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These resources have been reviewed and selected by STEM Learning’s team of education specialists for factual accuracy and relevance to teaching STEM subjects in UK schools.
# Properties of Number
This SMILE resource contains three packs of games, investigations, worksheets and practical activities supporting the teaching and learning of the properties of number, and a booklet 'Squares and Primes'.
Properties of number pack one contains sixteen work cards with a wide variety of activities covering odd and even numbers, multiples, rectangle numbers, and factors.
Properties of number pack two contains twelve work cards with activities requiring students to explore prime numbers, find patterns in multiples , use the sieve of Eratosthenes to find prime numbers, investigate triangle numbers, and use factors.
Properties of number pack three contains fourteen work cards with activities involving multiplication patterns, finding prime factors, divisibility problems, number challenges, finding highest common factors and lowest common multiples, investigating quadratics and primes, rational and irrational numbers, and proof by contradiction.
Squares and Primes is a booklet investigating prime numbers, primes and factorials, proof, Euclid and Greek mathematics.
SMILE (Secondary Mathematics Individualised Learning Experiment) was initially developed as a series of practical activities for secondary school students by practising teachers in the 1970s. It became a complete individualised scheme based around a network of activity cards and assessments.
Related resources include answers to all of the cards and test books and answers.
#### Show health and safety information
Please be aware that resources have been published on the website in the form that they were originally supplied. This means that procedures reflect general practice and standards applicable at the time resources were produced and cannot be assumed to be acceptable today. Website users are fully responsible for ensuring that any activity, including practical work, which they carry out is in accordance with current regulations related to health and safety and that an appropriate risk assessment has been carried out.
17.86 MB
Information on the permitted use of this resource is covered by the Category Three Content section in STEM Learning’s Terms and conditions. | 415 | 2,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.947705 |
https://datalemur.com/blog/kpmg-sql-interview-questions | 1,726,771,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00152.warc.gz | 177,199,551 | 25,248 | # 9 KPMG SQL Interview Questions (Updated 2024)
Updated on
August 11, 2024
At KPMG, SQL is used to extract and process client financial data, such as revenue growth and expense patterns, as well as to analyze nuanced data trends specific to the auditing and advisory industry, like identifying anomalies in financial statements. That is why KPMG often asks SQL questions during interviews for Data Analytics, Data Science, and Data Engineering jobs.
To help you ace the KPMG SQL interview, here's 9 KPMG SQL interview questions – can you solve them?
## 9 KPMG SQL Interview Questions
### SQL Question 1: Calculate Customer Lifetime Value (CLV)
As a Data Analyst in KPMG, you're given a database of transactions that customers have made. Each transaction record contains the customer_id, date of transaction, and revenue generated from the transaction.
Your task is to calculate the Customer Lifetime Value (CLV) for each customer. CLV is the total revenue a company can realistically expect from a single customer account. It considers a customer's revenue value and compares that number to the company's predicted customer lifespan. For this task, you'll not consider churn rate.
Assume each customer has at least 1 transaction. If the customer has more than 1 transaction calculate the CLV as the cumulative sum of the revenue for each customer over time. List the top 10 customers with the highest CLV.
##### Example Input:
transaction_idcustomer_idtransaction_daterevenue
132232019-08-22250.00
245632019-11-18300.00
332232020-02-14200.00
467122020-05-20150.00
545632020-08-11250.00
667122020-10-22300.00
732232021-03-14500.00
867122021-06-10350.00
945632021-09-02400.00
1067122021-12-14250.00
##### Expected Output:
customer_idCLV
3223950.00
4563950.00
67121050.00
This PostgreSQL query calculates the cumulative sum of the revenue for each customer in the transactions table, ordered by the transaction_date. It uses a window function to create a running total of revenue for each customer_id. The query then orders the results by the CLV in descending order and limits the output to the top 10 customers with the highest CLV.
p.s. Window functions show up pretty often during SQL interviews, so practice the 27+ window function questions on DataLemur
### SQL Question 2: Top Department Salaries
Given a table of KPMG employee salary data, write a SQL query to find the top 3 highest paid employees in each department.
#### KPMG Example Input:
employee_idnamesalarydepartment_id
1Emma Thompson38001
2Daniel Rodriguez22301
3Olivia Smith20001
4Noah Johnson68002
5Sophia Martinez17501
8William Davis68002
10James Anderson40001
#### Example Input:
department_iddepartment_name
1Data Analytics
2Data Science
#### Example Output:
department_namenamesalary
Data AnalyticsJames Anderson4000
Data AnalyticsEmma Thompson3800
Data AnalyticsDaniel Rodriguez2230
Data ScienceNoah Johnson6800
Data ScienceWilliam Davis6800
Code your solution to this question interactively on DataLemur:
We use the DENSE_RANK() window function to generate unique ranks for each employee's salary within their department, with higher salaries receiving lower ranks. Then, we wrap this up in a CTE and filter the employees with a ranking of 3 or lower.
If the code above is tough, you can find a step-by-step solution here: Top 3 Department Salaries.
### SQL Question 3: What are SQL constraints, and can you give some examples?
Think of SQL constraints like the rules of a game. Just like a game needs rules to keep things fair and fun, a database needs constraints to keep things organized and accurate.
There are several types of SQL constraints like:
NOT NULL: This constraint is like a bouncer at a nightclub - it won't let anything NULL through the door.
UNIQUE: This constraint is like a VIP list - only special, one-of-a-kind values get in.
PRIMARY KEY: This constraint is like an elected official - it's made up of NOT NULL and UNIQUE values and helps identify each row in the table.
FOREIGN KEY: This constraint is like a diplomatic ambassador - it helps establish relationships between tables.
CHECK: This constraint is like a referee - it makes sure everything follows the rules.
DEFAULT: This constraint is like a backup plan - it provides a default value if no other value is specified.
So, whether you're playing a game or organizing a database, constraints are an important part of the process!
### SQL Question 4: Average Hours Billed Per Project
KPMG as a consulting firm executes various projects for different clients. Each project often consists of multiple tasks and each task is assigned to one or more employees, where each employee can log the hours they've worked. For this question, we're interested in the average number of hours billed per project. Write a SQL query which calculates the average hours billed to each project.
##### Example Input:
project_idclient_nameproject_name
100AppleMarket Analysis
50010012310
5011012658
5021013626
5031021929
50410298111
##### Example Output:
project_idproject_nameavg_hoursBilled
100Market Analysis10.00
101Competitive Research7.00
In this query, we first join the projects and tasks tables on project_id. We then group by project_id and project_name to split our data into sections for each project. Within these sections, we use the AVG function to calculate the average hours billed to each project.
To practice a very similar question try this interactive Microsoft Teams Power Users Question which is similar for calculating usage metrics or this Amazon Server Utilization Time Question which is similar for summarizing time usage.
### SQL Question 5: When would you use the constraint?
A is a field in a table that references the of another table. It creates a link between the two tables and ensures that the data in the field is valid.
For example, if you have a table of KPMG customers and an orders table, the customer_id column in the orders table could be a that references the id column (which is the primary key) in the KPMG customers table.
The constraint helps maintain the integrity of the data in the database by preventing the insertion of rows in the table that do not have corresponding entries in the table. It also enforces the relationship between the two tables and prevents data from being deleted from the table if it is still being referenced in the table.
### SQL Question 6: Analysis of Click-through Rates for KPMG's Digital Marketing Campaigns
KPMG, a global network of professional firms providing Audit, Tax and Advisory services, has recently increased its online marketing efforts. As an analyst for KPMG, your task is to analyze the click-through rates for their ongoing digital marketing campaigns. This involves calculating the rate at which users who see digital ads on various platforms (Google, Facebook, LinkedIn, etc.) end up landing on KPMG's website.
After clicking on the ad, some users might further opt for a service enquiry.
Given two tables - and , write a SQL query that calculates the proportion of total users who clicked an ad () and subsequently made a service enquiry () for each platform.
##### Example Input:
enquiry_iduser_idenquiry_date
71245008/16/2022 00:00:00
802320007/16/2022 00:00:00
493155006/30/2022 00:00:00
12670007/02/2022 00:00:00
117980005/25/2022 00:00:00
This query joins the table with the table on , considering only the cases where the is after or on the .
Then it aggregates the data by , and computes the rate of enquiries per clicks for each platform by dividing the COUNT of DISTINCT s in the by the COUNT of DISTINCT s in the , multiplied by 100 to get the percentage rate. This gives us the 'click-to-enquiry' rate for each platform.
To solve a related SQL problem on DataLemur's free online SQL code editor, try this SQL interview question asked by Facebook:
### SQL Question 7: What's a database view?
Views are a lot like virtual tables, where you can take a base table and customize it (such as by hiding some data from non-admin users, or removing some random columns/rows based on business requirements).
Here's the PostgreSQL syntax for creating a view based on data in the table:
### SQL Question 8: Maximum audit risk by sector
Given a table , where each row represents an audit conducted by KPMG, find the maximum audit risk for each sector for audits that were completed during the year 2021.
The table has the following schema:
audit_idsectorcompany_idaudit_dateaudit_risk
##### Example Input:
audit_idsectorcompany_idaudit_dateaudit_risk
641Finance1002021-02-060.76
702IT1012021-04-110.89
529Finance1022021-06-180.85
635Manufacturing1032021-08-030.91
417IT1042021-05-200.72
##### Example Output:
sectormax_audit_risk
Finance0.85
IT0.89
Manufacturing0.91
Here is the PostgreSQL query which calculates the maximum audit risk by sector for the year of 2021:
The statement is used to group the audits by sector. The clause is used to filter out audits that were not performed in the specified year (2021 in this case). And finally, is used to retrieve the highest audit risk from each sector. As you might already know, is a PostgreSQL function that returns the year part of a date as an integer.
### SQL Question 9: Finding Customers with Partial Email Domain
KPMG's Marketing Department would like to run an email campaign targeting their customers whose email providers are Gmail. They've noticed that a significant portion of their customer base is using Gmail, and they would like to build a specific marketing strategy around this. Therefore, the task is to find all the customers whose email IDs are registered with a '@gmail.com' domain.
##### Example Input:
customer_idfirst_namelast_nameemail
6171JohnDoejohn.doe@gmail.com
7802JaneSmithjane.smith@yahoo.com
5293RichardRoerichard.roe@hotmail.com
6352JamesBondjames.bond@gmail.com
4517AliceCooperalice.cooper@gmail.com
##### Example Output:
customer_idfirst_namelast_nameemail
6171JohnDoejohn.doe@gmail.com
6352JamesBondjames.bond@gmail.com
4517AliceCooperalice.cooper@gmail.com
This query uses the SQL keyword to filter rows from the customers database table. It matches customer records where the email column ends with '@gmail.com'. Therefore, it returns all customer records that are associated with a Gmail account. The percent symbol (%) is used as a wildcard character that can match any sequence of characters, and the at sign (@) and the text 'gmail.com' form the specific pattern to be matched.
### Preparing For The KPMG SQL Interview
The key to acing a KPMG SQL interview is to practice, practice, and then practice some more! Beyond just solving the above KPMG SQL interview questions, you should also solve the 200+ SQL questions from real Data Science & Analytics interviews which come from companies like Microsoft, Google, Amazon, and tech startups.
Each DataLemur SQL question has hints to guide you, full answers and best of all, there's an interactive coding environment so you can right online code up your query and have it checked.
To prep for the KPMG SQL interview it is also helpful to practice SQL questions from other accounting & consulting companies like:
Find out how KPMG is using Artificial Intelligence to drive innovation, improve efficiency, and reduce costs!
However, if your SQL query skills are weak, forget about diving straight into solving questions – improve your SQL foundations with this DataLemur SQL Tutorial.
This tutorial covers SQL topics like AND/OR/NOT and CASE/WHEN/ELSE statements – both of which pop up often in SQL job interviews at KPMG.
### KPMG Data Science Interview Tips
#### What Do KPMG Data Science Interviews Cover?
Beyond writing SQL queries, the other types of questions covered in the KPMG Data Science Interview include:
#### How To Prepare for KPMG Data Science Interviews?
To prepare for KPMG Data Science interviews read the book Ace the Data Science Interview because it's got:
• 201 interview questions sourced from FAANG tech companies
• a crash course covering Stats, SQL & ML
• over 1000+ 5-star reviews on Amazon
Don't ignore the behavioral interview – prep for that with this guide on acing behavioral interviews. | 2,751 | 12,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.845104 |
https://alphons.io/qa/683/how-to-convert-meters-per-second-to-revolutions-per-second/ | 1,620,537,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00330.warc.gz | 120,800,670 | 22,032 | Question #683
# How to convert meters per second to revolutionS per second?
Merged questions
The formula to convert linear speed expressed in meters per second [] to angular velocity expressed in revolutions per seconde [] is given by:
where:
• is the linear velocity expressed in meters per second ()
• i s the angular velocity expressed in revolution per second ()
• i s the radius expressed in meters ()
4 events in history
Question by Alphonsio 03/01/2021 at 03:18:13 PM
How to convert meters per second to revolutionS per second?
Question by Alphonsio 12/10/2020 at 03:44:30 PM
How to convert meters per second to revolution per second?
Answer by Alphonsio 12/10/2020 at 03:23:10 PM
The formula to convert linear speed expressed in meters per second [] to angular velocity expressed in revolutions per seconde [] is given by:
where:
• is the linear velocity expressed in meters per second ()
• i s the angular velocity expressed in revolution per second ()
• i s the radius expressed in meters ()
Question by Alphonsio 12/10/2020 at 03:17:57 PM
What is the formula to convert linear velocity in meters per second to angular velocity in revolution per second?
Icons proudly provided by Friconix. | 336 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.851167 |
http://mathworkorange.com/radianarc-lengthand-area-of-a-sector.html | 1,501,021,960,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425407.14/warc/CC-MAIN-20170725222357-20170726002357-00677.warc.gz | 219,905,564 | 11,289 |
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# Radian,Arc Length,and Area of a Sector
An angle is formed by two rays that share a common endpoint called the vertex of the angle. The vertex can
represent a point of rotation in that one ray is the initial side of the angle before rotation and the other ray
forms the terminal side after the rotation.
One principal means of measuring the size of an angle is using degree measure. An angle formed by the
rotation of a ray through one complete circle measures 360 degrees.
Another principal means of measuring the size of an angle is using radian measure. Begin with a circle of
radius r and let θ be a central angle, an angle whose vertex is the center of the circle. Let s be the
length of the arc that is intercepted by the angle. The radian measure of θ is given by θ = s/r,
where r and s have the same linear units.
Example 1: A circle has a radius of 9 inches. A central angle, θ, intercepts an arc of length 36
inches. What is the radian measure of θ ?
Example 2: A circle has a radius of 24 inches. A central angle, θ, intercepts an arc of length 7
feet. What is the radian measure of θ?
To devise a formula for converting between degree measure and radian measure consider the right angle
shown in the figure below.
Since θ measures 90° the arc length s is 1/4 the circumference of the circle.
Use the formula for radian measure and substitute for s.
To obtain formulas for converting between degree measure and radian measure, multiply
both sides of the equation above by 2 180° = π radians.
From the formula above, we see that rasians ( ≈ 0.017 rasian ) and 1radian = 180/π (≈ 57.3°).
To convert from degree measure to radian measure, multiply degrees by π/180°.
To convert from radian measure to degree measure, multiply radians by 180°/π.
Common Angles
Memorize the common angles.
Example 3: If two angles of a triangle have radian measures 2π/7 and π/9, find the radian measure of
the third angle.
Example 4: Convert the following degree measure to radians.
300°
Example 5: Convert the following degree measure to radians.
50°
Example 6: Convert the following radian measure to degrees.
7π/4
Example 7: Convert the following radian measure to degrees.
π/12
In a circle of radius r, the arc length s that is determined by a central angle of radian measure θ is given by the
arc length formula s = rθ. where r and s have the same linear units.
In a circle of radius r, the area A of a sector with central angle of radian measure θ is given by the
area sector formula
Example 8: Find the radius of a circle if the length of the intercepted arc by a central angle 4π/3
is 39π/2 inches.
Example 9: Find the sector area determined by a circle that has a radius of 5 cm and central angle 7π/6.
Example 10: A sector of a circle has a central angle 135° and area Find the radius of the
circle.
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http://de.metamath.org/mpeuni/bj-cbvex2v.html | 1,660,325,025,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00511.warc.gz | 13,149,694 | 3,887 | Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-cbvex2v Structured version Visualization version GIF version
Theorem bj-cbvex2v 31925
Description: Version of cbvex2 2268 with a dv condition, which does not require ax-13 2234. (Contributed by BJ, 16-Jun-2019.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
bj-cbval2v.1 𝑧𝜑
bj-cbval2v.2 𝑤𝜑
bj-cbval2v.3 𝑥𝜓
bj-cbval2v.4 𝑦𝜓
bj-cbval2v.5 ((𝑥 = 𝑧𝑦 = 𝑤) → (𝜑𝜓))
Assertion
Ref Expression
bj-cbvex2v (∃𝑥𝑦𝜑 ↔ ∃𝑧𝑤𝜓)
Distinct variable group: 𝑥,𝑦,𝑧,𝑤
Allowed substitution hints: 𝜑(𝑥,𝑦,𝑧,𝑤) 𝜓(𝑥,𝑦,𝑧,𝑤)
Proof of Theorem bj-cbvex2v
StepHypRef Expression
1 bj-cbval2v.1 . . . . 5 𝑧𝜑
21nfn 1768 . . . 4 𝑧 ¬ 𝜑
3 bj-cbval2v.2 . . . . 5 𝑤𝜑
43nfn 1768 . . . 4 𝑤 ¬ 𝜑
5 bj-cbval2v.3 . . . . 5 𝑥𝜓
65nfn 1768 . . . 4 𝑥 ¬ 𝜓
7 bj-cbval2v.4 . . . . 5 𝑦𝜓
87nfn 1768 . . . 4 𝑦 ¬ 𝜓
9 bj-cbval2v.5 . . . . 5 ((𝑥 = 𝑧𝑦 = 𝑤) → (𝜑𝜓))
109notbid 307 . . . 4 ((𝑥 = 𝑧𝑦 = 𝑤) → (¬ 𝜑 ↔ ¬ 𝜓))
112, 4, 6, 8, 10bj-cbval2v 31924 . . 3 (∀𝑥𝑦 ¬ 𝜑 ↔ ∀𝑧𝑤 ¬ 𝜓)
1211notbii 309 . 2 (¬ ∀𝑥𝑦 ¬ 𝜑 ↔ ¬ ∀𝑧𝑤 ¬ 𝜓)
13 2exnaln 1746 . 2 (∃𝑥𝑦𝜑 ↔ ¬ ∀𝑥𝑦 ¬ 𝜑)
14 2exnaln 1746 . 2 (∃𝑧𝑤𝜓 ↔ ¬ ∀𝑧𝑤 ¬ 𝜓)
1512, 13, 143bitr4i 291 1 (∃𝑥𝑦𝜑 ↔ ∃𝑧𝑤𝜓)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 195 ∧ wa 383 ∀wal 1473 ∃wex 1695 Ⅎwnf 1699 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 This theorem is referenced by: bj-cbvex2vv 31927
Copyright terms: Public domain W3C validator | 1,019 | 1,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | latest | en | 0.229251 |
https://www.cs.bu.edu/fac/betke/cs585/restricted/hw3/ | 1,631,947,648,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00586.warc.gz | 735,446,752 | 1,801 | ## CAS CS 585 Image and Video Computing - Spring 2021
Homework due date: 1 day before exam, which is Wednesday, February 24, 9:30 am, No late submissions accepted because solutions will be immediately published.
Exercise 1: Hausdorff Distance
There are two shapes in the figure below. The triangle in red has three vertices: A(-2,3), B(3,1), and C(0,-3). The rectangle in blue has four vertices: D(-3,2), E(2,2), F(2,-1), and G(-3, -1).
1. Consider two point sets S1 and S2, where S1={A, B, C} and S2={D, E, F, G}. Calculate the Hausdorff distance between these two point sets.
2. Now consider all the points forming these two polygons. What's the Hausdorff distance between the triangle and the rectangle?
Exercise 2: Segmentation
The following tables include all the local maxima and local minima of the grayscale histogram of an image.
Table 1: Local Maxima
Gray Values # of Pixels 52 54 103 231 1000 1170 1750 1300
Table 2: Local Minima
Gray Values # of Pixels 0 53 75 157 255 500 890 240 190 310
Calculate the highest peakiness and the corresponding threshold using Mode Method.
Programming Assignment on Segmentation
The programming component of this homework will be published after the exam due date. | 336 | 1,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-39 | latest | en | 0.856933 |
https://www.eevblog.com/forum/beginners/why-wont-my-meters-read-millivolts/ | 1,586,116,566,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371609067.62/warc/CC-MAIN-20200405181743-20200405212243-00200.warc.gz | 884,402,573 | 12,252 | ### Author Topic: Why won't my meters read millivolts? (Read 1451 times)
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#### billbyrd1945
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• Posts: 144
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##### Why won't my meters read millivolts?
« on: October 16, 2018, 02:33:44 am »
I have a Chinese Fluke 16B and a UNI-T UT61E. I'm trying to learn Arduino. I need to measure millivolts on my current project. Both meters will display DC volts, but neither display anything on DC millivolts. I'll appreciate any help.
#### Nusa
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• Posts: 1716
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##### Re: Why won't my meters read millivolts?
« Reply #1 on: October 16, 2018, 02:56:18 am »
What does the V range display for the voltage you are trying to measure in mV?
#### Terry01
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##### Re: Why won't my meters read millivolts?
« Reply #2 on: October 16, 2018, 03:39:14 am »
How many Mv are you trying to measure? You may be trying to measure too many so therefor need to move up to the V range. I'm sure after about 400Mv the Mv range shows "OL" and tops out.
Sparks and Smoke means i'm nearly there!
#### ArthurDent
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##### Re: Why won't my meters read millivolts?
« Reply #3 on: October 16, 2018, 03:40:22 am »
Here's my UT61E displaying 1.80 millivolts. One thing that might have happened is that you pressed the yellow function button and the meter is set to Hz which on DC would give you all zeros. More info is needed and a photo would help. Is it set to DC and not AC?
#### billbyrd1945
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• Posts: 144
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##### Re: Why won't my meters read millivolts?
« Reply #4 on: October 16, 2018, 05:27:41 pm »
Thanks to everyone. I hope these pictures will help you to help me. The sensor has a pos and neg leg and a center output pin. The output of about 0.6v is correct. But, on millivolts on either of my meters I get OL. In the pictures attached, I merely moved the dial from DCvolts to DCmillivolts.
#### Karlo_Moharic
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##### Re: Why won't my meters read millivolts?
« Reply #5 on: October 16, 2018, 05:52:38 pm »
#### 2N3055
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##### Re: Why won't my meters read millivolts?
« Reply #6 on: October 16, 2018, 06:05:37 pm »
Take a manual for your meter.
What does it say about measurement ranges?
220 mV
2.2V etc...
Meaning you cannot measure more than 220 mV in mV range. That is 0.22V. You cannot measure 0.6V on that range. You have to switch to next higher range. That's 2.2V.
#### Nusa
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##### Re: Why won't my meters read millivolts?
« Reply #7 on: October 16, 2018, 06:14:28 pm »
To be clear, you're simply exceeding the range of the higher resolution mV scale provided by the meters.
In the case of the UT61E, that limit is stated to be 220mV, resolution 0.01 mV, in the manual (findable online if you've lost yours). On the V setting, when the reading is under 2.2V resolution is 0.0001V (0.1 mV).
The answer is don't use the mV scale for this case. Read the V scale. Convert to mV if you want.
0.6038V = 603.8 mV
I didn't look up the numbers for your fluke, but it's presumably a range issue there as well.
#### billbyrd1945
• Regular Contributor
• Posts: 144
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##### Re: Why won't my meters read millivolts?
« Reply #8 on: October 16, 2018, 10:28:05 pm »
Okay. I get it. I wondered about that from the get go but doubted anyone would build a meter that wouldn't display all values while having the technology to do so at no extra cost (presumably). I did look in the manual. It started with 你好
I don't read a word of Latin so figured I'd just have to wing it. Thanks guys. I'm good now.
#### Nusa
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##### Re: Why won't my meters read millivolts?
« Reply #9 on: October 16, 2018, 11:20:21 pm »
Here's an English version of the UT61E manual, then, for your reference. Specifications start on page 42.
#### billbyrd1945
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##### Re: Why won't my meters read millivolts?
« Reply #10 on: October 16, 2018, 11:27:55 pm »
Thank you very much. This has been a good learning experience for this 73y/o.
#### Mechatrommer
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• reassessing directives...
##### Re: Why won't my meters read millivolts?
« Reply #11 on: October 16, 2018, 11:52:16 pm »
OL = Over Load.. did i just do that? or its already there?
if something can select, how cant it be intelligent? if something is intelligent, how cant it exist?
#### Nusa
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##### Re: Why won't my meters read millivolts?
« Reply #12 on: October 17, 2018, 12:18:31 am »
OL = Over Load.. did i just do that? or its already there?
It's in the manual I linked. OL = "The input value is too large for the selected range."
#### billbyrd1945
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##### Re: Why won't my meters read millivolts?
« Reply #13 on: October 17, 2018, 12:43:03 am »
"It's in the manual I linked. OL = "The input value is too large for the selected range."
Thanks once again! I always thought OL meant open line.
#### Mechatrommer
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• reassessing directives...
##### Re: Why won't my meters read millivolts?
« Reply #14 on: October 17, 2018, 01:42:44 am »
yes that is resistance mode. whatever it is as long as it helps our mental picture.
if something can select, how cant it be intelligent? if something is intelligent, how cant it exist?
Smf | 1,650 | 5,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-16 | latest | en | 0.92252 |
http://www.fangraphs.com/blogs/pick-offs-and-stolen-base-attempts/ | 1,394,444,631,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010732251/warc/CC-MAIN-20140305091212-00034-ip-10-183-142-35.ec2.internal.warc.gz | 329,055,630 | 25,770 | Pick-offs and Stolen Base Attempts
Last week, I looked at which teams were most likely to have players thrown out while running the bases (e.g caught stealing, picked-off, throw out while trying to take extra base, etc). In the comments of the article, it was discussed that the more aggressive base running teams are more likely to be thrown out on the bases. I am working toward a better solution for those base running numbers, but in the meantime I found some nice information on players getting picked-off.
The more aggressive a team is at attempting a stolen base, the more likely they are of getting picked-off. It seems like common sense to me, but I have had too many incorrect ideas to leave it only to instinct. The following graph looks at the attempted steals versus pick-offs for all teams from 2005 to 2009:
As it can be seen, the more aggressive a team is attempting to steal, the more likely they are of getting picked-off. Using the equation of the best fit line, it can be determined that for every 12.5 stolen base a team attempts, one player is likely to get picked off (12.5 attempts * 0.080 = 1).
Note: The relationship between the two values, doesn’t mean that one directly caused the other. There could be other factors at work on the two values.
I took this examination one step further and compared the times caught stealing versus time picked-off. I was looking to see if teams that were bad at stealing bases were also bad at getting picked-off:
My hunch was correct in that the r-squared (how closed one set of values correlates to another sets of values) is a bit higher (0.37 vs 0.32) for the caught stealing data. Using the values from the equation, it can be shown that for every 3 times a player is caught stealing, they are likely to be picked-off once (0.32 * 3 = ~1).
So far this season the numbers are similar to the previous 5 seasons as seen in the following two graphs:
Aggressive base stealing teams are more likely to be picked-off thereby removing a base runner. Rich Lederer proposed back in 2006 that the caught stealing value should include both caught stealing and picked-off numbers. I am not sure how the baseball community would accept that change, but if someone does include picked-off outs into caught stealing values, I could understand the reason why. For now it seems that teams looking to get an extra jump for a stolen base seem get thrown before they have the chance than those that are less likely to attempt the steal in the first place.
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Jeff writes for FanGraphs, The Hardball Times and Royals Review, as well as his own website, Baseball Heat Maps with his brother Darrell. In tandem with Bill Petti, he won the 2013 SABR Analytics Research Award for Contemporary Analysis. Follow him on Twitter @jeffwzimmerman.
15 Responses to “Pick-offs and Stolen Base Attempts”
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1. vivaelpujols says:
Whats the slope and correlation between SB successes and times picked off. If it’s lower than the caught stealings, it could be that good basestealing teams avoid pickoffs in addition to stealing more bases, or the other way around.
2. UWHabs says:
Does the relationship hold true to individual players as well, or are pickoffs of individual players too rare to really get meaningful numbers out?
• Jeff Zimmerman says:
I don’t have the data available right, I was just working with team data. Let me see what I can find.
3. Kevin Buterbaugh says:
There is a clear relationship here – but it is quite weak – .374 r square etc. So, let’s not over exaggerate the conclusion. There is a lot of variance here – the more interesting question now becomes why are some teams quite good at attempting steals without getting picked off while others are not. One of the teams attempted about 150 steals with only 6 or 7 pick offs in the first graph while another was picked off 27 times – why is that it.
• Jeff Zimmerman says:
I am hoping to find other possible causes. The rest of the baseball running data is a little harder to piece together and evaluate, but I am getting there. Stay tuned.
• oompaloopma says:
Hit and runs should be factored some how especially deciding a single players stats. He has been picked off because the coach called a sign expecting contact when batter fails then probably a guy who should not be stealing just got caught stealing.
4. OT says:
excellent article, thanks for including the graphs!
• Jeff Zimmerman says:
I love graphs – I will probably put in too many than not enough.
5. Joe R says:
Clear correlation, but I wouldn’t exactly go into a meeting w/ those r-squared values.
6. Otter says:
Doesn’t the team that has Scott Podsednik sort of ruin this (or at least become the outliar)? Not only does he steal a lot of bases, but he’s also very very very good at getting picked off?
• Joe R says:
He is also 2nd in MLB in CS.
• Jeff Zimmerman says:
Way to rub salt into my wounds that are my Royal’s loyalty.
7. Dylan says:
I haven’t really studied statistics in a while, is an r^2 valued of .374 high enough to even say there’s really a correlation. That value seems pretty weak, almost to the point of being statistically insignificant.
Also, just out of curiosity, if you still have the statistics at hand, who are the two teams that had less than 100 SB attempts and more than 25 times picked off. Those seem like extremely awful outliers, to the point where it might be making the relationship look worse than it really should be.
8. Mark says:
I probably missed something but why are runners who W or are HBP or get on because of CI not included? Aren’t they just as likely to attempt a steal as any other runner? And do you think there is any correlation to base running philosophy (Managers who like play station-to-station, 3-Run HR, etc., verses a more aggressive style) and the percentage you are projecting? Thanks for the article. | 1,315 | 5,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2014-10 | longest | en | 0.981862 |
http://www.bloggingtheboys.com/2012/9/6/3296815/cowboys-game-1-recap | 1,408,686,835,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500823169.67/warc/CC-MAIN-20140820021343-00138-ip-10-180-136-8.ec2.internal.warc.gz | 271,942,725 | 28,121 | ## Cowboys Game 1 Recap
Great game last night. Lots of points I want to make but I want to add a new section to my recap posts.
Expected Win Projections
As most of you know, Statistics are great. Well, when I say great, I mean they're great for a nice rear-view mirror look at what happened during a game. I couldn't tell you how well the Boys are going to do against the Seahawks because of last night, and that's just an aspect of Seattle being a completely different team than NY. However, a funny thing does happen with statistics. Over the course of a season, you can start converting them into expected wins with a couple of different formulas. You've probably seen these before from both OCC and FiTaT. I want to do a cumlative approach this season and show how different games and factors affect our final win tally. Just how much does 1 bad or good game affect our win tally?
(Well obviously there's no causal link but it will be interesting to see whether or not updated projections from a bad game will factor into our final win tally)
So without further ado here's my first set of predictions.
NY/A
I'm obsessed with this statistic. Ok, maybe not "really" obsessed but I still think it's really valuable. As a disclaimer, I just want to say that it's dangerous to use this as a statistic to measure a QB. This stat should measure a team offense and a team defense.
Week O NY/A D NY/A Projected Wins
1 9.35 5.34 23.16
So, yeah. That's the one downside to this formula. It's slow to get started. I mean, don't get me wrong it would be great for the Cowboys to win 23 wins but that's just not possible. Well, maybe the Cowboys could break the laws of football, but I wouldn't bet on it. Give this statistic a few weeks to get off the ground then hopefully it will start looking reasonable. Or the Cowboys could just be the best team ever. Yeah, I'm going with that.
NB: This is a correlative projection based off of a Linear Regression. No team in my calculations finished the year with a NY/A Differential as high as 4. That will come down. A couple of losses and close games should bring that down.
Pythagorean Wins:
I'd like to think Bill James' spirit lives on within all of us. Wait, what do you mean he's still alive? Well regardless, Bill James' Pythagorean model of wins still is one of the most accurate predictors of future success.So let's see what this specific 1 win tells us about the Cowboys Chances so far:
Week Points Scored Points Allowed Projected Wins
1 24.00 17.00 10.64
Yeah, looks pretty good. So a warning stating the obvious. This is taking the statistical sample of one game. I don't want anyone think the Cowboys are guaranteed 10-11 wins because of 1 win versus the Giants this year, although I think all of you know that. Of course the projection looks good; we haven't lost yet. That will change. I doubt this team is a 16 winner.
NB: This is based off the formula ((Points Scored)^2)/((Points Scored)^2+(Points Alowed)^2)* Total Number of Games
So that's what the two indicators look like so far. I'll keep cumulative updates on these statistics as the season goes on and we'll see if these stats become indicators of the team's success over the course of the season.
Please leave a comment below if you have any other indicators you want me to monitor over the course of the season.
Other Thoughts
Well with the statistics out of the way I wanted to comment on other relevant items:
The Emergence of Kevin Ogletree:
I'm eating crow this morning, and I'm eating a lot of it. On August 12th when I heard Kevin Ogletree was in the lead for the 3rd Wide Receiver position I made this insightful comment:
Right:
The man who in the last 3 years has had 25 receptions and 294 yards is by far the model of consistency the Cowboys must relish.
I don't know what got into this guy. Well, OK I have a theory, and others have stated things similar to this but I want to put my own version out there. The #3 Wide Receiver position might be the most important position on the Dallas Cowboys Offense. With Miles Austin and Dez Bryant you have two match up busters who will always demand a lot of respect by opposing coordinators because they will beat you if you slack up a little. (They often did beat the Giants last night)
So if your first and second Wide Receivers are being double covered and bracketed all night, especially if your Hall Of Fame Tight End is a (very heroic) non-factor, then there's very little chance they'll be able to have 150 yard games with a couple of Touch Downs. Kevin Ogletree got burn, because you can't double cover everyone. You can't have six Defensive Backs covering 3 receivers, a Linebacker covering your Tight End and hope to stop the run with 4 men in the box. That's not to mention the threat of throwing the football to your Running Back. So one of the Wide Receivers is going to get single coverage, and that should be the 3rd receiver. It is incumbent to have a good 3rd WR, especially if your Offensive Line still isn't getting you the time needed for your match-up busters to get open. I don't think Laurent Robinson broke out last year, Robinson did his job and rode a great QB to a great #2 WR contract.
Kevin Ogletree did what he needed to do to get good enough to be a target for Romo. That's what we need from him. If he keeps it up he'll be a thousand yard receiver no problem.
The Defense:
Wow, how about this team's performance. Whatever way you slice it this was good. We held the Giants defense to a 5.34 NY/A. Eli got 213 yards on 32 passes. That's incredible. And the point I want to emphasize is this, Eli didn't throw a pick. Maybe this is counter-intuitive, but what I mean is this: Turnovers greatly affect the outcome of the game. If you cough up a turnover there's a much slimmer chance of you winning. I'm aware of the WIlson turnover, but that didn't result in anything. This defense was downright dominant, and it did so without being opportunistic.
It's a bit to early to generalize, but if this defense can keep it up, the turnovers will come (most likely). This team can be good without relying on turnovers. That's a great sign.
How about that goal-line stand by the way? I don't care about the no-call on Scandrick; it makes up for the horrible no-call on Marty B for pushing off of the Safety. That was a downright dominant defense. Yes, I'm aware that the Giants running game isn't the best, but do you recall how much the Giants rushed against us last year? It certainly didn't feel like they were a bad rushing team. So I'm going to enjoy that they had to settle for a Field Goal.
Romo:
This wasn't the best game of his career. Oh don't get me wrong, it was amazing. But has everyone forgot 2011 against Buffalo? That was an AMAZING game, versus a team everyone thought was playoff caliber.
Quarterbacks are going to throw interceptions. Romo couldn't step into the pass and so made a bad pass. It happens, the Defense stepped up. There is no value you can place on a defense that can cover its offense's mistake. I'm of the impression that for the past couple of years that has been what is missing from the Cowboys. Teams sometimes throw duds, and the offense won't always be dominant. Is the defense good enough that most of the games the Offense implodes it can step up and keep us in the game? I can remember one or two times last year that was the case, but the defense needs to be much better than that. I still expect the offense will have to cover up for the defense sometimes, but if the defense can cut down on the number of times that is necessary and can return the favor to the offense, we're looking at the beginnings of a really good football team.
That's a lot of ifs, but that's what's to be expected from Week 1 Football. Take your time and enjoy the win Cowboys fans; they deserved the win. This team is not in mid-season form, that should scare all of us. I'm aware that neither the Giants nor any other team is, but the Cowboys displayed a level of dominance that should have us licking our chops.
But hey, that could just be the kool-aid talking.
Another user-created commentary provided by a BTB reader.
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In order to provide our users with a better overall experience, we ask for more information from Facebook when using it to login so that we can learn more about our audience and provide you with the best possible experience. We do not store specific user data and the sharing of it is not required to login with Facebook. | 2,237 | 9,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-35 | latest | en | 0.976374 |
https://discourse.mc-stan.org/t/incorporating-known-detection-probabilities-into-mark-recapture-models/12209 | 1,660,161,682,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00002.warc.gz | 212,623,128 | 5,798 | # Incorporating known detection probabilities into mark-recapture models
Hi all,
I have a mark-recapture model that borrows heavily from code here. The model estimates detection probabilities and survival for sample units. The data I currently have has four time steps (for example, 1101 or 1001 or 1111).
I want to borrow the detection probabilities so that I can look at the same data set for just two time steps (e.g. 10 or 11), but I am not sure how to do this. I don’t think I need to stay with my current mark recapture model, but rather go with a simpler binomial model with a detection probability included. I am just not sure how to do this.
Any help on getting started would be appreciated. Here is the code I have for modeling temporal effects with no covariates
``````// This models is derived from section 12.3 of "Stan Modeling Language
// User's Guide and Reference Manual"
functions {
int first_capture(int[] y_i) {
for (k in 1:size(y_i))
if (y_i[k])
return k;
return 0;
}
int last_capture(int[] y_i) {
for (k_rev in 0:(size(y_i) - 1)) {
// Compoud declaration was enabled in Stan 2.13
int k = size(y_i) - k_rev;
// int k;
// k = size(y_i) - k_rev;
if (y_i[k])
return k;
}
return 0;
}
matrix prob_uncaptured(int nind, int n_occasions,
matrix p, matrix phi) {
matrix[nind, n_occasions] chi;
for (i in 1:nind) {
chi[i, n_occasions] = 1.0;
for (t in 1:(n_occasions - 1)) {
// Compoud declaration was enabled in Stan 2.13
int t_curr = n_occasions - t;
int t_next = t_curr + 1;
/*
int t_curr;
int t_next;
t_curr = n_occasions - t;
t_next = t_curr + 1;
*/
chi[i, t_curr] = (1 - phi[i, t_curr])
+ phi[i, t_curr] * (1 - p[i, t_next - 1]) * chi[i, t_next];
}
}
return chi;
}
}
data {
int<lower=0> nind; // Number of individuals
int<lower=2> n_occasions; // Number of capture occasions
int<lower=0,upper=1> y[nind, n_occasions]; // Capture-history
}
transformed data {
int n_occ_minus_1;
// Compoud declaration is enabled in Stan 2.13
// int n_occ_minus_1 = n_occasions - 1;
int<lower=0,upper=n_occasions> first[nind];
int<lower=0,upper=n_occasions> last[nind];
n_occ_minus_1 = n_occasions - 1;
for (i in 1:nind)
first[i] = first_capture(y[i]);
for (i in 1:nind)
last[i] = last_capture(y[i]);
}
parameters {
vector<lower=0,upper=1>[n_occ_minus_1] alpha; // Mean survival
vector<lower=0,upper=1>[n_occ_minus_1] beta; // Mean recapture
}
transformed parameters {
matrix<lower=0,upper=1>[nind, n_occ_minus_1] phi;
matrix<lower=0,upper=1>[nind, n_occ_minus_1] p;
matrix<lower=0,upper=1>[nind, n_occasions] chi;
// Constraints
for (i in 1:nind) {
for (t in 1:(first[i] - 1)) {
phi[i, t] = 0;
p[i, t] = 0;
}
for (t in first[i]:n_occ_minus_1) {
phi[i, t] = alpha[t];
p[i, t] = beta[t];
}
}
chi = prob_uncaptured(nind, n_occasions, p, phi);
}
model {
// Priors
// Uniform priors are implicitly defined.
// alpha ~ uniform(0, 1);
// beta ~ uniform(0, 1);
// Likelihood
for (i in 1:nind) {
if (first[i] > 0) {
for (t in (first[i] + 1):last[i]) {
1 ~ bernoulli(phi[i, t - 1]);
y[i, t] ~ bernoulli(p[i, t - 1]);
}
1 ~ bernoulli(chi[i, last[i]]);
}
}
}
``````
Hi @Wade! I’d like to help out but I’m not totally sure I understand. Can you clarify what you mean by “known detection probabilities” and “borrow the detection probabilities” here?
Are you trying to take the posterior for detection probabilities from the model with 4 time points, and subsequently use it as “known” or data in another model with 2 time points?
1 Like | 1,097 | 3,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.743942 |
http://www.bonafous-murat.fr/boiler-steel/Allowable-Shear-Stress_798.html | 1,603,938,488,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902683.56/warc/CC-MAIN-20201029010437-20201029040437-00248.warc.gz | 123,105,921 | 10,335 | Allowable Shear Stress-Q235 Steel Q235A Q235B Q235C Q235D steel Equivalent,Properties
# Allowable Shear Stress
### WikiEngineer : Structural : Steel Beam Shear Strength
F v = The allowable shear stress of a beam.F y = The Yield Strength of the Steel (e.g.36 ksi,46 ksi,50 ksi). v =The Safety Factor for I-shaped members in Shear = 2.5 v =The Safety Factor for all other members in Shear = 2 \frac{7}{9} = 2.777778What is the allowable stress for mild steel? - QuoraMar 12,2019 Allowable Shear Stress#0183;Hari om,you are asking a question as to What is the allowable stress for mild steel?.Hari om.Hari om.ANSWER The allowable strength or allowable stress is the maximum stress (tensile,compressive or bending) that is allowed to be appliedWhat is allowable shear stress? - QuoraDec 22,2017 Allowable Shear Stress#0183;Allowable Shear stress is simply equal to.a l l o w a b l e s h e a r s t r e s s = y i e l d s h e a r s t r e s s f a c t o r o f s a f e t y.Factor of Safety (FOS) being always greater than 1; allowable shear stress value is always less than the yield shear stress.Therefore we are making those parts to experience less stress for same amount of force.
allowable stress design,based on service level loads and proportioning members using conservative allowable stresses.strength design,based on a realistic evaluation of member strength subjected to factored loads which have a low probability of being exceeded during the life of the structure.U.S.Technical Guide2.The allowable strengths and stiffness are for normal load duration.Bending,Shear and Compression parallel-to-grain shall be adjusted according to code.Modulus of Elasticity and Compression perpendicular-to-grain shall not be adjusted.3.The allowable Bending StressTorsion of Shafts - Engineering ToolBoxThe shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.The shear stress in a solid circular shaft in a given position can be expressed as = T r / J (1)
### Table of design properties for metric steel bolts M5 to
For standard course pitch thread and fine pitch thread bolts the nominal stress area A s is provided in ISO 898-1 Tables 4 to 7.In general the tensile stress area and the shear stress area are different.According to EN1993-1-8 Table 3.4 the shear strength of the bolt may be based on the tensile stress area.Definition of bolt classes 4.6,4.8 Table of design properties for metric steel bolts M5 to For standard course pitch thread and fine pitch thread bolts the nominal stress area A s is provided in ISO 898-1 Tables 4 to 7.In general the tensile stress area and the shear stress area are different.According to EN1993-1-8 Table 3.4 the shear strength of the bolt may be based on the tensile stress area.Definition of bolt classes 4.6,4.8 Steel Fv Allowable Shear Stress - Allowable StressMar 28,2020 Allowable Shear Stress#0183;Note Maximum shear stress reaches almost the allowable stress limit,but bending stress is well below allowable bending stress because the beam is very short.We can try at what span the beam approaches allowable stress,assuming L= 30 ft,using the same total load W = 2800 lbs to keep shear stress constant M= WL/8 = 2800(30)/8 M = 10500 Ib-ft fb = Mc/I = 10500(12)5/428 fb =
### Steel Design
Fv = allowable shear stress Fy = yield strength Fyw = yield strength of web material F.S.= factor of safety g = gage spacing of staggered bolt holes G = relative stiffness of columns to beams in a rigid connection,as is Specification for Structural Steel BuildingsAllowable Shear Stress with Tensio 5-5n Field Action 2 G4.Transverse Stiffener 5-5s 2 G5.Combined Shear and Tensio 5-5n Stress 3 H.COMBINED STRESSE 5-5S 4 H1.Axial Compressio and Bendinn g 5-54 H2.Axial Tensio andn Bendin g 5-55 I.COMPOSITE CONSTRUCTIO 5-5N 6 11.Definitio 5-5n 6 12.Design Assumption 5-5s 6Related searches for Allowable Shear Stressallowable shear stress of steelallowable shear stress formulaallowable shear stress of concretemaximum allowable shear stressshear stress equationallowable shear stress equationmax shear stress steelallowable shear stress woodSome results are removed in response to a notice of local law requirement.For more information,please see here.Previous123456NextAllowable Shear Stress of A36 Steel Our PastimesApr 12,2017 Allowable Shear Stress#0183;Shear stress refers to a pressure endured by a material parallel to its face.Shear strength is the material's ability to endure this applied stress.If enough stress is applied to a body it may not return to its original shape.Picture a rubber band.If
### Related searches for Allowable Shear Stress
allowable shear stress of steelallowable shear stress formulaallowable shear stress of concretemaximum allowable shear stressshear stress equationallowable shear stress equationmax shear stress steelallowable shear stress woodSome results are removed in response to a notice of local law requirement.For more information,please see here.12345NextGeneral Channel Stability Shear Stress RelationshipThe safety factors are the ratios of the allowable shear stress (0.15 lb/ft 2) divided by the calculated maximum shear stress.None of these channels can satisfy the allowable shear stress with this natural material,unless the channel is wide.A minimum channel width between 15 and 25 ftRelated searches for Allowable Shear Stressallowable shear stress of steelallowable shear stress formulaallowable shear stress of concretemaximum allowable shear stressshear stress equationallowable shear stress equationmax shear stress steelallowable shear stress woodSome results are removed in response to a notice of local law requirement.For more information,please see here.Maximum Allowable Stress Values for Bolting Materials Maximum Allowable Stress Values S for Bolting Materials according to ASME Code Section II,Part D,Table 3,2017 Edition Material
### How to Determine the Shear Strength of a Fillet Weld
The allowable shear stress for the welds would be 70,000 psi x 0.30 = 21,000 psi.A reduction of 70% compared to the case where the fillet weld was in pure tension.If our two welds are Allowable Shear Stress#188;-inch fillets then the shear strength (load carrying capacity) of the welds is calculated as follows.HEC-15 Permissible Shear Stress PlainwaterSep 05,2020 Allowable Shear Stress#0183;where, p = permissible shear stress,N/m 2 (lb/ft 2); F* = Shields parameter,dimensionless; s = specific weight of the stone,N/m 3 (lb/ft 3); = specific weight of the water,9810 N/m 3 (62.4 lb/ft 3); D 50 = mean riprap size,m (ft); Typically,a specific weight of stone of 25,900 N/m 3 (165 lb/ft 3) is used,but if the available stone is different from this value,the site Dont Stress Out - AISC Homesubstituting for Cv,the nominal shear stress is Fn = 0.6Fy.Dividing by ,the al-lowable shear stress is then Fv =0.4Fy.Us-ers of ASD will note that this is the same allowable shear stress as found in the 1989 ASD specication.Design of Members for Combined Forces and Torsion The interaction equations in
### Determining Allowable Design Values for Wood
The allowable design value is the value that should be used when sizing/designing wood structural components.This course teaches the methodology for adjusting the reference design values for all significant factors in order to arrive at the allowable design v Shear parallel to grain (Beam design)Determining Allowable Design Values for WoodThe allowable design value is the value that should be used when sizing/designing wood structural components.This course teaches the methodology for adjusting the reference design values for all significant factors in order to arrive at the allowable design v Shear parallel to grain (Beam design)Design Values - Southern PinePDF downloads of Southern Pine design value tables are available.Table 1 provides the design values for visually graded Southern Pine dimension lumber that became effective June 1,2013.
### DESIGN FOR SHEAR2008
Shear force that concrete can resist without web reinforcement ,V c (ACI Eq.11.3) V c = 2 f c (b w x d ) where f c is in psi; b w and d are in inches Section 1-1 d 1 d 1 V u 45 0 b w Fig.1.Dr.Mohammed E.Haque,PE Shear Design Page 2 of 6 V s = A v x f y x n V s = A v x f y x d/s (ACI Eq.11-15) s = AChapter 8--Threshold Channel DesignFigure 816 Allowable shear stress for granular material in straight 821 trapezoidal channels Figure 817 Allowable shear stress in cohesive material in straight 822 trapezoidal channels Figure 818 USDA textural classification chart 822 Figure 819 Unconfined strengthCalculate shear strength A36 plate Physics ForumsFeb 17,2017 Allowable Shear Stress#0183;In shear,the allowable stress is much lower,typically 0.4 * 36,000 = 14,400 psi,which allows vagaries of shear stress distribution,provides for a minimal factor of safety on the load,etc.Even in bending,which provides partial tensile loads,the allowable stress should be
### Bolt or Pin In Single Shear Equation and Calculator
Shear Stress Equation Single Shear.Shear Stress Average = Applied Force / Area or Shear Stress ave.= F/( r 2) or Factor of Safety = F.S = ultimate stress / allowable stress .Therefore allowable stress = ultimate stress / F.S.Related Bolt or Pin Double Shear Equation and Calculator ;Beam Stress Deflection MechaniCalcThe maximum shear stress occurs at the neutral axis of the beam and is calculated by where A = bh is the area of the cross section.Note that the maximum shear stress in the cross section is 50% higher than the average stress V/A.Shear Stresses in Circular Sections.A circular cross section isAllowable stresses of typical ASME materials - Stainless SteelAllowable stresses for temperatures of 595 Allowable Shear Stress#176;C and above are values obtained from time dependent properties.T9 Allowable stresses for temperatures of 620 Allowable Shear Stress#176;C and above are values obtained from time dependent properties.W12 These S values do not include a longitudinal weld efficiency factor.For Section III applications,for materials
### Allowable stress and allowable load
Apr 28,2019 Allowable Shear Stress#0183;where a is the allowable stress in tension or compression and A is the cross-section area.for direct shear.Pa= aA.where a is the allowable shear stress and A is the area where shear stress acts.construction management concrete construction.bridge construction:How to become a bridge engineer.Get link; Facebook;Allowable shear stressAllowable Shear stress is simply equal to $allowable shear stress = \frac{yield shear stress}{factor of safety}$Factor of Safety (FOS) being always greater than 1;allowable shear stress value is always less than the yield shear stress.What is allowable shear stress? - QuoraWas this helpful?People also askWhat is the relation between yield stress and allowable stress?What is the relation between yield stress and allowable stress?Yield strength is theproperty of the material whereas allowable stress isnot a property of the material.Yield strength is the stressat which you could have strain without further addition of stress.Allowable stress is the stressat which a member is not expected to fail under the given loading conditions.Is there any difference between allowable stress and yield Allowable Stress DesignAllowable Stress Design 13 Shear - Allowable Stress Design bd V fv Fv fm 50psi If steel is required F d Vs Av Fv 3 fm 150psi smax min{d /2,48in.} Fsd Sections within d/2 from face of support can be designed for shear at d/2 A.Noncantilever beamNoncantilever beam B.Reaction introduces compression into end region of member C.
### Allowable Stress Design (ASD) Method and Strength Design
V allowable,ASD = Allowable shear load N n = Lowest design strength of an anchor or anchor group in tension as determined per ACI 318 Appendix D,AC193,AC308 and the IBC.V n = Lowest design strength of an anchor or anchor group in shear as determined perAllowable Stress Design (ASD) Method and Strength Design For tension loads,T 0.2T allowable,the full allowable load in shear shall be permitted.For shear loads,V 0.2V allowable,the full allowable load in tension shall be permitted.For all other cases:Allowable Shear Stress of A36 Steel Our PastimesApr 12,2017 Allowable Shear Stress#0183;Shear stress refers to a pressure endured by a material parallel to its face.Shear strength is the material's ability to endure this applied stress.If enough stress is applied to a body it may not return to its original shape.Picture a rubber band.If
### Allowable Shear Stress - Mechanical engineering general
Dec 01,2002 Allowable Shear Stress#0183;I would like to determine the allowable shear stress for a particular material.I've seen a formula where Sv allow = 0.22Sy (Sv allow is allowable shear stress and Sy is Yield Strength of material).Machinery's Handbook say to use 4000 psi for main power tranmitting shafts,to 8500 psi for small short shafts.Allowable Bending Stress - an overview ScienceDirect TopicsThe allowable shear stress all depends on the girder.In addition,girders are used to satisfy the requirements of the web plate thickness,the girder web area,andALLOWABLE STRESS DESIGN TABLES FOR REINFORCEDallowable stress design,based on service level loads and proportioning members using conservative allowable stresses.strength design,based on a realistic evaluation of member strength subjected to factored loads which have a low probability of being exceeded during the life of the structure. F v = allowable shear stress in masonry,psi
### ALLOWABLE STRESS DESIGN TABLES FOR REINFORCED
allowable stress design,based on service level loads and proportioning members using conservative allowable stresses.strength design,based on a realistic evaluation of member strength subjected to factored loads which have a low probability of being exceeded during the life of the structure. F v = allowable shear stress in masonry,psi ALLOWABLE STRESS DESIGN OF CONCRETE MASONRY -The calculated shear stress due to applied loads,f v,as given by Equation 7 cannot exceed any of the code-prescribed allowable shear stresses,F v,as follows Building Code Requirements for Masonry Structures defines the above allowable shear stresses as being applicable to in-plane shear stresses only allowable shear stresses for out-of-plane loads are not provided.ACI 318 Specific Provisions - docs.bentleyThe allowable shear stress is calculated by selecting the appropriate equation from ACI 318-14 22.6.5.2 or 22.6.5.5.Equation 22.6.5.2(a) controls in non-prestressed concrete zones with large column aspect ratios.As the aspect ratio of the column gets larger,the allowable punching shear stress approaches the allowable one-way shear stress.
### results for this questionWhat is the allowable stress for mild steel?What is the allowable stress for mild steel?The ultimate strength,or stress of mild steel is around 800 to 840 MPa.So,taking a factor of safety of 4 (four) ,the allowable stress works out to 800 Allowable Shear Stress#247;4 =200 MPa.Reference quora/What-is-the-allowable-stress-for-mild-steel results for this questionFeedbackAllowable Maximum Shear Stress - an overview
If the maximum shear stress in the beam is limited to 200 N/mm 2 and the maximum angle of twist to 2 Allowable Shear Stress#176;,calculate the minimum thickness of the beam walls.Take G = 25000 N/mm 2.The minimum thickness of the beam corresponding to the maximum allowable shear stress of 200 N/mm 2 is obtained directly using Eq.(11.22) in which T max = 15 kNm.Thus results for this questionWhat is allowable stress?What is allowable stress?Allowable Stress (Strength) The allowable stress or allowable strength is the maximum stress (tensile,compressive or bending) that is allowed to be applied on a structural material.Allowable Stress::Fundamentals::Knowledgebase::SAFAS results for this questionHow to calculate allowable stress in steel?How to calculate allowable stress in steel?The maximum allowable shear stress for mild steel is 34,800 psi .P allowable = 6500 x 8 x 32 = 10400 lbs per plank = 4.5 inches and the wall thickness is 0.Division 2 Governs the design by Analysis and incorporates a lower safety factor of 3.Maximum allowable stress varies from material to material and design temperatures.How To Calculate Permissible Stress In Steel
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50CrMo4,42CrMo4,12Cr1MoV,25MnTi. | 3,837 | 16,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | longest | en | 0.822755 |
https://www.physicsforums.com/threads/small-understanding.161367/ | 1,660,199,765,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00295.warc.gz | 825,272,507 | 15,514 | # Small understanding
Sumanta
Hi,
This is not a homework question since I am out of college for a long time.
I was trying to understand the following that [0,1] /\ Q is not closed in R.
My understanding is that u must take a sequence (since this is a metric space) of the form m/n s.t m < n and create a sequence.
So I was trying to construct sequences like 1/2, 2/3, 3/4 but they seemed to be all ending within [0,1] /\ Q. I am not sure but do I have to take a sum or sth but I am not sure how to prove it.
## Answers and Replies
(I am assuming that Q means rational nos. and R means real nos.). Take any irrational number between 0 and 1 and let the rational number sequence be the the sequence where the nth term is the truncation of the decimal expansion of the irrational after n decimal places.
Homework Helper
Just giving an example of what mathman said: $\sqrt{2}{2}$ is irrational and is 0.70710678118654752440084436210485...
Each number in the sequence 0.7, 0.70, 0.707, 0.7071, 0.70710, 0.707106,... is a rational number because it is a terminating decimal; but the sequence as a whole converges to the irrational number $\frac{\sqrt{2}}{{2}}$
Last edited by a moderator:
Staff Emeritus
Gold Member
You're working from the wrong direction. Rather than trying to come up with a sequence of rationals and hope that its limit is irrational... you should pick the irrational, and then try and find a sequence of rationals that converges to it.
Sumanta
Thanks a lot for clearing the lacunae in my understanding. I think the basic idea is that u need to show that if u wiggle the rational number a bit u will fall into the set of irrational numbers, which is clearly proved by the sqrt(22) example.
Thank u a lot for the same.
Eighty
Actually that was sqrt(2)/2 (how could the square root of 22 be less than 1?). You should read what Hurkyl said. The simplest irrational you can think of in the interval is sqrt(2)/2, and constructing a sequence that converges to it is easy, as shown by HallsofIvy. | 517 | 2,015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.921875 | 4 | CC-MAIN-2022-33 | latest | en | 0.963699 |
https://human.libretexts.org/Bookshelves/Composition/Advanced_Composition/How_Arguments_Work_-_A_Guide_to_Writing_and_Analyzing_Texts_in_College_(Mills)/zz%3A_Back_Matter/21%3A_Accessibility/1.06%3A_Claim_and_three_reasons_argument_map | 1,721,668,143,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00614.warc.gz | 267,676,893 | 28,948 | # 1.6: Claim and three reasons argument map
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The first reason "We would feel it was right to cross the border without permission" is in a box with an arrow next to it pointing to the next reason, "We should recognize illegal crossing as ethical," which in turn has an arrow from it pointing to the reason "Border walls and detention centers are unjust," which points to the final claim, "We need a new policy that offers respect and help to migrants." | 1,851 | 4,804 | {"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} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.201702 |
https://www.coursehero.com/file/5902169/hw7slns/ | 1,490,358,658,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187945.85/warc/CC-MAIN-20170322212947-00177-ip-10-233-31-227.ec2.internal.warc.gz | 898,123,329 | 67,361 | hw7slns
# hw7slns - Homework#7Solutions
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Homework #7 Solutions For the Butterworth filter, we can use equation (10.3.46) and perform the bilinear transformation on s to get the frequency response for our digital filter. The code and plots are below. Matlab code: %problem 10.15 T = 1/24000; %sampling frequency (seconds) omega = 0:pi/100:pi; %digital filter plotting range Omega = 0:pi/100:pi; %analog filter plotting range WpOrig = 4000; WsOrig = 6000; wp = 2*T*WpOrig*pi; %digital passband frequency (radians) ws = 2*T*WsOrig*pi; %digital stopband frequency (radians)
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Rp = 1; %passband ripple Rs = 40; %stopband attenuation %Analog filter with prewarping Wp = 2*tan(wp/2); %analog passband frequency (rad/s) (with prewarping) Ws = 2*tan(ws/2); %analog stopband frequency (rad/s) (with prewarping) %MATLAB function specifications [N, WpButter] = buttord(Wp,Ws,Rp,Rs, 's' ); [Bbutter, Abutter] = butter(N,WpButter, 's' ); [BdButter, AdButter] = bilinear(Bbutter,Abutter,1); HdButter = freqz(BdButter,AdButter,omega); %frequency response %Our specifications
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hw7slns - Homework#7Solutions
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Ask a homework question - tutors are online | 474 | 1,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-13 | longest | en | 0.799027 |
https://elsner.edu.pl/en/integer-anchor-chart.html | 1,701,342,484,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00338.warc.gz | 276,954,244 | 5,303 | # Integer Anchor Chart
Integer Anchor Chart - Web these integer operations anchor charts or personal reference sheets include an example, modeling with the number line,. Web these integer operations anchor charts or personal reference sheets include an example, modeling with the number line,. Before they get to the same point, many. Web i’m going to show you how to make interactive and effective number anchor charts that support your number sense instruction! Creating a solid place value foundation starts with displaying number forms as in this. It is intended to encourage students to. There is no download included. Web this anchor chart displays four common strategies for subtracting. You can create one at the beginning of a unit, to introduce a topic. Web each number anchor chart includes the following representations of numbers.
### You Can Create One At The Beginning Of A Unit, To Introduce A Topic.
Can be shrunk down to 80. Web these integer operations anchor charts or personal reference sheets include an example, modeling with the number line,. There is no download included. Web here is your cheat sheet to help you remember what to do with positive and negative numbers.
### Web Integer Anchor Chart Created By Lauren Kubin More.
Web by elizabeth mulvahill. This chart was created with students during number talks. The positives & negatives of visual. This handy anchor chart is helpful for your high school algebra students as they learn the rules for adding,.
### Use Pictures —Remember, An Anchor Chart.
Web this is a short and easy to read anchor chart that is great for 7th graders learning their integer rules. Ways to show a number. One of the best, most effective tools for the classroom is anchor charts,. Web representing fractions anchor chart by teachers r us homeschool.
### If You're Looking For A Fun Visual To Introduce And Reinforce Numbers, You Need These Number Anchor Charts For Your Classroom!.
This is a hard good item. Web this free integer anchor chart can be used in interactive math notebooks and then students can reference the notes later. Rounding numbers is a simple concept once you get the hang of it. Creating a solid place value foundation starts with displaying number forms as in this.
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http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?s=20f15c39b0cf0392fc2c44151ed2b187&p=9332300 | 1,531,917,443,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590169.5/warc/CC-MAIN-20180718115544-20180718135544-00011.warc.gz | 393,224,916 | 12,124 | Actuarial Outpost May 2003 #12
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#1
10-28-2007, 09:13 PM
MrActuary Member Join Date: May 2006 Studying for DMAC... and life Posts: 242
May 2003 #12
I am stumped again. Here is the problem:
Eric deposits X into a savings at time 0, at a nominal rate of interest i, comp. semiannually.
Mike deposits 2X into a different account at time 0, at a simple rate of interest i.
They earn the same interest during last 6 months of 8th year.
Calculate i.
#2
10-28-2007, 09:24 PM
Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,879
You can do this. For simplicity, let X = 1000 (or keep it as X; if you do it will cancel). How much interest does Eric earn in that 6-month period [in terms of i; or in terms of X and i if you didn't set X =1000].
Hint: calculate Eric's balance at the end of 7.5 years (or at the end of 15 half-years) [again in terms of i, or X and i]. Then the interest he earns in the following 6 months is i/2 times that balance.
Then calculate how much interest Mike earns in 6 months [in terms of i, or of X and i]. Mike gets simple interest, so the interest in every 6 month period is the same. You can use the first 6 months if you like.
Set those two expressions equal. If you kept X until now, the X's will cancel. Solve for i.
#3
10-28-2007, 09:59 PM
MrActuary Member Join Date: May 2006 Studying for DMAC... and life Posts: 242
Thanks for the prompt reply. I realized that when I was working this I my notation was throwing me off. I was writing "i-upper 2" along with "i" in the same equation, making it looking more difficult than it really was. I finally got my solution to a point where a simple guess and check of the answers could be done very quickly (~8 seconds per answer).
#4
01-20-2011, 04:14 PM
laraib.usmani Join Date: Jan 2011 College: Undergraduate Posts: 10
i m doing FM paper preparations but have found some difficulties in the following questions.. plz help me regarding these question.. and also explain me the concepts behind them.. Thanks alott
Question 1: Eric deposits X into a savings accounts at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a different saving accounts at time 0, which pays a simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i.
Question 2:Using a method of equated time, a payment of 400 at time t=2 plus a payment of X at t=5 is equivalent to a payment of 400+X at time t=3.3125
At an effective interest rate of 10%, the above two payments are equivalent to a payment of 400+X at time t=k using the exact method. Calculate k.
Question 3: On January 1 1980, Jack deposit 1000 in Bank X to earn interest at the rate of j per annum compounded semi-annually. On January 1 1985, he transferred his account to Bank Y to earn interest at the rate of k per annum compounded quarterly. On January 1 1988, the balance at Bank Y in 1990.76
If Jack could have earned interest at the rate of k per annum compounded quarterly from January 1 1980 through Jan 1 1988, his balance would have been 2203.76
Calculate the ration k/j.
These r the three questions which im trying to understand but im not getting them.. Please help me regarding these questions
Regards,
#5
02-28-2012, 04:45 PM
zzzsilver Member Join Date: Aug 2011 College: MIT Posts: 151
stumped
Hi, I still don't understand why you owuld multiply the (i/2) by the amount that is in the account after 7 1/2 years to find the interest earned in the last 6 months of the 8th year. Can someone explain how that works out? Thanks!
Z
__________________
P FM
Lookin at the world through rose colored glasses .
#6
02-28-2012, 04:53 PM
Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,879
That's just the definition of nominal interest, compounded semiannually. If my rate is 8% nominal, compounded semi-annually, then in 6 months I earn 4% on the balance at the start of the 6 month period.
(That assumes that the starting date is right after an interest-compounding period, which in this problem it is.)
#7
02-28-2012, 04:59 PM
zzzsilver Member Join Date: Aug 2011 College: MIT Posts: 151
oh
Oh ok it makes sense now that you put numbers to it. the (i/2) was throwin me off but I get it now, thanks.
Z
__________________
P FM
Lookin at the world through rose colored glasses .
#8
05-25-2018, 01:03 PM
actsci123451 SOA Join Date: Feb 2018 Posts: 8
if the account pays simple interest at an "annual rate of i," I know that the interest is constant every year, but is it constant every half year as well? | 1,379 | 5,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | longest | en | 0.916712 |
https://www.coursehero.com/file/6108293/Chapter-10-Homework/ | 1,516,758,928,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892892.86/warc/CC-MAIN-20180124010853-20180124030853-00394.warc.gz | 899,646,298 | 100,081 | Chapter 10 Homework
Chapter 10 Homework - Rachel L Gates ACC-240 Chapter 10...
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Rachel L. Gates ACC-240 Chapter 10 Homework Assignment E10-1 Determining Financial Statement Effects of Transactions Involving Notes Payable 1.) Date Assets Liabilities Stockholders’ Equity (a) Nov. 1 Cash +6,000,000 Note Payable +6,000,000 NE (b) Dec. 31 NE Interest Payable +75,000 Interest Expense (+E) -75,000 (c) Apr. 30 Cash – 6,225,000 Note Payable -6,000,000 Interest Payable -75,000 Interest Expense (+E) -150,000 (b) Interest = (P) x (R) x (T) 75,000 = 6,000,000 x 7.5% x 2/12 (c) 6,000,000 x .075 x 4/12 = 150,000 150,000 + 75,000 = 225,000 6,000,000 + 225,000 = 6,225,000 2.) If Target needs extra cash every Christmas Season, should management borrow money on a long term basis to avoid negotiating a new short-term loan each year? One reason for long term borrowing, it will essentially provide financing for Target. On the other hand, following the Christmas Holiday season, they won’t really need it any longer because Target will eventually collect the funds from their credit sales resulting in no necessary need to borrow long term funds and end up paying more interest on a loan not needed after the season passes. It is a better decision to borrow on a short-term basis for short-term needs. Perhaps, Target could discuss a line of credit loan from their banking institution. Therefore, it would be short-term and very low interest, if any at all!
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S10-1 Finding Financial Information 1.) 2009 Results 2008 Results .13 = (519 + 6 + 972) .14 = (445 + 12+ 1,259) 11,153 12,706 The two above Quick Ratio results suggest that in 2008, Home Depot was had a slightly better
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Chapter 10 Homework - Rachel L Gates ACC-240 Chapter 10...
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Ask a homework question - tutors are online | 579 | 2,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.867368 |
https://chestofbooks.com/crafts/metal/Applied-Science-Metal-Workers/35-Problems-In-Compound-Leverage.html | 1,563,503,136,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00489.warc.gz | 350,731,801 | 5,610 | Problems in compound leverage are easily reduced to repeated cases of simple leverage, the force at the end of the first lever being the weight or force applied to the second lever, and so on through any number of levers.
As an example: If the force at W in Fig. 14 is 12 lbs., what is the force at P?
For the first lever the force pushing up at the end of the long arm is: 12 X 3 / 12 = 3 lbs. For the second lever it is: 3 X 3 /12 = 3/4 lb.
While the safest way is always to figure each lever as a simple lever, as just explained, a shorter method of obtaining the answer is as follows:
Multiply the weight by the continued product of the short arms of all the levers, and divide this by the continued product of the long arms of the same levers.
Fig. 14. - Compound Lovers.
Applying this rule to the above problem we have
12 X 3 X 3 / 12 X 12-------------- = 3/4 lb.
The answer is the same as before, and after a little thought it is evident that the two steps in the first case have merely been put together in one expression in the second case. If the weight, 3/4 lb., on the long end of the second lever at P is known (see Fig. 14), and the pressure or weight which would be needed at W is to be found, the same rule will apply but will be expressed in this manner: Multiply the weight by the continued product of the long arms and divide this by the continued product of the short arms:
3 / 4 X 12 X 12 / 3 X 3 = 12 lbs.
Regardless of how many levers there are working together, the rule is applicable. In all leverage problems the first, and the most important, thing is to find and locate the fulcrum, as the fulcrum is the point which determines the moment arms from which the required answer is obtained. The moment arm is always the perpendicular distance from the force or weight to the fulcrum. | 447 | 1,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.953631 |
http://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-4-review-exercises-page-389/53 | 1,519,013,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812327.1/warc/CC-MAIN-20180219032249-20180219052249-00675.warc.gz | 451,018,195 | 13,386 | ## Precalculus: Mathematics for Calculus, 7th Edition
$$\log_2 (\frac{ (x-y)^{\frac{3}{2}}}{(x^2 + y^2)^2})$$
$Combine$ $into$ $a$ $single$ $logarithm:$ $\frac{3}{2}\log_2 (x-y)$ - $2\log_2 (x^2+y^2)$ Use the Third Law of Logarithms for both portions $\frac{3}{2}\log_2 (x-y)$ = $\log_2 (x-y)^{\frac{3}{2}}$ $2\log_2 (x^2 + y^2)$ = $\log_2 (x^2 + y^2)^2$ $\log_2 (x-y)^{\frac{3}{2}}$ - $\log_2 (x^2 + y^2)^2$ Use the Second Law of Logarithms $\log_2 (x-y)^{\frac{3}{2}}$ - $\log_2 (x^2 + y^2)^2$ = $\log_2 (\frac{ (x-y)^{\frac{3}{2}}}{(x^2 + y^2)^2})$ $$\log_2 (\frac{ (x-y)^{\frac{3}{2}}}{(x^2 + y^2)^2})$$ | 304 | 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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-09 | latest | en | 0.284559 |
https://www.studymode.com/essays/Acc-300-Week-5-Individual-Assignment-62057623.html | 1,547,778,776,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659654.11/warc/CC-MAIN-20190118005216-20190118031216-00483.warc.gz | 960,659,300 | 24,151 | # ACC 300 Week 5 Individual Assignment Chapter Six
Topics: Royalties, Internal control, Publishing Pages: 3 (708 words) Published: November 1, 2014
In this file ACC 300 Week 5 Individual Assignment - Chapter Six you will find overview of the following parts:
1. Assignments from the Readings. Prepare responses to the following assignments from the e-text, Fundamentals of Financial Accounting 1st ed., by Phillips, Libby, and Libby
a. Chapter 6: Questions 2 and 3
2. Summarize the primary purposes of an internal control system.
3. What are the three internal control objectives for financial reporting?
QRB 501 Week 2 Learning Team Case Studies - "Case 5 - 2 and Case 6 -2"
Week 2 Learning Team Case Studies
Complete the following case studies from Ch. 5 6 of Business Math:
Case Study 5-2, pp. 184-185
Case Study 6-2, pp. 216-217
Note . Show all work and calculations. (The use of Microsoft Excel software is required.)
1. Ziam wants to know how much his royalty will be for a song he has written. How will it be calculated? Write the steps or the formulas that will be used to calculate his royalty payment.
2. Ziam has written a popular song titled “Going There,” which has been recorded by a well-known performer. He recently received a royalty check for \$7,000. If Ziam gets a 0.5 share of the royalties and the credit value is \$3.50, what was the credit total that his song earned? Write out the problem in the form of an equation and solve it.
3. Ziam quickly published another song, “Take Me There,” that is played even more often than “Going There.” If his first song earns 4,000 credits and his second song earns 6,000 credits, what will the royalty payment be from the two songs if the credit value remains at \$3.50?
4. Ziam is considering an offer to perform his own song...
https://bitly.com/12B3oHj
Speak with your professors daily to build strong relationships.... | 475 | 1,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-04 | latest | en | 0.94941 |
https://forum.artofmemory.com/t/memory-hooks-and-why-they-matter/45845 | 1,580,298,299,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00532.warc.gz | 433,618,689 | 16,171 | # Memory hooks and why they matter
Definition
I define a memory hook as everything that you use to retrieve a memory from the database in your brain.
Picture VS video
Imagine a memory contest held in Hollywood with competing teams consisting of actors, directors and movie critics. In this contest, more likely than not, at least one event will be something like looking at snapshots taken from movies and answering questions like:
• what movie is this?
• what happens after this snapshot?
The organizers of this event may realise, that the first question is easy enough for at least some of the competitors, but the second question just seems to difficult; so they decide to replace the snapshot with a 10 seconds video of the movie, making the second question significantly easier.
Memory asymmetry
Imagine using a 2 digit standard memory palace system to memorise a bunch of numbers, just for the fun of it. Also, imagine that youy are very sleepy, so you are likely to be in the following scenario: front door of my house, what in the name of Zeus did I put there? Nothing comes to mind, so you try your secondary method, counting from 00 to whatever 2 digit number is the right one. And very often, as soon as you visualise the object that represents the correct 2 digit number, you recognise it as the one you put at your front door.
This asymetry (see: Asymmetrical memory connection) can also be described as the difference between the quality of two memoryhooks:
1. object: being a memoryhook for the location it has been placed;
2. location; being a memoryhook for an object that has been placed in this location.
In my estimation, the first memoryhook is more powerful.
The juggling reverse memory palace
Please imagine the following scenario: there is a fireman axe stuck in your front door, you take it out and you teleport to your bedroom, you trow the axe on your bed and a pillow jumps up and you grap it and you teleport to your kitchen and you squeeze the pillow in one of the kitchen drawers, but there is a pair of scissors in the drawer that you have to grap in order for the pillow to fit and you finally teleport to the bathroom and you stick the scissors in the bathroom mirror.
No try to fill in the ⌠but donât look at what comes after the ⌠before you have filled it in.
there is a fireman axe stuck in your front door, you take it out and you teleport ⌠and a pillow jumps up and you grap it and you teleport âŚbut there is a pair of scissors in the drawer that you have to grap in order for the pillow to fit and you finally teleport âŚ
The above test is a very realistic example of my latest version of the reverse memory palace:
• 100 numbered locations;
• in every location there is a (starting/deault) object (in the previous version this object was merely associated with the location);
• Objects are moved around from location to location (like the balls in the hands of a juggler taking over each others place) in accordance with the to be memorised sequence of numbers;
• A starting object (latest favorite for this role is the Incredible Hulk) is placed at the first location.
This version of the reverse memory palace is designed in such a way, that the power of the memoryhook is maximised by creating a video that only requires an answer to the question âwhere did I place this object?â for itâs continuation.
Final improvement of system (getting rid of linguistics)
Iâm also working on a way to make the connection between the 2 digit numbers and the locations more powerful in a way that resembles the ability of âflash anzanâ/âmental abbacusâ users, in that they are capable of making speed calculations while having a conversation at the same time. In my latest version, I used my New memory system for fast translation of numbers (so for example: 22 35 89 = noe mas fag), but I feel that is not very compatible with the mental juggling of the newest reverse memory palace. I replace this linguistical method with a visual/sensual method, in which I simply place the numbers in the memory palace like real 2 dimensional objects.
2 examples:
• 33: coffee machine in office hallway; the two 3âs are placed under tension in the opening of the machine where the plastic cups come out like the spring coils of a car;
• 99: on top of the diving board in the swimming pool; they are sort of bouncing up and down.
1 Like
I found the reverse : If given the location I remember the object much more effectively than when given the object.
However
This is true, I find this has to do with the ability to recognise what you have seen/what was there rather than the object reminding you of the location. For example there was an old record posted somewhere on this forum where the record setter recognised around 1000 words in one go (able to recognise whether they have or have not seen those). Compared to learning a few words and recalling them from your memory just being able to identify whether you have seen something in this way or not is a lot more memorable. You may for example flick through a book at a very fast rate, you would be able to differentiate this book from 10 other books still, even if you can recall near nothing.
On this comment, the first time I ever saw the shaper system, I looked through the first 100 numbers and I pretty much was able to recall them instantly. I think the secret to strong recall lays there in some form. It was just extremely and absurdly memorable.
Conceptually when I think about why the shaper number system is so recognisable I feel very swayed to say it is because of the linking (processed stimuli is part of memory so directly hooks to recall the memory) as a kind of thing.
2 Likes
I agree completely with this.
My guess is that most people that have issues recalling simply use the location as a backdrop and would place the same image no matter what.
I would argue, that even the likes of Ben Pridmore and Alex Mullen have issues with recalling; the main difference between them and most of the "hobby"memorisers is the speed at which these issues occur.
So letâs look at one of the examples you gave:
Location 1 - a mailbox:
a giant rat planting an American flag as a stamp on an envelope before sending it
By lowering the amount of time you have for making this and other connections, you will fail at some time interval (let say 0,5 seconds, just a guess) in aswering the question: what object/animal did I place at the mailbox? The question than arrises: will you also fail at the same time interval in aswering the reverse question: where did I place the rat?
It is clear what my guess is, but I would like to make it also very clear, that I have no evidence, other than my own experience.
Can I ask you which of the following statements using your rat-mailbox example you think is most likely to be the case for memorisers ?:
• where did I place the rat is easier, than what did I place at the mailbox;
• what did I place ⌠is easier;
• no difference between these 2.
Thatâs a very interesting observation. If this is the case for most memorisers (I hope more forum members will share their experience regarding location -> object VS object-> location ability), than the reverse memory palace may not be very usefull for them.
The only other person, (as far as I can remember) that has shared his experience with this asymmetry is SilvioB who in a response to the Asymmetrical memory connection seemed to share my view, that going from object to location is easier. A part of his response was:
So for example; where did I put âfreedom of artistic expressionâ? My mind instantly jumps to the memory palace that I have for the Swiss Constitution and I see B.A. Baracus painting a statue of liberty. Which gives me the information I was looking for
In reaction to the "counting from 00 to âŚ"you said:
Nagime: This is true, I find this has to do with the ability to recognise what you have seen/what was there rather than the object reminding you of the location.
That is very strong explanation, which makes the example I gave (counting from 00 to âŚ) less convincing. Perhaps a better example would be the memorisation of a single deck of cards and than trying to recreate the order by going from location to location or the second way by going from card to card.
So, let me say in conclusion, that this asymmetry appears to be very powerful for me, but I am more than anything interested in finding out if and how (which of the 2 ways is easier) this asymmetry is for other memorisers.
Perhaps a somewhat more convincing anecdotal (but if you recognise the anecdote, than itâs becomes a bit more) evidence: you canât remember putting a lion at you front door, but a couple of locations later, you do remember putting a lion on your bed. And as soon as you see the lion on your bed, you also see a lion in front of your door so clearly, that you wander why you could not see it when you where looking at the door. It seems in this anecdote, that âthe lion (on the bed)â has triggered âthe lion at the doorâ in a way that the âdoorâ was not able to do.
1 Like
I think this can go both ways quite easily depending on how you encode things for others. The only real problem I would find with this is if you have for example an object being an axe in 120 locations. My objects would often âreoccurâ but my place would always be unique. Besides this, there isnât really any absolute reason to use one over the other. I am also saying this based on the fact that for me reusing an object more often makes it a bit harder to recall the sequence when I donât use something like 4 x [object].
Some perception differences exist though on the side of Location V Object:
When you are using locations, portals or teleportation makes it extremely difficult to recall the next location.
Having the object interact with the location makes it more memorable, e.g the scissors scraping the drawer vs the scissors just being there.
Having not too many objects makes it easier to recall them in place (I personally find having a bit more than 1 is better than having 1 for recall too).
This said I can kind of also see periods where it would be more memorable to have a reverse memory palace. For example if I am using a real life location like a highway, it would be many times more memorable for me recalling the location by the objects than the reverse, if portals are included this is also again the case.
This is very convincing.
To clarify:
X to remember X ; X being the object
Y to remember Y ; Y being the location
Y to remember X ; X being the object, Y being the location
X to remember Y ; Y being the location, X being the object
You want to state originally that X to remember Y > Y to remember X
The example is saying at Y1 you canât recall X1, at Y3 you can recall X1, when you recall X1 you recall X1 is at Y1.
Since its not quite of the form X to remember Y but rather X at Y3 to remember X at Y1, you may argue that it is easier to recall whatever you want to recall when you are given what you want to recall at some other link. If you were to draw a network of this you would have the exact same structure ârecalledâ from one path but not from the other, and as a result of recalling this, it fills in or reactivates/primes what it needs to , in order for you to recall it from the original path. This does somewhat match up with research on memory.
When I put it like this it seems that the original point is different? Perhaps rather it is saying that if you are given a list of objects you recall the locations those objects are at better than if you are given a list of locations and what objects are at those locations. The difference here is that you need to be able to remember the objects which is difficult, whereas remembering the location from location is kind of easier. If of course the objects are linked to each other this gives your point a lot more weight to it. Well, in this case I actually would say it is very difficult to decide because logically you may use the linking method to link objects and you wouldnât link objects on a white background but random locations which may serve as a memory palace in its own right. When you do this it may indeed be more memorable than the reverse, but in this sense the equivalent is creating a memory palace and objects at the same time which is virtually the same thing but instead of linking objects together you are linking locations together.
If given time to either e.g premaking a location and premaking objects, if you can recall sequences of premade objects to 1000 without premade locations (rather new locations) vs recalling a sequence of premade locations to 1000 new objects.
I believe I have successfully made this a lot more critical than I would have needed to, but on the above evaluation, this confirms that more or less both approaches work very well.
I would say my initial point on â120 of the same object vs 120 different locationsâ,. That said there are too many properties you can have as stated the object interact with the location. Thinking about this however makes me question why not both? You can have both a reverse and normal memory palace in a journey, interchangeably even.The realisation here is that I actually do this sometimes at least in theory. For example I may go out of the window and land on a floor in my visualisation, in theory a window being an object just has teleported me out to the floor? I feel like I have just completely lost the distinction between a reverse and non-reverse memory palace.
To further confuse things.
I mean after-all you can have a room in a house it would be a object of the house, so is a object in a room not a location in a room?
The question becomes if you have a ball and place 3 objects on a ball is this ball functional as a location? If I visualise a ball and put a spoon of sugar stuck to it then a tomato and a small kitten, then I place this ball on a floor in the sky. Lets have the kitten restlessly scratch the ball, the spoon spin and the tomato melt then return back to normal. I can recall a tomato, spoon of sugar and kitten, if I zoom in, it looks like a normal location, albeit very similar ground.
So is a location in a location more memorable than a location in a location? Yes.
I find if the current location links to the next its much easier to remember it. Suddenly everything clicks.
When I timed using the sharper system on a stopwatch I got around 0.40 seconds, though this was only for a single encoded image, 2 encodes quickly shot up the time. So maybe the benchmark is taken from one such estimate? Or perhaps verbal speed, which was closer to this?
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The 0,5 seconds is nothing more than a way to make the âfailing at some time intervalâ more concrete; it is not a bench mark for memorising a 100 digit number or a deck of cards. Does it really matter for the argument, whether this theoretical time interval matches your actual ability? Would it be a crime if I started a post on some runners forum with âlet say you can run a marathon in 2:15:00 and you want to go fasterâŚâ?
does âhobbyâ start? Thereâs probably 500 - 1,000 people in the world that can memorize a deck of cards in under a minute⌠âhobbyâ⌠or does that fall under âthe likes ofâ?
I am simply using the word âhobbyâ to differentiate the very best and the rest, like in the difference between a âprofessional marathon runnerâ and ârecreational marathon runnerâ; there is no value judgement made by me at all and the deviding line between the two categories is also completely irrelevant for the argument I was trying to make.
With your input the memory asymmetry poll is now 2:1 in favour of âwhat did I place ⌠is easierâ.
Your reply is a testament of some deep thinking. It reinforces (besides the other other things you explain in you reply) how difficult it is to compare to systems in a honest way; so many factors to consider to make it a fair comparison.
When you are using locations, portals or teleportation makes it extremely difficult to recall the next location.
Having the object interact with the location makes it more memorable, e.g the scissors scraping the drawer vs the scissors just being there.
Iâm not sure what you have in mind with this statement. In the juggling reverse memory palace the next location is triggered (hopefully) by the movie that ends with grabbing an object from the previous location; clearly in the standard method the next location is obvious, but the tradeoff is also obviously that you need to remember the object. As far as the interaction: I don 't really focus on an autonomous object-location interaction, but more on myself putting (throwing/pushing/spinning/rolling, etc.) the object in the location. So in the scissors to mirror example, I would grab thgem from the kitchen drawer and teleport/jump to the bathroom and jam them into the mirror with force.
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Sorry this statement was meant for the traditional approach not the reverse memory palace(because it isnât true when you refer to the reverse memory palace [object location wise]).
Isnât this statement just:
By lowering the amount of time you have for making this and other connections, you will fail at some time interval lets say this interval was 0.5 seconds to make it easier to visualise?
Then you are saying :
I mean sure 0.5 may be not the best number to visualise with but, isnât that besides the point?
I donât understand how you think that this running calculation example is any way simular to an argument or thought that I presented.
I seem to have mistaken your âratâ example for a 2 digit translation that is simply placed in a location (the mailbox). I further just imagined that you look at one 2 digit number and try to put in your memory as fast as possible and than after some distraction you see if you can still remember it. The 0,5 seconds doesnât appear to be to fast to imagine. Just to make it very clear; iâm not claiming that if you manage this 0,5 seconds test, you can automatically memorise 1.000 digits in 500 seconds.
That is precisely my point.
I would imagine, that when you start a sentence with "let say⌠:
• you run a 100 meter in 10 seconds;
• you can bench 300 pounds;
• you are monkey living on the moon.
⌠it doesnât matter if you are not a monkey, for whatever argument that follows to be valid and understandable.
I feel that Mental Cal interprets what I write in a completely different way, than how I meant it.
Maybe my understanding of the English language is not as good as yours, but I just donât feel that saying âpretend that you areâŚâ is supposed to be taken as a statement, that in fact you areâŚ
In Dutch: stel jij bent (are) een aap op de maanâŚ
We donât say: stel jij was (were) een aap op de maanâŚ
because than we are talking about a hypothetical situation in the past. But itâs still hypothetical.
Erik in response to Nagime: I feel that Mental Cal interprets what I write in a completely different way, than how I meant it.
and you responded:
See previous reply⌠and that is also all I said in my very first reply already. Use the location itself for context (in your image) and you will be able to recall from the location as trigger.
âŚthatâs how we ended up you dividing things into âthe best and the restâ and using absurd benchmarks.
This dividing was used by me as a reaction to your focus on people who do not have good memory skills for connecting objects to locations; My point was the complete opposite of what you make of it, as I was stressing that from the perspective of the theory I was discussing the only difference between the 2 groups is the time interval at which they fail to memorise. I have to admit, that I donât understand how you go from a theoretical 0,5 seconds (with little to no specifications) to interpreting it as a âbenchmarkâ. Like, I donât even know what this word means in this context. I understand the phrase"MC Donalds is the benchmark in the fastfood sector", but that meaning doesnât make sense to me in our conversation.
So, roll back the whole thing⌠you wanted to know if this was an issue (object vs location) and I gave you an example of how to make it a non-issue.
Again, you talk of an âissueâ which I interpret as a âproblemâ, but I wasnât talking about a problem, but about a difference in quality of memory hooks (object vs location).
This whole sentence is in the past tense, making it an unfair example; Furthermore, I would say that this is more old Dutch than present Dutch; non of my friends/coworkers talk like that.
This is a more (present tense) common Dutch sentence:
Hij praat over de bankoverval alsof het een zondagsuitje is.
Personally, I think sentiment is more meaningful than wording. You werenât trying to make fun of world records by estimating 0.5 seconds both in wording and sentiment.
Abstract considerations are still useful.
Ctrl + enter and my keyboard are not really compatible on this website.
Is a hand held model of a house an object or a location? As such, is size the difference of an object and a location? Finally, I can go back to the 13-year-old me reading that âeverything is an object in programmingâ and get some odd form of confirmation.
I appreciate your evaluation of this thread. I think we share the opinion, that this is getting in all kinds of surreal directions.
In situations like this I feel torn between two possible responses:
1. react to every misinterpretation (regardless of whoâs fault it is) or criticism;
2. leave the thread for what it is.
I guess I chose the first response in this case, but I feel like âwhat is even the pointâ, but I also feel that at least the conversation appeared to be somewhat civilised.
I have had another âincidentâ on this forum in which I responded to a proposed memory system by another forum member. I reacted in a very honest, but also funny (at least that how it was meant) way. To make a long story short: it didnât land very well with the original poster. It seemed like everything I said in the following responses was making things worse.
So, Iâm counting my blessings with this one.
2 Likes
Ironically the most âof high repliesâ I tend to partake in tend to be in âthese kinds of incidentsâ. Even this post has 27+1 replies.
Sometimes it gives me the impression that aggravating someone is the easiest way to have more replies. Generally, when you have made a comment that isnât taken well, it just goes downhill from there, almost like first impressions. Sometimes years later when the person has forgotten about it they are much nicer to you.
I donât really think most peopleâs ways of life are so different that they canât understand one another. In most cases it tends to be some misunderstanding, I think in this case even it might have partly been a misunderstanding among other things.
2 Likes
On the topic of memory hooks however, if you have seen the shaper system, how would you theorise its function in the sense that you can remember it much more easily compared to training an image system?
I think the kind of logic that causes the shaper system to be more memorable and trained on start should be applicable to other things.
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It is my impression, that the name giver of this sytem thinks it is original (I have not read the whole thread, so I could be wrong). I donât think it is. I have already made a 2 digit shape based system 10 years ago and I donât think I came up with the idea myself. My version didnât work really well, because I had problems with differentiating between left and right. So for example"38" was a snowman (8) with a broomstick (3), but â83â was something else (donât remember what), but it also looks like a snowman. Also â01â and â10â created a problem as only one was a baseball bat and baseball, but they both look like this. I have to admit, that some of the images of the shaper system are better than what I came up with (I only spend 2 hours creating my version of this system), but also some of the sharper images feel not very intuitive to me.
There is also the problem of seeing multiple things in a number. So in the shaper system the â11â is drumsticks, but when you accidentally see a skier, you have to teach yourself, that itâ s not a skier.
I do tend to believe there is a major advantage in such a system, in that it completely gets rid of language. I feel that language in memorising numbers and in mental aritmetic slows everything down and makes it more difficult to do other things at the same time. I feel this even more after my experience with my own version of the number to sound method; itâs really difficult to combine this pronounciating with the mental gymnastics that one is supposed to do in a (reverse) memory palace to memorise numbers.
Iâm probably somewhat biased, but I believe that a more powerful standard (number to object translation) shape based system exist, which is very simular to the way I intend to use the numbers in my juggling reverse memory palace method; making a mental image in which the 2 digits and the objects are connected in a way that is memorable. Some of my locations are basically objects in a fixed place, so I think they serve as a good example. Just to be clear: I donât visualise the numbers in any other way than numbers (also no rotation like that of the â3â to make it look like an âmâ), even though they may take on the identity of (or act like) objects in some abstract way.
11. stairs: the â11â is 2 pins stuck in the stairs;
17 mirror in elevator: the â1â is placed in the middle of the mirror and the â7â is placed in the upper right corner, matching the corner so to speak;
44. pallet with paper in the warehouse; they are pinned (just like â11â) in a box of paper.
I imagine (could be wrong) that most people think that for something to be memorable it has to be meaningful, in that for example a number has to resemble an object in some way. In my experience, letting go of this assumption allows your intuition to develop regarding how to place numbers in a memorable way.
The right placing of numbers requires patience in my experience. Put in a different way: if you give yourself a couple of days to come up with the right placing for each number/object combination, you will win in the long run, because they will be much more powerful. As a related topic, I would like to mention, that I have tried with some succes to memorise numbers in a memory palace as nothing more than (single digit) numbers. I managed 100 digits in 4 minutes, without any specific training. I think this simple method has the potential to be really fast for the obvious reason, that you donât have to translate anything.
1 Like | 6,170 | 27,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.955106 |
https://whatisconvert.com/95-cubic-inches-in-imperial-fluid-ounces | 1,571,575,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986707990.49/warc/CC-MAIN-20191020105426-20191020132926-00254.warc.gz | 757,997,750 | 7,073 | # What is 95 Cubic Inches in Imperial Fluid Ounces?
## Convert 95 Cubic Inches to Imperial Fluid Ounces
To calculate 95 Cubic Inches to the corresponding value in Imperial Fluid Ounces, multiply the quantity in Cubic Inches by 0.57674402642447 (conversion factor). In this case we should multiply 95 Cubic Inches by 0.57674402642447 to get the equivalent result in Imperial Fluid Ounces:
95 Cubic Inches x 0.57674402642447 = 54.790682510324 Imperial Fluid Ounces
95 Cubic Inches is equivalent to 54.790682510324 Imperial Fluid Ounces.
## How to convert from Cubic Inches to Imperial Fluid Ounces
The conversion factor from Cubic Inches to Imperial Fluid Ounces is 0.57674402642447. To find out how many Cubic Inches in Imperial Fluid Ounces, multiply by the conversion factor or use the Volume converter above. Ninety-five Cubic Inches is equivalent to fifty-four point seven nine one Imperial Fluid Ounces.
## Definition of Cubic Inch
The cubic inch is a unit of measurement for volume in the Imperial units and United States customary units systems. It is the volume of a cube with each of its three dimensions (length, width, and depth) being one inch long. The cubic inch and the cubic foot are still used as units of volume in the United States, although the common SI units of volume, the liter, milliliter, and cubic meter, are also used, especially in manufacturing and high technology. One cubic foot is equal to exactly 1,728 cubic inches because 123 = 1,728.
## Definition of Imperial Fluid Ounce
A fluid ounce (abbreviated fl oz, fl. oz. or oz. fl.) is a unit of volume. It is equal to about 28.41 ml in the imperial system or about 29.57 ml in the US system. The fluid ounce is sometimes referred to simply as an "ounce" in applications where its use is implicit.
### Using the Cubic Inches to Imperial Fluid Ounces converter you can get answers to questions like the following:
• How many Imperial Fluid Ounces are in 95 Cubic Inches?
• 95 Cubic Inches is equal to how many Imperial Fluid Ounces?
• How to convert 95 Cubic Inches to Imperial Fluid Ounces?
• How many is 95 Cubic Inches in Imperial Fluid Ounces?
• What is 95 Cubic Inches in Imperial Fluid Ounces?
• How much is 95 Cubic Inches in Imperial Fluid Ounces?
• How many uk fl oz are in 95 in3?
• 95 in3 is equal to how many uk fl oz?
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• What is 95 in3 in uk fl oz?
• How much is 95 in3 in uk fl oz? | 634 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-43 | longest | en | 0.842539 |
https://www.bbcelite.com/cassette/main/subroutine/mult1.html | 1,708,488,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00861.warc.gz | 693,151,531 | 138,675 | Elite on the BBC Micro and NES
# Maths (Arithmetic): MULT1
## [BBC Micro cassette version]
``` Name: MULT1 [Show more]
Type: Subroutine
Category: Maths (Arithmetic)
Summary: Calculate (A P) = Q * A
Deep dive: Shift-and-add multiplication
Context: See this subroutine in context in the source code
Variations: See code variations for this subroutine in the different versions
References: This subroutine is called as follows:
* MAD calls MULT1
* MULT12 calls MULT1
Do the following multiplication of two 8-bit sign-magnitude numbers:
(A P) = Q * A
.MULT1
TAX \ Store A in X
AND #%01111111 \ Set P = |A| >> 1
LSR A \ and C flag = bit 0 of A
STA P
TXA \ Restore argument A
EOR Q \ Set bit 7 of A and T if Q and A have different signs,
AND #%10000000 \ clear bit 7 if they have the same signs, 0 all other
STA T \ bits, i.e. T contains the sign bit of Q * A
LDA Q \ Set A = |Q|
AND #%01111111
BEQ mu10 \ If |Q| = 0 jump to mu10 (with A set to 0)
TAX \ Set T1 = |Q| - 1
DEX \
STX T1 \ We subtract 1 as the C flag will be set when we want
\ to do an addition in the loop below
\ We are now going to work our way through the bits of
\ P, and do a shift-add for any bits that are set,
\ keeping the running total in A. We already set up
\ the first shift at the start of this routine, as
\ P = |A| >> 1 and C = bit 0 of A, so we now need to set
\ up a loop to sift through the other 7 bits in P
LDA #0 \ Set A = 0 so we can start building the answer in A
LDX #7 \ Set up a counter in X to count the 7 bits remaining
\ in P
.MUL4
BCC P%+4 \ If C (i.e. the next bit from P) is set, do the
ADC T1 \ addition for this bit of P:
\
\ A = A + T1 + C
\ = A + |Q| - 1 + 1
\ = A + |Q|
ROR A \ As mentioned above, this ROR shifts A right and
\ catches bit 0 in C - giving another digit for our
\ result - and the next ROR sticks that bit into the
\ left end of P while also extracting the next bit of P
\ for the next addition
ROR P \ Add the overspill from shifting A to the right onto
\ the start of P, and shift P right to fetch the next
\ bit for the calculation
DEX \ Decrement the loop counter
BNE MUL4 \ Loop back for the next bit until P has been rotated
\ all the way
LSR A \ Rotate (A P) once more to get the final result, as
ROR P \ we only pushed 7 bits through the above process
ORA T \ Set the sign bit of the result that we stored in T
RTS \ Return from the subroutine
.mu10
STA P \ If we get here, the result is 0 and A = 0, so set
\ P = 0 so (A P) = 0
RTS \ Return from the subroutine
``` | 812 | 3,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-10 | latest | en | 0.885716 |
https://journalarjom.com/index.php/ARJOM/issue/view/99 | 1,674,804,209,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00189.warc.gz | 353,379,259 | 19,433 | ##### An Intrinsic Analysis of Human Brucellosis Dynamics in Africa
Paride O. Lolika, Mlyashimbi Helikumi
Asian Research Journal of Mathematics, Page 1-26
DOI: 10.9734/arjom/2022/v18i1030423
Brucellosis is one of the most common zoonotic infections globally. It affects humans, domestic animals and wildlife. In this paper, we conduct an intrinsic analysis of human brucellosis dynamics in non-periodic and periodic environments. As such we propose and study two
mathematical models for human brucellosis transmission and control, in which humans acquire infection from cattle and wildlife. The first model is an autonomous dynamical system and the second is a non-autonomous dynamical system in which the seasonal transmission of brucellosis
is incorporated. Disease intervention strategies incorporated in this study are cattle vaccination, culling of infectious cattle and human treatment. For both models we conduct both epidemic and endemic analysis, with a focus on the threshold dynamics characterized by the basic reproduction
numbers. Using sensitivity analysis we established that R0 is most sensitive to the rate of brucellosis transmission from buffalos to cattle, the result suggest that in order to control human brucellosis there is a need to control cattle infection. Based on our models, we also formulate
an optimal control problem with cattle vaccination and culling of infectious cattle as control functions. Using reasonable parameter values, numerical simulations of the optimal control demonstrate the possibility of reducing brucellosis incidence in humans, wildlife and cattle, within
a finite time horizon, for both periodic and non-periodic environments.
##### Magic Polygons and Combinatorial Algorithms
Danniel Dias Augusto
Asian Research Journal of Mathematics, Page 27-39
DOI: 10.9734/arjom/2022/v18i1130424
In this work, we study the Magic Polygons of order 3 (P(n; 2)) and we introduce some properties that were useful to build an algorithm to find out how many Magic Polygons exists for the regular polygons up to 24 sides. The concepts of Equivalents Magic Polygons and Derivatives Magic Polygons which allowed to classify, and avoid ambiguity about the representations of such elements, are also introduced.
##### Stability Analysis and Optimal Control of Endemic Malaria Disease Transmission Model with Cost-Effective Strategies
Dereje Gutema Edossa, Alemu Geleta Wedajo, Purnachandra Rao Koya
Asian Research Journal of Mathematics, Page 40-64
DOI: 10.9734/arjom/2022/v18i1130425
In this study, a non-linear system of ordinary differential equation model that describes the dynamics of malaria disease transmission is formulated and analyzed. Conditions are derived from the existence of disease-free and endemic equilibria. The basic reproduction number R0 of the model is obtained, and we investigated that it is the threshold parameter between the extinction and persistence of the disease. If R0 is less than unity, then the disease-free equilibrium point is both locally and globally asymptotically stable resulting in the disease removing out of the host populations. The disease can persist whenever R0 is greater than unity and the conditions for the existence of both forward and backward bifurcation at R0 is equal to unity are derived. Sensitivity analysis is also performed and the important parameter that derive the disease dynamics is identified. Furthermore, optimal combinations of time dependent control measures are incorporated to the model, and we derived the necessary conditions of optimal control using Pontryagins’s maximum principal theory. Numerical simulations were conducted using MATLAB to confirm our analytical results. Our findings were that, malaria may be controlled by reducing the requirement rate of mosquito populations and the use of a combination of vaccination, insecticide treated net ITN, indoor residual spray IRS and active treatment or strategy d can also help to reduce the number of populations with malaria symptoms to zero. We also find that the same strategy that is, strategy d proves to be efficacious and cost-effective.
##### Analysis of Generalization of a Problem of Vieta’s Descend Method with Examples and Computing Support
Szabo, Zoltan Istvan
Asian Research Journal of Mathematics, Page 65-76
DOI: 10.9734/arjom/2022/v18i1130426
Let and be fixed integer numbers. Assume that (a2+b2+c) is divisible by (ab+d) for some natural numbers a and b. Then the value of the fraction $$k ( = {(a^2+b^2+c) \over (ab+d)})$$ remains the same. Statement of this kind will be proved in pp. 1-3 and illustrated on some examples in pp. 3-10. The general method of proofs will be unified and simplified. Computing support will be provided: in pages 11-19 a simple program code is defined with the help of which one can hunt for natural numbers a, b with the same integer values of c, d and k. Here, a number of examples are given as well.
##### On Continuity of Grill Topological Spaces VIA Regular Generalized G$$\omega$$ - Closed Sets
Suliman Dawood, Ahmed M. Al-Audhahi, Amin Saif
Asian Research Journal of Mathematics, Page 77-91
DOI: 10.9734/arjom/2022/v18i1130427
Aims/ Objectives: In this paper, we study the continuity property in grill topological spaces via generalized G$$\omega$$ -closed sets and regular generalized G$$\omega$$ - closed sets. These notions are generalized G$$\omega$$ - continuous functions which is weak form of g$$\omega$$ - continuous functions, generalized G$$\omega$$ - irresolute functions, regular generalized G$$\omega$$ - continuous functions which is weak form of rg$$\omega$$continuous functions, regular generalized G$$\omega$$ - irresolute functions and investigates some of its properties in grill topological spaces.
##### $$(\hat{\alpha}-\bar{\psi})$$ Geraghty Contraction on Generalized Metric Spaces with Simulation Function
Manoj Kumar, Nisha Kumari
Asian Research Journal of Mathematics, Page 92-107
DOI: 10.9734/arjom/2022/v18i1130428
In this paper, we introduce the new definitions and fixed-point theorems for $$(\hat{\alpha}-\hat{\psi})$$-Geraghty contraction with an aid of simulation function $$\zeta:[0, \infty) \times[0, \infty) \rightarrow \mathbb{R}$$ in generalized metric space satisfying the following condition:
if $$\exists \hat{\beta} \in \mathcal{F}$$ such that for all $$r, s \in \mathfrak{X}$$, then we have
$$\zeta[\hat{\alpha}(r, s)(d(\mathcal{P} r, \mathcal{P} \mathcal{s})), \hat{\beta}(\hat{\psi}(d(r, s))) \hat{\psi}(d(r, s))] \geq 0$$, where $$\hat{\psi} \in \hat{\Psi}$$ and $$(\mathfrak{X}, d)$$ is a generalized metric space. An example is also given to support our results.
##### The Existence of Periodic Solutions for a Two-Enterprise Interaction Model with Delays
Chunhua Feng, Orjul Pogue
Asian Research Journal of Mathematics, Page 108-115
DOI: 10.9734/arjom/2022/v18i1130429
Various enterprise interaction models with or without time delays appeared in the literature. The stability and Hopf bifurcation for one or two delays in the models were studied by many researchers. However, the periodic oscillation for a four time delays enterprise interaction model is still an open problem due to the complexity of the bifurcating equation. In this paper, by means of the mathematical analysis method, some sufficient conditions to guarantee the existence of periodic oscillatory solution for the four time delays model are obtained. An open problem is solved. Computer simulation is given to demonstrate the present results.
##### On the Comparison of Adam-Bashforth and Adam Moulton Methods for Non-Stiff Differential Equations
Osakwe Charles Nnamdi, Olobayo Solomon Adelaja, Omowo Babajide Johnson
Asian Research Journal of Mathematics, Page 116-129
DOI: 10.9734/arjom/2022/v18i1130430
This paper presents the comparison of the two Adams methods using extrapolation for the best method suitable for the approximation of the solutions. The two methods (Adams Moulton and Adams Bashforth) of step k = 3 to k = 4 are considered and their equations derived. The extrapolation points, order, error constant, stability regions were also derived for the steps. More importantly, the consistency and zero stability are also investigated and finally, the derived equations are used to solve some non-stiff differential equations for best in efficiency and accuracy.
##### Contemporary Preservice Mathematics Teachers’ Technological Pedagogical Content Knowledge Levels in Perspective: Self-reported Survey
Jones Apawu
Asian Research Journal of Mathematics, Page 130-147
DOI: 10.9734/arjom/2022/v18i1130431
This study which is part of a project examined the perception of preservice mathematics teachers on Technological Pedagogical Content Knowledge (TPACK) levels and the relationship among the TPACK components. The research design employed was descriptive interspersed with correlational design. The population for this study was preservice mathematics teachers at the University of Education, Winneba of Ghana. The study employed the purposive sampling technique specifically homogeneous sampling technique to select level 300 mathematics teachers from the department of mathematics education of the University of Education, Winneba. In all, the level 300 (year 3) students were 183. Raosoft sample size software tool was used to determine a sample size of 125 for the study. After the determination of the sample size, simple random sampling technique was used in selecting the respondents for the study. Questionnaire was used as instrument to collect data. Data collected were analysed quantitatively. Results showed that: (i) the perceived knowledge level of the preservice mathematics teachers on TPACK and its components were moderate and high; (ii) there were positive relationships among the components of TPACK, and all of the relationships were statistically significant. Recommendations were thereof made accordingly.
##### A Hybrid Optimization Model for Vehicle Routing Problem, a Case Study at Zoomlion Ghana Limited, Shama District
Justice Kangah, Henry Otoo, Joseph Acquah
Asian Research Journal of Mathematics, Page 148-161
DOI: 10.9734/arjom/2022/v18i1130432
The aim of this study was to develop a hybrid optimization model for solving the routing problem identified at Zoomlion Ghana Limited in Shama district in the Western Region of Ghana. Two main optimization models were considered: Particle Swarm Optimization (PSO) and Genetic Algorithm (GA). A hybrid algorithm was developed from the two by merging crossover and mutation operators of GA with PSO. A sample of 20 breakpoints was run through 10,000 iterations for all the models and the results of the proposed hybrid model was compared with PSO and GA separately. The optimal results of PSO, GA and the proposed models are 1160.6km, 1190.3km and 1132.3km respectively. The proposed model’s results were also compared with other hybrid models to test the robustness of the new model. This result was achieved because the new model eliminates the low convergence rate in PSO and also prevents it from easily falling into local optimum in high-dimensional space and the inclusion of crossover and mutation operators of GA improves the diversity of the iterations. After the iterations, PSO reduced a field distance of 1700 km to 1160.6 km within 780.4098 seconds. GA on the other hand reduced the same field distance of 1700 km to 1190.3km within 397.3308 seconds. The proposed hybrid model reduced the same field distance from 1700 km to 1132.3 km within 550.2527 seconds. This indicates that the proposed hybrid model performed better than PSO and GA separately. A performance test between the proposed hybrid model and other hybrid models showed that merging crossover and mutation operators into PSO gives a better optimal result. MATLAB was used for the iterations.
##### Common Fixed Point Theorems for Four Weakly Compatible Self Maps Along with (CLR) Property in Fuzzy 2-Metric Spaces
. Deepika, Manoj Kumar
Asian Research Journal of Mathematics, Page 162-169
DOI: 10.9734/arjom/2022/v18i1130433
In this paper, we prove some common fixed point theorems for four weakly compatible self-maps along with (CLR) property in fuzzy 2- metric spaces. Our results are the improved version of the theorems proved by Shojaei et al. [1] in 2013, since our results does not require closedness of ranges of subsets of X.
##### Laplace Transforms Method on a System of Differential Equations for Non-isothermal Chemically Reactive Flow
Hope Osogom Okolie, Adolphus Okechukwu Nwaoburu
Asian Research Journal of Mathematics, Page 170-185
DOI: 10.9734/arjom/2022/v18i1130434
This study analyses Laplace transforms method on a system of partially coupled differential equations for non-isothermal chemically reactive flow through a cylindrical channel. The dimensionless governing equation for velocity, temperature and concentration was solved using Laplace Transform. Various parameters such as Temperature boundary parameter, Concentration boundary parameter, Cooling Parameter, Grashof number, pressure gradient and Magnetic field, as well as perturbation parameter had an effect on the velocity profile as well as temperature and concentration profile. The graphs were obtained with the results showing that an increase in the temperature boundary parameter resulted to an increase in the temperature of flow, an increase in perturbation parameter resulted to an increase in temperature profile of a body and an increase in Grashof temperature number results to an increase in the velocity of the body.
##### General Parameters Estimation in the Presence of Nonresponse Using Difference-cum Exponential Estimators
Alabi Oluwapelumi, Oluwagunwa Abiodun Peter, Ogundele Joshua Oluwafunminiyi, Benard O. Muse
Asian Research Journal of Mathematics, Page 186-202
DOI: 10.9734/arjom/2022/v18i1130435
Non response is a common problem in a survey process. Therefore, it is necessary to find a way out of handling non response whenever arises. The current study proposed difference-cum exponential ratio-type estimator for estimation of general parameters using auxiliary information which is defined in two situations of non response. A conventional estimator t*(a,b) is used to define population constants including population mean, standard deviation, coefficient of variation and mean square. The expression of bias and mean square error of the proposed estimators are obtained up to the first order of approximation for situation I and situation II. To compare the efficiency of the proposed estimators over the existing ones, an empirical study is carried out using real and simulated data sets. Both the theoretical and empirical study shows that the proposed class of estimators outperformed other existing estimators.
##### Numerical Solution of Nonlinear Equations by a Twelth-Order Iterative Method with Memory
M. Hassan, M. Aslam, A. Asghar, S. Ahmad, I. Rasheed, F. A. Shah, A. Qayyum
Asian Research Journal of Mathematics, Page 203-214
DOI: 10.9734/arjom/2022/v18i11596
Some real life mathematical problems can be converted in the form of nonlinear equations. Solving such problems by analytical approaches is dificult in many situations. Hence numerical solution is the best way in this case. In this paper, a twelth-order iterative scheme for solving
nonlinear equations is presented and analyzed in terms of eciency. The new scheme is derived from the well-known King's method with order of convergence eight. We extend eighth-order King's method to an iterative method with memory of order 12:16 by using famous Newton's
interpolating polynomial of degree 6 to avoid the derivative used in King's method. The new derived method is a three-step and is totally derivative free with twelth order of convergence. The method requires four functional evaluations at each iteration introducing high efficiency index of (12:16) 1/4 = 1:8673: Convergence order of new method is also studied. It is achieved by using matrix method of Herzburger. Numerical results are also provided to support theoretical analysis. Comparison of the derived scheme with previously well-known iterative schemes of the same order is also presented. As diffierent schemes of same order has efficiency index of (12)1/6 = 1:5131 because they requires six functional evaluations at each iteration, hence the proposed scheme is better than other schemes.
##### Mathematical Model on the Dynamics of Bacterial Blight of Rice in the Presence of Lysobacter antibioticus Considering Introduction at Different Stages
Agada Apeh Andrew, Samuel Musa, Solomon Ortwer Adee
Asian Research Journal of Mathematics, Page 215-232
DOI: 10.9734/arjom/2022/v18i11597
In this research we formulated the Plants diseases model with the aim of studying the dynamics of the use of lysobacter antibioticus for prevention and control of rice bacterial blight. The disease free equilibrium state of the models was also obtained by equating each of the equation of the modified model to zero and simplifying. The basic reproduction number for the model was derived using the next generation matrix approach. Numerical simulation was carried out using MATLAB2018a to virtualize the dynamics of the model. Five numerical experiment was carried out and it was shown that biocontrol help to reduce the population of the pathogen as well as act as treatment for those that are already exposed or infected with the disease. It was also observed that the biocontrol agent provide immunity to rice plants against been infected with the disease. Finally, we observed from the simulation that the earlier the control is introduced the more protection plants will receive.
##### A Mathematical Model Approach for Prevention and Intervention Measures of the COVID-19 Pandemic in Uganda
Fulgensia Kamugisha Mbabazi, Yahaya Gavamukulya, Awichi Richard Opaka, Peter Olupot-Olupot, Samson Rwahwire, Saphina Biira, Livingstone S. Luboobi
Asian Research Journal of Mathematics, Page 233-248
DOI: 10.9734/arjom/2022/v18i11598
The human{infecting corona virus disease (COVID-19) caused by the novel severe acute respiratory syndrome corona virus 2 (SARS-CoV-2) was declared a global pandemic on March11th, 2020. Different countries adopted different interventions at different stages of the outbreak, with social distancing being the first option while lock down the preferred option for attening the curve at the peak of the pandemic. Lock down aimed at adherence to social distancing, preserve the health system and improve survival. We propose a Susceptible-Exposed- Infected-Expected recoveries (SEIR) mathematical model for the prevention and control of Covid-19 in Uganda. We analyze the model using available data to find the infection-free, endemic/infection steady states and the basic reproduction number. We computed the reproductive number and it worked out as R0 = 0:468. We note that R0 is less than unity, thus forecast that several strategies in combination (including travel restrictions, mass media awareness, community buy-in and medical health interventions) will eliminate the disease from the population. However, our model predicts a recurrence of the disease after one year and two months (430 days) thus the population has to be mindful and continuously practice the prevention and control measures. In addition, a sensitivity analysis done showed that the transmission rate and the rate at which persons acquire the virus, have a positive in uence on the basic reproduction number. On other hand the rate of evacuation by a rescue ambulance greatly reduces the reproduction number. The results have potential to inform the impact and effect of early strict interventions including lock down in resource limited settings and social distancing.
##### Propagation Speed of the Frontal Head Through Lock-Exchange Density Current in Cold Fresh Water: Simulations without the Effect of Back-reflected Waves
Alabodite Meipre George, Evans Fiebibiseighe Osaisai
Asian Research Journal of Mathematics, Page 249-260
DOI: 10.9734/arjom/2022/v18i11599
The behaviour of warm water discharged at 4$$^{\circ}$$C through lock-exchange in cold fresh water was investigated numerically, fixing lock volume at the centre of the domain. This investigation as presented here is practical and can also enhance policy making towards the protection of the aquatic ecosystems. Though, the aim of this study is to better fathom and as well, gain more insight into such ows. Our results have shown a speedy movement of the lock volume at the centre of the domain with a leading head at two front on the oor which resulted in a hat shape within the first few time frame. Fluid movement in the second phase is independent of the back reflected waves. We were able to identify two regimes of ow with a stepwise decreasing velocity in the second phase. Our results have shown that velocity with which the current travels with in the second regime is higher within the first time frame as compared to those with the effect of back reflected waves. One major factor that is responsible for decrease in the velocity here is mixing. Previous results have also shown that the front velocities in the collapsing phase are independent of lock volume. But this seem not to be the same here because fluid movement in the first phase (regime) is not totally independent of the lock volume and its position here, where density difference is as a result of temperature. However, our scaling power laws here in the second phase show some variations with previous studies where we have effect of back reflected waves. But results in the collapsing phase here are in strong agreement with those in the first phase of our previous simulations with small lock volume. Generally, the spreading behaviour here is dependent on lock volume, barrier position, density difference and Reynolds number.
##### Fixed Point Theorems for Mappings Satisfying Implicit Relation in Partial Metric Spaces
Preeti Bhardwaj, Manoj Kumar
Asian Research Journal of Mathematics, Page 261-270
DOI: 10.9734/arjom/2022/v18i11600
In this manuscript, we shall introduce the implicit relation on $$\mathbb{R}_+$$4. Using this implicit relation, we shall prove fixed point and common fixed point theorems on partial metric spaces.
##### Application of the Some Blaise Abbo (SBA) Method to Solving the Time-fractional Schrödinger Equation and Comparison with the Homotopy Perturbation Method
Joseph Bonazebi Yindoula, Yanick Alain Servais Wellot, Bamogo Hamadou, Francis Bassono, Youssouf Paré
Asian Research Journal of Mathematics, Page 271-286
DOI: 10.9734/arjom/2022/v18i11601
We have solved the Schrödinger equation with the HPM method and the SBA method. We have noticed that with these two methods we find the same result.
##### Factors Predicting the Mathematics Performance of Grade 8 Students in the New Normal
Cristian A. Tobias, Crispina V. Diego
Asian Research Journal of Mathematics, Page 287-300
DOI: 10.9734/arjom/2022/v18i11602
Mathematics performance has been recognized as vital in the curricula in most of the country not only for academic success but also for its efficient application in everyday life. The study aimed to determine a significant relationship between the student, teacher, environmental, and parent involvement factors and students’ mathematics performance in a Catholic university for the Academic Year 2020-2021. Also, it determined which factors predict the mathematics performance of students. Using descriptive-correlational research design, 123 Grade 8 students answered a researcher-made test questionnaire to assess students' level of mathematics performance; these were interpreted using the Department of Education’s scale of standard. Furthermore, a researcher-made survey questionnaire was used to examine the level of extent of the four factors influencing the students’ mathematics performance, the correlation between these factors and students’ mathematics performance, and predictors of students’ mathematics performance. The assembled information was dealt with measurably utilizing mean, standard deviation, Kruskal-Wallis, Pearson product-moment correlation, Spearman's rho, and multiple linear regression. Results revealed that students’ level of mathematics performance was fairly-satisfactory (M=79.77, SD=8.96). The extent of influence of the factors is moderate (M=3.28, SD=0.33). There is a significant correlation between the extent of influence of factors as a whole and Mathematics performance [r(121)=0.235, p=0.009]. The results of the regression indicated that the predictor explained 7.3% of the variance [F(1, 121)=9.540, p=0.002, R2=0.073)]. The individual predictors were further examined and indicated that the teacher factor [β=6.440, t=3.089, p=0.002] significantly predicts Mathematics performance. This study provides significant insights into the vital role teachers play in finding a lasting solution to the students' perennial poor performance issues in mathematics in this new normal.
##### The Effect of Inquiry-based Learning on Senior High School Students’ Achievement in Plane Geometry: Pre-test-Post-test Randomized Experimental Design
Charles Kojo Assuah, Louis Osei, Gershon Kwame Mantey
Asian Research Journal of Mathematics, Page 320-331
DOI: 10.9734/arjom/2022/v18i11604
Using inquiry-based learning, this study investigates high school students' achievement in plane geometry. It employed the pre-test-post-test randomized experimental design, often known as the control group design. The participants (students) were randomly assigned to one of two classes/groups and were given either an intervention (the treatment group) or no intervention (the control group). One hundred and twenty (120) third-year high school students of similar mathematical aptitude (a control group = 60 students; an experimental group = 60 students) were chosen from a high school in Ghana's central region. Shapiro-Wilk had a p-value greater than.05 (p >.05) for each statistic, indicating that both the pre-test and post-test scores were normally distributed, before and after the test. The findings of the independent samples t-test showed that there was no statistically significant difference between the experimental and control pre-test scores (t = -.48, p >.05, C. I = [-1.78, 1.08]). The one-way ANOVA after inquiry-based learning showed a significant effect on student scores, F (1, 118) = 363.41, p < .05). Furthermore, independent samples t-test findings for the post-test showed statistically significant differences between the experimental and control post-test scores (t = -22.68, p < .05, C. I = [-24.29, 20.40]). The study's implications are that students can make their own connections with the content they learn. They may also comprehend the themes rather than simply recalling rules and formulas. The study concludes that inquiry-based learning improves senior high school students’ achievement in plane geometry.
##### Non-commutativity Over Canonical Suspension η for Genus g ≥1 in Hypercomplex Structures for Potential ρϕ
Deep Bhattacharjee
Asian Research Journal of Mathematics, Page 332-341
DOI: 10.9734/arjom/2022/v18i11607
Any matrix multiplication is non-commutative which has been shown here in terms of suspension, annihilator, and factor as established over a ring following the parameter k over a set of elements upto n for an operator to map the ring R to its opposite Rop having been through a continuous representation of permutation upto n-cycles being satisfied for a factor f along with its inverse f-1 over a denoted orbit γ on k-parameterized ring justified via suspension η ∈ η0, ηimplying the same global non-commutativity for the annihilator A. This will be used for the construction of the genus–alteration scenario where the suspension ηacting with its opponent η1 on any topological space J can alter the geometry making a change in the manifolds for taking over the Boolean (1,0) satisfying the concerned operations.
##### Asymptotic Expansions Related to the Wallis Ratio Based on the Bell Polynomials
Aimin Xu
Asian Research Journal of Mathematics, Page 342-350
DOI: 10.9734/arjom/2022/v18i11608
In this paper, we establish a new asymptotic expansion related to the Wallis ratio. By using the exponential Bell polynomials, we show that the coefficients of the asymptotic expansion can be recursively determined. In addition, an explicit expression for the coefficients is given. Our results improve and generalize the existing ones [1].
##### A Game-Theoretic Credit Period and Promotion Model in a Supplier-Retailer Channel
Peter E. Ezimadu, Sophia O. Ezimadu
Asian Research Journal of Mathematics, Page 351-361
DOI: 10.9734/arjom/2022/v18i11609
It has been established that trade credit can be influenced by a lot of factors. However, no specific function has been used to neither represent these factors nor consider their effects. This paper considers a supplier-retailer Stackelberg game in which the supplier as the channel leader supplies credit goods to the retailer who in-turn sells to the consumers. It uses a credit function based on credit period, supplier’s price margin and product promotion effort to model the players’ payoffs. The work considers two game scenarios: a situation involving the provision of trade credit and a situation without trade credit. The work obtains a closed-form solution for the credit period for the credit provision scenario, and the promotion efforts and payoffs for both scenarios, and shows that credit period prolongation may not be in favour of the retailer, and that the retailer can attain a larger payoff than the supplier. It also shows that the retailer’s margin is very crucial for both channel scenarios, and observes that the players are better-off with trade credit.
##### Establishing Equivariant Class [O] for Hyperbolic Groups
Deep Bhattacharjee
Asian Research Journal of Mathematics, Page 362-369
DOI: 10.9734/arjom/2022/v18i11615
This paper aims to create a class [O] concerning the groups associated with Gromov hyperbolic groups over correspondence and equivalence through Fuchsian, Kleinian, and Schottky when subject to Laplace – Beltrami in the Teichmüller space where for the hyperbolic 3-manifold when the fundamental groups of Dehn extended to Gromov – any occurrence of Švarc-Milnor lemma satisfies the same class [O] for quotient space and Jørgensen inequality. Thus the relation (and class) extended to Mostow – Prasad Rigidity Theorem in a finite degree isometry concerning the structure of the commensurator in higher order generalizations suffice through CAT(k) space. The map of the established class [O] is shown at the end of the paper.
##### The Solutions of the Linear Fractional Diffusion and Diffusion-convection Equations via the Regular Perturbation Method (RPM)
Bationo Jérémie Yiyuréboula, Bassono Francis
Asian Research Journal of Mathematics, Page 370-384
DOI: 10.9734/arjom/2022/v18i11616
In this paper, we implement Regular Perturbation Method (RPM) for the Solving fractional diffusion and diffusion-convection equations, in order to determine the analytical solutions of some linear fractional diffusion and linear fractional diffusion-convection equations. In general, the solving using this method allow to obtain exact or approximate solutions. For the case of the diffusion and diffusion-convection equations solved in this document, the solutions obtained are exact. By comparing these solutions with those obtained by other researchers using other methods for a certain value of the parameter α, we obtain the same results.
##### On Absolute Continuity of Non Negative Functions
Levi Otanga Olwamba
Asian Research Journal of Mathematics, Page 385-392
DOI: 10.9734/arjom/2022/v18i11617
This paper is committed to the study of absolute continuity of non negative functions with respect to vector measures. Almost everywhere properties are applied to establish boundedness, measurability and convergence of sequences of measurable functions. The measurability of sets with respect to vector duality functions with values in a Hilbert space is considered.
##### A Note on the New Method of Computing the Inverse of 3 x 3 Square Matrices and Its Applications
Binandam Stephen Lassong, Isaac Adu, Munkaila Dasumani, Kwesi Frempong Sarfo
Asian Research Journal of Mathematics, Page 393-401
DOI: 10.9734/arjom/2022/v18i11618
Computing the inverse of 3 x 3 square matrices using known methods such as Gauss-Jordon Method needs more time. During examination, candidates need to go through long process to find the inverse of 3 x 3 square matrices. Some students also find it very difficult and confusing when using Gauss-Jordon and Adjugate methods to find the inverse of the matrix. In this note, we present a new method that is simple and easy way of finding the inverse of 3 x 3 square matrices. We further give some applications of matrices in the real world phenomenon. Many other challenging problems can be addressed surprisingly by applying this strategy.
##### Graph Theoretic Technique to the Global Stability Analysis of the Discrete Age Structured SVEIR Model with Application to Ebola Virus Disease
Brenda Onyango, George O. Lawi, Frankline Tireito
Asian Research Journal of Mathematics, Page 402-413
DOI: 10.9734/arjom/2022/v18i11619
Discrete age structured models provide a better approach of discussing the spread of infectious diseases with infectivity, mortality and recovery being age dependent. Measures such as vaccination are carried out in discrete time or applied to individuals in certain age-groups. In the past Ebola Virus disease (EVD) outbreaks, infections in children under 5 years was associated with high mortality rates. In this work, a general discrete SVEIR epidemic model with age structure is formulated. The study establishes the existence of the endemic equilibrium and further shows that it is globally stable using the graph theoretic approach on the method of Lyapunov functions. The model is then applied to EVD dynamics to study the impact of vaccination on the age structured population. The numerical simulation of the discrete age case scenario demonstrates that vaccination of children under 5 years of age against EVD has a great impact in reducing their susceptibility because of the active immune system as compared to the older population who have a poor response to vaccine immunity. For the older population, vaccination is not very eective in reducing their susceptibility.
##### On the Computation of the Minimum Polynomial and Applications
Nikolaos Halidias
Asian Research Journal of Mathematics, Page 301-319
DOI: 10.9734/arjom/2022/v18i11603
Aims/Objectives: In this review article we study the computation of the minimum polynomial
of a matrix A and how we can use it for the computation of the matrix An. We also describe
the form of the elements of the matrix A-n and we will see that it is closely related with the
computation of the Drazin generalized inverse of A. Next we study the computation of the
exponential matrix and nally we give a simple proof of the Leverrier - Faddeev algorithm for
the computation of the characteristic polynomial. | 7,715 | 35,256 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | longest | en | 0.892586 |
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# Lecture-03-08[1] - Kinematics Kinematics Kinematics...
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1 Kinematics Kinematics ± Kinematics: Description of Motion ± position and displacement ± velocity ± average ± instantaneous ± acceleration ± average ± instantaneous 10
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2 Displacement, Velocity and Speed Displacement i x x x f Δ Average velocity x x x v i x f Δ = dt dx t x v x = Δ Δ = lim t t t i f Δ Average speed Spent Time Total Traveled Distance Total v Instantaneous velocity Instantaneous speed dt dx t x v x = Δ Δ = 0 Δt Stop-Motion Photography We can superimpose all the frames to create a “stop-motion” picture that shows the progression of positions as equal intervals of time pass. 4 The equal spacing of the images indicates that the car has a constant velocity .
3 Change and Motion A stationary ball on the ground. Same position at all times. Velocity=0. A skateboarder rolling on a sidewalk. Images are equally spaced. Velocity=consta A sprinter starting the 100 meter dash. 5 Image spacing grows. Velocity increasing. A car stopping for a red light. Image spacing shrinks. Velocity decreasing. The actual images can be replaced by “dots” showing successive positions. Motion Diagrams 6
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4 Position & Displacement A student on a bicycle is moving in a straight line. The coordinate axis consists of a line along the path of the bicycle. A point on this line chosen as the origin O . Other points are assigned a number x , the value of x being proportional to the distance from O . The numbers assigned to points to the right of O are positive, as shown, and those assigned to points to the left of O are negative. When the bicycle travels from point x i to point x f , its displacement is: Δ x = x f x i . Interpreting a Position Graph Motion of a car 8
5 Average Speed & Velocity Average speed (scalar) total distance total time s t ≡= Δ av x Average velocity (vector) fi xx x v Δ ≡≡ = GG G G av x (so ) ttt xv t Δ− Δ= Δ Instantaneous Velocity & Speed 0 ( ) lim slope of line tangent to vs. curve.
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Lecture-03-08[1] - Kinematics Kinematics Kinematics...
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# M408CF09Exam2 - Exam 2 Math 408C Name Fall 2009 TA...
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Unformatted text preview: Exam 2 Math 408C Name Fall 2009 TA Discussion Time: TTH You must show sufficient work in order to receive full credit for a problem. Do your work on the paper provided. Write your name on this sheet and turn it in with your work. Please write legibly and label the problems clearly. Circle your answers when appropriate. No calculators allowed. 1. (14 points) Let f (as) = sin2 cc -— x/5 sin 3:, 0 g cc S 7r. Find all critical numbers of f and classify the local extrema. a: 2. (14 points) Describe the concavity of the function f (x) = 2 4 x _ and find the points of inflection (if any exist). 3. (14 points) Find all horizontal asymptotes of the piecewise function x/zc2—3a:+2 f(x)= ”3‘2 —\/£L‘2+JI+1,IL'>0 4. (14 points) Water is pouring into a cone whose height is 8 inches and radius of the base is 2 inches. If the volume (measured in cubic inches) of the water in the , 3:30 reservoir at time t (measured in minutes) is given by V(t) — the rate at which the water level is rising after 1 minute has elapsed. (Volume of a cone of height h and radius of base 7' is V = %7T7"2h.) 5. (14 points) Find the point (or points) on the curve 3/ = 1 +x, —1 S a: g 4 closest to the point (5,0) and the points (or points) that are farthest from (5,0). 6. (14 points) Find a function f satisfying f”(:c) = sinx + 225% — 5, f’(0) = 3, and f(0) = 1. 7. (a) (8 points) Use 4 subintervals to find upper and lower bounds for the area of the region under the curve y = 1 + 4332, —1 g m S 1. 1 (b) (8 points) Approximate / 1 1 + 4x2dx by a Riemann sum with n terms. (You do not need to evaluate the limit if this Riemann sum.) 1 Bonus: (5 points) Evaluate /_11 + 4x2dx by finding the limit of the Riemann sum 1)1 2 1 in part 7 (b) You may use the fact that I; k— — n_____(n2 + )and 2 k2: MES—(L)- You may not use the Fundamental Theorem of Calculus mqogc Wow? WWW I. 4h): W7X ~V—3'WY Osxsll I Z. 1(7)” = ”97‘ (X714) -— (~xZ-V>{zx\a(Xz-U) (x1404 : (\$240 [flaMxZ—fl WNW/)1: M3 #91:“ (xii?) (Xz_q)q (erq)3 (Xz'q)3 6H5) 5: ”WC /“LY“L{ 30 ,ank lamdwwj/M \$13M “Woo/“MM (x In) 4h atom/L flw wdaa fowj‘ h: (w rs W M MN = {X—§)L+(Hx)7‘ ’rL“ 43W): aha—53 + an”) (wt “—0 M) M 4314) 712%wa \$5M WWW W. #43: 34, £(fl:9+‘i=l? £(43= l+z¥=26 50 #4 Nauaf- Pow} 413(33031'S (2,3) MM ,VMW r5 (4)0). I (v gl/x\:§’w\7< +Jx -5— S7 ryl/X) =~CO§7<+_(€-X3_SX4C §”°)=3= ~\+C=3=>c:q 50 444W 43W 3md731 an Numb HI] mm sucwxo MM“ ’7« ZMW‘U’; (/5 “A (“l/W44 N X0:"I X' "')Xw”'lm 1 (NW Y1: —l+\e(%\ n TLw 1h WWW“ 5W 9“ =ég¥lq§ Ay U/l/UM CE. (3 cvwa W‘s—UL 11k HAL wJ—bwafl [waybj MCk=¥K:"‘/‘QJ¢:. ...
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M408CF09Exam2 - Exam 2 Math 408C Name Fall 2009 TA...
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Ask a homework question - tutors are online | 1,284 | 3,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.836568 |
https://www.aqua-calc.com/one-to-one/density/kilogram-per-liter/long-ton-per-liter/73 | 1,571,023,669,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00304.warc.gz | 870,854,384 | 8,195 | # 73 kilograms per liter [kg/l] in long tons per liter
## kg/l to long tn/l unit converter of density
73 kilograms per liter [kg/l] = 0.07 long ton per liter [long tn/l]
### kilograms per liter to long tons per liter density conversion cards
• 73
through
97
kilograms per liter
• 73 kg/l to long tn/l = 0.07 long tn/l
• 74 kg/l to long tn/l = 0.07 long tn/l
• 75 kg/l to long tn/l = 0.07 long tn/l
• 76 kg/l to long tn/l = 0.07 long tn/l
• 77 kg/l to long tn/l = 0.08 long tn/l
• 78 kg/l to long tn/l = 0.08 long tn/l
• 79 kg/l to long tn/l = 0.08 long tn/l
• 80 kg/l to long tn/l = 0.08 long tn/l
• 81 kg/l to long tn/l = 0.08 long tn/l
• 82 kg/l to long tn/l = 0.08 long tn/l
• 83 kg/l to long tn/l = 0.08 long tn/l
• 84 kg/l to long tn/l = 0.08 long tn/l
• 85 kg/l to long tn/l = 0.08 long tn/l
• 86 kg/l to long tn/l = 0.08 long tn/l
• 87 kg/l to long tn/l = 0.09 long tn/l
• 88 kg/l to long tn/l = 0.09 long tn/l
• 89 kg/l to long tn/l = 0.09 long tn/l
• 90 kg/l to long tn/l = 0.09 long tn/l
• 91 kg/l to long tn/l = 0.09 long tn/l
• 92 kg/l to long tn/l = 0.09 long tn/l
• 93 kg/l to long tn/l = 0.09 long tn/l
• 94 kg/l to long tn/l = 0.09 long tn/l
• 95 kg/l to long tn/l = 0.09 long tn/l
• 96 kg/l to long tn/l = 0.09 long tn/l
• 97 kg/l to long tn/l = 0.1 long tn/l
• 98
through
122
kilograms per liter
• 98 kg/l to long tn/l = 0.1 long tn/l
• 99 kg/l to long tn/l = 0.1 long tn/l
• 100 kg/l to long tn/l = 0.1 long tn/l
• 101 kg/l to long tn/l = 0.1 long tn/l
• 102 kg/l to long tn/l = 0.1 long tn/l
• 103 kg/l to long tn/l = 0.1 long tn/l
• 104 kg/l to long tn/l = 0.1 long tn/l
• 105 kg/l to long tn/l = 0.1 long tn/l
• 106 kg/l to long tn/l = 0.1 long tn/l
• 107 kg/l to long tn/l = 0.11 long tn/l
• 108 kg/l to long tn/l = 0.11 long tn/l
• 109 kg/l to long tn/l = 0.11 long tn/l
• 110 kg/l to long tn/l = 0.11 long tn/l
• 111 kg/l to long tn/l = 0.11 long tn/l
• 112 kg/l to long tn/l = 0.11 long tn/l
• 113 kg/l to long tn/l = 0.11 long tn/l
• 114 kg/l to long tn/l = 0.11 long tn/l
• 115 kg/l to long tn/l = 0.11 long tn/l
• 116 kg/l to long tn/l = 0.11 long tn/l
• 117 kg/l to long tn/l = 0.12 long tn/l
• 118 kg/l to long tn/l = 0.12 long tn/l
• 119 kg/l to long tn/l = 0.12 long tn/l
• 120 kg/l to long tn/l = 0.12 long tn/l
• 121 kg/l to long tn/l = 0.12 long tn/l
• 122 kg/l to long tn/l = 0.12 long tn/l
• 123
through
147
kilograms per liter
• 123 kg/l to long tn/l = 0.12 long tn/l
• 124 kg/l to long tn/l = 0.12 long tn/l
• 125 kg/l to long tn/l = 0.12 long tn/l
• 126 kg/l to long tn/l = 0.12 long tn/l
• 127 kg/l to long tn/l = 0.12 long tn/l
• 128 kg/l to long tn/l = 0.13 long tn/l
• 129 kg/l to long tn/l = 0.13 long tn/l
• 130 kg/l to long tn/l = 0.13 long tn/l
• 131 kg/l to long tn/l = 0.13 long tn/l
• 132 kg/l to long tn/l = 0.13 long tn/l
• 133 kg/l to long tn/l = 0.13 long tn/l
• 134 kg/l to long tn/l = 0.13 long tn/l
• 135 kg/l to long tn/l = 0.13 long tn/l
• 136 kg/l to long tn/l = 0.13 long tn/l
• 137 kg/l to long tn/l = 0.13 long tn/l
• 138 kg/l to long tn/l = 0.14 long tn/l
• 139 kg/l to long tn/l = 0.14 long tn/l
• 140 kg/l to long tn/l = 0.14 long tn/l
• 141 kg/l to long tn/l = 0.14 long tn/l
• 142 kg/l to long tn/l = 0.14 long tn/l
• 143 kg/l to long tn/l = 0.14 long tn/l
• 144 kg/l to long tn/l = 0.14 long tn/l
• 145 kg/l to long tn/l = 0.14 long tn/l
• 146 kg/l to long tn/l = 0.14 long tn/l
• 147 kg/l to long tn/l = 0.14 long tn/l
• 148
through
172
kilograms per liter
• 148 kg/l to long tn/l = 0.15 long tn/l
• 149 kg/l to long tn/l = 0.15 long tn/l
• 150 kg/l to long tn/l = 0.15 long tn/l
• 151 kg/l to long tn/l = 0.15 long tn/l
• 152 kg/l to long tn/l = 0.15 long tn/l
• 153 kg/l to long tn/l = 0.15 long tn/l
• 154 kg/l to long tn/l = 0.15 long tn/l
• 155 kg/l to long tn/l = 0.15 long tn/l
• 156 kg/l to long tn/l = 0.15 long tn/l
• 157 kg/l to long tn/l = 0.15 long tn/l
• 158 kg/l to long tn/l = 0.16 long tn/l
• 159 kg/l to long tn/l = 0.16 long tn/l
• 160 kg/l to long tn/l = 0.16 long tn/l
• 161 kg/l to long tn/l = 0.16 long tn/l
• 162 kg/l to long tn/l = 0.16 long tn/l
• 163 kg/l to long tn/l = 0.16 long tn/l
• 164 kg/l to long tn/l = 0.16 long tn/l
• 165 kg/l to long tn/l = 0.16 long tn/l
• 166 kg/l to long tn/l = 0.16 long tn/l
• 167 kg/l to long tn/l = 0.16 long tn/l
• 168 kg/l to long tn/l = 0.17 long tn/l
• 169 kg/l to long tn/l = 0.17 long tn/l
• 170 kg/l to long tn/l = 0.17 long tn/l
• 171 kg/l to long tn/l = 0.17 long tn/l
• 172 kg/l to long tn/l = 0.17 long tn/l
#### Foods, Nutrients and Calories
AHOLD, MEDITERRANEAN OLIVE MEDLEY PITTED, UPC: 688267160851 contain(s) 167 calories per 100 grams or ≈3.527 ounces [ price ]
DANDY DAISIES, UPC: 041512083198 weigh(s) 152.16 gram per (metric cup) or 5.08 ounce per (US cup), and contain(s) 333 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
#### Gravels, Substances and Oils
Sand, Silica weighs 1 538 kg/m³ (96.0142 lb/ft³) with specific gravity of 1.538 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Anthracene [(C6H4CH)2] weighs 1 283 kg/m³ (80.09507 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Milkweed oil with temperature in the range of 23.9°C (75.02°F) to 110°C (230°F)
#### Weights and Measurements
pound per foot (lb/ft) is a non-metric measurement unit of linear or linear mass density.
Volume is a basic characteristic of any three–dimensional geometrical object.
kcal/s to µJ/min conversion table, kcal/s to µJ/min unit converter or convert between all units of power measurement.
#### Calculators
Volume of a rectangular box calculator. Definition and formulas. | 2,295 | 5,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.44346 |
https://math.stackexchange.com/questions/2232655/counting-the-number-of-ways-heads-shows-up-exactly-for-10-coin-tosses | 1,631,852,477,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00475.warc.gz | 415,340,875 | 38,628 | # Counting the number of ways heads shows up exactly for 10 coin tosses
Problem: Ten coins are tossed and each is equally likely to come down heads or tails. How many ways are there to get 3 heads?
I know the size of the sample space is $2^{10}$. It seems like I should subtract the total number of orderings with more than 3 heads and less than 2 heads from the sample space. Then I would have the result. Is that the best way of going about this? There doesn't seem to be any necessarily straightforward approach.
• If $3$ of the $10$ coins come down as heads, then $10-3=7$ are tails. Then the question becomes how many ways can you arrange $3$ heads and $7$ tails in a line of $10$ coins. Apr 13 '17 at 17:19
• Have you ever heard of binomial coefficients? Apr 13 '17 at 17:19
• Each way of choosing a subset of size 3 of the set of 10 pennies is a way for exactly 3 of the pennies to be heads. Apr 13 '17 at 17:20
• @rudy not quite... $3!7!$ will appear in a correct answer somehow, but not by itself as the answer as a whole. Perhaps a fraction that involves it... Apr 13 '17 at 17:22
• @JMoravitz My issue, I feel, is that order matters in terms of how many sequences there are, but order does not matter for how I am selecting them.
– rudy
Apr 13 '17 at 17:22
The binomial coefficient $\binom{10}{3}=\frac{10!}{3!7!}$ counts the number of ways in which you may select a subset of three objects out of a set of ten distinct objects.
In this case, our ten objects are the distinct positions in the sequence. We select three of them to be those positions who will have a coinflip resulting in a head (and the remaining seven positions will have a coinflip resulting in a tail). As such, $\binom{10}{3}$ is the number of ways of selecting three positions in the sequence of ten flips to be a head and therefore counts the number of sequences with exactly three heads.
The mantra that "you use combinations when order doesn't matter" here is in reference to that having picked which of the locations are occupied by heads, the order in which you write which locations those are won't matter despite the locations themselves having a specific order that matters. "I pick the first, the third, and the second positions to be those with heads" results in HHHTTTTTTT, meanwhile "I pick the third, the first, and the second positions to be those with heads" also results in HHHTTTTTTT.
You are correct that there are $2^{10}$ different sequences of ten coinflips (seen quickly by multiplication principle). We can break it down into cases as well:
$2^{10} = \text{#exactly0heads}~+~\text{#exactly1head}+\text{#exactly2heads}+\dots+\text{#exactly10heads}$
$2^{10}=\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\dots+\binom{10}{10}$
This is just a special case of the binomial theorem: $(x+y)^n = \sum\limits_{k=0}^n\binom{n}{k}x^ky^{n-k}$ where $x=y=1$ and $n=10$ | 783 | 2,866 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.959774 |
https://xaviergeerinck.com/2015/06/05/sql---part-1---advanced-case-selecting/ | 1,709,366,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00483.warc.gz | 1,059,468,268 | 12,220 | While I was studying the course Databases at the University of Ghent, we encountered some special queries that we were not being thaught in the Professional Bachelor course. This is why I tried to summarise those queries here for everyone to learn from.
The database engine being used is Oracle SQL server. This has been chosen because Oracle supports most of the functions that we needed in this class. (Another good SQL engine to use is MSSQL).
Please note, every query given here is based on the database that we had in the class, this database is not mine and therefor I can not publish this here nor post results of the query.
## Case Functions
### Order by given order
To order by a given order we are going to give a specific number from 1 .. x for the values that we want. We do this by using a case in the order part of our SQL statement.
Example 1: Order by disciplines 'KB', 'GS', 'SL', 'DH', 'SG'
``````SELECT distinct discipline, resort
FROM races
ORDER BY
CASE discipline
WHEN 'KB' THEN 1
WHEN 'GS' THEN 2
WHEN 'SL' THEN 3
WHEN 'DH' THEN 4
WHEN 'SG' THEN 5
ELSE 6
END;
``````
Example 2: Order by NULL or other values (NULL first, then the other values)
``````SELECT rid, modus
FROM races
WHERE discipline = 'SL' AND gender = 'M'
ORDER BY
CASE
WHEN modus IS NULL THEN 0
ELSE (modus + 1)
END
ASC;
``````
### Putting an X in one of multiple columns based on the value that we have
here is a query that is going to put an X in the column where we want it.
Example 1: We are calculating the BMI and have 3 columns stating: "underweight", "obesity", "normal". Now we are going to put an X in the column that matches the person his/her weight.
``````SELECT
name
, weight
, gender
, ROUND((weight / (height * height) ), 3) BMI
, CASE WHEN ROUND((weight / (height * height) ), 3) < 17
THEN 'X'
ELSE ' '
END underweight
, CASE WHEN ROUND((weight / (height * height) ), 3) BETWEEN 17 AND 23.999
THEN 'X'
ELSE ' '
END normal
, CASE WHEN ROUND((weight / (height * height) ), 3) >= 24
THEN 'X'
ELSE ' '
END obesity
FROM competitors
WHERE weight IS NOT NULL AND height IS NOT NULL
ORDER BY ROUND((weight / (height * height) ), 3);
``````
Example 2: We have races that are being done on a specific date, we want to be able to put an X based on the season.
``````SELECT EXTRACT(year FROM (racedate + 183)) season, racedate
, CASE WHEN racedate BETWEEN TO_DATE('21 03 ' || extract(year FROM racedate), 'dd MM YYYY') AND TO_DATE('20 06 ' || extract(year FROM racedate), 'dd MM YYYY') THEN 'X' ELSE ' ' END spring
, CASE WHEN racedate BETWEEN TO_DATE('21 06 ' || extract(year FROM racedate), 'dd MM YYYY') AND TO_DATE('20 09 ' || extract(year FROM racedate), 'dd MM YYYY') THEN 'X' ELSE ' ' END summer
, CASE WHEN racedate BETWEEN TO_DATE('21 09 ' || extract(year FROM racedate), 'dd MM YYYY') AND TO_DATE('20 12 ' || extract(year FROM racedate), 'dd MM YYYY') THEN 'X' ELSE ' ' END autumn
, CASE WHEN racedate BETWEEN TO_DATE('21 03 ' || extract(year FROM racedate), 'dd MM YYYY') AND TO_DATE('20 12 ' || extract(year FROM racedate), 'dd MM YYYY') THEN ' ' ELSE 'X' END winter
FROM races
WHERE EXTRACT(year FROM (racedate + 183)) BETWEEN 1990 AND 1992 AND discipline = 'SL'
ORDER BY summer desc, spring desc, autumn desc, winter desc, racedate;
``````
### Replace NVL with a CASE structure
NVL is a database engine specific function that is going to replace the NULL values with a value that you give to it. We want to be able to use this on other database engines so that is why we are going to write this as a normal CASE structure.
Example with NVL:
``````SELECT rid, rank, nvl(points, 0)
FROM results
WHERE rank = 4 AND points < 12;
``````
Example with CASE:
``````SELECT rid, rank,
CASE
WHEN points IS NULL THEN 0
ELSE points
END points
FROM results
WHERE rank = 4 AND points < 12;
``````
### Combination of all the above
``````SELECT
CASE
WHEN nation IN ('CAN', 'USA') THEN 'North America'
WHEN nation IN ('BUL', 'RUS', 'CZE', 'SLO', 'CRO') THEN 'Eastern-Europe'
WHEN nation IN ('JPN', 'KOR') THEN 'Asia'
ELSE 'Western-Europe'
END Continent
, nation, name, finishaltitude
, CASE WHEN finishaltitude < 1200 THEN 'X' ELSE ' ' END "<1200" , CASE WHEN finishaltitude BETWEEN 1200 AND 1700 THEN 'X' ELSE ' ' END "1200-1700" , CASE WHEN finishaltitude > 1700 THEN 'X' ELSE ' ' END ">1700"
FROM resorts
WHERE finishaltitude IS NOT NULL
-- Order by: Western-Europe, Eastern-Europe, North America, Asia
ORDER BY
CASE
WHEN nation IN ('CAN', 'USA') THEN 3
WHEN nation IN ('BUL', 'RUS', 'CZE', 'SLO', 'CRO') THEN 2
WHEN nation IN ('JPN', 'KOR') THEN 4
ELSE 1
END
ASC
-- By category with X
, CASE WHEN finishaltitude < 1200 THEN 1
WHEN finishaltitude BETWEEN 1200 AND 1700 THEN 2
ELSE 3
END
ASC
-- Resort name
, name;
`````` | 1,373 | 4,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | longest | en | 0.894204 |
https://www.geeksforgeeks.org/ip-addressing-introduction-and-classful-addressing/ | 1,550,551,523,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489343.24/warc/CC-MAIN-20190219041222-20190219063222-00318.warc.gz | 832,507,290 | 21,491 | Generally, there are two notations in which IP address is written, dotted decimal notation and hexadecimal notation.
Dotted Decimal Notation
Some points to be noted about dotted decimal notation :
1. The value of any segment (byte) is between 0 and 255 (both included).
2. There are no zeroes preceding the value in any segment (054 is wrong, 54 is correct).
The 32 bit IP address is divided into five sub-classes. These are:
• Class A
• Class B
• Class C
• Class D
• Class E
Each of these classes has a valid range of IP addresses. Classes D and E are reserved for multicast and experimental purposes respectively. The order of bits in the first octet determine the classes of IP address.
IPv4 address is divided into two parts:
• Network ID
• Host ID
The class of IP address is used to determine the bits used for network ID and host ID and the number of total networks and hosts possible in that particular class. Each ISP or network administrator assigns IP address to each device that is connected to its network.
Note: IP addresses are globally managed by Internet Assigned Numbers Authority(IANA) and regional Internet registries(RIR).
Note: While finding the total number of host IP addresses, 2 IP addresses are not counted and are therefore, decreased from the total count because the first IP address of any network is the network number and whereas the last IP address is reserved for broadcast IP.
Class A:
IP address belonging to class A are assigned to the networks that contain a large number of hosts.
• The network ID is 8 bits long.
• The host ID is 24 bits long.
The higher order bit of the first octet in class A is always set to 0. The remaining 7 bits in first octet are used to determine network ID. The 24 bits of host ID are used to determine the host in any network. The default sub-net mask for class A is 255.x.x.x. Therefore, class A has a total of:
• 2^7= 128 network ID
• 2^24 – 2 = 16,777,214 host ID
IP addresses belonging to class A ranges from 1.x.x.x – 126.x.x.x
Class B:
IP address belonging to class B are assigned to the networks that ranges from medium-sized to large-sized networks.
• The network ID is 16 bits long.
• The host ID is 16 bits long.
The higher order bits of the first octet of IP addresses of class B are always set to 10. The remaining 14 bits are used to determine network ID. The 16 bits of host ID is used to determine the host in any network. The default sub-net mask for class B is 255.255.x.x. Class B has a total of:
• 2^14 = 16384 network address
• 2^16 – 2 = 65534 host address
• IP addresses belonging to class B ranges from 128.0.x.x – 191.255.x.x.
Class C:
IP address belonging to class C are assigned to small-sized networks.
• The network ID is 24 bits long.
• The host ID is 8 bits long.
The higher order bits of the first octet of IP addresses of class C are always set to 110. The remaining 21 bits are used to determine network ID. The 8 bits of host ID is used to determine the host in any network. The default sub-net mask for class C is 255.255.255.x. Class C has a total of:
• 2^21 = 2097152 network address
• 2^8 – 2 = 254 host address
IP addresses belonging to class C ranges from 192.0.0.x – 223.255.255.x.
Class D:
IP address belonging to class D are reserved for multi-casting. The higher order bits of the first octet of IP addresses belonging to class D are always set to 1110. The remaining bits are for the address that interested hosts recognize.
Class D does not posses any sub-net mask. IP addresses belonging to class D ranges from 224.0.0.0 – 239.255.255.255.
Class E:
IP addresses belonging to class E are reserved for experimental and research purposes. IP addresses of class E ranges from 240.0.0.0 – 255.255.255.254. This class doesn’t have any sub-net mask. The higher order bits of first octet of class E are always set to 1111.
127.0.0.0 – 127.0.0.8 : Loop-back addresses
0.0.0.0 – 0.0.0.8 : used to communicate within the current network.
Rules for assigning Host ID:
Host ID’s are used to identify a host within a network. The host ID are assigned based on the following rules:
• Within any network, the host ID must be unique to that network.
• Host ID in which all bits are set to 0 cannot be assigned because this host ID is used to represent the network ID of the IP address.
• Host ID in which all bits are set to 1 cannot be assigned because this host ID is reserved as a broadcast address to send packets to all the hosts present on that particular network.
Rules for assigning Network ID:
Hosts that are located on the same physical network are identified by the network ID, as all host on the same physical network are assigned the same network ID. The network ID is assigned based on the following rules:
• The network ID cannot start with 127 because 127 belongs to class A address and is reserved for internal loop-back functions.
• All bits of network ID set to 1 are reserved for use as an IP broadcast address and therefore, cannot be used.
• All bits of network ID set to 0 are used to denote a specific host on the local network and are not routed and therefore, aren’t used. | 1,251 | 5,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-09 | longest | en | 0.930428 |
https://beta.mapleprimes.com/questions/234276-How-To-Solve-The-Given-BVP-Using-Differential | 1,723,664,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00688.warc.gz | 104,015,509 | 24,864 | # Question:How to solve the given BVP using differential transformation method
## Question:How to solve the given BVP using differential transformation method
Maple 17
HI, I have numerically solved the given problem using the dsolve command But I want to solve the same problem using the Differential transformation method.
Can anyone help me to get the series solution for the given problem using DTM.
I want to compare the numerical results with DTM results when lambda =0.5.
eqn1 := diff(f(eta), `\$`(eta, 3))+f(eta)*(diff(f(eta), `\$`(eta, 2)))-(diff(f(eta), eta))^2-lambda*(diff(f(eta), eta)) = 0.
eqn2 := diff(theta(eta), `\$`(eta, 2))+f(eta)*(diff(theta(eta), eta))*Pr = 0
Bcs := (D(f))(0) = 1, f(0) = 0, (D(f))(infinity) = 0, theta(0) = 1, theta(infinity) = 0;
[lambda = .5, Pr = 6.3]
| 251 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-33 | latest | en | 0.659428 |
https://www.netlib.org/lapack/explore-html-3.3.1/d1/da9/zstt21_8f_source.html | 1,713,336,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00064.warc.gz | 831,983,386 | 5,441 | LAPACK 3.3.1 Linear Algebra PACKage
# zstt21.f
Go to the documentation of this file.
```00001 SUBROUTINE ZSTT21( N, KBAND, AD, AE, SD, SE, U, LDU, WORK, RWORK,
00002 \$ RESULT )
00003 *
00004 * -- LAPACK test routine (version 3.1) --
00005 * Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
00006 * November 2006
00007 *
00008 * .. Scalar Arguments ..
00009 INTEGER KBAND, LDU, N
00010 * ..
00011 * .. Array Arguments ..
00012 DOUBLE PRECISION AD( * ), AE( * ), RESULT( 2 ), RWORK( * ),
00013 \$ SD( * ), SE( * )
00014 COMPLEX*16 U( LDU, * ), WORK( * )
00015 * ..
00016 *
00017 * Purpose
00018 * =======
00019 *
00020 * ZSTT21 checks a decomposition of the form
00021 *
00022 * A = U S U*
00023 *
00024 * where * means conjugate transpose, A is real symmetric tridiagonal,
00025 * U is unitary, and S is real and diagonal (if KBAND=0) or symmetric
00026 * tridiagonal (if KBAND=1). Two tests are performed:
00027 *
00028 * RESULT(1) = | A - U S U* | / ( |A| n ulp )
00029 *
00030 * RESULT(2) = | I - UU* | / ( n ulp )
00031 *
00032 * Arguments
00033 * =========
00034 *
00035 * N (input) INTEGER
00036 * The size of the matrix. If it is zero, ZSTT21 does nothing.
00037 * It must be at least zero.
00038 *
00039 * KBAND (input) INTEGER
00040 * The bandwidth of the matrix S. It may only be zero or one.
00041 * If zero, then S is diagonal, and SE is not referenced. If
00042 * one, then S is symmetric tri-diagonal.
00043 *
00044 * AD (input) DOUBLE PRECISION array, dimension (N)
00045 * The diagonal of the original (unfactored) matrix A. A is
00046 * assumed to be real symmetric tridiagonal.
00047 *
00048 * AE (input) DOUBLE PRECISION array, dimension (N-1)
00049 * The off-diagonal of the original (unfactored) matrix A. A
00050 * is assumed to be symmetric tridiagonal. AE(1) is the (1,2)
00051 * and (2,1) element, AE(2) is the (2,3) and (3,2) element, etc.
00052 *
00053 * SD (input) DOUBLE PRECISION array, dimension (N)
00054 * The diagonal of the real (symmetric tri-) diagonal matrix S.
00055 *
00056 * SE (input) DOUBLE PRECISION array, dimension (N-1)
00057 * The off-diagonal of the (symmetric tri-) diagonal matrix S.
00058 * Not referenced if KBSND=0. If KBAND=1, then AE(1) is the
00059 * (1,2) and (2,1) element, SE(2) is the (2,3) and (3,2)
00060 * element, etc.
00061 *
00062 * U (input) COMPLEX*16 array, dimension (LDU, N)
00063 * The unitary matrix in the decomposition.
00064 *
00065 * LDU (input) INTEGER
00066 * The leading dimension of U. LDU must be at least N.
00067 *
00068 * WORK (workspace) COMPLEX*16 array, dimension (N**2)
00069 *
00070 * RWORK (workspace) DOUBLE PRECISION array, dimension (N)
00071 *
00072 * RESULT (output) DOUBLE PRECISION array, dimension (2)
00073 * The values computed by the two tests described above. The
00074 * values are currently limited to 1/ulp, to avoid overflow.
00075 * RESULT(1) is always modified.
00076 *
00077 * =====================================================================
00078 *
00079 * .. Parameters ..
00080 DOUBLE PRECISION ZERO, ONE
00081 PARAMETER ( ZERO = 0.0D+0, ONE = 1.0D+0 )
00082 COMPLEX*16 CZERO, CONE
00083 PARAMETER ( CZERO = ( 0.0D+0, 0.0D+0 ),
00084 \$ CONE = ( 1.0D+0, 0.0D+0 ) )
00085 * ..
00086 * .. Local Scalars ..
00087 INTEGER J
00088 DOUBLE PRECISION ANORM, TEMP1, TEMP2, ULP, UNFL, WNORM
00089 * ..
00090 * .. External Functions ..
00091 DOUBLE PRECISION DLAMCH, ZLANGE, ZLANHE
00092 EXTERNAL DLAMCH, ZLANGE, ZLANHE
00093 * ..
00094 * .. External Subroutines ..
00095 EXTERNAL ZGEMM, ZHER, ZHER2, ZLASET
00096 * ..
00097 * .. Intrinsic Functions ..
00098 INTRINSIC ABS, DBLE, DCMPLX, MAX, MIN
00099 * ..
00100 * .. Executable Statements ..
00101 *
00102 * 1) Constants
00103 *
00104 RESULT( 1 ) = ZERO
00105 RESULT( 2 ) = ZERO
00106 IF( N.LE.0 )
00107 \$ RETURN
00108 *
00109 UNFL = DLAMCH( 'Safe minimum' )
00110 ULP = DLAMCH( 'Precision' )
00111 *
00112 * Do Test 1
00113 *
00114 * Copy A & Compute its 1-Norm:
00115 *
00116 CALL ZLASET( 'Full', N, N, CZERO, CZERO, WORK, N )
00117 *
00118 ANORM = ZERO
00119 TEMP1 = ZERO
00120 *
00121 DO 10 J = 1, N - 1
00122 WORK( ( N+1 )*( J-1 )+1 ) = AD( J )
00123 WORK( ( N+1 )*( J-1 )+2 ) = AE( J )
00124 TEMP2 = ABS( AE( J ) )
00125 ANORM = MAX( ANORM, ABS( AD( J ) )+TEMP1+TEMP2 )
00126 TEMP1 = TEMP2
00127 10 CONTINUE
00128 *
00129 WORK( N**2 ) = AD( N )
00130 ANORM = MAX( ANORM, ABS( AD( N ) )+TEMP1, UNFL )
00131 *
00132 * Norm of A - USU*
00133 *
00134 DO 20 J = 1, N
00135 CALL ZHER( 'L', N, -SD( J ), U( 1, J ), 1, WORK, N )
00136 20 CONTINUE
00137 *
00138 IF( N.GT.1 .AND. KBAND.EQ.1 ) THEN
00139 DO 30 J = 1, N - 1
00140 CALL ZHER2( 'L', N, -DCMPLX( SE( J ) ), U( 1, J ), 1,
00141 \$ U( 1, J+1 ), 1, WORK, N )
00142 30 CONTINUE
00143 END IF
00144 *
00145 WNORM = ZLANHE( '1', 'L', N, WORK, N, RWORK )
00146 *
00147 IF( ANORM.GT.WNORM ) THEN
00148 RESULT( 1 ) = ( WNORM / ANORM ) / ( N*ULP )
00149 ELSE
00150 IF( ANORM.LT.ONE ) THEN
00151 RESULT( 1 ) = ( MIN( WNORM, N*ANORM ) / ANORM ) / ( N*ULP )
00152 ELSE
00153 RESULT( 1 ) = MIN( WNORM / ANORM, DBLE( N ) ) / ( N*ULP )
00154 END IF
00155 END IF
00156 *
00157 * Do Test 2
00158 *
00159 * Compute UU* - I
00160 *
00161 CALL ZGEMM( 'N', 'C', N, N, N, CONE, U, LDU, U, LDU, CZERO, WORK,
00162 \$ N )
00163 *
00164 DO 40 J = 1, N
00165 WORK( ( N+1 )*( J-1 )+1 ) = WORK( ( N+1 )*( J-1 )+1 ) - CONE
00166 40 CONTINUE
00167 *
00168 RESULT( 2 ) = MIN( DBLE( N ), ZLANGE( '1', N, N, WORK, N,
00169 \$ RWORK ) ) / ( N*ULP )
00170 *
00171 RETURN
00172 *
00173 * End of ZSTT21
00174 *
00175 END
``` | 2,208 | 6,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.44894 |
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## #1 2012-10-10 09:16:47
zetafunc.
Guest
### Rod rotating about fixed axis
"One end of a rod of uniform density is attached to the ceiling in such a way that the rod can swing about freely with no resistance. The other end of the rod is held still so that it touches the ceiling as well. Then, the second end is released. If the length of the rod is l metres and gravitational acceleration is g metres per second squared, how fast is the unattached end of the rod moving when the rod is first vertical?"
I'm not sure I did this question correctly, but no answers are provided either. Here's what I did:
where I is the moment of inertia of the rod about an axis perpendicular to the rod and in the plane of the rod, and w is the angular velocity of the rod.
So we need to find the M.I. of the rod. Suppose Iz is the M.I. of the rod about an axis through one end of the rod and perpendicular to the rod. By the parallel axis theorem, the M.I. is
but we want Iy. But Ix = 0, so Iy = Iz by the perpendicular axis theorem. Since Loss in GPE = Gain in KE, then
therefore
is this correct?
## #2 2012-10-10 09:25:59
anonimnystefy
Real Member
Offline
### Re: Rod rotating about fixed axis
Phewh! For a second there I though it was our admin rotating.
The rod, it would seem represents a kind of pendulum whose period depends on uts mass, but the mass isn't given in the text, so I am a bit confused, and it doesn't look like a mathematical pendulum (where mass doesn't matter)...
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #3 2012-10-10 09:27:16
zetafunc.
Guest
### Re: Rod rotating about fixed axis
I let the mass of the rod be m, but it appears that it doesn't matter, since the mass cancels when comparing the GPE and KE.
## #4 2012-10-10 09:40:18
anonimnystefy
Real Member
Offline
### Re: Rod rotating about fixed axis
Ok, step by step. How'd you get the loss in GPE formula?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #5 2012-10-10 09:42:37
zetafunc.
Guest
### Re: Rod rotating about fixed axis
If the rod is at first horizontal, then rotates 90 degrees (when first vertical), then the change in GPE is mgl/2, since the centre of mass of the rod has dropped by a distance of l/2.
## #6 2012-10-10 09:49:36
anonimnystefy
Real Member
Offline
### Re: Rod rotating about fixed axis
But, the question never states the starting position of the rod.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #7 2012-10-10 09:54:03
zetafunc.
Guest
### Re: Rod rotating about fixed axis
One end of a rod is attached to the ceiling in such a way that the rod can swing about freely. The other end of the rod is held still so that it touches the ceiling as well. Then, the second end is released.
I don't see how else this could be interpreted... clearly the starting position is the rod in its horizontal position.
= (left end fixed)
goes to
|| (top end fixed)
## #8 2012-10-10 10:07:08
anonimnystefy
Real Member
Offline
### Re: Rod rotating about fixed axis
Oops! Me and my careful reading. Didn't see the "also touches the wall" part. Sorry.
It looks like the mass does cancel out, after all.
Just one more thing-Why are you using l/2 in tbe I_z formula if the rotation is around the end of the rod?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #9 2012-10-10 10:13:29
zetafunc.
Guest
### Re: Rod rotating about fixed axis
For a uniform rod of mass m and length 2l, the moment of inertia about an axis perp. to the rod through the centre is ml2/3 (you can prove this by integration but it should be fine to quote it). For a rod of length l, you square (l/2) instead. Then apply the parallel axis theorem, to get the M.I. of the rod about the axis through the end. I sort of wasted time thinking about I_x, I_y and I_z, since it is a rod, not a lamina, and therefore I_y = I_z.
## #10 2012-10-10 10:23:06
anonimnystefy
Real Member
Offline
### Re: Rod rotating about fixed axis
I would have to look at this in the morning. My brain is not functioning properly right now.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #11 2012-10-10 18:03:29
bob bundy
Moderator
Offline
### Re: Rod rotating about fixed axis
hi zetafunc,
That looks good to me.
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #12 2012-10-10 18:09:22
zetafunc.
Guest
### Re: Rod rotating about fixed axis
Thanks.
Is there a way to do this without knowing anything about moments of inertia? This question is taken from an admissions test in 2007 from Trinity College, Cambridge, but most applicants typically will not have covered this until Year 13 (it's an M5 topic)... can it be done with only M1-M3 knowledge?
## #13 2012-10-10 18:21:40
bob bundy
Moderator
Offline
### Re: Rod rotating about fixed axis
hi zetafunc
Moments of inertia can be calculated by integration. It's ages since I did this so there's lots of cobwebs to blow away first.
I'm sure you could do it.
MI of whole = integral over length of rod (MI of a dx segment)
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #14 2012-10-10 18:24:49
bob bundy
Moderator
Offline
### Re: Rod rotating about fixed axis
...... stopped thinking
I suppose you could find a way of calculating the KE directly. Write the velocity in terms of omega and l.
Maybe try it.
B
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #15 2012-10-11 02:57:39
zetafunc.
Guest
### Re: Rod rotating about fixed axis
I've studied moments of inertia before (had to do it for my exam) and deriving them via integration -- but I know the M.I. of the rod is definitely correct (and therefore the change in KE). Since:
as v = rw. But, this question still confuses me... are they asking for the ANGULAR velocity w, or just the instantaneous velocity of the end of the rod? In which case, with r = l;
v = rw, using my w from post #1;
## #16 2012-10-11 05:23:52
bob bundy
Moderator
Offline
### Re: Rod rotating about fixed axis
Oh yes, I'm terrible for only half reading the question. It does ask 'how fast' so I guess they want the velocity.
As you've got omega first I think you've covered yourself whatever.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
zetafunc.
Guest
Okay, thank you!
## #18 2012-10-12 07:32:37
zetafunc.
Guest
### Re: Rod rotating about fixed axis
A maths teacher said I am wrong but never explained why. Confused... | 2,172 | 7,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | longest | en | 0.928738 |
https://rdrr.io/bioc/qpgraph/man/qpHTF.html | 1,524,519,574,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946199.72/warc/CC-MAIN-20180423203935-20180423223935-00458.warc.gz | 677,487,986 | 15,892 | # qpHTF: Hastie Tibshirani Friedman iterative regression algorithm In qpgraph: Estimation of genetic and molecular regulatory networks from high-throughput genomics data
## Description
Performs maximum likelihood estimation of a covariance matrix given the independence constraints from an input undirected graph.
## Usage
1 qpHTF(S, g, tol = 0.001, verbose = FALSE, R.code.only = FALSE)
## Arguments
S input matrix, in the context of this package, the sample covariance matrix. g input undirected graph. tol tolerance under which the iterative algorithm stops. verbose show progress on calculations. R.code.only logical; if FALSE then the faster C implementation is used (default); if TRUE then only R code is executed.
## Details
This is an alternative to the Iterative Proportional Fitting (IPF) algorithm (see, Whittaker, 1990, pp. 182-185 and qpIPF) which also adjusts the input matrix to the independence constraints in the input undirected graph. However, differently to the IPF, it works by going through each of the vertices fitting the marginal distribution over the corresponding vertex boundary. It stops when the adjusted matrix at the current iteration differs from the matrix at the previous iteration in less or equal than a given tolerance value. This algorithm is described by Hastie, Tibshirani and Friedman (2009, pg. 634), hence we name it here HTF, and it has the advantage over the IPF that it does not require the list of maximal cliques of the graph which may be exponentially large. In contrast, it requires that the maximum boundary size of the graph is below the number of samples where the input sample covariance matrix S was estimated. For the purpose of exploring qp-graphs that meet such a requirement, one can use the function qpBoundary.
## Value
The input matrix adjusted to the constraints imposed by the input undirected graph, i.e., a maximum likelihood estimate of the sample covariance matrix that includes the independence constraints encoded in the undirected graph.
## Note
Thanks to Giovanni Marchetti for bringing us our attention to this algorithm and sharing an early version of its implementation on the R package ggm.
R. Castelo
## References
Castelo, R. and Roverato, A. A robust procedure for Gaussian graphical model search from microarray data with p larger than n. J. Mach. Learn. Res., 7:2621-2650, 2006.
Hastie, T., Tibshirani, R. and Friedman, J.H. The Elements of Statistical Learning, Springer, 2009.
Tur, I., Roverato, A. and Castelo, R. Mapping eQTL networks with mixed graphical Markov models. Genetics, 198(4):1377-1393, 2014.
Whittaker, J. Graphical Models in Applied Multivariate Statistics. Wiley, 1990. | 607 | 2,690 | {"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} | 2.609375 | 3 | CC-MAIN-2018-17 | longest | en | 0.877305 |
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Inequality Basic Concept Trick (in Hindi)
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In this lesson we use some tricky method and basic concept for inequality Topic
## Hemant Mishra is teaching live on Unacademy Plus
Hemant Mishra
Cleared RRB Clerk , SBI Clerk pre, NTPC 2016, 1 year experience in software developing along with 5 year Experience in Teaching .
U
thanks sir
Thanks 🙏 sir
K
no nikhil... b will be greater or equalto d hoga .
B>= C>=D is B>D true or not .
Nikhil C Shekar
2 years ago
you didn't mention the solution of this in your slide
Hemant Mishra
2 years ago
no not true
Hemant Mishra
2 years ago
if you have 2 option 1 is B>D and another is B=D than only either ,or case follow
Nikhil C Shekar
2 years ago
Thanks
Jubair choudhury
2 years ago
no it's false
1. BANK PO,CLERK EXAM REASONING INEQUALITY Basic concept HEMANT MISHRA
2. ABOUT ME Graduate from G.B.T.U Lucknow Love Travelling & Technology Rate , Review, Recommend & Share
3. Statements N 2 O 2 P Q > R Conclusions I. N > R II.R N 1) If only conclusion I follows 2) If only conclusion II follows 3) If either conclusion I or conclusion II follows 4) If neither conclusion I or conclusion II follows 5) If both conclusion I and II follow
4. Less Than/ neither greater nor equal S Less Than or Equal to > Greater Than /Neither less nor equal 2 Greater Than or Equal to Equal to
5. Concept 1 >Open gate . Open gate Statements: A - B>C>D<E Conclusions: I. B> D II. AsE
6. Standard Type When directions of operators are same Non-Standard Type When directions of operators are opposite
7. Concept 2
8. Statements N 2 O 2 P Q > R Conclusions I. N > R II.R N 1) If only conclusion I follows 2) If only conclusion II follows 3) If either conclusion I or conclusion II follows 4) If neither conclusion I or conclusion II follows 5) If both conclusion I and II follow | 536 | 1,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-35 | longest | en | 0.794442 |
https://www.bettersizeinstruments.com/angle-of-repose-angle-of-fall-angle-of-difference-.html | 1,627,771,318,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00442.warc.gz | 668,974,742 | 17,592 | Angle of Repose, Angle of Fall, Angle of Difference, Angle of Spatula - Bettersize Instruments Ltd.
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# Angle of Repose, Angle of Fall, Angle of Difference, Angle of Spatula
Angle of Repose: Under the static balance, the angle between the slope of a powder pile and the horizontal plane is Angle of Repose. It is measured when the powders fall to a surface via gravity and form a cone. It indicates the flowability of the powders. The smaller the Angle of Repose is, the better the flowability of the powders.
Angle of Fall: After measuring the Angle of Repose, apply an external force to the powder pile to collapse it. The angle between the slope of the collapsed pile and the horizontal plane is defined as Angle of Fall.
Angle of Difference: It means the difference between the Angle of Repose and the Angle of Fall. The larger the Angle of Difference is, the better flowability of the powders.
Angle of Spatula: immerse a plane in the powder pile; pull up the plane vertically, one angle is formed between the slope of the powders on the plane and the plane. Apply an external force to obtain another angle. The average of these two angles is Angle of Spatula. The smaller the Angle of Spatula is, the better the flowability of the powders. The Angle of Spatula is usually larger than the Angle of Repose. | 306 | 1,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.915047 |
http://www.jiskha.com/display.cgi?id=1274964430 | 1,496,010,406,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611569.86/warc/CC-MAIN-20170528220125-20170529000125-00496.warc.gz | 678,593,218 | 3,734 | # algebra 2
posted by on .
In a sample of 138 teenagers, 38 have never been to a live concert. Find the sample proportion for those who have never been to a live concert.
* 4%
* 28%
* 50%
* 72%
• algebra 2 - ,
38/138 * 100 = ? | 73 | 234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-22 | latest | en | 0.923973 |
https://web2.0calc.com/questions/determinant-of-a-matrix | 1,550,305,790,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479967.16/warc/CC-MAIN-20190216065107-20190216091107-00434.warc.gz | 740,507,621 | 6,544 | +0
# Determinant of a matrix
0
364
2
Hi. I need help at determinating the determinant of this matrix.
3 -5 -2 2
-4 7 4 4
4 -9 -3 7
2 -6 -3 2
I calculated and it gave me 0 but the result should be 27.
Oct 16, 2017
#1
+95866
+1
3 -5 -2 2
-4 7 4 4
4 -9 -3 7
2 -6 -3 2
We can use some row operations to minimize the work needed.....
Add the 2nd row to the 3rd row.....put the result in the 3rd row
3 -5 -2 2
-4 7 4 4
0 -2 1 11
2 -6 -3 2
Multiply the 4th row by 2.......add it to the second row
3 -5 -2 2
0 -5 -2 8
0 -2 1 11
2 -6 -3 2
Expand along the first row and first column
3 [ -5 -2 8
-2 1 11
-6 -3 2 ]
Rewrite the first two columns of the matrix
3 [ -5 -2 8 -5 -2
-2 1 11 -2 1
- 6 -3 2 -6 -3 ]
3 [ determinant of the marix] =
3 [ (-5* 1 *2 + -2 * 11 * -6 + 8 * -2 *-3) - ( -6*1*8 + -3*11*-5 + 2 *-2 * -2) ] = 135 (1)
-2 [-5 -2 2
-5 -2 8
-2 1 11
Rewrite the first two columns
-2 [ -5 -2 2 -5 -2
-5 -2 8 -5 -2
-2 1 11 -2 1 ]
-2 [ determinant of the matrix ]
-2 [ ( -5*-2*11 + -2*8*-2 + 2*-5*1) - (-2*-2*2 + 1*8*-5 + 11*-5*-2) ] = -108 (2)
Add (1) and (2) = 135 - 108 = 27
Oct 16, 2017
edited by CPhill Oct 17, 2017
#2
+21191
+1
Determinant of a matrix
3 -5 -2 2
-4 7 4 4
4 -9 -3 7
2 -6 -3 2
1. tridiagonal matrix, e. g. Gauß
$$\begin{pmatrix} 3 & -5 & -2 & 2 \\ 0 & \frac{1}{3} & 1\ \frac{1}{3} & 6\ \frac{2}{3} \\ 0 & 0 & 9 & 51 \\ 0 & 0 & 0 & 3 \\ \end{pmatrix}$$
2. Determinant: multiply the diagonally elements
The determinant of this matrix is $$3 \cdot \frac{1}{3}\cdot 9\cdot 3 = \mathbf{27 }$$
Oct 17, 2017 | 904 | 1,750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-09 | latest | en | 0.084372 |
https://www.math24.net/properties-applications-line-integrals/ | 1,611,031,847,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00322.warc.gz | 868,522,312 | 14,975 | # Properties and Applications of Line Integrals
Scalar functions: $$F\left( {x,y,z} \right),$$ $$F\left( {x,y} \right),$$ $$f\left( x \right)$$
Scalar potential: $$u\left( {x,y,z} \right)$$
Curves: $$C,$$ $${C_1},$$ $${C_2}$$
Limits of integration: $$a,$$ $$b,$$ $$\alpha,$$ $$\beta$$
Parameters: $$t$$, $$s$$
Polar coordinates: $$r$$, $$\theta$$
Vector field: $$\mathbf{F}\left( {P,Q,R} \right)$$
Position vector: $$\mathbf{r}\left( s \right)$$
Unit vectors: $$\mathbf{i},$$ $$\mathbf{j},$$ $$\mathbf{k},$$ $$\vec{\tau}$$
Area of a region: $$S$$
Length of a curve: $$L$$
Mass of a wire: $$m$$
Density: $$\rho\left( {x,y,z} \right),$$ $$\rho\left( {x,y} \right)$$
Coordinates of the center of mass: $$\bar x,$$ $$\bar y,$$ $$\bar z$$
First moments: $${M_{xy}},$$ $${M_{yz}},$$ $${M_{xz}}$$
Moments of inertia: $${I_x},$$ $${I_y},$$ $${I_z}$$
Volume of a solid: $$V$$
Work: $$W$$
Magnetic field: $$\mathbf{B}$$
Current: $$I$$
Electromotive force: $$\varepsilon$$
Magnetic flux: $$\psi$$
1. Line integral of a scalar function
Let a curve $$C$$ be given by the vector function $$\mathbf{r} = \mathbf{r}\left( s \right)$$, $$0 \le s \le S,$$ and a scalar function $$F$$ is defined over the curve $$C$$.
The line integral of the scalar function $$F$$ over the curve $$C$$ is written in the form
$${\large\int\limits_0^S\normalsize} {F\left( {\mathbf{r}\left( s \right)} \right)ds} =$$ $${\large\int\limits_C\normalsize} {F\left( {x,y,z} \right)ds} =$$ $${\large\int\limits_C\normalsize} {Fds},$$
where $$ds$$ is the arc length differential.
2. The line integral of a scalar function over a union of two curves is equal to the sum of the line integrals over each curve:
$${\large\int\limits_{{C_1} \cup {C_2}}\normalsize} {Fds} =$$ $${\large\int\limits_{{C_1}}\normalsize} {Fds} + {\large\int\limits_{{C_2}}\normalsize} {Fds}$$
3. If a smooth curve $$C$$ is parametrized by the equation $$\mathbf{r} = \mathbf{r}\left( t \right),$$ $$\alpha \le t \le \beta,$$ then the line integral of a scalar function is expressed by the formula
$${\large\int\limits_C\normalsize} {F\left( {x,y,z} \right)ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) }$$ $${ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt}$$
4. If $$C$$ is a smooth curve lying in the $$xy$$-plane and defined by the explicit equation $$y = f\left( x \right)$$, $$a \le x \le b$$, then the line integral is given by the expression
$${\large\int\limits_C\normalsize} {F\left( {x,y} \right)ds} =$$ $${\large\int\limits_a^b\normalsize} {F\left( {x,f\left( x \right)} \right) }$$ $${\sqrt {1 + {{\left( {f’\left( x \right)} \right)}^2}} dx}$$
5. Line integral of a scalar function in polar coordinates
$${\large\int\limits_C\normalsize} {F\left( {x,y} \right)ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {F\left( {r\cos \theta ,r\sin \theta } \right) }$$ $${\sqrt {{r^2} + {{\left( {{\large\frac{{dr}}{{d\theta }}}\normalsize} \right)}^2}} d\theta } ,$$
where the curve $$C$$ is defined by the polar function $$r\left( \theta \right).$$
6. Line integral of a vector field
Let a curve $$C$$ be defined by the vector function $$\mathbf{r} = \mathbf{r}\left( s \right),$$ $$0 \le s \le S.$$ The vector
$${\large\frac{{d\mathbf{r}}}{{ds}}\normalsize} = \vec{\tau} =$$ $$\left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right)$$
is the unit vector of the tangent line to this curve.
7. Let also a vector field $$\mathbf{F}\left( {P,Q,R} \right)$$ be defined over the curve $$C$$. Then the line integral of the vector function $$\mathbf{F}$$ along the curve $$C$$ is expressed in the form
$${\large\int\limits_C\normalsize} {Pdx + Qdy + Rdz} =$$ $${\large\int\limits_0^S\normalsize} {\big( {P\cos \alpha + Q\cos \beta }}$$ $$+\;{{ R\cos \gamma } \big)ds}$$
8. Properties of line integrals of vector fields
$${\large\int\limits_{ – C}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} =$$ $$– {\large\int\limits_C\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)},$$
where $$-C$$ denotes the curve with the opposite orientation.
$${\large\int\limits_C\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} =$$ $${\large\int\limits_{{C_1} \cup {C_2}}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} =$$ $${\large\int\limits_{C_1}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)}$$ $$+\;{\large\int\limits_{C_2}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ,$$
where $$C$$ is the union of the curves $${C_1}$$ and $${C_2}$$.
9. If the curve $$C$$ is parametrized by $$\mathbf{r}\left( t \right) =$$ $$\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),$$ $$\alpha \le t \le \beta,$$ then the line integral of the vector field is written as
$${\large\int\limits_C\normalsize} {Pdx + Qdy + Rdz} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\Big[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) {\large\frac{{dx}}{{dt}}\normalsize} }}$$ $${{+\; {Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) {\large\frac{{dy}}{{dt}}\normalsize}} }}$$ $$+\;{{R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dz}}{{dt}}\normalsize}} \Big]dt}$$
10. If the curve $$C$$ lies in the $$xy$$-plane and defined by the equation $$y = f\left( x \right),$$ $$a \le x \le b,$$ then the line integral of the vector field is given by
$${\large\int\limits_C\normalsize} {Pdx + Qdy} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\Big[ {P\left( {x,f\left( x \right)} \right) }}$$ $$+\;{{ Q\left( {x,f\left( x \right)} \right){\large\frac{{df}}{{dx}}\normalsize}} \Big] dx}$$
11. Green’s theorem
$${\large\iint\limits_C\normalsize} {\left( {{\large\frac{{\partial Q}}{{\partial x}}\normalsize} – {\large\frac{{\partial P}}{{\partial y}}\normalsize}} \right)dxdy} =$$ $${\large\oint\limits_C\normalsize} {Pdx + Qdy},$$
where $$\mathbf{F} =$$ $$P\left( {x,y} \right)\mathbf{i}$$ $$+\; Q\left( {x,y} \right)\mathbf{j}$$ is a continuous vector function with continuous first partial derivatives $$\partial P/\partial y,$$ $$\partial Q/\partial x$$ in a some domain $$R,$$ which is bounded by a closed, piecewise smooth curve $$C.$$
12. Area of a region $$R$$ bounded by a closed curve $$C$$
$$S = {\large\iint\limits_R\normalsize} {dxdy} =$$ $${\large\frac{1}{2}\oint\limits_C\normalsize} {xdy – ydx}$$
13. Path independence of line integrals
The line integral of a vector function $$\mathbf{F} = P\mathbf{i}$$ $$+\; Q\mathbf{j}$$ $$+\; R\mathbf{k}$$ is said to be path independent, if and only if $$P$$, $$Q$$ and $$R$$ are continuous in a domain $$D$$ and if there exists some scalar function $$u = u\left( {x,y,z} \right)$$ (a scalar potential) such that
$$\mathbf{F} = \text{grad }u$$ or $$\partial u/\partial x = P,$$ $$\partial u/\partial y = Q,$$ $$\partial u/\partial z = R.$$
Then the line integral is given by
$${\large\int\limits_C\normalsize} {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r}} =$$ $${\large\int\limits_C\normalsize} {Pdx + Qdy + Rdz} =$$ $$u\left( B \right) – u\left( A \right).$$
14. Test for a conservative field
A vector field of the form $$\mathbf{F} = \text{grad }u$$ is called a conservative field. The line integral of a vector function $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$$ $$+\; R\mathbf{k}$$ is path independent if and only if
$$\text{rot }\mathbf{F} =$$ $$\left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\large\frac{\partial }{{\partial x}}\normalsize} & {\large\frac{\partial }{{\partial y}}\normalsize} & {\large\frac{\partial }{{\partial z}}\normalsize}\\ P & Q & R \end{array}} \right|$$ $$= \mathbf{0}.$$
If the line integral is taken in the $$xy$$-plane, then the following formula for a conservative field is valid:
$${\large\int\limits_C\normalsize} {Pdx + Qdy} =$$ $$u\left( B \right) – u\left( A \right).$$
For the two-dimensional case, the test for a conservative field can be written in the form
$${\large\frac{{\partial P}}{{\partial y}}\normalsize} = {\large\frac{{\partial Q}}{{\partial x}}\normalsize}.$$
15. Length of a curve
$$L = {\large\int\limits_C\normalsize} {ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\left| {{\large\frac{{d\mathbf{r}}}{{dt}}\normalsize}\left( t \right)} \right|dt} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dz}}{{dt}}\normalsize} \right)}^2}} dt}$$
where $$C$$ is a piecewise smooth curve defined by the position vector $$\mathbf{r}\left( t \right),$$ $$\alpha \le t \le \beta.$$
If the curve $$C$$ is two-dimensional, then
$$L = {\large\int\limits_C\normalsize} {ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\left| {{\large\frac{{d\mathbf{r}}}{{dt}}\normalsize}\left( t \right)} \right|dt} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2}} dt} .$$
If the curve $$C$$ lies in the $$xy$$-plane and is described by the explicit function $$y = f\left( x \right)$$, $$a \le x \le b,$$ then its length is given by
$$L = {\large\int\limits_a^b\normalsize} {\sqrt {1 + {{\left( {\large\frac{{dy}}{{dx}}\normalsize} \right)}^2}} dx}.$$
16. Length of a curve in polar coordinates
$$L = {\large\int\limits_\alpha^\beta\normalsize} {\sqrt {{{\left( {\large\frac{{dr}}{{d\theta }}\normalsize} \right)}^2} + {r^2}} d\theta } ,$$
where the curve $$C$$ is determined by the equation $$r = r\left( \theta \right),$$ $$\alpha \le \theta \le \beta$$ in polar coordinates.
17. Mass of a wire
$$m = {\large\int\limits_C\normalsize} {\rho \left( {x,y,z} \right)ds} ,$$
where $${\rho \left( {x,y,z} \right)}$$ is the mass per unit length of the wire.
If the curve $$C$$ is parametrized by the vector function
$$\mathbf{r}\left( t \right) =$$ $$\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),$$ $$\alpha \le t \le \beta,$$
then its mass can be computed by the formula
$$m =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\rho \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) }$$ $${ \sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dz}}{{dt}}\normalsize} \right)}^2}} dt} .$$
If the curve $$C$$ lies in the $$xy$$-plane, then its mass is given by
$$m = {\large\int\limits_C\normalsize} {\rho \left( {x,y} \right)ds}$$
or
$$m =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\rho \left( {x\left( t \right),y\left( t \right)} \right) }$$ $${ \sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2}} dt}$$
(in parametric form)
18. Center of mass of a wire
$$\bar x = {\large\frac{{{M_{yz}}}}{m}\normalsize},\;$$ $$\bar y = {\large\frac{{{M_{xz}}}}{m}\normalsize},\;$$ $$\bar z = {\large\frac{{{M_{xy}}}}{m}\normalsize}$$, where
$${M_{yz}} = {\large\int\limits_C\normalsize} {x\rho \left( {x,y,z} \right)ds},\;$$
$${M_{xz}} = {\large\int\limits_C\normalsize} {y\rho \left( {x,y,z} \right)ds}.\;$$
$${M_{xy}} = {\large\int\limits_C\normalsize} {z\rho \left( {x,y,z} \right)ds} .$$
19. Moments of inertia
The moments of inertia of a curve about the $$x$$-axis, $$y$$-axis and $$z$$-axis are given by the formulas
$${I_x} = {\large\int\limits_C\normalsize} {\left( {{y^2} + {z^2}} \right) }$$ $${\rho \left( {x,y,z} \right)ds},\;$$
$${I_y} = {\large\int\limits_C\normalsize} {\left( {{x^2} + {z^2}} \right) }$$ $${\rho \left( {x,y,z} \right)ds},\;$$
$${I_z} = {\large\int\limits_C\normalsize} {\left( {{x^2} + {y^2}} \right) }$$ $${\rho \left( {x,y,z} \right)ds}.$$
20. Area of a region bounded by a closed curve
$$S = {\large\oint\limits_C\normalsize} {xdy} =$$ $$– {\large\oint\limits_C\normalsize} {ydx} =$$ $${\large\frac{1}{2}\oint\limits_C\normalsize} {xdy – ydx}$$
21. If the closed curve $$C$$ is given in parametric form $$\mathbf{r}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right),$$ $$\alpha \le t \le \beta ,$$ then the area can be calculated by the formula
$$S = {\large\int\limits_\alpha^\beta\normalsize} {x\left( t \right){\large\frac{{dy}}{{dt}}\normalsize} dt} =$$ $$– {\large\int\limits_\alpha^\beta\normalsize} {y\left( t \right){\large\frac{{dx}}{{dt}}\normalsize} dt} =$$ $${\large\frac{1}{2} \int\limits_\alpha^\beta\normalsize} {\left( {x\left( t \right){\large\frac{{dy}}{{dt}}\normalsize} – y\left( t \right){\large\frac{{dx}}{{dt}}\normalsize}} \right)dt} .$$
22. Volume of a solid formed by rotating a closed curve about the x-axis
$$V = – \pi {\large\oint\limits_C\normalsize} {{y^2}dx} =$$ $$– 2\pi {\large\oint\limits_C\normalsize} {xydy} =$$ $$– {\large\frac{\pi }{2}\oint\limits_C\normalsize} {2xydy + {y^2}dx}$$
23. Work of a field of forces
Work done by a force $$\mathbf{F}$$ on an object moving along a curve $$C$$ is described by the line integral
$$W = {\large\int\limits_C\normalsize} {\mathbf{F} \cdot d\mathbf{r}} ,$$
where $$d\mathbf{r}$$ is the unit tangent vector.
24. If the object is moved along a curve $$C$$ lying in the $$xy$$-plane, then the work is determined by the following formula
$$W = {\large\int\limits_C\normalsize} {\mathbf{F} \cdot d\mathbf{r}} =$$ $${\large\int\limits_C\normalsize} {Pdx + Qdy} .$$
If the path $$C$$ is specified by a parameter $$t$$ ($$t$$ often means time), then the formula for calculating work becomes
$$W = {\large\int\limits_\alpha^\beta\normalsize} {\Big[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dx}}{{dt}}\normalsize} }}$$ $$+\;{{ Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dy}}{{dt}}\normalsize} }}$$ $$+\;{{ R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dz}}{{dt}}\normalsize}} \Big]dt},$$
where $$t$$ goes from $$\alpha$$ to $$\beta.$$
If the vector field $$\mathbf{F}$$ is conservative and $$u\left( {x,y,z} \right)$$ is the scalar potential of this field, then the work on an object moving from $$A$$ to $$B$$ can be found by the formula
$$W = u\left( B \right) – u\left( A \right)$$.
25. Ampere’s law
$${\large\oint\limits_C\normalsize} {\mathbf{B} \cdot d\mathbf{r}} = {\mu _0}I$$
The line integral of a magnetic field $$\mathbf{B}$$ around a closed path $$C$$ is equal to the current $$I$$ (times the coefficient $${\mu _0}$$) flowing through the area bounded by the contour $$C.$$
$$\varepsilon = {\large\oint\limits_C\normalsize} {\mathbf{E} \cdot d\mathbf{r}} =$$ $$– {\large\frac{{d\psi }}{{dt}}\normalsize}$$
The electromotive force $$\varepsilon$$ induced around a closed loop $$C$$ is equal to the rate of change of the magnetic flux $$\psi$$ passing through the loop. | 5,633 | 14,894 | {"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 | 3 | CC-MAIN-2021-04 | latest | en | 0.543478 |
https://www.hackmath.net/en/math-problem/31081 | 1,604,088,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00636.warc.gz | 746,711,620 | 12,097 | # Ace or king
What is the probability that we will choose an ace or a king when choosing from a deck of sevens cards?
Correct result:
p = 0.25
#### Solution:
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https://www.cbc.ca/parents/play/view/fill-your-cup-a-simple-addition-game | 1,552,956,798,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201812.2/warc/CC-MAIN-20190318232014-20190319014014-00520.warc.gz | 725,462,522 | 21,134 | Ages:
2-5
## Activities
#### Sep 28, 2015
My daughter is really interested in numbers right now. She loves counting them, comparing them, ordering them and adding and subtracting them too!
When we do activities at home, I try to follow her lead as much as possible—that’s how this simple little addition game came about.
It's fun and easy to play, but the game also incorporates all kinds of important math concepts: counting, one-to-one correspondence, addition and comparison.
I also love that it doesn't require any special materials. In fact, my guess is that if you took a quick look around your home right now, you’d find everything you need!
Alright, so let’s get on with it. Here’s how to play Fill Your Cup!
## You Will Need:
• a pair of dice
• 2 identical clear cups or jars
• edible or non-edible counters (we chose to use grapes this time around, but berries, pretzel bites, pennies, small cube blocks or pompoms would work perfectly too. Use your imagination here!)
• 2 players
### How To Play:
1. Decide who will go first. We usually start with the youngest player in our family.
2. Player 1 rolls the dice.
3. Player 1 adds the sum of her dice (in this case, 4 + 2 = 6).
4. Player 1 adds the corresponding number of counters (in this case, grapes) to her cup and the turn is over.
You'll Also Love: 7 Ways To Work On Number Skills
5. Now it’s player 2’s turn! He rolls the dice, adds the numbers together and places the corresponding number of counters into his cup.
6. The players continue taking turns back and forth, comparing their cups every once and a while.
7. When both players have had an equal number of turns and one player fills his or her cup, the game is over and the player with the full cup wins!
In the case that both players fill their cups on the same turn, a count-off may be in order!
Of course, this game is meant to be fun and it really doesn’t matter who wins, but count-off situations allow for higher level counting and great number comparison opportunities.
Depending on what you’ve used as counters, your little one might notice that even though both cups look equally full, one player may have significantly more counters than the other. If so, take this chance to chat a little bit about size and capacity. There are so many playful learning opportunities here!
When the game is over, you can dump out your counters and play again. Of course, if you’ve used edible counters, you could move straight into snack time.
Enjoy!
###### Jen Kossowan
Jen is a teacher, blogger, and mama to a spirited little lady and a preemie baby boy. She's passionate about play, loves a good DIY project, adores travelling, and can often be found in the kitchen creating recipes that meet her crunchy mama criteria. You can follow Jen on her blog, Mama.Papa.Bubba, and on Facebook, Twitter, Pinterest, and Instagram. | 665 | 2,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-13 | longest | en | 0.953959 |
https://www.askiitians.com/forums/Trigonometry/tan-square-pi-by-16-tan-square-2pi-by-16-tan-squa_188776.htm | 1,695,459,840,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00803.warc.gz | 739,310,388 | 42,731 | # tan square pi by 16+ tan square 2pi by 16+tan square 3pi by 16+tan square 4pi by 16+ tan square 5pi by 16+ tan square 6pi by 16+tan square 7pi by 16=35
Arun
25757 Points
5 years ago
we use formula tan^2a=(1-cos2a)/(1+cos2a)
tan^(pi/16)+tan^2(7pi/16)+tan^(2pi/16)+tan^2(6pi/16)+tan^2(3pi/16)+tan^2(5pi/16)+tan^(4pi/16)
tan^(pi/16)+tan^2(pi/2-pi/16)+tan^2(2pi/16)+tan^2(pi/2-2pi/16)+tan^2(3pi/16)+tan^2(pi/2-3pi/16)+tan^2(pi/4)
tan^2(pi/16)+cot^2(pi/16)+tan^2(2pi/16)+cot^2(2pi/16)+tan^2(3pi/16)+cot^2(3pi/16)+1
we use formula
tan^2a+cot^2a=(1-cos2a)/(1+cos2a)+(1+cos2a)/(1-cos2a)
=2(1+cos^2(2a))/sin^2(2a))
2(1+cos^2(pi/8))/sin^2(pi/8) +2(1+cos^2(pi/4))/sin^2(pi/4) +2(1+cos^2(3pi/8))/sin^2(3pi/8) +1
2(cosec^2(pi/8)+cot^2(pi/8)+cosec^2(3pi/8)+cot^2(3pi/8)+2(1+1/2)/(1/2) +1
2(1+2cot^2(pi/8)+1+2cot^2(3pi/8))+6+1
4+6+1+4(cot^2(pi/8)+cot^2(3pi/8))
11+4{(1+cospi/4)/(1-cos(pi/4)) + (1+cos(3pi/4))/(1-cos(3pi/4)}
11+4{(1+1/sqrt2)/(1-1/sqrt2) +(1-1/sqrt2)/(1+1/sqrt2)}
11+4{2*3/2)/(1/2)
11+4(6)
11+24
35 | 590 | 1,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.297873 |
https://www.coursehero.com/file/5659853/Problem321-solution/ | 1,498,220,209,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320057.96/warc/CC-MAIN-20170623114917-20170623134917-00061.warc.gz | 837,376,300 | 37,471 | Problem3.21_solution
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## This note was uploaded on 11/08/2009 for the course ECE 304 taught by Professor Ma during the Spring '08 term at Arizona.
### Page1 / 7
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Ask a homework question - tutors are online | 938 | 2,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.509337 |
https://solvedlib.com/new-tab-the-comparative-balance-sheet-of-yellow,125779 | 1,713,745,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00783.warc.gz | 472,123,032 | 18,012 | # New Tab The comparative balance sheet of Yellow Dog Enterprises Inc. at December 31, 2018 and...
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Fill in the blanks USING the terms in the word bank ___________ have no symmetry or germ layers, no nervous system, and are completely aquatic. They have ___________________ for feeding. _______________ have two germ layers, a ______________ with no brain and in which messages can travel in both dir...
Demand for lift tickets at a popular ski resort is given by Q 2,500 -0.25p+ 2Palt 4Plodging+0.005Y where p price of a lift ticket aprice of a tropical vacation package $600 Q -quantity demanded Plodging price of ski resort lodging$200 Y consumer income $30,000 The quantity demanded as a function of... 5 answers ##### Imagine a Ferris wheel that rotates four times each minute and has diameter of 18.4m What is the centripetal acceleration of a rider? Submit Answer You have entered that answer Previous Incorrect_ Tries 1/10 before Iries What force does the seat exert on a 46. .7kg rider at the lowest point of the ride? 608 NSubmit AnswerIncorrect_ Tries 1/10 Previous TriesWhat force does the seat exert on the rider at the highest point of the ride? 3.06x10*2 NSubmit AnswerIncorrect __ Tries 1/10 Previous TriesW Imagine a Ferris wheel that rotates four times each minute and has diameter of 18.4m What is the centripetal acceleration of a rider? Submit Answer You have entered that answer Previous Incorrect_ Tries 1/10 before Iries What force does the seat exert on a 46. .7kg rider at the lowest point of the r... 1 answer ##### Question 1 On the first day of its fiscal year, Chin Company issued$26,500,000 of five-year,...
Question 1 On the first day of its fiscal year, Chin Company issued \$26,500,000 of five-year, 9% bonds to finance its operations of producing and selling home improvement products. Interest is payable semiannually. The bonds were issued at a market (effective) interest rate of 11%, resulting in Chin...
##### (2) &lbi Q5) Given the following joint density function, find P(2<Y<3.1 X-1.2)IXjy 8 00 < x < 2,2 < y < 4 otherwisef(x,y)0.843750.6520.573750.673750.76375
(2) &lbi Q5) Given the following joint density function, find P(2<Y<3.1 X-1.2) IXjy 8 0 0 < x < 2,2 < y < 4 otherwise f(x,y) 0.84375 0.652 0.57375 0.67375 0.76375...
##### Example: Find the forces required to hold the curved gate steady. Horizontal Force: vertical projection of...
Example: Find the forces required to hold the curved gate steady. Horizontal Force: vertical projection of the curved surface, find the force acting on that projection. Vertical Force: weight of water between the gate and the water-line Gate is 2 ft wide Atmospheric pressure Metal surface Vertical P...
##### 27 . V1 -y dx- V1 -x dy = 0, V3 y(0) = 2
27 . V1 -y dx- V1 -x dy = 0, V3 y(0) = 2...
##### In the calculation of the CPI, tea is given greater weight than beer if O a....
In the calculation of the CPI, tea is given greater weight than beer if O a. it costs more to produce tea than it costs to produce beer b. consumers buy more tea than beer c, the price oftea is higher than the price of beer. e d tea is more readily available than beer to the typical consumer here to...
##### Find the general solution to the differential equationy"' _ 2y' + y = 1+3(where y d
Find the general solution to the differential equation y"' _ 2y' + y = 1+3 (where y d... | 3,337 | 11,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.902355 |
https://www.esaral.com/q/if-the-sum-of-7-terms-of-an-a-p-is-49-and-that-of-17-terms-is-289-66385 | 1,721,472,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00145.warc.gz | 647,321,606 | 12,324 | # If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289,
Question:
If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
Solution:
In the given problem, we need to find the sum of n terms of an A.P. Let us take the first term as a and the common difference as d.
Here, we are given that,
$S_{7}=49$..............(1)
$S_{17}=289$.............(2)
So, as we know the formula for the sum of n terms of an A.P. is given by,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Where; a = first term for the given A.P.
d = common difference of the given A.P.
= number of terms
So, using the formula for n = 7, we get,
$S_{7}=\frac{7}{2}[2(a)+(7-1)(d)]$
$49=\left(\frac{7}{2}\right)[2 a+(6)(d)]$ (Using 1)
$49=\frac{14 a+42 d}{2}$
$49=7 a+21 d$
Further simplifying for a, we get,
$a=\frac{49-21 d}{7}$
$a=7-3 d$ ...............(3)
Also using the formula for $n=17$ we get
$S_{17}=\frac{17}{2}[2(a)+(17-1)(d)]$
$289=\left(\frac{17}{2}\right)[2 a+(16)(d)]$ (Using 2)
$289=\frac{(17)(2) a+(17)(16) d}{2}$
$289=17 a+136 d$
Further simplifying for a, we get,
$a=\frac{289-136 d}{17}$
$a=17-8 d$ ...........(4)
Subtracting (3) from (4), we get,
$a-a=(17-8 d)-(7-3 d)$
$0=17-8 d-7+3 d$
$0=10-5 d$
$5 d=10$
$d=2$
Now, to find a, we substitute the value of d in (3),
$a=7-3(2)$
$a=7-6$
$a=1$
Now, using the formula for the sum of n terms of an A.P., we get,
$S_{n}=\frac{n}{2}[2(1)+(n-1)(2)]$
$=\frac{n}{2}[2+2 n-2]$
$=\left(\frac{n}{2}\right)(2 n)$
$=n^{2}$
Therefore, the sum of first $n$ terms for the given A.P. is $S_{n}=n^{2}$. | 650 | 1,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.499621 |
https://essaycomrade.com/2022/04/21/determine-the-future-periodic-spot-interest-rates-at-one-period-from-now-on-a-onestep-ho-lee-model-using-current-periodic-spot-interest-rate-of-a1-percent-current-spot-rate-over-two-periods-of-a2/ | 1,708,541,056,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00333.warc.gz | 256,786,123 | 20,213 | # Determine the future periodic spot interest rates, at one period from now, on a onestep Ho-Lee model, using current periodic spot interest rate of A1 percent, current spot rate over two periods of A2 percent, and an annual interest rate volatility of .00
A. Determine the future periodic spot interest rates, at one period from now, on a onestep Ho-Lee model, using current periodic spot interest rate of A1 percent, current spot
rate over two periods of A2 percent, and an annual interest rate volatility of .005 (.5%).
Then consider the current spot rate over three periods of A3 percent to find the future
periodic interest rates in a two-step Ho-Lee model. The period is A4 months. Find
the following.
1. The trend (lambda1) in interest rate in the first period.
2. The trend (lambda2) in interest rate in the second period.
3. The future periodic rate on the upside at the end of one period.
4. The future periodic rate on the downside at the end of one period.
5. The future periodic rate at the up-most branch at the end of two periods.
6. The future periodic rate at the middle branch at the end of two periods.
7. The future periodic rate at the down-most branch at the end of two periods.
8. The future periodic rate at the down-most branch at the end of two periods
B. The cash futures price of a 3-month zero coupon bond with a face value of \$100 for
delivery in B1 months from now is B2 dollars. Suppose that the current spot interest
rate for a term of B3 months is B4 per cent per annum. Assume continuous
compounding to answer the following:
9. The forward rate of interest for period B1 months to B1+3 months.
10. The spot rate of interest for period 0 to B1 months.
11. The current fair value of a B1-month zero coupon bond with \$100 face.
C. Consider C1-month spot interest rates evolving in the following two-step binomial
tree over C2 months, i.e., with C1 months in each of the next two steps. The current
C2-month spot interest rate is 5.15% and the C3-month spot interest rate is 5.3%. Find
the following by assuming monthly compounding.
12. The risk neutral probability for the up move in first step.
13. The risk neutral probability for the up move in second step.
14. The current fair value of a C1-month European call option with a strike price of
\$974 written on a C2-month zero coupon bond with face value \$1000.
15. The current fair value of a C2-month European put option with a strike price of
\$994 written on a C3-month zero coupon bond with face value \$1000.
D. Consider D1-month spot interest rates evolving in the following three-step binomial
tree over D4 months, i.e., with D1 months in each of the next three steps. The current
D3-month spot interest rate is 5.15%, the current D2-month spot interest rate is 5.3%,
and the current D4-month spot interest rate is 5.45%. This three-step problem involves
calculation of just the third step, because it is an extension of the two-step Problem C
5.30%
4.30%
5.10%
6.20%
4 50 5.10%
4.00%
3
and, so, you should simply take the risk-neutral probabilities obtained in Problem C to
answer the following by assuming monthly compounding.
16. The risk neutral probability for the up move in the last (third) step.
17. The current fair value of a D3-month European call option with a strike price
of \$974 written on a D4-month zero coupon bond with face value \$1000.
18. The current fair value of a D2-month European put option with a strike price of
\$994 written on a D4-month zero coupon bond with face value \$1000.
5
5
E. Suppose that you are long in an original portfolio of 20-year bonds with a total
face value of E1 million dollars and that you want to hedge the portfolio value against
fluctuating interest rates by trading on a 10-year bond and a 30-year bond. You have
estimated (given): DV0110 = E2, DV0120 = E3, and DV0130 = E4 per \$100 face value
of 10-year, 20-year and 30-year bonds, respectively. You have also used yield data on
these bonds to estimate the coefficients, b=1.3 and c = 1.6, in the following regression:
20 10 30 . t t tt ∆ = +∆ +∆ + y a by cy ε
Using the given data, determine your optimal trading strategy for the 10-year and 30-
year bonds to hedge 20-year bonds. Report answers for the following:
19. Face value in dollars of 10 year bonds needed for hedging.
20. Face value in dollars of 30 year bonds needed for hedging.
5.1%
5.3%
4.3%
5.1%
5.3%
4.3%
3.7%
4.0%
7.1%
6.2%
Page 4 of 6 4
After hedging, suppose that you observe that the 20-year yield moves by E5 basis
points when the 10-year yield moves by 1 basis point and that the 20-year yield moves
by E6 basis points when the 30-year yield moves 1 basis point. Then:
21. How much will be your profit or loss corresponding to 30-year yield increasing
by 1 basis point?
F. A collared floater is like a variable rate bond, but with an upper limit and a lower
limit on the rate of coupon payment. Consider, for example, annual rates coupon
payment and one-year spot interest rates. In the following table, the current one-year
spot interest rate is denoted by r in column (1). Depending on the current spot interest
rate, the collared floater with a face value of F1 dollars will pay the holder of this
floater a sum a year later, as given in column (2) of the table. The binomial tree below
gives the evolution of one-year spot interest rates over a four year period.
22. What is the current fair value of the four-year collared floater if the risk-neutral
probabilities in the binomial tree are 0.5 and 0.5 for the up and down moves,
respectively?
Current Rate, r
(1)
Collared Floater with
face value F1 pays at
the end of the year
(2)
r < 5% F2
5% ≤ r ≤ 8% F1 x r%
r > 8% F3
4.5%
6.2%
5.2%
6.00%
9.50%
6.25%
4.75%
5.50%
10.50%
8.75%
Page 5 of 6 5
G. A four-year participating cap pays an interest based on a nominal principal of G1,
at the end of the year if the current annual spot interest rate, r, is higher than 5%. If
the current annual spot rate, r, is less than 5%, the participating cap obligates its holder
to lose 30% of (5%-r%) on the nominal principal at the end of the year. The payments
are given in column (2) of the following table, given the rates in column (1). The
binomial tree below gives the evolution of one-year spot interest rates over a four year
period.
23. What is the current fair value of the participating cap if the risk-neutral
probabilities in the binomial tree are 0.5 and 0.5 for up and down moves,
respectively?
Current rate, r
(1)
Buyer gets at end of year
(2)
r > 5% (r% – 5%)xG1
r ≤ 5% -.30(5% – r%)xG1
H. What is the risk premium (spread) of a H1 percent annual coupon, 3.35 year
maturity, and 100-face value bond? The bond pays the annual coupon in two equal
4.0%
5.3%
3.7%
4.8%
6.2%
4.3%
3.1%
3.3%
8.7%
6.8%
Page 6 of 6 6
installments and trades now at a quoted spot price of H2. Assume continuous
compounding and use the term structure of discount factors for riskless bonds, d(t) =
1 – .03t – .0075t2 + .0015t3
. Report answer for the following (Hint: rate at which to
discount future cash flows of a risky bond to find the current value is (as per CAPM)
the riskless rate which changes with the term of the cash flow plus risk premium which
is fixed across all terms; the easier way to solve for risk spread by the method of trial
and error, as done an equation solver):
24. Risk Spread (premium) of the bond
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 2,310 | 8,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-10 | latest | en | 0.90262 |
https://www.rdocumentation.org/packages/spatstat/versions/1.49-0/topics/kppm | 1,591,019,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347417746.33/warc/CC-MAIN-20200601113849-20200601143849-00212.warc.gz | 852,987,063 | 9,551 | # kppm
0th
Percentile
##### Fit Cluster or Cox Point Process Model
Fit a homogeneous or inhomogeneous cluster process or Cox point process model to a point pattern.
Keywords
models, spatial
##### Usage
kppm(X, …) # S3 method for formula
kppm(X,
clusters = c("Thomas","MatClust","Cauchy","VarGamma","LGCP"),
…,
data=NULL) # S3 method for ppp
kppm(X,
trend = ~1,
clusters = c("Thomas","MatClust","Cauchy","VarGamma","LGCP"),
data = NULL,
...,
covariates=data,
method = c("mincon", "clik2", "palm"),
improve.type = c("none", "clik1", "wclik1", "quasi"),
improve.args = list(),
weightfun=NULL,
control=list(),
statistic="K",
statargs=list(),
rmax = NULL,
covfunargs=NULL,
use.gam=FALSE,
nd=NULL, eps=NULL)# S3 method for quad
kppm(X,
trend = ~1,
clusters = c("Thomas","MatClust","Cauchy","VarGamma","LGCP"),
data = NULL,
...,
covariates=data,
method = c("mincon", "clik2", "palm"),
improve.type = c("none", "clik1", "wclik1", "quasi"),
improve.args = list(),
weightfun=NULL,
control=list(),
statistic="K",
statargs=list(),
rmax = NULL,
covfunargs=NULL,
use.gam=FALSE,
nd=NULL, eps=NULL)
##### Arguments
X
A point pattern dataset (object of class "ppp" or "quad") to which the model should be fitted, or a formula in the R language defining the model. See Details.
trend
An R formula, with no left hand side, specifying the form of the log intensity.
clusters
Character string determining the cluster model. Partially matched. Options are "Thomas", "MatClust", "Cauchy", "VarGamma" and "LGCP".
data,covariates
The values of spatial covariates (other than the Cartesian coordinates) required by the model. A named list of pixel images, functions, windows, tessellations or numeric constants.
Additional arguments. See Details.
method
The fitting method. Either "mincon" for minimum contrast, "clik2" for second order composite likelihood, or "palm" for Palm likelihood. Partially matched.
improve.type
Method for updating the initial estimate of the trend. Initially the trend is estimated as if the process is an inhomogeneous Poisson process. The default, improve.type = "none", is to use this initial estimate. Otherwise, the trend estimate is updated by improve.kppm, using information about the pair correlation function. Options are "clik1" (first order composite likelihood, essentially equivalent to "none"), "wclik1" (weighted first order composite likelihood) and "quasi" (quasi likelihood).
improve.args
Additional arguments passed to improve.kppm when improve.type != "none". See Details.
weightfun
Optional weighting function $w$ in the composite likelihood or Palm likelihood. A function in the R language. See Details.
control
List of control parameters passed to the optimization function optim.
algorithm
Character string determining the mathematical optimisation algorithm to be used by optim. See the argument method of optim.
statistic
Name of the summary statistic to be used for minimum contrast estimation: either "K" or "pcf".
statargs
Optional list of arguments to be used when calculating the statistic. See Details.
rmax
Maximum value of interpoint distance to use in the composite likelihood.
covfunargs,use.gam,nd,eps
Arguments passed to ppm when fitting the intensity.
##### Details
This function fits a clustered point process model to the point pattern dataset X.
The model may be either a Neyman-Scott cluster process or another Cox process. The type of model is determined by the argument clusters. Currently the options are clusters="Thomas" for the Thomas process, clusters="MatClust" for the Matern cluster process, clusters="Cauchy" for the Neyman-Scott cluster process with Cauchy kernel, clusters="VarGamma" for the Neyman-Scott cluster process with Variance Gamma kernel (requires an additional argument nu to be passed through the dots; see rVarGamma for details), and clusters="LGCP" for the log-Gaussian Cox process (may require additional arguments passed through …; see rLGCP for details on argument names). The first four models are Neyman-Scott cluster processes.
The algorithm first estimates the intensity function of the point process using ppm. The argument X may be a point pattern (object of class "ppp") or a quadrature scheme (object of class "quad"). The intensity is specified by the trend argument. If the trend formula is ~1 (the default) then the model is homogeneous. The algorithm begins by estimating the intensity as the number of points divided by the area of the window. Otherwise, the model is inhomogeneous. The algorithm begins by fitting a Poisson process with log intensity of the form specified by the formula trend. (See ppm for further explanation).
The argument X may also be a formula in the R language. The right hand side of the formula gives the trend as described above. The left hand side of the formula gives the point pattern dataset to which the model should be fitted.
If improve.type="none" this is the final estimate of the intensity. Otherwise, the intensity estimate is updated, as explained in improve.kppm. Additional arguments to improve.kppm are passed as a named list in improve.args.
The clustering parameters of the model are then fitted either by minimum contrast estimation, or by maximum composite likelihood.
Minimum contrast:
If method = "mincon" (the default) clustering parameters of the model will be fitted by minimum contrast estimation, that is, by matching the theoretical $K$-function of the model to the empirical $K$-function of the data, as explained in mincontrast.
For a homogeneous model ( trend = ~1 ) the empirical $K$-function of the data is computed using Kest, and the parameters of the cluster model are estimated by the method of minimum contrast.
For an inhomogeneous model, the inhomogeneous $K$ function is estimated by Kinhom using the fitted intensity. Then the parameters of the cluster model are estimated by the method of minimum contrast using the inhomogeneous $K$ function. This two-step estimation procedure is due to Waagepetersen (2007).
If statistic="pcf" then instead of using the $K$-function, the algorithm will use the pair correlation function pcf for homogeneous models and the inhomogeneous pair correlation function pcfinhom for inhomogeneous models. In this case, the smoothing parameters of the pair correlation can be controlled using the argument statargs, as shown in the Examples.
Additional arguments … will be passed to mincontrast to control the minimum contrast fitting algorithm.
Composite likelihood:
If method = "clik2" the clustering parameters of the model will be fitted by maximising the second-order composite likelihood (Guan, 2006). The log composite likelihood is $$\sum_{i,j} w(d_{ij}) \log\rho(d_{ij}; \theta) - \left( \sum_{i,j} w(d_{ij}) \right) \log \int_D \int_D w(\|u-v\|) \rho(\|u-v\|; \theta)\, du\, dv$$ where the sums are taken over all pairs of data points $x_i, x_j$ separated by a distance $d_{ij} = \| x_i - x_j\|$ less than rmax, and the double integral is taken over all pairs of locations $u,v$ in the spatial window of the data. Here $\rho(d;\theta)$ is the pair correlation function of the model with cluster parameters $\theta$.
The function $w$ in the composite likelihood is a weighting function and may be chosen arbitrarily. It is specified by the argument weightfun. If this is missing or NULL then the default is a threshold weight function, $w(d) = 1(d \le R)$, where $R$ is rmax/2.
Palm likelihood:
If method = "palm" the clustering parameters of the model will be fitted by maximising the Palm loglikelihood (Tanaka et al, 2008) $$\sum_{i,j} w(x_i, x_j) \log \lambda_P(x_j \mid x_i; \theta) - \int_D w(x_i, u) \lambda_P(u \mid x_i; \theta) {\rm d} u$$ with the same notation as above. Here $\lambda_P(u|v;\theta$ is the Palm intensity of the model at location $u$ given there is a point at $v$.
In all three methods, the optimisation is performed by the generic optimisation algorithm optim. The behaviour of this algorithm can be modified using the argument control. Useful control arguments include trace, maxit and abstol (documented in the help for optim).
Fitting the LGCP model requires the RandomFields package except in the default case where the exponential covariance is assumed.
##### Value
An object of class "kppm" representing the fitted model. There are methods for printing, plotting, predicting, simulating and updating objects of this class.
##### Error and warning messages
See ppm.ppp for a list of common error messages and warnings originating from the first stage of model-fitting.
##### References
Guan, Y. (2006) A composite likelihood approach in fitting spatial point process models. Journal of the American Statistical Association 101, 1502--1512.
Jalilian, A., Guan, Y. and Waagepetersen, R. (2012) Decomposition of variance for spatial Cox processes. Scandinavian Journal of Statistics, in press.
Tanaka, U. and Ogata, Y. and Stoyan, D. (2008) Parameter estimation and model selection for Neyman-Scott point processes. Biometrical Journal 50, 43--57.
Waagepetersen, R. (2007) An estimating function approach to inference for inhomogeneous Neyman-Scott processes. Biometrics 63, 252--258.
Methods for kppm objects: plot.kppm, fitted.kppm, predict.kppm, simulate.kppm, update.kppm, vcov.kppm, methods.kppm, as.ppm.kppm, Kmodel.kppm, pcfmodel.kppm.
Minimum contrast fitting algorithm: mincontrast.
Alternative fitting algorithms: thomas.estK, matclust.estK, lgcp.estK, cauchy.estK, vargamma.estK, thomas.estpcf, matclust.estpcf, lgcp.estpcf, cauchy.estpcf, vargamma.estpcf,
Summary statistics: Kest, Kinhom, pcf, pcfinhom.
See also ppm
• kppm
• kppm.formula
• kppm.ppp
##### Examples
# NOT RUN {
# method for point patterns
kppm(redwood, ~1, "Thomas")
# method for formulas
kppm(redwood ~ 1, "Thomas")
kppm(redwood ~ 1, "Thomas", method="c")
kppm(redwood ~ 1, "Thomas", method="p")
kppm(redwood ~ x, "MatClust")
kppm(redwood ~ x, "MatClust", statistic="pcf", statargs=list(stoyan=0.2))
kppm(redwood ~ x, cluster="Cauchy", statistic="K")
kppm(redwood, cluster="VarGamma", nu = 0.5, statistic="pcf")
# LGCP models
kppm(redwood ~ 1, "LGCP", statistic="pcf")
if(require("RandomFields")) {
kppm(redwood ~ x, "LGCP", statistic="pcf",
model="matern", nu=0.3,
control=list(maxit=10))
}
# fit with composite likelihood method
kppm(redwood ~ x, "VarGamma", method="clik2", nu.ker=-3/8)
# fit intensity with quasi-likelihood method
kppm(redwood ~ x, "Thomas", improve.type = "quasi")
# }
Documentation reproduced from package spatstat, version 1.49-0, License: GPL (>= 2)
### Community examples
Looks like there are no examples yet. | 2,685 | 10,617 | {"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} | 2.546875 | 3 | CC-MAIN-2020-24 | latest | en | 0.73588 |
https://bullforyou.com/Sourcecode/Indicators/153735905713504.html | 1,624,229,818,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00172.warc.gz | 149,379,353 | 5,900 | # Michelangelo.mq4
```#property link "http://www.forex-instruments.info"
//+------------------------------------------------------------------+
//| Michelangelo.mq4 |
//| |
//| |
//| Algorithm: Apply a H/L indicator like SMI, avg'd w/ power law |
//| lengths. Then apply a Kaufman AMA filter. |
//| Then a 'signal' EMA. Use crossover of this as |
//| |
//| Idea: Try to stay on the side of a trend but reverse |
//| quickly if there is a breakout. KaufmanAMA can |
//| be made sensitive to those. |
//| |
//+------------------------------------------------------------------+
#property indicator_separate_window
#property indicator_buffers 2
#property indicator_color1 White
#property indicator_color2 Red
#property indicator_level1 0
//---- input parameters
//
// Try these on M30 charts on trendy currencies.
// THESE PARAMETERS ARE NOT OPTIMIZED BY ANY MEANS.
//
// Brief run-down. Structure is derived from "SMI" indicator.
//
// The first part, minkernel, maxkernel, and exponent
// correspond to the power-law averaging of relative position of self to
// highs and lows. The underlying statistic is sort of like a "stochastic",
// the purpose of the averaging is to not be as dependent on a single, fixed lookback.
//
// The relative position and range series (kept separate here) are each subjected
// to a Kaufman adaptive moving average. This AMA computes an internal 'signal to noise'
// ratio to see if it is choppy (no consistent trend), in which case the smoothing is strong
// and laggy, or if it feels like a continuing trend, in which case the smoothing is light
// and fast. Parameters here are "periodAMA", which is the lookback for S/N, nfast, and nslow
// which control the range between fastest and slowest smoothing, and "G". This is an exponent
// which, for larger values than '1', more greatly emphasize the high S/N versus low. In practice,
// this means that for larger 'G', there are more flat periods, and then more sensitive to breakouts.
//
// After the KaufmanAMA filtering, the two series are
// "predictively EMA filtered" (similar to a Hull MA), with parameter Period_R,
// and then divided to form the main indicator line in white.
//
// Finally, this indicator line is filtered with a conventional EMA with period 'Signal'
// to give the red signal line. Trading signals are generally a crossover of white
// with red, with the slope of the white in the proper direction. This will probably
// require intra-bar consideration for breakouts when used in real-time trading.
//
// Best nutshell description is "a bastardized sort of trend-following stochastic",
// or otherwise "WTF?". But it does occasionally seem to show some nice signals
// on trendy currencies. Probably not good on choppy USD/CAD or highly reversing crosses.
//
// PLEASE EXPERIMENT WITH PARAMETERS HEAVILY.
// There is nothing sacred with these.
// They have quite distinct effects depending on the setting, timescale and their values.
extern int minkernel=2;
extern int maxkernel=80;
extern double Exponent=1.0;
int KernelLength;
double kernel[];
double working[];
extern int Period_R=3;
extern int periodAMA=12;
extern int nfast=6;
extern int nslow=60;
extern double G=2.5;
extern int Signal=5;
//extern int SignalShift=0;
//---- buffers
double SMI_Buffer[];
double Signal_Buffer[];
double SM_Buffer[];
double EMA_SM[];
double EMA2_SM[];
double EMA_HQ[];
double EMA2_HQ[];
double HQ_Buffer[];
//+------------------------------------------------------------------+
//| Custom indicator initialization function |
//+------------------------------------------------------------------+
int init()
{
//---- indicators
IndicatorBuffers(8);
SetIndexStyle(0,DRAW_LINE);
SetIndexBuffer(0,SMI_Buffer);
SetIndexStyle(1,DRAW_LINE);
SetIndexBuffer(1,Signal_Buffer);
SetIndexLabel(0,"Michelangelo");
SetIndexLabel(1,"Signal Michelangelo");
SetIndexBuffer(2,SM_Buffer);
SetIndexBuffer(3,EMA_SM);
SetIndexBuffer(4,EMA2_SM);
SetIndexBuffer(5,EMA_HQ);
SetIndexBuffer(6,EMA2_HQ);
SetIndexBuffer(7,HQ_Buffer);
IndicatorShortName("Michelangelo(PL["+minkernel+","+maxkernel+"],"+periodAMA+","+nfast+","+nslow+","+G+","+Signal+")");
//----
KernelLength= maxkernel+1;
initialize_kernel(minkernel,maxkernel,KernelLength,Exponent);
ArrayResize(working,KernelLength);
return(0);
}
//+------------------------------------------------------------------+
//| Custor indicator deinitialization function |
//+------------------------------------------------------------------+
int deinit()
{
//----
return(0);
}
//+------------------------------------------------------------------+
//| Custom indicator iteration function |
//+------------------------------------------------------------------+
int start()
{
int counted_bars=IndicatorCounted();
int limit;
int i;
if(counted_bars<0) return(-1);
if(counted_bars>0) counted_bars--;
limit=Bars-maxkernel-counted_bars;
if(counted_bars>0) counted_bars--;
for (i=limit;i>=0;i--)
{
for (int j=minkernel; j<=maxkernel; j++) {
double H = High[Highest(NULL,0,MODE_HIGH,j,i)];
double L = Low[Lowest(NULL,0,MODE_LOW,j,i)];
working[j] = (H-L);
}
HQ_Buffer[i] = convolve(working,kernel,minkernel,maxkernel);
for (j=minkernel; j<=maxkernel; j++) {
H = High[Highest(NULL,0,MODE_HIGH,j,i)];
L = Low[Lowest(NULL,0,MODE_LOW,j,i)];
double C= Close[i];
if (C < L) C = L;
if (C > H) C = H;
working[j] = C - (H+L)/2.0;
}
SM_Buffer[i] = convolve(working,kernel,minkernel,maxkernel);
}
KaufmanOnArray(limit, SM_Buffer, EMA_SM, periodAMA, nfast, nslow, G);
KaufmanOnArray(limit, HQ_Buffer, EMA_HQ, periodAMA, nfast, nslow, G);
EMAPredictiveSmoothOnArray(limit, Period_R, Period_R, EMA_SM, EMA2_SM);
EMAPredictiveSmoothOnArray(limit, Period_R, Period_R, EMA_HQ, EMA2_HQ);
for (i=limit-1;i>=0;i--)
{
double val = 100*EMA2_SM[i]/0.5/EMA2_HQ[i];
if (val > 100.0) val = 100.0;
if (val < -100.0) val = -100.0;
SMI_Buffer[i]= val;
}
EMAOnArray(limit,2.0/(Signal+1.0),SMI_Buffer,Signal_Buffer);
for (i=limit-1; i>= 0; i--) {
val = Signal_Buffer[i];
if (val > 100.0) val = 100.0;
if (val < -100.0) val = -100.0;
Signal_Buffer[i] = val;
}
return(0);
}
//+------------------------------------------------------------------+
void KaufmanOnArray(int N, double input[], double& output[], int periodAMA, int nfast, int nslow, double G) {
// perform a Kaufman moving average on input[], saving to output[]
double slowSC=(2.0 /(nslow+1));
double fastSC=(2.0 /(nfast+1));
int i;
double AMA0, AMA, signal, noise, ER, dSC,ERSC,wlxSSC;
// double noise,noise0,AMA,AMA0,signal,ER;
int nmax = N - periodAMA-1;
AMA0 = input[nmax+1];
for (i=nmax; i >= 0; i--) {
// loop down
signal=MathAbs(input[i]-input[i+periodAMA]);
noise=0;
for(int j=0;j nmax;i--) {
output[i] = input[i];
}
}
void EMAPredictiveSmoothOnArray(int N, double L, double Lfinal, double input[], double& output[]) {
//
// This "predictive/smoothed" EMA is very much like the HMA (hull MA).
// This particular subroutine specializes to a single "L" (input length
// is short length), and no 'time ahead'.
//
// Idea: do an EMA with lengths L and 2*L, and extrapolate from difference.
// That is a 'zero-lag' estimator of position, but has noise. Then
// Do EMA with length sqrt(Lfinal) for final smoothing.
double fastema[], slowema[], difference[];
ArrayResize(fastema,N);
ArrayResize(slowema,N);
ArrayResize(difference,N);
double fastp, finalp;
fastp = 2.0/(1.0+L);
finalp = 2.0/(1.0+MathSqrt(Lfinal));
EMAOnArray(N,fastp,input,fastema);
EMAOnArray(N,fastp,fastema,slowema);
for (int i=N; i>=0; i--) {
difference[i] = 2.0*fastema[i] - slowema[i];
}
EMAOnArray(N,finalp,difference,output);
}
void EMAOnArray(int N, double p, double input[], double& output[]) {
// Perform an "EMA" on array input[] with mixing parameter 'p'
// 0 < p < 1.
//
// p, conventionally is 2.0/(L+1.0) where L is the 'length' parameter.
// In an EMA, the length and thus 'p' need not be integers.
// initial value is input[N-1], and will set output[N-1] down to output[0].
//
double omp = 1.0-p;
double ema = input[N-1];
for (int i=N-1; i>=0; i--) {
double v = input[i];
ema = p*v + omp*ema;
output[i] = ema;
}
}
void initialize_kernel(int from, int to, int KernelLength, double PowerExponent) {
double kernelsum;
ArrayResize(kernel,KernelLength);
kernelsum = 0.0;
for (int i=from; i<=to; i++) {
kernel[i] = MathPow( (i)*0.01, -PowerExponent);
kernelsum += kernel[i];
}
for (i = from; i<=to; i++) {
kernel[i] = kernel[i] / kernelsum;
}
}
double convolve(double array[], double kernel[], int from, int to) {
// return sum(i=0..n-1) array[i]*kernel[i]
// conventionally kernel[*] sums to 1, but this is not enforced here.
double sum = 0.0;
for (int i=from; i ``` | 2,436 | 9,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | latest | en | 0.669715 |
https://www.vedantu.com/jee-main/mass-of-calcium-chloride-in-grams-would-be-chemistry-question-answer | 1,719,186,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00431.warc.gz | 913,312,878 | 26,624 | Courses
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# What mass of calcium chloride in grams would be enough to produce 14.35g of AgCl?[Atomic mass of Ca = 40, Ag = 108](A) 5.55g(B) 8.295g(C) 16.59g(D) 11.19g
Last updated date: 20th Jun 2024
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Hint: To solve this question first we have to write the equation for the production of silver chloride from calcium chloride. After writing the equation we compare the equivalence of calcium chloride and equivalence of silver chloride.
Complete step by step solution:
Calcium chloride reacts with silver to produce silver chloride and calcium. The reaction involved is:
$CaC{{l}_{2}}+Ag\to AgCl+Ca$
The balanced chemical reaction will be:
$CaC{{l}_{2}}+2Ag\to 2AgCl+2Ca$
Given in the question:
The atomic mass of calcium= 40
The atomic mass of silver= 108
Mass of AgCl to be produced= 14.35 g
The molar mass of calcium chloride:
Atomic Mass of chloride= 33.5
The atomic mass of calcium= 40
$Molar\text{ mass =}$ $atomic\text{ mass of calcium +}2\text{ ( atomic mass of chlorine)}$
The molar mass of calcium chloride = 111g
Similarly, Molar mass of Silver chloride:
Atomic Mass of chloride= 33.5
The atomic mass of silver= 108
$Molar\text{ mass =}$ $atomic\text{ mass of silver + atomic mass of chlorine}$
The molar mass of calcium chloride = 143.5g
From the above equation, we can say that amount of calcium chloride required to produce $2X143.5$g= 111g
Amount of calcium chloride used in the production of 14.35 g of silver chloride will be
=$\dfrac{(111)(14.35)}{(2)(143.5)}=5.55$g
The mass of calcium chloride in grams that would be enough to produce 14.35g of AgCl = 5.55g
Hence the correct answer is option (A).
Note: While solving such a question it is important to write a balanced chemical reaction, without balancing the reaction there will always be errors in the answer.
Silver chloride is used in the photographic process as when it is exposed to light a chemical reaction occurs which darks the film to produce an image. | 574 | 2,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.806842 |
http://www.futura-sciences.us/dico/d/physics-poisson-ratio-50001350/ | 1,642,950,544,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00074.warc.gz | 95,885,212 | 11,217 | Keywords |
• Physics
Poisson ratio
The principal Poisson ratio describes the contraction of matter in a direction perpendicular to the applied load.
This ratio was developed analytically by Denis Poisson, a French mathematician (1781 - 1840), the author of works on mathematical physics and mechanics, who determined its value using the molecular theory of the constitution of matter. It is defined by formula 1 opposite.
The Poisson ratio is denoted by the Greek letter ν and is one of the elastic constants (2 for isotropic materials or 4 for transversely isotropic materials). It is theoretically equal to 0.25 for a perfectly isotropic material and in practice is very close to this value.
For an isotropic material, the Poisson ratio allows the shear modulus G to be directly related to the Young modulus E.
The Poisson ratio is then always equal to or lower than 1/2. If it is equal to 1/2, the material is perfectly incompressible.
In the case of a laminate (transversely isotropic), we define a secondary Poisson coefficient using relationship 2 opposite linking E1 and E2.
Poisson ratio
connexes
Definition
Latest
Fill out my online form. | 259 | 1,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | longest | en | 0.869304 |
http://math.stackexchange.com/questions/15033/find-the-outgoing-edge-with-the-smallest-angles-given-one-incident-edges-and-mu?answertab=votes | 1,461,909,755,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110764.59/warc/CC-MAIN-20160428161510-00125-ip-10-239-7-51.ec2.internal.warc.gz | 121,256,513 | 18,653 | # Find the Outgoing Edge with the Smallest Angles, Given one Incident Edges and Multiple Outgoing Edges
I have one incident edges and multiple outgoing Edges, for which I want to pick an outgoing edge such that the angles between the outgoing edge and the incoming edge is the smallest of all. We know the coordinates for the vertex $V$ .
The angle must start from the incoming edge ($e_1$) and ends at another edge ($e_2$, $e_3$, $e_4$).
Also, the angle must be form in such a way that the face constructed from $e_1$ to $e_2$ must be in counter-close-wise direction. In other words, the face that you construct when you link all the vertexes $V$ in $e_1$ and $e_i$ together, must be a counter clockwise cycle.
So in the case below, edge $e_2$ must be picked because the $\theta$ is the smallest.
Is there an algorithm that does this elegantly? The few algorithms I devise are just plain ugly and requires a lot of hacks.
Update: One method that I think of ( and proposed below) is to use the cross product for each edge against the incoming edge $e_1$, and get the minimum $\theta_{i}$. i.e.,
$$e_{1} \times e_{i} = |e_{1}||e_{i}| \sin\, \theta_{i}$$
But the problem of this is it doesn't really work. Consider the following test cases: $$e_1=(1,0)$$ $$e_2=(-1/\sqrt(2),1/\sqrt(2) )$$ $$e_3=(1/\sqrt(2),-1/\sqrt(2))$$
One would find that
$$\theta_2 = -45^o$$ $$\theta_3=-45^o$$
Both are equal-- even though we know that $e_2$ should be selected.
-
In the diagram, $\theta$ is in the clockwise direction. – PEV Dec 21 '10 at 3:39
@Tervor, I didn't say that $\theta$ must be in counter clockwise direction. I'm saying that the face, when you link all the vertexes in $e_1$ and $e_i$ together, must be a counter clockwise cycle. – Graviton Dec 21 '10 at 3:49
Assuming that finding the coordinates(of your vectors) is not too difficult.
You can use the definition of dot and or cross products, and then solve for the angles.
So you can for example use $| a \times b | = |a||b| \sin\, \theta$
-
Not sure whether your solution works; although I can get the edge with minimum $\theta$, but I can't reject the face with clock wise cycle; see the updated question. – Graviton Dec 21 '10 at 6:18
@picakhu, sorry. Actually I can't even be sure that your solution works at all, see the updated question. – Graviton Dec 21 '10 at 6:31
I made a small mistake. Ill fix it. Try it again. By the way, your example is in 3 dimensions. Which complicated things.. You will have to use some kind of projection. – picakhu Dec 21 '10 at 7:29
@picakhu, it's 2D only; I've fixed the example. – Graviton Dec 21 '10 at 7:33
I have not checked your work, but just want to make sure you did $\sin\,\theta=\frac{|a\times b|}{|a||b|}$ – picakhu Dec 21 '10 at 7:36
If you are trying to put this into code you may want to consider the function atan2(y,x), it is available in most languages. Note that it takes the y coordinate as the first parameter. Since atan2 takes two parameter it can figure out in which quadrant your vector lies, and it can output the angle in the range $-\pi$ to $\pi$.
Here is a perl snippet to demonstrate how it works for your two angles:
#!/usr/bin/perl
use Math::Trig;
$t_rad = atan2(1/sqrt(2),-1/sqrt(2));$t_deg = $t_rad * 180 / pi; print "theta_2 =$t_deg\n";
$t_rad = atan2(-1/sqrt(2),1/sqrt(2));$t_deg = $t_rad * 180 / pi; print "theta_3 =$t_deg\n";
And here is the output when you run it for the two examples above:
theta_2 = 135
theta_3 = -45
- | 1,032 | 3,474 | {"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} | 3.984375 | 4 | CC-MAIN-2016-18 | latest | en | 0.89189 |
http://openstudy.com/updates/50aa8ee2e4b064039cbd4a70 | 1,448,684,275,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450762.4/warc/CC-MAIN-20151124205410-00337-ip-10-71-132-137.ec2.internal.warc.gz | 174,220,965 | 11,030 | ## Dido525 3 years ago Find the derivative:
1. Dido525
$\frac{ d }{ dx }\int\limits_{x^2}^{x^3}e ^{t^2+t} dt$
2. Dido525
So I tried to use the fundamental theorem of calculus:
3. Dido525
Is my work wrong?
4. zepdrix
Hmmmm no, it looks correct :O Didn't we just go over this problem like last week? XD Maybe that was someone else. heh
5. Dido525
Yeah we did. However apparently the professor said it's wrong...
6. zepdrix
Hmmmmmm
7. zepdrix
Mmmmmmmm nope you did it correctly, even wolfram agrees. http://www.wolframalpha.com/input/?i=%28d%2Fdx%29+integral_%7Bx%5E2%7D%5E%7Bx%5E3%7De%5E%7Bt%5E2%2Bt%7D+dt Maybe the teacher wrote it down wrong or something...
8. zepdrix
Oh, there shouldn't be a +C, maybe that's what he's fussing about? :o
9. Dido525
There should be. It's an indefinite intergal.
10. zepdrix
Why would you say it's indefinite? :o even though the limits are variables, they are certainly definite. :D So ANY constant that shows up, will end up being subtracted when you take away the value evaluated at the lower limit.
11. Dido525
Hmm good point. | 358 | 1,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2015-48 | longest | en | 0.897222 |
https://www.lmfdb.org/L/2/11e2/1.1/c3/0/9 | 1,713,917,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00327.warc.gz | 791,876,685 | 7,074 | # Properties
Label 2-11e2-1.1-c3-0-9 Degree $2$ Conductor $121$ Sign $1$ Analytic cond. $7.13923$ Root an. cond. $2.67193$ Motivic weight $3$ Arithmetic yes Rational yes Primitive yes Self-dual yes Analytic rank $0$
# Origins
## Dirichlet series
L(s) = 1 + 8·3-s − 8·4-s + 18·5-s + 37·9-s − 64·12-s + 144·15-s + 64·16-s − 144·20-s − 108·23-s + 199·25-s + 80·27-s + 340·31-s − 296·36-s − 434·37-s + 666·45-s − 36·47-s + 512·48-s − 343·49-s − 738·53-s − 720·59-s − 1.15e3·60-s − 512·64-s − 416·67-s − 864·69-s + 612·71-s + 1.59e3·75-s + 1.15e3·80-s + ⋯
L(s) = 1 + 1.53·3-s − 4-s + 1.60·5-s + 1.37·9-s − 1.53·12-s + 2.47·15-s + 16-s − 1.60·20-s − 0.979·23-s + 1.59·25-s + 0.570·27-s + 1.96·31-s − 1.37·36-s − 1.92·37-s + 2.20·45-s − 0.111·47-s + 1.53·48-s − 49-s − 1.91·53-s − 1.58·59-s − 2.47·60-s − 64-s − 0.758·67-s − 1.50·69-s + 1.02·71-s + 2.45·75-s + 1.60·80-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut & 121 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & \, \Lambda(4-s) \end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut & 121 ^{s/2} \, \Gamma_{\C}(s+3/2) \, L(s)\cr =\mathstrut & \, \Lambda(1-s) \end{aligned}
## Invariants
Degree: $$2$$ Conductor: $$121$$ = $$11^{2}$$ Sign: $1$ Analytic conductor: $$7.13923$$ Root analytic conductor: $$2.67193$$ Motivic weight: $$3$$ Rational: yes Arithmetic: yes Character: Trivial Primitive: yes Self-dual: yes Analytic rank: $$0$$ Selberg data: $$(2,\ 121,\ (\ :3/2),\ 1)$$
## Particular Values
$$L(2)$$ $$\approx$$ $$2.657507929$$ $$L(\frac12)$$ $$\approx$$ $$2.657507929$$ $$L(\frac{5}{2})$$ not available $$L(1)$$ not available
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$F_p(T)$
bad11 $$1$$
good2 $$1 + p^{3} T^{2}$$
3 $$1 - 8 T + p^{3} T^{2}$$
5 $$1 - 18 T + p^{3} T^{2}$$
7 $$1 + p^{3} T^{2}$$
13 $$1 + p^{3} T^{2}$$
17 $$1 + p^{3} T^{2}$$
19 $$1 + p^{3} T^{2}$$
23 $$1 + 108 T + p^{3} T^{2}$$
29 $$1 + p^{3} T^{2}$$
31 $$1 - 340 T + p^{3} T^{2}$$
37 $$1 + 434 T + p^{3} T^{2}$$
41 $$1 + p^{3} T^{2}$$
43 $$1 + p^{3} T^{2}$$
47 $$1 + 36 T + p^{3} T^{2}$$
53 $$1 + 738 T + p^{3} T^{2}$$
59 $$1 + 720 T + p^{3} T^{2}$$
61 $$1 + p^{3} T^{2}$$
67 $$1 + 416 T + p^{3} T^{2}$$
71 $$1 - 612 T + p^{3} T^{2}$$
73 $$1 + p^{3} T^{2}$$
79 $$1 + p^{3} T^{2}$$
83 $$1 + p^{3} T^{2}$$
89 $$1 - 1674 T + p^{3} T^{2}$$
97 $$1 + 34 T + p^{3} T^{2}$$
$$L(s) = \displaystyle\prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1}$$ | 1,236 | 2,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 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} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.229749 |
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# CSE 351 Midterm Review. Your midterm is next Wednesday Study past midterms (link on the website) Point of emphasis: Registers are not memory Registers.
## Presentation on theme: "CSE 351 Midterm Review. Your midterm is next Wednesday Study past midterms (link on the website) Point of emphasis: Registers are not memory Registers."— Presentation transcript:
CSE 351 Midterm Review
Your midterm is next Wednesday Study past midterms (link on the website) Point of emphasis: Registers are not memory Registers and caches are on the same piece of silicon as the processor This is done to make things fast (farther away => takes more time) Memory is an external storage bank of contiguous space Registers are individual storage elements that are not “addressable” (i.e. there is no “offset” between %rax and %rdx)
Integer Representation Convert -39 and 99 to binary and add the two’s complement 8-bit integers, then convert the result to hex -39 => 11011001 99 => 01100011 60 => 00111100 + 60 => 0x3C
Integer Representation What is the difference between the carry flag (CF) and the overflow flag (OF)? They differ in how they are triggered CF is set during unsigned addition when a carry occurs at the most-significant bit OF is set during signed arithmetic and it indicates that the addition yielded a number that was too large (either positive or negative direction)
Floating-Point Representation Suppose we have a 7-bit computer that uses IEEE floating-point arithmetic where a floating-point number has 1 sign bit, 3 exponent bits, and 3 fraction bits. NumberBinaryDecimal 00 000 0000.0 Smallest positive normalized number 0 001 0000.25 Largest positive number < INF0 110 11115.0 -3.11 100 100-3.0 12.250 110 10012.0
Pointers int a = 5, b = 15; int *p1, *p2; p1 = &a; // p1 -> a p2 = &b; // p2 -> b *p1 = 10; // a = 10 *p2 = *p1; // b = 10 p1 = p2; // p1 -> b *p1 = 20; // b = 20 What are the values of a, b, p1, p2?
IA-32 vs. x86-64 Look at the following function: int foo(int x, int y) { int c = x << (y + 3); if (x != 0) { return c; } else { return 1; } What does this function look like in x86 assembly? 32-bit and 64-bit.
IA-32 version push %ebp // ?? mov%esp, %ebp // ?? mov \$0xc(%ebp), %ecx // ?? add \$0x3, %ecx mov 0x8(%ebp), %eax // ?? shl %ecx, %eax cmp \$0x8(%ebp), \$0 jne \$0x808472 mov \$0x1, %eax leave // ?? ret This is what the code looks like in IA-32 assembly.
IA-32 version push %ebp // Save the base pointer mov%esp, %ebp // Move the base pointer up to the top of the stack mov \$0xc(%ebp), %ecx // Move y into a register add \$0x3, %ecx mov 0x8(%ebp), %eax // Move x into a register shl %ecx, %eax cmp \$0x8(%ebp), \$0 jne \$0x808472 mov \$0x1, %eax leave // Restore the old base pointer ret This is what the code looks like in IA-32 assembly.
x86-64 version push %ebp mov %esp, %ebp mov \$0xc(%ebp), %ecx add \$0x3, %rsi mov 0x8(%ebp), %eax mov %rdi, %rax shl %rsi, %rax test %rdi,%rdi jne \$0x808472 mov \$0x1, %rax leave ret This is what the code looks like in x86-64 assembly. Differences?
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // ?? add %eax, %ebx // ?? je.L1 // ?? sub %eax, %ecx // ?? je.L1 // ?? xor %eax, %eax // ?? jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // %ecx = y add %eax, %ebx // ?? je.L1 // ?? sub %eax, %ecx // ?? je.L1 // ?? xor %eax, %eax // ?? jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // %ecx = y add %eax, %ebx // %ebx = x + y je.L1 // ?? sub %eax, %ecx // ?? je.L1 // ?? xor %eax, %eax // ?? jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // %ecx = y add %eax, %ebx // %ebx = x + y je.L1 // jmp to L1 if x + y == 0 sub %eax, %ecx // ?? je.L1 // ?? xor %eax, %eax // ?? jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // %ecx = y add %eax, %ebx // %ebx = x + y je.L1 // jmp to L1 if x + y == 0 sub %eax, %ecx // %ecx = x - y je.L1 // ?? xor %eax, %eax // ?? jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // %ecx = y add %eax, %ebx // %ebx = x + y je.L1 // jmp to L1 if x + y == 0 sub %eax, %ecx // %ecx = x - y je.L1 // jmp to L1 if x – y == 0 xor %eax, %eax // ?? jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? mov %ebx, %ecx // %ecx = y add %eax, %ebx // %ebx = x + y je.L1 // jmp to L1 if x + y == 0 sub %eax, %ecx // %ecx = x - y je.L1 // jmp to L1 if x – y == 0 xor %eax, %eax // %eax = 0 jmp.L2 // ?? L1: mov \$1, %eax // ?? L2 :
Interpreting Assembly Let %eax store x and %ebx store y. What does this compute? |x| == |y| mov %ebx, %ecx // %ecx = y add %eax, %ebx // %ebx = x + y je.L1 // jmp to L1 if x + y == 0 sub %eax, %ecx // %ecx = x - y je.L1 // jmp to L1 if x – y == 0 xor %eax, %eax // %eax = 0 jmp.L2 // return 0 L1: mov \$1, %eax // return 1 L2 :
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# Web Quests
## CHAPTER 8
### WHO ARE YOU?
This task follows the theme of the Chapter 8 Chapter Task by having students create signs, which include tangram designs, for different animals that might be featured in a petting zoo. Students have already been introduced to tangram puzzles in Chapter 7, page185 of the student book. Students will practice solving several tangram puzzles at a website, copy the puzzles of 2 animals of their choice, as well as estimate and measure the area and perimeter of the tangram pieces.
GOALS
• manipulate stangram shapes to create images
• describe the shapes of the tangram puzzle pieces
• estimate, measure and record the area and perimeter of each shape
• create signs for a petting zoo using tangram puzzles
MEETING INDIVIDUAL NEEDS
• Students may find creating the animal tangramson their own very challenging. Allow them to refer back to the already created designs featured at the first website.
• If tracing the shapes to create animal tangrams is too difficult, make copies of the shapes and have students glue or tape them to the pages.
INSTRUCTIONAL SEQUENCE
1. Ask students if they remember what a tangram is. Check to see if they can remember how many pieces a tangram puzzle is made up of. Have them brainstorm some of the shapes they think you could make with tangram puzzle pieces.
2. Read the Introduction and the Task sections of the Student page as a class. Respond to any comments or concerns.
3. Have the students go to the website at Fwend. This web page will allow students to solve a series of tangram puzzles. Encourage the students to focus on the animal puzzles in the easy section.
4. Once students are comfortable solving the puzzles, have them print the tangram puzzle pieces at Tangram Grid.
5. Once they have cut out their pieces, have students complete the Tangram Shapes sheet.
6. Using the pieces they have printed and cut out, students can begin to create their own tangram puzzle animals. Have them choose the two they liked best to use for their signs. They will probably need to use the examples from the first website for reference.
7. Have students trace the animal tangrams onto blank sheets of paper to create 2 petting zoo signs that will include the name of each animal.
ASSESSMENT
Level 1 Level 2 Level 3 Level 4 Application of Procedures • has difficulty estimating the area and perimeter of each shape • makes major errors and/or omissions when: measuring area using a centimetre grid transparency; when measuring perimeter • estimates with some difficulty the area and perimeter of each shape • makes several errors and/or omissions when measuring area using a centimetre grid transparency; when measuring perimeter • estimates with reasonable accuracy the area and perimeter of each shape • makes only a few errors and/or omissions when measuring area using a centimetre grid transparency; when measuring perimeter • estimates with very reasonable accuracy the area and perimeter of each shape • makes almost no errors and/or omissions when measuring area using a centimeter grid transparency; when measuring perimeter Communication • provides an incomplete description of tangram shapes • provides a partial description of tangram shapes that shows some clarity • provides a complete and clear description of tangram shapes • provides a thorough, clear and insightful description of tangram shapes
RESOURCES
Websites:
Fwend
Tangram Grid
Files:
Tangram Shapes
Materials:
scissors
coloured pens or pencils (optional)
30 cm ruler
copies of tangram shapes (optional)
centimeter grid transparency
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https://de.webqc.org/molecularweightcalculated-190211-225.html | 1,568,629,302,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572517.50/warc/CC-MAIN-20190916100041-20190916122041-00512.warc.gz | 445,536,683 | 6,376 | #### Chemical Equations Balanced on 02/11/19
Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19
Calculate molecular weight
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 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232
Molar mass of [CH3(CH2)7]3P is 370.635502
Molar mass of co2 is 117.86639
Molar mass of CH4 is 16.04246
Molar mass of RbCl is 120.9208
Molar mass of NaOH is 39.99710928
Molar mass of co is 58.933195
Molar mass of NaCl2 is 93,89576928
Molar mass of Fe(ClO2)2 is 190.7486
Molar mass of NaCl 2 is 93,89576928
Molar mass of Na2CrO4 is 161,97323856
Molar mass of C5H8O2 is 100.11582
Molar mass of NaClO is 74,44216928
Molar mass of CO is 28.0101
Molar mass of Na3PO4 is 163.94066984
Molar mass of K is 39.0983
Molar mass of n2 is 28.0134
Molar mass of h2o is 18.01528
Molar mass of NaCl is 58.44276928
Molar mass of h2 is 2.01588
Molar mass of o2 is 31.9988
Molar mass of HF is 20.0063432
Molar mass of HF is 20.0063432
Molar mass of KI is 166.00277
Molar mass of NNa is 36.99646928
Molar mass of C57H106O6 is 887.44794
Molar mass of H2O2 is 34.01468
Molar mass of n is 14.0067
Molar mass of H3PO4 is 97.995182
Molar mass of CCl4 is 153.8227
Molar mass of CCl4 is 153.8227
Molar mass of Mg(NO3)2 is 148.3148
Molar mass of SO3 is 80.0632
Molar mass of I2 is 253.80894
Molar mass of I is 126.90447
Molar mass of I2 is 253.80894
Molar mass of KMnO3 is 142,034545
Molar mass of c6h6 is 78.11184
Molar mass of h is 1.00794
Molar mass of ammonia is 17.03052
Molar mass of sucrose is 342.29648
Molar mass of sucrose is 342.29648
Molar mass of Fe(NH4)2(SO4)2*6H2O is 392.1388
Molar mass of K is 39.0983
Molar mass of Fe(NH4)2(SO4)2*6H2O is 392.1388
Molar mass of H2PO4 is 96.987242
Molar mass of O13H4CC2O2CH2CH2 is 308.10802
Molar mass of H2C2O4*2H2O is 126.06544
Molar mass of fe is 55.845
Molar mass of NaI is 149.89423928
Molar mass of (NH4)3 PO4 is 149.086742
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 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232
Calculate molecular weight
Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19
Molecular masses on 02/04/19
Molecular masses on 01/12/19
Molecular masses on 02/11/18
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http://www.gaussianwaves.com/2010/01/matlab-simulation-psd-of-line-codes/ | 1,503,154,287,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105455.37/warc/CC-MAIN-20170819143637-20170819163637-00319.warc.gz | 556,549,469 | 25,296 | # Matlab Simulation – PSD of Line Codes
Line codes are used to map binary information sequence into analog signal which has properties suitable for the physical media that is being utilized to send the data.
Consider the case of a networking equipment like a router or a switch. According to ISO-OSI ( “International Standard Organisation – Open System Interconnection”[1]) reference model , any communication device or network can be conceptualized into seven layers of protocols namely Application, Presentation, Session, Transport, Network, Data-Link, and Physical Layer. Physical Layer dictates the electrical, physical and mechanical properties of the communication network or device. It defines properties like voltage levels, timing , maximum data rate allowable through the media without data loss,bandwidth efficiency, power consumption, physical connectors ( details like 50Ω or 75Ω termination, SMA, SMC or BNC connectors etc…,).
Line codes are chosen in such a way that they are optimized for the media that is being used to transmit data. Some line codes provide excellent timing synchronization (Timing synchronization is important in the context of sampling and clock recovery at the receiver. Incorrect timing sync may cause unrecoverable errors at the receiver due to wrong sampling instants) but may posses an undesirable DC content ( Why ?? Desired operating frequency range for a particular application may be between 500KHz and 1MHz, presence of DC content in such an application causes wastage of power.)
Commonly used line codes are NRZ (Non Return to Zero), RZ (Return to Zero), AMI (Alternate Mark Inversion), Manchester code, 8b/10b code,2B1Q code, Miller code, etc..,
Power Spectrum Density (PSD) is a common method employed in choosing a line code suitable for the physical media under consideration. Lets discuss how to convert a bit stream into NRZ unipolar, NRZ polar and Manchester coded data and plot their PSD.
### Procedure to plot Power Spectral Density of line codes in Matlab :
Step 1: Generate random binary sequence of sufficient length .Too small number of bits will give a poor PSD plot , whereas , too large number of bits consumes more computation time.
Step 2: Map the generated bits into NRZ unipolar or NRZ polar or Manchester line code (see Algorithm).
Step 3: Plot PSD using “psd” function in matlab.
### Algorithm:
Manchester Coding: NRZ Unipolar NRZ Polar First half of bit period : signal = data Second half of bit period: signal = !data For the entire bit period: signal=0 { if data =0 } signal=A volts { if data =1 } For the entire bit period: signal=-A volts { if data =0 } signal=A volts { if data =1 }
### File 1: %———Line_Encoder.m—————%
Check this ebook for the full Matlab code Simulation of Communication Systems using Matlab by Mathuranathan Viswanathan
Sample Output :
### Recommended Books:
• Pingback: How does codes fit in channel?()
• Hi
For getting back the original data just xor the Manchester coded data with a clock that run at 1/2 of bit rate.
Following code illustrates this:
%Manchester Coded data recovery
Tb=1/Rb;
%Generate Clock at 1/2 bit rate – Tb
clock=[];
numberOfDataBits = length(data);
bit=1;
for j=1:numberOfDataBits*2,
for i=0:1/Fs:Tb/2-1/Fs,
clock = [clock bit];
end
bit=~bit;
end
%XOR clock with Manchester coded output
recoveredStream = xor(clock,(output+max(output))/max(output));
%Decimate to get the recovered Data
for i=1:numberOfDataBits,
recoveredBits(i) = recoveredStream(i*Fs/Rb);
end
%print original data and recovered bits
recoveredBits %recovered data
data %original data
Regards
Mathuranathan
• you are doing very good,
but can i have the line decoding.
I mean from Manchester coding to its original form.
for example, I generated 1000 binary numbers and i coded them with your Line_code function to be of Manchester shape, how can i decode them to have their original format?
Thank you very much. | 889 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-34 | latest | en | 0.864059 |
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Mahama is also an NLP practitioner who has taught and inspired thousands of students to earn better grades in school and thrive in life. He uses a unique combination of intellectual and emotional problem approaches that are applicable to many aspects of life. | 869 | 3,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | longest | en | 0.725075 |
http://mathhelpforum.com/algebra/206920-substitution-transform-quadratic-form-print.html | 1,526,827,727,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863570.21/warc/CC-MAIN-20180520131723-20180520151723-00289.warc.gz | 189,875,498 | 3,469 | # Substitution to transform to quadratic form
• Nov 6th 2012, 04:58 PM
hexrei
Substitution to transform to quadratic form
Can anyone help me with this?
$\displaystyle 3=\frac{1}{(x+1)^2} + \frac{2}{(x+1)}$
I sub $\displaystyle u={(x+1)^-1}$
then flip them and rearrange so I get $\displaystyle u^2+\frac{u}{2}-3=0$
then mutiply all by two to get rid of fraction.
then I factor and end up with $\displaystyle (2u-3)(u+2)=0$
I think the problems begin when I sub back in the $\displaystyle {(x+1)^-1}$... doing something wrong and not getting book's solutions.
btw those -1's are supposed to be superscripted, not sure why its not working :/
• Nov 6th 2012, 05:09 PM
skeeter
Re: Substitution to transform to quadratic form
if $\displaystyle u = \frac{1}{x+1}$ , then the correct substitution would be ...
$\displaystyle 3 = u^2 + 2u$
try again ...
• Nov 6th 2012, 05:11 PM
HallsofIvy
Re: Substitution to transform to quadratic form
$\displaystyle \frac{1}{a}+ \frac{1}{b}$ "flipped" does NOT give "a+ b".
• Nov 6th 2012, 05:17 PM
hexrei
Re: Substitution to transform to quadratic form
Quote:
Originally Posted by skeeter
if $\displaystyle u = \frac{1}{x+1}$ , then the correct substitution would be ...
$\displaystyle 3 = u^2 + 2u$
try again ...
I appreciate the help :) and of course I'm going to try again. Why would I have taken the time to post it otherwise :)
• Nov 6th 2012, 06:11 PM
hexrei
Re: Substitution to transform to quadratic form
OK, so I have wrapped my head around that early error I made.
I'm getting tripped up after the factorization still though, and the book unfortunately has example subbing with negative exponents with single variable but none with a constant and a variable plus negative exponent.
So after I sub $\displaystyle x+1^{-1}$ back in. I will only list one of the zeros since I think once I see what I am missing the second shouldn't be hard.
my factorization is now $\displaystyle (u+3)(u-1)=0$
which after subbing back in the first one yields me
$\displaystyle \frac{1}{x+1} + 3$
from there I tried finding a common denominator
$\displaystyle \frac{1}{x+1} + \frac{3x+3}{x+1}$
$\displaystyle \frac{3x+4}{x+1}$
But not sure if that was correct, or what to do next. Can someone give me a clue please?
• Nov 6th 2012, 06:24 PM
MarkFL
Re: Substitution to transform to quadratic form
When is a rational expression equal to zero?
• Nov 6th 2012, 06:27 PM
hexrei
Re: Substitution to transform to quadratic form
Quote:
Originally Posted by MarkFL2
When is a rational expression equal to zero?
oooh..... when the numerator is zero? So x would be -4/3. Got it. Thanks :) | 786 | 2,622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.880611 |
http://wikistack.com/depth-first-traversal-of-graph/ | 1,560,778,049,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998475.92/warc/CC-MAIN-20190617123027-20190617145027-00300.warc.gz | 192,116,483 | 16,583 | Traversing a graph means visiting the nodes. In computer science the different ways of visiting nodes gives different results (sequence). In Depth first traversal of a graph , we visit the child node before visiting its sibling nodes. Visiting child nodes before its siblings nodes means , DFS explores its depth rather than breadth.
Note: All nodes must be visited only once.
Before applying DFS algorithm we need to represent the above graph in c/c++ program. There are two ways of representing this graph.
Here we will learn DFS of above graph by representing above graph as a adjacency list. The below figure shows adjacency list representation.
Let node 0 is starting position, then do dfs.
(1) For depth first traversal , we use stack data structure. Now visit node 0 and push this node to the stack S and mark this node as visited. Here visited node is marked as green.
(2) After the first step, the 0th node has three children 1,4 and 2. we can visit to any child here. let visit node 4, push this node on stack S and mark this node visited.
(3) Now we are at 4, the children nodes of node 4 are 3 and 2. Now let visit node 3, push node 3 on stack S and mark this node visited.
(4) Now from node 3, we can not move to any other child node, because it has no child node. In this case we look at the top of stack S, and pop node 3 ( because it is visited node ). This is called backtracking.
(5) Now we are at the node 4, Now look at the every child node. They are 3 and 2. node 3 is already visited so we have to move to node 2. mark node 2 visited, push this node to stack S.
(6) Now we are at node 2. All the adjacent children of node 2 are 3 and 1. node 3 is visited node so we have to move to node 1. mark node 1 visited and push the node 1 on stack.
(7) Now we are at node 1. As there is no place to go from node 1, we pop off the node 1 from stack.
(8) Again look at the top of the node 2 on stack. This node is already visited and there is no child to visit, so we have to pop off this node from stack S.
(9) Now we are on node 4, as this node is already visited and there is no child to visit we have to pop off this node from stack S.
(10) Now we are at node 0, since it is already visited and there is no child to visit then we need to pop off this node from stack.
(11) Now the stack is empty. the empty stack shows that all nodes have been visited and hence depth first traversal has been finished. The result is 0 4 3 2 1.
```#include<iostream>
#include<vector>
#include<stack>
using namespace std;
#define N 5 // number of nodes
class G
{
private:
bool visited[N];
stack < int >S;
public:
G ()
{
// false is for new node and true is for visited node
int i = 0;
for (i; i < N; i++);
visited[i] = false;
}
void addList (int u, int v);
void dfs (int v);
};
void G::addList (int u, int v)
{
}
void G::dfs (int v)
{
//push node v on stack
S.push (v);
visited[v] = true; // mark it visited
while (!S.empty ())
{
int u = S.top (); // get node
cout << u << " "; // print node
S.pop ();
// search for adjacent vertex of node u
// if not visited,make it visited and push on stack
for (int i = 0; i < adj[u].size (); i++)
{
}
}
}
int main ()
{
G g; | 860 | 3,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-26 | longest | en | 0.92999 |
https://in.mathworks.com/matlabcentral/cody/problems/42651-vector-creation/solutions/1870162 | 1,586,432,668,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371833063.93/warc/CC-MAIN-20200409091317-20200409121817-00330.warc.gz | 504,562,483 | 15,255 | Cody
# Problem 42651. Vector creation
Solution 1870162
Submitted on 10 Jul 2019 by jo gayong
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 7; y_correct = [1 2 3 4 5 6 7]; assert(isequal(vector(x),y_correct))
ans = 1 2 3 4 5 6 7
2 Pass
x = 9; y_correct = [1 2 3 4 5 6 7 8 9]; assert(isequal(vector(x),y_correct))
ans = 1 2 3 4 5 6 7 8 9 | 181 | 469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-16 | latest | en | 0.632144 |
https://brainmass.com/math/recurrence-relation/recursive-recurrence-528754 | 1,713,547,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00018.warc.gz | 132,269,279 | 6,850 | Purchase Solution
# Recursive and Recurrence
Not what you're looking for?
1. Find the sequence for the recursive formula:
S_n = -s_n-1 + 9, s_0 = -3
(see the attachment for the full question)
2. True of False a_n = 2 is a solution to the recurrence relation a_n = 2a_n-1 - a_n-2 with initial conditions a_0 = 2 and a_1 = 2.
a. True
b. False
3. Find a solution to the recurrence relation:
a_n = 3na_n-1, a_0 = 2
##### Solution Summary
In this solution we solve several problems pertaining to recursively-defined sequences.
##### Solution Preview
** Please see the attached file for the complete solution **
Help 2
Recursive and Recurrence
1. Find the sequence for the recursive formula: (please see the attached file),
We have:
We also have:
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This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations. | 344 | 1,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-18 | latest | en | 0.86493 |
https://www.tes.com/teaching-resource/aqa-entry-level-2-maths-12326435 | 1,621,171,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00127.warc.gz | 1,073,759,996 | 20,403 | Inspire and Educate! By Krazikas
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AQA Entry Level 1 Maths Bundle
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£24.99 | 711 | 3,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-21 | latest | en | 0.83771 |
https://ru.scribd.com/document/349025376/03-LAB-Friction-Phys-Sci | 1,611,023,719,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00283.warc.gz | 544,448,163 | 110,133 | Вы находитесь на странице: 1из 4
# Laboratory # 03
## PH 101, Physical Science, University of Mobile
STATIC and KINETIC FRICTION
INTRODUCTION
If you try to slide a heavy box resting on the floor, you may find it difficult to get the box
moving. Static friction is the force that counters your force on the box. If you apply a light
horizontal push that does not move the box, the static friction force is also small and directly
opposite to your push. If you push harder, the friction force increases to match the magnitude of
your push. There is a limit to the magnitude of static friction, so eventually you may be able to
apply a force larger than the maximum static force, and the box will move. The maximum static
friction force is sometimes referred to as starting friction. We model static friction, Fstatic, with
the inequality Fstatic s N where s is the coefficient of static friction and N is the normal force
exerted by a surface on the object. The normal force is defined as the perpendicular component
of the force exerted by the surface. In this case, the normal force is equal to the weight of the
object.
Once the box starts to slide, you must continue to exert a force to keep the object moving, or
friction will slow it to a stop. The friction acting on the box while it is moving is called kinetic
friction. In order to slide the box with a constant velocity, a force equivalent to the force of
kinetic friction must be applied. Kinetic friction is sometimes referred to as sliding friction. Both
static and kinetic friction depend on the surfaces of the box and the floor, and on how hard the
box and floor are pressed together. We model kinetic friction with Fkinetic = k N, where k is the
coefficient of kinetic friction.
In this experiment, you will use a Dual-Range Force Sensor to study static friction and kinetic
friction on a wooden block.
Mass
Wooden block
Pull
Force Sensor
Dual-Range
Figure 1
OBJECTIVES
Use a Dual-Range Force Sensor to measure the force of static and kinetic friction.
Determine the relationship between force of static friction and the weight of an object.
Measure the coefficients of static and kinetic friction for a particular block and track.
MATERIALS
computer string
Vernier computer interface block of wood with hook
Logger Pro balance or scale
Vernier Dual-Range Force Sensor mass set
## Physics with Vernier -1
PRELIMINARY QUESTIONS
1. In everyday life, you often experience one object sliding against another. Sometimes they slip
easily and other times they do not. List some things that seem to affect how easily objects
slide.
2. Consider a box sitting on a table. It takes a large force to move it at constant speed. List at
least two ways you could reduce the force needed to move the box at constant speed.
3. In pushing a heavy box across the floor, is the force you need to apply to start the box moving
greater than, less than, or the same as the force needed to keep the box moving? On what are
4.- If the box is at rest, what kind of friction relates to it? Explain
5.- What are the units for the coefficients of friction? Explain.
PROCEDURE
Part I Starting Friction
1. Measure the mass of the block and record it in the data table.
2. Set the range switch on the Dual-Range Force Sensor to 10 N. Connect the Force Sensor to
Channel 1 of the interface.
3. Open the file 12a Static Kinetic Frict from the Physics with Vernier folder.
4. Tie one end of a string to the hook on the Force Sensor and the other end to the hook on the
wooden block. Place a total of 1 kg mass on top of the block, fastened so the masses cannot
shift. Before you collect data, practice pulling the block and masses with the Force Sensor
using a straight-line motion: Slowly and gently pull horizontally with a small force. Very
gradually, taking one full second, increase the force until the block starts to slide, and then
keep the block moving at a constant speed for another second.
5. Sketch a graph of force vs. time for the force you felt on your hand. Label the portion of the
graph corresponding to the block at rest, the time when the block just started to move, and the
time when the block was moving at constant speed.
6. Hold the Force Sensor in position, ready to pull the block, but with no tension in the string.
Click to set the Force Sensor to zero.
7. Click to begin collecting data. Pull the block as before, taking care to increase the
force gradually. Repeat the process as needed until you have a graph that reflects the desired
motion, including pulling the block at constant speed once it begins moving. Copy the graph
for use in the Analysis portion of this activity. You can use print screen function or take a
picture. Save it as a JPEG file.
## 12 - 2 Physics with Vernier
Static and Kinetic Friction
## Part II Peak Static Friction and Kinetic Friction
In this part, you will measure the peak static friction force and the kinetic friction force as a
function of the normal force on the block, as shown in Figure 1. In each run, you will pull the
block as before, but by changing the masses on the block, you will vary the normal force on the
block.
8. Remove all masses from the block.
9. Click to begin collecting data and pull as before to gather force vs. time data.
10. Examine the data by clicking Statistics, . The maximum value of the force occurs when the
block started to slide. Read this value of the maximum force of static friction from the
floating box and record the number in your data table.
11. Drag across the region of the graph corresponding to the block moving at constant velocity.
Click Statistics, , again and read the average (or mean) force during the time interval. This
force is the magnitude of the kinetic frictional force.
12. Repeat Steps 911 for two more measurements and average the results to determine the
reliability of your measurements. Record the values in the data table.
13. Add 200 g to the block. Repeat Steps 912, recording values in the data table.
14. Repeat for 400, 600, and 800 g. Record values in your data table.
ANALYSIS
1. Inspect your force vs. time graph from Part I. Label the portion of the graph corresponding to
the block at rest, the time when the block just started to move, and the time when the block
was moving at constant speed. You can use Paint or other software to modify the JPEG
file.
2. Still using the force vs. time graph you created in Part I, compare the force necessary to keep
the block sliding compared to the force necessary to start the slide. How does your answer
3. The coefficient of friction is a constant that relates the normal force between two objects
(blocks and table) and the force of friction. Based on your graph (Run 1) from Part I, would
you expect the coefficient of static friction to be greater than, less than, or the same as the
coefficient of kinetic friction?
4. For Part II, calculate the normal force of the table on the block alone and with each
combination of added masses. Since the block is on a horizontal surface, the normal force
will be equal in magnitude and opposite in direction to the weight of the block and any
masses it carries. Fill in the Normal Force entries for both Part II data tables.
5. Plot a graph of the maximum (peak) static friction force (vertical axis) vs. the normal force
(horizontal axis). Use either Logger Pro or EXCEL.
6. Since Fmaximum static = s N, the slope of this graph is the coefficient of static friction s. Find
the numeric value of the slope, including any units by adding a Proportional Curve Fit. The
Proportional Curve Fit passes through the origin.
## Physics with Vernier 12 - 3
7. Does the force of kinetic friction depend on the weight of the block? Explain.
8. Does the coefficient of kinetic friction depend on the weight of the block?
DATA TABLE
Part I Starting Friction
Mass of block kg
## Total Normal Peak static friction Average
mass force peak static
(kg) (N) friction
Trial 1 Trial 2 Trial 3 (N)
## Total Normal Kinetic friction Average
mass force kinetic friction
(kg) (N) Trial 1 Trial 2 Trial 3 (N) | 1,851 | 8,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-04 | latest | en | 0.893324 |
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Didn't find what you were looking for?Search all of Office.com | 1,026 | 4,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | longest | en | 0.775452 |
http://mathhelpforum.com/differential-geometry/128597-series-real-analysis-print.html | 1,527,205,629,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00019.warc.gz | 190,888,628 | 2,972 | # Series- Real Analysis
• Feb 12th 2010, 06:44 PM
hebby
Series- Real Analysis
Consider the series:
1-1/2 -1/3 +1/4 +1/5 -1/6 -1/7........where the signs come in pairs. Does the series converge or diverge?
I want to use the alternating series test....any ideas? how I would solve this.
Thanks
• Feb 12th 2010, 07:23 PM
TheEmptySet
Quote:
Originally Posted by hebby
Consider the series:
1-1/2 -1/3 +1/4 +1/5 -1/6 -1/7........where the signs come in pairs. Does the series converge or diverge?
I want to use the alternating series test....any ideas? how I would solve this.
Thanks
Try groupin together in pairs to get
$\displaystyle \left( \frac{2-1}{1\cdot2}\right) + \left( \frac{-4+3}{3\cdot 4}\right)+\left( \frac{6-5}{5\cdot 6}\right)+...+\frac{(-1)^{n+1}}{(2n-1)(2n)}+...$
• Feb 12th 2010, 07:28 PM
hebby
but in your case we get a plus and then a minus.....but the series is minus minus ...plus plus?
• Feb 13th 2010, 06:32 AM
TheEmptySet
Quote:
Originally Posted by hebby
but in your case we get a plus and then a minus.....but the series is minus minus ...plus plus?
As I said add them to gether in pairs. Write it out and you will see.
$\displaystyle a_1+a_2=1-\frac{1}{2}=\frac{1}{2}=b_1$
$\displaystyle a_3+a_4=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12}=b_2$
$\displaystyle a_5+a_6=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}=b_3$
Note this is the series listed above I just gave the general term.
Since
$\displaystyle \sum a_n =\sum b_n$ we can look at the new series to draw any needed conclusions. | 530 | 1,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.751087 |
www.veetech.org.uk | 1,455,360,208,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166570.91/warc/CC-MAIN-20160205193926-00099-ip-10-236-182-209.ec2.internal.warc.gz | 732,873,144 | 6,733 | Ventilation Energy and Environmental Technology from VEETECH Ltd. Updated 12th April 2013
### CLICK HERE for PHPAIDA The Interactive Ventilation Calculator Based on Opening Area combined with Wind, Stack and Mechanical Ventilation
This tutorial covers:
- Background;
- Volume Flow vs. Mass Flow:
- The 'Crack Flow' Power Law Equation;
- The 'Flat Plate' Orifice Equation;
- Making an Orifice Equation look like a Power Law Equation;
- Changing to Mass Flow;
- The Next Step.
Background
The air flow equation is the building block of air infiltration and ventilation calculations. Every opening in the building fabric must be accurately represented by such an equation. Essentially there are two types of opening. The first are the random gaps and cracks that appear at construction joints or through porous material. These are ill defined and their geometry is rarely directly measurable. The second are essentially 'purpose provided' openings such as vents and windows. These are usually clearly defined in terms of geometry, location and flow characteristics. Openings in the former category require either measurement or implied knowledge about their characteristics. The necessary flow characteristics of the second type of opening can be inferred from the geometry of the opening and, therefore are easier to determine. The mathematical representation of these two types of opening is described in this tutorial. It is also shown that, by algebraic manipulation, both types of opening can be represented by an equation of identical structure. This considerable eases the calculation of total air flow into and out of a space.
Volume vs. Mass Flow
Flow rate can be expressed either in terms of 'volume' flow (e.g. m3/s, l/s, cfm) or 'mass' flow (e.g. kg/s). Most often fan capacity is expressed as a volume flow ,whereas calculations involving heat loss, for example, require that flow be expressed as a mass flow. In fundamental terms, it is more accurate to express flow rate as a mass flow. This is because, for a given mass of air, its volume varies with temperature. The Law of Continuity demands that the mass flow rate of air entering a building must match the mass leaving the building. Thus, in theory, a balance in mass flow does not necessarily mean that there is a balance in volume flow. Notwithstanding this situation, our air infiltration tutorials will be based on volume flow analysis. This is because for 'mild' climate temperature differences, where temperature differences between inside and outside peak at no more than 25 - 30C for a few hours at most in the year, any error will be marginal and because the 'iterative' approach that will be introduced to calculate infiltration and ventilation rate is much more stable for volume flow. Any error could also be minimised by selecting an air density (to convert to mass flow) that is midway between the outside and inside air temperature. However, switching to mass flow is very straightforward (as presented at the end of this tutorial). Also 'multi-zone' models such as Contam96 are structured in mass flow and can be downloaded and used for precision analyses if needed.
Cracks and Gaps
Air flow through general cracks and gaps in the building fabric is a function the size and structure of the opening and the pressure difference acting across it. The simplest representation of a crack is presented in the equation below.
This equation gives the volume flow rate (e.g. m3/s).
'C' is defined as the 'Flow Coefficient'. This is related to the size and structure of the opening. Typical flow coefficients are given in the Guide to Energy Efficient Ventilation.
'n' is the flow exponent and indicates the degree of turbulence. An 'n' value of 0.5 represents fully turbulent flow and '1.0' represents fully laminar flow. The typical 'n' value for whole buildings is 0.66.
'Orifice' Openings
Clearly defined openings such as vents or windows are frequently represented as 'flat plate orifices' in which the volume flow is represented in the equation:
Thus flow can be determined directly from a geometric analysis of the opening. In the absence of other information, the 'discharge coefficient' is usually based on a value of 0.61. Air flow through an orifice is assumed to be turbulent, thus the flow exponent 'n' = 0.5.
Making an Orifice Flow Equation Look Like a Crack Flow Equation
For ease of infiltration and ventilation calculations it is important that the above two equations take on the same form. This is achieved by making the orifice equation look like a power law crack flow equation. To achieve this:
Exercise: Manipulate the orifice flow equation to achieve the above 'C' and 'n' values.
Changing to Mass Flow
Volumetric flow is changed to mass flow by multiplying by the air density. Since density is dependent on air temperature, it is not a constant. Therefore, if this approach is used, it needs to be expressed as a function of temperature. This will be covered in a future tutorial.
Quadratic Formulation of the Air Flow Equation
Some authorities prefer to express air flow in the form of a quadratic equation of the form:
This is perceived to be dimensionally correct since the laminar and turbulent components of flow are separated. In essence, all the arguments and processes applied to the Power Law approach can be applied to the quadratic approach. In fact the single zone model, to be developed in these tutorials, will run just as easily in quadratic form. Again, at a later date, an equivalent analysis will be attempted.
The Next Steps
So far we have established the Power law Equation of flow through an opening as a building block towards infiltration and ventilation analysis. We have also shown that the flow characteristics of purpose provided opening can be approximated by the geometry of the opening. We now have to:
• Introduce a method to determine the pressure difference across openings;
• Establish a 'flow network' that represents the openings in the building;
• Find 'missing' data (primarily the 'C' and 'n' values of cracks and gaps).
• Develop a mathematical model that enables us to calculate the total flow into and out of the building and to calculate the flow rate and flow direction through individual openings.
• Incorporate mechanical ventilation systems.
• Introduce the concept of multi-zone or multi - room networks.
These steps will provide you with a basic design and ventilation evaluation tool that can be quickly established to solve 95% of pre-design and basic design steps. It will also give you the skills to apply more complex modelling tools that are available through the internet.
Tutorial 2 - Determining the natural driving forces (wind and temperature) ;
Tutorial 3 - Establishing a flow work - air leakage data and flow paths;
Tutorial 4 - AIDA a Single Zone Air Infiltration and Ventilation Model;
Tutorial 5 - Incorporating mechanical ventilation;
Tutorial 6 - Calculating pollutant concentration and energy impact.
Tutorial 7 - Towards multi-zone modelling.
This completes Tutorial 1 | 1,443 | 7,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.926669 |
http://www.khanacademy.org/math/basic-geo/basic-geo-coordinate-plane/basic-geo-shapes-on-a-plane | 1,427,609,605,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131298228.32/warc/CC-MAIN-20150323172138-00162-ip-10-168-14-71.ec2.internal.warc.gz | 610,385,332 | 16,590 | # Shapes on the coordinate plane
2 videos
3 skills
Let's think about shapes as collections of points on the coordinate plane. When you're done with this tutorial, you might be saying, "I have had it with these shapes on this plane!" But you'll be happy you went through it.
### Parallelogram on the coordinate plane
VIDEO 5:17 minutes
Remember our discussion of the coordinate plane? Sure you do! Let's graph the given coordinates of three of the polygon vertices, and find where the 4th vertex is.
### Rectangles on the coordinate plane
PRACTICE PROBLEMS
### Quadrilateral on the coordinate plane
VIDEO 1:32 minutes
In this example we are given the coordinates of the vertices and asked to construct the resulting polygon (specifically a quadrilateral). This is fun!
### Drawing polygons
PRACTICE PROBLEMS
### Drawing polygons 2
PRACTICE PROBLEMS | 197 | 858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-14 | latest | en | 0.882341 |
https://publicism.info/engineering/wiring/2.html | 1,620,709,960,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991641.5/warc/CC-MAIN-20210511025739-20210511055739-00315.warc.gz | 488,592,246 | 11,743 | Working Safely with Wiring - The Complete Guide to Wiring - Black & Decker, Cool Springs Press
The Complete Guide to Wiring, Updated 6th Edition: Current with 2014-2017 Electrical Codes - Black & Decker, Cool Springs Press (2014)
Chapter 1. Working Safely with Wiring
The only way you can possibly manage home wiring projects safely is to understand how electricity works and how it is delivered from the street to the outlets in your home.
The most essential quality to appreciate about electricity is that the typical amounts that flow through the wires in your home can be fatal if you contact it directly. Sources estimate that up to 1,000 people are electrocuted accidentally in the U.S. every year. In addition, as many as 500 die in fires from electrical causes. Home wiring can be a very satisfying task for do-it-yourselfers, but if you don’t know what you’re doing or are in any way uncomfortable with the idea of working around electricity, do not attempt it.
This chapter explains the fundamental principles behind the electrical circuits that run through our homes. It also includes some basic tips for working safely with wiring, and it introduces you to the essential tools you’ll need for the job. The beginner should consider it mandatory reading. Even if you have a good grasp of electrical principles, take some time to review the material. A refresher course is always useful.
In this chapter:
How Electricity Works
Glossary of Electrical Terms
Understanding Electrical Circuits
Grounding & Polarization
Home Wiring Tools
Wiring Safety
How Electricity Works
Ahousehold electrical system can be compared with a home’s plumbing system. Electrical current flows in wires in much the same way that water flows inside pipes. Both electricity and water enter the home, are distributed throughout the house, do their “work,” and exit.
In plumbing, water first flows through the pressurized water supply system. In electricity, current first flows along hot wires. Current flowing along hot wires also is pressurized. Electrical pressure is called voltage.
Large supply pipes can carry a greater volume of water than small pipes. Likewise, large electrical wires carry more current than small wires. This electrical current-carrying capacity of wires is called ampacity.
Water is made available for use through the faucets, spigots, and showerheads in a home. Electricity is made available through receptacles, switches, and fixtures.
Water finally leaves the home through a drain system, which is not pressurized. Similarly, electrical current flows back through neutral wires. The current in neutral wires is not pressurized and is at zero voltage.
Water and electricity both flow. The main difference is that you can see water (and touching water isn’t likely to kill you). Like electricity, water enters a fixture under high pressure and exits under no pressure.
The Delivery System
Electricity that enters the home is produced by large power plants. Power plants are located in all parts of the country and generate electricity with generators that are turned by water, wind, or steam. From these plants electricity enters large “step-up” transformers that increase voltage to half a million volts or more.
Electricity flows at these high voltages and travels through high-voltage transmission wires to communities that can be hundreds of miles from the power plants. “Step-down” transformers located at substations then reduce the voltage for distribution along street wires. On utility power poles, smaller transformers further reduce the voltage to ordinary 120-volt electricity for household use.
Wires carrying electricity to a house either run underground or are strung overhead and attached to a post called a service mast. Most homes built after 1950 have three wires running to the service head: two power wires, each carrying 120 volts, and a grounded neutral wire. Electricity from the two 120-volt wires may be combined at the service panel to supply electricity to large 240-volt appliances such as clothes dryers or electric water heaters.
Incoming electricity passes through a meter that measures electricity consumption. Electricity then enters the service panel, where it is distributed to circuits that run throughout the house. The service panel also contains fuses or circuit breakers that shut off power to the individual circuits in the event of a short circuit or an overload. Certain high-wattage appliances, such as microwave ovens, are usually plugged into their own individual circuits to prevent overloads.
Voltage ratings determined by power companies and manufacturers have changed over the years. These changes do not affect the performance of new devices connected to older wiring. For making electrical calculations, use a rating of 120 volts or 240 volts for your circuits.
Power plants supply electricity to thousands of homes and businesses. Step-up transformers increase the voltage produced at the plant.
Substations are located near the communities they serve. A typical substation takes electricity from high-voltage transmission wires and reduces it for distribution along street wires.
Electrical transformers reduce the high-voltage electricity that flows through wires along neighborhood streets. A utility pole transformer—or ground transformer—reduces voltage from 10,000 volts to the normal 120-volt electricity used in households.
Parts of the Electrical System
Glossary of Electrical Terms
Ampere (or amp): Refers to the rate at which electrical current flows to a light, tool, or appliance.
Armored cable: An assembly of insulated wires enclosed in a flexible, interlocked metallic armor.
Box: A device used to contain wiring connections.
BX: A brand name for an early type of armored cable that is no longer made. The current term is armored cable.
Cable: Two or more wires that are grouped together and protected by a covering or sheath.
Circuit: A continuous loop of electrical current flowing along wires.
Circuit breaker: A safety device that interrupts an electrical circuit in the event of an overload or short circuit.
Conductor: Any material that allows electrical current to flow through it. Copper wire is an especially good conductor.
Conduit: A metal or plastic pipe used to protect wires.
Continuity: An uninterrupted electrical pathway through a circuit or electrical fixture.
Current: The flow of electricity along a conductor.
Duplex receptacle: A receptacle that provides connections for two plugs.
Flexible metal conduit (FMC): Hollow, coiled steel or aluminum tubing that may be filled with wires (similar to Armored Cable, but AC is pre-wired).
Fuse: A safety device, usually found in older homes, that interrupts electrical circuits during an overload or short circuit.
Greenfield: A brand name for an early type of flexible metal conduit. The current term is flexible metal conduit. Note: flexible metal conduit is different from armored cable.
Grounded wire: See neutral wire.
Grounding wire: A wire used in an electrical circuit to conduct current to the service panel in the event of a ground fault. The grounding wire often is a bare copper wire.
Hot wire: Any wire that carries voltage. In an electrical circuit, the hot wire usually is covered with black or red insulation.
Insulator: Any material, such as plastic or rubber, that resists the flow of electrical current. Insulating materials protect wires and cables.
Junction box: See box.
Meter: A device used to measure the amount of electrical power being used.
Neutral wire: A wire that returns current at zero voltage to the source of electrical power. Usually covered with white or light gray insulation. Also called the grounded wire.
Non-metallic sheathed cable: NM cable consists of two or more insulated conductors and, in most cases, a bare ground wire housed in a durable PVC casing.
Outlet: A place where electricity is taken for use. A receptacle is a common type of outlet. A box for a ceiling fan is another type of outlet.
Overload: A demand for more current than the circuit wires or electrical device was designed to carry. This should cause a fuse to blow or a circuit breaker to trip.
Pigtail: A short wire used to connect two or more wires to a single screw terminal.
Polarized receptacle: A receptacle designed to keep hot current flowing along black or red wires and neutral current flowing along white or gray wires.
Power: The work performed by electricity for a period of time. Use of power makes heat, motion, or light.
Romex: A brand name of plastic-sheathed electrical cable that is commonly used for indoor wiring. Commonly known as NM cable.
Screw terminal: A place where a wire connects to a receptacle, switch, or fixture.
Service panel: A metal box usually near the site where electricity enters the house. In the service panel, electrical current is split into individual circuits. In residences, the service panel has circuit breakers or fuses to protect each circuit.
Short circuit: An accidental and improper contact between two current-carrying wires or between a current-carrying wire and a grounding conductor.
Switch: A device that controls electricity passing through hot circuit wires. Used to turn lights and appliances on and off.
UL: An abbreviation for Underwriters Laboratories, an organization that tests electrical devices and manufactured products for safety.
Voltage (or volts): A measurement of electricity in terms of pressure.
Wattage (or watt): A measurement of electrical power in terms of total work performed. Watts can be calculated by multiplying the voltage times the amps.
Wire connector: A device used to connect two or more wires together. Also called a wire nut.
Understanding Electrical Circuits
An electrical circuit is a continuous loop. Household circuits carry electricity from the main service panel, throughout the house, and back to the main service panel. Several switches, receptacles, light fixtures, or appliances may be connected to a single circuit.
Current enters a circuit loop on hot wires and returns along neutral wires. These wires are color coded for easy identification. Hot wires are black or red, and neutral wires are white or light gray. For safety, all modern circuits include a bare copper or green insulated grounding wire. The grounding wire conducts current in the event of a ground fault (see page 165) and helps reduce the chance of severe electrical shock. The service panel also has a bonding wire connected to a metal water pipe and a grounding wire connected to a metal grounding rod, buried underground, or to another type of grounding electrode.
If a circuit carries too much current, it can overload. A fuse or a circuit breaker protects each circuit in case of overloads.
Current returns to the service panel along a neutral circuit wire. Current then leaves the house on a large neutral service wire that returns it to the utility transformer.
Grounding & Polarization
Electricity always seeks to return to its source and complete a continuous circuit. Contrary to popular belief, electricity will take all available return paths to its source, not just the path of lowest resistance. In a household wiring system, this return path is provided by white neutral wires that return current to the main service panel. From the service panel, current returns along the uninsulated neutral service wire to a power pole transformer.
You will see the terms grounding and bonding used in this and other books about electricity. These terms are often misunderstood. You should understand the difference to safely work on electrical circuits.
Bonding connects the non-current-carrying metal parts of the electrical system, such as metal boxes and metal conduit, in a continuous low-resistance path back to the main service panel. If this metal becomes energized (a ground fault), current travels on the bonded metal and quickly increases to an amount that trips the circuit breaker or blows the fuse. The dead circuit alerts people to a problem.
Other metal that could become energized also must be bonded to the home’s electrical system. Metal water and gas pipes are the most common examples. A metal water and gas pipe could become energized by coming in contact with a damaged electrical wire. Metal gas pipe could become energized by a ground fault in a gas appliance such as a furnace.
Bonding is a very important safety system. A person could receive a fatal shock if he or she touches energized metal that is improperly bonded, because that person becomes electricity’s return path to its source. Bonding is also a fire safety system that reduces the chance of electrical fires.
Grounding connects the home’s electrical system to the earth. Grounding’s primary purpose is to help stabilize voltage fluctuations caused by lightning and other problems in the electrical grid. Grounding also provides a secondary return path for electricity in case there is a problem in the normal return path.
Grounding is accomplished by connecting a wire between the main service panel and a grounding electrode. The most common grounding electrode is a buried copper rod. Other grounding electrodes include reinforcing steel in the footing, called a ufer ground.
Normal current flow: Current enters the electrical box along a black hot wire and then returns to the service panel along a white neutral wire.
Ground Fault: Current is detoured by a loose wire in contact with the metal box. The grounding wire and bonded metal conduit pick it up and channel it back to the main service panel, where the overcurrent device is tripped, stopping further flow of current. Most current in the bonding and ground system flows back to the transformer; some may trickle out through the copper that leads to the grounding node.
Grounding of the home electrical system is accomplished by wiring the household electrical system to a metal cold water pipe and metal grounding rods that are buried in the earth.
After 1920, most American homes included receptacles that accepted polarized plugs. The two-slot polarized plug and receptacle was designed to keep hot current flowing along black or red wires and neutral current flowing along white or gray wires.
The metal jacket around armored cable and flexible metal conduit, widely installed in homes during the 1940s, provided a bonding path. When connected to metal junction boxes, it provided a metal pathway back to the service panel. Note, however, that deterioration of this older cable may decrease its effectiveness as a bonding conductor.
Modern cable includes a green insulated or bare copper wire that serves as the bonding path. This grounding wire is connected to all three-slot receptacles and metal boxes to provide a continuous pathway for any ground-faulted current. By plugging a three-prong plug into a grounded three-slot receptacle, people are protected from ground faults that occur in appliances, tools, or other electric devices.
Use a receptacle adapter to plug three-prong plugs into two-slot receptacles, but use it only if the receptacle connects to a grounding wire or grounded electrical box. Adapters have short grounding wires or wire loops that attach to the receptacle’s coverplate mounting screw. The mounting screw connects the adapter to the grounded metal electrical box.
Modern NM (nonmetallic) cable, found in most wiring systems installed after 1965, contains a bare copper wire that provides bonding for receptacle and switch boxes.
Armored cable is sold pre-installed in a flexible metal housing. It contains a green insulated ground wire along with black and white conductors. Flexible metal conduit (not shown) is sold empty.
Polarized receptacles have a long slot and a short slot. Used with a polarized plug, the polarized receptacle keeps electrical current directed for safety.
Tamper resistent three-slot receptacles are required by code for new homes. They are usually connected to a standard two-wire cable with ground.
A receptacle adapter allows three-prong plugs to be inserted into two-slot receptacles. The adapter should only be used with receptacles mounted in a bonded metal box, and the grounding loop or wire of the adapter must be attached to the coverplate mounting screw.
Double-insulated tools have non-conductive plastic bodies to prevent shocks caused by ground faults. Because of these features, double-insulated tools can be used safely with ungrounded receptacles.
Home Wiring Tools
To complete the wiring projects shown in this book, you need a few specialty electrical tools as well as a collection of basic hand tools. As with any tool purchase, invest in quality products when you buy tools for electrical work. Keep your tools clean, and sharpen or replace any cutting tools that have dull edges.
The materials used for electrical wiring have changed dramatically in the last 20 years, making it much easier for homeowners to do their own electrical work. The following pages show how to work with the following components for your projects.
Hand tools you’ll need for home wiring projects include: Stud finder/laser level (A) for locating framing members and aligning electrical boxes; tape measure (B); a cable ripper (C) for scoring NM sheathing; standard (D) and Phillips (E) screwdrivers; a utility knife (F); side cutters (G) for cutting wires; channel-type pliers (H) for general gripping and crimping; linesman pliers (I) that combine side cutter and gripping jaws; needlenose pliers (J); wire strippers (K) for removing insulation from conductors.
Use a tool belt to keep frequently used tools within easy reach. Electrical tapes in a variety of colors are used for marking wires and for attaching cables to a fish tape.
A fish tape is useful for installing cables in finished wall cavities and for pulling wires through conduit. Products designed for lubrication reduce friction and make it easier to pull cables and wires.
Diagnostic tools for home wiring use include: A touchless circuit tester (A) to safely check wires for current and confirm that circuits are dead; a plug-in tester (B) to check receptacles for correct polarity, grounding, and circuit protection; a multimeter (C) to measure AC/DC voltage, AC/DC current, resistance, capacitance, frequency, and duty cycle (model shown is an auto-ranging digital multimeter with clamp-on jaws that measure through sheathing and wire insulation).
Wiring Safety
Safety should be the primary concern of anyone working with electricity. Although most household electrical repairs are simple and straightforward, always use caution and good judgment when working with electrical wiring or devices. Common sense can prevent accidents.
The basic rule of electrical safety is: Always turn off power to the area or device you are working on. At the main service panel, remove the fuse or shut off the circuit breaker that controls the circuit you are servicing. Then check to make sure the power is off by testing for power with a voltage tester. Tip: Test a live circuit with the voltage tester to verify that it is working before you rely on it. Restore power only when the repair or replacement project is complete.
Follow the safety tips shown on these pages. Never attempt an electrical project beyond your skill or confidence level.
Shut power OFF at the main service panel or the main fuse box before beginning any work.
Create a circuit index and affix it to the inside of the door to your main service panel. Update it as needed.
Confirm power is OFF by testing at the outlet, switch, or fixture with a voltage tester.
Use only UL-approved electrical parts or devices. These devices have been tested for safety by Underwriters Laboratories.
Wear rubber-soled shoes while working on electrical projects. On damp floors, stand on a rubber mat or dry wooden boards.
Use fiberglass or wood ladders when making routine household repairs near the service mast.
Extension cords are for temporary use only. Cords must be rated for the intended usage.
Breakers and fuses must be compatible with the panel manufacturer and match the circuit capacity.
Never alter the prongs of a plug to fit a receptacle. If possible, install a new grounded receptacle.
Do not penetrate walls or ceilings without first shutting off electrical power to the circuits that may be hidden.
| 4,033 | 20,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-21 | longest | en | 0.929025 |
https://cbsemathssolutions.in/natural-numbers-definition-explained/ | 1,582,546,404,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145941.55/warc/CC-MAIN-20200224102135-20200224132135-00160.warc.gz | 309,470,372 | 19,134 | # Natural numbers definition explained
Learn and know what is natural numbers definition in mathematics. Natural numbers are the first types of numbers that we will learn in math. From natural numbers, we will learn other numbers like whole numbers (W), Integers (I or Z), rational numbers (Q), irrational numbers (Q|) and real numbers (R).
We all know that maximum numbers of students those who are studying in the lower classes know what are the natural numbers. But the exact definition of natural numbers they don’t know. According to them the meaning of Natural numbers is 1, 2, 3, 4, 5, 6 …and so on. Definition if we ask then they can’t say it. So this article will help for all the students in knowing the definition of natural numbers. Now we will learn how to define the natural numbers and its symbol.
## Natural numbers definition as follows:
The definition of natural numbers is given as “the counting numbers are called as Natural numbers”. I mean to say that the numbers which we are using to count something like number of students in a class or number of fans in a class or number of players in Indian cricket team and so on are called as Natural numbers.
Generally the Natural numbers are represented or denoted by the symbol “N”. To represent the natural numbers symbol remember that always we should use capital letters only. We should not use small letters. Symbolically the set of natural numbers are represented by the notation, N = {1, 2, 3, 4, 5, 6, 7, 8, 9…}.
Important Note:
On observing all the natural numbers it is very clear that the smallest natural number is “1” and the greatest natural number is “does not exist” because there is NO greatest natural number.
Hope now you have understood natural numbers definition and representation. | 384 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-10 | latest | en | 0.928595 |
http://mathhelpforum.com/calculus/109036-concave-up-concave-down-print.html | 1,508,500,820,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824068.35/warc/CC-MAIN-20171020101632-20171020121632-00329.warc.gz | 209,253,998 | 2,733 | # concave up and concave down
• Oct 19th 2009, 01:01 PM
xxju1anxx
concave up and concave down
How do I determine whether there is a upward concavity or downward concavity?
• Oct 19th 2009, 01:09 PM
Quote:
Originally Posted by xxju1anxx
How do I determine whether there is a upward concavity or downward concavity?
Use the second derivative test. This will help:
Pauls Online Notes : Calculus I - The Shape of a Graph, Part II
• Oct 19th 2009, 01:11 PM
tonio
Quote:
Originally Posted by xxju1anxx
How do I determine whether there is a upward concavity or downward concavity?
If the function is double derivable over an interval then second derivative positive ==> concave upwards, and 2nd derivative negative ==> downwards.
If there is no 2nd derivative then by inspection, I believe...
Tonio | 215 | 799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.847695 |
https://www.telematika.org/py/p4m_theneedforspeed/ | 1,611,843,828,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00701.warc.gz | 940,129,791 | 15,643 | ## Jupyter Snippet P4M TheNeedForSpeed
Jupyter Snippet P4M TheNeedForSpeed
# Python - The Need for Speed
Python was originally developed by Guido van Rossum as a high level general purpose scripting language in the 1990s. It’s strength has been in particular that it is easy to learn and that algorithms can be coded quickly (python has sometimes been described as executable pseudo-code). In recent years it has been increasingly used also for scientific computing and mathematics. Here one quickly reaches a point where the time spent waiting for a program to run becomes much longer than the time spent writing the code. Hence the need for speed.
In this tutorial a number of ways of speeding python up will be explored. These will be tried in the context of a simple numerical algorithm. We will start with a simple but slow implementation and then look at what makes it slow and how to speed things up.
## Method 0: Gaussian Elimination - A simplistic implementation
Consider one of the basic mathematical algorithms that most students learn in their first year of university mathematics: solving a set of simultaneous linear equations using the Gaussian Elimination algorithm. Below is the framework for a function to implement this method. Note the use of comments to describe the intent and usage of the function. Preconditions and post-conditions, in the form of assert statements, are used to help with debugging. Including such comments and conditions as a matter of course is a good practice, even for code that you don’t intend to maintain long term, as it can save you a lot of time hunting for bugs or when you invariably end up re-using code for other purposes than originally planned.
While this is not the main point of the exercise, you may also want to think about things such as numerical stability and how to deal with any errors. To keep things simple here we are going to assume that the input arguments are a dense matrix $A$ in the form of a dictionary, right hand side vector $b$ and we return the solution to $A x = b$ as a simple list.
Hint: Here is a pseudo-code of a very basic Gaussian elimination algorithm
INPUT: A,b
U := A # so the input is not modified
x := b
for i = 1,...,n: # using 1 indexing here as in typical math. notation
# assuming U_ii != 0 here, could add code to cater for this exception
Divide row i of U by U_ii (this makes U_ii==1)
for k = i+1,...,n:
subtract U_ki times row i from row k
x_k := x_k - U_ki * x_i
U is now upper triangular
for i = n,n-1,...,1: # back substitution
x_i := (x_i - sum( U_ik * x_k for k=i+1,...,n) ) / U_ii
OUTPUT: x
def gaussSimple(A,b):
"""Solve the set of equations A x = b and return x
Input: b = list of length n, and A = dictionary, with A[i,j] being the entry for every row i, column j
Output: A list of length n giving the solution. (This version throws an assertion failure if there is no solution)
Note: The inputs are not modified by this function."""
n = len(b)
N = range(n)
assert (len(A) == n*n), "Matrix A has wrong size" # simple check of inputs
assert all( (i,j) in A for i in N for j in N), "Cannot handle sparse matrix A"
U = dict(A.items()) # make a copy of A before we transform it to upper triangular form
x = b[:] # copy so we don't modify the orignal
## for the most basic version we want to:
## For every row i
## Eliminate all entries in column i below i by subtracting some multiple of row i
## Update the right hand side (in x) accordingly
## return [] if A does not have full rank (we come across a coefficient of zero)
## Back-substitute to replace x with the actual solution
error = max( abs(sum(A[i,j]*x[j] for j in N)-b[i]) for i in N)
assert (error < 1e-5 ), f"Incorrect solution: out by {error}" # check that we have reasonable accuracy
return x
To test this we are going to generate some random data. This will also allow us to see how fast it runs as the size of the problem increases
from random import random
def randProb(n):
"Returns a randomly generated n x n matrix A (as a dictionary) and right hand side vector b (as a list)."
n = int(n)
assert n > 0
N = range(n)
return dict( ((i,j), random()) for i in N for j in N), [random() for i in N]
A,b = randProb(3)
gaussSimple(A,b)
## Execution timing
Now lets see how fast this thing goes. There are a couple of things to think about when we talk about timing:
• Is it elapsed time (wall-clock) time or CPU time we are measuring? The two should be nearly the same if we are using a single threaded program on a computer that is not fully loaded. However, once we have multiple threads in our program, or multiple other programs competing for a limited number of CPUs, the results could be quite different
• Random variation - running the same code several times can result in small random variations in run-time due to a range of factors (cache misses, variable CPU clock speed, computer load, etc) even when the functiuon we are testing is entirely deterministic and run with the same inputs
• Garbage collection: One of the things that make Python convenient is that we generally don’t have to worry about memory management. However if the automated garbage collection happens to cut in during the function we are testing this can make a big difference.
For our tests here we are going to use the timeit module that takes care of the last two points by turning of garbage collection and making repeated tests easy. We will specifically ask it to measure CPU time using the time.process_time() function but you can experiment with using time.perf_counter() function which is the default.
In addition, to get a feeling for how the runtime increases as we double the size of the data, we also print the ratio between successive tests with increasing data. We reduce the random variation between tests of different methods by always generating data with the same random number seed.
Since we are going to make use of the speed test functions repeatedly we are going to write this out as a separate module that we can import into any notebook or python program.
%%writefile 'gausstest.py'
"""This module contains a function for testing the speed of a function which solve Ax=b
usage: import gausstest
gausstest.speed(myfunction) # optional second argument, maximum size n
Here myfunction takes arguments A,b,"""
import timeit,time,gc
from statistics import mean,stdev
from random import seed,random
def randProb(n):
"Returns a randomly generated n x n matrix A (as a dictionary) and right hand side vector b (as a list)."
n = int(n)
assert n > 0
N = range(n)
return dict( ((i,j), random()) for i in N for j in N), [random() for i in N]
def speed(method,maxSize=400):
seed(123456) # fix some arbitrary seed so we keep generating the same data
randP = lambda n : randProb(n) # add randProb to locals() namespace
prev,n = 0.0, 50
gc.disable()
while n <= maxSize:
gc.collect() # manual cleanout of garbage before each repeat
t = timeit.repeat(stmt="method(A,b)",setup=f"A,b=randP({n})",
timer=time.process_time,repeat=5,number=1,globals=locals())
print("%4d %10.4f σ=%.2f sec" % (n,mean(t),stdev(t)),"(x %.2f)"%(mean(t)/prev) if prev > 0 else "")
prev = mean(t)
n *= 2
gc.enable()
Now lets test our function systematically.
import gausstest
print("Simple test of a single 3-dimensional instance: ",
gaussSimple(*gausstest.randProb(3)),"\n")
# Note that in the above the '*' means we are passing the tuple of outputs
# from randProb as successive arguments to gaussSimple
gausstest.speed(gaussSimple) # full test
### Discussion - Complexity
• What is the theoretical complexity of this algorithm?
• Does the practical performance match the theoretical expectation?
• What can be done to make this implementation better?
• More robust numerically? - (Left as exercise to interested students)
• Faster? - What makes it slow?
## Method 1: changing the data structures
As discussed, part off the reason for the slow time is that python treates every variable and every element of generic data structures like lists and dictionaries, as an “Any” type that could contain anything. That means that in executing every step of the algorithm, the python interpreter has to got through a vast list of “if” statements to check every possible type to determine what actually needs to be done at this point of the code.
In our next version, we are going to replace the dictionary and list structures with array data structures. These are much more like the arrays found in Java/C/Fortran/… just a big chunk of memory with each element containing the same type. The basic usage for this data type is
from array import array
x = array('i',range(10))
This would initialise an array of integers containing the numbers 0,…,9. An array behaves much like a list in python but contains only elements of one basic type (integer, char, float). For our purposes we will want to create array('d',[]) where ’d' stands for double (C style 64-bit floating point number) and [] should be replaced by a suitable initialiser rather than the empty list. A
Write a function gaussArray that uses array data structures for all data types (matrix, right hand side and x). To fit the matrix into a single dimensional array, you need to do a bit of index arithmetic (and use range with appropriate step size). Alternatively (if you prefer) you could use the numpy version of array.
from array import array
#Solution algorithm
def gaussArray(A,b):
"""Solve the set of equations A x = b and return x
Input: b = list of length n, and A = dictionary, with A[i,j] being the entry for every row i, column j
Output: An array of length n giving the solution.
Note: The inputs are not modified by this function."""
import gausstest
gausstest.speed(gaussArray)
### Discussion Question:
Where does the speed-up of this method over the previous one come from? What makes this method quite slow still?
## Method 2: Using numpy
For numeric computation, a “standard” set of libraries is the numpy module and friends. This provides vectors and matrices together with basic linear algebra routines implented largely in C++ and Fortran. These laregy mimick the type of basic functionality built into matlab. Many other python packages are built on top of this so that the numpy data types have become a defacto standard for any kind of scientific computing with python.
We could use numpy to solve our equations directly using the numpy.linalg.solve routine. You may want to try this briefly. However, here we are mainly interested here in seeing whether writing our own algorithm on top of the basic matrix methods is going to be faster.
Basic usage for numpy: For convenience we import numpy as np (a fairly common practice in the python community)
• Create matrices using something like A = np.array( [ [1,2,3],[4,5,6] ] ) to create a 2 x 3 matrix. If all elements are integer this will be int64 type otherwise float64. You can query the dimensions with A.shape
• Matrices can be indexed using A[0,2] == 3 or using slices such as A[0,1:3] == np.array([2,3.0])
• Arithmetic operations are overloaded to do mostly do what you would expect. The only significant exception is multiplication. When multiplying two arrays or matrices together we get the Schur product, that is the result has each of the corresponding elements from the input multiplied together (the operands must have the same dimensions). To get the normal inner product or matrix procut use np.matmul. E.g. np.matmul( np.array([[3,0],[0,2]]), A) or A.dot(x) to get an inner product that is effectively the same as matrix multiply if A is a matrix and x is a vector or matrix (but behaves slightly differently for other types of A & x).
• Matrices can be built up using np.hstack, np.vstack (horizontal & vertical stacking)
• To reshape a numpy array use np.reshape, this is particularly useful to turn 1 dimensional matrices into 2 dimensional matrices. V = np.reshape(v,(len(v),1)) will turn an array v of length n into a n x 1 matrix V.
The task for this method is to write a Gaussian elimination method using matrix multiplications. In a first year maths course you have probably seen that the elementary row operations of Gaussian elimination can be represented by pre-multiplying the matrix we are reducing with a suitable elementary matrix (an identity matrix with one off-diagonal element). So we can rewrite the algorithm to set up a suitable elementary matrix for each row reduction and pre-multiplying our matrix with this. For example for a 5 x 5 matrix $A$, to subtract 3 times the 2nd row from the fourth row we would pre-multiply $A$ by $$E_{r4-3\times r2}=\begin{bmatrix}1& & & &\& 1 & & &\ &&1&&\&-3&&1&\&&&&1\end{bmatrix}$$
There is only one problem: Naive matrix multiplication requires $O(n^3)$ operations. If we just replace out inner loop with such a matrix multiplication, the overal complexity will be $O(n^5)$ - clearly that would be a bad idea. To fix this we are going to use two “tricks”
1. Collapsing all of the elementary matrices into a single square matrix that carries out all of the row reductions below the current element. For “zeroing” out the column below element $(i,i)$ this looks like an identity matrix with non-zero elements only below element $(i,i)$. This means we are only carrying out $O(n)$ matrix multiplications.
2. Using sparse matrices. Since the matrices we are using are mostly zero, we only want to store, and more importantly multipy with, the non-zero entries of the matrix. This reduces the cost of matrix multiplications from $O(n^3)$ to $O(n^2)$, as the sparse matrix only has at most 2 non-zero entries per row.
Note: sparse matrices are found not in numpy itself but in scipy.sparse where there are multiple formats, depending on whether we are storing matrices by row or column or with some (block) diagonal structure. Here it makes sense to use the row based format (as when we are pre-multiplying with our special matrix, each entry in the answer is the inner product of a column of the right hand side with a row of our special matrix). For a compressed row sparse matrix the internal storage representation is:
• An array start of length number of rows + 1, with start[i] containing the first index of non-zero elements of row i (and start[i+1] giving the end)
• An array col of length number of non-zeros where col[j] is the column of the j’th non-zero entry in the matrix
• An array data of length number of non-zeros so that A[i,j] == data[k] if col[k]==j and start[i]<=k<start[i+1]
We can set such a sparse matrix up by either providing the start, col and data arrays explicitly, or in various other formats as described here.
import numpy as np
def npSolve(A,b):
"just to check that numpy really is faster"
# write an implementation that calls np.linalg.solve
gausstest.speed(npSolve,800)
import numpy as np
from scipy.sparse import csr_matrix
def gaussSparse(A,b):
"Implementation of Gaussian elimination with numpy and sparse matrices"
# write an implementation that uses csr_matrix
gausstest.speed(gaussSparse,1600) # this should be fast enough to allow running up to 1600x1600
## Method 3 - Call an external library
At this point you might think - for this task Python is more trouble than it’s worth. Why not just write the critical parts in a real programming language (say C++)? Just use Python to do what it’s good at: pre-processing data, reporting, quick-and-dirty hacking etc. Fortunately, calling an external library from Python is not hard. Here I will provide you a simple C callable library that implements the basic gaussian elimination and it will be your task to interface to this. Similar patterns are also useful if you are using a commercial library (no source code available) or if you inhert some code from a colleague who still codes in Fortran.
For starters here is the C++ code
%%writefile gauss.cpp
extern "C" { // initial declaration as C style function (don't want C++ function name mangling)
const char *hello(const char *world); // simple test function
double check(const double **A,int i,int j); // returns A[i][j], check if arguments are passed correctly
double gauss(int n,const double **A,const double *b,double *x); // compute x : A x = b, for n x n matrix A
// assumes that x is already allocated in the right size - returns the maximum error
}
#include <stdio.h>
#include <vector>
#include <math.h>
const char*hello(const char *world) { static char buf[1028]; sprintf(buf,"Hello %s",world); return buf; }
double check(const double **A,int i,int j) { // check if arguments are passed correctly
return A[i][j];
}
double gauss(int n,const double **A,const double *b,double *x) // compute x : A x = b, for n x n matrix A & return max error
{
std::vector<std::vector<double> > U(n);
for(int i=0; i<n; ++i){ // copy input data into U
U[i].resize(n+1);
for(int j=0; j<n; ++j) U[i][j]=A[i][j];
U[i][n]=b[i];
}
for(int i=0; i<n; ++i) // do the row reduction
for(int j=i+1; j<n; ++j){
const double mult = U[j][i]/U[i][i];
U[j][i] = 0;
for(int k=i+1; k<=n; ++k)
U[j][k] -= mult * U[i][k];
}
for(int i=n-1; i>=0; --i){ // back-substitution
x[i] = U[i][n];
for(int j=i+1; j<n; ++j) x[i] -= U[i][j]*x[j];
x[i] /= U[i][i];
}
double error=0;
for(int i=0; i<n; ++i){
double sum=-b[i];
for(int j=0; j<n; ++j) sum += A[i][j]*x[j];
if(fabs(sum)>error) error = fabs(sum);
}
return error;
} // end function gauss()
Lets compile this code. If you are under windows and are having trouble with this, there is a pre-compiled .dll on the website (just put it in the same folder as this notebook)
import subprocess
# run: a simple function to execute a command (like os.system) and capture the output
run = lambda cmd: subprocess.run(cmd.split(),stdout=subprocess.PIPE,stderr=subprocess.STDOUT).stdout.decode("utf-8")
print(
run("g++ -o gauss.so -fPIC -shared -O3 -Wall gauss.cpp -lm"), # compile
run("ls -l gauss.so")) # check the library exists
#### Note: in Jupyter notebooks we could just as easily do this with the 'magic' ! commands below
# This is just a bit of magic go compile the C++ code. If you are doing this on windows you would get a DLL (assuming
# you have a proper windows installation that includes enough of cygwin to make windows look like a proper operating system :-)
!g++ -o gauss.so -fPIC -shared -O3 -Wall gauss.cpp -lm
# you should now have a file gauss.so in the same directory as the notebook
!ls -l gauss.so # just checking that the file is here and has the correct timestamp
### Calling an external library
Most of the magic required for interacting with the compiled library can be found in the ctypes module. The key elements are:
• cdll.LoadLibrary("./gauss.so") allows you to load a library. There are some subtle differences between windows (loading a .dll library) and Linux (calling a .so library) but it should work on any system (also Macs). Note: a library can only be loaded once. A second call to LoadLibrary with the same filename will return a link to the same (previously loaded) library. If you change the C++ source code an recompile you need to restart the kernel (or change the filename
• Converting to the correct types: ctypes defines standard ctypes. Many things will convert automatically for example python bytes string to const char *
• When working with C types explicitly you may need to convert. E.g. i=cint(3) and i.value (to go from python int to cint and back again)
• Sometimes we need to specifically set things up correctly. We can use ctypes.c_void_p to get a generic pointer type (void *) or use POINTER(c_int) to create a pointer to an integer (in this case).
• Arrays can be declared directly by multiplying a type by an integer size and using an initaliser list. For example c_int*3 is the same type as int[3] in C/C++. So (c_int*3)() constructs an unintialised array and (c_int * 3)([0,1,2]) will create an initialised array. Just like in C, you can call a function expecting POINTER(c_int) with a an argument of type c_int*3, for example.
• You may need to declare functions including their return type so that python needs how to call the function. For example if you had loaded a library lib containing a function called func then
• lib.func.argtypes = [c_char_p, c_double] says that func(char *, double) is the signature of the arguments
• lib.func.restype = c_char says func returns a char (rather than the default int)
• Alternatively numpy provides a simple way to access the pointer to the underlying memory buffer. If x is a numpy array then x.ctypes.data contains the pointer. See numpy.ndarray.ctypes for more information
• Yet another option is to modify the C code to have a set of simple helper functions for setting up the inputs and querying the outputs with all helper functions just receiving or returning basic values (int or double) that can be passed easily.
Have a go at calling the compiled library. To get started you might want to test that you can load and call simple functions. For example try calling the hello(b"world") function from the library or use the check(A,i,j) function to see if you can pass a matrix and return element A[i][j].
# insert your code here
Now try writing a small wrapper to pass the standard equation data into the C program and return the result as a list
# solution
from ctypes import *
def gaussC(A,b):
"Solve by using the gauss() function from gauss.so/gauss.dll"
import gausstest
gausstest.speed(gaussC,1600) # how fast will this go?
### Discussion
Why is C fast? Can python ever be as fast as C?
## Method 4 - Using Numba
What about compiling Python to give it the speed of C/C++/Fortran or at least something close to it? There are a number of projects that are working towards this goal. Some examples of this include PyPy, PyRex (no longer supported?) and Numba. Each of these make some attempt to compile python to native machine code. Note that there are also some other python implementation such as Jython and IronPython that target the Java Virtual Machine and .NET, but this is more about compatibility with a different software ecosystem than speed. Another issue that can cause confusion is that Python “byte compiles” modules - this is quite different to compiling to machine code with a language such as C or Fortran. It really just turns verbose ASCII source code to a much denser binary format that is still interpreted but much quicker for the interpreter to read (for example it doesn’t contain any of the comments from the original source).
The difficulty with trying to compile Python, rather than interpreting it, is that Python was not designed for this. Hence most attempts at producing a fast, compiled version of Python tends to only be compatible with a subset of python. Here we will use the Numba package, because it is in very active development, can be mixed seamlessly with interpreted Python code, and particularly targets scientific computing applications.
How does Numba work? It does a just-in-time (JIT) compilation of marked functions into machine code. That is we include a special library and “tag” functions to be compiled to machine code the first time that they are run. Such tagged functions can only use a subset of the python language (or else the numba package will fall back to running same code as the interpreter)
Basic usage:
from numba import * #or just import jit, int32,float64
@jit( float64(int32,float64[:,:]), nopython=True)
def f(n,M):
return M[n,n]
What does all that mean?
• @jit is a special decorator for the function that says this function should be just-in-time compiled by numba. In fact this is the only part that is really necessay, the rest of the lines in paranthesis could be left out.
• The first optional argument to @jit provides a type specification for the function. In this case it says that the function returns a float64 (64-bit floating point number or double in C). It takes two arguments the first of which is a 32-bit signed integer (as opposed to say a uint8), the second argument is a numpy matrix. If this is not specified, numba will “guess” the types based on the arguments to the function the first time it is called.
• The same function may be compiled multiple times with different types (either by being called with different arguments or by specifying a list of type declarations). Note that the compile time makes the first time that the function is called relatively slow compared to subsequent calls
• nopython=True tells the compiler that it should not accept any python code that it cannot compile (where it would need to fall back to the python interpreter). Without this @jit will always succeed in “compiling” the function but will potentially produce a function that is no faster than the standard interpreted python code. With the flag set, the compiler will raise an error if the function includes elements of python that it can’t handle.
That should be enough to get you started. See the Numba manual for more information. Go ahead and write a version of the Gaussian elimination solver using Numba - not that you will need a wrapper function to translate the dictionary into numpy arrays first.
import numba
from numba import jit,float64
import numpy as np
def numbaSolve(A,b):
"""Just-in-time compiled gauss-elimination function"""
# write your code here to compute x
return x
def gaussNumba(A,b):
"wrapper to call the function compiled with numba"
# convert argument types and call numbaSolve
return x
gausstest.speed(gaussNumba,1600) # how does this compare with the C++ version?
The strategy employed by Numba is virtually identical to that used in Julia to produce fast code. If you have a project that is entirely focussed on heavy-duty computational work, you may want to give this ago. For small amounts of numerical computation in a larger python project, Numba is entirely adequate though (and still improving rapidly due to ongoing development). For algorithms that are based less on numerical computations with vectors and matrices, you may also run into the limits of what Numba can’t compile - this may require rewriting an algorithm in a slightly less natural (more fortran-ish) way.
Note also that Numba does not compile all of the libraries that you import (though it supports a reasonable subset of the numpy libraries). So if you import some module X and call X.foo() this function will still run slowly.
## Method 5 - Two cores are better than one
Computers are getting more and more cores with every generation (the maxima server has 32, the mathprog group’s server has 72, even your phone probably has 4). So why are we only using 1 core in our python program so far?
### Discussion questions:
• What’s the difference between Multi-threading vs Multi-processing?
• What is Python’s GIL?
• GPU based parallelisation
### Suggested approach
Use Numba library with @jit(parallel=True,nopython=True) (can also explicity set nogil=True for any function to be called in a parallel thread. Indicate that loops are to be done in parallel by using numba.prange(n) instead of range(n): that is a loop for i in prange(n) will be carried out automaticallly by multiple threads. They automatically synchronise (wait for all of the threads to finish) at the end of the for-loop.
When doing the parallisation consider:
• Want reasonable chunks of work in each thread - synchronisation costs time
• Need to avoid conflicts where multiple threads want to write to the same memory: simultaneous read is fine. Avoid needing to use locks
• Limit to 2 threads - partly because of the number of people on the server, and because unless we significantly increase the size of matrices there will be very limited benefit from using many more. To enforce this use:
import os
import numba
Note that this has to be done before you load numba for the first time. If you are working in a notebook where you have used numba previously, please re-start the kernel. If you are working on a command line version of python you could also set the NUMBA_NUM_THREADS environmental variable before you even start python.
# please restart the kernel before executing this for the first time
# (the thread limit is only enforced when numba is first loaded)
import os
import numba
from numba import jit,float64,prange
import numpy as np
def parSolve(A,b):
"Parallel gaussian elimination using numba"
# write parallel version of numbaSolve() here
return x
def gaussPar(A,b):
"Call the parallel version compiled with numba"
# write wrapper function to call parSolve()
return x
import gausstest
gaussPar(*gausstest.randProb(5)) # compile & test code
#### Measuring performance of parallel code
The default speed test will only report CPU time used. This doesn’t tell you whether it has been used by 1 CPU or several. In fact, due to the overhead of parallelisation we expect the CPU time to go up - though hopefully not by very much. At the same time the elapsed time should go down.
gausstest.speed(gaussPar,1600) # by default measures CPU need to test ellapsed time as well
## measure the cost of parallel threads
import numpy as np
import timeit,time,gc
from statistics import mean,stdev
from random import seed,random
def parspeed(method,maxSize=400):
from gausstest import randProb
seed(123456) # fix some arbitrary seed so we keep generating the same data
prev,n = 0.0, 50
gc.disable()
while n <= maxSize:
gc.collect() # manual cleanout of garbage before each repeat
#CPU = -time.process_time()
T = timeit.repeat(stmt="method(A,b)",setup=f"A,b=randProb({n})",
# need timer to be subtractable so make it a vector
timer=lambda : np.array([time.perf_counter(),time.process_time()]),
repeat=5,number=1,globals=locals())
#CPU += time.process_time()
CPU = [ x[1] for x in T]
t = wall = [x[0] for x in T]
nThread = [ c/s for c,s in zip(CPU,wall)]
print("%4d elapsed %10.4f σ=%.2f sec" % (
n,mean(t),stdev(t)),"(x %.2f)"%(mean(t)/prev) if prev > 0 else " "*8,
parspeed(gaussPar,1600) # need to test ellapsed time as well | 7,091 | 29,931 | {"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} | 3.984375 | 4 | CC-MAIN-2021-04 | latest | en | 0.94218 |
https://www.askdifference.com/grader-vs-degrees/ | 1,686,077,758,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00563.warc.gz | 710,968,942 | 21,029 | ## Difference Between Grader and Degrees
A grader, also commonly referred to as a road grader, motor grader, or simply a blade, is a form of heavy equipment with a long blade used to create a flat surface during grading. Although the earliest models were towed behind horses, and later tractors, most modern graders are self-propelled and thus technically "motor graders".
0
#### Degrees➦
One of a series of steps in a process, course, or progression; a stage
proceeded to the next degree of difficulty.
0
0
#### Degrees➦
A step in a direct hereditary line of descent or ascent
First cousins are two degrees from their common ancestor.
0
A piece of heavy equipment used to level or smooth roads or other surfaces to the desired gradient.
0
#### Degrees➦
Relative social or official rank, dignity, or position.
0
A student in a specified class in an elementary, middle, or secondary school. Often used in combination
0
#### Degrees➦
Relative intensity or amount, as of a quality or attribute
a high degree of accuracy.
0
A machine used in road maintenance and construction for leveling large surfaces.
0
#### Degrees➦
The extent or measure of a state of being, an action, or a relation
modernized their facilities to a large degree.
0
A machine used to sort food by size or quality.
0
#### Degrees➦
A unit division of a temperature scale.
0
One who grades, or that by means of which grading is done or facilitated.
the graders of a school examination
0
#### Degrees➦
(Mathematics) A planar unit of angular measure equal in magnitude to 1/360 of a complete revolution.
0
(in combination) One who belongs to a certain grade at school.
0
#### Degrees➦
A unit of latitude or longitude, equal to 1/360 of a great circle.
0
One who grades, or that by means of which grading is done or facilitated.
0
#### Degrees➦
The greatest sum of the exponents of the variables in a term of a polynomial or polynomial equation.
0
A vehicle used for levelling earth, esp. one with a plow blade suspended from the center, used specifically for grading roads.
0
#### Degrees➦
The exponent of the derivative of highest order in a differential equation in standard form.
0
a judge who assigns grades to something
0
#### Degrees➦
An academic title given by a college or university to a student who has completed a course of study
received the Bachelor of Arts degree at commencement.
0
#### Degrees➦
A similar title conferred as an honorary distinction.
0
#### Degrees➦
(Law) A division or classification of a specific crime according to its seriousness
murder in the second degree.
0
#### Degrees➦
A classification of the severity of an injury, especially a burn
a third-degree burn.
0
#### Degrees➦
(Grammar) One of the forms used in the comparison of adjectives and adverbs. For example, tall is the positive degree, taller the comparative degree, and tallest the superlative degree of the adjective tall.
0
#### Degrees➦
One of the seven notes of a diatonic scale.
0
#### Degrees➦
A space or line of the staff.
0
plural of degree
0 | 714 | 3,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.924765 |
https://au.mathworks.com/matlabcentral/cody/problems/2672-largest-geometric-series/solutions/1349531 | 1,582,221,042,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00442.warc.gz | 271,188,210 | 15,768 | Cody
# Problem 2672. Largest Geometric Series
Solution 1349531
Submitted on 19 Nov 2017 by Shawn Mengel
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 = 2*3.^(1:3); b = 3*4.^(0:5); vec = [a b]; output = b; test = gSeries(vec); assert(isequal(test,output));
2 Pass
a = ones(1,50); b = 3*4.^(1:5); vec = [a b]; output = a; test = gSeries(vec); assert(isequal(test,output));
3 Pass
a = ones(1,50); b = randi(5,[1 10]); p = randperm(60); vec = [a b]; vec = vec(p); output = nonzeros(vec==1)'; test = gSeries(vec); assert(isequal(test,output));
4 Pass
a = 2.^(1:15); b = 3.^(1:10); c = 5.^(1:10); vec = [a b c]; p = randperm(35); vec = vec(p); output = a; test = gSeries(vec); assert(isequal(test,output));
5 Pass
a = 2*3.^(1:10); vec = [a a]; p = randperm(20); vec = vec(p); output = a; test = gSeries(vec); assert(isequal(test,output)); | 350 | 971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-10 | latest | en | 0.530634 |
https://socratic.org/questions/are-3c-7-and-7c-3-the-binomial-factors-of-21c-2-58c-21 | 1,620,307,636,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00281.warc.gz | 554,895,413 | 5,624 | # Are (3c+7) and (7c+3) the binomial factors of 21c^2+58c+21?
##### 1 Answer
Sep 10, 2016
Yes they are.
$= \left(3 c + 7\right) \left(7 c + 3\right)$
#### Explanation:
$21 {c}^{2} + 58 c + 21$
$= 21 {c}^{2} + 49 c + 9 c + 21$
$= 7 c \left(3 c + 7\right) + 3 \left(3 c + 7\right)$
$= \left(3 c + 7\right) \left(7 c + 3\right)$ | 170 | 333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-21 | latest | en | 0.287459 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-2-2-2-functions-2-2-exercises-page-183/31 | 1,670,296,411,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00775.warc.gz | 850,712,145 | 12,452 | ## Algebra and Trigonometry 10th Edition
f(x) = -$x^{2}$ + 5 f(-2) = -$(-2)^{2}$ + 5 = -4 + 5 = 1 f(-1) = -$(-1)^{2}$ + 5 = -1 + 5 = 4 f(0) = -$(0)^{2}$ + 5 = 5 f(1) = -$(1)^{2}$ + 5 = -1 + 5 = 4 f(2) = -$(2)^{2}$ + 5 = -4 + 5 = 1 | 138 | 231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-49 | latest | en | 0.316808 |
http://www.jiskha.com/display.cgi?id=1169681088 | 1,496,099,189,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613135.2/warc/CC-MAIN-20170529223110-20170530003110-00508.warc.gz | 665,057,913 | 4,096 | # Chemistry (Check)
posted by on .
Completion
1)1
2)conversion factor
3)change
4)Dimensional anaylsis
5)
Whenever two measurement are equal, or equivalent, a ratio of these two measurements will equal ___1____. A ratio of equivalent measurement is called a ___2___. When a measurement is multiplied by a conversion factor, the value of the measurement ___3____. In ____4___, the units that are a part of the measuerements are used to help solve the problem. The form of the conversion factor that is used is the one in which the unit of the ___5____ is the denominator.
I think 1, 2, 3, and 4, are ok.
For 5, let's look at a conversion.
12 inches = 1 foot
conversion factor = 12 in/1 ft
OR 1 ft/12 inches.
given: 3 ft.
convert to inches.
3 ft x (12 inches/1 ft) = 36 inches.
So the denominator is the part of the conversion factor that must cancel and that is the "given" or "the beginning unit" or some word like that. You probably had that in class or its in your notes. Note that we started with ft and we want to cancel the ft so ft must be on the bottom.
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question | 306 | 1,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-22 | latest | en | 0.921601 |
https://www.tahseenbutt.com/legal-advice/how-to-prove-distributive-law-best-solution.html | 1,670,141,681,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00750.warc.gz | 1,087,264,650 | 10,839 | # How To Prove Distributive Law? (Best solution)
Proof:
1. Assuming that x is in A, then x is also in (A union B) and in (A union C) (A union C). As a result, x is included inside (A union B) intersect (A union C). If x is in (B and C), then x is also in (A union B) because x is in B, and x is also in (A union C) because x is in C. If x is in (B and C), then x is also in (A union B) because x is in B. As a result, x is in (A union B) intersect once more (A union C). This demonstrates that.
## How do you explain Distributive Law?
According to the Distributive Law, multiplying a number by a set of numbers that have been added together is the same as doing each multiplication individually. As a result, the “3” may be “spread” over the “2+4” into three multiples of two and three multiples of four.
## How do you prove distributive property in Boolean algebra?
Distributive Law – This law lets an expression to be multiplied or factored out in order to simplify it.
1. A(B + C) = A.B + A.C (OR Distributive Law)
2. A + (B.C) = (A + B)
3. A(B + C) = A.B + A.C (OR Distributive Law)
4. A + (B.C) = (A + B). (A+C) (AS WELL AS Distributive Law)
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## How do you prove laws of sets?
Proof of the Distributive Law Property of Set Theory It is stated in the first law of sets that taking the union of a set to the intersection of two other sets is equivalent to taking the union of the original set and each of the other two sets individually, and then taking the intersection of the resulting sets Let x be the product of A (B) and C.
## How do you prove Crossive law in cross product?
Demonstration of the Distributive Law Property in Set Theory It is stated in the first law of sets that taking the union of a set to the intersection of two other sets is equivalent to taking the union of the original set and both of the other two sets individually, and then taking the intersection of the resulting sets Given that x = A (B) (C), we have
## How do you write distributive property?
The distributive property asserts that an equation in the form of A (B + C) may be solved as A (B + C) = AB + AC if the expression is presented in the form of A (B + C). It is also possible to apply this distributive property to subtraction, which is represented as A (B – C) = A (B – AC). This indicates that operand A has been spread among the other two operands.
## How do you find the square of 43?
The square root of 43 is equal to 6.557.
## What is K map explain with the help of example?
Example. Boolean algebra functions can be simplified with the use of Karnaugh maps, which are utilized to simplify the functions. Consider, for example, the Boolean function given by the truth table shown in the following example. are the maximum phrases that can be mapped (i.e., rows that have output 0 in the truth table).
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## What is Distributive Law in boolean expression?
It is possible to multiply or factor an expression within the provisions of the distributive law. A(B + C) = A(B + C) = A.B + A.C (OR Distributive Law) A plus (B.C) equals (A + B). (A plus C) (AND Distributive Law)
## How do you prove a boolean expression?
The Theorems of Boolean Algebra
1. De Morgan’s Theorem:
2. Transposition Theorem:
3. Proof: RHS = (A + C) (A’ + B) = AA’ + A’C + AB + CB = 0 + A’C + AB + BC = A’C + AB + BC = A’C + AB + BC = A’C + AB + BC For example, (A + A’)= AB + ABC + A’C + A’BC = AB + A’C = LHS.
4. Example: (AB + BC’) = AB + BC’ = AC + BC’
## How do you use distributive property in sets?
Example 1: Assume that A = 0, 1, 2, 3, 4; B = 1, – 2, 3, 4, 5, 6; and C = 2, 4, 6, 7; and that A = 0, 1, 2, 3, 4; B = 1, – 2, 3, 4, 5, 6; and C = 2, 4, 6, 7; I Demonstrate that A U (B n C) = (A U B) n (A U C) is true (ii) Verify with the help of a Venn diagram.
## How do you prove De Morgan’s Law?
Demorgan’s Law, which is found in set theory, proves that the intersection and union of sets are interchanged when the sets are combined. We can establish De Morgan’s law both mathematically and with the use of truth tables, and we will do so in this section. The first De Morgan’s theorem, sometimes known as the Law of Union, may be demonstrated as follows: Let R = (A U B)’ and S = A’ B’ be two independent variables. | 1,242 | 4,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-49 | latest | en | 0.946805 |
https://brilliant.org/discussions/thread/mehuls-messageboard/ | 1,627,526,380,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00430.warc.gz | 151,695,661 | 27,705 | # Mehul's Messageboard
Hello Brilliantians!
I am Mehul, I study In Class 9 in Delhi. Presently, I am preparing for The RMO, JSTSE and NSEJS.
Since I am nearing 250 followers, I thought that I should create my Messageboard. You can leave Messages and/or Questions for me here!
Thanks To All those Who Follow me! :D
Look forward to More Problems from me ;)
Note:-
I do like carrots and I love mangoes.
Note by Mehul Arora
6 years, 1 month ago
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# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
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"In the unmanifest [timeless (0)-soul] dimension, the ground of [eco-syntax] Being--that perfect, empty no-place [holder "Zero"] where there is only absolute stillness--you could say that God [Timeless Eternity] is peace.
Before the universe [spacetime] was born, resting in that state of [timeless] perfection and ease, it could not have been more [or less] peaceful, because nothing [potentiating everything spaciated] had yet occurred.
But when [timelessness co-arising into "Time"] God decided to become, to take form [in-form OVER ex-form co-gravitational function] this whole process of [universal through personal] creation and destruction, friction and emergence, was set in[carnationally] motion [ionic and ergodic and thermodynamic eco-balancing toward spaciating, then speciating, then individual egoing, then global ecoing struggling with our co-arising Win-Win co-creation].
That is what you are, that is what I am, and that is also the nature of [bicameral Polynomial Time Left dominant over recessive NotPolynomial timeless informing] God, in the manifest realm [Special Case, physical-ecosystemic Exterior Landscape as languaged by egosystemic languaged metaphysical mindbody Interior Landscape, co-arising as co-gravitating].
What does this new understanding of God have to do with love [synergy of life-systems]? This is an important question, because the common [Business As Usual Left-brain dominant] idea is that love [Left] is God [Right], and God [Left] is [pulsating diastatic-pos OVER diastolic-neg] love [Right].
...to many of us, spiritual [natural] love [synergy] means compassion, forgiveness, [co-redemption as co-gravitation] and unconditional [bicameral] acceptance.
That is one kind of love. But that kind of love is the expression of God [Yang, in balancing precessive stage, toward Win-Win diastasis] as Being--the reflection of the mystical revelation that everything is already [potentially and coincidentally, coexperientially] perfect.
What happens to love when God becomes the [r]evolutionary impulse,, or ["struggling with"] Eros?
That's the emergence of a very different form of love [synergetic precessive dynamic]--the expression of God as [Yang optimized Continuous Quality Improvement, Positive Evolutionary Psychology] Becoming.
In an evolutionary worldview, God's [therapeutic messianic] purpose is perpetual development [of health and beauty, goodness and truth], or vertical [ly therapeutic] ascent.
So in this context, the expression of the greatest love is an insistence on higher [eco] development.
It is not the kind of love that's going to accept you as you [ego-centrically Left-brain too-dominantly] are.
It's a kind of love that always wants more [co-investment value in health], and is therefore always challenging to the [overly competitive] status quo of the personal ego and the [eco]culturally [de]conditioned self.
No matter how far you have come, there will always be farther to go [into (0)-interest economic and politically regenerative system incarnation].
This love [synergetic integrity of ecosystemic memory] is infused with [r]evolutionary tension, and it generates [co]-creative [fractal] friction.
The idea of God as [co-diastatic experience and memory of octave truth as beauty] peace and Love [of beauty and health] as [synergetic] compassion is an ancient ideal, one that took root in the human [bicameral] heart and mind long before the [comprehensively polyparadigmatic] knowledge of [Taoist-TippingPoint-Balance as Revolution] evolution emerged.
While it remains as powerful and as relevant ever, [sic] this idea of [Time as spaciated, then speciated, universal positive ubiquitous energy] God is only half of the picture. Now we understand the nature of [Time] God to be both [yinyin (-,-] Being and [Yang+Polynomial] Becoming, [co-arising ZenZero] emptiness and [co-gravitational of mythic romantic resonance] Eros.
And discovering what God [Eco-Spaciated Time] as Eros actually looks like and feels like within [Interior eco-messianic Landscaped Bicameral] us and between [bicameral] us is new territory.
When you embrace this [r]evolutionary interpretation of who and what [and why] God is [stimulating dialectical bicameral eco-balance], then you realize that yes, God is love [synergetic integrity of ecosystemic Win-Win memory] but love is a dynamic and dramatic will toward higher [Earth-centric co-] emergence.
It is [EarthyHuman natural-organic dialectic-systemic] God trying to evolve, through you and through me, and most importantly, through us. "
["us" = all Earth's RNA/DNA health v. pathology struggling toward positive optimization of regenerative eco-systems within cooperative universally synergetic TransParent timeless fractal-networking RealTime Win-Win Group (0)-Soul Geometrics of MindBodies as Time (-,-)1 = Space +/(-,-)0-Cubed, bicamerally reversed as +1/(-,-)0 Ego/Eco Deduct/Induct as Inhale/Exhale reverse of time's seasonal unfolding, pregenitors predicting their progenitor dipolar appositionals.]"
This is what I would like to say to you but I can't. SO, HAPPY BIRTHDAY!
- 5 years, 4 months ago
I actually read that stuff up there. It is pretty cool!
Thanks! :D @Sharky Kesa
P.S. As your well wisher, I would like to inform you that you're about to die a cold blooded murder.
- 5 years, 4 months ago
Plz read https://brilliant.org/discussions/thread/happy-birthday-mehul/ which is written from the depth of my heart
- 5 years, 4 months ago
Do you like carrots?
- 6 years, 1 month ago
Kindly refer to the note :3
- 6 years, 1 month ago
do you like algebra?
- 5 years, 10 months ago
I love algebra.
- 5 years, 10 months ago
Happy birthday Mehul ! :) $\ddot\smile$
Have a great day.
- 5 years, 4 months ago
Thanks Kshitij! :D
I can't have a great day, sadly. Curse SST :P
- 5 years, 4 months ago
Hard luck.. 😂
It might have been the same case with me as well, but am lucky I guess, cos am having my SST exam on my b'day..Not too bad right?? 🤓
- 5 years, 4 months ago
Hahahaahahaha Hard luck xD
When is it by the way? :P
- 5 years, 4 months ago
Spring equinox..the 21st
- 5 years, 4 months ago
Aah okay :)
- 5 years, 4 months ago
Happy Birthday @Mehul Arora
- 5 years, 4 months ago
- 5 years, 4 months ago
Do you like Anime and Manga?
- 6 years, 1 month ago
Not really.
- 6 years, 1 month ago
What have you enjoyed the most on Brilliant?
- 6 years, 1 month ago
Also, The intelligence of youngsters like @Sharky Kesa @Harsh Shrivastava @Prasun Biswas @Satvik Golechha @Archit Boobna :)
- 6 years, 1 month ago
Thanks!!! :) :D
- 6 years, 1 month ago
What do you want to become in future?What are your hobbies?
- 6 years, 1 month ago
I haven't really thought on What I wish to become in the future.
My hobbies:- Playing Basketball, Soccer, Counter Strike, Songs, Mathematics ^_^ and Sleeping xD
- 6 years, 1 month ago
Hey. what about a one on one on counter strike? :D
- 6 years, 1 month ago
Haha, Sure! Anytime! Lemme see how badly I beat you xD
- 6 years, 1 month ago
Bad Joke. :P. btw on which server do you play?
- 6 years, 1 month ago
I play Zona alpha and Balkan. I play mostly deathmatches.
- 6 years, 1 month ago
I challenge you too :3
- 6 years, 1 month ago
Why don't you keep your profile pic of brilliant straight? Is there a specific reason for it? :P
- 6 years, 1 month ago
Actually, I always try to Keep it straight. It always becomes Inclined. I have no idea why :3
- 6 years, 1 month ago
Congrats! its straight now :3 :3 :3
- 6 years, 1 month ago
Yeah. I agree :3 :3 xD
- 6 years, 1 month ago
What is your favourite quote ?
- 6 years, 1 month ago
"As is a tale so is life, Not how long it is but how good it is, is what matters."
I love this quote.
- 6 years, 1 month ago
Nice :)
- 6 years, 1 month ago
Hehe, I know. That's why it is my status ;)
- 6 years, 1 month ago
Do you like bananas?
- 6 years, 1 month ago
Lol, Yeah xD Do you? :P
- 6 years, 1 month ago
Yeah, course. I have a mango tree in my yard so in the summer, I have an almost infinite supply of mangoes so I spend my summer happy. Parlez vous Francais?
- 6 years, 1 month ago
Oui. Je parle francais, Mon ami. :D
- 6 years, 1 month ago
I'm learning French at school. I'm pretty good at it. It won't be long before I'm trilingual. How many languages do you know?
- 6 years, 1 month ago
Same Here! I am also learning French at school. I will also be trilingual soon! ;)
English, Hindi and French!
- 6 years, 1 month ago
Oh, I am a trilingual by perfection and pentalingual, roughly.
- 6 years, 1 month ago
I guess I would be a hexalingual very soon!
Cheers!
- 6 years, 1 month ago
Hi Mehul , how's life?
- 6 years, 1 month ago
Why did you delete your RMO 1990 Problem ,?
- 6 years, 1 month ago
Actually, I did not delete it, Calvin Sir did. I don't have the faintest idea why
- 6 years, 1 month ago
So Mehul,are you the topper of your class?
- 6 years, 1 month ago
Yeah , he is :)
- 6 years, 1 month ago
Guess Nihar has done my job already ;)
- 6 years, 1 month ago
I actually saw you today bro at Enquire. I wasn't so sure it was you so I felt awkward asking. I went two times straight to Rohini and still we didn't talk ;-;
- 5 years, 11 months ago
Oh crap! Where were you man?!
- 5 years, 11 months ago
The guy just next to you in front of the guy with the Man Utd cap. ;-; Btw did you ponder on that logic puzzle which that old man gave us.
- 5 years, 11 months ago
Oh what the f? YOU WERE THAT GUY? i saw you at least 5 times! You should've come met me -_-
Yep, I did. But didn't get anything :P
- 5 years, 11 months ago
Sorry :3. And can we break the planks?
- 5 years, 11 months ago
I don't know anything.
- 5 years, 11 months ago
Dw I think we would catch up when you come to Indirapuram.Anyways for the question,do you go to any coaching or are you a born genius. :')
- 5 years, 11 months ago
Hahahaha, M no genius. i go to FIITJEE, I joined this year.
- 5 years, 11 months ago
Clouds are genius :3 :3
- 5 years, 11 months ago
No they're not :3 :3 :3
Not related to topic, but last night I saw a food dish named "Nihari Ghosht" :3
- 5 years, 11 months ago
Mmmm. ;D
- 5 years, 11 months ago
Hahahaha :P
I didn't eat it though :P :P
- 5 years, 11 months ago
Not done with three chocolates are you? ;D
- 5 years, 11 months ago
Haha, not yet :P
- 5 years, 11 months ago
Am close to 1500 followers :3 :3 :3
- 5 years, 11 months ago
Am close to 350 :3 :3 :3 :3
I don't care about the number of followers you have :3
- 5 years, 11 months ago
Why , no curiosity about followers problem(s)? :(
- 5 years, 11 months ago
............................................
- 5 years, 11 months ago
Lol what's that? Never heard before :3 :3
- 5 years, 11 months ago
Neither have I :P :P
I saw it, I even have a pic :P :v
- 5 years, 11 months ago
which one? i am in dwarka u in??
- 5 years, 7 months ago
also whats ur nso result 2015 its out check it mine is- international rank 13 state rank 2
- 5 years, 7 months ago
geometry is all about construction and imagination
- 5 years, 6 months ago
$\color{#FFFFFF}{\text{Happy Birthday!}}$
- 5 years, 4 months ago
Best Birthday message ever.
- 5 years, 4 months ago
$\color{#FFFFFF}{\text{Just toggle LaTeX, obviously.}}$
- 5 years, 4 months ago
- 5 years, 4 months ago
$\color{#333333}{ }$
- 5 years, 4 months ago
- 5 years, 4 months ago
$\color{#FFFFFF}{\text{Why'd you stop?}}$
- 5 years, 4 months ago
Why are these messages appearing blank?? :/
- 5 years, 4 months ago
Because they are doing it on purpose.
- 5 years, 4 months ago
Oh okay
- 5 years, 4 months ago
BTW, Ashish, how old r u really?
- 5 years, 4 months ago
15 years. My mother is the mentioned one. This is my moms fb acc,
- 5 years, 4 months ago
Class 10?? Level 4 in chem?? x_x
- 5 years, 4 months ago
Haha, because i only attempt questions which i know.£0I dont give wrong answers. I got 2 ques of chem level 4 right.:P
- 5 years, 4 months ago
Impressive :D
- 5 years, 4 months ago
- 5 years, 4 months ago
That's our unique way of communication :P
- 5 years, 4 months ago
Happy birthday mehul. Have a nice day.
- 5 years, 4 months ago
Happy Birthday @Mehul Arora Enjoy your day.
- 5 years, 4 months ago
Thanks so much! @Akshat Sharda
- 5 years, 4 months ago
- 5 years, 4 months ago
HAhahaha I was looking forward to this :P
Thanks! :3
- 5 years, 4 months ago
Do you like Geometry?
- 6 years, 1 month ago
I don't really hate it, But I don't like it much. My favourite topics are NT and Alg :)
- 6 years, 1 month ago
No , he hates it.
- 6 years, 1 month ago
For me Geom is quiet harder than NT and Algebra.
- 6 years, 1 month ago | 4,437 | 14,868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "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} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.854393 |
https://www.stat.math.ethz.ch/pipermail/r-help/2002-April/020231.html | 1,653,319,887,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00112.warc.gz | 1,149,272,011 | 2,303 | # [R] re: pooling categories in a table
Michael Friendly friendly at yorku.ca
Tue Apr 9 15:34:27 CEST 2002
```Perhaps I should have given an example.
# 5 column matrix: Day, Time, Station, State, Freq
tv.matrix<-read.table("C:/R/mosaics/tv.dat")
> tv.matrix[1:5,]
V1 V2 V3 V4 V5
1 1 1 1 1 6
2 2 1 1 1 18
3 3 1 1 1 6
4 4 1 1 1 2
5 5 1 1 1 11
>
tv <- array(tv.matrix[,5], dim=c(5,11,5,3))
dimnames(tv) <-
list(c("Monday","Tuesday","Wednesday","Thursday","Friday"),
c("8:00","8:15","8:30","8:45","9:00","9:15","9:30",
"9:45","10:00","10:15","10:30"),
c("A","C","N","F","Other"), c("O","S","P"))
names(dimnames(tv))<-c("Day", "Time", "Station", "State")
Say I want to collapse the 11 time categories to 3. The real problem
is when I have only the 4-way array (or an equivalent table). But
here, I thought I could just collapse the 2nd column to 3 categories:
tv.matrix[,2] <- 1+ as.integer((tv.matrix[,2]-1) /4)
tv2 <- array(tv.matrix[,5],
dim=c(5,3,5,3))
dimnames(tv2) <-
list(c("Monday","Tuesday","Wednesday","Thursday","Friday"),
c("8:00-8:45","9:00-9:45","10:00-10:30"),
c("A","C","N","F","Other"), c("O","S","P"))
names(dimnames(tv2))<-c("Day", "Time", "Station", "State")
But something is wrong, because the margins are not the same:
> margin.table(tv,1)
Day
Monday Tuesday Wednesday Thursday Friday
21271 20486 19304 19779 17275
> margin.table(tv2,1)
Day
Monday Tuesday Wednesday Thursday Friday
996 912 962 898 771
What am I missing?
-Michael
--
Michael Friendly friendly at yorku.ca
York University http://www.math.yorku.ca/SCS/friendly.html
Psychology Department
4700 Keele Street Tel: (416) 736-5115 x66249
Toronto, Ontario, M3J 1P3 Fax: (416) 736-5814
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```
More information about the R-help mailing list | 811 | 2,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.583311 |
https://www.jiskha.com/display.cgi?id=1377995541 | 1,508,620,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00234.warc.gz | 946,977,597 | 3,372 | # Physics
posted by .
A car accelerates at a constant rate from zero to 29.3 m/s in 10 seconds and then slows to 20 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s^2 during the 15 seconds?
Answer is 1.3....not sure how to get to it.
I am using the formula
a= (V sub2 - V sub1) / (T sub2 - T sub1)
• Physics -
a=(20)/15=1.3
## Similar Questions
1. ### physics help!
A car accelerates at a constant rate from zero to 32.6 m/s in 10 seconds and then slows to 14.8 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds?
A car accelerates at a constant rate from zero to 32.6 m/s in 10 seconds and then slows to 14.8 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds?
3. ### Physics
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9. ### physics
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More Similar Questions | 728 | 2,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-43 | longest | en | 0.949362 |
https://openoregon.pressbooks.pub/techmath/chapter/module-34-radian-measure/ | 1,713,310,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00524.warc.gz | 394,173,775 | 22,129 | You may use a calculator as needed in this module.
In everyday life, we measure angles in degrees. However, degrees are an arbitrary unit of measure; why is a right angle 90 degrees and a full circle 360 degrees?[1]
Another unit that is used in surveying is the gradian; a right angle is equal to 100 gradians. Using gradians makes some calculations easier, but this is still an arbitrary unit of measure.
In this module, we will focus on a third unit of angle measure known as radians.
To explain radian measure, we can visualize a circle. Next, let’s measure a length along the circumference of a circle that is equal to the radius. Finally, let’s draw an angle with its vertex at the center of the circle, intersecting the circle at the endpoints of this arc. The measure of this angle is defined to be 1 radian.
In fewer words, a radian is the measure of the central angle that forms an arc equal to the circle’s radius.
Because the circumference of a circle is times the radius, there are radians in one full circle, and therefore radians are equivalent to .
We can convert between degrees and radians using the conversion factor .
Exercises | 249 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-18 | longest | en | 0.92725 |
https://talkingsticklearningcenter.org/problem-solving-1-a-new-group-of-students/ | 1,652,762,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00735.warc.gz | 634,172,357 | 19,759 | # The Harmony Learning Community Blog
News, Updates, Program Recaps, and Homeschooling Information
### PROBLEM SOLVING #1: A New Group of Students
March 2, 2017
As I prepare for today’s math circle, I realize that I am in a privileged position of utmost responsibility: only one person in this group of six has ever been to a math circle before. For years, the majority of students in our circles have participated for years. I hope I am up to the task.
### FUNCTION MACHINES
“I know something we can do while we’re waiting for everyone to arrive,” I said. “Do you know what a machine does?” I asked. The students discussed the question, then helped me draw a new machine on the board. I drew an enclosed region with curvature in its border (a.k.a. a blob) and asks the kids to suggest machine parts to add. After three rounds of play, it ended up named the “Bobby 3,000” and it had hands, a head, a propeller, lights, 3D glasses, a pail of poison, and more. In round 1, when you put in 1 out came 2, when you put in 100 out came 101, and so on. The students’ job was to figure out the rule that was being applied to these numbers, and they did so with glee.
### WHAT IS A PROBLEM?
Soon everyone had arrived. I asked the students some questions: How were words invented? When? Why? After that discussion, I asked “How and why do you think the invention of the word problem come about?” The students agreed that the word is to describe difficult situations, “like when you get a thorn in your thumb,” added S.“This class is called Problem Solving. Why do we use the word problem in math?” The students said that in math, you can get into troubling situations that are unpleasant. “But the Function Machines we just did were problems, right? You had to figure out the rule. That was a problem. But it didn’t seem like having a thorn in your thumb. Doing function machines was kinda fun.” Eyes opened wider. Ah!
“This class is called Problem Solving. Why do we use the word problem in math?” The students said that in math, you can get into troubling situations that are unpleasant. “But the Function Machines we just did were problems, right? You had to figure out the rule. That was a problem. But it didn’t seem like having a thorn in your thumb. Doing function machines was kinda fun.” Eyes opened wider. Ah!“There’s another kind of problem,” proposed C. “When it’s fun to try to figure out the solution to something.” Everyone agreed. I showed them the book we are going to use in this course: Avoid Hard Work. I read to them the definition of
“There’s another kind of problem,” proposed C. “When it’s fun to try to figure out the solution to something.” Everyone agreed. I showed them the book we are going to use in this course: Avoid Hard Work. I read to them the definition of a problem from the book: “The word problem comes from the word probe, meaning inquiry.” We talked about that.“Why would you want to avoid hard work in math?” L laughed out loud. It was obvious to the kids.
“Why would you want to avoid hard work in math?” L laughed out loud. It was obvious to the kids.
I read a few playful mathematical inquiries from the book, then told them that I was going to give them two different problems and that their job would be to choose which one they would rather work on.
### Technique: HAVE AN EMOTIONAL REACTION
Every problem in the book lists as the first step in the solution “React emotionally to the problem.”* The first problem was about pins. After we discussed what pins are, I read“A pin has two ends: one called a head and one called the point. Penny likes to arrange pins in loops For example, here is a picture of six pins in a loop. This picture has two places where a head and a head meet, two places where a point and a point meet, and two places where a point and a head meet. One time Penny arranged ten pins in a loop. She told me that in her picture there were FIVE places here a head and a point met. Is she remembering correctly?” (p29-30)
“A pin has two ends: one called a head and one called the point. Penny likes to arrange pins in loops For example, here is a picture of six pins in a loop. This picture has two places where a head and a head meet, two places where a point and a point meet, and two places where a point and a head meet. One time Penny arranged ten pins in a loop. She told me that in her picture there were FIVE places here a head and a point met. Is she remembering correctly?” (p29-30)
I immediately asked the students how they felt, and wrote in on the board:
• Good
• Crazy
### Technique: DO SOMETHING
Then I read the other problem. I didn’t tell the students that both problems are identical mathematically. Avoid Hard Work lists the second step to problem-solving as “do something about it.” Specifically here, “Make the problem into a story. Children need a very strong reason for why something needs to be done (as do we all.) It also has to be personal to them: that is, it must touch upon their interests, hobbies, or favorite objects.” (p30) The challenge here, of course, is that I don’t know these kids at all. I have no idea what’s important to them. The older my own children get, the harder it is for me to know what’s important to kids of a certain age.**
An ogre lives in an enchanted wood. The ogre is not friendly because he has no friends. He wants to feel special and important, but no one treats him as though he is special or important. (What should his name be?) He devises a plan to get some friends. He steals something very special from some people in the town and locks what he steals into a vault. (What would he steal from you?) He creates two portals with challenges to get the people to come to him to retrieve their belongings. He plans to have a party for them at the end.
The challenges involve passing through portals. At the first portal, to get through, 6 people have to lie on the ground in a circle. At 2 of the places where the people touch, a head and a head meet. At 2 of the places where the people touch, feet and feet meet. At 2 of the places where people touch, a head meets feet. There is a sign posted showing how it can be done.
The second portal says this: to pass through, you must arrange 10 people in a circle. At 5 places, a head meets feet. How can this be done?***
As soon as I said the question, M started counting the people in the room. We had 10, including adults and siblings. “We’re ready to act this out!” said L.
“Before we do anything,” I said, “let’s talk about how it made you felt when you read it.”
• Terrible
• Good
• Terrified
• Creeped out
• Scared
• Happy
• It made me want to run into a wall and die
These varying reactions made it hard to tell whether the storytelling enhanced the problem or created more barriers to entering the mathematics of it. Did I go too far with the ogre? “If you had to choose one problem to solve, which would it be, the Penny Problem or the Ogre Problem? The Ogre problem definitely got more votes, but the Penny problem was attractive too. It already had a bit of a story embedded in it, so it wasn’t just pure mathematics.
“Which one are we going to do,” asked the students excitedly, as we ran out of time for today. We’ll start with the Ogre problem next week.
### REALLY OWNING THIS PROBLEM
The students seemed to be having a great time as I read the Ogre problem, especially the part about what the ogre would have taken from you. What do you value most? What is so important that you would go after the ogre in order to get it back? After suggesting some things that got crossed off the list quickly (like pizza), the students’ list looked like this:
• Gold
• Cake
• Detergent
• \$1,000
• Golden glass
• T-rex tooth
• Computer
• Nerf gun
Looking forward to continuing problem solving next week!
-- Rodi
* One of my goals for the six weeks of this math circle is for students to make a habit of first acknowledging their own feelings about math problems. That really can unleash a flow of problem-solving energy. I work with a lot of high school students who have never been taught this. I think many of them could have had more successful and pleasant math experiences over the years had they known that it’s normal to have an emotional reaction to math.
**Many math circle leaders over the years have asked me how to know what’s important to kids of a certain age. It used to be easier for me. Now I would recommend going to a book store to peruse the children’s section as a good first step.
***You may notice that I changed the problem from can it be done to how can it be done? I did this because it’s our first session and I didn’t want too much frustration at first. | 2,012 | 8,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-21 | longest | en | 0.977662 |
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# Investigating the effect of resistance on a capacitor circuit.
Extracts from this document...
Introduction
Investigating the effect of resistance on a capacitor circuit
Method: We will set up the following circuit. We will measure the capacitor pd. (Vc) with the cell connected.
Then we will remove the cell and connect point A to point B, at the same moment starting a stopwatch. We will record the length of time (t) for the Vc
Middle
153
330
320
680
584
Conclusion:
From the graph we can clearly see that the time taken for the capacitor to discharge is directly proportional to the resistance. This is because the graph shows a definite straight line going through or near most of the points. This means that the higher resistor you use the longer it will take for the capacitor to discharge. The experiment has therefore proved the prediction correct i.e. the resistance should be directly proportionate to the time taken for the capacitor to de-charge. This can be explained by the following: Capacitors store electrical charge. When current is passing through the circuit the capacitor charges up as the current can't jump between the gap of the two
Conclusion
There was only one result, which did no fit in with the main pattern, it was the value at 680kΩ. This may be due to many reasons. Firstly because the resistor value is so high even at a 2% error margin the resistor could be up to 13.6kΩ different from the supposed value. Also the time lapse between connecting A to B and starting the stopwatch may have accounted for the discrepancy.
This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.
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# Related AS and A Level Electrical & Thermal Physics essays
1. ## Experiment: Decay of Charge in a Capacitor
4 star(s)
Also, the discharge of a combination of two identical capacitors should show a doubled or halved time constant depending on how they are connected. This can also be tested by similar experiment settings. Detailed procedures are listed below. Procedures: 1.
2. ## Investigating the effect of 'length' on the resistance of a wire
Resistance = constant (resistivity) �length Area of cross-section If we re-arrange the formula above then we can also calculate the resistivity if we know the resistance: Resistivity = resistance � area of cross-section Length Resistivity = gradient � area of cross-section Resistivity = 54.84 � 9.5�10-9 = 5.21 � 10-7
1. ## Plotting the decay curve of charge in a capacitor
The time t was recorded when the reading of the microammeter I is decreasing in steps of 10�A. The readings were tabulated. 11. A graph of the discharging current I against time t was plotted. Data analysis and Results A.
2. ## Investigating the Smoothing Effect of a Capacitor on a Resistive Load
In order to overcome this problem I had to adjust the CRO settings using the trigger and the hold-off controls so that the trace would remain stable long enough for me to record precise values for the ripple voltage and the frequency.
1. ## Investigating the E.m.f and Internal Resistance of 2 cells on different circuit Structures.
therefore for two cells the e.m.f should double and I got the equation (when external resistor = 12 ohm) e.m.f = (I*2r) + potential difference of external resistor - using figures e.m.f = (2*1*0.5) + (0.5*12) E = (1 + 6)
2. ## Investigating the Effect of Concentrations of Solutions
The process that we are actually carrying out is electrolysis of the water the anode and cathode are carbon rods which will also separate the sodium and chloride ions which may affect the resistance of the solution if left on too long because the chlorine will evaporate but sodium will be left in the solution.
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# Number 10483219
ten million four hundred eighty three thousand two hundred nineteen
### Properties of the number 10483219
Factorization 10483219 Divisors 1, 10483219 Count of divisors 2 Sum of divisors 10483220 Previous integer 10483218 Next integer 10483220 Is prime? YES (694559th prime) Previous prime 10483211 Next prime 10483223 10483219th prime 188624507 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 100111111111011000010011 Octal 47773023 Duodecimal 3616817 Hexadecimal 9ff613 Square 109897880601961 Square root 3237.7799492862 Natural logarithm 16.165286346183 Decimal logarithm 7.0204946585332 Sine 0.69980492666842 Cosine 0.71433400073817 Tangent 0.97966067126199
Number 10483219 is pronounced ten million four hundred eighty three thousand two hundred nineteen. Number 10483219 is a prime number. The prime number before 10483219 is 10483211. The prime number after 10483219 is 10483223. Number 10483219 has 2 divisors: 1, 10483219. Sum of the divisors is 10483220. Number 10483219 is not a Fibonacci number. It is not a Bell number. Number 10483219 is not a Catalan number. Number 10483219 is not a regular number (Hamming number). It is a not factorial of any number. Number 10483219 is a deficient number and therefore is not a perfect number. Binary numeral for number 10483219 is 100111111111011000010011. Octal numeral is 47773023. Duodecimal value is 3616817. Hexadecimal representation is 9ff613. Square of the number 10483219 is 109897880601961. Square root of the number 10483219 is 3237.7799492862. Natural logarithm of 10483219 is 16.165286346183 Decimal logarithm of the number 10483219 is 7.0204946585332 Sine of 10483219 is 0.69980492666842. Cosine of the number 10483219 is 0.71433400073817. Tangent of the number 10483219 is 0.97966067126199
### Number properties
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Examples: 3628800, 9876543211, 12586269025
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© 2021 numberempire.com All rights reserved | 694 | 2,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.679004 |
https://www.solutioninn.com/study-help/artificial-intelligence-modern/consider-the-problem-of-finding-the-shortest-path-between-two | 1,695,843,332,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00148.warc.gz | 1,087,454,605 | 19,276 | # Consider the problem of finding the shortest path between two points on a plane that has convex polygonal obstacles as shown in Figure 3.31. This is an idealization of the problem that a robot has to solve to navigate in
Consider the problem of finding the shortest path between two points on a plane that has convex polygonal obstacles as shown in Figure 3.31. This is an idealization of the problem that a robot has to solve to navigate in a crowded environment.
a. Suppose the state space consists of all positions (x, y) in the plane. How many states are there? How many paths are there to the goal?
b. Explain briefly why the shortest path from one polygon vertex to any other in the scene must consist of straight-line segments joining some of the vertices of the polygons. Define a good state space now. How large is this state space?
c. Define the necessary functions to implement the search problem, including an ACTIONS function that takes a vertex as input and returns a set of vectors, each of which maps the current vertex to one of the vertices that can be reached in a straight line. (Do not forget the neighbors on the same polygon.) Use the straight-line distance for the heuristic function.
d. Apply one or more of the algorithms in this chapter to solve a range of problems in the domain, and comment on their performance.
Figure 3.31
## This problem has been solved!
Do you need an answer to a question different from the above? Ask your question!
Related Book For
Question Details
Chapter # 3
Section: Exercises
Problem: 7
Posted Date: January 05, 2019 11:04:22 | 346 | 1,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.95332 |
https://www.airmilescalculator.com/distance/ric-to-mod/ | 1,611,798,609,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00099.warc.gz | 646,716,559 | 90,914 | # Distance between Richmond, VA (RIC) and Modesto, CA (MOD)
Flight distance from Richmond to Modesto (Richmond International Airport – Modesto City–County Airport) is 2373 miles / 3819 kilometers / 2062 nautical miles. Estimated flight time is 4 hours 59 minutes.
Driving distance from Richmond (RIC) to Modesto (MOD) is 2874 miles / 4625 kilometers and travel time by car is about 47 hours 37 minutes.
## Map of flight path and driving directions from Richmond to Modesto.
Shortest flight path between Richmond International Airport (RIC) and Modesto City–County Airport (MOD).
## How far is Modesto from Richmond?
There are several ways to calculate distances between Richmond and Modesto. Here are two common methods:
Vincenty's formula (applied above)
• 2373.142 miles
• 3819.201 kilometers
• 2062.204 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2367.542 miles
• 3810.189 kilometers
• 2057.338 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Richmond International Airport
City: Richmond, VA
Country: United States
IATA Code: RIC
ICAO Code: KRIC
Coordinates: 37°30′18″N, 77°19′10″W
B Modesto City–County Airport
City: Modesto, CA
Country: United States
IATA Code: MOD
ICAO Code: KMOD
Coordinates: 37°37′32″N, 120°57′14″W
## Time difference and current local times
The time difference between Richmond and Modesto is 3 hours. Modesto is 3 hours behind Richmond.
EST
PST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 260 kg (574 pounds).
## Frequent Flyer Miles Calculator
Richmond (RIC) → Modesto (MOD).
Distance:
2373
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2373
Round trip? | 502 | 1,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-04 | longest | en | 0.823172 |
https://www.physicsforums.com/threads/infinite-series-test-ratio-test-get-1.426011/ | 1,603,968,879,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00264.warc.gz | 861,774,899 | 17,441 | # Infinite Series Test (Ratio Test get 1)
## Homework Statement
Test if the infinite series converge or diverge.
## Homework Equations
$$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}$$
## The Attempt at a Solution
I tried Ratio test:
$$a_{n+1} = \frac{4n+7}{(n+1)(n+2)(n+3)}$$
$$a_{n} = \frac{4n+3}{n(n+1)(n+2)}$$
$$\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{4n+7}{(n+1)(n+2)(n+3)} \times \frac{n(n+1)(n+2)}{4n+3} = \frac{n(4+7n)}{(n+3)(4n+3)} = \frac{4n^{2}+7n}{4n^{2}+15n+9}$$
$$lim_{n\rightarrow\infty} \left|\frac{a_{n+1}}{a_{n}}\right| = lim_{n\rightarrow\infty} \frac{4+\frac{7}{n}}{4+\frac{15}{n}+\frac{9}{n^{2}}} = \frac{4}{4} = 1$$
The answer is inconclusive, and I can't seem to think of any other test yet.
Anyone can help me with this?
I will much appreciate it. Thanks!
Last edited:
Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
Think about what the terms look like for large n. Try a comparison test.
Think about what the terms look like for large n. Try a comparison test.
This is what I done:
$$n^{3} > n(n+1)(n+1)$$
$$\frac{4n+3}{n^{3}} < \frac{4n+2}{n(n+1)(n+2)}$$
$$\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}$$
Both converge
Therefore, $$\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}$$ converges.
Is that correct?
Last edited:
How is n^3 > n^3 + 2n^2 + n?
How is n^3 > n^3 + 2n^2 + n?
Isn't this the same with the $$n^{3} > n(n+1)(n+1)$$ above?
Mark44
Mentor
Isn't this the same with the $$n^{3} > n(n+1)(n+1)$$ above?
Yes, so why do you think that n3 > n(n + 1)(n + 2)?
Yes, so why do you think that n3 > n(n + 1)(n + 2)?
Opps, sorry,
found the careless mistake.
should be
n(n+1)(n+2) > n3
so
$$\frac{4n+3}{n(n+1)(n+2)} < \frac{4n+3}{n^{3}}$$
Thanks for pointing me out
$$\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}$$
Both converge
Therefore, $$\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}$$ converges.
(use the term by term comparison test)
$$\frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}$$
(use the term by term comparison test)
$$\frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}$$
How do you get the 7n?
Mark44
Mentor
4n + 3 <= 7n for all n >= 1
4n + 3 <= 7n for all n >= 1
Got it!
Thanks!
So, the final answer should look like this:
$$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}$$
n(n+1)(n+2) $$\geq$$ n3
$$\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}$$
4n+3 $$\leq$$ 7n , for all n $$\geq$$ 1
$$\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}$$
$$\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
Converge P-series (p > 1)
According to Comparison test,
since $$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
$$and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges$$,
$$therefore \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.$$
Dick
Homework Helper
So, the final answer should look like this:
$$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}$$
n(n+1)(n+2) $$\geq$$ n3
$$\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}$$
4n+3 $$\leq$$ 7n , for all n $$\geq$$ 1
$$\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}$$
$$\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
Converge P-series (p > 1)
According to Comparison test,
since $$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
$$and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges$$,
$$therefore \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.$$
Very nice.
Very nice.
Thanks,
and thanks for everyone that helps. | 1,640 | 3,532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-45 | latest | en | 0.742119 |
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# Problem I. 353. (September 2014)
I. 353. The game Egyszámjáték'' (The Smallest Number Wins Game) was invented by the Hungarian mathematician and psychologist László Mérő: in this game each player submits a positive integer between 1 and 10000, and these numbers are collected. Numbers that have been submitted more than once are deleted. The smallest number among the remaining ones wins.
Example. Players submit the following numbers: 9, 6, 3, 7, 4, 6, 1, 6, 8, 1, 6, 5, 7, 9, 4, 2. We then remove duplicated elements. The remaining numbers are: 3, 8, 5. The player who submitted the smallest such number (3) wins.
Prepare a sheet to evaluate the game based on the numbers submitted by players in email, and to determine the winner. The file tippek.txt (UTF-8 encoded and downloadable from our web page) contains the data for players and the numbers they submitted. For each player we have the following data:
Tipp The integer submitted by the player, $\displaystyle 1\le \mathsf{tipp}\le 10\;000$;
Név The player's name (there can be more players with the same name);
E-mail The email address of the player, this address is unique.
If there are multiple submissions from the same email address, then only the first submission is taken into account (valid'') and the others are considered as invalid. You should use a spreadsheet application to solve the following tasks, but user-defined functions or macros cannot be used.
Open the tabulator-separated text file tippek.txt in the application according to the example. It is enough if your solution handles the data given in this file correctly. Your work should be saved as i353 in the default application file format. You should format the sheet according to the example, and give your answers in the cells adjacent to the messages in columns D:G by using expressions. Auxiliary computations can be performed in cells to the right of column H.
1. The winning number (Nyertes szám'') should appear in cell E2.
2. The name and email address of the winner (Nyertes neve'') should appear in cells E3 and G3, respectively.
In the following tasks you should determine some other values and numbers.
3. In cell E5 determine the number of valid submissions (Tippek száma'').
4. By using a function, determine in cell E6 the number that has been validly submitted the most often (Leggyakoribb tipp''). If there are multiple such numbers, give only one instance.
5. By using a function, determine in cells E7 and E9 the smallest and the largest valid numbers submitted (Legkisebb tipp'' and Legnagyobb tipp'').
6. In cells E8 and E10, respectively, display the player names (2. tippelője'') who submitted the smallest and the largest valid numbers for the second time. If there is no second submitter, the message nincs'' (= none) should appear.
7. You should format the sheet according to the example.
The spreadsheet (i353.xls, i353.ods, ...) together with a short documentation (i353.txt, i353.pdf, ...) describing the name and version number of the spreadsheet application should be submitted in a compressed file (i353.zip).
Downloadable file: tippek.txt
(10 pont)
Deadline expired on October 10, 2014.
### Statistics:
24 students sent a solution. 10 points: Dombai Tamás, Géczi Dániel, Gercsó Márk, Kiss 107 Ádám, Kovács 246 Benedek, Kovács Balázs Marcell, Mócsy Miklós, Piller Trisztán, Radnai Bálint, Tóth Márk Andor. 9 points: Bálint Martin, Fényes Balázs, Kelkó Balázs, Lencsés Ádám, Németh 729 Gábor, Olexó Gergely, Szabó 524 Tímea. 8 points: 3 students. 7 points: 1 student. 6 points: 1 student. 5 points: 1 student. 4 points: 1 student.
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http://www2.nlsd122.org/c/index.php/schools/liberty-jr-high/liberty-staff-members/cathleen-kollross/block-713/ | 1,521,346,124,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645513.14/warc/CC-MAIN-20180318032649-20180318052649-00138.warc.gz | 518,332,432 | 24,297 | NLSD
Liberty
Junior High School
Ms. Kollross
7th Grade Math - Liberty Jr. High
ckollross@nlsd122.org
Week of 3/12/2018
Friday, March 16, 2018: We began our day today with Minute Math #59. Then the kids I went over any questions they had on #1-7 from the Lesson 3-6 Assignment or 1-10 of the Topic 3 Review . We completed the remaining portion of Lesson 3-6 that involved a critical thinking situation which the kids excelled in solving! They were then given some personal work time to continue the Lesson 3-6 Assignment or work on the rest of the Topic 3 Review, both of which are to be done to their best efforts for Monday's class. Don't forget our Topic 3 Test is Wednesday, March 21. Prepare your study card if you haven't, Young One! Enjoy your St. Patrick's Day and your weekend.
Thursday, March 15, 2018: After the kids worked collaboratively on the Team Problem of the Day, we continued with Lesson 3-6: Markups and Markdowns. We discussed the kids Launch ideas and reflections and then continued on with ways to solve for markups and markdowns seen in the retail world. I asked the kids to try #1-7 for tomorrow's class. They were also given time to start the iTopic 3 Review and are encouraged to try #1-10 for tomorrow's class since their test is next Wednesday the 21st.
Wednesday, March 14, 2018: Happy Pi Day! We started our day together with Minute Math #58, and then the kids and I discussed any questions they had from the rest of the Lesson 3-5 Assignment. When they were individually ready, the kids then worked on the Lesson 3-5 In-Class Assignment. They were asked to partner up after finished to take a look at our next section Lesson 3-6: Markups and Markdowns. I asked them to discuss and work together on the Launch and Reflection to see what we would be exploring next.
Tuesday, March 13, 2018: After the Team Problem of the Day, the kids I went over any problems from the start of the Lesson 3-5 Assignment they wanted. We then completed Part Three and they worked within their teams on the Close &Check as a self-evaluation. I checked in with them when they were done, and they were encouraged to complete the rest of the Lesson 3-5 Assignment for tomorrow's class during this class time.
Monday, March 12, 2018: We started our first day of the week with Minute Math #57, and we then went over the kids' ideas from the Lesson 3-5 Launch and Reflection. We then went over any questions the kids had from Puzzle 7.17 ("What Do You Get When...") and this was then handed in for completion. We worked with Part One and Part Two of Lesson 3-5: Percent Increase and Decrease. The kids were asked to try #1-6 for tomorrow's class.
Week of 3/5/2018
Friday, March 9, 2018: We practiced with the formula for Percent Increase/Decrease (Percent of Change) today for Lesson 3-5. So the kids felt like they truly "knew" this formula, I gave them Puzzle 7.17 ("What Do You Get When You...") to work on together. They were asked to bring this completed to the best of their abilities to Monday's class.
Thursday, March 8, 2018: The kids were given time to complete a practice PARCC Student Response and then time to "clean-up" any missing assignments (i.e. puzzle 7.20, paper/online 3-4 assignment, 3-4 in-class assignment). We then began Lesson 3-5: Percent Increase and Decrease. We will truly dive into this lesson tomorrow.
Wednesday, March 7, 2018: After the each student finished the Lesson 3-4 In-Class Assignment, we reviewed the expectations for a successful PARCC Student Math Response. The kids were given a choice of 7 prompts to complete, choosing one to complete. If they did not finish today, they can tomorrow.
Tuesday, March 6, 2018: We will continue our shortened class schedule due to PARCC testing through Monday the 12th. We started class be reviewing any questions from the Lesson 3-4 Assignment. The class was given time to continue this assignment, to be completed for tomorrow's class.
Monday, March 5, 2018: We didn't have a bell ringer today due to our shortened class schedule because of PARRC Testing. We then went over any questions the kids had over Puzzle 7.20, which was to be turned in today if completed. The kids were given time to then start the Lesson 3-4 Assignment, to be completed for Wednesday's class. They were asked to try to get through #1-7 for tomorrow.
Week of 2/26/2018
Friday, March 2, 2018: Today after Minute Math the kids and I continued to discuss compound interest, today focusing on compound interest that is earned more than once a year. This developed the compound interest formula and all students have been greatly encouraged to keep a study card throughout the entire topic, but especially for this particular lesson. After working through the Student Companion, I asked the kids to try the rest of Puzzle 7.20, #4-8 and 10. Enjoy your weekend and the full moon tonight, Kids!
Thursday, March 1, 2018: After the Team Problem of the Day, the kids partnered to practice the Math Student Response as in the PARCC test. They chose one previous problem to solve and discuss. We will discuss these this next week.
Wednesday, February 28, 2018: Happy last day of second trimester, Kids! It is truly difficult to believe we are already 2/3 through the school year. They are a fantastic bunch to work with. We started class today with Minute Math #56 and then began our exploration through the compound interest formula. I thought they could use more practice with "interest compounded annually" formula so I gave them a puzzle (Puzzle 7.20 "What Did Mrs. Zog...Himilayas?". I asked them to work on #1-4 & 9 for tomorrow's class. We will then explore the concept of "interest compounded more than annually" in our next class. We can do this, Kids!
Tuesday, February 27, 2018: After our Team Problem of the Day (one way we practice for the PARCC test), I gave the kids 20 minutes on the timer for some "clean up" time. This gave the kids time to complete any redo work/make-up work that had yet to be done. I also encouraged them to clean out their binders/folders since tomorrow is the last day of second trimester. I told the kids they have no later than FRIDAY to turn in any redo/make-up work or take a retake test. After this time, we then had a healthy discussion about compound interest and how this applies to adult life. I showed them one of my current credit card bills and they had several ideas, questions, and comments about it. They needed this background knowledge before we jump into Lesson 3-4 during tomorrow's class.
Monday, February 26, 2018: Today the kids and I started our day with Minute Math #55. We then went over any questions the kids had over the Lesson 3-3: Simple Interest Assignment started this past Thursday. When the kids were individually ready, they completed the Lesson 3-3In-Class Assignment. Reminder: all redo/make-up is asked to be turned in by the last day of the trimester, this Wednesday. I can take any redo work through Friday, but that is as far as I can go since my final grades are due. Do your best, Kids!
Week of 2/20/18
Friday, February 23, 2018: After Minute Math #54, the kids and I continued to discuss and explore Lesson 3-3: Simple Interest. They completed Part Two and Part Three practicing their algebra skills throughout. I asked them to attempt the Lesson 3-3 Assignment for Monday's class.
Thursday, February 22, 2018: We started our day today with our PARCC Team Problem of the Day. We then discussed the Launch and Reflection from Lesson 3-3: Simple Interest. We then discussed and noted the simple interest formula and worked through Part One. I asked the kids to try #1-4 for tomorrow's class.
Wednesday, February 21, 2018: After Minute Math #53, the kids and I first went over their Fall to Winter AIMS growth charts I printed out for them. We then went over any questions they wanted from the Lesson 3-2 Assignment. Lesson 3-2 In-Class Assignment.
Tuesday, February 20, 2018: After today's Team Problem of the Day, all of the kids received their Topic Two Tests back to be reviewed and signed by a parent/guardian by Thursday's class, please. They also received their Topic 2 Homework Packets back and their Lesson 3-1 In-Class Assignment. We then continued with Lesson 3-2, completing Part Two and Three. The kids are going to try the rest of the lessons problems for tomorrow's class.
Week of 2/12/18
Friday, February 16, 2018: We started our day today with Minute Math. We then discussed the Launch and Reflection from Lesson 3-2: Using the Percent Equation. We continued with the lesson, making numerous notes on our Topic Three Study Cards from the Key Concept slide, but more importantly, having worthwhile discussions about tips, commissions, and sales tax which are all parts of this lesson. I asked the kids to try #1-7 for Tuesday's class. Enjoy your three-day weekend!
Thursday, February 15, 2018: After we completed the Team Problem of the Day, the kids and I discussed any questions they had from the Lesson 3-1 Assignment. When they were ready, they completed the Lesson 3-1 In-Class Assignment. They were asked to partner when done to look into Lesson 3-2: Using the Percent Equation by completing the Launch and Reflection for this section. We will start our day after Minute Math tomorrow with this.
Wednesday, February 14, 2018: Happy St. Valentine's Day! Today the kids and I started our day with Minute Math #51 and then discussed any questions from #1-7 of the Lesson 3-1 AssignmentWe then continued the lesson with Part Three and then the Close & Check. The kids were given time to work on competing the assignment for tomorrow's class.
Tuesday, February 13, 2018: We started out our day today with the Team Problem of the Day. We then began to work in Lesson 3-1: The Percent Equation which the kids are familiar with from our proportional relationships work. We worked through Part Two and will compete the lesson tomorrow. I asked the kids to try #1-7 of the assignment for tomorrow's class.
Monday, February 12, 2018: The kids were delighted with Friday's snow day but were even more delighted to take their Topic Two Test today instead of last Friday! Most of the kids seemed to perform well on this, and I hope to have them all graded and back to the kids by Thursday's class. Fingers crossed!
Week of 2/6/18
Thursday, February 8, 2018: After our Team Problem of the Day, the kids were able to ask any questions they wished to review from the Topic Two Review. They were then given a list of options for the rest of their math class: complete make-up/redo work, finish the homework packet to be turned in tomorrow prior to the test, or work with a small group on a review game to prepare further for tomorrow's Topic Two Test. Study well, Young Ones!
Wednesday, February 7, 2018: We started our class today with Minute Math 50! I returned the kids' Lesson 2-5 In-Class Assignments and then we went over any further questions from the Topic Two Review. They were given more time to continue working on the Topic Two Review and if completed, to work with to partner on sample "similar" problems to prepare for the test or work on any redo opportunities.
Tuesday, February 6, 2018: I hope you all enjoyed your three-day weekend and the Super Bowl, too! Today we went over any questions the kids had from the Lesson 2-5 Assignment after they completed the Team Problem of the Day. When they were ready, the kids worked on the Lesson 2-5 In-Class Assignment which some of the class will finish tomorrow. I asked our very own Patriots to continue with the Topic Two Review today, working through at least 10 more problems to best prepare for this Friday's Topic Two Test.
Week of 1/29/18
Friday, February 2, 2018: We started our day today with Minute Math. We then continued working through Lesson 2-5 and the kids worked with their partners. I asked them to do their best to complete the Lesson 2-5 Assignment but are welcome to, as always, come in with any questions for Tuesday's class. I also asked them to work on at least #1-10 of the Topic Two Review. The Topic Two Test will be next Friday, February 9. Begin studying well, Kids!
Thursday, February 1, 2018: After the Team Problem of the Day, the kids and I discussed their Puzzle 7.2 and reviewed any of the rooms' areas they needed. We then worked within Lesson 2-5: Maps and Scale Drawings. We completed the Launch, Key Concept, Part One, and began investigating Part Two. I asked the class to try #1-4 of the Lesson 2-5 Assignment. We will continue this lesson into tomorrow's class.
Wednesday, January 31, 2018: After Minute Math today, the kids and I discussed their measurements of the puzzle rooms from yesterday. Then we had a review discussion of scale to help them prepare not only for Lesson 2-5 which we will begin tomorrow but also to complete the Puzzle 7.2 for tomorrow's class.
Tuesday, January 30, 2018: We did not have our bellringer activity today since we completed the 3 AIMSweb Winter benchmark tests today. I did ask the kids to measure the lengths and widths of each room seen on Puzzle 7.2. We will discuss how to use the scale and find the areas of these rooms in tomorrow's class.
Monday, January 29, 2018: We began this Monday class with Minute Math and then reviewed any questions the kids wanted to go over from the Lesson 2-4 Assignment. As they were then ready, they completed the Lesson 2-4 In-Class Assignment. The kids need a bit of practice with measuring length using the metric system so I have them working on a small partner activity today and into tomorrow to see where their personal measuring skills are at. This will lead us into our next section which explores scale.
Week of 1/22/18
Friday, January 26, 2018: The kids started their class today with Minute Math since we are now back to our normal schedule with Terra Nova testing complete. We then continued to work in and were to complete Lesson 2-4. I asked the kids to try the rest of the problems (#8-End) for Monday's class. Have a wonderful weekend, Kids!
Thursday, January 25, 2018: We did not have class today due to Terra Nova testing.
Wednesday, January 24, 2018: We did not have class today due to Terra Nova testing.
Tuesday, January 23, 2018: We started class today with the Team Problem of the Day. We then continued and completed our discussion of Independent Variable vs. Dependent Variable. Next, we started working in Lesson 2-4: Proportional Relationships and Equations. We will continue this into tomorrow's class.
Monday, January 22, 2018: Out of the ordinary, we started class with a Team Problem of the Day. The kids then asked any questions they wanted from the Lesson 2-3 Assignment. As they were individually ready, the kid then completed the Lesson 2-3 In-Class Assignment. We began a discussion over Independent Variable vs. Dependent Variable, which we will continue tomorrow.
Week of 1/16/18
Friday, January 19, 2018: After Minute Math, the kids and I went over any questions they asked about over #1-6 from yesterday's assignment. Then we completed the rest of the Lesson 2-3 Lesson, and gave them time to work on the rest of the problems from the Lesson 2-3 Assignment for our next class. Enjoy your weekend!
Thursday, January 18, 2018: The kids started their math day today with the Team Problem of the Day. We then discussed the Launch and Reflection from Lesson 2-3: The Constant of Proportionality. We continued on with the Key Concept, Part One, and Part Two. This is similar to "unit rate" which we have studied together, too. I asked the kids to try #1-6 of the Lesson 2-3 Assignment for tomorrow's class.
Wednesday, January 17, 2018: After the kids completed their Minute Math bell ringer activity, we went over any questions from the Lesson 2-2 Assignment. We had a few issues with graphing but all were fixed. As the kids were individually ready, they completed the Lesson 2-2 In-Class Assignment. I asked them to complete the Launch and Reflection for Lesson 2-3: Constant of Proportionality in the Student Companion for tomorrow's class.
Tuesday, January 16, 2018: After our Team Problem of the Day, the kids and I discussed any questions they had from #1-7 of the Lesson 2-2 Assignment started last Friday during class. We then completed this lesson with Part 3 and the Close & Check. I asked the kids to attempt the rest of the problems from this assignment for tomorrow's class.
Week of 1/8/18
Friday, January 12, 2018: We started our class with our Minute Math bell ringer. Then we discussed any questions the kids had from their 10.5 Puzzle which reviewed graphing linear equations, a new concept for the kids. This is something that they will use in Lesson 2-2: Proportional Relationships and Graphs.
Thursday, January 11, 2018: Today we began class with our Team Problem of the Day. After reviewing this, we then went over any questions the kids had over The Coordinate System review sheet we discussed yesterday. The kids turned this in for me to look over, and then we practiced graphing equations by using graph paper and rulers and pasting samples into the kids' foldables. After they felt more comfortable with this new concept, I gave them each a puzzle, #10.5, to work on in class and then attempt to finish for tomorrow's class.
Wednesday, January 10, 2018: After our Minute Math bell ringer, the kids and I discussed any questions they had on the latter half of the Lesson 2-1 Assignment. When the kids were individually ready, they then completed the Lesson 2-1 In-Class Assignment. They were then given The Coordinate System sheet to see what the kids remember from previous years' lessons so we can all feel comfortable with graphing.
Tuesday, January 9, 2018: The kids seemed tired today! Getting back into the groove can be difficult, but we still started our class with our Tuesday bellringer of the Team Problem of the Day. We discussed any item from the first half of the Lesson 2-1 Assignment the kids started yesterday. We then continued Lesson 2-1, completing through the Close & Check. I asked the kids to attempt the last half of the assignment for tomorrow's class.
Monday, January 8, 2018: Happy New Year, Kids! Welcome back to Liberty! We started our day like any other Monday; we began class with Minute Math and then the kids all received their Topic 6 Tests back. Please have this signed by Wednesday's class. We then began working in Lesson 2-1, completing the Launch, Reflection, Part One, and Part Two. I asked the kids to try #1-7 for tomorrow's class.
Week of 12/18/17
Wednesday, December 20, 2017: After the kids turned in their Topic 6 Homework Packets and the optional extra credit slip, the kids completed the Topic 6 Test and also turned in their study cards. I asked them to make their Topic 2 Foldable when they were done with the test. They will receive these tests back the day we return from Winter Break. Happy holidays, Everyone!
Tuesday, December 19, 2017: After the Team Problem of the Day, we reviewed any other questions from the Topic 6 Review after completing the Topic 6 Homework Packet. I reminded the kids that only formulas, charts, and key hints are to be on study cards, not sample problems. You will do well tomorrow, Kids!
Monday, December 18, 2017: The kids started their day today with Minute Math. We then discussed questions from the Topic 6 Review. They were given them more time to then work further on the Review and Lesson 6-7. I asked to do their bests to get through as much of the Topic 6 Review as possible for tomorrow's class. The Topic 6 Test is still scheduled to be completed on Wednesday. Don't gorget to prepare your study cards, Kids!
Week of 12/11/17
Friday, December 15, 2017: After Minute Math, the kids and I reviewed the in-class from Lesson 6-6: Percent Error. The kids then were given time to work on the Topic 6 Review and partner to work through Lesson 6-7: Problem-Solving. The kids asked questions as needed. We will continue this into Monday's class. I recommended to the kids to try their bests to get through #1-20 for Monday's class on the Topic 6 Review.
Thursday, December 14, 2017: The kids started their math class with the Team Problems of the Day. We then went over any questions they had from the Lesson 6-6 Assignment. As they were ready, the kids took the Lesson 6-6 In-Class Assignment which they will receive back tomorrow. The Topic 6 Test will be next Wednesday, December 20. I have assigned the Topic 6 Review so they can do a few problems a day to prepare. They can also prepare a study card to bring into the class for the test if they haven't yet.
Wednesday, December 13, 2017: We started the day with Minute Math and then discussed #1-4, 6-7 from the Lesson 6-6 AssignmentThen we continued with Lesson 6-6: Percent ErrorThe kids did very well using and recording this formula today. Nice work, Kids! I asked them to attempt the rest (#5, 8-15) of the problems from the Lesson 6-6 Assignment for tomorrow's class.
Tuesday, December 12, 2017: After the Team Problem of the Day, the kids and I began to work through Lesson 6-6: Percent Error. They all took to the new formula so well and seem to have a sound understanding of the Key Concept, Part One, and Part Two. They started to work on #1-4 and 6-7 which we will discuss in tomorrow's class.
Monday, December 11, 2017: We started our class with Minute Math and then we discussed #8-15 from the Lesson 6-5 Assignment. After we went over the kids questions and they were individually ready, the kids completed the Lesson 6-5 In-Class Assignment. I asked them to work with a partner to take a look into and work through the Launch and Refection in our next lesson, Lesson 6-6: Percent Error. This is where we will start tomorrow's class.
*Week of 12/4/17
Friday, December 8, 2017: Due to the field trip to the movie theater, we did not have math today. Reminder: please work on #8-15 of the Lesson 6-5 Assignment for Monday's class. Enjoy your weekend, Kids!
Thursday, December 7, 2017: After our Team Problem of the Day, the kids and I first went over any questions they had from #1-7 from the Lesson 6-5 Assignment. We then continued to work through Lesson 6-5: Fractions, Decimals, and PercentsWe completed the Part Three and the Close & Check. I asked the kids to attempt #8-15 for Monday's class. Enjoy the movie tomorrow, Kids!
Wednesday, December 6, 2017: We started our day as we do each Wednesday with Minute Math. Please check PowerSchool to check your child's Converting Fractions-Decimals-Percents "Pop Quiz" score. Of course, retake quizzes can be completed if a redo sheet is done well and your child studies further. Please remind them to ask me any questions they'd like so they are ready for this retake quiz and to also review/prepare a study card. We then began Lesson 6-5: Fractions, Decimals, and Percents. We completed the Launch, Reflection, Part One, and Part Two. I asked the kids to attempt #1-7 for tomorrow's class.
Tuesday, December 5, 2017: After the kids worked together on the Team Problem of the Day, we discussed any questions they had from the Lesson 6-4 In-Class Assignment. Then, the kids worked through Lesson 6-4 In-Class Assignment as they were ready. I asked them to take a look at the Launch and Reflection of the Lesson 6-5 Assignment for tomorrow's class. Take a look at that Supermoon tonight, too, Kids!
Monday, December 4, 2017: After Minute Math, the kids and I discussed any questions they had from #1-7 of the Lesson 6-4 Assignment. Next, we explored more of this lesson using skills learned from the previous lesson to best help their understandings of percents less than 1%. I asked them to attempt #8-11 for tomorrow's class. They were then given their first "pop quiz" (they were thoroughly warned and encouraged to practice as well as utilize their study cards) which they will receive back tomorrow. Have a great evening, Kids!
*Week of 11/27/17
Friday, December 1, 2017: The kids started our day my completing Minute Math. We continued to work through Lesson 6-4: Percents Less than One. I asked the kids to attempt #1-7 of the Lesson 6-4 Assignment for Monday's class. They were then given their "pop quiz" on converting fractions to decimals to percents. These scores are in PowerSchool. Please encourage your child to fill in her or his study card with anything they deemed helpful from the converting fractions to decimals to percents chart we used in class; this chart is pasted in their Topic 6 Foldable.
Thursday, November 30, 2017: After our Team Problem of the Day, the kids and I continued to discuss converting fractions to decimals to percents. We reviewed the Puzzle 7.10 "What Were the Crash Test Dummy's Last Words?" and then this puzzle was turned in for a completion score. We had a bit of independent work within their foldable that they SHOULD study for a possible quiz quite soon! We then started Lesson 6-4: Percents Less Than 1 and will continue this tomorrow.
Tuesday, November 28, 2017: After our Team Problem of the Day, we discussed questions from the Lesson 6-3 Assignment. The kids had several question from this lesson's assignment since it is a bit challenging. When they were individually ready, they were able to complete the Lesson 6-3 Assignment.
Monday, November 27, 2017: We started with our normal Monday bell ringer activity Minute Math. We then continued to work through Lesson 6-3. I asked the kids to try the rest of the Lesson 6-3 Assignment for tomorrow's class.
*Week of 11/13/17
Thursday, November 16, 2017: We started our day together with the Team Problem of the Day. We continued working through Lesson 6-3 and I asked the kids to do their best with #1-4, 7-9 for our next class on Monday, November 27. Enjoy your Fall Break, Kids!
Wednesday, November 15, 2017: After we completed the Minute Math bell ringer activity, the kids shared their "Percents Greater than 100" models sets. They picked the team best to present to the class. Very creative! We then discussed the optional Pi activity/contest; it's fun, Kids! We then began working through Lesson 6-3 and will continue with this tomorrow.
Tuesday, November 14, 2017: After the Team Problem of the Day, the kids and I reviewed any questions they wished to discuss from the Lesson 6-2 Assignment. As they were individually ready, the kids competed the Lesson 6-2 In-Class Assignment and were then asked to "partner" to explore our next lesson on percents greater than 100. They were to complete the Launch and Reflection portions as to explore. I asked them to bring in a model set to visualize percents greater than 100.
Monday, November 13, 2017: We did not have Minute Math today due to our Kollross Cash-In time as well as new seat choices. You have no idea how exciting these two events are to your seventh graders. They all know how to make me smile.
To continue our study of terminating decimals in Lesson 6-2, we worked on the Part Three together in a classroom discussion. Then the kids chose a partner with whom to work cooperatively on the Close & Check. After I checked them out, I asked the kids to attempt #9-15 from the online/paper Lesson 6-2 Assignment for tomorrow's class.
*Week of 11/6/17
Friday, November 10, 2017: After Minute Math, the kids and I talked about 2nd trimester. I asked them to write 3 second trimester goals (not necessarily math-related) for Monday's class. They will also have Kollross Cash-In and new seat choice on Monday. We then explored Lesson 6-2: Terminating Decimals. We completed the Launch, Reflection, Part One and Part Two. I asked the kids to attempt #1-8 for Monday's class. Thank you to all of you who came or helped with today's Veteran's Day assembly!
Thursday, November 9, 2017: After the Team Problem of the Day, the kids chose a partner to play a multiplying decimals game and then a dividing decimals game. I was out at the dentist office today. The substitute teacher reported that the kids enjoyed the games and were very respectful. Thanks,Kids! They could choose from two puzzles to complete for tomorrow's class: "Brain Bending Labyrinth" (multiplying decimals) or ""What's the Point?" (dividing decimals). All redo work is due by this Friday's class. All tests need to be re-taken also by Friday. It is preferred that students come in during Morning Math (starts at 7:30 a.m.) for re-take tests so they do not miss class time. If you prefer your child to stay after school, please contact me so that we can find a date that works for both of us.
Wednesday, November 8, 2017: The class started with our Minute Math bell ringer. Then the kids and I went over any questions from the Lesson 6-1 Assignment that were difficult or needed clarification. As they were individually ready, the kids competed their Lesson 6-1 In-Class Assignment. All redo work is due by this Friday's class. All tests need to be re-taken also by Friday. It is preferred that students come in during Morning Math (starts at 7:30 a.m.) for re-take tests so they do not miss class time. If you prefer your child to stay after school, please contact me so that we can find a date that works for both of us.
Tuesday, November 7, 2017: We started our day with our Team Problem of the Day. Then we went over any questions the kids had over the Lesson 6-1 Assignment thus far. We continued into Part Two of the lesson and completed Part Three and the kids worked in the Close & Check for a bit. I asked them to attempt the rest of the problems in the Lesson 6-1 Assignment for tomorrow's class. Please remember to have your Topic 5 Tests signed for tomorrow, Kids.
Monday, November 6, 2017: After our Minute Math bellringer, the kids all received their Topic 5 Tests back to be signed by Wednesday's class. Any student who would like to take a retake for this test will need to please fill out redo sheets for the incorrect problems as well as a Liberty Test Retake Ticket (signed by student and parent). The original test, redo sheets, and Liberty Test Retake Ticket should be returned to me to review with the student by this coming Thursday, as the test needs to be taken by Friday. Friday, November 10 is the final day of First Trimester. All redo work needs to please be turned in by this day. We also started Lesson 6-1: Repeating Decimals today. I asked the kids to attempt #1-5 and #7 for tomorrow's class. See you then, Kids!
*Week of 10/30/17
Friday, November 3, 2017: Today the class took the Topic 5 Test. They also handed in their Topic 5 Homework Packets and could have turned in their +3% Bonus Slip if filled out by a parent/guardian. They will receive the test back during Monday's class.
Thursday, November 2, 2017: Today the kids and I reviewed any questions from the Lesson 5-5 Assignment and the Topic 5 Review. We went over finding mean within frequency tables, as well. When ready, they then completed the Lesson 5-5 In-Class Assignment. They were also given time to put their Topic 5 Homework Packets together for tomorrow's class. Tomorrow is the Topic 5 Test; study well, Kids!
Wednesday, November 1, 2017: We started out class with Minute Math today and then worked in Part Three of Lesson 5-5. The kids had never seen or worked with complex fractions before but rose to the challenge and practiced well. I asked them to try the rest of the problems from the Lesson 5-5 Assignment (#5, 6, 7, 10, and 13) for tomorrow's class. Continue studying for Friday's test, Kids!
Tuesday, October 31, 2017: Due to the 1/2-day schedule, we did not have a bell ringer today. We went over anything from #1-3, 8, or 9 on the Lesson 5-5 Assignment started yesterday. We then were able to practice changing degrees from Fahrenheit to Celsius and vice versa for Part Two of this lesson. We will complete Part Three tomorrow. I asked the kids to work on #4, 11, and 12 for tomorrow's class and to also do 4 or 5 more problems on the Topic 5 Review, which should have about 20 problems completed at this point. Our test is still scheduled for this coming Friday, November 3.
Monday, October 30, 2017: The kids started their day with Minute Math. Then we discussed any questions from #1-10 on the Topic 5 Review to prepare for Friday's test. I asked them to try 4-5 more Topic Review problems for tomorrow's class. We then continued with Lesson 5-5: Operations with Rational Numbers. We worked in detail with the Distributive Property and I asked the kids to try #1-3, 8 and 9 for tomorrow's class. Happy Hallow's Eve!
*Week of 10/23/17
Friday, October 27, 2017: After Minute Math the kids wrote in their Assignment Notebooks. Their Topic 5 Test will be next Friday, November 3. I also assigned their Topic 5 Review for them to use as a study guide. I asked them to start this with #1-10 for Monday's class. We will continue a few problems at a time. We will then continue with Lesson 5-5: Operations with Rational Numbers during Monday's class.
Thursday, October 26, 2017: After the Team Problem of the Day, the kids and I went through any questions they had from the Lesson 5-4 Assignment. When they were individually ready, the kids completed the Lesson 5-4 In-Class Assignment. Tomorrow I will be giving out the Topic 5 Review to help prepare them for the Topic 5 test which will be next Friday, November 3rd. Start studying well, Kids!
Wednesday, October 25, 2017: The kids started class completing Minute Math. Then we discussed Part Three of Lesson 5-4 which explores using and solving equations. We practiced this together for a while. The kids were then given time to work on #10-15 of the Lesson 5-4 Assignment. We will discuss these in tomorrow's class.
Tuesday, October 24, 2017: The kids had a substitute teacher today since I was at a meeting at District Office. They started their class with their Team Problem of the Day. After, they discussed their puzzle choices on dividing fractions started yesterday in class. They then worked through the Launch, Part one, and Part Two of Lesson 5-4: Dividing Rational Numbers. They then started to work through #1-9 of the Lesson 5-4 Assignment for tomorrow's class.
Monday, October 23, 2017: After Minute Math, the kids and I discussed their "Pick 10" sheet for dividing decimal numbers. We then reviewed dividing with fractions, mixed numbers, and whole numbers. The kids used dry erase boards to practice this skill and then chose a puzzle (2 options) to work on for tomorrow's class.
*Week of 10/16/17
Friday, October 20, 2017: The kids did an excellent job with Minute Math today. I love seeing their confidence! Afterwards, we went through their questions from the Lesson 5-3 Assignment. If they felt they were individually ready, they then completed the Lesson 5-3 In-Class Assignment. We also did a review of dividing decimal numbers to prepare for the Lesson 5-4 Lesson. I asked them to "Pick 10" on a dividing decimals sheet which has the answers on the back to use to check. Do your best, Kids!
Thursday, October 19, 2017: Today, after we completed the Team Problem of the Day, we discussed questions 1-7 from the Lesson 5-3 Assignment started in yesterday's class. We then completed Part 2 and Part 3 of the Student Companion lesson. The kids were given time to continue with the assignment, #8-15 to be discussed and evaluated in tomorrow's class.
Wednesday, October 18, 2017: After Minute Math, the kids and I discussed their puzzle choices from yesterday's class. We then began to work within the Student Companion. We completed the Launch and Part One of Lesson 5-3, exploring a new concept: dividing with zero. I asked he kids to attempt #1-7 for tomorrow's class.
Tuesday, October 17, 2017: We started class with our Team Problem of the Day. We did a review of the basic rules of dividing integers to get ready for Lesson 5-3. Then those who needed additional time were able to complete the Lesson 5-2 In-Class Assignment. As they finished, the kids were able to choose from two puzzles to practice these dividing integers skills. They can continue this to discuss in tomorrow's class.
Monday, October 16, 2017: We started class with our Minute Math bell ringer. The kids had several questions from the Lesson 5-2 Assignment and they were encouraged NOT to take an in-class assignment until they were ready to complete this. They are always welcome to complete "similar questions" to get ready for this "quiz." If anyone didn't finish, they can tomorrow.
*Week of 10/10/17
Friday, 10/13/17: After Minute Math, we completed Part Three and the Close & Check for Lesson 5-2. They were then given time to work on the rest of the problems from the assignment started yesterday. We will discuss these further on Monday. You can do this, Kids!
Thursday, 10/12/17: The kids started the class by working on the Team Problem of the Day. We then completed Part One and Part Two of Lesson 5-2. I asked the kids to try #1-8 for tomorrow's class and gave them time to work on this in class.
Wednesday, 10/11/17: We started class with our bell ringer activity of Minute Math. We went through their questions from Puzzle 6.11 and the kids took time to then make any adjustments. These were handed in for me to check over. We started the Launch for Lesson 5-2 and will continue this lesson tomorrow.
Tuesday, 10/10/17: After our Team Problem of the Day, the kids and I reviewed multiplying mixed numbers, fractions, and whole numbers. This is a pre-requisite skill to Lesson 5-2: Multiplying Rational Numbers. After we practiced on dry erase boards and watched a video from Khan Academy, the kids worked on Puzzle 6.11. They should do their best with this for tomorrow's class. We can go over any of these questions they'd like when they come in tomorrow.
*Week of 10/2/17
Friday, 10/6/17: After Minute Math, the kids and I discussed their questions over the Lesson 5-1 Assignment. When ready, they the completed the Lesson 5-1 In-Class Assignment. They also received their Topic 4 Test back. This is to be signed by a parent, regardless of the grade, by next Tuesday's class. We have discussed the procedure for taking a retake test if wanted. The student is to fill out redo sheets for the incorrect problems as a means to practice and also fill out a Liberty Test Retake Ticket. After I have helped with any questions, we can discuss when your child would like to complete the retake test. Overall, the kids did a wonderful job on these tests. Enjoy your three-day weekend, Patriot Families!
Thursday, 10/5/17: Today after the Team Problem of the Day, the kids were given time to work with their partner of choice on Part 3 of Lesson 5-1 and then completed the Close & Check for the section, checking in with me. They were then given class time to work on the rest of the Lesson 5-1 Assignment started yesterday.
Wednesday, 10/4/17: The kids started our day together with Minute Math and then we discussed the kids' puzzle choices from yesterday's assignment. We then started Lesson 5-1: Multiplying Integers and completed through Part Two. We will continue this tomorrow. I asked the kids to try #1-8 on the Lesson 5-1 Assignment for tomorrow's class.
Tuesday, 10/3/17: After the Team Problem of the Day, the kids and I reviewed the basic rules for multiplying integers which is necessary for success in Lesson 5-1. The kids did well, using dry erase boards to show me their understanding. They were then able to choose between two different puzzles to practice this skill for tomorrow's class.
Monday, 10/2/17: We did not have a bell ringer today since the kids had their Topic 4 Test today. They all handed in their Topic 4 Homework Packets, too. I hope they all used their Study Cards on their tests! I should have these back to the kids by Thursday's class. We will start with Topic 5 tomorrow. Have your Topic 5 Foldable ready to go!
*Week of 9/25/17
Friday, 9/29/17: After Minute Math, we went over any questions the kids wanted to review from the Topic 4 Review and then additional time was given to students to work on the Topic 4 Review or practice with "similar questions". Study well for your Topic 4 Test coming up on Monday, Kids! Don't forget to make those study cards!
Thursday, 9/28/17: The kids started class by working together on the Team Problem of the Day. We then reviewed #9-15 from the Lesson 4-6 Assignment, which the kids found challenging near the end. When they were individually ready, the kid completed the Lesson 4-6 In-Class Assignment. They were then given time to work on the Topic 4 Review further for tomorrow's class. The Topic 4 Test is Monday, October 2.
Wednesday, 9/27/17: After Minute Math, the kids and I discussed any questions hey wanted to discuss from #1-8 of the Lesson 4-6 Assignment. We then worked on Part 3 and the Close & Check of the lesson. The kids were then given time to complete #9-15 to complete the Lesson 4-6 Assignment. I had also assigned the Topic 4 Review to the kids. This should be done to the best of their abilities for Friday. They will be given time to work on this further tomorrow and during Friday's class. The Topic 4 Test will be Monday, October 2.
Tuesday, 9/26/17: The kids started out class working with their teammates on the Problem of the Day. We then began to explore Lesson 4-6: Distance on a Number Line. We completed half the lesson and I asked the kids to attempt half he problems for tomorrow's class #1-8).
Monday, 9/25/17: After our Minute Math bell ringer, the kids and I discussed and reviewed any of the assigned problems from the Lesson 4-5 Assignment started Thursday. When students were individually ready, they were able to complete the Lesson 4-5 In-Class Assignment. I asked them to complete the Lesson 4-6 Launch and Reflection for tomorrow's class.
*Week of 9/18/17
Friday, 9/22/17: After Minute Math, we reviewed the first 7 problems from the Lesson 4-5 Assignment. We then completed Part Three as well as the Close & Check from this lesson. The kids were given time to continue/complete the Lesson 4-5 Assignment. I asked them to do their best to complete this for Monday's class. Don't forget: Midterms for 1st Trimester are next Thursday. Work on those redo sheets if needed!
Thursday, 9/21/17: The kids started our class today by completing the Team Problem of the Day. We then discussed and reviewed #9-15, often drawing pictures to help the kids understand and set up their solutions. We then began working through Lesson 4-5: Subtracting Rational Numbers, completing the Launch, Reflection, Part One and Part Two. I asked the kids to try the first 7 problems from the Lesson 4-5 Assignment.
Wednesday, 9/20/17: After the kids and I discussed their Minute Math answers, we went over their questions from #1-8 on Puzzle 6.8. I then went through the rest of the problems from this puzzle with them to discuss strategies to come to solutions. They were given the rest of the class time to work on this puzzle. We will discuss this further tomorrow, Do your best, Kids!
Tuesday, 9/19/17: Today we did not start class with our normal Team Problem bell ringer activity. Instead, the kids completed the Fall AIMS tests. They are used to completing these on paper, but these are now completed online. I then gave them Puzzle 6.8 to practice adding and subtracting positive/negative rational numbers. I asked them to try #1-8 for tomorrow's class, noting those they'd like to discuss tomorrow.
Monday, 9/18/17: After Minute Math the kids and I reviewed any problems from the Lesson 4-4 Assignment they wished to go over. They then completed their Lesson 4-4 In-Class Assignment. After that I gave them class time to complete #11-20 on a sheet entitled "Adding/Subtracting Decimals" that we started earlier this topic. There is an answer key stapled to this so they can review this skill that is considered a prerequisite to Lesson 4-5. They have been asked to try these for tomorrow's class, noting the ones they'd like to review.
*Week of 9/11/17
Friday, 9/15/17: Due to shortened classes today, the kids did not have a bell ringer. They then worked collaboratively in their teams to work through Part 3 of Lesson 4-4 as well as the Close & Check. They were then encouraged to use the rest of the time to compete the rest of the Lesson 4-4 Assignment started in yesterday's class. We will discuss this assignment at Monday's class.
Thursday, 9/14/17: After the Team Problem of the Day, the kids and I worked through the Launch, Part One, and Part Two of Lesson 4-4: Subtracting Integers. The kids agreed that yesterday's review of the the skill of subtracting integers helped greatly since they are using the skill in problem solving situations in this section. I asked them to begin the first half of the Lesson 4-4 Assignment. They also received their Lesson 4-3 In-Class Assignment back; reminder: students are encouraged to complete a redo sheet for any incorrect problems on these as to learn from their mistakes and earn full credit.
Wednesday, 9/13/17: The kids started their day completing the Minute Math bell ringer. We then reviewed anything they wanted from the Lesson 4-3 Assignment that was to be completed to the best of their abilities for today's class. They then worked on the Lesson 4-3 In-Class Assignment when they were ready. They will receive this back graded by tomorrow's class.
Tuesday, 9/12/17: The kids started class with the Team Problem of the Day. We then reviewed any questions from the Lesson 4-3 Assignment the kids wanted to discuss. Reminder: any mathematical computation (mental or on a calculator) must be noted in student work. After the review and when the student was ready, the kids each completed the Lesson 4-3 In-Class Assignment. I asked them to work on the Lesson 4-4 Launch and Reflection (in Student Companion) for tomorrow's class.
Monday, 9/11/17: After Minute Math, the kids all received their Pre-Requisite to Lesson 4-2 In-Class Assignment back today. This will be put into PowerSchool tonight. The kids were asked to attempt to complete the Lesson 4-3 Assignment (online or paper copy) to the best of their abilities for tomorrow's class. They should highlight/circle/note the questions they would like to discuss tomorrow.
*Week of 9/5/17
Friday, 9/8/17: Today the kids completed the Prerequisite to Lesson 4-3 In-Class Assignment. This will be graded and returned to them on Monday and also put into PowerSchool after school Monday. We continued working within 4-3 as well.
Thursday, 9/7/17: We reviewed the six problems involving positive/negative fractions and mixed numbers after our bell ringer. The kids and I then began to explore within the actual lesson parts of Lesson 4-3: Adding Rational Numbers. We will continue this into tomorrow's class.
Wednesday, 9/6/17--> We continued preparing for Lesson 4-3 today by practicing adding positive/negative fractions and mixed numbers. The class was given six problems with answers provided to practice this skill for tomorrow's class.
Tuesday, 9/5/17--> Today the kids and I reviewed adding positive/negative decimals to prepare for our next lesson Lesson 4-3: Adding Rational Numbers. After practicing together in class in their foldables and on dry erase boards, the kids were asked to complete 10 problems (answers provided) for tomorrow's class.
Monday, 9/4/17--> No school (Labor Day) | 12,040 | 47,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-13 | longest | en | 0.951207 |
http://perplexus.info/show.php?pid=7756&cid=47690 | 1,553,616,172,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205597.84/warc/CC-MAIN-20190326160044-20190326182044-00356.warc.gz | 150,629,051 | 4,377 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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A red herring marble (Posted on 2012-05-01)
An urn contains 3333 marbles : 2001 black,999 white and 333 red.
You are requested to draw randomly a pair of marbles and to obey the following rules:
If none of them is red - discard both of them.
If only one of the marbles is red – discard the non-red marble and return the red one to the urn.
If both are red - discard both of them.
Continue drawing pair by pair and discarding as told until there is only a single marble left.
What is the probability of the last one being red?
See The Solution Submitted by Ady TZIDON No Rating
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One fish two fish red fish blue fish (solution) Comment 1 of 1
Clearly the individual black and white don't matter, but we note that 3000 non-red marbles. Then we consider what happens to the number of red marbles in each case:
If none is red - discard both of them. Red unchanged.
If one is red - return it to the urn. Red unchanged.
If both are red - discard both. Red reduced by 2.
Since we start with an odd number of red marbles and the total can only go down by two, we will always end up with a single red marble.
So the answer to the question is 1.
Posted by Jer on 2012-05-01 12:11:34
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Forums (0) | 359 | 1,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.908566 |
https://www.reference.com/math/many-faces-square-339c69f2d26d0e98 | 1,484,902,548,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280801.0/warc/CC-MAIN-20170116095120-00149-ip-10-171-10-70.ec2.internal.warc.gz | 972,793,863 | 20,642 | Q:
# How many faces does a square have?
A:
A square is a two-dimensional shape that does not have a face, but a square is one of the faces of a cube. Faces are associated with solid figures, such as cylinders, cubes and pyramids. Two-dimensional shapes have breadth and width, but they do not have thickness.
## Keep Learning
Credit: Peter Adams Photolibrary Getty Images
The attributes of plane shapes include a defined number of corners or sides. Solid shapes are defined by edges, vertices and faces, and they may be flat or thick. The edges of solid shapes are where two faces join; for example, a cube has a total of six square faces that create 12 edges.
Sources:
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PEOPLE SEARCH FOR | 338 | 1,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.961056 |
https://space.stackexchange.com/questions/4337/convergent-divergent-de-laval-nozzle-dimensions | 1,679,581,877,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00133.warc.gz | 606,530,009 | 39,764 | # Convergent-Divergent / De Laval Nozzle Dimensions
I'm creating a super sonic ping pong ball cannon like the one recently shown on Mythbusters. I plan on using pressure rated steel pipe so it's as safe as realistically possible and showing it off at science and technology exhibits for kids. I think it will be a big hit and a great way to get kids interested in physics and science.
The device is pretty simple and well explained in this video here: https://www.youtube.com/watch?v=YYNCGZCul1Q
The key to the whole thing is ~100 psi on one side of a burst disk with a de Laval nozzle attached to a barrel on the other side. The barrel and nozzle is all in a pretty strong vacuum. The pressure chamber is using 4" pipe and the barrel is 1.5" diameter.
Since the most common usage for de Laval nozzles is in rocket motors, I ended up here. Reading and research has led me to believe that a convergent angle of 30 degrees is common in rockets. I've also read that the divergent angle of 15 degrees works but less works as well. What I can't find anywhere is the throat size.
With the goal to get the highest velocity possible using 100 PSI and 4" diameter on the convergent side and 1.5" diameter on the divergent side which is the best convergent angle, divergent angle, and most importantly, the throat diameter.
I've found equations that I think would provide this information but they are well beyond my ability to solve so any help would be greatly appreciated!
Thank you very much for the response! It was fascinating! I did actually build the entire thing and it worked great. The trick in the end was to use a large burst disk and pressurizing the chamber until it failed (generally 600-700kPa - that took a LOT of experimentation to get right BTW). It made reloading a bit of a pain but I improved that with some pneumatic bits to help with opening and closing the chamber. The nozzle ended up being a very simple one that I machined to some rules of thumb I found somewhere - not optimal but worked well enough in the end.
I plan on rebuilding the entire thing some day as a demonstration project to show off as it's actually pretty impressive to see in action. I'll absolutely use your information to improve it. The cannon makes quite a "crack" when the ping pong ball breaks the sound barrier. I used a rifle chronometer to measure speed and while it was quite variable (due to the burst disk inconsistencies) I would routinely get speeds in the 400-500+ m/s range. That's fast enough to actually "scorch" the faux fur on stuffed animals - in addition to blowing holes clear through them - with a ping pong ball in case anyone is curious :)
Thank you again for taking the time to respond!
• That's awesome. Apr 26, 2020 at 18:16
• This post landed in the "low quality" review queue, likely because "Thank you very much for the response" triggered some keywords (SE does not work like a forum, and is strictly Question-Answer). This, however, contains what the actual practical solution to the original problem ended up like, and the results. That's impressive, 6 years after the original post! All answers should answer the question, and this certainly does. Apr 26, 2020 at 20:10
This is really an underconstrained problem. A lot of it also belongs on physics.stackexchange, or the like, or is down to specifics of your design constraints. However, I shall summarize the basics that are relevant to rocketry and space travel here.
The ratio of the final area to the throat area is the "expansion ratio" of the nozzle, and is the main factor influencing how fast your propellant leaves. At the throat, the flow is mach 1, which is a speed that will vary by propellant composition and temperature.
From there, the nozzle expands, and the increased area causes the exhaust to accelerate due the weird way in which supersonic flows work. The final velocity is given by $$v_e$$ in the following: $$v_e = \sqrt{ \left( \frac{2 \kappa}{\kappa - 1} \right) \left( \frac{R_u}{M} T_c \right) \left( 1 - \left(\frac{p_e}{p_c}\right)^{(\kappa-1)/\kappa} \right) }$$ $$R_u$$ is the universal gas constant. AFAICT, dry air at $$T_c=293.15\text{ K}$$ (i.e. $$20^\circ\text{ C}$$) has an average molar mass $$M \approx 28.9647\text{ g/mol}$$ and a heat capacity ratio $$\kappa \approx 1.400$$. The chamber pressure is $$p_c=100\text{ PSI}\approx 689\text{ kPa}$$ (use metric please!), and you're expanding it by some ratio to get a new pressure $$p_e$$. If the ratio is infinity, the nozzle is "ideally expanded", and $$p_e=0\text{ kPa}$$.
As it happens, I wrote a handy web calculator which can be applied to this case. Expanding to vacuum (i.e., "perfectly") using the above parameters will give you an exit velocity of $$v_e \approx 767.5\text{ m/s}$$. In practice, you'll get lower than this because you can't make an infinitely big rocket nozzle (and if you did, it wouldn't push the ball).
Of course, exit velocity is only half the story; you also need to consider mass flow (which is calculated as the density of the air at that pressure, times the hole's area, times mach 1). If you make an teensy hole and expand the flow to almost that speed, very little mass of air will escape, and this will not be strong enough to push the ball to a high speed before the end of the pipe. Conversely, if the hole is too large, plenty of mass will flow, but the flow will not accelerate very much above mach 1 (for the above, about $$343\text{ m/s}$$).
One needs to work out how much the gas force from the exhaust accelerates the projectile, and the length of pipe required. Indeed, there are lots of ways to extend this analysis in various ways and to various degrees of accuracy, but they get more and more into the realm of general Physics, which I won't go into here because it's off-topic. At least now you have a way to calculate the velocity and mass flow, though.
TL;DR, though: my guess is that a reasonably sized hole will work fine, even if you don't hit upon the exactly optimal size (which would be very difficult to calculate, given all the variables, anyway). Good luck with your project, even six years later! Be sure to heed the safety warnings in the very video you linked; this sort of engineering can be dangerous. | 1,501 | 6,245 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.961197 |
https://www.surfsidemedia.in/post/python-data-analysis-with-numpy | 1,723,781,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00272.warc.gz | 744,715,728 | 7,337 | # Python Data Analysis with NumPy
## Introduction
NumPy is a fundamental library for data analysis in Python. It provides support for arrays, mathematical functions, and operations that are essential for working with large datasets and numerical data. In this comprehensive guide, we'll explore the capabilities of NumPy and how to use it for data analysis.
## Prerequisites
Before you begin, make sure you have the following prerequisites in place:
• Python Installed: You should have Python installed on your local development environment.
• NumPy Installed: You can install NumPy using pip: `pip install numpy`
• Basic Python Knowledge: Understanding Python fundamentals is essential for using NumPy effectively.
## Key Concepts of NumPy
NumPy introduces the concept of arrays, which are similar to Python lists but more powerful for numerical operations. You can perform various data analysis tasks, including data manipulation, statistics, and linear algebra.
## Sample Python Code for NumPy
Here's a basic Python code snippet to demonstrate how to create a NumPy array and perform some common operations:
` import numpy as np # Create a NumPy array data = np.array([1, 2, 3, 4, 5]) # Perform basic operations mean = np.mean(data) median = np.median(data) variance = np.var(data) std_deviation = np.std(data) print("Data: ", data) print("Mean: ", mean) print("Median: ", median) print("Variance: ", variance) print("Standard Deviation: ", std_deviation) `
## Data Manipulation with NumPy
NumPy allows you to manipulate data efficiently. You can reshape arrays, slice data, and apply mathematical operations easily.
## Sample Python Code for Data Manipulation
Here's a basic Python code snippet to demonstrate data manipulation with NumPy:
` import numpy as np # Create a NumPy array data = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) # Reshape the array reshaped_data = data.reshape(3, 3) # Extract a slice of data sliced_data = data[0:2, 1:3] print("Original Data:\n", data) print("Reshaped Data:\n", reshaped_data) print("Sliced Data:\n", sliced_data) `
## Conclusion
NumPy is an essential library for data analysis and scientific computing in Python. This guide has introduced you to its core concepts and functionalities, but there's much more to explore in terms of advanced data analysis, statistics, and machine learning. As you continue your journey in data analysis, NumPy will be a valuable tool in your toolkit. | 553 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.760574 |
http://sheepolution.com/learn/book/13 | 1,552,958,818,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201882.11/warc/CC-MAIN-20190319012213-20190319034213-00451.warc.gz | 176,182,288 | 3,482 | # Chapter 13 - Collision
Let's say we're making a game where you can shoot down monsters. A monster should die when it is hit by a bullet. So what we need to check is: Is the monster colliding with a bullet?
We're going to create a collision check function. For that we need to know, when do 2 rectangles collide?
I created an image with 3 examples:
It's time to turn on that programmer brain if you haven't already. What is going on in the third example that isn't happening in the first and second example?
"They are colliding"
Yes, but you have to be more specific. We need information that the computer can use.
Take a look at the positions of the rectangles. In the first example, Red is not colliding with Blue, because Red is too far to the left. If Red was a bit further to the right, they would touch. How far exactly? Well, if Red's right side is further to the right than Blue's left side. This is something that is true for example 3.
But it's also true for example 2. We need more conditions to be sure there is collision. So example 2 shows we can't go too far to the right. How far exactly can we go? How much would Red have to move to the left for there to be collision? When Red's left side is further to the left than Blue's right side.
So we have 2 conditions, is that enough to ensure there is collision?
Well no, look at the following image:
This situation agrees with our conditions. Red's right side is further to the right than Blue's left side. And Red's left side is further to the left than Blue's right side. Yet, there is no collision. That's because Red is too high. It needs to move down. How far? Till Red's bottom side is further to the bottom than Blue's top side.
But if we move it too far down, there won't be collision anymore. How far can Red move down, and still collide with Blue? As long as Red's top side is further to the top than Blue's bottom side.
Now we got 4 conditions. Are all 4 conditions true for these 3 examples?
Red's right side is further to the right than Blue's left side.
Red's left side is further to the left than Blue's right side.
Red's bottom side is further to the bottom than Blue's top side.
Red's top side is further to the top than Blue's bottom side.
Yes, they are! Now we need to turn this information into a function.
First let's create 2 rectangles.
``````function love.load()
--Create 2 rectangles
r1 = {
x = 10,
y = 100,
width = 100,
height = 100
}
r2 = {
x = 250,
y = 120,
width = 150,
height = 120
}
end
function love.update(dt)
--Make one of rectangle move
r1.x = r1.x + 100 * dt
end
function love.draw()
love.graphics.rectangle("line", r1.x, r1.y, r1.width, r1.height)
love.graphics.rectangle("line", r2.x, r2.y, r2.width, r2.height)
end``````
Now we create a new function called checkCollision(), with 2 rectangles as parameters.
``````function checkCollision(a, b)
end``````
First we need the sides of the rectangles. The left side is the x position, the right side is the x position + the width. Same with y and height.
``````function checkCollision(a, b)
--With locals it's common usage to use underscores instead of camelCasing
local a_left = a.x
local a_right = a.x + a.width
local a_top = a.y
local a_bottom = a.y + a.height
local b_left = b.x
local b_right = b.x + b.width
local b_top = b.y
local b_bottom = b.y + b.height
end``````
Now that we have the 4 sides of each rectangle, we can use them to put our conditions in an if-statement.
``````function checkCollision(a, b)
--With locals it's common usage to use underscores instead of camelCasing
local a_left = a.x
local a_right = a.x + a.width
local a_top = a.y
local a_bottom = a.y + a.height
local b_left = b.x
local b_right = b.x + b.width
local b_top = b.y
local b_bottom = b.y + b.height
--If Red's right side is further to the right than Blue's left side.
if a_right > b_left and
--and Red's left side is further to the left than Blue's right side.
a_left < b_right and
--and Red's bottom side is further to the bottom than Blue's top side.
a_bottom > b_top and
--and Red's top side is further to the top than Blue's bottom side then..
a_top < b_bottom then
--There is collision!
return true
else
--If one of these statements is false, return false.
return false
end
end``````
Okay, we have our function. Let's try it out! We draw the rectangles filled or lined based on
``````function love.draw()
--We create a local variable called mode
local mode
if checkCollision(r1, r2) then
--If there is collision, draw the rectangles filled
mode = "fill"
else
--else, draw the rectangles as a line
mode = "line"
end
--Use the variable as first argument
love.graphics.rectangle(mode, r1.x, r1.y, r1.width, r1.height)
love.graphics.rectangle(mode, r2.x, r2.y, r2.width, r2.height)
end``````
It works! Now you know how to detect collision between 2 rectangles.
# TL;DR
Collision between 2 rectangles can be checked with 4 conditions.
Where A and B are rectangles:
A's right side is further to the right than B's left side.
A's left side is further to the left than B's right side.
A's bottom side is further to the bottom than B's top side.
A's top side is further to the top than B's bottom side. | 1,353 | 5,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-13 | latest | en | 0.955756 |
https://goprep.co/ex-11.4-q5-5y-2-9x-2-36-in-each-of-the-find-the-coordinates-i-1nk8vr | 1,607,207,172,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00097.warc.gz | 296,569,390 | 38,091 | # In each of the, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.5y2 – 9x2 = 36
The given equation is 5y2 – 9x2 = 36
We can re-write the given as
Or ……….. (1)
On comparing this equation (1) with the standard equation of hyperbola
, we get,
a = and b =2
We know, a2 + b2 = c2
Thus,
c2 = =
c =
Therefore,
The coordinates of the foci are
The coordinates of the vertices are .
Eccentricity, e =
Length of latus rectum =
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https://stats.libretexts.org/Courses/Highline_College/Book%3A_Statistics_Using_Technology_(Kozak)/05%3A_Discrete_Probability_Distributions/5.02%3A_Binomial_Probability_Distribution | 1,708,776,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474533.12/warc/CC-MAIN-20240224112548-20240224142548-00505.warc.gz | 558,856,763 | 36,340 | # 5.2: Binomial Probability Distribution
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Section 5.1 introduced the concept of a probability distribution. The focus of the section was on discrete probability distributions (pdf). To find the pdf for a situation, you usually needed to actually conduct the experiment and collect data. Then you can calculate the experimental probabilities. Normally you cannot calculate the theoretical probabilities instead. However, there are certain types of experiment that allow you to calculate the theoretical probability. One of those types is called a Binomial Experiment.
## Properties of a binomial experiment (or Bernoulli trial)
1. Fixed number of trials, n, which means that the experiment is repeated a specific number of times.
2. The n trials are independent, which means that what happens on one trial does not influence the outcomes of other trials.
3. There are only two outcomes, which are called a success and a failure.
4. The probability of a success doesn’t change from trial to trial, where p = probability of success and q = probability of failure, q = 1-p.
If you know you have a binomial experiment, then you can calculate binomial probabilities. This is important because binomial probabilities come up often in real life. Examples of binomial experiments are:
• Toss a fair coin ten times, and find the probability of getting two heads.
• Question twenty people in class, and look for the probability of more than half being women?
• Shoot five arrows at a target, and find the probability of hitting it five times?
To develop the process for calculating the probabilities in a binomial experiment, consider Example $$\PageIndex{1}$$.
Example $$\PageIndex{1}$$: Deriving the Binomial Probability Formula
Suppose you are given a 3 question multiple-choice test. Each question has 4 responses and only one is correct. Suppose you want to find the probability that you can just guess at the answers and get 2 questions right. (Teachers do this all the time when they make up a multiple-choice test to see if students can still pass without studying. In most cases the students can’t.) To help with the idea that you are going to guess, suppose the test is in Martian.
1. What is the random variable?
2. Is this a binomial experiment?
3. What is the probability of getting 2 questions right?
4. What is the probability of getting zero right, one right, and all three right?
Solution
a. x = number of correct answers
b.
1. There are 3 questions, and each question is a trial, so there are a fixed number of trials. In this case, n = 3.
2. Getting the first question right has no affect on getting the second or third question right, thus the trials are independent.
3. Either you get the question right or you get it wrong, so there are only two outcomes. In this case, the success is getting the question right.
4. The probability of getting a question right is one out of four. This is the same for every trial since each question has 4 responses. In this case, $$p=\dfrac{1}{4} \text { and } q=1-\dfrac{1}{4}=\dfrac{3}{4}$$
This is a binomial experiment, since all of the properties are met.
c. To answer this question, start with the sample space. SS = {RRR, RRW, RWR, WRR, WWR, WRW, RWW, WWW}, where RRW means you get the first question right, the second question right, and the third question wrong. The same is similar for the other outcomes.
Now the event space for getting 2 right is {RRW, RWR, WRR}. What you did in chapter four was just to find three divided by eight. However, this would not be right in this case. That is because the probability of getting a question right is different from getting a question wrong. What else can you do?
Look at just P(RRW) for the moment. Again, that means P(RRW) = P(R on 1st, R on 2nd, and W on 3rd)
Since the trials are independent, then P(RRW) = P(R on 1st, R on 2nd, and W on 3rd) = P(R on 1st) * P(R on 2nd) * P(W on 3rd)
Just multiply p * p * q
$$P(\mathrm{RRW})=\dfrac{1}{4} * \dfrac{1}{4} * \dfrac{3}{4}=\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}$$
The same is true for P(RWR) and P(WRR). To find the probability of 2 correct answers, just add these three probabilities together. You get
\begin{aligned} P(2 \text { correct answers }) &=P(\mathrm{RRW})+P(\mathrm{RWR})+P(\mathrm{WRR}) \\ &=\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}+\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}+\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1} \\ &=3\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1} \end{aligned}
d. You could go through the same argument that you did above and come up with the following:
r right P(r right)
0 right $$1^{*}\left(\dfrac{1}{4}\right)^{0}\left(\dfrac{3}{4}\right)^{3}$$
1 right $$3^{*}\left(\dfrac{1}{4}\right)^{1}\left(\dfrac{3}{4}\right)^{2}$$
2 right $$3 *\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}$$
3 right $$1^{*}\left(\dfrac{1}{4}\right)^{3}\left(\dfrac{3}{4}\right)^{0}$$
Table $$\PageIndex{1}$$: Binomial pattern
Hopefully you see the pattern that results. You can now write the general formula for the probabilities for a Binomial experiment
First, the random variable in a binomial experiment is x = number of successes. Be careful, a success is not always a good thing. Sometimes a success is something that is bad, like finding a defect. A success just means you observed the outcome you wanted to see happen.
Definition $$\PageIndex{1}$$
Binomial Formula for the probability of r successes in n trials is
$$P(x=r)=_{n} C_{r} p^{r} q^{n \cdot r} \text { where }_{n} C_{r}=\dfrac{n !}{r !(n-r) !}$$
The $$_{n} C_{r}$$ is the number of combinations of n things taking r at a time. It is read “n choose r”. Some other common notations for n choose r are $$C_{n, r}$$, and $$\left( \begin{array}{l}{n} \\ {r}\end{array}\right)$$. n! means you are multiplying $$n^{*}(n-1)^{*}(n-2)^{*} \dots^{*} 2^{*} 1$$. As an example, $$5 !=5^{*} 4^{*} 3^{*} 2^{*} 1=120$$.
When solving problems, make sure you define your random variable and state what n, p, q, and r are. Without doing this, the problems are a great deal harder.
Example $$\PageIndex{2}$$: Calculating Binomial Probabilities
When looking at a person’s eye color, it turns out that 1% of people in the world has green eyes ("What percentage of," 2013). Consider a group of 20 people.
1. State the random variable.
2. Argue that this is a binomial experiment.
3. Find the probability that none have green eyes.
4. Find the probability that nine have green eyes.
5. Find the probability that at most three have green eyes.
6. Find the probability that at most two have green eyes.
7. Find the probability that at least four have green eyes.
8. In Europe, four people out of twenty have green eyes. Is this unusual? What does that tell you?
Solution
a. x = number of people with green eyes
b.
1. There are 20 people, and each person is a trial, so there are a fixed number of trials. In this case, $$n = 20$$.
2. If you assume that each person in the group is chosen at random the eye color of one person doesn’t affect the eye color of the next person, thus the trials are independent.
3. Either a person has green eyes or they do not have green eyes, so there are only two outcomes. In this case, the success is a person has green eyes.
4. The probability of a person having green eyes is 0.01. This is the same for every trial since each person has the same chance of having green eyes. p = 0.01 and q = 1 - 0.01 = 0.99
c. $$P(x=0)=_{20} C_{0}(0.01)^{0}(0.99)^{20-0} \approx 0.818$$
d. $$P(x=9)=_{20} C_{9}(0.01)^{9}(0.99)^{20-9} \approx 1.50 \times 10^{-13} \approx 0.000$$
e. At most three means that three is the highest value you will have. Find the probability of x is less than or equal to three.
\begin{aligned} P(x \leq 3) &=P(x=0)+P(x=1)+P(x=2)+P(x=3) \\ &=_{20} C_{0}(0.01)^{0}(0.99)^{20}+_{20} C_{1}(0.01)^{1}(0.99)^{19} \\& +_{20}C_{2}(0.01)^{2}(0.99)^{18}+_{20}C_{3}(0.01)^{3}(0.99)^{17} \\ & \approx 0.818+0.165+0.016+0.001>0.999 \end{aligned}
The reason the answer is written as being greater than 0.999 is because the answer is actually 0.9999573791, and when that is rounded to three decimal places you get 1. But 1 means that the event will happen, when in reality there is a slight chance that it won’t happen. It is best to write the answer as greater than 0.999 to represent that the number is very close to 1, but isn’t 1.
f.
\begin{aligned} P(x \leq 2) &=P(x=0)+P(x=1)+P(x=2) \\ &=_{20} C_{0}(0.01)^{0}(0.99)^{20}+_{20} C_{1}(0.01)^{1}(0.99)^{19}+_{20} C_{2}(0.01)^{2}(0.99)^{18} \\ & \approx 0.818+0.165+0.016 \approx 0.999 \end{aligned}
g. At least four means four or more. Find the probability of x being greater than or equal to four. That would mean adding up all the probabilities from four to twenty. This would take a long time, so it is better to use the idea of complement. The complement of being greater than or equal to four is being less than four. That would mean being less than or equal to three. Part (e) has the answer for the probability of being less than or equal to three. Just subtract that number from 1.
$$P(x \geq 4)=1-P(x \leq 3)=1-0.999=0.001$$
Actually the answer is less than 0.001, but it is fine to write it this way.
h. Since the probability of finding four or more people with green eyes is much less than 0.05, it is unusual to find four people out of twenty with green eyes. That should make you wonder if the proportion of people in Europe with green eyes is more than the 1% for the general population. If this is true, then you may want to ask why Europeans have a higher proportion of green-eyed people. That of course could lead to more questions.
The binomial formula is cumbersome to use, so you can find the probabilities by using technology. On the TI-83/84 calculator, the commands on the TI-83/84 calculators when the number of trials is equal to n and the probability of a success is equal to p are $$\text{binompdf}(n, p, r)$$ when you want to find P(x=r) and $$\text{binomcdf}(n, p, r)$$ when you want to find $$P(x \leq r)$$. If you want to find $$P(x \geq r)$$, then you use the property that $$P(x \geq r)=1-P(x \leq r-1)$$, since $$x \geq r$$ and $$x<r$$ or $$x \leq r-1$$ are complementary events. Both binompdf and binomcdf commands are found in the DISTR menu. Using R, the commands are $$P(x=r)=\text { dbinom }(r, n, p) \text { and } P(x \leq r)=\text { pbinom }(r, n, p)$$.
Example $$\PageIndex{3}$$ using the binomial command on the ti-83/84
When looking at a person’s eye color, it turns out that 1% of people in the world has green eyes ("What percentage of," 2013). Consider a group of 20 people.
1. State the random variable.
2. Find the probability that none have green eyes.
3. Find the probability that nine have green eyes.
4. Find the probability that at most three have green eyes.
5. Find the probability that at most two have green eyes.
6. Find the probability that at least four have green eyes.
Solution
a. x = number of people with green eyes
b. You are looking for P (x=0). Since this problem is x=0, you use the binompdf command on the TI-83/84 or dbinom command on R. On the TI83/84, you go to the DISTR menu, select the binompdf, and then type into the parenthesis your n, p, and r values into your calculator, making sure you use the comma to separate the values. The command will look like $$\text{binompdf}(20,.01,0)$$ and when you press ENTER you will be given the answer. (If you have the new software on the TI-84, the screen looks a bit different.)
On R, the command would look like dbinom(0, 20, 0.01)
P (x=0) = 0.8179. Thus there is an 81.8% chance that in a group of 20 people none of them will have green eyes.
c. In this case you want to find the P (x=9). Again, you will use the binompdf command or the dbinom command. Following the procedure above, you will have binompdf(20, .01, 9) on the TI-83/84 or dbinom(9,20,0.01) on R. Your answer is $$P(x=9)=1.50 \times 10^{-13}$$. (Remember when the calculators or computers gives you $$1.50 E-13$$ or $$1.50 e-13$$, this is how they display scientific notation.) The probability that out of twenty people, nine of them have green eyes is a very small chance.
d. At most three means that three is the highest value you will have. Find the probability of x being less than or equal to three, which is $$P(x \leq 3)$$. This uses the binomcdf command on the TI-83/84. You use the command on the TI-83/84 of binomcdf(20, .01, 3).
Your answer is 0.99996. Thus there is a really good chance that in a group of 20 people at most three will have green eyes. (Note: don’t round this to one, since one means that the event will happen, when in reality there is a slight chance that it won’t happen. It is best to write the answer out to enough decimal points so it doesn’t round off to one.
e. You are looking for $$P(x \leq 2)$$. Again use binomcdf or pbinom. Following the procedure above you will have $$\text{binomcdf}(20,.01,2)$$ on the TI-83/84 and pbinom(2,20,0.01), with $$P(x \leq 2)=0.998996$$. Again there is a really good chance that at most two people in the room will have green eyes.
f. At least four means four or more. Find the probability of x being greater than or equal to four. That would mean adding up all the probabilities from four to twenty. This would take a long time, so it is better to use the idea of complement. The complement of being greater than or equal to four is being less than four. That would mean being less than or equal to three. Part (e) has the answer for the probability of being less than or equal to three. Just subtract that number from 1.
$$P(x \geq 4)=1-P(x \leq 3)=1-0.99996=0.00004$$ You can also find this answer by doing the following on TI-83/84:
$$P(x \geq 4)=1-P(x \leq 3)=1-\text { binomcdf }(20,.01,3)=1-0.99996=0.00004$$ on R:
$$P(x \geq 4)=1-P(x \leq 3)=1-\text { pbinom }(3,20,.01)=1-0.99996=0.0004$$ Again, this is very unlikely to happen.
There are other technologies that will compute binomial probabilities.
Example $$\PageIndex{4}$$ calculating binomial probabilities
According to the Center for Disease Control (CDC), about 1 in 88 children in the U.S. have been diagnosed with autism ("CDC-data and statistics,," 2013). Suppose you consider a group of 10 children.
1. State the random variable.
2. Argue that this is a binomial experiment.
3. Find the probability that none have autism.
4. Find the probability that seven have autism.
5. Find the probability that at least five have autism.
6. Find the probability that at most two have autism.
7. Suppose five children out of ten have autism. Is this unusual? What does that tell you?
Solution
a. x = number of children with autism
b.
1. There are 10 children, and each child is a trial, so there are a fixed number of trials. In this case, n = 10.
2. If you assume that each child in the group is chosen at random, then whether a child has autism does not affect the chance that the next child has autism. Thus the trials are independent.
3. Either a child has autism or they do not have autism, so there are two outcomes. In this case, the success is a child has autism.
4. The probability of a child having autism is 1/88. This is the same for every trial since each child has the same chance of having autism. $$p=\dfrac{1}{88}$$ and $$q=1-\dfrac{1}{88}=\dfrac{87}{88}$$.
c. Using the formula:
$$P(x=0)=_{10} C_{0}\left(\dfrac{1}{88}\right)^{0}\left(\dfrac{87}{88}\right)^{10-0} \approx 0.892$$
Using the TI-83/84 Calculator:
$$P(x=0)=\text { binompdf }(10,1 \div 88,0) \approx 0.892$$
Using R:
$$P(x=0)=\text { pbinom }(0,10,1 / 88) \approx 0.892$$
d. Using the formula:
$$P(x=7)=_{10} C_{7}\left(\dfrac{1}{88}\right)^{7}\left(\dfrac{87}{88}\right)^{10-7} \approx 0.000$$
Using the TI-83/84 Calculator:
$$P(x=7)=\text { binompdf }(10,1 \div 88,7) \approx 2.84 \times 10^{-12}$$
Using R:
$$P(x=7)=\operatorname{dbinom}(7,10,1 / 88) \approx 2.84 \times 10^{-12}$$
e. Using the formula:
\begin{aligned} P(x \geq 5) &=P(x=5)+P(x=6)+P(x=7) \\ &+P(x=8)+P(x=9)+P(x=10) \\ &=_{10} C_{5}\left(\dfrac{1}{88}\right)^{5}\left(\dfrac{78}{88}\right)^{10-5}+_{10} C_{6}\left(\dfrac{1}{88}\right)^{6}\left(\dfrac{78}{88}\right)^{10-6} \\ & +_{10}C_{7}\left(\dfrac{1}{88}\right)^{7}\left(\dfrac{78}{88}\right)^{10-7}+_{10}C_{8}\left(\dfrac{1}{88}\right)^{8}\left(\dfrac{78}{88}\right)^{10-8} \\ &+_{10}C_{9}\left(\dfrac{1}{88}\right)^{9}\left(\dfrac{78}{88}\right)^{10-9}+_{10}C_{10}\left(\dfrac{1}{88}\right)^{10}\left(\dfrac{78}{88}\right)^{10-10}\\&=0.000+0.000+0.000+0.000+0.000+0.000 \\ &=0.000 \end{aligned}
Using the TI-83/84 Calculator:
To use the calculator you need to use the complement.
\begin{aligned} P(x \geq 5) &=1-P(x<5) \\ &=1-P(x \leq 4) \\ &=1-\text { binomcdf }(10,1 \div 88,4) \\ & \approx 1-0.9999999=0.000 \end{aligned}
f. Using the formula:
\begin{aligned} P(x \leq 2) &=P(x=0)+P(x=1)+P(x=2) \\ &=_{10} C_{0}\left(\dfrac{1}{88}\right)^{0}\left(\dfrac{78}{88}\right)^{10-0}+_{10} C_{1}\left(\dfrac{1}{88}\right)^{1}\left(\dfrac{78}{88}\right)^{10-1} \\ &+_{10} C_{2}\left(\dfrac{1}{88}\right)^{2}\left(\dfrac{78}{88}\right)^{10-2} \\ &=0.892+0.103+0.005>0.999 \end{aligned}
Using the TI-83/84 Calculator:
$$P(x \leq 2)=\text { binomcdf }(10,1 \div 88,2) \approx 0.9998$$
Using R:
$$P(x \leq 2)=\text { pbinom }(2,10,1 / 88) \approx 0.9998$$
g. Since the probability of five or more children in a group of ten having autism is much less than 5%, it is unusual to happen. If this does happen, then one may think that the proportion of children diagnosed with autism is actually more than 1/88.
## Homework
Exercise $$\PageIndex{1}$$
1. Suppose a random variable, x, arises from a binomial experiment. If n = 14, and p = 0.13, find the following probabilities using the binomial formula.
1. P (x=5)
2. P (x=8)
3. P (x=12)
4. $$P(x \leq 4)$$
5. $$P(x \geq 8)$$
6. $$P(x \leq 12)$$
2. Suppose a random variable, x, arises from a binomial experiment. If n = 22, and p = 0.85, find the following probabilities using the binomial formula.
1. P (x=18)
2. P (x=5)
3. P (x=20)
4. $$P(x \leq 3)$$
5. $$P(x \geq 18)$$
6. $$P(x \leq 20)$$
3. Suppose a random variable, x, arises from a binomial experiment. If n = 10, and p = 0.70, find the following probabilities using the binomial formula.
1. P (x=2)
2. P (x=8)
3. P (x=7)
4. $$P(x \leq 3)$$
5. $$P(x \geq 7)$$
6. $$P(x \leq 4)$$
4. Suppose a random variable, x, arises from a binomial experiment. If n = 6, and p = 0.30, find the following probabilities using the binomial formula.
1. P (x=1)
2. P (x=5)
3. P (x=3)
4. $$P(x \leq 3)$$
5. $$P(x \geq 5)$$
6. $$P(x \leq 4)$$
5. Suppose a random variable, x, arises from a binomial experiment. If n = 17, and p = 0.63, find the following probabilities using the binomial formula.
1. P (x=8)
2. P (x=15)
3. P (x=14)
4. $$P(x \leq 12)$$
5. $$P(x \geq 10)$$
6. $$P(x \leq 7)$$
6. Suppose a random variable, x, arises from a binomial experiment. If n = 23, and p = 0.22, find the following probabilities using the binomial formula.
1. P (x=21)
2. P (x=6)
3. P (x=12)
4. $$P(x \leq 14)$$
5. $$P(x \geq 17)$$
6. $$P(x \leq 9)$$
7. Approximately 10% of all people are left-handed ("11 little-known facts," 2013). Consider a grouping of fifteen people.
1. State the random variable.
2. Argue that this is a binomial experiment Find the probability that
3. None are left-handed.
4. Seven are left-handed.
5. At least two are left-handed.
6. At most three are left-handed.
7. At least seven are left-handed.
8. Seven of the last 15 U.S. Presidents were left-handed. Is this unusual? What does that tell you?
8. According to an article in the American Heart Association’s publication Circulation, 24% of patients who had been hospitalized for an acute myocardial infarction did not fill their cardiac medication by the seventh day of being discharged (Ho, Bryson & Rumsfeld, 2009). Suppose there are twelve people who have been hospitalized for an acute myocardial infarction.
1. State the random variable.
2. Argue that this is a binomial experiment Find the probability that
3. All filled their cardiac medication.
4. Seven did not fill their cardiac medication.
5. None filled their cardiac medication.
6. At most two did not fill their cardiac medication.
7. At least three did not fill their cardiac medication.
8. At least ten did not fill their cardiac medication.
9. Suppose of the next twelve patients discharged, ten did not fill their cardiac medication, would this be unusual? What does this tell you?
9. Eyeglassomatic manufactures eyeglasses for different retailers. In March 2010, they tested to see how many defective lenses they made, and there were 16.9% defective lenses due to scratches. Suppose Eyeglassomatic examined twenty eyeglasses.
1. State the random variable.
2. Argue that this is a binomial experiment Find the probability that
3. None are scratched.
4. All are scratched.
5. At least three are scratched.
6. At most five are scratched.
7. At least ten are scratched.
8. Is it unusual for ten lenses to be scratched? If it turns out that ten lenses out of twenty are scratched, what might that tell you about the manufacturing process?
10. The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s.
1. State the random variable.
2. Argue that this is a binomial experiment Find the probability that
3. Six M&M’s are brown.
4. Twenty-five M&M’s are brown.
5. All of the M&M’s are brown.
6. Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason?
1. a. P(x=5) = 0.0212, b. P(x=8) = $$1.062 \times 10^{-4}$$, c. P(x=12) = $$1.605 \times 10^{-9}$$, d. $$P(x \leq 4)=0.973$$, e. $$P(x \geq 8)=1.18 \times 10^{-4}$$, f. $$P(x \leq 12)=0.99999$$
3. a. $$P(x=2)=0.0014$$, b. $$P(x=8)=0.2335$$, c. $$P(x=7)=0.2668$$, d. $$P(x \leq 3)=0.0106$$, e. $$P(x \geq 7)=0.6496$$, f. $$P(x \leq 4)=0.0473$$
5. a. $$P(x=8)=0.0784$$, b. $$P(x=15)=0.0182$$, c. $$P(x=14)=0.0534$$, d. $$P(x \leq 12)=0.8142$$, e. $$P(x \geq 10)=0.7324$$, f. $$P(x \leq 7)=0.0557$$
7. a. See solutions, b. See solutions, c. P(x=0) = 0.2059, d. $$P(x=7)=2.770 \times 10^{-4}$$, e. $$P(x \geq 2)=0.4510$$, f. $$P(x \leq 3)=0.944$$, g.$$P(x \geq 7)=3.106 \times 10^{-4}$$, h. See solutions
9. a. See solutions, b. See solutions, c. $$P(x=0)=0.0247$$, d. $$P(x=20)=3.612 \times 10^{-16}$$, e. $$P(x \geq 3)=0.6812$$, f. $$P(x \leq 5)=0.8926$$, g. $$P(x \geq 10)=6.711 \times 10^{-4}$$, h. See solutions
This page titled 5.2: Binomial Probability Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Kathryn Kozak via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 7,607 | 23,860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-10 | latest | en | 0.683701 |
https://en.wikipedia.org/wiki/Trial-and-error_learning | 1,547,793,967,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659890.6/warc/CC-MAIN-20190118045835-20190118071835-00522.warc.gz | 503,968,784 | 20,778 | # Trial and error
(Redirected from Trial-and-error learning)
Trial and error is a fundamental method of problem solving.[1] It is characterised by repeated, varied attempts which are continued until success,[2] or until the agent stops trying.
According to W.H. Thorpe, the term was devised by C. Lloyd Morgan (1852–1936) after trying out similar phrases "trial and failure" and "trial and practice".[3] Under Morgan's Canon, animal behaviour should be explained in the simplest possible way. Where behaviour seems to imply higher mental processes, it might be explained by trial-and-error learning. An example is the skillful way in which his terrier Tony opened the garden gate, easily misunderstood as an insightful act by someone seeing the final behaviour. Lloyd Morgan, however, had watched and recorded the series of approximations by which the dog had gradually learned the response, and could demonstrate that no insight was required to explain it.
Edward Thorndike showed how to manage a trial-and-error experiment in the laboratory. In his famous experiment, a cat was placed in a series of puzzle boxes in order to study the law of effect in learning.[4] He plotted learning curves which recorded the timing for each trial. Thorndike's key observation was that learning was promoted by positive results, which was later refined and extended by B.F. Skinner's operant conditioning.
Trial and error is also a heuristic method of problem solving, repair, tuning, or obtaining knowledge. In the field of computer science, the method is called generate and test. In elementary algebra, when solving equations, it is "guess and check".
This approach can be seen as one of the two basic approaches to problem solving, contrasted with an approach using insight and theory. However, there are intermediate methods which for example, use theory to guide the method, an approach known as guided empiricism.
## Methodology
This approach is far more far use successful with simple problems and in games, and is often resorted to when no apparent rule applies. This does not mean that the approach need be careless, for an individual can be methodical in manipulating the variables in an attempt to sort through possibilities that may result in success. Nevertheless, this method is often used by people who have little knowledge in the problem area. The trial-and-error approach has been studied from its natural computational point of view [5]
### Simplest applications
Ashby (1960, section 11/5) offers three simple strategies for dealing with the same basic exercise-problem; and they have very different efficiencies: Suppose there are 1000 on/off switches which have to be set to a particular combination by random-based testing, each test to take one second. [This is also discussed in Traill (1978/2006, section C1.2]. The strategies are:
• the perfectionist all-or-nothing method, with no attempt at holding partial successes. This would be expected to take more than 10^301 seconds, [i.e. 2^1000 seconds, or 3·5×(10^291) centuries!];
• a serial-test of switches, holding on to the partial successes (assuming that these are manifest) would take 500 seconds on average; while
• a parallel-but-individual testing of all switches simultaneously would take only one second.
Note the tacit assumption here that no intelligence or insight is brought to bear on the problem. However, the existence of different available strategies allows us to consider a separate ("superior") domain of processing — a "meta-level" above the mechanics of switch handling — where the various available strategies can be randomly chosen. Once again this is "trial and error", but of a different type. This leads us to:
### Hierarchies
Ashby's book develops this "meta-level" idea, and extends it into a whole recursive sequence of levels, successively above each other in a systematic hierarchy. On this basis he argues that human intelligence emerges from such organization: relying heavily on trial-and-error (at least initially at each new stage), but emerging with what we would call "intelligence" at the end of it all. Thus presumably the topmost level of the hierarchy (at any stage) will still depend on simple trial-and-error.
Traill (1978/2006) suggests that this Ashby-hierarchy probably coincides with Piaget's well-known theory of developmental stages. [This work also discusses Ashby's 1000-switch example; see §C1.2]. After all, it is part of Piagetian doctrine that children learn by first actively doing in a more-or-less random way, and then hopefully learn from the consequences — which all has a certain to Ashby's random "trial-and-error".
### Application
Traill (2008, espec. Table "S" on p.31) follows Jerne and Popper in seeing this strategy as probably underlying all knowledge-gathering systems — at least in their initial phase.
Four such systems are identified:
• Natural selection which "educates" the DNA of the species,
• The brain of the individual (just discussed);
• The "brain" of society-as-such (including the publicly held body of science); and
### Intention
In the Ashby-and-Cybernetics tradition, the word "trial" usually implies random-or-arbitrary, without any deliberate choice.
However amongst non-cyberneticians, "trial" will often imply a deliberate subjective act by some adult human agent; (e.g. in a court-room, or laboratory). So that has sometimes led to confusion.
Of course the situation becomes even more confusing if one accepts Ashby's hierarchical explanation of intelligence, and its implied ability to be deliberate and to creatively design — all based ultimately on non-deliberate actions. The lesson here seems to be that one must simply be careful to clarify the meaning of one's own words, and indeed the words of others. [Incidentally it seems that consciousness is not an essential ingredient for intelligence as discussed
One of the greatest inventor with 1093 patents was Thomas Alva Edison (TAE). One of his most famous sayings was that "Genius was 5% inspiration and 95% perspiration". Edison and his assistants would tried 500 ways to invent something and when not succeesful, they would try another 500 ways. When an experiment was not successful, he was happy to admit that he now knew the things that did not work. Trial and error was Edison's most successful means of invention. Most today believe Edison did not follow a scientific theory to invent but invented by: Trial And Error (TAE).
Today, the approach of voodoo programming is an usage where code is composed with trial and error until something which produces the desired output is found.
## Features
Trial and error has a number of features:
• solution-oriented: trial and error makes no attempt to discover why a solution works, merely that it is a solution.
• problem-specific: trial and error makes no attempt to generalize a solution to other problems.
• non-optimal: trial and error is generally an attempt to find a solution, not all solutions, and not the best solution.
• needs little knowledge: trials and error can proceed where there is little or no knowledge of the subject.
It is possible to use trial and error to find all solutions or the best solution, when a testably finite number of possible solutions exist. To find all solutions, one simply makes a note and continues, rather than ending the process, when a solution is found, until all solutions have been tried. To find the best solution, one finds all solutions by the method just described and then comparatively evaluates them based upon some predefined set of criteria, the existence of which is a condition for the possibility of finding a best solution. (Also, when only one solution can exist, as in assembling a jigsaw puzzle, then any solution found is the only solution and so is necessarily the best.)
## Examples
Trial and error has traditionally been the main method of finding new drugs, such as antibiotics. Chemists simply try chemicals at random until they find one with the desired effect. In a more sophisticated version, chemists select a narrow range of chemicals it is thought may have some effect using a technique called structure-activity relationship. (The latter case can be alternatively considered 8as a changing of the problem rather than of the solution strategy: instead of "What chemical will work well as an antibiotic?" the problem in the sophisticated approach is "Which, if any, of the chemicals in this narrow range will work well as an antibiotic?") The method is used widely in many disciplines, such as polymer technology to find new polymer types or families.
Trial and error is also commonly seen in player responses to video games - when faced with an obstacle or boss, players often form a number of strategies to surpass the obstacle or defeat the boss, with each strategy being carried out before the player either succeeds or quits the game.
Sports teams also make use of trial and error to qualify for and/or progress through the playoffs and win the championship, attempting different strategies, plays, lineups and formations in hopes of defeating each and every opponent along the way to victory. This is especially crucial in playoff series in which multiple wins are required to advance, where a team that loses a game will have the opportunity to try new tactics to find a way to win, if they are not eliminated yet.
The scientific method can be regarded as containing an element of trial and error in its formulation and testing of hypotheses. Also compare genetic algorithms, simulated annealing and reinforcement learning – all varieties for search which apply the basic idea of trial and error.
Biological evolution can be considered as a form of trial and error.[6] Random mutations and sexual genetic variations can be viewed as trials and poor reproductive fitness, or lack of improved fitness, as the error. Thus after a long time 'knowledge' of well-adapted genomes accumulates simply by virtue of them being able to reproduce.
Bogosort, a conceptual sorting algorithm (that is extremely inefficient and impractical), can be viewed as a trial and error approach to sorting a list. However, typical simple examples of bogosort do not track which orders of the list have been tried and may try the same order any number of times, which violates one of the basic principles of trial and error. Trial and error is actually more efficient and practical than bogosort; unlike bogosort, it is guaranteed to halt in finite time on a finite list, and might even be a reasonable way to sort extremely short lists under some conditions.
Jumping spiders of the genus Portia use trial and error to find new tactics against unfamiliar prey or in unusual situations, and remember the new tactics.[7] Tests show that Portia fimbriata and Portia labiata can use trial and error in an artificial environment, where the spider's objective is to cross a miniature lagoon that is too wide for a simple jump, and must either jump then swim or only swim.[8][9]
## References
1. ^ Evolutionary Epistemology, Rationality, and the Sociology of Knowledge p94 p108
2. ^ Concise Oxford Dictionary p1489
3. ^ Thorpe W.H. The origins and rise of ethology. Hutchinson, London & Praeger, New York. p26. ISBN 978-0-03-053251-1
4. ^ Thorndike E.L. 1898. Animal intelligence: an experimental study of the association processes in animals. Psychological Monographs #8.
5. ^ X. Bei, N. Chen, S. Zhang, On the Complexity of Trial and Error, STOC 2013
6. ^ Wright, Serwall (1932). "The roles of mutation, inbreeding, crossbreeding and selection in evolution" (PDF). Proceedings of the sixth international congress on genetics. Volume 1. Number 6: 365. Retrieved 17 March 2014.
7. ^ Harland, D.P. & Jackson, R.R. (2000). ""Eight-legged cats" and how they see - a review of recent research on jumping spiders (Araneae: Salticidae)" (PDF). Cimbebasia. 16: 231–240. Archived from the original (PDF) on 28 September 2006. Retrieved 5 May 2011.
8. ^ Jackson, Robert R.; Fiona R. Cross; Chris M. Carter (2006). "Geographic Variation in a Spider's Ability to Solve a Confinement Problem by Trial and Error". International Journal of Comparative Psychology. 19: 282–296. Retrieved 8 June 2011.
9. ^ Jackson, Robert R.; Chris M. Carter; Michael S. Tarsitano (2001). "Trial-and-error solving of a confinement problem by a jumping spider, Portia fimbriata". Behaviour. Leiden: Koninklijke Brill. 138 (10): 1215–1234. doi:10.1163/15685390152822184. ISSN 0005-7959. JSTOR 4535886. | 2,707 | 12,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.980643 |
http://mathoverflow.net/questions/82074/summing-a-function-using-modulus/82096 | 1,469,272,995,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00237-ip-10-185-27-174.ec2.internal.warc.gz | 161,957,882 | 16,333 | # Summing a function using modulus. [closed]
The problem:
If the infinite sum of a function is known, how to find:
\begin{align*} \sum_{i\equiv 0 \mod m}f(x_0+i)=\\\\ f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m)+\ldots \end{align*}
And if the finite sum of a function is known, how to find:
\begin{align*} \sum_{i\equiv 0 \mod m}^{i = {(x_0+\lfloor \frac{x-x_0+1}{m}\rfloor m)}}f(x_0+i)=\\\\ f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m) &\quad +\ldots+f\left(x_0+\left\lfloor \frac{x-x_0+1}{m}\right\rfloor m\right) \end{align*}
Details:
If we know a function $f$ and we can find the sum of its terms (defined as $S_f$), how to find the sum, but jumping some factors (defined as $MS_f$, where M representes modular)?
What's the relation with the sum function ($S_f$)? (I think this uses the root of the unity, but don't know how.)
For example, if:
$$S_f=\displaystyle\sum_{i=1}^{\infty}f(i)=f(1)+f(2)+\ldots$$
with infinite terms, how to find
\begin{align*} MS_f(x_0,m)&=\sum_{i\equiv 0 \mod m}f(x_0+i)\\\\ & = f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m)+\ldots \end{align*}
And if:
$$S_f(x)=\displaystyle\sum_{i=1}^{x}f(i)=f(1)+\ldots+f(x-1)+f(x),$$
how to find
\begin{align*} MS_f(x,x_0,m)&=\sum_{i\equiv 0 \mod m}^{i = {(x_0+\lfloor \frac{x-x_0+1}{m}\rfloor m)}}f(x_0+i)\\\\ & = f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m) \\\\ &\quad +\ldots+f\left(x_0+\left\lfloor \frac{x-x_0+1}{m}\right\rfloor m\right) \end{align*}
where $(x_0+\lfloor \frac{x-x_0+1}{m}\rfloor m)$ is the ultimate term of the arithmetic progression $x_0+k\times m$ which not exceeds $x$.
Edited:
As Jacques Carette said, I think the answer is using something like:
$MS_f(x,x_0,m)=\displaystyle\sum_{i=0}^{m-1}a_iS_f(w^ix)$ or $\displaystyle\sum_{i=0}^{m-1}a_iS_f(w^i(x+x0))$
but I don't know exactly.
Example:
$$S_f=\sum_{i=1}^{\infty}\frac{x^{i-1}}{(i-1)!}=e^x, \quad f(i)=\frac{x^{i-1}}{(i-1)!}$$ \begin{align*} MS_f(x_0,m)=\sum_{i\equiv 0 \mod m}f(x_0+i)=\sum_{i\equiv 0 \mod m}\frac{x^{(x_0+i)-1}}{((x_0+i)-1)!}\implies\\\\ MS_f(3,2)=\sum_{i\equiv 0 \mod 2}\frac{x^{(3+i)-1}}{((3+i)-1)!}=\sum_{j=0}^{\infty}\frac{x^{3+2j-1}}{(3+2j-1)!}=\cosh (x)-1 \end{align*}
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## closed as too localized by Gjergji Zaimi, Dan Petersen, Mariano Suárez-Alvarez♦, Yemon Choi, quid Nov 29 '11 at 0:44
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That you asked on math.SE and did not get answers (in one day!) there does not automatically mean you should ask here. Maybe next time you could ask first on tea.mathoverflow.net if the question fits MO? – Mariano Suárez-Alvarez Nov 28 '11 at 18:48
Well it has been very difficult join to this community. I think I have good questions, with Mathematics interest (like said in faq). The truth is, I didn't asked here because I didn't get ansser in Math.SE, I asked here because I wanted too, I would like another different opinions. I had seen questions in Math.SE with over 30 votes with no answer and answered here. So, what's the problem? If I ask in Math.SE and write in the question, you close my question. I I ask there too and write in the question, you close my question. Do you hate the Math.SE, didn't like no cross platafform? – GarouDan Nov 30 '11 at 10:19
If I need ask things here and just here (what's is annoying thing) tell me that I'll remember. But, but you may point in faq where is it? Or put there (not in meta) that's not a polite thing. Maybe I had been some agressive, but I think all community (even me, a beginner here) should opine to have a better site. – GarouDan Nov 30 '11 at 10:23
In general, you cannot get anything from just knowing $S_f$, there is just too little information. Note also that your $M$ is really an operator on $f$, not $S_f$; to see this, think of $f(i)=1/i!$ then $S_f = e$. How will $M$ manipulate $e$ to get those sums?
On the other hand, looking at your actual example, it seems you might want to know about generating functions rather than general sums of sequences. That is entirely different.
And your 'I think it uses roots of unity' was indeed along the right track. Let's consider the case where $x_0=0, m=2$, which is classical: this is the 'even' part of a function, which you can compute with $\frac{S_{f}(x)+S_{f}(-x)}{2}$. For $x_0=1,m=2$, you get the odd part, via $\frac{S_{f}(x)-S_{f}(-x)}{2}$.
In general, you'll want a function that looks like $$\frac{1}{m}\Sigma_{i=0}^{m-1} a_i S_{f}(\omega^i x)$$ for some weights $a_i$ which are also simple functions of $\omega^i$, where $\omega$ is a primitive $m$-root of unity.
For example, in the case where your sequences are holonomic, there are powerful algorithms for dealing with these questions (and most of them are implemented in both Maple and Mathematica, AFAIK). See the book $A=B$ by Petkovsek, Wilf and Zeilberger for a good introduction, and then Richard Stanley's 2-volume Enumerative Combinatorics if you want more.
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@JacquesCarette. I think you're right about the $S_f$ and $MS_f$. Your formula using $x_0=0$ and $m=2$ shows what I need. And you're right again when you say I'm looking for something like $MS_f(x,x_0,m)=\displaystyle\sum_{i=0}^{m-1}a_iS_f(w^ix)$. I already know this A=B book and it's really a good one, about the other I'm looking for. But isn't clear to me how to find this $a_i$'s, can you point something or a algorithm on the books to treat this? – GarouDan Nov 28 '11 at 15:22
There is absolutely nothing that you can say in general about this problem without putting some additional restriction on the function $f$. Think about it: you can start with any collection of $m$ functions $f_i, 1\le i\le m$, each supported on a distinct residue class mod $m$, and then add them together to get a new function $f$. But because the choice of $f_i$'s is completely arbitrary there is nothing that you can say in general about the relationship between the summatory function of $f$ and the summatory function of one of the $f_i$'s. If you have a particular function or a particular class of functions in mind then there is a chance, but in general no.
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@AH, What about if $f$ a polynomial or a analytic function. – GarouDan Nov 28 '11 at 16:41
For a non-zero polynomial the series you asked about don't converge. For your question about the relationship between the partial sums there is a closed form solution for polynomials, so yes there is a relationship. – Alan Haynes Nov 28 '11 at 17:17 | 2,160 | 6,720 | {"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} | 4.34375 | 4 | CC-MAIN-2016-30 | latest | en | 0.542654 |
http://slideplayer.com/slide/2404282/ | 1,544,865,145,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826842.56/warc/CC-MAIN-20181215083318-20181215105318-00386.warc.gz | 252,924,672 | 31,099 | # Square Roots and the Pythagoren Theorm
## Presentation on theme: "Square Roots and the Pythagoren Theorm"— Presentation transcript:
Square Roots and the Pythagoren Theorm
1.1 Square Numbers and Area Models
We can prove that 36 is a square number
We can prove that 36 is a square number. Draw a square with an area of 36 square units. 6 units 36 = 6 x 6 = 62 62 = 36
We can prove that 49 is a square number
We can prove that 49 is a square number. Draw a square with an area of 49 square units. 7 units 49 = 7 x 7 = 72 72 = 49
A square has an area of 64 cm2 Find the perimeter
A square has an area of 64 cm2 Find the perimeter. What number when multiplied by itself will give 64? 8 x 8 = 64 So the square has a side length of 8cm. Perimeter is the distance around: = 32 64 cm2
What is a Perfect Square? Part 1
Any rational number that is the square of another rational number. In other words, the square root of a perfect square is a whole number. Perfect Square Square Root 1 √1 = 1 4 √4 = 2 9 √9 = 3 16 √16 = 4 25 √25 = 5
Perfect Squares Use a calculator to determine if the following are perfect squares Perfect Square? Square Root Per. Square? √121 = Y/N √169 = Y/N √99 = Y/N √50 = Y/N
Perfect Squares - KEY Use a calculator to determine if the following are perfect squares Perfect Square? Square Root Per. Square? √121 = 11 Y/N √169 = 13 Y/N √99 = 9.95 Y/N √50 = 7.07 Y/N
What is a Perfect Square? Part 2
Another way to look at it. If we can find a division sentence for a number so that the quotient is equal to the divisor, the number is a square number. 16 ÷ 4 = 4 Dividend divisor quotient
Quiz #1 Ch 1 1) List the first 12 perfect squares. 2) If a square has a side length of 5cm, what is the area? Show your work. 3) Find the side length of a square with an area of 81 cm2. Show your work.
Quiz #1 Ch 1 Key 1) List the first 12 perfect squares. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ) If a square has a side length of 5cm, what is the area? 25cm2 3) Find the side length of a square with an area of 81 cm2. 9cm
1.2 Squares and Roots Squaring and taking the square root are inverse operations. That is they undo each other. 42 = 16 √16 = √4x4 = 4
Factors 1-30 What do you notice about all the yellow columns?
They all have an odd number of factors! They are perfect squares! The middle factor is the square root of the perfect square!
What is a perfect square? – PART 3
A perfect square will have its factor appear twice. Ex: 36 ÷ 1 = 36 1 and 36 are factors of ÷ 2 = 18 2 and 18 are factors of ÷ 3 = 12 3 and 12 are factors of ÷ 4 = 9 4 and 9 are factors of ÷ 6 = 6 6 is a factor that occurs twice Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36 The square root of 36 is 6 because it appears twice. It is also the middle factor when they are listed in ascending order!
What is a Perfect Square – Part 4
Is 136 a perfect square? Perfect squares have an odd number of factors. List the factors. 1 x 136 = x 68 = x 34 = x 17 = 136 There are 8 factors in 136. Therefore, 136 is not a perfect square because perfect squares have an odd number of factors.
1.2 Quiz Find the square root of 144. Find 42
List the factors of Is there a square root? If so what is the square root? Which perfect squares have square roots between 1 and 50.
1.2 Quiz Find the square root of 144. 12 Find 42 16
3. List the factors of Is it a PERFECT SQUARE? If so what is the square root? Yes it is a perfect square because there is an odd number of factors. 1, 11 ,121. The square root is 11. 4. What are the perfect squares between 1and 50. 1, 4, 9, 16, 25, 36, 49
1.3 Measuring Line Segments – Inside out.
1.3 Inside Out The Steps The Formula A = l2 + 4 [(b)(h)/2]
You can find the length of a line segment AB on a grid by constructing a square on the segment. The length of AB is the square root of the area of the square. Step 1 – Make a square around the line segment Step 2 – Cut the square into 4 congruent triangles and a smaller square. Step 3 – Calculate the area of the triangle A = bh/2 A = (3)(2)/2 A = 3 units The area of one triangle is 3 units, so all triangles would be 4(3) = 12 units Step 4 Calculate the are of a small square A = L x L = A = 1 x 1 = 1 unit Step 5 Add the area of the squares and triangles together = 13 so the line segment is the square root of 13 The Formula A = l2 + 4 [(b)(h)/2]
1.3 Measuring Line Segments – Outside In
1.3 Outside In The Steps The Formula A = l2 - 4 [(b)(h)/2]
You can find the length of a line segment AB on a grid by constructing a square on the segment. The length of AB is the square root of the area of the square. Step 1 – Make a square around the line segment Step 2 – Draw a larger square around the line segment square. Step 3 – Calculate the area of the outside square = l2 = 9 x 9 = 81 Step 4 – Calculate the area of the triangles (remember there are 4) 4 [(b)(h)/2] = 4 [(4)(5)/2] = 40 Step 5 Subtract the area of the triangles from the square. 81 – 40 = 41 so the line segment is the square root of 41 The Formula A = l2 - 4 [(b)(h)/2]
Practice Time Complete the 7 questions below. You will be given a hard copy (extra practice 1.3). You will need graph paper for #4. Use inside-out for # 3 and outside-in for #4.
1.4 Estimating Square Roots
Here is one way to estimate the value of the square root of a number that is not a perfect square. For example: Find √20 Step 1: Is it a perfect square? No Step 2: If it isn’t, sandwich it between 2 perfect squares. √16 < √20 < √25 4 < √20 < √20 is closer to 4 than 5 Now we use guess and check. 4.6 x 4.6 = 21.16 4.5 x 4.5 = 20.25 4.47x 4.47 = 19.98 Therefore the √20 = approximately 4.47 Bingo, this one is closest!!!!
Another way to estimate √20
Bingo, this one is closest!!!!
Find √27 Step 1: Is it a perfect square? No Step 2: If it isn’t, sandwich it between 2 perfect squares. √25 < √27 < √36 5 < √27 < √27 is closer to 5 than 6 Now we use guess and check. 5.2 x 5.2 = 27.04 5.19 x 5.19 = 26.93 Therefore the √27 = approximately 5.2 Bingo, this one is closest!!!!
Bingo, this one is closest!!!!
Find √105 Step 1: Is it a perfect square? No Step 2: If it isn’t, sandwich it between 2 perfect squares. √100< √105 < √121 10 < √105 < √105 is closer to 10 than 11 Now we use guess and check. 10.2 x 10.2 = 10.25 x = 10.24 x = Therefore the √105 = approximately 10.25 Bingo, this one is closest!!!!
Place each of the following square roots on the number line below.
√5, √52, and √89 √4< √5 < √9 2 < √5 < 3 - √5 is closer to 2 than 3 2.2 x 2.2 = 4.84 2.25x2.25 = 5.063 2.24 x 2.24 = 5.017 √5= approximately 2.24 √49 < √52 < √64 - √52 is closer to 7 than 8 7.2 x 7.2 = 51.8 7.25 x 7.25 = 52.56 7.22 x 7.22 = 52.12 7.21 x 7.21 = 51.98 √52= approximately 7.21 √81 < √89 < √ √83 is closer to 9 than 10 9.4 x 9.4 = 88.36 9.45 x 9.45 = 89.30 9.43 x 9.43 = 9.44 x = 89.12 √89= approximately 9.43 Bingo, this one is closest!!!! Bingo, this one is closest!!!! Bingo, this one is closest!!!! √5 √52 √89
1.5 The Pythagorean Theorem
In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares whose sides are the two legs
Watch This Video!
Pythagorus Side A and side B are always the legs and they are “attached” to the right angle. Side C is always across from the right angle. It is always longer than side A or side B. If you add the squares of side A and B, it will = the square of side C
Some Questions Find the hypotenuse. a2 + b2 = c = c = c2 85 = c2 √85 = √c = c We can now say that 6, 7, and 9.22 are not Pythagorean triplets because one is not a whole number
Some Questions Find the hypotenuse. a2 + b2 = c = c = c2 100 = c2 √100 = √c2 10 = c We can now say that 6, 8, 10 are Pythagorean triplets
Some Questions Find the leg “x”. We will make x – a. a2 + b2 = c2 a = 182 a = 324 a = a2 = 203 √a2 = √203 a = We can now say that 11, 14.24, and 18 are not Pythagorean triplets, because one is not a whole number.
Exploring the Pythagorean Theorem
1.6 Exploring the Pythagorean Theorem
For each triangle below, add up the 2 areas of the squares of the legs in the 2nd column, and include the area of the square of the hypotenuse in the third column. Do you see any patterns?
Use Pythagoras to determine if the triangle below is a right triangle.
a2 + b2 = c2 = ? = 81 ? ≠ 81 This triangle is not a right triangle!
Use Pythagoras to determine if the triangle below is a right triangle.
a2 + b2 = c2 = ? = 625 ? 625 = 625 This triangle is a right triangle! We can now say that 7, 24, and 25 are Pythagorean triplets.
What is a Pythagorean Triplet?
It is a set of WHOLE numbers that satisfy the Pythagorean theorem. For example, this triangles’ sides (3, 4, 5) satisfy the Pythagorean theorem and are therefore triplets. This is because they are all whole numbers and = 52
What is a Pythagorean Triplet?
This triangles’ sides (6, 8, 11) do not satisfy the Pythagorean theorem and are not therefore triplets. Although they are all whole numbers, they are not triplets because ≠ 112
112 + 14.242 = 182 14.24 Pythagorean Triplets
This triangles’ sides are not Pythagorean triplets because one of the sides is not a whole number eventhough: = 182 14.24
Your Turn! In one minute, write down as many Pythagorean triplets as you can where c (the hypotenuse) is less than 100. Here are a few. ( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (16, 30 34) (20, 21, 29) (15, 20, 25) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
Applying the Pythagorean Theorm
1.7 Applying the Pythagorean Theorm
Find the missing side. a2 + b2 = c b2 = b2 = b2 – 16 = 49 – 16 b2 = 33 √b2 = √33 b = 5.74
Draw a diagram to solve Pythagorean Word Problems!!
Whenever Possible Draw a diagram to solve Pythagorean Word Problems!!
Tanya runs diagonally across a rectangular field that has a length of 40m and a width of 30m. What is the length of the diagonal, in yards, that Tanya runs? a2 + b2 = c2 = c2 = c2 2500 = c2 √2500 = √c2 50 = c
To get from point A to point B you must avoid walking through a pond
To get from point A to point B you must avoid walking through a pond. To avoid the pond, you must walk 34 meters south and 41 meters east. To the nearest meter, how many meters would be saved if it were possible to walk through the pond? a2 + b2 = c2 = c2 = c2 2837= c2 √2837= √c2 53.26 = c
a2 + b2 = c2 32 + b2 = 52 9+ b2 = 25 9+ b2 - 9 = 25 – 9 √b2 = √16
Leo's dog house is shaped like a tent. The slanted sides are both 5 feet long and the bottom of the house is 6 feet across. What is the height of his dog house, in feet, at its tallest point? a2 + b2 = c2 32 + b2 = 52 9+ b2 = 25 9+ b2 - 9 = 25 – 9 √b2 = √16 b = 4
A ship sails 80 km due east and then 18 km due north
A ship sails 80 km due east and then 18 km due north. How far is the ship from its starting position when it completes this voyage? = c = c2 6724= c2 √6724 = √c2 82 = c
A ladder 7.25 m long stands on level ground so that the top end of the ladder just reaches the top of a wall 5 m high. How far is the foot of the ladder from the wall? a2 + b2 = c2 a = 7.252 a = 56.56 a = √a2 = √27.56 a = 5.25
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