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://goodriddlesnow.com/riddles/by/funny-riddles/page:6/sort:rid/direction:desc | 1,611,626,036,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704795033.65/warc/CC-MAIN-20210126011645-20210126041645-00133.warc.gz | 368,014,272 | 9,323 | # Funny Riddles
##### Funny Riddle #802 (medium)
Question: Little Johnny is walking home. He has \$300 he has to bring home to his mom. While he is walking a man stops him and gives him a chance to double his money. The man says "I'll give you \$600 if you can roll 1 die and get a 4 or above, you can roll 2 dice and get a 5 or 6 on at least one of them, or you can roll 3 dice and get a 6 on at least on die. If you don't I get your \$300."
What does Johnny do to have the best chance of getting home with the money?
##### When is L greater than XL? (medium)
Question: How can "L" be greater in size than "XL"?
##### Funny Riddle #411 (medium)
Question: What is the only question you can't answer yes to?
##### When you don't know what I am, I am something (medium)
Question: When you do not know what I am, then I am something. But when you know what I am, then I am nothing. What am I?
##### Funny Riddle #247 (easy)
Question: Timmy's mother has three children. The first was named April. The next was named May.
What is the final one's name? | 281 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-04 | longest | en | 0.965855 |
https://brainly.in/question/112191 | 1,484,644,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279650.31/warc/CC-MAIN-20170116095119-00455-ip-10-171-10-70.ec2.internal.warc.gz | 813,954,030 | 9,469 | # Prove that 3√2 is irrational.
2
by dhruvivaja3
2015-05-15T20:31:53+05:30
Let 3√2 is rational number
therefore it is in the form p/q where p and q are co prime numbers
3√2=p/q
√2 = p/3q
therefore l.h.s. is irrational no. and rhs is rational no.
therefore our supposition is wrong
therefore it is a irrational number
2015-05-16T16:38:17+05:30
let us take contradiction that 3√2 is rational number.
so if 3√2 is a rational number then
let 3√2=a/b,where a and b are co primes
then
3
2=a/b
2=a/3b
here in the R.H.S side
that is a/3b is a rational number but in the L.H.S side
√2
is irratioanal
so 3√2
is a irrational number..................hope it helped u:)))))) | 238 | 664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.894866 |
https://rememberingsomer.com/how-much-was-a-pound-worth-in-1880/ | 1,652,987,328,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00088.warc.gz | 557,638,887 | 6,563 | In 1800 one British lb was precious \$4.44 (U.S.). In 1988 terms, ~ adjusting because that inflation, the 1800 Pound would certainly be precious \$30.62 (U.S.).
You are watching: How much was a pound worth in 1880
## How lot was 50 pounds 1912?
Buying power of £100 in 1912
Initial valueEquivalent value
£1 lb in 1912£117.17 pounds today
£5 pounds in 1912£585.83 pounds today
£10 pounds in 1912£1,171.67 pounds today
£50 pounds in 1912£5,858.34 pounds today
## What was a lb worth in 1910?
£1 in 1910 is worth £114.99 in 2018 £1 in 1910 is tantamount in purchasing power to around £114.99 in 2018, rise of £113.99 end 108 years. The pound had an typical inflation price of 4.49% every year between 1910 and 2018, developing a cumulative price increase of 11,399.19%.
## What to be a pound worth in 1950?
Why a pound this particular day is worth only 3% the a pound in 1950 £100 in 1950 is identical in purchasing power to around £3,515.00 today, boost of £3,415.00 end 71 years. The pound had an median inflation price of 5.14% every year in between 1950 and today, developing a accumulation price boost of 3,415.00%.
## What would 100 pounds buy 1880?
Inflation by nation For comparison, in the UK £100.00 in 1880 would be tantamount to £98.94 in 1881, one absolute readjust of £-1.06 and a cumulative change of -1.06%.
## What would a lb buy in 1850?
7736 = \$4.35 – the value of an English lb in 1850.
## How lot would 10000 pounds in Victorian time be precious today?
£10,000 in 1850 is indistinguishable in purchasing strength to about £1,279,642.86 in 2017, rise of £1,269,642.86 over 167 years. The pound had an mean inflation rate of 2.95% per year between 1850 and also 2017, developing a accumulation price increase of 12,696.43%.
## How lot was 100 pounds precious in victorian times?
£100 in 1850 is precious £13,808.94 this particular day £100 in 1850 is equivalent in purchasing power to about £13,808.94 today, an increase of £13,708.94 end 171 years. The pound had actually an average inflation price of 2.92% per year between 1850 and also today, developing a accumulation price boost of 13,708.94%.
## How lot was a lot of money in the 1800s?
\$100 in 1800 is identical in purchasing power to around \$2,136.47 today, rise of \$2,036.47 over 221 years. The dollar had an typical inflation rate of 1.40% per year in between 1800 and today, producing a cumulative price boost of 2,036.47%.
## What to be a million dollars precious in 1800?
\$1,000,000 in 1800 is precious \$21,364,682.54 this day \$1,000,000 in 1800 is tantamount in purchasing power to about \$21,364,682.54 today, rise of \$20,364,682.54 end 221 years. The dollar had actually an median inflation price of 1.40% per year in between 1800 and also today, creating a accumulation price increase of 2,036.47%.
## How lot was \$1000 precious in 1880?
\$1,000 in 1880 is precious \$26,391.67 this particular day \$1,000 in 1880 is tantamount in purchasing strength to around \$26,391.67 today, an increase of \$25,391.67 end 141 years. The dollar had actually an median inflation rate of 2.35% every year in between 1880 and today, producing a accumulation price increase of 2,539.17%.
## How lot did a house cost in 1860?
A four-room home in most eastern cities ran about \$4.50 every month. Outside of the city, floor cost roughly \$3 come \$5 an acre.
## What was the average wage in 1880?
Averages. If you functioned in manufacturing (as numerous did during this period of mechanization), you could have supposed to make roughly \$1.34 a job in 1880, i beg your pardon adds as much as \$345 yearly for an average 257 job of work-related in a offered year.
## What was minimum fairy in 1865?
YEARCommon LaborCarpenters
18651.502.50
18661.503.00
1867—-2.75
1868—-2.50
## How lot did workers obtain paid in the 1800s?
Pay was extremely low for typical workers throughout the industrial revolution. \$1.00 come \$1.50 to be the typical pay for men workers while ladies were payment less and also children the least. It was hardly enough to make a life and almost impossible to support a family.
\$674.96 a year
## Who began the minimum wage in America?
The first federal minimum fairy was created as component of the National commercial Recovery plot of 1933, signed right into law by chairman Franklin D. Roosevelt, but declared unconstitutional. In 1938 the same Labor criter Act developed it at \$0.25 an hour (\$4.60 in 2020 dollars).
\$3,300
## How to be minimum fairy created?
The nationwide minimum wage was produced by conference under the same Labor requirements Act (FLSA) in 1938. Others have said that the primary purpose was to aid the lowest payment of the nation’s working population, those that lacked adequate bargaining strength to secure for themselves a minimum subsistence wage.
## When walk Minimum Wage avoid being a life wage?
Key Takeaways. The federal minimum wage has stagnated at \$7.25 an hour because 2009. For most people, working for minimum fairy does not give them a life wage. Countless states and also cities have actually a higher minimum fairy in place—in some cases, more than double—but employees still battle to do ends meet.
## How lot money walk a solitary person must live comfortably?
This well-known general budgeting rule allocates 50% of annual income to necessities prefer housing, 30% to discretionary prices like travel, and also the continuing to be 20% to savings. The median crucial living wage across the whole US is \$67,690.
## Can a solitary person live ~ above minimum wage?
A Minimum fairy Budget. According to official government guidelines, solitary people supporting themselves top top minimum wage space not technically life in poverty. However, for single people v no dependents, the government insurance claims the federal minimum wage is an ext than sufficient to survive and stay the end of poverty.
## Can you live on 15 an hour?
yes, you deserve to live turn off \$15 one hour…as lengthy as you live in a city v a low cost of living index and do not have dependents. However, relying on how frugal you are, you likely won’t have much money left end at the end of the month for many ‘extras’.
## Is \$30 one hour good?
To someone who simply graduated highscool or also college, a \$30 one hour full-time position would be a good (possibly great) salary. Come an separation, personal, instance with two decades of experience and also multiple advanced degrees, \$30 one hour would most likely be far below market value.
See more: Are Selena Gomez And Becky G Sisters, Becky G And Selena Gomez Sisters
## Is obtaining paid 16 one hour good?
A task at \$16/hr for 40 hrs/wk will at best net girlfriend \$32,000 pre taxation dollars. In truth it’s most likely lower as hourly employees rarely job-related the identical of 52 permanent weeks per year.
## Is 20 dollars an hour good?
Through this dystopian prism, \$20 is fantastic hourly wage for unskilled labor and even entrance level liberal arts graduates. Where can I do 20 dollars an hour? | 1,819 | 7,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-21 | latest | en | 0.960191 |
https://dev.to/jamesshah/housie-tambola-ticket-generator-using-python-31m1 | 1,721,327,786,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00216.warc.gz | 177,049,582 | 25,926 | ## DEV Community
James Shah
Posted on • Originally published at bytetales.co
# Housie(Tambola) Ticket Generator using Python
Announcement: I recently developed my own blog at bytetales.co ππ Please check it out.
Housie aka Tambola aka Bingo (British Version) is probably the best game to play in Lockdown to get our minds off from Corona and all the negative news from all around the world and to have fun with friends and family. I've been playing it a lot, lately.
For those who don't know what Housie is, let me explain.
# What is Housie?
Housie is basically a game of probability in which players mark off numbers on cards as the numbers are drawn randomly by a caller. So at the beginning of the game, all the players get a ticket [3x9 matrix] in which there are total of 27 boxes and 15 of them are filled with a distinct random number from 1 to 90. Here is a sample housie ticket.
And then a caller calls a random number from 1 to 90 and the winner being the first person to mark off all their numbers. So, It's a pretty simple yet interesting and gripping game that one can play with friends and family.
# Housie Ticket Rules
Now there are some rules to generate housie tickets and you can't just randomly put 15 numbers in any boxes of your choice. So, let's see the rules and then create a Python script that will generate such a housie ticket for us.
RULE #1 - Each row cannot have more than 5 numbers.
RULE #2 - Each column is assigned a range of numbers only:
(ex. 1-10 can appear only in column 1)
RULE #3 - In a specific column, numbers must be arranged in ascending order from top to bottom.
Now, of course it's a cumbersome task(and a little biased, too) for a human to generate such tickets, and for the record, we are talking about at least 20-25 tickets per game. So yeah it's definitely not a fun task. So, I thought of creating a python script that would take the number of tickets that we want to generate as input and will return the tickets as an output.
I know, it's not at all a very fancy or difficult script to write but anyway I am free in this lockdown period so I thought why not?
Let's see how we are gonna build this script.
### Step 1 : Create a 3x9 matrix (array) of zeros
``````import numpy as np
# Create a 2D array[3x9] of 0s
ticket_array = np.zeros((3, 9), dtype=int)
``````
Here I've used `numpy` library of Python to create the `ticket_array` because while creating a single matrix, it may not be effective but when we want to create a very large number of lists, NumPy creates it very fast and therefore it is efficient to use numpy to create the array.
### Step 2: Randomly Generate 15 indices to fill values in a ticket.
Now, in numpy we can use tuple for indexing of an array. i.e. To get the first element of first row and first column from `ticket_array` , we can use `ticket_array[(0,0)]`.
And so, first we'll generate an array `total_indices` which contains all the indices of 3x9 array as tuples (i.e. (0,0), (0,1)), (0,2), ..., (2,8))
and then, we'll randomly generate 5 indices from each row.
``````total_indices = [(i, j) for i in range(3) for j in range(9)]
random_indices = []
first_row = random.sample(total_indices[:9], 5)
second_row = random.sample(total_indices[9:18], 5)
third_row = random.sample(total_indices[-9:], 5)
for i in first_row:
random_indices.append(i)
for i in second_row:
random_indices.append(i)
for i in third_row:
random_indices.append(i)
``````
### Step 3: Now Let's fill these indices with random values
So let's create a list ranging from 1 to 90. Then to satisfy the Rule #2, we'll check the column number from index and will fill the according value.
``````for num in random_indices:
if num[1] == 0:
number = random.choice(total_numbers[:10])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 1:
number = random.choice(total_numbers[10:20])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 2:
number = random.choice(total_numbers[20:30])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 3:
number = random.choice(total_numbers[30:40])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 4:
number = random.choice(total_numbers[40:50])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 5:
number = random.choice(total_numbers[50:60])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 6:
number = random.choice(total_numbers[60:70])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 7:
number = random.choice(total_numbers[70:80])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
elif num[1] == 8:
number = random.choice(total_numbers[80:89])
ticket_array[num] = number
total_numbers[total_numbers.index(number)] = 0
``````
### Step 4: Now let's sort the ticket_array column-wise
``````for col in range(9):
# if all the rows are filled with a random number
if(ticket_array[0][col] != 0 and ticket_array[1][col] != 0 and ticket_array[2][col] != 0):
for row in range(2):
if ticket_array[row][col] > ticket_array[row+1][col]:
temp = ticket_array[row][col]
ticket_array[row][col] = ticket_array[row+1][col]
ticket_array[row+1][col] = temp
# if 1st and 2nd row are filled by random number
elif(ticket_array[0][col] != 0 and ticket_array[1][col] != 0 and ticket_array[2][col] == 0):
if ticket_array[0][col] > ticket_array[1][col]:
temp = ticket_array[0][col]
ticket_array[0][col] = ticket_array[1][col]
ticket_array[1][col] = temp
# if 1st and 3rd row are filled by random number
elif(ticket_array[0][col] != 0 and ticket_array[2][col] != 0 and ticket_array[1][col] == 0):
if ticket_array[0][col] > ticket_array[2][col]:
temp = ticket_array[0][col]
ticket_array[0][col] = ticket_array[2][col]
ticket_array[2][col] = temp
# if 2nd and 3rd rows are filled with random numbers
elif(ticket_array[0][col] == 0 and ticket_array[1][col] != 0 and ticket_array[2][col] != 0):
if ticket_array[1][col] > ticket_array[2][col]:
temp = ticket_array[1][col]
ticket_array[1][col] = ticket_array[2][col]
ticket_array[2][col] = temp
``````
### Step 5: Wrap it up in a function and print the tickets.
To print the tickets I used tabulate package of python using which we can present array and matrix in a neat way. Learn more about Tabulate Here.
``````if __name__ == "__main__":
# Take number of tickets from user as system argument
numberOfTickets = sys.argv[1]
tickets = []
for i in range(int(numberOfTickets)):
ticket = getTickets()
tickets.append(ticket)
for ticket in tickets:
print(tabulate(ticket, tablefmt="fancy_grid", numalign="center"))
``````
And Done!! Now, Let's check by running the script from terminal.
``````\$ python ticket-generator.py 2
``````
Output:
You can check out the full code on Github :
## jamesshah / housie-tickets-generator
### Housie Tickets Generator Using Python
Originally Published At: bytetales.co | 1,920 | 7,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-30 | latest | en | 0.954596 |
https://www.physicsforums.com/threads/characteristics-of-a-particles-path-v-t-and-a-t.956151/ | 1,618,159,141,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00581.warc.gz | 1,056,371,993 | 19,337 | # Characteristics of a particles path- v(t) and a(t)
Gold Member
## Homework Statement
The position of a particle moving along a coordinate axis is given by ##s(t) = t^3-9t^2+24t+4## where t is greater than 0.
a) Find v(t)
b) At what time(s) is the particle at rest?
c) On what time intervals is the particle moving from left to right and at what times is it moving from right to left?
d) Sketch the path of the particle
## The Attempt at a Solution
My question is in regards to part d, in sketching the graph of the particle's travel.
To give you some information, I'll post the answers I got to parts a-c.
a) ##v(t) = 3t^2-18t+24##
b) The particle is at rest at t=2 second, and at t=4 seconds
c) The particle is moving from left to right on the intervals [0,2) and (4,∞).
The particle is moving from right to left on the interval (2,4)
Now how can I visually represent this? If I use a traditional X,Y coordinate plane, to move backwards it would look like I was going back in time.
#### Attachments
• Screen Shot 2018-09-24 at 11.30.14 PM.png
67.9 KB · Views: 257
Mark44
Mentor
## Homework Statement
The position of a particle moving along a coordinate axis is given by ##s(t) = t^3-9t^2+24t+4## where t is greater than 0.
a) Find v(t)
b) At what time(s) is the particle at rest?
c) On what time intervals is the particle moving from left to right and at what times is it moving from right to left?
d) Sketch the path of the particle
## The Attempt at a Solution
My question is in regards to part d, in sketching the graph of the particle's travel.
To give you some information, I'll post the answers I got to parts a-c.
a) ##v(t) = 3t^2-18t+24##
b) The particle is at rest at t=2 second, and at t=4 seconds
c) The particle is moving from left to right on the intervals [0,2) and (4,∞).
The particle is moving from right to left on the interval (2,4)
Now how can I visually represent this? If I use a traditional X,Y coordinate plane, to move backwards it would look like I was going back in time.
The problem states that the particle is moving along a coordinate axis. I would plot values of s(t) for t = 0, 1, 2, 3, 4, and label each point with the time value. For example, s(0) = 4, so the particle is 4 units to the right of 0, with t = 0. It will necessarily change directions when t = 2 and t = 4. You don't want to use a two-axis coordinate system.
opus and CWatters
Gold Member
Thanks Mark. I didn't realize that further into the text, it gives the graph.
The math laid out makes sense, but the graph has me confused- maybe because I'm used to only seeing X,Y coordinates.
So on the graph, it's clear that from time=0 seconds to time=2 seconds, the particle is traveling from left to right. Then from t=2 second to t=4 seconds, it's traveling from right to left. Then from t=4 seconds and forward, it's traveling from left to right.
But I don't get the numbers on the number line. From my math, these are the positions where the particle changes direction, and where it starts movement. For example, at t=2 seconds, the position function gives ##s(2) = (2)^3-9(2)^2+24(2)+4 = 24##. But how does a position function yield only one value, 24? We need an X and a Y for a position.
#### Attachments
• Screen Shot 2018-09-25 at 1.40.22 AM.png
11.3 KB · Views: 246
Ray Vickson
Homework Helper
Dearly Missed
Thanks Mark. I didn't realize that further into the text, it gives the graph.
The math laid out makes sense, but the graph has me confused- maybe because I'm used to only seeing X,Y coordinates.
So on the graph, it's clear that from time=0 seconds to time=2 seconds, the particle is traveling from left to right. Then from t=2 second to t=4 seconds, it's traveling from right to left. Then from t=4 seconds and forward, it's traveling from left to right.
But I don't get the numbers on the number line. From my math, these are the positions where the particle changes direction, and where it starts movement. For example, at t=2 seconds, the position function gives ##s(2) = (2)^3-9(2)^2+24(2)+4 = 24##. But how does a position function yield only one value, 24? We need an X and a Y for a position.
The problem did say "motion along a coordinate axis", so the motion is one-dimensional. For example, if you choose to have the motion going along the ##x##-axis, the position at ##t=2## is ##(24,0).##
opus
Gold Member
Ahh ok. That makes sense. Thank you guys. | 1,213 | 4,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-17 | longest | en | 0.914718 |
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Post subject: Math/Geometry question 2Posted: Sun Sep 25, 2011 10:44 am
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
Hello again,
I'm currently suck on another math question. Question 1 was asked here: http://twistypuzzles.com/forum/viewtopic.php?f=7&t=18998 and you were a great help then so I hope you can help again.
Everyone knows that the distance to the center of an equilateral triangle is 1/3 of its height in from the base. At least I know this holds for a plane but is it also true for equilateral triangles formed on a sphere?
Lets say I have an equilateral triangle cut by 3 great circles. Let's call one of these the equator and the base of this triangle. If the height of the triangle is A degrees above the equator then is the center of this triangle A/3 degrees above the equator? I don't think so... if A=90 isn't the center 45 degrees above the equator or A/2? Doing this in my head without a globe handy so I'm not certain. Anyways I'm looking for the general solution for any A (I think I need to say A is less then or equal to 90 degrees).
Help,
Carl
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Last edited by wwwmwww on Sun Sep 25, 2011 11:55 am, edited 1 time in total.
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 11:40 am
Joined: Mon Mar 30, 2009 5:13 pm
Hint: just think of the equilateral triangle that would lie on a plane connecting the 3 points, which would slice through the sphere, and project the centre of that triangle onto the surface of the sphere.
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If you want something you’ve never had, you’ve got to do something you’ve never done - Thomas Jefferson
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 11:54 am
Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
Kelvin Stott wrote:
Hint: just think of the equilateral triangle that would lie on a plane connecting the 3 points, which would slice through the sphere, and project the centre of that triangle onto the surface of the sphere.
NICE!!! I always over complicate things.
Thanks,
Carl
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 11:57 am
Joined: Mon Mar 30, 2009 5:13 pm
wwwmwww wrote:
NICE!!! I always over complicate things.
And I tend to over-simplify things, so I'm not even sure this will work.
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 12:36 pm
Joined: Thu Jan 06, 2005 8:53 pm
Location: Los Angeles
A couple thinks to think about:
Depending on the size of the spherical equilateral triangle, the angles change.
So, a great circle is technically a spherical equilateral triangle.
You can have an equilateral triangle that covers more than half of the sphere.
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 12:46 pm
Joined: Mon Mar 30, 2009 5:13 pm
TBTTyler wrote:
A couple thinks to think about:
Depending on the size of the spherical equilateral triangle, the angles change.
So, a great circle is technically a spherical equilateral triangle.
You can have an equilateral triangle that covers more than half of the sphere.
This is true, and in that case you can just project the centre of the planar triangle through the centre of the sphere to the surface on the opposite side of the sphere.
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 1:08 pm
Joined: Tue Jan 13, 2009 8:23 pm
Twelvety.
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For Jasmine Rose... Happy 2nd Birthday in Heaven, 2nd Dec 2013 xxx
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Post subject: Re: Math/Geometry question 2Posted: Sun Sep 25, 2011 2:27 pm
Joined: Thu Jan 06, 2005 8:53 pm
Location: Los Angeles
Kelvin Stott wrote:
TBTTyler wrote:
A couple thinks to think about:
Depending on the size of the spherical equilateral triangle, the angles change.
So, a great circle is technically a spherical equilateral triangle.
You can have an equilateral triangle that covers more than half of the sphere.
This is true, and in that case you can just project the centre of the planar triangle through the centre of the sphere to the surface on the opposite side of the sphere.
Quite true of course, but that wasn't my (completely moot ) point.
I swear I read these things, but It just never clicks what somebody is actually asking until after I answer
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Jump to: Select a forum ------------------ Announcements General Puzzle Topics New Puzzles Puzzle Building and Modding Puzzle Collecting Solving Puzzles Marketplace Non-Twisty Puzzles Site Comments, Suggestions & Questions Content Moderators Off Topic | 1,414 | 5,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2014-15 | latest | en | 0.923629 |
https://howafrica.com/checkout-7-outstanding-math-prodigies/ | 1,627,799,746,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00356.warc.gz | 307,399,633 | 33,980 | # Checkout 7 Outstanding Math Prodigies
1. The Chinese delivery man who discovered a simpler solution to a complex math problem
Yu Jianchun, a Chinese migrant worker from the Henan province, has no mathematical training and no college degree but is being hailed as a real-life version of Will Hunting, the character played by Matt Damon in the 1997 Oscar-winning filmGood Will Hunting, after finding an alternative method to verify Carmichael numbers.
Carmichael numbers, also known as “pseudoprimes,” are large numbers that only appear to be prime numbers, which are only divisible by one and themselves. They are used for credit card encryption and online payments, among other things. Some examinations can be done to find out which numbers are prime and which are Carmichael numbers, but it’s tricky work. Apparently, Yu has come up with a simpler way to verify Carmichael numbers.
The 33-year-old says he worked on a solution for the last eight years, relying solely on his intuition and his innate sensitivity to numbers—and he did it all in his free time because he had a full-time job working as a delivery man for a logistics company.
After verifying his solution to the Carmichael numbers, Yu apparently wrote several universities about it, but they wouldn’t give him the time of day, he was just a delivery man with no formal studies in mathematics, after all. But he recently got the chance to make his case at Zhejiang University, along with solutions to four other complex math problems. His way of verifying Carmichael numbers stunned those in attendance, including Professor Cai Tianxin, who later told reporters that Yu’s proof was much more efficient than traditional solutions.
2. The Nigerian professor who solved a 156-year-old mathematics problem
A Nigerian professor has reportedly solved a math problem which has eluded scholars for 156 years. Dr. Opeyemi Enoch, who teaches at the Federal University in the city of Oye Ekiti, might receive a \$1 million (£657,000) prize if his formula is correct.
Before you go thinking that this is a quick way to get rich, you should understand what exactly Enoch has (possibly) done. He believes he has found a solution to the Riemann Hypothesis—a mathematical problem first proposed by German mathematician Bernhard Riemann in 1859. It makes up one of the seven “millennium problems,” which are a set of questions proposed by the Clay Mathematics Institute in 2000.
3. The woman who solved a mathematical problem that was open for seven decades
Neena Gupta is another genius who has pulled off a remarkable feat. In 2014, she became a recipient of the prestigious Indian National Science Academy (INSA) Medal for Young Scientists for solving a math problem, which astonishingly enough, remained open for almost 70 years!
The problem that no one but Neena could solve is called the Zariski Cancellation Conjecture. INSA described her solution as, “one of the best works in algebraic geometry in recent years done anywhere.”
In addition to the INSA award, she was also awarded the Ramanujan Prize (2014), and the Saraswathi Cowsik Medal (2013) by the Tata Institute of Fundamental Research (TIFR) Alumni Association.
4. The math genius who solved a 100-year-old problem and refused a million dollar prize
The Poincare conjecture was a seemingly unsolvable theorem that was first proposed in 1904. But Russian mathematician Grigory Perelman published two proofs of the theorem back in 2002 and 2003, and according to The Utopianist, it wasn’t until last year that a team of advanced mathematicians at the Clay Mathematical Institute (CMI) finally proved his results valid.
His reward? One million dollars and the Fields Medal, or the math world’s equivalent of the Nobel Prize. But the private Perelman shrugged off the invite to accept the cash, saying that the knowledge he gained from proving the conjecture was more valuable than any monetary gain. (Source)
5. The woman who solved ‘The Monty Hall Problem’ and caused public hysteria
It was a simple brainteaser, but its solution made everyone go crazy. And when Marilyn vos Savant—a woman known for having the highest IQ recorded in the Guinness Book of Records—solved it, some people weren’t satisfied.
We’re talking about “The Monty Hall Problem,” made famous by the game show Let’s Make a Deal, hosted by Monty Hall. Here’s how it works: There are three doors. Behind one door is a car, and behind the other two, are goats. The player chooses a door, and then before being told what’s behind it, Hall opens one of the doors to reveal a goat, leaving the door the player chose, and a door they haven’t seen yet. Should the player switch their choice to increase chances of finding the car?
When a reader submitted that question to Savant’s newspaper column Ask Marilyn, she replied: “Yes, you should switch.” But her answer caused a firestorm. She received thousands of letters claiming she was wrong, Some of the mail was even laced with sexist comments like “There is such a thing as female logic.” All of this was shocking to Savant, mostly because the problem had actually been solved many times before—as far back as 1889, by French mathematician Joseph Bertrand. Regardless, Savant ended up convincing many of her readers that she was right.
Watch the video explaining everything:
6. The “human computer” who solved complicated cube roots by the age of 5
A human calculator that solved the most complex math problems with incredible ease, Shakuntala Devi was an Indian gem. But more than anything else, I see her as a genius of a woman who counters the absurd, yet popular, notion that says women are mathematically challenged.
Born in Bangalore on November 4, 1929, she was just 5, when, unlike other kids who were still trying to get a hang of counting, she traveled with her dad and solved complex cube roots in performances which fetched money for the family.
Devi grew up to become not just a math wizard who traveled the world giving it glimpses of the extraordinary, but a woman known as the world’s “human computer,“ who has her remarkable feats recorded in the 1982 edition of the Guinness Book of World Records. Some of her many arithmetic talents that left the world astonished:
• She added four complex numbers and multiplied the result by 9,878 in 20 seconds.
• She extracted the 23rd root of a 201-digit number in 50 seconds
• She solved the record-worthy multiplication problem of 13-digit numbers: 7,686,369,774,870 × 2,465,099,745,779, in 28 seconds
One problem randomly put before her by the Computer Department of Imperial College, London in 1980 had the gigantic answer of 18,947,668,177,995,426,462,773,730, which she took just 28 seconds to come up with and earned her a mention in the Guinness Book of World Records.
7. The British man who won \$720,000 for solving a 300-year-old math problem
In 1993, Andrew Wiles delivered the findings of his obsessive seven-year study on “Fermat’s Last Theorem” to Cambridge University. When the British mathematician wrote his proof on the blackboard at the end of his presentation, the 200 researchers attending the lecture sat in stunned silence and suddenly erupted into overwhelming applause.
Wiles’ work has since undergone changes—particularly after an error was noted in 1994, but 20-odd years after his feat, he was awarded the highly prestigious Abel Prize by the Norwegian Academy of Sciences and Letters in Oslo. The award is often referred to as the mathematician’s Nobel Prize, and also comes with 6 million Norwegian kroner (\$720,000) in prize money.
Pierre de Fermat proposed Fermat’s Last Theorem in 1637, which states “an + bn = cn. This equation has no solution in integers for n≥3.” In other words, n can never be more than 2 for the equation to work. It may seem simple enough, but definitive proof of the theory had alluded mathematicians throughout the centuries. You can learn more about the problem in the video below. | 1,753 | 7,957 | {"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} | 3.234375 | 3 | CC-MAIN-2021-31 | latest | en | 0.980504 |
https://pt.slideshare.net/DrRajuSingh/02numerical-method-assignment | 1,701,501,331,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00608.warc.gz | 529,153,258 | 82,915 | # 02-NUMERICAL METHOD (ASSIGNMENT)
STUDY INNOVATIONSEducator em Study Innovations
1. The number 3.14150 rounded to 3 decimals is (a) 3.14 (b) 3.141 (c) 3.142 (d) None of these 2. The number of significant digits in 0.003050 is (a) 7 (b) 6 (c) 4 (d) None of these 3. The number of significant digits in 20.035 is (a) 3 (b) 5 (c) 4 (d) None of these 4. The number of significant digits in 20340 is (a) 4 (b) 5 (c) 3 (d) None of these 5. The number 0.0008857 when rounded off to three significant digits yields (a) 0.001 (b) 0.000886 (c) 0.000885 (d) None of these 6. The number 3.68451 when rounded off to three decimal places becomes (a) 3.68 (b) 3.684 (c) 3.685 (d) None of these 7. The number of significant digits in the number 0.00452000 is (a) 3 (b) 5 (c) 8 (d) None of these 8. When a number is approximated to n decimal places by chopping off the extra digits, then the absolute value of the relative error does not exceed (a) (b) (c) (d) None of these 9. When the number 6.878652 is rounded off to five significant figures, then the round off error is (a) – 0.000048 (b) –0.00048 (c) 0.000048 (d) 0.00048 10. The number 0.0009845 when rounded off to three significant digits yields (a) 0.001 (b) 0.000987 (c) 0.000985 (d) None of these 11. A decimal number is chopped off to four decimal places, then the absolute value of the relative error is not greater tha (a) (b) (c) (d) None of these 12. If and are absolute errors in two numbers and respectively due to rounding or truncation, then (a) Is equal to (b) Is less then (c) Is less then or equal to (d) Is greater then or equal to 13. In general the ratio of the truncation error to that of round off error is (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) None of these 14. The equation is of the form (a) Algebraic (b) Linear (c) Quadratic (d) Transcendental 15. The root of the equation lies in the interval (a) (2, 3) (b) (3, 4) (c) (3, 5) (d) (4, 6) 16. The root of the equation in the interval (1, 2) is (a) 1.13 (b) 1.98 (c) 1.54 (d) No root lies in the interval (1, 2) 17. The equation has repeated root , if (a) (b) (c) (d) None of these 18. The root of the equation lies between (a) 3 and 3.5 (b) 2 and 3 (c) 3.5 and 4 (d) None of these 19. For the equation , if and then we will discard the value of the function at the point (a) a (b) b (c) c (d) Anyone out of a, b, c 20. The positive root of the equation lies in the interval (a) (0, 1) (b) (1, 2) (c) (2, 3) (d) (2, 4) 21. The positive root of the equation lies in the interval (a) (0, 1) (b) (1, 2) (c) (2, 3) (d) (3, 4) 22. One real root of the equation = 0 must lie in the interval (a) (0, 1) (b) (1, 2) (c) (–1, 0) (d) (–2, 0) 23. The number of positive roots of the equation is (a) 1 (b) 2 (c) 3 (d) None of these 24. Let be an equation and be two real numbers such that , then (a) At least one root of the equation lies in the interval (b) No root of the equation lies in the interval (c
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### 02-NUMERICAL METHOD (ASSIGNMENT)
• 1. 164 Numerical Methods 1. The number 3.14150 rounded to 3 decimals is (a) 3.14 (b) 3.141 (c) 3.142 (d) None of these 2. The number of significant digits in 0.003050 is (a) 7 (b) 6 (c) 4 (d) None of these 3. The number of significant digits in 20.035 is (a) 3 (b) 5 (c) 4 (d) None of these 4. The number of significant digits in 20340 is (a) 4 (b) 5 (c) 3 (d) None of these 5. The number 0.0008857 when rounded off to three significant digits yields (a) 0.001 (b) 0.000886 (c) 0.000885 (d) None of these 6. The number 3.68451 when rounded off to three decimal places becomes (a) 3.68 (b) 3.684 (c) 3.685 (d) None of these 7. The number of significant digits in the number 0.00452000 is (a) 3 (b) 5 (c) 8 (d) None of these 8. When a number is approximated to n decimal places by chopping off the extra digits, then the absolute value of the relative error does not exceed (a) n 10 (b) 1 10 n (c) 1 10 5 . 0 n (d) None of these 9. When the number 6.878652 is rounded off to five significant figures, then the round off error is (a) – 0.000048 (b) –0.00048 (c) 0.000048 (d) 0.00048 10. The number 0.0009845 when rounded off to three significant digits yields (a) 0.001 (b) 0.000987 (c) 0.000985 (d) None of these B Ba as si ic c L Le ev ve el l Significant Digit and Rounding off Numbers, Fundamental concepts 164
• 2. Numerical Methods 165 11. A decimal number is chopped off to four decimal places, then the absolute value of the relative error is not greater tha (a) 2 10 (b) 3 10 (c) 4 10 (d) None of these 12. If 1 e and 2 e are absolute errors in two numbers 1 n and 2 n respectively due to rounding or truncation, then 2 2 1 1 n e n e (a) Is equal to 2 1 e e (b) Is less then 2 1 e e (c) Is less then or equal to 2 1 e e (d) Is greater then or equal to 2 1 e e 13. In general the ratio of the truncation error to that of round off error is (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) None of these 14. The equation 0 1 sin 2 x e x is of the form (a) Algebraic (b) Linear (c) Quadratic (d) Transcendental 15. The root of the equation 0 1 6 3 x x lies in the interval (a) (2, 3) (b) (3, 4) (c) (3, 5) (d) (4, 6) 16. The root of the equation 0 5 3 3 x x in the interval (1, 2) is (a) 1.13 (b) 1.98 (c) 1.54 (d) No root lies in the interval (1, 2) 17. The equation 0 ) ( x f has repeated root ) , ( 2 1 x x a , if (a) 0 ) ( ' a f (b) 0 ) ( ' a f (c) 0 ) ( ' a f (d) None of these 18. The root of the equation 7 log 2 10 x x lies between (a) 3 and 3.5 (b) 2 and 3 (c) 3.5 and 4 (d) None of these 19. For the equation 0 ) ( x f , if 0 ) ( , 0 ) ( , 0 ) ( c f b f a f and c b then we will discard the value of the function ) (x f at the point (a) a (b) b (c) c (d) Anyone out of a, b, c 20. The positive root of the equation 0 3 x e x lies in the interval (a) (0, 1) (b) (1, 2) (c) (2, 3) (d) (2, 4) 21. The positive root of the equation 0 5 2 3 x x lies in the interval (a) (0, 1) (b) (1, 2) (c) (2, 3) (d) (3, 4) 22. One real root of the equation 1 5 3 x x = 0 must lie in the interval (a) (0, 1) (b) (1, 2) (c) (–1, 0) (d) (–2, 0) 23. The number of positive roots of the equation 0 5 3 3 x x is (a) 1 (b) 2 (c) 3 (d) None of these
• 3. 166 Numerical Methods 24. Let 0 ) ( x f be an equation and 2 1 , x x be two real numbers such that 0 ) ( ) ( 2 1 x f x f , then (a) At least one root of the equation lies in the interval ) , ( 2 1 x x (b) No root of the equation lies in the interval ) , ( 2 1 x x (c) Either no root or more than one root of the equation lies the interval ) , ( 2 1 x x (d) None of these 25. Let 0 ) ( x f be an equation let 2 1, x x be two real numbers such that 0 ) ( ) ( 2 1 x f x f , then (a) At least one root of the equation lies in ) , ( 2 1 x x (b) No root of the equation lies in ) , ( 2 1 x x (c) Either no root or an even number of roots lie in ) , ( 2 1 x x (d) None of these 26. If for 0 ) ( x f , 0 ) ( a f and 0 ) ( b f , then one root of 0 ) ( x f is (a) Between a and b (b) One of from a and b (c) Less than a and greater than b (d) None of these 27. If 0 ) ( ) ( b f a f , then an approximate value of a real root of 0 ) ( x f lying between a and b is given by (a) a b a bf b af ) ( ) ( (b) a b b af a bf ) ( ) ( (c) ) ( ) ( ) ( ) ( a f b f a bf b af (d) None of these 28. One root of 0 4 3 x x lies in (1, 2). In bisection method, after first iteration the root lies in the interval (a) (1, 1.5) (b) (1.5, 2.0) (c) (1.25, 1.75) (d) (1.75, 2) 29. A root of the equation 0 1 3 x x lies between 1 and 2. Its approximate value as obtained by applying bisection method 3 times is (a) 1.375 (b) 1.625 (c) 1.125 (d) 1.25 30. A root of the equation 0 4 3 x x lies between 1 and 2. Its approximate value, as obtained by applying bisection method 3 times, is (a) 1.375 (b) 1.750 (c) 1.975 (d) 1.875 31. Performing 3 iterations of bisection method, the smallest positive approximate root of equation 0 1 5 3 x x is B Ba as si ic c L Le ev ve el l Successive bisection method
• 4. Numerical Methods 167 (a) 0.25 (b) 0.125 (c) 0.50 (d) 0.1875 32. A root of the equation 0 5 3 3 x x lies between 2 and 2.5. Its approximate value, by applying bisection method 3 times is (a) 2.0625 (b) 2.3125 (c) 2.3725 (d) 2.4225 33. If for the function 0 ) ( x f , 0 ) ( a f and 0 ) ( b f , then the value of x in first iteration is (a) 2 b a (b) 2 a b (c) 2 2 b a (d) 2 2 a b 34. Using successive bisection method, a root of the equation 0 1 4 3 x x lies between 1 and 2, at the end of first interaction, it lies between (a) 1.62 and 1.75 (b) 1.5 and 1.75 (c) 1.75 and 1.87 (d) None of these 35. The nearest real root of the equation 0 2 x xe correct to two decimal places, is (a) 1.08 (b) 0.92 (c) 0.85 (d) 0.80 36. By the false position method, the root of the equation 0 1 9 3 x x lies in interval (2, 4) after first iteration. It is (a) 3 (b) 2.5 (c) 3.57 (d) 2.47 37. The formula [where ) ( 1 n x f and ) ( n x f have opposite sign at each step n ≥ 1] of method of False position of successive approximation to find the approximate value of a root of the equation 0 ) ( x f is (a) ) ( ) ( ) ( ) ( 1 1 1 n n n n n n n x x x f x f x f x x (b) ) ( ) ( ) ( ) ( 1 1 1 n n n n n n n x x x f x f x f x x (c) ) ( ) ( ) ( ) ( 1 1 1 n n n n n n n x x x f x f x f x x (d) ) ( ) ( ) ( ) ( 1 1 1 n n n n n n n x x x f x f x f x x 38. By false positioning, the second approximation of a root of equation 0 ) ( x f is (where 1 0, x x are initial and first approximations respectively) (a) ) ( ) ( ) ( 0 1 0 0 x f x f x f x (b) ) ( ) ( ) ( ) ( 0 1 0 1 1 0 x f x f x f x x f x (c) ) ( ) ( ) ( ) ( 0 1 1 1 0 0 x f x f x f x x f x (d) ) ( ) ( ) ( 0 1 0 0 x f x f x f x 39. A root of the equation 0 18 3 x lies between 2 and 3. The value of the root by the method of false position is (a) 2.526 (b) 2.536 (c) 2.546 (d) 2.556 40. The equation 0 4 3 3 x x has only one real root. What is its first approximate value as obtained by the method of false position in (–3, –2) (a) –2.125 (b) 2.125 (c) –2.812 (d) 2.812 B Ba as si ic c L Le ev ve el l Regula-Falsi method
• 5. 168 Numerical Methods 41. A root of equation 0 5 2 3 x x lies between 1 and 1.5. its value as obtained by applying the method of false position only once is (a) 3 4 (b) 27 35 (c) 25 23 (d) 4 5 42. If successive approximations are given by 1 3 2 1 , ,...... , , n n x x x x x , then Newton-Raphson formula is given as (a) ) ( ) ( 1 1 x f x f x x n n n (b) ) ( ) ( 1 n n n n x f x f x x (c) ) ( ) ( 1 n n n n x f x f x x (d) ) ( ) ( 1 n n n n x f x f x x 43. Newton-Raphson method is applicable only when (a) 0 ) ( x f in the neighbourhood of actual root x (b) 0 ) ( x f in the neighbourhood of actual root x (c) 0 ) ( x f in the neighbourhood of actual root x (d) None of these 44. Newton-Raphson processes has a (a) Linear convergence (b) Quadratic convergence (c) Cubic convergence (d) None of these 45. The condition for convergence of the Newton-Raphson method to a root is (a) 1 ) ( ) ( 2 1 f f (b) 1 ) ( ) ( f f (c) 1 ) ( ) ( 2 1 f f (d) None of these 46. The real root of the equation 0 5 3 x x lying between –1 and 2 after first iteration by Newton-Raphson method is (a) 1.909 (b) 1.904 (c) 1.921 (d) 1.940 47. A root of the equation 0 1 4 3 x x lies between 1 and 2. Its value as obtained by using Newton-Raphson method is (a) 1.775 (b) 1.850 (c) 1.875 (d) 1.950 48. The value of 0 x (the initial value of x) to get the solution in interval (0.5, 0.75) of the equation 0 3 5 3 x x by Newton-Raphson method, is (a) 0.5 (b) 0.75 (c) 0.625 (d) None of these 49. If a and a + h are two consecutive approximate roots of the equation 0 ) ( x f as obtained by Newtons method, then h is equal to B Ba as si ic c L Le ev ve el l Newton-Raphson method
• 6. Numerical Methods 169 (a) ) ( / ) ( a f a f (b) ) ( / ) ( a f a f (c) ) ( / ) ( a f a f (d) ) ( / ) ( a f a f 50. The Newton-Raphson method converges fast if ) ( f is ( is the exact value of the root) (a) Small (b) Large (c) 0 (d) None of these 51. If one root of the equation 0 ) ( x f is near to 0 x then the first approximation of this root as calculated by Newton-Raphson method is the abscissa of the point where the following straight line intersects the x-axis (a) Normal to the curve ) (x f y at the point )) ( , ( 0 0 x f x (b) Tangent to the curve ) (x f y at the point )) ( , ( 0 0 x f x (c) The straight line through the point )) ( , ( 0 0 x f x having the gradient ) ( 1 0 x f (d) The ordinate through the point )) ( , ( 0 0 x f x 52. A root of the equation 0 5 3 3 x x lies between 2 and 2.5. Its value as obtained by using Newton-Raphson method, is (a) 2.25 (b) 2.33 (c) 2.35 (d) 2.45 53. After second iteration of Newton-Raphson method, the positive root of equation 3 2 x is (taking initial approximation 2 3 ) (a) 2 3 (b) 4 7 (c) 56 97 (d) 200 347 54. If one root of the equation 0 1 2 3 x x is near to 1.0, then by Newton-Raphson method the first calculated approximate value of this root is (a) 0.9 (b) 0.6 (c) 1.2 (d) 0.8 55. The approximate value of a root of the equation 0 5 3 3 x x at the end of the second iteration by taking the initial value of the roots as 2, and by using Newton-Raphson method, is (a) 2.2806 (b) 2.2701 (c) 2.3333 (d) None of these 56. Newton-Raphson method is used to calculate 3 65 by solving 65 3 x . If 4 0 x is taken as initial approximation then the first approximation 1 x is (a) 65/16 (b) 131/32 (c) 191/48 (d) 193/48 57. Starting with 1 0 x , the next approximation 1 x to 3 / 1 2 obtained by Newton’s method is (a) 3 5 (b) 3 4 (c) 4 5 (d) 6 5 Advance Level
• 7. 170 Numerical Methods 58. Approximate value of nh x x dx y 0 0 by Trapezoidal rule, is [Where n i h x x y x y i i i i ..... , 2 , 1 , 0 , , ) ( 1 ] (a) )] ..... ( 2 [ 2 1 3 2 1 0 n n y y y y y y h (b) )] .... ( 2 ) ..... ( 4 [ 3 2 4 2 1 5 3 1 0 n n n y y y y y y y y y h (c) )] .... ( 4 ) ..... ( 2 [ 4 2 4 2 1 5 3 1 0 n n n y y y y y y y y y h (d) )] .... ( 2 ) ..... [ 2 1 5 3 1 4 2 0 n n y y y y y y y y h 59. Trapezoidal rule for evaluation of dx x f b a ) ( requires the interval (a, b) to be divided into (a) 2n sub-intervals of equal width (b) 2n + 1 sub-intervals of equal width (c) Any number of sub-intervals of equal width (d) 3n sub-intervals of equal width 60. The value of f(x) is given only at 1 , 3 2 , 3 1 , 0 x . Which of the following can be used to evaluate 1 0 ) ( dx x f approximately (a) Trapezoidal rule (b) Simpson rule (c) Trapezoidal as well as Simpson rule (d) None of these 61. A river is 80 metre wide. Its depth d metre and corresponding distance x metre from one bank is given below in table x : 0 10 20 30 40 50 60 70 80 y : 0 4 7 9 12 15 14 8 3 Then the approximate area of cross-section of river by Trapezoidal rule, is (a) 710 sq.m (b) 730 sq.m (c) 705 sq.m (d) 750 sq.m 62. A curve passes through the points given by the following table x : 1 2 3 4 5 y : 10 50 70 80 100 By Trapezoidal rule, the area bounded by the curve, the x-axis and the lines x = 1, x = 5, is (a) 310 (b) 255 (c) 305 (d) 275 63. From the following table, using Trapezoidal rule, the area bounded by the curve, the x-axis and the lines x = 7.47, x = 7.52, is x : 7.47 7.48 7.49 7.50 7.51 7.52 f(x) 1.93 1.95 1.98 2.01 2.03 2.06 B Ba as si ic c L Le ev ve el l Trapezoidal rule
• 8. Numerical Methods 171 : (a) 0.0996 (b) 0.0896 (c) 0.1096 (d) 0.0776 64. Let f (0) = 1, f(1) = 2.72, then the trapezoidal rule gives approximate value of 1 0 ) ( dx x f (a) 3.72 (b) 1.86 (c) 1.72 (d) 0.86 65. By Trapezoidal rule, the value of 1 0 3 dx x considering five sub-intervals, is (a) 0.21 (b) 0.23 (c) 0.24 (d) 0.26 66. The approximate value of 9 1 2 dx x by using Trapezoidal rule with 4 equal intervals is (a) 243 (b) 248 (c) 242.8 (d) 242.5 67. Taking n = 4, by trapezoidal rule, the value of 2 0 1 x dx is (a) 1.1125 (b) 1.1176 (c) 1.118 (d) None of these 68. With the help of trapezoidal rule for numerical integration and the following table x : 0 0.25 0.50 0.75 1 f(x) : 0 0.0625 0.2500 0.5625 1 The value of 1 0 ) ( dx x f is (a) 0.35342 (b) 0.34375 (c) 0.34457 (d) 0.33334 69. If for n = 3, the integral 10 1 3 dx x is approximately evaluated by Trapezoidal rule 10 1 3 3 3 7 2 10 1 3 dx x , then (a) 3 3 (b) 3 4 (c) 3 5 (d) 3 6 70. By trapezoidal rule, the value of dx x 2 1 1 , (using five ordinates) is nearly (a) 0.216 (b) 0.697 (c) 0.921 (d) None of these Advance Level B Ba as si ic c L Le ev ve el l Simpson’s one third rule
• 9. 172 Numerical Methods 71. The value of nh x x dx y 0 0 , n is even number, by Simpson’s one-third rule is (a) )] ...... ( 4 ) ..... ( 2 ) [( 3 2 4 2 1 3 1 0 n n n y y y y y y y y h (b) )] ...... ( 2 ) ..... ( 4 ) [( 3 2 4 2 1 3 1 0 n n n y y y y y y y y h (c) )] ...... ( 4 ) ..... ( 2 ) [( 3 2 4 2 1 3 1 0 n n n y y y y y y y y h (d) None of these 72. Simpson’s one-third rule for evaluation b a dx x f ) ( requires the interval [a, b] to be divided into (a) An even number of sub-intervals of equal width (b) Any number of sub-intervals (c) Any number of sub-intervals of equal width (d) An odd number of sub-intervals of equal width 73. Simpson rule for evaluation of b a dx x f ) ( requires the interval (a, b) to be divided into (a) 3n intervals (b) 2n + 1 intervals (c) 2n intervals (d) Any number of intervals 74. To calculate approximate value of by Simpson’s rule, the approximate formula is (a) 16 , 1 1 1 0 2 n dx x (b) 9 , 1 1 1 0 2 n dx x (c) 11 , 1 1 1 0 n dx x (d) 9 , 1 1 1 0 n dx x 75. In Simpson’s one-third rule, the curve y = f(x) is assumed to be a (a) Circle (b) Parabola (c) Hyperbola (d) None of these 76. A river is 80 feet wide. The depth d (in feet) of the river at a distance of x feet from one bank is given by the following table x : 0 10 20 30 40 50 60 70 80 y : 0 4 7 9 12 15 14 8 3 By Simpson’s rule, the area of the cross-section of the river is (a) 705 sq. feet (b) 690 sq. feet (c) 710 sq. feet (d) 715 sq. feet 77. A curve passes through the points given by the following table x : 1 1.5 2 2.5 3 3.5 4 y : 2 2.4 2.7 2.8 3 2.6 2.1 By Simpson’s rule, the area bounded by the curve, the x-axis and the lines x = 1, x = 4, is (a) 7.583 (b) 6.783 (c) 7.783 (d) 7.275 78. Using Simpson’s 3 1 rule, the value of 3 1 ) ( dx x f for the following data, is x : 1 1.5 2 2.5 3 f(x) : 2.1 2.4 2.2 2.8 3
• 10. Numerical Methods 173 (a) 55.5 (b) 11.1 (c) 5.05 (d) 4.975 79. By the application of Simpson’s one-third rule for numerical integration, with two subintervals, the value of 1 0 1 x dx is (a) 24 17 (b) 36 17 (c) 35 25 (d) 25 17 80. By Simpson’s rule, the value of 3 3 4 dx x by taking 6 sub-intervals, is (a) 98 (b) 96 (c) 100 (d) 99 81. If b a dx x f ) ( is numerically integrated by Simpson’s rule, then in any pair of consecutive sub-intervals by which of the following curves, the curve y = f(x) is approximated (a) Straight line (b) Parabola (c) Circle (d) Ellipse 82. If by Simpson’s rule )] ( 4 1 . 3 [ 12 1 1 1 1 0 2 b a dx x when the interval [0, 1] is divided into 4 sub-intervals and a and b are the values of 2 1 1 x at two of its division points, then the values of a and b are the following (a) 25 . 1 1 , 0625 . 1 1 b a (b) 5625 . 1 1 , 0625 . 1 1 b a (c) 1 , 25 . 1 1 b a (d) 25 . 1 1 , 5625 . 1 1 b a 83. If 09 . 20 , 39 . 7 , 72 . 2 , 1 3 2 1 0 e e e e and 60 . 54 4 e , then by Simpson’s rule, the value of 4 0 dx e x is (a) 5.387 (b) 53.87 (c) 52.78 (d) 53.17 84. If (2, 6) is divided into four intervals of equal region, then the approximate value of 6 2 2 1 dx x x using Simpson’s rule, is (a) 0.3222 (b) 0.2333 (c) 0.5222 (d) 0.2555 85. If h = 1 in Simpson’s rule, the value of 5 1 x dx is (a) 1.62 (b) 1.43 (c) 1.48 (d) 1.56
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4. ## function generator or polynomial block in PLC5
Thanks Michael, I plan to test it out tomorrow and hope the 5 breakpoints and the 2nd-order poly is accurate enough. much appreciated
5. ## function generator or polynomial block in PLC5
There was a post a few months ago, by a moderator, where he used Excel to do the same thing. I think it goes out to 5th order but I don’t have access to a computer right now (typing this in my phone, more than once :/ )
6. ## MSG alternative to acees BOOL type data?
Ok, then it’s easy, you can’t do that.
7. ## Absolute addressing on AB PLC
I’m not aware of anyway to see the level addressing that I think you want to see. Ie Tag Bool resides in some absolute memory address In the microprocessor and you want to see it. Even Mel’s reference to Local: for IO isn’t the actual physical level address of the microprocessor IO. It’s just a pointer. I think that knowing hardware level memory addressing would pose some level of security risk
8. ## function generator or polynomial block in PLC5
Thanks Mickey. Using the mycurvefit website, I was able to plot a 2nd order polynomial curve with five breakpoints, and then put the calculation into the CPT-compute instruction. Take a look at the pic, if you have time, and let me know if this is what you had in mind. Thanks!
9. ## function generator or polynomial block in PLC5
See link.. It is an online curve software to generate a formula for you. Then just use a compute instruction. https://mycurvefit.com/
10. ## function generator or polynomial block in PLC5
See link.. It is an online curve software to generate a formula for you. Then just use a compute instruction. https://mycurvefit.com/
11. ## function generator or polynomial block in PLC5
Thanks NevergoldMel. You are correct, the first curve was very linear, but this curve is more in line with actual characteristics. So, is there a way to create a poly in PLC5?
12. ## manoon DanW
I think you are a Modbus expert.
I need advice for communication modbus Tcp Fx5U to Fx5U
Thank you.
13. ## OMRON VS FACTORYTALK
You probably don't I'd suggest a OPC server.
14. ## Modbus Communication between Omron E5DC and Mitsubishi FX5U-64M
why you set analog Output ? You Can Setting Modbus RTU. Navigation>>>Parameter>>>Module Parameter>>>485 Serial Port >>Communication Protocol Type>>>>Modbus_RTU Communication
Local:
16. ## function generator or polynomial block in PLC5
With any margin of error that's a linear graph.
17. ## The fear of FOR TO loops
There's a reason QuickBASIC wasn't ported to windows.
18. ## Prv instruction in structured text
I've only used it for creating function blocks for alternate PID equations, odd pump curves, a rather odd pH vs flow equation, and such oddball things crazy P.E.'s come up with.
19. ## function generator or polynomial block in PLC5
How can I create a function generator or polynomial in PLC5/80? By this I mean I need a eleven point curve that changes the output based on the input. I do this all the time with DCS logic, but I don't see a "Polynomial" of "Function" block in RSLogix5. A simple curve might look like this. input output x1= 0 ...y1= 0 x2= 10 ...y2= 31 x3= 20 ...y3= 58 x4= 30 ...y4= 89 x5= 40 ...y5= 118 x6= 50 ...y6= 147 x7= 60 ...y7= 178 x8= 70 ...y8= 216 x9= 80 ...y9= 238 x10=90 ...y10= 269 x11=100 ...y11= 300
20. ## function generator or polynomial block in PLC5
How can I create a function generator or polynomial in PLC5/80? By this I mean I need a eleven point curve that changes the output based on the input. I do this all the time with DCS logic, but I don't see a "Polynomial" of "Function" block in RSLogix5. A simple curve might look like this. input output x1= 0 y1= 0 x2= 10 y2= 20 x3= 20 y3= 31 x4= 30 y4= 49 x5= 40 y5= 59 x6= 50 y6= 79 x7= 60 y7= 101 x8= 70 y8= 142 x9= 80 y9= 178 x10= 90 y10= 224 x11= 100 y11= 300
21. ## How I open file bin ihm fuji UG 530h ?
I need help, I want modification ihm fuji UG 530h , the upload file bin, I am use software VSF edit V6.0, but not open.
22. Yesterday
23. ## MSG alternative to acees BOOL type data?
Michael, your suggestion violates the initial condition: reading a BOOL tag should not require in the target controller any preprogrammed procedure specific to the tag name. pcmccartney1, the same, plus the MSG instruction still will not support boolean datatype, arrayed or not.
24. ## Absolute addressing on AB PLC
Good day. I'm using 1756 L71S GuardLogix Integrated Safety Controller with Studio 5000 v28. My controller is connected to a third party software through ethernet connection, it simulates my project hardware and provides all the signals to my PLC. To know all the hardware signallocation at PLC memory, it uses a Tag export file provided by Studio 5000. In other PLCs, when you declare a variable as hardware input, you have to select in which address from the memory you want this input to be read, something like "Bit Input 0.0". But in AB this step is skipped, that is all addresses are managed by Studio5000 and hidden from the user. Is there any configuration at Studio 5000 that let me assign or at least see this addresses? Normally I would prefer my tags addresses to be hidden, but in this specific case it is more useful to know memory addresses.
25. ## The fear of FOR TO loops
The biggest objection I have to loops in the PLC is that they're harder to diagnose later. I would say if you're using loops, be meticulous and over-zealous in your documentation. Explain every detail of what you're doing and why so a technician on 3rd shift won't have any issues figuring it out 5 years from now. Without having to call an engineer. From a programming standpoint, just be careful when designing your loops to properly limit check them so you don't run past the end of your array. Most production managers don't have much of a sense of humor when it comes to faulting processors. Writing the loop is often simpler to avoid typos and copy-paste errors than having a rung for each element. If you have 40 items, it may not be a huge deal to have a rung for each, but when your array is a couple hundred or a thousand items, you almost have to do a loop just to stay sane. As an end user (not an OEM) who has to support the code in the long term, I will sacrifice memory/performance every time for ease of troubleshooting except in a few limited scenarios where the loop is doing something that's not critical to the machine's operation and that's unlikely to require diagnostics later. Like looping through alarm messages to display on an HMI. War story: we had a contractor come in and create a very complex piece of code for us that has multiple nested loops that are almost impossible to untangle. It works great...until something goes wonky and we have to download an old copy of the program into the PLC to get it working again. Power-cycling the PLC doesn't fix it. There's some tag value somewhere that makes it not work. After several years of it happening every few months, we're no closer to identifying the pattern. I spent several days one time just going through the code and making a flowchart but wasn't able to reproduce the bug. Weirdly enough, we have the identical code running in 3 lines in 3 different processors. One of them (originally a ControlLogix -L55M12 at v16 and now an -L81E at v30) has never gone wonky, another one (ControlLogix -L71, v24) goes wonky every other month or so, and the third (CompactLogix -L33ER, v24) has only gone wonky twice in 1.5 years or so. I've run comparison after comparison between the systems and haven't been able to figure out why that would be. In my opinion, loops definitely have their place, but PLEASE document them so I don't get a phone call at 3AM.
26. ## The fear of FOR TO loops
Hi everybody,For a while I’ve been wondering, in every plc forum where the use of loops come to discussion I find most people answering that it’s best to stay away from them and that there are better, easier and mostly less scan time demanding solutions out there. Why is that?In my experience they’re the next best thing.Why would referencing data from an array of 40 Dwords using a FOR X:= 0 TO 40 DO loop be any more scan time demanding then the PLC scanning through 40 rungs of statments?Just curious if I’m missing something here. Cheers
27. ## Prv instruction in structured text
Oh and btw, thanks Michael, reading A 270 as a DINT made me recieve some numbers
28. ## Prv instruction in structured text
Maybe it’s a European thing Like Michael said, in certain cases it’s very useful. In my current project for example, I work with large amounts ascii strings and data set in Lwords, which need to be shifted or moved and referenced in various ways Also I have remote in- and outputs from 40 individual stations for which I use arrays and lots of FOR TO loops and CASE statements to make things easier to implement en modify. In my opinion, once you get the hang of it the possibilities are endless and it’s much less time consuming to write, especially when using loops. | 2,357 | 9,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.899756 |
https://thecooperschool.org/5th-grade-ambassadors/ | 1,582,384,630,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00545.warc.gz | 583,798,508 | 14,356 | # The Cooper School Daily
Science
When you want to move somewhere, you probably stand up and use those muscular legs to transport yourself! But what if you were a rock? How exactly would you get to a new location? After studying how large rocks break into smaller rocks for a couple weeks, your scientists have moved on to asking how these smaller rocks then move from one place to another. With the use of a stream table model, 5th graders investigated the movement of sand and silt under a normal rain condition. We noticed a couple of things! First, we noticed that the lighter material, silt, was picked up first by the water and moved farther than the sand. Then, we noticed that the water created valleys, canyons, and alluvial fans in our earth material. Next week we will look at how earth materials move during a flood situation!
Math
Your fifth grade mathematicians continue to work like middle schoolers as they expand their use of algebraic expressions. Now that they have a strong understanding of what an algebraic expression is, they applied this knowledge to real world situations in the form of rules and formulas. One formula we used was a way to predict when Old Faithful Geyser would erupt. After using the formula to calculate certain values, your bright students graphed these values on a two-coordinate grid. Creating a line graph allowed them to make predictions for a variety of scenarios. We practiced this skill over and over again using different formulas and graphing our results. Ask your child what they find interesting about algebra! | 350 | 1,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.919594 |
https://math.stackexchange.com/questions/711440/string-probability-with-conditional-prob-and-combinations | 1,560,814,854,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998581.65/warc/CC-MAIN-20190617223249-20190618005249-00075.warc.gz | 505,453,204 | 36,719 | # String probability (with conditional prob and combinations)
I'm having trouble with the questions below, all relating to string probability. I'll write the problem and then provide my work for my (incorrect) answer. Please help me figure out what I did wrong.
SET1: Assume a string, allowing repetition, is randomly selected from all strings of length 4 from the set A = {r, e, a, s, o, n}.
Q1. What is the probability that it contains exactly one letter that is a vowel?
A1: 6^4 total probability, thus one vowel should be equal to... (C(4,1) * C(4,3)) / (6^4)
Q2. What is the probability that such a string contains exactly two r's given that it contains exactly two o's?
A2: two r's and two o's in a string of length 4 have 6 permutations, so according to conditional probability, we must show P(two r's | two o's) = P(two r's and two o's) / P(two o's), thus we have... (6 / (6^4)) / (C(4,2) * ((1/6)^2) * C(4,2) * ((5/6)^2))
SET2: How many distinct permutations of the letters in "letters"...
Q1. Begin with two vowels?
A1: 2 * 1 * 5 * 4 * 3 * 2 * 1 = 2 * (5!), since there are two vowel choices for the first spot, one for the second spot, then five remaining letters, then four, etc...
Q2. Begin with two e's or end with two t's?
A2: 2(2 * (5!)) - (2 * 1 * 3 * 2 * 1 * 2 * 1), used similar logic to the logic explained above in A1.
Q3. Have the vowels together?
A3: 6!, since you group the vowels together as one entity then find a place for all 6 entities.
Thank you!
p.s. If someone could format this, I'm not familiar with latex formatting and didn't see anything about math formatting in the advanced formatting help link. :-/
• For the first, the vowel can be chosen in $3$ ways, its location in $4$ ways, and the remaining slots can be filled with consonants in $3^3$ ways, so the probability is $(3)(4)(3^3)/6^4$. For the second, your expression is hard to read, one should get $1/25$. – André Nicolas Mar 13 '14 at 22:43
• Thanks for the effort on the first one, could you explain it using combinations as I did so I can understand exactly where I went wrong? That would be great. An answer alone for the second doesn't do me much good though since I got it wrong and don't know how to find the 1/25 probability. :( I'm really looking to learn from my mistakes. – CODe Mar 13 '14 at 22:57
• My solution used combinations, the vowel can be chosen in $\binom{3}{1}$ ways, and its location can be chosen in $\binom{4}{1}$ ways. For every way of settling the kind and location of the vowel, there are $3^3$ permutations of the consonants that make up our $4$-letter word. – André Nicolas Mar 13 '14 at 23:03
• If you are going to use the conditional probability formula for the second, let's do probability of $2$ o's first. The $2$ o's can be place in $\binom{4}{2}$ ways. The remaining $2$ slots can be filled with non-o's in $5^2$ ways, so the probability is $\binom{4}{2}5^2/6^4$. For the probability of $2$ o's and $2$ r's, the only freedom we have is where the o's go, this can be chosen in $\binom{4}{2}$ ways, the probability of $2$ of each is $\binom{4}{2}/6^4$. – André Nicolas Mar 13 '14 at 23:08
For problem Set1:
I will use $v$ for vowel and $c$ for consonant. For (1), you have strings of length $4$. So $s = _ _ _ _$. You want a single vowel. Suppose $s = v _ _ _$. The remaining three spots are consonants. There are three consonants, with one consonant per slot. So $s = v c c c$. So the number of ways to get $v = \binom{3}{1} = 3$, and the number of ways to get each $c$ is also $3$. So there are $3 * 3^{3} = 3^{4}$ ways to form a string with the first letter a vowel.
Now consider if $v$ is in the second position. The number of such strings is the same as if $v$ was in the first position. So you can easily see that the answer to your question is $3^{4} * 4$.
For the problem Set2:
For your first problem with "letters", there are two e characters. So by the multinomial counting rule, you divide by 2. Thus, the number of ways to start with two vowels is 5!.
You are making the same mistake in the second problem. Since there are two t's, you have to divide by 2. Thus, you have 2∗5!−3! as your answer. Your thinking is correct, though. Count first the number of ways to get two e's upfront (which is the answer to your first problem), then count the number of ways to get two t's at the end (also the answer to your first problem), then subtract out the number of ways to start with two e's and end with two t's (ie., permute the middle 3 characters). | 1,311 | 4,510 | {"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} | 3.84375 | 4 | CC-MAIN-2019-26 | latest | en | 0.916735 |
http://www.answers.com/topic/cleto-1 | 1,544,655,443,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824180.12/warc/CC-MAIN-20181212225044-20181213010544-00492.warc.gz | 332,814,090 | 52,540 | ## Results for: Cleto-1
In Personal Finance
# What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow (MORE)
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# Square coin from Neatherland has flowers on 1 side and a 5c and no date.is it a coin or metal?
I have one as well....all I can tell you is that there is a date on the saide of the coin that is 5c it is in the corner of the flowers and the coin I have 1923 tests to be si (MORE)
In Acronyms & Abbreviations
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo (MORE)
In Coins and Paper Money
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
In Math and Arithmetic
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
In Numerical Analysis and Simulation
# What is the answer for 5c equals -75?
The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio (MORE)
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# How many pixels does the iPhone 5c have?
The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
In Temperature
# What is minus 5c in Fahrenheit?
(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
In iPhone 5
# How many inches is a iPhone 5c?
The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35" | 596 | 1,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.924785 |
https://www.physicsclassroom.com/mop/Work-and-Energy/Energy-Analysis/QG2help | 1,660,133,047,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00272.warc.gz | 826,798,520 | 22,709 | # Work and Energy - Mission WE8 Detailed Help
The path of a sledder gliding across the ice and snow is shown in the diagram below. Frictional forces can be assumed to be negligible. Perform an energy analysis and fill in all the blanks.
An energy analysis begins by first identifying which types of forces are doing net work upon the object. In the case of the sledder, frictional forces (friction and air resistance) are said to be negligible. The normal force acts perpendicular to the sledder's motion and does not do work upon the sledder. So gravity - a conservative force - is the only force doing work upon the sledder. Since the only work being done on the sledder is being done by a conservative force, the total mechanical energy will be conserved. See Know the Law section. The amount of total energy - KE plus PE - at the beginning of the motion will be the same amount at every location along the path of the sledder. And of course at location E, there is no potential energy since the sledder is at ground level (much like at location C). Using these principles, all blanks can be determined.
Energy Conservation: If the only forces doing net work upon an object are conservative forces (such as gravity and spring forces), then the mechanical energy of the object will be conserved. The energy may change from one form to another - potential to kinetic or vice versa; but the total amount of the two forms together will be unchanging. | 316 | 1,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.889804 |
http://www.numericalmethod.com/javadoc/suanshu/com/numericalmethod/suanshu/algebra/linear/matrix/doubles/factorization/eigen/qr/Deflation.html | 1,537,371,334,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156252.31/warc/CC-MAIN-20180919141825-20180919161825-00064.warc.gz | 383,556,339 | 3,805 | SuanShu, a Java numerical and statistical library
com.numericalmethod.suanshu.algebra.linear.matrix.doubles.factorization.eigen.qr
Class Deflation
• java.lang.Object
• com.numericalmethod.suanshu.algebra.linear.matrix.doubles.factorization.eigen.qr.Deflation
• public class Deflation
extends Object
A deflation found in a Hessenberg (or tridiagonal in symmetric case) matrix. Given a Hessenberg matrix, $\begin{bmatrix} H_{11} & H_{12} & H_{13}\\ 0 & H_{22} & H_{23}\\ 0 & 0 & H_{33} \end{bmatrix}$
• $$u_l$$ is the upper left hand corner index of H22;
• $$l_r$$ is the lower right hand corner index of H22
Deflation of an upper Hessenberg matrix splits it into multiple smaller upper Hessenberg matrices when the sub-diagonal entries are sufficiently small. For example, suppose $H = \begin{bmatrix} 1 & 2 & 3 & 4 & \\ 5 & 6 & 7 & 8 & \\ 0 & 9 & 10 & 11 & \\ 0 & 0 & 12 & 13 & \end{bmatrix}$ We can split H into H1 and H2, so that $H_1 = \begin{bmatrix} 1 & 2 \\ 5 & 6 \\ \end{bmatrix}$ $H_2 = \begin{bmatrix} 10 & 11 \\ 12 & 13 \\ \end{bmatrix}$
"Golub and van Loan, Section 7.5.1.", "G. H. Golub, C. F. van Loan, "7.5.2 Deflation," Matrix Computations, 3rd edition."
• Method Summary
All Methods
Modifier and Type Method and Description
int getLowerRight()
Gets the lower right corner of the deflation.
int getUpperLeft()
Gets the upper left corner of the deflation.
• Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• Method Detail
• getUpperLeft
public int getUpperLeft()
Gets the upper left corner of the deflation.
Returns:
the index of the upper left corner
• getLowerRight
public int getLowerRight()
Gets the lower right corner of the deflation.
Returns:
the index of the lower right corner | 551 | 1,799 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-39 | longest | en | 0.537187 |
https://worldbuilding.stackexchange.com/questions/101915/does-the-interior-surface-of-a-dyson-sphere-exert-gravity/101918 | 1,726,864,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00020.warc.gz | 555,865,115 | 46,331 | # Does the interior surface of a Dyson Sphere exert gravity?
The interior of this particular Dyson Sphere is so large that it encapsulates a star at its center. Otherwise it is filled with breathable air which creatures can breath and fly in. I imagine that for the most part it contains a zero gravity environment, except for the areas closest to the star and to the interior surface of the sphere itself. How would I calculate these?
Edit: Thank you for all the answers and insights.
• This has been asked a number of times on this site. I'll find a duplicate in a moment, but the answer is no. As you descend through a sphere, the only gravity you feel is the gravity below you (toward the center). Everything above you is balanced out by the mass "across the way" from you. Therefore, inside a Dyson Sphere there is no gravity due to mass (now... if you can get that sphere spinning...)
– JBH
Commented Jan 9, 2018 at 23:57
• Well... this isn't an exact duplicate but it goes into a ton of detail about the nature of dyson spheres.
– JBH
Commented Jan 10, 2018 at 0:02
• Nope. Gravity within a uniform hollow sphere is always zero. Check out this website for the more mathy version: grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/… Commented Jan 10, 2018 at 0:07
• The relevant science is The Shell Theorem and inside a symmetric shell, gravity is zero everywhere. More precisely the net force from the shell is zero everywhere inside the shell. Commented Jan 10, 2018 at 0:19
• I also have doubts about air. I think it would not be physically possible. Commented Jan 10, 2018 at 7:30
A Dyson sphere is a hypothetical megastructure that completely encompasses a star and captures most or all of its power output
I had to look up what a Dyson sphere was, but using this definition from Wikipedia, I'm assuming it's a spherical shell. If you can assume it is a uniform spherical shell, i.e. it has constant density throughout, then the sphere will exert no gravitational force on any object inside it. This can be easily proved using Gauss' law applied to gravity
$$\nabla \cdot g = -4 \pi G\rho$$
For thorough derivations, search 'gravity shell theorem'. This result was derived by Newton and can be easily obtained with only geometry, Newton's law of gravity and patience.
Things get trickier if you have a non-uniform density or do something else to break the symmetry of the problem, but that gets complicated quickly. If you need a gravitational-like force sticking people to the shell, there are much better solutions out there.
• Newton's law of gravity and patience. What goes up... must come down... but only if you wait long enough. Commented Jan 10, 2018 at 23:14
• Only when up and down are defined.
– Mary
Commented Dec 24, 2021 at 0:30
Using an equation derived from the Law of Universal Gravitation, you can calculate the force of gravity a distance $x$ away from the center of mass. The equation is here:
$$g = \frac{Gm}{x^2}$$
where $G$ is the gravitational constant, $m$ is the mass of the body, $x$ is the distance from the center of mass (for force at the surface of body, this is just the body's radius), and $g$ is the resultant acceleration due to gravity.
This equation is directly applicable to the question of the star's gravitational effects. Just based on some napkin math, I think you'll find that any point where the radiation from the star is non-lethal will also have negligible gravity from it.
However, this equation isn't very helpful for the Dyson sphere calculations. Sure, you could assume that the sphere is a point mass and do the same calculations, but a) this would be inaccurate as the sphere is hollow, and b) would imply that people on the inside of the sphere would be hanging on 'upside down', which doesn't seem right. So let's take a better look at it.
This research site has a good argument about this. Basically, it works out that for a uniform hollow sphere, all points within the sphere have an effective gravity of zero. This makes sense if you think about it: by being inside the sphere, you are automatically subjected to the net gravitational forces of all the surrounding sphere walls which must, intuitively, sum to zero. The differential equation on the link is the mathematical way of proving that intuition.
In summary
Your inhabitants would experience no gravity due to the sphere itself (while inside it, that is) no matter how massive the sphere's radius, thick the outer wall, or dense the unobtainium it is built out of.
The star, too, would exert little gravity on the inhabitants as (presumably) the inner surface of the sphere is very far away from the star in order to keep radiation at survivable levels. That said, if the radius of the sphere were to be very close to the radius of the star, then yes presumably there would be some gravity due to the star's mass, pulling towards the construction's center of gravity (the center of the star).
Side note...
One thing you could look into is spinning the sphere about a central axis in order to simulate gravity, much like hypothetical space ships/stations could. The acceleration of spinning would have a tendency to "push" people away from the center, resulting in the appearance of gravity on the inner surface.
• There is definitely gravity from the star and people on the sphere would need to be at least in an orbit (i.e. free-fall) assuming they want to stay in their nice habitable zone. In practice, the sphere should be rotating a bit faster than this orbit so that the rotating reference frame gives you the experience of a 'downward force' into the sphere wall. This only works in a narrow ring around the center for a true sphere, with one axis of rotation. It gets more complex if you have multiple rings that rotate independently. Commented Jan 10, 2018 at 3:12
• Ah, so a series of nested ringworlds could encapsulate the entire star and have 1g of gravity Commented Aug 13, 2022 at 17:57
The previous answers seem to assume a hollow sphere - but this is not the case here:
it encapsulates a star at its center. Otherwise it is filled with breathable air which creatures can breath and fly in.
So what we have here is a gas (and one star) filled balloon with a (presumably metal) outer skin. Leaving aside the fact that a star in an atmosphere would probably behave WILDLY different from one in a vacuum, this means we're looking at a body that has mass everywhere in it (although at different densities), leading to an overall gravity that points to the center... which means, anything inside your dyson sphere would be drawn towards said center, and thus the star.
If you want to avoid all your nice atmosphere and creatures falling to a fiery death (although it might be a long enough fall to starve on the way, not sure there^^) until you DO end up with a hollow shell (except for the star), I'd recommend rethinking where you want to place that atmosphere. If you handwave some force fields that limit the atmosphere to (some number of kilometers/miles/whatnot), you will have a hollow (except for the star, again) shell and can go with any of the other answers here :)
• +1. Not only the creatures etc. inside the sphere would fall into the star, but so would all the air, turning it into a much bigger, much hotter star than it was originally, surrounded by a near vaccum. Commented Jan 10, 2018 at 9:37
• @Nathaniel Well, normal stars spew out huge amounts of material. If you kept them inside a closed container, it might actually give out more atmosphere that it takes in. It might make for some very interesting behaviour overall, even if you just think about the inevitable "reflection" of massive amounts of radiation and matter back into the star... Commented Jan 10, 2018 at 12:39
• @Luaan that may be true - I'm not sure of the mass balance calculations but you might be right. Also: when it comes down to it, the only reason the Earth doesn't heat up to the temperature of the Sun is that it can radiate heat away into space. The inside of the sphere can't do that, so it will heat up to 6000 celsius or so. (Assuming a Sun-like star.) If that doesn't melt the sphere it will heat up even more because the star itself will not be able to cool down if it can't radiate into space without its radiation being reflected back. Commented Jan 10, 2018 at 13:12
• The star is not in an atmosphere. Imagine the air pressure of a 1 AU column of air!
– rek
Commented Jan 10, 2018 at 17:26
• Incidentally, a 1AU sphere of "breathable air", assuming a uniform air pressure of 0.4 bar, has over 1000 solar masses. The physics of thousand solar mass spheres of mostly nitrogen is not well studied, so whether the resulting ball of gas is hotter or cooler than the star that was there before isn't obvious. Commented Jan 10, 2018 at 18:01
Other answers address the question of gravity, so I'll just expand on the atmosphere topic.
### The Problem
The biggest physics issue with your proposed world is the atmosphere.
A star is formed when enough gas is present that the gravity from all of the gas is enough to collapse it down into a dense hot sphere. It would thus collapse any atmosphere around it into itself. First you may think, well maybe the star only collapses the heavy elements and the light ones could still form an atmosphere. However, most of a star is hydrogen, the lightest element. So this is not true.
The density and pressure of a gas grows exponentially towards a the source of gravity, so it is impossible to have an extremely thick atmosphere without reaching pressures where things form plasmas. The exponential growth is because each layer of gas must have a pressure that can support all of the weight above it.
Even if the star initially didn't have enough gravity by itself to suck down a giant atmosphere, even a low density atmosphere would have more mass and thus gravity than a typical star.
### The Solution
If you want to have a giant atmosphere, consider having it on the outside of a shell. So then you'd have a star, a large vacuum, a (possibly transparent (maybe even diamond)) inner shell, an atmosphere, and an optional outer shell.
With this arrangement your gravity in the atmosphere would be:
$$g=\frac{G\,(M_{star}+M_{shell}+ M_{atm})}{r^2}$$
Where $$r$$ is the distance to the center of the star, $$g$$ is your gravitational acceleration, $$G$$ is the universal constant of gravitation, and the $$M_{star}$$, $$M_{shell}$$, and $$M_{atm}$$ are the masses of the star, inner shell, and portion of the atmosphere closer to the star than $$r$$ respectively.
So now let's take a look at the equations for the atmosphere to see if we can come up with some numbers that will fit your criteria:
First the specific ideal gas law relating temperature $$T$$, density $$\rho$$, and pressure $$P$$:
$$\rho=\frac{P}{RT}$$
Where $$R$$ is the specific gas constant (for air = $$286.9\frac{J}{kg\,K}$$)
Let's say the temperature is constant to simplify our analysis, and keep our creatures comfortable.
The change in mass of atmosphere closer than $$r$$ as $$r$$ increases will just be the surface area of the sphere of size $$r$$ times the density at that $$r$$:
$$\frac{d\,M_{atm}}{dr}=4\,\pi\,r^2\,\rho=\frac{4\,\pi}{RT}\,r^2\,P$$
Then since the pressure of the atmosphere must support the weight of the gas above it, the rate of change is also related to the density:
$$\frac{dP}{dr}=\rho\,g=\frac{P}{RT}\frac{G\,(M_{star}+M_{shell}+ M_{atm})}{r^2}$$
To simplify things a little let's change our variables:
$$M=M_{star}+M_{shell}+ M_{atm}$$
$$\frac{dM}{dr}=\frac{dM_{atm}}{dr}$$
Now we almost have enough information to do a numerical integration; we just need our initial values, and our constants. So let's try:
$$T=25^\circ C$$
$$M_0=M_{sol}= 10^{30} kg$$
$$P_0=1 atm = 10^5 Pa$$
$$r_0 = 1 AU = 1.5\times 10^{11} m$$
Integrating numerically we can get a plot of pressure vs altitude:
As you can see, the pressure drops off to less than three quarters of the initial pressure (enough to cause altitude sickness) by about 5000 km. Certainly a thicker breathable atmosphere than the measly 2.4 km that earth has, but let's see if we can do better.
By increasing our starting radius and decreasing the mass of our star and shell we can decrease the rate of pressure drop off, so let's look at a start with near the minimum mass to still be a red dwarf, about a tenth of our sun, and let's start out 100 times as far away:
For this system it looks like the breathable atmosphere would extend out to 7000 km: not much of an improvement for the extremes it took to get there. At 100 AU out from the star, you'd probably need an outer shell to insulate and keep your atmosphere warm.
Other concerns are how you'd get an intershell with a curvature of 1 AU to withstand an atmosphere of pressure, even if it was 100 miles thick, it would need to withstand 100 GPa of stress (diamond breaks somewhere in the 70-300 GPa range) Of course 100 mile thick shell of diamond would have a mass of 160 solar masses, but of course with that much mass we'd need to support its own weight in addition to the atmosphere. Turns out we just can't do this, so maybe just hand wave it away? Weightless force field? | 3,170 | 13,231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.933715 |
https://www.brainkart.com/article/Probability-Distributions_41261/ | 1,726,097,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00061.warc.gz | 622,970,689 | 6,733 | Home | | Maths 12th Std | Probability Distributions
# Probability Distributions
Probability theory is nothing but common sense reduced to calculation -Laplace
Probability Distributions
Probability theory is nothing but common sense reduced to calculation -Laplace
The history of random variables and how they evolved into mapping from sample space to real numbers was a subject of interest. The modern interpretation certainly occurred after the invention of sets and maps (1900), but as Eremenko says, random variables were used much earlier. Mathematicians felt the need to interpret random variables as maps. In 1812, Laplace published his book on Theory analytique des probabilities in which he laid down many fundamental results in statistics. The first half of this treatise was concerned with probability methods and problems and the second half with statistical applications.
## Learning Objectives
Upon completion of this chapter, students will be able to
• define a random variable, discrete and continuous random variables
• define probability mass (density) function
• determine probability mass (density) function from cumulative distribution function
• obtain cumulative distribution function from probability mass (density) function
• calculate mean and variance for random variable
• identify and apply Bernoulli and binomial distributions.
## Introduction
The concept of a sample space that completely describes the possible outcomes of a random experiment has been developed in volume 2 of I year higher secondary course.
In this chapter, we learn about a function, called random variable defined on the sample space of a random experiment and its probability distribution.
Tags : Mathematics , 12th Maths : UNIT 11 : Probability Distributions
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Maths : UNIT 11 : Probability Distributions : Probability Distributions | Mathematics | 365 | 1,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-38 | latest | en | 0.890905 |
https://sangarema.com/simple-solutions-434 | 1,675,748,706,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00530.warc.gz | 492,606,304 | 5,301 | Keep reading to learn more about 6.2 homework answers and how to use it. Math can be difficult for some students, but with the right tools, it can be conquered.
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Proportions solver Solving multi step equation calculator How to get answers for math homework Y homework solver Exponential solver calculator | 850 | 4,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-06 | latest | en | 0.957581 |
https://ez.analog.com/search?q=AD8251 | 1,638,902,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00057.warc.gz | 321,456,194 | 29,913 | Hi,
I am attempting to use the AD825x devices as current sense and voltage sense instrumentation amplifiers in an analog control loop. I was unable to get the SPICE models for the AD825x to work with PSpice and LTSpice. While the circuit simulation was…
AD8250 has differential input impedance of 5.3Gohms/ 0.5 pf.
I have input signal whose differential input impedance which is driving AD8250 is 10Mohms & 5pf.
My customer is saying that i may not be able to give differential signal to AD8250 directly…
• ### Spice Model of AD8251
Hi
I simulated the AD8251 in LTspice and found strange input bias current.
This part's specification of input bias current is 30nA max but it show a much bigger value.
Would you tell me the reason?
Hi,
I wonder if it's enough for AD8253/AD8251/AD8250 to output ±10V when it is supplied with a power of ±12V ?
Thanks.
• ### Varying input impedance of AD8251
I am using an AD8251 to measure the amplitude of an AC signal across a sense resistor. The input impedance of the AD8251 is in parallel with my sense resistor. Under normal operation, the circuit works great, but when I use large amplitude or high frequency…
• ### AD8251 Input Bias Current Model
Spec sheet for AD8251 cites nA input bias currents.
I have a rather simple AD8251 simulation running in LTspice (wrapping the AD model) and it is modeling ~4.5uA of input bias current.
Any suggestions?
• ### Input capacitance measurement for AD8251
Hi All,
I am going to use programmable gain amplifier AD8251 in a data acquisition system.
I want to know how to measure input impedance and capacitance of AD8251 when it present in circuit.
Please anybody suggest me as soon as possible.
Thanks,
• ### AD8251 differential output voltage calculation question
I would like to use the AD8251 in-amp in the same circuit configuration shown in figure 61 of its datasheet. I have two questions...
1. I can calculate the in amp and the level shift output voltage equations separately but am having trouble coming up… | 485 | 2,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.8733 |
https://mathzsolution.com/what-would-base-11-be/ | 1,670,581,227,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00075.warc.gz | 427,291,522 | 17,880 | # What would base 11 be?
Base $10$ uses these digits: $\{0,1,2,3,4,5,6,7,8,9\};\;$ base $2$ uses: $\{0,1\};\;$ but what would base $1$ be?
Let’s say we define Base $1$ to use: $\{0\}$.
Because $10_2$ is equal to $010_2$, would all numbers be equal?
The way I have thought Base 1 might be represented is tally marks, $0_{10}$ would be represented by nothing. So, $5$ in Base 1 would be represented by $00000$? Or we could define Base 1 to use: $\{$|$\}$ and $5$ would be |||||? | 173 | 479 | {"found_math": true, "script_math_tex": 15, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.961825 |
http://usaco.org/index.php?page=viewproblem2&cpid=617 | 1,524,625,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00005.warc.gz | 337,095,478 | 3,873 | ## Problem 3. Load Balancing
Contest has ended.
Log in to allow submissions in analysis mode
Farmer John's $N$ cows are each standing at distinct locations $(x_1, y_1) \ldots (x_n, y_n)$ on his two-dimensional farm ($1 \leq N \leq 100$, and the $x_i$'s and $y_i$'s are positive odd integers of size at most $B$). FJ wants to partition his field by building a long (effectively infinite-length) north-south fence with equation $x=a$ ($a$ will be an even integer, thus ensuring that he does not build the fence through the position of any cow). He also wants to build a long (effectively infinite-length) east-west fence with equation $y=b$, where $b$ is an even integer. These two fences cross at the point $(a,b)$, and together they partition his field into four regions.
FJ wants to choose $a$ and $b$ so that the cows appearing in the four resulting regions are reasonably "balanced", with no region containing too many cows. Letting $M$ be the maximum number of cows appearing in one of the four regions, FJ wants to make $M$ as small as possible. Please help him determine this smallest possible value for $M$.
For the first five test cases, $B$ is guaranteed to be at most 100. In all test cases, $B$ is guaranteed to be at most 1,000,000.
#### INPUT FORMAT (file balancing.in):
The first line of the input contains two integers, $N$ and $B$. The next $n$ lines each contain the location of a single cow, specifying its $x$ and $y$ coordinates.
#### OUTPUT FORMAT (file balancing.out):
You should output the smallest possible value of $M$ that FJ can achieve by positioning his fences optimally.
#### SAMPLE INPUT:
7 10
7 3
5 5
9 7
3 1
7 7
5 3
9 1
#### SAMPLE OUTPUT:
2
Problem credits: Brian Dean
Contest has ended. No further submissions allowed. | 466 | 1,770 | {"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} | 3.21875 | 3 | CC-MAIN-2018-17 | longest | en | 0.925345 |
https://cryptonews.com/news/how-calculate-crypto-roi-5-tips-for-positive-return.htm | 1,701,599,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00779.warc.gz | 217,971,219 | 35,591 | # How to Calculate Crypto ROI and 5 Tips For a Positive Return
Disclaimer: The text below is an advertorial article that is not part of Cryptonews.com editorial content.
After investing in digital assets, users should know how to calculate return on investment when trading cryptocurrency. Since successful crypto traders tend to set targets before selling their holdings, it is crucial to calculate crypto gains.
This guide will discuss what is ROI in cryptocurrency and show users how to calculate crypto returns.
## How to Calculate ROI on Crypto – Quick Guide
How do you calculate crypto profit? Users can simply calculate crypto returns with a simple crypto ROI calculator formula. The steps below demonstrate how to do so using a Bitcoin ROI calculator formula:
• To calculate crypto return, users can reduce the original buying price from the current price and then divide the result with the original price again.
• For example, let’s assume that a user bought \$BTC at USD 20,000 and sold the token at USD 60,000.
• After reducing the original price from the current BTC price, users will get the value of USD 40,000. Then USD 40,000 must be divided by the original price, which is USD 20,000.
• Therefore, the crypto return on investment calculator assumes the ROI as 2, or 200%.
Check out IBAT – The Best Crypto with Highest ROI Potential in 2022
## What is ROI on Crypto?
Users must be wondering what is ROI in cryptocurrency? As described in the example above, the ROI crypto meaning is the return on investment on a particular asset.
This formula is used to calculate a user’s profit/loss. For example, a trader may want to exit a trade after making an ROI of 20%. This can be used to calculate a user’s crypto mining ROI, Bitcoin ROI or any other investment they choose to track.
## The Top 5 Tips for a Positive Return when Investing in Cryptocurrency
The following sections provide users with a few tips for crypto investing to potentially maximize profits.
1. Diversify Assets – Users should look at building their portfolio with multiple cryptocurrencies. Even when one of your assets is down, a diversified portfolio ensures that a single asset does not cause volatility to your investments.
2. Set a Stop-loss – A stop-loss is a pre-fixed order to sell a particular asset after it falls to a specific level. Successful traders implement this tool to ensure that they do not lose additional value, and exit the trade on time.
3. Pick High-Growth Tokens – Users can look to invest in upcoming tokens such as Battle Infinity and Lucky Block, which can potentially provide high gains in the long term.
4. Leverage Assets – Applying leverage on cryptocurrencies allows users to multiply the value of their position without depositing more funds. However, this is a high-risk high reward tip since users stand a chance to multiply their losses and profits.
5. Long-Term Investing – Investing for the long term is one of the popular tips for cryptocurrency trading. Users confident in a crypto project can simply invest in the token and hold it for multiple years.
Users looking for how to successfully day trade crypto can check out our guide on the best day trading cryptocurrency platforms in 2022.
## Which Crypto offers the Best ROI in 2022?
When looking for the best cryptos to invest in, users want to find out about the assets that provide the best ROI in 2022. After reviewing the available projects, we recommend Battle Infinity and Lucky Block as the best ROI projects in 2022.
Battle Infinity is a decentralized project that is currently in its presale phase. The platform offers its utility token (IBAT), which can be used to avail of several benefits within the Battle Infinity ecosystem. We will discuss more on this in the sections below.
Lucky Block is an NFT Competitions game that use LBLOCK, its native crypto, to distribute rewards. Users can participate in weekly main draws, and NFT draws and stand a chance to earn up to USD 50,0000 in rewards. After releasing in January 2022, LBLOCK soared by over 1,100% in the first two weeks.
Visit Lucky Block Now
## What is Battle Infinity?
As mentioned above, Battle Infinity is an upcoming crypto project aiming to revolutionize gaming technology by implementing various decentralized elements and blockchain protocols. Users can participate in different play-to-earn (P2E) platforms to earn rewards within the games.
IBAT is the native crypto of Battle Infinity, which is used to distribute rewards and handle the platform. One of the products of Battle Infinity is the Premier League P2E, the first NFT-based sports fantasy league. Users can also engage in the Battle Arena – a metaverse and virtual ecosystem on Battle Infinity.
Another feature is the Battle Market Players can access the Battle Market to customize and design their avatars minted as NFTs using ERC 721 protocols. Furthermore, the IBAT Battle Stake Arena lets users stake their IBAT tokens to receive a potentially high Annual Percentage Yield (APY) in return.
Battle Infinity even hosts global liquidity pools, where portions of IBAT tokens are deposited as transaction fees. Interested investors can join the Battle Infinity Telegram Group to stay updated with the latest developments in this decentralized ecosystem.
Cryptoassets are a highly volatile unregulated investment product. No UK or EU investor protection.
## Should I Buy Battle Infinity?
The sections below provide a few reasons and factors for why users may want to invest in Battle Infinity.
### Benefits of Buying Battle Infinity
• Innovative P2E Platform – Battle Infinity is the first ever blockchain protocol that lets users participate in an NFT-based sports league. Furthermore, users can participate in the platform’s metaverse, design their own NFT avatars and engage in events and concerts on a virtual ecosystem.
• Multi-Utility Token – IBAT is used to distribute in-game rewards, provide liquidity to Battle Infinity’s platform, used for monetization purposes within the metaverse and can be staked and swapped on Battle Infinity.
• Exciting Roadmap – Battle Infinity plans to expand its crypto project by implementing strategic NFT mainnent deployments, launch IBAT on PancakeSwap DEX and use several marketing campaigns to promote the brand.
## Battle Infinity Price
Battle Infinity (IBAT) launched for a presale price of USD 0.0015 per token on July 11th, 2022. The 90-day presale event is on course to sell out before the October 10th deadline due to the strong momentum the token has witnessed.
Within just over a week of the presale going live, IBAT has gathered over USD 600,000 in its seed funding round. Since the soft cap targets have been met, Battle Infinity is on course to begin with the game development and beta testing.
## Battle Infinity Price Prediction
While it is hard to predict when the crypto market will go up, Battle Infinity’s future price can be predicted by looking at its fundamentals.
According to a report published by McKinsey, the metaverse could potentially grow to hit a USD 5 trillion valuation. Battle Infinity has the potential to acquire a significant share due to the multiple features that it provides via the IBAT token.
Another boost for Battle Infinity is that it originated from India – a growing technology hotspot. With leading projects like Polygon emerging from this region, India is becoming a hub for tech-savvy investors and entrepreneurs.
IBAT is on course to provide significant returns after starting its presale launch with strong interest. Read our guide to see how IBAT could explode to deliver 100x gains
## How to Buy Battle Infinity – Detailed Tutorial
Now, let’s look at how to buy Battle Infinity.
### Step 2: Connect MetaMask to BSC
• Network Name: Smart Chain
• New RPC URL: https://bsc-dataseed.binance.org/
• ChainID: 56
• Symbol: BNB
• Block Explorer URL: https://bscscan.com
Users must buy Binance Coin before purchasing IBAT. Simply access a crypto exchange, purchase BNB and transfer the tokens to the MetaMask wallet.
Head over to the Battle Infinity presale website, and click on “Connect Wallet” to link MetaMask with Battle Infinity.
Users can then choose the MetaMask option and follow the instructions to link the wallet with the presale.
### Step 5: Buy IBAT (Battle Infinity)
Users can simply scroll down to the “Buy IBAT” open order position to continue.
Enter the amount of IBAT they wish to purchase and click on “Buy IBAT” to confirm the transaction. | 1,756 | 8,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.850936 |
ustsvs.com | 1,623,730,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00137.warc.gz | 47,335,992 | 6,075 | # 6/49 philippine lotto results
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https://www.snapxam.com/problems/40167986/1-10857-5000-2-x-151967-100 | 1,708,815,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00585.warc.gz | 992,097,806 | 14,650 | 👉 Try now NerdPal! Our new math app on iOS and Android
# Expand the expression $\frac{\frac{1}{10}\cdot 857}{5000}\cdot 10^{-8}x^4+2\left(x-\frac{51967}{100}\right)$
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## Final Answer
$1.71\times 10^{-10}x^4+2x+1039.34$
Got another answer? Verify it here!
## Step-by-step Solution
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Multiply $\frac{1}{10}$ times $857$
$\frac{\frac{857}{10}}{5000}\cdot 10^{-8}x^4+2\left(x-\frac{51967}{100}\right)$
Learn how to solve special products problems step by step online.
$\frac{\frac{857}{10}}{5000}\cdot 10^{-8}x^4+2\left(x-\frac{51967}{100}\right)$
Learn how to solve special products problems step by step online. Expand the expression (1/10*857)/500010^(-8)x^4+2(x+-51967/100). Multiply \frac{1}{10} times 857. Divide \frac{857}{10} by 5000. Divide -51967 by 100. Calculate the power 10^{-8}.
## Final Answer
$1.71\times 10^{-10}x^4+2x+1039.34$
## Explore different ways to solve this problem
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### Main Topic: Special Products
Special products is the multiplication of algebraic expressions that follow certain rules and patterns, so you can predict the result without necessarily doing the multiplication.
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Create an Account | 798 | 2,097 | {"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} | 3.9375 | 4 | CC-MAIN-2024-10 | longest | en | 0.544453 |
https://doc.sagemath.org/html/en/reference/algebras/sage/algebras/down_up_algebra.html | 1,722,683,417,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00400.warc.gz | 162,687,992 | 16,814 | # Down-Up Algebras#
AUTHORS:
• Travis Scrimshaw (2023-4): initial version
class sage.algebras.down_up_algebra.DownUpAlgebra(alpha, beta, gamma, base_ring)[source]#
Bases: CombinatorialFreeModule
The down-up algebra.
Let $$R$$ be a commutative ring, and let $$\alpha, \beta, \gamma \in R$$. The down-up algebra is the associative unital algebra $$DU(\alpha, \beta, \gamma)$$ generated by $$d, u$$ with relations
\begin{split}\begin{aligned} d^2u & = \alpha dud + \beta ud^2 + \gamma d, \\ du^2 & = \alpha udu + \beta u^2d + \gamma u. \end{aligned}\end{split}
The down-up algebra has a PBW-type basis given by
$\{ u^i (du)^j d^k \mid i,j,k \in \ZZ_{\geq 0} \}.$
This algebra originates in the study of posets. For a poset $$P$$, we define operators acting on $$R[P]$$ by
$d(y) = \sum_x x \qquad\qquad u(y) = \sum_z z,$
where $$y$$ covers $$x$$ and $$z$$ covers $$y$$. For $$r$$-differential posets we have $$du - ud = r 1$$, and thus it affords a representation of a Weyl algebra. This Weyl algebra is obtained as the quotient of $$DU(0, 1, 2r)$$ by the ideal generated by $$du - ud - r$$. For a $$(q, r)$$-differential poset, we have the $$d$$ and $$u$$ operators satisfying
\begin{split}\begin{aligned} d^2u & = q(q+1) dud - q^3 ud^2 + r d, \\ du^2 & = q(q+1) udu - q^3 u^2d + r u, \end{aligned}\end{split}
or $$\alpha = q(q+1)$$, $$\beta = -q^3$$, and $$\gamma = r$$. Specializing $$q = -1$$ recovers the $$r$$-differential poset relation.
Two other noteworthy quotients are:
• the $$q$$-Weyl algebra from $$DU(0, q^2, q+1)$$ by the ideal generated by $$du - qud - 1$$, and
• the quantum plane $$R_q[d, u]$$, where $$du = qud$$, from $$DU(2q, -q^2, 0)$$ by the ideal generated by $$du - qud$$.
EXAMPLES:
We begin by constructing the down-up algebra and perform some basic computations:
sage: R.<a,b,g> = QQ[]
sage: DU = algebras.DownUp(a, b, g)
sage: d, u = DU.gens()
sage: d * u
(d*u)
sage: u * d
u*d
sage: d^2 * u
b*u*d^2 + a*(d*u)*d + g*d
sage: d * u^2
b*u^2*d + a*u*(d*u) + g*u
>>> from sage.all import *
>>> R = QQ['a, b, g']; (a, b, g,) = R._first_ngens(3)
>>> DU = algebras.DownUp(a, b, g)
>>> d, u = DU.gens()
>>> d * u
(d*u)
>>> u * d
u*d
>>> d**Integer(2) * u
b*u*d^2 + a*(d*u)*d + g*d
>>> d * u**Integer(2)
b*u^2*d + a*u*(d*u) + g*u
We verify some examples of Proposition 3.5 in [BR1998], which states that the 0-th degree part is commutative:
sage: DU0 = [u^i * (d*u)^j * d^i for i,j in
....: cartesian_product([range(3), range(3)])]
sage: all(x.degree() == 0 for x in DU0)
True
sage: all(x * y == y * x for x, y in cartesian_product([DU0, DU0]))
True
>>> from sage.all import *
>>> DU0 = [u**i * (d*u)**j * d**i for i,j in
... cartesian_product([range(Integer(3)), range(Integer(3))])]
>>> all(x.degree() == Integer(0) for x in DU0)
True
>>> all(x * y == y * x for x, y in cartesian_product([DU0, DU0]))
True
We verify that $$DU(2, -1, \gamma)$$ can be described as the universal enveloping algebra of the 3-dimensional Lie algebra spanned by $$x,y,z$$ satisfying $$z = [x, y]$$, $$[x, z] = \gamma x$$, and $$[z, y] = \gamma y$$:
sage: R.<g> = QQ[]
sage: L = LieAlgebra(R, {('x','y'): {'z': 1}, ('x','z'): {'x': g}, ('z','y'): {'y': g}},
....: names='x,y,z')
sage: x, y, z = L.basis()
sage: (L[x, y], L[x, z], L[z, y])
(z, g*x, g*y)
sage: x, y, z = L.pbw_basis().gens()
sage: x^2*y - 2*x*y*x + y*x^2 == g*x
True
sage: x*y^2 - 2*y*x*y + y^2*x == g*y
True
sage: DU = algebras.DownUp(2, -1, g)
sage: d, u = DU.gens()
sage: d^2*u - 2*d*u*d + u*d^2 == g*d
True
sage: d*u^2 - 2*u*d*u + u^2*d == g*u
True
>>> from sage.all import *
>>> R = QQ['g']; (g,) = R._first_ngens(1)
>>> L = LieAlgebra(R, {('x','y'): {'z': Integer(1)}, ('x','z'): {'x': g}, ('z','y'): {'y': g}},
... names='x,y,z')
>>> x, y, z = L.basis()
>>> (L[x, y], L[x, z], L[z, y])
(z, g*x, g*y)
>>> x, y, z = L.pbw_basis().gens()
>>> x**Integer(2)*y - Integer(2)*x*y*x + y*x**Integer(2) == g*x
True
>>> x*y**Integer(2) - Integer(2)*y*x*y + y**Integer(2)*x == g*y
True
>>> DU = algebras.DownUp(Integer(2), -Integer(1), g)
>>> d, u = DU.gens()
>>> d**Integer(2)*u - Integer(2)*d*u*d + u*d**Integer(2) == g*d
True
>>> d*u**Integer(2) - Integer(2)*u*d*u + u**Integer(2)*d == g*u
True
Young’s lattice is known to be a differential poset. Thus we can construct a representation of $$DU(0, 1, 2)$$ on this poset (which gives a proof that Fomin’s growth diagrams are equivalent to edge local rules or shadow lines construction for RSK()):
sage: DU = algebras.DownUp(0, 1, 2)
sage: d, u = DU.gens()
sage: d^2*u == 0*d*u*d + 1*u*d*d + 2*d
True
sage: d*u^2 == 0*u*d*u + 1*u*u*d + 2*u
True
sage: YL = CombinatorialFreeModule(DU.base_ring(), Partitions())
sage: def d_action(la):
....: return YL.sum_of_monomials(la.remove_cell(*c) for c in la.removable_cells())
sage: def u_action(la):
....: return YL.sum_of_monomials(la.add_cell(*c) for c in la.addable_cells())
sage: D = YL.module_morphism(on_basis=d_action, codomain=YL)
sage: U = YL.module_morphism(on_basis=u_action, codomain=YL)
sage: for la in PartitionsInBox(5, 5):
....: b = YL.basis()[la]
....: assert (D*D*U)(b) == 0*(D*U*D)(b) + 1*(U*D*D)(b) + 2*D(b)
....: assert (D*U*U)(b) == 0*(U*D*U)(la) + 1*(U*U*D)(b) + 2*U(b)
....: assert (D*U)(b) == (U*D)(b) + b # the Weyl algebra relation
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(0), Integer(1), Integer(2))
>>> d, u = DU.gens()
>>> d**Integer(2)*u == Integer(0)*d*u*d + Integer(1)*u*d*d + Integer(2)*d
True
>>> d*u**Integer(2) == Integer(0)*u*d*u + Integer(1)*u*u*d + Integer(2)*u
True
>>> YL = CombinatorialFreeModule(DU.base_ring(), Partitions())
>>> def d_action(la):
... return YL.sum_of_monomials(la.remove_cell(*c) for c in la.removable_cells())
>>> def u_action(la):
... return YL.sum_of_monomials(la.add_cell(*c) for c in la.addable_cells())
>>> D = YL.module_morphism(on_basis=d_action, codomain=YL)
>>> U = YL.module_morphism(on_basis=u_action, codomain=YL)
>>> for la in PartitionsInBox(Integer(5), Integer(5)):
... b = YL.basis()[la]
... assert (D*D*U)(b) == Integer(0)*(D*U*D)(b) + Integer(1)*(U*D*D)(b) + Integer(2)*D(b)
... assert (D*U*U)(b) == Integer(0)*(U*D*U)(la) + Integer(1)*(U*U*D)(b) + Integer(2)*U(b)
... assert (D*U)(b) == (U*D)(b) + b # the Weyl algebra relation
Todo
Implement the homogenized version.
REFERENCES:
algebra_generators()[source]#
Return the algebra generators of self.
EXAMPLES:
sage: DU = algebras.DownUp(2, 3, 4)
sage: dict(DU.algebra_generators())
{'d': d, 'u': u}
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(2), Integer(3), Integer(4))
>>> dict(DU.algebra_generators())
{'d': d, 'u': u}
degree_on_basis(m)[source]#
Return the degree of the basis element indexed by m.
EXAMPLES:
sage: R.<a,b,g> = QQ[]
sage: DU = algebras.DownUp(a, b, g)
sage: I = DU.indices()
sage: DU.degree_on_basis(I([0, 3, 2]))
-2
sage: DU.degree_on_basis(I([2, 3, 0]))
2
sage: DU.degree_on_basis(I([2, 0, 3]))
-1
sage: DU.degree_on_basis(I([3, 10, 3]))
0
>>> from sage.all import *
>>> R = QQ['a, b, g']; (a, b, g,) = R._first_ngens(3)
>>> DU = algebras.DownUp(a, b, g)
>>> I = DU.indices()
>>> DU.degree_on_basis(I([Integer(0), Integer(3), Integer(2)]))
-2
>>> DU.degree_on_basis(I([Integer(2), Integer(3), Integer(0)]))
2
>>> DU.degree_on_basis(I([Integer(2), Integer(0), Integer(3)]))
-1
>>> DU.degree_on_basis(I([Integer(3), Integer(10), Integer(3)]))
0
gens()[source]#
Return the generators of self.
EXAMPLES:
sage: DU = algebras.DownUp(2, 3, 4)
sage: DU.gens()
(d, u)
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(2), Integer(3), Integer(4))
>>> DU.gens()
(d, u)
one_basis()[source]#
Return the index of the basis element of $$1$$.
EXAMPLES:
sage: DU = algebras.DownUp(2, 3, 4)
sage: DU.one_basis()
(0, 0, 0)
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(2), Integer(3), Integer(4))
>>> DU.one_basis()
(0, 0, 0)
product_on_basis(m1, m2)[source]#
Return the product of the basis elements indexed by m1 and m2.
EXAMPLES:
sage: R.<a,b,g> = QQ[]
sage: DU = algebras.DownUp(a, b, g)
sage: I = DU.indices()
sage: DU.product_on_basis(I([2,0,0]), I([4,0,0]))
u^6
sage: DU.product_on_basis(I([2,0,0]), I([0,4,0]))
u^2*(d*u)^4
sage: DU.product_on_basis(I([2,0,0]), I([0,0,4]))
u^2*d^4
sage: DU.product_on_basis(I([0,2,0]), I([0,4,0]))
(d*u)^6
sage: DU.product_on_basis(I([0,2,0]), I([0,0,4]))
(d*u)^2*d^4
sage: DU.product_on_basis(I([0,0,2]), I([0,0,4]))
d^6
sage: DU.product_on_basis(I([5,3,1]), I([1,0,4]))
u^5*(d*u)^4*d^4
sage: DU.product_on_basis(I([0,1,0]), I([1,0,0]))
b*u^2*d + a*u*(d*u) + g*u
sage: DU.product_on_basis(I([0,0,2]), I([1,0,0]))
b*u*d^2 + a*(d*u)*d + g*d
sage: DU.product_on_basis(I([0,0,1]), I([2,0,0]))
b*u^2*d + a*u*(d*u) + g*u
sage: DU.product_on_basis(I([0,0,1]), I([0,1,0]))
b*u*d^2 + a*(d*u)*d + g*d
sage: DU.product_on_basis(I([0,1,0]), I([3,0,0]))
(a^2*b+b^2)*u^4*d + (a^3+2*a*b)*u^3*(d*u) + (a^2*g+a*g+b*g+g)*u^3
sage: DU.product_on_basis(I([1,1,3]), I([0,1,1]))
(a^2*b^2+b^3)*u^3*d^6 + (a^3*b+a*b^2)*u^2*(d*u)*d^5 + (a^2*b*g+b^2*g)*u^2*d^5
+ (a^3+2*a*b)*u*(d*u)^2*d^4 + (a^2*g+a*g+b*g+g)*u*(d*u)*d^4
>>> from sage.all import *
>>> R = QQ['a, b, g']; (a, b, g,) = R._first_ngens(3)
>>> DU = algebras.DownUp(a, b, g)
>>> I = DU.indices()
>>> DU.product_on_basis(I([Integer(2),Integer(0),Integer(0)]), I([Integer(4),Integer(0),Integer(0)]))
u^6
>>> DU.product_on_basis(I([Integer(2),Integer(0),Integer(0)]), I([Integer(0),Integer(4),Integer(0)]))
u^2*(d*u)^4
>>> DU.product_on_basis(I([Integer(2),Integer(0),Integer(0)]), I([Integer(0),Integer(0),Integer(4)]))
u^2*d^4
>>> DU.product_on_basis(I([Integer(0),Integer(2),Integer(0)]), I([Integer(0),Integer(4),Integer(0)]))
(d*u)^6
>>> DU.product_on_basis(I([Integer(0),Integer(2),Integer(0)]), I([Integer(0),Integer(0),Integer(4)]))
(d*u)^2*d^4
>>> DU.product_on_basis(I([Integer(0),Integer(0),Integer(2)]), I([Integer(0),Integer(0),Integer(4)]))
d^6
>>> DU.product_on_basis(I([Integer(5),Integer(3),Integer(1)]), I([Integer(1),Integer(0),Integer(4)]))
u^5*(d*u)^4*d^4
>>> DU.product_on_basis(I([Integer(0),Integer(1),Integer(0)]), I([Integer(1),Integer(0),Integer(0)]))
b*u^2*d + a*u*(d*u) + g*u
>>> DU.product_on_basis(I([Integer(0),Integer(0),Integer(2)]), I([Integer(1),Integer(0),Integer(0)]))
b*u*d^2 + a*(d*u)*d + g*d
>>> DU.product_on_basis(I([Integer(0),Integer(0),Integer(1)]), I([Integer(2),Integer(0),Integer(0)]))
b*u^2*d + a*u*(d*u) + g*u
>>> DU.product_on_basis(I([Integer(0),Integer(0),Integer(1)]), I([Integer(0),Integer(1),Integer(0)]))
b*u*d^2 + a*(d*u)*d + g*d
>>> DU.product_on_basis(I([Integer(0),Integer(1),Integer(0)]), I([Integer(3),Integer(0),Integer(0)]))
(a^2*b+b^2)*u^4*d + (a^3+2*a*b)*u^3*(d*u) + (a^2*g+a*g+b*g+g)*u^3
>>> DU.product_on_basis(I([Integer(1),Integer(1),Integer(3)]), I([Integer(0),Integer(1),Integer(1)]))
(a^2*b^2+b^3)*u^3*d^6 + (a^3*b+a*b^2)*u^2*(d*u)*d^5 + (a^2*b*g+b^2*g)*u^2*d^5
+ (a^3+2*a*b)*u*(d*u)^2*d^4 + (a^2*g+a*g+b*g+g)*u*(d*u)*d^4
verma_module(la)[source]#
Return the Verma module $$V(\lambda)$$ of self.
EXAMPLES:
sage: R.<a,b,g> = QQ[]
sage: DU = algebras.DownUp(a, b, g)
sage: DU.verma_module(5)
Verma module of weight 5 of Down-Up algebra with parameters (a, b, g)
over Multivariate Polynomial Ring in a, b, g over Rational Field
>>> from sage.all import *
>>> R = QQ['a, b, g']; (a, b, g,) = R._first_ngens(3)
>>> DU = algebras.DownUp(a, b, g)
>>> DU.verma_module(Integer(5))
Verma module of weight 5 of Down-Up algebra with parameters (a, b, g)
over Multivariate Polynomial Ring in a, b, g over Rational Field
class sage.algebras.down_up_algebra.VermaModule(DU, la)[source]#
Bases: CombinatorialFreeModule
The Verma module $$V(\lambda)$$ of a down-up algebra.
The Verma module $$V(\lambda)$$ for the down-up algebra generated by $$d, u$$ is the span of $$\{v_n \mid n \in \ZZ_{\geq 0} \}$$ satisfying the relations
$d \cdot v_n = \lambda_{n-1} v_{n-1}, \qquad\qquad u \cdot v_n = v_{n+1},$
where $$\lambda_n = \alpha \lambda_{n-1} + \beta \lambda_{n-2} + \gamma$$ and we set $$\lambda_0 = \lambda$$ and $$\lambda_{-1} = 0$$.
By Proposition 2.4 in [BR1998], $$V(\lambda)$$ is simple if and only if $$\lambda_n \neq 0$$ for all $$n \geq 0$$. Moreover, a maximal submodule is spanned by $$\{ v_n \mid n > m \}$$, where $$m$$ is the minimal index such that $$\lambda_m = 0$$. Moreover, this is unique unless $$\gamma = \lambda = 0$$.
EXAMPLES:
sage: R.<a,b> = QQ[]
sage: DU = algebras.DownUp(0, b, 1)
sage: d, u = DU.gens()
sage: V = DU.verma_module(a)
sage: list(V.weights()[:6])
[a, 1, a*b + 1, b + 1, a*b^2 + b + 1, b^2 + b + 1]
sage: v = V.basis()
sage: d^2 * v[2]
a*v[0]
sage: d * (d * v[2])
a*v[0]
>>> from sage.all import *
>>> R = QQ['a, b']; (a, b,) = R._first_ngens(2)
>>> DU = algebras.DownUp(Integer(0), b, Integer(1))
>>> d, u = DU.gens()
>>> V = DU.verma_module(a)
>>> list(V.weights()[:Integer(6)])
[a, 1, a*b + 1, b + 1, a*b^2 + b + 1, b^2 + b + 1]
>>> v = V.basis()
>>> d**Integer(2) * v[Integer(2)]
a*v[0]
>>> d * (d * v[Integer(2)])
a*v[0]
The weight is computed by looking at the scalars associated to the action of $$du$$ and $$ud$$:
sage: d*u * v[3]
(b+1)*v[3]
sage: u*d * v[3]
(a*b+1)*v[3]
sage: v[3].weight()
(b + 1, a*b + 1)
>>> from sage.all import *
>>> d*u * v[Integer(3)]
(b+1)*v[3]
>>> u*d * v[Integer(3)]
(a*b+1)*v[3]
>>> v[Integer(3)].weight()
(b + 1, a*b + 1)
An $$U(\mathfrak{sl}_2)$$ example:
sage: DU = algebras.DownUp(2, -1, -2)
sage: d, u = DU.gens()
sage: V = DU.verma_module(5)
sage: list(V.weights()[:10])
[5, 8, 9, 8, 5, 0, -7, -16, -27, -40]
sage: v6 = V.basis()[6]
sage: d * v6
0
sage: [V.basis()[i].weight() for i in range(6)]
[(5, 0), (8, 5), (9, 8), (8, 9), (5, 8), (0, 5)]
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(2), -Integer(1), -Integer(2))
>>> d, u = DU.gens()
>>> V = DU.verma_module(Integer(5))
>>> list(V.weights()[:Integer(10)])
[5, 8, 9, 8, 5, 0, -7, -16, -27, -40]
>>> v6 = V.basis()[Integer(6)]
>>> d * v6
0
>>> [V.basis()[i].weight() for i in range(Integer(6))]
[(5, 0), (8, 5), (9, 8), (8, 9), (5, 8), (0, 5)]
Note that these are the same $$\mathfrak{sl}_2$$ weights from the usual construction of the irreducible representation $$V(5)$$ (but they are different as $$\mathfrak{gl}_2$$ weights):
sage: B = crystals.Tableaux(['A',1], shape=[5]) # needs sage.graphs
sage: [b.weight() for b in B] # needs sage.graphs
[(5, 0), (4, 1), (3, 2), (2, 3), (1, 4), (0, 5)]
>>> from sage.all import *
>>> B = crystals.Tableaux(['A',Integer(1)], shape=[Integer(5)]) # needs sage.graphs
>>> [b.weight() for b in B] # needs sage.graphs
[(5, 0), (4, 1), (3, 2), (2, 3), (1, 4), (0, 5)]
An example with periodic weights (see Theorem 2.13 of [BR1998]):
sage: # needs sage.rings.number_field
sage: k.<z6> = CyclotomicField(6)
sage: al = z6 + 1
sage: (al - 1)^6 == 1
True
sage: DU = algebras.DownUp(al, 1-al, 0)
sage: V = DU.verma_module(5)
sage: list(V.weights()[:8])
[5, 5*z6 + 5, 10*z6, 10*z6 - 5, 5*z6 - 5, 0, 5, 5*z6 + 5]
>>> from sage.all import *
>>> # needs sage.rings.number_field
>>> k = CyclotomicField(Integer(6), names=('z6',)); (z6,) = k._first_ngens(1)
>>> al = z6 + Integer(1)
>>> (al - Integer(1))**Integer(6) == Integer(1)
True
>>> DU = algebras.DownUp(al, Integer(1)-al, Integer(0))
>>> V = DU.verma_module(Integer(5))
>>> list(V.weights()[:Integer(8)])
[5, 5*z6 + 5, 10*z6, 10*z6 - 5, 5*z6 - 5, 0, 5, 5*z6 + 5]
class Element[source]#
Bases: IndexedFreeModuleElement
An element of a Verma module of a down-up algebra.
is_weight_vector()[source]#
Return if self is a weight vector.
EXAMPLES:
sage: DU = algebras.DownUp(2, -1, -2)
sage: V = DU.verma_module(5)
sage: V.zero().is_weight_vector()
False
sage: B = V.basis()
sage: [B[i].weight() for i in range(6)]
[(5, 0), (8, 5), (9, 8), (8, 9), (5, 8), (0, 5)]
sage: B[5].is_weight_vector()
True
sage: v = B[0] + B[1]
sage: v.is_weight_vector()
False
sage: DU = algebras.DownUp(2, -1, 0)
sage: V = DU.verma_module(0)
sage: B = V.basis()
sage: v = sum(i*B[i] for i in range(1,5))
sage: v.is_weight_vector()
True
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(2), -Integer(1), -Integer(2))
>>> V = DU.verma_module(Integer(5))
>>> V.zero().is_weight_vector()
False
>>> B = V.basis()
>>> [B[i].weight() for i in range(Integer(6))]
[(5, 0), (8, 5), (9, 8), (8, 9), (5, 8), (0, 5)]
>>> B[Integer(5)].is_weight_vector()
True
>>> v = B[Integer(0)] + B[Integer(1)]
>>> v.is_weight_vector()
False
>>> DU = algebras.DownUp(Integer(2), -Integer(1), Integer(0))
>>> V = DU.verma_module(Integer(0))
>>> B = V.basis()
>>> v = sum(i*B[i] for i in range(Integer(1),Integer(5)))
>>> v.is_weight_vector()
True
weight()[source]#
Return the weight of self.
For $$v_n$$, this is the vector with the pair $$(\lambda_n, \lambda_{n-1})$$.
EXAMPLES:
sage: R.<a,b,g> = QQ[]
sage: DU = algebras.DownUp(a, b, g)
sage: V = DU.verma_module(5)
sage: B = V.basis()
sage: B[0].weight()
(5, 0)
sage: B[1].weight()
(5*a + g, 5)
sage: B[2].weight()
(5*a^2 + a*g + 5*b + g, 5*a + g)
sage: V.zero().weight()
Traceback (most recent call last):
...
ValueError: the zero element does not have well-defined weight
sage: (B[0] + B[1]).weight()
Traceback (most recent call last):
...
ValueError: not a weight vector
>>> from sage.all import *
>>> R = QQ['a, b, g']; (a, b, g,) = R._first_ngens(3)
>>> DU = algebras.DownUp(a, b, g)
>>> V = DU.verma_module(Integer(5))
>>> B = V.basis()
>>> B[Integer(0)].weight()
(5, 0)
>>> B[Integer(1)].weight()
(5*a + g, 5)
>>> B[Integer(2)].weight()
(5*a^2 + a*g + 5*b + g, 5*a + g)
>>> V.zero().weight()
Traceback (most recent call last):
...
ValueError: the zero element does not have well-defined weight
>>> (B[Integer(0)] + B[Integer(1)]).weight()
Traceback (most recent call last):
...
ValueError: not a weight vector
highest_weight_vector()[source]#
Return the highest weight vector of self that generates self as a down-up module.
EXAMPLES:
sage: DU = algebras.DownUp(1, 2, 3)
sage: V = DU.verma_module(5)
sage: V.highest_weight_vector()
v[0]
>>> from sage.all import *
>>> DU = algebras.DownUp(Integer(1), Integer(2), Integer(3))
>>> V = DU.verma_module(Integer(5))
>>> V.highest_weight_vector()
v[0]
weights()[source]#
Return the sequence of weights $$(\lambda_n)_{n=0}^{\infty}$$.
EXAMPLES:
sage: R.<a,b,g> = QQ[]
sage: DU = algebras.DownUp(a, b, g)
sage: V = DU.verma_module(5)
sage: V.weights()
lazy list [5, 5*a + g, 5*a^2 + a*g + 5*b + g, ...]
sage: V = DU.verma_module(0)
sage: DU = algebras.DownUp(a, 1-a, 0)
sage: V = DU.verma_module(0)
sage: V.weights()
lazy list [0, 0, 0, ...]
>>> from sage.all import *
>>> R = QQ['a, b, g']; (a, b, g,) = R._first_ngens(3)
>>> DU = algebras.DownUp(a, b, g)
>>> V = DU.verma_module(Integer(5))
>>> V.weights()
lazy list [5, 5*a + g, 5*a^2 + a*g + 5*b + g, ...]
>>> V = DU.verma_module(Integer(0))
>>> DU = algebras.DownUp(a, Integer(1)-a, Integer(0))
>>> V = DU.verma_module(Integer(0))
>>> V.weights()
lazy list [0, 0, 0, ...]
We reproduce the Fibonacci numbers example from [BR1998]:
sage: R.<la> = QQ[]
sage: DU = algebras.DownUp(1, 1, 0, R)
sage: V = DU.verma_module(la)
sage: list(V.weights()[:11])
[la, la, 2*la, 3*la, 5*la, 8*la, 13*la, 21*la, 34*la, 55*la, 89*la]
>>> from sage.all import *
>>> R = QQ['la']; (la,) = R._first_ngens(1)
>>> DU = algebras.DownUp(Integer(1), Integer(1), Integer(0), R)
>>> V = DU.verma_module(la)
>>> list(V.weights()[:Integer(11)])
[la, la, 2*la, 3*la, 5*la, 8*la, 13*la, 21*la, 34*la, 55*la, 89*la] | 7,583 | 19,553 | {"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} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.730169 |
https://blogs.adelaide.edu.au/maths-learning/2016/08/05/where-the-complex-points-are-on-a-line/ | 1,720,850,210,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00260.warc.gz | 113,134,146 | 12,849 | # Where the complex points are: on a line
This is the second in what has turned out to be a series on Where the Complex Points Are.
In the introduction, I described the place where the complex points are: at each point (a,b) in the real plane, there is an iplane attached, which contains all the points (a+ci,b+di). These points are arranged with the x-axis showing the imaginary part added to the x-coordinate and the y-axis showing the imaginary part added to the y-coordinate. I imagine them as being a transparent sheet attached to the point itself and being able to be flattened out to sit on top of the real plane.
In this post, I want to talk about where the complex points are on a line. This is the first reason why this representation of complex points is so very cool.
A real line
Every real line has an equation like ax + by = c for some real a, b, and c. For example, 2x + 9y = 7, -3x + 5y=8, x – 2y = 3, x=4 and y=5. (I’ve used whole numbers here but there’s nothing stopping us using fractions or square roots or transcendental numbers like e.)
But we know that at every point there is an iplane attached which holds all the complex points. Which of these complex points are part of the line? First we need to remind ourselves that a point is on a line when it satisfies the equation of the line. That’s fundamentally what an equation is for: deciding if a point is part of a set or not.
Let’s consider the line x – 2y = 3…
• The point (3,0) is on this line because 3 – 2*0 = 3.
• The point (1,-1) is on this line because 1 – 2*(-1) = 1+2 = 3.
• The point (9,3) is on this line because 9 – 2*3 = 9-6 = 3.
• The point (3+2i,i) is on this line because 3+2i – 2*i = 3+2i-2i = 3.
• The point (1+4i,-1+2i) is on this line because 1+4i – 2*(-1+2i) = 1 + 4i +2 – 4i = 3.
• The point (9-10i,3-5i) is on this line because 9 -10i -2*(3-5i) = 9 – 10i – 6 + 10i = 3.
There is a pattern forming here: in order to get my complex point to be on the line, I need to make sure the imaginary parts are the right sizes so that they subtract away when I put them in the formula. The real parts have to create the number 3 I need as well, which means the real parts have to satisfy the original equation.
Let’s do this with algebra:
What does this mean? If the point (p+ri,q+si) is on the line with equation ax + by = c, then the point (p,q) has to be on the line. This means that all the complex points on the real line are in the iplanes at the points on the real line. That’s very pleasant — the complex points on this line are in some way attached to the line itself, because they’re in the iplanes attached to this line.
But it gets better! The complex point (p+ri,q+si) is the point with coordinates (r,s) in the iplane at (p,q). And the point (r,s) has to satisfy the equation ax + by = 0. That means the point (r,s) lies on a line parallel to our original line passing through the centre of the iplane. When you take the iplane at the point (p,q) and lay it flat on the real plane, the complex points in that iplane that are part of the line will lie on top of the existing real line! This is because they are on a line parallel to this line passing through a point on this line.
So in our representation, the complex points on a real line aren’t just in the iplanes attached to the real line, they look like they are actually on the line! Isn’t that cool?!!
You can play around with the iplanes attached to real lines in the following GeoGebra applet.
A line with real slope
There are a whole lot of lines in the complex plane that aren’t real lines. The first kind I want to deal with is those with real slope. That is, their equations are of the form ax + by = c + fi for some non-real number c+fi and real numbers a and b. This kind of line has no real points at all – it’s not possible to put real numbers into ax + by and get a non-real result. But where actually are all the non-real points on this line?
So, the points on the line with equation ax + by = c + fi are in the iplanes attached to the real line with equation ax + by = c, and in each iplane they form a line parallel to this real line. In fact, if you look at where they are across all the iplanes, they all form theĀ sameĀ line parallel to the real line.
You can investigate these kinds of lines using this GeoGebra applet.
Of particular interest for our purposes later on are the “vertical” complex lines with equations of the form x=fi. The points on this line are in the iplanes attached to the real line x=0 — that is, they’re in the iplanes attached to the y-axis. Inside each iplane they form a vertical line, which looks like it’s part of the real line x=f.
A line with unreal slope
The last kind of line is a line with unreal slope. It would have an equation like (a+di)x + (b+ei)y = c+fi. You’d need to make sure that a+di and b+ei weren’t related by a real multiple or you could divide by some complex number to produce real coefficients for the x and y — we’ve already covered lines like that. In fact, it must therefore be possible to arrange it so that the y-coefficient is 1. In that case, we might as well write it in “slope-intercept” form, which might be a bit easier to process. So let’s take a look at the line with equation y = (a+di)x + (c+fi).
Hmm. That’s not particularly edifying. What does that mean? Why don’t we pick an iplane and see what the points on our line look like in this iplane:
This means that given (p,q), we can find a unique value of r. Substituting this into the other equation gives us a unique value of s. That is, every iplane has exactly one point of this complex line! You can investigate what these points look like with this GeoGebra applet.
Note that there is in fact exactly one real point every line with complex slope. Using the notation above, it’s the point (-f/d, -af/d+c). It’s quite hard to find the point by trial and error in the GeoGebra applet!
Conclusion
So there you have it. Using iplanes, the lines with real slope actually look like lines, and the lines with unreal slope are a cloud of points with one in each iplane. In the next post, I’ll talk about parabolas…
This entry was posted in Isn't maths cool?, Thoughts about maths thinking and tagged , , . Bookmark the permalink.
## 3 Responses
1. […] Where the complex points are: on a line […]
2. […] Where the complex points are: on a line […]
3. […] Where the complex points are: on a line […] | 1,680 | 6,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-30 | latest | en | 0.924818 |
http://www.top-law-schools.com/forums/viewtopic.php?f=6&t=127903 | 1,527,053,689,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00598.warc.gz | 482,584,595 | 9,349 | ## target score: 176
Prepare for the LSAT or discuss it with others in this forum.
amkid100
Posts: 254
Joined: Thu Aug 12, 2010 8:02 pm
### target score: 176
do you think that -4 will more or less guarantee at least a 176? i don't think i can get much more accurate than that. here are the scales:
http://lsatblog.blogspot.com/2010/03/ls ... rsion.html
Hannibal
Posts: 2211
Joined: Mon Jul 12, 2010 12:00 pm
### Re: target score: 176
I don't think I've seen a test that has a harsher curve than -4 = 176.
Cromartie
Posts: 200
Joined: Wed Aug 18, 2010 12:27 pm
### Re: target score: 176
Hannibal wrote:I don't think I've seen a test that has a harsher curve than -4 = 176.
This is still fresh in my mind as I just took the PT this past Sunday. PT 48: -4 = 175. I was so pissed with the curve that I created a thread just to express my disgust.
amkid100
Posts: 254
Joined: Thu Aug 12, 2010 8:02 pm
### Re: target score: 176
Cromartie wrote:
Hannibal wrote:I don't think I've seen a test that has a harsher curve than -4 = 176.
This is still fresh in my mind as I just took the PT this past Sunday. PT 48: -4 = 175. I was so pissed with the curve that I created a thread just to express my disgust.
is that some sort of cosmic joke?
Posts: 1108
Joined: Fri Jul 23, 2010 4:18 am
### Re: target score: 176
-4 = 175? How is that even possible??
Cromartie
Posts: 200
Joined: Wed Aug 18, 2010 12:27 pm
### Re: target score: 176
Adjudicator wrote:-4 = 175? How is that even possible??
Well, the overall curve for the test was -8. Don't have the slightest clue how that happened. Granted the LG section was easier than in most PT's, the LR and RC sections were filled with subtleties and misdirections, making them pretty challenging.
Posts: 1108
Joined: Fri Jul 23, 2010 4:18 am
### Re: target score: 176
Cromartie wrote:
Adjudicator wrote:-4 = 175? How is that even possible??
Well, the overall curve for the test was -8. Don't have the slightest clue how that happened. Granted the LG section was easier than in most PT's, the LR and RC sections were filled with subtleties and misdirections, making them pretty challenging.
I did the LG section from that PT today and I didn't like it; I got -3. So I would have to have missed no more than 1 on the rest of that test to get a 175?
Outrageous!
Cromartie
Posts: 200
Joined: Wed Aug 18, 2010 12:27 pm
### Re: target score: 176
Cromartie wrote:
Adjudicator wrote:-4 = 175? How is that even possible??
Well, the overall curve for the test was -8. Don't have the slightest clue how that happened. Granted the LG section was easier than in most PT's, the LR and RC sections were filled with subtleties and misdirections, making them pretty challenging.
I did the LG section from that PT today and I didn't like it; I got -3. So I would have to have missed no more than 1 on the rest of that test to get a 175?
Outrageous!
Yup, it sucks. Missed 3 on the last LG (rock bands and folk bands) and 2 on RC. My reward for a -5? 174. I remember taking one of the earlier 40's where a -6 netted me a 176. | 915 | 3,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-22 | latest | en | 0.974854 |
http://slideplayer.com/slide/2576260/ | 1,508,274,065,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00662.warc.gz | 310,975,390 | 23,722 | # © Paradigm Publishing Inc. EXCEL Editing and Formatting Worksheets Section 2.
## Presentation on theme: "© Paradigm Publishing Inc. EXCEL Editing and Formatting Worksheets Section 2."— Presentation transcript:
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Skills Edit the content of cells Clear cells and cell formats Use proofing tools Insert and delete columns and rows Move and copy cells Use Paste Options to link cells Create formulas using absolute references Adjust column width and row height Change the font, size, style, and color of cells
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Skills Apply numeric formats and adjust the number of decimal places Use Undo, Redo, and Repeat Change cell alignment and indentation Add borders and shading Copy formats using Format Painter Apply cell styles Find and replace cell entries and formats Freeze and unfreeze panes Change the zoom percentage
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Open WBQuote-MarqueeProd.xlsx. Save workbook with Save As in ExcelS2 folder and name it ExcelS2-01.xlsx. ACTIVITY 2.1 ExcelS2-01.xlsx
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Double-click I21. Press Right or Left Arrow key to move insertion point between decimal point and 4, then press Delete. Type 3 then press Enter. ACTIVITY 2.1
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Make I27 active. Move pointer after 1 in Formula bar then click left mouse button. Press Backspace to delete 1, type 4, then click Enter on Formula bar. ACTIVITY 2.1
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Make A9 active then press Delete. Select range I17:I18. With Home tab active, click Clear then click Clear All. ACTIVITY 2.1
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Press Ctrl + Home to move active cell to A1. Click Review tab then click Spelling. Click Ignore All in Spelling box to skip all occurrences of Rivermist. ACTIVITY 2.1
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Click Change in Spelling box to replace Torontow with Toronto. Click Change in Spelling box to replace Persns with Persons. Complete spell check, change words as required. Click OK when spelling check is complete. ACTIVITY 2.1
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Editing and Clearing Cells; Using Proofing Tools Make A39 active. Click Thesaurus in Review tab. Point to transport in word list, click down-pointing arrow, then click Insert. Click Research in Proofing group to turn off Research task pane. Save workbook as ExcelS2-01.xlsx. ACTIVITY 2.1
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Inserting and Deleting Columns and Rows With ExcelS2-01.xlsx open, position cell pointer over row indicator 24, hold down left mouse button, then drag mouse down over 25. Click Home tab, click Insert arrow, then click Insert Sheet Rows. ACTIVITY 2.2
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Inserting and Deleting Columns and Rows Click A24, type Vegetable tray with dip, then press Enter. Type Seafood hors d'oeuvres then press Enter. ACTIVITY 2.2
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Inserting and Deleting Columns and Rows Select row 32. Click Delete arrow then click Delete Sheet Rows. ACTIVITY 2.2
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Inserting and Deleting Columns and Rows Select row 22. Hold down Ctrl key then select rows 30 and 35. Position pointer within any three selected rows, right-click to display shortcut menu and Mini toolbar, then click Delete. ACTIVITY 2.2
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Inserting and Deleting Columns and Rows Select column F then display shortcut menu and Mini toolbar by positioning cell pointer over column indicator letter F and right-clicking mouse. At shortcut menu, click Delete. ACTIVITY 2.2
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Inserting and Deleting Columns and Rows Click any cell to deselect column. Make F11 active, type November 5, 2009, then press Enter. Save ExcelS2-01.xlsx. ACTIVITY 2.2
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Moving and Copying Cells With ExcelS2-01.xlsx open, make E5 active. Click Cut in Clipboard group in Home tab. Make A9 active then click Paste in Clipboard group. ACTIVITY 2.3
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Moving and Copying Cells Select range A17:B18. Point at any of four borders surrounding selected range. Hold down left mouse button, drag top left corner of range to E15, then release mouse. ACTIVITY 2.3
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Moving and Copying Cells Make D28 active. Click Copy in Clipboard group. Make D32 active, click Paste arrow in Clipboard group, then click Paste Link. Press Esc to remove moving marquee from D28. ACTIVITY 2.3
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Moving and Copying Cells Make D28 active, edit value to 4.09. Make A23 active. Point at any of four borders surrounding A23 until pointer displays as white arrow with move icon attached, hold down Ctrl key, then drag mouse to A36. Release mouse button first then release Ctrl key. Save ExcelS2-01.xlsx. ACTIVITY 2.3
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Creating Formulas with Absolute Addressing With ExcelS2-01.xlsx open, make K20 active then type Food. Make L20 active then type Prep. Make M20 active then type Server. Type decimal values in K19, L19, and M19 as shown. ACTIVITY 2.4
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Creating Formulas with Absolute Addressing Make K21 active, type =(c21*e21*h21)*k19, then press F4. Press Enter. Make L21 active, type formula =(c21*e21*h21)*\$I\$19, then press Enter. Make M21 active then type formula =(c21*e21*h21)*\$m\$19 by typing dollar symbols or by pressing F4. ACTIVITY 2.4
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Creating Formulas with Absolute Addressing Select range K21:M21 then click Copy in Home tab. Select range K28:M28, hold down Ctrl key, then select range K32:M32. Click Paste in Clipboard group. Press Esc to remove moving marquee from selected range. ACTIVITY 2.4
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Creating Formulas with Absolute Addressing Click K28 then look at Formula bar to see formula that was pasted into cell: =(C28*E28*H28)*\$K\$19. Click L32 then look at Formula bar to see formula that was pasted into cell. Enter Sum function to calculate totals in K39, L39, and M39. Save ExcelS2-01.xlsx. ACTIVITY 2.4
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adjusting Column Width and Row Height With ExcelS2-01.xlsx open, make any cell in column C active. Click Format in Home tab then click Column Width. At Column Width box, type 14 then click OK or press Enter. ACTIVITY 2.5
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Position mouse pointer on boundary line in column indicator row between columns D and E until pointer changes to a vertical line with left- and right- pointing arrow. Hold down left mouse button, drag boundary line to right until Width: 15.00 (110 pixels) displays in ScreenTip, then release mouse button. Adjusting Column Width and Row Height ACTIVITY 2.5
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adjusting Column Width and Row Height Position mouse pointer on boundary line in column indicator row between columns E and F until pointer changes and double click left mouse button to Autofit width. Increase width of column B to 12 (89 pixels) using Column Width box or by dragging the column boundary. ACTIVITY 2.5
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Position mouse pointer on column indicator letter C, hold down left mouse button, then drag mouse right to column M. Position mouse pointer on any of right boundary lines to right of column E within selected range of columns until pointer changes. Drag boundary line right until Width: 15.00 (110 pixels) displays in ScreenTip then release mouse button. Adjusting Column Width and Row Height ACTIVITY 2.5
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adjusting Column Width and Row Height Click any cell to deselect columns. Move E15:F16 to A17:B18 then click any cell to deselect range. ACTIVITY 2.5
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Position mouse pointer on boundary line below row 20 until pointer changes to horizontal line with up- and down-pointing arrow. Drag boundary line down until Height: 19.50 (26 pixels) displays in ScreenTip then release mouse button. Save ExcelS2-01.xlsx. Adjusting Column Width and Row Height ACTIVITY 2.5
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Font, Size, Style, and Color of Cells With ExcelS2-01.xlsx open, make A9 active. Click Font arrow in Home tab, scroll down list of fonts to Impact. Click Impact at drop-down gallery. ACTIVITY 2.6
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Font, Size, Style, and Color of Cells Click Font Size arrow in Font group then click 18. With A9 still active, click Font Color arrow then click Blue color box in Standard Colors section. ACTIVITY 2.6
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Font, Size, Style, and Color of Cells Select A9:F9 then click Merge & Center in Alignment group. With merged cell A9 still selected, click Fill Color arrow in Font group then click Accent 5, Lighter 80% color box in Theme Colors section. ACTIVITY 2.6
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Font, Size, Style, and Color of Cells Make A39 active. Hold down Ctrl key then click F39. Click Bold and Italic in Font group. Click any cell to deselect A39 and F39. Save ExcelS2-01.xlsx. ACTIVITY 2.6
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Formatting Numeric Cells; Adjusting Decimal Places; Using Undo and Redo With ExcelS2-01.xlsx open, make F21 active. Hold down Ctrl, click I21, K21:M21, F39, I39, and K39:M39. Click Accounting Number Format in Home tab. ACTIVITY 2.7
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Click any cell to deselect the cells. Select F28:F37, I28:I37, and K28:M32. Click Comma Style in Number group. Formatting Numeric Cells; Adjusting Decimal Places; Using Undo and Redo ACTIVITY 2.7
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Formatting Numeric Cells; Adjusting Decimal Places; Using Undo and Redo Click any cell to deselect cells and review numeric values in worksheet. Select J21:J39. Click Percent Style in Number group. With range J21:J39 still selected, click Increase Decimal twice. With range J21:J39 still selected, click Decrease Decimal once. Click any cell to deselect range. ACTIVITY 2.7
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Formatting Numeric Cells; Adjusting Decimal Places; Using Undo and Redo Select D37 and H37 then click Increase Decimal twice to display two decimal places. Deselect cells. Click Undo two times. Click Redo two times. Save ExcelS2-01.xlsx. ACTIVITY 2.7
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Alignment and Indentation of Cells; Using Repeat With ExcelS2-01.xlsx open, edit column headings C20 and E20 to include a period after the abbreviation for number. Select C20:M20. Click Center in Home tab. Select C21, C28, and C32 then change alignment to center. Center entries in column E. ACTIVITY 2.8
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Alignment and Indentation of Cells; Using Repeat Select A22:A27. Click Increase Indent in Alignment group. Select A29:A31 then click Increase Indent. Select A33:A36 then click Increase Indent. ACTIVITY 2.8
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Alignment and Indentation of Cells; Using Repeat Select A20:M20 then bold the cells. Make F11 active then click Align Text Left in Alignment group. Select A20:M20. Click Middle Align in Alignment group and deselect range. ACTIVITY 2.8
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Changing the Alignment and Indentation of Cells; Using Repeat Select E1:F1 then click Merge & Center. Select E2:F2 then press Ctrl + Y. Select E3:F3 press Ctrl + Y. Select E1:E3 then click Align Text Right. Deselect range. Save ExcelS2-01.xlsx. ACTIVITY 2.8
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles With ExcelS2-01.xlsx open, change width of column A to 11.00 (82 pixels). AutoFit columns C–F, and H–M. Change width of column G to 4.00 (33 pixels). ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Select A20:M20. Click Bottom Border arrow in Font group in Home tab. Click Top and Bottom Border at drop-down list. Click any cell to deselect range and view border. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Select A21:B21, click Top and Bottom Border arrow, then click Outside Borders. Select A28:B28 then click Outside Borders button. (Do not click arrow.) Select A32:B32 then click Outside Borders. Deselect range. Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Make F39 active, click Outside Borders arrow, then click Top and Double Bottom Border. With F39 still active, double-click Format Painter in Clipboard group. Click I39 then drag cell pointer to select K39:M39. Click Format Painter to turn off feature. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Select E1:F8. Click Fill Color arrow in Font group then click white color box (Background 1). Deselect range. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Select A9. Click Cell Styles in Home tab. Move mouse over several cell style designs and watch Live Preview show style applied. Click Title style in Titles and Headings. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Select A20:M20, click Cell Styles in Styles group, then click Accent2 style in Themed Cell Styles. Select A21:B21, A28:B28, and A32:B32 and apply Accent1 style in Themed Cell Styles. Select K19:M19, apply Note style in Data and Model section. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Select A1:F42. Click Office then click Print. At Print box, click Selection in Print what section then click OK. Deselect range. Click Page Layout tab. Click Orientation in Page Layout tab then click Landscape. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Adding Borders and Shading; Copying Formats with Format Painter; Using Cell Styles Click Width arrow in Scale to Fit group then click 1 page. Click Height button arrow in Scale to Fit group then click 1 page. Click Quick Print button and save ExcelS2-01.xlsx. ACTIVITY 2.9
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Using Find and Replace With ExcelS2-01.xlsx open, press Ctrl + Home to make A1 active. Click Home tab. Click Find & Select in Editing group then click Find. ACTIVITY 2.10
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Using Find and Replace Type *tray* in Find what box then click Find Next. Click Find Next four times. Click Close to close Find and Replace box. ACTIVITY 2.10
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Click Find & Select in Editing group then click Replace. Drag to select *tray* in the Find what text box then type 40. Press Tab to move insertion point to Replace with box then type 43. Click Replace All, then click OK at message that Excel has completed search and made four replacements. Click Close button to close Find and Replace box. Using Find and Replace ACTIVITY 2.10 40 43
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Using Find and Replace Review No. of Days column in worksheet and note that four replacements were made in E21, E28, E32, and E37. Save ExcelS2-01.xlsx. ACTIVITY 2.10
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Freezing Panes; Changing the Zoom With ExcelS2-01.xlsx open, make C10 active. Click View tab. Click Freeze Panes in Window group. Click Freeze Panes at drop-down list. ACTIVITY 2.11
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Press Page Down to scroll down worksheet. Press Ctrl + Home. Note Excel returns to C10 instead of A1 since A1 is frozen. Scroll several screens to the right. Press Ctrl + Home. Scroll right until column H is immediately right of column B then scroll down until row 20 is immediately below row 9. Freezing Panes; Changing the Zoom ACTIVITY 2.11 Lines indicate which rows and columns are frozen.
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Freezing Panes; Changing the Zoom Click Freeze Panes then click Unfreeze Panes. Practice dragging Zoom slider at right end of Status bar and watch cells magnify and shrink as you drag right and left. Drag slider to halfway mark on slider bar to redisplay worksheet at 100%. ACTIVITY 2.11
© Paradigm Publishing Inc. Links to Activities 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 Click over 100% at left edge of slider bar to open Zoom box. Click 75% then click OK. Click Zoom In at right end of Zoom Slider bar. Click Zoom In until zoom percentage returns to 100%. Save and close ExcelS2-01.xlsx. Freezing Panes; Changing the Zoom ACTIVITY 2.11 | 6,689 | 20,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-43 | longest | en | 0.682643 |
https://nrich.maths.org/public/leg.php?code=-100&cl=1&cldcmpid=5676 | 1,508,719,534,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00239.warc.gz | 771,078,880 | 9,975 | # Search by Topic
#### Resources tagged with Trial and improvement similar to Getting the Balance:
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### There are 85 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Trial and improvement
### The Brown Family
##### Stage: 1 Challenge Level:
Use the information about Sally and her brother to find out how many children there are in the Brown family.
##### Stage: 1 Challenge Level:
There are three baskets, a brown one, a red one and a pink one, holding a total of 10 eggs. Can you use the information given to find out how many eggs are in each basket?
### Jumping Squares
##### Stage: 1 Challenge Level:
In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
##### Stage: 1 Challenge Level:
On a farm there were some hens and sheep. Altogether there were 8 heads and 22 feet. How many hens were there?
### Big Dog, Little Dog
##### Stage: 1 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Find the Difference
##### Stage: 1 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
### Spiders and Flies
##### Stage: 1 Challenge Level:
There were 22 legs creeping across the web. How many flies? How many spiders?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### A Shapely Network
##### Stage: 2 Challenge Level:
Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order.
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Domino Join Up
##### Stage: 1 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Amazing Alphabet Maze
##### Stage: 1 Challenge Level:
Can you go from A to Z right through the alphabet in the hexagonal maze?
### Fifteen Cards
##### Stage: 2 Challenge Level:
Can you use the information to find out which cards I have used?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Tricky Triangles
##### Stage: 1 Challenge Level:
Use the three triangles to fill these outline shapes. Perhaps you can create some of your own shapes for a friend to fill?
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### One Big Triangle
##### Stage: 1 Challenge Level:
Make one big triangle so the numbers that touch on the small triangles add to 10. You could use the interactivity to help you.
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Five Steps to 50
##### Stage: 1 Challenge Level:
Use five steps to count forwards or backwards in 1s or 10s to get to 50. What strategies did you use?
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### The Tall Tower
##### Stage: 1 Challenge Level:
As you come down the ladders of the Tall Tower you collect useful spells. Which way should you go to collect the most spells?
### Magic Squares 4x4
##### Stage: 2 Challenge Level:
Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square.
### Coded Hundred Square
##### Stage: 2 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### A Numbered Route
##### Stage: 2 Challenge Level:
Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### Plenty of Pens
##### Stage: 2 Challenge Level:
Amy's mum had given her £2.50 to spend. She bought four times as many pens as pencils and was given 40p change. How many of each did she buy?
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Let's Face It
##### Stage: 2 Challenge Level:
In this problem you have to place four by four magic squares on the faces of a cube so that along each edge of the cube the numbers match.
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### One Wasn't Square
##### Stage: 2 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### Four Colours
##### Stage: 1 and 2 Challenge Level:
Kate has eight multilink cubes. She has two red ones, two yellow, two green and two blue. She wants to fit them together to make a cube so that each colour shows on each face just once.
### Buckets of Thinking
##### Stage: 2 Challenge Level:
There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket.
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### Sliding Game
##### Stage: 2 Challenge Level:
A shunting puzzle for 1 person. Swop the positions of the counters at the top and bottom of the board.
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### Magic Matrix
##### Stage: 2 Challenge Level:
Find out why these matrices are magic. Can you work out how they were made? Can you make your own Magic Matrix?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Pizza Cut
##### Stage: 2 Challenge Level:
Using only six straight cuts, find a way to make as many pieces of pizza as possible. (The pieces can be different sizes and shapes). | 2,235 | 9,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.946685 |
https://www.nogrid.com/en/component/content/article/glass-rolling?catid=18&Itemid=230 | 1,653,066,833,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00655.warc.gz | 1,078,279,257 | 18,161 | # Simulation Glass Rolling
Figure 1: CAD model created with NOGRID's COMPASS
Figure 2: Setup glass rolling case in the GUI
Figure 3: Temperature at a certain time step
Figure 4: Animation of the glass rolling case during inlet temperature change
The simulation of glass rolling can perfectly be performed with our CFD software NOGRID points. Flat glass sheets are often produced by pouring the glass between two rollers, which are responsible to give the sheet the final thickness. Due to the cold rollers the glass viscosity is decreasing and the glass behind the rollers is almost inherently stable. The movie in figure 1 shows a glass rolling process (the geometry was published by G. Nölle: Technik der Glasherstellung, 1997). The color represents the temperature. The inlet temperature is increasing with time; thus it appears that the glass sheet becomes more instable after a while.
NOGRID points computes the real glass flow and temperature distribution in the complete forming area. It is also possible to perform the rolling process completely in 3D. Comparisons with various experimental data have shown the high precision of the simulation of this intricate flow process. The short computation time allows our customers incorporating the simulation into the design process to find an optimal design, thus saving time and effort in the design phase and increasing product quality.
Our software is able to predict such effects and helps designers to find the right speed for their processes. Beside the normal flow with Newtons viscous law NOGRID points is also able to run the computation with non-Newtonian materials (also viscoelastic material behavior is included).
• computation is in full 3D solving complete Navier-Stokes-Equations
• easy and intuitive setup of the FSI (Fluid-Structure-Interaction) case
• free definable material properties by equations or curves,
• moving parts with a lot of moving features
• open or closed domains including moving of additional parts
• computation of temperature transport in the solid rollers and in the fluid
## Nogrid's strengths
NOGRID's particular strengths are the rapid preprocessing (no fluid grid needs to be generated, only the boundary mesh, inner finite points are generated automatically depending on User setting initially and during computation) and the outstandingly short computation time even for complicated cavities.
As you can see in the image below, the boundary of the geometry still requires a mesh to allow the interior finite points to detect the boundary. The boundary must therefore be meshed and the finite points inside are automatically generated during the simulation controlled by User specifications.
## What is CFD from NOGRID?
CFD solves the fundamental equations that define the fluid flow process. With CFD software from NOGRID every engineer makes better decisions by predicting, analyzing and controlling fluid flow, heat and mass transfer or chemical reaction. By using NOGRID software you receive information on essential flow characteristics as for example flow distribution. Using it additional to testing and experimentation NOGRID software helps to improve the evaluation of your design – resulting in better construction and operation parameters, increasing planning security and money savings due to faster time to the marketplace for your product or process.
## Choose NOGRID
With NOGRID, you choose professional CFD software and services – our aim is helping you to be successful. When you decide to work with NOGRID you choose close cooperation with a dynamic, flat hierarchies-organization. Short information channels result in quick and accurate professional support and service. Our team consists of highly qualified employees, who are experts in fields such as numerical simulation or computational fluid dynamics. Based on our know-how, we are pleased to offer the following services, responding to your individual requirements:
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info@nogrid.com | 917 | 4,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | longest | en | 0.913716 |
https://socratic.org/questions/how-do-you-solve-2-2x-times4-4x-8-64 | 1,723,386,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00711.warc.gz | 416,801,368 | 5,912 | # How do you solve 2^(2x)times4^(4x+8)=64?
Sep 18, 2016
$x = - 1$
#### Explanation:
We have: ${2}^{2 x} \times {4}^{4 x + 8} = 64$
Let's express the numbers in terms of $2$:
$\implies {2}^{2 x} \times {\left({2}^{2}\right)}^{4 x + 8} = {2}^{6}$
Using the laws of exponents:
$\implies {2}^{2 x} \times {2}^{8 x + 16} = {2}^{6}$
$\implies {2}^{2 x + 8 x + 16} = {2}^{6}$
$\implies {2}^{10 x + 16} = {2}^{6}$
$\implies 10 x + 16 = 6$
$\implies 10 x = - 10$
$\implies x = - 1$
Therefore, the solution to the equation is $x = - 1$. | 245 | 540 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.647406 |
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Section7_7_review
# Section7_7_review - Section 7.7 Indeterminate Forms and...
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Section 7.7: Indeterminate Forms and L’Hopital’s Rule L’Hopital’s Rule: Let f and g be differentiable functions with g x (29 0 near a (except possibly at a ). If lim x a f x ( 29 g x ( 29 is an indeterminate quotient, then lim x a f x (29 gx (29 H lim x a f x (29 g x (29 where an indeterminate quotient is of the form 0 0 or ± ∞ ± ∞ . Comments: i) lim x a in L’Hopital’s Rule can be replaced with lim x a + , lim x a - , lim x →∞ , and lim x →-∞ . In other words, L’Hopital’s Rule is valid for one-sided limits as well as x to a constant, to infinity, and to negative infinity. ii) L’Hopital’s Rule can be used several times in succession, a long as you have an indeterminate quotient each time. iii) L’Hopital’s Rule is only valid for indeterminate quotients. If you have an indeterminate product (such as 0 ⋅∞ or 0 ⋅ -∞ ( 29 ), an indeterminate difference (such as ∞-∞ ), or an indeterminate power (such as 1 , 0 , or 0 0 ), the expression must first be converted to a quotient before L’Hopital’s Rule can be used.
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https://www.codeproject.com/Articles/6534/Convolution-of-Bitmaps?msg=1072381&PageFlow=FixedWidth | 1,496,014,689,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611569.86/warc/CC-MAIN-20170528220125-20170529000125-00474.warc.gz | 1,072,141,083 | 34,648 | 12,954,522 members (61,426 online)
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Posted 26 Mar 2004
# Convolution of Bitmaps
, 29 Mar 2006
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Article discussing how to convolve images, specificially the convolution of bitmaps.
## Introduction
Image convolution plays an important role in computer graphics applications. Convolution of images allows for image alterations that enable the creation of new images from old ones. Convolution also allows for important features such as edge detection, with many widespread uses. The convolution of an image is a simple process by which the pixel of an image is multiplied by a kernel, or masked, to create a new pixel value. Convolution is commonly referred to as filtering.
## Details
First, for a given pixel (x,y), we give a weight to each of the surrounding pixels. It may be thought of as giving a number telling how important the pixel is to us. The number may be any integer or floating point number, though I usually stick to floating point since floats will accept integers as well. The kernel or mask that contains the filter may actually be any size (3x3, 5x5, 7x7); however, 3x3 is very common. Since the process for each is the same, I will concentrate only on the 3x3 kernels.
Second, the actual process of convolution involves getting each pixel near a given pixel (x,y), and multiplying each of the pixel's channels by the weighted kernel value. This means that for a 3x3 kernel, we would multiply the pixels like so:
```(x-1,y-1) * kernel_value[row0][col0]
(x ,y-1) * kernel_value[row0][col1]
(x+1,y-1) * kernel_value[row0][col2]
(x-1,y ) * kernel_value[row1][col0]
(x ,y ) * kernel_value[row1][col1]
(x+1,y ) * kernel_value[row1][col2]
(x-1,y+1) * kernel_value[row2][col0]
(x ,y+1) * kernel_value[row2][col1]
(x+1,y+1) * kernel_value[row2][col2]```
The process is repeated for each channel of the image. This means that the red, green, and blue color channels (if working in RGB color space) must each be multiplied by the kernel values. The kernel position is related to the pixel position it is multiplied by. Simply put, the kernel is allocated in `kernel[rows][cols]`, which would be `kernel[3][3]` in this case. The 3x3 (5x5 or 7x7, if using a larger kernel) area around the pixel (x,y) is then multiplied by the kernel to get the total sum. If we were working with a 100x100 image, allocated as `image[100][100]`, and we wanted the value for pixel (10,10), the process for each channel would look like:
```float fTotalSum =
Pixel(10-1,10-1) * kernel_value[row0][col0] +
Pixel(10 ,10-1) * kernel_value[row0][col1] +
Pixel(10+1,10-1) * kernel_value[row0][col2] +
Pixel(10-1,10 ) * kernel_value[row1][col0] +
Pixel(10 ,10 ) * kernel_value[row1][col1] +
Pixel(10+1,10 ) * kernel_value[row1][col2] +
Pixel(10-1,10+1) * kernel_value[row2][col0] +
Pixel(10 ,10+1) * kernel_value[row2][col1] +
Pixel(10+1,10+1) * kernel_value[row2][col2] +```
Finally, each value is added to the total sum, which is then divided by the total weight of the kernel. The kernel's weight is given by adding each value contained in the kernel. If the value is zero or less, then a weight of 1 is given to avoid a division by zero.
The actual code to convolve an image is:
```for (int i=0; i <= 2; i++)//loop through rows
{
for (int j=0; j <= 2; j++)//loop through columns
{
//get pixel near source pixel
/*
if x,y is source pixel then we loop through and
get pixels at coordinates:
x-1,y-1
x-1,y
x-1,y+1
x,y-1
x,y
x,y+1
x+1,y-1
x+1,y
x+1,y+1
*/
COLORREF tmpPixel = pDC->GetPixel(sourcex+(i-(2>>1)),
sourcey+(j-(2>>1)));
//get kernel value
float fKernel = kernel[i][j];
//multiply each channel by kernel value, and add to sum
//notice how each channel is treated separately
rSum += (GetRValue(tmpPixel)*fKernel);
gSum += (GetGValue(tmpPixel)*fKernel);
bSum += (GetBValue(tmpPixel)*fKernel);
//add the kernel value to the kernel sum
kSum += fKernel;
}
}
//if kernel sum is less than 0, reset to 1 to avoid divide by zero
if (kSum <= 0)
kSum = 1;
//divide each channel by kernel sum
rSum/=kSum;
gSum/=kSum;
bSum/=kSum;```
The source code included performs some common image convolutions. Also included is a Convolve Image menu option that allows users to enter their own kernel. Common 3x3 kernels include:
```gaussianBlur[3][3] = {0.045, 0.122, 0.045, 0.122,
0.332, 0.122, 0.045, 0.122, 0.045};
gaussianBlur2[3][3] = {1, 2, 1, 2, 4, 2, 1, 2, 1};
gaussianBlur3[3][3] = {0, 1, 0, 1, 1, 1, 0, 1, 0};
unsharpen[3][3] = {-1, -1, -1, -1, 9, -1, -1, -1, -1};
sharpness[3][3] = {0,-1,0,-1,5,-1,0,-1,0};
sharpen[3][3] = {-1, -1, -1, -1, 16, -1, -1, -1, -1};
edgeDetect[3][3] = {-0.125, -0.125, -0.125, -0.125,
1, -0.125, -0.125, -0.125, -0.125};
edgeDetect2[3][3] = {-1, -1, -1, -1, 8, -1, -1, -1, -1};
edgeDetect3[3][3] = {-5, 0, 0, 0, 0, 0, 0, 0, 5};
edgeDetect4[3][3] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
edgeDetect5[3][3] = {-1, -1, -1, 2, 2, 2, -1, -1, -1};
edgeDetect6[3][3] = {-5, -5, -5, -5, 39, -5, -5, -5, -5};
sobelHorizontal[3][3] = {1, 2, 1, 0, 0, 0, -1, -2, -1 };
sobelVertical[3][3] = {1, 0, -1, 2, 0, -2, 1, 0, -1 };
previtHorizontal[3][3] = {1, 1, 1, 0, 0, 0, -1, -1, -1 };
previtVertical[3][3] = {1, 0, -1, 1, 0, -1, 1, 0, -1};
boxBlur[3][3] = {0.111f, 0.111f, 0.111f, 0.111f,
0.111f, 0.111f, 0.111f, 0.111f, 0.111f};
triangleBlur[3][3] = { 0.0625, 0.125, 0.0625,
0.125, 0.25, 0.125, 0.0625, 0.125, 0.0625};```
Last but not least is the ability to show a convoluted image as a grayscale result. In order to display a filtered image as grayscale, we just add a couple lines to the bottom of the `Convolve` function:
```//return new pixel value
if (bGrayscale)
{
int grayscale=0.299*rSum + 0.587*gSum + 0.114*bSum;
rSum=grayscale;
gSum=grayscale;
bSum=grayscale;
}
clrReturn = RGB(rSum,gSum,bSum);```
This means that the entire `Convolve` function now looks like:
```COLORREF CImageConvolutionView::Convolve(CDC* pDC, int sourcex,
int sourcey, float kernel[3][3], int nBias,BOOL bGrayscale)
{
float rSum = 0, gSum = 0, bSum = 0, kSum = 0;
COLORREF clrReturn = RGB(0,0,0);
for (int i=0; i <= 2; i++)//loop through rows
{
for (int j=0; j <= 2; j++)//loop through columns
{
//get pixel near source pixel
/*
if x,y is source pixel then we loop
through and get pixels at coordinates:
x-1,y-1
x-1,y
x-1,y+1
x,y-1
x,y
x,y+1
x+1,y-1
x+1,y
x+1,y+1
*/
COLORREF tmpPixel = pDC->GetPixel(sourcex+(i-(2>>1)),
sourcey+(j-(2>>1)));
//get kernel value
float fKernel = kernel[i][j];
//multiply each channel by kernel value, and add to sum
//notice how each channel is treated separately
rSum += (GetRValue(tmpPixel)*fKernel);
gSum += (GetGValue(tmpPixel)*fKernel);
bSum += (GetBValue(tmpPixel)*fKernel);
//add the kernel value to the kernel sum
kSum += fKernel;
}
}
//if kernel sum is less than 0, reset to 1 to avoid divide by zero
if (kSum <= 0)
kSum = 1;
//divide each channel by kernel sum
rSum/=kSum;
gSum/=kSum;
bSum/=kSum;
//add bias if desired
rSum += nBias;
gSum += nBias;
bSum += nBias;
//prevent channel overflow by clamping to 0..255
if (rSum > 255)
rSum = 255;
else if (rSum < 0)
rSum = 0;
if (gSum > 255)
gSum = 255;
else if (gSum < 0)
gSum = 0;
if (bSum > 255)
bSum = 255;
else if (bSum < 0)
bSum = 0;
//return new pixel value
if (bGrayscale)
{
int grayscale=0.299*rSum + 0.587*gSum + 0.114*bSum;
rSum=grayscale;
gSum=grayscale;
bSum=grayscale;
}
clrReturn = RGB(rSum,gSum,bSum);
return clrReturn;
}```
Last but not least, I did a little tweaking to get the program to load a default image from a resource (`IDB_BITMAP1`). Then, I added the ability to convolve this default image. The program will still load image from a file, the only difference is that it will now show a default image at startup.
Please note that this article is, by no means, an example of fast processing of pixels. It is merely meant to show how convolution can be done on images. If you would like a more advanced image processor, then feel free to email me with the subject "WANT CODE:ImageEdit Please". That is an unreleased image processor I have done, though parts are not implemented yet due to lack of time, that contains much more functionality, using the CxImage library as its basis for reading and saving images.
This article has no explicit license attached to it but may contain usage terms in the article text or the download files themselves. If in doubt please contact the author via the discussion board below.
A list of licenses authors might use can be found here
## About the Author
Web Developer United States
Programming using MFC and ATL for almost 12 years now. Currently studying Operating System implementation as well as Image processing. Previously worked on DSP and the use of FFT for audio application. Programmed using ADO, ODBC, ATL, COM, MFC for shell interfacing, databasing tasks, Internet items, and customization programs.
## Comments and Discussions
First PrevNext
24 bit RGB ñoqui4-Apr-06 4:00 ñoqui 4-Apr-06 4:00
Re: 24 bit RGB Fred Ackers13-Apr-06 4:03 Fred Ackers 13-Apr-06 4:03
Re: 24 bit RGB ñoqui13-Apr-06 4:40 ñoqui 13-Apr-06 4:40
Does not compile in MSVC 2003 Dark Alchemist6-Jan-06 8:27 Dark Alchemist 6-Jan-06 8:27
Re: Does not compile in MSVC 2003 Aqiruse29-Mar-06 9:08 Aqiruse 29-Mar-06 9:08
please , help ? dynamica1231-Oct-05 22:03 dynamica123 1-Oct-05 22:03
Terrible Christian Graus30-Mar-05 16:02 Christian Graus 30-Mar-05 16:02
Re: Terrible Georgi Petrov30-Mar-05 19:17 Georgi Petrov 30-Mar-05 19:17
Re: Terrible Christian Graus31-Mar-05 12:03 Christian Graus 31-Mar-05 12:03
Georgi Petrov wrote: This is not the problem, because can be fixed very easy. No, it can't. You need to fundamentally change how your code works. You need to either use GDI+, or use a DIBSection wrapper to create images in the first place, and you need to work independantly of any device context. Georgi Petrov wrote: The main problem in convolution is the proper filter selecion That depends. The 'proper' filter won't speed your code up, or make it useful to people wanting to work on images to save them again ( if you load a 24 bit image on an 8 bit display and use this code, you'll generate an 8 bit image to save ). Georgi Petrov wrote: FFT is still slow for 2.4GHz CPU, so 3x3 5x5 kernels are OK Where did fast fourier transforms enter the discussion ? Or does FFT mean something else here ? No matter what, your code is significantly slower than it could be. Christian I have several lifelong friends that are New Yorkers but I have always gravitated toward the weirdo's. - Richard Stringer
Re: Terrible Georgi Petrov1-Apr-05 1:19 Georgi Petrov 1-Apr-05 1:19
Re: Terrible Christian Graus1-Apr-05 8:49 Christian Graus 1-Apr-05 8:49
Re: Terrible Rick York29-Mar-06 12:21 Rick York 29-Mar-06 12:21
Re: Terrible Christian Graus29-Mar-06 12:23 Christian Graus 29-Mar-06 12:23
Re: Terrible Fred Ackers30-Mar-06 0:22 Fred Ackers 30-Mar-06 0:22
Re: Terrible JKaminski7-Apr-06 2:55 JKaminski 7-Apr-06 2:55
Re: Terrible Christian Graus1-Apr-05 8:50 Christian Graus 1-Apr-05 8:50
Re: Terrible Mohammad A Gdeisat31-Jul-06 9:32 Mohammad A Gdeisat 31-Jul-06 9:32
O! Yes it is excelent Georgi Petrov26-Aug-04 0:45 Georgi Petrov 26-Aug-04 0:45
Re: O! Yes it is excelent Christian Graus30-Mar-05 16:04 Christian Graus 30-Mar-05 16:04
Re: O! Yes it is excelent Georgi Petrov30-Mar-05 19:15 Georgi Petrov 30-Mar-05 19:15
Re: O! Yes it is excelent Christian Graus31-Mar-05 12:00 Christian Graus 31-Mar-05 12:00
Re: O! Yes it is excelent wittend22-Oct-07 10:02 wittend 22-Oct-07 10:02
Grayscale conversion Erik_Egsgard20-Aug-04 3:45 Erik_Egsgard 20-Aug-04 3:45
Re: Grayscale conversion Aqiruse20-Aug-04 13:58 Aqiruse 20-Aug-04 13:58
Re: Grayscale conversion Erik_Egsgard21-Aug-04 18:29 Erik_Egsgard 21-Aug-04 18:29
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https://www.shaalaa.com/textbook-solutions/c/balbharati-solutions-mathematics-and-statistics-2-commerce-11th-standard-hsc-maharashtra-state-board-chapter-7-probability_3290 | 1,652,688,258,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00028.warc.gz | 1,167,153,908 | 21,058 | Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board chapter 7 - Probability [Latest edition]
Chapter 7: Probability
Exercise 7.1Exercise 7.2Exercise 7.3Exercise 7.4Miscellaneous Exercise 7
Exercise 7.1 [Pages 99 - 100]
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board Chapter 7 Probability Exercise 7.1 [Pages 99 - 100]
Exercise 7.1 | Q 1. (a) | Page 99
State the sample space and n(S) for the following random experiment.
A coin is tossed twice. If a second throw results in a tail, a die is thrown.
Exercise 7.1 | Q 1. (b) | Page 99
State the sample space and n(S) for the following random experiment.
A coin is tossed twice. If a second throw results in head, a die thrown, otherwise a coin is tossed.
Exercise 7.1 | Q 2 | Page 99
In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).
Exercise 7.1 | Q 3. (a) | Page 99
A coin and a die are tossed. State sample space of following event.
A: Getting a head and an even number.
Exercise 7.1 | Q 3. (b) | Page 99
A coin and a die are tossed. State sample space of following event.
B: Getting a prime number.
Exercise 7.1 | Q 3. (c) | Page 99
A coin and a die are tossed. State sample space of following event.
C: Getting a tail and perfect square.
Exercise 7.1 | Q 4. (a) | Page 99
Find total number of distinct possible outcomes n(S) of the following random experiment.
From a box containing 25 lottery tickets any 3 tickets are drawn at random.
Exercise 7.1 | Q 4. (b) | Page 99
Find total number of distinct possible outcomes n(S) of the following random experiment.
From a group of 4 boys and 3 girls, any two students are selected at random.
Exercise 7.1 | Q 4. (c) | Page 99
Find total number of distinct possible outcomes n(S) of the following random experiment.
5 balls are randomly placed into 5 cells, such that each cell will be occupied.
Exercise 7.1 | Q 4. (d) | Page 99
Find total number of distinct possible outcomes n(S) of the following random experiment.
6 students are arranged in a row for a photograph.
Exercise 7.1 | Q 5. (a) | Page 99
Two dice are thrown. Write favourable Outcomes for the following event.
P: Sum of the numbers on two dice is divisible by 3 or 4.
Exercise 7.1 | Q 5. (b) | Page 99
Two dice are thrown. Write favourable outcomes for the following event.
Q: Sum of the numbers on two dice is 7.
Exercise 7.1 | Q 5. (c) (i) | Page 100
Two dice are thrown. Write favourable outcomes for the following event.
R: Sum of the numbers on two dice is a prime number.
Also, check whether Events P and Q are mutually exclusive and exhaustive.
Exercise 7.1 | Q 5. (c) (ii) | Page 100
Two dice are thrown. Write favourable outcomes for the following event.
R: Sum of the numbers on two dice is a prime number.
Also, check whether Events Q and R are mutually exclusive and exhaustive.
Exercise 7.1 | Q 6. (a) | Page 100
A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space if consideration of suits is not taken into account.
Exercise 7.1 | Q 6. (b) | Page 100
A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space if consideration of suits is taken into account.
Exercise 7.1 | Q 7 | Page 100
Box-I contains 8 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red(R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-I; if it turns up tails, a marble is chosen from Box-II. Describe the sample space.
Exercise 7.1 | Q 8. (a) | Page 100
Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample Space if cards are drawn with replacement.
Exercise 7.1 | Q 8. (b) | Page 100
Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample Space if cards are drawn without replacement.
Exercise 7.2 [Pages 102 - 103]
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board Chapter 7 Probability Exercise 7.2 [Pages 102 - 103]
Exercise 7.2 | Q 1. (a) | Page 102
A fair die is thrown two times. Find the chance that product of the numbers on the upper face is 12.
Exercise 7.2 | Q 1. (b) | Page 102
A fair die is thrown two times. Find the chance that sum of the numbers on the upper face is 10.
Exercise 7.2 | Q 1. (c) | Page 102
A fair die is thrown two times. Find the chance that Sum of the numbers on the upper face is at least 10.
Exercise 7.2 | Q 1. (d) | Page 102
A fair die is thrown two times. Find the chance that sum of the numbers on the upper face is at least 4.
Exercise 7.2 | Q 1. (e) | Page 102
A fair die is thrown two times. Find the chance that the first throw gives an odd number and second throw gives multiple of 3.
Exercise 7.2 | Q 1. (f) | Page 102
A fair die is thrown two times. Find the chance that both the times die shows same number (doublet).
Exercise 7.2 | Q 2. (a) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that both are black.
Exercise 7.2 | Q 2. (b) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that both are diamonds.
Exercise 7.2 | Q 2. (c) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that both are ace cards.
Exercise 7.2 | Q 2. (d) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that both are face cards.
Exercise 7.2 | Q 2. (e) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that one is spade and other is non-spade.
Exercise 7.2 | Q 2. (f) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that both are from same suit.
Exercise 7.2 | Q 2. (g) | Page 102
Two cards are drawn from a pack of 52 cards. Find the probability that both are from same denomination.
Exercise 7.2 | Q 3. (a) | Page 102
Four cards are drawn from a pack of 52 cards. Find the probability that 3 are king and 1 is jack.
Exercise 7.2 | Q 3. (b) | Page 102
Four cards are drawn from a pack of 52 cards. Find the probability that all the cards are from different suit.
Exercise 7.2 | Q 3. (c) | Page 102
Four cards are drawn from a pack of 52 cards. Find the probability that at least one heart.
Exercise 7.2 | Q 3. (d) | Page 102
Four cards are drawn from a pack of 52 cards. Find the probability that all cards are club and one of them a jack.
Exercise 7.2 | Q 4. (a) | Page 102
A bag contains 15 balls of three different colours: Green, Black and Yellow. A ball is drawn at random from the bag. The probability of green ball is 1/3. The probability of yellow ball is 1/5. What is the probability of blackball?
Exercise 7.2 | Q 4. (b) | Page 102
A bag contains 15 balls of three different colours: Green, Black and Yellow. A ball is drawn at random from the bag. The probability of green ball is 1/3. The probability of yellow ball is 1/5. How many balls are green, black, and yellow?
Exercise 7.2 | Q 5. (a) | Page 102
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that, number on ticket is divisible by 6?
Exercise 7.2 | Q 5. (b) | Page 102
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that, number on ticket is a perfect square?
Exercise 7.2 | Q 5. (c) | Page 102
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that, number on ticket is prime?
Exercise 7.2 | Q 5. (d) | Page 102
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that, number on ticket is divisible by 3 and 5?
Exercise 7.2 | Q 6. (a) | Page 102
From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains 3 boys and 2 girls.
Exercise 7.2 | Q 6. (b) | Page 102
From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains at least 3 boys.
Exercise 7.2 | Q 7 | Page 103
A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in socket. What is the probability that the room is lit?
Exercise 7.2 | Q 8. (a) | Page 103
The letters of the word LOGARITHM are arranged at random. Find the probability that vowels are always together.
Exercise 7.2 | Q 8. (b) | Page 103
The letters of the word LOGARITHM are arranged at random. Find the probability that vowels are never together.
Exercise 7.2 | Q 8. (c) | Page 103
The letters of the word LOGARITHM are arranged at random. Find the probability that Exactly 4 letters between G and H.
Exercise 7.2 | Q 8. (d) | Page 103
The letters of the word LOGARITHM are arranged at random. Find the probability that Begins with O and ends with T.
Exercise 7.2 | Q 8. (e) | Page 103
The letters of the word LOGARITHM are arranged at random. Find the probability that start with vowel and end with ends with consonant.
Exercise 7.2 | Q 9 | Page 103
The letters of the word SAVITA are arranged at random. Find the probability that vowels are always together.
Exercise 7.3 [Page 104]
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board Chapter 7 Probability Exercise 7.3 [Page 104]
Exercise 7.3 | Q 1 | Page 104
Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or number on the second die is greater than or equal to the number on the first die?
Exercise 7.3 | Q 2. (a) | Page 104
A card is drawn from a pack of 52 cards. What is the probability that, card is either red or black?
Exercise 7.3 | Q 2. (b) | Page 104
A card is drawn from a pack of 52 cards. What is the probability that card is either red or face card?
Exercise 7.3 | Q 3. (a) | Page 104
Two cards are drawn from a pack of 52 cards. What is the probability that, both the cards are of same colour?
Exercise 7.3 | Q 3. (b) | Page 104
Two cards are drawn from a pack of 52 cards. What is the probability that, both the cards are either black or queens?
Exercise 7.3 | Q 4. (a) | Page 104
A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is the probability that, number on the ticket is a perfect square or divisible by 4?
Exercise 7.3 | Q 4. (b) | Page 104
A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is the probability that, number on the ticket is a prime number or greater than 30?
Exercise 7.3 | Q 5. (a) | Page 104
Hundred students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed in both. Find the probability that student selected at random passed in at least one examination.
Exercise 7.3 | Q 5. (b) | Page 104
Hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that student selected at random passed in exactly one examination.
Exercise 7.3 | Q 5. (c) | Page 104
Hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that student selected at random failed in both the examinations.
Exercise 7.3 | Q 6. (a) | Page 104
If P(A) = 1/4, "P"("B") = 2/5 and "P"("A" ∪ "B") = 1/2 Find the value of the following probability: P(A ∩ B)
Exercise 7.3 | Q 6. (b) | Page 104
If P(A) = 1/4, P(B) = 2/5 and P(A ∪ B) = 1/2 Find the value of the following probability: P(A ∩ B')
Exercise 7.3 | Q 6. (c) | Page 104
If P(A) = 1/4, P(B) = 2/5 and P(A ∪ B) = 1/2 Find the value of the following probability: P(A' ∩ B)
Exercise 7.3 | Q 6. (d) | Page 104
If P(A) = 1/4, P(B) = 2/5 and P(A ∪ B) = 1/2 Find the value of the following probability: P(A' ∪ B')
Exercise 7.3 | Q 6. (e) | Page 104
If P(A) = 1/4, P(B) = 2/5 and P(A ∪ B) = 1/2 Find the value of the following probability: P(A' ∩ B')
Exercise 7.3 | Q 7 | Page 104
A computer software company is bidding for computer programs A and B. The probability that the company will get software A is 3/5, the probability that the company will get software B is 1/3, and the probability that the company will get both A and B is 1/8. What is the probability that the company will get at least one software?
Exercise 7.3 | Q 8 | Page 104
A card is drawn from a well shuffled pack of 52 cards. Find the probability of it being a heart or a queen.
Exercise 7.3 | Q 9 | Page 104
In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Exercise 7.4 [Pages 107 - 108]
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board Chapter 7 Probability Exercise 7.4 [Pages 107 - 108]
Exercise 7.4 | Q 1 | Page 107
Two dice are thrown simultaneously, If at least one of the dice show a number 5, what is the probability that sum of the numbers on two dice is 9?
Exercise 7.4 | Q 2 | Page 107
A pair of dice is thrown. If sum of the numbers is an even number, what is the probability that it is a perfect square?
Exercise 7.4 | Q 3 | Page 107
A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
Exercise 7.4 | Q 4 | Page 107
A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as
A: a club 6 card is drawn.
B: an ace card 18 drawn.
Determine whether the events A and B are independent or not.
Exercise 7.4 | Q 5. (a) | Page 107
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that, problem is solved?
Exercise 7.4 | Q 5. (b) | Page 107
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that, problem is not solved
Exercise 7.4 | Q 5. (c) | Page 107
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that, exactly two students solve the problem?
Exercise 7.4 | Q 6 | Page 107
The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive after 10 years?
Exercise 7.4 | Q 7. (a) | Page 108
In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random, what is the probability that the student has failed in subject I, if it is known that he is failed in subject II?
Exercise 7.4 | Q 7. (b) | Page 108
In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random, what is the probability that the student has failed in at least one subject?
Exercise 7.4 | Q 7. (c) | Page 108
In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random, what is the probability that the student has failed in exactly one subject?
Exercise 7.4 | Q 8 | Page 108
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third guns respectively. assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
Exercise 7.4 | Q 9. (a) | Page 108
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that, first is white and second is black?
Exercise 7.4 | Q 9. (b) | Page 108
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that, one is white and other is black?
Exercise 7.4 | Q 10 | Page 108
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, What is the probability that at least one ball is black?
Exercise 7.4 | Q 11 | Page 108
Two balls are drawn from an urn containing 5 green, 3 blue, and 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Exercise 7.4 | Q 12 | Page 108
A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball is drawn from each bag, what is the probability that two balls are of the same colour?
Exercise 7.4 | Q 13 | Page 108
Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards drawn are face cards?
Miscellaneous Exercise 7 [Pages 109 - 110]
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board Chapter 7 Probability Miscellaneous Exercise 7 [Pages 109 - 110]
Miscellaneous Exercise 7 | Q 1 | Page 109
From a group of 2 men (M1, M2) and three women (W1, W2, W3), two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E?
Miscellaneous Exercise 7 | Q 2 | Page 110
Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. What is the chance that three selected consists of 1 girl and 2 boys?
Miscellaneous Exercise 7 | Q 3 | Page 110
A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in socket. What is the probability that the room is lit?
Miscellaneous Exercise 7 | Q 4 | Page 110
There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.
Miscellaneous Exercise 7 | Q 5 | Page 110
A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product on the numbers is even?
Miscellaneous Exercise 7 | Q 6 | Page 110
A, B, and C are mutually exclusive and exhaustive events associated with the random experiment. Find P(A), given that P(B) = 3/2 P(A) and P(C) = 1/2 P(B).
Miscellaneous Exercise 7 | Q 7 | Page 110
An urn contains four tickets marked with numbers 112, 121, 122, 222 and one ticket is drawn at random. Let Ai (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A1, A2, and A3.
Miscellaneous Exercise 7 | Q 8 | Page 110
The odds against a certain event are 5: 2 and odds in favour of another independent event are 6: 5. Find the chance that at least one of the events will happen.
Miscellaneous Exercise 7 | Q 9. (a) | Page 110
The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that the couple will be alive 20 years hence.
Miscellaneous Exercise 7 | Q 9. (b) | Page 110
The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that at least one of them will be alive 20 years hence.
Miscellaneous Exercise 7 | Q 10 | Page 110
Two throws are made, the first with 3 dice and the second with 2 dice. The faces of each die are marked with the number 1 to 6. What is the probability that the total in first throw is not less than 15 and at the same time the total in the second throw is not less than 8?
Miscellaneous Exercise 7 | Q 11 | Page 110
Two-third of the students in a class are boys and rest are girls. It is known that the probability of girl getting first class is 0.25 and that of boy getting is 0.28. Find the probability that a student chosen at random will get first class.
Miscellaneous Exercise 7 | Q 12 | Page 110
A number of two digits is formed using the digits 1, 2, 3, ….., 9. What is the probability that the number so chosen is even and less than 60?
Miscellaneous Exercise 7 | Q 13 | Page 110
A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls are made without replacement. Find the probability that the first drawing will give 3 white balls and second drawing will give 3 red balls.
Miscellaneous Exercise 7 | Q 14. (a) | Page 110
The odds against student X solving a business statistics problem are 8: 6 and odds in favour of student Y solving the same problem are 14: 16 What is the chance that the problem will be solved, if they try independently?
Miscellaneous Exercise 7 | Q 14. (b) | Page 110
The odds against student X solving a business statistics problem are 8: 6 and odds in favour of student Y solving the same problem are 14: 16 What is the probability that neither solves the problem?
Chapter 7: Probability
Exercise 7.1Exercise 7.2Exercise 7.3Exercise 7.4Miscellaneous Exercise 7
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board chapter 7 - Probability
Balbharati solutions for Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board chapter 7 (Probability) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board solutions in a manner that help students grasp basic concepts better and faster.
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Concepts covered in Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board chapter 7 Probability are Introduction of Probability, Types of Events, Algebra of Events, Elementary Properties of Probability, Addition Theorem of Probability, Conditional Probability, Multiplication Theorem on Probability, Independent Events.
Using Balbharati 11th solutions Probability exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Balbharati Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board 11th prefer Balbharati Textbook Solutions to score more in exam.
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# 6th Grade Math Fractions Curriculum Unit Three
Common Core Standards
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In this NO PREP product you get an entire unit that covers: estimating fractions, adding fractions, subtracting fractions, adding and subtracting fractions word problems, multiplying fractions, using models for dividing fractions, dividing fractions using an algorithm, word problems for multiplication and division of fractions, and solving one-step equations using whole numbers and fractions. All you have to do is print and distribute the notes, worksheets, and assessments to your students.
ALL chapters include:
• Notes
• Practice Pages
Chapter 1 includes:
• Estimating Fractions
• Subtracting Fractions
• Wrapping Up Adding and Subtracting Fractions With Word Problems
Chapter 2 includes:
• Using Models for Multiplying Fractions
• Multiplying Fractions Less Than 1
• Multiplying Mixed Numbers
Chapter 3 includes:
• Using Models for Dividing Fractions
• Dividing Fractions Less Than 1
• Dividing With Mixed Numbers and Whole Numbers
• Word Problems Involving Multiplication and Division of Fractions
Chapter 4 includes:
• Addition and Subtraction Fact Families
• Solving One-Step Equations Using Addition and Subtraction
• Multiplication and Division Fact Families
• Solving One-Step Equations Using Multiplication and Division
There is also an assessment (quiz) over Chapters 1 and 2 and a test over the entire Unit!
The Common Core Content Standards that are included in this product are:
6. NS.A, 6.NS.A.1, 6.NS.B.3, 6.NS.B.4, 6.EE.A.2, 6.EE.A.2.a, 6.EE.A.2.b, 6.EE.A.2.c, 6.EE.A.3, 6.EE.B.6, 6.EE.B.7
*This is the same product as Fractions 6th Grade Curriculum Unit 3 using Google
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\$20.00 | 530 | 2,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.817893 |
https://www.sawaal.com/area-questions-and-answers/a-tank-is-25m-long-12m-wide-and-6m-deep-the-cost-of-plastering-its-walls-and-bottom-at-75-paise-per-_4235 | 1,718,620,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00742.warc.gz | 868,757,716 | 14,204 | 78
Q:
# A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is
A) Rs. 258 B) Rs. 358 C) Rs. 458 D) Rs. 558
Explanation:
Area to be plastered = $2l+b×h+l×b$
=$225+12×6+25×12=744sq.m$
Cost of plastering = $744×75100=Rs.558$
Subject: Area
Exam Prep: Bank Exams
Job Role: Bank PO
Q:
What is the area of this trapezoidal garden? (All measurements are in cm)
A) 60 sq.cm B) 180 sq.cm C) 210 sq.cm D) 240 sq.cm
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
3 3073
Q:
∆ABC is right angled at B. If cosA = 8/17, then what is the value of cotC ?
A) 15/8 B) 15/17 C) 8/17 D) 17/15
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
1 6343
Q:
The curved surface area and the diameter of a right circular cylinder are 660 sq.cm and 21 cm respectively. Find its height (in cm).
A) 9 B) 10 C) 12 D) 8
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
0 424
Q:
If the measure of the interior angle of a regular polygon is $1800$, then how many sides does it have?
A) 6 B) 8 C) 9 D) 5
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
1 583
Q:
The sum of the lengths of the edges of a cube is equal to the perimeter of a square. If the numerical value of the volume of the cube is equal to the numerical value of the area of the square, then the length of one side of the square is
A) 30 units B) 9 units C) 27 units D) 12 units
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
1 435
Q:
A square ground is to be covered by planting 100 saplings on each side. How many saplings are needed in all?
A) 400 B) 404 C) 396 D) 408
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
0 480
Q:
Find the curved surface area (in cm2) of a right circular cone of diameter 28 cm and slant height 12 cm.
A) 792 B) 728 C) 528 D) 493
Explanation:
Filed Under: Area
Exam Prep: Bank Exams
0 430
Q:
In ΔABC measure of angle B is $900$. If cotA = 8/15, and AB = 0.8cm, then what is the length (in cm) of side BC?
A) 1.7 B) 2 C) 1.5 D) 2.5 | 701 | 2,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.78734 |
https://www.quizover.com/trigonometry/terms/directrix-the-parabola-by-openstax | 1,527,031,369,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005116-00538.warc.gz | 823,565,894 | 16,242 | # 10.3 The parabola (Page 8/11)
Page 8 / 11
• Card 8 / 11:
directrix
a line perpendicular to the axis of symmetry of a parabola; a line such that the ratio of the distance between the points on the conic and the focus to the distance to the directrix is constant
• ## Keyboard Shortcuts
Previous Card ← Previous Card Button Next Card → Next Card Button Flip Card ↑ / ↓ / Return / Space
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
Is there any rule we can use to get the nth term ?
how do you get the (1.4427)^t in the carp problem?
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
hello
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
rotation by 80 of (x^2/9)-(y^2/16)=1
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
what is the standard form if the focus is at (0,2) ?
a²=4 | 571 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-22 | longest | en | 0.915248 |
https://tex.stackexchange.com/questions/454013/how-to-add-slopes-to-function-plotted-with-pgfplots | 1,653,458,607,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662580803.75/warc/CC-MAIN-20220525054507-20220525084507-00066.warc.gz | 613,093,956 | 66,308 | # How to add slopes to function plotted with pgfplots?
I'd like to add slopes to a function I've plotted that is similar to the following figure that I drew using asymptote. I'd like to remove asymptote dependency from my document by using pgfplots/tikz.
To get started I put a simple pgfplots script together (see below) but now need to figure out how to add the slopes (preferable by computing them, since if the function changes the slope can change).
In asymptote I did the following:
real dydt = 5*exp(-x); // derivative of f()
real theta = atan (dydt);
dx = 0.8*cos (theta);
draw (Scale((x-dx, f(x)-dydt*dx))--Scale((x+dx, f(x)+dydt*dx)), blue+linewidth(1.2));
I presume something similar can be done with pgfplots/tikz?
\documentclass{article}
\usepackage{tikz,pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
clip=false,
xmin = 0, xmax = 5,
ymin = 0, ymax = 5,
xlabel = Time,
ylabel = Concentration of Product
]
\node at (axis cs:3,3) {Slope = rate of change};
\draw[fill=blue](axis cs:0,{5 - 5*exp(-1.0*0)}) circle[blue, radius=6];
\draw[fill=blue](axis cs:1,{5*(1 - exp(-1.0*1))}) circle[blue, radius=6];
\draw[fill=blue](axis cs:2,{5*(1 - exp(-1.0*2))}) circle[blue, radius=6];
\end{axis}
\end{tikzpicture}
\end{document}
– user121799
Oct 5, 2018 at 17:53
• Thank you for that, I searched but hadn't found anything. Oct 5, 2018 at 18:14
I'd just take Jake's great answer and make minor modifications that allow one to place the tangent at various locations and control their appearance. Notice that with decorations you are automatically in the tangent space of the path, so all you really do is to draw a horizontal line in tangent space. That is, you do not have to do the same gymnastics that you did in asymptote.
\documentclass{article}
\usepackage{tikz,pgfplots}
% from https://tex.stackexchange.com/a/198046/121799
\usetikzlibrary{intersections}
\makeatletter
\def\parsenode[#1]#2\pgf@nil{%
\tikzset{label node/.style={#1}}
\def\nodetext{#2}
}
\tikzset{
add node at x/.style 2 args={
name path global=plot line,
/pgfplots/execute at end plot visualization/.append={
\begingroup
\@ifnextchar[{\parsenode}{\parsenode[]}#2\pgf@nil
\path [name path global = position line #1-1]
({axis cs:#1,0}|-{rel axis cs:0,0}) --
({axis cs:#1,0}|-{rel axis cs:0,1});
\path [xshift=1pt, name path global = position line #1-2]
({axis cs:#1,0}|-{rel axis cs:0,0}) --
({axis cs:#1,0}|-{rel axis cs:0,1});
\path [
name intersections={
of={plot line and position line #1-1},
name=left intersection
},
name intersections={
of={plot line and position line #1-2},
name=right intersection
},
label node/.append style={pos=1}
] (left intersection-1) -- (right intersection-1)
node [label node]{\nodetext};
\endgroup
}
}
}
\makeatother
\begin{document}
\begin{tikzpicture}
\begin{axis}[
clip=false,
xmin = 0, xmax = 5,
ymin = 0, ymax = 5,
xlabel = Time,
ylabel = Concentration of Product,
tangent/.style args={at #1 with style #2 and length #3}{
[
sloped,
append after command={(\tikzlastnode.west) edge [#2] (\tikzlastnode.east)},
minimum width=#3
]
}
}
] | 987 | 3,075 | {"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} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.72673 |
https://www.pirate-physics.net/at-physics/at-previous-cycles/ATc26 | 1,656,683,068,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00490.warc.gz | 1,022,818,816 | 23,637 | # AT Cycle 26
## 2/16 - 2/22
W 2/16
Th 2/17
F 2/18
T 2/22
### π΄β 1: W 2/16, π‘β 3: W 2/16 - circuits assessment
Circuits Unit Assessment TODAY!!!
Homework: Make sure you have reviewed all of the magnetism topics you should have learned last year on my AT Magnetism Review page. Upload pictures of your magnetism notes to βοΈ Google Classroom assignment by Wednesday at 10pm.
### π₯ 1: F 2/18 lab, π¨ 3: Th 2/17 lab - basic magnetism problems
We'll start the period by testing our understanding of force on moving charged particles in magnetic fields with a couple of demonstrations. We'll discuss applications including the mass spectrometer.
This hour, we'll apply our problem solving and collaboration skills by doing problems from AP problems for force by magnetic field. (MC ans: CADEEACDA)
Homework: Finish all of the problems and upload FRQ solutions to βοΈ Google Classroom assignment by Sunday at 10pm.
### *Z* 1: M 2/21, π 3: F 2/18 - β©οΈ Pivot magnetic force lab
Z-day for pd 1
1 - 7:40 - 8:24 (44)
2 - 8:29 - 9:13 (44)
3 - 9:18 - 10:02 (44)
4 - 10:06 - 10:50 (44)
L1 - 10:54 - 11:35 (41)
5a - 11:40 - 12:24 (44)
5b - 10:55 - 11:39 (44)
L2 - 11:43 - 12:
24 (41)
6 - 12:29 - 1:13 (44)
7 - 1:18 - 2:02 (44)
8 - 2:06 - 2:50 (44)
We'll continue to review magnetic force. We will make sure you understand the magnetic force equations. Then, we'll start a lab in β©οΈ Pivot Interactives "Force on a Current-Carrying Conductor in a Magnetic Field." This lab will verify the equation you learned last year of the magnetic force on a current carrying wire. Lab is due Tuesday at 10pm.
Homework: Finish AP problems for force by magnetic field and upload FRQ solutions to βοΈ Google Classroom assignment by Sunday at 10pm. β©οΈ Pivot Interactives "Force on a Current-Carrying Conductor in a Magnetic Field" is due on Tuesday at 10pm.
### β€οΈ 1: T 2/22, *Z* 3: M 2/21 - Black History Month βπΏβπΎβπ½
Z-day for pd 3
1 - 7:40 - 8:24 (44)
2 - 8:29 - 9:13 (44)
3 - 9:18 - 10:02 (44)
4 - 10:06 - 10:50 (44)
L1 - 10:54 - 11:35 (41)
5a - 11:40 - 12:24 (44)
5b - 10:55 - 11:39 (44)
L2 - 11:43 - 12:
24 (41)
6 - 12:29 - 1:13 (44)
7 - 1:18 - 2:02 (44)
8 - 2:06 - 2:50 (44)
Today, we'll check out the Black History Month programming provided by the school:
Homework: β©οΈ Pivot Interactives "Force on a Current-Carrying Conductor in a Magnetic Field" is due on Tuesday at 10pm.
Pd 1 - "Black Culture & Identity"
The Journey from Jersey City, NJ, West Windsor, NJ Howard University to Brooklyn, and NYC- Ms. Darnese Daniels is the Director of Equity Transformation and Culturally Responsive Environments (CRE) within the Office of Equity and Access for the New York City Department of Education.
Pd 3 - "Leaders of the Future"
Alumni will discuss how they got involved in their career of choice and what it was like for them to transition from college to a job. They will also talk about the racism and prejudice they experienced in their journey. | 1,015 | 3,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.841889 |
http://pdesolutions.com/help/lowvisc.html | 1,526,907,836,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864186.38/warc/CC-MAIN-20180521122245-20180521142245-00110.warc.gz | 228,105,314 | 4,557 | Sample Problems > Applications > Fluids > lowvisc
# lowvisc
Navigation: Sample Problems > Applications > Fluids >
# lowvisc
{ LOWVISC.PDE
This example is a modification of the VISCOUS.PDE problem, in which the
viscosity has been lowered to produce a Reynold's number of approximately
40. This seems to be the practical upper limit or Reynolds number for
steady-state solutions of Navier-Stokes equations with FlexPDE.
As the input pressure is raised, the disturbance in velocities propagates farther
down the channel. The channel must be long enough that the velocities
have returned to the open-channel values, or the P=0 boundary condition
at the outlet will be invalid and the solution will not succeed.
The problem computes half of the domain, with a reflective boundary at the bottom.
We have included four elevation plots of X-velocity, at the inlet, channel
center, obstruction center and outlet of the channel. The integrals presented
on these plots show the consistency of mass transport across the channel.
We have added a variable psi to compute the stream function for plotting stream lines.
}
title 'Viscous flow in 2D channel, Re > 40'
variables u(0.1) v(0.01) p(1) psi select ngrid = 40 definitions Lx = 5 Ly = 1.5 p0 = 2
speed2 = u^2+v^2
speed = sqrt(speed2)
dens = 1
visc = 0.04 vxx = -(p0/(2*visc*(2*Lx)))*(Ly^2-y^2) { open-channel x-velocity }
rball=0.4
cut = 0.1 { value for bevel at the corners of the obstruction }
penalty = 100*visc/rball^2
Re = globalmax(speed)*(Ly/2)/(visc/dens)
w = zcomp(curl(u,v)) ! vorticity is the source for streamline equation
initial values
u = 0.5*vxx v = 0 p = p0*(Lx+x)/(2*Lx)
equations
u: visc*div(grad(u)) - dx(p) = dens*(u*dx(u) + v*dy(u))
v: visc*div(grad(v)) - dy(p) = dens*(u*dx(v) + v*dy(v))
then
psi: div(grad(psi)) + w = 0 ! solve streamline equation separately from velocities
boundaries
region 1
start(-Lx,0)
load(u) = 0 value(v) = 0 load(p) = 0 value(psi) = 0
line to (Lx/2-rball,0)
value(u) = 0 value(v) = 0 load(p) = 0
mesh_spacing = rball/10 ! dense mesh to resolve obstruction
line to (Lx/2-rball,rball) bevel(cut)
to (Lx/2+rball,rball) bevel(cut)
to (Lx/2+rball,0)
mesh_spacing = 10*rball ! cancel dense mesh requirement
line to (Lx,0)
load(u) = 0 value(v) = 0 value(p) = p0 natural(psi) = 0
line to (Lx,Ly)
value(u) = 0 value(v) = 0 load(p) = 0 natural(psi) = normal(-v,u)
line to (-Lx,Ly)
load(u) = 0 value(v) = 0 value(p) = 0 natural(psi) = 0
line to close
monitors
contour(speed) report(Re)
contour(psi) as "Streamlines"
contour(max(psi,-0.003)) zoom(Lx/2-3*rball,0, 3*rball,3*rball) as "Vortex Streamlines"
vector(u,v) as "flow" zoom(Lx/2-3*rball,0, 3*rball,3*rball) norm
plots
contour(u) report(Re)
contour(v) report(Re)
contour(speed) painted report(Re)
vector(u,v) as "flow" report(Re)
contour(p) as "Pressure" painted
contour(dx(u)+dy(v)) as "Continuity Error"
elevation(u) from (-Lx,0) to (-Lx,Ly)
elevation(u) from (0,0) to (0,Ly)
elevation(u) from (Lx/2,0) to (Lx/2,Ly)
elevation(u) from (Lx,0) to (Lx,Ly)
contour(psi) as "Streamlines"
contour(max(psi,-0.003)) zoom(Lx/2-3*rball,0, 3*rball,3*rball) as "Vortex Streamlines"
vector(u,v) as "flow" zoom(Lx/2-3*rball,0, 3*rball,3*rball) norm
Transfer(u,v,p) ! write flow solution as initial values for Coupled_Contaminant.pde
end | 1,149 | 3,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | longest | en | 0.760832 |
https://boardgames.stackexchange.com/questions/18047/a-couple-of-hearts-probabilities/20315 | 1,580,085,178,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694071.63/warc/CC-MAIN-20200126230255-20200127020255-00139.warc.gz | 358,580,423 | 31,446 | # A Couple of Hearts Probabilities
Here's what I don't get. I've played a lot of Hearts for many years - thousands of games I'm sure. The number of times I've played a low heart, and the trick has gone 2H, 3H, 4H, 5H seems way out of probable reality. But the number of times I've seen all four players play the any of the same value card (e.g. 6C, 6D, 6H, 6S) is once. Why is the probability of 2H, 3H, 4H, 5H any lower than any 4 same value cards? In my naiveté it seems that in both cases, the probability calls for 4 players to have a specific card at the same time. Can anyone explain it to me like a 4 year old?
• xkcd.com/1364 – Tim Lymington Jun 30 '14 at 22:23
• The fact that you see 2H, 3H, 4H, 5H seems surprising. If the player stuck with the 5 is trying not to accumulate hearts, I'd expect that player to slap down the largest heart card he has. If he's trying to shoot the moon, then the 5 makes sense. – Ellesedil Jul 1 '14 at 0:42
• Sorry for not being clearer - the cards most times do not appear in that order but more likely 5H, 4H, 3H, 2H or some other combination started by the 4H or 5H. Thanks for your comments – Ricksx Jul 8 '14 at 16:00
• I'm pretty sure most 4 year olds don't understand probability either. – bwarner Oct 14 '14 at 12:46 | 398 | 1,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-05 | latest | en | 0.960811 |
https://us.metamath.org/nfegif/ssetpov.html | 1,702,275,968,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00541.warc.gz | 661,834,622 | 6,288 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > ssetpov Unicode version
Theorem ssetpov 5944
Description: The subset relationship partially orders the universe. (Contributed by SF, 12-Mar-2015.)
Assertion
Ref Expression
ssetpov S Po
Proof of Theorem ssetpov
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 ssetex 4744 . . . 4 S
21a1i 10 . . 3 S
3 vvex 4109 . . . 4
43a1i 10 . . 3
5 ssid 3290 . . . . 5
6 vex 2862 . . . . . 6
76, 6brsset 4758 . . . . 5 S
85, 7mpbir 200 . . . 4 S
98a1i 10 . . 3 S
10 sstr 3280 . . . . 5
11 vex 2862 . . . . . . 7
126, 11brsset 4758 . . . . . 6 S
13 vex 2862 . . . . . . 7
1411, 13brsset 4758 . . . . . 6 S
1512, 14anbi12i 678 . . . . 5 S S
166, 13brsset 4758 . . . . 5 S
1710, 15, 163imtr4i 257 . . . 4 S S S
18173ad2ant3 978 . . 3 S S S
19 eqss 3287 . . . . . 6
2011, 6brsset 4758 . . . . . . 7 S
2112, 20anbi12i 678 . . . . . 6 S S
2219, 21bitr4i 243 . . . . 5 S S
2322biimpri 197 . . . 4 S S
24233ad2ant3 978 . . 3 S S
252, 4, 9, 18, 24pod 5936 . 2 S Po
2625trud 1323 1 S Po
Colors of variables: wff setvar class Syntax hints: wa 358 w3a 934 wtru 1316 wcel 1710 cvv 2859 wss 3257 class class class wbr 4639 S csset 4719 Po cpartial 5891 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-13 1712 ax-14 1714 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 ax-nin 4078 ax-xp 4079 ax-cnv 4080 ax-1c 4081 ax-sset 4082 ax-si 4083 ax-ins2 4084 ax-ins3 4085 ax-typlower 4086 ax-sn 4087 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-nan 1288 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-eu 2208 df-mo 2209 df-clab 2340 df-cleq 2346 df-clel 2349 df-nfc 2478 df-ne 2518 df-ral 2619 df-rex 2620 df-reu 2621 df-rmo 2622 df-rab 2623 df-v 2861 df-sbc 3047 df-nin 3211 df-compl 3212 df-in 3213 df-un 3214 df-dif 3215 df-symdif 3216 df-ss 3259 df-pss 3261 df-nul 3551 df-if 3663 df-pw 3724 df-sn 3741 df-pr 3742 df-uni 3892 df-int 3927 df-opk 4058 df-1c 4136 df-pw1 4137 df-uni1 4138 df-xpk 4185 df-cnvk 4186 df-ins2k 4187 df-ins3k 4188 df-imak 4189 df-cok 4190 df-p6 4191 df-sik 4192 df-ssetk 4193 df-imagek 4194 df-idk 4195 df-iota 4339 df-0c 4377 df-addc 4378 df-nnc 4379 df-fin 4380 df-lefin 4440 df-ltfin 4441 df-ncfin 4442 df-tfin 4443 df-evenfin 4444 df-oddfin 4445 df-sfin 4446 df-spfin 4447 df-phi 4565 df-op 4566 df-proj1 4567 df-proj2 4568 df-opab 4623 df-br 4640 df-sset 4725 df-trans 5899 df-ref 5900 df-antisym 5901 df-partial 5902 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 1,467 | 2,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.11784 |
https://yourquickadvice.com/what-is-included-in-total-project-cost/ | 1,709,578,485,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00342.warc.gz | 1,093,518,995 | 14,914 | # What is included in total project cost?
## What is included in total project cost?
(1) The term “total project cost” in these Guidelines means all costs and expenses required for a construction project, which shall consist of construction cost, compensation cost, and incidental expenses for facilities.
What are the 3 main methods of cost estimating?
Methods of Cost Estimation in Projects – Tools and Techniques
• Expert Judgement Method.
• Analogous Estimating Method.
• Parametric Estimating Method.
• Bottom-up Estimating Method.
• Three-Point Estimating Method.
• Data Analysis Method.
• Project Management Information System Method.
• Decision-Making Method.
What are the components of project cost?
Project Costing: Fundamental Components of Project Cost
• Step 1: Resource planning. Resource planning is the process of ascertaining future resource requirements for an organization or a scope of work.
• Step 2: Cost estimating.
• Step 3: Cost budgeting.
• Step 4: Cost control.
### What are the 3 budget section of the overall project plan?
There are three types of budget estimates that occur during the project cycle, these estimates – rough order of estimate, contract and definitive, vary primarily on when they are done, how they are used and how accurate they are.
What is the cost of project?
Project Cost is the total funds needed to complete the project or work that consists of a Direct Cost and Indirect Cost. The Project Costs are any expenditures made or estimated to be made, or monetary obligations incurred or estimated to be incurred to complete the project which are listed in a project baseline.
What is total cost of a project?
Total Project Costs means all costs which have been or are estimated to be incurred by the Company with respect to the acquisition, design, development, construction, debt financing, leasing, and completion of a Project, which Total Project Costs (including without limitation tenant allowances) are initially estimated …
## How do you calculate cost?
The most common way to estimate costs is to make a list of items you need and add up their costs. Make sure you include all applicable costs, such as equipment and parts, materials and supplies, labor, financing, fees and licensing, transportation, and acquisition costs for land or facilities.
What is the formula to calculate cost?
The equation for the cost function is C = \$40,000 + \$0.3 Q, where C is the total cost. Note we are measuring economic cost, not accounting cost.
What are the 4 types of cost?
Direct Costs.
• Indirect Costs.
• Fixed Costs.
• Variable Costs.
• Operating Costs.
• Opportunity Costs.
• Sunk Costs.
• Controllable Costs.
• ### What is the cost of a project?
What is a budget for a project?
The Project Budget is a tool used by project managers to estimate the total cost of a project. A project budget template includes a detailed estimate of all costs that are likely to be incurred before the project is completed.
How do you cost a project?
5 Steps to Accurate Project Costing
1. Understand the scope of the work. First, you need to understand what it is the project is going to deliver.
2. Estimate the work. Next, estimate the work.
3. Include all other costs. This is the step many project managers miss out. | 678 | 3,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.910136 |
https://optimization.mccormick.northwestern.edu/index.php?title=Mixed-integer_cuts&diff=next&oldid=2582 | 1,653,469,376,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00030.warc.gz | 503,067,993 | 7,091 | # Difference between revisions of "Mixed-integer cuts"
Author: Tahir Kapoor (ChE 345 Spring 2015) Stewards: Dajun Yue and Fengqi You
# Introduction
Mixed-integer cuts or Cutting-plane method is an iterative approach used to simplify the solution of a mixed integer linear programming (MILP) problem. Cutting-plane methods work by first relaxing the MILP to a complementary linear programming problem and cutting the feasible region to narrow down the solution search space to only include feasible solutions. Unlike the branch and bound method, which subdivides the feasible region into multiple sub-regions, mixed-integer cuts result in one feasible region which can be solved using standard linear programming methods. Proper application of mixed-integer cuts will result in a convex hull reformulation of a MILP where every extreme point of the feasible region is an integer point. In practice, branch and bound methods are typically more efficient than cutting-plane methods, but cutting-plane methods was the first method proven that a MILP solution could be found in a finite number of steps.
# Gomory's Cuts
Cutting-plane methods were first developed by Ralph Gomory in the 1950s and Gomory Cuts remain among the basis of cutting-plane methods.
The Gomory Cut method of the above MILP problem is a multi-step process using the following steps:
1. Relax the MILP problem to it's complementary LP problem by dropping the requirement that xi must be integers.
2. Solve the linear programming problem to obtain a basic feasible solution.
3. If this vertex is not an integer point then apply a cut to find a hyperplane with the vertex on one side and all feasible integer points on the other and add it to the constraints
4. Repeat until a feasible integer solution is found
Using the simplex method to solve a linear program produces a set of equations of the form:
$x_i+\sum \bar a_{i,j}x_j=\bar b_i$
where xi is a basic variable and the xj's are the nonbasic variables. Rewrite this equation so that the integer parts are on the left side and the fractional parts are on the right side:
$x_i+\sum \lfloor \bar a_{i,j} \rfloor x_j - \lfloor \bar b_i \rfloor = \bar b_i - \lfloor \bar b_i \rfloor - \sum ( \bar a_{i,j} -\lfloor \bar a_{i,j} \rfloor) x_j.$
For any integer point in the feasible region the right side of this equation is less than 1 and the left side is an integer, therefore the common value must be less than or equal to 0. So the inequality
$\bar b_i - \lfloor \bar b_i \rfloor - \sum ( \bar a_{i,j} -\lfloor \bar a_{i,j} \rfloor) x_j \le 0$
must hold for any integer point in the feasible region. Furthermore, nonbasic variables are equal to 0s in any basic solution and if xi is not an integer for the basic solution x,
$\bar b_i - \lfloor \bar b_i \rfloor - \sum ( \bar a_{i,j} -\lfloor \bar a_{i,j} \rfloor) x_j = \bar b_i - \lfloor \bar b_i \rfloor > 0.$
So the inequality above excludes the basic feasible solution and thus is a cut with the desired properties. Introducing a new slack variable xk for this inequality, a new constraint is added to the linear program, namely
$x_k + \sum (\lfloor \bar a_{i,j} \rfloor - \bar a_{i,j}) x_j = \lfloor \bar b_i \rfloor - \bar b_i,\, x_k \ge 0,\, x_k \mbox{ an integer}.$ | 823 | 3,259 | {"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": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.897999 |
https://number.academy/8002360 | 1,679,439,628,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00251.warc.gz | 497,420,837 | 12,484 | # Number 8002360
Number 8,002,360 spell 🔊, write in words: eight million, two thousand, three hundred and sixty , 8.0 million. Ordinal number 8002360th is said 🔊 and write: eight million, two thousand, three hundred and sixtieth. The meaning of the number 8002360 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 8002360. What is 8002360 in computer science, numerology, codes and images, writing and naming in other languages
## What is 8,002,360 in other units
The decimal (Arabic) number 8002360 converted to a Roman number is (M)(M)(M)(M)(M)(M)(M)(M)MMCCCLX. Roman and decimal number conversions.
#### Weight conversion
8002360 kilograms (kg) = 17642002.9 pounds (lbs)
8002360 pounds (lbs) = 3629846.7 kilograms (kg)
#### Length conversion
8002360 kilometers (km) equals to 4972435 miles (mi).
8002360 miles (mi) equals to 12878555 kilometers (km).
8002360 meters (m) equals to 26254143 feet (ft).
8002360 feet (ft) equals 2439149 meters (m).
8002360 centimeters (cm) equals to 3150535.4 inches (in).
8002360 inches (in) equals to 20325994.4 centimeters (cm).
#### Temperature conversion
8002360° Fahrenheit (°F) equals to 4445737.8° Celsius (°C)
8002360° Celsius (°C) equals to 14404280° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
8002360 seconds equals to 3 months, 1 week, 1 day, 14 hours, 52 minutes, 40 seconds
8002360 minutes equals to 1 decade, 6 years, 6 months, 1 week, 6 days, 4 hours, 40 minutes
### Codes and images of the number 8002360
Number 8002360 morse code: ---.. ----- ----- ..--- ...-- -.... -----
Sign language for number 8002360:
Number 8002360 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 8002360
### Multiplications
#### Multiplication table of 8002360
8002360 multiplied by two equals 16004720 (8002360 x 2 = 16004720).
8002360 multiplied by three equals 24007080 (8002360 x 3 = 24007080).
8002360 multiplied by four equals 32009440 (8002360 x 4 = 32009440).
8002360 multiplied by five equals 40011800 (8002360 x 5 = 40011800).
8002360 multiplied by six equals 48014160 (8002360 x 6 = 48014160).
8002360 multiplied by seven equals 56016520 (8002360 x 7 = 56016520).
8002360 multiplied by eight equals 64018880 (8002360 x 8 = 64018880).
8002360 multiplied by nine equals 72021240 (8002360 x 9 = 72021240).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 8002360
Half of 8002360 is 4001180 (8002360 / 2 = 4001180).
One third of 8002360 is 2667453,3333 (8002360 / 3 = 2667453,3333 = 2667453 1/3).
One quarter of 8002360 is 2000590 (8002360 / 4 = 2000590).
One fifth of 8002360 is 1600472 (8002360 / 5 = 1600472).
One sixth of 8002360 is 1333726,6667 (8002360 / 6 = 1333726,6667 = 1333726 2/3).
One seventh of 8002360 is 1143194,2857 (8002360 / 7 = 1143194,2857 = 1143194 2/7).
One eighth of 8002360 is 1000295 (8002360 / 8 = 1000295).
One ninth of 8002360 is 889151,1111 (8002360 / 9 = 889151,1111 = 889151 1/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
8002360
#### Is Prime?
The number 8002360 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 8002360 are 2 * 2 * 2 * 5 * 37 * 5407
The factors of 8002360 are
1 , 2 , 4 , 5 , 8 , 10 , 20 , 37 , 40 , 74 , 148 , 185 , 296 , 370 , 740 , 1480 , 5407 , 10814 , 21628 , 27035 , 8002360 show more factors ...
Total factors 32.
Sum of factors 18495360 (10493000).
#### Powers
The second power of 80023602 is 64.037.765.569.600.
The third power of 80023603 is 512.453.253.683.544.260.608.
#### Roots
The square root √8002360 is 2828,844287.
The cube root of 38002360 is 200,019665.
#### Logarithms
The natural logarithm of No. ln 8002360 = loge 8002360 = 15,895247.
The logarithm to base 10 of No. log10 8002360 = 6,903218.
The Napierian logarithm of No. log1/e 8002360 = -15,895247.
### Trigonometric functions
The cosine of 8002360 is 0,585746.
The sine of 8002360 is 0,810495.
The tangent of 8002360 is 1,383697.
### Properties of the number 8002360
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 8002360 in Computer Science
Code typeCode value
8002360 Number of bytes7.6MB
Unix timeUnix time 8002360 is equal to Friday April 3, 1970, 2:52:40 p.m. GMT
IPv4, IPv6Number 8002360 internet address in dotted format v4 0.122.27.56, v6 ::7a:1b38
8002360 Decimal = 11110100001101100111000 Binary
8002360 Decimal = 120001120011201 Ternary
8002360 Decimal = 36415470 Octal
8002360 Decimal = 7A1B38 Hexadecimal (0x7a1b38 hex)
8002360 BASE64ODAwMjM2MA==
8002360 MD5f183765643d583670929becf46b8fc8b
8002360 SHA1aec33bab5aa0b9baecec4f6af8b1d5cd8c3ce70f
8002360 SHA256b8b485d21c6944d7edc2f9cca78b05db50b7b78922cc2dcb0be9ba93fd8e241b
8002360 SHA384340c502c4032b56e6035b44ccaf3b01e76326b4a870f51e4fd1be1ce77e197d3d1a240e323387ac2cf98d249393159b9
More SHA codes related to the number 8002360 ...
If you know something interesting about the 8002360 number that you did not find on this page, do not hesitate to write us here.
## Numerology 8002360
### Character frequency in the number 8002360
Character (importance) frequency for numerology.
Character: Frequency: 8 1 0 3 2 1 3 1 6 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 8002360, the numbers 8+0+0+2+3+6+0 = 1+9 = 1+0 = 1 are added and the meaning of the number 1 is sought.
## № 8,002,360 in other languages
How to say or write the number eight million, two thousand, three hundred and sixty in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 8.002.360) ocho millones dos mil trescientos sesenta German: 🔊 (Nummer 8.002.360) acht Millionen zweitausenddreihundertsechzig French: 🔊 (nombre 8 002 360) huit millions deux mille trois cent soixante Portuguese: 🔊 (número 8 002 360) oito milhões e dois mil, trezentos e sessenta Hindi: 🔊 (संख्या 8 002 360) अस्सी लाख, दो हज़ार, तीन सौ, साठ Chinese: 🔊 (数 8 002 360) 八百万二千三百六十 Arabian: 🔊 (عدد 8,002,360) ثمانية ملايين و ألفان و ثلاثمائةستون Czech: 🔊 (číslo 8 002 360) osm milionů dva tisíce třista šedesát Korean: 🔊 (번호 8,002,360) 팔백만 이천삼백육십 Danish: 🔊 (nummer 8 002 360) otte millioner totusinde og trehundrede og treds Dutch: 🔊 (nummer 8 002 360) acht miljoen tweeduizenddriehonderdzestig Japanese: 🔊 (数 8,002,360) 八百万二千三百六十 Indonesian: 🔊 (jumlah 8.002.360) delapan juta dua ribu tiga ratus enam puluh Italian: 🔊 (numero 8 002 360) otto milioni e duemilatrecentosessanta Norwegian: 🔊 (nummer 8 002 360) åtte million, to tusen, tre hundre og seksti Polish: 🔊 (liczba 8 002 360) osiem milionów dwa tysiące trzysta sześćdziesiąt Russian: 🔊 (номер 8 002 360) восемь миллионов две тысячи триста шестьдесят Turkish: 🔊 (numara 8,002,360) sekizmilyonikibinüçyüzaltmış Thai: 🔊 (จำนวน 8 002 360) แปดล้านสองพันสามร้อยหกสิบ Ukrainian: 🔊 (номер 8 002 360) вiсiм мiльйонiв двi тисячi триста шiстдесят Vietnamese: 🔊 (con số 8.002.360) tám triệu hai nghìn ba trăm sáu mươi Other languages ...
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There are no comments for this topic. | 2,784 | 7,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.686989 |
https://community.anaplan.com/t5/Anaplan-Platform/Working-Days-between-dates/m-p/35355 | 1,611,185,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00155.warc.gz | 281,049,959 | 30,388 | Contributor
## Working Days between dates
I'm looking to try and find away to calulate working days between two dates in a module; I know this has been addressed in a previous post, but it's from 2014 and I can't seem to access the supporting docs. that really show me what I need.
I'm a new model builder so any assitance would be appriciated, and apologies if this has been covered elsewhere and I've not found it.
5 REPLIES 5
Certified Master Anaplanner
## Re: Working Days between dates
Hi Ceri,
This requirement can be acheived in the following manner:
1) Create a module(say, "Time Mappings") to maintain the working day mappings with time scale set as "Day", as shown in below screenshot:
a) Working Day Check (Line item 1): A boolean formatted line item
b) Cumulative Working Days (Line item 2): A number formatted line item with formula "CUMULATE(IF Working Day Check THEN 1 ELSE 0)"
The first line should be maintained manually and it will be a yearly activity done by an admin (Working days as to be checked manually, as shown in screenshot below)
2) An example module to hold your start and end dates and to calculate number of working days:
Formula for calculating number of working days = "Time Mappings.Cumulative Working Days[LOOKUP: Date Input.End Date] - Time Mappings.Cumulative Working Days[LOOKUP: Date Input.Start Date] + 1"
Based on the inputs given in Step 1 and Step 2, you can see the number of working days as "5" in Jan 17 column.
Hope this example will solve your query. Let me know, if you need further explanation on this.
Regards,
CA Pavan Kumar
Contributor
## Re: Working Days between dates
That's great, thanks for your help.
Certified Master Anaplanner
## Re: Working Days between dates
Cheers!
Pavan
Occasional Contributor
## Re: Working Days between dates
https://help.anaplan.com/anapedia/Content/Calculation_Functions/Function%20Usage/Excel%20Equivalent%...
that will help translate from the excel NETWORKDAYS function to Anaplan. Hope this helpg
Certified Master Anaplanner
## Re: Working Days between dates
This is something my team has been trying to solve for as we needed to calculate the # of working days in each month for personnel planning. The provided solution is workable but had a bit more manual admin work than would be ideal for our situation. I was able take the outline provided and leverage the WEEKDAY() formula to automate the determination of which dates to include/exclude from working days.
See below for breakdown of my solution:
1. Create Date Input Module
1. Create 2 date formatted lines for start and end date
2. Create number formatted line item for # of cumulative Working Days as of the end date (this will reference the module created in the next step) Formula: Time Mappings.Cumulated Working Days[LOOKUP: End Date]
1. Create Time Mappings module with a timescale set to Day
1. Create a number formatted line item to pull in the day of the week for that specific date. Formula: WEEKDAY(START())
2. Create a number formatted line item to determine the number of Working Days. The formula will define which days of the week will be your non-working day (in the US that is generally Saturday (day 6) and Sunday (day 7). Formula: IF Day of Week = 6 OR Day of Week = 7 THEN 0 ELSE 1
3. Create Boolean formatted line item to determine which dates have working days between the two specified dates. Formula: START() >= Date Input.Start Date AND END() <= Date Input.End Date AND Working Day > 0
4. Create number formatted line it to calculate the cumulative number of working days where the Boolean is marked true. Formula: CUMULATE(IF Include in Cumulated? THEN 1 ELSE 0)
This solution allowed us to systematically calculate workday for all of our planning periods so wanted to share should it be helpful to anyone else in the future. | 897 | 3,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | longest | en | 0.895552 |
http://www.mywordsolution.com/question/the-rate-of-inflation-in-year-1-is-expected-to-be-14-year/93110254 | 1,696,186,676,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00409.warc.gz | 68,590,181 | 7,736 | +61-413 786 465
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## Statistics
The rate of inflation in year 1 is expected to be 1.4%, year two is 1.8%, and years three through five is expected to be 2%. Assume the real risk-free rate, r*, is 3% for all maturities. What should the yield to maturity on risk-free bonds that mature in (a) one year, (b) 2 years, (c) 5 years?
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M93110254
• Price:- \$10
Priced at Now at \$10, Verified Solution
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http://functionx.com/bcb/math/MinValue.htm | 1,369,329,291,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703635016/warc/CC-MAIN-20130516112715-00006-ip-10-60-113-184.ec2.internal.warc.gz | 116,535,128 | 2,319 | Statistics Functions: The Minimum Value of a Series
`double __fastcall MinValue(const double * Data, const int Data_Size);`
The MinValue() function gets a numeric value that represents the minimum value of the items of an array. This function takes two arguments. The first argument, Data, represents an array of integers or double-precision numbers. The second argument is the number-1 of the items of the array; for example, if the considered array has 4 members, the Data_Size argument would be 3. To use the MinValue() function, declare an array that involves the necessary numbers. You can initialize such a variable or request the values from the user. To calculate the minimum value of a range, supply the array and its size. If you do not know the dimension of the array, you can use the sizeof operator to find it out. Here is an example that uses the MinValue() function:
```//---------------------------------------------------------------------------
void __fastcall TForm1::Button1Click(TObject *Sender)
{
double Values[] = { 12.55, 10.15, 980.22, 50.50, 280.12 };
int Size = (sizeof(Values)/sizeof(double)) - 1;
double Minimum = MinValue(Values,Size);
Edit1->Text = Minimum;
}
//---------------------------------------------------------------------------``` | 272 | 1,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.572337 |
https://www.eng-tips.com/viewthread.cfm?qid=30500 | 1,696,105,847,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00157.warc.gz | 821,142,565 | 16,641 | ×
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# Min & Max velocity in pipes 5
## Min & Max velocity in pipes
(OP)
Hello
Do we have a maximum velocity for liquids in pipes,for preventing noise or errosion?
Do we have a minimum velocity for liquids in pipes? In relation of which problem?
Thanks for all replies
### RE: Min & Max velocity in pipes
Maximum velocity is a frequent topic on these forums. There must be some widely varying assumptions to get the range of numbers that I've seen here.
For "dry gas" flow I use an actual velocity (as opposed to "velocity" calculated as SCF/FlowArea) of 100 ft/second or 15 psi/mile pressure drop as my upper design condition. Design capacity will be the lower of the SCF/day you get from 100 ft/sec or the SCF/day you get from 15 psi/mile pressure drop. That choice stays way below errosional velocities, won't slam a slug of water into plant-inlet pipeworks, and gives me a makeup compression Hp requirement I can live with.
All of the vertical flow correlations show a critical velocity (for keeping water mobile) around 36 ft/sec. I figure that if it works in vertical flow, it should be quite conservative in horizontal flow. I've been using 36 ft/s as a minimum design condition for about 7 years now and see considerably less liquid accumulation in the piping I designed than I see in the piping I inherited.
Hope this helps
David
### RE: Min & Max velocity in pipes
2
There is a good thread on maximum velocity in this room with the title "what is the maximum velocity a pipe can take", posted on 8/20. A good rule of thumb that I have been using is the manufacturers warranty on valves. When I used certain valves on water service, it turned out that they were rated for no more than 11 fps at which point you were on your own with no support from the factory. Thats not to say you cant move water quicker, but why would you want to, the energy cost of doing this would be far greater that the cost of piping or equipment in the long run I think...
The minimum velocity I design around is that which will provide reynolds numbers in the sufficiently enough in the turbulent range that I would not find Re's in the transition or worst yet, laminar flow regeims when the system is working.
I hope this helped..
Bob
### RE: Min & Max velocity in pipes
I will first answer regarding the minimum flow velocity in a pipe.The minimum velocity is generally dictated by economic considerations; if you take a lesser velocity you end up with a bigger pipe size.The other considerations would be fouling characteristics of the fluid.
Regarding maximum velocity for erosion would depend on type of liquid and duty- short duration operation or continuous operation.Obviously, if it is a short duration/cycle operation, a higher velocity could be used. For pump suction lines and liquid near its vaporization point, maximum velocities are limited.For noise and especially for control valve applications, talk to the control valve suppliers fot the maximum recommended velocity.
Hope this helps.
Arun
### RE: Min & Max velocity in pipes
One thing to keep in mind is the effect of velocity of check valves, if applicable. To maintain a check valve fully open, without wear, a minimum velocity is required based on the specific valve type. Swing checks are the most susceptible to damage if the disc is constantly oscillating in the flow stream and require 12-18 feet per second to fully open. Nozzle Checks fully open as low as 3 feet per second.
### RE: Min & Max velocity in pipes
pumpvlvguy point is very important. Crane's TP #410 gives a formula for each tipe of checkvalve... the symbol U/L in crane's paper means : Underwriter's Laboratory (not defined in the Nomenclature page).
Saludos.
a.
### RE: Min & Max velocity in pipes
2
The 1997 ASHRAE Fundamentals Handbook page 33.3 table gives following information:
Max water velocity to minimise erosion
Normal operation Velocity
hours per year m/s
1500 4.6
2000 4.4
3000 4.0
4000 3.7
6000 3.0
Same page also says "velocities on the order of 3 to 5m/s lie within the range of allowable noise levels for residential and commercial builings"
### RE: Min & Max velocity in pipes
You can check the following threads for more valuable comments. I have made the same search recently..
Try advanced search/check "exact phrase" and write the below subjects, the threads will come up..Good luck
FORUM: Piping & fluid mechanics engineering
SUBJECT: Velocity limitation in pipe
FORUM: Chemical plant design & operations
SUBJECT: Maximum velocity of fluid in pipe
### RE: Min & Max velocity in pipes
A possible reason for a minimum flow not mentioned so far is that you may want to keep horizontal pipes full to to minimize pipewall corrosion or to preclude a stratified flow condition that would allow entrained air or gases to collect in the water void space and perhaps aggravate waterhammer effects on quick closures of valves. I found an old equation that might be used to estimate the flowrate to keep a horizontal pipe full. It's in a Design News reprint #RS528 from the June 12, 1963 issue of that magazine and is titled "Fluid Flow from Partially Filled Pipes" by D.P.Costa. It is:
q=7.3d^2.56*K^1.84 where K is the "fullness" fraction from 0.2 to 0.6, d is inner pipe diameter, in inches limited to 2 to 6 inch pipe and q is GPM. The equation came from a 1928 Purdue Univ. bulletin. I tried the equation on a much bigger pipe of 11.7 inch ID and ran the fraction up to 1.0. It didn't blow up and gave rather believable results for full flow. Has anybody got something better to calculate minimum flows for full horizontal pipes?
### RE: Min & Max velocity in pipes
To Pumpvlvguy and Abeltio,
Regarding swing check valve pivot hardware degradation from too low flowrates to backseat, there's an obscure report that deals with the influence of upstream pipebends on valve disk dynamics. It's Kalsi etal (1988), "Prediction of Check Valve Performance and Degradation in Nuclear Power Plant Systems", Kalsi Eng. Rpt. #1559 (NUREG/CR-5159). Among the findings:
"At proximities of 3 to 5 pipe diameters, upstraem elbows require an increase in flow velocity over baseline of 10 or 15% to fully open the valve disk. Proximities of 0 to 1 diameter will require up to 50% higher flow velocity over baseline to fully open the disk, except for clearway designs which will require velocities more than 100% higher than baseline [clearway disks lift nearly totally out of the flowstream and may never achieve stability under some flow conditions]."
"The amplitude of disk motion which occurs before fully seating the disk increases as the flow disturbance is brought closer to the valve. The maximum disk fluctuations in the case of a severe turbulence source can reach as high as 16 degrees and, for elbows, up to 9 degrees. The reducers have negligible effect on disk fluctuations."
"The onset of tapping begins at lower flow velocities when the elbow is oriented up than when oriented down. This results in wider tapping zones for elbow-up installations compared to elbow-down. This is particularly evident at 0d and 1d and less so at 3d and 5d (d is upstream pipe diameter distance)."
Kalsi etal appear to be talking about single upstream bends. Multiple, closely spaced bends upstream of swing check valves could have considerably worse consequences.
### RE: Min & Max velocity in pipes
(OP)
Hi Vanstoja,
Your formula is very useful,But "d" has dropped !!
And also give a small example with your formula.
Thanks
P.S. for full k=1 ?
### RE: Min & Max velocity in pipes
I don't understand what you mean by "d has dropped". Full is K=1.0. I ran 2, 6 and 11.688 inch diameters in an EXCEL spreadsheet and got the following results:
ID(in.) K Q(GPM)
2 0.2 2.23
2 0.4 7.98
2 0.6 16.82
6 0.2 37.09
6 0.4 132.79
6 0.6 280.02
11.688 0.2 204.47
11.688 0.4 723.03
11.688 0.6 1543.60
11.688 0.8 2620.74
11.688 1.0 3951.28
The increased flows for 11.688 pipe at K=0.8 and 1.0 are part of a progression of step increases of approximately 500, 800, 1100 and 1300 GPM for 0.2K steps. The 0.8 and 1.0 K flowrates are "reasonable" though I can't say for sure that they are correct. The test results were apparently only run up to 0.6 and the author was reluctant to extraapolate any further.
### RE: Min & Max velocity in pipes
There is an ASTM Standard on maximum velocity of hydrocarbons in pipelines, sorry i don't recall the number but a search on their web site will find it. The maximum velocity here is predicted by static electricity build up.
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Register now while it's still free! | 2,547 | 10,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.895517 |
https://www.experts-exchange.com/questions/28483499/Finding-the-second-largest-number-in-an-array.html | 1,531,842,655,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589752.56/warc/CC-MAIN-20180717144908-20180717164908-00262.warc.gz | 883,795,135 | 23,181 | # Finding the second largest number in an array
Hi experts,
Here is a quick question in excel.
We have a numerical array in A1:A100 and want to find out the second largest number. Is there a simple formula to do that?
Assume that we have found out the largest number and put it in B1 = Max(a1:a100), what can we do to use B1 too?
Thanks,
RDB
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I think I would find the largest, put it and its index somewhere, then make a(index)=0 and run max again. The result will be the second largest. I am assuming all positive numbers for this exercise.
0
Commented:
You find the second largest number using this formula
=LARGE(A1:A100,2)
0
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Have a better answer? Share it in a comment. | 409 | 1,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-30 | latest | en | 0.934956 |
https://brainmass.com/business/accounting/relevant-costing-81515 | 1,596,479,513,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735823.29/warc/CC-MAIN-20200803170210-20200803200210-00493.warc.gz | 241,303,411 | 11,304 | Explore BrainMass
# Relevant costing
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The Minnetonka Corporation, which produces and sells to wholesalers a highly successful line of water skis, has decided to diversify to stabilize sales throughout the year. The company is considering the production of cross-country skis.
After considerable research, a cross-country ski line has been developed. Because of the conservative nature of the company management, however, Minnetonka's president has decided to introduce only one type of the new skis for this coming winter. If the product is a success, further expansion in future years will be initiated.
The ski selected is a mass-market ski with a special binding. It will be sold to wholesalers for \$80 per pair. Because of availability capacity, no additional fixed charges will be incurred to produce the skis. A \$100,000 fixed charge will be absorbed by the skis, however, to allocate a fair share of the company's present fixed costs to the new product.
Using the estimated sales and production of 10,000 pair of skis as the expected volume, the accounting department has developed the following cost per pair of skis and bindings:
Direct Labor: \$35
Direct Material: \$30
Total Overhead: \$15
Total: \$80
Minnetonka has approached a subcontractor to discuss the possibility of purchasing the bindings. The purchase price of the bindings from the subcontractor would be \$5.25 per binding, or \$10.50 per pair. If the Minnetonka Corporation accepts the purchase proposal, it is predicted that direct-labor and variable-overhead costs would be reduced by 10% and direct-material costs would be reduced by 20%.
© BrainMass Inc. brainmass.com June 3, 2020, 6:58 pm ad1c9bdddf
https://brainmass.com/business/accounting/relevant-costing-81515
#### Solution Summary
Excel file contains solution of relevant costing problem.
\$2.19 | 428 | 1,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-34 | latest | en | 0.930882 |
http://www.lmfdb.org/ModularForm/GL2/TotallyReal/2.2.109.1/holomorphic/2.2.109.1-9.1-c | 1,571,875,400,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00456.warc.gz | 281,890,374 | 5,536 | # Properties
Base field $$\Q(\sqrt{109})$$ Weight [2, 2] Level norm 9 Level $[9, 3, 3]$ Label 2.2.109.1-9.1-c Dimension 2 CM no Base change no
# Related objects
• L-function not available
## Base field $$\Q(\sqrt{109})$$
Generator $$w$$, with minimal polynomial $$x^{2} - x - 27$$; narrow class number $$1$$ and class number $$1$$.
## Form
Weight [2, 2] Level $[9, 3, 3]$ Label 2.2.109.1-9.1-c Dimension 2 Is CM no Is base change no Parent newspace dimension 13
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{2}$$ $$\mathstrut +\mathstrut 2x$$ $$\mathstrut -\mathstrut 1$$
Norm Prime Eigenvalue
3 $[3, 3, -w + 6]$ $-1$
3 $[3, 3, w + 5]$ $\phantom{-}1$
4 $[4, 2, 2]$ $\phantom{-}e$
5 $[5, 5, -3w + 17]$ $\phantom{-}2e + 1$
5 $[5, 5, -3w - 14]$ $-1$
7 $[7, 7, w - 5]$ $\phantom{-}e + 1$
7 $[7, 7, w + 4]$ $-3e - 3$
29 $[29, 29, -w - 7]$ $-e - 6$
29 $[29, 29, -w + 8]$ $-3e - 4$
31 $[31, 31, -5w + 28]$ $-2e - 8$
31 $[31, 31, -5w - 23]$ $-2e$
43 $[43, 43, 6w + 29]$ $-e + 3$
43 $[43, 43, -6w + 35]$ $-5e - 1$
61 $[61, 61, 3w - 19]$ $\phantom{-}2e - 9$
61 $[61, 61, -3w - 16]$ $\phantom{-}2e + 7$
71 $[71, 71, -7w - 34]$ $\phantom{-}7e + 7$
71 $[71, 71, 7w - 41]$ $\phantom{-}3e + 7$
73 $[73, 73, 2w - 7]$ $-8e - 11$
73 $[73, 73, -2w - 5]$ $\phantom{-}4e + 9$
83 $[83, 83, -w - 10]$ $-e - 11$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
3 $[3, 3, -w + 6]$ $1$
3 $[3, 3, w + 5]$ $-1$ | 746 | 1,509 | {"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} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.222211 |
https://www.debateart.com/forum/topics/2103-the-bible-and-math-is-awesome | 1,726,194,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00723.warc.gz | 667,056,621 | 17,277 | # The Bible and Math is awesome
Author: Dr.Franklin
## Posts
Total: 53
Debates: 32
Posts: 10,666
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The Bible is so good and intelligent,it has to prove that there is a God, and Mathematics prove it
"The words of the LORD are pure words: as silver tried in a furnace of earth, purified seven times."PSALM 12:
The number 7 and multiples thereof play a significant role in biology (the gestation periods of mammals, the incubation periods of birds, and the development of insects can all be measured in multiples of seven days); in chemistry (seven periods in the periodic table of elements); in music (the seven steps of the octave); in our calendar (seven days in the week); and in light (the seven colours of the rainbow). After these discoveries were made, those scientists claimed that they were merely finding out the physical laws and patterns which our Creator had already designed (ROMANS 1:20).
"In the beginning God created the heaven and the earth." GENESIS 1:1
The Hebrew sentence consists of exactly 7 words, which have exactly 28 (4x7) letters. There are 3 nouns: God, heaven and earth. If we add together the numerical value of each of the letters in these three Hebrew nouns, we get exactly 777 (111x7). The numerical value of the Hebrew word "created" is 203 (29x7). The first three words contain the subject, with exactly 14 (2x7) letters, and the four remaining words contain the object with also exactly 14 letters. The Hebrew words for heaven and earth have 7 letters each. The value of the first, the two middle and the last letters in the sentence is 133 (19x7). The total value of the first and last letter of every word in the verse is 1,393 (199x7). The value of the first and last letter of the first and last word in this verse is 497 (71x7). The value of the first and last letter of each word in-between is 896 (128x7).
Trippy,right
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@Dr.Franklin
"In the beginning God created the heaven and the earth."
Fairy tale.
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@disgusted
you completely missed the point
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@Dr.Franklin
ahhhhhh nope, the whole point is that the silly book of fiction is meaningless.
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@disgusted
The point was that the Bible and mathematics support a very holy and brilliant book
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@Dr.Franklin
The point was that the Bible and mathematics support a very holy and brilliant book
The bible does not meet the math and science requirement of a 4th grader.
Christians struggle to explain what is the lengeth of a day in Genesis. Is it a 24 hour day, is a 1000 years equal to a day or a billion years equal to a day?
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@Harikrish
I don't like you, what you do on DDO is sick.
The bible does not meet the math and science requirement of a 4th grader.
Evidence? I have proved that is not the case
Christians struggle to explain what is the lengeth of a day in Genesis. Is it a 24 hour day, is a 1000 years equal to a day or a billion years equal to a day?
Ok and, people can believe in different things and be accepted by God, hence why there are Catholics, Protestants, Baptists,etc
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@Dr.Franklin
2 Samual 24:9
1 Chronicles 21:5
In the Bible:
3=1 (Trinity)
1=1000 (Days)
I just thought of this. If we ignore all of the times the Bible gets it wrong, it would be right 100% of the time.
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@croweupc
In the report in 2 Samuel, the number of men of valor who drew the sword was 800,000, but did not include the standing army of 288,000 described in 1 Chronicles 27:1–15, or the 12,000 specifically attached to Jerusalem described in 2 Chronicles 1:14. Including these figures gives the grand total of 1,100,000 men of valor who composed the entire army of the men of Israel. The figure of 470,000 in 1 Chronicles 21 did not include the 30,000 men of the standing army of Judah mentioned in 2 Samuel 6:1. This is evident from the fact that the Chronicler points out that Joab did not complete the counting of the men of Judah (1 Chron. 21:6). Both calculations are correct according to the groups which were included and excluded from each report. Please understand the context next time
In the Bible:
3=1 (Trinity)
1=1000 (Days)
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First I will point out that chapter 21 is about the census of able men who can fight and chapter 27 of 1 Chronicles is about active men month by month. Two separate numbers about two different groups, active versus able. 2 Chronicles 1 is under King Solomon, so this would be some time after King Davids census. Please understand the context next time!
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@croweupc
did not include the standing army of 288,000 described in 1 Chronicles 27:1–15, or the 12,000 specifically attached to Jerusalem
The figure of 470,000 in 1 Chronicles 21 did not include the 30,000 men of the standing army of Judah mentioned in 2 Samuel 6:1.
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The 30,000 in 2 Samuel 6:1 is how many men King David took to bring back the Ark. You talk about out of context, this is most certainly out of context. You cannot just take numbers from other parts of the book and add them together like that and call it good. These events didn’t happen at the same time. It contains years of King Davids reign into account. These numbers you are putting together are not connected.
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@croweupc
These numbers you are putting together are not connected.
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@Dr.Franklin
God sent Gad, to threaten David with how many years of famine?
2 Samuel 24:13: SEVEN years of famine.
1 Chron.: 21:12: THREE years of famine.
When David defeated the King of Zobah, how many horsemen did he capture?
1 Chron. 18:4: David took SEVEN THOUSAND horsemen
2 Samuel 8:4: David took ONE THOUSAND SEVEN HUNDRED horsemen
How many stalls for horses did Solomon have?
I Kings 4:26: FORTY THOUSAND
2 Chron. 9:25: FOUR THOUSAND
The Temple contained how many baths?
1 Kings 7:26: TWO THOUSAND baths.
2 Chron. 4:5: THREE THOUSAND baths
How old was Jehoiachin when he became king of Jerusalem?
2 Kings 24:8: EIGHTEEN years of age
2 Chron. 36:9: EIGHT years of age
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@Stronn
Samual and Chronicles are different look at point above
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@Stronn
2 Samuel 24:13: SEVEN years of famine.
1 Chron.: 21:12: THREE years of famine.
Regarding 2 Samuel 24:13, many English translations follow the Septuagint by using “three” in place of “seven.” If this were the original reading, then we would have an example of a copyist error. It is possible for copyist errors to have crept into some documents, and since the doctrine of inerrancy only applies to the original manuscripts, such errors would have no impact on this crucial doctrine.
Not surprisingly, some critics of biblical authority present this apparent incongruity as evidence confirming their pre-committed disbelief in the inerrancy and divine inspiration of Scripture. Others cite this to justify their claim that modern copies of the biblical texts insufficiently represent the original manuscripts.
In reality, these accusations of corruption are unwarranted, and there are at least a couple of plausible solutions that do not appeal to a copyist error.
The key lies in understanding the greater context of the account. Let us first consider a verse that precedes the account in 2 Samuel:
Now there was a famine in the days of David for three years, year after year; and David inquired of the Lord. And the Lord answered, “It is because of Saul and his bloodthirsty house, because he killed the Gibeonites.” (2 Samuel 21:1)
Clearly, Israel had already experienced three years of famine before David numbered the people of Israel and Judah—for reasons unrelated to the situation in question. 2 Samuel 24:1–7 record the initiation of the census, but we find in verse 8 that “when they had gone through all the land, they came to Jerusalem at the end of nine months and twenty days.”
So according to the text, numbering the people was nearly a year-long process, and there is no clear indication that God had suspended the initial three-year famine prior to the events in chapter 24. Now if God had combined three additional years of famine (1 Chronicles 21:12) with the three years of initialfamine, and a possible intervening year while the census was conducted, the resulting overall famine would have totaled about seven years (2 Samuel 24:13).
Some Christians have proposed another solution. They claim that these two passages describe the prophet Gad confronting David on two different occasions. According to this view, the “seven year” proposal was initially given four years prior to the “three year” proposal. Thus, the prophet would have confronted David and given him a few years to mull over his decision. During that time, David had repented of his actions so God reduced the time of punishment—something God definitely has the authority to do. A problem with this view is that if God reduced the seven years to three years because of David’s repentance, then why didn’t He reduce the length of the other options as well? So while this solution may seem less likely, it still provides another reasonable explanation.
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@Stronn
1 Chron. 18:4: David took SEVEN THOUSAND horsemen
2 Samuel 8:4: David took ONE THOUSAND SEVEN HUNDRED horsemen
We can take two steps to solve this mystery. The first step is to understand that horsemen and footmen were not exclusive categories, but that horsemen were a subset of footmen. Soldiers who were trained to ride horses were usually trained first as ground infantry just as armored vehicle operators of the USA Marines are all initially trained as ground infantry. There is a clear example of this double role elsewhere in scripture. 2 Samuel 10:18 describes a battle where David slew “forty thousand horsemen” of the Syrians (KJV, NASB, ESV). 1 Chronicles 19:18, describing the same event, says that David slew “forty thousand footmen” of the Syrians (KJV, NASB, ESV). 2 Samuel 10:18 and 1 Chronicles 19:18 read together suggest that the horsemen and the footmen were the same men just described differently. There were not 80,000 men in total but rather 40,000 men who took on double roles as horsemen and footmen. Hence in the case of the Syrians the subset (40,000 horsemen) occupied the entire set (40,000 footmen). Going back to the battle against Hadarezer, when 2 Samuel 8:4 says that there were “seven hundred horsemen, and twenty thousand footmen,” there were not 20,700 men in total, but rather 20,000 men of which 700 were considered horsemen by the author of 2 Samuel. Likewise, when 1 Chronicles 18:4 says that there were “seven thousand horsemen, and twenty thousand footmen,” there were not 27,000 men in total, but rather 20,000 men of which 7000 were considered horsemen by the author of 1 Chronicles. Listed categories that are connected by the conjunction "and" do not have to be exclusive categories. For example, the Bible often uses the phrase "Judah and Jerusalem" even though Jerusalem is part of Judah. Listed categories that are connected by "and" can overlap in what they refer to. Thus "x horsemen, and twenty thousand footmen" means that there were x number of entities that qualified as horsemen and twenty thousand entities that qualified as horsemen, not that there were two exclusive categories of horsemen and footmen.
The second step to solving the mystery is to see that the difference between 700 and 7000 horsemen is due to “horsemen” being a floating label. The designation, “horsemen” is a floating label because it attaches when a man is on a horse, and could detach when a man is no longer on a horse. Chariots can always be called “chariots” (thus 2 Samuel and 1 Chronicles both agree that there were a thousand chariots). Footmen can also always be called “footmen” since all chariot riders are trained with the basics of ground infantry (thus 2 Samuel and 1 Chronicles both agree that there were 20,000 footmen). However, horsemen are not always horsemen. If at the start of battle there were 7000 men on horses, one historian can say that David captured 7000 horsemen in battle. However, if at the end of the battle 6300 horses go out of commission and only 700 men remain on horses, then another historian can say that David captured 700 horsemen in battle. The discrepancy in numbers is due to the different perspectives of the historians. The author of 1 Chronicles still referred to the men who lost their horses as "horsemen" whereas the author of 2 Samuel only referred to the men still on horses as "horsemen." This discrepancy could have arisen if the historian of 1 Chronicles got his number from a headcount of horsemen prior to battle (as the two sides squared off against each other) and the historian of 2 Samuel got his number from a post-battle headcount. Both accounts are correct according to their own perspectives. There is evidence that different historical sources were used in the two books. For example, as mentioned earlier, the superficial difference between 2 Samuel 10:18 (“forty thousand horsemen”) and 1 Chronicles 19:18 (“forty thousand footmen”) demonstrates that one account is not a mere duplicate of the other.
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@Stronn
I Kings 4:26: FORTY THOUSAND
2 Chron. 9:25: FOUR THOUSAND
Typo, KJV fixed it
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@Stronn
1 Kings 7:26: TWO THOUSAND baths.
2 Chron. 4:5: THREE THOUSAND baths
The two verses do not contradict. What we have here are two descriptions that are not mutually exclusive. If the container held 3000 baths at a certain point in time, then surely there was a point during its filling when it held 2000 baths. Neither 1 Kings 7:26 nor 2 Chronicles 4:5 purport to provide the upper limit of the capacity. When the historian of 1 Kings 7:26 observed the container, it held 2000 baths. When the historian of 2 Chronicles 4:5 observed the container, it held 3000 baths.
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@Stronn
2 Kings 24:8: EIGHTEEN years of age
2 Chron. 36:9: EIGHT years of age
The KJV follows the Masoretic reading. Most modern translators speculate that the Masoretic text is in error, seeing that 2 Kings 24:8 says Jehoiachin was eighteen years old when he began to reign. However, there is no error in the Masoretic text. Jehoiachin became co-regent with his father Jehoiakim over Judah at age eight (2 Chronicles 36:9) and became the ruler “in Jerusalem” at age eighteen (2 Kings 24:8). The young age at which Jehoiachim became co-regent is not surprising, since his father’s interest would have been to secure an heir in the face of imminent Babylonian invasion. Jehoiachin’s co-regency of ten years corresponds perfectly with his father Jehoiakim’s reign of eleven years (2 Chronicles 36:5). Moreover, as soon as the Babylonian invasion looms into the picture, Chronicles begins to use the phrase, “king over Judah and Jerusalem” (2 Chronicles 36:4, 10). The phrase is never used in Kings or in Chronicles prior to the Babylonian invasion. Prior to the Babylonian invasion, there was no need to differentiate the king of Jerusalem from the king of Judah. However, as the Babylonians came and instituted their rule, the king of Jerusalem was no longer the default king of the rest of Judah. Thus, Chronicles begins to use the phrase “king over Judah and Jerusalem” to indicate a ruler who reigned over both Jerusalem and Judah. This point is significant in regards to ascertaining the total length of Jehoiachin’s time in office. Although 2 Kings 24:8 and 2 Chronicles 36:9 say that he reigned “in Jerusalem” for three months, that does not exclude the possibility of him co-reigning over Judah for the past ten years. Since his father was the sole regent over Jerusalem under Nebuchadnezzar’s regime (2 Kings 24:1), Jehoiachin did not have authority over Jerusalem despite having co-regency over Judah. The seeming contradiction between 2 Kings 24:8 and 2 Chronicles 36:9 is a testament of the confusing political scene of the time rather than an error in the Masoretic text.
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The Bible is full of these contradictions. It would seem no matter how many you show the Christian apologists will find some way to make it agree.
For instance:
Matt 27:6-8
This account has the chef priests buying the field with the money Judas threw at them. The field was named the field of blood because the money was used to arrest Jesus and was considered blood money. Judas hangs himself.
Acts 1:18-19
This account however has Judas buying the field and spilling his intestines and blood on it and that is why it is known as the field of blood.
When people start with the conclusions first, they will always end up at the answers they want. You have to start with the conclusion that the Bible is inerrant in order to make these verses non-contradictory. The lengths people go to try and keep it inerrant is astonishing.
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@Dr.Franklin
@croweupc
One should at least acknowlege the source.
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@keithprosser
good ol’ copy and paste!
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@Dr.Franklin
@croweupc
I note a great many Bible conrradictions concern Chronicles.
The theory I favour is that Chronicles is a re-write of the material in the earlier OT books. The writer of Chronicles (often called 'The Chronicler') seems to have been unable to resist the temptation to exaggerate some of the numbers to make the iconic and semi-legendary kings of Hebrew histrory (Solomon and David) appear even more glorious than in the old legends. His enlarged numbers and other changes never serve to dimish the picture of Hebrews or their great kings so they aren't random errors.
The differences in the portryal of David in Chronicles and the early books is also worth noting. For example In 2 Samuel David has the husband of Bathsheeba (Uriah the Hittite) by whom he has a child killed. The Chronicler completely omits that blemish on David's character.in his account.
Of course nobody knows exactly why the Chronicler wrote his book but there are plenty of theories. But whatever his reasons, he purposely 'tweaked' the old stories.
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@croweupc
It's ok to use the internet for research, but not to pass off entire posts as if they are out of your own head.
I think 'Dr' Franklin isn't worth engaging with - I doubt he has any original ideas of his own.
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@keithprosser
I was too lazy,
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@croweupc
Acts 1:18 describes what occurred after Judas hanged himself in Matthew 27:5. His body began to decay as it hung from the rope. Eventually, his corpse fell, and “burst asunder” when it hit the ground—he literally burst apart.
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@keithprosser
I agree.
I have no problem with people referencing, I do from time to time, but most people tend not to think about the issues independently first. I can think much deeper about a topic if I do not start by reading others opinions first. This, of course, is just my opinion.
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@croweupc
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@Dr.Franklin
You just proved my point. No matter what errors are in the Bible, you refuse to acknowledge them. There is no way you would cut this much slack for any other source. You have to bend over backwards and it still does not make sense. As I pointed out, the reason for it being called the field of blood is deferent. I don’t think apologists are not smart, quite the opposite, the smarter they are the more clever their narrative. What I do not think you understand is if you did not believe the Bible was inerrant, you would have automatically concluded that these two accounts were contradictions. It is only out of necessity that you have them agree. These accounts were written by two different authors, and they have two separate conclusions. | 5,420 | 20,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.913256 |
http://mathhelpforum.com/algebra/581-calculating-grade-print.html | 1,503,554,485,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133032.51/warc/CC-MAIN-20170824043524-20170824063524-00423.warc.gz | 260,895,696 | 2,957 | • Jul 16th 2005, 09:16 AM
Angel007
word problem
i am lost after i add 79,56,91,72 i divide by 4 after this is were i am stuck can you help me out i learn better if i can see the steps just once. here is the problem:
A student has made a 79, 56, 91 and 72 on four tests. Determine the possible scores on the fifth test that will result in a 75 or higher average (assume an average cannot go higher than 100)…Hint: set up an inequality to calculate the grade if the average would be between 75 and 100.
• Jul 16th 2005, 09:42 AM
MathGuru
A student has made a 79, 56, 91 and 72 on four tests. Determine the possible scores on the fifth test that will result in a 75 or higher average
The score on the fifth test we will call 'x'
so the average after the fifth test will be:
(79 + 56 + 91 + 72 + x)/5
or
(298 +x)/5
or
59.6 + x/5
so what we want is:
75 <= 59.6 + x/5
15.4 <= x/5
77 <= x
In order to get at least a 75, the student needs to score a 77 or higher.
• Jul 16th 2005, 10:43 AM
ticbol
Here is one way.
If there are 5 test in total, then the average grade is the sum of the 5 grades divided by 5.
75 <= (79 +56 +91 +72 +x)/5 <= 100
75 <= (298+x)/5 <= 100
Clear the fraction, multiply all sides by 5,
375 <= 298 +x <= 500
Isolate the x, subtract 298 from all sides,
77 <= x <= 202
That means on the 5th test, the student must score anywhere between 77 and 202.
Can a grade or test score be more than 100?
If not, then the 5th grade should be anywhere from 77 to 100. | 477 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.934868 |
https://www.askiitians.com/forums/Organic-Chemistry/21/3469/chemistry-quantum-theory.htm | 1,701,944,547,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00419.warc.gz | 735,519,435 | 44,440 | # A sample of hydrogen atoms and in each atom , the electron is excited to the energy level n.What is the? maximum no. of unique wavelengths this sample can emit? please give a simple explanation. Its really confusing.Also i would like to know the difference between the above and below question. There is a hydrogen atom in the ground state . i t is excited to a higher energy level n .When it comes back to ground state, it emits radiation .What is the max. no. of unique wavelength it can emit?
8 Points
14 years ago
As it seems to me both questions are one & same.You see whenever an electron is in higher energy level it loses exact quanta of energy in the form of radiation emitted to go in the lower energy level.For hydrogen atom this energy is inversely proportional to the square of level that electron occupy.
Now from level 'n' maximum transitions could be 'n-1' as electron could end up in any of lower 'n-1' levels.As you could verify that each transition is unique for the value of enery released as radiation.So you've 'n-1' unique wavelenths corresponding to them.
Similarly for electron in 'n-1'th level there are 'n-2' unique wavelengths & so on upto 1 for 'n=1' (one above ground level).
Adding all give max. no. of unique wavelengths=n*(n-1)/2 {or,nC2 ;i.e. choosing 2 levels out of n}
For 2nd version it may be explicitly said that no in-between transitions are taken into consideration,only transition to ground state is counted then you've distinct 'n-1' wavelengths,otherwise both Q's are same statements.
Here you must know that for the 1st version of the Q i neglect any interaction between radiation emitted by one atom & another atom.I presumed them to be single entities in behaviour,which couldn't be so.Because in a sample of hydrogen atoms series of radiation attacking some electrons may propel them to higher energy states than 'n' which in turn cause emission of more energetic radiations when these electrons fall back to lower levels which in turn excite more & more like an avalanche.You see it's impossible then to determine exact no. of possible transitions !!!!
Above argument is quite valid for low pressure & high temp. sample with infinite volume to occupy.
I hope you get this.Any further query in this is answered fully just make your Q to the point.(what u wanna know!) | 518 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.931365 |
https://wackerhardware.com/accessories/your-question-can-a-door-be-measured-in-meters.html | 1,627,656,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00713.warc.gz | 607,723,021 | 19,389 | # Your question: Can a door be measured in meters?
Contents
## How many meters is a door?
Standard height of front and back doors
Standard height sizes for external doors begin at 80 inches (2.03m) and go up as high as 8ft (2.44m) tall. However, the most common standard size for a front door is 80 inches (2.03m) in height and 36 inches (0.91m) wide.
## Is a door measured in cm or M?
The width of a door is about 1 meter (m). The millimeter, centimeter, decimeter, and meter are units of length.
## How do you measure the size of a door?
Take three measurements inside the door frame – at the top, middle and bottom. The widest measurement determines the width of the door. Use a tape measure to understand the width of the three areas of the door frame. Run the tape measure along the width of the frame from the left to the right and record these numbers.
## What unit would you measure a door?
Feet are commonly used in measurement for objects such as the length of a car, height of a door, or the distance between two objects. Another customary unit of measurement is yards.
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## What is standard door size?
The height for all passage doors must be a minimum of 80 inches and the standard width sizes for interior doors are 24”, 28”, 30”, 32” and 36”. The minimum recommended door width to allow persons with disabilities’ to pass through is 36 inches.
## What distance is a meter?
The metre is currently defined as the length of the path travelled by light in a vacuum in 1299 792 458 of a second. The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole along a great circle, so the Earth’s circumference is approximately 40000 km.
## How many cm means 1 meter?
100 centimeters equal to 1 meter or one centimeter equal to one-hundredth (i.e. 1/100 th) of meter. Centimeters are denoted by cm, whereas meters are denoted by m.
## How many cm are in me?
1 Meter (m) is equal to 100 centimeters (cm).
## What is about 1 meter long?
A meter (m) is about: a little more than a yard (1 yard is exactly 0.9144 meters) the width of a doorway (most doorways are about 0.8 to 0.9 m) half the length of a bed.
## What is the actual size of a 36 door?
3’0″ (36 Inch) Door Width (Actual Size 35-3/4″) by Door Closers USA.
## What is the standard height of a door?
For reference, standard interior door widths range from 24″-36″ (finished opening). While a standard door height is 80″ (finished opening). The difference between a rough opening and finished opening is usually 2-3 inches.
## What is standard door hinge size?
What size of door hinge do I need? Most homes have a standard sized hinge of 3.5″ x 3.5″ This measurement is from the bottom edge of the hinge to the top end of the hinge, and then from the outside edge to edge when the hinge is open. Exterior doors usually have a 4″ hinge on them, but not always.
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## What unit is weight measured in?
The unit of measurement for weight is that of force, which in the International System of Units (SI) is the newton. For example, an object with a mass of one kilogram has a weight of about 9.8 newtons on the surface of the Earth, and about one-sixth as much on the Moon.
## How do you measure a door jamb?
The jamb is the outermost portion of the frame on the sides and top of a door system. Measure the thickest part of the jamb. Add 2 1/2″ (Frames generally measure 1 1/4″ thick each x2). Add 3/4″ to unit width measurement.
## What is a door slab?
A slab door is a door without a frame or hinges, but usually includes a hole pre-cut for the doorknob. Once the door is hung in an existing frame, it just needs to be painted or stained (if necessary) and outfitted with knobs. Door slabs can be purchased new, handbuilt by a carpenter, or salvaged from antique stores. | 997 | 3,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-31 | longest | en | 0.93077 |
https://oeis.org/A105291 | 1,600,887,606,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212039.16/warc/CC-MAIN-20200923175652-20200923205652-00620.warc.gz | 494,630,647 | 4,070 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A105291 Triangle read by rows: T(m,n) = binomial(m!,n), m>=0, 0 <= n <= m!. 3
1, 1, 1, 1, 1, 2, 1, 1, 6, 15, 20, 15, 6, 1, 1, 24, 276, 2024, 10626, 42504, 134596, 346104, 735471, 1307504, 1961256, 2496144, 2704156, 2496144, 1961256, 1307504, 735471, 346104, 134596, 42504, 10626, 2024, 276, 24, 1, 1, 120, 7140, 280840, 8214570, 190578024 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,6 COMMENTS This is the number of nXm arrays with each row a permutation of 1..m, and rows in lexicographically strictly increasing order. For row 0, remember that 0!=1. LINKS Alois P. Heinz, Rows n = 0..6, flattened EXAMPLE Triangle begins: [1, 1], [1, 1], [1, 2, 1], [1, 6, 15, 20, 15, 6, 1], [1, 24, 276, 2024, 10626, 42504, 134596, 346104, 735471, 1307504, 1961256, 2496144, 2704156, 2496144, 1961256, 1307504, 735471, 346104, 134596, 42504, 10626, 2024, 276, 24, 1], ... MATHEMATICA Flatten[Table[Binomial[m!, n], {m, 0, 5}, {n, 0, m!}]] (* Harvey P. Dale, Apr 16 2013 *) CROSSREFS See A180397 for another version. Cf. A007318 (Pascal's triangle), A086687, A109892. Sequence in context: A039763 A094262 A123554 * A214631 A025270 A249450 Adjacent sequences: A105288 A105289 A105290 * A105292 A105293 A105294 KEYWORD nonn,tabf AUTHOR N. J. A. Sloane, Sep 03 2010, following a suggestion from R. H. Hardin, Aug 31 2010 STATUS approved
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Last modified September 23 14:29 EDT 2020. Contains 337310 sequences. (Running on oeis4.) | 724 | 1,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.641489 |
https://api.hypothes.is/search?q=tag%3Asymbology | 1,709,014,166,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00570.warc.gz | 104,533,122 | 10,237 | 4 Matching Annotations
1. Feb 2023
2. en.wikipedia.org en.wikipedia.org
URL
3. www.reddit.com www.reddit.com
1. Are there symbols for 'supported by' or 'contradicted by' etc. to show not quite formal logical relations in a short hand?
In addition to the other excellent suggestions, I don't think you'll find anything specific that that was used historically for these, but there are certainly lots of old annotation symbols you might be able to co-opt for your personal use.
Evina Steinova has a great free cheat sheet list of annotation symbols: The Most Common Annotation Symbols in Early Medieval Western Manuscripts (a cheat sheet).
More of this rabbit hole:
(Nota bene: most of my brief research here only extends to Western traditions, primarily in Latin and Greek. Obviously other languages and eras will have potential ideas as well.)
Tironian shorthand may have something you could repurpose as well: https://en.wikipedia.org/wiki/Tironian_notes
Some may find the auxiliary signs of the Universal Decimal Classification useful for some of these sorts of notations for conjoining ideas.
Given the past history of these sorts of symbols and their uses, perhaps it might be useful for us all to aggregate a list of common ones we all use as a means of re-standardizing some of them in modern contexts? Which ones does everyone use?
Here are some I commonly use:
Often for quotations, citations, and provenance of ideas, I'll use Maria Popova and Tina Roth Eisenberg's Curator's Code:
• ᔥ for "via" to denote a direct quotation/source— something found elsewhere and written with little or no modification or elaboration (reformulation notes)
• ↬ for "hat tip" to stand for indirect discovery — something for which you got the idea at a source, but modified or elaborated on significantly (inspiration by a source, but which needn't be cited)
Occasionally I'll use a few nanoformats, from the microblogging space, particularly
• L: to indicate location
For mathematical proofs, in addition to their usual meanings, I'll use two symbols to separate biconditionals (necessary/sufficient conditions)
• (⇒) as a heading for the "if" portion of the proof
• (⇐) for the "only if" portion
Some historians may write 19c to indicate 19th Century, often I'll abbreviate using Roman numerals instead, so "XIX".
Occasionally, I'll also throw drolleries or other symbols into my margins to indicate idiosyncratic things that may only mean something specifically to me. This follows in the medieval traditions of the ars memoria, some of which are suggested in Cornwell, Hilarie, and James Cornwell. Saints, Signs, and Symbols: The Symbolic Language of Christian Art 3rd Edition. Church Publishing, Inc., 2009. The modern day equivalent of this might be the use of emoji with slang meanings or 1337 (leet) speak.
URL
4. Oct 2020
5. en.wikipedia.org en.wikipedia.org
1. Early Christians used the ichthys, a symbol of a fish, to represent Jesus,[94][95] because the Greek word for fish, ΙΧΘΥΣ Ichthys, could be used as an acronym for "Ίησοῦς Χριστός, Θεοῦ Υἱός, Σωτήρ" (Iesous Christos, Theou Huios, Soter), meaning "Jesus Christ, Son of God, Saviour".
URL
6. Oct 2019
7. www.medievalists.net www.medievalists.net
1. side from hunting scenes, one may broadly divide ancient depictions of hares into two iconographical groups. In the first group, the hare embodies the love of a man for a woman. On numerous Greek vases, for example, we see a depiction of Eros, the God of Love, with his two symbols, the lyre and the hare; other scenes show Eros pursuing a hare as a symbol of love, which, as we know from Book 1 of Philostrates’ Imagines, has been given a special gift: fertility.
hares symbology | 891 | 3,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.932646 |
https://stats.stackexchange.com/questions/70801/how-to-normalize-data-to-0-1-range/71011 | 1,723,136,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00832.warc.gz | 427,312,296 | 49,301 | # How to normalize data to 0-1 range?
I am lost in normalizing, could anyone guide me please.
I have a minimum and maximum values, say -23.89 and 7.54990767, respectively.
If I get a value of 5.6878 how can I scale this value on a scale of 0 to 1.
• is this the way =(value-min)/(max-min) Commented Sep 23, 2013 at 15:31
• Explanation of protection: This question is attracting extra answers containing code solutions only. While these may be interesting or useful to some readers, it's not an aim of CV to provide repositories of code solutions. Commented May 27, 2015 at 8:42
• the solutions provided consider a linear contrast value - would you like a different normalization, for instance one that achieve an uniform probability for the output? Commented May 21, 2018 at 16:31
If you want to normalize your data, you can do so as you suggest and simply calculate the following:
$$z_i=\frac{x_i-\min(x)}{\max(x)-\min(x)}$$
where $$x=(x_1,...,x_n)$$ and $$z_i$$ is now your $$i^{th}$$ normalized data. As a proof of concept (although you did not ask for it) here is some R code and accompanying graph to illustrate this point:
# Example Data
x = sample(-100:100, 50)
#Normalized Data
normalized = (x-min(x))/(max(x)-min(x))
# Histogram of example data and normalized data
par(mfrow=c(1,2))
hist(x, breaks=10, xlab="Data", col="lightblue", main="")
hist(normalized, breaks=10, xlab="Normalized Data", col="lightblue", main="")
• I only wonder how the two quite different-looking histograms do illustrate the point of your (correct) answer? Commented Sep 23, 2013 at 16:21
• @ttnphns They look only different due to the binning of the histograms. My point however was to show that the original values lived between -100 to 100 and now after normalization they live between 0 and 1. I could have used a different graph to show this I suppose or just summary statistics.
– user25658
Commented Sep 23, 2013 at 16:23
• The gentle nudge by @ttnphns was meant to encourage you not only to use a less complicated means of illustrating a (simple) idea, but also (I suspect) as a hint that a more directly relevant illustration might be beneficial here. You could do both by finding a more straightforward way to graph the transformation when it is applied to the min and max actually supplied by the O.P.
– whuber
Commented Sep 23, 2013 at 17:12
• Is there a way to "normalize" to custom range instead of 0-1? Commented Oct 25, 2016 at 11:46
• @JohnDemetriou May not be the cleanest solution, but you can scale the normalized values to do that. If you want for example range of 0-100, you just multiply each number by 100. If you want range that is not beginning with 0, like 10-100, you would do it by scaling by the MAX-MIN and then to the values you get from that just adding the MIN. So scale by 90, then add 10. That should be enough for most of the custom ranges you may want. Commented Oct 29, 2017 at 18:54
The general one-line formula to linearly rescale data values having observed min and max into a new arbitrary range min' to max' is
newvalue= (max'-min')/(max-min)*(value-max)+max'
or
newvalue= (max'-min')/(max-min)*(value-min)+min'.
• This is correct, but not efficient. It is a linear transformation, so you would precalculate a and b constants, and then just apply newvalue = a * value + b. a = (max'-min')/(max-min) and b = max - a * max Commented Sep 23, 2013 at 19:18
• Do you know how to cite this? I mean, is there an "original" reference somewhere? Commented May 11, 2014 at 20:18
• @MarkLakata Slight (typo?) correction: b = max' - a * max or b = min' - (a * min)
– Nick
Commented Dec 29, 2014 at 18:09
• @Nick - yes. I'm missing a ' Commented Dec 30, 2014 at 5:33
• Can you please compare your normalisation here se.mathworks.com/matlabcentral/answers/… i.e. the equation u = -1 + 2.*(u - min(u))./(max(u) - min(u));. Commented Oct 24, 2016 at 21:41
Here is my PHP implementation for normalisation:
function normalize($value,$min, $max) {$normalized = ($value -$min) / ($max -$min);
return $normalized; } But while I was building my own artificial neural networks, I needed to transform the normalized output back to the original data to get good readable output for the graph. function denormalize($normalized, $min,$max) {
$denormalized = ($normalized * ($max -$min) + $min); return$denormalized;
}
$int = 12;$max = 20;
$min = 10;$normalized = normalize($int,$min, $max); // 0.2$denormalized = denormalize($normalized,$min, $max); //12 Denormalisation uses the following formula:$x (\text{max} - \text{min}) + \text{min}\$
• I don't think, that this is the only difference. In my code, I also showed, how to return a normalized value to the value it was before normalisation. I think, that makes it worth this answer. Commented May 27, 2015 at 9:02
• It's still true that you post only code: I think you need to emphasise any supposedly special virtues of code in commentary, as otherwise readers have to read the code to see what they are. Presumably inverting the scaling is of use only when (a) the original values have been overwritten but (b) the user has prudently remembered to save the minimum and maximum. My wider point, as commented above, is that CV does not aim to be a repository of code examples. Commented May 27, 2015 at 9:10
• There are some problems, whre you need to restore the value: Nueral Networks for example... But you're right, in manner of data analysis, this answer is very bad. Commented May 27, 2015 at 9:25
• @NickCox I found his answer to be more satisfactory than the accepted one. Commented Aug 30, 2015 at 18:40
• @Karl Morrison You can and should upvote it then. But code only answers strictly remain off-topic here. Commented Aug 31, 2015 at 9:08
### Division by zero
One thing to keep in mind is that max - min could equal zero. In this case, you would not want to perform that division.
The case where this would happen is when all values in the list you're trying to normalize are the same. To normalize such a list, each item would be 1 / length.
// JavaScript
function normalize(list) {
var minMax = list.reduce((acc, value) => {
if (value < acc.min) {
acc.min = value;
}
if (value > acc.max) {
acc.max = value;
}
return acc;
}, {min: Number.POSITIVE_INFINITY, max: Number.NEGATIVE_INFINITY});
return list.map(value => {
// Verify that you're not about to divide by zero
if (minMax.max === minMax.min) {
return 1 / list.length
}
var diff = minMax.max - minMax.min;
return (value - minMax.min) / diff;
});
}
### Example:
normalize([3, 3, 3, 3]); // output => [0.25, 0.25, 0.25, 0.25]
• This is a rescaling to a sum 1, not to a range 0-1. I just think the answer is off-topic therefore. Commented Oct 4, 2017 at 17:31
• Not so. normalize([12, 20, 10]) outputs [0.2, 1.0, 0.0], which is the same you would get with (val - min) / (max - min). Commented Jan 16, 2019 at 15:23
• @rodrigo-silveira I don't see why the all 0.25 output. Isn't it better all 0.5? All items are equal, so should be kept centered in the interval. Commented Apr 2, 2019 at 14:56
• If the variable is a constant, it won't be much use either as an outcome or as a a predictor. Either way, you should not want to standardize it. I suppose the main message here is "Watch out if you try to standardize every variable in sight, as your code will give puzzling results or even fail without a trap for this case", Commented May 16, 2020 at 8:27
• "will" in previous should be "may". Commented May 16, 2020 at 8:36
Try this. It is consistent with the function scale
normalize <- function(x) {
x <- as.matrix(x)
minAttr=apply(x, 2, min)
maxAttr=apply(x, 2, max)
x <- sweep(x, 2, minAttr, FUN="-")
x=sweep(x, 2, maxAttr-minAttr, "/")
attr(x, 'normalized:min') = minAttr
attr(x, 'normalized:max') = maxAttr
return (x)
}
The answer is right but I have a suggestion, what if your training data face some number out of range? you could use the squashing technique. it will be guaranteed never to go out of range. rather than this
I recommend using this
with squashing like this in min and max of the range
and the size of the expected out-of-range gap is directly proportional to the degree of confidence that there will be out-of-range values.
For more information, you can google: squashing the out-of-range numbers and refer to the data preparation book of "Dorian Pyle".
• Please edit your answer to use capitalisation as conventional. Consistent lower case may seem amusing or efficient, but it is more difficult for almost everyone to read. Commented Sep 25, 2013 at 12:12 | 2,382 | 8,561 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-33 | latest | en | 0.910619 |
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Ques:- A train of length 150 metres takes 40.5 seconds to cross a tunnel of length 300 metres. What is the speed of the train in km/hr?
A. 13.33
B. 26.67
C. 39.16
D. 40
1km = 1000m
Xkm = 300m
X=0.30km
1hr = 3,600s
X = 40.5s
0.01125hr
Therefore; 0.3km/0.1125hr= 26.6666=26.67km/hr
Ques:- Two trains 240 metres and 270 metres in length are running towards each other on parallel lines, one at the rate of 60 kmph and another at 48 kmph. How much time will they take to cross each other?
A. 14 sec.
B. 15 sec.
C. 16 sec.
D. 17 sec.
17 secs
Ques:- The HCF of two numbers is 11 and their LCM is 7700. If one of these numbers is 275, then the other one is
A. 918
B. 308
C. 283
D. 279
b
Ques:- Find out the wrong number in a given series.
263, 284, 393, 481, 482
A. 482
B. 481
C. 393
D. 284
B
Ques:- Two chords of lengths 2L1 and 2L2 are drawn in a circle. Their lengths are inversely proportional to the straight distance joining the centre.Find the radius of circle?
10m
Ques:- Are you sure you can join?
NO but i hope to you hire me
Ques:- If \$ stands for triple and # for negation, what is the value of #\$#3-#5
Ques:- Report of the Sachar Committee is related with?
Ques:- Why were you fired from last organisation?
Ques:- How much salary do you want to received?
Ques:- Why are u changing ur job | 588 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.910632 |
https://fwjustice.org/which-is-a-correct-statement-of-the-second-law-of-thermodynamics/ | 1,695,621,837,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00557.warc.gz | 300,192,042 | 10,541 | Which Is a Correct Statement of the Second Law of Thermodynamics?
Which Is a Correct Statement of the Second Law of Thermodynamics?
The second law of thermodynamics is a fundamental principle in physics that deals with the concept of entropy and the direction of natural processes. It is a fundamental law governing the behavior of energy and its conversion into work. In this article, we will explore the various statements of the second law of thermodynamics and discuss their implications.
Statement 1: The law of heat flow
One of the statements of the second law of thermodynamics is the law of heat flow. It states that heat naturally flows from a hotter object to a colder object until thermal equilibrium is reached. This statement emphasizes the irreversible nature of heat transfer and the tendency of energy to disperse and become more evenly distributed. For example, a cup of hot coffee will eventually cool down as heat dissipates into its surroundings.
Statement 2: The law of increasing entropy
Another correct statement of the second law of thermodynamics is the law of increasing entropy. Entropy is a measure of the disorder or randomness of a system. This law states that the entropy of an isolated system tends to increase over time. In other words, natural processes lead to a more disordered state. For instance, a perfectly organized deck of cards will eventually become shuffled, increasing the overall entropy of the system.
Statement 3: The law of energy conservation
The third statement of the second law of thermodynamics is the law of energy conservation. It states that energy cannot be created or destroyed but can only change from one form to another. This principle implies that the total amount of energy in a closed system remains constant. However, it also highlights that the quality of energy degrades over time. For example, when gasoline is burned in an engine, some of its energy is lost as waste heat, decreasing the overall usefulness of the energy.
FAQs:
Q1: Can entropy decrease in a system?
No, the law of increasing entropy states that the entropy of an isolated system tends to increase over time. While it is theoretically possible to decrease the entropy of a particular part of a system, the overall entropy of the system and its surroundings will always increase.
Q2: Does the second law of thermodynamics violate the law of energy conservation?
No, the second law of thermodynamics does not violate the law of energy conservation. Energy conservation states that the total energy in a closed system remains constant, but it does not specify the quality or usefulness of the energy. The second law explains how energy tends to disperse and degrade over time, leading to an increase in entropy.
Q3: Can the second law of thermodynamics be reversed?
No, the second law of thermodynamics is considered to be a fundamental law of nature and cannot be reversed. It is a statistical law that describes the behavior of energy and its conversion into work. While individual processes may temporarily appear to violate the second law, the overall trend of increasing entropy remains constant.
Q4: Are there any exceptions to the second law of thermodynamics?
There are no known exceptions to the second law of thermodynamics. It is a universally applicable principle that governs the behavior of energy and its conversion into work. While there may be temporary local decreases in entropy, the overall trend always leads to an increase in entropy.
In conclusion, the second law of thermodynamics is a fundamental principle that governs the behavior of energy and its conversion into work. It can be correctly stated as the law of heat flow, the law of increasing entropy, and the law of energy conservation. These statements highlight the irreversible nature of heat transfer, the tendency of systems to become more disordered over time, and the degradation of energy quality. The second law is a well-established principle with no known exceptions, and it plays a crucial role in understanding the behavior of natural processes. | 776 | 4,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.94085 |
https://www.doubtnut.com/question-answer/if-abc-are-the-angles-off-deltaabc-then-cotacotb-cotbcotc-cotccota-95421598 | 1,632,538,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00578.warc.gz | 741,450,845 | 79,078 | Login
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If A,B,C are the angles off De...
# If A,B,C are the angles off DeltaABC, then cotA*cotB+cotB*cotC+cotC*cotA=
Answer
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Text Solution
012-1
Answer :
B
Solution :
Given <br> A+B+C=pi <br> impliesA+B=pi-C <br> impliescot(A+B)=cot(pi-C) <br> cot(A+B)=cot(pi-C) <br> implies(cotAcotB-1)/(cotA+cotB)=-cotC <br> impliescotAcotB-1=-cotAcotC-cotBcotC <br> impliescotAcotB+cotBcotC+cotCcotA=1
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9:16 | 538 | 1,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-39 | latest | en | 0.397803 |
https://numberworld.info/331133212201 | 1,653,392,723,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00492.warc.gz | 481,540,594 | 3,898 | # Number 331133212201
### Properties of number 331133212201
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
4d1913d629
Base 32:
9kch7lh9
sin(331133212201)
-0.92184025226476
cos(331133212201)
-0.38757005728571
tan(331133212201)
2.378512568078
ln(331133212201)
26.52578658514
lg(331133212201)
11.52000274207
sqrt(331133212201)
575441.75396038
Square(331133212201)
1.0964920422255E+23
### Number Look Up
Look Up
331133212201 which is pronounced (three hundred thirty-one billion one hundred thirty-three million two hundred twelve thousand two hundred one) is a amazing figure. The cross sum of 331133212201 is 22. If you factorisate the number 331133212201 you will get these result 11 * 30103019291. The number 331133212201 has 4 divisors ( 1, 11, 30103019291, 331133212201 ) whith a sum of 361236231504. The figure 331133212201 is not a prime number. The number 331133212201 is not a fibonacci number. The number 331133212201 is not a Bell Number. The number 331133212201 is not a Catalan Number. The convertion of 331133212201 to base 2 (Binary) is 100110100011001000100111101011000101001. The convertion of 331133212201 to base 3 (Ternary) is 1011122201020120002000111. The convertion of 331133212201 to base 4 (Quaternary) is 10310121010331120221. The convertion of 331133212201 to base 5 (Quintal) is 20411130100242301. The convertion of 331133212201 to base 8 (Octal) is 4643104753051. The convertion of 331133212201 to base 16 (Hexadecimal) is 4d1913d629. The convertion of 331133212201 to base 32 is 9kch7lh9. The sine of the number 331133212201 is -0.92184025226476. The cosine of 331133212201 is -0.38757005728571. The tangent of 331133212201 is 2.378512568078. The square root of 331133212201 is 575441.75396038.
If you square 331133212201 you will get the following result 1.0964920422255E+23. The natural logarithm of 331133212201 is 26.52578658514 and the decimal logarithm is 11.52000274207. I hope that you now know that 331133212201 is impressive figure! | 726 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.656743 |
http://www.piclist.com/images/www/hobby_elec/e_ckt7_4.htm | 1,713,183,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00559.warc.gz | 55,502,030 | 1,970 | for A-stable multivibrator (TR type)
The period which repeats a blink is fixed by the value of capacitor (Cx) and resistor (Rx), and capacitor (Cy) and resistor (Ry). The half period time (t) is possible to calculate by the following formula. The half period is the time which is made H condition or an L condition. It uses a calculation formula on the side of TR1 as an example.
t : second, Cx : Farad, Rx : ohm
A repeat period is decided by the combination of Cx and Rx, and Cy and Ry.
In case of Cx, Cy=47 µF, Rx, Ry=22 k-ohm, it is as follows.
The time that TR1 becomes OFF condition (t1)
t1 = 0.69 * Cx * Rx = 0.69 * 47 * 10-6 * 22 * 103 = 0.713 seconds
TR2(t2) is similar.
The frequency (f)
f = 1 / ( t1+t2 ) = 1 / ( 0.7 + 0.7 ) = 0.71 Hz
The measurement frequency of the circuit which was made this time was to 0.78 Hz be. The error with the calculation value is for the error of the part and the characteristic of the transistor and so on. The measurement value of Cx, Cy was to approximately 47 µF be. The measurement value of Rx and Ry was to about 20 k-ohm be. When calculating with this value, the frequency becomes about 0.77 Hz. The calculation value and the measurement value are an near value. Don't expect a correct period with this circuit. The period changes with the temperature around. So, this circuit isn't possible to use for the circuit with strict period. | 388 | 1,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-18 | latest | en | 0.944378 |
https://www.sudokulovers.com/solution/100382040809000702000000000040200008003004005907001600000020010004000000200000590/ | 1,652,692,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00444.warc.gz | 1,164,064,837 | 17,529 | 1
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This Sudoku Puzzle has 61 steps and it is solved using Naked Single, Full House, Hidden Single, Locked Candidates Type 1 (Pointing), Locked Candidates Type 2 (Claiming), Locked Triple techniques.
Try To Solve This Puzzle
## Solution Steps:
1. Row 5 / Column 1 → 6 (Naked Single)
2. Row 1 / Column 7 → 9 (Naked Single)
3. Row 4 / Column 1 → 5 (Naked Single)
4. Row 1 / Column 9 → 6 (Naked Single)
5. Row 4 / Column 3 → 1 (Naked Single)
6. Row 1 / Column 3 → 5 (Naked Single)
7. Row 1 / Column 2 → 7 (Full House)
8. Row 4 / Column 7 → 3 (Naked Single)
9. Row 4 / Column 8 → 7 (Naked Single)
10. Row 6 / Column 8 → 2 (Naked Single)
11. Row 6 / Column 9 → 4 (Naked Single)
12. Row 5 / Column 8 → 2 (Naked Single)
13. Row 6 / Column 2 → 8 (Full House)
14. Row 5 / Column 2 → 8 (Full House)
15. Row 5 / Column 7 → 1 (Naked Single)
16. Row 6 / Column 4 → 5 (Naked Single)
17. Row 6 / Column 5 → 3 (Full House)
18. Row 3 / Column 7 → 8 (Naked Single)
19. Row 7 / Column 7 → 4 (Naked Single)
20. Row 8 / Column 7 → 2 (Full House)
21. Row 3 / Column 9 → 1 (Hidden Single)
22. Row 3 / Column 1 → 4 (Hidden Single)
23. Row 3 / Column 3 → 2 (Hidden Single)
24. Row 8 / Column 8 → 6 (Hidden Single)
25. Locked Candidates Type 1 (Pointing): 3 in b1 => r789c2<>3
26. Locked Candidates Type 1 (Pointing): 6 in b1 => r79c2<>6
27. Row 9 / Column 2 → 1 (Naked Single)
28. Locked Candidates Type 2 (Claiming): 9 in r4 => r5c45<>9
29. Row 5 / Column 4 → 7 (Naked Single)
30. Row 5 / Column 5 → 7 (Naked Single)
31. Row 3 / Column 6 → 7 (Hidden Single)
32. Row 9 / Column 9 → 7 (Hidden Single)
33. Row 7 / Column 9 → 3 (Full House)
34. Row 8 / Column 9 → 3 (Full House)
35. Row 7 / Column 1 → 7 (Full House)
36. Row 8 / Column 1 → 7 (Full House)
37. Row 9 / Column 6 → 3 (Hidden Single)
38. Locked Candidates Type 2 (Claiming): 8 in r8 => r7c46,r9c4<>8
39. Row 7 / Column 3 → 8 (Hidden Single)
40. Row 9 / Column 3 → 6 (Full House)
41. Row 9 / Column 4 → 4 (Full House)
42. Row 9 / Column 5 → 4 (Full House)
43. Row 8 / Column 6 → 8 (Hidden Single)
44. Locked Triple: 1,5,6 in r2c456 => r2c2,r3c45<>6, r2c8,r3c5<>5
45. Row 2 / Column 2 → 3 (Naked Single)
46. Row 3 / Column 2 → 6 (Full House)
47. Row 2 / Column 8 → 3 (Naked Single)
48. Row 3 / Column 4 → 9 (Naked Single)
49. Row 3 / Column 5 → 9 (Naked Single)
50. Row 3 / Column 8 → 5 (Naked Single)
51. Row 7 / Column 4 → 6 (Naked Single)
52. Row 8 / Column 4 → 1 (Full House)
53. Row 2 / Column 4 → 1 (Full House)
54. Row 4 / Column 5 → 6 (Naked Single)
55. Row 4 / Column 6 → 9 (Full House)
56. Row 8 / Column 5 → 5 (Naked Single)
57. Row 2 / Column 5 → 5 (Naked Single)
58. Row 7 / Column 6 → 5 (Naked Single)
59. Row 8 / Column 2 → 9 (Naked Single)
60. Row 2 / Column 6 → 6 (Naked Single)
61. Row 7 / Column 2 → 9 (Naked Single) | 1,170 | 2,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.470729 |
https://area.autodesk.com/blogs/the-3ds-max-blog/the-array-looping-cook-book/ | 1,580,318,213,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251801423.98/warc/CC-MAIN-20200129164403-20200129193403-00028.warc.gz | 307,259,804 | 18,735 | FORUMS
# The Array Looping Cook Book
By - - 3ds Max
Duration
14 mins
Last modification: 16 Sep, 2017
Note: The interface in this tutorial applies to MCG 2016/2017. The interface in MCG 2018 has been revised to a new node naming scheme.
If you’re coming from a scripting or a programming background, you may have spent some time in the MCG operator list searching for the “for loop”. While there is no specific operator called “for loop” in MCG by default, there is a wide variety of operators which behave like specialized versions of the for loop. In this post, we’ll be taking a tour of those operators, and we’ll be looking at common patterns you can apply to create and iterate on arrays.
In the screenshots below, the ParseIntArray, PrintWithLabel, PrintArrayWithLabel, ArrayOfPseudoRandomFloats, and PartialSum compounds were created during the development of this tutorial, so you won’t find them in your MCG operator list by default. To obtain these compounds, install the MCGArrayTests.mcg tool at the bottom of this post.
Note: The MaxScript code in the looping examples below only illustrates the behavior of the operators, and is not a direct translation of the implementation of those MCG operators. In fact, MCG graphs compile down to .NET bytecode, which in most cases runs faster than MaxScript.
Count
Let’s start with the simplest one of them all: Count. The Count operator returns the number of items in the array.
-- Count
A = #(1, 6, 12, 33, 54)
A.count -- => 5
ArrayOf
The ArrayOf operator creates an array of size “n” which contains repeated copies of the supplied value. In the example below, we’re using the ArrayOf operator to create an array containing five copies of the vector [0,0,1].
-- ArrayOf
function ArrayOf x n = (
result = #()
for i = 0 to (n-1) do (
append result x
)
return result
)
ArrayOf [0,0,1] 5 -- => #([0,0,1], [0,0,1], [0,0,1], [0,0,1], [0,0,1])
ArrayOfFunction - Creating an Array of Random Values
If you want to generate an array of random values, use the graph construction below, which makes use of the ArrayOfFunction operator. This operator creates an array by repeating the given function “n” times. The “Bind” operator in this graph forces the PseudoRandomFloat operator to use the same random number generator, instead of recreating a new generator on every loop. We explain the “Bind” operator in more detail in The Low Poly Modifier - Part 2.
-- ArrayOfFunction
function ArrayOfFunction f n = (
result = #()
for i = 0 to (n-1) do (
tmp = f()
append result tmp
)
return result
)
-- Set the random seed value and create a function which generates a new random
-- value on every call.
seed 1234.0
function myRandomFunction = (
return random 0.0 1.0 -- A random floating point number between 0.0 and 1.0
)
ArrayOfFunction (myRandomFunction) 5 -- => #(0.177433, 0.954723, 0.396885, 0.398289, 0.242126)
We condensed this construction into the ArrayOfPseudoRandomFloats compound, which you can use in your own graphs by installing the MCGArrayTests.mcg file linked at the end of this post.
Unit, Array2, Array3, Array4, and Concatenate
To create an array containing a set of hard-coded values (without the ParseIntArray compound we’ve been using so far), you can use Unit, Array2, Array3, and Array4. Use the Concatenate operator on two arrays to join them into a single array.
Range
Given a number “n”, the Range operator will produce an array of integers between 0 and (n-1). In many cases, the Range operator can act as a starting point in your graph to guide the overall behavior of your tool such as the number of iterations, the number of objects to scatter, etc. For example, we used the Range operator to drive the number of stairs in our Staircase building tool.
-- Range
function Range n = (
result = #()
for i = 0 to (n-1) do (
append result i
)
return result
)
Range 5 -- => #(0, 1, 2, 3, 4)
RangeInclusiveFloat / RangeExclusiveFloat
Similar to the Range operator, the RangeInclusiveFloat and RangeExclusiveFloat operators produce an array of “n” decimal values between 0.0 and 1.0. The “Inclusive” and “Exclusive” terms determine whether or not the value of 1.0 is included in the computed range of values.
When you connect them to a Map, these operators can be used to generate an array of proportional samples. For example, in the graph below, we’re using the RangeExclusiveFloat operator to sample four equidistant angles around a circle. Note that in MCG, trigonometric operators such as Sin and Cos require angular values in radians. In the graph below, we’ve chosen to convert these values into degrees to increase the readability of the resulting array.
-- RangeInclusiveFloat
function RangeInclusiveFloat n = (
step = 1.0 / (n-1)
result = #()
for i = 0 to (n-1) do (
append result (i*step)
)
return result
)
-- RangeExclusiveFloat
function RangeExclusiveFloat n = (
step = 1.0 / (n)
result = #()
for i = 0 to (n-1) do (
append result (i*step)
)
return result
)
RangeInclusiveFloat 5 -- => #(0, 0.25, 0.5, 0.75, 1)
RangeExclusiveFloat 5 -- => #(0, 0.2, 0.4, 0.6, 0.8)
Map
Map is one of the most fundamental operators in MCG. We covered it in our very first blog post to explain the function connector. The Map operator closely resembles the “for each” construct common to other programming languages. It iterates over all the items in an array, and it returns a new array of transformed values based on the function you provide.
-- Map
function Map xs fxn = (
result = #()
for item in xs do (
tmp = fxn item
append result tmp
)
return result
)
-- Initialize an array of integers.
A = #(0, 1, 2, 3, 4)
-- Declare the function to apply on each iteration
function myFunc x = (
return x + x
)
Map A myFunc -- => #(0, 2, 4, 6, 8)
Indices
The Indices operator creates an array containing all the valid indices of a given array. In contrast to MaxScript, array indexing in MCG is 0-based instead of 1-based. This means that the first item in an MCG array is indexed at 0 instead of at 1.
Note that we can achieve the same result as the graph above by connecting a Count operator to a Range operator.
-- Indices
function Indices A = (
result = #()
for i=0 to (A.count-1) do (
append result i
)
return result
)
-- Initialize an array of 8 values
A = #(5, 2, 6, 20, 4, 9, 16, 100)
Indices A -- => #(0, 1, 2, 3, 4, 5, 6, 7)
Filter
The Filter operator iterates over an array, and only keeps the items which fulfill the given condition function. For example, you can use the Filter operator to remove values below a certain threshold, or to keep a set of faces whose normals point up in +Z.
-- Filter
-- An underscore is added to "_Filter" because "filter" is a keyword in MaxScript.
function _Filter A fxn = (
result = #()
for item in A do (
cond = fxn item
if cond == true then
append result item
)
return result
)
-- Initialize an array of values.
A = #(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
-- Declare a filtering function which returns true if the input is even.
function filterFn x = (
return mod x 2 == 0
)
_Filter A filterFn -- => #(0, 2, 4, 6, 8)
At
The At operator returns the value at the specified index in the array. It behaves in the same way as the subscript operator “[ ]” in myArray[i]. Recall that array indexing in MCG is 0-based while in MaxScript, it is 1-based.
-- At
-- An underscore is added to "_At" because "at" is a keyword in MaxScript.
function _At A index = (
-- (!) MaxScript array indexing is 1-based so we’ll be adding 1 to the given index.
result = A[index+1]
return result
)
-- Initialize an array of values, and the index of the value we want to retrieve.
A = #(5, 2, 6, 20, 117, 9, 16, 100)
index = 4 -- 0-based index in A
_At A index -- => 117
Combine
The Combine operator acts like the Map operator, but requires two arrays instead of one. Each pair of items between these arrays is applied to the given function to produce a new array of values.
Keep in mind that the function you define for the Combine operator must expose two arguments (i.e. two unconnected connectors). Furthermore, the order of these arguments must match the order in which the arrays are connected into the Combine node. Specifically, the type of the first array “xs” must match the type of the first argument in the function. In the same way, the type of the second array “ys” must match the type of the second argument in the function.
We cover how to distinguish the order of function arguments in The Low Poly Modifier - Part 2, but if you want a quick summary, the idea is to traverse the function backwards in a “Depth-first Top-to-Bottom” manner. Begin your traversal at the end of the function, and continuously visit the incoming nodes prioritizing depth first (i.e. going backwards), then visiting the slots top-to-bottom. The order in which you encounter the unconnected connectors corresponds to the argument order of the function. During your traversal, ignore the arguments you have already discovered.
As a counterexample to the graph above, if we swapped the “xs” and “ys” inputs of the Combine node, the graph would not compile successfully because the type of the first array (IArray) would not match the type of the first argument (Int32). Likewise, the type of the second array (IArray) would not match the type of the second argument (Single).
-- Combine
function Combine A B fxn = (
result = #()
for i=0 to (A.count-1) do (
itemA = A[i+1] -- Adding 1 because MaxScript array indexing is 1-based.
itemB = B[i+1] -- Adding 1 because MaxScript array indexing is 1-based.
tmp = fxn itemA itemB
append result tmp
)
return result
)
-- Declare the function to apply on each iteration of Combine.
function myFunction x y = (
return x * y
)
-- Initialize an array of integer values and an array of random values
A = #(139, 462, 521, 206, 877, 99)
B = ArrayOfFunction myRandomFunction A.count
Combine A B myFunction -- => #(17.9539, 349.924, 510.956, 143.505, 314.525, 43.2008)
Creating an Indexed For Loop - Indices, Map, At
The Combine operator can be a bit tricky to work with, especially if you’re not familiar with functions containing two arguments. As an alternative, you can emulate an index-based for loop to access items in your arrays. To do this, use the Indices, Map and At operators as follows:
ZipToTuple
Another alternative to the Combine operator is to use ZipToTuple with a Map. The idea is to first “Zip” the items contained in the two arrays into an array of pairs (IArray). You can then use a Map to transform each Tuple2. Use the PairItem1 and PairItem2 operators to obtain the contents of the Tuple2, and continue your work from there.
Ultimately, the choice between Combine, ZipToTuple, or indexed for loops depends on your personal preference, with the caveat that in some corner cases, an indexed for loop might cause unexpected subgraph re-evaluations, which can be slower than the other alternatives.
GenerateN
The GenerateN operator creates an array of a fixed size by repeating the same function on the current value to generate the next value in the array. In the Horns and Transforms tutorial, we showed how to use the GenerateN operator to construct an evolving array of transformation matrices to define the shape of the horn.
-- GenerateN
function GenerateN first count nextFn = (
result = #()
current = first
for i=0 to (count-1) do (
-- Append the current value to the result.
append result current
-- Compute the next result based on the current value.
if (i > 0) then (
current = nextFn current
)
)
return result
)
-- Declare the function which GenerateN will use to create the next value in the array.
function myFunction x = (
return x + 10
)
-- The initial value of GenerateN will be 0
init = 0
-- Repeat GenerateN 5 times.
n = 5
GenerateN init n myFunction -- => #(0, 10, 20, 30, 40)
Generate
The Generate operator is very similar to GenerateN, with the exception that it stops building the array when the supplied condition function returns False. In the example below, the Generate operator will continue building the array as long as the current value is less than 50.
-- Generate
function Generate first conditionFn nextFn = (
result = #()
current = first
while true do (
-- Stop looping if the condition function returns false.
cond = conditionFn current
if cond == false then exit
-- Append the current value to the result.
append result current
-- Compute the next result based on the current value.
current = nextFn current
)
return result
)
-- Declare the condition function which Generate will use to continue iterating.
function myConditionFunction x = (
return x
)
-- Declare the function which Generate will use to create the next value in the array.
function myFunction x = (
return x + 10
)
-- The initial value of Generate will be 0
init = 0
Generate init myConditionFunction myFunction -- => #(0, 10, 20, 30, 40)
Aggregate
The Aggregate operator iterates over each item in the array and accumulates (or “aggregates”) a value by applying the given function. The function you connect to the Aggregate must expose two arguments. The first argument is the currently “accumulated” value, and the second argument is the current value in the array.
For example, if you open the Sum compound, you’ll notice it was implemented with an Aggregate.
-- Aggregate
function Aggregate xs init fxn = (
accumulated = init
for current in xs do (
accumulated = fxn accumulated current
)
return accumulated
)
-- Declare the function to apply on each iteration of the Aggregate.
function myFunction acc current = (
return acc + current
)
-- Initialize an array of integers.
A = #(139, 462, 521, 206, 877, 99)
Aggregate A 0 myFunction -- => 2304
Partial Sum with Aggregate
The Aggregate operator can also be used to implement a Partial Sum array, whereby each item in the Partial Sum array corresponds to the sum of the items up to that point in the original array. This can be useful to keep track of the current distance at each point along a path. In this case, the accumulated value is actually a growing array, to which we append the latest computed value.
-- PartialSum using Aggregate
-- Declare the function to apply on each iteration of the Aggregate.
function partialSumFn acc current = (
local lastItem
if acc.count > 0 then
lastItem = acc[acc.count]
else
lastItem = 0
tmp = (lastItem + current)
append acc tmp
return acc
)
-- Initialize an array of integers.
A = #(139, 462, 521, 206, 877, 99)
Aggregate A #() partialSumFn -- => [139, 601, 1122, 1328, 2205, 2304]
Flatten
The Flatten operator joins all the subarrays contained inside an array.
It is particularly useful when you want to “flatten” the result of nested iterations. For example, we required the Flatten operator in the Horns and Transforms tutorial to flatten the nested arrays of points into one continuous array of points for the QuadMeshStrip operator.
-- Flatten
function Flatten xss = (
result = #()
for subarray in xss do (
join result subarray
)
return result
)
-- Initialize an array of subarrays.
A = #( #(139, 139), #(462, 462), #(521, 521) )
Flatten A -- => #(139, 139, 462, 462, 521, 521)
Repeat
At first glance, the Repeat operator may seem like the closest implementation of a for loop. However, once you get familiar with MCG, you’ll realize that its usage is almost always eclipsed by one of the more efficient operators we listed above. To illustrate this point, we translated the following MaxScript code containing a standard for loop using the Repeat operator, and alternatively using the Map operator.
A = #(139, 462, 521, 206, 877, 99)
result = #()
for i=0 to (A.count-1) do (
x = A[i+1] -- add 1 because MaxScript array indexing is 1-based.
tmp = x + x
append result tmp
)
result -- => #(278, 924, 1042, 412, 1754, 198)
Using Repeat:
Using Map:
Here’s a more formal MaxScript translation of the Repeat operator.
-- Repeat
function Repeat init n bodyFn = (
current = init
for i=0 to (n-1) do (
current = bodyFn current i
)
return current
)
-- Define an array of values
A = #(139, 462, 521, 206, 877, 99)
-- Declare the function to repeat on each iteration.
function myFunction currentArray index = (
x = A[index+1] -- add 1 because MaxScript array indexing is 1-based
tmp = x + x
append currentArray tmp
return currentArray
)
Repeat #() A.count myFunction -- => #(278, 924, 1042, 412, 1754, 198)
While
The While operator acts like the Generate operator, however it does not necessarily generate an array, and acts much like a general-purpose while loop. In the graph below, we’re using a while operator to generate a limited Fibonacci sequence.
-- Fibonacci sequence using While
-- An underscore is added to "_While" because "while" is a keyword in MaxScript.
function _While init conditionFn bodyFn = (
current = init
while true do (
cond = conditionFn current
if cond == false then exit
current = bodyFn current
)
return current
)
-- Declare the condition function which will determine if the while operator should continue iterating.
function myConditionFunction x = (
lastItem = x[x.count]
return lastItem
)
-- Declare the body function which will be invoked on each iteration.
function myFunction x = (
-- Add the last and second to last items together to produce the next Fibonacci number.
f_n1 = x[x.count]
f_n2 = x[x.count-1]
tmp = f_n1 + f_n2
append x tmp
return x
)
-- Initialize an array containing the initial values of the Fibonacci sequence 0 and 1.
A = #(0,1)
_While A myConditionFunction myFunction -- => #(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233)
Instructions: Extract the file anywhere on your filesystem, then go to Scripting > Install Max Creation Graph (.mcg) Package, and select MCGArrayTests.mcg in the extracted location. Once the package is successfully installed, open the MaxScript listener (F11) and type MCGArrayTests() (note the “s” at the end of Tests). This will run the MCG array test cases we built for this tutorial, and will print the results to the Listener. Feel free to open the MCGArrayTests.maxtool file under C:\\Users\\\\Autodesk\\3ds Max 2016\\Max Creation Graph\\Tools\\Downloads to explore each construction in more detail.
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# A Brief Introduction To Structural Analysis
Structural Analysis is basically the study of the changes in the properties of physical structures and their elements due to the forces applied upon them. These forces are usually applied with regard to the movements of masses and their constituent parts.
Structural Analysis is an important part of construction and design. It takes into consideration the relationship of structure to its surroundings, its purpose, its integrity, its load-bearing features, its maintenance needs, and its future implications. This is because a construction project will inevitably be incomplete if the properties of the building are not well understood at the inception of the construction project. Moreover, without understanding the relationships of properties to one another, it is difficult to design, construct, maintain and repair a structure.
Structural analysis covers the design of buildings, bridges, dams, tunnels, and highways. It is used by architects, engineers, surveyors, engineers, inspectors and builders to determine the properties that determine the stability of the building or its component parts. Without understanding what properties constitute the building, it is difficult for the architect to design and construct the building in a way that can minimize its load bearing capacity. Likewise, it is difficult for the engineer to design a building in such a way that its integrity can be maintained even under high loads. It is also very difficult for the contractor to manage and maintain the building.
However, structural analysis is not only concerned with the properties of buildings. It also covers the relationships of the structures with their surroundings including climate, soil type and elevation, gravity, and air quality.
Structural analysis covers the study of structural equilibrium. This refers to a state where the properties of the building or its component parts are equal to or greater than those of their surroundings and are also symmetric about the direction of gravity and the orientation of the building relative to the earth’s surface. In order to determine whether the equilibrium is balanced or not, structural analysis of equilibrium is used in a design of a building.
Structural Analysis can be divided into several types and subtypes. It includes dynamic, static, elastic, equilibrium, dynamic equilibrium, structural equilibrium, and equilibrium stability. Dynamic equilibrium is a subtype of equilibrium stability and describes a state when the equilibrium of the building is achieved with respect to the external forces acting on the building such as wind, gravity and humidity.
Static analysis describes the equilibrium of a building when the building has been constructed and its characteristics remain unaffected by forces acting on it. It involves testing the construction stability at the beginning of a construction project and is necessary prior to its installation.
Elastic stability analysis and equilibrium analysis are the two main forms of structural analysis. Dynamic equilibrium and elastic stability analysis both involve balancing the equilibrium and the forces on the building.
The equilibrium stability of a building refers to the stability of the building following its installation. Equilibrium stability can only exist if the forces acting on the building are equal to or greater than those of its environment. In equilibrium stability, all the elements and the building itself are balanced. In other words, it is the best equilibrium stability.
A second form of structural analysis is called equilibrium stability of equilibrium. In this form, the components of the building are balanced after the installation of the building. The difference between equilibrium stability and dynamic stability is that in equilibrium stability, the balance of the building or components is determined at the time of installation, while in dynamic stability, it is determined at the time of the project’s inception. The building’s equilibrium is determined as a result of tests.
To determine the equilibrium stability, a variety of tests are used to identify the relationship of forces with the equilibrium. These tests include determining the relationship of forces on a building with the orientation of its components and the building’s axis. Also, tests can be used to determine the strength of the structure against the forces that are acting on it.
Structural analysis of equilibrium is one of the most important tools that are used in designing buildings. It involves the application of several methods and techniques. These methods include load-bearing, design stability, and load-bearing stability.
### Related posts:
A Brief Introduction To Structural Analysis
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posted by .
By considering a wire of radius 'r', lenght 'l' and resistivity 'e' through which a current I flows, show that
a) the rate of production of heat by it is I^2*P*l / pi*r^2
b) the rate of loss of heat from its surface is 2pi*r*t*h , where 'h' is the heat lost per unit area of surface per second.
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# Implicit Differentiation - Inverse Trigonometric Functions
If $\displaystyle \tan^{-1} \left(x^2y\right)=3x+6y,$ what is $\frac{dy}{dx} ?$
Given the function $\tan^{-1}(4xy)=5,$ What is $\frac{dy}{dx}?$
If $\displaystyle \cos^{-1} \left(\frac{6x-y}{3x+y}\right)=e^a,$ what is $\frac{dy}{dx} ?$
If $\displaystyle \sin^{-1} (7x^2-y^2)=\ln 6,$ what is $\frac{dy}{dx} ?$
Given the function $\cos^{-1}(3x+11y)=12,$ what is $\frac{dy}{dx} ?$
×
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find the Z-Score using the BMI data by calculating the Standard Deviation on the Sample and the Average BMI of the sample. Discuss briefly what this Z-Score reveals about the BMI data.
## Expert Solution Preview
Introduction:
The Z-Score is a commonly used statistical measurement that helps to understand how an individual measurement compares to the average measurement of a group. In the context of BMI data, calculating the Z-Score can provide insights into how a person’s BMI compares to the average BMI of a given sample population. By calculating the Standard Deviation and Average BMI of the sample, we can determine the Z-Score and analyze what it reveals about the BMI data.
To find the Z-Score, we first need to calculate the Standard Deviation of the BMI data on the sample and the Average BMI of the sample population. The Standard Deviation measures the dispersion of the BMI values, while the Average BMI provides the central tendency of the data.
Once we have these values, we can calculate the Z-Score using the formula:
Z-Score = (Individual BMI – Average BMI) / Standard Deviation
The Z-Score is essentially the number of standard deviations an individual BMI measurement is away from the average BMI of the sample population. It indicates whether a person’s BMI is above or below the average value and by how many standard deviations.
Interpreting the Z-Score can reveal valuable information about the BMI data. If the Z-Score is positive, it means the individual’s BMI is above the average of the sample population. Conversely, a negative Z-Score indicates a BMI below the average. Additionally, the magnitude of the Z-Score tells us how far the individual’s BMI is from the sample average in terms of standard deviations.
For example, if we find a Z-Score of 1.5 for a particular individual, it means their BMI is 1.5 standard deviations above the average BMI of the sample population. This suggests that the individual has a higher BMI compared to their peers.
On the other hand, if we calculate a Z-Score of -0.7, it signifies that the individual’s BMI is 0.7 standard deviations below the average BMI of the sample population. This implies that the individual has a lower BMI compared to others in the sample.
In summary, the Z-Score provides a standardized measure for comparing individual BMI values to the average BMI of a sample population. It offers insights into the relative position of an individual’s BMI value within the sample and helps identify whether it is below, above, or close to the average.
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
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Ineq. #3:
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Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
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• interval notation calculator | 975 | 4,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.97303 |
https://schoollearningcommons.info/question/find-the-area-of-metal-sheet-required-to-prepare-a-pipe-of-length-3cm-and-radius6-28-3-14-plzz-a-24485309-45/ | 1,653,504,742,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00227.warc.gz | 584,117,663 | 14,994 | ## find the area of metal sheet required to prepare a pipe of length 3cm and radius6.28. (π=3.14) plzz answer if anyone knows
Question
find the area of metal sheet required to prepare a pipe of length 3cm and radius6.28. (π=3.14)
plzz answer if anyone knows it’s correct answer
in progress 0
5 months 2022-01-06T00:53:26+00:00 1 Answer 0 views 0
## Answers ( )
118.3152 cm2
Step-by-step explanation:
length=3cm | 136 | 422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.703651 |
http://freesourcecode.net/socialtags/filter-theory | 1,524,626,617,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00236.warc.gz | 129,173,837 | 12,803 | # Least mean square for system identification in matlab
The following Matlab project contains the source code and Matlab examples used for least mean square for system identification. Least mean squares (LMS) algorithms are a class of adaptive filter used to mimic a desired filter by finding the filter coefficients that relate to producing the least mean squares of the error signal (difference between the desired and the actual signal).
# Adaptive degree polynomial filter (savitzky golay filter) in matlab
The following Matlab project contains the source code and Matlab examples used for adaptive degree polynomial filter (savitzky golay filter). function polynomial_degree=adpf(data_frame) Adaptive-Degree Polynomial Filter (Savitzky-Golay Filter) Commonly, the degree of the fitting polynomial for the Savitzky-Golay filter is fixed.
# The even length savitzky golay filter in matlab
The following Matlab project contains the source code and Matlab examples used for the even length savitzky golay filter. The even-length Savitzky-Golay filter (i.
# Z transform of 1d & 2d savitzky golay smoothing and differentiation filter in matlab
The following Matlab project contains the source code and Matlab examples used for z transform of 1d & 2d savitzky golay smoothing and differentiation filter. Compute and plot 1D and 2D Savitzky-Golay smoothing and differentiation filters.
# 2d & 3d spectra of savitzky golay smoothing and differentiation filters in matlab
The following Matlab project contains the source code and Matlab examples used for 2d & 3d spectra of savitzky golay smoothing and differentiation filters. This zip-file contains two m-files that generate 2D spectral plots of smoothing and differentiation filters.
# Image interpolation in matlab
The following Matlab project contains the source code and Matlab examples used for image interpolation. This function performs the interpolation of an image filtering it with the specified filter coefficients which are given as input. Any filter can be used as input.
# Fast susan feature detection and non linear filter in matlab
The following Matlab project contains the source code and Matlab examples used for fast susan feature detection and non linear filter. This is a fast implementation of the SUSAN feature detection algorithm described at: www.
# Non local means filter in matlab
The following Matlab project contains the source code and Matlab examples used for non local means filter. Implementation of the Non-Local Means Filter proposed by Buades et al. for robust image denoising. typical usage: fima=nlmeans(ima,5,2,sigma); being sigma the noise standard deviation
# This program designs a prototype filter for use in a quadrature mirror filter filterbank. in matlab
The following Matlab project contains the source code and Matlab examples used for this program designs a prototype filter for use in a quadrature mirror filter filterbank. . This program designs a prototype filter for use in a Quadrature Mirror Filter filterbank.
# Simulating cic filtering decimation without the filter design toolbox in matlab
The following Matlab project contains the source code and Matlab examples used for simulating cic filtering decimation without the filter design toolbox. Recently needed to see the effects of a CIC decimating filter but realized that with my new license of Matlab, I did not have access to the Filter Design Toolbox.
# Fast savitzky golay filter as multi-threaded c-mex in matlab
The following Matlab project contains the source code and Matlab examples used for fast savitzky golay filter as multi-threaded c-mex . fSGolayFilt is a fast polynomial smoothing filter for uniformly spaced signals.
# Gaussian filter, determination of integer parameters in matlab
The following Matlab project contains the source code and Matlab examples used for gaussian filter, determination of integer parameters. Within a sigma range separable filters with integer parameters are sought. The filters have a sum which is an integer power of 2 in order to divide the filter result by shifting the bits.
# 3d max min filter in matlab
The following Matlab project contains the source code and Matlab examples used for 3d max min filter. I=Maxin(I,n,f) I is input 3D image n is neighbourhood size. for example 3 for 3*3 f=1 for max filter f=0 for min filter
# Savitzky golay smooth differentiation filters and filter application in matlab
The following Matlab project contains the source code and Matlab examples used for savitzky golay smooth differentiation filters and filter application. Contains savitzkyGolay.
# First order statistics filter (sigma filter lee's filter) in matlab
The following Matlab project contains the source code and Matlab examples used for first order statistics filter (sigma filter lee's filter). fcnFirstOrderStatisticsFilter performs noise filtering on an image based on using First Order Local Statistics around a prespecified pixel neighborhood.
# Nth oct hand arm & ac filter tool box in matlab
The following Matlab project contains the source code and Matlab examples used for nth oct hand arm & ac filter tool box. The Nth Octave Hand Arm and AC Filter Tool Box is based on the octave toolbox by Christophe Couvreur which is on the Matlab File Exchange ID number 69.
# Generate coefficients for 1d savitzky golay smoothing filters in matlab
The following Matlab project contains the source code and Matlab examples used for generate coefficients for 1d savitzky golay smoothing filters. INPUT: order = order of the polynomial interval = the window size; *MUST* smaller than size(vIN) OUTPUT: c(ss,nn)= a matrix Rows corresponds to coefficients for the target point, g_t, at different location of the window (up >> down: g_t moves from left to right).
# Savitzky golay smoothing filter in matlab
The following Matlab project contains the source code and Matlab examples used for savitzky golay smoothing filter. The function smooths a input vector using Savitzky-Golay smoothing filter based on the degree of polynomial and the length of moving windows
# Savitzky golay smoothing filter for 3d data in matlab
The following Matlab project contains the source code and Matlab examples used for savitzky golay smoothing filter for 3d data. This function provide a Savitzky-Golay smoothing filter for 3D data.
# Savitzky golay smoothing filter for 2d data in matlab
The following Matlab project contains the source code and Matlab examples used for savitzky golay smoothing filter for 2d data. This function provide a Savitzky-Golay smoothing filter for 2D data.
# Uniform filter bank in matlab
The following Matlab project contains the source code and Matlab examples used for uniform filter bank. This function can be used to design a uniform filter bank (with M analysis and M synthesis filters) so that the perfect reconstruction is almost achieved.
# Image filter by spatial correlation in matlab
The following Matlab project contains the source code and Matlab examples used for image filter by spatial correlation. We propose a novel algorithm for removing noise in gray-levels and color images.
# H infinity optimal fractional delay filter in matlab
The following Matlab project contains the source code and Matlab examples used for h infinity optimal fractional delay filter. This is a code for designing H-infinity optimal fractional delay filters proposed first in the following paper: M.
# Designs a passive filter ( input filter ) for power applications in matlab
The following Matlab project contains the source code and Matlab examples used for designs a passive filter ( input filter ) for power applications. function [ stg_Ld, stg_Rd, att ] = filter_damping_design( stg_L, stg_C, fs, Zmax) Written by Dr.
# Filter characteristics in matlab
The following Matlab project contains the source code and Matlab examples used for filter characteristics. Using this Matlab script you get the amplitude response, the phase response,the frequency vector, the complex transfer function by knowing the filter coefficients:
# Filter for discrete time signal in matlab
The following Matlab project contains the source code and Matlab examples used for filter for discrete time signal. [y_fl, Y_fl, f, Y, f_fl, fl_function, filter]=f4ts(y,Fs,filter,plot1) filter that clean the signal from diffrent kind of noise.
# Minimum phase filter in matlab
The following Matlab project contains the source code and Matlab examples used for minimum phase filter. A system with all of its poles and zeroes strictly inside the unit circle is called a MINIMUM Phase filter.
# Linear extended whitening filters (demo) in matlab
The following Matlab project contains the source code and Matlab examples used for linear extended whitening filters (demo). Two functions that demonstrate computation of dual-function whitening filters termed Linear Extended Whitening Filters (EWF) such that: 1.
# Two dimensional trilateral filter in matlab
The following Matlab project contains the source code and Matlab examples used for two dimensional trilateral filter. TRILATERALFILTER Two-dimensional trilateral filter. outputImage = trilateralFilter(inputImage,sigmaC,epsilon) filters the data in inputImage with the 2-D trilateral filter by P. Choudhury and J. Tumblin, http://www.cs.northwestern.edu/~jet/publications.html This implementation is a conversion of the OpenCV implementation by Tobi Vaudrey
# Steerable filter in matlab
The following Matlab project contains the source code and Matlab examples used for steerable filter. This packages implements the steerable filter of the first derivative and the second of Gaussian function of the paper W.
## Pages
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http://www.algebra.com/algebra/homework/playground/lessons/test.faq?hide_answers=1&beginning=6885 | 1,369,175,001,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700795821/warc/CC-MAIN-20130516103955-00092-ip-10-60-113-184.ec2.internal.warc.gz | 325,914,462 | 11,041 | # Questions on test answered by real tutors!
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Question 484238: 1. Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus 8 is the sum of their squares. Thank you very much for the answer ! :) Click here to see answer by MathTherapy(1422)
Question 484461: The volume of a refrigerator is x^3-3x^2-16x-12. Its height is x + 2. What are the other two dimensions? Click here to see answer by ewatrrr(10682)
Question 484559: There is only one real number that is the sum of 4 and 6. Click here to see answer by richard1234(5390)
Question 484741: Hi. I was wondering if someone has a list of THE MOST IMPORTANT SAT questions. I really need to practice math section. Can anyone kindly send me the link or something. Thanks Click here to see answer by jim_thompson5910(28546)
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Question 486663: If 2a/a + 1/a = 4, then a = ? Click here to see answer by bucky(2189)
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Question 486659: A store charges \$49 for a pair of pants. This price is 40% more than the amount it costs the store to buy the pants. After a sale, any employee is allowed to purchase any remaining pairs of pants at 30% off the store's costs. How much would it cost an employee to purchase the pants after the sale? Click here to see answer by josmiceli(9668)
Question 486688: Find P ∩ Q. Write in correct set notation. In the diagram: circle P - inside clockwise p,e,a,l. circle Q - inside clockwise b,s,t. Outside the diagram ∪ Click here to see answer by chessace(471)
Question 486705: Use the definition of the derivation to show that (d/dx)(cos x)= -sin x. You can use the following 3 facts: lim h->0 (sin h)/h =1 lim h->0 (cos h -1) / h = 0 cos(θ + )= cosθcos - sinθsin Click here to see answer by richard1234(5390)
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Question 486985: What is the ordered pair for f(-5)=8 and identify the x and y values. Click here to see answer by jim_thompson5910(28546)
Question 487035: How to diferentiate a quadratic equation like x^2 + 5x + 6? How to solve the same equation if equeted to 0? Click here to see answer by richard1234(5390)
Question 487103: This spinner is spun 36 times. The spinner landed on A 6 times, on B 21 times, and on C 9 times. Compute the empirical probability that the spinner will land on B. Please help me!!Thanks a lot Click here to see answer by Theo(3458)
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Question 487775: A lies on mon, tues, wed and speak truths on other days, B lies on Thurs, Fri, Sat and speaks truths on other days. One day a said I lied today and B said I too lied today. What is the day? Click here to see answer by solver91311(16877) | 1,314 | 4,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2013-20 | latest | en | 0.912596 |
https://www.lambdatest.com/automation-testing-advisor/python/autotest_python-apply_constraint | 1,708,480,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00514.warc.gz | 911,215,288 | 46,398 | # How to use apply_constraint method in autotest
Best Python code snippet using autotest_python
singlepointcalc.py
Source:singlepointcalc.py
`1"""This test makes sure that the forces returned from a2SinglePointCalculator are immutable. Previously, successive calls to3atoms.get_forces(apply_constraint=x), with x alternating between True and4False, would get locked into the constrained variation."""5from ase.lattice.surface import fcc1116from ase.calculators.emt import EMT7from ase.io import read8from ase.constraints import FixAtoms9def check_forces():10 """Makes sure the unconstrained forces stay that way."""11 forces = atoms.get_forces(apply_constraint=False)12 funconstrained = float(forces[0, 0])13 forces = atoms.get_forces(apply_constraint=True)14 forces = atoms.get_forces(apply_constraint=False)15 funconstrained2 = float(forces[0, 0])16 assert funconstrained2 == funconstrained17atoms = fcc111('Cu', (2, 2, 1), vacuum=10.)18atoms[0].x += 0.219atoms.set_constraint(FixAtoms(indices=[atom.index for atom in atoms]))20# First run the tes with EMT and save a force component.21atoms.set_calculator(EMT())22check_forces()23f = float(atoms.get_forces(apply_constraint=False)[0, 0])24# Save and reload with a SinglePointCalculator.25atoms.write('singlepointtest.traj')26atoms = read('singlepointtest.traj')27check_forces()28# Manually change a value.29forces = atoms.get_forces(apply_constraint=False)30forces[0, 0] = 42.31forces = atoms.get_forces(apply_constraint=False)...`
## Automation Testing Tutorials
Learn to execute automation testing from scratch with LambdaTest Learning Hub. Right from setting up the prerequisites to run your first automation test, to following best practices and diving deeper into advanced test scenarios. LambdaTest Learning Hubs compile a list of step-by-step guides to help you be proficient with different test automation frameworks i.e. Selenium, Cypress, TestNG etc. | 474 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.630889 |
https://research.stlouisfed.org/fred2/series/MORG054LEIH | 1,440,997,970,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065534.46/warc/CC-MAIN-20150827025425-00118-ip-10-171-96-226.ec2.internal.warc.gz | 880,464,304 | 18,342 | # All Employees: Leisure and Hospitality in Morgantown, WV (MSA)
2015-07: 7.3 Thousands of Persons
Monthly, Seasonally Adjusted, MORG054LEIH, Updated: 2015-08-21 1:02 PM CDT
1yr | 5yr | 10yr | Max
The data services of the Federal Reserve Bank of St. Louis include series that are seasonally adjusted. To make these adjustments, we use the X-12 Procedure of SAS to remove the seasonal component of the series so that non-seasonal trends can be analyzed. This procedure is based on the U.S. Bureau of the Census X-12-ARIMA Seasonal Adjustment Program. More information on this program can be found at http://www.census.gov/srd/www/x12a/.
The seasonal moving average function used is that of the Census Bureau’s X-11-ARIMA program. This includes a 3x3 moving average for the initial seasonal factors and a 3x5 moving average to calculate the final seasonal factors. The D11 function is also used to output the entire seasonally adjusted series that is displayed. For specific information on the SAS X-12 procedure, please visit their website: http://support.sas.com/documentation/cdl/en/etsug/60372/HTML/default/viewer.htm#etsug_x12_sect001.htm.
Restore defaults | Save settings | Apply saved settings
Recession bars:
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(a) All Employees: Leisure and Hospitality in Morgantown, WV (MSA), Thousands of Persons, Seasonally Adjusted (MORG054LEIH)
Integer Period Range:
copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` Federal Reserve Bank of St. Louis, All Employees: Leisure and Hospitality in Morgantown, WV (MSA) [MORG054LEIH], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/MORG054LEIH/, August 31, 2015. ```
Retrieving data.
Graph updated. | 632 | 2,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-35 | latest | en | 0.862405 |
https://cp3.irmp.ucl.ac.be/projects/madgraph/wiki/MW_development?version=4 | 1,586,129,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00511.warc.gz | 422,618,637 | 8,289 | Version 4 (modified by omatt, 8 years ago) (diff)
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LowMassSMHiggs
## Ressource Material
1. Ongoing Paper: paper
## Code V2.5
After a new discussion (September 09) between Olivier and Pierre and inspired by a discussion to Fabio. We want to include the multi-channel techniques in MW. In the previous code (2.0), an overconstrained systems lead to unaligned peak, but also to different change of variable, different because they are not aligning the same peak. In consequence, the multi-channel techniques is possible to performed.
In principle, the multi-channel is pointless because the correct area of the PS should be selected by any of those change of variables (the constraints should be consistent) but MW computes also weights for different permutation/background where those constraints are in conflicts, causing slow integration. For those events/permutations, the multi-channel can contibutes to fastenize the integral (a lot?)
### Development of Code
In a first step, (in progress), we will keep the different changes of variables provides by the python supra-code and run in multi-channel in those cases. Some script are added in the Python supra-script in order to compute the weights associated to each channel. The 'MadEvent trick' is used in order to choose wisely the weights of each channel.
A second step, will be to check to modify the supra-script in order to create additional change of variables in order to have all peaks aligned in at least one change of variables
We could(should?) think to fuse the supra code with Madgraph5 in order to have a full multi-channel program (multi-channel on the square matrix element and on the peak position). But this will be MG5 dependant and has some drawback (impossibility to modify Transfer Function after the diagram generation)
### Test of V2.5
Test of V2.5 against 2.0:
Process param 2.0 2.5 Comment W production (->e-) W at 80.4 14.36273+-7.3e-05 14.36218 +-1.5e-05 OK W at 100 15.80653 +-8.8e-05 15.80328 +-2.0-05 OK W at 150 17.8815 +- 1.0e-04 17.87489 +-3.0e-05 OK (W>lnu)(W>jj) 80.4 48.27716+- 7e-05 48.255909+-6.6e-05 OK 100 46.64555+- 8e-05 46.632452+- 6.1e-05 OK 150 56.563823+- 6.5e-05 56.575452+-6.9e-05 OK semi-lept tt Mt=150 813.6807+-0.0015 813.6377+-0.0013 OK 152.5 813.3573+-0.0343 812.5640 +-0.0032 OK 155 812.6200+-0.0015 812.1788+-0.0024 OK 157.5 812.8469+-0.0013 812.8812+-0.0015 OK 160 814.1727+-0.0015 814.2527+-0.0047 OK 162.5 815.5738+-0.0013 816.0374+-0.0037 OK 165 819.1933+-0.0019 818.2402+-0.0014 more confident in 2.5 167.5 822.4768+-0.0026 821.9353+-0.0038 OK 170 827.7806+-0.0013 825.9346+-0.0019 more confident in 2.5
### Check of version 2.5
Phase-Space Check
process block V2.5 true value remarque pp>e+e-u A (6.2952+-0.0017)E-5 6.299E-5 bg>(t>b(W>jj))jj A+E 149301+-282 149748 jj(W>jj)jj A+D 3.2503+- 0.0046 3.254 jj(W>eve) A+C 3.244+- 0.008 3.254 jj(t>b(W>eve)) A+B 150800+-1900 149748 jj(W>bt) with (t>b(W>eve)) A+A (5.233+-0.004)E+19 5.22E9 bg>(t>b(W>jj))jj multi-channel 148500+- 1300 149748 (A+DD/A+E/A+11) (5e-17/1e-16/148500) bg>(t>b(W>jj))jj multi-channel 149748 (A+DD/A+E) (W>lve) B (1.32 +-0.08)E-8 1.28E-8 instability in the jacobian (if pz(l)=E(l)) (W>lve)j B (6.2927 +- 0.0039)E-5 6.299E-5 non sensitive to the jac=0 area uu~>(W+>b~(t>b(W+>e+ve)))(W->e-ve~) B+A 5.201 +-0.046 E9 5.22E9 B+B 149000+-500 149748 (W>lv)(W>lv) B+C 3.238+- 0.007 3.254 (W>lv)(Z>mu+mu-) B+D 3.24 +- 0.05 3.254 bg>(t>b(W+>jj))(W->e-ve~) B+E 148100+- 1100 149748 (W>lv)(Z>mu+mu-) B+fuse 3.24 +- 0.04 3.254 (W>lv)(Z>mu+mu-) multi-channel 3.254+-0.003 3.254 (W>svt~(sta->ta-n1)) B+C (both massif) (5.959+-0.006)E-5 5.967E-5 pp>(t>b(W+>e+ve)) C 1.28E-1 6.2982E-005 WRONG tt~ D 5.247 +- 0.021 E+9 5.227E+09 H>WW>e+e-veve~ E 531.567756+-121.365377 3.254 instability in jac ???? H>WWj>e+e-veve~j E 149753 WW>e+e-veve~ F 565.237354+-123.048905 3.254 instability in jac ????? WWj>e+e-veve~j E 149753 bu > td; t>benu C 0.0468 3.257 in progress bu > td; t>benu B 3.25 3.257 in progress
Cross-section Check
process block ratio ME/MW (W>lv) B 0.9991 +- 0.0014 (W>svt~(sta->ta-n1)) B+C 1.003 +- 0.006 SMU SMU F 1.020 +- 0.005 tt> fully leptonique D 1.00056 +- 0.025
## Code V2.0
After a discussion (18 jan 08) between Olivier and Pierre, we have planned to write a new version of the code, with particular emphase on the generic and modular characters of the algorithm. In this new code, a supra-code (python script) will produce a phase-space generator (in fortran) that is specific to a certain topology. This version is currently on CVS (see link above).
### Advantages of V2.0
The main advantage of this new organization of the code resides in the factorization of two difficulties:
• Given a topology, find a phase space generator that is appropriate to ME method
• write a code that select automatically the adequate phase-space parametrization for any topology
In the new code, the main code (which is almost the only topology-dependent file) has a very simple structure, that can be handled easily, even written explicitly for a given topology. Only the way to generate it automatically is intricate. If we choose the supra code to generate this file, it is still very easy to check its contents.
Also in this new version, it will be really easy to implement new blocks or new classes of ECS in the future.
### The supra-code (python script)
This is for sure the most difficult part. The supra code will
• read the topology and record it in an appropriate structure: MadWeightAnalyzer MadWeightDiagramClass
• determine the class of ECS as well as the block structure of each blob, and organize the blocks in the adequate order
• write the main code in fortran (see below) that is the topology-dependent part of the phase-space generator
### The phase-space generator
The associated fortran code will partially result from the supra-code. It will contain several things: PhaseSpaceGenerator
#### I. Generic subroutines
One subroutine for each block and each class of ECS. As discussed in the paper, a block (or a class of ECS) is nothing else than a specific local change of phase-space parametrization. Given the inputs (parameters+dynamical variables), the subroutine associated to a block will solve the change of variables and compute the jacobian. These subroutines are generic so that there is no need to generate them with a supra code.
#### II. The main code
This code will be specific to the topology at work, and generated by the supra-code. The aim of the main code is to generate a phase-space point and compute the associated phase-space weight and jacobian. This will be done according to the following steps:
• generation of invariant masses linked to factorized propagators. This generation will occur in an ascendant way so that we avoid unphysical regions as much as possible. This is particularly useful for topologies like H>WW*, where at least one of the W is off-shell.
• generation of momenta that are factorized (according to the decision made by the supra-code) according to the transfer function. We should generate points according to Gaussian distributions
• generation/resolution of the kinematics associated to the blobs (+ computation of the associated jacobian). These are made of blocks, in the sense discussed in the paper. The kinematics of any blob will be generated/resolved by calling (a certain number of) block subroutines defined in I.
• resolution of the change of variables corresponding to the ECS at work. This will be done by the call of a generic subroutine (see I.)
So the main code will be made of external calls to several blocks and one class of ECS. The number of calls, their ordering, the inputs that are passed through the subroutines depend on the topology. These issues will be handled in the supra-code.
#### III. Status and checks
Each block or ECS has been checked at least once by computing the phase space volume with a phase-space parametrization involving this block.
Block_name Toplology PS volume check Block A ECS B + bl A (massless) 3.78 +/- 0.008 3.89 Block B ECS B + bl B (massless) 0.0166 +/- 0.0001 0.0166 Block B ECS B + bl B (massive) 0.001407 +/- 9e-6 0.001401 Block C ECS B + bl C (massless) (6.28+/-0.014)e-5 6.30e-5 Block C ECS B + bl C (massive) (9.75+/-0.02)e-6 9.74e-6 Block D ECS B + bl D (massless) (6.27+/-0.02)e-5 6.30e-5 Block E ECS B + bl E (massless) 0.0165 +/-0.0001 0.0166 ECS A ECS A (3FP, massless) (6.301+/-0.01)e-5 6.299e-5 ECS A ECS A (3FP, massive) (9.741+/-0.01)e-6 9.755e-6 ECS B ECS B (3FP, massive) (9.740+/-?)e-6 9.755e-6 ECS C ECS C (3FP, massless) (6.293+/-0.003)e-5 6.299e-5 ECS C ECS C (3FP, massive) (1.716+/-0.001)e-5 1.715e-5 ECS D ECS D (massless) 43.73+/-12.41 44.60 ECS D ECS D ( massive) 692.29+/-17.88 694.28 ECS E ECS E (massless) 0.01662+/-? 0.01662 ECS E ECS E ( massive) 0.001627 0.001631 ECS F2 ECS E (massless) 0.01664+/-2e-5 0.01662 ECS F2 ECS E (massive) 0.001631+/-3e-6 0.001631
#### IV. Differential weight
A histrogram package (dbook) has been set up in MadWeight, in order to draw differential distributions of the weight. One first application is the reconstruction of top quark pair invariant mass in tt>2l events.
-
--- MadWeight: the general code
#### Status and checks
Checked processes (after the fuse of MW in MG trunk):
Process status gg> (h>(w+>e+ ve )(w- > mu- vm~ ) ) OK! Olivier gg8;gt;( h>(w+>e+ ve )(w- > e- ve~ ) ) -> event Olivier gg>( h>(w+>e+ ve (w- > e- ve~ ) ) -> event Olivier pp>(er+>e+ n1) (er->e- n1) OK! Pierre pp>(mur+>mu+ n1) (mur->mu- n1) ) not checked yet Pierre pp>(t>b(W+>e+ve))(t~>b~(W->mu-vm~)) Ok! Pierre pp>(t>b(W+>mu+vm)) (t~>b~(W->mu-vm~)) not checked yet pp>(t>b(W+>mu+vm)) (t~>b~(W->mu-vm~)) not checked yet pp>(t>b(W+>mu+vm)) (t~>b~(W->jj)) not checked yet pp>(t>b(W+>jj)) (t~>b~(W->jj)) not checked yet
#### Things to do for the next Release
1. Grid initialization.
2. relative weight between solution in Blob/ECS
#### Different Idea
1. Grid initialization:
The transfer functions coud(should?) have their own grid this could be achieved by integrating the one-dimensional transfer functions. If this is needed (probably only for some hard cases), this could be done on an event-by-event basis. In other case a typical events could do the job once.
2. Tau decay;
it could be possible to have a special change of variable for the tau>lepton decay, using the coliniarity of the decay. A starting grid for the neutrinos angles could be used if we want to stay completely general.
3. Transfer function:
we could use a supra code trick for the transfer functions. This will give gain of time (~1%) but the main goal is to simplify the way to add a transfer functions in special case. -> Done in Version 2.5
4. Permutation:
we could/should develop a python code treating the permutation (supressing the identical permutation). This could be achieved with a MG stand alone test.
5. grid for solution choices:
For the moment each solution (in change of variables) has the same probability to choose. Adding an adaptative grid on this could be interesting (->include those variable in VEGAS). An other idea will be to use the jacobian in order to improve our choices
6. Adaptative precision:
For the moment the weight in each SubProcess the weights are computed with the same precision, in a normalized mode or in order to have refine procedure, a adaptative precision could be cut the processing time by a significative factor.
-- Main.OlivierMattelaer - 24 Nov 2008 | 3,577 | 11,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-16 | latest | en | 0.901637 |
https://www.inchcalculator.com/height-converter/145cm/ | 1,725,721,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00619.warc.gz | 808,031,045 | 15,661 | # What is 145 cm in Feet and Inches?
Are you trying to figure out how tall 145 cm is in feet and inches?
## How tall is 145 cm in feet:
4 ft 9.1 in
(4' 9.1")
4.76 ft
57.09 in
145 centimeters is equal to 4' 9.1" in feet and inches.
## How to Convert 145 Centimeters to Feet and Inches
You can convert 145 centimeters to feet easily using a height converter or manually by following a few easy steps.
Step One: divide the height in centimeters by 2.54 to convert to inches.
145 cm ÷ 2.54 = 57.09 in
Step Two: divide the height in inches by 12 to find the number of feet. Take the whole number as the number of feet and note the remainder for the next step.
57.09 in ÷ 12 = 4.76 ft
whole feet = 4 ft
remainder = 0.76 ft
Step Three: multiply the remainder above by 12 to find the remaining inches.
0.76 ft × 12 = 9.09 in
Step Four: put the whole feet and remaining inches together.
4 ft 9.09 in
So, 145 centimeters is slightly taller than 4 ft 9 in. You can also use our centimeters to feet calculator for this.
## 145 Centimeters to Feet Height Conversion Chart
Chart showing the height conversion for 145 centimeters and other close measurements to feet and inches.
CentimetersFeet & InchesFeetInches
140 cm4' 7.1"4.59'55.12"
141 cm4' 7.5"4.63'55.51"
142 cm4' 7.9"4.66'55.91"
143 cm4' 8.3"4.69'56.3"
144 cm4' 8.7"4.72'56.69"
145 cm4' 9.1"4.76'57.09"
146 cm4' 9.5"4.79'57.48"
147 cm4' 9.9"4.82'57.87"
148 cm4' 10.3"4.86'58.27"
149 cm4' 10.7"4.89'58.66"
150 cm4' 11.1"4.92'59.06"
You can also refer to our height comparison chart for a more comprehensive list of height conversions. | 546 | 1,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.859556 |
https://handwiki.org/wiki/Fundamental_theorem_of_linear_algebra | 1,723,703,393,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00440.warc.gz | 223,006,908 | 11,124 | # Fundamental theorem of linear algebra
Short description: Name for certain results on linear maps between two finite-dimensional vector spaces
In mathematics, the fundamental theorem of linear algebra is a collection of statements regarding vector spaces and linear algebra, popularized by Gilbert Strang. The naming of these results is not universally accepted.
More precisely, let f be a linear map between two finite-dimensional vector spaces, represented by a m×n matrix M of rank r, then:
• r is the dimension of the column space of M, which represents the image of f;
• nr is the dimension of the null space of M, which represents the kernel of f;
• mr is the dimension of the cokernel of f.
The transpose MT of M is the matrix of the dual f* of f. It follows that one has also:
• r is the dimension of the row space of M, which represents the image of f*;
• mr is the dimension of the left null space of M, which represents the kernel of f*;
• nr is the dimension of the cokernel of f*.
The two first assertions are also called the rank–nullity theorem. | 236 | 1,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.910359 |
http://math.stackexchange.com/questions/238180/doubt-on-simplifying-logarithms | 1,455,485,106,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454702032759.79/warc/CC-MAIN-20160205195352-00039-ip-10-236-182-209.ec2.internal.warc.gz | 136,732,587 | 16,710 | # Doubt on simplifying logarithms
here's another doubt (sorry I'm a logarithms newbie)
Given: $\log_{10} 2 = a$ and $\log_{10} 3 = b$
Express $\log_{5} 10$ in terms of $a$ and $b$
I don't know from where to start yet. $2$ and $3$ doesn't seems to have much relation with $5$.
-
You don't need $\log_{10}3=b$. Here are some hints.
1. Factor 10 to get a different expression for $\log_510$.
2. Make use of the fact that $\log_ab=1/\log_ba$ to get an expression for $\log_210$ in terms of $a$.
Hint: what is the relation between $\log_5 10$ and $\log_{10} 5$? Between $\log_{10} 5$ and $\log_{10} 2$? | 207 | 603 | {"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} | 3.625 | 4 | CC-MAIN-2016-07 | longest | en | 0.90318 |
http://www.mathworks.com/matlabcentral/cody/problems/165-woodall-number/solutions/52015 | 1,485,094,375,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281426.63/warc/CC-MAIN-20170116095121-00340-ip-10-171-10-70.ec2.internal.warc.gz | 560,196,344 | 11,989 | Cody
# Problem 165. Woodall number
Solution 52015
Submitted on 26 Feb 2012 by Raphael Cautain
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
2 Pass
%% x = 2; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ```
3 Pass
%% x = 3; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ```
4 Pass
%% x = 7; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
5 Pass
%% x = 23; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
6 Pass
%% x = 63; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
7 Pass
%% x = 159; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
8 Pass
%% x = 383; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
9 Pass
%% x = 895; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
10 Pass
%% x = 1000; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ```
11 Pass
%% x = 2000; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ```
12 Pass
%% x = 2047; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
13 Pass
%% x = 3000; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ```
14 Pass
%% x = 3001; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ```
15 Pass
%% x = 4607; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
16 Pass
%% x = 10239; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
17 Pass
%% x = 22527; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
18 Pass
%% x = 7516192767; tf_correct = true; assert(isequal(woodall(x),tf_correct))
``` ```
19 Pass
%% x = 7516192766; tf_correct = false; assert(isequal(woodall(x),tf_correct))
``` ``` | 649 | 1,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-04 | latest | en | 0.364806 |
http://techkumar.com/standard-error/standard-error-of-estimate-formula.html | 1,511,482,567,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807044.45/warc/CC-MAIN-20171123233821-20171124013821-00079.warc.gz | 285,838,116 | 6,465 | Home > Standard Error > Standard Error Of Estimate Formula
# Standard Error Of Estimate Formula
## Contents
I usually think of standard errors as being computed as: $SE_\bar{x}\ = \frac{\sigma_{\bar x}}{\sqrt{n}}$ What is $\sigma_{\bar x}$ for each coefficient? Interpreting the variables using the suggested meanings, success in graduate school could be predicted individually with measures of intellectual ability, spatial ability, and work ethic. They are messy and do not provide a great deal of insight into the mathematical "meanings" of the terms. The rule of thumb here is that a VIF larger than 10 is an indicator of potentially significant multicollinearity between that variable and one or more others. (Note that a VIF http://techkumar.com/standard-error/what-is-the-standard-error-of-the-estimate.html
Coefficient of determination The great value of the coefficient of determination is that through use of the Pearson R statistic and the standard error of the estimate, the researcher can What is the most efficient way to compute this in the context of OLS? Sorry that the equations didn't carry subscripting and superscripting when I cut and pasted them. In a multiple regression model, the constant represents the value that would be predicted for the dependent variable if all the independent variables were simultaneously equal to zero--a situation which may useful source
## Standard Error Of Estimate Formula
When dealing with more than three dimensions, mathematicians talk about fitting a hyperplane in hyperspace. A low value for this probability indicates that the coefficient is significantly different from zero, i.e., it seems to contribute something to the model. In the residual table in RegressIt, residuals with absolute values larger than 2.5 times the standard error of the regression are highlighted in boldface and those absolute values are larger than Condidence Intervals for Regression Parameters A level C confidence interval for the parameter j may be computed from the estimate bj using the computed standard deviations and the appropriate critical value
A visual presentation of the scatter plots generating the correlation matrix can be generated using SPSS/WIN and the "Scatter" and "Matrix" options under the "Graphs" command on the toolbar. The estimated model ŷi = bo+b1xi1+b2xi2+….bpxip can be written as: + The expressions in the parentheses are standardized variables; b's; are unstandardized regression coefficients and s1, s2, …sp are the standard Loading... Standard Error Of The Regression Most multiple regression models include a constant term (i.e., an "intercept"), since this ensures that the model will be unbiased--i.e., the mean of the residuals will be exactly zero. (The coefficients
The discrepancies between the forecasts and the actual values, measured in terms of the corresponding standard-deviations-of- predictions, provide a guide to how "surprising" these observations really were. Standard Error Of Estimate Interpretation However, the regression equation itself should be reported in terms of the unstandardized regression coefficients so that prediction of y can be made directly from the x variables. Here is an example of a plot of forecasts with confidence limits for means and forecasts produced by RegressIt for the regression model fitted to the natural log of cases of http://www.psychstat.missouristate.edu/multibook/mlt06m.html THE ANOVA TABLE The ANOVA table output when both X1 and X2 are entered in the first block when predicting Y1 appears as follows.
If p is large relative to n, the model tends to fit the data very well. Standard Error Of Estimate Excel The standard errors of the coefficients are the (estimated) standard deviations of the errors in estimating them. Standardized regression coefficients The magnitude of the regression coefficients depends upon the scales of measurement used for the dependent variable y and the explanatory variables included in the regression equation. Like us on: http://www.facebook.com/PartyMoreStud...Link to Playlist on Regression Analysishttp://www.youtube.com/course?list=EC...Created by David Longstreet, Professor of the Universe, MyBookSuckshttp://www.linkedin.com/in/davidlongs...
## Standard Error Of Estimate Interpretation
When this happens, it often happens for many variables at once, and it may take some trial and error to figure out which one(s) ought to be removed. other Additional analysis recommendations include histograms of all variables with a view for outliers, or scores that fall outside the range of the majority of scores. Standard Error Of Estimate Formula Seasonal Challenge (Contributions from TeXing Dead Welcome) Does a long flight on a jet provide a headstart to altitude acclimatisation? Standard Error Of Estimate Calculator If partial correlation r12.34 is equal to uncontrolled correlation r12 , it implies that the control variables have no effect on the relationship between variables 1 and 2..
Sign in Share More Report Need to report the video? http://techkumar.com/standard-error/standard-error-of-estimate-multiple-regression.html Specifically, it is calculated using the following formula: Where Y is a score in the sample and Y’ is a predicted score. The figure below illustrates how X1 is entered in the model first. If the model's assumptions are correct, the confidence intervals it yields will be realistic guides to the precision with which future observations can be predicted. Standard Error Of Regression Coefficient
1. zedstatistics 324,055 views 15:00 How to Read the Coefficient Table Used In SPSS Regression - Duration: 8:57.
2. Since they have two categories, they manage to ‘trick' least squares, while entering into the regression equation as interval scale variables with just two categories.
3. There's not much I can conclude without understanding the data and the specific terms in the model.
When effect sizes (measured as correlation statistics) are relatively small but statistically significant, the standard error is a valuable tool for determining whether that significance is due to good prediction, or pxip + i for i = 1,2, ... However, a correlation that small is not clinically or scientifically significant. navigate here Therefore, the standard error of the estimate is a measure of the dispersion (or variability) in the predicted scores in a regression.
The computations derived from the r and the standard error of the estimate can be used to determine how precise an estimate of the population correlation is the sample correlation statistic. Standard Error Of Regression Calculator Specifically, although a small number of samples may produce a non-normal distribution, as the number of samples increases (that is, as n increases), the shape of the distribution of sample means That is, there are any number of solutions to the regression weights which will give only a small difference in sum of squared residuals.
## Usually the decision to include or exclude the constant is based on a priori reasoning, as noted above.
The predicted Y and residual values are automatically added to the data file when the unstandardized predicted values and unstandardized residuals are selected using the "Save" option. Entering X3 first and X1 second results in the following R square change table. It states that regardless of the shape of the parent population, the sampling distribution of means derived from a large number of random samples drawn from that parent population will exhibit How To Calculate Standard Error Of Regression Coefficient In the least-squares model, the best-fitting line for the observed data is calculated by minimizing the sum of the squares of the vertical deviations from each data point to the line
In theory, the t-statistic of any one variable may be used to test the hypothesis that the true value of the coefficient is zero (which is to say, the variable should Taken together with such measures as effect size, p-value and sample size, the effect size can be a very useful tool to the researcher who seeks to understand the reliability and The MINITAB "Regress" command produced the following results: Regression Analysis The regression equation is Rating = 61.1 - 3.07 Fat - 2.21 Sugars After fitting the regression line, it is important his comment is here The correlation between "Fat" and "Rating" is equal to -0.409, while the correlation between "Sugars" and "Fat" is equal to 0.271.
Consider for example, the relationship between income and gender y = a + bx where y = income of an individual, and x = a dichotomous variable, coded as 0 if The standard error statistics are estimates of the interval in which the population parameters may be found, and represent the degree of precision with which the sample statistic represents the population Designed by Dalmario. I.e., the five variables Q1, Q2, Q3, Q4, and CONSTANT are not linearly independent: any one of them can be expressed as a linear combination of the other four.
SEQUENTIAL SIGNIFICANCE TESTING In order to test whether a variable adds significant predictive power to a regression model, it is necessary to construct the regression model in stages or blocks. Matt Kermode 260,637 views 6:14 Statistics 101: Standard Error of the Mean - Duration: 32:03. Does this mean that, when comparing alternative forecasting models for the same time series, you should always pick the one that yields the narrowest confidence intervals around forecasts? Authors Carly Barry Patrick Runkel Kevin Rudy Jim Frost Greg Fox Eric Heckman Dawn Keller Eston Martz Bruno Scibilia Eduardo Santiago Cody Steele Multivariate Statistics: Concepts, Models, and Applications
See the beer sales model on this web site for an example. (Return to top of page.) Go on to next topic: Stepwise and all-possible-regressions Skip navigation UploadSign inSearch Loading... The standard error of the estimate is a measure of the accuracy of predictions. Predictor Coef StDev T P Constant 61.089 1.953 31.28 0.000 Fat -3.066 1.036 -2.96 0.004 Sugars -2.2128 0.2347 -9.43 0.000 S = 8.755 R-Sq = 62.2% R-Sq(adj) = 61.2% Significance Tests Figure 1.
Rating is available when the video has been rented. In words, the model is expressed as DATA = FIT + RESIDUAL, where the "FIT" term represents the expression 0 + 1x1 + 2x2 + ... However, in rare cases you may wish to exclude the constant from the model. The SPSS ANOVA command does not automatically provide a report of the Eta-square statistic, but the researcher can obtain the Eta-square as an optional test on the ANOVA menu.
In the case of simple linear regression, the number of parameters needed to be estimated was two, the intercept and the slope, while in the case of the example with two | 2,245 | 10,747 | {"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} | 3.65625 | 4 | CC-MAIN-2017-47 | latest | en | 0.882365 |
https://answers.launchpad.net/yade/+question/698392 | 1,631,867,406,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00338.warc.gz | 147,844,766 | 10,794 | # I calculated the kinetic energy by E=0.5mv^2+0.5Iw^2 but it is not equal to kineticEnergy() when O.cell.velGrad is set
I am conducting a simulation where the energy results are important for the study. So I want to study how the kinetic energy is calculated in YADE. In the documentation it shows kinetic energy is calculated by E=0.5*m*v^2+0.5*I*w^2, which is definitely normal and accurate. So I have tried to calculated kinetic energy by myself, as the particle mass m, velocity v, inertia I and angular velocity w can be all obtain by using O.bodies[]state.mass, O.bodies[].state.vel, O.bodies[].state.inertia and O.bodies[].state.angVel. But when I have set the value of O.cell.velGrad, the E=0.5*m*v^2+0.5*I*w^2 calculated by myself is much larger than utils.kineticEnergy(). And when I didn't use O.cell.velGrad, the E=0.5*m*v^2+0.5*I*w^2 calculated by myself is the same with utils.kineticEnergy()
I am thinking why this happened? Why setting the value of O.cell.velGrad will make E=0.5*m*v^2+0.5*I*w^2 larger than the utils.kineticEnergy()?
My code for calculating the E=0.5*m*v^2+0.5*I*w^2 is:
kineE=0
for i in bodies_id:
m=O.bodies[i].state.mass
v2=numpy.dot(O.bodies[i].state.vel, O.bodies[i].state.vel)
I=O.bodies[i].state.inertia
w2=numpy.dot(O.bodies[i].state.angVel, O.bodies[i].state.angVel)
kineE=kineE+0.5*m*v2+0.5*I*w2
Thank you very much!!
Yuxuan
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Revision history for this message Robert Caulk (rcaulk) said on 2021-08-16: #1
Hello,
Cheers,
Robert
Revision history for this message Yuxuan Wen (wenyuxuan) said on 2021-08-16: #2
Yes, I set the velGrad in the format as shown in the documentation: O.cell.velGrad=Matrix3(-0.5,0,0,0,-0.5,0,0,0,-0.5). And then the kinetic energy calculated by E=0.5*m*v^2+0.5*I*w^2 is different with the kineticEnergy(). I made a simplified code of the simulation, as shown follows. When you run this code directly, the kinetic energy is different, and when you comment the line of O.cell.velGrad, the kinetic energy will be the same.
#################################
from yade import pack, plot, qt, export, os
O.periodic=True
O.cell.hSize=Matrix3(0.03,0,0, 0,0.03,0, 0,0,0.03)
sp=pack.SpherePack()
sp.makeCloud(Vector3(0,0,0), Vector3(0.03,0.03,0.03), num=1000, rMean=0.001, rRelFuzz=0, distributeMass=False, periodic=True, seed=1)
O.engines=[
ForceResetter(),
InsertionSortCollider([Bo1_Sphere_Aabb(),Bo1_Box_Aabb()]),
InteractionLoop(
[Ig2_Sphere_Sphere_ScGeom(), Ig2_Box_Sphere_ScGeom()],
[Ip2_FrictMat_FrictMat_MindlinPhys(betan=0.3,betas=0.3)],
[Law2_ScGeom_MindlinPhys_Mindlin()] ),
NewtonIntegrator(gravity=(0,0,-9.81),damping=0.0),
PyRunner(command='check()',iterPeriod=1000),
]
O.dt=0.3*utils.PWaveTimeStep()
O.cell.velGrad=Matrix3(-0.5,0,0, 0,-0.5,0, 0,0,-0.5) # if comment this line, kineE will be the same with kineticEnergy()
def check():
kineE=0
for i in O.bodies:
m=i.state.mass
v2=numpy.dot(i.state.vel, i.state.vel)
I=sqrt(numpy.dot(i.state.inertia,i.state.inertia))
w2=numpy.dot(i.state.angVel, i.state.angVel)
kineE=kineE+0.5*m*v2+0.5*I*w2 # compare with utils.kineticEnergy()
print(kineE, kineticEnergy())
#################################
Thank you so much!
Kind Regards,
Yuxuan
Revision history for this message Jan Stránský (honzik) said on 2021-08-17 #3
Hello,
both values are computed differently.
E=0.5*m*v^2+0.5*I*w^2
of course still holds, but the values of "v" and "w" are different.
Have a look at source code [2]: conditions "if (scene->isPeriodic)", variables like "spin = scene->cell->getSpin()" and their usage, comments "Only take in account the fluctuation velocity" etc.
If still anything is not clear, please let us know.
If you think this discrepancy should be mentioned in the documentation, feel free to update it yourself or create an issue on gitlab pages of the project (not to be forgotten).
Cheers
Jan
Revision history for this message Yuxuan Wen (wenyuxuan) said on 2021-08-17: #4
Hello Jan,
Thank you for you help and sending me the position of the source code! I am much appreciated!
But I am still a little confused with the definition of the fluctuation velocity. As you said, the E=0.5*m*v^2+0.5*I*w^2 still holds, and "v" and "w" are the fluctuation velocity . In the source code, the fluctuation velocity is defined as:
bodyFluctuationVel(state->pos - state->vel * scene->dt, state->vel, scene->cell->velGrad)
which is a function of 1)updated position, 2) v, and 3) velGrad.
But I don't know how to calculate the fluctuation velocity by these 3 terms. I searched the documentation by "bodyFluctuationVel" and didn't find anything matched. I wonder if it is possible that you could tell me the expression of the fluctuation velocity? Thank you!
Kind Regards,
Yuxuan
Revision history for this message Yuxuan Wen (wenyuxuan) said on 2021-08-17: #5
Hello Jan,
Thanks for your suggestion, now I have created an issue on gitlab pages of the project [3]. Because I am still confused with the definition of fluctuation velocity, so I didn't update it myself but created an issue on the gitlab pages.
Kind Regards,
Yuxuan
Revision history for this message Jan Stránský (honzik) said on 2021-08-19: #6
Probably again only source code is relevant here [4].
The idea is this:
Let's have a particle at (0,0,0), i.e. one corner of periodic cell.
Properties of the particle should (must) be identical with its periodic images (in this case e.g. particles in other corners of the periodic cell). See [5] and pictures there for a reference.
Let's say the (0,0,0) particle has zero velocity.
Now consider the cell has some velGrad. E.g. it is stretching in x direction. Then he "image particles" in the "right face" of the cell has some non-zero positive velocity.
Now, from the periodicity point of view, it should not matter if you choose the (0,0,0) particle or any of its periodic images.
The (0,0,0) particle has zero velocity, the periodic images have non-zero velocity, which one should be chosen?? The "fluctuation part" of the velocity is chosen in Yade.
Similarly it is with angular velocity and cell spin [6]
Cheers
Jan
Revision history for this message Yuxuan Wen (wenyuxuan) said on 2021-08-19: #7
Thank you so much Jan!! Now I understand the definition of fluctuation velocity and why it is chosen in Yade!
Thanks again for you help!! I am much appreciated!
Yuxuan | 1,904 | 6,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-39 | latest | en | 0.869807 |
https://www.yaclass.in/p/mathematics-state-board/class-7/algebra-3102/introduction-to-identities-18198/re-0fa8eb42-9ce3-4b65-bc03-9f1db38c214c | 1,712,928,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00482.warc.gz | 961,158,639 | 13,011 | PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
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Identity $$2$$: ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$
Let us construct a figure of four regions. The two square shaped regions with the dimensions of $$3 × 3$$ (Blue) and $$2 × 2$$ (Yellow). Observe the remaining two rectangle shaped regions. Both are in $$3 × 2$$ (Green) dimension.
By observing the above rectangle, we can notice that:
$$\text{Area of the bigger square = Area of the two small square + Area of the two rectangles}$$
$$3 + 2$$$$2 + 3$$ $$=$$ $$(3 × 3) + (2 × 3) + (3 × 3) + (2 × 2)$$
Now, we simplify the LHS and RHS of the above expression.
LHS $$=$$ $$3 + 2$$$$2 + 3$$ $$=$$ $$5×5 = 25$$
RHS $$=$$ $$(3 × 3) + (2 × 3) + (3 × 3) + (2 × 2)$$
$$=$$ ${3}^{2}+\left(2×3\right)+\left(3×2\right)+{2}^{2}$ $$=$$ $$9+6+6+4 = 25$$
Therefore, LHS $$=$$ RHS
Similarly, if we use the variables in this case instead of number we get:
Assume the square of ABCD of side $$a + b$$. From the above figure, we can get that:
$$\text{The total area of the bigger square = The area of the two small squares × The are of the two rectangles}$$
That is, $$(a + b)^2$$$={a}^{2}+\mathit{ab}+\mathit{ba}+{b}^{2}$
Since, $$ba=ab$$; $$(a + b)^2$$ $={a}^{2}+\mathit{ab}+\mathit{ab}+{b}^{2}={a}^{2}+2\mathit{ab}+{b}^{2}$.
Therefore, ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$ is a identity.
Example:
Simplify the expression $\left(a+6\right)\left(a+6\right)$ using the identity ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$.
Now write the given expression $\left(a+6\right)\left(a+6\right)$ with respect to the given identity ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$.
$\begin{array}{l}\left(a+b\right)\left(a+b\right)=\left(a+6\right)\left(a+6\right)\\ \\ {\left(a+6\right)}^{2}=\left({a}^{2}+2\left(6×6\right)a+{6}^{2}\right)\end{array}$
Now simplify the expression.
$=\left({a}^{2}+72a+36\right)$.
Therefore, $\left(a+6\right)\left(a+6\right)$$=\left({a}^{2}+72a+36\right)$. | 931 | 2,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-18 | longest | en | 0.399862 |
https://documen.tv/8-the-equation-a-640s-gives-the-relationship-between-s-square-miles-and-a-acres-kevin-owns-8-5-s-21508396-28/ | 1,686,083,136,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00558.warc.gz | 231,589,661 | 15,045 | Question
8.
The equation a = 640s gives the relationship between s square miles and a acres. Kevin owns 8.5 square miles of farmland. How many acres does he own?
13.28 acres
1.33 acres
5,440 acres
544 acres
5440 acres
Step-by-step explanation:
Given: The equation a = 640s gives the relationship between s square miles and a acres. Kevin owns 8.5 square miles of farmland.
To Find: How many acres does he own.
Solution:
Area of farmland owned by kevin
The given equation is
here is area in acres, and is area in square miles.
putting values in equation
Hence kevin owns of farmland | 159 | 600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.930746 |
https://oeis.org/A332588 | 1,621,328,154,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00411.warc.gz | 452,620,065 | 4,557 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A332588 Let t_k denote the triangular number k*(k+1)/2. Suppose 0 < x < y < z are integers satisfying t_x + t_y = t_p, t_y + t_z = t_q, t_x + t_z = t_r, for integers p,q,r. Sort the triples [x,y,z] first by x, then by y. Sequence gives the values of x. 6
9, 14, 20, 23, 27, 35, 35, 41, 44, 51, 54, 54, 62, 65, 65, 66, 74, 76, 77, 77, 77, 83, 90, 104, 104, 105, 105, 105, 114, 119, 124, 131, 131, 135, 135, 152, 152, 161, 161, 165, 168, 170, 170, 174, 189, 189, 189, 209, 210, 210, 216, 228, 230, 230, 237, 245, 252, 252 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Ulas gives a table assuming 0 < x < y < z < 1000. Because of the assumption z < 1000, only the entries with x < 46 can be relied upon (above this it is possible that there are gaps in the table). LINKS Giovanni Resta, Table of n, a(n) for n = 1..162 Ulas Maciej, A note on Sierpinski's problem related to triangular numbers, arXiv:0810.0222 [math.NT], 2008. See Table 1. Ulas Maciej, A note on Sierpinski's problem related to triangular numbers, Colloq. Math. 117 (2009), no. 2, 165-173. See MR2550124. See Table 1. Giovanni Resta, Table of x, y, z, p, q, r for x < 1000 EXAMPLE The initial values of x, y, z, p, q, r are: x y z p q r -- --- --- --- ---- --- 9 13 44 16 46 45 14 51 104 53 116 105 20 50 209 54 215 210 23 30 90 38 95 93 27 124 377 127 397 378 35 65 86 74 108 93 35 123 629 128 641 630 41 119 285 126 309 288 44 245 989 249 1019 990 ... MATHEMATICA L = {}; t[n_] := n (n + 1)/2; Do[ syp = Solve[t[x] + t[y] == t[p] && p > 0 && y > x , {p, y}, Integers]; If[syp =!= {}, Do[{y1, p1} = {y, p} /. e; s = Solve[ t[y1] + t[z] == t[q] && t[x] + t[z] == t[r] && q > 0 && z > y1 && r > 0, {z, q, r}, Integers]; If[s =!= {}, L = Join[L, {x, y1, z, p1, q, r} /. s]], {e, syp}]], {x, 54}]; Sort[L][[All, 1]] (* Giovanni Resta, Mar 02 2020 *) CROSSREFS Cf. A000217, A332589, A332590, A332591, A332592, A332593. Sequence in context: A327896 A302056 A173792 * A034703 A006624 A184218 Adjacent sequences: A332585 A332586 A332587 * A332589 A332590 A332591 KEYWORD nonn AUTHOR N. J. A. Sloane, Feb 29 2020 EXTENSIONS Terms a(10) and beyond from Giovanni Resta, Mar 02 2020 STATUS approved
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Last modified May 18 04:22 EDT 2021. Contains 343994 sequences. (Running on oeis4.) | 1,135 | 2,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-21 | latest | en | 0.415767 |
https://dsp.stackexchange.com/questions/19799/periodicity-of-the-output-of-causal-time-invariant-systems-when-the-input-is-per/19807 | 1,627,584,295,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00210.warc.gz | 230,741,283 | 40,489 | # Periodicity of the output of causal time-invariant systems when the input is periodic
For time-invariant systems, if the input is periodic, the we can argue that the output is periodic:
If we shift the input signal one period, we come up with the same signal as before, and since the output must be shifted the same amount as a result of the system being TI, it should be periodic too.
For causal systems we know that the output signal at time $t$ depends on the input for times up to $t$. My question is, can we conclude that for causal TI systems if the input is periodic before $t_0$, the output will be periodic up to $t_0$ too?
Your line of reasoning is correct. For a (not necessarily linear) time-invariant system the following always holds: if $y(t)$ is the response to an input signal $x(t)$, then $y(t+T)$ must be the response to the shifted input signal $x(t+T)$. If the input is periodic with period $T$, this implies that $x(t)=x(t+T)$, and, consequently, $y(t)=y(t+T)$, i.e. the output signal is periodic with the same period as the input signal.
Furthermore, if the system is causal and the input up to time $t_0$ is periodic, the output up to time $t_0$ must be periodic too, because a causal system cannot know the (possibly non-periodic) input beyond time $t_0$.
This is (part of the) theory. But then comes @MBaz with his "counter-example" of an ideal integrator. The problem here is indeed, as already noted in a comment, the integrator's instability. Showing the time-invariance of a linear system involves manipulating the convolution integral. However, if for the given input signal that integral is divergent, the proof becomes invalid. And this is the case with systems having poles on the $j\omega$-axis (or, for discrete-time systems, on the unit circle). If an LTI system has a pole on the $j\omega$-axis and if you excite it with the corresponding pole frequency, you get an output signal with increasing amplitude, which is obviously non-periodic. You can also see it in the following way: the response of an LTI system to a complex exponential $e^{j\omega_0 t}$ is
$$y(t)=H(\omega_0)e^{j\omega_0 t}\tag{1}$$
Eq. (1) of course requires that $H(\omega_0)$ can be evaluated. For an ideal integrator this is not the case for $\omega_0=0$. Note that if the input signal to the integrator had no DC component, then the response to a periodic input signal would indeed be periodic.
• I'm not convinced. The integrator's linearity doesn't play a role in my example; I think the inegrator's time invariance can be proven without using the convolution integral. Furthermore, the question does not require the system to be stable. In my counter-example, the integrator is a time-invariant system with a monotonically increasing output, which (despite's @Dilip's handwaving) cannot be considered periodic, IMO. – MBaz Dec 31 '14 at 18:22
• @MBaz: The output of the integrator is infinite for a periodic input signal with a DC component. So there's not much sense in discussing its periodicity. The integrator is of course linear and time-invariant, and its time-invariance can be easily shown as long as its output doesn't become infinite. And it will certainly produce a periodic output for any periodic input with a zero DC component. In sum, for a periodic input signal, its response is either periodic or infinite. By the way, I didn't mean to criticize your example, I actually liked it because it's an interesting special case. – Matt L. Dec 31 '14 at 18:37
• That summary I can agree with. I didn't feel criticized; thanks for your comment. I was just worried about being misunderstood. – MBaz Dec 31 '14 at 18:57
I believe the answer is no: a periodic input does not imply a periodic output.
An example: let the input be a square wave with amplitudes 1 and 0, and let the time-invariant system be an integrator. The system's output is monotonic, and thus not periodic.
• Thanks for the answer. But the example system you provided is not BIBO stable, and I think can not be a legitimate example. – user215721 Dec 31 '14 at 6:38
• To echo the OP's comment, hand-wavingly, since the periodic square wave input has been there since $-\infty$, the output for all time is also infinite, and technically does satisfy the condition $y(t+T) = y(t)$ for all $t$. – Dilip Sarwate Dec 31 '14 at 14:56
• @user215721, you do not require the system to be BIBO stable in your question. – MBaz Dec 31 '14 at 18:25
• @Dilip, I don't agree that this can be hand-waved so easily: the output can be considered monotonically increasing, so not periodic. See also my comment to Matt's answer. – MBaz Dec 31 '14 at 18:26
• So what you are saying is that with $x(t)$ being your square wave with amplitudes $0$ and $1$, for $t_1 < t_2$, $$y(t_1)=\int_{-\infty}^{t_1}x(t)\,\mathrm dt \leq y(t_2) = \int_{-\infty}^{t_2}x(t)\,\mathrm dt\tag{1}$$ because $$y(t_2) = y(t_1) + \int_{t_1}^{t_2}x(t)\,\mathrm dt\tag{2}$$ and the integral on the right side of $2$ is $\geq 0$? Isn't this the same as the schoolboy assertion that "my infinity is bigger than your infinity because I added 1 to your infinity to get mine"? – Dilip Sarwate Jan 1 '15 at 15:59 | 1,372 | 5,160 | {"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} | 3.265625 | 3 | CC-MAIN-2021-31 | latest | en | 0.903942 |
https://whatisconvert.com/215-milliseconds-in-weeks | 1,591,353,116,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00132.warc.gz | 582,128,362 | 6,920 | What is 215 Milliseconds in Weeks?
Convert 215 Milliseconds to Weeks
To calculate 215 Milliseconds to the corresponding value in Weeks, multiply the quantity in Milliseconds by 1.6534391534392E-9 (conversion factor). In this case we should multiply 215 Milliseconds by 1.6534391534392E-9 to get the equivalent result in Weeks:
215 Milliseconds x 1.6534391534392E-9 = 3.5548941798942E-7 Weeks
215 Milliseconds is equivalent to 3.5548941798942E-7 Weeks.
How to convert from Milliseconds to Weeks
The conversion factor from Milliseconds to Weeks is 1.6534391534392E-9. To find out how many Milliseconds in Weeks, multiply by the conversion factor or use the Time converter above. Two hundred fifteen Milliseconds is equivalent to zero point zero zero zero zero zero zero three five five five Weeks.
Definition of Millisecond
A millisecond (from milli- and second; symbol: ms) is a thousandth (0.001 or 10−3 or 1/1000) of a second.
Definition of Week
A week (symbol: wk) is a time unit equal to seven days. It is the standard time period used for cycles of rest days in most parts of the world, mostly alongside—although not strictly part of—the Gregorian calendar. The days of the week were named after the classical planets (derived from the astrological system of planetary hours) in the Roman era. In English, the names are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.
Using the Milliseconds to Weeks converter you can get answers to questions like the following:
• How many Weeks are in 215 Milliseconds?
• 215 Milliseconds is equal to how many Weeks?
• How to convert 215 Milliseconds to Weeks?
• How many is 215 Milliseconds in Weeks?
• What is 215 Milliseconds in Weeks?
• How much is 215 Milliseconds in Weeks?
• How many wk are in 215 ms?
• 215 ms is equal to how many wk?
• How to convert 215 ms to wk?
• How many is 215 ms in wk?
• What is 215 ms in wk?
• How much is 215 ms in wk? | 490 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-24 | latest | en | 0.89383 |
https://en.unionpedia.org/Hypersphere | 1,579,607,482,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00365.warc.gz | 419,797,794 | 8,641 | Communication
Faster access than browser!
# Hypersphere
In geometry of higher dimensions, a hypersphere is the set of points at a constant distance from a given point called its center. [1]
## Circle
A circle is a simple closed shape.
## Codimension
In mathematics, codimension is a basic geometric idea that applies to subspaces in vector spaces, to submanifolds in manifolds, and suitable subsets of algebraic varieties.
## Curvature
In mathematics, curvature is any of a number of loosely related concepts in different areas of geometry.
## Dimension
In physics and mathematics, the dimension of a mathematical space (or object) is informally defined as the minimum number of coordinates needed to specify any point within it.
## Duncan Sommerville
Duncan MacLaren Young Sommerville (1879–1934) was a Scottish mathematician and astronomer.
## Euclidean space
In geometry, Euclidean space encompasses the two-dimensional Euclidean plane, the three-dimensional space of Euclidean geometry, and certain other spaces.
## Flat (geometry)
In geometry, a flat is a subset of n-dimensional space that is congruent to a Euclidean space of lower dimension.
## Geometry
Geometry (from the γεωμετρία; geo- "earth", -metron "measurement") is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space.
## Hyperplane
In geometry, a hyperplane is a subspace whose dimension is one less than that of its ambient space.
## Hypersurface
In geometry, a hypersurface is a generalization of the concepts of hyperplane, plane curve, and surface.
## Journal of Physics A
The Journal of Physics A: Mathematical and Theoretical is a peer-reviewed scientific journal published by IOP Publishing.
## Manifold
In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point.
## N-sphere
In mathematics, the n-sphere is the generalization of the ordinary sphere to spaces of arbitrary dimension.
## Three-dimensional space
Three-dimensional space (also: 3-space or, rarely, tri-dimensional space) is a geometric setting in which three values (called parameters) are required to determine the position of an element (i.e., point).
## University of Michigan
The University of Michigan (UM, U-M, U of M, or UMich), often simply referred to as Michigan, is a public research university in Ann Arbor, Michigan.
## 3-sphere
In mathematics, a 3-sphere, or glome, is a higher-dimensional analogue of a sphere.
## References
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https://www.opengl.org/discussion_boards/archive/index.php/t-181026.html | 1,531,835,506,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589726.60/warc/CC-MAIN-20180717125344-20180717145344-00200.warc.gz | 934,382,410 | 4,592 | PDA
View Full Version : Push/Pop Problem
Tolito
02-09-2013, 09:08 AM
For some reason, F is being placed relative to A instead of D even though E is placed relative to D. I don't see why this problem is occurring. If I remove E, F is placed relative to D (as it should be), but I need them both where they are.
// A
glPushMatrix();
glTranslatef(0,0,0);
glCallList(x);
// B
glPushMatrix();
glTranslatef(0,195,-245);
glCallList(x);
glPopMatrix();
// C
glPushMatrix();
glTranslatef(274,272,0);
glCallList(x);
// D
glPushMatrix();
glTranslatef(0,0,0);
// E
glPushMatrix();
glTranslatef(140,440,75);
glCallList(x);
glPopMatrix();
// F
glPushMatrix();
glTranslatef(298,323,59);
glCallList(x);
glPopMatrix();
Thank you! :)
Dark Photon
02-10-2013, 02:24 PM
I'm just going to re-post your source code with braces added for the Pushes and Pops. I think you'll see what the problem is. You'll notice that you're missing about 3 levels of Pops for this to be balanced:
// A
glPushMatrix();
{
glTranslatef(0,0,0);
glCallList(x);
// B
glPushMatrix();
{
glTranslatef(0,195,-245);
glCallList(x);
}
glPopMatrix();
// C
glPushMatrix();
{
glTranslatef(274,272,0);
glCallList(x);
// D
glPushMatrix();
{
glTranslatef(0,0,0);
// E
glPushMatrix();
{
glTranslatef(140,440,75);
glCallList(x);
}
glPopMatrix();
// F
glPushMatrix();
{
glTranslatef(298,323,59);
glCallList(x);
}
glPopMatrix();
Tolito
02-11-2013, 06:05 AM
Thank you for your response! I would like to add that, oddly enough, the problem occurs even with the pops. I should have posted that code first. I haven't the slightest idea why it is occurring. With or without the pops, F is being positioned relative to A instead of D. Everything else is positioned correctly except for this one, which is why I am so baffled.
Dark Photon
02-11-2013, 04:53 PM
Check for GL errors. If your pushes and pops are imbalanced, you should eventually trigger an error with pushing on a full stack or popping an empty one (GL_STACK_OVERFLOW, GL_STACK_UNDERFLOW). Also, all bets are off for interpreting what's happening. | 582 | 2,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.743409 |
https://es.mathworks.com/matlabcentral/cody/problems/7-column-removal/solutions/1101473 | 1,610,788,252,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703505861.1/warc/CC-MAIN-20210116074510-20210116104510-00060.warc.gz | 323,717,894 | 17,168 | Cody
# Problem 7. Column Removal
Solution 1101473
Submitted on 10 Jan 2017 by Devan Rao
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [1 2 3; 4 5 6]; n = 2; B_correct = [1 3; 4 6]; assert(isequal(column_removal(A,n),B_correct))
2 Pass
A = magic(4); n = 3; B = [16 2 13; 5 11 8; 9 7 12; 4 14 1]; B_correct = B; assert(isequal(column_removal(A,n),B_correct))
3 Pass
A = 1:10; n = 7; B_correct = [1 2 3 4 5 6 8 9 10]; assert(isequal(column_removal(A,n),B_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 256 | 724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.641737 |
http://www.numbersaplenty.com/67805 | 1,585,728,031,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505550.17/warc/CC-MAIN-20200401065031-20200401095031-00425.warc.gz | 267,951,745 | 3,391 | Search a number
67805 = 571191
BaseRepresentation
bin10000100011011101
310110000022
4100203131
54132210
61241525
7401453
oct204335
9113008
1067805
1146a41
12332a5
1324b2a
141a9d3
1515155
hex108dd
67805 has 8 divisors (see below), whose sum is σ = 82944. Its totient is φ = 53200.
The previous prime is 67801. The next prime is 67807. The reversal of 67805 is 50876.
It can be divided in two parts, 67 and 805, that multiplied together give a palindrome (53935).
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 67805 - 22 = 67801 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (67801) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 260 + ... + 450.
It is an arithmetic number, because the mean of its divisors is an integer number (10368).
267805 is an apocalyptic number.
It is an amenable number.
67805 is a deficient number, since it is larger than the sum of its proper divisors (15139).
67805 is a wasteful number, since it uses less digits than its factorization.
67805 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 267.
The product of its (nonzero) digits is 1680, while the sum is 26.
The square root of 67805 is about 260.3939323410. The cubic root of 67805 is about 40.7774978443. Note that the first 3 decimals are identical.
The spelling of 67805 in words is "sixty-seven thousand, eight hundred five".
Divisors: 1 5 71 191 355 955 13561 67805 | 497 | 1,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.88396 |
https://brilliant.org/problems/first-find-the-curve-then-area/ | 1,529,557,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00474.warc.gz | 562,308,410 | 16,980 | # First find the curve then area
Calculus Level pending
Let $$f\left( x \right)$$ be a differentiable function such that
$$f\left( x+y \right) ={ e }^{ x }f\left( y \right) +{ e }^{ y }f\left( x \right)$$
for all $$x,y$$
And Given that $$f^{ ' }\left( x \right) =1$$
Find the Area bounded by the curve and x axis
× | 111 | 321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.684168 |
https://genderen.org/unlock-the-secrets-of-percentages-with-this-step-by-step-guide | 1,696,211,938,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00382.warc.gz | 299,498,098 | 16,357 | # Unlock the Secrets of Percentages with this Step-by-Step Guide!
Introduction
Calculating percentages from a number is something that we all have to do from time to time. Whether you are a student, a business owner, or simply someone who wants to understand how to calculate percentages for their own personal reasons, it is important to understand the basics of how to calculate percentages from a number. This article will provide an overview of the basics of how to calculate percentages from a number, and answer some of the most commonly asked questions about the process.
1. What is a Percentage?
A percentage is a fraction of a whole expressed as a number between 0 and 100. It is a way to express a proportion of a total amount. For example, a percentage of 25% means that 25 out of every 100 of something is being referred to.
2. How do I Calculate a Percentage of a Number?
To calculate a percentage of a number, you first need to convert the percentage to a decimal. To do this, divide the percentage by 100. For example, 25% is the same as 0.25. Once you have the decimal, you can multiply it by the number you are trying to calculate the percentage of. For example, if you wanted to calculate 25% of 100, you would multiply 0.25 by 100 to get 25.
3. What is the Formula for Calculating a Percentage of a Number?
The formula for calculating a percentage of a number is as follows:
Percentage x Number = Result
For example, if you wanted to calculate 25% of 100, you would use the following formula:
25% x 100 = 25
4. What is the Percentage Increase of a Number?
The percentage increase of a number is the percentage by which a number has increased from its original value. For example, if a number has increased from 10 to 15, the percentage increase would be 50%. To calculate the percentage increase of a number, subtract the original number from the new number and then divide by the original number. The result is the percentage increase.
5. What is the Percentage Decrease of a Number?
The percentage decrease of a number is the percentage by which a number has decreased from its original value. For example, if a number has decreased from 10 to 5, the percentage decrease would be 50%. To calculate the percentage decrease of a number, subtract the new number from the original number and then divide by the original number. The result is the percentage decrease.
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6. How do I Calculate the Percentage Difference Between Two Numbers?
The percentage difference between two numbers is the percentage by which one number is greater or less than the other. To calculate the percentage difference between two numbers, subtract the smaller number from the larger number and then divide by the smaller number. The result is the percentage difference.
7. How do I Calculate the Percentage of a Whole?
To calculate the percentage of a whole, divide the part by the whole and then multiply by 100. For example, if you wanted to calculate what 25 is as a percentage of 100, you would divide 25 by 100 and then multiply by 100 to get 25%.
8. How do I Calculate the Percentage of a Number in Reverse?
To calculate the percentage of a number in reverse, divide the number by the percentage and then multiply by 100. For example, if you wanted to calculate what 25% of 100 is, you would divide 100 by 25 and then multiply by 100 to get 25.
9. How do I Calculate the Percentage of Increase or Decrease?
To calculate the percentage of increase or decrease, subtract the original number from the new number and then divide by the original number. The result is the percentage increase or decrease.
10. How do I Calculate the Percentage of a Number by Weight?
To calculate the percentage of a number by weight, divide the weight of the number by the total weight and then multiply by 100. For example, if you wanted to calculate what 25 kg is as a percentage of 100 kg, you would divide 25 by 100 and then multiply by 100 to get 25%.
11. How do I Calculate the Percentage of a Number by Volume?
To calculate the percentage of a number by volume, divide the volume of the number by the total volume and then multiply by 100. For example, if you wanted to calculate what 25 liters is as a percentage of 100 liters, you would divide 25 by 100 and then multiply by 100 to get 25%.
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12. How do I Calculate the Percentage of a Number by Time?
To calculate the percentage of a number by time, divide the amount of time of the number by the total amount of time and then multiply by 100. For example, if you wanted to calculate what 25 minutes is as a percentage of 100 minutes, you would divide 25 by 100 and then multiply by 100 to get 25%.
13. How do I Calculate the Percentage of a Number by Area?
To calculate the percentage of a number by area, divide the area of the number by the total area and then multiply by 100. For example, if you wanted to calculate what 25 square meters is as a percentage of 100 square meters, you would divide 25 by 100 and then multiply by 100 to get 25%.
14. How do I Calculate the Percentage of a Number by Length?
To calculate the percentage of a number by length, divide the length of the number by the total length and then multiply by 100. For example, if you wanted to calculate what 25 meters is as a percentage of 100 meters, you would divide 25 by 100 and then multiply by 100 to get 25%.
15. How do I Calculate the Percentage of a Number by Speed?
To calculate the percentage of a number by speed, divide the speed of the number by the total speed and then multiply by 100. For example, if you wanted to calculate what 25 kilometers per hour is as a percentage of 100 kilometers per hour, you would divide 25 by 100 and then multiply by 100 to get 25%.
16. How do I Calculate the Percentage of a Number by Temperature?
To calculate the percentage of a number by temperature, divide the temperature of the number by the total temperature and then multiply by 100. For example, if you wanted to calculate what 25 degrees Celsius is as a percentage of 100 degrees Celsius, you would divide 25 by 100 and then multiply by 100 to get 25%.
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17. How do I Calculate the Percentage of a Number by Force?
To calculate the percentage of a number by force, divide the force of the number by the total force and then multiply by 100. For example, if you wanted to calculate what 25 newtons is as a percentage of 100 newtons, you would divide 25 by 100 and then multiply by 100 to get 25%.
18. How do I Calculate the Percentage of a Number by Pressure?
To calculate the percentage of a number by pressure, divide the pressure of the number by the total pressure and then multiply by 100. For example, if you wanted to calculate what 25 kilopascals is as a percentage of 100 kilopascals, you would divide 25 by 100 and then multiply by 100 to get 25%.
19. How do I Calculate the Percentage of a Number by Energy?
To calculate the percentage of a number by energy, divide the energy of the number by the total energy and then multiply by 100. For example, if you wanted to calculate what 25 joules is as a percentage of 100 joules, you would divide 25 by 100 and then multiply by 100 to get 25%.
20. How do I Calculate the Percentage of a Number by Power?
To calculate the percentage of a number by power, divide the power of the number by the total power and then multiply by 100. For example, if you wanted to calculate what 25 watts is as a percentage of 100 watts, you would divide 25 by 100 and then multiply by 100 to get 25%.
Conclusion
Calculating percentages from a number is a straightforward process once you understand the basics. By understanding the formulas and the different types of calculations, you can easily calculate percentages from a number. Hopefully this article has answered some of the most commonly asked questions about how to calculate percentages from a number.
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#### Anthony Genderen
Hi there, I'm Anthony Genderen, a creative and passionate individual with a keen interest in technology, innovation, and design. With a background in computer science and a natural curiosity about how things work, I've always been drawn to the world of technology and its endless possibilities. As a lifelong learner, I love exploring new ideas and challenging myself to think outside the box. Whether it's through coding, graphic design, or other creative pursuits, I always strive to approach problems with a fresh perspective and find innovative solutions. In my free time, I enjoy exploring the great outdoors, trying new foods, and spending time with family and friends. I'm also an avid reader and love diving into books on topics ranging from science and technology to philosophy and psychology. Overall, I'm a driven, enthusiastic, and curious individual who is always eager to learn and grow. | 1,999 | 9,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-40 | latest | en | 0.93974 |
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### Multiply a Binomial by a Trinomial - Algebra I
In this video, you will learn how to multiply a binomial by a trinomial. A binomial consists of two terms and a trinomial consists of three terms. When you multiply a binomial by a trinomial, you distribute the first term of the binomial to all three terms of the trinomial, and then you distribute the second term of the binomial to the three terms as well. Then you just simply combine all like terms, and you have your product.
Multiply a Binomial by a Trinomial - Algebra I | 138 | 561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.910326 |
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# Question 14 In the following question, out of the four options, only one is correct. Write the correct answer: If x be any non-zero integer, then x−1 is equal to: (a) x (b) 1x (c) -x (d) −1x
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## Using law of exponents ,a−m=1am Similarly, x−1=1x
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https://www.scribd.com/document/356470308/jack4b-pdf | 1,540,360,677,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583519859.77/warc/CC-MAIN-20181024042701-20181024064201-00391.warc.gz | 1,063,545,213 | 57,821 | Solutions to Problems in Jackson
,
Classical Electrodynamics, Third Edition
Homer Reid
October 8, 2000
Chapter 4: Problems 8-13
Problem 4.8
A very long, right circular, cylindrical shell of dielectric constant /0 and inner and
outer radii a and b, respectively, is placed in a previously uniform electric field E0
with its axis perpendicular to the field. The medium inside and outside the cylinder
has a dielectric constant of unity.
(a) Determine the potential and electric fields in the three regions, neglecting end
effects.
(b) Sketch the lines of force for a typical case of b ≈ 2a.
(c) Discuss the limiting forms of your solution appropriate for a solid dielectric
cylinder in a uniform field, and a cylindrical cavity in a uniform dielectric.
We will take the axis of the cylinder to be the z axis and the electric field to
be aligned with the x axis: E0 = E0 î. Since the cylinder is very long and we’re
told to neglect end effects, we can ignore the z direction altogether and treat
this as a two-dimensional problem.
(a) The general solution of the Laplace equation in two dimensional polar co-
ordinates is
X
Φ(r, ϕ) = [An rn + Bn r−n ][Cn sin(nϕ) + Dn cos(nϕ)]
For the region inside the shell (r < a), the B coefficients must vanish to keep
the potential from blowing up at the origin. Also, in the region outside the shell
1
r<a X Φ(r. ϕ) = rn [Cn sin nϕ + Dn cos nϕ] + r−n [En sin nϕ + Fn cos nϕ]. −E0 r cos ϕ with An = 0 for n > 1. With these observations we may write expressions for the potential in the three regions: X rn [An sin nϕ + Bn cos nϕ]. i. −E0 r cos ϕ + r>b The normal boundary condition at r = a is .Homer Reid’s Solutions to Jackson Problems: Chapter 4 2 (r > b).e. a<r<b X r−n [Gn sin nϕ + Hn cos ϕ]. the only positive power of r in the sum must be that which gives rise to the external electric field.
.
∂Φ .
.
∂Φ .
.
0 = ∂r .
x=a− ∂r .
the tangential boundary condition at r = a is .x=a+ or 0 X n−1 na [An sin nϕ + Bn cos nϕ] = X nan−1 [Cn sin nϕ + Dn cos nϕ] − na−(n+1) [En sin nϕ + Fn cos nϕ] From this we obtain two equations: 0 An = Cn − En a−2n (1) 0 Bn = Dn − Fn a−2n (2) Next.
.
∂Φ .
.
∂Φ .
.
= ∂ϕ .
x=a+ ∂ϕ .
from the normal boundary condition at r = b we obtain 0 0 X −(n+1) − E0 cos ϕ − nb [Gn sin nϕ + Hn cos ϕ] = X nbn−1 [Cn sin nϕ + Dn cos nϕ] − nb−(n+1) [En sin nϕ + Fn cos ϕ] .x=a− or X X nan [An cos nϕ − Bn sin nϕ] = nan [Cn cos nϕ − Dn sin nϕ] + na−n [En cos nϕ − Fn sin nϕ] from which we obtain two more equations: An = Cn + En a−2n (3) −2n Bn = Dn + F n a (4) Similarly.
b2 ( + 0 )2 − a2 ( − 0 )2 . and (8) specify the same degenerate system of equations. the system of equations (2). (5). Next. for n 6= 1. we have the tangential boundary condition at r = b: X bE0 sin ϕ + nb−n [Gn cos nϕ − Hn sin nϕ] = X nbn [Cn cos nϕ − Dn sin nϕ] + nb−n [En cos nϕ − Fn sin nϕ] giving Gn = Cn b2n + En (7) 2 2n −b E0 δn1 + Hn = Dn b + Fn . (4). we have 0 1 0 B1 = D1 − F1 a−2 D1 = 1+ B1 2 ⇒ 1 2 0 B1 = D1 + F1 a−2 F1 = a 1− B1 . 2 and −H1 = b2 E0 + D1 b 2 − F 1 0 0 H1 = b 2 E 0 + D 1 b 2 + F 1 → 0 = 2b2 E0 + b2 1 + D1 + 1 − F1 0 0 Substituting from above. However. so Bn = Dn = Fn = Gn = 0 for n 6= 0. and (7) specify a degenerate system of linear equations.Homer Reid’s Solutions to Jackson Problems: Chapter 4 3 which leads to 0 Gn = Cn b2n − En − (5) 0 0 − b2 E0 δn1 − Hn = Dn b2n − Fn (6) Finally. (3). 1 2 −4b2 E0 = b ( + 0 )2 − a2 ( − 0 )2 B1 0 or −40b2 B1 = E0 . which can only be satisfied by taking An = Cn = En = Gn = 0 for all n. for n = 1. (8) The four equations (1). (6).
b ( + 0 )2 − a2 ( − 0 )2 The potential is −40 b2 · E0 rcos ϕ. (b) In Figure 4. as an appendix to this document I’ve included the C program I wrote to generate this plot. a cylindrical cavity in a uniform dielectric corresponds to . r<b E(r. Φ → −E0 r cos ϕ in all three regions. r<a b2 ( + 0 )2 − a2 ( − 0 )2 −20 b2 a2 Φ(r. = 50 . a<r<b b2 ( + 0 )2 − a2 ( − 0 )2 r −(b2 − a2 )(20 − 2 ) b2 2 2 2 2 · E0 cos ϕ − E0 rcos ϕ. In that case the field would look like 2 0 + 0 E0 î. ϕ) = ( + 0 )r + ( − 0 ) E0 cos ϕ.1 I’ve plotted the field lines for b = 2a. b < r. ϕ) = a − ( + 0 ) + ( − ) 0 2 E 0 sin ϕϕ̂ . (c) For a solid dielectric cylinder in a uniform field. Also.Homer Reid’s Solutions to Jackson Problems: Chapter 4 4 Then −20 ( + 0 )b2 D1 = E0 b2 ( + 0 )2 − a2 ( − 0 )2 −20 ( − 0 )a2 b2 F1 = 2 E0 b ( + 0 )2 − a2 ( − 0 )2 −b2 (b2 − a2 )(20 − 2 ) H1 = 2 E0 . a<r<b r 2 (b2 − a2 )(20 − 2 ) b − 2 · E0 [cos ϕr̂ + sin ϕϕ̂] 2 2 b ( + 0 ) − a ( − 0 ) 2 r +E0 [cos ϕr̂ − sin ϕϕ̂] . ϕ) = 2 (2 − 2 ) b E0 î − 0 E0 [cos ϕr̂ + sin ϕϕ̂]. b < r. we would have a → 0. b ( + 0 ) − a ( − 0 ) r As → 0 . r<a b ( + 0 )2 − a2 ( − 0 )2 2 20 b2 a2 ( + ) − ( − 0 2 E0 cos ϕr̂ ) 0 b2 ( + 0 )2 − a2 ( − 0 )2 r 2 E(r. The electric field is 40 b2 E0 [cos ϕr̂ − sin ϕϕ̂] . which is reassuring. r>b ( + 0 )2 r On the other hand.
8 for b = 2a. r > a.Homer Reid’s Solutions to Jackson Problems: Chapter 4 5 Figure 1: Field lines in Problem 4. r<a ( + 0 )2 E(r. = 50 . in which case the field becomes 40 E0 î. b → ∞. ( + 0 )2 ( + 0 ) r . ϕ) = 20 20 ( − 0 ) a 2 E0 î − E0 [cos ϕr̂ + sin ϕϕ̂].
while Φ2 comes from the external point charge. θ) = Bl r−(l+1) + l+1 Pl (cos θ). ∇·D = 0 there. (a) Find the potential at all points in space as an expansion in spherical harmonics. We will take the origin of coordinates at the center of the sphere. in the limit /0 → ∞. r<d 4π0 dl+1 Φ2 (r. the potential may be written as the sum of two components Φ1 and Φ2 . in the region r > a.9 A point charge q is located in free space a distance d away from the center of a dielectric sphere of radius a (a < d) and dielectric constant /0 . where Φ1 comes from the polarization charge on the surface of the sphere. Then the problem has azimuthal symmetry. whence X Φ(r. r<a l q r X Φ(r. we may expand it in Legendre polynomials: X Φ1 (r. 4π0 . 4π0 rl+1 Putting this all together we may write the potential in the three regions as X Al rl Pl (cos θ). θ) = Al rl Pl (cos θ) (r < a). so within the sphere the potential satisfies the normal Laplace equation. l Now. Since Φ1 satisfies the Laplace equation for r > a. (c) Verify that. Φ2 is just the potential due to a point charge at z = d: q X rl Pl (cos θ). r > d. (b) Calculate the rectangular components of the electric field near the center of the sphere. This means that polarization charge only exists on the surface of the sphere. θ) = Bl r−(l+1) Pl (cos θ) (r > a).Homer Reid’s Solutions to Jackson Problems: Chapter 4 6 Problem 4. But since the permittivity is uniform within the sphere. θ) = (9) X dl q Pl (cos θ). a<r<d 4π0 d X qdl Bl + r−(l+1) Pl (cos θ). (a) Since there is no free charge within the sphere. we may also write ∇·(D/) = ∇·E = 0 there. your result is the same as that for the conducting sphere. and put the point charge on the z axis at z = +h. r > d. l On the other hand.
Homer Reid’s Solutions to Jackson Problems: Chapter 4 7 The normal boundary condition at r = a is .
.
∂Φ .
.
∂Φ .
.
= 0 ∂r .
r=a− ∂r .
r=a+ lqal−1 → lAl al−1 = −(l + 1)Bl a−(l+2) + 0 4π0 dl+1 0 −(l + 1) −(2l+1) q → Al = Bl a + (10) l 4π0 dl+1 The tangential boundary condition at r = a is .
.
∂Φ .
.
∂Φ .
.
= ∂θ .
r=a− ∂θ .
as /0 → ∞ we have Al → 0 as must happen. . 4π0 d d rl+1 Comparing with (9) we see that this is just the potential of a charge −qa/d on the z axis at z = a2 /d. This is just the size and position of the image charge we found in Chapter 2 for a point charge outside a conducting sphere. since the field within a conducting sphere vanishes. we obtain 1 2l + 1 q Al = 0 + l+1 l l 4π0 dl+1 qa2l+1 1 Bl = 1− 0 + l+1 l 0 4π0 dl+1 In particular. and qa2l+1 Bl → − . the potential outside the sphere due to the polariza- tion charge at the sphere boundary is l 1 qa X a2 1 Φ1 (r. (12) 4π0 dl+1 With the coefficients (12). θ) = − Pl (cos θ).r=a+ q al → Al al = Bl a−(l+1) + 4π0 d(l+1) q a2l+1 → Bl = Al a2l+1 − (11) 4π0 dl+1 Combining (10) and (11).
respectively. That region is described by θ = π/2. We’ll orient this problem such that the boundary between the dielectric- filled space and the empty space is the xy plane. (a) Since the dielectric has uniform permittivity. so within its body we may take the potential to be a solution of the normal Laplace equation. 0<θ< 2 Φ(r. we have Φ(r. (a) Find the electric field everywhere between the spheres. and we . θ) = A1 rP1 (cos θ) + A2 r2 P2 (cos θ) + · · · q 30 1 50 2 2 2 = z+ (z − x − y ) + · · · 4π0 d2 ( + 20 ) 2 d3 (2 + 30 ) so the field components are q 50 x Ex = · +··· 4π0 d2 2 + 30 d q 50 y Ey = 2 · +··· 4π0 d 2 + 30 d q 30 50 z Ez = − + + · · · 4π0 d2 + 20 2 + 30 d Problem 4. carry charges ±Q. as shown in the figure. <θ<π 2 First let’s apply the boundary conditions at the interface between the di- electric and free space. 0 < θ < π/2. θ) = X π l −(l+1) [Cl r + Dl r ]Pl (cos θ). (b) Calculate the surface-charge distribution on the inner sphere. and the problem has azimuthal symmetry. The empty space between the spheres is half-filled by a hemi- spherical shell of dielectric (of dielectric constant /0 ). all the polarization charge exists on the boundary of the dielectric. a < r < b.10 Two concentric conducting spheres of inner and outer radii a and b.Homer Reid’s Solutions to Jackson Problems: Chapter 4 8 (b) Near the origin. (c) Calculate the polarization-charge density induced on the surface of the dielectric at r = a. Then the region occupied by the dielectric is the region a < r < b. The potential in the region between the spheres may then be written X π [Al rl + Bl r−(l+1) ]Pl (cos θ).
Homer Reid’s Solutions to Jackson Problems: Chapter 4 9 must have .
.
∂Φ .
.
∂Φ .
.
= 0 ∂θ .
θ=π/2+ ∂θ .
θ=π/2− .
.
∂Φ .
.
∂Φ .
.
= ∂r .
θ=π/2+ ∂r .
Similarly. since Pl0 (0) vanishes for even l.θ=π/2− which leads to X Al − Cl Pl0 (0)rl + Bl − Dl Pl0 (0)r−l+1 = 0 (13) 0 0 X l−1 l [Al − Cl ] P (0)r − (l + 1) [Bl − Dl ] Pl (0)r−l+2 = 0. the coefficients of each power of r must vanish identically. (14) is automatically satisfied for l odd. If the liquid rises an average height h between the electrodes when a potential difference V is established between them.13 Two long. There are actually two components of this charge. In (13). we can use Gauss’ law to determine the E field between the . cylindrical conducting surfaces of radii a and b are lowered vertically into a liquid dielectric. To begin. For other cases the vanishing of the coefficients must be brought about by taking Al = C l Bl = Dl . show that the susceptibility of the liquid is (b2 − a2 )ρgh ln(b/a) χe = 0 V 2 where ρ is the density of the liquid. First let’s work out what happens when a battery of fixed voltage V is con- nected between two coaxial conducting cylinders with simple vacuum between them. (16) Next let’s consider the charge at the surface of the inner sphere. and the susceptibility of air is neglected. this requirement is automatically satisfied for l even. l odd (15) 0 0 Al = C l Bl = Dl . (14) Since these equations must be satisfied for all r in the region a < r < b. coaxial. l even. and the other component comes from the bound polarization charge on the inner surface of the dielectric Problem 4. g is the acceleration due to gravity. one component comes from the surface distribution of the free charge +Q that exists on the sphere.
to establish a potential difference V between the conductors. So the . For our Gaussian pillbox we take a disk of thickness dz and radius r. the battery now has to establish a surface charge that is greater that it was before by a factor (/0 ). in order to establish this same E field in the presence of the retarding effects of the dielectric. then the E field must be just the same as it was in the no-dielectric case. Now suppose we introduce a dielectric material between the cylinders. With this greater charge on the electrodes. the battery has to flow enough charge to establish a surface charge of magnitude 0 V σ= (17) a ln(b/a) on the cylinder faces (the surface charges are of opposite sign on the two cylin- ders). the D field will now be bigger by a factor (/0 ) than it was in our above calculation. If the voltage between the cylinders is kept at V .Homer Reid’s Solutions to Jackson Problems: Chapter 4 10 cylinders. It is useful to figure out the energy per unit length stored in the electric field between the cylinder plates here. and the com- ponent normal to the side surfaces (the radial component) is uniform around the disc. a < r < b centered on the axis of the cylinders. since (18) is the energy per unit length stored in the field between the cylinders with just vacuum between them. Hence q 1 I E · dA = 2π r dzEρ = = (2π a dz)σ 0 0 aσ → Eρ (ρ) = 0 r where σ is the surface charge on the inner conductor. However. By symmetry there is no component of E normal to the top or bottom boundary surfaces. This is just 1 b 2π Z Z Wv = E · D ρ dρ dφ 2 a 0 Z b = π0 E 2 (ρ)ρ dρ a a2 σ 2 =π ln(b/a) 0 π0 V 2 = (18) ln(b/a) where the v subscript stands for ’vacuum’. This must integrate to give the correct potential difference between the conductors: Z b aσ b V =− Eρ (ρ)dρ = − ln a 0 a which tells us that. because this field integrated from a to b must still give the same potential difference.
Eg is easily calculated by noting that the area between the cylinders is π(b2 − a2 ). with a battery keeping a voltage V between the electrodes. With no potential between the cylinder plates. ln(b/a) On the other hand. The energy lost by the battery is twice that gained by the dielectric. The height at which we no longer gain by having more liquid between the cylinders is the height to which the system will settle. πV 2 Ee = −h( − 0 ) (20) ln(b/a) This must be balanced by the gravitational potential energy Eg of the excess liquid. the combined system of battery and di- electric can lower its energy by having more of the dielectric rise up between the cylinders. So suppose that. i. Turning now to the situation in this problem. Now suppose a battery of fixed potential V is connected between the two cylinder plates. and if this mass is at a height h above the liquid surface its excess gravitational energy is dEg = (dm)gh = πgρ(b2 − a2 )hdh. to get to this point the battery has had to flow enough charge to increase the surface charges to be of magnitude (/0 ) times greater than (17). .e. so the system with dielectric between the cylinders has lower overall energy than the system with vacuum between the cylinders by a factor πV 2 ∆W = ( − 0 ) (19) ln(b/a) (per unit length). However. We’ll take the boundary between the liquid and the air above it to be at z = 0. In doing this the internal energy of the battery decreases by an amount equal to the work it had to do to flow the excess charge. As we showed earlier. so the mass of liquid contained in a height dh between the cylinders is dm = ρπ(b2 − a2 )dh. The decrease in electrostatic energy this affords over the case with just vacuum filling that space is just (19) times the height. the liquid between the electrodes rises to a height h above the surface of the liquid outside the electrodes.Homer Reid’s Solutions to Jackson Problems: Chapter 4 11 energy per unit length stored in the field between the cylinders increases by a factor (/0 − 1) over the result (18): πV 2 ∆Wd = ( − 0 ) . we’ll take the axis of the cylinders as the z axis. at some point the energy win we get from this is balanced by the energy hit we take from the gravitational potential energy of having the excess liquid rise higher between the cylinders. namely 2πV 2 ∆Wb = −V dQ = V (2π a dσ) = ( − 0 ) ln(b/a) (per unit length). the liquid between the cylinders is at the same height as the liquid outside. so that the surface of the liquid is parallel to the xy plane.
But if the surface area of the vessel containing the liquid is sufficiently larger than the area between the cylinders. we find that the gravitational penalty of the excess liquid just counterbalances the electrostatic energy reduction when 2( − 0 )V 2 h= ρg(b2 − a2 ) ln(b/a) 2χe 0 V 2 = ρg(b2 − a2 ) ln(b/a) Solving for χe . ρgh(b2 − a2 ) ln(b/a) χe = .Homer Reid’s Solutions to Jackson Problems: Chapter 4 12 Integrating over the excess height of liquid between the cylinders. the difference layer will be thin and its energy shifts negligible. namely. Hence there are really two other contributions to the energy shift. h 1 Z Eg = πgρ(b2 − a2 ) h0 dh0 = πgρ(b2 − a2 )h2 . Actually we should note one detail here. . the surface of the liquid outside the cylinders must fall. When the surface of the liquid between the cylinders rises. since the total volume of the liquid is conserved. the change in gravitational and electrostatic energies of the thin layer of liquid outside the cylinders that falls away when the liquid rises between the cylinders. (21) 0 2 Comparing (20) to (21). 20 V 2 So I seem to be off by a factor of 2 somewhere.
of pts to plot for each line */ #define DELTAX (4.0 * B) / NUMPOINTS /* horiz spacing of pts */ #define DELTAY (4.phi).c" #define EZ 1.0 /* no.h> #include "/usr2/homer/include/GnuPlot.0 /* permittivity of cylinder */ #define E0 1.0 /* radius of inner cylinder */ #define B 8.Homer Reid’s Solutions to Jackson Problems: Chapter 4 13 Appendix Source code for field line plotting program used in Problem 4. double phi) { double Coeff.0 /* external field (irrelevant here) */ #define A 4. return Coeff*E0*cos(phi). /* * Program to draw field lines for Jackson problem 4. * Homer Reid October 2000 */ #include <stdio. if ( r < A ) Coeff=(4.8.0 /* number of field lines to draw */ #define NUMPOINTS 250. else Coeff=1.8. } .(EPS-EZ)*(A*A)/(r*r) ).0 /* radius of outer cylinder */ #define NUMLINES 25.0 * B) / NUMLINES /* vert spacing of initial pts */ #define DENOM (B*B*(EPS+EZ)*(EPS+EZ) .A*A*(EPS-EZ)*(EPS-EZ)) /* * Return r component of electric field at position (r. */ double Er(double r.0 /* permittivity of free space */ #define EPS 5.h> #include <math.A*A)*(EZ*EZ-EPS*EPS)*(B*B)/(r*r*DENOM)).0 .((B*B . else if ( r < B ) Coeff=(2*EPS*B*B/DENOM)*( (EPS+EZ) .0*EPS*EZ*B*B)/DENOM.
-2. else Coeff=1."%g %g\n". double phi) { double Coeff. fprintf(g."set size square\n"). FILE *g. return -Coeff*E0*sin(phi).y."set yrange [%g:%g]\n". g=GnuPlot("Field lines").0*B."plot ’-’ t ’’.0*B. .0*B). fprintf(g.phi)."set terminal postscript portrait color\n").x. else if ( r < B ) Coeff=(2*EPS*B*B/DENOM)*( (EPS+EZ) + (EPS-EZ)*(A*A)/(r*r) ). } void main() { double i. */ double Ephi(double r. fprintf(g."set multiplot \n").eps’\n"). phi<=2*M_PI. /* * Send basic GnuPlot configuration commands. phi+=(2*M_PI/100)) fprintf(g. for(phi=0.A*sin(phi)). fprintf(g."set output ’fig4."e\n").0 + ((B*B .-2."set xrange [%g:%g]\n".PhiComp.2.0*EPS*EZ*B*B)/DENOM. fprintf(g. */ fprintf(g.phi.2. /* * Draw circles at r=a and r=b.dy.r.1.j.A*A)*(EZ*EZ-EPS*EPS)*(B*B)/(r*r*DENOM)). ’-’ t ’’ with lines. fprintf(g.dx. */ fprintf(g.0*B). fprintf(g. if ( r < A ) Coeff=(4. ’-’ t ’’ with lines\n")."set noxtics\n").A*cos(phi). double RComp."set noytics\n"). fprintf(g.Homer Reid’s Solutions to Jackson Problems: Chapter 4 14 /* * Return phi component of electric field at (r.
0) phi=(y>0.x. and y * coordinate up or down by an amount depending on * the direction of the electric field at this point */ x+=DELTAX."%g %g\n". /* * compute rise and run of electric field */ RComp=Er(r. fprintf(g. else phi=atan(y/x).Homer Reid’s Solutions to Jackson Problems: Chapter 4 15 fprintf(g. .0) { /* * Compute starting x and y coordinates and initiate plot. i<=NUMLINES. phi+=(2*M_PI/100)) fprintf(g.phi).y).B*sin(phi)). /* * Draw field lines. if (x==0. /* * Plot NUMPOINTS points for this field line.0*i)/NUMLINES).sin(phi)*PhiComp.0. */ for (j=0."e\n"). dy=sin(phi)*RComp + cos(phi)*PhiComp.0*B. */ for (i=1.0) { /* * compute polar coordinates of present location */ r=sqrt(x*x + y*y). phi<=2*M_PI. y=2.2. j<NUMPOINTS."%g %g\n". dx=cos(phi)*RComp ."plot ’-’ t ’’ with lines\n"). */ x=-2. i+=1.0) ? M_PI/2. fprintf(g.0*B * ((NUMLINES ."e\n"). /* * bump x coordinate forward a fixed amount.B*cos(phi). j+=1.phi). y+=DELTAX * (dy/dx). for(phi=0.0 : -M_PI/2. fprintf(g.0.0. PhiComp=Ephi(r.
Homer Reid’s Solutions to Jackson Problems: Chapter 4 16 }. } .\n"). fprintf(g. printf("Thank you for your support. }."e\n"). | 7,294 | 20,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-43 | latest | en | 0.854776 |
http://www.physicsforums.com/showthread.php?s=680a06cd80ba23fa3a178e519f953af8&p=4215942 | 1,369,449,548,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705318091/warc/CC-MAIN-20130516115518-00070-ip-10-60-113-184.ec2.internal.warc.gz | 647,906,629 | 12,343 | ## How liquid pressure = dgh can be generalized
I've seen the standard derivation of the expression for liquid pressure
P = dgh where,
d = density of the liquid;
g = acceleration due to gravity;
h = height of liquid column
in many text books has been done by using a specific example of a cylindrical vessel.
In such a case, the geometry of the cylinder allows us to write Volume = Base Area x Height
So, the above result is trivial.
But in many cases where such a direct link between volume of the vessel and its area and the other dimension doesn't exist, like for a cone, its volume is V = $\pi$ r$^{2}$ x h x (1/3) , we get that extra constant 1/3 and liquid pressure is no more dgh I think.
How then can a specific example be used to speak for the general?
I've seen this kind of easy generalization in many other parts of physics such as torque on a plane coil in a magnetic field($\tau$=BINAcosx), where the actual derivation has been performed using a simple rectangular coil and said that the result is true for any shape of plane coil with the same area.
I understand that this has been done for simplifying things but is there some law in physics that I do not know of that allows for such bold assumptions to be made even just by watching patterns?
PhysOrg.com physics news on PhysOrg.com >> A quantum simulator for magnetic materials>> Atomic-scale investigations solve key puzzle of LED efficiency>> Error sought & found: State-of-the-art measurement technique optimised
Mentor
Liquid pressure is a function of depth only, the geometry does not matter (in equilibrium).
While you have a different amount of water in a cone, you also have (vertical) pressure on the walls there. A cylinder is easier to calculate, as all walls are vertical.
I understand that this has been done for simplifying things but is there some law in physics that I do not know of that allows for such bold assumptions to be made even just by watching patterns?
Sometimes there is. In the example of the coil, you can use vector calculus to see the identity.
But why is there the constant (1/3) in case of the cone as opposed to that in the case of a cylinder? Even if a cone and a cylinder are of the same height and base area, the pressure at the base of cone seems to be lesser by one-third than in cylinder according to this way of looking at liquid pressures. What is wrong here?
Mentor
## How liquid pressure = dgh can be generalized
The pressure is the same in the cone.
As I said, if you want to do the same derivation there, you would have to consider pressure from the walls, which is quite messy. It is easier to verify that the fluid has to have the same pressure at the same depth, and look at the interior of the fluid. This way, you get p=dgh independent of the geometry.
A more sophisticated derivation which works for any shape of container uses the ideas of work and energy. Imagine a small cylinder fitted with a piston of area A, which is inserted in the liquid so the piston is at depth h, at any orientation we choose, and in contact with the liquid on one side (of the piston). If we push the piston out, a small distance x into the liquid, we'll displace a mass ρAx of liquid, and an equal mass of liquid will, effectively, be displaced from depth h to the surface, gaining gravitational potential energy ρAxhg. We can equate this gain in GPE to the work p dV = p A dx done by the piston against the liquid pressure p. This gives p = ρgh. This is delightful because it shows the result to be totally independent of the container shape and of the orientation of the surface against which the liquid pushes. The piston is assumed to move slowly enough for viscous (dissipative) forces to be negligible; after all we are dealing with hydrostatics! I first saw this derivation in A Second Course of Mechanics by A E E McKenzie, a highly respected writer of British school Physics textbooks of some fifty years ago.
Recognitions:
Homework Help
Quote by s0ft in many text books has been done by using a specific example of a cylindrical vessel. In such a case, the geometry of the cylinder allows us to write Volume = Base Area x Height So, the above result is trivial.
Trivial? Hmm. You're assuming that the pressure on the base is uniform and that the net pressure from the sides of the container is zero. In the case of the cylinder, the assumptions hold. But in the case of a conical container, do they hold? No, that's why you can't use the same argument in that case.
There are several ways to generalise to find the pressure inside some arbitrary container. I think some people have already replied with good answers. Here is how I would do it. First, make some assumptions 1) the fluid is static (i.e. not moving) 2) there is zero shear forces within the fluid 3) the pressure in the fluid is the same in all direction (And therefore can be written as a scalar pressure) 4)gravity and density are constant with respect to time and space.
We can write the force on any arbitrary parcel of fluid as due to the force of pressure on the surface of that parcel, plus the force due to gravity on that parcel:
$$\vec{F} = - \oint P d \vec{A} - \int \rho g \hat{z} dV$$
Where P is the scalar pressure, $\rho$ is the density of the fluid and g is the absolute value of gravity, i.e. 9.81 m/s^2 We also know that because of Gauss' law, we can write:
$$\oint P d \vec{A} = \int \nabla P dV$$
So now we have:
$$\vec{F} = \int - (\nabla P) - \rho g \hat{z} dV$$
We have assumed that the fluid is static. This implies that the force on any arbitrary parcel of fluid is equal to zero, so the integral on the right hand side must be zero, over any arbitrary volume. To achieve this, we must make the integrand equal to zero at all points in the fluid. Therefore:
$$\nabla P = - \rho g \hat{z}$$
Now, since P is a scalar, we also have the equation:
$$dP = \nabla P \cdot d \vec{L}$$
Where $d \vec{L} = dx \hat{x} + dy \hat{y} + dz \hat{z}$ So we can use this, along with our equation to get:
$$dP = - \rho g \hat{z} \cdot d \vec{L}$$
And therefore, substituting in $d \vec{L}$ we get:
$$dP = - \rho g dz$$
And since we have assumed that density and gravity are constant,
$$P = - \rho g z + C$$
Where C is just a constant. If we now specify that h= -z (i.e. h measures distance downwards) and if we specify that C=0 when h=0, we finally get:
$$P= \rho g h$$
BruceW Wow! This certainly has the merit of sticking (through thick and thin) to the basic concepts of pressure and force (though I think the right hand bracket needs moving in your third equation).
Recognitions: Homework Help hehe, yes, I like the basic derivations. They make the most sense to me. Also, I think the 3rd equation is correct... All the right hand side is meant to be within the same volume integral.
That's why the bracket should, I think, be moved to between the $\widehat{z}$ and the dV, but this is the merest nitpick, and I wish I hadn't mentioned it. I thought your derivation was a tour de force, though the work-energy method delivers the result convincingly in a couple of lines with elementary mathematics - for someone willing to use a slightly less direct method.
Recognitions: Homework Help yeah. I try to use work-energy method most of the time. But when I am not sure about something, I go back to this kind of method. I think that means I need to get more familiar with work-energy methods. Lagrangian mechanics and Hamiltonian vector fields, I have had an introduction to, but I'm still not as comfortable with. Like for example, when the density and gravity are not constant, I know what to do with my method, since $$\frac{dP}{dz} = - \rho g$$ Then when $\rho$ and g are dependent on z, then you can just integrate with respect to z (where the limits are just the initial and final positions), then you will get the change in pressure between those two positions. But using the work-energy method, I would not be sure what to do.
Hadn't thought of extending the energy method to more difficult situations. Presumably its viability depends on whether or not ρg, in its dependency on position in the liquid, is derivable from a scalar potential function.
Quote by BruceW Now, since P is a scalar, we also have the equation: $$dP = \nabla P \cdot d \vec{L}$$ ]
If P is a scalar then wouldn't it's derivative be zero?
Mentor Scalar is not the same as constant. Temperature is a scalar, for example - it has a single (scalar) value for each point in time and space - but it can depend on time and space.
Recognitions: Homework Help That has helped me realise something that I missed. I should have said another assumption is that the pressure at some fixed point does not vary with time. When I used Gauss' law, $\nabla P$ must be the partial derivatives of the spatial dimensions, while holding time constant (If I'm correct... I think this is explained by the maths of differential forms). So therefore, $$dP = \nabla P \cdot d \vec{L} + \frac{\partial P}{\partial t} dt$$ We must have this extra term, because time is being held constant in the first term. So from here, if I assume that the second term is zero, we get $dP = \nabla P \cdot d \vec{L}$ and the rest of the derivation continues the same as I wrote above. Instead, we could not introduce this assumption, and keep the extra term. So for time-dependent problems, it gets a bit more complicated. You'd probably have to use Reynolds transport theorem. | 2,263 | 9,422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2013-20 | latest | en | 0.952367 |
https://www.jiskha.com/display.cgi?id=1225712082 | 1,503,552,031,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133032.51/warc/CC-MAIN-20170824043524-20170824063524-00317.warc.gz | 933,492,026 | 3,807 | # algebra
posted by .
need help simplifying 2x+3y+2(x-y)-3x
is there anyone that can help
• algebra -
Would it help to rearrange it like this:
2x+3y+2x-2y-3x
then collect x and y terms?
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