url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.prep4paper.com/quantitative-aptitude/Trigonometry-SSC-Shortcuts.html | 1,660,593,363,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00626.warc.gz | 82,510,631 | 5,350 | @ : Home > Quantitative Aptitude > Trigonometry > Trigonometry SSC Shortcuts
## SSC Shortcuts (Substitution method)
At times students find trigonometry questions in SSC paper very hard and they leave those questions unattempted. Whereas, there are other students who are well versed with shortcut methods. These students use shortcuts and solve such questions within seconds. Here we are discussing some shortcuts tips and tricks for our students so that next time you are in a position to solve complex questions within seconds.
Q19) If cos x + cos y = 2 then the value of sin x + sin y is
1)
0
2) 1
3) 2
4) -1
cos x + cos y = 2
cos x = 1; cos y = 1
x = y = 0
sin x + sin y = 0
Hence, option 1.
Q20) The value of is
1)
5
2) 4
3) 3
4) 2
Easiest way to solve such questions is to substitute value of θ.
Put θ = 45°
Hence, option 1.
Q21) If and then cos2α is
1)
2)
3)
4)
Full method to solve this question:
Shortcut:
Put α = 60°; β = 30
tan60° = n tan30° i.e. n = 3 or n2 = 9
sin60° = m sin30° i.e. m = √3 or m2 = 3
cos2α = cos260° = 1/4
For m2 = 3; n2 = 9
Option 1: 1/3 Eliminated.
Option 2: 1/4 Satisfied.
Option 3: 1/2 Eliminated.
Option 4: 3/10 Eliminated.
Hence, option 2.
Trigonometric Identities
In the last question we got a unique answer using substitution method. However, it may not always be true. At times using, substitution method we may get more than one option. So Lets discuss how to solve such questions.
Q22) If then the value of is
Use substitution method, put θ = 90° Answer is either option 3 or option 4.
As we did not get a unique answer so put θ=0° Answer is option 4.
Hence, option 4.
Q23) If then the value of
1)
9
2) 0
3) 1
4) 4
Easiest substitution for this question is
Put θ = 45°,Φ = 45°
x = a sec45° cos45° = a
y = b sec45° sin45° = b
z = c tan45° = c
Hence, option 3.
Q24) If sin θ + cosθ = x then the value of sin6θ + cos6θ is equal to
Use substitution method, put
Answer is either option 3 or option 4.
Now, put
Answer is either option 1 or option 3.
So, common answer from the above substitutions is option 3.
Hence, option 3.
Q25) If sinθ + cosθ = 1, what is the value of sinθ cosθ
1)
2
2) 0
3) 1
4) 1/2
sinθ + cosθ = 1
(sinθ + cosθ)2 = 1
sin2θ + cos2θ + 2sinθ cosθ = 1
1 + 2sinθ cosθ = 1 i.e. sinθ cosθ = 0
Shortcut:
sinθ + cosθ = 1 so, put θ = 0°
sinθ cosθ = sin0° cos0° = 0.
Hence, option 2.
Q26) The numerical value of is
1)
0
2) -1
3) 1
4) 2
Put θ = 45°
Hence, option 3.
Q27) If cosθ + secθ = 2, then the value of cos6θ + sec6θ is
1)
1
2) 2
3) 4
4) 8
cosθ + secθ = 2
cosθ + 1/cosθ = 2
So, cosθ = 1 i.e. θ =
0° cos6θ + sec6θ = cos60° + sec60° = 2
Hence, option 2.
Q28) If sinθcosθ = 1/2, then what is sin6θ + cos6θ equal to
1)
1
2) 2
3) 3
4) 1/4
sinθcosθ = 1/2
2sinθcosθ = 1
sin2θ = 1
2θ = 90°
θ = 45°
sin6θ + cos6θ = sin645° + cos645° = 1/4
Hence, option 4. | 1,130 | 2,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.780898 |
https://2fwww.politics-prose.com/book/9781681739267 | 1,701,743,132,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00462.warc.gz | 94,588,393 | 10,550 | Fluid Mechanics Experiments (Synthesis Lectures on Mechanical Engineering) (Paperback)
\$39.95
Special Order—Subject to Availability
Fluid mechanics is one of the most challenging undergraduate courses for engineering students. The fluid mechanics lab facilitates students' learning in a hands-on environment. The primary objective of this book is to provide a graphical lab manual for the fluid mechanics laboratory. The manual is divided into six chapters to cover the main topics of undergraduate-level fluid mechanics. Chapter 1 begins with an overview of laboratory objectives and the introduction of technical laboratory report content. In Chapter 1, error analysis is discussed by providing examples. In Chapter 2, fluid properties including viscosity, density, temperature, specific weight, and specific gravity are discussed. Chapter 3 revolves around the fluid statics include pressure measurement using piezometers and manometers. Additionally, hydrostatic pressure on the submerged plane and curved surfaces as well as buoyancy and Archimedes' Principle are examined in Chapter 3. In Chapter 4, several core concepts of fluid dynamics are discussed. This chapter begins with defining a control system based on which momentum analysis of the flow system is explained. The rest of the chapter is allotted to the force acting on a control system, the linear momentum equation, and the energy equation. Chapter 4 also covers the hydraulic grade line and energy grade line experiment. The effect of orifice and changing cross-sectional area by using Bernoulli's' equation is presented in Chapter 4. The application of the siphon is extended from Chapter 4 by applying Bernoulli's' equation. The last two chapters cover various topics in both internal and external flows which are of great importance in engineering design. Chapter 5 deals with internal flow including Reynolds number, flow classification, flow rate measurement, and velocity profile. The last experiment in Chapter 5 is devoted to a deep understanding of internal flow concepts in a piping system. In this experiment, students learn how to measure minor and major head losses as well as the impact of piping materials on the hydrodynamics behavior of the flow. Finally, open channels, weirs, specific energy, and flow classification, hydraulic jump, and sluice gate experiments are covered in Chapter 6.
Product Details ISBN: 9781681739267
ISBN-10: 1681739267
Publisher: Morgan & Claypool
Publication Date: September 16th, 2020
Pages: 101
Language: English
Series: Synthesis Lectures on Mechanical Engineering | 510 | 2,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.933364 |
http://globalred.com.au/contribution-of-euclid-in-geometry.html | 1,553,230,394,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202628.42/warc/CC-MAIN-20190322034516-20190322060516-00018.warc.gz | 83,559,909 | 5,120 | # Contribution of euclid in geometry. Greek Geometry 2019-03-04
Contribution of euclid in geometry Rating: 9,7/10 1244 reviews
## Who Is Euclid and What Did He Do?
Greek Geometry and Its Influence Greek geometry eventually passed into the hands of the great Islamic scholars, who translated it and added to it. Many of the advances in geometry and mathematics were used in what eventually became known as physics. In the section Ganita calculations of his astronomical treatise Aryabhatiya, he made the fundamental advance in finding the lengths of chords of circles, by using the half chord rather than the full chord method used by Greeks. The temporal axis can displace the object if the axis is itself curved; so the curvature of spacetime in a gravitational field must result from the curvature of time, not of space. Euclid As the Father of Geometry A common misconception is that Euclid invented all concepts of geometry.
Next
## Euclid and His Contributions to Mathematics Essay
He gathered the work of all of the earlier mathematicians and created his landmark work, 'The Elements,' surely one of the most published books of all time. He is an important historical figure because all of the rules we use in Geometry today are based on the writings of Euclid, specifically 'The Elements'. The whole is greater than a part. Geometry was revolutionized by Euclid, who introduced mathematical rigor and the axiomatic method still in use today. When I was a student of physics I was troubled by the difficulties presented and aired by Sachs in this book. Now we actually have two competing ways of understanding gravity, either through Einstein's geometrical method or through the interaction of virtual particles in quantum mechanics. For, as the remarkable Aristotle tells us, the same ideas have repeatedly come to men at various periods of the universe.
Next
## Contribution of euclid to geometry
At the heart of Western science, Greek geometers helped to build strong foundations of architecture, astronomy, mechanics, and optics. The compass that he designs in it is my favorite. Second, the compass was placed on one end of the line, and the width was adjusted to fit the whole line. However, he left behind a legacy that has survived almost two and a half millennia. Philosophy of Science Metaphysics Home Page Copyright c 1996, 1998, 1999 Kelley L. An object passing by the earth is accelerated towards the earth and thereby acquires a velocity along a vector where it previously may have had no velocity at all. These quotes about him are also not certain, since much history about scholars was in fact legend and myth in the times of Ancient Greece.
Next
## Contribution of euclid to geometry
If three dimensional space is not extrinsically curved into time according to the axiom of open ortho-curvature, then it must be time that is extrinsically curved into the dimensions of space. He was thepioneer in the field of hydrostatics without the use of Calculus. Some realization of this, unfortunately, leads people more easily to the conclusion that science is conventionalistic or a social construction than to the more difficult truth that much remains to be understood about reality and that philosophical questions and perspectives are not always useless or without meaning. That theory rests on the use of non-Euclidean geometry. Euclid was a part of that culture.
Next
## Euclid Biography
In free fall we are being displaced with space itself , and so we move with our entire frame of reference and would not be able to detect that locally. If our imagination is necessarily Euclidean, hard-wired into the brain as we might now think by analogy with computers, but Einstein found a way to apply non-Euclidean geometry to the world, then we might think that space does have a reality and a genuine structure in the world however we are able to visually imagine it. This book, therefore, is for the purification and training of the understanding, while the Elements contain the complete and irrefutable guide to the scientific study of the subject of geometry. Inside these ranges it is not common to find books dedicated to a certain concept. This is one of the most influential and successful textbook ever written. Ideally, in any axiomatic system, the assumptions are of such a basic and intuitive nature that their truth can be accepted without qualms. Nevertheless, there is still rarely a public word spoken about the philosophical intelligibility of Einstein's own theory: the Relativistic theory of gravitation.
Next
## What was the main contribution of Euclid in geometry?
It was he who discovered the subject of proportions and the construction of the cosmic figures. Euclid proved a sequence of theorems that marks the beginning of number theory as a mathematical endeavor versus a numerological one. This makes for a finite Big Bang regardless of the dynamical fate of the universe, where that fate is tied to the effect of the curvature of time, locally positively curved but globally possibly Lobachevskian or Euclidean. In light of the distinction between intrinsic and extrinsic curvature, we must consider all the kinds of ontological axioms that will cover all the possible spaces that Euclidean and non-Euclidean geometries can describe. Their approach was very pragmatic and aimed very much at practical uses. Euclid's book the Elements also contains the beginnings of number theory.
Next
## What is the contribution of Euclid to the development of geometry
He also showed that the volume of a sphere is two thirds the volume of a cylinder with the same height and radius. The new geometries were another one of the mathematicians' pretty toys until Einstein showed us that space was in fact curved. It owed its discovery to the practice of land measurement. He popularized the principle that things are flat and can be measured. Introduction Until recently, Albert Einstein's complaints in his later years about the intelligibility of Quantum Mechanics often led philosophers and physicists to dismiss him as, essentially, an old fool in his dotage. Lesson Summary Euclid was an ancient Greek mathematician in Alexandria, Egypt.
Next
## Who is Euclid?
Three dimensional space can still be conceived as having an inherent hetero-curvature apart from the gravitational fate of the universe: non-Euclidean without the need to regard time or anything else as a fourth dimension into which space needs to be extrinsically curved. Furthermore, since Kant believed that space was a form imposed by our minds on the world, he did not believe that space actually existed apart from our experience. Also we can only visualize a positively curved surface if this is embedded in a Euclidean volume with an explicit extrinsic curvature. Even today, this data is used in preparing Hindu calendars. One of his pupils, , took the development of geometry further. As such, although Euclid is often given credit for everything in his Elements, my statement of his contribution is writing the book.
Next | 1,400 | 7,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-13 | latest | en | 0.956841 |
https://kupdf.net/download/ch-02_596c40d7dc0d60ff73a88e76_pdf | 1,719,281,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00827.warc.gz | 314,237,071 | 26,923 | # Ch 02
July 17, 2017 | Author: Vincents Genesius Evans | Category: Multiplication, Prime Number, Arithmetic, Numbers, Physics & Mathematics
#### Short Description
Maths Quests Grade 7 Chapter 2...
#### Description
Multiples, factors and primes
2 A room measures 550 centimetres by 325 centimetres. What would be the side length of the largest square tile that can be used to tile the floor without any cutting? This chapter will provide you with skills to answer this question more easily.
42
Maths Quest 7 for Victoria
Multiples A multiple of a number is the answer obtained when that number is multiplied by another whole number. All numbers in the 5 times table are multiples of 5, so 5, 10, 15, 20, 25, . . . are all multiples of 5. The number 5 has been multiplied by one other number to find each of the numbers in the 5 times table, so they are all multiples of 5.
WORKED Example 1 List the first 5 multiples of 7. THINK
WRITE
1
First multiple is the number × 1 = 7 × 1.
2
Second multiple is the number × 2 = 7 × 2.
3
Third multiple is the number × 3 = 7 × 3.
4
Fourth multiple is the number × 4 = 7 × 4.
5
Fifth multiple is the number × 5 = 7 × 5.
7, 14, 21, 28, 35
WORKED Example 2 Write the numbers in the list that are multiples of 8. 18, 8, 80, 100, 24, 60, 9, 40 THINK
WRITE
1
The biggest number in the list is 100. List multiples of 8 using the 8 times table just past 100; that is, 8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40, etc.
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104
2
Write any multiples that appear in the list.
Numbers in the list that are multiples of 8 are 8, 24, 40, 80.
remember remember 1. A multiple of a number is the answer obtained when that number is multiplied by another whole number. For example, all numbers in the 5 times table are multiples of 5; that is, 5, 10, 15, 20, 25 . . . 2. If 2 or more numbers have the same multiple, it is called a common multiple.
Chapter 2 Multiples, factors and primes
2A WORKED
Example
1
WORKED
Example
2
43
Multiples
1 List the first 5 multiples of the following numbers. a 3 b 6 c 100 e 15 f 4 g 21 i 14 j 12 k 50 m 33 n 120 o 45
d h l p
2.1
11 25 30 72
2 Write the numbers in the following list that are multiples of 10. 10, 15, 20, 100, 38, 62, 70 3 Write the numbers in the following list that are multiples of 7. 17, 21, 7, 70, 47, 27, 35
Multiples
4 Write the numbers in the following list that are multiples of 16. 16, 8, 24, 64, 160, 42, 4, 32, 1, 2, 80 5 Write the numbers in the following list that are multiples of 35. 7, 70, 95, 35, 140, 5, 165, 105, 700 6 The numbers 16, 40 and 64 are all multiples of 8. Find 3 more multiples of 8 that are less than 100. 7 a List the first 10 multiples of 4. b List the first 10 multiples of 6. c In your lists, circle the multiples that 4 and 6 have in common (that is, circle the numbers that appear in both lists). d What is the lowest multiple that 4 and 6 have in common? This is the Lowest Common Multiple of 4 and 6, known as the LCM. 8 a b c d
Lowest common multiple
List the first 6 multiples of 3. List the first 6 multiples of 9. Circle the multiples that 3 and 9 have in common. What is the lowest common multiple of 3 and 9?
9 Find the LCM of each of the following pairs of numbers. a 3 and 6 b 5 and 7 c d 9 and 6 e 6 and 15 f g 7 and 10 h 9 and 12 i j 10 and 25 k 12 and 16 l
6 and 8 4 and 12 15 and 20 4 and 15
10 Answer true (T) or false (F) to each of the following statements. a The LCM of 4 and 5 is 20. b The LCM of 3 and 6 is 18. c 20 is a multiple of 10 and 2 only. d 15 and 36 are both multiples of 3. e 60 is a multiple of 2, 3, 6, 10 and 12. f 100 is a multiple of 2, 4, 5, 10, 12 and 25. g 30 is the LCM of 6 and 5.
Lowest common multiple
44
Maths Quest 7 for Victoria
11 multiple choice a The first 3 multiples of 9 are: A 1, 3, 9 B 3, 6, 9 b The first 3 multiples of 15 are: A 15, 30, 45 B 1, 3, 5, 15 c The LCM of 6 and 9 is: A 6 B 54 d The LCM of 7 and 12 is: A 21 B 84
C 9, 18, 27
D 9, 18, 81
E 1, 9, 18
C 30, 45, 60
D 1, 15, 30
E 45
C 36
D 9
E 18
C 48
D 1
E 12
12 Place the first 6 multiples of 3 into the triangle at right, so each line adds up to 27. Use each number once only.
number once only.
13 Kate goes to the gym every second evening while Ian goes every third evening. If they both attended the gym on Monday, how long will it be before they both attend again on the same day? What day of the week will this be? 14 Vinod and Elena are riding around a mountain bike trail. Each person completes one lap in the time shown on the stopwatches.
MIN
SEC
100/SEC
05:00:00
MIN
SEC
100/SEC
07:00:00
If they both begin cycling from the starting point at the same time, how long will it be before they pass this starting point again at exactly the same time?
Chapter 2 Multiples, factors and primes
45
15 Two smugglers, Bill Bogus and Sally Seadog have set up signal lights that flash continuously across the ocean. Bill’s light flashes every 5 seconds and Sally’s light flashes every 4 seconds. If they both start together, how long will it take for both lights to flash again at the same time? 16 Alex and Nadia were having races running down a flight of stairs. Nadia took the stairs 2 at a time while Alex took the stairs 3 at a time. In each case, they reached the bottom with no steps left over. a How many steps are there in the flight of stairs? List 3 possible answers. b What is the smallest number of steps there could be? c If Alex can also take the stairs 5 at a time with no steps left over, what is the smallest number of steps in the flight of stairs?
Multiples, factors and primes 01
GE
QUEST
EN
MAT H
S
GAME time
17 Twenty students in Year 7 were each given a different number from 1 to 20 and then asked to sit in numerical order in a circle. Three older girls, Milly, Molly and Mandy came to distribute jelly beans to the class. Milly gave red jelly beans to every 2nd student, Molly gave green jelly beans to every 3rd student and Mandy gave blue jelly beans to every 4th student. a Which student had jelly beans of all 3 colours? b How many students received exactly 2 jelly beans? c How many students did not receive any jelly beans?
CH
AL
L
1 I am a 2-digit number that can be divided by 3 with no remainder. The sum of my digits is a multiple of 4 and 6. My first digit is double my second digit. What number am I? 2 Find a 2-digit number such that if you subtract 3 from it, the result is a multiple of 3; if you subtract 4 from it, the result is a multiple of 4 and if you subtract 5 from it, the result is a multiple of 5. 3 In a class election with 3 candidates, the winner beat the other 2 candidates by 3 and 6 votes respectively. If 27 votes were cast, how many votes did the winner receive?
46
Maths Quest 7 for Victoria
In 1893, which country countr y was the first to let women women vote? vote? 24 44 27 48 10 60 40 14 65 28
Which people wer were e not allow allowed to vote until 1924? 18 36
6
8
12
15 100 9
34 30 45 20 22 42 21
Find the code to answer these questions by matching the letter beside each pair of numbers below with their lowest common multiple (LCM).
A
2, 7
D
5, 9
I
6, 4
N
8, 6
C 3, 15
D
7, 4
A
11, 2
N
3, 9
I
10, 4
L 8, 10
E
2, 5
N 10, 6
A 5, 12
M 4, 18
I
17, 2
N 6, 14
E
2, 3
R
4, 8
A
S
7, 3
W 3, 27
Z 16, 3
E
4, 11
N 13, 5
9, 2
A 4, 25
Chapter 2 Multiples, factors and primes
47
Factors A factor is a whole number that divides exactly into another whole number, with no remainder. If one number is divisible by another number, the second number divides exactly into the first number. For example, 8 is divisible by 4 because 8 ÷ 4 = 2. The number 4 is a factor of 8 because 4 divides into 8 twice with no remainder, or 8 ÷ 4 = 2. The number 2 is also a factor of 8 because 8 ÷ 2 = 4. The number 3 is a factor of 15 because 3 goes into 15, or 15 ÷ 3 = 5.
WORKED Example 3 Find all the factors of 14. THINK 1
2
3
WRITE
1 is a factor of every number and the number itself is a factor; that is, 1 × 14 = 14. 14 is an even number so 14 is divisible by 2 and is a factor. Divide the number by 2 to find the other factor (14 ÷ 2 = 7). Write a sentence placing the factors in order from smallest to largest.
1, 14
2, 7
The factors of 14 are 1, 2, 7 and 14.
Factor pairs It is often easiest to write factors in pairs called factor pairs. These are pairs of numbers which multiply to equal a certain number.
WORKED Example 4 List the factor pairs of 30. THINK 1 2 3
4 5
1 and the number itself are factors; that is, 1 × 30 = 30. 30 is an even number so 2 and 15 are factors; that is, 2 × 15 = 30. Divide the next smallest number into 30. Therefore, 3 and 10 are factors; that is, 3 × 10 = 30. 30 ends in 0 so 5 divides evenly into 30, that is, 5 × 6 = 30. List the factor pairs.
WRITE 1, 30 2, 15 3, 10
5, 6 The factor pairs of 30 are 1, 30; 2, 15; 3, 10 and 5, 6.
48
Maths Quest 7 for Victoria
Common factors A common factor is a number which is a factor of two or more given numbers. If the numbers 6 and 10 are given, then 2 is a common factor because 2 is a factor of 6 and 2 is a factor of 10. The Highest Common Factor or HCF is the largest of the common factors.
WORKED Example 5
a Find the common factors of 8 and 24 by: i listing the factors of 8 ii listing the factors of 24 iii listing the factors common to both 8 and 24. b State the highest common factor of 8 and 24. THINK
WRITE
a iii
a iii 1, 8 2, 4 Factors of 8 are 1, 2, 4, 8. iii 1, 24 2, 12 3, 8 4, 6 Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. iii Common factors are 1, 2, 4, 8.
Find the pairs of factors of 8. Write them in order. iii 1 Find the pairs of factors of 24. 2 Write them in order. iii Write the common factors. 1
2
b Find the highest common factor.
b HCF is 8.
Factors can make it easier to multiply numbers mentally. The following examples show the thought processes required.
WORKED Example 6
Find 12 × 15 using factors. THINK 1 2 3 4
WRITE
Write the question. Write one of the numbers as a factor pair. Multiply the first two numbers. Find the answer.
12 × 15 = 12 × 5 × 3 = 60 × 3 = 180
WORKED Example 7
Use factors to evaluate 32 × 25. THINK 1 2
3
WRITE
Write the question. Rewrite the question so that one pair of factors is easy to multiply. The number 4 is a factor of 32 and 4 × 25 = 100. Find the answer.
32 × 25 = 8 × 4 × 25 = 8 × 100 = 800
With practice this could be done in your head. Some useful products are 2 × 50 = 100, 8 × 125 = 1000, 4 × 250 = 1000.
Chapter 2 Multiples, factors and primes
49
remember remember 1. A factor is a whole number that divides exactly into another whole number, with no remainder. 2. The factors of a number can be found using factor pairs. One of the pairs may be a single number which multiplies by itself. The number 5 is a factor of 25 because 5 × 5 = 25. The number itself and 1 are always factors of a number. 3. If 2 or more numbers have the same factor, it is called a common factor. 4. The highest common factor or HCF of 2 or more numbers is the largest factor which divides into all of the given numbers.
2B WORKED
Example
3
WORKED
Example
4
Factors
1 Find all the factors of each of the following numbers. a 12 b 8 c 40 e 28 f 60 g 100 i 39 j 85 k 76 m 99 n 250 o 51
d h l p
2 List the factor pairs of: a 20 b 18
d 132
c
36
35 72 69 105
Factors
2.2
3 If 3 is a factor of 12, state the smallest number greater than 12 which has 3 as one of its factors. WORKED
Example
5
4 a Find the common factors of 15 and 35 by: i listing the factors of 15 ii listing the factors of 35 iii listing the factors common to both 15 and 35. b State the highest common factor of 15 and 35. 5 a List the factors of 21. b List the factors of 56. c Find the highest common factor of 21 and 56. 6 a List the factors of 27. b List the factors of 15. c Find the highest common factor of 27 and 15. 7 a List the factors of 7. b List the factors of 28. c Find the highest common factor of 7 and 28. 8 a List the factors of 48. b List the factors of 30. c Find the HCF of 48 and 30. 9 Find the highest common factor of 9 and 36. 10 Find the highest common factor of 26 and 65. 11 Find the highest common factor of 42 and 77.
Highest common factor
Highest common factor
50
Maths Quest 7 for Victoria
12 Find the highest common factor of 24 and 56. 13 Find the highest common factor of 36 and 64. 14 Find the highest common factor of 18 and 72. 15 Find the highest common factor of 45, 72 and 108. 16 multiple choice a A factor pair of 24 is: A 2, 4 B 4, 6 b A factor pair of 42 is: A 6, 7 B 20, 2 c The HCF of 12 and 30 is: A 2 B 3 d The HCF of 15 and 33 is: A 1 B 15
C 6, 2
D 2, 8
E 1, 14
C 21, 1
D 16, 3
E 0, 42
C 30
D 6
E 12
C 3
D 5
E 33
17 Which of the numbers 3, 4, 5 and 11 are factors of 2004? 18 Find the following using factors. a 12 × 25 b 12 × 35 6 d 11 × 16 e 11 × 14 g 20 × 15 h 20 × 18
c f i
12 × 55 11 × 15 30 × 21
19 Use factors to evaluate the following. a 36 × 25 b 44 × 25 7 d 72 × 25 e 124 × 25 g 56 × 50 h 48 × 125
c f i
24 × 25 132 × 25 52 × 250
WORKED
Example
WORKED
Example
20 Connie Pythagoras is trying to organise her Year 7 class into rows for their class photograph. If Ms Pythagoras wishes to organise the 20 students into rows containing equal numbers of students, what possible arrangements can she have? 21 multiple choice Tilly Tyler has 24 green bathroom tiles left over. If she wants to use them all on the wall behind the kitchen sink (without breaking any) which of the following arrangements would be suitable? I 4 rows of 8 tiles II 2 rows of 12 tiles III 4 rows of 6 tiles IV 6 rows of 5 tiles V 3 rows of 8 tiles A I and II B I, II and III C III, IV and V D II, IV and V E II, III and V 22 Lisa needs to cut tubing into the largest pieces of equal length that she can without having any offcuts left over. She has 3 sections of tubing; one of length 6 metres, another of length 9 metres and the third of length 15 metres. a How long should each piece of tubing be? b How many pieces of tubing will Lisa end up with?
GAM
me E ti
Multiples, factors and primes 02
23 Mario, Luigi, Dee Kong and Frogger are playing Nintendo. Mario takes 2 minutes to play a complete game, Luigi takes 3 minutes, Dee Kong takes 4 minutes and Frogger takes 5 minutes. They have 12 minutes to play. a If they play continuously, which of the players would be in the middle of a game as time ran out? b After how many minutes did this player begin the last game?
51
QUEST
GE
S
EN
MAT H
Chapter 2 Multiples, factors and primes
CH
AL
L
1 What number am I? I am a multiple of 5 with factors of 6, 4 and 3. The sum of my digits is 6. 2 My age is a multiple of 3 and a factor of 60. The sum of my digits is 3. How old am I? (There are two possible answers.) 3 Find the highest common factor of 462, 504 and 630.
How many tiles? At the start of this chapter, you may have considered the following situation. A room measures 550 centimetres by 325 centimetres. What would be the side length of the largest square tile that can be used to tile the floor without any cutting?
1. 2. 3. 4.
Try this question now. (Hint: Find the common factors of 550 and 325.) How many tiles would fit on the floor along the wall of length 550 centimetres? How many tiles would fit on the floor along the wall of length 325 centimetres? How many floor tiles would be needed for this room?
52 Desertt creatur Deser creatures es that are are sheltered sheltered by by spinife spinif ex in outback Australia Maths Quest 7 for Victoria
The highest common factor (HCF) of the pair of numbers given and the letter beside each gives the puzzle answer code.
A
20, 50
D
15, 25
L
42, 56
O
18, 30
S
12, 21
R
40, 16
T
70, 4
N
36, 54
P
27, 45
I
21, 35
M
60, 80
F
120, 90
H
13, 7
C
36, 24
E
30, 45
G
39, 26
B
12, 20
Z
75, 200
14
15
13
14
15
3
3
14
7
25
10
8
5
3
9
7
15
5
7
3
1
4
15
15
2
14
15
3
15
10
5
15
12 8
3 15
15 3
2
8
2
15 5
5
13 8 10 3
2 3
1
10
1
6
5 9
5 9 15
15 8
8 3
1 6 9 9 7 18 13 20 7 12 15 5 8 10 13 6 18 30 14 7 15 3 9 10 7 18 2 15 5 30 7 8 15 2 10 7 14 30 7 18 12 1 15 3
Chapter 2 Multiples, factors and primes
53
Prime numbers A prime number is a counting number which has exactly 2 factors, itself and 1. The number 3 is a prime number. Its only factors are 3 and 1. The number 7 is a prime number. Its only factors are 7 and 1. A composite number is one which has more than two factors. The number 10 is a composite number; its factors are 1, 10, 5 and 2. The number 16 is also a composite number; its factors are 1, 16, 2, 8 and 4. The number 1 is a special number. It is neither a prime number nor a composite number because it has only one factor, 1.
WORKED Example 8 Find 3 prime numbers that are greater than 10. THINK 1
2 3 4 5 6
WRITE
All even numbers except 2 are composite numbers because factors include 1, 2 and the number itself. Look at odd numbers only. Try 11. Factors of 11 are 11 and 1 only, so it is a prime number. Try 13. Factors of 13 are 1 and 13 only, so 13 is a prime number. Try 15. Factors are 1, 3, 5, 15. Not a prime number. Try 17. Factors are 1 and 17 only. Prime number. List the 3 prime numbers found.
11 13
17 Three prime numbers greater than 10 are 11, 13 and 17.
Sum of two prime numbers Every composite number which is odd, can be written as the sum of 2 prime numbers, one of which is 2. For example,
9 = composite
7 prime
+
2 prime
Other examples are: 15 = 13 + 2 21 = 19 + 2 25 = 23 + 2 A mathematician, Goldbach, believed that every even composite number could be written as the sum of two odd prime numbers. For example, 8 = 5 + 3 24 = 5 + 19 46 = 17 + 29 Is he right? Can you find an even composite number which is not the sum of two prime numbers?
54
Maths Quest 7 for Victoria
WORKED Example 9 Write each of the following as the sum of two odd prime numbers. a 14 b 28 THINK
WRITE
a
a 5 + 9 = 14 9 is not prime. 3 + 11 = 14 3 and 11 are prime.
Try a pair of odd numbers and check that they are prime. If they are not prime numbers, try other pairs of odd numbers until you have found a pair which are prime.
1 2
b Try a pair of odd numbers and check that they are prime.
b 5 + 23 = 28 5 and 23 are prime.
Often there is more than one solution. For 28, 5 + 23 = 28 and 11 + 17 = 28. The numbers 5, 23, 11 and 17 are all prime. Note that if the number is even, 2 will not be one of the prime numbers in the sum.
remember remember 1. A prime number is a counting number which has exactly two factors, itself and 1. 2. A composite number is one which has more than two factors. 3. The number one is not a prime number or a composite number.
2C
Prime numbers
1 a List the factors for each number between 1 and 20. b Use the factor lists to write down all the prime numbers up to 20. WORKED
Example Prime 8 numbers
2.3
2 Find 4 prime numbers which are between 20 and 40. 3 Can you find 4 prime numbers that are even? Explain.
4 Write each of the following as the sum of 2 odd prime numbers. a 8=_+_ b 12 = _ + _ 9 c 30 = _ + _ d 16 = _ + _ e 100 = _ + _ f 48 = _ + _ g 24 = _ + _ h 52 = _ + _ i 60 = _ + _ j 32 = _ + _
WORKED
Example
5 Answer true (T) or false (F) for each of the following. a All odd numbers are prime numbers. b No even numbers are prime numbers. c 1, 2, 3 and 5 are the first 4 prime numbers. d A prime number has 2 factors only.
Chapter 2 Multiples, factors and primes
55
6 multiple choice a The number of primes less than 10 is: A 4 B 3 C 5 D 2 E 6 b The first 3 prime numbers are: A 1, 2, 3 B 1, 3, 5 C 2, 3, 4 D 2, 3, 5 E 3, 5, 7 c The number 15 can be written as the sum of 2 prime numbers. These are: A 3 + 12 B 1 + 14 C 13 + 2 D 7+8 E 11 + 4 d Factors of 12 that are prime numbers are: A 1, 2, 3, 4 B 2, 3, 6 C 2, 3 D 2, 4, 6, 12 E 1 7 Twin primes are pairs of primes which are separated from each other by one even number. For example, 3 and 5 are twin primes. Find 2 more pairs of twin primes. 8 a Which of the numbers 2, 3, 4, 5, 6 and 7 cannot be the difference between 2 consecutive prime numbers? Explain. b For each of the numbers which can be a difference between 2 consecutive primes, give an example of a pair of primes less than 100 with such a difference.
QUEST
GE
S
EN
MAT H
9 The following numbers are not primes. Each of them is the product of 2 primes. Find the 2 primes in each case. a 365 b 187
CH
AL
L
1 What is the largest 3 digit prime number in which each digit is a prime number? 2 Find a prime number greater that 10 where the sum of the digits equals 11. 3 My age is a prime number. I am older than 50. The sum of my digits is also a prime number. If you add a multiple of 13 to my age the result is 100. How old am I? 4 I am a 3 digit number. I am divisible by 6. My middle digit is a prime number. The sum of my digits is 9. I am between 400 and 500. My digits are in descending order. 5 Angus is the youngest in his family and today he and his Dad share a birthday. Both their ages are prime numbers. Angus’s age has the same 2 digits as his Dad but in reverse order. In 10 years’ time, Dad will be three times as old as Angus. How old will each person be when this happens? 6 What is the largest 5-digit number you can write if each digit must be different and no digit may be prime?
2.4
2.1
56
Maths Quest 7 for Victoria
History of mathematics E R ATOS T H E N E S O F C Y R E N E ( c 2 7 0 B C t o 1 9 0 B C ) During his life . . . The Roman Empire introduces a minted coin, the Denarius. Hannibal crosses the Alps. The Great Wall of China is built. Terracotta Soldiers are buried in a tomb at Xian (China). Eratosthenes was a Greek astronomer and mathematician. He was born in Cyrene (now Shahhat, Libya) in approximately 270 BC. Some sources suggest he was born in 270 BC and died in 194 BC. Eratosthenes was first educated by his father, Aglaos, and was then educated in Athens. He eventually became the main librarian at the Alexandrian museum. Eratosthenes was the first person to work out an accurate measurement of the circumference of the Earth (when many people believed the Earth was flat!). He did this by comparing the angle of the shadows of sticks at two distant locations, Alexandria and Athens, at the same time of the day.
Eratosthenes was one of the most famous mathematicians of his day and was asked by the ruler of Egypt, Ptolemy III, to tutor his son Ptolemy Philadelphus. He also took on the job of being the chief librarian at the University of Alexandra where other mathematicians, such as Euclid, had held this position. As chief librarian he was able to borrow and copy books from the library.
Because he worked on many types of research, some people thought that he was not a specialist in any area. However, all the work that he did was always of the highest standard. Eratosthenes made many discoveries in many areas that still benefit us now. He tried to improve the accuracy of calendars and realised there was a need for leap years to keep calendars in time with the seasons. He published the first map of the world based on longitude and latitude. He compiled a star catalogue containing 675 stars. Eratosthenes discovered a way of finding prime numbers, called the Sieve of Eratosthenes (see page 57). The largest known prime is 21 398 269 − 1 which has 420 921 digits, but new primes are now being discovered regularly with the help of computers. When Eratosthenes was old he became blind. He was so upset by this disability that he committed suicide.
Questions 1. What was Eratosthenes’s main job in life? 2. What physical property was he the first person to measure accurately? 3. What was different about Eratosthenes’s map of the world? 4. Which ruler did he work for? 5. What is the largest known prime at the time of this publication? 6. Why did Eratosthenes become depressed and commit suicide? Research 1. Look on the Internet for information about primes and find the largest prime number known at present and compare it with the one reported above. How much bigger is it now? 2. Why do we need leap years (they occur once every 4 years normally) and what is the difference between the years 1900 and 2000 in terms of leap years?
Chapter 2 Multiples, factors and primes
57
The sieve of Eratosthenes An easy way to find prime numbers is to use the ‘Sieve of Eratosthenes’. Eratosthenes discovered a simple method of sifting out all of the composite numbers so that only prime numbers are left. You can follow the steps below to find all prime numbers between 1 and 100. Alternatively you can use the Excel file provided on the Maths Quest CD-ROM to simulate this process. Sieve of Eratosthenes (a) Copy the numbers from 1 to 100 in a grid as shown below. Use 1 centimetre square grid paper.
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
(b) Cross out 1 as shown. It is not a prime number. (c) Circle the first prime number, 2. Then cross out all of the multiples of 2. (d) Circle the next prime number, 3. Now cross out all of the multiples of 3 that have not already been crossed out. (e) The number 4 is already crossed out. Circle the next prime number, 5. Cross out all of the multiples of 5 that are not already crossed out. (f) The next number that is not crossed out is 7. Circle 7 and cross out all of the multiples of 7 that are not already crossed out. Continue until you have circled all of the prime numbers and crossed out all of the composite numbers.
58
Maths Quest 7 for Victoria
1. 2. 3. 4. 5.
Answer the following questions using the Sieve of Eratosthenes. How many prime numbers are there between 1 and 100? What is the largest prime number less than 100? How many prime numbers are there between 20 and 100? How many single digit prime numbers are there between 1 and 100? How many double digit prime numbers are there between 1 and 100?
1 1 Find all the factors of 64. 2 List the factor pairs of 24. 3 Find the highest common factor of 30 and 45. 4 List the first 5 multiples of 9. 5 Find the lowest common multiple of 4 and 9. 6 Find the lowest common multiple of 12 and 16. 7 List all the prime numbers between 20 and 50. 8 Write 20 as the sum of 2 prime numbers. 9 Write 50 as the sum of 2 prime numbers. 10 A school marching band is made up of 36 students. The band leader is trying to arrange them into rows containing equal numbers of students. What arrangements are possible?
Prime factors and factor trees A factor tree shows the prime factors of a composite 20 number. Each branch shows a factor of all numbers above it. Factors of 20 2 10 The last numbers are all prime numbers, therefore they are prime factors of the original number. From the factor tree shown, 2, 2 and 5 are prime factors Factors of 10 2 5 of 20. Note that the factor 2 need only be written once, so we 20 can write that the prime factors of 20 are 2 and 5. If we had chosen different factors of 20 to start with, would 4 5 we end up with different prime factors? Choosing 4 and 5 instead of 10 and 2 did not change the prime factors of 20. If all the prime factors are multiplied together, 2 2 the answer will be the original number. Prime factors are 2 and 5 2 × 2 × 5 = 20
59
Chapter 2 Multiples, factors and primes
WORKED Example 10
a Find the prime factors of 50 by drawing a factor tree. b Write 50 as a product of its prime factors. THINK
WRITE a
a
1
Find a factor pair of the given number and begin the factor tree (50 = 5 × 10).
2
If a branch is prime, no other factors can be found (5 is prime). If a branch is composite, find factors of that number; 10 is composite so 10 = 5 × 2. Continue until all branches end in a prime number then stop. Write the prime factors.
50
5
3 4
b Write 50 as a product of prime factors found in part (a).
10 50
5
10
5
2
2 and 5 are prime factors of 50. b 50 = 5 × 5 × 2
WORKED Example 11 Find the prime factors of 56. THINK 1
WRITE
Draw a factor tree. When all factors are prime numbers you have found the prime factors.
56
8
2
4
2 2
Write the prime factors.
7
2
The prime factors of 56 are 7 and 2.
remember remember 1. A factor tree shows the prime factors of a composite number. 2. The last numbers in the factor tree are all prime numbers, therefore they are prime factors of the original number. 3. Every composite number can be written as a product of prime factors. For example, 20 = 2 × 2 × 5.
20
4
2
5
2
60
Maths Quest 7 for Victoria
2D WORKED
Example
Prime factors
Prime factors
1
Prime factors and factor trees
i Find the prime factors of each of the following numbers by drawing a factor tree. ii Write each one as a product of its prime factors.
10
a d g j 2
b e h k
15 100 18 84
c f i l
30 49 56 98
24 72 45 112
i Find the prime factors of the following numbers by drawing a factor tree. ii Express the number as a product of its prime factors. a d g j
Prime factors
b e h k
40 121 3000 196
c f i l
35 110 64 90
32 150 96 75
3 multiple choice a A factor tree for 21 is: A 21
7
D
B
3
C
21
1
21
3
21
E
b A factor tree for 36 is: A 36
1
B
36
1
7
1
2
C
18
36
9
4
36
9
9
7
36
18
2
3
E
36
2
c
7
7×1 1
D
1
21
3 3×1
21
3
4
3
The prime factors of 16 are: A 1, 2 B 1, 2, 4 C 2, 4, 8
d The prime factors of 28 are: A 1, 28 B 2, 7 C 1, 2, 14
2
2
D 2
E 1, 2, 4, 8, 16
D 1, 2, 7
E 2, 7, 14
Chapter 2 Multiples, factors and primes
4 Find the prime factors of each of the following numbers. a 48 b 200 11 d 81 e 18 g 27 h 300 j 120 k 50
61
WORKED
Example
c f i l
42 39 60 80
QUEST
GE
S
EN
MAT H
5 State whether each of the following is true (T) or false (F). a The number 3 is the only prime factor of 9. b No two numbers can have the same prime factors. c The numbers 2, 3, 5 and 7 are the prime factors of 210. d The numbers 1, 2 and 5 are the prime factors of 40. e The prime factors of 220 are 2, 5 and 11. f All numbers have exactly 2 prime factors.
CH
AL
L
1 A whole number is ‘perfect’ if it equals the sum of all its proper factors. The proper factors of a number are all factors smaller than the number. So 6 is perfect since its proper factors are 1, 2 and 3 and 6 = 1 + 2 + 3. a Find all the proper factors of 28 and show that 28 is a perfect number. b Do the same for 496. 2 Find the 7 proper factors of the number 999. Is 999 a perfect number? 3 Use a calculator to find the 13 proper factors of 8128. Is 8128 a perfect number?
Index notation The product of factors can be written in a shorter form by using index notation or index form. A number in index form has two parts, the base and the index. The product of factors 3 × 3, can be written as 32. The 3 is the base and the 2 is the index. Two other examples are given below: 4 × 4 × 4 = 43 43 has a base of 4 and an index of 3. 5 2×2×2×2×2=2 25 is a number in which 2 is the base and 5 is the index. 5 2 is in index notation or index form. 2 × 2 × 2 × 2 × 2 is in expanded form. A composite number written as a product of prime factors can be written using index notation. So 50 = 2 × 5 × 5 = 2 × 52 and 56 = 2 × 2 × 2 × 7 = 23 × 7.
62
Maths Quest 7 for Victoria
WORKED Example 12 Write the following using index notation. a 5×5 b 2×2×6×6×6 THINK
WRITE
a
Write the multiplication. Write the number being multiplied as the base and the number of times it is multiplied as the index.
a 5×5 = 52
Write the multiplication. Write the number being multiplied as the base and the number of times it is multiplied as the index.
b 2×2×6×6×6 = 2 2 × 63
1 2
b
1 2
WORKED Example 13 Write 120 as a product of prime factors using index notation. THINK 1
WRITE
Find a factor pair and begin the factor tree. If the number on the branch is a prime number, stop. If not, continue until a prime number is reached.
120
12
4
2 2 3
Write the number as a product of prime factors. Write your answer using index notation.
10
3
5
2
120 = 2 × 2 × 2 × 3 × 5 120 = 23 × 3 × 5
WORKED Example 14
Write 53 in expanded form and then find the answer. THINK 1 2 3
4
Write the question in expanded form. Multiply the first 2 numbers. Multiply the answer by the next number and continue until all numbers have been multiplied. Write the answer.
2
WRITE 53 = 5 × 5 × 5 = 25 × 5 = 125 53 = 125
Chapter 2 Multiples, factors and primes
Graphics Calculator tip!
63
Calculating numbers in index form
To calculate a number in index form, first type in the base, then ^ , then the index and press ENTER . The screen opposite shows the calculation for 53 (from worked example 14).
remember remember 1. The product of prime factors can be written in a shorter form by using index notation. For example, 40 = 23 × 5. Using index notation, 23 is a number with base 2 and index of 3. 2. The 2 (or base number) is the number being multiplied and the 3 (or index) indicates how many times the base is being multiplied 23 = 2 × 2 × 2.
2E WORKED
Example
12a
WORKED
Example
12b
WORKED
Example
13 WORKED
Example
14
Index notation
1 Write the following using index notation. a 4×4×4 b 8×8 c 7×7×7×7 d 12 × 12 × 12 × 12 × 12 e 2×2×2×2×2×2×2 f 13 × 13 × 13 g 5×5×5 h 9×9×9×9×9×9×9×9 2 Write the following using index notation. a 2×2×3 b 3×3×3×3×2×2 c 5×5×2×2×2×2 d 7×2×2×2 e 5 × 11 × 11 × 3 × 3 × 3 f 13 × 5 × 5 × 5 × 7 × 7 g 2×2×2×3×3×5 h 3×3×2×2×5×5×5 3 Write the following as a product of prime factors using index notation. a 60 b 50 c 75 d 220 e 192 f 72 g 124 h 200 4 Write the following in expanded form and then find the answer. b 112 c 53 d 24 a 32 2 6 3 2 e 2×7 f 3×2 g 9 ×3 h 2 5 × 42 4 3 2 2 3 i 3 +7 j 2 +3 k 10 − 3 l 5 3 − 24 4 2 5 3 4 2 m 2 ÷2 n 3 ÷ 3 (Hint: Write 2 ÷ 2 as a fraction and simplify.) 5 Write one million using index notation. Use 10 as the base number. 6 multiple choice The largest number listed here is: A 150 B 26 C 34
D 53
E 72
7 multiple choice The smallest number listed here is: B 102 C 115 A 43
D 025
E 44
Index notation
64
Maths Quest 7 for Victoria
Divisibility Often it can be difficult to find all the factors of a number, or to work out if a number is prime or composite. The following tests will help you.
Divisibility tests — 2, 3 and 4 A number is divisible by a second number if it can be divided by the second number without a remainder. For example, 20 is divisible by 10, because 20 can be divided by 10, without a remainder; that is, 20 ÷ 10 = 2. There is a way to test whether a number is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10 or 11.
Divisible by 2 The number 2 goes into all even numbers. A number is divisible by 2 if it is even. Even numbers end in 0, 2, 4, 6 or 8. Divisible by 3 All of the numbers in the 3 times table listed below are divisible by 3. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39 . . . The sum of the digits in each number is 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 12 . . . Each of these sums is also divisible by 3. If the sum of the digits of a number is divisible by 3, then that number is also divisible by 3.
WORKED Example 15 State whether each of the following numbers is divisible by 3. a 45 b 92 c 66 d 5742 e 1 233 693 THINK To be divisible by 3 the sum of the digits must be divisible by 3. a For 45, the sum of digits is 4 + 5 = 9. Sum is divisible by 3.
WRITE
a 45 is divisible by 3.
b For 92, the sum of digits is 9 + 2 = 11. Sum is not divisible by 3.
b 92 is not divisible by 3.
c For 66, the sum of digits is 6 + 6 = 12. Sum is divisible by 3.
c 66 is divisible by 3.
d For 5742, the sum of digits is 5 + 7 + 4 + 2 = 18. Sum is divisible by 3.
d 5742 is divisible by 3.
e For 1 233 693, the sum of digits is 1 + 2 + 3 + 3 + 6 + 9 + 3 = 27. Sum is divisible by 3 (because 2 + 7 = 9 which is divisible by 3).
e 1 233 693 is divisible by 3.
Chapter 2 Multiples, factors and primes
65
Divisible by 4 If the last two digits of a number are divisible by 4, then the number is divisible by 4. The number 2032 is divisible by 4 because the last 2 digits, 32, are divisible by 4 (32 ÷ 4 = 8).
WORKED Example 16 State whether each of the following numbers is divisible by 4. a 48 b 3012 c 150 THINK
WRITE
Consider the last 2 digits and check whether this number is divisible by 4. a Last two digits are 48 which is divisible by 4 (48 ÷ 4 = 12).
a 48 is divisible by 4.
b Last two digits are 12 which is divisible by 4 (12 ÷ 4 = 3).
b 3012 is divisible by 4.
c Last two digits are 50 which is not divisible by 4 (50 ÷ 4 = 12.5).
c 150 is not divisible by 4.
remember remember 1. Even numbers are divisible by 2. 2. If the sum of the digits of a number is divisible by 3, then that number is also divisible by 3. 3. If the last 2 digits of a number are divisible by 4, then the number is divisible by 4.
2F WORKED
Example
15
WORKED
Example
16
Divisibility tests — 2, 3 and 4
For questions 1 to 4 write ‘yes’ if divisible and ‘no’ if not divisible. 1 State whether each of the following numbers is divisible by 3. a 27 b 51 c 38 d 126 e 4017 f 943 g 362 h 84 i 69 j 9462 k 523 l 763 836 2 State whether each of the following is divisible by 2. a 38 b 17 c 26 e 288 f 100 g 370 i 2020 j 719 k 4533
d 3093 h 4131 l 64 512
3 State whether each of the following numbers is divisible by 4. a 44 b 28 c 328 e 917 f 3041 g 6084 i 149 j 70 232 k 69 478
d 212 h 68 l 324 636
66
Maths Quest 7 for Victoria
4 State whether each of the following is divisible by 2, 3 and 4. a 12 b 30 c 48 e 180 f 236 g 3690 i 200 j 1116 k 92 472
d 312 h 552 l 75 148
5 Copy the following list and circle the numbers that are divisible by both 2 and 3. 87, 18, 108, 127, 12, 45, 3024, 96, 67, 429, 216 6 Copy the following list and circle the numbers that are divisible by 3 and 4. 12, 390, 420, 96, 880, 612, 264, 1038, 59, 2003 7 multiple choice A 250 gram block of Dairy Milk Chocolate has 60 pieces. It can be shared evenly between the following number of people: A 2, 5, 7
B 3, 5, 9
C 2, 5, 11
D 3, 5, 12
E 3, 10, 14
8 If there are 220 students in Year 7, could they form groups of: a 2 with no students left over? b 3 with no students left over? c 4 with no students left over? 9 A tennis team consists of 4 players. If a tennis club has 326 members who want to play in a team, will they be able to make teams of exactly 4 players, or will there be some players left over?
2.2 10 There are 294 students waiting to buy their lunch at the school canteen and there are 3 queues. Could the 3 queues be exactly the same size?
67
QUEST
GE
S
EN
MAT H
Chapter 2 Multiples, factors and primes
CH
AL
L
1 If you double my age and subtract 1, you will have a prime number. I am less than 20 years old. The sum of my digits is divisible by 2 and 3. How old am I? 2 Find the largest 8-digit number which is divisible by 2, 3 and 4. 3 A clock strikes at regular intervals. It strikes the number of hours on the hour, twice for every half hour and once for each of the remaining quarter hours. How many times does the clock strike in a 24-hour period?
2 1 List the first 3 multiples of 15. 2 List the factor pairs of 100. 3 Find the prime factors of 48 by drawing a factor tree. 4 Find the prime factors of 81. For questions 5 to 8, consider the following list of numbers: 36, 77, 108, 231, 459, 12 568. 5 Which numbers in the list are divisible by 2? 6 Which numbers in the list are divisible by 3? 7 Which numbers in the list are divisible by 4? 8 Which numbers in the list are divisible by 2, 3 and 4? 9 Jonathon has just won an enormous block of chocolate which is made up of 1344 little squares. Can he divide it evenly between himself and 3 friends? 10 At the swimming pool there are 3 water slides. On a really hot day there were 524 people queuing up for a turn. Is it possible that the 3 queues could be the same length?
68
Maths Quest 7 for Victoria
Divisibility tests — 5, 6, 7 and 8 Divisible by 5 The 5 times table shows all the numbers that 5 ‘goes into’. These include 5, 10, 15, 20, 25, 30, 35, 40, … All of these numbers end in 5 or 0. Numbers ending in 5 or 0 are divisible by 5.
WORKED Example 17 State whether each of the following is divisible by 5. a 2457 b 6730 THINK
WRITE
If the number ends in 0 or 5 it is divisible by 5. a The number ends in 7, not 0 or 5.
a 2457 is not divisible by 5.
b The number ends in 0.
b 6730 is divisible by 5.
Divisible by 6 A number is divisible by 6, if it is divisible by both 2 and 3.
WORKED Example 18 Find whether 324 is divisible by 6. THINK 1 2
3
WRITE
Check whether 324 is even and therefore divisible by 2. Check whether the sum of the digits is divisible by 3 so that the number is divisible by 3. Write the answer.
324 is even, so it is divisible by 2. 3 + 4 + 2 = 9, so 324 is divisible by 3.
324 is divisible by 6.
Divisible by 7 To test whether a number is divisible by 7 is more complicated. The following steps can be followed. 1. Write the number without the last digit. 2. Subtract 2 times the last digit from this number. 3. Repeat steps 1 and 2 until you are left with a 1 or 2 digit number. 4. If the 1 or 2 digit number is divisible by 7, the original number is divisible by 7. If the number you are left with is 0, the original number is divisible by 7.
Chapter 2 Multiples, factors and primes
69
WORKED Example 19 State whether the following are divisible by 7. a 2483 b 1967 THINK
WRITE
a
Write the number without the last digit. Subtract 2 times the last digit (2 × 3 = 6) from the number obtained in step 1. Repeat the process in order to obtain a 1 or 2 digit number. Write the number without the last digit. Subtract 2 times the last digit (2 × 2 = 4) from the number obtained in step 3. Check whether the 2 digit number obtained is divisible by 7. Write the answer.
a 248 248 − 6 = 242
Write the number without the last digit. Subtract 2 times the last digit (2 × 7 = 14). Repeat. Write without the last digit. Subtract 2 times the last digit (2 × 2 = 4). Check whether the 2-digit number is divisible by 7. Write the answer.
b 196 196 − 14 = 182 18 18 − 4 = 14 14 is divisible by 7.
1 2 3
4 5 6
b
1 2 3 4 5 6
24 24 − 4 = 20 20 is not divisible by 7. So 2483 is not divisible by 7.
So 1967 is divisible by 7.
Divisible by 8 If the last 3 digits of a number are divisible by 8 then the number is divisible by 8. The number 1032 is divisible by 8 because 032 is divisible by 8. The number 2804 is not divisible by 8 because 804 is not divisible by 8.
WORKED Example 20 State whether the following are divisible by both 5 and 6. a 270 b 4305 THINK
WRITE
a
a 270 ends in 0 so it is divisible by 5.
1 2
3
Check whether the number is divisible by 5. Check whether the number is divisible by 6 by checking if it is divisible by both 2 and 3. If the number is even, it is divisible by 2. If the sum of the digits is divisible by 3, the number is divisible by 3. Write the answer.
270 is even so it is divisible by 2. 2 + 7 + 0 = 9 which is divisible by 3. So 270 is divisible by 6.
270 is divisible by both 5 and 6.
70
Maths Quest 7 for Victoria
THINK
WRITE
b
b 4305 ends in 5 so it is divisible by 5.
1 2 3
Check whether the number is divisible by 5. Check whether the number is divisible by 6. Write the answer.
4305 is not divisible by 2 and so it is not divisible by 6. 4305 is not divisible by both 5 and 6.
remember remember 1. Numbers ending in 0 or 5 are divisible by 5. 2. Numbers which are divisible by both 2 and 3 are divisible by 6. 3. To test whether a number is divisible by 7, drop the last digit, and subtract twice the dropped digit from the others. Repeat this process until there are only 2 digits left and see if that number is divisible by 7. 4. If the last 3 digits are divisible by 8 then the number is divisible by 8.
2G
Divisibility tests — 5, 6, 7 and 8
For questions 1 to 5 write ‘yes’ if divisible and ‘no’ if not divisible. WORKED 1 State whether each of the following is divisible by 5. Example a 5 b 30 c 47 d 105 e 5032 17 f 1175 g 943 h 77 i 39 260 j 645 2 State whether each of the following is divisible by 6. a 42 b 63 c 54 d 124 18 f 312 g 4311 h 3106 i 9 246 000
e 96 j 72 144
3 State whether each of the following is divisible by 7. a 483 b 923 c 658 d 4235 19 f 1855 g 6007 h 9948 i 38 297
e 3126 j 64 582
WORKED
Example
WORKED
Example
4 State whether each of the following is divisible by 8. a 64 b 45 c 4064 d 4008 f 1020 g 112 h 55 112 i 42 391 016 5 State whether each of the following is divisible by both 5 and 6. a 30 b 250 c 18 000 d 630 20 f 2952 g 2040 h 960 i 197 460
e 71 045 j 3102
WORKED
Example
e 375 j 3 416 220
6 Copy the following list and circle the numbers that are divisible by both 5 and 8. 4040, 100, 240, 6024, 400, 10 000, 367 080
Chapter 2 Multiples, factors and primes
71
7 State true (T) or false (F) for each of the following. a 148 is divisible by 6. b 49 026 is divisible by 5. c 810 560 is divisible by 8. d 3510 is divisible by both 5 and 6. e 6080 is divisible by both 5 and 8. f 21 960 is divisible by both 6 and 8. g 410 is divisible by both 5 and 6. h 13 160 is divisible by both 6 and 8.
QUEST
GE
S
EN
MAT H
8 Could 142 students be broken up into volleyball teams, with no one left out, if each team has 6 players exactly?
CH
AL
L
1 How old am I? I am older than 9. The sum of my digits is 3. My age is divisible by 5 and 2. 2 My age is a multiple of the sum of the first 2 prime numbers. It is also divisible by 7 and has exactly 4 factors. How old am I? 3 My locker number has factors of 2, 5 and 9. Find my locker number if there are 9 other factors of this number. 4 What are the two numbers that are divisible by both 5 and 7 and have exactly 12 factors?
72
Maths Quest 7 for Victoria
Divisibility tests — 9, 10 and 11 Divisible by 9 A number is divisible by 9 if the sum of its digits is divisible by 9.
WORKED Example 21 State whether each of the following is divisible by 9. a 360 b 296 THINK
WRITE
a
Add the digits of the number. Check whether the sum of the digits is divisible by 9.
a 3 + 6 + 0 = 9. 9 is divisible by 9 so 360 is divisible by 9.
Add the digits of the number. Check whether the sum of the digits is divisible by 9.
b 2 + 9 + 6 = 17 The number 17 is not divisible by 9 so 296 is not divisible by 9.
1 2
b
1 2
Divisible by 10 The 10 times table shows all of the numbers 10 ‘goes into’. 10, 20, 30, 40, 50, . . . All of these numbers end in 0. Numbers ending in zero are divisible by 10. Divisible by 11 Divisibility by 11 is more complicated. To test whether a number is divisible by 11, the position of each digit must be considered as shown. For the number 6542, the 6 is in an odd position because it is in position 1 (1 is odd). The number 5 is in an even position because it is in position 2 (2 is even). The number 4 is in an odd position because it is in position 3 (3 is odd). The number 2 is in an even position because it is in position 4 (4 is even). Number 6 5 4 2 ↕ ↕ ↕ ↕ Position 1 2 3 4 If the sum of the odd-positioned digits equals the sum of the even-positioned digits or differs from the sum of the even-positioned digits by 11, then the number is divisible by 11. For example, 55 is divisible by 11 because 5 = 5. 6542 is not divisible by 11 because 6 + 4 ≠ 5 + 2 1364 is divisible by 11 because 1 + 6 = 3 + 4. 24 968 is not divisible by 11 because 2 + 9 + 8 ≠ 4 + 6. 913 is divisible by 11 because 9 + 3 ≠ 1 but the two sums differ by 11.
Chapter 2 Multiples, factors and primes
73
WORKED Example 22 State whether 495 is divisible by both 9 and 11. THINK 1
2
3
WRITE
The number is divisible by 9 if the sum of the digits is divisible by 9 (18 ÷ 9 = 2). The number is divisible by 11 if the sum of the odd-positioned digits (4 + 5) is equal to the sum of the evenpositioned digits (9). Write the answer.
4 + 9 + 5 = 18 so 495 is divisible by 9. 4 + 5 = 9 so 495 is divisible by 11.
495 is divisible by both 9 and 11.
Finding prime factors The divisibility tests can be used to find the prime factors of any number.
WORKED Example 23 Find the prime factors of 2160. THINK 1
2
3
4
WRITE
The number 2160 is divisible by 10 because 2160 ends in 0. Use 10 as a factor in the factor tree. The number 216 is divisible by 9 because 2 + 1 + 6 = 9 and 9 is divisible by 9. Use 9 as a factor in the factor tree. The number 24 is divisible by 3 because 2 + 4 = 6 and 6 is divisible by 3. Use 3 as a factor. The number 8 is even so use 2 as a factor and continue until all prime factors are found.
2160
10
5
216
9
2 3
24 3
3
8 2
4 2
5
Write the number as a product of prime factors using index notation.
2
2160 = 24 × 33 × 5
remember remember 1. A number is divisible by 9 if the sum of the digits is equal to 9. 2. A number is divisible by 10 if the number ends in 0. 3. A number is divisible by 11 if the sum of the odd-placed digits is equal to the sum of the even-placed digits or the two sums differ by 11.
74
Maths Quest 7 for Victoria
2H
Divisibility tests — 9, 10 and 11
For questions 1 to 3, write ‘yes’ if divisible and ‘no’ if not divisible. 1 State whether each of the following is divisible by 9. Example a 81 b 108 c 362 d 261 21 e 918 f 1602 g 34 901 h 60 201 i 7096 j 8 446 520 k 356 892 l 459 258 WORKED
WORKED
Example
2 State whether each of the following is divisible by 10. a 30 b 45 c 600 e 891 f 8910 g 365 i 77 602 j 714 980 k 1245
d 750 h 4120 l 3 205 400
3 Find whether each of the following is divisible by 11. a 44 b 99 c 132 e 931 f 721 g 241 i 6822 j 50 260 k 4 281 926
d 594 h 3102 l 8277
4 Copy the following list and circle the numbers that are divisible by both 9 and 11.
22
45, 99, 990, 1010, 32 313, 198, 297, 2970, 5167 5 State true (T) or false (F) for each of the following. a 628 010 is divisible by 10. b 36 256 is divisible by 11. c 2107 is divisible by 9. d 13 068 is divisible by 9 and 11. e 79 010 is divisible by 9 and 11. f 332 132 130 is divisible by 9, 10 and 11. g 349 160 is divisible by 9, 10 and 11. h 5 832 910 is divisible by 9, 10 and 11.
6 Write the following numbers as the product of prime factors using index notation. Use divisibility tests to help. 23 a 1575 b 616 c 1080 d 2646
WORKED
Example
7 Use a divisibility test to show that the following numbers are composite. a 7667 b 9229
2.3
QUEST
GE
S
EN
MAT H
8 Use divisibility tests to decide which of the following numbers are primes. a 119 b 209 c 103 d 91
CH
AL
L
1 My age is a 2 digit number. It is divisible by 2, 3, 6 and 9. I am older than 20 but younger than 50 years old. How old am I? 2 My age is greater than 10 and less than 20. It is a composite number that is divisible by 3 and 9. How old am I? 3 I am very old. My age is a 3 digit number that is divisible by 11. It is also a multiple of 4. How old am I? 4 My age is divisible by 11. It is also a multiple of 2 and has 6 factors. I am less than 50 years old. How old am I?
Chapter 2 Multiples, factors and primes
75
Square numbers and square roots Square numbers If a number has 2 equal factors it is called a square number. Some examples are given below: 4 is a square number because 2 × 2 = 4 9 is also a square number because 3 × 3 = 9 1 is also a square number because 1 × 1 = 1. The number 1 is the first square number, 4 is the second square number, 9 is the third square number . . . 1 = 1 × 1 = 12
4 = 2 × 2 = 22
9 = 3 × 3 = 32
Each of these numbers can be represented by dots in the following way. Can you see why they are called square numbers?
12 = 1
22 = 4
32 = 9
WORKED Example 24 Find the sixth square number. THINK 1
2 3
Write the number which shows the square number you are looking for. Multiply that number by itself. Find the answer. Write your answer in a sentence.
WRITE 6×6 = 36 So 36 is the sixth square number.
WORKED Example 25 Find the square numbers between 90 and 150. THINK 1 2
3
Use your knowledge of tables to find the first square number after 90. Find the square numbers which come after that one but before 150. Write the answer in a sentence.
WRITE 102 = 10 × 10 = 100 112 = 11 × 11 = 121 122 = 12 × 12 = 144 132 = 13 × 13 = 169 (too big) The square numbers between 90 and 150 are 100, 121 and 144.
76
Maths Quest 7 for Victoria
Graphics Calculator tip!
Calculating the square of a number
There are 3 ways to obtain the square of a number using a graphics calculator. 1. Type in the base, press ^ , then type in the index which is 2. Press ENTER to complete the calculation. x 2 , then 2. Type in the base, press the key marked press ENTER . 3. Type in the number, press the × sign then type in the same number again. Remember to press ENTER . All 3 methods are shown in the screen opposite.
Square roots Finding the square root of a number is the opposite of squaring a number. 42 = 16 means that 4 multiplied by itself is equal to 16. 16 = 4 means that we are finding a number that multiplies by itself to equal 16.
WORKED Example 26
Find: a
49
b
3600 .
THINK
WRITE
a Find a number which when multiplied by itself is equal to 49.
a 49 = 7 × 7 so
b Find a number which when multiplied by itself is equal to 3600. You may need to try a few different numbers until you find the right one.
b
Graphics Calculator tip!
So
49 = 7
3600 = 60 × 60 3600 = 60
Calculating the square root of a number
To calculate the square root of a number, first press 2nd [ ] then the number. Remember to press ENTER to complete the calculation. The operation x2 . is listed in yellow above the key marked This is why we need to first press the key marked 2nd . The screen shows the calculations needed for worked example 26.
remember remember 1. A square number is one which has 2 equal factors. 2. Finding the square root of a number is the opposite of squaring a number. 3. To find the square root of a number, find a number that multiplies by itself to equal the original number.
Chapter 2 Multiples, factors and primes
2I WORKED
Example
24 WORKED
Example
25
77
Square numbers and square roots
1 Find the eighth square number. 2 Find the following square numbers: a the 13th b the 15th c the 20th
d the 50th
e the 100th. Square numbers
3 a Find the square numbers between 50 and 100. b Find the square numbers between 160 and 200. 4 Find the even square numbers between 10 and 70.
2.3
5 Find the odd square numbers between 50 and 120. 6 a Find the prime numbers between 12 and 22. b Find the prime numbers between 62 and 72. 7 a b c d
WORKED
Example
26a WORKED
Example
26b
Find the sum of the first 2 odd numbers. Find the sum of the first 3 odd numbers. Find the sum of the first 4 odd numbers. The answers to a, b and c are all square numbers. Use this to find the sum of i the first 10 odd numbers ii the first 50 odd numbers iii the first 1000 odd numbers.
8 Find: 25 a 9 Find: 4900 a
b
64
b
14 400
c
b 92 − 36 e 3 2 − 22 ÷ 4 +
d
360 000
49
c f
160 000
52 × 22 × 49 9 × 42 − 144 ÷ 22
GE
QUEST
EN
MAT H
d
121
Square roots
10 Simplify the following. a 22 + 25 d 32 + 22 × 16
S
c
81
CH
AL
L
1 Megan has 3 game scores which happen to be square numbers. The first 2 scores have the same 3 digits. The total of the 3 scores is 590. What are the 3 scores? 2 The difference of the squares of two consecutive odd numbers is 32. What are the two odd numbers? 3 What is the number? It is greater than 152. It is not a multiple of 3 or 5. It is less than 202. It is not a multiple of 17. It is a square number. 4 Find a 2-digit number in which both digits are different and the difference between the square of the number and the square of the number with the digits reversed is a square number.
2.5
78
Maths Quest 7 for Victoria
Why did the woman Why woman leav leave the theatre theatre after Act 1? Determine the exact values of the terms below and match them up with the letter beside each to find the answer code.
16
A
32 =
D
43 =
G
49 =
I
24 =
M
62 =
O
25 =
R
25 =
C
36 =
E
33 =
H
52 =
L
9=
N
72 =
P
23 =
S
53 =
T
92 =
W
22 =
X
125 27
“ 12
144 =
6
49
32 36
5
125 3
125
36 64 125 27
5
81 8
12 36 25
16
8 8
12
12
7 7
125
9 4
6
5
Chapter 2 Multiples, factors and primes
79
Testing whether numbers are prime or composite You will need a calculator for this activity. Eratosthenes devised a way of testing whether numbers were prime or composite. He tested all the primes smaller than the square root of the number. If any of the primes divide evenly into the number then the number is composite. If none of the primes divide evenly into the number then it is prime. Let’s try this method with the number 1973. First find the square root of 1973 using a calculator, 1973 = 44.418 . . . The prime numbers less than 44 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 and 41. Test each prime to see whether the number 1973 is divisible by that prime. None of the primes up to 44 divide evenly into 1973. So 1973 is a prime number. Try this method with the following numbers. Classify each as prime or composite: 1. 1001 2. 3133 3. 8191.
Producing primes
80
Maths Quest 7 for Victoria
summary 1 2 3
4 5 6 7 8 9 10 11 12
Copy the sentences below. Fill in the gaps by choosing the correct word or expression from the word list below. A is a whole number that divides exactly into another whole number, with no remainder. Factors that are the same for 2 or more numbers are called . The largest of these is known as the (HCF). A is obtained by multiplying a number by another number. If two or more numbers have the same multiple, they are called common multiples. The smallest of these is known as the (LCM). A has two factors only, itself and 1. A has more than 2 factors. The number is neither a prime number nor a composite number. Composite numbers can be written as the product of factors. The product of factors can be written in a shorter form by using notation or index form. A number in index form can be written using notation. Numbers with 2 equal factors are called numbers. Taking the square root of a number is the opposite of a number. Tests of divisibility: A number is divisible by: 2 if it ends in an number. 3 if the sum of its digits is divisible by . 4 if the last digits are divisible by 4. 5 if it ends in 0 or . 6 if it is divisible by and 3. 8 if the last three digits are divisible by . 9 if the of the digits is divisible by 9. 10 if it ends in . 11 if the sum of the even-positioned digits equals the sum of the -positioned digits or the two sums differ by 11.
WORD
LIST
one prime number square squaring composite number 2 lowest common multiple even
common factors 3 highest common factor sum index two 8 5
multiple odd factor 0 prime expanded
Chapter 2 Multiples, factors and primes
81
CHAPTER review 1 List the first 5 multiples of each number. a 11 b 100 d 20 e 13
c f
5 35
2A
2 Find the lowest common multiple (LCM) of the following pairs of numbers. a 3 and 12 b 6 and 15 c 4 and 7 d 5 and 8
2A
3 In a race, one trail bike rider completes each lap in 40 seconds while another completes it in 60 seconds. How long after the start of the race will the 2 bikes pass the starting point together?
2A
4 Find all the factors of each of the following numbers. a 16 b 27 c 50 d 42 e 36 f 72
2B
5 Find the highest common factor of each of these pairs of numbers. a 8 and 20 b 15 and 35 c d 18 and 60 e 11 and 77 f
14 and 42 6 and 72
6 List the factor pairs of the following numbers. a 24 b 40 d 21 e 99
48 100
c f
2B 2B
7 Dhiba wants to cut equal lengths of streamers to decorate a hall. She wants them to be as long as possible. If she has a roll containing 15 metres and another containing 35 metres, what should be the length of each streamer to have no left over sections?
2B
8 List all of the prime numbers less than 30.
d 124
2C 2C 2C 2D
13 Write the following as prime factors using expanded form. a 44 b 132 c 150 d 360
2D 2E
9 How many single digit prime numbers are there? 10 Find the prime number which comes next after 50. 11 Find the prime factors of: a 99 b 63
c
125
12 Express 280 as a product of prime factors.
14 Write the following using index notation. a 2×2×2×2×2×2 b 5×3×3×7
c
2×2×2×3×3×5
2E
82
Maths Quest 7 for Victoria
15 Find the 2 smallest numbers which are divisible by both 2 and by 3.
2F 2F
16 State whether the following are divisible by both 3 and by 4. a 120 b 155 c 76 d 252
2F,G,H
17 a State the test of divisibility for: i 2 ii 5 iii 10. b What do these 3 tests have in common?
2G
18 Which of the following numbers are divisible by 6? a 65 b 121 c 90
d 294
2H
19 Which of the following numbers are divisible by 9? a 162 b 488 c 459
d 49 725
2H
20 Change the last digit to make each of the following numbers divisible by 11. a 654 321 b 764 320 c 111 337 772
2F,G,H
21 State true (T) or false (F) for each of the following. a 146 is divisible by 2. b 3100 is divisible by 5 and 10. c 435 is divisible by 2 and 5. d 144 is divisible by 3. e 7650 is divisible by 8. f 3124 is divisible by 4. g 24 264 is divisible by 3 and 4. h 6045 is divisible by 5 and 8. i 234 is divisible by 6. j 345 098 is divisible by 11. k 240 is divisible by 2, 3, 4, 5, 6 and 8.
2I
22 Simplify: a 112
2I
23 Find: a the 7th square number b the 40th square number.
2I
24 Simplify: 121 a
CHAPTER
test yourself
2
b 92
b
6400
c
c
302
250 000
d 702 | 19,558 | 61,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-26 | latest | en | 0.885557 |
https://nl.mathworks.com/matlabcentral/cody/problems/2019-dimensions-of-a-rectangle/solutions/469280 | 1,596,827,645,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737206.16/warc/CC-MAIN-20200807172851-20200807202851-00387.warc.gz | 426,353,331 | 15,735 | Cody
# Problem 2019. Dimensions of a rectangle
Solution 469280
Submitted on 10 Jul 2014 by Prateep Mukherjee
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1 length_correct = 3/sqrt(10); width_correct = 1/sqrt(10); [width, length] = rectangle(x); tolerance = 1e-12; assert( abs(length-length_correct)<tolerance && abs(width-width_correct)<tolerance )
x = 1
2 Pass
%% x = 2; width_correct = 2/sqrt(10); length_correct = 6/sqrt(10); [width, length] = rectangle(x); tolerance = 1e-12; assert(abs(length-length_correct)<tolerance && abs(width-width_correct)<tolerance) | 207 | 699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-34 | latest | en | 0.699408 |
https://pastebin.com/wHhwVQwE | 1,575,657,599,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00017.warc.gz | 493,342,404 | 8,829 | • API
• FAQ
• Tools
• Archive
SHARE
TWEET
# Skateboard
Bcadren Apr 18th, 2015 (edited) 313 Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
1. var newVertices : Vector3[];
2. var newUV : Vector2[];
3. var newTriangles : int[];
4. var triangles : Array;
5. var UVs : Array;
6. var vertices : Array;
7. var width : float;
8. var length : float;
9. var depth : float;
10. var duckHeight : float;
11. private var currentLength : int;
12.
13. function Start () {
14. UVs = new Array ();
15. newVertices = [Vector3(0,0,0), Vector3(0,0,width), Vector3(length,0,0), Vector3(length,0,width), Vector3(0,0,width/2),
16.
22.
23. Vector3(length,0,width/2),
29.
30. ]; currentLength = newVertices.Length;
31. vertices = Array(newVertices);
32. for (j=0; j<currentLength; j++)
34. newVertices = vertices.ToBuiltin(Vector3);
35. newTriangles = [0,1,3, 0,3,2, 0,9,4, 5,4,6, 6,4,7, 7,4,8, 8,4,9, 5,1,4,
36. 2,10,15, 15,10,14, 14,10,13, 13,10,12, 12,10,11, 11,10,3 ];
37. currentLength = (newTriangles.Length/3);
38. triangles = Array(newTriangles);
39. for (k=0; k< currentLength; k++){
44. 2,15,18, 18,15,31,
45. 15,14,31, 31,14,30,
46. 14,13,30, 30,13,29,
47. 13,12,28, 13,28,29,
48. 12,11,27, 12,27,28,
49. 11,3,19, 11,19,27,
50. 3,1,19, 1,17,19,
51. 1,5,21, 1,21,17,
52. 5,6,22, 21,5,22,
53. 6,23,22, 23,6,7,
54. 23,7,24, 24,7,8,
55. 8,25,24, 25,8,9,
56. 16,25,9, 16,9,0);
57. newTriangles = triangles.ToBuiltin(int);
58. for (i=0; i< newVertices.length; i++) | 672 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.222639 |
https://www.excelforum.com/excel-formulas-and-functions/637449-shift-differential.html | 1,624,047,770,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00430.warc.gz | 706,933,434 | 14,054 | # shift differential
1. ## shift differential
I am having trouble figuring out a formula for shift differential and I hope someone can help.
My shift diff. are base on the start of the shift. If I start at or after 2000 or before 0559 I will be paid non-premium rate. If I start at or after 1200 or before 1959 I will be paid a premium rate. These rates are for 8 hours or less of hours worked in one day.
My hours are start time E3 and end time F3
My premium shift start time is I3 and end time is J3
My non-premium start time is M3 and end time is N3
2. What happens if the shift starts between 06:00 and 12:00, is that a standard rate? Assuming that your standard rate is in G3, premium rate in K3 and non-premium in O3 you could try this formula
=IF(E3>=M3,O3,IF(E3>=I3,K3,IF(E3>N3,G3,O3)))*MOD( F3-E3,1)*24
although it would probably be easier if you could set up a small table like this in A1:B5
from rate
00:00 £10
08:00 £12
12:00 £15
20:00 £10
and then use
=LOOKUP(E3,A2:B5)*MOD(F3-E3,1)*24
3. Thank you but I don't think that is the formula I'm looking for. I'll try to explain it better so you or someone else can understand.
Here are my 3 shift:
Standard shift 0600-1400
For the hours to be premium hours the shift must start at or after 1200 and before 1959 hour and only up to 8 hours.
For the hours to be non-premium hours the shift must start at or after 2000 and before 0559 hour and only up to 8 hours.
On my timesheet I have the hours that I worked as E3 (start time) and F3 (end time). Then I have I3 as my premium start time and J3 as my premium end time. I have M3 as my non-premium start time and N3 as my non-premium end time.
If possible here is what I'd like to happen. Example: If I put that I worked from 1400-2200 in E3-F3. I would like it to also show up in I3-J3 under premium shift. Example: If I put that I work from 1800-0200 in E3-F3. I would like it to show up in I3-J3 under premium shift.
Thank you to anyone who can help.
There are currently 1 users browsing this thread. (0 members and 1 guests)
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
Search Engine Friendly URLs by vBSEO 3.6.0 RC 1 | 656 | 2,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-25 | latest | en | 0.944618 |
https://documen.tv/sarah-incorrectly-states-that-you-can-find-the-lcm-of-any-two-whole-numbers-by-multiplying-them-28277541-32/ | 1,680,315,568,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00778.warc.gz | 242,312,004 | 15,911 | Question
Sarah incorrectly states that you can find the LCM of any two whole numbers by multiplying them together. Provide a counterexample and use prime factorization to explain when her statement is true.
I need helpp!! :((((
1. RobertKer
The prime factorisation of 14 and 1 is the product of 2 and 7, therefore Sarah’s statement is correct for this particular example.
### What is prime factorisation?
Prime factorisation is defined as the prime numbers of an integer that divides the number without a reminder.
The Lowest common multiple of a number is also known as the smallest positive number that is a multiple of two or more numbers.
You can find the LCM of a whole number by multiplying the two smallest numbers that can divide the integer without reminder.
The prime factorisation of 14 is the product of 2 and 7,therefore Sarah’s statement is correct for this particular example. | 190 | 896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-14 | latest | en | 0.921778 |
https://www.jagranjosh.com/articles/cbse-class-12-physics-syllabus-latest-1587468949-1 | 1,619,018,559,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039544239.84/warc/CC-MAIN-20210421130234-20210421160234-00372.warc.gz | 940,458,399 | 37,531 | Check the new CBSE Class 12 Physics syllabus 2020-21 and download it in PDF format. CBSE Syllabus for Class 12 Physics is very important for the preparation of CBSE 12th Physics board exam 2020-21.
Created On: Sep 1, 2020 15:29 IST
Modified On: Sep 15, 2020 15:35 IST
CBSE Class 12 Physics Syllabus
Check the new CBSE Class 12 Physics syllabus 2020-21 and download it in PDF format. CBSE Syllabus for Class 12 Physics is very important for the preparation of CBSE 12th Physics board exam 2020-21.
CBSE Class 12 Physics Syllabus 2020-21: Topics Added/Deleted By CBSE On 7th July 2020
CBSE Class 12th Physics Notes: All Chapters & Important Resources
## CBSE Class 12 Physics Syllabus 2020-21
Time: 3 hrs. Max Marks: 70
No. of Periods Marks Unit–I Electrostatics 23 16 Chapter–1: Electric Charges and Fields Chapter–2: Electrostatic Potential and Capacitance Unit-II Current Electricity 15 Chapter–3: Current Electricity Unit-III Magnetic Effects of Current and Magnetism 16 17 Chapter–4: Moving Charges and Magnetism Chapter–5: Magnetism and Matter Unit-IV Electromagnetic Induction and Alternating Currents 19 Chapter–6: Electromagnetic Induction Chapter–7: Alternating Current Unit–V Electromagnetic Waves 2 18 Chapter–8: Electromagnetic Waves Unit–VI Optics 18 Chapter–9: Ray Optics and Optical Instruments Chapter–10: Wave Optics Unit–VII Dual Nature of Radiation and Matter 7 12 Chapter–11: Dual Nature of Radiation and Matter Unit–VIII Atoms and Nuclei 11 Chapter–12: Atoms Chapter–13: Nuclei Unit–IX Electronic Devices 7 7 Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits Total 118 70
Unit I: Electrostatics (23 Periods)
Chapter–1: Electric Charges and Fields
Electric Charges; Conservation of charge, Coulomb's law-force between two-point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field.
Electric flux, statement of Gauss's theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet
Chapter–2: Electrostatic Potential and Capacitance
Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field.
Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.
Unit II: Current Electricity (15 Periods)
Chapter–3: Current Electricity
Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm's law, electrical resistance, V-I characteristics (linear and nonlinear), electrical energy and power, electrical resistivity and conductivity; temperature dependence of resistance.
Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff's laws and simple applications, Wheatstone bridge, metre bridge(qualitative ideas only)
Potentiometer - principle and its applications to measure potential difference and for comparing EMF of two cells; measurement of internal resistance of a cell(qualitative ideas only)
Unit III: Magnetic Effects of Current and Magnetism (16 Periods)
Chapter–4: Moving Charges and Magnetism
Concept of magnetic field, Oersted's experiment.
Biot - Savart law and its application to current carrying circular loop.
Ampere's law and its applications to infinitely long straight wire. Straight and toroidal solenoids (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields
Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; moving coil galvanometer-its current sensitivity and conversion to ammeter and voltmeter.
Chapter–5: Magnetism and Matter
Current loop as a magnetic dipole and its magnetic dipole moment, magnetic dipole moment of a revolving electron, bar magnet as an equivalent solenoid, magnetic field lines; earth's magnetic field and magnetic elements.
Unit IV: Electromagnetic Induction and Alternating Currents (19 Periods)
Chapter–6: Electromagnetic Induction
Electromagnetic induction; Faraday's laws, induced EMF and current; Lenz's Law, Eddy currents. Self and mutual induction.
Chapter–7: Alternating Current
Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits
AC generator and transformer.
Unit V: Electromagnetic waves (2 Periods)
Chapter–8: Electromagnetic Waves
Electromagnetic waves, their characteristics, their Transverse nature (qualitative ideas only).
Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.
Unit VI: Optics (18 Periods)
Chapter–9: Ray Optics and Optical Instruments
Ray Optics: Refraction of light, total internal reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lensmaker's formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism.
Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.
Chapter–10: Wave Optics
Wave optics: Wavefront and Huygens principle, reflection and refraction of plane waves at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygens principle. Interference, Young's double slit experiment and expression for fringe width, coherent sources and sustained interference of light, diffraction due to a single slit, width of central maximum
Unit VII: Dual Nature of Radiation and Matter (7 Periods)
Chapter–11: Dual Nature of Radiation and Matter
Dual nature of radiation, Photoelectric effect, Hertz and Lenard's observations; Einstein's photoelectric equation-particle nature of light.
Experimental study of photoelectric effect
Matter waves-wave nature of particles, de-Broglie relation
Unit VIII: Atoms and Nuclei (11 Periods)
Chapter–12: Atoms
Alpha-particle scattering experiment; Rutherford's model of atom; Bohr model, energy levels, hydrogen spectrum.
Chapter–13: Nuclei Composition and size of nucleus Nuclear force
Mass-energy relation, mass defect, nuclear fission, nuclear fusion.
Unit IX: Electronic Devices (7 Periods)
Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits
Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Semiconductor diode - I-V characteristics in forward and reverse bias, diode as a rectifier; Special purpose p-n junction diodes: LED, photodiode, solar cell.
PRACTICALS [Total Periods: 32]
The record to be submitted by the students at the time of their annual examination has to include:
Record of at least 8 Experiments [with 4 from each section], to be performed by the students.
Record of at least 6 Activities [with 3 each from section A and section B], to be demonstrated by teacher
Evaluation Scheme
Time Allowed: Three hours, Max. Marks: 30
Two experiments one from each section 8+8 marks Practical record [experiments and activities] 7 marks Viva on experiments, and activities 7 marks Total 30 marks
SECTION–A Experiments
1. To determine resistivity of two / three wires by plotting a graph for potential difference versus current.
2. To find resistance of a given wire / standard resistor using a metre bridge.
OR
To verify the laws of combination (series) of resistances using a metre bridge.
OR
To verify the laws of combination (parallel) of resistances using a metre bridge.
3. To compare the EMF of two given primary cells using a potentiometer.
OR
To determine the internal resistance of a given primary cell using a potentiometer.
4. To determine resistance of a galvanometer by half-deflection method and to find its figure of merit.
5. To convert the given galvanometer (of known resistance and figure of merit) into a voltmeter of desired range and to verify the same.
OR
To convert the given galvanometer (of known resistance and figure of merit) into an ammeter of desired range and to verify the same.
6. To find the frequency of AC mains with a sonometer.
Activities
1. To measure the resistance and impedance of an inductor with or without iron core.
2. To measure resistance, voltage (AC/DC), current (AC) and check continuity of a given circuit using a multimeter.
3. To assemble a household circuit comprising three bulbs, three (on/off) switches, a fuse and a power source.
4. To assemble the components of a given electrical circuit.
5. To study the variation in potential drop with length of a wire for a steady current.
6. To draw the diagram of a given open circuit comprising at least a battery, resistor/rheostat, key, ammeter and voltmeter. Mark the components that are not connected in proper order and correct the circuit and also the circuit diagram.
Experiments
SECTION-B
1. To find the focal length of a convex lens by plotting graphs between u and v or between 1/u and 1/v.
2. To find the focal length of a convex mirror, using a convex lens.
OR
To find the focal length of a concave lens, using a convex lens.
3. To determine angle of minimum deviation for a given prism by plotting a graph between angle of incidence and angle of deviation.
4. To determine refractive index of a glass slab using a travelling microscope.
5. To find refractive index of a liquid by using convex lens and plane mirror.
6. To draw the I-V characteristic curve for a p-n junction diode in forward bias and reverse bias.
Activities
1. To identify a diode, an LED, a resistor and a capacitor from a mixed collection of such items.
2. Use of a multimeter to see the unidirectional flow of current in case of a diode and an LED and check whether a given electronic component (e.g., diode) is in working order.
3. To study the effect of intensity of light (by varying distance of the source) on an LDR.
4. To observe refraction and lateral deviation of a beam of light incident obliquely on a glass slab.
5. To observe polarization of light using two Polaroids.
6. To observe diffraction of light due to a thin slit.
7. To study the nature and size of the image formed by a (i) convex lens, (ii) concave mirror, on a screen by using a candle and a screen (for different distances of the candle from the lens/mirror).
8. To obtain a lens combination with the specified focal length by using two lenses from the given set of lenses. | 2,764 | 11,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-17 | latest | en | 0.758734 |
https://owenduffy.net/blog/?paged=2&m=201402 | 1,680,131,119,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00634.warc.gz | 506,231,528 | 15,306 | ## Cooling an IC2200H
I have an IC2200H mounted on my operating table with 25mm clearance above the radio and ample room for convection currents to assist in heat removal. It is concerning that the case temperature reaches temperatures that are not safe to touch, temperatures in excess of 75° (55° above ambient) have been measured and that has not triggered the internal temperature protection… so it could get hotter still!
Whilst it might take a while for the radio to reach high temperatures, in the long term, it must dissipate around 139W when transmitting on HIGH power setting and at ambient temperatures as high as 35° in the shack. (Rated input is 15A at 13.6V for 65W out, leaving 139W of heat to be dissipated.)
This is one of those high power mobile radios that advertises no fan as an advantage, but it is clearly not up to the task!
The objective of this change is to keep the external parts below 60°, the (ASTM standard C1055 1999) 5 second human skin burn threshold.
## Mag loop or radiating dummy load?
There is a seemingly endless series of articles on small transmitting loops on the cheap.
(eHam 2014) is another, it describes a so-called magnetic loop for transmitting on 14.2Mhz using 4.57m of 2.6mm copper wire for the main loop. The author reports the bandwidth of the finished antenna as 100kHz. One of the claimed benefits is that with such wide bandwidth, a variable tuning capacitor is not required.
## G/T
A metric that may be used to express the performance of an entire receive system is the ratio of antenna gain to total equivalent noise temperature, usually expressed in deciBels as dB/K. G/T is widely used in design and specification of satellite communications systems.
G/T=AntennaGain/TotalNoiseTemperature 1/K
Example: if AntennaGain=50 and TotalNoiseTemperature=120K, then G/T=50/120=0.416 1/K or -3.8 dB/K.
## VK2OMD ambient noise measurement 144MHz – 20140217
I made a measurement of ambient noise on 144MHz this morning using the technique described at (Duffy 2009).
First step is to recheck the NF of the receiver. The TS2000 is getting a little tired, NF=8.3dB.
The technique calculates ambient noise from the variation in receiver output noise of a receiver of known Noise Figure with the insertion of a known input attenuator. The receiver output noise was measured using NFM (Duffy 2007) which allowed integration over 20s for high resolution measurement.
## VK2OMD G5RV with tuned feeder – line loss
The G5RV Inverted V antenna system at VK2OMD is fed with 9m of home made open wire transmission line using 2mm diameter copper wires spaced 50mm giving a line with characteristic impedance of 450Ω. (Varney 1958) described the tuned feeder configuration of his popular G5RV antenna system.
## BLHeli v11.0 on APM X450 quad
I had some Skywalker 40A with BLHeli v11 spare from the tricopter project, so I have put them on the APM X450 quad for a trial.
The BLHeli beacon is really annoying, so I turned it off. Otherwise it is a pretty basic Multi BLHeli configuration.
The ESCs were throttle calibrated using APM’s all at once procedure… but it didn’t work as expected, the motors remained stopped when armed. A lot of fiddling around to find some parameter changes did not work until power was cycled. Set MOT_SPIN_ARMED=200, THR_MIN=220.
Initial tests in stable, alt hold and loiter modes have been good.
## Small transmitting loop review
I saw a recent ‘maker’ video describing a small transmitting loop for 40m.
The loop used a 3m length of 19mm copper pipe formed into a circle, and at the gap where the ends almost meet, a tuning capacitance is synthesised using coaxial cable.
Above is a screen shot from Reg Edwards loop design program. It calculates the radiation resistance at 0.005Ω, loss resistance of the loop at 0.035Ω, capacitance to resonate it of 206pF (Xc=108Ω), and a bandwidth of 3.2kHz.
## ATU voltage verification
I described a method for designing antenna systems to avoid excessive voltages in baluns and ATUs at (Duffy 2011) .
This article reports post implementation measurements of an antenna system designed using that method and using a G5RV Inverted V with tuned feeder and ATR-30 ATU with integral 1:1 current balun. The tuned feeder is a home-made line section of 2mm diameter copper conductors spaced 50mm, and 9m in length. An additional 0.5m of 135Ω line connects from the antenna entrance panel to the ATU.
## A look at internal losses in a typical ATU
This article explores the loss that may be encountered in an ATU in a practical setting.
The load is a G5RV with tuned feeders operating at 3.6MHz. The tuned feeder is 9m of open wire line of characteristic impedance 450Ω, and the impedance seen by the ATU is around 40-j150Ω, this is not a particularly onerous load.
## An inexpensive medium power tuner current balun for HF using Jaycar parts
This is a project to design and build a Guanella 1:1 (current) balun suited for up to 100W on HF with wire antennas and an ATU.
For use with a tuner, the most important design criteria are:
• high voltage withstand;
• high common mode impedance;
• power handling. | 1,247 | 5,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.904036 |
https://stackoverflow.com/questions/2234204/latitude-longitude-find-nearest-latitude-longitude-complex-sql-or-complex-calc | 1,571,108,930,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00547.warc.gz | 675,441,256 | 46,308 | # latitude/longitude find nearest latitude/longitude - complex sql or complex calculation
I have latitude and longitude and I want to pull the record from the database, which has nearest latitude and longitude by the distance, if that distance gets longer than specified one, then don't retrieve it.
Table structure:
id
latitude
longitude
place name
city
country
state
zip
sealevel
What you need is to translate the distance into degrees of longitude and latitude, filter based on those to bound the entries that are roughly in the bounding box, then do a more precise distance filter. Here is great paper that explains how to do all this:
http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL
• thanks, Nice paper!!! – BlaShadow Jan 13 '14 at 6:14
• Very nice doc, just to note: to use it in Km just convert earth radius unit from miles to Km (3956 -> 6371) – j.c Dec 27 '16 at 16:25
• Also, note that @Kaletha answer is a lot easier and works very well for relatively small areas (errors < 5 m on 20 Km distances due to earth curvature) – j.c Dec 27 '16 at 16:30
• very nice document – Techgeeks1 May 4 '17 at 7:54
• @j.c to be precise miles to km: 3956 -> 6366.564864. Why 6371? – ianaz Oct 25 '17 at 6:43
SELECT latitude, longitude, SQRT(
POW(69.1 * (latitude - [startlat]), 2) +
POW(69.1 * ([startlng] - longitude) * COS(latitude / 57.3), 2)) AS distance
FROM TableName HAVING distance < 25 ORDER BY distance;
where [starlat] and [startlng] is the position where to start measuring the distance.
• Just a performance note, it's best to not sqrt the distance variable but instead square the '25' test value... later sqrt the results that have passed if you need to show the distance – sradforth Jan 30 '12 at 13:46
• What would be the same query for the distance to be in meters ? (which is currently in miles, right?) – httpete Nov 28 '12 at 20:42
• What measurement is that 25? – Steffan Donal Aug 8 '13 at 21:30
• Just to clarify here 69.1 is the conversion factor for miles to latitude degrees. 57.3 is roughly 180/pi, so that's conversion from degrees to radians, for the cosine function. 25 is the search radius in miles. This is the formula to use when using decimal degrees and statute miles. – John Vance Nov 7 '13 at 22:00
• Additionally, it does not take into account the curvature of the earth. This would not be an issue for short search radii. Otherwise Evan's and Igor's answers are more complete. – John Vance Nov 8 '13 at 3:46
## Creating the Table
When you create the MySQL table, you want to pay particular attention to the lat and lng attributes. With the current zoom capabilities of Google Maps, you should only need 6 digits of precision after the decimal. To keep the storage space required for your table at a minimum, you can specify that the lat and lng attributes are floats of size (10,6). That will let the fields store 6 digits after the decimal, plus up to 4 digits before the decimal, e.g. -123.456789 degrees. Your table should also have an id attribute to serve as the primary key.
CREATE TABLE `markers` (
`id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`name` VARCHAR( 60 ) NOT NULL ,
`address` VARCHAR( 80 ) NOT NULL ,
`lat` FLOAT( 10, 6 ) NOT NULL ,
`lng` FLOAT( 10, 6 ) NOT NULL
) ENGINE = MYISAM ;
## Populating the Table
After creating the table, it's time to populate it with data. The sample data provided below is for about 180 pizzarias scattered across the United States. In phpMyAdmin, you can use the IMPORT tab to import various file formats, including CSV (comma-separated values). Microsoft Excel and Google Spreadsheets both export to CSV format, so you can easily transfer data from spreadsheets to MySQL tables through exporting/importing CSV files.
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Frankie Johnnie & Luigo Too','939 W El Camino Real, Mountain View, CA','37.386339','-122.085823');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Amici\'s East Coast Pizzeria','790 Castro St, Mountain View, CA','37.38714','-122.083235');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Kapp\'s Pizza Bar & Grill','191 Castro St, Mountain View, CA','37.393885','-122.078916');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Round Table Pizza: Mountain View','570 N Shoreline Blvd, Mountain View, CA','37.402653','-122.079354');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Tony & Alba\'s Pizza & Pasta','619 Escuela Ave, Mountain View, CA','37.394011','-122.095528');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Oregano\'s Wood-Fired Pizza','4546 El Camino Real, Los Altos, CA','37.401724','-122.114646');
## Finding Locations with MySQL
To find locations in your markers table that are within a certain radius distance of a given latitude/longitude, you can use a SELECT statement based on the Haversine formula. The Haversine formula is used generally for computing great-circle distances between two pairs of coordinates on a sphere. An in-depth mathemetical explanation is given by Wikipedia and a good discussion of the formula as it relates to programming is on Movable Type's site.
Here's the SQL statement that will find the closest 20 locations that are within a radius of 25 miles to the 37, -122 coordinate. It calculates the distance based on the latitude/longitude of that row and the target latitude/longitude, and then asks for only rows where the distance value is less than 25, orders the whole query by distance, and limits it to 20 results. To search by kilometers instead of miles, replace 3959 with 6371.
SELECT
id,
(
3959 *
) AS distance
FROM markers
HAVING distance < 28
ORDER BY distance LIMIT 0, 20;
This one is to find latitudes and longitudes in a distance less than 28 miles.
Another one is to find them in a distance between 28 and 29 miles:
SELECT
id,
(
3959 *
) AS distance
FROM markers
HAVING distance < 29 and distance > 28
ORDER BY distance LIMIT 0, 20;
• Should it be HAVING distance < 25 as we querying locations within a radius of 25 miles ? – Vitalii Elenhaupt Feb 1 '17 at 16:17
• Yes, you're right, I have fixed it, thanks. – Sviatoslav Oleksiv Feb 2 '17 at 10:19
• it should be > 25 , then it will search all records – vidur punj Mar 15 '18 at 12:07
• are you Sure @vidurpunj about > 25 ? – Amranur Rahman Jan 4 at 10:39
• @AmranurRahman give it a try. May be I was wrong. – vidur punj Jan 7 at 11:49
Here is my full solution implemented in PHP.
This solution uses the Haversine formula as presented in http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL.
It should be noted that the Haversine formula experiences weaknesses around the poles. This answer shows how to implement the vincenty Great Circle Distance formula to get around this, however I chose to just use Haversine because it's good enough for my purposes.
I'm storing latitude as DECIMAL(10,8) and longitude as DECIMAL(11,8). Hopefully this helps!
### showClosest.php
<?PHP
/**
* Use the Haversine Formula to display the 100 closest matches to \$origLat, \$origLon
* Only search the MySQL table \$tableName for matches within a 10 mile (\$dist) radius.
*/
include("./assets/db/db.php"); // Include database connection function
\$db = new database(); // Initiate a new MySQL connection
\$tableName = "db.table";
\$origLat = 42.1365;
\$origLon = -71.7559;
\$dist = 10; // This is the maximum distance (in miles) away from \$origLat, \$origLon in which to search
\$query = "SELECT name, latitude, longitude, 3956 * 2 *
ASIN(SQRT( POWER(SIN((\$origLat - latitude)*pi()/180/2),2)
+COS(\$origLat*pi()/180 )*COS(latitude*pi()/180)
*POWER(SIN((\$origLon-longitude)*pi()/180/2),2)))
as distance FROM \$tableName WHERE
and latitude between (\$origLat-(\$dist/69))
and (\$origLat+(\$dist/69))
having distance < \$dist ORDER BY distance limit 100";
\$result = mysql_query(\$query) or die(mysql_error());
while(\$row = mysql_fetch_assoc(\$result)) {
echo \$row['name']." > ".\$row['distance']."<BR>";
}
mysql_close(\$db);
?>
### ./assets/db/db.php
<?PHP
/**
* Class to initiate a new MySQL connection based on \$dbInfo settings found in dbSettings.php
*
* @example \$db = new database(); // Initiate a new database connection
* @example mysql_close(\$db); // close the connection
*/
class database{
function __construct(){
include "dbSettings.php";
\$this->database = \$dbInfo['host'];
\$this->mysql_user = \$dbInfo['user'];
\$this->mysql_pass = \$dbInfo['pass'];
\$this->openConnection();
}
function openConnection(){
}
}
}
?>
### ./assets/db/dbSettings.php
<?php
\$dbInfo = array(
'host' => "localhost",
'user' => "root",
);
?>
It may be possible to increase performance by using a MySQL stored procedure as suggested by the "Geo-Distance-Search-with-MySQL" article posted above.
I have a database of ~17,000 places and the query execution time is 0.054 seconds.
• mile * 1.609344 = km – Sinan Dizdarević Mar 30 '15 at 9:06
• WARNING. Excelent solution, but it has a bug. All the abs should be removed. There is no need to take the abs value when converting from degrees to radians, and, even if you did, you are doing it to just one of the latitudes. Please edit it so it fixes the bug. – Chango Jan 12 '16 at 21:59
• thanks @Chango, updated – circuitry Jan 13 '16 at 15:49
• And for anyone who wants this in meters: Convert 3956 miles to kilometers: Earth's radius; Convert 69 miles to kilometers: the aproxímate length of 1 degree latitude in km; And input the distance in kilometers. – Chango Jan 13 '16 at 16:15
• And replace 69 with 111,044736 (forgot that in above comment) – rkeet Apr 26 '16 at 10:31
Just in case you are lazy like me, here's a solution amalgamated from this and other answers on SO.
set @orig_lat=37.46;
set @orig_long=-122.25;
set @bounding_distance=1;
SELECT
*
,((ACOS(SIN(@orig_lat * PI() / 180) * SIN(`lat` * PI() / 180) + COS(@orig_lat * PI() / 180) * COS(`lat` * PI() / 180) * COS((@orig_long - `long`) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS `distance`
FROM `cities`
WHERE
(
`lat` BETWEEN (@orig_lat - @bounding_distance) AND (@orig_lat + @bounding_distance)
AND `long` BETWEEN (@orig_long - @bounding_distance) AND (@orig_long + @bounding_distance)
)
ORDER BY `distance` ASC
limit 25;
• what exactly does bounding_distance represent? does this value limit the results to a mile amount? so in this case it will return results within 1 mile? – james Nov 14 '12 at 14:43
• @ bounding_distance is in degrees here, and is used to speed up calculations by limiting the effective search region. For example, if you know your user is in a certain city, and you know you have a few points within that city, you can safely set your bounding distance to a few degrees. – Evan Nov 14 '12 at 16:22
• Which geographical distance formula is this using? – bbodenmiller Aug 19 '15 at 8:08
Easy one ;)
SELECT * FROM `WAYPOINTS` W ORDER BY
ABS(ABS(W.`LATITUDE`-53.63) +
ABS(W.`LONGITUDE`-9.9)) ASC LIMIT 30;
Just replace the coordinates with your required ones. The values have to be stored as double. This ist a working MySQL 5.x example.
Cheers
• No idea why upvote, OP wants to limit and order by certain distance, not limit by 30 and order by dx+dy – okm Oct 14 '12 at 12:58
• This did it for me. It's not what the OP wanted, but it's what I wanted, so thank you for answering! :) – Webmaster G Dec 9 '15 at 18:58
• how do this find nearest points ? ..for instance within 5km? – Prasobh.Kollattu Aug 9 at 10:17
• outermost ABS() is sufficient. – dzona Oct 11 at 20:01
You're looking for things like the haversine formula. See here as well.
There's other ones but this is the most commonly cited.
If you're looking for something even more robust, you might want to look at your databases GIS capabilities. They're capable of some cool things like telling you whether a point (City) appears within a given polygon (Region, Country, Continent).
Check this code based on the article Geo-Distance-Search-with-MySQL:
Example: find the 10 nearest hotels to my current location in a 10 miles radius:
#Please notice that (lat,lng) values mustn't be negatives to perform all calculations
set @my_lat=34.6087674878572;
set @my_lng=58.3783670308302;
SELECT dest.id, dest.lat, dest.lng, 3956 * 2 * ASIN(SQRT(POWER(SIN((@my_lat -abs(dest.lat)) * pi()/180 / 2),2) + COS(@my_lat * pi()/180 ) * COS(abs(dest.lat) * pi()/180) * POWER(SIN((@my_lng - abs(dest.lng)) * pi()/180 / 2), 2))
) as distance
FROM hotel as dest
having distance < @dist
ORDER BY distance limit 10;
#Also notice that distance are expressed in terms of radius.
Try this, it show the nearest points to provided coordinates (within 50 km). It works perfectly:
SELECT m.name,
m.lat, m.lon,
p.distance_unit
FROM <table_name> AS m
JOIN (
SELECT <userLat> AS latpoint, <userLon> AS longpoint,
50.0 AS radius, 111.045 AS distance_unit
) AS p ON 1=1
WHERE m.lat
BETWEEN p.latpoint - (p.radius / p.distance_unit)
AND p.latpoint + (p.radius / p.distance_unit)
ORDER BY distance_in_km
Just change <table_name>. <userLat> and <userLon>
simpledb.execSQL("CREATE TABLE IF NOT EXISTS " + tablename + "(id INTEGER PRIMARY KEY AUTOINCREMENT,lat double,lng double,address varchar)");
simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2891001','70.780154','craftbox');");
simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2901396','70.7782428','kotecha');");//22.2904718 //70.7783906
simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2863155','70.772108','kkv Hall');");
simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.275993','70.778076','nana mava');");
simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2667148','70.7609386','Govani boys hostal');");
double curentlat=22.2667258; //22.2677258
double curentlong=70.76096826;//70.76096826
double curentlat1=curentlat+0.0010000;
double curentlat2=curentlat-0.0010000;
double curentlong1=curentlong+0.0010000;
double curentlong2=curentlong-0.0010000;
try{
Cursor c=simpledb.rawQuery("select * from '"+tablename+"' where (lat BETWEEN '"+curentlat2+"' and '"+curentlat1+"') or (lng BETWEEN '"+curentlong2+"' and '"+curentlong1+"')",null);
Log.d("SQL ", c.toString());
if(c.getCount()>0)
{
while (c.moveToNext())
{
double d=c.getDouble(1);
double d1=c.getDouble(2);
}
}
}
catch (Exception e)
{
e.printStackTrace();
}
The original answers to the question are good, but newer versions of mysql (MySQL 5.7.6 on) support geo queries, so you can now use built in functionality rather than doing complex queries.
You can now do something like:
select *, ST_Distance_Sphere( point ('input_longitude', 'input_latitude'),
point(longitude, latitude)) * .000621371192
as `distance_in_miles`
from `TableName`
having `distance_in_miles` <= 'input_max_distance'
order by `distance_in_miles` asc
The results are returned in meters so if you want KM instead of miles use .0001 instead of .000621371192
• ST_Distance_Sphere doesn't exist in my host's install (mysql Ver 15.1 Distrib 10.2.23-MariaDB). I read somewhere to substitute ST_Distance but the distances are way off. – ashleedawg Jul 21 at 8:04
• @ashleedawg - From the version, I think you're using MariaDB, which is a fork of mysql. From this conversation it looks like MariaDB hasn't implemented ST_Distance_Sphere – Sherman Jul 22 at 20:46
It sounds like you want to do a nearest neighbour search with some bound on the distance. SQL does not support anything like this as far as I am aware and you would need to use an alternative data structure such as an R-tree or kd-tree.
Find nearest Users to my:
Distance in meters
Based in Vincenty's formula
i have User table:
+----+-----------------------+---------+--------------+---------------+
| id | email | name | location_lat | location_long |
+----+-----------------------+---------+--------------+---------------+
| 13 | xxxxxx@xxxxxxxxxx.com | Isaac | 17.2675625 | -97.6802361 |
| 14 | xxxx@xxxxxxx.com.mx | Monse | 19.392702 | -99.172596 |
+----+-----------------------+---------+--------------+---------------+
sql:
-- my location: lat 19.391124 -99.165660
SELECT
(ATAN(
SQRT(
)
,
) * 6371000) as distance,
users.id
FROM users
ORDER BY distance ASC
radius of the earth : 6371000 ( in meters)
Sounds like you should just use PostGIS, SpatialLite, SQLServer2008, or Oracle Spatial. They can all answer this question for you with spatial SQL.
• sounds like you should just NOT suggest that people switch their entire database platform and cause irrelevant results to show up in my google search when i explicitly search fro "Oracle"... – I wrestled a bear once. Jun 2 '16 at 20:29
In extreme cases this approach fails, but for performance, I've skipped the trigonometry and simply calculated the diagonal squared.
MS SQL Edition here:
DECLARE @SLAT AS FLOAT
DECLARE @SLON AS FLOAT
SET @SLAT = 38.150785
SET @SLON = 27.360249
SELECT TOP 10 [LATITUDE], [LONGITUDE], SQRT(
POWER(69.1 * ([LATITUDE] - @SLAT), 2) +
POWER(69.1 * (@SLON - [LONGITUDE]) * COS([LATITUDE] / 57.3), 2)) AS distance
FROM [TABLE] ORDER BY 3
This problem is not very hard at all, but it gets more complicated if you need to optimize it.
What I mean is, do you have 100 locations in your database or 100 million? It makes a big difference.
If the number of locations is small, get them out of SQL and into code by just doing ->
Select * from Location
Once you get them into code, calculate the distance between each lat/lon and your original with the Haversine formula and sort it.
• i would say 100 thousands – Basit Feb 10 '10 at 4:28 | 4,868 | 17,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-43 | latest | en | 0.793943 |
http://www.physicsforums.com/showthread.php?s=abaaeb5f3b3e2d0778193ccf2708bf5b&p=4513672 | 1,398,342,197,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00019-ip-10-147-4-33.ec2.internal.warc.gz | 782,842,896 | 10,416 | # Which is larger, Graham's number or
by MathJakob
Tags: graham, larger, number
P: 153 This isn't a homework problem or anything just simply curious. Which is a larger number, Graham's number or the estimated number of atoms in the universe factorial? So Graham's number or ##(10^{80})!##
PF Gold P: 161 Well, as an upper bound, $\left(10^{80}\right)! < \left(10^{80}\right)^{\left(10^{80}\right)} = 10^{80 * 10^{80}} = 10^{8*10^{81}}$ which, if I am not mistaken, is far less than $g_1$.
P: 153 So ##(10^{80})!## isn't even as large as g1... and Graham's number is g64... WOW
HW Helper
PF Gold
P: 12,016
## Which is larger, Graham's number or
The Arildno number is even greater:
Arildno number =Graham's number+1.
Can I also get a WOW! from you?
P: 153
Quote by arildno The Arildno number is even greater: Arildno number =Graham's number+1. Can I also get a WOW! from you?
No. Your number is meaningless, it's never been used in any meaningful way.
HW Helper
PF Gold
P: 12,016
Quote by MathJakob No. Your number is meaningless, it's never been used in any meaningful way.
P: 124
Sorry to link to wikipedia, but you may be interested in TREE(3), and I can't find too much info on it.
an extremely weak lower bound for TREE(3), is A(A(...A(1)...)), where the number of As is A(187196)
Graham's number, for example, is approximately A^64(4) which is much smaller than the lower bound A^(A(187196))(1).
A is the Ackermann function.
http://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem
P: 308
Quote by MathJakob This isn't a homework problem or anything just simply curious. Which is a larger number, Graham's number or the estimated number of atoms in the universe factorial? So Graham's number or ##(10^{80})!##
Tip: if you can write out the number in any "normal" way, it's waaaaaaaaaaaaaaaaay smaller than Graham's number.
P: 13 What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!
HW Helper
PF Gold
P: 12,016
Quote by Deedlit What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!
It seems the Arildno number is a dwarf, after all..
HW Helper
P: 3,436
Quote by arildno It seems the Arildno number is a dwarf, after all..
Poor thing, the Arildno number is coming under lots of fire in here!
Quote by Deedlit What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!
I haven't really been able to find any sources that explain some of the larger operators in the fast growing hierarchy. For example,
http://googology.wikia.com/wiki/Fast-growing_hierarchy
gives a list of growth rates, but there's a lot of details being left out. Sadly, I only understand how to calculate up to and including $f_{\epsilon_0}(n)$.
And to do so, I use a calculator of course
P: 13 Good to see you found the Googology Wiki! Beyond $\varepsilon_0$ we have: $\varepsilon_1 = \varepsilon_0 ^ {\varepsilon_0 ^ {\varepsilon_0 ^\ddots}}$ $\varepsilon_2 = \varepsilon_1 ^ {\varepsilon_1 ^ {\varepsilon_1 ^\ddots}}$ $\varepsilon_\omega = \sup \lbrace \varepsilon_0, \varepsilon_1, \varepsilon_2, \ldots \rbrace$ and so on. Eventually we get to $\zeta_0 = \varepsilon_{\varepsilon_{\varepsilon_\cdots}}$ We can set $\varphi(0, \alpha) = \omega^\alpha, \varphi(1, \alpha) = \varepsilon_\alpha, \varphi(2, \alpha) = \zeta_\alpha$. More generally, $\varphi(\alpha+1, \beta) =$ the $\beta$th fixed point of $f(\gamma) = \varphi(\alpha, \gamma)$. When $\alpha$ is a limit ordinal, $\varphi(\alpha, \beta)$ is the $\beta$th ordinal in the intersection in the ranges of $f(\delta) = \varphi(\gamma, \delta)$ for all $\gamma < \alpha$. So, for example, $\varphi(3, 0) = \varphi(2, \varphi(2, \varphi(2, \ldots)))$ $\varphi(\omega, 0) = \sup (\varphi(1, 0), \varphi (2, 0), \varphi (3, 0), \ldots)$ and so on. This takes us up to $\Gamma_0 = \varphi(\varphi(\varphi(\ldots, 0),0),0)$ or alternatively, $\Gamma_0$ is the smallest ordinal $\alpha$ such that $\alpha = \varphi(\alpha, 0)$. We can continue the notation with $\varphi(1, 0, 0) = \Gamma_0$, and $\varphi(1, 0, \alpha)$ is the $\alpha$th fixed point of $f(\beta) = \varphi(\beta, 0)$. The ordinals $\varphi (1, \alpha, \beta)$ are defined analogously to $\varphi(\alpha, \beta)$, i.e. each function $f(\beta) = \varphi (1, \alpha+1, \beta)$ enumerates the fixed points of $g(\beta) = \varphi(1, \alpha, \beta)$, and at limit ordinals you enumerate the intersection of the ranges of previous ordinals. Then $\varphi(2, 0, \alpha)$ is the $\alpha$th fixed point of $f(\beta) = \varphi(1, \beta, 0)$, and you can construct the hierarchy $\varphi(2, \alpha, \beta)$ similarly. At limit ordinals you take the intersection of the ranges again, and so we define $\varphi (\alpha, \beta, \gamma)$ for all ordinals $\alpha, \beta, \gamma$. Then we can define $\varphi (1, 0, 0, \alpha)$ to be the $\alpha$th fixed point of $f(\beta) = \varphi(\beta, 0, 0)$. We can then define $\varphi(\alpha, \beta, \gamma, \delta), \varphi(\alpha, \beta, \gamma, \delta, \epsilon)$, and so on. The general definition for the n-ary Veblen funciton is: $\varphi (\alpha) = \omega^{\alpha}$ $\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n + 1, 0, \ldots, 0, \beta)$ is the $\beta$th fixed point of the function $f(\gamma) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_n, \gamma, 0, \ldots, 0)$ When $\alpha_n$ is a limit ordinal, $\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n, 0, \ldots, 0, \beta)$ is the $\beta$th ordinal in the intersection of the ranges of $f(\delta) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_{n-1}, \gamma, \delta, 0, \ldots, 0)$ for all $\gamma < \alpha_n$. We define the Small Veblen Ordinal as $\sup (\varphi (1, 0), \varphi(1, 0, 0), \varphi(1, 0, 0), \ldots)$. TREE(n) is larger than the fast-growing hierarchy at the level of the Small Veblen Ordinal. (how much further is not known.). I'll go one step further: we can extend the n-ary Veblen function to transfinitely many places. Obviously, we can't write out transfinitely many variables, so we need to modify our notation: instead of $\varphi(\alpha, \beta, \gamma, \delta, \epsilon)$ we write $\varphi(\alpha @ 4, \beta @ 3, \gamma @ 2, \delta @ 1, \epsilon @ 0)$. So we append "@ n" to every variable, where n represents the index of the variable. This allows us to skip variables that are 0, and so we can notate things like $\varphi (1 @ \omega, \alpha @ 0)$. $\varphi(1 @ \omega, \alpha @ 0)$ is defined as the $\alpha$th ordinal that is a fixed point of $f(\beta) = \varphi (\beta @ n)$ for all $n < \omega$. More generally, we define $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n + 1, \gamma @ 0)$ is the $\gamma$th fixed point of $f(\delta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \delta @ \beta_n)$ When $\alpha_n$ is a limit ordinal, $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \gamma @ 0)$ is the $\gamma$th ordinal in the intersection of the ranges of $f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n + 1, \epsilon @ \beta_n)$ for all $\delta < \alpha_n$ When $\beta_n$ is a limit ordinal, $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n, \gamma @ 0)$ is the $\gamma$th ordinal in the intersection of the ranges of $f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \epsilon @ \delta)$ for all $\delta < \beta_n$ When $\alpha_n$ and $\beta_n$ are limit ordinals, $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \gamma @ 0$ is the $\gamma$th ordinal in the intersection of the ranges of $f(\zeta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n, \zeta @ \epsilon)$ for all $\delta < \alpha_n, \epsilon < \beta_n$ This defines $\varphi(\alpha_1 @ \beta_1, \ldots, \alpha_n @ \beta_n)$ for all $\alpha_i$ and $\beta_i$. This notation is known as Schutte's Klammersymbolen. The smallest ordinal $\alpha$ such that $\alpha = \varphi (1 @ \alpha)$ is known as the Large Veblen Ordinal. I would think that TREE(n) would not reach the Large Veblen Ordinal in the fast-growing hierarchy, but I don't think this is known. Phew! I hope this was at least somewhat comprehensible.
Related Discussions General Math 2 General Math 1 Linear & Abstract Algebra 5 General Math 2 General Math 16 | 2,959 | 9,264 | {"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} | 3.640625 | 4 | CC-MAIN-2014-15 | latest | en | 0.929885 |
goodparentsgreatkids.com | 1,516,310,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887621.26/warc/CC-MAIN-20180118210638-20180118230638-00212.warc.gz | 153,163,097 | 32,844 | # Math In Everyday Life
Parents frequently ask what they can do to help their child with math. I find that the most meaningful method is to seize upon those teachable moments in everyday life. The following are some examples that may or may not work for you and your child. It’s important to keep your child’s confidence level high. Pushing a child beyond what they are capable of will do more harm than good, so take your cues from them and have fun with math in everyday life.
In an elevator
Notice the buttons- Use the buttons as a number line, and ask
If the elevator stopped on the 4th floor and we hadto walk to the 6th floor, how many more floors would we have to walk?
Would we walk upstairs or down?
What is the biggest number?
Compare several numbers and put them in order from least to most and most to least
Walking in NYC
Looking at the street sign- we are on 14th street. How many blocks away is 10th street?
How many do you want?
What if I gave you
1 more/ less
2 more/less
3 more/less
Pretend Purchase
Pretend your child is buying something. Ask them how much they have? Then make up a price and ask if they have enough? Then ask how much more they need or how much left over money they would have.
Use actual pennies to work it out.
Sports
Compare points: who has more, how much more?
How much more does the losing team need to get to be equal or win?
Money
Teach the value of penny, nickel and dime.
Count by 1’s 5’s and 10’s.
Count by 5’s and 10’s and then add pennies.
In the Kitchen
Your child may not be ready to add fractions but exposing them to units of measure will be helpful.
Do they know the difference vetween a teaspoon and tablespoon? Show them how many teaspoons equal a tablespoon. If your child is ready see if they can figure out how many ways to get 4 teaspoons. Play with measuring.
Food
Pizza or pies – ask questions involving a certain number of people eating a slice and then how much will be leftover?
Throw in the word “each” as in if each person ate 2 pieces how much will be leftover. This requires three steps and understanding what “each” implies. First they have to count how many slices are in the pie, then add how much was eaten and finally subtract from the total. Using the real thing will teach this concept with relative ease.
Cut things in half and discuss equal parts.
Introduce halves and quarters.
Setting the Table
Ask your child to set the table but don’t give them enough. Then ask them how much more they need.
Games
Board and card games are wonderful opportunities to incorporate math. You probably do not want to inundate your child with questions and take out the play….. but once in a while ask
How much more does one person have than the other?
How many more spaces to get to the end?
What do the numbers on the dice add up to?
Sharing
When you share something with your child say: I have ___. If I give you ___ how many will I have left?
.….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….….…
Once you see the opportunities you will understand that there are endless ways to teach math in everyday life. Keep it REAL, keep it FUN and keep it going.
# Our High Tech Kids
Our High Tech Kids
Today I am writing about handwriting, but it is just one more ingredient in this high tech world that has negative implications for our children. Handwriting is eventually going to be a thing of the past. There are so many more high tech ways to communicate and schools are beginning to focus on those alternate methods.
As a student I have found that writing notes helped me to remember important details much better than if I had typed them. My own children found the same strategy to be true. Conversely, I can write more creatively when I type. I teach writing to young children to facilitate reading. It helps them to single out letter sounds and sight words in isolation and then apply it to text. Clearly, the act of writin,as well as typing, has an impact and now we know more about why that is.
In a recent New York Times article it was said that:
Children not only learn to read more quickly when they first learn to write by hand, but they also remain better able to generate ideas and retain information. In other words, it’s not just what we write that matters — but how.
When we write, a unique neural circuit is automatically activated,” said Stanislas Dehaene, a psychologist at the Collège de France in Paris. “There is a core recognition of the gesture in the written word, a sort of recognition by mental simulation in your brain.
And it seems that this circuit is contributing in unique ways we didn’t realize,” he continued. “Learning is made easier.”
2012 study led by Karin James, a psychologist at Indiana University, lent support to that view. Children who had not yet learned to read and write were presented with a letter or a shape on an index card and asked to reproduce it in one of three ways: trace the image on a page with a dotted outline, draw it on a blank white sheet, or type it on a computer. They were then placed in a brain scanner and shown the image again.
The researchers found that the initial duplication process mattered a great deal. When children had drawn a letter freehand, they exhibited increased activity in three areas of the brain that are activated in adults when they read and write: the left fusiform gyrus, the inferior frontal gyrus and the posterior parietal cortex.
By contrast, children who typed or traced the letter or shape showed no such effect. The activation was significantly weaker.
Dr. James attributes the differences to the messiness inherent in free-form handwriting: Not only must we first plan and execute the action in a way that is not required when we have a traceable outline, but we are also likely to produce a result that is highly variable.
In a study that followed children in grades two through five,Virginia Berninger, a psychologist at the University of Washington, demonstrated that printing, cursive writing, and typing on a keyboard are all associated with distinct and separate brain patterns — and each results in a distinct end product. When the children composed text by hand, they not only consistently produced more words more quickly than they did on a keyboard, but expressed more ideas. And brain imaging in the oldest subjects suggested that the connection between writing and idea generation went even further. When these children were asked to come up with ideas for a composition, the ones with better handwriting exhibited greater neural activation in areas associated with working memory — and increased overall activation in the reading and writing networks.
We live in an age where modern conveniences are changing at lightning speed. The way we communicate, watch TV, use computers……………………….all drastically different than just two years ago. Even the food we eat is less wholesome than in years past. Our brain and other organs are still the same and requires proper stimulation and nourishment. It scares me to think of how we may be hurting our children as a result of modern technology.
I urge all parents and teachers to be fully mindful as we prepare our children towards their future.
# Kindergarten in New York City
Kindergarten and Play
Last week, in yoga, the teacher innocently instructs us to be playful with our pose. Pretend that you are in kindergarten she relates. I fume. Kindergarten has very little to do with play these days…..though I wish it did. And why do people still think it does? There are schools that still value a less structured and academic curriculum but those are few and far between. Many preschools, as well as parents, encourage learning basic skills such as the alphabet and number recognition as a way to give their child a leg up. The value of pretend play, exercising fine and gross motor skills (small and large muscles,) and building a child’s capacity for complex learning tasks are significantly diinished. Instead, kindergarten curriculum goes straight to hard core academics with limitted time for a five year old to be emotionally and physically free. Easels are gone…too messy. Fingerpainting is out. Clay is rarely used and soon crayons will be substituted for iPads. Worksheets are frequently used.
Many educators believe that this push to teach kindergartners in this way is inappropriate and ineffective. Children who are allowed to explore and experiment are more likely to be motivated to learn. Critical thinking, which is at the heart of the new common core curriculum, is increased and children develop their social emotional skills in a supportive setting.
Parents who strive to give their children the best education may look for the most rigorous setting. In my opinion, a school that provides great stimulation for a child’s natural curiosity and then supports them with a flexible environment in which to explore will be most effective.
New York City provides many options in both the public and private arena. Tests such as the ERB for private schools are important to some schools but not others. The G&T program may be an option but is often unacceptable. Hunter Elementary School is a great school but your child must perform exceptionally on the Stanford Binet Exam which is an IQ test. There are many general public schools that are wonderful options if you live within the school’s boundaries.
As you explore kindergarten for your child’s next academic step be aware of the trapping of the so called “top ten schools.”There is more to look at than a school’s ranking. Observe well and seek the advice of educators you respect.
http://voices.washingtonpost.com/parenting/2007/05/kindergartens_not_just_play_an.html
# Enabling Our Children.….Why?
Why do we enable our children?
Are you a parent who thinks they are doing too much for their child?
Is your child lacking in motivation or responsibility?
1- Is my help truly necessary?
2- What is my motivation for helping?
Many times parents help or enable their child because they want their child to succeed. This will usually help in the short term as the child completes tasks, gets to appointments on time, is well prepared and turns in elevated assignments. Does the parent feel pride as their child excells? Would the child have excelled without their help? What message is the child gettinng from this help?
The practice of enabling becomes more apparent as the parent feels obliged to intervene for the child’s continued success. The suceess of their child becomes a goal for the parent when it should be a goal for the child. The continued assistance by the parent creates a system whreeby the child depends on the parent’s help and therefore becomes a passive participant. There is diminished motivation and often the child resents the parents for their input. In the long term the parent must decide whether to continue this support or demenstrate trust in their child.
Think about that for a moment or two. Would you ever consciously want to convey to your child that you don’t trust them? Demonstrating trust in your child is huge. If you are faced with the option of demonstrating to your child that you have faith in them and you choose to involve yourself to insure better results then you must look in the mirror and ask yourself if your motive is ultimately selfish. Yes, selfish. That may sound harsh but it is imperative that you separate your own ego from your child’s success. Parents naturally feel proud of their child. It is unhealthy for both the parent and child when the need to see a child succeed trumps the ability to do what is necessary to raise a responsible child.
Give your child the gift of independence, confidence and responsibility by showing them that you have faith in them.…and walk away.
As a parent educator and a tutor for early readers, I am pleased to announce the first book authored by my youngest student.
Once a child has a solid foundation of letter sounds then they can begin to use inventive spelling to represent words. Photographs or their own illustrations add the finishing touch.
As a seasoned and now retired kindergarten teacher, I am able to publish a book that has controlled vocabulary, repetitive and predictive text with pictures to provide a solid clue to the unknown word. The result is a book that the author is very proud of, as well as a book that can be an instpiration to young readers and writers in general.
I am proud of this first book and will soon be announcing his second in this series. Karate Is My Sport.
# The Myth of Gifted Education in New York City
As an educator I can fully appreciate the needs of highly gifted children. They see the world differently and often require unique strategies and opportunities for learning and reaching their potential.
In New York City children entering kindergarten are required to take a standardized test if they want to be part of the Gifted and Talented Program. Many parents see this as an opportunity for an elevated education. The reality is that the children who test into these programs are usually not the exceptional children educators consider gifted. They are bright, they are inquisitive, they learn more easily than some others, but they certainly do not require specialized instruction. What they do need is an attentive and nurturing teacher, a rich curriculum that respects their developmental stage and an environment that is safe and stimulating.
In California, where I taught for over thirty years, there were two levels of giftedness. The seminar program was solely for highly gifted students. The formula for the “cluster program,” the second tier of giftedness, was that 25% of the class needed to be identified as gifted while the rest of the class was not. This shifted the focus of the instruction to a higher level and was meant to raise the bar for the entire class of children. Differentiated teaching was still necessary and any decent teacher knows that differentiated teaching and individualized instruction must always be part of the plan. Children all learn at different speeds and modalities. Attention must be paid to these differences.
In my opinion, the major benefit of a gifted class is not the teaching style or the curriculum. It is the student. The benefits of a class that is comprised of high achievers are that a teacher can move quicker and delve deeper into all subjects. Class discussions are richer and students are motivated by their peers. In classes that lack bright students there is a greater chance that children will have learning difficulties as well as behavior problems. The teacher’s time may be skewed d to the most challenging students and a bright, well behaved child may get a poor academic experience.
In New York City, where I was born and raised, and now live, I am witnessing many frustrated parents who are unable to enroll their child in a good school. Their child may have tested high on the G&T exam but denied a slot due to lack of openings. Also, their child may not have received a high enough score and therefore their choices were limited. Some children may get into a gifted program but the commute may be so long that it would be unacceptable to even consider. Many children have no choice than go to a neighborhood school even though those schools are floundering. Teachers, parents as well as school administrators are frustrated and there is no solution in sight. Throwing money at schools has not been shown to make a difference. Testing has become a new source for corruption and neglect as schools find themselves being driven to raise test scores at the expense of quality instruction. Charter schools can be wonderful but, more often than not, are poorly managed and must report to a board who are overly interested in their financial investment.
Benjamin Disraeli once said: “Upon the education of the people of this country, the fate of this country depends.” Can we afford to continue to let our public schools flounder and die? We continue to explore and strategize solutions but what if the solution is not found within the school system but the child who enters it? What if that child came from a home where parents were involved, had elevated parenting skills and partnered with the child’s teacher? Early education for children and parents as well as ongoing parenting support is a path to change the fate of our public schools. Every classroom should have a passionate teacher and students who are eager and capable of learning. All schools should be good schools. Anything less is unacceptable.
The Failure of The American Schools:
http://www.theatlantic.com/magazine/archive/2011/06/the-failure-of-american-schools/308497/
No Rich Child Left Behind:
http://opinionator.blogs.nytimes.com/2013/04/27/no-rich-child-left-behind/
Whining (verb)
To give or make a long, high-pitched complaining cry or sound.
Complain in a feeble or petulant way.
If you are a parent there is no need to read the definition of whining. It is something children do naturally and parents seem to just get used to…. and yes, we often get annoyed or VERY annoyed. Some parents learn to deal with this and almost tune out the sound. Others find that it intolerable and can lose their temper and see things spiral downward.
Learning to control our reaction is not the only way to go. If your children learn to wine it can become an issue for them as teachers and friends may also find it an irritant. Whining becomes an important issue if it causes us, as parents, to behave in a less than acceptable way.
Recognizing the need to decrease whining is the first step and it is a giant step. This article will help you to understand how whining originates and will provide specific steps to diminish it. It will make a huge difference in your lives.
1. Evaluate. Is there anything specific that brings out the whiner in your child? Take notes about the time of the day, their physical and emotional state and the kinds of things they whine about. Don’t just take mental notes, write it down. You may gain a great deal of insight.
2. Understand. What does whining accomplish? It is usually a learned behavior from a child that is used to having to beg r make repeated requests for what they want. They are feeling powerless and defeated and therefore resort to whining. Pretty soon it can become their main way of communicating or requesting.
3. Model and practice. Your child is probably unaware of their whining and will require time to replace this behavior. Ask your child to practice asking for things in a regular voice and then when whining does occur, ask them to use that regular voice.
4. Praise. Catch them speaking in a non-whining voice and praise them by saying how nice and sweet their voice sounds. Be careful not to give a “back handed compliment,” by comparing it to the negative.
5. Awareness of YOUR behavior. Are you part of the problem or solution?
Child: They get frustrated and then they start to whine.
Adult: Be responsive to their frustration level and intervene before they start to whine. If they are asking you for something be aware of a tendency to ignore and only respond when they whine. Respond earlier, even if it is to say I heard you and you need to wait.
Child: Whines when they talk to you.
Adult: You ignore the whining and respond to them. If they whine ask them to please repeat their message in their regular voice.
Or…..
Child: Whines when they talk to you.
Adult: Gets angry. A child that can make a grown up lose control has learned a powerful tool. React calmly and your child will respond in kind.
Congratulations for reading this. It shows that you want to elevate your parenting skills. Stick with it and you will see the results you want. Remember, reversing bad habits may be a lot of work in the beginning but careful attention to these areas has big payoffs in the end. Teachers and other adults will have a more positive view of your child if they communicate well. Confidence is boosted and so is performance. It’s HUGE and therefore well worth your time.
# Science Inspired By Groundhog Day
Though Ground Hog day is basically a silly custom that makes little sense, it is a great time to teach children about shadows and have fun with them as well.
The philosophy behind Groundhog Day is that if the groundhog comes out of their burrow and sees their shadow it means a longer winter. Six more weeks to be exact. If there is no shadow then the groundhog feels comfortable hanging around in the field and then we know Spring is “just around the corner.” The premise is silly for so many reasons but most importantly, it implies that the presence of a bright, beautiful sun, which is necessary for a shadow, could predict cold weather ahead.
Then inside you can play with a flashlight which imitates the sun. If the flashlight is straight above your child’s head, or any other object, the shadow will be minimal. As the sun goes down or the flashlight makes an arc downward, the shadow gets longer. That is why shadows are longer in the morning and evening and shortest at high noon.
With that concept being demonstrated, take notice of shadows in your everyday life.
Bringing attention to the wonders of science stimulates your child’s curiosity and thirst for knowledge.
# Increase Responsibility Using A Checklist
There’s A Checklist For That
Everyday routines can be exhausting.
You know the kids are going to:
Ignore you
Put up a fuss
Do it slowly or poorly
You are tired of getting aggravated so you:
Yell
Nag
Give up and do it yourself
Or learn to let them do it it their own way
Many situations fall into this dance of parent and child:
Bedtime
Coming home from school
Homework
Helping with chores
Cleaning room
Though there are many things that we try to analyze and perfect, for many reasons parents often resign to the standard
practices that come naturally but may not be effective. When we take the time to evaluate we can make big changes. A checklist, combined with better understanding can provide harmony in the home and more responsible children.
Example:
Sarah was a mom of 3 darling girls ages 2, 4 and 6.
She frequently complained that the girls ganged up on her and it was especially terrible at bedtime.
Gymnastics in the bedroom including jumping on bed and using the window sill as a balance beam was their routine. Once in bed the party continued, starting with soft whispers and soon escalating into wild laugher. Often the girls would sneak out of bed and get massive amounts of food without the parents having a clue, only to find the remains under the bed the following day.
1- Examine and trouble shoot
What might be interfering with your intended goal?
When we examined the situation we saw that the girls had gotten into a routine that needed to be stopped in order to see change. The 6 year old, deprived of night time rest had resorted to taking a long nap in school which made it harder for her to be tired at night.
The plan was to create a new look to bedtime. The family, including the girls and I made a checklist of what needed to happen once the bedtime routine was initiated. Since one of the problems was the amount of time mom spent talking and reading at
bedtime we added a time component so cuddling and reading was reasonable. Talking would be saved for daylight hours, at least till things fell into place. I took pictures of them to further invite buy in. The necessity of sleep was explained. They knew that most fluids would stop after 6:00 and the restroom would be used prior to bedtime so access to the bathroom was not going to be granted. They also knew.…..and this was super important.…. that mom would be stationed outside their door and would know if there was any talking or whispering.
2- Create a checklist
Think about what it would look like if your child independently, or close toindependently, began and finished the chosen routine. A bare bones approach is just as effective as one that gives lots of details. Logic and the particulars of your situation will guide your decision. Do a run through to check for accuracy.
• Let your child be involved with the sequence of events
• Have your child pose for pictures for each step. iPads make inserting pics very easy
example for Going to Sleep:
Take a bath
Put on pajamas
Eat a snack
Brush teeth
Toilet
Story time:15 minutes
Cuddle time: 5 minutes each
Close eyes, think of something nice
The reward for a successful bedtime would be a prize that they had already picked out.
After many successful bedtimes the girls transitioned to a star chart which translated to a fun family trip. Now it’s just their routine and rewards are unnecessary.
The day we began mom and dad made sure the girls were tired out from the day. The checklist was brought out and the girls cooperated. Once they got into bed the lights in the entire apartment were turned off.
Mom made sure they could clearly see her right outside the room taking away the temptation of getting out of bed to assess the situation.
They had relatively few issues and after only one bathroom request (which was ignored) and one warning about whispers, they fell asleep
within minutes.
Mom kept expecting the drama to start but the evening was uneventful, with the exception of the silent cheers of a relieved mom and dad.
As the days unfolded the girls continued to embrace this new routine. The effect of harmony at bedtime had positive effects in the daytime as well including increased respect and cooperation. School time naps ended and the teacher reported better focus.
I know that this situation was extreme but the same process can be applied to more mild issues.
A mom I worked with was miserable about how her child kept coming home from school and dropped their stuff all over.
She said she’s tried everything but a conversation with her child which resulted in a jointly created checklist solved the problem.
# Plan For Misbehaving
Children will misbehave and parents will give warning, after warning , after warning. If this sounds too familiar, you are not alone. Parenting is nonstop but if we put effort into preparing for misbehavior and implement the plan consistently we can relax more and put aside the headache remedy.
When we make a plan that includes our children’s input we gain way more than peace of mind. Our children become more responsible, respectful and resilient.
The following gem from the beloved Jane Nelson illustrates how making a plan can gain the cooperation of our children.
1. make a plan
2. implement the plan consistently
3. ignore attention getting behavior that deviates from the plan
The Jones family is very excited. They have just finished planning a day at the beach. Seven-year-old Jason and five-year-old Jenny have promised that they won’t fight. Mr. Jones, has warned, “If you do, we’ll turn around and come back.” “We won’t, we won’t,” promise Jason and Jenny again.
The Jones family haven’t gone two miles when a loud wail is heard from the back seat, “Jason hit me.”
Mrs. Jones says, “What did we tell you kids about fighting?”
Jason defends himself, “Well, she touched me.”
Mr. Jones threatens, “You two had better cut it out, or we are going home.”
The children cry out it unison, “Nooooooo! We’ll be good.”
And they are — for about ten minutes. Then, another wail is heard, “He took my red crayon.”
Jason replies, “Well she was hogging it. It’s my turn.”
Mr. Jones says, “Do you want me to turn around and go home?”
Nooooooo. We’ll be good.”
All this misbehaving should be no surprise. It happens routinely but always met with an “I don’t know what to do with them” response.
And so the story goes. Throughout the day Jason and Jenny fight, and Mr. and Mrs. Jones make threats. At the end of the day, Mr. and Mrs. Jones are angry and threaten never to take the kids anywhere again. Jason and Jenny feel bad that they have made their parents so miserable. They are beginning to believe they really are bad kids—and they keep living up to their reputation.
Now let’s visit the Smith family. They have just planned their trip to the zoo during their weekly family meeting. Part of the planning included a discussion about limits and solutions. Mr. and Mrs. Smith have told Susan and Sam how miserable they feel when they fight. The kids promise they won’t. Mr. Smith said, “I appreciate that, and I think we should come up with a plan for what will happen if you forget.” The kids keep insisting they won’t fight. Mr. and Mrs. Smith know their children have good intentions, and they are also very familiar with the pattern of good intentions gone awry. So, they have decided what they will do and they will follow through.
Mrs. Smith says, “Well then, is it okay with you if we stop the car if you do forget? We don’t think it is safe to drive when you are fighting, so we’ll just pull over to the side of the road and wait for you to stop. You can let us know when you are ready for us to drive again. How do you feel about that solution?” Both kids agree with innocent enthusiasm.
Typically, it doesn’t take them long to forget their promise, and a fight begins. Mrs. Smith quickly and quietly pulls off to the side of the road. She and Mr. Smith take out magazines and start reading. Each child starts blaming the other while protesting his or her own innocence. Mr. and Mrs. Jones ignore them and just keep reading. It doesn’t take long for Susan to catch on that Mom and Dad must mean what they said. Susan says, “Okay, we are ready to keep driving.” Mr. Smith says, “We’ll wait until we hear it from both of you.” Sam says, “But, she hit me.”
Mom and Dad just keep reading. Susan hits Sam, “Tell them you are ready.” Sam cries, “She hit me again.” Mom and Dad just keep reading. Susan realizes that hitting Sam won’t help, so she tries to reason with him. “We’ll have to sit here forever if you don’t say you are ready.” Susan follows her parent’s lead and starts to color. Sam holds out for about three more minutes before saying, “I’m ready for you to start driving.” Mom says, “Thank you very much. I appreciate your cooperation.”
About 30 minutes later another fight starts. Mom starts to pull over to the side of the road. Both kids cry out in unison, “We’ll stop. We’re ready to keep driving.” There was no more fighting for the rest of the day, and the Smiths enjoyed a wonderful day at the zoo.
What is the difference between the Jones family and the Smith family? Are Jason and Jenny really “bad” kids?” No, the difference is that the Smith family is helping their children learn cooperation and problem solving skills while the Jones family is helping their children learn manipulation skills. Mr. and Mrs. Smith demonstrate that they say what they mean and mean what they say by using kind and follow through. Mr. and Mrs. Jones don’t. They used angry threats. This had a temporary effect, but the kids would soon be fighting again.
Mr. and Mrs. Smith stopped using words and instead followed through with kind and firm action. It took a little longer for the kids to catch on, but once they did it had a longer lasting effect. Because they are kids, they just had to test the waters one more time. When their parents started to follow through again the kids knew they meant what they said. They were left with the feeling, not that they were bad kids, but that they were clever enough to figure out a solution to the problem and that cooperation was the most effective alternative.
Misbehaving will happen. Plan for it. | 9,326 | 32,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-05 | longest | en | 0.92403 |
https://djangocentral.com/calculate-power-of-a-number/ | 1,716,772,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00649.warc.gz | 184,972,355 | 7,380 | # Support Our Site
To ensure we can continue delivering content and maintaining a free platform for all users, we kindly request that you disable your adblocker. Your contribution greatly supports our site's growth and development.
# Python Program to Calculate Power of a Number
## Problem Definition
Create a Python program to take two numbers from the user one being the base number another the exponent then calculate the power.
### Program
``````import math
base_number = float(input("Enter the base number"))
exponent = float(input("Enter the exponent"))
power = math.pow(base_number,exponent)
print("Power is =",power)``````
### Output
``````Enter the base number 2
Enter the exponent 4
Power is = 16.0``````
The built-in math module provides a number of functions for mathematical operations. The `pow()` method takes a base number and exponent as parameters and returns the power.
Since in Python, there is always more than one way of achieving things calculating power with the exponentiation operator is also possible. The exponentiation operator `x**y`evaluates to power.
### Program
``````base_number = int(input("Enter the base number"))
exponent = int(input("Enter the exponent"))
power = base_number ** exponentprint("Result is =",power)``````
### Output
``````Enter the base number 2
Enter the exponent 5
Result is = 32``````
PROGRAMS
# Latest Articles
Latest from djangocentral
## How to Use Subquery() in Django With Practical Examples
In the realm of web development, Django stands as a powerful and versatile framework for building robust applications. One of the key aspects of developing efficient and optimized web applications is handling database queries effectively. In this article…
## DRF Serializer: Handling OrderedDict and Converting It to a Dictionary or JSON
In Django Rest Framework (DRF) tests, when you access serializer.data, you might encounter an OrderedDict instead of a regular dictionary. This behavior is intentional and reflects the design of DRF's serialization process.Understanding the Problem The u…
## Django Rest Framework CheetSheet: Mastering API Development
Django Rest Framework (DRF) is a powerful toolkit that makes building robust and scalable web APIs with Django a breeze. Whether you're a seasoned Django developer or a newcomer, having a comprehensive cheat sheet at your disposal can be a game-changer. … | 506 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.774481 |
https://r4r.in/mcqs/learn.php?test_id=3444&subid=653&subcat=ICSE%20XII&test=MCQ%20Questions%20for%20Class%2012%20Accountancy%20set-9 | 1,679,638,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00657.warc.gz | 541,736,819 | 16,047 | # ICSE XII/MCQ Questions for Class 12 Accountancy set-9 Sample Test,Sample questions
## Question: ```A company’s net sales are ₹ 15,00,000; cost of sales is ₹ 10,00,000 and indirect expenses are ₹ 3,00,000, the amount gross profit will be: ```
1.₹ 13,00,000
2. ₹ 5,00,000
3.₹ 2,00,000
4. ₹ 12,00,000
Posted Date:-2022-10-26 16:39:45
## Question: ```A company’s shareholders fund was 7 8,00,000 in the year 2015. It because 7 12,00,000 in the year 2016. What is percentage of change ? ```
1.100%
2.25%
3. 50%
4.33.3%
Posted Date:-2022-10-26 16:39:16
## Question: ```Amount set aside to meet losses due to bad debts is called: ```
1. Reserve
2.Provision
3.Liability
4. None of these
Posted Date:-2022-10-26 16:08:05
## Question: ```An annual report is issued by company to its : ```
1. Directors
2. Auditors
3. Shareholders
4.Management
Posted Date:-2022-10-26 16:54:50
## Question: ```Analysis of financial statements involve : ```
2. Profit & Loss statement
3.Balance Sheet
4. All the above
Posted Date:-2022-10-26 16:44:37
## Question: ```Balance Sheet provides information about financial position of the enterprise : ```
1. At a Point of Time
2.Over a Period of Time
3.For a Period of Time
4. None of the above
Posted Date:-2022-10-26 16:55:21
## Question: ```Break-even Analysis shows: ```
1.Relationship between cost and sales
2. Relationship between production and purchases
3.Relationship between cost and revenue
4.None of these
Posted Date:-2022-10-26 16:53:04
## Question: ```Break-even point refers to that point where : ```
1.Total Costs are more than Total Sales
2. Total Costs are less than Total Sales
3.Total Costs are half of the Total Sales
4.Total Cost are equal to total sales
Posted Date:-2022-10-26 16:42:36
## Question: ```Common-size financial statements are mostly prepared: ```
1. In proportion
2. In percentage
3.(a) and (b) both
4.None of these
Posted Date:-2022-10-26 16:38:14
## Question: ```Common-size Statement are also known as: ```
1. Dynamic Analysis
2. Horizontal Analysis
3.Vertical Analysis
4.External Analysis
Posted Date:-2022-10-26 16:19:57
## Question: ```Comparative financial analysis process shows the comparison between the items of which statement: ```
1.Balance Sheet
2. Profit & Loss Statement
3.(a) and (b) both
4. None of these
Posted Date:-2022-10-26 16:37:04
## Question: ```Comparative Financial Statements show: ```
1.Financial position of a concern
2.Earning capacity of a concern
3.Both of them
4. None of these
Posted Date:-2022-10-26 16:36:30
## Question: ```Comparative Statements are also known as : ```
1.Dynamic Analysis
2. Horizontal Analysis
3. Vertical Analysis
4. External Analysis
Posted Date:-2022-10-26 16:19:05
## Question: ```Debentures are shown in the Balance Sheet under the head of: ```
1. Short-term Loan
2. Secured Loan
3.Current Liability
4. Share Capital
Posted Date:-2022-10-26 16:09:08
## Question: ```Divident is usually paid : ```
1.On Authorised Capital
2.On Ussued Capital
3.On Paid-up Capital
4. On Called-up Capital
Posted Date:-2022-10-26 16:08:37
## Question: ```Financial analysis is significant because it: ```
1. Ignores qualitative aspect
2. Judges operational efficiency
3.Suffers from the limitations of financial statements
4.It is affected by personal ability and bias of the analysis
Posted Date:-2022-10-26 16:45:27
## Question: ```Financial analysis is useful: ```
1. For Investors
2. For Shareholders
3. For Debenture holders
4. All the above
Posted Date:-2022-10-26 16:43:58
## Question: ```For calculating trend percentages any year is selected as: ```
1.Current year
2. Previous year
3.Base year
4.None of these
Posted Date:-2022-10-26 16:23:09
## Question: ```Goodwill of a company is shown on the assets side of the Balance Sheet under the head. ```
1.Current Assets
2.Non-current Assets
3. Miscellaneous Expenditure
4. None of these
Posted Date:-2022-10-26 16:11:57
## Question: ```Horizontal Analysis is also known as : ```
1. Dynamic Analysis
2. Structural Analysis
3.Static Analysis
4.None of these
Posted Date:-2022-10-26 16:18:06
## Question: ```If total assets of a firm are 7 10,00,000 and its non-current assets are 7 6,00,000, what will be the percentage of current assets on total assets ? ```
1. 60%
2.50%
3.40%
4.30%
Posted Date:-2022-10-26 16:41:22
## Question: ```If total assets of a firm are 7 12,00,000 and its non of non-current assets to total assets ? ```
1. 50%
2.75%
3. 25%
4. 80%
Posted Date:-2022-10-26 16:40:47
## Question: ```In a common-size Balance Sheet, total equity and liabilities are assumed to be equal to : ```
1. 1,000
2.100
3.10
4.1
Posted Date:-2022-10-26 16:41:51
## Question: ```In which meeting of company directors report is presented ? ```
1. Directors Meeting
2.Annual General Meeting
3.Manager’s Meeting
4. All of the above
Posted Date:-2022-10-26 16:57:44
## Question: ```Interpretation of Financial Statements includes: ```
1.Criticisms and Analysis
2. Comparison and Trend Study
3.Drawing Conclusion
4. All the above
Posted Date:-2022-10-26 16:13:38
## Question: ```On the basis of process, which of the following is the type of financial analysis ? ```
1. Horizontal Analysis
2. Vertical Analysis
3.Ratio Analysis
4. (a) and (b) both
Posted Date:-2022-10-26 16:58:27
## Question: ```Out of the following which parties are interested in financial statements ? ```
1.Managers
2.Financial Institutions
3. Creditors
4. All the these
Posted Date:-2022-10-26 16:51:57
## Question: ```Payment of Income Tax is considered as : ```
1.Direct Expenses
2.Indirect Expenses
3. Operating Expenses
4. None of these
Posted Date:-2022-10-26 16:43:06
## Question: ```Profit & Loss Account is also called : ```
1.Balance Sheet
2. Income Statements
3. Operating Profit
4. Investment
Posted Date:-2022-10-26 16:55:51
## Question: ```Sales less Cost of goods sold is called : ```
1. Operating Profit
2. Gross Profit
3.Net Profit
4. Total Profit
Posted Date:-2022-10-26 16:40:10
## Question: ```Securities Premium Account is shown on the liabilities side in the Balance Sheet Under heading ```
1.Reserves and Surplus
2.Current Liabilities and Provisions
3. Share Capital
4.Contingent Liabilities
Posted Date:-2022-10-26 16:09:46
## Question: ```Share capital is shown in Balance Sheet under. the head ? ```
1.Authorised Capital
2. Issued Capital
3.Paid-up Capital
4. Shareholders’ Funds
Posted Date:-2022-10-26 16:06:23
## Question: ```Tangible assets of company increased from T 4,00,000 to T 5,00,000. What is the percentage of change ? ```
1. 20%
2.25%
3. 33%
4.50%
Posted Date:-2022-10-26 16:38:46
## Question: ```The analysis of financial statement by a shareholder is an example of: ```
1. External Analysis
2.Internal Analysis
3.Vertical Analysis
4. Horizontal Analysis
Posted Date:-2022-10-26 16:21:28
## Question: ```The financial statements of a business enterprise include: ```
1.Balance Sheet
2.Profit & Loss Account
3.Cash Flow Statement
4.All the above
Posted Date:-2022-10-26 16:54:18
## Question: ```The form of Balance Sheet as per Companies Act, 2013 is: ```
1. Horizontal
2. Horizontal or Vertical
3.Vertical
4.None of these
Posted Date:-2022-10-26 16:11:22
## Question: ```The most commonly used tools for financial analysis are: ```
1.Comparative Statements
2.Common-size Statement
3.Accounting Ratios
4.All the above
Posted Date:-2022-10-26 16:20:57
## Question: ```The prescribe from the Balance Sheet has given in the Schedule: ```
1. VI Part I
2.VI Part II
3.III Part I
4.VII Part IV
Posted Date:-2022-10-26 16:06:55
## Question: ```Tools for comparison of financial statements are : ```
1.Comparative Balance Sheet
2.Comparative Income Statement
3.Common-size Statement
4.All the above
Posted Date:-2022-10-26 16:23:44
## Question: ```Trend ratios and trend percentage are used in : ```
1. Dynamic analysis
2. Static analysis
3.Horizontal analysis
4.Vertical Analysis
Posted Date:-2022-10-26 16:33:48
## Question: ```Vertical Analysis is also known as : ```
1. Static Analysis
2. Dynamic Analysis
3.Structural Analysis
4. None of these
Posted Date:-2022-10-26 16:18:34
## Question: ```What is shown by Balance Sheet ? ```
1. Accuracy of books of accounts
2.Profit or loss of a specific period
3. Financial position on a specific date
4.None of the above
Posted Date:-2022-10-26 16:50:46
## Question: ```What is shown by the Income Statement ? ```
1.Accuracy of books of accounts
2.Profit or loss of a certain period
3.Balance of Cash Book
4.None of these
Posted Date:-2022-10-26 16:50:16
## Question: ```Which of the following assets is not shown undeer the head ‘Fixed Asset’ in the Balance Sheet ? ```
1. Goodwill
2. Bills Receivable
3. Buildings
4. Vehicle
Posted Date:-2022-10-26 16:10:43
## Question: ```Which of the following is not a limitations of financial statement analysis ? ```
1. To measure the financial strength
2. Affected by window-dressing
3.Do not reflect changes in price level
4.Lack of Qualitative Analysis
Posted Date:-2022-10-26 16:52:34
## Question: ```Which of the following is the purpose or objective of financial analysis ? ```
1. To assess the current profitability of the firm
2.To measure the solvency of the firm
3. To assess the short-term and long-term liquidity position of the firm
4.All the above
Posted Date:-2022-10-26 16:51:23
## Question: ```Which of the following shows the actual financial position of n enterprise ? ```
1.Fund Flow
2. Balance Sheet
3.P & L A/c
4. Ratio Analysis
Posted Date:-2022-10-26 16:53:35
## Question: ```Which of the following statement is correct ? ```
1. Assets = Liabilities + Shareholders funds
2. Assets = Total funds
3.Assets = Funds of outsiders .
4. None of the above
Posted Date:-2022-10-26 16:57:09
## Question: ```Which of these are not the method of financial statement analysis ? ```
1.Ratio Analysis
2. Comparative Analysis
3.Trend Analysis
4. Capitalisation Method
Posted Date:-2022-10-26 16:37:38
## Question: ```Which Section of the Companies Act, 2013 requires that the Balance Sheet to be prepared in prescribed form ? ```
1.Section 128
2.Section 130
3. Section 129
4.Section 212
Posted Date:-2022-10-26 16:07:26
## More MCQS
##### R4R Team
R4Rin Top Tutorials are Core Java,Hibernate ,Spring,Sturts.The content on R4R.in website is done by expert team not only with the help of books but along with the strong professional knowledge in all context like coding,designing, marketing,etc! | 3,090 | 10,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.893682 |
https://www.transtutors.com/questions/here-is-information-related-to-schellhamer-company-for-2010-total-credit-sales-1-532-1131406.htm | 1,556,108,104,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578641278.80/warc/CC-MAIN-20190424114453-20190424140453-00520.warc.gz | 848,593,046 | 14,923 | # Here is information related to Schellhamer Company for 2010. Total credit sales $1,532,850 Accounts. 1 answer below » Here is information related to Schellhamer Company for 2010. Total credit sales$1,532,850 Accounts receivable at December 31 864,300 Bad debts written off 54,310 (a) What amount of bad debts expense will Schellhamer Company report if it uses the direct write-off method of accounting for bad debts? (b) Assume that Schellhamer Company decides to estimate its bad debts expense based on 3% of accounts receivable. What amount of bad debts expense will the company record if Allowance for Doubtful Accounts has a credit balance of $3,678? (c) Assume the same facts as in the question above, except that there is a$1,152 debit balance in Allowance for Doubtful Accounts. What amount of bad debts expense will Schellhamer record?
James C
Solution:- a) Computation of Bad Debt expense by the direct write-off method:- Bad Debt Expense = Bad Debt written = \$54,310 b) Computation of Bad Debt Expnese:- Closing... | 252 | 1,028 | {"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} | 2.9375 | 3 | CC-MAIN-2019-18 | latest | en | 0.913556 |
https://walkccc.me/LeetCode/problems/2415/ | 1,685,608,342,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00149.warc.gz | 680,220,434 | 65,087 | # 2415. Reverse Odd Levels of Binary Tree
• Time: $O(n)$
• Space: $O(h)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 class Solution { public: TreeNode* reverseOddLevels(TreeNode* root) { dfs(root->left, root->right, true); return root; } private: void dfs(TreeNode* left, TreeNode* right, bool isOddLevel) { if (left == nullptr) return; if (isOddLevel) swap(left->val, right->val); dfs(left->left, right->right, !isOddLevel); dfs(left->right, right->left, !isOddLevel); } };
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 class Solution { public TreeNode reverseOddLevels(TreeNode root) { dfs(root.left, root.right, true); return root; } private void dfs(TreeNode left, TreeNode right, boolean isOddLevel) { if (left == null) return; if (isOddLevel) { final int val = left.val; left.val = right.val; right.val = val; } dfs(left.left, right.right, !isOddLevel); dfs(left.right, right.left, !isOddLevel); } }
1 2 3 4 5 6 7 8 9 10 11 12 class Solution: def reverseOddLevels(self, root: Optional[TreeNode]) -> Optional[TreeNode]: def dfs(left: Optional[TreeNode], right: Optional[TreeNode], isOddLevel: bool) -> None: if not left: return if isOddLevel: left.val, right.val = right.val, left.val dfs(left.left, right.right, not isOddLevel) dfs(left.right, right.left, not isOddLevel) dfs(root.left, root.right, True) return root | 423 | 1,323 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-23 | latest | en | 0.374624 |
https://www.physicsforums.com/threads/velocity-of-efflux.868030/ | 1,618,189,811,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00366.warc.gz | 1,037,320,170 | 26,155 | # Velocity of efflux
## Homework Statement
A container of height H has a orifice at a distance h from the top, the ratio of the cross section area of container and orifice is. 1, find velocity of efflux
## Homework Equations
Using Bernoullis equation we get v=(2gh)^.5
Equation of continuity A1V1=A2V2
## The Attempt at a Solution
Used Bernoullis equation to get v=(2gh)^.5
I cant think of anything else, thanks in advance :) [/B]
haruspex
Homework Helper
Gold Member
## Homework Statement
A container of height H has a orifice at a distance h from the top, the ratio of the cross section area of container and orifice is. 1, find velocity of efflux
## Homework Equations
Using Bernoullis equation we get v=(2gh)^.5
Equation of continuity A1V1=A2V2
## The Attempt at a Solution
Used Bernoullis equation to get v=(2gh)^.5
I cant think of anything else, thanks in advance :) [/B]
Do you have some reason to doubt that answer?
That's not the answer, the answer has ratio of cross sections in it. It comes out to be $$\sqrt{2gh} / (1-(a/A)^2)$$ where a and A are cross section of orifice and Vessel respectively
Last edited:
haruspex
Homework Helper
Gold Member
That's not the answer, the answer has ratio of cross sections in it. It comes out to be $$\sqrt{2gh} / (1-(a/A)^2)$$ where a and A are cross section of orifice and Vessel respectively
That clearly is not right in general. As a tends to A, the answer must tend to √(2gh) - that's just free fall - and certainly not tend to infinity.
The question is not entirely clear about state. If this is shortly after the hole was made in the full tank, the velocity would be less than √(2gh) as the flow is still becoming established; if later, the water height is no longer h, so again it would be less than √(2gh). For it even to reach √(2gh), there would need to be a constant head arrangement, with water flowing in to maintain the h height.
AbhinavJ
That clearly is not right in general. As a tends to A, the answer must tend to √(2gh) - that's just free fall - and certainly not tend to infinity.
The question is not entirely clear about state. If this is shortly after the hole was made in the full tank, the velocity would be less than √(2gh) as the flow is still
becoming established; if later, the water height is no longer h, so again it would be less than √(2gh). For it even to reach √(2gh), there would need to be a constant head arrangement, with water flowing in to
maintain the h height.
Let me phrase the question directly,
Water is filled in a container upto 3 m. A small hole of Area a is punched im the wall of a container at a height 52.5 cm from bottom. The cross sectional area of container is A. If a/A=0.1 find velocity of water coming out of hole.
haruspex
Homework Helper
Gold Member
As I wrote, the velocity will not be constant. It will rise to a peak, then decline as the water level drops.
It can never exceed, or even quite reach, √(2gh) as that would violate energy conservation (which is the basis of Bernoulli).
Here's an analysis of what happens over time:
A tank of horizontal area A contains liquid density ##\rho## to an initial depth y0.
At one side near the base, there is a small hole area a.
Rate of conversion from GPE to KE:
##\rho Ay\dot y\ddot y-\frac 12{\dot y}^3\rho A \left(\frac{A^2}{a^2}-1\right)=-\rho A y g \dot y##
Writing ##k=(\frac Aa)^2-1##:
##y\ddot y-\frac 12k{\dot y}^2=-gy##
Writing ##v=\dot y## and ##v'=\frac{dv}{dy}##, we have ##\ddot y = \frac{d}{dt}v =\frac{dy}{dt}\frac{d}{dy}v= vv'##.
##yvv'-\frac 12kv^2=-gy##
Writing ##w=v^2##
##yw'=kw-2gy##
General solution: ##w=By^{k}##.
Particular Integral: ##w=cy##
##cy=kcy-2gy##
##c=\frac{2g}{k-1}##
Using initial condition:
##0=By_0^{k}+y_0\frac{2g}{k-1}##
##B=-\frac{2g}{(k-1)y_0^{k-1}}##
##w=\frac{2gy}{k-1}(1-\left(\frac{y}{y_0}\right)^{k-1})##
Peak velocity is at w'=0:
##w=\frac{2gy}k##
##\frac{k-1}{k}=(1-\left(\frac{y}{y_0}\right)^{k-1})##
##ky^{k-1}=y_0^{k-1}##
##y=y_0(k)^{-\frac 1{k-1}}##
##w_{max}=y_0(k)^{-\frac 1{k-1}}\frac {2g}k##
##v_{max}=\sqrt{y_0(k)^{-\frac 1{k-1}}\frac {2g}k}##
Edit: in the original version of the above I left out a factor of 1/2 in the first equation. This led to 2k appearing subsequently everywhere that should have been just k.
Last edited:
AbhinavJ
Chestermiller
Mentor
If I apply the Bernoulli equation (assuming that, because A>>a, the flow is in quasi-steady-state), I get:
$$gh+\frac{1}{2}v_1^2=\frac{1}{2}v_2^2$$where ##v_1## is the velocity at the upper fluid surface and ##v_2## is the velocity out the hole. From continuity, ##v_1A=v_2a##. So,
$$2gh=v_2^2\left(1-\left(\frac{a}{A}\right)^2\right)$$
This works well for A>>a.
haruspex
Homework Helper
Gold Member
If I apply the Bernoulli equation (assuming that, because A>>a, the flow is in quasi-steady-state), I get:
$$gh+\frac{1}{2}v_1^2=\frac{1}{2}v_2^2$$where ##v_1## is the velocity at the upper fluid surface and ##v_2## is the velocity out the hole. From continuity, ##v_1A=v_2a##. So,
$$2gh=v_2^2\left(1-\left(\frac{a}{A}\right)^2\right)$$
This works well for A>>a.
But, sadly, it is not valid. It illustrates the pitfalls of applying Bernoulli out of scope.
Bernoulli's equation assumes steady state. The problem statement asks for the flow when the tank is still full, so this is not steady state unless there is a source of water maintaining the level and, moreover, that water must enter with a downward velocity of v1. It is that velocity which provides the extra energy to exceed the √(2gh) limit at the bottom.
For a complete-ish analysis of how the flow varies over time as the tank empties (but still ignoring losses) please see my post #6.
If I apply the Bernoulli equation (assuming that, because A>>a, the flow is in quasi-steady-state), I get:
$$gh+\frac{1}{2}v_1^2=\frac{1}{2}v_2^2$$where ##v_1## is the velocity at the upper fluid surface and ##v_2## is the velocity out the hole. From continuity, ##v_1A=v_2a##. So,
$$2gh=v_2^2\left(1-\left(\frac{a}{A}\right)^2\right)$$
This works well for A>>a.
Thanks, this was what i needed :)
But, sadly, it is not valid. It illustrates the pitfalls of applying Bernoulli out of scope.
Bernoulli's equation assumes steady state. The problem statement asks for the flow
when the tank is still full, so this is not
But, sadly, it is not valid. It
illustrates the pitfalls of applying Bernoulli out of scope.
Bernoulli's equation assumes steady state. The problem statement asks for the flow
when the tank is still full, so this is not
steady state unless there is a source of water maintaining the level and, moreover, that water must enter with a downward velocity of v1. It is that velocity
which provides the extra energy to exceed
the √(2gh) limit at the bottom.
For a complete-ish analysis of how the flow varies over time as the tank empties (but still ignoring losses) please see my post #6.
steady state unless there is a source of water maintaining the level and, moreover,
that water must enter with a downward
velocity
of v1. It is that velocity which
provides the extra energy to exceed the √(2gh) limit at the bottom.
For a complete-ish analysis of how the flow varies over time as the tank empties (but s
t
ill ignoring losses) please see my post #6.
Agreed, but what high school physics demanded was what Chestermiller added. Highly appreciate your effort, I learned something new :)
Chestermiller
Mentor
But, sadly, it is not valid. It illustrates the pitfalls of applying Bernoulli out of scope.
Bernoulli's equation assumes steady state. The problem statement asks for the flow when the tank is still full, so this is not steady state unless there is a source of water maintaining the level and, moreover, that water must enter with a downward velocity of v1. It is that velocity which provides the extra energy to exceed the √(2gh) limit at the bottom.
For a complete-ish analysis of how the flow varies over time as the tank empties (but still ignoring losses) please see my post #6.
Thanks. I'll have to look over what you did more carefully. Certainly, this is an unsteady state situation, and a large fraction of the fluid in the tank is decelerating as time progresses, so the steady state version of Bernoulli is not strictly kosher to use. I had just thought that this approximation would be adequate, but maybe not. I also want to check out the unsteady state version of Bernoulli.
Chet
haruspex
Homework Helper
Gold Member
Agreed, but what high school physics demanded was what Chestermiller added. Highly appreciate your effort, I learned something new :)
Perhaps you miss my point. Even in high school physics one is supposed to know when equations apply. The answer given would have been a reasonable approximation if instead it had said that the water had been running out of the hole for some time and asked what the egress velocity is when the head height is h. But in the set-up given, your original answer is more accurate than the 'book' answer.
Perhaps you miss my point. Even in high school physics one is supposed to know when equations apply. The answer given would have been a reasonable approximation if instead it had said that the water had been running out of the hole for
some time and asked what the egress velocity is when the head height is h. But in the set-up given, your original answer is more
accurate than the 'book' answer.
So no matter how large the hole is, given the water level is same somehow the velocity will be (2gh)^.5 always?
Chestermiller
Mentor
Hi @haruspex
I have a bit of a problem with the analysis in post #6. For a continuum like this, in which the velocity at each location is changing with time, writing $$\frac{dv}{dt}=v\frac{\partial v}{\partial y}$$is not valid. The equation should read:
$$\frac{dv}{dt}=\left(\frac{\partial v}{\partial t}\right)_y+v\left(\frac{\partial v}{\partial y}\right)_t$$
I see what you are trying to do here, and it's admirable, but the partial time derivative with respect to t is critical. In addition, the term on the left side of the equation should be an integration over the volume of the tank, since the local kinetic energy per unit mass is varying with position (including horizontal position if the exit hole is vertical) and time.
I have some ideas on how to modify what you have done, but it's a work in progress.
Chet
haruspex
Homework Helper
Gold Member
So no matter how large the hole is, given the water level is same somehow the velocity will be (2gh)^.5 always?
No, that's not what I wrote.
Distinguish carefully these four descriptions:
1. A hole is made at distance h below the water level in the side of a tank.
In this arrangement, the water is initially static. As the water flows out of the hole, the water in the tank accelerates downwards. It takes a little time for the flow to become approximately steady, by which time the height of water above the hole will no longer be h. So the speed of the water jet out of the hole was less than √(2gh) while the flow was becoming established, and it will be less than √(2gh) later because the height has fallen.
Note that √(2gh/(1-(a/A)2)) > √(2gh), so is even less accurate than √(2gh).
2. At some point after the flow has become reasonably steady, we note that the height of the water level above the hole is x. We can now use Bernoulli's equation to find the speed of the jet, and that gives √(2gx/(1-(a/A)2)). But note x, not h.
3. As in (1), but a constant head device keeps the tank full. E.g. water flows in from a pipe at the top of the tank, and overfows the tank if it comes in too fast. A while after the hole is made, the flow becomes established, so we can apply Bernoulli. But the water entering the tank at the top has no velocity along the flow line, so does not contribute to the energy. Bernoulli's equation now gives √(2gh).
4. As in (3), but the water enters the tank from an adjacent reservoir with a level just above the top of the tank. Note that this will not keep the tank completely full. Since the water flows into the tank with a purely horizontal component, it has to fall a short distance to reach the descending surface in the tank. Using h as the distance from hole to top of tank, Bernoulli gives √(2gh). If instead we use x, the distance of the water level in the tank above the hole, Bernoulli gives √(2gx/(1-(a/A)2)).
Now, I see Chet has some reservations about my analysis in post #6, and he may well be right, but that is largely independent from my much simpler analysis in this post.
Chestermiller
Mentor
We're back!!! Haruspex and I have been collaborating offline during the past few days to develop a better quantitative understanding of the transient flow behavior of this system. We have developed a time-dependent model of the flow, and have performed calculations using this model. We are here to report on the details of the flow analysis, the results of our calculations, and our conclusions.
INTRODUCTION
The usual steady state Bernoulli equation does not correctly describe the effect of the area ratio a/A (where a is the hole area and A is the tank cross sectional area) on the effluent velocity. This is because the Bernoulli equation applies only to steady state flow, and the flow in this system is transient. Because the level of fluid is changing, the fluid velocity at any constant elevation within the tank is varying with time. But the usual Bernoulli equation does not take this part of the fluid acceleration into account. It includes only the advective part of the acceleration. As the ratio of the areas a/A gets higher, the error in the prediction from the usual Bernoulli equation prediction gets worse. For the case where a/A = 1, the Bernoulli equation totally fails to predict the required free fall. To determine the solution to this problem correctly, a time dependent modification to the Bernoulli equation must be used, which properly includes the missing part of the acceleration.
MODEL DESCRIPTION
In the present development, we assume that the exit hole is situated in the bottom of the tank. This allows us to assume that the flow velocity in the tank is essentially vertical and 1D, rather than having to contend with a complicated 2D fluid flow approaching the exit hole. This substantially simplifies the determination of the kinetic energy of the fluid in the tank, as well as the rate at which gravitational work is being done on the fluid in the tank.
Let
##v_x(t)## = downward vertical exit velocity from tank
##v(t)## = downward vertical velocity of fluid in tank
##a##=cross sectional area of exit hole
##A##=cross sectional area of tank
##h(t)##=location of upper water surface at time t
From the continuity equation: $$v(t)=\frac{av_x(t)}{A(z)}\tag{1}$$
From the kinematics, $$\frac{dh}{dt}=-v(t)=-\frac{av_x(t)}{A}\tag{2}$$
We shall next perform a mechanical energy balance on the system, setting the rate at which gravitational work is being done on the tank contents equal to the rate of change of kinetic energy of the fluid in the tank plus the rate at which kinetic energy is leaving the tank in the exit stream.
The rate at which gravitational work is being done on the contents of the tank at time t is obtained by multiplying the rate of doing gravitational work per unit volume by the volume of the tank:
$$\rho g v(t) Ah(t)=\rho g v_x(t)ah(t)\tag{3}$$
The total kinetic energy of the fluid in the tank at time t is obtained by multiplying kinetic energy per unit volume by the volume of the tank:
$$\rho \frac{v^2(t)}{2}Ah=\rho \frac{v_x^2(t)}{2}\frac{a^2}{A}h(t)\tag{4}$$
The rate of change of fluid kinetic energy within the tank is obtained by evaluating the time derivative of the expression in Eqn. 4:
$$\rho \frac{v_x^2(t)}{2}\frac{a^2}{A}\frac{dh}{dt}+\rho v_x(t)\frac{a^2}{A}h(t)\frac{dv_x(t)}{dt}\tag{5}$$
Substituting Eqn. 2 into this expression yields:
$$-\rho \frac{v_x^3(t)}{2}\frac{a^3}{A^2}+\rho v_x(t)\left(\frac{a^2}{A}\right)h(t)\frac{dv_x(t)}{dt}\tag{6}$$
The rate at which kinetic energy is exiting the tank at time t is given by:
$$\rho v_x(t)a\frac{v_x^2(t)}{2}\tag{7}$$
If we now perform a transient mechanical energy balance on the system by setting the rate of doing gravitational work on the fluid in the tank equal to the rate of kenetic energy generation within the tank plus the rate of kinetic energy leaving in the exit stream, we obtain:
$$\rho g v_x(t)ah(t)=-\rho \frac{v_x^3(t)}{2}\frac{a^3}{A^2}+\rho v_x(t)\left(\frac{a^2}{A}\right)h(t)\frac{dv_x(t)}{dt}+\rho v_x(t)a\frac{v_x^2(t)}{2}\tag{8}$$
If we divide Eqn. 8 by ##\rho v_x(t)a##, we obtain:
$$g h(t)=\frac{v_x^2(t)}{2}\left[1-\left(\frac{a}{A}\right)^2\right]+h(t)\frac{a}{A}\frac{dv_x(t)}{dt}\tag{9}$$
Eqn. 9 is fully consistent with the standard form of the transient Bernoulli equation presented in the literature.
The time t can be replaced as the independent variable in this equation by the depth h, by combining Eqn. 9 with Eqn. 2 to yield:
$$2g h=v_x^2\left[1-\left(\frac{a}{A}\right)^2\right]-h\left(\frac{a}{A}\right)^2\frac{dv_x^2}{dh}\tag{10}$$
Note the comparison between Eqn. 10 and the form of the Bernoulli equation written in previous posts of this thread. The key difference is the second term on the right hand side, involving the derivative of ##v_x^2## with respect to h. This term captures the effects of the portion of the fluid acceleration that is omitted from the steady flow version of the Bernoulli equation.
I think I'll stop here for now. I'll be back tomorrow with the analytic solution to Eqn. 10, together with some graphs of computational results obtained using the analytic solution, and a discussion of these results.
Chet
Last edited:
cheapstrike
Chestermiller
Mentor
MODEL SOLUTION
Equation Eqn. 10 of the previous post tells us that the velocity of the effluent stream from the tank vx will be a function of the area ratio a/A, the initial depth of the (inviscid) fluid in the tank h0, and the fluid depth at any arbitrary time t, h(t). The general analytic solution to this equation, subject to the initial condition ##v_x=0## at h=h0 is given by:$$v_x=\sqrt{2gh\frac{\left[1-(h/h_0)^{\frac{1-2r}{r}}\right]}{1-2r}}\tag{11}$$where ##r=(a/A)^2##. For the special limiting cases in which ##r=1/\sqrt{2}## and ##r=1##, this solution reduces to:
$$v_x=\sqrt{-4gh\ln(h/h_0)}\tag{for r=1/√2}$$
$$v_x=\sqrt{2g(h_0-h)}\tag{for r = 1}$$
For the case of r =1 (i.e., the case in which the exit hole area is equal to the tank area), the above equation for the efflux velocity vx is, as expected, just that predicted for free fall.
MODEL RESULTS
The results calculated from Eqn. 11 for the efflux v_x (normalized by ##\sqrt{2gh_0}## as a function of the (dimensionless) fluid depth ratio h/h0 and the (dimensionless) area ratio a/A are shown in the figure below.
In all cases, the efflux velocity is equal to zero initially (i.e., when h = h0), and then rises rapidly as the fluid, both inside the tank and in the efflux, accelerates. However, as the depth of fluid in the tank continues to decrease, the efflux velocity passes through a maximum and then begins to decrease. Eventually, as the fluid depth approaches zero, the efflux velocity, of course, also drops to zero. (In the case of a/A = 1, the maximum velocity is attained just as the tank reaches empty.)
The results in the above figure are fully consistent with what Haruspex has been saying all along, to wit, the efflux velocity vxis always less than the Torricelli velocity based on the initial depth of fluid in the tank ##\sqrt{2gh_0}##. However, it is also of interest to compare vx with the Torricelli velocity calculated on the basis of the instantaneous depth of fluid in the tank h(t), rather than on the depth at time zero. The figure below shows the efflux velocity ##v_x## normalized in terms of the Torricelli velocity calculated on the basis of the instantaneous depth h(t) plotted as a function of dimensionless depth ##h/h_0## at a selection of values of the area ratio a/A.
According to the results in the figure, for (small) values of the area ratio a/A less than 0.5, the dimensionless efflux velocity ##v_x/\sqrt{2gh(t)}## levels off to a constant value as the depth of fluid in the tank decreases. The smaller the value of a/A, the more rapidly the dimensionless velocity levels off. From our analytic solution (Eqn. 11), the value to which the dimensionless velocity levels off is given by:
$$(v_x)_{level}=\sqrt{\frac{2gh(t)}{[1-2(a/A)^2]}}\tag{12}$$
For the case of a/A = 0.1 in the OP of this thread, we see from the figure that, once the depth h(t) has decreased to about 90% of the initial depth h0, the velocity has already leveled off.
For small values of a/A, Eqn. 12 for ##(v_x)_{level}## can be expressed, to linear terms in ##(a/A)^2## by:
$$(v_x)_{level}~\approxeq \frac{\sqrt{2gh}}{[1-(a/A)^2]}$$
By an strange coincidence, this is the same relationship for the efflux velocity that our high schooler OP had written in his second post of this thread. He was able to obtain this result by the fortuitous cancellation of two errors: 1. incorrectly assuming that the steady state Bernoulli equation could be applied this transient problem and 2. failing to take the square root of the denominator in obtaining ##v_x## from his equation for ##v_x^2##.
This pretty much completes what haruspex and I wanted to cover.
Chet
cheapstrike, AbhinavJ and haruspex
By an strange coincidence, this is the same relationship for the efflux velocity that our high schooler OP had written in his second post of this thread. He was able to obtain this result by the fortuitous cancellation of two errors: 1.
incorrectly assuming that the steady state Bernoulli equation could be applied this transient problem and 2. failing to take the square root of the denominator in obtaining
vxvxv_x from his equation for v2xvx2v_x^2
.
Wow, I am a lucky guy :p | 5,924 | 21,807 | {"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} | 3.84375 | 4 | CC-MAIN-2021-17 | longest | en | 0.933731 |
http://tech.ctkschool.org/archive/class-of-2016/geometrypuzzles | 1,701,564,632,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100476.94/warc/CC-MAIN-20231202235258-20231203025258-00481.warc.gz | 44,277,775 | 8,691 | Archive > Class of 2016 >
### Geometry Puzzles!
posted Feb 10, 2015, 9:33 AM by mdreyfus@ctkschool.org [ updated Sep 11, 2016, 3:03 PM by kmcmillan@ctkschool.org ]
Here are some geometry activities you can try out next:
At KhanAcademy.org you will find some recommended skills to practice.
### Proofs
Proofs are like puzzles. To prove something in geometry, you can only use other geometry facts you already know. You can't use measuring in a proof!
1) Angles in a trapezoid
• Draw a pair of parallel lines
• Draw TWO transversals. Do not connect them.
• Pairs of supplementary angles
• What's the sum of the interior angles? How can you know this WITHOUT MEASURING THEM?
2) Angles in a triangle
• Using lines (not line segments), draw a triangle
• Select a vertex and the line opposite that vertex. Construct a parallel line.
• Using what you know about parallel lines cut by a transversal, find the two ? angles.
• What is the sum of the interior angles of the triangle? How can you know this without measuring them? | 273 | 1,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-50 | latest | en | 0.89558 |
https://www.thestudentroom.co.uk/showthread.php?t=1277560 | 1,547,957,550,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583690495.59/warc/CC-MAIN-20190120021730-20190120043730-00144.warc.gz | 961,755,134 | 40,493 | You are Here: Home
# Chemistry understanding help. watch
Announcements
1. I was doing some exercises, i have a question here.
If the answer for a certain ppm question is, say, 3000. But if i were to write if at 0.003 ppm, is it acceptable?
Does the term ppm indicates 10^6 or it's only a unit?
I'm confused now Because some practices that i did, when the answer is 2.16 X 10^6, they just write it as 2.16 ppm.
2. ppm means parts per million. 2 ppm means '2 particles in every million'.
If you get an answer of 10^6 you can write it using SI prefixes- 10^6 is mega or M. So 2x10^6g is 2Mg. http://en.wikipedia.org/wiki/SI_prefix
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: May 21, 2010
Today on TSR
### Four simple steps to get an A*
Here's all you have to do
### University open days
• University of East Anglia
All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
Wed, 30 Jan '19
• Aston University
Wed, 30 Jan '19
• Solent University | 347 | 1,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.909799 |
https://people.math.harvard.edu/~knill/pedagogy/algebra/index.html | 1,716,857,565,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00298.warc.gz | 380,075,368 | 3,739 | This one-day workshop was organized by the Texas Valley Communities Foundation and the Texas Graduate Center, which is affiliated with the Math for teaching degree program at the Harvard extension school. Thanks to Mary Alice Reyes and Adriana Lopez from the Texas Graduate Center for arranging and organizing that (and dinner). It was an inspiring workshop with amazing contributions from the class which I still have to digest. There were almost 30 teachers present. (Photos by the center: pic1, pic2, pic3.) Some handouts are to the right. In the wake of the preparations, I also mixed in a bit of algebra in my current passion for geometry on graphs. Feb 3, 2017: Hidden figures shows the importance of algebra skills: More about the math. See also 10-2=20 [Jun 2, 2017] and Percentages [Jun 13, 2017]. Leonard Euler who lived from 1707 to 1783 is the grand master of Pedagogy in the realm of algebra. Euler also invented graph theory (Koenigsberg Bridge Problem), seeds of topology (Euler characteristic etc) and so many other things. He is probably the most inspiring mathematician ever, not only because of his theorems and formulas v-e+f=2,exp(i π)+1=0,1+1/4+1/9+1/16+...=π2/6 etc), but also because of his outreach and his passion for making it accessible. Euler walked the talk, like many of the teachers who throw their energy into the noble cause of teaching. Euler's contributions to algebra pedagogy was not only in writing his textbook in Algebra but producing a gold standard in clarity which is hard to surpass. It is one of the most successful textbooks of all times. Update 9/24/2022:
I have here PDF files of Euler's Algebra textbook:
English translation [PDF] (with notes of Bernoulli and Lagrange) German Part I [PDF], German Part II [PDF]
The textbook is quite small as can be seen on this page from 2006. Some blog notes about addition and multiplication, on networks. This topic got mentioned at the end of this workshop illustrating that basic discoveries in arithmetic are not over (we have barely scratched the surface!)
The picture to the right was taken in my Office. It is part of this panorama. To the left (click for more photos), we see the "Fraction lab".
Some photos from the workshop. I hope to link later to more, done by the graduate center.
There is also great food in the Texas valley. Here is a photo from a restaurant (thanks to Claudia from the Texas Valley Communities Foundation) for that lunch:
The Renaissance hotel
Some pictures from an early morning run:
And from the McAllen airport featuring a Brother Wright type flyer (a replica close to Flyer III from 1905 or Model AB from 1910)
And while flying back, the magical Chicago O'Hare Airport tunnel connecting B and C. Oliver Knill, January, 20 2017, | 632 | 2,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.964788 |
http://www.thephysicsforum.com/homework-help/5681-need-help-circuit-diagram.html | 1,558,492,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256724.28/warc/CC-MAIN-20190522022933-20190522044933-00504.warc.gz | 347,664,949 | 10,074 | # Thread: Need help with this circuit diagram.
1. Hello everyone, I'm new to this site. I'm studying Bachelors of Computer Science in Pakistan. I'm not really good at physics. I have got a compulsory subject to study in this semester, "Basic Electronics". We are following 4-5 books for this course.
I need help with this question. Can anyone explain this to me?
What is the voltage across R1 and R2 when the switch is in position 1
I know that it's very easy but I never took interest in physics in high school nor I got any good teacher which can help me make my concepts stronger.
MODERATOR NOTE: Moved to "Homework Help".
2. Kirchhoff's circuit laws
3. Originally Posted by aimi95
Hello everyone, I'm new to this site. I'm studying Bachelors of Computer Science in Pakistan. I'm not really good at physics. I have got a compulsory subject to study in this semester, "Basic Electronics". We are following 4-5 books for this course.
I need help with this question. Can anyone explain this to me?
What is the voltage across R1 and R2 when the switch is in position 1
I know that it's very easy but I never took interest in physics in high school nor I got any good teacher which can help me make my concepts stronger.
MODERATOR NOTE: Moved to "Homework Help".
The voltage across R1 is zero as in switch position one. There is no applied potential (voltage) from the battery. The voltage across R2 is dependent on the battery voltage and the resistance value of the resistor. (You have shown the schematic in position two, not one)
Ohms law is you divide the battery voltage by the resistor value. This will give the current flow value in amperage. (I =V/R)
Then step two is to use the equation V = IxR - This means to know the voltage drop across the resistor (in the case of this circuit it will be a negative voltage drop) you multiply the current in amps times the resistance in ohms.
I have used V to indicate voltage - your school may use E but same thing. The voltage is in volts. I means current and as measured in Amps. R meant resistance and is measured in ohms.
4. Originally Posted by aimi95
Hello everyone, I'm new to this site. I'm studying Bachelors of Computer Science in Pakistan. I'm not really good at physics. I have got a compulsory subject to study in this semester, "Basic Electronics". We are following 4-5 books for this course.
I need help with this question. Can anyone explain this to me?
What is the voltage across R1 and R2 when the switch is in position 1
I know that it's very easy but I never took interest in physics in high school nor I got any good teacher which can help me make my concepts stronger.
MODERATOR NOTE: Moved to "Homework Help".
The voltage across R2 is zero as in switch position shown there is no applied potential (voltage) from the battery. The voltage across R1 is dependent on the battery voltage and the resistance value of the resistor.
Ohms law is you divide the battery voltage by the resistor value. This will give the current flow value in amperage. (I =V/R)
Then step two is to use the equation V = IxR - This means to know the voltage drop across the resistor you multiply the current in amps times the resistance in ohms.
I have used V to indicate voltage - your school may use E but same thing. The voltage is in volts. I means current and as measured in Amps. R meant resistance and is measured in ohms.
Posting Permissions
You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Forum Rules | 850 | 3,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-22 | latest | en | 0.963288 |
http://chronux.org/chronuxFiles/Documentation/chronux/spectral_analysis/continuous/mtspecgramtrigc.html | 1,695,451,873,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00870.warc.gz | 7,521,694 | 3,758 | Home > chronux > spectral_analysis > continuous > mtspecgramtrigc.m
# mtspecgramtrigc
## PURPOSE
Multi-taper event triggered time-frequency spectrum - continuous process
## SYNOPSIS
function [S,t,f,Serr]=mtspecgramtrigc(data,E,win,movingwin,params)
## DESCRIPTION
``` Multi-taper event triggered time-frequency spectrum - continuous process
Usage:
[S,t,f,Serr]=mtspecgramtrigc(data,E,win,movingwin,params)
Input:
Note units have to be consistent. Thus, if movingwin is in seconds, Fs
data (single channel data) -- required
E (event times) -- required
win (in the form [winl winr] i.e window around each event)
required
movingwin (in the form [window winstep] i.e length of moving
window and step size) -
required
Note that units for the windows have
to be consistent with
units of Fs
params: structure with fields tapers, pad, Fs, fpass, err, trialave
- optional
tapers : precalculated tapers from dpss or in the one of the following
forms:
(1) A numeric vector [TW K] where TW is the
time-bandwidth product and K is the number of
tapers to be used (less than or equal to
2TW-1).
(2) A numeric vector [W T p] where W is the
bandwidth, T is the duration of the data and p
is an integer such that 2TW-p tapers are used. In
this form there is no default i.e. to specify
the bandwidth, you have to specify T and p as
well. Note that the units of W and T have to be
consistent: if W is in Hz, T must be in seconds
and vice versa. Note that these units must also
be consistent with the units of params.Fs: W can
be in Hz if and only if params.Fs is in Hz.
The default is to use form 1 with TW=3 and K=5
Note that T has to be equal to movingwin(1).
pad (padding factor for the FFT) - optional (can take values -1,0,1,2...).
to the next highest power of 2 etc.
Defaults to 0.
Fs (sampling frequency) - optional. Default 1.
fpass (frequency band to be used in the calculation in the form
[fmin fmax])- optional.
Default all frequencies between 0 and Fs/2
err (error calculation [1 p] - Theoretical error bars; [2 p] - Jackknife error bars
[0 p] or 0 - no error bars) - optional. Default 0.
trialave (average over events when 1, don't average when 0) - optional. Default 0
Output:
S (triggered spectrum in form time x frequency x events for trialave=0;
or in the form time x frequency trialave=1)
t (times)
f (frequencies)
Serr (error bars) only for err(1)>=1```
## CROSS-REFERENCE INFORMATION
This function calls:
• createdatamatc Helper function to create an event triggered matrix from univariate
• mtspecgramc Multi-taper time-frequency spectrum - continuous process
This function is called by:
## SOURCE CODE
```0001 function [S,t,f,Serr]=mtspecgramtrigc(data,E,win,movingwin,params)
0002 % Multi-taper event triggered time-frequency spectrum - continuous process
0003 %
0004 % Usage:
0005 %
0006 % [S,t,f,Serr]=mtspecgramtrigc(data,E,win,movingwin,params)
0007 % Input:
0008 % Note units have to be consistent. Thus, if movingwin is in seconds, Fs
0009 % has to be in Hz. see chronux.m for more information.
0010 % data (single channel data) -- required
0011 % E (event times) -- required
0012 % win (in the form [winl winr] i.e window around each event)
0013 % required
0014 % movingwin (in the form [window winstep] i.e length of moving
0015 % window and step size) -
0016 % required
0017 % Note that units for the windows have
0018 % to be consistent with
0019 % units of Fs
0020 % params: structure with fields tapers, pad, Fs, fpass, err, trialave
0021 % - optional
0022 % tapers : precalculated tapers from dpss or in the one of the following
0023 % forms:
0024 % (1) A numeric vector [TW K] where TW is the
0025 % time-bandwidth product and K is the number of
0026 % tapers to be used (less than or equal to
0027 % 2TW-1).
0028 % (2) A numeric vector [W T p] where W is the
0029 % bandwidth, T is the duration of the data and p
0030 % is an integer such that 2TW-p tapers are used. In
0031 % this form there is no default i.e. to specify
0032 % the bandwidth, you have to specify T and p as
0033 % well. Note that the units of W and T have to be
0034 % consistent: if W is in Hz, T must be in seconds
0035 % and vice versa. Note that these units must also
0036 % be consistent with the units of params.Fs: W can
0037 % be in Hz if and only if params.Fs is in Hz.
0038 % The default is to use form 1 with TW=3 and K=5
0039 % Note that T has to be equal to movingwin(1).
0040 %
0041 % pad (padding factor for the FFT) - optional (can take values -1,0,1,2...).
0043 % to the next highest power of 2 etc.
0044 % e.g. For N = 500, if PAD = -1, we do not pad; if PAD = 0, we pad the FFT
0045 % to 512 points, if pad=1, we pad to 1024 points etc.
0046 % Defaults to 0.
0047 % Fs (sampling frequency) - optional. Default 1.
0048 % fpass (frequency band to be used in the calculation in the form
0049 % [fmin fmax])- optional.
0050 % Default all frequencies between 0 and Fs/2
0051 % err (error calculation [1 p] - Theoretical error bars; [2 p] - Jackknife error bars
0052 % [0 p] or 0 - no error bars) - optional. Default 0.
0053 % trialave (average over events when 1, don't average when 0) - optional. Default 0
0054 % Output:
0055 % S (triggered spectrum in form time x frequency x events for trialave=0;
0056 % or in the form time x frequency trialave=1)
0057 % t (times)
0058 % f (frequencies)
0059 % Serr (error bars) only for err(1)>=1
0060
0061 if nargin < 4; error('Need data, events and parameters for the windows'); end;
0062 if nargin < 5; params=[]; end;
0063
0064 if length(params.tapers)==3 & movingwin(1)~=params.tapers(2);
0065 error('Duration of data in params.tapers is inconsistent with movingwin(1), modify params.tapers(2) to proceed')
0066 end
0067
0069 clear tapers pad fpass trialave
0070 if nargout > 3 && err(1)==0;
0071 % Cannot compute error bars with err(1)=0. change params and run again.
0072 error('When Serr is desired, err(1) has to be non-zero.');
0073 end;
0074 data=change_row_to_column(data);
0075 data=createdatamatc(data,E,Fs,win);
0076 if nargout==4;
0077 [S,t,f,Serr]=mtspecgramc(data,movingwin,params);
0078 else
0079 [S,t,f]=mtspecgramc(data,movingwin,params);
0080 end;```
Generated on Fri 28-Sep-2012 12:34:30 by m2html © 2005 | 1,923 | 7,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.827235 |
http://mathhelpforum.com/algebra/10258-find-k-print.html | 1,505,938,726,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687447.54/warc/CC-MAIN-20170920194628-20170920214628-00207.warc.gz | 210,410,422 | 3,330 | # Find k
• Jan 18th 2007, 06:33 PM
symmetry
Find k
An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?
(1) 128 (3) 8
(2) 64 (4) 4
• Jan 18th 2007, 06:37 PM
topsquark
Quote:
Originally Posted by symmetry
An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?
(1) 128 (3) 8
(2) 64 (4) 4
Max height is when the derivative of the height function goes to 0. So:
$h'(t) = -32t + k = 0$
Thus $k = 32t = 32 \cdot 4 = 128$
-Dan
• Jan 18th 2007, 06:41 PM
symmetry
ok
Derivative is a calculus word.
I am not asking calculus questions at this point in the study book.
Thanks anyway!
• Jan 18th 2007, 06:54 PM
topsquark
All right. In that case we need to look at the height function a bit more carefully. It is an inverted parabola. The maximum height will be at the vertex of the parabola, which is on the axis of symmetry.
Given a parabola $y = ax^2 + bx + c$ the axis of symmetry will be the line $x = -\frac{b}{2a}$.
We have the parabola $h = -16t^2 + kt + 3$. We know that the location of the max height is the line t = 4, which is our axis of symmetry. So:
$t = - \frac{k}{2 \cdot -16} = 4$
So
$k = 128$.
-Dan
• Jan 20th 2007, 04:56 AM
symmetry
ok
Dan,
Thanks for breaking the question some other way.
I am not ready for calculus questions at this point in the course of my review for the June state test.
Thanks!
• Jan 20th 2007, 09:32 AM
CaptainBlack
Quote:
Originally Posted by symmetry
Dan,
Thanks for breaking the question some other way.
I am not ready for calculus questions at this point in the course of my review for the June state test.
Thanks!
I think we would all be interested to know what test you are preparing for.
Its always nice to have some context when subjected to an avalache of
questions:)
RonL | 644 | 2,038 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-39 | longest | en | 0.905959 |
https://metanumbers.com/58658 | 1,621,119,620,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00239.warc.gz | 410,069,102 | 7,483 | ## 58658
58,658 (fifty-eight thousand six hundred fifty-eight) is an even five-digits composite number following 58657 and preceding 58659. In scientific notation, it is written as 5.8658 × 104. The sum of its digits is 32. It has a total of 3 prime factors and 8 positive divisors. There are 28,980 positive integers (up to 58658) that are relatively prime to 58658.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 32
• Digital Root 5
## Name
Short name 58 thousand 658 fifty-eight thousand six hundred fifty-eight
## Notation
Scientific notation 5.8658 × 104 58.658 × 103
## Prime Factorization of 58658
Prime Factorization 2 × 139 × 211
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 58658 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 58,658 is 2 × 139 × 211. Since it has a total of 3 prime factors, 58,658 is a composite number.
## Divisors of 58658
1, 2, 139, 211, 278, 422, 29329, 58658
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 89040 Sum of all the positive divisors of n s(n) 30382 Sum of the proper positive divisors of n A(n) 11130 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 242.194 Returns the nth root of the product of n divisors H(n) 5.27026 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 58,658 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 58,658) is 89,040, the average is 11,130.
## Other Arithmetic Functions (n = 58658)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 28980 Total number of positive integers not greater than n that are coprime to n λ(n) 4830 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5922 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 28,980 positive integers (less than 58,658) that are coprime with 58,658. And there are approximately 5,922 prime numbers less than or equal to 58,658.
## Divisibility of 58658
m n mod m 2 3 4 5 6 7 8 9 0 2 2 3 2 5 2 5
The number 58,658 is divisible by 2.
## Classification of 58658
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
• Sphenic
## Base conversion (58658)
Base System Value
2 Binary 1110010100100010
3 Ternary 2222110112
4 Quaternary 32110202
5 Quinary 3334113
6 Senary 1131322
8 Octal 162442
10 Decimal 58658
12 Duodecimal 29b42
20 Vigesimal 76ci
36 Base36 199e
## Basic calculations (n = 58658)
### Multiplication
n×i
n×2 117316 175974 234632 293290
### Division
ni
n⁄2 29329 19552.7 14664.5 11731.6
### Exponentiation
ni
n2 3440760964 201828156626312 11838836011386209296 694442442755892264884768
### Nth Root
i√n
2√n 242.194 38.8546 15.5626 8.98805
## 58658 as geometric shapes
### Circle
Diameter 117316 368559 1.08095e+10
### Sphere
Volume 8.45416e+14 4.32379e+10 368559
### Square
Length = n
Perimeter 234632 3.44076e+09 82954.9
### Cube
Length = n
Surface area 2.06446e+10 2.01828e+14 101599
### Equilateral Triangle
Length = n
Perimeter 175974 1.48989e+09 50799.3
### Triangular Pyramid
Length = n
Surface area 5.95957e+09 2.37857e+13 47894.1
## Cryptographic Hash Functions
md5 be9b6cec65c863e1b55538d2c7fcc9b0 57c1542690146b8c3f2d6e65469cf3dbd283e21b e75ed3aaf1c826d246f9af3d11c8d2b3fbc5aae34050880ea1f55f02e6fb2827 5fe15860977e5a42bf1f7cdd226258b13bf42ed0ef91060caa3d2670acec00ed5a47eea158f1858f0690cb9ac7dd0dfb3c659a2821fea04a4b37fb3916ae4986 a2760e20e6a6f38378663f3cf7274687cf56c863 | 1,477 | 4,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.814819 |
https://www.slideshare.net/RaynaCody/parallelogram-story-12121614 | 1,498,161,440,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00348.warc.gz | 925,322,576 | 33,765 | Upcoming SlideShare
×
# Parallelogram story
1,489 views
Published on
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total views
1,489
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
9
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Parallelogram story
1. 1. Parallelogram Story By: Rayna Cody
2. 2. IntroductionHello! my name is Rayna. I am sixteen yearsold and I go to New Heights Academy CharterSchool. I am currently taking a geometry classwhich is taught by my teacher Ms.Bush. Andthe goal of this project is to describe andexplain the characteristics of a rectangle.
3. 3. Basic PropertiesA rectangle has four sides and two pairs havethe same length. And, in order to find thediagonals of a rectangle, one must use thePythagorean theorem.
4. 4. Specific PropertiesA rectangle is a plane figure with four rightangles and congruent diagonals. Also, itsopposite sides are parallel and congruent.The consecutive angles are supplementary.The length of opposite sides are equal. Andthe diagonals bisect each other.
5. 5. ComparisonAll of the parallelograms are similar to oneanother because, each of their angles mustadd up to a total of 360°. Therefore eachangle equals 90°. Rectangles opposite sidesare all parallel. Plus, they each have foursides. In addition to the fact that, they eachare quadrilaterals.
6. 6. ContrastThis shape is unlike other parallelogramsbecause this one has adjacent angles that areequal. A rhombus is a quadrilateral with fourequal sides and a rectangle is a quadrilateralwith four right angles. A rectangle isespecially unlike others because all angles addup to ninety degrees.
7. 7. ProblemFrancis constructed a rectangular path. The distancearound the rectangular path is 42m. The length of thepath is 3m less than twice the width. Find the lengthand width of the path.
8. 8. SolveProperty L=2w-32L+2W=422(2W-3)+2W=424W-6+2W=426W=42+66W=48 =8 rectangular path 6(W=48 6=8)W=8 ÷L=2x8-3L=16-3=13L=13Proof2x13+2x8=4226+16=4242=42
9. 9. ○ ExamplesExamples of this rectangle in everyday life isof the rectangle inreal life are a laptop, a Scott Foresmandictionary, and a McDougal Littell geometrytextbook. These four all have four rightangles, parallel sides, 90° angles. Plus all fourshapes have four sides.
10. 10. Conclusion Overall, this project was a learning experience for me. For example I learnedthat; in a rectangle there at least have to be two congruent sides, and Ilearned that a rectangle is a plane figure, the adjacent angles are equal andall of the angles added up are 360°. The most surprising thing I found outduring this project was how much each of the parallelograms have incommon. I also enjoyed Photo Story a lot because the special effects areamazing plus it had many unique components which allowed me to create aneven more interesting story about my shape, it allows me to animate mywork, insert the pictures I needed when necessary and it makes your workpresentable. | 802 | 3,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-26 | latest | en | 0.847185 |
https://1linux.info/understanding-order-of-operations-in-programming/ | 1,620,317,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00389.warc.gz | 91,084,371 | 30,923 | # Understanding Order of Operations in Programming
### Introduction
As a coder, you’re probably used to telling computers what to do. Type up some code, run it, and the computer gets to work executing whatever command you gave it.
Even though we have this powerful reign over computers, there’s still a lot of magic constantly occurring in our code that we tend to overlook. This is especially true if you’re working with high-level languages with pre-built functions, as most of us are. And of course while there’s no real reason to reinvent the wheel or try to implement these helpful functions on your own, it’s still fun to take a peek under the hood and see what’s going on!
Today we’re going to take a closer look at one of these seemingly obvious concepts that we’ve probably all used at one point or another: the order of operations.
Say you want to evaluate this expression: `5 + 10 * 3`. According to the mathematical order of operations, you should multiply `10 by 3` first and then add `5` to the product of that, but how exactly would you tell a computer to do this?
There are a few different ways we can parse this equation, but some require a little more background than others.
In this tutorial, you’ll see how to go about solving this.
This method requires that we first convert our equation into the correct format. Once it’s in a more machine readable form, then we can go ahead and feed it through our parsing algorithm which will actually calculate it.
First I’ll show you how to get the correct format so we can see what the end result should look like, then I’ll walk you through the actual algorithm used to evaluate the expression. Just to keep things simple, we’ll only be dealing with four operators in this example: addition, subtraction, multiplication, and division.
## Infix to Postfix
Even though you may not realize it yet, most of you are probably already familiar with infix notation. The above expression, `5 + 10 * 3`, is written in infix notation. It just means the operators fall in between the operands that they’re acting upon.
### What is Postfix Notation?
As mentioned earlier, we need to convert our equation into a format that the computer can easily understand. This format is called postfix notation.
Expressions written in postfix notation will have all operators following their operands.
This is important because when the machine is reading expressions in this format, it will never encounter an operator before the operands it’s acting on, which means it won’t have to go back and forth.
So in the previous example, `5 + 10 * 3` becomes `5 10 3 * +`.
This may look unusual, but there’s a methodology to arrive at this.
### Human-Friendly Way to Convert from Infix to Postfix
1. Add in parentheses in order of precedence
`` (5 + (10 * 3)) ``
1. Move every operator to the right, directly before its closing parenthesis
`` (5 ( 10 3 *)+) ``
1. Now just drop the parentheses altogether, which leaves us with our expression in postfix notation
`` 5 10 3 * + ``
Another example, just to show that the operators won’t necessarily always be at the very end:
`` 8 * 4 + 2 ((8 * 4) + 2) ((8 4 *) 2 +) 8 4 * 2 + ``
Again, this isn’t really ideal for the computer to do. It still wouldn’t know where to put the parentheses. Luckily for us, we have a an algorithm to produce the same results.
## Shunting Yard Algorithm
The Shunting Yard Algorithm was developed by Dijkstra as a means to convert infix notation to postfix notation.
Before we go any further, let’s just quickly review the two data structures we’re going to be using here: a stack and a queue. We can use an array to hold both of these sets of data. The main difference comes from the order we’re adding and removing the data.
Queue — When we add data to a queue, we’re pushing it onto the back. Just imagine you’re getting in line for an event and every person in line is an element in the queue. When you walk up to the line, you’re automatically inserted into the back of the line. As the event starts letting people in (removing elements from the queue), they pull from the front of the line since those people have been there longer. You can remember this with the acronym FIFO: first in, first out.
Stack — Every time we add a new element to the stack, it will be put on top (or at the front) instead of in the back. When we want to remove an item from the stack, we’ll pop off the top item. Because new elements always go on top, those new ones will always be popped off first when we need to remove something. This can be remembered with the acronym LIFO: last in, first out.
For this algorithm, assume we have one temporary stack to hold the operators (we’ll call it “operator stack”) and one queue that will hold the final result.
### How It Works
The Shunting Yard Algorithm follows four basic steps:
1. Parse the expression from left to right.
2. If we see an operand, output it to the results queue immediately.
3. If we see an operator:
a. If the operator stack is empty, push the incoming operator onto the operator stack.
b. If the incoming operator has higher precedence than what’s currently at top of the operator stack, push the incoming operator onto the top of the stack.
c. If the incoming operator has equal precendence, pop the top operator from the stack, output it to the queue, and push the incoming operator onto the stack.
d. If the incoming operator has lower precedence pop the top operator from the stack, output it to the queue, and test the incoming operator with the new top of the stack.
4. Once we have parsed the whole expression, pop all remaining tokens from the operator stack.
### Evaluating Our Expression with the Shunting Yard Algorithm
It’s hard to make sense of those steps without seeing it in action, so let’s walk through our previous example and try to format it with the algorithm!
Convert this equation from infix notation to postfix notation: `5 + 10 * 3`
Let’s set up our two arrays: one for the results output and one for the temporary operator stack.
`` expression = 5 + 10 * 3 output = [] operator stack = [] ``
First we start reading our expression from left to right. So first up we have `5`. Since this is an operand, we can output it immediately.
`` expression = + 10 * 3 output = [5] operator stack = [] ``
Next we see the `+`. The operator stack is empty, so we can just push it there
`` expression = 10 * 3 output = [5] operator stack = [+] ``
Next up is `10`, so we’ll output immediately.
`` expression = * 3 output = [5, 10] operator stack = [+] ``
Now we hit another operator, `*`. Since the operator stack isn’t empty, we have to compare it to the current top of the operator stack to see which has higher precedence.
If we look above, we see the current top of the stack is `+`. So comparing the two, we know multiplication has higher precedence than addition.
This means we can just push it onto the top of the stack, which gives us:
`` expression = 3 output = [5, 10] operator stack = [*, +] ``
Now we hit our final value, `3`. Since this isn’t an operator, we can just output it immediately.
`` expression is now empty output = [5, 10, 3] operator stack = [*, +] ``
Since the expression is now empty, all we need to do is pop all tokens from the operator stack and output them immediately. When we pop from the stack, we’re grabbing from the top, so first we’ll take the `*` to push to the end of the queue and then we’ll take the `+`.
`` output = [5, 10, 3, *, +] ``
And that’s it! As you can see, it matches the above method where we just add parentheses, but this way is much easier for a computer to do.
### Precedence Rules
You may have noticed there was one point where instead of using the algorithm to decide, we relied on our own knowledge to make a choice between what to do next: determining which operator had higher precedence.
It’s not important right now while you understand the concepts behind the algorithm, but when we’re writing the actual code to solve this, we’re going to have to build in some precedence rules.
All we have to do is create an object that will essentially rank each operator. We’ll give the multiplication and division operators a rank of 2 and the addition and subtraction operators a rank of 1.
When we code it up, we’ll just compare two operators by comparing their numerical rank. The actual numbers 1 and 2 here are arbitrary, so don’t get too caught up in that. Just know that multiplication ranks higher than addition, so it has a higher number.
`` const precedence = { "*":2, "/":2, "+":1, "-":1 }; ``
## Evaluating our Postfix Expression
We finally have our expression in Postfix Notation. Now we can use this format to evaluate it.
### Algorithm to Evaluate Our Expression
Here’s how we’ll do it:
2. Parse the first token in the expression.
3. If it’s an operand, push it onto the stack.
4. If it’s an operator, pop off the appropriate number of operands from the stack into temporary variables. (For example, multiplication is a binary operator, so if you are parsing and you hit a multiplication operator, then you know to pop off two operands.)
5. Evaluate that expression using the current operator and the two operands that were popped.
6. Push that result to the top of the stack.
7. Repeat 2-7 until the expression is empty.
In our example, we’re only dealing with binary operators, so we can just always pop off two operands when we see an operator. If we wanted to expand our example to handle all operators, we’d have to handle unary operators such as `!`.
### Walking Through the Algorithm
Let’s walk through some pseudo-code where we use the algorithm to evaluate the above postfix notation expression:
`5 10 3 * +`.
First we start by pushing every operand onto the stack until we hit an operator.
`` expression = [5, 10, 3, *, +] - push 5 - push 10 - push 3 stack = [3, 10, 5] ``
So now we get to our first operator, `*`, which means it’s time to start popping. We pop until we have two values.
`` - pop 3 - pop 10 ``
Alright now we have our two operands, `3` and `10`, so we will combine this with our operator, `*`, leaving us with `10 * 3`.
`` expression = [+] stack = [5] tempOperand1 = 3 tempOperand2 = 10 tempOperator = * eval(tempOperand1 + tempOperator + tempOperand2) // 3 * 10 ``
We evaluate that, get `30`, and then push this back onto the stack. We now have the following:
`` expression = [+] stack = [30, 5] ``
So we start parsing the expression again and we immediately hit an operator. Again, we have to pop from the stack until we have two operands.
`` expression = [] stack = [] tempOperand1 = 30 tempOperand2 = 5 tempOperator = + eval(tempOperand1 + tempOperator + tempOperand2) // 30 + 5 ``
We pop `30` and the `5` and we are ready to evaluate again. `5 + 30` gives us `35` and we can now push this back onto the stack.
Going back to our original expression to parse for the next token, we find that it’s empty!
`` expression = [] stack = [35] ``
This either means that we are done or that the original expression was malformed.
Let’s check by looking at our stack. Thankfully it only has one value in it, so this means we are done and `35` is the final output of the original expression, `5 + 10 * 3`.
## Prefix Notation
The algorithm for evaluating an expression in prefix notation is essentially the same as above, except this time you will read from right to left. All we need is a small modification to the code and we can also evaluate for prefix notation.
If we go back to our original method of adding parentheses and moving operators, we can convert to prefix notation in the same way we did postfix. Instead of moving the operators to the end of their operands, we’ll move them to the beginning. Once we’ve done that, we can drop the parentheses altogether and then we have our prefix notation expression!
`` 5 + 10 * 3 (5 + (10 * 3)) (+ 5 (* 10 3)) + 5 * 10 3 ``
If you want to put your knowledge to the test, try to figure out how you’d do this algorithmically with a small modification to the shunting yard algorithm.
## Conclusion
In this tutorial, you’ve created an algorithm for converting expressions to postfix notation and tested it by evaluating an expression. | 2,903 | 12,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-21 | latest | en | 0.937423 |
https://en.wikibooks.org/wiki/Real_analysis/Dedekind's_construction | 1,529,816,847,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00049.warc.gz | 591,768,191 | 15,166 | # Real Analysis/Dedekind's construction
(Redirected from Real analysis/Dedekind's construction)
## Contents
Historically Dedekind gave the first construction of the real numbers. Like Cantor's construction, Dedekind's method constructs the real numbers from the set rational numbers. Dedekind's construction gives a more geometric picture of the real numbers.
The idea of the construction is that every real number ${\displaystyle r}$ should cut the number line into two subsets, the numbers less than ${\displaystyle r}$ and the numbers greater than or equal to ${\displaystyle r}$ . These two sets are different for every real number, and given these sets we could determine ${\displaystyle r}$ . In fact, because each real number divides the number line, it also divides the rational numbers into two. And if we knew these two sets of rational numbers, then we could also determine ${\displaystyle r}$ . Hence when looking for a way to define the real numbers, we could define them as the collection of ways to partition the rational numbers into two sets. Today when discussing Dedekind cuts one usually only keeps track of one of these two sets. Our definition of a cut could informally be thought of as the numbers less than ${\displaystyle r}$ .
## Definition
We say that ${\displaystyle \alpha \subset \mathbb {Q} }$ is a cut if and only if one of these is true:
The Dedekind Cut ${\displaystyle x^{2}}$
1. If ${\displaystyle p\in \alpha }$ and ${\displaystyle q}$ is a rational number so that ${\displaystyle q , then ${\displaystyle q\in \alpha }$ . So for any element of the cut, every rational number smaller than it is also an element of the cut.
2. There exists a rational number ${\displaystyle a}$ such that ${\displaystyle p for all ${\displaystyle p\in \alpha }$ . So there's a rational number that's larger than any element in a particular cut.
3. For every ${\displaystyle p\in \alpha }$ , there exists an ${\displaystyle r\in \alpha }$ such that ${\displaystyle p . So every element of the cut has yet another element larger than it. If that seems impossible, imagine a series of rational numbers that approximate the real number we're trying to define. For example, the cut ${\displaystyle x^{2}<2}$ contains a series of countably infinitely many rational numbers like 1, 1.2, 1.4, 1.41, 1.414, 1.4142, etc. each of which is larger than every previous one and a bit closer to ${\displaystyle {\sqrt {2}}}$ .
## The Construction
We now outline how to make the set of Dedekind cuts form an ordered field that satisfies the the least upper bound axiom.
#### Order
First we discuss how to compare the relative size of two cuts.
Definition Given two cuts ${\displaystyle \alpha ,\beta }$ , we say ${\displaystyle \alpha =\beta }$ if they are the same subset of the rational numbers. That is if ${\displaystyle \alpha \subseteq \beta }$ and ${\displaystyle \beta \subseteq \alpha }$ .
Exercise Show equality defined as in the previous definition is an equivalence relation.
Definition Given two cuts ${\displaystyle \alpha ,\beta }$ , we say ${\displaystyle \alpha <\beta }$ if ${\displaystyle \alpha \subset \beta }$ .
Exercise Show the above order satisfies the order axioms and that ${\displaystyle \mathbb {R} }$ is totally ordered.
We now also assume the usual definitions to ${\displaystyle \geq ,<,>}$ , we leave this to the reader to supply the correct definitions.
Exercise Given two cuts ${\displaystyle \alpha ,\beta }$ , show that if ${\displaystyle \alpha <\beta }$ then there is a rational number ${\displaystyle p\in \beta }$ so that ${\displaystyle p\notin \alpha }$ .
Exercise (Trichotomy) Show that with this order, every cut satisfies exactly one of the following three inequalities: ${\displaystyle \alpha <0,\alpha =0,\alpha >0}$ .
#### Field Axioms
Let ${\displaystyle \alpha ,\beta }$ be two cuts
Define addition as ${\displaystyle \alpha +\beta =\{p+q:p\in \alpha ,q\in \beta \}}$ .
Define negation as ${\displaystyle -\alpha =\{r\in \mathbb {Q} :r<-p{\text{ for some }}p\not \in \alpha \}}$
Let 0 be the cut defined by ${\displaystyle \mathbf {0} =\{r:r<0\}}$ .
Exercise For two cuts α, β verify that α+β, −α, and 0 are cuts. Furthermore show that the the set of cuts forms an abelian group with this definition of addition. What is the identity element?
Defining multiplication requires a bit more care. We define
${\displaystyle \alpha \cdot \beta ={\begin{cases}\{p\cdot q\mid p\in \alpha ,p>0,q\in \beta \}&{\text{if }}\alpha \geq \mathbf {0} ,\beta \geq \mathbf {0} ;\\\{p\cdot q\mid p\notin \alpha ,q\in \beta ,q>0\}&{\text{if }}\alpha \geq \mathbf {0} ,\beta <\mathbf {0} ;\\\{p\cdot q\mid p\in \alpha ,q\notin \beta ,q>0\}&{\text{if }}\alpha <\mathbf {0} ,\beta \geq \mathbf {0} ;\\\{r\mid r
We comment that in the definition of multiplication we needed to be careful to choose elements that had a particular sign. This is possible by the exercises involving the order on Dedekind cuts.
We shall define the cut 1 by 1={r|r<1}.
For a cut α ≠ 0 we define
${\displaystyle \alpha ^{-1}={\begin{cases}\{r\mid r<{\frac {1}{p}},p\not \in \alpha \}&{\text{if }}\alpha >\mathbf {0} ;\\\{r\mid r<{\frac {1}{p}},p\not \in \alpha ,{\text{ and }}p<0\}&{\text{if }}\alpha <\mathbf {0} .\end{cases}}}$
Exercise For two cuts α, β verify that α·β, α-1 (assuming α ≠ 0), and 1 are cuts. Show that with these definitions the set of Dedekind cuts forms an ordered field.
#### Least Upper Bound Property
We will now show the set of Dedekind cuts satisfies the least upper bound axiom.
Let A be a non-empty collection of cuts, and suppose that there is a cut β such that α<β for all α. In other words, A is a non-empty collection of cuts that has an upper bound given by β.
Now we define a subset of the rational numbers, δ, by taking the unions of all of the subsets given by the cuts in A. That is ${\displaystyle \delta =\displaystyle \bigcup _{\alpha \in A}\alpha }$
Now we wish to show that δ is a cut. Suppose p∈δ, if follows that p∈α0 for some α0 in A. Then if q<p then q∈α0 and hence q∈δ. This shows the first property. Also, there is some r∈α0 with r>p, but then r∈δ. Hence the third property holds. To see that the second property holds, notice that δ⊆β because each α⊆β. By the cut properties for β there is a rational number b so that b>p for all p∈β. Therefore, b>p for all p∈δ. Hence δ is a cut.
Now, by definition of the order, δ is an upper bound of A. If η any cut with η<δ, then there is a rational number p∈δ and p∉η. But p∈α0 for some α0 in A. Hence η is not greater then α0, and therefore η is not an upperbound for A. Therefore δ is the least upper bound. This shows that the set of Dedekind cuts satisfies the least upper bound axiom.
## Reference
• W. Rudin, Principles of Mathematical Analysis, McGraw-Hill International | 1,899 | 6,811 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 41, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-26 | latest | en | 0.885329 |
https://www.texasgateway.org/resource-index/?f%5B0%5D=im_field_resource_subject%3A1&f%5B1%5D=im_field_resource_subject%3A2&f%5B2%5D=sm_field_resource_grade_range%3A8&f%5B3%5D=sm_field_resource_grade_range%3A6&f%5B4%5D=sm_field_resource_grade_range%3A10&f%5B5%5D=sm_field_resource_grade_range%3A3&f%5B6%5D=sm_field_resource_grade_range%3A11&f%5B7%5D=sm_field_resource_audience%3Ateacher&%3Bf%5B1%5D=sm_field_resource_grade_range%3A12&%3Bf%5B2%5D=im_field_resource_subject%3A2&%3Bf%5B3%5D=sm_field_resource_grade_range%3A2&%3Bf%5B4%5D=sm_field_resource_type%3Apro_dev | 1,576,253,388,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540564599.32/warc/CC-MAIN-20191213150805-20191213174805-00496.warc.gz | 769,463,186 | 16,505 | • Resource ID: TEA001
• Grade Range: K–8
• Subject: Math
### TXRCFP: Texas Response to Curriculum Focal Points for K-8 Mathematics Revised 2013
The Texas Response to Curriculum Focal Points Revised 2013 was created from the 2012 revision of the TEKS as a guide for implementation of effective mathematics instruction by identifying critical areas of content at each grade level.
• Resource ID: TEKS12_MATH_03_001
• Grade Range: 3
• Subject: Math
### Determining the Perimeter of a Polygon (Series and Activity 1)
This activity provides an opportunity for students to investigate the perimeter of polygons.
• Resource ID: TEKS12_MATH_06_002
• Grade Range: 6
• Subject: Math
### Area of Triangles, Parallelograms, and Trapezoids
These activities provide an opportunity for students to explore the area formulas for triangles, trapezoids, and parallelograms.
• Resource ID: TEKS12_MATH_07_001
• Grade Range: 8
• Subject: Math
### Reflections
These activities provide an opportunity for students to explore reflections both on and off of a coordinate plane.
• Resource ID: TEKS12_MATH_07_002
• Grade Range: 8
• Subject: Math
### Translations
These activities provide an opportunity for students to explore translations both on and off of a coordinate plane.
• Resource ID: TEKS12_MATH_08_001
• Grade Range: 8
• Subject: Math
### Dilations
These activities provide an opportunity for students to explore dilations both on and off of a coordinate plane.
• Resource ID: K2KA103
• Grade Range: 8–10
• Subject: Math
### Kid2Kid: Determining the Meaning of Slope and Intercepts
Kid2Kid videos on determining the meaning of slope and intercepts in English and Spanish
• Resource ID: MATH_IMG_001
• Grade Range: K–8
• Subject: Math
### Interactive Math Glossary
The Interactive Math Glossary is provided by the Texas Education Agency to help teachers explore and understand mathematics vocabulary.
• Resource ID: M8M4L3*
• Grade Range: 8
• Subject: Math
### Determining the Effects of Proportional Change on Area
Given pictorial representations and problem situations involving area, the student will describe the effects on area when dimensions are changed proportionally.
• Resource ID: M8M4L2*
• Grade Range: 8
• Subject: Math
### Determining the Effects of Proportional Change on Perimeter
Given pictorial representations and problem situations involving perimeter, the student will describe the effects on perimeter when dimensions are changed proportionally.
• Resource ID: M8M1L2*
• Grade Range: 8
• Subject: Math
### Approximating the Value of Irrational Numbers
Given problem situations that include pictorial representations of irrational numbers, the student will find the approximate value of the irrational numbers.
• Resource ID: M8M1L3*
• Grade Range: 8
• Subject: Math
### Expressing Numbers in Scientific Notation
Given problem situations, the student will express numbers in scientific notation.
• Resource ID: M8M2L15*
• Grade Range: 8
• Subject: Math
### Determining if a Relationship is a Functional Relationship
The student is expected to gather and record data & use data sets to determine functional relationships between quantities.
• Resource ID: A1M1L3a
• Grade Range: 9–12
• Subject: Math
### Writing Verbal Descriptions of Functional Relationships
Given a problem situation containing a functional relationship, the student will verbally describe the functional relationship that exists.
• Resource ID: A1M1L6
• Grade Range: 9–12
• Subject: Math
### Writing Inequalities to Describe Relationships (Graph → Symbolic)
Given the graph of an inequality, students will write the symbolic representation of the inequality.
• Resource ID: A1M1L7
• Grade Range: 9–12
• Subject: Math
### Writing Inequalities to Describe Relationships (Symbolic → Graph)
Describe functional relationships for given problem situations, and write equations or inequalities to answer questions arising from the situations.
• Resource ID: M8M2L2*
• Grade Range: 8
• Subject: Math
### Graphing Dilations, Reflections, and Translations
Given a coordinate plane, the student will graph dilations, reflections, and translations, and use those graphs to solve problems.
• Resource ID: M8M2L3*
• Grade Range: 8
• Subject: Math
### Graphing and Applying Coordinate Dilations
Given a coordinate plane or coordinate representations of a dilation, the student will graph dilations and use those graphs to solve problems.
• Resource ID: A1M1L8
• Grade Range: 9–11
• Subject: Math
### Connecting Multiple Representations of Functions
The student will consider multiple representations of linear functions, including tables, mapping diagrams, graphs, and verbal descriptions.
• Resource ID: A1M2L2
• Grade Range: 9–12
• Subject: Math
### Determining Reasonable Domains and Ranges (Verbal/Graph)
Given a graph and/or verbal description of a situation (both continuous and discrete), the student will identify mathematical domains and ranges and determine reasonable domain and range values for the given situations. | 1,162 | 5,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.784508 |
https://www.airmilescalculator.com/distance/ybr-to-gst/ | 1,620,364,372,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.96/warc/CC-MAIN-20210507025943-20210507055943-00634.warc.gz | 543,593,562 | 41,354 | # Distance between Brandon (YBR) and Gustavus, AK (GST)
Flight distance from Brandon to Gustavus (Brandon Municipal Airport – Gustavus Airport) is 1543 miles / 2483 kilometers / 1341 nautical miles. Estimated flight time is 3 hours 25 minutes.
Driving distance from Brandon (YBR) to Gustavus (GST) is 2080 miles / 3347 kilometers and travel time by car is about 47 hours 45 minutes.
## Map of flight path and driving directions from Brandon to Gustavus.
Shortest flight path between Brandon Municipal Airport (YBR) and Gustavus Airport (GST).
## How far is Gustavus from Brandon?
There are several ways to calculate distances between Brandon and Gustavus. Here are two common methods:
Vincenty's formula (applied above)
• 1543.157 miles
• 2483.471 kilometers
• 1340.967 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
• 1538.571 miles
• 2476.090 kilometers
• 1336.981 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 Brandon Municipal Airport
City: Brandon
IATA Code: YBR
ICAO Code: CYBR
Coordinates: 49°54′36″N, 99°57′6″W
B Gustavus Airport
City: Gustavus, AK
Country: United States
IATA Code: GST
ICAO Code: PAGS
Coordinates: 58°25′31″N, 135°42′25″W
## Time difference and current local times
The time difference between Brandon and Gustavus is 3 hours. Gustavus is 3 hours behind Brandon.
CDT
AKDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 182 kg (402 pounds).
## Frequent Flyer Miles Calculator
Brandon (YBR) → Gustavus (GST).
Distance:
1543
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1543
Round trip? | 501 | 1,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-21 | latest | en | 0.819192 |
http://gmatclub.com/forum/p-can-hit-a-target-4-times-in-5-shots-q-3-times-in-4-shots-3604.html | 1,484,599,892,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279248.16/warc/CC-MAIN-20170116095119-00256-ip-10-171-10-70.ec2.internal.warc.gz | 118,484,457 | 40,173 | P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 16 Jan 2017, 12:51
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots
Author Message
Intern
Joined: 29 Aug 2003
Posts: 29
Location: Mumbai
Followers: 0
Kudos [?]: 0 [0], given: 0
P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots [#permalink]
### Show Tags
10 Dec 2003, 05:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit?
Director
Joined: 13 Nov 2003
Posts: 790
Location: BULGARIA
Followers: 1
Kudos [?]: 47 [0], given: 0
### Show Tags
10 Dec 2003, 06:03
5/6 ...i suppose..
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8
Kudos [?]: 245 [0], given: 0
### Show Tags
10 Dec 2003, 06:29
at least 2 means that 2 or 3
P(all 3)=4/5*3/4*2/3=2/5
P(2)=[4/5*3/4*1/3]+[3/4*2/3*1/5]+[4/5*2/3*1/4]=1/5+1/10+2/15=13/30
P=2/5+13/30=25/30=5/6
Intern
Joined: 29 Aug 2003
Posts: 29
Location: Mumbai
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
10 Dec 2003, 07:41
Right.
5/6 it is...
Manager
Joined: 22 Nov 2003
Posts: 54
Location: New Orleans
Followers: 1
Kudos [?]: 0 [0], given: 0
### Show Tags
10 Dec 2003, 13:48
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67
Kudos [?]: 861 [0], given: 781
### Show Tags
13 Dec 2003, 00:27
4128851 wrote:
P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit?
prob of hitting zero shots = 1/5 * 1/4 *1/3 = 1/60
prob of hitting exactly one shot = 4/5 * 1/4*1/3 + 1/5* 3/4 * 2/3 + 1/5 * 1/4 * 2/3 = 9/60
P(0) + P(1) = 1/60 + 9/60 = 10/60 =1/6
required prob = 1- 1/6 = 5/6
Re: probability!! [#permalink] 13 Dec 2003, 00:27
Display posts from previous: Sort by | 971 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-04 | latest | en | 0.841152 |
https://byjus.com/cbse/projectile-motion-problems/ | 1,550,516,280,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487624.32/warc/CC-MAIN-20190218175932-20190218201932-00517.warc.gz | 497,403,044 | 94,991 | # Projectile Motion- Jumping off Cliffs
Projectile Motion Problems:
The problems based on the projectile motion can be of following types:
1. The initial height of the projectile is lower than the final height of the projectile. In the given figure, the initial height of projectile is A and final height is B, which is a higher point relative to A.
2. The initial height of the projectile is greater than the final height of the projectile. The initial height of projectile i.e. A is higher with respect to the final height of projectile i.e. B.
3. The initial and final height of the projectile is same. The initial height A of the projectile is same as the final height B of the projectile as shown in the figure.
Let us now solve a few projectile motion problems based on these cases.
Problem 1: An angry bird is launched with an initial velocity of $10√2 m/s$ at an angle of $45^{\circ}$ as shown. Determine the time in which it will hit the pig on the wooden platform. Assume the acceleration due to gravity is $10 m/s^2$.
Solution: From the data given in the problem,
Component of velocity in x direction,
$v_{ox}~=~v_o~cos~θ~=~10√2~cos~45^{\circ}~=~10 ~m/s$
Component of velocity in y-direction,
$v_{oy}~=~v_o~sin~θ~=~10√2~sin~45^{\circ}~=~10~ m/s$
As there is no acceleration in the horizontal direction and the velocity in the horizontal direction is constant. Also, the distance traveled in horizontal direction or range is given as $10 m$.
Since, $Range~=~velocity~×~time$
$⇒~10~=~10t$
$⇒~t~=~1~ sec$
This is the time required for the bird to reach the wooden platform, which is at a height of $15 ~m$ from the ground as given.
Now, will the bird even hit the pig? Let us try to figure that out.
The distance traveled by the projectile in the y-direction is at the end of 1 sec will be:
$s_y~=~ v_{oy}~ t~ +~\frac{1}{2}~ a_y t^2$
$⇒~s_y~=~10(1)~-~\frac{1}{2}~ 10(1)^2$
$⇒~s_y~=~5 m$
So, the bird is at a height of 5m at the end of 1sec and the pig is sitting at a height of 15 m. Therefore, it will miss the pig and he is going to leave. The motion of the bird will be as shown in figure.
Problem 2: A missile is being launched from a cliff of a given height as shown in the figure given below. Calculate the range of the projectile. Assume the acceleration due to gravity is $10 ~m/s^2$.
Solution: Such projectile motion problems can be solved easily by resolving the initial velocity into vertical and horizontal components. This can be considered as the sum of two one-dimensional motions.
Resolving the initial velocity, we have
Since the time of flight depends upon the component of velocity in the vertical direction. Hence, using the equations of motion in y-direction and considering upward displacement as positive in y-direction and forward displacement positive in x-direction, we have
$s_y~=~ v_{oy}~ t~ +~\frac{1}{2}~ a_y t^2$
$⇒-45~=~40t~+~\frac{1}{2}~ (10) ~(t)^2$
$⇒~5t^2~-~40t~-~45~=~0$
$⇒~t^2~-~8t~-~9~=~0$
$⇒~t~=~9~ sec$ and $t~=~-1~ sec$
$t~=~-1$ implies that the time taken for a ball launched at position A to reach at position B with the velocity of $40~ m/s$ is $1 ~sec$.
The time, taken by the projectile to reach back to the same height as point A is $9 ~sec$.
Now, $Range~=~Velcoity~×~time$
$⇒~R~=~(40~×~9)~m$
$⇒~R~=~360 ~m$
Thus, the range of projectile is $360~ m$.
Problem 3: In a cricket match, batsman hits the ball at a velocity of $10√2 ~ m/s$ at an angle of $45^{\circ}$ in order to score a six. A bowler can catch a ball, which is at a height of 3 m with respect to the ground. Determine the time in, which bowler will catch the ball.
Solution: The initial velocity can be resolved as
According to the given problem:
$v_{oy}~=~10$
$a~=~-10 ~m/s$
$s_y~=~3 m$
According to equations of motion in a plane:
$s_y~= ~v_{oy}~ t~ +~\frac{1}{2}~ a_y ~t^2$
Substituting the values we have;
$3~=~10(t)~-~\frac{1}{2} ~(10) ~t^2$
$5t^2~-~10t~+~3~=~0$
Solving for $t$ we get,
$t~=~1~±~\frac{\sqrt{40}}{10} ~sec$
Thus, the projectile motion problems may seem to appear as different but they can be easily solved using the correct approach.’
#### Practise This Question
Choose the correctly matched pair:
NCERT Related Articles NCERT Solutions for Class 12 NCERT Solutions for Class 11 NCERT Solutions for Class 10 NCERT Solutions for Class 12 Maths NCERT Solutions for Class 11 Maths NCERT Solutions for Class 10 Maths NCERT Solutions for Class 12 Physics NCERT Solutions for Class 11 Physics NCERT Solutions for Class 10 Science NCERT Solutions for Class 12 Chemistry NCERT Solutions for Class 11 Chemistry NCERT Solutions for Class 6 NCERT Solutions for Class 12 Biology NCERT Solutions for Class 11 Biology NCERT Solutions for Class 6 Maths NCERT Solutions for Class 9 NCERT Solutions for Class 8 NCERT Solutions for Class 6 Science NCERT Solutions for Class 9 Maths NCERT Solutions for Class 8 Maths NCERT Solutions for Class 4 NCERT Solutions for Class 9 Science NCERT Solutions for Class 8 Science NCERT Solutions for Class 4 Maths NCERT Solutions for Class 5 NCERT Solutions for Class 7 NCERT Solutions for Class 7 Maths NCERT Solutions for Class 4 Science NCERT Solutions for Class 5 Maths NCERT Solutions for Class 7 Science NCERT Solutions NCERT Books NCERT Solutions for Class 5 Science | 1,492 | 5,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.53125 | 5 | CC-MAIN-2019-09 | latest | en | 0.884458 |
https://planetmath.org/allsolutionofthelorenzequationenteranellipsoid | 1,618,916,945,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00085.warc.gz | 542,304,510 | 3,908 | # all solution of the Lorenz equation enter an ellipsoid
If $\sigma,\tau,\beta>0$ then all solutions of the Lorenz equation
$\displaystyle\dot{x}$ $\displaystyle=$ $\displaystyle\sigma(y-x)$ $\displaystyle\dot{y}$ $\displaystyle=$ $\displaystyle x(\tau-z)-y$ $\displaystyle\dot{z}$ $\displaystyle=$ $\displaystyle xy-\beta z$
will enter an ellipsoid centered at $(0,0,2\tau)$ in finite time. In addition the solution will remain inside the ellipsoid once it has entered. To observe this we define a Lyapunov function
$V(x,y,z)=\tau x^{2}+\sigma y^{2}+\sigma(z-2\tau)^{2}.$
It then follows that
$\displaystyle\dot{V}$ $\displaystyle=$ $\displaystyle 2\tau x\dot{x}+2\sigma y\dot{y}+2\sigma(z-2\tau)\dot{z}$ $\displaystyle=$ $\displaystyle 2\tau x\sigma(y-x)+2\sigma y(x(\tau-z)-y)+2\sigma(z-2\tau)(xy-% \beta z)$ $\displaystyle=$ $\displaystyle-2\sigma(\tau x^{2}+y^{2}+\beta(z-r)^{2}-b\tau^{2}).$
We then choose an ellipsoid which all the solutions will enter and remain inside. This is done by choosing a constant $C>0$ such that the ellipsoid
$\tau x^{2}+y^{2}+\beta(z-r)^{2}=b\tau^{2}$
is strictly contained in the ellipsoid
$\tau x^{2}+\sigma y^{2}+\sigma(z-2\tau)^{2}=C.$
Therefore all solution will eventually enter and remain inside the above ellipsoid since $\dot{V}<0$ when a solution is located at the exterior of the ellipsoid.
Title all solution of the Lorenz equation enter an ellipsoid AllSolutionOfTheLorenzEquationEnterAnEllipsoid 2013-03-22 15:15:28 2013-03-22 15:15:28 Daume (40) Daume (40) 4 Daume (40) Result msc 34-00 msc 65P20 msc 65P30 msc 65P40 msc 65P99 | 532 | 1,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 23, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-17 | longest | en | 0.628208 |
http://www.algebra.com/algebra/homework/equations/Equations.faq?hide_answers=1&beginning=11385 | 1,369,379,660,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704288823/warc/CC-MAIN-20130516113808-00070-ip-10-60-113-184.ec2.internal.warc.gz | 323,105,098 | 13,085 | # Questions on Algebra: Equations answered by real tutors!
Algebra -> Algebra -> Equations -> Questions on Algebra: Equations answered by real tutors! Log On
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Algebra: Equations Solvers Lessons Answers archive Quiz In Depth
Question 218935: 3(x-1)=30 Click here to see answer by rfer(12662)
Question 218945: given the equation 2x + 3y =270 solve for y if x = 15 solve for x if y = 15 Click here to see answer by rfer(12662)
Question 219030: -6b+1+4b=3b+6 solve for b please Click here to see answer by jim_thompson5910(28595)
Question 219031: 9x-(6x-1)=2 Click here to see answer by jim_thompson5910(28595)
Question 219044: how do i solve for c? c-8=2+5c Click here to see answer by jim_thompson5910(28595)
Question 219083: how to solve homework problem 2w-4w=-10 Click here to see answer by chibisan(131)
Question 219114: I need your help, can you please tell me what the answer to 1/2(6-z)=z please and thank you for your time. Click here to see answer by jim_thompson5910(28595)
Question 219141: what are the steps to figure out: -14=5-G? I know that G=19, but I can't get the steps. Click here to see answer by jim_thompson5910(28595)
Question 219169: 7-2n=n-14 Click here to see answer by jim_thompson5910(28595)
Question 219183: what does n equal to in: -6n-9=-2n+53 Click here to see answer by jim_thompson5910(28595)
Question 219183: what does n equal to in: -6n-9=-2n+53 Click here to see answer by rfer(12662)
Question 219187: What is the wavelength of radiation with a frequency of 1.50X10^13 Hz? Does this radiation have a longer or shorter wavelength than red light? Click here to see answer by Alan3354(30993)
Question 219105: 40-8x/5=-2x Click here to see answer by checkley77(12569)
Question 219264: 3(n-1)-2=4(n-4)-(n-11) Click here to see answer by checkley77(12569)
Question 219246: five times the sum of a number and nine is thirty five Click here to see answer by ankor@dixie-net.com(15656)
Question 219409: convert the rate using dimensional analysis. round to the nearest tenth, if necessary. 25cm/s=________m/h a.225 b.900 c.9000 d.90 Click here to see answer by stanbon(57361)
Question 219458: Please help with the steps to solve . I'm not sure how to isolate the x. Click here to see answer by ichudov(499)
Question 219464: Can someone explain to me how this is done: Profit = -0.25Q^2 + 10Q - 4 Profit = -0.25(Q^2-40Q) - 4 Profit = -0.25 (Q^2-40Q +400) - 4 +100 Profit = -0.25(Q-20)^2 + 96 Therefore Profit(max) = 96 @ Q = 20 How do you get 40Q? How do you get 400? How do you get (Q-20)? Also - is this algebra? Since I do not understand the rules here on how to work through this problem, what should I be studying here? Thanks, SJ Click here to see answer by stanbon(57361)
Question 219562: Can you show me how to work this problem. It's equations containing decimals. 0.24+0.18x=0.08x-0.06 Click here to see answer by jim_thompson5910(28595)
Question 219587: if sally paints a house in 4 hours and john paints the same house in 6 hours how long will it take for them both to paint the house Click here to see answer by solver91311(16897)
Question 219605: Y+7Y=104 What is Y= to? Click here to see answer by jim_thompson5910(28595)
Question 219618: 5x-6=3x+6 Click here to see answer by jim_thompson5910(28595)
Question 219622: Translate to an algebraic expression. the product of 10% and some number. here is what I got. 0.10y,is that right and Solve. 4x - (3x + 3) =2 please explain step by step how you got the answer. and find the slope, if it exists x= -8, m = ? Click here to see answer by jim_thompson5910(28595)
Question 219681: 3v-9=7+2v Click here to see answer by jim_thompson5910(28595)
Question 219678: Translate To An Equation, then solve: If the larger of two numbers is 5 less than twice the smaller number and the sum of the two numbers is 28. What are the numbers? Click here to see answer by rfer(12662)
Question 219683: 12(2-x)=6 Click here to see answer by jim_thompson5910(28595)
Question 219699: 5/6x+2/3=-1/6x-5/3 Click here to see answer by jim_thompson5910(28595)
Question 219715: solve the equation x2-x-6=0 Click here to see answer by checkley77(12569)
Question 219734: Please help me solve this equation and explain to me how you did it: 6n-3n=12 Click here to see answer by likaaka(51)
Question 219729: describe what steps you would take to solve 8x+5=2x-7 Click here to see answer by likaaka(51)
Question 219822: What are the steps to solving this problem? 905=5a Click here to see answer by checkley77(12569)
Question 219821: 4050 meters divided by 1/3 Click here to see answer by checkley77(12569)
Question 219818: 12=a/36+17 Click here to see answer by checkley77(12569)
Question 219857: i need a answer. the question is that i need a 7th grade algebra equation with the answer 29. but the trick is that you can only use the numbers 1,2,3,4 each ONCE. YOU CAN USE EXPONETS ALSO. FOR EXAMPLE: 4-2x3+1=7 Click here to see answer by stanbon(57361)
Question 219877: i would like to know how to figure aout this formula, the formula for the volume of a right circular cylinder is v=3.14 r(squared)h. the value of h can be expressed as. can't figure the answer Click here to see answer by rfer(12662)
Question 219890: i am really struggeling e=wtih my homework. i am in 7th grade and in pre-algebra my question is that i need a algebra equation that EQUALS 29 and only using A,B,C,D EACH ONLY ONCE. A=1 B=2 C=3 D=4 FOR EXAMPLE: (bc)+dxa 2x3=6+4=10x1= 10 so the answer would be 10 Click here to see answer by josmiceli(9686)
Question 219943: i need help with algebraic equations using fractions such as 1/3w+6=-2 Click here to see answer by vleith(2825)
Question 219943: i need help with algebraic equations using fractions such as 1/3w+6=-2 Click here to see answer by rfer(12662)
Question 220062: What does Q equal in this equation? 300-3Q=50+15*(1.1)^1 Click here to see answer by rfer(12662)
Question 220045: how do i solve this equation 22+4y-14=0 shown in steps please and thank you Click here to see answer by rfer(12662)
Question 219921: Please explain how to solve this equation and the way you did it: -3(2x+8)-8=-26 Click here to see answer by chibisan(131)
Question 220116: 0.05z+0.2=0.15z-10.5 Click here to see answer by susanaramya(12)
Question 220115: 0.12x+3-0.8x=0.22x-0.6 Click here to see answer by vleith(2825)
Question 220100: c-9.02=-8.6 Click here to see answer by susanaramya(12)
Question 220162: what are the steps to solving this equation? It involves fractions so I hope I write this out for you correctly: [(-4.57x-3.61)/2.78]-[(3.11x-2.46)/3.44]=1.77x Click here to see answer by stanbon(57361)
Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250, 20251..20295, 20296..20340, 20341..20385, 20386..20430, 20431..20475, 20476..20520, 20521..20565, 20566..20610, 20611..20655, 20656..20700, 20701..20745, 20746..20790, 20791..20835, 20836..20880, 20881..20925, 20926..20970, 20971..21015, 21016..21060, 21061..21105, 21106..21150, 21151..21195, 21196..21240, 21241..21285, 21286..21330, 21331..21375, 21376..21420, 21421..21465, 21466..21510, 21511..21555, 21556..21600, 21601..21645, 21646..21690, 21691..21735, 21736..21780, 21781..21825, 21826..21870, 21871..21915, 21916..21960, 21961..22005, 22006..22050, 22051..22095, 22096..22140, 22141..22185, 22186..22230, 22231..22275, 22276..22320, 22321..22365, 22366..22410, 22411..22455, 22456..22500, 22501..22545, 22546..22590, 22591..22635, 22636..22680, 22681..22725, 22726..22770, 22771..22815, 22816..22860, 22861..22905, 22906..22950, 22951..22995, 22996..23040, 23041..23085, 23086..23130, 23131..23175, 23176..23220, 23221..23265, 23266..23310, 23311..23355, 23356..23400, 23401..23445, 23446..23490, 23491..23535, 23536..23580, 23581..23625, 23626..23670, 23671..23715, 23716..23760, 23761..23805, 23806..23850, 23851..23895, 23896..23940, 23941..23985, 23986..24030, 24031..24075, 24076..24120, 24121..24165, 24166..24210, 24211..24255, 24256..24300, 24301..24345, 24346..24390, 24391..24435, 24436..24480, 24481..24525, 24526..24570, 24571..24615, 24616..24660, 24661..24705, 24706..24750, 24751..24795, 24796..24840, 24841..24885, 24886..24930, 24931..24975, 24976..25020, 25021..25065, 25066..25110, 25111..25155, 25156..25200, 25201..25245, 25246..25290, 25291..25335, 25336..25380, 25381..25425, 25426..25470, 25471..25515, 25516..25560, 25561..25605, 25606..25650, 25651..25695, 25696..25740, 25741..25785, 25786..25830 | 6,098 | 14,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2013-20 | latest | en | 0.887424 |
https://mcqseries.com/logic-gates-145/ | 1,721,651,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00851.warc.gz | 338,582,196 | 15,780 | Logic Gates – 145
5555555555
145. One possible output expression for an AND-OR-Invert circuit having one AND gate with inputs A, B, and C and one AND gate with inputs D and E is ________.
(a)
(b)
(c)
(d)
Explanation
Explanation : No answer description available for this question. Let us discuss.
Subject Name : Electronics Engineering Exam Name : IIT GATE, UPSC ESE, RRB, SSC, DMRC, NMRC, BSNL, DRDO, ISRO, BARC, NIELIT Posts Name : Assistant Engineer, Management Trainee, Junior Engineer, Technical Assistant
Electronics & Communication Engineering Books Sale Question Bank On Electronics & Communication Engineering Book - question bank on electronics & communication engineering; Language: english; Binding: paperback ₹ 287 Sale A Handbook for Electronics Engineering(Old Edition) Made Easy Editorial Board (Author); Marathi (Publication Language); 620 Pages - 01/01/2015 (Publication Date) - Made Easy Publications (Publisher) ₹ 320 Basic Electronic Devices and Circuits Patil (Author); English (Publication Language) Electronic Circuits: Analysis and Design (SIE) | 3rd Edition Neamen, Donald (Author); English (Publication Language); 1351 Pages - 08/25/2006 (Publication Date) - McGraw Hill Education (Publisher) ₹ 1,265
Related Posts
• Logic Gates » Exercise - 114. The output of a ......... gate is only 1 when all of its inputs are 1. (a) NOR (b) XOR (c) AND (d) NOT
Tags: logic, gates, output, gate, inputs, electronics, engineering
• Logic Gates » Exercise - 15. In the circuit shown the value of inputs A and B for F = 0 is : & (a) 00 (b) 01 (c) 10 (d) 11
Tags: logic, gates, circuit, inputs, electronics, engineering
• 5555555555127. When does the output of a NOR gate = 0? (a) Whenever a 0 is present at an input (b) Only when all inputs = 0 (c) Whenever a 1 is present at an input (d) Only when all inputs = 1
Tags: inputs, logic, gates, output, gate, electronics, engineering
• 555555555558. The output of an AND gate with three inputs, A, B, and C, is HIGH when ________. (a) A = 1, B = 1, C = 0 (b) A = 0, B = 0, C = 0 (c) A = 1, B = 1, C = 1 (d) A = 1, B = 0, C = 1
Tags: logic, gates, output, gate, inputs, electronics, engineering
• 5555555555124. When does the output of a NAND gate = 1? (a) Whenever a 0 is present at an input (b) Only when all inputs = 0 (c) Whenever a 1 is present at an input (d) Only when all inputs = 1
Tags: inputs, logic, gates, output, gate, electronics, engineering | 699 | 2,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.697978 |
https://bibbase.org/network/publication/feldman-vondrak-tightboundsonlowdegreespectralconcentrationofsubmodularandxosfunctions-2015 | 1,627,443,686,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00058.warc.gz | 151,895,515 | 4,160 | Tight Bounds on Low-degree Spectral Concentration of Submodular and XOS functions. Feldman, V. & Vondrak, J. arXiv:1504.03391 [cs], April, 2015. arXiv: 1504.03391
Submodular and fractionally subadditive (or equivalently XOS) functions play a fundamental role in combinatorial optimization, algorithmic game theory and machine learning. Motivated by learnability of these classes of functions from random examples, we consider the question of how well such functions can be approximated by low-degree polynomials in \${\textbackslash}ell_2\$ norm over the uniform distribution. This question is equivalent to understanding of the concentration of Fourier weight on low-degree coefficients, a central concept in Fourier analysis. We show that 1. For any submodular function \$f:{\textbackslash}\{0,1{\textbackslash}\}{\textasciicircum}n {\textbackslash}rightarrow [0,1]\$, there is a polynomial of degree \$O({\textbackslash}log (1/{\textbackslash}epsilon) / {\textbackslash}epsilon{\textasciicircum}\{4/5\})\$ approximating \$f\$ within \${\textbackslash}epsilon\$ in \${\textbackslash}ell_2\$, and there is a submodular function that requires degree \${\textbackslash}Omega(1/{\textbackslash}epsilon{\textasciicircum}\{4/5\})\$. 2. For any XOS function \$f:{\textbackslash}\{0,1{\textbackslash}\}{\textasciicircum}n {\textbackslash}rightarrow [0,1]\$, there is a polynomial of degree \$O(1/{\textbackslash}epsilon)\$ and there exists an XOS function that requires degree \${\textbackslash}Omega(1/{\textbackslash}epsilon)\$. This improves on previous approaches that all showed an upper bound of \$O(1/{\textbackslash}epsilon{\textasciicircum}2)\$ for submodular and XOS functions. The best previous lower bound was \${\textbackslash}Omega(1/{\textbackslash}epsilon{\textasciicircum}\{2/3\})\$ for monotone submodular functions. Our techniques reveal new structural properties of submodular and XOS functions and the upper bounds lead to nearly optimal PAC learning algorithms for these classes of functions.
@article{feldman_tight_2015,
title = {Tight {Bounds} on {Low}-degree {Spectral} {Concentration} of {Submodular} and {XOS} functions},
url = {http://arxiv.org/abs/1504.03391},
abstract = {Submodular and fractionally subadditive (or equivalently XOS) functions play a fundamental role in combinatorial optimization, algorithmic game theory and machine learning. Motivated by learnability of these classes of functions from random examples, we consider the question of how well such functions can be approximated by low-degree polynomials in \${\textbackslash}ell\_2\$ norm over the uniform distribution. This question is equivalent to understanding of the concentration of Fourier weight on low-degree coefficients, a central concept in Fourier analysis. We show that 1. For any submodular function \$f:{\textbackslash}\{0,1{\textbackslash}\}{\textasciicircum}n {\textbackslash}rightarrow [0,1]\$, there is a polynomial of degree \$O({\textbackslash}log (1/{\textbackslash}epsilon) / {\textbackslash}epsilon{\textasciicircum}\{4/5\})\$ approximating \$f\$ within \${\textbackslash}epsilon\$ in \${\textbackslash}ell\_2\$, and there is a submodular function that requires degree \${\textbackslash}Omega(1/{\textbackslash}epsilon{\textasciicircum}\{4/5\})\$. 2. For any XOS function \$f:{\textbackslash}\{0,1{\textbackslash}\}{\textasciicircum}n {\textbackslash}rightarrow [0,1]\$, there is a polynomial of degree \$O(1/{\textbackslash}epsilon)\$ and there exists an XOS function that requires degree \${\textbackslash}Omega(1/{\textbackslash}epsilon)\$. This improves on previous approaches that all showed an upper bound of \$O(1/{\textbackslash}epsilon{\textasciicircum}2)\$ for submodular and XOS functions. The best previous lower bound was \${\textbackslash}Omega(1/{\textbackslash}epsilon{\textasciicircum}\{2/3\})\$ for monotone submodular functions. Our techniques reveal new structural properties of submodular and XOS functions and the upper bounds lead to nearly optimal PAC learning algorithms for these classes of functions.},
urldate = {2016-03-10TZ},
journal = {arXiv:1504.03391 [cs]},
author = {Feldman, Vitaly and Vondrak, Jan},
month = apr,
year = {2015},
note = {arXiv: 1504.03391},
keywords = {Computer Science - Data Structures and Algorithms, Computer Science - Learning}
} | 1,200 | 4,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.726634 |
https://whatisconvert.com/190-inches-in-nautical-miles | 1,628,196,583,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00599.warc.gz | 612,727,126 | 7,640 | # What is 190 Inches in Nautical Miles?
## Convert 190 Inches to Nautical Miles
To calculate 190 Inches to the corresponding value in Nautical Miles, multiply the quantity in Inches by 1.3714902807775E-5 (conversion factor). In this case we should multiply 190 Inches by 1.3714902807775E-5 to get the equivalent result in Nautical Miles:
190 Inches x 1.3714902807775E-5 = 0.0026058315334773 Nautical Miles
190 Inches is equivalent to 0.0026058315334773 Nautical Miles.
## How to convert from Inches to Nautical Miles
The conversion factor from Inches to Nautical Miles is 1.3714902807775E-5. To find out how many Inches in Nautical Miles, multiply by the conversion factor or use the Length converter above. One hundred ninety Inches is equivalent to zero point zero zero two six one Nautical Miles.
## Definition of Inch
An inch (symbol: in) is a unit of length. It is defined as 1⁄12 of a foot, also is 1⁄36 of a yard. Though traditional standards for the exact length of an inch have varied, it is equal to exactly 25.4 mm. The inch is a popularly used customary unit of length in the United States, Canada, and the United Kingdom.
## Definition of Nautical Mile
The nautical mile (symbol M, NM or nmi) is a unit of length, defined as 1,852 meters (approximately 6,076 feet). It is a non-SI unit used especially by navigators in the shipping and aviation industries, and also in polar exploration.
## Using the Inches to Nautical Miles converter you can get answers to questions like the following:
• How many Nautical Miles are in 190 Inches?
• 190 Inches is equal to how many Nautical Miles?
• How to convert 190 Inches to Nautical Miles?
• How many is 190 Inches in Nautical Miles?
• What is 190 Inches in Nautical Miles?
• How much is 190 Inches in Nautical Miles?
• How many nmi are in 190 in?
• 190 in is equal to how many nmi?
• How to convert 190 in to nmi?
• How many is 190 in in nmi?
• What is 190 in in nmi?
• How much is 190 in in nmi? | 520 | 1,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.894395 |
https://www.easycalculation.com/trigonometry/trigonometric-expressions-to-summation-tan-and-cot.php | 1,575,907,651,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00179.warc.gz | 683,795,789 | 6,717 | English
# Transformation of Tangent and Cotangent Summation into a Product
Simple trigonometric calculator which is used to transform the difference of tangent and cotangent function into product.
## Transform Summation of Tangent and Cotangent into a Product
°
°
°
Simple trigonometric calculator which is used to transform the difference of tangent and cotangent function into product.
#### Formula:
tan(x)+cot(y)=cos((x-y)) / (cos(x)*sin(y)) Where, x = Angle (α) y = Angle (β) | 117 | 486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.783916 |
https://www.thestudentroom.co.uk/showthread.php?t=4143437 | 1,518,993,315,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.30/warc/CC-MAIN-20180218212626-20180218232626-00734.warc.gz | 992,121,052 | 38,594 | You are Here: Home >< Maths
# Hard vector question! watch
1. Find coordinates where line through (-5,-1) in the direction (2/3) (this is the direction vector) meets circle at x^2 + y^2 = 65?
I'll tag Zacken
2. (Original post by mil88)
Find coordinates where line through (-5,-1) in the direction (2/3) (this is the direction vector) meets circle at x^2 + y^2 = 65?
I'll tag Zacken
Conver the line into cartesian form y = mx + c and then plug it into the circle equation to get a quadratic in x.
What does the direction vector mean and how does it relate to the gradient of a line...? Think about it a bit.
3. (Original post by Zacken)
Conver the line into cartesian form y = mx + c and then plug it into the circle equation to get a quadratic in x.
What does the direction vector mean and how does it relate to the gradient of a line...? Think about it a bit.
Oh snap, I did exactly what you said before, but I wrote gradient to be 2/3 rather than 3/2 ! Hence, I got incorrect values.
By the way, can you actually work it out using vectors, or do you have to convert it to y= mx + c?
4. (Original post by mil88)
Oh snap, I did exactly what you said before, but I wrote gradient to be 2/3 rather than 3/2 ! Hence, I got incorrect values.
By the way, can you actually work it out using vectors, or do you have to convert it to y= mx + c?
You probably could in some perverted non A-Level way. Pretty sure you should convert it to y = mx + c.
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: June 5, 2016
Today on TSR
### Three reasons you may feel demotivated right now
...and how to stay positive
### Can I get A*s if I start revising now?
Discussions on TSR
• Latest
Poll
Useful resources
### Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
### How to use LaTex
Writing equations the easy way
### Study habits of A* students
Top tips from students who have already aced their exams
## Groups associated with this forum:
View associated groups
Discussions on TSR
• Latest
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 626 | 2,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-09 | latest | en | 0.924867 |
http://msgroups.net/sqlserver.programming/decimal-problem/98932 | 1,591,477,798,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348519531.94/warc/CC-MAIN-20200606190934-20200606220934-00413.warc.gz | 84,237,818 | 11,647 | #### decimal problem
```Hi all
I a very strange effect when using different decimal settings
and calculating some values:
DECLARE @Decimal1 decimal(5, 2)
DECLARE @Decimal2 decimal(30, 2)
SET @Decimal1 = 0.2
SET @Decimal2 = 0.2
SELECT (CONVERT(decimal(15, 15), 0.000000000322900) * (1 - @Decimal1))
SELECT (CONVERT(decimal(15, 15), 0.000000000322900) * (1 - @Decimal2))
The first result is correct (0.00000000025832000),
the second is wrong, why ???
Best regards
Frank Uray
```
0
Utf
4/28/2010 1:16:13 PM
sqlserver.programming 1873 articles. 0 followers.
1 Replies
915 Views
Similar Articles
[PageSpeed] 25
```Hi Frank
This is what happens when you multiple these precision&scales
Precision
15 + 30 = 45 + 1 = 46
Scale
2+15= 17
so (46,17)
http://msdn.microsoft.com/en-us/library/ms190476.aspx?ppud=4
http://blogs.msdn.com/sqlprogrammability/archive/2006/03/29/564110.aspx
* The result precision and scale have an absolute maximum of 38. When a
result precision is greater than 38, the corresponding scale is reduced to
prevent the integral part of a result from being truncated.
Jon
"Frank Uray" wrote:
> Hi all
>
> I a very strange effect when using different decimal settings
> and calculating some values:
>
> DECLARE @Decimal1 decimal(5, 2)
> DECLARE @Decimal2 decimal(30, 2)
>
> SET @Decimal1 = 0.2
> SET @Decimal2 = 0.2
>
> SELECT (CONVERT(decimal(15, 15), 0.000000000322900) * (1 - @Decimal1))
> SELECT (CONVERT(decimal(15, 15), 0.000000000322900) * (1 - @Decimal2))
>
> The first result is correct (0.00000000025832000),
> the second is wrong, why ???
>
>
> Best regards
> Frank Uray
```
0
Utf
4/28/2010 1:32:05 PM
Similar Artilces:
SOP NonInventory Unit Cost 6 Decimal Places
Is there any simple way to increase the currency decimal places to greater than the current highest value of 5? Thanx Scott, I don't believe there is any easy or even moderately difficult way to do this, as the field types used to store amounts in GP only go up to 5 decimal places. -- Victoria Yudin Dynamics GP MVP Flexible Solutions - home of GP Reports http://www.flex-solutions.com/gpreports.html blog: http://www.victoriayudin.com "Scott Rutledge" <sersoar@hotmail.com> wrote in message news:E102B81B-8555-400E-843C-240C58794294@microsoft.com... > Is there any...
Another Standby & System Stanby Problem
Up until about a week ago my wife was able to put her desktop PC into Standby Mode (Start/Turn Off Computer/Standby), and the "Power Schemes" worked so that the computer would go into "System Standby" after 1 hour of non use. Now only the screen saver comes on correctly but the computer will not go into System Standby after 1 hour of non use nor can the computer be put into Standby manually (Start/Turn Off Computer/Standby). When she tried to put the computer into Standby mode manually, it turns off the monitor, starts to turn off the computer but it comes r...
Excel Search Problem
I have five columns (A,B,C,D,E) of which I have a varying number of row per time. Column A will contain numbers from 1+ and Column E wil contain numbers and text strings. What I want to do is have a colum E that has a formula that for each row use its corrisponding value in and search All values in column E for the same value and if that A contains a value listed in column E somewhere return the label "True and if not found return "False". Example A B C D E F 1 - - - 2000 FALSE 2 - - - 2001 FALSE 2000 - - - STORM TRUE 200...
Need solution to formula problem
Help please! I have two rows of numbers with some cell that will contain 'N/A". If the top row is highest I need it highlighted yellow if it is not highest or is "N/A" the cell needs to remain with no fill. I can do a condtional formatting that works great until a cell contains the "N/A". Is there a formula that I can do in conditional formatting or some other options that will solve this problem. As suggested in your multi-post in .programming (please do not multi-post) Try this .. Select rows 1 & 2, and apply cond format Cond1's formula: =A\$1=&q...
Fetch XML problem
Hi, all. I have a problem that is caused by a limitation of the fetch XML schema: I want to calculate the prices for a list products according to a special pricelist by the default unit of measure. I found no way to select unique product and priceleveldetail details for the products because I can only give one join attribute (productid or uomid) for the link-entity. Example: <fetch mapping='logical'> <entity name='product'> <attribute name='productid'/> <attribute name='name'/> <attribute name='price'/>...
Reduce currency decimals
Hi, Is there any tool which will reduce the currency decimals on our Great Plains system. We want to reduce our system currency decimals. Thank you. KT I'm not sure if this is what you mean, but there's Tools >> Utilities >> Inventory >> Change Decimal Places This utility can change the decimal places for items. I don't know of any system-wide change for decimal places. Eleni "KT" wrote: > Hi, > > Is there any tool which will reduce the currency decimals on our Great > Plains system. We want to reduce our system currency decimals....
Outlook 2000 "Out of memory or system resources" problems
OS: Win XP Pro CPU/Ram: 1.3M Presario Laptop 256MB RAM I run Win XP Pro on Pentium 1.3M with 256MB Ram, I'm getting "out of memory or low in system resources" error while using OUTLOOK 2000. Some folders don't open and and I can't move items between folders. This started just 1 week ago. I've tried updating Office to with SR1 update. It was successful though I got one error "cannot apply patch to C:\config.msi\pt472.tmp....." saying patch may have been applied by other setup... I've tried reinstalling my McAfee virusscan, repairing office(whic...
Connectivity Problems
Sorry about the length of this post, but I have no idea what I'm talking about, let alone explain it! We have two sites (Head Office and Remote) with a Netgear FVS338 VPN firewall at each end. The remote site connects via a Terminal Server and runs a couple of EPOS systems. Our setup in the Head Office is thus: Workstation -> SBS 2003 R2 Standard -> FVS338 VPN Firewall -> Internet Router Recently, the VPN keeps dropping and if I set up a permanent ping I can see that several times a day we get 7 or 8 pings timing out. This is enough to kill the EPOS connect...
Font dialog Sample Window problem
Am using a CFontDialog derived class (CInwinFontDlg) to allow user to select new font/size etc. But for some reason cannot get the text in Sample window to either show or change when font attribute selected. The following is how my font dlg is constructed. CInwinFontDlg dlg(&lf, CF_SCREENFONTS| CF_INITTOLOGFONTSTRUCT| CF_FIXEDPITCHONLY| CF_SHOWHELP | CF_ENABLETEMPLATE); Does one have to control the look and appearance or does the Common Windows' font dialog do this with the which CF flag? -- Kevin Kohler INWIN Development Manager Response Technology, Inc. Kevin, How you constru...
sd reader problem with optima notebook
i am using a 32 bit version of vista ultimate. the inbuilt sd reader is completley unresponsive. I do also have a problem with no drivers installed for PCI Flash Memory . I am wondering how do i get my hands on some drivers to fix this problem? I dont have a manufacturer drivers disk and the site was of no help. L40 Optima Centoris V662. "Insane" <Insane@discussions.microsoft.com> wrote in message news:F04AC29F-ECD6-410A-BBD4-DA401B06DD3E@microsoft.com... >i am using a 32 bit version of vista ultimate. the inbuilt sd reader is > completley unresponsive. Your questi...
SBS Site Problems w/ CRM
After upgrading from 1.2 to 3.0, we have major problems with our SBS websites. 1) CRM is setup to use the IP address of the server and this site runs fine. 2) Upgrading knocked out our "companyweb" site installed by SBS ("service unavailable") 3) SQL Reporting doesn't work because it's trying to use the default address which is already used by CRM, RWW, companyweb, etc. I'm really confused about how to sort out all the websites. How can all these sites run on one IP address? So much for MS making this install nice and trouble free for SBS users! -- Bra...
decimal problem
Hi all I a very strange effect when using different decimal settings and calculating some values: DECLARE @Decimal1 decimal(5, 2) DECLARE @Decimal2 decimal(30, 2) SET @Decimal1 = 0.2 SET @Decimal2 = 0.2 SELECT (CONVERT(decimal(15, 15), 0.000000000322900) * (1 - @Decimal1)) SELECT (CONVERT(decimal(15, 15), 0.000000000322900) * (1 - @Decimal2)) The first result is correct (0.00000000025832000), the second is wrong, why ??? Thanks for any comments Best regards Frank Uray Hi Frank This is what happens when you multiple these precision&scales Precision 15 + 3...
table problems
when i try to open an attachment...my publisher freezes up with a mesage that reads... the table is too large for the workspace because of a printer or style change, your table has become larger than publisher's workspace. to solve this problem, you can delete your table, or remove some text from your table. note: publisher may have locked the table size as a resuilt of this action. if your table does not expand to accommodate the text, click grow to fit text on the table menu. What are they talking about??? do i need to take the excel spreadsheet out?? or is it soemthing els...
I am using a Microsoft Jet Engine database. While testing an ADO implementation, I found that I ran into trouble if I repeatedly tried to update the field with the 'too large' data; eventually ADO became convinced that I had never issued an Edit or AddNew. Of course, Edit does not exist in ADO; the edit is implied when you try to set a field. To make sure the error wasn't elsewhere, I have removed all wrapper functions and created this test case that reproduces the error using primitives. I have discovered that issuing a CancelUpdate() clears the error, but then any other edits...
Problem with paths with spaces SUDO
I have this function... function sud([scriptblock]\$SomethingToExecuteInElevation,[switch]\$NoExit){ \$location ="`"`"`""+ (Get-Location) +"`"`"`""; if(!\$NoExit){ \$command = "-command &{set-Location \$location; & \$SomethingToExecuteInElevation;}" Start-Process powershell -Verb Runas -ArgumentList \$command } else{ \$command = "-NoExit -command &{set-Location \$location; & \$SomethingToExecuteInElevation;}" Start-Process powershell -Verb Runas -ArgumentList \$command }} the problem i...
registry problem
I was trying to create a registry key for my dialog project and I added SetRegistryKey(_T("Simple Alarm Clock")); in BOOL CAlarm_ClockApp::InitInstance() , but nothing gets written to the registry. Is there anything else I need to do in order to get my dialog project to write to the registry? Z.K. That simply sets the registry key for the application. You still have to write something there. You can then use WriteProfileInt/WriteProfileString to write something at that key. -- Ajay Kalra [MVP - VC++] ajaykalra@yahoo.com "Z.K." <nospam@nospam.net> wro...
problem in mail
We are facing some problem in moving the users from one domain to other .For example the below user is moved from FMGNT domain to VPNT domain . When we moved his mailbox It was showing correctly mail box is moved to VPNT domain But once his account has been moved when we checked his mailbox his mail box is shown in FMG NT any solution Please It was showing correctly like the below screen But once his account has been moved when we checked his mailbox is at FMGNT please find the below screen shot Make sure that the new domain has ha...
Decimal
In a laboratory test we do the temperature has to be reported to the nearest 0.5�C. Can anyone let me know how to format it so that when the result is a whole number, it leaves the decimal point out? eg 22�C or 22.5�C (not 22.�C or 22.0�C) Alan "Alan Cocks" <alandrob@netspace.net.au> wrote in message news:btlqhd\$19fi\$1@otis.netspace.net.au... > In a laboratory test we do the temperature has to be reported to the > nearest 0.5�C. > Can anyone let me know how to format it so that when the result is a whole > number, it leaves the decimal point out? &g...
Synchronisation problem
I have a HP iPAQ 2200 PDA and I synchronise it with Outlook. If I create an appointment in my iPAQ it synchronises correctly and the appointment appears in Outlook. However, if I create an appointment in Outlook even after synchronisation the appointment will not appear in my iPAQ. Has anyone else experienced this problem? Any ideas how to fix it? I appreciate and help or advice given. Thanks in advance! Probably best to post this in m.p.pocketpc or ..pocketpc.activesync as it isn't really an Outlook issue.... Neil Hindry wrote: > I have a HP iPAQ 2200 PDA and I synchronise ...
I am running 10.4.6 on a Mac G5 dual 2 gig. Upon launcing a MS Office 2004 application today I took notice of the available upgrade to 11.2.6 and proceded with it. This is where things get wacky. I did the upgrade and promptly encountered an error message stating I was missing a file and installation could not continue. Great. Turns out that all of Office apps were wiped out due to this non completed upgrade. Not a biggy. I uninstalled MS OFFICE and rebuilt permissions. Reinstalled MS OFFICE 2004 and tried to get my user data (email and contacts) back into Entourage. No such luck. Er... | 3,396 | 13,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.695685 |
https://mathandmovement.com/activity/super-squares/ | 1,656,813,154,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00093.warc.gz | 442,152,502 | 55,019 | # Super Squares
Created by:
Math & Movement
## Activity Instructions
Put black construction paper down on the numbers 1, 2, 11, and 12. This makes a 2 x 2 square and the square number 4. Have students count the four squares to discover that 2 x 2 is actually 4. Ask, “What is the next square number? Who can guess?” Students can find the answer by covering the 3 x 3 square (covering the numbers 1, 2, 3, 11, 12, 13, 21, 22, and 23). Have the children skip count by threes to discover that 3 x 3 is 9.
Continue laying down pieces of black construction paper to make additional square numbers. | 167 | 597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-27 | latest | en | 0.902108 |
http://docplayer.net/22779657-I-perimeter-area-learning-goals-304.html | 1,542,072,922,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741176.4/warc/CC-MAIN-20181113000225-20181113022225-00552.warc.gz | 96,410,254 | 28,941 | # I Perimeter, Area, Learning Goals 304
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 U N I T Perimeter, Area, Greeting cards come in a variety of shapes and sizes. You can buy a greeting card for just about any occasion! Learning Goals measure and calculate perimeter estimate, measure, and calculate area explore how the areas of rectangles, parallelograms, and triangles are related develop formulas for the area of a parallelogram and the area of a triangle sketch polygons, given perimeter and area measurements 304
2 and Volume Key Words variable base height What measurements are needed to make the envelope for a greeting card? Why is the area of paper used important when calculating costs? 305
3 L E S S O N Perimeter How can you find the perimeter of this square? Remember, a formula is a shortcut to state a rule. 3 cm You will need square dot paper and triangular dot paper. Share the work. Draw 15 different polygons. Make sure there are at least two of each of these types of polygons: square rectangle parallelogram rhombus triangle Find the perimeter of each polygon. For which types of polygons can you find a shortcut to calculate the perimeter? Write the shortcut as a formula. Show and Share Share your formulas with another group of students. Compare your formulas. Discuss any differences. For which types of polygons is it possible to write more than one formula? Explain. 306 LESSON FOCUS Measure and calculate perimeter.
4 For any polygon, we can find its perimeter by adding the side lengths. For this hexagon: 27 mm 15 mm 38 mm 31 mm You could also write the perimeter of this hexagon in centimetres. 182 mm cm 18.2 cm 9 mm 62 mm Perimeter The perimeter of this hexagon is 182 mm. We can find shortcuts for calculating the perimeters of some polygons. A parallelogram has two pairs of equal sides. Here is one rule for the perimeter of a parallelogram. Add the measures of a longer side and a shorter side. Then multiply by 2. 6 cm Formula 11 cm Perimeter (Longer side Shorter side) 2 P ( w) 2 (11 6) The perimeter of this parallelogram is 34 cm. We use to represent the longer side, w for the shorter side, and P for perimeter. These letters are called variables. Unit 9 Lesson 1 307
5 1. Which unit would you use to measure the perimeter of each item? Explain your choice. a) a football field b) your teacher s desk c) a swimming pool d) Manitoba e) a floor tile 2. Measure to find the perimeter of each parallelogram. Which unit did you use? Why? Write each perimeter using a different unit. a) b) 3. Refer to question 2. Suppose the lengths of the longer sides of each parallelogram increase by 2 cm. What would happen to each perimeter? Explain. 4. Find the perimeter of this parallelogram. 5 cm 3 cm Describe how you could change the parallelogram so it has perimeter 20 cm. Give 2 different answers. 5. One formula for calculating the perimeter of a parallelogram is shown in Connect. Write a different formula for the perimeter of a parallelogram. Explain your reasoning. Numbers Every Day Number Strategies Is each number prime or composite? How do you know? Unit 9 Lesson 1
6 6. a) Find the perimeter of each polygon. A B b) Suppose the side lengths of each polygon are doubled. What would happen to each perimeter? Explain. 7. The perimeter of a triangular field is 1.4 km. How long might each side be? 8. Your teacher will give you a large copy of these regular polygons. A C G E D B F H a) Find and record the perimeter of each polygon. b) How is the perimeter of a regular polygon related to the number of its sides? Write a formula to find the perimeter of a regular polygon. How are the side lengths of a figure and its perimeter related? Use examples to explain. Choose one room in your home. Find the perimeter of the floor. Which units did you use to measure? ASSESSMENT FOCUS Question 6 Unit 9 Lesson 1 309
7 L E S S O N Exploring Rectangles You will need 1-cm grid paper. Draw a 2-cm by 3-cm rectangle on grid paper. Find the area of the rectangle. Suppose the length of the rectangle doubles. Predict the area of the new rectangle. Check your prediction. Suppose the width of the original rectangle doubles. Predict the area of the new rectangle. Check your prediction. Suppose both the length and the width double. Predict the area of the new rectangle. Check your prediction. How does the area of each new rectangle compare to the area of the original rectangle? Show and Share Share your work with another pair of students. Describe any patterns. What do you think happens to the area of a rectangle when the length triples? The width triples? Both the length and the width triple? Edmond built a crate for his dog. The dimensions of the floor of the crate are 0.8 m by 1.2 m. 310 LESSON FOCUS Relate the lengths of the sides to the perimeter and the area of a rectangle.
8 You can use formulas to find the perimeter and the area of the floor of the crate. Perimeter (Longer side + Shorter side) 2 P ( w) 2 ( ) Area Length Width A w The perimeter of the floor of the crate is 4 m and the area is 0.96 m 2. You can write the floor area of the crate in square centimetres. Use: 1 m 100 cm A ( ) ( ) The floor area of the crate is 9600 cm 2. So, 0.96 m 2 and 9600 cm 2 are equal areas. In the formula for the area of a rectangle, the length and the width are the factors and the area is the product. When you convert a measure of area from square metres to square centimetres, each dimension is multiplied by 100, so the area is multiplied by , or Edmond doubled the width of the crate. How do the perimeter and the area of the floor of the new crate compare to those of the original crate? The width of the new crate is: m 1.6 m P ( w) 2 A w ( ) The perimeter of the floor of the new crate is 5.6 m and the area is 1.92 m 2. The perimeter of the new crate is 5.6 m 4 m, or 1.6 m greater than the original. The area of the new crate is 1.92 m 2, which is m 2. So, the area of the new crate is double the original area. Unit 9 Lesson 2 311
9 1. Which unit would you use to express each perimeter and area? Explain your choice. a) a park close to your home b) the cover of your notebook c) the library floor in your school d) your bedroom closet ceiling e) the top of a chalkbrush 2. Find the perimeter and the area of each rectangle. Write each measure in two different units. a) b) c) 31 m 44 cm 30 m 36 mm 28 mm 14 cm 3. For each area in question 2, which numbers are factors and which are products? 4. Rectangle A has area 40 cm 2 and length 8 cm. The area of rectangle B is one-half the area of rectangle A. The rectangles have the same length. What is the width of rectangle B? 5. Which has the greater area? Explain. a) a rectangle with area 0.84 m 2 b) a rectangle with area 8400 cm 2 Math Link Your World When we measure very large areas, we use square kilometres (km 2 ) or hectares (ha). One hectare is the area of a square with side length 100 m. 1 ha m m 1 ha 312 Unit 9 Lesson 2
10 6. A rectangle has dimensions 3 cm by 5 cm. What happens to the perimeter of the rectangle in each case? a) The width is doubled. b) The length is doubled. c) Both the length and the width are doubled. Show your work. 7. The perimeter of a square is 20 cm. What is the area of the square? How do you know? 8. Lena used 34 m of fencing to enclose a rectangular section of her backyard. What might the area of the enclosed section be? Give 3 different answers. 9. A square room has area 16 m 2. What is the perimeter of the room? 10. Pat is making rectangular placemats each with area 1200 cm 2. She wants to put braid trim around the edges. How much trim does Pat need for each placemat? Is there more than one answer? Explain. What happens to the perimeter and the area of a rectangle if the length and the width are multiplied by the same number? Use diagrams to explain. Numbers Every Day Number Strategies Find each answer. Order the answers from least to greatest ASSESSMENT FOCUS Question 6 Unit 9 Lesson 2 313
11 L E S S O N Christine plans to frame her Grade 4 saxophone certificate with a 5-cm wide mat. The certificate measures 30 cm by 25 cm. What is the area of the glass that covers Christine s certificate and the mat? Show and Share Share your work with another pair of students. Describe the strategy you used to solve the problem. Strategies Best Ride Sports sells snowboard equipment and clothing. The store measures 15 m by 13 m. The checkout desk measures 3 m by 1 m. A display shelf measures 11 m by 2 m. The floor around the desk and shelf is tiled. What area of the floor is tiled? What do you know? The store measures 15 m by 13 m. The checkout desk is 3 m by 1 m and the shelf is 11 m by 2 m. The floor that is not covered by the desk and shelf is tiled. Make a table. Use a model. Draw a diagram. Solve a simpler problem. Work backward. Guess and check. Make an organized list. Use a pattern. Draw a graph. Use logical reasoning. What strategy will you use to solve the problem? You could draw a diagram. Sketch the store, the desk, and the shelf. Mark the dimensions on the sketch. 314 LESSON FOCUS Interpret a problem and select an appropriate strategy.
12 Find the total area of the store. Then find the area covered by the desk and the shelf. Subtract the area covered by the desk and the shelf from the total floor area. Check your work. Is the sum of the tiled area and the areas covered by the desk and the shelf equal to the total area? 1. The Wongs are putting a flagstone deck around their pool. The pool is rectangular. Its dimensions are 8 m by 4 m. The deck will surround the pool. It will have a width of 2 m. a) What is the area of the deck? b) How much security fencing is required around the deck? Choose one of the Strategies 2. Scalene ABC has related side lengths. Side AB is 7 cm shorter than side BC. Side AC is 4 cm longer than side BC. The perimeter of ABC is 48 cm. What is the length of each side? Choose one of the problems in this lesson. Could you have solved the problem without drawing a diagram? Explain. Unit 9 Lesson 3 315
13 L E S S O N Area of a Parallelogram Which of these figures are parallelograms? How do you know? A B C D You will need scissors and 1-cm grid paper. Copy parallelogram A on grid paper. Estimate, then find, the area of the parallelogram. A B C Cut out the parallelogram. Cut along the broken line segment. Arrange the two pieces to form a rectangle. What is the area of the rectangle? How does the area of the rectangle compare to the area of the parallelogram? Repeat the activity for parallelograms B and C. Show and Share Share your work with another pair of students. Can every parallelogram be changed into a rectangle by cutting and moving one section? Explain. How is the area of a parallelogram related to the area of the rectangle? Explain. Numbers Every Day Number Strategies List 5 numbers that are divisible by both 4 and 6. What do the numbers have in common? 316 LESSON FOCUS Develop a formula to find the area of a parallelogram.
14 To estimate the area of this parallelogram, count the whole squares and the part squares that are one-half or greater. 8 5 There are: 33 whole squares 8 part squares that are one-half or greater The area of this parallelogram is about 41 square units. Any side of a parallelogram is a base. The height of the parallelogram is the length of a line segment that joins parallel sides and is perpendicular to the base. A rectangle is a parallelogram. The length is the base. The width is the height. The area of a parallelogram is the same as the area of a rectangle with the same base and height. height base length and base width and height 5 5 Use b for base. Use h for height. 8 8 Area Base Height Area Base Height A b h A b h The area of the rectangle The area of the parallelogram is 40 square units. is 40 square units. 1. Copy each parallelogram onto 1-cm grid paper. Draw a rectangle with the same base and height. a) b) c) Unit 9 Lesson 4 317
15 2. Estimate, then find the area of each parallelogram. a) b) c) 3. Find the area of each parallelogram. a) b) c) 24 cm 13 mm 16 km 24 km 13 cm 15 mm 4. For each parallelogram, find the length of the base or the height. a) Area 60 m 2 b) Area 6 mm 2 c) Area 30 cm 2 height 2 mm 5 cm base 12 m base 5. Which has the greatest area? The least area? a) a rectangle with base 10 cm and height 5 cm b) a parallelogram with base 7 cm and height 8 cm c) a square with sides 7 cm long 6. Sketch a parallelogram with area 24 square units. How many different ways can you do this? Explain. 7. A student says the area of this parallelogram is 20 cm 2. Explain the student s error. 4 cm 5 cm 8. Which parallelogram has the greater area? Explain. a) a parallelogram with base 15 cm and height 6 cm b) a parallelogram with base 15 cm and side length 6 cm 4 cm How is finding the area of a parallelogram like finding the area of a rectangle? How is it different? 318 ASSESSMENT FOCUS Question 6 Unit 9 Lesson 4
16 L E S S O N Exploring Triangles and Rectangles You will need scissors and 1-cm grid paper. Draw a rectangle on grid paper. Find and record the area of the rectangle. Draw a triangle inside the rectangle so that: At least one side of the triangle coincides with one side of the rectangle. The third vertex is on the opposite side of the rectangle. Shade the triangle. Estimate the area of the triangle. Cut out the rectangle. Then cut out the triangle. Arrange the unshaded pieces to cover the triangle. What is the area of the triangle? How do you know? Repeat the activity 2 more times. Use a different side of the rectangle for one side of the triangle. Use different points for the third vertex of the triangle. Show and Share Share your work with another student. Compare the area of the rectangle and the area of the triangle inside it. What do you notice? Explain. Any side of a triangle can be its base. The height of a triangle is the perpendicular line segment that joins the base to the opposite vertex. height base LESSON FOCUS Relate the area of a triangle to the area of a rectangle. 319
17 Here is one way to find the area of this triangle. 4 cm 15 cm Draw a rectangle with the same base and height. Each square represents 1 cm 2. I estimate the area of the triangle by counting squares and part squares. The area of this triangle is about 30 cm 2. 4 cm 15 cm Find the area of the rectangle. Area 15 cm 4 cm 60 cm 2 The pieces of the rectangle outside the triangle can be rearranged to make a congruent triangle. The areas of the congruent triangles are equal. So, the area of one triangle is one-half the area of the rectangle. One-half of 60 is 30. The area of the triangle is 30 cm 2. The area of a triangle is one-half the area of the rectangle with the same base and height. We can write a formula for the area of a triangle: Area (Base Height) 2 A (b h) 2 1. Estimate, then find, the area of each triangle. a) b) c) d) 320 Unit 9 Lesson 5
18 2. What is the area of the sail on Matthew s windsurfer? 3.5 m 2 m 3. Each triangle has area 12 m 2. Find each base or height. a) base b) c) height 4 m 12 m height 8 m 4. Use 0.5-cm grid paper. a) Draw as many different triangles as you can with area 16 cm 2. b) How do you know the triangles are different? c) Have you drawn all possible triangles? Explain. 5. Which triangle has the greater area? How do you know? a) Triangle A has base 13 m and height 5 m. b) Triangle B has base 9 m and height 8 m. 6. A rectangle and a triangle have the same area. The dimensions of the rectangle are 6 m by 8 m. The base of the triangle is 8 m. What is the height of the triangle? How do you know? 7. A triangle has area 21 cm 2. Its base is 7 cm long. Would all triangles with base 7 cm have area 21 cm 2? Explain. How are the areas of triangles and rectangles related? Use examples to explain. Numbers Every Day Number Strategies Write each decimal in words and in expanded form ASSESSMENT FOCUS Question 4 Unit 9 Lesson 5 321
19 L E S S O N Exploring Triangles and Parallelograms You will need scissors, tape, and 1-cm grid paper. B A C Make 2 copies of triangle A on grid paper. Cut out the triangles. Arrange the triangles to make a parallelogram. Find the area of the parallelogram. What is the area of triangle A? Repeat the activity for triangles B and C. Show and Share Share your work with another pair of students. Did you make the same parallelogram each time? Can two congruent triangles always be joined to make a parallelogram? Explain. How is the area of a triangle related to the area of the parallelogram? Explain. Numbers Every Day Number Strategies Write each measure in metres. 344 mm 1045 cm km 322 LESSON FOCUS Relate the area of a triangle to the area of a parallelogram.
20 Here is one way to find the area of this triangle. 4 cm 11 cm Draw a congruent triangle. Cut out both triangles. Arrange the triangles to make a parallelogram. 4 cm 11 cm Find the area of the parallelogram. A b h Two congruent triangles were used to make the parallelogram. So, the area of the triangle is one-half the area of the parallelogram. One-half of 44 is 22. The area of a triangle is one-half the area of the parallelogram with the same base and height. The area of the triangle is 22 cm 2. We can use the formula for the area of a triangle: Area (Base Height) 2 Or, A (b h) 2 Unit 9 Lesson 6 323
21 1. Copy each triangle on 1-cm grid paper. Draw a related parallelogram. a) b) c) 2. Find the area of each triangle. a) b) c) d) e) f) 3. Find the base or height of each triangle. a) Area 18 cm 2 b) Area 32 m 2 c) Area 480 mm 2 height 4 m base height 30 mm 9 cm 4. Mark is designing a skateboard ramp. He has sketched the side view. What is the area of this face of the ramp? 324 Unit 9 Lesson 6
22 5. Draw a parallelogram on 1-cm grid paper. Draw a diagonal to divide the parallelogram into 2 triangles. Find the area of each triangle. 6. Use 1-cm grid paper. Draw 3 different triangles with base 5 cm and height 6 cm. Find the area of each triangle. How do the areas compare? Explain. 7. Draw each triangle on 1-cm grid paper. Then find the area of the triangle. a) base 12 cm and height 4 cm b) base 6 cm and height 4 cm c) base 12 cm and height 2 cm d) base 6 cm and height 2 cm Compare the bases, the heights, and the areas of the 4 triangles. 8. Use 1-cm grid paper. How many different triangles can you draw that have base 5 cm and height 3 cm? Sketch each triangle. Find the area of each triangle. 9. Find the area of this triangle. a) Suppose the base is doubled. Explain what happens to the area. b) Suppose both the height and the base of the original triangle are doubled. Explain what happens to the area. c) What do you think happens to the area when both the height and the base are tripled? Explain. Show your work. Use examples to explain how the area of a triangle and the area of a parallelogram are related. ASSESSMENT FOCUS Question 9 Unit 9 Lesson 6 325
23 Show What You Know LESSON 1 1. Measure to find the perimeter of each polygon. Which unit did you use? Why? a) b) A B 2. Calculate the perimeter of a regular hexagon with side length 4.8 cm. Show your work Tania dug this rectangular garden. 3 m 8 m a) Find the area of the garden in square metres and in square centimetres. b) Find the perimeter in metres and centimetres For each parallelogram, identify a base and a height. Then find its area. a) b) c) 5. Which has the greater area? a) a rectangle with base 7 cm and height 3 cm b) a parallelogram with base 6 cm and height 4 cm 326 Unit 9
24 LESSON Sketch a rectangle with perimeter 18 cm that has the least area possible. The length and width are whole numbers of centimetres. What is the area of the rectangle? 7. A rectangle and a triangle have the same area. The rectangle has base 5 cm and height 4 cm. What might the base and height of the triangle be? Give two different answers. 8. Find the area of each triangle. a) b) c) 9. Sketch 3 different triangles with area 32 cm Some oil spilled on the ground during the loading of an oil truck. The clean-up crew marked a triangular section on the ground that contained the spill. The base of the triangular section is 12 m. The height is 5 m. What area of the ground was covered by the spill? U N I T Learning Goals measure and calculate perimeter estimate, measure, and calculate area explore how the areas of rectangles, parallelograms, and triangles are related develop formulas for the area of a parallelogram and the area of a triangle sketch polygons, given perimeter and area measurements Unit 9 327
25 Stationery Design You will need a ruler, coloured pencils or markers, 1-cm grid paper, scissors, and tape. 1. Design and create a greeting card and its envelope. On grid paper, record a pattern for the card and for the envelope. 2. What is the area of the card? The envelope? What is the perimeter of the card? The envelope? How are the areas of rectangles, parallelograms, and triangles related? How is the perimeter of a figure related to its side lengths? Use examples to explain. 328 Unit 9
26 Check List 3. Suppose you have a rectangular piece of paper that measures 28 cm by 43 cm. How many cards could you cut from one piece of paper? How many envelopes? 4. Write about your design. How did you choose the shape of your card? The envelope? How did you choose the dimensions of the card and figure out the dimensions of the envelope? Your work should show an appropriate greeting card design on grid paper the numbers of cards and envelopes that would fit on one piece of paper correct calculations of area and perimeter clear explanations of your design and your procedures Unit 9 329
### What You ll Learn. Why It s Important
These students are setting up a tent. How do the students know how to set up the tent? How is the shape of the tent created? How could students find the amount of material needed to make the tent? Why
### Length, perimeter and area 3.1. Example
3.1 Length, perimeter and area Kno and use the names and abbreviations for units of length and area Be able to measure and make a sensible estimate of length and area Find out and use the formula for the
### Convert between units of area and determine the scale factor of two similar figures.
CHAPTER 5 Units of Area c GOAL Convert between units of area and determine the scale factor of two. You will need a ruler centimetre grid paper a protractor a calculator Learn about the Math The area of
### Area and Perimeter. Practice: Find the perimeter of each. Square with side length of 6 cm. Rectangle with side lengths of 4 cm and 7 cm
Area and Perimeter Perimeter: add up all the sides (the outside of the polygon) Practice: Find the perimeter of each Square with side length of 6 cm Rectangle with side lengths of 4 cm and 7 cm Parallelogram
### *1. Derive formulas for the area of right triangles and parallelograms by comparing with the area of rectangles.
Students: 1. Students understand and compute volumes and areas of simple objects. *1. Derive formulas for the area of right triangles and parallelograms by comparing with the area of rectangles. Review
### Archdiocese of Washington Catholic Schools Academic Standards Mathematics
5 th GRADE Archdiocese of Washington Catholic Schools Standard 1 - Number Sense Students compute with whole numbers*, decimals, and fractions and understand the relationship among decimals, fractions,
### Identify relationships between and among linear and square metric units. area = length x width = 60 cm x 25 cm = 1500 cm 2
1 Unit Relationships Identify relationships between and among linear and square metric units. 1. Express each area in square centimetres. a) 8 m 2 80 000 cm 2 c) 3.5 m 2 35 000 cm 2 b) 12 m 2 120 000 cm
### Lesson 6. Unit 3. Building with Blocks. Area
Math 5 Lesson 6 Area Building with Blocks Lian is making a fort for her action figures out of building blocks. She has 42 cm blocks. She wants to have the largest area possible. How can she arrange her
### Math 6: Unit 7: Geometry Notes 2-Dimensional Figures
Math 6: Unit 7: Geometry Notes -Dimensional Figures Prep for 6.G.A.1 Classifying Polygons A polygon is defined as a closed geometric figure formed by connecting line segments endpoint to endpoint. Polygons
### Model area formulas for parallelograms, trapezoids, and triangles.
Answers Teacher Copy Lesson 23-3 Area of Triangles, Trapezoids, and Polygons Learning Targets p. 297 Model area formulas for parallelograms, trapezoids, and triangles. Write equations that represent problems
### CALCULATING PERIMETER. WHAT IS PERIMETER? Perimeter is the total length or distance around a figure.
CALCULATING PERIMETER WHAT IS PERIMETER? Perimeter is the total length or distance around a figure. HOW DO WE CALCULATE PERIMETER? The formula one can use to calculate perimeter depends on the type of
### Grade 7 Area of triangle
Grade 7 Area of triangle 7.SS.2 Develop and apply a formula for determining the area of triangles parallelograms circles 1. Illustrate and explain how the area of a rectangle can be used to determine the
### Working in 2 & 3 dimensions Revision Guide
Tips for Revising Working in 2 & 3 dimensions Make sure you know what you will be tested on. The main topics are listed below. The examples show you what to do. List the topics and plan a revision timetable.
### The area of a figure is the measure of the size of the region enclosed by the figure. Formulas for the area of common figures: square: A = s 2
The area of a figure is the measure of the size of the region enclosed by the figure. Formulas for the area of common figures: square: A = s 2 s s rectangle: A = l w parallelogram: A = b h h b triangle:
### Perimeter and Area. Chapter 11 11.1 INTRODUCTION 11.2 SQUARES AND RECTANGLES TRY THESE
PERIMETER AND AREA 205 Perimeter and Area Chapter 11 11.1 INTRODUCTION In Class VI, you have already learnt perimeters of plane figures and areas of squares and rectangles. Perimeter is the distance around
### GAP CLOSING. 2D Measurement. Intermediate / Senior Student Book
GAP CLOSING 2D Measurement Intermediate / Senior Student Book 2-D Measurement Diagnostic...3 Areas of Parallelograms, Triangles, and Trapezoids...6 Areas of Composite Shapes...14 Circumferences and Areas
### Squares and Square Roots
Squares and Square Roots Focus on After this lesson, you will be able to... determine the square of a whole number determine the square root of a perfect square Literacy Link A square number is the product
### Functional Skills Mathematics
Functional Skills Mathematics Level Learning Resource Perimeter and Area MSS1/L.7 Contents Perimeter and Circumference MSS1/L.7 Pages 3-6 Finding the Area of Regular Shapes MSS1/L.7 Page 7-10 Finding the
### CHAPTER. Symmetry and Surface Area GET READY 2. Math Link Warm Up Line Symmetry Warm Up 15
CHAPTER 1 Symmetry and Surface Area GET READY Math Link 4 1.1 Warm Up 5 1.1 Line Symmetry 6 1. Warm Up 15 1. Rotation Symmetry and Transformations 16 1.3 Warm Up 5 1.3 Surface Area 6 Graphic Organizer
### 43 Perimeter and Area
43 Perimeter and Area Perimeters of figures are encountered in real life situations. For example, one might want to know what length of fence will enclose a rectangular field. In this section we will study
### Areas of Polygons. Goal. At-Home Help. 1. A hockey team chose this logo for their uniforms.
-NEM-WBAns-CH // : PM Page Areas of Polygons Estimate and measure the area of polygons.. A hockey team chose this logo for their uniforms. A grid is like an area ruler. Each full square on the grid has
### Grade 9 Mathematics Unit 3: Shape and Space Sub Unit #1: Surface Area. Determine the area of various shapes Circumference
1 P a g e Grade 9 Mathematics Unit 3: Shape and Space Sub Unit #1: Surface Area Lesson Topic I Can 1 Area, Perimeter, and Determine the area of various shapes Circumference Determine the perimeter of various
5 th Grade Mathematics Instructional Week 20 Rectilinear area with fractional side lengths and real-world problems involving area and perimeter of 2-dimensional shapes Paced Standards: 5.M.2: Find the
### b = base h = height Area is the number of square units that make up the inside of the shape is a square with a side length of 1 of any unit
Area is the number of square units that make up the inside of the shape of 1 of any unit is a square with a side length Jan 29-7:58 AM b = base h = height Jan 29-8:31 AM 1 Example 6 in Jan 29-8:33 AM A
### Covering & Surrounding Notes Problem 1.1
Problem 1.1 Definitions: Area - the measure of the amount of surface enclosed by the sides of a figure. Perimeter - the measure of the distance around a figure. Bumper-cars is one of the most popular rides
### Lesson 21. Chapter 2: Perimeter, Area & Volume. Lengths and Areas of Rectangles, Triangles and Composite Shapes
ourse: HV Lesson hapter : Perimeter, rea & Volume Lengths and reas of Rectangles, Triangles and omposite Shapes The perimeter of a shape is the total length of its boundary. You can find the perimeter
### Developing Conceptual Understanding of Number. Set J: Perimeter and Area
Developing Conceptual Understanding of Number Set J: Perimeter and Area Carole Bilyk cbilyk@gov.mb.ca Wayne Watt wwatt@mts.net Perimeter and Area Vocabulary perimeter area centimetres right angle Notes
### PERIMETERS AND AREAS
PERIMETERS AND AREAS 1. PERIMETER OF POLYGONS The Perimeter of a polygon is the distance around the outside of the polygon. It is the sum of the lengths of all the sides. Examples: The perimeter of this
Grade 6 Rectangle area 6.SS.3 Develop and apply a formula for determining the perimeter of polygons area of rectangles volume of right rectangular prisms 1. Explain, using models, how the perimeter of
### Volume of Right Prisms Objective To provide experiences with using a formula for the volume of right prisms.
Volume of Right Prisms Objective To provide experiences with using a formula for the volume of right prisms. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game
### MODULE FRAMEWORK EN ASSESSMENT SHEET
MODULE FRAMEWORK EN ASSESSMENT SHEET LEARNING OUTCOMES (LOS) ASSESSMENT STANDARDS (ASS) FORMATIVE ASSESSMENT ASs Pages and (mark out of 4) LOs (ave. out of 4) SUMMATIVE ASSESSMENT Tasks or tests Ave for
### CHAPTER 27 AREAS OF COMMON SHAPES
EXERCISE 113 Page 65 CHAPTER 7 AREAS OF COMMON SHAPES 1. Find the angles p and q in the diagram below: p = 180 75 = 105 (interior opposite angles of a parallelogram are equal) q = 180 105 0 = 35. Find
### Polygon of the Day Discovering Area Formulas
Polygon of the Day Discovering Area Formulas Utilizing Technology and Manipulatives Cuisenaire pop-cubes Geometry tiles GSP triangles tool Trapezoid magnets Sixth Grade 6-day Unit Plan (40 minute classes)
### Perimeter, Area, and Volume
Perimeter is a measurement of length. It is the distance around something. We use perimeter when building a fence around a yard or any place that needs to be enclosed. In that case, we would measure the
### Covering and Surrounding: Homework Examples from ACE
Covering and Surrounding: Homework Examples from ACE Investigation 1: Extending and Building on Area and Perimeter, ACE #4, #6, #17 Investigation 2: Measuring Triangles, ACE #4, #9, #12 Investigation 3:
### Formulas for Area Area of Trapezoid
Area of Triangle Formulas for Area Area of Trapezoid Area of Parallelograms Use the formula sheet and what you know about area to solve the following problems. Find the area. 5 feet 6 feet 4 feet 8.5 feet
### Unit 8 Angles, 2D and 3D shapes, perimeter and area
Unit 8 Angles, 2D and 3D shapes, perimeter and area Five daily lessons Year 6 Spring term Recognise and estimate angles. Use a protractor to measure and draw acute and obtuse angles to Page 111 the nearest
### ME7-18, ME7-19, ME7-20, ME7-21
1 Measurement Grade 7 Part 1 Introduction In this unit, students will determine the relationships between units of length, area, volume, and capacity. They will develop and use formulas for the areas of
### AUTUMN UNIT 3. first half. Perimeter. Centimetres and millimetres. Metres and centimetres. Area. 3D shapes PART 3 MEASURES AND PROPERTIES OF SHAPES
PART AUTUMN first half MEASURES AND PROPERTIES OF SHAPES SECTION Perimeter SECTION Centimetres and millimetres SECTION Metres and centimetres SECTION Key Stage National Strategy CROWN COPYRIGHT 00 Area
### MATH STUDENT BOOK. 6th Grade Unit 8
MATH STUDENT BOOK 6th Grade Unit 8 Unit 8 Geometry and Measurement MATH 608 Geometry and Measurement INTRODUCTION 3 1. PLANE FIGURES 5 PERIMETER 5 AREA OF PARALLELOGRAMS 11 AREA OF TRIANGLES 17 AREA OF
### Integrated Algebra: Geometry
Integrated Algebra: Geometry Topics of Study: o Perimeter and Circumference o Area Shaded Area Composite Area o Volume o Surface Area o Relative Error Links to Useful Websites & Videos: o Perimeter and
### Sect 8.3 Quadrilaterals, Perimeter, and Area
186 Sect 8.3 Quadrilaterals, Perimeter, and Area Objective a: Quadrilaterals Parallelogram Rectangle Square Rhombus Trapezoid A B E F I J M N Q R C D AB CD AC BD AB = CD AC = BD m A = m D m B = m C G H
### G6-9 Area of Composite Shapes
G6-9 Area of Composite Shapes 1. a) Calculate the area of each figure. b) Draw a line to show how Shape C can be divided into rectangles A and. i) ii) A C A C Area of A = Area of A = Area of = Area of
### Surface Area of a Prism
Surface Area of a Prism Focus on After this lesson, you will be able to... φ link area to φ surface area find the surface area of a right prism Most products come in some sort of packaging. You can help
### 2.4. Determining Square Roots of Rational Numbers. Explore Square Roots of Rational Numbers. Focus on
2.4 Focus on After this lesson, you will be able to determine the square root of a perfect square rational number determine an approximate square root of a non-perfect square rational number Determining
### Circumference and area of a circle
c Circumference and area of a circle 22 CHAPTER 22.1 Circumference of a circle The circumference is the special name of the perimeter of a circle, that is, the distance all around it. Measure the circumference
### Family Letters. Assessment Management. Playing Fraction Of
Formula for the Area of a Parallelogram Objectives To review the properties of parallelograms; and to guide the development and use of a formula for the area of a parallelogram. www.everydaymathonline.com
### Imperial Measures of Length
SI system of measures: Imperial Measures of Length is an abbreviation for a system of units based on powers of 10; the fundamental unit: of length is the ; of mass is the ; and of time is the. Imperial
### PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures.
PERIMETER AND AREA In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. Perimeter Perimeter The perimeter of a polygon, denoted by P, is the
### Unit 2, Activity 2, One-half Inch Grid
Unit 2, Activity 2, One-half Inch Grid Blackline Masters, Mathematics, Grade 8 Page 2-1 Unit 2, Activity 2, Shapes Cut these yellow shapes out for each pair of students prior to class. The labels for each
Geometry Progress Ladder Maths Makes Sense Foundation End-of-year objectives page 2 Maths Makes Sense 1 2 End-of-block objectives page 3 Maths Makes Sense 3 4 End-of-block objectives page 4 Maths Makes
### Shape, space and measures 1
Shape, space and measures 1 contents There are three lessons in this unit, Shape, space and measures 1. S1.1 Lines, length and perimeter 3 S1.2 Area of a rectangle 6 S1.3 Solving problems 9 Resource sheets
### Signs, Signs, Every Place There Are Signs! Area of Regular Polygons p. 171 Boundary Lines Area of Parallelograms and Triangles p.
C H A P T E R Perimeter and Area Regatta is another word for boat race. In sailing regattas, sailboats compete on courses defined by marks or buoys. These courses often start and end at the same mark,
### Inv 1 5. Draw 2 different shapes, each with an area of 15 square units and perimeter of 16 units.
Covering and Surrounding: Homework Examples from ACE Investigation 1: Questions 5, 8, 21 Investigation 2: Questions 6, 7, 11, 27 Investigation 3: Questions 6, 8, 11 Investigation 5: Questions 15, 26 ACE
### Surface Area and Volume
1 Area Surface Area and Volume 8 th Grade 10 days by Jackie Gerwitz-Dunn and Linda Kelly 2 What do you want the students to understand at the end of this lesson? The students should be able to distinguish
### Geometry: Chapter Questions. 1. How is the formula for area of a parallelogram related to area of a rectangle?
Geometry: Chapter Questions. How is the formula for area of a parallelogram related to area of a rectangle?. How is the formula for area of a triangle related to area of a rectangle?. How do you find the
### Geometry Non Calculator
Geometry Non Calculator Revision Pack 35 minutes 35 marks To use alongside mymaths.co.uk and livemaths.co.uk to revise for your GCSE exam Page 1 of 14 Q1. Diagram NOT accurately drawn Work out the size
### Chapter 6. Similarity of Figures
Chapter 6 Similarity of Figures 6.1 Similar Polygons 6.2 Determining if two Polygons are Similar 6.3 Drawing Similar Polygons 6.4 Similar Triangles 21 Name: 6.1 Similar Polygons A. What makes something
### Lateral and Surface Area of Right Prisms
CHAPTER A Lateral and Surface Area of Right Prisms c GOAL Calculate lateral area and surface area of right prisms. You will need a ruler a calculator Learn about the Math A prism is a polyhedron (solid
### Area of Parallelograms and Triangles
Context: Grade 7 Mathematics. Area of Parallelograms and Triangles Objective: The student shall derive the formulas for the area of the triangle and the parallelogram and apply these formulas to determine
### GAP CLOSING. 2D Measurement GAP CLOSING. Intermeditate / Senior Facilitator s Guide. 2D Measurement
GAP CLOSING 2D Measurement GAP CLOSING 2D Measurement Intermeditate / Senior Facilitator s Guide 2-D Measurement Diagnostic...4 Administer the diagnostic...4 Using diagnostic results to personalize interventions...4
### 11-4 Areas of Regular Polygons and Composite Figures
1. In the figure, square ABDC is inscribed in F. Identify the center, a radius, an apothem, and a central angle of the polygon. Then find the measure of a central angle. Center: point F, radius:, apothem:,
### Sum of the interior angles of a n-sided Polygon = (n-2) 180
5.1 Interior angles of a polygon Sides 3 4 5 6 n Number of Triangles 1 Sum of interiorangles 180 Sum of the interior angles of a n-sided Polygon = (n-2) 180 What you need to know: How to use the formula
### Q1. Lindy has 4 triangles, all the same size. She uses them to make a star. Calculate the perimeter of the star. 2 marks.
Q1. Lindy has 4 triangles, all the same size. She uses them to make a star. Calculate the perimeter of the star. Page 1 of 16 Q2. Liam has two rectangular tiles like this. He makes this L shape. What is
### Geometry Chapter 9 Extending Perimeter, Circumference, and Area
Geometry Chapter 9 Extending Perimeter, Circumference, and Area Lesson 1 Developing Formulas for Triangles and Quadrilaterals Learning Target (LT-1) Solve problems involving the perimeter and area of triangles
### TEKS TAKS 2010 STAAR RELEASED ITEM STAAR MODIFIED RELEASED ITEM
7 th Grade Math TAKS-STAAR-STAAR-M Comparison Spacing has been deleted and graphics minimized to fit table. (1) Number, operation, and quantitative reasoning. The student represents and uses numbers in
### 5E Lesson Plan Perimeter and Area
5E Lesson Plan Perimeter and Area teachhouston Student Name: Mentor Teacher Name: Grade Level: 4th Lesson Teaching Date: Concept(s): Perimeter is the distance around the outside of a shape. Area is the
### You can use the postulates below to prove several theorems.
Using Area Formulas You can use the postulates below to prove several theorems. AREA POSTULATES Postulate Area of a Square Postulate The area of a square is the square of the length of its side, or s.
### Formula for the Area of a Triangle Objective To guide the development and use of a formula for the area of a triangle.
Formula for the rea of a Triangle Objective To guide the development and use of a formula for the area of a triangle. www.everydaymathonline.com epresentations etoolkit lgorithms Practice EM Facts Workshop
### A. 32 cu ft B. 49 cu ft C. 57 cu ft D. 1,145 cu ft. F. 96 sq in. G. 136 sq in. H. 192 sq in. J. 272 sq in. 5 in
7.5 The student will a) describe volume and surface area of cylinders; b) solve practical problems involving the volume and surface area of rectangular prisms and cylinders; and c) describe how changing
Unit 6: POLYGONS Are these polygons? Justify your answer by explaining WHY or WHY NOT??? a) b) c) Yes or No Why/Why not? Yes or No Why/Why not? Yes or No Why/Why not? a) What is a CONCAVE polygon? Use
### WHAT IS AREA? CFE 3319V
WHAT IS AREA? CFE 3319V OPEN CAPTIONED ALLIED VIDEO CORPORATION 1992 Grade Levels: 5-9 17 minutes DESCRIPTION What is area? Lesson One defines and clarifies what area means and also teaches the concept
### CONNECT: Volume, Surface Area
CONNECT: Volume, Surface Area 1. VOLUMES OF SOLIDS A solid is a three-dimensional (3D) object, that is, it has length, width and height. One of these dimensions is sometimes called thickness or depth.
### = = 4 + = 4 + = 25 = 5
1 4 1 4 1 4 = 4 6+ 1 = 4 + = 4 + = 25 = 5 5. Find the side length of each square as a square root. Then estimate the square root. A B C D A side length B side length C side length D side length = 4 2+
### Three daily lessons. Year 5
Unit 6 Perimeter, co-ordinates Three daily lessons Year 4 Autumn term Unit Objectives Year 4 Measure and calculate the perimeter of rectangles and other Page 96 simple shapes using standard units. Suggest
### 10.1: Areas of Parallelograms and Triangles
10.1: Areas of Parallelograms and Triangles Important Vocabulary: By the end of this lesson, you should be able to define these terms: Base of a Parallelogram, Altitude of a Parallelogram, Height of a
### Lesson 6. Identifying Prisms. Daksha and his brother and sister are playing a dice game. When a die is rolled, one number is displayed on the top.
Math 4 Lesson 6 Identifying Prisms Dice Games Daksha and his brother and sister are playing a dice game. When a die is rolled, one number is displayed on the top. When you roll the dice they will only
### Postulate 17 The area of a square is the square of the length of a. Postulate 18 If two figures are congruent, then they have the same.
Chapter 11: Areas of Plane Figures (page 422) 11-1: Areas of Rectangles (page 423) Rectangle Rectangular Region Area is measured in units. Postulate 17 The area of a square is the square of the length
### Geometry. Geometry is the study of shapes and sizes. The next few pages will review some basic geometry facts. Enjoy the short lesson on geometry.
Geometry Introduction: We live in a world of shapes and figures. Objects around us have length, width and height. They also occupy space. On the job, many times people make decision about what they know
### CONNECT: Areas, Perimeters
CONNECT: Areas, Perimeters 1. AREAS OF PLANE SHAPES A plane figure or shape is a two-dimensional, flat shape. Here are 3 plane shapes: All of them have two dimensions that we usually call length and width
### SGS4.3 Stage 4 Space & Geometry Part A Activity 2-4
SGS4.3 Stage 4 Space & Geometry Part A Activity 2-4 Exploring triangles Resources required: Each pair students will need: 1 container (eg. a rectangular plastic takeaway container) 5 long pipe cleaners
### Study Guide. 6.g.1 Find the area of triangles, quadrilaterals, and other polygons. Note: Figure is not drawn to scale.
Study Guide Name Test date 6.g.1 Find the area of triangles, quadrilaterals, and other polygons. 1. Note: Figure is not drawn to scale. If x = 14 units and h = 6 units, then what is the area of the triangle
### SHAPE, SPACE AND MEASURES
SHPE, SPCE ND MESURES Pupils should be taught to: Understand and use the language and notation associated with reflections, translations and rotations s outcomes, Year 7 pupils should, for example: Use,
### Pythagorean Theorem. Inquiry Based Unit Plan
Pythagorean Theorem Inquiry Based Unit Plan By: Renee Carey Grade: 8 Time: 5 days Tools: Geoboards, Calculators, Computers (Geometer s Sketchpad), Overhead projector, Pythagorean squares and triangle manipulatives,
### Activity Set 3. Trainer Guide
Geometry and Measurement of Plane Figures Activity Set 3 Trainer Guide GEOMETRY AND MEASUREMENT OF PLANE FIGURES Activity Set 3 Copyright by the McGraw-Hill Companies McGraw-Hill Professional Development
### CC Investigation 4: Measurement
Content Standards 6.G., 6.G. CC Investigation : Measurement Mathematical Goals At a Glance PACING 3 days DOMAIN: Geometry Find the volume of a right rectangular prism with fractional edge lengths by packing
### Shape and Space. General Curriculum Outcome E: Students will demonstrate spatial sense and apply geometric concepts, properties and relationships.
Shape and Space General Curriculum Outcome E: Students will demonstrate spatial sense and apply geometric concepts, properties and relationships. Elaboration Instructional Strategies/Suggestions KSCO:
### Dyffryn School Ysgol Y Dyffryn Mathematics Faculty
Dyffryn School Ysgol Y Dyffryn Mathematics Faculty Formulae and Facts Booklet Higher Tier Number Facts Sum This means add. Difference This means take away. Product This means multiply. Share This means
### Area of Parallelograms, Triangles, and Trapezoids (pages 314 318)
Area of Parallelograms, Triangles, and Trapezoids (pages 34 38) Any side of a parallelogram or triangle can be used as a base. The altitude of a parallelogram is a line segment perpendicular to the base
### MATH STUDENT BOOK. 7th Grade Unit 9
MATH STUDENT BOOK 7th Grade Unit 9 Unit 9 Measurement and Area Math 709 Measurement and Area Introduction 3 1. Perimeter 5 Perimeter 5 Circumference 11 Composite Figures 16 Self Test 1: Perimeter 24 2.
### 3. If AC = 12, CD = 9 and BE = 3, find the area of trapezoid BCDE. (Mathcounts Handbooks)
EXERCISES: Triangles 1 1. The perimeter of an equilateral triangle is units. How many units are in the length 27 of one side? (Mathcounts Competitions) 2. In the figure shown, AC = 4, CE = 5, DE = 3, and
### Geometry Chapter 9 Extending Perimeter, Circumference, and Area
Geometry Chapter 9 Extending Perimeter, Circumference, and Area Lesson 1 Developing Formulas for Triangles and Quadrilaterals Learning Targets LT9-1: Solve problems involving the perimeter and area of
### Build your skills: Perimeter and area Part 1. Working out the perimeter and area of different shapes
Working out the perimeter and area of different shapes This task has two parts. Part 1 In this part, you can brush up your skills and find out about perimeter and area. Part 2 In the second part, you can
### Squares and Square Roots
Squares and Square Roots Focus on After this lesson, you will be able to... φ determine the φ square of a whole number determine the square root of a perfect square The Pythagoreans were members of an
### Algebra Geometry Glossary. 90 angle
lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example:
### Measurement. Learning Goals U N I T
At U N I T Measurement the Zoo Learning Goals measure length in millimetres select referents for units of measure relate units of measure draw different rectangles for a given perimeter or area estimate
### What You ll Learn. Why It s Important
What is a circle? Where do you see circles? What do you know about a circle? What might be useful to know about a circle? What You ll Learn Measure the radius, diameter, and circumference of a circle. | 12,029 | 47,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-47 | latest | en | 0.852952 |
https://www.coursehero.com/file/5973841/Onesampletestsconfidint/ | 1,498,569,463,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321410.90/warc/CC-MAIN-20170627115753-20170627135753-00061.warc.gz | 827,266,871 | 24,753 | Onesampletestsconfidint
# Onesampletestsconfidint - PSY 2801 Summer 2010 One Sample...
This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: PSY 2801: Summer 2010 One Sample Tests/Confidence Intervals Jeff Jones University of Minnesota Jeff Jones Hypothesis Tests for Means The first several statistical tests have the same form: H : μ = μ 1 H 1 : μ 6 = μ 1 That is for the two-sided test. We want to test whether a mean is equal or not equal to a hypothesized value. This is why we call these tests “one sample test” -- we are making inferences about the population value from one group of people , and we are testing this against a known μ parameter . Jeff Jones Hypothesis Tests for Means So, for instance: 1 Is the mean IQ for women equal to 105? 2 Is the average height of men equal to 5 feet 8 inches? 3 Do 2009 U of M Psych freshmen spend 4 hours, on average, on homework each night? 4 Is the average number of minutes bored (in my lectures) 90 minutes? So for each of those, we are only estimating the mean of one group, and we know the population mean to which we are comparing that group. Jeff Jones Notes Notes Notes Steps for One Sample Hypothesis Tests Here are the basic steps for a one sample hypothesis test: 1 Form null and alternative hypothesis (as always). 2 Choose α level (as always). 3 Choose the test to carry out 4 Divide the mean difference by the standard error to obtain the test statistic (perform the test) 5 Check the probability of the test statistic occurring (given what?) 6 Choose to reject or not reject H . Jeff Jones z-score Revisited Remember the z-score? z i = x i- μ σ If x i comes from a Normal Distribution, and we perform this operation on every x i , then the z i ’s have a Standard Normal Distribution. In this case, we can look in the z-table to find the probability of scoring greater than this x-- it’s just the probability or area greater than this z value in the z-table. Jeff Jones z-test Now, instead of wanting something like the probability of scoring a value greater than or equal to an x , we want to find something like the probability of scoring a sample mean greater than or equal to an ¯ x . So, by the CLT: if x is Normally Distributed, ¯ x is Normally Distributed with a mean of μ and a standard deviation of σ √ N . So, now ¯ x is our observation , and instead of looking at a distribution of x , we’re looking at a distribution of ¯ x . Jeff Jones Notes Notes Notes z-test This leads to the natural extension of the z-score, which is the z-test (or a z-score for means and NOT individual values): z = ¯ x- μ σ √ N This equation is a specific example of most of the parametric inferential statistics we will encounter in this class: test statistic = statistic- parameter standard error Jeff Jones Pieces of Information The pieces of information you need to perform a z-test: 1 A sample mean ( ¯ x ) 2 The number of people in your sample ( N ) Both of the above you can get from the data that you collect -- they are data specific pieces of information....
View Full Document
## This note was uploaded on 10/08/2010 for the course PSY 2801 taught by Professor Guyer during the Summer '08 term at Minnesota.
### Page1 / 17
Onesampletestsconfidint - PSY 2801 Summer 2010 One Sample...
This preview shows document pages 1 - 4. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 877 | 3,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.879412 |
https://discuss.pytorch.org/t/linear-algebra-question-inverse/87058 | 1,653,828,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00284.warc.gz | 256,693,910 | 7,590 | # [LINEAR ALGEBRA] [QUESTION] Inverse
Quick question about an implementation choice (didn’t feel appropriate to raise on github)
Psuedo-inverses are proper inverses when a matrix is square + full rank.
It seems like the implementation in torch isn’t doing anything to discover rank and so can’t fall back on classical inverse algorithms for efficiency.
I am playing with this example:
``````import torch
nrows = 10000
ncols = 10000
torch.inverse(torch.rand(nrows,ncols).cuda())
``````
``````
CPU times: user 3.8 s, sys: 141 ms, total: 3.94 s
Wall time: 927 ms
``````
swapping that last line for `torch.pinverse(torch.rand(nrows,ncols).cuda())` yields
``````CPU times: user 2min 17s, sys: 1.95 s, total: 2min 19s
Wall time: 20.7 s
``````
So, even with added overhead to discover rank with something like Cholesky (for which it seems an implementation exists), large speedups can be gained.
What’s going on under the hood with pseudo-inverse and why is it taking so much longer? mat-mult is sub-second, classical inversion for full-rank operators is sub-second, so even manual-construction wins out:
``````A = torch.rand(nrows,ncols).cuda()
torch.mm(torch.inverse(torch.mm(A.t(), A)), A.t())
``````
``````CPU times: user 3.67 s, sys: 167 ms, total: 3.83 s
Wall time: 1.09 s
``````
Moreover, sometimes `pinverse` runs out of memory, but the latter won’t (a kernel restart allows `pinverse` to run again), and the solutions do seem to agree on average:
Hey,
One thing to be careful of here is that the CUDA api is asynchronous. So without the appropriate `torch.cuda.synchronize()` calls, you shoud not read too much into these timings.
Still you can see here that we use LU factorization for the inverse and delegate all the heavy lifting to magma.
The pinverse is here on the other hand and uses svd (done by magma as well). Note that it needs to handle batches of inputs.
Quick question about an implementation choice
Definitely open to discussions!
1 Like
thanks so much for the clarification. I’m a maths person who has only used CPUs to date, finally got a GPU to play around with. So, I’m not exactly sure where those calls should go.
The differences in algos though does explain a lot of the difference between my one-liner and the psuedo-inverse method; I really appreciate the linking to the source files, as it’s a large library to sort through!
A cholesky-based inverse may provide some major gains; c.f matlab’s benchmarks in this (admittedly decade old, but still relevant I think) paper: https://arxiv.org/pdf/0804.4809.pdf
correct me if I’m wrong, but as far as I understand, deep-learning networks induce operators that are very ill-conditioned. So, my assumption is that `pinverse` is a often-called function during training. Is that the case? If not, then I’m not so sure how helpful work on improving it would be.
I wouldn’t say it is a very commonly used operator in general for neural nets. But it is useful and we do want make sure we’re not making obvious mistake here.
That being said, advanced users do have access to all the “low level” functions like lu, svd, getr etc and can use the one they know fits best their use case.
Considering that we have both usecases, what would be the overhead here of checking the rank for people that have known non-invertible matrices?
good question. out of curiosity, how does the development team think of overhead? Would numerical tests suffice (perhaps calling `.cpu()` to collect results at the end of my operations to account for the async?), or are you looking for a Big-O analysis for rank reveal + LU over straight SVD? I’d need to look into the specifics of the magma implementation for that (well, frankly, I have some friends in the department who’d be better suited for that).
perhaps calling `.cpu()` to collect results at the end of my operations to account for the async?
You can do something like:
``````torch.cuda.synchronize()
start = time.time()
# your_op
torch.cuda.synchronize()
elapsed = time.time() - start
``````
how does the development team think of overhead?
I am not working on perfs explicitly so the definitive answer will definitely be given by the relevant people on github
But even though the Big-O notation is a good hint, I think actual runtime will be the most important.
Also such choice can depend on the input size, batch size, etc.
cc @VitalyFedyunin what do you think?
Thanks!
Pretty similar benchmark.
Card: RTX2080S, don’t know how this may differ on better hadware. But the stark difference is a clue for an opportunity to improve performance out of the box. I can do some more numerical studies and see if I can get insight to the theoretical cost question
That sounds good.
After a quick offline chat with Vitaly, I think we would be happy to add some heuristics here.
Unfortunately, my linear algebra is a bit rusty but if you could open an issue on github with some details on the heuristic you want to use (and a quick idea why). And a script that can be used to get a first idea of the impact on runtime both for full rank, not full rank matrices and ill conditioned matrices.
I think we can then build on that to make sure that the general penalty in time is small enough for the benefit we get.
1 Like
You can quite update pinverse implementation as it is easy to extract rank from `svd` calls results here https://github.com/pytorch/pytorch/blob/6debc28964872e26724764e213c40db366327f95/aten/src/ATen/native/LinearAlgebra.cpp#L94
that sounds good. if we can agree on a good benchmarking test, I can devote a little time to experimenting with it.
while I don’t often code in c++, I can totally follow along with that snippet based on the math (also a testament to well-written code).
however, while SVD is rank-revealing, I think what I’d propose is to avoid the cost of computing it altogether potentially, since that seems to be the bottleneck. My demo was just on full-rank operators, and that SVD implementation you linked is more stable for ill-conditioned matrices. | 1,435 | 6,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.889769 |
https://datascience.stackexchange.com/questions/11167/algorithm-suggestion-for-a-specific-problem | 1,708,912,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00251.warc.gz | 204,379,096 | 41,469 | # Algorithm Suggestion For a Specific Problem
I'm working on a problem where in I have some data sets about some power generating units. Each of these units have been activated to run in the past and while activation, some units went into some issues. I now have all these data and I would like to come up with some sort of Ranking for these generating units. The criteria for ranking would be pretty simple to start with. They are:
1. Maximum number of times a particular generating unit was activated
2. How many times did the generating unit ran into problems during activation
Later on I would expand on this ranking algorithm by adding more criteria. I will be using Apache Spark MLIB library and I can already see that there are quite a few algorithms already in place.
http://spark.apache.org/docs/latest/mllib-guide.html
I'm just not sure which algorithm would fit my purpose. Any suggestions?
• It sounds like you are not trying to predict anything but rather characterize the performance. True?
– Pete
Apr 13, 2016 at 14:28
• Yes, I just want to use the dense data and come up with a ranking order for all the power generating units. Apr 13, 2016 at 17:39
• Do you want an exact count of number of activations etc. or approximate counts? If approximate then there are interesting algorithms if exact then it's just a group by operation for which you can look up the algo in spark Apr 19, 2016 at 14:13
You can use a clustering algorithm such as k-means to divide the generators into groups. You never know what kind of groups you'll get until you try it. Try and assess the character of each group of generators as you increase the number of clusters. At some point you should find a meaningful division of generators. The inputs to your k-means algorithm will be the criteria you mentioned in your post: the number of times it was activated, the number of activation problems, and so forth. When you are finished, the group a generator belongs to is its ranking. This method will not generate a ranking of 1-1000 if you have 1000 generators. Rather it will give you, for example with k=3: a group of 243 outstanding generators, 320 average generators, and 446 terrible generators.
• Cool! That seems to be a good starting point. I will explore more on it! Thanks for the suggestion. Apr 15, 2016 at 5:05
With a few exceptions, you can pretty much use any machine learning algorithm for your model. The beauty of most machine learning packages is that the interface for each model is mostly the same (although the tuning parameters will differ), and it takes just a few lines of code to try out each model. There is no reason you should artificially constrain yourself to trying certain models.
Some exceptions to this rule are algorithms that might only work for classification or only work for regression. It sounds like you're trying to predict a continuous target variable that you'll then use for ranking. If that's the case, then you won't be able to use an algorithm called Naive Bayes because it can only output probabilities. In other rare cases like Deep Learning models, the run time can be very long (hours or days) and in those cases you wouldn't want to use an algorithm like that unless you had a good reason to do so (e.g., face recognition in images). You should be able to use nearly every algorithm in MLlib though: gradient boosting, random forests, etc.
• Thanks for the post. Could you point me to some examples? Apr 12, 2016 at 14:24
• Sure. Are you looking for MLlib examples specifically, or any type of machine learning library? What's your preferred language and what is the file size of your data? Apr 12, 2016 at 14:29
• Yes, I'm only looking for MLib examples and I use Scala. I want to do this on a historical data set that would be say a hundred of MB in size to start with. Apr 12, 2016 at 14:31
• Unfortunately I haven't personally used MLlib in Scala, only python, so maybe another poster can show you some of their own code (I would just end up pulling something from Google). Also I think you're trying to say MLlib instead of Mlib. There is supposed to be an extra L in there so that when fully expanded it shows: Machine Learning (ML) library (lib): MLlib Apr 12, 2016 at 14:38
• I'm referring to the Spark-MLib library. Sorry for not being explicit! Apr 12, 2016 at 14:40 | 987 | 4,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.961586 |
https://www.assignmentessays.com/stats-a-demographer-using-a-random-sample-of-n-500-people/ | 1,596,825,378,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737206.16/warc/CC-MAIN-20200807172851-20200807202851-00090.warc.gz | 593,015,044 | 10,484 | # STATS – A demographer, using a random sample of n = 500 people
December 12, 2016
Question
A demographer, using a random sample of n = 500 people, obtained a 95 percent confidence interval for mean age at marriage (m) in years for US adults. The CI was (26.4, 27.3). Suppose another demographer obtained a separate independent sample of people and also constructed a 95% confidence interval for the same quantity m, what can we say about this new interval as compared to the previous one?
narrower and would be less likely to contain m.
wider and would be more likely to contain m.
narrower and would be more likely to contain m.
It may be wider or narrower, but it has same chance to contain m.
narrower but we can’t be sure if it would be more or less likely to contain m.
Order your essay today and save 20% with the discount code: ESSAYHELP | 211 | 850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.958511 |
http://mathematica.stackexchange.com/users/50/j-m?tab=activity | 1,461,903,725,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110372.12/warc/CC-MAIN-20160428161510-00123-ip-10-239-7-51.ec2.internal.warc.gz | 133,507,042 | 15,552 | J. M.
Reputation
56,855
146/100 score
Apr 26 comment NSolve won't act on very large powers @Jeff, did you try increasing the setting of WorkingPrecision instead? Apr 26 revised Wolfram Cloud and reCAPTCHA edited tags Apr 26 revised How dangerous is 'Overflow occurred in computation' message? edited tags Apr 26 comment How dangerous is 'Overflow occurred in computation' message? …and by sharing your code, we just might be able to propose reformulations that avoid your overflow problems altogether. Apr 26 revised Kernel crashes when computing finite difference mixed derivative with respect to y & z but works fine when computing with respect to x & y or x & z? edited tags Apr 26 comment Wolfram Cloud and reCAPTCHA Maybe you can use their API with the built-in functionality for APIs? Apr 26 comment The Disc and Washer Methods Have you looked at RevolutionPlot3D[] already? Apr 26 awarded Nice Answer Apr 26 revised Calculate $\pi$ using Monte Carlo methods edited tags Apr 25 comment Construct certain univariate functions over bounded interval having given set of moments This seems more like a math question than a Mathematica question. Apr 25 comment Find the number of $n$ such that $n!$ is a sum of three squares Huh, would've thought keeping everything in integers would be faster. Oh well. Apr 25 comment Find the number of $n$ such that $n!$ is a sum of three squares It's annoying that BitShiftRight[] is still uncompilable. Otherwise, nice! BTW: I'd use Quotient[m, 2] instead of Floor[m/2]. Apr 25 revised Use Map on a Function with two variables edited tags Apr 25 comment NSolve won't act on very large powers Huh. If you set WorkingPrecision to anything other than the default value, it works. Apr 25 comment Checking if input is a function BTW, you might also be interested in PolynomialQ[]. Apr 25 comment Checking if input is a function Variables[#^2 - # + 7] will work, tho. Apr 25 revised Plot values of an $m\times n$ matrix on the complex plane with color varying along $m$ added 27 characters in body; edited tags; edited title Apr 25 revised Find the number of $n$ such that $n!$ is a sum of three squares deleted 7 characters in body; edited tags Apr 25 answered Return a sentence with numerical values Apr 25 revised FindInstance function edited tags | 527 | 2,284 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-18 | latest | en | 0.880675 |
https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9e81a4f98389efdf:conics-intro/v/introduction-to-conic-sections | 1,685,715,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00323.warc.gz | 905,715,787 | 95,512 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Precalculus
### Course: Precalculus>Unit 5
Lesson 1: Introduction to conic sections
# Intro to conic sections
Sal introduces the four conic sections and shows how they are derived by intersecting planes with cones in certain ways. Created by Sal Khan.
## Want to join the conversation?
• Will two same parabolas, placed side to side ( like this <>) , make a perfect ellipse ?
• Nope. They will have a sharp angle in the joint. If you see, for them to fit, a parabola must be vertical at some point, which is actually not true, it gets always steeper but don't get vertival until reaching infinity.
However, If you considered possible to reach infinity, then placing them infinitely separated could make them actually fit, but I doubt that shape would form an ellipse. (By the way, all shapes are either ellipses or aren't ellipses at all. There is no such a thing as a perfect ellipse : )
• So these conic sections are 2D shapes that intersect a 3D cone? Are there 3 dimensional shapes that intersect 4 dimensional shapes?
• Camerxn, you're right that time is a temporal dimension, not a spatial one, although ANY method of distinguishment (color, musical note, shape, etc.) could be a dimension, however '4D' is considered to be 4 spatial dimensions (x axis, y axis, z axis, w axis).
The more dimensions (particularly spatial ones) the more fun you can have. You can define the foci for an ellipse by using a cone, two spheres and zero math. In four dimensions, I'm willing to bet that there are ways of defining 3-dimensional shapes in a similar manner (I can't think of any off the top of my head). For the cone, make a sphere just big enough to touch the desired ellipse at one point inside the cone, and the other sphere just small enogh to touch the same ellipse in a second point, nestled on top of the cone (think of an Ice cream cone), those two points are the foci.
Three dimensional knots wouldn't but four dimensional knots would (think of a bow knot with 3 bows instead of two, something like that I think).
• Can't you also get a triangle by going perpendicular to the base of the cone and going through the tip?
• Good catch! It's not quite a triangle, because the theoretical cones in this example have infinite height, but it is two intersecting straight lines. Other "weird" examples are a single point, a single straight line, and two parallel straight lines. (See if you can figure out how to get those!) All of these examples are called degenerate conic sections, and we will never speak of them again. ^_^
• Dose the Hyperbola ever touch the asmpotote?
• No, hyperbolas never reach the asymptotes, which is why they are called asymptotes. As the hyperbola gets further and further away from the center, the hyperbola approaches the asymptotes, but is unable to touch it.
• So I understand how the circle, ellipse, parabola, and the hyperbola are related, but I still don't understand what the meaning of conic sections is.
• To 'section' something is to cut it... like in biology class when you 'dissect' a frog.
Imagine a cone... an ice cream cone (with no ice cream) or one of those orange cones they put around utility vehicles in the street.
If you hold such a cone so its central axis is vertical, then cut it with a horizontal plane, the cut edge will be a circle. If you tilt the cutting plane a bit, but not so much that it is parallel to the outer edge of the cone and section the cone the cut edge will be an ellipse. If you section the cone with a plane that is parallel to the outer surface of the cone the cut edge will be a parabola and if you tilt the cutting plane past that point and on to vertical you will get a hyperbola.
So the 'conic sections' are literally the shapes you get when you section a cone.
See http://en.wikipedia.org/wiki/Conic_section for a lot more detail.
• From 2009, what a classic.
• Can I differentiate or integrate conic section (circle, parabola, ellipse, hyperbola) equations?
• Yes, absolutely, if you want to learn partial differentiation early! Since there are two variables in these equations, you can take two separate derivatives. When you take the partial derivative of the function for the X variable, consider Y a constant (and vice versa). For example, given `x^2 + y^2` then the partial derivative with respect to X would be just 2x + 0. The result is the change in the X direction when Y is not changing at all. This concept is built upon in Multivariable Calculus and is really fun :)
• Why only circle,ellipse,parabola and hyperbola be conic sections ,others can't?
• Because those are the only shapes you can get from slicing two cones(one on top, point touching the other).
(1 vote)
• At , Sal draws a "cone", however it appears to be two cones on top of each other (tip to tip). Why is this? | 1,170 | 5,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.926861 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2001/Whitmire/Wednesday/anothernew8.html | 1,542,722,243,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746398.20/warc/CC-MAIN-20181120130743-20181120152743-00253.warc.gz | 187,291,177 | 1,401 | Paula Whitmire
Assignment 8
Altitudes and Orthocenters
Triangle BCD below has orthocenter L constructed from the intersection of the three lines containing the altitudes. The altitudes are perpendicular segments from the vertices to the opposite sides.
Now we will find the orthocenter of triangle LDC.
The orthocenter will be point B.
Now we will find the orthocenter of triangle BLD. The orthocenter will be point C.
Now we will find the orthocenter of triangle BLC and it is point D.
Now we construct the circumcenter of triangle BCD. It is point T.
Now we construct the circumcircle of triangle BCD with center T.
After much investigating, it was determined that the circumcenters for the smaller triangles was outside this triangle. Plus it is very late. Thanks again Dr. Wilson, I never dreamed I would learn so much.
RETURN | 189 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-47 | latest | en | 0.906643 |
https://education.ti.com/en/84activitycentral/us/detail?id=C3BDB13D578F4E8AB3617AEFD29F25E2&t=E4E7B037162E40BFA0393F1D8A1C6D42 | 1,685,839,765,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00403.warc.gz | 256,017,652 | 16,292 | Education Technology
• Device
• TI-83 Plus Family
• TI-84 Plus
• TI-84 Plus Silver Edition
• TI-84 Plus C Silver Edition
• TI-84 Plus CE
• TI-84 Plus CE Python
• Software
TI Connect™
TI Connect™ CE
Science: Walk the Line
by Texas Instruments
Overview
In this activity, students' will create constant-speed motion plots and develop linear equations to describe them mathematically.
Key Steps
• Distribute the pages to your class.
Follow the Activity procedures:
• Place the CBR on a table and stand 0.5 m away from it
• Walk away from the CBR at a slow and steady pace
• Examine the distance versus time graph
• Record the coordinates and calculate the slope and the starting y- intercept
• Perform a linear regression and compare the slope and the intercept values
• Students create a distance-time graph made by walking at a constant rate in order to find the mathematical equation that duplicates the plot.
• Students use the guess-and-test method to find the equation of the line that matches their data.
• Students use linear regression to gauge the accuracy of their calculations of the slope using the coordinates of two data points. | 256 | 1,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.808818 |
http://www.research.stlouisfed.org/fred2/series/LNS12600000 | 1,416,805,886,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400380368.73/warc/CC-MAIN-20141119123300-00084-ip-10-235-23-156.ec2.internal.warc.gz | 461,794,475 | 19,042 | # Employed, Usually Work Part Time
2014-10: 27,693 Thousands of Persons (+ see more)
Monthly, Seasonally Adjusted, LNS12600000, Updated: 2014-11-07 9:07 AM CST
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
The series comes from the 'Current Population Survey (Household Survey)'
The source code is: LNS12600000
Release: Employment Situation
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Employed, Usually Work Part Time, Thousands of Persons, Seasonally Adjusted (LNS12600000)
The series comes from the 'Current Population Survey (Household Survey)'
The source code is: LNS12600000
Employed, Usually Work Part Time
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` US. Bureau of Labor Statistics, Employed, Usually Work Part Time [LNS12600000], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/LNS12600000/, November 23, 2014. ```
Retrieving data.
Graph updated.
#### Recently Viewed Series
Subscribe to our newsletter for updates on published research, data news, and latest econ information.
Name: Email: | 506 | 1,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-49 | latest | en | 0.790252 |
https://discourse.processing.org/t/lerp-unintentional-offset-help-me-understand/40090 | 1,675,889,025,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00148.warc.gz | 224,631,430 | 6,774 | # Lerp unintentional offset: help me understand
I honestly feel someone else will look at this and understand the issue right away, like I feel like it is glaringly obvious but somehow I don’t get it. I am simply trying to get a set of evenly set points that fall along a line between two points (b.x, b.y) and (c.x, c.y)
What am I missing here? The points are always offset from the line and they should fall on top. Why aren’t they aligned?
``````PVector b, c;
void setup() {
size(480, 710);
noLoop();
}
void draw() {
background(255);
b = new PVector(random(10, 300), random(200,500));
c = new PVector(width, random(0,(b.y-10)));
for (float i=b.x; i <c.x; i=i+3) {
float amt = i/(c.x-b.x);
float bc = lerp(b.y, c.y, amt);
line(c.x,c.y, b.x, b.y);
point(i, bc);
}
}
``````
The value of `amt` is thrown off by starting `i` at `b.x`.
Printing out `amt` will show some values above 1.
Adding `b.x` to `point()` will put the dots on the line but not at the start.
To account for the offset, set `i` from zero to `range` and `float range = c.x - b.x`.
That will bring `amt` between 0 and 1.
Then add the starting x: `point(i+b.x, bc);`.
Also, since the solid line is drawn across the entire range, you could bring that outside of the loop.
1 Like
``````
PVector b, c;
void setup() {
size(480, 710);
noLoop();
}
void draw() {
background(255);
b = new PVector(random(10, 300), random(200, 460));
c = new PVector(width-33, random(0, (b.y-10)));
line(c.x, c.y, b.x, b.y);
for (float i=0; i <= 10; i+=1) {
float amt = i/10;
float bcX = lerp(b.x, c.x, amt);
float bcY = lerp(b.y, c.y, amt);
stroke(255, 0, 0);
ellipse(bcX, bcY, 4, 4); // or point
}
}
``````
1 Like
Hello @i-draw-monkeys,
I encourage you as a beginner to use the resources here:
There is an example of lerping here:
https://processing.org/reference/lerp_.html
Once you understand that you can advance to:
https://processing.org/reference/PVector_lerp_.html
This helps to understand linear interpolation:
`:)`
1 Like
This is the answer. It is the “i” starting at b.x as the problem because the “i” is reused in the numerator of amt! Totally makes sense, was the thing I was overlooking. Changing this line from:
`float amt = i/(c.x-b.x);`
to
`float amt = (i-b.x)/(c.x-b.x);`
makes it so the points overlay the line as they should.
Thanks!
I am big into the reference section! I definitely read those multiple times before asking the question. Unfortunately, I still wasn’t able to find my error.
1 Like
This is basically what I ended up doing to build my file a couple days ago when I had the problem. But now I also see the problem with the original code. Many ways to solve an issue! Thanks!
1 Like | 784 | 2,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | latest | en | 0.82716 |
http://computer-programming-forum.com/44-ada/3d35f2906172d240.htm | 1,601,325,387,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401614309.85/warc/CC-MAIN-20200928202758-20200928232758-00023.warc.gz | 28,545,679 | 5,435 | Multidimensional array vs. array of array
Author Message
Multidimensional array vs. array of array
How does a skilled Ada programmer define a return type containing a
multidimensional array? Of course, the simple solution works:
type Equations is array (Rows range <>, Columns range <>) of Number;
type Workspace (row : Rows; col : Columns) is record
eqs : Equations (Rows'First .. row, Columns'First .. col);
... some more stuff ...
end record;
function Init_Workspace return Workspace;
But this approach fails miserably on extracting a single row from the result
as well as on construction the result in a recursive function.
I do not found a way to define this type as an array of arrays despite
writing a generic package over the maximum values:
generic
type Columns is range <>;
package
type Equation is arry (Columns) of Number;
type Equations is array (Rows range <>) of Equation;
type Workspace (row : Rows) is record
eqs : Equations (Rows'First .. row);
... some more stuff ...
end record;
function Init_Workspace return Workspace;
end package;
But this approach does not work for a situation where determining the
resulting constraints is equivalent to simulating the Init_Workspace
recursion. This should be avoided: Error-prone, duplicated and slightly
modified code.
Any idea?
Sun, 30 May 2004 06:22:00 GMT
Multidimensional array vs. array of array
Quote:
> How does a skilled Ada programmer define a return type containing a
> multidimensional array? Of course, the simple solution works:
> type Equations is array (Rows range <>, Columns range <>) of Number;
> type Workspace (row : Rows; col : Columns) is record
> eqs : Equations (Rows'First .. row, Columns'First .. col);
> ... some more stuff ...
> end record;
> function Init_Workspace return Workspace;
> But this approach fails miserably on extracting a single row from the
result
> as well as on construction the result in a recursive function.
Would it help you to just define a distinct type to represent a row in
isolation, but keep your matrix as defined above (i.e. not in terms of
composition of the row type)?
-- mark
Sun, 30 May 2004 06:45:20 GMT
Multidimensional array vs. array of array
Quote:
> How does a skilled Ada programmer define a return type containing a
> multidimensional array?
I suspect the answer is : you don't. Define a procedure with an 'in
out' parameter instead. Or one 'in' and one 'out'.
Quote:
> Of course, the simple solution works:
> type Equations is array (Rows range <>, Columns range <>) of Number;
> type Workspace (row : Rows; col : Columns) is record
> eqs : Equations (Rows'First .. row, Columns'First .. col);
> ... some more stuff ...
> end record;
> function Init_Workspace return Workspace;
> But this approach fails miserably on extracting a single row from the result
> as well as on construction the result in a recursive function.
Yes, this is a fundamental limitation. If you think about a reasonable
implementaion of multi-dimensional arrays, and what you are asking the
compiler to do, it will make sense that it is a limitation.
--
-- Stephe
Sun, 30 May 2004 07:03:51 GMT
Multidimensional array vs. array of array
Quote:
>> How does a skilled Ada programmer define a return type containing a
>> multidimensional array?
>I suspect the answer is : you don't. Define a procedure with an 'in
>out' parameter instead. Or one 'in' and one 'out'.
Parameters of unconstraint type always have the constraints as in parameters.
In order to obtain the constraints from the calling procedure, you have to
return an unconstraint type.
Returning a multidimensional array is easy.
Returning an array of arrays is impossible.
A multidimensional array is often unusable in further processes. There a
array of arrays is necessary. Of course, I can use Unchecked_Conversion,
but this is not portable and not a good style.
Quote:
>> But this approach fails miserably on extracting a single row from the
>> result as well as on construction the result in a recursive function.
>Yes, this is a fundamental limitation. If you think about a reasonable
>implementaion of multi-dimensional arrays, and what you are asking the
>compiler to do, it will make sense that it is a limitation.
I read all the discussions about Ada83/95 multidimensional arrays and know
where the limitation comes from. But I still ask: How does a skilled Ada
programmer return a array of arrays?
Sun, 30 May 2004 16:39:28 GMT
Multidimensional array vs. array of array
Quote:
> I read all the discussions about Ada83/95 multidimensional arrays and know
> where the limitation comes from. But I still ask: How does a skilled Ada
> programmer return a array of arrays?
You can do it two different ways:
1) Put the array of arrays in a record and return the record
2) Create an access type for the array of arrays. Dynamically
allocate the array of arrays, and return its access value.
Jim Rogers
Colorado Springs, Colorado USA
Mon, 31 May 2004 08:26:19 GMT
Multidimensional array vs. array of array
Quote:
>> I read all the discussions about Ada83/95 multidimensional arrays and know
>> where the limitation comes from. But I still ask: How does a skilled Ada
>> programmer return a array of arrays?
>You can do it two different ways:
>1) Put the array of arrays in a record and return the record
I'm unable to do this. Example?
Quote:
>2) Create an access type for the array of arrays. Dynamically
> allocate the array of arrays, and return its access value.
*gna* No way. *drawing anti-satan symbols*
Mon, 31 May 2004 17:02:39 GMT
Multidimensional array vs. array of array
Quote:
> >You can do it two different ways:
> >1) Put the array of arrays in a record and return the record
> I'm unable to do this. Example?
I don't think this can be done.
Quote:
> >2) Create an access type for the array of arrays. Dynamically
> > allocate the array of arrays, and return its access value.
> *gna* No way. *drawing anti-satan symbols*
Anyway that solves nothing, in itself.
I suspect the only neat solution is for the next revision to permit
multi-dimensional slices, and a new kind of 'dimension reducing' array
conversion.
In the meantime, I can only suggest that you fudge by:
(1) using an appropriate multi-dimensional array; AND
(1a) redefine all your row/column functions so that they accept this
multidimensional array (and simply check that they have been given just one
row or column); OR
(1b) define a set of 'wrapper' functions that do this check, extract the row
or column into an appropriate array type, and then call the proper function,
perhaps doing the reverse conversion on the way back; OR
(1c) redefine all your row/column functions so that they accept this
multidimensional array plus an identification of the row/column to be
operated upon.
Horrid, yes, but I can't think of anything better. E.g.:
type Matrix is array (Positive range <>,
Positive range <>) of Float;
...
function Sum_Row (M: in Matrix) return Float is
R: Float := 0.0;
begin
if M'Length(1) /= 1 then
raise Constraint_Error; -- or another exception
end if;
for i in M'Range(2) loop
R := R + M(M'First(1),i); -- M'First(1)=M'Last(1)=row
end loop;
return R;
end;
An alternative with a wrapper function would be:
type Vector is array (Positive range <>) of Float;
...
function Sum (V: in Vector) return Float is
R: Float := 0.0;
begin
for i in V'Range loop
R := R + V(i);
end loop;
return R;
end;
...
function Sum_Row (M: in Matrix) return Float is
V: Vector(M'Range(2));
begin
if M'Length(1) /= 1 then
raise Constraint_Error; -- or another exception
end if;
for i in M'Range(2) loop
V(i) := M'(M'First(1),i);
end loop;
return Sum(V);
end;
Finally, you could pass the whole matrix (presumably this would actually be
done by reference), and simply identify the row/column to be acted upon:
function Sum_Row (M: in Matrix;
Row: in Positive) return Float is
R: Float := 0.0;
begin
for i in M'Range(2) loop
R := R + M(Row,i);
end loop;
return R;
end;
It may be that the last solution, while ugly, is actually the fastest.
Does this help at all?
--
Best wishes,
Nick Roberts
Tue, 01 Jun 2004 05:49:35 GMT
Multidimensional array vs. array of array
Quote:
>Does this help at all?
Yes, but I perfer a differnt approach: Returning a record containing a
multidimensional array, define an constraint array of array type from the
constraint result (ws : Workspace := Init_Workspace; type ... ws.length;),
converting (copy) the multidimensional array to a variable of the new type
and work with this new type. This is reasonable, because the following
computations are much slower, than the conversion itself.
Tue, 01 Jun 2004 17:45:21 GMT
Multidimensional array vs. array of array
Quote:
>> How does a skilled Ada programmer define a return type containing a
>> multidimensional array? Of course, the simple solution works:
>> type Equations is array (Rows range <>, Columns range <>) of Number;
>> type Workspace (row : Rows; col : Columns) is record
>> eqs : Equations (Rows'First .. row, Columns'First .. col);
>> ... some more stuff ...
>> end record;
>> function Init_Workspace return Workspace;
>> But this approach fails miserably on extracting a single row from the
>> result as well as on construction the result in a recursive function.
>Would it help you to just define a distinct type to represent a row in
>isolation, but keep your matrix as defined above (i.e. not in terms of
>composition of the row type)?
After returning the Workspace in order to define a unconstrained variables,
it is possible.
Fri, 04 Jun 2004 23:56:28 GMT
Page 1 of 1 [ 9 post ]
Relevant Pages
Powered by phpBB® Forum Software | 2,406 | 9,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.755678 |
http://www.helpteaching.com/questions/Place_Value/Grade_1 | 1,435,872,384,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095671.53/warc/CC-MAIN-20150627031815-00293-ip-10-179-60-89.ec2.internal.warc.gz | 509,922,397 | 11,400 | Tweet
Place Value questions are available in the following grade levels:
##### Browse Questions
You can create printable tests and worksheets from these Grade 1 Place Value questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.
1 2 3
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2b
Count the pictures.
Which is correct?
1. 1 ten + 4 ones = 14
2. 4 tens + 1 one = 14
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2b
Count the pictures.
Which is correct?
1. 2 tens + 1 one = 12
2. 1 ten + 2 ones = 12
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2a
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2a
Count the bunnies.
Which is correct?
1. 1 ten + 4 ones = 14
2. 4 tens + 1 one = 14
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2a
Grade 1 Place Value CCSS: 3.NBT.A.1
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2c
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2a
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2b
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2b
Which shows 17?
1. 1 ten and 1 one
2. 7 tens and 7 ones
3. 1 ten and 7 ones
4. 7 tens and 1 one
Grade 1 Place Value CCSS: 1.NBT.B.2
Grade 1 Place Value CCSS: 1.NBT.B.3
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2b
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2b
Grade 1 Place Value CCSS: 1.NBT.B.2
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2c
Grade 1 Place Value CCSS: 1.NBT.B.2
Grade 1 Place Value CCSS: 1.NBT.B.2, 1.NBT.B.2a
Grade 1 Place Value CCSS: 1.NBT.B.2
Grade 1 Place Value CCSS: 1.NBT.B.2
1 2 3
You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges. | 682 | 1,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2015-27 | longest | en | 0.697013 |
https://interviewmania.com/machine-design/machine-design-miscellaneous/1/7 | 1,721,515,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00787.warc.gz | 274,926,211 | 10,555 | ## Machine Design Miscellaneous
#### Machine Design
1. The bolts in a rigid flanged coupling connecting two shafts transmitting power are subjected to
1. NA
##### Correct Option: A
NA
1. The bolts in a rigid flanged coupling connecting two shafts transmitting power are subjected to
1. NA
##### Correct Option: A
NA
1. Stress concentration in a machine component of a ductile material is not so harmful as it is in a brittle material because
1. Local yielding reduces stress concentration effect.
##### Correct Option: A
Local yielding reduces stress concentration effect.
1. In a bolted Joint two smooth members are connected with an axial tightening force of 2200 N. If the bolt used has metric threads of 4 mm pitch, the torque required for achieving the tightening force is
1. Tightening force = 2200 N
Pitch = 4 mm = 0.004 m
Torque = Tightening force × 0.004 = 1.4 N-m 2π
##### Correct Option: C
Tightening force = 2200 N
Pitch = 4 mm = 0.004 m
Torque = Tightening force × 0.004 = 1.4 N-m 2π
1. A bolted Joint is shown below. The maximum shear stress, in MPa, in the bolts at A and B, respectively are
1. FA = FB = FC = 10 kN 3
FA = P.e.r = 10 × 103 × 150 × 40 rA² + rB² + rC² 40² + 0² + 40²
F'A = 18.75 kN = F'C
Shear load at A and C
FR = √(FA)² + (F'A = √(3.33)² + (18.75)²
FR = 19.04 KN
τmax = 19.04 × 103 (π / 4) × 102
( τmax )A,C = 242.5 MPa
( τmax )B = 42.44 MPa
##### Correct Option: A
FA = FB = FC = 10 kN 3
FA = P.e.r = 10 × 103 × 150 × 40 rA² + rB² + rC² 40² + 0² + 40²
F'A = 18.75 kN = F'C
Shear load at A and C
FR = √(FA)² + (F'A = √(3.33)² + (18.75)²
FR = 19.04 KN
τmax = 19.04 × 103 (π / 4) × 102
( τmax )A,C = 242.5 MPa
( τmax )B = 42.44 MPa | 613 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-30 | latest | en | 0.792425 |
http://www.sqlservercentral.com/Forums/Topic1359289-391-1.aspx | 1,386,354,131,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052338/warc/CC-MAIN-20131204131732-00067-ip-10-33-133-15.ec2.internal.warc.gz | 535,669,322 | 24,662 | Recent PostsRecent Posts Popular TopicsPopular Topics
Home Search Members Calendar Who's On
Stored Procedure help Rate Topic Display Mode Topic Options
Author
Message
Posted Friday, September 14, 2012 7:53 AM
Forum Newbie Group: General Forum Members Last Login: Tuesday, September 18, 2012 10:23 AM Points: 4, Visits: 12
Hi, I need help in a stored procedure that, counts the number of Saturdays in a month, returns 12 rows, each row containing number of Saturdays for each month of the year using a single parameter as the specification for the year.
Post #1359289
Posted Friday, September 14, 2012 8:07 AM
SSCoach Group: General Forum Members Last Login: Friday, November 01, 2013 1:55 PM Points: 15,442, Visits: 9,579
Like this:`DECLARE @StartDate DATE = '20120101';WITH Seeds(Seed) AS (SELECT * FROM ( VALUES ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1) ) AS V (C)), Numbers(Number) AS (SELECT ROW_NUMBER() OVER (ORDER BY S1.Seed) - 1 FROM Seeds AS S1 CROSS JOIN Seeds AS S2), Calendar(Date) AS (SELECT DATEADD(DAY, Number, @StartDate) FROM Numbers WHERE DATEADD(DAY, Number, @StartDate) < DATEADD(YEAR, 1, @StartDate)) SELECT DATEPART(MONTH, Date), COUNT(*) FROM Calendar WHERE DATEPART(weekday, Date) = 7 GROUP BY DATEPART(MONTH, Date) ORDER BY DATEPART(MONTH, Date);`I use the Seed and Numbers CTEs to build a table of numbers from 0 - 400, then use the Calendar CTE to build a table of all dates in a year, from the Numbers CTE. It can work even better if you have a persisted Calendar table (those have a lot of good uses). Assuming you don't have one, this will work.This solution depends on features from SQL 2008 and later. Based on the forum the question was posted in, that should be okay. If you're actually using a prior version of SQL Server (2005 or earlier), you'll need to change the Seeds CTE so that it uses Union All statements instead of a Table Value Constructor. - Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETCProperty of The Thread"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
Post #1359306
Posted Friday, September 14, 2012 9:29 AM
SSCommitted Group: General Forum Members Last Login: Today @ 10:58 AM Points: 1,869, Visits: 4,117
It also depends on @@DATEFIRST = 7.Otherwise you might be counting Sundays or another day. Luis C.Please don't trust me, test the solutions I give you before using them.Forum Etiquette: How to post data/code on a forum to get the best help
Post #1359418
Posted Friday, September 14, 2012 10:18 AM
SSCommitted Group: General Forum Members Last Login: Today @ 3:51 AM Points: 1,566, Visits: 2,206
Code below has fewer calcs and does not depend on any SQL date settings.`DECLARE @year intSET @year = 2012 --<<-- chg as neededSELECT month_start, DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, DATEADD(MONTH, 1, month_start))) / 7 - DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, month_start)) / 7FROM ( SELECT CAST(CAST(@year * 10000 + 0101 AS char(8)) AS datetime) AS month_start UNION ALL SELECT CAST(@year * 10000 + 0201 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0301 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0401 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0501 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0601 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0701 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0801 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0901 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1001 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1101 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1201 AS char(8))) AS months_of_the_yearORDER BY month_start` SQL DBA,SQL Server MVP('07, '08, '09)One man with courage makes a majority. Andrew Jackson
Post #1359486
Posted Monday, September 17, 2012 8:50 AM
Forum Newbie Group: General Forum Members Last Login: Tuesday, September 18, 2012 10:23 AM Points: 4, Visits: 12
Thanks!
Post #1360254
Posted Monday, September 17, 2012 9:08 AM
SSCrazy Group: General Forum Members Last Login: Wednesday, November 20, 2013 10:00 AM Points: 2,719, Visits: 4,724
a bit shorter version...`DECLARE @year intSET @year = 2015SELECT MONTH(mfd) AS MonthNo ,DATENAME(MONTH,mfd) AS MonthName ,DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, DATEADD(MONTH, 1, mfd))) / 7 - DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, mfd)) / 7 AS NoOfSaturdaysFROM ( SELECT CAST(CAST(@year * 100 + m.m AS CHAR(6)) + '01' AS datetime) mfd FROM (VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) m(m) ) q ORDER BY q.mfd ` _____________________________________________"The only true wisdom is in knowing you know nothing""O skol'ko nam otkrytiy chudnyh prevnosit microsofta duh!"(So many miracle inventions provided by MS to us...)How to post your question to get the best and quick help
Post #1360270
Posted Monday, September 17, 2012 9:28 AM
SSCoach Group: General Forum Members Last Login: Friday, November 01, 2013 1:55 PM Points: 15,442, Visits: 9,579
ScottPletcher (9/14/2012)Code below has fewer calcs and does not depend on any SQL date settings.`DECLARE @year intSET @year = 2012 --<<-- chg as neededSELECT month_start, DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, DATEADD(MONTH, 1, month_start))) / 7 - DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, month_start)) / 7FROM ( SELECT CAST(CAST(@year * 10000 + 0101 AS char(8)) AS datetime) AS month_start UNION ALL SELECT CAST(@year * 10000 + 0201 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0301 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0401 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0501 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0601 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0701 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0801 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0901 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1001 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1101 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1201 AS char(8))) AS months_of_the_yearORDER BY month_start`Looks like you're picking the start date to be the day of the week you want. Correct? 6 Jan 1900 as the seed because it's a Saturday, right?If you're doing that calculation in the script or as a parameter, and concerned about @@DateFirst, you can modify mine like this:`DECLARE @StartDate DATE = '20120101', @SeedDate DATE = '19000106';WITH Seeds(Seed) AS (SELECT * FROM ( VALUES ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1) ) AS V (C)), Numbers(Number) AS (SELECT ROW_NUMBER() OVER (ORDER BY S1.Seed) - 1 FROM Seeds AS S1 CROSS JOIN Seeds AS S2), Calendar(Date) AS (SELECT DATEADD(DAY, Number, @StartDate) FROM Numbers WHERE DATEADD(DAY, Number, @StartDate) < DATEADD(YEAR, 1, @StartDate)) SELECT DATEPART(MONTH, Date), COUNT(*) FROM Calendar WHERE DATEPART(weekday, Date) = DATEPART(weekday, @SeedDate) GROUP BY DATEPART(MONTH, Date) ORDER BY DATEPART(MONTH, Date);`Not that it gives any execution-time advantage either way. Both run in 0 milliseconds on my test server.Would come down to readability for each. Unsurprisingly, I find mine more readable. Easier to maintain. But since I wrote it, that's about as unfair a test as is possible. - Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETCProperty of The Thread"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
Post #1360280
Posted Monday, September 17, 2012 9:34 AM
Forum Newbie Group: General Forum Members Last Login: Tuesday, September 18, 2012 10:23 AM Points: 4, Visits: 12
sorry last request... if Sunday were to be added on there? So both Saturday and Sunday.
Post #1360285
Posted Monday, September 17, 2012 9:39 AM
SSCoach Group: General Forum Members Last Login: Friday, November 01, 2013 1:55 PM Points: 15,442, Visits: 9,579
Eugene Elutin (9/17/2012)a bit shorter version...`DECLARE @year intSET @year = 2015SELECT MONTH(mfd) AS MonthNo ,DATENAME(MONTH,mfd) AS MonthName ,DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, DATEADD(MONTH, 1, mfd))) / 7 - DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, mfd)) / 7 AS NoOfSaturdaysFROM ( SELECT CAST(CAST(@year * 100 + m.m AS CHAR(6)) + '01' AS datetime) mfd FROM (VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) m(m) ) q ORDER BY q.mfd `I like it.I'd move the math for generating day-1 of the desired year into a variable at the top of the script, just for readability, but it works as-is.`DECLARE @Year CHAR(4) = '2012'; -- input parameter if procDECLARE @StartDate DATE = @Year + '0101';SELECT DATEADD(MONTH, [month], @StartDate) AS MonthStartFROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11)) AS TVC([month])` - Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETCProperty of The Thread"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
Post #1360287
Posted Monday, September 17, 2012 10:31 AM
SSCommitted Group: General Forum Members Last Login: Today @ 3:51 AM Points: 1,566, Visits: 2,206
GSquared (9/17/2012)ScottPletcher (9/14/2012)Code below has fewer calcs and does not depend on any SQL date settings.`DECLARE @year intSET @year = 2012 --<<-- chg as neededSELECT month_start, DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, DATEADD(MONTH, 1, month_start))) / 7 - DATEDIFF(DAY, '19000106', DATEADD(DAY, -1, month_start)) / 7FROM ( SELECT CAST(CAST(@year * 10000 + 0101 AS char(8)) AS datetime) AS month_start UNION ALL SELECT CAST(@year * 10000 + 0201 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0301 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0401 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0501 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0601 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0701 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0801 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 0901 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1001 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1101 AS char(8)) UNION ALL SELECT CAST(@year * 10000 + 1201 AS char(8))) AS months_of_the_yearORDER BY month_start`Looks like you're picking the start date to be the day of the week you want. Correct? 6 Jan 1900 as the seed because it's a Saturday, right?If you're doing that calculation in the script or as a parameter, and concerned about @@DateFirst, you can modify mine like this:`DECLARE @StartDate DATE = '20120101', @SeedDate DATE = '19000106';WITH Seeds(Seed) AS (SELECT * FROM ( VALUES ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1), ( 1) ) AS V (C)), Numbers(Number) AS (SELECT ROW_NUMBER() OVER (ORDER BY S1.Seed) - 1 FROM Seeds AS S1 CROSS JOIN Seeds AS S2), Calendar(Date) AS (SELECT DATEADD(DAY, Number, @StartDate) FROM Numbers WHERE DATEADD(DAY, Number, @StartDate) < DATEADD(YEAR, 1, @StartDate)) SELECT DATEPART(MONTH, Date), COUNT(*) FROM Calendar WHERE DATEPART(weekday, Date) = DATEPART(weekday, @SeedDate) GROUP BY DATEPART(MONTH, Date) ORDER BY DATEPART(MONTH, Date);`Not that it gives any execution-time advantage either way. Both run in 0 milliseconds on my test server.Would come down to readability for each. Unsurprisingly, I find mine more readable. Easier to maintain. But since I wrote it, that's about as unfair a test as is possible. And unsuprisingly, I find my version more readable.I mean, seriously, three levels of CTEs with a GROUP BY "more readable" than two DATEDIFF functions??I should have added a comment about the date seed, just to be clear.My code as originally written also works in earlier versions of SQL (I think it would even work in 7.0). SQL '08-specific features are great, when needed, but I don't use them just for the sake of using them. We still have '05 instances where I work, and I think some other people do too.And, yes, the code for the month generation can be shortened, but I think it's clearer the longer way, and that bit won't affect execution time. I have no doubt that the two DATEDIFFs will have much less overhead than the cross joins, etc., although it may not be signficant really, because SQL is so fast at doing cross joins. SQL DBA,SQL Server MVP('07, '08, '09)One man with courage makes a majority. Andrew Jackson
Post #1360316
Permissions | 3,959 | 12,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-48 | longest | en | 0.764425 |
https://efxa.org/2011/02/09/ | 1,685,294,167,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644309.7/warc/CC-MAIN-20230528150639-20230528180639-00779.warc.gz | 256,729,394 | 33,128 | ## Genetic Algorithm (Evolutionary Algorithm).
Taxonomy
The Genetic Algorithm is an Adaptive Strategy and a Global Optimization technique. It is an Evolutionary Algorithm and belongs to the broader study of Evolutionary Computation. The Genetic Algorithm is a sibling of other Evolutionary Algorithms such as Genetic Programming, Evolution Strategies, Evolutionary Programming, and Learning Classifier Systems. The Genetic Algorithm is a parent of a large number of variant techniques and sub-fields too numerous to list.
## Iterated Local Search (Stochastic Algorithm).
Taxonomy
Iterated Local Search is a Metaheuristic and a Global Optimization technique. It is an extension of Multi Start Search and may be considered a parent of many two-phase search approaches such as the Greedy Randomized Adaptive Search Procedure and Variable Neighborhood Search.
## Random Search (Stochastic Algorithm).
Taxonomy
Random search belongs to the fields of Stochastic Optimization and Global Optimization. Random search is a direct search method as it does not require derivatives to search a continuous domain. This base approach is related to techniques that provide small improvements such as Directed Random Search, and Adaptive Random Search.
## Adaptive Random Search (Stochastic Algorithm).
Taxonomy
The Adaptive Random Search algorithm belongs to the general set of approaches known as Stochastic Optimization and Global Optimization. It is a direct search method in that it does not require derivatives to navigate the search space. Adaptive Random Search is an extension of the Random Search and Localized Random Search algorithms.
## Stochastic Hill Climbing (Stochastic Algorithm).
Taxonomy
The Stochastic Hill Climbing algorithm is a Stochastic Optimization algorithm and is a Local Optimization algorithm (contrasted to Global Optimization). It is a direct search technique, as it does not require derivatives of the search space. Stochastic Hill Climbing is an extension of deterministic hill climbing algorithms such as Simple Hill Climbing (first-best neighbor), Steepest-Ascent Hill Climbing (best neighbor), and a parent of approaches such as Parallel Hill Climbing and Random-Restart Hill Climbing. | 411 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-23 | latest | en | 0.884446 |
https://archive.guildofarchivists.org/wiki/Reference:RAWA/D%27ni_symbol_for_25 | 1,722,761,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00197.warc.gz | 78,850,516 | 25,307 | # Reference:RAWA/D'ni symbol for 25
## Part 1
RAWA, is it true that the "X" is a D'ni 25? And, if so, it that method preferred over writing a Dni "10"? (I've sort of assumed that a blank box is the number 0, correct me if I'm wrong.)
Yes, 25 can be written in two different ways in Dni, the X on the Book of D'ni is the number 25 written as a single digit. It is the only digit that has this dual representation. The single digit 25 is rarely used, it is far more common to see "10" for 25 instead.
0 is not an empty box, but is a box with a single dot in the center. The map included with the Book of Ti'ana has at least two zeroes in it. (I didn't look very long to find those two, there may be more.)
Shorah,
RAWA
--- _ -----------------------------------------------------------------
/ * Richard A. Watson * mailto:rawa@cyan.com *
\_/ * Programmer / D'ni Historian * http://www.cyan.com *
C Y A N ------------------------------------------------------------
## Part 2
I know this was discussed immediately upon the opening of the spoiler lyst... and I may have missed a message or two. Did anyone uncover the meaning of the '/' number? That's the only missing piece in my number knowledge.
Just to help curb the confusion here, the single diagonal slash was an error in the original D'ni number font we made. The only place that I know the old font was used by mistake is on the watch in Gehn's bedroom.
Sorry for the confusion.
RAWA
--- _ -----------------------------------------------------------------
/ * Richard A. Watson * mailto:rawa@cyan.com *
\_/ * Programmer / D'ni Historian * http://www.cyan.com *
C Y A N ------------------------------------------------------------
## Part 3
In the original game design, 25 was meant to be figured out by process of elimination. But later we decided to just not use it at all. I just double-checked the prototype to make sure that I was remembering correctly, and when generating the dome combo, it doesn't use 1 or 25.
It is possible that Broderbund (Mac/PC) or Sunsoft (Game machines) didn't implement this limitation in their versions. Out of the dozens of times I've played each of their versions, I never got 1 or 25 in the dome combo, but that may just have been coincidence.
A Fan wrote:
If the D'ni numbering system is base 25 (and it is), then 25 would be two digits. The first would be the symbol for 1, followed by a symbol for 0. I believe the symbol for 0 is a square with a dot in the middle. (25 x 1 ) + 0 = 25 We use the same logic for Gehn's Age 233. (25 X 9) + 8 + 233
For mathematical purposes you are correct. In these cases, twenty-five is always represented as two digits ("10"). But the D'ni also has a single symbol for 25 used in other instances (like comparisons, where they judge things on a scale from 1 to 25).
One Fan wrote:
BTW, the symbol for zero is a square with a single diagonal line. Check Gehn's timepiece. When it's open, there are number symbols on the upper area of the inner cylinder. They begin with this symbol, and continue with the symbols for one, two, three, etc.
Another Fan wrote:
I disagree. I believe the square with the diagonal line is an error. I remember a post many months ago stating the square with a dot inside is "0". I believe it was RAWA who posted it. But I may be wrong. Maybe both are correct.
There are two ways of looking at this:
1) The number on the timepiece was meant to be 25, but there was an error in the font that was used.
2) The single slash is a special symbol representing the combination of 0 (a dot) and 25 (an "x") only used in cases like the clock where the number sequence is cyclical, and can be thought of a wapping around upon itself.
#1 is the truth, but I like #2 better. ;)
Shorah,
RAWA
--- _ -----------------------------------------------------------------
/ * Richard A. Watson * mailto:rawa@cyan.com *
\_/ * Programmer / D'ni Historian * http://www.cyan.com *
C Y A N ------------------------------------------------------------
## Part 4
(Note: this was not posted on the Riven Lyst, but as a personal message on the Myst Online forums)
Message subject: Re: ConScript Unicode Registry
From: RAWA
Sent: Fri 18 Oct 2013, 19:17 (UTC+1)
To: korovev
(...)
The "Alternative 0" represents both 0 and 25 simultaneously, and is used for cyclical representations where 0 and 25 would overlap. Examples from our culture (kind of): a 24-clock where either 00:00 and 24:00 could be used to represent midnight; or a compass where 0 and 360 degrees "overlap".
The "Alternative 25" that looks like an asterisk in a box (D'ni numerals 6 and 25 in the same box) - is the D'ni symbol for the concept of "infinity" - something that cannot be numbered.
Hope that helps,
RAWA | 1,178 | 4,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.900119 |
https://www.helpingwithmath.com/printables/tables_charts/4oa4-prime-numbers01.htm | 1,555,849,565,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578531462.21/warc/CC-MAIN-20190421120136-20190421142136-00192.warc.gz | 706,978,801 | 11,782 | # Prime Numbers up to 100
You will find help with determining whether numbers are prime or composite here. There is also a printable chart that lists all the prime numbers up to 251 here. Note: This is a 2-page document with the first page intended for color printing and the second for black and white. print only the pages required.
------ Note: The Information above this point will not be sent to your printer --------
--------- Page Break----End of Page 1 of 2 ---------
------------------------End of Page 2 of 2 ---------
------ Note: The Information below this point will not be sent to your printer --------
- By HelpingWithMath.com
### Related Resources
The Factors and multiples Chart above is aligned, either partially or wholly, with the standard 4OA04 from the Common Core Standards For Mathematics (see the shortened extract below). The resources below are similarly aligned.
Find all factor pairs for a whole number in the range 1 to 100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1 to 100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1 to 100 is prime or composite. | 260 | 1,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | longest | en | 0.800769 |
http://codeforces.com/topic/58857/en2 | 1,601,033,901,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400223922.43/warc/CC-MAIN-20200925084428-20200925114428-00795.warc.gz | 33,514,802 | 14,831 | Rating changes for the last round are temporarily rolled back. They will be returned soon. ×
Explain the problem
Revision en2, by plehem, 2018-03-24 17:43:38
Problem
I might have misunderstood the problem. Please explain the following case
3 12 2
5 9 1
Author's code output: 30, but there are 48 non-decreasing subsequences, e.g.
5 9 1 5 9 1 5 9 1 5 9 1 --> b
1 2 3 4 5 6 7 8 9 10 11 12 --> positions
length 1 subsequence 12
length 2:
1-4,1-5,1-7,1-8,1-10,1-11,2-5,2-8,2-11,3-4,3-5,3-6,3-7,3-8,3-9,3-10,3-11,3-12,
4-7,4-8,4-10,4-11,5-8,5-11,6-7,6-8,6-9,6-10,6-11,6-12
7-10,7-11,8-11,9-10,9-11,9-12
total 36+12 = 48
What's wrong here?
#### History
Revisions
Rev. Lang. By When Δ Comment
en11 plehem 2018-03-26 12:46:01 3 Tiny change: ' 48 but it's 30:\n\n~' -> ' 48 but it is 30:\n\n~'
en10 plehem 2018-03-26 10:32:57 31 Tiny change: 'preciated!\n\nSorry, for my poor English.' -> 'preciated!'
en9 plehem 2018-03-26 02:48:20 5 Tiny change: 'y, for my English.' -> 'y, for my poor English.'
en8 plehem 2018-03-25 23:42:40 2 Tiny change: 'already>\nSince, I' -> 'already>\n\nSince, I'
en7 plehem 2018-03-25 23:41:55 23 Tiny change: 'eryone! \n\nSince, I' -> 'eryone! \n<fucking reply already>\nSince, I'
en6 plehem 2018-03-25 21:08:28 4 Tiny change: 'weak at **dp** tagged ' -> 'weak at **DP** tagged '
en5 plehem 2018-03-25 19:53:01 12
en4 plehem 2018-03-25 19:50:16 411 Tiny change: ' [problem:588D]. I found' -> ' [problem:<b>588D<b>]. I found'
en3 plehem 2018-03-24 20:31:19 2 Tiny change: '[Problem](http://c' -> '[Problem D](http://c'
en2 plehem 2018-03-24 17:43:38 4 Tiny change: 'problem/587/B)\n\nI mig' -> 'problem/588/D)\n\nI mig'
en1 plehem 2018-03-24 14:08:18 614 Initial revision (published) | 770 | 1,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.481446 |
https://math.stackexchange.com/questions/2708553/easy-way-to-determine-if-a-matrix-is-positive-semidefinite | 1,558,329,814,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255562.23/warc/CC-MAIN-20190520041753-20190520063753-00356.warc.gz | 534,610,647 | 31,001 | # Easy way to determine if a matrix is positive-semidefinite?
What is an easy way of checking if this matrix on $R^2_{++}$ is positive semidefinite ? I saw this thread Checking if a matrix is positive semidefinite using Sylvester's criterion but don't understand what principal minors are. My class works ONLY on 2x2 matrices and did not mention anything about minors.
For 2x2 matrices, without using the word minors/principals/etc, do I just have to find out if the 3 conditions are satisfied ?
$H[0][0] \ge 0$
$H[1][1] \ge 0$
$H[0][0] \times H[1][1] - H[0][1] \times H[1][0] \ge 0$
H = $$\begin{bmatrix}\frac{2x_2^2}{(x_1x_2)^3} & \frac{1}{(x_1x_2)^2} \\\frac{1}{(x_1x_2)^2} & \frac{2x_1^2}{(x_1x_2)^3}\end{bmatrix}$$
What about positive definite ? | 268 | 757 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-22 | latest | en | 0.714983 |
https://lvbet.com/casino-blog/roulette/romanosky-roulette-system/ | 1,716,336,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00734.warc.gz | 329,703,796 | 24,069 | # ROMANOSKY ROULETTE SYSTEM
Roulette is a game that has wowed players from the very beginning. And when online roulette was on the rise, this love for roulette only spread farther and wider. But, with the increasing popularity of online roulette comes an increased number of roulette variants. Don’t get us wrong, having more roulette games to play is excellent, but more variants also come with different rules, which may not be the easiest to pick up.
That said, this is where roulette strategies can truly shine. Different players use different roulette strategies, methods and betting techniques — there are TONNES to choose from. While betting systems do not guarantee success, as roulette is a game of chance, they can help you allocate the funds in your bankroll in the most efficient way possible.
Some of the best roulette strategies include the Martingale betting system, the Paroli roulette strategy and the Fibonacci strategy. However, with this article, we will shed some light on the Romanosky betting system (also called Romanovsky roulette system).
## WHAT IS THE ROMANOSKY ROULETTE SYSTEM?
If you’re one of the more experienced players in the online casino scene, the Romanosky betting system is probably a strategy you’ve come by before. Here’s a quick breakdown:
The Romanosky system is a roulette betting technique that involves covering 32 numbers on the roulette wheel with eight different units, leaving only five uncovered numbers. It does this by playing two dozen bets and two corner bets to cover as much of the roulette table as possible.
Unlike betting on a single number, this strategy is often used by high rollers with a large bankroll. But it can also work for players who prefer to play it safe.
The majority of roulette strategies out there were designed to work solely on outside bets, but this strategy involves betting on several number combinations.
## HOW DOES THE ROMANOSKY ROULETTE STRATEGY WORK?
The Romanosky system works best on a French or European roulette table. It can work on an American roulette game, but the double zero would really throw off the math, so it’s best to stick to single-zero tables.
There are six variations of a Romanosky bet but don’t worry; it’s not as daunting as it might sound. No matter what variation you choose to use, each wager will cover a large part of the roulette wheel. The only difference is that different bets cover different parts of the wheel.
### ROMANOSKY BET 1
The first bet provides you with one chip on your win. It involves placing three units on the first and second dozens bet, one unit on the square that covers numbers ranging from 25-29 and another unit on the square that covers numbers ranging from 32-36. Your uncovered numbers will be zero, 27, 30, 31, and 34. This stake offers a general probability of winning of 86%.
### ROMANOSKY BET 2
The second bet selection involves placing three units on dozens, one unit on the square covering numbers ranging from 26-30 and one unit on the square covering 31-35. With this stake, players will receive one betting chip profit on each win. Newbie players can begin by betting on a single number. All uncovered numbers include zero, 25, 28, 33, and 36.
It is important to note that the dozens bets must not cover the numbers of the corner bets.
### ROMANOSKY BET 3
Just like the first two bets, the same strategy can be applied to bet three, four, five and six, only this time, the units placed on the dozens and corners bets are a little different.
For the third bet in the Romanosky strategy, players must place three units on the first and third dozen, and one unit on the corners that cover 13-17 and 20-24. Uncovered numbers include the zero, 5, 18, 19, and 22.
### ROMANOSKY BET 4
This fourth stake is slightly different from the previous bet. Here, three units must be placed on the first and third dozen, and two corner bets must be placed with one betting unit on 14-18 and 19-23. Uncovered numbers include the zero, 13, 16, 21, and 24. With this bet, you will wager a total of eight chips with a profit of one single chip.
### ROMANOSKY BET 5
The fifth bet uses three units on the second and third dozen bets, and one for the corner bets that cover numbers ranging from one to five and eight to 12. Thus, the uncovered numbers include zero, three, six, seven, and 10.
### ROMANOSKY BET 6
For bet six, players must place three units on the second and third dozen, and one for the corners two through six and seven through 11, leaving zero, one, four, nine and 12 as uncovered numbers.
## HOW TO USE THE ROMANOSKY ROULETTE STRATEGY
It’s pretty safe to say that the Romanosky strategy is quite straightforward — once you’ve gotten down one bet, you can pretty much discard the rest, as they all lead to the same outcome. That said, while this may be true for casino experts, some roulette rookies may not be as convinced. Check out these tips for using the Romanosky strategy:
• Determine the amount you want to bet per round and split it into eight equal units. Doing so will help protect your funds. Each unit has to be as large as the minimum stake on the table.
• The minimum bet for the Romanosky strategy must be 8x the minimum bet of the table.
• By placing chips on dozens and corner bets, you will be effectively covering the majority of the wheel, allowing you to cover previous losses with low risk.
• This combination of bets offers an 86% odds of winning despite the one-chip profit.
## PROS AND CONS OF ROMANOSKY ROULETTE SYSTEM
Like all betting systems, the Romanosky strategy comes with both benefits and pitfalls. It is always important to remember that no matter what strategy you choose to use, the game of roulette is a game of chance, so there is never a guarantee that you’ll actually hit any of the numbers you bet on.
This is also true due to the house edge, which describes the mathematical advantage the house has over the player — the house will always win in the long run. That is not to say that the Romanosky strategy is not worth using. Let’s take a look at the pros and cons:
One of the main reasons why players should take advantage of the Romanoksy strategy is that it’s super easy to understand and learn. Sure, there are six different types of bets they can place, but all players need to do is learn one, and they’re good to go!
On top of that, the fact that players are covering 32 numbers, including the middle column, offering a whopping 86% odds of winning, makes this betting technique extremely attractive — especially to new players.
The best part is that the Romanosky system is not only tailored to rookies. Players with more knowledge of the game and larger budgets looking to make a profit can make use of this strategy by combining it with negative progression systems. While raising bets after losing a round can be risky, it might help cover losses.
So, with that, the two most prominent advantages of using this wagering system include high odds of winning and versatility.
The disadvantages of using this method are quite plain to see. This strategy banks off of betting high amounts and earning low profit in return as players are required to bet eight units with one in return.
So, if players are counting on recovering their losses, those losses will be recovered rather slowly. This system is a low-risk-low-gain type of game strategy, which some players might not find worth using if they’re looking for a larger profit margin.
## ROMANOSKY ROULETTE STRATEGY TIPS AND TRICKS
Are you all set and ready to try out this strategy? Here are a few tips and tricks to keep in mind to make the most out of your roulette gaming experience!
### USE ONE VARIANT OR SHUFFLE THROUGH DIFFERENT ONES
If you enjoy chasing hot and cold numbers, shuffling through the Romanosky stake variants will do just the trick! Remember, each variant covers different sets of numbers but all cover 32 numbers, so if you see one particular number trending, or one falling off a little, switch up the variant to cover the numbers you prefer.
### MAXIMISE THE STRATEGY’S VERSATILITY
This is a super great tip when playing on tables with number multipliers. Evolution’s Lightning Roulette, for example, features lightning multipliers that appear on the table at the beginning of every round. These multipliers help boost your payout if you’re lucky to bet on a number featuring one of these zapping bonuses.
By using this strategy to cover the majority of the board, you’ll be effectively chasing one of these numbers.
### COMBINE THE ROMANOSKY WITH OTHER WAGERING METHODS
While the Romanosky offers a pretty low payout, the brilliant thing about this technique is that it can be combined with other betting systems. Many players use the Romanosky in conjunction with the Martingale system, but you can use any other progression system you like!
### THE HOUSE ALWAYS WINS
This is probably the most crucial tip we can give you. While the Romanosky method brags about an 86% chance of winning, it is not full-proof. The house will win at some point, as depicted by the house edge. But, to increase your chances, you can try playing at tables with a lower edge, like French roulette, for instance.
### PRACTISE IN THE DEMO MODE
You will notice a vast array of First-Person roulette games when playing online casinos. These tables work with a computer, so you’ll play against an algorithm rather than a live dealer. And this means that they also come with a free demo.
The demo is a free-play version of the game that allows you to use virtual funds to spin the wheel rather than your actual cash. You won’t be able to win anything, but you can use the demo to give the Romanosky method a test drive before using it for real.
## TRY THE ROMANOSKY SYSTEM WHILE PLAYING ROULETTE AT LV BET
Roulette is a game that is packed with entertainment, and an equally entertaining casino experience is precisely what will take your gaming to the next level.
In an exciting industry full of online casinos, LV BET is among the best of them. Our enthusiastic members dedicate time to posting potent and insightful content to spread knowledge to a wider audience of casino players. At LV BET, inexperienced players gain a vast array of tips and tricks, as well as fun casino bonuses and sweet deals.
Watch the roulette wheel spin as you have a whack at all the best roulette strategies on amazing games like Lightning Roulette and Immersive Roulette. If you’re looking for casino quality — you’ll definitely get it at LV BET!
## FAQ
### ✅ IS THE ROMANOSKY ROULETTE STRATEGY LEGAL?
Despite its effectiveness in boosting your odds of winning with low stakes, the Romanosky strategy is totally legal to use at casinos. Do not forget, however, that no strategy is full-proof. At one point or another, you will wrack up a few losing bets — but that’s all part of the experience!
### ✅ CAN I USE ROMANOSKY SYSTEM FOR DIFFERENT CASINO GAMES?
No, this strategy can only be used on roulette as it involves placing bets on specific numbers on the wheel.
### ✅ HOW TO MEMORISE ROMANOSKY CHARTS?
You don’t have to! Once you figure out one, you’ll practically know them all. The only difference between each chart is that they place units on different combinations of numbers. | 2,488 | 11,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.933247 |
http://www.let.rug.nl/~gosse/termpedia2/termpedia.php?language=dutch_general&density=7&link_color=000000&termpedia_system=perl_db&url=http%3A%2F%2Fen.wikipedia.org%2Fw%2Findex.php%3Ftitle%3DSearch_game%26amp%3Baction%3Dedit%26amp%3Bsection%3D1 | 1,563,518,676,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526064.11/warc/CC-MAIN-20190719053856-20190719075856-00070.warc.gz | 237,821,047 | 14,806 | # Search game
A search game is a two-person zero-sum game which takes place in a set called the search space. The searcher can choose any continuous trajectory subject to a maximal velocity constraint. It is always assumed that neither the searcher nor the hider has any knowledge about the movement of the other player until their distance apart is less than or equal to the discovery radius and at this very moment capture occurs. As mathematical models, search games can be applied to areas such as hide-and-seek games that children play or representations of some tactical military situations. The area of search games was introduced in the last chapter of Rufus Isaacs' classic book "Differential Games"[1] and has been developed further by Shmuel Gal[2][3] and Steve Alpern.[3] The princess and monster game deals with a moving target.
## Strategy
A natural strategy to search for a stationary target in a graph is to find a minimal closed curve L that covers all the arcs of the graph. (L is called a Chinese postman tour). Then, traverse L with probability 1/2 for each direction. This strategy seems to work well if the graph is Eulerian. In general, this random Chinese postman tour is indeed an optimal search strategy if and only if the graph consists of a set of Eulerian graphs connected in a tree-like structure.[4] A misleadingly simple example of a graph not in this family consists of two nodes connected by three arcs. The random Chinese postman tour (equivalent to traversing the three arcs in a random order) is not optimal, and the optimal way to search these three arcs is complicated.[2]
## Unbounded domains
In general, the reasonable framework for searching an unbounded domain, as in the case of an online algorithm, is to use a normalized cost function (called the competitive ratio in Computer Science literature). The minimax trajectory for problems of these types is always a geometric sequence (or exponential function for continuous problems). This result yields an easy method to find the minimax trajectory by minimizing over a single parameter (the generator of this sequence) instead of searching over the whole trajectory space. This tool has been used for the linear search problem, i.e., finding a target on the infinite line, which has attracted much attention over several decades and has been analyzed as a search game.[5] It has also been used to find a minimax trajectory for searching a set of concurrent rays. Optimal searching in the plane is performed by using exponential spirals.[2][3][6] Searching a set of concurrent rays was later re-discovered in Computer Science literature as the 'cow-path problem'.[7]
## References
1. ^ Rufus Isaacs, Differential Games, John Wiley and Sons, (1965),
2. ^ a b c S. Gal, Search Games, Academic Press, New York (1980)
3. ^ a b c S. Alpern and S. Gal, The Theory of Search Games and Rendezvous, Springer (2003).
4. ^ S. Gal, On the optimality of a simple strategy for searching graphs, Int. J. Game Theory (2000).
5. ^ A. Beck and D.J. Newman. Yet More on the linear search problem. Israel J. Math. (1970).
6. ^ M. Chrobak, A princess swimming in the fog looking for a monster cow, ACM Sigact news, 35(2), 74–78 (2004).
7. ^ MY Kao, JH Reif and SR Tate, Searching in an unknown environment: an optimal randomized algorithm for the cow-path problem, SODA 1993. | 754 | 3,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.931101 |
https://www.hindawi.com/journals/ads/2015/645746/ | 1,656,310,517,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00459.warc.gz | 843,852,754 | 150,091 | / / Article
Research Article | Open Access
Volume 2015 |Article ID 645746 | https://doi.org/10.1155/2015/645746
S. M. Hatefi, S. A. Torabi, "A Common Weight Linear Optimization Approach for Multicriteria ABC Inventory Classification", Advances in Decision Sciences, vol. 2015, Article ID 645746, 11 pages, 2015. https://doi.org/10.1155/2015/645746
# A Common Weight Linear Optimization Approach for Multicriteria ABC Inventory Classification
Revised22 Oct 2014
Accepted11 Dec 2014
Published01 Jan 2015
#### Abstract
Organizations typically employ the ABC inventory classification technique to have an efficient control on a huge amount of inventory items. The ABC inventory classification problem is classification of a large amount of items into three groups: A, very important; B, moderately important; and C, relatively unimportant. The traditional ABC classification only accounts for one criterion, namely, the annual dollar usage of the items. But, there are other important criteria in real world which strongly affect the ABC classification. This paper proposes a novel methodology based on a common weight linear optimization model to solve the multiple criteria inventory classification problem. The proposed methodology enables the classification of inventory items via a set of common weights which is very essential in a fair classification. It has a remarkable computational saving when compared with the existing approaches and at the same time it needs no subjective information. Furthermore, it is easy enough to apply for managers. The proposed model is applied on an illustrative example and a case study taken from the literature. Both numerical results and qualitative comparisons with the existing methods reveal several merits of the proposed approach for ABC analysis.
#### 1. Introduction
Inventory classification using ABC analysis, which is based on the Pareto principle, is one of the most widely employed inventory control techniques in practice [1, 2]. ABC analysis classifies the inventory items or stock keeping units (SKUs) into three classes, namely, A (very important), B (moderately important), and C (relatively unimportant) requiring different level of control for each class and, at the same time, setting the same service level for all SKUs in a class. More details on inventory control policies for these classes of items can be found in Silver et al. [3].
Traditional ABC analysis considers only a single measure, most often annual dollar usage, to classify inventory items. However, in addition to this criterion, inventory holding unit cost, part criticality, length and variability of replenishment lead time, commonality, obsolescence, substitutability, number of requests per year, scarcity, durability, reparability, order size requirement, stockability, demand distribution, and stock-out unit penalty cost are recently recognized as other important criteria which affect inventory classification [1, 4, 5]. Thus, it can be realized that the traditional ABC analysis may be an inefficient method for appropriate classification of inventory items in practice [68].
Several methods have been suggested to solve the problem of multiple criteria inventory classification (MCIC) in the literature. Bhattacharya et al. [9] and Rezaei and Dowlatshahi [10] provide comprehensive reviews of various methods introduced in the literature for MCIC issue. The implementation of analytic hierarchy process (AHP) to study this issue is addressed by Flores et al. [11] and Partovi and Burton [12]. More recently, a fuzzy AHP-DEA approach is proposed by Hadi-Vencheh and Mohamadghasemi [13] to solve the MCIC problem. However, when applying the AHP method, it is often a difficult task for the decision maker to accurately assign exact values to pairwise comparisons. Also, expert’s opinion and judgment may play important role in deriving the criteria weights which consequently can affect ABC classification results considerably.
Bhattacharya et al. [9] propose a distance-based multicriteria consensus framework based on the concepts of positive-ideal and negative-ideal solutions for the ABC analysis. The authors also affirm that constructing fuzzy models such as fuzzy TOPSIS and neurofuzzy hybrid models would be suitable when considering the ambiguity of attribute values. An application of the case-based distance model to solve the MCIC problem can be seen in Chen et al. [14]. Recently, a new approach based on loss profit is proposed to deal with ABC analysis [15].
Application of artificial intelligence methods such as genetic algorithms, particle swarm optimization, and artificial neural networks to solve the MCIC problem can be found in Guvenir and Erel [6], Tsai and Yeh [16], and Partovi and Anandarajan [8], respectively. Artificial intelligence based classification techniques are addressed by Yu [17] to resort MCIC problem. Also, a fuzzy classification approach is proposed in the case where there exists nominal and nonnominal attributes [2]. Similarly, Rezaei and Dowlatshahi [10] present a rule-based method for classifying inventory items in a multicriteria setting. However, these methods are too complicated to be applied in practice so that they may not be easily understood by inventory managers.
A number of optimization based methods have also been developed to solve the MCIC problem. Ramanathan [1] proposes a weighted linear optimization model (hereafter R-model) for classifying inventory items with multiple criteria. The spirit of R-model is based upon the concept of data envelopment analysis (DEA) which has no subjectivity in determining the weights of criteria; that is, the weights are endogenously and repeatedly generated by a DEA-like model. Zhou and Fan [18] extend the R-model by incorporating some balancing features for MCIC using two sets of weights that are most favorable and least favorable for each item. Then, the final performance score of each item is obtained by aggregating the best and worst performance scores and using a control parameter called whose value is determined by the decision maker subjectively. In a similar way, Chen [19] proposes the peer estimation approach in which the performance score of each inventory item is first measured in the most favorable and least favorable senses and then they are aggregated without any subjectivity. Chen [20] proposes another alternative model that resorts to virtual items and the concepts of TOPSIS for ABC analysis. The proposed model provides a unique inventory classification without any subjectivity.
Stanford and Martin [21] introduce a general ABC inventory classification system as the foundation for a normative model of the maintenance cost structure and stock turnover characteristics of a large, multi-item inventory system with constant demand. Ng [5] proposes a weighted linear optimization model for ABC inventory classification. The author also introduces a transformation technique to simplify the classification procedure (hereafter NG-model) which aids the inventory managers with obtaining the aggregated scores of inventory items without a linear optimizer. Hadi-Vencheh [22] provides an improved version of the NG-model which is a nonlinear programming model (hereafter HV-model). Both the NG and HV models require prior assumption on the importance order of the criteria which is subjectively determined by the decision maker. More recently, Teunter et al. [23] present an alternative approach for using multicriteria methodologies based on multiple way classifications. The authors propose a new cost criterion for ranking SKUs from an inventory cost perspective that accounts for demand volume, holding cost (purchase price), shortage cost (criticality), and average order quantity.
Park et al. [24] proposed a cross-evaluation-based weighted linear optimization model for the MCIC problem. Their proposed method incorporates a cross-efficiency evaluation approach into a weighted linear optimization model for finer classification of inventory items. For classification of inventory items, Soylu and Akyol [25] incorporated the preference of the decision-maker into the decision making process. They applied two utility-function-based sorting methods to solve the MCIC problem. Bacchetti et al. [26] proposed a hierarchical multicriteria classification method for inventory management purposes and applied it in a case study of the spare parts business of a household appliance manufacturer. Rezaei and Salimi [27] developed an interval programming model for ABC inventory classification. Their proposed model provides optimal results instead of an expert-based classification and it does not require precise values of item parameters. Torabi et al. [28] introduced an imprecise data envelopment analysis model to classify inventory items in the case where there exist both quantitative and qualitative criteria. Furthermore, Hatefi et al. [29] developed an iterative DEA-like model to solve the MCIC problem under quantitative and qualitative criteria. Lolli et al. [30] introduces a new hybrid method based on the AHP and the K-means algorithm to solve the MCIC problem.
This paper proposes an alternative optimization-based model in which the composite performance scores of all inventory items are calculated simultaneously via a set of common weights. The proposed common weight linear optimization model has a notable computational saving in terms of the number of required LP models that must be solved and therefore can considerably reduce the processing time when controlling a large number of inventory items. Also, no subjective information is required to run the proposed model which is essential in an accurate and fair decision environment.
The rest of the paper is organized as follows. In Section 2, a brief discussion is provided about the R-model as the first optimization-based ABC classifier. Then, the proposed common weight linear optimization model is explained in detail. An application of the proposed model for ABC analysis is shown by a small numerical example as well as a case study taken from the literature in Sections 3 and 4, respectively. Section 5 provides a comprehensive comparison and discussion about the proposed approach and the most relevant existing methods in the literature, that is, the R-model, the ZF-model, Chen’s [20] model, the NG-model, and the HV-model. Finally, concluding remarks are reported in Section 6.
#### 2. An Alternative Common Weights MCIC Approach
Suppose that there are inventory items that are being classified as A, B, and C classes based on incommensurable criteria. Let denote the value of criterion with respect to inventory item . For simplicity, it is assumed that all the criteria are of benefit type, that is, positively related to the importance of an item. It is noteworthy to mention that the cost type criteria (i.e., those criteria that are negatively related to the importance level of items) can easily be converted into the benefit type ones by considering their reciprocal values [1]. Furthermore, in order to avoid any problem arising from the criteria scale differences, all measures are first normalized by using the following linear normalization procedure [5, 31]:
Notably, the above transformation formula converts all measurements into the interval. Hereafter for simplicity, we suppose is the normalized value of criterion with respect to the inventory item . Normalization of criteria has two main advantages. First, it avoids any problem arising from the criteria scale differences. Second, normalized data are fairly easy to interpret [32]. It is worthy to mention that there exist several normalization methods in the literature. Some of the well-known normalization methods are listed below [33].(i)Ranking Method. It is the simplest normalization technique, which is not affected by outliers.(ii)Standardization (or -Scores) Method. It converts indicators to a common scale with a mean of zero and standard deviation of one.(iii)Min-Max Method. It normalizes criteria to have an identical range by subtracting the minimum value and dividing by the range of the indicator values.(iv)Distance to a Reference Method. It measures the relative position of a given indicator vis-à-vis a reference point.
The concerned problem is aggregating the performances of an inventory item in terms of different criteria into a single score (called aggregated performance or composite index) for the subsequent ABC classification. The R-model introduced by the Ramanathan [1] is as follows: where for indicates the aggregated performance of inventory item with respect to all of underlying criteria and is the weight of criterion which is generated endogenously by the R-model when evaluating item . Objective function of the R-model is similar to the simple additive weighting (SAW) aggregation method, while the weights of criteria are changeable for each item and calculated endogenously by the model. Consequently, a set of indices for all items are provided by solving the R-model repeatedly for each inventory item. It should be noted that the R-model is very similar to an output maximizing multiplier DEA model with multiple outputs and one constant input [1]. It is noted that there are several weighting and aggregation methods in the literature. Some are taken from statistical models, such as factor analysis, data envelopment analysis, and unobserved components models (UCM), or from participatory methods like budget allocation processes (BAP), analytic hierarchy processes (AHP), and conjoint analysis (CA). Equal weighting (EW) is one of the simplest weighting methods. Furthermore, three simple but popular aggregation methods are the simple additive weighting (SAW) method, the weighted product (WP) method, and the weighted displaced ideal (WDI) method [33].
In spite of popularity of the R-model, Zhou and Fan [18] mathematically proved that if an item has a value dominating other items in terms of a certain criterion, it would always obtain an aggregated performance score of 1 even if it has severely bad values with respect to other criteria. This may lead to the situation where an item with a high value in an unimportant criterion but with low values in other important criteria is inappropriately classified as class A, which may not reflect the real position of this inventory item. Furthermore, the obtained weights provided by the R-model are often unrealistic and lead to poor discriminatory power among SKUs. That is, for each inventory item, the criteria which have good performance may receive extremely high weights and those having bad performance may receive extremely low weights leading to extreme weights that are often unrealistic and impractical because of ignoring the impact of criteria with extremely low weight values for ABC classification [31, 34]. Accordingly, this paper proposes a new MCIC model in which all inventory items are evaluated using a set of common weights to enable a fair comparison among them that differs with the R-model in which the inventory items are evaluated by different sets of weights. For doing so, let denote the performance score deviation of item from the unity when it is under evaluation. Now, by considering , the R-model can be reformulated as follows when evaluating item : where denotes the performance score deviation of item , . Notably, two separate indices and are used in Model (2) that both of them are bounded as , . The first index is related to the th linear programming model that should be solved for measuring the score of item . The first constraints of Model (2) indicate that when evaluating item , the composite performance score of each item, that is, for , plus the related performance deviation from the unity must be equal to 1.
It is worthwhile to point out when item is under evaluation, the resulting weight vector is similar for both Models (1) and (2) that provides the most favorable weights for this item. Also, provided by Model (1) is equal to that is provided by Model (2). Notably, Model (2) should also be solved times to minimize the performance score deviation of each item. However, solving this model for each item separately provides totally different weight vectors for criteria, each of which might be unrealistic as mentioned before.
To overcome this deficiency, we propose a common weight DEA-like model as follows. The objective function of Model (2) only involves the performance score deviation of item . However, to maximize the aggregated performance scores of all inventory items simultaneously as it is needed in a common weight framework, we need to reformulate this model via a set of common weights. For doing so, this paper uses the minimax approach to minimize the maximum efficiency deviation among all items. It is noteworthy to mention that the minimax efficiency score is employed to avoid yielding unrealistic weight dispersion and improve the discriminating power in the context of DEA models [35]. In this manner, the proposed common weight linear model can be written as follows: where denotes the common weight with respect to criterion among all inventory items which is endogenously generated by solving Model (3). Notably, the constraints ; assure that is the maximum of values while they do not change the feasible region of decision variables as discussed in Li and Reeves [35]. When Model (3) is solved, the composite index of all inventory items can be calculated by for . After determining the item scores, we sort them in a nonincreasing order and then classify them by the Pareto rule based principle of ABC analysis. The classifying distribution of inventory items performs based on their scores. For example, the classifying distribution, that is, 20% of the inventory items with best scores in class A, 30% in class B, and 50% in class C, can be considered.
According to the literature of common weights DEA models [31, 34], all inventory items are concurrently evaluated using a set of common weights to enable a fair comparison among them which highly differs with the R-model in which each inventory item is evaluated by its own favored weights. Furthermore, the common weight structure of the proposed method improves the discriminatory power among all inventory items. Notably, unrealistic weight distribution is a main difficulty of some DEA models where some input-output weights achieve the extreme or zero values which in turn leads to wrong assessment of some DMUs as efficient units. However, as discussed by Li and Reeves [35] and Bal et al. [36], the minimax approach provides realistic input-output weights and improves the discrimination power among DMUs. Accordingly, the main reasons for applying the minimax approach to derive the final scores of inventory items are as follows.
The minimax approach does not lead to the most favorable weights for each inventory item under evaluation as the R-model does. Therefore, the items’ scores generated by this approach are tighter than those provided by the R-model. In particular, if inventory item achieves score of 1 by employing the minimax approach, it must also achieve score of 1 by employing the R-model, since the minimax score requires . However, if inventory item achieves score of 1 by employing the R-model, it may receive a score less than 1 by applying the minimax approach, because or does not necessarily means that is being minimized. According to this discussion, it can be affirmed that the minimax approach generally results in fewer efficient items with score of 1 leading to discrimination power improvement of Model (3). Moreover, since is defined as a function of all deviation variables and each deviation variable is associated with a constraint, therefore, minimizing is equivalent to imposing rigorous constraints on common weight variables. In this way, the range of criteria weights is effectively restricted leading to moderating the homogeneity of weights dispersion on a common base and hence Model (3) provides a realistic and reasonable set of weights for the ABC analysis.
In spite of above descriptive advantages of proposed Model (3), its attractiveness can be more declared through clarifying its interesting mathematical properties as follows.
(a) Model (3) Is always Feasible
Proof. Suppose that all required data (i.e., the values) in Model (3) are first normalized. So, we would have , . Then, the following solution is feasible for Model (3).
Consider for and ; .
(b) ,
Proof. Taking into account the nonnegativity of the variables and for each and and the fact that , , it would be clear that , . Therefore, according to the constraints , we can conclude that , .
(c) The Common Weights Generated by Proposed Model (3) Do Not Exceed 1 (i.e., , )
Proof. Assume that for at least criterion for which we have , Model (3) yields . Then, the constraint enforces the corresponding deviation variable, that is, must be negative, a contradiction. Therefore, the common weights generated by proposed Model (3) do not exceed 1 and we have , .
(d) Regarding Proposed Model (3). If th item be minimax efficient, that is, , then it is recognized as an efficient item by the R-model as well; that is, . However, if the th item be efficient by the R-model, it may be minimax inefficient by employing Model (3), since it may not be assured that is being minimized. Also, it can be concluded that if by using Model (1), then Model (3) yields and consequently it will be inefficient since . On the other hand, if a specific item achieves a score smaller than 1 in terms of Model (1), then it also gives a score less than 1 when applying Model (3).
#### 3. Numerical Example
In this section, we provide a numerical example to illustrate the advantages of the proposed approach. The example presented in Table 1 is part of the data set used as a case study in the literature [1, 5, 11, 18, 20, 22]. More details about this data set are reported in the next section. The data set contains 7 inventory items with three benefit-type criteria, that is, average unit cost (AUC), annual dollar usage (ADU), and lead time (LT). After normalization, Models (1) to (3) are applied and related results are given in Table 2.
Item number Average unit cost (\$) Annual dollar usage (\$) Lead time (week) 1 Item S41 19.8 79.2 2 2 Item S42 37.7 75.4 2 3 Item S43 29.89 59.78 5 4 Item S44 48.3 48.3 3 5 Item S45 34.4 34.4 7 6 Item S46 28.8 28.8 3 7 Item S47 8.46 25.38 5
Model (1) Model (2) Model (3) Item score Item score Item score Item 41 1 0 1 0 1 0 1 0 0.38861 Item 42 1 0.1504 0.9572 0 1 0.1504 0.9572 0 0.58458 Item 43 1 0 1 0.6014 1 0 1 0.6014 0.80642 Item 44 1 1 0 0 1 0.6747 0.543 0.4696 0.71328 Item 45 1 0 0 1 1 0 0.7426 0.8755 1 Item 46 0.5498 0.9198 0 0.4011 0.5498 0.9198 0 0.4011 0.38861 Item 47 0.6 0 0 1 0.6 0 0 1 0.38861
To achieve inventory item scores, that is, , Model (1) is separately solved for each inventory item by changing the objective function accordingly. In a similar way, Model (2) is repeatedly solved for each inventory item. For example, when evaluating inventory item 44, Models (1) and (2) can be written as follows.
Model (1). Consider
Model (2). Consider where , , and denote the weights of AUC, ADU, and LT criteria, respectively. After solving the above Models (1) and (2), the respective weight vectors and are found, respectively. Notably, item 44 dominates all items in terms of AUC and hence Model (1) generates high weight value of 1 for this criterion and assigns for other criteria weights, which is clearly unrealistic because of ignoring the impact of ADU and LT criteria completely and at the same time assigning an extreme weight value for the AUC. The obtained score for this item is 1 that is calculated as by Model (1) and () by Model (2). It is worth pointing out that the solution generated by using Models (1) and (2) for evaluating this item is not unique. It is actually software dependent. For example, the solutions of and are optimal for Model (1) when evaluating item 44. According to the results of Table 2, seven different weight vectors are produced by employing Models (1) or (2) to calculate final scores of all items. However, these items can be more fairly and confidently evaluated by just a unique set of common weights by solving proposed Model (3) that is written as follows.
Model (3). Consider where , , and denote the common weights of AUC, ADU, and LT, respectively. Proposed Model (3) provides the score of all items by a more distributed weight vector that is both unique and more reasonable than those generated by Models (1) and (2). Furthermore, when comparing the items’ score columns presented in Table 2, it can be affirmed that Model (3) gives more discriminated scores than those obtained by Models (1) and (2).
#### 4. Case Study
The proposed common weight linear optimization model is applied for the same multicriteria inventory classification problem as discussed in the literature [1, 5, 11, 18, 20, 22]. The data include 47 inventory items (SKUs) with four criteria, namely, annual dollar usage (ADU), average unit cost (AUC), critical factor (CF), and lead time (LT). However, we follow Ng [5], Zhou and Fan [18], Chen [20], and Hadi-Vencheh [22] studies and consider just ADU, AUC, and LT criteria for ABC classification, since CF is a categorical and discontinuous criterion and not suitable for linear models. Note that all the three criteria that are presented in Table 3 are positively related to the importance level of inventory items. The same classifying distribution, that is, 10 class A, 14 class B, and 23 class C, is adopted for illustration and comparison purposes as considered in the earlier similar works.
Item # Average unit cost (\$) Annual dollar usage (\$) Lead time Proposed performance scores ABC classification Proposed model Chen modela ZF-model R-modelb HV-model NG-model S2 210 5670 5 1 A A A A A A S29 134.34 268.68 7 0.7002 A A A A A A S10 160.5 2407.5 4 0.6817 A A A B A A S13 86.5 1038 7 0.6027 A A A A A A S9 73.44 2423.52 6 0.5648 A A A A A A S14 110.4 883.2 5 0.5454 A A A B A B S28 78.4 313.6 6 0.4950 A A A A B B S18 49.5 594 6 0.4288 A B A A B B S45 34.4 34.4 7 0.4240 A B B A B B S3 23.76 5037.12 4 0.4199 A A A A A A S31 72 216 5 0.4165 B B B B B B S40 51.68 103.36 6 0.4153 B B B B B B S8 55 2640 4 0.4089 B A B B B B S1 49.92 5840.64 2 0.4063 B A A A A A S23 86.5 432.5 4 0.4062 B B B C B B S5 57.98 3478.8 3 0.3924 B B B B A A S39 59.6 119.2 5 0.3795 B B B B B B S19 47.5 570 5 0.3649 B B B B B B S34 7.07 190.89 7 0.3571 B C B A B B S33 49.48 197.92 5 0.3556 B B B B B B S22 65 455 4 0.3496 B B B C C C S20 58.45 467.6 4 0.3326 B C B C C C S15 71.2 854.4 3 0.3243 B B C C C C S12 20.87 1043.5 5 0.3124 B C B B B B S37 30 150 5 0.3016 C C B B C C S6 31.24 2936.67 3 0.2996 C B C C B A S43 29.89 59.78 5 0.2978 C C C B C C S7 28.2 2820 3 0.2869 C B C C B B S38 67.4 134.8 3 0.2859 C C C C C C S35 60.6 181.8 3 0.2695 C C C C C C S16 45 810 3 0.2526 C C C C C C S4 27.73 4769.56 1 0.2473 C B C B A A S21 24.4 463.6 4 0.2415 C C C C C C S47 8.46 25.38 5 0.2392 C C C B C C S44 48.3 48.3 3 0.2314 C C C C C C S17 14.66 703.68 4 0.2249 C C C C C C S27 84.03 336.12 1 0.2231 C C C C C C S36 40.82 163.28 3 0.2160 C C C C C C S24 33.2 398.4 3 0.2048 C C C C C C S26 33.84 338.4 3 0.2042 C C C C C C S32 53.02 212.08 2 0.1929 C C C C C C S46 28.8 28.8 3 0.1785 C C C C C C S42 37.7 75.4 2 0.1466 C C C C C C S30 56 224 1 0.1438 C C C C C C S11 5.12 1075.2 2 0.0989 C C C C C C S41 19.8 79.2 2 0.0989 C C C C C C S25 37.05 370.5 1 0.0989 C C C C C C
Source: Chen [20].
bSource: Zhou and Fan [18].
Before applying our proposed model, all measures are normalized. Proposed Model (3) generates the common weights 0.2290, 0.5474, and 0.3453 for ADU, AUC, and LT criteria, respectively. The fifth column of Table 3 shows the optimal aggregated performance scores of inventory items by solving proposed Model (3). The ABC classification with the above three criteria and by using the proposed model, the Chen [20] model, the R-model, the ZF-model, the NG-model, and the HV-model [22] has also been listed in Table 3. As our proposed model and the Chen-model [20] and ZF-model are various extensions of the R-model, therefore, our quantitative comparisons are considerably focused on the results of these models.
Zhou and Fan [18] discuss that the ZF model which is an extended version of the R-model provides a more reasonable and encompassing index for ABC classification. When our model is compared to the ZF-model, 4 out of 47 inventory items do not have the same classification. For class A items identified by our proposed model, nine items are classified as group A items in both models. Also, 12 out of 14 class B items and 22 out of 23 class C items are matched in both models. For instance, item S34 is classified as class A by the R-model as it has the highest lead time and this is the only criterion taken into account in the model. However, it is classified as class B in both our model and the ZF model since, in addition to the high lead time, they also take into consideration its low average unit cost.
When comparing our proposed model with the Chen-model [20], 10 out of 47 items have different classifications. Items 18 and 45 are classified as class B by the Chen-model [20], while they are moved up to class A by applying Model (3). Two class A items and three class C items resulting from the Chen-model [20] exchanged their classifications and shifted to class B when applying proposed Model (3). Furthermore, three class B items sorted by the Chen-model [20] are moved down to class C by using the proposed approach. In particular, item 15 with the normalized scores of 0.323 (AUC), 0.143 (ADU), and 0.333 (LT) dominates item 44 with the normalized scores of 0.211 (AUC), 0.003 (ADU), and 0.333 (LT). Consequently, as item 44 is rated as class C by all methods, it is more appropriate that item 15 is shifted to a better class like B which coincides with the results of proposed Model (3) and the Chen-model [20]. As another comparison, item 9 has the normalized data as 0.333 (AUC), 0.412 (ADU), and 0.833 (LT), which is classified as class A by applying all models, and the normalized data of item 8 are as 0.243 (AUC), 0.445 (ADU), and 0.5 (LT). Item 9 dominates item 8 in terms of the AUC and LT criteria. Also, both items are relatively similar in terms of the ADU. Therefore, it is expected that item 8 might be inferior when compared to item 9. Therefore, it is more appropriate that item 8 be moved down to a second-rate class like B which coincides with classification of Model (3) as well as four other models presented in Table 3. However, item 8 is rated as class A when employing the Chen-model [20].
According to the results, the classification results reported by the Chen-model [20] are different for 3 out of 47 items, that is, items 8, 34, and 12, from other classifications while in the proposed classification, just item 1 has different classification from those provided by the other models. The reasons for this difference can be justified as follows. Consider the normalized data, , for item 22 which is classified as class B by the ZF-model, the Chen-model [20], and our proposed model, and for item 1 which is evaluated as class A by five models presented in Table 3. It is clear that item 22 dominates item 1 in terms of AUC and LT criteria. Furthermore, proposed Model (3) assigns larger values for the weights of AUC and LT than the weight of ADU. In this way, item 22 is considerably superior to item 1 and hence it is more appropriate to put item 1 in an inferior class like class B which matches with our proposed classification. However, the main difference between ours and other models is that our composite index is calculated based on a unique set of common weights among all items.
When the proposed model is compared with the HV-model [22] that is an extended version of the NG-model, 11 out of 47 inventory items are not coincident. Among the class A items recognized in our proposed model, 7 out of 10 items are classified as class A in both models. Furthermore, 9 out of 14 class B items and 20 out of 23 class C items are similarly classified in both models.
#### 5. Discussion
To alleviate deficiencies of the R-model, Zhou and Fan [18] proposed another DEA-based model that aggregates the best and worst performance scores of each item using a control parameter called to obtain the ZF-index for ABC classification. The poor discrimination power in the R-model is also improved in the ZF-model. However, choosing an appropriate value for this subjective parameter is crucial and depends on the decision maker’s preference on two extreme cases. This flexibility may also sometimes cause difficulty for decision makers to make a subjective choice in specifying the value of . Furthermore, different values of may lead to distinct, misleading, and nonunique results. It is noteworthy to mention that linear optimization models must be solved in order to calculate the ZF-index for all SKUs. Furthermore, Chen [20] proposed an alternative model for ABC analysis whose structure is similar to the ZF-model but without any subjectivity.
Ng [5] proposes a novel and easy-to-use method that does not require a linear optimizer for ABC analysis. Besides its many advantages, Hadi-Vencheh [22] discusses that the Ng-model leads to a situation where the NG-index for each item is independent from the obtained weights. Consequently, Hadi-Vencheh [22] improves the NG-model and constructs a nonlinear programming model that keeps the impacts of weights when calculating the final indices. The HV-model is solved for each inventory item repeatedly (i.e., it requires to run nonlinear programming models) and a different set of weights is obtained for calculating the final index of each item. Moreover, both the NG and HV models need prior assumption on the importance order of criteria which is determined subjectively by the decision maker. It should be noted that when the number of criteria is large, it is an overwhelming task for the decision maker to rank all criteria.
Relying on the above discussions, the proposed method generates a set of common weights for evaluating all items simultaneously that obviously improves the discrimination power among them. Also, the proposed method removes any subjectivity which may lead to confusing interpretation of ABC classification results and hence provides unique ABC classification results. Table 4 shows the several merits of our proposed method in summary when comparing it with the recently developed methods for ABC inventory classification.
Qualitative attributes Proposed model Chen-model [20] ZF-model [18] R-model [1] HV-model [22] NG-model [5] Type of optimization model Linear Linear Linear Linear Nonlinear — Type of weights for evaluating each item common changeable changeable changeable changeable Independent Number of models which must be solved 1 At least 2M 2M M M — Requiring subjective information No No Yes No Yes Yes
As highlighted by Chen et al. [37], common weights strategy to DEA has several features. First, the common weights strategy diminishes computational complexity/time compared with the traditional model. Second, common weights models show higher discrimination power than other classical models. However, for a typical MCIC problem, the proposed common weights strategy incorporated in the minimax approach in this paper has the following merits.(1)Despite other existing methods that generate several sets of weights for the criteria, the proposed method obtains a unique set of common weights for evaluating all inventory items which is very essential for a fair classification of items and provides unique results for ABC analysis.(2)The proposed model like the models presented by Chen [19, 20] requires no subjective information leading to provide a unique classification results for inventory managers while some recent MCIC approaches do. Requiring any subjective information may lead to nonunique and misleading results for the ABC analysis, making difficulty in reaching a final decision.(3)The proposed method is very efficient from the computational point of view since only a LP model needs to be solved while the other methods need that several linear or even nonlinear models to be solved.(4)The common weight structure incorporated in our proposed method improves the poor discrimination power of other methods especially the R-model by providing a full-ranked vector of items based upon their composite performance scores.
Relying on the aforementioned merits, it can be concluded that the proposed model is an easy-to-understand and easy-to-use method that helps inventory managers and practitioners to manage inventory items in reality. The proposed method can be applied to industry. Furthermore, it is easy to apply for human managers. The only background required for implementing our proposed model is to know how to solve linear programming models via a commercial software like LINGO. In summary, inventory managers can perform the following steps to classify their inventory items.(1)Provide normalized measures by the transformation procedure presented in the paper.(2)Solve proposed Model (3) on normalized data by the commercial software.(3)Provide item scores by formula for and then sort them in a nonincreasing order.(4)Classify inventory items by the Pareto rule based principle of ABC analysis.
#### 6. Concluding Remarks
This paper addresses the ABC inventory classification problem through the multiple criteria inventory classification (MCIC) approach and proposes a common weight linear optimization method that enables us to classify inventory items using a set of common weights in an efficient and effective way. The numerical results along with some qualitative assessments confirm the superiority of the proposed method when compared with the previously developed relevant techniques.
Recently, finding common weights based on the DM’s preference information is addressed by Jahanshahloo et al. [38] in the DEA literature that measures the preferences of decision maker when generating the common weights. Accordingly, incorporating the inventory manager’s preferences in a new model to derive common weights is proposed for future research in the context of MCIC. Furthermore, all above-mentioned DEA-based methods in this paper can only take into account quantitative criteria for dealing with MCIC problem.
#### Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
#### Acknowledgment
The authors are grateful to the respected reviewers for their valuable comments in preparation of the revised paper.
#### References
1. R. Ramanathan, “ABC inventory classification with multiple-criteria using weighted linear optimization,” Computers and Operations Research, vol. 33, no. 3, pp. 695–700, 2006. View at: Publisher Site | Google Scholar
2. C.-W. Chu, G.-S. Liang, and C.-T. Liao, “Controlling inventory by combining ABC analysis and fuzzy classification,” Computers & Industrial Engineering, vol. 55, no. 4, pp. 841–851, 2008. View at: Publisher Site | Google Scholar
3. E. A. Silver, D. F. Pyke, and R. Peterson, Inventory Management and Production Planning and Scheduling, Wiley, New York, NY, USA, 1998.
4. B. E. Flores and D. C. Whybark, “Implementing multiple criteria ABC analysis,” Journal of Operations Management, vol. 7, no. 1-2, pp. 79–85, 1987. View at: Publisher Site | Google Scholar
5. W. L. Ng, “A simple classifier for multiple criteria ABC analysis,” European Journal of Operational Research, vol. 177, no. 1, pp. 344–353, 2007. View at: Publisher Site | Google Scholar
6. H. A. Guvenir and E. Erel, “Multicriteria inventory classification using a genetic algorithm,” European Journal of Operational Research, vol. 105, no. 1, pp. 29–37, 1998. View at: Publisher Site | Google Scholar
7. J. Huiskonen, “Maintenance spare parts logistics: special characteristics and strategic choices,” International Journal of Production Economics, vol. 71, no. 1–3, pp. 125–133, 2001. View at: Publisher Site | Google Scholar
8. F. Y. Partovi and M. Anandarajan, “Classifying inventory using an artificial neural network approach,” Computers and Industrial Engineering, vol. 41, no. 4, pp. 389–404, 2001. View at: Google Scholar
9. A. Bhattacharya, B. Sarkar, and S. K. Mukherjee, “Distance-based consensus method for ABC analysis,” International Journal of Production Research, vol. 45, no. 15, pp. 3405–3420, 2007. View at: Publisher Site | Google Scholar
10. J. Rezaei and S. Dowlatshahi, “A rule-based multi-criteria approach to inventory classification,” International Journal of Production Research, vol. 48, no. 23, pp. 7107–7126, 2010. View at: Publisher Site | Google Scholar
11. B. E. Flores, D. L. Olson, and V. K. Dorai, “Management of multicriteria inventory classification,” Mathematical and Computer Modelling, vol. 16, no. 12, pp. 71–82, 1992. View at: Google Scholar
12. F. Y. Partovi and J. Burton, “Using the analytic hierarchy process for ABC analysis,” International Journal of Operations & Production Management, vol. 13, no. 9, pp. 29–44, 1993. View at: Publisher Site | Google Scholar
13. A. Hadi-Vencheh and A. Mohamadghasemi, “A fuzzy AHP-DEA approach for multiple criteria ABC inventory classification,” Expert Systems with Applications, vol. 38, no. 4, pp. 3346–3352, 2011. View at: Publisher Site | Google Scholar
14. Y. Chen, K. W. Li, D. Marc Kilgour, and K. W. Hipel, “A case-based distance model for multiple criteria ABC analysis,” Computers and Operations Research, vol. 35, no. 3, pp. 776–796, 2008. View at: Publisher Site | Google Scholar
15. Y.-Y. Xiao, R.-Q. Zhang, and I. Kaku, “A new approach of inventory classification based on loss profit,” Expert Systems with Applications, vol. 38, no. 8, pp. 9382–9391, 2011. View at: Publisher Site | Google Scholar
16. C.-Y. Tsai and S.-W. Yeh, “A multiple objective particle swarm optimization approach for inventory classification,” International Journal of Production Economics, vol. 114, no. 2, pp. 656–666, 2008. View at: Publisher Site | Google Scholar
17. M.-C. Yu, “Multi-criteria ABC analysis using artificial-intelligence-based classification techniques,” Expert Systems with Applications, vol. 38, no. 4, pp. 3416–3421, 2011. View at: Publisher Site | Google Scholar
18. P. Zhou and L. Fan, “A note on multi-criteria ABC inventory classification using weighted linear optimization,” European Journal of Operational Research, vol. 182, no. 3, pp. 1488–1491, 2007. View at: Publisher Site | Google Scholar
19. J.-X. Chen, “Peer-estimation for multiple criteria ABC inventory classification,” Computers & Operations Research, vol. 38, no. 12, pp. 1784–1791, 2011. View at: Publisher Site | Google Scholar | MathSciNet
20. J.-X. Chen, “Multiple criteria ABC inventory classification using two virtual items,” International Journal of Production Research, vol. 50, no. 6, pp. 1702–1713, 2012. View at: Publisher Site | Google Scholar
21. R. E. Stanford and W. Martin, “Towards a normative model for inventory cost management in a generalized ABC classification system,” Journal of the Operational Research Society, vol. 58, no. 7, pp. 922–928, 2007. View at: Publisher Site | Google Scholar
22. A. Hadi-Vencheh, “An improvement to multiple criteria ABC inventory classification,” European Journal of Operational Research, vol. 201, no. 3, pp. 962–965, 2010. View at: Publisher Site | Google Scholar
23. R. H. Teunter, M. Z. Babai, and A. A. Syntetos, “ABC Classification: service levels and inventory costs,” Production and Operations Management, vol. 19, no. 3, pp. 343–352, 2010. View at: Publisher Site | Google Scholar
24. J. Park, H. Bae, and J. Bae, “Cross-evaluation-based weighted linear optimization for multi-criteria ABC inventory classification,” Computers & Industrial Engineering, vol. 76, pp. 40–48, 2014. View at: Publisher Site | Google Scholar
25. B. Soylu and B. Akyol, “Multi-criteria inventory classification with reference items,” Computers and Industrial Engineering, vol. 69, no. 1, pp. 12–20, 2014. View at: Publisher Site | Google Scholar
26. A. Bacchetti, F. Plebani, N. Saccani, and A. A. Syntetos, “Empirically-driven hierarchical classification of stock keeping units,” International Journal of Production Economics, vol. 143, no. 2, pp. 263–274, 2013. View at: Publisher Site | Google Scholar
27. J. Rezaei and N. Salimi, “Optimal ABC inventory classification using interval programming,” International Journal of Systems Science, 2013. View at: Publisher Site | Google Scholar
28. S. A. Torabi, S. M. Hatefi, and B. Saleck Pay, “ABC inventory classification in the presence of both quantitative and qualitative criteria,” Computers and Industrial Engineering, vol. 63, no. 2, pp. 530–537, 2012. View at: Publisher Site | Google Scholar
29. S. M. Hatefi, S. A. Torabi, and P. Bagheri, “Multi-criteria ABC inventory classification with mixed quantitative and qualitative criteria,” International Journal of Production Research, vol. 52, no. 3, pp. 776–786, 2014. View at: Publisher Site | Google Scholar
30. F. Lolli, A. Ishizaka, and R. Gamberini, “New AHP-based approaches for multi-criteria inventory classification,” International Journal of Production Economics, vol. 156, pp. 62–74, 2014. View at: Publisher Site | Google Scholar
31. E. E. Karsak and S. S. Ahiska, “A common-weight MCDM framework for decision problems with multiple inputs and outputs,” in Computational Science and Its Applications—ICCSA 2007, vol. 4705 of Lecture Notes in Computer Science, pp. 779–790, Springer, Berlin, Germany, 2007. View at: Publisher Site | Google Scholar
32. P. Zhou, B. W. Ang, and K. L. Poh, “Comparing aggregating methods for constructing the composite environmental index: an objective measure,” Ecological Economics, vol. 59, no. 3, pp. 305–311, 2006. View at: Publisher Site | Google Scholar
33. M. Nardo, M. Saisana, A. Saltelli, S. Tarantola, A. Hoffman, and E. Giovannini, “Handbook on constructing composite indicators: methodology and user guide,” OECD Statistics Working Paper 2005/3, 2005. View at: Google Scholar
34. E. E. Karsak and S. S. Ahiska, “Practical common weight multi-criteria decision-making approach with an improved discriminating power for technology selection,” International Journal of Production Research, vol. 43, no. 8, pp. 1537–1554, 2005. View at: Publisher Site | Google Scholar
35. X.-B. Li and G. R. Reeves, “Multiple criteria approach to data envelopment analysis,” European Journal of Operational Research, vol. 115, no. 3, pp. 507–517, 1999. View at: Publisher Site | Google Scholar
36. H. Bal, H. H. Örkcü, and S. Çelebioǧlu, “Improving the discrimination power and weights dispersion in the data envelopment analysis,” Computers and Operations Research, vol. 37, no. 1, pp. 99–107, 2010. View at: Publisher Site | Google Scholar
37. Y.-W. Chen, M. Larbani, and Y.-P. Chang, “Multiobjective data envelopment analysis,” Journal of the Operational Research Society, vol. 60, no. 11, pp. 1556–1566, 2009. View at: Publisher Site | Google Scholar
38. G. R. Jahanshahloo, M. Zohrehbandian, A. Alinezhad, S. A. Naghneh, H. Abbasian, and R. K. Mavi, “Finding common weights based on the DM's preference information,” Journal of the Operational Research Society, vol. 62, no. 10, pp. 1796–1800, 2011. View at: Publisher Site | Google Scholar
Copyright © 2015 S. M. Hatefi and S. A. Torabi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 11,355 | 47,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.860537 |
https://byjus.com/physics/scalars-and-vectors/ | 1,669,524,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00318.warc.gz | 177,358,014 | 217,460 | # Scalar And Vector
In Physics, we often use the terms force, speed, velocity and work, and these quantities are classified as a scalar or vector quantities. A scalar quantity is a physical quantity with only magnitudes, such as mass and electric charge. On the other hand, a vector quantity is a physical quantity that has both magnitudes and directions like force and weight. In this article, let us familiarize ourselves with vectors and scalars.
## What is a Scalar Quantity?
A scalar quantity is defined as the physical quantity with only magnitude and no direction. Such physical quantities can be described just by their numerical value without directions. The addition of these physical quantities follows the simple rules of algebra, and here, only their magnitudes are added.
### Examples of Scalar Quantities
Some examples of scalar include:
• Mass
• Speed
• Distance
• Time
• Volume
• Density
• Temperature
## What is a Vector Quantity?
A vector quantity is defined as the physical quantity that has both directions as well as magnitude.
A vector with the value of magnitude equal to one is called a unit vector and is represented by a lowercase alphabet with a “hat” circumflex i.e. “û“.
Visualize unit vector with the help of the video given below:
### Examples of Vector Quantities
Examples of vector quantity include:
• Linear momentum
• Acceleration
• Displacement
• Momentum
• Angular velocity
• Force
• Electric field
• Polarization
## Difference Between Scalars and Vectors
The difference between Scalars and Vectors is crucial to understand in physics learning.
We have listed the various differences between a scalar and vector in the table below:
Vector Scalar A physical quantity with both the magnitude and direction. A physical quantity with only magnitude. A number (magnitude), direction using unit cap or arrow at the top and unit. A number (magnitude) and unit Quantity symbol in bold and an arrow sign above Quantity symbol Yes No Velocity and Acceleration Mass and Temperature
### Vector Addition and Subtraction
After understanding what is a vector, let’s learn vector addition and subtraction. The addition and subtraction of vector quantities do not follow the simple arithmetic rules. A special set of rules are followed for the addition and subtraction of vectors. Following are some points to be noted while adding vectors:
• Addition of vectors means finding the resultant of a number of vectors acting on a body.
• The component vectors whose resultant is to be calculated are independent of each other. Each vector acts as if the other vectors were absent.
• Vectors can be added geometrically but not algebraically.
• Vector addition is commutative in nature, i.e.,
$$\begin{array}{l}\underset{A}{\rightarrow}+\underset{B}{\rightarrow}=\underset{B}{\rightarrow}+\underset{A}{\rightarrow}\end{array}$$
Now, about vector subtraction, it is the same as adding the negative of the vector to be subtracted. To better understand, let us look at the example given below.
Let us consider two vectors, A and B, as shown in the figure below. We need to subtract vector B from vector A. It is just the same as adding vector B and vector A. The resultant vector is shown in the figure below.
### Vector Notation
For vector quantity usually, an arrow is used on the top as shown below, which represents the vector value of the velocity and also explains that the quantity has both magnitudes as well as direction.
$$\begin{array}{l}Vector\, Notation= \vec{v}\end{array}$$
## Scalar and Vector Solved Problems
Q1: Given below is a list of quantities. Categorize each quantity as being either a vector or a scalar.
20 degrees Celsius 5 mi, North 256 bytes 5 m 30 m/sec, East 4000 Calories
20 degrees Celsius Scalar 5 mi, North Vector 256 bytes Scalar 5 m Scalar 30 m/sec, East Vector 4000 Calories Scalar
Q2: Ashwin walks 10 m north, 12 m east, 3 m west and 5 m south and then stops to drink water. What is the magnitude of his displacement from his original point?
Answer: We know that displacement is a vector quantity; hence the direction Ashwin walks will be positive or negative along an axis.
To find the total distance travelled along the y-axis, let us consider the movement towards the north to be positive and the south to be negative.
$$\begin{array}{l}\sum y=10\,m-5\,m=5\,m\end{array}$$
He moved a net of 5 meters to the north along the y-axis.
Similarly, let us consider his movement towards the east to be positive and the west to be negative.
$$\begin{array}{l}\sum y=-3\,m+12\,m=9\,m\end{array}$$
He moved a net of 9 m to the east.
Using Pythagoras theorem, the resultant displacement can be found as follows:
$$\begin{array}{l}D^2=(\sum x^2)+(\sum y^2)\end{array}$$
Substituting the values, we get
$$\begin{array}{l}D^2=(9^2)+(5^2)\end{array}$$
$$\begin{array}{l}D^2=(106)^2\end{array}$$
$$\begin{array}{l}\sqrt{D^2}=\sqrt{(106)^2}\end{array}$$
$$\begin{array}{l}D=10.30\,m\end{array}$$
Q3. What is the magnitude of a unit vector?
Answer: The magnitude of a unit vector is unity. A unit vector has no units or dimensions.
## Frequently Asked Questions – FAQS
### What is vector and scalar quantity in Physics?
A scalar quantity is defined as the physical quantity that has only magnitude. On the other hand, a vector quantity is defined as the physical quantity that has both magnitude as well as direction.
### How are vector and scalar different?
Vectors have both magnitude and direction but scalars have only magnitude.
### How are vectors and scalars quantities alike?
Scalars and vectors both have specific unit and dimension. Both of these quantities are measurable. Moreover, both possess magnitude.
### What are the examples of scalar?
Mass and electric charge are examples of scalars.
### What are the examples of vectors?
Displacement and angular velocity are examples of vectors.
Stay tuned to BYJU’S and Fall in Loev with Learning!
Test your Knowledge on Scalars And Vectors
1. Why Kinetic Energy is a scalar quantity while velocity is a vector quantity?
• Kinetic energy is a scalar quantity because it doesn’t have a direction, unlike velocity. The kinetic energy of an object is completely described by magnitude alone.
its velocity square and the dot product of two vectors is a scalar quantity .hence,
kinetic energy isa scalar quantity
2. Pratibha Menda
How weight is vector quantity
• The gravitational acceleration is a vector quantity which has magnitude and direction. As a result, the weight of the object is a vector quantity because of its gravitational acceleration.
3. Kaif Sufi
Why do we need vector and scalar?
• Many quantities in Physics are vectors and when we try to add two vectors without considering their directions we usually end up with wrong results. Hence, vectors are important for accurate results.
Even though scalars signify just the magnitude, they are used all the time in Physics.
4. Sonu mishr khan patel singh
What is use in daily life of scalar and vector quantities | 1,619 | 7,032 | {"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-2022-49 | longest | en | 0.909953 |
https://www.physicsforums.com/threads/help-with-friction-forces-box-resting-on-a-board-on-a-frictionless-surface.1062280/ | 1,720,798,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00144.warc.gz | 703,310,198 | 17,574 | # Help with Friction Forces: Box resting on a board on a frictionless surface
• alomari
alomari
Homework Statement
A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1). The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is μs. The coefficient of kinetic friction between the board and the box is, as usual, less than μs.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.
Figure1 of 1
The figure shows a box of mass m 1 lying on a board of mass m 2 and length L. The board lies on a horizontal surface. Force F acts to the right on the board.
Part A
Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).Express your answer in terms of some or all of the variables μs, m1, m2, g, and L. Do not include Ff in your answer.
Relevant Equations
Static friction force : Fs = μs N
Hello, I'm struggling understanding why my answer to this question is incorrect. I know the right answer, but why is this way wrong? I have attached my solution and would really appreciate the help.
Thank you!
#### Attachments
• Ex = Its _ m.g Ms.pdf
231.9 KB · Views: 9
• 1714477940604.png
10.6 KB · Views: 14
Imagine that the mass ##m_2## is huge, say, the mass of Earth. According to your answer, pulling the Earth with the force ##F_{min}=m_1 g \mu_s## would move it out from under the box ##m_1## sitting on it...
Hill said:
Imagine that the mass ##m_2## is huge, say, the mass of Earth. According to your answer, pulling the Earth with the force ##F_{min}=m_1 g \mu_s## would move it out from under the box ##m_1## sitting on it...
I understand, and the answer (m1 + m2)g μs makes logical sense, but I don't understand why my working is not right? The force of friction acting on the box on top would be the same as the force of friction acting on the box on the bottom, and wouldn't we just have to apply a minimum force of the same magnitude to overcome it? Am I missing a force vector here?
What will be the accelerations of the two bodies if ##F=m_1\mu_s g##?
alomari said:
Am I missing a force vector here?
No, you are not missing a force, but
alomari said:
we just have to apply a minimum force of the same magnitude to overcome it
is incorrect. We rather need to prevent them from moving together.
alomari said:
I understand, and the answer (m1 + m2)g μs makes logical sense, but I don't understand why my working is not right? The force of friction acting on the box on top would be the same as the force of friction acting on the box on the bottom, and wouldn't we just have to apply a minimum force of the same magnitude to overcome it? Am I missing a force vector here?
Start by considering that the force of static friction ##F_f## is the net force on ##m_1## in the horizontal direction. Remember that the force of static friction has an upper limit that it cannot exceed.
• Introductory Physics Homework Help
Replies
5
Views
710
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
736
• Introductory Physics Homework Help
Replies
9
Views
951
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
669
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
29
Views
2K | 955 | 3,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.925598 |
https://www.calculateme.com/compare-fractions/2-20/5-32 | 1,669,766,738,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710712.51/warc/CC-MAIN-20221129232448-20221130022448-00306.warc.gz | 753,413,755 | 2,912 | # What's Bigger 2/20 or 5/32?
Is two twentieths greater than five thirty-seconds? Use this calculator to quickly compare the size of two fractions.
Fraction 1
/
Fraction 2
/
220
is smaller than
532
Fraction Decimal Value
220 0.1
532 ≈ 0.156
How much less?
220
is 36% smaller than
532 | 88 | 285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-49 | latest | en | 0.879633 |
https://autoinsurancecompaniesranked.com/is-poor-credit-worse-than-bad-credit/ | 1,701,221,386,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00690.warc.gz | 147,277,558 | 13,222 | # Is poor credit worse than bad credit?
Is poor credit worse than bad credit?
Generally, having no credit is better than having bad credit, though both can hold you back. People with no credit history may have trouble getting approved for today’s best credit cards, for example — while people with bad credit may have trouble applying for credit, renting an apartment and more.
Can a single parent with a child go to University on Universal Credit?
If you are a lone parent student you are eligible for UC if you are responsible for a child under 16, or a young person aged 16-19 in full-time non-advanced education. You no longer count as responsible for a child once they reach 1 September after their 19th birthday.
How do you prove financial dependency to your parents?
Confirmation of residency in a shared household; Shared bank accounts or investments; A loan or mortgage in joint names; Wills naming each other as the main beneficiary.
What is a high household income UK?
In the year ending March 2019, the average (median) annual household income in each quintile before housing costs were paid was: top quintile: £54,000. second highest quintile: £35,700. middle quintile: £26,800.
Can an individual charge interest on a loan UK?
Yes, the maximum rate that can be charged by any business or individual over a loan is set at 25% for England. By law, you cannot charge any rate above this.
How much interest can you charge on a personal loan?
The average interest rate on a personal loan is 10.71% as of March 2023, and well-qualified borrowers can get rates of 7–8%. 5 You can get a better deal if you have a strong income and credit history or can offer collateral to secure your loan.
How much interest can you legally charge UK?
The interest you can charge if another business is late paying for goods or a service is ‘statutory interest’ – this is 8% plus the Bank of England base rate for business to business transactions. You cannot claim statutory interest if there’s a different rate of interest in a contract.
How do I charge interest on a loan?
Assume you borrow \$100 at 6% for one year. How much interest will you pay? The simple interest formula is: Interest = Principal x rate x time 4.
Can you give a family member an interest free loan UK?
You do not have to charge interest for the loan and in the majority of family situations loans are made interest-free. If you do charge interest, the interest payments received by you will be taxable income in your hands and must be declared to HMRC.
What is a disadvantage from obtaining a loan from a family member?
Under certain circumstances, a family loan might introduce tax issues for the lender, leading to unwelcome stress and an additional financial and administrative burden. While lenders are allowed to charge a reasonable interest rate, they also have to pay tax on the income earned from the loan.
Why the student is independent?
What Is an Independent Student? For the FAFSA, an independent student is someone who will not receive any financial support for their education from their parents or guardians. This means the Federal Student Aid office does not use parents’ or guardians’ financial information to calculate independent students’ EFC.
What is the maximum student maintenance loan UK?
What are the minimum and maximum Maintenance Loans in England? The minimum Maintenance Loan on offer for students from England is £3,698. This is paid to students with a household income of £58,291 or more who will live at home during their time at uni. The maximum Maintenance Loan is £13,022.
Does inheritance count as income for student loans UK?
Additionally, if the student has already graduated and is repaying their student loan, an inheritance may affect the amount they repay each month. The amount of student loan repayments are based on the borrower’s income, so if the inheritance increases the borrower’s income, their repayments may increase as well.
What is the parental income?
This is the value of any payments paid or provided to a parent (from any source) that are: used by the parent to replace lost or diminished income of the parent or the parent’s family, or. used to meet usual living expenses of the parent or the parent’s family, and.
Can I let people borrow money and charge interest?
Can I Legally Lend Money to a Friend and Charge Interest? You can lend money at interest, provided that the interest rate falls within the appropriate legal guidelines. Most states have usury laws that limit the maximum amount of interest that a lender can charge.
Do I have to charge interest on a loan to a family member?
The IRS mandates that any loan between family members be made with a signed written agreement, a fixed repayment schedule, and a minimum interest rate. (The IRS publishes Applicable Federal Rates (AFRs) monthly.)
How do you charge interest on borrowed money?
For example, if you take out a five-year loan for \$20,000 and the interest rate on the loan is 5 percent, the simple interest formula would be \$20,000 x .05 x 5 = \$5,000 in interest.
What is the average interest rate on a loan UK?
In June 2022, the average annual percentage rate (APR) on a £10,000 personal loan was 4.11%, which is the highest this figure has been since 2016. In June 2022, the average APR on a £5,000 personal loan was 8.2%, up from 8.14% the previous month.
Can I lend my daughter money to buy a house?
Yes, you can loan a family member money to buy a house. It is very important to get the terms of the loan set out legally as if you get it wrong, the loan could give the Lender an unintended beneficial interest (that’s a type of ownership) over the property.
What is charging more interest than is legally allowed considered?
An interest rate that exceeds the legal rate of interest is classified as usury. There are usually stiff penalties for usury in most states, such as fines or even the forfeiture of principal and/or interest. | 1,276 | 5,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-50 | latest | en | 0.967545 |
http://www.clubdetirologrono.com/2nd-grade-addition-and-subtraction-worksheets/free-math-worksheets-and-printouts-second-grade-addition-subtraction-additiondrills25-le/ | 1,566,803,234,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00512.warc.gz | 236,430,506 | 10,845 | ## Clubdetirologrono
Category : Math Worksheet.
Author : Diaz Firtha.
Published : Thu, Jul 18th 2019 09:11 AM.
Format : jpg/jpeg.
Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child grasp of a subject is deeper and comprehensive. There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
Now that we know the benefits of mathematics what is needed to conquer this hard to tackle and arduous subject. Answer is quite simple actually, practice, a lot of practice from early childhood. But in order to perform any task or practice any task, one requires resources. Obtaining a resource is not difficult in today modern world but affordable resource is definitely a rare commodity. Today education is a sector which gets very little funding from the federal government.
What's Popular?
##### Pattern worksheets for kindergarten
###### Printable 2nd grade math worksheets
Today we all know that benefits of math are considerable. Math is not a subject one learns by reading the problems and solutions. American children have very little practice with multi-step problems, and very few opportunities to think their way in to and through problems that do not look like all the others. With a packed curriculum and the increased emphasis on testing, our children are taught tons of procedures - but procedures disconnected from when to use them, and why. Sustained thinking - the key ingredient to math success - is painfully absent in too many math classes.
## Ccss 2 nbt 5 worksheets two digit addition and subtraction within 2nd grade pdf ccss2n
#### 3 digit subtraction worksheets 2nd grade addition and with regrouping second math sheets column digits no
##### Free math worksheets and printouts threedigitadditionnoregro
###### 2nd grade subtraction worksheets subtracting 1 and 10
People have to start from the ground, then first step, second, third and so on to reach their destination floor. Exactly the same way students have to start from Kindergarten, then grade one, grade two and three and so on to reach their math destination. Also, if some of the steps are broken in the staircase, it is still hard to reach the desired floor using those steps. Same way, if you are missing some of the basic concepts from elementary grades, math for you is still hard. Now, the kindergarten, first grade and second grade are like first couple of the steps of the stairs. You can learn this level of math easily, as you can jump enough to take yourself to second or third step of the stairs easily. As it is very hard to reach sixth or seventh step of a stairs by jumping from the ground, exactly the same way to learn grade five or higher grade math is very hard (or impossible) without having the good knowledge of the kindergarten to grade three or grade four math. | 689 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-35 | latest | en | 0.949964 |
https://www.elitedigitalstudy.com/11605/how-many-elements-has-p-a-if-a-f | 1,670,019,337,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00510.warc.gz | 792,968,167 | 10,699 | How many elements has P (A), if A = Φ?
Asked by Pragya Singh | 11 months ago | 78
##### Solution :-
If A is a set with m elements
n (A) = m then n [P (A)] = 2m
If A = Φ we get n (A) = 0
n [P(A)] = 20 = 1
Therefore, P (A) has one element.
Answered by Abhisek | 11 months ago
### Related Questions
#### If A = {x : x ϵ R, x < 5} and B = {x : x ϵ R, x > 4}, find A ∩ B.
If A = {x : x ϵ R, x < 5} and B = {x : x ϵ R, x > 4}, find A ∩ B.
#### Prove that A – B = A ∩ B.’
Prove that A – B = A ∩ B.’
#### Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}.
Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}. | 285 | 662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.715083 |
http://www.drumtom.com/q/at-1-atm-how-much-energy-is-required-to-heat-71-0-g-of-h2o-s-at-12-0-c-to-h2o-g-at-159-0-c | 1,485,102,689,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281450.93/warc/CC-MAIN-20170116095121-00474-ip-10-171-10-70.ec2.internal.warc.gz | 430,697,780 | 8,322 | # At 1 atm, how much energy is required to heat 71.0 g of H2O(s) at –12.0 °C to H2O(g) at 159.0 °C?
• At 1 atm, how much energy is required to heat 71.0 g of H2O(s) at –12.0 °C to H2O(g) at 159.0 °C?
... / At 1 atm, how much energy ... Ask ... how much energy is required to heat 77.0 g of H2O(s) ... 0 degrees C, q1=m CP dt=77.0g(2.087 J/g*C)(0+12.0 C)=1 ...
Positive: 67 %
At 1 atm, how much energy is required to heat 83.0 g ... required to heat 83.0 g of H2O(s) at –12.0 °C to ... take energy as well. (83 g ice)(12 °C) ...
Positive: 64 %
### More resources
Physical Chemistry by P Bahadur ... — (in C.G.S.) nh u 1 for H = 2.1847 x 108 cm sec"1, ... Calculate the energy required for the process; He + (g) ...
Positive: 67 %
At 1 atm, how much energy is required to heat 55.0 g of H2O(s) ... how much energy is required to heat 37.0g of H2O at -24.0 C to ... Chemistry - At 1 atm, ...
Positive: 62 %
... Chemisty Ch. 11 & 12 Don't ... How much energy is required to heat 36.0 g H2O from a ... 114.6 °C at 1 atm. | 383 | 1,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-04 | longest | en | 0.883797 |
https://origin.geeksforgeeks.org/nfa-machines-accepting-all-strings-that-ends-or-not-ends-with-substring-ab/ | 1,685,588,223,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00086.warc.gz | 495,771,189 | 37,419 | GFG App
Open App
Browser
Continue
# NFA machines accepting all strings that ends or not ends with substring ‘ab’
Prerequisite: Finite Automata Introduction
Problem-1:
Construction of a minimal NFA accepting a set of strings over {a, b} in which each string of the language ends with ‘ab’.
Explanation:
The desired language will be like:
`L1 = {ab, abbab, abaab, ...........}`
Here as we can see that each string of the above language ends with ‘ab’ but the below language is not accepted by this NFA because some of the string of below language does not end with ‘ab’.
`L2 = {bba, abb, aaabbbb, .............}`
The state transition diagram of the desired language will be like below:
In the above NFA, the initial state ‘X’ on getting ‘a’ as the input it either remains in the state of itself or transit to a state ‘Y’ and on getting ‘b’ as the input it remains in the state of itself. The state ‘Y’ on getting ‘b’ as the input it transmits to a final state ‘Z’.
Problem-2:
Construction of a minimal NFA accepting a set of strings over {a, b} in which each string of the language is not ending with ‘ab’.
Explanation: The desired language will be like:
`L1 = {bba, abb, aaabbbb, .............}`
Here as we can see that each string of the above language is not ending with ‘ab’ but the below language is not accepted by this NFA because some of the string of below language is ending with ‘ab’.
`L2 = {ab, abab, ababaab..............}`
The state transition diagram of the desired language will be like below:
In the above NFA, the initial state ‘X’ on getting ‘a’ as the input it remains in the state of itself and on getting ‘b’ as the input it transmits to a state ‘Y’. The state ‘Y’ on getting ‘b’ as the input it either remains in the state of itself or transmits to a final state ‘Z’. The final state ‘Z’ on getting ‘a’ as the input it remains in the state of itself.
My Personal Notes arrow_drop_up | 486 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.906698 |
kunstschule-jever.de | 1,652,687,588,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00542.warc.gz | 29,503,106 | 5,702 | # Bode plot single pole
• Ответил Роберт, делая несколько шагов.
• Understanding Bode Plots, Part 4: Complex Systems - Video - MATLAB
• Безусловно, только разведывательный отряд, высланный За завтраком состоялся семейный совет.
• Wie flirtet der jungfraumann
• Kosten single tageskarte
• Partnervermittlung ukraine preise
• Lose bekanntschaft bedeutung
So far, we have looked at the asymptotic behavior of first order constructs, like pure integrators or single bode plot single pole and zeros. Once you start working with typical dynamic systems, it is very likely that you will have to deal with higher order polynomial expressions. The trick to dealing with those is remembering that any polynomial, no matter the order, can always be factored into a bunch of first order constructs which will correspond to the real roots, and a number of second order constructs which will correspond to complex conjugate pairs of roots.
Typical examples of second order systems are mass spring dampers and RLC circuits.
### Regelungstechnische Begriffe: deutsch–english
Both of this, depending on the ratio bode plot single pole either the damping to the mass or the resistance to the inductance, will have a pair of complex conjugate roots.
Anyways, if we calculate the roots of that quadratic polynomial using the standard formula-- minus b plus minus square root of whatever-- we find that the complex conjugate pair of roots will have this form.
Note that these roots will only be a complex conjugate pair as long as the damping ratio zeta is less than 1. Anything greater than 1, and both roots will become real numbers, which means that the system will behave as the product of two first order poles. As we did before to calculate the frequency response, we replace s by jw in the transfer function.
### Understanding Bode Plots, Part 3: Simple Systems
So both the magnitude and the phase will be approximately 0. Finally, when the frequency w is much larger than the natural frequency the quadratic term will dominate. When taking the log, the square will come out and multiply the 20, so the magnitude will asymptotically approach a straight line with a slope of dBs per decade.
The phase will go to degrees because G will now fall bode plot single pole the negative real axis. This adjusted value is what is called a damped natural frequency. And note that, in that case, the magnitude of the resonant peak will go towards infinity.
A small value of the damping ratio means a higher and sharper resonant peak as well as a sharper shift in the phase. You can see that as we increase zeta, the magnitude of the resonant peak comes down and the phase transition becomes smoother. Here I want to highlight the damping ratio of 0.
This damping makes the magnitude -3 dBs at the natural frequency. At this point let me go back to our interactive design tool because there are a couple of additional things I want to highlight.
### Understanding Bode Plots, Part 4: Complex Systems
First, let me bring in a pair of complex conjugate poles, and I am going to place them close to 10 radians per second. Let me just make sure that I set the natural frequency to exactly I notice that, because I am starting with a damping ratio of 1, my polynomial is the product of two real roots.
As soon as I change the damping to any value lower than 1, let's say 0. If I choose a smaller damping ratio, what you're going to see is a sharper and higher magnitude peak.
We just saw how a function like Bode in MATLAB can quickly and easily create a frequency response plot directly from the dynamic equations of, or the input, output transfer functions of our system. The key as control engineers is not just to be able to create those plots. The important thing is having a good understanding of what those magnitude and phase traces are telling us about our system behavior and stability. Bode plots were originally developed by Dr. Hendrik Bode, hence the name, while he was working for Bell Labs in the s, just before World War II.
In this particular case, I picked 0. | 886 | 4,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.772437 |
https://stackoverflow.com/questions/1727669/contruct-3d-array-in-numpy-from-existing-2d-array/68082251 | 1,716,405,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00744.warc.gz | 464,548,582 | 44,357 | # Contruct 3d array in numpy from existing 2d array
During preparing data for NumPy calculate. I am curious about way to construct:
``````myarray.shape => (2,18,18)
``````
from:
``````d1.shape => (18,18)
d2.shape => (18,18)
``````
I try to use NumPy command:
``````hstack([[d1],[d2]])
``````
but it looks not work!
Just doing `d3 = array([d1,d2])` seems to work for me:
``````>>> from numpy import array
>>> # ... create d1 and d2 ...
>>> d1.shape
(18,18)
>>> d2.shape
(18,18)
>>> d3 = array([d1, d2])
>>> d3.shape
(2, 18, 18)
``````
• I have one question similar. If I have already got the d3 with shape(2,18,18) and I want to add another 2-d array d4 (18x18) into d3 to make 3-d array(3,18,18). What should I do? Dec 30, 2015 at 3:40
• You simply `vstack(d3, d4[np.newaxis,...])`, as in my answer. Mar 25, 2016 at 20:56
• It must be vstack((d3, d4[np.newaxis,...])). Two '(( ))' are needed. Oct 12, 2017 at 8:01
hstack and vstack do no change the number of dimensions of the arrays: they merely put them "side by side". Thus, combining 2-dimensional arrays creates a new 2-dimensional array (not a 3D one!).
You can do what Daniel suggested (directly use `numpy.array([d1, d2])`).
You can alternatively convert your arrays to 3D arrays before stacking them, by adding a new dimension to each array:
``````d3 = numpy.vstack([ d1[newaxis,...], d2[newaxis,...] ]) # shape = (2, 18, 18)
``````
In fact, `d1[newaxis,...].shape == (1, 18, 18)`, and you can stack both 3D arrays directly and get the new 3D array (`d3`) that you wanted.
• `np.vstack([a[np.newaxis,...],b[np.newaxis,...]])` worked like charm! Thanks. Nov 24, 2017 at 14:11
``````arr3=np.dstack([arr1, arr2])
``````
arr1, arr2 are 2d array `shape (256,256)`, arr3: `shape(256,256,2)`
• Up for this, because it works for joining RGB channels on images too, as the final shape must be `(height,width,3)`. Jun 26, 2020 at 14:30
A lot of versatility is provided by the `np.stack()` function. You can use it like this:
``````>>> d3 = np.stack([d1, d2])
>>> d3.shape
(2, 18, 18)
``````
However you can also specify the axis along which the arrays get joined. So if you wanted to join channels of a RGB image, you would use:
``````>>> rgb = np.stack([r, g, b], axis=-1) # r, g and b each have shape (18, 18)
>>> rgb.shape
(18, 18, 3)
`````` | 783 | 2,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.817518 |
https://www.experts-exchange.com/questions/27637302/Which-math-operator-is-faster.html | 1,506,081,991,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688940.72/warc/CC-MAIN-20170922112142-20170922132142-00064.warc.gz | 785,109,897 | 28,330 | Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17
x
Solved
# Which math operator is faster?
Posted on 2012-03-17
Medium Priority
409 Views
Which line should process faster and why?
``````if (x == y) { }
``````
OR
``````if (x > y) { }
``````
OR
``````if (x < y) { }
``````
OR
``````if (x >= y) { }
``````
Which operator should process faster (+, -, * or /)?
0
Question by:Mohamed Abowarda
[X]
###### Welcome to Experts Exchange
Add your voice to the tech community where 5M+ people just like you are talking about what matters.
• Help others & share knowledge
• Earn cash & points
• Learn & ask questions
• 2
LVL 84
Accepted Solution
Dave Baldwin earned 532 total points
ID: 37734190
The 'if' lines should all be the same because subtraction is normally used and the results are used to determine the response. The first checks for a 0 result, the second checks for a positive result, negative result, and positive or 0 result.
For the operators, + and - are the fastest for integers because they are a single step operation, * and / are multiple steps. For floating point, they are all multiple steps.
0
LVL 12
Author Comment
ID: 37734234
Can you give me an example on how the computer process +, -, *, and /?
I want detailed information.
Thanks,
0
LVL 51
Assisted Solution
Gustav Brock earned 268 total points
ID: 37734321
You've got it:
http://msdn.microsoft.com/en-us/library/aa309387(v=vs.71).aspx
That's for .Net 1.1. For 4.0 it is comes with the Microsoft Windows SDK Tools and is typically located in:
C:\Program Files\Microsoft SDKs\Windows\v7.0A\bin\NETFX 4.0 Tools\ildasm.exe
/gustav
0
LVL 84
Assisted Solution
Dave Baldwin earned 532 total points
ID: 37734324
For integers and characters that fit in the main CPU registers (32 or 64 bit), + and - can simply be done in a single step by the main ALU. All other operations are passed to the math coprocessor. This page http://www.top500.org/2007_overview_recent_supercomputers/intel_xeon gives a good example of the blocks in an Intel CPU. An earlier CPU would be easier to understand since there isn't as much detail. Here's one http://www.laynetworks.com/Block%20diagram%20of%20the%20Pentium.htm of an early Pentium CPU. But they keep adding things to speed up the instructions. Now all of this is hidden from you by the C# compiler and the .NET framework. The compiler determines what kind of data you are using and selects the appropriate .NET CLR routines to process it.
0
## Featured Post
Question has a verified solution.
If you are experiencing a similar issue, please ask a related question
Creating an analog clock UserControl seems fairly straight forward. It is, after all, essentially just a circle with several lines in it! Two common approaches for rendering an analog clock typically involve either manually calculating points with…
Wouldn’t it be nice if you could test whether an element is contained in an array by using a Contains method just like the one available on List objects? Wouldn’t it be good if you could write code like this? (CODE) In .NET 3.5, this is possible…
Want to learn how to record your desktop screen without having to use an outside camera. Click on this video and learn how to use the cool google extension called "Screencastify"! Step 1: Open a new google tab Step 2: Go to the left hand upper corn…
We’ve all felt that sense of false security before—locking down external access to a database or component and feeling like we’ve done all we need to do to secure company data. But that feeling is fleeting. Attacks these days can happen in many w…
###### Suggested Courses
Course of the Month8 days, 18 hours left to enroll
#### 721 members asked questions and received personalized solutions in the past 7 days.
Join the community of 500,000 technology professionals and ask your questions. | 994 | 3,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-39 | longest | en | 0.833894 |
https://123deta.com/document/zwv4o1vl-the-number-of-extensions-of-latin-rectangles.html | 1,685,595,366,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00783.warc.gz | 97,506,491 | 49,844 | # the number of extensions of Latin rectangles
## Full text
(1)
### the number of extensions of Latin rectangles
B. D. McKay and I. M. Wanless Department of Computer Science Australian National University Canberra ACT 0200 Australia
bdm@cs.anu.edu.au / imw@cs.anu.edu.au
Submitted: February 18, 1997; Accepted: March 15, 1997; Received in final form: February 8, 1998.
AMS Classifications 15A15, 05C70, 05B15.
Let k ≥ 2, m ≥ 5 and n = mk be integers. By finding bounds for certain rook polynomials, we identify thek×nLatin rectangles with the most extensions to (k+ 1)×n Latin rectangles. Equivalently, we find the (n−k)-regular subgraphs of Kn,n which have the greatest number of perfect matchings, and the (0,1)-matrices with exactlyk zeroes in every row and column which maximise the permanent. Without the restriction on nbeing a multiple ofk we solve the above problem (and the corresponding minimisation problem) for k = 2. We also provide some computational results for small values of nand k.
Our results partially settle two open problems of Minc and conjectures by Merriell, and Godsil and McKay.
§1. The problem
Letk andnbe positive integers with k≤n. A k×nLatin rectangle is a k×n matrix of entries from{1,2, . . . , n}such that no entry is duplicated within any row or any column.
We useL(k, n) for the set ofk×nLatin rectangles. ForR∈L(k, n) defineE(R) to be the number ofR0 ∈L(k+1, n) such that the firstkrows ofR0are identical to the corresponding rows of R. We say that E(R) is the number of extensions of R. We call R1 ∈ L(k, n) a maximising rectangle ifE(R1)≥E(R) for everyR∈L(k, n). We defineMk,n =E(R1) for a maximising R1. Similarly we call R2 ∈L(k, n) a minimising rectangle ifE(R2)≤E(R) for every R∈L(k, n) and define mk,n = E(R2) for a minimising R2. We are interested in identifying maximising and minimising rectangles and in finding estimates for Mk,n and mk,n. In particular, we concentrate on maximising rectangles in the case when n = mk for some integer m.
The problem has (at least) two other guises which are fruitful to consider. With each R∈L(k, n) we associate G(R), a subgraph of the complete bipartite graph Kn,n, defined as follows. Let {u1, u2, . . . , un} and {v1, v2, . . . , vn} be the two vertex sets. We put an
(2)
edge (ui, vj) in G(R) precisely when symbol i occurs in column j of R. For any spanning subgraph G of Kn,n we use G to denote the complement within Kn,n of G. Note that G(R) isk-regular, G(R) is (n−k)-regular andE(R) is the number of perfect matchings in G(R). Finding a maximisingk×nLatin rectangle is equivalent to maximising the number of perfect matchings in an (n−k)-regular subgraph of Kn,n.
The other incarnation of the problem is in (0,1)-matrices. Let Λkn denote the set of (0,1)-matrices of order n in which the row and column sums are all equal to k. With R and G(R) we associateA(R)∈Λkn defined by
A(R)
ij =
1, if ui is adjacent tovj inG(R);
0, otherwise.
We call A(R) the biadjacency matrix ofG(R). Note that E(R) is the permanent of A(R), the biadjacency matrix of G(R). Hence the question of finding a maximising k×n Latin rectangle relates to maximising the permanent of (0,1)-matrices of order n with all line sums equal to n−k.
The association between R,G(R) andA(R) is so strong that we will tend to blur any distinction and think of them as a single object. It should be apparent that we are only interested in the structure ofG(R) up to isomorphism, orA(R) up to permutations of the rows and columns.
The principal result of the paper (Theorem 10) is that ifm≥5 then every maximising R∈L(k, mk) has G(R) isomorphic to m copies of Kk,k. This partially answers problems 4 and 12 of Minc [12].
§2. What is known
The literature on bounds for permanents is quite extensive. Minc [11], [12] and Schri- jver [13] are recommended starting points. Of particular interest to us are the Egorychev- Falikman Theorem (formerly the van der Waerden conjecture) which yields
mk,n ≥n!(1−k/n)n, (1)
and the Br`egman bound,
Mk,n ≤ (n−k)!n/(nk)
. (2)
Br`egman proved (2) in [3], which also contains the following theorem.
Theorem 1. Let k, m ≥ 2 be integers and R ∈ L (m−1)k, mk
be maximising. Then G(R) consists ofm copies of Kk,k.
We note the following corollary.
(3)
Corollary 1. R∈L(k,2k)is maximising if and only if G(R) is disconnected.
There has been substantial effort towards enumerating Latin squares (n×n Latin rectangles), often by counting the extensions of Latin rectangles. The best asymptotic estimates to date are contained in [7], which employs similar tools to the present paper.
Let R ∈ L(k, n). An i-matching in G(R) is a set of i vertex-disjoint edges in G(R).
Let mi(R) denote the number of i-matchings in G(R) and adopt the convention that m0(R) = 1. We define the rook polynomial ρ(R, x) by
ρ(R, x) =ρ G(R), x
= Xn i=0
(−1)imi(R)xni.
The two features of rook polynomials which we exploit most are demonstrated in the following two results. The first is a consequence of the work of Heilmann and Lieb [8], while the second is due to Joni and Rota [9].
Theorem 2. For anyR∈L(k, n) wherek ≥2, the roots ofρ G(R), x
are real and lie in the open interval (0,4k−4). For R∈L(1, n),ρ G(R), x
= (x−1)n.
Theorem 3. The number of extensions of R∈ L(k, n) is given by E(R) = I0 ρ(R, x) , where the linear operator Iab(·)is defined by
Iab f(x)
= Z b
a
e−xf(x)dx.
The integral defined in Theorem 3 is the fundamental tool in this paper, as it was in [7]. We use I(·) as shorthand for I0(·).
Two other well known properties of the rook polynomial are worth noting. Firstly, it is multiplicative on components. That is, if {Ci}i is the set of components of a graph G then ρ(G, x) =Q
iρ(Ci, x). Secondly, for an arbitrary integera, ρ(Ka,a) =La(x) = (−1)aa!
Xa i=0
a i
(−x)i
i! . (3)
That is, the rook polynomial of a complete bipartite graph is a Laguerre polynomial, normalised to be monic.
§3. The k = 2 case
Every R ∈ L(1, n) satisfies E(R) = n!Pn
i=0(−1)i/i!, that being the number of de- rangements of{1,2, . . . , n}. Hence, the smallest case for which the question of identifying maximising rectangles is interesting is the case k = 2.
(4)
LetUn,tdenote the set of (0,1)-matrices of orderncontaining exactlytzeroes (without restriction on row or column sums). In [4] the matrices maximising the permanent inUn,t
are identified for t ≤ 2n. When t = 2n the answer turns out to be an element of Λnn2, except in the case n = 5. The maximising rectangles in L(2, n) are thereby found for all n6= 5. In Theorem 4 (below) we present a new way of obtaining this result.
Every component of G(R) for R ∈ L(2, n) is a cycle of even length. We use Ca to denote a cycle of length a, and define pi =pi(x) = ρ(C2i, x) for each i≥ 2. By extension we define p0 = 2 andp1 =x−2 so that pi(4x2) = 2T2i(x) for each i≥ 0, where Tn(x) is the nth Chebyshev polynomial of the first kind. This leads [14] to
papb =pa+b +pab for 0≤b≤a. (4) Formula (4) is the key to the next two theorems, because it shows us when it is profitable to split long cycles.
Theorem 4. When 2 ≤ n ≤ 4 or n ≥ 8 the maximising R ∈ L(2, n) are those which maximise the number of components in G(R). For 5 ≤ n ≤ 7 the maximising 2 ×n rectangles are those Rfor which
G(R)∼=
(C10 for n= 5, C6+C6 for n= 6, C10+C4 orC6+ 2C4 for n= 7.
Here + denotes disjoint union and rGis shorthand for G| +G+{z. . .+G}
rtimes
.
Proof: The theorem is easily established forn≤7 so we assumen≥8. LetR∈L(k, n) be maximising and suppose G(R) consists of c cycles C2a1, C2a2, . . . , C2ac. Clearly n= P
ai and ρ G(R), x
= Q
pai, and we may suppose for convenience that the ai are arranged in non-increasing order. We first show that a1 ≤ 5. Suppose this were not the case and consider the rectangle R0 formed from R by ‘splitting’ the C2a1 into C4+C2a14. Then by (4)
ρ G(R0), x
=p2pa12
Y
i2
pai =ρ G(R), x
+pa14
Y
i2
pai. Now
E(R0) =I
ρ G(R0), x
=E(R) +I
pa14
Y
i2
pai
. (5)
Our assumptions thatn >6 anda1 >5 mean thatI pa14
Q
i2pai
>0 by (1) because it is the number of extensions of some rectangle inL(2, n−4). Thus (5) breaches our choice ofR, proving that a1 ≤5.
(5)
We next examine the case when a1 = 5. Let R0 be the rectangle obtained fromR by splitting C2a1 into C6+C4. Then E(R0) =E(R) +I p1
Q
i2pai
. If a2 ≥3 then I
p1Y
i2
pai
=I
pa2+1Y
i3
pai
+I
pa21Y
i3
pai
which is positive because the first term on the right is positive and the second non-negative, again by considering the integrals as counts of extensions of certain Latin rectangles. Thus we may assume that ai= 2 for i≥2. Now
I p1pc21
=I p3pc22
+I p1pc22 which by induction yields that I p1pc−12
is zero when c = 2 and positive for c ≥ 3. As n≥8 it follows that there must be at leastc≥3 cycles, and hence a1 = 5 is contradictory.
Now we eliminate the possibility that a1 = 4. Let R0 be the rectangle obtained from R by splitting C2a1 into C4+C4. Then
E(R0) =E(R) +I
p0Y
i2
pai
=E(R) + 2I Y
i2
pai
> E(R).
Which means that G(R) consists entirely of C4’s and C6’s. To complete the proof of the theorem it suffices to show that (for n ≥ 8) replacing 2C6 by 3C4 will increase the number of extensions. Consider
I
p32Y
i3
pai
−I
p23Y
i3
pai
= 3I
p2
Y
i3
pai
−2I Y
i3
pai
.
This is clearly positive since for anyν ≥2, appending aC4 to an element ofL(2, ν) always increases the number of extensions. To see this note the injection which takes
α1 α2 α3 . . . αν β1 β2 β3 . . . βν
e1 e2 e3 . . . eν
to
α1 α2 α3 . . . αν ν+ 1 ν + 2 β1 β2 β3 . . . βν ν+ 2 ν + 1 ν+ 1 ν+ 2 e3 . . . eν e1 e2
.
(3 similar injections are obtained by swapping e1 ↔e2 and/orν1 ↔ν2 in the image.) Theorem 5. The minimisingR∈L(2, n) are precisely those for which
G(R)∼=
C2n for n≤4,
C2ν+2+C for odd n= 2ν+ 1≥5, C12 or C8+C4 or 3C4 for n= 6,
C2n or C2ν+2+C2 for even n= 2ν ≥8.
Proof: Similar to Theorem 4. Equation (4) tells us when replacing two cycles by a single
cycle reduces E(R). We omit the details.
Having completely solved the k = 2 case, we may assume for the remainder of the paper that k ≥3.
(6)
Define Sm,k ∈ L(k, mk) to be such that G(Sm,k) ∼= mKk,k. In [7] the following conjecture was made.
Conjecture 1. If R∈L(k, mk) is maximising then G(R)∼=G(Sm,k).
This paper represents an effort to resolve this conjecture. We will show that it is substantially (though not without exception) correct. We know already from Corollary 1 that the conjecture is true for all k when m= 2. We also know by Theorem 4 that there exists a counterexample when k = 2 andm= 3. Specifically,
E(S3,2) =E
1 2 3 4 5 6
2 1 4 3 6 5
= 80<82 =E
1 2 3 4 5 6
2 3 1 5 6 4
.
The only other case where we know Conjecture 1 fails is for k = m = 3. It is an easy matter to check that
E(S3,3) = 12096<12108 =E
1 2 3 4 5 6 7 8 9
2 3 4 1 6 7 8 9 5
3 4 1 2 7 9 6 5 8
.
Curiously, in both the above examples Sm,k is in fact minimising among rectangles for which G(R) is disconnected. This is particularly interesting in light of our main result.
A more general attempt to identify the matrices in Λkn which maximise the permanent was made by Merriell [10]. Merriell completely solved the k = 2 and k = 3 cases and conjectured a partial answer for larger values.
Let Jr and Zr denote r ×r blocks of ones and zeroes respectively. We also use J without a subscript to denote a (not necessarily square) block of ones of unspecified, but implied dimensions. Finally, let Dr = Ir denote the complement of the order r identity matrix,Ir. Merriell’s conjectures can then be stated as:
Conjecture 2. Suppose k ≤ n ≤ 2k and that either k ≥ 5 or n is even. The maximum permanent in Λkn is achieved by a matrix with block structure
A J J B
where Aand B are square matrices with orders that differ by at most 1. Furthermore, A and B should be chosen to maximise their individual permanents.
Conjecture 3. Let n= tk+r for integers k ≥ 5, t ≥1 and r ≥0. Then the maximum permanent in Λkn is achieved by
(t−r)Jk+rDk+1 when r≤min{t, k−3}, (t−1)Jk+Xk,r when r=k−2 or r =k−1,
(7)
where
Xk,k2 =
J Ik1
Ik1 J
and Xk,k1 =
J Zk1
Ik J
.
Conjecture 3 was shown to fail for n = 14, k = 5 in [16], and it follows that the conjecture fails for n = 9 + 5t, k = 5 for every positive integer t. Also Conjecture 2 is known [2] to fail for n = 9, k = 7. However, Merriell himself acknowledged that his pattern broke down in certain small cases (all of which he hoped to have excluded). The experience of this paper shows that isolated counterexamples do not render a conjecture on maximising the permanent in Λkn worthless. The primary issue is whether the pattern holds for sufficiently large k and n.
In fact there is a serious flaw in Conjecture 2. For any positive integer a, it implies that there are maximising rectangles R ∈ L(2,4a+ 2) and R1, R2 ∈ L(2,2a+ 1) such that G(R) ∼= G(R1) +G(R2), which contradicts Theorem 4 for all a ≥ 2. A similar observation applied to Theorem 10 will furnish another infinite family of counterexamples to Conjecture 2. Conjecture 3 remains unresolved for k ≥6.
The question of finding the maximum permanent in Λkn when k does not divide n is problem 4 in [11] and [12]. Problem 12 of [12] asks whether this maximum permanent is achieved by a circulant. A circulant is a square matrix which is a linear combination of powers of the permutation matrix corresponding to the full cycle (123. . . n). It is well known that in the cases covered by Theorem 1, the maximum permanent is achieved by a circulant. Since the complement of a circulant is also a circulant, our main result will furnish another set of examples where the maximum is achieved by a circulant.
In Table 1 below we identify maximising R ∈L(k, n) for some small values of k and n. In the process we get more data relating to Minc’s questions and Conjectures 1 to 3.
For example, despite failing when (m, k) = (3,2) or (3,3), we see that Conjecture 1 is true for (3,4), (4,3), (5,3) and probably also for (3,5) and (4,4). Note also, by Theorem 4, that the conjecture holds for (m,2) whenever m >3.
In the light of Table 1 we propose the following research problem.
Research problem. When are the following statements true of maximisingRinL(k, n)?
(a) G(R) is unique up to isomorphism.
(b) G(R)contains exactly bn/kc components (that being the greatest possible number of components). Similarly,G(R) contains bn/(n−k)c components.
(c) G(R)∼=G(R0) for a maximising R0 ∈L(n−k, n).
(d) A(R)can be constructed (up to permutation of the rows and columns) from copies of J1 by recursive use of the direct sum and complement operations.
Properties (a), (b), (c) and (d) seem to commonly but not universally hold. Can this observation be formalised? Note that for each property, Table 1 provides at least one counterexample. See also the forthcoming paper, [15].
(8)
Table 1 (part 1): A(R) for maximising R∈L(k, n).
n\k 3 4 5 6 7
7 Figure 1 J3⊕D4 2J2⊕D3 D7 J7
(148) (54) (8) [1] [0]
8 2D4 2J4 2D4 4J2 D8
[1313] [576] [81] [16] [1]
9 D4⊕J2⊕D3 Figure 2 J4⊕D5 3J3 3J2⊕D3
(12108) (2916) (1056) [216] (16)
10 2J3⊕D4 2D5 2J5 J4⊕2D3 2J3⊕D4
(127044) [32826] [14400] (1968) (324)
11 2J3⊕J2⊕D3 D5⊕3J2 J5⊕D6 J5⊕D6 D5⊕2D3
(1448640) (373208) (86400) (31800) (3608)
12 4J3 3J4 2D6 2J6 2D6
[17927568] [4783104] [1181737] [518400] [70225]
13 3J3⊕D4 2J4⊕D5 D6⊕2J2⊕D3
(238673088) (65641536) (15950816)
14 3J3⊕J2⊕D3 2J4⊕3J2 2(2J2⊕D3) 2J7
(3410776944) (961491456) (241119120) [25401600]
15 5J3 2J4⊕J3⊕D4 3J5
[52097831424] (14992781184) [3891456000]
16 4J3⊕D4 4J4
(846230552208) [248341303296]
Key (also see notes on next page) A = complement of A
Jr =r×r block of 1s
Dr=Ir, whereIr is the order r identity
⊕ = direct sum rA=A| ⊕A⊕{z. . .⊕A}
rtimes
(9)
Table 1 (part 2): A(R) for maximising R∈L(k, n).
n\k 8 9 10 11 12
10 5J2 D10 J10 − −
[32] [1] [0]
11 J3⊕2D4 4J2⊕D3 D11 J11
(486) (32) [1] [0]
12 3J4 4J3 6J2 D12 J12
[13824] [1296] [64] [1] [0]
13 J5⊕2D4 2J4⊕D5 3J3⊕D4 5J2⊕D3 D13
(157560) (25344) (1944) (64) [1]
14 J5⊕Figure 2 2J4⊕2D3 2J3⊕2D4 7J2
(349920)† (47232) (2916) [128]
15 3J5 J4⊕D5⊕2D3 5J3
[1728000] (86592) [7776]
16 2J8 4J4
[1625702400] [331776]
Notes:
• The table shows A(R) for maximising R ∈ L(k, n). To maximise the permanent in Λnnk use the complement, A(R).
• In each caseMk,n is given belowA(R). Values of Mk,n which exceed those predicted by Conjecture 2 are listed in bold.
• The sole value of Mk,n which breaches Conjecture 3 is marked with a †. Note that this value exceeds that of the counterexample provided in [16].
• Values of Mk,n which are achieved by circulant matrices are given in [brackets], whereas other values appear in (parentheses).
• The results were found by computer enumeration of graphs, except for those which follow from Theorem 1, and the casen= 15, k = 3 which follows from Theorem 10.
• Some of the results presented here were previously known from [10].
• Results marked * are provisional because not all graphs could be enumerated. All disconnected graphs and graphs with disconnected complement were generated in these cases. In addition, connected graphs containing at least 115, 421, 42 and 1212 4-cycles respectively were generated for (n, k) = (12,5), (12,7), (13,4) and (13,9).
(10)
0 0 1 0 1 1 0
0 1 0 0 1 0 1
1 0 0 0 0 1 1
0 0 0 0 1 1 1
1 1 0 1 0 0 0
1 0 1 1 0 0 0
0 1 1 1 0 0 0
or
0 0 0 0 1 1 1
0 0 0 0 1 1 1
0 0 0 1 0 1 1
0 1 1 0 1 0 0
1 0 1 1 0 0 0
1 1 0 1 0 0 0
1 1 1 0 0 0 0
Figure 1: A(R) for maximising R∈L(3,7).
0 0 0 0 0 1 1 1 1
0 0 0 0 0 1 1 1 1
0 0 0 0 1 1 0 1 1
0 0 0 0 1 0 1 1 1
0 0 1 1 0 1 1 0 0
1 1 1 0 1 0 0 0 0
1 1 0 1 1 0 0 0 0
1 1 1 1 0 0 0 0 0
1 1 1 1 0 0 0 0 0
Figure 2: A(R) for maximising R∈L(4,9).
There are only two cases where both G(R) and G(R) are connected. These cases do not fit easily into the table, so they are dealt with separately in Figures 1 and 2. Figure 1 shows the only case covered by Table 1 where G(R) is not unique up to isomorphism.
Another case (n= 7, k = 2) appeared in Theorem 4.
§5. Above the roots
We begin the proof of our main result by investigating the behaviour of the rook polynomial above its largest root. Let R ∈ L(k, n) and suppose v is a vertex of G(R).
Imitating [6] we define a tree T(R, v) as follows. The vertices of T(R, v) correspond to paths in G(R) which start at v. Two vertices are adjacent if, of the two paths they correspond to, one is a maximal proper subpath of the other. The root of T(R, v) is the vertex corresponding to the empty path. Let ηv,r be the number of closed walks of length r inT(R, v) starting atv and define
wr(R) = 12X
v
ηv,2r.
The following properties of wr(R) are known ([6], [7]) (a) wr(R) =P
iλri where{λ1, λ2, . . . , λn} are the roots of ρ(R, x).
(11)
(b) Let s be the number of 4-cycles in G(R). Then w1 =nk,
w2 =nk(2k−1), w3 =nk(5k2−6k+ 2),
w4 =nk(14k3−28k2+ 20k−5)−4s,
w5 =nk(42k4−120k3+ 135k2−70k+ 14)−40(k−1)s.
(6)
(c) The rook polynomial ρ(R, x) is given by the power series ρ(R, x) =xnexp
− X r=1
wr(R) rxr
(7) which is convergent provided x lies above the greatest root of ρ(R, x).
Theorem 6. Suppose S = Sm,k and R∈ L(k, mk). Let λS and λR be the largest roots ofρ(S, x) and ρ(R, x) respectively. Then λR≥λS and wr(R)≥wr(S) for all r.
Proof: Let v be a vertex in G(A) for some A ∈ L(k, mk). Consider a vertex u of T(A, v) which is a distance d ≥1 from the root. Let P be the set of vertices in the path corresponding to u and eu ∈ P the final vertex in that path. Then the degree of u in T(A, v) is given by deg(u) = 1 +|N(eu)\P|where N(eu) is the set of neighbours of eu in G(A). SinceG(A) is bipartite and k-regular we have
deg(u) = 1 +k− |N(eu)∩P| ≥1 +k− d
2
. (8)
Now inG(S) every component is complete which means that the bound (8) is achieved in T(S, v) for every vertex u(except the root, which is necessarily of degreek). It follows that T(S, vS) is isomorphic to a subgraph of T(R, vR) for arbitrary vertices vS andvR in G(S) and G(R) respectively. Hence ηvS,r ≤ηvR,r for every r which means that wr(S)≤wr(R).
Now since the rth moment of the roots of ρ(R, x) dominates the rth moment of the roots of ρ(S, x) we conclude that λR ≥ λS, otherwise taking r sufficiently large yields a
The original reasoning behind Conjecture 1 is embodied in the following result.
Theorem 7. Suppose R∈L(k, mk) is not isomorphic toS =Sm,k. Then for x≥4k−4, ρ(S, x)−ρ(R, x)≥ρ(S, x) 2(k−1)2x4 + 15(k−1)3x5
.
Proof: By applying (7) and Theorem 6 we see that forx≥ 4k−4, ρ(R, x)
ρ(S, x) = exp
−X
r≥1
wr(R)−wr(S) rxr
≤exp
− w4(R)−w4(S)
4x4 − w5(R)−w5(S) 5x5
.
(12)
Next we use (6), which shows that ρ(R, x) ≤ ρ(S, x) exp −(s−t)(x4+ 8(k−1)x5) where s, t are the number of 4-cycles in G(S) andG(R) respectively. If we can show that s−t ≥2(k−1)2 then by applying Taylor’s Theorem to exp(·) we will get
ρ(R, x)
ρ(S, x) ≤1− 2(k−1)2
x4 − 16(k−1)3 x5 +
2(k−1)2
x4 + 16(k−1)3 x5
2
. (9)
Since 2(k −1)2x−4 + 16(k −1)3x−52
≤ (k−1)3x−5 for x ≥ 4(k−1), the theorem is proved once we have (9).
It remains to shows−t≥2(k−1)2. Letvbe a vertex in G(A) for someA∈L(k, mk).
Define Bv to be the subgraph induced by the ball of radius 2 around v in G(A). Suppose that the vertices at distance 2 from v are v1, v2, . . . , vl for some l≥ k−1. Let the degree ofvi inBv be di, and relabel if necessary so that di ≥di+1 for each i. Callfv the number of 4-cycles in G(A) which involve v. We have
fv = Xl i=1
di
2
. (10)
Note that since G(A) is k-regular bipartite, we must have Pl
i=1di =k(k−1) and di ≤k for each i. With these restrictions it is easily calculated that fv ≤ (k−1) k2
by noting that a+12
+ b21
> a2 + b2
provideda≥b. The maximum for fv is achieved only when l =k−1 and eachdi =k, which means thatv is contained in a complete componentKk,k. It follows that s= 12n(k−1) k2
> t, where n=mk.
It remains to find the maximum possible value of t. Take a copy of S and perform the following surgery. Remove edges (x, y) and (x0, y0) from different components ofS and replace them with edges (x, y0) and (x0, y) to get a new graph S0. The surgery destroys 2(k−1)2 of the 4-cycles in S, and does not create any new 4-cycles in S0. We claim that t ≤ 12n(k −1) k2
− 2(k− 1)2. First note that R must have at least 4k vertices which are not in complete components. Of these vertices, unless there is a vertex v satisfying fv > (k −1) k2
−(2k−3) then we immediately have that t ≤ 12n(k−1) k2
−k(2k −3) which is sufficient for k ≥ 2. Hence, (10) tells us the only remaining possibility is that l =k and d1 =d2 = . . .=dk2 =k, dk1 =k−aand dk =afor some a≤2. The a= 1 case when Rhas only 4k vertices not in complete components is now easily seen to be the
best of the remaining options.
§6. Between the roots
We study the behaviour of the rook polynomial below its largest root.
(13)
Theorem 8. Let w≈ 0.27846satisfy w+ log(w) + 1 = 0. Let β = 4(k−1) and suppose λn < β is the largest root of ρ(R, x) for a rectangle R∈L(k, n). Then
(a) |ρ(R, x)| ≤(β−x)nw−φn for all k < x < λn, where φ=w(β−k)/((w+ 1)(β−x)).
(b) ρ(R, x)≤(x−k)n2(β−x)2 ≤(x−k)n whenever (β+kw)/(1 +w)≤x≤λn
(c) |ρ(R, x)| ≤xnwϕn for all0< x≤k, where ϕ=kwx1/(w+ 1).
(d) |ρ(R, x)| ≤(k−x)n for all x≤kw/(1 +w).
Proof: We prove only (a) and (b); the proofs of (c) and (d) being similar. Let {λi}ni=1
be the roots of ρ(R, x), labelled in non-decreasing order. Suppose x is in the interval (k, λn) and choose a so that λa ≤ x < λa+1. We consider moving the λi in order to maximiser(x) =Q
|x−λi|, while preservingP
λi =nk. First we moveλa+1, λa+2, . . . , λn
so that they coincide at (λa+1 + λa+2 + . . .+ λn)/(n−a), and move λ1, λ2, . . . , λa so they are all equal to (λ12 +. . .+λa)/a. The arithmetic/geometric mean inequality ensures that r(x) will not be decreased by this process. Next we move λa+1, λa+2, . . . , λn
to β, and at the same time move the lower group of roots λ1, λ2, . . . , λa to α, where α= (nk−(n−a)β)/a. This further adjustment clearly does not decrease r(x). Now we have r(x, a) = (x−α)a(β−x)na. If we define θ by
θ = ∂
∂alog(r) = log
x−α β−x
− n(β−k) a(x−α) then
∂θ
∂a =−n2(β−x)2 a3(x−α)2 ≤0.
From this we conclude that for x fixed, r has a single maximum when θ = 0 at a= w(β−k)n
(w+ 1)(β−x). (11)
Substituting (11) into r(x, a) = (x−α)a(β−x)na yields (a). Note that we can do better when x ≥ (β+kw)/(1 +w), meaning that the maximum (11) occurs above the greatest feasible value ofa. In this caser(x, a) increases monotonically witha. By choicea≤n−1, and note that if a=n−1 then ρ(R, x) is negative. Part (b) of the theorem follows.
Theorem 9. |ρ(R, x)| ≤(x2−2kx+ 2k2−k)n/2 for allR∈L(k, n) and x≥0.
Proof: Suppose ρ(R, x) =Q
(x−λi). A standard inequality of means gives
|ρ(R, x)|1/n
1 n
X(x−λi)2 1/2
. (12)
The required bound follows from (12) and knowledge of the first two moments, (6).
(14)
§7. The ‘large’ cases
We present two simple lemmas which will help identify maximisingk×mk rectangles for large m and k.
Lemma 1. Let τ = 32n and m≥5. ThenIτ ρ(R, x)
133 eτ(τ −k)n for R∈L(k, n).
Proof: Suppose ρ(R, x) = Q
i(x−λi). By the arithmetic/geometric mean inequality we have ρ(R, x)≤ n1 P
(x−λi)n
= (x−k)n provided x≥max{λi}. Hence Iτ ρ(R, x)
≤ Z
τ
ex(x−k)ndx=eτ Xn i=0
n!(τ −k)i
i! ≤eτ Xn i=0
nni(τ −k)i. Since τ = 32n > n+k we see immediately that,
Iτ ρ(R, x)
≤eτ(τ −k)n X i=0
n τ −k
i
=eτ(τ −k)n+1/(τ −n−k).
Finally, (τ −k)/(τ −n−k) = 3 + 4/(m−2)≤13/3 for m≥5.
Lemma 2. Suppose that R ∈ L(k, n) where n = mk. Define m0 = min{m,6}. Then I04k ρ(R, x)≤4ke(2m0)k(12m0k)n.
Proof: It was proved in [7] that
I04k ρ(R, x)≤2ne2k Z 6k
2k
exxndx. (13)
Since dxd (exxn) = exxn1(n−x) we can bound the integrand in (13) by its value at x=m0k, giving Z 6k
2k
exxndx≤4kem0k(m0k)n.
In what follows we supposeS =Sm,k andR∈L(k, n) do not have isomorphic graphs.
Then by combining Lemma 1, Lemma 2 and (1), I4kτ ρ(S, x)
=I0 ρ(S, x)
−I04k ρ(S, x)
−Iτ ρ(S, x)
≥n!
m−1 m
n
− 4k(12m0k)n
e(m02)k133 eτ(τ −k)n. (14) Now 2(k−1)2x4 is a decreasing function of x for x >0, so by Theorem 7,
I4k ρ(S, x)−ρ(R, x)
≥I4kτ ρ(S, x)−ρ(R, x)
2(k−1)2 τ4
I4kτ ρ(S, x)
. (15)
(15)
Also Lemma 2 tells us that
I04k ρ(S, x)−ρ(R, x)≤I04k ρ(S, x)+I04k ρ(R, x)≤ 8k(12m0k)n e(m02)k . Combining with (14) and (15) we see that if
n!
m−1 m
n
− 4k(12m0k)n
e(m02)k133 eτ(τ −k)n− 4k(12m0k)nτ4
e(m02)k(k−1)2 >0 (16) then I4k ρ(S, x)−ρ(R, x)
+I04k ρ(S, x)−ρ(R, x)
>0 and so E(S)> E(R).
Define q5 = 51, q6 = 15, q7 = 8, q8 = 5, q9 = 4 and qi = 3 for i ≥ 10. It is a simple matter to establish that (16) holds for 5≤m≤10 and k =qm. We use this as a basis for induction.
In the notation of [1] we use Γ and ψ to denote the gamma and digamma functions respectively. Note that Γ(n+ 1) = n! and ψ(x) = dxd log Γ(x). Suppose that we make the following definitions
f1 = Γ(n+ 1) mm1n
f2 = 4k(12m0k)ne(2m0)k
f3 = 133 eτ(τ −k)n f4 = 4k(12m0k)ne(2m0)kτ4(k−1)2
with the aim of showing that f1 dominates the inequality (16). We shall prove that the ratios f1/f2, f1/f3 and f1/f4 are increasing functions of k for any fixed m≥ 5, provided k ≥qm. However, first we must show that (16) holds for k = 3 and m≥10. To that end, we fix m0 = 6 and observe that
1 kf1
∂f1
∂m =ψ(n+ 1) + log
m−1 m
+ 1
m−1 >log(k) + log(m−1) + 1
m−1 (17) because ψ(n+ 1)>logn for n >0. Meanwhile,
1 kf4
∂f4
∂m = log(k) + log(3) + 4 n
and log(m−1) > log(3) + 1 for m ≥ 10 so we conclude that log(f1/f4) is an increasing function of m in this range. Immediately we get that f1/f2 also increases with m for m≥10 because f4/f2= (3mk/2)4(k−1)−2 trivially increases with m. In addition,
1 kf3
∂f3
∂m = log(k) + log(32m−1)− 12 + 2
3m−2. (18)
Now 2/(3m−2) < 1/(m−1) for positive m and log(32m−1)− 12 < log(m−1) for all m >(√
e−1)/(√
e−3/2)≈4.362. Hence by (17) we see that log(f1/f3) increases withm
(16)
in the required range. Since (16) holds when k = 3 and m= 10 we conclude that it must hold for k = 3 and m≥10.
Next we fix m≥5 and show that (16) holds for all k ≥qm, using the knowledge that it holds when k =qm. We have,
1 mf1
∂f1
∂k =ψ(n+ 1) + log
m−1 m
>log(k) + log(m−1).
By comparison, 1 mf4
∂f4
∂k = log(k) + log(12m0) + 1 + 2k2+k−5 mk(k−1) − m0
m. Now (2k2 +k−5)/(k2−k) is a decreasing function for k ≥ (5 +√
10)/3≈ 2.721, so for our purposes we may bound it by its value when k =qm. It is then established by an easy case analysis that ∂k log(f1)> ∂k log(f4) form≥5 andk ≥qm, so we see that f1/f4 does indeed increase with k in this range. Moreover f4/f2 = (3n/2)4/(k−1)2 is an increasing function of k provided k≥2, so f1/f2 must also increase withk in the required range. It remains to use the same approximation used on (18) to show that
1 mf3
∂f3
∂k = log(k) + log(32m−1)− 12 <log(k) + log(m−1)< 1 mf1
∂f1
∂k .
We conclude thatf1/f2,f1/f3 andf1/f4are increasing functions ofk, providedm≥5 and k ≥ qm. Therefore inequality (16) holds for all m≥ 5 and k ≥ qm. We are left with only finitely many cases to check; namelyk = 3,4, . . . ,(qm−1) for m= 5,6,7,8,9. These cases will be checked in the final section.
§8. The ‘small’ cases
We turn our attention to the cases left unresolved by the preceding section, namely 3≤k ≤qm−1 for 5 ≤m≤9. Since ρ(S, x)≥0 for λS ≤x ≤λR we see from Theorem 7 that
Iλ4k4
S ρ(S, x)
≥Iλ4k4
R ρ(R, x) and hence
E(S)−E(R)≥I4k4 ρ(S, x)−ρ(R, x)
−I0λR ρ(R, x)
+I0λS ρ(S, x)
. (19) Notice that by Theorem 7,
I4k4 ρ(S, x)−ρ(R, x)
≥I4k4 ρ(S, x)(2(k−1)2x4+ 15(k−1)3x5)
. (20)
(17)
Now for specific values of m and k, the bound in (20) can be explicitly calculated, as can I0λS ρ(S, x)
, because we know thatρ(S, x) = (Lk)m whereLk is defined by (3). Hence the only term in (19) we need to work on isI0λR ρ(R, x)
. We use Theorem 8 and Theorem 9.
Define the cutoffs and functions
c5 = 4k−4, f5 =ex(x−k)n2(c5−x)2, c4 = c5+kw
1 +w , f4 =ex(c5−x)nw(w/(w+1))n(c5k)/(c5x)
, c3 = min{3k, c4}, f3 =ex(x2−2kx+ 2k2−k)n/2,
c2 =k−1, f2 =exxnwc1n/x, c1 = kw
1 +w, f1 =ex(k−x)n. so that
I0λR ρ(R, x)
≤ X5
i=1
Z ci ci1
fidx (21)
where we assumec0 = 0. Note thatf5 is positive betweenλR andc5. We consider the last integral in (21) first. We have,
df5
dx =ex(x−k)n3(c5−x) x2 −(n+k+c5)x+c5(n+k−2) + 2k
. (22) Notice that the discriminant ∆ = (n+k−c5)2+ 8(c5−k) of the quadratic term in (22) satisfies (n+k−c5)2 <∆<(n+k−c5+ 3k)2. We conclude thatf5 achieves its maximum in the interval [k, c5] when x equals
x0 = 12(n+k+c5)− 12 (n+k−c5)2+ 8(c5 −k)1/2
. This certainly means that
Z c5
c4
f5dx≤(c5−c4)e−x0(x0−k)n−2(c5−x0)2.
We bound the other four integrals in (21) by noticing that the integrand is concave in each case (although the integral of f1 may be explicitly calculated if preferred). To begin, suppose l=ax+band f =e−xlnwc/l wherea, b, c and w are independent of x. Then
l2 f
d2f dx2 =
caln(w)
l +l+a−an 2
+ (n−1)a2−2al≥(n−1)a2−2al.
It is now a simple matter to check thatf1,f2andf4 are concave in their required intervals.
Next we show that f3 is concave for x ∈[c2, c3]. We claim that in this interval, and for integers k ≥ 3, m≥ 5 and n = mk it can easily be checked that n−2(x−k) > √
n.
Then d2f3
dx2 = f3
(x2−2kx+ 2k2−k)2
n(x−k)−(x−k)2−(k2−k)2
+n (k2−k)−(x−k)2 (23)
(18)
which is clearly positive unless we assume (x−k)2 ≥k2−k ≥6. Sincex−k≥c2−k =−1 it follows thatx > k and hencen(x−k)−(x−k)2−(k2−k)≥ n−2(x−k)
(x−k)>0.
But now
n(x−k)−(x−k)2−(k2−k)>√
n(x−k)>0 which means that (23) is positive, as required.
Now the integral of a concave function can be bounded above by taking a simple polygonal approximation to the curve. Specifically, in (21) we can subdivide each interval into σ subintervals each of width δi = (ci−ci1)/σ, giving,
I0λR ρ(R, x)
≤(c5−c4)f5(x0) + X4
i=1
Xσ j=1
δi
2
fi ci1+ (j−1)δi
+fi ci1+jδi
. (24)
Takingσ = 10 the bound in (24) was computed form= 5,6,7,8,9 andk = 3,4, . . . , qm−1.
Together with (20) this allowed confirmation that E(S) > E(R) in each of these cases, except when m= 5 and k ≥ 12. For this subcase (20) becomes too slow to compute for large k, but it is sufficient to use (24) as a substitute for Lemma 2 in the derivation of (16). Specifically, ifB is the bound computed in (24) and
n!(4/5)n−B− 133 eτ(τ −k)n− Bτ4
(k−1)2 >0 (25)
thenE(S)> E(R). It can quickly be confirmed that (25) holds fork = 12,13, . . . ,50 with m= 5, which completes the proof of the ‘small’ cases. The entire calculation was checked independently by numerical integration. Combined with the results of the preceding sec- tion, we get the main result.
Theorem 10. Letm≥5,k ≥2andn=mkbe integers. IfR∈L(k, n)is maximising then G(R)∼=mKk,k. Equivalently, ifM is a(0,1)-matrix with exactlyk zeroes in each row and in each column then the permanent per(M) is maximised (uniquely, up to permutations of the rows and columns) by the matrix with block structure
Zk Jk Jk · · · Jk
Jk Zk Jk · · · Jk
Jk Jk Zk · · · Jk
... ... ... . .. ...
Jk Jk Jk · · · Zk
(26)
where Zk, Jk are k×k blocks of zeroes and ones respectively.
Corollary 2. For integersm≥5, k ≥2 andn=mk Mk,n =
Z
0
ex Lk(x)m
dx.
(19)
Also note that [5] cites an inclusion-exclusion formula of Kaplansky for the permanent of (26). However, to find Mk,n it is just as easy to calculate the integral in Corollary 2.
In closing, we observe that it is quite possible that Theorem 10 can be extended to show that Conjecture 1 holds with only a finite number of exceptions when m= 3. In fact we conjecture that there are no exceptions other than the two discussed in§4. However the techniques presented in this paper are not yet strong enough to apply when m < 5.
Hope of proving there are finitely many exceptions to Conjecture 1 is bolstered by the observation that Lk(x)≤k!ex/2 for x≥0 (c.f. inequality 22.14.12 of [1]) and hence
Z 4k−4
0
exρ(Sm,k)dx≤(k!)m
Z 4k−4
0
e(m2)x/2dx≤ 2(k!)m
m−2e2k(m2). (27) This means by (1) that for fixed m ≥ 3 and n = mk → ∞ the initial segment of the integral I(Sm,k) is asymptotically insignificant compared to E(Sm,k), because
2(k!)m
m−2e2k(m2) .
n!(1−k/n)n
=O k(m1)/2 e2m4 (m−1)m
k
=o(1).
IfI04k4(R) could be similarly handled for otherR∈L(k, n) then Theorem 7 would suffice.
(20)
§9. References
[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, Dover pub- lications, New York, 1965.
[2] V. I. Bolshakov, The spectrum of the permanent on Λkn, Proceedings of the All-Union Seminar on Discrete Mathematics and its Applications (Russian) ed. O. B. Lupanov, Moskov. Gos. Univ., Moscow, 1986, 65-73.
[3] L. M. Br`egman, Some properties of nonnegative matrices and their permanents,Soviet Math. Dokl. 14:945-949 (1973).
[4] R. A. Brualdi, J. L. Goldwasser and T. S. Michael, Maximum permanents of matrices of zeroes and ones, J. Comb. Th. A47:207-245 (1988).
[5] F. R. K. Chung, P. Diaconis, R. L. Graham and C. L. Mallows, On the permanents of complements of the direct sums of identity matrices, Adv. in Appl. Math. 2:121-137 (1981).
[6] C. D. Godsil, Matchings and walks in graphs, J. Graph Th. 5:285-297 (1981).
[7] C. D. Godsil and B. D. McKay, Asymptotic enumeration of Latin rectangles,J. Comb.
Th. B 48:19-44 (1990).
[8] O. J. Heilmann and E. H. Lieb, Theory of monomer-dimer systems, Comm. Math.
Physics 25:190-232 (1972).
[9] S. A. Joni and G.-C. Rota, A vector space analog of permutations with restricted position, J. Comb. Th. A 29:59-73 (1980).
[10] D. Merriell, The maximum permanent in Λkn, Linear and Multilinear Algebra 9:81-91 (1980).
[12] H. Minc, Theory of permanents 1978-1981,Linear and Multilinear Algebra 12:227-263, (1983).
[13] A. Schrijver, Bounds on permanents, and the number of 1-factors and 1-factorizations of bipartite graphs, London Math. Soc. Lecture Note Ser. 82:107-134 (1983).
[14] I. M. Wanless, The Holens-–Dokovi´c Conjecture on permanents fails, (submitted for publication).
[15] I. M. Wanless, Maximising the permanent and complementary permanent of (0,1)- matrices with constant line sum, (submitted for publication).
[16] N. Zagaglia-Salvi, Permanents and determinants of circulant (0,1)-matrices, Matem- atiche (Catania) 39:213-219 (1984).
Updating...
## References
Related subjects : | 13,627 | 36,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-23 | longest | en | 0.826941 |
https://id.scribd.com/document/348551337/Unit-3-Review-on-Forces | 1,568,942,416,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573801.14/warc/CC-MAIN-20190920005656-20190920031656-00030.warc.gz | 546,448,740 | 72,867 | Anda di halaman 1dari 4
Review Problems for Unit 3
For each problem draw a force diagram that would be accepted on the AP exam, show all work, and write out
the sum of force/equation that is needed.
1. A 1240 kg wrecking ball is pulled up and over by a wire. (a) What is the tension in the wire and (b) how
hard is the wall pushing on the ball?
10
(a) T = 12,591.29 N (b) Fwall = 2,186.45 N
2. A 9.34 kg object slides down the hill. (a) What is the normal force on the object? (b) What is the frictional
force on the object if it moves at a constant velocity? (c) If the surface has a coefficient of friction equal to
0.3, what is the new frictional force and what is the acceleration of the object?
27
(a) FN = 83.22 N (b) Ff = -42.40 N (c) Ff = -24.97 N a = 1.87 m/s2
Unit 3: Newtons Laws of Motion Page 1 of 4
3. (a) What is the normal force and (b) what is the force of friction for the situation below if the box isnt
moving? (c) What would be the coefficient of friction is the box accelerates at 2 m/s2 to the right?
F = 100 N
42
m =25kg
(a) FN = 183.07 N (b) Ff = - 74.31 N (c) = .13
4. A mass of 15 kg hangs as shown below with 3 strings attached. The tension is string 1 is 283 N, find the
tension is the other two strings.
41o
String 2
String 1
String 3 being
pulled down
T2 = 374.98 N T3 = -96.01 N
5. A 50 kg ball rests motionless on the incline because of the tension in the rope. Assuming there is no
friction, (a) what is the tension in the rope and (b) the normal force acting on box?
String
19o
(a) T = -162.78 N (b) FN = 472.76 N
Unit 3: Newtons Laws of Motion Page 2 of 4
6. A 5.0-kilogram monkey hangs initially at rest from two
vines, A and B, as shown above. Each of the vines has
length 10 meters and negligible mass. Determine the
tension in vine B while the monkey is at rest.
TA = 25N TB = 43.3 N
7. A 10-kilogram block rests initially on
a table as shown in cases I and II above.
The coefficient of sliding friction
between the block and the table is 0.2.
The block is connected to a cord of
negligible mass, which hangs over a
massless frictionless pulley. In case I a
force of 50 N is applied to the cord.
In case II an object of mass 5 kilograms is hung on the bottom of the cord. (a) Calculate the acceleration of the
10-kilogram block in case I. (b) Calculate the acceleration of the 10-kilogram block in case II. (c) Find the
tension in the cable for Case II.
(a) 3 m/s2 (b) 2 m/s2 (c) 40 N
Unit 3: Newtons Laws of Motion Page 3 of 4
8. A 55 kg women is in an elevation at rest. (a) What is the normal force acting on the women if the elevator
starts to accelerate down at 4 m/s2? (b) What is the normal force acting on the women when the elevator starts
to slow down at 2 m/s2?
(a) 330 N (b) 660 N
9. A rotten pumpkin that has a mass of 9 kg is dropped and is moving at a speed of 17 m/s right before it hits
the ground. (a) What is the acceleration of the pumpkin during impact if it stops in 0.22 s? (b) What force does
the ground exert on the pumpkin during impact with the ground? (c) What force does the pumpkin apply to the
ground during impact? | 966 | 3,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-39 | latest | en | 0.919679 |
http://www.econometricsbysimulation.com/2012/07/jensens-inequality.html | 1,397,938,188,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00610-ip-10-147-4-33.ec2.internal.warc.gz | 413,429,171 | 30,403 | ## Monday, July 2, 2012
### Jensen's inequality
* Jensen's inequality
* If f is convex then the Expected value of f(x) is greater that f of the expected value of x.
* If f'(x) >= 0 E(f(x)) >= f(E(x))
* The reverse with concave.
* Likewise if f'(x) <= 0 E(f(x)) <= f(E(x))
* This is easy to show:
clear
set obs 100
gen x=runiform()*-10
* This will make it so that x is always positive making derivatives of functions easy to interpret.
* f(x)=x^2 -> f''(x)=2 > 0 thus convex
gen fx = x^2
sum x
* The expected value is approximated by the mean of x
local fEx = (r(mean))^2
di "So: f(E(x)) ~ " string(`fEx',"%9.2f")
sum fx
di "E(f(x)) ~ " string(r(mean),"%9.2f") " is greater than f(E(x)) ~ " string(`fEx',"%9.2f")
* Thus Jensen's inequality works!
* It is a very helpful property in probability theory.
* However, the medians function passes through monotonic functions (completely unrelated to Jensen's inequality)
qui sum fx, detail
local medfx = r(p50)
qui sum x, detail
local medx = r(p50)
di "med(f(x)) ~ " string(`medfx',"%9.2f") " is approximately the same as f(med(x)) ~ " string(`medx'^2,"%9.2f") | 366 | 1,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2014-15 | longest | en | 0.785775 |
https://listenwithothers.com/2010/06/page/2/ | 1,529,443,588,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00170.warc.gz | 651,307,521 | 17,822 | # Archive for June, 2010
## Pandigital Squares by Oyler
Posted by Listen With Others on 17 June 2010
Ten double sided cards each have a different single digit printed on each side. When the cards are arranged in a row a pandigital square, P, is formed. When the cards are turned over and kept in the same order the result is a different pandigital square Q. In the clues the subscripts refer to the cards in positions 1 to 10 respectively. For example if P was 6154873209 then P25 would be the four digit string 1548. In order for solvers to identify P and Q, the grid, which has 180° rotational symmetry, should be completed. In the grid no entry starts with zero and all are different. P and Q should be written underneath the grid.
________________________________
Across Down 1 P13 + P89 1 P3 x P6 3 P10 x P10 = Q12 2 Q10 x Q34 5 Q47 3 Q8 x Q23 7 Q3 x Q4 x Q5 4 P6 (P7 + P8) 8 P4 (Q12 – P12 ) / Q9 6 P46 + Q46 + P34 + Q67 9 P36 7 P24 + Q68 – Q10 12 P2 x P7 8 Q10 x Q12 13 P79 10 Q4 x Q4 = Q34 11 P1 x P2 x P3 x P4
## Pandigital Squares by Oyler: Solution
Posted by Listen With Others on 16 June 2010
From the clue lengths and numbering we can deduce the grid to be as follows.
10d. Q4 is 5 or 6 so Q34 is 25 or 36. So 5a starts with a 5 or 6.
2d. Multiple of 25 or 36 with the multiple being 4, 5, 6 or 9. This yields
25 36 4 100 144 5 125 180 6 150 216 9 225 324
The only fit is with 150 so Q10 is 6 and Q34 is 25. Entering these in the grid gives up P4 as 2. So we have P as ???2?????? and Q as **25*****6.
8a. Q12 > P12. So Q1 is not 1.
3a. P10 is 4, 5, 6 or 9. But from 8a it is not 4 and from the fact that P4 is 2 it is not 5. So P10 is 6 or 9. If it is 6 then Q12 is 36 but Q10 is 6 so P10 is 9 and 3a is 81. Q1 is 8 and Q2 is 1. Using the 2 digit termini of square numbers tells us that P9 is even but not 2 which already appears in P. Also Q9 is odd and will be 3, 7 or 9.
8d. 6 x 81 = 486 so P6 is 8 and P8 is 6.
12a. P2 or P5 must be 5.
11d. 1 must be in P1 to P4 inclusive.
3d. This is a multiple of 12 so must be 84 and Q8 is 7.
5a. Q6 is 4.
7d. Starts with at least a 4 so 7a must be 90. Therefore Q5 is 9, Q9 is 3 and Q7 is 0. Thus Q is 8125940736 which is 901442.
5a is 5940.
6d ends in 4 so P5 is 4. Now P9 is 0.
13a is ?60.
11d. P2 must be 5.
4d is a multiple of 8 and is 104 so P7 is 7.
12a is 35.
13a is 760.
11d is 30.
1d is 24 so P3 is 3 and P1 is 1.
P is thus 1532487609 which is 391472.
The remaining entries are
7d is 933, 9a is 3248, 1a is 213 and 6d is 914 which check out.
The final grid is as follows:
## Double Shuffling and Dealing by Auctor
Posted by shirleycurran on 11 June 2010
The Numpty team gazed for a good hour at Auctor’s Double Shuffling and Dealing and got nowhere at all. Part of the trouble was that phrase in parenthesis ‘… the answer (which is also the grid entry) with one letter omitted’. The answer is usually the ‘grid entry’, so was this something special? Were we entering the answer with one letter omitted?
Our first red herring seemed to confirm this Numpty diversion. If we used CHARMING for ‘Fetching’, (Fetching up special seat ahead of new government) then the clue led to an anagram of CHAIR plus NG (CHARING) and we had the letter M in the margin. Great! Except that we couldn’t cope with that little word ‘up’ in the clue.
Fortunately, remembering what a bunch of oenophiles these setters are, we rethought and came up with SPEWING (much less fetching as a word) and SHERRY to go with it on the top line. And we were on our way.
Not being particularly competent with wordplay played into our hands with this crossword. It wasn’t long before we had a complete grid, often including the only word that the Chambers CD Rom produced to fill the light, even if we could see no link at all to the wordplay. We wondered about the proliferation of moles (and suspected that we were heading for a Hamlet quotation – ‘Canst work in the ground so fast, old mole?’) and muttered seditiously about the definitions appearing in the middle of clues, but we lacked the sophistication that would have led more competent solvers to grasp what was going on – until EMICATED appeared at 17dn and we suddenly twigged that we had seen its definition (sparkled) at 26dn.
Suddenly the TALPIDAE burrowed their way to the moles and we saw what was going on.
Meanwhile ‘I HAVE DECEIVED YOU BOTH, I HAVE DIRECTED YOU TO …’ emerged from our extra letters and Google gave us two more words: ‘WRONG PLACES’. We were told what we had just realized! The Host at the Garter was our guilty party and we were left with WGHEHMETVOOODSL to convert into four words. Easy! GODS and LOVE leapt out at us, leaving ‘WHOM THE’ – so ‘Whom the gods love’. We had already found MENANDER in the usual diagonal slot and Google confirmed the link.
All that was left to do was to sort out those extra words we had happily ignored and return them to their own wordplay. Here is my ultimate solution with a couple of problems at 24 and 29 across that someone will perhaps explain to me? I wonder, too, whether I have spotted the correct definition for EVOKE. ‘Cause’ seemed to be the only remaining wandering word.
Auctor certainly challenged us – thank you!
Posted in Solving Blogs | Tagged: , | 1 Comment »
## Back Gate by Tiburon
Posted by linxit on 5 June 2010
Well, having been one of the first volunteers when Chris originally planned Listen With Others, I suppose I’d better finally produce my first blog entry. Unfortunately about the time LWO got started my job changed and I stopped being single, so my weekend time was much more restricted than before. And I’m quite lazy…
Anyway, back to Tiburon. I met him for the first time at the Magpie party a few weeks earlier, and he mentioned a Listener in the pipeline, so I thought I’d better have a crack at it at least. Misprints in every definition, with messages in the misprints and the corrected letters. Must have been difficult to write the clues!
I printed it out at work on the Friday and had my first look on the train without any solving aids. First read through the clues didn’t produce too much – RAG WEEK was the first one I got, followed by WET ROT (I often find multi-word answers easier to spot), then managed to complete most of the SE corner and put in a wrong answer at 17A. Next chance to look at it was Saturday morning, with Chambers, Bradfords, TEA and the Internet at my disposal. About three hours later I emerged victorious with a completed grid.
Strangely enough, up to that time I’d not bothered to even look at the two columns of letters I’d been neatly writing in next to each completed clue as I went along, so it came as quite a surprise when I instantly saw DEOX/YRIB/ONUC/LEIC/DCID (oops) in the misprints and its four component chemicals in the corrections. But where did that rogue D come from?
36D Following Peru’s appeal for help dash for Colombia (5)
“dash” becomes “cash” and it’s PE + SOS, surely! Took me another couple of minutes to see that it could also work with “dash” -> “dosh”.
DNA stands out in the middle, and I can see a double helix of names winding around it. I’d heard of WATSON and CRICK, but had to look up FRANKLIN and WILKINS on Wikipedia to confirm them. I also (rarely for me) saw the significance of the title, which gave me fond memories of Ploy’s Signal Boxes (Listener 4004).
Posted in Solving Blogs | 4 Comments »
## Back Gate by Tiburon
Posted by shirleycurran on 4 June 2010
This was the numpty team’s first Tiburon and we approached it with some trepidation. Rule number one; read the preamble carefully several times and highlight some instructions! What a good thing that we noticed that we were looking for misprints in certain clues and corrections in the remainder! That instruction was slightly daunting; how were we going to separate them?
Solving progress was very slow. A friend suggested that this puzzle would fit into the Magpie A or B category. I found the wordplay very challenging – clearly there’s a long uphill climb to the D/E level. However, almost every time that I solved a clue, I commented ‘That was brilliant!’ or ‘That was magic!’ or ‘What a splendid clue!’ Well, they were, weren’t they?
Take 22d. ‘Nit treatments limited to prime locations’ I wonder how long Tiburon pondered before he spotted the fact that RETE (net) occupied the ‘prime’ (i.e. 2,3,5 and 7) locations of ‘treatments’.
Misprints are so often fairly obvious but not so with 37ac. ‘Ocarina formed of pipes etc.’ The surface reading is magic – an ocarina is formed of pipes isn’t it? The obvious misprint would have been ‘acarina’, so, for far too long, I tried to find some mites that would be anagrammed to ‘pipes etc.’ Oh no! AARONIC finally appeared and the new magic surface reading of ‘popes etc.’
Then there was the ARAL SEA. It took me a while to spot that ALS in AREA fulfilled the needs of the clue, ‘Lar(G)e lake once further in extent’. It took me even longer to recognise the sheer genius and relevance of the surface reading concerning this shrinking lake.
My favourite of all these fabulous clues has to be, ‘Exclude from li(S)t in newspaper limo having disembarked’. The thought of the SUN ROLLS is delightful. I have a vision of Murdoch swanning around in his Rolls – then we disembark him (remove the SS) and are left with UNROLL.
There was such pleasure in these wonderful clues that the team wasn’t particularly troubled by our slow rate. The four corners, one by one, slowly filled (that’s like saying soccer is a game of two halves, and we either win or we lose, isn’t it?) until we suddenly saw THYMINE in our corrected misprints. What a give-away! At once, the others resolved themselves into CYTOSINE, GUANINE and ADENINE and we were left with DEOXYRIBONUCLEIC ACID. Of course, that resolved a few of our remaining problems with the clues. 19d. ‘Pool buddy in bright costume’. I had been struggling to spot the misprint, and having visions of some dazzling orange and turquoise Speedo-clad bronzed Adonis at the poolside – but no, it was not to be. The L in Pool was the misprint and, of course, Kanga was Pooh’s buddy.
SALVE had given me Ice in the place of Ace – most unsatisfactory! However, the C was now needed as the misprint in 5d. ‘Ace possibly, except straddling line’, and here was another brilliant clue with an Ave as a Salve!
Finding WILKINS, WATSON, CRICK and FRANKLIN spiralling their way down the grid in symmetrical curves and encapsulating the DNA theme was the final beautiful touch of this magic crossword. What a triumph for Tiburon!
Posted in Solving Blogs | Tagged: , | 1 Comment » | 2,896 | 10,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-26 | latest | en | 0.889503 |
https://cyberchalky.wordpress.com/2010/08/ | 1,606,222,825,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176256.21/warc/CC-MAIN-20201124111924-20201124141924-00671.warc.gz | 247,122,991 | 20,420 | ## Archive for August 2010
### Bayesian Probability
August 29, 2010
For those of you who wish to try and understand Bayesian Probability, Congratulations and Commiserations. Congratulations because you are doing the right thing, commiseration because I know what sort of headache you are going to have in about 2 hours….
Bayesian Theory
See you in Class.
### Troublesome, tricky triangles of terror!
August 8, 2010
Triangles are the simplest of geometrical shapes – three connected straight lines form a triangle. Of course, we need ways to classify the different possible types of triangle (right-angled or not, equilateral, isoceles or scalene, etc.)
But that is old – you learned that information in primary school – early in secondary school you learned about the angles (summing to 180, acute, obtuse, complementary, etc). This year we want to be able to identify if two triangles are similar (same internal angles, different side lengths) or congruent (exactly the same – i.e. same sidelengths and same angles.)
### Mathematical limbo – or getting under the line!
August 8, 2010
Have you ever tried to limbo? Bending over backwards, contorting yourself to get under a low stick? You know you have – if you think you haven’t you’re probably just repressing the memories!
Calculus is just the same, except the limbo stick is a function (possibly on fire, and definitely covered in sharp spikes). Instead of bending over backwards and ruining your spine, you have to twist, torment and contort your brain in order to force the function to reveal the area between it and the axis. But don’t worry – you can do this – after all, you’ve survived Maths Methods this far! | 379 | 1,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-50 | latest | en | 0.912318 |
http://www.mrob.com/pub/muency/jordancurvemethod.html | 1,718,756,412,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00871.warc.gz | 46,033,430 | 4,734 | mdbtxt1
mdbtxt2
Proceed to Safety
# Jordan Curve Method
Robert P. Munafo, 2023 Jun 21.
The "Jordan curve method" (named for the Jordan curve theorem which is the basis of the Even-odd rule) is a way to find the period of a mu-atom or mu-molecule whose location is known only to be within some bounding rectangle. It is also described in the Period article.
Given a rectangle containing some or all of the Mandelbrot set, one can perform iterations on the rectangle's corners to determine the period of the lowest-period mu-atom within the rectangle, and its approximate location within the rectangle.
1) Start with the four corners of the rectangle as complex numbers, and treat each as the parameter C in the standard iteration algorithm. Start with iteration count N=0.
2) Perform one iteration for each point (making sure that each point is iterated with its own C value; there is a different C value for each of the four iterations). Add one to N.
3) Using the Jordan curve theorem, determine if the current four points define the boundary of a quadrilateral that surrounds the origin. If not, go back to step (2).
If so, then N is the period of a mu-atom within the initial rectangle. The location of the origin in relation to the current positions of the four points corresponds to the position of the mu-atom relative to the original corners of the rectangle.
To find the (approximate) coordinates of the nucleus of the mu-atom, use the method described in the article Inverse Interpolation for a Quadrilateral, with the (iterated) corners as the coordinates (a, b), (c, d), (e, f), and (g, h); and set the values of v and w both to 0. Calculate p and q as described in that article, and they give the position of the mu-atom relative to the original rectangle.
Program
The following program sets up an example and prints the calculations as it goes; you can use the example values to verify that your implementation is working. It uses coordinates of a square enclosing the period-5 mu-molecule R2F(1/2(1/3B1)B1)S (see that article for a picture).
NOTE: The variable names ja, jb, ..., jh correspond to the the variables a, b, ..., h seen at the start of the program in the Jordan Curve Theorem article. See that article for the program listing for the "subprogram jct_test" referenced by this program.
NOTE: The variable names ca, cb, ..., ch correspond to the the variables a, b, ..., h seen at the start of the program in the Inverse Interpolation for a Quadrilateral article. See that article for the program listing for the "subprogram inverse_2d_interpolate" referenced by this program.
// Set C real/imag parameters for 4 complex numbers // We want the period of R2F(1/2(1/3B1)B1)S, the island // located at -1.255 + 0.382i +- 0.005 k1r = -1.250 k1i = 0.387 k2r = -1.260 k2i = 0.387 k3r = -1.260 k3i = 0.377 k4r = -1.250 k4i = 0.377 // Initial Z values are all 0 z1r = 0 z1i = 0 z2r = 0 z2i = 0 z3r = 0 z3i = 0 z4r = 0 z4i = 0 // other setup and loop start go here for iter=1 to 100 // iterate 1st corner zr2 = z1r*z1r - z1i*z1i zi2 = 2*z1r*z1i z1r = zr2 + k1r z1i = zi2 + k1i // iterate 2nd corner zr2 = z2r*z2r - z2i*z2i zi2 = 2*z2r*z2i z2r = zr2 + k2r z2i = zi2 + k2i // iterate 3rd corner zr2 = z3r*z3r - z3i*z3i zi2 = 2*z3r*z3i z3r = zr2 + k3r z3i = zi2 + k3i // iterate 4th corner zr2 = z4r*z4r - z4i*z4i zi2 = 2*z4r*z4i z4r = zr2 + k4r z4i = zi2 + k4i // show current iterated coordinates print "iter ";iter;": (";z1r;", ";z1i; print "), (";z2r;", ";z2i;"), (";z3r;", ";z3i; print "), (";z4r;", ";z4i;")" // set parameters to JCT subroutine ja = z1r jb = z1i jc = z2r jd = z2i je = z3r jf = z3i jg = z4r jh = z4i // Perform Jordan Curve test, same algorithm as // shown in "Jordan Curve Theorem" article: run subprogram jct_test // if detected, print result and end if (jct_answer) then print "The lowest period mu-atom nucleus within" print "this rectangle has period ";iter // now use reverse interpolation to estimate the nucleus location ' set parameters to inverse2Dinterpolate subroutine ' vertex coordinates (different order from Jordan) ca = z1r cb = z1i cc = z2r cd = z2i ce = z4r cf = z4i cg = z3r ch = z3i ' and point of interest (origin) cv = 0 cw = 0 ' call subroutine for inverse interpolation run subprogram inverse_2d_interpolate if (i2i_error < 0) then print "Could not reverse-interpolate to find a nucleus." else ' compute coordinates of located point with respect ' to original corners kwid = k2r - k1r; khei = k1i - k3i; nr = k1r + pp*kwid; ni = k1i - qq*khei; print "Nucleus is at approximately (";nr;", ";ni;")" end if end program end if // loop continue, test itmax, end without success next iter print "Iterated ";iter;" times without finding a period." end program
To verify proper operation, run the program and check for the following values to be printed out:
iter 1: (-1.25, 0.387), (-1.26, 0.387), (-1.26, 0.377), (-1.25, 0.377) iter 2: (0.1627, -0.5805), (0.1778, -0.5882), (0.1855, -0.5730), (0.1704, -0.5655) iter 3: (-1.5605, 0.1981), (-1.5744, 0.1778), (-1.5540, 0.1644), (-1.5408, 0.1843) iter 4: (1.1459, -0.2312), (1.1871, -0.1728), (1.1278, -0.1341), (1.0900, -0.1910) iter 5: (0.0097, -0.1428), (0.1194, -0.0233), (-0.0060, 0.0746), (-0.0984, -0.0393) The lowest period mu-atom nucleus within this rectangle has period 5
Example Implementations
A period finder using this method is implemented in Color MANDELZOOM and in Power MANDELZOOM; the latter article shows it being used (in the embedded Julia set image).
revisions:
20230218 first version; | 1,760 | 5,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.882706 |
https://www.lixingqiu.com/2020/03/11/%E5%8B%87%E9%97%AF%E9%BB%91%E6%9A%97%E8%BF%B7%E5%AE%AB/ | 1,725,845,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00660.warc.gz | 837,522,351 | 14,502 | # 勇闯黑暗迷宫
### 勇闯黑暗迷宫
```"""
勇闯黑暗迷宫,迷宫会停电,仅凭记忆去碰小花!
"""
from sprites import *
from winsound import PlaySound,SND_ASYNC
PlaySound('勇闯.wav',SND_ASYNC)
screen = Screen()
screen.setup(640,480)
screen.title('勇闯黑暗迷宫') # 设定标题
ball = Sprite(1) # 新建小球角色
ball.color('black')
ball.bk(300)
ball.pensize(20)
ball.pendown()
ball.fd(400)
ball.penup()
ball.goto(0,250)
ball.pendown()
ball.goto(0,100)
ball.penup()
ball.goto(0,-250)
ball.pendown()
ball.goto(0,-100)
ball.penup()
ball.goto(160,100)
ball.pendown()
ball.goto(160,-100)
ball.penup()
ball.goto(-310,230) # 定位到这开始画黑色边框
ball.pendown()
for _ in range(2):
ball.fd(610)
ball.rt(90)
ball.fd(460)
ball.rt(90)
ball.penup()
ball.goto(-200,-150) # 最终定位到这
ball.say('按方向箭头\n操作我碰到小花',10,False)
flower = Sprite('res/flower.png') # 新建小花角色
flower.goto(-200,150) # 小花定位到这里
w = Sprite(visible=False) # 写字的角色
w.goto(-100,30) # 定位到(-100,30)
info = "马上要停电,请立即记住迷宫布局"
ft = ('',23,'normal')
w.write(info) # 写提示信息
w.wait(4) # 等待4秒钟
w.goto(0,30) # 定位到(0,30)
# 下面是写倒计时3,2,1
for x in range(3):
w.clear()
w.write(3-x,align='center',font=ft)
w.wait(1)
w.clear()
screen.bgcolor('black') # 把背景设为黑色
leftkey = Key('Left') # 新建左方向箭头
rightkey = Key('Right') # 新建右方向箭头
upkey = Key('Up') # 新建上方向箭头
downkey = Key('Down') # 新建下方向箭头
screen.listen() # 监听屏幕按键
flag = None
```
(会员专属：能浏览所有文章，下载所有带链接的Python资源。) | 611 | 1,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.189413 |
https://nbviewer.ipython.org/gist/uncommoncode/f2fbd405e1f219660130/ | 1,653,309,088,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00662.warc.gz | 486,228,973 | 19,325 | In [1]:
import math
import random
import matplotlib.pyplot as plt
import matplotlib
%matplotlib inline
In [181]:
def random_sum(sampler, count):
return sum(sampler() for i in range(count))
def random_xor(sampler, count):
return reduce(lambda a,b: a^b, (sampler() for i in range(count)))
def sample_sum(samples):
return sum(samples)
def sample_xor(samples):
return reduce(lambda x, y: x^y, samples)
In [182]:
count = 100000
n = 10
max_value = 2**8-1
sampler = lambda: random.randint(0, max_value)
samples = [[sampler() for i in range(n)] for i in range(count)]
standard_samples = [sample_sum(sample) for sample in samples]
mod_samples = [sample_sum(sample) % max_value for sample in samples]
xor_samples = [sample_xor(sample) % max_value for sample in samples]
In [183]:
plt.hist(standard_samples);
The sum of independent random variables approaches a normal distribution due to the central limit theorem.
In [184]:
plt.hist(mod_samples);
The distribution of the modulo sum of random variables can become uniform again due to boundary conditions of a sort that are removed due to the modulo operation. See http://emmettmcquinn.com/blog/ for more.
In [185]:
plt.hist(xor_samples);
XOR is a common simple technique to combine entropy (see here for more).
In [ ]: | 321 | 1,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.547041 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=12905 | 1,369,386,310,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704392896/warc/CC-MAIN-20130516113952-00039-ip-10-60-113-184.ec2.internal.warc.gz | 309,916,827 | 1,005 | ```Question 24271
Y=X^4
LET US CONSIDER THIS IN OUR FIELD OF REAL NUMBERS.
TWO CONITIONS ARE TO BE SATISFIED FOR THIS RELATION TO BE AFUNCTION
1.FOR ALL X IN REAL NUMBERS ,THERE SHOULD BE A VALUE OF Y IN REAL NUMBERS.....THIS IS SATISFIED.
2.SAME VALUE OF X SHOULD NOT GIVE RAISE TO MORE THAN ONE VALUE OF Y ...THIS IS ALSO TRUE.
HENCE THIS IS A FUNCTION.
NOTE.....WE CAN HAVE SAME Y VALUE FOR MORE THAN ONE X VALUE AS IS THE CASE HERE.X=+1 GIVES Y=+1...AND X=-1 ALSO GIVES Y=+1..THIS IS OK.BUT THE OTHER WAY ROUND IS NOT ALLOWED FOR THE RELATION TO BE A FUNCTION.FOR EXAMPLE Y=SQRT.OF X IS NOT A FUNCTION SINCE X=+1 CAN GIVE RAISE TO 2 DIFFERENT VALUES OF Y NAMELY +1 AND -1.HENCE THIS IS NOT A FUNCTION.``` | 221 | 708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-20 | latest | en | 0.524077 |
https://elmhurstskiclub.org/and-pdf/2469-handbook-of-univariate-and-multivariate-data-analysis-and-interpretation-pdf-183-85.php | 1,627,598,738,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00333.warc.gz | 248,356,146 | 9,277 | # Handbook Of Univariate And Multivariate Data Analysis And Interpretation Pdf
File Name: handbook of univariate and multivariate data analysis and interpretation .zip
Size: 11481Kb
Published: 13.05.2021
We have enough money handbook of univariate and multivariate data analysis with ibm spss second edition and numerous books collections from fictions to scientific research in any way. Logistic regression can handle both categorical and continuous variables, and the predictors do not have to be normally distributed, linearly related, or of equal variance within each group Tabachnick and Fidell Fast Download speed and ads Free!
Descriptive statistics can be used to summarize the data. If your data is categorical, try the frequencies or crosstabs procedures. If your data is scale level, try summaries or descriptives. If you have multiple response questions, use multiple response sets.
Report : Case Summaries. Shoukri and Mohammad A. Identify where multivariate methods fit in the grand scheme of the empirical research process Student Learning Outcomes: 1. From the menu bar, click File, then New, and then Syntax.
Demonstrate an understanding of the principles of probability theory in basic multivariate statistical analyses including Multiple Regression, Analysis of Variance, Analysis of Covariance in association with the SPSS Program ; and 4. Specific details on this assignment will be distributed in class. Prior to testing for Multicollinearity or indeed any statistical test, it is good practice to examine each variable on its own; this is called univariate analysis.
Silahkan WhatsApp: Biaya ribu sd ribu Sesuai Beban. Proses 1 sd 3 Hari Tergantung Antrian. By Juana Sanchez. Univariate, Bivariate, and Multivariate Statistics Using R offers a practical and very user-friendly introduction to the use of R software that covers a range of statistical methods featured in data analysis and data science.
The author— a noted expert in quantitative teaching —has written a quick go-to reference for performing essential Part I. The Basics of Multivariate Design Chapter 1. An Introduction to Multivariate Design Chapter 2. Data Screening Chapter 3B. Comparisons of Means Chapter 4A. Univariate Comparison of Means Chapter 4B. Multivariate Analysis of Variance The documents include the data, or links to the data, for the analyses used as examples.
Voelkl and S. Springer, Sage, Multivariate analysis was performed on factors found to be significant in the univariate analysis. The following variables were included: animal fat, animal protein, vitamin D, riboflavin, niacin, vitamin B 12 , pantothenic acid, sodium, animal iron, zinc, selenium, cholesterol, fiber, vitamin K, vitamin C, and plant calcium. Compound miter saw stand plans. Extjs layout Dr israel manon deaths.
Data view Variable view. Make sure to save both if you want to save your changes in data or analysis. Manova Correlation Top dresser spreader for sale.
The fully revised Tenth Edition offers step-by-step instruction on data analysis using the latest version Organized to parallel most introductory research methods texts, this text starts with an introduction to computerized data analysis and the social research process, then takes readers step-by-step through univariate, bivariate, and multivariate analysis using SPSS Statistics.
Save with MyShopping. Students learn how to use the SPSS statistical program to apply these techniques to analysing data for research.
Rucker , John W. Suttie , Donald B. McCormick Many statistics texts tend to focus more on the theory and mathematics underlying statistical tests than on their applications and interpretation. In order to use the software you will need for this book, you will either have to purchase SPSS or find a computer laboratory in which SPSS has been installed.
With the step-by-step tutorial on the new features, students and empirical researchers can use it as a handbook when they conduct data analysis. Hard disk external 1tb terbaik. Seeburg jukebox. A frame cabin kits utah. Gauss seidel method application in real life.
Kalyan mein aaj ka open ank. Ariens 42 inch deck belt diagram. Inbox yahoo mail. Word layout greyed out. Could not connect to mail plugin altserver. Tzhaar slayer task rs3 Can cengage mindtap detect copy and paste Mpa vs mph reddit. Vts tradefed. Portable buildings antlers ok.
Old nbme exams. Optimus prime original toy price. Federal 22lr ammo. Peterbilt inside air cleaner lights. Fnaf vincent x male reader lemon. Wrecked ss for sale. Smithfly update. The sum of 3 consecutive even numbers is Microsoft icons. Herbology morton. Lennox 14acx reviews. Kahulugan at kahalagahan ng ekonomiks lesson plan. Teton leef vs altos. How long is pennsylvania dutch egg nog good for unopened.
Waves reading and questions. Openhab2 bindings. Faulty alignment dollar bills. Best yupoo clothing sellers. Maricopa county superior court clerk records. Echolink node list. Suzuki outboard reviews. Dap weldwood contact cement cleaner and thinner msds. Determining spinner probability answer key. Pfsense change log. Fat daddy wheels. How do i stop deleted emails from coming back on my mac. Frigidaire lp conversion kit instructions.
Toyota sequoia ecu location. N vs n How to find dupe stashes 2b2t. Stock tank swimming pool cover. Open baffle active crossover. Samsung rstdbp water filter. Ngauge mustang. Nightbot mod only commands. Stimulus and response worksheet. B3 rt24 tugboat pushing barges.
Weimaraner puppies for sale texas. Google chrome wonpercent27t open on mac Airgun nation hunting.
## Handbook of univariate and multivariate data analysis with ibm spss
As the name implies, multivariate regression is a technique that estimates a single regression model with more than one outcome variable. When there is more than one predictor variable in a multivariate regression model, the model is a multivariate multiple regression. Please Note: The purpose of this page is to show how to use various data analysis commands. It does not cover all aspects of the research process which researchers are expected to do. In particular, it does not cover data cleaning and checking, verification of assumptions, model diagnostics and potential follow-up analyses.
Getting, cleaning, analyzing and visualizing raw data is the main job responsibility of industry data scientists. Analyses were unchanged when patients who died within It is a useful data analysis software for introducing students to statistical analysis. Ib math applications and interpretation textbook pearson pdf. The fully revised Tenth Edition offers step-by-step instruction on data analysis using the latest version
E-mail: mgrootveld dmu. Herein, multicomponent nuclear magnetic resonance NMR analysis is used as a model to delineate how advanced statistical tools, both univariate and multivariate, can be implemented to effectively perform complex spectral dataset analyses in metabolomic applications, and to provide valuable, validated conclusions therein. Computational techniques are now embedded into spectral interpretation from an analytical chemist's perspective. However, there are challenges to applying such advanced statistical probes, which will be explored throughout this chapter.
We have enough money handbook of univariate and multivariate data analysis with ibm spss second edition and numerous books collections from fictions to scientific research in any way. Logistic regression can handle both categorical and continuous variables, and the predictors do not have to be normally distributed, linearly related, or of equal variance within each group Tabachnick and Fidell Fast Download speed and ads Free! Descriptive statistics can be used to summarize the data.
Enter your mobile number or email address below and we'll send you a link to download the free Kindle App. Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. To get the free app, enter your mobile phone number.
### Multivariate analysis
In the situation where there multiple response variables you can test them simultaneously using a multivariate analysis of variance MANOVA. This article describes how to compute manova in R. For example, we may conduct an experiment where we give two treatments A and B to two groups of mice, and we are interested in the weight and height of Univariate and multivariate logistic regression analysis was used to find factors associated with PMI. In the univariate analysis, PMI was associated with com-plex lesion characteristics, such as the lesion length, lesion angle, calcification, and Medina classification.
Haynes ManualsThe Haynes Author : Robert Ho Description:Many statistics texts tend to focus more on the theory and mathematics underlying statistical tests than on their applications and interpretation. This can leave readers with little understanding of how to apply statistical tests or how to interpret their findings. While the SPSS statistical software has done much to alleviate the frustrations of social science professionals and students who must analyze data, they still face daunting challenges in selecting the proper tests, executing the tests, and interpreting the test results. With emphasis firmly on such practical matters, this handbook sets forth clear guidelines for performing specific statistical tests with SPSS and interpreting the output. The author clearly explains the purpose of each test and the research designs for which they are relevant, demonstrates the execution of the tests, and explains how to interpret the results obtained.
Ho, Robert. Handbook of univariate and multivariate data analysis and interpretation with SPSS / Robert Ho. p. cm. Includes bibliographical references and index.
## Nuil V.
PDF | Abstracts not available for BookReviews | Find, read and cite all the Handbook of Univariate and Multivariate Data Analysis and Interpretation with SPSS.
## Generoso D.
PDF | On Sep 3, , Shahram Vahedi published Handbook of univariate and multivariate data analysis and interpretation with SPSS | Find. | 2,055 | 10,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-31 | latest | en | 0.830838 |
http://newton.ex.ac.uk/handbook/PHY/modules/PHY2023.html | 1,531,759,257,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00232.warc.gz | 266,978,152 | 5,034 | PHY2023 Thermal Physics 2017-18
Dr J. Anders
Delivery Weeks: T2:01-11
Level: 5 (NQF)
Credits: 15 NICATS / 7.5 ECTS
Enrolment: 162 students (approx)
### Description
This module builds on the discussion of thermal properties in the Stage 1 PHY1024 Properties of Matter module, introduces classical thermodynamics and shows how its laws arise naturally from the statistical properties of an ensemble. Real-world examples of the key ideas are presented and their application in later modules such as PHY2024 Condensed Matter I and PHY30703 Stars from Birth to Death is stressed. The concepts developed in this module are further extended in the PHYM001 Statistical Physics module.
### Module Aims
The aim of Classical thermodynamics is to describe the states and processes of of systems in terms of macroscopic directly measurable properties. It was largely developed during the industrial revolution for practical purposes, such as improving the efficiency the steam-engines, and its famous Three Laws are empirically based.
The aim of statistical mechanics, which had major contributions from Maxwell, Boltzmann and Gibbs, is to demonstrate that statistical methods can predict the bulk thermal properties of a system from an atomistic description of matter. The theory provides the only tractable means of analysing the almost unimaginable complexity of an N-body system containing 1023 particles. The classical Second Law of Thermodynamics finds a natural explanation in terms of the evolution of a system from the less probable to the more probable configurations.
### Intended Learning Outcomes (ILOs)
A student who has passed this module should be able to:
• Module Specific Skills and Knowledge:
1. explain the nature of classical entropy, and its relationship to the second law of thermodynamics;
2. determine the maximum efficiency of simple heat-engines and heat pumps;
3. calculate the equilibrium energy distribution of a system using the Boltzmann distribution;
4. explain the origin of the second law from a statistical viewpoint;
5. describe the significance of various thermodynamic potentials and deduce relations between them;
6. demonstrate, by calculating certain properties of real gases, an understanding of the limitations of the ideal gas law;
7. calculate bulk thermodynamic properties such as heat capacity, entropy and free energy from the partition function;
8. predict whether a gas constitutes a classical or a quantal gas, and explain key differences in the behaviour of these;
• Discipline Specific Skills and Knowledge:
1. use calculus to calculate maximum and minimun values of constrained multivariable systems;
2. use graphs and diagrams to illustrate arguments and explanations;
• Personal and Key Transferable / Employment Skills and Knowledge:
1. use a range of resources to develop an understanding of topics through independent study;
2. solve problems;
3. apply general concepts to a wide range of specfic systems and situations;
4. meet deadlines for completion of work for problems classes and develop appropriate time-management strategies.
### Syllabus Plan
1. Introduction
Brief historical survey.
2. Classical Thermodynamics
1. Zeroth, first and second laws of thermodynamics
2. Temperature scales, work, internal energy and heat capacity
3. Entropy, free energies and the Carnot Cycle
4. Changes of state
5. Heat engines and heat pumps
6. The Fundamental Thermodynamic Relationship
7. Thermodynamic potentials and Maxwell relations
8. Real gases
3. Statistical Physics
1. Maxwell-Boltzmann distribution
2. Boltzmann energy sharing
3. Microscopic / statistical interpretation of entropy
4. Statistical Thermodynamics
1. Density of states
2. The partition function Z
3. Macroscopic functions of state in terms of Z.
4. Equation of state for an ideal monatomic gas
5. The equipartition theorem
6. Quantum statistical mechanics; the Bose-Einstein and Fermi-Dirac distributions
### Learning and Teaching
#### Learning Activities and Teaching Methods
Description Study time KIS type
22×1-hour lectures 22 hours SLT
5×6-hour self-study packages 30 hours GIS
8×2-hour problems sets 16 hours GIS
Problems class support 8 hours SLT
Tutorial support 3 hours SLT
Reading, private study and revision 71 hours GIS
#### Assessment
Weight Form Size When ILOS assessed Feedback
0% Exercises set by tutor 3×1-hour sets (typical) Scheduled by tutor 1-14 Discussion in tutorials
0% Guided self-study 5×6-hour packages Fortnightly 1-14 Discussion in tutorials
10% 8 × Problems sets 2 hours per set Weekly 1-14 Marked in problems class, then discussed in tutorials
15% Mid-term Test 30 minutes Weeks T2:06 1-13 Marked, then discussed in tutorials
75% Examination 120 minutes May/June assessment period 1-13 Mark via MyExeter, collective feedback via ELE and solutions.
### Resources
The following list is offered as an indication of the type & level of information that students are expected to consult. Further guidance will be provided by the Module Instructor(s).
Core text:
Supplementary texts:
ELE:
### Further Information
#### Prior Knowledge Requirements
Pre-requisite Modules Properties of Matter (PHY1024) and Mathematics for Physicists (PHY1026) Mathematics with Physical Applications (PHY2025)
#### Re-assessment
Re-assessment is not available except when required by referral or deferral.
Original form of assessment Form of re-assessment ILOs re-assessed Time scale for re-assessment
Whole module Written examination (100%) 1-13 August/September assessment period
Notes: See Physics Assessment Conventions.
#### KIS Data Summary
Learning activities and teaching methods
SLT - scheduled learning & teaching activities 33 hrs
GIS - guided independent study 117 hrs
PLS - placement/study abroad 0 hrs
Total 150 hrs
Summative assessment
Coursework 10%
Written exams 90%
Practical exams 0%
Total 100%
#### Miscellaneous
IoP Accreditation Checklist TD-01 Zeroth, first and second laws of thermodynamics TD-02 Temperature scales, work, internal energy and heat capacity TD-03 Entropy, free energies and the Carnot Cycle TD-04 Changes of state SM-02 Statistical basis of entropy SM-03 Maxwell-Boltzmann distribution SM-04 Bose-Einstein and Fermi-Dirac distributions SM-05 Density of states and partition function Availability unrestricted Distance learning NO Keywords Physics; Thermodynamic; Properties; Heat; Energy; System; State; Distribution; Boltzmann; Entropy; Functions. Created 01-Oct-10 Revised 01-Oct-11 | 1,399 | 6,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-30 | latest | en | 0.895169 |
http://bio-aromatica.com/background-comorbidity-adjustment-can-be-an-important-component-of-health-services/ | 1,652,997,804,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00178.warc.gz | 6,303,285 | 27,899 | # Background Comorbidity adjustment can be an important component of health services
Background Comorbidity adjustment can be an important component of health services research and clinical prognosis. using a SEER-Medicare data example. We examined the ability of summary comorbidity Lenvatinib steps to adjust for confounding using simulations. Results We devised a numerical proof that discovered that the comorbidity overview procedures work prognostic or modification mechanisms in success analyses. Once one understands the comorbidity rating no other information regarding the comorbidity factors used to make the rating is generally required. Our data example and simulations confirmed this acquiring. Conclusions Overview comorbidity procedures like the Charlson Comorbidity Index and Elixhauser ratings are commonly employed for scientific prognosis and comorbidity modification. We have provided a theoretical justification that validates the use of such scores under many conditions. Our simulations generally confirm the power of the summary comorbidity steps as substitutes for use of the individual comorbidity variables in health services research. One caveat is usually that a summary measure may only be as good as the variables used to produce it. Introduction Baseline comorbidity adjustment is an important component of health services research and clinical prognosis. Researchers have widely used summary steps for comorbidity Lenvatinib adjustment in outcome studies that use administrative health data.[1][2][3] Lenvatinib When adjusting for comorbidities researchers may consider comorbidities individually or through the use of summary steps such as the Charlson Comorbidity Index [4][5][6] or the Elixhauser comorbidity steps [7][8]. In statistical models investigators might ROM1 incorporate comorbidities such as diabetes or heart disease by including indication covariates to denote whether the condition is present (the indication equals 1 if the condition is present 0 normally). In contrast summary steps such as the Charlson Comorbidity Index attach weights to each condition and then sum the weights of those conditions which are present in an individual.[4] The Charlson Comorbidity Index is based on a number of conditions that are each assigned an integer weight from one to six with a weight of six representing the most severe morbidity. The summation of the weighted comorbidity scores results in a summary score. In this paper we use the Charlson Comorbidity Index as the main example of a comorbidity summary measure due to its common use. A Web of Science search finds that the original and derivative papers concerning the Charlson Comorbidity Index have been cited over 8 800 occasions. While initially developed for use with medical records data the Charlson Comorbidity Index has been adapted for use with health claims data.[5][6][9] The validity of the Charlson Comorbidity Index as well as its adaptations have been investigated in multiple studies.[10][11][12] The success of the index has Lenvatinib prompted inquiry into further adaptations of the Charlson Comorbidity Index using questionnaire and physician claims based indices.[13][14][15] While the Charlson Comorbidity Index is commonly used competitor comorbidity measures have been developed. As an additional example we also investigate properties of the more recently developed Elixhauser score.[8] Like the Charlson score the Elixhauser score was derived using regression estimates. Whether it is better to use the Charlson Comorbidity Index or the individual comorbidities separately in statistical models is an open question. For example using ICD-10 data from a multinational group of patients Sundararajan represent the success time its possibility thickness function and represents the threat while is certainly a vector of covariates is certainly a vector from the understood beliefs represent a comorbidity rating produced from a threat rate; is certainly a function of = = predicated on a model. While we frequently suppose that the estimator of converges to the reality Lenvatinib as the test size increases (an asymptotic result) there could be some bias or performance ramifications of using the estimation in small examples. We explore this in the simulation section. Data example For the info example we utilized Security Epidemiology and FINAL RESULTS (SEER) data that were associated with Medicare promises data.[23] The SEER data source is maintained with the National. | 874 | 4,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.870076 |
http://www.monroecc.edu/etsdbs/MCCatPub.nsf/Web+Course+Descriptions/C95C854E997104F985258193005E17A6?OpenDocument | 1,527,127,878,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00064.warc.gz | 412,311,859 | 7,569 | # Course Descriptions
## MTH 175 - Precalculus
4 Credits
A study of the properties and graphs of functions, including polynomial, absolute value, power, piecewise, radical, rational, exponential, logarithmic, trigonometric, and inverse trigonometric. Topics also include a study of analytic trigonometry and an introduction to vectors. This course is intended to prepare students for the study of calculus. Four class hours. (SUNY-M)
Prerequisite: MTH 165 with a grade of C or higher, or MTH 141 with a grade of C or higher, or MCC Level 9 Mathematics placement or higher.
Course Learning Outcomes
1. Rewrite a logarithmic expression using properties of logarithms.
2. Perform operations on functions, which may include compositions or arithmetic combinations.
3. Determine simplified difference quotients.
4. Determine the inverse of a one-to-one function.
5. Construct graphs of equations, functions, or their transformations.
6. Identify appropriate features of the graph of an equation such as intercepts, asymptotes, maximum values, minimum values, domain, range, or symmetry.
7. Determine the values (in exact form when possible) of the trigonometric functions of any angle.
8. Verify a trigonometric identity.
9. Solve a trigonometric equation.
10. Evaluate expressions involving inverse trigonometric functions.
11. Use the Law of Sines and/or Law of Cosines to solve an oblique triangle.
12. Construct the graph of y = a f (bx + c) + d where f is a trigonometric function.
13. Identify appropriate features of the graph of a trigonometric function, such as amplitude, period, phase shift, vertical shift, domain, or range.
14. Perform operations on vectors.
Course Offered Fall and Spring
Use links below to see if this course is offered:
Fall Semester 2018
Summer Session 2018 | 400 | 1,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-22 | latest | en | 0.83391 |
https://swagasoft.com.ng/how-to-make-a-frequency-counter-algorithm/ | 1,638,023,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00580.warc.gz | 620,046,053 | 14,521 | November 27, 2021
# How to make a frequency counter algorithm
The frequency counter algorithm measures the occurrence of a value in data. It uses objects or sets to collects the value/frequency of data.
## Frequency counter algorithm example javascript
Write a function called frequencyCounter, which accepts a string. The function should return the frequency of each character in the given string.
## Frequency counter pseudocode
• Write a function that accepts a string.
• Declare an empty object to hold your frequency.
• Loop over the the given string with K = 0 and less than the length of the string.
• Check if the object declared already have the value in loop, increment object value by 1.
• Otherwise set object value = 1.
• Finally return the object.
## Write a function that accepts a string.
``````const frequencyCounter = (str) => {
}``````
## Declare an empty object to hold your frequency.
``````const frequencyCounter = (str) => {
let textFrequency = {}
}``````
## Loop over the given string
``````const frequencyCounter = (str) => {
let textFrequency = {}
for (let i = 0; i < str.length; i++) {
const element = str[i];
}
}``````
## Check if the object declared already has the element in the loop, increment by 1 or set by 1.
``````const frequencyCounter = (str) => {
let textFrequency = {}
for (let i = 0; i < str.length; i++) {
const element = str[i];
if(textFrequency[element]){
textFrequency[element] += 1;
}else{
textFrequency[element] = 1
}
}
}``````
## Finally return the frequency object.
``````const frequencyCounter = (str) => {
let textFrequency = {}
for (let i = 0; i < str.length; i++) {
const element = str[i];
if(textFrequency[element]){
textFrequency[element] += 1;
}else{
textFrequency[element] = 1
}
}
return textFrequency;
}
console.log(frequencyCounter('madam')); // { m: 2, a: 2, d: 1 }``````
``````const frequencyCounter = (str) => {
let textFrequency = {}
for (let i = 0; i < str.length; i++) {
const element = str[i];
if(textFrequency[element]){
textFrequency[element] += 1;
}else{
textFrequency[element] = 1
}
}
return textFrequency;
}
console.log(frequencyCounter(' Pseudopseudohypoparathyroidism'));
*/
{
P: 1,
s: 3,
e: 2,
u: 2,
d: 3,
o: 4,
p: 3,
h: 2,
y: 2,
a: 2,
r: 2,
t: 1,
i: 2,
m: 1
} /*``````
Below is a code example I wrote removing multiple users from a list using a frequency counter.
``````const getLeaderBorad = async (req, res) => {
await GameRecord.find({correct_qst : 15}).sort({minutes: 1}).sort({seconds: 1}).limit(50).then((leaders)=> {
let uniqueCount = {};
for (let i = 0; i < leaders.length; i++) {
if(uniqueCount[element.user] == element.user){
}else{
uniqueCount[element.user] = element.user;
}
}
})
}``````
With the given example in this article, you should identify where the frequency counter algorithm is used.
You can also view other algorithms examples.
If you want to improve your skill in algorithms, use any of the links below.
If you find this helpful share.
#### Simon Jerry
View all posts by Simon Jerry → | 817 | 3,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-49 | latest | en | 0.670973 |
http://list.seqfan.eu/pipermail/seqfan/2008-November/000225.html | 1,669,612,340,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00331.warc.gz | 30,996,930 | 2,561 | # [seqfan] Differences Between Some Primes (Stupid Question?)
Leroy Quet q1qq2qqq3qqqq at yahoo.com
Sun Nov 23 19:43:14 CET 2008
```I just posted these sequences.
%I A152073
%S A152073 2,3,5,7,11,13,17,19,13,29,29,37,41
%N A152073 a(n) = the largest prime < p(n) such that p(n) - a(n) is a power of 2, where p(n) is the nth prime.
%C A152073 Does every odd prime differ from at least one lower prime by a power of 2? Or is this sequence not defined for some terms?
%e A152073 Checking over the primes less than the 10th prime = 29: 29 - 23 = 6, not a power of 2. 29-19 = 10, not a power of 2. 29-17 = 12, not a power of 2. But 29-13 = 16, a power of 2. Since p = 13 is the largest prime p such that 29 - p = a power of 2, then a(10) = 13.
%Y A152073 A139758
%K A152073 more,nonn
%O A152073 2,1
(A139758 considers the primes {a(n)} where each a(n) - p(n) is a power of 2.)
%I A152075
%S A152075 3,5,7,13,13,19,19,29,29,31,37,43,43
%N A152075 a(n) = the smallest prime p > p(n) such that p - p(n) is squarefree, where p(n) is the nth prime.
%Y A152075 A152076
%K A152075 more,nonn
%O A152075 1,1
%S A152076 2,3,5,5,11,11,17,17,23,29,31,31
%N A152076 a(n) = the largest prime p < p(n) such that p(n) - p is squarefree, where p(n) is the nth prime. a(n) = 0 if no such prime p exists.
%C A152076 Does every odd prime differ from some smaller prime by a squarefree integer? Or is there at least one term of this sequence equal to 0?
%Y A152076 A152075
%K A152076 more,nonn
%O A152076 2,1
Is there always a prime that satisfies the conditions of these sequences? Or are the sequences sometimes undefined (or equal to 0, in the case of A152076)?
In the case of A152075 it seems to be *VERY* likely that every term exists.
(Because
p(n) + product{q = primes < some Q, q not = p(n)} q
is very likely to be prime for SOME positive value of Q, I would think.)
As for sequences A152073 and A152076, it seems that a simple check of terms would soon bring up a case where each sequence is not defined or is 0, if there are any such cases. (Perhaps if I just checked a couple more primes by hand I would have come across such a case.)
Thanks,
Leroy Quet
``` | 747 | 2,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.862463 |
https://www.cs.ox.ac.uk/teaching/courses/automatalogicgames/ | 1,656,917,719,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00734.warc.gz | 747,511,177 | 8,449 | # Automata, Logic and Games: 2021-2022
Lecturer Mikolaj Bojanczyk Degrees MSc in Mathematics and Foundations of Computer Science Term Michaelmas Term 2021 (24 lectures)
## Overview
The lectures for this course will be held in person.
To introduce the mathematical theory underpinning the Computer-Aided Verification of computing systems. The main ingredients are:
• Automata (on infnite words and trees) as a computational model of state-based systems
• Logical systems (such as temporal and modal logics) for specifying operational behaviour
• Two-person games as a conceptual basis for understanding interactions between a system and its environment
## Learning outcomes
At the end of the course students will be able to:
• Describe in detail what is meant by a Buchi automaton, and the languages recognised by simple examples of Buchi automata.
• Use linear-time temporal logic to describe behaviourial properties such as recurrence and periodicity, and translate LTL formulas to Buchi automata.
• Use S1S to define omega-regular languages, and give an account of the equivalence between S1S definability and Buchi recognisability.
• Explain the intuitive meaning of simple modal mu-calculus formulas, and describe the correspondence bewteen property-checking games and modal mu-calculus model checking.
## Prerequisites
Knowledge of first-order predicate calculus will be assumed. Familiarity with the basics of Finite Automata Theory and Computability (for example, as covered by the course, Models of Computation) and Complexity Theory would be very helpful. Candidates who do not have the required background are expected to have taken the course Introduction to the Foundations of Computer Science.
## Synopsis
• Automata on infinite words. Buchi automata: Closure properties. Determinization and Rabin automata.
• Nonemptiness and Nonuniversality problems for Buchi automata.
• Linear temporal logic and alternating Buchi automata.
• Modal mu-calculus: Fundamental Theorem, decidability and finite model property. Parity Games and the Model-Checking Problem: memoryless determinacy, algorithmic issues.
• Monadic Second-order Logic and its relationship with the modal mu-calculus.
Selected parts from:
• J. Bradfield and C. P. Stirling. Modal logics and mu-calculi. In J. Bergstra, A. Ponse, and S. Smolka, editors, Handbook of Process Algebra, pages 293{332. Elsevier, North-Holland, 2001.
• B. Khoussainov and A. Nerode. Automata Theory and its Applications. Progress in Computer Science and Applied Logic, Volume 21. Birkhauser, 2001.
• C. P. Stirling. Modal and Temporal Properties of Processes. Texts in Computer Science. Springer-Verlag, 2001.
• W. Thomas. Languages, Automata and Logic. In G. Rozenberg and A. Salomaa, editors, Handbook of Formal Languages, volume 3. Springer-Verlag, 1997.
• M. Y. Vardi. An automata-theoretic approach to linear temporal logic. In Logics for Concurrency: Structure versus Automata, ed. F. Moller and G. Birtwistle, LNCS vol. 1043, pp. 238-266, Springer-Verlag, 1996.
## Feedback
Students are formally asked for feedback at the end of the course. Students can also submit feedback at any point here. Feedback received here will go to the Head of Academic Administration, and will be dealt with confidentially when being passed on further. All feedback is welcome. | 768 | 3,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.844009 |
https://www.crazy-numbers.com/en/66 | 1,685,766,126,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00145.warc.gz | 789,068,341 | 6,098 | Discover a lot of information on the number 66: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 66
Is 66 a prime number? No
Is 66 a perfect number? No
Number of divisors 8
List of dividers 1, 2, 3, 6, 11, 22, 33, 66
Sum of divisors 144
Prime factorization 2 x 3 x 11
Prime factors 2, 3, 11
## How to write / spell 66 in letters?
In letters, the number 66 is written as: Sixty-six. And in other languages? how does it spell?
66 in other languages
Write 66 in english Sixty-six
Write 66 in french Soixante-six
Write 66 in spanish Sesenta y seis
Write 66 in portuguese Sessenta e seis
## Decomposition of the number 66
The number 66 is composed of:
2 iterations of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
Other ways to write 66
In letter Sixty-six
In roman numeral LXVI
In binary 1000010
In octal 102
In US dollars USD 66.00 (\$)
In euros 66,00 EUR (€)
Some related numbers
Previous number 65
Next number 67
Next prime number 67
## Mathematical operations
Operations and solutions
66*2 = 132 The double of 66 is 132
66*3 = 198 The triple of 66 is 198
66/2 = 33 The half of 66 is 33.000000
66/3 = 22 The third of 66 is 22.000000
662 = 4356 The square of 66 is 4356.000000
663 = 287496 The cube of 66 is 287496.000000
√66 = 8.124038404636 The square root of 66 is 8.124038
log(66) = 4.1896547420264 The natural (Neperian) logarithm of 66 is 4.189655
log10(66) = 1.8195439355419 The decimal logarithm (base 10) of 66 is 1.819544
sin(66) = -0.026551154023967 The sine of 66 is -0.026551
cos(66) = -0.99964745596635 The cosine of 66 is -0.999647
tan(66) = 0.026560517776039 The tangent of 66 is 0.026561 | 607 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-23 | longest | en | 0.806046 |
https://iq.opengenus.org/defining-2d-array-in-python/ | 1,726,266,817,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00568.warc.gz | 282,703,616 | 21,023 | ×
Search anything:
# Defining 2D array in Python
#### Software Engineering Python
Open-Source Internship opportunity by OpenGenus for programmers. Apply now.
In this article, we have explored the different ways of defining a 2D array in Python. We have explored three approaches:
• Creating a List of Arrays
• Creating a List of Lists
• Creating 2D array using numpy
Some terminologies in Python:
Array: An array is a collection of homogeneous elements (i.e. belonging to the same data type) that are stored in contiguous memory locations. In Python, there is a module âarrayâ that needs to be imported to declare/use arrays.
List: A list in Python is collection of elements that can belong to different data types. So, a List can be an array if the user chooses to store only homogeneous elements in it. There is no need to explicitly import a module for declaration of Lists.
So, what are 2D arrays/lists?
2D arrays/lists can be defined as an array/list of arrays/lists. 2D arrays are also called 'Matrices'.
# Creating a List of Arrays:
Syntax of defining an array in python:
``````import array as arr #imports array module
a = arr.array(âiâ,[1,2,3,4]) #defines a 1D array of integer type
``````
NOTE: You need to mention the data type of the elements and if an element of different data type is inserted, an exception âIncompatible data typesâ is thrown.
Letâs create a 3*3 matrix(i.e. 2D array)
``````import array as arr #imports module
a = [] #defining an empty list
for i in range(0,3): #i iterates from 0 to 3
row = arr.array(âiâ) #defining an array
for j in range(0,4): #j iterates from 0 to 4
row.append(i*4+j) #row-wise position is appended
a.append(row) #row array is from the end added to the list 'a'
print(a)
``````
Output: [array('i', [0, 1, 2, 3]), array('i', [4, 5, 6, 7]), array('i', [8, 9, 10, 11])]
# Creating a List of Lists:
Now, here is a better and easier way to create a 2D array.
There are different ways in which this can be done with Lists and here I will be showing 3 of them.
Method 1:
This is the most simple way.
``````rows = 3 #value of rows is defined 3
cols = 3 #value of columns is defined 3
mat = [[0]*cols]*rows #2D list initialized with value 0 is defined
print(mat) #contents of the 2D array 'mat' is displayed
``````
Output:
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
Method 2:
``````rows = 3 #value of rows is defined 3
cols = 3 #value of columns is defined 3
mat = [[0 for i in range(cols)] for j in range(rows)] #2D stored in 'mat'
print(mat) #contents of the 2D array 'mat' is displayed
``````
Output:
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
Method 3:
``````rows = 3 #value of rows is defined 3
cols = 3 #value of columns is defined 3
mat=[] #empty List is defined
for i in range(cols): #i iterates from 0 to value 'cols'
col = [] #empty List is defined
for j in range(rows): #j iterates from 0 to value 'rows'
col.append(0) #0 is added to list 'col'
mat.append(col) #List 'col' is added to List 'mat'
print(mat) #contents of List 'mat' displayed
``````
Output:
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
# Creating 2D array using numpy:
Though we can use the List for the same purpose but itâs slow. Numpy is a pre-defined package in Python used for performing powerful mathematical operations. It provides an array object much faster than traditional Python lists.
• Using numpy.array()
``````import numpy as np
#initializing 3 different Lists
row1 = [1,2,3]
row2 = [4,5,6]
row3 = [7,8,9]
mat = np.array([row1,row2,row3]) #initializing array with 3 Lists
print(mat) #display contents of List 'mat'
``````
Output:
[[1 2 3]
[4 5 6]
[7 8 9]]
• Using numpy.asarray()
``````#initializing 3 different Lists
import numpy as np
row1 = [1,2,3]
row2 = [4,5,6]
row3 = [7,8,9]
mat = np.asarray([row1,row2,row3]) #initializing array with 3 Lists
print(mat) #display contents of List 'mat'
``````
Output:
[[1 2 3]
[4 5 6]
[7 8 9]]
In numpy arrays there are many functions that can be applied for mathematical computations.So, if you wish to explore more on numpy arrays you can refer to this link.
# Applications:
The applications of 2D arrays are:
• To represent images and manipulate them
• To represent any 2-D grid
• Used in programming techniques like Dynamic Programming
# Exercise Problem:
Write a program that takes the number of rows and the number of columns of the 2D array as input and then takes input row-wise and displays the 2D array as output.
# Solution:
``````rows = int(input("Enter the no. of rows:")) #user input number of rows
cols = int(input("Enter the no. of columns:")) #user input number of columns
mat = [] #defining empty List
print("Enter the elements row-wise:")
for i in range(rows): #i iterates from 0 to value of 'rows'
row = [] #defining empty List
for j in range(cols): #j iterates from 0 to value of 'cols'
row.append(int(input())) #user input is added in List 'row'
mat.append(row) #List 'row' is added in List 'mat'
for i in range(rows): #i iterates from 0 to value of 'rows'
for j in range(cols): #j iterates from 0 to value of 'cols'
print(mat[i][j], end = " ") #display value at ith row and jth column
print() #line change
``````
Output:
Enter the no. of rows: 2
Enter the no. of columns: 4
Enter the elements row-wise:
1
2
3
4
5
6
7
8
1 2 3 4
5 6 7 8
# Time Complexity:
Worst case time complexity: O(RC)
Average case time complexity: Î(R
C)
Best case time complexity: Ί(RC)
Space complexity: Î(R
C)
Since we are iterating from column to column in every row hence the complexity is rows*columns.
NOTE: âRâ is the number of rows and âCâ is the number of columns.
Defining 2D array in Python | 1,742 | 5,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.816715 |
https://www.electrondepot.com/electrodesign/square-root-circuit-769018-.htm | 1,679,979,215,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00238.warc.gz | 838,000,745 | 13,311 | Square root circuit
• posted
Hi all, I may need to make a square root circuit to linearize the gain in a temperature control loop with resistors as heaters. (Speed is not very important.) For ~\$7 I can use an analog multiplier (AD633) or something with matched transistors. ala B. Pease page 30 here.
I'm wondering if there is something else (in nature) that naturally takes a square root. For instance a silly idea would be to look at the temperature rise of a resistor driven by a voltage.... (kinda the inverse of the problem I'm trying to solve.)
(Oh sure the obvious answer is a computer.)
George H.
• posted
Yup, a 40 cent uC is probably the answer here.
• posted
I recently did this, using a BJT with another one (diode-connected) as emitter degeneration, and another two diodes (bjts) from base to ground. To make it square-root, you put a bias resistor from the midpoint of the diodes to the positive supply, enough so that the forward voltage of the bottom one stays nearly constant with drive.
Not brilliant, but good enough for linearizing a heater.
Cheers
Phil Hobbs
```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted
Here's a weird idea, don't know how practical it is...
Build an analog computer using op amps that solves the differential equation dy/dx - 1/(2x) = 0 using x as your input. Then run the solution through an exponential converter and the output of the exponentiator will be proportional to sqrt(x)...
• posted
Another way you maybe already know is to use a differential amplifier, an ordinary amplifier, and the analog multiplier as a squarer
The input is applied to the differential amplifier along with the output of the squaring circuit, so at the output of the differential amplifier you have K1(ae_2^2 - e_1) where e_1 is the input and e_2 is the square root output.
Then run that output through another amplifier with gain K2 and feed it to the input of the squarer so you have K1K2(ae_2^2 - e_1) = e_2
and if the gains K1 and K2 are large you then have ae_2^2 - e^1 = 0
• posted
One can also make an inverting opamp and add a diode across the feedback resistor, with maybe a little extra resistance in series with the diode. Like you say, not brilliant, but good enough. That's the first step towards a piecewise breakpoint thing.
You can get a single breakpoint (slope change) by using two opamps, summing their outputs, and scale so that one hits its rails. Or three. Or four.
High precision (like, 0.02%) math can be done with PWM multipliers, which can be cheap but take more thinking. Process controllers used to do that before uPs came along.
```--
John Larkin Highland Technology, Inc
picosecond timing laser drivers and controllers ```
• posted
Two transistors an a few resistors can make a passable squaring circuit. An output amp and some feedback can make that a square root. I'll leave the details to the student.
All square roots get crazy near zero, which is a good reason to not do it too well.
```--
John Larkin Highland Technology, Inc
picosecond timing laser drivers and controllers ```
• posted
I drew that up. "Scratching head..." I feed the base (of the transistor) with a current ? and take the output (voltage) from it's emitter.
No.. I feed the base with a voltage.
George H.
• posted
Yeah thanks, Once I have a multiplier it's fairly easy.
George H.
• posted
Try something really tacky like this:
Version 4 SHEET 1 880 680 WIRE -208 144 -240 144 WIRE -160 144 -208 144 WIRE -96 144 -160 144 WIRE 128 144 -16 144 WIRE 272 144 128 144 WIRE 368 144 272 144 WIRE 432 144 368 144 WIRE 272 176 272 144 WIRE -160 224 -160 144 WIRE 128 256 128 144 WIRE 272 288 272 256 WIRE -160 336 -160 304 WIRE 128 400 128 336 WIRE 272 400 272 352 WIRE 272 400 128 400 WIRE 272 432 272 400 WIRE 272 560 272 496 FLAG -160 336 0 FLAG 272 560 0 FLAG -208 144 IN FLAG 368 144 OUT SYMBOL res 288 272 R180 WINDOW 0 -53 70 Left 2 WINDOW 3 -53 38 Left 2 SYMATTR InstName R2 SYMATTR Value 0 SYMBOL diode 256 288 R0 WINDOW 0 64 13 Left 2 WINDOW 3 48 46 Left 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL res 0 128 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R3 SYMATTR Value 20K SYMBOL voltage -160 208 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 0 -67 106 Left 2 WINDOW 3 -65 174 Left 2 SYMATTR InstName V3 SYMATTR Value PULSE(0 10 0 2) SYMBOL diode 256 432 R0 WINDOW 0 64 13 Left 2 WINDOW 3 48 46 Left 2 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL res 112 240 R0 WINDOW 0 -66 31 Left 2 WINDOW 3 -66 65 Left 2 SYMATTR InstName R1 SYMATTR Value 5K TEXT -96 24 Left 2 !.tran 3 TEXT 64 24 Left 2 ;WORSE SQUARE ROOT TEXT 112 56 Left 2 ;JL July 2015 TEXT 320 392 Left 2 ;BAV23 DUAL
```--
John Larkin Highland Technology, Inc
picosecond timing laser drivers and controllers ```
• posted
Oh boy, I've opened a can of worms now. (The can is my brain, and the worms my lack of understanding.) I get a Log function from diode-opamp thing.. (but how do I get the square root?) I could do a log anti-log, with a factor of 2 (or 1/2) somewhere.
Oh I'd never heard of that... V1 = pulse height, V2 = pulse width. Average is ~V1*V2... thanks
George H.
• posted
Oh sure once I have a multiplier, I make V^2 and just feed that back in an opamp.
Yeah.. that's why it's a bit scary for me to think about doing it with transistors. I'l make some silly mistake and the circuit will blow up when the temperature changes.
George H.
• posted
What you're trying to do is similar to what is done on the front end of some photodiode transimpedance amplifiers. There you've got to squeeze a big range into an ADC. Here is an example of Log TIA at Linear Tech.
It's not a square root circuit but that's where I would start.
• posted
That sort of thing is wild overkill to tweak the dynamics of a PID driving a resistive heater.
This is interesting because I now have a requirement to build two temperature controllers, on opposite sides of a roughly 1 square inch pcb, using 0805 resistors as the heaters. But maybe I should use mosfets as the heaters, and keep things linear. Heck, I may not need an opamp!
```--
John Larkin Highland Technology, Inc
picosecond timing laser drivers and controllers ```
• posted
• posted
It just occurred to me that zener diodes might make nice heaters (but probably not for your app).
The opamp current and thermal limits, so it probably doesn't blow up. Some artistic copper pours could help spread the heat.
It may need R9 to tame the loop gain, but it could drop something small, one volt maybe.
I need two heaters, one on each side of a tiny round PCB. The two Things have a nasty phase change sort of effect around room temp, and have to be heated to around 70C.
I guess resistors as heaters and PWM would work here too.
```--
John Larkin Highland Technology, Inc
picosecond timing precision measurement ```
• posted
Ahh.. got it, I drew Phil H's circuit like that the first time, but then didn't like the collector resistor.
Thanks Simon
George H.
• posted
Yeah well no one (but me) liked the FET as heater idea. (It still seems like the way to go... There's no room temperature pass element to get hot for one.)
George H.
• posted
I've heard of using zeners for heaters... don't you want to control current? Or is that the purpose of R9? Maybe split R9 in half and stick one piece on each side to balance the heat. (Maybe the matching is not that critical. The zener voltages won't match either.)
George H.
• posted
I've used mosfets as heaters several times. Works great, but you do need a current limit of some sort. Resistors limit themselves.
A 3-terminal voltage regulator makes a neat heater. It protects itself.
```--
John Larkin Highland Technology, Inc
picosecond timing precision measurement ```
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners. | 2,150 | 8,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-14 | latest | en | 0.924079 |
https://cs.stackexchange.com/questions/54800/how-to-give-an-upper-bound-on-this-bin-packing-problem/54806 | 1,566,638,960,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00248.warc.gz | 424,749,635 | 30,476 | # How to give an upper bound on this bin packing problem?
In the bin packing with fragile objects (BPFO) problem one is given a set of objects $\{1,\ldots,n\}$ where each object $i$ has a weight $w_i$ and a fragility $f_i$ for all $i$ in the set $\{1,\ldots,n\}$. (We assum that $w_i\le f_i$.) Also, we have a set of bins of infinite capacity (each) in which we want to place the objects.
The objective is to assign the objects to the minimum number of bins such that the sum of the weights in each bin is at most the minimum fragility. That is, if we assign the set of objects $S$ to bin $B$ than we must have:
$$\sum_{i\in S}w_i\le\min\limits_{i\in S}f_i.\quad\quad\quad (1)$$
The BPFO problem is given in this paper. Let us denote the optimal value of BPFO by $OPT$.
I am trying to relax BPFO by modifying equation$~(1)$ as follows.
$$\sum_{i\in S}w_i\le f_i,\forall i\in S$$
Let us denote the optimal value of the relaxed problem by $OPTR$.
Now, I would like to upper bound $OPT$ by $OPTR$. How can I achieve this?
It is clear that a solution to BPFO is a solution to the relaxed problem and then $OPT\ge OPTR$.
Is there a way to say that $OPT\le \alpha(n)OPTR$ where $\alpha(n)\ge 1$ ?
Your relaxation is same as original problem.
$$\sum_{i\in S}w_i\leq f_j,\forall j\in S$$
is same condition as
$$\sum_{i\in S}w_i\leq\min\limits_{j\in S}f_j$$.
because
$$\min\limits_{i\in S}f_i \leq f_j,\forall j\in S$$.
• A solution to the original problem implies a solution to the relaxed problem but a solution to the relaxed problem does not imply a solution to the original problem. Am I confusing? Are you saying that both problems are the same? – drzbir Mar 23 '16 at 1:46
• Yes, $Z < x_1$ and $Z < x_2$ if and only if $Z < \min(x_1,x_2)$. – Shreesh Mar 23 '16 at 1:48
• So, $OPTR=OPT$? – drzbir Mar 23 '16 at 1:53
• Yes, that is exact. – Shreesh Mar 23 '16 at 1:59 | 617 | 1,880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-35 | latest | en | 0.888726 |
https://www.assignmenthippo.com/operational-amplifiers-assignment-help | 1,695,470,419,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00475.warc.gz | 719,333,207 | 10,340 | # Operational Amplifiers Assignment Help
There is no need to worry about achieving HD grades this semester in Operational Amplifiers! Let's accomplish your grades together with just one click! Though how boring the topic may sound, our Operational Amplifiers Assignment Help experts are there to make it interesting. So, let's begin to learn this with full curiosity.
What do you mean by Operational Amplifiers? Operations amplifiers amplify the voltage and are DC-linked with one output and differential input. The resistor and capacitor are outer feedback parts.
Where were they found initially? They were found in frequency-dependent circuits, mathematical functions in non-linear and linear criteria and analogue computers.
We understand how difficult it is to complete an excellent assignment within the given time. But, this problem is solved through Assignment Hippo. Our services provide assistance not only to Australian native students but to international students too who are pursuing operational amplifiers in any of the Australian universities! Yes, you read it right! You can reach the peak of success with HD grades this semester only!
## What Is The Role of An Operational Amplifier?
According to the experts of Online Operational Amplifiers Assignment Help in Australia, operational amplifiers are connected with other circuits to work. These amplifiers are devices that are linear and exhibit all the characteristics of an ideal DC amplifier.
Operational Amplifiers are utilised in the way of signal conditioning, addition, subtraction, and differentiation and integration of mathematical functions.
## Categories of Operational Amplifiers
Our Online Operational Amplifiers Assignment Help in Australia helps thoroughly to understand the categories of operational amplifiers.
There are three categories of operational amplifiers:
1. In and out of voltage
2. In and out of the current
3. Current in and voltage out in Trans resistance.
4. Voltage in and current out in Transconductance.
You can always explore and get an ample amount of understanding through our well-structured Operational Amplifiers Assignment Sample.
## Terms of Operational Amplifiers That Would Excite You
According to our Operational Amplifiers Assignment Help experts, the following are the terms you should know:
1. Differential Amplifier
The difference that lies between two input signals at the input phase gives output voltage is known as a differential amplifier.
2. Common Mode Of operation
In this mode, input and output signals are in the equilibrium stage. Collector voltages work in the same or opposite directions as output voltage. This makes a proportional circuit and creates a zero difference between collector voltages.
3. Differential Input Voltage
When an operational amplifier creates an impact on the difference between the input voltages and not about a common potential, the differential input voltage is created.
4. Open-loop gain
Open-loop gain is the gain without any kind of positive and negative feedback. The primary role of the operational amplifier is to maximise the amplification of the input signal.
There are more terms to know, for that you can always take help from our Operational Amplifiers Assignment Samples available on our website.
Experts of Online Operational Amplifiers Assignment Help in Australia are well familiar with the advantages. Let's have a look at them:
1. Operational Amplifiers are smaller in size.
2. They are more useful and reliable compared to other amplifiers.
3. Operational amplifiers consume less power.
4. Reuse of operational amplifiers is possible.
5. Cost is less as compared to other parts.
## FAQs Answered By Our Operational Amplifiers Assignment Help Experts
### Q.1 What are the uses of Operational Amplifiers?
When we need to amplify the weak current in a circuit, operational amplifiers come to the rescue. Operational Amplifiers are incorporated in headphones, stereo systems, and radios.
### Q.2 Where are Operational Amplifiers applied?
Operational Amplifiers are applied as a voltage follower, an integrator, and an effective rectifier.
### Q.3 What do you mean by Inverting an Operational Amplifier?
Inverting an operational amplifier gives an output of its input by about 180. If the output is positive, then input is negative and vice versa.
## Get Benefit from Our Operational Amplifiers Assignment Help And Secure HD Grades This Semester At Affordable Prices
Our sole concept of providing an operational amplifiers assignment help is to encourage students to solve/write their dissertations/thesis, essays, case studies, or any other kind of assignment and guide students to understand key concepts of Operational Amplifiers Assignment topics through top-quality mentorship. We delegate our Online Operational Amplifiers Assignment Help in Australia to help students and scholars based on their subject matter expertise to provide high-quality homework support. So, make sure to get HD grades with the help of our Assignment Help provider
• Proper citation and referencing styles, containing APA, Chicago, MLA, etc.
• Proofreading and copy-editing assistance for each assignment and dissertations at your doorstep.
• Our Rates are student-friendly and very much affordable.
• You get in-depth research on the allocated topic.
• Arguments herein are backed by high-quality proof.
• Proper development of arguments.
• 100% plagiarism-free valid assignments and dissertations.
• Free Turnitin checks and reports.
• Regular discounts
• You can make your assignment interesting through the guidance of the Operational Amplifiers Assignment Sample available on our website.
• Live tracking based on web and app.
• 100% confidentiality of the students
• We have an excess of resources available from our Operational Amplifiers Assignment Help experts and are there to flourish your success.
What are you waiting for? Grab these affordable and attractive offers now! Prove to us that you are the best. Success is in our hands!
## Buy Operational Amplifiers Assignment Help Online
Talk to our expert to get the help with Operational Amplifiers Assignment Help Answers from Assignment Hippo Experts to complete your assessment on time and boost your grades now
The main aim/motive of the finance assignment help services is to get connect with a greater number of students, and effectively help, and support them in getting completing their assignments the students also get find this a wonderful opportunity where they could effectively learn more about their topics, as the experts also have the best team members with them in which all the members effectively support each other to get complete their diploma assignment help Australia. They complete the assessments of the students in an appropriate manner and deliver them back to the students before the due date of the assignment so that the students could timely submit this, and can score higher marks. The experts of the Assignment Help services at assignmenthippo.com are so much skilled, capable, talented, and experienced in their field and use our best and free Citation Generator and cite your writing assignments, so, for this, they can effectively write the best economics assignment help services.
### Get Online Support for Operational Amplifiers Assignment Help Assignment Help Online
Want to order fresh copy of the Sample Operational Amplifiers Assignment Help Answers? online or do you need the old solutions for Sample Operational Amplifiers Assignment Help, contact our customer support or talk to us to get the answers of it. | 1,403 | 7,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.93627 |
https://nl.mathworks.com/matlabcentral/cody/problems/44080-construct-a-diagadiag-matrix/solutions/1710137 | 1,606,419,241,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00166.warc.gz | 413,531,081 | 17,069 | Cody
# Problem 44080. Construct a "diagAdiag" matrix
Solution 1710137
Submitted on 19 Jan 2019 by HH
• Size: 9
• This is the leading solution.
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 = 3; y_correct = [1 2 6; 3 5 7;4 8 9]; assert(isequal(diagAdiag(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In diagAdiag (line 2) In ScoringEngineTestPoint1 (line 3) In solutionTest (line 3)]
2 Pass
x = 4; y_correct = [1 2 6 7; 3 5 8 13;4 9 12 14;10 11 15 16]; assert(isequal(diagAdiag(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In diagAdiag (line 2) In ScoringEngineTestPoint2 (line 3) In solutionTest (line 5)]
3 Pass
x = 5; y_correct = [1 2 6 7 15;3 5 8 14 16;4 9 13 17 22;10 12 18 21 23;11 19 20 24 25]; assert(isequal(diagAdiag(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In diagAdiag (line 2) In ScoringEngineTestPoint3 (line 3) In solutionTest (line 7)]
4 Pass
x = 6; y_correct = [ 1 2 6 7 15 16; 3 5 8 14 17 26; 4 9 13 18 25 27; 10 12 19 24 28 33; 11 20 23 29 32 34; 21 22 30 31 35 36]; assert(isequal(diagAdiag(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In diagAdiag (line 2) In ScoringEngineTestPoint4 (line 8) In solutionTest (line 9)]
5 Pass
x = 7; y_correct = [ 1 2 6 7 15 16 28; 3 5 8 14 17 27 29; 4 9 13 18 26 30 39; 10 12 19 25 31 38 40; 11 20 24 32 37 41 46; 21 23 33 36 42 45 47; 22 34 35 43 44 48 49]; assert(isequal(diagAdiag(x),y_correct))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In diagAdiag (line 2) In ScoringEngineTestPoint5 (line 9) In solutionTest (line 11)]
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 846 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | latest | en | 0.613683 |
https://diffzi.com/14-2-wire-amps-2/ | 1,685,595,193,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00576.warc.gz | 241,023,225 | 13,983 | # 14/2 Wire Amps
In the world of electrical wiring, there are a variety of different types of wire that are used for different purposes. One of the most common types of wire is 14/2 wire, which is used for a variety of different applications. In this article, we will discuss what 14/2 wire is, what it is used for, and how to calculate the amps that it can handle.
What is 14/2 wire?
14/2 wire is a type of electrical wire that is used for a variety of different applications. The “14” in the name refers to the gauge of the wire, which is a measurement of the wire’s diameter. The “2” refers to the number of conductors, or wires, that are inside the wire. 14/2 wire is typically made from copper, although it can also be made from aluminum.
What is 14/2 wire used for?
14/2 wire is commonly used in residential electrical wiring applications. It is often used to power lighting fixtures, outlets, and other devices. It can also be used for low-voltage applications such as doorbells or thermostats.
How do you calculate the amps that 14/2 wire can handle?
The amps that 14/2 wire can handle is dependent on a few factors. The first factor is the wire’s resistance. The resistance of the wire is determined by its length and gauge. The longer the wire, the greater its resistance. The gauge of the wire also affects its resistance – the lower the gauge, the lower the resistance.
The second factor that affects the amps that 14/2 wire can handle is the temperature of the wire. When electrical current flows through a wire, it generates heat. The higher the current, the more heat is generated. If the wire gets too hot, it can become damaged or even start a fire. The temperature rating of the wire is determined by the insulation that surrounds it. The insulation is rated for a specific temperature range – if the wire gets too hot, the insulation can melt or break down.
To calculate the amps that 14/2 wire can handle, you can use the following formula:
Amps = (Watts / Volts) x .80
Watts: The total power consumption of the device that will be connected to the wire.
Volts: The voltage of the circuit that the wire will be connected to (.e.g. 120 volts for a household circuit)
.80: This is a safety factor that accounts for the wire’s resistance and temperature.
For example, let’s say that you want to install a lighting fixture that consumes 100 watts and will be connected to a 120-volt circuit. Using the formula above, you can calculate the amps that the 14/2 wire can handle:
Amps = (100 watts / 120 volts) x .80 = .67 amps
In this scenario, 14/2 wire would be sufficient because it can handle up to 15 amps of current.
What are some other factors to consider when working with 14/2 wire?
When working with 14/2 wire, there are a few other factors that you should consider:
– Always use wire that is rated for the specific application. Different types of wire have different ratings for things like temperature, moisture resistance, and more. Make sure that the wire you are using is appropriate for the task at hand.
– Be mindful of the wire’s length. As we mentioned earlier, the longer the wire, the greater its resistance. This means that if you are running wire over a long distance, you may need to use a wire with a higher gauge to handle the additional resistance.
– Follow all safety guidelines. Electrical work can be dangerous if not done properly. Make sure that you understand all safety guidelines and take appropriate precautions when working with electrical wiring.
– Don’t overload the circuit. Just because the wire can handle a certain amount of current doesn’t mean that the circuit can handle more than that. If you overload the circuit, you can cause damage to the wire, start a fire, or even electrocute someone.
Conclusion
14/2 wire is a common type of electrical wire that is used for a variety of different applications. It is important to understand how to calculate the amps that the wire can handle and to follow all safety guidelines when working with electrical wiring. By taking the appropriate precautions, you can ensure that your electrical system is safe and reliable for years to come. | 916 | 4,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-23 | latest | en | 0.944643 |
https://math.stackexchange.com/questions/1266074/composition-of-probability-distribution-functions | 1,561,396,901,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999620.99/warc/CC-MAIN-20190624171058-20190624193058-00081.warc.gz | 520,285,381 | 35,498 | # composition of probability distribution functions
Suppose we are given $X \sim \mathcal{N}(\mu,\Sigma)$. Then, we define the random variable $Y$ as follows:
$Y_i = 1 + X_i$ if $X_i \ge 0$
$Y_i = \exp(X_i)$ if $X_i \lt 0$.
How do I go about calculating the probability density of Y? And $E[Y_i]$ for all $i$? (not a homework problem - it is taken from a paper where the Y model the prior distribution of surface emissions of methane...)
Is the derivation in @martini's reply correct if the $X$ are correlated ($\Sigma$ is not diagonal)?
For the density: Note that $X_i \sim N(\mu_i, \Sigma_{ii})$. Hence, for $t \in \mathbf R^+$, $t \le 1$, we have \begin{align*} \def\P{\mathbf P}\P(Y_i \le t) &= \P(e^{X_i} \le t)\\ &= \P(X_i \le \log t)\\ &= \frac 1{\sqrt{2\pi\Sigma_{ii}}} \int_{-\infty}^{\log t} \exp\bigl(-(x-\mu_i)^2/2\Sigma_{ii}\bigr)\, dx \end{align*} For $t \ge 1$, we have \begin{align*} \P(Y_i \le t) &= \P(Y_i \le 1) + \P(1 \le Y_i \le t)\\ &= \P(Y_i \le 1) + \P(0 \le X_i \le t-1)\\ &= \P(X_i \le 0) + \P(0 \le X_i \le t-1)\\ &= \P(X_i \le t-1)\\ &= \frac 1{\sqrt{2\pi\Sigma_{ii}}} \int_{-\infty}^{t-1} \exp\bigl(-(x-\mu_i)^2/2\Sigma_{ii}\bigr)\, dx \end{align*} Taking derivatives, we say that $Y_i$'s density is given by $$f_i(t) = \begin{cases} 0 & t \le 0\\ \frac 1t \cdot (2\pi\Sigma_{ii})^{-1/2} \exp\bigl(-(\log t - \mu_i)^2/2\Sigma_{ii}\bigr) & 0< t \le 1\\ (2\pi \Sigma_{ii})^{-1/2}\exp\bigl(-(t-1 - \mu_i)^2/2\Sigma_{ii}\bigr) & t > 1 \end{cases}$$ For the expectation, compute $\int_{\mathbf R} tf_i(t)\, dt$.
• great, thanks. one comment: you seem to have left out the "-" sign inside the $exp$ functions. – qazxswedc May 4 '15 at 10:19
• i'm not sure if this derivation works if $\Sigma$ is not diagonal. If the $X_i$ are correlated, should there not be some correlation between the $Y_i$ as well? – qazxswedc Oct 22 '15 at 8:27 | 737 | 1,862 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-26 | latest | en | 0.665406 |
https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/Calculate-month-end-dates-between-two-dates/td-p/2773116 | 1,725,778,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00457.warc.gz | 162,995,025 | 50,041 | cancel
Showing results for
Did you mean:
Find everything you need to get certified on Fabric—skills challenges, live sessions, exam prep, role guidance, and more. Get started
Frequent Visitor
## Calculate month end dates between two dates
I need to count the number of month end dates between two dates. I've tried using DATEDIFF but cannot quite make this work with the added requirement of counting month end dates.
In the example below I've shown the months count results I would expect to see by calculating the month end dates crossed between the two dates. If the start or end date lands on the month end then they count as being crossed.
Start date End Date Months count
01/08/22 30/08/22 0
20/08/22 31/08/22 1
31/08/22 01/09/22 1
31/08/22 30/09/22 2
20/08/22 01/09/22 1
20/08/22 30/09/22 2
20/08/22 31/07/23 12
Can anyone suggest a solution to this?
1 ACCEPTED SOLUTION
Frequent Visitor
I found a slightly different solution by creating an end of month column for the start date, and then then using an end of month calc for the end date but adding a day to the end date to push it into the next month if it fell on the month end date. This allowed me to accruately count month ends.
This was the dax I used below and 'StartDate_EndofMonth' iis the end of month column I created for the start date.
Months count = DATEDIFF(EOMONTH('table'[StartDate_EndofMonth],0), EOMONTH('table'[End Date]+1,0),MONTH)
2 REPLIES 2
Frequent Visitor
I found a slightly different solution by creating an end of month column for the start date, and then then using an end of month calc for the end date but adding a day to the end date to push it into the next month if it fell on the month end date. This allowed me to accruately count month ends.
This was the dax I used below and 'StartDate_EndofMonth' iis the end of month column I created for the start date.
Months count = DATEDIFF(EOMONTH('table'[StartDate_EndofMonth],0), EOMONTH('table'[End Date]+1,0),MONTH)
Super User
You could add a End of Month column to your date table and then use
Months count =
VAR StartDate =
SELECTEDVALUE ( 'Table'[Start date] )
VAR EndDate =
SELECTEDVALUE ( 'Table'[End date] )
VAR Result =
CALCULATE (
COUNTROWS ( VALUES ( 'Date'[End of month] ) ),
DATESBETWEEN ( 'Date'[Date], StartDate, EndDate )
)
RETURN
Result
Announcements
#### Europe’s largest Microsoft Fabric Community Conference
Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data.
#### Power BI Monthly Update - August 2024
Check out the August 2024 Power BI update to learn about new features.
#### Fabric Community Update - August 2024
Find out what's new and trending in the Fabric Community.
Top Solution Authors
Top Kudoed Authors | 755 | 2,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.78931 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.