that material that provides the equivalent shielding of one metre (≈39.4 in) of water.
This unit is commonly used in underground science to express the extent to which the overburden (usually rock) shields an underground space or laboratory from cosmic rays. The actual thickness of overburden through which cosmic rays must traverse to reach the underground space varies as a function of direction due to the shape of the overburden, which may be a mountain, or a flat plain, or something more complex like a cliff side. To express the depth of an underground space in mwe (or kmwe for deep sites) as a single number, the convention is to use the depth beneath a flat overburden at sea level that gives the same overall cosmic ray muon flux in the underground location.
The strontium unit, formerly known as the Sunshine Unit (symbol S.U.), is a unit of biological contamination by radioactive substances (specifically strontium-90). It is equal to one picocurie of Sr-90 per gram of body calcium. Since about 2% of the human body mass is calcium, and Sr-90 has a half-life of 28.78 years, releasing 6.697+2.282 MeV per disintegration, this works out to about 1.065×10−12 grays per second. The permissible body burden was established at 1,000 S.U.
### Banana equivalent dose
Bananas, like most organic material, naturally contain a certain amount of radioactive isotopes—even in the absence of any artificial pollution or contamination. The banana equivalent dose, defined as the additional dose a person will absorb from eating one banana, expresses the severity of exposure to radiation, such as resulting from nuclear weapons or medical procedures, in terms that would make sense to most people. This is approximately 78 nanosieverts - in informal publications one often sees this estimate rounded up to 0.1 μSv.
## Other
### ASTA pungency unit
ASTA (American Spice Trade Association) pungency unit is based on a scientific method of measuring chili pepper 'heat'. The technique utilizes high performance liquid chromatography to identify and measure the concentrations of the various compounds that produce a heat sensation. Scoville units are roughly 15 times higher than Pungency units while measuring capsaicin, so a rough conversion is to multiply Pungency by 15 to obtain Scoville heat units.[50]
### Bibles, Encyclopaediae, and the Library of Congress: data volume
A CD-ROM can easily store the entirety of a paper encyclopedia's words and images, plus audio and video clips
When the Compact Disc began to be used as a data storage device, the CD-ROM, journalists had to compare the disc capacity (650 megabytes) to something everyone could imagine. Since many Western households had a Christian Bible, and the Bible is a comparatively long book, it was often chosen for this purpose. The King James Version of the Bible in uncompressed plain 8-bit text contains about 4.5 million characters,[51] so a CD-ROM can store about 150 Bibles.
The print version of the Encyclopædia Britannica is another common data volume metric. It contains approximately 300 million characters,[52] so two copies would fit onto a CD-ROM and still have 50 megabytes left over.
The term Library of Congress is often used as an unusual unit of measurement to represent an impressively large quantity of data when discussing digital storage or networking technologies. It refers to the US Library of Congress. Information researchers have estimated that the entire print collections of the Library of Congress represent roughly 10 terabytes of uncompressed textual data.[53]
Diagrams and photographs are not included in the total. Since these take up considerably more space than text, this would be an important consideration for digital storage of large book collections. In practice, diagrams can often be expressed more compactly using vector graphics, and data compression software can pack more text into the available space. English text can be compressed to about 11% of its original size.[54] Thus, one might claim that the entire print collections of the Library of Congress could be packed onto a single 2.2 terabyte storage unit, so long as one excluded images, ignored foreign language holdings, and did not count the growth of the print collections by at least 6 million volumes since the time of the study that produced the 10 terabyte number.[55][56] Furthermore, the modern Encyclopædia Britannica contains over 35000 image, audio, and video assets and fits on a single DVD-ROM.[57]
Though some sources have suggested that 10 terabytes represents the total quantity of data stored at the Library of Congress,[58][59][60][61][62][63][64][65][66] this is a significant underestimate, given that the Web Archiving program[67] had by itself collected 385 terabytes of data as of January 2013.[68] A slide from a September 2012 presentation by a Library of Congress storage engineer furthermore noted institutional storage capacity in excess of 27 petabytes,[69] casting further doubt on the accuracy of the 10 terabyte number as applied to the entire holdings.
The Library of Congress itself has made no official claims of how much data is represented by its entire holdings, and several employees have expressly disavowed any of the specific numbers that have been attributed.[70][71]
A similar unit of measure for early computers was phone numbers and phone books.
### Big Mac Index: purchasing power parity
A McDonald's Big Mac hamburger, as bought in the United States
The Economist's Big Mac Index compares the purchasing power parity of countries in terms of the cost of a Big Mac hamburger.[72] This was felt to be[citation needed] a good measure of the prices of a basket of commodities in the local economy including labour, rent, meat, bread, cardboard, advertising, lettuce, etc.
A similar system used in the UK is the 'Mars bar'. Tables of prices in Mars Bars have intermittently appeared in newspapers over the last 20 years, usually to illustrate changes in wages or prices over time without the confusion caused by inflation.[73]
### Centipawn
Chess software frequently uses centipawns internally or externally as a unit measuring how strong each player's situation position is, and hence also by how much one player is beating the other, and how strong a possible move is. 100 centipawns = the value of 1 pawn - more specifically, something like the average value of the pawns at the start of the game, as the actual value of pawns depends on their position. Loss of a pawn will therefore typically lose that player 100 centipawns (which is usually enough to lose the game). The centipawn is a convenient size for choosing between different possible moves, as in a given position, chess software will often rate the better of two moves within a few centipawns of each other.
### FLOPS
In computing, FLOPS (FLoating point Operations Per Second) is a measure of a computer's computing power. It is also common to see measurements of kilo-, mega-, and teraFLOPS.
It is also used to compare the performance of algorithms in practice.[citation needed]
### Garn
The Garn is NASA's unit of measure for symptoms resulting from space adaptation syndrome, the response of the human body to weightlessness in space, named after US Senator Jake Garn, who became exceptionally spacesick during an orbital flight in 1985. If an astronaut is completely incapacitated by space adaptation syndrome, he or she is under the effect of one Garn of symptoms.[74]
### KLOC: computer program length
A computer programming expression, the K-LOC or KLOC, pronounced kay-lok, standing for "kilo-Lines of Code", i.e., thousand lines of code. The unit was used, especially by IBM managers,[75] to express the amount of work required to develop a piece of software. Given that estimates of 20 lines of functional code per day per programmer were often used, it is apparent that 1 K-LOC could take one programmer as long as 50 working days, or 10 working weeks. This measure is no longer in widespread use because different computer languages require different number of lines to achieve the same result.
Error rates in programming are also measured in "Errors per K-LOC", which is called the defect density. NASA's SATC is one of the few organisations to claim zero defects in a large (>500K-LOC) project, for the space shuttle software.
A rare multiple of K-LOC is GLOC, pronounced jee-lok, standing for "giga-lines of code".
### Micromort
A micromort is a unit of risk measuring a one-in-a-million probability of death (from micro- and mortality). Micromorts can be used to measure riskiness of various day-to-day activities. A microprobability is a one-in-a million chance of some event; thus a micromort is the microprobability of death. For example, smoking 1.4 cigarettes increases one's death risk by one micromort, as does traveling 230 miles by car.
### Mother Cow Index
Formerly used in real estate transactions in the American Southwest, it was the number of pregnant cows an acre of a given plot of land could support. It acted as a proxy for the agricultural quality, natural resource availability, and arability of a parcel of land.[76]
### Nibble
A measure of quantity of data or information, the "nibble" (sometimes spelled "nybble" or "nybl") is normally equal to 4 bits, or one half of the common 8-bit byte. The nibble is used to describe the amount of memory used to store a digit of a number stored in binary-coded decimal format, or to represent a single hexadecimal digit. Less commonly, 'nibble' may be used for any contiguous portion of a byte of specified length, e.g. "6-bit nibble"; this usage is most likely to be encountered in connection with a hardware architecture in which the word length is not a multiple of 8, such as older 36-bit minicomputers.
### Nines
Numbers very close to, but below one are often expressed in "nines" (N - not to be confused with the unit newton), that is in the number of nines following the decimal separator in writing the number in question. For example, "three nines" or "3N" indicates 0.999 or 99.9%, "four nines five" or "4N5" is the expression for the number 0.99995 or 99.995%.[citation needed]
Typical areas of usage are:
• the reliability of computer systems, that is the ratio of uptime to the sum of uptime and downtime. "Five nines" reliability in a continuously operated system means an average downtime of no more than approximately five minutes per year.
• the purity of materials, such as gases and metals.
### Pain
The dol (from the Latin word for pain, dolor) is a unit of measurement for pain.
James D. Hardy, Herbert G. Wolff, and Helen Goodell of Cornell University proposed the unit based on their studies of pain during the 1940s-1950s. They defined one dol to equal to "just noticeable differences" (jnd's) in pain. The unit never came into widespread use and other methods are now used to assess the level of pain experienced by patients.
### Proof: alcohol concentration
A (nearly empty) bottle of 151 proof rum.
Up to the 20th century, alcoholic spirits were assessed in the UK by mixing with gunpowder and testing the mixture to see if it would still burn; spirit that just passed the test was said to be at 100° proof. The UK now uses percentage alcohol by volume at 20 °C (68 °F), where spirit at 100° proof is approximately 57.15% ABV; the US uses a "proof number" of twice the ABV at 60 °F (15.5 °C).[citation needed]
### Savart
The Savart is an 18th century unit for measuring the frequency ratio of two sounds, it is equal to 1/300 of an octave, or 1/25 of a semitone. Still used in some programs, but considered too rough for most purposes. Cent is preferred.
### Schmidt sting pain index and Starr sting pain index
These are pain scales rating the relative pain caused by different hymenopteran stings. Schmidt has refined his Schmidt Sting Pain Index (scaled from 1 to 4) with extensive anecdotal experience, culminating in a paper published in 1990 which classifies the stings of 78 species and 41 genera of Hymenoptera. The Starr sting pain scale uses the same 1-to-4 scaling. In practice, the Comparative Pain Scale is used by doctors when working with patients.[77]
### Scoville heat unit
The Scoville scale is a measure of the hotness of a chili pepper. It is the degree of dilution in sugar water of a specific chili pepper extract when a panel of 5 tasters can no longer detect its 'heat'.[78] Pure capsaicin (the chemical responsible for the 'heat') has 1.6×107 Scoville heat units.[79]
The number of people that met a condition is often measured with the capacity of major sports and arts facilities, usually the largest ones in a region. For example, Uruguay's Estadio Centenario is often used in Uruguay,[80][81] while in parts of the United States, Michigan Stadium is used in this manner.[citation needed]
### Telecommunications traffic volume
The erlang, named after A. K. Erlang, as a dimensionless unit is used in telephony as a statistical measure of the offered intensity of telecommunications traffic on a group of resources. Traffic of one erlang refers to a single resource being in continuous use, or two channels being at fifty percent use, and so on, pro rata. A lot of telecommunications management and forecasting software uses this unit on a day to day basis—but strictly speaking it is a telecom sector specific network stress measurement unit.
### X-ray intensity
The crab is defined as the intensity of X-rays emitted from the Crab Nebula at a given photon energy up to 30 kiloelectronvolts. The Crab Nebula is often used for calibration of X-ray telescopes. For measuring the X-ray intensity of a less energetic source, the milliCrab (mCrab) may be used.
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5. ^ NZ fishermen land colossal squid
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72. ^ "Big MacCurrencies". The Economist. 1998-04-09. Retrieved 2007-07-24.
73. ^ Mars Bar, Nico Colchester Fellowship, FT.com (Financial Times website). Article dated 2001-01-26. Retrieved 2007-01-13.
74. ^ pg 35, Johnson Space Center Oral History Project, interview with Dr. Robert Stevenson
75. ^ "Triumph of the Nerds". PBS. | 13,537 | 55,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.959799 |
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2013-10-06, 16:08 #1 Tony Reix Oct 2002 3·11 Posts 500€ Reward for a proof for the Wagstaff primality test conjecture Hi, I've decided to offer a 500€ reward for the first guy/lady who will be able to provide a proof for the conjecture that Anton Vrba and myself have found, years ago, about proving the primality of a Wagstaff number. I've just started the process of defining the rules and putting in place some pages on the Web. I have much more work to do. So, the first version is very raw: http://tony.reix.free.fr/WagstaffReward/ . Anyway, I plan to improve these pages. I have to refresh my memory about HTML, Lyx, and LaTeX stuff. Such a candidate proof will have to be reviewed by peers and to be published in a well-known Mathematics journal. I also have to take decisions about if proving one of the two other conjectures (Mersenne and Fermat) should deserve a reward. I also need approval from Anton Vrba, and provide thanks to other people who provided help to this. I am also looking for help and support. Anyone who would like to read candidate proofs, or anyone who would like to sponsor the reward (give money !), is welcome. One important help would come from Number Theory experts to give their opinion about the importance of these conjectures and to write it down. In my opinion, it is worth to give light on these conjectures due to several reasons: 1) Number Theory books are wrong ! Yes, LLT test is not only for Mersenne numbers and N+1 world is not for LLT and N-1 world for other stuff. As an example, 4 years after me, someone published a proof that LLT test for Fermat numbers is equivalent to Pépin's test. 2) Having a primality test for Wagstaff numbers will extend the GIMPS search and it will speed up finding a 100 millions digits prime, because there are more Wagstaff primes than Mersenne primes; 3) it's fun ! 4) There is a powerful and magic Math world in these DiGraph cycles. I'll work on this during the next months since, in November, I'll be in Asia (Singapour, Bali, Angkor, Luang Prabang) for fun and photography, and I will not be able to do anything. Moreover, I'm now involved in preparing my trip, plus many other stuff dealing with my photographer activity. My goals are: 1) get famous ! (more seriously, I would be VERY happy that all the time I spent working on this damned and hard Math stuff would lead to something real and useful) and thus appear in Number Theory books, with Anton Vrba and the prover of the conjecture; 2) let my children be proud of their damned father; 3) bring attention to Math theory more than to prime records; 4) extend our knowledge about this area and bring Mathematicians to explore this LLT magic world ! Also, I know that Paul Erdős would have liked this reward ! Though my Erdős number is probably close to infinity ! More seriously, I DO believe that studying the properties of the Cycles of a Digraph under $S^2-2 \ mod(N)$ will bring very powerful theorems. And, if you would read Lucas papers, as I did, you would know that he found a lot of things (but without valid proofs ! ). Any comment or suggestions are welcome ! Regards Tony Reix
2013-10-06, 19:29 #2 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 25·3·101 Posts Tony, you may want to post to NMBRTHRY list. It is a slow list (posts are approved once a week or so), but there are many people there who will pass their professional opinion. (The intersection of that list with this forum is in single digits, ever wondered why?) When you will post there, you will do better by dropping the second half that starts with "In my opinion, it is worth to give light on these conjectures due to several reasons: 1) Number Theory books are wrong" etc. Firstly, this bumps the crank score so that you will lose half of your audience right away. Secondly, all of it is already in the conjecture. Post it in plain text without any fancy attachments. A 500 Eur Conjecture: Let: $N_q = 2^q + 1\ and\ W_q = N_q / 3$ (a Wagstaff number), and $S_0 = 1/4 \ (mod\ N_q),\ S_{i+1} = {S_i}^2-2\ (mod\ N_q)$ for 0` This (above) is basically all you need to post.
2013-10-06, 20:48 #3
xilman
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
2·5,573 Posts
Quote:
Originally Posted by Batalov NMBRTHRY"] (The intersection of that list with this forum is in single digits, ever wondered why?).
Are you sure of that? You may be correct but I already know of you, me, Bob, Bruce, Tom W, Phil Carmody and there are probably a few others I forget right now. Perhaps other NMBRTHRY lurkers could make themselves known.
2013-10-06, 21:15 #4 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 25×3×101 Posts To be honest, I didn't count myself. I am a lurker there. I agree that the count would be somewhat higher if you included people who essentially left this forum, like, e.g. maxal.
2013-10-06, 22:06 #5 only_human "Gang aft agley" Sep 2002 2·1,877 Posts I lurk there. I've given up thinking about Zhi-Wei Sun's conjectures though. Before I can think about one, it seems there are a dozen more.
2013-10-07, 00:48 #6
CRGreathouse
Aug 2006
3·1,993 Posts
Quote:
Originally Posted by xilman Are you sure of that? You may be correct but I already know of you, me, Bob, Bruce, Tom W, Phil Carmody and there are probably a few others I forget right now. Perhaps other NMBRTHRY lurkers could make themselves known.
I'm one.
2013-10-09, 19:34 #7 Tony Reix Oct 2002 3·11 Posts Review of the Conjectures I've asked Hugh Williams his opinion about the conjectures. He accepted and he first wanted to look at the conjectures and the proofs of necessity. He is not really sure that they are correct or provable. So, I"ll wait for him to give me his opinion and recommendations. It is very kind of him to take time looking at this. However, he spent so many hours, days, weeks, months, and years on this subject that his opinion does count. Wait & See. If these conjectures are nuts, I'll be free to spend more time on taking photographs ! ;) and I'll buy a new Nikon lens instead !
2013-10-10, 01:23 #8 maxal Feb 2005 11×23 Posts It may be worth to point out the previous discussion on this topic inspired by Anton Vrba's proposed proof: http://www.mersenneforum.org/showthread.php?t=10737
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≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔 | 2,348 | 7,423 | {"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": 5, "/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-2022-05 | latest | en | 0.937944 |
http://nrich.maths.org/2381/note | 1,505,983,756,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687711.44/warc/CC-MAIN-20170921082205-20170921102205-00020.warc.gz | 254,204,162 | 4,660 | ### Rain or Shine
Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry.
### Knock-out
Before a knockout tournament with 2^n players I pick two players. What is the probability that they have to play against each other at some point in the tournament?
### Squash
If the score is 8-8 do I have more chance of winning if the winner is the first to reach 9 points or the first to reach 10 points?
# FA Cup
##### Stage: 5 Challenge Level:
This requires an understanding of the simple rules of probability and a lot of calculating with fractions. | 139 | 645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-39 | latest | en | 0.950485 |
http://rosa.unipr.it/fsda/mdpdReda.html | 1,642,819,662,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00607.warc.gz | 50,973,195 | 8,399 | mdpdReda
mdpdReda allows to monitor Minimum Density Power Divergence criterion to parametric regression problems.
Syntax
• out=mdpdReda(y, X)example
• out=mdpdReda(y, X,Name,Value)example
Description
out =mdpdReda(y, X) Call of mdpdReda with all default options.
out =mdpdReda(y, X, Name, Value) Example of use of option alpha.
Examples
expand all
Call of mdpdReda with all default options.
Simulate a regression model.
n=100;
p=3;
sig=0.01;
eps=randn(n,1);
X=randn(n,p);
bet=3*ones(p,1);
y=X*bet+sig*eps;
% Contaminate the first 10 observations.
y(1:10)=y(1:10)+0.05;
[out] = mdpdReda(y, X,'plots',1);
Example of use of option alpha.
n=100;
p=3;
sig=0.01;
eps=randn(n,1);
X=randn(n,p);
bet=3*ones(p,1);
y=X*bet+sig*eps;
% Contaminate the first 10 observations.
y(1:10)=y(1:10)+0.05;
[out] = mdpdReda(y, X,'plots',1,'alphaORbdp','alpha','tuningpar',[1 0.8 0.2 0]);
Related Examples
expand all
mdpdReda applied to example 1 of Durio and Isaia (2011).
600 points generated according to the model Y=0.5*X1+0.5*X2+eps and n2 = 120 points (outliers), drawn from the model X1,X2~U(0,1) eps~N(0,0.1^2)
n=600;
p=2;
sig=0.1;
eps=randn(n,1);
X=rand(n,p);
bet=0.5*ones(p,1);
y=X*bet+sig*eps;
[out] = mdpdReda(y,X ,'plots',1);
mdpdReda applied to Forbes data.
load('forbes.txt');
y=forbes(:,2);
X=forbes(:,1);
[outalpha0] = mdpdReda(y, X, 'plots',1);
mdpdReda applied to multiple regression data.
load('multiple_regression.txt');
y=multiple_regression(:,4);
X=multiple_regression(:,1:3);
[out] = mdpdReda(y, X, 'plots',1);
Input Arguments
y — Response variable. Vector.
Response variable, specified as a vector of length n, where n is the number of observations. Each entry in y is the response for the corresponding row of X.
Missing values (NaN's) and infinite values (Inf's) are allowed, since observations (rows) with missing or infinite values will automatically be excluded from the computations.
Data Types: double
X — Predictor variables. Matrix.
Matrix of explanatory variables (also called 'regressors') of dimension n x (p-1) where p denotes the number of explanatory variables including the intercept.
Rows of X represent observations, and columns represent variables. By default, there is a constant term in the model, unless you explicitly remove it using input option intercept, so do not include a column of 1s in X. Missing values (NaN's) and infinite values (Inf's) are allowed, since observations (rows) with missing or infinite values will automatically be excluded from the computations.
Data Types: double
Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside single quotes (' '). You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'tuningpar',[1 0.8 0.5 0.4 0.3 0.2 0.1] , 'alphaORbdp','bdp' , 'modelfun', modelfun where modelfun = @(beta,X) X*beta(1).*exp(-beta(2)*X); , 'beta0',[0.5 0.2 0.1] , 'intercept',true , 'conflev',0.99 , 'plots',0
tuningpar —tuning parameter.scalar | Vector.
tuningpar may refer to $\alpha$ (default) or to breakdown point (depending on input option alphaORbdp.
As the tuning parameter $\alpha$ (bdp) decreases the robustness of the Minimum Density Power Divergence estimator decreases while its efficiency increases (Basu et al., 1998). For $\alpha=0$ the MDPDE becomes the Maximum Likelihood estimator, while for $\alpha=1$ the divergence yields the $L_2$ metric and the estimator minimizes the $L_2$ distance between the densities, e.g., Scott (2001), Durio and Isaia (2003). The sequence is forced to be monotonically decreasing, e.g. alpha=[1 0.9 0.5 0.01]. The default for tuningpar is a sequence from 1 to 0 with step -0.01.
Example: 'tuningpar',[1 0.8 0.5 0.4 0.3 0.2 0.1]
Data Types: double
alphaORbdp —ctuning refers to $\alpha$ or to breakdown point.character.
Character which specifies what are the values in input option tuningpar. If this option is not specified or it is equal to 'alpha' then program assumes that the values of tuningpar refer to 'alpha', elseif this option is equal to 'bdp', program assumes that the values of tuninpar refer to breakdownpoint.
Example: 'alphaORbdp','bdp'
Data Types: char
modelfun —non linear function to use.function_handle | empty value (default).
If modelfun is empty the link between $X$ and $\beta$ is assumed to be linear else it is necessary to specify a function (using @) that accepts two arguments, a coefficient vector and the array X and returns the vector of fitted values from the non linear model y. For example, to specify the hougen (Hougen-Watson) nonlinear regression function, use the function handle @hougen.
Example: 'modelfun', modelfun where modelfun = @(beta,X) X*beta(1).*exp(-beta(2)*X);
Data Types: function_handle or empty value
theta0 —empty value or vector containing initial values for the coefficients (beta0 and sigma0) just in case modelfun is non empty.if modelfun is empty this argument is ignored and LMS solution will be used as initial solution for the minimization.
Example: 'beta0',[0.5 0.2 0.1]
Data Types: double
intercept —Indicator for constant term.true (default) | false.
If true, and modelfun is empty (that is if the link between X and beta is linear) a model with constant term will be fitted (default), else no constant term will be included.
This argument is ignored if modelfun is not empty.
Example: 'intercept',true
Data Types: boolean
conflev —Confidence level.scalar.
Confidence level which is used to declare units as outliers.
Usually conflev=0.95, 0.975 0.99 (individual alpha) or 1-0.05/n, 1-0.025/n, 1-0.01/n (simultaneous alpha).
Default value is 0.975.
Example: 'conflev',0.99
Data Types: double
plots —Plot on the screen.scalar.
If plots = 1, generates a plot of the monitoring of residuals against alpha.
Example: 'plots',0
Data Types: single | double
Output Arguments
out — description Structure
A structure containing the following fields
Value Description
Beta
matrix containing the mpdp estimator of regression coefficients for each value of alpha
Scale
vector containing the estimate of the scale (sigma) for each value of alpha.
RES
n x length(alpha) matrix containing the robust scaled residuals for each value of bdp
Outliers
Boolean matrix containing the list of the units declared as outliers for each value of alpha using confidence level specified in input scalar conflev
conflev
confidence level which is used to declare outliers.
alpha
vector which contains the values of alpha which have been used. To each value of alpha corresponds a value of bdp (see out.bdp).
bdp
vector which contains the values of bdp which have been used. To each value of bdp corresponds a value of alpha (see out.alpha).
y
response vector y. The field is present if option yxsave is set to 1.
X
data matrix X. The field is present if option yxsave is set to 1.
class
'MDPDReda'
Fval
Value of the objective function and reason fminunc or fminsearch stopped and v. Matrix.
length(alpha)-by-3 matrix.
The first column contains the values of alpha which have been considered.
The second column contains the values of the objective function at the solution.
The third column contins the details about convergence. A value greater then 0 denotes normal convergence. See help of functions fminunc or fminsearch for further details.
We assume that the random variables $Y|x$ are distributed as normal $N( \eta(x,\beta), \sigma_0)$ random variable with density function $\phi$.
Note that if the model is linear $\eta(x,\beta)= x^T \beta$. The estimate of the vector $\theta_\alpha=(\beta_1, \ldots, \beta_p)^T$ (Minimum Density Power Divergence Estimate) is given by:
$\mbox{argmin}_{\beta, \sigma} \left[ \frac{1}{\sigma^\alpha \sqrt{ (2 \pi)^\alpha(1+\alpha)}}-\frac{\alpha+1}{\alpha} \frac{1}{n} \sum_{i=1}^n \phi^\alpha (y_i| \eta (x_i, \beta), \sigma) \right]$ As the tuning paramter $\alpha$ increases, the robustness of the Minimum Density Power Divergence Estimator (MDPDE) increases while its efficieny decreases (Basu et al. 1998). For $\alpha=0$ the MDPDE becomes the Maximum Likelihood Estimator, while for $\alpha=1$ the estimator minimizes the $L_2$ distance between the densities (Durio and Isaia, 2003),
References
Basu, A., Harris, I.R., Hjort, N.L. and Jones, M.C., (1998), Robust and efficient estimation by minimizing a density power divergence, "Biometrika", Vol. 85, pp. 549-559.
Durio,A.,Isaia,E.D.(2003), A parametric regression model by minimum L2 criterion: a study on hydrocarbon pollution of electrical transformers, "Developments in Applied Statistics, Metodoloski Zvezki", Vol. 19, pp. 69-83.
Durio A., Isaia E.D. (2011), The Minimum Density Power Divergence Approach in Building Robust Regression Models, "Informatica", Vol. 22, pp. 43-56. | 2,497 | 8,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | longest | en | 0.554569 |
https://de.mathworks.com/matlabcentral/cody/problems/2017-side-of-an-equilateral-triangle/solutions/1107448 | 1,579,450,534,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594662.6/warc/CC-MAIN-20200119151736-20200119175736-00549.warc.gz | 402,022,120 | 15,531 | Cody
# Problem 2017. Side of an equilateral triangle
Solution 1107448
Submitted on 19 Jan 2017 by Angela Labianca-Campbell
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = 1; x_correct = 2/sqrt(sqrt(3)); tolerance = 1e-12; assert(abs( side_length(A) - x_correct ) < tolerance)
2 Pass
A = sqrt(3); x_correct = 2; tolerance = 1e-12; assert(abs(side_length(A) - x_correct) < tolerance)
3 Pass
A = 2; x_correct = 2*sqrt(2)/sqrt(sqrt(3)); tolerance = 1e-12; assert(abs(side_length(A) - x_correct) < tolerance) | 198 | 636 | {"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} | 2.875 | 3 | CC-MAIN-2020-05 | latest | en | 0.638901 |
https://www.cleariitmedical.com/2020/10/position-of-roots-basic.html | 1,674,885,865,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00559.warc.gz | 697,006,014 | 119,354 | ## Position of the roots - Basic
In algebra, a quadratic equation is any equation that can be rearranged in standard form as where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no term.
Q1. If a, b, c are real numbers such that , then the quadratic equation 3ax2+2bx+c has
• At least one root in [0, 1]
• At least one root in [1, 2]
• At least one root in [–1, 0]
• None of these
At least one root in [0, 1]
Q2. The number of values of k for which the equation x2-3x+k=0 has two real and distinct roots lying in the interval (0, 1), are
• 0
• 2
• 3
• Infinitely many
0
Q3. The value of k for which the equation (k-2)x2+8x+k+4=0 has both real, distinct and negative is
• 0
• 2
• 3
• -4
3
Q4. If x2-1 is a factor of x4+ax3+3x-b , then
• a=3,b=-1
• a=-3, b= 1
• a=-3,b=-1
• None of these
a=-3, b= 1
Q5. If (x-1)3 is factor of x4+ax3+bx2+cx-1 then the other factor is
• x+2
• x+1
• x+3
• None of these
x+1
Q6. The complete solution of the inequation x2-4x<12 is
• x greater than 6 and x less than -2
• x greater than -6 and x less than 2
• x greater than 2 and x less than 6
• x greater than -2 and x less than 6
x greater than -2 and x less than 6
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UTF-8 | 908 | 2,344 | {"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-2023-06 | longest | en | 0.793453 |
https://analystnotes.com/cfa_question.php?p=P6L0B2W3Y | 1,716,676,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00813.warc.gz | 78,065,317 | 6,350 | ### CFA Practice Question
There are 985 practice questions for this topic.
### CFA Practice Question
We want to test the claim that a steady diet of wolfsbane will cause an 18-year-old werewolf to lose EXACTLY 10 lbs. over 5 months. A random sample of 49 werewolf was taken, yielding an average weight loss over 5 months of 12.5 lbs with S = 7 lbs. Let alpha = .02. What is the calculated value suitable for testing the above hypothesis?
A. 12.5
B. 7 x 2.5
C. 2.5
t(calculated) = (12.5-10)/(7/SQRT(49)) = 2.5
User Comment
isida test statistic question: (sample statistic - null value) / standard error standard error = standard deviation / sqrt N
Masterkang Obviously, he read to many Harry Potter Books...
achu we know it's t-test because S sample sd, not population std dev.
steved333 Yes, but sample size is large, so it doesn't matter. T-stat is the same as Z in this case.
poomie83 Was alpha there as useless information?
ybavly @poomie83
yes
johntan1979 Agree with steved333. n>30, so Z.
sgossett86 the equation for variance is single bernoulli variable
Shaan23 You guys need to read the notes. One we dont know if this is normal or not. Population variance is unknown. Whenever population variance is unknown always jump to t-test.
But remember t-test approaches the z test when n becomes big. | 353 | 1,306 | {"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 | 0.877548 |
https://aeropuertosevilla.info/craps-rolls/ | 1,618,279,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00194.warc.gz | 196,220,205 | 10,864 | ## Craps Rolls
Once a point is made on the first roll or a come point on a succeeding roll, you may take the. In order to tell how many numbers youve hit during a roll at the casino craps table, the easiest way is to put chips aside as you roll. You use one-dollar chips (usually white) for one through four, then a red chip for a five, add white for six through nine, then two reds for 10 and so on. To roll a 2, 3, or 12 on the come out roll. A player betting on the Pass line or Come loses on crap out, but the roll does not lose when a point is established. Don't Pass and Don't Come wins if a 2 or 3 craps is rolled on come out, but ties (pushes) if a 12 is rolled on come out.
vert1276
do you think an average craps games gets. No how many points are passed, but just total number of rolls. Now im sure it depends how many players are at the table because the dealer needs to take and pay bets after every roll..But lets say the table is a little over half full and there are 10 players. How many rolls do you think there would be per hour?
MathExtremist
### Craps Rolls Names
With 10 players, between 80 and 100.
'In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice.' -- Girolamo Cardano, 1563
vert1276
With 10 players, between 80 and 100.
Craps, of course, is a male-dominated game, so we hear the roll of 2-3 is also known as the “waitress roll,” because it’s a “pair and a tray.” Naturally, our list isn’t complete. Names like “boxcars” for 12 have sometimes been replaced with colorful counterparts. But if you're undaunted and really want to test your system, read on. The wrong way to test your system is to gather random roulette spins, baccarat rounds, or craps rolls and then apply your system to it. The reason this fails is that it's impossible for you to hand-check several thousand rounds.
WOW thats it? I wonder how they rate comps? They must just have a universal formula they always use? and dont change their EV based on how many people are at the table. For example they always assume 120 rolls per hour no matter how many players are at the table. So if you are playing at a table with 3 players you are getting ripped off on comps? But if you are at a full table(16 players) you are getting over comped?
### Craps Rolls Log
FleaStiff
The 'official' rate of rolls per hour or something is probably a rounded off figure easy to remember and easy to multiply.
One reason these darned dice setters are so hated is the house just wants to keep the dice moving. I've never seen dice move faster than on those day boats in Florida. The Strip is fast in Vegas, the downtown and locals casinos are thought to be generally slower sometimes of necessity due to inexperienced dealers.
Its the same way with blackjack. Those boats deal very fast.
Even in a place like Hawaiian Gardens in Los Angeles, bingo numbers are called out every 4.2 seconds. Speed is the essence.
The house edge is zero if the dice are stationary. The house edge is zero if the blackjack dealer is shuffling. Its 'action' times 'house edge' but its decisions per hour that count.
As for the casinos just picking a number and going with it for Comp purposes: that is fine with me. I'd rather wind up on the wrong side of a rough rule of thumb than have them take time to be precisely accurate. Those blackjack dealers may soak their wrists in liniment every night. I don't know, but they sure make money for themselves and for the house only when dealing, so they deal very fast or they get fired.
Its also one reason why I just pick up two dice and throw them. No song, no dance, no hem, no haw. No incantations about chicken dinners, No Baby Needs New Shoes. No This, No That. I pick the dice up and lob 'em against the back wall. I absolutely loathe all these superstitious types who do anything else but pick up the dice and roll them. If the Stick pushed the dice out to you, its time. You pick them up and you roll them. If hands are in the way, curse and yell if you want to but roll the bones!! Craps is not like Baccarat wherein everyone tries to slow the game down a bit in a campaign against the House so as to score a little 'extra' high class booze or something. Craps is Action. Want to rub a bald head for luck? Sure, Lady but do it on your own time! Want to have chicken for dinner? Go ahead, Jerk, I sure ain't stopping you from it. Just pick up the darned dice and hit the back wall with them.
AZDuffman
With 10 players, between 80 and 100.
I will second that I asked a casino manager the number he wanted per hour in class and this was exactly what he stated.
All animals are equal, but some are more equal than others
DJTeddyBear
The Wiz answered this question, for Craps, BJ and Roulette. It's the first question in Ask The Wizard # 136
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
NowTheSerpent
do you think an average craps games gets. No how many points are passed, but just total number of rolls. Now im sure it depends how many players are at the table because the dealer needs to take and pay bets after every roll..But lets say the table is a little over half full and there are 10 players. How many rolls do you think there would be per hour?
The per-hour goal I've heard is 102 at a full table.
AlanMendelson
I was at a table once when they were lucky to get one roll in five minutes. No joking. The problem was the shooter had already held the dice for about an hour and there was a lot of money on the table. The players were not experienced or even regular players. And as the dealers went around the table paying bets, the players were shouting out instructions and then changing instructions and then changing them again and again. It got to the point that two floor people had to come over to try to restore order.
Now obviously in the early going the rolls came quickly. But when 'real money' was on the table? It was a mess. When one player said same bet another player said 'take my bets off,' and then the player who said same bet would say reduce me to half. And then another player would say off, and another would say press, and then everyone would change again.
Making matters worse, most of the players were friends and their wives were at the table telling their husbands what to do and the whole process would start over. And the shooter would throw another number..
FleaStiff
There will always be extreme situations on each side of the table, but in general casinos know that it is 'Action' time 'House Edge' and that House Edge only applies when Dice are on the way to the back wall and when black jack cards are actually being dealt.
House Edge of BJ during a shuffle is zero and of craps during all the hoopla of chicken dinner. Thats why stickmen are so important in setting the pace of the game. Don't run over the base dealers but do give them the 'call' clear and crisp and a patter that tells them what to do in what order.
The Florida Boats are fast because the crew gets constantly reminded of be fast or be gone.
Strip casinos tend to be fast, downtown is a bit slower and local casinos can be fairly relaxed, particularly if its a break in dealer learning his trade. Remember, casinos do have surveys in progress and MBA types with stop watches from time to time.
on
If you consider yourself to be a serious craps player, you probably know the details on your personal record for longest consecutive roll.
In a highly volatile game of chance like craps – which affords the average shooter just 8.5 rolls before they “seven out” to end the table’s fun – going on an extended streak of success as the shooter can be an unforgettable experience.
There you are playing baccarat, taking center stage while an entire looks on with bated breath as you prepare to roll. You’ve already hit a few point numbers to cash in Pass Line bets for your fellow players, so excitement is in the air. The dice keep tumbling and dodging the dreaded 7 with a point number set, landing on every alternative number a time or two to produce winners for the exotic bettors.
The clock keeps ticking and you keep rolling winners, and before you know it, you’ve just set a new highwater mark for your longest stretch as a shooter without sevening out. Eventually, the party ends when the dice show 4-3 on a 9 point, but no bother – you nearly managed to eclipse the one-hour plateau with a 53-minute roll.
In most Las Vegas casinos, a roll like that would cause an immediate buzz across the gaming floor, both among players and dealers alike. And for good reason, as the average craps roll tends to last for just about 20 minutes.
So what if I told you a tourist from Honolulu, Hawaii once held the dice in hand for three hours and six minutes without ever sevening out?
That incredible craps session lasting 118 straight seven-less rolls might sound like one of Sin City’s infamous myths, like pure oxygen pumped into the air supply to keep losing gamblers blissfully unaware. But if you pay a visit to the classic California casino in Downtown Las Vegas, you can still find a full-fledged shrine to the original “Golden Arm” himself – the late Stanley Fujitake.
## Fujitake Sets a World Record for Longest Consecutive Craps Roll
Back on May 28, 1989, Fujitake and his wife Satsuko took advantage of a Hawaii-focused travel promotion offered by the California Hotel & Casino to make one of their regular visits to Las Vegas.
An avid craps enthusiast herself, Satsuko Fujitake taught her husband the game during their courtship. As she told Hawaii News Now in 2009, shortly after Fujitake’s record finally fell (more on this to come), Satsuko soon suspected her husband had been bitten hard by the craps bug:
“Mom, where is Dad going out all the time? I said, ‘Well, he must have trouble with his stomach, he’s going to the bathroom.’
I didn’t believe that, I knew what he was doing – he was on the table every time he went out.”
So it was that Satsuko woke up one morning to find her husband’s side of the bed hadn’t even been slept in. That’s because Fujitake has spent the night putting on one of the greatest gambling shows Las Vegas has ever witnessed.
The action started around midnight when Fujitake – a mild-mannered man of diminutive stature who looked every part the average Las Vegas tourist – placed a simple \$5 bet on the Pass Line.
Over the next 3+ hours, Fujitake could do no wrong with the dice, rolling over and over again without sevening out. As the epic rolling session progressed, onlookers crowded the table and wagered everything they had to get in on Fujitake’s good fortune.
Guido Metzger – who worked as a dealer at the California back then before rising to become director of casino operations for parent company Boyd Gaming – recalled the frantic crush of bettors surrounding Fujitake in an interview with Boyd’s Buzz:
“They had trouble keeping up with the chip payouts that night.
My table was empty. But there were at least 30 to 40 people trying to place bets at his table.
They couldn’t get fills to the table fast enough and had to start issuing scrip [casino credit] because not enough people were going to the cage and cashing in their chips.”
What races make up the triple crown in horse racing. With winners coming on every roll, the California’s coffers were soon being drained for six-figure sums. The outlays became so onerous that John Repetti, casino manager for the California at the time, was called in from home to supervise the situation.
As he told the Los Angeles Times in a 2017 retrospective on Fujitake’s record-setting roll. Repetti was literally roused from his slumber in order to personally monitor the increasingly expensive craps game:
“The first call came and he’d been shooting for an hour, and we were losing a couple hundred thousand dollars at the time. I said if he continued, to call me at every \$100,000 loss interval.
Well, the calls kept coming every 15 minutes. Another \$100,000. And another \$100,000.
After the fourth call and fifth call, I decided I’d better get some clothes on and get downtown.”
A seasoned veteran of the casino gambling industry, Repetti knew instantly that he was witnessing a historic run of good luck, as he told News at the Cal a few months afterward:
“Half an hour is average, over an hour is amazing, but more than three hours is totally astounding.”
In the end, Fujitake held the dice for 118 consecutive rolls without sevening out, a feat which earned him \$30,000 in winnings.
But according to David Strow, who serves as vice president of corporate communications for Boyd Gaming, Fujitake was hardly the biggest winner to benefit from the legendary roll. As Strow remembered it in a 2017 interview with PokerNews, Fujitake’s fellow players placed larger bets along the way and wound up winning upwards of \$1 million:
“That was one of the ironic things about his roll – the other players at the table ended up winning a lot more money than Stanley did!”
## Wife Remembers the Late Legend and Love of Her Life
Stanley Fujitake passed away in 2000 at the age of 77, but he was survived by his wife Satsuko and their sons Dennis, Lester, and Kevin.
And while the children may have wondered where Dad was during those late nights at the California’s craps tables, Satsusko told Hawaii News Now that she is glad Stanley was able to enjoy the game he loved so dearly:
“It was a miracle, because it’s impossible to hold the dice.
It doesn’t happen all the time, maybe it’s only once in a lifetime deal.”
Satsuko was there that night, but after wandering the casino floor for a while, she couldn’t find the small of stature Stanley amidst the crowd. Later on, as she played video poker in another area of the casino, Satsuko found herself surrounded by well-wishers celebrating her husband’s new crown as the King of Craps:
“People came up to congratulate me and I thought, geez, I didn’t do anything, I didn’t even hit a royal, why are they congratulating me?
Then I realized, he was the one with the dice.”
## Fujitake’s Record Gets Smashed in the Garden State
For 20 years following his world record roll, nobody could top Fujitake’s mark of three hours and six minutes without turning over the dice.
The record stood until 2009, when a craps rookie named Patricia Demauro visited the Borgata casino in Atlantic City, New Jersey on a whim. Bored with the penny slots, her pal invited her to take a crack at craps, leading to one of the more improbable feats in gambling history.
DeMauro rolled 154 times consecutively without sevening out, a session which lasted four hours and 18 minutes altogether – or a full hour longer than Fujitake’s previous record.
When asked about her late husband’s historic feat falling into second place, Satsuko Fujitake told Hawaii News Now that Stanley’s record run will always be number one in her heart:
“As my husband of 54 years, in my heart, he is still the champ to me and will be forever.”
## The “Golden Arm” Club Carries on Fujitake’s Legacy in Fine Fashion
In 1992, the California Hotel & Casino commemorated Fujitake’s record roll by creating the “Golden Arm” award.
Ever since then, any craps player at the California who can roll for one hour or more without sevening out earns entrance to the Golden Arm club. Admission comes with a plaque memorializing the date and length of session, while members are given a snazzy blue shirt proclaiming them to be Golden Arms.
The name comes straight from Fujitake himself, after the proud craps player told Repetti that “this arm is golden” upon receiving a check for \$30,000. Fujitake went on to top the 60-minute mark without sevening out on three other occasions, proving that his proficiency with the dice was no fluke.
You can learn more about the Golden Arm club – and the California’s annual craps tournament held in Fujitake’s honor – in this profile by the L.A. Times.
## Conclusion
Managing to beat the average of 8.5 rolls without sevening out is enough to get most craps players’ heart’s pumping, so just imagine what Fujitake was feeling as the hours passed by. Runs like that are the stuff of gambling lore, but for one unforgettable night back in 1989, a tourist in Sin City simply refused to lose. The next time you’re in Downtown Las Vegas, make sure to pay homage to Fujitake and his record-setting roll by visiting the California and its Golden Arm “wall of fame.” | 3,779 | 16,638 | {"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} | 3.25 | 3 | CC-MAIN-2021-17 | latest | en | 0.965489 |
https://www5.in.tum.de/wiki/index.php?title=Algorithms_for_Scientific_Computing_-_Summer_17&diff=prev&oldid=22721 | 1,638,527,036,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00362.warc.gz | 1,136,676,932 | 13,105 | # Difference between revisions of "Algorithms for Scientific Computing - Summer 17"
Term
Summer 2017
Lecturer
Time and Place
Lecture: Mon 8:30-10:00, Fri 10:15-11:45, MI Hörsaal 2 (1st lecture: Mon, Apr 24)
Tutorial: Wed 10:15-11:45, MI 00.13.09A
Audience
see module description (IN2001) in TUMonline
Tutorials
Emily Mo-Hellenbrand, M.Sc., Jean-Matthieu Gallard, M.Sc.
Exam
Mon, Aug 7, 10.30 in lecture hall MI HS 1 (F.L. Bauer Hörsaal)
Semesterwochenstunden / ECTS Credits
6 SWS (4V + 2Ü) / 8 Credits
TUMonline
https://campus.tum.de/tumonline/wbLv.wbShowLVDetail?pStpSpNr=950290914
## News & Announcements
• The tutorial on Wednesday 12.07 will include the beginning of the lecture on space-filling curve
• The Mock Exam and its solution is now posted in the "Worksheets and Solutions" table. Please note:
• Disclaimer: this mock exam merely serves the purpose of giving you some ideas/hints on what to expect in the actual exam (e.g., exam format, possible questions, difficulty levels). Please do NOT assume that you will get the same (or very similar) questions in the actual exam, as there are many ways to ask a question on the same subject!
• Exam coverage: You should prepare for all 4 topics, i.e., FFT, Hier. methods, Sparse grids, SFC. And you should expect questions from all lecture slides (except for Red parts) and worksheet exercises. Pseudo code questions are possible to appear.
• Preparation hint: Try to solve & understand all the exercises in the worksheets and the mock exam.
• The supplement material of transforming the regularization formula into a linear system (Lecture July 10, slide 18) is uploaded.
• Worksheet 9 code template is updated (fixed compatibility issues with Python 3). Please re-download the template zip. NOTE: you need the files supplied in the template zip to run the Worksheet 9 code solution.
• Worksheet 7 solution is updated (mistake in ex4 corrected). Please re-check the solution!
• please re-check the solution of exercise 1 on worksheet 4; this has been corrected!
• as an exception, the lecture on Fri, May 19, will start at 10.30 (until 12.00)
Many applications in computer science require methods of (numerical) mathematics - especially in science and engineering, of course, but also in surprisingly many areas that one might suspect to be directly at the heart of computer science:
Consider, for example, Fourier and wavelet transformations, which are indispensable in image processing and image compression. Similar, numerical methods for approximation have become essential techniques for high-dimensional classification problems in data science. Essentially, these methods come down to the question of how to represent and process information or data as (multi-dimensional) continuous functions. "Algorithms for Scientific Computing" thus provides an algorithmically oriented introduction to the foundations of such mathematical methods.
Topics include:
• The fast Fourier transformation (FFT) and some of its variants:
• FCT (Fast Cosine Transform), real FFT, Application for compression of video and audio data
• Hierarchical and recursive methods in scientific computing
• From Archimedes' quadrature to the hierarchical basis
• Classification problems
• From the hierarchical basis to wavelets
• High-demonsional problems
• Sparse grids and the sparse-grid combination technique
• Octrees and Space filling curves (SFCs):
• Construction and properies of SFCs
• Application for parallelization and to linearize multidimensional data spaces in data bases
## Lecture Slides and Supplementary Materials
Lecture slides are published here successively. For future lectures, the respective slides from summer 2016 will be linked.
## Worksheets and Solutions
Number Topic Worksheet Tutorial Solution
1 Discrete Fourier Transform I Worksheet 1Python Introduction Apr. 26
2 Discrete Fourier Transform II Worksheet 2 Worksheet 2 Notebook template May 3
- - - May 10 tutorial cancelled due to student assembly
3 Discrete Cosine Transform Worksheet 3 Worksheet 3 Notebook template Template Exercise 1 May 17
4 Discrete Fourier Transform III Worksheet 4 May 24
5 Numerical Quadrature 1D Worksheet 5 Worksheet 5 Notebook template May 31
6 Hierarchical Basis Worksheet 6 Jun. 07
7-Part1 Function Approximation and Wavelet Ex1-3: Worksheet 7 Worksheet 7 Notebook template Jun. 14
7-Part2 Function Approximation and Wavelet Ex4-5: See above Jun. 21 See above
8 Multi-dimensional Quadrature Worksheet 8 Worksheet 8 template Jun. 28
9 Multi-dimensional hierarchization and adaptive sparse grids Worksheet 9 Worksheet 9 code template Jul. 05
- Mock Exam Mock exam -
10 Grammars for space-filling curves Worksheet 10 Worksheet 10 code template Notebook template Jul. 12
11 Arithmetization of space-filling curves Worksheet 11 code template Notebook template Jul. 19
## Exam
• type: written exam, duration: 100 min
• time, date, room: Mon, Aug 7, 2017, 10.30-12.10 (MI HS 1, Friedrich L. Bauer Hörsaal)
• note that the exam will start precisely on 10.30; please be in the exam room by 10.15, at the latest!
• helping material:
• you may use one hand-written sheet of paper (size A4, front and back may be used)
• no other helping material of any kind is allowed
• extra session for questions: t.b.a.
Books that are labeled as "available as e-book" can be accessed as e-book via the TUM library - see the ebooks website of the library for details on how to access the books.
### Fast Fourier Transform:
The lecture is oriented on: | 1,302 | 5,488 | {"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-2021-49 | latest | en | 0.842002 |
http://maxen.pro/functions-study-material-for-iit-jee/ | 1,529,271,755,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859817.15/warc/CC-MAIN-20180617213237-20180617233237-00222.warc.gz | 204,717,700 | 14,253 | Functions Study Material for IIT JEE
Functions Study Material for IIT JEE
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