url
stringlengths 6
1.61k
| fetch_time
int64 1,368,856,904B
1,726,893,854B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int32 9.6k
1.74B
| warc_record_length
int32 664
793k
| text
stringlengths 45
1.04M
| token_count
int32 22
711k
| char_count
int32 45
1.04M
| metadata
stringlengths 439
443
| score
float64 2.52
5.09
| int_score
int64 3
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://aufdercouch.net/how-much-does-one-dollar-bill-weigh/
| 1,653,499,746,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00289.warc.gz
| 172,885,762
| 4,795
|
Our simple calculator permits you to calculation the load of money with ease. If you"ve ever wondered what \$1,000,000 weighs in quarters or \$100 bills, ours calculator will certainly tell you. Simply pick any amount you like and also you have the right to calculate how much it weighs in receipt or coins of any type of value.
How to use the calculator
Fill in the "Amount" ar with the dissension amount friend would choose to calculation the weight of, select the invoice or coin denomination you"re interested in, and also tap "Calculate Weight" to receive your results.
How our calculator works
If you"ve ever before had \$1,000,000 in a suitcase (us neither) you will understand that the weight of an lot of money deserve to differ greatly depending on the denominations provided to make up this amount.Let"s take it an example. The bureau of Engraving and also Printing claims that all us bills sweet a single gram. This means that \$1,000,000 in \$100 bills weighs around 10 kilograms (22.046 pounds). However, if you wanted your million in single dollar bills, that same amount the money would certainly weigh a metric ton (2,204.623 pounds).Coins are a small more complicated as castle weigh various amounts. The U.S. Mint tells us that those soldier in your pocket each weigh 5.7 grams, an interpretation that \$1,000,000 in soldier weighs a colossal 22.68 metric tons, the indistinguishable of 22,680 kilograms, 25 us tons, or 50,000.841 pounds.
You are watching: How much does one dollar bill weigh
Money load CalculatorAmount (\$):Denomination:100 dissension bills50 dissension bills20 dollar bills10 dissension bills5 dissension bills1 dollar bills1 disagreement coins50 cent coins (half dollars)25 cent coins (quarters)10 cent coins (dimes)5 cent coins (nickels)1 cent coins (pennies)
| 416
| 1,806
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.15625
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.905127
|
https://web2.0calc.com/questions/what-is-the-probability-that-the-outcome-of-the-roll
| 1,566,507,830,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00201.warc.gz
| 710,400,942
| 5,795
|
+0
# What is the probability that the outcome of the roll is an even sum or a sum that is a multiple of 3?
0
183
1
+390
What is the probability that the outcome of the roll is an even sum or a sum that is a multiple of 3?
May 28, 2018
#1
+1
All EVEN sums =
2=1/36
4 =3/36
6=5/36
8=5/36
10=3/36
12=1/36
Add up all the probabilities and you get: 18/36 = 1/2
All multiples of 3=
3=2/36
6=5/36
9=4/36
12=1/36
Add up all the probabilities and you get =12/36 = 1/3
May 29, 2018
| 192
| 478
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2019-35
|
latest
|
en
| 0.935269
|
http://naturalunits.blogspot.com/2013/10/smoothing-polygon.html
| 1,508,396,380,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00170.warc.gz
| 250,724,765
| 18,193
|
## Oct 9, 2013
### Smoothing a polygon
Ref. [1] has an interesting observation. For any closed polygon, if one takes the midpoint repeatedly, the polygon will be eventually smoothed out. The authors observed that the curves will become an elliptic.
The mapping is in fact independent in x and y directions. Therefore, after large enough iteration, the x and y coordinate will be "in phase", i.e. belong to the same eigensubspace.
randline[n_Integer, a_: - 1, b_: 1] := RandomReal[{a, b}, {n, 2}]
mpt[pts_, n_: 3] :=
Table[Sum[pts[[Mod[i + j - 1, Length@pts, 1]]], {j, 1, n}]/n, {i, 1,
Length[pts]}]
nl[npts_, nitr_, n_: 2, a_: - 1, b_: 1] :=
NestList[mpt[#, n] &, randline[npts, a, b], nitr];
randplot[npt_, nit_, nst_: 10, na_: 2, pt_: False, init_: False, a_: - 1,
b_: 1] :=
Module[{n, i0 = Min[Max[1, nst], nit + 1]},
n = nl[npt, nit, na, a, b];
Show[
If[pt,
Graphics[
Flatten[Table[{Hue[(i - i0)/(nit + 1 - i0)],
Line[closeline[n[[i]]]], PointSize[Medium],
Point[n[[i]]]}, {i, nst, nit + 1}], 1]],
Graphics[
Flatten[Table[{Hue[(i - nst)/(nit + 1 - i0)],
Line[closeline[n[[i]]]]}, {i, nst, nit + 1}], 1]]
], If[init,
Graphics[{Opacity[0.6], Line[closeline[n[[1]]]]}],
Graphics[]
]
]
]
I here also present the spectrum
reference:
[1]:http://www.cs.cornell.edu/cv/ResearchPDF/PolygonSmoothingPaper.pdf
| 474
| 1,312
|
{"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}
| 2.953125
| 3
|
CC-MAIN-2017-43
|
longest
|
en
| 0.645476
|
https://cpep.org/mathematics/556760-what-is-perimeter-and-area-of-an-equallateral-triangle.html
| 1,721,335,930,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00557.warc.gz
| 161,327,401
| 7,504
|
3 July, 05:33
# What is perimeter and area of an equallateral triangle?
+4
1. 3 July, 05:55
0
For perimeter just add all the sides
For area the formula is
1/2 * b*h*sin60
H = height
| 64
| 186
|
{"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-2024-30
|
latest
|
en
| 0.837752
|
http://www.gabormelli.com/RKB/Marginal_Probability_Function
| 1,695,519,782,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00392.warc.gz
| 72,348,231
| 23,719
|
# Marginal Probability Function
A marginal probability function is a multivariate probability function that reports the marginal probability value for an event (without aposteriori knowledge of the events for the other random variables).
## References
### 2009
• (Wikipedia, 2009) ⇒ http://en.wikipedia.org/wiki/Marginal_distribution
• In probability theory and statistics, the marginal distribution of a subset of a collection of random variables is the probability distribution of the variables contained in the subset. The term marginal variable is used to refer to those variables in the subset of variables being retained. These terms are dubbed "marginal" because they used to be found by summing values in a table along rows or columns, and writing the sum in the margins of the table. The distribution of the marginal variables (the marginal distribution) is obtained by "marginalising" over the distribution of the variables being discarded.
• The context here is that the theoretical studies being undertaken, or the data analysis being done, involves a wider set of random variables but that attention is being limited to a reduced number of those variables. In many applications an analysis may start with a given collection of random variables, then first extend the set by defining new ones (such as the sum of the original random variables) and finally reduce the number by placing interest in the marginal distribution of a subset (such as the sum). Several different analyses may be done, each treating a different subset of variables as the marginal variables.
• Given two random variables $\displaystyle{ X }$ and $\displaystyle{ Y }$ whose joint distribution is known, the marginal distribution of $\displaystyle{ X }$ is simply the probability distribution of $\displaystyle{ X }$ averaging over information about Y. This is typically calculated by summing or integrating the joint probability distribution over Y.
• For discrete random variables, the marginal probability mass function can be written as Pr(X = x).
| 395
| 2,040
|
{"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}
| 3.734375
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.883153
|
https://studyadda.com/question-bank/handling-data_q30/2650/234123
| 1,571,723,483,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987803441.95/warc/CC-MAIN-20191022053647-20191022081147-00434.warc.gz
| 705,615,304
| 20,012
|
• question_answer Direction: The incomplete pictograph shows the amount of money Dinesh spent on four days. Day Amount of money spent Wednesday Thursday Friday Saturday Sunday Each stands for Rs. 20 On which two days did Dinesh spend the same amount of money? A) Wednesday and Thursday B) Thursday and Friday C) Wednesday and Friday D) Thursday and Saturday
Solution :
The no. of symbols for the amount of money spent on Wednesday and Friday are the same. So, Dinesh spent the same amount of money on Wednesday and Friday.
You need to login to perform this action.
You will be redirected in 3 sec
| 138
| 603
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.796875
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.920974
|
https://nrich.maths.org/public/leg.php?code=71&cl=3&cldcmpid=2194
| 1,511,107,864,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934805687.20/warc/CC-MAIN-20171119153219-20171119173219-00049.warc.gz
| 665,142,897
| 10,370
|
# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Smith and Jones:
Filter by: Content type:
Stage:
Challenge level:
### There are 176 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Doodles
##### Stage: 4 Challenge Level:
Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Three Frogs
##### Stage: 4 Challenge Level:
Three frogs hopped onto the table. A red frog on the left a green in the middle and a blue frog on the right. Then frogs started jumping randomly over any adjacent frog. Is it possible for them to. . . .
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### A Biggy
##### Stage: 4 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
### Sixational
##### Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Archimedes and Numerical Roots
##### Stage: 4 Challenge Level:
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Tri-colour
##### Stage: 3 Challenge Level:
Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### Russian Cubes
##### Stage: 4 Challenge Level:
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Stage: 4
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Mod 3
##### Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Go Forth and Generalise
##### Stage: 3
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Cycle It
##### Stage: 3 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
### Long Short
##### Stage: 4 Challenge Level:
What can you say about the lengths of the sides of a quadrilateral whose vertices are on a unit circle?
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### Proof: A Brief Historical Survey
##### Stage: 4 and 5
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Picture Story
##### Stage: 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
##### Stage: 4 Challenge Level:
Four jewellers share their stock. Can you work out the relative values of their gems?
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
### Cosines Rule
##### Stage: 4 Challenge Level:
Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.
### Con Tricks
##### Stage: 3
Here are some examples of 'cons', and see if you can figure out where the trick is.
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Why 24?
##### Stage: 4 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Mouhefanggai
##### Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Composite Notions
##### Stage: 4 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
### Unit Fractions
##### Stage: 3 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
| 2,398
| 9,857
|
{"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}
| 4.125
| 4
|
CC-MAIN-2017-47
|
longest
|
en
| 0.849148
|
https://convertoctopus.com/75-5-grams-to-pounds
| 1,611,648,220,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610704799711.94/warc/CC-MAIN-20210126073722-20210126103722-00737.warc.gz
| 275,762,140
| 7,662
|
## Conversion formula
The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:
1 g = 0.0022046226218488 lb
To convert 75.5 grams into pounds we have to multiply 75.5 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:
1 g → 0.0022046226218488 lb
75.5 g → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 75.5 g × 0.0022046226218488 lb
M(lb) = 0.16644900794958 lb
The final result is:
75.5 g → 0.16644900794958 lb
We conclude that 75.5 grams is equivalent to 0.16644900794958 pounds:
75.5 grams = 0.16644900794958 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 6.0078459602649 × 75.5 grams.
Another way is saying that 75.5 grams is equal to 1 ÷ 6.0078459602649 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that seventy-five point five grams is approximately zero point one six six pounds:
75.5 g ≅ 0.166 lb
An alternative is also that one pound is approximately six point zero zero eight times seventy-five point five grams.
## Conversion table
### grams to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from grams to pounds
grams (g) pounds (lb)
76.5 grams 0.169 pounds
77.5 grams 0.171 pounds
78.5 grams 0.173 pounds
79.5 grams 0.175 pounds
80.5 grams 0.177 pounds
81.5 grams 0.18 pounds
82.5 grams 0.182 pounds
83.5 grams 0.184 pounds
84.5 grams 0.186 pounds
85.5 grams 0.188 pounds
| 508
| 1,722
|
{"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}
| 4.03125
| 4
|
CC-MAIN-2021-04
|
latest
|
en
| 0.81784
|
https://www.shaalaa.com/question-bank-solutions/the-separation-between-plates-parallel-plate-capacitor-0-500-cm-its-plate-area-100-cm2-0-400-cm-thick-metal-plate-inserted-gap-its-faces-parallel-plates-capacitors-capacitance_68930
| 1,585,765,049,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00435.warc.gz
| 1,136,903,053
| 11,407
|
Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
Share
# The Separation Between the Plates of a Parallel-plate Capacitor is 0⋅500 Cm and Its Plate Area is 100 Cm2. a 0⋅400 Cm Thick Metal Plate is Inserted into the Gap with Its Faces Parallel to the Plates. - Physics
ConceptCapacitors and Capacitance
#### Question
The separation between the plates of a parallel-plate capacitor is 0⋅500 cm and its plate area is 100 cm2. A 0⋅400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.
#### Solution
Given:
Area of the plate = 100 cm2
Separation between the plates = 0.500 "cm" = 5 xx 10^-3 "m"
Thickness of the metal, t = 4 xx 10^-3 "m"
therefore C = (∈_0A)/(d-t+t/k)
Here,
k = Dielectric constant of the metal
d = Separation between the plates
t = Thickness of the metal
For the metal, k = ∞
therefore C = (∈_0A)/(d-t) = ((8.85 xx 10^-12) xx 10^-12)/((5-4) xx 10^-3) = 88 "pF"
Here, the capacitance is independent of the position of the metal. At any position, the net separation is (d − t).
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [2]
Solution The Separation Between the Plates of a Parallel-plate Capacitor is 0⋅500 Cm and Its Plate Area is 100 Cm2. a 0⋅400 Cm Thick Metal Plate is Inserted into the Gap with Its Faces Parallel to the Plates. Concept: Capacitors and Capacitance.
S
| 451
| 1,536
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.15625
| 4
|
CC-MAIN-2020-16
|
longest
|
en
| 0.817302
|
https://ru.scribd.com/presentation/260785377/quadrilaterals-foldables
| 1,581,925,972,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00294.warc.gz
| 551,638,668
| 79,203
|
Вы находитесь на странице: 1из 23
## Polygons with four sides
Foldable
1. Take out a piece of
notebook paper and
make a hot dog fold
over from the right
side over to the pink
line.
Foldable
2. Now, divide the right
hand section into 5
sections by drawing 4
evenly spaced lines.
3. Use scissors to cut
but ONLY to the crease!
## The fold crease
Foldable
4. Write
down the left hand side
Foldable
## 5. Fold over the top cut
section and write
PARALLELOGRAM
on the outside.
6. Reopen the fold.
## The fold crease
Foldable
1. Opposite angles are congruent.
2. Consecutive angles are supplementary.
3. Opposite sides are congruent.
4. Diagonals bisect each other.
5. Diagonals make 2 congruent triangles.
## 7. On the left hand
section, draw a
parallelogram.
8. On the right hand
side, list all of the
properties of a
parallelogram.
Foldable
## 1. Opposite angles are congruent.
2. Consecutive angles are supplementary.
3. Opposite sides are congruent.
4. Diagonals bisect each other.
5. Diagonals make 2 congruent triangles.
## * Fold over the second
cut section and write
RECTANGLE on the
outside.
* Reopen the fold.
Foldable
1. Opposite angles are congruent.
2. Consecutive angles are supplementary.
3. Opposite sides are congruent.
4. Diagonals bisect each other.
5. Diagonals make 2 congruent triangles.
## * On the left hand
section, draw a
rectangle.
* On the right hand
side, list all of the
properties of a
rectangle.
## 2. Has 4 right angles
3. Diagonals are congruent.
Foldable
1. Opposite angles are congruent.
2. Consecutive angles are supplementary.
3. Opposite sides are congruent.
4. Diagonals bisect each other.
5. Diagonals make 2 congruent triangles.
## * Fold over the third
cut section and write
RHOMBUS on the
outside.
* Reopen the fold.
## 2. Has 4 right angles
3. Diagonals are congruent.
Foldable
## 1. Opposite angles are congruent.
2. Consecutive angles are supplementary.
3. Opposite sides are congruent.
section, draw a
rhombus.
## 4. Diagonals bisect each other.
5. Diagonals make 2 congruent triangles.
## 1.Is a special type of parallelogram.
2. Has 4 right angles
3. Diagonals are congruent.
## * On the right hand
side, list all of the
properties of a
rhombus.
## 1. Is A Special type of Parallelogram
2. Has 4 Congruent sides
3. Diagonals are perpendicular.
4. Diagonals bisect opposite angles
Foldable
## 1. Opposite angles are congruent.
2. Consecutive angles are
supplementary.
3. Opposite sides are congruent.
4. Diagonals bisect each other.
5. Diagonals make 2 congruent
triangles.
## 1.Is a special type of parallelogram.
2. Has 4 right angles
3. Diagonals are congruent.
## * Fold over the third
cut section and write
SQUARE on the
outside.
* Reopen the fold.
## 1. Is A Special type of Parallelogram
2. Has 4 Congruent sides
3. Diagonals are perpendicular.
4. Diagonals bisect opposite angles
Foldable
1. Opposite angles are congruent.
2. Consecutive angles are supplementary.
3. Opposite sides are congruent.
4. Diagonals bisect each other.
5. Diagonals make 2 congruent triangles.
## * On the left hand
section, draw a square.
## 1.Is a special type of parallelogram.
2. Has 4 right angles
3. Diagonals are congruent.
## * On the right hand
side, list all of the
properties of a square.
## 2. Has 4 Congruent sides
3. Diagonals are perpendicular.
4. Diagonals bisect opposite angles
## 1. Is a parallelogram, rectangle, and
rhombus
2. 4 congruent sides and 4 congruent
* Place in your
notebook and save for
tomorrow.
Foldable
(right) angles
## The last Box will be
Regular
Irregular trapezoid
Trapezoids
Kites
Characteristics Of Other
All trapezoids
## Exactly 1 pair of parallel sides
360 degrees
Regular Trapezoids:
## Exactly 1 pair of parallel sides
360 degrees
Consecutive angles total 180 degrees
Base angles are congruent
Characteristics Of Other
4 sides
360 degrees
Kites
Diagonals perpendicular
opposite sides that are parallel and of
equal length and opposite angles are
equal
Indicates equal sides
pairs of equal sides and four right
angles (90 degrees)
Indicates equal sides
## Rhombus: Parallelogram with four
equal sides and opposite angles equal
## Indicates equal sides
equal sides and four right angles
(90 degrees)
Indicates equal sides
## Box indicates 900 angle
pair of parallel sides
## Parallel sides never
meet.
with no equal sides and no equal
angles
1
rectangle
4
parallelogram
irregular
trapezoid
3
rhombus
square
Interior Angles
## Interior angles: An interior angle
(or internal angle) is an angle
formed by two sides of a simple
polygon that share an endpoint
## Interior angles of a quadrilateral
always equal 360 degrees
| 1,309
| 4,691
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875
| 4
|
CC-MAIN-2020-10
|
latest
|
en
| 0.770907
|
https://www.teachoo.com/2305/595/Ex-5.3--8---Chapter-5-Class-11---Solve-equation/category/Ex-5.3/
| 1,723,066,729,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00233.warc.gz
| 792,816,358
| 21,357
|
Chapter 4 Class 11 Complex Numbers
Serial order wise
Transcript
Ex5.3, 8 Solve the equation 3x2 2x + 3 3 = 0 3x2 2x + 3 3 = 0 The above equation is of the form 2 + + = 0 Where a = 3 , b = 2, and c = 3 3 x = ( ( ^2 4 ))/2 = ( (" " 2 ) ((" " 2 )^2 4 (3 ) 3 3) )/(2 3) = ( 2 (2 4 3 3))/(2 3) = ( 2 (2 36))/(2 3) = ( 2 ( 34))/(2 3) = ( 2 ( 1 34))/(2 3) = ( 2 ( 1 ) (34 ))/(2 3) = ( 2 34 )/(2 3) Thus, = ( 2 34 )/(2 3)
| 228
| 415
|
{"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-2024-33
|
latest
|
en
| 0.752162
|
https://studyaffiliates.com/the-maximum-displacement-of-the-object-is-a0-5-m-and-its-maximum-acceleration-is-amax-3-m-s2-2/
| 1,624,427,603,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623488534413.81/warc/CC-MAIN-20210623042426-20210623072426-00292.warc.gz
| 466,091,040
| 11,882
|
# The maximum displacement of the object is A=0.5 m and its maximum acceleration is a(max)= 3 m/s^2.
An object attached to a horizontal spring is oscillating back and forth along a frictionless table. The maximum displacement of the object is A=0.5 m and its maximum acceleration is a(max)= 3 m/s^2.
How much time elapses between an instant when the object’s displacement is at a maximum and the next time when its speed is at a maximum?
| 106
| 439
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.484375
| 3
|
CC-MAIN-2021-25
|
latest
|
en
| 0.905465
|
https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.APPROX/RANDOM.2023.19
| 1,716,773,329,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00752.warc.gz
| 183,446,130
| 16,298
|
Document
# Tighter Approximation for the Uniform Cost-Distance Steiner Tree Problem
## File
LIPIcs.APPROX-RANDOM.2023.19.pdf
• Filesize: 0.78 MB
• 16 pages
## Cite As
Josefine Foos, Stephan Held, and Yannik Kyle Dustin Spitzley. Tighter Approximation for the Uniform Cost-Distance Steiner Tree Problem. In Approximation, Randomization, and Combinatorial Optimization. Algorithms and Techniques (APPROX/RANDOM 2023). Leibniz International Proceedings in Informatics (LIPIcs), Volume 275, pp. 19:1-19:16, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2023)
https://doi.org/10.4230/LIPIcs.APPROX/RANDOM.2023.19
## Abstract
Uniform cost-distance Steiner trees minimize the sum of the total length and weighted path lengths from a dedicated root to the other terminals. They are applied when the tree is intended for signal transmission, e.g. in chip design or telecommunication networks. They are a special case of general cost-distance Steiner trees, where different distance functions are used for total length and path lengths. We improve the best published approximation factor for the uniform cost-distance Steiner tree problem from 2.39 [Khazraei and Held, 2021] to 2.05. If we can approximate the minimum-length Steiner tree problem arbitrarily well, our algorithm achieves an approximation factor arbitrarily close to 1+1/√2. This bound is tight in the following sense. We also prove the gap 1+1/√2 between optimum solutions and the lower bound which we and all previous approximation algorithms for this problem use. Similarly to previous approaches, we start with an approximate minimum-length Steiner tree and split it into subtrees that are later re-connected. To improve the approximation factor, we split it into components more carefully, taking the cost structure into account, and we significantly enhance the analysis.
## Subject Classification
##### ACM Subject Classification
• Theory of computation → Approximation algorithms analysis
• Theory of computation → Routing and network design problems
##### Keywords
• cost-distance Steiner tree
• approximation algorithm
• uniform
## Metrics
• Access Statistics
• Total Accesses (updated on a weekly basis)
0
## References
1. Sanjeev Arora. Polynomial time approximation schemes for euclidean traveling salesman and other geometric problems. Journal of the ACM, 45(5):753-782, 1998.
2. Marcelo P.L. Benedito, Lehilton L.C. Pedrosa, and Hugo K.K. Rosado. On the inapproximability of the cable-trench problem. Procedia Computer Science, 195:39-48, 2021.
3. Jarosław Byrka, Fabrizio Grandoni, Thomas Rothvoss, and Laura Sanità. Steiner tree approximation via iterative randomized rounding. Journal of the ACM, 60(1):article 6, 2013.
4. Chandra Chekuri, Sanjeev Khanna, and Joseph Naor. A deterministic algorithm for the cost-distance problem. In Proc. ACM-SIAM symposium on Discrete Algorithms (SODA '01), 2001, pages 232-233. SIAM, 2001.
5. Miroslav Chlebík and Janka Chlebíková. The steiner tree problem on graphs: Inapproximability results. Theoretical Computer Science, 406(3):207-214, 2008.
6. Julia Chuzhoy, Anupam Gupta, Joseph Naor, and Amitabh Sinha. On the approximability of some network design problems. ACM Transactions on Algorithms, 4(2):article 23, 2008.
7. Siad Daboul, Stephan Held, Bento Natura, and Daniel Rotter. Global interconnect optimization. ACM Transactions on Design Automation of Electronic Systems, 2023. (to appear, conference paper in Proc. ICCAD '19). URL: https://doi.org/10.1145/3587044.
8. Fabrizio Grandoni and Thomas Rothvoß. Network design via core detouring for problems without a core. In International Colloquium on Automata, Languages, and Programming, pages 490-502, 2010.
9. Sudipto Guha, Adam Meyerson, and Kamesh Munagala. A constant factor approximation for the single sink edge installation problem. SIAM Journal on Computing, 38(6):2426-2442, 2009.
10. Longkun Guo, Nianchen Zou, and Yidong Li. Approximating the shallow-light steiner tree problem when cost and delay are linearly dependent. In Proc. International Symposium on Parallel Architectures, Algorithms and Programming, pages 99-103, 2014.
11. Stephan Held, Dirk Müller, Daniel Rotter, Rudolf Scheifele, Vera Traub, and Jens Vygen. Global routing with timing constraints. IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems, 37(2):406-419, 2018.
12. Stephan Held and Daniel Rotter. Shallow-light steiner arborescences with vertex delays. In Proc. International Conference on Integer Programming and Combinatorial Optimization (IPCO '13), pages 229-241, 2013.
13. Stephan Held and Yannik.K.D. Spitzley. Further improvements on approximating the uniform cost-distance steiner tree problem. Technical report, Research Institute for Discrete Mathematics, University of Bonn, 2022. URL: https://arxiv.org/abs/2211.03830.
14. Raja Jothi and Balaji Raghavachari. Improved approximation algorithms for the single-sink buy-at-bulk network design problems. Journal of Discrete Algorithms, 7(2):249-255, 2009.
15. Ardalan Khazraei and Stephan Held. An improved approximation algorithm for the uniform cost-distance Steiner tree problem. In Proc. Workshop on Approximation and Online Algorithms (WAOA 2020), pages 189-203. Springer, 2021.
16. Samir Khuller, Balaji Raghavachari, and Neal E. Young. Balancing minimum spanning trees and shortest-path trees. Algorithmica, 14(4):305-321, 1995.
17. Adam Meyerson, Kamesh Munagala, and Serge Plotkin. Cost-distance: Two metric network design. SIAM Journal on Computing, 38(4):1648-1659, 2008.
18. Daniel Rotter. Timing-constrained global routing with buffered Steiner trees. PhD thesis, Universitäts-und Landesbibliothek Bonn, 2017.
19. Kunal Talwar. The single-sink buy-at-bulk lp has constant integrality gap. In International Conference on Integer Programming and Combinatorial Optimization, pages 475-486, 2002.
20. Vera Traub and Rico Zenklusen. Local search for weighted tree augmentation and steiner tree. In Proc. ACM-SIAM Symposium on Discrete Algorithms (SODA '22), pages 3253-3272, 2022.
X
Feedback for Dagstuhl Publishing
| 1,575
| 6,081
|
{"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
|
longest
|
en
| 0.805521
|
http://www.weegy.com/?ConversationId=QUOCUZHZ
| 1,444,501,405,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-40/segments/1443737961332.92/warc/CC-MAIN-20151001221921-00234-ip-10-137-6-227.ec2.internal.warc.gz
| 1,080,140,665
| 8,465
|
Not a good answer? Get an answer now. (Free)
Rating
There are no new answers.
There are no comments.
Questions asked by the same visitor
A furniture store has a chair originally priced at \$76 on sale for \$58. What is the percent of decrease rounded to the nearest tenth?
Weegy: 76-58=18 18/76=23.7% is the percent of decrease rounded to the nearest tenth. (More)
Question
Asked 8/28/2013 8:56:17 AM
A video, originally priced at \$15, is on sale for \$5.50. What is the percent of decrease rounded to the nearest whole number? A. 63% B. 10% C. 17% D. 47%
Weegy: A video, originally priced at \$15, is on sale for \$5.50. What is the percent of decrease rounded to the nearest whole number? The answer is A. 63%. [smile] (More)
Question
Asked 8/28/2013 8:58:34 AM
The electric bill in May was \$248.00. In June the bill was \$265.00. What is the percent of increase rounded to the nearest tenth?
Weegy: (265-248)/248*100=6.9. The percent increase in the electric bill is 6.9%. (More)
Question
Asked 8/28/2013 8:59:17 AM
A video, originally priced at \$15, is on sale for \$5.50. What is the percent of decrease rounded to the nearest whole number?
Weegy: The percent of decrease rounded to the nearest whole number is 63% less from the original price. Solution: divide \$5.50 by \$5.50 which is equals to 1. Then, divide \$5.50 by \$15 equals 0.37. 0.37 multiply by 100% equals 37%. [ You've paid only 37% of the original price of the video and you got 63% less. ] (More)
Question
Asked 8/28/2013 8:58:10 AM
51 is 85% of what number?
Question
Updated 8/29/2013 9:03:24 AM
51 is 85% of 60
Added 8/29/2013 9:03:24 AM
Popular Conversations
(-5a 2)3·a 5
Weegy: (-5a^2)3 a^5 =(- 5) * a^2 * 3 * a^5 = - 15 * a^7 = - 15a^7 User: (y 2)5 · y 8 Weegy: What can I do for ...
10/9/2015 7:48:52 AM| 8 Answers
(fg 2)4 User: Evaluate m 0 - n 2 for m = 2 and n = -1.
Weegy: m^0 - n^2 for m = 2 and n = -1 (2)^0 - (-1)^2; = 1 - 1; = 0 User: 1.5m 7(-4m 5)2 Weegy: pass to other ...
10/9/2015 7:58:29 AM| 3 Answers
(-7p)^3
Weegy: (-7p)^3 = (-7p)(-7p)(-7p) = -343p^3 User: (a 2 b 3)4 User: (k^3)5 = k^8
10/9/2015 7:42:44 AM| 2 Answers
Ophelia's death occurs in which part of Shakespeare's ...
Weegy: Ophelia tragically drowns after going insane following the death of her father
10/9/2015 10:21:47 AM| 2 Answers
In “By the Waters of Babylon,” when John sees the city “as it had ...
Weegy: B. New York City as it had been. User: The question below refers to the selection "By the Waters of Babylon." ...
10/9/2015 11:02:29 AM| 2 Answers
Solve the inequality. 2
Weegy: y 2 User: Solve the inequality. x + 5 > 3 A. x > 2 B. x > 2 C. x Weegy: ?y 2
10/9/2015 11:01:13 AM| 2 Answers
Change 1.4 to a fraction. 1/4 1 1/4 1 1/2 1 2/5
10/9/2015 11:14:53 AM| 2 Answers
Find the quotient. Simplify if possible. q/7,q/26
Weegy: (q/7)/(q/26)= (q/7)*(26/q)=26q/7q=26/7 User: Paul needs 2 1/4 yards of fabric to make a tablecloth. How many ...
10/9/2015 8:34:21 PM| 2 Answers
Weegy Stuff
S
L
Points 244 [Total 1784] Ratings 0 Comments 244 Invitations 0 Offline
S
Points 139 [Total 815] Ratings 0 Comments 139 Invitations 0 Offline
S
1
L
L
P
R
P
Points 94 [Total 9852] Ratings 0 Comments 94 Invitations 0 Offline
S
L
Points 60 [Total 3922] Ratings 1 Comments 50 Invitations 0 Offline
S
Points 24 [Total 24] Ratings 0 Comments 24 Invitations 0 Offline
S
Points 11 [Total 62] Ratings 1 Comments 1 Invitations 0 Offline
S
Points 10 [Total 10] Ratings 1 Comments 0 Invitations 0 Offline
S
R
R
Points 10 [Total 346] Ratings 1 Comments 0 Invitations 0 Offline
S
Points 8 [Total 439] Ratings 0 Comments 8 Invitations 0 Offline
S
R
Points 7 [Total 290] Ratings 0 Comments 7 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
| 1,392
| 3,707
|
{"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}
| 3.828125
| 4
|
CC-MAIN-2015-40
|
longest
|
en
| 0.92321
|
https://allhomeworksolutions.com/states-naval-power/
| 1,718,862,853,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00377.warc.gz
| 77,861,666
| 38,509
|
# State’s Naval Power
In Canvas you will find a dataset labeled RAP_naval_rep.dta. This is a slightly different version of the Crisher and Souva (2014) naval data set from a paper I published in 2017.
The variables are:
• “navycat” Naval Categories. A 4-tier categorical variable of naval power.
• “democ” A binary measure of democracy. 1 = Democracy, 0 = Non-democracy .
• “advecon” A binary measure of advanced economy using energy consumption. 1 = Yes, 0 = no.
• “lrgstate” A binary measure of large state using urban population. 1 = yes, 0 = no.
• “ton10” Total tonnage (in ten thousands). So, 10,000 tons would be shown as 1 ton.
• “energy” Continuous measure of energy consumption.
• “urbprop” Continuous measure of portion of total population in urban areas.
• “polity2” Continuous measure of democracy. Higher numbers means more democratic.
Here’s the assignment: I want you to evaluate some potential explanations for a state’s naval power. In other words, we are going to use the tonnage measure (ton10) as a dependent variable. So, we have a simple research question – what factors impact a state’s naval power. This will require you to conduct the three different types of bivariate hypotheses tests we’ve been using in class.
1. Is there a relationship between regime type and naval category? How did you come up with your answer?
2. Is there a relationship between advanced economies and naval category? How did you come up with your answer?
3. Is there a relationship between large states and naval category? How did you come up with your answer?
4. Do democracies have more tonnage than non-democracies? How did you come up with your answer?
6. Do large states have more tonnage than small states? How did you come up with your answer?
7. Is there a correlation between tonnage, energy consumption, urban population, and the continuous measure of democracy? How is this different from what you found in the previous tests?
8. What is the main weakness for these different tests?
9. In a paragraph, sum up your findings.
The above questions can be answered without seeking outside academic resources (you should cite the textbook and the Crisher & Souva article). The Stata codes necessary for the above questions have all been discussed in class or were used in the in-class assignment.
Your paper is to be 2-3 pages long with typical formatting – double spacing, 12 point font, and 1 inch margins. Please include any graphs (you should include at least one scatter plot) as a separate page at the end of your paper with a proper title (Figure 1: XXXX). Graphs/figures do not count toward the 2-3 page count. You should include at least two tables as well.
Do you need help with this assignment or any other? We got you! Place your order and leave the rest to our experts.
Quality Guaranteed
| 644
| 2,821
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.53125
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.948273
|
http://gmatclub.com/forum/if-jim-earns-x-dollars-per-hour-it-will-take-him-4-hours-to-116483.html
| 1,485,039,390,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00512-ip-10-171-10-70.ec2.internal.warc.gz
| 122,073,401
| 61,657
|
If Jim earns x dollars per hour, it will take him 4 hours to : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 21 Jan 2017, 14:56
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# If Jim earns x dollars per hour, it will take him 4 hours to
Author Message
TAGS:
### Hide Tags
Manager
Status: Its Wow or Never
Joined: 11 Dec 2009
Posts: 205
Location: India
Concentration: Technology, Strategy
GMAT 1: 670 Q47 V35
GMAT 2: 710 Q48 V40
WE: Information Technology (Computer Software)
Followers: 11
Kudos [?]: 192 [0], given: 7
If Jim earns x dollars per hour, it will take him 4 hours to [#permalink]
### Show Tags
05 Jul 2011, 13:23
4
This post was
BOOKMARKED
00:00
Difficulty:
45% (medium)
Question Stats:
59% (02:06) correct 41% (00:58) wrong based on 291 sessions
### HideShow timer Statistics
If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost?
(1) Tom makes 20% less per hour than Jim does.
(2) x + y = $43.75 [Reveal] Spoiler: OA _________________ --------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right. Manager Joined: 14 Mar 2011 Posts: 203 Followers: 6 Kudos [?]: 118 [0], given: 21 Re: Price of jacket [#permalink] ### Show Tags 05 Jul 2011, 14:43 It is given tom earns X$ per hour and it takes him 4 hrs to purchase the jacket. so price of jacket 4x $. Now jim earns Y and it takes him 5hrs to earn enough to buy the jacket. So jacket's price = 5y. so 4x= 5y thus x= 5y/4. (1) Insufficient. - It is a redundant information. The question stem already gives the relation between x and y. (2) sufficient - x + y = 43.75, so substitute the value of x for y. 5y/4 + y =43.75 >> 9y= 43.75 * 4 >> the price of jacket is 5y , so price = 43.75* 5*4 /9 = 875/9$
Senior Manager
Joined: 03 Mar 2010
Posts: 440
Schools: Simon '16 (M)
Followers: 5
Kudos [?]: 271 [1] , given: 22
### Show Tags
06 Jul 2011, 00:05
1
KUDOS
Jim's rate x dollar/hour.
Tom's rate y dollar/hour
Money made by Jim in 4 hours = Money made by Tom in 5 hours
=> 4x=5y----(1)
Stmt1: y=x-0.2x => y=0.8x
Put in (1) => 4x=4x. Same equation. Not sufficient.
Stmt2: x + y = $43.75 Also, 4x = 5y. Two equation , two unknown. Sufficient. OA B _________________ My dad once said to me: Son, nothing succeeds like success. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13680 [2] , given: 222 Re: Price of jacket [#permalink] ### Show Tags 06 Jul 2011, 02:00 2 This post received KUDOS Expert's post mojorising800 wrote: If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost? (1) Tom makes 20% less per hour than Jim does. (2) x + y =$43.75
You can also use ratios here to arrive at answer (B) in moments.
Time taken by Jim : Time take by Tom = 4:5
Hourly rate of Jim : Hourly rate of Tom = 5:4 = x:y
(Since Money Earned = Hourly Rate * Time and Money to be earned is the same for both Jim and Tom)
Statement 1: We already know that hourly rate of Tom is 20% less than the hourly rate of Jim (because it is in the ratio 5:4). No information about actual dollar amount so not sufficient.
Statement 2: On the ratio scale, x+y is 9 but it is actually 43.75. So we can get the values of x and y and hence can find the cost of the jacket. Sufficient.
Note: You don't need to calculate the cost of the jacket. You just need to say whether you can calculate it.
For more on ratios, see
http://www.veritasprep.com/blog/2011/03 ... of-ratios/
http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/
http://www.veritasprep.com/blog/2011/03 ... -problems/
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: ==GMAT Ninja== Joined: 08 Jan 2011 Posts: 247 Schools: ISB, IIMA ,SP Jain , XLRI WE 1: Aditya Birla Group (sales) WE 2: Saint Gobain Group (sales) Followers: 5 Kudos [?]: 78 [0], given: 46 Re: Price of jacket [#permalink] ### Show Tags 08 Jul 2011, 09:02 mojorising800 wrote: If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost? (1) Tom makes 20% less per hour than Jim does. (2) x + y =$43.75
B For me
as A and the information given in the question are same
B give an additional equation which is required to solve for the values
I wish these types of questions come in my real GMAT.......
_________________
WarLocK
_____________________________________________________________________________
The War is oNNNNNNNNNNNNN for 720+
see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html
do not hesitate me giving kudos if you like my post.
Director
Joined: 01 Feb 2011
Posts: 755
Followers: 14
Kudos [?]: 115 [0], given: 42
### Show Tags
08 Jul 2011, 19:38
Jack - $x/hr has to work 4 hours to be able to buy a jacket =$4x spent on the jacket
Tom - $y/hr has to work 5 hours to be able to buy the same jacket =$5y spent on the jacket
=>4x=5y=?
1. Not sufficient
y =(80/100)x => y =4x/5=>5y =4x
which is exact same equation
2. Sufficient
we can solve this equation and equation in the stem to get the jacket price.
VP
Joined: 24 Jul 2011
Posts: 1126
GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Followers: 124
Kudos [?]: 541 [0], given: 19
Re: If Jim earns x dollars per hour, it will take him 4 hours [#permalink]
### Show Tags
19 Nov 2011, 11:44
4x = 5y
=> x/y = 4/5
Statement (1) just restates this fact, so nothing new here. Insufficient.
Statement (2) gives us another equation in x and y which we can solve to find x (or y) and calculate 4x or 5y. Sufficient.
(B) it is.
_________________
GyanOne | Top MBA Rankings and MBA Admissions Blog
Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching
Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738
Manager
Joined: 27 Oct 2011
Posts: 191
Location: United States
Concentration: Finance, Strategy
GMAT 1: Q V
GPA: 3.7
WE: Account Management (Consumer Products)
Followers: 5
Kudos [?]: 150 [0], given: 4
Re: If Jim earns x dollars per hour, it will take him 4 hours [#permalink]
### Show Tags
11 Feb 2012, 16:10
+1 B
solve for x and solve for y. x = price of jacket / 4 and y = p/5.
when you know the relative rates of x and y they only tell you what p equals each other...
X + Y will you the rates and setting p/4 + p/5 you can solve what p is equal to.
_________________
DETERMINED TO BREAK 700!!!
Intern
Joined: 08 Nov 2012
Posts: 22
Followers: 0
Kudos [?]: 7 [0], given: 6
Re: If Jim earns x dollars per hour, it will take him 4 hours [#permalink]
### Show Tags
08 May 2013, 15:15
I dont understand why A is not sufficient. If tom earns 80% of what Jim makes, then i get the equation .8x(5)=x(4) where I got x=6.25 plug that back into the givens and I get the answer for J. I understand how you can say that this is not giving us new information but when you plug it back in you get an answer for J.
Here was my math
.8x(5)=x(4)
5=.2x(4)
1.25=.2x
x=6.25
Can someone help me understand where I went off into the wrong direction? Ive got this problem wrong 3x in the course of my studies and cannot understand where I am losing it. Thanks
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 630
Followers: 80
Kudos [?]: 1120 [0], given: 136
Re: If Jim earns x dollars per hour, it will take him 4 hours [#permalink]
### Show Tags
08 May 2013, 20:38
Richard0715 wrote:
I dont understand why A is not sufficient. If tom earns 80% of what Jim makes, then i get the equation .8x(5)=x(4) where I got x=6.25 plug that back into the givens and I get the answer for J. I understand how you can say that this is not giving us new information but when you plug it back in you get an answer for J.
Here was my math
.8x(5)=x(4)
5=.2x(4)
1.25=.2x
x=6.25
Can someone help me understand where I went off into the wrong direction? Ive got this problem wrong 3x in the course of my studies and cannot understand where I am losing it. Thanks
Jim earns 4*x dollars in 4 hours = price of the jacket = 4x.
Similarly, Tom earns 5*y dollars in 5 hours = price of the same jacket = 5y.
Thus as 4x = 5y, we have y = 0.8x
Now from F.S 1, we know that Tom makes 80% of what Jim earns per hour--> Thus, y = 0.8x. This is exactly what we had, without the aid of any Fact Statements. Thus Insufficient.
_________________
Intern
Joined: 20 Sep 2012
Posts: 22
Followers: 0
Kudos [?]: 42 [0], given: 7
IF Jim earns x dollars per hour [#permalink]
### Show Tags
12 Oct 2013, 09:22
If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost?
(1) Tom makes 20% less per hour than Jim does.
(2) x + y = $43.75 Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93358 [0], given: 10557 Re: IF Jim earns x dollars per hour [#permalink] ### Show Tags 12 Oct 2013, 10:20 skrishnakarthik wrote: If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost? (1) Tom makes 20% less per hour than Jim does. (2) x + y =$43.75
Merging similar topics. Please refer to the solutions above.
_________________
Manager
Joined: 10 Jun 2015
Posts: 128
Followers: 1
Kudos [?]: 25 [0], given: 0
Re: If Jim earns x dollars per hour, it will take him 4 hours to [#permalink]
### Show Tags
13 Aug 2015, 22:11
mojorising800 wrote:
If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost?
(1) Tom makes 20% less per hour than Jim does.
(2) x + y = \$43.75
Jim x dollars 4 hours
Tom y dollars 5 hours
Since 4x = 5y, x/y 5/4; it implies y is 20% less than x
Therefore, statement (1) is redundant.
state(2)
x+y =43.75
4x=5y
therefore, x and y can be found
It is sufficient to answer the question
Re: If Jim earns x dollars per hour, it will take him 4 hours to [#permalink] 13 Aug 2015, 22:11
Similar topics Replies Last post
Similar
Topics:
Bill started walking with a speed of X miles per hour. 2 27 Apr 2016, 23:41
1 If Machine X can produce 80 widgets in 4 hours, how long would it take 4 17 Feb 2016, 02:39
10 On a certain nonstop trip, Marta averaged x miles per hour for 2 hours 6 26 Oct 2015, 08:48
2 Each week John earns X dollars per hour for the first 40 3 25 Nov 2010, 04:56
3 For a week Raymond is paid at the rate of x dollars per hour 7 10 Oct 2008, 09:47
Display posts from previous: Sort by
| 3,615
| 12,072
|
{"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
| 4
|
CC-MAIN-2017-04
|
latest
|
en
| 0.859189
|
https://39thstreetgourmet.com/qa/what-is-the-difference-between-table-spoon-and-desert-spoon.html
| 1,627,243,791,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00637.warc.gz
| 92,848,332
| 8,490
|
# What Is The Difference Between Table Spoon And Desert Spoon?
## Which spoon is a tablespoon?
A typical large dinner spoon is about 1 tablespoon in size.
It is not often the case, but some might consider the dinner spoon as the one used towards a normal bowl of soup or cereal..
## How many dessert spoons make a tablespoon Australia?
two dessert spoonsThere are approximately one and a half dessert spoons per tablespoon. One dessert spoon holds 10 milliliters, and 1 tablespoon holds 15 milliliters. In Australia, the measurement is two dessert spoons per 20-milliliter tablespoon.
## What does tsp mean?
Thrift Savings PlanTSPAcronymDefinitionTSPTeaspoonTSPThrift Savings PlanTSP3-(Trimethylsilyl)-Propionic-2,2,3,3-D4 AcidTSPTraveling Salesman Problem141 more rows
## What are spoon sizes?
Size/No.1. … Spoons vary in length (11″, 13″, 15″, 18″, 21″) for ease of use in cooking or serving. … A thumb notch on a server or spoon handle prevents the spoon from slipping into the pan and prevents hands from sliding into the food.More items…
## How much does a plastic spoon hold?
The disposable spoon is available in two sizes, 2.5 ml (similar to a teaspoon) and 10 ml (similar to a tablespoon), and is ideal for sampling powders, granulates and fluids.
## What can I use if I don’t have a tablespoon?
Method 1 of 2: The most simple conversion is tablespoons to teaspoons. If you are missing a tablespoon, simply measure out three level teaspoons instead. Measure 1/16 of a cup. A tablespoon is equivalent to 1/16 of a cup, which will allow you to easily measure out that amount without a measuring spoon.
## Is a tablespoon the big spoon?
A tablespoon is a large spoon. In many English-speaking regions, the term now refers to a large spoon used for serving; however, in some regions, it is the largest type of spoon used for eating. By extension, the term is also used as a cooking measure of volume.
## What is a rounded tablespoon?
SPOON MEASURES usually means a rounded tablespoon. A rounded tablespoon means there is as much of the product you are measuring above the top edge of the spoon as there is in the “bowl” of the spoon. A heaped tablespoon means as much as you can get on the spoon without it falling off.
## Does T stand for teaspoon or tablespoon?
Teaspoon = t. or tsp. Tablespoon = T. or tbsp.
## How many grams is a tablespoon?
17.07 gramsBaking Conversion TableU.S.Metric1/2 teaspoon2.84 grams1 teaspoon5.69 grams1/2 tablespoon8.53 grams1 tablespoon17.07 grams82 more rows
## What’s bigger a tablespoon or dessert spoon?
A teaspoon is the smallest, a tablespoon is the largest, and then a DESSERT spoon falls in between. … A tablespoon (in the UK and South Africa) is accepted to mean 15ml of powder or liquid, while a teaspoon clocks in at 5ml.
## What is a 10mL spoon?
10mL equals two teaspoons (2tsp). A tablespoon is three times bigger than a teaspoon and three teaspoons equal one tablespoon (1Tbsp or 1Tb). One tablespoon also equals 15mL.
## What spoon do you eat cereal with?
table spoonThe most common spoon is the table spoon. We use these for soups, ice cream and cereal.
## How can I measure a teaspoon without a teaspoon?
3. Hand Comparisons1/8 teaspoon = 1 pinch between thumb, index and middle fingers.1/4 teaspoon = 2 pinches between thumb, index and middle fingers.1/2 teaspoon = Cup your hand, pour a quarter sized amount in your palm.1 teaspoon = Top joint of index finger.1 tablespoon = Entire thumb.More items…•Sep 2, 2015
## Is a dessert spoon the same as a tablespoon?
Worldwide, there are many different units of measure. … In the UK, dessertspoons are a level of measurement similar to a US Tablespoon. One level dessertspoon (Also known as dessert Spoon or abbreviated as dstspn) is equal to two teaspoons (tsp), 10 milliliters (mLs). A US tablespoon (tbls) is three teaspoons (15mL).
| 954
| 3,865
|
{"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-2021-31
|
latest
|
en
| 0.904847
|
https://math.stackexchange.com/questions/3802279/examples-and-counterexamples-of-relations-which-satisfy-certain-properties/3802280
| 1,726,513,380,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00085.warc.gz
| 346,388,651
| 45,169
|
Examples and Counterexamples of Relations which Satisfy Certain Properties
Definition: Given a set $$X$$, a relation $$R$$ on $$X$$ is any subset of $$X\times X$$. A relation $$R$$ on $$X$$ is said to be
1. reflexive if $$(x,x) \in R$$ for all $$x \in X$$,
2. irreflexive if $$(x,x) \not\in R$$ for all $$x \in X$$,
3. transitive if $$(x,y) \in R$$ and $$(y,z) \in R$$ implies that $$(x,z)\in R$$,
4. intransitive (or antitransitive) if $$(x,y) \in R$$ and $$(y,z) \in R$$ implies that $$(x,z)\not\in R$$,
5. symmetric if $$(x,y) \in R$$ implies that $$(y,x) \in R$$,
6. antisymmetric if $$(x,y) \in R$$ and $$(y,x) \in R$$ implies that $$x=y$$.
Given any combination of the properties listed above, is there a nontrivial (i.e. nonempty) relation which satisfies that combination of properties?
• This question and answer are meant to be a canonical answer to similar questions. See this meta question. I invite comments and critiques. Commented Aug 25, 2020 at 0:56
• @HereToRelax Note the context given in the comment right above yours. Based on moderate community demand, this question was posted to act as a duplicate for all similar questions. Without that context, I agree that this would generally be considered a low-quality question (though not a PSQ, as the PSQ version would be "Give an example of a relation which is transitive, but not reflexive"). Commented Apr 28, 2021 at 17:56
• I see, apologies Commented Apr 28, 2021 at 18:07
1. Reflexivity and Irreflexivity
A relation on a nonempty set cannot be both reflexive and irreflexive. This follows almost immediately from the definitions: a reflexive relation on a nonempty set $$X$$ must contain every pair of the form $$(x,x) \in X\times X$$, while an irreflexive relation cannot contain any such pair. Reflexivity and irreflexivity are mutually exclusive properties.
2. Transitivity and Intransitivity
A relation may be vacuously transitive and intransitive: if there is no $$y$$ such that $$(x,y),(y,z) \in R$$ for some $$x$$ and $$z$$, then the hypotheses of both transitivity and intransitivity fail. Any conclusion is implied by a false hypothesis, so such a relation is both transitive and intransitive. For example, let $$R$$ be the relation on the three element set $$X = \{1,2,3\}$$ given by $$R = \{ (1,2), (1,3) \}.$$ This relation is (trivially) both transitive and intransitive, as there is no $$y$$ which appears in the first slot of one pair, and in the second slot of another.
Aside from such vacuous examples (vacuous in the sense that the hypotheses are false, not in the sense that they are "easy"), a relation cannot be both transitive and intransitive: if $$(x,y), (y,z) \in R$$, then either $$(x,z) \in R$$ (and $$R$$ is not intransitive), or $$(x,z) \not\in R$$ (and $$R$$ is not transitive). Aside from vacuous examples, these two properties are mutually exclusive.
3. Intransitivity and Irreflexivity
A nontrivial relation which is intransitive must also be irreflexive. The essential idea here is that reflexive relations "build in" transitive relations. More formally, consider a proof by contraposition: suppose that $$R$$ is a nontrivial relation which is not irreflexive. Then there is some $$x$$ such that $$(x,x) \in R$$. Taking $$x=y=z$$, this implies that $$(x,y), (y,z), (x,z) \in R,$$ which contradicts the definition of intransitivity. Thus $$R$$ is not intransitive. Therefore a relation which is not irreflexive is not intransitive.
By contraposition, an intransitive relation must be irreflexive.
4. Symmetry and Antisymmetry
Perhaps counterintuitively, a nontrivial relation can be both symmetric and antisymmetric. Suppose that $$R$$ is a nontrivial relation which is both symmetric and antisymmetric. As $$R$$ is nontrivial, it contains some pair $$(x,y)$$. The symmetry of $$R$$ implies that $$(y,x)$$ is also in $$R$$. The antisymmetry of $$R$$ then implies that $$x=y$$. Hence a relation on a set $$X$$ which is both symmetric and antisymmetric must be a subset of the diagonal $$\{(x,x) : x \in X\}$$. Any such relation is vacuously transitive, and can be reflexive if it is the entire diagonal (this is the equality relation). There is no nontrivial irreflexive relation which is both symmetric and antisymmetric.
5. Examples on a Set with Three Elements
The remainder of this answer is structured as follows: the set $$X$$ is the three element set $$X = \{1,2,3\}$$. Each of the items below gives an example of a relation $$R$$ on $$X$$ which satisfies various combinations of the properties listed in the question. The examples are labeled with a string such as "[RT-]".
• The first character may be R for a reflexive relation, I for an irreflexive relation, or - for a relation which is neither reflexive nor irreflexive.
• The second character may be T for a transitive relation, I for an intransitive relation, or - for a relation which is neither transitive nor intransitive.
• The third character may be S for a symmetric relation, A for an antisymmetric relation, or - for a relation which is neither symmetric nor antisymmetric.
Commentary is given in cases where it might be illuminating.
• [RTS] $$R = \{(1,1), (1,2), (2,2), (2,1), (3,3)\}$$
• [RTA] $$R = \{(1,1), (1,2), (1,3), (2,2), (2,3), (3,3)\}$$
• [RT-] $$R = \{ (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (3,3) \}$$
• [RIS] No example exists, see 3.
• [RIA] No example exists, see 3.
• [RI-] No example exists, see 3.
• [R-S] $$R = \{(1,1), (1,2), (2,1), (2,2), (2,3), (3,2), (3,3)\}$$
• [R-A] $$R = \{(1,1), (1,2), (2,2), (2,3), (3,3)\}$$
• [R--] $$R = \{(1,1), (2,2), (3,3), (1,2), (2,3) \}$$
• [ITS] No nontrivial example exists.
Suppose that $$R$$ is some nontrivial, irreflexive, transitive relation. If $$R$$ is not antisymmetric, then there exist pairs $$(x,y)$$ and $$(y,x)$$ which are both elements of $$R$$. But $$R$$ is transitive, so $$(x,x)$$ and $$(y,y)$$ must also be elements of $$R$$. In other words, a nontrivial, irreflexive, transitive relation must be antisymmetric.
• [ITA] $$R = \{(1,2), (1,3), (2,3)\}$$
The usual order relations ($$\le$$, $$<$$, $$\ge$$, $$>$$) on $$\mathbb{R}$$ are more interesting examples of relations which are transitive and antisymmetric. Weak inequalities are reflexive, while strict inequalities are irreflexive.
• [IT-] No nontrivial example exists, see [ITS].
• [IIS] $$\{(1,2), (2,1)\}$$.
• [IIA] $$R = \{(1,2)\}$$
• [II-] $$\{(1,2), (1,3), (2,1)\}$$.
• [I-S] $$R = \{(1,2), (2,1), (2,3), (3,2) \}$$.
Transitivity and intransitivity can be a little hard to see by inspection. This relation is not intransitive, as every intransitive relation must be antisymmetric; and it is not transitive, as $$(1,2),(2,3) \in R$$ but $$(1,3)\not\in R$$.
• There is no example of an irreflexive and antisymmetric relation on $$X$$ which is neither transitive nor intransitive. However, if $$R$$ is a relation on as set $$Y = \{a,b,c,d\}$$, then an example exists:
[I-A] $$R = \{ (a,b), (a,c), (b,c), (c,d) \}$$
This relation is not transitive, because $$(a,c), (c,d) \in R$$, but $$(a,d)\not\in R$$; and is not intransitive, because $$(a,b), (b,c), (a,c) \in R$$.
• [I--] $$R = \{(1,2), (1,3), (2,1), (2,3)\}$$
• [-TS] $$R = \{(1,1), (1,2), (2,1), (2,2)\}$$
Note that the above relation is not reflexive on the three element set $$X = \{1,2,3\}$$ because it does not contain the pair $$(3,3)$$. However, thought of as a relation on the two element set $$\{1,2\}$$, this relation is reflexive.
• [-TA] $$R = \{(1,1), (1,2), (2,3), (3,1)\}$$
• [-T-] $$R = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3) \}$$
• [-IS] No nontrivial example exists, see 3.
• [-IA] No nontrivial example exists, see 3.
• [-I-] No nontrivial example exists, see 3.
• [--S] $$R = \{(1,2), (2,1), (2,2) \}$$
• [--A] $$R = \{(1,1), (1,2), (2,3) \}$$
• [---] $$R = \{ (1,1), (1,2), (2,1), (2,3) \}$$
Abstractly, it is good to have simple examples and counterexamples to different permutations of relational properties. However, it is also useful to have in mind more interesting models—every one of these properties comes from something in the world. The arbitrary permutations of properties may not have any useful meaning, but the properties themselves are interesting.
• An equivalence relation is any relation which is reflexive, transitive, and symmetric. The most basic such relation is equality ($$=$$): $$x=y$$ if and only if $$x$$ and $$y$$ are, in fact, the same object. Abusing notation a bit, this means that $$=$$, though of as an equivalence relation on some arbitrary set $$X$$, is the diagonal of $$X\times X$$. That is, $$= \quad\text{is the set}\quad \{ (x,x) : x \in X\}.$$ There are other important equivalence relations, and many important properties in mathematics hold only "up to equivalence" with respect to some equivalence relation.
For example, $$1/2$$ and $$2/4$$ are not really the same object—ask any second grader. If I have a package of two cookies, then I can have one cookie, and give another to a friend. We each get one of the two cookies, or $$1/2$$ of the package. If I have a package of four cookies, then I can have two and give two to a friend. We each get two cookies, or $$2/4$$ of the package. Two is not one! These things are different. However, from the point of view of addition and multiplication, $$1/2$$ and $$2/4$$ behave in essentially the same way—they are equivalent with respect to a relation which ultimately gives us the rational numbers. Hence we can treat them as though they are the same object (and typically do!).
• Order relations are examples of transitive, antisymmetric relations. For example, $$\le$$, $$\ge$$, $$<$$, and $$>$$ are examples of order relations on $$\mathbb{R}$$—the first two are reflexive, while the latter two are irreflexive. Set containment relations ($$\subseteq$$, $$\supseteq$$, $$\subset$$, $$\supset$$) have simililar properties.
In general, I think that it is reasonable to think of transitive, antisymmetric relations as those relations which "rank" or "order" things in some rough way. Inequalities order numbers, set containment relations order sets, taxonomies classify and order living organisms, etc.
• Intransitive relations are kind of an odd duck, and it is not immediately obvious how they might come up in the real world. However, they do! My favorite example is the two-player game "Rock-Paper-Scissors". Rock beats scissors, scissors beats paper, paper beats rock. The relation "beats" is intransitive. Parenthood is also (generally speaking—one can always find exceptions once human behaviour is involved) an intransitive relation: I am the parent of my daughter, and my mother is my parent, but my mother is not my daughter's parent.
To start, it's worth pointing out that, as defined in the question above, none of pairs of "opposite" properties (reflexive / irreflexive, transitive / intransitive or symmetric / antisymmetric) are actually antonyms. Not only is it possible for a relation to satisfy none of these properties (and indeed most random relations on sufficiently large sets won't satisfy any of them), but it's also possible for a relation to satisfy two seemingly opposite properties at once, with some restrictions:
• There is only one relation which is both reflexive and irreflexive, as defined above: the empty relation $$R = \emptyset$$ on the empty set $$S = \emptyset$$. For relations on non-empty sets, reflexivity and irreflexivity are mutually exclusive.
• If no element $$x \in X$$ appears on both the left side and the right side of a relation, it is vacuously both transitive and intransitive. In other words, $$R$$ is both transitive and intransitive if and only if $$R \subset A \times B$$ for some disjoint subsets $$A$$ and $$B$$ of $$X$$. Such relations are always irreflexive and antisymmetric, never symmetric unless $$R = \emptyset$$, and never reflexive unless $$R = X = \emptyset$$.
• A relation is both symmetric and antisymmetric if all its elements are of the form $$(x, x)$$ for some $$x \in X$$. In other words, $$R$$ is both symmetric and antisymmetric if and only if $$(x,y) \in R \implies x = y$$. Such relations are always transitive, and never intransitive unless $$R = \emptyset$$; they can be either reflexive (if $$R = \{(x,x): x \in X\}$$) or irreflexive (if $$R = \emptyset$$) or neither.
Except for the "paradoxical" combinations described above, the three properties of (ir)reflexivity, (in)transitivity and (anti)symmetry are mostly independent of each other. The only further restrictions are that:
• An intransitive relation must also be irreflexive (and thus cannot be reflexive unless the underlying set is empty): setting $$x = y = z$$ in the definition of intransitivity leads to a contradiction unless $$(x,x) \notin R$$ for all $$x \in X$$.
• An irreflexive transitive relation must be antisymmetric (and cannot be symmetric unless empty): if $$R$$ contains both $$(x,y)$$ and $$(y,x)$$, then transitivity implies that $$R$$ must also contain $$(x,x)$$ and $$(y,y)$$.
For all other combinations of the six properties, examples of relations with (only) those properties exist on the four-element set $$X = \{a,b,c,d\}$$. Here is a complete list of them, generated using a simple Python script:
• reflexive, symmetric, antisymmetric, transitive: $$R_d = \{(a,a), (b,b), (c,c), (d,d)\}$$
• reflexive, symmetric, transitive: $$\{(a,b), (b,a)\} \cup R_d$$
• reflexive, symmetric: $$\{(a,b), (a,c), (b,a), (c,a)\} \cup R_d$$
• reflexive, antisymmetric, transitive: $$\{(a,b)\} \cup R_d$$
• reflexive, antisymmetric: $$\{(a,b), (b,c)\} \cup R_d$$
• reflexive, transitive: $$\{(a,b), (b,a), (c,d)\} \cup R_d$$
• reflexive: $$\{(a,b), (a,c), (b,a)\} \cup R_d$$
• irreflexive, symmetric, antisymmetric, transitive, intransitive: $$\emptyset$$
• irreflexive, symmetric, intransitive: $$\{(a,b), (b,a)\}$$
• irreflexive, symmetric: $$\{(a,b), (a,c), (b,a), (b,c), (c,a), (c,b)\}$$
• irreflexive, antisymmetric, transitive, intransitive: $$\{(a,b)\}$$
• irreflexive, antisymmetric, transitive: $$\{(a,b), (a,c), (b,c)\}$$
• irreflexive, antisymmetric, intransitive: $$\{(a,b), (b,c)\}$$
• irreflexive, antisymmetric: $$\{(a,b), (a,c), (b,c), (b,d)\}$$
• irreflexive, intransitive: $$\{(a,b), (a,c), (b,a)\}$$
• irreflexive: $$\{(a,b), (a,c), (b,a), (b,c)\}$$
• symmetric, antisymmetric, transitive: $$\{(a,a)\}$$
• symmetric, transitive: $$\{(a,a), (a,b), (b,a), (b,b)\}$$
• symmetric: $$\{(a,a), (a,b), (b,a)\}$$
• antisymmetric, transitive: $$\{(a,a), (a,b)\}$$
• antisymmetric: $$\{(a,a), (a,b), (b,c)\}$$
• transitive: $$\{(a,a), (a,b), (b,a), (b,b), (c,d)\}$$
• none: $$\{(a,a), (a,b), (a,c), (b,a)\}$$
Each relation in the list above satisfies all the properties named in that entry and no others out of the six properties listed in the question. Each example relation on the list has the smallest possible number of pairs among all relations on $$X$$ with that combination of properties, and is the first in lexicographical order among those with the same number of pairs.
The only combination of properties that requires a four-element set is "irreflexive and antisymmetric (and neither transitive nor intransitive)," for which the minimal example is $$R = \{(a,b), (a,c), (b,c), (b,d)\}$$. All other combinations on the list above can also be exhibited with relations on a three-element set, although in some cases an example with only three elements may require more pairs than if a fourth element is allowed.
The only technically possible combination of properties missing from the list above is "reflexive, irreflexive, symmetric, antisymmetric, transitive, intransitive", which, as noted above, is only possible in the vacuous case when $$R = X = \emptyset$$.
| 4,685
| 15,689
|
{"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": 176, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.328125
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.838208
|
https://www.infogalactic.com/info/Set_(mathematics)
| 1,627,729,295,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00410.warc.gz
| 847,670,040
| 21,083
|
# Set (mathematics)
A set of polygons in a Venn diagram
In mathematics, a set is a collection of distinct objects, considered as an object in its own right. For example, the numbers 2, 4, and 6 are distinct objects when considered separately, but when they are considered collectively they form a single set of size three, written {2,4,6}. Sets are one of the most fundamental concepts in mathematics. Developed at the end of the 19th century, set theory is now a ubiquitous part of mathematics, and can be used as a foundation from which nearly all of mathematics can be derived. In mathematics education, elementary topics such as Venn diagrams are taught at a young age, while more advanced concepts are taught as part of a university degree. The German word Menge, rendered as "set" in English, was coined by Bernard Bolzano in his work The Paradoxes of the Infinite.
## Definition
Passage with a translation of the original set definition of Georg Cantor
A set is a well defined collection of distinct objects. The objects that make up a set (also known as the elements or members of a set) can be anything: numbers, people, letters of the alphabet, other sets, and so on. Georg Cantor, the founder of set theory, gave the following definition of a set at the beginning of his Beiträge zur Begründung der transfiniten Mengenlehre:[1]
A set is a gathering together into a whole of definite, distinct objects of our perception [Anschauung] or of our thought—which are called elements of the set.
Sets are conventionally denoted with capital letters. Sets A and B are equal if and only if they have precisely the same elements.[2]
Cantor's definition turned out to be inadequate for formal mathematics; instead, the notion of a "set" is taken as an undefined primitive in axiomatic set theory, and its properties are defined by the Zermelo–Fraenkel axioms. The most basic properties are that a set has elements, and that two sets are equal (one and the same) if and only if every element of each set is an element of the other.
## Describing sets
There are two ways of describing, or specifying the members of, a set. One way is by intensional definition, using a rule or semantic description:
A is the set whose members are the first four positive integers.
B is the set of colors of the French flag.
The second way is by extension – that is, listing each member of the set. An extensional definition is denoted by enclosing the list of members in curly brackets:
C = {4, 2, 1, 3}
D = {blue, white, red}.
One often has the choice of specifying a set either intensionally or extensionally. In the examples above, for instance, A = C and B = D.
There are two important points to note about sets. First, a set can have two or more members which are identical, for example, {11, 6, 6}. However, we say that two sets which differ only in that one has duplicate members are in fact exactly identical (see Axiom of extensionality). Hence, the set {11, 6, 6} is exactly identical to the set {11, 6}. The second important point is that the order in which the elements of a set are listed is irrelevant (unlike for a sequence or tuple). We can illustrate these two important points with an example:
{6, 11} = {11, 6} = {11, 6, 6, 11} .
For sets with many elements, the enumeration of members can be abbreviated. For instance, the set of the first thousand positive integers may be specified extensionally as
{1, 2, 3, ..., 1000},
where the ellipsis ("...") indicates that the list continues in the obvious way. Ellipses may also be used where sets have infinitely many members. Thus the set of positive even numbers can be written as {2, 4, 6, 8, ... }.
The notation with braces may also be used in an intensional specification of a set. In this usage, the braces have the meaning "the set of all ...". So, E = {playing card suits} is the set whose four members are ♠, ♦, ♥, and ♣. A more general form of this is set-builder notation, through which, for instance, the set F of the twenty smallest integers that are four less than perfect squares can be denoted
F = {n2 − 4 : n is an integer; and 0 ≤ n ≤ 19}.
In this notation, the colon (":") means "such that", and the description can be interpreted as "F is the set of all numbers of the form n2 − 4, such that n is a whole number in the range from 0 to 19 inclusive." Sometimes the vertical bar ("|") is used instead of the colon.
## Membership
If B is a set and x is one of the objects of B, this is denoted xB, and is read as "x belongs to B", or "x is an element of B". If y is not a member of B then this is written as yB, and is read as "y does not belong to B".
For example, with respect to the sets A = {1,2,3,4}, B = {blue, white, red}, and F = {n2 − 4 : n is an integer; and 0 ≤ n ≤ 19} defined above,
4 ∈ A and 12 ∈ F; but
9 ∉ F and green ∉ B.
### Subsets
If every member of set A is also a member of set B, then A is said to be a subset of B, written AB (also pronounced A is contained in B). Equivalently, we can write BA, read as B is a superset of A, B includes A, or B contains A. The relationship between sets established by ⊆ is called inclusion or containment.
If A is a subset of, but not equal to, B, then A is called a proper subset of B, written AB (A is a proper subset of B) or BA (B is a proper superset of A).
Note that the expressions AB and BA are used differently by different authors; some authors use them to mean the same as AB (respectively BA), whereas other use them to mean the same as AB (respectively BA).
A is a subset of B
Example:
• The set of all men is a proper subset of the set of all people.
• {1, 3} ⊆ {1, 2, 3, 4}.
• {1, 2, 3, 4} ⊆ {1, 2, 3, 4}.
The empty set is a subset of every set and every set is a subset of itself:
• ∅ ⊆ A.
• AA.
An obvious but useful identity, which can often be used to show that two seemingly different sets are equal:
• A = B if and only if AB and BA.
A partition of a set S is a set of nonempty subsets of S such that every element x in S is in exactly one of these subsets.
### Power sets
The power set of a set S is the set of all subsets of S. Note that the power set contains S itself and the empty set because these are both subsets of S. For example, the power set of the set {1, 2, 3} is {{1, 2, 3}, {1, 2}, {1, 3}, {2, 3}, {1}, {2}, {3}, ∅}. The power set of a set S is usually written as P(S).
The power set of a finite set with n elements has 2n elements. This relationship is one of the reasons for the terminology power set[citation needed]. For example, the set {1, 2, 3} contains three elements, and the power set shown above contains 23 = 8 elements.
The power set of an infinite (either countable or uncountable) set is always uncountable. Moreover, the power set of a set is always strictly "bigger" than the original set in the sense that there is no way to pair every element of S with exactly one element of P(S). (There is never an onto map or surjection from S onto P(S).)
Every partition of a set S is a subset of the powerset of S.
## Cardinality
The cardinality | S | of a set S is "the number of members of S." For example, if B = {blue, white, red}, | B | = 3.
There is a unique set with no members and zero cardinality, which is called the empty set (or the null set) and is denoted by the symbol ∅ (other notations are used; see empty set). For example, the set of all three-sided squares has zero members and thus is the empty set. Though it may seem trivial, the empty set, like the number zero, is important in mathematics; indeed, the existence of this set is one of the fundamental concepts of axiomatic set theory.
Some sets have infinite cardinality. The set N of natural numbers, for instance, is infinite. Some infinite cardinalities are greater than others. For instance, the set of real numbers has greater cardinality than the set of natural numbers. However, it can be shown that the cardinality of (which is to say, the number of points on) a straight line is the same as the cardinality of any segment of that line, of the entire plane, and indeed of any finite-dimensional Euclidean space.
## Special sets
There are some sets that hold great mathematical importance and are referred to with such regularity that they have acquired special names and notational conventions to identify them. One of these is the empty set, denoted {} or ∅. Another is the unit set {x}, which contains exactly one element, namely x.[2] Many of these sets are represented using blackboard bold or bold typeface. Special sets of numbers include
• P or ℙ, denoting the set of all primes: P = {2, 3, 5, 7, 11, 13, 17, ...}.
• N or ℕ, denoting the set of all natural numbers: N = {1, 2, 3, . . .} (sometimes defined containing 0).
• Z or ℤ, denoting the set of all integers (whether positive, negative or zero): Z = {..., −2, −1, 0, 1, 2, ...}.
• Q or ℚ, denoting the set of all rational numbers (that is, the set of all proper and improper fractions): Q = {a/b : a, bZ, b ≠ 0}. For example, 1/4 ∈ Q and 11/6 ∈ Q. All integers are in this set since every integer a can be expressed as the fraction a/1 (ZQ).
• R or ℝ, denoting the set of all real numbers. This set includes all rational numbers, together with all irrational numbers (that is, numbers that cannot be rewritten as fractions, such as √2, as well as transcendental numbers such as π, e and numbers that cannot be defined).
• C or ℂ, denoting the set of all complex numbers: C = {a + bi : a, bR}. For example, 1 + 2iC.
• H or ℍ, denoting the set of all quaternions: H = {a + bi + cj + dk : a, b, c, dR}. For example, 1 + i + 2jkH.
Positive and negative sets are denoted by a superscript - or +. For example ℚ+ represents the set of positive rational numbers.
Each of the above sets of numbers has an infinite number of elements, and each can be considered to be a proper subset of the sets listed below it. The primes are used less frequently than the others outside of number theory and related fields.
## Basic operations
There are several fundamental operations for constructing new sets from given sets.
### Unions
The union of A and B, denoted AB
Two sets can be "added" together. The union of A and B, denoted by A ∪ B, is the set of all things that are members of either A or B.
Examples:
• {1, 2} ∪ {1, 2} = {1, 2}.
• {1, 2} ∪ {2, 3} = {1, 2, 3}.
• {1, 2, 3} ∪ {3, 4, 5} = {1, 2, 3, 4, 5}
Some basic properties of unions:
• AB = BA.
• A ∪ (BC) = (AB) ∪ C.
• A ⊆ (AB).
• AA = A.
• A ∪ ∅ = A.
• AB if and only if AB = B.
### Intersections
A new set can also be constructed by determining which members two sets have "in common". The intersection of A and B, denoted by AB, is the set of all things that are members of both A and B. If AB = ∅, then A and B are said to be disjoint.
The intersection of A and B, denoted AB.
Examples:
• {1, 2} ∩ {1, 2} = {1, 2}.
• {1, 2} ∩ {2, 3} = {2}.
Some basic properties of intersections:
• AB = BA.
• A ∩ (BC) = (AB) ∩ C.
• ABA.
• AA = A.
• A ∩ ∅ = ∅.
• AB if and only if AB = A.
### Complements
The relative complement
of B in A
The complement of A in U
The symmetric difference of A and B
Two sets can also be "subtracted". The relative complement of B in A (also called the set-theoretic difference of A and B), denoted by A \ B (or AB), is the set of all elements that are members of A but not members of B. Note that it is valid to "subtract" members of a set that are not in the set, such as removing the element green from the set {1, 2, 3}; doing so has no effect.
In certain settings all sets under discussion are considered to be subsets of a given universal set U. In such cases, U \ A is called the absolute complement or simply complement of A, and is denoted by A′.
Examples:
• {1, 2} \ {1, 2} = ∅.
• {1, 2, 3, 4} \ {1, 3} = {2, 4}.
• If U is the set of integers, E is the set of even integers, and O is the set of odd integers, then U \ E = E′ = O.
Some basic properties of complements:
• A \ BB \ A for AB.
• AA′ = U.
• AA′ = ∅.
• (A′)′ = A.
• A \ A = ∅.
• U′ = ∅ and ∅′ = U.
• A \ B = AB.
An extension of the complement is the symmetric difference, defined for sets A, B as
$A\,\Delta\,B = (A \setminus B) \cup (B \setminus A).$
For example, the symmetric difference of {7,8,9,10} and {9,10,11,12} is the set {7,8,11,12}.
### Cartesian product
A new set can be constructed by associating every element of one set with every element of another set. The Cartesian product of two sets A and B, denoted by A × B is the set of all ordered pairs (a, b) such that a is a member of A and b is a member of B.
Examples:
• {1, 2} × {red, white} = {(1, red), (1, white), (2, red), (2, white)}.
• {1, 2} × {red, white, green} = {(1, red), (1, white), (1, green), (2, red), (2, white), (2, green) }.
• {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}.
Some basic properties of cartesian products:
• A × = ∅.
• A × (BC) = (A × B) ∪ (A × C).
• (AB) × C = (A × C) ∪ (B × C).
Let A and B be finite sets. Then
• | A × B | = | B × A | = | A | × | B |.
For example,
• {a,b,c}×{d,e,f}={(a,d),(a,e),(a,f),(b,d),(b,e),(b,f),(c,d),(c,e),(c,f)}.
## Applications
Set theory is seen as the foundation from which virtually all of mathematics can be derived. For example, structures in abstract algebra, such as groups, fields and rings, are sets closed under one or more operations.
One of the main applications of naive set theory is constructing relations. A relation from a domain A to a codomain B is a subset of the Cartesian product A × B. Given this concept, we are quick to see that the set F of all ordered pairs (x, x2), where x is real, is quite familiar. It has a domain set R and a codomain set that is also R, because the set of all squares is subset of the set of all reals. If placed in functional notation, this relation becomes f(x) = x2. The reason these two are equivalent is for any given value, y that the function is defined for, its corresponding ordered pair, (y, y2) is a member of the set F.
## Axiomatic set theory
Although initially naive set theory, which defines a set merely as any well-defined collection, was well accepted, it soon ran into several obstacles. It was found that this definition spawned several paradoxes, most notably:
• Russell's paradox—It shows that the "set of all sets that do not contain themselves," i.e. the "set" { x : x is a set and xx } does not exist.
• Cantor's paradox—It shows that "the set of all sets" cannot exist.
The reason is that the phrase well-defined is not very well defined. It was important to free set theory of these paradoxes because nearly all of mathematics was being redefined in terms of set theory. In an attempt to avoid these paradoxes, set theory was axiomatized based on first-order logic, and thus axiomatic set theory was born.
For most purposes however, naive set theory is still useful.
## Principle of inclusion and exclusion
This principle gives us the cardinality of the union of sets. \begin{align} \left|A_{1}\cup A_{2}\cup A_{3}\cup\ldots\cup A_{n}\right|= & \left(\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|+\ldots\left|A_{n}\right|\right)- \\ & \left(\left|A_{1}\cap A_{2}\right|+\left|A_{1}\cap A_{3}\right|+\ldots\left|A_{n-1}\cap A_{n}\right|\right)+ \\ &\ldots+ \\ &\left(-1\right)^{n-1}\left(\left|A_{1}\cap A_{2}\cap A_{3}\cap\ldots\cap A_{n}\right|\right) \end{align}
## De Morgan's Law
De Morgan stated two laws about Sets.
If A and B are any two Sets then,
• (A ∪ B)′ = A′ ∩ B′
The complement of A union B equals the complement of A intersected with the complement of B.
• (A ∩ B)′ = A′ ∪ B′
The complement of A intersected with B is equal to the complement of A union to the complement of B.
| 4,417
| 15,660
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "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-2021-31
|
latest
|
en
| 0.957582
|
https://open.library.ubc.ca/cIRcle/collections/ubctheses/831/items/1.0080783
| 1,477,617,207,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988721415.7/warc/CC-MAIN-20161020183841-00002-ip-10-171-6-4.ec2.internal.warc.gz
| 860,782,202
| 143,200
|
# Open Collections
## UBC Theses and Dissertations
### A mathematical model for cement kilns Darabi, Pirooz 2007-12-31
Media
[if-you-see-this-DO-NOT-CLICK]
ubc_2007-0071a.pdf [ 12.01MB ]
[if-you-see-this-DO-NOT-CLICK]
JSON: 1.0080783.json
JSON-LD: 1.0080783+ld.json
RDF/XML (Pretty): 1.0080783.xml
RDF/JSON: 1.0080783+rdf.json
Turtle: 1.0080783+rdf-turtle.txt
N-Triples: 1.0080783+rdf-ntriples.txt
Original Record: 1.0080783 +original-record.json
Full Text
1.0080783.txt
Citation
1.0080783.ris
#### Full Text
`A MATHEMATICAL MODEL FOR CEMENT KILNS By PIROOZDARABI B.Sc, Sharif University of Technology, Iran, 2004 A THESIS SUBMITTED IN P A R T I A L F U L F I L L M E N T OF THE R E Q U I R E M E N T FOR THE DEGREE OF M A S T E R OF APPLIED SCIENCE in •THE F A C U L T Y OF G R A D U A T E STUDIES (Mechanical Engineering) The University of British Columbia January 2007 Pirooz Darabi, 2006 ABSTRACT Rotary kilns have numerous industrial applications including cement production. Frequent operational problems such as low thermal efficiency, refractory failure, and poor product quality have prompted extensive efforts to improve and optimize their design. Mathematical modeling and Computational Fluid Dynamics constitute effective tools recently used for these purposes. A cement kiln consists of three major parts: the hot flow, the bed, and the wall. A CFD code which had the capability of simulating the hot gas was developed further to simulate the kiln. In the present work, two 1-D mathematical models are proposed and implemented in the existing CFD code. The first model consists of the steady-state solution for the material and temperature evolution within the bed. The second one simulates tire combustion in the kiln. The tire burning model assumes that tire combustion occurs in two major successive steps, devolatization and char combustion. For the devolatization model, external heat and mass transfer, three parallel reactions, and enthalpy effects are considered the dominant phenomena. The char combustion model considers the enthalpy effect and the external mass transfer. With the aid of the developed model, full-scale industrial cement kilns under steady-state and realistic operational conditions are simulated. In addition, cement kilns with combustion of full scrap tires in the middle of them are mathematically modeled. The limits and feasibility of tire combustion are further explored by running numerical simulations with different tire flow rates and different injector locations. The flow field, temperature distribution and species distribution are presented. Analysis of the results indicates that, with the help of the proposed model, a better understanding of the important processes within cement kilns can be obtained. The model can be used for addressing operational problems and optimizing designs. It is also concluded that successful firing of tires can lead to a cheaper, longer lasting, and less polluting kiln. TABLE OF CONTENTS Abstract ii Table of Contents iii List of Tables viii List of Figures xi Nomenclature xiii Acknowledgements xvii Dedication xviii Chapter 1. Introduction 1 1.1. Cement Production Process 1 1.2. Process o f Cement Rotary K i l n s 5 1.3. Operational Problems o f Cement Rotary K i l n s 6 1.4. Multi-Dimensional Modeling 9 1.5. Motivation for this Work 14 1.6. Objectives 16 Chapter 2. Physical Model and Governing Equations - iii - 18 2.1. Hot F l o w M o d e l 19 2.1.1. Conservation Equations 19 2.1.2. Solution Methodology 21 2.1.3. Boundary Conditions 22 2.2. Wall/Refractories M o d e l 27 2.2.1. Governing Equation 27 2.2.2. Solution Methodology 27 2.2.3. Boundary Conditions 30 2.3. Other Models 31 2.4. Post-Processing 31 Chapter 3. Clinker Formation in Cement Kilns 33 3.1. Introduction 33 3.2. Clinker Formation Process 34 3.3. Objectives 36 3.4. Literature Survey 36 3.5. Mathematical M o d e l 38 3.5.1. Clinker Formation Model 39 3.5.2. Formation of Thermal NO 49 3.5.3. Modified Parent Code 50 x - iv - 3.6. Grid Generation 52 3.6.1. Geometry and Refractory Lining Details for Kiln #1 52 3.6.2. Geometry and Refractory Lining Details for Kiln #2 54 3.7. Boundary Conditions 55 3.8. Solution Methodology 61 3.9. Results and Discussion 62 3.9.1. Validation on Clinker Formation Model 63 3.9.2. Mass and Energy Balance 66 3.9.3. Clinker Temperature and Composition 68 3.9.4. Gas Temperature Distribution 70 3.9.5. NO Emissions 75 3.9.6. Refractory Temperature Distribution 76 x 3.10. Summary and Conclusions 77 3.11. Future W o r k 79 Chapter 4. Mid-Kiln Firing of Tires in Cement Kilns 80 4.1. Introduction 80 4.2. Objectives 83 4.3. Literature Survey 84 4.4. Physical M o d e l 86 - v- 4.4.1. 4.5. Assumptions 86 Mathematical M o d e l 87 4.5.1. Tire Combustion Model 4.5.2. Formation of Thermal NO 92 4.5.3. Modified Parent Code 92 4.6. 87 x G r i d and Boundary Conditions 95 4.6.1. Test Kiln 95 4.6.2. Tire Combustion Model 99 4.7. Solution Methodology 100 4.8. Results and Discussion 100 4.8.1. Mass and Energy Balance 4.8.2. Clinker Temperature and Composition 103 4.8.3. Gas Temperature Distribution 105 4.8.4. Refractory Temperature Distribution 106 4.8.5. NO Emissions 10 7 4.8.6. Overall Impact of Mid-Kiln Firing of Tires on Kiln Performance 108 4.8.7. Sensitivity Analysis 110 4.9. 101 x Summary and Conclusions Chapter 5. 117 S u m m a r y , Conclusions, a n d Recommendations - vi - 120 Bibliography 125 Appendix A 131 - vii - LIST OF TABLES Table 2-1: Components of equation (2-1) 20 Table 2-2: Curve fit data for thermal conductivities of refractory layers 29 Table 3-1: Raw material components 33 Table 3-2: Clinker phases 34 Table 3-3: Clinker formation process 34 Table 3-4: Clinker formation reactions [3] 34 Table 3-5: Simplified clinker formation reactions 41 Table 3-6: Thermal data for clinker formation reactions 43 Table 3-7: Summary of production rates 44 Table 3-8: Summary of reaction rates 45 Table 3-9: Summary of the pre-exponential factors and activation energies for clinker formation reactions 45 Table 3-10: Material information for Kiln #1 57 Table 3-11: Fuel input information for Kiln #1 57 Table 3-12: A i r input information for Kiln #1 58 Table 3-13: Material information for Kiln #2 60 Table 3-14: Fuel input information for Kiln #2 60 - viii - Table 3-15: Air input information for Kiln #2 61 Table 3-16: Loss-free composition of Kiln #1 64 Table 3-17: Loss-free composition of Kiln #2 64 Table 3-18: Comparison between the model prediction and Bogue calculation for Kiln #1 65 Table 3-19: Comparison between the model prediction and Bogue calculation for Kiln #2 65 Table 3-20: Mass and energy balance for Kiln #1 66 Table 3-21: Mass and energy balance for Kiln #2 67 Table 3-22: Clinker temperature at the kiln exit 70 Table 3-23: Summary of the average gas and clinker temperature at the exit and the peak 72 Table 3-24: Dust information for Kiln #1 74 Table 4-1: Composition and kinetic parameters 90 Table 4-2: Proximate, ultimate analysis and lower heating value for the tire rubber [39] 91 Table 4-3: Material information for test kiln 97 Table 4-4: Fuel input information for test kiln 97 Table 4-5: Air input information for test kiln 98 Table 4-6: Input data for tire combustion sub-model for the case with 20% co-combustion 99 Table 4-7: Mass and energy balance: without co-combustion of tire 101 Table 4-8: Mass and energy balance: with 10% co-combustion of tire 102 - ix - Table 4-9: Mass and energy balance: with 20% co-combustion of tire 102 Table 4-10: Summary of important parameters with tire injection at x=50 m and different tire flow rates 109 Table 4-11: Summary of important parameters with 20% co-combustion of tire and different tire injection points 114 Table 4-12: Summary of results for three different grids -x- 116 LIST OF FIGURES Figure 1-1: Cement production line (reproduced form [1]) 4 Figure 1-2: The kiln system 8 Figure 1-3: Heat transfer paths (taken from [2]) 8 Figure 2-1: Coupling of the models (taken from [2]) 18 Figure 2-2: Boundary conditions of the hot flow 26 Figure 2-3: Refractory lining details 29 Figure 3-1: Kiln hood details of Kiln #1 53 Figure 3-2: Refractory lining details of Kiln #1 53 Figure 3-3: Kiln hood details of Kiln #2 54 Figure 3-4: Refractory lining details of Kiln #2 55 Figure 3-5: Boundary conditions for Kiln #1 56 Figure 3-6: Boundary conditions for Kiln #2 59 Figure 3-7: Axial profde of species in Kiln #1 69 Figure 3-8: Axial profde of species in Kiln #2 69 Figure 3-9: Gas temperature distribution in Kiln #1 71 Figure 3-10: Gas temperature distribution in Kiln #2 71 - xi - Figure 3-11: N O distribution in Kiln #1 75 Figure 3-12: N O distribution in Kiln #2 76 Figure 3-13: Refractory temperature distribution in Kiln #1 77 Figure 3-14: Refractory temperature distribution in Kiln #2 77 Figure 4-1: Boundary conditions for test kiln 96 Figure 4-2: Axial profde of species: without co-combustion of tire 104 Figure 4-3: Axial profde of species: with 10% co-combustion of tire 104 Figure 4-4: Axial profde of species: with 20% co-combustion of tire 105 Figure 4-5: Gas temperature distribution 106 Figure 4-6: Refractory temperature distribution 106 Figure 4-7: N O concentration distribution 108 Figure 4-8: Axial profde of species: with 50% co-combustion of tire 111 Figure 4-9: Gas temperature distribution: with 50% co-combustion of tire 112 Figure 4-10: N O distribution: with 50% co-combustion of tire 112 Figure 4-11: Axial profde of species: 20% co-combustion of tire with tire injection point x=80 m 115 Figure 4-12: Axial gas temperature profde for three different grids 117 - xii - NOMENCLATURE A area, m -<4 pre-exponential factor, 1/s B body force vector per unit volume, N/m c molecular concentration, mol/m Cj Constant in k - s turbulence model C 2 Constant in k - £ turbulence model C 3 Constant in k - E turbulence model C p specific heat capacity, JV(kg.K) 2 3 3 C^ Constant in k - s turbulence model d kiln diameter, m E activation energy, J/mol F, tire flow rate, 1/s F additional viscous term vector due to the turbulence, N/m G turbulent energy generation rate, kg/(m.s ) h enthalpy, J/kg visc 3 h convective heat transfer coefficient, W/(m .K) k thermal conductivity, W/(m.K) k turbulent kinetic energy per unit mass, m /s 2 c 2 - xiii - 2 k rate constant, 1/s /, turbulent length scale, m L kiln length, m Lf us m latent heat of fusion, J/kg mass, kg mass flow rate, kg/s m mj mass fraction of species, wt% M molecular weight, kg ^' tire flow rate, tire/revolution P pressure, N/m 2 Pr ; q Constant in k - £ turbulence model heat flux, W/m heat flow rate per unit volume, W/m 2 3 heat flow rate, W r reaction rate, 1/s r kiln radius, m R universal gas constant, JV(kg.K) R production rate, mol/(m .s) R thermal heat resistance, K/W 3 Constant in k - E turbulence model - xiv - * source terms in equation 1-1 t time, s T temperature, K u velocity in coordinate direction x, m/s U mean velocity, m/s velocity, m/s v v v velocity in coordinate direction y, m/s st0 stoichiometric matrix V reactor volume, m ^ fluid velocity vector ^' additional viscous term vector due to the turbulence, N/m w velocity in coordinate direction z, m/s W weight, kg x axial position, m x mass fraction, wt% Y mass fraction, wt% 3 - xv - Greek: r, * diffusion coefficients in equation 1-1 AH reaction heat per unit mass, J/kg AH reaction heat per unit volume, J/m ^ bed angle, degree s turbulent energy dissipation rate per unit mass, m /s 3 2 emissivity £ K Von-Karman constant ^"8 effective viscosity, kg/(m.s) ^' laminar viscosity, kg/(m.s) turbulent viscosity, kg/(m.s) ^ unknown in equation 1-1 P density, kg/m CT Stephan-Boltzmann constant, W/(m .K ) 3 2 a k Constant in k - s turbulence model a e Constant in k - s turbulence model 0 3 4 kiln rotational speed, rpm - xvi - ACKNOWLEDGEMENTS I would not have completed this thesis without the significant help of many great people who have made my life at U B C a pleasant and productive experience. First and foremost, I would like to express my sincere gratitude to my outstanding supervisors, Dr. Martha Salcudean and Dr. Ian Gartshore, for their continuous support and patience during my stay at U B C . I consider myself very fortunate for having the opportunity to work with them. I would also like to thank my co-supervisor, Dr. Steven Rogak, for his guidance. For their unlimited technical support, I gratefully acknowledge, Dr. Jerry Yuan, who always had time for discussion no matter how busy he was and without whom I could not have accomplished this work, and Dr. David Stropky, who generated the complex grids with the software developed by him. I am very glad to have had the chance to work and collaborate with great companies such as Process Simulations Ltd. (PSL) and Lafarge, through PSL. I would also like to thank Maureen Phillips for all her help and support. To my wonderful friends at U B C , I send a special thank you, especially Niusha Javid, Hassan Rivaz, Alireza Forghani, A l i Asadkarami, Amir Nejat, Bijan Azadi, Massimo Diciano, A l i Kashani, Amin Karami, Hamed Mahmudi, Mohammad Sepasi, A l i Soltanzade, and Tingwen L i . They have all contributed to my happy and memorable life in Vancouver. Finally, this work is dedicated to my wonderful parents and my dear sisters for all their love and support. I always felt them near me, even though they were living far away during my period of study in Canada. They have been my main motivation throughout my studies. - xvii - To my - xviii - parents Chapter 1. INTRODUCTION Kilns are often regarded as the heart of cement manufacturing plants. Plant profits are affected by the efficiency of these machines. One of the most important phases of cement production takes place in kilns. In order to produce cement with acceptable quality, it is important to make sure that the material has been properly burned. For these reasons, the utmost importance has been given to kiln operation, and efforts have been made to improve their design in order to get high-quality products with the lowest cost. Rotary kilns, due to having the capability of handling varied feedstock and long residence time, are widely employed by industry in processes such as calcination of limestone and production of cement. Similar to many other industrial processes their operation is not free of problems, among which dust generation, emission of pollutants, low thermal efficiency, low product quality, and refractory failure are considered as the most important ones. This chapter starts with a brief introduction to the cement production process and cement kilns; next, the problems associated with the operation of cement kilns and the need for modeling are addressed. Then there is a literature survey on the models available for rotary kilns, and the motivations leading to the present work are described. At the end, the objectives of this work are presented. 1.1. C E M E N T PRODUCTION PROCESS Cement is a fine, gray substance and is made from a mixture of natural elements such as limestone, clay, sand, and/or shale. Cement is the final product of a cement plant, and is packed and shipped for further usage. Most commonly, cement is used to produce concrete. When cement is mixed with water, it can form a hard, solid material called concrete. Concrete is inert, waterproof, fire resistant, and is the most common construction material in the world. The production of cement, schematically shown in Figure 1-1, consists of three fundamental stages: 1. Preparation of the raw material 2. Production of the clinker 3. Preparation of the cement The first step in the cement manufacturing process is the preparation of the raw materials. Generally, the raw mixture for cement production consists of limestone (the major ingredient), shale, clay, sand, and iron ore. After being quarried from local rocks, the material is loaded into trucks for transportation to the crushing plant. Through a series of crushers and screens, the raw materials are reduced to an acceptable size for clinker production. Then the raw materials are mixed to meet a desired composition and are fed into a mill where the raw materials are ground. The material exiting the mill is called kiln feed and is stored in silos until required for further processing in the plant. After being prepared, the raw mixture goes into the pyroprocessing system. This system either consists of a preheater tower, a calciner, and a dry rotary kiln or only a wet process kiln. In either of them, the same chemical reactions take place: evaporation of the moisture content of the raw mixture, calcination of the limestone, and the reactions between the calcium oxide with the other materials. This results in a final nodular product known as "clinker". The hot clinker will be discharged into a clinker cooler where some of the heat is recovered. The recovered heat will be returned to the pyroprocessing system; this, to some extent, reduces the fuel consumption and improves the energy efficiency of the system. The cooled clinker, which can now be handled on standard conveying equipments, will be stored in clinker silos until needed for cement production. Clinker, with some gypsum and other additives, will be ground together to form cement. Cement will be stored in cement silos and ready for distribution. It will be shipped in bulk or in bags and distributed to customers. Figure 1-1: Cement production line (reproduced form [1]) ,2. PROCESS OF C E M E N T R O T A R Y KILNS Rotary kilns are usually considered to be the most important part of any cement manufacturing plant. They are used by industry to heat the solid material to the point where a required chemical reaction(s) can take place. This heating process needs efficient timing. The residence time of the solid material in the kiln is an important design component and is set by factors such as diameter, length, rotational speed, and slope. A rotary cement kiln, as shown in Figure 1-2, is a very large rotating sloped cylinder 3-5 meters in diameter and 45-165 meters in length, depending on the type of process and production rate. The raw material is fed from the elevated cold end and, due to rotation and gravity, moves down to the far end of the kiln. The other end, also known as fire-end, is hot whose temperature is maintained by combustion of fuel. In the hot zone of a cement rotary kiln, the temperature of the solid material can go up to or beyond 1480 °C. Physical/chemical phenomena that occur inside a rotary kiln include three-dimensional turbulent flow with fuel combustion in the gas phase, three-dimensional flow plus the chemical evolution of species in the bed of material (mud) along with three modes of heat transfer (radiative, convective and conductive) between and within the three components of the kiln, i.e. the hot gas, the mud and the rotating wall. In addition, the hot wall loses some of heat via radiation and convection (either free or forced) to its surroundings. A l l the above-mentioned heat transfer paths are schematically depicted in Figure 1-3. There are two distinct flow phases within a cement rotary kiln: the hot gas and the solid material. The overall flow of the solid material is simple; it enters the kiln from the cold end and flows downward to the other end. On the other hand, the flow within the hot gas is very complex and usually more than one source of air flow exists in the kiln system. Typically, there are three sources for the air inside a kiln system (primary, secondary and in-leakage) and one for the fuel. As mentioned before, at the fire-end of the kiln, fuel combustion occurs where fuel reacts with the oxygen in the air and burns. Fuel is usually delivered via the burner system, which can be a simple pipe or a more sophisticated system; the primary air is also supplied through the burner system. The secondary air comes from the bottom of the cooler which after recovering the energy of hot clinker and passing through the kiln hood, enters the flame area. The third one is the in-leakage air which mostly comes through the air gaps in the kiln hood. ,3. O P E R A T I O N A L P R O B L E M S O F C E M E N T R O T A R Y KILNS Cement production is an energy intensive process; it requires high temperatures and large energy amounts. In addition, the cement industry operates at low thermal efficiencies [1] and due to having very high-temperature environment (especially in the kiln system) emits NO . This suggests the need x to seek opportunities to improve energy efficiency and reduce the emissions of a cement plant. Pyroprocessing is the main step for cement production, during which the necessary reactions leading to cement formation occur. The pyroprocessing step dominates the energy consumption and environmental impacts related to the cement industry, and accounts for approximately 75% of the energy consumption of the plant [1]. Therefore, by improving the pyroprocessing step and more specifically the cement rotary kiln, the greatest reductions in the energy consumption and pollutant emissions of the cement industry can be achieved. As mentioned above, the high temperature and chemical processes that take place in kilns lead to formation of pollutants such as N O . The emission of these pollutants and keeping them under certain x limits (imposed by regulations) are among the important issues for cement kilns operators. Refractory failure is another issue related to rotary kilns, primarily due to the expense of replacing bricks. In addition for replacing brick, the kiln system and plant production, will be shut down and this leads to loss of production. Dust generation from the raw material is an important problem associated with these machines. However, by applying new technologies such as bag fdters and special add-ons, the dust particles are continuously collected and recirculated into the kiln system. This allows for a more effective control on this issue. Figure 1-3: Heat transfer paths (taken from [2]) .4. MULTI-DIMENSIONALMODELING The physical/chemical processes inside a rotary kiln are complicated and poorly characterized and understood. In addition, the operation of rotary kilns is accompanied by problems such as low thermal efficiency, poor product quality and emission of pollutants. Based on the source and the nature of the problem, some of them are easily fixed while the other are sophisticated and require additional investigations. The existence of such problems brings the need for having a good understanding and a good physical basis for the effective control and optimization of the system. In addition, it will be very helpful to develop a tool that can predict the kiln performance with a new firing system, and assess the effects of changes in different parameters on the process, etc. In order to obtain a better understanding of the processes inside a kiln and reach enhanced operational conditions and designs, numerous studies have been performed. These studies have been conducted in two general paths: experimental/empirical and mathematical/analytical. As mentioned earlier, rotary kilns are very large high-temperature cylinders in which complex processes occur. These aspects produce technical and economical difficulties for conducting experiments on full-scale rotary kilns. For instance, it will not be easy or inexpensive to obtain measurement data; in addition, they are often limited. Moreover, the measurement data come with some extent of uncertainty and error. These data are associated with certain operational conditions which will reduce their applicability range. Testing new burners or new operating conditions is generally costly and risks compromising the production that is expected from the kiln. Therefore, it would be very useful to have a computational model with predictive capabilities to guide the experiments, reduce their number, and diminish the risk associated with them. There are some empirical or semi-empirical studies conducted on rotary kiln. With the aid of these studies, the operational condition of a rotary kiln (most generally, the heat and mass balance) or the impact of one key element on the overall process, such as the relation between the kiln rotational speed and residence time of the material can be obtained, see [4] for more detail. However, as with all empirical or semi-empirical correlations, they are limited to certain geometries and process parameters and their extrapolation outside the range used to establish them is risky. Besides experimental and empirical studies, analytical and computational methods can be useful in understanding the design and operation of industrial systems such as rotary kilns. Rotary kilns, due to the complexity of the processes within them, do not lend themselves to analytical solutions. Therefore, the governing equations of the process have been solved numerically using computational methods. Computational fluid dynamics (CFD) is one of these tools and is mainly employed to simulate the processes associated with fluid flow. For this purpose, initially the governing partial differential equations (PDE) of the system are established and then with the aid of a C F D code, are numerically solved. In addition to CFD, mathematical modeling has been found to be a feasible approach to understanding and optimizing complex systems such cement rotary kilns. In a cement rotary kiln, in addition to the fluid flow, other phenomena such as heat transfer, turbulence and chemical evolution should be taken into account as well. After modeling a problem using C F D , detailed useful information, such as velocity, temperature, and species concentration distributions are obtained. Analysis of the C F D results can increase our insight and understanding - 10- over the investigated process; lead us to improved designs with better operational conditions and product qualities and, even increase the speed of the required modifications, if any, made on the process and design of the system. It is important to validate a C F D model (as much as possible) in order to ascertain that it is a reasonably good representation of the system. It is also important to mention that many of the problems related to the kiln operation, such as poor product quality and refractory failure can be associated with the operation and design of the burner which can be improved by a better understanding of the combustion process in the flame area. Therefore, it will be crucial to simulate the fluid flow and fuel combustion in the flame area as detailed and reliable as possible. Over the past years, the physical components of rotary kilns have been separately investigated leading to an improved understanding of the processes occurring in kilns. For instance, the heat transfer [5-8] and the bed motion [9-12] have been the subject of many numerical and experimental studies. There can be many different reasons for conducting separate investigations of only one component of a rotary kiln. The lack of understanding of the other process components and the impossibility of representing and solving the governing equations for the whole system can be considered as the most important reasons for this matter. However, as time passed, numerical tools and methods for simulating such complicated systems became available and hence the understanding of the process within rotary kilns was improved. In addition, the components in a rotary kiln (the hot flow, the bed and the wall) act together. Therefore, in order to develop a complete model for rotary kilns, all the interactions between the components including all the important phenomena have to be considered. This goal can be achieved by developing a fully-coupled model including all the important components of a rotary kiln. - 11 - One-dimensional models can be useful for obtaining a better understanding and improving the kiln operation and design. This goal has been achieved by studying the impact of key variables such as kiln speed or production rate on the kiln operation. However, such models have never been suitable for detailed investigation. For example, they do not have the capability to optimize the burner design. This deficiency is mainly because of the assumption made in one-dimensional models which is the uniform distribution of any parameter in the cross-sections. Although these models do not provide detailed information and lack some essential capabilities, they represent a considerable progress over previous completely empirical practices. M . A . Martins et al. [13], for example, developed a 1-D mathematical model for the simulation of petroleum coke calcination in rotary kilns. They predicted the axial temperature and composition profiles within the bed and the gas and showed better agreement with measured data in comparison to other simulations in the literature. H . A . Spang [14] developed a 1-D dynamic model for cement kilns in which the axial temperature and material evolution within the gas and solid phase were obtained. He indicated that with the applied approach the qualitative behavior of an actual kiln could be simulated well. He was also able to model the impact of chain section on the heat transfer between the solid material and the gas. However, he predicted a too high peak temperature in the burning zone, which is mainly caused by the instability associated with the positive feedback from the endothermic reactions during the formation of clinker in the solid material. In addition, the mass transfer between the solid and the gas as well as the liquid formation within the bed is missing in this work. Similar to other 1-D models, no detailed description regarding the burner area and flame shape can be obtained from this work. - 12- In addition to one-dimensional models, a number of multi-dimensional models have been developed for rotary furnaces. For instance, F. Marias [15] has simulated an incinerator consisting of a 1-D model for the burning bed of solid waste and a 3-D model for the gas phase. He has been able to predict the combustion of the volatile matter within the incinerator as well as the thermal and chemical species evolutions. However, the wall is not included in the model and this, in fact, reduces the accuracy of this work. It should be stressed that, although investigating the quantitative results of this or similar works can improve the understanding on the flow and heat transfer behavior within rotary kilns, multi-dimensional models with more emphasis on clinker formation in the bed or similar processes are still required. In 1995, Bui et al. [16] developed a mathematical model for coke calcining kilns which includes 3-D calculations of material transport along with the fuel combustion in the gas phase. In addition, a 3-D model is proposed for the bed motion assuming two layers of material in the bed: active layer and plug flow layer. Both layers are assumed to behave like Newtonian fluids. However, since the bed chemistry of a cement kiln makes it different from coke calcining kilns, models with cement bed characteristics are still required. Lime and cement kilns, to some extent, have similar bed behaviours. They both have the calcination step in common. But the chemical reactions occurring in the bed of cement kilns are more complicated than lime kilns. M . Georgallis [2] recently developed a three-dimensional steady-state model to predict the flow and heat transfer in the pre-heat and calcination zones of rotary lime kilns. His model consists of three full-coupled sub-models: the hot flow, the bed, and the rotary wall. Similar to Bui's work, his bed model consists of two Newtonian regions in which the calcination - 13 - reaction is modeled and is assumed to be dominated by heat transfer. The model is validated against some experimental data and shows the potential for optimizing the process and operation of a lime kiln. However, the author points out that for solving the equations related to the bed motion, a better computational scheme is needed. He also suggests a 1-D approach for solving the wall model in order to reduce the computation time. In addition, the preheating zone of the lime kiln is missing in his work and this, in fact, reduces the applicability range of the proposed model. His work was related to lime kilns and not cement kilns. The most complete and recent work on cement kilns, to the author's knowledge, was done by Mastorakos et al. [17] in which a 2-D axisymmetric gas model is coupled with 1-D models of the wall and the bed. They claim that clinker chemistry, shell temperature, and the composition of the exhaust gas are well predicted. In the bed model of this work, the clinker chemistry is included and this makes the model different from lime kilns. Similar to any other model their work had uncertainties such as insufficient information given in the heat and temperature range of clinkering reaction, and reaction rates that were set by trial and error to match the expected composition at the exit. ,5. M O T I V A T I O N F O R Tins WORK The literature review revealed the capability of modeling for complex system such as cement rotary kilns. However, all the above models on cement kilns had some shortcomings; for example, Spang had simulated the whole cement kiln in 1-D and had missed some important aspects of the system such as liquid formation and mass transfer between the bed and the gas. In the other article, Mastorakos et al. proposed a model with insufficient information on its clinker formation model. The proposed clinker model in this work is a combination of what was achieved by Spang [14] and Mastorakos et al. [17]. The information regarding the clinker formation reactions such as heat of - 14- reaction, reaction rates, and specific heat of bed material were taken from Spang's work while the liquid formation model, which was missing in Spang's work, was completely inspired by the other work. Another important aspect related to hot furnaces such as cement rotary kilns is the formation of N O , x which was also missing in the above studies. Cement kilns are expensive to run and emit pollutants especially nitrogen oxides. These two negative aspects of them led researchers and plant operators to try alternative fuels, such as waste derived fuels, to save energy and reduce the emissions of pollutants. A literature survey revealed that no mathematical studies on the combustion of scrap tires in cement kilns have been performed before. In this work, combustion of full scrap tires in the middle of the kiln is numerically simulated and analyzed for the first time. Refractory failure is another important problem associated with rotary kilns, and has to be avoided as much as possible. Flame impingement on the wall is one of the major causes. Therefore, by detailed investigation of the processes in a cement kiln, especially in the flame area, an improved understanding of the characteristics of the hot flow and its impact on the refractory lining of the kiln can be obtained. The primary goal of this work was to develop a model to simulate cement rotary kilns, including the clinker formation process and formation of N O . Flexibility and completeness can be considered the x major advantages of the present work over the previous ones. With the aid of this study, the important processes within a cement rotary kiln are quantitatively simulated. Co-combustion of tire in the middle of cement kilns along with its impacts on N O emissions and product quality are investigated x as well. - 15 - Three different industrial kilns are targeted for the simulations carried out in this work: Kiln #1, Kiln #2 and test kiln. Kilns #1 and #2 are used for the simulation related to the clinker production and test kiln, which is basically the cylindrical section of Kiln #1, is used for mid-kiln firing of tires. These kilns were investigated through collaboration with PSL and Lafarge. For this work, essential information for full-scale models consisting of geometry details and boundary conditions were extracted and developed by Jerry Yuan. Based on these details, complex grids were generated by David Stropky, using a grid generation software developed by him. .6. OBJECTIVES The need for development of a fully-coupled three-dimensional cement rotary kiln was described earlier. It was also mentioned that emissions of pollutants, low thermal efficiency, and high operational costs are the major concerns associated with cement rotary kilns. The overall objective of this work is to model mathematically cement rotary kilns to provide a scientific tool for obtaining an improved understanding and ultimately use this mathematical model as a tool to optimize their design and operation. The objectives of the current research can be summarized as, 1. To develop a fully-coupled model capable of simulating the production of clinker in cement rotary kilns 2. To validate the developed CFD model against available plant data 3. To obtain a better understanding of the important phenomena within cement rotary kilns 4. To investigate tire combustion in the middle of cement kilns in order to achieve a better understanding of its possibilities and limitations - 16- Chapter 2 describes the C F D code developed in our research group and used for the simulations in this work. In Chapter 3, full-scale cement rotary kilns are modeled and investigated. The combustion of full scrap tires inside the cement kilns is modeled in Chapter 4. The summary of this work, along with its conclusions and recommendations, are presented in Chapter 5. - 17- Chapter 2. PHYSICAL MODEL AND GOVERNING EQUATIONS The model for cement rotary kilns must include the hot flow in the gas, the process within the bed of material, and the wall. In order to develop a fully-coupled model, these three models should be coupled by exchanging the necessary information such as temperature, heat flux, and flow rate of C 0 2 release from the bed, as shown in Figure 2-1. This coupling is essential and cannot be ignored because the gas flow and temperature computations need the inner wall and the bed surface temperatures as boundary conditions. Similarly the heat flux from the gas is applied as a boundary condition for the bed and the wall. Figure 2-1: Coupling of the models (taken from [2]) In the present work, a steady-state model to simulate the physical/chemical processes inside a cement rotary kiln is presented. The temperature distribution and species concentrations inside the cement kiln at different operating conditions are visualized and/or tabulated. Analysis of these results can - 18- lead to a better understanding of the processes inside cement rotary kilns and can, in fact, aid us in obtaining improved design and operating conditions for a given cement kiln. In this chapter, the governing equations, the solution methodology, and the boundary conditions the gas and the wall are described. The model for the bed of material will be explained in Chapter 3. 2.1. H O T FLOW M O D E L In order to simulate the gas, the governing conservation equations for the fluid flow, heat transfer, and chemical evolutions are written and solved. 2.1.1. Conservation Equations A l l the simulations in this work were basically performed with a C F D solver originally developed by Dr. Nowak in our research group at the University of British Columbia [18] and developed further by Jerry Yuan for lime kilns [19]. The model is based on steady-state assumptions for incompressible fluids which, in addition to solving for Navier-Stokes equations, includes the thermal energy, the turbulent energy and its dissipation (k - e model), and the conservation equations for the mass fractions of chemical species. In addition to the laminar and turbulent steady-state flows, more complex flows, including buoyancy effects and fuel combustion, can also be simulated with this code. The general steady-state conservation equation for the flow under consideration can be written as v-[/#7-r>vf] = s where % are the unknowns, f (2-1) are the diffusion coefficients and 5^ are the source terms of 17 partial differential equations, as given in Table 2-1. - 19- Table 2-1: Components of equation (2-1) Equation Continuity 1 Momentum (3) V Energy h Turbulent kinetic energy k Dissipation rate of turbulent energy e Species (10) mi 0 0 -VP + B + F visc Veff S h cAG-c p^-+c &-<y) 2 k Veff iP k s m Sm t whereV = (u,v,w) is the fluid velocity vector (m/s), p is the fluid density (kg/m ), jU is the effective 3 eff viscosity (kg/m.s), P is the pressure (N/m ), B - (B , B , B ) is the body force vector per unit volume 2 x (N/m ), F 3 W j c = (F ,F ,F )\s vx vy vz y z the additional viscous terms due to the turbulence per unit volume (N/m ), h is the enthalpy (J/kg), k is the kinetic energy of turbulence per unit mass (m /s ), e is the 3 2 2 dissipation rate of turbulence per unit mass (m /s ), G is the turbulence energy generation rate 2 3 (kg/m.s ) and mi is the mass fraction of species (wt%). 3 -20- The species considered in the transport equations consists of seven elementary, whose transport equations are always solved ( 0 , CH4, C 0 , CO, H 0 , H and N ) and three secondary species (NO, 2 2 2 2 2 N H and HCN) which are considered only for N O concentrations. 3 The effective viscosity (/Ll ) is given by eff Me =M,+Mi (2-2) ff where is the laminar viscosity and /x is the turbulent viscosity evaluated from the relation t u, = pC^k le. 2 hi addition, the turbulence energy generation rate (G) is given by 2 {dxj 2 + fdw} 2 + {dz) + (du dv] 2 (dw du^ + dx —- + v. dx + — 2 (dv + —+ dw} y -V + pk)V-V (2-3) The usual values ofthe constants are: C, = 1.44, C = 1.92, C = 1.0, 2 3 0.09, a = 1.0, a = K /[(C 2 k E 2 Ci) Cn" ] = 1.22 (where K = 0.42 is the Von Karman constant), Pr, = 1.0 and 5m, = 1.0. 2 .1.2. Solution Methodology The conservation equations are discretized by the first order Hybrid scheme [20] and are solved with a C F D code based on a finite volume method [21]. In addition, pressure-correction and multisegment techniques are applied. Applying these techniques results in an efficient code, that has the capability -21 - to calculate the steady-state flows in complex three-dimensional regions. A l l the codes are written in F O R T R A N algorithmic language. In the pressure-correction method (for incompressible problems), a time derivative of pressure is added to the continuity equation, which tightly couples the continuity and momentum equations. This coupling makes the system of equations easier to solve. Furthermore, the strategy of dividing the domain into segments results in the capability of local refinement of the mesh to capture the important interactions and processes with more detail and with better numerical efficiency. The code has been validated through a number of problems with complex geometries and has proven to be a reliable solver [22]. .1.3. Boundary Conditions The code allows for dividing the boundary conditions into patches. This patching system allows for separating different boundary conditions from each other. The boundary conditions of the hot gas for a cement rotary kiln are shown in Figure 2-2. In addition to the cylindrical section, the kiln hood, the burner, and the cooler are also depicted in this figure. Boundary conditions for various patches are described as below: aa) Flow rate condition: This condition allows for setting convective flow rates and sets the diffusive terms to zero. -22- For primary air, secondary air, in-leakage air, and the fuel, a mass flow rate condition is imposed. This condition allows for prescribing the uniform distribution of the normal component of velocity on that patch, according to the following formula: v =m/(pA) (2-4) n where v„ is the normal component of velocity (m/s), m is the mass flow rate (kg/s), p is the density (kg/m ) and A is the area of the patch (m ). 3 2 The fuel is assumed to be a combination of coke and coal and the air is assumed to be consisting of 23.13% oxygen, 76.15% nitrogen, and the rest moisture. In addition to the above conditions for the inlets, the values of the turbulent energy and its dissipation rate are calculated with the following equations (2-5) k = \.5U' 3 £ = C ' 4 — (2-6) where k is the turbulent kinetic energy, U is the mean velocity at the inlet, -= is the turbulence intensity (set to 5%), e is the dissipation rate of turbulent energy, is constant (set to 0.09) and /, is the turbulent length scale and is set to 10% of a characteristic geometric length. -23- bb) Wall condition: different types of wall conditions are used for this work: regular wall, refractory, and bed of material. For regular walls, no-slip and no-penetration conditions are used. A no-slip boundary condition says that the fluid in contact with a wall will have the same velocity as the velocity of the wall. A nopenetration boundary condition means no fluid flow, either convective or diffusive, is allowed through the wall. A regular wall condition is set for the hood and cooler wall. In addition, in order to avoid heat loss and concentrate on the process within the cylindrical section of the model, the wall temperature of these two regions is set to the temperature of the secondary air, and therefore no heat transfer is allowed through them. Two other types of wall conditions are used: the bed of material and the refractory which are both employed for the rotary kiln (see Figure 2-2). For refractory, similar to regular walls, no-slip and nopenetration conditions are set. However, in order to calculate for the heat losses, conduction is allowed through it. In addition, due to having negligible rotation speeds in comparison to the velocities in the flow field, the velocity of the rotary wall is set to zero. For the bed of material, again a no-slip condition is assumed. However, because of the chemical evolution within the bed and the release of C 0 and H 0 , a no-penetration condition can no longer be 2 2 used. The mass flow rate of these two components is set as a source term for their transport equations in the gas region close to the surface of the bed. In addition, similar to the refractory wall, the bed has energy interaction with the hot flow, i.e. it is assumed that the hot gas is the heat source ofthe bed. -24- For all the walls, the tangential stress and the values of k and e are imposed according to the wall function method proposed by Launder and Spalding [23]. The emissivity of the refractory and the bed surface, for radiation calculation, is set to a default value of 0.8. cc) Outlet condition: it is assumed that the kiln exit is sufficiently far away and not affected by the downstream conditions; therefore, a zero-gradient boundary condition, for velocity and scalars, is imposed at the exit. dv ^ =0 ^ =0 dn an (2-7) (2-8) hi addition to the above boundary conditions, a constant laminar viscosity of //, = 2*10 kgl(m.s) is 5 considered in the calculations; turbulent Prandtl and Schmidt numbers are also assumed to be constant and equal to 1.0 throughout the whole field. -25- 2.2. WALL/REFRACTORIES M O D E L In addition to the model for the hot gas, the existing C F D code includes the wall model. The wall model is coupled with the hot flow model in order to calculate the inner wall temperature as a boundary condition for the hot flow and also to get the heat loss through the refractory wall. 2.2.1. Governing Equation The conduction equation in cylindrical coordinates is written as follows [24] i a „ ar, 1 a ., ar. a .. ar, . —) +—(k — ) = 0 dO dx dx (kr—) + —(k rdr dr r d0 2 (2-9) Comparing the order of magnitude of the temperature gradients in the radial direction with those in the angular and axial direction and considering the values of the kiln length and diameter, by performing a simple order of magnitude analysis, it can be concluded that, the angular and axial terms can be neglected. Therefore, the above equation is reduced to the following equation 1 3 r dr <i (kr—)\ =ft 0 3 7 dr (2-10) 2.2.2. Solution Methodology The wall of rotary kilns is usually multi-layered with the following arrangement, from inside to outside: coating, brick and steel shell. Therefore, the 1-D conduction equation for the multi-layered refractory wall is written as follows -27- where Q is the conductive heat flow rate (W), T cond T w oul w in is the inner temperature of the wall (K), is the outer temperature of the wall and /?, is the heat resistance of the i * layer (KAV), which is calculated as follows [24] ln out,l (2-12) V 1'".' J 2nk L ; where r , is the outer radius of the i layer (m), r , is the inside radius of the i layer (m), L is the oM in height of the cylinder (m) and k is the thermal conductivity of the i" layer (W/m.K). 1 t The refractory detail of kiln #1, for instance, is shown in Figure 2-3. As illustrated in this figure, at each cross-section the wall of this kiln consists of three different layers, which are specified by different names and have different thicknesses and thermal conductivities. The conductivity of each refractory layer varies with its temperature; this effect is included in the model by writing the thermal conductivity in the form of jfc,. =a,.+fcT + c,T 2 (2-13) where a , b and c, are constants and can be derived by fitting the thermal conductivity data (provided { t by the manufacturer) in curves. Table 2-2 summarises the value of these constants for all the refractory layers in this work. -28- Table 2-2: Curve fit data for thermal conductivities of refractory layers a Type [W/(m.K)] [W/(m.K )] [W/(m.K )] Stainless Steel 14.7 0.016 -0.504x10' Magnel RS 5.23 -0.0019 - Aladin 80 2.00 - - Magnel R S V 5.23 -0.0019 - TZ40 3.06 -0.0003 Z_Dol 3.06 -0.0003 PERMAL7 5.23 Alumina 2 3 -0.0019 - 2.00 - - RT 150 0.4 - - Castable MC22 1.00 - - Coating 4.13 -0.00458 0.154x10 Kricon 32 1.10 - - AS90S 3.60 -0.001 6" Thick Oogjjgg 1' Long, 1.7" Thick Magnal RS bi ; 12" Thick Coating 0' Lang, 1.7- Thick Magnel RSV 6' Long, t.7" Thick Alodin-oO 6' Long, 8.7" Thick TZ40 12' Lang, 1.7" Thick ZDol ' 2" Thick Coating 5 s 2" Thick Coating ag' Long, 1.7" Thick TOSAIumnla II' Lang, t.7" Thick 1.6" Thick StalnlaaaStac Caatabl. MC22 125' Long, B.r Thick RT1S0 27' Long, 1.7' Thick PERMAL7 Figure 2-3: Refractory lining details -29- 160' U n g , 8.7" Thick RT 160 .2.3. Boundary Conditions Two different boundary conditions are considered for the wall model: one as a net heat flux on the inner side and the other one as a heat loss on the outer side. For the inner wall, a net heat flux is calculated and then imposed on each control volume of the refractory, according to the following equation. This net heat flux consists of three parts: convection from the hot gas to the inner wall, radiation from the hot gas to the inner wall, and radiation from the wall to the hot gas. The first two parts are provided by the CFD code and are accounted for by the first term on the right hand side of this equation; while the third part is a function of the inner wall temperature. q =q -oeX,in net where q (2-14) in is the net heat flux passing through the wall (W/m ), q is the sum of convective and 2 net in radiative heat flux taken from the hot gas (W/m ), o~ is Stefan-Boltzmann constant (<x = 5.67xl0~ 2 8 W/(m .K )), Ei is the emissivity of the inner wall (which is set to 0.8 as default) and T 2 4 n w in is the temperature of the inner wall (K). As mentioned before, the outer wall shell loses heat to the ambient air through convection and radiation, which can be formulated as follows lout = K ( W,out - am) T T + °~ o e ( W,ou, ~ am ) T M -30- T ( "1 ) 2 5 where q is the heat flux lost from the shell to the ambient air (W/m ), h is the convective heat 2 out c transfer coefficient (W/m .K), T 2 w oul (K), and e out is the outer wall temperature (K), T is the ambient temperature am is the emissivity of the outside wall (which is also set to 0.8 as default). It is assumed that the kiln shell loses heat by free convection. This heat transfer coefficient, for turbulent cases (Gr Pr^ > 10 ), is calculated with the following equation [24] 9 f (2-16) K=c(T -T J" w a where h is the convective heat transfer coefficient (W/m .K), T is the kiln shell temperature (K), 2 c T am w is the ambient temperature (K), and constant c is equal to 1.24 for horizontal cylinders (W/m .K ). 2 2.3. 4/3 OTHER MODELS A l l the above equations are the general conservation equations for the hot flow and the wall, and are solved by the original C F D code. However, whenever a new process is added to the model, such as clinker formation (Chapter 3) or tire combustion (Chapter 4), the code is modified accordingly. A l l the necessary modifications, such as solution of the new conservation equations and coupling considerations are explained in their relevant chapters and were not included in this chapter. 2.4. POST-PROCESSING After solving the model equations with the fully-coupled C F D code, the results are written in a series of data and diagnostic files. The diagnostic files are usually investigated to make sure the code has -31 - reached a satisfactory convergence and the data files are used for analysis and post-processing purposes. -32- Chapter 3. C L I N K E R F O R M A T I O N IN C E M E N T K I L N S INTRODUCTION 3.1. Cement kilns are utilized by industries to produce clinker. In order to produce clinker, a raw material with a specific mixture is fed into the kiln, and after a series of chemical reactions, known as "clinkering reactions", clinker is formed. The whole process leading to the production of clinker is called "clinker formation" [25]. The hot clinker leaves the kiln and is subjected to a series of postprocesses until cement is produced. These processes are summarized as follows: the hot clinker after being cooled down, partially crunched and homogenized is thrown into the cement mill for final grinding. During the grinding, a small amount of additives (gypsum and limestone) is combined with the clinker, and finally cement with the desired composition is produced. Clinker is different from cement, and these two terms cannot be interchangeably used. Cement is the final product of a cement plant while clinker is the final product of a cement kiln. The raw material of cement kilns, as shown in Table 3-1, is mainly a mixture of calcium carbonate, silicon dioxide and aluminium, and iron oxides [25]. Due to the chemical reactions taking place within cement kilns, the chemical composition of clinker is different from that of the raw material (see Table 3-2). Table 3-1: Raw material components Chemical Formula Name CaC0 Calcium Carbonate 3 Si0 Silicon dioxide (Silica) 2 Al 0 2 Aluminium oxide (shale) 3 Fe 0 2 Iron oxide 3 -33- Table 3-2: Clinker phases Name Shorthand formula Typical percentage Tricalcium silicate (Alite) C S (3CaO.Si0 ) 45-65 Dicalcium silicate (Belite) C S (2CaO.Si0 ) 10-25 Tricalcium aluminate C A (3CaO.Al 0 ) 7-12 3 2 2 2 3 Tetracalcium aluminoferrite 2 3 C AF (ACaO.Al 0 .Fe 0 ) 4 2 3 2 5-11 3 CLINKER FORMATION PROCESS 3.2. In a cement rotary kiln, the solid material with the aid of a 3-4 degree inclination and the rotational motion of the kiln (with an approximate speed of 1-5 rpm), flows toward the burner end. Simultaneously, the hot gas flows above the solid material in the opposite direction. The hot gas is the major heat source for the bed of material. As the solid material flows, it gains heat, and its temperature increases. After a certain temperature, a complex series of exothermic and endothermic chemical reactions (as summarized in Table 3-3 and Table 3-4) occur. Table 3-3: Clinker formation process Temperature [°C] Process 100 Evaporation of free water 900 -1200 Reaction between calcium carbonate or calcium oxide and alumino-silicates 1250 - 1280 Beginning of liquid formation > 1280 Further liquid formation & completion of formation of compounds Table 3-4: Clinker formation reactions [3] No. Standard enthalpy of reaction [kj/kg] Reaction 1 CaC0 ^>CaO + C0 T +1780 2 MgC0 ^>MgO+C0 +1395 3 CaO + Al 0 -> CA 4 2CaO + Fe 0 -> C F 3 2 3 2 2 -100 3 2 3 -114 2 -34- 5 2CaO + Si0 ->C2S 6 CA + C F + CaO -> C AF -732 +25 7 CA + 2CaO->C A +25 8 C S + CaO-^C S +59 2 2 4 3 2 3 The length of a rotary kiln depends mainly on the amount of time required to heat up the raw material to a certain temperature. In cement kilns, based on Table 3-3, the objective is to reach a very high bed temperature, at least 1280 ° C . The kiln length is also dependent on the type of the kiln process: wet and dry. In wet kilns the feed material has high moisture content and therefore, longer kilns are needed. Dry kilns have a preheater tower installed before them. The preheating towers dry and partially calcinate the solid material and as a result, dry kilns are much shorter than wet ones. The highly endothermic reaction of CaC0 — > CaO + C0 3 2 T , the first step in clinker formation, is called calcination. Partially calcinated raw material means that the calcium carbonate, to some extent but not completely, has been decomposed into CaO and C 0 . 2 A cement kiln consists of three distinct temperature zones: the preheating zone, the calcining zone and the burning zone. Each zone is named based on its process. These zones are bounded by the temperature range and the chemical reactions taking place within them [25]. In the preheating zone, the solid material is dried and heated to the point that calcination can begin (900 ° C ) . The calcination zone is self-explanatory; in this zone calcination is initiated and completed. During calcination the C 0 content of the bed is released and transferred to the hot gas. Because of this, the solid mass flow 2 rate will gradually decrease. The end of calcination is the beginning of the burning zone. The burning zone is the hottest part of the kiln, where the temperature of the bed may exceed 1280°C. In this zone, the solid material is melted and liquid is formed. The liquid is principally composed of alumina and iron oxides. It acts as flux in the formation of the calcium silicates of clinker, C S and C S . The 3 2 amount of the molten phase is sufficient enough to cause the material to cohere into small balls of -35 - clinker. Finally, the clinker consisting of C S and C S crystals embedded in a matrix of the aluminates 3 2 and ferrites ( C A and C A F ) is formed. 3 4 This chapter is organized as follows: first, the objectives are mentioned and it is followed by a discussion on the literature related to this work. Then, a 1-D bed model, including the governing material and energy balance equations, and the coupling with the existing C F D code is described. After that, the grids for two full-scale cement kilns including their boundary conditions are presented. Then, the simulations results are analyzed and validated against limited plant data. Finally, summary, conclusions, and future recommendations are presented. 3.3. OBJECTIVES The primary objectives of this work are to develop a scientific tool for simulating cement rotary kilns and to obtain an improved understanding of the important processes inside them. In order to fulfill the above objectives, a simplified one-dimensional model based on the ODEs describing the axial evolution of the temperature and material composition inside the bed was developed. Then, this model was implemented in an existing U B C C F D code developed in our research group [18, 21]. With the aid of this code, two full-scale cement kilns were simulated and limited available plant data were used for validation. 3.4. LITERATURE SURVEY For this work detailed studies on simulation of cement rotary kilns were required. Unfortunately, a literature survey revealed that only a few papers were available. The scarcity of literature is possibly -36- exacerbated by the desire of the companies engaged in this research area to keep their data and experience confidential and therefore gain a competitive advantage. An earlier paper was written by H . A . Spang [14] in which a dynamic model of cement kilns was developed. In this work, a coupled 1-D mathematical model for coal/oil-fired cement kilns including the gas, the solid, and the wall was developed and investigated through a series of plots. The author concluded that with the aid of the proposed model, qualitative behaviour of actual cement kilns can be predicted properly. Although modeling the cement kilns in only one dimension can be too much of a simplification, but this work was a good starting point for future work. In particular, Spang showed that the two exothermic reactions of clinker formation make the numerical solution unstable and, in fact, no truly steady-state solution was obtained by him. This instability was one of the main deficiencies of Spang's work. Many years later, E. Mastorakos et al. [17] simulated a coal-fired rotary cement kiln. Similar to Spang's work, the bed equations were solved in a one-dimensional domain. However, in this work, the flow field, the temperature and mass distribution in the hot gas along with the wall temperature were obtained in 3-D. They revealed satisfactory predictions compared to both the limited measurements in a full-scale cement kiln, and the trends based on experience. However, their clinker model was not complete; for instance, some major thermo-chemical data, such as temperature ranges and the heat of clinker formation reactions were missing. The kinetics data (pre-exponential factors and activation energies) were significantly different from that of Spang [14]. As a matter of fact, E. Mastorakos et al. have mentioned that the pre-exponential factors and the activation energies were chosen by trial and error to get the expected clinker composition at the kiln exit. This, of course, limits predictive capabilities of the model. Despite some deficiencies, the introduction of a closure equation for the liquid formation, based on physical insight, was one of the important contributions of -37- their work and was used in our model. A similar equation for the liquid formation was written by G . Locher [3]. In this work, a 1-D steady-state mathematical model, including the axial material and temperature evolution within the bed, was developed and then implemented in the existing code. It is important to mention that the bed model in this work is a combination of E. Mastorakos et al. and Spang's work. To be specific, the original clinker formation model was based on Spang's model and it was further developed by including the liquid formation model, inspired by what was achieved by E . Mastorakos et al, and the mass transfer between the gas and the bed. The major advantages of the developed bed model in the present work lie in its flexibility and completeness. With the aid of the developed code, industrial-size cement rotary kilns are simulated and analyzed. Based on the quantitative and qualitative analysis of the results, it is ultimately concluded that significant progress has been achieved in the development of a predictive scientific tool for the simulation of cement rotary kilns. Also, the developed model can be used to address operational problems and optimize the design and operation of these complicated machines. ,5. MATHEMATICAL MODEL The formulation and coding of the mathematical model consists of two separate steps: the development of the bed model including the clinker formation process, and its implementation in the existing code. -38- 3.5.1. C l i n k e r F o r m a t i o n M o d e l Initially a set of assumptions and simplifications used in the development of the model are described. Then, the conservation equations for the mass and energy of the bed are derived. 3.5.1.1. Assumptions and Simplifications As mentioned earlier, a cement kiln is a counter-current system in which the solid material flows in one direction and the gas flows in the opposite direction. The C F D code used for the simulations in this work already had the capability of calculating the steady-state flows in complex threedimensional geometries. Therefore, no additional considerations were required for the gas phase. The clinker formation process is complicated and it is not possible to model in all details. So efforts were made to develop a simple model for the clinker formation process, while the essential characteristics and features were maintained. The solid material of the bed flows by a combination of the gravity and rotational forces, both leading to a flow with diffusive and convective movement of particles. Mathematical equations for the particular motions inside the bed are complex [9-11] and are beyond the scope of this work. Therefore, the flow of the solid material required significant simplification. For example, all the mixing and gravity effects were neglected and, similar to the previous works [14, 17], it was assumed that the movement of the solid material is governed by the plug flow rules. A constant axial velocity was chosen for the plug flow of the bed. In addition to this, the bed chemistry was significantly simplified. In reality, chemical reactions in the solid phase take place at the interface between crystals and are limited by both diffusion and kinetics. However, due to the high extent of mixing in the bed and following the path of other researchers, the chemical reactions within the bed were assumed to be only limited by kinetics [14, 17]. Furthermore, -39- the reaction rates were determined by Arrhenius' law. Heats of clinkering reactions, specific heat and latent heat of the solid material were assumed to be constant. It was also assumed that the bed temperature was controlled by the external heat transfer from the hot gas, the heat released due to the clinkering reactions and convection and conduction with the wall. However, the heat transfer between the solid material and the wall was neglected in this work. For a real kiln, due to the mixing of particles and rotation of the wall, the bed surface is not flat and can have different irregular shapes [9]. In addition, the bed height is not necessarily constant throughout the whole kiln. However, for simplicity and avoiding grid generation difficulties, the bed height was assumed to be constant throughout the whole kiln and the bed surface was assumed to be flat. For modeling the clinker formation process, all the reactions listed in Table 3-4 should be considered. This will not be an easy task because it needs dealing with 14 components and eight reactions for which eight rate constants - i.e. eight activation energies and pre-exponential factors - are to be known. These eight rate constants (&,.) should be derived either experimentally or by trial and error, which is difficult and time-consuming. However, H . A . Spang [14] proposed a simpler model with five arbitrary reactions. In this system only the main components of the clinker formation process, both from the raw material mixture and the clinker phases, are involved and are summarized in Table 3-5. This simplified system of reactions is the base for our clinker formation model. -40- Table 3-5: Simplified clinker formation reactions Rate constant Reaction No. CaC0 ^>CaO+C0 3 2 2 2CaO + 3 CS + 2 Si0 ^C S 2 CaO^C S 3 3CaO + Al 0 4 2 2 -^C A 3 3 ACaO + Al 0 + Fe 0 -> C AF 5 2 3.5.1.2. 3 2 3 4 Energy Balance Equations For rotary kilns, heat transfer plays an important role and therefore efforts have been directed to improving its understanding [8]. It consists of radiation and convection between the hot gas, the bed, and the refractory wall, particle-particle conduction and radiation among bed material particles, conduction through the wall and convection and radiation between the kiln shell and the ambient air. In this work, all the above listed terms are accounted for except the particle-particle conduction and radiation in the bed (that are accounted for in a global way in the one-dimensional approximation and are not individually evaluated) and heat transfer between the wall and the bed. The heat transfer mechanisms between and within the gas and the wall were already included in the existing code. However, for coupling the code the ones related to the bed needed to be taken care of. There are two main heat transfer/generation mechanisms for the bed: the first one is the heat transfer between the hot gas and the bed, which occurs through a combination of convection and radiation. The second one is the summation of the heat of the clinkering reactions. The final temperature of the bed depends on the total amount of heat transferred and/or generated via the above mechanisms. The energy balance equation, assuming plug flow condition for the bed and neglecting pressure losses, can be written as -41 - p ^r dV 1 (3-D * Two equations, equations (3-2) and (3-3), were derived form the above equation mC AT = Q (3-2) p (3-3) where m is the mass flow rate of material (kg/s), C is the specific heat of the bed material p (C =1088 J/kg.K) [14], v is the axial velocity of the bed material (m/s), T is the temperature of the p bed (K), Q is the net heat flow to the bed (W), AH is the heat of reaction (J/kg) and r is the rate of r t reaction (1/s). In our modeling, the following strategy was applied to get the bed temperature: first, the net amount of heat flux (radiation and convection) from the gas to the bed was found and multiplied by the area of the bed in each control volume to get the net heat flow. Then, based on the amount of this heat, with equation (3-2) the bed temperature at the exit of the control volume was calculated. The average temperature of that control volume was calculated and, then reaction rates and the total heat generated due to clinkering were calculated. With the aid of equation (3-3), the bed temperature was updated. If the temperature difference between the new one, calculated by equation (3-3), and the old one, calculated by equation (3-2), was less than one degree, it was acceptable; if not, the new temperature was considered an old temperature, and the above procedure was repeated until the above mentioned temperature difference went under one degree in that control volume. -42- It should be mentioned that each clinkering reaction can take place only in a given range of temperature, out of which it is ignored. Table 3-6 contains a summary of the temperature ranges and standard enthalpies ofthe clinkering reactions, taken from [3]. Table 3-6: Thermal data for clinker formation reactions Reaction No. CaC0 ->CaO 1 + 3 2CaO + Si0 2 CS + 6 Clinker 2 873 to 1573 -1.124e6 [J/kgCaO] 1473 to 1553 +8.01e4 [J/kgCaO] 1473 to 1553 -4.34e4 1473 to 1553 -2.278e5 [J/kg CaO] 2 3 2 ACaO + Al 0 +1.782e6 [J/kgCaO] 2 3CaO + Al 0 5 823 to 1233 C0 1 CaO^>C S 2 4 Standard enthalpy of the reaction -*C S 2 3 Temperature range [K] ->C A 3 3 + Fe 0 3 2 Jo/ 3 -> C A F 4 —> Clinker > 1553 /l<? 1 [J/kg CaO] +6.00e5 [J/kg] Positive sign indicates an endothermic reaction and vice versa. 3.5.1.3. Material Balance Equations As mentioned earlier, a cement rotary kiln behaves approximately like a plug flow reactor, where the bed material moves with a constant axial velocity and there is no axial diffusion. Therefore, the general plug flow reactor formula is used for the material balance in the bed: dc. —L dt = d(c.v) - ± - + R. dx (3-4) Assuming a constant axial velocity ( v ) and steady-state condition, the above equation is reduced to dc. v~^ = Rj dx (3-5) -43- where C. represents the concentration of component j with production rate R. The above equation was separately written and solved for each species involved in the simplified clinker formation reactions (as in Table 3-5) excepting for C0 . In order to complete the material 2 balance equations, the production rate (Rj) of each species is required. The production rate matrix (R) can be determined by the equation (3-6) and are summarized in Table 3-7. Table 3-7: Summary of production rates Component Production rate CaC0 = -1 3 CaO R Si0 R>= ~h 2 2 CS C 5 *5 3 . Al 03 r 2 - r 3 - r 4 ~ r 5 = 3 r = -r -r 4 2 5 R =h CA 7 3 Fe 0 = 3 C AF 4 \ - r = h~h 2 2 = R, ~5 r = 5 r According to Table 3-7, the production rate of each component is a function of reaction rates. Therefore, for calculating the production rates, reaction rates should be determined. Reaction rates can be written in different forms such as r. = A'.cJ' cJ ...orr. = 2 1 2 k.Y^Y^... (whereYj represents the mass fraction of component j). Based on Spang's work, the reaction rates were written in the latter form and are summarized in Table 3-8. -44- Table 3-8: Summary of reaction rates No. Reaction rate Reaction l CaC0 ^>CaO + C0 t 3 2 2CaO 3 CS 4 + Si0 -^C S 2 + 2 1 2 '2 r 3 + Al 0 2 3 + Fe 0 2 -> 3 C AF — k Y 5 Y 3 4 - i-y r 4 Y ^l^CaO^CiS 4 ACaO ± 2 r 5 2 ± '3 ^>C A 3 V *'2 Si0 CaO r = kY 3 2 —u v r 2 CaO^C S 3CaO + Al 0 1 CaCOi 4 /1/JOJ CaO y Y 5 CaO Al 0 x 2 3 Fe 0 2 3 As mentioned earlier, the reaction rates are determined by Arrhenius' law; therefore, the reaction constants can be written as below k =A exp(-E /RT) i i (3-7) i where A, is the pre-exponential factor (1/s), E is the activation energy (J/mol) and R is the universal i gas constant A and and is equal to/? = 8.314 (J/g.mol.K)=l.9872 (Callg.mol.R). The values of were taken from Spang [14] and are summarized in Table 3-9. i Table 3-9: Summary of the pre-exponential factors and activation energies for clinker formation reactions Rate constant Pre-exponential factor [1/s] (A ) Activation energy [J/mol] (E ) K 4.55e31 7.81e5 k 4.11e5 1.93e5 k 1.33e5 2.56e5 K 8.33e6 1.94e5 K 8.33e8 1.85e5 2 3 i -45- { Finally, the production rates were plugged into equation (3-5) and nine ordinary differential equations (normalized with respect to the mass of CaO) for the axial evolution of the components involved in clinker formation, excepting for C 0 , were obtained and are as follows 2 dY M CaCO] _ V dx ^ ^ C a r M Ca0 ^i sio cao Y ° y C Y Y Ca0 2 8 Ca dY _ dY ^ M F = <£c . ^ L v 4M 5 CiS = Y Al 0 (3 9) Fe rq -^ 2 M . IT^^c C a O A , A F e A ( } 0 y o Y Cat? dY a f 0-12) 3 F e A C 2 S -J7^K CaO C,S i W Y 2 V y 5 C ° " ° M 2 ~^Tr~ 2 Si0 CaO k 4 Cfl C0 M cs ~TT F n Ca0 (3-10) 4 M A Ca0 v 5 Ca0 3M * ° >>°> dx kY Y ^Y ^ M -^P2M -^VL = dx V k^Y Y ^ ^i cao c s Y '^T - /"} Q\ CaCOs j y dx Y Cat? 0"13) r = f l 0 -46- ^ (3-14) dY,C,A M 3 M CaO dx , dY k.4 Y~„„Y. CaO i A M CtAF y CfAF ^ ' (3-15) l 0 } Y n 16. 4M Ca0 where M , is the molecular weight. The amount of C0 , 2 being released into the gas phase, was determined by writing a proportional equation between the calcium carbonate and the carbon dioxide and is as follows M co = m 2 C<h M Caeo, m (3-17) CaCO, M Equations (3-8) to (3-17) mathematically describe the axial evolution of the composition inside the bed of solid material and were solved to simulate the clinker formation process. 3.5.1.4. Closure Equations and Parameters In addition to the mass and energy balance equations, there are some additional process aspects, such as liquid formation, and parameters, such as the specific heat of solid material, bed angle etc., that should be addressed. * Reaction HI (C3S formation), which is the final step in the clinker formation, occurs only in the presence of liquid phase. In this work, the involvement of reaction IU was judged according to the -47- availability of liquid; if liquid was available, reaction HI was taken into account and the fraction of the related components were updated accordingly. Otherwise, the rate of this reaction was set to zero. The liquid melting inside the bed was modeled by including an additional equation taking into account a new parameter, liquid fraction ( ) . If the bed temperature exceeded a temperature (called the fusion temperature T fus fusion = 1553 A!" [3]), based on the excess amount of heat and the latent heat of = 600U I kg [3], the fusion fraction was calculated; Otherwise, no liquid was formed. A n upper limit of Y fus = 0.3 [3] was set, above which the fusion was neglected. Fusion fractions more than zero meant the liquid was available and C S was forming. 3 It was assumed that the solid material flows with a constant velocity, which was set as v = 0.0127 m/ s [14] for all the cases regardless of the size of the kiln. In addition, the specific heat of the solid material was assumed to be constant ( C = 1 0 8 8 J/kg.K [14]) for all the cases. The bed angle (6) was another important parameter, which cannot be measured and because of its direct impact on the heat flow to the bed was required to be chosen carefully. The higher the bed angle, the bigger the bed surface area will be. The surface area is multiplied by the heat flux from the gas phase, and any increase in its value will increase the heat flow to the bed of material. In summary, by increasing the bed angle, the solid phase temperature increases; this relationship was one of the main challenges we encountered in the modeling procedure. In literature [14, 17], a bed angle of 90 degrees was chosen; however, we noticed that not all kilns can have the same bed angles. This conclusion was reached after very high clinker temperatures were -48- obtained for kilns with bed angle of 0 = nil. This problem was addressed by repetitive adjustment of the bed angle until relatively good results (considering the clinker temperature, composition and exhaust gas temperature) were achieved. Not only was this task hard but also time-consuming; each time the bed angle was changed, the whole grid was regenerated and the entire code execution process, which usually takes at least a few days, was repeated. For instance, for simulating Kiln #1, a Pentium HJ computer with a 1.3 GHz C P U and 2.1 G B of R A M was used. First, the cold flow was computed in 18 hours (2000 iterations). Then, the hot flow including the bed model was obtained in 130 hours (after 10000 iterations). Finally, the post-processing code for N O calculations was x executed and was completed in 6.5 hours. In total, the simulation of Kiln #1 one was completed in 154.5 hours (6.4 days). The issue of bed angle and its influence on the process remains a difficult issue and constitutes a limitation for the model predictive capabilities. For existing kilns there is information available on gas temperature at the exit, clinker composition and temperature at the exit that one can use to tune the model. It is necessary to perform a mass and energy balance together with the detailed C F D computations to make sure that the assumptions made are reasonable. i.5.2. Formation of Thermal N O x "Nitrogen oxides" (NO ) is a general term used for gases which contain nitrogen and oxygen in x varying amounts. Thermal N O refers to the nitrogen oxides formed at very high temperatures, x usually above 2200° F, and is the dominant formation mechanism in hot furnaces such as cement kilns. The formation rate of thermal N O is primarily a function of temperature and the residence time x of nitrogen at that temperature, which can mathematically be predicted by the following equation. -49- d[NO] ^(kMOiW-k^lNOf) —= 2[0]: dt (M^J + MWO]) i ; 3 (gmolIm .s) U> For the simulations in this chapter, the rate data (ki) were selected from the work done by Hanson and Salimian and the partial equilibrium approach was used for determining the concentration of oxygen radicals (see Appendix A for more detail). The source term, due to the formation of thermal N O , is x calculated with following equation S , th where M is wNO N0 =M, w NO kg l{m\s) at the molecular weight of N O , and ^[NO] ^ ^ dt (3-19) formation rate, is computed as mentioned above. 5.3. M o d i f i e d P a r e n t C o d e It is apparent that no clinker formation model can be sufficient on its own and should be incorporated in a model consisting of the gas and the wall as well. This is because the wall, the gas, and the clinker model all interact and impact on each other. For instance, the gas temperature depends on the wall temperature and the clinker temperature. The solid phase absorbs energy from the gas and the wall, and its temperature will depend on the local temperature of both of them. These interactions are important and have to be included in final model. The existing C F D code had the capability of 3-D simulation of the hot flow, and the multi-layered wall. However, a model for the bed was missing. Therefore, the developed bed model was implemented in the existing code. For coupling the interactions between the hot flow and the bed, the -50- mass and heat transfer between the existing code and the bed model were considered. This will also result in the conservation of the mass and energy of the whole system. For including the energy interactions, only one modification was required. This was done by considering the heat transfer between the hot gas and the bed. The heat flux equation, equation (3-20), was written such that a portion of the heat flux was absorbed by the bed and the rest was reflected back to the hot gas. <ln=q»,-°e,A (3-20) where q^, is the net heat flux into the bed (W/m ), q is the sum of convective and radiative heat 2 in flux taken from the hot gas (W/m ), cris Stefan-Boltzmann constant (a = 5.67xl0~ W/(m .K )), £;•„ is 2 8 the emissivity of the bed (which is set to 0.8 as default) and T bin 2 4 is the temperature of the bed (K). For conserving the mass of the system, the mass transfer from the bed to the gas was taken into account. It consisted of two parts: release of H 0 during drying and release of C 0 during the 2 2 calcination. The amount of H 0 and C 0 (in the form of mass flow rate) was treated as an additional 2 2 source term in the transport equation of these components in the gas flow field. For a real kiln, in addition to the release of these two components, the solid material constantly generates dust. The dust particles will eventually become a part of the gas phase; they will drift and leave the kiln at the exit. Including dust circulation is complicated and needs some basic knowledge. i Unfortunately, in the literature no discussions were available on this matter. Modeling dust formation was beyond the scope of this thesis and therefore it was neglected. -51 - 3.6. GRID GENERATION First, the grid for two full-scale cement kilns, including the rotary kiln, the hood, and a length of the cooler, was generated. Then, the boundary conditions were imposed. In addition, the wall of each kiln was implemented according to the refractory lining details provided by the plant. Kiln characteristic data for two industrial kilns were extracted and supplied by Dr. Jerry Yuan. The grids including the proper boundary conditions were generated with a special software developed by Dr. David Stropky. 3.6.1. Geometry and Refractory Lining Details for Kiln #1 The smaller kiln, referred to as Kiln #1, had a cylindrical section with a length of L , =154.65 m and a diameter ofd =3.1394 m . A l l the geometry details were extracted from the related AutoCAD fdes i of this kiln and are depicted in Figure 3-1. In addition, the refractory details of this kiln are exhibited in Figure 3-2. The gird for Kiln #1 was generated based on the given dimensions and considering a bed angle of<9 = 78°. -52- L 2.1082m 7' - 4 1/2" 2.2479m Figure 3-1: Kiln hood details of Kiln #1 E 2 S m .1 £s 6' Thick Coaling Y long, 6.7" Thick Ma grid RS 12* Thick Coating V Long, 8.7- Thick Magn.l RSV 6' Long, 8.7" Thick Aladln-SO 2- Thick Coating W Long, 8.7" Thick 70% Alumnia S3' Lena. 8.7' Thick ZDol 8' Long, 8.7" Thick TZ40 2' Thick Coating 16' Long, 8.7" Thick Caatabla MC22 125' Long, 8.7" Thick RT150 27' Long, 8.7' Thick PERMAL7 Figure 3-2: Refractory lining details of Kiln #1 -53- 1.5" Thick Stainlaaa Steal 160* Long, 8.7" Thick RT 150 3.6.2. Geometry and Refractory Lining Details for Kiln #2 The other kiln, Kiln #2, had a cylindrical section with a length of L =154.82 m and a diameter 2 of d = 4.3586 m . Similar to the other kiln, the dimensions and the detail required for the model were 2 read off AutoCAD files and are depicted in Figure 3-3 and Figure 3-4. Finally, based on the characteristic data of Kiln #1 and considering a bed angle of 6 = 84°, its grid was generated. 7' 5-1/4" 2.267m T 11-1/4" 2.4194m 2.6353m 3 2.7432m 20' 0" 6.096m E 19' 4-9/16" 5.9071m Figure 3-3: Kiln hood details of Kiln #2 -54- 8.1' Long, 8.7" Thick Aladln-M 15' Long, 8.7" Thick TZ40 15' Long, 8.7" Thick AS90S 120' Long, 8.7" Thick RT 150 18' Long, 8.7" Thick Aladln-80 125' Long, 8.7" Thick RT 150 F i g u r e 3-4: Refractory lining details of Kiln #2 After generating the grid, the boundary conditions based on the real conditions of the kilns were applied. The boundary conditions of each model were imposed according to its raw material, fuel, and air input data, and are summarized in the following tables. 3.7. BOUNDARY CONDITIONS Boundary conditions are needed for the primary and secondary air, fuel, refractory, wall and bed. A l l the above boundary conditions are schematically depicted in Figure 3-5 and Figure 3-6 and tabulated in the following tables. Two additional conditions were considered for the cylindrical section and are not depicted in these figures: it is assumed that the kiln exit is not affected by the downstream conditions and therefore, zero-gradient boundary condition is imposed for it. It also assumed that the kiln shell loses heat to the ambient air (withT ami = 3 0 ° C ) through convection and radiation. Further detail for the above mentioned boundary conditions is available in Chapter 2. -55- Table 3-10: Material information for Kiln #1 Inlet Material Information Flow rate [kg/s] 20.788 Temperature [K\ 338 Mass Fraction (% wt) Component CaC0 77.23 CaO 0.00 Si0 13.69 Al 03 , 3.36 Fe 03 1.70 Inert + others 3.85 Moisture 0.17 Total 100 3 2 2 2 Clinker Information Flow rate [kg/s] 13.52 Temperature [K] 1633 Table 3-11: Fuel input information for Kiln #1 Coal/coke Fuel type Flow rate [kg/s] 2.7167 Temperature [K\ 333 Proximate Analysis (%) Moisture 1.24 Volatile 21.86 Fixed Carbon 67.35 Ash 9.55 Total 100 Elemental Analysis (%) -57- 1.24 Moisture Carbon 71.81 Hydrogen 3.73 Nitrogen 1.08 Sulphur 5.45 Oxygen 7.14 Chlorine 0.00 Ash 9.55 Total 100 High heating value (HHV) [U/kg] 27566 Low heating value (LHV) [kJ/kg] 26720 Table 3-12: A i r input information for Kiln #1 Primary Air Flow rate [kg/s] 6.516 Temperature [K] 365 Secondary Air Flow rate [kg/s] 14.346 Temperature [K] 1122 In-leakage Air Flow rate [kg/s] 6.954 Temperature [K] 298 Total air flow rate [kg/s] 27.816 -58- F i g u r e 3-6: Boundary conditions for Kiln #2 -59- Table 3-13: Material information for Kiln #2 Inlet Material Information Flow rare [kg/s] 30.60 Temperature [K] 338 Mass Fraction (% wt) Component CaC0 77.66 CaO 0.00 Si0 13.52 Al 03 3.39 Fe 03 1.98 Inert + others 3.45 Moisture 0.17 Total 100 3 2 2 2 Clinker Information Flow rate [kg/s] 20.014 Temperature [K] 1698 Table 3-14: Fuel input information for Kiln #2 Fuel type Coal/coke Flow rate [kg/s] 4.1467 Temperature [K] 333 Proximate Analysis (%) Moisture 1.09 Volatile 21.36 Fixed Carbon 62.40 Ash 15.15 Total 100 Elemental Analysis (%) Moisture 1.09 Carbon 73.51 -60- Hydrogen 3.74 Nitrogen 1.13 Sulphur 5.26 Oxygen 0.13 Chlorine 0.00 Ash 15.15 Total 100 High heating value (HHV) [kJ/kg] 27696 Low heating value (LHV) [kJ/kg] 26858 Table 3-15: Air input information for Kiln #2 Primary Air 9.7315 Flow rate [kg/s] 361 Temperature [K] Secondary Air 24.678 Flow rate [kg/s] Temperature [K] 1049 In-leakage Air 11.1661 Flow rate [kg/s] 296 Temperature [K] 45.5756 Total air flow rate [kg/s] 3.8. SOLUTION M E T H O D O L O G Y After finishing the initial set-up of the problem, which consisted of the development of the bed model, its implementation in the parent code, generating the grid, and imposing the boundary conditions, it was time to numerically solve the system of the equations for the whole model. The final code was executed in three steps: first, the cold flow field, in order to get only the velocity contours, was created. Then, the code for the whole model, including all the sub-models, was -61 - executed until a good convergence was achieved. The convergence behaviour was judged by continuous monitoring of the maximum heat flux residual of the refractories, the maximum temperature of the solid material, and the other residuals of the enthalpy, mass, etc. Finally, after simulating the problem, including all the primary parameters, the post-processing was performed to obtain the concentration of secondary species especially NO. Due to having very small concentrations, it can be assumed that the reactions leading to formation of NO have no practical influence on the aerodynamics/combustion of the model. Therefore, the computations related to the formation of nitrogen oxides can be decoupled from those of the main code and handled in a post-processor. The possibility of having this decoupling will allow us to freeze the whole flow field and execute only the post-processing in order to calculate for the species related to the formation of nitrogen oxides. This, in fact, will save significant amount of computational time. Due to the strong interactions between the sub-models, an iterative procedure between the predictions for the gas, the wall, and the bed was used. In addition, due to the counter-current flow within cement rotary kilns, a single iteration (in the second step of the code execution) was divided into two subiterations. First, the equations related to the gas phase and the wall were solved; then, moving in the opposite direction, the equations related to the bed were solved. ,9. R E S U L T S AND DISCUSSION Two full-scale cement kilns were simulated with the developed code. In order to obtain an estimate on the performance of the developed model and also improve the understanding on the processes taking place inside a cement rotary kiln, the predictions and results were validated, compared and analyzed. Particularly, the clinker temperature at the kiln exit and the flue gas temperature were -62- compared against limited measurement data provided by the plant. The simulation results are summarized and discussed as follows. 3.9.1. Validation on Clinker Formation Model One of the main objectives of this chapter was to improve the understanding of the clinker formation process inside cement rotary kilns. The clinker composition from the simulation was validated against the composition calculated by the Bogue formula [26]. 3.9.1.1. Bogue Calculation Bogue calculations are the most important and frequently used indicators of the chemical composition of clinker. For this calculation, it is assumed that the final clinker will be composed only of the four main clinker components (as in Table 3-2). According to the Bogue calculation, the major components of the clinker can be calculated by the following equations: C S = 4.07CaO-(7.6Si0 +6J2Al 0 3 2 2 +1.43Fe 0 ) 3 2 C S = 2.87Si0 - 0.754C 5 2 2 C A = 2.65 Al 0 3 2 3 2 C AF = 3.04Fe O 4 2 3 (3-21) (3-22) 3 -1.69Fe 0 3 3 (3-23) (3-24) The components calculated by Bogue are only potential compositions and any change in the cooling rate or burning temperature can change the true composition considerably. However, Bogue results become close to the true composition if the un-combined or free lime content of the clinker is considered in the calculations. -63- 3.9.1.2. Modified Kiln Feed Composition In Bogue calculation, it is assumed that clinker is a composition of the four main components of the kiln feed (CaO ,Si0 ,Al 0 2 K 0,S0 , 2 3 2 3 and Fe 0 ) 2 3 plus inert (a combination of chemicals such as Na 0, 2 etc.). Therefore, the kiln feed composition was modified such that it consisted only of the above components and all the losses (components that leave the bed such as CO2 and moisture) were excluded. This re-calculated composition, containing only the major components plus inert, is called the "loss-free composition" (see Table 3-16 and Table 3-17). Loss-free composition is the one used in Bogue calculations. In addition, clinker analyses usually contain some information regarding un-combined or free lime content. This lime represents the amount of CaO that has not participated in the clinkering reactions and hence its amount should be subtracted from the total lime content in the loss-free composition. As reported from the plant, Kiln #1 and #2 contained 0.89% and 0.95% free lime, respectively. Table 3-16: Loss-free composition of Kiln #1 Component M a s s F r a c t i o n (% wt) CaO 65.68 Si0 20.79 Al 03 5.10 Fe 03 2.58 Inert + others 5.85 Total 100 2 2 2 -64- Table 3-17: Loss-free composition of Kiln #2 Component Mass Fraction (% wt) CaO 66.06 Si0 20.54 Al 02> 5.15 Fe 03 3.01 Inert + others 5.24 Total 100 2 2 2 3.9.1.3. Comparison between Model Prediction and Bogue Calculation Table 3-18 and Table 3-19 contain a summary of results for both kilns. Table 3-18 shows a fairly good agreement between the model and Bogue calculations for Kiln #1 while for Kiln #2, as seen in Table 3-19, the predictions for the first two components of clinker are not that good. Table 3-18: Comparison between the model prediction and Bogue calculation for Kiln #1 CS 63.65 Calculated by Bogue (% wt) 67.78 CS 11.40 8.47 CA 9.15 9.15 C AF 7.86 7.86 CaC0 0.00 0.00 CaO 2.05 0.89 Si0 0.06 0.00 0.00 0.00 Fe 0 0.00 0.00 Inert + others 5.83 5.85 Total 100 100 Component Model Predictions (% wt) 3 2 3 4 3 2 Al 0 2 3 2 3 -65 - Table 3-19: Comparison between the model prediction and Bogue calculation for Kiln #2 CS 53.93 Calculated by Bogue (% wt) 70.09 CS 2 18.15 6.00 CA 8.55 8.56 C,AF 9.15 9.15 CaC0 0.00 0.00 CaO Si0 4.97 0.95 0.01 0.00 Al 0 3 0.00 0.00 Fe 0 0.00 0.00 Inert + others 5.24 5.24 Total 100 100 Component Model Predictions (% wt) 3 3 3 2 2 2 3 3.9.2. M a s s a n d E n e r g y Balance Table 3-20 and Table 3-21 contain a summary of the mass and energy balance for both kilns. The balances were preformed by considering the main sources of mass and energy. As summarized in Table 3-20 and Table 3-21, the mass and energy are well balanced in both kilns. The acceptable balance indicates that the coupling between the main code and the bed model is correctly established; it also means that the code has reached a satisfactory level of convergence. The following observations concerning the energy distribution in studied cement rotary kilns can be reported: about 70% of total heat input is provided by fuel combustion and the rest is from the sensible heat in the gas and feed material. About 20% of total energy is consumed by heating the solid material and formation of clinker, and about 4% of it is lost through the refractories to the atmosphere. Approximately 25% of the heat is carried away with the hot clinker leaving the kiln, and most of the remaining energy is lost by exhaust gases (50%). -66- Table 3-20: Mass and energy balance for Kiln #1 Input Sensible heat in air Mass flow rate [kg/s] 27.8 Temperature \K\ 1122 Enthalpy 23.2 22.18 \MW\ Percent [%] Sensible heat in feed material 20.788 338 7.645 7.3 Sensible heat in fuel 2.7167 333 1.135 1.12 72.59 69.4 104.57 100 Fuel combustion Total 51.304 Mass flow rate \kg/s\ 37.1 Temperature Enthalpy [K] [MW] 1224 50 47.55 1661 27.206 25.87 0.70 0.67 Heat absorbed by clinkerization 21.273 20.23 Shell heat losses 4.265 4.06 Heat of unburned carbon 1.705 1.62 105.149 100 Output Exit gas losses Clinker discharge 13.691 Ash discharge 0.444 Total 51.2347 Table 3-21: Mass and energy balance for Kiln #2 Input Sensible heat in air Mass flow rate [kg/s] 46.2 Temperature Enthalpy [K] [MW] 1049 37.1 22.98 Percent [%] Sensible heat in feed material 30.596 338 11.251 6.97 Sensible heat in fuel 4.1467 333 1.733 1.07 111.372 68.98 161.456 100 Fuel combustion Total 80.942 Temperature Enthalpy [K] [MW] 1267 81.1 50.8 1562 37.867 23.72 2.06 1.29 Heat absorbed by clinkerization 31.455 19.7 Shell heat losses 6.572 4.12 Heat of unburned carbon 0.528 0.37 Output Exit gas losses Mass flow rate [kg/s] 59.4 Clinker discharge 20.148 Ash discharge 0.856 Total 80.404 159.582 -67- 100 .9.3. C l i n k e r T e m p e r a t u r e a n d C o m p o s i t i o n The beginning zone of the kiln is known as the preheating zone where only the water content of the solid material evaporates. As depicted in Figure 3-7 and Figure 3-8, almost half of the kiln (80 m) is occupied by this zone. The next zone is the calcination zone and is self-explanatory; in this zone, the first step of the clinker formation, the calcination takes place. During calcination, calcium carbonate decomposes into calcium oxide and carbon dioxide; the carbon dioxide will travel from the solid phase to the gas phase. This phenomenon is one of the main losses of the clinker formation process and will result in a decrease in the mass flow rate of the solid material as it moves down. Based on Figure 3-7 and Figure 3-8, about 60 meters of the kiln are occupied by the calcination zone. Since calcination is highly endothermic, we assumed that it is controlled only by heat transfer, and the clinker temperature remains constant at 1089 K during the process. After the completion of calcination, the temperature of the solid phase starts rising again and, as summarized in Table 3-5, the rest of the clinkering reaction, along with the formation of liquid will take place. These reactions consist of formation of the four phases of the clinker (C S, C S , C A and C4AF); as listed in Table 2 3 3 3-6, each clinkering reaction can take place in a given range of temperature and will either consume or generate heat. The most important of them is the formation of C2S, due to it being highly exothermic. Referring to Figure 3-7 and Figure 3-8 clinker temperature peaks approximately where the highest flame temperature is attained. After this peak, the clinker cools down and its temperature drops to around 1600 K . The hot clinker leaves the kiln and is transported to the other sections of the cement plant for further processing. -68- Predicted Axial Profile in Kiln #1 Distance from Kiln Hood [m] F i g u r e 3-7: Axial profde of species in Kiln #1 -69- 3.9.3.1. Validation on Clinker Temperature at the Exit A comparison between model result and plant measurement showed that the clinker temperature for Kiln #1 was well predicted. However, for Kiln #2 it was poorly predicted. Further investigation revealed that since inadequate heat has been absorbed by the bed of Kiln #2, its clinker temperature is lower than expected, its free-lime content is relatively high and its product quality is poor. Comparing the predicted clinker temperature at the exit with the plant data (Table 3-22) confirms that more heat should have been transferred to the bed of Kiln #2. Two possible modifications are recommended to compensate this matter: the first solution is to build a new grid with a bigger bed angle. This will increase the surface area of the bed and therefore, the heat flow to it. The second one is including the heat transfer between the wall and the bed in the model. In areas where the inner wall temperature is higher that the bed, such as burning zone, the direction of the heat flow will be from the wall to the bed, which results in a clinker with higher temperature in those areas temperature. Table 3-22: Clinker temperature at the kiln exit Reported from plant Model prediction [K] Kiln #1 1661 [K] 1633 Kiln #2 1562 1698 3.9.4. G a s T e m p e r a t u r e D i s t r i b u t i o n Figure 3-9 and Figure 3-10 show the contour plot of the temperature within the gas phase. According to these figures, as combustion takes place in the flame area, the flame temperature peaks, 2200 K at x=25m from the burner. As the hot gas moves downstream, it loses heat and its temperature decreases, reaching around 1250 K at the exit. Elevated gas temperatures, especially in the flame area, indicates the high potential of cement kilns for formation of thermal N O . x -70- 75 50 125 150 Gas Temperature Distribution in Kiln #1 • ••• Tga«[K] 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 9 z=40tn z=30m z=20m z=10m z=50m Figure 3-9: Gas temperature distribution in Kiln #1 25 75 SO 100 125 150 Gas Temperature Distribution in Klin #2 TgttlKI I •Liasi • I; 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 z=10m z=20m z=30m z=40m Figure 3-10: Gas temperature distribution in Kiln #2 -71 - z=50m 3.9.4.1. Validation on Gas Temperature at the Exit Table 3-23 contains a summary of the maximum and exit temperatures of the gas and the bed. As showed in this table, the predicted gas temperature is around 1250 K at the kiln exit while the measured gas temperature was reported to be around 1100 K . After investigation, we concluded that this difference is mainly due to ignoring dust circulation in the system for both kilns. By considering dust in the gas phase, some heat will be absorbed and carried away by these particles. In addition, dust particles contain calcium carbonate and are calcinated as they move downstream. Calcination consumes heat and this will reduce the gas temperature as well. Therefore, by considering dust circulation in the system the gas temperature will be reduced. The coupling between the gas peak temperature and the clinker temperature is one of the important observations through Table 3-23. As mentioned earlier, we assumed that convection and radiation from the gas are the main heat sources for the bed. Due to very high gas temperatures, it can be concluded that the radiative term is the dominant one. The kiln with the higher gas peak temperature (i.e. Kiln #1) has a higher clinker peak temperature. Another important observation the interaction between the solid phase and the gas at the kiln exit; the kiln with the lower flue gas temperature (i.e. Kiln #1) has a higher clinker temperature at the exit and vice versa. Table 3-23: Summary of the average gas and clinker temperature at the exit and the peak K i l n #1 K i l n #2 Gas temperature at the kiln exit [K] 1224 1260 Gas temperature at its peak [K] 2088 2033 Clinker temperature at the kiln exit [K] 1661 1562 Clinker temperature at its peak [K] 1862 1741 -72- Here is an approximate calculation on how including dust will impact the gas temperature at the exit. Table 3-24, for instance, contains the dust information for Kiln #1. There are two energy aspects that have to be considered: difference of sensible heat of dusts, and required heat for dust calcination. The enthalpy different between the return, material, and exit dust will be: AH = H -H =H -H -H 2 l e r m = (3.944 fcg/s)(l .08 kJ/kg.K)(l063K) -(2.639kg/s)(l.08 -(\.10kgls)(\m kJ I kg.K)(361 K) kJ/kg.K)(33\$K) = 4.53-1.05-0.62 = 2.86MW In addition to this, dust carries some C a C 0 . Calcination consumes heat and this heat has to be 3 accounted for as well. The amount of available C 0 in dust can be calculated based on given loss of 2 ignition (LOI) and mass flow rate of each part. Therefore, the amount of net generated C 0 due to 2 dust calcination can be calculated: m COi = ( 2 . 6 3 9 ^ / ^ X 0 . 0 9 7 ) + (U 0 kg / s)(03434) - (3.944kg / * ) ( 0 . 1 1 3 2 ) = 0.256 + 0.584 - 0.446 = 0394kg I s The amount of required heat for dust calcination will be E = (0394kg I s)(—)(1679 44 Ulkg) = \ .5MW -73- Therefore, a total amount of 2.86+1.5=4.36 M W of energy will be carried away by dust which has to be subtracted from the total flue gas enthalpy. The equivalent reduction in gas temperature at the exit will be: = AT 8 - = = s m.C. \A5K (21Mgls)(\mkJ Ikg.K) Considering this, the flue gas temperature will drop to 1224-145=1079 K . This estimated flue gas temperature is now close to measured flue gas temperature at the plant (1100 K ) . So it can be concluded that considering kiln dust can bring the gas temperature down to values comparable to plant data. Same calculations can be performed for the other kiln. Table 3-24: Dust information for Kiln #1 Return Dust Information Flow rate [kg/s] 2.639 Temperature [K] 367 LOI 0.097 Exit Dust Information Flow rate [kg/s] 3.944 Temperature [AT, 1063 LOI 0.1132 Feed Material Dust Information Flow rate [kg/s] 1.7 Temperature [K] 338 LOI 0.3434 -74- 3.9.5. N O Emissions x Nitrogen oxides are hazardous pollutants. Their emissions have to be reduced and controlled as much as possible. Therefore, it is essential to estimate their emissions for any system that emits these pollutants, such as cement kilns, and try to reduce them by applying different technologies. As depicted in Figure 3-11 and Figure 3-12, due to having a very high temperature (over \800°K), the NO concentrations in the flame area can be very high, 2600 ppm and 3600 ppm for Kiln #1 and #2, respectively. Moving downstream, the gas temperature decreases and hence the N O concentrations reduce and are 2020 and 2100 ppm at the exit of Kiln #1 and #2, respectively. K (I 25 SO 75 MM NOx Distribution in Kiln#1 Figure 3-11: NO distribution in Kiln #1 -75- 125 150 0 25 50 75 100 125 L, 150 NOx Distribution in Kiln #2 M z =i0m z=20m 9 Mf z=30m z=40m z=50m Figure 3-12: NO distribution in Kiln #2 3.9.6. Refractory Temperature Distribution As explained earlier, the hot gas, the wall, and the bed interact and have impacts on each other. The inner wall temperature receives heat from the gas; some of this heat is absorbed and travels through the shell, and the rest will be reflected. In addition, the inner wall interacts with the bed, which as described before is neglected in this thesis. Due to the close interaction between the hot gas and the inner wall, similar trends can be observed in both of them. This fact can be confirmed by comparing the inner wall temperature distribution of these two kilns, Figure 3-13 and Figure 3-14, with the temperature distribution within the hot gas, Figure 3-9 and Figure 3-10. The inner wall temperature peaks where the maximum gas temperature is attained and will drop as the gas flows downstream. Due to the low ambient temperature, the outer wall loses heat and has a much lower temperature compared to that of the inner wall. -76- 50 100 I I mar Refractory | Temperatoro |K] 2 0 Q m 4 Q 0 5 0 Q I ^ 7 M m 150 ggg 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 Kiln #1: Refractory Temperature Distribution 20 40 60 80 100 120 140 160 180 200*220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540 Figure 3-13: Refractory temperature distribution in Kiln #1 Kiln #2: Refractory Temperature Distribution Terra>erature [K] 20 40 60 80 100 120140160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540 560 Figure 3-14: Refractory temperature distribution in Kiln #2 3.10. S U M M A R Y A N D C O N C L U S I O N S A one-dimensional mathematical model including all the main phenomena associated with the formation of clinker in the bed of material was developed and then implemented in a CFD code. With -77- the aid of this code, two full-scale cement rotary kilns under realistic operational conditions were targeted and simulated. The modeling results were analyzed and compared against limited available data. Kiln #1 showed good agreement with the experimental evidence of clinker quality; on the other hand, Kiln #2 had low clinker quality and temperature which were not observed in practice. The simulation result indicated a tight coupling between the amount of heat transfer to the bed and the clinker formation process. The likely reason for poor prediction for Kiln #2 is an incorrect evaluation of the heat flux to the bed as a result of an inadequate estimate of the bed angle. An overall mass and energy balance was performed for both kilns. The analysis of the energy balance revealed that, for the cement kilns studied in this work: the combustion of the fuel accounts for 70 % of the total heat input and the rest is compensated by the sensible heat of the feed material and air. The heat lost via flue gases accounts for 50% of the total heat output, the solid material absorbs about 40-45% of the total energy and around 5% of the heat is lost to the ambient air through refractories. Moreover, for further investigation of the simulation results, a series of plots were developed. These plots included the contour plots of temperature, N O distribution within the gas phase, and temperature distribution in the wall along with the axial plots of the average concentration of important species in the gas and solid phase. Finally, it can be concluded that considerable progress has been achieved to develop a predictive tool for simulating the important processes in full-scale cement rotary kilns. By investigating the simulation results of two industrial kilns under realistic operational conditions and quantifying the energy and mass flows in various parts of the system, a better understanding of the cement kiln operation was achieved. It was shown that a reasonable estimate of the bed angle is necessary in order -78- to obtain good predictions. Also, the developed model can be used for simulating and studying cement kilns with different conditions, such as with a new firing system, a new type of fuel, a new energy distribution layout, or even a combination of them. The subject of the next chapter of this thesis is "Mid-Kiln Firing of Tires in Cement Kilns" in which the combustion of full scrap tires in the middle of the kiln is modeled and its impacts on the kiln operation are studied. 3.11. F U T U R E W O R K hi this work, some phenomena including dust circulation, heat transfer between the bed and the wall, and rotation of the kiln were neglected. Considering these phenomena can result in a more complete model with improved results and can contribute to a better understanding of the whole process. More specifically, considering dust circulation can lower flue gas temperature. In addition, including the heat transfer between the bed and the wall will have direct impact on the bed temperature and might bring the model predictions closer to the plant data. The kiln rotation is also important and should be implemented when heat transfer between the bed and the wall is considered. In addition, more plant data are required for the validation of the simulation results such as the gas flow field, temperature distribution of the wall, and the bed height. -79- Chapter 4. 4.1. M I D - K I L N FIRING OF TIRES IN C E M E N T K I L N S INTRODUCTION Approximately 2.5x10 tones of tire are produced in North America each year [27]. Due to an 6 increase in vehicle usage, this rate is increasing. Scrap tires are not easily recycled and traditionally for many years, have been disposed of in stockpiles or landfills. Dumping tires has potential health and environmental hazards. For instance, tire stockpiles can catch fire (which are often hard to extinguish) and emit pollutants. In addition, wet stockpiles of tire can be breeding sites for mosquitoes. Some countries are planning to ban landfdling of tires in the near future. A l l this brings challenges for thinking of possible ways to reuse them. Tires are made of rubbery materials and have high carbon content and high heating values. Scrap tires are cheap and available. These factors make tires a good choice for being utilized as fuel. Fuel from tires is also known as tire-derived fuel (TDF). Since for most of the cases, tires are burned as secondary fuel, the term "co-combustion of tire" is also frequently used. Tire-derived fuel is the largest market for scrap tire utilization. It accounts for about 82% of scrap tire recycling [28]. Cement kilns have excellent combustion characteristics for burning scrap tires. They operate at very high temperatures and have long residence times. With an adequate supply of oxygen, complete burnout of tires can be achieved inside them. They also have the potential to use 100% of scrap tire in a completely efficient manner [29]; the ash from tire combustion is combined into clinker and does not require any disposal. In addition, tire combustion in cement kilns seems to be environmentally sound; based on a report [30], co-combustion of scrap tires results in either a decrease or no change in the emissions of the pollutants such as C O , N O , S 0 and particulate matters. Tire-derived fuel in x cement kilns is experiencing an increasing trend. -80- 2 One of the main objectives of this work is to simulate the combustion of tires inside cement kilns (which is also known as "mid-kiln firing of tires"). It should be mentioned that the expression "midkiln" does not necessarily mean the middle point of the kiln. In mid-kiln firing of tires in cement kilns, full scrap tires (in one piece without any pre-processing), via a special conveyor, will be elevated and thrown (injected) somewhere inside the kiln. Originally, there is no opening on the kiln shell foreseen for this matter; hence, a special opening will be created at a designated axial location. Through this opening a number of tires will be thrown into the kiln. Tires will fall and rest on top of the solid material and move with it to the downstream of the kiln (where the main fuel is burned). As they move down, they gain heat and bum. In co-combustion, the amount of heat being released from tire combustion will compensate the reduction in the mass flow rate of the original fuel. It was mentioned that full tires are injected into the middle of the kiln, but it was not explained why tires are in one piece and not shredded, and why in the middle of the kiln rather than in a hot area such as the flame area. The supply of tire for cement kilns can be done in two different ways: whole and chipped. Both have their own advantages and disadvantages [31]. Combustion of chipped tire particles has been employed at many facilities worldwide. Shredded tires can be introduced into the burner zone of the kiln and can be consistently delivered to the fuel system. But the high cost of shredding often makes it uneconomical. On the other hand, burning the whole tire is usually cheap; there is no shredding cost involved. However, in order to inject full tires in the middle of the kiln, a new fuel entry point and a special fuel delivery system will be required. This, in fact, imposes some additional cost. Some kilns have chain curtains (which act as a heat exchanger) at the feed end. The chain section permits only the flow of fine materials and acts as a barrier to the movement of large tires. Therefore, for kilns with chain section, large tires cannot be injected at the feed end and have to be injected somewhere else. -81 - One of the main reasons for burning the tires in the middle of the kiln is its influence on the energy distribution of the system. Before introduction of the tire, the fuel is burned in one zone (flame area). By co-combustion of tires in the middle of the kiln, the energy required for the processes within the kiln will not be concentrated in one spot and will be distributed. This is known as fuel stagedcombustion. Staged-combustion is effective in reducing N O emissions because it lowers the x concentration of oxygen and nitrogen in the primary combustion zone, it reduces the peak flame temperature by allowing for gradual combustion of the fuel, and reduces the amount of time the fuel and air mixture is exposed to at high temperatures. However, the amount of co-combustion of tires is limited (10-20% energy-based) for different reasons. First, as mentioned earlier, tires will be thrown into the middle of the kiln through an opening which is sized based on the size of the tires. Its size is also optimized such that the amount of energy and mass lost through it is minimized. In addition, tires are fed into the kiln discontinuously; whenever the opening meets the end of the conveyor, it will be opened to let the tires in, which takes some time. Based on the size of the opening and the available time for having the entrance open, there will be an upper limit for the number of the tires that can be thrown into the kiln. Thus, the percentage of co-combustion of tires is practically bounded by an upper limit. In addition to this practical obstacle, because of the process and the flow of the cement kilns, at least a certain amount of energy should be released in the flame area to trigger the clinkering reactions. "... Field studies have also confirmed that T D F (Tire-Derived Fuel) can be used successfully as a 10-20% supplementary fuel in properly designed solid-fuel combustors with good combustion control and add-on particulate controls ..."[32]. Another good question is " i f the energy distribution is good and can reduce the emissions of pollutants, why not have as many openings as possible?" This is mainly because of their disadvantages. Significant amount of heat and mass can be lost through an opening, and each opening -82- will need its own conveyor and imposes some costs. In order to maintain a plant at profitable and efficient conditions, it is essential to minimize the losses and operating cost. Also, the location of the opening is important and cannot be at any axial position. Tires should be injected at locations where complete combustion of them can be ensured (where gas temperature is high enough) and is practical. For example, due to excessive gas temperatures and thick coating in the burning zone, placement of the injector in that section would not be practical. Because of the very high temperature within the burning zone, some of the solid material (around 20-30%) will melt and form liquid; as the kiln rotates and the material flows some of this molten material will stick to the inner wall and coat it [25]. In some areas of the burning zone, the coating layers can be thick which will cause practical difficulties in placing the opening. In addition, as mentioned earlier, tires cannot be injected into the chain section either. For these reasons, tires are usually introduced at the beginning of the transition zone, between the burning and calcining zone where high gas temperatures, not excessive, are encountered. Because the transition zone is usually short, it would not be practical to have more than one opening. This chapter is organized as follows: first the objectives of this work are mentioned which are followed by literature survey. After this, the physical and mathematical model for tire combustion is described. Then, the geometry and boundary conditions for a test kiln are summarized. Later, the results, including a parametric study along with some tables and figures, are presented and analyzed. Finally, summary and conclusions are explained. 4.2. OBJECTIVES The objectives of this chapter are to develop a tool to simulate scientifically the tire combustion phenomena, improve the understanding of combustion of tire in rotary kilns, study its impacts on the -83- product quality and process, explore the limitations on the flow rate and axial location of tire injection, and finally quantify the stack emissions mainly nitrogen oxides. In order to fulfill the above objectives, a simplified one-dimensional model for tire combustion was developed and then implemented in the U B C C F D code developed in our research group [18, 21]. A test cement kiln with different operational conditions was simulated and the final results were compared and analyzed. ,3. L I T E R A T U R E SURVEY A literature review showed that no investigations regarding the combustion of full tires were done let alone on their combustion in the middle of the cement kilns. The combustion of tire takes place in two steps: the first step is called "devolatization" or "pyrolysis," during which the volatile content changes phase (from solid to gas); the second step, known as "char combustion," occurs as the fixed carbon of tire reacts with oxygen and releases heat [33]. The occurrence of devolatization is the key to the availability of the fixed carbon and no char combustion occurs if no volatile matter has been released before. It is also important to mention that in an inert atmosphere (in the absence of oxygen) only pyrolysis without char combustion takes place. The combination of these two processes is known as "tire combustion" phenomenon and any mathematical model proposed for tire combustion should include both of them. Different authors have investigated the pyrolysis and combustion of shredded (small) tire particles [34-37]; however, for our work, literature with emphasis on the devolatization and combustion of large tire particles was required. To the author's knowledge, there were no articles on combustion of large tire particles; however, two articles on the pyrolysis of large tire particles were found. The first one was by J. Yang et al. [38] in -84- which a two-dimensional transient pyrolysis model for large tire particles (cylindrical samples with a diameter of 40 mm and a length of 60 mm) was introduced. Their model included a combination of heat transfer, mass transfer, and pyrolysis kinetics. They concluded that the slowest processes are the conduction and mass transfer in the reaction zone. Therefore, the rates of these processes control the overall rate of the tire pyrolysis. Dealing with a large size of tire particle samples with a high content of volatile matters (VM) differentiates their work from other pyrolysis models. The model was tested against experimental data and good agreement was found. The second article was recently published by M . B . Larsen et al. [39]. They performed an experimental study in order to investigate the effects of particle size and surrounding temperature on the devolatilization rate of large tire particles in which cylindrical tire particles with diameters between 7.5 and 22 mm were tested. They found a significant effect of particle size and surrounding temperature on the rate of devolatilization, i.e. larger particles and lower temperatures increase the devolatilization time. They also developed a mathematical model for the devolatilization process including internal and external heat transfer, three parallel devolatilization reactions, and reaction enthalpy effects. They also revealed good agreement with experimental data. The two latter articles were both on large tire particles; but they both lacked the combustion step, and the particles used by them were still not big enough for our work. Although none of the above literature was giving adequate information and data for our case, their process insight and thermal and kinetics data were inspiring and useful. With the aid of the above articles and a basic knowledge on char combustion, a one-dimensional mathematical model for the tire combustion (including devolatization and char combustion) was developed and implemented in our C F D code. With this model, co-combustion of full scrap tires inside a test cement kiln with different operational conditions was simulated and investigated. -85- 1.4. PHYSICAL M O D E L In mid-kiln firing of tires, full tires are thrown in the middle of the kiln and burned. This description is rather general and needs more explanation. Further explanation on the whole phenomenon is given as follows: full tires, after being injected through an opening into the kiln, rest on top of the bed and move with it towards the flame area. Meanwhile, the tire gains heat from its vicinity, mainly the hot gas flowing on top of it. When the surface layer becomes hot enough, the devolatization reactions begin and some volatile matter is released into the atmosphere. As the heat is transferred towards the core of the tire, the devolatization reactions develop deeper into it. At the same time, the fixed carbon content of the tire reacts with the available oxygen, and some heat will be released; this step is known as "char combustion". This whole process goes on until nothing but ash is left. .4.1. Assumptions For mid-kiln firing, full tires are thrown into the kiln through an opening. Practically, this opening decreases the efficiency of the kiln by allowing the outflow of mass and heat. However, in order to facilitate the modeling, this opening was not included in the physical model. In this work, tire injection part (from the point they are thrown until they rest on top of the bed) was neglected, and tires were introduced on top of the bed. From this point it was assumed that tires move downstream with the same velocity as the bed has until they are completely burned and nothing but ash is left. The remaining ash leaves the kiln at the exit. It also assumed that the tires are continuously fed to the kiln. The rate is calculated based on the number of tires injected per revolution and kiln angular velocity. Only one heat source for tires is considered and that is the radiative term from the hot gas on top of them. -86- In addition, the effect of tire on the gas flow field is considered negligible and no physical thickness is accounted for tires. However, the surface area of tires is a key parameter that has a direct impact on the heat flow calculations to tire and is addressed when the heat transfer equation is introduced. 4.5. MATHEMATICAL MODEL In order to complete the code and include tire combustion in it, two general steps were taken. First, the 1-D model for combustion of tires was developed; then, this model was coupled with the existing code for cement rotary kiln (which was developed in the previous chapter). 4.5.1. T i r e C o m b u s t i o n M o d e l Mathematically, tire combustion consists of two steps: a set of devolatilization reactions and a final step of char combustion. Both steps are governed by internal and external mass and heat transfer, and kinetics of chemical reactions. Modeling tire combustion including all the above mentioned mechanisms would not be possible because of the lack of data and knowledge. Therefore, a simplified model based on the governing mechanisms, that limit the rate for each step, was developed. 4.5.1.1. Heat Transfer First of all, according to the following formula, the net heat flux absorbed by the tire was calculated. Q^o^cTt-T?) -87- (4-D where g , i s the net heat flow rate (W),(7 = 5.671*10 W / ( m . K ) i s the Stephan-Boltzmann 8 2 4 ne constant, A, is the area of tire (m ), e is the tire emissivity, T is the gas average temperature (K) and 2 t T is the tire temperature (K). Tire surface area was calculated based on the diameter of tire and tire t rim. The area was assumed to be constant. It should be mentioned that tires do not necessarily occupy the full surface of the mud; the amount of coverage depends on the ratio of the tire surface and the mud surface in each control volume. Knowing the net heat flux and tire mass flow rate, the new tire temperature at the exit of each control volume was calculated with the following equation T , =T +Q /(m.C ) l new where T is tnew lMd nel (4-2) p the new tire temperature (at the end of the control volume) in Kelvin, T told is the old tire temperature (at the beginning of the control volume) in Kelvin, Q is the net heat flow rate (W), net m is the tire mass flow rate (kg/s), C is the specific heat of tire (C =2000 J/kg.K [39]). p p The tire mass flow rate is calculated as follows m = (N W .(O)/60 r (4-3) t where TV, is the number of tires injected per revolution of the kiln (tire/rotation), W is the weight of t one tire (kg/tire) and co is the rotational speed of the kiln (rotation/min). -88- It is also assumed that no temperature gradient exists within the tire and the whole particle is at a T single average temperature. This average temperature ( ' ' +T new ''" ) was used for calculating the rate of devolatization reactions in each control volume. '.5.1.2. Devolatization Process Devolatization is the process of transformation of volatile matter content into gas. In the present work, inspired by others such as M . B. Larsen et al. [39] and J. Yang et al. [38], it is assumed that tire devolatization consists of three independent parallel reactions and takes place in a single step according to the following equation Solid —> t gaSj, i = 1,3 (4-4) We assumed that the external heat transfer and chemical reaction kinetics were the controlling parameters for the devolatilization process. Mass transfer and internal heat transfer were neglected. As explained by M . B . Larsen et al. [39], these assumptions have been successfully used by other authors in the devolatization modeling. Mathematically, the devolatization process was formulated with a first-order Arrhenius equation as [39] dx(-EA —'- - - A exp —'- x, , i = 1,3 dt ^ {RT J ' ; where x is the mass fraction of the volatile component. t -89- (4-5) After being initiated, the devolatization goes on until no volatile matter is left. Table 4-1 contains a summary of the initial mass fraction for each volatile component and the kinetic parameters for each reaction, taken from M . B . Larsen et al. [39]. The summation of the initial fractions of three volatile components is equal to 64%, which indicates the total volatile matter content of the sample tire used in this work. Table 4-1: Composition and kinetic parameters Parameter Value Unit Initial mass fraction •*i,o 0.15 kg/kg of volatile specie i X 2,0 0.13 kg/kg 0.36 kg/kg Pre-exponential factor and £, 49.1 kJ/mol activation energies A 100 1/s E 207 kJ/mol 2 3.93xl0 £ 3 3 A 4.5.1.3. 14 1/s 212 kJ/mol 1.05x10° 1/s Char Combustion Based on the proximate analysis, the fixed carbon content of tires is usually high. For instance, as indicated in Table 4-2, the sample tire for this work contains 29.7% fixed carbon. This means that char combustion is important and cannot be neglected. For char combustion, the amount of available char at each control volume was required. As explained earlier, carbon becomes available only if the devolatization has already happened; in other words, there will be no carbon to burn if no volatile matter has been released before. In order to quantify this, -90- we assumed that the amount of exposed carbon for combustion, based on the proximate analysis, is proportional to the amount of volatile matter released in each control volume. After being exposed to the air, the carbon reacts with oxygen and burns. The reaction of fixed carbon with oxygen is char combustion. For simplicity, we assumed that oxygen availability is the only controlling parameter for this phenomenon. If enough oxygen is available, all the exposed carbon reacts with the oxygen molecules and CO2 is formed. This C 0 is released to the gas. If the amount of available oxygen is not 2 adequate, a portion of it burns and the rest combines with the unburned carbon content (if any) transported from the previous control volumes and is transferred to the next control volume. Char combustion proceeds until no fixed carbon is left. The end of char combustion is the end of tire combustion and after that, nothing but tire ash is left. The char combustion reaction ( C + C> —> C0 ) 2 2 is exothermic and as a result, the gas temperature increases in the adjacent area. The heating value of this reaction is 32.79 MJ/kg. Table 4-2: Proximate, ultimate analysis and lower heating value for the tire rubber [39] Proximate analysis Ultimate analysis Truck tire rubber Unit Volatiles 64.1 wt% Fixed C 29.7 wt% Moisture 1.2 wt% Ash 5.0 wt% C 82.0 wt% H 6.71 wt% S 1.35 wt% N 0.32 wt% 0 3.42 wt% 35.0 MJ/kg Lower heating value -91 - 1.5.2. Formation of Thermal NO x In the existing code, the formation rate of thermal NO* was mathematically predicted by the following equation WO] — dt = 2[0]- (kA[0 ][N ]^_ [N0f) ; - (gmollnf.s) [k [0 ] + k_ [NO]) 2 2 2 2 2 (4-6) x For the simulations in this chapter, the rate data (k ) were selected from the work done by Baulch et ; al. and the equilibrium approach was used for determining the concentration of oxygen radicals (see Appendix 1 for more detail). The source term due to thermal N O formation is calculated with the x following equation V - v o = ,NO kg/(m\s) M W where M wNO at is the molecular weight of N O , and ^INO] ^ ^ dt (4-7) formation rate, is computed as mentioned above. 1.5.3. Modified Parent Code In order to simulate the combustion of tire in the middle of the cement kilns, the developed code for modeling cement kilns (as in previous chapter) was modified. The modifications included the interactions between the hot flow and tires and consisted of heat and mass transfer between them. For including the mass transfer, the gaseous component being released or absorbed due to tire combustion were accounted for. After calculating the extent of devolatization, based on the -92- composition of the tire and knowing the tire mass flow rate, the flow rate of each species (0 , CH4, 2 C 0 , H 0, etc.) was calculated. 2 2 As seen in Table 4-2, tire composition in elemental form (the mass fractions of elements such as C, H, O, etc.) is given. If this tire composition was going to be considered in the calculations, their own transport equations had to be written and solved. This, in fact, increases the complexity of the problem and requires a significant amount of time for development. Therefore, with a pre-processing code, the elemental analysis of the tire was converted into molecular composition (mass fractions of molecules 0 , CH4, C 0 , H 0 , etc. as in Table 4-6). Similar to many other CFD codes, this form was 2 2 2 suitable for our code and there was no need for considering any additional transport equations. In addition to the release of gaseous components due to the devolatization, the mass transfer due to the char combustion was also considered. The amount of consumed oxygen was treated as a sink term in the transport equation of oxygen in the gas flow field; similarly, the formed C 0 was added as a 2 source term in its transport equation. The source terms for the conservation equation of the species (5^ ) as in Table 1-1 were calculated with the following equations 5 = -F,.m .32/T2 0z c ^0/4 co s = t F 2 _ ( voi m (4-8) - CH (4-9) - co + m .44 /12) (4-10) Pf vol m X A x c 2 -93- F .m .x t SH 0 ~ ^ 2 co (4-11) -H0 (4-12) vol vol m X 2 (4-13) (4-14) where S is the source term of each component (kg/s), F is the tire flow rate (1/s), ;t is the mass m t ( fraction of each component (wt%), m is the mass of fixed carbon (kg) and m is the mass of c vol volatile matter (kg). In order to conserve energy, in addition to accounting for the heat transfer between tire and the hot gas (which was explained earlier), the enthalpy effects from tire combustion were also included. Equation (4-15) includes all these effects: the first term accounts for the pyrolysis heat, the second term accounts for the heat release due to the char combustion, and the last term accounts for the sensible heat of gaseous components. S was treated as a source term for the energy equation of Table n 1-1. Sh = t F ( vol m -Mlpyro + m C ^ch + ( vol m + m C ) t 'p ) T C (4-15) where S is the additional energy considered as a source term in energy equations of the gas phase h (W), F,is the tire flow rate (1/s), m is the mass of carbon (kg), AH is c pyro the pyrolysis heating value (J/kg), m is the mass of volatile matter (kg), AH is the heat of char combustion (J/kg), T is the tire vol ch temperature (K) and C is the specific heat of tire. p -94- t 4.6. G R I D AND BOUNDARY CONDITIONS For simulating tire combustion in cement kilns, the grid including the proper boundary conditions for all its sub-models (the hot flow, the wall, the clinker and tire combustion) was required. 4.6.1. Test Kiln For the simulations in this chapter, the cylindrical section of Kiln #1, referred to as test kiln, was targeted and then proper boundary conditions were prescribed. The developed grid included 23, 37 and 213 nodal points, in r, 6 and z direction respectively. The boundary conditions for the hot flow including the primary air, secondary air, fuel and wall conditions are schematically shown in Figure 4-1. In addition, the imposed boundary conditions for the hot flow and the bed model are tabulated in the following tables. It should be stressed that the kiln characteristic data were extracted and supplied by Dr. Jerry Yuan and the grid with the proper boundary conditions was generated with a special software developed by Dr. David Stropky. -95- -96- Table 4-3: Material information for test kiln Inlet Material Information How rare [kg/s] 20.8 Temperature [K] 338 Mass Fraction (% wt) Component CaC0 77.23 CaO 0.00 Si0 13.69 Al 03 3.36 Fe 03 1.7 3 2 2 2 Inert + others 3.85 Moisture 0.17 Total 100 Clinker Information Flow rate [kg/s] 13.52 Temperature [K] 1633 Table 4-4: Fuel input information for test kiln Coal/coke Fuel type Flow rate [kg/s] 2.72 Temperature [K] 333 Proximate Analysis (%) Moisture 1.24 Volatile 21.86 Fixed Carbon 67.35 Ash 9.55 Total 100 Elemental Analysis (%) 1.24 Moisture 71.81 Carbon -97- Hydrogen 3.73 Nitrogen 1.08 Sulphur 5.45 Oxygen 7.14 Chlorine 0.00 Ash 9.55 Total 100 High heating value (HHV) [kJAg] 27566 Low heating value (LHV) [kJ/kg] 26720 Table 4-5: Air input information for test kiln Primary Air Flow rate [kg/s] 6.52 Temperature [K] 365 Secondary Air Flow rate [kg/s] 21.3 Temperature [K~] 1122 Total air flow rate [kg/s] 27.82 In addition to the above conditions, it is assumed that the kiln shell loses heat, through natural convection and radiation, to the ambient atmosphere with T^,, =30 °C . Moreover, a zero-gradient boundary condition was imposed for the kiln exit. Chapter 2 contains more detail about the boundary conditions. The above boundary conditions are for test kiln with basic conditions, i.e. without any co-combustion of tire. For simplicity, we assumed that the total amount of heat needed for the whole process within the kiln is constant for all the cases. We also assumed that the airflow and distribution (flow rates, -98- inlet temperatures, and injection points) would remain unchanged and only the fuel mass flow rate changes for cases with tire combustion. Whenever co-combustion of tire existed, based on the heating values of the original fuel, tire and the amount of co-combustion, the fuel mass flow rate was modified and the rest of boundary conditions remained unaffected. For instance, for the case with 20% co-combustion of tire, the flow rate of the original flow was reduced to 2.72 x 0.8 = 2.176 kg/s . 4.6.2. T i r e Combustion M o d e l The boundary conditions for tire combustion model were handled separately. Table 4-6, for instance, contains a summary for the boundary conditions for the case with 20% co-combustion of tire. Similar to the mass flow rate of the original, the tire flow rate was adjusted for different co-combustion percentages. This setting was done by changing proportionally the number of tires dropped per rotation for different co-combustion rates. T a b l e 4-6: Input data for tire combustion sub-model for the case with 20% co-combustion Number of tires dropped in a rotation 6 Kiln rotation speed, rpm 1 Axial location of tire drop in, m 50 Tire temperature when dropped in, K 300 Weight of one whole tire, kg 4.15 Tire diameter, m 0.5776 Tire rim diameter, m 0.3556 2000 Tire specific heat, J/kg.K C h e m i c a l composition of the volatile matter content (WJ) CH4 content, wt% 0 C 0 content, wt% 0 CO content, wt% 9.34 2 -99- 0 H 0 content, wt% 2 H content, wt% 3.99 C H„ content, wt% 83.63 2 m 40.217 Lower heating value of C H , MJ/kg m 4.7. n SOLUTION METHODOLOGY After finishing the initial set-up of the problem, which consisted of the development of the tire combustion model, its implementation in the existing code, and grid generation along with the assignment of the appropriate boundary conditions, the system of the equations for the whole model was solved. The final code was executed in three steps: first, the cold flow field was created. Then, the code including all the sub-models was executed until a good convergence was achieved. After simulating the problem, including all the primary parameters, in order to get the concentration of secondary species (such as NO), the post-processing code was used. The computations related to formation of nitrogen oxides were decoupled from those of the main code and handled in a postprocessor. We stress again that the decoupling allows freezing the whole flow field and only executing the post-processing in order to calculate for the species related to the formation of nitrogen oxides. This, in fact, will save a lot of computational time. 4.8. RESULTS A N D DISCUSSION Originally, three different test cases were simulated: the first case had no tire combustion, and the second and third cases were with 10% and 20% co-combustion of tire, respectively. In all the above simulations, tires were injected at x=50 m; this distance is measured from the burner end. - 100- 4.8.1. Mass and Energy Balance Table 4-7 to Table 4-9 contain a summary of the mass and energy balance for the above mentioned cases. Both terms are well balanced for all cases. This indicates that the coupling between the main code and tire combustion model is properly done; it also indicates that the code has reached a satisfactory level of convergence. A detailed discussion on the important parameters is presented later. Table 4-7: Mass and energy balance: without co-combustion of tire Input Sensible heat in air Mass flow rate [kg/s] Average temperature [«] 27.8 Enthalpy [MW] Percent [%] 28.0 25.60 Sensible heat in feed material 20.788 338 7.645 6.99 Sensible heat in fuel 2.717 333 1.135 1.04 Sensible heat in tire - - - - Heat form fuel combustion 72.59 69.4 Heat from tire combustion - - 109.37 100 51.305 Total 37.1 Average temperature [K\ 1225 Clinker discharge 13.691 1811 Fuel ash discharge 0.448 Tire ash discharge - Output Mass flow rate [kg/s] Enthalpy [MW] 50.1 45.50 29.441 26.72 0.68 0.62 - - 21.291 19.34 Shell heat losses 7.20 6.54 Heat of unbumed carbon 1.328 1.21 Heat of water evaporation 0.076 0.07 Exit gas losses - Heat absorbed by clinkerization Total Imbalance 51.239 110.116 0.066 -0.746 - 101 - 100 Table 4-8: Mass and energy balance: with 10% co-combustion of tire Input Sensible heat in air Mass flow rate [kg/s] Average temperature [K] 27.8 Enthalpy [MW] Percent [%] 28.0 25.60 Sensible heat in feed material 20.788 338 7.645 6.99 Sensible heat in fuel 2.445 333 1.022 0.93 Sensible heat in tire 0.208 300 0.125 0.11 Heat form fuel combustion 65.33 59.73 Heat from tire combustion 7.263 6.64 109.384 100 51.241 Total 37.1 Average temperature [K] 1228 Clinker discharge 13.691 1740 Fuel ash discharge 0.401 Tire ash discharge 0.01 Output Mass flow rate [kg/s] Enthalpy [MW] 50.1 45.37 28.383 25.69 0.600 0.54 0.025 0.02 Heat absorbed by clinkerization 22.943 20.77 Shell heat losses 7.205 6.52 Heat of unburned carbon 1.141 1.03 Heat of water evaporation 0.068 0.06 57.202 110.466 100 0.039 -1.082 Exit gas losses Total Imbalance 1210 Table 4-9: Mass and energy balance: with 20% co-combustion of tire Input Sensible heat in air Mass flow rate [kg/s] Average temperature [K] 27.8 Enthalpy [MW] Percent [%] 28.0 25.59 Sensible heat in feed material 20.788 338 7.645 6.99 Sensible heat in fuel 2.173 333 0.908 0.83 Sensible heat in tire 0.415 300 0.249 0.23 Heat form fuel combustion 58.071 53.08 Heat from tire combustion 14.525 13.28 109.397 100 51.176 Total Output Mass flow rate [kg/s] - 102- Average temperature [K] Enthalpy [MW] 37.1 1239 50.7 45.82 Clinker discharge 13.691 1740 28.383 25.65 Fuel ash discharge 0.351 0.537 0.49 Tire ash discharge 0.021 0.045 0.04 Heat absorbed by clinkerization 22.943 20.73 Shell heat losses 7.200 6.51 Heat of unburned carbon 0.79 0.71 Heat of water evaporation 0.061 0.05 110.66 -1.263 100 Exit gas losses Total Imbalance 1095 51.163 0.13 4.8.2. Clinker Temperature and Composition Comparing Figure 4-2 to Figure 4-4, it is observed that co-combustion of tires has no effect on the clinker formation process. This conclusion is based on the fact that the species evolutions in the bed are the same for all three cases. This is the key to tire utilization in cement kilns because the main function of a cement kiln is clinker production. As long as the clinker composition is maintained, the co-combustion of tire is assumed to be successful. Another important factor is the production rate, which may slightly decrease due to tire firing [40]; based on this report, the reduction in the kiln production rate is because of the increased dust pickup and the loss in the stability of the kiln. However, possible production rates that can occur with tire injection were not considered and the production rate for all cases was assumed to be unchanging. In the area close to the tire injector (x=50 m), a slight increase in the gas temperature is observed (see Figure 4-3 and Figure 4-4). This is mainly because the tire devolatization process is exothermic and some heat will be released to the gas in its vicinity. The higher the tire feed rate, the bigger this increase will be. A quantitative discussion on all the important parameters is presented later on. - 103- Predicted Axial Profile in Test Kiln Distance from Kiln Hood [m] Figure 4-2: Axial profile of species: without co-combustion of tire Predicted Axial Profile in Test Kiln Distance from Kiln Hood [m] Figure 4-3: Axial profde of species: with 10% co-combustion of tire - 104- Predicted Axial Profile in Test Kiln Distance from Kiln Hood [m] Figure 4-4: Axial profile of species: with 20% co-combustion of tire 4.8.3. Gas Temperature Distribution The contour plots for the gas temperature for all three cases were similar but not completely the same; as tire was burned in the middle of the kiln, there was a slight decrease in its temperature in the flame area (see Table 4-10) and a slight increase in the area close to the surface of the tire. According to Table 4-10, the more co-combustion of tire, the bigger the reduction in the peak temperature will be. However, this difference was not noticeable in the contour plots of temperature of these three cases; therefore, only one of the contour plots (that of without tire combustion) is depicted as in Figure 4-5. - 105- 0 25 50 75 100 125 150 • • • • • • • • • • • W W II.IL L B B B S . . •• Gas Temperature Distribution Tg«IK] i 400 ^ i ^ i ^ H H H H K 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 4 4 1=3 Om z=20m z=40m 9 \$ Z=50m Figure 4-5: Gas temperature distribution 4.8.4. Refractory Temperature Distribution It was also observed that the mid-kiln firing of tires had no effect on refractory temperature distribution and therefore only the case without any co-combustion of tire is represented in Figure 4-6. ) | 50 1 1 1 Inner Refractory Temperature [K] 1 100 1 1 1 1 1 1 Wall Temperature Distribution 1 1 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 Outer She! Temperature |K] 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 Figure 4-6: Refractory temperature distribution - 106- 1 1 150 1— 4.8.5. NO X Emissions Any change in the gas temperature, specifically for temperatures beyond 1800 K , has a direct impact on N O formation rate [41]. As mentioned earlier, one of the desired impacts of the mid-kiln firing of tires is the reduction in the flame temperature and consequently a decrease in N O concentrations x throughout the whole kiln. More specifically, for the case without tire combustion (Figure 4-7a), N O concentration reaches a peak around 2500 ppm in the flame area and goes down to 1908 ppm at the exit. For the case with 10% tire (Figure 4-7.b), the maximum N O concentration goes down to around 1900 ppm in the flame area and reduces to 1591 ppm (17% reduction) at the exit. Similarly, with 20% co-combustion of tire, the peak and the exit concentrations are around 1600 and 1283 ppm (33% decrease), respectively. - 107- a) Without Co-Combustion of Tire 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 b) With 10% Co-Combustion of Tire 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 c) With 20% Co-Combustion of Tire 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 Figure 4-7: NO concentration distribution 8.6. Overall Impact of Mid-Kiln Firing of Tires on Kiln Performance Table 4-10 contains a summary of all the important parameters that are used to analyze the overall impact of tire utilization on test kiln. As tire is burned, the maximum gas average-temperature drops in the flame area. It has decreases from 2118 K to 2091 and 2059, respectively, for 10% and 20% co-combustion of tire. This finally contributes to a reduction in NO concentrations at the kiln exit (17% and 3 3 % , for 10% and 20% tire - 108- combustion). The maximum clinker temperature and the clinker temperature at the kiln exit decrease as the maximum gas temperature reduces (see Table 4-10). Because of tire combustion, the gas average-temperature at the kiln exit increases. The gas temperature at the kiln exit directly indicates the exit gas losses. The higher this temperature, the more these losses will be. Therefore, this parameter is important and should be investigated to make sure that the heat losses are still within a reasonable range. More specifically, the energy distribution within the kiln should be such that all the necessary processes and reactions are successfully achieved and not too much heat is lost via flue gasses. Referring to the energy balance tables (Table 4-7 to Table 4-9), it is observed that the enthalpy lost through the exit gas is slightly increased (for the worst case, it has increased from 50.1 M W to 50.7). On the other hand, since the clinker composition at the exit for all three cases was the same, it is concluded that the additional heat loss via flue gases is acceptable. More detailed discussion on this issue is presented in the sensitivity analysis section. Table 4-10: Summary of important parameters with tire injection at x=50 m and different tire flow rates Kiln without cocombustion of tire Kiln with 10% cocombustion of tire Kiln with 20% cocombustion of tire Gas average temperature at kiln exit [K] 1225 1228 1239 Maximum gas average temperature [K] 2118 2091 2059 N O concentration at kiln exit [ppm] 1908 1591 1283 Relative N O reduction [%] - 17 33 Maximum clinker average temperature [K] 2008 1939 1925 Clinker average temperature at kiln exit [K] 1811 1740 1740 - 109- 4.8.7. Sensitivity Analysis So far we have learned that co-combustion of tires (up to 20%) significantly reduces N O emissions. x It was also shown that mid-kiln firing (up to 20%) of tires had no negative impact on clinker quality. It was highlighted that the total enthalpy carried out by the exit flue gasses is an important parameter and should be investigated to ensure that not too much heat is lost via exit gases. This parameter is tightly coupled with the bed processes. If the total enthalpy of exit flue gasses, for example, increases, the total heat absorbed by the bed reduces. A n excessive reduction results in clinker with poor composition. So not all mid-kiln firing of tires can be successful and each kiln has proper rates, for a specific injection point. The whole process is successful as long as N O emissions at the kiln exit are x reduced, while clinker quality at the exit is still acceptable. There are mainly two parameters that can be investigated: the tire feed rate and the location of the tire injector. 4.8.7.1. Tire Feed Rate So far it has been shown that for test kiln co-combustion of tires up to 20% is successful and N O x concentration at the kiln exit reduces without any negative impact on the product quality. More reduction might be achieved if more tires are burned. So limits of the tire flow rate have to be explored. Despite the existence of practical barriers, such as opening size and the available time for throwing tires in the kiln, a case with higher flow rates in order to improve our understanding regarding this issue was simulated. We simulated a rather extreme case with 50% co-combustion where tires were still injected at x=50 m. We found that the gas average temperature in the flame area was reduced and the N O emission at x the kiln exit was reduced as well, a 70% reduction was observed. However, since not enough heat was available in the flame area, the clinker temperature was significantly reduced (see Figure 4-8) - 110- and clinker composition at the kiln exit was poor and not acceptable. The fact that the total heat lost by the exit flue gasses was increased from 50.1 MW to 54.3 MW confirms that more heat should have been absorbed by the bed. Because of the high tire flow rates, an excessive amount of heat was released in the tire burning area, and as a result the gas temperature in that area was even higher than that of the flame area (see Figure 4-9). Accordingly, more N O was formed in that area than in the flame area (see Figure;, 4-10); x however, the NO concentration in that area was still low and NO concentration at the kiln exit was low too. Predicted Axial Profile in test Kiln with 50% TDF Distance from Kiln Hood [m] Figure 4-8: Axial profile of species: with 50% co-combustion of tire - Ill - 25 50 75 100 125 150. Gas Temperature Distribution Toa«N 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 4 z=40m 7=10m Figure 4-9: Gas temperature distribution: with 50% co-combustion of tire 25 50 75 100 125 150 NOx Distribution N O [ppm] | 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 • z-nom # z=20m • z=30m f z=46m z=50m Figure 4-10: N O distribution: with 50% co-combustion of tire 112 • 4.8.7.2. Location of Tire Injector In order to optimize the location of the injection point, the thermal profde and the production rate (provided that the clinker quality is satisfactory) should be simulated as a function of injector location. The injection point should be such that the heat losses via flue gasses are minimized while complete combustion of tires can be ensured. In order to investigate the impact of the injection point, the test kiln with three new injection points, x=30, 40, and 80 m, was simulated. For all these simulations, the tire rate was fixed at 20%. Table 4-11 contains a summary of important parameters for these three cases. The original case (with tire injector at x=50 m) is included in this table as well. As the tire injector is placed closer to the flame area, less heat is lost via the flue gasses at the exit. This is actually good, and means that considering a constant heat loss through the wall, more heat is absorbed by the bed. The more heat absorbed by the bed, the higher its temperature will be (see Table 4-11). This guarantees that, if a clinker with an acceptable composition has been produced by throwing tire at x=50, for cases with injection points less than 50 (here, x=30 and 40 m) acceptable product quality will be obtained as well. This was confirmed by comparing the axial concentration profiles of clinker components of these cases with the original case. However, for the case of x=80 m, the heat lost by the flue gasses was too high (see the gas temperature at the exit) and not enough heat was absorbed by the bed. Therefore, the product quality was poor (see Figure 4-11). - 113- For the first three cases (x=30, 40, and 50 m), the maximum gas temperature is almost the same, and as a result, N O concentration at the kiln exit was the same. On the other hand, for the last case (x=80), the maximum flame temperature was low and N O concentrations at the exit were significantly reduced. Based on the above observation, one can conclude that varying the injection point from 50 to 30 m would result in acceptable clinker quality and unchanged N O emissions while putting the injection point somewhere farther downstream, e.g. at a distance of 80 m, would result in a poor quality product and reduced N O emissions. Table 4-11: Summary of important parameters with 20% co-combustion of tire and different tire injection points Injection point x=30 Injection point x=40 Injection point x=50 Injection point x=80 Tire combustion finished at x= ... [m] 2.7 4.2 6.5 10.9 Gas average temperature at kiln exit [K] 1232 1235 1239 1278 Maximum gas average temperature [K] 2058 2059 2059 2047 N O concentration at kiln exit [ppm] 1281 1280 1283 1217 Maximum clinker average temperature [K] 1912 1921 1925 1801 Clinker average temperature at kiln exit [K] 1756 1751 1740 1591 - 114- Predicted Axial Profile in test kiln 4.8.7.3. Grid Sensitivity Analysis The original grid was compared against two finer grids, one 50% finer and the other one 100% finer. For this comparison, the test kiln with the same geometry and boundary conditions without any tire combustion was simulated. Referring to the trend in Figure 4-12 and the data in Table 4-12, as the nodal points increase, the gas temperature profiles tend to become the same and it is concluded that the grid convergence is achieved. Based on the results of this simulation it seems that the second grid should be chosen for our simulations; however, due to having significantly smaller computational times and showing a better convergence behavior especially on enthalpy imbalance, the original grid (the coarsest one in Table 4-12) was preferred to the other two. - 115- Table 4-12: Summary of results for three different grids Case 1 Case 2 Case 2 23x37x213 29 x 37 x 249 29 x 49 x 249 181263 267177 353829 1.0 -1.5 -2.0 Maximum local temperature of the gas[K] 2190 2163 2159 Maximum axial temperature of the gas [K] 2064 2042 2033 Enthalpy imbalance 0.80e+04 -.19e+06 -.10e+06 Mass imbalance -.86e-04 0.27e-04 0.14e-04 Grid points in r, 8 and z direction Number of nodes Nodes ratio (N/N,) - 116- Grid Sensetivity Analysis 2500 -r=1.0 -r=1.5 -r=2.0 0.00 20.00 40.00 60.00 80.00 1 00.00 120.00 140.00 160.00 Axial position (m) Figure 4-12: Axial gas temperature profile for three different grids 9. SUMMARY AND CONCLUSIONS A one-dimensional mathematical model for tire combustion was developed and implemented in an existing code. With the aid of this code, three different cases (one without tire combustion and the other two with co-combustion of tire, 10% and 20% respectively) were simulated and analyzed through a series of contour plots and tables. The primary results of these simulations revealed that the co-combustion of tire, up to 20% at x=50 m for the text kiln, has no negative impact on the product quality and causes significant reductions in - 117- N O emissions at the exit (up to 33% for the case with 20% co-combustion). Finally, it is concluded x that for this kiln co-combustion of tires up to 20% at x=50 is successfully utilized. Therefore, tire combustion can contribute to a cement kiln with less pollutant and less operational costs. In order to explore the limits and possibilities of tire combustion, in addition to above operational conditions, test kiln with different tire flow rates and injector locations was simulated. We particularly investigated their impact on two important parameters: the clinker composition and N O x emissions at the exit. The summary of our observations along with conclusions is presented in the following: For studying the impact of tire flow rate, we increased the co-combustion rate up to 50%. The simulation results revealed that N O emissions were significantly reduced (70%), but the clinker x composition was not acceptable. Based on this observation, we concluded that the high tire flow rates, in addition to having practical limitations, may also impair the clinker quality. Therefore, an upper limit should be prescribed for the amount of co-combustion in the middle of the kiln. Based on our analysis, we recommend 20% co-combustion of tires. On other hand, the test kiln with three new tire injection points was simulated. The results showed that whenever the tire injection point was less than 50 m, the clinker composition was still acceptable and N O emissions were almost the same. This means that for this kiln, there is actually more than one suitable point for injecting tires, although some of these points may not be practically accessible (they may be in the burning zone). We also simulated a test case with tire injector at x=80 m. For this case, because of injecting the tire in an inappropriate area, too much heat was lost through the exit flue gasses and the clinker did not absorb adequate heat; therefore, its quality was poor. It also did not - 118- have that much of an effect on N O emissions at the exit. Therefore, we concluded that placing the tire injector somewhere far from the burner can result in a kiln operation with unacceptable product. We recommend a tire injector at around x=50 m. - 119- Chapter 5. SUMMARY, CONCLUSIONS, AND RECOMMENDATIONS A fully-coupled mathematical model, including the hot flow, the bed, and the wall of rotary kilns, was developed. With the aid of this model, full-scale industrial cement kilns under steady-state and realistic operational conditions were simulated. The results were analyzed and validated against limited measurement data, hi addition, a 1-D model for tire combustion was added to the existing code. This new code was used for modeling cement rotary kilns with combustion of full scrap tires in the middle of them. The visualized solutions such as the temperature distribution and species concentrations in the gas flow-field, the temperature distribution within the wall, and the material and temperature evolution in the bed were analyzed and compared. The results indicated that significant progress has been made to create a predictive tool for cement rotary kilns. In addition, with the aid of the developed model, a better understanding of cement kilns with various operational conditions, kiln geometries, and burner designs can be obtained. The model can be used to diagnose the operational problems associated with this complex equipment. A brief summary of the present work, some concluding remarks, and recommendations for future work, are given as follows. Summary and Contributions: • A n existing C F D code, which has the capability of simulating rotary furnaces including the hot gas and the wall, was used for further development. - 120- A 1-D model for the transport phenomena in the bed, including the species and thermal evolution, was developed. This model included ten elements and five chemical reactions between them. For simulating cement rotary kilns, the bed model was implemented into the existing C F D code; coupling was achieved by implementing the heat and mass transfer between the gaseous phase and the bed. A new 1-D model for combustion of full tires in the middle of rotary kilns was developed and implemented in a cement kiln model. The implementation of tire combustion model was achieved by including the mass and heat transfer between different sub-models. With the aid of this code, cement kilns with different tire combustion conditions (different rates and different injector locations) were simulated and analyzed. The tire combustion model consisted of two sub-models: devolatization and char combustion. Attempts were made to include all the important physical and chemical phenomena associated with the combustion of tire in this model. For devolatization, the external heat transfer and chemical reaction kinetics were assumed to be dominant. Char combustion was assumed to be controlled by oxygen availability. A mathematical model for mid-kiln firing of tires was developed. The newly developed model was applied to a test cement kiln with different tire combustion rates and different locations of tire injector and the results of the simulations were analyzed. These simulations were performed to investigate and explore the limitations and feasibility of tire combustion in cement rotary kilns. The product quality and NO* concentration at the kiln exit were chosen as the main performance indicators of co-combustion of tires in the middle of cement kilns. - 121 - • It was shown that 20% co-combustion of tire can reduce the N O emissions of test kiln x substantially without any negative impact on its product quality. Conclusions: • Analysis and investigation of simulation results revealed that a useful tool in advancing the understanding of cement kilns has been developed. The developed model can be used to address operational problems and improve kiln design. • The simulation result indicated that the clinker formation process is very sensitive to the heat transfer to the bed, and a good estimate of the bed angle is a key step for gaining high quality products. • The results indicated that almost half of a wet process kiln is dedicated to the preheating of the raw material. This means that by introducing a drying system before the kiln system, the kiln length can be significantly reduced. That proves again the usefulness of the modem practice of using preheaters. • A heat balance of the kiln indicated that for a wet process kiln, the combustion of the fuel accounts for 70% of the inlet heat and the rest (30%) is accounted for by the sensible heat of the air, the raw material, and the fuel. About 50% of the total heat is lost by flue gases; the bed material absorbs 40-45% of this heat and the rest of it (5-10%) is lost through the refractory wall to the ambient air. The fact that almost half of the kiln energy is lost through the flue gasses is one the major reasons for low thermal efficiency of cement kilns. • Tire combustion simulations have shown that tires were completely burned. However, for certain firing rates and axial locations the product quality deteriorated. The model can be used to establish the desired firing rate and axial location for tire burning. - 122- • For the cases without any negative impact on the product quality, a significant reduction in NO* emissions at the exit was observed. This, in fact, was one of the desired impacts from tire combustion. • Based on the simulation results, it was concluded that the best performance can be achieved when 20% of the primary fuel is replaced with tires somewhere in the middle of the calcination zone. • It was concluded that successful utilization of tires, as expected, can save energy and money, and lower the emissions of pollutants such as N O . It is also likely to expand refractory x lifetime because of the lower gas temperature especially in the burner area. • By burning tires with different rates and different axial location, it was confirmed that a certain flame temperature is required for the clinker formation process. Low flame temperatures or inappropriate energy distributions (by burning too much tire inside the kiln) can result in poor product quality. Therefore, the energy distribution within the system has to be chosen carefully. This is one of the main reasons for limiting the amount of tire cocombustion. Recommendations: • Including the heat transfer between the wall and the bed of material would lead to more accurate model predictions • Modeling the dust circulation in the system would contribute to a more complete model and is expected to bring the gas temperature closer to plant measurement data. • Considering the rotational term in the wall model would result in better temperature predictions. - 123 - Improving the bed model, for example expanding the computational domain to three dimensions could be pursued. However, this would increase the complexity of the model and computational time. Also the bed physics is very complex and not well understood. Therefore, it is possible that the increase in complexity might not result in commensurate model improvement. Validating the simulation results for cases with tire combustion inside the kiln against some measurement data would be very useful. - 124- REFERENCES [1] Choate, W.T., 2003, "Energy and Emission Reduction Opportunities for the Cement Industry," US Department of Energy, Energy Efficiency and Renewable Energy. [2] Mike Georgallis, 2004, "Mathematical Modelling of Lime Kilns," The University of British Columbia. [3] Locher, G., 2002, "Mathematical Models for the Cement Clinker Burning Process- Part 1: Reactions and Unit Operations," Z K G International, 55pp. 29-38. [4] Peray, K . E . , 1979, "Cement Manufacturer's Handbook," Chemical Publishing Co. Inc. [5] Brimacombe, J. K., and Watkinson, A . P., 1978, "Heat Transfer in a Direct Fired Rotary Kiln-I. Pilot Plant and Experimentation," Met. Trans. B , 9Bpp. 201-208. [6] Watkinson, A. P., and Brimacombe, J. K., 1978, "Heat Transfer in a Direct Fired Rotary Kiln-II. Heat Flow Results and their Interpretation," Met. Trans. B , 9Bpp. 209-219. [7] Gorog, J. P., Adams, T. N . , and Brimacombe, J. K., 1983, "Heat Transfer from Flames in a Rotary Kiln," Met. Trans. B , 14Bpp. 411-424. [8] Boateng, A . A . , and Barr, P. V . , 1996/7, " A Thermal Model for the Rotary Kiln Including Heat Transfer within the Bed," International Journal of Heat and Mass Transfer, 39(10) pp. 2131-2143. [9] Mellmann, J., 2001/8/28, "The Transverse Motion of Solids in Rotating Cylinders—Forms of Motion and Transition Behaviour," Powder Technology, 118(3) pp. 251-270. - 125- [10] Ding, Y . L., Seville, J. P. K., and Forster, R., 2001/3, "Solids Motion in Rolling Mode Rotating Drums Operated at Low to Medium Rotational Speeds," Chemical Engineering Science, 56(5) pp. 1769-1780. [11] Ding, Y . L., Forster, R., and Seville, J. P. K., 2002/4, "Segregation of Granular Flow in the Transverse Plane of a Rolling Mode Rotating Drum," International Journal of Multiphase Flow, 28(4) pp. 635-663. [12] Van Puyvelde, D. R., 2006/7, "Comparison of Discrete Elemental Modelling to Experimental Data Regarding Mixing of Solids in the Transverse Direction of a Rotating Kiln", Chemical Engineering Science, 61(13) pp. 4462-4465. [13] Martins, M . A . , Oliveira, L . S., and Franca, A . S., 2001/9, "Modeling and Simulation of Petroleum Coke Calcination in Rotary Kilns," Fuel, 80(11) pp. 1611-1622. [14] Spang, H . A . , 1972, " A Dynamic Model of a Cement Kiln," Automatica, 8pp. 309. [15] Marias, F., 2003/6/15, " A Model of a Rotary Kiln Incinerator Including Processes Occurring within the Solid and the Gaseous Phases," Computers & Chemical Engineering, 27(6) pp. 813-825. [16] R.T. Bui, G . Simard, A . Charette, Y . Kocaefe, and J. Perron, 1995, "Mathematical Modeling of the Rotary Coke Calcining Kiln," Canadian Journal of Chemical Engineering, 73pp. 534-544. [17] Mastorakos, E., Massias, A., and Tsakiroglou, C. D., 1999/1/1, "CFD Predictions for Cement Kilns Including Flame Modelling, Heat Transfer and Clinker Chemistry," Applied Mathematical Modelling, 23(1) pp. 55-76. - 126- [18] Nowak, Z. P. and Salcudean, ML, 1996, "Turbulent Flow Calculations by the Nonlinear MultiGrid Method," Journal of Applied Mathematics and Mechanics, 76pp. 463. [19] Yuan J., Stropky D., Bibeau E.L., 2003, "Using Process Modelling to Improve Lime Kiln Operations," Pulp and Paper Canada, 104(11) pp. 41-44. [20] Patankar, S.V., 1980, "Numerical Heat Transfer and Fluid Flow," Hemisphere Publishing Corp. [21] He, P., Salcudean, M . , and Gartshore, I. S., 1996/5, "Multigrid Calculation of Fluid Flows in Complex 3D Geometries using Curvilinear Grids," Computers & Fluids, 25(4) pp. 395-419. [22] HE, P., 1995, "Numerical Simulation of Film Cooling of a Turbine Blade using Block-Structured Curvilinear Grids," University of British Columbia. [23] B.E. Launder and D.B. Spalding, 1974, "The Numerical Computation of Turbulent Flows," Computer Methods in Applied Mechanics and Engineering, 3pp. 269-289. [24] Holman, J.P., 2002, "Heat transfer," McGraw-Hill, Boston. [25] Peray, K . E . and Waddell, J.J., 1972, "The Rotary Cement Kiln," Chemical Publishing Co., Inc., New York, pp. 194. [26] Bogue, R. H., 1929, "Calculation of the Compounds in Portland Cement," Ind. Eng. Chem. Anal. Ed., 1pp. 192-197. [27] de Marco Rodriguez, I., Laresgoiti, M . F., and Cabrero, M . A., 2001/8, "Pyrolysis of Scrap Tyres," Fuel Processing Technology, 72(1) pp. 9-22. - 127 - [28] Yamaguchi, E., 2000, "Waste Tire Recycling," 2006(June/29) available from: http://www.p2pavs.org/ref711/10504/. [29] Barlaz, M . A . , Eleazer, I.,William E., and Whittle, D. J., 1993/12, "Potential to use Waste Tires as Supplemental Fuel in Pulp and Paper M i l l Boilers, Cement Kilns and in Road Pavement," Waste Management & Research, 11(6) pp. 463-480. [30] Karell, M . , 2000, "Regulation Impacts on Scrap Tire Combustion: Part II," 2006(June/29), available from: http://www.pirnie.com/resources pubs air febOO 3.html.. [31] Stillwagon, M . A . , and Wahlquist, C. J., 1999, "Building A Successful Tire-Derived Fuel Program," 2006(June/29), available from: http://cementamericas.com/mag/cement building successful tirederived/. [32] Reisman, J. I., October 1997, "Air Emissions from Scrap Tire Combustion," 2006(June/29) pp. 117, available from: http://www.epa.gov/ttn/catc/dir1 /tire eng.pdf. [33] Conesa, J. A., Font, R., and Fullana, A., 1998/10, "Kinetic Model for the Combustion of Tyre Wastes," Fuel, 77(13) pp. 1469-1475. [34] Williams, P. T., Besler, S., and Taylor, D. T., 1990/12, "The Pyrolysis of Scrap Automotive Tyres: The Influence of Temperature and Heating Rate on Product Composition," Fuel, 69(12) pp. 1474-1482. [35] Conesa, J. A., and Marcilla, A., 1996/8, "Kinetic Study of the Thermogravimetric Behavior of Different Rubbers," Journal of Analytical and Applied Pyrolysis, 37(1) pp. 95-110. - 128- [36] Conesa, J. A . , Font, R., and Marcilla, A., 1997/8, "Mass Spectrometry Validation of a Kinetic Model for the Thermal Decomposition of Tyre Wastes," Journal of Analytical and Applied Pyrolysis, 43(1) pp. 83-96. [37] Leung, D . Y . C , and Wang, C. L., 1998/5, "Kinetic Study of Scrap Tyre Pyrolysis and Combustion," Journal of Analytical and Applied Pyrolysis, 45(2) pp. 153-169. [38] Yang, J., Tanguy, P. A., and Roy, C , 1995/6, "Heat Transfer, Mass Transfer and Kinetics Study of the Vacuum Pyrolysis of a Large used Tire Particle," Chemical Engineering Science, 50(12) pp. 1909-1922. [39] Larsen, M . B., Schultz, L., and Glarborg, P., 2006/0, "Devolatilization Characteristics of Large Particles of Tyre Rubber under Combustion Conditions," Fuel, 85(10-11) pp. 1335-1345. [40] Ukrainetz, P. and Denizeau, P., 1995, "Mid-Kiln Injection of Tire-Derived Fuel (TDF) at the Lafarge Canada Inc. Cement Plant at St. Constant, Quebec". [41] Fluent Inc., 2003, "User's Guide Index,"Fluent Inc., Chap. 18. [42] Westenberg, A., 1971, Combustion Science and Technology, 4pp. 59-64. [43] Warnatz, J., 1991, "NOx Formation in High Temperature Processes," Proc. European Gas Conference, Anonymous pp. 303-320. [44] Baulch, D. L . et al., 1992, "Evaluated Kinetic Data for Combustion Modelling," Journal of Physical and Chemical Reference Data, 21(3) pp. 411-734. - 129- [45] Flower, W . L., Hanson, R.K., and Kruger, C H . In 15th Symp. (Int'l.) on Combustion, pp. 823. The Combustion Institute, 1975. [46] Baulch, D.L., Drysdall, D.D., and Home, D.G., 1973, "Evaluated Kinetic Data for High Temperature Reactions," Butterworth. - 130- Appendix A. FORMATION OF THERMAL NO x The principal reactions governing the formation of thermal N O , known as the extended Zeldovich x mechanism, are as follows: 0 +N< **• >N + NO (A-l) N +0< **t W + NO (A-2) 2 2 N + OH<r^-^H + NO (A-3) Based on the above reactions, the formation rate of NO can be calculated [41] ^~p- = *, [0][N ] + k [N][0 ] + k [N][OH] 2 2 2 3 ( A 4 ) -fc_,[/V][/V0] - k_ [0][NO] - k_ [H][NO] 2 3 where all concentrations are in gmol/m . 3 In order to calculate the formation rates of N O , the concentrations of N , N , O, H , and O H are 2 required. A.l. T H E QUASI-STEADY ASSUMPTION FOR [N] When sufficient oxygen is available, the rate of consumption of nitrogen atoms will be equal to the rate of its formation, and a quasi-steady state can be assumed. Therefore, the N O formation rate can be modified to [41] - 131 - k,[N ]-d[NO] dt 2 k [0 ] 2 = 2[0]- 2^ k,k_ [NO] 2 2 (gmol/m .s) (A-5) k,[NO] 1+- k^O^ + k^OH] A.2. DETERMINING O RADICAL CONCENTRATION There are three approaches used by F L U E N T [41] to determine the concentration of oxygen radicals: the equilibrium approach, the partial equilibrium approach, and the predicted concentration approach. A.2.1. Method 1: Equilibrium Approach The thermal N O formation rate is much slower than the main hydrocarbon oxidation rate; therefore, x the formation process of thermal N O can often be decoupled from the main combustion reactions. x Using this approach, according to Westenberg [42], the equilibrium oxygen atom concentration can be calculated from the following expression: [O] = 3 . 9 7 x l 0 r 5 1 / 2 e ( 3 1 0 9 0 / 7 ' [O ] ) 1 / 2 2 (gm //m ) 3 O (A-6) A.2.2. Method 2: Partial Equilibrium Approach The first method can be improved by accounting for third-body reactions as 0 +M <r*0 + 0 + M 2 Considering this reaction, the oxygen atom concentration can be calculated by the following expression [43]: [O] = 36.647/ ' 1 2 ( e - 2 7 1 2 3 / r ) [0 ] 2 1 / 2 (gmol I m ) 3 This method generally leads to a higher oxygen atom concentration compared to Method 1. - 132- (A-7) A.2.3. Method 3: Predicted O Approach When the concentration ofthe oxygen atom is calculated using an advanced chemistry model, [O] can be taken simply from it. A.3. D E T E R M I N I N G O H R A D I C A L C O N C E N T R A T I O N There are three approaches used by F L U E N T [41] to determine the O H radical concentration: exclusion of O H approach, the partial approach, and the predicted O H concentration approach. A.3.1. Method 1: Exclusion of O H Approach In this approach, it is assumed that the k [0 ] 2 2 eq k [OH] » 3 eq and therefore the third reaction in the extended Zeldovich mechanism is neglected. A.3.2. Method 2: Partial Equilibrium Approach In this approach, the O H radicals are not ignored anymore and their concentration is calculated with the following expression [44]: [OH] = 2] 2.9 7 ^ 5 7 E ( - 4 5 9 5 / 7 ' [O] ) 1/2 [H 0] U2 2 (gmol I m ) 3 (A-8) A.3.3. Method 3: Predicted O H Approach Similar to the predicted O approach, when the O H radical concentration is calculated using an advanced chemistry model, [OH] can be taken from the local O H mass fraction. - 133- A.4. T H E R M A L N O X REACTION RATES The expressions for the rate coefficients for the extended Zeldovich mechanism based on the evaluation of Hanson and Salimian [45] and Baulch et al. [46] are summarized in the following table. Table A - l : Rate data (m /gmol.s) for N O kinetics 3 Baulch et al. Rate constant Hanson and Salimian A Reaction [m 3 KT ) gmol" s" 1 fi 1 fi E/R [K] 38000 A [m 3 fi E/R [K] 1.82E+8 0 38370 3.8E+7 0 425 gmor s" K ^ ] 1 1 *i 0 + N ->N + NO 7.6E+7 0 *-i N + NO^>0 + N 2 1.6E+7 0 Ic^ N +0 ^ 0 + NO 6.4E+3 1 3150 1.8E+4 1 4680 19500 3.8E+3 1 20820 7.1E+7 0 450 1.7E+8 0 24560 2 2 O + NO^N + 0 1.5E+3 1 *3 N + OH ->// + NO 4.1E+7 0 *-3 H + NO-+N + OH 2.0E+8 0 2 23650 Note: reaction rate constants (k ) are in the form k = AT p t A.5. e\p(-E/RT). SUMMARY The thermal N O , formation rate can be predicted by equation (A-5). The expressions for the rate coefficients for the extended Zeldovich mechanism are often selected based on the evaluation of Baulch et al. [46] . The O-atom concentration needed in equation (A-5) is often computed according to equation (A-6), the equilibrium assumption. With the exclusion of the O H approach, equation (A-5) will be simplified to: ^\[mP2W-k. k^NOf) d[NO] — x = 2[0]- dt ; r (*2[O ] + 2 lfc_i[iV0j) - 134- (emollm .s) ^ ^ A Finally, the N O source term due to formation of thermal N O can be calculated x d[NO] S,h,NO= w,NO . where M w N 0 3 . kg/(m.s) M d is the molecular weight of NO, and -135- dt l s c o m (A- p t d from equation (A-9). U e `
#### Cite
Citation Scheme:
Citations by CSL (citeproc-js)
#### Usage Statistics
United States 69 19
Nigeria 28 26
India 24 34
United Kingdom 16 8
China 16 23
Germany 15 5
Malaysia 13 0
Bulgaria 12 0
Ukraine 11 0
Brazil 9 28
Australia 9 4
Iran 8 12
Unknown 157 178
Ashburn 21 0
San Mateo 9 2
Montreal 8 2
San Jose 7 0
Shenzhen 6 22
Beijing 6 1
London 6 1
Vancouver 5 1
Saint Petersburg 5 0
Mountain View 5 4
Magdeburg 4 1
Shenyang 4 0
{[{ mDataHeader[type] }]} {[{ month[type] }]} {[{ tData[type] }]}
#### Embed
Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
``` ```
<div id="ubcOpenCollectionsWidgetDisplay">
<script id="ubcOpenCollectionsWidget"
src="{[{embed.src}]}"
data-item="{[{embed.item}]}"
data-collection="{[{embed.collection}]}"
`http://iiif.library.ubc.ca/presentation/dsp.831.1-0080783/manifest`
| 49,702
| 184,344
|
{"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}
| 2.59375
| 3
|
CC-MAIN-2016-44
|
longest
|
en
| 0.852876
|
https://askmycalculator.com/mutual-fund-calculator
| 1,723,551,280,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00358.warc.gz
| 87,930,886
| 12,190
|
# Mutual Fund Calculator
Published: 4/25/2021 Last updated: 2/16/2023
## Mutual Fund Calculator Definition
A mutual fund calculator is a financial tool used to estimate the potential returns of investing in a mutual fund. Mutual funds are investment vehicles that pool money from multiple investors to purchase a diversified portfolio of stocks, bonds, or other securities.
A mutual fund calculator takes into account factors such as the initial investment amount, the expected rate of return, and the length of the investment term to calculate the potential future value of the investment.
These calculators are useful for investors looking to compare the potential returns of different mutual funds or to estimate how much they would need to invest in order to reach a specific financial goal. However, it is important to note that mutual fund calculators are only estimates and actual returns may vary depending on market conditions and other factors.
### Mutual Fund Calculator formula
A mutual fund calculator formula is a mathematical equation used to calculate the potential returns of a mutual fund investment. The formula typically takes into account the initial investment amount, the expected rate of return, and the length of the investment term.
The most commonly used formula for calculating mutual fund returns is the compound interest formula. This formula calculates the future value of an investment by multiplying the initial investment amount by one plus the rate of return raised to the power of the investment term.
For example, if an investor invests \$10,000 in a mutual fund with an expected rate of return of 7% for 10 years, the future value of the investment can be calculated using the compound interest formula as follows: Future Value = \$10,000 x (1 + 0.07) ^ 10 = \$19,671.56. It is important to note that this is only an estimate and actual returns may vary depending on market conditions and other factors.
### Mutual Fund Calculator examples 20000\$
A mutual fund calculator can be a useful tool for investors looking to estimate the potential returns on a \$20,000 investment in a mutual fund. Using a mutual fund calculator, an investor can input the initial investment amount, expected rate of return, and investment term to generate an estimated future value.
For example, if an investor invests \$20,000 in a mutual fund with an expected rate of return of 8% for a 10-year investment term, the mutual fund calculator can estimate that the future value of the investment will be approximately \$46,610.
Alternatively, if an investor has a specific financial goal, such as saving for a child's education or a down payment on a home, they can use a mutual fund calculator to determine the required investment amount and investment term to reach their goal. However, it's important to keep in mind that mutual fund calculators are only estimates and actual returns may vary depending on market conditions and other factors. Therefore, investors should also consult with a financial advisor before making any investment decisions.
### Mutual Fund Calculator explanation
A mutual fund calculator is a financial tool that helps investors estimate the potential returns on their mutual fund investments. The calculator typically requires input of the initial investment amount, expected rate of return, and investment term to generate an estimated future value.
The expected rate of return is an estimate of the average annual return that the mutual fund is likely to provide over the investment term. The investment term is the period of time during which the investor plans to hold the mutual fund investment. The mutual fund calculator uses these inputs to estimate the potential future value of the investment based on a compound interest formula.
This formula takes into account the initial investment amount, the expected rate of return, and the investment term to calculate the future value of the investment. However, it's important to remember that mutual fund calculators are only estimates and actual returns may vary depending on market conditions and other factors. Therefore, it's always a good idea to consult with a financial advisor before making any investment decisions.
### The Anatomy of Mutual Funds: Components and Structure
Mutual funds are often hailed as one of the most popular investment vehicles, providing an avenue for investors to pool their resources together and invest in a diversified portfolio. However, to fully appreciate the utility and potential of mutual funds, one needs to understand its intricate components and structure.
Selection of Securities:
At the heart of every mutual fund lies the pool of securities it invests in. This can range from equities and bonds to money market instruments and commodities. The selection is based on the fund's objectives. For instance, an equity fund will predominantly invest in stocks, while a bond fund will focus on corporate or government bonds. The portfolio manager, with the assistance of a research team, meticulously selects these securities, aiming to generate maximum returns in line with the fund's strategy and risk tolerance.
Management Strategies:
Different funds employ varied management strategies. Two of the most common are active and passive management. In active management, the fund manager makes deliberate investment decisions to outperform a benchmark index. This involves continuous research, analysis, and adjustments to the portfolio. Conversely, in passive management, the fund simply aims to replicate the performance of a benchmark index, like the S&P 500, leading to lower transaction costs and management fees.
Net Asset Value (NAV):
An essential component of mutual funds is the Net Asset Value (NAV). This represents the value of one share in a mutual fund, and it's calculated daily based on the total value of the fund's assets minus its liabilities. The NAV provides a clear picture of the fund's performance and is the price at which investors buy and sell mutual fund shares.
Expense Ratios:
All mutual funds incur costs, including management fees, administrative expenses, and distribution fees. These costs are aggregated and represented as a percentage of the fund's average assets, known as the expense ratio. It's crucial for investors to be aware of these fees, as they can significantly impact returns over time.
Share Classes:
Many mutual funds offer different share classes, each with its own fee structure and minimum investment requirement. Common share classes include Class A (front-end load), Class B (back-end load), and Class C (level-load). The choice between these depends on the investor's investment horizon and the amount they wish to invest.
In conclusion, mutual funds are a complex blend of strategies, decisions, and structures. By understanding each component, investors can make informed choices that align with their financial goals and risk appetite.
Risk vs. Reward: Evaluating Mutual Fund Investments In the world of investments, mutual funds are often perceived as a balanced option that offers diversification. However, like all investment avenues, they come with their own set of risks and rewards. Striking the right balance between these two facets is crucial for investors aiming to achieve their financial objectives.
### Understanding the Risk-Reward Spectrum:
The risk-reward spectrum is a fundamental concept in investing. Essentially, it dictates that the potential return on an investment is directly proportional to the risk associated with it. Mutual funds that invest predominantly in equities or stocks, for instance, tend to be on the higher end of the risk spectrum but also offer the possibility of higher returns. Conversely, debt or money market funds are generally perceived as lower risk but may provide comparatively modest returns.
### Metrics to Consider:
• Standard Deviation: This metric offers insights into a fund's volatility. A higher standard deviation indicates that the fund's returns have fluctuated more significantly over a given period, signaling higher risk.
• Beta: It measures a fund's sensitivity to market movements. A beta greater than 1 indicates that the fund is more volatile than the market, while a beta less than 1 signifies less volatility.
• Sharpe Ratio: This ratio calculates the average return earned in excess of the risk-free rate per unit of volatility. A higher Sharpe ratio indicates better risk-adjusted performance.
### Other Indicators:
Beyond quantitative metrics, it's essential to consider other factors, such as the fund's investment objective, its portfolio composition, and the track record of the fund manager. These qualitative indicators can provide valuable context about the fund's risk profile.
### Diversification as a Strategy:
One of the primary reasons investors turn to mutual funds is the inherent diversification they offer. By investing in a broad array of securities, mutual funds can mitigate some of the risks associated with individual assets. This doesn't eliminate risk but spreads it across various investments.
### The Essence of Diversification:
At its core, diversification is the practice of spreading investments across multiple securities or asset classes to reduce the risk associated with any single asset's adverse movement. Think of it as the financial equivalent of the adage "don't put all your eggs in one basket."
### Key Benefits of Diversification:
• Risk Reduction: The primary benefit of diversification is risk mitigation. By holding a mix of assets, the negative performance of a few investments can be counterbalanced by the positive performance of others.
• Potential for Enhanced Returns: Diversified portfolios can tap into multiple sectors, industries, or regions, some of which might outperform others, leading to potentially enhanced overall returns.
• Access to Multiple Asset Classes: Mutual funds offer investors an opportunity to invest in a blend of equities, bonds, commodities, and more, all within a single fund.
• Protection Against Volatility: A well-diversified portfolio tends to be less susceptible to market volatility than a concentrated one, offering more stability to investors.
### Diversification in Mutual Funds:
Mutual funds inherently offer diversification since they pool together money from various investors to invest in a broad spectrum of securities. Depending on the fund's objective, it might invest across sectors, market caps, regions, or even countries. This makes mutual funds especially suitable for investors seeking diversification without the need to handpick numerous individual securities.
### Limits to Diversification:
While diversification is a powerful tool, it's crucial to recognize its limitations. Diversification reduces company-specific or unsystematic risk, but it cannot eliminate market-wide or systematic risk. Events affecting the entire market, like broad economic downturns, can impact diversified portfolios as well.
### Mutual Fund Calculator FAQ
Mutual fund calculators are financial tools that help investors estimate the potential returns on their mutual fund investments. Here are some frequently asked questions about mutual fund calculators:
1. What is a mutual fund calculator? A mutual fund calculator is a tool that helps investors estimate the potential future value of their mutual fund investment based on the initial investment amount, expected rate of return, and investment term.
2. How do I use a mutual fund calculator? To use a mutual fund calculator, you need to input the initial investment amount, expected rate of return, and investment term. The calculator will then estimate the potential future value of the investment based on these inputs.
3. Are mutual fund calculators accurate? Mutual fund calculators are only estimates and actual returns may vary depending on market conditions and other factors. However, they can be useful in providing investors with an idea of the potential returns on their investments.
4. What are the advantages of using a mutual fund calculator? The main advantage of using a mutual fund calculator is that it helps investors estimate the potential future value of their investments, which can help with investment planning and decision-making.
5. Can I use a mutual fund calculator for any type of mutual fund? Yes, mutual fund calculators can be used for any type of mutual fund, whether it's an equity mutual fund, debt mutual fund, or hybrid mutual fund.
6. Should I consult with a financial advisor before using a mutual fund calculator? It's always a good idea to consult with a financial advisor before making any investment decisions, including using a mutual fund calculator. A financial advisor can help you understand your investment goals and risk tolerance, and can provide personalized investment advice.
| 2,382
| 12,886
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.4375
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.913666
|
https://arend-lang.github.io/arend-lib/v1.10.0/arend-html-files/src/Set/Subset.html
| 1,721,535,586,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00383.warc.gz
| 87,972,986
| 15,584
|
```\import Data.Array
\import Function.Meta
\import Logic
\import Logic.Meta
\import Meta
\import Order.Lattice
\import Order.PartialOrder
\import Paths
\import Paths.Meta
\import Topology.Locale
\open Bounded(top,top-univ)
\func Set (X : \hType) => X -> \Prop
\where {
\func Union {X : \hType} (S : Set X -> \hType) : Set X
=> \lam a => ∃ (U : S) (U a)
\lemma Union-cond {X : \hType} {S : Set X -> \hType} {U : Set X} (SU : S U) : U ⊆ Union S
=> \lam Ux => inP (U,SU,Ux)
\func Total {X : \Type} (U : Set X) => \Sigma (x : X) (U x)
\func restrict {X : \Type} {U : Set X} (V : Set X) : Set (Total U)
=> \lam s => V s.1
\func Prod {X Y : \hType} (U : Set X) (V : Set Y) : Set (\Sigma X Y)
=> \lam s => \Sigma (U s.1) (V s.2)
\func CoverInter {X : \hType} (C : Set (Set X)) (D : Set (Set X)) : Set (Set X)
=> \lam W => ∃ (U : C) (V : D) (W = U ∧ {SetLattice X} V)
\func CoverInterBig {X : \hType} (Cs : Array (Set (Set X))) : Set (Set X)
=> Big CoverInter (single (top {SetLattice X})) Cs
}
\func single {X : \Set} (a : X) : Set X
=> a =
\lemma single_<= {X : \Set} {a : X} {U : Set X} (Ua : U a) : single a ⊆ U
=> \lam p => rewriteI p Ua
\func \infix 8 ^-1 {X Y : \hType} (f : X -> Y) (S : Set Y) : Set X
=> \lam a => S (f a)
\type Subset \alias \infix 4 ⊆ {X : \hType} (U V : Set X) => \Pi {x : X} -> U x -> V x
\func Compl {X : \hType} (U : Set X) : Set X
=> \lam x => Not (U x)
\instance SetLattice (A : \hType) : Locale (Set A)
| <= => ⊆
| <=-refl p => p
| <=-transitive p q u => q (p u)
| <=-antisymmetric p q => ext \lam x => ext (p,q)
| meet U V a => \Sigma (U a) (V a)
| meet-left => __.1
| meet-right => __.2
| meet-univ p q c => (p c, q c)
| top _ => \Sigma
| top-univ _ => ()
| Join {J} f a => ∃ (j : J) (f j a)
| Join-cond j c => inP (j,c)
| Join-univ d (inP (j,c)) => d j c
| Join-ldistr>= (d, inP (j,c)) => inP (j,(d,c))
\func Refines {X : \hType} (C D : Set (Set X)) : \Prop
=> ∀ {U : C} ∃ (V : D) (U ⊆ V)
\lemma Refines-cover {X : \hType} {C D : Set (Set X)} (r : Refines C D) {x : X} (p : ∃ (U : C) (U x)) : ∃ (V : D) (V x) \elim p
| inP (U,CU,Ux) => \case r CU \with {
| inP (V,DV,U<=V) => inP (V, DV, U<=V Ux)
}
\lemma Refines-single_top {X : \hType} {C : Set (Set X)} : Refines C (single top)
=> \lam _ => inP (top, idp, top-univ)
\lemma Refines-refl {X : \hType} {C : Set (Set X)} : Refines C C
=> \lam CU => inP (_, CU, <=-refl)
\lemma Refines-trans {X : \hType} {C D E : Set (Set X)} (r1 : Refines C D) (r2 : Refines D E) : Refines C E
=> \lam CU => \case r1 CU \with {
| inP (V,DV,U<=V) => \case r2 DV \with {
| inP (W,EW,V<=W) => inP (W, EW, U<=V <=∘ V<=W)
}
}
\lemma Refines-inter-left {X : \hType} {C D : Set (Set X)} : Refines (Set.CoverInter C D) C
=> \lam {_} (inP (U,CU,_,_,idp)) => inP (U, CU, meet-left)
\lemma Refines-inter-right {X : \hType} {C D : Set (Set X)} : Refines (Set.CoverInter C D) D
=> \lam {_} (inP (_,_,V,DV,idp)) => inP (V, DV, meet-right)
\lemma Refines-inter-big {X : \hType} (Cs : Array (Set (Set X))) (j : Fin Cs.len) : Refines (Set.CoverInterBig Cs) (Cs j) \elim Cs, j
| C :: Cs, 0 => Refines-inter-left
| C :: Cs, suc j => Refines-trans Refines-inter-right (Refines-inter-big Cs j)
\lemma Refines-inter {X : \hType} {C D C' D' : Set (Set X)} (r : Refines C D) (r' : Refines C' D') : Refines (Set.CoverInter C C') (Set.CoverInter D D')
=> \lam {W} (inP (U,CU,U',C'U',W=UU')) => \case r CU, r' C'U' \with {
| inP (V,DV,U<=V), inP (V',D'V',U'<=V') => inP (V ∧ V', inP (V, DV, V', D'V', idp), rewrite W=UU' \$ MeetSemilattice.meet-monotone U<=V U'<=V')
}
\lemma CoverInterBig-char {X : \hType} {Cs : Array (Set (Set X))} : Set.CoverInterBig Cs = \lam W => ∃ (Us : Array (Set X) Cs.len) (\Pi (j : Fin Cs.len) -> Cs j (Us j)) (W = Big {Set X} (∧) top Us) \elim Cs
| nil => exts \lam W => ext (\lam p => inP (nil, \case __ \with {}, inv p), \lam (inP t) => inv t.3)
| C :: Cs => pmap (Set.CoverInter C) CoverInterBig-char *> exts \lam W => ext (
\lam (inP (U, CU, V, inP (Us,CsUs,q), p)) => inP (U :: Us, \case \elim __ \with {
| 0 => CU
| suc j => CsUs j
}, p *> pmap (U ∧) q),
\lam (inP (Us,CsUs,p)) => inP (Us 0, CsUs 0, _, inP (\lam j => Us (suc j), \lam j => CsUs (suc j), idp), p))
\func extend {X : \Type} {U : Set X} (V : Set (Set.Total U)) : Set X
=> \lam x => \Sigma (Ux : U x) (V (x,Ux))
\lemma extend-sub {X : \Type} {U : Set X} {V : Set (Set.Total U)} : extend V ⊆ U
=> __.1
\lemma extend-mono {X : \Type} {U : Set X} {V W : Set (Set.Total U)} (p : V ⊆ W) : extend V ⊆ extend W
=> \lam {x} (Ux,Vx) => (Ux, p Vx)
\lemma extend_meet {X : \Type} {U : Set X} {V W : Set (Set.Total U)} : extend (V ∧ W) = extend V ∧ extend W
=> <=-antisymmetric (\lam e => ((e.1, e.2.1), (e.1, e.2.2))) (\lam e => (e.1.1, (e.1.2, transport W (ext idp) e.2.2)))
\lemma restrict_extend {X : \Type} {U : Set X} {V : Set (Set.Total U)} : Set.restrict (extend V) = V
=> <=-antisymmetric (\lam r => transport V (ext idp) r.2) (\lam {x} Vx => (x.2,Vx))
\lemma extend_restrict {X : \Type} {U : Set X} {V : Set X} : extend (Set.restrict {X} {U} V) ⊆ V
=> __.2
\lemma ^-1_<= {A B : \hType} (f : A -> B) {U V : Set B} (p : U ⊆ V) : f ^-1 U ⊆ f ^-1 V
=> p __
\func ^-1_FrameHom {A B : \hType} (f : A -> B) : FrameHom (SetLattice B) (SetLattice A) \cowith
| func => ^-1 f
| func-<= p {_} => p
| func-top>= {_} _ => ()
| func-meet>= c => c
| func-Join>= c => c```
| 2,297
| 5,319
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.416456
|
http://mathhelpforum.com/differential-equations/140095-using-seperation-variables.html
| 1,480,724,395,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00243-ip-10-31-129-80.ec2.internal.warc.gz
| 176,082,204
| 9,871
|
# Thread: Using seperation of variables?
1. ## Using seperation of variables?
What is the procedure or what are the steps to be carried out here:
$\frac{X\prime}{X}=\frac{1}{4}(\frac{Y}{Y\prime})=\ lambda$
The answer is given as $X=e^{\lambda\alpha}$ and $Y=e^{\frac{\beta}{4\lambda}}$
I don't know how to get to the answer, i have done seperation of variables, but not when i have a derivative as a denominator!
2. Originally Posted by punkstart
What is the procedure or what are the steps to be carried out here:
$\frac{X\prime}{X}=\frac{1}{4}(\frac{Y}{Y\prime})=\ lambda$
The answer is given as $X=e^{\lambda\alpha}$ and $Y=e^{\frac{\beta}{4\lambda}}$
I don't know how to get to the answer, i have done seperation of variables, but not when i have a derivative as a denominator!
This produces 2 equations
$X^{ \prime } - \lambda X = 0$
and
$\lambda Y^{ \prime } - \frac{Y}{4} = 0$
This can become
$Y^ { \prime } - \frac{ Y }{ \lambda 4} = 0$
To solve both,
let $X=e^{rx}$
let $Y=e^{qy}$
Therefore the first equation becomes
$re^{rx} - \lambda e^{rx} = 0$
$r= \lambda$
Hence,
$X=C_1 e^{ \lambda x }$
Of course the co-efficient is just a constant to represent the general solution. Which can be re-written in the form that they gave you.
The same follows for the Y equation!
| 398
| 1,293
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875
| 4
|
CC-MAIN-2016-50
|
longest
|
en
| 0.868698
|
https://englishlangkan.com/question/dianna-went-to-the-hardware-store-and-bought-3-identical-copper-pipes-when-diana-lined-up-the-pi-14105051-78/
| 1,631,974,838,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00181.warc.gz
| 296,168,962
| 14,411
|
## Dianna went to the hardware store and bought 3 identical copper pipes. When Diana lined up the pipes end – to -end, the line was 0.3 meters
Question
Dianna went to the hardware store and bought 3 identical copper pipes. When Diana lined up the pipes end – to -end, the line was 0.3 meters long. How long was each pipe?
in progress 0
1 week 2021-09-10T14:27:11+00:00 1 Answer 0
| 112
| 382
|
{"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-2021-39
|
latest
|
en
| 0.961591
|
https://www.scienceforums.net/topic/77819-is-there-a-quick-way-to-determine-if-this-is-a-prime-number/page/2/#comment-778566
| 1,660,827,698,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00557.warc.gz
| 829,836,889
| 36,522
|
# Is there a quick way to determine if this is a prime number?
## Recommended Posts
Collatz Theory is a newer form of mathematics(recent). I developed it originally to try to solve the Collatz conjecture, but I found more than I bargained for with Collatz Theory. For example, I have developed algorithms for the factorizations of composites consisting of two primes, the Mersenne hailstone sequences, hailstone series equations, and Collatz-Matrix equations.
The most helpful for this problem are the Mersenne hailstone sequences, hailstone series equations, and Collatz-Matrix equations. I am getting really close to finding some equation to easily prove a Mersenne number to be a prime. It will just take some more analysis.
Hopefully you get one and learn more from you, this will help me further if I take up post graduate degrees.
• Replies 63
• Created
#### Popular Days
Well one can try to find common relationships of each product when two Mersenne primes are added to 1 and divided by each other.
$U = 4^{10^{8}}L + 4^{10^{8}} -1$
Yeah, I got to that as well, but it doesn't seem like it could do anything. Because one requires that "+1" for both primes, the relation becomes pretty trivial. We're simply looking at powers of 2 separated arithmetically by a relatively large number, so I don't see how it could shed light on your candidate's primality. However, I may be thinking too superficially, and one could use more indirect logic and make use of it in a proof.
##### Share on other sites
Yeah, I got to that as well, but it doesn't seem like it could do anything. Because one requires that "+1" for both primes, the relation becomes pretty trivial. We're simply looking at powers of 2 separated arithmetically by a relatively large number, so I don't see how it could shed light on your candidate's primality. However, I may be thinking too superficially, and one could use more indirect logic and make use of it in a proof.
Well, a proof could be derived from it. Since the two variables of U and L are prime(assuming), then it could try to prove that after a certain amount of numbers that are derived from the Mersenne formula that it will be prime or not prime.
For example, I used the new formula to get more primes than the Mersenne formula, however it took the increase of the exponent n for the 4 values to do so. After a certain while, you have to increase n in order to get more primes as you increase the size of the primes.
##### Share on other sites
Well, a proof could be derived from it. Since the two variables of U and L are prime(assuming), then it could try to prove that after a certain amount of numbers that are derived from the Mersenne formula that it will be prime or not prime.
For example, I used the new formula to get more primes than the Mersenne formula, however it took the increase of the exponent n for the 4 values to do so. After a certain while, you have to increase n in order to get more primes as you increase the size of the primes.
As I was reviewing I remembered that the exponent of 2 is prime so maybe I can create a simple program from visula basic C++ to run a non terminating exponent input on two and so on and forth. The program is relatively easy to make but the problem the processing capability of current computers. I'll test it this upcoming monday and ask some help from my younger brother to create the program.
It goes like these: (2^2)-1 = 3 which is prime by law and then substitute it back to (2^3)-1 =7 prime again, and so on and so forth until I reach the maximum possible prime number.
Edited by gabrelov
##### Share on other sites
As I was reviewing I remembered that the exponent of 2 is prime so maybe I can create a simple program from visula basic C++ to run a non terminating exponent input on two and so on and forth. The program is relatively easy to make but the problem the processing capability of current computers. I'll test it this upcoming monday and ask some help from my younger brother to create the program.
It goes like these: (2^2)-1 = 3 which is prime by law and then substitute it back to (2^3)-1 =7 prime again, and so on and so forth until I reach the maximum possible prime number.
If I'm understanding your process correctly, you're basically doing: $p_{n+1}=2^{p_{n}}-1$ for $p_1=2$?
The statement is that "For $M_n=2^n-1$, if $M_n$ is prime, then $n$ itself is also prime." However, the converse is not generally true, and so the primality of the resulting numbers in your sequence will be uncertain. Your sequence will get very large very quickly, so you'll be testing extremely large numbers (though I may be misinterpreting your idea).
##### Share on other sites
If I'm understanding your process correctly, you're basically doing: $p_{n+1}=2^{p_{n}}-1$ for $p_1=2$?
The statement is that "For $M_n=2^n-1$, if $M_n$ is prime, then $n$ itself is also prime." However, the converse is not generally true, and so the primality of the resulting numbers in your sequence will be uncertain. Your sequence will get very large very quickly, so you'll be testing extremely large numbers (though I may be misinterpreting your idea).
This is the equation I am referring to:
$U = 4^{n}L + 4^{n} - 1$
As the variable L gets larger, so will the variable n over time. For example, once U equals 71 n must be increased to 2 in order to get more prime numbers than when it is equal to 1.
##### Share on other sites
If I'm understanding your process correctly, you're basically doing: $p_{n+1}=2^{p_{n}}-1$ for $p_1=2$?
The statement is that "For $M_n=2^n-1$, if $M_n$ is prime, then $n$ itself is also prime." However, the converse is not generally true, and so the primality of the resulting numbers in your sequence will be uncertain. Your sequence will get very large very quickly, so you'll be testing extremely large numbers (though I may be misinterpreting your idea).
I'm sorry but i just got my idea from the theorem that if (2^n)-1 results in a prime then n is also a prime number.
Since Its just a theorem it may hold true for small values here are examples below and to note that n should always be prime to get also prime. Maybe its not true for bigger numbers so I'm gonna test it for bigger numbers.
22-1 = 3
23-1 = 7
25-1 = 31
27-1 = 127
The problem with my equation is that it may be simple but I think computer will not be able to run it when the number reaches so high such as thousands digits. If the computer stops processing there is nothing left doing but back to drawing board.
I think I will resort to mathematical equations again and derive and also to be able to get accurate answer.
Edited by gabrelov
##### Share on other sites
I have now been using a prime factorization algorithm to determine if the candidate is a prime number(it is going to take a while). However, it most likely is a prime number and(if my calculations are correct), it is over 18,000,000 digits long. However, I am still doing some calculations to find the precise size of the number.
UPDATE: The number is 77,631,169 digits long.
Well, I hope here is some good news.
I have used Fermat's Little Theorem to try to predict that it was a prime number, and it seems to be true. You can check my work(if anyone knows Mathematica, here is the code).
Simplify[Mod[2^257885161, 257885161] == Mod[2, 257885161],
Element[2, Integers] && Element[257885161, Primes]]
And the result turned out to be true. Also, I found another prime candidate that is over 100,000,000(167,940,168 digits long more exactly) digits long:
Simplify[Mod[2^557885161, 557885161] == Mod[2, 557885161],
Element[2, Integers] && Element[557885161, Primes]]
Though Fermat's little theorem may not be proof enough that they are prime, it still gives a hopeful chance that they are.
The first prime candidate(with a need of confirmation) is now proven to be a prime using Fermat's Little Theorem and Wilson's Theorem.
FullSimplify[Mod[((2^(257885161) - 1) - 1)!, (2^(257885161) - 1)],
Element[(2^(257885161) - 1), Primes]]
Due to some limitations with Mathematica, I will have to work around with the second prime candidate, but it has been proven with Fermat's Little Theorem to be prime.
Edited by Unity+
##### Share on other sites
Here is the sufficient proof of the primality of the first candidate using Wilson's theorem.
If this is true, then the candidate is prime.
EDIT: The first candidate has been proven a prime, a Mathematica document will be uploaded to prove as such.
Edited by Unity+
##### Share on other sites
hmm... are you sure that you are maintaining precision. For a start Mathematica relies on GMP which maxes out at 2^32 - and your number has (let alone its factorial) has more digits than that. How long is it taking mathematica to crunch an answer?
##### Share on other sites
hmm... are you sure that you are maintaining precision. For a start Mathematica relies on GMP which maxes out at 2^32 - and your number has (let alone its factorial) has more digits than that. How long is it taking mathematica to crunch an answer?
Mathematica's memory for amount of digits depends on the memory of the computer and Mathematica doesn't calculate the value in one try, but relies on an algorithm to calculate all the values(I can tell you that it takes a long time for an output to come out). I know this because I have programmed specific mathematical algorithms involving(which have existed before) to carry out the product larger than the 2^32 limit(though, I have a 64 bit computer, so it would max out at 2^64).
Well, for the actual proving of the prime number as it would be confirmed Daedalus is using some program using the NVidia GPU, which will still take some time in order for it to confirm the prime number. I simply used Fermat's Little theorem(which is still not enough to prove it, but states that it most likely is prime).
I did recheck Wilson's theorem document and realized there was something wrong. Mathematica did stop it's calculation there(unfortunately). Right now, I am waiting for Daedalus to report whether CUDALucas has determined it to be a prime or not. It will take some time. I am keeping my fingers crossed. I will give him credit if it is prime.
Since I am calculating this with Mathematica 9, there is much precision within the process. Though you could argue that computers do make rounding mistakes, which may be relevant to calculation in this circumstance, I don't see any evidence as to this conclusion.
Edited by Unity+
##### Share on other sites
even a 64 bit computer would top out at 2^37. Note I am not saying that GMP cannot handle numbers bigger than 2^32 - it can handle numbers with 2^32 digits!! But your number has close to that - and its factorial will have far more than that.
Mathematica's memory for amount of digits depends on the memory of the computer and Mathematica doesn't calculate the value in one try, but relies on an algorithm to calculate all the values(I can tell you that it takes a long time for an output to come out). I know this because I have programmed specific mathematical algorithms involving(which have existed before) to carry out the product larger than the 2^32 limit(though, I have a 64 bit computer, so it would max out at 2^64).
Well, for the actual proving of the prime number as it would be confirmed Daedalus is using some program using the NVidia GPU, which will still take some time in order for it to confirm the prime number. I simply used Fermat's Little theorem(which is still not enough to prove it, but states that it most likely is prime).
I did recheck Wilson's theorem document and realized there was something wrong. Mathematica did stop it's calculation there(unfortunately). Right now, I am waiting for Daedalus to report whether CUDALucas has determined it to be a prime or not. It will take some time. I am keeping my fingers crossed. I will give him credit if it is prime.
Since I am calculating this with Mathematica 9, there is much precision within the process. Though you could argue that computers do make rounding mistakes, which may be relevant to calculation in this circumstance, I don't see any evidence as to this conclusion.
If you look back at my initial response you will notice that I gave you how long it took for the largest mersenne prime to be checked on a GPU. It took 3 and a bit days.
Your number is almost exactly 2^200,000,000 times bigger than that number. How long do you think it is gonna take to run that check?
BTW Fermat Pseudoprimes satisfy the fermat little and yet are composite.
BTW 2 Lucas - Lehmer is a probabilistic test - and not deterministic. The deterministic are much much longer
##### Share on other sites
even a 64 bit computer would top out at 2^37. Note I am not saying that GMP cannot handle numbers bigger than 2^32 - it can handle numbers with 2^32 digits!! But your number has close to that - and its factorial will have far more than that.
If you look back at my initial response you will notice that I gave you how long it took for the largest mersenne prime to be checked on a GPU. It took 3 and a bit days.
Your number is almost exactly 2^200,000,000 times bigger than that number. How long do you think it is gonna take to run that check?
BTW Fermat Pseudoprimes satisfy the fermat little and yet are composite.
BTW 2 Lucas - Lehmer is a probabilistic test - and not deterministic. The deterministic are much much longer
Well...I don't know then. I guess that ends my search. I don't really have the resources to do anything of this magnitude at this moment.
##### Share on other sites
take a look at the great internet mersenne prime search - and take heart! They use distributed computing - maybe whilst you wait for inspiration you can join in
##### Share on other sites
take a look at the great internet mersenne prime search - and take heart! They use distributed computing - maybe whilst you wait for inspiration you can join in
##### Share on other sites
I just noticed that there is a test feature in the Prime95 program to test specific primes. This will come in handy.
Since the iteration is at 12678/257885161(approximately, I altered some iteration counts so that it only shows it for a certain amount). It will take a while, but with Prime95 it should work(unless they are also not reliable).
Edited by Unity+
##### Share on other sites
• 2 weeks later...
I am very sorry sir I cannot help you yet since my exams are approaching near and I am very nervous already and really spending all my time studying. I hope you understand.
##### Share on other sites
Apparently the prime was already tested before(supposedly the error Prime95 gave me indirectly states this).
Well, at least things were learned.
##### Share on other sites
Sorry - I never thought to check if it had already been factored
http://www.mersenne.org/report_exponent/?exp_lo=257885161&exp_hi=257885171&B1=Get+status
What strikes me is the fact that it says that even the highest number that the software can do is already factored, which means no more tests can be done until they update the software to do higher calculations.
##### Share on other sites
What strikes me is the fact that it says that even the highest number that the software can do is already factored, which means no more tests can be done until they update the software to do higher calculations.
mmm - Unless they went straight to the end of the book and read the last page to the largest number they could test and did that as a proof of concept - but there are still oodles of numbers to test in between the ceiling (ie every possibility tested below) and the maximum. I must admit if I had a new program and lots of run time I would probably "shoot for the moon" and go for the highest possibility first - when that failed I would be sensible and start marking off from the already established ceiling.
##### Share on other sites
• 2 months later...
Sir Unity may I ask what happened?
Hve you found out if that is a prime?
##### Share on other sites
Sir Unity may I ask what happened?
Hve you found out if that is a prime?
It turned out that the number was already tested not to be a prime.
##### Share on other sites
It turned out that the number was already tested not to be a prime.
##### Share on other sites
Oh I don't know about that... I learnt a lot from the thread and I hope Unity did as well; that seems a pretty good (even if not optimal) outcome
## Create an account
Register a new account
| 3,858
| 16,506
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625
| 4
|
CC-MAIN-2022-33
|
latest
|
en
| 0.966186
|
http://www.sparknotes.com/math/geometry2/theorems/section5.rhtml
| 1,511,549,248,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00753.warc.gz
| 491,667,030
| 14,140
|
# Geometry: Theorems
## Contents
page 1 of 2
Page 1
Page 2
#### Tangent Segments
Given a point outside a circle, two lines can be drawn through that point that are tangent to the circle. The tangent segments whose endpoints are the points of tangency and the fixed point outside the circle are equal. In other words, tangent segments drawn to the same circle from the same point (there are two for every circle) are equal.
Figure %: Tangent segments that share an endpoint not on the circle are equal
#### Chords
Chords within a circle can be related many ways. Parallel chords in the same circle always cut congruent arcs. That is, the arcs whose endpoints include one endpoint from each chord have equal measures.
Figure %: Arcs AC and BD have equal measures
When congruent chords are in the same circle, they are equidistant from the center.
Figure %: Congruent chords in the same circle are equidistant from the center
In the figure above, chords WX and YZ are congruent. Therefore, their distances from the center, the lengths of segments LC and MC, are equal.
A final word on chords: Chords of the same length in the same circle cut congruent arcs. That is, if the endpoints of one chord are the endpoints of one arc, then the two arcs defined by the two congruent chords in the same circle are congruent.
#### Intersecting Chords, Tangents, and Secants
A number of interesting theorems arise from the relationships between chords, secant segments, and tangent segments that intersect. First of all, we must define a secant segment. A secant segment is a segment with one endpoint on a circle, one endpoint outside the circle, and one point between these points that intersects the circle. Three theorems exist concerning the above segments.
#### Theorem 1
PARGRAPH When two chords of the same circle intersect, each chord is divided into two segments by the other chord. The product of the segments of one chord is equal to the product of the segments of the other chord.
Page 1
Page 2
| 451
| 2,011
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.65625
| 5
|
CC-MAIN-2017-47
|
latest
|
en
| 0.907828
|
https://converter.ninja/volume/us-tablespoons-to-uk-tablespoons/58-ustablespoon-to-brtablespoon/
| 1,606,233,230,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00650.warc.gz
| 240,873,440
| 5,457
|
58 US tablespoons in UK tablespoons
Conversion
58 US tablespoons is equivalent to 57.1754904875 UK tablespoons.[1]
Conversion formula How to convert 58 US tablespoons to UK tablespoons?
We know (by definition) that: $1\mathrm{ustablespoon}\approx 0.98578431875\mathrm{brtablespoon}$
We can set up a proportion to solve for the number of UK tablespoons.
$1 ustablespoon 58 ustablespoon ≈ 0.98578431875 brtablespoon x brtablespoon$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{brtablespoon}\approx \frac{58\mathrm{ustablespoon}}{1\mathrm{ustablespoon}}*0.98578431875\mathrm{brtablespoon}\to x\mathrm{brtablespoon}\approx 57.17549048750001\mathrm{brtablespoon}$
Conclusion: $58 ustablespoon ≈ 57.17549048750001 brtablespoon$
Conversion in the opposite direction
The inverse of the conversion factor is that 1 UK tablespoon is equal to 0.0174900117423326 times 58 US tablespoons.
It can also be expressed as: 58 US tablespoons is equal to $\frac{1}{\mathrm{0.0174900117423326}}$ UK tablespoons.
Approximation
An approximate numerical result would be: fifty-eight US tablespoons is about fifty-seven point one seven UK tablespoons, or alternatively, a UK tablespoon is about zero point zero two times fifty-eight US tablespoons.
Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
| 396
| 1,454
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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}
| 4.03125
| 4
|
CC-MAIN-2020-50
|
latest
|
en
| 0.725019
|
https://www.termpaperwarehouse.com/essay-on/Accounting-Ratios/492965
| 1,556,113,870,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-18/segments/1555578643556.86/warc/CC-MAIN-20190424134457-20190424160457-00504.warc.gz
| 816,231,816
| 16,040
|
# Accounting Ratios
Submitted By Aayushipandey
Words 6484
Pages 26
CHAPTER 3 – ANALYZING FINANCIAL STATEMENTS
Questions
LG1-LG5 1. Classify each of the following ratios according to a ratio category (liquidity ratio, asset management ratio, debt management ratio, profitability ratio, or market value ratio).
a. Current ratio – liquidity ratio
b. Inventory turnover ratio – asset management ratio
c. Return on assets – profitability ratio
d. Accounts payable period – asset management ratio
e. Times interest earned – debt management ratio
f. Capital intensity ratio – asset management ratio
g. Equity multiplier – debt management ratio
h. Basic earnings power ratio – profitability ratio
LG1 2. For each of the actions listed below, determine what would happen to the current ratio. Assume nothing else on the balance sheet changes and that net working capital is positive.
a. Accounts receivable are paid in cash – Current ratio does not change
b. Notes payable are paid off with cash – Current ratio increases
c. Inventory is sold on account – Current ratio does not change
d. Inventory is purchased on account– Current ratio decreases
e. Accrued wages and taxes increase – Current ratio decrease
f. Long-term debt is paid with cash – Current ratio decreases
g. Cash from a short-term bank loan is received – Current ratio decreases
LG1-LG5 3. Explain the meaning and significance of the following ratios
a. Quick ratio - Inventories are generally the least liquid of a firm’s current assets. Further, inventory is the current asset for which book values are the least reliable measures of market value. In practical terms, what this means is that if the firm must sell inventory to pay upcoming bills, the firm is most likely to have to discount inventory items in order to liquidate them, and so therefore they are the assets on which losses are most likely to occur. Therefore, the quick (or acid-test) ratio measures a…...
### Similar Documents
#### Ratio
...Ratio Formula Current Assets Current Liabilities Short term Investments Current Receivables Current Liabilities Net sales Average accounts receivable Cost of goods sold Average inventory 365 365 Net Sales Average total assets Total liabilities Total assets Total equity Total assets Total liabilities Total equity Income before interest expense and income taxes Interest Expense Net Income Net Sales Cost of goods sold Net Sales Net Income Average total assets Net income Preferred dividents Average common stockholders equity Net Sales Measure of Short‐term debt‐paying ability (2:1 guideline) Immediate short‐term debt‐paying ability (1:1 guideline) Efficiency of collection (bigger is better) Efficiency of inventory management (higher is better) Liquidity of receivables (not exceeding 1 1/3 times the days Liquidity of inventory Efficiency of assets in producing sales Creditor financing and leverage (1 is all debt, .50 means half of the assets are through debt) Owner financing Debt versus equity financing Protection in meeting interest payments (large ratio means less risky to creditors) Net income in each sales dollar (10‐15% for appliance and 1% or 2% for supermarket) Gross margin in each sales dollar Overall profitability of assets Profitability of owner investment Liquidity and Efficiency Current Ratio Acid‐test ratio Accounts receivable turnover Inventory turnover Days’ sales uncollected ......
Words: 982 - Pages: 4
#### Vodafone Accounting Ratios
...Accounting Project FAC210 / Dr.Doaa Abdu * Liquidity Ratios: 1. Current Ratio =Current AssetsCurrent Liabilites = 13029/27947 = 0.47 (2009) = 14219/28616 = 0.50 (2010) Comment: Current Ratio shows the company's ability to pay its short term obligations by its current assets. The ratio is better when higher and it improved from 2009 to 2010 2. Quick Ratio =Current Assets-InventoryCurrent Liabilites = (13029-412)/27947 = 0.45 (2009) = (14219-433)/28616 = 0.48 (2010) Comment: Quick Ratio shows the company's ability to pay its short term obligations by its easy to convert to cash assets, which the inventory isn't. The ratio is better when higher and it improved from 2009 to 2010 * Activity Ratios 1. Inventory Turnover =Cost of goods soldInventory = 25842/412 = 63 times (2009) = 29439/433 = 68 times (2010) Comment: A ratio showing how many times a company's inventory is sold and replaced over a period. The higher the better and it improved from 2009 to 2010 2. Average Collection Period =Accounts RecievableSales per day = 7662/ (41017/365) = 68 days (2009) = 8784/ (44472/365) = 72 days (2010) Comment: This is a ratio that shows how much time it is taking the company to collect its payments owed from customers. The lower the better which means that it deteriorated from 2009 to 2010 3. Asset Turnover =SalesTotal Assets = 41017/152699 = 0.27 (2009) = 44472/156985 = 0.28......
Words: 571 - Pages: 3
#### Ratio
...[pic] ANALYSIS OF FINANCIAL STATEMENTS OF HOTEL LEELA VENTURES TABLE OF CONTENTS :: 1) INTRODUCTION TO HOTEL INDUSTRY 2) PROFILE OF HOTEL LEELA VENTURES LTD. 3) OBJECTIVE OF ANALYSIS AND METHODOLOGY 4) FINANCIAL ANALYSIS USING RATIO ANALYSIS 5) INTERPRETATIONS OF THE RATIOS 6) RECOMMENDATIONS 7) REFERENCES INTRODUCTION TO HOTEL INDUSTRY Over the last decade and half the mad rush to India for business opportunities has intensified and elevated room rates and occupancy levels in India. Even budget hotels are charging USD 250 per day. The successful growth story of 'Hotel Industry in India' seconds only to China in Asia Pacific. 'Hotels in India' have supply of 110,000 rooms. According to the tourism ministry, 4.4 million tourists visited India last year and at current trend, demand will soar to 10 million in 2010 – to accommodate 350 million domestic travelers. 'Hotels in India' has a shortage of 150,000 rooms fueling hotel room rates across India. With tremendous pull of opportunity, India is a destination for hotel chains looking for growth. The World Travel and Tourism Council, India, data says, India ranks 18th in business travel and will be among the top 5 in this decade. Sources estimate, demand is going to exceed supply by at least 100% over the next 2 years. Five-star hotels in metro cities allot same room, more than once a day to different guests, receiving......
Words: 20390 - Pages: 82
#### Accessing the Financial Performance Using Accounting Ratios of Barclays Bank Ghana Limited
...1.0 INTRODUCTION 1 1.2 OVERVIEW OF INDUSTRY 2 2.0 COMPANY PROFILE 14 2.1 WHAT IS IMPRESSIVE ABOUT BBGL 15 3. ACCOUNTING POLICIES 17 4.0 FINANCIAL ANALYSIS 20 4.1 RATIO ANALYSIS 20 4.2 COMMON SIZE ANALYSIS 24 4.3 TREND ANALYSIS 25 5.0 EVALUATION AND CONCLUSION 27 6.0 WORKINGS 28 7.0 REFERENCES 34 1.0 INTRODUCTION With encouragements from the Ghana Stock Exchange for companies to be listed, it is very expedient that companies who show interest in being listed provide a profitable and an efficient view of the company to prospective investors. This is achieved in its financial statements. The decision to invest or not to invest in a company depends on the effectiveness and efficiency of the firm under consideration. Using various financial statement analysis tools, the potential investor may be able to make a decision to invest. The decision to invest does not only affect the investor but the firm as a whole. The firm will be able to raise enough capital to finance its operations. For firms whose capital requirements do not meet the requirement set by the bank of Ghana in February 2008 with deadline being December 2012, this is an avenue where such firms can fulfil this new requirement. With the current developments in the banking industry, banks of which Barclays bank is of no exception, should work at not only being profitable but also given investors value for their money. 1.2 OVERVIEW OF INDUSTRY In anticipation of the expected economic growth......
Words: 7280 - Pages: 30
#### Ratio
... | |Working Capital is more a measure of cash flow than a ratio. The result of this calculation must be a positive number. It is calculated | |as shown below: | |Working Capital = Total Current Assets - Total Current Liabilities | |Bankers look at Net Working Capital over time to determine a company's ability to weather financial crises. Loans are often tied to | |minimum working capital requirements. | |Accounting Ratios and its utility | |A relationship between various accounting figures, which are connected with each other, expressed in mathematical terms, is called | |accounting ratios. | |According to Kennedy and Macmillan, "The relationship of one item to another expressed in simple mathematical form is known as ratio." | |Robert Anthony defines a ratio as – "simply one number expressed in terms of another." | |Accounting ratios are very useful as they briefly summarise the result of......
Words: 3169 - Pages: 13
#### Ratio and Financial Ratio Analyisis
...Ratio and Financial Statements Analysis Kimberly Y. Gruber University of Maryland University College Dr. Sunando Sengupta 07/25/2013 Turnitin Score: 23% Executive Summary The purpose of this paper is to examine ratio and financial statement analysis. Such analysis is a useful tool for managers and stakeholders to evaluate a company’s financial health in order to identify opportunities for growth and areas of weakness so as to institute corrective measures. Financial statements are used in order to predict trends of cash flow within the business as well as predict the potential of a business and if they are capable of financial growth. Ratio analysis examines the probability of a company’s profit or a company’s loss. This paper will examine the benefits and limitations of ratio analysis, explain what factors impact the meaningfulness of such measures and what new practices or theories may be emerging regarding the application of ratio and financial statement. The paper concludes that ratio and financial statements is an essential tool used in analyzing a company’s profit. Introduction to Ratio and Financial Statement Analysis Though there are various methods for monitoring the liquidity of businesses the most common has been the use of financial ratios. Ratio analysis is an established technique—involving the relationship between two or more variables--that is used to conduct qualitative analysis of information contained in a company’s financial statement......
Words: 1722 - Pages: 7
Free Essay
#### Ratios
...a. Explain HOW accounting ratios can be used to monitor the financial performance of the organisation. Ensure you provide examples and use the correct data from the case study To Pass this criteria students must: * Explain what specific areas each individual ratio measures providing specific examples and data from the case study * Identify at least 2 things that the ratios can be measured against to assess a business’s performance. * Explain why it is valuable for a business to measure each ratio in improving performance, * | SCI-FI COLLECTORS year ending March 2011 | SCI-FI COLLECTORS year ending March 2012 | Industry averages | SCI-FI COLLECTORS year ending March 2013 | Return on capital employed | 95% | 95% | 80% | 90% | Gross profit percentage | 44% | 45% | 39% | 41% | Net profit percentage | 28% | 27% | 30% | 16% | Stock turnover | 44 days | 56 days | 62 days | 77 days | Debtor payment period | 1 day | 1 day | 1 day | 1 day | Credit payment period | 6 days | 13 days | 12 days | 16 days | Current ratio | 4.8 : 1 | 4.9 : 1 | 4.5 : 1 | 5 : 1 | Acid test | 1.1:1 | 0.89:1 | 1:1 | 0.6 : 1 | Summary of Ratios for Sci Fi Collectors * providing specific examples and data from the case study A partir de la informacion reflejada en el balance de una empresa, podremos realizer un diagnostic mediante los differentes tipos de ratio para monitorear el desempeno financiero de esta organizacion, ya que estas tecnicas nos ayudaran a......
Words: 1137 - Pages: 5
#### Ratio
... RATIO ANALYSIS: Fundamental Analysis has a very broad scope. One aspect looks at the general (qualitative) factors of a company. The other side considers tangible and measurable factors (quantitative). This means crunching and analyzing numbers from the financial statements. If used in conjunction with other methods, quantitative analysis can produce excellent results. Ratio analysis isn't just comparing different numbers from the balance sheet, income statement, and cash flow statement. It's comparing the number against previous years, other companies, the industry, or even the economy in general. Ratios look at the relationships between individual values and relate them to how a company has performed in the past, and might perform in the future. MEANING OF RATIO: A ratio is one figure express in terms of another figure. It is a mathematical yardstick that measures the relationship two figures, which are related to each other and mutually interdependent. Ratio is express by dividing one figure by the other related figure. Thus a ratio is an expression relating one number to another. It is simply the quotient of two numbers. It can be expressed as a fraction or as a decimal or as a pure ratio or in absolute figures as “ so many times”. As accounting ratio is an expression relating two figures or accounts or two sets of account heads or group contain in the financial statements. MEANING OF RATIO ANALYSIS: Ratio......
Words: 12924 - Pages: 52
#### D2 - the Adequacy of Accounting Ratios
Words: 1607 - Pages: 7
#### Ratio Analysis Under Different Accounting Rules
...Summary Comparing different companies under different rules, or comparing one company in different time period can be complicating and often misleading when differences in accounting methods are not captured. In this essay, I will start the analysis by examining each firm’s change in accounting methods compared to their previous year. Then I will move on to comparing the three companies’ ratios to conduct analysis of each company based on the DuPont Method. In addition, I will also look at how changes in accounting methods affected Seven and I holdings’ results in 2013 when compared to 2012. Finally, I will conclude my analysis on how comparable it was under different accounting methods based on above analysis. _____ Three Companies Chosen for Analysis For this ratio analysis assignment, I chose three companies from the UK, US and Japan: Walmart Country: US Industry: Supermarket Retail Accounting Standard: GAAP TESCO Country: UK Industry: Supermarket Retail Accounting Standard: IFRS Seven and I Holdings Country: Japan Industry: Supermarket Retail Accounting Standard: Japan Standard _____ Change/Amendments in Accounting Methods Walmart Even though there is a mention about recent accounting pronouncements and future adoption of those policies, there is and will be no effect in the firm’s net income, financial position or cash flows. Sainsbury Although there were two amendments effective from this annual reports, the firm has......
Words: 1255 - Pages: 6
#### Accounting Ratio
...Liquidity Ratios • Measures the short-term ability of the company to pay its maturing obligations and to meet unexpected needs for cash. • Ratios include the current ratio, the acid-test ratio, receivables turnover, and inventory turnover. Liquidity ratio = current asset___ current liabilities = 360042949 266476991 = 1.35 Asid test ratio = cash + short - long term investment + receivable (net) Current liabilities = 238446962+26656729+76692918____________________ 266476991 = 1.28 Receivable turnover = __Net credit sales____ Average net receivable = 574273012 (238446962+138220476)/2 = 3.049 : 1 365 days / 3.049 times = 119.71 days This means that receivables are collected on average every 217 days Inventory turnover = cost of goods sold Average inventory = 461246689 (18246340+2574857)/2 = 44.31 Profitability Ratios • Measures the income or operating success of a company for a given period of time. • Income, or lack of it, affects the company’s ability to obtain debt and equity......
Words: 391 - Pages: 2
#### Accounting Ratios
...History Friedrich Fischer designs the ball grinder. This machine allows steel balls to be ground to an absolutely round state for the first time – and in large volumes. Thanks to this innovation, he lays the foundation for the entire rolling bearing industry. Thus, the worldwide success story of the ball bearing begins in Schweinfurt.Later, 1883 is officially declared the year in which the company was founded. 1896Friedrich Fischer applies for permission to build a new plant near the train station in Schweinfurt – a step towards a new industrial dimension. The new plant produces 10 million balls per week. The company is incorporated one year later. On 29 july 1905, the FAG brand was registered with the patent office in Berlin. The registered trademark FAG, which stands for Fischers Aktien-Gesellschaft, is protected in over 100 countries today. In 1909 Georg Schäfer takes over the “First Automated Cast Steel Factory, previously Friedrich Fischer, AG“ (“Erste Automatische Gußstahlkugelfabrik, vormals Friedrich Fischer, AG“) The Schaeffler Technologies AG & Co. KG (also known as Schaeffler Group respectively Schaeffler Gruppe inGerman) is a privately owned major manufacturer of rolling element bearings for automotive, aerospace and industrial uses. In Germany, the main brands of the Schaeffler Group – INA, FAG and LuK – are marketed by Schaeffler Technologies AG & Co. KG and LuK GmbH & Co. oHG. Business of the Company Rolling And Plain Bearings * Deep...
Words: 620 - Pages: 3
#### Accounting Ratios and Monitoring Business Performance
...Task 11 (D2) Accounting ratios and monitoring business performance Ratio analysis can be used as a management tool to monitor and improve the performance of HSBC as well as being used by those outside of the organisation such as bank regulators, potential shareholders and suppliers to look at the performance of HSBC and compare it with other similar organisations. Information used for comparison must be accurate - otherwise the results will be misleading. There are four main methods of ratio analysis - liquidity, solvency, efficiency and profitability. If ratios of companies are to be compared it is important that the companies are in the same industry. It would be appropriate to compare HSBC ratios with other the ratios of other banks but not for example a construction company. Liquidity ratios These ratios should be used on a daily basis by management to monitor performance and manage cash flow risks. There are three types of liquidity ratio: * Current ratio - current assets divided by current liabilities. This assesses whether you have sufficient assets to cover your liabilities. A ratio of two for example shows you have twice as many current assets as current liabilities. * Quick or acid-test ratio - current assets (excluding inventory) divided by current liabilities. A ratio of one shows liquidity levels are high - an indication of solid financial health. * Defensive interval - liquid assets divided by daily operating expenses. This measures how......
Words: 1053 - Pages: 5
#### Ratios
...The use of Financial Ratios for Research: Problems Associated with and Recommendations for Using Large Databases Introduction The use of financial ratio analysis for understanding and predicting the performance of privately owned business firms is gaining in importance in published research. Perhaps the major problem faced by researchers is the difficulty of obtaining an adequate sample of representative financial statements with many studies using 50 or fewer firms for analysis. However, when larger databases are used, it is important to know that they have problems as well and that adjustments to these samples must be made to permit the use of multivariate analysis techniques. Understanding how to properly use large databases for ratio analysis will become of importance now that the Kauffman Center for Entrepreneurial Leadership (KCEL) has developed a financial statement database of more than 400,000 privately owned firms with a significant number of these including a base year and three operating years of financial statements. This database is currently available to a team of scholars working closely with the KCEL on selected internal studies. It is expected that this database will become generally available to researchers and this source of financial statement information is likely to become the standard for financial performance research in the future. For the first time, scholars will have a large commonly available database of privately owned firm......
Words: 1788 - Pages: 8
#### Ratios
...The Use of Ratio Analysis Ratio analysis is a tool used by individuals to conduct a quantitative analysis of information in a company's financial statements. Ratios are calculated from current year numbers and are then compared to previous years, other companies, the industry, or even the economy to judge the performance of the company. Ratio analysis is predominately used by proponents of fundamental analysis to judge the performance of the company. Analyzing ratios is used to evaluate a company's present performance and its possible future performance. In a fact, interpretation of different accounting ratio lets the researcher fully understand the financial condition and performance of a business concern. Ratio itself is the comparison of one figure to another relevant figure. (http://www.investopedia.com/terms/r/ratioanalysis.asp) There are many ratios that you can use to analyze the financial health of a business. In this paper I will discuss four financial performance areas that I think are worth analyzing: Liquidity, profitability, solvency, and efficiency. I will discuss the strengths and weaknesses of using these ratios. First of all, Liquidity is the ability of the firm to convert assets into cash. It is also called marketability or short-term solvency. The liquidity of a business firm is usually of particular interest to its short-term creditors since the liquidity of the firm measures its ability to pay those creditors. Several financial ratios measure the......
Words: 798 - Pages: 4
| 5,002
| 22,281
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2019-18
|
longest
|
en
| 0.905219
|
http://www.encyclopediaofmath.org/index.php/Appell_polynomials
| 1,371,533,929,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368706933615/warc/CC-MAIN-20130516122213-00093-ip-10-60-113-184.ec2.internal.warc.gz
| 428,151,851
| 9,424
|
# Appell polynomials
A class of polynomials over the field of complex numbers which contains many classical polynomial systems. The Appell polynomials were introduced by P.E. Appell [1]. The series of Appell polynomials is defined by the formal equality
(1)
where is a formal power series with complex coefficients , and . The Appell polynomials have an explicit expression in terms of the numbers as follows:
The condition is tantamount to saying that the degree of the polynomial is .
There is another, equivalent, definition of Appell polynomials. Let
be a differential operator, generally of infinite order, defined over the algebra of complex polynomials in the variable . Then
i.e. is the image of the function under the mapping , .
The class of Appell polynomials is defined as the set of all possible systems of polynomials with generating functions of the form (1). To say that a system of polynomials (of degree ) belongs to the class amounts to saying that the relationships
are valid.
The Appell polynomials of class are sometimes defined by
which, apart from normalization, are equivalent to those given above.
Appell polynomials of class are used to solve equations of the form:
(2)
The formal equality for
makes it possible to write the solution of (2) in the form
where are Appell polynomials with the generating function . In this connection the expansion of analytic functions into Appell polynomials is of special interest. Appell polynomials also find use in various problems connected with functional equations, including differential equations other than (2), in interpolation problems, in approximation theory, in summation methods, etc. (cf. [1][6]). For a more general account of the theory of Appell polynomials of class , and a number of applications, see [6].
The class contains, as special cases, a large number of classical sequences of polynomials. Examples, up to a normalization, are the Bernoulli polynomials
etc. For many other examples, see [2] and [3], Vol. 3.
There exist various generalizations of Appell polynomials, which are also known as systems of Appell polynomials. These include the Appell polynomials with generating functions of the form
(3)
and the Appell polynomials with the more general generating functions:
(4)
(see, for example, [2], [3], Vol. 3). If is the inverse function to the function , then the fact that the system of polynomials belongs to the class of sequences of Appell polynomials with a generating function of type (3) is equivalent to the validity of the relations
There are only five weighted orthogonal systems of sequences of Appell polynomials on the real axis with generating functions of the type (3); these include only one orthogonal system with generating functions of the type (1), which consists of Hermite polynomials with the weight on the real axis (cf. [7]).
For the expansion in series by Appell polynomials with generating functions of the types (3) and (4), and interconnections of these polynomials by various functional equations see [2], [7], [8].
The class , where is an integer, of Appell polynomials is defined as follows: It is the set of all systems of polynomials for each of which the (formal) representation
is valid. Here, , and , , are formal power series, the free terms of which are such that the degree of the polynomial is . To say that a sequence of polynomials of degree belongs to amounts to saying that the relations
are valid. For problems on the expansion of analytic functions in series by Appell polynomials of class , see . They are closely connected with the problem of finding analytic solutions of functional equations of the type
Appell polynomials in two variables were introduced by P. Appell [10]. They are defined by the equations:
in which it is assumed that , for ; these Appell polynomials are analogues of the Jacobi polynomials. The Appell polynomials are orthogonal with the weight
(5)
to any polynomials in two variables of degree lower than , over the domain , where is the triangle: , , ; however, they do not form a system of orthogonal functions with the weight in (see, for example, [3], Vol. 2).
#### References
[1] P.E. Appell, Ann. Sci. École Norm. Sup. , 9 (1880) pp. 119–144 [2] R.P. Boas, R.C. Buck, "Polynomial expansions of analytic functions" , Springer & Acad. Press (U.S.A. & Canada) (1958) [3] H. Bateman (ed.) A. Erdélyi (ed.) , Higher transcendental functions , 1–3 , McGraw-Hill (1953–1955) [4] B. Wood, "Generalized Szász operators for the approximation in the complex domain" SIAM J. Appl. Math. , 17 (1969) pp. 790–801 [5] G. Szegö, "Orthogonal polynomials" , Amer. Math. Soc. (1975) [6] N. Bourbaki, "Elements of mathematics. Functions of a real variable" , Addison-Wesley (1976) (Translated from French) [7] J. Meixner, "Orthogonale Polynomsysteme mit einer besonderen Gestalt der erzeugenden Funktion" J. London Math. Soc. (1) , 9 (1934) pp. 6–13 [8] Ch.A. Anderson, "Some properties of Appell-like polynomials" J. Math. Anal. Appl. , 19 (1967) pp. 475–491 [9a] Yu.A. Kaz'min, "Expansions in series of Appell polynomials" Math. Notes , 5 : 5 (1969) pp. 304–311 Mat. Zametki , 5 : 5 (1969) pp. 509–520 [9b] Yu.A. Kaz'min, "On Appell polynomials" Math. Notes , 6 : 2 (1969) pp. 556–562 Mat. Zametki , 6 : 2 (1969) pp. 161–172 [10] P.E. Appell, Arch. Math. Phys. (1) , 66 (1881) pp. 238–245
| 1,365
| 5,383
|
{"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}
| 3.53125
| 4
|
CC-MAIN-2013-20
|
longest
|
en
| 0.932593
|
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-4-section-4-4-trigonometric-functions-of-any-angle-exercise-set-page-575/38
| 1,529,905,820,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867493.99/warc/CC-MAIN-20180625053151-20180625073151-00563.warc.gz
| 841,221,437
| 14,303
|
## Precalculus (6th Edition) Blitzer
$30^o$
RECALL: The following are the means on how to find the reference angle of an angle $\theta$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $180^o-\theta$ (3) Quadrant III: $\theta - 180^o$ (4) Quadrant IV: $360^o-\theta$ $210^o$ is in Quadrant III. Thus, using forrmula (3) above gives: reference angle of $210^o$ = $210^o-180^o=30^o$
| 146
| 406
|
{"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}
| 3.765625
| 4
|
CC-MAIN-2018-26
|
longest
|
en
| 0.740453
|
https://www.vskills.in/certification/tutorial/excel-formulas/
| 1,716,570,588,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00353.warc.gz
| 926,316,916
| 21,731
|
# Excel Formulas
A formula is a set of mathematical instructions that can be used in Excel to perform calculations. Formals are started in the formula box with an = sign.
There are many elements to and excel formula.
• References: The cell or range of cells that you want to use in your calculation
• Operators: Symbols (+, -, *, /, etc.) that specify the calculation to be performed
• Constants: Numbers or text values that do not change
• Functions: Predefined formulas in Excel
To create a basic formula in Excel:
• Select the cell for the formula
• Type = (the equal sign) and the formula
• Click Enter
Calculate with Functions
A function is a built in formula in Excel. A function has a name and arguments (the mathematical function) in parentheses. Common functions in Excel:
• Sum: Adds all cells in the argument
• Average: Calculates the average of the cells in the argument
• Min: Finds the minimum value
• Max: Finds the maximum value
• Count: Finds the number of cells that contain a numerical value within a range of the argument
To calculate a function:
• Click the cell where you want the function applied
• Click the Insert Function button
• Choose the function
• Click OK
• Complete the Number 1 box with the first cell in the range that you want calculated
• Complete the Number 2 box with the last cell in the range that you want calculated
Function Library
The function library is a large group of functions on the Formula Tab of the Ribbon. These functions include:
• AutoSum: Easily calculates the sum of a range
• Recently Used: All recently used functions
• Financial: Accrued interest, cash flow return rates and additional financial functions
• Logical: And, If, True, False, etc.
• Text: Text based functions
• Date & Time: Functions calculated on date and time
• Math & Trig: Mathematical Functions
Relative, Absolute, and Mixed References
Calling cells by just their column and row labels (such as “A1”) is called relative referencing. When a formula contains relative referencing and it is copied from one cell to another, Excel does not create an exact copy of the formula. It will change cell addresses relative to the row and column they are moved to. For example, if a simple addition formula in cell C1 “=(A1+B1)” is copied to cell C2, the formula would change to “=(A2+B2)” to reflect the new row. To prevent this change, cells must be called by absolute referencing and this is accomplished by placing dollar signs “\$” within the cell addresses in the formula. Continuing the previous example, the formula in cell C1 would read “=(\$A\$1+\$B\$1)” if the value of cell C2 should be the sum of cells A1 and B1. Both the column and row of both cells are absolute and will not change when copied. Mixed referencing can also be used where only the row OR column fixed. For example, in the formula “=(A\$1+\$B2)”, the row of cell A1 is fixed and the column of cell B2 is fixed.
Linking Worksheets
You may want to use the value from a cell in another worksheet within the same workbook in a formula. For example, the value of cell A1 in the current worksheet and cell A2 in the second worksheet can be added using the format “sheetname!celladdress”. The formula for this example would be “=A1+Sheet2!A2” where the value of cell A1 in the current worksheet is added to the value of cell A2 in the worksheet named “Sheet2”.
### Get industry recognized certification – Contact us
keyboard_arrow_up
| 793
| 3,459
|
{"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}
| 3.234375
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.805982
|
https://ncatlab.org/nlab/show/formal+topology
| 1,675,243,054,000,000,000
|
application/xhtml+xml
|
crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00501.warc.gz
| 433,305,419
| 13,975
|
# nLab formal topology
Formal topology
This entry is about a generalized notion of topology. For the notion of formal space in the sense of rational homotopy theory, see formal dg-algebra.
# Formal topology
## Idea
Formal topology is a programme for doing topology in a finite, predicative, and constructive fashion.
It is a kind of pointless topology; in the context of classical mathematics, it reproduces the theory of locales rather than topological spaces (although of course one can recover topological spaces from locales).
The basic definitions can be motivated by an attempt to study locales entirely through the posites that generate them. However, in order to recover all basic topological notions (particularly those associated with closed rather than open features) predicatively, we need to add a ‘positivity’ relation to the ‘coverage’ relation of sites.
## Definitions
A formal topology or formal space is a set $S$ together with
• an element $\top$ of $S$,
• a binary operation $\cap$ on $S$,
• a binary relation $\lhd$ between elements of $S$ and subsets of $S$, and
• a unary relation $\Diamond$ on $S$,
such that
1. $a = b$ whenever $a \lhd \{b\}$ and $b \lhd \{a\}$,
2. $a \lhd U$ whenever $a \in U$,
3. $a \lhd V$ whenever $a \lhd U$ and $x \lhd V$ for all $x \in U$,
4. $a \cap b \lhd U$ whenever $a \lhd U$ or $b \lhd U$,
5. $a \lhd \{ x \cap y \;|\; x \in U,\; y \in V \}$ whenever $a \lhd U$ and $a \lhd V$,
6. $a \lhd \{\top\}$,
7. $\Diamond x$ for some $x \in U$ whenever $\Diamond a$ and $a \lhd U$, and
8. $a \lhd U$ whenever $a \lhd U$ if $\Diamond a$,
for all $a$, $b$, $U$, and $V$.
We interpret the elements of $S$ as basic opens in the formal space. We call $\top$ the entire space and $a \cap b$ the intersection of $a$ and $b$. We say that $a$ is covered by $U$ or that $U$ is a cover of $a$ if $a \lhd U$. We say that $a$ is positive or inhabited if $\Diamond a$. (For a topological space equipped with a strict topological base $S$, taking these intepretations literally does in fact define a formal space; see the Examples.)
Some immediate points to notice:
• If we drop (1), then the hypothesis of (1) defines an equivalence relation on $S$ which is a congruence for $\top$, $\cap$, $\lhd$, and $\Diamond$, so that we may simply pass to the quotient set. In appropriate foundations, we can even allow $S$ to be a preset originally, then use (1) as a definition of equality.
• We can prove that $(S,\cap,\top)$ is a bounded semilattice; if (as the notation suggests) we interpret this as a meet-semilattice, then $a \leq b$ if and only if $a \lhd \{b\}$. Conversely, we could require that $(S,\cap,\top)$ be a semilattice originally, then let (1) say that $a \leq b$ whenever $a \lhd \{b\}$.
• We can prove that $\Diamond a$ holds iff every cover of $a$ is inhabited and that $\Diamond a$ fails iff $a \lhd \empty$. Accordingly, this predicate is uniquely definable (in two equivalent ways, one impredicative and one nonconstructive) in a classical treatment; only in a treatment that is both predicative and constructive do we need to include it in the axioms. See positivity predicate.
## Examples
Let $X$ be a topological space, and let $S$ be the collection of open subsets of $X$. Let $\top$ be $X$ itself, and let $a \cap b$ be the literal intersection of $a$ and $b$ for $a, b \in S$. Let $a \lhd U$ if and only $U$ is literally an open cover of $a$, and let $\Diamond a$ if and only if $a$ is literally inhabited. Then $(S,\top,\cap,\lhd,\Diamond)$ is a formal topology.
The above example is impredicative (since the collection of open subsets is generally large), but now let $S$ be a base for the topology of $X$ which is strict in the sense that it is closed under finitary intersection. Let the other definitions be as before. Then $(S,\top,\cap,\lhd,\Diamond)$ is a formal topology.
More generally, let $B$ be a subbase for the topology of $X$, and let $S$ be the free monoid on $B$, that is the set of finite lists of elements of $B$ (so this example is not strictly finitist), modulo the equivalence relation by which two lists are identified if their intersections are equal. Let $\top$ be the empty list, let $a \cap b$ be the concatenation of $a$ and $b$, let $a \lhd U$ if the intersection of $a$ is contained in the union of the intersections of the elements of $U$, and let $\Diamond a$ if the intersection of $a$ is inhabited. Then $(S,\top,\cap,\lhd,\Diamond)$ is a formal topology.
Let $X$ be an accessible locale generated by a posite whose underlying poset $S$ is a (meet)-semilattice. Let $\top$ and $\cap$ be as in the semilattice structure on $S$, and let $a \lhd U$ if $U$ contains a basic cover (in the posite structure on $S$) of $a$. Then we get a formal topology, defining $\Diamond$ in the unique way.
The last example is not predicative, and this is in part why one studies formal topologies instead of sites, if one wishes to be strictly predicative. (It still needs to be motivated that we want $\Diamond$ at all.)
## In dependent type theory
In dependent type theory, if one doesn’t have a type of all propositions, then one usually cannot define a relation between a type $S$ and the type of all subtypes of $S$, which is necessary for defining the coverage $\lhd$. Instead, one has to use a Tarski universe to define the type of all locally $U$-small subtypes of $S$, and use that to define a locally $U$-small formal topology on $S$.
Let $(U, T)$ be a Tarski universe, and let $\mathrm{Prop}_U$ be the type of all propositions in $U$
$\mathrm{Prop}_U \coloneqq \sum_{A:U} \mathrm{isProp}(T(A))$
which comes with an embedding $\mathrm{asType}_U:\mathrm{Prop}_U \hookrightarrow U$. In addition, we define the relation $a \in_S^U A$ for $a:S$ and $A:S \to \mathrm{Prop}_U$ as
$a \in_S^U A \coloneqq \mathrm{isContr}(T(\mathrm{asType}(A(a)))$
The singleton subtype function is a function $\{-\}:S \to (S \to \mathrm{Prop}_U)$, such that for all elements $a:S$, $p(a):a \in_S^U \{a\}$, and for all functions $R:S \to (S \to \mathrm{Prop}_U)$ such that for all elements $a:S$, $p_R(a):a \in_S^U R(a)$, the type of functions $(b \in_S^U \{a\}) \to (b \in_S^U R(a))$ is contractible for all $a:A$ and $b:A$.
Given a Tarski universe $(U, T)$, a locally $U$-small formal topology or locally $U$-small formal space is a type $S$ together with
• an element $\top:S$,
• a binary operation $\cap:S \times S \to S$,
• a binary relation $a \lhd A$ between elements of $S$, $a:S$, and locally $U$-small subsets of $S$, $A:S \to \mathrm{Prop}_U$, and
• a unary relation $\Diamond a$ on $S$,
such that
1. For all elements $a:S$ and $b:S$, the canonical function
$\mathrm{idToFormalTop}(a, b):(a =_S b) \to ((a \lhd \{b\}) \times (b \lhd \{a\}))$
is an equivalence of types.
$\prod_{a:S} \prod_{b:S} \mathrm{isEquiv}(\mathrm{idToFormalTop}(a, b))$
2. For all elements $a:S$ and subsets $A:S \to \mathrm{Prop}_U$, if $a \in_S^U A$, then $a \lhd A$
$\prod_{a:S} \prod_{A:S \to \mathrm{Prop}_U} (a \in_S^U A) \to (a \lhd A)$
3. For all elements $a:S$ and subsets $A:S \to \mathrm{Prop}_U$ and $B:S \to \mathrm{Prop}_U$, if $a \lhd A$ and for all elements $x:S$, $x \lhd B$ and $x \in_S^U A$, then $a \lhd B$
$\prod_{a:S} \prod_{A:S \to \mathrm{Prop}_U} \prod_{B:S \to \mathrm{Prop}_U} \left((a \lhd A) \times \prod_{x:S} (x \in_S^U A) \times (x \lhd B)\right) \to (a \lhd B)$
4. For all elements $a:S$ and $b:S$ and subsets $A:S \to \mathrm{Prop}_A$, if $a \lhd A$ or $b \lhd A$, then $a \cap b \lhd A$.
$\prod_{a:S} \prod_{b:S} \prod_{A:S \to \mathrm{Prop}_U} ((a \lhd A) \times (b \lhd A)) \to (a \cap b \lhd A)$
5. For all elements $a:S$ and subsets $A:S \to \mathrm{Prop}_U$ and $B:S \to \mathrm{Prop}_U$, if $a \lhd A$ and $a \lhd B$, then
$a \lhd \sum_{z:S} \prod_{x:S} \prod_{y:S} (x \in_S^U A) \times (y \in_S^U B) \times (z =_S x \cap u)$
$\prod_{a:S} \prod_{A:S \to \mathrm{Prop}_U} \prod_{B:S \to \mathrm{Prop}_U} (a \lhd A) \times (a \lhd B) \to \left(a \lhd \sum_{z:S} \prod_{x:S} \prod_{y:S} (x \in_S^U A) \times (y \in_S^U B) \times (z =_S x \cap y)\right)$
6. For all elements $a:S$, $a \lhd \{\top\}$,
$\prod_{a:S} a \lhd \{\top\}$
7. For all elements $a:S$ and subsets $A:S \to \mathrm{Prop}_U$, if $\Diamond a$ and $a \lhd A$, there exists an element $x:S$ such that $x \in_S A$ and $\Diamond x$.
$\prod_{a:S} \prod_{A:S \to \mathrm{Prop}_U} (\Diamond a) \times (a \lhd A) \to \sum_{x:S} (x \in_S A) \times (\Diamond x)$
8. For all elements $a:S$ and subsets $A:S \to \mathrm{Prop}_U$, if $\Diamond a$ implies $a \lhd A$, then $a \lhd A$.
$\prod_{a:S} \prod_{A:S \to \mathrm{Prop}_U} (\Diamond a \to a \lhd A) \to (a \lhd A)$
## References
• Mike Fourman and Grayson (1982); Formal Spaces. This is the original development, intended as an application of locale theory to logic.
• Giovanni Sambin (1987); Intuitionistic formal spaces; pdf.
• This is the probably the main reference on the subject.
• Warning: you can ignore the material about foundations and type theory, but if you do read any of it, the term ‘category’ means (roughly) proper class; this was common in type theory in the 1980s.
• Giovanni Sambin (2001); Some points in formal topology; pdf. This has newer results, alternative formulations, etc.
• Erik Palmgren, From Intuitionistic to Point-Free Topology: On the Foundation of Homotopy Theory, Logicism, Intuitionism, and Formalism Volume 341 of the series Synthese Library pp 237-253, 2005 (pdf)
Last revised on November 17, 2022 at 22:01:31. See the history of this page for a list of all contributions to it.
| 3,052
| 9,549
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 198, "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-2023-06
|
longest
|
en
| 0.820313
|
http://www.chegg.com/homework-help/questions-and-answers/a-road-between-fairbanks-and-nome-alaska-will-have-a-most-likely-construction-of-4-million-q3260060
| 1,369,480,512,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368705939136/warc/CC-MAIN-20130516120539-00092-ip-10-60-113-184.ec2.internal.warc.gz
| 388,799,151
| 7,633
|
## expected values
a road between fairbanks and Nome, alaska, will have a most likely construction of \$4 million per mile. Doubling this cost is considered to have a probability of 30%, and cutting it by 25% is considered to have a probability of 10%. the states interst rate is 8% and the road should last 40 years. what is the probability distribution of the equivalent annual construction cost per mile? calculate the expected value of the equivalent annyual construction cost per mile
| 106
| 490
|
{"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-2013-20
|
latest
|
en
| 0.94905
|
https://math.stackexchange.com/questions/1195739/resolvent-decay-behavior?noredirect=1
| 1,726,507,896,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00705.warc.gz
| 353,711,140
| 37,498
|
# Resolvent: Decay Behavior
Given a Hilbert space $\mathcal{H}$.
Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
Denote resolvent set: $$\rho(H):=\{z\in\mathbb{C}:(z-H)^{-1}\in\mathcal{B}(\mathcal{H})\}$$
Define the ratio: $$\eta_z:\sigma(H)\to\mathbb{C}:\lambda\mapsto\tfrac{|z-\lambda|}{1+|\lambda|}$$
Then one has estimates: $$z\in\rho(H):\quad\delta_-(z)\leq\eta_z\leq\delta_+(z)$$
Moreover one has: $$z\in\rho(H):\quad\delta_\pm(z)=\delta_\pm(\overline{z})$$
How to prove these?
• I'm not sure what you are asking. Presumably $m_z$ and $M_z$ are constants that depend on $z$ but not on $\lambda$, but what is $\lambda$? When you ask "how to check these," are you asking about a proof? a computation? Commented Mar 26, 2015 at 18:40
• @hardmath: Oh yes these are supposed to be constants w.r.t. $\lambda$ that still depend on $z$. $\lambda$ is meant to lie within the spectrum. (I've added that.) I'm asking for a quick 'proof'. Hope it is clearer now. Thanks for pointing me out on this!!! Commented Mar 27, 2015 at 17:24
This has very little to do with Hilbert spaces or Hamiltonians. If $\sigma(H)$ is a closed set in the complex plane and $z \notin \sigma(H)$, then $f(\lambda) = \dfrac{|z - \lambda|}{1 + |\lambda|}$ is a continuous positive function on $\mathbb C$, with limit $1$ as $|\lambda| \to \infty$, and its only zero is at $\lambda = z$. Therefore it is bounded away from $0$ on $\{\lambda: |\lambda - z| \ge d(z, \sigma(H))\}$.
| 498
| 1,479
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.34375
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.846512
|
http://www.revisemri.com/questions/artefacts/motion_phase
| 1,544,978,253,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376827769.75/warc/CC-MAIN-20181216143418-20181216165418-00150.warc.gz
| 471,166,486
| 5,023
|
This site will look better in a browser that supports web standards, but it is accessible to any browser or Internet device.
### QUESTIONS» Image Artefacts
These questions are concerned with MRI image features which do not represent the object in the field-of-view.
# Motion Artefact
## Why is motion artefact only seen in the phase encoding direction?
Although the encoding methods in each direction are mathematically equivalent, a frequency-encode step takes much less time (of the order of milliseconds) than a phase encoding step (of the order of seconds). In the frequency-encode direction, all the samples of a signal are acquired in the time of a single echo, whereas in the phase-encode direction all the lines of k-space must be collected to obtain a complete data set for Fourier reconstruction—each line is separated by the time interval TR. This means that most motions that occur during clinical MRI are much slower than the rapid sampling process along the frequency encoding axis. For this reason although motion artefacts along the frequency encoding axis may occur, they are usually insignificant and the most effect they have is a slight blurring.
In the following images, the motion artefact (blurring in the phase encoding direction—up-down—across the whole image) is present in spite of the direction of the motion which occured during the scan.
Motion in the phase encoding direction (up-down). The strong contrast of the bottle-phantom simply represents where the phantom was when the central k-space lines were acquired (high up; some aliasing has occured). The edge information is in the centre of the image because this is where the phantom was when the peripheral k-space lines were acquired. The blurring of signal across the whole image in the phase encoding direction (up-down) is motion artefact.
Motion in the frequency encoding direction (left-right). The strong contrast of the bottle-phantom simply represents where the phantom was when the central k-space lines were acquired (to the right). The edge information is in the centre of the image because this is where the phantom was when the peripheral k-space lines were acquired. The blurring of signal across the whole image in the phase encoding direction (up-down) is motion artefact.
Motion artefacts are also considered in the K-Space Tool.
| 462
| 2,341
|
{"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-2018-51
|
longest
|
en
| 0.947692
|
https://www.yumpu.com/en/document/view/59996491/us6506148/19
| 1,529,922,242,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867644.88/warc/CC-MAIN-20180625092128-20180625112128-00331.warc.gz
| 949,869,096
| 20,345
|
Views
2 months ago
# US6506148
## 15 f D(r) = eV(r) =
15 f D(r) = eV(r) = (1-(r/R)), (12) directed perpendicular to the Screen. Note that the plate Separation 6 has dropped out. This means that the precise location of the polarization charges is not critical in the present model, and further that Ö may be taken as Small as desired. Taking 6 to Zero, one thus arrives at the mathemati cal model of pulsed dipoles distributed over the circular CRT Screen. The field due to the charge distribution qo will be calculated later. The electric field induced by the distributed dipoles (12) can be calculated easily for points on the centerline of the Screen, with the result V(0) where V(0) is the pulse voltage (11) at the Screen center, p the distance to the rim of the Screen, and Z the distance to the center of the screen. Note that V(0) pulses harmonically with frequency J, because in (11) the sinusoidal part J of the beam current varies in this manner. The electric field (13) due to the dipole distribution causes a potential distribution V(r)/2 over the Screen and a potential distribution of-V(r)/2 over the polarization disc, where V(r) is nonuniform as given by (11). But Since the polarization disc is a perfect conductor it cannot Support Voltage gradients, and therefore cannot have the potential distribu tion -V(r)/2. Instead, the polarization disc is at ground potential. This is where the charge distribution q(r) comes in; it must be such as to induce a potential distribution V(r)/2 over the polarization disc. Since the distance between polar ization disc and Screen Vanishes in the mathematical model, the potential distribution V(r)/2 is induced over the screen as well. The total potential over the monitor screen thus becomes V(r) of (11), while the total potential distribution over the polarization disc becomes uniformly zero. Both these potential distributions are as physically required. The electric charges q are moved into position by polarization and are partly drawn from the earth through the ground connection of the CRT. In our model the charge distribution qo is located at the Same place as the dipole distribution, viz., on the plane Z=0 within the circle with radius R. At points on the center line of the Screen, the electric field due to the monopole distri bution qo is calculated in the following manner. AS discussed, the monopoles must be Such that they cause a potential (po that is equal to V(r)/2 over the disc with radius R centered in the plane Z=0. Although the charge distribution do(r) is uniquely defined by this condition, it cannot be calculated easily in a Straightforward manner. The difficulty is circumvented by using an intermediate result derived from Excercise 2 on page 191 of Kellogg (1953), where the charge distribution over a thin disc with uniform potential is given. By using this result one readily finds the potential (p(Z) on the axis of this disc as 2 (14) s" (z) = V f(R), where f3(R) is the angle Subtended by the disc radius R, as viewed from the point Z on the disc axis, and V* is the disc potential. The result is used here in an attempt to construct the potential (p(Z) for a disc with the nonuniform potential US 6,506,148 B2 15 25 35 40 45 50 55 60 65 16 V(r)/2, by the ansatz of writing the field as due to a linear combination of abstract discs with various radii R and potentials, all centered in the plane Z=0. In the ansatz the potential on the Symmetry axis is written where W is chosen as the function 1-R,°/R, and the constants a and b are to be determined Such that the potential over the plane z=0 is V(r)/2 for radii r ranging from 0 to R, with V(r) given by (11). Carrying out the integration in (15) g|Ves In order to find the potential over the disc r-R in the plane Z=0, the function (po(z) is expanded in powers of Z/R for 0
17 experimental data points 103 for the 15- computer monitor and for the 30" TV tube. FIG. 18 shows that the developed theory agrees fairly well with the experimental results. From the best fit one can find the center-Screen Voltage pulse amplitudes. The results, normalized as discussed above, are V(0)=266.2 volt for the 15" computer monitor and V(0)= 310.1 volt for the 30" TV tube. With these amplitudes in hand, the emitted pulsed electric field along the center line of the monitors can be calculated from the Sum of the fields (13) and (19). For instance, for the 15" computer monitor with 1.8% RGB pulse modulation used in the /3 Hz sensory resonance experiments mentioned above, the pulsed electric field at the center of the subject, located at Z=70 cm on the Screen center line, is calculated as having an amplitude of 0.21 V/m. That Such a pulsed electric field, applied to a large portion of the Skin, is Sufficient for exciting the /2 HZ Sensory resonance is consistent with experimental results discussed in the 874 patent. In deriving (11), the dimensionless number 2 t?cAm was said to be much smaller than unity. Now that the values for V(0) are known, the validity of this statement can be checked. Eq. (11) implies that V(0) is equal to m/4t. The Sum of the beam currents in the red, green, and blue electron guns for 100% intensity modulation is estimated to have pulse amplitudes J of 0.5 mA and 2.0 mA respectively for the 15" computer monitor and the 30" TV tube. Using the derived values for V(0), one arrives at estimates for the screen resistivity m as 6.7 MS.2/square and 1.9 MG2/square respectively for the 15" computer monitor and the 30" TV tube. Estimating the Screen capacity cA as 7 pf and 13 pf, 27t?cAm is found to be 148x10 and 78x10, respectively for the 15" computer monitor and the 30" TV tube. These numbers are very Small compared to unity, So that the Step from (10) to (11) is valid. The following procedures were followed in preparing pulsed images for the field measurements. For the 15" computer monitor the images were produced by running the VB6 program discussed above. The pulsed image comprised the full screen with basic RGB values chosen uniformly as R=G=B=127, with the exception of an on/off button and a few data boxes which together take up 17% of the screen area. The image intensity was pulsed by modifying the R, G, and B values by integer-rounded Sine functions AR(t), AG(t), and AB(t), uniformly over the image, except at the button and the data boxes. The measured electric field pulse ampli tudes were normalized to a pulsed image that occupies all of the screen area and has 100% intensity modulation for which the image pulses between black and the maximum intensity, for the fixed RGB ratios used. The image intensity depends on the RGB values in a nonlinear manner that will be be discussed. For the measurements of the pulsed electric field emitted by 30" TV tube, a similar image was used as for the 15" computer monitor. This was done by playing back a camcorder recording of the computer monitor display when running the VB6 program, with 40% pulse modulation of R, G, and B. In front of the monitor, i.e., for Z>0, the parts (13) and (19) contribute about equally to the electric field over a practical range of distances Z. When going behind the monitor where Z is negative the monopole field flipS Sign So that the two parts nearly cancel each other, and the resulting field is very Small. Therefore, in the back of the CRT, errors due to imperfections in the theory are relatively large. Moreover our model, which pretends that the polarization charges are all located on the polarization disc, fails to account for the electric field flux that escapes from the outer regions of the back of the Screen to the earth or whatever conductors US 6,506,148 B2 15 25 35 40 45 50 55 60 65 18 happen to be present in the vincinity of the CRT. This flaw has relatively more Serious consequences in the back than in front of the monitor. Screen emissions in front of a CRT can be cut dramati cally by using a grounded conductive transparent Shield that is placed over the Screen or applied as a coating. Along the lines of our model, the shield amounts to a polarization disc in front of the Screen, So that the latter is now Sandwiched between to grounded discs. The Screen has the pulsed potential distribution V(r) of (11), but no electric flux can escape. The model may be modified by choosing the polar ization disc in the back Somewhat Smaller than the Screen disc, by a fraction that Serves as a free parameter. The fraction may then be determined from a fit to measured fields, by minimizing the relative Standard deviation between experiment and theory. In each of the electron beams of a CRT, the beam current is a nonlinear function of the driving Voltage, i.e., the Voltage between cathode and control grid. Since this function is needed in the normalization procedure, it was measured for the 15" computer monitor that has been used in the /3 Hz Sensory resonance experiments and the electric field mea Surements. Although the beam current density can be determined, it is easier to measure the luminance, by reading a light meter that is brought right up to the monitor Screen. With the RGB values in the VB6 program taken as the same integer K, the luminance of a uniform image is proportional to the image intensity I. The luminance of a uniform image was measured for various values of K. The results were fitted with I=cK, (20) where c is a constant. The best fit, with 6.18% relative standard deviation, was obtained for Y=2.32. Screen emissions also occur for liquid crystal displayS (LCD). The pulsed electric fields may have considerable amplitude for LCDs that have their driving electrodes on opposite sides of the liquid crystal cell, for passive matrix as well as for active matrix design, Such as thin film technology (TFT). For arrangements with in-plane switching (IPS) however, the driving electrodes are positioned in a Single plane, So that the Screen emission is very Small. For arrange ments other than IPS, the electric field is closely approxi mated by the fringe field of a two-plate condenser, for the Simple case that the image is uniform and extends over the full Screen. For a circular LCD Screen with radius R, the field on the center line can be readily calculated as due to pulsed dipoles that are uniformly distributed over the screen, with the result where E(z) is the amplitude of the pulsed electric field at a distance Z from the Screen and V is a voltage pulse amplitude, in which the aperture ratio of the LCD has been taken into account. Eq. (21) can be used as an approximation for Screens of any shape, by taking R as the radius of a circle with the same area as the Screen. The result applies to the case that the LCD does not have a ground connection, So that the top and bottom electrodes are at opposite potential, i.e., V/2 and -V/2. If one Set of LCD electrodes is grounded, monopoles are needed to keep these electrodes at Zero potential, much as in the case of a CRT discussed above. The LCD situation is Simpler however, as there is no charge injection by electron beams, So that the potentials on the top and bottom plates of the condenser in the model are spatially uniform. From (14)
| 2,580
| 11,143
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.71875
| 3
|
CC-MAIN-2018-26
|
longest
|
en
| 0.93025
|
https://sciencebyjason.com/properties-of-logarithms-2.html
| 1,723,391,986,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00446.warc.gz
| 410,240,635
| 5,600
|
# Properties of Logarithms
These are my notes and worked examples of properties of logarithms.
Ankr Store on Amazon, keep your electronics charged by the best! f you buy something, I get a small commission and that makes it easier to keep on writing. Thank you in advance if you buy something.
Applications involving logarithms play an important role in modern day computation. One reason for this is that logarithms possess several important properties. For example, the loudness of a sound can be measured in decibels by the formula $$f(x) = 10\log (10^{16}x)$$ where x is the intensity of the sound in watts per square centimeter.
Basic Properties of Logarithms
Logarithms possess several important properties. One property of logarithms states that the sum of the logarithms of two numbers equals the logarithm of their product.
$\log 5 + \log 2 = \log 10$
$\log 5 + \log 25 = \log 100$
Property 1
This property is a direct result of the inverse property $$\log_{a} a^{x} = x$$
$\log_{a} 1 = \log_{a} a^{0} = 0$
$\log_{a} a = \log_{a} a^{1} = 1$
So:
$\log 1 = 0$
$\ln e = 1$
Property 2
If m and n are positive numbers, then we can write $$m = a^{x}$$ and $$n = a^{d}$$ for real numbers.
$\log_{a} m + \log_{a} n = \log_{a} a^{x} + \log_{a} a^{d} = c + d$
$\log_{a}(mn) = \log_{a} (a^{x}a^{d}) = \log_{a} (a^{c + d}) = c + d$
So:
$\log_{a} m + \log_{a} n = \log_{a} (mn)$
Example: Let $$m = 100 \text{ and } n = 1000$$
$\log m \log n = \log 100 + \log 1000 = \log 10^{2} + \log 10^{3} = 2 + 3 = 5$
$\log (mn) = \log (100 * 1000) = \log 100000 = \log 10^{5} = 5$
Property 3
Let $$m = a^{c}$$ and $$n = a^{d}$$ for real numbers c and d
$\log_{a} m - \log_{a} n = \log_{a} a^{c} - \log_{a} a^{d} = c - d$
$\log_{a} \frac{m}{n} = \log_{a} \frac{a^{c}}{a^{d}} = \log_{a} (a^{c - d}) = c - d$
So:
$\log_{a} m - \log_{a} n = \log_{a} \frac{m}{n}$
Example: Let m = 100 and n = 1000
$\log m - \log n = \log 100 - \log 1000 = \log 10^{2} - \log 10^{3} = 2 - 3 = -1$
$\log \frac{m}{n} = \log \frac{100}{1000} = \log \frac{1}{10} = \log 10^{-1} = -1$
Property 4
Let $$m = a^{c}$$ and r be any real number
$\log_{a} m^{r} = \log_{a} (a^{c})^{r} = \log_{a} (a^{cr}) = cr$
$r \log_{a} m = r \log_{a} a^{c} = rc$
So:
$\log_{a} m^{r} = r \log_{a} m$
Example: Let m= 100 and r = 3
$\log m^{r} = \log 100^{3} = \log 1,000,000 = \log 10^{6} = 6$
$r \log m = 3 \log 100 = 3 \log 10^{2} = 3 * 2 = 6$
Example
Expand $$\log xy$$
$\log xy = \log x + \log y$
Example
Expand $$\ln \frac{6}{z}$$
$\ln \frac{6}{z} = \ln 6 - \ln z$
Example
Expand $$\log 2x^{4}$$
$\log 2x^{4} = \log 2 + \log x^{4} = \log 2 + 4 \log x$
Example
Expand $$\ln \frac{7x^{3}}{k}$$
$\ln \frac{7x^{3}}{k} = \ln 7x^{3} - \ln k = \ln 7 + \ln x^{3} - \ln k = \ln 7 + 3 \ln x - \ln k$
Example
Analyzing sound with decibels
Sound levels in decibels can be computed by $$D(x) = 10 \log (10^{16})x$$
Use properties of logarithms to simplify the formula
$D(x) = 10 \log (10^{16})x$
$10(\log 10^{16}) + \log x$
$10(16 + \log x)$
$160 + 10 \log x$
Example
Write the expression as the logarithm of a single expression
$\ln 2e + \ln \frac{1}{e}$
$\ln(2e*\frac{1}{e})$
$= \ln 2$
Example
Write the expression as the logarithm of a single expression
$\log_{2} 27 + \log_{2} x^{3}$
$= \log_{2}(27x^{3})$
Example
Write the expression as the logarithm of a single expression
$\log x^{3} - \log x^{2}$
$\log \frac{x^{3}}{x^{2}}$
$= \log x$
Example
Write the expression as the logarithm of a single expression
$\log 5 + \log 15 - \log 3$
$\log (5 * 15) - \log 3$
$\log \frac{5 * 15}{3}$
$= \log 25$
Example
Write the expression as the logarithm of a single expression
$2 \ln x - \frac{1}{2}\ln y - 3\ln z$
$\ln x^{2} - \ln y^{\frac{1}{2}} - \ln z^{3}$
$\ln \frac{x^{2}}{y^{\frac{1}{2}}} - \ln z^{3}$
$\ln \frac{x^{2}}{(y^{\frac{1}{2})}(z^{3})}$
$\ln \frac{x^{2}}{(z^{3})(\sqrt{y})}$
Example
Write the expression as the logarithm of a single expression
$5 \log_{3} x + \log_{3} 2x - \log_{3} y$
$\log_{3} x^{5} + \log_{3} 2x - \log_{3} y$
$\log_{3} (x^{5} * 2x) - \log_{3} y$
$\log_{3} \frac{2x^{6}}{y}$
| 1,670
| 4,125
|
{"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}
| 4.375
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.730586
|
https://www.math24.net/newtons-second-law-motion
| 1,618,914,971,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00254.warc.gz
| 987,589,841
| 14,995
|
# Newton’s Second Law of Motion
Newton’s second law establishes a relationship between the force $$\mathbf{F}$$ acting on a body of mass $$m$$ and the acceleration $$\mathbf{a}$$ caused by this force.
The acceleration $$\mathbf{a}$$ of a body is directly proportional to the acting force $$\mathbf{F}$$ and inversely proportional to its mass $$m,$$ that is
${\mathbf{a} = \frac{\mathbf{F}}{m}\;\;\text{or}\;\;}\kern-0.3pt{\mathbf{F} = m\mathbf{a} = m\frac{{{d^2}\mathbf{r}}}{{d{t^2}}}.}$
This formulation is valid for systems with constant mass. When the mass changes (for example, in the case of relativistic motion), Newton’s second law takes the form
$\mathbf{F} = \frac{{d\mathbf{p}}}{{dt}},$
where $$\mathbf{p}$$ is the impulse (momentum) of the body.
In general, the force $$\mathbf{F}$$ can depend on the coordinates of the body, i.e., the radius vector $$\mathbf{r},$$ its velocity $$\mathbf{v},$$ and time $$t:$$
$\mathbf{F} = \mathbf{F}\left( {\mathbf{r},\mathbf{v},t} \right).$
Below we consider the special cases where the force $$\mathbf{F}$$ depends only on one of these variables.
### Force Depends on Time: $$\mathbf{F} = \mathbf{F}\left( t \right)$$
Assuming that the motion is one-dimensional, Newton’s second law is written as the second order differential equation:
$m\frac{{{d^2}x}}{{d{t^2}}} = F\left( t \right).$
Integrating once, we find the velocity of the body $$v\left( t \right):$$
${v\left( t \right) }={ {v_0} + \frac{1}{m}\int\limits_0^t {F\left( \tau \right)d\tau } .}$
Here we assume that the body begins to move at time $$t = 0$$ with the initial velocity $$v\left( {t = 0} \right) = {v_0}.$$ Integrating again, we get the law of motion $$x\left( t \right):$$
$x\left( t \right) = {x_0} + \int\limits_0^t {v\left( \tau \right)d\tau } ,$
where $${x_0}$$ is the initial coordinate of the body, $$\tau$$ is the variable of integration.
### Force Depends on the Velocity: $$\mathbf{F} = \mathbf{F}\left( {\mathbf{v}} \right)$$
When a solid body moves in a liquid or gaseous environment it experiences a drag force (or a frictional force). At low velocities $$\mathbf{v},$$ this force is proportional to the velocity $$\mathbf{v}:$$
$\mathbf{F} = – k\mathbf{v}.$
The coefficient $$k$$ in turn is proportional to the viscosity $$\eta.$$ In particular, if the body has a spherical shape, the drag force is described by the Stokes’ law:
$\mathbf{F} = – 6\pi \eta R\mathbf{v},$
where $$R$$ is the radius of the ball, $$\eta$$ is the viscosity of the environment.
In this mode of motion Newton’s second law is written (in one-dimensional approximation) as the following differential equation:
$m\frac{{{d^2}x}}{{d{t^2}}} = m\frac{{dv}}{{dt}} = – kv.$
Integrating this equation with the initial condition $$v\left( {t = 0} \right) = {v_0}$$ gives
${\frac{{dv}}{v} = – \frac{k}{m}dt,\;\;}\Rightarrow {\int\limits_{{v_0}}^v {\frac{{du}}{u}} = – \frac{k}{m}\int\limits_0^t {d\tau } .}$
Here $$u$$ and $$\tau$$ are integration variables. The velocity of the body varies from $${v_0}$$ to $$v$$ as the time changes from $$0$$ to $$t.$$ Consequently,
${\ln v – \ln {v_0} = – \frac{k}{m}t,\;\;}\Rightarrow {\ln \frac{v}{{{v_0}}} = – \frac{k}{m}t,\;\;}\Rightarrow {v\left( t \right) = {v_0}{e^{ – {\large\frac{k}{m}\normalsize}t}}.}$
Thus, if the drag force is proportional to the velocity of the body, its speed will decrease exponentially.
The law of motion $$x\left( t \right)$$ can be easily found by repeated integration:
${x\left( t \right) = {x_0} + \int\limits_0^t {v\left( \tau \right)d\tau } } = {{x_0} + \int\limits_0^t {{v_0}{e^{ – {\large\frac{k}{m}\normalsize}\tau }}d\tau } } = {{x_0} – \frac{{m{v_0}}}{k}\left( {{e^{ – {\large\frac{k}{m}\normalsize}t}} – 1} \right) } = {{x_0} + \frac{{m{v_0}}}{k}\left( {1 – {e^{ – {\large\frac{k}{m}\normalsize} t}}} \right).}$
The last formula shows that the path traversed by the body to a complete stop, is equal to $$\large\frac{{m{v_0}}}{k}\normalsize,$$ i.e., proportional to the initial momentum of the body $$m{v_0}.$$
As the velocity of a body increases, the physics of the process changes. The kinetic energy of the body begins to be spent not only on the friction between the layers of liquid, but also on the movement of the fluid in front of the body. In this mode, the drag force becomes proportional to the square of the velocity:
$F = – \mu \rho S{v^2},$
where $$\mu$$ is the coefficient of proportionality, $$S$$ is the cross-sectional area of the body, $$\rho$$ is the density of the medium.
The nonlinear regime described above appears on the conditions
$\mathbf{\text{Re}} = \frac{{\rho vL}}{\eta } > 100,$
where $$\mathbf{\text{Re}}$$ is the dimensionless Reynolds number, $$\eta$$ is the viscosity of the medium, $$L$$ is a characteristic cross-sectional size, for example, the radius of the body.
Considering one-dimensional motion, we write Newton’s second law for this case in the form
${m\frac{{{d^2}x}}{{d{t^2}}} = m\frac{{dv}}{{dt}} }={ – \mu \rho S{v^2}.}$
Integrating, we find the velocity of the body:
${\frac{{dv}}{{{v^2}}} = – \frac{{\mu \rho S}}{m}dt,\;\;}\Rightarrow {\int\limits_{{v_0}}^v {\frac{{du}}{{{u^2}}}} = – \frac{{\mu \rho S}}{m}\int\limits_0^t {d\tau } .}$
Here $$u$$ and $$\tau$$ again denote the integration variables. For the time $$t,$$ the velocity of the body will decrease from an initial value $${v_0}$$ to the final value $$v.$$ As a result, we obtain
${- \left( {\frac{1}{v} – \frac{1}{{{v_0}}}} \right) = – \frac{{\mu \rho S}}{m}t,\;\;}\Rightarrow {\frac{1}{v} = \frac{1}{{{v_0}}} + \frac{{\mu \rho S}}{m}t,\;\;}\Rightarrow {v\left( t \right) = \frac{1}{{\frac{1}{{{v_0}}} + \frac{{\mu \rho S}}{m}t}} }={ \frac{{{v_0}}}{{1 + \frac{{\mu \rho S{v_0}}}{m}t}}.}$
Integrate again to find the law of motion $$x\left( t \right):$$
$\require{cancel} {x\left( t \right) = \int\limits_0^t {\frac{{{v_0}}}{{1 + \frac{{\mu \rho S{v_0}}}{m}\tau }}d\tau } } = {\int\limits_0^t {\frac{{\cancel{v_0}}}{{1 + \frac{{\mu \rho S{v_0}}}{m}\tau }}\frac{{d\left( {1 + \frac{{\mu \rho S{v_0}}}{m}\tau } \right)}}{{\frac{{\mu \rho S\cancel{v_0}}}{m}}}} } = {\frac{m}{{\mu \rho S}}\int\limits_0^t {\frac{{d\left( {1 + \frac{{\mu \rho S{v_0}}}{m}\tau } \right)}}{{1 + \frac{{\mu \rho S{v_0}}}{m}\tau }}} } = {\frac{m}{{\mu \rho S}}\left[ {\left. {\ln \left( {1 + \frac{{\mu \rho S{v_0}}}{m}\tau } \right)} \right|_0^t} \right] } = {\frac{m}{{\mu \rho S}}\ln \left( {1 + \frac{{\mu \rho S{v_0}}}{m}t} \right).}$
It is important to bear in mind that these formulas are valid for sufficiently large values of the velocity: at lower velocities, this model is physically incorrect, since the drag force begins to depend on the velocity linearly (this case was considered previously).
### Force Depends on the Position: $$\mathbf{F} = \mathbf{F}\left( x \right)$$
Examples of forces that depend only on the coordinate are, in particular:
• Elastic force $$F = -kx;$$
• Force of gravitational attraction $$F = – G\large\frac{{{m_1}{m_2}}}{{{x^2}}}\normalsize.$$
The motion of a body of mass $$m$$ connected to a spring under the force of elasticity is determined by the differential equation
${m\frac{{{d^2}x}}{{d{t^2}}} = – kx\;\;\text{or}\;\;}\kern-0.3pt{\frac{{{d^2}x}}{{d{t^2}}} + \frac{k}{m}x = 0.}$
This equation describes the undamped periodic oscillations with a period
$T = 2\pi \sqrt {\frac{m}{k}} .$
In the case of gravitational attraction, the body motion is described by the nonlinear differential equation
$\frac{{{d^2}x}}{{d{t^2}}} = – G\frac{M}{{{x^2}}},$
where $$M$$ is the mass of the attracting body (for example, the mass of the Earth or the Sun), $$G$$ is the universal gravitational constant.
The solution of this equation is given on the page Newton’s Law of Universal Gravitation.
In the case where the force depends on the coordinate, the acceleration is conveniently represented in the form:
${a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}\frac{{dx}}{{dt}} }={ v\frac{{dv}}{{dx}}.}$
Then the differential equation can be written as
${m\frac{{{d^2}x}}{{d{t^2}}} = m\frac{{dv}}{{dt}} }={ mv\frac{{dv}}{{dx}} }={ F\left( x \right).}$
Separating the variables $$v$$ and $$x,$$ we have
${mvdv = F\left( x \right)dx,\;\;}\Rightarrow {m\int\limits_{{v_0}}^v {udu} = \int\limits_0^L {F\left( x \right)dx} ,\;\;}\Rightarrow {\frac{{m{v^2}}}{2} – \frac{{mv_0^2}}{2} }={ \int\limits_0^L {F\left( x \right)dx} .}$
The last equation expresses the law of conservation of energy. The left side describes the change in kinetic energy, and the right side corresponds to the work of a variable force $${F\left( x \right)}$$ when the body is moved by a distance $$L.$$
The subsequent integration of the function $${v\left( t \right)}$$ allows to find the law of motion $${x\left( t \right)}.$$ Unfortunately, this is not always possible because of the cumbersome analytical expressions for $${v\left( t \right)}.$$
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
A body begins to fall from a height $$H$$ under the action of gravity. While falling it experiences resistance proportional to the velocity. Determine the time of the drop.
### Example 2
At the initial moment, a chain of length $$L$$ hangs over the edge of the table so that the force of gravity is balanced by the friction force (Figure $$3$$). As a result of a small shift $$\varepsilon$$ the chain starts sliding. Determine the time $$T,$$ for which the chain completely slips off the table. The coefficient of friction between the chain and the surface of the table is equal to $$\mu.$$
Page 1
Concept
Page 2
Problems 1-2
| 3,170
| 9,582
|
{"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}
| 4.28125
| 4
|
CC-MAIN-2021-17
|
latest
|
en
| 0.731809
|
https://codingwithnick.in/bitwise-and-30-days-of-code-solution/
| 1,721,655,654,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00505.warc.gz
| 150,921,844
| 47,813
|
Home30 Days Of CodeHackerRank Day 29: Bitwise AND 30 days of code solution
# HackerRank Day 29: Bitwise AND 30 days of code solution
Today we are going to solve HackerRank Day 29: Bitwise AND 30 days of code solutions in C, C++, Java, Javascript & Python.
## Objective
Welcome to the last day! Today, we’re discussing bitwise operations.
Given set S = {1, 2, 3, . . . ,N}. Find two integers, A and B (where A < B), from set S such that the value of A&B is the maximum possible and also less than a given integer, K. In this case, & represents the bitwise AND operator.
Function Description
Complete the bitwiseAnd function in the editor below.
bitwiseAnd has the following paramter(s):
– int N: the maximum integer to consider
– int K: the limit of the result, inclusive
Returns
– int: the maximum value of A&B within the limit.
## Input Format
The first line contains an integer, T, the number of test cases.
Each of the T subsequent lines defines a test case as 2 space-separated integers, N and K, respectively.
## Constraints
• 1 <= T <= 103
• 2 <= N <= 103
• 2 <= K <= N
Sample Input
``````STDIN Function
----- --------
3 T = 3
5 2 N = 5, K = 2
8 5 N = 8, K = 5
2 2 N = 8, K = 5``````
Sample Output
``````1
4
0``````
## Bitwise AND Solution in C
``````#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int maxvalue(int n, int k) {
int i, j;
int res = 0, max_res = 0;
for (i = 1; i <= n; i++) {
for (j = i+1; j <=n ; j++) {
int val = i &j;
if (val > max_res && val < k) {
max_res = val;
}
}
}
return max_res;
}
int main(){
int t;
int a0;
scanf("%d",&t);
for(a0 = 0; a0 < t; a0++){
int n;
int k;
scanf("%d %d",&n,&k);
printf("%d\n", maxvalue(n, k));
}
return 0;
}``````
## Bitwise AND Solution in C++
``````#include <iostream>
#include <vector>
using namespace std;
int main(){
int ncases, n, k, max = 0, tmp = 0;
vector<int> range;
range.reserve(1000);
cin >> ncases;
for(int i = 0; i < ncases; ++i){
cin >> n >> k;
for(int j = 0; j < n; ++j)
range.push_back(j + 1);
for(int x = 0; x < range.size() - 1; ++x){
for(int y = x + 1; y < range.size(); ++y){
tmp = range[x] & range[y];
if(tmp < k)
max = (tmp > max ? tmp : max);
}
}
cout << max << '\n';
range.clear();
max = 0;
}
}``````
## Bitwise AND Solution in Java
``````import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner in = new Scanner(System.in);
int q = in.nextInt();
for (int i = 0; i < q; i++) {
int n = in.nextInt();
int k = in.nextInt();
int maxed = 0;
for (int b = 2; b <= n; b++) {
for (int a = 1; a < b; a++) {
if (a == b) continue;
int ab = a&b;
if (ab > maxed && ab < k) maxed = ab;
}
}
System.out.println(maxed);
}
}
``````
## Bitwise AND Solution in Python 2
``````#!/bin/python
import sys
t = int(raw_input().strip())
for a0 in xrange(t):
n,k = raw_input().strip().split(' ')
n,k = [int(n),int(k)]
maxValue = 0
check = 0
for i in range(n-1,0,-1):
for j in range(n,i,-1):
tmp = i&j
if ((tmp > maxValue)&(tmp < k)):
maxValue = tmp
if (maxValue + 1 == k):
check = 1
break
if (check == 1):
break
print maxValue``````
## Bitwise AND Solution in Python 3
``````#!/bin/python3
import math
import os
import random
import re
import sys
def max_bit(n,k):
maximum = 0
for i in range(1,n+1):
for j in range(1,i):
h = i & j
if maximum < h < k:
maximum = h
if maximum == k-1:
return maximum
return maximum
if __name__ == '__main__':
t = int(input())
for t_itr in range(t):
nk = input().split()
n = int(nk[0])
k = int(nk[1])
print(max_bit(n,k))``````
## Bitwise AND Solution in Javascript
``````process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function main() {
var t = parseInt(readLine());
for(var a0 = 0; a0 < t; a0++){
var n_temp = readLine().split(' ');
var n = parseInt(n_temp[0]);
var k = parseInt(n_temp[1]);
function findMaxPoss(arr) {
var res = 0;
for(var i = 0; i < arr.length; i++){
for(var j = i + 1; j < arr.length; j++){
var ans = arr[i] & arr[j];
if((ans > res) && (ans < k)){
res = ans;
}
}
}
return res;
}
console.log(findMaxPoss(range(n)));
}
function range(n){
return Array.apply(null, Array(n)).map(function (_, i) {return i + 1;});
}
}``````
RELATED ARTICLES
| 1,532
| 4,726
|
{"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}
| 3.390625
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.386088
|
https://discuss.codecademy.com/t/sum-of-digits-of-0303-is-15-why/152198
| 1,537,743,469,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267159820.67/warc/CC-MAIN-20180923212605-20180923233005-00260.warc.gz
| 496,360,923
| 4,173
|
# Sum of digits of 0303 is 15 - Why?
#1
Hi guys.
I have some question about this code. It works properly but if I try to pass over an argument that equals (0303) the function returns "15". Does anybody know why it is happening? ))
Digit_sum question
#2
The leading `0` in `0303` indicates to the Python interpreter that it is to be evaluated as an octal (base `8`) number. Expressed as a decimal number, it is `195`. The sum of the digits of `195` is `15`. Tricky, but very interesting.
#3
Thanks a lot for your explanation. I didn't have it on my mind )
#4
| 154
| 564
|
{"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-2018-39
|
latest
|
en
| 0.938607
|
htsrec.com
| 1,686,320,770,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00200.warc.gz
| 346,838,567
| 14,055
|
# What is the definition of ROI in poker?
In poker, the return on investment, or ROI, is a figure that indicates how lucrative a player’s participation in tournaments is. It helps you to analyze if participation in online activities is beneficial or detrimental to your bankroll. To do this, it is required to examine this figure and use the facts to additional successes in online poker.
What is ROI in poker?
This word translates as “return on investment” and refers to the ratio of net profit to money invested. ROI is a percentage return on investment that is solely used in MTT and SNG poker. Cache employs the idea of victory rate.
Before beginning the calculations, newcomers to poker need grasp the difference between a tournament and a traditional cash game:
The blinds in MTT and SNG increase at a certain frequency; the precise strategy to use against opponents depends on their position at the table;
To play, a certain amount of money, known as the buy-in, must be paid; the quantity of chips at the table does not match the cash deposit.
If a player remains in the event for an extended period of time, he will be able to join the reward area. The higher his position, the greater the payment.
How to Do It
The statistic is used to calculate the profit from participation in online or offline activities. How it is calculated:
ROI = (total tournament wins minus total buy-ins) – 1.
Sample calculation: Ivan completed 100 MTTs in a month. Each buy-in was \$40. The total wins at the end of the month are \$5,000. Using the formula:
(5,000 / 4,000) – 1 = 25%
According to the figures, the gamer profited \$0.25 for every dollar invested.
| 365
| 1,651
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.96875
| 3
|
CC-MAIN-2023-23
|
longest
|
en
| 0.943364
|
https://forums.wolfram.com/mathgroup/archive/1997/Aug/msg00339.html
| 1,685,900,393,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00573.warc.gz
| 290,802,520
| 7,841
|
Re: inverse command of Series[]
• To: mathgroup at smc.vnet.net
• Subject: [mg8054] Re: inverse command of Series[]
• From: weber at math.uni-bonn.de (Matthias Weber)
• Date: Mon, 4 Aug 1997 01:47:37 -0400
• Organization: RHRZ - University of Bonn (Germany)
• Sender: owner-wri-mathgroup at wolfram.com
```In article <5s0v3n\$q7n\$1 at dragonfly.wolfram.com>, "Nguyen N. Anh"
<anh at chm.ulaval.ca> wrote:
> Hello,
>
> I've written a program in Mathematica and it works quite well. However, after
> convergence it gave as result the following polinomial:
>
> 1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120 + x^6/720 + x^7/5040
>
> I know that this is the power series expansion of function E^x but I want that
> my program recognizes it automatically and outputs the generating function of
> the polinomial calculated ( E^x in this case).
>
> Is there any command or package which can do that ?
>
> Any help will be appreciated
>
> Regards
>
Here is a suggestion how to 'recognize' a polynomial as a
truncated power series:
Let us assume that the function you want to recognize
satisfies some linear ode of order n. You can get its coefficients approximately
by differentiating the polynomial n times and finding the coefficients
of a linear combination which makes this zero.
Then you can solve this ode explicitly, giving you the required function.
For other types of functions, you might be able to modify this
approach.
However, you should be aware that this procedure will not
prove anything.
Good luck,
Matthias Weber
```
• Prev by Date: Re: Re: How to select unique elements in a list?
• Next by Date: Re: inverse command of Series[]
• Previous by thread: Re: inverse command of Series[]
• Next by thread: Re: inverse command of Series[]
| 478
| 1,751
|
{"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}
| 2.6875
| 3
|
CC-MAIN-2023-23
|
longest
|
en
| 0.895043
|
http://www.sigmainfy.com/blog/leetcode-permutation-sequence.html
| 1,487,958,497,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-09/segments/1487501171629.84/warc/CC-MAIN-20170219104611-00095-ip-10-171-10-108.ec2.internal.warc.gz
| 587,021,546
| 3,379
|
# LeetCode Permutation Sequence: Math Approach Rather Than DFS
## Overview
To solve LeetCode Permutation Sequence, math could be adopted to identify the correct number in the corresponding position rather than do DFS simulation.
## LeetCode Permutation Sequence
The set `[1,2,3,…,<i>n</i>]` contains a total of n! unique permutations.
By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):
1. `"123"`
2. `"132"`
3. `"213"`
4. `"231"`
5. `"312"`
6. `"321"`
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
## Solution Analysis: Math Approach Rather Than DFS
(1) DFS and enumerate all the sequences by the increasing order, when enumerate the k-th sequence, return it. This would be TLE for large test case
(2) Actually there is “O(n)” solution. From pure mathematical view, for example we want to find k-th sequence of length n, basically let a[1]…a[n] denote the k-th sequence, think about this: except a[1] there are totally (n-1)! different permutations for a[2]…a[n], then a[1] = k / (n-1)! and we need to find k % (n-1)! -th element is the a[2]…[an] sequence, this is recursively the same problem. Just be careful when dealing with some border cases like k is exactly (n-1)! or 3 * (n-1)! sort of these cases. The following code is accepted by LeetCode OJ to pass this Permutation Sequence problem:
```string getPermutation(int n, int k) {
string sol;
vector<char> nums;
int per[10];
per[0] = per[1] = 1;
for (int i = 1; i <= n; ++i) {
per[i] = i * per[i-1];
nums.push_back('0' + i);
}
int v = 0;
for (int i = n; i >= 1; --i) {
v = (k - 1) / per[i - 1];
sol.push_back(nums[v]);
nums.erase(nums.begin() + v);
k = (k - 1) % per[i-1] + 1;
}
return sol;
}
```
And not that the real time complexity is O(N^2) because erase operation takes O(N) time, I say this is a “O(N)” approach in a sense of that it just linear scan the permutation to generate the final result.
## Summary
To solve LeetCode Permutation Sequence, math could be adopted to identify the correct number in the corresponding position rather than do DFS simulation.
Written on June 15, 2013
| 613
| 2,173
|
{"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}
| 4.34375
| 4
|
CC-MAIN-2017-09
|
latest
|
en
| 0.764354
|
http://www.jiskha.com/display.cgi?id=1186244610
| 1,493,401,523,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917123046.75/warc/CC-MAIN-20170423031203-00182-ip-10-145-167-34.ec2.internal.warc.gz
| 555,829,313
| 3,630
|
# Geometry
posted by on .
If M is the midpoint of line AB.How do i Find the coordinates of B given A(4,-2) and M(1,3).
M = (A + B)/2 ---->
B = 2 M - A = 2(1,3) - (4,-2) =
(2-4,6+2) = (-2,8)
| 85
| 194
|
{"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-2017-17
|
latest
|
en
| 0.773569
|
https://centregalilee.com/why-am-i-taller-than-my-parents/
| 1,638,909,257,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964363418.83/warc/CC-MAIN-20211207201422-20211207231422-00128.warc.gz
| 228,690,677
| 5,152
|
depends entirely on her genetics, diet and also environment so no one hear have the right to say noþeles definitively
You are watching: Why am i taller than my parents
(Original article by fishy789) I"m a 5"6 15 year old male and also my parents space both 5"2. Ns don"t feel as if I"ve had actually any significant growth spurts however I to be wondering exactly how much taller than my parents I can get. My grandparents to be 4"8 and 5"3 (dad"s side) and also 5"5 and also 5"7.5 (mum"s side). I additionally had a 6"3 good grandfather (mum"s side). Is it also late for me to acquire a development spurt and also how much taller than my parents have the right to I get?
Average of both her parents" height then +5 customs if you"re a guy and also -5 if you"re a girl is a way to calculation your height.That would placed you in ~ 5"7 but obviously, no everyone is the same and some are lot taller and some space much much shorter than this. As you are only 15, girlfriend still have actually a lot of of farming to do.
There"s no proven method to guess a child"s adult height. However, number of formulas can administer a reasonable guess: v for child growth. Here"s a renowned example: include the mother"s height and the father"s elevation in either inches or centimeters. Include 5 customs (13 centimeters) for guys or subtract 5 inch (13 centimeters) because that girls. Division by two.Not guaranteed. And also if it"s best you"re pretty much as tall together your going centregalilee.comme get!
(Original short article by fishy789) I"m a 5"6 15 year old male and also my parents room both 5"2. Ns don"t feel together if I"ve had any significant growth spurts however I was wondering how much taller 보다 my parental I deserve to get. My grandparents to be 4"8 and also 5"3 (dad"s side) and 5"5 and also 5"7.5 (mum"s side). I likewise had a 6"3 great grandfather (mum"s side). Is it too late because that me to acquire a development spurt and also how much taller than my parents have the right to I get?
at 15 ns was 5ft7 climate randomly over article GCSE holiday i prospered steadily till now ( 17) to 6ft2 therefore theres quiet a possibility
(Original short article by glad-he-ate-her) at 15 i was 5ft7 then randomly over short article GCSE holiday i grew steadily till now ( 17) to 6ft2 therefore theres quiet a opportunity
This is a an extremely uncentregalilee.commmon i wouldn"t get your wishes up. Height isn"t all its cracked approximately be therefore don"t worry around it.
(Original short article by cbreef) Yup very same kind of point here. My dad is perhaps 5ft 9, yet I"m 6ft 3 You"re a girl.
(Original short article by cbreef) Yup exact same kind of point here. Mine dad is possibly 5ft 9, mum 5ft 2, however I"m 6ft 3 You"re a girl.
(Original post by Goaded) however my mum is 5"9 and all the other girls in my household are 5"7 and above except one.
Height doesn"t issue as much if you"re a girl to it is in honest. Males don"t really treatment so it"s only really relevant if you carry out some sport or something,
See more: What Is In The Middle Of An Olive, A Quick Guide To Olives And Olive Oil
(Original article by cbreef) elevation doesn"t matter as much if you"re a girl to it is in honest. Guys don"t really care so it"s only really appropriate if you carry out some sports or something,
| 886
| 3,327
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.78125
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.967383
|
http://mathhelpforum.com/advanced-statistics/105711-random-variables.html
| 1,526,950,365,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794864572.13/warc/CC-MAIN-20180521235548-20180522015548-00278.warc.gz
| 185,767,118
| 10,707
|
1. ## Random Variables
Given $\displaystyle (\Omega, \mathcal{A}, P)$, suppose X is a random variable with $\displaystyle X\geq0$ and E{X}=1. Definite $\displaystyle Q:\mathcal{A}->R$ by Q(A)=$\displaystyle E{(X1_A)}$. Show that if P(A) = 0, then Q(A)=0. Give an example that shows that Q(A)=0 does not in general imply P(A)=0.
I'm pretty sure that $\displaystyle E{(1_A)}=P(A)$, so I was thinking that maybe we could break Q(A) up into $\displaystyle E{(X)}E{(1_A)}$. However, this would be assuming that X and 1_A are independent random variables? Would this be true? If so, then it's very obvious why, if P(A)=0, then Q(A)=0. Anyone have any hints? Thanks!
2. Originally Posted by azdang
Given $\displaystyle (\Omega, \mathcal{A}, P)$, suppose X is a random variable with $\displaystyle X\geq0$ and E{X}=1. Definite $\displaystyle Q:\mathcal{A}->R$ by Q(A)=$\displaystyle E{(X1_A)}$. Show that if P(A) = 0, then Q(A)=0. Give an example that shows that Q(A)=0 does not in general imply P(A)=0.
I'm pretty sure that $\displaystyle E{(1_A)}=P(A)$, so I was thinking that maybe we could break Q(A) up into $\displaystyle E{(X)}E{(1_A)}$. However, this would be assuming that X and 1_A are independent random variables? Would this be true? If so, then it's very obvious why, if P(A)=0, then Q(A)=0. Anyone have any hints? Thanks!
It is false (without further assumption) that $\displaystyle Q(A)=E{(X)}E{(1_A)}$. However, if $\displaystyle P(A)=0$, then $\displaystyle X 1_A=0$ almost-everywhere (it is 0 outside $\displaystyle A$, and $\displaystyle A$ is negligible), hence $\displaystyle Q(A)=0$.
As for the example, you must choose $\displaystyle X$ in such a way that $\displaystyle X 1_A$ is zero everywhere, i.e. $\displaystyle X$ is 0 on $\displaystyle A$, and it is chosen outside A in such a way that $\displaystyle E[X](=E[X 1_{A^c}])=1$).
3. Ooh, the first part seems so obvious. Thank you, Laurent.
As for the second part, I'm still a little lost. My first question is: Is it correct that $\displaystyle X1_A$ = X if x is in A and 0 if x is not in A?
4. Originally Posted by azdang
Ooh, the first part seems so obvious. Thank you, Laurent.
As for the second part, I'm still a little lost. My first question is: Is it correct that $\displaystyle X1_A$ = X if x is in A and 0 if x is not in A?
No. What is true is that $\displaystyle X(\omega) 1_A(\omega)=X(\omega)$ if $\displaystyle \omega\in A$ and $\displaystyle X(\omega) 1_A(\omega)=0$ else. Remember $\displaystyle A$ is an event, not a subset of $\displaystyle \mathbb{R}$.
For instance (you should write this more formally), suppose we toss a fair coin, and define the event $\displaystyle A=\{\text{the outcome is heads}\}$ and $\displaystyle X$ is defined to be 0 in case heads shows up, and to be 2 if tails shows up. Then what are $\displaystyle P(A)$, $\displaystyle E[X]$ and $\displaystyle E[X 1_A]$?
5. Well, P(A) should be .5, correct? So, wouldn't $\displaystyle EX1_A$ be 0 in either case (if we roll a heads or a tales)? So in this case, Q(A) = 0, but P(A) does not equal 0.
6. Originally Posted by azdang
Well, P(A) should be .5, correct? So, wouldn't $\displaystyle EX1_A$ be 0 in either case (if we roll a heads or a tales)? So in this case, Q(A) = 0, but P(A) does not equal 0.
Yes. (More correctly, $\displaystyle X1_A$ is 0 in both cases, hence $\displaystyle E(X1_A)=0$. )
| 1,040
| 3,371
|
{"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}
| 3.90625
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.885919
|
https://bedtimemath.org/fun-math-soccer-for-your-car/
| 1,726,850,727,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00549.warc.gz
| 104,014,983
| 19,140
|
It’s easy to kick a soccer ball, since it’s so much smaller than you. But try kicking one as tall as your bedroom ceiling! This great video shows Trackhoe Soccer, where cars and trucks play giant soccer with an 8-foot-high ball. Cars and trucks can’t kick, of course, so they drive, skid and turn to push the ball into the goal – while an excavator goalie tries to block them!
Wee ones: What shape is a soccer ball?
Little kids: What do you call the 5-sided black shapes on the soccer ball? Bonus: If the car team scores 1 goal, then the truck team scores 2 goals, then cars score 1, then trucks score 2…which team scores the 10th goal to keep the pattern?
Big kids: Trackhoe soccer is played once a year, and this 2018 game is the 5th one! In what year was the 1st game? Bonus: If there are 18 players out there, and there are twice as many white vehicles as colored ones, how many white vehicles are there?
The sky’s the limit: If there are equal numbers of cars and trucks, and 1/2 the cars and 2/3 the trucks are red, and there are 2 more red trucks than red cars…how many trucks are there?
Wee ones: A circle, or in 3D, a “sphere.”
Little kids: Pentagons. Bonus: The car team, since trucks will score goals 8 and 9.
Big kids: In 2014. Remember, you can’t just subtract 3, since that would bring you to year “zero” when there was no game. Bonus: 12 white and 6 colored. If there are twice as many whites, it’s like a set of colored cars and 2 more sets the same size, making 3 “sets” total.
The sky’s the limit: 2/3 of the number is 2 more than 1/2 of the same number. And the difference between 2/3 and 1/2 is 1/6. So 2 is 1/6 of the total, and there are 12 trucks.
| 468
| 1,682
|
{"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}
| 3.921875
| 4
|
CC-MAIN-2024-38
|
latest
|
en
| 0.961115
|
https://www.freebasic.net/forum/viewtopic.php?p=233121
| 1,591,105,777,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00297.warc.gz
| 702,018,159
| 8,582
|
## Associative Memory
User projects written in or related to FreeBASIC.
sean_vn
Posts: 283
Joined: Aug 06, 2012 8:26
### Associative Memory
There where a lot of papers in the 1970s and 1980s about associative memory. It actually turns out there is an easy solution to the problem:
Code: Select all
`type associativememory veclen as ulong density as ulong hash as ulong '32 bit unsigned integer weights as single ptr 'pointer to 32 bit floats binarize as boolean ptr work as single ptr declare sub init(veclen as ulong,density as ulong,hash as ulong) declare sub free() declare sub train(target as single,inVec as single ptr) declare function recall(inVec as single ptr) as single declare sub signflip(vec as single ptr,h as long) declare sub wht(vec as single ptr)end typesub associativememory.init(veclen as ulong,density as ulong, hash as ulong) this.veclen=veclen this.density=density this.hash=hash weights=callocate(veclen*density,sizeof(single)) 'allocate zeroed memory binarize=callocate(veclen*density,sizeof(boolean)) work=callocate(veclen,sizeof(single))end subsub associativememory.free() deallocate(weights) deallocate(binarize) deallocate(work)end subfunction associativememory.recall(inVec as single ptr) as single dim as ulong i,j dim as single x=0f dim as single ptr wtidx=weights dim as boolean ptr bidx=binarize for i=0 to veclen-1 work[i]=inVec[i] next for i=0 to density-1 signflip(work,hash+i) 'sign flip and wht=random projection wht(work) for j=0 to veclen-1 if work[j]>0f then x+=wtidx[j] bidx[j]=true else x-=wtidx[j] bidx[j]=false end if next wtidx+=veclen bidx+=veclen next return xend functionsub associativememory.train(target as single,inVec as single ptr) dim as ulong i,j dim as single ptr wtidx=weights dim as boolean ptr bidx=binarize dim as single e=target-recall(inVec) 'get the prediction error e/=veclen*density 'scale the error correctly so that it will be fully corrected for i=0 to density-1 for j=0 to veclen-1 if bidx[j] then wtidx[j]+=e else wtidx[j]-=e end if next wtidx+=veclen bidx+=veclen nextend sub' Pseudorandomly flip the sign of the values in vector using h as a seed' h is a 32 bit signed intergersub associativememory.signflip(vec as single ptr,h as long) dim as ulong i for i=0 to veclen-1 h*=1664525 h+=1013904223 if h<0 then vec[i]=-vec[i] nextend sub'Fast Walsh Hadamard transform. Leaves vector length unchangedsub associativememory.wht(vec as single ptr) dim as ulong i,j,hs=1 dim as single a,b,scale=1.0/sqr(veclen) while hs<veclen i=0 while i<veclen j=i+hs while i<j a=vec[i] b=vec[i+hs] vec[i]=a+b vec[i+hs]=a-b i+=1 wend i+=hs wend hs+=hs wend for i=0 to veclen-1 vec[i]*=scale nextend subscreenres 300,300,32print "Please wait!"dim as associativememory netdim as single ptr vec=callocate(256,sizeof(single)) 'allocate zeroednet.init(256,3,1234567)for i as ulong=0 to 99 for j as ulong=0 to 254 vec[j]=1f net.train(sin(j*.06)*100,vec) vec[j+1]=1f net.train(cos(j*.2)*100,vec) vec[j]=0f vec[j+1]=0f nextnextclsfor i as ulong=0 to 254 vec[i]=1f pset (i,150-sin(i*.06)*100),rgb(0,255,0) pset (i,150-net.recall(vec)),rgb(255,255,0) vec[i]=0fnextfor i as ulong=0 to 254 vec[i]=1f vec[i+1]=1f pset (i,150-cos(i*.2)*100),rgb(0,255,0) pset (i,150-net.recall(vec)),rgb(255,0,255) vec[i]=0f vec[i+1]=0fnextdeallocate(vec)net.free()getkey`
It's based on this observation:
"There has been a lack of discussion about binarization in neural networks. Multiplying those +1/-1 values by weights and summing allows you to store values with a high degree of independence. For a given binary input and target value you get an error. You divide the error by the number of binary values and then you simply correct each of the weights by the reduced error taking account of the binary sign. That gives a full correction to get the correct target output. In higher dimensional space most vectors are orthogonal. For a different binary input the adjustments you made to the weights will not align at all. In fact they will sum to Gaussian noise by the central limit theorem. The value you previously stored for a second binary input will now be contaminated by a slight amount of Gaussian noise which you can correct for. This will now introduce an even smaller amount of Gaussian noise on the value for the first binary input. Iterating back and forth will get rid of the noise entirely for both binary inputs.
This has high use in random projection,reservoir and extreme learning machine computing. And in fact turns a simple locality sensitive hash formed by random projection followed by binarization into a useful single layer neural network. "
https://software.intel.com/en-us/forums ... pic/734095
If you use excess memory (to get an over-determined solution) it will provide a manner of error correction.
Last edited by sean_vn on Jun 10, 2017 3:33, edited 1 time in total.
sean_vn
Posts: 283
Joined: Aug 06, 2012 8:26
### Re: Associative Memory
I wrote the code in such a way that it can be easily translated into C.
sean_vn
Posts: 283
Joined: Aug 06, 2012 8:26
### Re: Associative Memory
There are better constants you can use in the signflip sub.
Code: Select all
`' h is a 32 bit signed intergersub associativememory.signflip(vec as single ptr,h as long) dim as ulong i for i=0 to veclen-1 h*=&h9E3779B9 h+=&h6A09E667 if h<0 then vec[i]=-vec[i] nextend sub`
Anyway you don't need to use all the input dimensions. However I did find it best if using say x in the range 0 to 1, to use 1-x as an input as well. It's interesting to see the sort of crystalline (polytope) approximation you get. The example is kinda slow, however if someone created a specialized chip (or you just used a GPU) then massive speedup are possible.
Code: Select all
`screenres 512,256,32print "Please wait!"dim as associativememory netdim as single ptr vec=callocate(256,sizeof(single)) 'allocate zeroednet.init(256,10,1234567)for i as ulong=0 to 100 for xi as ulong=0 to 99 for yi as ulong=0 to 99 dim as single x=xi*0.01f,y=yi*0.01f vec[0]=x vec[1]=y vec[2]=1-x vec[3]=1-y net.train(sin(x*6)*sin(y*6),vec) next nextnext clsfor xi as ulong=0 to 249 for yi as ulong=0 to 249 dim as single x=xi*0.004f,y=yi*0.004f vec[0]=x vec[1]=y vec[2]=1-x vec[3]=1-y dim as long c1=255*sin(x*6)*sin(y*6) dim as long c2=255*net.recall(vec) if c1>255 then c1=255 if c1<0 then c1=0 if c2>255 then c2=255 if c2<0 then c2=0 pset (xi,yi),rgb(c1,50,0) pset (xi+256,yi),rgb(50,c2,0) nextnextdeallocate(vec)net.free()getkey`
dafhi
Posts: 1334
Joined: Jun 04, 2005 9:51
### Re: Associative Memory
Code: Select all
`' .. paste the udt from abovetype float as singletype myint as integerfunction clamp(in as float, hi as float=1, lo as float=0) as float if in < lo then return lo if in > hi then return hi return inEnd Functionsub main var w = 800 var h = 400 screenres 800,400,32 print "Please wait!" var network_area = 30 '' small as possible for post-training performance var dens = 9 dim as myint c = network_area \ dens dim as associativememory net(c-1) dim as float ptr vec=callocate(256,sizeof(float)) 'allocate zeroed for i as long = 0 to ubound(net) net(i).init(256,dens,i*3+11) next var siz = 8 ' recommended minimum: 8 var inc = 1.2 / siz for i as long=1 to 60 '' good baseline for xi as long=0 to siz-1 for yi as long=0 to siz-1 dim as float x=xi*inc,y=yi*inc vec[0]=x vec[1]=y vec[2]=1-x vec[3]=1-y for j as long = 0 to ubound(net) net(j).train(sin(x*6)*sin(y*6),vec) next next next next cls var scale = 22 siz *= scale inc /= scale for xi as ulong=0 to siz-1 for yi as ulong=0 to siz-1 dim as float x=xi*inc,y=yi*inc vec[0]=x vec[1]=y vec[2]=1-x vec[3]=1-y dim as long c1=clamp(255*sin(x*6)*sin(y*6), 255) dim as single c2 for j as long = 0 to ubound(net) c2 += net(j).recall(vec) Next: c2 = clamp(255*c2/c, 255) pset (xi,yi),rgb(c1,c1,c1) pset (xi+siz,yi),rgb(c2,c2,c2) next if inkey<>"" then exit for sleep 1 next deallocate(vec) for i as long = 0 to ubound(net) net(i).free() next getkeyend submain`
| 2,986
| 9,020
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.765625
| 3
|
CC-MAIN-2020-24
|
latest
|
en
| 0.409359
|
http://xavierdupre.fr/app/onnxcustom/helpsphinx/_sources/api/onnxops/onnx__GatherElements.rst.txt
| 1,669,723,326,000,000,000
|
text/plain
|
crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00247.warc.gz
| 111,464,675
| 4,877
|
.. _l-onnx-doc-GatherElements: ============== GatherElements ============== .. contents:: :local: .. _l-onnx-op-gatherelements-13: GatherElements - 13 =================== **Version** * **name**: `GatherElements (GitHub) `_ * **domain**: **main** * **since_version**: **13** * **function**: False * **support_level**: SupportType.COMMON * **shape inference**: True This version of the operator has been available **since version 13**. **Summary** GatherElements takes two inputs `data` and `indices` of the same rank r >= 1 and an optional attribute `axis` that identifies an axis of `data` (by default, the outer-most axis, that is axis 0). It is an indexing operation that produces its output by indexing into the input data tensor at index positions determined by elements of the `indices` tensor. Its output shape is the same as the shape of `indices` and consists of one value (gathered from the `data`) for each element in `indices`. For instance, in the 3-D case (r = 3), the output produced is determined by the following equations: :: out[i][j][k] = input[index[i][j][k]][j][k] if axis = 0, out[i][j][k] = input[i][index[i][j][k]][k] if axis = 1, out[i][j][k] = input[i][j][index[i][j][k]] if axis = 2, This operator is also the inverse of ScatterElements. It is similar to Torch's gather operation. Example 1: :: data = [ [1, 2], [3, 4], ] indices = [ [0, 0], [1, 0], ] axis = 1 output = [ [1, 1], [4, 3], ] Example 2: :: data = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], ] indices = [ [1, 2, 0], [2, 0, 0], ] axis = 0 output = [ [4, 8, 3], [7, 2, 3], ] **Attributes** * **axis**: Which axis to gather on. Negative value means counting dimensions from the back. Accepted range is [-r, r-1] where r = rank(data). Default value is ``0``. **Inputs** * **data** (heterogeneous) - **T**: Tensor of rank r >= 1. * **indices** (heterogeneous) - **Tind**: Tensor of int32/int64 indices, with the same rank r as the input. All index values are expected to be within bounds [-s, s-1] along axis of size s. It is an error if any of the index values are out of bounds. **Outputs** * **output** (heterogeneous) - **T**: Tensor of the same shape as indices. **Type Constraints** * **T** in ( tensor(bfloat16), tensor(bool), tensor(complex128), tensor(complex64), tensor(double), tensor(float), tensor(float16), tensor(int16), tensor(int32), tensor(int64), tensor(int8), tensor(string), tensor(uint16), tensor(uint32), tensor(uint64), tensor(uint8) ): Constrain input and output types to any tensor type. * **Tind** in ( tensor(int32), tensor(int64) ): Constrain indices to integer types **Examples** **_gather_elements_0** :: axis = 1 node = onnx.helper.make_node( 'GatherElements', inputs=['data', 'indices'], outputs=['y'], axis=axis, ) data = np.array([[1, 2], [3, 4]], dtype=np.float32) indices = np.array([[0, 0], [1, 0]], dtype=np.int32) y = gather_elements(data, indices, axis) # print(y) produces # [[1, 1], # [4, 3]] expect(node, inputs=[data, indices.astype(np.int64)], outputs=[y], name='test_gather_elements_0') **_gather_elements_1** :: axis = 0 node = onnx.helper.make_node( 'GatherElements', inputs=['data', 'indices'], outputs=['y'], axis=axis, ) data = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]], dtype=np.float32) indices = np.array([[1, 2, 0], [2, 0, 0]], dtype=np.int32) y = gather_elements(data, indices, axis) # print(y) produces # [[4, 8, 3], # [7, 2, 3]] expect(node, inputs=[data, indices.astype(np.int64)], outputs=[y], name='test_gather_elements_1') **_gather_elements_negative_indices** :: axis = 0 node = onnx.helper.make_node( 'GatherElements', inputs=['data', 'indices'], outputs=['y'], axis=axis, ) data = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]], dtype=np.float32) indices = np.array([[-1, -2, 0], [-2, 0, 0]], dtype=np.int32) y = gather_elements(data, indices, axis) # print(y) produces # [[7, 5, 3], # [4, 2, 3]] expect(node, inputs=[data, indices.astype(np.int64)], outputs=[y], name='test_gather_elements_negative_indices') **Differences** .. raw:: html
`0` `0` `GatherElements takes two inputs data and indices of the same rank r >= 1` `GatherElements takes two inputs data and indices of the same rank r >= 1` `1` `1` `and an optional attribute axis that identifies an axis of data` `and an optional attribute axis that identifies an axis of data` `2` `2` `(by default, the outer-most axis, that is axis 0). It is an indexing operation` `(by default, the outer-most axis, that is axis 0). It is an indexing operation` `3` `3` `that produces its output by indexing into the input data tensor at index` `that produces its output by indexing into the input data tensor at index` `4` `4` `positions determined by elements of the indices tensor.` `positions determined by elements of the indices tensor.` `5` `5` `Its output shape is the same as the shape of indices and consists of one value` `Its output shape is the same as the shape of indices and consists of one value` `6` `6` `(gathered from the data) for each element in indices.` `(gathered from the data) for each element in indices.` `7` `7` `8` `8` `For instance, in the 3-D case (r = 3), the output produced is determined` `For instance, in the 3-D case (r = 3), the output produced is determined` `9` `9` `by the following equations:` `by the following equations:` `10` `10` `::` `::` `11` `11` `12` `12` ` out[i][j][k] = input[index[i][j][k]][j][k] if axis = 0,` ` out[i][j][k] = input[index[i][j][k]][j][k] if axis = 0,` `13` `13` ` out[i][j][k] = input[i][index[i][j][k]][k] if axis = 1,` ` out[i][j][k] = input[i][index[i][j][k]][k] if axis = 1,` `14` `14` ` out[i][j][k] = input[i][j][index[i][j][k]] if axis = 2,` ` out[i][j][k] = input[i][j][index[i][j][k]] if axis = 2,` `15` `15` `16` `16` `This operator is also the inverse of ScatterElements. It is similar to Torch's gather operation.` `This operator is also the inverse of ScatterElements. It is similar to Torch's gather operation.` `17` `17` `18` `18` `Example 1:` `Example 1:` `19` `19` `::` `::` `20` `20` `21` `21` ` data = [` ` data = [` `22` `22` ` [1, 2],` ` [1, 2],` `23` `23` ` [3, 4],` ` [3, 4],` `24` `24` ` ]` ` ]` `25` `25` ` indices = [` ` indices = [` `26` `26` ` [0, 0],` ` [0, 0],` `27` `27` ` [1, 0],` ` [1, 0],` `28` `28` ` ]` ` ]` `29` `29` ` axis = 1` ` axis = 1` `30` `30` ` output = [` ` output = [` `31` ` [` `32` `31` ` [1, 1],` ` [1, 1],` `33` `32` ` [4, 3],` ` [4, 3],` `34` ` ],` `35` `33` ` ]` ` ]` `36` `34` `37` `35` `Example 2:` `Example 2:` `38` `36` `::` `::` `39` `37` `40` `38` ` data = [` ` data = [` `41` `39` ` [1, 2, 3],` ` [1, 2, 3],` `42` `40` ` [4, 5, 6],` ` [4, 5, 6],` `43` `41` ` [7, 8, 9],` ` [7, 8, 9],` `44` `42` ` ]` ` ]` `45` `43` ` indices = [` ` indices = [` `46` `44` ` [1, 2, 0],` ` [1, 2, 0],` `47` `45` ` [2, 0, 0],` ` [2, 0, 0],` `48` `46` ` ]` ` ]` `49` `47` ` axis = 0` ` axis = 0` `50` `48` ` output = [` ` output = [` `51` ` [` `52` `49` ` [4, 8, 3],` ` [4, 8, 3],` `53` `50` ` [7, 2, 3],` ` [7, 2, 3],` `54` ` ],` `55` `51` ` ]` ` ]` `56` `52` `57` `53` `**Attributes**` `**Attributes**` `58` `54` `59` `55` `* **axis**:` `* **axis**:` `60` `56` ` Which axis to gather on. Negative value means counting dimensions` ` Which axis to gather on. Negative value means counting dimensions` `61` `57` ` from the back. Accepted range is [-r, r-1] where r = rank(data). Default value is 0.` ` from the back. Accepted range is [-r, r-1] where r = rank(data). Default value is 0.` `62` `58` `63` `59` `**Inputs**` `**Inputs**` `64` `60` `65` `61` `* **data** (heterogeneous) - **T**:` `* **data** (heterogeneous) - **T**:` `66` `62` ` Tensor of rank r >= 1.` ` Tensor of rank r >= 1.` `67` `63` `* **indices** (heterogeneous) - **Tind**:` `* **indices** (heterogeneous) - **Tind**:` `68` `64` ` Tensor of int32/int64 indices, with the same rank r as the input.` ` Tensor of int32/int64 indices, with the same rank r as the input.` `69` `65` ` All index values are expected to be within bounds [-s, s-1] along` ` All index values are expected to be within bounds [-s, s-1] along` `70` `66` ` axis of size s. It is an error if any of the index values are out of` ` axis of size s. It is an error if any of the index values are out of` `71` `67` ` bounds.` ` bounds.` `72` `68` `73` `69` `**Outputs**` `**Outputs**` `74` `70` `75` `71` `* **output** (heterogeneous) - **T**:` `* **output** (heterogeneous) - **T**:` `76` `72` ` Tensor of the same shape as indices.` ` Tensor of the same shape as indices.` `77` `73` `78` `74` `**Type Constraints**` `**Type Constraints**` `79` `75` `80` `76` `* **T** in (` `* **T** in (` `77` ` tensor(bfloat16),` `81` `78` ` tensor(bool),` ` tensor(bool),` `82` `79` ` tensor(complex128),` ` tensor(complex128),` `83` `80` ` tensor(complex64),` ` tensor(complex64),` `84` `81` ` tensor(double),` ` tensor(double),` `85` `82` ` tensor(float),` ` tensor(float),` `86` `83` ` tensor(float16),` ` tensor(float16),` `87` `84` ` tensor(int16),` ` tensor(int16),` `88` `85` ` tensor(int32),` ` tensor(int32),` `89` `86` ` tensor(int64),` ` tensor(int64),` `90` `87` ` tensor(int8),` ` tensor(int8),` `91` `88` ` tensor(string),` ` tensor(string),` `92` `89` ` tensor(uint16),` ` tensor(uint16),` `93` `90` ` tensor(uint32),` ` tensor(uint32),` `94` `91` ` tensor(uint64),` ` tensor(uint64),` `95` `92` ` tensor(uint8)` ` tensor(uint8)` `96` `93` ` ):` ` ):` `97` `94` ` Constrain input and output types to any tensor type.` ` Constrain input and output types to any tensor type.` `98` `95` `* **Tind** in (` `* **Tind** in (` `99` `96` ` tensor(int32),` ` tensor(int32),` `100` `97` ` tensor(int64)` ` tensor(int64)` `101` `98` ` ):` ` ):` `102` `99` ` Constrain indices to integer types` ` Constrain indices to integer types`
.. _l-onnx-op-gatherelements-11: GatherElements - 11 =================== **Version** * **name**: `GatherElements (GitHub) `_ * **domain**: **main** * **since_version**: **11** * **function**: False * **support_level**: SupportType.COMMON * **shape inference**: True This version of the operator has been available **since version 11**. **Summary** GatherElements takes two inputs `data` and `indices` of the same rank r >= 1 and an optional attribute `axis` that identifies an axis of `data` (by default, the outer-most axis, that is axis 0). It is an indexing operation that produces its output by indexing into the input data tensor at index positions determined by elements of the `indices` tensor. Its output shape is the same as the shape of `indices` and consists of one value (gathered from the `data`) for each element in `indices`. For instance, in the 3-D case (r = 3), the output produced is determined by the following equations: :: out[i][j][k] = input[index[i][j][k]][j][k] if axis = 0, out[i][j][k] = input[i][index[i][j][k]][k] if axis = 1, out[i][j][k] = input[i][j][index[i][j][k]] if axis = 2, This operator is also the inverse of ScatterElements. It is similar to Torch's gather operation. Example 1: :: data = [ [1, 2], [3, 4], ] indices = [ [0, 0], [1, 0], ] axis = 1 output = [ [ [1, 1], [4, 3], ], ] Example 2: :: data = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], ] indices = [ [1, 2, 0], [2, 0, 0], ] axis = 0 output = [ [ [4, 8, 3], [7, 2, 3], ], ] **Attributes** * **axis**: Which axis to gather on. Negative value means counting dimensions from the back. Accepted range is [-r, r-1] where r = rank(data). Default value is ``0``. **Inputs** * **data** (heterogeneous) - **T**: Tensor of rank r >= 1. * **indices** (heterogeneous) - **Tind**: Tensor of int32/int64 indices, with the same rank r as the input. All index values are expected to be within bounds [-s, s-1] along axis of size s. It is an error if any of the index values are out of bounds. **Outputs** * **output** (heterogeneous) - **T**: Tensor of the same shape as indices. **Type Constraints** * **T** in ( tensor(bool), tensor(complex128), tensor(complex64), tensor(double), tensor(float), tensor(float16), tensor(int16), tensor(int32), tensor(int64), tensor(int8), tensor(string), tensor(uint16), tensor(uint32), tensor(uint64), tensor(uint8) ): Constrain input and output types to any tensor type. * **Tind** in ( tensor(int32), tensor(int64) ): Constrain indices to integer types
| 4,131
| 12,123
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2022-49
|
latest
|
en
| 0.724718
|
https://gmatclub.com/forum/it-takes-printer-a-4-more-minutes-than-printer-b-to-print-60752.html?oldest=1
| 1,508,543,515,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187824471.6/warc/CC-MAIN-20171020230225-20171021010225-00677.warc.gz
| 701,565,382
| 40,271
|
It is currently 20 Oct 2017, 16:51
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# It takes printer A 4 more minutes than printer B to print 40
Author Message
VP
Joined: 22 Nov 2007
Posts: 1079
Kudos [?]: 679 [0], given: 0
It takes printer A 4 more minutes than printer B to print 40 [#permalink]
### Show Tags
02 Mar 2008, 09:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages?
12
18
20
24
30
Kudos [?]: 679 [0], given: 0
Director
Joined: 10 Sep 2007
Posts: 934
Kudos [?]: 338 [0], given: 0
### Show Tags
02 Mar 2008, 09:55
Assume that B prints 40 Pages in X mins, then B prints 40/X pages in 1 min. So total pages it prints in 6 minutes is 240/X.
As A is taking 4 more minutes than B in printing the pages so A print 40/(X+4) pages in 1 min. So total pages it prints in 6 minutes is 240/(X+4).
So Total pages printed by A + B in 6 mins = 240/X + 240/(X+4) = 50
Solving this Quarditic quation will give X=8 as only real value.
There by A takes 12 mins to print 40 pages, hence for printing 80 pages it will take 24 mins.
Ans D
Kudos [?]: 338 [0], given: 0
SVP
Joined: 28 Dec 2005
Posts: 1545
Kudos [?]: 179 [0], given: 2
### Show Tags
02 Mar 2008, 11:56
assume B takes x minutes to do 40 pages. B's rate is 40/x pages per minute, and A's rate is 40/x+4 minutes.
together, we have 40/x + 40/x+4 = 50/6 pages per minute. solve for x
Kudos [?]: 179 [0], given: 2
Senior Manager
Joined: 19 Nov 2007
Posts: 459
Kudos [?]: 218 [0], given: 4
### Show Tags
02 Mar 2008, 17:52
pmenon wrote:
assume B takes x minutes to do 40 pages. B's rate is 40/x pages per minute, and A's rate is 40/x+4 minutes.
together, we have 40/x + 40/x+4 = 50/6 pages per minute. solve for x
which gives D.
_________________
-Underline your question. It takes only a few seconds!
-Search before you post.
Kudos [?]: 218 [0], given: 4
Re: Work-rate [#permalink] 02 Mar 2008, 17:52
Display posts from previous: Sort by
| 850
| 2,689
|
{"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}
| 4.4375
| 4
|
CC-MAIN-2017-43
|
latest
|
en
| 0.826714
|
https://www.nagwa.com/en/videos/543134526190/
| 1,725,953,211,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00244.warc.gz
| 860,600,869
| 35,952
|
Question Video: Finding the Magnitude of the Displacement Vector | Nagwa Question Video: Finding the Magnitude of the Displacement Vector | Nagwa
# Question Video: Finding the Magnitude of the Displacement Vector Mathematics • Second Year of Secondary School
## Join Nagwa Classes
A particle moves such that its position vector π« = (4π‘ β 2)π’ + (3π‘ + 1)π£. The magnitude of its displacement till π‘ = 2 equals οΌΏ length units. [A] 7 [B] 10 [C] 13 [D] β85
03:02
### Video Transcript
A particle moves such that its position vector π« is equal to four π‘ minus two π’ plus three π‘ plus one π£. The magnitude of its displacement till π‘ equals two equals what length units. Is it (A) seven, (B) 10, (C) 13, or (D) root 85?
In this question, weβre given the position vector of a particle, and we need to calculate the magnitude of its displacement at π‘ equals two. We recall that the displacement vector π¬ of π‘ is equal to π« of π‘ minus π« of zero, where π« of zero is the initial position of the particle. As we need to find the displacement when π‘ equals two, π¬ of two will be equal to π« of two minus π« of zero. We can calculate π« of two and π« of zero by substituting two and zero into our equation for the position vector. π« of two is equal to four multiplied by two minus two π’ plus three multiplied by two plus one π£. This simplifies to six π’ plus seven π£.
π« of zero is equal to four multiplied by zero minus two π’ plus three multiplied by zero plus one π£, where π’ and π£ are the standard perpendicular unit vectors. This simplifies to negative two π’ plus π£. The displacement at π‘ equals two is therefore equal to six π’ plus seven π£ minus negative two π’ plus π£. Six π’ minus negative two π’ is equal to eight π’, and seven π£ minus π£ is equal to six π£. π¬ of two is equal to eight π’ plus six π£.
We are asked to calculate the magnitude of this. And we know that the magnitude of any vector is equal to the square root of the sum of the squares of its individual components. The magnitude of the displacement is equal to the square root of eight squared plus six squared. Eight squared is equal to 64, and six squared is 36. So we have the square root of 100. This is equal to 10. We can therefore conclude that the magnitude of the displacement of the particle till π‘ equals two is equal to 10 length units. The correct answer is option (B).
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions
| 763
| 2,699
|
{"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}
| 4.53125
| 5
|
CC-MAIN-2024-38
|
latest
|
en
| 0.799462
|
http://www.chegg.com/homework-help/suppose-y-gamma-distribution-parameters-c-0-constant-derive-chapter-6-problem-12e-solution-9780495385080-exc
| 1,469,633,722,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257826908.63/warc/CC-MAIN-20160723071026-00190-ip-10-185-27-174.ec2.internal.warc.gz
| 373,891,250
| 20,212
|
View more editions
# TEXTBOOK SOLUTIONS FOR Mathematical Statistics with Applications 7th Edition
• 1713 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
100% (3 ratings)
Suppose that Y has a gamma distribution with parameters α and β and that c > 0 is a constant.
a Derive the density function of U = cY .
b Identify the density of U as one of the types we studied in Chapter 4. Be sure to identify any parameter values.
c The parameters α and β of a gamma-distributed random variable are, respectively, “shape” and “scale” parameters. How do the scale and shape parameters for U compare to those for Y ?
STEP-BY-STEP SOLUTION:
Chapter: Problem:
100% (3 ratings)
• Step 1 of 4
Let Y be a random variable as follows Gamma distribution with parameter and
That is the probability density function of Y is
a) Define U= CY
The probability density function of U is
• Chapter , Problem is solved.
Corresponding Textbook
Mathematical Statistics with Applications | 7th Edition
9780495385080ISBN-13: 0495385085ISBN:
Alternate ISBN: 9780495110811
| 299
| 1,209
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.78125
| 3
|
CC-MAIN-2016-30
|
latest
|
en
| 0.762998
|
https://www.coursehero.com/file/6385421/1990AB1/
| 1,524,236,965,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00167.warc.gz
| 771,917,858
| 25,591
|
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
1990AB1
1990AB1 - “mm g i1 is E 5“ E E lci‘iO—W‘HBI 1 A...
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: “mm g i1 is E:- 5“: E'- E. lci‘iO—W‘HBI 1. A particle, initialiy at rest, moves along the x -axis so that its acceieration at an}r time t 2 0 is given by a(t) = 1212 — 4. The position ofthe particle when t 2 l is x0) = 3. (a) Find the values of t for which the particle is at rest. (33) Write an expression for the position x(t) of the particle at any time t 2 0. (c) Find the total distance traveled by the particle from t = 0 to I = 2. at) v(t\= moat = +6—th + c, tzo chfl=O=> c.=o «ML vac): Lit3—‘lt = 4t(t2—|)7tzo wt): 0 when 12:0}! . Bud: 1:20 VLt3=O when 13:0 I VGA: -- + ‘2': O l 52 b) x Lt): Enema = t*-— 2t“+—Q. xm=3=> :—a+c,=3 OE c,=‘+ TB]:= [Skits—Molt] + I S [741:3- titldt I =[2t‘- 5*]; + Et“- 2151131 .2 -I + (9—(—o)=/o to H T) . v(t) 1990 A131 ...
View Full Document
{[ snackBarMessage ]}
| 445
| 1,084
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2018-17
|
latest
|
en
| 0.684749
|
https://stackoverflow.com/questions/13416894/calculate-the-number-of-months-between-two-dates-in-php/65849578
| 1,721,070,711,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00327.warc.gz
| 460,687,845
| 59,872
|
# Calculate the number of months between two dates in PHP?
Without using PHP 5.3's date_diff function (I'm using PHP 5.2.17), is there a simple and accurate way to do this? I am thinking of something like the code below, but I don't know how to account for leap years:
\$days = ceil(abs( strtotime('2000-01-25') - strtotime('2010-02-20') ) / 86400);
\$months = ???;
I'm trying to work out the number of months old a person is.
• Does that also rule out the DateTime class? Commented Nov 16, 2012 at 12:50
• any reason you didnt like the solution on php.net? php.net/manual/en/datetime.diff.php#107434 this one Commented Nov 16, 2012 at 12:50
• What exactly do you mean with number of months? What's the difference of months between 5 October and 3 November of the same year? And what about 31st of October and 1 November? Commented Nov 16, 2012 at 12:52
• Count the number of years, multiply it by 12, subtract the beginning month from the end month, then do the same with the days. -edit- or do what deceze suggested ;) Commented Nov 16, 2012 at 12:52
• Thanks all, I've updated the question now which should answer your queries. @NappingRabbit, its extremely long! Commented Nov 16, 2012 at 12:57
\$date1 = '2000-01-25';
\$date2 = '2010-02-20';
\$ts1 = strtotime(\$date1);
\$ts2 = strtotime(\$date2);
\$year1 = date('Y', \$ts1);
\$year2 = date('Y', \$ts2);
\$month1 = date('m', \$ts1);
\$month2 = date('m', \$ts2);
\$diff = ((\$year2 - \$year1) * 12) + (\$month2 - \$month1);
You may want to include the days somewhere too, depending on whether you mean whole months or not. Hope you get the idea though.
• Actually this approach works better than others. For example date_diff() produces a diff of "7 months and 30 days" between 2009-09-01 and 2010-05-01 while this solutions produces "8 months" always. Commented Jun 29, 2017 at 7:14
• You can use date('n', \$ts1) to get numbers without leading zeros....
– Čamo
Commented Jan 4 at 10:09
Here is my solution. It checks both years and months of dates and find difference.
\$date1 = '2000-01-25';
\$date2 = '2010-02-20';
\$d1=new DateTime(\$date2);
\$d2=new DateTime(\$date1);
\$Months = \$d2->diff(\$d1);
\$howeverManyMonths = ((\$Months->y) * 12) + (\$Months->m);
• the cleanest solution in my opinion BUT set up the time zone parameter \$d1=new DateTime(\$date2, new DateTimeZone('Europe/Paris') ) Commented Sep 6, 2020 at 9:10
• the perfect solution, way cleanner Commented Oct 12, 2020 at 18:33
This is a simple method I wrote in my class to count the number of months involved into two given dates :
public function nb_mois(\$date1, \$date2)
{
\$begin = new DateTime( \$date1 );
\$end = new DateTime( \$date2 );
\$end = \$end->modify( '+1 month' );
\$interval = DateInterval::createFromDateString('1 month');
\$period = new DatePeriod(\$begin, \$interval, \$end);
\$counter = 0;
foreach(\$period as \$dt) {
\$counter++;
}
return \$counter;
}
• Instead of looping over the period, you can simply use iterator_count(\$period) Commented Dec 21, 2015 at 19:12
• I really like this solution, especially with @Koen. 's replacement for the loop. Commented Sep 23, 2019 at 14:09
• I changed this to \$end = \$end->modify( '+1 day' ); in my code and worked great Commented Jul 18, 2020 at 12:55
This is how I ended up solving it. I know I'm a little late but I hope this saves someone a lot time and lines of code.
I used DateInterval::format to display a human readable countdown clock in years, months, and days. Check https://www.php.net/manual/en/dateinterval.format.php for the format table to see your options on how to modify your return values. Should give you what you're looking for.
\$origin = new DateTime('2020-10-01');
\$target = new DateTime('2020-12-25');
\$interval = \$origin->diff(\$target);
echo \$interval->format('%y years, %m month, %d days until Christmas.');
Outputs: 0 years, 2 month, 24 days
Like this:
\$date1 = strtotime('2000-01-25');
\$date2 = strtotime('2010-02-20');
\$months = 0;
while ((\$date1 = strtotime('+1 MONTH', \$date1)) <= \$date2)
\$months++;
echo \$months;
If you want to include days to, then use this:
\$date1 = strtotime('2000-01-25');
\$date2 = strtotime('2010-02-20');
\$months = 0;
while (strtotime('+1 MONTH', \$date1) < \$date2) {
\$months++;
\$date1 = strtotime('+1 MONTH', \$date1);
}
echo \$months, ' month, ', (\$date2 - \$date1) / (60*60*24), ' days'; // 120 month, 26 days
• This would tell you that between 2012-01-01 and 2012-02-01 there are two months. Even if you changed <= to < it would yield two months for the Jan-1 --> Feb-2 timespan, which might not be the best solution. Commented Nov 16, 2012 at 12:58
• Using the dates you specified I get 1 returned because it updates \$date1 before it checks if it is less and adds a month.
Commented Nov 16, 2012 at 13:01
• All those using this, do take care of the while statement with some limiting value for its iteration. I'm not saying if the code is wrong, it is just bit safe with bit precautions with iterative statements. Commented Sep 6, 2018 at 10:35
my function to resolve issue
function diffMonth(\$from, \$to) {
\$fromYear = date("Y", strtotime(\$from));
\$fromMonth = date("m", strtotime(\$from));
\$toYear = date("Y", strtotime(\$to));
\$toMonth = date("m", strtotime(\$to));
if (\$fromYear == \$toYear) {
return (\$toMonth-\$fromMonth)+1;
} else {
return (12-\$fromMonth)+1+\$toMonth;
}
}
Here is my solution. It only checks years and months of dates. So, if one date is 31.10.15 and other is 02.11.15 it returns 1 month.
function get_interval_in_month(\$from, \$to) {
\$month_in_year = 12;
\$date_from = getdate(strtotime(\$from));
\$date_to = getdate(strtotime(\$to));
return (\$date_to['year'] - \$date_from['year']) * \$month_in_year -
(\$month_in_year - \$date_to['mon']) +
(\$month_in_year - \$date_from['mon']);
}
\$d1 = new DateTime("2009-09-01");
\$d2 = new DateTime("2010-09-01");
\$months = 0;
while (\$d1 <= \$d2){
\$months ++;
}
print_r(\$months);
I recently needed to calculate age in months ranging from prenatal to 5 years old (60+ months).
Neither of the answers above worked for me. The first one I tried, which is basically a 1 liner for deceze's answer
\$bdate = strtotime('2011-11-04');
\$edate = strtotime('2011-12-03');
\$age = ((date('Y',\$edate) - date('Y',\$bdate)) * 12) + (date('m',\$edate) - date('m',\$bdate));
. . .
This fails with the set dates, obviously the answer should be 0 as the month mark (2011-12-04) hasn't been reached yet, how ever the code returns 1.
The second method I tried, using Adam's code
\$bdate = strtotime('2011-01-03');
\$edate = strtotime('2011-02-03');
\$age = 0;
while (strtotime('+1 MONTH', \$bdate) < \$edate) {
\$age++;
\$bdate = strtotime('+1 MONTH', \$bdate);
}
. . .
This fails and says 0 months, when it should be 1.
What did work for me, is a little expansion of this code. What I used is the following:
\$bdate = strtotime('2011-11-04');
\$edate = strtotime('2012-01-04');
\$age = 0;
if(\$edate < \$bdate) {
//prenatal
\$age = -1;
} else {
//born, count months.
while(\$bdate < \$edate) {
\$age++;
\$bdate = strtotime('+1 MONTH', \$bdate);
if (\$bdate > \$edate) {
\$age--;
}
}
}
My solution was mix of a few answers. I did not want to do a loop especially when I have all the data in the \$interval->diff, I just did the math, and the number of months could be negative if in my case so this was my approach.
/**
* Function will give you the difference months between two dates
*
* @param string \$start_date
* @param string \$end_date
* @return int|null
*/
public function get_months_between_dates(string \$start_date, string \$end_date): ?int
{
\$startDate = \$start_date instanceof Datetime ? \$start_date : new DateTime(\$start_date);
\$endDate = \$end_date instanceof Datetime ? \$end_date : new DateTime(\$end_date);
\$interval = \$startDate->diff(\$endDate);
\$months = (\$interval->y * 12) + \$interval->m;
return \$startDate > \$endDate ? -\$months : \$months;
}
Follow up on the answer of @deceze (I've upvoted on his answer). Month will still count as a whole even if the day of the first date didn't reached the day of the second date.
Here's my simple solution on including the day:
\$ts1=strtotime(\$date1);
\$ts2=strtotime(\$date2);
\$year1 = date('Y', \$ts1);
\$year2 = date('Y', \$ts2);
\$month1 = date('m', \$ts1);
\$month2 = date('m', \$ts2);
\$day1 = date('d', \$ts1); /* I'VE ADDED THE DAY VARIABLE OF DATE1 AND DATE2 */
\$day2 = date('d', \$ts2);
\$diff = ((\$year2 - \$year1) * 12) + (\$month2 - \$month1);
/* IF THE DAY2 IS LESS THAN DAY1, IT WILL LESSEN THE \$diff VALUE BY ONE */
if(\$day2<\$day1){ \$diff=\$diff-1; }
The logic is, if the day of the second date is less than the day of the first date, it will reduce the value of \$diff variable by one.
\$date1 = '2000-01-25';
\$date2 = '2010-02-20';
\$ts1 = strtotime(\$date1);
\$ts2 = strtotime(\$date2);
\$year1 = date('Y', \$ts1);
\$year2 = date('Y', \$ts2);
\$month1 = date('m', \$ts1);
\$month2 = date('m', \$ts2);
\$diff = ((\$year2 - \$year1) * 12) + (\$month2 - \$month1);
If Month switched from Jan to Feb above code will return you the \$diff = 1 But if you want to consider next month only after 30 days then add below lines of code along with above.
\$day1 = date('d', \$ts1);
\$day2 = date('d', \$ts2);
if(\$day2 < \$day1){ \$diff = \$diff - 1; }
To calculate the number of CALENDAR MONTHS (as also asked here) between two dates, I usually end up doing something like this. I convert the two dates to strings like "2020-05" and "1994-05", then get their respective result from the below function, then run a subtraction of those results.
/**
* Will return number of months. For 2020 April, that will be the result of (2020*12+4) = 24244
* 2020-12 = 24240 + 12 = 24252
* 2021-01 = 24252 + 01 = 24253
* @param string \$year_month Should be "year-month", like "2020-04" etc.
*/
static private function calculate_month_total(\$year_month)
{
\$parts = explode('-', \$year_month);
\$year = (int)\$parts[0];
\$month = (int)\$parts[1];
return \$year * 12 + \$month;
}
function date_duration(\$date){
\$date1 = new DateTime(\$date);
\$date2 = new DateTime();
\$interval = \$date1->diff(\$date2);
if(\$interval->y > 0 and \$interval->y < 2){
return \$interval->y.' year ago';
}else if(\$interval->y > 1){
return \$interval->y.' years ago';
}else if(\$interval->m > 0 and \$interval->m < 2){
return \$interval->m.' month ago';
}else if(\$interval->m > 1){
return \$interval->y.' months ago';
}else if(\$interval->d > 1){
return \$interval->d.' days ago';
}else{
if(\$interval->h > 0 and \$interval->h < 2){
return \$interval->h.' hour ago';
}else if(\$interval->h > 1){
return \$interval->h.' hours ago';
}else{
if(\$interval->i > 0 and \$interval->i < 2){
return \$interval->i.' minute ago';
}else if(\$interval->i > 1){
return \$interval->i.' minutes ago';
}else{
return 'Just now';
}
}
}
}
Returns 11 Months ago, 2 years ago, 5 minutes ago style of date differentiation
Example:
echo date_duration('2021-02-28 14:59:00.00');
Will return '1 Month ago' depending on your current month
I just want to share the function I wrote. You can modify it to get both months and years by entering the related datetime format to define the modifier aka modify("+1 something").
/**
* @param DateTimeInterface \$start a anything using the DateTime interface
* @param DateTimeInterface|null \$end to calculate the difference to
* @param string \$modifier for example: day or month or year.
* @return int the count of periods.
*/
public static function elapsedPeriods(
DateTimeInterface \$start,
DateTimeInterface \$end = null,
string \$modifier = '1 month'
): int
{
// just an addition, in case you just want the periods up untill now
if (\$end === null ) {
\$end = new DateTime();
}
// we clone the start, because we dont want to change the actual start
// (objects are passed by ref by default, so if you forget this you might
\$cloned_start = clone \$start;
// we create a period counter, starting at zero, because we want to count
// periods, assuming the first try, makes one period.
// (week, month, year, et cetera)
\$period_count = 0;
// so while our \$cloned_start is smaller to the \$end (or now).
// we will increment the counter
while (\$cloned_start < \$end) {
// first off we increment the count, for the first iteration could end
// the cycle
\$period_count++;
// now we modify the cloned start
\$cloned_start->modify(sprintf("+1 %s", \$modifier));
}
return \$period_count; // return the count
}
cheers
\$d1 = date('Y-m-d', strtotime('2015-02-13'));
\$d2 = date('Y-m-d');
\$d1 = new DateTime(\$d1); \$d2 = new DateTime(\$d2);
\$interval = \$d2->diff(\$d1);
\$interval->m = (\$interval->m + (12 * \$interval->y));
\$months = \$interval->format('%m');
– Community Bot
Commented May 21, 2022 at 19:17
You can make the calculations manually, for which I created a simple class. You input the start and end dates, and get the amount of months between those two dates.
The formulate converts the days into decimal point values taking into account the amount of days the corresponding month had in that specific year.
Here is the class:
class TimeHelper
{
private \$years = 0;
private \$months = 0;
private \$days = 0;
public function getMonthsDiff(\$start, \$end) : float
{
\$this->resetDifValues();
\$start = new \DateTime(\$start);
\$end = new \DateTime(\$end);
\$this->getYears(\$end, \$start);
\$this->getMonths(\$end, \$start);
\$this->getDays(\$end, \$start);
\$totalMonths = (\$this->years * 12) + \$this->months;
\$totalMonths = floatval(\$totalMonths) + (floatval(\$this->days) / floatval(\$this->getDaysInStartMonth(\$start)));
return \$totalMonths;
}
/**
* @return void
*/
private function resetDifValues(): void
{
\$this->years = 0;
\$this->months = 0;
\$this->days = 0;
}
/**
* @param \DateTime \$end
* @param \DateTime \$start
* @return void
*/
private function getYears(\DateTime \$end, \DateTime \$start): void
{
\$this->years = intval(\$end->format('Y')) - intval(\$start->format('Y'));
if (\$this->years < 0) {
throw new \InvalidArgumentException('The end date must be greater than the start date');
}
}
/**
* @return void
*/
private function convertYearToMonths(): void
{
if (\$this->years <= 0) {
throw new \InvalidArgumentException('The end date must be greater than the start date');
}
\$this->years--;
\$this->months += 12;
}
/**
* @param \DateTime \$end
* @param \DateTime \$start
* @return void
*/
private function getMonths(\DateTime \$end, \DateTime \$start): void
{
\$this->months = intval(\$end->format('m')) - intval(\$start->format('m'));
if (\$this->months < 0) {
\$this->convertYearToMonths();
}
}
/**
* @param \DateTime \$end
* @return void
*/
private function convertMonthInDays(\DateTime \$end, \DateTime \$start): void
{
if (\$this->months <= 0) {
\$this->convertYearToMonths();
}
\$this->months--;
\$daysInMonthBeforeEndDate = \$this->getDaysInStartMonth(\$start);
\$this->days += \$daysInMonthBeforeEndDate;
}
/**
* @param \DateTime \$end
* @param \DateTime \$start
* @return void
*/
private function getDays(\DateTime \$end, \DateTime \$start): void
{
\$this->days = intval(\$end->format('d')) - intval(\$start->format('d'));
if (\$this->days < 0) {
\$this->convertMonthInDays(\$end, \$start);
}
}
/**
* @param \DateTime \$start
* @return int
*/
private function getDaysInStartMonth(\DateTime \$start): int
{
\$monthBeforeEndDate = intval(\$start->format('m'));
return cal_days_in_month(CAL_GREGORIAN, \$monthBeforeEndDate, \$start->format('Y'));
}
}
To get the result you just have to instantiate the class and call the method getMonthsDiff.
\$timeHelper = new TimeHelper();
\$monthsDiff = \$timeHelper->getMonthsDiff(\$project->dt_start, \$project->dt_end);
This will give you the amount of months in float format like 1.0 or 10.935...
| 4,597
| 15,822
|
{"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-2024-30
|
latest
|
en
| 0.886854
|
https://goprep.co/q5-y-axis-and-line-x-4-are-parallel-lines-what-is-the-i-1nkchv
| 1,604,133,951,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00624.warc.gz
| 338,023,279
| 28,727
|
# Y-axis and line x = -4 are parallel lines. What is the distance between them?
The equation of line parallel to Y-axis is x = a.
x = -4
a = -4
Since the distance between and the line is 4 units.
The graph is shown below:
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Basic Understanding of Coordinate Geometry33 mins
Champ Quiz | 2-Dimension( Coordinate geometry )36 mins
Quiz | Imp. Qs. on Coordinate Geometry39 mins
Know How to Solve Complex Geometry Problems!27 mins
Coordinate Geometry45 mins
Euclid's Geometry51 mins
Quiz | Euclid's Geometry44 mins
Basics of Coordinate Geometry43 mins
Quiz | Imp. Qs. of Coordinate Geometry46 mins
A Solid Grip on Basics of Coordinate Geometry49 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
| 244
| 1,027
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.765625
| 3
|
CC-MAIN-2020-45
|
longest
|
en
| 0.838876
|
www.olliewilliams.org
| 1,550,915,745,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-09/segments/1550249495888.71/warc/CC-MAIN-20190223082039-20190223104039-00288.warc.gz
| 395,927,939
| 15,257
|
# Infix notation: Pratt parsers and building a calculator
Welcome to the third installment (part 1, part 2) of my series about building a parser framework in Apple's Swift language. In this post I'll show how we parse infix notation and build up a simple calculator.
At this point we've got a pretty solid recursive descent parser which makes it easy to describe parsers for data which are presented largely sequentially: where what we parse earlier in the stream dictates how we parse what comes later. Formally, our parser is good for LL(k) grammars and, as we saw in the previous post, it's pretty easy to build a parser for JSON, which is LL(1).
## Infix notation
We humans tend to read and write sequentially; hence, many representations of data have this sequential characteristic. One important exception to this is mathematical notation. Consider a calculator application in which we want to parse the input:
``+ 3 * 4 2``
This is prefix notation. Like S-expressions and JSON our parser is very happy to process this type of input because it's sequential: we first read the `+` which tells us to "recursively parse two more expressions and add them". The operators `*` and `+` come before their operands. Sadly, no one wants to write a formula like this. We'd far rather use infix notation in which the operators come between the operands:
``3 + 4 * 2``
Here we have to read the `3` and put it in a safe place before we get to the `+`. Now at least the parser knows that it needs to read one more expression. But should it read just the `4`, or the complete remainder of the string `4 * 2`? At school I was taught BIDMAS, which means that I know that the multiplication should happen before the addition. That's the hard bit about infix notation: our parser must take precedence (and associativity1) into account. These are properties of the operators which dictate how terms are grouped; they provide implicit parentheses in our string. BIDMAS tells us that `*` has higher precedence than `+`, so our example should be grouped as
``3 + 4 * 2 = 3 + (4 * 2)``
Now we know that after parsing `3` and `+` we should parse the rest of the string. But the parser won't know this until it has parsed the `4` (also stashing that somewhere safe) and got to the `*`.
If we want this parser framework to be a serious tool with which a real programming language could be parsed, we'll need to be able to handle infix notation. It is possible to build an infix parser with our current parsing tools with precedence climbing. But this would be painful to implement, hard to debug, and inefficient. It would also be more static than I want, in that it would only offer a fixed number of precedence levels. I would like to make it relatively easy for new operators to be added dynamically to the parser. After all, Swift has this feature, and I have benefitted from it greatly in making this library.
## Pratt parser
The recursive descent approach does almost everything we want from a parser succinctly and cleanly. I therefore want to make only small changes to the framework to accommodate operator precedence. To our rescue comes the Pratt parser (Wikipedia article). The introduction to this paper does a nice job of summing up the raison d'être of this little project
...we find some language implementers willing to incorporate as much syntax as possible provided they do not have to work hard at it...
I won't give a tutorial of how the Pratt parser works. You can either read the paper, or check out some other online resources: Wikipedia, blog, blog. Instead I'll sketch out what you need to know to drive this part of framework.
## Calculator demo
To demonstrate how to handle infix (and prefix, and postfix) notation, let's build a simple calculator demo. There are lots of static parts to this so, like the JSON parser, I will house them as static members of a `struct`:
``````struct Calculator : Parser {
static let skip = regex("\\s*")
...
}``````
To add infix parsing, we create an instance of `OperatorPrecedence`:
``````class OperatorPrecedence<
T, O:Parser, P:Parser
where P.Target==T, O.Target==String > : Parser
{
typealias Target = T
private let primary : P
...
init(opFormat:O, primary:P) {
self.primary = primary
...
}
...
}``````
I sure do love me some strong typing. Swift occupies a middle ground between more functional languages (F#, ML, Haskell), and more C-like languages (C, C++, C#). I think that, on the whole, Swift does a grand job of bridging these worlds; but there are a few warts, generic signatures like this being one of them. As you can see, `OperatorPrecedence` is constructed with two sub-parsers. `opFormat` is used to parse `String`s which represent operator symbols. The `primary` parser is used to parse primary expressions: those things which the operators stick together. For example, in the expression:
``3 + y * 5``
`+` and `*` are the operators; `3`, `y`, and `5` are the primary expressions. Primary elements are the most tightly-bound in an expression and have higher precedence than any other operator we might add.
Let's add an `OperatorPrecedence` parser to `Calculator`:
``````struct Calculator : Parser {
...
static let opFormat = (regex("[+*-/%\\^]")) ~> skip
static let primary = LateBound<Double>()
static let opp = OperatorPrecedence(opFormat:opFormat, primary:primary)
...
}``````
For this simple tool, the `opFormat` looks for one of the single-character symbols `+`, `*`, `%`, `\`, or `^`. These have their usual meanings for anyone familiar with C, with the exception of `^`, which I'm using to represent exponentiation: i.e., we represent 23 as `2^3`.
The `primary` parser is more complicated. Whenever `LateBound` appears we know there's going to be some recursion. In this case it's because we want our calculator to be able to handle expressions containing parentheses. We do this by treating parenthesized subexpressions as primary, capturing their maximum precedence2.
``````struct Calculator : Parser {
...
static let oparen = const("(") ~> skip
static let cparen = const(")") ~> skip
static let brackets = oparen >~ opp ~> cparen
...
}``````
The other types of primary expression are function calls and, of course, literal floating-point constants:
``````struct Calculator : Parser {
...
static let arg1 = oparen >~ opp ~> cparen
static let sinfunc = const("sin") >~ arg1 |> sin
static let cosfunc = const("cos") >~ arg1 |> cos
static let tanfunc = const("tan") >~ arg1 |> tan
static let expfunc = const("exp") >~ arg1 |> exp
static let logfunc = const("log") >~ arg1 |> log
static let sqrtfunc = const("sqrt") >~ arg1 |> sqrt
static let funcs = sinfunc | cosfunc | tanfunc | expfunc | logfunc | sqrtfunc
static let flt = FloatParser(strict:false) ~> skip
static let primaryImpl = funcs | brackets | flt
...
}``````
I am pleased how simple it is to define the function call parsers. The `sin`, `cos`, etc. at the end of each line are the actual runtime functions provided by Swift, and we're using the pipe operator `|>` to incorporate them into a parser.
So much for static components. As I mentioned above, one of my desiderata for the parser was that it would be easy to add operators dynamically. This is done by adding `OperatorHandler`s. `OperatorHandler` is an `enum`:
``````enum OperatorHandler<T> {
typealias Binary = (T, T) -> T?
typealias Unary = T -> T?
case Prefix(Unary, Int)
case LeftInfix(Binary, Int)
case RightInfix(Binary, Int)
case Postfix(Unary, Int)
...
}``````
where each `case` holds a "builder" method which implements the operation, and an integer precedence. `LeftInfix` and `RightInfix` are left- and right-associative infix operators, which take two arguments. `Prefix` and `Postfix` operators are unary, and can be thought of as infix operators which only have a right or left hand side, respectively.
In our calculator there's now a one-time initialization (no static constructors in Swift) to setup the things I couldn't fit into declarative statements. This includes adding the operators and their handlers:
``````struct Calculator : Parser {
...
private static func initialize() -> Void {
if primary.inner == nil {
opp.addOperator("+", .LeftInfix({return \$0 + \$1}, 50))
opp.addOperator("-", .LeftInfix({return \$0 - \$1}, 50))
opp.addOperator("*", .LeftInfix({return \$0 * \$1}, 70))
opp.addOperator("/", .LeftInfix({return \$0 / \$1}, 70))
opp.addOperator("%", .LeftInfix({return \$0 % \$1}, 70))
primary.inner = primaryImpl.parse
}
}
...
}``````
A higher precedence value indicates that an operator binds tighter. So with the precedences and associativities I've used here, we'll get
``3 + 4 ^ 6 * 8 + 2 = (3 + ((4 ^ 6) * 8)) + 2``
Because it's common for some symbols to be used both as infix and prefix, with different meanings, I've made sure `OperatorPrecedence` can handle this. The most common examples of this are unary `+` and `-`. In the expression
``9 * -(4 - 2)``
the first `-` character is a prefix operator indicating unary negation, whereas the second `-` is the infix symbol for binary subtraction.
And—modulo a bit of plumbing to honor the `Parser` protocol—that's pretty much it (see the full `struct` here). As always, feel free to pull the code from GitHub and have a go!
## Next
As listed in the previous post, the last piece of this puzzle is high-quality error handling. Actually, any error handling would be an improvement.
1. Associativity tells us how operators of like precedence are grouped. If we say that `+` is left associative, and `*` is right associative we get the groupings:
``````7 + 9 + 3 = (7 + 9) + 3
7 * 9 * 3 = 7 * (9 * 3)``````
I wasn't taught about this at school because we only dealt with associative algebras where it didn't matter... at least, I'm pretty sure I didn't do the cross product until University.
2. This could also be achieved by making the `(` and `)` symbols be pre- and postfix operators with appropriate precedences and semantics. I think this way is clearer.
| 2,431
| 10,005
|
{"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}
| 3.390625
| 3
|
CC-MAIN-2019-09
|
latest
|
en
| 0.938141
|
https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-11-section-11-2-the-cosine-ratio-and-applications-exercises-page-511/5
| 1,722,944,862,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640484318.27/warc/CC-MAIN-20240806095414-20240806125414-00867.warc.gz
| 620,309,919
| 12,337
|
## Elementary Geometry for College Students (7th Edition)
cos α = $\frac{sqrt of 2}{sqrt of 3}$ , cos β = $\frac{sqrt of 3}{sqrt of 5}$
Step 1: By Pythagoras theorem $(sqrt of 3)^{2}$ + $(sqrt of 2)^{2}$ = $c^{2}$ 3 + 2 = $c^{2}$ c = $\sqrt 5$ Step 2: cos α = $\frac{length of adjacent}{length of hypotenuse}$ cos α = $\frac{sqrt of 2}{sqrt of 3}$ Similarly cos β = $\frac{length of adjacent}{length of hypotenuse}$ cos β = $\frac{sqrt of 3}{sqrt of 5}$ Therefore cos α = $\frac{sqrt of 2}{sqrt of 3}$ , cos β = $\frac{sqrt of 3}{sqrt of 5}$
| 201
| 542
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.28125
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.690859
|
https://finnolux.com/lab-report-125/
| 1,591,460,361,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590348517506.81/warc/CC-MAIN-20200606155701-20200606185701-00003.warc.gz
| 338,845,786
| 14,122
|
Rock Street, San Francisco
What is the Identity of this Hydrate? Date performed: May 30, 12 Date Submitted: June 6, 2012 Name: 2968 Instructor: Reid A hydrate was given to our group and the identity of the hydrate was unknown. The lab workers were told to determine the identity of the unknown hydrate. The identity of the hydrate could be determined by calculating the hydrate’s percent of water. So the lab workers set out to determine the water percent of the unknown hydrate. The percent of any compound or element can be found by using a certain formula. This formula is: % of element = Mass of element or compound/Total mass of compound axes.
In order to use this formula the mass of the water and the total mass of the hydrate had to be found. The lab workers were first given a test tube, test tube clamp, top loader electronic balance, Bunsen burner and unknown hydrate. First the lab workers inserted a small portion of the hydrate into the test tube. The test tube with the hydrate in it was inserted into the beaker. The beaker was used because it allowed the lab workers to weigh the test tube and the hydrate with ease. The weight of the test tube, beaker and hydrate were all found. The weight of the test tube and beaker were subtracted out of the total weight.
We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!
order now
The left over weight belonged to the hydrate. The hydrate was then burned by using the Bunsen burner; until there was no remaining water left. Once the water was thoroughly burned off, the left over anhydrous salt was weighted. The Bunsen burner was used because it provided heat, which evaporated the water of the hydrate. The weight of the anhydrous salt was subtracted from the total weight of Water Percent I Hydrate Weight Anhydrous salt weight I salt percent 16. 30% | . Egg 1 0. Egg | 83. 70% | the hydrate. This calculation left the lab workers with the weight of the water.
The total weight of the hydrate and the weight of the water were plugged into the percent formula; which determined the percent. Water Percent I Hydrate Weight I Anhydrous Salt Weight I Water Weight I 14. 50% | . Gill I. Go 1. Gig The hydrate is Barium chloride dehydrate. This conclusion was drawn because the water percent of the unknown hydrate was extremely close to the water percent of barium chloride dehydrate. The little difference between the water percent of the unknown hydrate and the water percent of barium chloride dehydrate can be attributed to human error.
### Post Author: admin
x
Hi!
I'm Eric!
Would you like to get a custom essay? How about receiving a customized one?
Check it out
| 616
| 2,657
|
{"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-2020-24
|
latest
|
en
| 0.963948
|
https://doczz.net/doc/7272856/slide-1
| 1,632,616,537,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00106.warc.gz
| 273,404,893
| 7,254
|
Slide 1
Transcription
Slide 1
```A Semi-Analytic Model of Type Ia
Supernova Turbulent Deflagration
Kevin Jumper
Advised by Dr. Robert Fisher
May 3, 2011
Review of Concepts
• Type Ia supernovae may
be “standard candles”
• Progenitor is a white
dwarf in a singledegenerate system
• Accretion causes carbon
ignition and deflagration
• Fractional burnt mass is
important for describing
deflagration
Credit: NASA, ESA, and A. Field (STScI), from Briget Falck.
“Type Ia Supernova Cosmology with ADEPT.“ John Hopkins
University. 2007. Web.
The Semi-Analytic Model
• One dimensional – a single flame bubble
expands and vertically rises through the star
• The Morison equation governs bubble motion
t = time
R = bubble radius
ρ1 = bubble (ash) density
ρ2 = background star (fuel) density
• Proceeds until breakout
V = bubble volume
g = gravitational acceleration
CD = coefficient of drag
The Semi-Analytic Model (Continued)
• The coefficient of
drag depends on the
Reynolds Numbers
(Re).
•Higher Reynolds
numbers indicate
greater fluid turbulence.
3.0
Coefficient of Drag
• Δx is grid resolution
Coefficient of Drag vs. Reynolds
Number
2.5
2.0
1.5
1.0
0.5
0.0
0
20 40 60 80 100 120 140
Reynolds Number
The Three-Dimensional Simulation
• Used by a graduate student
in my research group
• Considers the entire star
• Proceeds past breakout
• Grid resolution is limited to
8 kilometers
• Longer execution time than
semi-analytic model
Credit: Dr. Robert Fisher, University of Massachusetts Dartmouth
Project Objectives
• Analyze the evolution of the flame bubble.
• Determine the fractional mass of the progenitor
burned during deflagration.
• Compare the semi-analytic model results against
the 3-D simulation.
• Add the physics of rotation to the semi-analytic
model.
Comparison with 3-D Simulations
(Updated)
Log Speed vs. Position
•The model’s bubble
rise speed is increased
due to a lower
coefficient of drag.
3
Log [Speed (km/s)]
• There is still good
initial agreement
between the model
(blue) and the
simulation (black).
2
1
0
0
400
800
1200
Position (km)
1600
Comparison with 3-D Simulations
(Updated)
•Now the model
and simulation
begin to diverge at
about 200 km.
8
Log [Area (km^2)]
•The bubble’s area
is decreased in the
model, as it has
less time to expand.
Log Area vs. Position
7
6
5
4
3
0
400
800
Position (km)
1200
1600
Comparison with 3-D Simulations
(Updated)
•The early
discrepancy
between the
volume of the
model and
simulation is much
smaller.
Log Volume vs. Position
12
11
Log [Volume (km^3)]
•The model has
greater volume until
an offset of about
600 km.
10
9
8
7
6
5
4
0
400
800
Position (km)
1200
1600
Comparison with 3-D Simulations
(Updated)
•The simulation
predicts about 1% at
breakout.
Fractional Burnt Mass vs. Position
0.040
Fractional Burnt Mass
•As predicted, the
model’s fractional
burnt mass is higher
(about 3%).
0.035
0.030
0.025
0.020
0.015
0.010
0.005
0.000
•We still need to
refine the model.
0
400
800
1200
Position (km)
1600
Adding Rotation to the Model
Spherical Coordinates
• Cartesian coordinates are
inconvenient for rotation
problems.
• r = radius from origin
• θ = inclination angle
(latitude)
• Φ = azimuth angle
(longitude)
• The above conventions may
vary by discipline.
Weisstein, Eric W. "Spherical Coordinates." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/SphericalCoordinates.html
Image Credit: Wikipedia
Adding Rotation to the Model
Force Equation
The rotating star is a noninertial reference
frame, which causes several “forces” to act
upon the bubble.
F’ = Fphysical + F’Coriolis + F’transverse + F’centrifugal – mAo
All forces except Fphysical depend on the motion
of the bubble relative to the frame.
Credit: Fowles and Cassiday. “Analytical Mechanics.” 7th ed. Thomson: Brooks/Cole.
2005. Print.
Adding Rotation to the Model
Summary of Forces
• Fphysical: forces due to matter acting
on the bubble
• F’Coriolis: acts perpendicular to the
velocity of the bubble in the
noninertial system
• F’transverse: acts perpendicular to
radius in the presence of angular
acceleration
• F’centrifugal: acts perpendicular and
out from the axis of rotation
• mAo: inertial force of translation
Credit: Fowles and Cassiday, page 199
Credit: Fowles and Cassiday. “Analytical Mechanics.” 7th ed. Thomson: Brooks/Cole.
2005. Print.
Future Work
• Try to narrow the discrepancy so that the
model and simulation agree within a factor of
two
• Program the effects of rotation into the semianalytic model
A Semi-Analytic Model of Type Ia
Supernovae
Questions?
```
| 1,304
| 4,518
|
{"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-2021-39
|
latest
|
en
| 0.797708
|
https://www.homeworkmarket.com/content/demorgan%E2%80%99s-laws-using-venn-diagram
| 1,571,459,235,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986688826.38/warc/CC-MAIN-20191019040458-20191019063958-00282.warc.gz
| 944,587,262
| 40,660
|
# DEMORGAN’S LAWS USING A VENN DIAGRAM
Part I
Demonstrate DeMorgan’s Laws using a Venn diagram.
1. Draw a Venn diagram showing the elements of sets A, B, and the universe for all 4 regions.
2. Draw a second diagram showing only the elements of the complement of set A.
3. Draw a third diagram showing only the elements of the complement of set B.
4. Draw a fourth diagram showing the union of steps 2 and 3 for the first DeMorgan's law or the intersection of steps 2 and 3 for the second DeMorgan's law.
5. Draw a fifth diagram showing the elements of the intersection of sets A and B for the first DeMorgan's law or the union of sets A and B for the second DeMorgan's law.
6. Finally, draw the last diagram showing the complement of step 5. Compare the results from step 4 against those in step 6 to prove both DeMorgan's laws. In the absence of data elements, you can use 4 different colors to clearly indicate the regions of the universe and follow all of the steps.
Part II
Define two propositions (simple statements that can be either true or false). Give a real-world example of 2 propositions that r and can represent. Call them r and s. Create a truth table that shows all values of the following:
Proposition Definition Your example explained r s ¬ r ¬s r ∧ s ¬ r v s r ∧¬s
Interpret the columns of the truth table for those proposition examples. Interpret the operations on the propositions and the values in the table based on the operations and the values of r and s.
• Posted: 5 years ago
DEMORGAN’S LAWS USING A VENN DIAGRAM
Purchase the answer to view it
Save time and money!
Our teachers already did such homework, use it as a reference!
• Rated 2 times
### Demonstrate DeMorgan’s Laws using a Venn diagram
PART I : Demonstrate DeMorgan's Laws using a Venn diagram
1.Draw a Venn diagram showing the elements of sets A, B, and the universe for all 4 regions.
2.Draw a second diagram showing only the …
• Rated 2 times
### MATHS-Draw a Venn diagram showing the elements of sets A, B, and the universe for all 4 regions
1.Draw a Venn diagram showing the elements of sets A, B, and the universe for all 4 regions.
2.Draw a second diagram showing only the elements of the complement of set A.
3.Draw a third diagram showing …
• Not rated
### melissa_morgan only
Phase 1 Individual Project Deliverable Length: 2 Parts: See Assignment Details Details: Weekly tasks or assignments (Individual or Group Projects) will be duebyMonday and late …
• Not rated
### can you plz help me
Demonstrate DeMorgan’s Laws using a Venn diagram.
1. Draw a Venn diagram showing the elements of sets A, B, and the universe for all 4 regions.
2. Draw a second diagram showing only the elements of the …
• Not rated
### Logic and Set Theory
Logic and Set Theory
Mon, 1/11/16
Numeric
100
2 Parts: See Assignment Details
Part I
Demonstrate DeMorgan’s Laws using a Venn diagram.
1. Draw a Venn diagram showing the elements of …
| 733
| 2,971
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.25
| 4
|
CC-MAIN-2019-43
|
latest
|
en
| 0.790316
|
https://www.convertunits.com/from/foot+%5Bsurvey%5D/to/sagene
| 1,685,328,559,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00257.warc.gz
| 791,459,494
| 12,702
|
## Convert foot [survey] to sagene
foot [survey] sagene
Did you mean to convert foot foot [Egypt] foot [France] foot [iraq] foot [Netherlands] foot [pre-1963 Canada] foot [Rome] foot [survey] to sagene
How many foot [survey] in 1 sagene? The answer is 6.999986. We assume you are converting between foot [survey] and sagene. You can view more details on each measurement unit: foot [survey] or sagene The SI base unit for length is the metre. 1 metre is equal to 3.2808333333333 foot [survey], or 0.4686914135733 sagene. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between feet [survey] and sagene. Type in your own numbers in the form to convert the units!
## Quick conversion chart of foot [survey] to sagene
1 foot [survey] to sagene = 0.14286 sagene
5 foot [survey] to sagene = 0.71429 sagene
10 foot [survey] to sagene = 1.42857 sagene
20 foot [survey] to sagene = 2.85715 sagene
30 foot [survey] to sagene = 4.28572 sagene
40 foot [survey] to sagene = 5.7143 sagene
50 foot [survey] to sagene = 7.14287 sagene
75 foot [survey] to sagene = 10.71431 sagene
100 foot [survey] to sagene = 14.28574 sagene
## Want other units?
You can do the reverse unit conversion from sagene to foot [survey], or enter any two units below:
## Enter two units to convert
From: To:
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
| 515
| 1,829
|
{"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
| 0.767399
|
http://clay6.com/qa/48710/if-the-coefficient-of-x-7-in-the-expansion-of-a-2x-b-x-is-equal-to-the-coef
| 1,513,608,138,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948617816.91/warc/CC-MAIN-20171218141805-20171218163805-00469.warc.gz
| 54,419,629
| 27,347
|
# If the coefficient of $x^7$ in the expansion of $(a^2x+b^{-1}x^{-1})^{11}$ is equal to the coefficient of $x^{-7}$ in $(ax-b^{-1}x^{-2})^{11}$ then $ab=$
$\begin{array}{1 1}(A)\;1\\(B)\;2\\(C)\;3\\(D)\;4\end{array}$
## 1 Answer
Toolbox:
• $T_{r+1}=nC_r a^{n-r} b^r$
$T_{r+1}$ in $(ax^1+b^{-1}x^{-1})^{11}$
$\Rightarrow 11C_r(ax^2)^{11-r}(\large\frac{1}{bx})^r$
$\Rightarrow 11C_r a^{11-r} b^{-r} x^{22-3r}$
Let $T_{r+1}$ contains $x^7$
$\therefore 22-3r=7$ or $5$
$\therefore T_{r+1}=T_{5+1}=11C_5a^{11-5}b^{-5} x^7$
$\therefore$ Coefficient of $x^7=11C_5a^6b^{-5}$
$T_{r+1}$ in $(ax-b^{-1}x^{-2})^{11}$
$\Rightarrow 11C_r (ax) ^{11-r} (-b^{-1}x^{-2})^r$
$\Rightarrow 11C_r(-1)^ra^{11-r}b^{-r} x^{11-3r}$
Let $T_{r+1}$ contains $x^{-7}$
$\therefore 11-3r=-7$ or 6
$\therefore T_{r+1}=T_{6+1}=11C_6 (-1)^6a^{11-6} b^{-6}x^{-7}$
$\therefore$ Coefficient of $x^{-7}=11C_6a^5b^{-5}$
$\therefore 11C_5a^6b^{-5}=11C_6a^5b^{-6}$
$\Rightarrow a=b^{-1}\Rightarrow ab=1$
Hence (A) is the correct answer.
answered Jun 25, 2014
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
| 528
| 1,090
|
{"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-2017-51
|
longest
|
en
| 0.355724
|
https://forum.upcase.com/t/code-review-spiral-traverse/7064
| 1,660,943,251,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882573760.75/warc/CC-MAIN-20220819191655-20220819221655-00508.warc.gz
| 260,251,516
| 4,628
|
# Code review - Spiral Traverse
Here’s my solution to the classic “traverse a matrix in a spiral pattern” problem in Ruby!
After finishing this up, I took a look at other solutions and realized you can do cool things with the Matrix class, which I was unaware of. Also, other solutions tend to dismantle the original matrix & rebuild it into a new one-dimensional array, whereas I leave the original intact.
Does anyone know how I might refactor the repetitive `if` statements in my private methods `traverse_top(i)`, etc? They are necessary to check whether or not the matrix has been iterated through. I feel like there is something obvious that would render them redundant, but it’s escaping me.
All thoughts / constructive feedback welcome!
``````class SpiralTraverse
def initialize
@matrix = []
@width = ""
@height = ""
@num_elements = ""
end
def traverse(matrix)
@matrix = matrix
@width = matrix.first.length
@height = matrix.length
@num_elements = matrix.flatten.length
run
end
def run
@x = 0
(0...@width).each do |i|
traverse_top(i)
traverse_right(i)
traverse_bottom(i)
traverse_left(i)
end
end
private
def end_of_matrix?
@x < @num_elements
end
def traverse_top(i)
if end_of_matrix?
(i...@width - i).each do |top|
puts "#{@matrix[i][top]}"
@x += 1
end
else
return
end
end
def traverse_right(i)
if end_of_matrix?
(i+1..@height-1-i).each do |right|
puts "#{@matrix[right][@width - 1 - i]}"
@x += 1
end
else
return
end
end
def traverse_bottom(i)
if end_of_matrix?
(i+1..@width-2-i).reverse_each do |bottom|
puts "#{@matrix[@height-1-i][bottom]}"
@x += 1
end
else
return
end
end
def traverse_left(i)
if end_of_matrix?
(i+1..@height-1-i).reverse_each do |left|
puts "#{@matrix[left][i]}"
@x += 1
end
else
return
end
end
end
``````
| 488
| 1,754
|
{"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
| 3
|
CC-MAIN-2022-33
|
latest
|
en
| 0.734725
|
http://hplgit.github.io/wavebc/doc/pub/._wavebc_cyborg002.html
| 1,685,237,063,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224643388.45/warc/CC-MAIN-20230527223515-20230528013515-00514.warc.gz
| 21,775,713
| 4,843
|
$$\newcommand{\tp}{\thinspace .}$$
# Mathematical model and solution method
We solve a one-dimensional, linear, constant-coefficient wave equation by an explicit finite difference method.
## The wave equation problem
The standard, linear wave equation in a homogeneous one-dimensional medium reads $$$$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}, \quad x\in (0,L),\ t\in (0,T]\tp \tag{1}$$$$ The unknown function $$u$$ depends on space $$x$$ and time $$t$$: $$u=u(x,t)$$.
The need for boundary conditions in the wave equation. Four initial and boundary conditions must be specified to have a unique solution:
• Initial condition for $$u(x,0)$$
• Initial condition for $$u_t(x,0)$$
• Boundary condition at $$x=0$$
• Boundary condition at $$x=L$$
## Initial conditions
Most demonstrations will start with an initial profile of $$u$$, $$u(x,0) = I(x),$$ being at rest, i.e., $$\frac{\partial}{\partial t}u(x,0) = 0\tp$$ Two initial profiles will be considered:
• a plug as to left in Figure 1
• a Gaussian peak as to the right in Figure 1
Figure 1: Examples on initial conditions.
## Boundary conditions
### Fixed $$u$$
At $$x=0$$ we will sometimes use the condition $$u=0$$, often known as a homogeneous Dirichlet condition. This condition will mirror the wave.
### Reflecting condition
At $$x=0$$ and/or $$x=L$$ we will apply a reflecting or no-flux condition: $$$$\frac{\partial u}{\partial x}=0\tp \tag{2}$$$$ This condition reflects the wave into the domain again, as a surface wave hits a vertical wave, runs up to the double amplitude, and propagates back into the domain again. This type of boundary condition is also referred to as a Neumann condition.
### Feeding a wave from the boundary
We shall demonstrate the effect of moving $$u$$ at the boundary $$x=0$$ to feed the domain with an incoming wave. The boundary condition then reads $$u(0,t) = U_0(t),$$ for some given function $$U_0(t)$$. A particular choice in a later demonstration is a sine function that is active in three different time intervals: $$U_0 (t) = \left\lbrace\begin{array}{ll} \frac{1}{4}\sin(6\pi t),& t\in T_1\hbox{ or } t\in T_2\hbox{ or } t\in T_3\\ 0,& \hbox{otherwise} \end{array}\right.$$ where $$T_1=[0, \frac{1}{6}]$$, $$T_2=[\frac{3}{4}, \frac{5}{6}]$$, and $$T_3=[\frac{3}{2},\frac{11}{6}]$$. The movement of $$u$$ at the boundary will produce a wave that is by the PDE transported to the right into the domain.
### Open boundary condition
Very often one wants to let a wave travel through the boundary without being disturbed. Such a condition is called an open boundary condition, or a radiation condition, or an artificial boundary condition: \begin{align} \frac{\partial u}{\partial t} - c\frac{\partial u}{\partial x} &= 0,\quad x=0, \tag{3}\\ \frac{\partial u}{\partial t} + c\frac{\partial u}{\partial x} &= 0,\quad x=L\tp \tag{4} \end{align} These conditions work exactly in 1D, but are challenging to generalize and implement in 2D and 3D.
### Periodic boundary condition
When following a wave motion over large distances, it is desireable to let a wave travel out of the right domain and at the same time feed the wave back into the domain from the left. This approach avoids a very large domain where nothing happens in the majority of the domain. A periodic boundary condition at $$x=0$$ can be used to feed the signal traveling out at $$x=L$$ into the domain: $$$$u(0,t) = u(L,t)\tp \tag{5}$$$$ The condition at $$x=L$$ is then an open boundary condition (4).
## Numerical solution method
The wave equation is solved by an explicit finite difference method, which is of second-order in space and time. A uniform mesh with spacing $$\Delta x$$ and $$\Delta t$$ is used in space and time, respectively. The no-flux or Neumann boundary conditions are implemented by modifying the computational stencil at the boundary. The open boundary conditions are implemented by forward in time, upstream in space finite differences, which exactly let the wave out of the boundary. More details are found in Appendix: Numerical solution method. Parts of the computer code are explained in Appendix: Computer code.
| 1,103
| 4,153
|
{"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": 2, "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": 3, "x-ck12": 0, "texerror": 0}
| 4.28125
| 4
|
CC-MAIN-2023-23
|
latest
|
en
| 0.851357
|
https://www.physicsforums.com/threads/matrix-norms-and-proving-their-characteristics.88960/
| 1,675,122,788,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00673.warc.gz
| 946,711,312
| 16,942
|
Matrix Norms, and proving their characteristics
pinodk
I have a couple of questions on this matter...
QUESTION 1.)
I am to show a certain Lemma, that the definition of a matrix norm
||A|| = max(||x||<=1) ||Ax||
obides with the definitions of a standard vector norm
a.) ||A||>=0 for all A and ||A||=0 <=> A=0
b.) ||tA|| = |t| ||A|| for any real number t
c.) ||A+B|| <=||A||+||B|| for all A,B
Ive shown b and c, although b is confusing me somewhat:
b.) ||tA|| must satisfy the following
||tA|| = max(||x||<=1) ||tAx||
= |t| max(||x||<=1) ||Ax||
= |t| ||A||
I thinkthis proves it, but still I am not quite sure what would happen if t is negative from the beginning.
Could someone tell me if ||tA|| is then equal to |t| ||A|| and why?
The really important part of of question 1 is getting a hint of how to show a.), since i have no idea how to...
Question 2
Show that
||A|| = max(||x||!=0) ||Ax||/||x||
From a definition in the book, I know that ||Ax||<=||A|| ||x||
then i must have that
||A|| <= max(||x||!=0) ||A|| ||x||/||x||
giving me that
||A|| <= max(||x||!=0) ||A||
which proves it.
Is this proof ok?
Question 3
The definition i use above is proven in the book, the first line of this proof says
||Ax|| = ||A (x/||x||) x||
why is this so?
Is it cause x/||x|| is a unit vector, and thus makes no difference when multiplied on A?
Anyone who can help me, will receive an appropriate e-hug :-)
Cheers
Daniel
Homework Helper
Question 1:
You know Ax is a vector right, and you know that the norm of a vector is greater than or equal to 0, with equality iff the vector is the zero-vector, right? So this is enough to show that ||A|| > 0. It remains to show that ||A|| = 0 iff A = 0.
Assume ||A|| = 0
Then max||x|| < 1||Ax|| = 0
And then ||Ax|| = 0 for all x such that ||x|| < 1
From here, you need to prove that A = 0, and the above should be enough to make that easy.
On the other hand, assume A is 0 and prove that ||A|| = 0. That should be easy.
Question 2.
Your proof is not right. For one, you end up proving an inequality, and you're asked to prove an equality. But the overall approach is wrong. I assume that you are still working with the norm from the previous question, so:
||A|| = max||x|| < 1||Ax||
You want to prove that ||A|| = max||x|| != 0||Ax||/||x||, so you want to prove that:
max||x|| < 1||Ax|| = max||x|| != 0||Ax||/||x||
Let m = ||x||, and define v as x/m, so x = mv
max||x|| < 1||Ax||
= maxm < 1m||Av||
= m x maxm < 1||Av||
max||x|| != 0||Ax||/||x||
= maxm != 0m||Av||/m
= maxm != 0||Av||
= 1 x maxm != 0||Av||
All you have to do is show:
m x maxm < 1||Av|| = 1 x maxm != 0||Av||
Question 3.
Doesn't make any sense. On the right side of the equation, you seem to be "multiplying" two vectors because you have:
(x/||x||) x
Do you mean the dot product of (x/||x||) with x, or the cross product of those two vectors or, as I suspect, was this a typo?
pinodk
typo
It was a typo :-)
it should have been
(x/||x||)||x||
should make more sense now :-D
I have a very basic question also, I would like the following definition to be described in as plain words as possible, I want to be absolutely sure what it means...
||A|| = max(||x||<=1) ||Ax||
I was trying to explain to a mate of mine, and found out I couldn't answer his following questions with certainty
What i told him, was that u multiply the mxn matrix with a n-vector, and get a new m vector, then take the max value in this.
Why should we do the Ax multiplication? Is x to be chosen somehow, what is the idea of this?
||x||<=1 - this is based on ||x|| being the 2 norm right? the length of the vector?
Homework Helper
Certainly you can see that (x/||x||)||x|| = x.
||x|| is the length of x, yes. If x is a 2-vector, then the set of x satisfying ||x|| < 1 is just the unit circle, so you are basically seeing how A transforms the unit circle, then finding the point in the image of the circle under the transformation A that is farthest from the origin.
Staff Emeritus
Gold Member
If it helps, one of the more important properties of an operator norm is that it satisfies:
||Tx|| <= ||T|| ||x||
The definition of ||T|| is chosen so that ||T|| is large enough for this inequality to be valid, but no larger than necessary.
In fact, I've seen the definition written in three different ways (I'm not sure if I've seen the one you wrote or not): (click the image to see how to make them)
$$||T|| := \sup_{x \neq 0} \frac{||Tx||}{||x||}$$
$$||T|| := \sup_{||x|| \leq 1} ||Tx||$$
$$||T|| := \sup_{||x|| = 1} ||Tx||$$
Exercise 1: prove that all three are equivalent.
Exercise 2: explain why these definitions capture the notion I gave above for why we choose the operator norm the way we do.
pinodk
About the (x/||x||)||x|| being equal to x...
Doh! Of course i can see that... sorry :-S
| 1,477
| 4,795
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.46875
| 3
|
CC-MAIN-2023-06
|
latest
|
en
| 0.905568
|
http://www.mapleprimes.com/tags/conjecture
| 1,498,367,749,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00677.warc.gz
| 568,793,462
| 17,001
|
## Modular Arithmetic...
Asked by:
How do I use msolve to solve y^2 + y - 11=0 in Zp for all primes p with 41=< p =<107 ?
Also, using the results make a conjecture describing the primes p for which there are solutions to y^2 + y - 11 = 0 in Zp
This was what I did.
41<=p<=107
msolve(y^2 + y - 11=0, p)
but I received this error, no implementation of msolve matches the arguments in call, msolve(y^2 + y - 11=0, p)
Any help is appreciated. Thanks
## Calculating Collatz's Conjecture ...
Asked by:
Hi everyone, I'm trying to print out Collatz's Conjecture's steps for any given value with the following code but it takes forever and prints nothing. Any idea on how I can get it working ?
checkCollatzValue:=proc(val) local res, remaining;
while res <> 1 do
remaining = irem(val, 2); remaining;
if remaining = 0 then res = val / 2; else res = val * 3 + 1; fi;
res;
od;
end proc;
## A list of small graphs, invariants and conjectured...
by: Maple
A list of small graphs with associated pictures and tables of
values of various graph invariants.
The graph invariants were made using Maple programs which uses
the networks and GraphTheory packages.
A picture presents some inequality conjectures between the graph invariants.
http://www.msci.memphis.edu/~speeds/
Sam Speed August 29, 2011
## MRB constant rational?A
Page 1 of 1
| 369
| 1,347
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.046875
| 3
|
CC-MAIN-2017-26
|
latest
|
en
| 0.876161
|
http://www.jiskha.com/display.cgi?id=1287672830
| 1,498,505,695,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320863.60/warc/CC-MAIN-20170626184725-20170626204725-00652.warc.gz
| 542,589,497
| 4,006
|
# Chemistry
posted by .
In this case, the inflection point, and
equivalence, occurs after 23.25mL of 0.40 M NaOH has been delivered.
Moles of base at the equivalence point can be determined from the volume of base delivered to
reach the equivalence point and the concentration of NaOH.
1. Calculate the moles of NaOH added to reach the equivalence point (i.e. after the addition
of 23.25 mL of 0.40 M NaOH).
2. Using the fact that the acid and base are in exact stoichiometric proportions at the
equivalence point, calculate the moles of citric acid in the solution. The balanced
equation is:
H3C6H5O7(aq) + 3 NaOH(aq) -----> Na3C6H5O7(aq) + 3H2O(l)
3. calculate the mass of citric acid in the drink mix.
• Chemistry -
1) moles = moles x L.
2) moles NaOH x (3 moles citric acid/1 mole NaOH = moles citric acid.
3). mass citric acid = moles x molar mass.
| 259
| 860
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.65625
| 3
|
CC-MAIN-2017-26
|
latest
|
en
| 0.821571
|
https://www.b4x.com/android/forum/threads/sd-xui_view3d-mesh-from-a-dem.148417/
| 1,723,307,016,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00688.warc.gz
| 524,418,349
| 18,357
|
# Android QuestionSD XUI_View3D--mesh from a dem--
#### PABLO2013
##### Well-Known Member
Longtime User
Greetings, I hope you are well, I want to know how I could create a surface from a dem 4x4, the cell width is 5, this library adapts very well but I don't know how to do it, thanks, I'll give a similar example, thanks
+1000 tks Star-Dust
example:
`````` Case "Function 3D"
Dim ID As Int = 0
For z=-120 To 120 Step 10
For x=-120 To 120 Step 10
Dim NewPointList(2) As Point3D_Type
Dim y1 As Int =50 * CosD(z*2) * SinD(x*2)
Dim y2 As Int =50* CosD(z*2) * SinD(x*2+20)
NewPointList(0)=(P3D.CtP(X,y1,z))
NewPointList(1)=(P3D.CtP(X+10,y2,z))
ID=ID+1
Next
Next
For x=-120 To 120 Step 10
For z=-120 To 120 Step 10
Dim NewPointList(2) As Point3D_Type
Dim y1 As Int =50 * CosD(z*2) * SinD(x*2)
Dim y2 As Int =50* CosD(z*2+20) * SinD(x*2)
NewPointList(0)=(P3D.CtP(X,y1,z))
NewPointList(1)=(P3D.CtP(X,y2,z+10))
ID=ID+1
Next
Next``````
dem:
`````` Dim dem(4, 4) As Double
dem(0, 0) = 100.0
dem(0, 1) = 50.0
dem(0, 2) = 70.0
dem(0, 3) = 10.0
dem(1, 0) = 10.0
dem(1, 1) = 10.0
dem(1, 2) = 10.0
dem(1, 3) = 10.0
dem(2, 0) = 10.0
dem(2, 1) = 220.0
dem(2, 2) = 10.0
dem(2, 3) = 10.0
dem(3, 0) = 10.0
dem(3, 1) = 10.0
dem(3, 2) = 10.0
dem(3, 3) = 10.0``````
#### Star-Dust
##### Expert
Longtime User
Forgive me but I don't quite understand what you are asking
#### PABLO2013
##### Well-Known Member
Longtime User
Hi Star D. Tks for write. I want to plot a Dem (matrix 4x4) . It have 0.0 ....0.1....0.2 positions and values = heights i want plot it like a terrain or mesh like a picture
#### Star-Dust
##### Expert
Longtime User
If you see example 1 for B4A, it achieves a similar thing by drawing horizontal and vertical lines passing through the vertices
If I can find some time, I'll give you an example
Last edited:
#### Star-Dust
##### Expert
Longtime User
Sample for this
B4X:
`````` Dim dem(4, 4) As Double
dem(0, 0) = 100.0
dem(0, 1) = 50.0
dem(0, 2) = 70.0
dem(0, 3) = 10.0
dem(1, 0) = 10.0
dem(1, 1) = 10.0
dem(1, 2) = 10.0
dem(1, 3) = 10.0
dem(2, 0) = 10.0
dem(2, 1) = 220.0
dem(2, 2) = 10.0
dem(2, 3) = 10.0
dem(3, 0) = 10.0
dem(3, 1) = 10.0
dem(3, 2) = 10.0
dem(3, 3) = 10.0``````
#### Attachments
• dem.zip
14.4 KB · Views: 118
#### PABLO2013
##### Well-Known Member
Longtime User
Thank you very much, I appreciate your time and your code, I have reviewed many other codes in other languages and yours is the best, it is light, and for applications with limited resources it works very well, I understand that the expectations were not as expected, but I think that these are very specific codes and the interest is more limited, I think, in any case, thanks a lot.
#### Star-Dust
##### Expert
Longtime User
That's the simplest, but better implementations could be done. Instead of drawing lines, trapezoids could be created which would also give a more realistic light effect.
It also depends on the distance between the dots, maybe 50dip is too little to get the correct ide of the image. Try with 300dip and reduce the zoom from 1 to 0.2 you will have a better effect.
It all depends on what you want to achieve with such a small grid
#### Attachments
• Dem.zip
3.4 KB · Views: 123
Last edited:
#### PABLO2013
##### Well-Known Member
Longtime User
Thank you very much, I appreciate your time.
I am going to review what you indicate and I would like to know three additional things:
1. when you talk about trapezoids, how would at least one line of code be.
2. when I add an image to mark a specific point since it is in 2d when rotating In some positions it looks deformed, it is possible to improve this and I still do not.
3. understand roughly the mechanism of data entry to the library. In this case, two "for" statements are added and NewPointList(0) and NewPointList(1) are added. .. and they are added as polygons (I don't understand this two-point polygon...)
3. And to finish... I'm doing this... I draw a 3d figure in autocad (based on triangle blocks), then I go convert with qgis to geojeson (i make a parser) and from there to a 3d object from your library, the process is not very efficient...but it works well in 2d, in 3d it's complicated...is there any idea to do that better...I need this to be selectable parts (individual) to get its data (x,y,z)
many tks.
Last edited:
#### Star-Dust
##### Expert
Longtime User
1. When I was talking about trapezoids I was referring to the vertices you indicated in the first post. In general, 3D objects are represented in the files as many triangles which then make up the objects. In the case of the vertices that you indicated to me by grouping them by 4 you can draw trapezoids which are filled objects, with respect to lines.
2. If the image is warped it is obviously off center and rotated. If it is perfectly aligned parallel to the X axis it is straight. If it is far from the PIVOT point it is distorted from the display trajectory.
3. In reality they are lines, I used the command which easily allowed me to obtain a line with 2 points even if the method is used to draw polygons. If I entered all the points together it would create a polygon with many points and the result would not be correct. The correct solution in terms of logic is the one I posted in the previous post
4. I need to understand better what you want to do
#### PABLO2013
##### Well-Known Member
Longtime User
Ok, sorry for making a lot of confusion.
The issue is that I have an electric tower drawn in 3d in autocad) by means of the line between points are around 150 lines, I want to use your library to present it but I also want to select or deselect a particular line and know its x, y, Z values
I did it as indicated in the previous post (I invented a method--autocad--draw triangles as blocks--parse in qgis to geojeson---and convert to P3D.LoadObiectj3 file, it works well for 2d, but for 3d it is very laborious, ...
I used stl format, and it works, but stl format makes the figure a single one as a solid and I want is to be able to touch its parts...I'll make a diagram to see you... ..thank you.
Last edited:
#### Star-Dust
##### Expert
Longtime User
With my library it is possible to click only one element because it handles the touch event. With the stl format the object is divided into triangles so it will be possible to click triangles not lines.
When you load an STL file all elements acquire the same ID and therefore raise the same event any triangle. So you should modify the reading of the file to assign a different id to each element.
I don't know if the loading and processing speed is sufficient for what you need.
Replies
11
Views
6K
Android Question Wifi Mesh
Replies
7
Views
1K
Replies
6
Views
1K
Replies
26
Views
10K
Android Example BitmapMeshView
Replies
1
Views
3K
| 2,129
| 6,791
|
{"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-2024-33
|
latest
|
en
| 0.721832
|
https://teaching.ncl.ac.uk/bms/wiki/index.php?title=Nernst_Equation&diff=prev&oldid=971
| 1,596,888,532,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00099.warc.gz
| 509,161,584
| 7,189
|
# Nernst Equation
(Difference between revisions)
Jump to: navigation, search
Revision as of 15:07, 15 November 2010 (view source)← Older edit Revision as of 15:29, 15 November 2010 (view source)Newer edit → Line 49: Line 49: F is the Faraday's constant; F = 96,485.3415 C mol-1 F is the Faraday's constant; F = 96,485.3415 C mol-1 − [A-] is the concentration of ion outside the membrane (in this case is anion, negative charge ion)
+ [A-]o is the concentration of ion outside the membrane (in this case is anion, negative charge ion)
− [A-] is the concentration of ion inside the membrane (in this case is anion, negative charge ion) + [A-]i is the concentration of ion inside the membrane (in this case is anion, negative charge ion) == Application == == Application ==
## Revision as of 15:29, 15 November 2010
Nernst Equation is an equation used to calculate the electrical potential of a chemical reaction. In its equilibrium state, the Nernst equation should be zero. It also shows the direct relation between energy or potential of a cell and its participating ions. The equation is proposed by a German chemist, Walther H. Nernst (1864-1941).[1]
## Equation
Nernst equation can be expressed as follows:
where
Ecell is the half-cell potential difference
Eθcell is the standard half-cell potential
R is the universal gas constant; R = 8.314471 J K-1 mol-1
T is the thermodynamics temperature, in Kelvin; 0 K = -273.15oC
z is the number of moles of electrons transferred between cells (defined by the valency of ions)
F is the Faraday's constant; F = 96,485.3415 C mol-1
[red] is the concentration of ion that gained electrons (reduction)
[oxi] is the concentration of ion that lost electrons (oxidation)
## Membrane potential
Main article: Membrane potential
Nernst equation is also can be used to calculate the potential of an ion across the membrane. For potential difference of a membrane, we can manipulate the Nernst Equation as follows:
or
where
Em is the potential difference of an ion between membranes
R is the universal gas constant; R = 8.314471 J mol-1
T is the thermodynamics temperature, in Kelvin; 0 K = -273.15oC
z is the number of moles of electrons transferred between membranes (defined by the valency of ion)
F is the Faraday's constant; F = 96,485.3415 C mol-1
[A-]o is the concentration of ion outside the membrane (in this case is anion, negative charge ion)
[A-]i is the concentration of ion inside the membrane (in this case is anion, negative charge ion)
## Application
### Ussing study of frog skin
In biochemistry, Nernst equation can be used to calculate the potential difference of ion between membranes. Hans H. Ussing, a Danish scientist, used a frog skin to measure the potential difference of sodium and potassium ions across the membranes with his famous invention, the Ussing chamber.
[2]Ussing model of transepithelial ions absorption.
For example at the standard condition and temperature of 25oC (298K), the above sodium ion membrane potential can be calculated as:
### Goldman equation
Main article: Goldman equation
In presence of more than one ion, the Nernst equation can be modified into Hodgkin-Katz-Goldman equation or is commonly known as Goldman equation. Goldman equation is proposed by David E. Goldman of Columbia University together with Alan L. Hodgkin and Bernard Katz.
## References & Notes
1. http://nobelprize.org/nobel_prizes/chemistry/laureates/1920/nernst-bio.html, The Nobel Prize in Chemistry 1920; Walther Nernst
2. Diagram based on CMB2003: Cell and Membrane Transport lecture note (2010).
| 913
| 3,597
|
{"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}
| 3.265625
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.921286
|
http://wwprojekt.eu/Mar/23238/
| 1,556,106,632,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-18/segments/1555578641278.80/warc/CC-MAIN-20190424114453-20190424140453-00075.warc.gz
| 193,463,086
| 7,334
|
# Calculation Of Ball Mill Residence Time
### residence time distribution in cement mills
calculation of ball mill residence time . calculation of ball mill residence time » cement ball mill residence time » how to calculate motor kw of ball mill » what (Residence Time Distribution) 120 toneladas por ventas de la planta de trituración horas
### hammer mill efficiancy calculation
how to calculate residence time of a hammer mill » ball mill calculations to improve . Hammer Mill, Sigma Improving the efficiency of fine grinding How do you calculate ball mill residence time >> Get .
### An overfilling indicator for wet overflow ball mills .
The equations for the residence time calculation can be implemented in a spread sheet, or coded in toolbox software. Once the volume-based residence time is determined for a particular ball mill, it can be compared with the database to see what band of residence time the
### MATERIAL RETENTION TIME IN A BALL MILL & VRM
Aug 11, 2010 · material retention time in a ball mill & vrm. dear all can anybody suggest me how to calculate the material retention time in a . also called residence time.
### Training Courses based on the Grinding Software
Training Courses based on the Grinding Software . Ball charges Composition Calculator for Monochamber Mills 2. Ball charges . Residence time inside the mill
### Ball Mill Design/Power Calculation
The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are; material to be ground, characteristics, Bond Work Index, bulk density, specific density, desired mill tonnage capacity DTPH, operating % solids or pulp density, feed size as F80 and maximum 'chunk size', product size as P80 and .
### residence time in vertical roller mill - rock-island
how to calculate ball mill residence time - rlmc. how to calculate ball mill residence time. COMPARTMENTAL MODELLING OF FLOW IN A CLINKER .
### calculation of ball mill residence time - rrcser
RTD, Holdup, Ball mill, Ball load, Slurry. estimated and the calculated values compared reasonably with the measured ones within the limits of . and Ball Loading on Mill Residence Time Distribution.
### Ball Mill Power Calculation Pdf - Hansie en Grietjie ECD .
What is the formula to calculate mill motor power of ball mill How do you calculate ball mill residence time? mass of mill charge divided by mass flow rate. >>Chat Online; ball mill power calculation pdf Crushing Equipments SAM stone crushing equipment is designed to achieve maximum productivity and high reduction ratio From .
### How To Calculate Ball Mill Residence Time
how to calculate ball mill residence time. More About cement ball mill residence time, calculate ball mill retention time calculate ball mill retention time The influence of mill speed and pulp density on. calculate residence time through ball mill
### ball mill residence time - The English Academy
how to calculate residence time in a ball mill. 16 822 USD. How to calculate residence time in a ball mill . How do you calculate ball mill residence time .
### Mill Speed - Critical Speed - Paul O. Abbe
Talk with the Experts at Paul O. Abbe about your process requirements and Ball Mill . and Mill Speed - Critical Speed . individually or several at a time.
### material retention time in cement ball mill
cement ball mill residence time - miningmalaysia . MATERIAL RETENTION TIME IN A BALL MILL & VRM – CemNet. CemNet the home of International Cement Review CALCULATE THE MATERIAL RETENTION TIME IN A CLOSED CIRCUIT BALL .
### Calculate and Select Ball Mill Ball Size for Optimum .
In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do.
How do you calculate ball mill residence time. mass of mill charge divided by mass flow rate.. Engineering How do you calculate ball mill residence.
### An overfilling indicator for wet overflow ball mills .
The equations for the residence time calculation can be implemented in a spread sheet, or coded in toolbox software. Once the volume-based residence time is determined for a particular ball mill, it can be compared with the database to see what band of residence time the
### How To Calculate Residence Time In A Ball Mill
Material retention time in a ball MILL & VRM. Dear all can anybody suggest me how to calculate the material retention time in a closed circuit ball MILL in each compartment?The retention time, also called residence time, is the ratio (M/Q) [h] .
### How To Calculate Ball Mill Residence Time
calculate residence time through ball mill - prabodhana. ball sizes, and residence time, a continuous ball mill is studied for optimizing cement raw. calculate optimal raw material feed ratio.
### calculate ball mill retention time - crusher-mill.site
calculate ball mill retention time – Grinding . calculation of ball mill residence time - SAMAC Crusher Screen Plate, SAMAC beneficiation equipment -SAM.
### cement ball mill residence time - marketingpower
How To Calculate Ball Mill Dense For Cement Grinding. how to calculate ball mill dense for cement grinding, How do you calculate ball mill residence time - WikiAnswers The ball mill grinder is going to be more efficient because the pr, The volume of a sphere (or ball, Read more
### How To Calculate Ball Mill Residence Time
how to calculate residence time in a ball mill. how to calculate residence time in a ball mill. how to calculate ball mill residence time volume of the pulp in the mill is then calculated using the mill discharge .
The residence time distribution (RTD) of mineral slurry and slurry holdup volume in an industrial ball mill has been successfully studies using tracer tests. Six different conditions of solids concentration and three levels of ball loading were assessed.
### Ball Mill Application and Design - Paul O. Abbe
The residence time in ball mills is long enough that all particles . at Paul O. Abbe to help you determine which design and size ball mill would be best for .
### how to calculate ball mill residence time - diavista
calculate residence time through ball mill. How do you calculate ball mill residence time - Answers. mass of mill charge divided by mass flow rate. .
### cement ball mill residence time - spskhanapur
how to calculate ball mill residence time Documents. BinQ Mining Equipment how to calculate ball mill residence time, How To Calculate Residence Time In A Ball How To Calculate Residence Time In A Ball
### Residence time distributions in a stirred ball mill and .
Residence time distributions in a . Calculation of moments E(t) = c1 . In contrast to the behaviour of conventional ball mills, where the mean residence time t is .
### how to calculate residence time of a hammer mill
formula 39 s to calculate efficiency of ball mill . classifying hammer mill; lection function S, residence time distribution To calculate lining for a cylindrical mill
### formula to calculate grinding media wear rate for cement mill
Feb 14, 2016 · SAM can provide the proper grinding measure to all kinds of methods how to calculate grinding media sizes for ball mill grinding media charge calculation in finish cement ball mill . calculate ball mill grinding media in cement. calculate ball mill grinding media in mill The ball Many factors are involved in the selection of .
### How To Calculate Ball Mill Residence Time
calculate residence time through ball mill - prabodhana. ball sizes, and residence time, a continuous ball mill is studied for optimizing cement raw. calculate optimal raw material feed ratio.
### Residence Time Ball Mill - totalspace
Residence Time Distribution In Cement Mills. Leave your quotation. calculation of ball mill residence time - calculation of ball mill residence time » cement ball mill residence time » how to calculate motor .
• Previous Page: Quary Machines In Kenya
• Next Page: Equipment For Ore Concentrate
• #### related information
• Mill In Liners Plate Design Calculation In Cement Plant
• Cement Ball Mill Efficiency Calculation
• Calculation Copper Ore Price
• Information Of Calculation For Ball Mills
• Ton Capacity Grinding Mill Media Charge Calculation
• Hammer Mill Crusher Calculation
• Calculation Of Ball Mill Design
• Calculation Method Of Vibrating Screen
• Calculation Of Crusher For Captive Power Plant
• Cone Crusher Utilisation Capacity Calculation
• Calculation Of Ball Mill Residence Time
• Yield Calculation For Iron Processing Plant
• Hammer Crusher Materials And Production Capacity Calculation
• Code For Ball Mill Sizing Calculation
• Coal Mill Design Calculation
• Calculation For Arb In Coal
• Calculation Impact Crusher Capacity
• Small Ball Mill Design Calculation
• Jaw Stone Crusher Calculation Capacity
• Calculation Of Payload For Mining Equipment
| 1,773
| 8,957
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2019-18
|
latest
|
en
| 0.778062
|
https://www.physicsforums.com/threads/calculations-for-shooting-ions.832665/
| 1,511,376,659,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806620.63/warc/CC-MAIN-20171122175744-20171122195744-00289.warc.gz
| 842,371,774
| 16,926
|
# Calculations for shooting ions
1. Sep 15, 2015
### sander2798
Hello everyone,
the problem I have to solve is as following.
You're in space and you're moving with a constant speed. Now, you want to accelerate. For this you will be using Newton's third law.
Xenon 1+ iones will be shot from the back of the spacecraft with high speeds, the avarage speed an ion will be accelerated to is 30.000 m/s. This will result in a gain in velocity of the spacecraft, according to Newton's 3rd law. Calculate how many velocity a spacecraft with a mass of 1000 kg can gain by shooting away 50kg of Xenon 1+ iones.
My attempt:
Mass of one Xenon +1 ion:
= 131.30 x 1.660538921 x 10^-27 = 2.18 x 10^-25 kg
Kinetic energy gain of ion:
1/2 mv^2 = 1/2* 2.18*10^-25 *30000^2 = 9.81*10^-17 Joule
According to newton's third law the gain in kinetic energy of the ion should be equal to the gain in kinetic energy of the spacecraft, right? (not sure about this)
1/2 mv^2 = 1/2 * 1000 * v^2 = 9.81*10^-17
v = 4.43 * 10^-10 m/s for every ion shot
Calculating how many iones we have:
100000 / 131.30u = 761.614 mol.
761.614 * 6,022 * 10^23 = 4.586 * 10^26 iones.
Shooting away all those iones will create a gain in speed of:
4.586 * 10^26 * 4.43 * 10^-10 = 2.03 * 10^17 m/s
But since this is about a few thousand times the speed of light, I assume I did something wrong somewhere. If anyone could help me I'd be very thankful.
Sander.
2. Sep 15, 2015
### paisiello2
This is wrong. You should look up the definition on Wikipedia or something.
What you actually did here is apply the principle of conversation of energy. However. conservation of energy doesn't work the way you put it because you neglected the work of the magnetic field that accelerated the ions to begin with.
Can you think of something else that is conserved here instead?
Good for you, though, that you did a reality check at the end and realized you did something wrong.
3. Sep 15, 2015
### sander2798
Thanks, now have the correct answer! (I hope :P)
| 600
| 2,019
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375
| 4
|
CC-MAIN-2017-47
|
longest
|
en
| 0.932098
|
https://math.stackexchange.com/questions/2136729/maximun-distance-between-a-geodesic-and-another-curve
| 1,558,870,317,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232259126.83/warc/CC-MAIN-20190526105248-20190526131248-00014.warc.gz
| 544,045,613
| 31,448
|
# Maximun distance between a geodesic and another curve
Consider a Riemannian manifold $M$ and two points $x,y \in U \subset M$ and a local diffeomorphic chart $\phi:U \rightarrow \mathbb{R}^n$.
Let $l_{uv}:[0,a] \rightarrow \mathbb{R}^n$ be the straight line segment that connects $u$ and $v$ in $\mathbb{R}^n$.
Hence $\gamma := \phi^{-1}(l_{\phi(x) \phi(y)})$ is a curve on $M$ that connects $x$ and $y$
Now consider $\psi:[0,T] \rightarrow \mathbb{R}^n$ the shortest path on $M$ that connects $x$ and $y$.
My questions is: What a is upper bound for the maximum distance between $\gamma$ and $\psi$?
| 201
| 606
|
{"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}
| 2.671875
| 3
|
CC-MAIN-2019-22
|
latest
|
en
| 0.741442
|
http://www.instructables.com/id/How-to-Measure-and-Mark-with-Precision/?ALLSTEPS
| 1,397,752,016,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00066-ip-10-147-4-33.ec2.internal.warc.gz
| 497,545,540
| 36,068
|
# How to Measure and Mark with Precision
Measuring is the first step in wood-working. The second step is to make a mark on the measured point. The combination of the two processes—measuring and marking—forms the foundation of accurate work. If either is done incorrectly, much effort and material are wasted. Here are a few pointers that can improve the quality of your work by reducing the chance for error.
This project was originally published in the January 2001 issue of Popular Mechanics. You can find more great projects at Popular Mechanics DIY Central.
Remove these ads by Signing Up
## Step 1: Ruler Selection & Marking
A wood folding ruler, a tape measure and a few steel bench rulers (6 in., 12 in. and 24 in.) meet most measuring needs (Photo 1). To mark from any one of these rulers, use a sharp 2H pencil, the best general marking device for woodworking. These pencils have a medium-hard lead and when properly sharpened leave a fine line that’s readily visible on most surfaces. When marking a line with a ruler, be sure to hold the pencil at an angle so that its point meets the workpiece at the same place the ruler does (Photo 2, left pencil). Holding the pencil perpendicular to the work surface results in the center of the pencil point being inaccurately positioned.
## Step 2: Basic Measurement
When measuring from the edge of a workpiece, hold a stopblock next to the edge, then press the ruler against the block (Photo 1). For utmost precision, hold the ruler on its edge against the work surface (Photo 2). This way, the ruler’s markings meet the workpiece surface and you avoid mismarking the workpiece due to an off-center line-of-sight created by the ruler’s thickness. A similar distortion occurs when transferring the measurement from a tape measure. To avoid this, hook the tape’s end clip firmly to the edge of the workpiece, then tip the blade so its edge curves down to meet the surface. Then mark the dimension on the workpiece (Photo 3).
## Step 3: Measuring Circles
Use a pair of blocks and a ruler of sufficient length when measuring the outside diameter of a circular workpiece. Read the measurement at the inside edges of the blocks.
## Step 4: Middle Marking and Finding Center
Dividing a board into any number of equal spaces can be done easily without calculation. Simply set a ruler diagonally across the board with the desired number of inch graduations divided between the board’s edges. Then make a mark at each graduation (Photo 1).
A similar method is used to find the board’s center. Angle a ruler across the board’s width so that a whole number on the ruler is at each edge of the workpiece. Find the midway point between the two numbers, and mark there (Photo 2). With either method, it doesn’t matter if you place the end of the ruler at the board’s edge or if you use the middle of the ruler.
## Step 5:
To take the inside measurement of an assembly, use a folding ruler with a sliding extension. Hold the ruler against one end of the assembly, and slide its brass extension so that it butts against the other end of the assembly (Photo 1). The measurement of the sliding extension, added to the ruler’s dimension, equals the inside dimension of the assembly. Without a folding ruler, use a wood block and a ruler that is shorter than the interior dimension. Overlap the ruler and block, then add the block’s dimension to the dimension at which it overlaps the ruler (Photo 2).
## Step 6: The Penny Trick
A last tip: you can use a penny to mark a 3⁄8-in. radius. Hold the penny as shown above and mark along its edge.
maurifix3 months ago
So easy and yet so important. Great Instructtable. Awesome. Some I use, but two of them are so great. Thank you!
3mrkamel4 months ago
I am sorry but I don't understand the "Step 4: Middle Marking and Finding Center" part... why I have to put the ruler diagonally? Thanks for this great instructable :-)
russ_hensel4 months ago
Old school was to use a knife to make the marks, you can still see these in old furniture. The knife mark also helps to first make the cut, and on a circular saw it can help avoid chip out. That said I did learn from this instructable.
mhmcgrath4 months ago
More importantly, if striking a line against reference (measurement) marks, first place the point of the pencil on the reference mark, then incline it, move the straightedge to the inclined pencil point, then make the pencil line. This insures that the center of the pencil line actually aligns to the reference mark center. When I asked my first carpentry boss if he wanted me to cut to leave the line, or take the line - his response. "You take half and I'll take half!" This man measured twice, cut once - and you better d**n sure do the same! (BTW, in wood work we prefer a striking knife/scribe to a pencil when accuracy really counts!)
saints754 months ago
Thank you
jimmysymo4 months ago
I worked in an envelpoe making factory.We had over a dozen machines.English and German.made .After 15 years working on most of them,I decided to make the IS0 2000
standard for the whole factory.The first thing I did was to buy a metal ruler That was tested and certified to be true..From there all the rulers I bought were tested against this one,and signed that I had tested.that way.. All our envelopes now are the right size.
4 months ago
I have just made my 2nd Law ( Always check what I have written before I post to Instucts)
ENVELOPE and NOT envelpoe. BUT then I know I:m not perfect. :-)
glassgiant4 months ago
Wow. I feel like a complete idiot. All these years of calculating the centre of boards. Just... wow.
Phil B6 months ago
A man gave a presentation on measurement for woodworking. He made the point that the same ruler should be used throughout a project because they can and do vary from one another a little. I had never thought about that before.
6 months ago
I heard from a colleague that he once compared folding meters at a building site. He found two - which were being used in construction - that differed by 4cm over 2 meters... So I completely agree.
4 months ago
You think that's bad, I once bought a \$500 paper cutter that was not square. I took it back and traded it for another, which also was not square! The third one was. That was back in college. Since then, I have boughten two drywall squares which were not square! It says "square" in the name, for God's sake!! lol
4 months ago
what they need is....... GIANT CALIPERS
Pra_ga4 months ago
Thanks for sharing these great tips
And the photographs are simply amazing!
The Fettler4 months ago
Great tips, I'll be putting them straight into practice.
Thanks very much indeed :-)
darman124 months ago
Thank you for sharing this. Really great tips! My favorite tip is the one for finding the center of a board without calculation.
Marina L.4 months ago
Favorited! Thanks for sharing.
I'm the queen of "Measure twice, screws-up once".
4 months ago
That's really funny :)
Satrek6 months ago
Thanks for sharing this!
I always told myself that slight offsets and mistakes in the stuff I made (especially concerning woodworking which I want to get into more) would give my stuff an authentic or rustic note - now I am runnning out of excuses.
| 1,669
| 7,210
|
{"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-2014-15
|
latest
|
en
| 0.924727
|
http://vunivere.ru/work23378
| 1,502,912,934,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886102393.60/warc/CC-MAIN-20170816191044-20170816211044-00535.warc.gz
| 371,258,289
| 12,479
|
What is Mathematics? Things to remember about the Faculty of Applied Mathematics, Computer and Information Science
Содержание работы
Faculty of Applied Mathematics, Computer and
Information Science
Introduction in Speciality
What is Mathematics?
Mathematics is a living subject that is constantly changing and evolving. This evolution is driven by our need to solve a new problem, or the development of some abstract line of logical thought. The power of mathematics is such that the solution to a new problem often lies in Mathematics conceived using abstract thought many years, sometimes centuries, ago.
Why study Mathematics and Computer Science? The huge increase in the availability and power of computers has led to them being a necessary part of any modern direction. Often, the problems that arise, say, in the business world must be first formulated using Mathmatics before the computer can be of use. For instance, the aerodynamic design of cars and aeroplanes is carried out using computers. However, the computer -and the engineer- need to know the mathematical equations that govern the airflow around the car. The Mathematics and Computer science will give you a blend of skills required to tackle these types of problems in the business environment, (business — 1. Бизнес, дело. 2. Экономическая деятельность, деловая жизнь. 3. Компания, производственное предприятие).
Information Systems and Information Technologies. The first are: the knowledge and skilles necessary to design and develop software systems that allow a wide range of users to easily interrogate large sources of information. An information systems professional will combine soft systems development knowledge, strong inrterface design skills, a sound understanding of a system's business context, and an ability to communicate and empathise (придавать особое значение) with potential users of a system. The second are: the Information technologies such as: Visual-basic, Microsoft Access and Excel, Internet Development Tools, PC-based Networking, Numerical Methods, IT Tools, Introduction to Programming, Database Management Systems, Modern Aspects of Data Processing etc.
What are Mathematics and Statistics?
Cosider the following facts:
- the display you see on your mobile phone was developed using Mathematics
- The Software and Hard Ware on your computer were desined using Mathematics
- Mathematics and Statistics lie behind the weather prediction on TV
- Mathematics is used to make data transfer on the Internet secure
- The banking, investment and financial operations of large companies and the stockmarket rely on sophisticated Mathematics and Statistics
- Forensic procedure (судебный) use Probability and Statistics to assess the value of evidence when presenting it to a court
- Medicines are licensed for use only after the most rigorous testing in statistically-designed and controlled experiments
- Marine scientists use Statistics in their analysis of catch and survey data to estimate the size of fish population and inform political decisions about how to halt the dwindling fish stocks etc.
Mathematical and Statistical Techniques are being used to study:
- Electric and Mechanical activity in the heart to help diagnose and prevent coronary disease
- Growth of cancer cells in the body so that possible drug therapies can undergo trials before being given to patients
- Spread of HIV to inform us of its future impact on society
- Gene mapping of the human body to help uncover the complex interplay between genes.
Why study Mathematics, Statistics & Business?
There is a strong emphasis on statistical techniques in data analysis and on use of mathematical models.
What are Computer Science and Artificial Intelligence? Computer Science and Artificial Intelligence are at the forefront of development of technological tools to improve the quality of life. Computer-based interactive systems are used in all kinds of areas: administrative and management jobs, education, and research in other disciplines.
In computer science it is programming, technical communication and human-computer interaction.
In Artificial Intelligence, the automation of all aspects of human cognition from its foundation to approaches to symbolic and non-symbolic Al, natural languagies processing, computer vision, evolutionary and adaptive systems.
Things to remember about the Faculty of Applied Mathematics, Computer and Information Science
Тип:
Статьи
Размер файла:
19 Kb
Скачали:
0
| 854
| 4,558
|
{"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-2017-34
|
longest
|
en
| 0.825868
|
https://fr.mathworks.com/matlabcentral/answers/508910-doubt-in-matlab-coding
| 1,632,542,182,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00672.warc.gz
| 309,685,031
| 32,687
|
# Doubt in MATLAB coding.
3 views (last 30 days)
Joy Salomi on 4 Mar 2020
Edited: Joy Salomi on 6 Mar 2020
Hello. I'm trying to plot a differntial equation using MATLAB. When I run my program, I'm getting different graph.
I've attached the differential equation and my MATLAB coding below.
Can anyone tell me how to use the boundary conditions correctly?
function abcd
ex4init=bvpinit(linspace(0.00001,100,21),[0,0]);
sol=bvp4c(@ex4ode,@ex4bc,ex4init)
plot(sol.x,sol.y(1,:),'blue');
end
function f=ex4ode(r,y)
a=0.001391304348;
b=19.31459475;
f=[y(2)
-(2/r)*y(2)-a*(1-exp(b*y(1)))
];
end
function res=ex4bc(ya,yb)
res=[
ya(1)+1
yb(2)
]
end
This is the graph which I got.
But this is the exact graph.
##### 2 CommentsShowHide 1 older comment
Joy Salomi on 4 Mar 2020
Hello Sir,
If I change x axis range to 1e-9 m, I didn't get any graph. Plot is empty.
What shall I do?
Robert U on 5 Mar 2020
Edited: Robert U on 6 Mar 2020
Hi Joy Salomi,
Thanks for the paper. The differential equation should be written in spherical coordinates (edit: I think there is a mistake to assume R to be constant). Since R is assumed to be constant there is no singularity. I adjusted the code and tested it in Matlab 2019b. Depending on chosen xmesh results might differ. You have to try a bit to get a stable solution. I attached an example for the D = 200 nm graph.
abcd.m
function abcd(ah,D,phi_s)
ex4init=bvpinit(linspace(0,D/2,3e3),[phi_s 0]);
sol=bvp4c(@ex4ode,@ex4bc,ex4init);
plot(ah,sol.x*1e9,sol.y(1,:));
ah.XLabel.String = 'r [nm]';
ah.YLabel.String = 'phi [V]';
function dphidr=ex4ode(r,phi)
a = -1.602176634e-19*1e17*(1e-2)^(-3)/(8.8541878125e-12*11.5);
b = 1.602176634e-19/(physconst('Boltzmann')*600);
dphidr = [ phi(2)
a * ( 1-exp( b*phi(1) ) ) ];
end
function res=ex4bc(ya,yb)
res=[ ya(1)+ phi_s
yb(2) ];
end
end
Example call:
fh = figure;
ah = axes(fh);
hold(ah,'on')
arrayfun(@(dIn) abcd(ah,200e-9,dIn),[0.1,0.3,0.7,1])
Result:
Kind regards,
Robert
Joy Salomi on 6 Mar 2020
Hello Sir,
The code you sent is working.
Thank you so much for your help Sir. :)
This is a bit hard to solve without knowing more of the problem. What I would do is to break down the problem and analyse each step. For example, look at the values you have in the solution "sol". I plot separately:
>> plot(sol.x);
>> plot(sol.y);
And also
>> plot(sol.y');
Is anything here remotely connected to what you are expecting? My guess is that no, sol.y has values between 0 and 12e4 whilst the graph you want is between -1 and 0. So the problem is not really about Matlab but about the equations that you are using.
Hope this helps.
Joy Salomi on 4 Mar 2020
Thank you for your reply Sir. But my doubt is somewhat different. I'm trying to get the exact curve which is already available.
This is the curve I want.
Robert U on 4 Mar 2020
Hi Joy Salomi,
the implementation of a boundary value problem with singular term has to be done a bit different than you did.
function abcd
ex4init=bvpinit(linspace(0,100,5),[-1 0]);
S = [0 0; 0 -2];
options = bvpset('SingularTerm',S);
sol=bvp4c(@ex4ode,@ex4bc,ex4init,options);
plot(sol.x,sol.y(1,:),'blue');
end
function f=ex4ode(r,y)
a=0.001391304348;
b=19.31459475;
f = [ y(2)
-a*(1-exp(b*y(1))) ];
end
function res=ex4bc(ya,yb)
res=[ ya(1)+1
yb(2) ];
end
Using the values "a" and "b" you supplied still leads to a singular Jacobian matrix which provoces an error. Changing both these values to one leads to a solution that qualitatively corresponds to the curves drawn.
There are two more points that hinder me to test my solution:
1. I don't know how "a" and "b" translate to "phi_s".
2. I still assume that r is given in SI units. If r is in unit [m] the interval needs to be adjusted to [nm] as shown in your graph.
Kind regards,
Robert
Joy Salomi on 5 Mar 2020
Hello Sir,
I have attached the PDF for your reference.
a and b are combination of some parametrs.
By subtituting the values of the parameters, I have found the values of a and b.
| 1,264
| 3,950
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.96875
| 3
|
CC-MAIN-2021-39
|
latest
|
en
| 0.685009
|
https://espanol.libretexts.org/Negocio/Gerencia/Libro%3A_Una_introducci%C3%B3n_a_la_cooperaci%C3%B3n_y_al_mutualismo_(Boland)/02%3A_Las_cooperativas_y_Mutuales_son_Firmas/2.06%3A_Referencias
| 1,718,818,317,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00426.warc.gz
| 209,542,058
| 29,706
|
Saltar al contenido principal
# 2.6: Referencias
$$\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}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
La discusión sobre el diseño de organizaciones y gobierno corporativo se basa en conceptos presentados en Economía Gerencial y Arquitectura Organizacional, por James Brickley, Clifford Smith y Jerold Zimmerman, y sobre The Ownership of Enterprise de Henry Hansmann. El Árbol de Decisiones de Hacer o Comprar está adaptado de David Besanko, David Dranove, Mark Shanley y Scott Schaefer, Economics of Strategy. La decisión de hacer o comprar fue planteada por primera vez por Ronald Coase y publicada en “La naturaleza de la firma”, Economica 4, 16 (nov. 1937) :386—405. El Teorema de Coase fue publicado en “El problema del costo social” Revista de Derecho y Economía 3,1 (1960) :1—44. Otros, sobre todo Oliver Williamson, han estudiado y escrito más a fondo sobre este concepto. Dos artículos de Williamson que son accesibles para un lector laico son “La teoría de la firma como estructura de gobierno: de la elección al contrato”, Journal of Economic Perspectives 16, 3 (2002): 171—195, y “The Economics of Governance”, The American Economic Review 95,2 (2005) :1—18. Tanto Coase como Williamson han sido reconocidos con el Premio Sveriges Riksbank en Ciencias Económicas en Memoria de Alfred Nobel por su trabajo en esta área. La información sobre Coase y Williamson y los resúmenes de estos artículos se pueden encontrar en http://tinyurl.com/yaa7n3h9 y http://tinyurl.com/ybq272yu.
La exposición sobre diseño de empleos es modificada de Economía Gerencial y Arquitectura Organizacional por James Brickley, Clifford Smith y Jerold Zimmerman.
La prueba 1.5 se basa en una cifra anterior realizada por Tom Pierson.
El trabajo de Eugene Fama y Michael Jensen se utiliza como base para la discusión sobre gestión de decisiones y derechos en “Problemas de agencia y reclamos residuales”, Journal of Law and Economics 26 (1983): 327—349. Fama fue reconocida con el Premio Sveriges Riksbank de Ciencias Económicas en Memoria de Alfred Nobel; la información sobre él se puede encontrar en http://tinyurl.com/y7zyxb9y
La discusión sobre artículos y estatutos proviene de una variedad de conversaciones con abogados cooperativos a lo largo de los años, pero mi colega, el abogado David Swanson, ha tenido el mayor impacto en mí en esta discusión.
La información sobre los estatutos legales de las cooperativas se puede encontrar en tinyurl.com/y8e8crw2
Hay una serie de fuentes para los roles y responsabilidades de los directores. Probablemente el más sucinto se puede encontrar en tinyurl.com/y74c86x2.
Una amplia introducción a las cooperativas está disponible en tinyurl.com/y8lyhgw8.
This page titled 2.6: Referencias is shared under a CC BY-NC license and was authored, remixed, and/or curated by Michael Boland (University of Minnesota Libraries ) .
| 1,392
| 4,459
|
{"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}
| 2.59375
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.136348
|
https://www.fixya.com/support/t132234-canon_p23_dhv_calculator_need_some_info
| 1,680,036,147,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00518.warc.gz
| 866,795,164
| 40,856
|
# Canon P23 DHV Calculator / I need some info on it
I'm trying to do my payroll with the calculator, and i cannot figure out how to put it in time format. could you please let me know how i could do this?
Posted by on
### Anonymous
• Level 1:
An expert who has achieved level 1.
Governor:
An expert whose answer got voted for 20 times.
Corporal:
An expert that has over 10 points.
Mayor:
An expert whose answer got voted for 2 times.
• Contributor
What I do is just base the time on quarter hour and the 24 hour clock.
Example
6:30am = 6.5
3:45pm = 15.75
every 15 min = .25
Clock in at 11:45 am and clock out at 5:15 pm
17.25-11.75=5.5
Posted on Apr 14, 2009
×
my-video-file.mp4
×
## Related Questions:
### Order tape for printing
Lorena, Staples will have it in stock, Check the attached links,
"I hope this helped you out, if so let me know by pressing the helpful button. Check out some of my other posts if you need more tips and info."
Amazon com Calculator Plain Paper 2 25 150 Pack of 3 Calculator And Cash...
Canon 1904B001AA P23 DH Printing Calculator Staples
How to Load Paper Into Canon P23 DH eHow
Mar 05, 2016 • Canon P23-DHV Calculator
### Instructions
P23-DHV:
P23-DHV G:
http://www.usa.canon.com/app/pdf/calculator/P23-DHV_G_USA_GB.pdf
Aug 26, 2011 • Canon P23-DH Calculator
### WHAT INK ROLLER MODEL NUMBER FITS CANNON P23-DHV
Canon P23-DHVInk roller Part# IR40T
http://www.gorillapaper.com/ink-roller-ir40t-b-r.html
Thank you
Sep 30, 2010 • Canon P23-DHV Calculator
### I have a Canon P23-DHV G. It rounds up my numbers, and for payroll I need the whole number. Please help.
On the function switches, one is called the decimal point system (it has the numbers 0,2,4 & F for floating) The higher the number, the more digits are shown after the decimal point. You can download the manual by using the following webpage: http://consumer.usa.canon.com/cusa/office/products/hardware/calculators/printing_calculators/palm_printing/p23_dh_v?selectedName=Specifications#BrochuresAndManuals
Aug 10, 2010 • Canon P23-DH Calculator
### How to change the paper on canon p23 dhv calculator
Follow the illustrated instructions in the manual available at the manufacturer's web site
May 20, 2010 • Canon P23-DHV Calculator
### How do I set my blue tax buttons on the Canon P23-DHV calculator? I lost my manual.
http://www.usa.canon.com/cusa/support/office/calculators/portable_palm_printers/p23_dh_v#BrochuresAndManuals
May 20, 2010 • Canon P23-DHV Calculator
### I need a manual for the Canon P23-DHV calculator.
http://www.devicemanuals.com/guide/Calculator/Canon/Canon-Portable-Palm-Printer-P23-DH-V-User-Guide-TkRFeE5UZz.html
Jan 19, 2010 • Canon P23-DHV Calculator
152 views
Level 3 Expert
Level 3 Expert
| 800
| 2,757
|
{"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}
| 3.234375
| 3
|
CC-MAIN-2023-14
|
latest
|
en
| 0.806941
|
https://vustudents.ning.com/group/cs101introductiontocomputing/forum/topics/cs-101-introduction-to-computing-assignment-no-04-fall-due-date
| 1,582,646,806,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00000.warc.gz
| 600,265,244
| 22,092
|
We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
www.vustudents.ning.com
www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More
Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion
# CS 101 Introduction to Computing Assignment No. 04 Fall Due Date 24-1-2013
Assignment No. 04
Semester: Fall 2012
Introduction to Computing-CS101
Total Marks: 15
Due Date:24/01/2013
Instructions:
• You will submit your assignment before or on due date on VU-LMS.
• Assignment should be completed by your own efforts it should not be copied from internet, handouts or books.
• You should submit your MS Word File via assignment interface at VU-LMS.
• Assignment sent via Email will not be replied and accepted/graded in any case.
• If the submitted assignment does not open or file is corrupt, it will not be graded.
• You will submit solution only in MS Word file (.doc or .docx) file.
Objectives:
• To understand the fundamental concepts of Spread sheet and Network topologies
For any query about the assignment, contact at cs101@vu.edu.pk
Assignment Questions
Question: 1 Marks 10
Suppose given is a worksheet of MS Excel. (Figure 1)
Apply the Most appropriate formulas in MS Excel using the same data given in the below diagram, to find the result (same as shown in the diagram) of the below mentioned requirements and just write down the formulas in the solution file (MS word file).
Total sale Total sale of Asif Average sale of Asif Total Number of Products sold by Asif Maximum Price
(Figure 1)
Notes for question # 1: 1) While writing formulas; only use the given range in formulas, other range should not be accepted 2) Do not send excel file, send us the word file only 3) Do not paste the excel file in the word file. 4) You are to only write down formulas in the word file.
Question: 2 Marks 5
The “HaseebSoft” organization launches a project for 1 year. Now the Project Manger wants to build a computer network for the project smooth running of the project. The project details are given below:
Project Duration: 1 year (Extendable)
Required Resources: 12 computers+ 1 printer+1 scanner
Future enhancements: 8 to 10 computers may be added
For such type of scenario, which topology is most suitable for network and why? Justify your answer.
BEST OF LUCK
+ How to Join Subject Study Groups & Get Helping Material?
+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?
+ VU Students Reserves The Right to Delete Your Profile, If?
Views: 2781
.
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
### Replies to This Discussion
Our main purpose here discussion not just Solution
Q1 solution
Total Sale: 129000 ==>>
=SUM(B6:B15)
Total sale of Asif: 78000 ==>> =SUM(B6,B8,B10,B12)
Average sale of Asif: 195000 ==>> =AVERAGE(B6,B8,B10,B12)
Total Number of Product sold by Asif: 4 ==>> =COUNT(B6,B8,B10,B12)
Maximum Price: 55000 ==>> =MAX(B6:B14)
its not logically solution yaar,,,,,think a while... if teacher give you an assignment which contains over one million records and you have to fulfill above requirement then "would you perform said action" Of course not otherwise it would take couple of days to find where is asif and its product sold-out figures:::::try sumif equal to formulas by giving specific range....Am sorry Its my suggestion not discouraging you all....Its looking very simple solution but professionlizum required to fill it with future references,,,,, try it like this (=SUMIF(K8:K16,"=Asif",J8:J16)
sir i think star topology is the best for asked setup. on switch is required
All students must share points and also your problems.... so it can discussed and solution can prepare.
Don’t wait for solutions just participate in Discussions because after discussions a correct solution will prepare...
2nd: star topology.
qus1:
total number of product of asif =COUNTIF(C6:C14,"=Asif")
u just write formulas in word format.
Question: 01
Solution:
• Total sale: =SUM(B6:B14)
• Total sale of Asif: =SUM(B6,B8, B10,B12)
• Average sale of Asif: =AVERAGE(B6,B8,B10,B12)
• Total number of products sold by Asif: =COUNT(B6,B8,B10,B12)
• Maximum Price: =MAX(B6:B14)
I am a bit confuse about the 4th one, ap logon me sy kisi ne ye likha he k, =countif(c6:c14,"=asif") ka function use hoga lekin meri knowledge k mutabiq, ye function to is liay use hona chahiye k data men name Asif kitni bar aya he? answer to same hoga lekin formula men, I think kuch garbar he.. Kyunk hmen to no. of products batani he..nai?
Please somebody guide in this area and correct me if i am wrong...
Aur 2nd question men I think star hoga....
Uper wali diagram sy bhe yehi prove hota...
Thanks...
for solution of second question (those don't have enough knowledge about networking) make your house electricity board planning on simple paper, and you have to give connection to all rooms, all appliances and may be extended on upper portion in future.....This typically paper work lead you how to fll gap in networking by focusing extension in future...it means you have to focus on "Main boards" where all electricity connections having plug in and out with some extra switches for extension.....
thaaaaaaaaaaaaaaaaaaaaaaaaaaanx a lot for sharing samara...
Ist question ma mja b doubt ha plz any one guide correct solution,
## Latest Activity
+ ! ! ! ❣ maho ❣ ! + added a discussion to the group MTH603 Numerical Analysis
### mth603 past paper
1 minute ago
3 minutes ago
+ ! ! ! ❣ maho ❣ ! + joined + M.Tariq Malik's group
### MTH603 Numerical Analysis
7 minutes ago
+ ! ! ! ❣ maho ❣ ! + posted a discussion
### Ary kia krti ho??????????????????
9 minutes ago
11 minutes ago
+ıllıllı \$µǥąя ǥ€ɲɨµ\$ ıllıllı+ liked Isha Chuhdary's discussion paka
14 minutes ago
15 minutes ago
17 minutes ago
Alisha Noor replied to Isha Chuhdary's discussion Bari
18 minutes ago
Isha Chuhdary replied to Isha Chuhdary's discussion Mera sher
20 minutes ago
Isha Chuhdary replied to Isha Chuhdary's discussion paka
21 minutes ago
+ıllıllı \$µǥąя ǥ€ɲɨµ\$ ıllıllı+ liked Isha Chuhdary's discussion Mera sher
27 minutes ago
1
2
3
| 1,727
| 6,769
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.09375
| 3
|
CC-MAIN-2020-10
|
latest
|
en
| 0.834638
|
https://hunterriverpei.com/how-much-is-7-8-of-a-cup/
| 1,653,348,214,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00260.warc.gz
| 373,691,324
| 4,572
|
To usage this cups to tablespoons calculator, just form some worth in the very first left text box. It will certainly be converted automatically.
You are watching: How much is 7/8 of a cup
## Calculator Use
To usage this converter, just choose a unit to convert from, a unit to convert to, then form the value you desire to convert. The result will be displayed immediately.
This converter accepts decimal, integer and also fractional values as input, therefore you can input values like: 1, 4, 0.5, 1.9, 1/2, 3 1/2, etc.
Maximum approximation error because that the fractions:Exact fraction 1% 2% 5%10%15%
teaspoonstablespoonsfluid ouncescupspintsquartsgallonsimperial teaspoonsimperial tablespoonsimperial liquid ouncesimperial cupsimperial pintsimperial quartsimperial gallonsmilliliterscubic centimeters (cc)metric teaspoonsmetric tablespoonsmetric cupsliterscubic inchescubic feetbarrels that oilcubic meters⇨teaspoonstablespoonsfluid ouncescupspintsquartsgallonsimperial teaspoonsimperial tablespoonsimperial liquid ouncesimperial cupsimperial pintsimperial quartsimperial gallonsmilliliterscubic centimeters (cc)teaspoonsmetric tablespoonsmetric cupsliterscubic inchescubic feetbarrels the oilcubic meters=
See more: How Many Days Is 168 Hours To Days Conversion: Plus Calculator (Hr To D)
## The definitive liquid volume counter table
US CustomaryImperialMetric
1 teaspoon = 4.929 milliliters = 1/3 tablespoon
1 tablespoon = 14.79 milliliters = 3 teaspoons = 1/2 ounce = 1/16 cup = 1/32 pint = 1/64 quart
1 oz = 29.57 milliliters = 6 teaspoons = 2 tablespoons = 1/8 cup = 1/16 pint = 1/32 quart = 1/128 gallon
1 cup = 236.6 milliliters = 48 teaspoons = 16 tablespoons = 8 ounces = 1/2 pint = 1/4 quart = 1/16 gallon
1 pint = 473.2 milliliters = 96 teaspoons = 32 tablespoons = 16 ounces = 2 cup = 1/2 quart = 1/8 gallon
1 quart = 946.4 milliliters = 192 teaspoons = 64 tablespoons = 32 ounces = 4 cups = 2 pints = 1/4 gallon
1 gallon = 3785 milliliters = 768 teaspoons = 256 tablespoons = 128 ounces = 16 cups = 8 pints = 4 quarts
1 teaspoon = 4.4395 milliliters = 1/4 tablespoon = 1/8 oz = 1/64 cup = 1/128 pint
1 tablespoon = 17.758 milliliters = 4 teaspoons = 1/2 oz = 1/16 cup = 1/32 pint = 1/64 quart
1 liquid oz = 35.516 milliliters = 8 teaspoons = 2 tablespoons = 1/8 cup = 1/16 pint = 1/32 quart = 1/128 gallon
1 cup = 284.13 milliliters = 64 teaspoons = 16 tablespoons = 8 ounces = 1/2 pint = 1/4 quart = 1/16 gallon
1 pint = 568.26 milliliters = 128 teaspoons = 32 tablespoons = 16 ounces = 2 cups = 1/2 quart = 1/8 gallon
1 quart = 1136.5 milliliters = 256 teaspoons = 64 tablespoons = 32 ounces = 4 cup = 2 pints = 1/4 gallon
1 gallon = 4546.1 milliliters = 1024 teaspoons = 256 tablespoons = 128 ounces = 16 cup = 8 pints = 4 quarts
1 tespoon = 5 milliliters = 5 cc = 1/3 metric tablespoon
1 tablespoon = 15 milliliters = 15 cc = 3 metric teaspoons
1 liter = 1000 milliliters = 1000 cc = 200 metric teaspoons = 66 2/3 metric tablespoons = 4 metric cups
| 944
| 2,982
|
{"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-2022-21
|
latest
|
en
| 0.629556
|
https://www.physicsforums.com/threads/very-stuck-on-electrostatics-question.308634/
| 1,685,644,843,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00617.warc.gz
| 1,047,391,762
| 14,148
|
# Very stuck on electrostatics question
• latentcorpse
#### latentcorpse
State Gauss' Law
A spherical volume carries a uniform charge density $\rho_0$. A particle of mass m and charge -q is placed inside the sphere at radial distance r. (This additional charge does not distort the field arising from the $\rho_0$ charge density.) Show that the particle oscillates harmonically and find its' oscillation frequency.
don't really know what to do here.
obviously gauss' law is $\nabla \cdot \vec{E}=\frac{\rho_0}{\epsilon}$
and the lorentz force will probably come into play, $F=\int_V \rho_0 \vec{E} dV$ seeing as ther's no magnetic field.
but i can't put it together.
thanks guys.
## Answers and Replies
What is the equation of motion for a harmonic oscillator? Compare the equation of motion of a harmonic oscillator to the equation of motion you have for this given problem. It should strike you as very similar.
yeah. just got it there actually. thanks though. took me flippin ages!
| 241
| 995
|
{"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}
| 2.96875
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.922571
|
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3/17/b/a/
| 1,656,287,480,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00787.warc.gz
| 924,143,987
| 76,911
|
# Properties
Label 3.17.b.a Level $3$ Weight $17$ Character orbit 3.b Analytic conductor $4.870$ Analytic rank $0$ Dimension $4$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$3$$ Weight: $$k$$ $$=$$ $$17$$ Character orbit: $$[\chi]$$ $$=$$ 3.b (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$4.86973631570$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\mathbb{Q}[x]/(x^{4} + \cdots)$$ Defining polynomial: $$x^{4} + 3814x^{2} + 2981440$$ x^4 + 3814*x^2 + 2981440 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$2^{6}\cdot 3^{8}$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \beta_1 q^{2} + (\beta_{2} + \beta_1 - 513) q^{3} + (\beta_{3} + 3 \beta_{2} - \beta_1 - 3116) q^{4} + (15 \beta_{3} - 45 \beta_{2} + 95 \beta_1) q^{5} + (81 \beta_{3} - 21 \beta_{2} - 2532 \beta_1 - 100764) q^{6} + (217 \beta_{3} + 651 \beta_{2} - 217 \beta_1 - 785386) q^{7} + (264 \beta_{3} - 792 \beta_{2} + 50528 \beta_1) q^{8} + ( - 729 \beta_{3} + 135 \beta_{2} - 154413 \beta_1 + 4654665) q^{9}+O(q^{10})$$ q + b1 * q^2 + (b2 + b1 - 513) * q^3 + (b3 + 3*b2 - b1 - 3116) * q^4 + (15*b3 - 45*b2 + 95*b1) * q^5 + (81*b3 - 21*b2 - 2532*b1 - 100764) * q^6 + (217*b3 + 651*b2 - 217*b1 - 785386) * q^7 + (264*b3 - 792*b2 + 50528*b1) * q^8 + (-729*b3 + 135*b2 - 154413*b1 + 4654665) * q^9 $$q + \beta_1 q^{2} + (\beta_{2} + \beta_1 - 513) q^{3} + (\beta_{3} + 3 \beta_{2} - \beta_1 - 3116) q^{4} + (15 \beta_{3} - 45 \beta_{2} + 95 \beta_1) q^{5} + (81 \beta_{3} - 21 \beta_{2} - 2532 \beta_1 - 100764) q^{6} + (217 \beta_{3} + 651 \beta_{2} - 217 \beta_1 - 785386) q^{7} + (264 \beta_{3} - 792 \beta_{2} + 50528 \beta_1) q^{8} + ( - 729 \beta_{3} + 135 \beta_{2} - 154413 \beta_1 + 4654665) q^{9} + ( - 3130 \beta_{3} - 9390 \beta_{2} + 3130 \beta_1 - 4661640) q^{10} + ( - 7341 \beta_{3} + 22023 \beta_{2} + 618715 \beta_1) q^{11} + ( - 2187 \beta_{3} + 367 \beta_{2} - 463277 \beta_1 + 143123652) q^{12} + (10972 \beta_{3} + 32916 \beta_{2} - 10972 \beta_1 - 395182606) q^{13} + (57288 \beta_{3} - 171864 \beta_{2} - 3365950 \beta_1) q^{14} + (49005 \beta_{3} - 10275 \beta_{2} + 6442185 \beta_1 + 1707489720) q^{15} + (59304 \beta_{3} + 177912 \beta_{2} - 59304 \beta_1 - 3640317152) q^{16} + ( - 189948 \beta_{3} + 569844 \beta_{2} - 10252764 \beta_1) q^{17} + ( - 161838 \beta_{3} + 56214 \beta_{2} + \cdots + 10576244520) q^{18}+ \cdots + ( - 374876825841 \beta_{3} + \cdots + 47\!\cdots\!40) q^{99}+O(q^{100})$$ q + b1 * q^2 + (b2 + b1 - 513) * q^3 + (b3 + 3*b2 - b1 - 3116) * q^4 + (15*b3 - 45*b2 + 95*b1) * q^5 + (81*b3 - 21*b2 - 2532*b1 - 100764) * q^6 + (217*b3 + 651*b2 - 217*b1 - 785386) * q^7 + (264*b3 - 792*b2 + 50528*b1) * q^8 + (-729*b3 + 135*b2 - 154413*b1 + 4654665) * q^9 + (-3130*b3 - 9390*b2 + 3130*b1 - 4661640) * q^10 + (-7341*b3 + 22023*b2 + 618715*b1) * q^11 + (-2187*b3 + 367*b2 - 463277*b1 + 143123652) * q^12 + (10972*b3 + 32916*b2 - 10972*b1 - 395182606) * q^13 + (57288*b3 - 171864*b2 - 3365950*b1) * q^14 + (49005*b3 - 10275*b2 + 6442185*b1 + 1707489720) * q^15 + (59304*b3 + 177912*b2 - 59304*b1 - 3640317152) * q^16 + (-189948*b3 + 569844*b2 - 10252764*b1) * q^17 + (-161838*b3 + 56214*b2 + 8793279*b1 + 10576244520) * q^18 + (-642849*b3 - 1928547*b2 + 642849*b1 - 14029279090) * q^19 + (156720*b3 - 470160*b2 + 38786240*b1) * q^20 + (-474579*b3 - 29575*b2 - 100640323*b1 + 31113859266) * q^21 + (2197030*b3 + 6591090*b2 - 2197030*b1 - 43386453000) * q^22 + (269922*b3 - 809766*b2 + 160696274*b1) * q^23 + (4819824*b3 - 1206816*b2 - 10320936*b1 + 25128893088) * q^24 + (-2677100*b3 - 8031300*b2 + 2677100*b1 - 2018512175) * q^25 + (2896608*b3 - 8689824*b2 - 525661630*b1) * q^26 + (-12852999*b3 + 6926958*b2 + 381455082*b1 - 79495916913) * q^27 + (-1461558*b3 - 4384674*b2 + 1461558*b1 + 186713000264) * q^28 + (-19387467*b3 + 58162401*b2 - 67056155*b1) * q^29 + (6845310*b3 - 15563430*b2 + 1435642290*b1 - 440582279400) * q^30 + (-1406435*b3 - 4219305*b2 + 1406435*b1 + 617945289062) * q^31 + (32957760*b3 - 98873280*b2 - 1034157312*b1) * q^32 + (29898801*b3 - 8940783*b2 - 4837111995*b1 - 902674487880) * q^33 + (30586056*b3 + 91758168*b2 - 30586056*b1 + 680315403168) * q^34 + (32370030*b3 - 97110090*b2 + 8406238750*b1) * q^35 + (-38531295*b3 + 149915907*b2 + 1279373103*b1 - 304913531628) * q^36 + (-10171068*b3 - 30513204*b2 + 10171068*b1 + 92640803474) * q^37 + (-169712136*b3 + 509136408*b2 - 6384518782*b1) * q^38 + (-23995764*b3 - 356967130*b2 - 5444069098*b1 + 1755542556846) * q^39 + (-200036240*b3 - 600108720*b2 + 200036240*b1 - 2948821773120) * q^40 + (117316386*b3 - 351949158*b2 - 5574907390*b1) * q^41 + (-114870798*b3 + 39061974*b2 + 34045248720*b1 + 6896970921960) * q^42 + (337207927*b3 + 1011623781*b2 - 337207927*b1 - 7016255515666) * q^43 + (98916144*b3 - 296748432*b2 - 28965427520*b1) * q^44 + (546017355*b3 + 1636632135*b2 - 13469720445*b1 + 4698831610800) * q^45 + (102663044*b3 + 307989132*b2 - 102663044*b1 - 10998644876208) * q^46 + (279906036*b3 - 839718108*b2 + 47334571156*b1) * q^47 + (-129697848*b3 - 3433761320*b2 - 30929705096*b1 + 10260489838752) * q^48 + (-340857524*b3 - 1022572572*b2 + 340857524*b1 + 7369565634291) * q^49 + (-706754400*b3 + 2120263200*b2 + 29817561025*b1) * q^50 + (-1353590676*b3 + 320159340*b2 - 58664684772*b1 - 20710393805664) * q^51 + (-429371358*b3 - 1288114074*b2 + 429371358*b1 + 10548272280104) * q^52 + (-1217891637*b3 + 3653674911*b2 + 58830403419*b1) * q^53 + (614286747*b3 + 10193718537*b2 - 18807312078*b1 - 26765915189076) * q^54 + (-771928300*b3 - 2315784900*b2 + 771928300*b1 + 72563413309200) * q^55 + (3368575056*b3 - 10105725168*b2 - 16497051200*b1) * q^56 + (1405910763*b3 - 16268322157*b2 + 281784759599*b1 - 83782277122086) * q^57 + (4101249250*b3 + 12303747750*b2 - 4101249250*b1 + 2199105495720) * q^58 + (-723473769*b3 + 2170421307*b2 - 709479233185*b1) * q^59 + (3573292320*b3 - 901020480*b2 - 28385648880*b1 + 14031610223040) * q^60 + (-688871180*b3 - 2066613540*b2 + 688871180*b1 + 90567448270802) * q^61 + (-371298840*b3 + 1113896520*b2 + 634670614082*b1) * q^62 + (-8281674009*b3 + 32517007929*b2 + 294488024733*b1 - 66674590946586) * q^63 + (-4233528768*b3 - 12700586304*b2 + 4233528768*b1 - 163487435694848) * q^64 + (-3695375970*b3 + 11086127910*b2 + 391268209150*b1) * q^65 + (-4804904610*b3 - 35734197510*b2 - 1054373974830*b1 + 333192237511320) * q^66 + (-5788376453*b3 - 17365129359*b2 + 5788376453*b1 - 6693601340866) * q^67 + (-4373713344*b3 + 13121140032*b2 - 355339116288*b1) * q^68 + (13759763382*b3 - 3523618698*b2 - 286628665938*b1 + 14705793322704) * q^69 + (1446682300*b3 + 4340046900*b2 - 1446682300*b1 - 573090571544400) * q^70 + (14167417398*b3 - 42502252194*b2 + 1183517442630*b1) * q^71 + (1703148120*b3 + 31551116760*b2 + 192537358992*b1 + 599412438619200) * q^72 + (33109695792*b3 + 99329087376*b2 - 33109695792*b1 + 79407221884994) * q^73 + (-2685161952*b3 + 8055485856*b2 + 213595144130*b1) * q^74 + (5854817700*b3 - 11342851475*b2 + 1229878500925*b1 - 377841466256625) * q^75 + (-12026161606*b3 - 36078484818*b2 + 12026161606*b1 - 502160550127096) * q^76 + (22266543222*b3 - 66799629666*b2 - 6353070111850*b1) * q^77 + (-34601333598*b3 + 9439966614*b2 + 2621457560280*b1 + 384544861463880) * q^78 + (-39564406019*b3 - 118693218057*b2 + 39564406019*b1 + 1198504291296230) * q^79 + (-42538765440*b3 + 127616296320*b2 + 1971904217600*b1) * q^80 + (21122653986*b3 - 101573452278*b2 - 1983465554670*b1 - 1611048213013599) * q^81 + (-30797930380*b3 - 92393791140*b2 + 30797930380*b1 + 397278120330000) * q^82 + (-65301927471*b3 + 195905782413*b2 + 8311670714089*b1) * q^83 + (3196427346*b3 + 181622393750*b2 + 859264991102*b1 - 302630975549784) * q^84 + (62226401520*b3 + 186679204560*b2 - 62226401520*b1 + 1999998523143360) * q^85 + (89022892728*b3 - 267068678184*b2 - 11026332183550*b1) * q^86 + (-58824632673*b3 + 12110061039*b2 - 8467621176045*b1 - 2212542398810520) * q^87 + (93752159600*b3 + 281256478800*b2 - 93752159600*b1 - 842572473526080) * q^88 + (100743455694*b3 - 302230367082*b2 + 17987355800910*b1) * q^89 + (131111284230*b3 - 471182776110*b2 - 1778617653030*b1 + 887283661326840) * q^90 + (-94371880694*b3 - 283115642082*b2 + 94371880694*b1 + 2332134557914252) * q^91 + (44792651808*b3 - 134377955424*b2 - 1688122782592*b1) * q^92 + (3075873345*b3 + 613046675957*b2 + 1265131825097*b1 - 516051849190446) * q^93 + (-12845226584*b3 - 38535679752*b2 + 12845226584*b1 - 3214899032416992) * q^94 + (-341233243890*b3 + 1023699731670*b2 - 26456773272290*b1) * q^95 + (6998928768*b3 + 3524620032*b2 + 17301662760384*b1 + 3876907639949568) * q^96 + (-66747558308*b3 - 200242674924*b2 + 66747558308*b1 - 3458448750650446) * q^97 + (-89986386336*b3 + 269959159008*b2 + 11423043309699*b1) * q^98 + (-374876825841*b3 - 763573764357*b2 + 12441440722815*b1 + 4735794274334640) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 2052 q^{3} - 12464 q^{4} - 403056 q^{6} - 3141544 q^{7} + 18618660 q^{9}+O(q^{10})$$ 4 * q - 2052 * q^3 - 12464 * q^4 - 403056 * q^6 - 3141544 * q^7 + 18618660 * q^9 $$4 q - 2052 q^{3} - 12464 q^{4} - 403056 q^{6} - 3141544 q^{7} + 18618660 q^{9} - 18646560 q^{10} + 572494608 q^{12} - 1580730424 q^{13} + 6829958880 q^{15} - 14561268608 q^{16} + 42304978080 q^{18} - 56117116360 q^{19} + 124455437064 q^{21} - 173545812000 q^{22} + 100515572352 q^{24} - 8074048700 q^{25} - 317983667652 q^{27} + 746852001056 q^{28} - 1762329117600 q^{30} + 2471781156248 q^{31} - 3610697951520 q^{33} + 2721261612672 q^{34} - 1219654126512 q^{36} + 370563213896 q^{37} + 7022170227384 q^{39} - 11795287092480 q^{40} + 27587883687840 q^{42} - 28065022062664 q^{43} + 18795326443200 q^{45} - 43994579504832 q^{46} + 41041959355008 q^{48} + 29478262537164 q^{49} - 82841575222656 q^{51} + 42193089120416 q^{52} - 107063660756304 q^{54} + 290253653236800 q^{55} - 335129108488344 q^{57} + 8796421982880 q^{58} + 56126440892160 q^{60} + 362269793083208 q^{61} - 266698363786344 q^{63} - 653949742779392 q^{64} + 13\!\cdots\!80 q^{66}+ \cdots + 18\!\cdots\!60 q^{99}+O(q^{100})$$ 4 * q - 2052 * q^3 - 12464 * q^4 - 403056 * q^6 - 3141544 * q^7 + 18618660 * q^9 - 18646560 * q^10 + 572494608 * q^12 - 1580730424 * q^13 + 6829958880 * q^15 - 14561268608 * q^16 + 42304978080 * q^18 - 56117116360 * q^19 + 124455437064 * q^21 - 173545812000 * q^22 + 100515572352 * q^24 - 8074048700 * q^25 - 317983667652 * q^27 + 746852001056 * q^28 - 1762329117600 * q^30 + 2471781156248 * q^31 - 3610697951520 * q^33 + 2721261612672 * q^34 - 1219654126512 * q^36 + 370563213896 * q^37 + 7022170227384 * q^39 - 11795287092480 * q^40 + 27587883687840 * q^42 - 28065022062664 * q^43 + 18795326443200 * q^45 - 43994579504832 * q^46 + 41041959355008 * q^48 + 29478262537164 * q^49 - 82841575222656 * q^51 + 42193089120416 * q^52 - 107063660756304 * q^54 + 290253653236800 * q^55 - 335129108488344 * q^57 + 8796421982880 * q^58 + 56126440892160 * q^60 + 362269793083208 * q^61 - 266698363786344 * q^63 - 653949742779392 * q^64 + 1332768950045280 * q^66 - 26774405363464 * q^67 + 58823173290816 * q^69 - 2292362286177600 * q^70 + 2397649754476800 * q^72 + 317628887539976 * q^73 - 1511365865026500 * q^75 - 2008642200508384 * q^76 + 1538179445855520 * q^78 + 4794017165184920 * q^79 - 6444192852054396 * q^81 + 1589112481320000 * q^82 - 1210523902199136 * q^84 + 7999994092573440 * q^85 - 8850169595242080 * q^87 - 3370289894104320 * q^88 + 3549134645307360 * q^90 + 9328538231657008 * q^91 - 2064207396761784 * q^93 - 12859596129667968 * q^94 + 15507630559798272 * q^96 - 13833795002601784 * q^97 + 18943177097338560 * q^99
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 3814x^{2} + 2981440$$ :
$$\beta_{1}$$ $$=$$ $$6\nu$$ 6*v $$\beta_{2}$$ $$=$$ $$( -3\nu^{3} + 132\nu^{2} - 6690\nu + 251724 ) / 22$$ (-3*v^3 + 132*v^2 - 6690*v + 251724) / 22 $$\beta_{3}$$ $$=$$ $$( 9\nu^{3} + 396\nu^{2} + 20202\nu + 755172 ) / 22$$ (9*v^3 + 396*v^2 + 20202*v + 755172) / 22
$$\nu$$ $$=$$ $$( \beta_1 ) / 6$$ (b1) / 6 $$\nu^{2}$$ $$=$$ $$( \beta_{3} + 3\beta_{2} - \beta _1 - 68652 ) / 36$$ (b3 + 3*b2 - b1 - 68652) / 36 $$\nu^{3}$$ $$=$$ $$( 11\beta_{3} - 33\beta_{2} - 3356\beta_1 ) / 9$$ (11*b3 - 33*b2 - 3356*b1) / 9
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/3\mathbb{Z}\right)^\times$$.
$$n$$ $$2$$ $$\chi(n)$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
2.1
− 52.1196i − 33.1293i 33.1293i 52.1196i
312.717i −5369.70 3770.02i −32256.2 276758.i −1.17895e6 + 1.67920e6i −7.10881e6 1.04072e7i 1.46206e7 + 4.04878e7i 8.65472e7
2.2 198.776i 4343.70 + 4917.21i 26024.2 482304.i 977423. 863422.i 5.53804e6 1.81999e7i −5.31128e6 + 4.27178e7i −9.58704e7
2.3 198.776i 4343.70 4917.21i 26024.2 482304.i 977423. + 863422.i 5.53804e6 1.81999e7i −5.31128e6 4.27178e7i −9.58704e7
2.4 312.717i −5369.70 + 3770.02i −32256.2 276758.i −1.17895e6 1.67920e6i −7.10881e6 1.04072e7i 1.46206e7 4.04878e7i 8.65472e7
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3.17.b.a 4
3.b odd 2 1 inner 3.17.b.a 4
4.b odd 2 1 48.17.e.b 4
5.b even 2 1 75.17.c.d 4
5.c odd 4 2 75.17.d.b 8
12.b even 2 1 48.17.e.b 4
15.d odd 2 1 75.17.c.d 4
15.e even 4 2 75.17.d.b 8
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
3.17.b.a 4 1.a even 1 1 trivial
3.17.b.a 4 3.b odd 2 1 inner
48.17.e.b 4 4.b odd 2 1
48.17.e.b 4 12.b even 2 1
75.17.c.d 4 5.b even 2 1
75.17.c.d 4 15.d odd 2 1
75.17.d.b 8 5.c odd 4 2
75.17.d.b 8 15.e even 4 2
## Hecke kernels
This newform subspace is the entire newspace $$S_{17}^{\mathrm{new}}(3, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4} + 137304 T^{2} + \cdots + 3863946240$$
$3$ $$T^{4} + 2052 T^{3} + \cdots + 18\!\cdots\!41$$
$5$ $$T^{4} + 309212805600 T^{2} + \cdots + 17\!\cdots\!00$$
$7$ $$(T^{2} + 1570772 T - 39368833865900)^{2}$$
$11$ $$T^{4} + \cdots + 12\!\cdots\!00$$
$13$ $$(T^{2} + 790365212 T + 53\!\cdots\!60)^{2}$$
$17$ $$T^{4} + \cdots + 28\!\cdots\!60$$
$19$ $$(T^{2} + 28058558180 T - 15\!\cdots\!64)^{2}$$
$23$ $$T^{4} + \cdots + 32\!\cdots\!40$$
$29$ $$T^{4} + \cdots + 52\!\cdots\!00$$
$31$ $$(T^{2} - 1235890578124 T + 38\!\cdots\!44)^{2}$$
$37$ $$(T^{2} - 185281606948 T - 79\!\cdots\!60)^{2}$$
$41$ $$T^{4} + \cdots + 10\!\cdots\!00$$
$43$ $$(T^{2} + 14032511031332 T - 47\!\cdots\!00)^{2}$$
$47$ $$T^{4} + \cdots + 26\!\cdots\!60$$
$53$ $$T^{4} + \cdots + 12\!\cdots\!60$$
$59$ $$T^{4} + \cdots + 11\!\cdots\!00$$
$61$ $$(T^{2} - 181134896541604 T + 77\!\cdots\!04)^{2}$$
$67$ $$(T^{2} + 13387202681732 T - 28\!\cdots\!20)^{2}$$
$71$ $$T^{4} + \cdots + 29\!\cdots\!00$$
$73$ $$(T^{2} - 158814443769988 T - 92\!\cdots\!60)^{2}$$
$79$ $$(T^{2} + \cdots + 10\!\cdots\!96)^{2}$$
$83$ $$T^{4} + \cdots + 20\!\cdots\!60$$
$89$ $$T^{4} + \cdots + 56\!\cdots\!00$$
$97$ $$(T^{2} + \cdots + 81\!\cdots\!20)^{2}$$
| 7,597
| 15,568
|
{"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}
| 2.859375
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.296381
|
https://fornoob.com/which-groups-fought-against-each-other-in-civil-war-of-1642-1645/
| 1,638,144,780,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00543.warc.gz
| 350,152,147
| 23,260
|
# Which groups fought against each other in civil war of 1642-1645
Which groups fought against each other in civil war of 1642-1645
Hey there, you are talking about the English Civil War, the groups were the Parliamentarians and the Royalists.
61
Step-by-step explanation:
16 + 45
Convert 16 like this!
10 + 6 = 16
Convert 45 like this!
40 + 5 = 45
6 + 5 = 11
10 + 40 = 50
50 + 11 = 61
alright now listen fe203(s)=567.66666
Explanation:
a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.
b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .
The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.
c. CI [6583.336 ,6816.336]
d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.
Step-by-step explanation:
Population mean = u= 6500 hours.
Population standard deviation = σ=500 hours.
Sample size =n= 50
Sample mean =x= 6,700 hours
Sample standard deviation=s= 600 hours.
Critical values, where P(Z > Z) =∝ and P(t >) =∝
Z(0.10)=1.282
Z(0.05)=1.645
Z(0.025)=1.960
t(0.01)(49)= 1.299
t(0.05)= 1.677
t(0.025,49)=2.010
Let the null and alternate hypotheses be
H0: u = 6500 against the claim Ha: u ≠ 6500
Applying Z test
Z= x- u/ s/√n
z= 6700-6500/500/√50
Z= 200/70.7113
z= 2.82=2.82
Applying t test
t= x- u /s/√n
t= 6700-6500/600/√50
t= 2.82
a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.
Yes we reject H0 for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645
For t test we reject H0 as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.
b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .
The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.
c. The 95 % confidence interval of the population mean life is estimated by
x ± z∝/2 (σ/√n )
6700± 1.645 (500/√50)
6700±116.336
6583.336 ,6816.336
d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.
Can I talk to you please I need to.
Explanation:
Even though it is a little bit complicated to distinguish the different options, these are the answers:
1. Oliver Cromwell was Lord Protector of the Commonwealth of England, Scotland, and Ireland from 1653 until his death (1658).
2. King Charles I was executed by Puritans in January 1649, since he rejected the demand of the English Parliament for the establishment of a constitutional monarchy.
3. The Commonwealth of England was created after the end of the Second Civil War and the execution of Charles I, in 1649, and it lasted until 1660. During that time, England, Wales, Ireland, and Scotland were ruled as a republic.
4. Theaters in London closed between 1642 and 1660.
5. The office of the Lord Protector was created in 1653.
6. The English Civil Wars (there were three of them in this period) took place between 1642 and 1651. The First Civil War took place between 1642 and 1646. It was a conflict between supporters of the monarchy of Charles I and those who rejected it.
nah
Step-by-step explanation:
i just wanna listen music
it makes me calm
Parliamentarians ("Roundheads") and Royalists ("Cavaliers") found against each other in the civil war of 1642-1645.
Roundheads supported the Parliament of England during the English Civil War which lasted from 1641 to 1652. Roundheads were Parliamentarians, who fought against King Charles I of England and his supporters, called the Cavaliers or Royalists, who claimed rule by absolute monarchy and the principle of the 'divine right of kings'. The Roundhead party wanted the Parliament to have supreme control over executive administration of the country.
Hottest videos
Related Posts
| 1,205
| 4,015
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.4375
| 4
|
CC-MAIN-2021-49
|
latest
|
en
| 0.834431
|
https://www.keyword-suggest-tool.com/search/odds+ratio+with+zero+cell/
| 1,611,733,659,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610704821253.82/warc/CC-MAIN-20210127055122-20210127085122-00287.warc.gz
| 845,721,074
| 8,832
|
Odds ratio with zero cell
Odds ratio with zero cell keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
Keyword Suggestions
Odds ratio with zero cell
( Please select at least 2 keywords )
Most Searched Keywords
Websites Listing
We found at least 10 Websites Listing below when search with odds ratio with zero cell on Search Engine
R - Odds ratio with zero cell counts - Cross Validated
I am trying to run a binomial GLM but have some zero cell counts resulting in estimates of +/- infinity. In the text book "An Introduction to Categorical Data Analysis" Agresti A. (1996) it suggests to add a small constant to the zero cells - which I also came across in the more accessible paper here.. However, when I try this in R I get a warning message of:
https://stats.stackexchange.com/questions/69844/odds-ratio-with-zero-cell-counts
DA: 23 PA: 49 MOZ Rank: 72
16.9.2 Studies with zero-cell counts - Cochrane
Odds ratio and risk ratio methods require zero-cell corrections more often than difference methods, except for the Peto odds ratio method, which only encounters computation problems in the extreme situation of no events occurring in all arms of all studies.
https://handbook-5-1.cochrane.org/chapter_16/16_9_2_studies_with_zero_cell_counts.htm
DA: 25 PA: 50 MOZ Rank: 77
Odds ratio - Wikipedia
An odds ratio greater than 1 indicates that the condition or event is more likely to occur in the first group. And an odds ratio less than 1 indicates that the condition or event is less likely to occur in the first group. The odds ratio must be nonnegative if it is defined. It is undefined if p 2 q 1 equals zero, i.e., if p 2 equals zero or q ...
https://en.wikipedia.org/wiki/Odds_ratio
DA: 16 PA: 16 MOZ Rank: 32
How to estimate odds ratios with zeros when running binary ...
And another model, estimated using forward stepwise (likelihood ratio), produced odds ratio of 274.744 with sig. 0.000. Total N is 180, missing 37. The model is fitted based on Omnibus and Hosmer ...
https://www.researchgate.net/post/How_to_estimate_odds_ratios_with_zeros_when_running_binary_logistic_regression_in_SPSS
DA: 20 PA: 50 MOZ Rank: 50
How to calculate OR (odd ratio) if one of groups is "0" in ...
Just adding 0.5 to each of the cells and then calculate the odds ratio over these adjusted cell counts (known as Haldane-Anscombe correction) is indeed a common practice, and it also remove some ...
https://www.researchgate.net/post/How_to_calculate_OR_odd_ratio_if_one_of_groups_is_0_in_a_case-control_study
DA: 20 PA: 50 MOZ Rank: 50
Principles of Epidemiology | Lesson 3 - Section 5
The odds ratio is sometimes called the cross-product ratio because the numerator is based on multiplying the value in cell “a” times the value in cell “d,” whereas the denominator is the product of cell “b” and cell “c.” A line from cell “a” to cell “d” (for the numerator) and another from cell “b” to cell “c ...
https://www.cdc.gov/csels/dsepd/ss1978/lesson3/section5.html
DA: 11 PA: 41 MOZ Rank: 52
An Estimate of the Odds Ratio That Always Exists
Title: An Estimate of the Odds Ratio That Always Exists Created Date: 10/31/2006 7:00:37 AM
https://www.keyword-suggest-tool.com
DA: 28 PA: 28 MOZ Rank: 28
PMean: Is my odds ratio zero or infinity? | PMean
The odds ratio of 10 and the odds ratio of 0.1 are the flip sides of the same coin. These are complementary ratios. Get used to it. Half of the software packages out there will compute an odds ratio of 10 and the other half will compute an odds ratio of 0.1. Some people try to standardize the notation for odds ratios.
http://blog.pmean.com/zero-or-infinity/
DA: 14 PA: 18 MOZ Rank: 32
The odds ratio | The BMJ
For the example, the log odds ratio is log e (4.89)=1.588 and the confidence interval is 1.588±1.96×0.103, which gives 1.386 to 1.790. We can antilog these limits to give a 95% confidence interval for the odds ratio itself,2 as exp(1.386)=4.00 to exp(1.790)=5.99. The observed odds ratio, 4.89, is not in the centre of the confidence interval ...
https://www.bmj.com/content/320/7247/1468.1
DA: 11 PA: 24 MOZ Rank: 35
| 1,117
| 4,257
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.015625
| 3
|
CC-MAIN-2021-04
|
latest
|
en
| 0.880881
|
https://www.gradegorilla.com/microIB/4waves/M_polarisation.php
| 1,696,232,049,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00042.warc.gz
| 843,258,229
| 4,512
|
10 minutes maximum! Can you do it in 5?
1. Which of these can be used to polarise light?
• A. Polaroid filters and reflection from plane surfaces.
• B. Coloured filters and reflection from plane surfaces.
• C. Polaroid filters and transmission through a vacuum.
• D. Coloured filters and transmission through a vacuum.
2. Which of the following waves cannot be polarised?
• A. microwave.
• B. sound wave.
• C. X-ray
• D. gamma ray.
3. When a wave of amplitude A0 and intensity I0 passes through a polarising filter the intensity of the transmitted wave is ½I0. This means the transmitted wave amplitude is:
• A. A0
• B. ½ A0
• C. 1/√2 A0
• D. ¼ A0
4. Which of these units can be used to measure the intensity of electromagnetic radiation?
• A. W
• B. W m-2
• C. J
• D. J m-2
5 -8. A light ray passes through 2 polarising filters. The first filter polarises the wave in a vertical plane. The second filter is rotated and the transmitted light I2 observed.
5. Which of these gives the correct terms for the first and second filters?
First Filter Second Filter A initial stage filter polariser B analyser polariser C polariser amplitude moderator D polariser analyser
6. The intensity of light incident on the first filter is I0. Which of these describes the intensity of light after the second filter I2 when it is held vertically or horizontally?
Second filter horizontal Second filter vertical A I2 = 0 I2 < I0 B I2 = I0 I2 = I0 C I2 = 0 I2 = I0 D I2 < I0 I2 < 0
7. The final light intensity I2 is compared to I1 , the intensity after the first filter. What is the intensity of I2 when the angle between the two filters is 450, given that cos(450) = 1/√2?
8. Which of the 4 graphs below correctly shows the variation in intensity of the final observed beam I2 as the second filter is slowly rotated through 900 from its initial position? It is initially vertically aligned.
9. Which of the following rely on polarisation to function?
• A. CD discs.
• B. Mobile phone signals.
• C. LED TV screens.
• D. LCD displays.
10. When fishing, people often wear sunglasses to reduce the polarised glare from the light reflected from the water surface.
In which direction is the reflected light polarised, and in which orientation must the polarising filters in the sunglasses be aligned?
Polarisation of reflected light Alignment of filters in sunglasses A vertically vertically B vertically horizontally C horizontally vertically D horizontally horizontally
| 599
| 2,471
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.0625
| 3
|
CC-MAIN-2023-40
|
longest
|
en
| 0.868218
|
https://www.physicsforums.com/threads/quick-question-about-ratio-test-for-series-convergence.808350/
| 1,508,741,618,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825700.38/warc/CC-MAIN-20171023054654-20171023074654-00404.warc.gz
| 990,807,851
| 16,155
|
# Quick question about Ratio Test for Series Convergence
1. Apr 13, 2015
### ColtonCM
1. The problem statement, all variables and given/known data
This is the question I have (from a worksheet that is a practice for a quiz). Its a conceptual question (I guess). I understand how to solve ratio test problems.
"Is this test only sufficient, or is it an exact criterion for convergence?"
2. Relevant equations
Recall the ratio-test: If {an}n∈N is a positive sequence and there is a number a < 1 such that eventually an+1 ≤ a then the series is convergent. If, eventually, an+1 ≥ 1 then the series is divergent.
3. The attempt at a solution
I would assume that it would be considered "only sufficient," since if the result yields a ratio of one, convergence cannot be determined, thus it is not an absolute criterion.
Would this line of reasoning be correct?
Thanks,
Colton
2. Apr 13, 2015
### fourier jr
that sounds right. it's sufficient but not necessary because there are other ways to determine whether or not a series converges eg $\sum_{n} \frac{1}{n^{2}}$ is known to converge but the ratio test doesn't give any information about it.
3. Apr 13, 2015
### ColtonCM
Sounds good, thanks!
| 303
| 1,207
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.125
| 3
|
CC-MAIN-2017-43
|
longest
|
en
| 0.936455
|
https://www.muffwiggler.com/forum/viewtopic.php?f=17&t=138926&start=625
| 1,618,234,142,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00532.warc.gz
| 984,250,257
| 19,190
|
## DIY Mutable successfull builds hall of fame
From circuitbending to homebrew stompboxes & synths, keep the DIY spirit alive!
Moderators: Kent, luketeaford, Joe.
EvanLang
Learning to Wiggle
Posts: 12
Joined: Wed Apr 10, 2019 12:13 am
Location: Denver, CO
It's alive!!!
collumn2016
Learning to Wiggle
Posts: 47
Joined: Mon Aug 29, 2016 4:42 am
Location: in orbit
### Marbles Calibration
Hi, built marbles awhile back and eventually sorted out my problems with your guys help!
Reason for post is my uncertainty in whether or not i have calibrated my marbles correctly. I used the guide provide by forestcaver https://github.com/forestcaver/Marbles-panel
I tried the method but i am not very confident it has worked correctly. I was confused by the process and wonder if anyone could do a video or talk me through what i should be looking for at each step. I am okay with the volt/oct setup using 1 and 3volts.
When i turn the pots looking for close to 1 or close to 3 i am unsure if over 1 or over 3 counts as these values are not seen in the files provided. It seems kind of arbitrary at times (obviously my lack of understanding of the process) where the 1 volt or 3 on the volt/oct is quite discrete and easy to achieve.
Could anyone explain to me why the calibration by forestcaver takes the form it does? Or how the spreadsheet formula is derived.
Thanks
forestcaver
Ultra Wiggler
Posts: 925
Joined: Tue Mar 14, 2017 4:41 pm
Location: UK
Contact:
### Re: Marbles Calibration
collumn2016 wrote: Reason for post is my uncertainty in whether or not i have calibrated my marbles correctly.
When i turn the pots looking for close to 1 or close to 3 i am unsure if over 1 or over 3 counts as these values are not seen in the files provided. It seems kind of arbitrary at times (obviously my lack of understanding of the process) where the 1 volt or 3 on the volt/oct is quite discrete and easy to achieve.
Could anyone explain to me why the calibration by forestcaver takes the form it does? Or how the spreadsheet formula is derived.
Thanks
Outputting 1v and 3v - those knob and button settings tell marbles to output octaves so you set the knobs to get marbles to think it is outputting exactly 1v and 3v. You then measure the error.
The calibration takes the form it does for several reasons:
1. You need to calibrate the module, otherwise it is pretty useless as a quantised sequencer
2, MI didnt expose the calibration in the user interface. I believe because there were a lot of support calls and very few people have voltmeters of the required precision and accuracy.
3. I’m lazy and only really care about diying a few personal modules and this was a quick easy hack to get my modules calibrated
4. A more general purpose calibration routine would be to either use the serial port and drive the factory calibration code directly via some wrapper harness program (I actually did this when experimenting as well but it’s not usable by anyone else and I lost interest in tidying it up). Or you could edit the firmware to present a ui within the module for calibration. Originally I just hacked the firmware to directly output the voltages I wanted (which is what I did for stages and tides 2), then realised I could do it by adjusting the knobs and buttons. I have various hacks that output values on the serial ports for testing....
5. The original notes on how I calibrated it were very terse. Bu Ma took those notes and wrote the detailed pdf I placed on the github. I have no interest in writing long notes or providing support! I just shared what I did.
6. These modules aren’t really designed for diy so you are on your own a bit really.
7. Most of the fun in diying the MI stuff lies in figuring out what it does and why it does it. The learning bit is the most important part for me... Without that element, I think it’s better to just buy the factory module - far less cost in time and risk! I dunno - having a video to walk through all this seems just wrong to me :-)
8. Obviously these are just my personal views - but you did ask :-)
The spreadsheet is just used to derive the linear dac offset and scale from the two point values (each of which has a coded dac value that corresponds to that value) you have measured (derived from what the firmware does). It’s pretty self-explanatory, isnt it?
Ps. Should this be in the *unsuccessful* builds thread ?
collumn2016
Learning to Wiggle
Posts: 47
Joined: Mon Aug 29, 2016 4:42 am
Location: in orbit
Hi Forestcaver, been a few weeks now. Got a new meter and it is a 50,000 count one and 5 digits. Tried the calibration again and managed x1 through x3 okay but couldn't get the y to behave like the rest. When i set it up as described i tried changing the bias but the voltage would not change. I played around with the rate as suggested but i couldn't get a change in voltage as i had had with the others. The module seems to play in tune but i am not 100 percent sure, i compared some of the note sequences from ben's very thorough video and it seems to behave the same. Thanks for your explicit explanation above and appreciate your time. I don't know how to move it to unsuccessful builds :-). I agree yes we have to try and work out what is going on when we build these modules but my leaning is more towards using them creatively. Especially this module and similar ones too. I love generative music and they are very important to me and my system i built for this. Take it easy and thanks again.
forestcaver
Ultra Wiggler
Posts: 925
Joined: Tue Mar 14, 2017 4:41 pm
Location: UK
Contact:
That’s odd you couldnt get quantised output from Y. Good news it’s working well...
steviet
Wiggling with Experience
Posts: 435
Joined: Wed Oct 17, 2018 9:47 am
Location: Montréal
Contact:
Hate to bump this bad boy, but I'm super happy this one turned out, thanks to forestcaver for the calibration instructions! One of my new favourite modules:
Soundcloud
Website - DIY/Euro/Gear Blog updated every Sunday
proonjooce
Learning to Wiggle
Posts: 22
Joined: Wed Sep 04, 2019 2:41 am
Branches and Ripples done!
Braids and Clouds up next
belzrebuth
Common Wiggler
Posts: 80
Joined: Mon Aug 19, 2013 5:47 pm
Location: Greece
Question for those with working DIY Stages:
Does the "B" buttons take only 1 sec to put segments in looping mode?!
Mine take a lot longer (4-5 sec) but apart from that the module works fine
Cycling thru green, yellow and red modes is instant so no worries there it's just the cycling mode that takes longer.
forestcaver
Ultra Wiggler
Posts: 925
Joined: Tue Mar 14, 2017 4:41 pm
Location: UK
Contact:
belzrebuth wrote:Question for those with working DIY Stages:
Does the "B" buttons take only 1 sec to put segments in looping mode?!
Mine take a lot longer (4-5 sec) but apart from that the module works fine
Cycling thru green, yellow and red modes is instant so no worries there it's just the cycling mode that takes longer.
Built three - all 1s. Just checked the latest mi fw - it should be 1s as well. Have a look in ui.cc and see what your duration is set to....
IIRC (vaguely) someone released a hacked fw that took 5s as they wanted it a bit longer....
belzrebuth
Common Wiggler
Posts: 80
Joined: Mon Aug 19, 2013 5:47 pm
Location: Greece
I thought that there must be a clock issue since the "1000" might not mean anything to my MCU and presto!I just found what my problem is..
My crystal does not oscillate at all..
Tried a different crystal, same thing..
Tried a thru hole crystal on the pads that are not gnd same thing too.:')
I entirely removed the xtal, turned on the module and behaves exactly like before.
So I have a working stages without a working clock
edit : It was the orientation of the 100R resistors..Now it works fine.
KittenVillage
Wiggling with Experience
Posts: 283
Joined: Sun Feb 17, 2019 10:30 am
Location: Glenshaw, PA
Contact:
### Re: DIY Mutable successfull builds hall of fame
Here's a nice pair of shades!
20200124_104543.jpg (219.09 KiB) Viewed 678 times
I've had these in the build queue for 9 months. So proud of these babies! Very easy build!
Eurorack is an exercise in delayed desire.
Hales
Common Wiggler
Posts: 212
Joined: Mon Apr 30, 2018 11:22 am
Location: Strasbourg, France
### Re: DIY Mutable successfull builds hall of fame
Here is my current setup, I've build Veils and Ripples that sit at the upper right of the case… first time that I do SMD soldering.
Out of frame, there is a digitakt for drums.
Which modules would you add in the free space?
I eventually look for an eurorack sequencer because I don't always feel comfortable handling the beatstep pro, the digitakt sequencer is good, but I find it painful not to have scales mode. I hear about the Hermod, this one seems to be a good companion of the digitakt…
IMG_1296.JPG (2.48 MiB) Viewed 649 times
Super Deluxe Wiggler
Posts: 2933
Joined: Wed Dec 25, 2013 6:24 am
### Re: DIY Mutable successfull builds hall of fame
peaks is good for drums, warps as a folder etc for your other vco's
Hashtag Octothorpe
Learning to Wiggle
Posts: 16
Joined: Wed May 24, 2017 2:39 pm
Location: Grand Rapids
### Re: DIY Mutable successfull builds hall of fame
Maybe the most unusual Braids build.....
It's inspired by and I referenced this project.
squarewavesurfer
Wiggling with Experience
Posts: 278
Joined: Tue Jan 27, 2015 2:43 pm
### Re: DIY Mutable successfull builds hall of fame
Hashtag Octothorpe wrote:
Sun Feb 09, 2020 8:43 pm
Maybe the most unusual Braids build.....
That is awesome! Braids looks pretty cool in a can.
Really unique take on a modular synth.
WTB: Verbos Scan & Pan; Sputnik modules (EF/Preamp, 6 channel mixer, 5 step voltage source); Mutable Instruments Factory panels
epijdemic
Learning to Wiggle
Posts: 10
Joined: Mon Dec 23, 2019 5:13 pm
### Re: DIY Mutable successfull builds hall of fame
Done:
2x Braids (Antumbra Knot)
2x Plaits (Antumbra Knit)
Clouds (Smog)
Elements (Atom)
Marbles (Cara)
Ripples
Grids
Rings
Tides (2018)
2x Stages
Streams
In Progress:
Warps, Veils, Shelves, Jinx
eljay
Common Wiggler
Posts: 143
Joined: Sun Jan 10, 2016 4:50 pm
Location: Dorset, UK
### DIY Yarns
My first Mutable build, really pleased with the end result. Board and plain metal front panel from Amazing Synths, great service, fast delivery and useful supporting advice. Many thanks Adam. Only my third SMD project, it really stretched my abilities but I like a challenge. I don't think I will get into anything that uses smaller components than this but I am looking forward to building a Tides module.
The front panel cover was printed on Vinyl Matte self adhesive paper, I think it works quite well. I couldn't get Green display panels so I got yellow instead, they are quite bright but are actually OK. I had some frustration identifying a suitable encoder, in the end I went for a Bourns PEC11R-4220F-S0024, the bush length on this is just too short to get the front panel nut on, but it works with a little squeezing of the PCB toward the front panel, enough to get the nut on a couple of threads and hold it firm. The attachment to the front panel with the nut is essential to stop the PCB flexing when pressing the encoder switch.
Setting up the Mutable development environment on my PC was a lot easier in practice than it appeared at first sight and I invested in an ST-LinkV2 and the Olimex JTAG-20-10 converter cable. After a couple of false starts, I needed to build the Yarns bootloader before running the overall Yarns makefile and provide the module power from a Eurorack supply, it ran without error and downloaded the image to the micro-controller. A quick re-cycle of the module power and there it was!
Calibration is straight forward and is now done. My only negative observation is that the Mutable documentation doesn't appear to accurately reflect the functionality of the version 1.5 software that I've installed, some of the menu items and options have changed, I assume for the better.
IMG_0721 (2).JPG (1.12 MiB) Viewed 526 times
Super Deluxe Wiggler
Posts: 2933
Joined: Wed Dec 25, 2013 6:24 am
### Re: DIY Mutable successfull builds hall of fame
Looks great
Staticcharge
Common Wiggler
Posts: 245
Joined: Tue May 05, 2015 2:02 pm
Location: Somewhere between yesterday and tomorrow.
### Re: DIY Yarns
eljay wrote:
Thu Feb 20, 2020 5:02 pm
My first Mutable build, really pleased with the end result. Board and plain metal front panel from Amazing Synths, great service, fast delivery and useful supporting advice. Many thanks Adam. Only my third SMD project, it really stretched my abilities but I like a challenge. I don't think I will get into anything that uses smaller components than this but I am looking forward to building a Tides module.
The front panel cover was printed on Vinyl Matte self adhesive paper, I think it works quite well. I couldn't get Green display panels so I got yellow instead, they are quite bright but are actually OK. I had some frustration identifying a suitable encoder, in the end I went for a Bourns PEC11R-4220F-S0024, the bush length on this is just too short to get the front panel nut on, but it works with a little squeezing of the PCB toward the front panel, enough to get the nut on a couple of threads and hold it firm. The attachment to the front panel with the nut is essential to stop the PCB flexing when pressing the encoder switch.
Setting up the Mutable development environment on my PC was a lot easier in practice than it appeared at first sight and I invested in an ST-LinkV2 and the Olimex JTAG-20-10 converter cable. After a couple of false starts, I needed to build the Yarns bootloader before running the overall Yarns makefile and provide the module power from a Eurorack supply, it ran without error and downloaded the image to the micro-controller. A quick re-cycle of the module power and there it was!
Calibration is straight forward and is now done. My only negative observation is that the Mutable documentation doesn't appear to accurately reflect the functionality of the version 1.5 software that I've installed, some of the menu items and options have changed, I assume for the better.
IMG_0721 (2).JPG
Where did you find the DIN sockets, Mouser don't stock the ones on the BOM.
Super Deluxe Wiggler
Posts: 2933
Joined: Wed Dec 25, 2013 6:24 am
### Re: DIY Mutable successfull builds hall of fame
eljay
Common Wiggler
Posts: 143
Joined: Sun Jan 10, 2016 4:50 pm
Location: Dorset, UK
### Re: DIY Mutable successfull builds hall of fame
I searched for SD-50BV on Mouser, there are plenty in stock. These fit and work perfectly.
Kylep357
1-Post Wiggler
Posts: 1
Joined: Thu Dec 05, 2019 1:28 pm
Location: New York
### Re: DIY Mutable successfull builds hall of fame
Recently finished all 7 of these modules! I’ve been wanting to get into Eurorack for a while, and decided that I should start with all DIY and save some \$\$. Panels are etched aluminum filled with enamel, came out pretty well I think.
Attachments
eljay
Common Wiggler
Posts: 143
Joined: Sun Jan 10, 2016 4:50 pm
Location: Dorset, UK
### Re: DIY Mutable successfull builds hall of fame
Second successful Mutable build, a version 1 Tides, board and bare metal panel from Amazing Synths. I've used Vinyl Matte self adhesive paper to print out and attach a front panel cover (not as impressive as the etched aluminium above but it works Ok for me and cost about 50p). Having set up the software build environment to build the Yarns the build for this was a simple step and repeat, everything worked first time, calibration took a couple of minutes. Again, really pleased with the end result, I do take my time to assemble the board, I hand solder the components and I'm still developing my SMD technique. For the processor I use a pin to apply solder paste to the legs in a thin strip and then use a hot air workstation to solder, if necessary I clean any bridges with de-solder braid. The rest is soldered with an iron, 0.7mm solder and a flux pen.
Super Deluxe Wiggler
Posts: 2933
Joined: Wed Dec 25, 2013 6:24 am
### Re: DIY Mutable successfull builds hall of fame
looks nice, I'm glad it went ok, have you tried drag soldering?
eljay
Common Wiggler
Posts: 143
Joined: Sun Jan 10, 2016 4:50 pm
Location: Dorset, UK
### Re: DIY Mutable successfull builds hall of fame
I have tried drag soldering and generally use the technique on all the ICs except the processor chip. I tried unsuccessfully to drag solder a processor chip and have since adapted a solder paste technique for these.
| 4,180
| 16,510
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.53125
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.931916
|
http://www.physicsforums.com/showthread.php?t=314301
| 1,386,940,789,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386164943590/warc/CC-MAIN-20131204134903-00094-ip-10-33-133-15.ec2.internal.warc.gz
| 496,626,123
| 10,106
|
# acceleration of the lower block
by brunettegurl
Tags: acceleration, block
P: 138 1. The problem statement, all variables and given/known data In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 58.6 N. The coefficient of kinetic friction between the lower block and the surface is 0.204. The coefficient of kinetic friction between the lower block and the upper block is also 0.204. What is the acceleration of the lower block, if the mass of the lower block is 4.68 kg and the mass of the upper block is 2.11 kg? 2. Relevant equations F=ma / friction = $$\mu$$mg 3. The attempt at a solution so i wrote the forces that affect each box For the upper box ::$$\sum$$ F= T-$$\mu$$m1g for the lower box :: $$\sum$$ F= F-T- friction1- friction2 i solved the first equation for tension and then substituted it into the 2nd equation and i solved for acceleration...can u pls. tell me what i'm doin wrong thanx :)) Attached Thumbnails
Mentor
P: 40,279
Quote by brunettegurl so i wrote the forces that affect each box For the upper box ::$$\sum$$ F= T-$$\mu$$m1g for the lower box :: $$\sum$$ F= F-T- friction1- friction2
OK, but what did you use for the friction forces on the lower box?
i solved the first equation for tension and then substituted it into the 2nd equation and i solved for acceleration...can u pls. tell me what i'm doin wrong thanx :))
I assume you applied Newton's 2nd law to each box to get two equations. Show exactly what you did and then we can see what went wrong.
P: 138 for friction 1 i used (coeffecient of frictionxmass1xgravity) and for friction 2 i use (coeffecient of frictionxmass2xgravity) and the equation i got before isolating for acceleration is :: a(m2+m1)= Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass2xgravity) so then a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2)
Mentor
P: 40,279
## acceleration of the lower block
Quote by brunettegurl for friction 1 i used (coeffecient of frictionxmass1xgravity)
OK.
and for friction 2 i use (coeffecient of frictionxmass2xgravity)
Careful: What's the normal force between the bottom block and the floor?
P: 138 is it f 2 = coeffecient of frictionxmass2+ mass1xgravity...???
Mentor
P: 40,279
Quote by brunettegurl is it f 2 = coeffecient of frictionxmass2+ mass1xgravity...???
Yes, the normal force is the combined weight of both blocks, thus f2 = μ(m1 + m2)g.
P: 138 ok so i changed that and my answer is still coming out wrong is there anything else wrong i cld have done in the initial equation??
Mentor
P: 40,279
Quote by brunettegurl ok so i changed that and my answer is still coming out wrong is there anything else wrong i cld have done in the initial equation??
Show me your final equation.
P: 138 my final equation wld look like :: m2a = [Fapplied-m1a-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2) Fapplied is the force given in the question
Mentor
P: 40,279
Quote by brunettegurl my final equation wld look like :: m2a = [Fapplied-m1a-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2) Fapplied is the force given in the question
Your equations are a bit confusing, due to what may be a few typos. (The '=', for one.) Can you rewrite using F, m1, m2, μ, and g? Try to simplify as much as possible.
P: 138 m2a= [F -m1a- (μm1g)- (μg(m1+m2)) - (μm2g)] a = [F - (μm1g) - (μg(m1+m2)) - (μm2g)]/(m1+m2) iss that clearer ??
Mentor
P: 40,279
Quote by brunettegurl m2a= [F -m1a- (μm1g)- (μg(m1+m2)) - (μm2g)] a = [F - (μm1g) - (μg(m1+m2)) - (μm2g)]/(m1+m2) iss that clearer ??
Yes, much clearer. You have an extra m2 term in there for some reason. If you write the original Newton's law equations for m1 and m2, perhaps we can see where it got snuck in.
P: 138 this is what i have as my equations for m1 and m2 m1:: m1a= T- μm1g m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g i dont see the extra m2 in my equations can u be a bit more specific
Mentor
P: 40,279
Quote by brunettegurl this is what i have as my equations for m1 and m2 m1:: m1a= T- μm1g m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g
That second friction term in the m2 equation is the problem--you're using the wrong mass.
P: 138 so this m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g wld be this m2 :: m2a = Fapp - T- μ(m1+m2)g- μm1g.... then the final equation wld look like a = [Fapp - (μm1g) - (μg(m1+m2)) - (μm1g)]/(m1+m2) is that correct??
Mentor
P: 40,279
Quote by brunettegurl then the final equation wld look like a = [Fapp - (μm1g) - (μg(m1+m2)) - (μm1g)]/(m1+m2) is that correct??
That looks good. But consolidate some of those terms.
Related Discussions Advanced Physics Homework 1 Advanced Physics Homework 1 Introductory Physics Homework 8 Introductory Physics Homework 1 Introductory Physics Homework 3
| 1,678
| 5,327
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625
| 4
|
CC-MAIN-2013-48
|
longest
|
en
| 0.864073
|
https://socratic.org/questions/how-would-you-use-the-kinetic-theory-to-explain-why-a-helium-filled-balloon-shri
| 1,579,754,246,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00167.warc.gz
| 632,784,253
| 6,298
|
How would you use the kinetic theory to explain why a helium filled balloon "shrinks" when it is taken from a warm room to the outside on a cold day?
Jun 6, 2017
From kinetic theory, the pressure can be interpreted as, 2/3 the kinetic energy of molecules per unit volume.
$p = \frac{2 N E}{3} V$
Also, from kinetic interpretation of temperature, the ideal gas molecules have their energy proportional to the absolute temperature.
Thus, $E = \frac{3 k T}{2}$
Combining the two relations,
$p V$ is proportional to $T$.
Then for a decrease in temperature $T$, then $V$ and $p$ decrease.
This explains why the balloon shrinks when the temperature decreases.
| 163
| 662
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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}
| 3.28125
| 3
|
CC-MAIN-2020-05
|
latest
|
en
| 0.844753
|
http://dl.begellhouse.com/jp/journals/61fd1b191cf7e96f,6a4cc29505086840,1c6eda411d03872e.html
| 1,579,562,376,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00177.warc.gz
| 51,475,626
| 7,704
|
ライブラリ登録: Guest
インパクトファクター: 1.016 5年インパクトファクター: 1.194 SJR: 0.554 SNIP: 0.68 CiteScore™: 1.18
ISSN 印刷: 1543-1649
ISSN オンライン: 1940-4352
# International Journal for Multiscale Computational Engineering
DOI: 10.1615/IntJMultCompEng.v6.i4.30
pages 309-325
## Scaling Up of an Underground Waste Disposal Model with Random Source Terms
Olivier Gipouloux
Université de Lyon, CNRS, Institut Camille Jordan, Batiment ISTIL; and Département de Mathématiques, Faculté des Sciences et Techniques, Université de Saint Etienne, France
Farid Smai
Université de Lyon, CNRS, Institut Camille Jordan, Batiment ISTIL, France
### 要約
In this paper, we study the global behavior of an underground waste disposal in order to obtain an accurate upscaled model suitable for the computations involved in safety assessment processes. The disposal is made of a high number of modules containing the waste material. One supposes that the modules or disposal units are periodicaly distributed and that each disposal unit leaks randomly. Recent theoretical works, by use of a combination of classical homogenization and probability theories, allow to derive a global model where only one deterministic leaking source is considered. We propose, using these theoretical results, to give a numerical approach for a such multiscale problem to obtain solutions of the deterministic problem and the first-order statistical moments.
### Articles with similar content:
SHORT REVIEW OF LAGRANGIAN STOCHASTIC MODELS FOR SIMULATION OF ATMOSPHERIC POLLUTANT DISPERSION
Hybrid Methods in Engineering, Vol.2, 2000, issue 3
Domenico Anfossi
Multiscale Modeling for Planar Lattice Microstructures with Structural Elements
International Journal for Multiscale Computational Engineering, Vol.4, 2006, issue 4
Ken Ooue, Isao Saiki, Kenjiro Terada, Akinori Nakajima
Discretization and Solver Methods with Analytical Characteristic Methods for Advection-Diffusion Reaction Equations and 2D Applications
Journal of Porous Media, Vol.12, 2009, issue 7
Juergen Geiser
Iterative Algorithms for Computing the Averaged Response of Nonlinear Composites under Stress-Controlled Loadings
International Journal for Multiscale Computational Engineering, Vol.4, 2006, issue 4
| 564
| 2,218
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2020-05
|
latest
|
en
| 0.61193
|
https://pvpmc.sandia.gov/modeling-steps/2-dc-module-iv/effective-irradiance/
| 1,675,078,315,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00640.warc.gz
| 464,589,753
| 16,028
|
Effective irradiance is total plane of array (POA) irradiance adjusted for angle of incidence losses, soiling, and spectral mismatch. In a general sense it can be thought of as the irradiance that is “available” to the PV array for power conversion.
In the context of the Sandia PV Array Performance Model (SAPM), effective irradiance ($E_{e}$) is defined specifically as:
$E_{e}=\frac{I_{sc}}{I_{sc0}\left&space;\{&space;1+\alpha&space;_{Isc}\left&space;(&space;T_{c}-T_{0}&space;\right&space;)&space;\right&space;\}}$, where $I_{sc}$ is the short circuit current. It can be calculated from Eq. 1 described in the SAPM, from measured irradiance, air mass, angle of incidence, and several empirically determined module coefficients.
Alternatively, effective irradiance can be measured directly by obtaining IV curves from a matched reference module.
$E_{e}=\frac{I_{scr}}{I_{sc0r}\left&space;\{&space;1+\alpha&space;_{Iscr}\left&space;(&space;T_{cr}-T_{0}&space;\right&space;)&space;\right&space;\}}\times&space;SF$
• $SF$ is the soiling factor (=1 when clean)
A simplified approach using a single irradiance sensor has also been suggested, shown below. Caution should be used when applying this method because irradiance sensors (pyranometers) require careful calibration and frequent cleaning. These calibrations may not account for all instrument responses (e.g., angular), and instruments can vary in their spectral response and acceptance angles, etc. (thermopile vs. photodiode)
$E_{e}=\frac{E_{POA}}{E_{0}}\times&space;SF$
• $E_{POA}$ is the plane of array irradiance.
• $E_{0}$ is the reference irradiance ($1000&space;W/m^{2}$)
| 455
| 1,648
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.015625
| 3
|
CC-MAIN-2023-06
|
longest
|
en
| 0.801568
|
http://freetofindtruth.blogspot.com/2016/09/13-39-52-86-113-167-colonial-pipeline.html
| 1,586,151,505,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00125.warc.gz
| 75,188,158
| 28,770
|
## Thursday, September 22, 2016
### 13 39 52 86 113 167 | Colonial Pipeline leak in Alabama, a Super Bowl 51 tribute? (Gold King Mine & Super Bowl 50 de ja vu)
This pipeline runs from Houston to New York. Why am I thinking of Super Bowl 51 and NRG Stadium, or en-er-gy? *Please recall that last year there was the Gold King Mine River Disaster that was a tribute to the Broncos Super Bowl to be won. That river disaster was mainly in Colorado, and began 187-days before the Super Bowl was played, August 5, 2015. Peyton Manning, age 39, played for the Broncos. This disaster is connected to New York, where Eli Manning, plays QB for the Giants. Eli is also in his 13th season and this leaking pipe has a lot to do with '13' as you will see.
The pipeline that burst is the 'Colonial Pipeline'.
'39' is the New York number, remember, every Giants Super Bowl connects to '39', and the Jets have only won one, 16-7; that was Super Bowl 3.
7/26/1788 = 7+2+6+1+7+8+8 = 39 (13x3 = 39)
Recall the gematria of 'NRG'.
NRG = 14+18+7 = 39 (Construction began March 9, or 3/9)
As I've been saying, there are a lot of signs of the Giants or a different New York team being in the Super Bowl this year, in Houston, at NRG Stadium.
For one last '39', let's examine 'pipe leak'.
In Pythagorean Gematria, 'Colonial Pipeline' is also interesting.
The word 'pipeline' also has gematria of '86'.
Right now the New England team in the NFL is missing its 39-year old QB due to 'Deflate Gate'.
I bring up New England because they have a major '99' connection, like the September 9 start date of this pipeline leak.
The number 'thirteen' also connects to '99'.
The leak began in 'Alabam', having gematria of '13'.
September 9 is also the day that leaves 113-days left in the year.
Where the leak began, Shelby County, has a connection to the Super Bowl date, February 5, written 5/2 or 2/5.
This year Eli Manning earned his 106th win of his career to start the season.
He earned his 52nd regular season win in New York against his father's team, the Saints, also from an energy city, not far from Houston.
Eli won his last Super Bowl in Lucas Oil Stadium, where his big brother once played, the home of the Indianapolis Colts.
http://kfor.com/2016/09/16/state-of-emergency-declared-after-crucial-oil-pipeline-leaks-250000-gallons-in-alabama/
The other thing to consider is that this leak began September 9, three-days after the river leak in Russia that turned the river to 'Crimson Tide', the name of the football team that plays in Alabama. See my work on that river leak, a leak that was first reported in the media the day of the season opener for the NFL, September 8, one day before this leak began.
From the date of the incident, September 9, until the upcoming Super Bowl, is 149-days. That is relevant because this is the 47th Super Bowl of the modern era, and the first Super Bowl was in the 47th season of the league, at the end of '66.
1. Here in Atl it has been contrived inconvenience and borderline hysteria at one point. All over pretty much nothing. I feel like the "shortage" was resolved before it even began. Just an excuse to scare the masses and drive up demand/prices.
1. That sounds about right. Order out of chaos (or made up chaos).
2. That's exactly how it was, Matt. Up on Lake Lanier it was Gasageddon, but not to keep ot or store it, just to have enough to fill up the boats and other entertainment. Now, it settles again. These things almost always have a weekend run then fizzle.
2. The word "shortage" reminds me that the rapper "Shawty Lo" supposedly died in car crash in Atlanta during this same period of supposed gas shortages, 9.21.2016.
So, his name reflects both Short.y and Low as in short, at the same time we are being told there is a fuel shortage. What a coincidence!
Yeah right, clearly this is NLP reinforcement. Layering stories representing the main idea you want to put out there to the masses, and reinforcing it over and over.
1. Prunella, I Love Your Posts!!!! That is really being alert, keep on posting!!
| 1,062
| 4,070
|
{"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-2020-16
|
latest
|
en
| 0.972766
|
https://raw.githubusercontent.com/rlabbe/Kalman-and-Bayesian-Filters-in-Python/master/05-Multivariate-Gaussians.ipynb
| 1,660,448,974,000,000,000
|
text/plain
|
crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00444.warc.gz
| 458,661,725
| 757,791
|
{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "[Table of Contents](./table_of_contents.ipynb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Multivariate Gaussians\n", "\n", "Modeling Uncertainty in Multiple Dimensions" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "%matplotlib inline" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "scrolled": true }, "outputs": [ { "data": { "text/html": [ "\n", " \n", " " ], "text/plain": [ "" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#format the book\n", "import book_format\n", "book_format.set_style()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Introduction" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The techniques in the last chapter are very powerful, but they only work with one variable or dimension. They provide no way to represent multidimensional data, such as the position and velocity of a dog in a field. Position and velocity are related to each other, and as we learned in the g-h chapter we should never throw away information. In this chapter we learn how to describe this relationship probabilistically. Through this key insight we will achieve markedly better filter performance." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Multivariate Normal Distributions\n", "\n", "We've been using Gaussians for a scalar random variable, expressed as $\\mathcal{N}(\\mu, \\sigma^2)$. A more formal term for this is *univariate normal*, where univariate means 'one variable'. The probability distribution of the Gaussian is known as the *univariate normal distribution*.\n", "\n", "What might a *multivariate normal distribution* be? *Multivariate* means multiple variables. Our goal is to be able to represent a normal distribution with multiple dimensions. I don't necessarily mean spatial dimensions - if we track the position, velocity, and acceleration of an aircraft in (x, y, z) that gives us a nine dimensional problem. Consider a two dimensional case. It might be the *x* and *y* coordinates of a robot, it might be the position and velocity of a dog on the x-axis, or milk production and feed rate at a dairy. It doesn't really matter. We can see that for $N$ dimensions, we need $N$ means, which we will arrange in a column matrix (vector) like so:\n", "\n", "$$\n", "\\mu = \\begin{bmatrix}\\mu_1\\\\\\mu_2\\\\ \\vdots \\\\\\mu_n\\end{bmatrix}\n", "$$\n", "\n", "Let's say we believe that $x = 2$ and $y = 17$. We would have\n", "\n", "$$\n", "\\mu = \\begin{bmatrix}2\\\\17\\end{bmatrix} \n", "$$\n", "\n", "The next step is representing our variances. At first blush we might think we would also need N variances for N dimensions. We might want to say the variance for x is 10 and the variance for y is 4, like so. \n", "\n", "$$\\sigma^2 = \\begin{bmatrix}10\\\\4\\end{bmatrix}$$ \n", "\n", "This is incomplete because it does not consider the more general case. In the **Gaussians** chapter we computed the variance in the heights of students. That is a measure of how the heights vary relative to each other. If all students are the same height, then the variance is 0, and if their heights are wildly different, then the variance will be large. \n", "\n", "There is also a relationship between height and weight. In general, a taller person weighs more than a shorter person. Height and weight are *correlated*. We want a way to express not only what we think the variance is in the height and the weight, but also the degree to which they are correlated. In other words, we want to know how weight varies compared to the heights. We call that the *covariance*. \n", "\n", "Before we can understand multivariate normal distributions we need to understand the mathematics behind correlations and covariances." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Correlation and Covariance\n", "\n", "*Covariance* describes how much two variables vary together. Covariance is short for *correlated variances*. In other words, *variance* is a measure for how a population vary amongst themselves, and *covariance* is a measure for how much two variables change in relation to each other. For example, as height increases weight also generally increases. These variables are *correlated*. They are *positively correlated* because as one variable gets larger so does the other. As the outdoor temperature decreases home heating bills increase. These are *inversely correlated* or *negatively correlated* because as one variable gets larger the other variable lowers. The price of tea and the number of tail wags my dog makes have no relation to each other, and we say they are *uncorrelated* or *independent*- each can change independent of the other.\n", "\n", "Correlation allows prediction. If you are significantly taller than me I can predict that you also weigh more than me. As winter comes I predict that I will be spending more to heat my house. If my dog wags his tail more I don't conclude that tea prices will be changing.\n", "\n", "For example, here is a plot of height and weight of students on the school's track team. If a student is 68 inches tall I can predict they weigh roughly 160 pounds. Since the correlation is not perfect neither is my prediction. " ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "data": { "image/png": "iVBORw0KGgoAAAANSUhEUgAAAwUAAAFzCAYAAABvgaWMAAAAOXRFWHRTb2Z0d2FyZQBNYXRwbG90bGliIHZlcnNpb24zLjQuMywgaHR0cHM6Ly9tYXRwbG90bGliLm9yZy/MnkTPAAAACXBIWXMAAA9hAAAPYQGoP6dpAABeCUlEQVR4nO3dd3hUBdr+8e+kE0JHEiKhrFQBWUREEGmSRARCEZEeQuasrsquCru+6rrgFn3Xta34c3XPpBCKYAEEQSAoRQgqRUSKSEekRAQTAiGZZM7vD15mjbQEkzkp9+e6vC7nmZM5zzxMMnPPaQ7LsixERERERKTK8rO7ARERERERsZdCgYiIiIhIFadQICIiIiJSxSkUiIiIiIhUcQoFIiIiIiJVnEKBiIiIiEgVp1AgIiIiIlLFKRSIiIiIiFRxAXY3UB54PB6OHDlCjRo1cDgcdrcjIiIiIvKLWZbF6dOniYyMxM/vytsCFAqAI0eOEBUVZXcbIiIiIiKl7ttvv6VRo0ZXXEahAKhRowZwfmA1a9b0+frdbjfLly8nJiaGwMBAn6+/qtCcfUez9g3N2Xc0a9/RrH1Dc/YdO2ednZ1NVFSU97PulSgUgHeXoZo1a9oWCkJDQ6lZs6Z+McuQ5uw7mrVvaM6+o1n7jmbtG5qz75SHWRdn93gdaCwiIiIiUsUpFIiIiIiIVHEKBSIiIiIiVZxCgYiIiIhIFadQICIiIiJSxSkUiIiIiIhUcQoFIiIiIiJlyLLs7uDqdJ0CEREREZFSlpNXwAvLdpG+4xg5Z/z559driL4xgsmxrQgLLn8fwctfRyIiIiIiFVhOXgFDX1/HnswcPBaAg6wfz5G2/gAZe08w78Hby10w0O5DIiIiIiKl6IVlu34SCP7LY8GezBxeXLbLnsauQKFARERERKQUrdh5/KJAcIHHgvSdx33bUDEoFIiIiIiIlBLLsnAXXvnI4oJCC6ucHX2sUCAiIiIiUkocDgeB/o4rLhPg78DhuPIyvqZQICIiIiJSivq2CcfvMp/5/RwQ3Sbctw0Vg0KBiIiIiEgpmhzbiuYNwi4KBn4OaN4gjEmxrexp7AoUCkRERERESlFYcADzHryd+K5NaVQ7hFqBFo1qhxDftWm5PB0p6DoFIiIiIiKlLiw4gClxbXmyX0sWL15C//49CAwMtLuty9KWAhERERGRMlTOjim+JIUCEREREZEqTqFARERERKSKUygQEREREaniFApERERERKo4hQIRERERkSpOoUBEREREpIpTKBARERERqeIUCkREREREqjiFAhERERGRKk6hQERERESkilMoEBERERGp4hQKRERERETKUGFhod0tXJVCgYiIiIhIKcvOzubNN9/ktttuY/bs2Xa3c1W2hoLnnnuOzp07U6NGDRo0aMDgwYPZtWtXkWUsy2Lq1KlERkZSrVo1evXqxfbt24ssk5eXx8SJE6lfvz7Vq1cnLi6Ow4cP+/KpiIiIiEgVZ1kW69evZ8KECTRs2JAHHniAzZs3s2bNGizLsru9K7I1FKxevZqHHnqITz/9lPT0dAoKCoiJieHMmTPeZZ5//nleeuklXnvtNTZs2EBERATR0dGcPn3au8wjjzzC/PnzmTNnDmvXriUnJ4cBAwZUiE01IiIiIlKxnTx5kn/961+0b9+ebt26kZKSwtmzZ2ndujXPP/88L774Ig6Hw+42ryjAzpUvXbq0yO2UlBQaNGjApk2b6NGjB5Zl8corr/DUU08xdOhQAKZPn054eDizZ8/m/vvvJysri6SkJGbMmEHfvn0BmDlzJlFRUaxYsYLY2FifPy8RERERqdwsy2LVqlWYpsm8efPIy8sDoFq1atx7770YhsHtt99OQUEBS5Yssbnbq7M1FPxcVlYWAHXr1gVg//79HDt2jJiYGO8ywcHB9OzZk4yMDO6//342bdqE2+0uskxkZCTt2rUjIyPjkqEgLy/P+w8H5/f5AnC73bjd7jJ5bldyYZ12rLsq0Zx9R7P2Dc3ZdzRr39GsfUNzvnbHjx8nLS2NlJQU9uzZ46136NCBxMRERowYQe3atQEoKCiwddYlWWe5CQWWZfHYY4/RvXt32rVrB8CxY8cACA8PL7JseHg4Bw8e9C4TFBREnTp1Llrmws//3HPPPcczzzxzUX358uWEhob+4udyrdLT021bd1WiOfuOZu0bmrPvaNa+o1n7huZcPIWFhWzZsoX09HQ2bNjg3UU9JCSEHj16EBMTww033IDD4SAjI+OSj2HHrM+ePVvsZctNKHj44YfZunUra9euvei+n++DZVnWVffLutIyTzzxBI899pj3dnZ2NlFRUcTExFCzZs1r6P6XcbvdpKenEx0dTWBgoM/XX1Vozr6jWfuG5uw7mrXvaNa+oTkXz7fffktqairTp0/n0KFD3nqXLl1ITExk2LBhhIWFXfEx7Jz1hb1hiqNchIKJEyeycOFC1qxZQ6NGjbz1iIgI4PzWgIYNG3rrmZmZ3q0HERER5Ofnc+rUqSJbCzIzM+nWrdsl1xccHExwcPBF9cDAQFt/Mexef1WhOfuOZu0bmrPvaNa+o1n7huZ8MbfbzeLFizFNk6VLl+LxeACoU6cOY8eOxel00r59+xI/rh2zLsn6bD37kGVZPPzww8ybN4+PP/6YZs2aFbm/WbNmREREFNnckp+fz+rVq70f+Dt16kRgYGCRZY4ePcq2bdsuGwpERERERH5q7969PPHEEzRu3JghQ4awZMkSPB4PvXr1YtasWRw5csR7hqHKyNYtBQ899BCzZ8/m/fffp0aNGt5jAGrVqkW1atVwOBw88sgjPPvss7Ro0YIWLVrw7LPPEhoayqhRo7zLJiYmMmnSJOrVq0fdunWZPHky7du3956NSERERETk5/Ly8pg/fz6mafLxxx976w0aNGD8+PE4nU5atGhhY4e+Y2so+Pe//w1Ar169itRTUlIYP348AH/84x/Jzc3lwQcf5NSpU3Tp0oXly5dTo0YN7/Ivv/wyAQEBDB8+nNzcXO68805SU1Px9/f31VMRERERkQpix44dmKZJWloaJ0+eBM4fwxobG4thGAwcOLDK7VZlaygozpXdHA4HU6dOZerUqZddJiQkhGnTpjFt2rRS7E5EREREKoszZ87wzjvvYJpmkTMENWrUiAkTJjBhwgSaNGliY4f2KhcHGouIiIiIlIXNmzdjmiazZ8/2no3H39+fgQMHYhgGsbGx2rsEhQIRERERqWSys7OZPXs2pmmyefNmb/1Xv/oVTqeT8ePHFzmzpSgUiIiIiEglYFkW69evxzRN3n77be+Fu4KCghg6dCiGYdCrVy/8/Gw9+Wa5pVAgIiIiIhXWDz/8wIwZMzBNkx07dnjrbdq0wTAMxo4dS/369W3ssGJQKBARERGRCsXj8bBq1SpM02TevHnk5+cDUK1aNe677z6cTifdunXD4XDY3GnFoVAgIiIiIhXCsWPHSE1NxeVysXfvXm+9Y8eOGIbBqFGjqFWrlo0dVlwKBSIiIiJSbhUWFrJs2TJM02TRokUUFhYCUKNGDUaNGoVhGHTq1MnmLis+hQIRERERKXcOHTpEcnIyycnJfPvtt956165dMQyD4cOHU716dRs7rFwUCkRERESkXHC73SxatAiXy8XSpUu9F7qtW7cuY8eOxel00q5dO5u7rJwUCkRERETEVnv27MHlcpGamsrx48e99d69e2MYBkOGDCEkJMTGDis/hQIRERER8blz584xb948XC4XK1eu9NbDw8NJSEggMTGR5s2b29hh1aJQICIiIiI+s337dkzTZMaMGZw8eRIAh8PBXXfdhWEYDBgwgMDAQJu7rHoUCkRERESkTJ05c4a3334b0zRZv369tx4VFcWECROYMGECjRs3trFDUSgQERERkTKxadMmTNNk9uzZnD59GoCAgAAGDhyIYRjExMTg7+9vc5cCCgUiIiIiUoqysrKYPXs2pmnyxRdfeOs33HADTqeT8ePHExERYWOHcikKBSIiIiLyi1iWRUZGBqZp8vbbb5ObmwtAUFAQ99xzD4Zh0LNnT/z8/GzuVC5HoUBERERErsmJEyeYMWMGLpeLHTt2eOs33ngjhmEwduxY6tWrZ2OHUlwKBSIiIiJSbB6Ph5UrV2KaJvPnzyc/Px+A0NBQ7rvvPgzD4LbbbsPhcNjcqZSEQoGIiIiIXNXRo0dJTU3F5XKxb98+b/3mm2/GMAxGjhxJrVq1bOxQfgmFAhERERG5pMLCQpYuXYppmnzwwQcUFhYCULNmTUaPHo3T6eTmm2+2uUspDQoFIiIiIlLEwYMHSU5OJjk5mcOHD3vr3bp1wzAM7r33XqpXr25jh1LaFApEREREBLfbzcKFCzFNk+XLl2NZFgD16tVj3LhxJCYm0rZtW5u7lLKiUCAiIiJShe3evRuXy0VqaiqZmZneep8+fTAMgyFDhhAcHGxjh+ILCgUiIiIiVcy5c+dYvXo1L730EmvWrPHWw8PDSUhIIDExkebNm9vYofiaQoGIiIhIFbFt2zZM02TGjBmcOnUKAD8/P+666y4Mw6B///4EBgba3KXYQaFAREREpBLLyclh7ty5uFwuPv30U2+9fv36/Pa3v8UwDKKiomzsUMoDhQIRERGRSsayLDZt2oRpmrz11lucPn0agICAAOLi4khISCA/P5+BAwdqy4AACgUiIiIilcaPP/7IrFmzcLlcbNmyxVtv3rw5hmEQHx9PeHg4brebJUuW2NeolDsKBSIiIiIVmGVZrFu3DtM0eeedd8jNzQUgODiYe+65B8Mw6NmzJw6Hw+ZOpTxTKBARERGpgL7//nvS0tJwuVx8/fXX3nrbtm0xDIOxY8dSt25dGzuUikShQERERKSC8Hg8fPzxx5imyfz583G73QCEhoYyYsQIDMOgS5cu2iogJaZQICIiIlLOHTlyhJSUFJKSkti/f7+33qlTJwzDYOTIkdSsWdPGDqWiUygQERERKYcKCgpYunQppmmyePFiCgsLAahZsyZjxozB6XTSsWNHm7uUykKhQERERKQcOXDgAMnJySQnJ/Pdd99567fffjuGYXDvvfcSGhpqY4dSGSkUiIiIiNgsPz+fhQsXYpom6enpWJYFQL169YiPj8fpdNKmTRubu5TKTKFARERExCa7du0iKSmJ1NRUvv/+e2+9b9++OJ1OBg8eTHBwsI0dSlWhUCAiIiLiQ7m5ubz33nuYpsmaNWu89YYNG5KQkEBiYiK/+tWvbOxQqiKFAhEREREf2Lp1K6ZpMnPmTH788UcA/Pz86NevH4Zh0L9/fwICfPfR7P/2UBIBwM/Ola9Zs4aBAwcSGRmJw+FgwYIFRe7Pycnh4YcfplGjRlSrVo02bdrw73//u8gyeXl5TJw4kfr161O9enXi4uI4fPiwD5+FiIiIyKXl5OTgcrno0qULHTp04LXXXuPHH3+kSZMm/OUvf+HgwYN88MEHDBo0yCeBICevgKkLt9PrxTVM2eRPrxfXMHXhdnLyCsp83VK+2bql4MyZM3To0IGEhATuueeei+5/9NFHWblyJTNnzqRp06YsX76cBx98kMjISAYNGgTAI488wqJFi5gzZw716tVj0qRJDBgwgE2bNuHv7+/rpyQiIiJVnGVZbNiwAdM0mTNnDjk5OQAEBAQwaNAgDMOgb9++Pv+ckpNXwNDX17EnMwePBeAg68dzpK0/QMbeE8x78HbCgrUTSVVl6798v3796Nev32XvX79+PfHx8fTq1QuA3/zmN7z55pts3LiRQYMGkZWVRVJSEjNmzKBv374AzJw5k6ioKFasWEFsbKwvnoaIiIgIp06dYtasWZimydatW731Fi1aYBgG8fHxNGjQwLb+Xli26yeB4L88FuzJzOHFZbuYEtfWnubEduU6Dnbv3p2FCxcyYcIEIiMjWbVqFd988w3/+te/ANi0aRNut5uYmBjvz0RGRtKuXTsyMjIuGwry8vLIy8vz3s7OzgbA7XZ7LxfuSxfWace6qxLN2Xc0a9/QnH1Hs/adijZry7JYu3YtycnJvPfee5w7dw6A4OBghg4dSmJiInfccQcOhwOw93ml7zh2USC4wGOdv//Jfi1921QVYOdruiTrLNeh4NVXX8UwDBo1akRAQAB+fn64XC66d+8OwLFjxwgKCqJOnTpFfi48PJxjx45d9nGfe+45nnnmmYvqy5cvt/ViIOnp6batuyrRnH1Hs/YNzdl3NGvfKe+zzsrKYuXKlaSnpxe5wFiTJk2Ijo6mZ8+e1KhRg5ycHD788EMbOz3PsiDnjD/guOwyp8/ksnjxEhyXX0R+ATte02fPni32suU+FHz66acsXLiQJk2asGbNGh588EEaNmzo3V3oUizL8ibyS3niiSd47LHHvLezs7OJiooiJiaGmjVrlupzKA632016ejrR0dEEBgb6fP1VhebsO5q1b2jOvqNZ+055nrXH4+Gjjz4iKSmJRYsWeb+FrV69OsOHDycxMZHOnTtf8TOInf759Rqyfjx32fvDqlejf/8ePuyoarDzNX1hb5jiKLehIDc3lyeffJL58+fTv39/AG666Sa2bNnCCy+8QN++fYmIiCA/P59Tp04V2VqQmZlJt27dLvvYwcHBl7wQSGBgoK1/gOxef1WhOfuOZu0bmrPvaNa+U55m/d1335GSkkJSUhIHDhzw1jt37ozT6WTkyJHUqFHDvgaLKfrGCNLWH7jkLkR+Doi5MaLczLwysuM1XZL12XpK0iu5sH+/n1/RFv39/fF4PAB06tSJwMDAIptjjh49yrZt264YCkRERESupKCggIULFzJw4EAaN27M008/zYEDB6hVqxYPPfQQW7Zs4fPPP+c3v/lNhQgEAJNjW9G8QRh+P9uQ4eeA5g3CmBTbyp7GpFywdUtBTk4Oe/bs8d7ev38/W7ZsoW7dujRu3JiePXvyhz/8gWrVqtGkSRNWr15NWloaL730EgC1atUiMTGRSZMmUa9ePerWrcvkyZNp3779FXcvEhEREbmU/fv3k5SUREpKCkeOHPHW77jjDgzD4J577rH1+MNfIiw4gHkP3s6Ly3aRvuMYp8/kUqN6NaJvjGBSbCudjrSKs/Vff+PGjfTu3dt7+8J+/vHx8aSmpjJnzhyeeOIJRo8ezcmTJ2nSpAl///vfeeCBB7w/8/LLLxMQEMDw4cPJzc3lzjvvJDU1VdcoEBERkWLJz89nwYIFuFyuInsf1K9fn/j4eJxOJ61bt7axw9ITFhzAlLi2PNmvJYsXL6F//x7aZUgAm0NBr169sK5wje2IiAhSUlKu+BghISFMmzaNadOmlXZ7IiIiUol9/fXXuFwupk+fzokTJ7z16OhoDMMgLi7ukscgVhbl9HhosYm2E4mIiEiVkZuby7vvvotpmnzyySfeemRkJAkJCSQmJtKsWTMbOxSxh0KBiIiIVHpffvklpmkyc+ZMsrKyAPDz86N///44nU7uvvtuAgL0sUiqLr36RUREpFI6ffo0b731FqZpsnHjRm+9adOmJCYmkpCQwPXXX29jhyLlh0KBiIiIVBqWZfH5559jmiZz5szhzJkzwPnztQ8ePBin00nfvn0vOuW5SFWnUCAiIiIV3smTJ5k5cyYul4uvvvrKW2/ZsiWGYTBu3DgaNGhgY4ci5ZtCgYiISCVlWRaOSnyKGcuyWLNmDaZp8u6775KXlwecPzPhsGHDMAyDO+64o1LPQKS0KBSIiIhUIjl5BbywbBcrdh7HXWgR6O+gb5twJleii1NlZmYyffp0XC4X33zzjbd+0003YRgGo0ePpk6dOjZ2KFLxVI6/DiIiIkJOXgFDX1/HnswcPD+5DFDa+gNk7D3BvAdvr7DBwOPxkJ6ejmmavP/++xQUFABQvXp1Ro4ciWEYdO7cWVsFRK5RxfzLICIiIhd5YdmuiwIBgMeCPZk5vLhsF1Pi2trT3DU6fPgwKSkpJCUlcfDgQW/91ltvxTAM7rvvPmrUqGFjhyKVg0KBiIhIJbFi5/GLAsEFHgvSdx6vEKGgsLCQhQsXkpKSwocffojH4wGgdu3ajB07FqfTyU033WRzlyKVi0KBiIhIJWBZFu7CyySC/1NQaJXrg4/37dvHm2++iWmanDp1ylvv0aMHhmFwzz33UK1aNRs7FKm8FApEREQqAYfDQaD/lT/sB/g7yl0gyMvLY8GCBbzx5n9YtfJjbz2gem06Rw/mtb9M5ub25X/rhkhFp1AgIiJSSfRtE07a+gOX3IXIzwHRbcJ939Rl7Ny5E9M0SUtL44cffvi/qoOQpr8mrEMsoS26cCwgkP9ZcYJ5LQsq7AHSIhWFfsNEREQqicmxrcjYe+Kig439HNC8QRiTYlvZ1xxw9uxZ3nnnHUzTZN26dd56jboN8G/dm+o3xRBQ67/BpSIfIC1S0SgUiIiIVBJhwQHMe/B2Xly2i/SdxykotAjwdxDdJpxJNl6n4IsvvsDlcjFr1iyysrIA8Pf3p3///hiGwXNbg/kuO/+SP1uRDpAWqcgUCkRERCqRsOAApsS1ZUpcW1sPKs7Ozuatt97CNE02bdrkrTdr1ozExEQSEhKIjIzEsiz+uvXjKzxS+T9AWqQyUCgQERGppHz9IdqyLD777DNM02TOnDmcPXsWgMDAQIYMGYJhGPTp0wc/P78iPVbEA6RFKhuFAhEREflFTp48yYwZM3C5XGzbts1bb9WqFYZhMG7cOK677rrL/nxFOkBapLJSKBAREZESsyyLVatW4XK5eO+998jLywMgJCSE4cOH43Q66d69e7G+4S/vB0iLVAUKBSIiIlJsx48fJzU1FZfLxZ49e7z1Dh06YBgGo0ePpnbt2iV6zCIHSO84xukzudSoXo3oGyNsPUBapCq5pt+yb7/9lgMHDnD27Fmuu+462rZtS3BwcGn3JiIiIuVAYWEhy5cvx+VysXDhQgoKCgAICwtj1KhRGIZBp06dftF+/xcOkH6yX0sWL15C//49CAwMLK2nICJXUexQcPDgQd544w3eeustvv32Wyzrv9v3goKCuOOOO/jNb37DPffcU+QAIhEREamYvv32W5KTk0lOTubQoUPeepcuXTAMg/vuu4+wsLBSX6+OKRbxvWJ9ev/9739P+/bt2b17N3/5y1/Yvn07WVlZ5Ofnc+zYMZYsWUL37t15+umnuemmm9iwYUNZ9y0iIiJlwO12s2DBAvr370/Tpk2ZOnUqhw4dok6dOvzud79j69atfPrppyQmJpZJIBARexRrS0FQUBB79+695JkDGjRoQJ8+fejTpw9TpkxhyZIlHDx4kM6dO5d6syIiIlI29u7di8vlIjU1lWPHjnnrPXv2xDAMhg4dSrVq1WzsUETKUrFCwT//+c9iP+Ddd999zc2IiIiI7+Tl5TF//nxM0+Tjj/97AbEGDRowfvx4EhMTadmypY0dioivlPhA49zcXCzLIjQ0FDh/rMH8+fNp06YNsbGxpd6giIiIlK4dO3ZgmiZpaWmcPHkSOH8RsZiYGAzDYODAgQQFBdncpYj4UolDwaBBgxg6dCgPPPAAP/74I126dCEwMJATJ07w0ksv8dvf/rYs+hQREZFf4MyZM7zzzjuYpklGRoa33qhRIyZMmEBCQgJNmza1r0ERsVWJTxO0efNm7rjjDgDeffddwsPDOXjwIGlpabz66qul3qCIiIhcu82bN/Pb3/6WyMhIEhISyMjIwN/fn0GDBvHBBx9w4MABnnnmGQUCkSquxFsKzp49S40aNQBYvnw5Q4cOxc/Pj9tuu42DBw+WeoMiIiJSMtnZ2cyePRvTNNm8ebO33qxZM5xOJ+PHjycyMtLGDkWkvClxKGjevDkLFixgyJAhLFu2jEcffRSAzMxMatasWeoNioiIyNVZlsX69esxTZO3336bs2fPAufPIDhkyBAMw6B37966lpCIXFKJQ8Gf//xnRo0axaOPPkqfPn3o2rUrcH6rQceOHUu9QREREbm8H374gRkzZmCaJjt27PDW27Rpg9PpZNy4cdSvX9/GDkWkIihxKBg2bBjdu3fn6NGjdOjQwVu/8847GTJkSKk2JyIiIhfzeDysWrUK0zSZN28e+fn5AFSrVo3hw4djGAbdunXDoUsDi0gxlTgUAERERBAREcG3336Lw+GgUaNG3HrrraXdm4iIiPzEsWPHSE1NxeVysXfvXm+9Y8eOGIbByJEjqV27tn0NikiFVeJQUFBQwDPPPMOrr75KTk4OAGFhYUycOJEpU6YQGBhY6k2KiIhUVYWFhSxbtgzTNFm0aBGFhYUA1KhRg1GjRmEYBp06dbK5SxGp6EocCh5++GHmz5/P888/7z2eYP369UydOpUTJ07wxhtvlHqTIiIiVc2hQ4dITk4mOTmZb7/91lvv2rUrTqeT4cOHExYWZmOHIlKZlDgUvPXWW8yZM4d+/fp5azfddBONGzdmxIgRCgUiIiLXyO12s2jRIlwuF0uXLsWyLADq1q3L2LFjcTqdtGvXzuYuRaQyKnEoCAkJueQFTpo2bapLoouIiFyDPXv24HK5SE1N5fjx49567969MQyDIUOGEBISYmOHIlLZlfhkxQ899BB//etfycvL89by8vL4+9//zsMPP1yix1qzZg0DBw4kMjISh8PBggULLlpm586dxMXFUatWLWrUqMFtt93GoUOHiqx74sSJ1K9fn+rVqxMXF8fhw4dL+rRERER86ty5c8yePZs+ffrQokUL/vGPf3D8+HHCw8N5/PHH+eabb/j4448ZOXKkAoGIlLlibSkYOnRokdsrVqygUaNG3lOSfvnll+Tn53PnnXeWaOVnzpyhQ4cOJCQkcM8991x0/969e+nevTuJiYk888wz1KpVi507dxb54/jII4+waNEi5syZQ7169Zg0aRIDBgxg06ZN+Pv7l6gfERGRsrZ9+3ZM02TGjBmcPHkSAIfDwV133YVhGAwYMEAn7RARnytWKKhVq1aR2z//AB8VFXVNK+/Xr1+RYxN+7qmnnuLuu+/m+eef99Z+9atfef8/KyuLpKQkZsyYQd++fQGYOXMmUVFRrFixgtjY2GvqS0REpDSdOXOGt99+G9M0Wb9+vbceFRXFhAkTmDBhAo0bN7axQxGp6ooVClJSUsq6j4t4PB4WL17MH//4R2JjY/niiy9o1qwZTzzxBIMHDwZg06ZNuN1uYmJivD8XGRlJu3btyMjIuGwoyMvLK7L7U3Z2NnD+AC+32112T+oyLqzTjnVXJZqz72jWvqE5+861znrz5s0kJSUxZ84cTp8+DYC/vz8DBgxgwoQJxMTEeLdq69/xPL2ufUNz9h07Z12SdTqsC6c2sJnD4WD+/PneD/zHjh2jYcOGhIaG8re//Y3evXuzdOlSnnzySVauXEnPnj2ZPXs2CQkJRT7gA8TExNCsWTPefPPNS65r6tSpPPPMMxfVZ8+eTWhoaKk/NxERqTrOnDnD6tWrWbFiBfv27fPWIyIiiI6Opk+fPtSpU8fGDkWkqjh79iyjRo0iKyuLmjVrXnHZYm0p6NixY7Evlb558+ZiLXc1Ho8HgEGDBvHoo48C8Otf/5qMjAzeeOMNevbsedmftSzriv0+8cQTPPbYY97b2dnZREVFERMTc9WBlQW32016ejrR0dHaj7QMac6+o1n7hubsO1ebtWVZrF+/nqSkJN59911yc3MBCAoKYsiQISQmJtKjRw/8/Ep8fo8qR69r39CcfcfOWV/YG6Y4ihUKLnx770v169cnICCAG2+8sUi9TZs2rF27Fjj/rUt+fj6nTp0q8q1LZmYm3bp1u+xjBwcHExwcfFE9MDDQ1l8Mu9dfVWjOvqNZ+4bm7Ds/n/WJEydIS0vD5XKxc+dOb/3GG2/EMAzGjh1LvXr17Gi1wtPr2jc0Z9+xY9YlWV+xQsGUKVOuuZlrFRQUROfOndm1a1eR+jfffEOTJk0A6NSpE4GBgaSnpzN8+HAAjh49yrZt24ocnCwiIlJaPB4PK1euxDRN5s+fT35+PgChoaHcd999GIbBbbfdVuwt7CIi5UGJL15WHFfbfeeCnJwc9uzZ4729f/9+tmzZQt26dWncuDF/+MMfuO++++jRo4f3mIJFixaxatUq4PxZkRITE5k0aRL16tWjbt26TJ48mfbt23vPRiQiIlIaTp48yT/+8Q9SUlKKHCtw8803YxgGI0eOvOhsfSIiFUWxQkGbNm14+umnGTZs2BWvWrx7925eeuklmjRpwv/8z/9c9XE3btxI7969vbcv7OcfHx9PamoqQ4YM4Y033uC5557jd7/7Ha1ateK9996je/fu3p95+eWXCQgIYPjw4eTm5nLnnXeSmpqqaxSIiMgvVlhYyNKlS3nzzTdZvHix93i3mjVrMnr0aJxOJzfffLPNXYqI/HLFCgX/7//9Px5//HEeeughYmJiuOWWW4iMjCQkJIRTp06xY8cO1q5dy44dO3j44Yd58MEHi7XyXr16cbWTH104f/PlhISEMG3aNKZNm1asdYqIiFzNwYMHSU5OJjk5mcOHD3vr3bp1wzAM7r33XqpXr25jhyIipatYoaBPnz5s2LCBjIwM5s6dy+zZszlw4AC5ubnUr1+fjh07Mm7cOMaMGUPt2rXLuGUREZHS53a7WbhwIaZpsnz5cu+XVvXq1WPMmDE0b96c+++/XwdlikilVKJjCrp163bFs/qIiIhUNLt378blcpGamkpmZqa33qdPHwzDYMiQIfj5+bFkyRIbuxQRKVtlcqCxiIhIeXbu3Dnee+89TNNk9erV3npERAQJCQkkJiZyww03eOu66quIVHYKBSIiUmVs27YN0zSZMWMGp06dAsDPz4+77roLwzDo37+/dg8SkSpJoUBERCq1nJwc5s6di2mafPbZZ95648aNSUxMJCEhgaioKBs7FBGxn0KBiIhUOpZlsXHjRlwuF7NnzyYnJweAgIAA4uLiMAyD6Ohonb5aROT/KBSIiEil8eOPPzJr1ixM0+TLL7/01ps3b45hGMTHxxMeHm5jhyIi5VOJQ4G/vz9Hjx6lQYMGReo//PADDRo0oLCwsNSaExERuRrLsli3bh2mafLOO++Qm5sLQHBwMPfccw+GYdCzZ08cDofNnYqIlF8lDgWXu9hYXl7eFa92LCIiUpq+//570tLScLlcfP311956u3btMAyDMWPGULduXRs7FBGpOIodCl599VUAHA4HLpeLsLAw732FhYWsWbOG1q1bl36HIiIi/8fj8fDxxx9jmibz58/3nio0NDSUESNGYBgGXbp00VYBEZESKnYoePnll4HzWwreeOONIgdnBQUF0bRpU954443S71BExEaWZekDZjlw5MgRUlJSSEpKYv/+/d76LbfcgmEYjBgxgpo1a9rYoYhIxVbsUHDhj3Dv3r2ZN28ederUKbOmRETslJNXwAvLdrFi53HchRaB/g76tglncmwrwoJ1fgZfKSgoYOnSpZimyeLFi73HrNWqVYvRo0djGAa//vWv7W1SRKSSKPG728qVK8uiDxGRciEnr4Chr69jT2YOnp8cQpW2/gAZe08w78HbFQzK2IEDB0hOTiY5OZnvvvvOW+/evTuGYTBs2DBCQ0Nt7FBEpPIp8TtbYWEhqampfPTRR2RmZuLxeIrc//HHH5dacyIivvbCsl0XBQIAjwV7MnN4cdkupsS1tae5Siw/P5+FCxdimibp6enek1rUq1eP+Ph4nE4nbdq0sblLEZHKq8Sh4Pe//z2pqan079+fdu3aaV9bEalUVuw8flEguMBjQfrO4woFpWjXrl0kJSWRmprK999/76337dsXp9PJ4MGDCQ4OtrFDEZGqocShYM6cObz99tvcfffdZdGPiIhtLMvCXXiZRPB/CgotHXz8C+Xm5vLee+9hmiZr1qzx1hs2bEhCQgKJiYn86le/srFDEZGqp8ShICgoiObNm5dFLyIitnI4HAT6X/nDfoC/Q4HgGm3duhXTNJk5cyY//vgjAH5+fvTr1w/DMOjfvz8BATpeQ0TEDiX+6ztp0iT+9a9/8dprr+mNUUQqnb5twklbf+CSuxD5OSC6Tbjvm6rAcnJymDNnDqZp8vnnn3vrTZo0ITExkYSEBBo1amRjhyIiAsUMBUOHDi1y++OPP+bDDz+kbdu2BAYGFrlv3rx5pdediIiPTY5tRcbeExcdbOzngOYNwpgU28q+5ioIy7LYsGEDpmkyZ84ccnJyAAgICGDQoEEYhkHfvn2LXO9GRETsVaxQUKtWrSK3hwwZUibNiIjYLSw4gHkP3s6Ly3aRvvM4BYUWAf4OotuEM0nXKbiiU6dOMWvWLEzTZOvWrd56y5YtcTqdxMfH06BBAxs7FBGRyynWu1tKSkpZ9yEiUm6EBQcwJa4tU+La6qDiq7Asi08++QSXy8U777zDuXPnAAgODubee+/F6XTSo0cPzVBEpJzTV14iIlegD7OX9v333zN9+nRcLhe7du3y1tu3b49hGIwZM0ZXvhcRqUBKHAo6dux4yTdJh8NBSEgIzZs3Z/z48fTu3btUGhQRkfLB4/Hw0UcfYZomCxYswO12A1C9enVGjhyJ0+nk1ltvVZASEamA/Er6A3fddRf79u2jevXq9O7dm169ehEWFsbevXvp3LkzR48epW/fvrz//vtl0a+IiPjYd999x9/+9jduuOEGYmJieOedd3C73XTu3Jn//Oc/HD16FNM06dKliwKBiEgFVeItBSdOnGDSpEk8/fTTRep/+9vfOHjwIMuXL2fKlCn89a9/ZdCgQaXWqIiI+E5BQQEffvghpmmyePFiPB4PcP7EE2PGjMEwDDp06GBzlyIiUlpKHArefvttNm3adFF9xIgRdOrUCdM0GTlyJC+99FKpNCgiIr6zf/9+kpKSSElJ4ciRI976HXfcgWEYDBs2jGrVqtnYoYiIlIUSh4KQkBAyMjIuuqpxRkYGISEhwPn9ToODg0unQxERKVP5+fksWLAAl8tFenq6t16/fn3i4+NxOp20bt3axg5FRKSslTgUTJw4kQceeIBNmzbRuXNnHA4Hn3/+OS6XiyeffBKAZcuW0bFjx1JvVkRESs/XX3+Ny+Vi+vTpnDhxwluPjo7GMAzi4uL0BY+ISBVR4lDwpz/9iWbNmvHaa68xY8YMAFq1aoVpmowaNQqABx54gN/+9rel26mIiPxiubm5vPvuu5imySeffOKtR0ZGkpCQQGJiIs2aNbOxQxERscM1Xadg9OjRjB49+rL3a39TEZHy5csvv8Q0TWbOnElWVhYAfn5+9O/fH6fTyd13301AgC5dIyJSVekdQESkkjp9+jRz5szBNE02bNjgrTdt2pTExEQSEhK4/vrrbeywYrEsuzsQESk7xQoFdevW5ZtvvqF+/frUqVPniuehPnnyZKk1JyIiJWNZFp9//jmmaTJnzhzOnDkDQGBgIIMHD8YwDO688078/Ep8mZoqKSevgBeW7SJ9xzFyzvjzz6/XEH1jBJNjWxEWrO/VRKTyKNZftJdffpkaNWoA8Morr5RlPyIicg1OnTrFzJkzMU2Tr776yltv1aoVTqeTcePG0aBBAxs7rHhy8goY+vo69mTm4LEAHGT9eI609QfI2HuCeQ/ermAgIpVGsf6axcfHX/L/RUTEPpZlsW3bNubOncu8efPIy8sDzp86+t5778UwDLp3766rDF+jF5bt+kkg+C+PBXsyc3hx2S6mxLW1pzkRkVJ2TV9x7N27l5SUFPbu3cu//vUvGjRowNKlS4mKiqJtW/2BFBEpS5mZmUyfPh3TNNm9e7e3ftNNN2EYBqNHj6ZOnTo2dlg5rNh5/KJAcIHHgvSdxxUKRKTSKPFOpatXr6Z9+/Z89tlnzJs3j5ycHAC2bt3KlClTSr1BERE5f1HIZcuWce+999KoUSP++Mc/snv3bkJCQkhMTOTzzz9ny5YtPPzwwwoEpcCyLNyFVz6yuKDQwtLRxyJSSZR4S8H//M//8Le//Y3HHnvMe5wBQO/evfnXv/5Vqs2JiFR1hw8fJiUlhaSkJA4ePOit33rrrUyYMIFatWpxzz33EBgYaGOXlY/D4SDQ/8q7XQX4O7RrlohUGiUOBV999RWzZ8++qH7dddfxww8/lEpTIiJVWUFBAYsXL8Y0TT788EM8Hg8AtWvXZuzYsTidTm666SbcbjdLliyxudvKq2+bcNLWH7jkLkR+DohuE+77pkREykiJdx+qXbs2R48evaj+xRdflPh812vWrGHgwIFERkbicDhYsGDBZZe9//77cTgcF539KC8vj4kTJ1K/fn2qV69OXFwchw8fLlEfIiLlwb59+3jqqado3LgxgwcPZvHixXg8Hnr06MGMGTM4cuQIr776KjfddJPdrVYJk2Nb0bxBGH4/2xjg54DmDcKYFNvKnsZERMpAiUPBqFGjePzxxzl27BgOhwOPx8O6deuYPHky48aNK9FjnTlzhg4dOvDaa69dcbkFCxbw2WefERkZedF9jzzyCPPnz2fOnDmsXbuWnJwcBgwYQGFhYYl6ERGxQ15eHnPnzqVv377ccMMNPPvssxw9epTrrruOyZMn8/XXX7N69WrGjBmjq8X7WFhwAPMevJ34rk1pVDuEWoEWjWqHEN+1qU5HKiKVTon/ov39739n/PjxXH/99ViWxY033khhYSGjRo3iT3/6U4keq1+/fvTr1++Ky3z33Xc8/PDDLFu2jP79+xe5Lysri6SkJGbMmEHfvn0BmDlzJlFRUaxYsYLY2NiSPTkRER/ZuXMnLpeLtLQ0Tpw4AZzfjz06OhrDMIiLiyMoKMjmLiUsOIApcW15sl9LFi9eQv/+PXT8hohUSsUOBXv27KF58+YEBgYya9Ys/vKXv/DFF1/g8Xjo2LEjLVq0KPXmPB4PY8eO5Q9/+MMlT3W6adMm3G43MTEx3lpkZCTt2rUjIyPjsqEgLy/Pez5vgOzsbADcbjdut7uUn8XVXVinHeuuSjRn39GsL+3s2bO8++67pKSksG7dOm/9+uuvJz4+nvHjx9O0aVNv/Wrz05x9x+1243Bo1r6g17VvaM6+Y+esS7LOYoeCli1bcv3119O7d2/69OlD7969GTZs2DU1WFz/+Mc/CAgI4He/+90l7z927BhBQUEXnX4vPDycY8eOXfZxn3vuOZ555pmL6suXLyc0NPSXNf0LpKen27buqkRz9h3N+rx9+/aRnp7O6tWrOXv2LAB+fn7ccsstREdHc/PNN+Pv78+OHTvYsWNHiR9fc/Ydzdp3NGvf0Jx9x45ZX3jPKY5ih4LVq1ezevVqVq1axUMPPcS5c+do3LixNyD07t27xAcaX8mmTZv417/+xebNm0t8yjfLsq74M0888QSPPfaY93Z2djZRUVHExMRQs2bNa+75WrndbtLT04mOjtZm6TKkOfuOZn3+78rcuXNJTk5m06ZN3nqzZs1ISEhg3LhxlzxOqiQ0Z9/RrH1Hs/YNzdl37Jz1hb1hiqPYoeCOO+7gjjvu4E9/+hNut5v169ezatUqVq1axVtvvUVeXh7Nmzdn165d19T0z33yySdkZmbSuHFjb62wsJBJkybxyiuvcODAASIiIsjPz+fUqVNFthZkZmbSrVu3yz52cHAwwcHBF9UDAwNt/cWwe/1VhebsO1Vt1pZl8dlnn2GaJnPnzuXMmTPA+TkMGTIEwzDo06cPfn4lPsfDFVW1OdtJs/Ydzdo3NGffsWPWJVnfNZ06ITAwkB49etC5c2e6du3KsmXLME2TPXv2XMvDXdLYsWO9Bw9fEBsby9ixY0lISACgU6dOBAYGkp6ezvDhwwE4evQo27Zt4/nnny+1XkREruTkyZPMnDkT0zTZtm2bt966dWucTifjxo3juuuus7FDERGRKytRKDh37hwZGRmsXLmSVatWsWHDBpo1a0bPnj3597//Tc+ePUu08pycnCJBYv/+/WzZsoW6devSuHFj6tWrV2T5wMBAIiIiaNXq/Lmha9WqRWJiIpMmTaJevXrUrVuXyZMn0759+4sChYhIabIsi9WrV2OaJu+995735AUhISEMHz4cwzC4/fbbdcVbERGpEIodCnr27MmGDRu44YYb6NGjBxMnTqRnz56Eh1/7FR03btxI7969vbcv7OcfHx9PampqsR7j5ZdfJiAggOHDh5Obm8udd95Jamoq/v7+19yXiMjlHD9+nOnTp+Nyudi9e7e33qFDBwzDYPTo0dSuXdu+BkVERK5BsUNBRkYGDRs2pHfv3vTq1YsePXpQv379X7TyXr16YVmXuH78ZRw4cOCiWkhICNOmTWPatGm/qBcRkcspLCwkPT0d0zRZuHAhBQUFAISFhTFq1CgMw6BTp07aKiAiIhVWsUPBjz/+yCeffMKqVav4xz/+wciRI2nZsiU9e/akV69e9OzZU/vMikilcvjwYZKTk0lKSuLQoUPeepcuXTAMg/vuu4+wsDAbOxQRESkdxQ4F1atX56677uKuu+4C4PTp06xdu5aVK1fy/PPPM3r0aFq0aFHkIDsRkYrG7XazePFiTNNk6dKleDweAOrUqcPYsWNxOp20b9/e5i5FRERK1zWdfQjOh4S6detSt25d6tSpQ0BAADt37izN3kREfGbv3r0kJSWRkpJS5OKHvXr1wul0MnToUKpVq2ZjhyIiImWn2KHA4/GwceNGVq1axcqVK1m3bh1nzpzxXuX4//2//1fkoGERkfIuLy+P+fPnY5omH3/8sbfeoEEDxo8fT2JiIi1btrSxQxEREd8odiioXbs2Z86coWHDhvTq1YuXXnqJ3r17c8MNN5RlfyIipW7Hjh24XC7S0tL44YcfAHA4HMTExGAYBgMHDiQoKMjmLkVERHyn2KHgn//8J71799a3ZiJSIZ09e5a3334b0zTJyMjw1q+//nomTJjAhAkTaNq0qX0NioiI2KjYoeD+++8vyz5ERMrEF198gWmazJo1i+zsbAD8/f0ZMGAAhmFw11136bomIiJS5V3zgcYiIuVVdnY2s2fPxjRNNm/e7K3/6le/wul0Eh8fT2RkpI0dioiIlC8KBSJSKViWxaeffoppmsydO5ezZ88CEBQUxJAhQzAMg969e+Pn52dzpyIiIuWPQoGIVGg//PADM2bMwOVysX37dm+9TZs2GIbB2LFjf/HV10VERCo7hQIRqXAsy2LVqlWYpsm8efPIy8sDoFq1agwfPhzDMOjWrRsOh8PmTkVERCoGhQIRqTCOHTvG9OnTcblc7Nmzx1vv2LEjhmEwcuRIateubV+DIiIiFZRCgYiUa4WFhSxfvhzTNFm0aBEFBQUA1KhRg1GjRmEYBp06dbK5SxERkYpNoUBEyqVDhw6RkpJCcnIyhw4d8ta7du2K0+lk+PDhhIWF2dihiIhI5aFQICLlhtvt5oMPPsA0TZYuXYplWQDUqVOHcePG4XQ6adeunc1dioiIVD4KBSJiuz179uByuUhNTeX48ePeeq9evTAMg6FDhxISEmJjhyIiIpWbQoGI2OLcuXPMnz8f0zRZuXKltx4eHs748eNJTEykRYsWNnYoIiJSdSgUiIhPbd++HZfLRVpaGidPngTA4XAQGxuLYRgMHDiQwMBAm7sUERGpWhQKRKTMnTlzhrfffhvTNFm/fr233qhRIxITE5kwYQKNGze2sUMREZGqTaFARMrM5s2bSU1NZfbs2WRnZwPg7+/PwIEDMQyD2NhY/P39be5SREREFApEyjnLsirUlXmzsrKYMWMGL7/8Mvv27fPWb7jhBpxOJ/Hx8TRs2NDGDkVEROTnFApEyqGcvAJeWLaLFTuP4y60CPR30LdNOJNjWxEWXP5+bS3LYv369Zimydtvv83Zs2cBCAoKYujQoRiGQa9evfDz87O5UxEREbmU8vfpQqSKy8krYOjr69iTmYPH+m89bf0BMvaeYN6Dt5ebYPDDDz+QlpaGy+Vix44d3nqbNm3o2rUrf//734mIiLCxQxERESkOfW0nUs68sGzXRYEAwGPBnswcXly2y57GLvTh8fDxxx8zcuRIIiMjeeyxx9ixYwfVqlVj/PjxrFu3ji1bthAXF0e9evVs7VVERESKp3x83SgiXit2Hr8oEFzgsSB953GmxLX1bVPA0aNHSU1NJSkpib1793rrN998M4ZhMHLkSGrVqgWcvzKxiIiIVBwKBSLliGVZuAsvkwj+T0Gh5bODjwsLC1m2bBmmabJo0SIKCwsBqFGjBqNHj8YwDG6++eYy70NERETKlkKBSDnicDgI9L/yh/0Af0eZB4JDhw6RlJREcnIyhw8f9ta7deuGYRjce++9VK9evUx7EBEREd9RKBApZ/q2CSdt/YFL7kLk54DoNuFlsl63282iRYswTZNly5ZhWecbqFu3LuPGjcPpdNK2re93WxIREZGyp1AgUs5Mjm1Fxt4TFx1s7OeA5g3CmBTbqlTXt3v3blwuF6mpqWRmZnrrffr0wel0MmTIEEJCQkp1nSIiIlK+KBSIlDNhwQHMe/B2Xly2i/SdxykotAjwdxDdJpxJpXSdgnPnzjFv3jxM02TVqlXeenh4OAkJCSQmJtK8efNfvB4RERGpGBQKRMqhsOAApsS1ZUpc21I9qHjbtm2YpsmMGTM4deoUcP44hn79+uF0OhkwYACBgYGlsi4RERGpOBQKRMq5XxoIzpw5w9y5czFNk08//dRbj4qKIjExkQkTJhAVFfVL2xQREZEKTKFApBKyLItNmzZhmiZvvfUWp0+fBiAgIICBAwdiGAYxMTH4+/vb3KmIiIiUBwoFIpVIVlYWs2bNwjRNtmzZ4q03b94cp9NJfHw8ERER9jUoIiIi5ZJCgUgFZ1kW69atw+Vy8fbbb5ObmwtAcHAw99xzD06nk549e+Ln52dzpyIiIlJeKRSIVFAnTpwgLS0Nl8vFzp07vfW2bdtiGAZjxoyhXr16NnYoIiIiFYVCgUgF4vF4WLlyJaZpMn/+fPLz8wEIDQ1lxIgROJ1ObrvttjK/4rGIiIhULgoFUqVYl7hKcEVw9OhRUlJSSEpKYt++fd56p06dMAyDkSNHUrNmTRs7FBERkYrM1p2M16xZw8CBA4mMjMThcLBgwQLvfW63m8cff5z27dtTvXp1IiMjGTduHEeOHCnyGHl5eUycOJH69etTvXp14uLiOHz4sI+fiZRnOXkFTF24nV4vrmHKJn96vbiGqQu3k5NXYHdrV1RYWMjixYsZPHgwUVFRPPXUU+zbt4+aNWvy29/+ls2bN7Nx40buv/9+BQIRERH5RWwNBWfOnKFDhw689tprF9139uxZNm/ezNNPP83mzZuZN28e33zzDXFxcUWWe+SRR5g/fz5z5sxh7dq15OTkMGDAAAoLC331NKQcy8krYOjr60hbf4DvfjxHltvBdz+eI239AYa+vq5cBoODBw/y5z//mSZNmjBgwADef/99CgsLuf3220lNTeXo0aO8/vrrdOzY0e5WKx2rom5KEhER+YVs3X2oX79+9OvX75L31apVi/T09CK1adOmceutt3Lo0CEaN25MVlYWSUlJzJgxg759+wIwc+ZMoqKiWLFiBbGxsWX+HKR8e2HZLvZk5uD52Wc9jwV7MnN4cdkupsS1tae5n8jPz2fRokWYpsny5cu9H07r1avHuHHjcDqd3HjjjTZ3WTnl5BXwwrJdrNh5HHehRaC/g75twpkc24qwYO1hKSIiVUOFesfLysrC4XBQu3ZtADZt2oTb7SYmJsa7TGRkJO3atSMjI+OyoSAvL4+8vDzv7ezsbOD8Lktut7vsnsBlXFinHeuu7NJ3HLsoEFzgsc7f/2S/lr5t6ie++eYbkpOTmTlzJpmZmd56nz59mDBhAoMGDSI4OBioWK+PivKazskrYPh/PmPv92eKvE7S1h9g3Z7vefs3Xcp1MKgoc64MNGvf0ax9Q3P2HTtnXZJ1Oqxysr3c4XAwf/58Bg8efMn7z507R/fu3WndujUzZ84EYPbs2SQkJBT5gA8QExNDs2bNePPNNy/5WFOnTuWZZ565qD579mxCQ0N/2RORcsOyYMomf7Lclz8TT61Ai2c6FeLLk/Xk5eXx6aefsnz5crZv3+6t16lThz59+tC3b18aNmzou4aqsPf2+/HJMQcWF78AHFjcEWFxTzOPDZ2JiIj8cmfPnmXUqFFkZWVd9fjD8vsV2E+43W5GjBiBx+Ph9ddfv+rylmVd8ZSMTzzxBI899pj3dnZ2NlFRUcTExNhywKbb7SY9PZ3o6GgCAwN9vv7K7J9fryHrx3OXvT+sejX69+/hk16++uorkpOTmT17NqdOnQLAz8+Pu+66iwkTJtCvX79K8+9fUV7Tz7+4BotLvz4sHOw7V4277/bN6+NaVJQ5Vwaate9o1r6hOfuOnbO+sDdMcZT7UOB2uxk+fDj79+/n448/LvKhPSIigvz8fE6dOkWdOnW89czMTLp163bZxwwODvbukvFTgYGBtv5i2L3+yij6xgjS1h+45C5Efg6IuTGiTGeek5PD3LlzMU2Tzz77zFtv3LgxiYmJJCQkEBUVVWbrt1t5fk1blkXBVTYCFHggICCg3F/3oTzPubLRrH1Hs/YNzdl37Jh1SdZn69mHruZCINi9ezcrVqy46OqsnTp1IjAwsMgByUePHmXbtm1XDAVSdUyObUXzBmH4/ewznZ8DmjcIY1Jsq1Jfp2VZbNiwgd/85jc0bNgQp9PJZ599RkBAAPfccw8ffvgh+/bt489//nOlDgTlncPhIND/yh/2A/wd5T4QiIiIlAZbtxTk5OSwZ88e7+39+/ezZcsW6tatS2RkJMOGDWPz5s188MEHFBYWcuzYMQDq1q1LUFAQtWrVIjExkUmTJlGvXj3q1q3L5MmTad++vfdsRFK1hQUHMO/B23lx2S7Sdxzj9JlcalSvRvSNEUwq5bPL/Pjjj8yaNQvTNPnyyy+99RYtWuB0OomPjyc8PLzU1ie/XN824VfckhTdRv9eIiJSNdgaCjZu3Ejv3r29ty/s5x8fH8/UqVNZuHAhAL/+9a+L/NzKlSvp1asXAC+//DIBAQEMHz6c3Nxc7rzzTlJTU/H39/fJc5DyLyw4gClxbXmyX0sWL15C//49Sm3znWVZrF27FtM0eeeddzh37vz+6cHBwQwbNgzDMOjRo4e+bS6nJse2ImPviYtOW1uWW5JERETKI1tDQa9eva54saDinBgpJCSEadOmMW3atNJsTSqp0vps/v3335OWlobL5eLrr7/21tu1a4dhGIwZM4a6deuWzsqkzBTZkrTzOAWFFgH+DqLbhJf6liQREZHyTO94IsXk8Xj46KOPcLlczJ8/33vu3+rVqzNixAgMw+DWW2/VVoEK5sKWpClxba965jIREZHKSqFA5CqOHDlCSkoKSUlJ7N+/31u/5ZZbMAyDESNG2HIqWyl9CgQiIlJVKRSIXEJBQQEffvghLpeLxYsXU1hYCECtWrUYM2YMTqfzomNdRERERCoqhQKRnzhw4ABJSUkkJydz5MgRb7179+4YhsGwYcN01WsRERGpdBQKpMrLz8/n/fffxzRNVqxY4T3AvV69esTHx+N0OmnTpo3NXYqIiIiUHYUCqbJ27dqFy+Vi+vTpfP/999563759MQyDQYMGXfLK1yIiIiKVjUKBVCl5eXnMnDmTlJQUPvnkE2+9YcOGJCQkkJiYyK9+9SsbOxQRERHxPYUCqRK2bt3Km2++yfTp0zlz5gwAfn5+3H333RiGwd13301AgH4dREREpGrSpyCptE6fPs2cOXNwuVx8/vnn3nqTJk1wOp2MHz+eRo0a2dihiIiISPmgUCCVimVZbNiwAdM0mTNnDjk5OQAEBgYycOBA2rdvzxNPPKFjBURERER+QqFAKoVTp04xa9YsTNNk69at3nrLli0xDINx48ZRp04dlixZgp+fn42dioiIiJQ/CgVSYVmWxSeffIJpmrz77rucO3cOgJCQEIYNG4ZhGNxxxx3eq9S63W472xUREREptxQKpMLJzMwkLS0Nl8vFrl27vPX27dtjGAZjxoyhTp06NnYoIiIiUrEoFEiF4PF4WLFiBaZp8v7773u/9a9evTojR47EMAw6d+7s3SogIiIiIsWnUCDl2nfffUdycjJJSUkcPHjQW+/cuTOGYTBixAhq1KhhY4ciIiIiFZ9CgZQ7BQUFLFmyBNM0WbJkCR6PB4DatWszZswYnE4nHTp0sLlLERERkcpDoUDKjf3795OUlERKSgpHjhzx1nv06IHT6WTYsGFUq1bNxg5FREREKieFArFVXl4e77//PqZpsmLFCm/9uuuuIz4+HqfTSatWrWzsUERERKTyUygQW3z99de4XC6mT5/OiRMnAHA4HERHR+N0Ohk0aBBBQUE2dykiIiJSNSgUiM+cPXuWd999F9M0Wbt2rbceGRnJhAkTmDBhAs2aNbOxQxEREZGqSaFAytyXX36JaZrMnDmTrKwsAPz8/Ojfvz+GYdCvXz8CAvRSFBEREbGLPolJmTh9+jRvvfUWpmmyceNGb71p06Y4nU7Gjx/P9ddfb2OHIiIiInKBQoGUGsuy+Oyzz3C5XMyZM4czZ84AEBgYyODBgzEMgzvvvBM/Pz+bOxURERGRn1IokF/s5MmTzJw5E9M02bZtm7feqlUrDMNg3LhxXHfddTZ2KCIiIiJXolAg18SyLFavXo3L5eLdd98lLy8PgJCQEIYPH47T6aR79+44HA6bOxURERGRq1EokBI5fvw406dPx+VysXv3bm+9Q4cOGIbB6NGjqV27tn0NioiIiEiJKRTIVRUWFrJixQpM0+T999+noKAAgLCwMEaOHIlhGNxyyy3aKiAiIiJSQSkUyGUdPnyY5ORkkpOTOXjwoLfepUsXDMPgvvvuIywszMYORURERKQ0KBRIEQUFBSxevBjTNPnwww/xeDwA1K5dm7Fjx2IYBu3bt7e5SxEREREpTQoFAsC+fftwuVykpqZy9OhRb71nz54YhsHQoUOpVq2ajR2KiIiISFlRKKjC8vLyWLBgAaZp8tFHH3nrDRo0YPz48SQmJtKyZUsbOxQRERERX1AoqIJ27tyJaZqkpaXxww8/AOBwOIiJicEwDAYOHEhQUJDNXYqIiIiIrygUVBFnz57lnXfewTRN1q1b5603atSICRMmkJCQQNOmTe1rUERERERso1BQyX3xxReYpsmsWbPIzs4GwN/fnwEDBmAYBnfddRf+/v42dykiIiIidlIoqISys7N56623ME2TTZs2eevNmjXD6XQyfvx4IiMjbexQRERERMoThYJKwrIsPv30U0zTZO7cuZw9exaAoKAghgwZgmEY9O7dGz8/P5s7FREREZHyRqGggjt58iQzZszANE22b9/urbdu3RrDMBg3bhz169e3sUMRERERKe9s/dp4zZo1DBw4kMjISBwOBwsWLChyv2VZTJ06lcjISKpVq0avXr2KfPCF86fVnDhxIvXr16d69erExcVx+PBhHz4L37Msi5UrVzJq1CgiIyN55JFH2L59O9WqVSM+Pp61a9eyY8cOHnvsMQUCEREREbkqW0PBmTNn6NChA6+99tol73/++ed56aWXeO2119iwYQMRERFER0dz+vRp7zKPPPII8+fPZ86cOaxdu5acnBwGDBhAYWGhr56Gzxw/fpx//OMftGzZkj59+vDWW2+Rl5dHx44def311zly5AipqancfvvtOBwOu9sVERERkQrC1t2H+vXrR79+/S55n2VZvPLKKzz11FMMHToUgOnTpxMeHs7s2bO5//77ycrKIikpiRkzZtC3b18AZs6cSVRUFCtWrCA2NtZnz6WsFBYWsnz5clwuFwsXLqSgoACAGjVqMGrUKAzDoFOnTjZ3KSIiIiIVWbk9pmD//v0cO3aMmJgYby04OJiePXuSkZHB/fffz6ZNm3C73UWWiYyMpF27dmRkZFToUHDo0CFSUlJITk7m0KFD3nrXrl1xOp0MHz6csLAwGzsUERERkcqi3IaCY8eOARAeHl6kHh4ezsGDB73LBAUFUadOnYuWufDzl5KXl0deXp739oXz97vdbtxud6n0XxIX1nn27FmWL19OcnIyy5Ytw7IsAOrUqcOYMWNISEigXbt2F/2cFM+FeWluZU+z9g3N2Xc0a9/RrH1Dc/YdO2ddknWW21Bwwc/3jbcs66r7y19tmeeee45nnnnmovry5csJDQ29tkZ/gaNHj7JixQoSEhI4deqUt96uXTuio6Pp2rUrQUFBHDp0qMhWA7k26enpdrdQZWjWvqE5+45m7TuatW9ozr5jx6wvnKK+OMptKIiIiADObw1o2LCht56ZmendehAREUF+fj6nTp0qsrUgMzOTbt26Xfaxn3jiCR577DHv7ezsbKKiooiJiaFmzZql/VSuaOrUqTz77LPe2+Hh4YwdO5aEhARatGjh014qO7fbTXp6OtHR0QQGBtrdTqWmWfuG5uw7mrXvaNa+oTn7jp2zvrA3THGU21DQrFkzIiIiSE9Pp2PHjgDk5+ezevVq/vGPfwDQqVMnAgMDSU9PZ/jw4cD5b923bdvG888/f9nHDg4OJjg4+KJ6YGCgz/+xunbtisPhoGPHjjz++OMMGTJEv5xlzI5/56pKs/YNzdl3NGvf0ax9Q3P2HTtmXZL12RoKcnJy2LNnj/f2/v372bJlC3Xr1qVx48Y88sgjPPvss7Ro0YIWLVrw7LPPEhoayqhRowCoVasWiYmJTJo0iXr16lG3bl0mT55M+/btvWcjKu/uuusuvvnmG7Zv387dd9+tX0wRERER8TlbQ8HGjRvp3bu39/aFXXri4+NJTU3lj3/8I7m5uTz44IOcOnWKLl26sHz5cmrUqOH9mZdffpmAgACGDx9Obm4ud955J6mpqfj7+/v8+VyLgIAAmjRpctFF2UREREREfMXWUNCrVy/vGXYuxeFwMHXqVKZOnXrZZUJCQpg2bRrTpk0rgw5FRERERCo/W69oLCIiIiIi9lMoKCeusMFERERERKRMlduzD1UFOXkFvLBsF+k7jpFzxp9/fr2G6BsjmBzbirBg/dOIiIiIiG/ok6dNcvIKGPr6OvZk5uCxABxk/XiOtPUHyNh7gnkP3q5gICIiIiI+od2HbPLCsl0/CQT/5bFgT2YOLy7bZU9jIiIiIlLlKBTYZMXO4xcFggs8FqTvPO7bhkRERESkylIosIFlWbgLr3xkcUGhdcXTtYqIiIiIlBaFAhs4HA4C/R1XXCbA34HDceVlRERERERKg0KBTfq2CcfvMp/5/RwQ3Sbctw2JiIiISJWlUGCTybGtaN4g7KJg4OeA5g3CmBTbyp7GRERERKTKUSiwSVhwAPMevJ34rk1pVDuEWoEWjWqHEN+1qU5HKiIiIiI+pU+eNgoLDmBKXFue7NeSxYuX0L9/DwIDA+1uS0RERESqGG0pKCd0TLGIiIiI2EWhQERERESkilMoEBERERGp4hQKRERERESqOIUCEREREZEqTqFARERERKSKUygQEREREanidJ0CwLIsALKzs21Zv9vt5uzZs2RnZ+s6BWVIc/Ydzdo3NGff0ax9R7P2Dc3Zd+yc9YXPthc+616JQgFw+vRpAKKiomzuRERERESkdJ0+fZpatWpdcRmHVZzoUMl5PB6OHDlCjRo1cNhwFbHs7GyioqL49ttvqVmzps/XX1Vozr6jWfuG5uw7mrXvaNa+oTn7jp2ztiyL06dPExkZiZ/flY8a0JYCwM/Pj0aNGtndBjVr1tQvpg9ozr6jWfuG5uw7mrXvaNa+oTn7jl2zvtoWggt0oLGIiIiISBWnUCAiIiIiUsUpFJQDwcHBTJkyheDgYLtbqdQ0Z9/RrH1Dc/Ydzdp3NGvf0Jx9p6LMWgcai4iIiIhUcdpSICIiIiJSxSkUiIiIiIhUcQoFIiIiIiJVnEKBiIiIiEgVp1DgQ9999x1jxoyhXr16hIaG8utf/5pNmzZ577csi6lTpxIZGUm1atXo1asX27dvt7HjiulKc3a73Tz++OO0b9+e6tWrExkZybhx4zhy5IjNXVdMV3tN/9T999+Pw+HglVde8W2TlUBx5rxz507i4uKoVasWNWrU4LbbbuPQoUM2dVxxXW3WOTk5PPzwwzRq1Ihq1arRpk0b/v3vf9vYccXUtGlTHA7HRf899NBDgN4PS8uV5qz3w9J1tdf0T5XX90OFAh85deoUt99+O4GBgXz44Yfs2LGDF198kdq1a3uXef7553nppZd47bXX2LBhAxEREURHR3P69Gn7Gq9grjbns2fPsnnzZp5++mk2b97MvHnz+Oabb4iLi7O38QqoOK/pCxYsWMBnn31GZGSk7xut4Ioz571799K9e3dat27NqlWr+PLLL3n66acJCQmxr/EKqDizfvTRR1m6dCkzZ85k586dPProo0ycOJH333/fvsYroA0bNnD06FHvf+np6QDce++9gN4PS8uV5qz3w9J1tdf0BeX6/dASn3j88cet7t27X/Z+j8djRUREWP/7v//rrZ07d86qVauW9cYbb/iixUrhanO+lM8//9wCrIMHD5ZRV5VTcWd9+PBh6/rrr7e2bdtmNWnSxHr55ZfLvrlKpDhzvu+++6wxY8b4qKPKqzizbtu2rfWXv/ylSO3mm2+2/vSnP5Vla5Xe73//e+uGG26wPB6P3g/L0E/nfCl6Pyw9l5p1eX8/1JYCH1m4cCG33HIL9957Lw0aNKBjx46Ypum9f//+/Rw7doyYmBhvLTg4mJ49e5KRkWFHyxXS1eZ8KVlZWTgcjkt+wy2XV5xZezwexo4dyx/+8Afatm1rU6cV29Xm7PF4WLx4MS1btiQ2NpYGDRrQpUsXFixYYF/TFVRxXtPdu3dn4cKFfPfdd1iWxcqVK/nmm2+IjY21qeuKLz8/n5kzZzJhwgQcDofeD8vIz+d8KXo/LB2XmnWFeD+0O5VUFcHBwVZwcLD1xBNPWJs3b7beeOMNKyQkxJo+fbplWZa1bt06C7C+++67Ij9nGIYVExNjR8sV0tXm/HO5ublWp06drNGjR/u404qvOLN+9tlnrejoaO83JeXxm5Hy7mpzPnr0qAVYoaGh1ksvvWR98cUX1nPPPWc5HA5r1apVNndfsRTnNZ2Xl2eNGzfOAqyAgAArKCjISktLs7Hrim/u3LmWv7+/9/1P74dl4+dz/jm9H5aeS826IrwfKhT4SGBgoNW1a9citYkTJ1q33XabZVn//SN45MiRIss4nU4rNjbWZ31WdFeb80/l5+dbgwYNsjp27GhlZWX5qsVK42qz3rhxoxUeHl7kj2J5/CNY3l1tzt99950FWCNHjiyyzMCBA60RI0b4rM/KoDh/P/75z39aLVu2tBYuXGh9+eWX1rRp06ywsDArPT3d1+1WGjExMdaAAQO8t/V+WDZ+Puef0vth6fr5rCvK+6F2H/KRhg0bcuONNxaptWnTxnt2kIiICACOHTtWZJnMzEzCw8N902QlcLU5X+B2uxk+fDj79+8nPT2dmjVr+rLNSuFqs/7kk0/IzMykcePGBAQEEBAQwMGDB5k0aRJNmza1oeOK6Wpzrl+/PgEBAcV63cuVXW3Wubm5PPnkk7z00ksMHDiQm266iYcffpj77ruPF154wY6WK7yDBw+yYsUKnE6nt6b3w9J3qTlfoPfD0nWpWVeU90OFAh+5/fbb2bVrV5HaN998Q5MmTQBo1qwZERER3qPV4fw+aatXr6Zbt24+7bUiu9qc4b9/AHfv3s2KFSuoV6+er9usFK4267Fjx7J161a2bNni/S8yMpI//OEPLFu2zI6WK6SrzTkoKIjOnTtf9XUvV3e1WbvdbtxuN35+Rd86/f398Xg8PuuzMklJSaFBgwb079/fW9P7Yem71JxB74dl4VKzrjDvh3ZvqqgqPv/8cysgIMD6+9//bu3evduaNWuWFRoaas2cOdO7zP/+7/9atWrVsubNm2d99dVX1siRI62GDRta2dnZNnZesVxtzm6324qLi7MaNWpkbdmyxTp69Kj3v7y8PJu7r1iK85r+ufK4ubS8K86c582bZwUGBlr/+c9/rN27d1vTpk2z/P39rU8++cTGziue4sy6Z8+eVtu2ba2VK1da+/bts1JSUqyQkBDr9ddft7HziqmwsNBq3Lix9fjjj190n94PS8/l5qz3w9J3pdf0z5XH90OFAh9atGiR1a5dOys4ONhq3bq19Z///KfI/R6Px5oyZYoVERFhBQcHWz169LC++uorm7qtuK405/3791vAJf9buXKlfU1XUFd7Tf9cefwjWBEUZ85JSUlW8+bNrZCQEKtDhw7WggULbOi04rvarI8ePWqNHz/eioyMtEJCQqxWrVpZL7744mVP8SiXt2zZMguwdu3addF9ej8sPZebs94PS9+VXtM/Vx7fDx2WZVk2bKAQEREREZFyQscUiIiIiIhUcQoFIiIiIiJVnEKBiIiIiEgVp1AgIiIiIlLFKRSIiIiIiFRxCgUiIiIiIlWcQoGIiIiISBWnUCAiIiXStGlTXnnllWIvf+DAARwOB1u2bCm1Hnr06MHs2bO9tx0OBwsWLCj2z3/wwQd07NgRj8dTaj2JiFRkCgUiIlXA+PHjGTx48EX1VatW4XA4+PHHH4v9WBs2bOA3v/lN6TUHpKamUrt27WIt+8EHH3Ds2DFGjBjhrR09epR+/foVe30DBgzA4XAUCRYiIlWZQoGIiJTIddddR2hoqG3rf/XVV0lISMDP779vYREREQQHB5focRISEpg2bVpptyciUiEpFIiISBEZGRn06NGDatWqERUVxe9+9zvOnDnjvf/nuw99/fXXdO/enZCQEG688UZWrFhxyd159u3bR+/evQkNDaVDhw6sX78eOL+1IiEhgaysLBwOBw6Hg6lTp16ytxMnTrBixQri4uKK1H+6vgu7K82bN++S67sgLi6Ozz//nH379l3boEREKhGFAhER8frqq6+IjY1l6NChbN26lblz57J27VoefvjhSy7v8XgYPHgwoaGhfPbZZ/znP//hqaeeuuSyTz31FJMnT2bLli20bNmSkSNHUlBQQLdu3XjllVeoWbMmR48e5ejRo0yePPmSj7F27VpCQ0Np06bNVZ/L5dZ3QZMmTWjQoAGffPJJMSYjIlK5BdjdgIiI+MYHH3xAWFhYkVphYWGR2//85z8ZNWoUjzzyCAAtWrTg1VdfpWfPnvz73/8mJCSkyPLLly9n7969rFq1ioiICAD+/ve/Ex0dfdH6J0+eTP/+/QF45plnaNu2LXv27KF169bUqlULh8PhfYzLOXDgAOHh4UV2HbqcK63vguuvv54DBw5c9bFERCo7bSkQEakievfuzZYtW4r853K5iiyzadMmUlNTCQsL8/4XGxuLx+Nh//79Fz3mrl27iIqKKvJh/tZbb73k+m+66Sbv/zds2BCAzMzMEj2H3Nzci4LJ5RRnfdWqVePs2bMl6kFEpDLSlgIRkSqievXqNG/evEjt8OHDRW57PB7uv/9+fve73130840bN76oZlkWDoejWOsPDAz0/v+FnynpKUHr16/PqVOnSm19J0+e5LrrritRDyIilZFCgYiIeN18881s3779ovBwOa1bt+bQoUMcP36c8PBw4PwpS0sqKCjool2ZLqVjx44cO3aMU6dOUadOnRKv56fOnTvH3r176dix4y96HBGRykC7D4mIiNfjjz/O+vXreeihh9iyZQu7d+9m4cKFTJw48ZLLR0dHc8MNNxAfH8/WrVtZt26d90Dj4m5BgPNnNMrJyeGjjz7ixIkTl92lp2PHjlx33XWsW7eu5E/uZz799FOCg4Pp2rXrL34sEZGKTqFARES8brrpJlavXs3u3bu544476NixI08//bR3n/yf8/f3Z8GCBeTk5NC5c2ecTid/+tOfAIq97z9At27deOCBB7jvvvu47rrreP755y+7vgkTJjBr1qySP7mfeeuttxg9erSt11wQESkvHJZlWXY3ISIilce6devo3r07e/bs4YYbbij1xz9+/Dht27Zl06ZNNGnS5Joe4/vvv6d169Zs3LiRZs2alXKHIiIVj0KBiIj8IvPnzycsLIwWLVqwZ88efv/731OnTh3Wrl1bZut8//33qVu3Lnfcccc1/fznn3/O/v37ue+++0q5MxGRikmhQEREfpG0tDT++te/8u2331K/fn369u3Liy++SL169exuTUREikmhQERERESkitOBxiIiIiIiVZxCgYiIiIhIFadQICIiIiJSxSkUiIiIiIhUcQoFIiIiIiJVnEKBiIiIiEgVp1AgIiIiIlLFKRSIiIiIiFRxCgUiIiIiIlXc/werVn4Xv4CzugAAAABJRU5ErkJggg==\n", "text/plain": [ "
| 23,932
| 37,788
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.25
| 3
|
CC-MAIN-2022-33
|
latest
|
en
| 0.764732
|
https://ru.scribd.com/document/336143883/MELJUN-CORTES-OPERATING-SYSTEM-Lesson3-updates
| 1,563,672,406,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00027.warc.gz
| 522,745,270
| 63,248
|
You are on page 1of 7
# Operating System
TOPIC OBJECTIVE:
## No. Of Hours: 3HRS
At the end of the topic discussion the students should be able to:
1. Compute/Solve the Efficiency and Reliability of CPU.
--
CPU UTILIZATION
It determines how efficient and reliable a computer system. To determine the efficiency of the
CPU, it must be equal to 50%, the card reader must be equal to 25% and line printer must be equal to
25%. If the result of the computation will not meet the required percentage then, it is less efficient and
reliable.
FORMULAS:
CR Time
= number of cards
CR speed
LP Time
= number of lines
LP speed
CPU Time
= number of instruction * CPU speed
TOTAL Time = CPU Time + CR Time + LP Time
To get the efficiency used the following formula:
% Efficiency of Line Printer
% Efficiency of CPU
_______________________
## = (CR Time / Total Time) * 100
= (LP Time / Total Time) * 100
= (CPU Time / Total Time) * 100
100 %
Example:
Given the following characteristics:
1. % Eff. Of CR
= 48.78%
TT
= 0.41 sec
CPU Time
= 10 ms
No. of instructions = 10,000
Required:
a. CPU speed in s
b. CR Time
c. %Eff.of LP
d. % Eff. Of CPU
2. % Eff. Of CPU
TT
LP Speed
= 19.85%
= .001525 hr
= 30 lines / min
Required:
a. LP Time
a. CPU Time
b. CR Time
e. %Eff.of CR
## Professor: MELJUN P. CORTES
e. %Eff. Of TT
c.. %Eff. Of LP
Operating System
SOLUTIONS TO PROBLEM GIVEN ABOVE:
1. a) CPU speed
= ___CPU Time__
no. of instruction
CPU speed
= 10 ms__________
10,000 instructions
10ms * 1 x 10-3 secs.
1 ms
10,000 instructions
= .01
10,000 seconds
= 1 x 10-6 sec
= 1s
b) CR Time
CR Time
## = Total Time * %Eff.of CR
100
= .41 * 48.78
100
= .199 sec or .2sec
= 48.78%
d) %Eff. Of CPU
= 100% - 97.56%
= 2.44%
## Professor: MELJUN P. CORTES
Operating System
III. PROCESS MANAGEMENT
PROCCES MANAGEMENT allows time sharing of CPU in allocating different processes to the
processor.
PCB is a data structure, which contains important information about certain process. It is also called a
task control block. The information or content of PCB are process state, program counter, CPU
Registers, CPU Scheduling information, Memory-management Information, Accounting Information
and I/O information.
ILLUSTRATION of PCB
Pointer
Process
State
Process number
Program Counter
Registers
Memory limits
List of Open Files
:
Process State. The state may: be new, ready, running, and waiting, halted and so on.
Program Counter. The counter indicates the address of the next instruction to be
executed for this process.
CPU Registers. The registers vary in number and type, depending on the computer
architecture. They include accumulators, index registers, stack pointers, and generalpurpose registers, plus any condition-code information. Along with the program counter,
this state information must be saved when an interrupt occurs, to allow the process to
be continued correctly afterward.
CPU Scheduling Information. This information includes a process priority, pointers to
scheduling queues, and any other scheduling parameters.
Memory-Management information. This information may include such information as
the value of the base and limit registers, the page tables or the segment tables
depending on the memory system used by the operating system.
Accounting information. This information includes the amount of CPU and real time
used, time limits, account numbers, job or process numbers and so on.
I/O status information. The information includes the list of I/O devices (such as tape
drives) allocated to this process, a list of open files, and so on.
PROCESS The designer of Multics system was the first who introduced the word or term process in
the mid 60s. Since that time, process used somewhat interchangeably with task or job, has been
given a lot of definitions. Some of these are:
A program in execution.
An asynchronous activity
Operating System
## The animated spirit of a procedure in execution.
The locus of control of a procedure in execution.
That which is manifested by the existence of a process control block in the operating
system.
That entity to which processors are assigned.
The dispatch able unit.
A process is more than a program code (sometimes known as text section). It also includes the
current activity, as represented by the value of the program counter and the contents of the
processors registers. A process generally includes the process stack, containing temporary data such
as subroutines, parameters, return address, and temporary variables and data section containing
global variables.
A resource of Process includes CPU Time, memory, files, and I/O devices to complete the task.
These resources are either given to the process when it is created, or allocated to it while it is running.
PROCESS TRANSITION DIAGRAM
Run
Subm
it
Hold
y
Compl
ete
Wait
A process goes through a series of discrete process states. Various events can cause a
process to change states.
Submit state, a state where a process is being submitted to the system and the system will
respond to the request.
Eventually, when a job is submitted to the system or CPU, a corresponding process is created
and normally inserted at the back of the ready list. The assignment of the CPU to the first process on
the ready list is called dispatching, and is performed by a system hardware called dispatcher.
Hold state, a state where a process is being transformed into a machine readable form.
Ready state a state where a process submitted is ready for execution, but there are more
processes running than the processor available.
To prevent any process from monopolizing the system, the operating system sets a hardware
interrupting clock or interval timer (idle) to allow particular user to run for a specific time interval called
quantum time. The interrupting clock generates an interrupt, causing the operating system to regain
control. The operating system then makes the previously running process ready, and makes the first
process on the ready list running.
A running state a currently procedure of processing takes place were from the ready list
provided that the English like language used by a specific users the operating system handles the
operation to transformed the English like language into a machine readable form. After a thorough
procedure of transformation of language and when it is successfully done it is immediately scheduled
to the processor for processing.
A process is said to be running if it is currently has the CPU. A process is said to be ready if it
is used a CPU if one were available. A process is said to be blocked if it is waiting for some event to
happen before it can proceed.
## Professor: MELJUN P. CORTES
Operating System
A wait state was some of the processes are waiting for some event. Once, the CPU reset the
resource it means that it is ready to accept another process. Timer or interrupt clock queue the waiting
jobs to dispatcher or ready state to prepare for next execution.
Complete State the process is already finished and the resources used are reclaimed for the
next process in the queue.
Before a process will be submitted (next) to the system, TRAFFIC CONTROLLER will check
the Available resources. It will keep track the status of all hardware and software resources.
There are two types of scheduler in operating system that will manage the flow of the system.
SCHEDULER. A process migrates between the various scheduling queues throughout its lifetime. The
operating system must select, for scheduling purposes, processes from these queues in some
fashion. The selection process is carried out by the appropriate scheduler.
1. JOB SCHEDULER also known as Long Term Scheduler. It will select jobs being submitted to
the system and will put all the jobs selected in the physical memory of the system.
There are several criteria for job selection:
a. capacity of the job
b. priority of process
c. expected runtime
d. amount of input and output
e. special resources required
2. CPU SCHEDULER or also known as Short Term Scheduler. This scheduler will select process
in the memory and put it on time of the queue.
There are types of queue.
execution. The process that resides in main memory and are ready and waiting to
execute are kept on a list are called ready queue.
b. Device Queue will hold processes that are listed for a particular input output deices.
In CPU scheduler, there are different algorithms used for handling processes, which are
categorized into two:
1. Non Deterministic Scheduling Algorithms
2. Deterministic Scheduling Algorithms
Deterministic Scheduling Algorithms is an algorithm wherein all characteristics of a process
are known in advance or before execution. These characteristics are:
a. execution time (ti)
b. priority time (wi)
d. finishing time
There are three algorithms under this class.
1. Minimum Mean Response Time (MMRT) all processes are arranged according to increasing
execution time. Execution time is denoted as ti. In case of the same execution time, first come
first serve is supplied.
2. Minimum Weighted Response Time (MWRT) all processes are arranged according to
increasing execution time divided by the priority. Priority is denoted as wi.
## Professor: MELJUN P. CORTES
Operating System
3. Maximum/Minimum Lateness (MML) all processes are arranged according to increasing
the lower or smallest execution time will be loaded first.
Example: Considering the following execution time, deadline and priority time. Schedule the given
process and arranged them according to MMRT, MWRT and MML algorithm.
PROCESS
P1
P2
P3
Execution Time
10 ms
5ms
15 ms
Priority Time
2
1
3
12
6
16
SOLUTION:
MMRT ti
CPU
P2
5 ms
P1
P3
10 ms
15 ms
MWRT ti / wi
P1 10 / 2 = 5
P2 5 / 5 = 1
P3 15 / 3 = 5
CPU
P2
1ms
P1
5 ms
P3
5 ms
MML di
CPU
P2
6 ms
P1
12 ms
P3
16 ms
EXERCISES:
Schedule the following jobs using MMRT, MWRT and MML where:
A = 5, B= 3 C = 2.
PROCESS
J
U
L
Y
Execution Time
A*A
AB*2
B * A/ 3
A*C
Priority Time
C*C
BC
A*C
B*3
| 2,388
| 9,974
|
{"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}
| 3.71875
| 4
|
CC-MAIN-2019-30
|
latest
|
en
| 0.785987
|
http://www.brightstorm.com/qna/question/7918/
| 1,369,560,481,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368706794379/warc/CC-MAIN-20130516121954-00073-ip-10-60-113-184.ec2.internal.warc.gz
| 369,759,526
| 9,842
|
Quick Homework Help
# The results of an exam are distributed with a standard deviation of 6.54%. If 14% of the students scored below 55%, then the value of the mean correct to the nearest 10th of a percent is _____? ⚑ Flag
by Montanna002 at January 16, 2011
Montanna -You did not say, but assuming a standard normal distribution, the value of z corresponding with an area of 0.14 is equal to -1.0803. Now all you have to do is solve this equation for the mean:Hope this helped
Steve204 January 17, 2011
| 138
| 507
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.125
| 3
|
CC-MAIN-2013-20
|
latest
|
en
| 0.94826
|
http://en.wikipedia.org/wiki/Talk:Situation_calculus
| 1,416,838,919,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-49/segments/1416400380638.30/warc/CC-MAIN-20141119123300-00236-ip-10-235-23-156.ec2.internal.warc.gz
| 102,676,952
| 12,709
|
# Talk:Situation calculus
## NonEncylopedic tone?
Does anyone else feel the tone of this artical is wrong?--Oxinabox (talk) 11:33, 19 June 2012 (UTC)
## Origin
Situation calculus was first introduced by McCarthy solo in 1963. Original Source:
``` @TechReport{mccar-1963:situaactio:TR,
author = "John McCarthy",
title = "Situations, actions, and causal laws",
institution = "Artificial Intelligence Project, Stanford University",
year = "1963",
number = "AIM-2",
}
```
Also supported by Reiter himself in his book: Knowledge in Action
``` @Book{Reiter01,
title = "Knowledge in Action: Logical Foundations for Describing and Implementing Dynamical Systems",
publisher = "MIT Press, Bradford Books",
year = "2001",
author = "R. Reiter",
}
```
-Rob
## Type of second order logic?
Is situation calculus a kind of "second order logic"? Maybe not. It just reified first order logic. -Reiter calls SitCalc a 'dialect of FOL'.
I was also wondering this, Russel & Norvig (1995, AI: A Modern Approach, p 204) and Luciano Serafini (Trento, slides, http://sra.itc.it/people/serafini/teaching/dottorato-dit/2005/situation-calculus.pdf) both explicitly state that it is first order, even though the cited articles indeed call it second order (Wouter van Atteveldt, 128.237.249.187)
AI: A "Modern" Approach is such a terrible book. Do yourself a favor and burn it. Fresheneesz (talk) 00:25, 20 November 2007 (UTC)
It can be viewed either way: It's really an issue of whether you view reification as the conceptual technique by which the situation calculus is constructed, or merely a backend implementation method for reasoning. Consider the first-order language that contains just the fluents as predicates, and no change over time; e.g. you have things like On(Plate, Table) which either hold or don't hold. To go from that to the situation calculus, you now need to make statements about the situations in which these erstwhile predicates hold, like Holds(On(Plate, Table), s). But what just happened when you did that? One view is that you created a new first-order language, with predicates such as Holds(), in which the predicates of your original first-order language are now function symbols; this is called reification (and indeed was invented by McCarthy for the situation calculus). Another view is that you left your original language as is, and just added new predicates like Holds(), which take the original predicates as arguments; therefore your language is now second order. If you were using an automated-reasoning system, you might still do your reasoning by reifying the fluents to convert it to a first-order language, but in this second view you consider reification an implementation detail. Neither of these views is really inherently correct or incorrect, and which an author prefers depends on whether they consider it more sensible to view fluents as reified entities or not, from a knowledge-representation point of view. I personally see it as more sensible to consider reification inherent, and therefore situation calculus as a first-order logic, because fluents are not really predicates anymore (it makes no real sense to use a fluent outside one of the situation-calculus predicates). --Delirium (talk) 10:20, 8 September 2008 (UTC)
You often want a second order component to get rid of non-standard models; in particular you want every situation to have a corresponding term and the situations to form a nice tree with transitivity properties, and some kind of predicate minimization or circumscription is conventional for doing this; I guess you could also just use some semantics which doesn't have non-standard models which violate the transitivity relationship. Some kind of sorts/types are also convenient for just writing the stuff down. 124.170.245.172 (talk) 12:09, 6 October 2008 (UTC)
Ah yeah, I was bracketing circumscription in the above discussion, and just discussing whether having predicates that take fluents as parameters makes the language second-order or not. The reified situation calculus with circumscription added is second order, since the circumscription of first-order formulas is second-order (though some cases can be converted to first-order formulas, through fairly complex translation schemes). I've sometimes seen this distinction clarified by using a phrase like "first-order logic augmented with circumscription" (I believe Shanahan uses that phrase in his writing on the event calculus), to emphasize that everything except the circumscription, which is usually considered sort of separate (in that you don't write normal formulas in it, but just consider it part of the background machinery) is first-order. --Delirium (talk) 02:22, 12 January 2009 (UTC)
## Better than first-order logic?
How is situation calculus more useful than first-order or second-order logic? It seems to me that a situation is simply another variable one can define in first-order logic - in which case situation calculus is simply a special case. Fresheneesz (talk) 00:25, 20 November 2007 (UTC)
It depends in what sense you're talking about "useful". It is indeed a special case in the sense that it's a particular theory about change over time formalized within classical logic (although often a non-monotonic version that uses circumscription), and doesn't require any changes to the underlying logic. That's actually the entire point of the fluent approach to reasoning about dynamical situations: encoding dynamical situations via a system of predicates within classical logic, rather than via specialized temporal logics that have new logic-level operators. The other widely used such approach is the event calculus. Using such a system is "more useful" than developing an ad-hoc theory of change over time within the base logic only in that there has already been considerable work done in putting together the system of predicates and knowledge-representation strategies in ways that are aimed at resulting in the expected inferences and not producing unwanted inferences (i.e. tackling issues like the frame problem and the ramification problem). --Delirium (talk) 09:56, 8 September 2008 (UTC)
## Clarification of "Dynamical Domains"
The term "dynamical domain" has been used multiple times in logic articles. I can't find its definition anywhere, and so I'm guessing it means something along the lines of "constantly changing/variable world" from the independent definitions of the two words. Based on how it is used, however, I do think it needs some definition somewhere - either short definitions in the articles that use it, or a full article that describes the concept (or perhaps just a Wiktionary entry). 209.173.109.104 (talk) 23:00, 20 February 2009 (UTC)
## Why the focus on Reiter?
Why is there so much focus on Reiter's version of the Situation Calculus? I do think his version has some interesting features, and I'm not suggesting that any information in the article needs to be removed, but it was McCarthy that came up with the calculus originally, and he does have a new (2001) version which may be useful to readers (with new features like the "Occurs(e,s)" predicate [I think it's a predicate...]).
The article states that "The main version of the situational calculus that is presented in this article is based on that introduced by Ray Reiter in 1991" in the intro, but it doesn't say (or cite) what makes it "the main version", nor does it say why it is the focus of the article. I suspect it just happens to be the version of the calculus that the article writer learned first.
In any case, the new formalism by McCarthy can be found at [1] if anyone wants to add information about it. His original article on the subject (by John McCarthy and Pat Hayes in 1969) can be found at the same site with many of his other published papers. 209.173.109.104 (talk) 22:57, 6 March 2009 (UTC)
From my understanding, McCarthy's original version did not receive that much attention because it was lacking a solution to the frame problem. This was Reiter's major contribution and lead to a revived interest. Reiter---together with many others from the cognitive robotics group at Univ. of Toronto and elsewhere---then developed Golog on top of it, which was successfully used in systems and keeps being researched. As a side product of this, there are Prolog implementations of Reiter's situation calculus, included in the Golog systems. Christian.fritz (talk) 05:31, 10 September 2009 (UTC)
### Removed reference
The peer-reviewed paper "Learning With C4.5 in a Situation Calculus Domain" (removed by Christian.fritz) introduces automatically discovering temporal rules, and use of Prolog to express them in a Situation Calculus problem domain. It is an example of practical use of Situation Calculus, something missing from the current page. The notion that this page should only concern theory is arbitrary and reduces the usefulness of the page.
CF: Anonymous user 136.159.4.38: a lot of what you seem to be doing on Wikipedia (anonymously; https://en.wikipedia.org/wiki/Special:Contributions/136.159.4.38) is promoting work by Kamran Karimi and Howard J. Hamilton -- since we don't know your name we can only speculate that you are one of these two persons. This is not what Wikipedia is for. Please stop doing this. — Preceding unsigned comment added by Christian.fritz (talkcontribs) 16:33, 16 July 2014 (UTC)
Christian.fritz, you did not address any of the points raised in support of adding that reference. — Preceding unsigned comment added by 136.159.4.38 (talk) 17:48, 16 July 2014 (UTC)
| 2,172
| 9,612
|
{"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-2014-49
|
latest
|
en
| 0.926225
|
https://ignitegirls.com/nutrition/how-many-calories-do-you-burn-carrying-groceries.html
| 1,632,661,958,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00538.warc.gz
| 368,219,935
| 16,281
|
# How many calories do you burn carrying groceries?
Contents
The average person burns 165-220 calories per hour while shopping.
## How many calories do you burn putting away groceries?
Once home, putting away the groceries for fifteen minutes adds about 22 calories.
## How many extra calories do you burn carrying weight?
The more weight you’re carrying, the more calories you’ll expand. Carrying a light load (like a day pack) burns approximately 50 more calories per hour on the trail. Supplies for a long backpacking trip add roughly 100 calories or more to your hourly burn rate.
## How many calories do we burn while doing nothing?
This means that, at rest, they’ll burn approximately 1,829.8 calories in a day (equation: 66 + (6.2 x 180) + (12.7 x 72) – (6.76 x 40) = 1,829.8).
## How many calories do you burn walking with baby in carrier?
If you carry your baby with you on a walk, you’ll burn about 238 calories per hour depending on your weight and your infant’s size.
IT IS INTERESTING: Do slimming vests really work?
## Does walking in the mall burn calories?
Calorie burn: about 250 calories for an hour of power (mall) walking.
## Does pushing a buggy burn more calories?
They found that pushing the stroller with two hands increases the amount of calories burned by about 5 percent, pushing with one hand increases it by about 6 percent, and the push-and-chase method increases it by about 8 percent.
## What exercise burns the most calories?
Running is the winner for most calories burned per hour. Stationary bicycling, jogging, and swimming are excellent options as well. HIIT exercises are also great for burning calories. After a HIIT workout, your body will continue to burn calories for up to 24 hours.
## How many calories should I eat a day to lose 2 pounds a week?
It is generally not advisable to lose more than 2 pounds per week as it can have negative health effects, i.e. try to target a maximum daily calorie reduction of approximately 1000 calories per day.
## How can I lose a lb a day?
You need to burn 3500 calories a day to lose one pound a day, and you need anywhere between 2000 and 2500 calories in a day if you are doing your routine activities. That means you need to starve yourself the whole day and exercise as much as to lose the remaining calories.
## How many calories do you naturally burn in a day?
To lose a pound, you need to have a good idea of how many calories you burn (use for energy) on an average day. According to the U.S. Department of Health and Human Services, the average adult woman expends roughly 1,600 to 2,400 calories per day, and the average adult man uses 2,000 to 3,000 calories per day.
IT IS INTERESTING: Question: Can dementia cause weight loss?
## Do you burn calories when you fart?
Experts say farting is a passive activity — so it probably doesn’t burn any calories at all. When you fart, your muscles relax and the pressure in your gut pushes the gas out without effort. You burn calories when your muscles work, not relax.
## How do I calculate how many calories I burn a day?
Calculating daily calorie burn
1. For men: 66 + (6.2 x weight) + (12.7 x height) – (6.76 x age)
2. For women: 655.1 + (4.35 x weight) + (4.7 x height) – (4.7 x age)
| 783
| 3,253
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2021-39
|
latest
|
en
| 0.911216
|
https://www.solvedanswer.com/hideki-spends-all-of-his-money-on-movie-tickets-and-groceries-suppose-hideki-is-presented-with-two-options-a-and-b-option-a-gives-him-a-voucher-of-x-that/
| 1,696,016,900,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00207.warc.gz
| 1,089,359,510
| 24,167
|
# Hideki spends all of his money on movie tickets and groceries. Suppose Hideki is presented with two options: A and B. Option A gives him a voucher of \$X that
Hideki spends all of his money on movie tickets and groceries. Suppose Hideki is presented with two options: A and B. Option A gives him a voucher of \$X that he can only spend on groceries. Option B gives him \$X that he can spend on either or both of the goods. Which of the following is correct?
A) Hideki always prefers Option A to Option B.
B) Hideki can never be indifferent between Option A and Option B.
C) Hideki is always indifferent between Option A and Option B.
D) Hideki can prefer Option B to Option A but cannot prefer Option A to Option B.
CategoriesUncategorized
| 171
| 743
|
{"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-2023-40
|
latest
|
en
| 0.963694
|
http://mathhelpforum.com/algebra/197526-exponential-growth-decay-print.html
| 1,529,799,282,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267865438.16/warc/CC-MAIN-20180623225824-20180624005824-00438.warc.gz
| 204,096,655
| 2,489
|
Exponential growth or decay
• Apr 18th 2012, 06:12 PM
vaironxxrd
Exponential growth or decay
Is the following function growth or decay?
problem:
\$\displaystyle f(x) = (4)^-^x\$
Answer:Growth because a>0 and b >1. Now the graph is actually decaying(or just falling down), this gives me the notion I'm wrong.
• Apr 18th 2012, 07:40 PM
highvoltage
Re: Exponential growth or decay
I concur. This is decay. The negative in the exponent reflects the graph about the y-axis changing it from growth to decay.
| 148
| 504
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.824222
|
http://slideplayer.com/slide/3153035/
| 1,513,389,938,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00146.warc.gz
| 248,122,537
| 18,739
|
State Assignment using Rules Jacob Boles Ece 572 Fall 99.
Presentation on theme: "State Assignment using Rules Jacob Boles Ece 572 Fall 99."— Presentation transcript:
State Assignment using Rules Jacob Boles Ece 572 Fall 99
Introduction In this presentation I will show an example of state assignment by heuristic rules and compare it to the assignment down by partition pairs.
So that my example is more relevant and unique, I will use the simplified state machine for this machine A F B E C D X=0 X=1 X=0 X=1 X=0 A
State Assignment by Rules Rule 1 –States with most incoming branches should be assignment least number of 1’s in code. –This implies that state A which has the most incoming branches by far should be zero. All the other states have about the same number of incoming branches so we take no precedence A <= 000
State Assignment by Rules Rule 2 –State with common next state on the same input condition should be assigned adjacent codes. –In my example this only occurs for E&C&A E & C & A should be adjacent to each other A F B E C D X=0 X=1 X=0 X=1 X=0 STATES A,C and E transit to the same state A under x=0
State Assignment by Rules Rule 3 –Next state of same state should be adjacent codes according to adjacency of branch conditions. –This is a little harder to see but implies … A adj. B A adj. D D adj. E F adj. C Impossible to do all these with 3 bits! A F B E C D X=0 X=1 X=0 X=1 X=0
State Assignment by Rules Rule 4 –States that form a chain on same branch should be adjacent codes. A F B E C D X=0 X=1 X=0 X=1 X=0 Two chains: Chain A->B->F->C->D->E Chain B->C->A X=0
State Assignment by Rules Our assignment … Rule 2: E&C&A adjacent A 000 B 001 C 010 D 100 E 101 F 011 Rule 1: A <=000 Violates only 1 rules Rule 3 A adj. B A adj. D D adj. E F adj. C Rule 4: Chain A->B->F->C->D->E Chain B->C->A Impossible w/o violating Rule 1 Violates this rule
State Assignment by Rules Q0 = XC + X’D + [XD] Q1 = X’B + XF + [X’F + XB] Q2 = XA + [XD] + [X’F + XB] Assuming sharing of common logic: # gates = 5+4+3 = 12 A 000 B 001 C 010 D 100 E 101 F 011 A F B E C D X=0 X=1 X=0 X=1 X=0 x=0 Symbolic In this example partition pair method does not give a good solution. Check with more advanced partition methods if this statement is valid
Comparison of results Easy to do Fast Efficient for small problems with limited number of variables Advantages Disadvantages Rules and heuristics Partitioning More complex Can be slow if problem is large or bad partition Will always find best solution if given time Better than trying every possibility Rules may not always hold true Inefficient for large variable problems.
Similar presentations
| 724
| 2,656
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.0625
| 3
|
CC-MAIN-2017-51
|
latest
|
en
| 0.805712
|
http://forums.wolfram.com/mathgroup/archive/2005/Jul/msg00542.html
| 1,585,445,051,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00221.warc.gz
| 76,217,741
| 8,336
|
Re: Empirical CDF
• To: mathgroup at smc.vnet.net
• Subject: [mg58926] Re: Empirical CDF
• From: Mark Fisher <mark at markfisher.net>
• Date: Sun, 24 Jul 2005 01:21:48 -0400 (EDT)
• References: <dbt3oh\$skn\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```In the meantime, Piecewise has arrived, which Mathematica *does* know how to
integrate. Here is an example of how to use it to compute an emprical
cummulative distribution function given a list of observations.
MakeEDFPiecewise[list_] :=
With[{d = 1/Length[list]},
Block[{x},
Function @@ {x,
Piecewise[
Transpose[{Range[1, d, -d], Thread[x >= Reverse[Sort[list]]]}]
]}
]]
rand = Table[Random[], {100}];
fun = MakeEDFPiecewise[rand];
Plot[fun[x], {x, -.25, 1.25}, PlotPoints -> 300];
fun2 = Block[{x}, Function @@ {x, Integrate[fun[x], x]}];
Plot[fun2[x], {x, -.25, 1.25}, PlotPoints -> 300, PlotRange -> All]
For faster execution one can replace Function with Compile in the
definition of MakeEDFPiecewise. (I use an Option to control the choice.)
However, there are three costs: It takes longer to compute the function
itself, the resulting object is larger, and it Mathematica complains when one
tries to integrate it analytically (and returns an uncompiled result).
--Mark
P.S. Question to Mathematica developers: Why does fun2'[x] sometimes (but not
always) produce the error message Reduce::ratnz?
David Kahle wrote:
> MathGroup -
>
> swidrygiello posted on Fri, 13 Sep 2002 a question on how to create the
> empirical cumulative distribution function (ECDF) with Mathematica. Since
> then, few others (Mark Fisher, , for example) have posted
> responses utilizing the Interpolation[] command. However, similar results
> can be achieved using the simple code :
>
> For[i=1,i<(Length[rand]+1),preF[x_,i_]:=UnitStep[x-rand[[i]]],i++]
> F[x_]:=Sum[preF[x,i],{i,Length[rand]}]/Length[rand]
>
> Where 'rand' is the vector containing the random observations. The
> resulting function F is right continuous, limits to 0 and 1 as we would
> like it to, and has the correct step sizes. Even better, Mathematica is
> more comfortable manipulating UnitStep functions than it is piecewise
> functions defined with the Which[] command. For example, if we would like
> to integrate the ECDF to test for certain orderings (Stochastic dominance,
> etc.), Mathematica understands integrating the F as defined above, but if
> we use the Which[] command it resorts to numerical techniques which
> typically fail for various reasons. Hope it helps.
>
> David Kahle
> david.kahle at richmond.edu
>
```
• Prev by Date: Re: Follow-on: StyleForm and font selection
• Next by Date: Re: limit problem
• Previous by thread: Re: Empirical CDF
• Next by thread: Interpolation problem, optimization algorithms
| 768
| 2,774
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2020-16
|
longest
|
en
| 0.771972
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.