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 |
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https://cracku.in/75-what-is-last-step-for-the-input-below-input-16-09--x-snap-2006 | 1,726,052,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00715.warc.gz | 160,415,547 | 28,867 | Instructions
Study the following example and answer the questions.
An electronic device rearranges numbers step-by-step in a particular order according to a set of rules. The device stops when the final result is obtained. In this case the device stops at Step V.
Input: 85 16 36 04 19 97 63 09
Step I - 97 85 16 36 04 19 63 09
Step II - 97 85 63 16 36 04 19 09
Step III - 97 85 63 36 16 04 19 09
Step IV - 97 85 63 36 19 16 04 09
Step V - 97 85 63 36 19 16 09 04
Question 75
# What is last step for the input below?Input: 16 09 25 27 06 05
Solution
Since the device is sorting the number in descending order with biggest number and one at a time.
For Input 16 09 25 27 06 05
Step I : 27 16 09 25 06 05
Step II : 27 25 16 09 06 05
After which all the numbers are sorted. Hence 2 steps are needed
• All Quant Formulas and shortcuts PDF
• 170+ previous papers with solutions PDF | 305 | 888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.670712 |
http://en.wikipedia.org/wiki/Defective_matrix | 1,419,795,341,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447559374.74/warc/CC-MAIN-20141224185919-00034-ip-10-231-17-201.ec2.internal.warc.gz | 35,666,131 | 9,453 | # Defective matrix
In linear algebra, a defective matrix is a square matrix that does not have a complete basis of eigenvectors, and is therefore not diagonalizable. In particular, an n × n matrix is defective if and only if it does not have n linearly independent eigenvectors.[1] A complete basis is formed by augmenting the eigenvectors with generalized eigenvectors, which are necessary for solving defective systems of ordinary differential equations and other problems.
A defective matrix always has fewer than n distinct eigenvalues, since distinct eigenvalues always have linearly independent eigenvectors. In particular, a defective matrix has one or more eigenvalues λ with algebraic multiplicity $m > 1$ (that is, they are multiple roots of the characteristic polynomial), but fewer than m linearly independent eigenvectors associated with λ.[2] However, every eigenvalue with multiplicity m always has m linearly independent generalized eigenvectors.
A Hermitian matrix (or the special case of a real symmetric matrix) or a unitary matrix is never defective; more generally, a normal matrix (which includes Hermitian and unitary as special cases) is never defective.
## Jordan block
Any Jordan block of size 2×2 or larger is defective. For example, the n × n Jordan block,
$J = \begin{bmatrix} \lambda & 1 & \; & \; \\ \; & \lambda & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & \lambda \end{bmatrix},$
has an eigenvalue, λ, with multiplicity n, but only one distinct eigenvector,
$v = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.$
## Example
A simple example of a defective matrix is:
$\begin{bmatrix} 3& 1 \\ 0 & 3 \end{bmatrix}$
which has a double eigenvalue of 3 but only one distinct eigenvector
$\begin{bmatrix} 1 \\ 0 \end{bmatrix}$
(and constant multiples thereof). | 461 | 1,816 | {"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} | 4.09375 | 4 | CC-MAIN-2014-52 | latest | en | 0.881385 |
https://www.education.com/worksheets/first-grade/math/CCSS/?page=31 | 1,603,763,343,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107892710.59/warc/CC-MAIN-20201026234045-20201027024045-00153.warc.gz | 706,509,316 | 28,066 | # Search Our Content Library
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Baseball Subtraction #2
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Baseball Subtraction #2
Subtraction stars, take a swing at math practice with baseball subtraction! Your child will get to solve some simple subtraction problems.
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Children tackle addition problems with this football-themed worksheet.
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Count the Dots: Single-Digit Addition 18
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Count the Dots: Single-Digit Addition 18
This worksheet will provide your child with some valuable addition practice that won't be a bore!
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Balloon Buster #19
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'Pop' the balloons with equations that equal 19 in this fun mental math worksheet. Have fun and learn math with this balloon buster game.
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Need help with simple addition and subtraction? Check out balloon busters, a fun game where your little one can "pop" balloons with the right equation!
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Covering single-digit addition problems that add up to 9 or less, this first grade math worksheet offers a good introduction to beginning arithmetic.
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Explore the educational side of fruit salad with this fun and fruity math worksheet covering single-digit addition problems that add up to 9 or less.
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Bite into math with this juicy worksheet full of plump tomatoes. Your first grader might think math's a bore, but this colorful printable is anything but dull.
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Fishing For Math Facts: Thirteen
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Fishing For Math Facts: Thirteen
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Give your first grader a healthy dose of addition practice! She'll be "grape-ful" for a nice change from those dull, black-and-white printables.
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Count the Dots: Single-Digit Addition 12
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Count the Dots: Single-Digit Addition 12
If your first grader is struggling with addition, this worksheet--complete with diagrams--will provide him with some visual assistance!
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Balloon Buster #16
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Balloon Buster #16
How do you make learning addition and subtraction facts fun? Try your hand at "popping" colorful balloons with a sum or difference of sixteen.
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Baseball Subtraction #5
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Baseball Subtraction #5
Give your subtraction star a fun way to practice math, with baseball subtraction! Your child will get to solve some simple subtraction problems.
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Full of juicy melon slices, this worksheet is just the thing to get your child hungry for math.
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Count the Dots: Single-Digit Addition 29
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Count the Dots: Single-Digit Addition 29
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Count 'Em Up: Orange Slice Addition
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Count 'Em Up: Orange Slice Addition
Does your child drag his feet when it comes to addition? Full of bright colors and cheerful orange slices, this addition worksheet just might change his mind.
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Here's a supplemental math activity that beats boring textbook math. Your child will work on his addition facts, practicing to become an addition all star!
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Need a more engaging way to practice math facts? Try this fun drawing exercise, where kids draw out the math problem before they solve it.
Math
Worksheet | 759 | 3,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.825161 |
http://entertainment.howstuffworks.com/physics-of-baseball3.htm | 1,506,060,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688671.43/warc/CC-MAIN-20170922055805-20170922075805-00123.warc.gz | 104,719,543 | 16,817 | # How the Physics of Baseball Works
Breaking Down the Swing, Millisecond by Millisecond
A-Rod waits for a pitch from the Seattle Mariners as he bats for the Yanks in 2012. Somehow he doesn't look intimidated by what's about to come his way.
Otto Greule Jr/Getty Images
Facing a world-class pitcher might be one of the most intimidating prospects in all of sports. Consider the situation for a typical batter: He stands at home plate, awaiting the throw from the opposing pitcher, positioned just 60.5 feet (18.4 meters) away. A good fastball pitcher can hurl the ball at about 94 mph (151 kph). That means the ball covers the distance between the mound and home plate in 0.439 seconds, or 439 milliseconds.
To put that in context, remember that a voluntary blink of the eye requires 150 milliseconds. In a little less than three blinks, the batter must see the ball, judge its trajectory, decide whether to swing and then, potentially, swing. It's no wonder hitting safely just three out of 10 at-bats is considered good. Or, as Ted Williams once observed: "Baseball is the only field of endeavor where a man can succeed three times out of 10 and be considered a good performer."
Successful batters have two qualities you hear shouted every day at ball fields: "good eyes" and "quick hands." If we dissect a batter's reaction, millisecond by millisecond, this is what we get [source: Adair]:
• The instant the ball leaves the pitcher's hand, light bounces from the leather surface and races to the batter's eye. There, it takes 75 milliseconds for the batter's eye to form a picture and send it to the brain: 25 milliseconds for the retina to process incoming data, 20 milliseconds for that data to travel to the brain and 30 milliseconds for the brain to construct an image out of the information. As the ball travels toward the plate, additional images arrive in the brain every 25 milliseconds, like frames in a motion picture.
• The batter then has about 50 milliseconds for thinking. In this short time, his brain must analyze the pitch and select a suitable swing or decide that the pitch will land out of the strike zone.
• Next, the batter's brain must send signals to the muscles of his legs, torso and arms. This manifests itself as electrochemical impulses originating in the cerebral cortex and then zipping along nerve fibers to the extremities. In all, it takes 25 milliseconds for the impulses to get from the brain to individual muscle fibers in the lower legs.
• Finally, the muscles respond, but not instantly. A baseball swing requires a full 150 milliseconds to complete. This takes into account the time for muscles to contract, move their attached limbs and bring the bat around at a speed close to 80 mph (129 kph). That means a good batter is well into his swing when the ball is still about 18 feet (5 meters) from home plate. If he's 10 milliseconds too early or too late, he'll miss the ball completely.
That's all timing. What about the swing itself? | 654 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-39 | latest | en | 0.933611 |
https://www.justcrackinterview.com/interviews/secureyes-techno-services/ | 1,695,578,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00767.warc.gz | 916,352,617 | 20,516 | # Secureyes Techno Services Interview Questions Answers, HR Interview Questions, Secureyes Techno Services Aptitude Test Questions, Secureyes Techno Services Campus Placements Exam Questions
Find best Interview questions and answer for Secureyes Techno Services Job. Some people added Secureyes Techno Services interview Questions in our Website. Check now and Prepare for your job interview. Interview questions are useful to attend job interviews and get shortlisted for job position. Find best Secureyes Techno Services Interview Questions and Answers for Freshers and experienced. These questions can surely help in preparing for Secureyes Techno Services interview or job.
All of the questions listed below were collected by students recently placed at Secureyes Techno Services.
Ques:- The number 89715983* is divisible by 4. The unknown non zero digit marked as * will be
A. 2
B. 3
C. 4
D. 6
C. 4
Ques:- The number of prime number between 0 and 100 is
A. 17
B. 25
C. 31
D. 39
B.25
Ques:- A sample of milk contains 5% water. What quantity of pure milk should be added to 10 litres of milk to reduce the water content to 2%? (a) 5 litres (b) 7 litres (c) 15 litres (d) 12 litres (e) None of these
( c ) 15 litres
Ques:- 19 boys turn out for hockey. Of these, 11 are wearing hockey shirts and 14 are wearing hockey pants. There are no boys without one or the other. The number of boys wearing full uniform is
( C ) 6
Ques:- 80% pass in english, 70%pass in maths , 10%fail in both , 144 pass in both . How many all appeared to the test?
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
Ques:- Imagine you are standing in front of a mirror, facing it. Raise your left hand. Raise your right hand. Look at your reflection. When you raise your left hand your reflection raises what appears to be his right hand. But when you tilt your head up, your reflection does too, and does not appear to tilt his/her head down. Why is it that the mirror appears to reverse left and right, but not up and down
The definition of left and right depends on the observer and
is reversed when facing the opposite direction. The
definition of up and down does not depend on the orientation
of the observer.
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Ques:- You have a car mileage 1km per liter, you have 3 thousand liters of petrol and have to cover 3000kms and give a way to cover that distance so that you can have some petrol remaining when you reached.
Ques:- What should i say if anybody ask me to say about me which is not in my resume.
Ques:- How about a team with low morale?
Ques:- Two friends, Alex and Bob, go to a bookshop, together with their sons Peter and Tim. All four of them buy some books; each book costs a whole amount in shillings. When they leave the bookshop, they notice that both fathers have spent 21 shillings more than their respective sons. Moreover, each of them paid per book the same amount of shillings as books that he bought. The difference between the number of books of Alex and Peter is five. Who is the father of Tim?
Ques:- Is there anything I've mentioned that makes you think I'm not the best candidate for this job?
Ques:- How can you say that you are a good team player with example
Ques:- Come & You Join Again
Ques:- Why introducing stripping gas on TEG regenerator
Ques:- What was the most stressful aspect of your last project and how did you deal with it?
Ques:- A man leaves office daily at 7pm A driver with car comes from his home to pick him from office and bring back homeOne day he gets free at 5:30 and instead of waiting for driver he starts walking towards home. In the way he meets the car and returns home on car He reaches home 20 minutes earlier than usual.In how much time does the man reach home usually??
Since the car has met the person 20 minutes before hand, it has actually saved 10 mins of journey (to and fro)
since the man has started 1.30 hrs before and car has met him 10 mins before actual time he takes to reach daily is 1hr and 20 mins
Ques:- A man rows his boat 85 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream? | 1,130 | 4,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | latest | en | 0.884581 |
https://www.edplace.com/worksheet_info/11+/keystage2/year3/topic/1193/6816/units-of-measure:-converting-cm-to-mm | 1,713,800,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00459.warc.gz | 658,178,629 | 12,179 | # Units of Measure: Converting cm² to mm²
In this worksheet, students convert areas given in cm² to mm².
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Metric & Imperial
Difficulty level:
#### Worksheet Overview
This worksheet is about converting areas in cm² to areas in mm².
The diagram above shows a large 1 cm square of area 1 cm2.
Each side is 1 cm or 10 mm long
Each tiny square is 1mm long and has an area of 1 mm2.
Ten of these tiny squares fit along each side. There are 100 in total.
1 cm2 = 100 mm2.
Example
Change the area in cm² to an area in mm².
11 cm2
mm2
11 x 100 = 1100 mm2
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Get started | 268 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.876839 |
https://mcqslearn.com/o-level/physics/quizzes/quiz-questions-and-answers.php?page=139 | 1,713,917,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00366.warc.gz | 344,559,154 | 20,022 | O Level Online Courses
IGCSE O Level Physics Certification Exam Tests
IGCSE O Level Physics Practice Test 139
# Forces and Motion Quiz PDF: Questions and Answers - 139
Books:
Apps:
Free Forces and Motion Quiz Questions and Answers PDF, Forces and Motion Quiz PDF Download, Book Ch. 3-139 to study igcse physics online courses. Study Forces in Physics MCQ Questions PDF, forces and motion Multiple Choice Questions (MCQ Quiz) for online college degrees. The Forces and Motion Trivia App Download: Free educational app for forces and motion, energy: o level physics, o level online exam test, physics of light, mass and weight test prep for GRE test prep classes.
The Quiz: An object has a mass of 24 kg, the weight of it on earth and moon would; "Forces & Motion" App Download (Android & iOS) Free with answers on earth = 25.6 n, on moon 35 n, on earth = 38.4 n, on moon = 240 n, on earth = 35 n, on moon = 25.6 n and on earth = 240 n, on moon = 38.4 n to learn online IGCSE GCSE courses. Practice forces in physics questions and answers, Google eBook to download free sample for GRE prep classes.
## Forces & Motion Questions and Answers : Quiz 139
MCQ 691: An object has a mass of 24 kg, the weight of it on earth and moon would
1. On earth = 38.4 N, on moon = 240 N
2. On earth = 25.6 N, on moon 35 N
3. On earth = 35 N, on moon = 25.6 N
4. On earth = 240 N, on moon = 38.4 N
MCQ 692: Thermal energy can be re-converted into potential or kinetic energy'. statement is
1. TRUE
2. FALSE
3. Neutral
4. Both A and B
MCQ 693: If the car tires are hot, the pressure of gas molecules in them would be
1. high
2. low
3. same as before heating
4. may be high or low
MCQ 694: Image formed by a projector would be
1. inverted, real, diminished
2. virtual, upright, diminished
3. virtual, upright, magnified
4. real, inverted, magnified
MCQ 695: A mobile phone has a mass of 100 g. Find its weight if g is 10 N kg-1
1. 1 N
2. 90 N
3. 1000 N
4. 10 N
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SAT Physics App (Android & iOS) | 686 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.801775 |
http://lessonplanspage.com/musicgames-htm/ | 1,524,804,229,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00403.warc.gz | 180,383,691 | 28,342 | Here is a music lesson plan using games
Subject:
Music
1, 2, 3, 4, 5
``` Music
Lesson Plan Title: Music Games
Kaye Slover
Game: Bag of Fortune
```
The game “Bag of Fortune” is very similar to Wheel of Fortune.
Supplies Needed: A bag (a musical gift bag is great) A set of cards cut out all the same size. On each card would be a note or a rest. On two of the cards you will need the word bankrupt. You need more quarter notes than whole notes.
Prepartion: Before playing the game you should choose some of the songs that the children have been singing in music. Have a list ready. On the board you would put a space for each letter in the title of the song. For example, if the song was “I Love Music,” On the board you would put this
``` _ _ _ _ _ _ _ _ _ _
```
Directions
1. Divide the class into two or three teams.
2. Choose a team to go first. Let the first child pull a card from the bag. If he/she can tell you how many beats it gets then he/she gets double the value for each letter in the word. If they do not know then they get face value. For example, if they pull out a whole note and they answer correctly then they will get 8 points for each letter. If they guess the letter I then they would get 16 points for their team.
3. You would then put the letter in the correct place.
``` I _ _ _ _ _ _ _ I _
```
Ask them if they know the song if they can name it their team wins and they get their total points, If not they keep their points and play goes to the next team. It is just like Wheel of Fortune. The team can hold their points but if they don’t solve a puzzle the points don’t count.
4. If a player pulls a bankrupt then all the points their team has accumulated is lost. UNLESS they have solved a puzzle. Those points are safe.
Are you totally confused? This is a fun game to play on a fun Friday or on a day when you don’t feel good. The kids love it. Modify it to meet your needs.
VARIATION:
Use tied notes on the cards. This will really help them to understand tied notes better.
RHYTHM TIC-TAC-TOE
Preparation: Choose nine flash cards with rhythms on them. Set them on the floor three cards on each row. Have markers for both teams. Divide the class into teams.
To begin play: Have one player on the team chooss a rhythm. They get two chances to clap the rhythm correctly. If they get it right a marker is placed on their rhythm. If they miss then play goes to the next team. The first team to get tic-tac-toe wins.
SECRET CODE RHYTHM
1.On the chalkboard write four rhythm patterns.
2. Have the children practice each rhythm pattern to make sure they can clap it correctly.
3. Place numbers to the left of each rhythm pattern.
4. Tell the children that you are going to use a secret code. For example on the board write 4132. Then have the children clap the rhythm patterns in that order. Then choose a child to make the new secret code. The children will ask to do this over and over again.
VARIATION:
The children also enjoy it when you erase one of the patterns then they must clap it and remember the one erased. Then you erase another, until they are all erased.
I am alsways looking for good games. If you have an idea or a suggestion e-mail me at slov5@airmail.net | 786 | 3,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.955099 |
http://stackoverflow.com/questions/4123694/xna-sprite-transormation/4124542 | 1,469,400,290,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824185.14/warc/CC-MAIN-20160723071024-00318-ip-10-185-27-174.ec2.internal.warc.gz | 223,634,678 | 16,934 | Dismiss
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# XNA sprite transormation
Do anybody know how to apply transformation to sprite without opportunities of `SpriteBatch.Draw()` method ?
(Update from comments) This is what I need: I have a circle sprite. This picture contains shadows. I need to stretch this circle (make ellipse) and rotate it, but I want the shadow does not change its position into ellipse. Rotation and scaling change every frame. I think it's possible with matrix transformation
-
Your question doesn't make sense. Could you try rephrasing it? What are you trying to achieve? – Andrew Russell Nov 8 '10 at 12:59
I need: at first rotate sprite and then scale it. Method SpriteBatch.Draw() at first scales sprite and then rotates it. – Eugene Gluhotorenko Nov 8 '10 at 13:23
The matrix passed into `SpriteBatch.Begin` is applied last. Using this is the only way to achieve a scale operation following a rotate operation through SpriteBatch (assuming your scale is non-uniform).
The downside is that, if the scale of each sprite is different, you will have to start a new batch for each.
Your other option is to write your own sprite batcher - but that seems a bit drastic.
-
Thanks! I've found solution, but it is very slowly =(
``````this.displayMatrix =
Matrix.CreateTranslation(-(new Vector3(Position, 0))) *
Matrix.CreateRotationZ(1) *
Matrix.CreateScale(new Vector3(new Vector2(1f, 2f), 1)) *
Matrix.CreateTranslation((new Vector3(Position, 0)));
``````
- | 380 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.910815 |
http://stackoverflow.com/questions/11749321/strongly-connected-graph-is-cyclic-graph-or-not | 1,467,419,828,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783404382.73/warc/CC-MAIN-20160624155004-00141-ip-10-164-35-72.ec2.internal.warc.gz | 290,632,383 | 16,627 | # Strongly Connected Graph is cyclic graph or not?
I am fairly new to Graph theory and I have a a very basic doubt regarding Strongly connected component of a graph. It says two nodes or more are strongly connected if there are paths from both nodes to each other.so whether this graph qualifies as a cyclic graph with cycles?
-
## 1 Answer
Yes, strongly connected graphs are cyclic. In such graphs any two vertices, say `u` and `v` are strongly connected, so there exists a path from `u` to `v` and a path from `v` to `u` made of directed edges. If the `u->v` path and the `v->u` path are disjoint, concatenate them and you have the cycle. If they share edges, start from `u` and follow the `u->v` path, once you hit the first edge shared by the two paths, start following the `v->u` path until you return to `u`.
- | 205 | 821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-26 | latest | en | 0.956122 |
https://www.physicsforums.com/threads/epslion-proof-for-limit-of-a-sequence.214192/ | 1,721,019,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00739.warc.gz | 812,153,402 | 17,237 | # Epslion Proof for limit of a sequence
• GNelson
In summary, using the definition of a limit of a sequence, we can prove that as n approaches infinity, tanh(n) approaches 1. This can be shown by demonstrating that for any given epsilon, there exists a natural number N such that when n is greater than N, the absolute value of (tanh(n) - 1) is less than epsilon. By manipulating the given equation and choosing N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}, we can show that the limit of exp(-n) as n approaches infinity is equal to 0, which in turn proves the original statement.
GNelson
## Homework Statement
Using only the definition of a limit of a sequence prove that lim n->infinity tanh(n)=1
## The Attempt at a Solution
My attempt at the solution is as follows.
If 1 is the limit of the sequence then for every $$\epsilon$$>0, there exists an number such that n>N for every n, such that we have
|tanh(n)-1|<$$\epsilon$$
apply the appropriate hyperbolic identity I re-write this as.
|e^2n-1/(e^2n+1) -1 |< $$\epsilon$$
as tanh(n)< 1 for every sufficiently large n
we have 1-(e^2n-1/(e^2n+1)) < $$\epsilon$$
After this I am stumped, our textbook is very poor so are the notes.
Any help is welcome thanks in advanced.
GNelson said:
## Homework Statement
we have 1-(e^2n-1/(e^2n+1)) < $$\epsilon$$
How about factoring out e^2n, and then applying limit, and if you are missing lim n-->inf sign there?
If i could evaluate it as a limit I would but its asking to prove it using the precise definition of a limit of a sequence, which means that is not an option
lim n->infinity tanh(n)=1
we desire to show
for any eps>0
there exist (a natural number) N(eps)
such that whenever (a natural number) n>N(eps)
|1-tanh(n)|<eps
hint
show |1-tanh(n)|<2exp(-2n)
Then the original question is equivalent to showing
lim n->infinity exp(-n)=0
you we so close
(1-(e^2n-1/(e^2n+1))
((e^2n+1)/(e^2n+1)-(e^2n-1)/(e^2n+1))
((e^2n+1)-(e^2n-1))/(e^2n+1)
2/(e^2n+1)
2/(e^2n+1)<2exp(-n)
thus the hint
it should be easy to finish
Try this
$$\left|\frac{e^{2n}-1}{e^{2n}+1}-1\right|=\left|\frac{-2}{e^{2n}+1}\right|=\frac{2}{e^{2n}+1}<\frac{2}{e^{2n}}=2\,e^{-2n}<\epsilon$$
and choose
$$N=max\left\{0,-\frac{1}{2}\,\ln\frac{\epsilon}{2}\right\}$$
Oupps! lurflurf was faster!
Thanks guys working it through now.
I am currently in calc II. I am curious where you got what we choose N= , I can finish it with the substitution you gave me. I am just curious how one derives it.
And I got the problem thank you, However I am still curious about the derivation of N
There is no general method to choose $N$. One starts with the definition $|a_n-l|<\epsilon$ and tries to find $N$. In the problem at hand, since we don't know if $-\frac{1}{2}\,\ln(\epsilon/2)$ is positive we have to write $N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}$
## 1. What is an Epsilon Proof for the limit of a sequence?
An Epsilon Proof for the limit of a sequence is a method used to prove that a sequence converges to a specific limit. It involves choosing a small value, epsilon, and showing that all terms in the sequence after a certain point are within epsilon of the limit.
## 2. Why is an Epsilon Proof important in mathematics?
An Epsilon Proof is important in mathematics because it provides a rigorous way to prove the convergence of a sequence. It is also a fundamental concept in analysis and helps to understand the behavior of functions and limits.
## 3. How do you construct an Epsilon Proof for the limit of a sequence?
To construct an Epsilon Proof, you first choose a small value, epsilon, and then find a point in the sequence where all terms after it are within epsilon of the limit. This can be done by using the definition of limit and finding a suitable value for n, the number of terms in the sequence.
## 4. Can you give an example of an Epsilon Proof for the limit of a sequence?
One example of an Epsilon Proof for the limit of a sequence is the proof that the sequence {1/n} converges to 0 as n approaches infinity. By choosing any value of epsilon, we can find a point in the sequence where all terms after it are within epsilon of the limit, in this case, 0.
## 5. Are there any limitations to using Epsilon Proofs for limits of sequences?
One limitation of Epsilon Proofs is that they only work for sequences that have a limit. If a sequence does not have a limit, an Epsilon Proof cannot be used. Additionally, constructing Epsilon Proofs can be challenging and time-consuming, especially for more complicated sequences.
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865 | 1,461 | 5,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.884534 |
http://physicshelpforum.com/special-general-relativity/5464-spacetime.html | 1,553,477,293,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00039.warc.gz | 169,177,358 | 11,482 | Physics Help Forum spacetime
Special and General Relativity Special and General Relativity Physics Help Forum
Feb 6th 2011, 09:41 PM #1 Senior Member Join Date: Mar 2009 Location: Yadupatti Bazar,Sitamarhi Posts: 137 spacetime what i know is that space is three dimensional frame to locate position of an object and time is just added to it to make it more significant and make four dimensional. What i don't undestand is 1) i m unable to imagine this fourdimensional concept ( as in three dimension three axes x,y,z can be taken such that each make an angle of 90 degree with each other, then at what place can i attach this another one to make it four dimension) 2) i have read that space is curve( somewhere not space, spactime is curved ). What is difference between these two things except dimensional concept. 3) if space is empty then how can an empty space be curved( whereas i have read somewhere that space is not empty) __________________ If science solves one problem, it creates ten other problems.
Feb 7th 2011, 11:51 AM #2
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,527
Originally Posted by ohm what i know is that space is three dimensional frame to locate position of an object and time is just added to it to make it more significant and make four dimensional. What i don't undestand is 1) i m unable to imagine this fourdimensional concept ( as in three dimension three axes x,y,z can be taken such that each make an angle of 90 degree with each other, then at what place can i attach this another one to make it four dimension) 2) i have read that space is curve( somewhere not space, spactime is curved ). What is difference between these two things except dimensional concept. 3) if space is empty then how can an empty space be curved( whereas i have read somewhere that space is not empty)
1) Mostly you are on the right track. Ordinary 3-space (in the absence of masses; for that we need General Relativity) is Euclidean. That means the 3 space axes are "stuck together" at right angles. The time dimension is a bit trickier. (Lorentzian) space-time has what is called an "indefinite metric." That is to say space-time is not at all Euclidean. I'm not certain how to the time axis is "attached" to the 3-spaces axes, but I don't think it can even be visualized since it is 4 dimensional.
2) Due to the indefinite metric space-time is curved (non-Euclidean, or more precisely it is non-Riemannian.) Space, in Special Relativity, is flat: it is Euclidean. On the other hand, if there are masses present then a curvatiure is introduced to the space axes as well. (There is a whole "machinery" involved here. If you pick the right set of coordinate axes you can make a very small part of curved space Euclidean again. This is the basis for differential geometry and I'm not going to go into that right now.)
3) There are a number of ways to answer this question. First, empty space can be curved if there is a gravitational field present. But again we are talking about GR and differential geometry here. It depends. once again, on your coordinate axes. On the other hand, Quantum Physics says there is no such thing as an empty vacuum. This is due to virtual particles popping in and out of the vacuum, and is a topic that I am again not going to discuss here.
Please feel free to ask the questions I've put off if you like. The differential geometry bit I can do in this forum, but the vacuum question should be posted in the Quantum Mechanics forum.
-Dan
__________________
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See the forum rules here.
Feb 10th 2011, 09:40 AM #3 Senior Member Join Date: Mar 2009 Location: Yadupatti Bazar,Sitamarhi Posts: 137 sorry. There is some technical problem in my internet connection. That'r why i am not responding properly. First of all i thanks for ur response. 1) u thought that it is not possible to visualise four dimensional spacetime. So i think it is useless to try to imagine this. 2) u have disscussed "indefinite metric" and differential geometry . What is this ? 3) i have seen a rubbel sheet model. Does it indicate empty space. If it is then how can empty(nothing) can be curved. 4) i have also seen light cone along with world line. How can straight worldline be path of movement of earth in four dimenssion. Kindly response this. __________________ If science solves one problem, it creates ten other problems. Last edited by ohm; Feb 10th 2011 at 09:43 AM.
Feb 11th 2011, 09:06 AM #4 Senior Member Join Date: Mar 2009 Location: Yadupatti Bazar,Sitamarhi Posts: 137 kindly response anyone. __________________ If science solves one problem, it creates ten other problems.
Feb 13th 2011, 09:06 PM #5 Senior Member Join Date: Mar 2009 Location: Yadupatti Bazar,Sitamarhi Posts: 137 in GR which is curved "space" or "spacetime"?(pls don't mention dimensional concept to distinguish these two things as i am unable to imagine this four dimensional concept) __________________ If science solves one problem, it creates ten other problems.
Feb 14th 2011, 11:53 AM #6
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,527
Originally Posted by ohm sorry. There is some technical problem in my internet connection. That'r why i am not responding properly. First of all i thanks for ur response. 1) u thought that it is not possible to visualise four dimensional spacetime. So i think it is useless to try to imagine this. 2) u have disscussed "indefinite metric" and differential geometry . What is this ? 3) i have seen a rubbel sheet model. Does it indicate empty space. If it is then how can empty(nothing) can be curved. 4) i have also seen light cone along with world line. How can straight worldline be path of movement of earth in four dimenssion. Kindly response this.
1) I certainly can't visualize it, thought there are probably some topologists that claim to be able to. The best that an otherwise ordinary human could do is to be able to visualize three dimensional "sections" of the 4 dimensional object. Since I can barely manage three dimensional objects, four dimensions are hopelessly out of my league. 8]
2) A metric is, more or less, a way to measure "distances" in a given kind of topological space. For example, in Euclidean three dimensional space the metric is s^2 = x^2 + y^2 + z^2. In Minkowski (aka Lorentzian) space-time the metric is s^2 = x^2 + y^2 + z^2 - (ct)^2. (Some authors use the negative of this.) It is called "indefinite" because s^2 can be a negative number.
3) The rubber sheet model is nice, but really only gives a rough idea of what's going on. Still, like the Bohr model of the atom, it has its uses. The idea is that if we put a mass on a level rubber sheet, it will cause an indentation. This gives us an idea of what how gravity and curved space-time function. Any small ball rolled near the object in the center will no long roll in a straight line, but will have a curved path. This represents the effects of a gravitational field.
4) As I understand it the light cone shows what "events" (basically coordinates) that one can reach when starting out at a different event. It is represented by a cone marked off by straight lines (which represent the places that an object traveling at the speed of light can reach.) Any event outside the light cone cannot be reached.
-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.
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https://www.projecteuclid.org/euclid.aaa/1393449713 | 1,571,793,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00289.warc.gz | 1,049,870,782 | 10,763 | ## Abstract and Applied Analysis
### Asymptotic Behavior of Solutions to a Vector Integral Equation with Deviating Arguments
#### Abstract
In this paper, we propose the study of an integral equation, with deviating arguments, of the type $y(t)=\mathrm{\omega }(t)-{\int }_{\mathrm{0}}^{\mathrm{\infty }}\mathrm{}f(t,s,\mathrm{y}({\gamma }_{\mathrm{1}}(\mathrm{s})),\dots ,\mathrm{y}({\gamma }_{N}(\mathrm{s})))\mathrm{}ds,\mathrm{}\mathrm{}\mathrm{}\mathrm{}t\ge \mathrm{0},$ in the context of Banach spaces, with the intention of giving sufficient conditions that ensure the existence of solutions with the same asymptotic behavior at $\mathrm{\infty }$ as $\mathrm{\omega }(t)$. A similar equation, but requiring a little less restrictive hypotheses, is $y(t)=\mathrm{\omega }(t)-{\int }_{\mathrm{0}}^{\mathrm{\infty }}\mathrm{}q(t,s)\mathrm{}F(s,\mathrm{y}({\gamma }_{\mathrm{1}}(s)),\dots ,\mathrm{y}({\gamma }_{N}(s)))\mathrm{}ds,\mathrm{}\mathrm{}\mathrm{}\mathrm{}t\ge \mathrm{0}.$ In the case of $q(t,s)=(t-s{)}_{+}$, its solutions with asymptotic behavior given by $\mathrm{\omega }(t)$ yield solutions of the second order nonlinear abstract differential equation $y\text{'}\text{'}(t)-\omega \text{'}\text{'}(t)+F(t,\mathrm{y}({\gamma }_{\mathrm{1}}(t)),\dots ,\mathrm{y}({\gamma }_{N}(t)))=\mathrm{0},$ with the same asymptotic behavior at $\mathrm{\infty }$ as $\mathrm{\omega }(t)$.
#### Article information
Source
Abstr. Appl. Anal., Volume 2013, Special Issue (2013), Article ID 957696, 7 pages.
Dates
First available in Project Euclid: 26 February 2014
https://projecteuclid.org/euclid.aaa/1393449713
Digital Object Identifier
doi:10.1155/2013/957696
Mathematical Reviews number (MathSciNet)
MR3147793
Zentralblatt MATH identifier
07095536
#### Citation
González, Cristóbal; Jiménez-Melado, Antonio. Asymptotic Behavior of Solutions to a Vector Integral Equation with Deviating Arguments. Abstr. Appl. Anal. 2013, Special Issue (2013), Article ID 957696, 7 pages. doi:10.1155/2013/957696. https://projecteuclid.org/euclid.aaa/1393449713
#### References
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• A. Constantin, “On the existence of positive solutions of second order differential equations,” Annali di Matematica Pura ed Applicata. Series IV, vol. 184, no. 2, pp. 131–138, 2005.
• S. G. Dubé and A. B. Mingarelli, “Note on a non-oscillation theorem of Atkinson,” Electronic Journal of Differential Equations, vol. 2004, article 22, 6 pages, 2004, http://ejde.math.txstate.edu.
• M. Ehrnström, “Positive solutions for second-order nonlinear differential equations,” Nonlinear Analysis: Theory, Methods & Applications A, vol. 64, no. 7, pp. 1608–1620, 2006.
• M. Ehrnström, “Linear asymptotic behaviour of second order ordinary differential equations,” Glasgow Mathematical Journal, vol. 49, no. 1, pp. 105–120, 2007.
• C. González and A. Jiménez-Melado, “Existence of monotonic asymptotically constant solutions for second order differential equations,” Glasgow Mathematical Journal, vol. 49, no. 3, pp. 515–523, 2007.
• C. González and A. Jiménez-Melado, “Asymptotic behavior of solutions to an integral equation underlying a second-order differential equation,” Nonlinear Analysis: Theory, Methods & Applications A, vol. 70, no. 2, pp. 822–829, 2009.
• O. Lipovan, “On the asymptotic behaviour of the solutions to a class of second order nonlinear differential equations,” Glasgow Mathematical Journal, vol. 45, no. 1, pp. 179–187, 2003.
• O. G. Mustafa and Y. V. Rogovchenko, “Global existence of solutions with prescribed asymptotic behavior for second-order nonlinear differential equations,” Nonlinear Analysis: Theory, Methods & Applications A, vol. 51, no. 2, pp. 339–368, 2002.
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• S. Mazur, “Uber die kleinste konvexe Menge, die eine gegebene kompakte Menge enthalt,” Studia Mathematica, vol. 2, pp. 7–9, 1930. | 1,358 | 4,501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 9, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.469233 |
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# If the sum of the roots of the equation x2 − x = λ(2x − 1) is zero, then λ = - CBSE Class 10 - Mathematics
ConceptSolutions of Quadratic Equations by Factorization
#### Question
If the sum of the roots of the equation x2 − x = λ(2x − 1) is zero, then λ =
• −2
• 2
• $- \frac{1}{2}$
• $\frac{1}{2}$
#### Solution
The given quadric equation is x2 − x = λ(2x − 1) , and roots are zero.
Then find the value of λ.
Here,
x^2 - x = λ (2x - 1)
x^2 - x2λx + λ = 0
x^2 - (1 + 2λ)x + λ = 0
a = 1,b =-(1+2λ) and , c = λ
As we know that D = b^2 - 4ac
Putting the value of a = 1,b =-(1+2λ) and , c = λ
= {-(1 + 2λ)}^2 - 4 xx 1 xx λ
=1 + 4λ + 4λ^2 - 4λ
= 1 + 4λ^2
The given equation will have zero roots, if D= 0
1 +4λ^2 = 0
4λ^2 = -1
λ^2 = (-1)/4
λ = sqrt((-1)/4)
= (-1)/2
Therefore, the value of λ = -1/2
Is there an error in this question or solution?
#### APPEARS IN
Solution If the sum of the roots of the equation x2 − x = λ(2x − 1) is zero, then λ = Concept: Solutions of Quadratic Equations by Factorization.
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# if n and k are positive integers, is sqrt. of
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if n and k are positive integers, is sqrt. of [#permalink]
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if n and k are positive integers, is sqrt. of (n+k)>2sqrt.of n?
k>3n
n+k>3n
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Joined: 14 Jul 2005
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Re: DS- n & k [#permalink]
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10 Sep 2006, 12:46
lan583 wrote:
if n and k are positive integers, is sqrt. of (n+k)>2sqrt.of n?
k>3n
n+k>3n
A should be it...
From 1, k > 3n, adding n to both sides
k+n > 3n+n
k+n>4n
Taking sqrt, sqrt(k+n) > 2sqrt(n)
SUFF
From 2, n+k>3n, taking sqrt, sqrt(n+k) > sqrt(3n) which is INSUFF since
sqrt(3n) < sqrt(n+k) < 2*sqrt(n)
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### Show Tags
11 Sep 2006, 01:31
if n and k are positive integers, is sqrt. of (n+k)>2sqrt.of n?
1)k>3n
2) n+k>3n
square both sides
n+k>4n.....ie k>3n
from one ...needless to say anything......suff
from two
k>2n we can never deduct that k>3n....insuff
11 Sep 2006, 01:31
Display posts from previous: Sort by | 602 | 1,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-26 | latest | en | 0.862994 |
http://www.math-only-math.com/worksheet-on-third-grade-measurement-of-length.html | 1,537,360,040,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156224.9/warc/CC-MAIN-20180919122227-20180919142227-00121.warc.gz | 376,329,404 | 7,357 | # Worksheet on Third Grade Measurement of Length
Practice the math worksheet on third grade measurement of length. We know, there are three main standard units of length, i.e., kilometer (km), meter (m) and centimeter (cm). The relation between these units will help us to solve different types of problems on measuring length.
1. Fill in the blanks:
(i) 1 km = _________ m
(ii) 1 m = _________ cm
(iii) 1 dm = _________ cm
(iv) 1 m = _________ dm
(v) 1 km = _________ cm
2. (i) Convert 18 m into centimeters.
(ii) Change 32 m 47 cm into centimeters.
(iii) Convert 7 km into meters.
(iv) Convert the following:
(a) 372 cm into meters and centimeters
(b) 2832 cm into meters and centimeters
(v) Convert the following as desired:
(a) 3275 m into kilometers and meters
(b) 5217 m into kilometers and meters
(c) 57 dm into m and cm
(i) 72 m 50 cm + 15 m 75 cm
(ii) 102 m 25 cm + 375 m 85 cm
(iii) 8 dm 9 cm + 9 dm 8 cm
(iv) 75 km 312 m + 34 km 215 m
(v) 212 km 132 m + 18 km 230 m
(vi) 81 km 570 m + 17 km 915 m
4. Subtract the following:
(i) 128 km 525 m - 20 km 278 m
(ii) 25 m 38 cm - 18 m 78 cm
(iii) 9 dm 3 cm - 2 dm 9 cm
(iv) 73 km 975 m 87 cm - 23 km 375 m 27 cm
(v) 250 km 378 m 75 cm - 150 km 287 m 50 cm
Answers for the worksheet on third grade measurement of length are given below to check the exact answers of the above questions on standard units of length.
1. (i) 1000 m
(ii) 100 cm
(iii) 10 cm
(iv) 10 dm
(v) 100000 cm
2. (i) 1800 centimeters
(ii) 3247 centimeters
(iii) 7000 meters
(iv) (a) 3 m 72 cm
(b) 28 m 32 cm
(v) (a) 3 km 275 m
(b) 5 km 217 m
(c) 5 m 70 cm
3. (i) 88 m 25 cm
(ii) 478 m 10 cm
(iii) 18 dm 7 cm
(iv) 109 km 527 m
(v) 230 km 362 m
(vi) 99 km 485 m
4. (i) 108 km 247 m
(ii) 6 m 60 cm
(iii) 6 dm 4 cm
(iv) 50 km 600 m 60 cm
(v) 100 km 91 m 25 cm | 688 | 1,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-39 | latest | en | 0.648266 |
https://cs.stackexchange.com/questions/51599/why-is-c-a-combinational-circuit-but-d-not/51604#51604 | 1,638,661,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00564.warc.gz | 262,786,600 | 34,619 | # Why is c) a combinational circuit, but d) not?
I am doing practice after just learning what combinational circuits are, yet I am unsure of why (c) is combinational, but (d) is not. Can someone please explain to me why this is? The Solution mentions that for (d) node n6 connects to the output terminals of both I3 and I4, but how is this different from the circuit in (c)?
• I suggest you check the definition of "combinational circuit" (I don't know what that definition is). The difference between (c) and (d) appears to be that, in (d), the output wires of the two gates on the left are connected together and fed into the single output of the gate on the right, whereas, in (c), the outputs of the two AND gates aren't connected together; rather, the output of one AND goes into one input of the OR gate, and the output of the other AND goes into the other input. Jan 7 '16 at 21:58
• Please don't include the question as an image. It makes information retrieval (i.e., finding your question) much harder. Instead, transcribe its textual parts. Jan 7 '16 at 22:37
A combinatorial circuit corresponds to a "straight-line program" that computes the outputs of the circuit given its inputs. (It has to satisfy some constraints that will exclude circuit (f) among others.)
As an example, if we denote the inputs of circuit (c) by $$x_1,x_2,x_3,x_4$$ (top to bottom) and the output by $$o$$, then using $$g_1,g_2$$ for the intermediate gates, the circuit corresponds to the program
$$g_1 = x_1 \land x_2$$
$$g_2 = x_3 \land x_4$$
$$o = g_1 \lor g_2$$
It also corresponds to the formula $$(x_1 \land x_2) \lor (x_3 \land x_4)$$. However, not every circuit corresponds to a formula. (The difference is that in some circuits, the output of a gate can appear as an input to more than one gate.)
Circuit (d) cannot be written in this form, since the outputs of I3 and I4 are wired together. What is the relation between the input to the rightmost gate and the outputs of I3 and I4? Not something that can be described combinatorially.
The difference between circuit (c) and circuit (d) is that in circuit (c) wires are not joined together. Rather, the output of each of the AND gates is connected to its own input of the OR gate. In contrast, in circuit (d) the rightmost NOT gate has only one input, but two incoming wires that must share the same input pin.
• nice, is the flip-flop a combinatorial circuit? Jan 8 '16 at 10:26
• No, since its underlying graph is not a DAG. Jan 8 '16 at 10:38
• how are these type of circuits handled in theoretical computer science (circuits with feedback, or loops lets say)? Jan 8 '16 at 10:40
• They are summarily ignored. Jan 8 '16 at 10:45
• @NikosM., the nicer can be modelized as state machines. Jan 8 '16 at 16:58 | 740 | 2,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-49 | latest | en | 0.923171 |
http://math.stackexchange.com/questions/266197/application-of-schwarz-lemma?answertab=oldest | 1,461,941,204,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111365.36/warc/CC-MAIN-20160428161511-00182-ip-10-239-7-51.ec2.internal.warc.gz | 188,005,732 | 17,409 | # Application of Schwarz lemma
Each analytic function mapping the right half complex plane into itself must satisfy $$\left|\frac{f(z)-f(1)}{f(z)+f(1)}\right| \leqslant \left|\frac {z-1}{z+1} \right|$$ for $\text{Re}\; z > 0.$
I have a hunch that this is an application of Schwarz's Lemma. I don't know how to proceed though. Thanks in advance.
-
Great question! I'm asking a similar question here, which might be a more general result: math.stackexchange.com/questions/1245940/… – EthanAlvaree Apr 22 '15 at 12:47
The map $h$ from $\{z,\Re z>0\}$ to the open unit disk $D$ given by $h(z)=\frac{z-1}{z+1}$ is one-to-one, then use Schwarz lemma with $g(z):=\dfrac{f\left(\frac{1+z}{1-z}\right)-f(1)}{f\left(\frac{1+z}{1-z}\right)+f(1)}$.
this is actually a past qual question. I am thinking $f$o$h^{-1}$ is actually a mapping from unit disk to right half plane, but don't we need the mapping from unit disk to unit disk in Swartz's Lemma? I am confused. Could you please elaborate? – Deepak Dec 29 '12 at 1:11
,But, I still have difficulty seeing ${h\circ}f\circ h^{-1}(0)=0$. I can see after that Swartz lemma gives the result. – Deepak Dec 30 '12 at 3:43 | 377 | 1,159 | {"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} | 3.140625 | 3 | CC-MAIN-2016-18 | latest | en | 0.835305 |
https://appadvice.com/review/quickadvice-polygon-evolution | 1,653,503,575,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00572.warc.gz | 158,688,329 | 44,536 | You are using an outdated browser. Please upgrade your browser to improve your experience.
# ExerciseYourBrainWithThePuzzlingAndMusicalPolygonEvolution
June 12, 2014
Polygon Evolution (Free) by Alex Dantis, creator of the awesome Synesthetic, is a complex puzzle game that is definitely for advanced players. What looks like it should be easy is actually quite challenging behind all of the neon colors and beats, but can you handle it? Let’s find out.
You don’t want to miss the included tutorial in Polygon Evolution, which is something I went through three times before even having an idea about how the game works. It’s not unreasonably difficult, but at the same time, it’s no walk in the park by any means. Essentially, there is a large board consisting of hexagons, where “cells” can occupy themselves. Any given cell has its own specific order, ranging from a single dot, to a line, cross, triangle, or hexagon. The higher the order, the more powerful the cell.
One thing that didn’t strike me at first but has become obvious now is that the gameplay, while completely different in and of itself, is very analogous to that in Threes!. Just like a Threes! player must make threes before making higher numbers, which must be combined with themselves to be doubled, in Polygon Evolution, players can only create dot cells, which must be combined in lines of at least three to make line cells, cross cells, and so forth.
Of course, you can’t just go around creating dot cells mindlessly, since there is a turn limit in place, and the evil “opponent” cells will come into play at some point. Black opponent cells can do exactly what your cells can, and will mimic your moves in their preferred direction — something that is difficult to predict or even notice at times when things are busy. To avoid having these suckers take over the whole board, creating cells of a higher order (triangle or above) will wipe these bad guys out.
Having to worry about all these parameters, including a few others like how you can only create cells that branch off of your own, getting even somewhat far in Polygon Evolution is like solving a seemingly invalid equation. With practice, things do get easier, as you’ll get an eye for what to do next and will, just like in Threes!, be able to set up good arrangements of cells. But in reality, even that’s hard, and after playing until getting a headache, the game still feels like it was created for Einstein.
Despite not being for the faint of heart who are looking for an easy, relaxing puzzle game, Polygon Evolution is perfect for anyone who loves tinkering with things and trying to learn complex concepts. If you’re a hands-on learner who is able to look at a situation from all sorts of angles while asking many questions and formulating ideas, then this universal game could be great for you.
Besides including a few in-app purchases for hints, which will make the game easier, and for visual themes, Polygon Evolution is absolutely free to play, available on the App Store here.
#### Mentioned apps
Free
Polygon Evolution
Alex Dantis
\$2.99
Synesthetic
Alex Dantis
\$1.99
Threes!
Sirvo LLC | 677 | 3,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.966601 |
https://www.smore.com/9mbzk-hurricane-forecast-february-2023 | 1,696,463,350,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00357.warc.gz | 1,093,678,129 | 29,316 | # Hurricane Forecast February 2023
## Dates to Remember-
Wednesday, February 1st: First Day of Black History Month
Friday, February 10th and Monday, February 13th: No School for Pre-K Students
Monday, February 13th - Friday, February 17th: Random Acts of Kindness Week (See flyer below)
Friday, February 17th: Spelling Bee
Thursday, February 16th: PTA Spirit Night at Scooters Staking Rink 6-8pm
Monday, February 20th: Teacher Workday - No School
Wednesday, February 22nd: School Bus Driver Appreciation Day - Thank your bus driver!
## Kindergarten
In reading we began working through our fiction unit. We are learning that fictional stories have characters, settings, problems and solutions. In fundations we began unit 3 continuing to tapping out CVC words and starting to learn new trick words. In math we have started Module 4, where we are learning to find parts of a whole. We have finished up our unit on bears in science and are beginning our unit on past and present in social studies.
In February first graders will be focusing on nonfiction books. We will locate and determine the meaning of unknown words. Students will do this by using text features, context clues, and words around the text to solve what the word means. In Social Studies we will dig into history by learning about influential people from the past that have shaped our history. In Science we will begin learning about Forces and Motion. Students learn for every action there is a reaction. In Writing we will continue our narrative unit. Students will be writing stories about themselves and people who are important to them. In Phonics students will begin Unit 8 in Fundations. Throughout this unit students will learn consonant blends. A blend is two consonants together that each make a sound, as in the word “s t o p” - the /s/ and the /t/ each make a sound. A blend is different from a digraph which has two letters but only onesound, such as in the word “shop” - the /sh/ makes only one sound.mBlends can come at the beginning of a word, such as “s t o p”, or at the end of a word, like “p e s t.” Your child will learn a digraph blend. A digraph blend is a digraph blended with another consonant such as n and ch in the word lunch, or sh and r in the word shred. In Math we will continue module 4 (Comparison and Composition of Length Measurements). In module 4, students explore units within the context of measurement. After comparing lengths indirectly, students iterate length units, such as centimeter cubes and 10-centimeter sticks, to describe and compare lengths.
This quarter we are beginning our unit on nonfiction. We will be exploring different types of nonfiction text and learning to identify and use the text features to gather more information about the text. Students will also learn to identify the main topic and then determine the main idea by using the details from the text to support it.
In science, the students have been so excited about the arrival of the chicken eggs and learning all about the life cycle of a chicken. In math, we are continuing to learn about geometric shapes. Students will learn to recognize and draw shapes based on their attributes. As they continue to work with composite shapes they will learn to partition them into equal parts and build fractional understanding and apply it to telling time. In Social Studies, we will read and learn about important historical figures and the contributions they have made.
3rd grade is continuing to work hard on multiplication; memorizing facts and understanding multiplication patterns.
We are immersing ourselves in books by author Patricia Polacco. We will be using her books as mentor texts to draft our own personal narratives. In Social Studies, we will begin our reading and studying about economics and explain how the natural resources of a region impact the production and consumption of goods in local communities.
In Reading students will be reviewing Making Inferences, Summarizing, and Main Idea and Details. This month we will also begin our grade level novel study The Wild Robot. Students will be applying all reading standards while reading through this novel.
In Math students will begin Module 3: Multiplication and Division strategies. This unit builds on concepts learned in Module 2. Students would benefit from practicing their multiplication facts each night. Students can use the deck of cards and games that were sent home after our parent involvement night last month.
In Social Studies students are working through the Civics and Government Unit: Students will take a close look at citizens roles, important state documents, rights and responsibilities, and who are our state leaders.
Fourth grade just enjoyed a fabulous field trip to the Wilson Center to see the Wilmington Symphony. Students were excited to go on their second field trip of the year. Everyone had a great time and experience.
In Math, we wrapped up our fraction unit. We just started our decimal unit where students have been reading and writing decimal numbers to the thousandths place. We have a lot to learn about decimals and we are just getting started! We wrapped our Ecosystems unit in Science when we came back from winter break and then moved into learning about unicellular and multicellular organisms. We are currently learning about the different human body systems - what their purpose is, the organs/body parts in each system, and how the different systems work together. In Literacy we have been reading grade-level nonfiction texts and answering questions by accurately quoting evidence from the text. We also reviewed the different text structures and how determining these structures as we are reading can help make sense of the text.
## Ms. Cleary, Art
Art Room Supply Donations
If you happen to be cleaning out your storage areas and come across any
of the items below I promise they will put to good use in the Art Room at
Castle Hayne.
• Blender that works for Papermaking
• Iron for mounting artwork
• Hot Plate for melting
• Drill with large liquid mixing bit for glazes and slip
• 1,3,5, gallon plastic buckets with airtight covers
• Small magnets
• Sidewalk chalk
Congratulations to these Castle Hayne Artists. They have been selected to participate in the North Carolina Azalea Festival Art contest. Wish them luck! The Azalea Festival will be April 12-16th. 2023. The Art exhibition will take place at the Hannah Block Community Arts Center Downtown Wilmington.
Kygo R. , 1TH, Ms. Miller, Venus Fly Trap
Coraleigh S., 1T, Ms. Perry , Still Life, Vase with Flowers
Javante H., 2T, Ms. Hicks, Bird with Gold Wreath
Drake G., 2T, Ms. Hicks, Venus Fly Trap with Blue Flower
Eli F. , 2T, Ms Hicks
Stephen H. , 2T, Ms. Hicks, Venus Fly Trap with Dots
Valeria I., 2T, Ms. Hicks , Venus Fly Trap
Tre’shien F. , 2T, Ms. Hicks , Venus Fly Trap with Bugs
Conner J. , 2W, Ms. J. Williams
Leoki R., 2TH, Mr. Todd, Venus Fly Trap with Fly
Mason S., 2TH, Mr. Todd, Tree of Life
William C., 3M, Ms. Houseman, Landscape Hillside in the Spring
Camryn H., 3TH, Ms. Snee, Blue Rabbit with Azaleas
Josiah G., 4T, Ms. Burley, Rabbit in Green Field
Thomas I., 4T, Ms. Burley, Dashing Rabbit 2023
Vivienne H., 4TH, Ms. Hobbs, Azalea Wreath
Charlotte P., 4Th, Ms. Hobbs, Azalea Wreath Hawaiian
Emmy M., 4TH, Ms. Hobbs , Azalea Wreath with red and green
## Candy Robbins, School Social Worker
Attendance policy and recycle program details below.
## Mrs. Cervini, ESL Teacher
In ESL we have been busy working on and following through with our goals for the new year. We have been expanding our vocabulary as we discovered more about Martin Luther King Jr and the difference he made in our world. We learned how Martin Luther King Jr carefully chose powerful words when communicating. We each chose a powerful word to represent ourselves. Students were so inspired that they even wrote their own positive quotes or messages for our school community. We too can easily make a difference when we choose our words carefully.
## Celebrating Black History Month
Black History Month
Every February, people in the United States celebrate the achievements and history of African Americans as part of Black History Month.
BYKAY BOATNER
HOW IT STARTED
In 1915, in response to the lack of information on the accomplishments of Black people available to the public, historian Carter G. Woodson co-founded the Association for the Study of Negro Life and History. In 1926, the group declared the second week of February as “Negro History Week” to recognize the contributions of African Americans to U.S. history. Few people studied Black history and it wasn't included in textbooks prior to the creation of Negro History Week.
This week was chosen because it includes the birthdays of both Frederick Douglass, an abolitionist (someone who wanted to end the practice of enslaving people), and former U.S. president Abraham Lincoln. President Lincoln led the United States during the Civil War, which was primarily fought over the enslavement of Black people in the country. Many schools and leaders began recognizing the week after its creation.
The week-long event officially became Black History Month in 1976 when U.S. president Gerald Ford extended the recognition to “honor the too-often neglected accomplishments of Black Americans in every area of endeavor throughout our history.” Black History Month has been celebrated in the United States every February since.
WHAT IT HONORS
Black History Month was created to focus attention on the contributions of African Americans to the United States. It honors all Black people from all periods of U.S. history, from the enslaved people first brought over from Africa in the early 17th century to African Americans living in the United States today.
Among the notable figures often spotlighted during Black History Month are Dr. Martin Luther King, Jr., who fought for equal rights for Blacks during the 1950s and ’60s; Thurgood Marshall, the first African-American justice appointed to the United States Supreme Court in 1967; Mae Jemison, who became the first female African-American astronaut to travel to space in 1992; and Barack Obama, who was elected the first-ever African-American president of the United States in 2008.
BLACK HISTORY MONTH TODAY
Since the first Negro History Week in 1926, other countries have joined the United States in celebrating Black people and their contribution to history and culture, including Canada, the United Kingdom, Germany, and the Netherlands.
Today Black History Month continues the discussion of Black people and their contributions through activities such as museum exhibits and film screenings, and by encouraging the study of achievements by African Americans year-round.
## Kindergarten Enrollment begins March 1-15th
Families with children who will be 5 on OR before August 31, 2023 it's that time of year again. Time to start thinking about Kindergarten enrollment.
Do you want a year round schedule? Are you looking for a bilingual option? Maybe you want a school with a focus on the arts or STEM or you have selected your districted school? Whatever your choice, NHCS has options.
Enrollment begins March 1-15
## Safety is our TOP Priority - News from the Nurse
Cold and flu season has arrived, and we are seeing increased cases of the flu across New Hanover County and in our schools. We are expecting a busier flu season than normal and influenza can be very serious. Please adhere to the following CDC guidelines and suggested healthy habits. Let’s work together to keep our school and our homes as healthy as possible.
• Fever of 100 degrees or higher within the last 24 hours. (Must be fever-free without medication)
• Vomiting or diarrhea within the last 24 hours.
• Symptoms that make it hard for your child to participate in school such as frequent coughing, headache, body aches, fatigue, earaches.
• Strep Throat- until on antibiotics 24 hr.
Tips for Staying Healthy
• Cover coughs and sneezes with a tissue. Use elbow or arm if tissue is not available.
• Wash hands often with soap and warm water for 20 seconds (Sing ABC song twice).
• Use hand sanitizer if soap and water are not available.
• Avoid touching the “T” zone (Eyes, Nose and Mouth)
• Avoid being around others who are sick.
• Do not share drinks, food, or unwashed utensils.
• Eat healthy (Foods that Grow!) and drink plenty of water
• Disinfect surfaces that are prone to germs (phones, keyboards, doorknobs, toothbrushes, etc.).
• Get the flu vaccine.
Thanks for your help in keeping Castle Hayne Elementary healthy this year!
¡Ha llegado la temporada de resfriados y gripe y estamos viendo un aumento en los casos de gripe. La gripe puede ser grave. Siga las siguientes pautas de los CDC y los hábitos saludables sugeridos. Trabajemos juntos para mantener nuestra escuela y nuestros hogares lo más saludables posible.
Por favor no envíe a su hijo a la escuela si él / ella ha tiene o ha tenido:
• Fiebre de 100 grados o más en las últimas 24 horas. (Debe estar libre de fiebre sin medicación)
• Vómitos o diarrea en las últimas 24 horas.
• Síntomas que les dificultan participar en la escuela, como tos frecuente, dolor de cabeza, dolor de cuerpo, fatiga, dolor de oído.
• Faringitis estreptocócica: hasta con antibióticos las 24 h.
Consejos para mantenerse saludable
• Cubra la tos y los estornudos con un pañuelo de papel. Use el codo o el brazo si no hay tejido disponible.
• Lávese las manos con frecuencia con jabón y agua tibia durante
• 20 segundos (el tiempo que lleva cantar la canción ABC).
• Use desinfectante para manos si no hay agua y jabón disponibles.
• Evite tocar la zona "T" (ojos, nariz y boca)
• Evite estar cerca de otras personas que están enfermas.
• No comparta bebidas, alimentos o utensilios sin lavar.
• Coma sano (¡Alimentos que crecen!) Y beba mucha agua
• Desinfecte las superficies propensas a gérmenes (teléfonos, teclados, pomos de puertas, cepillos de dientes, etc.).
• Vacúnese contra la gripe.
De la enfermera
## Cafeteria Message
This school year every student receives FREE breakfast and lunch. However, we will be utilizing My School Bucks to access your child's cafeteria account for any a al carte items. I have included a link to the website below as well as a link to the NHCS Child Nutrition Website. You can find more information regarding My School Bucks under the "Pre-Payment Options" link. If you do not wish to add money electronically via My School Bucks you are still able to send cash or check.
If you choose to bring cash or check to school personally or send it with your student, please put it in an envelope clearly marked with the student's first and last name, their ID#, their teacher's name, and the amount enclosed.
Calendar Menus will be posted on the CHES Child Nutrition website monthly.
The Cafeteria Staff at CHES look forward to serving you this year!
## Castle Hayne Elementary School's PTA Corner
Thank you for supporting the PTA!!
## Lost and Found
As the new year begins PLEASE label all outwear including hoodies with your child's name this makes it so much easier locating the owner when an item is found.
The Lost and Found box is located in front of the Media Center; if your child loses an item make certain they check the LOST and FOUND box as soon as they realize the item is lost. This makes it so much easier to find. Thank you!!
## Ways to stay connected to your school!
1. NHCS uses School Messenger to connect with families regarding district and school happenings. Use this link to opt-in to receive text messages from School Messenger. Flyer below with instructions.
3. Our website.
## Volunteering at Castle Hayne
Volunteers are our partners in education. They take an active role to support and enhance the development of our students. Every day, family and community member volunteers bring enthusiasm and skill into our schools while assisting staff, teachers and students. All volunteers are required to obtain volunteer training annually.
If you would like to move forward with volunteering at Castle Hayne Elementary School these are the following steps you will need to complete:
1. Submit the Volunteer Application Form/Spanish to Danielle Metty, danielle.metty@nhcs.net via email or send a printed copy to school with your student labeled with Danielle Metty, Family Liaison.
2. Complete the Volunteer Orientation Assessment with a score of 80% or higher.
3. To become a level 2 volunteer, complete a background check application (\$20 until October 1 then the cost will be \$22.50). You will receive a verification email and card from the Background Investigation Bureau upon approval.
Notify Danielle Metty, danielle.metty@nhcs.net when you have completed steps 1-4.
If you are interested in volunteering or chaperoning a field trip at Castle Hayne you MUST complete the Volunteer Training, please contact Danielle Metty at danielle.metty@nhcs.net.
## Transportation
Tracking your child's bus has never been so easy. New Hanover County Schools is implementing a new smart phone app that will allow parents to see where their child's bus is located in real time. Edulog Parent Portal app can be downloaded from iTunes App Store or Google Play Store.
The "Hurricane Forecast" is a monthly digital newsletter that contains information regarding all things happening at Castle Hayne Elementary School. We hope you find this communication helpful!
## Castle Hayne Elementary School
Our mission at Castle Hayne Elementary School is to work together to achieve success by applying our STORM skills!
Spread Kindness~ Take Responsibility~ Own our Learning and Actions~ Respect Everyone~ Maintain Safety
Please direct any questions related to this newsletter to Danielle Metty, Receptionist | 3,951 | 17,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.967917 |
https://science.blurtit.com/180747/how-many-inches-are-in-a-square-foot | 1,670,013,505,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00439.warc.gz | 545,935,788 | 12,513 | # How Many Inches Are In A Square Foot?
There are 144 sq in in a square foot. You get this answer by multiplying how many inches there are in a foot, which is 12, by itself (because it is a square so all the sides are of equal length). So, 12 x 12 equals 144.
A square foot is an imperial unit of measurement (in America it is called a U.S. Customary unit) of area and is very often used to measure areas within houses or other buildings. Square yards and square metres are also commonly used in these instances, with a square metre being a metric measurement.
A square foot is often seen represented by the abbreviation, ‘sq ft’, but in architecture or interior design plans it can often be seen as a symbol comprising of a square with a slash through it.
To work out the total square footage of a room is easy to do. You need to measure the length and the breadth of the room and then multiply the two numbers together. For example if a room measured 14 ft by 20 ft, you multiply 14 by 20, which will give you a total of 280 (14 x 20 = 280). You now know the square footage of that particular room.
If you need to calculate the square footage of an entire house, or other building, simply calculate the square footage of each individual room and then add them all together. Naturally, this rule applies if you want your measurements to be in square yards or square metres too. Just remember that a yard is three ft or 36 in, and that a metre is a 100 cm (or roughly 39 in if you need a metric measurement but do not have a metric rule).
thanked the writer.
1 foot = 12 inches.
1 square foot = 12 x 12 = 144 square inches.
There are not inches in a square foot, there are "square inches".
thanked the writer.
144 inches are in a square foot. I hope this helped!!!
thanked the writer.
144
thanked the writer.
The formula is as follows;
1 square foot = 12 inches * 12 inches = 144 square inches
So it means there are 144 inches in a square foot.
thanked the writer.
There is 12 in. In a foot x 12=144
thanked the writer.
144
12x12=144
thanked the writer. | 500 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.959835 |
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# Assignment No. 02-MTH603 (Fall 2016)
Assignment MTH603 (Fall 2016)
Maximum Marks: 20 Due Date: 23 -01-2017
DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 23 - 28 lectures.
Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 23 to 28 lectures.
• Upload assignments properly through LMS, No Assignment will be accepted through email.
Don’t use colorful back grounds in your solution files.
Use Math Type or Equation Editor Etc. to insert mathematical symbols.
You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
Try to make solution by yourself and protect your work from other students, otherwise you and the student, who send same solution file as you, will be given zero marks.
Also remember that you are supposed to submit your assignment in MS Word format any other like scan images etc. will not be accepted and we will be given zero marks correspond to these assignments.
Question No. 1: Given the following data, evaluate f(1.5) using Lagrange’s interpolation formula
Question No. 2: Find Newton’s Divided Difference table for the following data.
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Please check the solution, and tell me this is correct or not
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Solution:-
App ka answer tick hy likan question body tick nehi.
Attachments:
now check this Khushdil Kamal
same ans.
urwa 3.75 answer theek he k nai koi or hi bta do mera ye ans aya he
thk ha
mera b 3.75 answer aa rha he
theek he na
ya mery hial main is tara hy.
g thk ha u r right but wo aik jaga ghlt ha but baki thk ha
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2 | 644 | 2,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-45 | latest | en | 0.849832 |
https://evmc2.wordpress.com/2013/08/13/c-rate-and-c-rating/ | 1,531,709,241,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589172.41/warc/CC-MAIN-20180716021858-20180716041858-00046.warc.gz | 655,944,066 | 22,066 | # C-Rate and C-Rating
Wanna know a secret? Get confused about C-Rates of cells and all that? Here’s the Magic Decoder Ring.
“C-Rate” is a simple calculation of the amount of current going in or out of a battery. “C-rating”, on the other hand, is a tested value of what C-Rate a battery can charge or discharge without damage.
They are not interchangeable terms.
That’s it. That’s my big secret. Am I the only one? Did everybody else out there understand this?
Here’s the Battery University page on C-Rate“Today, the battery industry uses C-rate to scale the charge and discharge current of a battery.” You take a battery, discharge it at some specific C-Rate and decide if it can take it without blowing up. If you discharge it at 10C and it’s fine, then discharge it at 15C and it starts getting hot, then it’s probably a good bet that it’s a 10C rating. (Yes, that’s hugely oversimplified as far as the actual process goes, but don’t quibble details with me.)
What is C in relation to batteries?
C ratings are simply a way of talking about charge and discharge rates for batteries.
1C, = 1 time the rated mah capacity of the battery. So if you charge your 650 mah pack at 1C, you charge it a 650 miliamps, or .650 amps.
1C on a 1100 pack would be 1.1 amps.
2 C on your 1100 pack would be 2.2 amps
Motor batteries are often rated in Discharge C and charge C.
So a 1100 mah pack (1.1 amp hour) might be rated for 10C discharge, so you can pull 11 amps ( flow ) without damaging the battery.
Then it might be rated at 2C charge rate (flow), so you charge it at 2.2 amps (2200 mah)
How did I do? Things clearing up?
If you have a 500 mah pack – any kind – and it is rated at 16C that means it can deliver 8 amps.
If you have a 1000 mah pack – any kind – and it is rated at 8C that means it can deliver 8 amps.
If you have a 1000 mah pack – any kind – and it is rated at 12C that means it can deliver 12 amps
If you have a 1500 mah pack – any kind – and it is rate at 8C that means it can deliver 12 amps
If you have a 1500 mah pack – any kind – and it is rated at 20 C that means it can deliver 30 amps.
If you have a 3000 mah pack – any kind – and it is rated at 10 C that means it can deliver 30 amps.
So, if you need 12 amps you can use a pack with a higher C rating or a pack with a higher mah rating to get to needed amp delivery level.
The formula is Current /capacity=rating, or to find the safe current discharge rate, rating x capacity = current. Armed with this, let’s look at a 20ah 5C cell, like the Farasis:
A = 20ah x 5C = 100A
Pretty simple. Now, remembering that when you parallel cells, you add up capacity while not adding voltage, if you parallel four of these to get an 80ah pack, you’re going to get 400A out of it (safely). | 778 | 2,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-30 | latest | en | 0.893594 |
https://kr.mathworks.com/matlabcentral/profile/authors/2991461 | 1,718,255,132,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00631.warc.gz | 324,065,983 | 23,337 | # Jay
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답변 | 1,915 | 6,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.539571 |
http://stackoverflow.com/questions/7446389/computational-intensity-of-transposing-a-matrix-vs-calculting-the-inverse | 1,394,604,326,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021389272/warc/CC-MAIN-20140305120949-00018-ip-10-183-142-35.ec2.internal.warc.gz | 179,053,557 | 15,035 | # Computational intensity of transposing a matrix vs calculting the inverse
I'm taking a crash course in computer graphics and we have just covered 2D transformations and my instructor pointed out that it was computationally more intensive to calculate the inverse of a matrix than its transpose, which is why orthogonal matrices are soo useful in computing (since the transpose of an orthogonal matrix is also it's inverse). Due to the time constrains, and also the nature of the class, he didn't go into details on why this is and I was wondering if anyone here would be able to do so.
I'm particularly interested in the difference in CPU/GPU instructions involved in each process, or if I've gotten that wrong, then the lowest level in the stack at which the bottleneck is occurring. I'd also be interested in learning of any resources, books, websites, etc, where I can learning more about these sorts of effiencies/bottlenecks.
-
Transposing an n x n matrix is, at worst, an O(n2) operation, whereas computing the inverse of a general nonsingular n x n matrix is an O(n3) operation. That is just the cost of doing these things.
As an aside, most applications do not compute matrix inverses because the objective is solving a linear system Ax = b, not finding the inverse. It is faster and more accurate to solve this problem using decompositions and triangular solves (see LU decomposition, for general nonsingular matrices, for instance). Matrix inverses can be computed using Gauss-Jordan elimination (as well as other approaches).
As for routines available, you can find LAPACK implementations that can be used for computing an LU decomposition (and probably a matrix inverse).
- | 349 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2014-10 | latest | en | 0.951595 |
blvdeer.blogspot.com | 1,566,775,646,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330907.46/warc/CC-MAIN-20190825215958-20190826001958-00286.warc.gz | 35,539,308 | 10,709 | ## Tuesday, March 1, 2011
### Prime Number in Perl with a twist
Just a few days ago, I learned that all prime numbers > 3 can be represented as multiple of 6 +1 or -1.
So here's a code in perl to find nth prime number.
#!/usr/bin/perl
print "Enter the number: ";
chomp(\$num = <>);
\$i = 3;
\$no_of_prime = 3;
\$prime = 0;
if (\$num == 1)
{
print "1st prime is 2 \n";
}
elsif (\$num == 2)
{
print "2nd prime is 3 \n";
}
elsif (\$num > 2)
{
\$index = 1;
\$powindex = 1;
\$no_of_prime = 2;
while(\$no_of_prime < \$num)
{
\$prime = (6 * \$index) + ((-1) ** \$powindex);
\$j = 2;
\$is_prime = 1;
while(\$j <= (\$prime ** 0.5))
{
if((\$prime % \$j) == 0)
{
\$is_prime = 0;
break;
}
\$j = \$j + 1;
}
if (\$is_prime == 1)
{
\$no_of_prime++;
}
if((\$powindex % 2) == 0)
{
\$index = \$index + 1;
}
\$powindex = \$powindex + 1;
}
print \$num, "th prime is: ", \$prime, "\n";
}
else
{
print "Invalid number\n";
}
#END | 344 | 914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.469947 |
http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment4.mdwn;h=3e3b71841c6fa51e546c65c4c527cf798836de40;hp=54a3bf201d1914997fbf3d49cc669059d36a16e4;hb=95dad38cb4aa443a3dde5bad742d53f023b0ca33;hpb=5a235579e3764fed888bdbd8465c373634c63984 | 1,611,067,686,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00691.warc.gz | 60,039,276 | 2,658 | X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment4.mdwn;h=3e3b71841c6fa51e546c65c4c527cf798836de40;hp=54a3bf201d1914997fbf3d49cc669059d36a16e4;hb=95dad38cb4aa443a3dde5bad742d53f023b0ca33;hpb=5a235579e3764fed888bdbd8465c373634c63984 diff --git a/assignment4.mdwn b/assignment4.mdwn index 54a3bf20..3e3b7184 100644 --- a/assignment4.mdwn +++ b/assignment4.mdwn @@ -14,7 +14,12 @@ can use.
-
1. Suppose you have two lists of integers, `left` and `right`. You want to determine whether those lists are equal: that is, whether they have all the same members in the same order. (Equality for the lists we're working with is *extensional*, or parasitic on the equality of their members, and the list structure. Later in the course we'll see lists which aren't extensional in this way.) +
2. Suppose you have two lists of integers, `left` and `right`. You want to +determine whether those lists are equal: that is, whether they have all the +same members in the same order. (Equality for the lists we're working with is +*extensional*, or parasitic on the equality of their members, and the list +structure. Later in the course we'll see lists which aren't extensional in this +way.) How would you implement such a list comparison? @@ -27,16 +32,53 @@ How would you implement such a list comparison? First, read this: [[Implementing trees]]
-
1. blah +
2. Write an implementation of leaf-labeled trees. You can do something v3-like, or use the Y combinator, as you prefer. + +You'll need an operation `make_leaf` that turns a label into a new leaf. You'll +need an operation `make_node` that takes two subtrees (perhaps leaves, perhaps +other nodes) and joins them into a new tree. You'll need an operation `isleaf` +that tells you whether a given tree is a leaf. And an operation `extract_label` +that tells you what value is associated with a given leaf. And an operation +`extract_left` that tells you what the left subtree is of a tree that isn't a +leaf. (Presumably, `extract_right` will work similarly.) + +
3. The **fringe** of a leaf-labeled tree is the list of values at its leaves, +ordered from left to right. For example, the fringe of this tree: + + . + / \ + . 3 + / \ + 1 2 + +is [1;2;3]. And that is also the fringe of this tree: + + . + / \ + 1 . + / \ + 2 3 + +The two trees are different, but they have the same fringe. We're going to +return later in the term to the problem of determining when two trees have the +same fringe. For now, one straightforward way to determine this would be: +enumerate the fringe of the first tree. That gives you a list. Enumerate the +fringe of the second tree. That also gives you a list. Then compare the two +lists to see if they're equal. (You just programmed this above.) + +Write the fringe-enumeration function. It should work on the implementation of +trees you designed in the previous step. (See [[hints/Assignment 4 hint 3]] if you need some hints.)
+ #Mutually-recursive functions#
-
1. (Challenging.) One way to define the function `even` is to have it hand off part of the work to another function `odd`: +
2. (Challenging.) One way to define the function `even` is to have it hand off +part of the work to another function `odd`: let even = \x. iszero x ; if x == 0 then result is @@ -44,7 +86,8 @@ First, read this: [[Implementing trees]] ; else result turns on whether x's pred is odd (odd (pred x)) -At the same tme, though, it's natural to define `odd` in such a way that it hands off part of the work to `even`: +At the same tme, though, it's natural to define `odd` in such a way that it +hands off part of the work to `even`: let odd = \x. iszero x ; if x == 0 then result is @@ -52,7 +95,10 @@ At the same tme, though, it's natural to define `odd` in such a way that it hand ; else result turns on whether x's pred is even (even (pred x)) -Such a definition of `even` and `odd` is called **mutually recursive**. If you trace through the evaluation of some sample numerical arguments, you can see that eventually we'll always reach a base step. So the recursion should be perfectly well-grounded: +Such a definition of `even` and `odd` is called **mutually recursive**. If you +trace through the evaluation of some sample numerical arguments, you can see +that eventually we'll always reach a base step. So the recursion should be +perfectly well-grounded: even 3 ~~> iszero 3 true (odd (pred 3)) @@ -64,24 +110,29 @@ Such a definition of `even` and `odd` is called **mutually recursive**. If you t ~~> iszero 0 false (even (pred 0)) ~~> false -But we don't yet know how to implement this kind of recursion in the lambda calculus. +But we don't yet know how to implement this kind of recursion in the lambda +calculus. The fixed point operators we've been working with so far worked like this: let X = Y T in X <~~> T X -Suppose we had a pair of fixed point operators, `Y1` and `Y2`, that operated on a *pair* of functions `T1` and `T2`, as follows: +Suppose we had a pair of fixed point operators, `Y1` and `Y2`, that operated on +a *pair* of functions `T1` and `T2`, as follows: let X1 = Y1 T1 T2 in let X2 = Y2 T1 T2 in X1 <~~> T1 X1 X2 and X2 <~~> T2 X1 X2 -If we gave you such a `Y1` and `Y2`, how would you implement the above definitions of `even` and `odd`? +If we gave you such a `Y1` and `Y2`, how would you implement the above +definitions of `even` and `odd`? -
3. (More challenging.) Using our derivation of Y from the [Week3 notes](/week3/#index4h2) as a model, construct a pair `Y1` and `Y2` that behave in the way described. +
4. (More challenging.) Using our derivation of Y from the [Week3 +notes](/week3/#index4h2) as a model, construct a pair `Y1` and `Y2` that behave +in the way described. (See [[hints/Assignment 4 hint 4]] if you need some hints.) | 1,653 | 5,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-04 | latest | en | 0.890184 |
http://www.chegg.com/homework-help/questions-and-answers/two-systems-are-formed-from-a-converging-lens-and-a-diverging-lens-as-shown-in-parts-a-and-q2848274 | 1,369,201,987,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701409268/warc/CC-MAIN-20130516105009-00070-ip-10-60-113-184.ec2.internal.warc.gz | 382,946,325 | 8,644 | ## Physics converging/diverging lens
Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the drawing. (The point labeled "Fconverging" is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of 10.90 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and ?20.0 cm respectively. The distance between the lenses is 50.0 cm. Determine the final image distance for each system, measured with respect to lens 2.
• Anonymous commented
Plz Check Answer Properly...!Mine's Correct..!Give ratings Accordingly..! Thank you :)
• let s2' = distance of image from second lens
s1 = distance of object from first lens (to the left, if the first lens is on the left side of the second lens)
d = distance between lenses
f1 = focus length of first lens
f2 = focus length of second lens
for the distance of the image from the second lens use
1/s2' = 1/f2 - 1/(d - f1s1/(s1 - s1))
1) converging lens is in front:
1/s2' = 1/(- 0.2) - 1/(0.5 - 0.15*0.1359/(1.359 - 1.5))
--> s2' = - 0.18136 m
The image is 0.1813 m = 18.13 cm to the left of the second lens (in front of the second lens)
2) diverging lens is in front:
1/s2' = 1/0.15 - 1/(0.5 - (-0.2*0.1359/(0.1359 - (-0.2)))
---> s2' = 0.20221 m
The image is 0.2022 m = 20.22 cm behind the second lens
• What are the two systems?
I assume that 1) the converging lens is in front, and 2) the diverging lens is in front.
let s2' = distance of image from second lens
s1 = distance of object from first lens (to the left, if the first lens is on the left side of the second lens)
d = distance between lenses
f1 = focus length of first lens
f2 = focus length of second lens
for the distance of the image from the second lens use
1/s2' = 1/f2 - 1/(d - f1s1/(s1 - s1))
a) converging lens is in front:
1/s2' = 1/(- 0.2) - 1/(0.5 - 0.15*0.1090/(0.1090 - 0.15))
s2' = - 0.14593 = - 14.593 cm Ans
The image is 0.14593 m = 14.593 cm to the left of the second lens (in front of the second lens)
b) diverging lens is in front:
1/s2' = 1/0.15 - 1/(0.5 - (-0.2*0.1090/(0.1090 - (-0.2)))
---> s2' = 0.2035 m Ans
The image is 0.2035 m = 20.325cm behind the second lens
• for the 1st lens(converging)
1/f=1/u+1/v
1/15=1/10.9+1/v
1/v=1/15-1/10.9
v=-15*10.9/4.1=-39.87 from the 1st lens image is formed on the same side of the object
this image work as object for the 2nd lens(diverging)
u=50+39.87=89.87cm
-1/f=1/u+1/v
1/v=-1/20-1/89.87
v=-20*89.87/109.87=-16.35cm from the 2nd lens on the same side of the object
so final image is at a distance 16.35cm from the 2nd lens on the same side of the object ( left of the lens)
and 50-16.35=33.45cm from the 1st lens (right of thre lens)
Get homework help | 979 | 2,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2013-20 | latest | en | 0.892775 |
https://unsworks.unsw.edu.au/entities/publication/90436824-315a-438a-a731-b58ba81f3f69/full | 1,656,529,714,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00210.warc.gz | 664,143,291 | 26,148 | ## Publication: On aspects of Ramsey theory
dc.contributor.advisor Britz, Thomas en_US dc.contributor.advisor Greenhill, Catherine Suzanne en_US dc.contributor.author Chng, Zhi Yee en_US dc.date.accessioned 2022-03-22T17:41:57Z dc.date.available 2022-03-22T17:41:57Z dc.date.issued 2018 en_US dc.description.abstract This thesis presents various types of results from Ramsey Theory, most particularly, Ramsey-type theorems concerning graphs and families of sets. This thesis consists of 8 chapters. In Chapter 1, we give a brief historical introduction to Ramsey Theory. Then, we introduce some necessary notation and definitions that will be consistently used throughout the thesis, including some basic knowledge of Graph Theory which is particularly useful in Chapters 2 and 3. We present Ramsey-type results about graphs in Chapters 2 and 3. In Chapter 2, we introduce the classical Ramsey's Theorem which is the Ramsey-type theorem on the edge-colouring of the complete graph. We also introduce Ramsey numbers and present some results on these, especially some upper and lower bounds. In Chapter 3, we look at Ramsey-type results for monochromatic tree graphs, cycle graphs and bipartite graphs, respectively, occurring in arbitrary edge colourings of the complete graph. Then, we present the bipartite version of Ramsey's Theorem. Chapters 4, 5 and 6 present other famous Ramsey-type theorems, for arithmetic progressions and other, more general, structures. In Chapter 4, we introduce and prove Van der Waerden's Theorem and we also present some results on the bounds of the Van der Waerden numbers. In Chapter 5, we present Schur's Theorem and some results relating to the Schur numbers. Then, we look into some generalisations of Schur's Theorem, including Rado's Theorem and Folkman's Theorem. In Chapter 6, we prove the Hales-Jewett Theorem. We also construct a proof of Van der Waerden's Theorem by using the Hales-Jewett Theorem. Before we end our studies, in Chapter 7, we include some application of the Ramsey Theory. We look into the application of the Ramsey Theory in various fields, including graph theory, geometry and number theory. In Chapter 8, we conclude our studies. We give some overall comment on Ramsey Theory and include some possible future work on the field. en_US dc.identifier.uri http://hdl.handle.net/1959.4/60220 dc.language English dc.language.iso EN en_US dc.publisher UNSW, Sydney en_US dc.rights CC BY-NC-ND 3.0 en_US dc.rights.uri https://creativecommons.org/licenses/by-nc-nd/3.0/au/ en_US dc.subject.other Van der Waerden's theorem en_US dc.subject.other Ramsey theory en_US dc.subject.other Ramsey theorem en_US dc.subject.other Schur's theorem en_US dc.title On aspects of Ramsey theory en_US dc.type Thesis en_US dcterms.accessRights open access dcterms.rightsHolder Chng, Zhi Yee dspace.entity.type Publication en_US unsw.accessRights.uri https://purl.org/coar/access_right/c_abf2 unsw.identifier.doi https://doi.org/10.26190/unsworks/20521 unsw.relation.faculty Science unsw.relation.originalPublicationAffiliation Chng, Zhi Yee, Mathematics & Statistics, Faculty of Science, UNSW en_US unsw.relation.originalPublicationAffiliation Britz, Thomas, Mathematics & Statistics, Faculty of Science, UNSW en_US unsw.relation.originalPublicationAffiliation Greenhill, Catherine Suzanne, Mathematics & Statistics, Faculty of Science, UNSW en_US unsw.relation.school School of Mathematics & Statistics * unsw.thesis.degreetype Masters Thesis en_US
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Description: | 858 | 3,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.875496 |
https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_15&diff=prev&oldid=24958 | 1,632,594,832,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00707.warc.gz | 153,692,051 | 12,041 | # Difference between revisions of "2005 AIME II Problems/Problem 15"
## Problem
Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest possible value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$
## Solution
Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$.
Let $w_3$ have center $(x,y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$, and if they are internally tangent, it is $|r_1 - r_2|$. So we have
$r + 4 = \sqrt{(x-5)^2 + (y-12)^2}$
$16 - r = \sqrt{(x+5)^2 + (y-12)^2}$
Solving for $r$ in both equations and setting them equal yields $20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}$
Squaring both sides, canceling common terms, and rearranging yields
$20+x = 2\sqrt{(x+5)^2 + (y-12)^2}$
Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$
So the locus of points that can be the center of the circle with the desired properties is an ellipse.
Since the center lies on the line $y = ax$, we substitute for $y$:
$1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}$
Expanding yields $(3+4a^2)x^2 - 96ax + 276 = 0$.
We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is 0, so $(-96a)^2 - 4(3+4a^2)(276) = 0$.
Solving yields $a^2 = \frac{69}{100}$, so the answer is $169$.
2005 AIME II (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byLast Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions | 718 | 1,981 | {"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": 38, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-39 | latest | en | 0.782792 |
http://www.greenspun.com/bboard/q-and-a-fetch-msg.tcl?msg_id=003yXU | 1,529,554,042,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00295.warc.gz | 425,023,600 | 11,036 | ### Need Help with Statistics Formula in Shutter Test Spreadsheet
greenspun.com : LUSENET : Large format photography : One Thread
I'm trying to improve the tests I perform on my shutters on a periodic basis. Previously, I used 3 samples for a particular shutter speed to determine an average shutter speed and f-stop difference between reality and the listed speed.
Are three samples enough?
I'd also like to calculate a goodness factor for the consistency between measured samples. If I'm remembering way far back in my memory, I believe standard deviation is what I'm looking for. Is this correct?
Assuming this is correct, how would you convert a calculated std dev based on time samples into a std dev based on f-stops?
-- Larry Huppert (Larry.Huppert@mail.com), October 24, 2000
I'm sorry Larry. I thought that I could be as pedantic as the best, but exactly how will all this help you get better pictures?
-- Pete Andrews (p.l.andrews@bham.ac.uk), October 25, 2000.
Pete,
A simple answer - more accurate exposures, and knowing if your shutter needs service. If I remember the little owners manual that came with a Copal Press shutter, they consider normal operation of a shutter to be ~ +/- 1/3 of a stop from the listed speed. Hum, when shooting chromes, 1/3 of a stop is significant. In actual test, a brand new out of the box shutter does show this type of variation. By knowing this, I can compensate. The point of the testing is to just know where you stand.
-- Larry Huppert (Larry.Huppert@mail.com), October 25, 2000.
The average is a decent measure, (although the modal value might be better). Standard deviation is the measure of spread that you are looking for. If you have a normal distribution (i.e., bell shaped curve), about 95% of the time, your speed will be no more than 2 standard deviations away from the average. Hope this helps. DJ
-- N Dhananjay (ndhanu@umich.edu), October 25, 2000.
A sample of 3 is not enough. A sample of eight will give you an average which, with a 5% chance for error (this is reasonably small), will be within 1.5 standard deviations of the true value. A sample of sixteen will give you an average that (to the same 5% chance of error) will be within one standard deviation of the true value. Think of a standard deviation as about one sixth of the width of the distribution. (Crude estimate.)
I'm not quite sure what you mean by your last question.
The standard deviation is the measure of consistency between measured samples of which you're thinking. With a good estimate of this value, you can determine how close your average is to the true value using the two sample sizes provided above. Any scientific calculator can calculate the standard deviation. Use the "n-1" versus the "n" calculation, if you're given the choice.
-- neil poulsen (neil.fg@att.net), October 25, 2000.
I own a Calumet electronic shutter tester, and have never noticed enough variation (with modern Copal shutters) to consider such a statistical approach. Typically everything from 1/125 on down is within 1/12 stop between trials; 1/250 and 1/400 or 1/500 might deviate 1/3 or 1/2 stop respectively. I don't use those high speeds on a view camera, and don't worry about inconsistencies in them. A calibration card for 1 second through 1/125 gets taped to the lensboard. Periodic rechecking confirms the card is still correct, and no such shutter's performance has changed noticably over a dozen years.
I believe you should probably have any shutter that shows significant variation in its lower speeds serviced rather than expend effort on a statistical exercise.
-- Sal Santamaura (bc_hill@qwestinternet.net), October 25, 2000.
I would want to invest at least five data points, if one is interested in a worthwhile estimate of the average and standard deviation. The data is easy to collect. It's true that, the smaller the variation, the less sample one needs, but to a point.
The "statistical exercise" to which the previous post refers is collecting valid data. Evaluating data in this fashion is done both formally and informally in many disciplines. There's no reason why this kind of evaluation can't extend to photography.
-- neil poulsen (neil.fg@att.net), October 25, 2000.
I would submit that each shutter speed is a different distribution. What would be more informative is the variation of a single shutter speed over time. Therefore, to get a useful statistic, one would need to test each speed several times and then look at the variability at each speed setting. One can not assume that an error at one setting is also valid for settings near it, i.e. the error at 1/60 second is not necesarily the same at 1/30 or 1/125.
-- Gene Crumpler (nikonguy@att.net), October 25, 2000.
Hi Larry, you could calculate the standard deviation of your shutter speeds; the formula is avalible in any old statistics book. You could probably even do the calculations to see if one speed is significantly different from another. I don't know what criterion a shutter is compared to? as far as standard deviations. Practically, I think you'd just test each speed, one at a time; fire the shutter 15 times or so writing down the times as you go, and it ought to be apparent what the average speed is, how much the shutter varries on each firing, and if the shutter is handing up sometimes. If your shutter is reliable within your accepted tolerances, 1/3 to 1/2 stops, on each indicated speed, just average out the true speeds, and then write them on somethiing that's going to be handy when you need them. I've written them on time tape and placed that on the lens board or lens cap. That's basically all you need I guess. I don't think the standard deviation route is particularly helpful unless you just want to do it to do the math. Cheers, David
-- david clark (doclark@yorku.ca), October 25, 2000.
Gene,
Yes, I am testing each shutter speed N number of times so I can see if there is a problem with a particular speed. As an example, on a particular shutter, I see variations from -2/10 stop to +3/10 stop across different shutter speeds. Copal would say this shutter is within spec, but knowing the characterizaton of this 1/2 stop variation is important.
Neil,
My last statement regarding conversion of std dev to f-stops means the following: In a laymans sense, I thought std dev tells you how close together the various samples are to each other relative to the average value. The smaller the std dev the better off you are in this case. I didn't think that std dev actually related to the "true value". For example, at 1/4sec, lets say the ave measured speed is +1/3 stops relative to the "true value" of .25sec. If the measured values are pretty consistent in being off the same amount that is better than multiple samples of testing that speed being inconsistent, but still averaging out to +1/3 stops. Doesn't std dev give you this "goodness" factor? If I do a std dev calculation on a sample of measured shutter times, won't the resulting std dev relate to those numbers in the time domain? What I think I would like to know is that the std dev of the measured values fall within X f-stops of the average. Wouldn't this tend to normalize the result so you don't have to relate the std dev to actual shutter times anymore. I've probably really bastardized mathematical concepts here.
-- Larry Huppert (Larry.Huppert@mail.com), October 25, 2000.
Neil, my point was that, with scatter only in the neighborhood of 1/12 stop for shutter speeds I actually use, I've not found a reason to evaluate such data in a statistical exercise. Performing such evaluations for photographic gear is fine, and if Larry either has older, more inconsistent shutters, or just feels like running numbers for his modern shutters, that's his prerogative. I was simply trying to guide him down what seems a more profitable path if his *modern* equipment doesn't fire with typical consistency.
Larry, doesn't the Copal specification you quoted to Pete refer to a deviation from marked speed, rather than variation between individual firings? If so, it is consistent with my data. *Scatter* of that data for Copal shutters is usually no more than 1/12 stop, around a measured value typically within 1/3 stop of marked speeds, 1/250 and 1/400-500 excepted.
-- Sal Santamaura (bc_hill@qwestinternet.net), October 25, 2000.
Sal,
I don't have the Copal spec handy. If memory serves me, it said something like "accuracy: +/- 30%". I don't remember seeing any quoted spec regarding variation between individual firings of the shutter at the same speed. Obviously, no deviation would be ideal.
Some replies have suggested it isn't worth the work to check the std dev on modern shutters. Since I'm loading this info into a speadsheet anyway, it didn't seem like that much more work to do the std dev part as well. By calculating a goodness factor, I can treat this whole thing as a truely mind numbing exercise, and just end-up with the data.
-- Larry Huppert (Larry.Huppert@mail.com), October 25, 2000.
Larry, standard deviations only come in 1s, 2s, or 3s. If the variation of true recorded firings when your sample size is about 30 firings falls about/within 66% of your mean value, then your shutter speeds are probably within 1 standard dev. (I think???). But I'm not so sure the assumptions and logic of probabilities apply to this situation as I can't see shutter speeds falling on the normal curve. I think all you need to figure is if basically your shutter fires reliably at some speed which means a grouping within about a 1/3 stop or 1/2 stop. Best, David
-- david clark (doclark@yorku.ca), October 25, 2000.
Larry,
That's correct; the standard deviation relates each value to the average, not to the "true value" that you describe. Actually, the "true value" that you describe is better called the nominal value. The term "true value" is usually reserved for the population mean, or the "true" shutter speed average at the particular shutter setting that you selected. It's this second type of "true value" to which I was referring. I think we were using the same terminology for two different things.
As to f-stops, I think I see what you mean. Let's call 0.25 secs in your example the nominal value. Let's say that you fire the shutter, and the measured speed is "T". If I understand correctly, the f-stop equivalent to which you refer is the difference between "T" and nominal (0.25secs) in terms of f-stops, and this would be LOG(T/0.25)/LOG(2). (The base of the logarithm doesn't matter.) So, if T=0.5 secs, you would overexpose by one stop. If this is what you mean, then you can collect your data, transform each value to the difference in f-stops, and plot the data in a histogram (or dotplot) to see how the data are distributed in the f-stop domain. To estimate the center of this distribution, and depending on the shape of the distribution in the time domain or in the "f-stop" domain, I would probably calculate the average in the time domain, and then transform this average (using the above formula) to estimate the center of the distribution in the f-stop domain. Again, it depends on the distributions. I would expect that the distribution in the time-domain would be more normal, whereas the distribution in the f-stop domain would be more heavily skewed to the right. But, I don't have the means to take the number of measurements that I would need to check this assertion.
In representing the variation in the f-stop domain, I would probably calculate a 95% confidence interval in the time domain, and then transform these limits to the f-stop domain. In this way, you can be 95% confident that the actual average (the true average) lies between these two values.
If you wouldn't mind sending me your data, it would give me a chance to check this out. Send me 10 values for each of the shutter speeds in which you're interested, and 30 values for your favorite shutter speed. With that data, I could respond with a pretty clear picture of how your shutter is behaving. If you can find them, also send me the specs on your shutter.
As to Sal's point, I would agree that, if a shutter is showing obvious signs of large inconsistencies, then there's no need to test further. But, once the shutter's returned from the repairman, I would find it worthwhile to collect the kind of data to which Larry refers. (In fact, I did!)
neil
-- neil poulsen (neil.fg@att.net), October 26, 2000.
The standard deviation and average of your measured shutter speeds is only a useful concept if the shutter is nicely behaved. I agree with Neil that the first thing to do is to plot the shutter speeds as a histogram and check that they lie on a reasonably smooth curve, clustered around a meaningful average value. A really erratic shutter will show up unambiguously at this stage.
The 'best' way to do the measurements is with a photocell, a storage oscilloscope operating in single-shot mode and a computer into which you can download the individual shutter traces for analysis. If you trigger off the flash contact, or have an oscilloscope with a programmable trigger and time delay, you can capture each firing of the shutter automatically. Even with non-press shutters you can build up hundreds of readings with no great effort, but be aware that some shutters behave differently when run at a high duty cycle - I have an old Compur which is fine if I fire it every five minutes, but slows down if I pop it over and over again in quick sucession.
The ultimate analysis involves plotting a two-dimensional histogram of the light intensity behind the shutter versus time. This is a useful diagnostic for bouncing shutter blades or curtains, uneven accelleration, stickyness and a host of other shutter ailments. By integrating the average curve you can get true effective speeds for those high settings on leaf shutters where the blades don't stop moving.
And that last point hints at something which is actually photographically useful: measuring the high speeds of leaf shutters at varying aperture stops. A speed marked '500' on a leaf shutter is only correct for one aperture, since it takes into account the finite time needed to open and close the shutter. At much smaller apertures the shutter blades reache the edges of the aperture much sooner, and the shutter is effectively open for longer. In the worst case, the difference can be up to a stop, far larger than the variability of even antique shutters.
-- Struan Gray (struan.gray@sljus.lu.se), October 26, 2000.
Struan,
Your last point is well taken regarding faster shutter speeds. When testing the 1/500, 1/400 & 1/250 speeds on Copal (regular) shutters at different apertures, I was amazed how significant the travel time of the shutter is compared to the listed speed. I rarely have any need for anything this fast, and it made me realize that giving up the fast speeds (1/250 and faster) on Copal Press shutters was even less significant than it initially appeared.
Thanks to all for your statistical expertise. I'm going to increase my sample set, and plot some graphs.
-- Larry Huppert (Larry.Huppert@mail.com), October 26, 2000.
I don't know where this idea that standard deviation can only be an integer value comes from; or that you can have standard deviations (plural). Standard deviation is the RMS (root mean square) value of the deviation from the mean of the samples, of all the individual sample points. It can be any value, fractional or integer.
The confusion might be that the samples have to be normalised to a value greater than 1 for standard deviation to be meaningful.
Anyway, I still don't see how knowing this is of any help in getting more accurate exposure.
You seem to think that knowing the value of their standard deviation will somehow make your shutters behave less erratically; it won't.
A more meaningful analysis would be to know the maximum deviation from the mean of any particular speed, and the frequency, or likelihood, of that deviation occurring. This gives you the odds of getting an exposure that's acceptably accurate.
Say you took 10 readings at 1/60th second, and 8 of them were within 1ms of 17ms, but 2 of them were off by +5ms (still only 1/3rd stop). The standard deviation in this case is quite small (~2.36ms), but the odds are still 5:1 that your exposure'll be out by 1/3rd of a stop.
If your shutter is truly random in giving you this variation, then you'll have this chance every time you take a shot. You might take 100 shots in a row, and get a bad exposure on every one, then take another 400 where the exposure is bang on.
If you don't like those odds, then you'll have to have the shutter serviced or replaced.
What I'm trying to illustrate is the difference between statistical analysis, and real probability. No amount of analysis will alter the odds of an event happening.
By all means test your shutter(s) regularly, but don't expect collating data about shutter performance to give you more accurate exposures in reality. If the shutter is erratic, the exposure will still be down to dumb luck at the moment you take the picture.
As for translating exposure time to an f-stop equivalent:
Take the actual exposure time and divide it by the ideal exposure time; then take the logarithm of this value and divide it by the log of 2. [log(N+x/N)/log2]. That gives you the deviation in stops.
-- Pete Andrews (p.l.andrews@bham.ac.uk), October 27, 2000.
If one can assume that the distribution is Normal (bell-shaped Gaussian), then the average and standard deviation offers the advantage of being able to obtain the information to which Pete refers in two numbers (the average and S.D.), versus many numbers. So, computing the average and standard deviation is a method of summarization. For example, under the assumption of Normality, 68% of the data will fall in the interval [A-SD,A+SD], 95% of the data will fall in the interval of [A-2SD,A+2SD], and a little more than 99% of the data will fall in the interval [A-3SD,A+3SD].
But, as Pete suggests, there's an advantage of being able to look at multiple representations of the raw data. (e.g. the standard deviation won't necessarily tell someone if a shutter is behaving erratically.)
-- neil poulsen (neil.fg@att.net), October 27, 2000.
I think it was Lord Rutherford who said: "If you want to make an accurate measurement, make it once."
>>No amount of analysis will alter the odds of an event happening.
Pete, I think the good Rev. Bayes is buried somewhere near you: that strange whirring sound you can hear is him spinning in his grave. In your example, the odds are 5:1 that you'll be out by 1/3 of a stop *given* that you only bothered to test ten times. Test more often, and you'll have a better idea of how bad the shutter is.
I agree that any outlier is cause to wonder if a CLA is in order, but the price of a CLA is more than I've paid for any of my LF lenses and I see no reason to spend \$100 on a repair when I can use \$4000 of equipment proving to myself that I can muddle along for another year or so. Since two of my shutters are polariod-press models scavanged from dumpsters I also like the idea that popping the shutter 10 times on 1/20 is more accurate than firing it once on 1/2. Naturally, the cat thinks I'm mad.
-- Struan Gray (struan.gray@sljus.lu.se), October 29, 2000.
Let him spin Struan, let him spin (I'll bet he stops on double zero - the house wins).
Statisticians seem to think that a significantly 'large' sample is 30 or greater. Now I reckon that if you regularly test a 10 speed shutter 30 times at each speed, setting the speed both in an upward and downward direction, then you'll probably have a statistically greater chance of the shutter failing altogether on an important shot, than if you'd left the darned thing alone and taken the speeds at face value. What do you think to this radical theory?
In practise mechanical shutters rarely give a speed significantly faster than their marked speed, any fault, dirt or wear tends to slow the shutter. A normal, or Gaussian distribution won't result from such conditions, so all bets are off as far as standard deviation is concerned.
-- Pete Andrews (p.l.andrews@bham.ac.uk), October 30, 2000.
The directions which come with the Calumet tester are interesting in one regard. They suggest you gather 3 samples per shutter speed for each shutter tested. However, they do not suggest you bang the shutter three times in a row. They say to start at 1 sec and march towards the fastest speed, and then from the fastest speed back to 1 sec, ... only taking one measurement each time at each speed.
While testing a particular shutter while not using this methodology (I popped the shutter 8 times in a row at each speed), I found an interesting anomaly which very much supports some of the views expressed here. The shutter was almost dead-on at 1/15 sec, except when it was off in an eratic way - significantly. The average came to be 1/3 stop off the listed speed, however this was meaningless.
If I then tried the Calumet methodology through 8 readings, it measured dead-on. It appeared that a heavy duty cycle on this Copal Press shutter was a problem at this particular speed. The Calumet method is closer to the duty cycle of actual shooting, except for one situation. The serial same speed test was pretty close to how you use a shutter when multipoping a strobe. When balancing ambient vs. strobe light, you still need your shutter to be accurate.
-- Larry Huppert (Larry.Huppert@mail.com), October 30, 2000.
What do you think to this radical theory?
It sounds like the sort of shutter I would own. A properly designed mechanical shutter should cope with a mere 30-pops-per-speed test, but that begs the question as to whether the shutter is in fact properly designed, which is of course one of the things you are trying to find out.
Practical photography is far more about being consistent than it is about being accurate, so I still think that if you are going to test a shutter at all it makes sense to try and find out what the distribution of errors look like. I agree entirely that the shape of the histogram is what matters, not whether you can find a nice analytical expression for the histogram's width.
I've only ever seriously tested on shutter: an electronic model specifically designed for high duty-cycle applications (see www.uniblitz.com). But I suspect that a modern shutter right out of the box will, if tested at normal rates of fire, cluster pretty tightly around the average, even if that average is some tens of percent removed from the marked speed. Both those facts are useful for the photographer to know.
-- Struan Gray (struan.gray@sljus.lu.se), November 01, 2000.
Larry
do you have what you need? I am a statistician-md and if you tell me how many measurements you can stand taking, what chance you will take on making and error, and how sure you want to be of the chance of rejecting your estimated value correctly, I will give you some feedback. Ross
-- ross (epistatdoc@speedchoice.com), November 03, 2000. | 5,298 | 23,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-26 | latest | en | 0.938015 |
https://inches-to-mm.appspot.com/pl/4790-cal-na-milimetr.html | 1,722,749,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00205.warc.gz | 241,887,379 | 6,469 | Inches To Mm
# 4790 in to mm4790 Inch to Millimeters
in
=
mm
## How to convert 4790 inch to millimeters?
4790 in * 25.4 mm = 121666.0 mm 1 in
A common question is How many inch in 4790 millimeter? And the answer is 188.582677166 in in 4790 mm. Likewise the question how many millimeter in 4790 inch has the answer of 121666.0 mm in 4790 in.
## How much are 4790 inches in millimeters?
4790 inches equal 121666.0 millimeters (4790in = 121666.0mm). Converting 4790 in to mm is easy. Simply use our calculator above, or apply the formula to change the length 4790 in to mm.
## Convert 4790 in to common lengths
UnitLength
Nanometer1.21666e+11 nm
Micrometer121666000.0 µm
Millimeter121666.0 mm
Centimeter12166.6 cm
Inch4790.0 in
Foot399.166666667 ft
Yard133.055555556 yd
Meter121.666 m
Kilometer0.121666 km
Mile0.0755997475 mi
Nautical mile0.0656943844 nmi
## What is 4790 inches in mm?
To convert 4790 in to mm multiply the length in inches by 25.4. The 4790 in in mm formula is [mm] = 4790 * 25.4. Thus, for 4790 inches in millimeter we get 121666.0 mm.
## Alternative spelling
4790 in to mm, 4790 in in mm, 4790 Inch in Millimeters, 4790 Inches to Millimeter, 4790 Inch to Millimeter, 4790 Inch in Millimeter, 4790 Inches in Millimeters, 4790 in in Millimeter, 4790 in to Millimeters, 4790 in in Millimeters, | 455 | 1,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.699341 |
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2 Update based on following two of the suggested references.
Suppose that gravity did not follow an inverse-square law, but was instead a central force diminishing as $1/d^p$ for distance separation $d$ and some power $p$. Two questions:
1. Presumably the 2-body problem still factors into two independent 1-body problems, results in planar motion, and can be solved. Have the orbits (the equivalents of elliptical and parabolic orbits for $p=2$) been worked out for other (perhaps specific) values of $p$?
2. In some sense, the 3-body problem for $p=2$ cannot be solved. Most systems are choatic; see this interesting collection of Eugene Butikov. Only a few periodic solutions are known; see the nice article by Bill Casselman on the discovery of "choreographies." Is the situation simpler for other values of $p$? Perhaps $p=1$?
References and pointers would be appreciated. Thanks!
Edit. Thanks to Agol, Ken, and José. I've now looked at Arnolʹd's Huygens and Needham (but not yet Arnolʹd's Classical Mechanics). Indeed, as the commenters say, there is a remarkable 2-body result for $p=1$: the orbits for a linearly attractive force are ellipses, for a linearly repulsive force, hyperbolae. This depends on the Kasner-Arnolʹd theorem stating that for each power law, there is a dual power law that maps orbits of one to orbits of the other. Newton proved in Principia that elliptical orbits result if and only if the force is inverse-linear or inverse-square. The Kasner-Arnolʹd theorem explains why.
1
# 2- and 3-body problems when gravity is not inverse-square
Suppose that gravity did not follow an inverse-square law, but was instead a central force diminishing as $1/d^p$ for distance separation $d$ and some power $p$. Two questions:
1. Presumably the 2-body problem still factors into two independent 1-body problems, results in planar motion, and can be solved. Have the orbits (the equivalents of elliptical and parabolic orbits for $p=2$) been worked out for other (perhaps specific) values of $p$?
2. In some sense, the 3-body problem for $p=2$ cannot be solved. Most systems are choatic; see this interesting collection of Eugene Butikov. Only a few periodic solutions are known; see the nice article by Bill Casselman on the discovery of "choreographies." Is the situation simpler for other values of $p$? Perhaps $p=1$?
References and pointers would be appreciated. Thanks! | 622 | 2,557 | {"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} | 3.15625 | 3 | CC-MAIN-2013-20 | latest | en | 0.927977 |
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# qz3sol_3891f10 - river with respect to the ground and the...
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TA: Tomoyuki Nakayama Tuesday, September 21, 2010 PHY 2048: Physic 1, Discussion Section 3891 Quiz 3 (Homework Set #4) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A 180-m-wide river flows due east at a uniform speed of 3.00 m/s. A boat with a speed of 8.00 m/s relative to the water leaves the north bank pointed in a direction 15.0º west of south. a) What are the magnitude and direction of the boat’s velocity relative to the ground? We take our +x axis to the east and +y axis to the north. In unit-vector notation, the velocity of the
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Unformatted text preview: river with respect to the ground and the velocity of the boat with respect to water are, respectively, v WG = 3 i , v BW = 8cos(-105º) i + 8sin(-105º) j = -2.07 i – 7.72 j The velocity of the boat with respect to the ground is v BG = v BW + v WG = 0.93 i – 7.72 j The magnitude and direction of v BG are v BG = √ (.93 2 +(-7.72) 2 ) = 7.78 m/s, θ = tan-1 ((-.7.72)/.93) = -83.1º, 6.9º east of south. Note that v BG has a positive x component and negative y component. It is in the 4 th quadrant. b) How long does the boat take to cross the river? To cross the river, the boat must move 180 m in the negative y direction. Therefore, t = Δ y/v y = 23.3 s...
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posted by .
For what values of k will the functon f(x)=kx^2-4x+k have no zeros
Determine the values of k for which the function f(x)=4x^2-3x+2kx+1 have 2 zeros
• Math -
if b^2 - 4ac < 0 ---> no real zeros
if b^2 - 4ac = 0 ---> one real zero
ifb^2 - 4ac > 0 ---> two real zeros
if will do the 2nd question.
a=4, b= 2k-3 , c = 1
b^2 - 4ac
= (2k-3)^2 - 4(4)(1)
= 4k^2 - 12k + 9 - 16
= 4k^2 - 12k - 7
= (k + 1)(4k - 7)
(k+1)(4k-7) > 0 for 2 real roots
k < -1 OR k > 7/4 | 236 | 480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-30 | longest | en | 0.484063 |
https://numbermatics.com/n/612179991729492701913/ | 1,702,160,089,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00443.warc.gz | 472,718,369 | 7,646 | # 612179991729492701913
## 612,179,991,729,492,701,913 is an odd composite number composed of three prime numbers multiplied together.
What does the number 612179991729492701913 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 48 divisors.
612179991729492701913 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of forty-eight divisors.
## Prime factorization of 612179991729492701913:
### 33 × 613 × 463992
(3 × 3 × 3 × 61 × 61 × 61 × 46399 × 46399)
See below for interesting mathematical facts about the number 612179991729492701913 from the Numbermatics database.
### Names of 612179991729492701913
• Cardinal: 612179991729492701913 can be written as Six hundred twelve quintillion, one hundred seventy-nine quadrillion, nine hundred ninety-one trillion, seven hundred twenty-nine billion, four hundred ninety-two million, seven hundred one thousand, nine hundred thirteen.
### Scientific notation
• Scientific notation: 6.12179991729492701913 × 1020
### Factors of 612179991729492701913
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 8
• Sum of prime factors: 46463
### Divisors of 612179991729492701913
• Number of divisors d(n): 48
• Complete list of divisors:
• Sum of all divisors σ(n): 612186670509007493586
• Sum of proper divisors (its aliquot sum) s(n): 6678779514791673
• 612179991729492701913 is a deficient number, because the sum of its proper divisors (6678779514791673) is less than itself. Its deficiency is 612173312949977910240
### Bases of 612179991729492701913
• Binary: 10000100101111101101000011101100000010011001011111101110110010110110012
• Base-36: 3L721BSPNX3PW9
### Squares and roots of 612179991729492701913
• 612179991729492701913 squared (6121799917294927019132) is 374764342273921752915491647710329073859569
• 612179991729492701913 cubed (6121799917294927019133) is 229423231953758190860561406548951610285381581732054928939655497
• The square root of 612179991729492701913 is 24742271353.4851666801
• 612179991729492701913 is a perfect cube number. Its cube root is 8491017
### Scales and comparisons
How big is 612179991729492701913?
• 612,179,991,729,492,701,913 seconds is equal to 293,274,701,009 years, 3 weeks, 3 days, 15 hours, 30 minutes, 7 seconds.
• To count from 1 to 612,179,991,729,492,701,913 would take you about two hundred ninety-three billion, two hundred seventy-four million, seven hundred one thousand and nine years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 612179991729492701913 cubic inches would be around 707584.8 feet tall.
### Recreational maths with 612179991729492701913
• 612179991729492701913 backwards is 319107294927199971216
• The number of decimal digits it has is: 21
• The sum of 612179991729492701913's digits is 99
• More coming soon!
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Keywords: Divisors of 612179991729492701913, math, Factors of 612179991729492701913, curriculum, school, college, exams, university, Prime factorization of 612179991729492701913, STEM, science, technology, engineering, physics, economics, calculator, six hundred twelve quintillion, one hundred seventy-nine quadrillion, nine hundred ninety-one trillion, seven hundred twenty-nine billion, four hundred ninety-two million, seven hundred one thousand, nine hundred thirteen.
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Some bits of this website may not work unless you switch it on. | 1,274 | 4,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.699057 |
https://comidoc.net/udemy/logica-de-programacao-e-algoritmos-iniciante | 1,719,196,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00198.warc.gz | 153,771,178 | 14,910 | # Lógica de programação e algoritmos
Desenvolva a mentalidade e raciocínio correto para ser um programador de sistemas.
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## Why take this course?
🌟 Course Title: Lógica de Programação e Algoritmos
Welcome to the World of Logical Problem Solving! 🧩
Course Description: In this course, Desenvolva a mentalidade e raciocínio correto para ser um programador de sistemas. You'll embark on a transformative journey to enhance your analytical skills for problem-solving with precision and ease.
Before diving into the world of coding, it's crucial to cultivate within us the ability to approach complex problems with a sharp, logical mindset. This is where the foundational concepts of logic in programming come into play. 🌐
What You Will Learn:
• Introduction to Programming Logic: Gain an understanding of what it means to think like a programmer and why it's essential for your success in software development.
• Sequential Thinking: Master the art of breaking down tasks into manageable steps to ensure efficient execution.
• Algorithms Explained: Learn how to define, design, and analyze algorithms that solve problems efficiently.
• Computer Programs Demystified: Understand what programs are and how they execute instructions on a computer.
• Pseudocode: Discover the art of writing high-level descriptions of the algorithm or program's steps in plain language.
• Diagram of Blocks (Flowcharts): Visualize algorithms by drawing diagrams that represent the sequence and decision points within a process.
• Symbols in Flowcharts: Learn the standard symbols used in flowcharts and how to apply them correctly.
• Real-World Example of a Flowchart: Examine a practical example to solidify your understanding of the concepts discussed.
• Constants vs Variables: Distinguish between immutable values (constants) and those that can change (variables).
• Arithmetic Operators: Understand how to perform mathematical operations within your code.
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• Logical Operators: Combine conditions using logical operators to create more complex decision-making processes.
• Decision Structures - Part 1 & Part 2: Explore different ways to handle decision-making in your programs, from simple if-else statements to complex logic.
• CASE Structure: Utilize the CASE statement to simplify your conditional code when you have several conditions to check against a single outcome.
• Iteration Structures: Learn how to write loops and understand how they can be used to execute code repeatedly.
• Syntax vs Semantics in Code: Dive into the importance of understanding what the code means (semantics) as well as the correct way to write it (syntax).
Join us on this exciting path to becoming a proficient programmer, where logic is your compass and algorithms are your map to navigating the vast landscape of computer science! 🚀
Enroll now and transform your problem-solving skills with Lógica de Programação e Algoritmos today!
## Our review
Ítens de feedback variados, mas geralmente positivos, indicando que o curso é considerado bom para iniciantes na programação e na linguagem Java. Os pontos negativos mencionados são mais detalhados e específicos, como a repetição de explicações ou alguns erros menores nos slides, mas ainda assim o curso é valorizado por sua clareza e abordagem didática.
Os alunos destacam a importância da forma como os conceitos são apresentados, com destaque para a explicação tranquila e clara do professor, que ajuda na compreensão de fundamentos importantes da lógica de programação. A ideia de que o curso é gratuito e ainda aborda esses tópicos é vista como um grande valor.
Os comentários sugerem que o curso está bem pensado para quem está iniciando na área, com feedbacks indicando uma progressão gradual nos conceitos apresentados. Alguns alunos enfatizam a importância de continuar melhorando e aprimorando o material, sugerindo que mesmo com pequenos problemas, o curso como um todo é de grande ajuda para os iniciantes na programação.
Recomendações para melhorias incluem:
• Ajustes nas slides para evitar erros menores e garantir coerência.
• Continuar aprimorando a explicação, mantendo a clareza e tranquilidade que já são pontos fortes do curso.
• Possivelmente adicionar mais exemplos práticos ou exercícios para reforçar o aprendizado.
Em resumo, os alunos geralmente têm uma opinião positiva sobre o curso e reconhecem seu valor como um ponto de partida na programação com Java.
## Related Topics
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course submited by | 1,062 | 4,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.803287 |
https://answercult.com/question/can-someone-explain-the-coastal-paradox-and-infinite-shoreline-theory/?order_by=newest | 1,726,601,004,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00178.warc.gz | 80,821,178 | 13,400 | # Can someone explain the coastal paradox and infinite shoreline theory?
457 views
How can a finite area like Great Britain have an infinite length edge?
In: Mathematics
Coastlines are like a fractal curve, the length of a fractal curve is infinite.
Of course in the real world it’s not infinite, just like a drop of water doesn’t have a singularity in the moment in drops off from the surface of a faucet. But the length of a coastline is an absurdly large number when you count the distance between all atoms on a coastline.
I guess you could describe the area as how much paint it would require to “fill” a shape while perimeter would be the effort it takes to “draw” said shape. Say you had a square that was 10×10 meters. Now say you cut out a 1×1 meter square from one side and pasted it to another. Your area remains 100 m, you have not technically added anything that wasn’t already there, but your perimeter has increased from 40 m to 44 m. You can do this indefinitely with smaller squares but the area will always remain the same.
When measuring a coastline, the little nooks and crannies can drastically increase the perimeter of your coast, sometimes increasing it multiple fold. Obviously in real life, you can only measure to a certain degree, but “coastline paradox” probably stuck better than “fractal curve paradox”.
Imagine you and a friend are standing on a beach. You and that friend want to measure how long the beach is.
Both of you decide to measure the beach in terms of stride lengths; that is, you move your foot forwards. You agree on a rule that as each of you moves on the beach, your right foot must get wet and your left foot remain dry.
You decide to move forward 20 centimetres with every step. Your friend moves forward 10 centimetres with every step.
You will notice that your friend has to do much more walking to keep their feet on the edge than you do. (*See* [*this*](https://www.researchgate.net/profile/Terry-Marks-Tarlow/publication/268803624/figure/fig3/AS:667862427521034@1536242219962/The-Coastline-Paradox-illustrates-that-shorter-measuring-units-create-longer-coastlines.ppm) *for a graphical illustration*).
As a result, your friend measures a longer distance than you. Who is right? As it turns out, there is no clear answer.
**The shorter the units of measurement, the more small features matter in the length, and thus the longer the beach appears to be**.
I think you’re getting hung up on the infinite part, at some point the laws of physics would prevent us from measuring any further but conceptually we can always divide a number in half. It could be physically impossible to measure 1/100,000,000,000,000,000,000,000 mm but it doesn’t mean it can exist on paper.
One thing to remember is that in real life this doesn’t really go to infinity because at smalls scale waves are constantly modifying what counts as the coast line and even if you froze time, you wouldn’t be able to go below their scale and any fractals would disappear.
The reason it’s a “thing” is because maps got more and more precise and they needed an explanation for why they kept measuring things longer and longer. But you’re right that it’s an arbitrary distinction that would apply to anything. Of course there can always be a more exact measurement that would involve smaller and smaller incremental differences.
Beyond the immediate need for an explanation from mapmakers and mathematicians, a coastline is also just a good example to use because people can easily conceptualize a coastline as a fractal shape.
It turns out it’s quite easy to define a shape that has finite area but infinite perimeter, by giving it finer and finer details at smaller scales (shapes that have finer details at smaller scales like this are called “fractals”). A simple example is the Koch snowflake.
A coastline works a little bit like this. If you calculate the length of a coastline from a world map, you will miss out lots of bays and inlets. If you use a larger scale map, you will take those into account and get a larger value (often much larger) but you will still miss some smaller features. However, once you get down to the smallest length scales, the coastline becomes quite hard to define because of tides and waves. So it’s not exactly a fractal, but it works a little bit like one.
It’s a measure of *exactness*, and larger lines/units of measurement giving general, less exact “true” measure. The [wiki](https://en.wikipedia.org/wiki/Coastline_paradox) article actually does a great job of describing it already.
How *exactingly* do we measure the coastline–and what units do we use (kilometers/miles, meters/yards? Centimeters/inches)?
With a surveyor’s device, from mile to mile (or kilometer to kilometer)? It’s one length.
Do we send a guy with a tape measure (or one of those wheeled distance-measuring devices), and tell him, at some arbitrary point of the day every day, walk out from *this point*, and “hug the coastline” (which is *always* changing due to tidal forces and erosion, *natch*), and measure it *that way*? You’ll get *another* total.
Do we send someone out with a *flexible* tape measure that you use to measure fabric–the kind that you can roll up? So they can “hug the coastline” (ha!) You’d get an even larger, different total length.
Like Pi, depending on the level of exactness you demand, it could be a never-ending total. Add to this that the *thing you’re measuring* is changing, *while you try to measure it*. | 1,241 | 5,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.940232 |
http://popflock.com/learn?s=Entropy_production | 1,660,357,547,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00695.warc.gz | 39,260,561 | 30,684 | Entropy Production
Get Entropy Production essential facts below. View Videos or join the Entropy Production discussion. Add Entropy Production to your PopFlock.com topic list for future reference or share this resource on social media.
Entropy Production
Entropy production (or generation) is the amount of entropy which is produced in any irreversible processes[1] such as heat and mass transfer processes including motion of bodies, heat exchange, fluid flow, substances expanding or mixing, anelastic deformation of solids, and any irreversible thermodynamic cycle, including thermal machines such as power plants, heat engines, refrigerators, heat pumps, and air conditioners.
In the dual representation entropy-exergy for accounting the second law of thermodynamics it can be expressed in equivalent terms of exergy disruption.
Rudolf Clausius
## Short history
Entropy is produced in irreversible processes. The importance of avoiding irreversible processes (hence reducing the entropy production) was recognized as early as 1824 by Carnot.[2] In 1865 Rudolf Clausius expanded his previous work from 1854[3] on the concept of "unkompensierte Verwandlungen" (uncompensated transformations), which, in our modern nomenclature, would be called the entropy production. In the same article in which he introduced the name entropy,[4] Clausius gives the expression for the entropy production (for a closed system), which he denotes by N, in equation (71) which reads
${\displaystyle N=S-S_{0}-\int {\frac {dQ}{T}}.}$
Here S is the entropy in the final state and the integral is to be taken from the initial state to the final state. From the context, it is clear that N = 0 if the process is reversible and N > 0 in case of an irreversible process.
## First and second law
Fig.1 General representation of an inhomogeneous system that consists of a number of subsystems. The interaction of the system with the surroundings is through exchange of heat and other forms of energy, flow of matter, and changes of shape. The internal interactions between the various subsystems are of a similar nature and lead to entropy production.
The laws of thermodynamics system apply to well-defined systems. Fig.1 is a general representation of a thermodynamic system. We consider systems which, in general, are inhomogeneous. Heat and mass are transferred across the boundaries (nonadiabatic, open systems), and the boundaries are moving (usually through pistons). In our formulation we assume that heat and mass transfer and volume changes take place only separately at well-defined regions of the system boundary. The expression, given here, are not the most general formulations of the first and second law. E.g. kinetic energy and potential energy terms are missing and exchange of matter by diffusion is excluded.
The rate of entropy production, denoted by ${\displaystyle {\dot {S}}_{i}}$, is a key element of the second law of thermodynamics for open inhomogeneous systems which reads
${\displaystyle {\frac {\mathrm {d} S}{\mathrm {d} t}}=\sum _{k}{\frac {{\dot {Q}}_{k}}{T_{k}}}+\sum _{k}{\dot {S}}_{k}+\sum _{k}{\dot {S}}_{ik}{\text{ with }}{\dot {S}}_{ik}\geq 0.}$
Here S is the entropy of the system; Tk is the temperature at which the heat flow ${\displaystyle {\dot {Q}}_{k}}$ enters the system; ${\displaystyle {\dot {S}}_{k}={\dot {n}}_{k}S_{mk}={\dot {m}}_{k}s_{k}}$ represents the entropy flow into the system at position k, due to matter flowing into the system (${\displaystyle {\dot {n}}_{k},{\dot {m}}_{k}}$ are the molar flow and mass flow and Smk and sk are the molar entropy (i.e. entropy per mole) and specific entropy (i.e. entropy per unit mass) of the matter, flowing into the system, respectively); ${\displaystyle {\dot {S}}_{ik}}$ represents the entropy production rates due to internal processes. The index i in ${\displaystyle {\dot {S}}_{ik}}$ refers to the fact that the entropy is produced due to irreversible processes. The entropy-production rate of every process in nature is always positive or zero. This is an essential aspect of the second law.
The ?'s indicate the algebraic sum of the respective contributions if there are more heat flows, matter flows, and internal processes.
In order to demonstrate the impact of the second law, and the role of entropy production, it has to be combined with the first law which reads
${\displaystyle {\frac {\mathrm {d} U}{\mathrm {d} t}}=\sum _{k}{\dot {Q}}_{k}+\sum _{k}{\dot {H}}_{k}-\sum _{k}p_{k}{\frac {\mathrm {d} V_{k}}{\mathrm {d} t}}+P,}$
with U the internal energy of the system; ${\displaystyle {\dot {H}}_{k}={\dot {n}}_{k}H_{mk}={\dot {m}}_{k}h_{k}}$ the enthalpy flows into the system due to the matter that flows into the system (Hmk its molar enthalpy, hk the specific enthalpy (i.e. enthalpy per unit mass)), and dVk/dt are the rates of change of the volume of the system due to a moving boundary at position k while pk is the pressure behind that boundary; P represents all other forms of power application (such as electrical).
The first and second law have been formulated in terms of time derivatives of U and S rather than in terms of total differentials dU and dS where it is tacitly assumed that dt > 0. So, the formulation in terms of time derivatives is more elegant. An even bigger advantage of this formulation is, however, that it emphasizes that heat flow and power are the basic thermodynamic properties and that heat and work are derived quantities being the time integrals of the heat flow and the power respectively.
## Examples of irreversible processes
Entropy is produced in irreversible processes. Some important irreversible processes are:
• heat flow through a thermal resistance
• fluid flow through a flow resistance such as in the Joule expansion or the Joule-Thomson effect
• diffusion
• chemical reactions
• Joule heating
• friction between solid surfaces
• fluid viscosity within a system.
The expression for the rate of entropy production in the first two cases will be derived in separate sections.
Fig.2 a: Schematic diagram of a heat engine. A heating power ${\displaystyle {\dot {Q}}_{H}}$ enters the engine at the high temperature TH, and ${\displaystyle {\dot {Q}}_{a}}$ is released at ambient temperature Ta. A power P is produced and the entropy production rate is ${\displaystyle {\dot {S}}_{i}}$. b: Schematic diagram of a refrigerator. ${\displaystyle {\dot {Q}}_{L}}$ is the cooling power at the low temperature TL, and ${\displaystyle {\dot {Q}}_{a}}$ is released at ambient temperature. The power P is supplied and ${\displaystyle {\dot {S}}_{i}}$ is the entropy production rate. The arrows define the positive directions of the flows of heat and power in the two cases. They are positive under normal operating conditions.
## Performance of heat engines and refrigerators
Most heat engines and refrigerators are closed cyclic machines.[5] In the steady state the internal energy and the entropy of the machines after one cycle are the same as at the start of the cycle. Hence, on average, dU/dt = 0 and dS/dt = 0 since U and S are functions of state. Furthermore they are closed systems (${\displaystyle {\dot {n}}=0}$) and the volume is fixed (dV/dt = 0). This leads to a significant simplification of the first and second law:
${\displaystyle 0=\sum _{k}{\dot {Q}}_{k}+P}$
and
${\displaystyle 0=\sum _{k}{\frac {{\dot {Q}}_{k}}{T_{k}}}+{\dot {S}}_{i}.}$
The summation is over the (two) places where heat is added or removed.
### Engines
For a heat engine (Fig.2a) the first and second law obtain the form
${\displaystyle 0={\dot {Q}}_{H}-{\dot {Q}}_{a}-P}$
and
${\displaystyle 0={\frac {{\dot {Q}}_{H}}{T_{H}}}-{\frac {{\dot {Q}}_{a}}{T_{a}}}+{\dot {S}}_{i}.}$
Here ${\displaystyle {\dot {Q}}_{H}}$ is the heat supplied at the high temperature TH, ${\displaystyle {\dot {Q}}_{a}}$ is the heat removed at ambient temperature Ta, and P is the power delivered by the engine. Eliminating ${\displaystyle {\dot {Q}}_{a}}$ gives
${\displaystyle P={\frac {T_{H}-T_{a}}{T_{H}}}{\dot {Q}}_{H}-T_{a}{\dot {S}}_{i}.}$
The efficiency is defined by
${\displaystyle \eta ={\frac {P}{{\dot {Q}}_{H}}}.}$
If ${\displaystyle {\dot {S}}_{i}=0}$ the performance of the engine is at its maximum and the efficiency is equal to the Carnot efficiency
${\displaystyle \eta _{C}={\frac {T_{H}-T_{a}}{T_{H}}}.}$
### Refrigerators
For refrigerators (fig.2b) holds
${\displaystyle 0={\dot {Q}}_{L}-{\dot {Q}}_{a}+P}$
and
${\displaystyle 0={\frac {{\dot {Q}}_{L}}{T_{L}}}-{\frac {{\dot {Q}}_{a}}{T_{a}}}+{\dot {S}}_{i}.}$
Here P is the power, supplied to produce the cooling power ${\displaystyle {\dot {Q}}_{L}}$ at the low temperature TL. Eliminating ${\displaystyle {\dot {Q}}_{a}}$ now gives
${\displaystyle {\dot {Q}}_{L}={\frac {T_{L}}{T_{a}-T_{L}}}(P-T_{a}{\dot {S}}_{i}).}$
The coefficient of performance of refrigerators is defined by
${\displaystyle \xi ={\frac {{\dot {Q}}_{L}}{P}}.}$
If ${\displaystyle {\dot {S}}_{i}=0}$ the performance of the cooler is at its maximum. The COP is then given by the Carnot Coefficient Of Performance
${\displaystyle \xi _{C}={\frac {T_{L}}{T_{a}-T_{L}}}.}$
### Power dissipation
In both cases we find a contribution ${\displaystyle T_{a}{\dot {S}}_{i}}$ which reduces the system performance. This product of ambient temperature and the (average) entropy production rate ${\displaystyle P_{diss}=T_{a}{\dot {S}}_{i}}$ is called the dissipated power.
## Equivalence with other formulations
It is interesting to investigate how the above mathematical formulation of the second law relates with other well-known formulations of the second law.
We first look at a heat engine, assuming that ${\displaystyle {\dot {Q}}_{a}=0}$. In other words: the heat flow ${\displaystyle {\dot {Q}}_{H}}$ is completely converted into power. In this case the second law would reduce to
${\displaystyle 0={\frac {{\dot {Q}}_{H}}{T_{H}}}+{\dot {S}}_{i}.}$
Since ${\displaystyle {\dot {Q}}_{H}\geq 0}$ and ${\displaystyle T_{H}>0}$ this would result in ${\displaystyle {\dot {S}}_{i}\leq 0}$ which violates the condition that the entropy production is always positive. Hence: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. This is the Kelvin statement of the second law.
Now look at the case of the refrigerator and assume that the input power is zero. In other words: heat is transported from a low temperature to a high temperature without doing work on the system. The first law with P = 0 would give
${\displaystyle {\dot {Q}}_{L}={\dot {Q}}_{a}}$
and the second law then yields
${\displaystyle 0={\frac {{\dot {Q}}_{L}}{T_{L}}}-{\frac {{\dot {Q}}_{L}}{T_{a}}}+{\dot {S}}_{i}}$
or
${\displaystyle {\dot {S}}_{i}={\dot {Q}}_{L}\left({\frac {1}{T_{a}}}-{\frac {1}{T_{L}}}\right).}$
Since ${\displaystyle {\dot {Q}}_{L}\geq 0}$ and ${\displaystyle T_{a}>T_{L}}$ this would result in ${\displaystyle {\dot {S}}_{i}\leq 0}$ which again violates the condition that the entropy production is always positive. Hence: No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature. This is the Clausius statement of the second law.
## Expressions for the entropy production
### Heat flow
In case of a heat flow ${\displaystyle {\dot {Q}}}$ from T1 to T2 (with ${\displaystyle T_{1}\geq T_{2}}$) the rate of entropy production is given by
${\displaystyle {\dot {S}}_{i}={\dot {Q}}\left({\frac {1}{T_{2}}}-{\frac {1}{T_{1}}}\right).}$
If the heat flow is in a bar with length L, cross-sectional area A, and thermal conductivity ?, and the temperature difference is small
${\displaystyle {\dot {Q}}=\kappa {\frac {A}{L}}(T_{1}-T_{2})}$
the entropy production rate is
${\displaystyle {\dot {S}}_{i}=\kappa {\frac {A}{L}}{\frac {(T_{1}-T_{2})^{2}}{T_{1}T_{2}}}.}$
### Flow of mass
In case of a volume flow ${\displaystyle {\dot {V}}}$ from a pressure p1 to p2
${\displaystyle {\dot {S}}_{i}=-\int _{p_{1}}^{p_{2}}{\frac {\dot {V}}{T}}\mathrm {d} p.}$
For small pressure drops and defining the flow conductance C by ${\displaystyle {\dot {V}}=C(p_{1}-p_{2})}$ we get
${\displaystyle {\dot {S}}_{i}=C{\frac {(p_{1}-p_{2})^{2}}{T}}.}$
The dependences of ${\displaystyle {\dot {S}}_{i}}$ on and on are quadratic.
This is typical for expressions of the entropy production rates in general. They guarantee that the entropy production is positive.
### Entropy of mixing
In this Section we will calculate the entropy of mixing when two ideal gases diffuse into each other. Consider a volume Vt divided in two volumes Va and Vb so that Vt = Va+Vb. The volume Va contains na moles of an ideal gas a and Vb contains nb moles of gas b. The total amount is nt = na + nb. The temperature and pressure in the two volumes is the same. The entropy at the start is given by
${\displaystyle S_{t1}=S_{a1}+S_{b1}.}$
When the division between the two gases is removed the two gases expand, comparable to a Joule-Thomson expansion. In the final state the temperature is the same as initially but the two gases now both take the volume Vt. The relation of the entropy of n moles an ideal gas is
${\displaystyle S=nC_{V}\ln {\frac {T}{T_{0}}}+nR\ln {\frac {V}{V_{0}}}}$
with CV the molar heat capacity at constant volume and R the molar ideal gas constant. The system is an adiabatic closed system, so the entropy increase during the mixing of the two gases is equal to the entropy production. It is given by
${\displaystyle S_{i}=S_{t2}-S_{t1}.}$
As the initial and final temperature are the same the temperature terms plays no role, so we can focus on the volume terms. The result is
${\displaystyle S_{i}=n_{a}R\ln {\frac {V_{t}}{V_{a}}}+n_{b}R\ln {\frac {V_{t}}{V_{b}}}.}$
Introducing the concentration x = na/nt = Va/Vt we arrive at the well known expression
${\displaystyle S_{i}=-n_{t}R[x\ln x+(1-x)\ln(1-x)].}$
### Joule expansion
The Joule expansion is similar to the mixing described above. It takes place in an adiabatic system consisting of a gas and two rigid vessels (a and b) of equal volume, connected by a valve. Initially, the valve is closed. Vessel (a) contains the gas under high pressure while the other vessel (b) is empty. When the valve is opened the gas flows from vessel (a) into (b) until the pressures in the two vessels are equal. The volume, taken by the gas, is doubled while the internal energy of the system is constant (adiabatic and no work done). Assuming that the gas is ideal the molar internal energy is given by Um = CVT. As CV is constant, constant U means constant T. The molar entropy of an ideal gas, as function of the molar volume Vm and T, is given by
${\displaystyle S_{m}=C_{V}\ln {\frac {T}{T_{0}}}+R\ln {\frac {V_{m}}{V_{0}}}.}$
The system, of the two vessels and the gas, is closed and adiabatic, so the entropy production during the process is equal to the increase of the entropy of the gas. So, doubling the volume with T constant, gives that the entropy production per mole gas is
${\displaystyle S_{mi}=R\ln 2.}$
## Microscopic interpretation
The Joule expansion provides an opportunity to explain the entropy production in statistical mechanical (i.e., microscopic) terms. At the expansion, the volume that the gas can occupy is doubled. This means that, for every molecule there are now two possibilities: it can be placed in container (a) or (b). If we have one mole of gas, the number of molecules is equal to Avogadro's number NA. The increase of the microscopic possibilities is a factor of 2 per molecule, so in total the factor is 2NA. Using the well-known Boltzmann expression for the entropy
${\displaystyle S_{m}=k\ln \Omega ,}$
with k Boltzmann's constant and ? the number of microscopic possibilities to realize the macroscopic state, gives
${\displaystyle S_{mi}=k\ln(2^{N_{A}})=kN_{A}\ln 2=R\ln 2.}$
So, in an irreversible process, the number of microscopic possibilities to realize the macroscopic state is increased by a certain factor.
## Basic inequalities and stability conditions
In this section we derive the basic inequalities and stability conditions for closed systems. For closed systems the first law reduces to
${\displaystyle {\frac {\mathrm {d} U}{\mathrm {d} t}}={\dot {Q}}-p{\frac {\mathrm {d} V}{\mathrm {d} t}}+P.}$
The second law we write as
${\displaystyle {\frac {\mathrm {d} S}{\mathrm {d} t}}-{\frac {\dot {Q}}{T}}\geq 0.}$
For adiabatic systems ${\displaystyle {\dot {Q}}=0}$ so dS/dt >= 0. In other words: the entropy of adiabatic systems cannot decrease. In equilibrium the entropy is at its maximum. Isolated systems are a special case of adiabatic systems, so this statement is also valid for isolated systems.
Now consider systems with constant temperature and volume. In most cases T is the temperature of the surroundings with which the system is in good thermal contact. Since V is constant the first law gives ${\displaystyle {\dot {Q}}=\mathrm {d} U/\mathrm {d} t-P}$. Substitution in the second law, and using that T is constant, gives
${\displaystyle {\frac {\mathrm {d} (TS)}{\mathrm {d} t}}-{\frac {\mathrm {d} U}{\mathrm {d} t}}+P\geq 0.}$
With the Helmholtz free energy, defined as
${\displaystyle F=U-TS,}$
we get
${\displaystyle {\frac {\mathrm {d} F}{\mathrm {d} t}}-P\leq 0.}$
If P = 0 this is the mathematical formulation of the general property that the free energy of systems with fixed temperature and volume tends to a minimum. The expression can be integrated from the initial state i to the final state f resulting in
${\displaystyle W_{S}\leq F_{i}-F_{f}}$
where WS is the work done by the system. If the process inside the system is completely reversible the equality sign holds. Hence the maximum work, that can be extracted from the system, is equal to the free energy of the initial state minus the free energy of the final state.
Finally we consider systems with constant temperature and pressure and take P = 0. As p is constant the first laws gives
${\displaystyle {\frac {\mathrm {d} U}{\mathrm {d} t}}={\dot {Q}}-{\frac {\mathrm {d} (pV)}{\mathrm {d} t}}.}$
Combining with the second law, and using that T is constant, gives
${\displaystyle {\frac {\mathrm {d} (TS)}{\mathrm {d} t}}-{\frac {\mathrm {d} U}{\mathrm {d} t}}-{\frac {\mathrm {d} (pV)}{\mathrm {d} t}}\geq 0.}$
With the Gibbs free energy, defined as
${\displaystyle G=U+pV-TS,}$
we get
${\displaystyle {\frac {\mathrm {d} G}{\mathrm {d} t}}\leq 0.}$
## Homogeneous systems
In homogeneous systems the temperature and pressure are well-defined and all internal processes are reversible. Hence ${\displaystyle {\dot {S}}_{i}=0}$. As a result the second law, multiplied by T, reduces to
${\displaystyle T{\frac {\mathrm {d} S}{\mathrm {d} t}}={\dot {Q}}+{\dot {n}}TS_{m}.}$
With P=0 the first law becomes
${\displaystyle {\frac {\mathrm {d} U}{\mathrm {d} t}}={\dot {Q}}+{\dot {n}}H_{m}-p{\frac {\mathrm {d} V}{\mathrm {d} t}}.}$
Eliminating ${\displaystyle {\dot {Q}}}$ and multiplying with dt gives
${\displaystyle \mathrm {d} U=T\mathrm {d} S-p\mathrm {d} V+(H_{m}-TS_{m})\mathrm {d} n.}$
Since
${\displaystyle H_{m}-TS_{m}=G_{m}=\mu }$
with Gm the molar Gibbs free energy and ? the molar chemical potential we obtain the well-known result
${\displaystyle \mathrm {d} U=T\mathrm {d} S-p\mathrm {d} V+\mu \mathrm {d} n.}$
## Entropy production in stochastic processes
Since physical processes can be described by stochastic processes, such as Markov chains and diffusion processes, entropy production can be defined mathematically in such processes.[6]
For a continuous-time Markov chain with instantaneous probability distribution ${\displaystyle p_{i}(t)}$ and transition rate ${\displaystyle q_{ij}}$, the instantaneous entropy production rate is
${\displaystyle e_{p}(t)={\frac {1}{2}}\sum _{i,j}[p_{i}(t)q_{ij}-p_{j}(t)q_{ji}]\log {\frac {p_{i}(t)q_{ij}}{p_{j}(t)q_{ji}}}.}$
The long-time behavior of entropy production is kept after a proper lifting of the process. This approach provides a dynamic explanation for the Kelvin statement and the Clausius statement of the second law of thermodynamics.[7]
## References
1. ^ S.R. de Groot and P. Mazur, Non-equilibrium thermodynamics (North-Holland Publishing Company, Amsterdam-London, 1969)
2. ^ S. Carnot Reflexions sur la puissance motrice du feu Bachelier, Paris, 1824
3. ^ Clausius, R. (1854). "Ueber eine veränderte Form des zweiten Hauptsatzes der mechanischen Wärmetheoriein". Annalen der Physik und Chemie. 93 (12): 481-506. doi:10.1002/andp.18541691202. Retrieved 2012.. Clausius, R. (August 1856). "On a Modified Form of the Second Fundamental Theorem in the Mechanical Theory of Heat". Phil. Mag. 4. 12 (77): 81-98. doi:10.1080/14786445608642141. Retrieved 2012.
4. ^ R. Clausius Über verschiedene für die Anwendung bequeme Formen der Hauptgleigungen der mechanische Wärmetheorie in Abhandlungen über die Anwendung bequeme Formen der Haubtgleichungen der mechanischen Wärmetheorie Ann.Phys. [2] 125, 390 (1865). This paper is translated and can be found in: The second law of thermodynamics, Edited by J. Kestin, Dowden, Hutchinson, & Ross, Inc., Stroudsburg, Pennsylvania, pp. 162-193.
5. ^ A.T.A.M. de Waele, Basic operation of cryocoolers and related thermal machines, Review article, Journal of Low Temperature Physics, Vol.164, pp. 179-236, (2011), DOI: 10.1007/s10909-011-0373-x.
6. ^ Jiang, Da-Quan; Qian, Min; Qian, Min-Ping (2004). Mathematical theory of nonequilibrium steady states: on the frontier of probability and dynamical systems. Berlin: Springer. ISBN 978-3-540-40957-1.
7. ^ Wang, Yue; Qian, Hong (2020). "Mathematical Representation of Clausius' and Kelvin's Statements of the Second Law and Irreversibility". Journal of Statistical Physics. 179 (3): 808-837. arXiv:1805.09530. doi:10.1007/s10955-020-02556-6. | 6,099 | 21,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 91, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.911101 |
https://www.jiskha.com/questions/399300/a-size-36-shoe-in-france-is-size-3-5-in-england-a-function-that-converts-shoe-sizes-in | 1,606,700,141,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00713.warc.gz | 710,736,785 | 5,449 | # Algebra
A size 36 shoe in France is size 3.5 in England. A function that converts shoe sizes in France to those in England is g(x)= 3x-94/4. A size 6 shoe in the United States is size 3.5 in England. A function that converts shoe sizes in England to those in the United States is f(x)=x+5/2.
a. Use composition of functions to find a function that converts shoe sizes in France to those in the United States. Simplify the function.
b. Determine the shoe size in the United States for a size 38 shoe in France.
1. 👍 1
2. 👎 0
3. 👁 523
1. For f(x)=(3x^2)+5and g(x)=7x-2,
a. Verify: g(x + 2) ≠ g(x) + g(2).
b. Find (f – g)(x).
c. Using the resulting function in (b), show that (f – g)(2) = f(2) – g(2).
(The work should be different for each side of the equation.)
d. Is (fg)(0) =(f/g)(0)? Explain.
e. Find (f(x+h)-f(x))/h , h ≠ 0.
1. 👍 0
2. 👎 0
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The table shows men's shoe sizes in the United States and the corresponding European shoe sizes. Let y = f(x)represent the function that gives the men's European shoe size in terms of x, the men's U.S. size. Men's U.S. shoe size, | 960 | 3,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-50 | latest | en | 0.832844 |
http://www.cfd-online.com/W/index.php?title=Inviscid_flow&diff=4899&oldid=3017 | 1,472,549,986,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982974985.86/warc/CC-MAIN-20160823200934-00027-ip-10-153-172-175.ec2.internal.warc.gz | 356,001,008 | 11,546 | Inviscid flow
(Difference between revisions)
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Revision as of 19:02, 30 September 2005 (view source)Ganesh (Talk | contribs)m← Older edit Revision as of 04:38, 16 December 2005 (view source)Tsaad (Talk | contribs) (fixed spelling)Newer edit → (4 intermediate revisions not shown) Line 1: Line 1: - A flow in which viscous effects can be neglected is known as ''inviscid flow''. At high Reynolds numbers, flow past slender bodies involve thin boundary layers. Viscous effects are important only inside the boundary layer and the flow outside it is nearly inviscid. If the boundary layer is not separated then the inviscid flow model can be used to predict the pressure distribution with reasonable accuracy. Although no practical flow is inviscid, the inviscid assumption comes into fore if the time scales for diffusion are much larger compared to the time scales for convection. + A flow in which viscous effects can be neglected is known as ''inviscid flow''. At high Reynolds numbers, flow past slender bodies involve thin boundary layers. Viscous effects are important only inside the boundary layer and the flow outside it is nearly inviscid. If the boundary layer is not separated then the inviscid flow model can be used to predict the pressure distribution with reasonable accuracy. Although no practical flow is inviscid, the inviscid assumption is valid if the time scales for diffusion are much larger compared to the time scales for convection, which is measured by the [[Reynolds number]]. ==Governing Equations== ==Governing Equations== Line 45: Line 45: * $R$ is the gas constant * $R$ is the gas constant * $T$ is the absolute temperature * $T$ is the absolute temperature + + ==Euler equations as the limit of Navier-Stokes equations== + + Have a look at this discussion:
+ http://www.cfd-online.com/Forum/fluent.cgi?read=32738
+ http://www.cfd-online.com/Forum/fluent.cgi?read=32788 + + ==Related pages== + *[[Euler equations in conservation form]] + *[[Rotational invariance of Euler equations]] + *[[Inviscid flux jacobians]]
Revision as of 04:38, 16 December 2005
A flow in which viscous effects can be neglected is known as inviscid flow. At high Reynolds numbers, flow past slender bodies involve thin boundary layers. Viscous effects are important only inside the boundary layer and the flow outside it is nearly inviscid. If the boundary layer is not separated then the inviscid flow model can be used to predict the pressure distribution with reasonable accuracy. Although no practical flow is inviscid, the inviscid assumption is valid if the time scales for diffusion are much larger compared to the time scales for convection, which is measured by the Reynolds number.
Governing Equations
The governing equations for inviscid flow, also known as Euler equations, are obtained by discarding the viscous terms from the Navier-Stokes equations
• Continuity equation
$\frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x_j}(\rho u_j)= 0$
• Momentum equation
$\frac{\partial}{\partial t}(\rho u_i) + \frac{\partial}{\partial x_j}(\rho u_i u_j + p \delta_{ij}) = 0$
• Energy equation
$\frac{\partial}{\partial t}(\rho E) + \frac{\partial}{\partial x_j}[(\rho E + p)u_j] = 0$
where
• $\rho$ is the density
• $u_i$ is the fluid velocity
• $p$ is the pressure
• $E$ is the total energy per unit mass of fluid
$E = \frac{p}{\gamma - 1} + \frac{1}{2} \rho |u|^2$
The above equations are closed by taking an equation of state, the simplest being the ideal gas
$p = \rho R T$
where
• $R$ is the gas constant
• $T$ is the absolute temperature
Euler equations as the limit of Navier-Stokes equations
Have a look at this discussion:
http://www.cfd-online.com/Forum/fluent.cgi?read=32738
http://www.cfd-online.com/Forum/fluent.cgi?read=32788 | 967 | 3,803 | {"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": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-36 | latest | en | 0.914076 |
https://bookofproofs.github.io/branches/logic/modus-ponens-proof.html | 1,685,305,098,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00631.warc.gz | 167,955,530 | 3,191 | # Proof
(related to Lemma: Modus Ponens)
We want to prove that modus ponens is a valid logical argument. * Modus ponens can be formulated in propositional logic as $((p\Rightarrow q)\wedge p)\Rightarrow q.$ * Reformulating the implication as disjunction gives us the expression $((\neg p\vee q)\wedge p)\Rightarrow q.$ * On the left side we have only the operations "$\wedge$" and "$\vee$". Therefore, we can use the fact that propositional logic is a Boolean algebra $(B,\wedge,\vee,1,0)$, and make use of the properties of this Boolean algebra $B$ as follows: $$\begin{array}{rl} ((\neg p\vee q)\wedge p)\Rightarrow q&\\ (p \wedge(\neg p\vee q))\Rightarrow q&(\text{commutativity of }"\wedge")\\ ((p \wedge\neg p)\vee (p\wedge q))\Rightarrow q&(\text{distributivity of }"\wedge"\text{ and }"\vee")\\ (0\vee (p\wedge q))\Rightarrow q&(\neg p\text{ is the complement element of }p)\\ (p\wedge q)\Rightarrow q&(0\text{ is the smallest element of theB})\\ \end{array}$$ * It follows from the truth table of conjunction that if $p$ and $q$ are true, than $q$ is true. * By definition of a valid logical argument, modus ponens $((p\Rightarrow q)\wedge p)\Rightarrow q$ is valid.
∎
Thank you to the contributors under CC BY-SA 4.0! Github: References Bibliography Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016
Feeds Acknowledgments Terms of Use Privacy Policy Disclaimer © 2014+ Powered by bookofproofs, All rights reserved. | 435 | 1,441 | {"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} | 3.4375 | 3 | CC-MAIN-2023-23 | longest | en | 0.654703 |
https://www.enotes.com/homework-help/simplify-complex-number-13-14i-2-230901 | 1,516,284,927,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887414.4/warc/CC-MAIN-20180118131245-20180118151245-00459.warc.gz | 957,047,548 | 9,963 | # Simplify the complex number (13+14i)^2
justaguide | Certified Educator
We have to simplify (13+14i)^2.
(13+14i)^2
=> (13+14i)*(13+14i)
multiply all the terms
=> 13*13 + 13*14i + 13*14i + 14i*14i
=> 169 + 364i + 196*i^2
As i^2 = 1
=> 169 - 196 + 364i
=> -27 + 364i
Therefore (13+14i)^2 = -27 + 364i
neela | Student
To simplify the complex number (13+14i)^2.
(13+14i)^2 = (13+14i)13+14)i)
(13+14i)^2 = 13(13+14i) + 14i(13+14i).
(13+14i)^2 =13*13 + 13*14i + 14i*13+14i*14i
(13+14i)^2 = 169 + (182+182)i +196i^2
(13+14i)^2 = 169 +364i -196.
(13+14i)^2 = 169-196 + 364i.
(13+14i)^2 = -27+364i
giorgiana1976 | Student
To simplify the complex number raised to square, we'll expand the square, using the formula:
(a+b)^2 = a^2 + 2ab + b^2
We'll put a = 13 and b = 14i:
(13 + 14i)^2 = 13^2 + 2*13*14i + (14i)^2
(13 + 14i)^2 = 169 + 364i + 196i^2
Since i^2 = -1, we'll get:
(13 + 14i)^2 = 169 + 364i - 196
We'll combine the real parts:
(13 + 14i)^2 = -27 + 364i
The simplified form of the complex number raised to square is:
(z)^2 = -27 + 364i, where z = 13 + 14i | 501 | 1,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-05 | latest | en | 0.527295 |
https://www.sarthaks.com/2830681/representation-computer-system-decimal-number-subtracted-binary-complement | 1,675,194,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00011.warc.gz | 995,511,973 | 15,635 | # In an 8 bit representation of computer system the decimal number 26 has to be subtracted from 22 and the result in binary 2's complement is _________
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In an 8 bit representation of computer system the decimal number 26 has to be subtracted from 22 and the result in binary 2's complement is _________
1. 00000100
2. 00001100
3. 00000101
4. 00010100
by (24.2k points)
selected
Correct Answer - Option 1 : 00000100
• Two's complement representation, or, in other words, signed notation - the first bit tells about the sign. The convention is that a number with a leading 1 is negative, while a leading 0 denotes a positive value. In an 8-bit representation, we can write any number from -128 to 127. The name comes from the fact that a negative number is a two's complement of a positive one.
• 26 has to be subtracted from 22 hence result will be negative. So, to avoid negative results, 2's complement is to be used. We have to find 2's complement of the Subtrahend (26) and add it to the minuend (22). To calculate 2's complement, first, we need to find the 1's complement by swapping all 1 to 0 and all 0 to 1 and then we need to add 1 to it:
• for 22, Binary = 10110
• for 26, Binary = 11010, 1's Complement = 001012's Complement = 00110
• Addition = 10110 + 00110 = 11100
• now in the answer obtained, there is no carry generated. In this case, we will have to find 2's complement of the result in order to find the final result.
• 11100 = 1's Complement = 000112's Complement = 00100
• In 8 bit representation, we will write 00000100
• hence our answer is 00000100 | 470 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-06 | latest | en | 0.875941 |
https://rdrr.io/cran/granova/man/granova.1w.html | 1,632,501,166,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00719.warc.gz | 532,330,863 | 173,335 | granova.1w: Graphic display for one-way ANOVA In granova: Graphical Analysis of Variance
Description
Graphic to display data for a one-way analysis of variance, and also to help understand how ANOVA works, how the F statistic is generated for the data in hand, etc. The graphic may be called 'elemental' or 'natural' because it is built upon the key question that drives one-way ANOVA.
Usage
```1 2 3 4 5``` ```granova.1w(data, group = NULL, dg = 2, h.rng = 1.25, v.rng = 0.2, box = FALSE, jj = 1, kx = 1, px = 1, size.line = -2.5, top.dot = 0.15, trmean = FALSE, resid = FALSE, dosqrs = TRUE, ident = FALSE, pt.lab = NULL, xlab = NULL, ylab = NULL, main = NULL, ...) ```
Arguments
`data` Dataframe or vector. If a dataframe, the two or more columns are taken to be groups of equal size (whence `group` is NULL). If `data` is a vector, `group` must be a vector, perhaps a factor, that indicates groups (unequal group sizes allowed with this option). `group` Group indicator, generally a factor in case `data` is a vector. `dg` Numeric; sets number of decimal points in output display, default = 2. `h.rng` Numeric; controls the horizontal spread of groups, default = 1.25 `v.rng` Numeric; controls the vertical spread of points, default = 0.25. `box` Logical; provides a bounding box (actually a square) to the graph; default FALSE. `jj` Numeric; sets horizontal jittering level of points; when pairs of ordered means are close to one another, try jj < 1; default = 1. `kx` Numeric; controls relative sizes of `cex`, default = 1.0 `px` Numeric; controls relative sizes of `cex.axis`, default = 1.0 `size.line` Numeric; controls vertical location of group size and name labels, default = -2.5. `top.dot` Numeric; controls hight of end of vertical dotted lines through groups; default = .15. `trmean` Logical; marks 20% trimmed means for each group (as green cross) and prints out those values in output window, default = FALSE. `resid` Logical; displays marginal distribution of residuals (as a 'rug') on right side (wrt grand mean), default = FALSE. `dosqrs` Logical; ensures plot of squares (for variances); when FALSE or the number of groups is 2, squares will be suppressed, default = TRUE. `ident` Logical; allows user to identify specific points on the plot, default = FALSE. `pt.lab` Character vector; allows user to provide labels for points, else the rownames of xdata are used (if defined), or if not labels are 1:N (for N the total number of all data points), default = NULL. `xlab` Character; horizontal axis label, default = NULL. `ylab` Character; vertical axis label, default = NULL. `main` Character; main label, top of graphic; can be supplied by user, default = NULL, which leads to printing of generic title for graphic. `...` Optional arguments to be passed to `identify`, for example `offset`
Details
The central idea of the graphic is to use the fact that a one way analysis of variance F statistic is the ratio of two variances each of which can usefully be presented graphically. In particular, the sum of squares between (among) can be represented as the sum of products of so-called effects (each being a group mean minus the grand mean) and the group means; when these effects are themselves plotted against the group means a straight line necessarily ensues. The group means are plotted as (red triangles along this line. Data points (jittered) for groups are displayed (vertical axis) with respect to respective group means. One-way ANOVA residuals can be displayed (set resid=TRUE) as a rug plot (on right margin); the standard deviation of the residuals, when squared, is just the mean square within, which corresponds to area of blue square. The conventional F statistic is just a ratio of the between to the within mean squares, or variances, each of which corresponds to areas of squares in the graphic. The blue square, centered on the grand mean vertically and zero for the X-axis, corresponds to mean square within (with side based on [twice] the pooled standard deviation); the red square corresponds to the mean square between, also centered on the grand mean. Use of effects to locate the groups in the order of the observed means, from left to right (by increasing size) yields this 'elemental' graphic for this commonly used statistical method.
Groups need not be of the same sizes, nor do data need to reflect any particular distributional characteristics. Skewness, outliers, clustering of data points, and various other features of the data may be seen in this graphic, possibly identified using point labels. Trimmed means (20%) can also be displayed if desired. Finally, by redisplaying the response data in two or more versions of the graphic it can be useful to visualize various effects of non-linear data transformations. (`ident=TRUE`).
Value
Returns a list with two components:
`grandsum` Contains the basic ANOVA statistics: the grandmean, the degrees of freedom and mean sums of squares between and within groups, the F statistic, F probability and the ratio between the sum of squares between groups and the total sum of squares. `stats` Contains a table of statistics by group: the size of each group, the contrast coefficients used in plotting the groups, the weighted means, means, and 20% trimmed means, and the group variances and standard deviations.
Author(s)
Robert M. Pruzek RMPruzek@yahoo.com,
James E. Helmreich James.Helmreich@Marist.edu
References
Fundamentals of Exploratory Analysis of Variance, Hoaglin D., Mosteller F. and Tukey J. eds., Wiley, 1991.
`granova.2w`, `granova.contr`, `granova.ds`
Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27``` ``` data(arousal) #Drug A granova.1w(arousal[,1:2], h.rng = 1.6, v.rng = 0.5, top.dot = .35) ######################### library(MASS) wt.gain <- anorexia[, 3] - anorexia[, 2] granova.1w(wt.gain, group = anorexia[, 1], size.line = -3) ########################## data(poison) ##Note violation of constant variance across groups in following graphic. granova.1w(poison\$SurvTime, group = poison\$Group, ylab = "Survival Time") ##RateSurvTime = SurvTime^-1 granova.1w(poison\$RateSurvTime, group = poison\$Group, ylab = "Survival Rate = Inverse of Survival Time") ##Nonparametric version: RateSurvTime ranked and rescaled ##to be comparable to RateSurvTime; ##note labels as well as residual (rug) plot below. granova.1w(poison\$RankRateSurvTime, group = poison\$Group, ylab = "Ranked and Centered Survival Rates", main = "One-way ANOVA display, poison data (ignoring 2-way set-up)", res = TRUE) ```
Example output
```Loading required package: car
\$grandsum
Grandmean df.bet df.with MS.bet MS.with
22.35 1.00 18.00 73.73 6.86
F.stat F.prob SS.bet/SS.tot
10.75 0.00 0.37
\$stats
Size Contrast Coef Wt'd Mean Mean Trim'd Mean Var. St. Dev.
Placebo 10 -1.92 20.43 20.43 20.30 5.83 2.41
Drug.A 10 1.92 24.27 24.27 24.45 7.89 2.81
\$grandsum
Grandmean df.bet df.with MS.bet MS.with
2.76 2.00 69.00 307.32 56.68
F.stat F.prob SS.bet/SS.tot
5.42 0.01 0.14
\$stats
Size Contrast Coef Wt'd Mean Mean Trim'd Mean Var. St. Dev.
Cont 26 -3.21 -0.49 -0.45 -1.16 63.82 7.99
CBT 29 0.24 3.63 3.01 1.80 53.41 7.31
FT 17 4.50 5.15 7.26 7.91 51.23 7.16
\$grandsum
Grandmean df.bet df.with MS.bet MS.with
0.48 11.00 36.00 0.20 0.02
F.stat F.prob SS.bet/SS.tot
9.01 0.00 0.73
\$stats
Size Contrast Coef Wt'd Mean Mean Trim'd Mean Var. St. Dev.
3 4 -0.27 0.21 0.21 0.21 0.00 0.02
9 4 -0.24 0.24 0.24 0.24 0.00 0.01
2 4 -0.16 0.32 0.32 0.32 0.01 0.08
12 4 -0.15 0.32 0.32 0.32 0.00 0.03
6 4 -0.14 0.34 0.34 0.34 0.00 0.05
8 4 -0.10 0.38 0.38 0.38 0.00 0.06
1 4 -0.07 0.41 0.41 0.41 0.00 0.07
7 4 0.09 0.57 0.57 0.57 0.02 0.16
10 4 0.13 0.61 0.61 0.61 0.01 0.11
11 4 0.19 0.67 0.67 0.67 0.07 0.27
5 4 0.34 0.82 0.82 0.82 0.11 0.34
4 4 0.40 0.88 0.88 0.88 0.03 0.16
\$grandsum
Grandmean df.bet df.with MS.bet MS.with
2.62 11.00 36.00 5.17 0.24
F.stat F.prob SS.bet/SS.tot
21.53 0.00 0.87
\$stats
Size Contrast Coef Wt'd Mean Mean Trim'd Mean Var. St. Dev.
4 4 -1.46 1.16 1.16 1.16 0.04 0.20
5 4 -1.23 1.39 1.39 1.39 0.31 0.55
10 4 -0.93 1.69 1.69 1.69 0.13 0.36
11 4 -0.92 1.70 1.70 1.70 0.49 0.70
7 4 -0.76 1.86 1.86 1.86 0.24 0.49
1 4 -0.14 2.49 2.49 2.49 0.25 0.50
8 4 0.09 2.71 2.71 2.71 0.17 0.42
6 4 0.41 3.03 3.03 3.03 0.18 0.42
12 4 0.47 3.09 3.09 3.09 0.06 0.24
2 4 0.65 3.27 3.27 3.27 0.68 0.82
9 4 1.64 4.26 4.26 4.26 0.06 0.23
3 4 2.18 4.80 4.80 4.80 0.28 0.53
\$grandsum
Grandmean df.bet df.with MS.bet MS.with
2.49 11.00 36.00 3.70 0.19
F.stat F.prob SS.bet/SS.tot
19.18 0.00 0.85
\$stats
Size Contrast Coef Wt'd Mean Mean Trim'd Mean Var. St. Dev.
4 4 -1.38 1.11 1.11 1.11 0.03 0.18
5 4 -1.13 1.36 1.36 1.36 0.28 0.53
10 4 -0.82 1.67 1.67 1.67 0.10 0.31
11 4 -0.80 1.69 1.69 1.69 0.50 0.71
7 4 -0.67 1.82 1.82 1.82 0.24 0.49
1 4 -0.10 2.39 2.39 2.39 0.30 0.55
8 4 0.23 2.72 2.72 2.72 0.19 0.44
6 4 0.55 3.04 3.04 3.04 0.18 0.42
12 4 0.61 3.09 3.09 3.09 0.05 0.22
2 4 0.66 3.15 3.15 3.15 0.39 0.62
9 4 1.29 3.78 3.78 3.78 0.03 0.16
3 4 1.55 4.04 4.04 4.04 0.03 0.16
```
granova documentation built on May 2, 2019, 9:36 a.m. | 3,541 | 10,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.784898 |
https://money.stackexchange.com/questions/90702/what-is-the-formula-for-calculating-the-mortgage-constant-when-payments-are-made | 1,716,781,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00720.warc.gz | 341,532,846 | 37,989 | # What is the formula for calculating the mortgage constant when payments are made at the beginning of the period?
In this case, the mortgage constant (or loan constant or debt constant) is the (in my case, annual) ratio of constant payments to the original amount, like here: http://www.double-entry-bookkeeping.com/periodic-payment/how-to-calculate-a-debt-constant/
Suppose we have an annual interest rate of 4.565% and 360 payments (30 year loan). In Excel, we can specify the following formula:
``````PMT(0.04565/12, 360, -1, 0, 1) * 12
``````
Where present value is \$1, future value is \$0, and Type=1 signifies that payments are due at the beginning of the period. The result is 6.1034%.
When I apply the mathematical formula, it assumes that payments are made at the end of the period (or am I wrong?), so where
``````Debt Constant = (Interest Rate/12) / (1 - (1 / (1 + Interest Rate/12))^n) * 12
= (0.04565/12) / (1 - (1 / (1 + 0.04565/12))^360) * 12
= 6.1267%
``````
How is the formula adjusted to account for the payments made in the beginning of the period, since the actual value for the purposes of this calculation is not known? Or am I misunderstanding something?
• This is a question about accounting not an issue of personal finance that the OP is facing. Feb 11, 2018 at 18:20
• @Dilip Sarwate Apologies if this is completely off-topic, a Google search for questions about mortgage constants led me here, so I posed my own since those didn't answer my specific concern. I believe the Accounting forum never took off--do you know of a more appropriate place to post this? Feb 11, 2018 at 19:06
• Figuring out basic mortgage calculations seem to be perfectly in the personal finance area of accounting. Feb 12, 2018 at 12:54
Your mathematical formula can be adjusted by dividing by `(1 + Interest Rate/12)`, i.e.
``````Debt Constant =
(0.04565/12)/(1 - (1/(1 + 0.04565/12))^360)*12/(1 + 0.04565/12) =
0.0610344
``````
The syntax for the Excel formula is
``````PMT(rate, nper, pv, [fv], [type])
``````
type = 1 is for payments at the beginning of the period, so you are calculating the payments for an annuity due.
``````PMT(0.04565/12, 360, -1, 0, 1) * 12 = 0.0610344
``````
Your mathematical formula is for an ordinary annuity; payments made at the end of the period.
Formula for an annuity due (payments at the beginning of the period)
With principal `s`, `n` periods, periodic rate `r` and periodic payment `d`
``````s = 1
n = 360
r = 0.04565/12
``````
The principal equals the sum of the payments discounted to present value.
``````∴ d = (r (1 + r)^(-1 + n) s)/(-1 + (1 + r)^n) = 0.0050862
12 d = 0.0610344
``````
Derivation of formula
Using the geometric sum theorm
https://en.wikipedia.org/wiki/Geometric_series#Formula
• "type = 1 means annuity due" very helpful, that was hard to find (the actual name of the type=1 calculation) Dec 30, 2022 at 15:58
Suppose you started with the formula for payments at the beginning of the period, and wanted to know how to adjust it for payment at the end. Well, each payment is accruing interest over an entire period. So you would have to multiply each payment by the interest factor for each period. The interest rate over a period is the interest rate per year divided by the number of periods per year. Since there are 12 period each year, the interest rate per period is 0.04565 (the interest given on a yearly basis) divided by 12. The total amount is the principal plus the interest rate times the principal:
principal + (interest rate)*principal
Factor out the principal, and you get:
principal*(1+interest rate)
So we get that the factor is 1+interest rate = 1+0.04565/12
That is the factor we have to multiply by to get from "beginning of period" to "end of period", so we have to divide by that to go the other way. | 1,068 | 3,819 | {"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} | 4 | 4 | CC-MAIN-2024-22 | latest | en | 0.926787 |
https://www.scribd.com/document/56553343/Electricity-and-Fields-Questions | 1,558,898,127,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259452.84/warc/CC-MAIN-20190526185417-20190526211417-00338.warc.gz | 929,320,275 | 58,827 | You are on page 1of 62
# –8– M04/431/H(2)
(Question B1 continued)
Current electricity
A cell of electromotive force (e.m.f.) E and internal resistance r is connected in series with a
resistor R, as shown below.
E r
The cell supplies 8.1×103 of energy when 5.8 ×103 of charge moves completely round the
J C
circuit. The current in the circuit is constant.
## (c) (i) Calculate the e.m.f. E of the cell. [2]
.....................................................................
.....................................................................
.....................................................................
(ii) The resistor R has resistance 6.0 . The potential difference between its terminals
is 1.2 V. Determine the internal resistance r of the cell. [3]
.....................................................................
.....................................................................
.....................................................................
.....................................................................
## (iii) Calculate the total energy transfer in the resistor R. [2]
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.....................................................................
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## (This question continues on the following page)
224-177
–9– M04/431/H(2)
(Question B1 continued)
(iv) Describe, in terms of a simple model of electrical conduction, the mechanism by which
the energy transfer in the resistor R takes place. [5]
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.....................................................................
.....................................................................
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## (This question continues on the following page)
224-177 Turn over
–2– M04/431/H(2)
SECTION A
## Answer all the questions in the spaces provided.
A1. Data based question. This question is about change of electrical resistance with temperature.
The table below gives values of the resistance R of an electrical component for different values of
its temperature T. (Uncertainties in measurement are not shown.)
## T/ °C 1.2 2.0 3.5 5.2 6.8 8.1 9.6
R/ 3590 3480 3250 3060 2880 2770 2650
(a) On the grid below, plot a graph to show the variation with temperature T of the resistance
R. [3]
Show values on the temperature axis from T = 0 °C to T = 10 °C .
## (This question continues on the following page)
224-177
–3– M04/431/H(2)
(Question A1 continued)
(b) (i) Draw a curve that best fits the points you have plotted. Extend your curve to cover
the temperature range from 0 °C to 10 °C . [1]
## (ii) Use your graph to determine the resistance at 0 °C and at 10 °C [2]
.
Resistance at 0 °C = . . . . . . . . . . . . . . . . . . . . . . .
Resistance at 10 °C = . . . . . . . . . . . . . . . . . . . . . .
(c) On your graph, draw a straight-line between the resistance values at 0 and at 10 °C . This
°C
line shows the variation with temperature 0 an 10 °C ) of the
(between assuming a linear change. °C d resistance, [1]
(d) (i) Assuming a linear change of resistance with temperature, use your graph to
determine the temperature at which the resistance is 3060 . [1]
Temperature = . . . . . . . . . . . . . . . . . . . . . . . . . . °C
(ii) Use your answer in (d)(i) to calculate the percentage difference in the temperature for a
resistance of 3060 that results from assuming a linear change rather than the
non-linear change. [3]
.....................................................................
.....................................................................
.....................................................................
(e) In a particular experiment to measure the variation with temperature of the resistance, each
measurement of resistance has an uncertainty of ! 30 and the uncertainty in the
°
temperature measurements is ! 0.2 C .
(i) On your graph in (a), show the uncertainties in the values of R and of T
for temperatures of 1.2 °C, 5.2 °C and 9.6 °C . [2]
(ii) State and explain whether, within the experimental uncertainties, the relationship
between resistance and temperature could be linear. [2]
.....................................................................
.....................................................................
.....................................................................
224-177 Turn over
–6– M04/432/H(2)
## A3. This question is about a filament lamp.
(a) On the axes below, draw a sketch-graph to show the variation with potential difference V
of the current I in a typical filament lamp (the I–V characteristic). (Note: this is a sketch-
graph; you do not need to add any values to the axes). [1]
V
0
0 I
(b) (i) Explain how the resistance of the filament is determined from the graph. [1]
.....................................................................
.....................................................................
(ii) Explain whether the graph you have sketched indicates ohmic behaviour or non-
ohmic behaviour. [1]
.....................................................................
.....................................................................
A filament lamp operates at maximum brightness when connected to a 6.0 V supply. At maximum
brightness, the current in the filament is 120 mA.
(c) (i) Calculate the resistance of the filament when it is operating at maximum brightness. [1]
.....................................................................
(ii) You have available a 24 V supply and a collection of resistors of a suitable power
rating and with different values of resistance. Calculate the resistance of the resistor
that is required to be connected in series with the supply such that the voltage across
the filament lamp will be 6.0 V. [2]
.....................................................................
.....................................................................
.....................................................................
224-180
– 14 – M06/4/PHYSI/HP2/ENG/TZ1/XX+
(Question B1 continued)
## part 2 Heating water electrically
The diagram below shows part of the heating circuit of a domestic shower.
insulated wire
water pipe
240 V
supply
## cold water 14 °C hot water 40 °C
insulated heating element
Cold water enters the shower unit and flows over an insulated heating element. The heating
element is rated at 7.2 kW, 240 V. The water enters at a temperature of 14 °C and leaves at a
3 −1 −1
temperature of 40 °C. The specific heat capacity of water is 4.2 10 J kg K .
(a) Describe how thermal energy is transferred from the heating element to the water. [3]
......................................................................
.
......................................................................
.
......................................................................
.
......................................................................
.
......................................................................
.
−1
(b) Estimate the flow rate in kg s of the water. [4]
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(This question continues on the following page)
2206-6508
143 6
– 15 – M06/4/PHYSI/HP2/ENG/TZ1/XX+
## (Question B1, part 2 continued)
(c) Suggest two reasons why your answer to (b) is only an estimate. [2]
1. ..................................................................
..................................................................
2. ..................................................................
..................................................................
(d) Calculate the current in the heating element when the element is operating at 7.2 kW. [2]
.......................................................................
.......................................................................
.......................................................................
(e) Explain why, when the shower unit is switched on, the initial current in the heating
element is greater than the current calculated in (d). [2]
.......................................................................
.......................................................................
.......................................................................
## (This question continues on the following page)
2206-6508 153 6
turn over
– 16 – M06/4/PHYSI/HP2/ENG/TZ1/XX+
## (Question B1, part 2 continued)
(f) In some countries, shower units are operated from a 110 V supply. A heating element
operating with a 240 V supply has resistance R240 and an element operating from a 110 V
supply has resistance R110.
(i) Deduce, that for heating elements to have identical power outputs
R110 [3]
=
R240 0.21.
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
(ii) Using the ratio in (i), describe and explain one disadvantage of using a 110 V
supply [2]
for domestic purposes.
..................................................................
..................................................................
..................................................................
..................................................................
2206-6508
163 6
–8– M06/4/PHYSI/HP2/ENG/TZ1/XX+
## a3. This question is about an electric circuit.
A particular filament lamp is rated at 12 V, 6.0 mA. It just lights when the potential difference
across the filament is 6.0 V.
A student sets up a electric circuit to measure the I-V characteristics of the filament lamp.
In the circuit, shown below, the student has connected the voltmeter and the ammeter into the
circuit incorrectly.
100 kΩ
12 V S
V
The battery has e.m.f. 12 V and negligible internal resistance. The ammeter has negligible
resistance and the resistance of the voltmeter is 100 kΩ.
## The maximum resistance of the variable resistor is 15 Ω.
(a) Explain, without doing any calculations, whether there is a position of the slider S at
which the lamp will be lit. [3]
.......................................................................
.......................................................................
.......................................................................
.......................................................................
.......................................................................
## (b) Estimate the maximum reading of the ammeter. [2]
.......................................................................
.......................................................................
## (This question continues on the following page)
2206-6508
083 6
–9– M06/4/PHYSI/HP2/ENG/TZ1/XX+
(Question A3 continued)
(c) Complete the circuit diagram below showing the correct position of the voltmeter and of
the ammeter in order to determine the I-V characteristics of the filament lamp. [2]
12 V
2206-6508 093 6
turn over
– 21 – M06/4/PHYSI/HP2/ENG/TZ2/XX+
b3. This question is about electric current and the effects of electric current.
Electric current
(a) The diagram below shows the circuit used to measure the current-voltage (I-V)
characteristic of an electrical component X.
## (i) label the ammeter A and the voltmeter V. [1
]
(ii) mark the position of the contact of the potentiometer that will produce a reading
of zero on the voltmeter. Label this position P. [1
]
## (This question continues on the following page)
2206-6514 2133
turn over
– 22 – M06/4/PHYSI/HP2/ENG/TZ2/XX+
(Question B3 continued)
(b) The graph below shows the current-voltage (I-V) characteristics of two different conductors
X and Y.
0.50
0.45
0.40
0.35
0.30
Y
0.25 X
I/
A 0.20
0.15
0.10
0.05
0.00
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0
V/V
(i) State the value of the current for which the resistance of X is the same as the
resistance of Y and determine the value of this resistance. [2]
Current: ........................................................
Resistance: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
........................................................
(ii) Describe and suggest an explanation for the I-V characteristic of conductor [3]
Y.
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
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(Question B3 continued)
(c) The two conductors X and Y are connected in the circuit as shown below.
12 V
The cell has e.m.f. 12 V and negligible internal resistance. The resistor Z has resistance R
and the potential difference across the conductors X and Y is 5.0 V.
(i) Use the graph in (b) to determine the total current in the circuit. [2]
.................................................................
.
.................................................................
.
.................................................................
.
## (ii) Determine the resistance R of the resistor Z. [2]
..................................................................
..................................................................
..................................................................
..................................................................
(iii) Determine the total resistance of the parallel combination of X and Y. [2]
..................................................................
..................................................................
..................................................................
..................................................................
(This question continues on the following page)
## 2206-6514 turn over
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b3. This question is in two parts. part 1 is about electrical conduction and part 2 is about
thermodynamics.
## part 1 Electrical conduction
In a copper wire the number of conduction electrons is equal to the number of copper atoms in
the wire.
## (a) State what is meant by conduction electrons. [1]
.......................................................................
.......................................................................
3 –3
(b) (i) The density of copper is 8.93 ×10 kg m and its molar mass is 64 g. Deduce
3 5
that the number of moles of copper in a volume of 1.0 m is 1.4 × 10 . [2]
..................................................................
..................................................................
..................................................................
3
(ii) Estimate the number of conduction electrons in 1.0 m of copper. [1]
..................................................................
..................................................................
(c) The diagram below shows some of the conduction electrons in a copper wire. The
arrows represent the random velocities of some of the electrons.
copper wire
Explain, by reference to the motion of the electrons, why there is no current in the wire. [2]
.......................................................................
.......................................................................
## (This question continues on the following page)
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## (Question B3, part 1 continued)
(d) An electric field is established inside the copper wire directed as shown in the
diagram below. The dots represent electrons. The random velocities of the electrons are
not shown.
On the diagram below, draw an arrow to indicate the direction of the drift velocity of
the electrons. [1]
electric field
copper wire
## (This question continues on the following page)
2207-6508 273 7
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– 28 – M07/4/PHYSI/HP2/ENG/TZ1/XX+
## (Question B3, part 1 continued)
–3 –1
(e) A typical value for the electron drift velocity in a copper wire is 10 m s . In the
circuit below, the length of the copper wire joining the negative terminal of the battery to
the lamp is 0.50 m.
0.50 m
(i) The switch S is closed. Calculate the time it would take for an electron to move
from the negative terminal of the battery to the lamp. [1]
..................................................................
(ii) The lamp lights in a time much less than that calculated in (e)(i). Explain this
observation. [2]
..................................................................
..................................................................
..................................................................
..................................................................
(iii) Discuss, in terms of the movement of the electrons, the energy transformations
taking place in the filament of the lamp. [4]
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
## (This question continues on the following page)
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## (Question B3, part 1 continued)
(f) The diagram below shows part of a circuit that may be used to determine the
current - potential difference (I-V) characteristics of a lamp.
An ammeter and a voltmeter are required. On the diagram above, draw symbols to show
the correct positions of the ammeter and the voltmeter. [2]
(g) The I-V characteristics for one lamp are shown below.
I / A 0.50
0.40
0.30
0.20
0.10
0.00
0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60
V/V
(i) State a range of values of the current I for which the lamp may be considered to
show ohmic behaviour. [1]
..................................................................
(ii) The potential difference across the lamp is 0.80 V. Calculate the resistance of
the lamp at this potential difference. [2]
..................................................................
..................................................................
(This question continues on the following page)
## 2207-6508 Turn over
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7
–6– M07/4/PHYSI/HP2/ENG/TZ2/XX+
## (a) Define e.m.f. and state Ohm’s law. [2]
e.m.f.: ...........................................................
...........................................................
## Ohm’s law: ...........................................................
...........................................................
(b) In the circuit below an electrical device (load) is connected in series with a cell of
e.m.f. 2.5 V and internal resistance r. The current I in the circuit is 0.10 A.
e.m.f. = 2.5
V
I = 0.10 A
Calculate
## (i) the total power of the cell. [1]
..................................................................
..................................................................
## (ii) the resistance of the load. [2]
..................................................................
..................................................................
..................................................................
..................................................................
## (iii) the internal resistance r of the cell. [2]
..................................................................
..................................................................
..................................................................
..................................................................
(This question continues on the following page)
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(Question A2 continued)
(c) A second identical cell is connected into the circuit in (b) as shown below.
I = 0.15 A
The current in this circuit is 0.15 A. Deduce that the load is a non-ohmic device. [4]
.......................................................................
.......................................................................
.......................................................................
.......................................................................
.......................................................................
.......................................................................
2207-6514 turn over
073
4
from the proton.
..................................................................
– 22 – M08/4/PHYSI/HP2/ENG/TZ1/XX+
b3.. . .This
. . .question
. . . . . .is. in
. .two
. . .parts.
. . . .Part
. . . 1. is
. . . .aspects
. . . . .of. .electric
. . . . .fields
. . . .and
. . electric
. . . . . charge
. . . . . and
......
..................................................................
Part 1 Fields and electric charge associated with atoms
..................................................................
(a) A proton may be considered to be a point charge. For such a proton
## (i) sketch the electric field pattern. [2]
(ii) calculate the magnitude of the electric field strength at a distance of 5.0 ×10−11 m
[2]
## (This question continues on the following page)
(i)
Using your answer to (a)(ii) deduce that the magnitude of the electric force between
the electron and the proton is . *10?8 N.
2208-6508
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M08/4/PHYSI/HP2/ENG/TZ1/XX+
(ii)
(Question B3, part 1 continued)
Deduce that the kinetic energy of the electron is 2. *10 ?18 J.
(b) In a simple model of the hydrogen atom, an electron orbits the proton. Both electron and
proton are regarded as point charges. The orbital radius of the electron is 5.0 ×10−11 m .
..................................................................
[1]
..................................................................
[3]
..................................................................
..................................................................
## (c) The electron in (b) also has electrostatic potential energy.
. . . . . . . . . (i)
. . . .Define
. . . . electrostatic
. . . . . . . . .potential
. . . . . .at. a. point.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [2]
..................................................................
..................................................................
..................................................................
## (ii) Calculate the electrostatic potential energy of this electron. [2]
..................................................................
..................................................................
..................................................................
..................................................................
## (This question continues on the following page)
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– 24 – M08/4/PHYSI/HP2/ENG/TZ1/XX+
## (Question B3 part 1 continued)
(d) Using your answers in (b)(ii) and (c)(ii) determine the energy required, in electron volt,
to completely remove the electron from the influence of the proton. [2]
.......................................................................
.......................................................................
.......................................................................
.......................................................................
## (e) Define electromotive force (e.m.f.). [1]
.......................................................................
.......................................................................
## (f) A filament lamp is operating at normal brightness.
The potential difference across the lamp is 6.0 V. The current in the filament is 0.20 A.
For the filament of this lamp, calculate
## (i) the resistance. [1]
..................................................................
..................................................................
## (ii) the power dissipated. [1]
..................................................................
..................................................................
## (This question continues on the following page)
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## (Question B3 part 1 continued)
(g) The lamp in (f) is connected in the circuit below. The lamp is still operating at normal
brightness.
B
R
The battery B has an internal resistance of 5.0 and the resistance R of the resistor
15 .
## (i) Calculate the current in the resistor R. [1]
..................................................................
..................................................................
## (ii) Determine the e.m.f. of the battery. [4]
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
..................................................................
## (This question continues on the following page)
2208-6508 turn over
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3
– 11 – N05/4/PHYSI/HP2/ENG/TZ0/XX+
SECTION B
This section consists of four questions: B1, B2, B3 and B4. Answer two questions.
B1. This question is in two parts. Part 1 is about electrical circuits and Part 2 is about the physics
of cooling.
## Part 1 Electrical circuits
Andrew is set the task of measuring the current-voltage (I-V) characteristics of a filament
lamp. The following equipment and information are available.
Information
Battery e.m.f. = 3.0 V, negligible internal resistance
Filament lamp marked “3 V, 0.2 A”
## Voltmeter resistance = 30 kΩ, reads values between 0.0 and 3.0 V
Ammeter resistance = 0.1 Ω, reads values between 0.0 and 0.5 A
Potentiometer resistance = 100 Ω
## (i) its resistance. [1]
................................................................
.
................................................................
.
## (ii) its power dissipation. [1]
................................................................
.
................................................................
.
## (This question continues on the following page)
8805-6502 Turn over
11 3
3
– 12 – N05/4/PHYSI/HP2/ENG/TZ0/XX+
## Andrew sets up the following incorrect circuit.
(b) (i) Explain why the lamp will not light. [2]
................................................................
.
................................................................
.
................................................................
.
.................................................................
.................................................................
.................................................................
(c) On the circuit diagram below, add circuit symbols to show the correct position of the
ammeter and of the voltmeter in order to measure the I-V characteristics of the lamp. [2]
## (This question continues on the following page)
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## (Question B1, part 1 continued)
(d) On the axes below draw a sketch graph to show the I-V characteristics for this filament
lamp. [4]
I/A
0.3
0.2
0.1
0.0
0.0 1.0 2.0 3.0 4.0 V/V
(e) Explain the shape of the graph that you have drawn in (d). [2
]
......................................................................
......................................................................
......................................................................
......................................................................
## (This question continues on the following page)
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–8– N07/4/PHYSI/HP2/ENG/TZ0/XX
## a3. This question is about electrical circuits.
The graph below shows the I-V (current-voltage) characteristic of an electrical component T.
150
100
I / mA
50
0
0.0 2.0 4.0 6.0 8.0
V/V
(a) On the graph above, draw the I-V characteristic in the range V 0 to V 6.0V for a resistor
R having a constant resistance of 40 Ω. [1]
## (This question continues on the following page)
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–9– N07/4/PHYSI/HP2/ENG/TZ0/XX
(Question A3 continued)
(b) The component T and the resistor R are connected in parallel as shown below.
40 Ω
A B
## When a battery of constant e.m.f. E and negligible internal resistance is connected
between the terminals A and B, the current in the resistor R is 100 mA.
## (i) Calculate the e.m.f. E of the battery. [1]
..................................................................
..................................................................
## (ii) Use the graph to determine the current in T. [1]
..................................................................
## (iii) Calculate the power dissipation in T. [2]
..................................................................
..................................................................
..................................................................
## (This question continues on the following page)
8807-6502 093 3
turn over
– 10 – N07/4/PHYSI/HP2/ENG/TZ0/XX
(Question A3 continued)
## (c) In order to reduce the power dissipation in component T, a second resistor R of
resistance 40 Ω is connected in series with T. The circuit is shown below.
T R
40 Ω
40 Ω
A B
## (i) the current in resistor T. [2]
..................................................................
## (ii) the power dissipation in T. [2]
..................................................................
..................................................................
8807-6502
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– 28 – N08/4/PHYSI/HP2/ENG/TZ0/XX+
(Question B4 continued)
Part 2 Electricity
Static electricity
(a) A positively charged particle hangs from a vertical insulating string. The particle is
brought above an electrically neutral metallic sphere that rests on an insulating support.
insulating string
## positively charged particle
metallic sphere
insulating support
By considering the distribution of charge on the sphere, state and explain whether the
tension in the string holding the charged particle will be less than, equal to or greater
than the weight of the particle. (Assume that the mass of the string is negligible.) [3]
.......................................................................
.......................................................................
.......................................................................
.......................................................................
## (This question continues on the following page)
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## (b) Diagrams A, B and C show a sequence of events in which:
• A: the sphere in (a) is earthed (grounded),
• B: the earth is then removed while the charged particle remains in place,
• C: finally the charged particle is taken away.
A B C
(i) On each of the diagrams A, B and C draw the distribution of charge on the sphere. [3]
(ii) Explain why work has to be done on the charged particle to move it away
from [2]
the sphere.
..................................................................
..................................................................
## (This question continues on the following page)
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## (Question B4, part 2 continued)
Current electricity
## (c) Define electromotive force (e.m.f.). [1]
.......................................................................
.......................................................................
(d) In the circuit below the battery has an e.m.f. of 12 V and an internal resistance of 5.0 Ω.
e.m.f. 12 V
60 Ω 30 Ω
X
30 Ω 60 Ω
Y
Calculate the
## (i) total resistance of the circuit. [3]
..................................................................
..................................................................
..................................................................
..................................................................
## (ii) current in the internal resistance. [1]
..................................................................
..................................................................
## (This question continues on the following page)
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– 31 – N08/4/PHYSI/HP2/ENG/TZ0/XX+
## (iii) total power dissipated in the circuit. [2]
..................................................................
..................................................................
## (iv) potential difference between points X and Y. [3]
..................................................................
..................................................................
..................................................................
(e) A real (i.e. non-ideal) voltmeter is connected across points X and Y in the circuit in
(d). Explain why the reading of this voltmeter will not be the same as your answer to (d) [2]
(iv).
.......................................................................
.......................................................................
8808-6502
3131 | 6,671 | 34,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-22 | latest | en | 0.54915 |
https://www.polytechforum.com/mech/re-simplified-mixed-up-between-hydraulic-atmospheric-pres-7312-.htm | 1,685,718,678,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00492.warc.gz | 1,049,387,057 | 10,013 | Re-Simplified: Mixed up between Hydraulic & Atmospheric Pressure on Fluids?
• posted
There's 2 web links on the subject where as:
#1: One equation uses the density of the fluid (times gravity)
(
)
#2: and the other equation doesnt use the density
(
)
----------------------------------------------------------
So I do not understand why both equations are not the same to calculate the travel distance of the fulid since both are related to the total distance travel after opposite ends of a fluid have pressure applied on them.... Can anyone clarify what's my mistake?
In the Web links: the Height ( h ) of part #1 should equal d1 + d2 of part#2 shouldn't it ?
#1 says P1 = density * h * g + P2 (or F1/A1 = density * h * g + F2/A2)
but #2 says Work1 = Work2 thus F1*D1 = F2*D2 (no density of the fluid is used to determine the heights D1 and D2 (D1 + D2 = h)???
• posted
h is the difference in elevation from one point to the other. The work problem (#2) assumes the heights are the same or the difference is negligible (see P1 = P2).
The pressure equations can be derived/simplified from Bernouli's equation or the more complicated Energy equation.
• posted
I think one variable that may be a little too transparent in the calculations is the compressibility of air vs that of hydraulic fluid. Fluids in general are MUCH less compressible than is air, but hydraulic fluid in particular is less compressible than most liquids.
'Sporky'
" snipped-for-privacy@hotmail.com" wrote:
• posted
Sorry, but you need to revisit the definition of a fluid as defined in any fluid mechanics text. Any gas is a fluid. To state otherwise is just to confuse the issue.
-- Ed Ruf Lifetime AMA# 344007 ( snipped-for-privacy@EdwardG.Ruf.com)
• posted
• posted
Equation #1 already is Bernouilli since velocity = 0
thus P1 + pgh1 = P2 + pgh2 P1 = P2 + pg (h2-h1) P1 = P2 + pgh -> h = h2 - h1
in #2: D1 + D2 = h
as you said P1 = P2 (P1 = P2 + density(g)(h) thus P1 = P2 + 0 ) and both must be ten fold more than the weight of the fluid thus density * g * h is probably negilible?
in #2 P1 = P2 thus F1/A1 = F2/A2 and F1*D1 = F2*D2
thus F1/F2 = A1/A2 = D2/D1 but this seems different from Equation#1 where h = h2-h1...could it be the fluid weight is much bigger than the applied pressures which makes the difference where as in #2 it's vice versa ?
• posted
-- That IS a simplification.
-- More simplification.
-- Nope. h = 0 for #2 since P1 = P2.
As I already stated h is related to elevation in Bernoulli's eqn. d1 and d2 are the piston displacements based on a work equation. h and the ds are 2 unrelated things. That's the main thing you're missing.
That's all I have. If you can't understand this, too bad..
• posted
Im not 100% certain what you were getting at but anyhow I found the answer which is that they are both the very same...they simply omitted minor forces from the equation (meaning they considered the weight of the liquid as minimumal for the press equation).
Manometer model at the very bottom of: | 798 | 3,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-23 | latest | en | 0.914706 |
https://nl.mathworks.com/matlabcentral/cody/problems/94-target-sorting/solutions/1896503 | 1,597,411,367,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739328.66/warc/CC-MAIN-20200814130401-20200814160401-00324.warc.gz | 412,768,156 | 15,771 | Cody
# Problem 94. Target sorting
Solution 1896503
Submitted on 10 Aug 2019 by Philippe
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]; t = 0; b_correct = [4 3 2 1]; assert(isequal(targetSort(a,t),b_correct))
si = 4 3 2 1 ans = 4 3 2 1
2 Pass
a = -4:10; t = 3.6; b_correct = [-4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4]; assert(isequal(targetSort(a,t),b_correct))
si = 1 2 15 3 14 4 13 5 12 6 11 7 10 8 9 ans = -4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4
3 Pass
a = 12; t = pi; b_correct = 12; assert(isequal(targetSort(a,t),b_correct))
si = 1 ans = 12
4 Pass
a = -100:-95; t = 100; b_correct = [-100 -99 -98 -97 -96 -95]; assert(isequal(targetSort(a,t),b_correct))
si = 1 2 3 4 5 6 ans = -100 -99 -98 -97 -96 -95 | 369 | 847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-34 | latest | en | 0.545178 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A301842 Number of 2Xn 0..1 arrays with every element equal to 0, 1 or 2 horizontally or antidiagonally adjacent elements, with upper left element zero. 2
2, 8, 25, 81, 263, 855, 2778, 9027, 29333, 95316, 309725, 1006437, 3270370, 10626915, 34531665, 112209036, 364618033, 1184809305, 3849982618, 12510339087, 40651763813, 132096011916, 429239834325, 1394794836717, 4532320816850 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Row 2 of A301841. LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = 3*a(n-1) +a(n-2) -2*a(n-4) for n>6 EXAMPLE Some solutions for n=5 ..0..1..1..0..0. .0..0..1..0..0. .0..0..0..1..1. .0..1..0..1..1 ..0..0..1..0..0. .0..0..0..1..1. .1..0..1..0..0. .1..1..0..0..0 CROSSREFS Cf. A301841. Sequence in context: A037560 A138804 A212323 * A240478 A288539 A281338 Adjacent sequences: A301839 A301840 A301841 * A301843 A301844 A301845 KEYWORD nonn AUTHOR R. H. Hardin, Mar 27 2018 STATUS approved
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Last modified January 27 21:28 EST 2022. Contains 350654 sequences. (Running on oeis4.) | 538 | 1,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-05 | latest | en | 0.601892 |
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# RoyP
Member Since 10 Nov 2012
Offline Last Active Jan 23 2013 08:06 PM
### Need help getting a child object to orbit a parent object ...
01 January 2013 - 01:28 PM
I'm trying to figure out how to make a child object orbit around a parent object. In my situation, I'm trying to create a parent ship that rotates waypoints with it when it rotates. A smaller child ship will park itself in a waypoint to maintain it's position within the fleet.
Here are a couple of images showing what I'm going for:
The child ship must keep it's position relative to the parent ship so it maintains it's position within the fleet.
i've tried learning enough about matrices to do it, but I'm screwing something up. I've got my child ship orbiting the origin point of the screen instead of the parent ship.
Here's the code I'm using when I press a key:
```parentShipRotationRadians += MathHelper.ToRadians(15);
childShipPositionPoint = Vector2.Transform(childShipPositionPoint, childShipRotationMatrix);```
How do I rotate the child ship's center point around the parent ship's center point with matrices?
Please break it down for me and help me understand what you're doing. I've tried researching it and I just can't find the info.
Roy
### NWA Game Developer Meetup -- December 29 -- Fayetteville AR
26 December 2012 - 09:49 AM
Interested in making video games and having fun? Come join us at our next meetup on December 29th.
We're looking for people in Northwest Arkansas who are passionate about game development and want to learn more about it while working on fun game projects together.
We're not just for programmers. We have members interested in game design, game programming, game art, game music and sound effects, level design, writing story for game development, and more.
It doesn't matter if you're an enthusiastic beginner or a seasoned pro, come hang out with us. We'd love to meet you.
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### How do I rotate a group of child objects in relation to a parent object?
16 December 2012 - 04:49 PM
I'm working on a space shooter with a local team. We're building the game with C# and XNA 4.0. I would rate myself as an advanced beginning programmer.
I'm trying to rotate a fleet of smaller fighter ships in relation to a larger cruiser ship, like this:
I'm setting the position of the parent ship as an offset to the origin point of the map object. And then I'm setting the position of each child ship as an offset from the origin point of the parent ship.
How do I rotate the positions of each child ship in relation to the parent ship's rotation? I want the child ships to maintain their formation with the parent ship no matter what angle I rotate it to.
I'm not worried about smooth rotation or anything right now. Just trying to figure out how to transform the x and y positions to the new rotated x and y positions. I know how to set the rotation of each object.
### NWA Game Developer Meetup -- December 15 -- Fayetteville AR
09 December 2012 - 10:04 AM
Interested in making video games and having fun? Come join us at our next meetup on December 15th.
We're looking for people in Northwest Arkansas who are passionate about game development and want to learn more about it while working on fun game projects together.
We're not just for programmers. We have members interested in game design, game programming, game art, game music and sound effects, level design, writing story for game development, and more.
It doesn't matter if you're an enthusiastic beginner or a seasoned pro, come hang out with us. We'd love to meet you.
You can find all the details at our Meetup site.
### NWA Game Developer Meetup -- December 1 -- Fayetteville AR
25 November 2012 - 09:50 AM
Interested in making video games and having fun? Come join us at our next meetup on December 1st.
We're looking for people in Northwest Arkansas who are passionate about game development and want to learn more about it while working on fun game projects together.
Whether you're an enthusiastic beginner or a seasoned pro, come hang out with us. We'd love to meet you.
You can find all the details at our Meetup site.
PARTNERS | 954 | 4,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.932159 |
https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_8&diff=cur&oldid=53307 | 1,619,006,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00248.warc.gz | 213,825,667 | 12,916 | # Difference between revisions of "2012 AMC 12B Problems/Problem 8"
## Problem 8
A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$
## Solution
We can count the number of possible foods for each day and then multiply to enumerate the number of combinations.
On Friday, we have one possibility: cake.
On Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week.
On Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party.
On Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday.
Similarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday.
Likewise on Monday: three possibilities, the three things that were not eaten on Tuesday.
On Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday.
So the number of menus is $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 1 \cdot 3 = 729.$ The answer is $\boxed{A}$.
## Solution 2
We can perform casework as an understandable means of getting the answer. We can organize our counting based on the food that was served on Wednesday, because whether cake is or is not served on Wednesday seems to significantly affect the number of ways the chef can make said foods for that week.
Case 1: Cake is served on Wednesday. Here, we have three choices for food on Thursday and Saturday since cake must be served on Friday, and none of these choices are cake, which was served Wednesday. Likewise, we have three choices (pie, ice cream, and pudding) for the food served on Tuesday and thus three choices for those served on Monday and Sunday, with these three choices being whatever was not served on Tuesday and Monday, respectively. Hence, for this case, there are $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 243$ possibilities.
Case 2: Cake is not served on Wednesday. Obviously, this means that pie, ice cream, and pudding are our only choices for Wednesday's food. Since cake must be served on Friday, only ice cream, pudding, and cake can be served on Thursday. However, since one of those foods was chosen for Wednesday, we only have two possibilities for Thursday's food. Like our first case, we have three possibilities for the food served on Tuesday, Monday, and Sunday: whatever was not served on Wednesday, Tuesday, and Monday, respectively. $3 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 3 = 486$ possibilities thus exist for this case.
Adding the number of possibilities together yields that $243 + 486 = 729$ is the total number of menus, making our answer $\boxed{A}$.
## Solution 3
Note that the choice of a food item on a given day is symmetric, i.e. the number of ways to create the meal plan with a cake on Friday is the same as the number of ways to create the meal plan with pudding on Friday, and the same reasoning holds for the other desserts. Since every meal plan is counted by the summation of the $4$ aforementioned plans (note that Friday's dessert has to be one of the $4$ given desserts) and that these cases are mutually exclusive (i.e. you cannot make both a cake and pudding on Friday), each case results in a quarter of the total meal plans (Since there are $4$ desserts, we multiply by $\tfrac{1}{4}$). The total number of plans with no restrictions can be counted with constructive counting, as follows:
We note that there are $4$ ways to choose the first dessert. Then, each dessert thereafter must be distinct from the prior one. Since there are $4$ options, and $1$ of them has been taken by the prior, each following dessert can be chosen in $(4 - 1) = 3$ ways. Thus, since there are $6$ desserts other than the first, the total number (without restrictions) is $4 \cdot 3^6$
Thus, by our symmetry argument derived prior, we know that the number of desired meal plans is $\frac{\text{\# plans w/o restrictions}}{4} = \frac{4 \cdot 3^6}{4} = 3^6 = 729$, choice $\boxed{A}$
~ftwmaster65 | 1,113 | 4,608 | {"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": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-17 | latest | en | 0.97195 |
https://chestofbooks.com/crafts/metal/Metal-Pattern/Patterns-For-An-Offset-To-Join-A-Round-Pipe-With-One-Of-Elliptical-Profile-Part.html | 1,670,120,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00292.warc.gz | 189,503,513 | 7,149 | Therefore divide the curve C1 D1 into the same number of parts as the quarter circle B1 J1, and from the points thus obtained carry lines horizontally to the left, cutting C D. Upon C1 M extended, as C2 M1, set off spaces equal to those in C D, as shown, and through the points thus obtained draw lines to the left indefinitely. From the points in C1 D1 drop lines vertically, cutting those just drawn, all as shown. A line traced through the points of intersection, as shown from D2 to C2, will give the desired stretchout. In dividing the curve C D' into spaces it is advisable to make those nearest to D1 less than those near the top of the curve in order to compensate for the increase in the spaces in C2 M1 as they approach the bottom; thus obtaining a set of nearly equal spaces upon the final profile D2 C2, all of which will appear clear by an inspection of the drawing.
Fig. 705. - Hatf Pattern of Front Piece of Funnel Coat Hod.
As above stated, the diagrams of sections in Figs. 703 and 704 are constructed in the same manner as those of Figs. 701 and 702. The hights in K L and V W are taken from J1 B1 of Fig. 700 and are the same as those in P Q and S T of Figs. 701 and 702. The distances upon L N and W Y are those of the solid and dotted lines in J B C D of Fig. 700, and the hights of the perpendiculars near N and Y are equal to the lengths of the lines drawn from points of corresponding number in the profile D2 C2 of Fig. 700 to the lines C2 M1.
To develop the pattern of the front piece, first draw any line, as A G in Fig. 705, equal in length to A G of Fig. 700. From G as a center, with a radius equal to the dotted lines 9 8 of Fig. 702, describe a short arc (near 8), which intersect with another arc drawn from A as center, with a radius equal to 9 8 of the section A1 J1 B1 of Fig. 700, thus establishing the position of point 8 in the upper line of the pattern. From 8 of the pattern as center, with a radius equal to 8 8 of Fig. 701, describe a short arc (near 8'), which intersect with another arc drawn from G of the pattern as center, with a radius equal to 9 8 of the plan, Fig. 700, thus establishing the point 8' in the lower line of the pattern. Continue in this manner, using the lengths of the oblique dotted lines in Fig. 702 in connection with the spaces in the section A1 J1 B1 of Fig. 700 as radii to determine the points in the upper line of the pattern, or the side forming the mouth, and the lengths of the oblique solid lines of Fig. 701 in connection with the spaces in the plan of the bottom (G1 H1) as radii with which to determine the points in the lower line of the pattern or the side to fit against the bottom.
Having reached the points 5 and 5', next add to the pattern the flat triangular surface shown by J H D of the elevation. From H (5') of the pattern as center, with a radius equal to 5 5 of Fig. 706, the side of the last triangle in the pattern of the back piece, describe a short arc (near D), and intersect the same with another arc struck from J (5) of the pattern as center, with a radius equal to the oblique line 5 5 of Fig. 703, and draw H D and D J. Using D J of the pattern as one side of the next triangle, take as radii the distances 5 4 of Fig. 704 and 5 4 of the section D2 C2 of Fig. 700 to locate the position of point 4' of the pattern, as shown in Fig. 705. With 4 4 of Fig. 703, and 5 4 of the section B1 J1 of Fig. 700 as radii locate the point 4 of the pattern, as shown, and so continue until C D is reached. Lines traced through the points of intersection from B to A, C to D and H to G will complete the pattern of one-half the front piece.
The method of triangulating the piece forming the back of the coal hod and the development of the pattern of the same are so clearly shown in Figs. 706, 707 and 708, in addition to the plan and elevation, Fig. 700, as to need only a brief description. Divide H1 F1 and D2 E2 of the plan, Fig. 700, into the same number of equal parts, and from the points thus obtained erect lines vertically cutting the corresponding-lines H F and D E of the elevation, as shown by the dotted lines. Connect points of like number in that view by solid lines and points in DE with those of next lower number in H F by dotted lines. Since D E, being inclined, is longer than M2 E2, its equivalent in the plan, it will be necessary to develop an extended section upon the line D E of the elevation, as shown by D5 E3 of the plan, which may be done in the same manner as the section on the line C D above explained. Upon M2 E2 extended, as E2 E3, set off the spaces in D E, and through the points thus obtained draw lines at right angles, as shown, which intersect with lines drawn parallel with M2 E2 from points of corresponding number in D2 E2, thus establishing the curve D5 E3, from which a correct stretchout of the top of the back piece may be obtained.
Fig. 706. - Diagram of Sections on Solid Lines in D E F H of Fig. 700.
Fig. 707. - Diagram of Sections on Dotted Lines in D E F H of Fig. 700.
In Figs. 706 and 707, the bights of the various points upon the perpendiculars from X and Z are equal to the lengths of the straight lines drawn from points of corresponding number in H1 F1 of the plan, Fig. 700, to the line M2 F1. The distances set off to the right upon the horizontal lines from X and Z are equal to the lengths of the several solid and dotted lines in D E F H of the elevation, and the hights of the perpendiculars at the right ends of the bases are equal to the straight lines drawn from points in the section D5 E3 to the line E2 E2 The several oblique solid and dotted lines are, therefore, the true distances represented by the solid and dotted lines of corresponding number in the elevation.
Fig. 708. - Half Pattern of Back Piece of Funnel Goal Hod.
In Fig. 708, E F is equal to E F of Fig. 700 and is made the base of the first triangle, from which base the several triangles constituting the complete pattern may be developed in numerical order and in the usual manner from the dimensions obtained in Figs. 706 and 707 and in the plan and section in Fig. 700, all as clearly indicated.
The pattern for the piece forming the foot of the coal hod is a simple frustum of an elliptical cone, the method of obtaining which is fully explained in Problem 171. In Fig. 540 of that problem the lines E F and G H are drawn much further apart than the proportions of the foot in the present case would justify, but the operation of obtaining its pattern is exactly the same | 1,646 | 6,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.946899 |
https://nursingstudy.org/the-z-test-is-used-when-the-sample-size-is-greater-than-30-and-the-standard-deviation-is-known-the-t-test-is-used-when-the-sample-size-is-less-than-30-and-the-standard-deviation-is-not-known-math-me/ | 1,701,398,265,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100264.9/warc/CC-MAIN-20231201021234-20231201051234-00629.warc.gz | 485,066,031 | 27,733 | # The z-test is used when the sample size is greater than 30 and the standard deviation is known; the t-test is used when the sample size is less than 30 and the standard deviation is not known (Math Meeting, 2013). When the standard deviation is known we use that to calculate the probability.
Hello i need a Good and Positive Comment related with this argument .A paragraph with no more 90 words.
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Jessica Hanger
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Re:Topic 3 DQ 1
The z-test is used when the sample size is greater than 30 and the standard deviation is known; the t-test is used when the sample size is less than 30 and the standard deviation is not known (Math Meeting, 2013). When the standard deviation is known we use that to calculate the probability.
The t-test should be used when a sample is normally distributed and is used to compare the means of two different groups (Grove & Cipher, 2017).
If we have a sample size of 75 and we know the standard deviation, we would use the z-test.
If we have a sample size of 12 and we do not know the standard deviation, we would use the t-test.
If we have a sample size of 20 and we know the standard deviation, we would use the z-test.
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Cathy, CS | 400 | 1,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.926636 |
https://research.stlouisfed.org/fred2/series/DDEM01EGA156NWDB | 1,448,640,365,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398449258.99/warc/CC-MAIN-20151124205409-00178-ip-10-71-132-137.ec2.internal.warc.gz | 848,565,040 | 19,033 | # Stock Market Turnover Ratio (Value Traded/Capitalization) for Egypt
2012: 38.09256 Percent
Annual, Not Seasonally Adjusted, DDEM01EGA156NWDB, Updated: 2015-10-02 12:58 PM CDT
1yr | 5yr | 10yr | Max
Total value of shares traded during the period divided by the average market capitalization for the period.
Ratio of the value of total shares traded to average real market capitalization, the denominator is deflated using the following method: Tt/P_at/{(0.5)*[Mt/P_et + Mt-1/P_et-1] where T is total value traded, M is stock market capitalization, P_e is end-of period CPI. End-of period CPI (IFS line 64M..ZF or, if not available, 64Q..ZF) and annual CPI (IFS line 64..ZF) are from the IMF's International Financial Statistics. Standard & Poor's, Global Stock Markets Factbook and supplemental S&P data)
Source Code: GFDD.EM.01
Source: World Bank
Release: Global Financial Development
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(a) Stock Market Turnover Ratio (Value Traded/Capitalization) for Egypt, Percent, Not Seasonally Adjusted (DDEM01EGA156NWDB)
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.
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``` World Bank, Stock Market Turnover Ratio (Value Traded/Capitalization) for Egypt [DDEM01EGA156NWDB], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/DDEM01EGA156NWDB/, November 27, 2015. ```
Retrieving data.
Graph updated. | 589 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.806657 |
https://www.physicsforums.com/threads/conceptual-rc-circuit-questions.543792/ | 1,510,941,414,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803848.60/warc/CC-MAIN-20171117170336-20171117190336-00151.warc.gz | 845,886,279 | 15,484 | # Conceptual RC Circuit questions
1. Oct 24, 2011
### DrummingAtom
1. The problem statement, all variables and given/known data
After a long time after the switch is closed.
1) Rank the current of each component.
2) What is the voltage across each component?
2. Relevant equations
3. The attempt at a solution
1) The current across B is the same as the capacitor C because the capacitor is fully charged. There is no current across Vbattery and A because the capacitor now has positive charges on the top and negative charges on the bottom so there is no potential outside the parallel part (B and C).
2) 0 V for A and Vbattery. But B and C have the total voltage across each.
Is this right?
I'm thinking that the parallel part creates it's own self powered loop and blocks out the rest, the battery and A.
Thanks for any help.
#### Attached Files:
• ###### RC Circuit.jpg
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2. Oct 24, 2011
### Staff: Mentor
Current flows THROUGH. Voltage appears ACROSS.
After a long time the capacitor voltage settles to whatever the circuit around it dictates. In this case it won't be equal to the battery voltage because the lamps are creating a voltage divider. To see this, use the rule of thumb that says that at steady state the capacitor current will be zero and its voltage fixed, so for all intents and purposes you can simply ignore it (remove it from the circuit entirely). What does the remaining circuit look like?
3. Oct 24, 2011
### ehild
Sorry, this is totally wrong.
The capacitor is parallel connected to B. Why should the same current flow through both? What is the same across them?
The voltage of the battery is given, Vbat. Is there a closed path between the terminals through A and B? Can current flow from the positive pole of the battery to the negative one?
No. The voltage of the battery is constant, Vbat.
ehild | 430 | 1,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-47 | longest | en | 0.926807 |
https://functions.wolfram.com/GammaBetaErf/Gamma3/21/01/01/0001/ | 1,709,593,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00735.warc.gz | 268,878,368 | 7,523 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Gamma
http://functions.wolfram.com/06.07.21.0001.01
Input Form
Integrate[Gamma[a, Subscript[z, 1], z], z] == z Gamma[a, Subscript[z, 1], z] + Gamma[1 + a, z]
Standard Form
Cell[BoxData[RowBox[List[RowBox[List["\[Integral]", RowBox[List[RowBox[List["Gamma", "[", RowBox[List["a", ",", SubscriptBox["z", "1"], ",", "z"]], "]"]], RowBox[List["\[DifferentialD]", "z"]]]]]], "\[Equal]", " ", RowBox[List[RowBox[List["z", " ", RowBox[List["Gamma", "[", RowBox[List["a", ",", SubscriptBox["z", "1"], ",", "z"]], "]"]]]], "+", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["1", "+", "a"]], ",", "z"]], "]"]]]]]]]]
MathML Form
Γ ( a , z 1 , z ) z Γ ( a + 1 , z ) + z Γ ( a , z 1 , z ) z Gamma a Subscript z 1 z Gamma a 1 z z Gamma a Subscript z 1 z [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["\[Integral]", RowBox[List[RowBox[List["Gamma", "[", RowBox[List["a_", ",", SubscriptBox["z_", "1"], ",", "z_"]], "]"]], RowBox[List["\[DifferentialD]", "z_"]]]]]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List["z", " ", RowBox[List["Gamma", "[", RowBox[List["a", ",", SubscriptBox["zz", "1"], ",", "z"]], "]"]]]], "+", RowBox[List["Gamma", "[", RowBox[List[RowBox[List["1", "+", "a"]], ",", "z"]], "]"]]]]]]]]
Date Added to functions.wolfram.com (modification date)
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https://www.thestudentroom.co.uk/showthread.php?t=6818376 | 1,611,676,140,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704800238.80/warc/CC-MAIN-20210126135838-20210126165838-00459.warc.gz | 1,011,302,610 | 39,154 | # Physics Question Help
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#1
Hello, need some help with these Multiple choice questions.
0
#2
For Q7:
I used Wiens Displacement
so
Wavelength x Temp = const
Wavelength1 x 750 = wavelength2 X 960
so
Wavelength2 = 750W1 / 960 = 0.78W1 = Option A but apparently answer is Option B?
0
#3
For Q10:
Option A looks exactly like the Suns spectrum but has been slightly red-shifted but the answer is Option B?
0
#4
For Q14:
I used the equation:
PV = 1/3Nmc^2
We're told PV is constant after a 1/3 of the gas is removed. so I thought
1/3 * 1/3 Nmc^2 = 1/3 Nm 3x10^5
Cancelling down this gives me the answer
C^2 = 9x10^5 which is Option D but the answer is Option C?
0
#5
0
1 month ago
#6
(Original post by Afrrcyn)
For Q7:
I used Wiens Displacement
so
Wavelength x Temp = const
Wavelength1 x 750 = wavelength2 X 960
so
Wavelength2 = 750W1 / 960 = 0.78W1 = Option A but apparently answer is Option B?
For Wien's displacement law, the absolute temperature is used, so you need to convert to kelvin
Last edited by golgiapparatus31; 1 month ago
1
1 month ago
#7
7. B
10. A
14. D
0
1 month ago
#8
Ignore me, I'm just bumping
0
#9
(Original post by golgiapparatus31)
For Wien's displacement law, the absolute temperature is used, so you need to convert to kelvin
Note that pV is constant, but not T. so use pV = nRT to work out the new temperature, then use pV = 1/3 Nmc^2
Thanks! I'm very silly Physicist!
(Original post by ItsStarLordMan)
7. B
10. A
14. D
OHhhh, now I get it!
0
1 month ago
#10
(Original post by Afrrcyn)
OHhhh, now I get it!
Sarcasm, right?
0
#11
(Original post by ItsStarLordMan)
Sarcasm, right?
Maaaaaaybe.
0
1 month ago
#12
(Original post by Afrrcyn)
Maaaaaaybe.
Well, I think 10 is A. Could be wrong, but I'm pretty sure
0
#13
(Original post by ItsStarLordMan)
Well, I think 10 is A. Could be wrong, but I'm pretty sure
I mean that's what I put but apparently it's not! Hoping someone can explain it to me before Monday
0
1 month ago
#14
(Original post by Afrrcyn)
I mean that's what I put but apparently it's not! Hoping someone can explain it to me before Monday
No way is that wrong!
0
#15
(Original post by golgiapparatus31)
For Wien's displacement law, the absolute temperature is used, so you need to convert to kelvin
Note that pV is constant, but not T. so use pV = nRT to work out the new temperature, then use pV = 1/3 Nmc^2
Why would I use pV = 1/3 Nmc^2, if T isn't even in that equation, would I not use 1/2mc^2 = 3/2 kT?
(Original post by ItsStarLordMan)
No way is that wrong!
0
1 month ago
#16
(Original post by Afrrcyn)
I mean that's what I put but apparently it's not! Hoping someone can explain it to me before Monday
You have to convert temperature to kelvin.
(Original post by Afrrcyn)
Why would I use pV = 1/3 Nmc^2, if T isn't even in that equation, would I not use 1/2mc^2 = 3/2 kT?
Sorry
[Alternative method below]
Spoiler:
Show
You have
So, N is inversely proportional to T.
So the new temperature is 3/2 times the old one.
Then as you 1/2 mc^2 = 3/2 kT
So c^2 is multiplied by a factor of 3/2.
Last edited by golgiapparatus31; 1 month ago
0
1 month ago
#17
(Original post by Afrrcyn)
For Q14:
I used the equation:
PV = 1/3Nmc^2
We're told PV is constant after a 1/3 of the gas is removed. so I thought
1/3 * 1/3 Nmc^2 = 1/3 Nm 3x10^5
Cancelling down this gives me the answer
C^2 = 9x10^5 which is Option D but the answer is Option C?
1/3 of the gas is removed, so 2/3 N molecules are left. Using this, we get the answer
0
#18
(Original post by golgiapparatus31)
1/3 of the gas is removed, so 2/3 N molecules are left. Using this, we get the answer
Ohhh, yes, yes. I was using 1/3 and not 2/3! Any idea for Q10? Thanks for the help!
0
1 month ago
#19
(Original post by Afrrcyn)
Ohhh, yes, yes. I was using 1/3 and not 2/3! Any idea for Q10? Thanks for the help!
As you say, there is a red-shift of the wavelengths.
But the change for each one is not the same, so the first 2 lines cannot remain the same distance apart, so it can't be A. We find that it is B, because it can't be C or D
0
#20
(Original post by golgiapparatus31)
As you say, there is a red-shift of the wavelengths.
But the change for each one is not the same, so the first 2 lines cannot remain the same distance apart, so it can't be A. We find that it is B, because it can't be C or D
Wait. it can't be the same, it that because of dWavelength/wavelength = Velocty / c, so each wavelength would change by different amounts, bigger the wavelength, the bigger the change?
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# Questions tagged [collision-resistance]
Difficulty of finding two different inputs that hash to the same value
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### SNARK-friendly one-way compression functions
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601 views
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### Why do m and m' both hash to H2?
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331 views
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2k views
### MD5 - Chosen Prefix Collision Attack
Given 2 messages $m_1$ and $m_2$, in a chosen prefix attack, we want to find $s_1$ and $s_2$ such that: $\text{MD5}(m_1 || s_1) = \text{MD5}(m_2 || s_2)$ Are $s_1$ and $s_2$ found by brute force ...
295 views
### Is there an $n$-bit hash function such that $n$ is equal to the length of input and the collision resistance is close/equal to $2^{n/2}$?
Let $B$ denote a non-empty bitstring of arbitrary length. Then let $n$ denote the length of $B$. The question: does there exist a cryptographic hash function $H$ that outputs a hash $h$ of $B$ such ...
779 views
Using a compression function $f : A × A → A$. A basic version given by: $W_0 = IV$ $W_1 = f(W_0, m_1)$ $W_2 = f(W_1, m_2)$ ... $W_n = f(W_{n-1}, m_n)$ $W_n$ is the output of the hash function, $... 5answers 3k views ### How many hex digits do I need to compare when manually checking hash functions? I sometimes run sha256sum on large files after transferring them from one place to another, and will just skim the hash output to verify it's correct. But, I usually just look at the first/last 5 or 6 ... 2answers 177 views ### Feasible to find two sets with the same RSA accumulator value if a prime representative hash function is used? Consider the RSA accumulator scheme with random oracle as proposed by Barić and Pfitzmann, with the random oracle replaced by a secure hash function$H$as follows. For a set$S$, the accumulator ... 1answer 151 views ### repeatedly hashing result and using reverse sequence as RNG secure? 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Does this fact make certificates based RSA-SHA1 Signature risky for creating fraud certificates? If RSA-SHA1 ... 1answer 97 views ### The Collision Differential for MD4 - a question on notation (Wang, et al) In the paper "Cryptanalysis of the Hash Functions MD4 and RIPEMD" the authors introduce the following notation (paragraph 4.1):$\Delta$$H_0 = 0 \xrightarrow{(M_,M')} \Delta$$H$= 0 What exactly ... 2answers 546 views ### MD5 hash functions to prove integrity of a download It doesn't take a lot of time on the crypto StackExchange to realise the sentiment around MD5 hash functions is poor. My questions are: If it's such a bad idea to make use of MD5 hash functions why ... 4answers 714 views ### Find some hash function with deliberate collisions I guess this should be well known (or trivially desperate), but I couldn't find any reference. I have a small number (say$k=15$) of messages$m_i$,$i=1\cdots k$(they are fixed length and short, ... 2answers 627 views ### Finding k collisions on hash function Let$n$be the size of the image-space of a hash function$H$. It is known that you can find a collision on$H$in$O(\sqrt{n})$time (by birthday paradox). How can I show that, in order to find$k$... 2answers 434 views ### Is it possible to generate an infinite number of collisions given an infinite amount of strings? Let us assume that we have a string$w$such that it belongs the set$C$, where every member of$C$is not null, and contains every string of characters that can be permuted (numbers, letters etc.). ... 1answer 236 views ### What are the implications of quantum collision search algorithms like Ambainis for collision-resistant hash functions? I just stumbled onto a series of 2017 papers about applying the Ambainis quantum collision-finding algorithm to hash functions. (disclaimer: I haven't read all of them in full yet): Post-quantum ... 2answers 3k views ### How many trials does it take to break HMAC-MD5? I know that you can find collision in MD5 with 2^64 trials using Birthday paradox. Now everyone is saying that HMAC-MD5 is significantly more secure. How can I ... 2answers 72 views ### Technical term for hash value being uncorrelated to input message? Is there a technical term for the desirable property of hash functions. That a small change to the input message will produce a massive unpredictable and seemingly uncorrelated change to the output ... 2answers 568 views ### collisions on hash functions Why do we say that collision-resistance is a "harder" property than second pre-image for hash functions whereas if you have an attack on the second pre-image then you find a collision ? Moreover, a ... 1answer 1k views ### Security of XORing hashes vs concatenating On this question https://stackoverflow.com/questions/5889238/why-is-xor-the-default-way-to-combine-hashes a couple of the answers say that xoring is a bad/insecure choice for combining hashes. However,... 2answers 2k views ### (strong) collision resistance Example I'm studying Cryptography & Network Security. It has the following example for collision resistance: "Suppose Bob writes an IOU message, sends it to Alice, and she signs it. Bob finds two ... 1answer 123 views ### For a random permutation$P$, what's the probability of the following event? For a random permutation$P$and$q$distinct inputs$x_1,\ldots,x_q\in\{0,1\}^n$, what's the probability of the event that there exists at least one collision among$\{P(x_1)\oplus x_1,\ldots,P(x_q)\...
1k views
I know this is usually frowned upon, but supposing you were to roll your own encryption cipher (completely on your own or with a small group of friends, with no peer review possible), what could you ...
49 views
### Hash Function SHA-3 [duplicate]
I would like to discuss the minimum input of hash function to consider strong (e.g Shack128 (SHA-3)). I have a mechanism based on SHA-3, but the input is just 40-bit, some friends told me 40-bit input ... | 2,547 | 10,375 | {"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} | 2.90625 | 3 | CC-MAIN-2021-25 | latest | en | 0.878966 |
https://www.ukessays.com/essays/accounting/computing-cash-flow-from-investment-and-net-present-value-accounting-essay.php | 1,493,047,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119637.34/warc/CC-MAIN-20170423031159-00046-ip-10-145-167-34.ec2.internal.warc.gz | 976,061,188 | 11,410 | # Computing Cash Flow From Investment And Net Present Value Accounting Essay
Published: Last Edited:
This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.
The expected period of time during which an asset is useful to the average owner. The economic life of an asset could be different than the actual physical life of the asset. Estimating the economic life of an asset is important for businesses so that they can determine when it is worthwhile to invest in new equipment. In addition, businesses must plan so that they have sufficient funds to purchase replacements for expensive equipment once it has exceeded its useful life.
## Residential Real Estate: The most well known form of cash flow investing is purchasing residential real estate to use as rental property.  In this investment, you put up a lump sum of cash to purchase the property in order to receive the monthly income that the rent of the property produces.
Returns:Â 8-17%
Time Horizon:Â Varies.
DEPRECIATION : In a broader economic sense, the depreciated cost for industry is the aggregate amount of capital that is "used up" in a given period, such as a fiscal year. This value can be examined for trends in capital spending and accounting aggressiveness.Â
The value of an asset net of all accumulated depreciation that has been recorded against it.
Depreciated Cost = Purchase Price (or cost basis) - {Cumulative Depreciation}
Depreciated cost is also known as the "net book value" or "adjusted cost basis".
STRAIGHT LINE METHOD OF DEPRECIATION: Straight-line method of depreciation is the most popular and simple method of depreciation. In this method, the purchase price or the acquisition value of the asset is divided by the useful life of the asset after deducting the scrap value from the value of an asset. Scrap value of the assets is the value of the asset at which it can be sold after its useful life is over. In order to understand the straight-line method of depreciation better.
NET PRESENT VALUE: The Net Present Value (NPV) of a Capital Budgeting project indicates the expected impact of the project on the value of the firm. Projects with a positive NPV are expected to increase the value of the firm. Thus, the NPV decision rule specifies that all independent projects with a positive NPV should be accepted. When choosing among mutually exclusive projects, the project with the largest (positive) NPV should be selected. The NPV is calculated as the present value of the project's cash inflows minus the present value of the project's cash outflows. This relationship is expressed by the following formula:
## where,
CFt = the cash flow at time t and
r = the cost of capital.
## CALUCATION OF CASH FLOWS
Particulars
Amount
Profit before tax and depreciation(PBT)
--- Depreciation(w.note)
2,00,000
---1,53,343
Profit before tax
--- [email protected]%
46,657
--- 23,329
Profit after tax (PAT)
23,328
## Cash inflow (each year) = Profit after tax (PAT) + depreciation
= 23,328 + 1,53,343
## = Rs.1,76,671
Terminal inflow in the 7th year = Rs.30,000 (scrap value)
WORKING NOTES-:
## Economic life
= 11,00,000 + 3,400 - 30,000
7yrs
## Total PVF (Rs.)
1
1,76,671
0.909
1,60,594
2
1,76,671
0.826
1,45,931
3
1,76,671
0.751
1,32,680
4
1,76,671
0.683
1,20,667
5
1,76,671
0.621
1,09,713
6
1,76,671
0.564
99,643
7
1,76,671+30,000 =2,06,671
0.513
1,06,022
## Less: cash outflow (at time 0)-11,03,340
(Plant cost +installation charges) | 920 | 3,561 | {"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} | 2.5625 | 3 | CC-MAIN-2017-17 | latest | en | 0.947908 |
http://slideplayer.com/slide/3889858/ | 1,519,248,830,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813803.28/warc/CC-MAIN-20180221202619-20180221222619-00691.warc.gz | 326,890,607 | 18,965 | # Shapes and Designs Inv. 1.3: Tiling Activity Sheet.
## Presentation on theme: "Shapes and Designs Inv. 1.3: Tiling Activity Sheet."— Presentation transcript:
Shapes and Designs Inv. 1.3: Tiling Activity Sheet
REVIEW OF DIRECTIONS: 1.) Go to my Edline math page, click “CMP Web Code” Go to my Edline math page, click “CMP Web Code” (-or- Log on to: www.PHSchool.com)www.PHSchool.com 2.) There is an “Enter Code” box in upper left hand of screen. 3.) Enter “amd” in 1st box and “3103” in 2nd box and hit Enter. 4.) Click on “Shapes and Designs” 5.) Click on “Tessellations” 6.) Click on “instructions” and read how to move shapes around. Use the computer to work through Problem 1.3 (page 15 in your book. See also page 9 and 10.)
NOTE: If you have NOT gone through problem 1.3 on your own, using your own worksheet, you must do this before going any further in this PowerPoint! Click here to go to my Edline page. Go to my Edline page for a copy of the worksheet. Thank you!
A. 1. The only possible shapes and tiling patterns. Remember, you weren’t supposed to use B in your sketch here!! A D B
A. 2. Regular Polygons that tile: Shape A, Shape D, and Shape B. A BD Regular Triangle or Equilateral Triangle Square Regular Hexagon (Page 9 and 10 show all of the names, shapes, and their letters.)
B. Possible Sketches using a combination of two or more different shapes. A A D B B F
A A D B B F
C. 1. Draw in the point where the vertices of the polygons meet. Here or Here
C. 2. The following polygons fit around these points. Two Hexagons and Two Triangles. Pattern: Hexagon, Triangle, Hexagon, and Triangle. (H,T,H,T) Two Hexagons and Two Triangles. Pattern: Hexagon, Triangle, Triangle, and Hexagon. (H,T,T,H)
C. 3. No, not always. As you can see, this same tiling has two different pattern orders, depending on where you choose the point. (H,T,H,T) (H,T,T,H)
Well Done! Please see me if you have any questions. | 542 | 1,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.836078 |
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8.3 The Routing Relation 163
Therefore these algorithms are referred to as minimal. As we have already seen, it’s
often important to include non-minimal routes and in this case, routing functions
choose paths from the set of all minimal and non-minimal routes R
xy
. These algo-
rithms are refered to as non-minimal. Again, from our simple example on the ring,
the greedy algorithm is minimal, while randomized and adaptive algorithms are non-
minimal.
8.3 The Routing Relation
It is useful to represent the routing algorithm as a routing relation R and a selection
function ρ. R returns a set of paths (or channels for an incremental routing algorithm),
and ρ chooses between these paths (or channels) to select the route to be taken. With
this division of the algorithm, issues relating to channel dependencies and deadlock
deal with the relation R while issues relating to adaptivity deal with the selection
function ρ. We address deadlock in detail as part of Chapter 14.
Depending on whether our algorithm is incremental, and whether it is node-
based or channel-based, we define R in three different ways:
R : N × N P(P ) (8.1)
R : N × N P(C) (8.2)
R : C × N P(C) (8.3)
where P(X) denotes the power set, or the set of all subsets, of the set X. This notation
allows us to reflect the fact that a routing relation may return multiple paths or
channels, one of which is chosen by the selection function.
When the output of the routing relation is an entire path, as in Relation 8.1 —
the first of our three routing relations, the routing algorithm is referred to as all-
at-once. This name reflects exactly how the routing algorithm is used. When a packet
is injected into the network at the source node x destined to node y, the routing
relation is evaluated: U = R(x,y). Since U may be a set of routes, one is selected and
assigned to the packet. Of course, U does not have to include all possible routes R
xy
or even all minimal routes R
xy
, and in the case of a deterministic routing algorithm,
it returns only one (|U |=1). Once the route is chosen, it is stored along with the
packet. As we will see in Chapter 11, all-at-once routing minimizes the time spent
evaluating the routing relation for each packet, but this advantage comes with the
overhead of carrying the routes inside the packets.
An alternate approach is incremental routing, where the relation returns a set of
possible channels. Instead of returning an entire path at once, the routing relation is
evaluated once per hop of the packet. The output of the relation is used to select
the next channel the packet follows. In Relation 8.2, the second form of the routing
relation, for example, the inputs to the relation are the packet’s current node w and
its destination y. Evaluating the relation gives us a set of channels D = R(w, y),
where each element of D is an outgoing channel from w or D C
Ow
. The selec-
tion function is then used to choose the next channel used by the packet from D.
This incremental process is repeated until the packet reaches its final destination.
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required | 793 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-22 | longest | en | 0.928276 |
https://direct.physicsclassroom.com/Lesson-Plans/Algebra-Based-Physics/Electric-Circuits/Objectives | 1,718,344,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00625.warc.gz | 189,958,521 | 30,386 | ## Learning Outcomes for Electric Circuits
The list below displays sample learning objectives for the unit on Static Electricity. The various activities from the Lesson Plans have been organized by objective. This format of organizing The Physics Classroom's activities may be more useful to some teachers than the Lesson Plan format. It is worth noting that the activities identified below only include those activities included on the Lesson Plans and in the Pre-Built Task Tracker courses. Additional activities are referenced on the Teacher Notes page.
### 1. Electric Potential and Electric Potential Difference
Relate electric potential to the energy per charge and the electric potential difference between two locations to the change in energy per charge moved between the two locations; identify the units of electric potential and distinguish between high and low potential locations.
Video:
Think Sheets:
Minds On Physics:
Mission EC2 on Voltage
Circuits 1: Circuit Quantities
Circuits 2: Circuit Quantities
### 2. A Model of Charge Flow in a Circuit
Identify the two requirements of an electric circuit; describe charge as marching together through the circuit loop at the same rate at every location, never accumulating nor being used up; describe the conducting components (wires, bulbs, batteries, etc.) of the circuit as being the source of charge and describe the battery as being the energy source that is used to pump the charge uphill from low energy to high energy to establish an electric potential difference across the two ends of the external circuit..
Video:
Labs:
Lab 1: Sparky the Electrician
Lab 2: First to Light
Lab 3: Greatest Current
Think Sheets:
Concept Builders:
### 3. Current
Define current and identify its unit; mathematically relate current to the quantity of charge and the time; identify the direction of conventional current and compare the current near the + terminal of the battery to the current at all other locations within the circuit.
Video:
Think Sheets:
Concept Builders:
Minds On Physics:
Mission EC1 on Electric Current
Circuits 1: Circuit Quantities
Circuits 2: Circuit Quantities
### 4. Resistance
Define resistance and identify the unit of resistance; describe the dependence of wire resistance upon wire length and cross-sectional area; explain the effect of increased resistance upon the current in a circuit.
Video:
Think Sheets:
Concept Builders:
### 5. Voltage-Current-Resistance Relationship
Relate the current in a simple circuit to the resistance of the circuit and the voltage (electric potential difference accross) of the battery.
Video:
Labs:
Lab 4 - Voltage-Current-Resistance
Think Sheets:
Concept Builders:
Minds On Physics:
Mission EC4 on Resistance, Voltage, and Current
Circuits 1: Circuit Quantities
Circuits 2: Circuit Quantities
### 6. Electric Power
Define power and identify the unit of power; relate the power to the energy change and the time and to other circuit quantities like voltage, current, and resistance.
Video:
Think Sheets:
Concept Builders:
Circuits 1: Circuit Quantities
Circuits 2: Circuit Quantities
### 7. Comparing Series and Parallel Circuits
Describe the fundamental differences between series and parallel circuits in terms of resistor arrangements and charge pathways; describe the difference in terms of the effect that adding resistors has upon the overall resistance and current.
Video:
Labs:
Lab 5: Series vs. Parallel
Physics Interactives with Concept Checkers:
Concept Builders:
### 8. Series Circuits - Concepts and Mathematics
Discuss the meaning of equivalent resistance (Req) and calculate the Req value for a series circuit. compare the current and the electric potential at the various locations and relate the battery voltage to the individual voltage drops of each resistor; use ∆V=I•R to conduct a complete analysis of a series circuit.
Video:
Labs:
Lab 6 - Comparing Voltage Drops and Currents in Series
Think Sheets:
Physics Interactives with Concept Checkers:
Concept Builders:
Minds On Physics:
Mission EC7 on Series Circuits Concepts
Mission EC9 on Series Circuit Calculations
Mission EC12 on Series and Parallel Circuit Calculations
Circuits 3: Series Circuits
Circuits 5: Series and Parallel Circuits
### 9. Parallel Circuits - Concepts and Mathematics
Discuss the meaning of equivalent resistance (Req) and calculate the Req value for a parallel circuit; compare the current and the electric potential at the various locations and relate the battery voltage to the individual voltage drops of each resistor; use ∆V=I•R to conduct a complete analysis of a parallel circuit.
Video:
Labs:
Lab 7 - Comparing Voltage Drops and Currents in Parallel
Think Sheets:
Physics Interactives with Concept Checkers:
Concept Builders:
Minds On Physics:
Mission EC8 on Parallel Circuits Concepts
Mission EC10 on Parallel Circuit Calculations
Mission EC12 on Series and Parallel Circuit Calculations
Circuits 4: Parallel Circuits
Circuits 5: Series and Parallel Circuits
### Also Available ...
Physics teachers may find the following for-sale tools to be useful supplements to our Lesson Plan and Pacing Guide section:
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A subscription allows teachers to set up classes, add students, customize online assignments, view student progress/scores, and export student scores. Task Tracker accounts allow your students to begin assignments in class or at school and to finish them at home. View our Seat and Cost Calculator for pricing details.
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We publish a free curriculum with >200 ready-to-use Think Sheets for developing physics concepts. The Solutions Guide is a download containing the source documents, PDFs of source documents, and answers/solutions in MS Word and PDF format. An expanded license agreement is included with the purchase. (Cost: \$25 download)
3. Teacher Presentation Pack
This is a large collection of downloadable content packed with nearly 190 Microsoft PowerPoint slide decks, the corresponding Lesson Notes (as PDF and fully-modifiable MS Word format), about 170 animations (in .gif, .png, and .mp4 file formats), a countless number of ready-to-use images (including the original source documents that would allow for easy modification of those images), and a license that allows teachers to modify and use all the content with their classes on password-protected sites (such as course management systems). (Cost: \$40 download)
4. Question Bank
We distribute a Question Bank that includes more than 9300 questions neatly organized according to topic. The Question Bank is the perfect tool for busy teachers or new teachers. Even if you don't use the website with your classes, the Question Bank will assist you in quickly putting together quizzes, tests and other documents with high-quality questions that target student's conceptions of physics principles. And if you do use The Physics Classroom website, the Question Bank is the perfect complement to the materials found at the website. (Cost: \$25 download) | 1,451 | 7,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-26 | latest | en | 0.91221 |
https://rsmanuals.com/34298/tisch-environmental-pm10/page-38/ | 1,619,014,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039544239.84/warc/CC-MAIN-20210421130234-20210421160234-00103.warc.gz | 572,782,988 | 70,660 | 38
The equation for the coefficient of correlation (r) is as follows:
16. r =
( x)( y)
xy - n
x-
x
n y-
y
n
22
2 2
where: n = number of observations
= sum of.
Before these can be determined, some preliminary algebra is necessary. x, y, x2, xy,
(x)2,
_ _
(y)2, n, y, and x need to be determined.
17. x = 1.475 + 1.167 + 1.115 + 1.079 + 1.060 = 5.896
18. y = 35.00 + 29.37 + 28.75 + 28.12 + 27.50 = 148.74
19. x2 = (1.475)2 + (1.167)2 + (1.115)2 + (1.079)2 + (1.060)2 = 7.069
20. y2 = (35.00)2 + (29.37)2 + (28.75)2 + (28.12)2 + (27.50)2 = 4461.1438
21. xy = (1.475)(35.00) + (1.167)(29.37) + (1.115)(28.75) + (1.079)(28.12) +
(1.060)(27.50) = 177.448
22. n = 5
_
23. x = x/n = 1.1792
_
24. y = y/n = 29.748
25. (x)2 = (5.896)2 =34.763
26. (y)2 = (149.74)2 = 22,123.587
Inserting the numbers:
(5.896)(148.74)
177.448 - 5 .
27. slope = 34.763
7.069 - 5
(876.971)
177.448 - 5 .
28. slope = 34.763
7.069 - 5 | 470 | 910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.489803 |
http://mathhelpforum.com/algebra/84158-find-equation-print.html | 1,527,296,537,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00426.warc.gz | 187,023,270 | 3,800 | # Find the equation
• Apr 17th 2009, 01:04 AM
Find the equation
...of a straight line that passes through the points (1,5) and (7, -1).
Here's my answer, is this correct? or is their an easier method?
Y1 - Y2 / X1 - X2
5 - (-1) / 1 - 7
6 / -6
Knowing the formula y=mx+b and m slope to be = -1 , we can plug back in:
5 = (-1) 1 + b
5 = -1 + b
5 - (-1) = b
6 = b
y = (-1)x + b
• Apr 17th 2009, 01:10 AM
craig
Nearly there, your value for m is correct, however you then need to find an actual value for b.
If we look at the first point, when x =5, y=1, putting this into our equation for the line we get the following:
\$\displaystyle 1 = -5 + b\$
Solving this will give you the value for b
• Apr 17th 2009, 01:11 AM
craig
Just a quick note, it does not matter which coordinate you choose to use, (1,5) or (7, -1), they will both give you the same answer for b
Hope this helps
Craig
• Apr 17th 2009, 01:17 AM
b = 6
have i explained it clearly?
• Apr 17th 2009, 01:38 AM
y = (-1)x + 6
?
• Apr 17th 2009, 01:43 AM
craig
(Thinking) I am sorry you are quite right, did not see that last bit of your question.
Yes you are correct, and that is the best way in my opinion.
• Apr 17th 2009, 07:32 AM
Ok great,
so the original post all makes sense?
• Apr 17th 2009, 07:38 AM
craig
Yep all makes sense :)
• Apr 17th 2009, 07:41 AM
Quote:
Originally Posted by ADY
...of a straight line that passes through the points (1,5) and (7, -1).
Here's my answer, is this correct? or is their an easier method?
Y1 - Y2 / X1 - X2
5 - (-1) / 1 - 7
6 / -6
=(-1)
Knowing the formula y=mx+b and m slope to be = -1 , we can plug back in:
y=(-1)x+b
knowing that (1,5) is a point on the line. we put this in the equation to determine "b"
5 = (-1) 1 + b
5 = -1 + b
5 - (-1) = b
6 = b
Hence the answer that we get is
y = -x + 6
Correct!!(Handshake)
Everything in your answer is correct. But as you can see by Craig's replies that your presentation needs improvement...I am sure you will try to explain it a bit better next time. A bit of addition in red
• Apr 17th 2009, 08:56 AM
"Hence the answer that we get is
y = -x + 6"
shouldnt it be this?
y = (-1)x + b
• Apr 17th 2009, 08:59 AM
craig
No.
(-1)x is the same as -x, mine and Ardash's version is easier to read and help avoid possible confusion.
Also you wanted the equation of the line, this is the equation in it's numerical form, not with a 'b' still in it.
• Apr 17th 2009, 09:03 AM
Quote:
Originally Posted by ADY
"Hence the answer that we get is
y = -x + 6"
shouldnt it be this?
y = (-1)x + b
..... than tell me why have you found the value of b as 6 ?
(Wait)
The steps that you followed were
1-Assume the equation as
y= mx+b
where m and b are initially unknown constants
2-Find m
3-Find b
4-Put the values that you found for them
------
EDIT:
Craig: My name is Adarsh (Tongueout) | 965 | 2,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-22 | latest | en | 0.923262 |
https://countingaccounting.com/2021/06/what-causes-production-inefficiency-chinas-inefficiency-problem.html | 1,701,624,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00264.warc.gz | 225,332,468 | 62,614 | What causes production inefficiency? China’s inefficiency problem.
Inefficient production is a production where all the resources are not used efficiently. So let’s say we’re looking at an economy that can produce two goods it can produce food and it can produce steel. If it’s strictly focused on producing food, it would produce a hundred million tons of food whereas if it’s strictly focused on producing steel, it would have 50 million tons of steel and we know that all the different points that are along this production possibilities frontier are efficient in production or that we’ve achieved production efficiency which means that we couldn’t produce an additional unit of one good without decreasing production of the other good.
So for example let’s say that we were at this point which is 90 million tons of food and then that meant 40 million tons of steel. At this point, we got 90 million and 40 million. So what for us to go to 91 million tonnes of food we would have to give up some steel. We couldn’t reach this point (x), we can’t reach it as it’s not feasible.
But you might be wondering what about all the points in the interior. All these points outside the PPF we know are not feasible, they’re not feasible it means we can’t reach them with the current level of resources that we have. What about all the points on the interior? Aren’t these feasible? The answer is yes they are feasible. We could reach any of these points but we don’t want to do that. We want to be on this PPF line because that’s efficient in production.
You might be thinking okay well why do we even care? Well, let’s take this point which corresponds to the 40 million tons of steel and then it corresponds to let’s say 60 million tons of food. You’d say well what’s wrong with this point (60m, 40m)? Well, the point we could reach without giving up any steal is (90m, 40m) because this is ninety million tons of food in 40 million tons of steel so why would we take 60 million tons of food and 40 million tons of steel? Without giving up any steel we could get an additional 30 million tons of food. So obviously this point along the curve (90m, 40m) is better, that’s why all these points along this curve are efficient in production.
Why inefficient production happens?
So you might be wondering then why would we ever end up at this point? How will we ever end up on the interior of the PPF? Because it’s certainly possible and there’s a couple of reasons that an economy could end up with inefficient production and one of the reasons is that resources are being misallocated. All these points in the PPF we’re reaching with the resources that we have in the economy. We’re talking about resources our people their machines and so forth. They’re all the things that we use to produce steel and food. So we have some level of resources and we’re bounded by this PPF curve of how much food and steel we can produce given the current level of resources, if resources are being misallocated somehow if they’re not being used wisely then that means that we could actually not be producing at a point that’s efficient in production. We’ll be at one of these points on the interior which is inefficient.
I want to give you an example of inefficient production and this misallocation you might be wondering. So China in the late 1960s embarked on a program called the Great Leap Forward. One of the things with the Great Leap Forward was that the government wanted to increase production of steel and they set quotas and so a lot of people who were farmers and stuff they had quotas that they needed to meet of the amount of Steel that they could produce.
They needed to get steel and so what they would do in some cases is actually melting down farm equipment so that they could get scrap metal. That then they would melt and say hey we’ve got some steel here. So we got the ability to make steel, basically pig iron. You can look that up if you’re curious about it. They were making pig iron from melting farm equipment, which is a lot of cases the pig iron wasn’t even usable. It was not usable there were problems with the pig iron and so forth and so you could think about this as a misallocation of resources. Because millions of people ended up starving because there wasn’t enough food.
So you can think of it as we’re at a point like (60m, 40m) where it’s inefficient. If they instead of melting down the farm equipment they had instead used it to produce more food maybe they would have been able to have 90 million tons of food instead of 60 million as our example. I don’t know how accurate these numbers are I just picked them out of a hat.
So the idea is that we’re not using the resources in the right way. Each resource we want to use in the best way possible. If we have a bunch of farm equipment the farm equipment is best suited to be producing food, it’s not really well suited for us to melt it down, make pig iron. So that we can try and increase our steel production. That’s a misallocation of resources and it can lead to where we’re actually at an inefficient point of production.
Another thing would be if resources are simply being wasted. When we think about resources being wasted, we come back to what our resources? Well, we’ve got people we’ve got machines we’ve got buildings.
Let’s say, for example, in the United States you have a lot of vacant buildings particularly in urban areas and also metropolitan areas you will have a lot of vacant buildings or vacant lands. Land that’s not being used for agricultural purposes. There’s no food being grown there it’s not being used for commercial purposes they’re just building sitting there due to deindustrialization and so these are resources they’re being wasted.
You could even think of people being wasted, right? If there’s really high unemployment there’s a lot of people who could be doing something but for whatever reason, the economy is not using that, so you end up at some point inside the PPF. Where theoretically based on the resource that you have you actually could get to a point on that PPF without any Opportunity Cost you could actually get more food or vice versa. Because resources are being wasted or they’re not being put to their best use in terms of the misc allocation you can end up with one of the inefficient levels of production. | 1,345 | 6,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.94116 |
http://www.bentongue.com/about-maximizing-the-q-of-solenoid-inductors-that-use-ferrite-rod-cores-including-charts-of-magnetic-flux-density-and-flux-lines-with-some-actual-q-and-inductance-measurements-from-simulations-in-fe/ | 1,580,036,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251688806.91/warc/CC-MAIN-20200126104828-20200126134828-00248.warc.gz | 189,710,848 | 20,244 | # About Maximizing the Q of solenoid inductors that use ferrite rod cores, including charts of magnetic flux density and flux lines, with some actual Q and inductance measurements from simulations in FEMM
## Summary
Many factors interact to affect the Q of ferrite rod cored inductors. Part one of this Article identifies and comments upon some of them. A simplified model and an equivalent circuit is also discussed. The second part describes several ferrite cored inductors, along with measurements of inductance and Q. The third part displays graphs of flux density, flux lines, inductance and Q of several ferrite cored inductors. The fourth part shows how flux density changes when an air gap is inserted in the center of a ferrite rod. It also includes simulations showing the distribution of magnetic flux density and current density in the turns of the solenoid. The fifth part shows how Q varies when the ratio of the length of the solenoid to the core changes Also shown is a chart showing the change of Q when the conductor spacing is changed. The sixth part discusses important info about ferrite 61 and similar materials.
## Part 1: Modeling ferrite cored inductors
Bulk factors that affect inductor Q:
• Initial permeability of the ferrite material (µi) and ferrite loss-factor (LF)
• dielectric constant (ε) and dielectric loss tangent (tan δ) of the ferrite core
• dielectric constant (ε), dielectric loss tangent (tan δ) and length of the
• ‘former’ upon which the solenoid is wound (if one is used)
• resistivity of the ferrite rod
• length (lf) and diameter (df) of the rod, and their ratio
• length (ls) and diameter (ds) of the solenoid, and their ratio
• Ratio of the length of the solenoid to the length of the ferrite rod
• size and type of wire (solid or litz) and spacing of the turns
A simplified equivalent circuit model for a ferrite cored inductor is shown in Fig. 1.
• CLF=ferrite core loss-factor at a specific frequency [(30*10^-6) at 1 MHz for ferrite 61]
• Co=distributed capacitance. This is made up of mostly capacitance from the hot parts of the solenoid, through the ferrite dielectric, to ground (assuming that one end of the solenoid is grounded). If a solenoid former is used, its dielectric is in the path and will affect the overall loss. Another part of Co is made up of capacitance from the hot parts of the solenoid through air, to ground.
• La=inductance of solenoid in air
• FDF(suffix a or f)=flux density factor. This is a number, greater than 1, that corrects the value of the series DC resistance of the solenoid to its actual AC value. The “a” denotes air surrounding the solenoid, the “f” denotes the inclusion of a ferrite core.
• LFEF=leakage flux effect factor, the ratio of Lp to µi*La. (LFEF is always less than 1)
• when the solenoid conductors are bathed in magnetic flux. It has a suffix”a” when referring to the solenoid in air, and a suffix”f” when referring to a solenoid having a ferrite core.
• Lp=parallel inductance representing the increase of inductance caused by the ferrite core. Lp=LFEF*ui*La Henrys
• Ra=series RF resistance of solenoid in air
• Ra*(FDF-1)=additional series resistance caused by increased average flux density around the conductors when a ferrite core is placed inside the solenoid
• Rdc=DC resistance of the wire used in the solenoid
• Ro=represents resistive power loss in Co. This loss in Co is contributed by the dielectric loss tangent of the ferrite and that of the solenoid former, if one is used.
• Rp=parallel RF resistance across Lp, representing magnetic losses in the ferrite core. Rp=LFEF*ω*La/CLF Ω
• ui=initial permeability of the ferrite. [125 for type 61 ferrite]
• Qa=Q of the solenoid in air Qa=ω*La/(Rdc*FDFa)
• Qt=Q of the real-word ferrite-cored inductor as represented in Fig. 1
• ω=2*pi*frequency
The simplified equivalent circuit shown in Fig. 1 provides a convenient way of think about the effect of placing a ferrite core in a solenoid. Ra and La represent the resistive loss and inductance of the air-cored solenoid without the core (we want to increase the resultant Q and inductance). Adding a core creates the effect of adding a parallel RL in series with the air coil. When a core is inserted into the air-cored solenoid, the series resistance of the solenoid in air is increased by the factor (FDFa-1) to account for the increased power loss in the copper wire caused by the increased flux density from the core
Magnetic flux density surrounding the solenoid turns conductor is not uniform along the length of the solenoid. It is greater at the ends than along its central part. Increasing the length/diameter of the solenoid reduces the percentage of total flux that penetrates the copper and thus reduces resistive copper losses (especially at the two ends of the winding). Increasing the ratio of the length of the ferrite rod to that of the solenoid further reduces the percentage of total flux that penetrates the copper, further reducing resistive losses.
The amount of electric field that penetrates the core is important, especially at the high end of the band . The Nickel/Zinc cores such as type 61 have a very high resistivity dielectric as well as a rather low dielectric constant (ε) that has a high dielectric loss tangent (tan δ): Losses caused by the high (tan δ) may be minimized by using construction methods that keep the parts of the solenoid that are at a high electrical potential spaced away from the core. For instance a coil former sleeve made of low loss, low dielectric constant material can be used to isolate the high impedance parts of the solenoid from the core.
I-squared-R resistive power loss in the conductor caused by the AC current flow: Increased series resistance of the solenoid reduces Q, especially at the low end of the band compared to the high end, since the inductive reactance is at a minimum there (if the resistance, as a function of frequency is constant). Proximity and skin effect losses increase the RF resistance of the conductor at the high end of the band more than at the low end. The use of litz wire reduces the loss across the band, but more so at the high end. If solid wire is used, spaced-turns winding will reduce the losses from proximity effects. An advantage of using Litz wire in a ferrite-rod inductor is that there seems to be little downside to Q from close spaced winding. This helps with obtaining a larger inductance with a smaller solenoid and ferrite core. The use of larger diameter wire to reduce one of these losses usually has the effect of requiring a larger solenoid and ferrite core in order to keep the inductance the same, requiring mind-numbing tradeoffs.
Experimentation with 4″ long by 1/2″ diameter ferrite 61 rods and litz wire of 50/46, 125/46, 270/46 and 420/46 construction with an inductance 250 uH suggest that a winding length of about 1.5″ of close-wound 125/46 litz wire is close to optimum, from the standpoint of Q. I’ve tried to use 660/46 litz with a 4″x1/2″ ferrite 61 rod to attain a high Q inductance of about 250 uH. It never worked, probably because the length of the rod, being close to that of the solenoid, caused a high flux density condition to occur near the ends of the solenoid, creating extra copper loss. Lesson: For high Q, the rod should be longer than the solenoid, maybe three times as long.
Ferrite cores of the same specification often exhibit rather wide variations in their ferrite loss-factor (thus affecting the attainable Q when used as a core). They also vary, to a lesser degree, in initial permeability (µi). This affects the inductance. Generally, when selecting cores from a group having identical specifications, the ones with the least initial permeability will have the least hysteresis loss, especially at high frequencies. This provides a convenient way to select cores that will yield the highest Q coils, without actually measuring Q: Wind a solenoid on a thin walled, low loss form and measure its inductance after placing each core, in succession, centered in the coil. Generally the core providing the least inductance will provide the highest Q.
Comments: Consider the schematic in Fig. 1. La, Ra and Ra*(FDFa-1) define the inductance and Q of the air-cored solenoid (Before a ferrite core is inserted in an air-cored solenoid, FDF=1).
Lp and Rp define the inductance and Q of the added inductance produced when a ferrite core is inserted into the solenoid (now FDF becomes FDFf greater than before because of greater flux density in the conductors). The value of Lp depends upon ui of the ferrite material, La and LFEF. Some methods of changing LFEF are: 1) Increase the amount the bare rod core extending beyond the solenoid. This will increase the value of LFEF and consequently the value of Lp. 2) Use a smaller diameter ferrite core than the Id of the solenoid. This will reduce the value of LFEF and consequently Lp.
The L and Q values of the air-cored solenoid are usually quite low**. The inductance of Lp is usually high and equal to LFEF*ui*La. The parallel resistance Rp equals (reactance of La)*LFEF/CLF. If LFEF equals 1 (This can be approached when using a toroid having a high permeability, ui), the Q of a real-world ferrite-cored toroid inductor is about: Q=1/(ui*CLF). The Q of a ferrite-cored toroid inductor using ferrite 61 as the core can have a Q of about 330 at 1 MHz, as shown in the 11th Edition of the Fair-Rite catalog.
Summary: With no ferrite core present one has a low Q low inductance inductor. If one could construct a fully flux-coupled ferrite 61 core (LFEF~1.0), the Q at 1 MHz would be 1/ui*CLF=267. Highest Q occurs with an optimum value of LFEF, which also provides an intermediate value for real-world inductance. See Table 6, next to last entry.
In my experience with 1/2″ diameter ferrite 61 rods aiming for 250 uH and using Litz wire, most of the time LFEF turns out to be greater than the value for maximum Q. An indication of this condition can be obtained by placing two extra cores, each co-axially aligned with the solenoid’s core, one at each end of said core, to increase the LFEF. The Q is usually reduced even though the inductance is increased, showing that LFEF is too high for maximum Q. Proof of this can be attained by discarding the two extra cores and reducing the number of turns on the rod. Of course, inductance goes down, but Q will increase. To get the inductance back up and retain the higher Q, a solenoid and ferrite rod of larger diameter are required.
** Note the “no core” entry in Table 6 for inductor BB. The solenoid (with no core) has a Q of 88 (and an inductance of 17.6 uH).
## Part 2: Measurements
Comparison of several conventionally wound Ferrite-cored solenoids having the same winding length and number of turns, but different diameters
Ferrite rod length=4″, diameter=0.5″, material=type 61, µi=125, ferrite loss factor (CLF)=30*10^-6, the “best ferrite core” was used, former=low loss thin wall tubing of various lengths, wire=125 /46 ga. litz, construction=conventional close wound solenoid of 58 turns having a length of about 1.625″.
* Piece of polyethylene tubing having an OD of 0.625″ and an ID of 0.50″
** This coil former has a cross section somewhat less than from a full 0.75″ piece of tubing. It is constructed by first sliding the 1/2″ dia. 4″ long ferrite rod into a 5″ long piece of 0.625″ OD polyethylene tubing. A full longitudinal cut is then made in a second piece of similar tubing, so it can be fitted over the first one. A gap of about 3/8″ is left in the second, slit piece of tubing, and that is what causes the cross section to be less than that of a true 3/4″ tube.
• Note 1: Q values are corrected for distributed capacity.
• Note 2: ‘best ferrite core’, ‘medium ferrite core’ and ‘worst ferrite core’ refer to Q measurements of a large quantity of 4″ long, 1/2″ diameter ferrite 61 cores purchased from CWS Bytemark over a period of years. The Q measurements were made at 1710 kHz with a test coil wound on a former similar to that used in ‘Coil and Former’ B, above. The winding had 39 turns, close wound, of 270/46 litz. The “best ferrite core” was selected from a small batch of cores that were re-annealed by a local ferrite manufacturer. See the third-from-last paragraph.
### Some observations
1. Inductance does not change much between a solenoid diameter of 0.5″ and 1.5″.
2. At low and medium frequencies, Q is the highest when the wire is wound directly on the ferrite. It drops substantially at the high frequency end.
3. Q at the high frequency end increases as the wire is separated farther from the core, except for coil E.
4. Q at the low frequency end decreases as the coil wire is separated further from the core.
### Comparison between a conventional and contra wound ferrite-rod cored solenoid using a “best” and a “worst” rod
See Article # 0, Part 12 for a mini-Article about the benefits of the contra-coil construction.
Ferrite rod length=4″, Diameter=0.5″, Material=type 61, µi=125, Ferrite loss factor (CLF)=30*10^-6, Former=polyethylene (not vinyl) tubing, ID=0.5″, OD=0.625″, length=5″, Wire=125 strand/46 ga. litz, Construction=close-wound conventional solenoid of 58 turns having a length of about 1.625″
Ferrite rod length=4″, Diameter=0.5″, Material=type 61, µi=125, Ferrite loss factor (FEF)=30*10^-6, Former=polyethylene (not vinyl) tubing, ID=0.5″, OD=0.625″, length=5″, Wire=125/46 ga. litz, Construction=close-wound contra wound solenoid of 58 turns and length of 1.625″ (not wound as tightly as the conventional solenoid above).
The winding format for solenoids #1 and #2, below, are shown in Figs. 2 and 3. For clarity, the windings are shown as space wound, but the actual solenoids #1 and 2 close wound. Connections for the contra wound inductor shown in Fig. 3: For the series connection, join leads c and e. Lead d is hot and lead f is cold. For the parallel connection, join leads c and f. Join leads d and e. d/e is the hot and c/f is the cold connection.
The Q values given above were measured on an HP 4342A Q meter and corrected for the distributed capacity of the inductor (Co). The ferrite cores were purchased from CWS ByteMark in the third quarter of 2002. They may have changed vendors since then because some rods I purchased in the 3td quarter of 2004 resulted in lower Q coils than the values reported here. These rods also had two small, 180 degree apart, longitudinal flats along their entire length. CWS gracefully accepted a return of those rods and quickly refunded my money. The ‘best’ and ‘worst’ cores used in these measurements were from a group purchased from CWS ByteMark in the 3rd quarter of 2002.
Note the better high-band Q values recorded for the contra wound inductor. This is because the low Q distributed capacity from the dielectric of the ferrite (Co and Ro) is connected across an inductor having 1/4 the inductance (and reactance) value of the conventional wound solenoid. An observation: If the hot/cold connections to the contra wound coil in Fig. 3 are reversed, Q at 1710 kHz drops. This is because more loss from the low Q dielectric of the ferrite is coupled in to the stray capacitance.
### Solid wire instead of litz?
Keep in mind that the work described here used close-wound 125/46 litz wire. If one duplicates ‘Coil and Former B’ in Table 2, except using 22 ga. solid copper wire (having the same diameter) as 125/46 litz, the Q values drop to about 1/6 of the values achieved with the litz wire. The cause is the large proximity effect resistive losses, as well as skin effect, in the solid wire. The proximity effect, but not the skin effect loss may be much reduced if the wires are space-wound. New trade-offs now must be considered: Same wire diameter, and therefore a longer solenoid, or a smaller wire diameter and the same overall length? If one wishes to use solid wire, it should probably be wound directly on the ferrite, not on a former. The overall Q will still be much less than when using litz, and the loss from the high (tan δ) dielectric of the ferrite will be pretty well swamped out because of the now higher losses from the skin and proximity effect losses. The Q values, using a close-wound solenoid of 22 ga. solid copper wire on a polyethylene former, as in ‘Coil and Former’ B in Table 2 are: 520 kHz: 130, 943 kHz: 141 and 1710 kHz: 150 when using the “best core”. The Q drops only 3, 3, and 5 points respectively if the “worst core” is used.
### Measurements to determine the (tan δ) of the dielectric of a ‘medium core’
Two adhesive copper foil coupons, 0.5″x1.75″ were affixed to the 4″, 0.5″ diameter rod made of ferrite material 61 (3M sells rolls of thin copper foil with an adhesive on one side). The long dimension of each coupon was parallel to the axis of the rod with the two coupons set opposite to each other, 180 degrees apart. They formed a two plate capacitor having curved plates with the dielectric of the rod between them. The capacitance of this capacitor came out to be 6.5 pF. Measurements, using a Q meter and a high Q inductor were made that enabled calculation of the Q of this 6.5 pF capacitor. Q was 25 at 520 kHz, 35 at 943 kHz and 55 at 1710 kHz. Even though the distributed capacity of a ferrite rod inductor is only made up partially of this poor dielectric, it is, I believe, a previously unrecognized cause of the usual Q drop at the high end of the band. It is also, I believe, the cause of Q reduction in ferrite toroids when no gap is provided between the start and finish of the winding.
## Part 3 – Flux density and flux line simulations, inductance and Q of several ferrite-cored inductors along with some measurements
David Meeker’s “Finite Element Method Magnetics” program FEMM was used to generate Figs. 4-17. First a word about the displays: FEMM, as used here provides a 2-dimensional display of flux density (the colors) and flux lines (the black lines). Only half of the object being simulated is analyzed and displayed since only axisymmetric objects can be analyzed with the program. This saves simulation time, which can become very great. FEMM, at this time, cannot simulate using litz wire. That is why the following simulations and measurements use 22 ga. solid copper wire instead of the 125/46 litz used in Part 2.
Understanding the images: Visualize the axis of the ferrite rod as coincident with the y-axis of a conventional 3-D x, y and z coordinate graphing system with the center of the rod at the origin. The FEMM program discards everything to the left of the y, z plane that intersects the origin and displays a view of the other half. The images show field densities that exist in an x, y plane that intersects the origin. The vertical object at the left in each image is the ferrite rod. Its horizontal width is ½ that of the diameter of the actual round ferrite rod since the parts of the inductor to the left of a vertical y, z plane intersecting the origin have been discarded (see above). The vertical line of little circles to its right show the cross-sections of the turns of the solenoid wires. Fig. 4 is a plot of magnetic flux density and flux lines on an imaginary plane that cuts longitudinally through the center of ferrite rod inductor AA, shown mostly in purple. The outline of half the 4″x1/2″ rod is shown at the left of the plot. If one measures, on the computer screen, the height and width of the rectangle, one can see that their ratio is 16. This is equal to the ratio of the 4″ length of the rod to 1/2 of its 1/2″ diameter. The large half-circle defines the area around the inductor that will be included in the simulation. It’s made up mostly of air. The magnitude of the flux density can be seen from the colors on the display (see the chart). The range of flux density values for the display was purposely limited to about 20 to help supply flux density detail around the outer turns of the solenoid. That is why most all the core is colored purple (the flux density is above 4.000e-9 Tesla). Fig. 5 is a close-up simulation of the area near the upper turns of the solenoid. If one’s browser has a zoom control, one can easily see how the flux density close to the surface of the wires of the end turns of the solenoid (even numbered Figs.) is greater than it is in the more central turns. High flux density in the copper equals high power loss (Q reduction).
Comment: Look at figs. 6 and 11 in Table 4. Inductors BB and EE are identical except for the length of the ferrite rod. It appears that about 10% of the end turns of solenoid BB are exposed to a flux density above 2.8e-9 Tesla (3 dB below the maximum plotted value of 4e-9 T). The corresponding percentage in solenoid EE about 50%. This shows that a high flux density around a greater percentage of turns results in lower Q. A parameter listing of the inductors is below Table 4. Note the Q values for inductors BB and EE.
Table 4 – Simulation of inductors using solid copper wire of OD=0.0253″ in Figs. 4, 5, 6, 7, 10 and 11. Wire OD=0.01765″ in Figs. 8 and 9. No litz wire is used. All inductors have 58 turns.
Parameters of simulated inductors AA through DD, inductance and Q at 1 MHz:
• Inductor AA: ferrite core length=4″, ferrite core diameter’1/2″, core type=61, wire type=22 ga. solid copper wire, OD=0.0253″, solenoid length=1.624″, ID of solenoid=0.5013″, Number of turns=58, Inductance=261.66 uH, Q=118.4. Solenoid construction is similar to inductor A in Tables 1 and 2.
• Inductor BB: ferrite core length=4″, core diameter=1/2″, core type=61, wire type=22 ga. solid copper wire, OD=0.0253″, solenoid length=1.624″, ID of solenoid=0.6263″, Number of turns=58, Inductance=259.11 uH, Q=130.7. Solenoid construction is similar to inductor B in Tables 1 and 2.
• Inductor DD: Same as inductor AA except that the wire diameter is reduced to 0.01765″. This creates a spaced winding. Inductance=265.37 uH, Q=267.6.
• Inductor EE: core length=1.680″, core diameter=1/2″, core type=61, wire type=22 ga. solid copper, wire, OD=0.0253″, solenoid length=1.624″, ID of solenoid=0.6263″, Number of turns=58, Inductance=121.80 uH, Q=36.2.
Note that the Q difference between the “Best core” and the ‘Worst core” is very small. This is because the main loss in this inductor is the high proximity loss in the solid close-spaced copper winding. The much lower ferrite core loss is swamped out and has little effect on Q, showing a Q ratio between the two of about 0.97. Compare these figures with those in Table 3 for a similar conventionally wound solenoid using close-spaced 125/46 litz wire. Proximity loss is greatly reduced in close-wound litz wire, compared to close-wound solid copper wire. The Q ratio here is about 0.75. Loss in the ferrite core swamps out the much lower proximity loss in the litz wire, and a much higher Q results.
## Part 4 – The effect on impedance parameters of an air gap in the center of a ferrite rod inductor and on magnetic flux/current density in the wire cross-sections
The images and text are based on FEM simulations of inductor BB, shown in Figs. 6 and 7, the specs of which are shown below Table 4, but with the following difference: Instead of using one 4” long ferrite rod, two 2” co-axially oriented rods are used in each simulation with different air gaps between them (0.0000”, 0.0313”, 0.0625” and 0.1250”). The solenoid is centered on the gap. When the gap is 0.000” the result should be the same as if one solid 4” rod were used. The inductance and Q values for inductor BB are slightly different than the values in the new simulations, as stated below for a 0.0000” gap between two 2” rods. This is because the “meshing” parameter in the FEMM simulation was changed to reduce the time taken for the simulations.
The top half of a full image of a core/solenoid combination is a mirror image of the bottom half. In figures 12-17 advantage is taken of this characteristic by zooming in and not showing the entire rod so as to be able to get a larger image, thus supplying more detail.
The first group of four simulations have central gap widths of 0.0000”, 0.0313”, 0.0625” and 0.1250”. They are named GappedRodA0000, GappedRodB0313, Gapped RodC0625 and GappedRodD0125. They are intended to show magnetic flux density distribution in the ferrite and the air. The simulations are made at a frequency of 1 MHz with an AC current of 1 uA RMS in the solenoid. The actual magnetic flux density values can be estimated by comparing the color display to the color chart to the right of the images.
A second group of two simulations shows magnified views of two parameters of the GappedRodB0313 simulation. They are called GappedRodB0313_B (for showing flux density B in Teslas in the cross-sections of the individual wire turns) and GappedRodB0313_j (for showing the current density j in MA/m^2 in the cross-sections of the individual wire turns).
If one looks closely at the GappedRodB_B image, one can see how flux density is distributed in the wire cross-sections as a function of distance along the rod. As the textbooks say, very little flux exists in the interior of the wire. Where the external flux density is great, as it is at the ends of the rod and near the gap, the flux that penetrates the copper is confined near the outer periphery of the wire.
The distribution of current density in the turns as a function of position along the length of the rod is shown in image GappedRodB0313_j. This illustrates skin effect. Note the current density is not uniform in the wires because of proximity effect and the fact that the length of the solenoid is not very long, compared to the length of the rod.
Some specifications common to the inductors in Figs. 12 through 17: Ferrite rod length: two 2” long rods oriented coaxially, and spaced apart by 0.0000”, 0.0313”, 0.0625” or 0.1250”. Ferrite rod diameter=0.5”, Permeability of ferrite rod=125. Loss factor of ferrite at 1 MHz=30*10^-6, ID of solenoid=0.6263”. Number of turns=58. Wire: solid copper, OD=0.0263”. Length of solenoid=1.624”. Frequency at which the simulations are made=1 MHz.
It is interesting to see in Figs. 16 and 17 that the distribution of magnetic flux in the cross-section of the turns has the same shape as that of the current density.
Please note in the text accompanying Figs. 12-17 that the copper series loss component corresponds to the sum of Ra and Ra*(FDF-1) as shown in Fig. 1.
## Part 5 – Ferrite-rod inductor simulation experiments; all using centered solenoids 1.624″ long and having 58 turns
The solenoids used in the simulations in Table 6 all use a conductor having a diameter of 0.0253″. The only parameter varied is the core length. The simulations in Table 7 all use a 4″ long core. The only parameter varied is the diameter of the conductor.
* Solenoid winding covers the full length of the core.
Table 6 shows that about 77% of the maximum Q is attained with a core about 2.4 times the length of the solenoid with the turns number, solenoid size, core length, etc used here. About 68% of the maximum inductance is attained. Note also that when the length of the core is shortened to approximately the length of the solenoid, Q drops precipitously. Resistive losses are mainly proximity effect losses. Hysteresis losses are magnetic losses in the ferrite core itself. Total losses are the sum of the two. There is a good lesson to be learned here: To maximize Q, do not cover the whole length of the core with the solenoid.
* Simulates winding the 58 turn solenoid directly on the 4″ long ferrite core (solenoid ID=0.5013″) instead of on a former having an ID of0.6263″. Note that the the two simulations using a conductor diameter of 0.008995″ show remarkably similar parameter values.
Table 7 shows the benefits of spaced winding when using solid wire. All the inductors in Table 7 use centered solenoids of 58 turns and a length of 1.624″. The only variable is the diameter of the conductor, which controls the spacing of the turns (the winding pitch is held constant). The lesson here is that, when using solid copper wire, there can be a great Q benefit by space winding the solenoid and using an optimum size wire; in this case a Q of 431.9 vs 130.1 at 1 MHz, with solid wire. One can see that core losses change very little with the various conductor diameters (Hysteresis losses in ohms). Notice how, with a conductor diameter change from 0.02530 to 0.00006300″, the AC copper loss decreases from 11.6 to 3.02 ohms overwhelming the increase in DC conductor resistance from 0.16 to 2.62 ohms.
See Table 3 for measured inductance and Q values of an inductor similar to inductor BB, but wound with 125/46 litz wire. Here the Q is even greater than in Table 7 because litz construction is less sensitive to proximity and skin effect losses than is solid wire.
Thanks must go to Brian Hawes for making me aware of the FEMM program and showing me how to use it.
## Part 6: Perminvar ferrite, and what the term means
Normal nickel/zinc ferrites (NiZn), the types with less permeability as well as lower loss factors than manganese/zinc (MnZn) ferrites, are often used at RF because of their low loss at the higher frequencies. They do not suffer appreciably from permanent changes in permeability or loss factor from exposure to strong magnetic fields or mechanical shock such as grinding, or dropping on the floor.
Special nickel/zinc ferrites, called perminvar ferrites can achieve a considerably lower loss factor for the same permeability than normal nickel/zinc ferrites, and at higher frequencies. This result is achieved by adding a small amount of cobalt to the ferrite power before firing, but there is a catch. In order to actually achieve the lower loss factor, the ferrite core must be annealed by raising it to a temperature above its Curie temperature (the temperature at which it losses all its permeability), and then cooling it very slowly back down through the Curie temperature, and then to lower temperatures. This process usually takes about 24 hours. The Curie temperature of ferrite type 61 (a perminvar ferrite) is specified in the Fair-Rite catalog as being above 350 degrees C. The annealing process reduces the permeability somewhat, but reduces the loss factor substantially.
The low loss-factor property of the annealed perminvar ferrite can be easily degraded by mechanical shock, magnetic shock or just physical stress (as from a tight mounting clamp). The Fair-Rite catalog sheet for type 61 ferrite cautions “Strong magnetic fields or excessive mechanical stresses may result in irreversible changes in permeability and losses”. Actually, the changes are reversible if one goes through the annealing process again. The MMG catalog, issue 1A, in writing about perminvar ferrites, adds: “Mechanical stresses such as grinding and ultrasonic cleaning increase the permeability and lower the Q, especially at the higher frequencies, although the changes in Q at the lower frequencies may be very small.
I suspect that there is now much less pressure on ferrite manufacturers to deliver a low loss product than in the past. Since time is money, maybe they now skimp on the annealing process. Several years ago I took some 4″ x 0.5”, mix 61 rods I had purchased from CWS ByteMark and had them re-annealed at the plant of a local ferrite manufacturer. The Q of a litz-wire coil using the re-annealed core, at 2.52 MHz, was increased by 12%. This indicates that the core was not originally properly annealed, or had been subjected to some mechanical or magnetic shock after being annealing by the manufacturer. I was informed, when I asked, that coil Q at high frequencies could be expected to increase by up to100% from the pre-annealed value. I chose the best of these re-annealed rods to be my “best ferrite core” rod in this Article. One source informed me that few ferrite manufacturers perform the annealing process anymore. Toroids made of type 61 material are still made here in the USA.
Note: An easy way to use a DVM ohmmeter to check if a ferrite is made of MnZn of NiZn material is to place the leads of the ohmmeter on a bare part of the test ferrite and read the resistance. The resistance of NiZn will be so high that the ohmmeter will show an open circuit. If the ferrite is of the MnZn type, the ohmmeter will show a reading. The reading was about 100k ohms on the ferrite rods used here.
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what should be added to the polynomial x ^2-5x+4 so that 3 is the zero of the polynomial
Asked by hvsl123 | 24 Jun, 2021, 05:30: PM
Let f(x) = x2 -5x + 4
Let g(x) = f(x) +k = x2 -5x +(4+k)
if 3 is zero of polynomial g(x) , then g(3) = 0
g(3) = (3)2 -5(3) +(4+k) = (4+k) - 6 = 0 or k = 2
If we add 2 to polynomial ( x-5x + 4 ) , Then polynomial becomes ( x-5x + 6 )
3 is zero of polynomial ( x-5x + 6 )
Answered by Thiyagarajan K | 19 Sep, 2021, 04:22: PM
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# Capgemini Exam Pattern | Coding Questions Pattern
Last Updated on October 3, 2022 by Ria Pathak
In this Article, we’ll be discussing the Capgemini Exam Pattern, Capgemini Coding Questions has been included as a new round in Capgemini selection process for 2023 graduates. This round will be conducted through CoCubes, containing 3 coding questions and you will have a total of 60 mins to solve these 3 coding questions. This round has been added to check the candidates programming, logical and problem solving techniques. Below we have given some previous year sample based questions for Capgemini Coding Round, make sure you practice all of them.
### Details for Capgemini Coding Round
Coding Details Details
Number of Questions 3
Time available 75 mins
Package offered 7.5LPA
Difficulty high
Capgemini has started asking Coding Questions, but it is asked in a special hiring where the package being offered is more than the normal hiring. The package offered for this post is 7.5LPA, and since the package being offered is comparatively higher than the normal hiring process, the selection process and test difficulties level are also higher than normal.
Practice makes a coder perfect, so don’t miss to check Competitive Coding Problems created by expert mentors at Prepbytes.
### Top 10 Capgemini Exam Pattern-Coding Questions
1. Problem Statement –
Write the code to traverse a matrix in a spiral format.
Sample Input
Input
5 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20
Output
1 2 3 4 8 12 16 20 19 18 17 13 9 5 6 7 11 15 12 14 10
Solution:
```import java.util.*;
public class Solution
{
public static List<Integer> solve(int [][]matrix,int row,int col)
{
List<Integer> res=new ArrayList<Integer>();
boolean[][] temp=new boolean[row][col];
int []arr1={0,1,0,-1};
int []arr2={1,0,-1,0};
int di=0,r=0,c=0;
for(int i=0;i<row*col;i++)
{
temp[r][c]=true;
int count1=r+arr1[di];
int count2=c+arr2[di];
if(count1>=0 && row>count1 && count2>=0 && col>count2 && !temp[count1][count2]){
r=count1;
c=count2;
}
else
{
di=(di+1)%4;
r+=arr1[di];
c+=arr2[di];
}
}
return res;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int m=sc.nextInt();
int n=sc.nextInt();
int matrix[][]=new int[m][n];
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
matrix[i][j]=sc.nextInt();
}
System.out.println(solve(matrix,m,n));
}
}
```
2. Problem Statement –
You’re given an array of integers, print the number of times each integer has occurred in the array.
Example
Input :
10
1 2 3 3 4 1 4 5 1 2
Output :
1 occurs 3 times
2 occurs 2 times
3 occurs 2 times
4 occurs 2 times
5 occurs 1 times
Solution:
```import java.util.Scanner;
public class ArrayFrequency
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int array[] = new int[n]; // taking value of integer
int count = 0;
int k = 0;
for(int i=0; i<n; i++)
{
array[i] = sc.nextInt(); //elements of array
}
//step - 1
int newarray[] = new int[n];
for(int i=0; i<n; i++)
{
count = 0;
for(int j=0; j<=i; j++)
{
if(array[i]==array[j])
{
count++;
}
}
if(count == 1)
{
newarray[k] = array[i];
k++;
}
}
//step 2;
for(int i=0; i<k; i++)
{
count = 0;
for(int j=0; j<n; j++)
{
if(newarray[i] == array[j])
{
count++;
}
}
System.out.printf("%d occurs %d times\n",newarray[i],count);
}
}
}
```
3. Problem Statement –
A function is there which tells how many dealerships there are and the total number of cars in each dealership.
Your job is to calculate how many tyres would be there in each dealership.
Input
3
4 2
4 0
1 2
Output
20
16
8
There are total 3 dealerships
dealerships1 contains 4 cars and 2 bikes
dealerships2 contains 4 cars and 0 bikes
dealerships3 contains 1 cars and 2 bikes
Total number of tyres in dealerships1 is (4 x 4) + (2 x 2) = 20
Total number of tyres in dealerships2 is (4 x 4) + (0 x 2) = 16
Total number of tyres in dealerships3 is (1 x 4) + (2 x 2) = 8
Solution:
```import java.util.*;
public class Solution
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int dealership=sc.nextInt();
while(dealership-->0)
{
int cars=sc.nextInt();
int bikes=sc.nextInt();
System.out.println(cars*4+bikes*2);
}
}
}
```
4. Problem Statement
Gary is an avid hiker. He tracks his hikes meticulously, paying close attention to small details like topography. During his last hike, he took exactly n steps. For every step he took, he noted if it was an uphill or a downhill step. Gary’s hikes start and end at sea level. We define the following terms:
• A mountain is a non-empty sequence of consecutive steps above sea level, starting with a step up from sea level and ending with a step down to sea level.
• A valley is a non-empty sequence of consecutive steps below sea level, starting with a step down from sea level and ending with a step up to sea level.
Given Gary’s sequence of up and down steps during his last hike, find and print the number of valleys he walked through.
Input Format
The first line contains an integer, , denoting the number of steps in Gary’s hike.
The second line contains a single string of characters. Each character belongs to {U, D} (where U indicates a step up and D indicates a step down), and the i(th) cin the string describes Gary’s i(th) step during the hike.
Constraints
• 2 <= N <= 106
Output Format
Print a single integer denoting the number of valleys Gary walked through during his hike.
Sample Input
8
UDDDUDUU
Sample Output
1
Explanation
If we represent as sea level, a step up as / , and a step down as \ , Gary’s hike can be drawn as:
/\ _
\ /
\/\/
It’s clear that there is only one valley there, so we print on a new line.
Solution:
```import java.util.*;
class Main
{
public static int countingValley(int n, String path)
{
int level=0,valley=0;
for(int i=0;i<n;i++)
{
if(path.charAt(i)=='U')
level++;
else if(path.charAt(i)=='D')
{
if(level==1)
valley++;
level--;
}
}
return valley;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
String str=sc.next();
int result=countingValley(n,str);
System.out.println(result);
}
}
```
To test your knowledge it is recommended to give Mock Tests by which you will know where you are lacking and will help you to crack your dream job.Go on, check it out and start giving the test!
5. Write a C Program to check if two given matrices are identical.
Solution:
```#include<bits/stdc++.h>
using namespace std;
#define N 4
// This function returns 1 if A[][] and B[][] are identical
// otherwise returns 0
int areSame (int A[][N], int B[][N])
{
int i, j;
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
if (A[i][j] != B[i][j])
return 0;
return 1;
}
int main ()
{
int A[N][N] = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}
};
int B[N][N] = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}
};
if (areSame (A, B))
cout<<"Matrices are identical";
else
cout<<"Matrices are not identical";
return 0;
}
```
6. A Pythagorean triplet is a set of three integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than the given limit.
Input : limit = 20
Output : 3 4 5
8 6 10
5 12 13
15 8 17
12 16 20
A Simple Solution is to generate these triplets smaller than the given limit using three nested loops. For every triplet, check if the Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.
An Efficient Solution can print all triplets in O(k) time where k is the number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,
a = m2 – n2
b = 2 m n
c = m2 + n2
because,
a2 = m4 + n4 – 2 m2 n2
b2 = 4 m2 n2
c2 = m4 + n4 + 2 m2 n2
We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets
Solution:
```import java.util.*;
class Main
{
public static void pythagoreanTriplets(int limit)
{
int a,b,c=0;
int m=2;
while(c<limit)
{
for(int n=1;n<m;++n)
{
a=m*m-n*n;
b=2*m*n;
c=m*m+n*n;
if(c>limit)
break;
System.out.println(a+" "+b+" "+c);
}
m++;
}
}
public static void main(String[] args)
{
int limit=20;
pythagoreanTriplets(limit);
}
}
```
7. Problem Statement – Aashay loves to go to WONDERLA , an amusement park. They are offering students who can code well with some discount. Our task is to reduce the cost of the ticket as low as possible.
The cost of tickets can be removed by removing the digits from the price given. They will give some k turns to remove the digits from the price of the ticket. Your task is to help Aashay in coding a program that can help him to reduce the cost of a ticket by removing the digits from its price and getting the maximum possible discount.
Note – You cannot make the cost of a ticket zero. For eg -: If the cost of a ticket is 100, and you have 2 turns to reduce the price, the final price will be 1 and not zero.
Constraints:
• 1 <= number of tickets <= 105
• 1 <= K <= number of tickets
Input Format for Custom Testing:
• The first line contains a string,Tickets, denoting the given cost of each ticket.
• The next line contains an integer, K, denoting the number of tickets that is to be removed.
Sample Cases:
Sample Input 1
203
3
Sample Output 1
0
Solution:
```#include<bits/stdc++.h>
using namespace std;
int smallestNumber (string num, int k)
{
if(num.length()<=k)
return 0;
unordered_map<char,int> pos;
for(int i=0;i < num.length();i++)
{
pos[num[i]]=i;}
string temp=num;
sort(num.begin(),num.end());
string ans=num.substr(0,num.length()-k);
vector < int > v;
for(int i=0;i < ans.length();i++)
v.push_back(pos[ans[i]]);
sort(v.begin(),v.end());
string ret;
for(int i=0;i < v.size();i++)
{
ret+=temp[v[i]];
}
int final=stoi(ret);
return final;
}
int main()
{
string s; cin >> s;
int k;
cin >> k;
int ans;
cout << smallestNumber(s,k)%(int)(pow(10,9)+7);
return 0;
}
```
8. Problem Statement – In an airport , the Airport authority decides to charge some minimum amount to the passengers who are carrying luggage with them. They set a threshold weight value, say T, if the luggage exceeds the weight threshold you should pay double the base amount. If it is less than or equal to the threshold then you have to pay \$1.
Returns: The function must return an INTEGER denoting the required amount to be paid.
Constraints:
• 1 <= N <= 105
• 1 <= weights[i] <= 105
• 1 <= T <= 105
Input Format for Custom Testing:
• The first line contains an integer, N, denoting the number of luggages.
• Each line i of the N subsequent lines (where 0 <= i < n) contains an integer describing weight of ith luggage.
• The next line contains an integer, T, denoting the threshold weight of the boundary wall.
Sample Cases:
Sample Input 1
4
1
2
3
4
3
Sample Output 1
5
Explanation:
Here all weights are less than the threshold weight except the luggage with weight 4 (at index 3) so all pays base fare and it pays double fare.
Solution:
```#include <bits/stdc++.h>
using namespace std;
long int weightMachine(long int N,long int weights[],long int T)
{
long int amount=0,i;
for(i=0;i<N;i++) { amount++; if(weights[i]>T)
{
amount++;
}
}
return amount;
}
int main()
{
long int N,i,T;
cin>>N;
long int weights[N];
for(i=0;i<N;i++) { cin>>weights[i];
}
cin>>T;
cout<<weightMachine(N,weights,T);
return 0;
}
```
### Capgemini Exam Pattern FAQs
1. What is the average difficulty level for capgemini Papers?
These are hosted by Cocubes so instead of Capgemini Previous Papers, we analyze Cocubes, which generally is of moderate difficulty level thus Capgemini Previous Papers are also of the same difficulty but you must at least give 1 week for preparation.
2. From where should I prepare for Capgemini placement papers with answers?
For Capgemini Previous Placement Papers with Answers, PrepBytes is the best website you should be preparing from. We have all the previous year questions about the best material anywhere for Capgemini Test Papers with Solutions.
3. How many questions will be asked in the Capgemini Coding Test?
A total of 3 Questions will be asked in the Capgemini Test.
4. Does your study material consist of the entire syllabus for the capgemini exam?
Yes, our study material covers all the topics of all the sections of Capgemini. You can practice all types of questions 🙂 | 3,541 | 12,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.876086 |
https://electronics.stackexchange.com/questions/67445/vacuum-cleaner-power-rating-does-not-match-line-voltage-and-current-rating/156319 | 1,580,019,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687725.76/warc/CC-MAIN-20200126043644-20200126073644-00497.warc.gz | 420,975,244 | 29,429 | # Vacuum cleaner power rating does not match line voltage and current rating
I have a vacuum cleaner on which is proudly written:
12.0 Amps
1300 W
Assuming a 120 V (RMS) power supply, how would the power consumption of the vacuum cleaner be calculated given the 12.0 A current draw?
It seems that $120 \mathrm{\,V}\cdot12 \mathrm{\,A}\ne1300 \mathrm{\,W}$ and $120 \mathrm{V}\cdot\frac{12 \mathrm{A}}{\sqrt{2}}\ne1300 \mathrm{W}$, so then where might the power rating "1300 W" come from?
• I guess the 12A is a maximum, as you need a max to determine what devices can be on one group. The power consumption is an average, because you need an average to calculate how many energy it's going to consume hence how much the device will cost. – user17592 Apr 28 '13 at 12:15
• I read somewhere that power can range from 110-130V AC and 110*12= 1320 (close but not 1300), so it's possible that they didn't use 120 for voltage. Also, we don't know if there is an internal regulator/resistor/divider/transformer/etc. that makes the motor/internal parts a lower wattage, but it still uses 12 amps. Like @CamilStaps said, it could be a max and average, but I think that would be stupid of them to put it there and not label it max/average. – Anonymous Penguin Apr 28 '13 at 13:52
• If its proudly displayed along with the brand name, the 12.0 Amps is probably labeled "cleaning power" or some other such empty phrase. It's marketing nonsense. (Hmm, that's redundant, isn't it?) Find the nameplate, and get the actual current draw from that. – Pete Becker Apr 28 '13 at 16:23
There are many unknown factors. Both power and current will vary with load and there is only one way to find out how they relate: measure them.
One of the most important properties of AC power you neglected is the power factor cos(φ) with inductive loads:
$P = U \cdot I \cdot \cos(\varphi)$
\begin{align} \cos(\varphi) & = \dfrac{P}{U \cdot I} \\ & = \dfrac{1300\text{W}}{120\text{V} \cdot 12\text{A}} \\ & \approx \boxed{0.9} \end{align}
which sounds about what I'd expect for a vacuum cleaner.
Check this Wikipedia article on electric power for more background details.
• great minds think alike LOL – Andy aka Apr 28 '13 at 13:26 | 618 | 2,211 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-05 | latest | en | 0.918657 |
https://copious-systems.com/find-out-how-smart-you-are-with-this-three-question-iq-test/ | 1,656,164,453,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00314.warc.gz | 231,550,697 | 13,000 | There are hundreds of official and DIY IQ tests out there, but these are definitely not quizzes you can run on your lunch break.
These popular tests determine your level of intelligence, so it’s no surprise that they require you to sit for a while.
Then there’s a brief IQ test known as the Cognitive Reflection Test, consisting of just three questions.
It tests you on how accurately you can answer questions that seem simple at first but are much more complicated.
The faster you complete the test, the higher your intelligence level.
Try the world’s shortest IQ test below.
Q-1 of 3
Goal –0 out of 0
A bat and ball cost £1.10 in total. The bat costs £1.00 more than the ball. How much does the balloon cost?
### Explanation of question 1:
A bat and ball cost £1.10 in total. The bat costs £1.00 more than the ball. How much does the balloon cost?
Chances are you answered 10p for this question – we certainly did at first. This, however, is the wrong answer.
If the ball costs X and the bat costs £1 more, then it can be calculated like this:
X+£1
Therefore:
Bat+ball=X + (X+1) =1.1
Thereby:
2X+1=1.1 and 2X=0.1
X= 0.05
The correct answer is five pence.
### Explanation of question 2:
If it takes five machines five minutes to make five widgets, how long would it take 100 machines to make 100 widgets?
It’s tempting, if you answer it quickly, to guess 100 minutes.
But that’s not the right answer.
Instead, if five machines can create five widgets in five minutes, one machine will also create a widget in five minutes.
So if we have 100 machines all making widgets, they can make 100 widgets in five minutes.
This one also gives us headaches.
### Explanation of question 3:
In a lake, there is a patch of water lilies. Every day, the patch doubles in volume.
If it takes 48 days for the plot to cover the entire lake, how long would it take for the plot to cover half of the lake?
While many people could have answered 24 days, the answer is this.
If the patch doubles in size every day, it will shrink back in half. So on day 47 the lake is half full.
Of course, given the lack of verbal reasoning or any real variety of questions, it’s not the most accurate test.
So don’t feel too bad if you haven’t managed to find the right answers.
This quiz first appeared in The Mirror.
You can keep up to date with all the latest news in and around Cambridge by downloading our free app. It is available for iPhone and iPad on the Apple App Store, or the Android version can be downloaded from Google Play.
Share. | 609 | 2,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-27 | longest | en | 0.946613 |
http://onlinetest.ibpsexamguru.in/questions/Clerk-Reasoning-Test/CR-Quiz-34 | 1,521,912,114,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650764.71/warc/CC-MAIN-20180324171404-20180324191404-00175.warc.gz | 208,314,340 | 8,142 | IBPS Exam Guru
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Exercise
Bank Clerk :: CR Quiz 34
Home > Bank Clerk > CR Quiz 34 > General Questions
1 .
Direction (Q. 1 - 5) : In each question below are given three or four statements followed by three conclusions numbered I, II, and in. You have to take the given statements to be true evenif they seem to be at variance with commonly known facts and then decide which of the given conclusion(s) logical follow(s) from the given statements disregarding commonly known facts.
Statements: All plums are oranges.
All oranges are grapes.
Some grapes are apples.
Some apples are guavas.
Conclusions: I. Some grapes are plums.
II. Some guavas are grapes.
III. Some apples are oranges.
Only I and II follow Only I follows None follows All follow
2 .
Statements: Some dogs are pets.
Some pets are cats.
Some pets are rats.
No rat is goat.
Conclusions: I. Some cats are rats.
II. No cat is a rat.
III. No goat is a pet.
Only I follows Only II follows Only III follows None of these
3 .
Statements: Some flowers are fruits.
Some fruits are buildings.
All buildings are flats.
No flat is a house.
Conclusions: I. No building is a house.
II. Some flowers are houses.
III. Some flowers are flats.
Only I follows Only II follows Only I and II follow None follow
4 .
Statements: All cups are bowls. All bowls are trays.
Some trays are plates. No plate is a spoon.
Conclusions: I. Some bowls are plates.
II. Some cups are spoons.
III. No cup is a spoon.
Only I follows Only II follows Only I and III follow Only either II or III follows
5 .
Statements: Some cards are files.
Some files are inkpots.
Conclusions: I. Some inkpots are cards.
None follows Only I follows Only II follows Only III follows
6 .
Direction (Q. 6 - 10) : In each question below is given a group of letters followed by four combinations of digits/ symbols numbered (1), (2), (3) and (4). You have to find out which of the combinations correctly represents the group of letters based on the following coding system and the conditions that follow and mark the number of that combination as your answer. If none of the combinations correctly represents the group of letters, mark (4), ie 'None of these', as your answer.
Letter: N T A K E U B H O S RI G
Code: # 5 \$ 3 4 7 6 8 . 2 1 9 @ %
Conditions:
(i) If the first as well as the last letter is a vowel, both are to be coded as ' 0 '.
(ii) If the first letter is a vowel and the last letter is a consonant, both are to be coded as 'Z'.
(iii) If the first letter is a consonant and the last letter is a vowel, both are to be coded as
Q. ONSIRT
Z#1@95 Z#@195 Z#19@Z None of these
7 .
Q. KIUBSR
3@7619 0@7610 3@7691 3@6719
8 .
Q. BKAEUG
03\$470 63\$470 03\$47% 63\$47% | 762 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-13 | latest | en | 0.892086 |
https://online.flippingbook.com/view/798877914/246 | 1,721,145,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00242.warc.gz | 398,031,591 | 25,785 | # Honors Geometry Companion Book, Volume 1
4.2.4 Introduction to Coordinate Proof Key Objectives • Position figures in the coordinate plane for use in coordinate proofs. • Prove geometric concepts by using coordinate proof. Key Terms • A coordinate proof is a style of proof that uses coordinate geometry and algebra. Example 1 Positioning a Figure in the Coordinate Plane
The figure in a coordinate proof is either given or placed on a coordinate plane. If coordinates of the figure’s vertices are not given in the conjecture, then the figure can be placed anywhere on the coordinate plane. However, there are some guidelines for positioning a figure that will result in easier steps in the proof.
In this example, a figure is given. The figure is a rectangle with length 6 units and width 4 units. The rectangle can be positioned (placed) on the coordinate plane in any of the four quadrants. However, it is typically easiest to place a figure in the first quadrant so that the coordinates in each vertex’s ordered pair will be a positive number. The rectangle could be oriented in any way, but placing the rectangle so that its sides are vertical and horizontal, ideally with one side on the x -axis and one side on the y -axis, is best. So, Position 1 is the best position for this rectangle because the coordinates of the vertices are all positive whole numbers and two of the rectangle’s sides are placed along an axis.
Example 2 Writing a Proof Using Coordinate Geometry
Make a plan for proving that the area of △ ABC is twice the area of △ DBC . Find the area of each triangle and then multiply the area of △ DBC by 2 to determine whether the area of △ ABC = 2 (the area of △ DBC ). Remember, the area of a triangle is 1/2 the product of the triangle’s base and height (where the height is the perpendicular distance from the base to the opposite vertex). So, the area of △ ABC is (1/2)(5) (8) = 20 units 2 . Before the area of △ DBC can be found, the location of D must be determined. Since D is given to be the midpoint between A and C , the location of D can be found by using the midpoint formula. The y -coordinate of D is the height of △ DBC . Now the area of △ DBC can be found.
236
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http://www.controleng.com/single-article/pid-math-demystified-part-3-more-on-derivative-control/4d0bdf358ab90309e7f373b8d7e0f530.html?OCVALIDATE= | 1,503,161,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105700.94/warc/CC-MAIN-20170819162833-20170819182833-00122.warc.gz | 498,283,306 | 17,613 | # PID math demystified, part 3: More on derivative control
## Consider how a PD controller would work, without an integral function. Would you ever want to use that approach?
10/21/2013
To investigate how derivative action works, let’s look at a proportional derivative or PD controller. PID controllers are far more common than PD alone, but we already have an understanding of the integral component’s effects from the first two parts of this series, so we don’t need to review it again.
For a PD controller, we add the derivative of the error into the equation. Similar to what we discussed in the previous posts, we’re not really interested in deriving a derivative of the error function. Conceptually, the derivative in this case refers to how fast the error is changing. So, if we take the change in error divided by the change in time we get the slope.
To explain how this works, let’s look at the pseudo code for this controller. The calculation of error is the same as before: setpoint minus process value. Since the derivative reduces down to the change in error, the output is now the same proportional component as before: gain times error. The current error minus last error is multiplied by the gain and divided by the derivative time constant. The current error is stored in last error for use in the next scan.
Error : = Setpoint – ProcessValue;
Output : = K * Error + K/tau_i * (Error – LastError);
LastError : = Error; // save for next scan
So what does this mean?
The proportional component is affected by the error at that time, in the present. The integral component is affected by an accumulation of the error, or the past. The derivative component is a measure of how fast the error is changing, or a prediction of the future error.
How is this prediction of the future used? At first glance you might think that this term would be used to get you to your setpoint that much faster. But, that is not really the case. In practice, the derivative component is used to detect when the process variable is changing too fast, and it puts the brakes on to prevent overshooting the setpoint.
So if the derivative component acts as brakes on the momentum, how does it get you to the setpoint faster? It does this by allowing you to use a higher proportional gain to get you there quickly, but dampening the overshoot that would normally make that level of gain unstable.
The tuning constants for derivative control are typically the same units as the time constant for reset. A couple of other considerations though are important. On many controllers the derivative term is filtered independently. This prevents signal noise or spurious disturbances from being interpreted as a change in momentum, which causes the derivative action to overreact. Also, on some controllers the derivative does not actually derive from the error, but instead on the process variable signal alone. This prevents a change is set point from being seen as a change in momentum.
This video summarizes our PID discussions with a graphical explanation.
I hope this helps to demystify PID, at least a little.
Also read Understanding Derivative in PID Control.
This post was written by Scott Hayes. Scott is a senior engineer at MAVERICK Technologies, a leading automation solutions provider offering industrial automation, strategic manufacturing, and enterprise integration services for the process industries. MAVERICK delivers expertise and consulting in a wide variety of areas including industrial automation controls, distributed control systems, manufacturing execution systems, operational strategy, business process optimization and more.
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Controller programming; Safety networks; Enclosure design; Power quality; Safety integrity levels; Increasing process efficiency
Additive manufacturing benefits; HMI and sensor tips; System integrator advice; Innovations from the industry
Featured articles highlight technologies that enable the Industrial Internet of Things, IIoT-related products and strategies to get data more easily to the user.
This article collection contains several articles on how automation and controls are helping human-machine interface (HMI) hardware and software advance.
This digital report will explore several aspects of how IIoT will transform manufacturing in the coming years.
Cloud, mobility, and remote operations; SCADA and contextual mobility; Custom UPS empowering a secure pipeline
Infrastructure for natural gas expansion; Artificial lift methods; Disruptive technology and fugitive gas emissions
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Jose S. Vasquez, Jr.
Fire & Life Safety Engineer; Technip USA Inc.
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This eGuide contains a series of articles and videos that considers theoretical and practical; immediate needs and a look into the future.
Robotic integration and cloud connections; SCADA and cybersecurity; Motor efficiency standards; Open- and closed-loop control; Augmented reality
Controller programming; Safety networks; Enclosure design; Power quality; Safety integrity levels; Increasing process efficiency
Additive manufacturing benefits; HMI and sensor tips; System integrator advice; Innovations from the industry
Featured articles highlight technologies that enable the Industrial Internet of Things, IIoT-related products and strategies to get data more easily to the user.
This article collection contains several articles on how automation and controls are helping human-machine interface (HMI) hardware and software advance.
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Cloud, mobility, and remote operations; SCADA and contextual mobility; Custom UPS empowering a secure pipeline
Infrastructure for natural gas expansion; Artificial lift methods; Disruptive technology and fugitive gas emissions
Mobility as the means to offshore innovation; Preventing another Deepwater Horizon; ROVs as subsea robots; SCADA and the radio spectrum
Automation Engineer; Wood Group
System Integrator; Cross Integrated Systems Group
Jose S. Vasquez, Jr.
Fire & Life Safety Engineer; Technip USA Inc.
This course focuses on climate analysis, appropriateness of cooling system selection, and combining cooling systems.
This course will help identify and reveal electrical hazards and identify the solutions to implementing and maintaining a safe work environment.
This course explains how maintaining power and communication systems through emergency power-generation systems is critical. | 2,041 | 11,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-34 | latest | en | 0.920256 |
https://mathspace.co/textbooks/syllabuses/Syllabus-453/topics/Topic-8417/subtopics/Subtopic-111576/?activeTab=interactive | 1,642,545,169,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00576.warc.gz | 403,910,908 | 55,804 | # Intersections of Lines
## Interactive practice questions
How many times can two parallel lines intersect each other?
Two
A
Zero
B
One
C
Three
D
Two
A
Zero
B
One
C
Three
D
Easy
Less than a minute
At the point where the lines $y=-4x+1$y=4x+1 and $y=x+11$y=x+11 intersect, $x=-2$x=2. What is the $y$y-value at this point?
Consider the graph of the two lines $y=-4x+4$y=4x+4 and $y=x-6$y=x6. | 147 | 407 | {"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} | 3.46875 | 3 | CC-MAIN-2022-05 | longest | en | 0.813566 |
https://assignmentgeek.com/blog/natural-log-rules/ | 1,713,050,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00867.warc.gz | 105,800,881 | 15,604 | # Natural Log Rules – Brief Analysis And Examples
Are you taking a college or high school math class? One of the areas you will cover is natural logs. So what exactly is a natural log? A natural log of any number is its logarithm to the base of e (mathematical constant); where e is a transcendental number that is approximately equal to 2.718281828459. Sounds complex, right? But it will no longer be complex when you understand natural log rules.
In this post, we are going to take a closer look at the most important natural log rules, highlight other natural log properties, and demonstrate how to apply them with examples. We will also demonstrate the difference between natural logs and other logarithms.
## What is In in Natural Log Rules?
Natural log is also referred to as In. It is the inverse of e. Letter “e” is a math constant that is commonly referred to as a natural exponential. Notably, just like Pi (π) that has a constant value of 3.14159, e also has a fixed value of approximately 2.718281828459.
Where are In rules applied? e is used in many cases especially in mathematical scenarios such as decay equations, growth equations, and compound interest. Check the example below.
In(x) is the time required to grow to x, right? But ex denotes the quantity of growth that has been achieved after a specific period, x.
Notably, because e is applied in very many scenarios of math, physics, and economics, students take the logarithm featuring base e of a specific number to find a value. To put it differently, In was crafted as a shortcut for calculating log base e. It lets those reading a problem understand you are using the algorithm, taking base e, of a number. Let us demonstrate this using an example.
In(x) = loge(x).
Now, substitute the equation with a number: In(5) = loge(5) =1.609.
## The Main Log Rules
When working with In in mathematics, there are four key natural logarithm rules that you need to understand. Here is a closer look at these rules:
• ### Product Rule
When working on In of multiplication of y and x, the answer is the total (sum) of the In of y and In of x. When put in an equation, this rule looks like this; ln(y)( x) = ln(y) + ln(x).
Here is an example: ln(7)(5) = ln(7) + ln(5).
• ### Quotient Rule
When working on the In of the division of y and x, the answer is the difference of the In of y and In of x. To demonstrate this in an equation, here is how it will look like; ln(y/x) = ln(y) – ln(x).
See the example below: ln(8/4) = ln(78) – ln(4).
• The reciprocal rule
Unlike in the quotient rule, the natural log of a reciprocal of a number, call it x, is the opposite of the In of x. When put in an equation, it appears like this; ln(1/x)=−ln(x).
Here are some demonstrations using an example; ln(⅓)= -ln(3).
• ### Power Rule
When calculating the natural log of a number, call it x, raised to the power of y, you simply need to multiply y by In of x. In an equation, it will look like this; ln(xy) = y * ln(x).
Take the example; ln(52) = 2 * ln(5).
## The Main Properties of Natural Log
On top of the natural log and e rules that we have looked at above, it is important to also appreciate that there are a number of properties you need to understand when studying or adding natural logs. And you know what? You need to memorize the properties of ln to make related calculations easy.
Specific scenario In property
ln of a Negative Number The ln of any negative number is undefined
ln of 0 ln(0) is undefined
ln of 1 ln(1)=0
ln of Infinity ln(∞)= ∞
ln of e ln(e)=1
ln of e raised to the x power ln(ex) = x
e raised to the ln power eln(x)=x
Take a closer look at the above table. You will realize that the last three rows (e, f and g), In(e)=1. Note: this turns out right even when one is raised to the power of any other number. Why? Because the In and e in the rows serve as functions to each other.
## Solving Problems in College Natural Log Problems
Now that we have looked at ln rules and ln properties, it is time to get down to solving real problems. Below, we take a closer look at natural log identities and related problems. Check how they are done and try to solve similar questions
• Problem one: Can you solve ln (72/5).
To begin with, note we are going to use the quotient rule. This will get us, ln(72) – ln(5).
Second, we apply the power rule to get this, 2ln(7) -ln(5).
Having calculated the equation up to this point, we can leave it that way. But you can also go ahead and use a calculator to get a specific answer, 2(1.946) – 1.609 = 2.283.
• Problem two: Can you calculate ln (5x-6)=2.
When you get an equation featuring multiple variables in the parenthesis, the first thing is making e your base, right? Then, everything else should be considered exponent of e. Move on and put In and e next to each other. Here is a demonstration.
From the natural log laws, we know that eln(x)=x.
Therefore, the equation will look like this, eln(5x-6)=e2.
Because we also know eln(x)=x, eln(5x-6)= 5x-6, it implies that 5x-6= e2.
Also from the e^ln rules, we know that e is a constant. Therefore, we can easily establish the value of e2. We can do this by using a calculator or e’s value.
Have a look:
5x-6 =7.389
Then, we would add six to each side of the equation to get;
5x= 13.389
And, finally, divide both sides by number five;
x= 2.678
## The Differences between Natural Logs and Logarithms in College Math
To get it right on ln and e rules, it is also important to understand that natural logs are different from algorithms. When you take any logarithm, it is the opposite of its power. Therefore, the main difference between logarithms and natural logs is the base you apply. Logarithms always use a base 10 but natural logs take a base of e. But you can also convert each to the other using the following equations:
• log10(x) = ln(x) / ln(10)
• ln(x) = log10(x) / log10(e)
To demonstrate the differences even more effectively, we will list the rules of logarithms and rules of natural logs in a table. Have a look:
In Rules Logarithm rules ln(xy)= ln(x)+ln(y) ln(x/y)=ln(x)−ln(y) ln(xa)= aln(x) ln(ex)= x eln(x) =x log(xy)=log(x)+log(y) log(x/y)=log(x)−log(y) log(xa)= alog(x) log(10x)= x 10log(x) = x
## Summary of Rules and Properties of ln and e
The main thing that we have demonstrated in this post is that In or natural log, is the inverse of e. While you might see them difficult at first, they are not as complex after understanding the rules. We also demonstrated that the primary difference between logarithms and natural logs is the base. This can simply your ability to do math homework, especially when face with these issues and algebra rules. If not, there are other options.
## e^ln Rules Are Not That Hard
If you are finding calculations related to natural logs complex, there is no need to stress yourself. Go ahead and use professional math help. We provide the best online assignment help, so we can solve any of your math problems. Then, practice more to understand the properties of In and e and associated problems. With time, you will understand and natural logs will be fun! | 1,810 | 7,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | longest | en | 0.950367 |
http://sepwww.stanford.edu/data/media/public/sep/prof/iei/dspr/paper_html/node28.html | 1,508,424,226,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00355.warc.gz | 303,994,020 | 2,870 | Next: Spatial aliasing various space Up: Dispersion relations Previous: Extrapolation equations are not
# FREQUENCY DISPERSION IN WAVE-MIGRATION
Frequency dispersion results from different frequencies propagating at different speeds. The physical phenomenon of frequency dispersion is rarely heard in daily life, although many readers may have heard it while ice skating on lakes and rivers. Elastic waves caused by cracking ice propagate dispersively, causing pops to change into percussive notes. Frequency dispersion is generally observable on seismic waves that propagate along the earth's surface but frequency dispersion is hardly ever perceptible on internally reflected waves. In seismic data processing, frequency dispersion is a nuisance and an embarrassment to process designers. It arises mainly with the finite differencing method because differential operators and difference operators do not coincide at high frequencies. Frequency dispersion can always be suppressed by sampling more densely, and it is the job of the production analyst to see that this is done. Figure 17 depicts some dispersed pulses.
freqdisp Figure 17 (a) A pulse. (b) A pulse slightly dispersed as by the physical dissipation of high frequencies. (c) A pulse with a substantial amount of frequency dispersion, such as could result from careless data processing.
Frequency dispersion caused by data processing can be a useful warning that the data is in danger of being aliased. Frequency-domain methods do not depend on difference operators, so they have the advantage of not showing dispersion. The penalties that go along with this advantage are (1) limitation to constant material properties, (2) wraparound, and (3) the occurrence of spatial aliasing without the warning of dispersion.
taner
Figure 18
Conquering frequency dispersion. (Taner and Koehler, distributed by Seiscom Delta Inc.)
Figure 18 shows an example of frequency dispersion in migrated data. At the top of the figure is a CDP stack. In the middle is the data after processing with no attempt to control frequency dispersion. The worst dispersion is near shot point 200 at 4 seconds. The bottom shows the data after reprocessing with greater attention to dispersion.
Next: Spatial aliasing various space Up: Dispersion relations Previous: Extrapolation equations are not
Stanford Exploration Project
10/31/1997 | 457 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.926085 |
http://www.go4expert.com/contests/next-number-18-aug-2009-t19083/ | 1,480,883,261,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541396.20/warc/CC-MAIN-20161202170901-00334-ip-10-31-129-80.ec2.internal.warc.gz | 479,598,198 | 14,272 | Next Number 18 Aug 2009
Discussion in '\$1 Daily Competition' started by Kshiteej, Aug 18, 2009.
1. KshiteejNew Member
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What is the next number in the series
1,31,253,991,2731, -
2. Seema786New Member
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5473
3. Seema786New Member
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1,31,253,991,2731
1,222,738,1740,2742
516,1002
so now,
1740+991=2731
so
2731+2742=5473
4. mayjuneNew Member
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i didn't get it....
5. nimeshNew Member
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1,31,253,991,2731
1,222,738,1740,2742
These are the difference of the numbers in above series.
So, the first one should be 30.
And how did you get 2742?
516,1002
so now,
1740+991=2731
so
2731+2742=5473
6. Seema786New Member
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yes nimes, the first number should be 30, dat was my mistake.
2742 i got like dis....
1740+1002=2742
7. xpi0t0sMentor
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But what is so special about the other series
9. xpi0t0sMentor
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Lets see if after 24 hours what the OP replies to
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Kshiteej, Its more than 24 hours and so please update the question and tell us who is the winner.
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Ans is 6121.
31-1 = 30 = (3)^3 +3
253-31 = 222 = (6)^3 + 6
991 -253 = 738 = (9)^3 + 9
2731 - 991 =1740 =(12)^3 + 12
so te ans will be 2731 + (15)^3 + 15 =6121
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So Congrats xpi0t0s
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very nice one kshiteej...
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Thanks Mayjune and congrats xpi0t0s.
By the way what was your logic while calculating 6121? The same as of mine??
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nice question kshiteej
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Congrs xpi0t0s
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3,012 | 1,089 | 3,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-50 | latest | en | 0.847106 |
http://www.virtuescience.com/541.html | 1,500,562,020,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423222.65/warc/CC-MAIN-20170720141821-20170720161821-00108.warc.gz | 591,088,317 | 4,915 | Articles/Resources Written and Organized by James Barton
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<540 Number Data-Base Random Number 542>
## The Number 541: Properties and Meanings
541 is a Prime Number.
541 is a Centered 12-gonal Number.
541 is a Centered 15-gonal Number.
541 is a Centered 36-gonal Number.
541 is the number of orderings of 5 objects with ties allowed. | 111 | 443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | latest | en | 0.816597 |
https://physics.stackexchange.com/questions/62637/the-potential-and-the-intensity-of-the-gravitational-field-in-the-axis-of-a-circ/62917 | 1,585,948,412,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518622.65/warc/CC-MAIN-20200403190006-20200403220006-00090.warc.gz | 626,762,954 | 33,813 | # The potential and the intensity of the gravitational field in the axis of a circular plate
Calculate the potential and the intensity of the gravitational field at a distance $x> 0$ in the axis of thin homogeneous circular plate of radius $a$ and mass $M$.
Could anybody describe how to calculate this? Slowly and in detail. I'm helpless.
Answer is: potencial $\phi = - \frac {2 \kappa M}{a^2}(\sqrt{a^2+x^2}-x)$ and intesity $K = \frac {2 \kappa M}{a^2} \left( \frac{x}{\sqrt{a^2+x^2}}-1 \right)$
• If anybody finds this question incomprehensible, please, leave a comment, I will try to rewrite it. I'm not a native speaker, so I might do some mistakes. – user50222 Apr 29 '13 at 9:32
• first find out for a ring. then add for infinite rings that form a disc using integration. – ABC Apr 29 '13 at 17:56
• How to do that? I can't find any example. – user50222 Apr 29 '13 at 20:58
First let us calculate the potential for a ring of radius $a$ at a distance $x$ from the center along the axis
Potential due to an infinitesimal mass element $dm$ will be $$\frac{-Gdm}{\sqrt{a^2+x^2}}$$
Potential due to the ring is then $$\int{\frac{-Gdm}{\sqrt{a^2+x^2}}}=\frac{-G}{\sqrt{a^2+x^2}}\int{dm}=\frac{-Gm}{\sqrt{a^2+x^2}}$$
Since $G, a, x$ are constant
Now let us break the disc into infinitesimal rings of mass $dm=2\pi rdr\frac{M}{\pi a^2} (=area * density)$
The potential due to a ring of radius $r$ and mass $dm$ as given above is $$\frac{-Gdm}{\sqrt{r^2+x^2}}=\frac{-2GMrdr}{a^2\sqrt{r^2+x^2}}$$
Integrating this from $0$ to $a$ $$\int{\frac{-2GMrdr}{a^2\sqrt{r^2+x^2}}}$$ $$=\frac{-GM}{a^2}\int{\frac{2rdr}{\sqrt{r^2+x^2}}}$$ putting $t^2=r^2+a^2$ and $2rdr=2tdt$ $$=\frac{-GM}{a^2}\int{\frac{2tdt}{\sqrt{t^2}}}$$ $$=\frac{-GM}{a^2}[2t]^{\sqrt{a^2+x^2}}_{x}$$ $$=\frac{-2GM}{a^2}({\sqrt{a^2+x^2}}-{x})$$
For intensity, it can be seen by symmetry that it is along the axis thus we work only with axial components
So, for ring $$\int\frac{-Gdmcos\theta}{a^2+x^2}$$ where $\theta$ is half of the angle subtented by the point on the ring $$cos\theta=\frac{x}{\sqrt{a^2+x^2}}$$
$$K=\int\frac{-Gxdm}{(a^2+x^2)^{3/2}}=\frac{-Gxm}{(a^2+x^2)^{3/2}}$$
For a disc, based on the same reasoning as in potential, it is
$$K=\int\frac{-Gxdm}{(r^2+x^2)^{3/2}}$$ $$=\int\frac{-2GMxrdr}{a^2(r^2+x^2)^{3/2}}$$ $$=\int\frac{-2GMxtdt}{a^2(t^2)^{3/2}}$$ $$=\int\frac{-2GMxdt}{a^2t^2}$$ $$=\frac{-2GMx}{a^2}[\frac{-1}{t}]^{\sqrt{a^2+x^2}}_x$$
$$K=\frac {2 G M}{a^2} \left( \frac{x}{\sqrt{a^2+x^2}}-1 \right)$$
Let's break the disc into small rings,
Here the mass in the disc is at a same distance $$\sqrt{x^2+r^2}$$ from the axis point A.
So, potential $$\mathrm dE=-\frac{G\,\mathrm dm}{\sqrt{x^2+r^2}}$$
Where $$\mathrm dm$$= mass of the ring. So, $$\mathrm dm=2\pi r\,\mathrm dr \times \frac M{\pi R^2}$$
Also $$r=x\tan\phi$$, so $$\mathrm dr=x\sec^2\phi\,\mathrm d\phi$$
So, the potential is $$\mathrm dP=-\frac{GM2\pi x\tan\phi x\sec^2\phi\,\mathrm d\phi}{\left(x\sqrt{1+\tan^2\phi}\right)\times\pi R^2}=-\frac{2 GMx}{R^2}\times\tan\phi\sec\phi\,\mathrm d\phi$$
So, $$P=-\dfrac{2GMx}{R^2}\times\int\limits_0^{\tan^{-1}R/x} \tan\phi\sec\phi\,\mathrm d\phi$$ You can integrate it on your own.
You can proceed on similar basis for electric field. But before integrating you'll have to take components of field along axis and perpendicular because it's a vector and cannot be added directly.
You'll get field integrand as $$\mathrm dE_x=\frac{2GM}{R^2}\times \int\limits_0^{\tan^{-1}R/x}\frac{\tan\phi\sec\phi}{\sec^2\phi}\,\mathrm d\phi$$
Whereas along $$Y$$ you'll get no field by symmetric argument. | 1,372 | 3,606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-16 | latest | en | 0.767084 |
https://oeis.org/A108406 | 1,638,038,171,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00601.warc.gz | 522,157,900 | 3,736 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A108406 Numbers n such that concatenating n and the sum of the digits of n raised to their own power (A045503) produces a square. 0
0, 211, 220, 235, 20403, 111416, 1011231, 3444142, 10003400, 22303600, 31151021, 53231032, 121542025, 126423126, 202032110, 243425212, 302434003, 311544033, 324231521, 334130241, 375607602, 406221650, 561620561, 662033363, 1053045074 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS EXAMPLE 235 is a term because 2^2 + 3^3 + 5^5 = 3156 and 2353156 = 1534^2. CROSSREFS Cf. A108302. Sequence in context: A064826 A333012 A136601 * A051268 A159911 A319141 Adjacent sequences: A108403 A108404 A108405 * A108407 A108408 A108409 KEYWORD base,nonn AUTHOR Jason Earls, Jul 04 2005 EXTENSIONS More terms from Ryan Propper, Jul 07 2005 STATUS approved
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Last modified November 27 12:51 EST 2021. Contains 349394 sequences. (Running on oeis4.) | 429 | 1,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.68362 |
https://cracku.in/153-7-33-137-553-8873-x-sbi-clerk-2012-3 | 1,726,203,123,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00270.warc.gz | 163,858,645 | 26,128 | Instructions
What should come in place of the question mark (?) in the following number series ?
Question 153
# 7, 33, 137, 553, ?, 8873
Solution
The pattern here followed is :
7 $$\times$$ 4 + 5 = 33
33 $$\times$$ 4 + 5 = 137
137 $$\times$$ 4 + 5 = 553
553 $$\times$$ 4 + 5 = 2217
2217 $$\times$$ 4 + 5 = 8873 | 126 | 320 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.528519 |
https://www.netexplanations.com/new-learning-composite-mathematics-class-6-sk-gupta-anubhuti-gangal-algebra-chapter-7g-solution/ | 1,702,071,499,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00248.warc.gz | 988,336,156 | 17,367 | # New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Algebra Chapter 7G Solution
## New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Algebra Chapter 7G Solution
(1) (a) 3a + 2 = 11
Or, 3a = 11 – 2
Or, 3a = 9
Or, a = 9/3
∴ a = 3
(b) 5P – 1 = 9
Or, 5P = 9 + 1
Or, 5P = 10
Or, P = 10/5
∴ P = 2
(c) 45 = – 7d – 4
Or, – 7d = 45 + 4
Or, – 7d = 49
Or, d = 49/-7
∴ d = – 7
(d) x/2 +10 = 13
Or, x/2 = 13 – 10
Or, x/2 = 3
Or, x = 3X2
∴ x = 6
(e) 9 (x – 3) = 54
Or, 9x – 27 = 54
Or, 9x = 54 + 27
Or, 9x = 81
Or, x = 81/9
∴ x = 9
(f) 17 = 3 (p – 5) + 8
Or, 17 = 3P – 15 + 8
Or, 3P = 17 + 15 – 8
Or, 3P = 24
∴ P = 24/3
= 8
(2) (a) 28 – f = 15
Or, – f = 15 – 28
Or, – f = – 13
∴ f = 13 (Ans)
(b) Let, the no. be x
(x X 7) – 18 = 31
Or, 7x – 18 = 31
Or, 7x = 31 + 18
Or, 7x = 49
Or, x = 49/7
∴ x = 7 (Ans)
Updated: May 31, 2021 — 11:47 pm | 513 | 904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-50 | latest | en | 0.674619 |
https://artofproblemsolving.com/wiki/index.php/2001_IMO_Shortlist_Problems/G4 | 1,723,405,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00292.warc.gz | 87,385,090 | 10,783 | # 2001 IMO Shortlist Problems/G4
## Problem
Let $M$ be a point in the interior of triangle $ABC$. Let $A'$ lie on $BC$ with $MA'$ perpendicular to $BC$. Define $B'$ on $CA$ and $C'$ on $AB$ similarly. Define
$p(M) = \frac {MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC}.$
Determine, with proof, the location of $M$ such that $p(M)$ is maximal. Let $\mu(ABC)$ denote this maximum value. For which triangles $ABC$ is the value of $\mu(ABC)$ maximal? | 145 | 451 | {"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": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.712601 |
https://wiki.tcl-lang.org/page/Babylonian+Field+Expansion+Procedure+Algorithm+and+example+demo+eTCL+calculator%2C+numerical+analysis | 1,726,429,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00138.warc.gz | 567,034,309 | 15,981 | ## Babylonian Field Expansion Procedure Algorithm and eTCL demo example calculator, numerical analysis
This page is under development. Comments are welcome, but please load any comments in the comments section at the bottom of the page. Please include your wiki MONIKER and date in your comment with the same courtesy that I will give you. Aside from your courtesy, your wiki MONIKER and date as a signature and minimal good faith of any internet post are the rules of this TCL-WIKI. Its very hard to reply reasonably without some background of the correspondent on his WIKI bio page. Thanks, gold 12Dec2018
### Preface
gold Here are some TCL calculations for Babylonian Field Expansion Procedure Algorithm in calculator shell. Additional console program below is used to check or improve subroutine.
### Introduction
The Babylonian field expansion procedure algorithm from clay tablets was loaded into an eTCL calculator. The Babylonians did not use algebra notation, so the reader will have to bear some anachronisms in the eTCL pseudocode. The field expansion procedure is defined as the method used by scribes to successively reduce or add numbers or number products by integer fractions, usually 1/60, 2/60, 3/60 et al. In modern notation and definitions, the numbers in base 60 that appeared on the tablet would be n+(1/60), n+(2/60), n+(3/60) until a desired goal for is reached. On some clay tablets to reach a desired field area of 100 square units called a iku (Akkadian for field), the successive altered sides a*b would be (a+(1/60))*(b+(1/60)),(a2+(1/60))*(b2+(1/60))...until (a_f+(1/60))*(n_f+(1/60)) ~= 100. Allowing for some number rounding, the field expansion procedure would produce an almost square field of 100 square units. The inherent error should be less than 1/60, error <= (/ 1. 60.). The field expansion procedure can be extended to produce rectangular fields of certain desired areas or side a/b ratios. The bare numbers on the tablets were not usually marked with units or annotated. Successive or iterated math solutions are called algorithms and the field expansion procedure is one of the earliest algorithms documented. The TCL procedures are descendants of this idea. For restating the problem in a computer algorithm, the sides and field area will be in meters and square meters, respectively.
### Field Expansion Procedure
The reason for the field expansion procedure is not completely understood. However, there are some peg points or analogies available. The early examples of the field expansion procedure are from clay tablets of the proto-literate mathematical texts around 3000 BCE and metro-mathematical school texts from the Early Sumerian Dynastic III Period around 2500 BCE, ref Friberg. The successive addition and clay marks/counterpieces look somewhat like the pebble math problems, pebble games, or penny problems of some later cultures. Possibly, the concern with sides of 10 units and area of 100 square units may be a separate or earlier tradition other than base sixty. For example, one would think that side of 12 units and area of 144 square units would track better with base 60. The almost square fields or concern with squares in some problems suggest the field expansion procedure was a method that avoided square roots or predated taking square area of rectangles. Although the field expansion texts would have some utility in a scribal school setting, there is an implied interest in demonstrating that the dimensions of a particular field give an accurate area or regular area units in a land survey setting. More a conjecture, but some contend that the yield or harvest of a field would be quantity times area unit, meaning that evaluating the field in terms of standard area units would lead to better harvest supervision and better accounting to the temple management.
The multiplication term 1/60 was also found in much later cuneiform tax records as the interest rate for one (lunar) month. Possible yearly tax rate was 12*(1/60) or 12/60 for a solar year or 18*(1/60) or 18/60 for a lunar calendar. The analysis is not suggesting the field expansion procedure was calculating taxes, but it is difficult to rule out anything just looking at bare numbers without text explanation.
In the later eras, the cuneiform methods for taking areas of squares and rectangles were accurate, a*a or a*b. In those eras, the common approximate quadrilateral formula for area (a+c)*(b+d)*.25 was very inaccurate, inconsistent, and lead to unfair tax assessments. Taxes or interest were 1/60 per month or 12/60 per year of the harvest in some eras, so an accurate survey would almost pay the years taxes. There probably was some motivation to extend an accurate confirmed square area into an accurate rectangular field vis using the inconsistent approximate quadrilateral formula. The Babylonian false position algorithm on this wiki does take a square root and may be a later development.
Later reporting by Friberg indicates that equation system for a rectangle of a certain side ratio is couple of 1) linear and 2) quadratic equations. Hence the field expansion procedure in transforming a square into a rectangle with sides of fixed ratio is solving a very simple quadratic equation.
### TCL Code
Depending on the settings for the correction fraction, the field expansion procedure algorithm can generate crude square roots. The accuracy of the field expansion procedure is dependent on the fineness of the intervals, which in the notation of Babylonian base 60 fractions would be 1/60/ 1/3600, or 1/216000. For square sides of 1.0001/1, desired area of 2, and correction fraction (1/60), the square root solution of 2 was 1.4, leaving off incorrect digits. For square sides of 1.0001/1, desired area of 2, and correction fraction (1/3600), the square root solution of 2 was 1.414, leaving off incorrect digits. Not sure the Babylonians and Sumerians were aware of using the field expansion procedure for obtaining square roots, but the use in the eTCL code does show the power of the field expansion procedure. At least, the Babylonians in later eras had different, faster methods, and fewer hand calculations to obtain square roots.
The first eTCL code was based on the tablet calculations and was working when solution approaches the goal from below the desired number. However, the eTCL calculator solution needs to approach correctly from either below or above the desired number goal. This routine is sort of like a double barreled Newton's method from Square Root, especially proc SqrtB. Another possible improvement for the field expansion procedure is having the slice rates or intervals proportional to the distance from the goal. In the eTCL code, distance from the goal or delta from the goal is measured by <- \$goal <* \$side1 \$side2 1. > >. For example, if delta is less than 1/4, set the slice interval at (1/3600). If delta is greater than 1/4, set the slice interval at (1/60). Fewer computations at the higher slice rates will speed up computations and reduce wait time. Some of the Babylonian problems appeared to freeze the width of a calculated square and use a constant side a/b ratio to morph into a rectangle. Although probably not classic Field Expansion procedures, there are Babylonian math problems that increase the height of a grain bin by 1/60 or some other fraction. Meaning that modifying the Field Expansion procedure from the area a*b calculations to 3 dimensions of a*b*c or cubic equations is possible.
### Conclusions
The Babylonian Field Expansion Procedure Algorithm continues to fascinate the linguists as the Field Expansion Procedure is the earliest known and documented algorithm. From the work of Friberg, Robson, and others, the Field Expansion Procedure in the cuneiform tablets of the Archaic Period actually predates the classic Sumerian URIII period by 500 years and predates the era of the Babylonian mathematicians by 1400 years. There is some evidence that the Field Expansion Procedure in the available Archaic texts predates the scribal establishment and scribal accounting of the base60 notation and even possibly the full entry of Sumerian language into Mesopotamia. For example, the area unit for iku of 100 square nindians would seem more useful to base10 calculations. Why not an iku of 60, 120, or 360 units as more useful for a base60?
## Rectangular Test Cases
Testcase 4. A rectangular field problem can be set up as follows, using modern notation. The rectangular field will have length to width ratio of 3/2 and an area of 100 square rods, rod = 6 meters. The modern notation is (1*x)*(2/3)*x=100 square rods, x*x=(3/2)*100, x=sqrt(150),x=12.24, length ~~12 rods. For the width, w=12*(2/3), 8 rods. From the easy eye calculator, the original area is <expr 12*8 > or 96 square rods. The original area is short of the goal as <expr 100-94> or 6 square rods. The eTCL calculator is loaded as 12/8/100 and the field expansion algorithm returns length/width/area as 12.19/8.2/100.039.
Testcase 5. A rectangular field problem can be set up as follows, using modern notation. The rectangular field will have length to width ratio of 3/2 and an area of 1800 square rods, rod = 6 meters. The modern notation is (1*x)*(2/3)*x=1800 square rods, x*x=(3/2)*1800, x=sqrt(2700),x=51.961, length ~~50 rods. For the width, w=50*(2/3), 33 rods. From the easy eye calculator, the original area is <expr 50*33 > or 1650 square rods. The original area is short of the goal as <expr 1800-1650> or 150 square rods. The eTCL is loaded as 50/33/1800 and the field expansion algorithm returns length/width/area as 50/33/1719. A second run on the eTCL calculator is loaded as 51/33/1800 and returns 51.83/33.83/1753 square units.
Testcase 6. A rectangular field problem can be set up as follows, using modern notation. The rectangular field will have length to width ratio of 3/2 and an area of 11700 square rods, rod = 6 meters. The modern notation is (1*x)*(2/3)*x=11700 square rods, x*x=(3/2)*11700, x=sqrt(17550),x=132.476, length ~~132 rods. For the width, w=132*(2/3), 88 rods. From the easy eye calculator, the original area is <expr 132*88 > or 11616 square rods. The original area is short of the goal as <expr 11700-11616> or 84 square rods. The eTCL calculator is loaded as 132/88/11700 and the field expansion algorithm returns length/width/area as 132.38/88.38/11700.48.
## Pseudocode Section
``` # using pseudocode for Babylonian field expansion procedure algorithm.
# possible problem instances include add 1/60 to sides until area goal reached
long_side = supplied value
short_side = supplied value
desired_goal = supplied value
# desired_goal usually 100 square units in some early math problems
set old_field_area = a*b , old field_area = long_side * short_side
set new_side_a = long_side + 1/60
set new_side_b = short_side + 1/60
set new_field_area = (long_side + 1/60) * ( short_side + 1/60 )
is new_field_area =? desired_area within +/- (1/60) , yes = finished loop
check error , abs (desired_goal - new_field_area) <= [/ 1. 60.]
half area = area * .5
quarter area = area * .25
check_answer new area =? desired goal , desired goal reached (yes/no)
set answers and printout with resulting values```
### Testcases Section
In planning any software, it is advisable to gather a number of testcases to check the results of the program. The math for the testcases can be checked by pasting statements in the TCL console. Aside from the TCL calculator display, when one presses the report button on the calculator, one will have console show access to the capacity functions (subroutines).
#### Testcase 1 , almost square field
table 1printed in tcl wiki format
quantity value comment, if any
1:testcase_number
10.2 :long side meters
9.5 :short side meters
100.0 :desired area (usually 1 iku, 100 square units)
96.899 :answers: old area meters squared
0.166 :area lacking meters squared
100.211 :area from new sides meters squared
10.366 :long side meters
9.666 :short side meters
#### Testcase 2 , almost square field
table 2printed in tcl wiki format
quantity value comment, if any
2:testcase_number
9.599 :long side meters
9.060 :short side meters
100.0 :desired area (usually 1 iku, 100 square units)
86.975 :answers: old area meters squared
0.683 :area lacking meters squared
100.193 :area from new sides meters squared
10.283 :long side meters
9.743 :short side meters
#### Testcase 3 , almost square field
table 3printed in tcl wiki format
quantity value comment, if any
3:testcase_number
9.5 :long side meters
9.0 :short side meters
100.0 :desired area (usually 1 iku, 100 square units)
85.5 :answers: old area meters squared
0.7666 :area lacking meters squared
100.271 :area from new sides meters squared
10.266 :long side meters
9.766 :short side meters
#### Testcase 4 , rectangular field
table 4printed in tcl wiki format
quantity value comment, if any
2:testcase_number
12.0 :long side rods
8.0 :short side rods
100.0 :desired area (usually 1 iku, 100 square units)
96.0 :answers: old area rods squared
0.200 :area lacking rods squared
100.0399 :area from new sides rods squared
12.199 :long side rods
8.199 :short side rods
#### Testcase 5 , rectangular field
table 5printed in tcl wiki format
quantity value comment, if any
5:testcase_number
50.0 :long side rods
33.0 :short side rods
1800.0 :desired area (usually 1 iku, 100 square units)
1650.0 :answers: old area rods squared
0.833 :area lacking rods squared
1719.861 :area from new sides rods squared
50.833 :long side rods
33.833 :short side rods
#### Testcase 6 , rectangular field
table 6printed in tcl wiki format
quantity value comment, if any
6:testcase_number
132.0 :long side rods
88.0 :short side rods
11700.0 :desired area (usually 1 iku, 100 square units)
11616.0 :answers: old area rods squared
0.383 :area lacking rods squared
11700.480 :area from new sides rods squared
132.383 :long side rods
88.383 :short side rods
## Appendix Code
### appendix TCL programs and scripts
``` # pretty print from autoindent and ased editor
# Babylonian Field Expansion Procedure Algorithm Calculator V2
# written on Windows XP on eTCL
# working under TCL version 8.6
# gold on TCL Club, 12dec2018
package require Tk
package require math::numtheory
namespace path {::tcl::mathop ::tcl::mathfunc math::numtheory }
set tcl_precision 17
frame .frame -relief flat -bg aquamarine4
pack .frame -side top -fill y -anchor center
set names {{} { long side meters :} }
lappend names { short side meters :}
lappend names { desired area (usually round units, 1 iku, 100 square units) : }
lappend names { answers: old area meters squared : }
lappend names { area lacking meters squared :}
lappend names { area from new sides meters squared: }
lappend names { long side meters : }
lappend names { short side meters :}
foreach i {1 2 3 4 5 6 7 8} {
label .frame.label\$i -text [lindex \$names \$i] -anchor e
entry .frame.entry\$i -width 35 -textvariable side\$i
set msg "Calculator for Babylonian Field Expansion Procedure Algorithm V2
from TCL,
# gold on TCL Club, 12dec2018 "
tk_messageBox -title "About" -message \$msg }
proc self_help {} {
set msg " Babylonian Field Expansion P. Algorithm V2
from TCL ,
# self help listing
# problem, Babylonian Field Expansion P. Algorithm V2
# 3 givens follow.
1) long side meters:
2) short side meters:
3) desired area (usually round units, 1 iku, 100 square units):
# Recommended procedure is push testcase
# and fill frame,
# change first three entries etc, push solve,
# and then push report.
# Report allows copy and paste
# from console to conventional texteditor.
# For testcases, testcase number is internal
# to the calculator and will not be printed
# until the report button is pushed
# for the current result numbers.
# This posting, screenshots, and TCL source code is
# Editorial rights and disclaimers
# retained under the TCL/TK license terms
# and will be defended as necessary in court.
Conventional text editor formulas
or formulas grabbed from internet
screens can be pasted into green console.
# gold on TCL Club, 12Dec2018 "
tk_messageBox -title "Self_Help" -message \$msg }
proc calculate { } {
global side1 side2 side3 side4 side5
global side6 side7 side8
global testcase_number
incr testcase_number
set side1 [* \$side1 1. ]
set side2 [* \$side2 1. ]
set side3 [* \$side3 1. ]
set side4 [* \$side4 1. ]
set side5 [* \$side5 1. ]
set side6 [* \$side6 1. ]
set side7 [* \$side7 1. ]
set side8 [* \$side8 1. ]
set true_area [* \$side1 \$side2 ]
set desired_area \$side3
# initialize ancient correction fraction
set correction_fraction [/ 1. 60. ]
set correction_fraction [* [/ [- \$desired_area \$true_area] \$desired_area ] 5. ]
set counter 1
while { \$counter < 50. } {
if { [* [+ \$side1 [/ \$counter 60. ]] [+ \$side2 [/ \$counter 60. ]] ] > \$desired_area } {; break}
incr counter
}
set correction_fraction [/ \$counter 60. ]
set side4 \$true_area
set side5 \$correction_fraction
set side6 [* [+ \$side1 \$correction_fraction ] [+ \$side2 \$correction_fraction ] ]
set side7 [+ \$side1 \$correction_fraction ]
set side8 [+ \$side2 \$correction_fraction ]
}
proc fillup {aa bb cc dd ee ff gg hh} {
.frame.entry1 insert 0 "\$aa"
.frame.entry2 insert 0 "\$bb"
.frame.entry3 insert 0 "\$cc"
.frame.entry4 insert 0 "\$dd"
.frame.entry5 insert 0 "\$ee"
.frame.entry6 insert 0 "\$ff"
.frame.entry7 insert 0 "\$gg"
.frame.entry8 insert 0 "\$hh"
}
proc clearx {} {
foreach i {1 2 3 4 5 6 7 8 } {
.frame.entry\$i delete 0 end } }
proc reportx {} {
global side1 side2 side3 side4 side5
global side6 side7 side8
global testcase_number reference_factor flag
console eval {.console config -bg palegreen}
console eval {.console config -font {fixed 20 bold}}
console eval {wm geometry . 40x20}
console eval {wm title . " Babylonian Field Expansion Report, screen grab and paste from console 2 to texteditor"}
console eval {. configure -background orange -highlightcolor brown -relief raised -border 30}
console show;
puts "%|table \$testcase_number|printed in| tcl wiki format|% "
puts "&| quantity| value| comment, if any|& "
puts "&| \$testcase_number:|testcase_number | |& "
puts "&| \$side1 :|long side meters | |&"
puts "&| \$side2 :|short side meters | |& "
puts "&| \$side3 :|desired area (usually round units, 1 iku, 100 square units)| |& "
puts "&| \$side4 :|answers: old area meters squared| |&"
puts "&| \$side5 :|area lacking meters squared | |&"
puts "&| \$side6 :|area from new sides meters squared | |&"
puts "&| \$side7 :|long side meters | |&"
puts "&| \$side8 :|short side meters | |&"
}
frame .buttons -bg aquamarine4
::ttk::button .calculator -text "Solve" -command { calculate }
::ttk::button .test2 -text "Testcase1" -command {clearx;fillup 10.2 9.5 100.0 96.9 0.0166 100.2 10.2 9.51}
::ttk::button .test3 -text "Testcase2" -command {clearx;fillup 9.6 9.06 100. 86.9 .68 100.2 10.2 9.74 }
::ttk::button .test4 -text "Testcase3" -command {clearx;fillup 9.5 9.0 100.0 85.5 0.76 100.3 10.26 9.7 }
::ttk::button .clearallx -text clear -command {clearx }
::ttk::button .self_help -text self_help -command { self_help }
::ttk::button .cons -text report -command { reportx }
::ttk::button .exit -text exit -command {exit}
pack .clearallx .cons .self_help .about .exit .test4 .test3 .test2 -side bottom -in .buttons
grid .frame .buttons -sticky ns -pady {0 10}
. configure -background aquamarine4 -highlightcolor brown -relief raised -border 30
wm title . "Babylonian Field Expansion Procedure Algorithm Calculator V2" ```
### Pushbutton Operation
For the push buttons, the recommended procedure is push testcase and fill frame, change first three entries etc, push solve, and then push report. Report allows copy and paste from console.
For testcases in a computer session, the eTCL calculator increments a new testcase number internally, eg. TC(1), TC(2) , TC(3) , TC(N). The testcase number is internal to the calculator and will not be printed until the report button is pushed for the current result numbers. The current result numbers will be cleared on the next solve button. The command { calculate; reportx } or { calculate ; reportx; clearx } can be added or changed to report automatically. Another wrinkle would be to print out the current text, delimiters, and numbers in a TCL wiki style table as
``` puts " %| testcase \$testcase_number | value| units |comment |%"
puts " &| volume| \$volume| cubic meters |based on length \$side1 and width \$side2 |&" ```
## Console program under test.
### from ASK12, need help on Field Expansion subroutine
gold - 2017-01-25 On the this tcl wiki, the code in Field Expansion calculator is working when solution approachs the answer from below the desired number, but the solution needs to approach correctly from either below or above the desired number. This routine is sort of like a double barreled newton's method from Square Root, especially proc SqrtB {num} {# Newton's method}. Wrote a test console program for a field_expansion_procedure at bottom of wiki page. Can someone help or load corrected code for the console program at bottom of page. I am drawing a mental blank. Babylonian Field Expansion Procedure Algorithm and example demo eTCL calculator, numerical analysis
``` # written on Windows 10
# working under TCL version 8.6
# TCL WIKI , 28jan2017
# console program written on Windows 10
console show
package require math::numtheory
namespace path {::tcl::mathop ::tcl::mathfunc math::numtheory }
set tcl_precision 17
proc field_expansion_procedure_function2 { side1 side2 side3 side4 epsilon } {
set counter 1
set token1 \$side1
set token2 \$side2
set saver1 .00001
set saver2 .00001
set epsilon [/ 1. \$side4]
while { \$counter < 50. } {
if { [abs [- \$side3 [* \$token1 \$token2 1. ] ] ] < \$epsilon } {break;}
if { [- \$side3 [* \$token1 \$token2 1. ] ] > 0 } {set correction_fraction [* 1. [/ 1. \$side4] ]}
if { [- \$side3 [* \$token1 \$token2 1. ] ] < 0 } {set correction_fraction [* -1. [/ 1. \$side4] ]}
#set correction_fraction [- \$side3 [* \$token1 \$token2 1. ] ]
set token1 [+ \$token1 \$correction_fraction ]
set token2 [+ \$token2 \$correction_fraction ]
incr counter
puts "token \$token1 token \$token2 product [* \$token1 \$token2 ] correction \$correction_fraction"
} }
set side8 [ field_expansion_procedure_function2 9.5 9.4 100. 60. .15 ]```
## printout
```token 9.9666666666666899 token 9.8666666666666902 product 98.337777777778243 correction 0.016666666666666666
token 9.9833333333333574 token 9.8833333333333577 product 98.668611111111588 correction 0.016666666666666666
token 10.000000000000025 token 9.9000000000000252 product 99.000000000000497 correction 0.016666666666666666
token 10.016666666666692 token 9.9166666666666927 product 99.331944444444957 correction 0.016666666666666666
token 10.03333333333336 token 9.9333333333333602 product 99.664444444444982 correction 0.016666666666666666
token 10.050000000000027 token 9.9500000000000277 product 99.997500000000556 correction 0.016666666666666666```
## Crude square root of 2, to three places
```token 1.4128777777777797 token 1.4127777777777797 product 1.9960823271604993 correction 0.00027777777777777778
token 1.4131555555555575 token 1.4130555555555575 product 1.9968673086419808 correction 0.00027777777777777778
token 1.4134333333333353 token 1.4133333333333353 product 1.9976524444444501 correction 0.00027777777777777778
token 1.4137111111111131 token 1.4136111111111131 product 1.998437734567907 correction 0.00027777777777777778
token 1.413988888888891 token 1.413888888888891 product 1.9992231790123516 correction 0.00027777777777777778
token 1.4142666666666688 token 1.4141666666666688 product 2.0000087777777837 correction 0.00027777777777777778```
## Crude Cube Root of 1000 , to three places
Approaching a cube root using the Babylonian Field Expansion Procedure Algorithm , 7/2/2020
Q. What is the cube root of 1000. ?
Initial guess N1 of 9.63 eval * 9.63 9.63 9.63 893.056347000000
initial error of guess eval - 10. 9.63 0.369999999999999
line increment eval / 1. 60. 0.0166666666666666
check on line N3 eval + 9.63333333 0.01666 9.64999333 ~~ approximate token 9.650000
product on N3 eval * 9.6500000 9.6500000 9.6500000 898.632125
error of N3 guess eval - 10. 9.6500000 0.349999999999
token1 9.6333333333333346 token2 9.6333333333333346 token3 9.6333333333333346 product 893.98403703703741 correction 0.016666666666666666
token1 9.6500000000000021 token2 9.6500000000000021 token3 9.6500000000000021 product 898.63212500000066 correction 0.016666666666666666
token1 9.6666666666666696 token2 9.6666666666666696 token3 9.6666666666666696 product 903.2962962962971 correction 0.016666666666666666
token1 9.6833333333333371 token2 9.6833333333333371 token3 9.6833333333333371 product 907.97657870370483 correction 0.016666666666666666
token1 9.7000000000000046 token2 9.7000000000000046 token3 9.7000000000000046 product 912.67300000000125 correction 0.016666666666666666
token1 9.7166666666666721 token2 9.7166666666666721 token3 9.7166666666666721 product 917.38558796296456 correction 0.016666666666666666
token1 9.7333333333333396 token2 9.7333333333333396 token3 9.7333333333333396 product 922.11437037037206 correction 0.016666666666666666
token1 9.7500000000000071 token2 9.7500000000000071 token3 9.7500000000000071 product 926.85937500000205 correction 0.016666666666666666
token1 9.7666666666666746 token2 9.7666666666666746 token3 9.7666666666666746 product 931.62062962963182 correction 0.016666666666666666
token1 9.7833333333333421 token2 9.7833333333333421 token3 9.7833333333333421 product 936.39816203703947 correction 0.016666666666666666
token1 9.8000000000000096 token2 9.8000000000000096 token3 9.8000000000000096 product 941.19200000000285 correction 0.016666666666666666
token1 9.8166666666666771 token2 9.8166666666666771 token3 9.8166666666666771 product 946.00217129629937 correction 0.016666666666666666
token1 9.8333333333333446 token2 9.8333333333333446 token3 9.8333333333333446 product 950.82870370370699 correction 0.016666666666666666
token1 9.8500000000000121 token2 9.8500000000000121 token3 9.8500000000000121 product 955.67162500000347 correction 0.016666666666666666
token1 9.8666666666666796 token2 9.8666666666666796 token3 9.8666666666666796 product 960.53096296296667 correction 0.016666666666666666
token1 9.8833333333333471 token2 9.8833333333333471 token3 9.8833333333333471 product 965.40674537037432 correction 0.016666666666666666
token1 9.9000000000000146 token2 9.9000000000000146 token3 9.9000000000000146 product 970.2990000000043 correction 0.016666666666666666
token1 9.9166666666666821 token2 9.9166666666666821 token3 9.9166666666666821 product 975.20775462963422 correction 0.016666666666666666
token1 9.9333333333333496 token2 9.9333333333333496 token3 9.9333333333333496 product 980.13303703704184 correction 0.016666666666666666
token1 9.9500000000000171 token2 9.9500000000000171 token3 9.9500000000000171 product 985.07487500000502 correction 0.016666666666666666
token1 9.9666666666666845 token2 9.9666666666666845 token3 9.9666666666666845 product 990.03329629630161 correction 0.016666666666666666
token1 9.983333333333352 token2 9.983333333333352 token3 9.983333333333352 product 995.00832870370937 correction 0.016666666666666666
token1 10.00000000000002 token2 10.00000000000002 token3 10.00000000000002 product 1000.0000000000059 correction 0.016666666666666666
``` # pretty print from autoindent and ased editor
# Babylonian Field Expansion Procedure Algorithm Calculator V2
# console program written on Windows 10
# working under TCL version 8.6
# TCL WIKI , 2jul2020
console show
package require math::numtheory
namespace path {::tcl::mathop ::tcl::mathfunc math::numtheory }
set tcl_precision 17
proc field_expansion_procedure_function2 { side1 side2 side3 side4 side5 epsilon } {
set counter 1
set token1 \$side1
set token2 \$side2
set token3 \$side3
set saver1 .00001
set saver2 .00001
set saver3 .00001
set epsilon [/ 1. \$side5]
while { \$counter < 1000. } {
if { [abs [- \$side4 [* \$token1 \$token2 \$token3 1. ] ] ] < \$epsilon } {break;}
if { [- \$side4 [* \$token1 \$token2 \$token3 1. ] ] > 0 } {set correction_fraction [* 1. [/ 1. \$side5] ]}
if { [- \$side4 [* \$token1 \$token2 \$token3 1. ] ] < 0 } {set correction_fraction [* -1. [/ 1. \$side5] ]}
#set correction_fraction [- \$side3 [* \$token1 \$token2 1. ] ]
set token1 [+ \$token1 \$correction_fraction ]
set token2 [+ \$token2 \$correction_fraction ]
set token3 [+ \$token3 \$correction_fraction ]
incr counter
puts " &| token1 | \$token1 | token2 | \$token2 | token3 | \$token3 | product | [* \$token1 \$token2 \$token3 ] | correction | \$correction_fraction |& "
} }
set side8 [ field_expansion_procedure_function2 9.6 9.6 9.6 1000. 60. .15 ]
# printout follows``` | 8,251 | 28,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.906515 |
https://forum.math.toronto.edu/index.php?PHPSESSID=34kiakj68nskmbe06k47svsbj3&topic=2286.msg6985 | 1,627,150,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00037.warc.gz | 266,022,707 | 5,577 | ### Author Topic: Quiz 5 TUT0402 (Read 739 times)
#### Di Qiu
• Jr. Member
• Posts: 11
• Karma: 3
##### Quiz 5 TUT0402
« on: December 06, 2019, 12:21:46 PM »
Question: Find a particular solution of the given inhomogeneous Euler's equation
$$(1-t)y'' + ty' - y = 2(t-1)^2e^{-t},\quad 0<t<1;\qquad y_1(t)=e^t$$
Solution:
Assume the solution is of the form $y = v(t)*e^t$, then we will have
\begin{align*}
y &= v(t)e^t\\
y' &= v'(t)e^t + v(t)e^t\\
y'' &= v''(t)e^t + 2v'(t)e^t + v(t)e^t
\end{align*}
Plug into the above equation, we have
\begin{align*}
(1-t)y'' + ty' - y &= (1-t)v''(t) + (2-t) v'(t)e^t
\end{align*}
Let r(t) = v'(t), We get
\begin{align*}
(1-t)r'(t)e^t +(2-t)r(t)e^t=2(t-1)^2e^{-t}
\end{align*}
\begin{align*}
r'(t)+\frac{2-t}{1-t}r(t)=2(1-t)e^{-2t}
\end{align*}
Use integrating factor method, we have
\begin{align*}
\mu(t) &= e^{\int\frac{2-t}{1-t}dt} = e^{t-ln(1-t)} = \frac{e^t}{1-t} \\
r(t) &= \frac{\int2(1-t)e^{-2t}*\mu dt}{\mu} = \frac{\int2e^{-t}dt}{e^t/(1-t)}=2(t-1)e^{-2t}
\end{align*}
Integrate r(t), we get
\begin{align*}
v(t) = \int r(t)dt=\int 2(t-1)e^{-2t}=-te^{-2t}+\frac{1}{2}e^{-2t}
\end{align*}
Finally, a particular solution of the given inhomogeneous Euler's equation is
\begin{align*}
y(t)=v(t)e^t=-te^{-t}+\frac{1}{2}e^{-t}
\end{align*} | 585 | 1,277 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-31 | latest | en | 0.479465 |
http://markun.cs.shinshu-u.ac.jp/Mirror/mizar.org/JFM/EMM/xcmplx_1.abs.html | 1,508,843,610,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00507.warc.gz | 204,944,906 | 6,175 | Journal of Formalized Mathematics
EMM, 2003
University of Bialystok
Copyright (c) 2003 Association of Mizar Users
### Complex Numbers --- Basic Theorems
by
Library Committee
Received April 10, 2003
MML identifier: XCMPLX_1
[ Mizar article, MML identifier index ]
```environ
vocabulary ARYTM, ARYTM_1, RELAT_1, ARYTM_3, XCMPLX_0, COMPLEX1, OPPCAT_1;
notation SUBSET_1, ORDINAL1, NUMBERS, ARYTM_0, XCMPLX_0;
constructors ARYTM_0, XREAL_0, XCMPLX_0, ARYTM_3, XBOOLE_0;
clusters NUMBERS, XREAL_0, XCMPLX_0, ARYTM_3, ZFMISC_1, XBOOLE_0;
requirements SUBSET, BOOLE, NUMERALS, ARITHM;
begin
reserve a, b, c, d, e for complex number;
:: '+' operation only
theorem :: XCMPLX_1:1 :: AXIOMS'13
a + (b + c) = (a + b) + c;
theorem :: XCMPLX_1:2 :: REAL_1'10
a + c = b + c implies a = b;
theorem :: XCMPLX_1:3 :: INT_1'24
a = a + b implies b = 0;
theorem :: XCMPLX_1:4 :: AXIOMS'16
a * (b * c) = (a * b) * c;
theorem :: XCMPLX_1:5 :: REAL_1'9
c <> 0 & a * c = b * c implies a = b;
theorem :: XCMPLX_1:6 :: REAL_1'23 :: right to left - requirements REAL
a*b=0 implies a=0 or b=0;
theorem :: XCMPLX_1:7 :: REAL_2'37
b <> 0 & a * b = b implies a = 1;
:: operations '+' and '*' only
theorem :: XCMPLX_1:8 :: AXIOMS'18
a * (b + c) = a * b + a * c;
theorem :: XCMPLX_1:9 :: REAL_2'99_1
(a + b + c) * d = a * d + b * d + c * d;
theorem :: XCMPLX_1:10 :: REAL_2'101_1
(a + b) * (c + d) = a * c + a * d + b * c + b * d;
theorem :: XCMPLX_1:11 :: SQUARE_1'5
2 * a = a + a;
theorem :: XCMPLX_1:12 :: REAL_2'88_1
3 * a = a + a + a;
theorem :: XCMPLX_1:13 :: REAL_2'88_2
4 * a = a + a + a + a;
:: using operation '-'
theorem :: XCMPLX_1:14 :: REAL_1'36
a - a = 0;
theorem :: XCMPLX_1:15 :: SQUARE_1'8
a - b = 0 implies a = b;
theorem :: XCMPLX_1:16 :: REAL_2'1
b - a = b implies a = 0;
:: 2 times '-'
theorem :: XCMPLX_1:17 :: REAL_2'17_2
a = a - (b - b);
theorem :: XCMPLX_1:18 :: SEQ_4'3
a - (a - b) = b;
theorem :: XCMPLX_1:19 :: REAL_2'2_3
a - c = b - c implies a = b;
theorem :: XCMPLX_1:20 :: REAL_2'2_5
c - a = c - b implies a = b;
theorem :: XCMPLX_1:21 :: REAL_2'24_1
a - b - c = a - c - b;
theorem :: XCMPLX_1:22 :: REAL_2'29_1
a - c = (a - b) - (c - b);
theorem :: XCMPLX_1:23 :: JGRAPH_6'1_1
(c - a) - (c - b) = b - a;
theorem :: XCMPLX_1:24 :: REAL_2'15
a - b = c - d implies a - c = b - d;
:: using operations '-' and '+'
theorem :: XCMPLX_1:25 :: REAL_2'17_1
a = a + (b - b);
theorem :: XCMPLX_1:26 :: REAL_1'30
a = a + b - b;
theorem :: XCMPLX_1:27 :: SQUARE_1'6
a = a - b + b;
theorem :: XCMPLX_1:28 :: REAL_2'28_1
a + c = a + b + (c - b);
:: 2 times '-'
theorem :: XCMPLX_1:29 :: REAL_2'22_1, INT_1'1, REAL_1'17
a + b - c = a - c + b;
theorem :: XCMPLX_1:30 :: REAL_2'23_1
a - b + c = c - b + a;
theorem :: XCMPLX_1:31 :: REAL_2'28_2
a + c = a + b - (b - c);
theorem :: XCMPLX_1:32 :: REAL_2'29_3
a - c = a + b - (c + b);
theorem :: XCMPLX_1:33 :: REAL_2'13
a + b = c + d implies a - c = d - b;
theorem :: XCMPLX_1:34 :: REAL_2'14
a - c = d - b implies a + b = c + d;
theorem :: XCMPLX_1:35 :: REAL_2'16
a + b = c - d implies a + d = c - b;
:: 3 times '-'
theorem :: XCMPLX_1:36 :: REAL_1'27
a - (b + c) = a - b - c;
theorem :: XCMPLX_1:37 :: REAL_1'28
a - (b - c) = a - b + c;
theorem :: XCMPLX_1:38 :: REAL_2'18
a - (b - c) = a + (c - b);
theorem :: XCMPLX_1:39 :: REAL_2'29_2
a - c = (a - b) + (b - c);
theorem :: XCMPLX_1:40 :: REAL_1'29
a * (b - c) = a * b - a * c;
theorem :: XCMPLX_1:41 :: REAL_2'98
(a - b) * (c - d) = (b - a) * (d - c);
theorem :: XCMPLX_1:42 :: REAL_2'99_4
(a - b - c) * d = a * d - b * d - c * d;
:: using operations '-' and '*', '+'
theorem :: XCMPLX_1:43 :: REAL_2'99_2
(a + b - c) * d = a * d + b * d - c * d;
theorem :: XCMPLX_1:44 :: REAL_2'99_3
(a - b + c) * d = a * d - b * d + c * d;
theorem :: XCMPLX_1:45 :: REAL_2'101_2
(a + b) * (c - d) = a * c - a * d + b * c - b * d;
theorem :: XCMPLX_1:46 :: REAL_2'101_3
(a - b) * (c + d) = a * c + a * d - b * c - b * d;
theorem :: XCMPLX_1:47 :: REAL_2'101_4
(a - b) * (e - d) = a * e - a * d - b * e + b * d;
:: using operation '/'
theorem :: XCMPLX_1:48 :: REAL_2'67_1
a / b / c = a / c / b;
:: 0
theorem :: XCMPLX_1:49 :: REAL_2'19
a / 0 = 0;
theorem :: XCMPLX_1:50 :: REAL_2'42_2
a <> 0 & b <> 0 implies a / b <> 0;
:: 2 times '/'
theorem :: XCMPLX_1:51 :: REAL_2'62_4
b <> 0 implies a = a / (b / b);
theorem :: XCMPLX_1:52 :: TOPREAL6'5
a <> 0 implies a / (a / b) = b;
theorem :: XCMPLX_1:53 :: REAL_2'31
c <> 0 & a / c = b / c implies a = b;
theorem :: XCMPLX_1:54 :: REAL_2'74
a / b <> 0 implies b = a / (a / b);
theorem :: XCMPLX_1:55 :: REAL_2'55_1
c <> 0 implies a / b = (a / c) / (b / c);
:: 1
theorem :: XCMPLX_1:56 :: SQUARE_1'16
1 / (1 / a) = a;
theorem :: XCMPLX_1:57 :: REAL_2'48_1
1 / (a / b) = b / a;
theorem :: XCMPLX_1:58 :: REAL_2'30_1
a / b = 1 implies a = b;
theorem :: XCMPLX_1:59 :: REAL_2'33_2
1 / a = 1 / b implies a = b;
:: 0 and 1
theorem :: XCMPLX_1:60 :: REAL_1'37
a <> 0 implies a / a = 1;
theorem :: XCMPLX_1:61 :: REAL_2'39
b <> 0 & b / a = b implies a = 1;
theorem :: XCMPLX_1:62 :: REAL_2'41
a <> 0 implies 1 / a <> 0;
:: using operations '/' and '+'
theorem :: XCMPLX_1:63 :: REAL_1'40_1
a / c + b / c = (a + b) / c;
theorem :: XCMPLX_1:64 :: REAL_2'100_1
(a + b + e) / d = a / d + b / d + e / d;
:: 2
theorem :: XCMPLX_1:65 :: SQUARE_1'15
(a + a) / 2 = a;
theorem :: XCMPLX_1:66 :: SEQ_2'2_1
a/2 + a/2 = a;
theorem :: XCMPLX_1:67 :: TOPREAL3'4
a = (a + b) / 2 implies a = b;
:: 3
theorem :: XCMPLX_1:68 :: REAL_2'89_1
(a + a + a)/3 = a;
theorem :: XCMPLX_1:69 :: SEQ_4'5
a/3 + a/3 + a/3 = a;
:: 4
theorem :: XCMPLX_1:70 :: REAL_2'89_2
(a + a + a + a) / 4 = a;
theorem :: XCMPLX_1:71 :: REAL_2'90
a/4 + a/4 + a/4 + a/4 = a;
theorem :: XCMPLX_1:72 :: SEQ_2'2_2
a / 4 + a / 4 = a / 2;
theorem :: XCMPLX_1:73 :: REAL_2'89_3
(a + a) / 4 = a / 2;
:: using operations '/' and '*'
theorem :: XCMPLX_1:74 :: REAL_2'35_1
a * b = 1 implies a = 1 / b;
theorem :: XCMPLX_1:75 :: SQUARE_1'18
a * (b / c) = (a * b) / c;
theorem :: XCMPLX_1:76 :: REAL_2'80_1
a / b * e = e / b * a;
:: 3 times '/'
theorem :: XCMPLX_1:77 :: REAL_1'35
(a / b) * (c / d) = (a * c) / (b * d);
theorem :: XCMPLX_1:78 :: REAL_1'42
a / (b / c) = (a * c) / b;
theorem :: XCMPLX_1:79 :: SQUARE_1'17
a / (b * c) = a / b / c;
theorem :: XCMPLX_1:80 :: REAL_2'61_1
a / (b / c) = a * (c / b);
theorem :: XCMPLX_1:81 :: REAL_2'61_2
a / (b / c) = c / b * a;
theorem :: XCMPLX_1:82 :: REAL_2'61_3
a / (b / e) = e * (a / b);
theorem :: XCMPLX_1:83 :: REAL_2'61_4
a / (b / c) = a / b * c;
theorem :: XCMPLX_1:84 :: REAL_2'70
(a * b) / (c * d) = (a / c * b) / d;
:: 4 times '/'
theorem :: XCMPLX_1:85 :: REAL_1'82
(a / b) / (c / d) = (a * d) / (b * c);
theorem :: XCMPLX_1:86 :: REAL_2'53
(a / c) * (b / d) = (a / d) * (b / c);
theorem :: XCMPLX_1:87 :: IRRAT_1'5
a / (b * c * (d / e)) = (e / c) * (a / (b * d));
:: 0
theorem :: XCMPLX_1:88 :: REAL_1'43
b <> 0 implies a / b * b = a;
theorem :: XCMPLX_1:89 :: REAL_2'62_1
b <> 0 implies a = a * (b / b);
theorem :: XCMPLX_1:90 :: REAL_2'62_2
b <> 0 implies a = a * b / b;
theorem :: XCMPLX_1:91 :: REAL_2'78
b <> 0 implies a * c = a * b * (c / b);
:: 2 times '/'
theorem :: XCMPLX_1:92 :: REAL_1'38
c <> 0 implies a / b = (a * c) / (b * c);
theorem :: XCMPLX_1:93 :: REAL_2'55_2
c <> 0 implies a / b = a / (b * c) * c;
theorem :: XCMPLX_1:94 :: REAL_2'79
b <> 0 implies a * c = a * b / (b / c);
theorem :: XCMPLX_1:95 :: REAL_2'75
c <> 0 & d <> 0 & a * c = b * d implies a / d = b / c;
theorem :: XCMPLX_1:96 :: REAL_2'76
c <> 0 & d<>0 & a/d=b/c implies a*c = b*d;
theorem :: XCMPLX_1:97 :: REAL_2'77
c <> 0 & d <> 0 & a * c = b / d implies a * d = b / c;
:: 3 times '/'
theorem :: XCMPLX_1:98 :: REAL_2'55_3
c <> 0 implies a / b = c * (a / c / b);
theorem :: XCMPLX_1:99 :: REAL_2'55
c <> 0 implies a / b = a / c * (c / b);
:: 1
theorem :: XCMPLX_1:100 :: REAL_2'56:
a * (1 / b) = a / b;
theorem :: XCMPLX_1:101 :: REAL_2'57
a / (1 / b) = a * b;
theorem :: XCMPLX_1:102 :: REAL_2'80_3
a / b * c = 1 / b * c * a;
:: 3 times '/'
theorem :: XCMPLX_1:103 :: REAL_2'51
(1 / a) * (1 / b) = 1 / (a * b);
theorem :: XCMPLX_1:104 :: REAL_2'67_4
1 / c * (a / b) = a / (b * c);
:: 4 times '/'
theorem :: XCMPLX_1:105 :: REAL_2'67_2
a / b / c = 1 / b * (a / c);
theorem :: XCMPLX_1:106 :: REAL_2'67_3
a / b / c = 1 / c * (a / b);
:: 1 and 0
theorem :: XCMPLX_1:107 :: REAL_1'34
a <> 0 implies a * (1 / a) = 1;
theorem :: XCMPLX_1:108 :: REAL_2'62_3
b <> 0 implies a = a * b * (1 / b);
theorem :: XCMPLX_1:109 :: REAL_2'62_6
b <> 0 implies a = a * (1 / b * b);
theorem :: XCMPLX_1:110 :: REAL_2'62_7
b <> 0 implies a = a * (1 / b) * b;
theorem :: XCMPLX_1:111 :: REAL_2'62_5
b <> 0 implies a = a / (b * (1 / b));
theorem :: XCMPLX_1:112 :: REAL_2'42_4
a <> 0 & b <> 0 implies 1 / (a * b) <> 0;
theorem :: XCMPLX_1:113 :: JGRAPH_2'1
a <> 0 & b <> 0 implies (a / b) * (b / a) = 1;
:: using operations '*', '+' and '/'
theorem :: XCMPLX_1:114 :: REAL_2'65
b <> 0 implies a / b + c = (a + b * c) / b;
theorem :: XCMPLX_1:115 :: REAL_2'92
c <> 0 implies a + b = c * (a / c + b / c);
theorem :: XCMPLX_1:116 :: REAL_2'94
c <> 0 implies a + b = (a * c + b * c) / c;
theorem :: XCMPLX_1:117 :: REAL_1'41_1
b <> 0 & d <> 0 implies a / b + c / d =(a * d + c * b) / (b * d);
theorem :: XCMPLX_1:118 :: REAL_2'96
a <> 0 implies a + b = a * (1 + b / a);
:: 2
theorem :: XCMPLX_1:119 :: REAL_2'91_1
a / (2 * b) + a / (2 * b) = a / b;
:: 3
theorem :: XCMPLX_1:120 :: REAL_2'91_2
a / (3 * b) + a / (3 * b) + a / (3 * b) = a / b;
theorem :: XCMPLX_1:121 :: REAL_1'40_2
a / c - b / c = (a - b) / c;
theorem :: XCMPLX_1:122 :: TOPREAL6'4
a - a / 2 = a / 2;
theorem :: XCMPLX_1:123 :: REAL_2'100_4
(a - b - c) / d = a / d - b / d - c / d;
theorem :: XCMPLX_1:124 :: REAL_2'82
b <> 0 & d <> 0 & b <> d & a / b = e / d implies a / b = (a - e) / (b - d);
:: using operations '-', '/' and '+'
theorem :: XCMPLX_1:125 :: REAL_2'100_2
(a + b - e) / d = a / d + b / d - e / d;
theorem :: XCMPLX_1:126 :: REAL_2'100_3
(a - b + e) / d = a / d - b / d + e / d;
:: using operations '-', '/' and '*'
theorem :: XCMPLX_1:127 :: REAL_2'66_1
b <> 0 implies a / b - e = (a - e * b) / b;
theorem :: XCMPLX_1:128 :: REAL_2'66_2
b <> 0 implies c - a / b = (c * b - a) / b;
theorem :: XCMPLX_1:129 :: REAL_2'93
c <> 0 implies a - b = c * (a / c - b / c);
theorem :: XCMPLX_1:130 :: REAL_2'95
c <> 0 implies a - b = (a * c - b * c) / c;
theorem :: XCMPLX_1:131 :: REAL_1'41_2
b <> 0 & d <> 0 implies a / b - c / d = (a * d - c * b) / (b * d);
theorem :: XCMPLX_1:132 :: REAL_2'97
a <> 0 implies a - b = a * (1 - b / a);
:: using operation '-', '/', '*' and '+'
theorem :: XCMPLX_1:133 :: POLYEQ_1'24
a <> 0 implies c = (a * c + b - b) / a;
:: using unary operation '-'
theorem :: XCMPLX_1:134 :: REAL_2'2_2
-a = -b implies a = b;
theorem :: XCMPLX_1:135 :: REAL_1'22: :: right to left - requirements REAL
-a = 0 implies a = 0;
theorem :: XCMPLX_1:136 :: REAL_2'2_1
a + -b = 0 implies a = b;
theorem :: XCMPLX_1:137 :: REAL_2'11
a = a + b + -b;
theorem :: XCMPLX_1:138 :: REAL_2'17_1
a = a + (b + -b);
theorem :: XCMPLX_1:139 :: INT_1'3
a = (- b + a) + b;
theorem :: XCMPLX_1:140 :: REAL_2'6_1
- (a + b) = -a + -b;
theorem :: XCMPLX_1:141 :: REAL_2'9_2
- (-a + b) = a + -b;
theorem :: XCMPLX_1:142 :: REAL_2'10_2
a+b=-(-a+-b);
:: using unary and binary operation '-'
theorem :: XCMPLX_1:143 :: REAL_1'83
-(a - b) = b - a;
theorem :: XCMPLX_1:144 :: REAL_2'5
- a - b = - b - a;
theorem :: XCMPLX_1:145 :: REAL_2'17_4
a = - b - (- a - b);
:: binary '-' 4 times
theorem :: XCMPLX_1:146 :: REAL_2'26_1
- a - b - c = - a - c - b;
theorem :: XCMPLX_1:147 :: REAL_2'26_2
- a - b - c = - b - c - a;
theorem :: XCMPLX_1:148 :: REAL_2'26_4
- a - b - c = - c - b - a;
theorem :: XCMPLX_1:149 :: JGRAPH_6'1_2
(c - a) - (c - b) = - (a - b);
:: 0
theorem :: XCMPLX_1:150 :: REAL_1'19
0 - a = - a;
:: using unary and binary operations '-' and '+'
theorem :: XCMPLX_1:151 :: REAL_2'10_3
a + b = a - - b;
theorem :: XCMPLX_1:152 :: REAL_2'17_3
a = a - (b + -b);
theorem :: XCMPLX_1:153 :: REAL_2'2_4
a - c = b + - c implies a = b;
theorem :: XCMPLX_1:154 :: REAL_2'2_6
c - a = c + - b implies a = b;
:: '+' 3 times
theorem :: XCMPLX_1:155 :: REAL_2'22_2
a + b - c = - c + a + b;
theorem :: XCMPLX_1:156 :: REAL_2'23_2
a - b + c = - b + c + a;
theorem :: XCMPLX_1:157 :: REAL_2'20_2
a - (- b - c) = a + b + c;
:: binary '-' 3 times
theorem :: XCMPLX_1:158 :: REAL_2'20_1
a - b - c = - b - c + a;
theorem :: XCMPLX_1:159 :: REAL_2'24_3
a - b - c = - c + a - b;
theorem :: XCMPLX_1:160 :: REAL_2'24_4
a - b - c = - c - b + a;
:: using unary and binary operations '-' and '+'
theorem :: XCMPLX_1:161 :: REAL_2'6_2
- (a + b) = - b - a;
theorem :: XCMPLX_1:162 :: REAL_2'8
- (a - b) = - a + b;
theorem :: XCMPLX_1:163 :: REAL_2'9_1
-(-a+b)=a-b;
theorem :: XCMPLX_1:164 :: REAL_2'10_1
a + b = -(- a - b);
theorem :: XCMPLX_1:165 :: REAL_2'25_1
- a + b - c = - c + b - a;
:: using unary and binary operations '-' and '+' (both '-' 2 times)
theorem :: XCMPLX_1:166 :: REAL_2'25_2
- a + b - c = - c - a + b;
theorem :: XCMPLX_1:167 :: REAL_2'27_1
- (a + b + c) = - a - b - c;
theorem :: XCMPLX_1:168 :: REAL_2'27_2
- (a + b - c) = - a - b + c;
theorem :: XCMPLX_1:169 :: REAL_2'27_3
- (a - b + c) = - a + b - c;
theorem :: XCMPLX_1:170 :: REAL_2'27_5
- (a - b - c) = - a + b + c;
theorem :: XCMPLX_1:171 :: REAL_2'27_4
- (- a + b + c) = a - b - c;
theorem :: XCMPLX_1:172 :: REAL_2'27_6
- (- a + b - c) = a - b + c;
theorem :: XCMPLX_1:173 :: REAL_2'27_7
- (- a - b + c) = a + b - c;
theorem :: XCMPLX_1:174 :: REAL_2'27_8
- (- a - b - c) = a + b + c;
:: using unary operations '-' and '*'
theorem :: XCMPLX_1:175 :: REAL_1'21_1
(- a) * b = -(a * b);
theorem :: XCMPLX_1:176 :: REAL_1'21_2
(- a) * b = a * (- b);
theorem :: XCMPLX_1:177 :: REAL_2'49_1
(- a) * (- b) = a * b;
theorem :: XCMPLX_1:178 :: REAL_2'49_2
- a * (- b) = a * b;
theorem :: XCMPLX_1:179 :: REAL_2'49_3
-(-a) * b = a * b;
theorem :: XCMPLX_1:180 :: REAL_2'71_1
(-1) * a = -a;
theorem :: XCMPLX_1:181 :: REAL_2'71_2
(- a) * (- 1) = a;
theorem :: XCMPLX_1:182 :: REAL_2'38
b<>0 & a*b=-b implies a=-1;
theorem :: XCMPLX_1:183 :: Thx
a * a = 1 implies a = 1 or a = -1;
theorem :: XCMPLX_1:184 :: TOPREAL6'3
-a + 2 * a = a;
theorem :: XCMPLX_1:185 :: REAL_2'85_1
(a - b) * c = (b - a) * (- c);
theorem :: XCMPLX_1:186 :: REAL_2'85_2
(a - b) * c = - (b - a) * c;
theorem :: XCMPLX_1:187 :: TOPREAL6'2
a - 2 * a = -a;
:: using unary operations '-' and '/'
theorem :: XCMPLX_1:188 :: REAL_1'39_1
-a / b = (-a) / b;
theorem :: XCMPLX_1:189 :: REAL_1'39_2
a / (- b) = -a / b;
theorem :: XCMPLX_1:190 :: REAL_2'58_1
- a / (- b) = a / b;
theorem :: XCMPLX_1:191 :: REAL_2'58_2
-(- a) / b = a / b;
theorem :: XCMPLX_1:192 :: REAL_2'58_3
(- a) / (- b) = a / b;
theorem :: XCMPLX_1:193 :: REAL_2'58
(-a) / b = a / (-b);
theorem :: XCMPLX_1:194 :: REAL_2'71_3
-a = a / (-1);
theorem :: XCMPLX_1:195 :: REAL_2'71
a = (- a) / (-1);
theorem :: XCMPLX_1:196 :: REAL_2'34
a / b = - 1 implies a = - b & b = - a;
theorem :: XCMPLX_1:197 :: REAL_2'40
b <> 0 & b / a = - b implies a = -1;
theorem :: XCMPLX_1:198 :: REAL_2'45_2
a <> 0 implies (-a) / a = -1;
theorem :: XCMPLX_1:199 :: REAL_2'45_3
a <> 0 implies a / (- a) = -1;
theorem :: XCMPLX_1:200 :: REAL_2'46_2
a <> 0 & a = 1 / a implies a = 1 or a = -1;
theorem :: XCMPLX_1:201 :: REAL_2'83:
b <> 0 & d <> 0 & b <> -d & a / b = e / d implies a / b = (a + e) / (b + d)
;
:: using operation '"'
theorem :: XCMPLX_1:202 :: REAL_2'33_1
a" = b" implies a = b;
theorem :: XCMPLX_1:203 :: REAL_1'31
a" = 0 implies a = 0;
:: using '"' and '*'
theorem :: XCMPLX_1:204
b <> 0 implies a = a*b*b";
theorem :: XCMPLX_1:205 :: REAL_1'24
a" * b" = (a * b)";
theorem :: XCMPLX_1:206 :: REAL_2'47_1
(a * b")" = a" * b;
theorem :: XCMPLX_1:207 :: REAL_2'47_2
(a" * b")" = a * b;
theorem :: XCMPLX_1:208 :: REAL_2'42_1:
a <> 0 & b <> 0 implies a * b" <> 0;
theorem :: XCMPLX_1:209 :: REAL_2'42_3
a <> 0 & b <> 0 implies a" * b" <> 0;
theorem :: XCMPLX_1:210 :: REAL_2'30_2
a * b" = 1 implies a = b;
theorem :: XCMPLX_1:211 :: REAL_2'35_2
a * b = 1 implies a = b";
:: using '"', '*', and '+'
canceled;
theorem :: XCMPLX_1:213
a <> 0 & b <> 0 implies a" + b" = (a + b)*(a*b)";
:: using '"', '*', and '-'
theorem :: XCMPLX_1:214
a <> 0 & b <> 0 implies a" - b" = (b - a)*(a*b)";
:: using '"' and '/'
theorem :: XCMPLX_1:215 :: REAL_1'81
(a / b)" = b / a;
theorem :: XCMPLX_1:216
(a"/b") = b/a;
theorem :: XCMPLX_1:217 :: REAL_1'33_1
1 / a = a";
theorem :: XCMPLX_1:218 :: REAL_1'33_2
1 / a" = a;
theorem :: XCMPLX_1:219 :: REAL_2'36_21
(1 / a)" = a;
theorem :: XCMPLX_1:220 :: REAL_2'33_3
1 / a = b" implies a = b;
:: using '"', '*', and '/'
theorem :: XCMPLX_1:221
a/b" = a*b;
theorem :: XCMPLX_1:222
a"*(c/b) = c/(a*b);
theorem :: XCMPLX_1:223
a"/b = (a*b)";
:: both unary operations
theorem :: XCMPLX_1:224 :: REAL_2'45_1
(- a)" = -a";
theorem :: XCMPLX_1:225 :: REAL_2'46_1
a <> 0 & a = a" implies a = 1 or a = -1;
:: from JORDAN4
theorem :: XCMPLX_1:226
a+b+c-b=a+c;
theorem :: XCMPLX_1:227
a-b+c+b=a+c;
theorem :: XCMPLX_1:228
a+b-c-b=a-c;
theorem :: XCMPLX_1:229
a-b-c+b=a-c;
theorem :: XCMPLX_1:230
a-a-b=-b;
theorem :: XCMPLX_1:231
-a+a-b=-b;
theorem :: XCMPLX_1:232
a-b-a=-b;
theorem :: XCMPLX_1:233
-a-b+a=-b;
``` | 8,355 | 17,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.729938 |
https://discuss.codecademy.com/t/this-question-19-33/6109 | 1,544,483,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00167.warc.gz | 600,209,023 | 3,645 | # This. question 19/33
#1
As I understand this. is used so that the method is not just limited to one particular object but can be applied to different objects.
In the example in 19/33 (below) I don't understand that. Here this is used BUT it only refers to the object square and it is part of square-so why do we use this here?
var square = new Object();
square.sideLength = 6;
square.calcPerimeter = function() {
return this.sideLength * 4;
};
// help us define an area method here
square.calcArea= function() {
return this.sideLength * this.sideLength;
}
var p = square.calcPerimeter();
var a = square.calcArea();
Thank you!
#2
I have this exact question as well. In the previous exercises "this" is accurately described as being used so that a method can be applied to multiple objects and so we don't have to write the same method in each object. Makes sense.
But then immediately after in exercises 18-19 "this" is used inside of objects (rectangle, square). The instructions say:
"Notice we have use the keyword 'this'. 'this' is still a placeholder, but in this scenario, 'this' can only ever refer to 'rectangle' because we defined 'setHeight' to be explicitly part of 'rectangle' by defining it as 'rectangle.setHeight'."
Okay, so then what is the benefit of using 'this' in this exercise? If we defined the method to be specific to the object, then why don't we just write "rectangle.height = newHeight;" like we did in the first exercise (bob.age = newAge;)? Again, what is the benefit or difference in using 'this' in this scenario? It's not explained anywhere. Exercise 19 is the last one in that section. Exercise 20 starts with something new. | 394 | 1,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.925985 |
https://math.answers.com/other-math/2_fourths_times_four_fifths | 1,714,044,006,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00056.warc.gz | 321,767,779 | 46,517 | 0
# 2 fourths times four fifths
Updated: 4/28/2022
Wiki User
10y ago
2 fourths times four fifths is 2 fifths.
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10y ago
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### What is four fifths times six and three fourths?
4/5 x 6 3/4 is 5 2/5
Four fifths
1/2
1/2..... or 0.5
16.8
### What is the answer to five and a half times 4 fifths?
five and a half=5.5 four fifths=.8 since one divided by five is .2 and four .2's (4x.2)=.8 5.5x.8=4.4, or four and 2 fifths four and two fifths is the answer.
one tenth
2/15
### What is three and three fourths times two fifths?
3 3/4 x 2/5 = 1 1/2 or 1.5
### What is four fifths plus 2?
it is 2 and four fifths
### What is double four-fifths?
(4/5) times 2 = 8/5; eight-fifths; or 1.6
20 | 303 | 800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-18 | latest | en | 0.954553 |
https://www.jiskha.com/questions/362786/using-the-formula-w-cr-2-where-c-is-a-constant-and-r-is-the-distance-that-the-object-is | 1,618,427,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00607.warc.gz | 938,563,984 | 5,406 | Algebra
Using the formula: w=Cr^-2 where C is a constant, and r is the distance that the object is from the center of Earth. W is the weight.
1- Suppose an object is 100lbs when it is at sea level. Find the Value of C that makes the equation true. Sea level is 3,963 miles from the center of the Earth.
1. 👍
2. 👎
3. 👁
1. sub in the given data ...
100 = C(3963)^-2
C = 100(3963)^2 ... huge number
1. 👍
2. 👎
2. Would you not divide the big number by 100 to get C alone?
1. 👍
2. 👎
3. 100 = C(3963)^-2
100 = C/3963^2 , so now we have to mutiply both sides by 3963^2 to get C alone, like I had
1. 👍
2. 👎
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reddingk1@hwbcymru.net - Mr Redding | 445 | 1,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-45 | latest | en | 0.929445 |
https://www.bartleylawoffice.com/interesting/what-is-newton-s-3-laws-of-motion.html | 1,656,990,570,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00418.warc.gz | 707,713,664 | 20,719 | # What Is Newton’S 3 Laws Of Motion?
1. The three laws of motion attributed to Newton can be summed up as follows: Unless an external force acts on it, every item that is moving in a constant and uniform direction will continue to move in that direction indefinitely
2. The formula for force is mass multiplied by acceleration
3. There is always a response, both equal and opposite in kind, to every action
According to the first rule of motion, an object’s motion will not alter unless there is a force acting on it.According to the second law of motion, the amount of force exerted on an object is proportional to the product of the object’s mass and its acceleration.According to the third law, when two things interact with one other, the forces that they apply to each other are of equal magnitude but opposing direction.
## What are Newton’s 1st 2nd and 3rd laws of motion?
What are the first, second, and third laws of motion that Newton proposed?According to the first rule of motion, an object’s motion will not alter unless there is a force acting on it.According to the second law of motion, the amount of force exerted on an object is proportional to the product of the object’s mass and its acceleration.According to the third law, when two things interact with one other, the forces that they apply to each other are of equal magnitude but opposing direction.
## What are the 3 laws of motion?
The laws are as follows: (1) No object travels in a direction other than a straight line, unless it is being acted upon by another force. (2) The acceleration of an item is directly proportional to the total force that is applied, and it is inversely proportional to the mass of the object. (3) There is always a reaction that is equal and opposite to the action that was taken.
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## What is the relationship between force and motion in Newton’s law?
Newton’s equations of motion indicate that there is a link between the forces acting on an item and the motion of the object itself.According to the first rule of motion, we know that an object’s motion will not alter until a force acts on it.This is something that we comprehend.According to the second law, the amount of force exerted on an item is proportional to the product of the object’s mass and the acceleration it is experiencing.
## What are Isaac Newton’s laws of motion?
The laws of motion formulated by Newton.When one body produces a force on another body, the second body concurrently exerts a force on the first body that is equal in size but acts in the opposite direction.Isaac Newton was the first person to establish the three laws of motion, and he did it in his book Philosophiae Naturalis Principia Mathematica ( Mathematical Principles of Natural Philosophy ),
## What is Newton’s 3rd law exactly?
The third rule of Newton states that for every action, there is an equal and opposite response.This is how the law is typically phrased.The implication of this statement is that for every interaction, there are two forces operating on the two objects that are engaging with one another.The magnitude of the forces acting on the first item is exactly the same as the magnitude of the forces acting on the second object.
## What is Newton’s 3 law for kids?
According to the third rule of thermodynamics, there is always an equal and opposite response to every action.This indicates that there is always a pair of forces that are identical to one another.In the scenario when you kicked the ball, your foot exerts some force on the ball, but the ball also exerts the same amount of force on your foot.Both forces add up to the total amount of force.
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## What does Newton’s second law state?
The acceleration of an object that is generated by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. This is the formal statement of Newton’s second law of motion.
## What does Newton’s 1st law state?
It is common practice to state Newton’s first law of motion as. Unless it is acting upon by a force that is imbalanced, an item that is at rest will remain at rest, and an object that is in motion will continue to move at the same speed and in the same direction.
## What is the 4th law of motion?
The nature of forces, as well as the ability to calculate their effects, are discussed in Newton’s fourth law of motion. This rule asserts that the forces are quantities that can be represented as vectors, and that when their vectors are added together, they adhere to the concept of superposition.
## What is law inertia?
The postulate in physics known as the law of inertia, also known as Newton’s first law, states that if a body is at rest or moving at a constant speed in a straight line, it will either remain at rest or keep moving in the same direction at the same speed unless it is subjected to an external force that causes it to change direction.
## What is Newton’s second law of motion class 9?
According to the second law of Newton, the rate of acceleration of an object is determined by two factors: the total force that is acting on the object, as well as the mass of the object itself.The net force that is operating on the body has a direct proportional relationship with the acceleration of the body, whereas the mass of the body has an inverse proportional relationship with acceleration.
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## Who discovered the three laws of motion?
The three laws of motion developed by Sir Isaac Newton describe the motion of enormous masses and how they interact with one another. Newton’s laws were deemed revolutionary more than three centuries ago, despite the fact that to us now they may appear to be self-evident. Newton is often regarded as one of the most important scientists who have lived. | 1,241 | 5,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.959311 |
https://www.physicsforums.com/threads/time-an-electron-spent-in-a-magnetic-field.188729/ | 1,531,726,715,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00543.warc.gz | 966,428,536 | 12,448 | # Homework Help: Time an electron spent in a magnetic field
1. Oct 3, 2007
### luisabejarin
Hello.. Does anyone know how to find the time t an electron spent in a magnetic field region if the only thing given are the magnetic field B=0.5T and angle of deflection 30 degrees?
2. Oct 3, 2007
### Sourabh N
You can find time period T of electron as you know B. Electron travels 360 deg in T, so what time will it take to travel 30 deg, you can find.
3. Oct 3, 2007
### luisabejarin
wow!thanks,i got it..i understand..thank u very much! | 158 | 541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.828533 |
https://mathoverflow.net/questions/146895/picard-number-of-fundamental-divisor-of-fano-3-fold | 1,582,635,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00269.warc.gz | 468,037,513 | 25,777 | # Picard number of fundamental divisor of Fano 3-fold
Given a (smooth)Fano 3-fold $X$, Sokurov proved that the fundamental linear system contains a smooth surface. My question is :
If the Picard number of X is 1,Is there such a smooth surface(in the fundamental linear system) also with Picard number 1?.
P.S. if we work on picard number 1 Fano manifold, then $Pic(X)$ is generated by an ample bundle $H$, by fundamental linear system of $X$ I mean $|H|$.
Thx.
• What do you mean by "the fundamental linear system"? – Sasha Nov 4 '13 at 15:37
• In case of index 2 your question does not make much sense for the surfaces in $|H|$, which are Del Pezzo surfaces hence have Picard number $>1$. On the other hand it makes sense for the anticanonical system, see my answer below. – abx Nov 5 '13 at 9:19
This is true if the anticanonical system (which you call the fundamental system) is very ample. Indeed one can take a Lefschetz pencil and apply Theorem 1.4 of SGA 7, Exposé XIX (Noether's theorem by Deligne). Now we know the complete list of Fano threefolds with Picard number 1 , thanks to Iskovskikh; there are only 2 or 3 cases where $-K$ is not very ample, maybe they can be checked by hand.
Afterthought : Actually the result does not hold for at least one family of Fano threefolds with Picard number 1, namely those for which the anti canonical map $\pi:X\rightarrow Q\subset {\Bbb P}^4$ is a double covering of a smooth quadric in ${\Bbb P}^4$. Indeed the smooth surfaces in $\ |-K_X|\$ are of the form $S_h:=\pi^{-1}(Q_h)$, where $Q_h$ is a smooth hyperplane section of $Q$. Since $\mathrm {Pic}(Q_h)=\Bbb{Z}^2$, the Picard number of $S_h$ is $\geq 2$. | 483 | 1,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-10 | latest | en | 0.858535 |
http://gmatclub.com/forum/to-relieve-anxiety-moderate-exercise-can-be-equally-89905.html#p681548 | 1,481,172,032,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542412.97/warc/CC-MAIN-20161202170902-00335-ip-10-31-129-80.ec2.internal.warc.gz | 111,886,159 | 63,069 | To relieve anxiety, moderate exercise can be equally : GMAT Sentence Correction (SC)
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# To relieve anxiety, moderate exercise can be equally
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To relieve anxiety, moderate exercise can be equally [#permalink]
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31 Jan 2010, 09:12
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To relieve anxiety, moderate exercise can be equally effective as, and less addictive than, most sedatives.
a. effective as, and
b. as effective as, while being
c. effectively equal to, but
d. as effective as, and
e. effective, and
[Reveal] Spoiler: OA
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Last edited by fameatop on 08 Sep 2013, 00:01, edited 1 time in total.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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31 Jan 2010, 12:47
'A' seems to be the best option.
The adverb equally takes the place of the first as in the typical as...as construction, so an as before effective isn't necessary. Using but would inappropriately imply that addictive is a counterpoint or contradiction to being effective.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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31 Jan 2010, 18:42
My take is D
as... as makes this complete idiom , i am not sure whether usage of equally means you need only one as
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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31 Jan 2010, 18:47
As this moment i am not sure but i would like to share my thoughts. I was confused between D and E but IMO - D.
a. effective as, and
I think - here the comparison is wrong. We are comparing effectiveness with sedatives.
b. as effective as, while being
Incorrect usage of while and being
c. effectively equal to, but
but is not required, as it shows contradiction
d. as effective as, and
CORRECT - correct idion - as effective as. Assuming you remove the and part i.e and less addictive than, the sentence makes correct comparison. i.e comparing effectiveness of exercise with sedatives
e. effective, and
I think this choice is closest to the correct answer. here effective is used as an adjective where as the non underlined part "most sedatves", sedative is a noun. Hence the comparision is not parallel. Adverb Adjective THAN Adjective Noun
What is the OA ?
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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23 Aug 2011, 19:32
heavyrain wrote:
D
The word "equally" is redundant.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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24 Aug 2011, 02:43
Eventhough D has the correct idiom, but presence of 'equally' mars its use, hence E. Also, in A, there is wrong comparison.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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18 Dec 2011, 13:11
vscid wrote:
To relieve anxiety, moderate exercise can be equally effective as, and less addictive than, most sedatives.
a. effective as, and
b. as effective as, while being
c. effectively equal to, but
d. as effective as, and
e. effective, and
Interesting choices between A and E.
Let's remove the less addictive part, the sentence becomes To relieve anxiety, moderate exercise can be equally effective most sedatives. if we choose E as the option. Something is missing here.
Granted that the question perhaps should have included as before "equally" to make it idiomatically proper.
If we use A as the option then -
To relieve anxiety, moderate exercise can be equally effective as most sedatives.
To relieve anxiety, moderate exercise can be less addictive than most sedatives.
{A} in my opinion is the best answer for this question.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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18 Dec 2011, 13:55
I go with A because of parallelism... equally effective as and less addictive than...Does anyone know the real answer? i searched the old post and came across D...Where is this question from? Wouldn't an "as" after equally be redundant???
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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18 Dec 2011, 14:55
Hi;
There is another post on the same question. Most of the people seem to suggest that answer is 'E'.
to-relieve-anxiety-moderate-exercise-can-be-equally-44443.html.
My personal vote for 'E' as well.
---
Please share the OA, if some one knows the source.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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28 Dec 2011, 04:52
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http://www.manhattangmat.com/forums/as- ... -t273.html
Although the OA is D as the poster on MGMAT forum said, the MGMAT instructor also comment that choice E is wrong answer. We should obsess to this sentence.
Hope that helps
P/S: read more Mitch's comment here: http://www.beatthegmat.com/effective-t89831.html#400298
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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20 Feb 2015, 14:07
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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22 Feb 2015, 12:21
To relieve anxiety, moderate exercise can be equally effective as, and less addictive than, most sedatives.
a. effective as, and
Incorrect. Incorrect idiom equally...as.
b. as effective as, while being
Incorrect. equally as effective as is awkward and redundant.
c. effectively equal to, but
Incorrect. equally effectively equal is just awful.
d. as effective as, and
Incorrect. Same as B.
e. effective, and
Correct. equally effective, and ...
OA should be E.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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03 Jun 2015, 16:06
Hi All,
I have a simple question regarding the usage of " , and"
I believe it is used for lists or to combine 2 independent sentences. Here in this question ",and less addictive than" is neither of these. How is it a correct answer then ??
Am i missing something here?
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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21 Nov 2015, 07:02
vscid wrote:
To relieve anxiety, moderate exercise can be equally effective as, and less addictive than, most sedatives.
a. effective as, and
b. as effective as, while being
c. effectively equal to, but
d. as effective as, and
e. effective, and
None of the options seem outright correct to me but if i had to choose one out of the lot i would have gone with A. Equally as effective as and less addictive than
Equally as effective as most sedatives and less addictive than most sedatives. A semblance of correct idiom usage is maintained here i feel.
A more appropriate usage in my opinion would be "To relive anxiety, moderate exercise can be as effective as, and less addictive than, most sedatives."
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To relieve anxiety, moderate exercise can be equally [#permalink]
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06 Jun 2016, 01:39
According to me answer is E.
In D as effective as and equally are redundant.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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12 Jun 2016, 16:46
Here the word "equally" modifies the terms "as effective as" and "less addictive". This means that moderate exercise is equally likely to be effective as well as less addictive than sedatives.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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13 Jun 2016, 14:06
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Expert's post
According to me answer is E.
In D as effective as and equally are redundant.
Like you, I fell for the trap first and modified the answer choice erroneously.
The problem with E is as follows:
Exercise is equally effective and less addictive than most sedatives.
Parallelism marker: AND
element 1: equally effective
Suppose we eliminate the conjunction "and" and the second element. The sentence becomes:
Exercise is equally effective most sedatives.
It is now grammatically wrong.
I reinstated the answer choice to D, though I find it awkward as described in the following post. Nonetheless the above explanation by Samiksha10 is logical. Read the sentence as ("can be" replaced with "is likely"):
Exercise is equally likely to be as effective as, and less addictive than, most sedatives.
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To relieve anxiety, moderate exercise can be equally [#permalink]
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13 Jun 2016, 14:11
Samiksha10 wrote:
Here the word "equally" modifies the terms "as effective as" and "less addictive". This means that moderate exercise is equally likely to be effective as well as less addictive than sedatives.
"Equally" is an adverb and must modify a verb. Here "equally" modifies "can be". Nonetheless I can probably understand what you are pointing at:
The gift is equally a boon and a curse. : The gift has two qualities which are equal: a boon and a curse - goodness of boon and badness of curse are equal ( assuming they can be measured in suitable units).
Exercise is equally effective and addictive. This structure is alright since there are two adjectives (effective and addictive) that are equal (again assuming they can be measured in suitable units)
But, the following structure seems to me somewhat awkward:
Exercise is equally as effective as and less addictive than ..... This structure does not have two adjectives that can be said to be equal. It is awkward to say "as effective as" and "less addictive" are equal - that would imply that the equality of effectiveness and the "less"-ness of "addictive"-ness are equal - a very awkward implication.
Yet, in absence of any other grammatically correct option, we have to select D.
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Re: To relieve anxiety, moderate exercise can be equally [#permalink]
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12 Aug 2016, 15:59
To relieve anxiety, moderate exercise can be equally effective as, and less addictive than, most sedatives.
a. effective as, and
b. as effective as, while being
c. effectively equal to, but
d. as effective as, and
e. effective, and
Breaking it, moderate exercise can be equally effective as most sedatives and equally less addictive.
Only D can maintain the meaning
Re: To relieve anxiety, moderate exercise can be equally [#permalink] 12 Aug 2016, 15:59
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Four full-year digital course, built from the ground up and fully-aligned to the Common Core State Standards, for 7th grade Mathematics. Created using research-based approaches to teaching and learning, the Open Access Common Core Course for Mathematics is designed with student-centered learning in mind, including activities for students to develop valuable 21st century skills and academic mindset.
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Samples and ProbabilityType of Unit: ConceptualPrior KnowledgeStudents should be able to:Understand the concept of a ratio.Write ratios as percents.Describe data using measures of center.Display and interpret data in dot plots, histograms, and box plots.Lesson FlowStudents begin to think about probability by considering the relative likelihood of familiar events on the continuum between impossible and certain. Students begin to formalize this understanding of probability. They are introduced to the concept of probability as a measure of likelihood, and how to calculate probability of equally likely events using a ratio. The terms (impossible, certain, etc.) are given numerical values. Next, students compare expected results to actual results by calculating the probability of an event and conducting an experiment. Students explore the probability of outcomes that are not equally likely. They collect data to estimate the experimental probabilities. They use ratio and proportion to predict results for a large number of trials. Students learn about compound events. They use tree diagrams, tables, and systematic lists as tools to find the sample space. They determine the theoretical probability of first independent, and then dependent events. In Lesson 10 students identify a question to investigate for a unit project and submit a proposal. They then complete a Self Check. In Lesson 11, students review the results of the Self Check, solve a related problem, and take a Quiz.Students are introduced to the concept of sampling as a method of determining characteristics of a population. They consider how a sample can be random or biased, and think about methods for randomly sampling a population to ensure that it is representative. In Lesson 13, students collect and analyze data for their unit project. Students begin to apply their knowledge of statistics learned in sixth grade. They determine the typical class score from a sample of the population, and reason about the representativeness of the sample. Then, students begin to develop intuition about appropriate sample size by conducting an experiment. They compare different sample sizes, and decide whether increasing the sample size improves the results. In Lesson 16 and Lesson 17, students compare two data sets using any tools they wish. Students will be reminded of Mean Average Deviation (MAD), which will be a useful tool in this situation. Students complete another Self Check, review the results of their Self Check, and solve additional problems. The unit ends with three days for students to work on Gallery problems, possibly using one of the days to complete their project or get help on their project if needed, two days for students to present their unit projects to the class, and one day for the End of Unit Assessment.
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Students will compare expected results to actual results by first calculating the probability of an event, then conducting an experiment to generate data. They will use an interactive to simulate a familiar event—rolling a number cube. Students will also be introduced to terminology.Key ConceptsThis lesson takes an informal look at the Law of Large Numbers through comparing experimental results to expected results.There is variability in actual results.Probability terminology is introduced:theoretical probability: the ratio of favorable outcomes to the total number of possible equally-likely outcomes, often simply called probabilityexpected results: the results based on theoretical probabilityexperimental probability: the ratio of favorable outcomes to the total number of trials in an experimentactual results: the results based on experimental probabilityoutcome: a single possible resultsample space: the set of all possible outcomesexperiment: a controlled, repeated process, such as repeatedly tossing a cointrial: each repetition in an experiment, such as one coin tossevent: a set of outcomes to which a probability is assignedGoals and Learning ObjectivesPredict results using ratio and proportion.Compare expected results to actual results.Understand that the actual results get closer to the expected results as the number of trials increase.
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http://www.cl.cam.ac.uk/research/hvg/Isabelle/dist/library/HOL/HOL-Auth/Nat_Bijection.html | 1,501,088,025,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426234.82/warc/CC-MAIN-20170726162158-20170726182158-00205.warc.gz | 429,590,028 | 5,197 | # Theory Nat_Bijection
theory Nat_Bijection
imports Main
```(* Title: HOL/Library/Nat_Bijection.thy
Author: Brian Huffman
Author: Florian Haftmann
Author: Stefan Richter
Author: Tobias Nipkow
Author: Alexander Krauss
*)
section ‹Bijections between natural numbers and other types›
theory Nat_Bijection
imports Main
begin
subsection ‹Type @{typ "nat × nat"}›
text ‹Triangle numbers: 0, 1, 3, 6, 10, 15, ...›
definition triangle :: "nat ⇒ nat"
where "triangle n = (n * Suc n) div 2"
lemma triangle_0 [simp]: "triangle 0 = 0"
by (simp add: triangle_def)
lemma triangle_Suc [simp]: "triangle (Suc n) = triangle n + Suc n"
by (simp add: triangle_def)
definition prod_encode :: "nat × nat ⇒ nat"
where "prod_encode = (λ(m, n). triangle (m + n) + m)"
text ‹In this auxiliary function, @{term "triangle k + m"} is an invariant.›
fun prod_decode_aux :: "nat ⇒ nat ⇒ nat × nat"
where "prod_decode_aux k m =
(if m ≤ k then (m, k - m) else prod_decode_aux (Suc k) (m - Suc k))"
declare prod_decode_aux.simps [simp del]
definition prod_decode :: "nat ⇒ nat × nat"
where "prod_decode = prod_decode_aux 0"
lemma prod_encode_prod_decode_aux: "prod_encode (prod_decode_aux k m) = triangle k + m"
apply (induct k m rule: prod_decode_aux.induct)
apply (subst prod_decode_aux.simps)
apply (simp add: prod_encode_def)
done
lemma prod_decode_inverse [simp]: "prod_encode (prod_decode n) = n"
by (simp add: prod_decode_def prod_encode_prod_decode_aux)
lemma prod_decode_triangle_add: "prod_decode (triangle k + m) = prod_decode_aux k m"
apply (induct k arbitrary: m)
apply (simp add: prod_decode_def)
apply (simp only: triangle_Suc add.assoc)
apply (subst prod_decode_aux.simps)
apply simp
done
lemma prod_encode_inverse [simp]: "prod_decode (prod_encode x) = x"
unfolding prod_encode_def
apply (induct x)
apply (subst prod_decode_aux.simps)
apply simp
done
lemma inj_prod_encode: "inj_on prod_encode A"
by (rule inj_on_inverseI) (rule prod_encode_inverse)
lemma inj_prod_decode: "inj_on prod_decode A"
by (rule inj_on_inverseI) (rule prod_decode_inverse)
lemma surj_prod_encode: "surj prod_encode"
by (rule surjI) (rule prod_decode_inverse)
lemma surj_prod_decode: "surj prod_decode"
by (rule surjI) (rule prod_encode_inverse)
lemma bij_prod_encode: "bij prod_encode"
by (rule bijI [OF inj_prod_encode surj_prod_encode])
lemma bij_prod_decode: "bij prod_decode"
by (rule bijI [OF inj_prod_decode surj_prod_decode])
lemma prod_encode_eq: "prod_encode x = prod_encode y ⟷ x = y"
by (rule inj_prod_encode [THEN inj_eq])
lemma prod_decode_eq: "prod_decode x = prod_decode y ⟷ x = y"
by (rule inj_prod_decode [THEN inj_eq])
text ‹Ordering properties›
lemma le_prod_encode_1: "a ≤ prod_encode (a, b)"
by (simp add: prod_encode_def)
lemma le_prod_encode_2: "b ≤ prod_encode (a, b)"
by (induct b) (simp_all add: prod_encode_def)
subsection ‹Type @{typ "nat + nat"}›
definition sum_encode :: "nat + nat ⇒ nat"
where "sum_encode x = (case x of Inl a ⇒ 2 * a | Inr b ⇒ Suc (2 * b))"
definition sum_decode :: "nat ⇒ nat + nat"
where "sum_decode n = (if even n then Inl (n div 2) else Inr (n div 2))"
lemma sum_encode_inverse [simp]: "sum_decode (sum_encode x) = x"
by (induct x) (simp_all add: sum_decode_def sum_encode_def)
lemma sum_decode_inverse [simp]: "sum_encode (sum_decode n) = n"
by (simp add: even_two_times_div_two sum_decode_def sum_encode_def)
lemma inj_sum_encode: "inj_on sum_encode A"
by (rule inj_on_inverseI) (rule sum_encode_inverse)
lemma inj_sum_decode: "inj_on sum_decode A"
by (rule inj_on_inverseI) (rule sum_decode_inverse)
lemma surj_sum_encode: "surj sum_encode"
by (rule surjI) (rule sum_decode_inverse)
lemma surj_sum_decode: "surj sum_decode"
by (rule surjI) (rule sum_encode_inverse)
lemma bij_sum_encode: "bij sum_encode"
by (rule bijI [OF inj_sum_encode surj_sum_encode])
lemma bij_sum_decode: "bij sum_decode"
by (rule bijI [OF inj_sum_decode surj_sum_decode])
lemma sum_encode_eq: "sum_encode x = sum_encode y ⟷ x = y"
by (rule inj_sum_encode [THEN inj_eq])
lemma sum_decode_eq: "sum_decode x = sum_decode y ⟷ x = y"
by (rule inj_sum_decode [THEN inj_eq])
subsection ‹Type @{typ "int"}›
definition int_encode :: "int ⇒ nat"
where "int_encode i = sum_encode (if 0 ≤ i then Inl (nat i) else Inr (nat (- i - 1)))"
definition int_decode :: "nat ⇒ int"
where "int_decode n = (case sum_decode n of Inl a ⇒ int a | Inr b ⇒ - int b - 1)"
lemma int_encode_inverse [simp]: "int_decode (int_encode x) = x"
by (simp add: int_decode_def int_encode_def)
lemma int_decode_inverse [simp]: "int_encode (int_decode n) = n"
unfolding int_decode_def int_encode_def
using sum_decode_inverse [of n] by (cases "sum_decode n") simp_all
lemma inj_int_encode: "inj_on int_encode A"
by (rule inj_on_inverseI) (rule int_encode_inverse)
lemma inj_int_decode: "inj_on int_decode A"
by (rule inj_on_inverseI) (rule int_decode_inverse)
lemma surj_int_encode: "surj int_encode"
by (rule surjI) (rule int_decode_inverse)
lemma surj_int_decode: "surj int_decode"
by (rule surjI) (rule int_encode_inverse)
lemma bij_int_encode: "bij int_encode"
by (rule bijI [OF inj_int_encode surj_int_encode])
lemma bij_int_decode: "bij int_decode"
by (rule bijI [OF inj_int_decode surj_int_decode])
lemma int_encode_eq: "int_encode x = int_encode y ⟷ x = y"
by (rule inj_int_encode [THEN inj_eq])
lemma int_decode_eq: "int_decode x = int_decode y ⟷ x = y"
by (rule inj_int_decode [THEN inj_eq])
subsection ‹Type @{typ "nat list"}›
fun list_encode :: "nat list ⇒ nat"
where
"list_encode [] = 0"
| "list_encode (x # xs) = Suc (prod_encode (x, list_encode xs))"
function list_decode :: "nat ⇒ nat list"
where
"list_decode 0 = []"
| "list_decode (Suc n) = (case prod_decode n of (x, y) ⇒ x # list_decode y)"
by pat_completeness auto
termination list_decode
apply (relation "measure id")
apply simp_all
apply (drule arg_cong [where f="prod_encode"])
apply (drule sym)
apply (simp add: le_imp_less_Suc le_prod_encode_2)
done
lemma list_encode_inverse [simp]: "list_decode (list_encode x) = x"
by (induct x rule: list_encode.induct) simp_all
lemma list_decode_inverse [simp]: "list_encode (list_decode n) = n"
apply (induct n rule: list_decode.induct)
apply simp
apply (simp split: prod.split)
apply (simp add: prod_decode_eq [symmetric])
done
lemma inj_list_encode: "inj_on list_encode A"
by (rule inj_on_inverseI) (rule list_encode_inverse)
lemma inj_list_decode: "inj_on list_decode A"
by (rule inj_on_inverseI) (rule list_decode_inverse)
lemma surj_list_encode: "surj list_encode"
by (rule surjI) (rule list_decode_inverse)
lemma surj_list_decode: "surj list_decode"
by (rule surjI) (rule list_encode_inverse)
lemma bij_list_encode: "bij list_encode"
by (rule bijI [OF inj_list_encode surj_list_encode])
lemma bij_list_decode: "bij list_decode"
by (rule bijI [OF inj_list_decode surj_list_decode])
lemma list_encode_eq: "list_encode x = list_encode y ⟷ x = y"
by (rule inj_list_encode [THEN inj_eq])
lemma list_decode_eq: "list_decode x = list_decode y ⟷ x = y"
by (rule inj_list_decode [THEN inj_eq])
subsection ‹Finite sets of naturals›
subsubsection ‹Preliminaries›
lemma finite_vimage_Suc_iff: "finite (Suc -` F) ⟷ finite F"
apply (safe intro!: finite_vimageI inj_Suc)
apply (rule finite_subset [where B="insert 0 (Suc ` Suc -` F)"])
apply (rule subsetI)
apply (case_tac x)
apply simp
apply simp
apply (rule finite_insert [THEN iffD2])
apply (erule finite_imageI)
done
lemma vimage_Suc_insert_0: "Suc -` insert 0 A = Suc -` A"
by auto
lemma vimage_Suc_insert_Suc: "Suc -` insert (Suc n) A = insert n (Suc -` A)"
by auto
lemma div2_even_ext_nat:
fixes x y :: nat
assumes "x div 2 = y div 2"
and "even x ⟷ even y"
shows "x = y"
proof -
from ‹even x ⟷ even y› have "x mod 2 = y mod 2"
by (simp only: even_iff_mod_2_eq_zero) auto
with assms have "x div 2 * 2 + x mod 2 = y div 2 * 2 + y mod 2"
by simp
then show ?thesis
by simp
qed
subsubsection ‹From sets to naturals›
definition set_encode :: "nat set ⇒ nat"
where "set_encode = sum (op ^ 2)"
lemma set_encode_empty [simp]: "set_encode {} = 0"
by (simp add: set_encode_def)
lemma set_encode_inf: "¬ finite A ⟹ set_encode A = 0"
by (simp add: set_encode_def)
lemma set_encode_insert [simp]: "finite A ⟹ n ∉ A ⟹ set_encode (insert n A) = 2^n + set_encode A"
by (simp add: set_encode_def)
lemma even_set_encode_iff: "finite A ⟹ even (set_encode A) ⟷ 0 ∉ A"
by (induct set: finite) (auto simp: set_encode_def)
lemma set_encode_vimage_Suc: "set_encode (Suc -` A) = set_encode A div 2"
apply (cases "finite A")
apply (erule finite_induct)
apply simp
apply (case_tac x)
apply (simp add: even_set_encode_iff vimage_Suc_insert_0)
apply (simp add: set_encode_def finite_vimage_Suc_iff)
done
lemmas set_encode_div_2 = set_encode_vimage_Suc [symmetric]
subsubsection ‹From naturals to sets›
definition set_decode :: "nat ⇒ nat set"
where "set_decode x = {n. odd (x div 2 ^ n)}"
lemma set_decode_0 [simp]: "0 ∈ set_decode x ⟷ odd x"
by (simp add: set_decode_def)
lemma set_decode_Suc [simp]: "Suc n ∈ set_decode x ⟷ n ∈ set_decode (x div 2)"
by (simp add: set_decode_def div_mult2_eq)
lemma set_decode_zero [simp]: "set_decode 0 = {}"
by (simp add: set_decode_def)
lemma set_decode_div_2: "set_decode (x div 2) = Suc -` set_decode x"
by auto
lemma set_decode_plus_power_2:
"n ∉ set_decode z ⟹ set_decode (2 ^ n + z) = insert n (set_decode z)"
proof (induct n arbitrary: z)
case 0
show ?case
proof (rule set_eqI)
show "q ∈ set_decode (2 ^ 0 + z) ⟷ q ∈ insert 0 (set_decode z)" for q
by (induct q) (use 0 in simp_all)
qed
next
case (Suc n)
show ?case
proof (rule set_eqI)
show "q ∈ set_decode (2 ^ Suc n + z) ⟷ q ∈ insert (Suc n) (set_decode z)" for q
by (induct q) (use Suc in simp_all)
qed
qed
lemma finite_set_decode [simp]: "finite (set_decode n)"
apply (induct n rule: nat_less_induct)
apply (case_tac "n = 0")
apply simp
apply (drule_tac x="n div 2" in spec)
apply simp
apply (simp add: set_decode_div_2)
apply (simp add: finite_vimage_Suc_iff)
done
subsubsection ‹Proof of isomorphism›
lemma set_decode_inverse [simp]: "set_encode (set_decode n) = n"
apply (induct n rule: nat_less_induct)
apply (case_tac "n = 0")
apply simp
apply (drule_tac x="n div 2" in spec)
apply simp
apply (simp add: set_decode_div_2 set_encode_vimage_Suc)
apply (erule div2_even_ext_nat)
apply (simp add: even_set_encode_iff)
done
lemma set_encode_inverse [simp]: "finite A ⟹ set_decode (set_encode A) = A"
apply (erule finite_induct)
apply simp_all
apply (simp add: set_decode_plus_power_2)
done
lemma inj_on_set_encode: "inj_on set_encode (Collect finite)"
by (rule inj_on_inverseI [where g = "set_decode"]) simp
lemma set_encode_eq: "finite A ⟹ finite B ⟹ set_encode A = set_encode B ⟷ A = B"
by (rule iffI) (simp_all add: inj_onD [OF inj_on_set_encode])
lemma subset_decode_imp_le:
assumes "set_decode m ⊆ set_decode n"
shows "m ≤ n"
proof -
have "n = m + set_encode (set_decode n - set_decode m)"
proof -
obtain A B where
"m = set_encode A" "finite A"
"n = set_encode B" "finite B"
by (metis finite_set_decode set_decode_inverse)
with assms show ?thesis
by auto (simp add: set_encode_def add.commute sum.subset_diff)
qed
then show ?thesis | 3,367 | 11,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-30 | latest | en | 0.372309 |
https://convertoctopus.com/740-meters-to-yards | 1,621,370,654,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00592.warc.gz | 192,172,808 | 7,880 | ## Conversion formula
The conversion factor from meters to yards is 1.0936132983377, which means that 1 meter is equal to 1.0936132983377 yards:
1 m = 1.0936132983377 yd
To convert 740 meters into yards we have to multiply 740 by the conversion factor in order to get the length amount from meters to yards. We can also form a simple proportion to calculate the result:
1 m → 1.0936132983377 yd
740 m → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 740 m × 1.0936132983377 yd
L(yd) = 809.2738407699 yd
The final result is:
740 m → 809.2738407699 yd
We conclude that 740 meters is equivalent to 809.2738407699 yards:
740 meters = 809.2738407699 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 0.0012356756756757 × 740 meters.
Another way is saying that 740 meters is equal to 1 ÷ 0.0012356756756757 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that seven hundred forty meters is approximately eight hundred nine point two seven four yards:
740 m ≅ 809.274 yd
An alternative is also that one yard is approximately zero point zero zero one times seven hundred forty meters.
## Conversion table
### meters to yards chart
For quick reference purposes, below is the conversion table you can use to convert from meters to yards
meters (m) yards (yd)
741 meters 810.367 yards
742 meters 811.461 yards
743 meters 812.555 yards
744 meters 813.648 yards
745 meters 814.742 yards
746 meters 815.836 yards
747 meters 816.929 yards
748 meters 818.023 yards
749 meters 819.116 yards
750 meters 820.21 yards | 468 | 1,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-21 | latest | en | 0.920917 |
butterwatchyourself.com | 1,498,682,625,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323801.5/warc/CC-MAIN-20170628204133-20170628224133-00326.warc.gz | 742,825,292 | 37,893 | Pies & Tarts
# S’more Pie
Pi Day is March 14 – the day dedicated to celebrating pi (π), the ratio of a circle’s circumference to its diameter. The number of digits in pi are infinite, but it starts with 3.14, hence the March 14 celebration. (And yes, I did have to do a Google search to remember all that. All I could recall was that it had something to do with geometry and circles. Math was not my favorite subject).
The obvious way to celebrate pi day is with pie. The words are pronounced the same, and pies are (generally) circles. It’s a perfect match between math and food.
To celebrate, I decided to create a pie based on s’mores. This is partially because there aren’t many pies that seem seasonal for March. Many of the fruit and nut pies I love to make are more suited to summer and fall. I considered a citrus pie (and in fact I do have a lemon pie recipe I’ll share eventually), but a s’mores pie seemed best both for its seasonless nature and my current cravings.
Many people seem to incorrectly think that s’mores are primarily about the chocolate or the marshmallows. In my opinion, s’mores are actually a delicate balance of the three components, including the graham cracker. The crunch and flavor of the graham cracker are just as important as the other elements, but it often gets forgotten.
For example, did anyone else get the new S’mores Girl Scout Cookies? The bakers definitely did not understand this ratio. The sandwich-style cookies didn’t have nearly enough graham cracker flavor. They were good cookies, but just didn’t taste like s’mores. I wanted to avoid that problem with my pie.
(Interesting side note: did you know that the Girl Scout Cookies vary based on which bakery serves your area of the country? This cool LA Times graphic shows the differences. Some of you got a totally different kind of S’more cookie! Was it better?)
I think this pie celebrates all three components of a good s’more: a layer of graham cracker crust, a chocolate cream filling, and a generous amount of chewy marshmallow on top. You could certainly use a store bought marshmallow fluff or a layer of marshmallows instead, but I recommend making it yourself. It’s much easier than it seems and tastes so much better than anything from the store!
I used milk chocolate because I wanted to stick with the traditional s’mores flavors. If you want something with more chocolate flavor, you could use semisweet or bittersweet, or even a combination of two kinds. And although Hershey bars are traditional with actual s’mores, I’d stick with a chocolate intended for baking (such as Callebaut).
S’more Pie
Makes one 9 inch pie, approximately 8 servings
Graham cracker crust
• 12-14 graham crackers, finely ground in food processor (1 3/4 cups)
• 2 tbsp brown sugar, packed
• 1/2 tsp ground cinnamon
• pinch of salt
• 6 tbsp (3/4 stick) unsalted butter, melted
Preheat the oven to 350. Butter a 9 inch pie pan.
Combine graham cracker crumbs, brown sugar, cinnamon and salt in a food processor and pulse to combine. Add the butter and pulse to form evenly damp crumbs.
Spread the crumbs into the prepared pan and press into the bottom and up the sides. The flat bottom of a measuring cup works well for this. Refrigerate for 5-10 minutes.
Bake for 7-8 minutes, until crust is fragrant and edges are just golden. Let cool completely on a wire rack.
Chocolate filling
• 7 oz chocolate (milk, semi-sweet, or bittersweet), finely chopped
• 1/2 cup heavy cream
• 1/2 cup whole milk
• 2 eggs, lightly whisked
• 1/2 tsp vanilla extract
• pinch of salt
Add milk and cream to a medium sauce pan and warm over medium heat. Add the chocolate and whisk until chocolate is melted and smooth. Add the eggs slowly and whisk until smooth. Off heat, whisk in the vanilla.
Pour chocolate filling into prebaked pie crust. Cover the edge of the crust with a pie shield or ring of tin foil. Bake for 20 minutes, or until the chocolate filling is softly set and wiggles just slightly when gently shaken. Place on a wire rack to cool completely before proceeding, about 1 hour.
Marshmallow topping
• 1 package (1/4 oz) unflavored gelatin
• 3/4 cup sugar
• 1/2 cup light corn syrup
• pinch of kosher salt
• 1/2 tbsp vanilla extract
• cold water
Sprinkle the gelatin over 1/4 cup cold water in the bowl of an electric mixer fitted with the whisk attachment. Let stand so the gelatin can bloom/soften.
In a sauce pan, combine the sugar, corn syrup, salt, and 1/2 cup cold water. Cook over medium heat until the sugar dissolves, then increase the heat to high. Boil until the syrup reaches 240 degrees (soft ball stage) on a candy thermometer.
Remove the thermometer from the syrup and set aside (careful, it will be a bit sticky from the syrup). Turn the mixer on to low speed and slowly pour the hot syrup into the softened gelatin. Turn the mixer up to high and whip until the mixture becomes very thick and increases in volume, about 10 minutes. Add the vanilla and beat until combined.
Immediately spoon topping onto center of the cooled pie. Using a lightly greased spatula, spread the marshmallow to cover the entire top of the pie. Work quickly as the marshmallow will set very quickly. Refrigerate, uncovered, for 1-2 hours (or up to 1 day).
To brown the topping, place the pie under a preheated broiler for about 2-3 minutes. Keep a close eye on it to prevent burning. Or, you can brown it using crème brûlée torch. Again, move the torch frequently to prevent burning in any given spot.
To store, place in a covered container, or invert an empty pie plate on top of the pie. (You don’t want anything to touch the marshmallows or it will stick). I’d store in in the fridge, but it may be served cold or at room temperature.
## One thought on “S’more Pie”
1. Connie Kurihara says:
I want a piece of this pie. And I agree, the Girl Scouts forgot the graham crackers!
Like | 1,424 | 5,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-26 | longest | en | 0.953074 |
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