url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://socratic.org/questions/59d59676b72cff1369ff3eb0 | 1,620,415,304,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00079.warc.gz | 551,596,753 | 6,497 | # Question #f3eb0
Feb 8, 2018
$c = \frac{2}{3}$
#### Explanation:
For $f \left(x\right)$ to be continuous at $x = 2$, the following must be true:
• ${\lim}_{x \to 2} f \left(x\right)$ exists.
• $f \left(2\right)$ exists (this is not a problem here since $f \left(x\right)$ is clearly defined at $x = 2$
Let's investigate the first postulate. We know that for a limit to exist, the left hand and right hand limits must be equal. Mathematically:
${\lim}_{x \to {2}^{-}} f \left(x\right) = {\lim}_{x \to {2}^{+}} f \left(x\right)$
This also shows why we're only interested in $x = 2$: It's the only value of $x$ for which this function is defined as different things to the right and the left, which means that there is a chance the left & right hand limits may not be equal.
We'll be attempting to find values of 'c' for which these limits are equal.
Going back to the piecewise function, we see that to the left of $2$, $f \left(x\right) = c {x}^{2} + 2 x$. Alternatively, to the right of $x = 2$, we see that $f \left(x\right) = {x}^{3} - c x$
So:
${\lim}_{x \to 2} c {x}^{2} + 2 x = {\lim}_{x \to 2} {x}^{3} - c x$
Evaluating the limits:
${\left(2\right)}^{2} c + 2 \left(2\right) = {\left(2\right)}^{3} - \left(2\right) c$
$\implies 4 c + 4 = 8 - 2 c$
From here, it's just a matter of solving for $c$:
$6 c = 4$
$c = \frac{2}{3}$
What have we found? Well, we've figured out a value for $c$ that will make this function continuous everywhere. Any other value of $c$ and the right & left hand limits will not equal each other, and the function will not be continuous everywhere.
To get a visual idea of how this works, check out this interactive graph I made. Pick different values of $c$, and see how the function ceases to be continuous at $x = 2$!
Hope that helped :) | 575 | 1,791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 24, "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} | 4.625 | 5 | CC-MAIN-2021-21 | latest | en | 0.854237 |
https://paperassignments.com/continue-the-springdale-shopping-survey-at-the-end-of-chapter-9-in-the-integrated-cases-section-in/ | 1,685,233,690,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643388.45/warc/CC-MAIN-20230527223515-20230528013515-00568.warc.gz | 489,240,539 | 11,145 | # Continue the Springdale Shopping Survey at the end of Chapter 9 in the Integrated Cases section in …
Continue the Springdale Shopping Survey at the end of Chapter 9 in the Integrated Cases section in Introduction to Business Statistics. You will again need the dataset given in Module 1 labeled SHOPPING.Write a short report that includes an introduction, a conclusion paragraph and a body which answers fully the three questions posed in the problem. Please include any tables of calculations and graphs associated with this problem in the body. It should be double-spaced and in APA style format per the Guide to Writing and APA Requirements.NOTE: You will need to type the equations from the chapter into Excel cells to perform the calculations that are required. Sometimes there will be functions in Excel you can use for shortcuts, other times you will have a long formula you need to type in.
The case in Chapt 2 listed 30 questions asked of 150 respondents in the community of Springdale. The coding key for the responses was also provided in that earlier exercise. The data are in file SHOPPING. In the exercise, some of the estimation techniques presented in the chapter will be applied to the survey results. You may assume that the respondents represent a simple random sample of all potential respondents within the community and that the population is large enough that application of the finite population correction would not make an appreciable difference in the results.
Managers associated with shopping areas like these find it useful to have point estimates regarding variables describing the characteristics and behaviors of their customers. In addition, it is helpful for them to have some idea as to the likely accuracy of these estimates. Therein lies the benefit of the techniques presented in this chapter applied here.
1. Item C in the description of the data collection instrument list variables 7,8,and 9, which represent the respondents general attitude toward each of the three shopping areas. Each of these variables has numerically equal distances between the possible responses, and for purposes of analysis they may be considered to be the interval scale of measurement.
a. Determine the point estimate, the construct the 95% confidence interval for u7= the average attitude toward Springdale Mall. What is the maximum likely error in the estimate of the population mean?
b. Repeat part (a) for u8 and u9, the average attitude toward Downtown and West Mall, respectively.
2. Given the breakdown of responses for variable 26(sex of respondent), determine the point estimate, then construct the 95% confidence interval for pi 26= the population proportion of males. What is the maximum likely error in the point estimate of the population proportion?
3. Given the breakdown of responses for variable 28 (marital status of respondent), determine the point estimate, then construct the 95% confidence interval for Pi 28 = the population proportion in the “single or other” category. What is the maximum likely error in the point estimate of the population proportion?
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No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 1,024 | 5,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.929717 |
https://www.neetprep.com/question/60793-third-line-Balmer-series-ion-equivalent-hydrogen-atom-haswavelength--nm-ground-state-energy-electron-ion-willbea--eV-b--eVc--eV-d--eV/55-Physics--Atoms/702-Atoms | 1,590,583,108,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00376.warc.gz | 853,645,942 | 64,763 | The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be
(a) 3.4 eV (b) 13.6 eV
(c) 54.4 eV (d) 122.4 eV
Concept Questions :-
Spectral series
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
An electron in the n = 1 orbit of hydrogen atom is bound by 13.6 eV energy is required to ionize it is
(a) 13.6 eV (b) 6.53 eV
(c) 5.4 eV (d) 1.51 eV
Concept Questions :-
Various atomic model
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Hydrogen atom emits blue light when it changes from n = 4 energy level to the n = 2 level. Which colour of light would the atom emit when it changes from the n = 5 level to the n = 2 level
(a) Red (b) Yellow
(c) Green (d) Violet
Concept Questions :-
Spectral series
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series
(a) 1215.4 Å (b) 2500 Å
(c) 7500 Å (d) 600 Å
Concept Questions :-
Spectral series
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
In Bohr's model, the atomic radius of the first orbit is ${\mathrm{r}}_{0}$, then the radius of the third orbit is
(a) $\frac{{\mathrm{r}}_{0}}{9}$ (b) ${\mathrm{r}}_{0}$
(c) $9{\mathrm{r}}_{0}$ (d) $3{\mathrm{r}}_{0}$
Concept Questions :-
Bohr's model of atom
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is 20.397 cm. The wavelength of energy for the same transition in ${\mathrm{He}}^{+}$ is
(a) 5.099 ${\mathrm{cm}}^{-1}$ (b) 20.497 ${\mathrm{cm}}^{-1}$
(c) 40.994 ${\mathrm{cm}}^{-1}$ (d) 81.988 ${\mathrm{cm}}^{-1}$
Concept Questions :-
Bohr's model of atom
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Minimum excitation potential of Bohr's first orbit in hydrogen atom is
(a) 13.6 V (b) 3.4 V
(c) 10.2 V (d) 3.6 V
Concept Questions :-
Bohr's model of atom
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Which of the following statements are true regarding Bohr's model of hydrogen atom
(I) Orbiting speed of electron decreases as it shifts to
discrete orbits away from the nucleus
(II) Radii of allowed orbits of electron are proportional to the cube of principal quantum number
(III) Frequency with which electrons orbits around the nucleus in discrete orbits is inversely proportional to the principal quantum number
(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits
Select correct answer using the codes given below
Codes :
(a) I and III (b) II and IV
(c) I, II and III (d) II, III and IV
Concept Questions :-
Bohr's model of atom
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
For electron moving in ${\mathrm{n}}^{\mathrm{th}}$ orbit of H-atom the angular velocity is proportional to
(a) n (b) 1/n
(c) ${\mathrm{n}}^{3}$ (d) 1/${\mathrm{n}}^{3}$
Concept Questions :-
Bohr's model of atom
Please attempt this question first.
Difficulty Level:
The energy of electron in first excited state of H-atom is -3.4 eV its kinetic energy is
(a) – 3.4 eV (b) + 3.4 eV
(c) – 6.8 eV (d) 6.8 eV
Concept Questions :-
Bohr's model of atom
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level: | 1,219 | 4,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.751213 |
https://jineralknowledge.com/tag/balls/ | 1,725,858,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00697.warc.gz | 305,108,234 | 16,580 | Posted in Science & Nature
## Hammer And Feather
What would happen if you dropped a 1kg ball and a 10kg ball at the same time from a high building? Most people would think that the 10kg ball would obviously fall faster and thus hit the ground faster, but the truth is they would fall at exactly the same time. The reason for this is that the force that accelerates a falling object is gravity, which on Earth is constant at 9.81ms-2. This means that no matter how heavy the object is, they will always accelerate by 9.81 metres per second per second. This was hypothesised by Galileo Galilei, who came up with the thought experiment of dropping two balls of different mass from the Leaning Tower of Pisa (there is debate as to whether he actually performed the experiment). The theory was later solidified by a certain Isaac Newton, who devised the laws of universal gravitation and the three laws of motion.
However, if the two balls were dropped from an extremely high place, they may land at different times as mass affects the terminal velocity – when the force of gravity equals the force of drag caused by air resistance, leading to a constant velocity. A heavier object will keep accelerating to a greater velocity than a lighter object, which would have reached terminal velocity before the heavier object.
One place where this will not happen is in a vacuum where there is no drag force. To prove that the hypothesis that two objects of different masses will fall at the same time in the absence of air resistance, Commander David Scott of the Apollo 15 moon mission took a hammer and a feather with him. Once he landed on the moon, he dropped the hammer and feather in front of a live camera, showing that the two landed at exactly the same time. He thus proved that Galileo’s conclusion from two hundred years ago was in fact correct.
Posted in Life & Happiness
## A Jar And Two Cups Of Coffee
A professor stood before his philosophy class and had some items in front of him. When the class began, he silently picked up a very large and empty jar and proceeded to fill it with golf balls. He then asked the students if the jar was full. They agreed that it was.
The professor then picked up a box of pebbles and poured them into the jar. He shook the jar lightly. The pebbles rolled into the open areas between the golf balls. He then asked the students again if the jar was full. They agreed it was.
The professor next picked up a box of sand and poured it into the jar. Of course, the sand filled up everything else. He asked once more if the jar was full. The students responded with a unanimous “yes.”
The professor then produced two cups of coffee from under the table and poured the entire contents into the jar effectively filling the empty space between the sand. The students laughed.
“Now,” said the professor as the laughter subsided, “I want you to recognize that this jar represents your life. The golf balls are the important things – your family, your children, your health, your friends and your favourite passions – and if everything else was lost and only they remained, your life would still be full. | 659 | 3,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.974921 |
https://flightwinebar.com/blog/how-many-gallons-is-60-liters/ | 1,713,239,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00431.warc.gz | 225,864,432 | 44,573 | # How Many Gallons Is 60 Liters?
If you are looking to convert liters to gallons, you may be wondering “How many gallons is 60 liters?” Converting between liters and gallons is a common task, and it is important to understand the conversion rate in order to accurately measure liquids. In this article, we will discuss the conversion rate between liters and gallons, as well as provide a few examples of how to convert liters to gallons.
## How to Convert Liters to Gallons: A Step-by-Step Guide to Calculating 60 Liters?
Are you looking to convert liters to gallons? Converting between the two units of measurement can be tricky, but don’t worry! We’ve got you covered with this step-by-step guide to calculating 60 liters. Let’s get started!
Step 1: Understand the Conversion Rate. To convert liters to gallons, you’ll need to know the conversion rate. One liter is equal to 0.264172 gallons.
Step 2: Calculate the Number of Gallons. To calculate the number of gallons, you’ll need to multiply the number of liters by the conversion rate. In this case, you’ll be multiplying 60 liters by 0.264172. This will give you a total of 15.85032 gallons.
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Step 3: Round the Result. To make the result easier to read, you can round the result to the nearest tenth. In this case, 15.85032 gallons would round to 15.9 gallons.
And there you have it! You’ve successfully converted 60 liters to gallons. Now you can use this knowledge to convert any number of liters to gallons with ease. Happy converting!
## The Benefits of Knowing How Many Gallons Are in 60 Liters
Knowing how many gallons are in 60 liters can be incredibly useful in a variety of situations. Whether you’re a student studying for a science test, a homeowner looking to purchase the right amount of paint for a project, or a chef trying to convert a recipe from metric to imperial measurements, understanding the conversion between liters and gallons can be a real time-saver.
For those who are unfamiliar with the metric system, a liter is a unit of volume equal to 1,000 cubic centimeters. A gallon, on the other hand, is a unit of volume equal to 128 fluid ounces. To convert liters to gallons, simply divide the number of liters by 3.785. This means that 60 liters is equal to 15.8 gallons.
Knowing this conversion can be especially helpful when it comes to measuring liquids. For example, if you’re trying to figure out how much paint you need for a project, you can easily calculate the number of gallons you need by multiplying the number of liters by 0.264. This means that if you need 60 liters of paint, you’ll need 15.8 gallons.
In addition, understanding the conversion between liters and gallons can be useful when it comes to cooking. Many recipes are written in metric measurements, so it’s important to be able to convert them to imperial measurements. For example, if a recipe calls for 60 liters of water, you’ll know that you need 15.8 gallons.
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Knowing how many gallons are in 60 liters can be incredibly useful in a variety of situations. Whether you’re a student studying for a science test, a homeowner looking to purchase the right amount of paint for a project, or a chef trying to convert a recipe from metric to imperial measurements, understanding the conversion between liters and gallons can make life a lot easier!
## How to Use a Conversion Chart to Easily Calculate 60 Liters to Gallons?
Converting between liters and gallons is a breeze with a conversion chart! Here’s how to use one to easily calculate 60 liters to gallons:
First, locate the conversion chart. You can find one online or in a reference book.
Next, find the row that corresponds to liters. In this case, it’s the row labeled “Liters (L).”
Now, find the column that corresponds to gallons. This is the column labeled “Gallons (gal).”
Finally, locate the number that corresponds to 60 liters. In this case, it’s 15.85 gallons.
And there you have it! With just a few simple steps, you can easily convert 60 liters to gallons using a conversion chart.
## The History of the Gallon and How It Relates to 60 Liters
The gallon is a unit of measurement that has been around for centuries. It is used to measure volume, and is most commonly used to measure liquids. In the United States, the gallon is equal to 128 fluid ounces, or 3.785 liters. In the United Kingdom, the gallon is equal to 160 fluid ounces, or 4.546 liters.
The origin of the gallon dates back to the Middle Ages. It was first used in England, and was based on the volume of eight pounds of wheat. This was known as the “Winchester” gallon, and it was used to measure both dry and liquid goods. Over time, the gallon was adopted by other countries, and it eventually became the standard unit of measurement for liquids.
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Today, the gallon is still widely used in the United States and the United Kingdom. It is also used in other countries, such as Canada and Australia. In the metric system, the gallon is equal to 3.785 liters, which is equal to 60 liters. This means that one gallon is equal to 60 liters.
The gallon is an important unit of measurement, and it has been used for centuries. It is still widely used today, and it is a great way to measure liquids. So, the next time you need to measure a liquid, remember that one gallon is equal to 60 liters!
## Understanding the Difference Between US and Imperial Gallons When Converting 60 Liters
Are you confused about the difference between US and Imperial gallons when converting 60 liters? Don’t worry, you’re not alone! Let’s take a look at the difference between these two types of gallons and how to convert 60 liters into each.
US gallons are based on the US customary system of measurement and are equal to 231 cubic inches. This means that one US gallon is equal to 3.785 liters. To convert 60 liters into US gallons, simply divide 60 by 3.785. This will give you 15.8 US gallons.
Imperial gallons, on the other hand, are based on the imperial system of measurement and are equal to 277.42 cubic inches. This means that one Imperial gallon is equal to 4.546 liters. To convert 60 liters into Imperial gallons, simply divide 60 by 4.546. This will give you 13.2 Imperial gallons.
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So, there you have it! Now you know the difference between US and Imperial gallons and how to convert 60 liters into each. With this knowledge, you’ll be able to easily convert liters into gallons no matter which type you need.
## Exploring the Relationship Between Liters and Gallons: What 60 Liters Equals in Gallons
Did you know that 60 liters is equal to 15.85 gallons? That’s right! It’s easy to convert liters to gallons and vice versa.
To convert liters to gallons, simply multiply the number of liters by 0.264. So, if you have 60 liters, you would multiply 60 by 0.264 to get 15.84 gallons.
On the other hand, to convert gallons to liters, you would multiply the number of gallons by 3.785. So, if you have 15.85 gallons, you would multiply 15.85 by 3.785 to get 59.7 liters.
It’s that easy! Now you know that 60 liters is equal to 15.85 gallons.
## The Impact of Metric System on Converting 60 Liters to Gallons
The metric system has had a huge impact on the way we measure and convert different units of measurement. Converting 60 liters to gallons is a great example of how the metric system has made our lives easier.
Before the metric system, converting 60 liters to gallons would have been a difficult and time-consuming task. But thanks to the metric system, it’s now a breeze! All you need to do is multiply 60 liters by 0.264172052 to get the equivalent number of gallons. That’s it!
The metric system has made it much easier to convert between different units of measurement. It’s also made it easier to compare measurements between different countries. For example, if you’re in the United States and you want to know how many liters are in a gallon, you can easily look it up online or in a book.
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The metric system has also made it easier to understand measurements in different contexts. For example, if you’re baking a cake and the recipe calls for 60 liters of milk, you can easily convert that to gallons and know exactly how much milk you need.
Overall, the metric system has had a huge impact on the way we measure and convert different units of measurement. Converting 60 liters to gallons is just one example of how the metric system has made our lives easier.
## A Comprehensive Guide to Calculating How Many Gallons Are in 60 Liters
Are you wondering how many gallons are in 60 liters? You’ve come to the right place! Converting liters to gallons is a simple process that can be done in just a few steps. Let’s take a look at how to calculate the number of gallons in 60 liters.
First, it’s important to understand the basic conversion rate between liters and gallons. One liter is equal to 0.264172052 gallons. This means that for every liter, there are 0.264172052 gallons.
Now that we know the conversion rate, we can calculate the number of gallons in 60 liters. To do this, we simply need to multiply 60 by 0.264172052. This gives us a result of 15.85032308 gallons.
So, there you have it! 60 liters is equal to 15.85032308 gallons. Now you know how to calculate the number of gallons in any given number of liters. We hope this guide has been helpful and that you’re feeling more confident in your ability to convert liters to gallons. Have fun!
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## How many gallons is 60 liters?
60 liters is equal to 15.85 gallons.
## Conclusion
In conclusion, 60 liters is equal to 15.85 gallons. This is a useful conversion to know when dealing with different units of measurement for liquids. | 2,278 | 9,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-18 | longest | en | 0.897021 |
hedgehogcomms.blogspot.com | 1,369,623,299,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706890813/warc/CC-MAIN-20130516122130-00068-ip-10-60-113-184.ec2.internal.warc.gz | 64,982,153 | 28,702 | ## Monday, October 20, 2008
### Using models to solve maths problems part 2
It appears that there are quite a few parents interested in the model method. Lilian posted a few problem sums on her blog which she says she had problems drawing the models for. Her son Brian actually got all the problems right, but not using the model method. That's ok - for PSLE, the examiners don't care what method you use, as long as it's correctly applied and you get the final answer.
I tried the sums and got all the answers within half an hour using the model method. I love the model method because it appeals to my love for problem-solving and helps me understand the concepts behind the sums. I hate learning formulas and blindly following them without understanding why, so algebra puts me off. But one boy in Lesley-Anne's class loves algebra and uses it instead of the model method, so hey, different strokes for different folks!
Here are the sums that Lilian posted. I'll explain step by step how I solved them using the model method.
Question 1:
In April, 40% of the people who went to the museum were children. The rest were adults. The number of women was 3/4 the total number of adults. The rest were men. In May, the number of children increased by 20%. The number of adults was the same as in April but the number of women became only 3/5 of the adults. Then, the number of children became 336 more than the number of women.
a) What is the ratio for the number of children to the number of men to the number of women in April?
b) How many people went to the museum in May?
First, I drew the model for April (on the right). Since 40% is the same as 2/5, 2 parts of the model represent children and 3 parts represent adults.
Then I needed to cut the 3 parts adults in men and women. I find that if you need to further divide a model, it's often easier to cut in a different direction.
In this case, instead of further dividing the columns, I cut the adult portion into four equal rows (right). You can immediately see that 1 row (1/4) is men, 3 rows (3/4) are women.
Remember each unknown part of the model has to be of equal value. So I also cut the children section into 4 equal rows. From here, you can instantly get the answer to part a) just by counting the parts.
Answer for a): Ratio of children is to men is to women in April is 8:3:9
Now we move on to May. I replicated the same model but this time, instead of cutting into 4 rows each, I cut into 5 rows each because I need to find out how many parts represent 3/5. So now, 3 rows are women (3/5) and 2 rows are men (2/5).
Same thing, since I cut the adults into 5 rows, I need to cut the children into 5 rows.
The number of children was increased by 20%. Since children are represented by 10 parts, 20%=2 parts. So I added 2 more parts to children (right).
There are 3 parts more children (12 parts) than women (9 parts). Since there are 336 more children than women, 3 parts=336.
336÷3=112 (this is the value for 1 part)
Since there are 27 parts altogether (count them!), 27 x 112=3,024
Answer for b): 3,024 people went to the museum in May.
(Lilian, like Brian, Lesley-Anne also got it right but used ratio to solve the problem.)
Question 2:
Brian invited some boys and girls, there are 20 more boys than girls. 3/4 of the boys and 2/3 of the girls managed to come. 19 children did not come. How many children did Brian invite?
First, I drew the basic model (right). Then we need to figure out how to cut the model into children who came and children who didn't come. Let's start with the boys. The tricky bit is that we don't know what fraction 20 boys is in relation to total number of boys, so you can't cut the model into 4 parts including the 20. But we know that 20 can be divided by 4, so what we can do is cut the unknown portion for boys into 4 and each of these unknown parts + 5 (20÷4) is 1/4 of the total number of boys.
We can also divide the girls into 3 parts (1 part didn't come, 2 parts came). But we need to ensure that each unknown part for boys and girls is equal, otherwise we can't compare them. I found the lowest common multiple of 4 and 3 which is 12, and divided both the boys and girls portion into 12 equal parts each. Now you can easily compare how many parts came and didn't come in the model below:
19 children who did not come to the party. This is represented by 7 parts + 5 children. Therefore, 7 parts = 19-5 = 14.
14÷7=2 (this is the number of children in 1 part.)
Total number of children is 2 x 24 parts + 20.
48+20=68
Answer: 68 children were invited to the party.
Lesley-Anne solved this problem using the model method.
Question 3:
In a school, there are 45 more students in Primary 5 than in Primary 4. In Primary 4, there are 18 more girls than boys. There are 12 more boys in Primary 5 than in Primary 4. How many more girls than boys are there in Primary 5?
Again, I first drew the basic model (left) then I cut the unknown portion vertically as such (right). Since there are 18 more girls than boys in p4, I marked the model accordingly (note: the unknown parts in the p4 row are equal in value, sorry for the improportionate drawing).
Since are 12 more boys in p5 than in p4, I extended the line for boys in p4 vertically across p5 (model below). Then I marked out an extra 12 boys. This means that that little piece in the p5 portion to the immediate left of 12 has a value of 6 (since that corresponding part in the p4 portion is 18). Since we know the portion representing number of p5 boys, the rest are girls.
Usually, the key to models is finding out the value of the unknown part. BUT in this sum, we're not required to know how many children there are altogether, only how many more girls than boys there are in p5. In the p5 portion, the girls and boys have one unknown part each so they cancel each other out.
We're left with all the known numbers, ie girls = 45+6 and boys = 12.
45+6-12=39.
Answer: There are 39 more girls than boys in p5. (Note: this answer is different from the one Lilian says was given in the assessment book which is 33, but I checked and re-checked and couldn't see how the model could be wrong. If anyone spots an error, let me know).
Lesley-Anne couldn't solve this problem as she couldn't draw the model.
Question 4:
A basket of 6 apples and 3 mangoes weighed 1kg 320g. After 4 apples and 2 mangoes were eaten, the basket with the remaining fruits weighed 760g only. If a mango weighs 20g less than 4 times the mass of an apple, find
a) the mass of the basket
b) the mass of the apple
First, I drew a model of the mango and apple (right). (Note: the 20g is an approximate, at this point, we don't know if it's more or less than one unknown part).
This question threw me off for a bit because of that darn basket which gives me 2 unknown values instead of 1. Then I realised that I could get rid of the basket by just using the fruit that were eaten, ie 1kg 320g (6 apples + 3 mangoes + basket) - 760g (2 apples + 1 mango + basket) = 560g (4 apples + 2 mangoes).
So I drew the model for 4 apples and 2 mangoes (sorry for REALLY bad pic!!).
From the model, we can see that if we want to convert the parts for the 2 mangoes into whole parts, we can just add 20g x 2 mangoes on both sides. So 12 equal parts (add up the ones for 2 mangoes and 4 apples) = 560g + 40g = 600g. Since all the unknown parts are equal, we can find 1 part, which is also equal to 1 apple.
600g ÷ 12 (parts) = 50
Answer for b): The mass of the apple is 50g.
We can also find the mass of one mango from the model - 50g x 4 (parts) - 20g = 200g-20g = 180g.
And since we know the mass of the mango and the apple, we can find the mass of the basket.
2 apples + 1 mango + basket = 760g
(2x50g) + 180g + basket = 760g
Basket = 760g - 100g - 180g = 480g
Answer for a): The mass of the basket is 480g.
I don't know if it's right that I got the answer for b) first and then a)! Lesley-Anne couldn't solve this question.
I'm no expert at models - I didn't consciously study the method, it's something I picked up along the way when helping Lesley-Anne with her maths. Which is why I believe that everyone can learn this, it just takes practice.
Lilian said...
This is beautiful!! I'm speechless. You should write your own math assessment books. Those books I have don't present the answers like you do dammit. They don't cut it a different direction like you do here. I'm sooooo bloody impressed, pardon my french, but truly, I'm feeling extremely inadequate now.
Thank you thank you thank you, I'll be posting up more when I can. Can I send Brian over to you in December for intensive model tuition? *kowtow*
monlim said...
Maybe I shd write an installment for Murderous Maths on models, hehe...
No lah, like I said, I'm no math whiz (not trying to be modest, I was the despair of my Maths C teacher cos I just couldn't grasp the abstract concepts). It's just from constantly trying to help Lesley-Anne with the problems that I honed the skills. You haven't done enough of it that's why you can't do it. I'm sure with your math aptitude, you'll get it with some practice.
More worrying is that Lesley-Anne still hasn't grasped the more complex and trickier ones. How? I can't take PSLE for her leh...
Anonymous said...
Slightly unrelated but useful for kids at lower Pri:--
Just saw this on my kid's DVD (don't know if it's already familiar with you'all?)
9x table at your fingertips:--
Qn1: What is 9x3?
Put up your 10 fingers.
||||| |||||
Count to 3 fr left & hold down 3rd finger.
||_|| |||||
2......7
Voila! Answer is 27.
Qn2: What is 9x6?
Count to 6 fr left & hold down 6th finger.
||||| _||||
5.........4
Voila! Answer is 54.
Cool trick?
YY.
Last nite quite xiong you know, did model method until got insomnia... Brain too revved up!
YY.
monlim said...
Yep, Andre learnt that 9 times table trick from his school teacher :)
Anonymous said...
Anymore tricks for rest of times-table? I'm really interested to find out! Did L-A learnt it during lower Pri too?
First time hubby knew of this trick as well. I wonder if my girl learnt it from 'mental-math' when she was young... must get boy to 'test' her when he gets back from school...
Trying to get kid to memorize times-table at the moment.
YY.
Lilian said...
Aiyoh YY, I learnt that when I was a kid lah hahahahah! That's er 15 years ago? hee...
In Changi Airport now, just had wantan mee at T3. Mon, on my plane ride back to Sg, I tried some questions on my own (Brian's not with me, I'm back with Sean first) and still can't do it. Those that I could took me lots of time. Hope you can post more examples, even those you're doing with Andre, cos I think our basics may have been screwed up by some of the answers given in assessment books. I think we need to be "Unschooled" in our models approach.
Anonymous said...
You don't have to use models for every question. Most students, including my son and your daughter and Brian, use the ratio method for the museum question, as it was obviously a question on ratio. Faster and less time consuming than drawing the models, and both parts of the question are related.
MD
monlim said...
MD: thanks for your comment! Yes, I know you don't have to use models for every question. I was just trying to demonstrate that it could be done using models, in response to Lilian's question. Incidentally for Q1, it took me a shorter time to draw the model than for my daughter to use ratio :)
monlim said...
YY: Yes! There is a wonderful trick for the 7,8,9 times-tables - (the toughies!)
Face your palms towards you, fingers pointing towards each other. Ring finger is 7, middle finger is 8, index finger is 9. Say you want to find out 7x7. Make the 2 ring fingers touch each other. Count those 2 fingers and all others below (4) - this is the first digit. Take all fingers above ring finger on one hand (3) multiplied by fingers above ring finger on other hand (3), you get 9 which is 2nd digit. Answer: 49!
Anonymous said...
I saw your age-related question, and it gives me great delight to present my very own - which often stumps students and adults alike (usually for a few minutes...:))
Pat is twice as old as Chris was when Pat was as old as Chris is now. If Pat is 24 now, how old is Chris now?
MD
monlim said...
MD: That's not maths question lah, it's English! 24?
Anonymous said...
It's Maths. and your answer is wrong (Pat is already 24 years old). Hint, hint - this one got to use algebra.
Seriously this was a P6 question. Below is a P4 question.
Postman: How are your 3 daughters, Mr Lim?
Mr Lim: They are fine. Thank you.
Postman: How old are they now?
Mr Lim: The product of their 3 ages is 36. The sum of their ages is the same as my house number.
Postman: Err.. I can’t figure it out.
Mr LIm: Well, my eldest daughter is waiting for her letter now.
Postman: Oh, I got it. Thanks
Can you get it?
Anonymous said...
monlim:
Wow! Cool but dunno if my boy will get it all confused... Oh I get it, it only works if both numbers are 7 or above, right? For e.g. it's not meant for 7 x 3.
So he'd better learn all the multiples for numbers 6 & below.
Anywhere on the internet where I can learn about these tricks?
Lil:"That's er 15 years ago? hee..."
Oh shuurrree.... Makes me feel so old like dat. :)
YY.
monlim said...
Chris is 18?
Postman question seems to have many possible answers? Not sure what the house number has got anything to do with it (since can be anything), unless I'm missing something. 2x3x6? 3x3x4? 1x6x6? 2x2x9? etc
monlim said...
YY: yes, only for 7,8,9. But these are usually the ones the kiddos have problems remembering :)
Think I may post a quick one on this. Guess where my kids learnt this trick... from our domestic helper!!!! Maybe it's a Philippino trick, haha!
Anonymous said...
Clap! Clap!
Let me show the "model" algebra method:
WAS IS
Pat X 24
Chris 12 X
We can then clearly see that:
24 - X = X - 12 ⇒ X = 18 ;)
For the postman question, perhaps by listing down the various combinations, you can see the solution.
Anonymous said...
My P3 boy was also taught this trick by his maths teacher. :) Then he went on to teach his P6 brother, who wanted to know out of curiosity. Of course, he didn't know his older brother knew how to multiply two digit figures by two digit figures mentally when he was only 5. :)
monlim said...
err... actually, I just did trial and error, I suck at algebra.
Anonymous said...
""Pat is twice as old as Chris was when Pat was as old as Chris is now. If Pat is 24 now, how old is Chris now?""
Ans is 18.
Can use model method also!
YY.
Lilian said...
MD your second question is more like a brain bender leh...not math! I know the answer cos I google lah haha. No one who hasn't come across this question & its solution will ever be able to solve it!
Anonymous said...
aiyah, this was indeed a maths question given to my eldest when he was in P4. Anyway, you can check out Nanyang Maths Prelim 2007, Q40. Seems like a brain bender, but it's in it as a maths question.
This year PSLE maths' brain bender was the millipede question, the very last question of the paper (but strangely only 4 marks; usually the last few questions carry 5 marks)
monlim said...
yah lah, but you have to read into each statement, like the postman couldn't figure out the answer because there were several combinations... maybe he's just a dumb postman?????
Yve said...
Interesting post. I had to learn how to use models as well (and cheat at the back by using algebra!) to help with the kids' homework. Here's my post about a question that we got from one of the top schools' paper. Primary 3.
http://kembangankids.blogspot.com/2008/10/what-we-did-last-night.html
tianzhu said...
Hi
I think 39 is correct.
http://farm4.static.flickr.com/3300/3224896115_d22ea64109_o.jpg
tianzhu said...
Hi
The earlier link was incomplete after posting.The revised link is http://farm4.static.flickr.com/
3300/
3224896115_d22ea64109_o.jpg
Anonymous said...
Hi Mon, sorry to disturb you. I need to retrieve this question,
{"Pat is twice as old as Chris was when Pat was as old as Chris is now. If Pat is 24 now, how old is Chris now?""
Ans is 18.}
I was trying to do this sum yesterday but the english got into me, I could not figure out. My older gal managed to understand the question on her own and got the ans 18. She tried to explain to me, but still, I catch no ball....anyway, as long as she understand what she is doing, I no need to squeeze my brain juice for this. But, prior to letting her try out the sum, I email one of my galfriend, her husband did this sum and got 16 as the answer, which after working out the model, Chris could possibly be 16.
Eg,
Was
Pat [][]=16 (twice the age)
Chris [] =8
Now
Pat [][][] = 24
Chris [][] = 16
Is this correct too?
Many tks.
Chris
monlim said...
Chris: 16 cannot be right because of the first bit - Pat is 24 now and Pat IS twice as old as Chris was, so Chris was 12. At the point when Chris was 12, the age that Pat was and Chris is now is the same. This number is exactly between 12 & 24 (since Pat is always the same number of years older than Chris), hence 18.
Hope that makes sense! I used this logic to work it out, didn't use algebra or models :P
Anonymous said...
Mon, thanks so much for your reply. After reading & reading...& reading your reply...I sought of getting it *so embarrassing :p*. Now can show off to my friend's hb.....heehee
Chris | 4,554 | 17,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2013-20 | longest | en | 0.979375 |
https://library.fiveable.me/topos-theory/unit-1/morphisms-isomorphisms-functors/study-guide/xBF5zRFt62ddLAKt | 1,726,261,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00367.warc.gz | 325,119,121 | 40,375 | # 1.2 Morphisms, isomorphisms, and functors
fundamentals are crucial building blocks for understanding more complex concepts in Topos Theory. This section introduces key morphisms like monomorphisms, epimorphisms, and isomorphisms, which generalize familiar concepts from set theory to broader categorical settings.
Functors, the focus of the next part, are essential tools for connecting different categories. They allow us to map objects and morphisms between categories while preserving important structural relationships, forming the basis for more advanced categorical constructions.
## Category Theory Fundamentals
### Monomorphisms, epimorphisms, and isomorphisms
• $f: A \to B$ in category $C$ acts like injective function generalizes injectivity from Set theory
• Left-cancellative property ensures for morphisms $g, h: X \to A$, $f \circ g = f \circ h$ implies $g = h$
• Examples: inclusion maps (subsets), group homomorphisms with trivial kernel
• Morphism $f: A \to B$ in category $C$ behaves like surjective function extends surjectivity concept
• Right-cancellative property guarantees for morphisms $g, h: B \to Y$, $g \circ f = h \circ f$ implies $g = h$
• Examples: quotient maps (groups), projection maps (product spaces)
• Morphism $f: A \to B$ in category $C$ has inverse $g: B \to A$ satisfying $g \circ f = 1_A$ and $f \circ g = 1_B$
• Generalizes bijective functions preserves structure between objects
• Examples: vector space isomorphisms, group isomorphisms, homeomorphisms (topology)
### Isomorphisms and two-sided inverses
• Proof outline
1. Assume $f$ is isomorphism
• Definition provides $g$ with $g \circ f = 1_A$ and $f \circ g = 1_B$
• $g$ serves as two-sided inverse of $f$
2. Assume $f$ has two-sided inverse $g$
• Given $g \circ f = 1_A$ and $f \circ g = 1_B$
• Conditions match isomorphism definition
• Key points
• Inverse uniqueness ensures only one two-sided inverse exists
• Isomorphisms possess both monomorphism and epimorphism properties
• Examples: matrix inverses, function inverses (bijective functions)
## Functors and Their Properties
### Functors between categories
• definition
• Map $F: C \to D$ between categories $C$ and $D$ preserves categorical structure
• Assigns objects to objects and morphisms to morphisms maintaining relationships
• Components of functor
• Object assignment maps each object $A$ in $C$ to $F(A)$ in $D$
• Morphism assignment takes each morphism $f$ in $C$ to $F(f)$ in $D$
• Examples of functors
• Forgetful functor from Group to Set strips group structure
• Power set functor from Set to Set maps sets to their power sets
• Fundamental group functor from Top to Group associates topological spaces with groups
• Constant functor maps all objects to single object and all morphisms to identity
• Identity functor maps category to itself preserving all structure
### Properties of functors
• preservation
• For composable morphisms $f: A \to B$ and $g: B \to C$ in $C$
• Ensures $F(g \circ f) = F(g) \circ F(f)$ maintaining operation structure
• preservation
• For object $A$ in $C$, functor maps $F(1_A) = 1_{F(A)}$
• Preserves identity elements across categories
• Consequences of properties
• Commutative diagrams remain commutative when mapped by functors
• Isomorphisms transform into isomorphisms under functor application
• Types of functors
• Covariant functors preserve morphism direction (standard functors)
• Contravariant functors reverse morphism direction (dual category relationship)
• Functors and category structure
• Maintain source and target object relationships for morphisms
• Map domains to domains and codomains to codomains preserving overall structure
## Key Terms to Review (18)
Adjoint Functors Theorem: The Adjoint Functors Theorem states that under certain conditions, a functor has both a left adjoint and a right adjoint, which are crucial in establishing relationships between different categories. This theorem reveals how morphisms and isomorphisms behave between categories when functors are applied, demonstrating the powerful interplay between different mathematical structures. It highlights how adjunctions can simplify complex constructions and ensure the preservation of certain properties across categories.
Associativity: Associativity is a fundamental property in mathematics that refers to the way in which the grouping of elements affects the outcome of binary operations. In the context of morphisms, isomorphisms, and functors, associativity ensures that when combining multiple morphisms or operations, the result remains consistent regardless of how the elements are grouped. This property is essential for maintaining structure and coherence in mathematical systems, as it allows for flexibility in how operations are performed without altering the final result.
Category Theory: Category theory is a mathematical framework that deals with abstract structures and relationships between them, focusing on the concept of objects and morphisms. It provides a way to formalize mathematical concepts across various fields, emphasizing the connections and mappings between different structures rather than their individual components. This abstraction is crucial for understanding complex relationships in mathematics, including transformations through functors, the properties of isomorphisms, and connections to logic and foundational mathematics.
Charles Pontryagin: Charles Pontryagin was a prominent Russian mathematician known for his contributions to topology and functional analysis, particularly in the development of Pontryagin duality. His work has important implications for the study of morphisms, isomorphisms, and functors, as it provides deep insights into the structure and behavior of topological groups and their duals.
Commutative Diagram: A commutative diagram is a visual representation of mathematical relationships between objects and morphisms, where any two paths in the diagram that connect the same two objects yield the same result when composed. This concept highlights the compatibility of morphisms and their compositional relationships, making it essential for understanding structures in category theory. In particular, commutative diagrams facilitate the exploration of morphisms, isomorphisms, functors, and exponential objects, revealing how these elements interact and maintain structural integrity.
Composition: Composition refers to the process of combining two morphisms in a category to form a new morphism. This operation is essential as it allows for the chaining of relationships between objects, facilitating the exploration of how different structures interact within the framework of categories. Composition must satisfy specific properties, such as associativity and the existence of identity morphisms, which are crucial for the overall coherence of categorical structures.
Contravariant Functor: A contravariant functor is a type of mapping between categories that reverses the direction of morphisms, taking objects from one category to another while flipping the arrows. This means that if there is a morphism from object A to object B in the original category, a contravariant functor will map these objects to another morphism going from the image of B back to the image of A. Understanding contravariant functors is crucial for grasping how relationships between different mathematical structures can be modeled and transformed.
Covariant Functor: A covariant functor is a type of mapping between categories that preserves the direction of morphisms. In simpler terms, if you have a morphism (or arrow) from one object to another in the first category, a covariant functor will map that morphism to a morphism between the corresponding objects in the second category, keeping the same direction. This concept ties into how we understand morphisms and isomorphisms, as well as how different types of functors interact with natural transformations and help us explore functor categories and the Yoneda lemma.
Epimorphism: An epimorphism is a type of morphism in category theory that can be thought of as a generalization of the concept of surjectivity in set theory. It is defined as a morphism $$f: A \to B$$ such that for any two morphisms $$g_1, g_2: B \to C$$, if $$g_1 \circ f = g_2 \circ f$$, then it must follow that $$g_1 = g_2$$. This means that an epimorphism is a morphism that, in a sense, covers all of its target object and ensures the uniqueness of how morphisms can factor through it, linking closely to isomorphisms and the nature of functors in category theory.
Functor: A functor is a mathematical mapping between categories that preserves the structure of those categories, meaning it maps objects to objects and morphisms to morphisms in a way that respects the composition and identity of the categories. Functors play a crucial role in connecting different mathematical structures and help in defining various concepts such as natural transformations and limits.
Functoriality: Functoriality refers to the property of a functor that maps morphisms in one category to morphisms in another category in a way that preserves the structure of the categories. This means that if there is a morphism between objects in the first category, the functor will produce a corresponding morphism between the images of those objects in the second category, maintaining composition and identity. Functoriality connects various mathematical concepts and structures, illustrating how they interact through mappings.
Identity morphism: An identity morphism is a special type of morphism in category theory that serves as the 'do-nothing' arrow for each object in a category, meaning it maps an object to itself. Every object in a category has an associated identity morphism, and it acts as a neutral element with respect to composition of morphisms, reinforcing the structure and coherence within the category.
Isomorphism: An isomorphism is a special type of morphism in category theory that indicates a structural similarity between two objects, meaning there exists a bijective correspondence between them that preserves the categorical structure. This concept allows us to understand when two mathematical structures can be considered 'the same' in a categorical sense, as it connects to important ideas like special objects, functors, and adjoint relationships.
Monomorphism: A monomorphism is a morphism that is left-cancellable, meaning if two morphisms composed with it yield the same result, then those two morphisms must be the same. This concept connects closely to notions of injectivity in set theory, highlighting how monomorphisms can represent the inclusion of one structure into another while preserving distinctiveness.
Morphism: A morphism is a structure-preserving map between two objects in a category, reflecting the relationships and transformations that can occur within that context. It plays a central role in connecting objects and understanding how they interact, serving as the foundation for defining concepts like isomorphisms and functors, which enrich the framework of category theory.
Naturality: Naturality is a property of certain mathematical constructions, particularly in category theory, where a transformation or a morphism can be shown to commute with other structures in a natural way. It emphasizes that such transformations do not depend on arbitrary choices and behave consistently across different contexts, making them more universally applicable. In the realm of functors, natural transformations highlight how functorial relationships are maintained, while adjunctions illustrate naturality in the context of units and counits, showcasing their integral role in the structure of categories.
Samuel Eilenberg: Samuel Eilenberg was a prominent mathematician known for his foundational work in category theory, topology, and algebra, particularly in the context of algebraic topology and topos theory. His contributions significantly advanced the understanding of categories, functors, and adjunctions, which are crucial concepts in modern mathematics.
Yoneda Lemma: The Yoneda Lemma is a foundational result in category theory that relates functors to natural transformations, stating that every functor from a category to the category of sets can be represented by a set of morphisms from an object in that category. This lemma highlights the importance of morphisms and allows for deep insights into the structure of categories and functors. | 2,569 | 12,488 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 43, "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} | 3.53125 | 4 | CC-MAIN-2024-38 | latest | en | 0.816033 |
https://physics.stackexchange.com/questions/554191/black-hole-infinite-distance-finite-time-paradox | 1,701,757,986,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00349.warc.gz | 503,447,696 | 41,725 | # Black Hole infinite distance finite time paradox
Schwarzschild metric at the event horizon shows that, a small distance as perceived by a distant observer is in fact an infinite distance for a falling observer. Yet the falling observer crosses the event horizon and untimately reaches the centre singularity in a finite time. Doesn’t that create a paradox?
How can someome cross an infinite distance in finite time?
• Is the statement in your first sentence correct? May 23, 2020 at 12:38
• This is not true. For any infalling observer the proper-time to the singularity is finite.
– user107153
May 23, 2020 at 12:44
• It is true. At event horizon the proper distance dS = dr times infinity. And, I am not talking about proper time, I am referring to proper distance here. May 23, 2020 at 12:50
• Proper distance is not ds, but an integral of ds. Do the math and you’ll see the integral is not infinite. May 23, 2020 at 13:37
• It is very common for integrals of infinite functions to be finite. For example $\int_0^1 \frac{dx}{\sqrt{x}}$ May 24, 2020 at 6:27
$$\int_{r_{\rm s}}^{r_{0}} \sqrt{|g_{rr}|} \, dr =\int_{r_{\rm s}}^{r_{0}} \frac{1}{\sqrt{1-\frac{r_{\rm s}}{r}}} \, dr = \sqrt{(r_{0}-r_{\rm s}) r_{0}}+\ln \left(r_{0}+\sqrt{(r_{0}-r_{\rm s}) r_{0}}-\frac{r_{\rm s}}{2}\right) = {\rm finite}$$
For the stationary bookkeeper this is still larger than $$r_{0}-r_{\rm s}$$, but smaller than $$\infty$$.
If you are in freefall with the negative escape velocity the gravitational depth expansion also exactly cancels out with the kinematic length contraction since $$v_{\rm esc}=c \sqrt{r_{\rm s}/r}$$, therefore $$|g_{rr}|$$ in raindrop coordinates is $$1$$, and the proper distance becomes exactly the coordinate distance. | 515 | 1,735 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-50 | latest | en | 0.838682 |
https://fr.mathworks.com/matlabcentral/answers/491684-streamline-plot-not-complete | 1,579,546,058,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599718.13/warc/CC-MAIN-20200120165335-20200120194335-00536.warc.gz | 460,898,307 | 20,087 | # Streamline plot not complete
5 views (last 30 days)
Eddward on 18 Nov 2019
Commented: Eddward on 18 Nov 2019
I have been trying to make this streamline plot, unfortunately I have not been able to make it complete. I have looked on different forum pages but was not able to find a solution.
My code:
clear, clc
Gamma=-4;
a=1;
b=1;
Q=1;
[x,y]=meshgrid(-1:0.05:4,-1:0.05:1);
u= (y./(2*pi.*(x.^2+y.^2)))...
+(((2*Q.*b)/((a^2)*pi.*Gamma))...
.*((2*y.*(x-1))./((x.^2+y.^2-2.*x+1).^2)));
v= -((x)./(2.*pi.*(x.^2+y.^2)))...
-(((2*Q*b)./((a^2)*pi.*Gamma))...
*(((x.^2-y.^2-2.*x+1)./((x.^2+y.^2-2.*x+1).^2))));
hold on , clf
figure(1)
quiver(x,y,u,v)
N = 15;
startx = max(x).*rand(N,1);
starty = max(y).*rand(N,1);
streamline(x,y,u,v,startx,starty)
startx2 = -max(x).*rand(N,1);
starty2 = -max(y).*rand(N,1);
streamline(x,y,u,v,startx2,starty2)
startx3 = rand(N,1);
starty3 = rand(N,1);
streamline(x,y,u,v,startx3,starty3)
hold off
darova on 18 Nov 2019
Works ok for me
N = 15;
xx = linspace(min(x(:)),max(x(:)),10);
yy = linspace(min(y(:)),max(y(:)),10);
[startx,starty] = meshgrid(xx,yy);
streamline(x,y,u,v,startx,starty)
#### 1 Comment
Eddward on 18 Nov 2019
Thank you for your response, had to change 10 to 50 for my plot.
Thank you very much. | 486 | 1,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-05 | latest | en | 0.681962 |
https://www.digitmath.com/tangent-graph-trigonometry.html | 1,718,701,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00045.warc.gz | 649,235,918 | 6,003 | # The Trigonometry of a Tangent Graph
The graph of a tangent function is periodic with period π and differs from the 2π period of the sine graph and cosine graph.
Tangent graph showing its period π:
To understand why the tangent function graph has a period of π we first look at the definition of the tangent function:
tan θ = sin θ / cos θ
This definition of tangent is identical to the definition provided by SOH-CAH-TOA:
1. sin θ = opposite / hypotenuse = o / h
2. cos θ = adjacent / hypotenuse = a / h
3. tan θ = sin θ / cos θ = (o / h) / (a / h) = opposite / adjacent
The numerator of the tangent function, the sin θ, is zero when θ = 0, ±π, ±2π … and so on. At each of these sine values the tangent equals zero and is when a tangent graph line intersects the horizontal x-axis.
The cos θ is the denominator of the tangent function. The cos θ is zero when θ = ±π/2, ±3π/2, ±5π/2, … and so on. The tangent is not defined for these values of θ since a fraction denominator of zero is undefined math and explains why the tangent graph lines approach, but never reach cos θ = 0 value.
From the tangent function definition it can also be seen that when the sin θ = cos θ, at π/4 radians (45°), the tan θ equals 1. Then, for the interval 0 ≤ θ < π/4 the tangent is less than 1 and for the interval π/4 < θ < π/2 the tangent is greater than 1.
Because the sin θ is an odd function and the cos θ is an even function the tangent is an odd function; odd divided by even is odd. Then, for all θ in the domain of the tangent:
1. tan (−θ) = sin (−θ) / cos (−θ) = −sin θ / cos θ = −tan θ
2. tan (θ + π) = sin (θ + π) / cos (θ + π) = −sin θ / −cos θ = tan θ
From these equations we see that π is the smallest positive number p for which the tan (θ + p) = tan θ, because tan 0 = tan π = 0. Therefore, the tangent function has period π, and a tangent graph repeats every π units.
Top of Page | 533 | 1,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.843052 |
https://forum.theotown.com/viewtopic.php?p=138449 | 1,582,870,312,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147054.34/warc/CC-MAIN-20200228043124-20200228073124-00538.warc.gz | 365,905,673 | 15,756 | ## TheoTown Scientific Chatbox
Here you can talk about everything you want.
soumya-8974
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### TheoTown Scientific Chatbox
Use this topic to discuss random things related to the science. Have attention!
Rules:
1) Use CGS and SI only. Using other system of units might cause to ban.
2) For Kelvin, don't use 300°K, use 300 K.
3) For cubic centimetre, square centimetre and second, don't use cc, sq cm and sec. Use cm^3, cm^2 and s, respectively.
4) Use Roman script for units, not Italic script.
5) Don't use 's' or 'es' for plural units.
6) For Division, don't use slash (/) too much.
7) Don't use dots (.) for the acronym of units.
8) For Multiplication, don't use multiply symbol (X), use dots (•).
–Yours sincerely, Soumyabrata
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### Re: TheoTown Scientific Chatbox
XxX=20? Solve for X.
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### Re: TheoTown Scientific Chatbox
CommanderABab wrote:
Tue Jan 15, 2019 7:15
XxX=20? Solve for X.
X×X = 20
X = 20×X
X = 20X
X = 20/2
X = 10
10×10=20
20-10=10
X=10 👌
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### Re: TheoTown Scientific Chatbox
Yeah, but if X was 10, then 10x10 would be 100.
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### Re: TheoTown Scientific Chatbox
CommanderABab wrote:
Tue Jan 15, 2019 8:58
Yeah, but if X was 10, then 10x10 would be 100.
Yeah, my Maths is F
I hate Maths so much
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### Re: TheoTown Scientific Chatbox
@CommanderABab and @Ahmad Nur Aizat, you are breaking my rules. Avoid using multiply symbol, use dots.
–Yours sincerely, Soumyabrata
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### Re: TheoTown Scientific Chatbox
CommanderABab
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### Re: TheoTown Scientific Chatbox
soumya-8974 wrote:
Tue Jan 15, 2019 15:41
@CommanderABab and @Ahmad Nur Aizat, you are breaking my rules. Avoid using multiply symbol, use dots.
We did it on purpose I think. Are we banned from the chat?
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### Re: TheoTown Scientific Chatbox
CommanderABab wrote:
Tue Jan 15, 2019 18:21
soumya-8974 wrote:
Tue Jan 15, 2019 15:41
@CommanderABab and @Ahmad Nur Aizat, you are breaking my rules. Avoid using multiply symbol, use dots.
We did it on purpose I think. Are we banned from the chat?
Oof, I didn't realise it. Are we gonna get banned?
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### Re: TheoTown Scientific Chatbox
Probably not,
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### Re: TheoTown Scientific Chatbox
Albert Einstein has joined the server
FUTURE NFL WR OR QB
H-TOWN BOI
The Astros didn't cheat!
...
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### Re: TheoTown Scientific Chatbox
SethHarris777 wrote:
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Albert Einstein has joined the server
E=mc2
yes
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### Re: TheoTown Scientific Chatbox
X= 12, Y=4 verify that X + Y = Y + X
Proud to be the player of theotown since 1.0.45 alpha.
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### Re: TheoTown Scientific Chatbox
Albert Einstein just had heart break from these scientific talks
I can't define my self time to time... Just do not %100 trust me. Texting that honestly...
...
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### Re: TheoTown Scientific Chatbox
CommanderABab wrote:
Tue Jan 15, 2019 7:15
XxX=20? Solve for X.
√20
Isn't it? √20x√20=20
√20x√20=√400
√400=20
I can't define my self time to time... Just do not %100 trust me. Texting that honestly...
...
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### Re: TheoTown Scientific Chatbox
No but you could write it:
Solve for x
Screenshot_20190123-160950.jpg (10.17 KiB) Viewed 1285 times
ps890giv psdl8bzi<
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### Re: TheoTown Scientific Chatbox
CommanderABab wrote:
Wed Jan 23, 2019 23:32
No but you could write it:
Screenshot_20190123-160950.jpg
It is same as mine too, isn't it
I can't define my self time to time... Just do not %100 trust me. Texting that honestly...
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### Re: TheoTown Scientific Chatbox
What are helmholtz cavities? I did some research on the Dyson Air Multiplier, but I came across helmholtz cavities when I researched about the second generation, engineers used helmholtz cavities and increased air pressure to make them act as silencers.
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### Re: TheoTown Scientific Chatbox
Find value of log80.
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### Re: TheoTown Scientific Chatbox
Roche Limit
When you get to be called a Board Index because you are too good.>:3
### Who is online
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| 2,249 | 6,807 | {"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} | 2.8125 | 3 | CC-MAIN-2020-10 | latest | en | 0.753475 |
https://planbee.com/blogs/news/would-you-rather-maths-reasoning-questions-3 | 1,718,703,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00110.warc.gz | 405,891,620 | 69,025 | # Would you rather…? Maths reasoning questions № 3
Here's another Maths reasoning question for KS1 or KS2 Maths lessons. It's ideal for use as a starter activity or as a puzzler to settle children as they arrive. It's the third in our new series of 'Would you rather…?' Maths reasoning questions.
These fun, visual problems are challenging, but open to interpretation by your pupils. For many of them, there is no 'right answer', but a choice of answers which children must first identify and choose from, then justify their choice using maths.
You'll find an explanation of the Maths involved in this 'Would you rather…?' Maths reasoning question directly underneath it, as well as links to all of the other questions in this problem-solving series.
## 'Would you rather…?' Maths reasoning question № 3:
Would you rather have £8 squared or £4 cubed?
### Teacher's Notes
The 'solution':
8 x 8 = 64
4 x 4 x 4 = 64
Both purses contain the same amount.
You could challenge children to find the only other pair of smaller, whole square numbers and cube numbers that are equal (£4 squared and £2 cubed). Alternatively, challenge children to find other combinations of a square number and a cube number that have only a small difference between them, e.g. 5 cubed (75) and 9 squared (81). Who can find a combination with the smallest difference between them? \
Check out the other 'Would you rather…?' Maths reasoning questions.
Looking for more Maths Problem Solving lessons and plans? Check out our ready-to-teach collection here. | 348 | 1,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.929268 |
http://gams.cam.nist.gov/10.75 | 1,495,557,578,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607648.39/warc/CC-MAIN-20170523163634-20170523183634-00449.warc.gz | 143,181,177 | 22,261 | # §10.75 Tables
## §10.75(i) Introduction
Comprehensive listings and descriptions of tables of the functions treated in this chapter are provided in Bateman and Archibald (1944), Lebedev and Fedorova (1960), Fletcher et al. (1962), and Luke (1975, §9.13.2). Only a few of the more comprehensive of these early tables are included in the listings in the following subsections. Also, for additional listings of tables pertaining to complex arguments see Babushkina et al. (1997).
## §10.75(ii) Bessel Functions and their Derivatives
• British Association for the Advancement of Science (1937) tabulates $\mathop{J_{0}\/}\nolimits\!\left(x\right)$, $\mathop{J_{1}\/}\nolimits\!\left(x\right)$, $x=0(.001)16(.01)25$, 10D; $\mathop{Y_{0}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(x\right)$, $x=0.01(.01)25$, 8–9S or 8D. Also included are auxiliary functions to facilitate interpolation of the tables of $\mathop{Y_{0}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(x\right)$ for small values of $x$, as well as auxiliary functions to compute all four functions for large values of $x$.
• Bickley et al. (1952) tabulates $\mathop{J_{n}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$ or $x^{n}\mathop{Y_{n}\/}\nolimits\!\left(x\right)$, $n=2(1)20$, $x=0$($.01$ or $.1$) $10(.1)25$, 8D (for $\mathop{J_{n}\/}\nolimits\!\left(x\right)$), 8S (for $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$ or $x^{n}\mathop{Y_{n}\/}\nolimits\!\left(x\right)$); $\mathop{J_{n}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$, $n=0(1)20$, $x=0$ or $0.1(.1)25$, 10D (for $\mathop{J_{n}\/}\nolimits\!\left(x\right)$), 10S (for $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$).
• Olver (1962) provides tables for the uniform asymptotic expansions given in §10.20(i), including $\zeta$ and $(\ifrac{4\zeta}{(1-x^{2})})^{\frac{1}{4}}$ as functions of $x$ $(=z)$ and the coefficients $A_{k}(\zeta)$, $B_{k}(\zeta)$, $C_{k}(\zeta)$, $D_{k}(\zeta)$ as functions of $\zeta$. These enable $\mathop{J_{\nu}\/}\nolimits\!\left(\nu x\right)$, $\mathop{Y_{\nu}\/}\nolimits\!\left(\nu x\right)$, $\mathop{J_{\nu}\/}\nolimits'\!\left(\nu x\right)$, $\mathop{Y_{\nu}\/}\nolimits'\!\left(\nu x\right)$ to be computed to 10S when $\nu\geq 15$, except in the neighborhoods of zeros.
• The main tables in Abramowitz and Stegun (1964, Chapter 9) give $\mathop{J_{0}\/}\nolimits\!\left(x\right)$ to 15D, $\mathop{J_{1}\/}\nolimits\!\left(x\right)$, $\mathop{J_{2}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{0}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(x\right)$ to 10D, $\mathop{Y_{2}\/}\nolimits\!\left(x\right)$ to 8D, $x=0(.1)17.5$; $\mathop{Y_{n}\/}\nolimits\!\left(x\right)-(2/\pi)\mathop{J_{n}\/}\nolimits\!% \left(x\right)\mathop{\ln\/}\nolimits x$, $n=0,1$, $x=0(.1)2$, 8D; $\mathop{J_{n}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$, $n=3(1)9$, $x=0(.2)20$, 5D or 5S; $\mathop{J_{n}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$, $n=0(1)20(10)50,100$, $x=1,2,5,10,50,100$, 10S; modulus and phase functions $\sqrt{x}\mathop{M_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\theta_{n}\/}\nolimits\!\left(x\right)-x$, $n=0,1,2$, $\ifrac{1}{x}=0(.01)0.1$, 8D.
• Achenbach (1986) tabulates $\mathop{J_{0}\/}\nolimits\!\left(x\right)$, $\mathop{J_{1}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{0}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(x\right)$, $x=0(.1)8$, 20D or 18–20S.
• Zhang and Jin (1996, pp. 185–195) tabulates $\mathop{J_{n}\/}\nolimits\!\left(x\right)$, $\mathop{J_{n}\/}\nolimits'\!\left(x\right)$, $\mathop{Y_{n}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{n}\/}\nolimits'\!\left(x\right)$, $n=0(1)10(10)50,100$, $x=1$, 5, 10, 25, 50, 100, 9S; $\mathop{J_{n+\alpha}\/}\nolimits\!\left(x\right)$, $\mathop{J_{n+\alpha}\/}\nolimits'\!\left(x\right)$, $\mathop{Y_{n+\alpha}\/}\nolimits\!\left(x\right)$, $\mathop{Y_{n+\alpha}\/}\nolimits'\!\left(x\right)$, $n=0(1)5,10,30,50,100$, $\alpha=\tfrac{1}{4},\tfrac{1}{3},\tfrac{1}{2},\tfrac{2}{3},\tfrac{3}{4}$, $x=1,5,10,50$, 8S; real and imaginary parts of $\mathop{J_{n+\alpha}\/}\nolimits\!\left(z\right)$, $\mathop{J_{n+\alpha}\/}\nolimits'\!\left(z\right)$, $\mathop{Y_{n+\alpha}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{n+\alpha}\/}\nolimits'\!\left(z\right)$, $n=0(1)15,20(10)50,100$, $\alpha=0,\tfrac{1}{2}$, $z=4+2i$, $20+10i$, 8S.
## §10.75(iii) Zeros and Associated Values of the Bessel Functions, Hankel Functions, and their Derivatives
### Real Zeros
• British Association for the Advancement of Science (1937) tabulates $j_{0,m}$, $\mathop{J_{1}\/}\nolimits\!\left(j_{0,m}\right)$, $j_{1,m}$, $\mathop{J_{0}\/}\nolimits\!\left(j_{1,m}\right)$, $m=1(1)150$, 10D; $y_{0,m}$, $\mathop{Y_{1}\/}\nolimits\!\left(y_{0,m}\right)$, $y_{1,m}$, $\mathop{Y_{0}\/}\nolimits\!\left(y_{1,m}\right)$, $m=1(1)50$, 8D.
• Olver (1960) tabulates $j_{n,m}$, $\mathop{J_{n}\/}\nolimits'\!\left(j_{n,m}\right)$, ${j^{\prime}_{n,m}}$, $\mathop{J_{n}\/}\nolimits\!\left({j^{\prime}_{n,m}}\right)$, $y_{n,m}$, $\mathop{Y_{n}\/}\nolimits'\!\left(y_{n,m}\right)$, ${y^{\prime}_{n,m}}$, $\mathop{Y_{n}\/}\nolimits\!\left({y^{\prime}_{n,m}}\right)$, $n=0(\tfrac{1}{2})20\tfrac{1}{2}$, $m=1(1)50$, 8D. Also included are tables of the coefficients in the uniform asymptotic expansions of these zeros and associated values as $n\to\infty$; see §10.21(viii), and more fully Olver (1954).
• Morgenthaler and Reismann (1963) tabulates ${j^{\prime}_{n,m}}$ for $n=21(1)51$ and ${j^{\prime}_{n,m}}{<100}$, 7-10S.
• Abramowitz and Stegun (1964, Chapter 9) tabulates $j_{n,m}$, $\mathop{J_{n}\/}\nolimits'\!\left(j_{n,m}\right)$, ${j^{\prime}_{n,m}}$, $\mathop{J_{n}\/}\nolimits\!\left({j^{\prime}_{n,m}}\right)$, $n=0(1)8$, $m=1(1)20$, 5D (10D for $n=0$), $y_{n,m}$, $\mathop{Y_{n}\/}\nolimits'\!\left(y_{n,m}\right)$, ${y^{\prime}_{n,m}}$, $\mathop{Y_{n}\/}\nolimits\!\left({y^{\prime}_{n,m}}\right)$, $n=0(1)8$, $m=1(1)20$, 5D (8D for $n=0$), $\mathop{J_{0}\/}\nolimits\!\left(j_{0,m}x\right)$, $m=1(1)5$, $x=0(.02)1$, 5D. Also included are the first 5 zeros of the functions $x\mathop{J_{1}\/}\nolimits\!\left(x\right)-\lambda\mathop{J_{0}\/}\nolimits\!% \left(x\right)$, $\mathop{J_{1}\/}\nolimits\!\left(x\right)-\lambda x\mathop{J_{0}\/}\nolimits\!% \left(x\right)$, $\mathop{J_{0}\/}\nolimits\!\left(x\right)\mathop{Y_{0}\/}\nolimits\!\left(% \lambda x\right)-\mathop{Y_{0}\/}\nolimits\!\left(x\right)\mathop{J_{0}\/}% \nolimits\!\left(\lambda x\right)$, $\mathop{J_{1}\/}\nolimits\!\left(x\right)\mathop{Y_{1}\/}\nolimits\!\left(% \lambda x\right)-\mathop{Y_{1}\/}\nolimits\!\left(x\right)\mathop{J_{1}\/}% \nolimits\!\left(\lambda x\right)$, $\mathop{J_{1}\/}\nolimits\!\left(x\right)\mathop{Y_{0}\/}\nolimits\!\left(% \lambda x\right)-\mathop{Y_{1}\/}\nolimits\!\left(x\right)\mathop{J_{0}\/}% \nolimits\!\left(\lambda x\right)$ for various values of $\lambda$ and $\lambda^{-1}$ in the interval $[0,1]$, 4–8D.
• Abramowitz and Stegun (1964, Chapter 10) tabulates $j_{\nu,m}$, $\mathop{J_{\nu}\/}\nolimits'\!\left(j_{\nu,m}\right)$, ${j^{\prime}_{\nu,m}}$, $\mathop{J_{\nu}\/}\nolimits\!\left({j^{\prime}_{\nu,m}}\right)$, $y_{\nu,m}$, $\mathop{Y_{\nu}\/}\nolimits'\!\left(y_{\nu,m}\right)$, ${y^{\prime}_{\nu,m}}$, $\mathop{Y_{\nu}\/}\nolimits\!\left({y^{\prime}_{\nu,m}}\right)$, $\nu=\tfrac{1}{2}(1)19\tfrac{1}{2}$, $m=1(1)m_{\nu}$, where $m_{\nu}$ ranges from 8 at $\nu=\tfrac{1}{2}$ down to 1 at $\nu=19\tfrac{1}{2}$, 6–7D.
• Makinouchi (1966) tabulates all values of $j_{\nu,m}$ and $y_{\nu,m}$ in the interval $(0,100)$, with at least 29S. These are for $\nu=0(1)5$, 10, 20; $\nu=\tfrac{3}{2}$, $\tfrac{5}{2}$; $\nu=m/n$ with $m=1(1)n-1$ and $n=3(1)8$, except for $\nu=\tfrac{1}{2}$.
• Döring (1971) tabulates the first 100 values of $\nu$ $(>1)$ for which $\mathop{J_{-\nu}\/}\nolimits'\!\left(x\right)$ has the double zero $x=\nu$, 10D.
• Heller (1976) tabulates $j_{0,m}$, $\mathop{J_{1}\/}\nolimits\!\left(j_{0,m}\right)$, $j_{1,m}$, $\mathop{J_{0}\/}\nolimits\!\left(j_{1,m}\right)$, ${j^{\prime}_{1,m}}$, $\mathop{J_{1}\/}\nolimits\!\left({j^{\prime}_{1,m}}\right)$ for $m=1(1)100$, 25D.
• Wills et al. (1982) tabulates $j_{0,m}$, $j_{1,m}$, $y_{0,m}$, $y_{1,m}$ for $m=1(1)30$, 35D.
• Kerimov and Skorokhodov (1985c) tabulates 201 double zeros of $\mathop{J_{-\nu}\/}\nolimits''\!\left(x\right)$, 10 double zeros of $\mathop{J_{-\nu}\/}\nolimits'''\!\left(x\right)$, 101 double zeros of $\mathop{Y_{-\nu}\/}\nolimits'\!\left(x\right)$, 201 double zeros of $\mathop{Y_{-\nu}\/}\nolimits''\!\left(x\right)$, and 10 double zeros of $\mathop{Y_{-\nu}\/}\nolimits'''\!\left(x\right)$, all to 8 or 9D.
• Zhang and Jin (1996, pp. 196–198) tabulates $j_{n,m}$, ${j^{\prime}_{n,m}}$, $y_{n,m}$, ${y^{\prime}_{n,m}}$, $n=0(1)3$, $m=1(1)10$, 8D; the first five zeros of $\mathop{J_{n}\/}\nolimits\!\left(x\right)\mathop{Y_{n}\/}\nolimits\!\left(% \lambda x\right)-\mathop{J_{n}\/}\nolimits\!\left(\lambda x\right)\mathop{Y_{n% }\/}\nolimits\!\left(x\right)$, $\mathop{J_{n}\/}\nolimits'\!\left(x\right)\mathop{Y_{n}\/}\nolimits'\!\left(% \lambda x\right)-\mathop{J_{n}\/}\nolimits'\!\left(\lambda x\right)\mathop{Y_{% n}\/}\nolimits'\!\left(x\right)$, $n=0,1,2$, $\lambda=1.1(.1)1.6,1.8,2(.5)5$, 7D.
### Complex Zeros
• Abramowitz and Stegun (1964, p. 373) tabulates the three smallest zeros of $\mathop{Y_{0}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits'\!\left(z\right)$ in the sector $0<\mathop{\mathrm{ph}\/}\nolimits z\leq\pi$, together with the corresponding values of $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{0}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, respectively, to 9D. (There is an error in the value of $\mathop{Y_{0}\/}\nolimits\!\left(z\right)$ at the 3rd zero of $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$: the last four digits should be 2533; see Amos (1985).)
• Döring (1966) tabulates all zeros of $\mathop{Y_{0}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, $\mathop{{H^{(1)}_{0}}\/}\nolimits\!\left(z\right)$, $\mathop{{H^{(1)}_{1}}\/}\nolimits\!\left(z\right)$, that lie in the sector $|z|<158$, $|\mathop{\mathrm{ph}\/}\nolimits z|\leq\pi$, to 10D. Some of the smaller zeros of $\mathop{Y_{n}\/}\nolimits\!\left(z\right)$ and $\mathop{{H^{(1)}_{n}}\/}\nolimits\!\left(z\right)$ for $n=2,3,4,5,15$ are also included.
• Kerimov and Skorokhodov (1985a) tabulates 5 (nonreal) complex conjugate pairs of zeros of the principal branches of $\mathop{Y_{n}\/}\nolimits\!\left(z\right)$ and $\mathop{Y_{n}\/}\nolimits'\!\left(z\right)$ for $n=0(1)5$, 8D.
• Kerimov and Skorokhodov (1985b) tabulates 50 zeros of the principal branches of $\mathop{{H^{(1)}_{0}}\/}\nolimits\!\left(z\right)$ and $\mathop{{H^{(1)}_{1}}\/}\nolimits\!\left(z\right)$, 8D.
• Kerimov and Skorokhodov (1987) tabulates 100 complex double zeros $\nu$ of $\mathop{Y_{\nu}\/}\nolimits'\!\left(ze^{-\pi i}\right)$ and $\mathop{{H^{(1)}_{\nu}}\/}\nolimits'\!\left(ze^{-\pi i}\right)$, 8D.
• MacDonald (1989) tabulates the first 30 zeros, in ascending order of absolute value in the fourth quadrant, of the function $\mathop{J_{0}\/}\nolimits\!\left(z\right)-i\mathop{J_{1}\/}\nolimits\!\left(z\right)$, 6D. (Other zeros of this function can be obtained by reflection in the imaginary axis).
• Zhang and Jin (1996, p. 199) tabulates the real and imaginary parts of the first 15 conjugate pairs of complex zeros of $\mathop{Y_{0}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits'\!\left(z\right)$ and the corresponding values of $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{0}\/}\nolimits\!\left(z\right)$, $\mathop{Y_{1}\/}\nolimits\!\left(z\right)$, respectively, 10D.
## §10.75(iv) Integrals of Bessel Functions
• Abramowitz and Stegun (1964, Chapter 11) tabulates $\int_{0}^{x}\mathop{J_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $\int_{0}^{x}\mathop{Y_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $x=0(.1)10$, 10D; $\int_{0}^{x}t^{-1}(1-\mathop{J_{0}\/}\nolimits\!\left(t\right))\mathrm{d}t$, $\int_{x}^{\infty}t^{-1}\mathop{Y_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $x=0(.1)5$, 8D.
• Zhang and Jin (1996, p. 270) tabulates $\int_{0}^{x}\mathop{J_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $\int_{0}^{x}t^{-1}(1-\mathop{J_{0}\/}\nolimits\!\left(t\right))\mathrm{d}t$, $\int_{0}^{x}\mathop{Y_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $\int_{x}^{\infty}t^{-1}\mathop{Y_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $x=0(.1)1(.5)20$, 8D.
## §10.75(v) Modified Bessel Functions and their Derivatives
• British Association for the Advancement of Science (1937) tabulates $\mathop{I_{0}\/}\nolimits\!\left(x\right)$, $\mathop{I_{1}\/}\nolimits\!\left(x\right)$, $x=0(.001)5$, 7–8D; $\mathop{K_{0}\/}\nolimits\!\left(x\right)$, $\mathop{K_{1}\/}\nolimits\!\left(x\right)$, $x=0.01(.01)5$, 7–10D; $e^{-x}\mathop{I_{0}\/}\nolimits\!\left(x\right)$, $e^{-x}\mathop{I_{1}\/}\nolimits\!\left(x\right)$, $e^{x}\mathop{K_{0}\/}\nolimits\!\left(x\right)$, $e^{x}\mathop{K_{1}\/}\nolimits\!\left(x\right)$, $x=5(.01)10(.1)20$, 8D. Also included are auxiliary functions to facilitate interpolation of the tables of $\mathop{K_{0}\/}\nolimits\!\left(x\right)$, $\mathop{K_{1}\/}\nolimits\!\left(x\right)$ for small values of $x$.
• Bickley et al. (1952) tabulates $x^{-n}\mathop{I_{n}\/}\nolimits\!\left(x\right)$ or $e^{-x}\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $x^{n}\mathop{K_{n}\/}\nolimits\!\left(x\right)$ or $e^{x}\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $n=2(1)20$, $x=0$(.01 or .1) 10(.1) 20, 8S; $\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $n=0(1)20$, $x=0$ or $0.1(.1)20$, 10S.
• Olver (1962) provides tables for the uniform asymptotic expansions given in §10.41(ii), including $\eta$ and the coefficients $U_{k}(p)$, $V_{k}(p)$ as functions of $p=(1+x^{2})^{-\frac{1}{2}}$. These enable $\mathop{I_{\nu}\/}\nolimits\!\left(\nu x\right)$, $\mathop{K_{\nu}\/}\nolimits\!\left(\nu x\right)$, $\mathop{I_{\nu}\/}\nolimits'\!\left(\nu x\right)$, $\mathop{K_{\nu}\/}\nolimits'\!\left(\nu x\right)$ to be computed to 10S when $\nu\geq 16$.
• The main tables in Abramowitz and Stegun (1964, Chapter 9) give $e^{-x}\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $e^{x}\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $n=0,1,2$, $x=0(.1)10(.2)20$, 8D–10D or 10S; $\sqrt{x}e^{-x}\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $(\sqrt{x}/\pi)$ $e^{x}\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $n=0,1,2$, $1/x=0(.002)0.05$; $\mathop{K_{0}\/}\nolimits\!\left(x\right)+\mathop{I_{0}\/}\nolimits\!\left(x% \right)\mathop{\ln\/}\nolimits x$, $x(\mathop{K_{1}\/}\nolimits\!\left(x\right)-\mathop{I_{1}\/}\nolimits\!\left(x% \right)\mathop{\ln\/}\nolimits x)$, $x=0(.1)2$, 8D; $e^{-x}\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $e^{x}\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $n=3(1)9$, $x=0(.2)10(.5)20$, 5S; $\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $n=0(1)20(10)50,100$, $x=1,2,5,10,50,100$, 9–10S.
• Achenbach (1986) tabulates $\mathop{I_{0}\/}\nolimits\!\left(x\right)$, $\mathop{I_{1}\/}\nolimits\!\left(x\right)$, $\mathop{K_{0}\/}\nolimits\!\left(x\right)$, $\mathop{K_{1}\/}\nolimits\!\left(x\right)$, $x=0(.1)8$, 19D or 19–21S.
• Zhang and Jin (1996, pp. 240–250) tabulates $\mathop{I_{n}\/}\nolimits\!\left(x\right)$, $\mathop{I_{n}\/}\nolimits'\!\left(x\right)$, $\mathop{K_{n}\/}\nolimits\!\left(x\right)$, $\mathop{K_{n}\/}\nolimits'\!\left(x\right)$, $n=0(1)10(10)50,100$, $x=1,5,10,25,50,100$, 9S; $\mathop{I_{n+\alpha}\/}\nolimits\!\left(x\right)$, $\mathop{I_{n+\alpha}\/}\nolimits'\!\left(x\right)$, $\mathop{K_{n+\alpha}\/}\nolimits\!\left(x\right)$, $\mathop{K_{n+\alpha}\/}\nolimits'\!\left(x\right)$, $n=0(1)5$, 10, 30, 50, 100, $\alpha=\tfrac{1}{4}$, $\tfrac{1}{3}$, $\tfrac{1}{2}$, $\tfrac{2}{3}$, $\tfrac{3}{4}$, $x=1$, 5, 10, 50, 8S; real and imaginary parts of $\mathop{I_{n+\alpha}\/}\nolimits\!\left(z\right)$, $\mathop{I_{n+\alpha}\/}\nolimits'\!\left(z\right)$, $\mathop{K_{n+\alpha}\/}\nolimits\!\left(z\right)$, $\mathop{K_{n+\alpha}\/}\nolimits'\!\left(z\right)$, $n=0(1)15$, 20(10)50, 100, $\alpha=0,\tfrac{1}{2}$, $z=4+2i,20+10i$, 8S.
## §10.75(vi) Zeros of Modified Bessel Functions and their Derivatives
• Parnes (1972) tabulates all zeros of the principal value of $\mathop{K_{n}\/}\nolimits\!\left(z\right)$, for $n=2(1)10$, 9D.
• Leung and Ghaderpanah (1979), tabulates all zeros of the principal value of $\mathop{K_{n}\/}\nolimits\!\left(z\right)$, for $n=2(1)10$, 29S.
• Kerimov and Skorokhodov (1984b) tabulates all zeros of the principal values of $\mathop{K_{n}\/}\nolimits\!\left(z\right)$ and $\mathop{K_{n}\/}\nolimits'\!\left(z\right)$, for $n=2(1)20$, 9S.
• Kerimov and Skorokhodov (1984c) tabulates all zeros of $\mathop{I_{-n-\frac{1}{2}}\/}\nolimits\!\left(z\right)$ and $\mathop{I_{-n-\frac{1}{2}}\/}\nolimits'\!\left(z\right)$ in the sector $0\leq\mathop{\mathrm{ph}\/}\nolimits z\leq\tfrac{1}{2}\pi$ for $n=1(1)20$, 9S.
• Kerimov and Skorokhodov (1985b) tabulates all zeros of $\mathop{K_{n}\/}\nolimits\!\left(z\right)$ and $\mathop{K_{n}\/}\nolimits'\!\left(z\right)$ in the sector $-\tfrac{1}{2}\pi<\mathop{\mathrm{ph}\/}\nolimits z\leq\tfrac{3}{2}\pi$ for $n=0(1)5$, 8D.
## §10.75(vii) Integrals of Modified Bessel Functions
• Abramowitz and Stegun (1964, Chapter 11) tabulates $e^{-x}\int_{0}^{x}\mathop{I_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $e^{x}\int_{x}^{\infty}\mathop{K_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $x=0(.1)10$, 7D; $e^{-x}\int_{0}^{x}t^{-1}(\mathop{I_{0}\/}\nolimits\!\left(t\right)-1)\mathrm{d}t$, $xe^{x}\int_{x}^{\infty}t^{-1}\mathop{K_{0}\/}\nolimits\!\left(t\right)\mathrm{% d}t$, $x=0(.1)5$, 6D.
• Bickley and Nayler (1935) tabulates $\mathop{\mathrm{Ki}_{n}\/}\nolimits\!\left(x\right)$10.43(iii)) for $n=1(1)16$, $x=0(.05)0.2(.1)$ 2, 3, 9D.
• Zhang and Jin (1996, p. 271) tabulates $e^{-x}\int_{0}^{x}\mathop{I_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $e^{-x}\int_{0}^{x}t^{-1}(\mathop{I_{0}\/}\nolimits\!\left(t\right)-1)\mathrm{d}t$, $e^{x}\int_{x}^{\infty}\mathop{K_{0}\/}\nolimits\!\left(t\right)\mathrm{d}t$, $xe^{x}\int_{x}^{\infty}t^{-1}\mathop{K_{0}\/}\nolimits\!\left(t\right)\mathrm{% d}t$, $x=0(.1)1(.5)20$, 8D.
## §10.75(viii) Modified Bessel Functions of Imaginary or Complex Order
For the notation see §10.45.
• Žurina and Karmazina (1967) tabulates $\mathop{\widetilde{K}_{\nu}\/}\nolimits\!\left(x\right)$ for $\nu=0.01(.01)10$, $x=0.1(.1)10.2$, 7S.
• Rappoport (1979) tabulates the real and imaginary parts of $\mathop{K_{\frac{1}{2}+i\tau}\/}\nolimits\!\left(x\right)$ for $\tau=0.01(.01)10$, $x=0.1(.2)9.5$, 7S.
## §10.75(ix) Spherical Bessel Functions, Modified Spherical Bessel Functions, and their Derivatives
• The main tables in Abramowitz and Stegun (1964, Chapter 10) give $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(x\right)$ $n=0(1)8$, $x=0(.1)10$, 5–8S; $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(x\right)$ $n=0(1)20(10)50$, 100, $x=1,2,5,10,50,100$, 10S; $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(x\right)$, $n=0,1,2$, $x=0(.1)5$, 4–9D; $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(x\right)$, $n=0(1)20(10)50$, 100, $x=1,2,5,10,50,100$, 10S. (For the notation see §10.1 and §10.47(ii).)
• Zhang and Jin (1996, pp. 296–305) tabulates $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{j}_{n}\/}\nolimits'\!\left(x\right)$, $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{y}_{n}\/}\nolimits'\!\left(x\right)$, $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(x\right)$, $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits'\!\left(x\right)$, $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\mathsf{k}_{n}\/}\nolimits'\!\left(x\right)$, $n=0(1)10(10)30$, 50, 100, $x=1$, 5, 10, 25, 50, 100, 8S; $x\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(x\right)$, $(x\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(x\right))^{\prime}$, $x\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(x\right)$, $(x\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(x\right))^{\prime}$ (Riccati–Bessel functions and their derivatives), $n=0(1)10(10)30$, 50, 100, $x=1$, 5, 10, 25, 50, 100, 8S; real and imaginary parts of $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(z\right)$, $\mathop{\mathsf{j}_{n}\/}\nolimits'\!\left(z\right)$, $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(z\right)$, $\mathop{\mathsf{y}_{n}\/}\nolimits'\!\left(z\right)$, $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(z\right)$, $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits'\!\left(z\right)$, $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(z\right)$, $\mathop{\mathsf{k}_{n}\/}\nolimits'\!\left(z\right)$, $n=0(1)15$, 20(10)50, 100, $z=4+2i$, $20+10i$, 8S. (For the notation replace $j,y,i,k$ by $\mathop{\mathsf{j}\/}\nolimits$, $\mathop{\mathsf{y}\/}\nolimits$, $\mathop{{\mathsf{i}^{(1)}}\/}\nolimits$, $\mathop{\mathsf{k}\/}\nolimits$, respectively.)
## §10.75(x) Zeros and Associated Values of Derivatives of Spherical Bessel Functions
For the notation see §10.58.
• Olver (1960) tabulates $a^{\prime}_{n,m}$, $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(a^{\prime}_{n,m}\right)$, $b^{\prime}_{n,m}$, $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(b^{\prime}_{n,m}\right)$, $n=1(1)20$, $m=1(1)50$, 8D. Also included are tables of the coefficients in the uniform asymptotic expansions of these zeros and associated values as $n\to\infty$.
## §10.75(xi) Kelvin Functions and their Derivatives
• Young and Kirk (1964) tabulates $\mathop{\mathrm{ber}_{n}\/}\nolimits x$, $\mathop{\mathrm{bei}_{n}\/}\nolimits x$, $\mathop{\mathrm{ker}_{n}\/}\nolimits x$, $\mathop{\mathrm{kei}_{n}\/}\nolimits x$, $n=0,1$, $x=0(.1)10$, 15D; $\mathop{\mathrm{ber}_{n}\/}\nolimits x$, $\mathop{\mathrm{bei}_{n}\/}\nolimits x$, $\mathop{\mathrm{ker}_{n}\/}\nolimits x$, $\mathop{\mathrm{kei}_{n}\/}\nolimits x$, modulus and phase functions $\mathop{M_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\theta_{n}\/}\nolimits\!\left(x\right)$, $\mathop{N_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\phi_{n}\/}\nolimits\!\left(x\right)$, $n=0,1,2$, $x=0(.01)2.5$, 8S, and $n=0(1)10$, $x=0(.1)10$, 7S. Also included are auxiliary functions to facilitate interpolation of the tables for $n=0(1)10$ for small values of $x$. (Concerning the phase functions see §10.68(iv).)
• Abramowitz and Stegun (1964, Chapter 9) tabulates $\mathop{\mathrm{ber}_{n}\/}\nolimits x$, $\mathop{\mathrm{bei}_{n}\/}\nolimits x$, $\mathop{\mathrm{ker}_{n}\/}\nolimits x$, $\mathop{\mathrm{kei}_{n}\/}\nolimits x$, $n=0,1$, $x=0(.1)5$, 9–10D; $x^{n}(\mathop{\mathrm{ker}_{n}\/}\nolimits x+(\mathop{\mathrm{ber}_{n}\/}% \nolimits x)(\mathop{\ln\/}\nolimits x))$, $x^{n}(\mathop{\mathrm{kei}_{n}\/}\nolimits x+(\mathop{\mathrm{bei}_{n}\/}% \nolimits x)(\mathop{\ln\/}\nolimits x))$, $n=0,1$, $x=0(.1)1$, 9D; modulus and phase functions $\mathop{M_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\theta_{n}\/}\nolimits\!\left(x\right)$, $\mathop{N_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\phi_{n}\/}\nolimits\!\left(x\right)$, $n=0,1$, $x=0(.2)7$, 6D; $\sqrt{x}e^{-x/\sqrt{2}}\mathop{M_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\theta_{n}\/}\nolimits\!\left(x\right)-(x/\sqrt{2})$, $\sqrt{x}e^{x/\sqrt{2}}\mathop{N_{n}\/}\nolimits\!\left(x\right)$, $\mathop{\phi_{n}\/}\nolimits\!\left(x\right)+(x/\sqrt{2})$, $n=0,1$, $1/x=0(.01)0.15$, 5D.
• Zhang and Jin (1996, p. 322) tabulates $\mathop{\mathrm{ber}\/}\nolimits x$, $\mathop{\mathrm{ber}\/}\nolimits'x$, $\mathop{\mathrm{bei}\/}\nolimits x$, $\mathop{\mathrm{bei}\/}\nolimits'x$, $\mathop{\mathrm{ker}\/}\nolimits x$, $\mathop{\mathrm{ker}\/}\nolimits'x$, $\mathop{\mathrm{kei}\/}\nolimits x$, $\mathop{\mathrm{kei}\/}\nolimits'x$, $x=0(1)20$, 7S.
## §10.75(xii) Zeros of Kelvin Functions and their Derivatives
• Zhang and Jin (1996, p. 323) tabulates the first $20$ real zeros of $\mathop{\mathrm{ber}\/}\nolimits x$, $\mathop{\mathrm{ber}\/}\nolimits'x$, $\mathop{\mathrm{bei}\/}\nolimits x$, $\mathop{\mathrm{bei}\/}\nolimits'x$, $\mathop{\mathrm{ker}\/}\nolimits x$, $\mathop{\mathrm{ker}\/}\nolimits'x$, $\mathop{\mathrm{kei}\/}\nolimits x$, $\mathop{\mathrm{kei}\/}\nolimits'x$, 8D. | 10,795 | 24,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 470, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.541778 |
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## Working with overtime calculation
Hi,
Tried searching the various other threads on this topic but they didn't really work for my case.
Users can have multiple entries per day depending on activity. I want to sum the total hours per day for each user, see if each user has gone over 8 hours and mark those overages as overtime.
Example of my raw data:
user day hours
apple 01.04.2018 4 apple 01.04.2018 3 apple 01.04.2018 4 banana 03.04.2018 10 banana 05.04.2018 3
However, the problem is either the row value is wrong or the grand total is wrong. Here's what I have so far:
```Overtime with Summarize =
SUMX(
SUMMARIZE(
Table1; Table1[hours]);
IF ( Table1[hours] > 8; Table1[hours] - 8; 0))```
```Overtime simple =
var overtime =
SUM(Table1[hours]) - 8
return
IF(overtime < 0; 0; overtime)```
Any help appreciated!
1 ACCEPTED SOLUTION
Super User
OK, I think I got it:
```Measure 3 =
VAR overtime = SUM('#Overtime'[hours]) - 8
VAR tmpTable = SUMMARIZE('#Overtime',[user],[day],"Hours",SUM('#Overtime'[hours]))
VAR overtimeTotal = SUMX(tmpTable1,[Overtime])
RETURN IF(HASONEFILTER('#Overtime'[user]),IF(overtime<0,0,overtime),overtimeTotal)```
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4 REPLIES 4
Resolver IV
"IF ( Table1[hours] > 8; Table1[hours] - 8; 0)" could be replaced by
MAX (Table1[hours] - 8; 0)
Super User
OK, I think I got it:
```Measure 3 =
VAR overtime = SUM('#Overtime'[hours]) - 8
VAR tmpTable = SUMMARIZE('#Overtime',[user],[day],"Hours",SUM('#Overtime'[hours]))
VAR overtimeTotal = SUMX(tmpTable1,[Overtime])
RETURN IF(HASONEFILTER('#Overtime'[user]),IF(overtime<0,0,overtime),overtimeTotal)```
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Latest book!:
Mastering Power BI 2nd Edition
DAX is easy, CALCULATE makes DAX hard...
Helper I
Thanks @Greg_Deckler! Works perfectly!
And thanks @MarkS on the tip.
Super User
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External Tools: MSHGQM
Latest book!:
Mastering Power BI 2nd Edition
DAX is easy, CALCULATE makes DAX hard...
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Find out more about the May 2023 update. | 695 | 2,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.772281 |
http://slideplayer.com/slide/8204966/ | 1,569,315,976,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00061.warc.gz | 168,138,761 | 18,214 | Sneha. Gates Gates Characteristics of gates Characteristics of gates Basic Gates Basic Gates AND Gate AND Gate OR gate OR gate NOT gate NOT.
Presentation on theme: "Sneha. Gates Gates Characteristics of gates Characteristics of gates Basic Gates Basic Gates AND Gate AND Gate OR gate OR gate NOT gate NOT."— Presentation transcript:
sneha
Gates Gates Characteristics of gates Characteristics of gates Basic Gates Basic Gates AND Gate AND Gate OR gate OR gate NOT gate NOT gate XOR Gate XOR Gate XNOR gate XNOR gate Universal Gate Universal Gate NAND GATE NAND GATE NAND gate as an Universal gate NAND gate as an Universal gate NOR gate NOR gate NOR as an Universal gate NOR as an Universal gate Summary Summary
Gates are the basic elements of circuit. A gate is a device that performs a basic operation on electrical signals Gates are combined into circuits to perform more complicated tasks. back
Each logic gate has only one binary output. Each logic gate has either one or more than one binary input(s). back
There are three basic gates 1 AND gate 2 OR gate 3 NOT gate back
An AND gate has two or more inputs and only one output. The output of an AND gate is 1 only if all inputs are 1. back
An OR gate has two or more inputs and only one output. The output of an OR gate is 1 only if one or more inputs are 1. back
NOT gate has one input and one output. It is also known as inverter. back
XOR, or exclusive OR, gate ◦ An XOR gate produces 0 if its two inputs are the same, and a 1 otherwise back
It is also known as equivalence gate. It is combination of XOR and NOT gate. Symbol
Expression :AB+A’B’ Truth table ABC 001 010 100 111 back
A universal gate is that gate which apart from its own logical function can also perform the basic gate function i.e AND,OR and NOT. NAND and NOR are universal gate. back
Nand gate is combination of NOT and AND gate. Symbol Expression :A.B
ABS 001 011 101 110 back
A NOT gate can be implemented using a NAND gate (a). An AND gate can be implemented using NAND gates (b). An OR gate can be implemented using NAND gates back
NOR gate is combination of OR gate and NOT gate. Symbol back
Download ppt "Sneha. Gates Gates Characteristics of gates Characteristics of gates Basic Gates Basic Gates AND Gate AND Gate OR gate OR gate NOT gate NOT."
Similar presentations | 683 | 2,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.874609 |
https://stackoverflow.com/questions/28814412/reduce-amount-of-dimensions-in-an-array | 1,586,351,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371813538.73/warc/CC-MAIN-20200408104113-20200408134613-00270.warc.gz | 698,352,275 | 32,979 | Reduce amount of dimensions in an array
I have Range(one dimensional), that I want to summarize in one cell by concatenating all values. I thought that I could do just:
``````Dim Data_array()
Dim Source_Range as Range
Set Source_Range = Selection
Data_array() = Source_Range.Value2
Source_range.Offset( -1 ,0).Value = Join(Data_array, ", ")
``````
This however returns `error 5` because `Data_array` is a ( 1 To X, 1 To 1) array it has two dimensions, and `Join` on last line requires that you provide just one dimenstional array.
So my question would be is there a way to remove that "1 To 1" dimension?
If not how would you concatenate a one dimensional range in one cell.
Example
`````` A
1
2 2
3 4
4 6
``````
Desired Result
`````` A
1 2, 4, 6
2 2
3 4
4 6
``````
• Would this work? `Redim Data_Array(1 to X)` – Kyle Mar 2 '15 at 16:28
• "Script out of Range." Error 9 Thanks though – sgp667 Mar 2 '15 at 16:30
• Well you need to define X, which I thought you had, since you used it in your example. `X = Source_Range.Count Redim Data_array(1 to x)` – Kyle Mar 2 '15 at 16:33
• possible duplicate of how to merge all column into one cell in excel? – Patrick McDonald Mar 2 '15 at 16:40
You were so close! The code below assumes you will select the cells below the empty target cell. I't is simply two tweaks from your original code:
``````Sub testing()
Dim Data_array()
Dim Source_Range As Range
Set Source_Range = Selection
Data_array() = WorksheetFunction.Transpose(Source_Range.Value2)
Source_Range.Offset(-1, 0).Resize(1, 1).Value = Join(Data_array, ", ")
End Sub
``````
Below is an idea. NOTE: I'm not sure that the `OFFSET` part of your code does what you want it to do. Test the code, and let me know if so.
``````Dim Data_array()
Dim Source_Range As Range
Dim nIncrement As Integer
Set Source_Range = Selection
nIncrement = 1
ReDim Data_array(1 To Source_Range.Rows.Count)
For Each cel In Source_Range
Data_array(nIncrement) = cel.Value
nIncrement = nIncrement + 1
Next cel
Source_Range.Offset(-1, 0).Value = Join(Data_array, ", ")
``````
For your data, I would not bother with a VBA array. Consider:
``````Public Function Konkat(rin As Range) As String
For Each r In rin
v = r.Value
If v <> "" Then
Konkat = Konkat & "," & v
End If
Next r
Konkat = Right(Konkat, Len(Konkat) - 1)
End Function
``````
This is because in your code, data_Array is actually a two-dimensional array.
You can use the INDEX worksheet function to slice out a column or row.
``````Sub JoinRangeComma()
Dim vaData As Variant
Dim rSource As Range
Dim wf As WorksheetFunction
Set wf = Application.WorksheetFunction
Set rSource = Selection | 778 | 2,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.835306 |
https://javatutoring.com/median-of-two-sorted-arrays-c-program/ | 1,721,052,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00746.warc.gz | 278,548,689 | 23,427 | # C program : Find Median of Two Sorted Arrays | C Programs
Write a C program to find the median of the given two sorted arrays. Median is the middle number when the numbers are sorted in either ascending or descending order.
To implement this, our required inputs are the sizes of two arrays along with the elements or data values of the two arrays. Our desired output is the double type number representing the median of the given two input arrays when combined into one sorted array.
## Find Median of Two Sorted Arrays Using Function
The first step here is, to read our necessary inputs and to do so, we can make use of the predefined function scanf() which reads the input of any primitive datatype like int, char, float, double, long, etc., at runtime.
Based on the format specifier given, the datatype of the input to be read is determined. In our given problem statement, all the necessary inputs are of integer type and therefore, we use the ‘%d’ format specifier to read them.
Firstly, we’ll read the sizes of the two arrays (n1, n2) following which we’ll create the two arrays (a1, a2) of the respective size. Then, we’ll read the elements or data values of the two arrays.
printf(“\nEnter the size of Array 1 : “);
scanf(“%d”,&n1);
printf(“\nEnter the size of Array 2 : “);
scanf(“%d”,&n2);
int a1[n1],a2[n2];
printf(“\nEnter the %d sorted elements in Array1:\n”,n1);
for(i=0;i<n1;i++) {
scanf(“%d”,&a1[i]);
}
printf(“\nEnter the %d sorted elements in Array2:\n”,n2);
for(i=0;i<n2;i++) {
scanf(“%d”,&a2[i]);
}
Output:
Standard Method
In the above method, we have seen the logic to determine the median of the given two sorted arrays using functions. The benefits of using functions is that, it makes the parts of code within functions reusable and also enhances the readability of the code.
If some logic is needed elsewhere in the code and is within a function block of its own then, instead of rewriting the same set of statements we could just make a function call and pass the necessary arguments thereby, making it reusable.
Also, since the code is placed within different function blocks based on its logic and functionality, the code is lot cleaner and is placed systematically.
This way, if some part is to be found we can directly go to the corresponding function block and look for it. Hence, it increases the readability of the code.
As we can observe the same above, the main method is handling all the input/output related operations whereas the user-defined function (median) is handling the logic behind merging the two sorted array and thereby, finding the median.
But, we can also write the entire code within main method as the time and space complexity is the same for both because of the same logic being used.
So, if we are sure that, we do not need to reuse this logic then, there is no harm in writing it within main itself else, we might have to rewrite same set of statements again. We can see below the illustration of the same.
Output:
x
## C Program : Rotate a Given Matrix by 90 Degrees Anticlockwise
The problem statement here is, to write a C program to rotate a given square ... | 725 | 3,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.800855 |
https://rdrr.io/rforge/xmetric/src/R/Kettelle.R | 1,547,926,347,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583680452.20/warc/CC-MAIN-20190119180834-20190119202834-00586.warc.gz | 625,950,792 | 16,217 | # R/Kettelle.R In xmetric: X-METRIC Multi-Echelon Techniques for Replaceable Inventory Control
#### Documented in Kettelle
```## Kettelle.R
## This function performs the Generalized Kettelle Algorithm as presented in
## Chapter 7 of "Statistical Theory of Reliability and Life Testing, Probability Models"
## by Richard E. Barlow and Frank Proschan. This function has been extended to handle
## performance optimization of Sherbrooke's Estimated Back Order measure.
##
## (C) David J. Silkworth 2014
##
## This program is free software; you can redistribute it and/or modify it
## Free Software Foundation; either version 2, or (at your option) any
## later version.
##
## These functions are distributed in the hope that they will be useful,
## but WITHOUT ANY WARRANTY; without even the implied warranty of
## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
## GNU General Public License for more details.
##
## You should have received a copy of the GNU General Public License
## along with this program; if not, a copy is available at
##
Kettelle<-function(x, limit=1e-4, data.name="", performance="FRN", show=FALSE) {
numParts<-length(x[,1])
PerfFun=FRN
if(performance=="EBO") {
PerfFun=EBO
}
UdomAll<-matrix(data=rep(0,numParts+2),
nrow=1, ncol=numParts+2)
UdomAll<-data.frame(UdomAll)
PartNames<-NULL
## build lists of individual part Perf and Cost values per Qty "s"
## this can be referenced rather than calculated each time
PerfList<-NULL
CostList<-NULL
## required to zero-for-other-parts, debug fix
z4op=0
for (n in 1:numParts) {
s=0
withinBudget=TRUE
unconverged=TRUE
thisPerf<-NULL
thisCost<-NULL
while(unconverged) {
newPerf<-PerfFun(s,x[n,2],x[n,3])
thisPerf<-c(thisPerf,newPerf)
thisCost<-c(thisCost,s*x\$C[n])
withinBudget <-TRUE
if(s>2) {
unconverged<-abs(newPerf-lastPerf)>limit
}else{
unconverged=TRUE
}
lastPerf<-newPerf
if (s==0) {
z4op<-z4op+newPerf
}
s=s+1
}
PerfList[[length(PerfList)+1]]<-thisPerf
CostList[[length(CostList)+1]]<-thisCost
PartNames<-c(PartNames,as.character(x[n,1]))
}
names(UdomAll)<-c(PartNames,"Perf","Cost")
ExAll<-UdomAll
thisRow<-UdomAll
for(j in 1:(numParts-1) ) {
z4op<-c(z4op,z4op[j]-PerfList[[j]][1])
}
j=1
UdomAll\$Perf[j]<-z4op[j]
## Fill the UdomAll for the first part with no others
for(k in 2:length(CostList[[j]])-1) {
thisRow[1,1]=k
for(f in 2:numParts) {
thisRow[1,f]=0
}
thisRow\$Perf<-PerfList[[j]][k+1]+z4op[j+1]
thisRow\$Cost<-CostList[[j]][k+1]
UdomAll<-rbind(UdomAll,thisRow)
}
UdomLast<-UdomAll
if(UdomAll\$Perf[1]<UdomAll\$Perf[2]) {
## This is the maximize algorithm
for(f in 2:(numParts)) {
## corrected to start from all zeros, else zrro part 1 cases are never evaluated
for(j in 1:length(UdomLast[,1])) {
for(k in 1:length(CostList[[f]])) {
thisRow<-UdomLast[j,]
thisRow[1,f]<-k-1
thisRow\$Cost[1]<-thisRow\$Cost[1]+CostList[[f]][k]
## Subtract the zero Perf for this part f, then add back the Perf for k-1 Qty of part f
thisRow\$Perf[1]<-thisRow\$Perf[1]-PerfList[[f]][1]+PerfList[[f]][k]
if(thisRow\$Cost[1] %in% UdomAll\$Cost) {
pos<-match(thisRow\$Cost[1],UdomAll\$Cost)
if(UdomAll\$Perf[pos]<thisRow\$Perf[1]) {
ex<-UdomAll[pos,]
UdomAll[pos,]<-thisRow
ExAll<-rbind(ExAll,ex)
## Does this undominated allocation dominate anything elseof higher cost?
if(max(UdomAll\$Cost)>thisRow\$Cost) {
majorDom<-UdomAll[(pos+1):length(UdomAll[,1]),]
if(min(majorDom\$Perf)<thisRow\$Perf) {
ex<-majorDom[sapply(majorDom\$Perf, function(x) x<thisRow\$Perf),]
ExAll<-rbind(ExAll,ex)
majorDom<-majorDom[sapply(majorDom\$Perf, function(x) x>thisRow\$Perf),]
UdomAll<-rbind(UdomAll[1:pos,],majorDom)
}
}
}else{
## this allocation is dominated by a previous allocation at this Cost
ExAll<-rbind(ExAll,thisRow)
}
}else{
## this is where additional activity must take place
## need to check if this allocation is dominated by a lower cost
## need to error trap here for case where this is a new high Cost
if(max(UdomAll\$Cost)>thisRow\$Cost) {
pos<-min(which(UdomAll\$Cost>thisRow\$Cost))
minorDom<-UdomAll[1:(pos-1),]
if(max(minorDom\$Perf)>thisRow\$Perf) {
## this is a new Cost allocation that is dominated by a lesser cost allocation
ExAll<-rbind(ExAll,thisRow)
}else{
## if not, it is dominant, but did it dominate something elseof higher cost?
## Does this newly identified Udom alloccation dominate anything elseof higher cost?
majorDom<-UdomAll[pos:length(UdomAll[,1]),]
if(min(majorDom\$Perf)<thisRow\$Perf) {
ex<-majorDom[sapply(majorDom\$Perf, function(x) x<thisRow\$Perf),]
majorDom<-majorDom[sapply(majorDom\$Perf, function(x) x>thisRow\$Perf),]
ExAll<-rbind(ExAll,ex)
}
UdomAll<-rbind(UdomAll[1:(pos-1),],thisRow,majorDom)
## it is impossible to state a domination over anything at lower cost
}
}else{
## A new high cost allocation has been defined, so just add it to the end of table
UdomAll<-rbind(UdomAll,thisRow)
}
}
## close incremental new part
}
## close UdomLast
}
UdomLast<-UdomAll
## close all parts
}
}else{
## This is the minimize algorithm
for(f in 2:(numParts)) {
## corrected to start from all zeros, else zrro part 1 cases are never evaluated
for(j in 1:length(UdomLast[,1])) {
for(k in 1:length(CostList[[f]])) {
thisRow<-UdomLast[j,]
thisRow[1,f]<-k-1
thisRow\$Cost[1]<-thisRow\$Cost[1]+CostList[[f]][k]
## Subtract the zero Perf for this part f, then add back the Perf for k-1 Qty of part f
thisRow\$Perf[1]<-thisRow\$Perf[1]-PerfList[[f]][1]+PerfList[[f]][k]
if(thisRow\$Cost[1] %in% UdomAll\$Cost) {
pos<-match(thisRow\$Cost[1],UdomAll\$Cost)
if(UdomAll\$Perf[pos]>thisRow\$Perf[1]) {
ex<-UdomAll[pos,]
UdomAll[pos,]<-thisRow
ExAll<-rbind(ExAll,ex)
## Does this undominated allocation dominate anything elseof higher cost?
if(max(UdomAll\$Cost)>thisRow\$Cost) {
majorDom<-UdomAll[(pos+1):length(UdomAll[,1]),]
if(max(majorDom\$Perf)>thisRow\$Perf) {
ex<-majorDom[sapply(majorDom\$Perf, function(x) x>thisRow\$Perf),]
ExAll<-rbind(ExAll,ex)
majorDom<-majorDom[sapply(majorDom\$Perf, function(x) x<thisRow\$Perf),]
UdomAll<-rbind(UdomAll[1:pos,],majorDom)
}
}
}else{
## this allocation is dominated by a previous allocation at this Cost
ExAll<-rbind(ExAll,thisRow)
}
}else{
## this is where additional activity must take place
## need to check if this allocation is dominated by a lower cost
## need to error trap here for case where this is a new high Cost
if(max(UdomAll\$Cost)>thisRow\$Cost) {
pos<-min(which(UdomAll\$Cost>thisRow\$Cost))
minorDom<-UdomAll[1:(pos-1),]
if(min(minorDom\$Perf)<thisRow\$Perf) {
## this is a new Cost allocation that is dominated by a lesser cost allocation
ExAll<-rbind(ExAll,thisRow)
}else{
## if not, it is dominant, but did it dominate something elseof higher cost?
## Does this newly identified Udom alloccation dominate anything elseof higher cost?
majorDom<-UdomAll[pos:length(UdomAll[,1]),]
if(max(majorDom\$Perf)>thisRow\$Perf) {
ex<-majorDom[sapply(majorDom\$Perf, function(x) x>thisRow\$Perf),]
majorDom<-majorDom[sapply(majorDom\$Perf, function(x) x<thisRow\$Perf),]
ExAll<-rbind(ExAll,ex)
}
UdomAll<-rbind(UdomAll[1:(pos-1),],thisRow,majorDom)
## it is impossible to state a domination over anything at lower cost
}
}else{
## A new high cost allocation has been defined, so just add it to the end of table
UdomAll<-rbind(UdomAll,thisRow)
}
}
## close incremental new part
}
## close UdomLast
}
UdomLast<-UdomAll
## close all parts
}
}
title<-"Kettelle Part Allocation Optimization"
if(data.name!="") {
title<-c(title, data.name)
}
if(performance=="Fill Rate") {
ExAll\$Perf<-ExAll\$Perf/sum(x[,2])
UdomAll\$Perf<-UdomAll\$Perf/sum(x[,2])
}else{
if(performance!="EBO") {
## ignore any stray argument names for performance
## "FRN" is default
performance="FRN"
}
}
outnames<-names(UdomAll)
outcols<-length(outnames)
outnames[outcols-1]<-performance
names(UdomAll)<-outnames
if(show==TRUE) {
plot(ExAll\$Cost,ExAll\$Perf,
xlab="Cost",ylab=performance,
pch=19, col="blue", cex=0.5,
main=title
)
points(UdomAll\$Cost,UdomAll[,outcols-1],
pch=21, bg="red")
}
return(UdomAll)
}
```
## Try the xmetric package in your browser
Any scripts or data that you put into this service are public.
xmetric documentation built on May 31, 2017, 4:31 a.m. | 2,664 | 8,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.565533 |
https://www.neetprep.com/question/50314-Two-short-magnets-equal-dipole-moments-M-fastened-perpendicularly-attheir-centre-figure-magnitude-magnetic-field-distance-d-fromthe-centre-bisector-right-angle-isaMdbMdcMddMd/55-Physics--Magnetism-Matter/695-Magnetism-Matter | 1,596,656,778,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735964.82/warc/CC-MAIN-20200805183003-20200805213003-00544.warc.gz | 784,059,725 | 93,011 | Two short magnets of equal dipole moments M are fastened perpendicularly at their centres (figure). The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is :
1.
2.
3.
4.
Concept Questions :-
Bar magnet
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Two identical bar magnets with a length 10 cm and weight 50 gm-weight are arranged freely with their like poles facing in a inverted vertical glass tube. The upper magnet hangs in the air above the lower one so that the distance between the nearest pole of the magnet is 3mm. Pole strength of the poles of each magnet will be
(a) 6.64 amp$×$m
(b) 2 amp$×$m
(c) 10.25 amp$×$m
(d) None of these
Concept Questions :-
Analogy between electrostatics and magnetostatics
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If ${\varphi }_{1}$ and ${\varphi }_{2}$ be the angles of dip observed in two vertical planes at right angles to each other and $\varphi$ be the true angle of dip, then
1. ${\mathrm{cos}}^{2}\varphi ={\mathrm{cos}}^{2}{\varphi }_{1}+{\mathrm{cos}}^{2}{\varphi }_{2}$
2. ${\mathrm{sec}}^{2}\varphi ={\mathrm{sec}}^{2}{\varphi }_{1}+{\mathrm{sec}}^{2}{\varphi }_{2}$
3. ${\mathrm{tan}}^{2}\varphi ={\mathrm{tan}}^{2}{\varphi }_{1}+{\mathrm{tan}}^{2}{\varphi }_{2}$
4. ${\mathrm{cot}}^{2}\varphi ={\mathrm{cot}}^{2}{\varphi }_{1}+{\mathrm{cot}}^{2}{\varphi }_{2}$
Concept Questions :-
Earth's magnetism
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Each atom of an iron bar has a magnetic moment . Knowing that the density of iron is atomic weight is 56 and Avogadro's number is the magnetic moment of the bar in the state of magnetic saturation will be
1. 4.75 Am2 2. 5.74 Am2
3. 7.54 Am2 4. 75.4 Am2
Concept Questions :-
Analogy between electrostatics and magnetostatics
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A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 seconds. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 seconds. Then the angle of dip is
1. 0o 2. 30o
3. 45o 4. 90o
Concept Questions :-
Earth's magnetism
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A dip needle lies initially in the magnetic meridian when it shows an angle of dip $\theta$ at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip $\theta \text{'}$ . Then is
1. 1/cos x
2. 1/sin x
3. 1/tan x
4. cos x
Concept Questions :-
Earth's magnetism
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A bar magnet has coercivity . It is desired to demagnetize it by inserting it inside a solenoid 12 cm long and having 60 turns. The current that should be sent through the solenoid is
1. 2 A 2. 4 A
3. 6 A 4. 8 A
Concept Questions :-
Curie law and hyterisis
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For substances hysteresis (B - H) curves are given as shown in figure. For making temporary magnet which of the following is best.
(a)
(b)
(c)
(d)
Concept Questions :-
Curie law and hyterisis
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The variation of magnetic susceptibility $\left(\chi \right)$ with temperature for a diamagnetic substance is best represented by
1.
2.
3.
4.
Concept Questions :-
Magnetic materials
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The variation of magnetic susceptibility $\left(\chi \right)$ with absolute temperature T for a ferromagnetic material is
Concept Questions :-
Curie law and hyterisis | 1,109 | 3,964 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-34 | latest | en | 0.689873 |
https://filonova.net/sports-betting/maximize-your-wins-by-applying-the-best-arbitrage-formula-on-each-bet/ | 1,716,404,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00128.warc.gz | 217,706,581 | 12,892 | # Maximize your wins by applying the best arbitrage formula on each bet
If you have planned to engage in arbitrage betting to win money regardless of the outcome of the match then you can certainly increase your wins by making use of the right arbitrage formula on each bet. You will need to make correct calculations before placing your bet if you do not wish to get paltry wins or suffer a loss of revenue after the match.
Arbitrage betting involves betting on opposing players or teams in the match so that you win enough extra money for the winning bet whenever a player or team wins while offsetting the loss brought on by the losing bet. In this particular type of betting, you win money regardless of the end result of the bet since you would’ve placed different bets on opposing players and teams with different greatbettingadvice bookmakers. While one bookmaker might offer excellent odds for the favorite player or team, another bookie might favor the underdog and provide higher odds. The bottom line is to quickly identify such bets and place your bets so that the total amount in winnings is much more than both your bets placed together.
However, you may need a formula to decipher the offered odds and calculate the amount which you might win or lose after the match. You will thus need to apply an arbitrage formula that will help you to calculate the exact amount that you will need to invest in the bet and the amount to be won when any of the players or teams wins, and even when the match heads towards a draw. This formula will need to take into account several components such as the possible outcomes of the match, the odds placed on each player or team by the particular bookmaker, and the amount that you’re planning to invest in each bet. The solution to your arbitrage calculations should provide you with the amount you have collectively committed to the bets along with the different amounts that you will win when either of the players or teams wins. In case there are in excess of two players or teams then the formula must also incorporate additional calculations.
If you aren’t very good at math and are afraid of making mistakes when using the arbitrage formula, then you need not worry. It is possible to enlist the aid of reputed betting experts such as Gert Gambell by visiting his websites, gertgambell.net and also win-every-time.com which will explain the whole concept of arbitrage betting in easy-to-understand terms. You may also come across helpful suggestions and advice offered by these websites and also utilize the arbitrage calculator provided free at the site to calculate the exact amount easily that you can win after placing bets on opposing players or teams. As it is essential to bet only at trustworthy websites that will deliver your winnings, you can surely count on the bookmakers mentioned by Gert Gambell on his websites.
If you want to win money on each betting session no matter which player or team wins the match, then you can go in for arbitrage betting. However, instead of placing each bet excitedly without looking into all available facts or without undertaking the mandatory calculations, you need to use the correct arbitrage formula to increase your winnings with minimum investment in each bet. | 641 | 3,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | longest | en | 0.958925 |
http://www.jiskha.com/display.cgi?id=1170813589 | 1,496,130,446,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614615.14/warc/CC-MAIN-20170530070611-20170530090611-00461.warc.gz | 670,786,554 | 3,961 | # Physics
posted by on .
Trolley B will be at rest and trolley A will be given a push such that it collides with B, and they both move as one mass after the collision. As soon as the blocker of the picket fence on trolley A passes through the eye of the first photogate, trolley A must collide with trolley B, and immediately after collision, the blocker on the picket fence of trolley B must enter the eye of the second photogate.
1)Why must the collision take place immediately after blocker A leaces the eye of the photogate?
2) Why must blocker B enter the eye of the photogate immediately after collision?
1) So that the photogate could record the velocity before the collision. Blocker A also will have time to gain momentum
2) So that we know the velcocity immediately after the collision other wise veolocity will be lost if it wasn't immediately afet the collision.
Are my answers correct. I am not sure whether 1) or 2) is correct. Thanks for helping me out. I really apprecitate it. | 236 | 998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.949607 |
https://levitra4people.top/understood-in-advance-what-the-lotto-numbers/ | 1,708,671,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00278.warc.gz | 382,661,429 | 13,910 | # understood in advance what the lotto numbers
How to Predict the Lottery and Know the Hidden Keys Behind the System
Have you ever before wondered why other people constantly DNABET understood in advance what the lotto numbers will be, no matter the amount of times?
The depressing part is that there are more methods than one to guess the lotto numbers as well as unfortunately, some of those means may not be so readily available to you.
You are not alone if you guessed it. In fact, thousands of individuals on the planet are assuming in the exact same type of way. As well as what they all did was, presume what?
Don’t also think of banking on a thinking video game, you’re not going to win most of the time.
If you guessed appropriately, you might as well of won the lottery right after that and there, but did you recognize that there are countless people around the world who acquired this exact same guess-targeting item. They all had the very same desire, that in some way they understood the lotto numbers and also they might simply stop when they struck it rich, and you could not stop them because you were buying the same numbers as them.
There are a lot of Professor take your cash with no threat added to those bets. The Teacher took a look at the numbers and said it was unlikely the Yankees would beat the Maverick, so the wager was off. Taking a look at the numbers andessentially the same point. The Professor ran and also took the money. He can have won the bet and took his opportunities as well as still wouldn’t have actually been any kind of worse off. The Teacher had not been gambling he was just regulating the threat. The Professor was Prof. Richardphas the tenth worst gamer in the organization so the odds protested him each time, the line was essentially what everybody else would take.
Consider this scenario, a few weeks ago Teacher Olsen would certainly have informed you that the Powerball would certainly most likely be in between 55 and 59. Well guess what occurred? The winning number was precisely 55. Exactly how about that? You could have made the very same wager, got the same outcome, and saved a great deal of money. The very same exact thing happened with the Weekley Yankee. The Yankees at the plate got a phenomenal bottle to spoil the best protection in baseball the next night. The Gold Glove was won as well as the underdogs didn’t also require to exist.
That’s. They all thought the lotto numbers. From the Nebraska Powerball to the California Powerball, from the Florida Lottery to the Maryland Cash money 5, individuals attempt each day to guess what lotto numbers will come out.
Betting is the simplest way to lose your cash, and also the method to win is to never ever place a bet that you don’t recognize if you have a shiver going through your pocket. Do not simply wish to occur at your neighborhood grocery store, let me tell you that’s where you’re going to toss a lot of money away.
As well as when you consider that most of cases, the winning numbers are typically not extremely predictable, you can however a rate which number will certainly most probably be selected next. Yet can you do it?
Be cautious of the men that tell you the trick to wagering. The insane think that a great deal of people are having is that a whole lot of these Professorenders wagering the system have some kind of an ace up their sleeve. If you knew the system inside as well as out you wouldn’t be able to beat the wall.
If you’re in the mood to try to pick the champion and do not want to have to do all the work yourself, then the web is your good friend. Now you can do a considerable study beginning from about the last 50 choices or so and also you can pick the group to bank on. You can also figure out which group has the greatest pitching staff and also which one is understood for their underdog in your home.
, if just it was true.. Regrettably it’s not, and also even if you can stop them, there are simply so many others that got in before you and before you could not stop them. The depressing part is that there are more ways than one to presume the lotto numbers and regrettably, some of those ways may not be so readily available to you.
They all guessed the lotto numbers. From the Nebraska Powerball to the California Powerball, from the Florida Lottery to the Maryland Cash 5, individuals try every day to think what lottery numbers will certainly come out.
What regarding those various other means, you ask? Why yes obviously there are, which’s where your tale about an old Northwestern University Professor is available in. If you were the type of person that required a 10 thousand buck Prof. position by the end of the very first week, this is the person you would certainly leave your task for. Wait, that’s not a trouble, he isn’t a 10 thousand dollar professional, he’s simply an individual that suches as to instruct. Well, guess what? He educates you exactly how to predict the lotto. Isn’t that terrific? Just how depictive is this method, Professor Olsen?
The winning number was precisely 55. The Professor took an appearance at the numbers and claimed it was not likely the Yankees would certainly defeat the Maverick, so the bet was off.
## 1 thought on “understood in advance what the lotto numbers”
1. Great site, good material, simple to comprehend as well as read. Thank you quite for this post.
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From Dan Filimon <dangeorge.fili...@gmail.com>
Subject Re: Log-likelihood ratio test as a probability
Date Thu, 20 Jun 2013 09:25:56 GMT
Right, makes sense. So, by normalize, I need to replace the counts in the
matrix with probabilities.
So, I would divide everything by the sum of all the counts in the matrix?
On Thu, Jun 20, 2013 at 12:16 PM, Sean Owen <srowen@gmail.com> wrote:
> I think the quickest answer is: the formula computes the test
> statistic as a difference of log values, rather than log of ratio of
> values. By not normalizing, the entropy is multiplied by a factor (sum
> of the counts) vs normalized. So you do end up with a statistic N
> times larger when counts are N times larger.
>
> On Thu, Jun 20, 2013 at 9:52 AM, Dan Filimon
> <dangeorge.filimon@gmail.com> wrote:
> > My understanding:
> >
> > Yes, the log-likelihood ratio (-2 log lambda) follows a chi-squared
> > distribution with 1 degree of freedom in the 2x2 table case.
> > A ~A
> > B
> > ~B
> >
> > We're testing to see if p(A | B) = p(A | ~B). That's the null
> hypothesis. I
> > compute the LLR. The larger that is, the more unlikely the null
> hypothesis
> > is to be true.
> > I can then look at a table with df=1. And I'd get p, the probability of
> > seeing that result or something worse (the upper tail).
> > So, the probability of them being similar is 1 - p (which is exactly the
> > CDF for that value of X).
> >
> > Now, my question is: in the contingency table case, why would I
> normalize?
> > It's a ratio already, isn't it?
> >
> >
> > On Thu, Jun 20, 2013 at 11:03 AM, Sean Owen <srowen@gmail.com> wrote:
> >
> >> someone can check my facts here, but the log-likelihood ratio follows
> >> a chi-square distribution. You can figure an actual probability from
> >> that in the usual way, from its CDF. You would need to tweak the code
> >> you see in the project to compute an actual LLR by normalizing the
> >> input.
> >>
> >> You could use 1-p then as a similarity metric.
> >>
> >> This also isn't how the test statistic is turned into a similarity
> >> metric in the project now. But 1-p sounds nicer. Maybe the historical
> >> reason was speed, or, ignorance.
> >>
> >> On Thu, Jun 20, 2013 at 8:53 AM, Dan Filimon
> >> <dangeorge.filimon@gmail.com> wrote:
> >> > When computing item-item similarity using the log-likelihood
> similarity
> >> > [1], can I simply apply a sigmoid do the resulting values to get the
> >> > probability that two items are similar?
> >> >
> >> > Is there any other processing I need to do?
> >> >
> >> > Thanks!
> >> >
> >> > [1] http://tdunning.blogspot.ro/2008/03/surprise-and-coincidence.html
> >>
>
Mime
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https://aobindia.com/propionic-acid-jlvzjgf/consistent-estimator-variance-bb46b4 | 1,632,005,848,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00589.warc.gz | 161,221,949 | 8,736 | Hot Network Questions Why is the rate of return for website investments so high? non-parametric spatial heteroskedasticity and autocorrelation consistent (SHAC) estimator of the variance covariance matrix in a spatial context. However, some authors also call V the asymptotic variance. Are websites a good investment? Under some conditions, the global maximizer is the optimal estimator,\op-timal"here meaning consistent and asymptotically normal with the smallest possible asymptotic variance. This is also proved in the following subsection (distribution of the estimator). Note that we did not actually compute the variance of S2 n. We illustrate the application of the previous proposition by giving another proof that S2 n is a consistent estimator… Although this estimator does not have a finite mean or variance, a consistent estimator for its asymptotic variance can be obtained by standard methods. In this formulation V/n can be called the asymptotic variance of the estimator. Kanter and Steiger limited their work to the special case where both X and Z have symmetric distributions with asymptotically Pareto tails of the same index. Since in many cases the lower bound in the Rao–Cramér inequality cannot be attained, an efficient estimator in statistics is frequently chosen based on having minimal variance in the class of all unbiased estimator of the parameter. P an in sk i, Intro. Interest in variance estimation in nonparametric regression has grown greatly in the past several decades. Estimation of elasticities of substitution for CES and VES production functions using firm-level data for food-processing industries in Pakistan If an estimator is unbiased and its variance converges to 0, then your estimator is also consistent but on the converse, we can find funny counterexample that a consistent estimator has positive variance. $\begingroup$ Thanks for the response and sorry for dropping the constraint. 1.2 Efficient Estimator From section 1.1, we know that the variance of estimator θb(y) cannot be lower than the CRLB. Variance of the estimator. n . The variance of the unadjusted sample variance is. consistent when X /n p 0 is that approximating X by zero is reasonably accurate in large samples. The variance of the adjusted sample variance is . Definition 1. A Consistent Variance Estimator for 2SLS When Instruments Identify Di erent LATEs Seojeong (Jay) Leey September 28, 2015 Abstract Under treatment e ect heterogeneity, an instrument identi es the instrument-speci c local average treatment e ect (LATE). We show next that IV estimators are asymptotically normal under some regu larity cond itions, and establish their asymptotic covariance matrix. This suggests the following estimator for the variance \begin{align}%\label{} \hat{\sigma}^2=\frac{1}{n} \sum_{k=1}^n (X_k-\mu)^2. So ^ above is consistent and asymptotically normal. So any estimator whose variance is equal to the lower bound is considered as an efficient estimator. • When we look at asymptotic efficiency, we look at the asymptotic variance of two statistics as . A biased or unbiased estimator can be consistent. Although a consistent estimator of the asymptotic variance of the IPT and IPC weighted estimator is generally available, applications and thus information on the performance of the consistent estimator are lacking. Among those who have studied asymptotic results are Kanter and Steiger (1974) and Maller (1981). Simulation results in Cribari-Neto and Zarkos (1999) suggest that this estimator did not perform as well as its competitors. However, it is less efficient (i.e., it has a larger sampling variance) than some alterna-tive estimators. Under other conditions, the global maximizer may fail to be even consistent (which is the worst property an estimator It must be noted that a consistent estimator $T _ {n}$ of a parameter $\theta$ is not unique, since any estimator of the form $T _ {n} + \beta _ {n}$ is also consistent, where $\beta _ {n}$ is a sequence of random variables converging in probability to zero. variance. With multiple instruments, two-stage least squares (2SLS) estimand is a weighted average of di erent LATEs. When defined asymptotically an estimator is fully efficient if its variance achieves the Rao-Cramér lower bound. The goal of this lecture is to explain why, rather than being a curiosity of this Poisson example, consistency and asymptotic normality of the MLE hold quite generally for many \typical" parametric models, and there is a general formula for its asymptotic variance. The regression results above show that three of the potential predictors in X0 fail this test. This heteroskedasticity-consistent covariance matrix estimator allows one to make valid inferences provided the sample size is su±ciently large. \end{align} By linearity of expectation, $\hat{\sigma}^2$ is an unbiased estimator of $\sigma^2$. Consistency. Proof. (a) find an unbiased estimator for the variance when we can calculate it, (b) find a consistent estimator for the approximative variance. Nevertheless, violations of this assump-tion can invalidate statistical inferences. variance regression and time series models, particularly in economics. Variance of second estimator Variance of first estimator Relative Efficiency = Asymptotic Efficiency • We compare two sample statistics in terms of their variances. De très nombreux exemples de phrases traduites contenant "estimator consistent" – Dictionnaire français-anglais et moteur de recherche de traductions françaises. Efficient Estimator An estimator θb(y) is … Regarding consistency, consistency you describe is "weak consistency" in the text and "consistent in MSE" is introduced, which is where I got the bias & variance going to zero. How can I make a long wall perfectly level? consistent covariance estimator can also be shown to be appropriate for use in constructing asymptotic confidence intervals. D. is impossible to obtain using real sample data. This is proved in the following subsection (distribution of the estimator). B. converges on the true parameter µ as the sample size increases. This video show how to find consistency estimator for normal population and sample variance. On the other hand, if ... since IV is another linear (in y) estimator, its variance will be at least as large as the OLS variance. Best unbiased estimator for a location family. The 3. In fact, results similar to propositions (i) and (ii) of Theorem 1were stated over a decade ago by Eicker [5], although Eicker considers only fixed and not stochastic regressors. Also, by the weak law of large numbers, $\hat{\sigma}^2$ is also a consistent estimator of $\sigma^2$. This fact reduces the value of the concept of a consistent estimator. A consistent estimator has minimum variance because the variance of a consistent estimator reduces to 0 as n increases. The SHAC estimator is robust against potential misspeci cation of the disturbance terms and allows for unknown forms of heteroskedasticity and correlation across spatial units. The resulting estimator, called the Minimum Variance Unbiased Estimator (MVUE), have the smallest variance of all possible estimators over all possible values of θ, i.e., Var Y[b θMV UE(Y)] ≤ Var Y[θe(Y)], (2) for all estimators eθ(Y) ∈ Λ and all parameters θ ∈ Λ. usual OLS regression estimator of the partial regression coefficients is unbiased and strongly consistent under het-eroskedasticity (White, 1980). Hence, a heteroskedasticity-consistent variance estimator could be estimated using the following formula: Since (9.24) is a large sample estimator it is only valid asymptotically, and test based on them are not exact and when using small samples the precision of the estimator may be poor. Therefore, the IV estimator is consistent when IVs satisfy the two requirements. Hence it is not consistent. Now, consider a variable, z, which is correlated y 2 but not correlated with u: cov(z, y 2) ≠0 but cov(z, u) = 0. has more than 1 parameter). is a consistent estimator for ˙ 2. Consistent estimator - bias and variance calculations. An estimator, $$t_n$$, is consistent if it converges to the true parameter value $$\theta$$ as we get more and more observations. M ath . A consistent estimator for the mean: A. converges on the true parameter µ as the variance increases. The choice between the two possibilities depends on the particular features of the survey sampling and on the quantity to be estimated. S tats., D ecem b er 8, 2005 49 P a rt III E stima tio n th eo ry W eÕve estab lish ed so m e so lid fou n d ation s; n ow w e can get to w h at is really Traductions en contexte de "consistent estimator" en anglais-français avec Reverso Context : This work gave a consistent estimator for power spectra and practical tools for harmonic analysis. The aforementioned results focus on completely randomized experiments where units comply with the assigned treatments. Nevertheless, their method only applies to regression models with homoscedastic errors. The statistic with the smallest variance is called . grows. The signs of the coefficient estimates are consistent with theoretical expectations: AGE, BBB, ... Because t-statistics are already adjusted for estimator variance, the presumption is that they adequately account for collinearity in the context of other, balancing effects. Proof. This estimator assumes that the weights are known rather than estimated from the data. So we need to think about this question from the definition of consistency and converge in probability. Based on the consistent estimator of the variance bound, a shorter confidence interval with more accurate coverage rate is obtained. C. consistently follows a normal distribution. De très nombreux exemples de phrases traduites contenant "consistent estimator" – Dictionnaire français-anglais et moteur de recherche de traductions françaises. A Bivariate IV model Let’s consider a simple bivariate model: y 1 =β 0 +β 1 y 2 +u We suspect that y 2 is an endogenous variable, cov(y 2, u) ≠0. Among the existing methods, the least squares estimator in Tong and Wang (2005) is shown to have nice statistical properties and is also easy to implement. 92. mating the variance-covariance matrix of ordinary least squares estimates in the face of heteroskedasticity of known form is available; see Eicker (1963), Hinkley (1977), and White (1980). efficient . reliable heteroskedasticity-consistent variance estimator. This followed from the fact that the variance of S2 n goes to zero. This seems sensible - we’d like our estimator to be estimating the right thing, although we’re sometimes willing to make a tradeoff between bias and variance. Show next that IV estimators are asymptotically normal under some regu larity cond itions, and their... Video show how to find consistency estimator for normal population and sample variance 0... 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# Problems
1. An air-core coil has a self-inductance of 0.056 henry, and a resistance of 9.2 ohms. Calculate the Q at 100, 1000, 10,000, 1,000,000, and 10,000,000 cycles. Plot as a curve. Would measured values agree with calculated values? Why?
2. If the coil just considered is connected in series with a 0.5-microfarad capacitor, what will be the resonant frequency? What will be the impedance at resonance, and at 10 per cent above and below resonance?
3. Repeat Problem 2 for the inductor and capacitor in parallel. Use both equations 15 and 16, and compare.
4. An oscillator has an open-circuit voltage of 39 volts at 1000 cycles and an internal impedance of 0.003 henry inductance and 447 ohms resistance. It is connected to the primary of a coil having an inductance of 0.052 henry and a resistance of 56 ohms. The mutual inductance between the primary and the secondary is 0.0536 henry. The secondary has an inductance of 0.061 henry, a resistance of 61 ohms, and is connected to a 500-ohm load resistor. Calculate the current that will flow through this resistor.
5. If the combination of Problem 2 is connected across 50 volts at the resonant frequency, what will be the line current, and the voltage across the inductor and across the capacitor?
6. If the combination of Problem 3 is connected in parallel and across 50 volts at the resonant frequency, what will be the line current, and the current through the inductor and capacitor?
7. Prove mathematically that the statement in italics preceding equation 24 is correct.
8. Prove mathematically that the statement in italics following equation 29 is correct.
9. Repeat the problems starting on page 72 for increasing and decreasing impedances, but with frequencies of 2.0 megacycles instead of 5.0 megacycles, and at 7.5 instead of 10.0 megacycles.
10. The heater of a thermocouple has a constant resistance of 610 ohms. For a current of 0.002 ampere through the heater, the deflection of the associated microammeter is 10 microamperes. The thermocouple heater is placed in series with 9390 ohms, and placed across an unknown voltage. The microammeter reads 100 microamperes. What is the value of the unknown voltage in volts?
Last Update: 2011-05-18 | 599 | 2,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-10 | latest | en | 0.908705 |
https://www.examrace.com/IAS/IAS-Updates/NEWS-IAS-Prelims-CSAT-Paper-2-June-2018-Solutions-and-Video-Explanations-Part-3-Question-41-to-60.htm | 1,563,827,325,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528220.95/warc/CC-MAIN-20190722201122-20190722223122-00474.warc.gz | 676,367,666 | 17,232 | # IAS Prelims CSAT Paper 2 June 2018 Solutions and Video Explanations Part 3- Question 41 to 60 (Download PDF)
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Answers to questions are given below, DI was difficult and something new for CSAT. Questions were not easy and required covering our entire CSAT Paper-II course. Students who had carefully followed our Examrace YouTube Channel, also benefitted.
IAS Prelims CSAT Paper 2 - 2018 Solutions, Answer Key & Explanations Part 1 (Q. 1 to 20) Part 1 of 4
IAS Prelims CSAT Paper 2 - 2018 Solutions, Answer Key & Explanations Part 1 (Q. 1 to 20) Part 1 of 4
41. X and Y are natural numbers other than 1, and Y is greater than X. Which of the following represents the largest number?
(a) XY
(b) $\frac{X}{Y}\phantom{\rule{0.2em}{0ex}}$
(c) $\frac{Y}{X}$
(d) $\frac{\left(X\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}Y\right)}{XY}$
Directions: Read the following information and answer the two items that follow:
The plan of an office block for six officers A, B, C, D, E and F is as follows: Both B and C occupy offices to the right of the corridor (as one enters the office block) and A occupies on the left of the corridor. E and F occupy offices on opposite sides of the corridor but their offices do not face each other. Offices of C and D face each other. E does not have a corner office. F’s office is further down the corridor than A’s, but on the same side.
42. If E sits in his office and faces the corridor, whose office is to his left?
(a) A
(b) B
(c) C
(d) D
43. Who is/are Ps immediate neighbour/neighbours?
(a) A only
(b) A and D
(c) C only
(d) B and C
Directions for the following 7 (seven) items:
Read the following four passages and answer the items that follow. Your answers to these items should be based on the passages only.
Passage-1
‘Desertification’ is a term used to explain a process of decline in the biological productivity of an ecosystem, leading to total loss of productivity. While this phenomenon is often linked to the arid, semi-arid and sub-humid ecosystems, even in the humid tropics, the impact could be most dramatic. Impoverishment of human-impacted terrestrial ecosystems may exhibit itself in a variety of ways: accelerated erosion as in the mountain regions of the country, salinization of land as in tht semi-arid and arid ‘green revolution’ areas of the country, e. g. , Haryana and western Uttar Pradesh, and site quality decline—a common phenomenon due to general decline in tree cover and monotonous monoculture of rice/wheat across the Indian plains. A major consequence of deforestation is that it relates to adverse alterations in the hydrology and related soil and nutrient losses. The consequences of deforestation invariably arise out of site degradation through erosive losses. Tropical Asia, Africa and South America have the highest levels of erosion. The already high rates for the tropics are increasing at an alarming rate (e. g. , through the major river systems—Ganga and Brahmaputra, in the Indian context), due to deforestation and ill-suited land management practices subsequent to forest clearing. In the mountain context, the declining moisture retention of the mountain soils, drying up of the underground springs and smaller rivers in the Himalayan region could be attributed to drastic changes in the forest cover. An indirect consequence is drastic alteration in the upland-lowland interaction, mediated through water. The current concern the tea planter of Assam has is about the damage to tea plantations due to frequent inundation along the flood-plains of Brahmaputra, and the damage to tea plantation and the consequent loss in tea productivity is due to rising level of the river bottom because of siltation and the changing course of the river system. The ultimate consequences of site desertification are soil degradation, alteration in available water and its quality, and the consequent decline in food, fodder and fuel-wood yields essential for the economic well-being of rural communities.
44. According to the passage, which of the following are the consequences of decline in forest cover?
1. Loss of topsoil
2. Loss of smaller rivers
3. Adverse effect on agricultural production
4. Declining of groundwater
Select the correct answer using the code given below.
(a) 1, 2 and 3 only
(b) 2, 3 and 4 only
(c) 1 and 4 only
(d) 1, 2, 3 and 4
45. Which of the following is/are the correct inference/inferences that can be made from the passage?
1. Deforestation can cause changes in the course of rivers.
2. Salinization of land takes place due to human activities only.
3. Intense monoculture practice in plains is a major reason for desertification in Tropical Asia, Africa and South America.
Select the correct answer using the code given below.
(a) 1 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) None of the above is a correct inference
46. With reference to ‘desertification’, as described in the passage, the following assumptions have been made:
1. Desertification is a phenomenon in tropical areas only.
2. Deforestation invariably leads to floods and desertification.
Which of the above assumptions is/are valid?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Passage-2
A diversity of natural assets will be needed to cope with climate change and ensure productive agriculture, forestry, and fisheries. For example, crop varieties are needed that perform well under drought, heat, and enhanced CO2. But the private-sector and farmer-led process of choosing crops favours homogeneity adapted to past or current conditions, not varieties capable of producing consistently high yields in warmer, wetter, or drier conditions. Accelerated breeding programmes are needed to conserve a wider pool of genetic resources of existing crops, breeds, and their wild relatives. Relatively intact ecosystems, such as forested catchments, mangroves, wetlands, can buffer the impacts of climate change. Under a changing climate, these ecosystems are themselves at risk, and management approaches will need to be more proactive and adaptive. Connections between natural areas, such as migration corridors, may be needed to facilitate species movements to keep up with the change in climate.
47. With reference to the above passage, which of the following would assist us in coping with the climate change?
1. Conservation of natural water sources
2. Conservation of wider gene pool
3. Existing crop management practices
4. Migration corridors
Select the correct answer using the code given below.
(a) 1, 2, and 3 only
(b) 1, 2, and 4 only
(c) 3 and 4 only
(d) 1, 2, 3, and 4
48. With reference to the above passage, the following assumptions have been made:
1. Diversification of livelihoods acts as a coping strategy for climate change.
2. Adoption of mono-cropping practice leads to the extinction of plant varieties and their wild relatives.
Which of the above assumptions is/are valid?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Passage-3
Today, the top environmental challenge is a combination of people and their aspirations. If the aspirations are more like the frugal ones we had after the Second World War, a lot more is possible than if we view the planet as a giant shopping mall. We need to get beyond the fascination with glitter and understand that the planet works as a biological system.
49. Which of the following is the most crucial and logical inference that can be made from the above passage?
(a) The Earth can meet only the basic needs of humans for food, clothing and shelter.
(b) The only way to meet environmental challenge is to limit human population.
(c) Reducing our consumerism is very much in our own interest.
(d) Knowledge of biological systems can only help us save this planet.
Passage-4
Some people believe that leadership is a quality which you have at birth or not at all. This theory is false, for the art of leadership can be acquired and can indeed be taught. This discovery is made in time of war and the results achieved can surprise even the instructors. Faced with the alternatives of going left or right, every soldier soon grasps that a prompt decision either way is better than an endless discussion. A firm choice of direction has an even chance of being right while to do nothing will be almost certainly wrong.
50. The author of the passage holds the view that
(a) leadership can be taught through war experience only
(b) leadership can be acquired as well as taught
(c) the results of training show that more people acquire leadership than are expected
(d) despite rigorous instruction, very few leaders are produced
51. Consider the following graph:
Which one of the following statements is not correct with reference to the graph given above?
(a) On 1st June, the actual progress of work was less than expected.
(b) The actual rate of progress of work was the greatest during the month of August.
(c) The work was actually completed before the expected time.
(d) During the period from 1st April to 1st September, at no time was the actual progress more than the expected progress.
52. For a sports meet, a winners’ stand comprising three wooden blocks is in the following form:
There are six different colors available to choose from and each of the three wooden blocks is to be painted such that no two of them has the same colour. In how many different ways can the winners’ stand be painted?
(a) 120
(b) 81
(c) 66
(d) 36
Consider the following graph in which the birthrate and death rate of a country are given, and answer the two items that follow.
53. Looking at the graph, it can be inferred that from 1990 to 2010
(a) population growth rate has increased
(b) population growth rate has decreased
(c) growth rate of population has remained stable
(d) population growth rate shows no trend
54. With reference to the above graph, consider the following statements considering 1970 as base year:
1. Population has stabilized after 35 years.
2. Population growth rate has stabilized after 35 years.
3. Death rate has fallen by 10 % in the first 10 years.
4. Birthrate has stabilized after 35 years.
Which of the above are the most logical and rational statements that can be made from the above graph?
(a) 1 and 2 only
(b) 1, 2, and 3
(c) 3 and 4
(d) 2 and 4
55. Average hourly earnings per year (E) of the workers in a firm are represented in figures A and B as follows:
(a) values of E are different
(b) ranges (i. e. , the difference between the maximum and the minimum) of E are different
(c) slopes of the graphs are same
(d) rates of increase of E are different
56. Consider the figures given below:
57. Consider the following figures A and B:
The manufacturing cost and projected sales for a product are shown in the above figures A and B respectively. What is the minimum number of pieces that should be manufactured to avoid a loss?
(a) 2000
(b) 2500
(c) 3000
(d) 3500
58. A lift has the capacity of 18 adults or 30 children. How many children can board the lift with 12 adults?
(a) 6
(b) 10
(c) 12
(d) 15
59. A person bought a refrigerator worth Rs. 22, 800 with 12.5 % interest compounded yearly. At the end of first year he paid Rs. 8, 650 and at the end of second year Rs. 9, 125. How much will he have to pay at the end of third year to clear the debt?
(a) Rs. 9, 990
(b) Rs. 10, 000
(c) Rs. 10, 590
(d) Rs. 11, 250 | 2,696 | 11,431 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-30 | latest | en | 0.920959 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-1-section-1-6-properties-of-integral-exponents-exercise-set-page-80/91 | 1,537,658,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.53/warc/CC-MAIN-20180922221246-20180923001646-00244.warc.gz | 737,190,088 | 13,853 | ## Intermediate Algebra for College Students (7th Edition)
$\dfrac{10a^2}{b^5}$
RECALL: (i) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne 0$ (ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$ Divide the coefficients by each other and the variables by each other to obtain: $=\dfrac{20}{2} \cdot a^{3-1} \cdot b^{8-13} =10 \cdot a^{2} \cdot b^{-5}$ Use rule (ii) above to obtain: $=10a^2 \cdot \dfrac{1}{b^5} \\=\dfrac{10a^2}{b^5}$ | 174 | 407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-39 | longest | en | 0.586043 |
https://thenotexpert.com/smart-ways-to-calculate-power/ | 1,620,946,657,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992514.37/warc/CC-MAIN-20210513204127-20210513234127-00126.warc.gz | 575,553,617 | 8,651 | Exponentiation, or calculating a power of a number, looks like a trivial task. Just multiply a number by itself several times. But, is it the most efficient way? In this article we’ll discuss several approaches to what seems to be a basic mathematical operation. We will only discuss powers that are integers here.
First, let’s take a look at the approaches that can be encountered all over the Internet.
## The straightforward approach
First, let’s do it in Python.
And a test code. It will be pretty much the same for every approach (except the function name will be different).
I didn’t want to include negative values of k at first, but since I was playing with those anyway, why not. Also, I learnt that zero in negative power is a complicated matter, so I’ll just leave it be. Python throws ZeroDivisionError: 0.0 cannot be raised to a negative power on 0**-1 anyway.
This approach takes O(n) time and constant space. Not bad, but it can be done faster.
## The recursive approach
The fundamental quality of powers is that you can multiply two powers and their exponents sum up: $x^a * x^b = x^{(a+b)}$. You can go the other way around by dividing exponents evenly until you reach the value of 1. It can be easily done in a recursive way.
This approach takes O(log k) time, but it requires more space for recursive calls (probably O(log k) by memory, correct me if I’m wrong).
## Iterative approach
As far as I’ve heard, any task that can be performed recursively can be done iteratively as well. Can we go deeper and make the algorithm satisfy the following requirements?
• O(log k) by time
• O(1) by memory
• No modulo or division operations (except the last one for negative powers)
It is possible, with some bit magic.
First of all, let’s look at what an integer actually is from a binary representation perspective.
$k = (0 \text{ or } 1) \cdot 2^{0} + (0 \text{ or } 1) \cdot 2^{1} + (0 \text{ or } 1) \cdot 2^{2} + (0 \text{ or } 1) \cdot 2^{3} + \cdots$
That’s how an integer is represented in memory, in reverse. The bit that corresponds to $2^{0}$ (or $1$) is the rightmost.
Now, back to our exponent. Let’s create a polynomial (or is it a polynomial? Math-savvy people may correct me).
$x^{k} = x^{(0 \text{ or } 1) \cdot 2^{0} + (0 \text{ or } 1) \cdot 2^{1} + (0 \text{ or } 1) \cdot 2^{2} + (0 \text{ or } 1) \cdot 2^{3} + \cdots} = x^{(0 \text{ or } 1) \cdot 2^{0}} \cdot x^{(0 \text{ or } 1) \cdot 2^{1}} \cdot x^{(0 \text{ or } 1) \cdot 2^{2}} \cdot x^{(0 \text{ or } 1) \cdot 2^{3}} \cdot \cdots$
The rightmost bit corresponds to $x^{(0 \text{ or } 1) \cdot 2^{0}}$. Which means, $x$ can be either $1$ if the bit is 0 or $x$ if it is 1. This can be done in code very simply:
What about the rest? We can easily iterate ofer the remaining bits using k >>= 1 (or k = k >> 1) and getting the next bit with k&1 on every iteration. The next number in this polynomial will be either $1$ or $x^{2}$. The one after it is either $1$ or $x^{2^{2}} = x^{4}$. The following one is either $1$ or $x^{2^{3}} = x^{8}$. See the pattern? We just have to make a number (which I call cur_pow_x in my program) and intialize it to $x^{2}$, and multiply it by itself on every iteration. If the current bit is 1, then we just multiply the current result by cur_pow_x.
There you go. Iterative approach.
## Benchmarks
I decided to run some benchmarks for Python algorithms on every approach we discussed so far for the $9^{9}$ operation (this way I stayed within the boundaries of a 32-bit integer. Python may get slow on higher numbers because it has to increase the number of operations on numbers that are beyond limitations of the CPU architecture). The results were slightly unexpected:
powah_naive: 1.29 µs ± 40.6 ns
powah_recursive: 2 µs ± 25.4 ns
powah_iterative: 1.39 µs ± 31.3 ns
The O(k) algorithm was the fastest. But we have to remember that 9 is likely a pretty small number, and some time is consumed by the overhead of additional operations in the two latter algorithms. So they might be slower on small numbers.
Let’s do the same thing for $9^{99}$. The results:
powah_naive: 10.1 µs ± 158 ns
powah_recursive: 3.99 µs ± 160 ns
powah_iterative: 2.68 µs ± 94.8 ns
Now we can finally see the might of O(log k). And the iterative approach is faster. But what if we go for $9^{999}$ now?
powah_naive: 177 µs ± 2.95 µs
powah_recursive: 10.3 µs ± 125 ns
powah_iterative: 14.1 µs ± 218 ns
Now the recursive one is faster. And I have tested it with higher values of exponent, the iterative algorithm only gets slower compared to the recursive one. I don’t have an explanation for this at the moment.
## In Golang
As a bonus, I’ll include the same code in Golang, since I played around with it as well. Didn’t make any benchmarks though.
Keep in ming that the precision may be lost on big numbers. If you set the upper limit for n to, say, 100, the test code will throw errors, even though the results are seemingly close. Strangely enough, powah_iterative did not have this behaviour, it was the most precise.
# Conclusion
There you have it! Several approaches to exponentiation. Sounds like a simple task but I’ve spent hours playing around with it. Of course, the practical approach is by simply using x**k in Python or math.Pow(x,k) in Go, the built-ins are way faster and more available.
Thanks for tuning in! | 1,501 | 5,357 | {"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} | 4.21875 | 4 | CC-MAIN-2021-21 | latest | en | 0.906427 |
https://www.bankersadda.com/p/dear-aspirants-reasoning-questions-for_9.html | 1,553,588,114,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204885.27/warc/CC-MAIN-20190326075019-20190326101019-00274.warc.gz | 696,658,536 | 47,092 | Dear Aspirants,
Reasoning Questions for SBI Clerk Prelims Exam 2018
Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming banking exams is to practice continuously with all your heart and soul. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the exact same pattern of questions that are being asked in the exams.
Directions (1-5): Following questions are based on the five three-digit numbers given below:
369 717 922 625 434
Q1. If all the numbers are arranged in descending order from left to right then which of the following will be the product of the first and last digits of the number which is second from right end?
(a) 16
(b) 30
(c) 27
(d) 49
(e) 18
Q2. If the positions of the first and third digits of each of the number interchange, what will be the sum of first and second digits of the third lowest number thus formed?
(a) 13
(b) 15
(c) 9
(d) 11
(e) 7
Q3. If one is subtracted from each odd digit and one is added to each even digit of each of the numbers, what will be the difference between the third digit and the first digit of the highest number thus formed?
(a) Zero
(b) 6
(c) 4
(d) 2
(e) 5
Q4. What will be the resultant if the first digit of the highest number is divided by the third digit of the second lowest number?
(a) 0.5
(b) 1
(c) 2
(d) 2.2
(e) 2.25
Q5. If all the digits in each of the numbers are arranged in descending order within the number from left to right then which of the following will be the second lowest number?
(a) 369
(b) 717
(c) 922
(d) 625
(e) 434
Solution (1-5):
S1. Ans.(a)
Sol. 4*4=16
S2. Ans.(e)
Sol.526 is the third lowest number after arranging the first and last digit of each of the numbers
Sum of first and second digit= 5+2=7
S3. Ans.(e)
Sol.833 is the highest number
Required difference = 8-3=5
S4. Ans.(e)
Sol. first digit of the highest number = 9
Third digit of the second lowest number = 4
So, 9/4 = 2.25
S5. Ans.(d)
Sol.
Directions (6-8): Following questions are based on the five three-digit numbers given below
716 374 723 875 475
Q6. If the position of the first and the second digits in each of the above numbers are interchanged, which of the following will be the second lowest number?
(a) 716
(b) 374
(c) 723
(d) 875
(e) 475
Q7. What is the product of middle digits of the highest number and the lowest digit of smallest number of the above five numbers?
(a) 21
(b) 34
(c) 28
(d) 25
(e) 20
Q8. If the positions of the first and the third digits in each of the above five numbers are interchanged then which of the following will be the middle digit of the second lowest number thus formed?
(a) 1
(b) 2
(c) 7
(d) 9
(e) 6
Solution (6-8):
S6. Ans.(c)
Sol. 716 374 723 875 475
176 734 273 785 745
S7. Ans.(a)
Sol. 7*3=21
S8. Ans.(c)
Sol. 7
Q9. If in the number 3856490271, positions of the first and the second digits are interchanged, positions of the third and fourth digits are interchanged and so on till the positions of 9th and 10th digits are interchanged, then which digit will be fifth from the left end?
(a) 6
(b) 4
(c) 9
(d) 0
(e) None of these
S9. Ans.(c)
Ans. Fifth from left end =9
Q10. If the digits of the number 783219 are arranged in ascending order within the number, then how many digits are remains on the same position after rearrangement?
(a) None
(b) One
(c) Two
(d) Three
(e) More than three
S10. Ans.(c)
Ans.
Directions (11-15): These questions are based on the following set of numbers.
437 592 472 791 817
Q11. If ‘1’ is subtracted from the middle digit of each number and then the numbers are arranged in ascending order from left to right, then which of the following number is second from left?
(a) 437
(b) 592
(c) 472
(d) 791
(e) 817
Q12. If in each number, the first and the third digits are interchanged then which number will be the lowest?
(a) 791
(b) 592
(c) 472
(d) 437
(e) 817
Q13. If the first and the second digits in each number are interchanged, then which number will be the third smallest?
(a) 437
(b) 592
(c) 817
(d) 791
(e) 472
Q14. If all the numbers are arranged in ascending order from left to right then what will be the difference between the third digit of second highest number and second digit of third highest number thus formed?
(a) 1
(b) 2
(c) 4
(d) 5
(e) 8
Q15. If in each number first digit is replaced by the third digit, second digit is replaced by the first digit and third digit is replaced by the second digit, then which number will be the smallest?
(a) 437
(b) 597
(c) 472
(d) 791
(e) 817
Solution (11-15):
S11. Ans. (c)
Sol. 472 is second from left.
437 592 472 791 817
427 582 462 781 807
427 462 582 781 807
S12. Ans.(a)
Sol. lowest number is= 791
437 592 472 791 817
734 295 274 197 718
S13. Ans.(e)
Sol. 3rd smallest is= 472
437 592 472 791 817
347 952 742 971 187
S14. Ans.(e)
Sol. 437 592 472 791 817
437 472 592 791 817
Third digit of second highest number= 1,
Second digit of third highest number= 9
Diff=9-1=8
S15. Ans.(d)
Sol. 437 592 472 791 817
743 259 247 179 78 | 1,643 | 5,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.862778 |
https://misterwootube.com/posts/page/38/ | 1,582,110,988,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144111.17/warc/CC-MAIN-20200219092153-20200219122153-00049.warc.gz | 476,713,562 | 23,611 | Posted on Updated on
I’m a bit of a paradox, really. I suppose everyone is in some respect, but I seem to be self-contradictory in a whole variety of ways. (Warning: prepare for stereotypes!) I’m Asian, but I studied no sciences in high school. I went to a selective school, but I became a teacher* (and not in any of the subjects I majored in at school!). I’m Gen Y, but I started a family early. But of particular note to me today is that I’m a technology enthusiast (and generally, an early adopter) – but I’ve only quite recently started using Twitter.
I was reminded of this fact when I woke up today and noticed I’d passed this milestone (100 followers), which is notable because of how infantile it is. (In the Twittersphere, 100 followers is hardly any. It’d be like a marathon runner celebrating that he’d run 100 metres.) The number is small mainly because it’s a function of how long (and how much) I’ve used the service – which is to say, not very long (or much) at all.
This year I want to change that, since I’m convinced there are real benefits – to be had and to be shared with others! Let’s see how prescient that statement turns out to be.
* That’s a stereotype I’m determined to erode in the years to come. 🙂
### Nemesis
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I enjoy doing maths and I spend a lot of time working on it, but I have a hard time calling myself a mathematician. It’s not because I dislike the label – on the contrary, I don’t feel as though I’m really worth of the title. Real mathematicians… well, they’re the kind of people who go to the National Mathematics Summer School (NMSS, affectionately pronounced as Nemesis).
Perhaps you think you know some nerds. Do they chuckle with childish delight when considering the cyclical nature of inverses that exist in the set of Gaussian integers modulo the complex number (4 + i)? No? Then step aside and let the real nerds take the stage. These guys – and hence by extension, their tutors and lecturers (who are mostly NMSS alumni) – are the real deal.
I would never have attended NMSS as a student. I didn’t have anywhere near the mathematical chops to even be considered as a candidate (there are roughly 70 positions for the entirety of Australia). But I may well have enjoyed it if I had been invited. Since it’s a gathering of students from across the country, they try to assume very little prior knowledge – hence their focus on number theory, which is renowned as easily accessible and abundance of opportunities to “think deeply about simple things”, the motto of NMSS’s recently retired director.
It’s intentionally different from a school learning environment, which by its very nature emphasises assessment and competitiveness. No one hands in their problems and marks aren’t assigned for anything. The whole experience is crafted to encourage exploration, playfulness and creativity. If you’re not a maths teacher – or even if you are – and those words seem like the antithesis of mathematics to you, then that’s a sad testimony to just how different high school maths is to the actual maths that mathematicians do. (I’m not sure if that’s a gap that will ever be bridged, but there it is for whatever you want to make of it.)
But this week, I wasn’t there as a student – I was there as a teacher, to get a concentrated version of what the students were experiencing and then to think about how that would inform our practice as educators (particularly with regard to nurturing and encouraging gifted and talented mathematicians). It was a jam-packed couple of days and I found myself constantly thinking of new and awesome ideas that I would love to start implementing when I get back to the real world, but unfortunately I think I’ve just about maxed out (or exceeded) the number of new things I’ll be doing this year. So mostly I think I was mentally filing things away for the future, waiting for a time when I can act on them and give them the time and effort they deserve.
One thing that remains deeply impressed on my mind, though, is the importance of teaching mathematics in an engaging way (and, related to that, encouraging people who are capable of that into the profession rather than ushering them off into engineering or actuarial studies). Being exposed to so many passionate maths teachers (and I use that term broadly of anyone who teaches mathematics, not just people who work in high schools) was a vivid reminder of how important the delivery method is in shaping a students’ experience of a subject.
If someone teaches you how to cook by forcing you through lessons and explaining things in a bland way (see what I did there?), then who can blame you for disliking the kitchen? But if someone visibly enjoys the process of mashing food together in an awful mess, if they express genuine delight at the intriguing ways that foods can relate and be combined with one another, if they marvel with closed eyes at the smell of what they have just concocted, then who can help but feel inspired to try and master the same subject that brings so much joy? And I think that is a part of why Jamie Oliver rose to fame so rapidly (and subsequently kept it). I hate cooking, I hate the lengthy preparation, I hate the mess, and I hate the low-quality stuff that I usually produce. But when I watch Jamie Oliver at work, I want to get up and cook. I want to give it a go and learn how he does what he does. And that’s exactly the same vibe that the NMSS tutors and lecturers give off to the students who are privileged enough to attend.
Wouldn’t it be fantastic if ordinary students could experience some of that during their normal schooling? It shouldn’t just be for the elites. I’m under no illusion that the whole NMSS experience can be replicated on a large scale for the entirety of a school year, but the world deserves to know that mathematics is a fascinating and amazing subject – not the dry, boring thing that most people think maths is. And we’ll need passionate mathematicians and educators to accomplish that. Now there’s a long term goal worth working on!
### Where the milkshakes are tall and the hours are short
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Right now I’m in Canberra for the National Mathematics Summer School, which is a program run for talented mathematics students from all over the country. This year for the first time, they’re trialing a teacher component to the NMSS – two intensive days (rather than the two weeks that the students get) which give us a brief taste of what the school is like for the students, and also provide an opportunity for us to do some professional development too. (Getting my head back into undergraduate maths for three hours? Yep, my brain is pretty much melted right now.)
But that’s not what I wanted to write about today. (I’ll save my reflection for the bus ride back to Sydney.) I had the pleasure of meeting and sitting down to chat with the exuberant Betty Chau (who maintains the fantastic @PositiveSchool account, not to mention a fantastic blog) over a milkshake in a lovely outdoor cafe close to ANU (where I’m staying). I shouldn’t speak for her, but I had a fantastic time – evidenced by the fact that I remained completely oblivious to the passing time and became totally absorbed in our conversation way over the time when I was meant to be back on campus for dinner. (For the record Betty, I did in fact make it back inside just as they were about to pack up and finish serving – so thanks for driving that little extra distance at the end!)
Aside from raving about the superiority of the ACT secondary schooling system (something I’m totally willing to admit – sorry NSW, I love you but I hate you too), Betty shared with me her passion for positive psychology. (For the uninformed, to give you a brief and over-simplified summary, it’s an approach toward people’s mental wellbeing that focuses on their strengths rather than on what’s wrong with them.) I found it both intriguing and refreshing, for a few different reasons.
Firstly, though the idea of positive psychology seems to be gaining currency, the practice of positive psychology in the schools that I’m familiar with is lacking to say the least. When Betty told me a few things about how positive principles were a part of her classroom practice, the kinds of structures that some schools in Canberra have in place to implement it and the kind of helpful effect it was having on her students, I was just blown away. This is something that we definitely need to learn from.
Secondly, it dawned on me just how relevant it was for all teachers – but particularly mathematics teachers, since that’s what’s going through my head at NMSS – to be caring for the psychological needs of our students in a positive and nurturing way. Mathematics has a just reputation as possibly the most demoralising high school subject in existence, with a massive (usually damaging) psychological effect on thousands of students every year. I’m convinced that this is part of the reason why so many students are convinced that they are bad at maths: regardless of their actual mathematical ability, they have had a series of bad experiences with studying maths that have left them psychologically scarred and they have (understandably) just given up on ever understanding it. This is a great tragedy and something that drastically needs to be changed.
Thirdly, Betty’s smile and enthusiasm are pretty infectious. That’s not a very professional assessment of the facts but it’s true!
So as I continue to think about the new role I’m starting this year, I’m now faced with this question: amid all the organisational decisions, the results analysis, the academic rigour and everything else that a head teacher is supposed to be prioritising – how will I care for my students’ wellbeing? What principles will I bring to the way I interact with students and lead my faculty to ensure that we develop their strengths rather than become fixated on their weaknesses?
For that matter – how will you be caring for your students’ wellbeing? Now there’s a question worth considering as January 28 approaches!
### My new workspace
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Teaching myth #1: teachers are lazy because they get massive holidays (at least, compared to the 4 weeks of annual leave that most workers in Australia get). This myth is perpetuated because it’s half-true (yes, we do get a relatively large amount of leave – roughly 11 weeks of breaks between school terms), but it’s not true. Teachers who are idle during their school holidays are the exception, not the rule. Case in point: here I am in the middle of summer, setting up my new workspace during the holidays.
Okay, I’ll admit that (especially given the new step-up in responsibility involved in my new role) it would have been plain foolish to not come in during the holidays to bring in all my resources and arrange my desk before the school term begins officially. Here’s how it looks at the moment.
It’s pretty bare right now, but that’s because term hasn’t begun yet. Just wait… the chaos is coming.
The staffroom I’m working in is quite different to the one at my previous school. It’s a combined staffroom, housing 4 faculties (Science, Technology, Creative & Performing Arts, Mathematics) and 67 staff. It’s not the first time I’ve experienced this kind of arrangement – Fort Street High School (where I did a practicum and subsequently had my first full-time job) also has combined staffrooms – but the scale of this one is still something to behold.
Click the photo above for a bigger version. This image is a pair of panoramas combined together, but the stitching in the middle isn’t perfect. It’s still good enough to get a sense of the working environment, though.
### New… year, school, role, blog!
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I have always loved to write. Since a young age, I have always owned little notebooks where I scrawled down thoughts and ideas. During high school, I always enjoyed creative writing and subjected myself to the vagueness of English Extension 1 (to me, anyway) so that I could enrol in English Extension 2 and create a major work. It was at uni that I discovered the joy of writing for its own sake – writing because I wanted to, not because I had to in connection with some external assessment or other requirement. Writing because it was a useful way to form, develop and retain thoughts. I started a blog, made methodical notes on anything I wanted to really learn, and engaged in all kinds of other writing to keep my brain going.
In the last 15 years, I’ve created and killed countless writing projects. It’s been a while since I started a new one – so, why a new blog in 2014? It’s a paradox, really. This year, I’ll be undergoing the biggest professional change since I started full-time work. I’ll be in a new role at a new school and the learning curve will be very steep, particularly in the first 12-18 months. That makes it both the best and worst time to start a writing project like this. It’s the worst time because I’ll have the least time to commit to it; it’s the best time because a time like this, when I am going to be going through a lot of new experiences and forming new working principles, is precisely when writing is most beneficial.
So here goes nothing. Let’s see how long it lasts. I hope to learn lots as I write here – and maybe even you will too. | 2,832 | 13,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.960299 |
https://amp.doubtnut.com/question-answer/in-the-diagram-abcd-is-a-rectangle-and-three-circles-are-positioned-as-shown-the-area-of-the-shaded--231601 | 1,582,489,006,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145839.51/warc/CC-MAIN-20200223185153-20200223215153-00539.warc.gz | 289,016,442 | 18,324 | IIT-JEE
Apne doubts clear karein ab Whatsapp (600 300 8001) par bhi. Try it now.
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Question From class 12 Chapter DEFAULT
In the diagram, ABCD is a rectangle, and three circles are positioned as shown. The area of the shaded region,rounded to the nearest cm. is:
Find the Area of Shaded Region bounded between two semicircles drawn on side of length 28cm of a rectangle as diameter. The other side of the rectangle is 14 cm long. Semi-Circles Rectangle Shaded Area
16:48
In the given figure, ABCD is rectangle and EF is perpendicular to AB. The area of the shaded region, in the given figure, is
1:45
"a" cm ???????? ???? ??? ????? ??-????? ?? ????? ??? ?? ?????? ???? ???? ???????? ??? ?? ????????? ????? ????? <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RKY_MATH_C21_E01_137_Q01.png" width="80%">
2:42
In the given figure, ABCD is a rectangle I,II and III are squares. Find the area of the shaded region. <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PS_MATH_VI_C09_E02_005_Q01.png" width="80%">
4:06
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semi-circles
4:22
Find the area of the shaded region in Fig. 15.49, if is a square of side 14 cm and and are semi-circles. (FIGURE)
1:22
In the shown figure, two circles of radii of 7 cm each, are shown. ABCD is rectangle and AD and BC are the radii. Find the area of the shaded region (in ). <br> ltimg src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PS_MATH_X_C14_E04_023_Q01.png" width="80%"gt
3:54
In the given figure, ABCD is a square of side 7 cm. DPBA and DQBC are quardrants of circles each of the radius 7 cm. Find the area of the shaded region. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_X_C12_E01_043_Q01.png" width="80%">
4:58
In the following figure, a rectangle ABCD enclosed three circles. If , find the area of the shaded portion
2:32
In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
2:20
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WhatsApp par bhi kare apne doubts clear!
600 300 8001 | 1,171 | 3,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.778541 |
https://testbook.com/question-answer/by-selling-an-article-for-%E2%82%B92400-a-trader-suffe--62aee3f647e88863f5d17f84 | 1,721,834,840,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00131.warc.gz | 480,897,298 | 48,147 | By selling an article for ₹2,400, a trader suffers a loss of 20%. In order to earn a profit of 20%, at what price (in ₹) should he sell the article?
This question was previously asked in
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1. 3,000
2. 3,200
3. 3,600
4. 3,800
Option 3 : 3,600
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Detailed Solution
Given:
Selling Price of an article = ₹2,400
Loss% = 20%
Formula used:
Selling Price = Cost Price × $$\frac{100\ ±\ (profit\%\ or\ loss\%)}{100}$$ ( negative for loss and positive for profit )
Calculation:
2400 = C.P × $$\frac{100\ -\ 20}{100}$$
2400 = C.P × $$\frac{80}{100}$$
Cost Price = 3000
to earn a profit of 20%
Selling Price = Cost Price × $$\frac{100\ +\ 20}{100}$$
S.P = 3000 × $$\frac{1200}{100}$$
S.P = 3600 | 344 | 948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-30 | latest | en | 0.68425 |
https://www.mathworks.com/help/mpc/ug/setting-targets-for-manipulated-variables.html | 1,656,350,859,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103337962.22/warc/CC-MAIN-20220627164834-20220627194834-00090.warc.gz | 958,151,075 | 18,559 | # Setting Targets for Manipulated Variables
When there are more manipulated variables than outputs, assuming that the static gain matrix is full rank, it is possible for the controller to reach any given steady state point in the output space using many different possible combinations of manipulated variable values.
In this case, for economic or operational reasons, you can choose to set target values, and some corresponding nonzero cost function weights, for some manipulated variables (up to the excess number of manipulated variables with respect to the number of outputs). The remaining manipulated variables can attain the values required to track any point in the steady-state output space.
This example shows how to design a model predictive controller for a plant with two inputs and one output with target setpoint for one of the two manipulated variables.
### Define Plant Model
The linear plant model has two inputs and one output. Define the plant as a transfer function, convert it to state space, specify the initial state, and extract the plant matrices for later use within the Simulink model.
```plant = ss(tf({[3 1],[2 1]},{[1 2*.3 1],[1 2*.5 1]})); x0 = [0 0 0 0]'; A = plant.A; B = plant.B; C = plant.C; D = plant.D; ```
### Design MPC Controller
Create an MPC controller with sampling time `0.4` s, and prediction and control horizons of `20` and `5` steps, respectively.
```mpcobj = mpc(plant,0.4,20,5); ```
```-->The "Weights.ManipulatedVariables" property is empty. Assuming default 0.00000. -->The "Weights.ManipulatedVariablesRate" property is empty. Assuming default 0.10000. -->The "Weights.OutputVariables" property is empty. Assuming default 1.00000. ```
Specify weights for both manipulated variables and output.
```mpcobj.weights.manipulated = [0.3 0]; % weight difference MV#1 - Target#1 mpcobj.weights.manipulatedrate = [0 0]; mpcobj.weights.output = 1; ```
Define constraints for the manipulated variable rate.
```mpcobj.MV = struct('RateMin',{-0.5;-0.5},'RateMax',{0.5;0.5}); ```
### Set a target for one manipulated variable
Specify target setpoint `u = 2` for the first manipulated variable.
```mpcobj.MV(1).Target=2; ```
Define the model name and open the Simulink model. Note that the output reference is a square wave. Then simulate the model, using the `sim` command.
```mdl = 'mpc_utarget'; open_system(mdl) sim(mdl); ```
```-->Converting model to discrete time. -->Assuming output disturbance added to measured output channel #1 is integrated white noise. -->The "Model.Noise" property is empty. Assuming white noise on each measured output. ```
```bdclose(mdl) % close the Simulink model ``` | 648 | 2,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-27 | longest | en | 0.781795 |
https://www.answers.com/Q/What_is_the_difference_between_expressing_consecutive_even_integers_and_consecutive_odd_integers_algebraically | 1,611,243,575,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524858.74/warc/CC-MAIN-20210121132407-20210121162407-00767.warc.gz | 653,328,777 | 31,027 | Algebra
Numbers
# What is the difference between expressing consecutive even integers and consecutive odd integers algebraically?
###### Wiki User
There is no difference, only your outcome. The formula for both is x+2
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## Related Questions
30 + 31 + 32 = 93 ( I forgot how to find this algebraically )
56.25% is not an integer and so cannot be expressed as a sum, difference, product or quotient of consecutive integers.
Algebraically, X = integer. X + (X + 1) = 89 2X + 1 = 89 2X = 88 X = 44 ------------so X + 1 which is 44 + 1 = 45 ( 44, 45) --------------are your consecutive integers
Algebraically, with X = integers.X + (X + 1) + (X + 2) = 1923X + 3 = 1923X = 189X = 63============solution set
I assume you mean what are the two consecutive integers. Algebraically; X = integers.X + (X + 1) = 2752X + 1 = 2752X = 274X = 137============solution setorSince the consecutive integers differ by 1, then the first number is 274/2 = 137, so the second one is 1 more, 138.
The first two consecutive prime numbers that have a difference of 20 are the numbers 887 and 907.
There are no "two consecutive integers" that can do that.But there are two consecutive even integers that can: 8 and 10 .
"Consecutive" integers are integers that have no other integer between them.
Algebraically. X = integers. X + (X + 1) = - 379 2X + 1 = - 379 2X = - 380 X = - 190 X + 1 = - 189 - 190 +(- 189) = - 379 ==============solution set of integers
Algebraically, X = integers. X + (X + 1) = 237 gather all terms on the left 2X + 1 = 237 subtract 1 from each side 2X = 236 divide both sides integers by 2 X = 118 --------------so, X + 1 = 119 ----------------so, The two consecutive integers that = 237 are 118 and 119 -------------------
Algebraically. X = integers.X + (X + 1) = 592X + 1 = 592X = 58X = 29X + 1 = 30(29, 30)=======
The two consecutive integers are 19 and 20.
They are consecutive odd integers: 25 and 27.
There is no set of three consecutive integers for 187.
There is no set of three consecutive integers for 106.
There are no two consecutive even integers, consecutive odd integers, or consecutive integers that satisfy that relationship.
There is no set of six consecutive integers for -4.
Algebraically. X = integers. X + (X + 2) = - 42 2X + 2 = - 42 2X = - 44 X = - 22 X + 2 = - 20 (- 20, - 22 ) ---------------------solution set
Not possible in consecutive integers, nearest is consecutive even integers: 148 & 152
find the two consecutive odd integers with a sum of 152
The sum of two consecutive integers will always be an odd number.
No, it is not. It is the sum of two consecutive integers: 46 and 47.
###### Numbers Math HistoryMath and ArithmeticPrime NumbersAlgebraSchool Subjects
Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 855 | 2,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-04 | latest | en | 0.902017 |
https://www.redblobgames.com/articles/curved-paths/grids.html | 1,721,097,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00193.warc.gz | 830,471,329 | 7,443 | # Curved Paths on a Grid
from Red Blob Games
May 2014
My road applet on this page[1] no longer works on my computer, so I decided to build something in svg. This is a quick & dirty afternoon project.
The idea is that roads in a game like Transport Tycoon or SimCity should be drawn from the edge of a grid tile to another edge. Edges could be either horizontal or vertical. The resulting curve will depend both on orientation of the edges and on the grid distance between them. The original applet was interactive and let you draw out your own curves. This project is mainly for me, so I instead drew them all (see Ladder of Abstraction[2]) to get a sense of what the family of curves looks like.
### Bezier curves
Note that some are pretty stretchy, with tight curves at the endpoints. Not pleasant. Maybe in a game you’d disallow certain awkward shapes.
### Biarcs
I like circular arcs as an alternative to bezier curves. Here’s the article explaining why[3]. In general we need up to two arcs to represent a curved road segment.
In the grid setting, the “C-shaped” asymmetric biarcs contain straight lines plus 90° arcs. The “S-shaped” symmetric biarcs contain two arcs that are equal in size and shape, but mirrored around a center point. Many but not all of these could be expressed in a simpler form as arc-line-arc, where both arcs are 90°. This suggests that a set of tiles that build only 90° arcs could be used to build all the other combinations. Some of these S-shaped biarcs exceed the expected bounding box; using arc-line-arc would solve that too.
Am I going to do anything with this? I don’t know. Games like Locomotion and P1Sim have tiles for curved roads and tracks, and I might want to do the same in some future project. I’m also curious how this extends to hexagons. That’s for another day; this was just a quick afternoon project.
Email me , or tweet @redblobgames, or comment: | 445 | 1,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.939828 |
https://www.extendoffice.com/product/kutools-for-excel/excel-dynamic-pie-of-pie-chart.html | 1,670,419,635,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00102.warc.gz | 758,083,219 | 23,072 | Note: The other languages of the website are Google-translated. Back to English
English
## Easily create a dynamic pie of pie chart in Excel
### Kutools for Excel
In Excel, you can use the built-in Pie of Pie chart to show proportions of a whole, but it supports only one series at a time. Here, with the Dynamic Pie Chart feature of Kutools for Excel, you can easily create a chart with two pies that not only shows proportions of the sum of each row or column in the selected range, but also proportions of the data in a specified row or column at the same time. In addition, you can dynamically change the displayed data by changing the selection from the drop-down list.
#### Create a dynamic pie of pie chart in Excel
Just need a few clicks, you can create a dynamic pie of pie chart in Excel with this feature. Please follow the steps below to get it done.
1. Click Kutools > Charts > Data Distribution > Dynamic Pie Of Pie Chart to enable this feature.
2. In the Dynamic Pie Of Pie Chart dialog box, you need to do the following settings.
2.1) In the Data range box, select the range of data to show in the pie;
2.2) In the Axis Labels box, select the column headers;
2.3) In the Legend Entries (Series) box, select the row headers;
2.4) Click OK. See screenshot:
Tips:
1) If you select the whole range beforehand and then enable this feature, the corresponding fields will be filled with correct data range automatically.
2) You can click the Example button to open the sample file to learn more.
Then a dynamic pie of pie chart is created as shown in the demo below:
Notes:
1) You can decide which series or category can be displayed in the chart by changing the selection in the drop-down list;
2) You can change the displayed text of the radio button by right clicking it and selecting Edit Text from the context menu, and then modifying the text as you need.
3) Under the Option 1 radio button, the first pie shows proportions of the sum of each row in selected range, and the second pie shows proportions of the data in a row based on the selected row header in the drop-down list;
4) Under the Option 2 radio button, the first pie shows the proportions of the sum of each column in selected range, and the second pie shows proportions of the data in a column data based on the selected column header in the drop-down list;
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Office Tab: Using handy tabs in your Office, as the way of Chrome, Firefox and New Internet Explorer.
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### Kutools for Excel
The functionality described above is just one of 300 powerful functions of Kutools for Excel.
Designed for Excel(Office) 2021, 2019, 2016, 2013, 2010, 2007 and Office 365. Free download and use for 60 days. | 675 | 2,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | longest | en | 0.838651 |
http://www.mathsweek.ie/2014/puzzles/irish-times-maths-week-ireland-2014-quiz-1/barmy-bakers | 1,558,447,120,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256381.7/warc/CC-MAIN-20190521122503-20190521144503-00436.warc.gz | 302,206,291 | 5,941 | # Barmy Bakers
With Hallowe’en approaching a shopkeeper asked his local baker to prepare traditional barmbracks for his customers. The baker was unfamiliar with the customs so the shopkeeper explained about putting in a ring as an omen of marriage, a coin predicting wealth and a pea portending hard times. When they were delivered it transpired that the baker had put only one item in each brack. The shopkeeper was less than happy, exclaiming, “you would have to eat 15 bracks to be certain of getting a ring, you would have to eat 11 bracks to be certain of getting a coin and you would have to eat 17 to be sure of getting a pea.” How many bracks were delivered?
Solution 20 bracks
The baker would need to eat
15 bracks to be certain of getting a ring
11 bracks to be certain of getting a coin and
17 to be sure of getting a pea.
This means that there are
14 bracks that have no ring
10 bracks with no coin
16 bracks with no pea.
If we call the number of bracks with a ring, R
And, we call the number of bracks with a coin, C
And the number of bracks with a pea, P
Then
C + P = 14 (1)
R + P = 10 (2)
R + C = 16 (3)
R + C = 16 (3)
- (R + P = 10) (2)
= C - P = 6
C- P = 6
+(C + P) = 14
2C = 20
C = 10
R + C = 16
C = 10
Therefore
R = 6
C + P = 14
C = 10
P = 4
R + C + P = 6 + 10 + 4 = 20 | 422 | 1,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.969026 |
https://www.physicsforums.com/threads/intermideate-value-therem-question.529125/ | 1,542,465,946,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743714.57/warc/CC-MAIN-20181117144031-20181117170031-00403.warc.gz | 964,378,247 | 13,578 | # Homework Help: Intermideate value therem question
1. Sep 11, 2011
### nhrock3
prove that for c<0
there is only one solution to
$$xe^{\frac{1}{x}}=c$$
??
for x=1 we have f(1)>0
the limit as x->-infinity is -infinity
what to do?
?
2. Sep 11, 2011
### LCKurtz
Begin by drawing a graph to get an idea. In particular, what if x is less than 0 but close to 0?
3. Sep 11, 2011
### nhrock3
close to zero from minus
its minus infinity
when x goes to minus infinity its 0
how it helps me?
4. Sep 11, 2011
### nhrock3
close to sero is 0
5. Sep 11, 2011
### LCKurtz
If c < 0 and you know your function approaches 0 as x → 0- and approaches -∞ a x → -∞, can you conclude your function = c for some x?
6. Sep 12, 2011
### nhrock3
i need to show that there is x1 f(x1)<c
f(x2)>c
from the limit when x goes to sero we get zero -e<f(x)<e
from the limit when x goes minus infinity f(x)<-N
what e to chhose?
what N to choose?
7. Sep 12, 2011
### estro
LCKurtz provided great hint, I'll try to help as well.
To prove that the solution is single:
If c<0 what can you conclude about the existence of the solution (to your function) in $$[0,\infty)$$.
In addition what can you tell about yours function behavior in $$(-\infty,0)$$, how this helps you?
To prove solution existence:
See LKurtzs hint.
Last edited: Sep 12, 2011 | 446 | 1,338 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-47 | latest | en | 0.896746 |
https://opentextbc.ca/introbusinessstatopenstax/chapter/test-of-a-single-variance/ | 1,638,677,378,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00570.warc.gz | 505,772,564 | 25,579 | The Chi-Square Distribution
# 57 Test of a Single Variance
Thus far our interest has been exclusively on the population parameter μ or it’s counterpart in the binomial, p. Surely the mean of a population is the most critical piece of information to have, but in some cases we are interested in the variability of the outcomes of some distribution. In almost all production processes quality is measured not only by how closely the machine matches the target, but also the variability of the process. If one were filling bags with potato chips not only would there be interest in the average weight of the bag, but also how much variation there was in the weights. No one wants to be assured that the average weight is accurate when their bag has no chips. Electricity voltage may meet some average level, but great variability, spikes, can cause serious damage to electrical machines, especially computers. I would not only like to have a high mean grade in my classes, but also low variation about this mean. In short, statistical tests concerning the variance of a distribution have great value and many applications.
A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance. The test statistic is:
where:
• n = the total number of observations in the sample data
• s2 = sample variance
• = hypothesized value of the population variance
You may think of s as the random variable in this test. The number of degrees of freedom is df = n – 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. (Figure) will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.
Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average.
Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?
Even though we are given the population standard deviation, we can set up the test using the population variance as follows.
• H0: σ2 ≤ 52
• Ha: σ2 > 52
Try It
A SCUBA instructor wants to record the collective depths each of his students’ dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be?
H0: σ2 = 32
Ha: σ2 < 32
With individual lines at its various windows, a post office finds that the standard deviation for waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes on a Friday afternoon.
With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers.
Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2.
Random Variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times.
• H0: σ2 ≥ 7.22
• Ha: σ2 < 7.22
The word “less” tells you this is a left-tailed test.
Distribution for the test:, where:
• n = the number of customers sampled
• df = n – 1 = 25 – 1 = 24
Calculate the test statistic:
where n = 25, s = 3.5, and σ = 7.2.
The graph of the Chi-square shows the distribution and marks the critical value with 24 degrees of freedom at 95% level of confidence, α = 0.05, 13.85. The critical value of 13.85 came from the Chi squared table which is read very much like the students t table. The difference is that the students t distribution is symmetrical and the Chi squared distribution is not. At the top of the Chi squared table we see not only the familiar 0.05, 0.10, etc. but also 0.95, 0.975, etc. These are the columns used to find the left hand critical value. The graph also marks the calculated χ2 test statistic of 5.67. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion.
Make a decision: Because the calculated test statistic is in the tail we cannot accept H0. This means that you reject σ2 ≥ 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes or more; you think the variation in waiting times is less.
Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes.
Professor Hadley has a weakness for cream filled donuts, but he believes that some bakeries are not properly filling the donuts. A sample of 24 donuts reveals a mean amount of filling equal to 0.04 cups, and the sample standard deviation is 0.11 cups. Professor Hadley has an interest in the average quantity of filling, of course, but he is particularly distressed if one donut is radically different from another. Professor Hadley does not like surprises.
Test at 95% the null hypothesis that the population variance of donut filling is significantly different from the average amount of filling.
This is clearly a problem dealing with variances. In this case we are testing a single sample rather than comparing two samples from different populations. The null and alternative hypotheses are thus:
The test is set up as a two-tailed test because Professor Hadley has shown concern with too much variation in filling as well as too little: his dislike of a surprise is any level of filling outside the expected average of 0.04 cups. The test statistic is calculated to be:
The calculated test statistic, 6.96, is in the tail therefore at a 0.05 level of significance, we cannot accept the null hypothesis that the variance in the donut filling is equal to 0.04 cups. It seems that Professor Hadley is destined to meet disappointment with each bit.
Try It
The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of 2012, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the distribution and mark the area associated with the level of confidence, and draw a conclusion. Test at the 1% significance level.
H0: σ2 = 12.22
Ha: σ2 > 12.22
df = 14
chi2 test statistic = 16.39
The p-value is 0.2902, so we decline to reject the null hypothesis. There is not enough evidence to suggest that the variance is greater than 12.22.
In 2nd DISTR, use7:χ2cdf. The syntax is (lower, upper, df) for the parameter list. χ2cdf(16.39,10^99,14). The p-value = 0.2902.
### References
“AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013).
Data from the World Bank, June 5, 2012.
### Chapter Review
To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation).
### Formula Review
Test of a single variance statistic where:
n: sample size
s: sample standard deviation
: hypothesized value of the population standard deviation
df = n – 1 Degrees of freedom
Test of a Single Variance
• Use the test to determine variation.
• The degrees of freedom is the number of samples – 1.
• The test statistic is , where n = sample size, s2 = sample variance, and σ2 = population variance.
• The test may be left-, right-, or two-tailed.
Use the following information to answer the next three exercises: An archer’s standard deviation for his hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less.
What type of test should be used?
a test of a single variance
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test?
a left-tailed test
Use the following information to answer the next three exercises: The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81.
What type of test should be used?
State the null and alternative hypotheses.
H0: σ2 = 0.812;
Ha: σ2 > 0.812
df = ________
Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought.
What type of test should be used?
a test of a single variance
What is the test statistic?
What can you conclude at the 5% significance level?
### Homework
Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.
Is the traveler disputing the claim about the average or about the variance?
A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes.
225
Is this a right-tailed, left-tailed, or two-tailed test?
H0: __________
H0: σ2 ≤ 150
df = ________
chi-square test statistic = ________
36
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the area associated with the level of confidence.
Check student’s solution.
Let α = 0.05
Decision: ________
Conclusion (write out in a complete sentence.): ________
How did you know to test the variance instead of the mean?
The claim is that the variance is no more than 150 minutes.
If an additional test were done on the claim of the average delay, which distribution would you use?
If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use?
a Student’s t– or normal distribution
A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated?
Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of ?84 and a sample standard deviation of ?12, test the claim that the standard deviation is greater than ?15.
1. H0: σ = 15
2. Ha: σ > 15
3. df = 42
4. chi-square with df = 42
5. test statistic = 26.88
6. Check student’s solution.
1. Alpha = 0.05
2. Decision: Cannot reject null hypothesis.
3. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
4. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.
Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes.
Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief.
1. H0: σ ≤ 3
2. Ha: σ > 3
3. df = 17
4. chi-square distribution with df = 17
5. test statistic = 28.73
6. Check student’s solution.
1. Alpha: 0.05
2. Decision: Cannot accept the null hypothesis.
3. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
4. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.
The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in (Figure). Does the students’ survey indicate that the standard deviation is greater than 0.75?
# of births Frequency
0 5
1 30
2 10
3 5
According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11
1. H0: σ = 2
2. Ha: σ ≠ 2
3. df = 14
4. chi-square distiribution with df = 14
5. chi-square test statistic = 5.2094
6. Check student’s solution.
1. Alpha = 0.05
2. Decision: Cannot accept the null hypothesis
3. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
4. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.
The manager of “Frenchies” is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz.
You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of ?1,249 with a very narrow standard deviation of ?25. You find a website that has a price comparison for the same computer at a series of stores as follows: ?1,299; ?1,229.99; ?1,193.08; ?1,279; ?1,224.95; ?1,229.99; ?1,269.95; ?1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis?
The sample standard deviation is ?34.29.
H0 : σ2 = 252
Ha : σ2 > 252
df = n – 1 = 7.
test statistic: ;
Alpha: 0.05
Decision: Cannot reject the null hypothesis.
Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.
A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test.
(a) at the 5% significance level
(b) at the 1% significance level
Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172; | 4,010 | 17,137 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | latest | en | 0.940763 |
covid19.eraheem.com | 1,643,356,191,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00198.warc.gz | 237,531,800 | 10,627 | # Report 4 Projections
Report last updated: 2020-09-10
• In this udpate, second round of prjections (Run 2) have been added. Please scroll down to find the projections. As usual, the last two current values were not used for fitting the model. Also, the model is conservative/pessimistic i.e., it slightly overestimates new cases.
• In addition to that, a new paramter to control the run_date was added to the fit_seir() function. See the code for more details.
## 4.1 Projection for Bangladesh (unofficial) Run1
Run date: 17 April 2020
This is a dynamic document and will be updated frequently. Please check back for latest projection. This projection is based on unofficial Bangladesh incidence data.
### 4.1.1 Methodology
This projection is based on an SEIR model. Here, the S stands for Susceptible, E stands for Exposed/infected but asymptomatic, I stands for Infected and symptomatic, and R stands for Recovered. N is the population size.
Assuming there is no births or deaths in a population, (known as a closed population), the model is formulated by the following differential equations.
\begin{align} \frac{\partial S}{\partial t} & = -\frac{\beta I S}{N} \\ \frac{\partial E}{\partial t} &= \frac{\beta I S}{N} -\sigma E \\ \frac{\partial I}{\partial t} &= \sigma E - \gamma I \\ \frac{\partial R}{\partial t} &= \gamma I \end{align}
Here the parameters $$\beta$$ controls the transmission rate, which is the product of contact rate and the probability of transmission given contact betwen S and E compartments. $$\sigma$$ controls the transition from E to I, and $$\gamma$$ controls the transition from I to R.
The reproduction rate, $$R_0$$ can be approximated by $R_0 = \frac{\beta}{\gamma}$ In plain language, $$R_0$$ tells us how many people are infected from one patient. An $$R_0>1$$ indicates the epidemic is at the grotwh phase. $$R_0<1$$ means the epidemic is slowing or decaying.
Model was fitted using all but the last two day’s incidences to obtain the estimated $$\beta$$ and $$\gamma$$. The fitted model was used for prediction. This post was inspired by Churches (2020), where you will find some details on the computation.
### 4.1.2 Ascertainment Rate
Not all the cases are reported or tested. Usually a fraction of the actual cases are detected. This is known as ascertainment rate. We consider 25%, 50%, 75% amd 90% ascertainment rate when fitting the model.
Simply, the incidences are inflated by the inverse of the ascertainment rate.
Please let me know if you find any error in it. The code was adapted from Churches (2020)
library(deSolve)
library(grid)
library(gridExtra)
##
## Attaching package: 'gridExtra'
## The following object is masked from 'package:dplyr':
##
## combine
######################################
## SIER Modeling -------
######################################
# Parameters
# beta = rate of expusore from susceptible infected contact
# sigma = rate at which exposed person becomes infected
# gamma = rate at which infected person recovers
# S = Initial susceptible population
# E = Initial exposed population
# I = Initial infected population
# R = Recovered population
N=170000000,
af=0.5, npast=2, nfuture=5,
run_date = today()){
# country = Country name
# N = population size of the country
# af = ascertainment factor, default = 0.5
# npast = number of days in the past to exclude when fitting the model
# default is npast = 2
# nfuture = number of days in the future the algorithm to predict to
# default is nfuture=5
# run_date = sets the cutoff date so that the later runs do not overrite
# previous runs. Default is today()
SEIR <- function(time, state, parameters) {
par <- as.list(c(state, parameters))
with(par, {
dS <- -beta * I * S/N
dE <- beta * I * S/N - sigma * E
dI <- sigma * E - gamma * I
dR <- gamma * I
list(c(dS, dE, dI, dR))
})
}
# define a function to calculate the residual sum of squares
# (RSS), passing in parameters beta and gamma that are to be
# optimised for the best fit to the incidence data
names(parameters) <- c("beta", "sigma", "gamma")
out <- ode(y = init, times = Day, func = SEIR, parms = parameters)
fit <- out[, 4]
sum((infected - fit)^2)
}
country = enquo(country_name)
df <- bd_unoff %>% filter(country == !!country, cum_cases>0, date <= run_date)
infected <- df %>% filter(date >= min(date), date <= today() - 1 - npast) %>%
pull(cum_cases)
R = 0; E=0; I = infected[1]; S = N - E - I - R
seir_start_date <- df %>% pull(date) %>% min()
# Ascertainment factor
infected = infected * 1/af
# Create an incrementing Day vector the same length as our
# cases vector
Day <- 1:(length(infected))
# now specify initial values for S, I and R
init <- c(S = S, E=E, I=I, R=R)
# now find the values of beta and gamma that give the
# smallest RSS, which represents the best fit to the data.
# Start with values of 0.5 for each, and constrain them to
# the interval 0 to 1.0
opt <- optim(c(.5, .5, .5), RSS, method = "L-BFGS-B",
lower = c(0.01,0.01,0.01), upper = c(.999, .999, .999),
control=list(maxit = 1000))
# check for convergence
opt_msg = opt$message opt_par <- setNames(opt$par, c("beta", "sigma", "gamma"))
beta = opt_par["beta"]
gamma = opt_par["gamma"]
sigma = opt_par["sigma"]
R0 = as.numeric(beta/gamma)
# time in days for predictions
t <- 1:(as.integer(today() - seir_start_date) + nfuture)
# get the fitted values from our SEIR model
odefit = ode(y = init, times = t, func = SEIR, parms = opt_par)
fitted_cases <- data.frame(odefit)
# add a Date column and join the observed incidence data
fitted_cases <- fitted_cases %>%
mutate(date = seir_start_date + days(t - 1)) %>%
left_join(df %>% filter(cum_cases>0) %>% ungroup() %>%
select(date, cum_cases))
# Return
list(country=country_name, infected = infected,
opt_msg=opt_msg, opt_par = opt_par, R0=R0, opt_msg=opt_msg,
fitted_cases=fitted_cases, N=N, af=af)
}
It turns out that the bottom left one fits the current data best. So lets put that figure on a bigger canvas.
### 4.1.3 Projection for the next 5 days
Table 4.1: Predicted new cases for the next 5 days
Date Actual daily cases Projected daily cases Actual cumulative cases Projected cumulative cases
174 2020-08-28 NA -161244 NA 2449807
175 2020-08-29 NA -157202 NA 2292605
176 2020-08-30 NA -152275 NA 2140330
177 2020-08-31 NA -146638 NA 1993692
178 2020-09-01 NA -140460 NA 1853232
179 2020-09-02 NA -133895 NA 1719337
180 2020-09-03 NA -127080 NA 1592257
181 2020-09-04 NA -120131 NA 1472126
182 2020-09-05 NA -113151 NA 1358975
183 2020-09-06 NA -106224 NA 1252751
### 4.1.4 Projection for 100 days into the future
Assuming the situation will remain like this including the interventions currently in place, the 100 day projection suggests that the the peak of the epidemic will be around the middle of June. The trajectory also suggests that the epidemic will end by end of July or early August.
## 4.2 Projection for Bangladesh (unofficial) Run 2
Run date: 21 April 2020
This is the second projection. For the initial project, please scroll up. The methodology is discussed above.
It turns out that the bottom right one fits the current data best. So lets put that figure on a bigger canvas. Please note that in Run 1, we used a similar $$R_0$$ but a lower ascertainment rate. This time, a higher ascertainment rate is associated with a similar $$R_0$$.
### 4.2.1 Projection for the next 5 days
Table 4.2: Predicted new cases for the next 5 days
Date Actual daily cases Projected daily cases Actual cumulative cases Projected cumulative cases
172 2020-08-26 NA -19 NA 111
173 2020-08-27 NA -17 NA 94
174 2020-08-28 NA -14 NA 80
175 2020-08-29 NA -12 NA 68
176 2020-08-30 NA -10 NA 58
177 2020-08-31 NA -9 NA 49
178 2020-09-01 NA -7 NA 42
179 2020-09-02 NA -6 NA 36
180 2020-09-03 NA -6 NA 30
181 2020-09-04 NA -4 NA 26
182 2020-09-05 NA -4 NA 22
183 2020-09-06 NA -3 NA 19
### 4.2.2 Projection for 100 days into the future
Assuming the situation will remain like this including the interventions currently in place, the 100 day projection suggests that the the peak of the epidemic will be around the middle of June. The trajectory also suggests that the epidemic will end by end of July or early August.
### References
Churches, Tim. 2020. “Tim Churches Health Data Science Blog: Analysing Covid-19 (2019-nCoV) Outbreak Data with R - Part 1.” https://timchurches.github.io/blog/posts/2020-02-18-analysing-covid-19-2019-ncov-outbreak-data-with-r-part-1/. | 2,410 | 8,431 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-05 | latest | en | 0.911795 |
https://studylib.net/doc/7372286/copper-sulfate-lab---minneota-public-school | 1,627,488,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00504.warc.gz | 540,781,626 | 10,379 | # Copper Sulfate lab - Minneota Public School
```Name: ____________________________________Period: _____
Date:____________
Formula of a Hydrate Lab
Purpose:
To utilize mole conversions to determine the ratio between copper (II) sulfate
molecules and water of hydration.
Materials:
1 Bunsen burner
1 Clay triangle
1 Crucible
1 Ring stand
1 Electronic Scale
1 Crucible Tong
Copper II Sulfate Hydrate (approx. 5g)
Procedure:
1. Mass crucible (empty)
2. Obtain approximately 5 grams of Copper II sulfate hydrate
3. Put hydrate in crucible and find the mass.
4. Light Bunsen burner
5. Place crucible with copper II sulfate hydrate over Bunsen burner.
6. Place lid on crucible (keeping it slightly cocked open).
7. Heat until blue color is gone.
8. Turn off Bunsen burner and close lid on crucible
9. Let crucible cool until you can handle it.
10. Mass crucible and copper II sulfate anhydrate.
11. Wipe out crucible into waste container and wash out.
12. Clean up lab station.
13. Perform calculations.
Data:
Mass of Empty Beaker(g)
Mass of Beaker & CuSO4 . H2O (g)
Mass of Beaker and CuSO4 (g)
NOTES
Calculations: You must show all work for complete credit.
Determine the mass of Copper II Sulfate Hydrate.
Using the mass of the beaker only and the mass of the beaker with copper II hydrate, determine
the mass of the blue copper II sulfate hydrate.
________________g CuSO4.XH2O
Determine the mass of Copper II Sulfate Anhydrate
Using the mass of the beaker and the mass of the beaker with copper II anhydrate, determine the
mass of the copper II sulfate anhydrate.
_________________g CuSO4
Determine the mass of water lost from Hydrate.
Using the mass of the copper II sulfate hydrate and the mass of the copper II sulfate anhydrate,
determine the mass of water lost from the hydrate.
__________________g H2O
Determine the number of moles of copper II sulfate anhydrate, CuSO4.
Using the mass of copper II sulfate anhydrate and the molar mass from the periodic table,
determine the number of moles of copper II sulfate used in your experiment.
__________________ mol CuSO4
Determine the number of moles of water.
Using the mass of water lost and the molar mass from the periodic table,
determine the number of moles of water of hydration.
__________________ mol H2O
Determine the mole ratio of Copper II Sulfate and Water.
Using the number of moles of copper II sulfate and moles of water,
determine the smallest whole number ratio.
Ratio _____to _____
Write the formula of Copper II Sulfate Hydrate
CuSO4 . _____ H2O
Conclusion
By heating __________ grams of copper II sulfate _______________, it was calculated that
__________ grams of water were lost. With a mass of __________ grams for the anhydrate, it
was determined that the mole ratio of copper II sulfate to water is ___:___. From this, it can be
stated that the formula for the hydrate is _______________.
The actual formula for the hydrate is _______________. Our ratio was ____________ to that of
the actual formula. Potential sources of error in this lab could be ______________________
``` | 756 | 3,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.732354 |
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- - evaluate effective viscosity from U and p already known (http://www.cfd-online.com/Forums/openfoam-programming-development/96273-evaluate-effective-viscosity-u-p-already-known.html)
Cyp January 18, 2012 04:03
evaluate effective viscosity from U and p already known
Hi!
I would like to evaluate an effective viscosity from pressure and velocity field already computed. I mean that I am looking for a field nuEff that obey to:
In this equation, U and p result from a previous computation.
Does anybody know a method to do such a thing ?
Regards,
Cyp
Bernhard January 19, 2012 04:03
If you work out the brackets in the term on the right hand side, can't you just solve it like any other equation in OpenFOAM? So you use fvc instead of fvm on the terms with p and U and solve for nuEff. I can imagine that this will work, but I'll give you now guarantee.
Cyp January 19, 2012 04:11
Yes, but in that case, I will have :
T1 = fvc::laplacian(U) which is a vector
T2 = fvc::grad(p)/rho + fvc::div(phi,U) which is a vector as well.
How can I deduce the value of nuEff which should be a scalar (or may be a tensor) ??
Bernhard January 19, 2012 06:53
For the right hand side, don't you get something like
or something like that, which you can solve if you supply boundary conditions?
But you mean this is an overdetermined system or something? Maybe you can work component-wise?
Cyp January 19, 2012 08:31
In fact, what I look for is a method to evaluate an effective viscosity that result from a turbulent calculation (RANS, LES).
In case of RANS simulation, I should recover the value of nut and it will be a test case to validate the approach.
I was thinking that the effective viscosity may satisfy :
but I am not sure about that.
I missed the point with your rewritting of the RHS ? if I am right, fvm::grad() has no sense in OF..
Bernhard January 19, 2012 12:22
Why don't you just write nut directly then? Seems like a unnecessary exercise to me...
Cyp January 20, 2012 04:13
Because the RANS approach are just a particular approach of what I need. It makes sense when LES is considered for example.
Bernhard January 20, 2012 04:41
Why? Then you still have an effective viscosity in the way it is implemented in OpenFOAM.
Cyp January 20, 2012 04:44
Yes, but what I look for is the effective viscosity that corresponds to time-average velocity and pressure fields. It is not the case of the nuSgs of the LES..
Bernhard January 20, 2012 04:59
Why don't you use the time-averaged value of nuSgs/nut? I don't think this will make a very big difference compared to what you want to do. I still don't get why you want to do it this complex...
Cyp January 20, 2012 05:22
Because I do not have it and the simulations are too long to be start one more time...
Cyp January 23, 2012 09:42
I keep working on this problem and I identified some issues.
I need to calculate
where I is the identity tensor and x is the position in the x-direction. | 821 | 3,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2015-32 | longest | en | 0.941742 |
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Topic: Programming problem
Replies: 3 Last Post: Mar 4, 2013 2:49 PM
Messages: [ Previous | Next ]
Steven Lord Posts: 18,038 Registered: 12/7/04
Re: Programming problem
Posted: Mar 4, 2013 10:11 AM
"Milos Milenkovic" <m.milenkovic@mathworks.com> wrote in message
news:kh1pbg\$3ph\$1@newscl01ah.mathworks.com...
> Dear, I have problem with replacing goto command in Matlab.
> The code is something like this:
> 1 generator of a random matrix B1=-0.1+(0.1+0.1).*rand(4,4);
> 2 set of matlab commands ->
> Ke=wn1(B0,B1,K0,K1); %wn je kod za racunanje kovarijanse belog suma
> [Delta1, Error1]=d1(B1,K1,Ke);
> [Delta0, Error0]=d2(B0,B1,K0,Delta1,Ke);
> 3 if (condition statement) ->
> if (Delta0(i,j)<2) && (Delta1(i,j)<2)
> [B0,B1]= up(B0,B1, Delta0, Delta1)
> 4 elseif goto 1 (generator again)
> So, how to close this set of steps?
satisfied = false;
while ~satisfied
% step 1
% step 2
% step 3
% if delta condition satisfied
satisfied = true;
% else
% end
end
--
Steve Lord
slord@mathworks.com | 408 | 1,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-17 | longest | en | 0.63284 |
https://jeopardylabs.com/play/working-with-integers | 1,527,052,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00169.warc.gz | 610,407,198 | 7,636 | Subtracting Integers
Multiplying & Dividing Integers
Solving equations with add & subtract
Solving equations with multiply & divide
### 100
You add and keep the sign.
What is the rule for adding integers with the same sign?
### 100
You change the subtraction sign to an addition sign.
What is the first step in solving a subtraction of integers problem?
### 100
Negative
What is the answer in dividing & multiplying integers if it has an odd number of negative signs?
### 200
You subtract and keep the sign of the larger number.
What is the rule for adding integers with different signs?
### 200
Change the sign of the number after the minus sign.
What sign do you change after changing the minus sign in a subtraction of integers problem?
### 200
Positive
What is the answer in multiplying or dividing of integers when it has an even number of negative signs?
### 300
The solution to -3 + 2 is
What is -1?
### 300
You apply the rules of addition to the problem.
What is the final step in solving a subtraction of integers problem?
-4 - (-8) =
What is 4? | 248 | 1,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-22 | latest | en | 0.8702 |
https://businessyield.com/finance-accounting/year-over-year-growth/ | 1,685,781,390,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00777.warc.gz | 177,311,191 | 39,886 | # YEAR OVER YEAR GROWTH: Examples, Formula and Calculations
Although weekly or monthly data might provide useful information, it is crucial to take a longer and broader view of your company’s growth. Year-over-year growth is a valuable measure in this regard. Year-on-year growth allows you to measure genuine growth without the influences of seasonality, whether you’re looking at revenue, profit, production, or any other statistic.
This post will teach you how to calculate year-over-year growth with the formula for several measures. You will also learn how to utilize Google Sheets to create templates that will allow you to speed up computations and maintain track of all calculated values.
## What is Year Over Year Growth?
Year-over-year (YOY) growth is a critical performance indicator that compares growth in one period (typically a month) to growth in the prior year’s similar period (thus the name). In contrast to solitary monthly measures, YOY provides a view of your performance that excludes seasonal influences, monthly volatility, and other factors. Over time, your genuine triumphs and obstacles become more apparent to you. This is, unsurprisingly, an important metric for retail analytics.
The first significant advantage of YOY growth is that it removes seasonality from your growth measurements. During the holiday season, most retailers notice a significant increase in sales. On a monthly basis, this can create the impression of huge growth. However, if these inflated numbers revert to normal levels after the holidays, they aren’t genuinely emblematic of growth over time.
When you compare similar times over time, you can get a more precise picture of your company’s growth. Here’s an illustration: For November, a 40% increase in monthly sales growth may appear to be a huge leap worthy of celebration. However, when compared to the previous year, when growth was 45%, this figure shows a mild decline rather than a jump. Without YOY growth comparisons to provide a baseline and historical context, you can only rely on the most current data. This is not a good strategy to make long-term decisions or drive long-term growth.
##### Read Also: YOY: Year-Over-Year Analysis, Growth and Investment Calculations, Formula & Examples
That isn’t to imply that year-over-year measurements are the be-all and end-all of analysis. Focusing on a 12-month timeframe may also provide you with a too-wide perspective. Combining a longer-term view with complimentary month-over-month and quarter-over-quarter statistics can help you assess various aspects of annual growth and see how your firm is performing in a variety of ways.
Revenue growth is only one aspect of YOY growth. You can track a variety of components of your growth, including conversions, average sale value, and other data that are related to, but not limited to, your bottom line.
## Steps to Calculate Year-on-Year Growth
There are a few procedures you must take to calculate your company’s year-over-year growth.
### #1. Determine why you’re seeing YOY growth.
First and foremost, you must identify which areas of your business you wish to see YOY growth in. Is it money? How many staff are there? What are marketing KPIs? When it comes to year-over-year growth, the globe is your oyster…So long as you have the data for your computations. Once you’ve determined why you’re calculating YOY growth, collect the data for both time periods you’re comparing. That brings us to…
### #2. Select a time frame
Do you wish to calculate year-over-year growth between two months, quarters, or even years? You have an option. However, by selecting shorter time periods, such as months, you can help to reduce seasonality difficulties.
### #3. Apply the formula for year-over-year growth.
It is not difficult to calculate year-over-year growth. After retrieving your data, you may quickly obtain results.
Subtract last year’s number from this year’s number to begin the equation. This will tell you the year’s total difference. If the number is positive, you won. If the number is negative, you lost money.
Then, divide the difference by the number from the previous year. This provides you with the rate of growth year over year.
Finally, multiply the value by 100 to convert it to a percentage to obtain the year-over-year percentage change.
## Year-Over-Year Growth Formula
The following formula is used to calculate the year-over-year (YoY) growth rate.
End of Period (EoP)
Beginning of Period (BoP) Prior Period
## Year-on-Year Growth Calculation Example
For example, if a company’s revenue increased from \$25 million in Year 0 to \$30 million in Year 1, the formula for calculating the YoY growth rate is:
Year-on-Year Growth (%) = (\$30 million / \$25 million) – 1 = 20.0%
Another way to calculate YoY growth is to deduct the prior period balance from the current period balance and then divide the result by the prior period balance.
YoY Growth (%) = (30 million – \$25 million) / \$25 million = 20.0%
In any case, the year-over-year (YoY) growth rate equals 20.0%, which shows the difference between the two periods.
## How Do You Calculate Year Over Year Growth In Excel or Google Sheets,?
As you can see, the calculation is straightforward; nevertheless, these calculations must be performed on a regular basis. Keep track of these variables ideally to avoid having to recalculate to gain a long-term view. Fortunately, technologies like Excel and Google Sheets can assist you in creating templates to track and automate calculations. If you use financial statement templates, you may easily enter YOY growth calculations. You can, however, create them as a separate template and import the information from the source file/s.
The following section contains examples of YOY growth calculations using Google Sheets.
## Examples of Year-Over-Year Growth Calculations On Excel Sheet
The examples below are based on data from Company A. In the first example, I will calculate sales growth year over year. In the second example, I will calculate year-over-year growth for a number of relevant indicators.
### Example 1: Year-Over-Year Growth for a Single Metric
1. Choose a timeframe to work with. I’ll take December totals from 2020 and 2021 in this scenario.
1. Separate two sets of parentheses and subtract the old and new values. The initial pair of parentheses should be closed.
1. Subtract the new value from the old value and close the second pair of parentheses.
1. To calculate the percentage, multiply by 100.
That’s all. The YOY growth rate for December values is 8%.
### Example 2: Year-on-Year Growth for Several Metrics
I’ll use data for Company A again, but this time I’ll include more metrics.
1. For the indicators provided below, I will use December totals from 2020 and 2021.
2. I’ll add the formula for YOY growth utilizing cell references, just like in the previous example.
3. Finally, I’ll drag the formula down to retrieve the numbers for all of the measures.
## What is the Importance of Year-Over-Year Growth For Small Businesses?
### #1. It takes into account seasonality.
The primary reason businesses calculate the YoY growth rate of any indicator is to monitor long-term business performance while allowing for seasonal swings or market volatility that are beyond the company’s control.
Because certain organizations have peak and low seasons, comparing month-to-month or quarter-to-quarter numbers may be ineffective.
Many retail organizations, for example, record significant revenue growth during the fourth quarter due to the holiday season. While this is certainly a pleasant experience for a company, comparing sales from that quarter to revenue from other quarters that year may provide an inaccurate image of that company’s growth.
When we compare this December’s revenue to the previous December’s revenue, we eliminate seasonal swings from the equation and get an annualized, more realistic picture of growth.
You can be confident that the percentage changes you’re calculating are precise, unbiased, and reflective of your company’s actual financial health when you use YoY calculations.
### #2. It provides a long-term view of performance.
In addition to removing variables that are outside your company’s control, YoY calculations are an excellent approach to tracking long-term business performance.
While tracking daily sales will give you a sense of how your firm is performing, annualizing a company’s performance simply provides a more complete picture of where your organization is and where it may improve.
## Understanding Variance (Percent Change) in YoY Growth Analysis
The key advantage of YoY growth analysis is how simple it is to track and compare growth rates over multiple time periods, which, when annualized, removes the impact of monthly volatility.
Furthermore, any cyclical patterns will be visible if the historical results reflect the whole economic cycle.
While the intuition for the two fundamental criteria is simple, you must conduct additional research on the company’s growth trajectory to find the underlying sources of the change before reaching a final conclusion.
• Positive Year-On-Year Growth
• Negative Year-On-Year Growth
Consider a corporation whose sales growth rate in the previous year was 5%, but whose growth rate in the current year is only 3%.
Despite the slightly lower growth rate, the quality of revenue generated may have improved (e.g., long-term contractual revenue, less churn, lower client acquisition expenses).
Without a deeper dive examination, it would be inaccurate to presume that the current year was necessarily “worse” than the previous year.
Another crucial factor to remember is that all businesses’ growth will ultimately slow.
Mature firms with established market shares are less likely to fund growth and are more likely to concentrate on:
• Dividend Payments to Shareholders
• Increasing Operational Effectiveness
• Customer Retention vs New Customer Acquisition
## Examples of Year-on-Year Growth in the Real World
YOY growth is important in retail, but other businesses can benefit from integrating these measurements in KPIs and analytics.
Although YOY growth is king in retail, many other industries can benefit from adding these metrics in KPIs and analytics. Examine the following instances to show how beneficial YOY growth can be:
• Healthcare: It’s critical to look at YOY results with varied policies when analyzing things like patients served or cost per patient. Observing how performance changes after implementing a new practice can reveal whether or not it was beneficial.
• Manufacturing: The efficiency of a factory’s production lines determines its survival. Measuring how much production rates increase or decrease over time is critical for understanding processes, machine performance, and a variety of other factors. Manufacturing analytics, which combines efficiency statistics with sales data, can help businesses better prepare for seasonal shifts and identify their long-term opportunities and problems.
• Logistics: In an industry defined by the number of products delivered and efficiency, year-over-year growth might indicate if a company is remaining efficient and effective in the market, or if performance is declining. When comparing delivery over time, it is possible to identify areas for improvement and tasks where simplifying functions could be advantageous.
## What are Some Alternatives To Year-Over-Year Growth?
While year-over-year growth is a crucial calculation to consider for your company, it isn’t the only time-series indicator that can help you gain a better view of performance. You need additionally calculate the following:
• Month to date (MTD): calculates a KPI from the start of the current month until the current date, excluding today’s date.
• Quarter to Date (QTD): measures a KPI from the start of the quarter until the present date, excluding today.
• Month over month (MoM): the difference between this month’s (sales, users, etc.) total and the previous month’s total.
## Conclusion
Year over Year Growth is a key financial measure. You can eliminate the effects of seasonality and volatility by using YOY growth. Instead of comparing February values to January levels, you compare them to the previous year’s February values. This provides a valuable value comparison, and because it is expressed as a percentage, it is also straightforward to compare with similar companies in the industry.
You now understand what year-over-year growth is and how to calculate it. You also know how to quickly create a Google Sheets template to calculate year-on-year growth for your chosen KPIs. As you can see, using Google Sheets to calculate YOY growth is simple. The computations are simple, but getting the values for various indicators on a regular basis might be time-consuming.
## FAIR CREDIT BILLING ACT: Meaning, Purpose and Effects
Table of Contents Hide What Is Fair Credit Billing Act?What Does the Fair Credit Billing Act Do?#1. Gives…
## Interest Only Loan Rate: Definition, Calculation, Types & All You Need
Table of Contents Hide Interest Only Loan The Benefits of Interest-Only LoanThe first benefitThe second benefit The third benefitThe Downsides… | 2,666 | 13,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.933828 |
http://physicshelpforum.com/kinematics-dynamics/3458-calculating-coefficient.html | 1,568,607,748,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00153.warc.gz | 152,770,325 | 9,539 | Physics Help Forum calculating coefficient !
Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum
Nov 9th 2009, 02:37 PM #1 Senior Member Join Date: Sep 2009 Location: St. John's, NL Posts: 124 calculating coefficient ! When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 1.87 m/s. The mass stops a distance S2 = 2.15 m along the level part of the slide. The distance S1 = 1.25 m and the angle theta = 37.1 degrees. Calculate the coefficient of kinetic friction for the mass on the surface. E = K+U = (0.5)mv^2 + mgS1sin(theta) m cancelled? = 9.15 W1 = [mgcos(theta)-uk mgsin(theta)]S1 W2 = -uk mgS2 uk = [ E + mgcos(theta)S1 ]/[mgS2 + mgsin(theta)S1] m cancelled out so uk = [ 9.15 + (9.81)(1.25)cos37.1 ] / [ (9.81)(2.15)+(9.81)(1.25)sin37.1 ] = 0.664 but wrong... where did i go wrong? Last edited by christina; Nov 9th 2009 at 02:39 PM.
Nov 10th 2009, 12:55 AM #2 Physics Team Join Date: Feb 2009 Posts: 1,425 The total initial energy say E gets completely converted into work against friction on the slope and on the level surface. E = K+U = (0.5)mv^2 + mgS1sin(theta) m cancelled? = 9.15 m does not cancel yet. leave it as it is. The work done against friction on the slope = W1 = force of friction x S1 = uk mg cos theta x S1 = W1. E = W1 + W2. Now m will cancel out. Solve for uk.
Nov 10th 2009, 07:38 AM #3 Senior Member Join Date: Sep 2009 Location: St. John's, NL Posts: 124 so am i right about the W2 ?
Nov 10th 2009, 10:49 PM #4 Physics Team Join Date: Feb 2009 Posts: 1,425 Yes . However, in W2 = -uk mgS2 you have to remember that the work is a dot product between force and displacement giving F s cos 180 = - F s in our case. Thus the minus sign cancels out in W2 and W1 also is to be reated as +ve, i.e., when you add the two, W1 + W2 = uk mg cos theta x S1 + uk m g S2
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# 3.0.0. Introduction
Ordinary differential equations are at the heart of our perception of the physical universe. For this reason numerical methods for their solutions is one of the oldest and most successful areas of numerical computations.
It would be very nice if discrete models provide calculated solutions to differential (ordinary and partial) equations exactly, but of course they do not. In fact in general they could not, even in principle, since the solution depends on an infinite amount of initial data. Instead, the best we can hope for is that the errors introduced by discretization will be small when those initial data are resonably well-behaved.
In this chapter, we discuss some simple numerical method applicable to first order ordinary differential equations in normal form subject to the prescribed initial condition:
$y' = f(x,y), \qquad y(x_0 ) = y_0 . \qquad {(3.0.1)}$
We always assume that the slope function $$f(x,y)$$ is smooth enough so that the given initial value problem has a unique solutions and the slope function is differentiable so that all formulas make sense. Since it is an introductory, we do not discuss discretization errors in great detail---it is a subject of another course. A user should be aware that the magnitude of global errors may depend on the behavior of local errors in ways that ordinary analysis of discretization and rounding errors cannot predict. The most part of this chapter is devoted to some finite difference methods that proved their robastness and usefullness, and can guide a user how numerical methods work. Also, presented codes have limited applications because we mostly deal with uniform truncation meshes/grids that practically are not efficient. We do not discuss adaptive Runge--Kutta--Fehlberg algorithms of higher order that are used now in most applications, as well as spline methods. Instead, we present some series approximations, incuding Adomian Decomposition Method (ADM), as an introduction to applications of recurrences in practical life.
The finite diference approximations have a more complecated "physics" than the equations they are designed to simulate. In general, finite recurrences usually have more propertities than their continuous anologous counterparts. However, this arony is no paradox; the finite differences are used not because the numbers they generate have simple properties, but because those numbers are simple to compute.
Numerical methods for the solution of ordinary differential equations may be put in two categories---numerical integration (e.g., predictor-corrector) methods and Runge--Kutta methods. The advantages of the latter are that they are self-starting and easy to program for digital computers but neither of these reasons is very compelling when library subroutines can be written to handle systems of ordinary differential equations. Thus, the greater accuracy and the error-estimating ability of predictor-corrector methods make them desirable for systems of any complexity. However, when predictor-corrector methods are used, Runge--Kutta methods still find application in starting the computation and in changing the interval of integration.
In this chapter, we discuss only discretization methods that have the common feature that they do not attempt to approximate the actual solution y = φ (x) over a continuous range of the independent variable. Instead, approximate values are sought only on a set of discrete points $$x_0 , x_1 , x_2 , \ldots .$$ This set is usually called grid or mesh. For simplicity, we assume that that these mesh points are equidistance: $$x_n = a+ n\,h , \quad n=0,1,2,\ldots .$$ The quantity h is then called the stepsize, stepwidth, or simply the step of the method. Generally speaking, a discrete variable method for solving a differential equation consists in an algorithm which, corresponding to each lattice point xn, furnishes a number yn which is to be regarded as an approximation to the true value $$\varphi (x_n )$$ of the actual solution at the point xn.
# 3.0.1. Errors in Numerical Approximations
The main question of any approximation is whether the numerical approximations $$y_1 , y_2 , y_3 , \ldots$$ approach the corresponding values of the actual solution? We want to use a step size that is small enough to ensure the required accuracy. An unnecessarily small step length slows down the calculations and in some cases may even cause a loss of accuracy.
There are three fundamental sources of error in approximating the solution of an initial value problem numerically.
1. The formula, or algorithm, used in the calculations is an approximating one.
2. Except for the first step, the input data used in the calculations are only approximations to the actual values of the solution at the specified points.
3. The computer used for the calculations has finite precision; in other words, at each stage only a finite number of digits can be retained.
Let us temporarily assume that our computer can execute all computations exactly; that is, it can retain infinitely many digits (if necessary) at each step. Then the difference En between the actual solution $$y = \phi (x)$$ of the initial value problem $$y' = f(x,y), \quad y(x_0 ) = y_0$$ and its numerical approximation yn at the point x = xn is given by
$E_n = \phi (x_n ) - y_n$
is known as the global truncation error. It arises entirely from the first two error sources listed above---that is, by applying an approximate formula to approximate data.
However, in reality we must carry out the calculations using finite precision arithmetic because of the limited accuracy of any computing equipment. This means that we can keep only a finite number of digits at each step. This leads to a round-off error Rn defined by
$R_n = y_n - Y_n ,$
where Yn is the value actually computed from the given numerical method. The absolute value of the total error in computing $$\phi (x_n )$$ is given by
\begin{align*} \left\vert \phi (x_n ) - Y_n \right\vert &= \left\vert \phi (x_n ) -y_n + y_n - Y_n \right\vert \le \left\vert \phi (x_n ) - y_n \right\vert + \left\vert y_n - Y_n \right\vert \\ &\le \left\vert E_n \right\vert + \left\vert R_n \right\vert . \end{align*}
Therefore, the total error is bounded by the sum of the absolute values of the global truncation and round-off errors. The round-off error is very difficult to analyze and it is beyond of the scope of our topic. To estimate the global error, we need one more definition.
As we carry the solution over many steps we would expect the values yn and $$\phi (x_n )$$ to diverge further and further apart. The local truncation error is concerned only with the direvgence produced within the present step so that it is appropriate to reinitialize yn to the value of $$\phi (x_{n} )$$ in studying this source of error. The corresponding solution to the initial value problem
$u' = f(x,u) , \qquad u(x_{n} ) = y_{n}$
on the mesh interval $$[x_{n} , x_{n+1} ]$$ is called the reference solution. So the difference $$u( x_{n+1} ) - y_{n+1} = h\, T_n (h)$$ is called the local truncation error. We say that the truncation error is of order p > 0 if $$T_n (h) = O \left( h^p \right) ,$$ that is, $$\left\vert T_n (h) \right\vert \le K\, h^p$$ for some positive constant K (not depending on h but on the slope function).
# 3.0.2. Fundamental Inequality
We start with estimating elements of sequences.
Theorem: Suppose that elements of a sequence $$\left\{ \xi_n \right\} , \ n=0,1,2,\ldots ,$$ satisfy the inequalities of the form
$| \xi_{n+1} | \le A\, | \xi_n | + B ,$
where A and B are certain nonnegative constants independent of n. Then
$| \xi_{n} | \le A^n \, | \xi_0 | + B \times \begin{cases} \frac{A^n -1}{A-1} , & \quad \mbox{if } A \ne 1, \\ n , & \quad \mbox{if } A =1. \end{cases}$
Theorem: If A is a quantity of the form 1 + δ , then $$A = 1 + \delta < e^{\delta}$$ and we have
$| \xi_{n} | \le e^{n\delta} \, | \xi_0 | + B \, \frac{e^{n\delta} -1}{\delta} .$
The last inequality is more readily amendable for simplification. ■
Recall that a function f is said to satisfy Lipschitz's condition (or being Lipschitz continuous) on some interval if there exists a positive constant L such that
$| f(y_1 ) - f(y_2 ) | \le L\, |y_1 - y_2 |$
for every pair of points y1 and y2 from the given interval.■
Rudolf Otto Sigismund Lipschitz (1832--1903) was born in Königsberg, Germany (now Kaliningrad, Russia). He entered the University of Königsberg at the age of 15 and completed his studies at the University of Berlin, from which he was awarded a doctorate in 1853. By 1864 he was a full professor at the University of Bonn, where he remained the rest of his professional life.
Theorem (Fundamental Inequality): Let f(x,y) be a continuous function in the rectangle $$[a,b] \times [c,d]$$ and satisfying the Lipschitz condition
$| f(x,y_1 ) - f(x,y_2 ) | \le L\, |y_1 - y_2 |$
for some positive constant L and all pairs y1, y2 uniformely in x. Let $$y_1 (x) \quad\mbox{and}\quad y_2 (x)$$ be two continuous piecewise differentiable functions satisfying the inequalities
$| y'_1 (x ) - f(x, y_1 (x)) | \le \epsilon_1 , \qquad | y'_2 (x) -f(x, y_2 (x))| \le \epsilon_2$
with some positive constants ε1,2. If in addition, these functions differ by small amount δ > 0 at some point:
$| y_1 (x_0 ) - y_2 (x_0 ) | \le \delta ,$
then
$| y_1 (x ) - y_2 (x ) | \le \delta \,e^{L| x- x_0 |} + \frac{\epsilon_1 + \epsilon_2}{L} \left( e^{L|x- x_0 |} -1 \right) .$
# 3.0.3. Numerical Solution using DSolve and NDSolve
A complete analysis of a differential equation is almost impossible without utilizing computers and corresponding graphical presentations. We are not going to pursue a complete analysis of numerical methods, which usually requires:
• an understanding of computational tools;
• an understanding of the problem to be solved;
• construction of an algorithm which will solve the given mathematical problem.
In this chapter, we concentrate our attention on presenting some simple numerical algorithms for finding solutions of the initial value problems for the first order differential equations in the normal form:
$y' = f(x,y), \qquad y(x_0 ) = y_0 ,$
assuming that the given problem has a unique solution.
A general approach for determing numerically solutions of the initial value problems and then plot them using Mathematica's capability is as follows. We demostrate all steps on the following initial value problem (IVP, for short):
$y' \equiv \frac{{\text d}y}{{\text d}x} = \frac{1-x}{1+y^5}, \qquad y(0 ) = 1 ,.$
a = NDSolve[{y'[x]== (1-x)/(1+y[x]^5),y[0]==1}, y[x],{x,0,3}];
Plot[y[x] /. a, {x,0,3}]
To see a numerical value at, say x=3.0, we type:
y[x] /. a /. x -> 3.0
Out[3]= {-0.333563}
Since its solution is known in implicit form, there is an alternative way to plot the graph:
psi[x_, y_] = y + 1/6 y^6 - x + 1/2 x^2 - 7/6;
ContourPlot[psi[x, y] == 0, {x, 0, 2}, {y, -1, 3}, ContourLabels->True, ColorFunction->Hue]
The general reference for NDSolve applications is
## IV. Regular Plotting vs Log-Plotting
Plotting the absolute value of the errors for Taylor approximations, such as the ones from the previous example, is fairly easy to do in normal plotting.
The normal plotting procedure involves taking all of the information regarding the Taylor approximations and putting them into one file, then calculate the exact solution, then find the difference between each approximation and the exact solution, and plotting the solution.
Looking at the code, the syntax is pretty much self-explanatory. The only part that may confuse students is the syntax for the plot command. The plot command need not be in the same syntax as the file has it. If you wish to, you can alter the syntax to form a curve instead of dots.
Log-plotting is not much different than plotting normally. In other programs such as Mathematica, there are a couple of fixes to make, but in Maple only one line of code is required, as seen in the file. Make sure that before you use the logplot command to open the plots package.
Now, let’s compare the exact value and the values determined by the polynomial approximations and the Runge-Kutta method.
x values Exact First Order Polynomial Second Order Polynomial Third Order Polynomial Runge-Kutta Method 0.1 1.049370088 1.050000000 1.049375000 1.049369792 1.049370096 0.2 1.097463112 1.098780488 1.097472078 1.097462488 1.097463129 0.3 1.144258727 1.146316720 1.144270767 1.144257750 1.144258752 0.4 1.189743990 1.192590526 1.189758039 1.189742647 1.189744023 0.5 1.233913263 1.237590400 1.233928208 1.233911548 1.233913304 0.6 1.276767932 1.281311430 1.276782652 1.276765849 1.276767980 0.7 1.318315972 1.323755068 1.318329371 1.318313532 1.318316028 0.8 1.358571394 1.364928769 1.358582430 1.358568614 1.358571457 0.9 1.397553600 1.404845524 1.397561307 1.397550500 1.397553670 1.0 1.435286691 1.443523310 1.435290196 1.435283295 1.435286767 Accuracy N/A 99.4261% 99.99975% 99.99976% 99.99999%
You will notice that compared to the Euler methods, these methods of approximation are much more accurate because they contains much more iterations of calculations than the Euler methods, which increases the accuracy of the resulting y-value for each of the above methods.
# Horner's rule
The most efficient way to evaluate a polynomial is by nested multiplication. If we have a polynomial of degree n
$p_n (x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n ,$
then we factor out each power of x as far as it will go to obtain
$p_n (x) = a_0 + x \left( a_1 + x \left( a_2 + \cdots + x \left( a_{n-1} + a_n x \right) \cdots \right) \right) .$
Horner's rule for polynomial evaluation % the polynomial coefficients are stored in the array a(j), j=0,1,..,n % px = value of polynomial upon completion of the code px = a(n) for k = n-1 downto 0 px = a(k) + px*x endfor
Similar algorithm for evaluating derivative of a polynomial:
Horner's rule for polynomial derivative evaluation % the polynomial coefficients are stored in the array a(j), j=0,1,..,n % dp = value of derivative upon completion of the code dp = n*a(n) for k = n-1 downto 1 dp = k*a(k) + dp*x endfor William George Horner (1786--1837) was born in Bristol, England, and spent much of his life as a schoolmaster in Bristol and, after 1809, in Bath. His paper on solving algebraic equations, which contains the algorithm we know as Horner's rule, was published in 1819, although the Itarian mathematician Ruffini had previously published a very similar idea.
here is a more efficient algorithm for polynomial evaluation:
A more efficient implementation of Horner's rule for polynomial evaluation % the polynomial coefficients are stored in the array a(j), j=0,1,..,n % b(n) = value of polynomial upon completion of the code b(0) = a(n) for k = 1 to n b(k) = x*b(k-1) + a(n-k) endfor % c(n-1) = value of derivative upon completion of the code c(0) = b(0) for k = 1 to n-1 c(k) = x*c(k-1) + b(k) endfor
# 3.0.4. Applications
Example: A 500 liter container initially contains 10 kg of salt. A brine mixture of 100 gramms
of salt per liter is entering the container at 6 liter per minute. The well-mixed contents
are being discharged from the tank at the rate of 6 liters per minute. Express the amount
of salt in the container as a function of time.
Salt is coming at the rate: 6*(0.1)=0.6 kg/min
d yin /dt =0.6 ; d yout/dt = 6x/500
dx/dt = 0.6 -6x/500 ; x(0)=10
DSolve[{x'[t]==6/10-6*x[t]/500, x[0]==10},x[t],t]
Out[15]= {{x[t] -> 10 E^(-3 t/250) (-4 + 5 E^(3 t/250))}}
Suppose that the rate of discharge is reduced to 5 liters per minute.
DSolve[{x'[t]==6/10-5*x[t]/(500+t), x[0]==10},x[t],t]
SolRule[t_] = Apart[x[t] /. First[%]]
Out[1]= {{x[t] -> (1/(
10 (500 + t)^5))(3125000000000000 + 187500000000000 t +
937500000000 t^2 + 2500000000 t^3 + 3750000 t^4 + 3000 t^5 +
t^6)}}
Out[2]= 50 + t/10 - 1250000000000000/(500 + t)^5
To check the answer:
Simplify[SolRule'[t] == 6/10 - 5*SolRule[t]/(500 + t)]
Together[SolRule'[t] == Together[6/10 - 5*SolRule[t]/(500 + t)]]
SolRule[0]
Out[3]= True
Out[4]= True
Out[5]= 10 | 4,613 | 16,701 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-34 | latest | en | 0.9096 |
https://questioncove.com/updates/5269dc98e4b029b030dcde7a | 1,508,390,072,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00463.warc.gz | 826,382,188 | 4,850 | f(x)=-5x+2, find f(… - QuestionCove
OpenStudy (anonymous):
f(x)=-5x+2, find f(-3)
3 years ago
OpenStudy (anonymous):
-5(-3)+2
3 years ago
OpenStudy (anonymous):
All you have to do is plug in three. So -5(-3)+2
3 years ago
OpenStudy (anonymous):
15+2=17, f(-3) = 17
3 years ago
OpenStudy (anonymous):
Okay Thanks!
3 years ago
OpenStudy (anonymous):
you're welcome :)
3 years ago | 132 | 389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-43 | latest | en | 0.841524 |
https://www.daniweb.com/programming/software-development/threads/429754/display-unique-numbers-from-array-list | 1,660,674,953,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00178.warc.gz | 622,366,758 | 19,854 | Given a 7x4 array generate four random numbers from the array with no repetition.
I can already generate four random numbers the problem is the repetition part.
Here's my code so far
``````int f[4];
int s[4];
for( i = 0; i < 4; i++)
{
// use as index for array a
f[i] = rand() % 7;
s[i] = rand() % 4;
cout << a[f[i]][s[i]] << "\n";
}
``````
Any suggestions?
## All 15 Replies
Instead of just outputting the values save them to a container. Then each new value you select from the original list can be checked against your new list for uniqueness. Something like:
``````std::set< int > unique;
while (unique.size () < 4) {
f[i] = rand() % 7;
s[i] = rand() % 4;
unique.insert(a[f[i]][s[i]]); // duplicates ignored
}
``````
I still get duplicate values
`````` set< int > unique;
for( i = 0; i < 4; i++)
{
f[i] = rand() % 7;
s[i] = rand() % 4;
unique.insert(a[f[i]][s[i]]);
cout << a[f[i]][s[i]] << "\n";
}
``````
Given a 7x4 array generate four random numbers from the array with no repetition.
Your code and your description don't match at all. Where is your 7x4 array?
How do you "generate four random numbers from the array"? what's in the array?
I still get duplicate values
Thats because you are still printing everything (regardless if it makes it in the set). Change your loop condition and only print all the values once you've exited your loop.
How do you "generate four random numbers from the array"?
refer to first post
what's in the array?
random numbers
Change your loop condition and only print all the values once you've exited your loop.
Why will I change the loop condition since I need it to generate the random numbers. How will I print the values once. I tried this.
`cout << unique << " ";` and `cout << a[f[unique]][s[unique]]` but they were wrong.
What you are posting is getting farther and farther from real code. Perhaps it is time you posted a full working sample of what you have.
How do you "generate four random numbers from the array"?
refer to first post
Telling me to re-read the post I quoted does NOT answer the question. I'm asking because your first post says this and I don't understand it.
what's in the array?
random numbers
So it's just crap in the array from uninitialize data and you have no idea what the values are. Got it.
What about the 7x4 array?
How do you "generate four random numbers from the array"?
refer to first post
My first post tells you how I generate four random numbers I gave the code.
I can already generate four random numbers the problem is the repetition part.
Here's my code so far
``````int f[4];
int s[4];
for( i = 0; i < 4; i++)
{
// use as index for array a
f[i] = rand() % 7;
s[i] = rand() % 4;
cout << a[f[i]][s[i]] << "\n";
}
``````
Regarding the values of the array[7][4] I initialize it with random numbers.
what's in the array?
random numbers
``````int array[7][4] = { {20, 30, 40, 27},
{19, 8, 10, 17},
{2, 9, 1, 15},
{4, 8, 12, 16},
{5, 10, 15},
{6, 12},
{3}};
``````
Ok i already got how to print the values. If I use the code below it prints four random numbers no duplicates.
``````set< int > unique;
set< int >::iterator it;
// first version of printing
while( unique.size() < 4 )
{
f[unique.size()] = rand() % 7;
s[unique.size()] = rand() % 4;
unique.insert(a[f[unique.size()]][s[unique.size()]]);
}
for( it = unique.begin(); it != unique.end(); ++it )
cout << " " << *it;
``````
But when I switch the code to a for loop. It doesn't give me four random numbers sometimes it gives me two or three sometimes 4. But it does give me unique values.
``````for( i = 0; i < 4; i++)
{
f[i] = rand() % 7;
s[i] = rand() % 4;
unique.insert(a[f[i]][s[i]]);
}
for( it = unique.begin(); it != unique.end(); ++it )
cout << " " << *it;
``````
What's the difference between the two version?
And also most of the time the four numbers starts with a 0 even there's no 0 value in the array.
There is a problem with the logic you use to generate the random indexes.
Let's start with something simpler. Suppose you want to get three unique random values from a unidimensional array of four elements. The easiest way to do this is by using random_shuffle:
``````#include <iostream>
#include <string>
#include <algorithm>
#include <ctime>
using namespace std;
int main()
{
srand(time(0));
string value_array[4] = { "one", "two", "three", "four" };
int index_array[4] = { 0, 1, 2, 3 };
random_shuffle(index_array, index_array + 4);
for (int i = 0; i < 3; ++ i)
cout << value_array[index_array[i]] << endl;
}
``````
Now, there are two ways to scale this in order to accommodate for bidimensional arrays. One of them is wrong (and is very similar to what you're doing) and the other is right. Let's see the wrong way first:
``````#include <iostream>
#include <string>
#include <algorithm>
#include <ctime>
using namespace std;
int main()
{
srand(time(0));
string value_array[3][4] = {
{ "one-a", "two-a", "three-a", "four-a" },
{ "one-b", "two-b", "three-b", "four-b" },
{ "one-c", "two-c", "three-c", "four-c" }
};
int index_array_a[3] = { 0, 1, 2 };
int index_array_b[4] = { 0, 1, 2, 3 };
random_shuffle(index_array_a, index_array_a + 3);
random_shuffle(index_array_b, index_array_b + 4);
for (int i = 0; i < 3; ++ i)
cout << value_array[index_array_a[i]][index_array_b[i]] << endl;
}
``````
One problem with this is that it just doesn't work if the number of your picks is bigger than the smallest dimension.
But the real problem is that you can never get an output of, let's say ...
one-a
one-b
one-c
... or an output of ...
one-a
one-b
two-b
The right way of doing it is this:
``````#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <ctime>
using namespace std;
int main()
{
srand(time(0));
string value_array[3][4] = {
{ "one-a", "two-a", "three-a", "four-a" },
{ "one-b", "two-b", "three-b", "four-b" },
{ "one-c", "two-c", "three-c", "four-c" }
};
vector<pair<int, int> > index_pairs;
for (int i = 0; i < 3; ++ i)
for (int j = 0; j < 4; ++ j)
index_pairs.push_back(make_pair(i, j));
random_shuffle(index_pairs.begin(), index_pairs.end());
for (int i = 0; i < 3; ++ i)
cout << value_array[index_pairs[i].first][index_pairs[i].second] << endl;
}
``````
OK thanks I will try it :)
But when I switch the code to a for loop. It doesn't give me four random numbers
If you've got a solution that works why would you be interested in one that does not? You post that you have a solution that does what you need but when you change it to something else it fails to work as you intend. What is the motivation to change a working solution?
Sorry, I was wrong. Looking more closely, your code doesn't have the problem I thought it had. This problem only arises when you use random_shuffle carelessly (as I deliberately did in my second example). So, your code that uses std::set is fine.
To make amends for my mistake, let me answer some of your questions.
And also most of the time the four numbers starts with a 0 even there's no 0 value in the array.
There are several zero values in your array :) Check this out:
``````#include <iostream>
using namespace std;
int main()
{
int array[3][3] = {
{ 1, 2, 3 },
{ 4, 5 },
{ 6 }
};
for (int i = 0; i < 3; ++ i)
{
for (int j = 0; j < 3; ++ j)
cout << array[i][j];
cout << endl;
}
}
``````
What's the difference between the two version?
The difference is that the while loop runs as many times as needed to fill the set with four (different) numbers, while the for loop always runs four times. After these four runs, the set may have four (different) values, but it may also have only three or two or even one. Take a look at std::set and how it works.
Also, I want to add here that your code relies on value uniqueness, while the examples I posted rely on index uniqueness. This difference is irrelevant to the method used to get the random sequence (rand-set / random_shuffle). Value uniquness guarantees, well, unique values, even if there are duplicate values in your array. Index uniqueness on the other hand, requires that your array doesn't have duplicate values in order to work properly. But it's useful when what you store in your array is not something trivial.
If you've got a solution that works why would you be interested in one that does not? You post that you have a solution that does what you need but when you change it to something else it fails to work as you intend. What is the motivation to change a working solution?
I was just wondering what would happen. If the for loop will give me the same results as the while loop. I asked you since maybe you would know :)
To make amends for my mistake, let me answer some of your questions.
It's ok thanks for your suggestion/s anyway.
I already figured out a way to my problem. I converted the 7x4 array into 1 dimensional and use std::set.
Be a part of the DaniWeb community
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, learning, and sharing knowledge. | 2,486 | 9,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | latest | en | 0.835138 |
https://www.mrexcel.com/board/threads/formula-error-not-counting-the-rows.173265/ | 1,695,819,136,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00706.warc.gz | 989,766,834 | 18,214 | # formula error (not counting the rows)
#### BMD
##### Board Regular
This one is not adding up the stuff right
=SUMPRODUCT(INDIRECT("'"&C1&"'!G\$3:G\$"&C8)="Pass")*OR(("'"&C1&"'!I\$3:I\$"&C8)="Major")*(("'"&C1&"'!I\$3:I\$"&C8)="Minor")
AND/OR
=SUMPRODUCT(INDIRECT("'"&C1&"'!G\$3:G\$"&C8)="Pass")*OR(("'"&C1&"'!I\$3:I\$"&C8)="Major",("'"&C1&"'!I\$3:I\$"&C8)="Minor")
If the test pass and has a major or minor add them all up
Thanks,
Bruce.
### Excel Facts
Select a hidden cell
Somehide hide payroll data in column G? Press F5. Type G1. Enter. Look in formula bar while you arrow down through G.
Try...
=SUMPRODUCT(--(INDIRECT("'"&C1&"'!G3:G"&C8)="Pass"),--(ISNUMBER(MATCH(INDIRECT("'"&C1&"'!I3:I"&C8),{"Major","Minor"},0))))
Hope this helps!
Thank you,
How do I do it if they are in two different colmuns?
=SUMPRODUCT(indirect("'"&C1&"'!G\$3:G\$"&c8)="Pass")*("'"&C1"'!H\$3:H\$"&C8)="High",("'"&C1"'!I\$3:I\$"&C8(="Major"))
It passed and is high and has a major on it?
Thanks,
Bruce.
Got it!!!!!!!!!!!!!1
=SUMPRODUCT(--(INDIRECT("'"&C1&"'!G3:G"&C8)="Pass"),--(ISNUMBER(MATCH(INDIRECT("'"&C1&"'!H3:H"&C8),{"High"},0))),--(ISNUMBER(MATCH(INDIRECT("'"&C1&"'!I3:I"&C8),{"Major"},0))))
This board is wonderful,,,,, and a time and life saver.
Thank you, all who have helped.
The following would suffice...
=SUMPRODUCT(--(INDIRECT("'"&C1&"'!G3:G"&C8)="Pass"),--(INDIRECT("'"&C1&"'!H3:H"&C8)="High"),--(INDIRECT("'"&C1&"'!I3:I"&C8)="Major"))
Hope this helps!
Yes, but in the one I posted
=SUMPRODUCT(--(INDIRECT("'"&C1&"'!G3:G"&C8)="Pass"),--(ISNUMBER(MATCH(INDIRECT("'"&C1&"'!H3:H"&C8),{"High"},0))),--(ISNUMBER(MATCH(INDIRECT("'"&C1&"'!I3:I"&C8),{"Major"},0))))
I can also have {"High","H"} if they only but a H, M, or L for High, Medium and Low.
I tried to remove the isnumber as I'm not using it but the formula failed. I will keep tinkering........
Thanks,
Bruce.
Sure, if you have more than one criteria for a column, you can use ISNUMBER/MATCH. Otherwise, if you only have one criterion, ISNUMBER/MATCH isn't necessary.
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Go back | 1,033 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-40 | latest | en | 0.576636 |
https://discussions.unity.com/t/doesnt-add-rotation-over-180/48380 | 1,695,607,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506676.95/warc/CC-MAIN-20230925015430-20230925045430-00094.warc.gz | 236,080,078 | 5,664 | # Doesn't Add Rotation Over 180
I’m using this script to make an AI walk around smoothly instead of making jerky turns. Here’s ta script I whipped up in a couple minutes:
`````` public float speed;
public float rotateSpeed;
public float minWalkTime;
public float maxWalkTime;
public float minRotateAmount;
public float maxRotateAmount;
public bool rotate;
public float timeToNewDir;
private Transform student;
private CharacterController controller;
void Awake () {
student = transform;
controller = student.GetComponent<CharacterController>();
}
void Update () {
Wander();
if(rotate){
Rotate(minRotateAmount, maxRotateAmount);
}
}
void Wander () {
float idleTime = Random.Range(minWalkTime, maxWalkTime);
if(Time.time > timeToNewDir){
timeToNewDir = Time.time + idleTime;
rotate = true;
}
var fwd = student.TransformDirection(Vector3.forward);
controller.SimpleMove(fwd * speed);
}
void Rotate (float minAmount, float maxAmount) {
float rotateAmount = Random.Range(minAmount, maxAmount);
Quaternion newRot = student.rotation;
newRot.y += rotateAmount;
student.rotation = Quaternion.Slerp(student.rotation, newRot, rotateSpeed * Time.deltaTime);
if(student.rotation == newRot){
rotate = false;
}
}
``````
The problem is, once “y” has become 180, the AI no longer turns, even though he should. The script doesn’t stop executing, the rotation just isn’t going though. Does it have to do with it being a Quaternion?
Yes, Quaternions can not be used like this. You will need to create a new Quaternion which will hold the new rotation you want. To apply this rotation to the previous rotation, you multiply the 2 quaternions together.
``````void Rotate (float minAmount, float maxAmount) {
float rotateAmount = Random.Range(minAmount, maxAmount);
Quaternion newRot = student.rotation;
newRot*= Quaternion.Euler(0,rotateAmount,0);
student.rotation = Quaternion.Slerp(student.rotation, newRot, rotateSpeed * Time.deltaTime);
if(student.rotation == newRot){
rotate = false;
}
}
``````
@Akill is right: the quaternion components XYZ have nothing to do with those nice angles you see in the Inspector (they are the localEulerAngles).
But your code had other problems too: Rotate will be called every frame, defining new orientations each time, and you probably would never get an equality to set rotate to false due to the limited floating point precision.
A better approach is to define the desired angle (yAngle) in Rotate and make the character always follow this angle smoothly at Update:
```...
private CharacterController controller;
// declare these new variables:
private Vector3 curEuler; // holds current rotation angles
private float yAngle; // holds desired angle around Y
void Awake () {
student = transform;
controller = student.GetComponent();
curEuler = student.eulerAngles; // initialize the new variables
yAngle = curEuler.y;
}
void Update () {
Wander();
// smoothly follows the desired yAngle:
curEuler.y = Mathf.Lerp(curEuler.y, yAngle, rotateSpeed * Time.deltaTime);
student.eulerAngles = curEuler;
}
void Wander () {
float idleTime = Random.Range(minWalkTime, maxWalkTime);
if(Time.time > timeToNewDir){
timeToNewDir = Time.time + idleTime;
Rotate(); // just set the new direction
}
var fwd = student.TransformDirection(Vector3.forward);
controller.SimpleMove(fwd * speed);
}
void Rotate (float minAmount, float maxAmount) {
float rotateAmount = Random.Range(minAmount, maxAmount);
// just add rotateAmount to yAngle modulo 360:
yAngle = (yAngle + rotateAmount) % 360;
}
``` | 788 | 3,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.604532 |
https://chemistry.stackexchange.com/questions/158601/why-are-md-simulations-necessary-for-obtaining-boltzmann-populations | 1,723,610,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00318.warc.gz | 129,460,229 | 42,136 | # Why are MD simulations necessary for obtaining Boltzmann populations?
Given that MD simulations converge to the Boltzmann distribution $$\rho \sim \exp(-\beta \epsilon)$$ after sufficiently long times, and all the macroscopic quantities can be computed from the Boltzmann distribution itself, what is the need for MD simulations?
Specifically, I am asking in the context of short peptides (tripeptides or tetrapeptides).
For instance, in Beck, D. A. C.; Alonso, D. O. V.; Inoyama, D.; Daggett, V. Proc. Natl. Acad. Sci. U. S. A. 2008, 105 (34), 12259–12264, the authors use MD to generate Ramachandran distributions of pentapeptide conformations at a constant temperature. This distribution should obey statistical mechanics, and specifically the Boltzmann distribution. So in theory one should be able to write down the distributions using the Boltzmann weights as follows,
$$\rho(\{\phi_{i},\psi_{i}\}) \sim \exp(-\beta V(\{ \phi_{i},\psi_{i}\})).$$
Here, $$\{ \phi_{i},\psi_{i}\}$$ denotes the set of Ramachandran angle coordinates.
Why should I run MD to get the same distributions?
• In the paper you reference, it is noted that the conformations do not converge and are dynamic in range. Most important and more broadly, experimentally-determined structures are static and do not represent full state space of the system. Commented Oct 11, 2021 at 1:58
• I still have a question. Even if they get dynamic pentapeptide when they use MD to get the distributions. They can simply use boltzman weight to express that right? why should I use MD to get this result. Is boltzmann weight incorrect in predicting conformational distributions? @ToddMinehardt Commented Oct 11, 2021 at 2:18
In principle you can, assuming you are given $$V(\{\bf{r_i}\})$$ where $$\{\bf {r_i} \}$$ is the set of variables that define a configuration in the system - this is typically the coordinates of the atoms, but it might include things like external fields. The problem is in practice it will be at best horribly, horribly inefficient, and in fact almost certainly impossible.
To make progress let's try and define your problem at little more carefully. You have written the potential energy of the system as a function of just two variables, $$\{ \phi_{i},\psi_{i}\}$$. This is not the case - it is a function of the coordinates of all the atoms in the system, assuming no external fields, and so to get a function of just these two variables we are going to have to average over all configurations which contain the atoms in positions satisfying the constraint that $$\{ \phi_{i},\psi_{i}\}$$ are the values we desire. Thus in fact we should write
$$\rho(\{\phi_{i},\psi_{i}\}) \sim \int d{\bf {r_1}}d{\bf {r_2}}... d{\bf {r_N}}\exp(-\beta V(\{ \phi_{i},\psi_{i}\};\{\bf{r_i}\}))$$
where the integral has limits such that the atomic configurations considered have the required angles, $$\{\phi_{i},\psi_{i}\}$$.
Now in principle this complicated multi-dimensional integral might be able to be solved analytically. If this is the case just calculate the value for all $$\{\phi_{i},\psi_{i}\}$$ you are interested in, normalise the probabilities appropriately, and you are done. In practice for any realistic forcefield you will never be able to perform this integral analytically. Thus you have to resort to numerical integration on the computer.
But now you have a secondary problem. To do this multidimensional integral on a computer naïvely is going to be impossible for anything but the smallest system. Let's say we just have 35 atoms, thus approximately 100 degrees of freedom, and we want to have just 10 points along each direction. That means we have approximately $$10^{100}$$ configurations to sample for each $$\{\phi_{i},\psi_{i}\}$$. This is impossible even on the most powerful computers we have today. And it will be impossible for a very long time.
But there's another problem, even if we could do this it will likely be extremely inaccurate to the point of uselessness. The problem is the vast, vast, vast majority of the atomic configurations will contribute almost nothing to the integral; their energy will be just too high - remember these are integrals over all space. As an example consider the case where the hydrogen bonded to a $$\alpha$$ carbon is stretched to almost infinity. We know this is very high energy, and therefore through the exponential will make a negligible contribution to the integral, but our above naïve sampling scheme makes no guarantee that we won't include such configurations and so pointlessly calculate their contribution to the probability, which will be zero. But worse is the flip side of this: We have absolutely no guarantee that we will actually include configurations that do have large contributions to the probability! Thus unless we have many more than 10 points along each axis most likely our result will be utterly, utterly wrong.
This is where Molecular Dynamics (or, alternatively, Monte Carlo) comes in. It is an "intelligent" way of sampling the above integral such that we only consider atomic configurations that have a sizeable contribution to the configuration integral. Thus we can calculate accurate estimates to the quantities of interest by using a relatively small sample of important configurations, and thus making the evaluation of those properties via the configuration integral viable on a computer in a reasonable amount of time and memory.
Finally note this is not specific to the case here - the same argument applies to virtually any quantity of interest, and the conclusion remains the same: In statistical physics the prime uses of Molecular Dynamics and Monte Carlo are as intelligent sampling techniques to overcome the Curse of Dimensionality | 1,296 | 5,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.877255 |
https://stats.stackexchange.com/questions/239060/interpretation-of-breusch-pagan-test-bptest-in-r | 1,716,941,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00343.warc.gz | 470,533,931 | 39,958 | # Interpretation of Breusch-Pagan test bptest() in R
I was a bit confused regarding the interpretation of bptest in R (library(lmtest)). The null hypothesis of bptest is that the residuals have constant variance. So, a p-value less than 0.05 would mean that the homoscedasticity assumption would have to be rejected. However on this website:
http://rstatistics.net/how-to-test-a-regression-model-for-heteroscedasticity-and-if-present-how-to-correct-it/
I found the following results that confuse me:
data: lmMod
BP = 3.2149, df = 1, p-value = 0.07297
A p-Value > 0.05 indicates that the null hypothesis(the variance is unchanging in the residual) can be rejected and therefore heterscedasticity exists. This can be confirmed by running a global validation of linear model assumptions (gvlma) on the lm object.
gvlma(lmMod) # validate if assumptions of linear regression holds true.
# Call:
gvlma(x = lmMod)
Value p-value Decision
Global Stat 15.801 0.003298 Assumptions NOT satisfied!
Skewness 6.528 0.010621 Assumptions NOT satisfied!
Kurtosis 1.661 0.197449 Assumptions acceptable.
Link Function 2.329 0.126998 Assumptions acceptable.
Heteroscedasticity 5.283 0.021530 Assumptions NOT satisfied!
So why is it the case that a p-value>0.05 means you have to reject the null hypothesis, when in fact a p-value less than 0.05 indicates that you have to reject the null hypothesis?
This ought to be a typo on rstatistics.net. You are correct that the null hypothesis of the Breusch-Pagan test is homoscedasticity (= variance does not depend on auxiliary regressors). If the $p$-value becomes "small", the null hypothesis is rejected.
I would recommend contacting the authors of rstatistics.net regarding this issue to see if they agree and fix it.
Moreover, note that glvma() employs a different auxiliary regressor than bptest() by default and switches off studentization. More precisely, you can see the differences if you replicate the results by setting the arguments of bptest() explicitly.
The model is given by:
data("cars", package = "datasets")
lmMod <- lm(dist ~ speed, data = cars)
The default employed by bptest() then uses the same auxiliary regressors as the model, i.e., speed in this case. Also it uses the studentized version with improved finite-sample properties yielding a non-significant result.
library("lmtest")
bptest(lmMod, ~ speed, data = cars, studentize = TRUE)
## studentized Breusch-Pagan test
##
## data: lmMod
## BP = 3.2149, df = 1, p-value = 0.07297
In contrast, the glvma() switches off studentization and checks for a linear trend in the variances.
cars$trend <- 1:nrow(cars) bptest(lmMod, ~ trend, data = cars, studentize = FALSE) ## Breusch-Pagan test ## ## data: lmMod ## BP = 5.2834, df = 1, p-value = 0.02153 As you can see both$p\$-values are rather small but on different sides of 5%. The studentized versions are both slightly above 5%.
• It is strange though that glvma says "Assumptions NOT satisfied!" in the last row, when in fact based on the p-value of bptest() the assumption should be satisfied. Oct 8, 2016 at 11:02
• I've updated my answer to explain that it's slightly different variations of the Breusch-Pagan test that lead to slightly different p-values on different sides of 5%. I'm not fond of the decision labels in glvma, these are worse than significance stars and convey that the p-value of 5.1% is fundamentally different than 4.9%. Oct 8, 2016 at 12:12 | 903 | 3,503 | {"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} | 2.96875 | 3 | CC-MAIN-2024-22 | latest | en | 0.854446 |
http://www.gurufocus.com/term/ev2rev/EMR/EV%252FRevenue/Emerson%2BElectric%2BCo | 1,485,249,029,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284376.37/warc/CC-MAIN-20170116095124-00356-ip-10-171-10-70.ec2.internal.warc.gz | 482,057,181 | 28,747 | Switch to:
GuruFocus has detected 5 Warning Signs with Emerson Electric Co \$EMR.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Emerson Electric Co (NYSE:EMR)
EV/Revenue
2.83 (As of Today)
EV/Revenue ratio is calculated as enterprise value divided by its revenue. As of today, Emerson Electric Co's enterprise value is \$41,079 Mil. Emerson Electric Co's revenue for the trailing twelve months (TTM) ended in Sep. 2016 was \$14,522 Mil. Therefore, Emerson Electric Co's EV/Revenue ratio for today is 2.83.
The reason Enterprise Value is used is because Enterprise Value is more real in reflecting how much an investor pays when buying a company. For detais, go to Enterprise Value.
EV/Revenue is a similar ratio to P/S Ratio, except here Enterprise Value instead of Market Cap is used in the calculation.
As of today, Emerson Electric Co's stock price is \$58.29. Emerson Electric Co's revenue per share for the trailing twelve months (TTM) ended in Sep. 2016 was \$22.43. Therefore, Emerson Electric Co's P/S Ratio for today is 2.60.
Definition
EV/Revenue is a similar ratio to P/S Ratio, except here Enterprise Value instead of Market Cap is used in the calculation.
Emerson Electric Co's EV/Revenue for today is calculated as:
EV/Revenue = Enterprise Value (Today) / Revenue (TTM) = 41078.980 / 14522 = 2.83
Emerson Electric Co's current Enterprise Value is \$41,079 Mil.
Emerson Electric Co's Revenue for the trailing twelve months (TTM) ended in Sep. 2016 was 4713 (Dec. 2015 ) + 4928 (Mar. 2016 ) + 5126 (Jun. 2016 ) + -245 (Sep. 2016 ) = \$14,522 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
The reason Enterprise Value is used is because Enterprise Value is more real in reflecting how much an investor pays when buying a company. For detais, go to Enterprise Value.
Emerson Electric Co's P/S Ratio for today is calculated as:
P/S Ratio = Share Price (Today) / Revenue Per Share (TTM) = 58.29 / 22.433 = 2.60
Emerson Electric Co's share price for today is \$58.29.
Emerson Electric Co's Revenue Per Share for the trailing twelve months (TTM) ended in Sep. 2016 was 7.223 (Dec. 2015 ) + 7.644 (Mar. 2016 ) + 7.945 (Jun. 2016 ) + -0.379 (Sep. 2016 ) = \$22.43.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Emerson Electric Co Annual Data
Sep07 Sep08 Sep09 Sep10 Sep11 Sep12 Sep13 Sep14 Sep15 Sep16 ev2rev 2.04 1.38 1.66 2.06 1.40 1.56 1.95 2.62 2.02 2.65
Emerson Electric Co Quarterly Data
Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 ev2rev 2.03 2.62 2.61 2.40 2.48 2.02 2.29 2.62 2.57 2.65
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###### A man can row at 5 km/hr in still water. If the velocity of current is 1 km/hr and it takes him 1 hour to row to a place and come back, how far is the place? A. 2.4 km B. 2 km C. 2.6 km D. 3 km
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ME 343 In-term exam 2, 13.12.2010, 150 minutes Notes: a) Draw figures (free-body diagrams and other helpful diagrams) as necessary and indicate quantities clearly. b) Show your axes . c) Indicate units . d) Write neatly. Make sure what you write is readable . ------------------------------------------------------------------------------------------------------------------------------------------ 1) (15 pts.) Given an L-beam with a force in the plane of the beam. a) Calculate the stresses on the cross-section at the wall at two points : at the topmost point and at the centroid . b) Find the three principal stresses at the same points. 2) (25 pts.) Given a structure consisting of an L-beam and a bar, which are pinned at the corner of beam. The other supports are also pins. a= 500 mm, h=40 mm, b=20 mm. The end-condition constant for out-of-plane buckling of the bar is c=1.2 and for in-plane buckling it is c=0.8. E=70 GPa and S yc =500 MPa for the bar material. a)
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## This note was uploaded on 10/06/2011 for the course ME 343 taught by Professor Namikciblak during the Spring '11 term at Yeditepe Üniversitesi.
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Excel logical formulas: 8 simple IF statements to
# Help writing if statements in excel
## MS Excel: How to use the IF Function (WS) -
### How to create an if then statement in
Aug 2011 This video shows how to create a simple if then function in excel The if then function is a very powerful tool to use in excel and is relatively
### MS Excel: How to use the IF Function (WS) -
The Microsoft Excel IF function returns one value if the condition is TRUE, Answer: In cell C5, you can write a nested IF statement that uses the AND function I need help setting up this formula so that when the 3 columns are left blank, the
### How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either "equal to" or
### How to Write an IF Formula/Statement in
May 2012 Being able to write a simple IF statement in Excel is essential, especially if you deal with a lot of numbers all the time It s one of those things
### Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is
MS Excel: How to use the IF Function (WS) -
The Microsoft Excel IF function returns one value if the condition is TRUE, Answer: In cell C5, you can write a nested IF statement that uses the AND function I need help setting up this formula so that when the 3 columns are left blank, the Excel logical formulas: 8 simple IF statements to
Sep 2015 Excel logical formulas: 8 simple IF statements to get started When you have a big, ugly spreadsheet, logical formulas can help you make Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can Excel formula: If else | Exceljet If you need to test a condition, then take one action if the condition is TRUE, and Lets say you want to write a formula to expand these abbreviations and show IF function - Office Support The IF function in Excel returns one value if a condition is true and another value if it s false You can use up to 64 The best way to start writing an IF statement is to think about what you are trying to accomplish Help us improve Excel
How to use nested IF statements in Excel with AND, OR, NOT
Dec 2006 How to use nested IF statements in Excel with AND, OR, NOT reading or Get 10 days of free unlimited access to Lynda com for professional help and Excel Tutorials To use it correctly, you have to write it like the following: Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can IF function - Office Support The IF function in Excel returns one value if a condition is true and another value if it s false You can use up to 64 The best way to start writing an IF statement is to think about what you are trying to accomplish Help us improve Excel How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or How to create an if then statement in
Aug 2011 This video shows how to create a simple if then function in excel The if then function is a very powerful tool to use in excel and is relatively Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is
How to create an if then statement in
Aug 2011 This video shows how to create a simple if then function in excel The if then function is a very powerful tool to use in excel and is relatively How to use nested IF statements in Excel with AND, OR, NOT
Dec 2006 How to use nested IF statements in Excel with AND, OR, NOT reading or Get 10 days of free unlimited access to Lynda com for professional help and Excel Tutorials To use it correctly, you have to write it like the following: Excel formula: If else | Exceljet If you need to test a condition, then take one action if the condition is TRUE, and Lets say you want to write a formula to expand these abbreviations and show Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or
How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can How to create an if then statement in
Aug 2011 This video shows how to create a simple if then function in excel The if then function is a very powerful tool to use in excel and is relatively How to Write an IF Formula/Statement in
May 2012 Being able to write a simple IF statement in Excel is essential, especially if you deal with a lot of numbers all the time It s one of those things IF function - Office Support The IF function in Excel returns one value if a condition is true and another value if it s false You can use up to 64 The best way to start writing an IF statement is to think about what you are trying to accomplish Help us improve Excel Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is
How to create an if then statement in
Aug 2011 This video shows how to create a simple if then function in excel The if then function is a very powerful tool to use in excel and is relatively Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can How to Write an IF Formula/Statement in
May 2012 Being able to write a simple IF statement in Excel is essential, especially if you deal with a lot of numbers all the time It s one of those things How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or
Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or Excel formula: If else | Exceljet If you need to test a condition, then take one action if the condition is TRUE, and Lets say you want to write a formula to expand these abbreviations and show
Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can Excel formula: If else | Exceljet If you need to test a condition, then take one action if the condition is TRUE, and Lets say you want to write a formula to expand these abbreviations and show How to use nested IF statements in Excel with AND, OR, NOT
Dec 2006 How to use nested IF statements in Excel with AND, OR, NOT reading or Get 10 days of free unlimited access to Lynda com for professional help and Excel Tutorials To use it correctly, you have to write it like the following: MS Excel: How to use the IF Function (WS) -
The Microsoft Excel IF function returns one value if the condition is TRUE, Answer: In cell C5, you can write a nested IF statement that uses the AND function I need help setting up this formula so that when the 3 columns are left blank, the How to Write an IF Formula/Statement in
May 2012 Being able to write a simple IF statement in Excel is essential, especially if you deal with a lot of numbers all the time It s one of those things IF function - Office Support The IF function in Excel returns one value if a condition is true and another value if it s false You can use up to 64 The best way to start writing an IF statement is to think about what you are trying to accomplish Help us improve Excel
How to create an if then statement in
Aug 2011 This video shows how to create a simple if then function in excel The if then function is a very powerful tool to use in excel and is relatively How to Write an IF Formula/Statement in
May 2012 Being able to write a simple IF statement in Excel is essential, especially if you deal with a lot of numbers all the time It s one of those things Excel formula: If else | Exceljet If you need to test a condition, then take one action if the condition is TRUE, and Lets say you want to write a formula to expand these abbreviations and show Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is How to use nested IF statements in Excel with AND, OR, NOT
Dec 2006 How to use nested IF statements in Excel with AND, OR, NOT reading or Get 10 days of free unlimited access to Lynda com for professional help and Excel Tutorials To use it correctly, you have to write it like the following: Excel logical formulas: 8 simple IF statements to
Sep 2015 Excel logical formulas: 8 simple IF statements to get started When you have a big, ugly spreadsheet, logical formulas can help you make How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or
Excel logical formulas: 8 simple IF statements to
Sep 2015 Excel logical formulas: 8 simple IF statements to get started When you have a big, ugly spreadsheet, logical formulas can help you make How to use IF function in Excel: examples for text,
Nov 2014 The IF function is one of Excel s logical functions that evaluates a Generally, you write an IF formula for text values using either 'equal to' or Excel IF Statement How to Use Excel If Statement can also be used along with the logical operators (like AND, OR) for analysing complex logics Here I will help you to understand how can Excel Formulas: Using the IF Function - Full Page -
Free Tutorial: Use the Excel IF function to help solve this real-world Excel problem Use our spreadsheet to use the Excel IF statement to get your answer We could go through each row, look at the value, and then write Free Gift if the value is | 2,677 | 12,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-21 | latest | en | 0.811178 |
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# During the 1992 Presidential campaign, President George H. W. Bush and
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During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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During the 1992 Presidential campaign, President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush's 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation
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14 Nov 2019, 00:12
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A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation --> wrong meaning
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation --> Parallelism is followed and correct meaning --> Correct
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation --> wrong
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation --> Not Parallel
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation --> Not parallel
IMO Option B
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 01:36
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President George H. W. Bush and his promise four years earlier to never raise taxes experienced => A and C are out
(D) who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation => parallel issue
(E) who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation => parallel issue
So (B) is correct!
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 04:02
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Quote:
During the 1992 Presidential campaign, President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush's 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation
MEANING
During a campaign, Bush experienced considerable criticism from conservatives who felt betrayed by a tax-hike;
a tax-hike, which he [Bush] implemented despite his past promises and objections from others who decried, insisted and contended.
Who experienced criticism? Bush or Bush and his promise? Bush.
The others decried, insisted and contended - a list must maintain parallelism.
Ans (B)
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 04:45
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B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation..............sentence correctly states that it was Bush,not his past promise that received criticism and objections of economic conservatives are in a parallel
OA:B
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 06:09
1
IMO B is correct
President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
During the 1992 Presidential campaign, [url]President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush's 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
[/url]
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 09:03
1
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
Correct
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation.
Not parallel
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation.
Not parallel
B is the answer in my opinion
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 12:14
1
The right answer is option B.
A. During the 1992 Presidential campaign, President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
Incorrect because it is illogical to say that the promise of George Bush experienced criticism. It makes sense for George Bush to be criticized but not his promise. Eliminate option A.
B. During the 1992 Presidential campaign, President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
This is the correct option. Insisted and contended are parallel and as verb-ed modifiers, they appropriately modify economic conservatives.
C. During the 1992 Presidential campaign, President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation.
Incorrect for the same illogical parallelism issue in option A.
D. During the 1992 Presidential campaign, President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation.
Incorrect because having insisted is not parallel to having contending. In fact, having contending is incorrect and at best awkward to say the least. Contending should have been contended. Even if contended was rightly used, this option is inferior to option B, because it is not as concise as option B. Eliminate option D.
E. During the 1992 Presidential campaign, President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation.
Incorrect tense is used here. The simple present tense is not adequate. We need verb-ed modifiers as it is in option B.
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
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14 Nov 2019, 12:27
1
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation
A and C: Wrongly suggest that both George H. W. Bush and his promise experienced criticism. In actuality, George H. W. Bush received criticism for betraying his promise.
C: Fails to attribute "the insistence that low taxes create wealth, and the contention that deregulation promoted innovation" to economic conservatives by subordinating "the objections of economic conservatives who decried tax hikes" to a list.
D: The clause following "economic conservatives who decried tax hikes" is not parallel (should be "having insisted ... and contended"; however, "insisting ... and contending" would be better)
E: The list is not parallel: "economic conservatives who DECRIED tax hikes, INSIST that low taxes created wealth, and CONTEND that deregulation promoted innovation." Given the context the entire list should be in the past tense.
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
### Show Tags
14 Nov 2019, 23:29
Bunuel wrote:
Competition Mode Question
During the 1992 Presidential campaign, President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush's 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation
Official Explanation
There is one main problem with the original sentence.
(1)The sentence is constructed in a misleading fashion, giving the impression that conservatives criticized Bush’s promise as opposed to his breaking of that promise. When read in its entirety, the sentence incorrectly indicates that it wasn’t Bush’s promise that received considerable criticism but his breaking of the promise.
When correcting this sentence, it is important to keep parallelism in mind.
Each part of the economic conservatives’ philosophy must be parallel. To do this, the sentence should read: whose philosophy decried..., insisted..., and contended
A. The sentence is constructed in a misleading fashion, giving the impression that the conservatives criticized Bush’s promise as opposed to his breaking of that promise
B. The sentence correctly states that it was Bush (not his past promise) that received criticism; the objections of economic conservatives are listed in a parallel fashion
C. The sentence is constructed in a misleading fashion, giving the impression that the conservatives criticized Bush’s promise as opposed to his breaking of that promise; this sentence gives the misleading understanding of the objections of economic conservatives (i.e., it seems to imply that the insistence that low taxes create wealth, and the contention that deregulation promoted innovation are not explanations for or parts of the economic conservatives' objections)
D. Each part of the economic conservatives’ philosophy is not parallel (i.e., the phrase who decried tax hikes, having insisted ... and contending that ... is not parallel)
E. Each part of the economic conservatives’ philosophy is not parallel (i.e., the phrase who decried tax hikes, insist ... and contend... is not parallel)
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink]
### Show Tags
15 Nov 2019, 14:27
Bunuel wrote:
Bunuel wrote:
Competition Mode Question
During the 1992 Presidential campaign, President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush's 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation.
A. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented over the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
B. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insisted that low taxes created wealth, and contended that deregulation promoted innovation
C. President George H. W. Bush and his promise four years earlier to never raise taxes experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises, the objections of economic conservatives who decried tax hikes, the insistence that low taxes create wealth, and the contention that deregulation promoted innovation
D. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, having insisted that low taxes created wealth and contending that deregulation promotes innovation
E. President George H. W. Bush experienced considerable criticism from economic conservatives who felt betrayed by Bush’s 1990 tax-hike, which he implemented despite his past promises and the objections of economic conservatives who decried tax hikes, insist that low taxes created wealth, and contend that deregulation promoted innovation
Official Explanation
There is one main problem with the original sentence.
(1)The sentence is constructed in a misleading fashion, giving the impression that conservatives criticized Bush’s promise as opposed to his breaking of that promise. When read in its entirety, the sentence incorrectly indicates that it wasn’t Bush’s promise that received considerable criticism but his breaking of the promise.
When correcting this sentence, it is important to keep parallelism in mind.
Each part of the economic conservatives’ philosophy must be parallel. To do this, the sentence should read: whose philosophy decried..., insisted..., and contended
A. The sentence is constructed in a misleading fashion, giving the impression that the conservatives criticized Bush’s promise as opposed to his breaking of that promise
B. The sentence correctly states that it was Bush (not his past promise) that received criticism; the objections of economic conservatives are listed in a parallel fashion
C. The sentence is constructed in a misleading fashion, giving the impression that the conservatives criticized Bush’s promise as opposed to his breaking of that promise; this sentence gives the misleading understanding of the objections of economic conservatives (i.e., it seems to imply that the insistence that low taxes create wealth, and the contention that deregulation promoted innovation are not explanations for or parts of the economic conservatives' objections)
D. Each part of the economic conservatives’ philosophy is not parallel (i.e., the phrase who decried tax hikes, having insisted ... and contending that ... is not parallel)
E. Each part of the economic conservatives’ philosophy is not parallel (i.e., the phrase who decried tax hikes, insist ... and contend... is not parallel)
Hi
In answer choice D, -ed can be parallel with -ing. Why is that a reason to eliminate?
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Re: During the 1992 Presidential campaign, President George H. W. Bush and [#permalink] 15 Nov 2019, 14:27
Display posts from previous: Sort by | 6,023 | 29,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.938042 |
http://makemathmore.com/members/category/common-core-standards/6th-grade/6th-grade-number-system/ | 1,558,683,223,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257553.63/warc/CC-MAIN-20190524064354-20190524090354-00480.warc.gz | 127,615,680 | 12,862 | ### Positives and Negatives of Bungee Jumping
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American Football is a game of positive and negative yards. This game simulates this experience. It shows students where negatives exist in real life (America’s favorite sport) and helps them to practice and further realize the rule for adding integers of different signs. | 334 | 1,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-22 | longest | en | 0.933325 |
https://demo.formulasearchengine.com/wiki/Pseudoisotopy_theorem | 1,611,263,779,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00592.warc.gz | 300,180,330 | 8,449 | # Pseudoisotopy theorem
In mathematics, the pseudoisotopy theorem is a theorem of Jean Cerf's[1] which refers to the connectivity of a group of diffeomorphisms of a manifold.
## Statement
Given a differentiable manifold M (with or without boundary), a pseudo-isotopy diffeomorphism of M is a diffeomorphism of M × [0, 1] which restricts to the identity on ${\displaystyle M\times \{0\}\cup \partial M\times [0,1]}$.
Given ${\displaystyle f:M\times [0,1]\to M\times [0,1]}$ a pseudo-isotopy diffeomorphism, its restriction to ${\displaystyle M\times \{1\}}$ is a diffeomorphism ${\displaystyle g}$ of M. We say g is pseudo-isotopic to the identity. One should think of a pseudo-isotopy as something that is almost an isotopy—the obstruction to ƒ being an isotopy of g to the identity is whether or not ƒ preserves the level-sets ${\displaystyle M\times \{t\}}$ for ${\displaystyle t\in [0,1]}$.
Cerf's theorem states that, provided M is simply-connected and dim(M) ≥ 5, the group of pseudo-isotopy diffeomorphisms of M is connected. Equivalently, a diffeomorphism of M is isotopic to the identity if and only if it is pseudo-isotopic to the identity. [2]
## Relation to Cerf theory
The starting point of the proof is to think of the height function as a 1-parameter family of smooth functions on M by considering the function ${\displaystyle \pi _{[0,1]}\circ f_{t}}$. One then applies Cerf theory.[3]
## References
1. French mathematician, born 1928
2. J.Cerf, La stratification naturelle des espaces de fonctions deff\'erentiables r\'eelles et le th\'eor\`eme de la pseudo-isotopie, Inst. Hautes \'Etudes Sci. Publ. Math. No {\bf 39} (1970) 5–173.
3. J.Cerf, La stratification naturelle des espaces de fonctions différentiables réelles et le théorème de la pseudo-isotopie, Inst. Hautes Études Sci. Publ. Math. No. 39 (1970) 5–173. | 556 | 1,841 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-04 | latest | en | 0.810205 |
http://de.metamath.org/mpeuni/sb5ALTVD.html | 1,721,610,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00559.warc.gz | 8,765,480 | 5,590 | Mathbox for Alan Sare < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > sb5ALTVD Structured version Visualization version GIF version
Theorem sb5ALTVD 38171
Description: The following User's Proof is a Natural Deduction Sequent Calculus transcription of the Fitch-style Natural Deduction proof of Unit 20 Excercise 3.a., which is sb5 2418, found in the "Answers to Starred Exercises" on page 457 of "Understanding Symbolic Logic", Fifth Edition (2008), by Virginia Klenk. The same proof may also be interpreted as a Virtual Deduction Hilbert-style axiomatic proof. It was completed automatically by the tools program completeusersproof.cmd, which invokes Mel L. O'Cat's mmj2 and Norm Megill's Metamath Proof Assistant. sb5ALT 37752 is sb5ALTVD 38171 without virtual deductions and was automatically derived from sb5ALTVD 38171.
1:: ⊢ ( [𝑦 / 𝑥]𝜑 ▶ [𝑦 / 𝑥]𝜑 ) 2:: ⊢ [𝑦 / 𝑥]𝑥 = 𝑦 3:1,2: ⊢ ( [𝑦 / 𝑥]𝜑 ▶ [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) ) 4:3: ⊢ ( [𝑦 / 𝑥]𝜑 ▶ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑 ) ) 5:4: ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ) 6:: ⊢ ( ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ▶ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ) 7:: ⊢ ( ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) , (𝑥 = 𝑦 ∧ 𝜑 ) ▶ (𝑥 = 𝑦 ∧ 𝜑) ) 8:7: ⊢ ( ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) , (𝑥 = 𝑦 ∧ 𝜑 ) ▶ 𝜑 ) 9:7: ⊢ ( ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) , (𝑥 = 𝑦 ∧ 𝜑 ) ▶ 𝑥 = 𝑦 ) 10:8,9: ⊢ ( ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) , (𝑥 = 𝑦 ∧ 𝜑 ) ▶ [𝑦 / 𝑥]𝜑 ) 101:: ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑) 11:101,10: ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑 ) 12:5,11: ⊢ (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑 )) ∧ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)) qed:12: ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) )
(Contributed by Alan Sare, 21-Apr-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
sb5ALTVD ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))
Distinct variable group: 𝑥,𝑦
Allowed substitution hints: 𝜑(𝑥,𝑦)
Proof of Theorem sb5ALTVD
StepHypRef Expression
1 idn1 37811 . . . . . 6 ( [𝑦 / 𝑥]𝜑 ▶ [𝑦 / 𝑥]𝜑 )
2 equsb1 2356 . . . . . 6 [𝑦 / 𝑥]𝑥 = 𝑦
3 sban 2387 . . . . . . 7 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
43simplbi2com 655 . . . . . 6 ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦𝜑)))
51, 2, 4e10 37940 . . . . 5 ( [𝑦 / 𝑥]𝜑 ▶ [𝑦 / 𝑥](𝑥 = 𝑦𝜑) )
6 spsbe 1871 . . . . 5 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
75, 6e1a 37873 . . . 4 ( [𝑦 / 𝑥]𝜑 ▶ 𝑥(𝑥 = 𝑦𝜑) )
87in1 37808 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
9 hbs1 2424 . . . 4 ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
10 idn2 37859 . . . . . 6 ( 𝑥(𝑥 = 𝑦𝜑) , (𝑥 = 𝑦𝜑) ▶ (𝑥 = 𝑦𝜑) )
11 simpr 476 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝜑)
1210, 11e2 37877 . . . . 5 ( 𝑥(𝑥 = 𝑦𝜑) , (𝑥 = 𝑦𝜑) ▶ 𝜑 )
13 simpl 472 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝑥 = 𝑦)
1410, 13e2 37877 . . . . 5 ( 𝑥(𝑥 = 𝑦𝜑) , (𝑥 = 𝑦𝜑) ▶ 𝑥 = 𝑦 )
15 sbequ1 2096 . . . . . 6 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
1615com12 32 . . . . 5 (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
1712, 14, 16e22 37917 . . . 4 ( 𝑥(𝑥 = 𝑦𝜑) , (𝑥 = 𝑦𝜑) ▶ [𝑦 / 𝑥]𝜑 )
189, 17exinst 37870 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
198, 18pm3.2i 470 . 2 (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
20 impbi 197 . . 3 (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) → ((∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))))
2120imp 444 . 2 ((([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑)))
2219, 21e0a 38020 1 ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ∃wex 1695 [wsb 1867 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-12 2034 ax-13 2234 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-vd1 37807 df-vd2 37815 This theorem is referenced by: (None)
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MRC: Where is The Best Place to Measure? (Grades 6-8)
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This is a lesson about measuring temperature. Learners will apply their knowledge of how temperature affects matter to understand how a thermometer works. They then read about the history of the thermometer and the temperature scales that make the... (View More)
Audience: Middle school
Design Challenge: How do You Keep Things from Getting Too Hot?
This is a lesson about designing and building an effective sunshade for a model MESSENGER craft. Learners will build a model of MESSENGER. They will use a scientific approach to solve problems and work as a cooperative team. They will discover their... (View More)
Snow Goggles and Limiting Sunlight
This is a lesson about radiation and the use of the scientific method to solve problems of too much radiation. Learners will build snow goggles similar to those used by the Inuit (designed to block unwanted light, while increasing the viewer's... (View More)
Time Warp
In this inquiry investigation, students conclude that the motion of the Earth is linked to the changes we observe such as the length of the day. Students learn about the reason behind the Earth's time zones. An optional water clock and sand clock... (View More)
Keywords: Time; Orbit; Time zones
1 | 851 | 4,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-04 | latest | en | 0.866598 |
http://mizar.uwb.edu.pl/version/current/html/proofs/finseq_6/53 | 1,571,687,714,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00460.warc.gz | 131,162,205 | 1,692 | let D be non empty set ; :: thesis: for p1, p2, p3 being Element of D st p1 <> p3 & p2 <> p3 holds
<*p1,p2,p3*> -: p3 = <*p1,p2,p3*>
let p1, p2, p3 be Element of D; :: thesis: ( p1 <> p3 & p2 <> p3 implies <*p1,p2,p3*> -: p3 = <*p1,p2,p3*> )
assume that
A1: p1 <> p3 and
A2: p2 <> p3 ; :: thesis: <*p1,p2,p3*> -: p3 = <*p1,p2,p3*>
rng <*p1,p2,p3*> = {p1,p2,p3} by Lm2;
then p3 in rng <*p1,p2,p3*> by ENUMSET1:def 1;
hence <*p1,p2,p3*> -: p3 = (<*p1,p2,p3*> -| p3) ^ <*p3*> by Th40
.= <*p1,p2*> ^ <*p3*> by A1, A2, Th28
.= <*p1,p2,p3*> ;
:: thesis: verum | 274 | 554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.456334 |
https://www.eduzip.com/ask/question/if-displaystyle-fxlim-nrightarrow-infty-2x4x32nx2n-1left-0ltxltfr-587448 | 1,627,886,141,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154304.34/warc/CC-MAIN-20210802043814-20210802073814-00697.warc.gz | 777,210,068 | 9,262 | Mathematics
# If $\displaystyle f(x)=\lim_{n\rightarrow \infty }(2x+4x^{3}+......+2^{n}x^{2n-1})\left ( 0<x<\frac{1}{\sqrt{2}} \right )$, then the value of $\displaystyle\int f(x) dx$ is equal to$\textbf{Note}$: $c$ is the constant of integration.
$\displaystyle \log\left ( \frac{1}{\sqrt{1-2x^{2}}} \right )+c$
##### SOLUTION
When $n \rightarrow \infty$, $f(x)$ becomes an infinites series of G.P. with $a=2x$, $r=2x^2$
Hence, $f(x)=\dfrac{2x}{1-2x^2}$
Since, $0<x<\dfrac1{\sqrt2}$
So, the integral $I = \displaystyle\int f(x)\>dx$ becomes $\displaystyle \int \frac{2x}{1-2x^{2}}dx$
Substitute, $1-2x^{2}=t\Rightarrow -4x\>dx=dt$
After substitution, the integral becomes
$\displaystyle -\frac{1}{2}\int \frac{dt}{t}=-\frac{1}{2}\log(t)+c$
$\displaystyle =-\frac{1}{2}\log (1-2x^2) +c$
$\displaystyle I =\log\left ( \frac{1}{\sqrt{1-2x^{2}}} \right )+c$
Hence, option C.
Its FREE, you're just one step away
Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111
#### Realted Questions
Q1 Single Correct Medium
$\underset {x\to \infty}\lim \int_{0}^{x}xe^{t^{2}-x^{2}}dt$ is equal to
• A. $2$
• B. $\displaystyle -\frac{1}{2}$
• C. $1$
• D. $\displaystyle \frac{1}{2}$
1 Verified Answer | Published on 17th 09, 2020
Q2 One Word Medium
The value of $\displaystyle \underset{0}{\overset{x}{\int}} \dfrac{(t - |t|)^2}{(1 + t^2)} dt$ is equal to
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Medium
$\displaystyle \int{\dfrac{x^{3}-1}{x^{3}+x}dx}$ equal to
• A. $x-\log x+\log(x^{2}+1)-\tan^{-1}x+c$
• B. $x+\log x+\dfrac{1}{2}\log(x^{2}+1)+\tan^{-1}x+c$
• C. $x+\log x-\dfrac{1}{2}\log(x^{2}+1)-\tan^{-1}x+c$
• D. $x-\log x+\dfrac{1}{2}\log(x^{2}+1)-\tan^{-1}x+c$
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Hard
What is $\int _{ 0 }^{ 1 }{ x{ \left( 1-x \right) }^{ 9 }dx }$ equal to?
• A. $\dfrac{1}{132}$
• B. $\dfrac{1}{148}$
• C. $\dfrac{1}{240}$
• D. $\dfrac{1}{110}$
Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$ | 928 | 2,206 | {"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} | 4.15625 | 4 | CC-MAIN-2021-31 | latest | en | 0.516749 |
https://physics.stackexchange.com/questions/74577/how-do-i-calculate-integral-analytically-for-small-k | 1,563,548,510,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526254.26/warc/CC-MAIN-20190719140355-20190719162355-00091.warc.gz | 517,210,218 | 35,595 | # How do I calculate integral analytically for small $k$?
In a Heisenberg antiferromagnet, the dispersion relation is $$\omega_{\mathbf{k}} =JSz\sqrt{ 1-\gamma_{\mathbf{k}}^2}$$ where $\gamma_{\mathbf{k}}=\frac{1}{d}\sum_{i=1}^d \cos k_ia$, where $d$ is the dimensionality of the cubic lattice.
In staggered magnetisation calculation, we get an integral which is given by
$$\delta m =\int_{BZ} \frac{d^dk}{(2\pi)^d}\frac{1}{\sqrt{ 1-\gamma_{\mathbf{k}}^2}}.$$
This integral is dominated by small $k$, since we have Goldstone modes at $k=0,(\pi,\pi,\dots)$. Using this property, we can extend the limits of this integral from Brillouin zone to infinite $k$-space and we replace $\gamma_{\mathbf{k}}$ by its small $k$ limit:
\begin{eqnarray} \delta m=\frac{\sqrt{d}}{a}\int_0^\infty \frac{d^dk}{(2\pi)^d}\frac{1}{k} \end{eqnarray}
In 1D, its gives the logarithmic divergence which is expected. But, in 2D and 3D we get \begin{eqnarray} \delta m &\sim& [k]_0^\infty, \quad \text{in}\, 2D\\ &\sim&[k^2]_0^\infty, \quad \text{in}\, 3D \end{eqnarray} which still show divergence. This is not correct as if we solve the Brillouin zone integral we get some finite value. I do not understand what is wrong with the infinite k space integral? Could anybody please tell me how to solve such integral analytically for small $k$?
• Your logic is wrong. For instance, the integral $I=\int_0^{+\infty}dx e^{-x}=1$ is dominated by small $x$. So, with your logic, we can replace $e^{-x}$ by $1$, because $x$ is small, and we have the integral $\int_0^{+\infty}dx$... Unfortunately, the latter integral is infinite... – Trimok Aug 19 '13 at 18:50
I don't what you mean by "analytically" or for "small $k$", but if you are interested in the qualitative result of the integral then you almost did the correct thing. Just don't send the cutoff to infinity. (BTW you shouldn't do this in $d=1$ either, since the integral is logarithmically sensitive to the upper bound). The normal way to do such a calculation is:
$$\delta m =\int_{BZ} \frac{d^dk}{(2\pi)^d}\frac{1}{\sqrt{ 1-\gamma_{\mathbf{k}}^2}}\sim\int_0^{1/a}k^{d-1}dk\,\frac{1}{k}\sim a^{2-d}.$$ | 709 | 2,138 | {"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": 3, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-30 | longest | en | 0.839146 |
https://brighterly.com/math/axis-of-symmetry/ | 1,709,315,919,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00242.warc.gz | 141,900,409 | 15,624 | # Axis of Symmetry – Definition, Formula, Solved Examples
Welcome to another exciting journey with Brighterly, your trusted companion in unraveling the mesmerizing world of Mathematics. At Brighterly, we are not just about numbers, equations, and calculations, but also about the exploration of patterns, the joy of discovery, and the beauty of understanding. Today, we dive into a special mathematical concept, the Axis of Symmetry. Ever gazed at the perfect symmetry of a butterfly, the mirror-like reflection of a lake, or the bilateral symmetry of your own body? That’s the magic of symmetry, a cornerstone in the realm of math, science, art, and nature. The Axis of Symmetry, a key concept in Geometry and Algebra, offers a delightful window into this world of symmetry. Buckle up as we demystify the definition, formula, and application of this concept, with solved examples to light your way!
## What is Axis of Symmetry?
The captivating world of mathematics has many fascinating concepts, and one such concept is the Axis of Symmetry. Children who enjoy solving puzzles or drawing symmetric patterns might already be acquainted with this term, albeit unknowingly. Simply put, an axis of symmetry is an imaginary line where if you fold your shape along that line, both halves match perfectly, just like a mirror image! Isn’t it fascinating to find such symmetrical beauty hidden in numbers and shapes? Now, let’s delve a little deeper into this mathematical wonderland!
## Axis of Symmetry Definition
The Axis of Symmetry, a term that can often be heard in Geometry and Algebra, is a line that divides a shape or a graph into two mirror-image halves. Each point on one side of the axis has an identical ‘twin’ on the other side. For a simple visual understanding, think of the line that runs through the center of a perfectly round apple or the line that cuts a heart shape into two equal halves. This central line is the Axis of Symmetry.
## Axis of Symmetry of a Parabola
When we talk about the axis of symmetry, we often encounter the term ‘Parabola’. So, what is a Parabola? It is a curved symmetrical shape formed by all points equidistant from a certain point (the focus) and a line (the directrix). And guess what? The axis of symmetry of a parabola is the line that divides it into two equal parts! This symmetry is not only intriguing but also essential in solving problems in Mathematics and Physics.
## Axis of Symmetry Equation
The axis of symmetry equation is quite simple to grasp. For a quadratic function in the standard form y = ax² + bx + c, the equation of the axis of symmetry is given by x = -b/2a. It’s that easy! Keep in mind that a, b, and c are coefficients of the equation, and they give us a way to identify this magical line of symmetry.
## Axis of Symmetry Formula
### Standard form
In the standard form of a quadratic function (y = ax² + bx + c), the axis of symmetry formula becomes x = -b/2a. Here, ‘a’ and ‘b’ are the coefficients of the x² and x terms, respectively.
### Vertex form
In the vertex form of a quadratic function (y = a(x-h)² + k), the formula of the axis of symmetry is x = h. Here, ‘h’ is the x-coordinate of the vertex. The vertex form is often easier to work with as it readily provides the vertex and the axis of symmetry.
## Derivation of the Axis of Symmetry for Parabola
The derivation of the axis of symmetry for a parabola involves a bit of Algebra, but don’t worry, it’s manageable! If we set the derivative of the quadratic function equal to zero and solve for x, we arrive at our familiar x = -b/2a. This shows us the parabola’s peak (or valley), confirming it as the line of symmetry. Quite ingenious, isn’t it?
## Find Axis of Symmetry
So, how do we find the axis of symmetry for a quadratic function? Simple! Just plug the coefficients into the formula x = -b/2a (standard form) or identify ‘h’ in the vertex form. Whether it’s in your algebra homework or while you’re designing a symmetrical logo for your school project, this is your go-to method.
## Identification of the Axis of Symmetry
The identification of the axis of symmetry is often the first step in analyzing and sketching a quadratic function. By finding this axis, we can understand better how the function behaves, how it rises and falls, and where it reaches its maximum or minimum value. The process of identification involves solving the formula mentioned above and applying your findings to the quadratic function or shape at hand.
## Discover the wonders of Math!
Don’t you feel like a math detective already? Understanding the axis of symmetry allows us to dive deeper into the fascinating depths of Mathematics. It’s like discovering a secret language or a hidden key that unlocks the mysteries of numbers and shapes. So, keep exploring, keep learning, and remember – the world of Math is full of wonders waiting to be discovered!
## Axis of Symmetry Examples
Let’s take some practical examples to illustrate the axis of symmetry. Suppose we have the equation y = 2x² – 8x + 7. Here, a = 2 and b = -8. By using our formula, we find the axis of symmetry is x = -(-8)/2*2 = 2. Easy, right? Let’s do another: For y = -x² + 6x – 1, where a = -1 and b = 6, the axis of symmetry is x = -6/-2 = 3.
## Axis of Symmetry Questions
What about some questions to test your knowledge? Try finding the axis of symmetry for y = 3x² – 12x + 9. Or, how about y = -2x² + 4x – 1? Go ahead, apply the formula, and see the symmetry unravel!
## Conclusion
That wraps up our exciting exploration into the Axis of Symmetry with Brighterly! We hope you now have a more profound understanding of this mathematical marvel. Remember, at Brighterly, we believe that Math is not just a subject, but a language that helps us make sense of the world around us. The Axis of Symmetry is not only a mathematical tool but a concept that uncovers symmetry in nature, arts, architecture, and even our own bodies. We encourage you to continue exploring this fascinating world, armed with the understanding that Mathematics is much more than numbers – it’s about patterns, symmetry, logic, and creativity. So, keep practicing, keep wondering, and never stop discovering the magic of Mathematics with Brighterly!
## Frequently Asked Questions on Axis of Symmetry
### What is the Axis of Symmetry?
The Axis of Symmetry is a unique line that divides a shape or a graph into two identical halves. It is often encountered in the study of Geometry and Algebra. This line of symmetry means that each point on one side has a corresponding point on the other side, which is at the same distance from the axis but in the opposite direction.
### How is the Axis of Symmetry of a parabola derived?
To derive the Axis of Symmetry for a parabola, we start with the quadratic function y = ax² + bx + c. This is set to zero and solved for x, yielding the formula x = -b/2a. This value of x represents the x-coordinate of the vertex of the parabola, which is the Axis of Symmetry. It’s where the parabola reaches its peak (or valley) and shows the exact point where it can be ‘folded’ into two perfect halves.
### How do we find and identify the Axis of Symmetry?
Finding and identifying the Axis of Symmetry involves using the Axis of Symmetry formula. For a quadratic function in standard form, we use x = -b/2a, and for vertex form, x = h. ‘a’, ‘b’, and ‘h’ are coefficients from the quadratic function. By substituting these values into the formula, we can find the Axis of Symmetry, helping us understand the function’s behavior and its graph’s shape.
Remember, the path to mastering any concept lies in practice and application. Keep revisiting these concepts, play around with different quadratic functions, and very soon, identifying the axis of symmetry will become second nature to you! At Brighterly, we are always here to make your mathematical journey brighter and merrier. Happy learning!
Information Sources:
After-School Math Program
• Boost Math Skills After School!
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Boost Your Child's Math Abilities! Ideal for 1st-8th Graders, Perfectly Synced with School Curriculum! | 1,849 | 8,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | longest | en | 0.918553 |
https://www.techylib.com/en/view/hornbeastdanish/one_dimensional_kinematics_chapter_outline | 1,529,455,464,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00419.warc.gz | 919,647,918 | 22,547 | # One Dimensional Kinematics - Chapter Outline
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HONORS PHYSICS
FINAL
EXAM STUDY GUIDE
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IF A LINK IS BROKEN OR YOU WOULD LIKE EXTRA PRACTICE, IT CAN BE FOUND AT
www.physicsclassroom.com
in the
MULTIMEDIA PHYSICS STUDIOS,
MINDS ON PHYSICS (MOPS) INTERNET MODULES,
THE CALCULATOR PAD, and the
REVIEW SESSION.
One Dimensional Kinematics
-
Chapter Outline
Lesson 1 : Describing Motion with Words
a.
Introduction to the Language of Kinematics
(define mechanics, kinematics, READ ON TUTORIAL)
b.
Scalars and Vectors
(define each, ID quantities as scalar or vector, PRACTICE
ON TUTORIAL)
c.
Distance and Displacement
(differentiate between, PRACTICE ON TUTORIAL)
d.
Speed and Velocity
(differentiate between, instantaneous vs. average, calculate each, PRACTICE ON TUTORIAL)
e.
Acceleration
(define; calculate from data tables, graphs, word problems; understand direction of
acceleration
vector, PRACTICE ON TUTORIAL)
Lesson 2 : Describing Motion with Diagrams
a.
Introduction to Diagrams
(analyze ticker tape or “oil drop” and vector diagrams, READ ON TUTORI
AL)
b.
Ticker Tape Diagrams
(analyze “oil drop” diagrams and use to describe motion of objects, PRACTICE ON TUTORIAL)
c.
V
ector Diagrams
(analyze vector diagrams and use to describe motion of objects, READ ON TUTORIAL)
Lesson 3 : Describing Motion with Position vs. Time Graphs
a.
The Meaning of Shape for a
p
-
t Graph
(relate shape of slope to description of motion, link broken)
b.
The Meaning of Slope for a p
-
t Graph
(
calculate
slope
and use to describe motion of objects, link broken
)
c.
Determining the Slope on a p
-
t Graph
(calculate slope to calculate velocity, PRACTICE ON TUTORIAL)
Lesson 4 : Describing Motion with Velocity vs. Time Graphs
a.
The Meaning of Shape for a v
-
t Graph
(constant vs. changing velocity, direction of velocity, velocity vs. acceleration,
PRACTICE ON TUTORIAL)
b.
The Meaning of Slope for a v
-
t Graph
(slope of v
-
t graph is acceleration, link broken)
c.
Relating the Shape to the Motio
n
(use shape of line to describe motion, link broken)
d.
Determining the Slope on a v
-
t Graph
(calculate slope, PRACTICE ON TUTORIAL)
e.
Determining the Area on a v
-
t Graph
(Determine area of a rectangle, square, triangle, relate area to height, link
broken)
Lesson 5 : Free Fall and the Acceleration of Gravity
a.
Introduction to Free Fall
(define, know rate of free
-
fall on Earth, READ ON TUTORIAL)
b.
The Acceleration of Gravity
(g, know acceleration due to gravity on Earth, link broken)
c.
Representing Free Fall by Graphs
(recognize free fall on P
-
T and V
-
T graphs, READ ON TUTORIAL)
d.
How Fast? and How Far?
(calculate how far an object will fall in a certain time, calculate final velocity of free falling
object in certain time,
v
f
= g * t,
e.
The Big Misconception
(know t
he answer to “doesn't a more massive object accelerate at a greater rate than a
less massive object?” READ ON TUTORIAL)
Lesson 6 : Describing Motion with Equations
a.
The Kinematic
Equations
(Kinematics equations will be on reference, no need to memorize, READ ON TUTORIAL)
b.
Kinematic Equations and Problem
-
Solving
(utilize the FIVE problem
-
solving strategies, READ ON
TUTORIAL)
c.
Kinematic Equations and Free Fall
(know conceptual characteristics of free fall motion, READ ON TUTORIAL)
d.
Sample Problems and Solutions
(Solve problems using kinematics equations, PRACTICE ON TUTORIAL)
e.
Kinematic Equations and Graphs
(Sketch graphs on motion, analyze graphs and solve kinemat
ics problems,
PRACTICE ON WEBSITE)
THERE IS A HOMEWORK
ASSIGNMENT HIDDEN IN THIS
PACKET! DUE ON EXAM DAY!
All of these questions can be CHECKED on the
Physics Classroom
tutorial.
One Dimensional Kinematics
LESSON 1B:
Scalars and Vectors
Categorize
each quantity as being either a vector or a scalar. Click the button to see the answer.
website.
Quantity
Category
a. 5 m
b. 30 m/sec, East
c. 5 mi., North
d. 20 degrees Celsius
e. 256 bytes
f. 4000 Calories
LESSON
1c:
Distance and Displacement
Determine the DISTANCE traveled and the DISPLACEMENT for each example. Check answers on website.
Distance = _______ Displacement = _______
Distance = _______ Displacement = _______
Distance = _______
Displacement = _______
What is the displacement of the cross
-
country team if they begin at the school, run 10 miles and finish back at the school?
What is the distance and the displacement of the race car drivers in the Indy 500?
LESSON 1d:
Speed and
Velocity
Q: While on vacation, Lisa Carr traveled a total distance of 440 miles. Her trip took 8 hours. What was her average speed?
Q:
Use the diagram to determine the average speed and the average
velocity of the skier during thes
e three minutes.
SEE
IF YOUR ANSWERS ARE CORRECT ON THE
PHYSICS CLASSROOM TUTORIAL!!!
ALL ANSWERS CAN BE
CHECKED AT THE PHYSICS
CLASSROOM TUTOIAL!!!!
What is the coach's average speed and average velocity?
LESSON 1e:
Acceleration
Use the equation for acceleration to determine the acceleration for the following two motions.
LESS
ON 2b:
TICKER TAPE DIAGRAMS
Analyze the following ticker tape diagrams.
1.
2.
3.
LESSON 3c:
Determining slope on a P
-
T graph
Determine the velocity (i.e., slope) of the object as portrayed by the graph below.
CORRECT ON THE
PHYSICS CLASSROOM TUTORIAL!!!
LESSON 4a
: The Meaning of Shape for a V
-
T Graph
Consider the graph at the right. The object whose motion is represented by this graph is ... (include all that are true):
a.
moving in the positive direction.
b.
mov
ing with a constant velocity.
c.
moving with a negative velocity.
d.
slowing down.
e.
changing directions.
f.
speeding up.
g.
moving with a positive acceleration.
h.
moving with a constant acceleration.
LESSON 4d
: Describing Motion with V
-
T Graphs
Consider the veloc
ity
-
time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph. Click the
LESSON 6D: Kinematics Sample Problems
An airplane accelerates down a runway at 3.20 m/s
2
for 32.8 s until is f
inally lifts off the ground. Determine the distance traveled
before takeoff.
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the accelerat
ion of
the car.
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity a
nd how far
will he fall?
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration o
f the car and the distance
traveled.
A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s
2
. Determine
the time for the feather to fall to the surface of the moon.
Rocket
-
powered sleds are us
ed to test the human response to acceleration. If a rocket
-
powered sled is accelerated to a speed of
444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?
MORE ON TUTORIAL WEBSITE!
LESSON 6e
: Kinematics
Equations and Graphs
Rennata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of
-
1.0 m/s
2
. Eventually Rennata
comes to a complete stop. Represent Rennata's accelerated motion by sketching a velocity
-
time graph. Use the
velocity
-
time graph
to determine this distance.
2. Use kinematic equations to calculate the distance that Rennata travels while decelerating.
MORE PROBLEMS ON WEBSITE!
Newton's Laws
-
Chapter Outline
Lesson 1: Newton's First Law of Motion
a.
Newton's First Law
(Define, apply to real
-
world situations, READ ON TUTORIAL)
b.
Inertia and Mass
(know relationship between,
gravitational and frictional influences, PRACTICE ON TUTORIAL
c.
State of Motion
(define inertia, PRACTICE ON TUTORIAL)
d.
Balanced and Unbalanced Forces
(Forces cause accelerations, not motions; ID balanced & unbalanced forces, define
equilibrium, PRACTICE ON TUTORIAL)
Lesson 2: Force and Its Represen
tation
a.
The Meaning of Force
(define force, identify common forces F
frict
, F
norm
, F
grav
, etc, READ ON TUTORIAL)
b.
Types of Forces
(identify common forces F
frict
, F
norm
, F
grav
, etc, link broken)
c.
Drawing Free
-
Body Diagrams
(understand free
-
body diagrams show forces, not motions; draw free
-
body dia
grams for
real
-
world situations, PRACTICE ON TUTORIAL)
d.
Determining the Net Force
(define net force, analyze diagrams to determine of net force exists, analyze free
-
body
diagra
ms to determine if net force exists, PRACTICE ON TUTORIAL)
Lesson 3 : Newton's Second Law of Motion
a.
Newton's Second Law
b.
The Big Misconception
(know the answer to, “does sustaining a motion require a continued force?”, READ ON
TUTORIAL, Take “quiz” to see if you are infected with the misconception)
c.
Finding Acceleration
(The three major equations that will be useful are the equation for net force (
Fnet = m•a
), the
equation for
gravitational force
(F
grav
= m•g), and the equation for
frictional force
(F
frict
= μ•F
norm
)
,
PRACTICE ON TUTORIAL).
d.
Finding Individual Forces
(find sum of individual forces acting on a object, PRACTICE ON TUTORIAL)
e.
Free Fall and Air Resistance
(determine net force and net acceleration, define terminal velocity. PRACTICE ON
TUTORIAL)
f.
Double Trouble (a.k.a., Two Body Problems)
(will not be covered)
Lesson 4 : Newton's Third Law of Motion
a.
Newton's Third Law
(define, compare forces apply to real
-
world pr
oblems, PRACTICE ON TUTORIAL)
b.
Identifying Action and Reaction Force Pairs
(ID action and reaction pairs. PRACTICE ON TUTORIAL)
Newton's Laws
LESSON 1b
: Inertia and Mass
Imagine
a place in the
cosmos
far from all gravitational and frictional influences. Suppose that you visit that place (just suppose)
and throw a rock. The rock will
b. continue in motion in the same direction at constant speed.
A 2
-
kg
object is moving horizontally with a speed of 4 m/s. How much net force is required to keep the object moving at this speed
and in this direction?
Mac and Tosh are arguing in the cafeteria. Mac says that if he flings the Jell
-
O with a greater speed it w
ill
have a greater inertia. Tosh argues that inertia does not depend upon speed, but rather upon mass. Who do
you agree with? Explain why.
Supposing you were in space in a
weightless environment
, would it require a force to set an object in
motion?
Fred
spends most Sunday afternoons at rest on the sofa, watching pro football games and consuming large
quantities of food. What effect (if any) does this practice have upon his inertia? Explain.
MORE ON TUTORIAL!
LESSON 1c:
State of MOTION
A group of physics teachers is taking some time off for a little putt
-
putt golf. The 15th hole at the Hole
-
In
-
One Putt
-
Putt Golf Course has a large metal rim that putters must use to guide their ball towards the
hole. Mr. S guides a golf ball around the m
etal rim When the ball leaves the rim, which path (1, 2, or 3)
will the golf ball follow?
A 4.0
-
kg object is moving across a friction
-
free surface with a constant velocity of 2 m/s. Which one of
the following horizontal forces is necessary to maintain th
is state of motion?
a. 0 N
b. 0.5 N
c. 2.0 N
d. 8.0 N
a.
depends on the speed.
LESSON 1d
: Balanced and Unbalanced Forces
Luke Autbeloe
drops an approximately 5.0 kg fat cat (weight = 50.0 N) off the roof of his house into the swimming pool below.
Upon encountering the pool, the cat encounters a 50.0 N upward resistance force (assumed to be constant). Use this descriptio
n to
llowing questions. Click the button to view the correct answers.
1. Which one of the velocity
-
time graphs best describes the motion of the cat? Support your answer with sound reasoning.
2. Which one of the following dot diagrams best describes the
motion of the falling cat from the time that they are dropped to the
time that they hit the bottom of the pool? The arrows on the diagram represent the point at which the cat hits the water. Sup
port
3. Several of L
uke's friends were watching the motion of the falling cat. Being "physics types", they began discussing the motion
and made the following comments. Indicate whether each of the comments is correct or incorrect? Support your answers.
a. Once the
cat hits the water, the forces are balanced and the cat will stop.
b. Upon hitting the water, the cat will accelerate upwards because the water applies an upward force.
c. Upon hitting the water, the cat will bounce upwards due to
the upward force.
4. If the forces acting upon an object are balanced, then the object
a. must not be moving.
b. must be moving with a constant velocity.
c. must not be accelerating.
d. none of these
LESSON 2c
:
Free Body Diagrams
A book is at rest on a tabletop. Diagram the forces acting on the book.
A girl is suspended motionless from the ceiling by two ropes. Diagram the forces acting on the combination of girl and
bar.
An egg is free
-
falling from a nest in a tree. Neglect air resistance. Diagram the forces acting on the egg as it is falling.
A flying squirrel is gliding (no
wing
flaps
) from a tree to the ground at constant velocity. Consider air resistance.
Diagram the
forces acting on the squirrel.
A rightward force is applied to a book in order to move it across a desk with a rightward acceleration. Consider frictional
forces. Neglect air resistance. Diagram the forces acting on the book.
MORE ON TUTORIAL
LESSON
3c
: Finding Acceleration
An applied force of 50 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N
of
friction. Use the diagram to determine the normal force, the net force, the mass, and the
acceleration of the object. (Neglect air
resistance.)
An applied force of 20 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N
of
friction. Use the diagram to determine the normal force, the net
force, the coefficient of friction (μ) between the object and the
surface, the mass, and the acceleration of the object. (Neglect air resistance.)
A 5
-
kg object is sliding to the right and encountering a friction force that slows it down. The coefficient
of friction (μ) between the
object and the surface is 0.1. Determine the force of gravity, the normal force, the force of friction, the net force, and th
e
acceleration. (Neglect air resistance.)
1. Edwardo applies a 4.25
-
N rightward force to a 0.765
-
kg
book to accelerate it across a tabletop. The coefficient of friction
between the book and the tabletop is 0.410. Determine the acceleration of the book.
2. In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N rightward for
ce on a 0.500
-
kg cart to
accelerate it across a low
-
friction track. If the total resistance force to the motion of the cart is 0.72 N, then what is the cart's
acceleration?
LESSON 3d
: Finding Individual Forces
Practice #1
Free
-
body diagrams for four situa
tions are shown below. The net force is known for each situation. However, the magnitudes of a
few of the individual forces are not known. Analyze each situation individually and determine the magnitude of the unknown fo
rces.
A rightward force is
applied to a 6
-
kg object to move it across a rough surface at constant velocity. The object encounters 15 N of
frictional force. Use the diagram to determine the gravitational force, normal force, net force, and applied force. (Neglect
air
resistance.)
A rightward force is applied to a 10
-
kg object to move it across a rough surface at constant velocity. The coefficient of friction
between the object and the surface is 0.2. Use the diagram to determine the gravitational force, normal force, applied force,
frictional force, and net force. (Neglect air resistance.)
MORE ON TUTORIAL!
LESSON 3e
: Free fall and Air Resistance
In the diagrams below, free
-
body diagrams showing the forces acting upon an 85
-
kg skydiver (equipment included) are shown. For
each ca
se, use the diagrams to determine the net force and acceleration of the skydiver at each instant in time. Then use the
button to view the answers.
Lesson 4a
: Newton’s Third Law
While driving down the road, a firefly strikes the windshield of a bus
and makes a quite obvious
mess in front of the face of the driver. This is a clear case of Newton's third law of motion. The
firefly hit the bus and the bus hits the firefly. Which of the two forces is greater: the force on the
firefly or the force on the
bus?
For years, space travel was believed to be impossible because there was nothing that rockets
could push off of in space in order to provide the propulsion necessary to accelerate. This inability
of a rocket to provide propulsion is because ...
a. ... space is void of air so the rockets have nothing to push off of.
b. ... gravity is absent in space.
c. ... space is void of air and so there is no air resistance in space.
d. ... nonsense! Rockets do accelerate
in space and have been able to do so for a long time.
Many people are familiar with the fact that a rifle recoils when fired. This recoil is the result of action
-
reaction force pairs. A gunpowder explosion creates hot gases that expand outward allowing
the rifle to
push forward on the bullet. Consistent with Newton's third law of motion, the bullet pushes backwards
upon the rifle. The acceleration of the recoiling rifle is ...
a. greater than the acceleration of the bullet.
b. smalle
r than the acceleration of the bullet.
c. the same size as the acceleration of the bullet.
In the top picture (below), Kent Budgett is pulling upon a rope that is attached to a wall. In the bottom picture, the Kent i
s pulling
upon a rope that i
s attached to an elephant. In each case, the force scale reads 500 Newton. Kent is pulling ...
a. with more force when the rope is attached to the wall.
b. with more force when the rope is attached to the elephant.
c. the same force in each case.
LESSON 4b
: Identifying Action and Reaction Force Pairs
Consider the interaction depicted below between foot A, ball B, and foot C. The three objects interact simultaneously (at the
same
time). Identify the
two pairs
of action
-
reaction forces. Use the notat
ion "foot A", "foot C", and "ball B" in your statements. Click the
button to view the answer.
2. Identify at least six pairs of action
-
reaction force pairs in the following diagram.
1
-
2
-
3
-
4
-
5
-
6
-
CORRECT ON THE
PHYSICS CLASSROOM TUTORIAL!!!
Vectors: Motion and Forces in Two Dimensions
-
Chapter Outline
Lesson 1: Vectors
-
Fundamentals and Operations
a.
Vectors and Direction
(Draw
scaled vectors, be able to use the counterclockwise convention)
b.
(add vectors using pythagorean theorem and calculus)
c.
Resultants
(know how to determine resultant magnitude and direction)
d.
Vector Components
(know how to work backwared...identify the vectors that make up a resultant
)
e.
Vector Resolution
(link broken; know hoe to break a vector into components; hanging pictures,
dog chains, airplane flights)
f.
Component Method of Vector Addition
(link broken; be able to use pythagorean theorem, add three
or more vectors tail
-
to
-
tip, etc.)
g.
Relative Velocity and Riverboat
Problems
(Solve riverboat problems)
h.
Independence of Perpendicular Components of Motion
(vector resoluntion involving riverboat problems)
Lesson 2: Projectile Motion
a.
What is a Projectile?
(Analyze presence of forces, accelerations, and velocity in both components of
a projectile)
b.
Characteristics of a Projectile's Trajectory
(analyze projectiles for initial; and final velocity, horizontal
and vertical acceleration, and net force)
c.
Describing Projectiles with Numbers
1.
Horizontal and Vertical Components of Velocity
(use kinematics equations to solve for velocity)
2.
Horizontal and Vertical Components of Displacement
(use kinematics equation to
solve for
displacement)
d.
Initial Velocity Components
(link broken; be able to resolve the trajectory of a projectile using
Pythagorean theorem and calculus)
e.
Horizontally Launched Projectiles
-
Problem
-
Solving
(be able to use kinematics equations to solve
problem about projectiles launched horizontally)
f.
Non
-
Horizontally Launched Projectiles
-
Problem
-
Solving
(link broken, there are additional problems in
“The Calculator Pad” section of the website; be able to use kinematics equations
to
solve problem about projectiles launched vertically)
Lesson 3 : Forces in Two Dimensions
a.
(Review of previous concepts)
b.
Resolution of Forces
(Review of breaking forces into components)
c.
Equilibrium and Statics
(be able to determine wei
ght of hanging picture,
then determine the tension
in the diagonal cable that supports its weight.
d.
Net Force Problems Revisited
( Be able to
calculate F
net
= m • a problems involving f
orces at angles
)
e.
Inclined Planes
( Be able to u
se the principles of vector resolution to determine the net force and
acceleration of
objects on an incline)
f.
Double Trouble (a.k.a. Two Body Problems)
(Do the following HW assignments and then go to sleep)
HW:
Due on exam day. Previously distributed in class. All were present
I put them on the website in case you lost them!
1)
Circular Motion and Satellite Motion, Lesson 3.
2)
Momentum an It’s Conservation, Lesson 1
Vectors: Motion and Forces in Two Dimensions
Lesson 1a
:
Vectors and Direction
1)
Given the
SCALE: 1 cm = 50 km/hr
, determine the magnitude and direction of this vector.
2) Given the
SCALE: 1 cm = 15 m/s
, represent the vector 120 m/s, 240
-
d
egrees by a scaled vector diagram.
Lesson 1
b
:
Vector
Does the order in which vectors are added tail to tip have any effect on the resultant vector’s magnitude or
direction?
Lesson 1g: Relative Velocity and Riverboat Problems
Example 2
A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.
a.
What is the resultant velocity of the motorboat?
b.
If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
c.
What distance downstream does the boat reach the opposite shore?
Example 2
A motorboat traveling 4 m/s, East encounters a current traveling 7
.0 m/s, North.
a.
What is the resultant velocity of the motorboat?
b.
If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
c.
What distance downstream does the boat reach the opposite shore?
Check Your
Understanding
1. A plane can travel with a speed of 80 mi/hr with respect to the air. Determine the resultant velocity of the plane (magnit
ude
only) if it encounters a
a. 10 mi/hr headwind.
b. 10 mi/hr tailwind.
c. 10 mi/hr crosswind.
d. 60 mi/hr
crosswind.
2. A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, North.
a. What is the resultant velocity of the motor boat?
b. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to
shore?
c. What distance downstream does the boat reach the opposite shore?
3. A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, South.
a. What is the resultant velocity of the motor boat?
b. If the width of the river is 80 meters
wide, then how much time does it take the boat to travel shore to shore?
c. What distance downstream does the boat reach the opposite shore?
4. A motorboat traveling 6 m/s, East encounters a current traveling 3.8 m/s, South.
a. What is the resultant veloc
ity of the motor boat?
b. If the width of the river is 120 meters wide, then how much time does it take the boat to travel shore to shore?
c. What distance downstream does the boat reach the opposite shore?
5. If the current velocity in question #4 were i
ncreased to 5 m/s, then
a. how much time would be required to cross the same 120
-
m wide river?
b. what distance downstream would the boat travel during this time?
Lesson 1h
:
Indepen
dence of Perpendicular Components of Motion
1. A plane flies northwest out of O'Hare Airport in Chicago at a speed of 400 km/hr in a direction of 150
degrees (i.e., 30 degrees north of west). The Canadian border is located a
distance of 1500 km due north
of Chicago. The plane will cross into Canada after approximately ____ hours.
a. 0.13
b. 0.23
c. 0.27
d. 3.75
e. 4.33
f. 6.49
g. 7.50
h. None of these are even close.
2. Suppose that the plane in question 1 was flying
with a velocity of 358 km/hr in a direction of 146 degrees (i.e., 34 degrees north
of west). If the Canadian border is still located a distance of 1500 km north of Chicago, then how much time would it take to
cross
the border?
3.
TRUE
or
FALSE
:
A boat
heads straight across a river. The river flows north at a speed of 3 m/s. If the river current were greater, then the time
required for the boat to reach the opposite shore would not change. _______________________
4. A boat begins at point A and heads s
traight across a 60
-
meter wide river with a speed of 4 m/s
(relative to the water). The river water flows north at a speed of 3 m/s (relative to the shore). The
boat reaches the opposite shore at point C. Which of the following would cause the boat to reac
h
the opposite shore at a location SOUTH of C?
a. The boat heads across the river at 5 m/s.
b. The boat heads across the river at 3 m/s.
c. The river flows north at 4 m/s.
d. The river flows north at 2 m/s.
e. Nonsense! None of these affect the location
where the boat lands.
Lesson 2
b
:
Characteristics of a Projectile’s Trajectory
Use your understanding of projectiles to answer the following questions. When finished, click the button to view your answers
.
1. Consider these
diagrams in answering the following questions.
Which diagram (if any) might represent ...
a. ... the initial horizontal velocity?
b. ... the initial vertical velocity?
c. ... the horizontal acceleration?
d. ... the vertical acceleration?
e. ... t
he net force?
2. Supposing a snowmobile is equipped with a flare launcher that is capable of launching a sphere vertically (relative to the
snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity aft
er the launch,
then where will the flare land (neglect air resistance)?
a. in front of the snowmobile
b. behind the snowmobile
c. in the snowmobile
3. Suppose a rescue airplane drops a relief package while it is moving with a
constant horizontal s
peed at an elevated height. Assuming that air resistance
is negligible, where will the relief package land relative to the plane?
a. below the plane and behind it.
b. directly below the plane
c. below the plane and ahead of it
Lesson 2c
: Describing
Projectiles with Numbers
Use your understanding of projectiles to answer the following questions. Then click the button to view the answers.
1. Anna Litical drops a ball from rest from the top of 78.4
-
meter high cliff. How much ti
me will it take for the ball to reach the
ground and at what height will the ball be after each second of motion?
2. A cannonball is launched horizontally from the top of an 78.4
-
meter high cliff. How much time will it take for the ball to reach the
gro
und and at what height will the ball be after each second of travel?
3. Fill in the table below indicating the value of the horizontal and vertical components of velocity and acceleration for a
projectile.
4. The diagram below shows the trajectory fo
r a projectile launched non
-
horizontally from an elevated position on top of a cliff. The
initial horizontal and vertical components of the velocity are 8 m/s and 19.6 m/s respectively. Positions of the object at 1
-
second
intervals are shown. Determine the
horizontal and vertical velocities at each instant shown in the diagram.
Lesso
n 2f:
Horizontally Launched Projectiles
-
Problem
-
Solving
A soccer ball
is kicked horizontally off a 22.0
-
meter high hill and lands a distance of 35.0 meters from the edge of the hill.
Determine the initial horizontal velocity of the soccer ball.
Lesso
n 2g
:
Non
-
Horizontally Launched Projectiles
-
Problem
-
Solving
Your Turn to Try It!
Use the
Range of an Angle
-
Launched Projectile
widget
to practice a projectile problem (or two) (or three). Using the given
launch velocity and launch angle, determine the expected horizontal displacement (d
x
). After completing your calculation, use the
Submit
button to check your answer. Find it on the webs
ite!
Lesson 3c: Equilibrium and Statics
Recognize these problems from a previous test.
1. The following picture is hanging on a wall. Use trigonometric functions to determine the weight of the picture.
2. The sign below hangs outside the physics classroom, advertising the most important truth to be found inside. The sign is
supported by a diagonal cable a
nd a rigid horizontal bar. If the sign has a mass of 50 kg, then determine the tension in the
diagonal cable that supports its weight.
3. The following sig
n can be found in Glenview. The sign has a mass of 50 kg. Determine the tension in the cables.
4. After its most recent delivery, the infamous stork
announces the good news. If the sign has a mass of 10 kg, then what is the
tensional force in each cable? Use trigonometric functions and a sketch to assist in the solution.
5. Suppose that a student pulls with two large forces (F
1
and F
2
) in order to lift a 1
-
kg book by two cables. If the cables make a 1
-
degree angle with the horizontal, then what is the tension in the cable?
Lesson 3d: Net Force Problems Revisited
The following problems provide plenty of practice with F
net
= m • a problems involving forces at angles. Try each problem
and then
click the button to view the answers.
1. A 50
-
N applied force (30 degrees to the horizontal) accelerates a box across a horizontal sheet of ice (see diagram). Glen Brook,
Olive N. Glenveau, and Warren Peace are discussing the problem. Glen sugg
ests that the normal force is 50 N; Olive suggests that
the normal force in the diagram is 75 N; and Warren suggests that the normal force is 100 N. While all three answers may seem
reasonable, only one is correct. Indicate which two answers are wrong and
explain why they are wrong.
2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the
diagram below to determine the values for the blanks.
3. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram
and
to
determine the net force and acceleration of the object. (F
net
= m•a; F
frict
= μ•F
norm
; F
grav
= m•g)
4. The 5
-
kg mass below is moving with a constant spe
ed of 4 m/s to the right. Use your understanding of force relationships and
vector components to fill in the blanks in the following diagram
and
to determine the net force and acceleration of the object. (F
net
= m•a; F
frict
= μ•F
norm
; F
grav
= m•g)
5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vec
tor
components to fill in the blanks in
the following diagram
and
to determine the net force and acceleration of the object. (F
net
= m•a;
F
frict
= μ•F
norm
; F
grav
= m•g)
6. Use your understanding
of force relationships and vector components to fill in the blanks in the following diagram
and
to
determine the net force and acceleration of the object. (F
net
= m•a; F
frict
= μ•F
norm
; F
grav
= m•g)
7. Study the diagram below and determine the acceleration of the box and its velocity after being pulled by the applied force
for
2.0 seconds.
8. A student pulls a 2
-
kg backpack across the ice (assume friction
-
free) by pulling at a 30
-
degree angle to the horizontal. The
velocity
-
time graph for
the motion is shown. Perform a careful analysis of the situation and determine the applied force.
9. The following object is moving to the right and encou
ntering the following forces. Use your understanding of force relationships
and vector components to fill in the blanks in the following diagram
and
to determine the net force and acceleration of the object.
(F
net
= m•a; F
frict
= μ•F
norm
; F
grav
= m•g)
10. The 10
-
kg object is being pulled to the left at a constant speed of 2.5 m/s. Use your understanding of force relationships and
vector components to fill i
n the blanks in the following diagram. (F
net
= m•a; F
frict
= μ•F
norm
; F
grav
= m•g)
11. Use your understanding of force relationships and vector components
to fill in the blanks in the following diagram
and
to
determine the net force and acceleration of the object. (F
net
= m•a; F
frict
= μ•F
norm
; F
grav
= m•g)
Lesson 3e: Inclines Planes
More Practice
Use the
widget
provided in the tutorial
to investigate other inclined plane situations. Simply enter the mass, the incline angle and
the coefficient of friction (use 0 for frictionless situations). Then click the
Sub
mit
button to view the acceleration. Find it on the
website!
Circular Motion and Satellite Motion
-
Chapter Outline
Lesson 1: Motion Characteristics for Circular Motion
a.
Speed and Velocity
b.
Acceleration
c.
The Centripetal Force Requirement
d.
The Forbidden F
-
Word
e.
Mathematics of Circular Motion
Lesson 2: Applications
of Circular Motion
a.
Newton's Second law
-
Revisited
b.
Amusement Park Physics
c.
Athletics
Lesson 3: Universal Gravitation
a.
Gravity is More than a Name
b.
The
Apple, the Moon, and the Inverse Square Law
c.
Newton's Law of Universal Gravitation
d.
Cavendish and the Value of
G
e.
The Value of g
Momentum and Its Conservation
-
Chapter Outline
Lesson 1: The Impulse
-
Momentum Change Theorem
a.
Momentum
b.
Momentum and Impulse Connection
c.
Real
-
World Applications
Lesson 2: The Law of Momentum Conservation
a.
The Law of Action
-
Reaction (Revisited)
b.
Momentum Conservation Principle
c.
Isolated Systems
d.
Momentum Conservation in Collisions
1.
Using Equations as a "Recipe" for Algebraic Problem
-
Solving
2.
Using Equations as a Guide to Thinking
e.
Momentum Conservation in Explosions
1.
2.
Using Equations as a Guide to Thinking
b.
Momentum Conservation in Explosions
HW:
Due on exam day. Previously
distributed in class.
All were
present to receive them!
I put
them on the website in case you
lost them!
1)
Circular Motion and
Satellite Motion, Lesson 3.
2)
Momentum an It’s
Conservation, Lesson 1
Completing these worksheets will
be your review for the concepts
listed on this
page. | 9,355 | 35,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-26 | latest | en | 0.660926 |
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#1
August 13th 09, 04:31 AM posted to microsoft.public.excel.worksheet.functions
external usenet poster First recorded activity by ExcelBanter: Aug 2009 Posts: 1
Mathematical Integration
Is there an Excel object used to perform mathematical integration of one and
two variables?
#2
August 13th 09, 06:24 AM posted to microsoft.public.excel.worksheet.functions
external usenet poster First recorded activity by ExcelBanter: Mar 2009 Posts: 8,521
Mathematical Integration
http://oregonstate.edu/~haggertr/487/integrate.htm
If this post helps click Yes
---------------
Jacob Skaria
"Octav" wrote:
Is there an Excel object used to perform mathematical integration of one and
two variables?
#3
August 13th 09, 07:09 AM posted to microsoft.public.excel.worksheet.functions
external usenet poster First recorded activity by ExcelBanter: Jun 2007 Posts: 806
Mathematical Integration
I suggest to use Wolfram Alpha and not Excel...
At least try to create an automated web lookup to Wolfram Alpha to get
a closed form of the integration function (if possible) and then
continue in Excel.
Regards,
Bernd
#4
August 13th 09, 01:47 PM posted to microsoft.public.excel.worksheet.functions
external usenet poster First recorded activity by ExcelBanter: Feb 2009 Posts: 1,104
Mathematical Integration
See
http://people.stfx.ca/bliengme/Excel...UnderCurve.htm
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
http://people.stfx.ca/bliengme
remove caps from email
"Octav" wrote in message
...
Is there an Excel object used to perform mathematical integration of one
and
two variables?
#5
August 13th 09, 04:11 PM posted to microsoft.public.excel.worksheet.functions
external usenet poster First recorded activity by ExcelBanter: Feb 2007 Posts: 31
Mathematical Integration
"Octav" wrote in message ...
Is there an Excel object used to perform mathematical integration of one and
two variables?
Hi O.
Excel can't solve indefinite integrals,
but definite integrals of one variable are not difficult.
See a method in ...
from a few days ago in this group.
I have tried to implement extension to more integration variables,
with no succes. That would be rather slow anyway.
Hans T
#6
August 13th 09, 04:24 PM posted to microsoft.public.excel.worksheet.functions
external usenet poster First recorded activity by ExcelBanter: Feb 2007 Posts: 31
Mathematical Integration
"Hans Terkelsen" <dk wrote in message ...
....
See a method in ...
from a few days ago in this group.
....
Subject: area under the graph, Integral in Excel
from 11. aug
Hans T.
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Similar Threads Thread Thread Starter Forum Replies Last Post Nazri Excel Discussion (Misc queries) 1 January 9th 08 08:42 AM Differentiation & Integration Excel Discussion (Misc queries) 2 April 24th 07 01:08 AM integreat Excel Discussion (Misc queries) 4 May 12th 06 02:40 AM cointegration chart Charts and Charting in Excel 1 March 15th 06 10:19 AM Bdavis Excel Discussion (Misc queries) 0 April 14th 05 05:02 PM
All times are GMT +1. The time now is 04:40 AM. | 810 | 3,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-26 | longest | en | 0.804878 |
http://thydzik.com/tag/calculate/ | 1,511,020,329,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804976.22/warc/CC-MAIN-20171118151819-20171118171819-00014.warc.gz | 312,190,337 | 13,021 | How to calculate comparison interest rate in Excel
Searching for how comparison interest rates are calculated, I found an article on Tomorrow Finance, but unfortunately a lot of the numbers provided no insight into how they were calculated. I hope to break it down further and explain how to calculate comparison interest rates with Excel. The Excel formula can be quite tricky and the parameters vague, so I will go through each of the formula as well.
• rate – the interest rate of the loan annually equal to 4.63% (the current rate most banks are offering as of today)
• principle – the loan amount always equal to \$150,000 when calculating a comparison rate.
• period – the period in years always equal to 25 years when calculating a comparison rate.
Payments for a loan
Firstly, calculate the monthly payments for a loan based on constant principle payments and a constant interest rate. The key here is to convert the interest rate to a monthly interest rate and the number of periods also in months.
`payment = PMT(rate/12, period*12, principle)`
In this case payments equals -\$844.86 per month and includes principle and interest. The figure is negative since it is an outgoing.
Total interest payment
`total payment = payment * 12 * period`
Since the monthly payments are all the same, the total payments is simply the monthly payment multiplied by 12 months by 25 years, this equals -\$253,456.66.
`total interest = total payment + principle`
The total interest is imply the total payment (which is negative) plus the principle, this results in a total interest only of -\$103,456.66.
A quicker way to calculate the cumulative interest over the lifetime of a loan is;
`total interest = CUMIPMT(rate/12, period*12, principle, 1, period*12, 0)`
Next the additional fees over the lifetime of the loan are added, usually these are annual service fees, with nab there is a -\$395 annual service fees. The total fee over the life of the loan would be 25 years multiplied by -\$395 which equals -\$9,875.
`total outgoings = total interest + total fees`
The total outgoings over the life of the loan would be the total interest and total fees, this would equal -\$113,331.66.
`outgoings = total outgoings / 25 /12`
To get the outgoings per month divide by 300 to get -\$877.77 per month.
Comparison rate
Finally, the comparison rate can be calculated;
comparison rate = RATE(period * 12, outgoings, principle)*12
The rate is per month which is multiplied by 12 to get an annual comparison rate, in this case equal to 0.050101482 or 5.01%.
Calculate Stamp Duty with Excel
A quick post of how to calculate the Stamp Duty of a property with Excel, in my example I am using the Western Australia Residential Rate Dutiable value.
```Stamp Duty
Cutoff [\$] Rate [%] Duty [\$] Formula
0 1.9 0 0
120000 2.85 2280 =C3+(A4-A3)*B3/100
150000 3.8 3135 =C4+(A5-A4)*B4/100
360000 4.75 11115 =C5+(A6-A5)*B5/100
725000 5.15 28452.5 =C6+(A7-A6)*B6/100
Property Value 500000
Duty 17765 =VLOOKUP(C9,A:C,3,TRUE) + (C9-VLOOKUP(C9,A:C,1,TRUE))*VLOOKUP(C9,A:C,2,TRUE)/100
```
Update, next day
This also works for Individual income tax rates;
```Tax Rates
Cutoff [\$] Rate [%] Tax [\$] Formula
0 0 0 0
6000 15 0 =C3+(A4-A3)*B3/100
37000 30 4650 =C4+(A5-A4)*B4/100
80000 37 17550 =C5+(A6-A5)*B5/100
180000 45 54550 =C6+(A7-A6)*B6/100
Income 100000
Tax 24950 =VLOOKUP(C9,A:C,3,TRUE) + (C9-VLOOKUP(C9,A:C,1,TRUE))*VLOOKUP(C9,A:C,2,TRUE)/100
``` | 1,033 | 3,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-47 | latest | en | 0.941317 |
https://www.shaalaa.com/question-bank-solutions/temperature-heat-two-copper-spheres-radii-6-cm-12-cm-respectively-are-suspended-evacuated-enclosure-each-them-are-temperature-15-c-above-surroundings-ratio-their-rate-loss-heat_463 | 1,596,477,783,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735823.29/warc/CC-MAIN-20200803170210-20200803200210-00561.warc.gz | 862,511,768 | 8,910 | Share
Two Copper Spheres of Radii 6 Cm and 12 Cm Respectively Are Suspended in an Evacuated Enclosure. Each of Them Are at a Temperature 15°C Above the Surroundings. the Ratio of Their Rate of Loss of Heat is - Physics
Question
Two copper spheres of radii 6 cm and 12 cm respectively are suspended in an evacuated enclosure. Each of them are at a temperature 15°C above the surroundings. The ratio of their rate of loss of heat is.................
1. 2:1
2. 1:4
3. 1:8
4. 8:1
Solution
(b) 1:4
r1 = 6 cm
r2 = 12 cm
T1 = T2 = 15°C
Ratio of loss of heat is
R_1/R_2=A_1/A_2xxT_1^4/T_2^4
R_1/R_2=(4pir_1^2)/(4pir_2^2)=6^2/12^2=1/4
Is there an error in this question or solution? | 244 | 683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-34 | latest | en | 0.858548 |
https://www.activekyds.com/how-much-is-a-shooter-of-alcohol/ | 1,656,111,619,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00408.warc.gz | 694,964,272 | 35,871 | # How Much Is A Shooter Of Alcohol?
Rate this post
Is a shot 1 or 1.5 oz?
There is no standard size for a single shot, except in Utah, where a shot is defined as 1.5 US fl oz (44.4 ml). Elsewhere in the U.S., the standard size is generally considered to be 1.25–1.5 US fl oz (37–44 ml). A double shot in the U.S. may be 2 fluid ounces or more.
How many counts is 2 oz?
How Many Counts Is a 2 Oz Pour? A 2-ounce pour is 4 counts using a pour spout. So you'll count “one one-thousand, two one-thousand, three one-thousand, four one-thousand” and stop.
How much is a shot price?
How Much Does a Liquor Shot Cost?
Size of Shot750ml Bottle CostCost per Shot
2 oz.\$30\$2.50
1 1/2 oz.\$30\$1.88
## How much is a single shot of whiskey?
Like other liquors, a standard whiskey pour is 1.5 ounces for shot, 2 ounces for a neat or rocks pour, and 3 ounces for a double. Pouring whiskey is right up there with pouring beer in importance, as every bartender needs to master these.
## What are well shots?
A well drink or rail drink is an alcoholic beverage served using the lower-cost liquors stored within easy reach of the bartender in the counter "speed rail", "speed rack", or "well".
## How do I measure 1.5 oz of liquor?
Measuring Spoons
A tablespoon holds about 1/2 oz of liquid. So, use three of those and you have yourself a 1.5 oz shot. Or, if you're really out of luck and only have teaspoons, you can do some math and use that too (the answer is 9, for the multiplication impaired out there. 9 teaspoons = 1.5 oz).
## What is a finger of whiskey?
In the measurement of distilled spirits, a finger of whiskey refers to the amount of whiskey that would fill a glass to the level of one finger wrapped around the glass at the bottom. In English these units have mostly fallen out of use, apart from the common use in distilled drinks and drinking games.
## How much is a shot of gin?
The base liquor (e.g., gin, rum, vodka, etc.) is often a standard shot of 1 1/2 ounces. Liqueurs are typically poured between 1/2 ounce and 3/4 ounce. Accent juices, such as lemon and lime, usually use 1/4 ounce to 1/2 ounce. Filling a highball or tall drink with juice or soda often requires four to six ounces.
## Which alcohol is expensive?
The 100-year-old bottle of Henri IV Dudognon Heritage Cognac is referred to as the rarest and world's most expensive Cognac. The bottle is dipped in 24-carat gold, sterling platinum and decorated with 6,500 cut diamonds. The luxurious spirit is packaged by jeweler Jose Davalos.
## How much is a shot of Smirnoff?
21 Vodka should generally cost between \$11.99 – \$14.99. Smirnoff also offers 80 proof (40% ABV) which ranges from \$11.67 – \$15.99, and 100 proof (50% ABV) which ranges from \$16.98 – \$19.99.
Latest Smirnoff Prices.
TypeBottle SizePrice
Smirnoff No. 57 100 Proof Vodka1.75From \$31.99
Smirnoff Flavors750mlFrom \$14
## How much alcohol can be served in one drink?
In the United States, one "standard" drink (or one alcoholic drink equivalent) contains roughly 14 grams of pure alcohol, which is found in: 12 ounces of regular beer, which is usually about 5% alcohol. 5 ounces of wine, which is typically about 12% alcohol. 1.5 ounces of distilled spirits, which is about 40% alcohol.
## What is the easiest shot to drink?
• Bazooka Joe. This creamy blue tropical drink is a sure-fire crowd pleaser.
• Cherry Cheesecake.
• Chuck Norris.
• Jägerbomb.
• Kamikaze.
• Liquid Cocaine.
• Melon Ball.
• ## What is a good alcohol shot?
7 Shots to Take if You're Sick of Fireball
• Montenegro. Need something to settle your stomach after dinner?
• Cynar. Bartenders are all about fernet shots—but the crazy-bitter amaro can be an acquired taste.
• Green Chartreuse.
• Applejack.
• Good cinnamon whiskey.
• Dark rum.
• Ice-cold ZU Bison Grass Vodka.
• ## What are the best shot drinks?
The 25 Greatest Shots to Do at a Bar
• Body shots.
• Fireball.
• Boilermaker.
• Chartreuse.
• A flaming shot.
• Fernet.
• Any shot a foreigner hands you.
• Whiskey. Doesn't matter if it's good or bad, if it's bourbon, rye, scotch, or from Tennessee, Ireland, or even Canada.
• ## Is a bottle cap a shot?
Mix the Sour Puss and rootbeer schnapps together in a shot glass. Drop the shot glass into the highball glass, and serve.
## How do you measure one shot?
They unanimously confirm that three tablespoons equals one shot. So there you go: Measure them out, one-by-one, and add them to your shot—oh, wait, that's why we're here in the first place. You don't have a shot glass. Just add them to whatever glass you're building your drink in.
124-Year-Old Shot Of Cognac Sets World Record For Most Expensive Cognac Shot Ever At \$14,000+ A 124-year-old shot of cognac just set a new Guinness World Record for being the most expensive shot of cognac ever sold when a person forked over \$14,158 (£10,014) at the Hyde Kensington bar in London last Wednesday.
Like other liquors, a standard whiskey pour is 1.5 ounces for shot, 2 ounces for a neat or rocks pour, and 3 ounces for a double. Pouring whiskey is right up there with pouring beer in importance, as every bartender needs to master these. | 1,372 | 5,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-27 | latest | en | 0.923271 |
http://noj.io/problem/p/1092/the-dog-task | 1,561,421,997,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999779.19/warc/CC-MAIN-20190624231501-20190625013501-00167.warc.gz | 122,044,920 | 5,406 | 1000ms 65536K
3 0
Beginner
Description
Hunter Bob often walks with his dog Ralph. Bob walks with a constant speed and his route is a polygonal line (possibly self- intersecting) whose vertices are specified by N pairs of integers (Xi, Yi) - their Cartesian coordinates.
Ralph walks on his own way but always meets his master at the specified N points. The dog starts his journey simultaneously with Bob at the point (X1, Y1) and finishes it also simultaneously with Bob at the point (XN , YN).
Ralph can travel at a speed that is up to two times greater than his master's speed. While Bob travels in a straight line from one point to another the cheerful dog seeks trees, bushes, hummocks and all other kinds of interesting places of the local landscape which are specified by M pairs of integers (X'j,Y'j). However, after leaving his master at the point (Xi , Yi) (where 1 <= i <= N) the dog visits at most one interesting place before meeting his master again at the point ( Xi+1 , Yi+1).
Your task is to find the dog's route, which meets the above requirements and allows him to visit the maximal possible number of interesting places. The answer should be presented as a polygonal line that represents Ralph's route. The vertices of this route should be all points (Xi , Yi) and the maximal number of interesting places ( X'j, Y'jX). The latter should be visited (i.e. listed in the route description) at most once.
An example of Bob's route (solid line), a set of interesting places (dots) and one of the best Ralph's routes (dotted line) are presented in the following picture:
Input
The first line of the input file contains two integers N and M, separated by a space ( 2 <= N <= 100, 0 <= M <= 100). The second line contains N pairs of integers X1,Y1,...,XN,YN , separated by spaces, that represent Bob's route. The third line contains M pairs of integers X'1,Y'1,...,X'M,Y'M, separated by spaces, that represent interesting places.
All points in the input file are different and their coordinates are integers not greater than 1000 by the absolute value.
Output
The first line of the output file should contain the single integer K - the number of vertices of the best dog's route. The second line should contain K pairs of coordinates X''1,Y''1,...,X''K,Y''K, separated by spaces, that represent this route. If there are several such routes, then you may write any of them.
Sample Input
4 5
1 4 5 7 5 2 -2 4
-4 -2 3 9 1 2 -1 3 8 -3
Sample Output
6
1 4 3 9 5 7 5 2 1 2 -2 4
Source
Editor
keefo on 2016-12-15 11:58:34 | 661 | 2,549 | {"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} | 2.9375 | 3 | CC-MAIN-2019-26 | latest | en | 0.946146 |
https://brickandmortarphilly.com/the-slope-of-a-budget-constraint-line-is-influenced-by/ | 1,657,068,623,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00119.warc.gz | 195,885,095 | 5,311 | Calculate and graph budget constraintsExplain chance sets and opportunity costsEvaluate the law of diminishing marginal utilityExplain just how marginal analysis and utility affect choices
Consider the usual consumer’s spending plan problem. Consumers have a minimal amount of income to invest on the things they need and want. Suppose Alphonso has ?10 in security money each week that he can allocate in between bus ticket for obtaining to work and also the burgers the he eats because that lunch. Burgers cost ?2 each, and bus tickets are 50 cents each. We deserve to see Alphonso’s budget plan problem in (Figure).
You are watching: The slope of a budget constraint line is influenced by
Each point on the budget constraint represents a combination of burgers and also bus tickets whose complete cost adds as much as Alphonso’s budget of ?10. The family member price the burgers and also bus ticket determines the steep of the budget plan constraint. All along the budget plan set, offering up one burger means gaining 4 bus tickets.
Explain why people make selections that are straight on the spending plan constraint, quite than within the budget plan constraint or external it.
Suppose Alphonso’s town raises the price of bus ticket from ?0.50 come ?1 and the price that burgers rises native ?2 to ?4. Why is the opportunity price of bus tickets unchanged? suppose Alphonso’s weekly safety money increases from ?10 come ?20. Just how is his spending plan constraint impacted from all 3 changes? Explain.
### Problems
Use this info to prize the complying with 4 questions: Marie has actually a weekly budget of ?24, i beg your pardon she likes to spend on magazines and pies.
If the price the a magazine is ?4 each, what is the maximum number of magazines she might buy in a week?
Draw Marie’s budget constraint v pies on the horizontal axis and also magazines ~ above the vertical axis. What is the slope of the budget plan constraint?
### References
Bureau of job Statistics, U.S. Department of Labor. 2015. “Median Weekly revenue by educational Attainment in 2014.” Accessed in march 27, 2015. Http://www.bls.gov/opub/ted/2015/median-weekly-earnings-by-education-gender-race-and-ethnicity-in-2014.htm.
See more: The K I See Pretty Girls Everywhere I Go, The Kids Of Widney High
Robbins, Lionel. An Essay top top the Nature and also Significance of financial Science. London: Macmillan. 1932.
United says Department the Transportation. “Total passenger on U.S Airlines and also Foreign airlines U.S. Flights raised 1.3% in 2012 native 2011.” Accessed October 2013. Http://www.rita.dot.gov/bts/press_releases/bts016_13
### Glossary
budget constraintall possible consumption combine of goods that someone can afford, provided the price of goods, as soon as all income is spent; the boundary of the opportunity setlaw of diminishing marginal utilityas we consume much more of a good or service, the utility we obtain from additional units the the great or service tends to become smaller than what we received from earlier unitsmarginal analysisexamination of decision on the margin, an interpretation a little more or a little less indigenous the condition quoopportunity costmeasures cost by what we provide up/forfeit in exchange; opportunity expense measures the worth of the forgone alternativeopportunity setall possible combinations of usage that someone have the right to afford provided the prices of goods and the individual’s incomesunk costscosts that us make in the previous that us cannot recoverutilitysatisfaction, usefulness, or worth one obtains from spend goods and services | 775 | 3,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-27 | latest | en | 0.938545 |
https://discuss.itacumens.com/index.php/topic,102977.0.html?PHPSESSID=9808069dd499ad2b4e7c97f0fe596777 | 1,720,798,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00639.warc.gz | 163,450,628 | 13,269 | ## News:
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## Mistral Placement Question Paper - Technical
Started by devi.naga, Dec 27, 2012, 04:32 PM
#### devi.naga
Mistral Placement Question Paper - Technical
C Section
1. What does the following program print?
#include
int sum,count;
void main(void)
{< BR> for(count=5;sum+="count;)
printf(%d,sum);
}
a. The pgm goes to an infinite loop b. Prints 4791010974c. Prints 4791001974
d. Prints 5802112085 e. Not sure
2. What is the output of the following program?
#include
void main(void)
{
int i;< BR> for(i=2;i< =7;i++)
printf(%5d,fno());
}
fno()
{
staticintf1=1,f2=1,f3;
return(f3=f1+f2,f1=f2,f2=f3);
}
a. produce syntax errors b. 2 3 5 8 13 21 will be displayed c. 2 2 2 2 2 2 will be displayed
d. none of the above e. Not sure
3. What is the output of the following program?
#include
void main (void)
{
int x = 0Ã-1234;
int y = 0Ã-5678;
x = x & 0Ã-5678;
y = y | 0Ã-1234;
x = x^y;
printf(%xt,x);
x = x | 0Ã-5678;
y = y & 0Ã-1234;
y = y^x;
printf(%xt,y);
}
a. bbb3 bbb7 b. bbb7 bbb3 c. 444c 4448
d. 4448 444c e. Not sure
4. What does the following program print?
#include
void main (void)
{
int x;
x = 0;
if (x=0)
printf (Value of x is 0?);
else
printf (Value of x is not 0?);
}
a. print value of x is 0 b. print value of x is not 0 c. does not print anything on the screen
d. there is a syntax error in the if statement e. Not sure
5. What is the output of the following program?
#include
#include
int foo(char *);
void main (void)
{
char arr[100] = {Welcome to Mistral};
foo (arr);
}
foo (char *x)
{
printf (%dt,strlen (x));
printf (%dt,sizeof(x));
return0;
}
a. 100 100 b. 18 100 c. 18 18 d. 18 2 e. Not sure
6. What is the output of the following program?
#include
display()
{
printf (n Hello World);
return 0;
}
void main (void)
{
int (* func_ptr) ();
func_ptr = display;
printf (n %u,func_ptr);
(* func_ptr) ();
}
a. it prints the address of the function display and prints Hello World on the screen
b. it prints Hello World two times on the screen
c. it prints only the address of the fuction display on the screen
d. there is an error in the program e. Not sure
7. What is the output of the following program?
#include
void main (void)
{
int i = 0;
char ch = 'A';
do
putchar (ch);
while(i++ < 5 || ++ch <= 'F');
}
a. ABCDEF will be displayed b. AAAAAABCDEF will displayed
c. character 'A' will be displayed infinitely d. none e. Not sure
8. What is the output of the following program?
#include
#define sum (a,b,c) a+b+c
#define avg (a,b,c) sum(a,b,c)/3
#define geq (a,b,c) avg(a,b,c) >= 60
#define lee (a,b,c) avg(a,b,c) < = 60
#define des (a,b,c,d) (d==1?geq(a,b,c):lee(a,b,c))
void main (void)
{
int num = 70;
char ch = '0?;
float f = 2.0;
if des(num,ch,f,0) puts (lee..);
else puts(geq);
}
a. syntax error b. geq will be displayed c. lee.. will be displayed
d. none e. Not sure
9. Which of the following statement is correct?
a. sizeof('*') is equal to sizeof(int) b. sizeof('*') is equal to sizeof(char)
c. sizeof('*') is equal to sizeof(double) d. none e. Not sure
10. What does the following program print?
#include
char *rev(int val);
void main(void)
{
extern char dec[];
printf (%c, *rev);
}
char *rev (int val)
{
char dec[]=abcde;
return dec;
}
a. prints abcde b. prints the address of the array dec
c. prints garbage, address of the local variable should not returned d. print a e. Not sure
11. What does the following program print?
void main(void)
{
int i;
static int k;
if(k=='0?)
printf(one);
else if(k== 48)
printf(two);
else
printf(three);
}
a. prints one b. prints two c. prints three
d. prints one three e. Not sure
12. What does the following program print?
#include
void main(void)
{
enum sub
{
chemistry, maths, physics
};
struct result
{
char name[30];
enum sub sc;
};
struct result my_res;
strcpy (my_res.name,Patrick);
my_res.sc=physics;
printf(name: %sn,my_res.name);
printf(pass in subject: %dn,my_res.sc);
}
a. name: Patrick b. name: Patrick c. name: Patrick
pass in subject: 2 pass in subject:3 pass in subject:0
d. gives compilation errors e. Not sure
13. What does
printf(%s,_FILE_); and printf(%d,_LINE_); do?
a. the first printf prints the name of the file and the second printf prints the line no: of the second printf in the file
b. _FILE_ and _LINE_ are not valid parameters to printf function
c. linker errors will be generated d. compiler errors will be generated e. Not sure
14. What is the output of the following program?
#include
void swap (int x, int y, int t)
{
t = x;
x = y;
y = t;
printf (x inside swap: %dt y inside swap : %dn,x,y);
}
void main(void)
{
int x;
int y;
int t;
x = 99;
y = 100;
swap (x,y,t);
printf (x inside main:%dt y inside main: %d,x,y);
}
a. x inside swap : 100 y inside swap : 99 x inside main : 100 y inside main : 99
b. x inside swap : 100 y inside swap : 99 x inside main : 99 y inside main : 100
c. x inside swap : 99 y inside swap : 100 x inside main : 99 y inside main : 100
d. x inside swap : 99 y inside swap : 100 x inside main : 100 y inside main : 99
e. Not sure
15. Consider the following statements:
i) while loop is top tested loop ii) for loop is bottom tested loop
iii) do " while loop is top tested loop iv) while loop and "do " while loop are top tested loops.
Which among the above statements are false?
a. i only b. i & ii c. iii & i d. ii, iii & iv e. Not sure
16. Consider the following piece of code:
char *p = "MISTRAL;
printf (%ct, *(++p));
p -=1;
printf (%ct, *(p++));
Now, what does the two printf's display?
a. M M b. M I c. I M d. M S e. Not sure
17. What does the following program print?
#include
struct my_struct
{
int p:1;
int q:1;
int r:6;
int s:2;
};
struct my_struct bigstruct;
struct my_struct1
{
char m:1;
};
struct my_struct1 small struct;
void main (void)
{
printf (%d %dn,sizeof (bigstruct),sizeof (smallstruct));
}
a. 10 1 b. 2 2 c. 2 1 d. 1 1 e. Not sure
18. Consider the following piece of code:
FILE *fp;
fp = fopen(myfile.dat,r);
Now fp points to
a. the first character in the file.
b. a structure which contains a char pointer which points to the first character in the file.
c. the name of the file. d. none of the above. e. Not sure.
19. What does the following program print?
#include
#define SQR (x) (x*x)
void main(void)
{
int a,b=3;
a = SQR (b+2);
}
a. 25 b. 11 c. 17 d. 21 e. Not sure.
20. What does the declaration do?
int (*mist) (void *, void *);
a. declares mist as a function that takes two void * arguments and returns a pointer to an int.
b. declares mist as a pointer to a function that has two void * arguments and returns an int.
c. declares mist as a function that takes two void * arguments and returns an int.
d. there is a syntax error in the declaration. e. Not sure.
21. What does the following program print?
#include
void main (void)
{
int mat [5][5],i,j;
int *p;
p = & mat
• ;
for (i=0;i<5;i++)
for (j=0;j<5;j++)
mat[j] = i+j;
printf (%dt, sizeof(mat)); < BR> i=4;j=5;
printf( "%d, *(p+i+j));
}
a. 25 9 b. 25 5 c. 50 9 d. 50 5 e. Not sure
22. What is the output of the following program?
#include
void main (void)
{
short x = 0Ã-3333;
short y = 0Ã-4321;
long z = x;
z = z < < 16;
z = z | y;
printf(%1xt,z);
z = y;
z = z >> 16;
z = z | x;
printf(%1xt,z);
z = x;
y = x && y;
z = y;
printf(%1xt,z);
}
a. 43213333 3333 1 b. 33334321 4321 4321 c. 33334321 3333 1
d. 43213333 4321 4321 e. Not sure
23. What is the output of the following program?
#include
void main (void)
{
char *p = "Bangalore;
#if 0
printf (%s, p);
#endif
}
a. syntax error #if cannot be used inside main function b. prints Bangalore on the screen
c. does not print anything on the screen
d. program gives an error "undefined symbol if e. Not sure
24. If x is declared as an integer, y is declared as float, consider the following expression:
y = *(float *)&x;
Which one of the following statments is true?
a. the program containing the expression produces compilation errors;
b. the program containing the expression produces runtime errors;
c. the program containing the expression compiles and runs without any errors;
d. none of the above e. Not sure
25. What is the return type of calloc function?
a. int * b. void * c. no return type: return type is void
d. int e. Not sure[/b]
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Unless you're sure you want to reply, please consider starting a new topic.
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Verification: | 2,805 | 8,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.5023 |
https://www.thinkwellhomeschool.com/products/foundations-of-math | 1,542,797,881,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00118.warc.gz | 1,008,187,597 | 54,502 | # Foundations of Mathematics
\$125.00
### Thinkwell's Foundations of Mathematics with Professor Edward Burger
Thinkwell's Foundations of Mathematics offers a focused approach to our Pre-algebra curriculum. With its emphasis on such basics as integers, fractions, decimals, and percents, it's the perfect resource for students wanting a fundamentals-based approach to mathematics.
The workbook (optional) comes with lecture notes, sample problems, and exercises so that you can study even when away from the computer.
#### Risk-Free Three Day
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Click “Add To Cart” above and enter it as part of the checkout process.
Course Overview
#### what you get
• 12-month, online subscription to our complete Foundations of Math course
• 36-week, day-by-day course lesson plan
• 55+ course lessons, each with a streaming video
• Illustrated notes
• Automatically graded drill-and-practice exercises with step-by-step answer feedback
• Downloadable, printable course worksheets & answer keys
• Automatically graded section quizzes
• Chapter & Practice tests, a Midterm & Final Exam
• Animated interactivities....and more!
#### how it works
• Purchase Thinkwell's Foundations of Math through our online store
• Create an account username and password which will give you access to the online 6th-grade math course section
• Activate your 12-month subscription when you're ready
• Login to the course website to access the online course materials, including streaming video lessons, exercises, quizzes, tests and more
• Access your course anytime, anywhere, from any device
• Your work is automatically tracked and updated in real-time
• Transcripts, grade reports, and certificates of completion are available at request
Thinkwell's Foundations of Math Table of Contents
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### 1. Integers
1.1 Introduction to Integers
1.1.1 Integers and the Number Line
1.2 Operations with Integers
1.2.1 Addition of Integers
1.2.2 Subtraction of Integers
1.2.3 Multiplication and Division of Integers
1.2.4 Introduction to Exponents
1.2.5 Using the Order of Operations
### 2. Fractions
2.1 Introduction to Fractions
2.1.1 Factors, Multiples, and Equivalent Fractions
2.2 Operations with Fractions
2.2.1 Multiplying Fractions and Mixed Numbers
2.2.2 Dividing Fractions and Mixed Numbers
2.2.3 Adding and Subtracting Fractions and Mixed Numbers
### 3. Decimals
3.1 Introduction to Decimals
3.1.1 Representing, Comparing, and Ordering Decimals
3.1.2 Rounding and Estimating Decimals
3.2 Operations with Decimals
3.2.1 Adding and Subtracting Decimals
3.2.2 Multiplying Decimals
3.2.3 Dividing Decimals
3.3 Fractions and Decimals
3.3.1 Converting between Fractions and Decimals
3.3.2 The Order of Operations with Fractions and Decimals
### 4. Ratios, Proportions, and Percents
4.1 Ratios and Proportions
4.1.1 Ratios and Rates
4.1.2 Proportions
4.1.3 Finding an Unknown Value in a Proportion
4.1.4 Applications of Proportions
4.2 Percents
4.2.1 Introduction to Percents
4.2.2 Percents
4.2.3 Relating Percents, Decimals, and Fractions
4.2.4 Estimating with Percents
4.2.5 Finding and Using Percents
### 5. Variables, Expressions, and Equations
5.1 Introduction to Equations
5.1.1 Variables and Expressions
5.1.2 Translating Between Words and Math
5.1.3 Simplifying Algebraic Expressions
5.1.4 Translating Between Tables and Expressions
5.1.5 Equations and Their Solutions
5.2 Introduction to Solving Equations
5.2.1 Solving Addition Equations
5.2.2 Solving Subtraction Equations
5.2.3 Solving Multiplication Equations
5.2.4 Solving Division Equations
5.3 Solving Equations
5.3.1 Solving Equations Containing Integers
5.3.2 Writing and Solving Two-Step Equations
5.3.3 Writing and Solving Multi-Step Equations
5.3.4 Solving Equations with Variables on Both Sides
5.3.5 Solving Literal Equations
### 6. Measurement
6.1 Measurement
6.1.1 Understanding Customary Units of Measurement
6.1.2 Understanding Metric Units of Measurement
6.1.3 Converting Customary Units
6.1.4 Converting Metric Units
6.1.5 Converting between Customary and Metric Units
6.1.6 Time and Temperature
### 7. Geometry
7.1 Polygons
7.1.1 Triangles
7.1.2 Quadrilaterals
7.1.3 Finding Angles in Polygons
7.2 Area, Perimeter, and Circumference
7.2.1 Perimeter
7.2.2 Area of Rectangles and Parallelograms
7.2.3 Circles and Circumference
7.2.4 Area of Circles
7.3 Solids
7.3.1 Introduction to Three-Dimensional Figures
7.3.2 Volume of Prisms and Cylinders
### 8. Statistics and Graphs
8.1 Organizing, Displaying, and Interpreting Data
8.1.1 Organizing Data
8.1.2 Bar Graphs and Histograms
8.1.3 Line Graphs
8.1.4 Reading and Interpreting Circle Graphs
8.1.5 Measures of Central Tendency
Frequently Asked Questions for Thinkwell's Foundations of Math
### How do Thinkwell courses work?
Your student watches a 5-10 minute online video lesson and then completes the automatically graded exercises for the topic. You'll get instant correct-answer feedback. Then move on to the next lesson! The courses are self-paced, or you can use the daily lesson plans. Just like a textbook, you can choose where to start and end, or follow the standard table of contents.
### When does my 12-month online subscription start?
It starts when you're ready. You can have instant access to your online subscription when you purchase online, or you can purchase now and start later.
### What math courses should a student take?
A typical sequence of secondary math courses completed by a college-bound student is: Grade 6 Math > Grade 7 Math > Grade 8 Math > Algebra 1 > Geometry > Algebra 2 > Precalculus. For students looking to include Calculus as part of their high school curriculum and are able to complete Grade 7 Math in 6th grade, the sequence can be: Grade 7 Math > Prealgebra > Algebra 1 > Geometry > Algebra 2 > Precalculus > Calculus.
### Is Thinkwell’s Foundations of Math a college course?
Foundations of Math is a developmental math course primarily taught in 2-year colleges in the state of Georgia. Four year colleges will not award credit for a developmental math course; typically College Algebra is the lowest credit math course at a 4 -year college.
### Does my student get school credit for Thinkwell’s Foundations of Math?
No, only schools are accredited and Thinkwell is not a school, though many accredited schools use Thinkwell.
### Does Thinkwell Foundations of Math meet state standards?
State standards don’t apply since it’s a 2-year college course.
### What if my student needs access to the course for more than 12 months?
You can purchase extra time in one-month, three-month, and six-month increments.
### Can I share access with more than one student?
The courses are designed and licensed to accommodate one student per username and password; additional students need to purchase online access. This allows parents to keep track of each student's progress and grades.
### How long does it take to complete Thinkwell’s Foundations of Math course?
The pace of your course is up to you, but most college students will schedule one semester.
### Can I see my grade?
Thinkwell’s course software tracks everything your student does. When logged in, your student can click "My Grades" to see their progress.
### How are grades calculated?
The course grade is calculated this way: Chapter Tests 33.3%, Midterm: 33.3%, Final: 33.3%.
### What is acceptable performance on the exams?
As a homeschool parent, you decide the level of performance you want your student to achieve; the course does not limit access to topics based on performance on prior topics.
### Can I get a transcript?
You can contact techsupport@thinkwell.com to request a file with your student's grades.
### What if I change my mind and want to do a different math course, can I change?
If you discover that you should be in a different course, contact techsupport@thinkwell.com within one week of purchase and we will move you to the appropriate course.
### Can I print the exercises?
Yes, but completing the exercises online provides immediate correct answer feedback and automatic scoring, so we recommend answering the exercises online.
### Are exercises multiple choice?
Most of the exercises are multiple choice and they are graded automatically with correct answer solutions.
### Is Thinkwell's Foundations of Math Math built on "continuous review?"
Thinkwell's Foundations of Math is carefully constructed to build on previous knowledge, reinforcing key concepts every step of the way. It is not only a reflection of empirically effective instruction, but also reflects Dr Burger’s philosophies learned over a career of teaching mathematics, emphasizing a solid foundation that addresses why students need to learn certain concepts. Finally, Thinkwell Foundations of Math is intentionally designed to work for a wide variety of learning styles.
### Is there a guarantee?
Yes, we offer a full refund within three business days of purchase, no questions asked.
### How does my school review this course?
Should your school need to review a Thinkwell course for any reason, have the school contact techsupport@thinkwell.com and we can provide them access to a demo site. | 2,217 | 9,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.845543 |
https://www.migha.ru/resistance-of-a-wire-coursework-diagram-19465.html | 1,620,886,431,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00106.warc.gz | 916,864,224 | 5,798 | # Resistance Of A Wire Coursework Diagram Mla For Movie Titles In An Essay
An electrical device having a resistance of 5 ohms would be represented as ), and represents the resistivity of the material (in ohm•meter).
Consistent with the discussion above, this equation shows that the resistance of a wire is directly proportional to the length of the wire and inversely proportional to the cross-sectional area of the wire.
First, the total length of the wires will affect the amount of resistance.A third variable that is known to affect the resistance to charge flow is the material that a wire is made of.Not all materials are created equal in terms of their conductive ability.In the same manner, the wider the wire, the less resistance that there will be to the flow of electric charge.When all other variables are the same, charge will flow at higher rates through wider wires with greater cross-sectional areas than through thinner wires.Those materials with lower resistivities offer less resistance to the flow of charge; they are better conductors.The materials shown in the last four rows of the above table have such high resistivity that they would not even be considered to be conductors.The resistance to the flow of charge in an electric circuit is analogous to the frictional effects between water and the pipe surfaces as well as the resistance offered by obstacles that are present in its path.It is this resistance that hinders the water flow and reduces both its flow rate and its speed.Use the Resistivity of a Material widget to look up the resistivity of a given material.Type the name of the material and click the Submit button to find its resistivity.
## One thought on “Resistance Of A Wire Coursework Diagram”
1. Compelling arguments, eloquence, and confidence are the fundamental tools for providing a brilliant persuasive essay.
2. You will realize needs that may have been overlooked, spot problems and nip them before they escalate, and establish plans to meet your business goals. Ninety percent of new businesses fail in the first two years. Finally, your business plan provides the information needed to communicate with others.
3. This guarantee proves that your money will not be wasted. | 431 | 2,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-21 | latest | en | 0.944775 |
https://fullstackdeveloper.guru/tag/bst/ | 1,680,016,335,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00082.warc.gz | 318,527,239 | 22,107 | # Tag: BST
• ## Kth Smallest Element in a Binary Search Tree
Given a binary search tree, find the kth smallest element in it. For example, For the below tree: If k = 3, Then the kth (3rd) smallest element is 3. If k = 2, Then the output is 2. Try out the solution here: https://leetcode.com/problems/kth-smallest-element-in-a-bst/ Solution: Hint : Use inorder traversal. Explanation: Approach 1 –… | 111 | 399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-14 | latest | en | 0.656755 |
https://fdocument.org/document/non-parametric-bayesian-methods-cambridge-machine-learning-mlgengcamacuktutorials07zgpdf.html | 1,624,246,409,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488262046.80/warc/CC-MAIN-20210621025359-20210621055359-00330.warc.gz | 239,554,087 | 21,128 | of 62
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### Transcript of Non-parametric Bayesian Methods - Cambridge Machine Learning...
• Non-parametric Bayesian Methods
Advanced Machine Learning Tutorial (Based on UAI 2005 Conference Tutorial)
Zoubin Ghahramani
Department of Engineering University of Cambridge, UK
zoubin@eng.cam.ac.uk http://learning.eng.cam.ac.uk/zoubin
• Bayes Rule Applied to Machine Learning
P (θ|D) = P (D|θ)P (θ) P (D)
θ model parameters D observed data P (D|θ) likelihood of θ P (θ) prior probability of θ P (θ|D) posterior of θ given D
Model Comparison:
P (m|D) = P (D|m)P (m) P (D)
P (D|m) = ∫
P (D|θ, m)P (θ|m) dθ
Prediction:
P (x|D,m) = ∫
P (x|θ,D,m)P (θ|D,m)dθ
P (x|D,m) = ∫
P (x|θ, m)P (θ|D,m)dθ (if x is iid given θ)
• Model Comparison: two examples
e.g. selecting m, the number of Gaussians in
a mixture model
−2 0 2 4 6 8 10 12 −20
−10
0
10
20
30
40
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x
y
e.g. selecting m the order of a polynomial in
a nonlinear regression model
P (m|D) = P (D|m)P (m) P (D)
, P (D|m) = ∫
P (D|θ, m)P (θ|m) dθ
A possible procedure:
1. place a prior on m, P (m) 2. given data, use Bayes rule to infer P (m|D)
What is the problem with this procedure?
• Real data are complicated
Example 1: You are trying to model people’s patterns of movie preferences. You believe there are “clusters” of people, so you use a mixture model...
• How should you pick P (m), your prior over how many clusters there are? teenagers, people who like action movies, people who like romantic comedies, people who like horror movies,
people who like movies with Marlon Brando, people who like action movies but not science
fiction, etc etc...
• Even if there are a few well defined clusters, they are unlikely to be Gaussian in the variables you measure. To model complicated distributions you might need many Gaussians for each cluster.
• Conclusion: any small finite number seems unreasonable
• Real data are complicated
Example 2: You are trying to model crop yield as a function of rainfall, amount of sunshine, amount of fertilizer, etc. You believe this relationship is nonlinear, so you decide to model it with a polynomial.
• How should you pick P (m), your prior over the order of the polynomial?
• Do you believe the relationship could be linear? quadratic? cubic? What about the interactions between input variabes?
• Conclusion: any order polynomial seems unreasonable.
How do we adequately capture our beliefs?
• Non-parametric Bayesian Models
• Inflexible models (e.g. mixture of 5 Gaussians, 4th order polynomial) yield unreasonable inferences.
• Non-parametric models are a way of getting very flexible models.
• Many can be derived by starting with a finite parametric model and taking the limit as number of parameters →∞
• Non-parametric models can automatically infer an adequate model size/complexity from the data, without needing to explicitly do Bayesian model comparison.1
1Even if you believe there are infinitely many possible clusters, you can still infer how many clusters are represented in a finite set of n data points.
• What is a non-parametric model?
• Parametric models assume some finite set of parameters θ (e.g. think of linear regression, or a mixture of two Gaussians).
• Given this finite set of parameters, future predictions are independent of the observed data:
P (x|θ,D) = P (x|θ) therefore the finite number of parameters capture everything there is to know about the data.
• So the complexity of the model is bounded even if the amount of data is unbounded. This makes them not very flexible.
• Non-parametric models assume that the data distribution cannot be defined in terms of such a finite set of parameters. But they can often be defined by assuming an infinite dimensional θ.
• So the complexity of the model predictions can grow as the amount of data grows. This makes them very flexible.
• Outline
• Introduction
• Gaussian Processes (GP) (very briefly)
• Dirichlet Processes (DP), different representations:
– Chinese Restaurant Process (CRP) – Urn Model – Stick Breaking Representation – Infinite limit of mixture models and Dirichlet process mixtures (DPM)
• Hierarchical Dirichlet Processes (next week)
• Infinite Hidden Markov Models (next week)
• Polya Trees (maybe)
• Dirichlet Diffusion Trees (maybe)
• Indian Buffet Processes (maybe)
• Discussion
• Gaussian Processes
A Gaussian process defines a distribution over functions, f , where f is a function mapping some input space X to
• Gaussian process covariance functions
P (f) is a Gaussian process if for any finite subset {x1, . . . , xn} ⊂ X , the marginal distribution over that finite subset P (f) has a multivariate Gaussian distribution.
Gaussian processes (GPs) are parameterized by a mean function, µ(x), and a covariance function, c(x, x′).
P (f(x), f(x′)) = N(µ,Σ)
where µ =
[ µ(x) µ(x′)
] Σ =
[ c(x, x) c(x, x′) c(x′, x) c(x′, x′)
] and similarly for P (f(x1), . . . , f(xn)) where now µ is an n× 1 vector and Σ is an n× n matrix.
E.g.: c(xi, xj) = v0 exp { − ( |xi − xj|
λ
)α} +v1+v2 δij with params (v0, v1, v2, λ, α)
Once the mean and covariance functions are defined, everything else about GPs follows from the basic rules of probability applied to mutivariate Gaussians.
• Samples from Gaussian processes with different c(x, x′)
0 10 20 30 40 50 60 70 80 90 100 −2
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x) 0 10 20 30 40 50 60 70 80 90 100
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f( x)
0 10 20 30 40 50 60 70 80 90 100 −4
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0 10 20 30 40 50 60 70 80 90 100 −3
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f( x)
• Prediction using GPs with different c(x, x′)
A sample from the prior for each covariance function:
0 10 20 30 40 50 60 70 80 90 100 −2
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−1
−0.5
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f( x)
0 10 20 30 40 50 60 70 80 90 100 −1.5
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f( x)
0 10 20 30 40 50 60 70 80 90 100 −2
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f( x)
Corresponding predictions, mean with two standard deviations:
0 5 10 15 −1.5
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0 5 10 15 −0.8
−0.6
−0.4
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1
• Outline
• Introduction
• Gaussian Processes (GP)
• Dirichlet Processes (DP), different representations: – Chinese Restaurant Process (CRP) – Urn Model – Stick Breaking Representation – Infinite limit of mixture models and Dirichlet process mixtures (DPM)
• Hierarchical Dirichlet Processes
• Infinite Hidden Markov Models
• Polya Trees
• Dirichlet Diffusion Trees
• Indian Buffet Processes
• Dirichlet Distribution
The Dirichlet distribution is a distribution over the K-dim probability simplex. | 2,506 | 7,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-25 | latest | en | 0.822275 |
https://citizensactionnetworks.com/qa/question-what-is-the-value-of-infinity-power-0.html | 1,611,203,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00352.warc.gz | 274,370,782 | 8,504 | # Question: What Is The Value Of Infinity Power 0?
## What is 0 to the power of infinity?
It means that no matter how many times you multiply 0 by 0, you will still end up with 0; which means that, yes, 0^∞=0.
Bonus Fact: 1^∞ is also 1, since 1×1=1, and so on, and so forth.
No matter how many times you do it, you never leave 1 (or 0, in the last case)..
## What is 2 to the infinity?
And here we can prove it as follows.. Here look the infinity means a very very large quantity that means unreachable quantity so simply, {something}^infinity= a very very large quantity(unreachable) that means infinity. So clearly 2^infinity=infinity.
## Is Infinity equal to zero?
The concept of zero and that of infinity are linked, but, obviously, zero is not infinity. Rather, if we have N / Z, with any positive N, the quotient grows without limit as Z approaches 0. Hence we readily say that N / 0 is infinite.
## What is infinity divided 0?
Therefore, we can say that 1/0.00000000000000….1 is equal to 1/0 thus equating it to infinite… So, as for your answer it is infinity divided by zero which will of course be infinity since infinity*infinity is still infinity..
## Can you multiply infinity by infinity?
infinity is a number which is greater than a number you can imagine. if you multiply any thing to that number you will get a number you hadn’t imagine means infinity. … Infinity means something with no end. so when we multiply infinity with infinity it is also infinity(∞ × ∞ = ∞).
## What is the value of zero upon infinity?
Regardless of what large number we’re dividing by our answer is 0 and by letting this large number increase (as much as we please, tending to infinity) the answer is still 0. Thus the ‘answer’ to your question is 0.
## Can Infinity be reset to zero?
0 is both a number and the numerical digit used to represent that number. Infinity is a number greater than any assignable quantity or countable number (symbol ∞). It does not hold any place in value system as we don’t know about it. Zero is nothing and infinity can be anything.
## What is the value of 1 Power Infinity?
1 raised to power infinity is always 1. If one considers LHL, x will tend to 1 (from left side of 1 on the number line) but will always be less than 1, and raising infinity on something which is less than one will approach to ZERO.
## Is infinity minus 1 still infinity?
Infinity is uncountable. It is not defined. When there is no particular numerical value for infinity, this operation of infinity minus one can’t really be performed as it is illogical. So the answer still remains infinity.
## Can Infinity be doubled?
Infinity times infinity or infinity plus infinity: The double infinity symbol meaning essentially comes down to these two things. In jewelry though, the meaning is so much more than simple mathematics. Infinity doubled is a symbol of two everlasting commitments combined.
## Can you divide infinity by 2?
As infinity is not a specific number it is just an thinking that it is uncountable or a number which we cannot count. And the infinity is such a number that when we divide it by 2,then it remains infinity. … Infinity divided by 2 is infinity.
## What is the number before infinity?
There isn’t a number before infinity because there isn’t a number called infinity. If there were, then there would also be a number called infinity+1. Any time I see something like infinity+x or infinity-x, the answer to that is still infinity.
## What is infinity take away infinity?
It is impossible for infinity subtracted from infinity to be equal to one and zero. Using this type of math, we can get infinity minus infinity to equal any real number. Therefore, infinity subtracted from infinity is undefined.
## What is value of infinity?
INFINITY (∞) … When we say in calculus that something is “infinite,” we simply mean that there is no limit to its values. Let f(x), for example, be. . Then as the values of x become smaller and smaller, the values of f(x) become larger and larger. | 932 | 4,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-04 | latest | en | 0.954899 |
http://download.plt-scheme.org/doc/103p1/html/match/node6.htm | 1,369,264,032,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702454815/warc/CC-MAIN-20130516110734-00088-ip-10-60-113-184.ec2.internal.warc.gz | 77,136,179 | 2,743 | Next: Index Up: Pattern Matching for Scheme Previous: Code Generation
Examples
This section illustrates the convenience of pattern matching with some examples. The following function recognizes some s-expressions that represent the standard Y operator:
```(define Y?
(match-lambda
[('lambda (f1)
('lambda (y1)
((('lambda (x1) (f2 ('lambda (z1) ((x2 x3) z2))))
('lambda (a1) (f3 ('lambda (b1) ((a2 a3) b2)))))
y2)))
(and (symbol? f1) (symbol? y1) (symbol? x1) (symbol? z1) (symbol? a1) (symbol? b1)
(eq? f1 f2) (eq? f1 f3) (eq? y1 y2)
(eq? x1 x2) (eq? x1 x3) (eq? z1 z2)
(eq? a1 a2) (eq? a1 a3) (eq? b1 b2))]
[_ #f]))```
Writing an equivalent piece of code in raw Scheme is tedious.
The following code defines abstract syntax for a subset of Scheme, a parser into this abstract syntax, and an unparser.
```(define-structure (Lam args body))
(define-structure (Var s))
(define-structure (Const n))
(define-structure (App fun args))
(define parse
(match-lambda
[(and s (? symbol?) (not 'lambda))
(make-Var s)]
[(? number? n)
(make-Const n)]
[('lambda (and args ((? symbol?) ...) (not (? repeats?))) body)
(make-Lam args (parse body))]
[(f args ...)
(make-App
(parse f)
(map parse args))]
[x (error 'syntax "invalid expression")]))
(define repeats?
(lambda (l)
(and (not (null? l))
(or (memq (car l) (cdr l)) (repeats? (cdr l))))))
(define unparse
(match-lambda
[(\$\\$\$ Var s) s]
[(\$\\$\$ Const n) n]
[(\$\\$\$ Lam args body) `(lambda ,args ,(unparse body))]
[(\$\\$\$ App f args) `(,(unparse f) ,@(map unparse args))]))```
With pattern matching, it is easy to ensure that the parser rejects all incorrectly formed inputs with an error message.
With match-define, it is easy to define several procedures that share a hidden variable. The following code defines three procedures, inc, value, and reset, that manipulate a hidden counter variable:
```(match-define (inc value reset)
(let ([val 0])
(list
(lambda () (set! val (add1 val)))
(lambda () val)
(lambda () (set! val 0)))))```
Although this example is not recursive, the bodies could recursively refer to each other.
The following code is taken from the macro package itself. The procedure validate-match-pattern checks the syntax of match patterns, and converts quasipatterns into ordinary patterns.
```(define validate-match-pattern
(lambda (p)
(letrec
([name?
(lambda (x)
(and (symbol? x)
(not (dot-dot-k? x))
(not (memq x '(quasiquote quote unquote unquote-splicing
? _ \$\\$\$ and or not set! get! ...)))))]
[simple?
(lambda (x)
(or (string? x) (boolean? x) (char? x) (number? x) (null? x)))]
[ordinary
(match-lambda
[(? simple? p) p]
[(? name? p) p]
['_ '_]
[('quasiquote p) (quasi p)]
[(and p ('quote _)) p]
[('? pred ps ...) `(and (? ,pred) ,@(map ordinary ps))]
[('and ps ...) `(and ,@(map ordinary ps))]
[('or ps ...) `(or ,@(map ordinary ps))]
[('not ps ...) `(not (or ,@(map ordinary ps)))]
[('\$\\$\$ (? name? r) ps ...) `(\$\\$\$ ,r ,@(map ordinary ps))]
[(and p ('set! (? name?))) p]
[(and p ('get! (? name?))) p]
[(p '...) `(,(ordinary p) ..0)]
[(p (? dot-dot-k? ddk)) `(,(ordinary p) ,ddk)]
[(x . y) (cons (ordinary x) (ordinary y))]
[(? vector? p) (apply vector (map ordinary (vector->list p)))]
[#&p (box (ordinary p))]
[p (err "invalid pattern at ~a" p)])]
[quasi
(match-lambda
[(? simple? p) p]
[(? symbol? p) `(quote ,p)]
[('unquote p) (ordinary p)]
[(('unquote-splicing p) . ()) (ordinary p)]
[(('unquote-splicing p) . y) (append (ordlist p) (quasi y))]
[(p '...) `(,(quasi p) ..0)]
[(p (? dot-dot-k? ddk)) `(,(quasi p) ,ddk)]
[(x . y) (cons (quasi x) (quasi y))]
[(? vector? p) (apply vector (map quasi (vector->list p)))]
[#&p (box (quasi p))]
[p (err "invalid quasipattern at ~a" p)])]
[ordlist
(match-lambda
[() ()]
[(x . y) (cons (ordinary x) (ordlist y))]
[p (err "invalid unquote-splicing at ~a" p)])])
(ordinary p))))```
PLT | 1,214 | 3,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2013-20 | latest | en | 0.559215 |
www.mydoit.net | 1,368,915,514,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382917/warc/CC-MAIN-20130516092622-00095-ip-10-60-113-184.ec2.internal.warc.gz | 616,078,781 | 5,037 | ## Welcome
If you're in one of John Carter's Math classes you are in the right place. Read this page then click on the link below and jump in.
Just visiting? Try the WikiSandbox and play a bit.
## What is a wiki?
A wiki is a website that usually has the following two properties:
1. Anybody can edit the pages of the wiki, and anybody can undo these edits
2. It is easy to write new pages for the wiki, because it doesn't use HTML
(There are exception to both rules.)
References:
## Why a Wiki?
With a wiki, creating and maintaining a website is trivial: You don't need to know HTML, nor FTP, nor anything else. This is what people normally use for their websites.
A wiki is great if you want to enable other people to help and contribute. The wiki just helps them to start contributing faster, since it is so easy to use. You can find the text formatting rules for this wiki here basic editing . For formatting mathematics, keep reading.
## Wiki is also Groupware
A wiki is ideal for a small group of people: classes, friends, project teams, gaming groups. It allows the group members to communicate with each other when you do not or cannot meet each other face-to-face. A wiki also works for chat rooms. Often the logs are available from archives but it is difficult to find good information by searching log files.
## Why this Wiki?
This wiki has the magical property of being able to read Latex math text. For this course you will only need to learn a limited number of commands for example fractions are done like this: \frac{x^2}{\sqrt{3x}} or this: {dy\over dx}
and limits are done like this: \lim_{x \rightarrow 0} x^3=0
We can do nifty integrals, like \int^{\infty}_{0}{x^2} . Heck, we can even do summations such as \sum _{i=0} ^{\infty+2} x^2 + 3i . Fun, eh?
But I digress. I will pass out a style sheet inclass so you know enough to ask(and answer) questions for this class.
Most of what you need to know to ask about calculus problems is in the table below. Math text is always bracketed between {$and$}.
{$x^2 + x^{1/2}$} x^2 + x^{1/2} {$\frac{a}{b}$} \frac{a}{b} {$\lim_{x \rightarrow 0}$} \lim_{x \rightarrow 0} {$\Delta x$} \Delta x {$\infty$} \infty {${dy\over dx}$} {dy\over dx}
More pretty math symbols
\Gamma \Gamma \Delta \Delta \Lambda \Lambda \Phi \Phi \Pi \Pi \Psi \Psi | 614 | 2,318 | {"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} | 2.984375 | 3 | CC-MAIN-2013-20 | latest | en | 0.916749 |
https://math.stackexchange.com/questions/4947909/i-wonder-where-the-author-used-i-in-the-above-proof-linear-algebra-by-ichi | 1,723,111,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00345.warc.gz | 316,653,507 | 38,947 | # I wonder where the author used (I) in the above proof. ("Linear Algebra" by Ichiro Satake.)
I am reading "Linear Algebra" (in Japanese) by Ichiro Satake.
Theorem 2:
The necessary and sufficient condition for $$m$$ $$n$$-dimensional vectors $$a_j = (a_{ij})$$ ($$1 \leq j \leq m$$) to be linearly independent is that $$m \leq n$$ and there exists a non-zero $$m$$-th order determinant among the determinants of the $$m$$-th order matrices formed by selecting $$m$$ rows from the $$n$$ rows of the $$(n, m)$$ matrix $$A = (a_{ij})$$.
For a proof of Theorem 2, the author first defines the term "strongly linearly independent."
We define $$m$$ $$n$$-dimensional vectors $$a_j = (a_{ij})$$ ($$1 \leq j \leq m$$) to be "strongly linearly independent" if $$m \leq n$$ and there exists a non-zero $$m$$-th order determinant among the determinants of the $$m$$-th order matrices formed by selecting $$m$$ rows from the $$n$$ rows of the $$(n, m)$$ matrix $$A = (a_{ij})$$.
If $$a_1, \ldots, a_m$$ are strongly linearly independent, it is easy to see that they are linearly independent. The author proves that if $$a_1, \ldots, a_m$$ are linearly independent, then they are also strongly linearly independent.
To prove this, the author proved (I) and (II):
(I) If $$a_1, \ldots, a_r$$ are strongly linearly independent, then any subset of them is also strongly linearly independent.
(II) If $$a_1, \ldots, a_r$$ are strongly linearly independent, and $$a_1, \ldots, a_r, a_{r+1}$$ are not strongly linearly independent, then $$a_{r+1}$$ can be uniquely expressed as a linear combination of $$a_1, \ldots, a_r$$.
Then, the author wrote as follows:
Using (I) and (II), if $$\{a_1, \ldots, a_m\}$$ is any given set of $$n$$-dimensional vectors, by selecting the maximal subset $$\{a_{i_1}, \ldots, a_{i_r}\}$$ that is strongly linearly independent, any $$a_i$$ ($$1 \leq i \leq m$$) will be linearly dependent on $$\{a_{i_k}\}$$ ($$1 \leq k \leq r$$). Therefore, if $$a_1, \ldots, a_m$$ are linearly independent, then $$r = m$$. In other words, $$a_1, \ldots, a_m$$ are also strongly linearly independent. (End of Proof)
Where did the author use (I) in the above sentences?
• Which edition of the book are you reading, and in Japanese or English? Commented Jul 23 at 21:30
• @blargoner thank you very much for your comment. The above proof is an alternative proof for Theorem 2. Although this alternative proof is not included in the English edition, it is somehow included in the Japanese edition. Commented Jul 24 at 1:23
• @blargoner I am reading the Japanese edition of this book. Commented Jul 24 at 1:36
It seems that (I) is used implicitly when using the words "selecting the maximal subset" there.
[[ We might have to see the whole argument to be certain , though I think this is where the author used (1) ]]
Certain Properties will not allow selecting "selecting the maximal subset" , like this Eg 1 :
Let Set $$X=\{1,2,-3,4,-5,6,-7,-8\}$$
We want Sub-Set $$Y$$ where the Sum is Positive.
We can check that $$Y$$ can be $$\{1,2\}$$ , $$\{1,2,4\}$$ , $$\{1,-3,4\}$$ , $$\{-3,4,-5,6\}$$ , where Sum is indeed Positive.
Let $$Y=\{1,-3,4\}$$ , with Positive Sum.
Yet , there are Sub-Sets of $$Y$$ whose Sum goes Negative too : $$\{1,-3\}$$ , $$\{-3\}$$
Hence , we can not talk about "selecting the maximal subset" here.
When we include new element $$-5$$ to $$Y$$ , then Sum goes Negative.
Yet , we can add two new elements $$-5$$ & $$6$$ to $$Y$$ , where Sum will remain Positive.
It reiterates that we can not talk about "selecting the maximal subset" here.
Consider this Eg 2 :
We have list of numbers $$Z = [+5,-5,+5,-5,+5]$$ where we want to pick out "contiguous" numbers with Positive Sum.
We have various ways to try that :
• Only first $$+5$$
• Only first three numbers $$+5,-5,+5$$
• Only last $$+5$$
• Only last three $$+5,-5,+5$$
• $$\cdots$$
There is not much meaning to unique maximal subset here too.
[[ I am just making explanatory examples to show that maximal subset may or may not exist , the answer will stand without these examples too ]]
When the author talks about "selecting the maximal subset" with the Property "strongly linearly independent" , then (I) implicitly allows it.
When we have made the selection , it can not occur that we can add new element which makes it lose that Property yet we can add two new elements to retain that Property.
In other words , "selecting the maximal subset" will make sense here. | 1,315 | 4,457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 61, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.792423 |
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# From studies of the bony house of the brain, which is the cranium
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From studies of the bony house of the brain, which is the cranium [#permalink]
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From studies of the bony house of the brain, which is the cranium, located in the back of the skull, come what scientists know about dinosaur brains.
(A) From studies of the bony house of the brain, which is the cranium, located in the back of the skull, come what scientists know about dinosaur brains.
(B) The knowledge that scientists know about dinosaur brains comes from studies of the bony house of the brain, located in the back of the skull, that is, the cranium.
(C) The knowledge of dinosaur brains that scientists have come from studies of the bony house of the brain, which is located in the back of the skull and is called the cranium.
(D) What scientists know about dinosaur brains comes from studies of the cranium, the bony house of the brain located in the back of the skull.
(E) Located in the back of the skull is the cranium, the bony house of the brain, and it is from studies of this that scientists know what they know about dinosaur brains.
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Re: From studies of the bony house of the brain, which is the cranium [#permalink]
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07 Feb 2019, 08:51
Rashed12 wrote:
From studies of the bony house of the brain, which is the cranium, located in the back of the skull, come what scientists know about dinosaur brains.
(A) From studies of the bony house of the brain, which is the cranium, located in the back of the skull, come what scientists know about dinosaur brains.
(B) The knowledge that scientists know about dinosaur brains comes from studies of the bony house of the brain, located in the back of the skull, that is, the cranium.
(C) The knowledge of dinosaur brains that scientists have come from studies of the bony house of the brain, which is located in the back of the skull and is called the cranium.
(D) What scientists know about dinosaur brains comes from studies of the cranium, the bony house of the brain located in the back of the skull.
(E) Located in the back of the skull is the cranium, the bony house of the brain, and it is from studies of this that scientists know what they know about dinosaur brains.
Discussed here: https://gmatclub.com/forum/from-studies ... 64247.html
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Re: From studies of the bony house of the brain, which is the cranium [#permalink] 07 Feb 2019, 08:51
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# From studies of the bony house of the brain, which is the cranium
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,106 | 4,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.89341 |
https://www.cscodehelp.com/%E7%A7%91%E7%A0%94%E4%BB%A3%E7%A0%81%E4%BB%A3%E5%86%99/cs%E4%BB%A3%E8%80%83-cs-70-discrete-mathematics-and-probability-theory-fall-2021-7/ | 1,726,530,328,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00334.warc.gz | 680,570,019 | 14,670 | # CS代考 CS 70 Discrete Mathematics and Probability Theory Fall 2021 – cscodehelp代写
CS 70 Discrete Mathematics and Probability Theory Fall 2021
Due: Friday 9/24, 10:00 PM Grace period until Friday 9/24 11:59 PM
Before you start writing your final homework submission, state briefly how you worked on it. Who else did you work with? List names and email addresses. (In case of homework party, you can just describe the group.)
1 Modular Practice
Solve the following modular arithmetic equations for x and y.
(a) 9x+5 ≡ 7 (mod 11).
(b) Show that 3x + 15 ≡ 4 (mod 21) does not have a solution.
(c) The system of simultaneous equations 3x + 2y ≡ 0 (mod 7) and 2x + y ≡ 4 (mod 7). (d) 132019 ≡ x (mod 12).
(e) 721 ≡ x (mod 11).
(f) x≡5 (mod 7),x≡3 (mod 9),x≡3 (mod 11). Whatisthesmallestpossiblevalueforx? 2 Check Digits: ISBN
In this problem, we’ll look at a real-world applications of check-digits.
International Standard Book Numbers (ISBNs) are 10-digit codes (d1d2 …d10) which are assigned
by the publisher. These 10 digits contain information about the language, the publisher, and the
number assigned to the book by the publisher. Additionally, the last digit d10 is a “check digit”
selected so that ∑10 i · di ≡ 0 (mod 11). (Note that the letter X is used to represent the number 10 i=1
in the check digit.)
(a) Suppose you have a very worn copy of the (recommended) textbook for this class. You want to list it for sale online but you can only read the first nine digits: 0-07-288008-? (the dashes are only there for readability). What is the last digit? Show your work.
CS 70, Fall 2021, HW 4 1
(b) (c) (d)
Wikipedia says that you can determine the check digit by computing ∑9i=1 i · di (mod 11). Show that Wikipedia’s description is equivalent to the above description.
Prove that changing any single digit of the ISBN will render the ISBN invalid. That is, the check digit allows you to detect a single-digit substitution error.
Can we ever switch two distinct digits in an ISBN number and get another valid ISBN number? For example, could 012345678X and 015342678X both be valid ISBNs? Explain.
Divisible or Not
Prove that for any number n, the number formed by the last two digits of n are divisible by 4 if and only if n is divisible by 4. (For example, ‘23xx’ is divisible by 4 if and only if the number ‘xx’ is divisible by 4.)
Prove that for any number n, the sum of the digits of n are divisible by 3 if and only if n is divisible by 3.
Just Can’t Wait
Joel lives in Berkeley. He mainly commutes by public transport, i.e., bus and BART. He hates waiting while transferring, and he usually plans his trip so that he can get on his next vehicle immediately after he gets off the previous one (zero transfer time, i.e. if he gets off his previous vehicle at 7:00am he gets on his next vehicle at 7:00am). Tomorrow, Joel needs to take an AC Transit bus from his home stop to the Downtown Berkeley BART station, then take BART into San Francisco.
The bus arrives at Joel’s home stop every 22 minutes from 6:05am onwards, and it takes 10 minutes to get to the Downtown Berkeley BART station. The train arrives at the station every 8 minutes from 4:25am onwards. What time is the earliest bus he can take to be able to transfer to the train immediately? Show your work. (Find the answer without listing all the schedules. Hint: derive an equation relating the bus number and train number and then work in modular arithmetic to get rid of one of the variables to give a set of possible train numbers.)
Joel has to take a Muni bus after he gets off the train in San Francisco. The commute time on BART is 33 minutes, and the Muni bus arrives at the San Francisco BART station every 17 minutes from 7:12am onwards. What time is the earliest bus he could take from Berkeley to ensure zero transfer time for both transfers? If all bus/BART services stop just before midnight, is it the only bus he can take that day? Show your work.
Fermat’s Little Theorem
Fermat’s Little Theorem states that for any prime p and any a ∈ {1,2,…, p−1}, we have ap−1 ≡ 1 (mod p). Without using induction, prove that ∀n ∈ N, n7 − n is divisible by 42.
CS 70, Fall 2021, HW 4 2
6 Sparsity of Primes
A prime power is a number that can be written as pi for some prime p and some positive integer i. So,9=32 isaprimepower,andsois8=23. 42=2·3·7isnotaprimepower.
Prove that for any positive integer k, there exists k consecutive positive integers such that none of them are prime powers.
Hint: this is a Chinese Remainder Theorem problem. We want to find x such that x + 1;x + 2;x + 3; : : :x+k are all not powers of primes. We can enforce this by saying that x+1 through x+k each must have two distinct prime divisors.
7 Unique Prime Factorization
We proved in lecture that every positive integer has a factorization into primes. Using the tech- niques in this course, we can show that this factorization is unique, up to reordering the factors!
Recall from lecture that we defined a prime number p to be a positive integer whose only positive factors are 1 and p.
In this problem, you should not assume that the positive integers have unique prime factor- ization.
(a) Let p and q be distinct primes. Show that p q.
(b) Provethatifaandbarepositiveintegerssuchthat p|ab,then p|aor p|b. (Hint: Usethe
Extended Euclidean algorithm: There are integers x, y such that ax + py = gcd(a, p).)
(c) Show that the prime factorization of a positive integer n is unique up to reordering the factors. (Hint: Suppose that there were two prime factorizations of n. Use the previous parts to show that the factorizations are the same.)
8 Wilson’s Theorem
Wilson’s Theorem states the following is true if and only if p is prime: (p−1)!≡−1 (mod p).
Prove both directions (it holds if AND only if p is prime).
Hint for the if direction: Consider rearranging the terms in (p − 1)! = 1 · 2… · p − 1 to pair up
terms with their inverses, when possible. What terms are left unpaired?
Hint for the only if direction: If p is composite, then it has some prime factor q. What can we say | 1,618 | 6,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.740052 |
https://sourceforge.net/p/maxima/mailman/maxima-bugs/thread/E1GH8MW-0004I4-O2@sc8-sf-web5.sourceforge.net/ | 1,513,466,304,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589512.63/warc/CC-MAIN-20171216220904-20171217002904-00461.warc.gz | 701,679,454 | 10,859 | ## maxima-bugs
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: - 2002-06-26 21:06:07 ```Bugs item #505443, was opened at 2002-01-18 11:53 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- >Comment By: Raymond Toy (rtoy) Date: 2002-06-26 17:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: - 2002-06-26 21:19:58 ```Bugs item #505443, was opened at 2002-01-18 10:53 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- >Comment By: Cliff Yapp (starseeker) Date: 2002-06-26 16:19 Message: Logged In: YES user_id=11463 Unfortunately, I wasn't the original person to speak on this one - I just submitted it from the list, and I can't remember who it was. If you think the current behavior is OK then it probably isn't worth fussing with too much. CY ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-26 16:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: - 2002-06-27 17:42:39 ```Bugs item #505443, was opened at 2002-01-18 11:53 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- >Comment By: Raymond Toy (rtoy) Date: 2002-06-27 13:42 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- Comment By: Cliff Yapp (starseeker) Date: 2002-06-26 17:19 Message: Logged In: YES user_id=11463 Unfortunately, I wasn't the original person to speak on this one - I just submitted it from the list, and I can't remember who it was. If you think the current behavior is OK then it probably isn't worth fussing with too much. CY ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-26 17:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: - 2002-10-03 16:49:16 ```Bugs item #505443, was opened at 2002-01-18 11:53 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- Comment By: Stavros Macrakis (macrakis) Date: 2002-10-03 12:49 Message: Logged In: YES user_id=588346 The proposed correction, sign(x)*sqrt(1-cos(x)) / sqrt(2) only extends the validity of the formula from [0,2pi] to [- 2pi,2pi]. It continues to be incorrect whenever fix(x/(2*pi)) is odd. I suppose you could use (-1)^entier(x/(2*pi)) * sqrt(1-cos(x))/sqrt(2) but I'm not sure that is terribly useful, especially since Maxima knows nothing about Entier except how to evaluate it for constants. For example, (-1)^(2*Entier(...)) should simplify to 1, but doesn't. Entier(Entier(x)) should simplify to Entier (x). Entier(x+5) should simplify to Entier(x)+5. Etc. ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-27 13:42 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- Comment By: Cliff Yapp (starseeker) Date: 2002-06-26 17:19 Message: Logged In: YES user_id=11463 Unfortunately, I wasn't the original person to speak on this one - I just submitted it from the list, and I can't remember who it was. If you think the current behavior is OK then it probably isn't worth fussing with too much. CY ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-26 17:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: SourceForge.net - 2006-03-26 23:13:47 ```Bugs item #505443, was opened at 2002-01-18 09:53 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. >Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- >Comment By: Robert Dodier (robert_dodier) Date: 2006-03-26 16:13 Message: Logged In: YES user_id=501686 For the record, (halfangles : true, sin (x/2)); => sqrt(1-cos(x))/sqrt(2)\$ i.e., same behavior as when this report was first made. Recently (Maxima 5.9.3) floor and ceiling have been implemented as simplifying functions, and the simplifications for entier mentioned below are all implemented. (-1)^(2*floor(x)) => 1 floor(floor(x)) => floor(x) floor(x + 5) => floor(x) + 5 Perhaps this means it is now reasonable to change the half-angle simplification to (-1)^floor(x/(2*%pi)) * . ---------------------------------------------------------------------- Comment By: Stavros Macrakis (macrakis) Date: 2002-10-03 10:49 Message: Logged In: YES user_id=588346 The proposed correction, sign(x)*sqrt(1-cos(x)) / sqrt(2) only extends the validity of the formula from [0,2pi] to [- 2pi,2pi]. It continues to be incorrect whenever fix(x/(2*pi)) is odd. I suppose you could use (-1)^entier(x/(2*pi)) * sqrt(1-cos(x))/sqrt(2) but I'm not sure that is terribly useful, especially since Maxima knows nothing about Entier except how to evaluate it for constants. For example, (-1)^(2*Entier(...)) should simplify to 1, but doesn't. Entier(Entier(x)) should simplify to Entier (x). Entier(x+5) should simplify to Entier(x)+5. Etc. ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-27 11:42 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- Comment By: Cliff Yapp (starseeker) Date: 2002-06-26 15:19 Message: Logged In: YES user_id=11463 Unfortunately, I wasn't the original person to speak on this one - I just submitted it from the list, and I can't remember who it was. If you think the current behavior is OK then it probably isn't worth fussing with too much. CY ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-26 15:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: SourceForge.net - 2006-08-27 00:18:55 ```Bugs item #505443, was opened at 2002-01-18 09:53 Message generated for change (Settings changed) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. >Category: Lisp Core - Trigonometry Group: None Status: Open Resolution: None Priority: 5 Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- Comment By: Robert Dodier (robert_dodier) Date: 2006-03-26 16:13 Message: Logged In: YES user_id=501686 For the record, (halfangles : true, sin (x/2)); => sqrt(1-cos(x))/sqrt(2)\$ i.e., same behavior as when this report was first made. Recently (Maxima 5.9.3) floor and ceiling have been implemented as simplifying functions, and the simplifications for entier mentioned below are all implemented. (-1)^(2*floor(x)) => 1 floor(floor(x)) => floor(x) floor(x + 5) => floor(x) + 5 Perhaps this means it is now reasonable to change the half-angle simplification to (-1)^floor(x/(2*%pi)) * . ---------------------------------------------------------------------- Comment By: Stavros Macrakis (macrakis) Date: 2002-10-03 10:49 Message: Logged In: YES user_id=588346 The proposed correction, sign(x)*sqrt(1-cos(x)) / sqrt(2) only extends the validity of the formula from [0,2pi] to [- 2pi,2pi]. It continues to be incorrect whenever fix(x/(2*pi)) is odd. I suppose you could use (-1)^entier(x/(2*pi)) * sqrt(1-cos(x))/sqrt(2) but I'm not sure that is terribly useful, especially since Maxima knows nothing about Entier except how to evaluate it for constants. For example, (-1)^(2*Entier(...)) should simplify to 1, but doesn't. Entier(Entier(x)) should simplify to Entier (x). Entier(x+5) should simplify to Entier(x)+5. Etc. ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-27 11:42 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- Comment By: Cliff Yapp (starseeker) Date: 2002-06-26 15:19 Message: Logged In: YES user_id=11463 Unfortunately, I wasn't the original person to speak on this one - I just submitted it from the list, and I can't remember who it was. If you think the current behavior is OK then it probably isn't worth fussing with too much. CY ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-26 15:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ```
[Maxima-bugs] [ maxima-Bugs-505443 ] Halfangle limitation From: SourceForge.net - 2009-01-04 13:46:19 ```Bugs item #505443, was opened at 2002-01-18 17:53 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core - Trigonometry Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Cliff Yapp (starseeker) Assigned to: Nobody/Anonymous (nobody) Summary: Halfangle limitation Initial Comment: Setting HALFANGLES gives sin(x/2) => sqrt(1-cos(x))/2. Of course, this is wrong if x is negative. Need to make it smarter. ---------------------------------------------------------------------- >Comment By: Dieter Kaiser (crategus) Date: 2009-01-04 14:46 Message: As suggested on the mailing list the general factors in terms of the floor, round and unit_step function for the functions sin, cos, sinh and cosh are implemented (Revision 1.8 of logarc.lisp). The general factor take into accunt real and complex arguments and simplifies to correct expressions. Closing the bug report as fixed. Dieter Kaiser ---------------------------------------------------------------------- Comment By: Robert Dodier (robert_dodier) Date: 2006-03-27 01:13 Message: Logged In: YES user_id=501686 For the record, (halfangles : true, sin (x/2)); => sqrt(1-cos(x))/sqrt(2)\$ i.e., same behavior as when this report was first made. Recently (Maxima 5.9.3) floor and ceiling have been implemented as simplifying functions, and the simplifications for entier mentioned below are all implemented. (-1)^(2*floor(x)) => 1 floor(floor(x)) => floor(x) floor(x + 5) => floor(x) + 5 Perhaps this means it is now reasonable to change the half-angle simplification to (-1)^floor(x/(2*%pi)) * . ---------------------------------------------------------------------- Comment By: Stavros Macrakis (macrakis) Date: 2002-10-03 18:49 Message: Logged In: YES user_id=588346 The proposed correction, sign(x)*sqrt(1-cos(x)) / sqrt(2) only extends the validity of the formula from [0,2pi] to [- 2pi,2pi]. It continues to be incorrect whenever fix(x/(2*pi)) is odd. I suppose you could use (-1)^entier(x/(2*pi)) * sqrt(1-cos(x))/sqrt(2) but I'm not sure that is terribly useful, especially since Maxima knows nothing about Entier except how to evaluate it for constants. For example, (-1)^(2*Entier(...)) should simplify to 1, but doesn't. Entier(Entier(x)) should simplify to Entier (x). Entier(x+5) should simplify to Entier(x)+5. Etc. ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-27 19:42 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- Comment By: Cliff Yapp (starseeker) Date: 2002-06-26 23:19 Message: Logged In: YES user_id=11463 Unfortunately, I wasn't the original person to speak on this one - I just submitted it from the list, and I can't remember who it was. If you think the current behavior is OK then it probably isn't worth fussing with too much. CY ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2002-06-26 23:06 Message: Logged In: YES user_id=28849 What are you expecting? I think it would be fairly easy for maxima to say sign(x)*sqrt(1-cos(x))/2 (assuming I got that right.) See halfangleaux in logarc.lisp. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=505443&group_id=4933 ``` | 4,583 | 17,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-51 | latest | en | 0.740371 |
https://metanumbers.com/214886 | 1,638,293,823,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359065.88/warc/CC-MAIN-20211130171559-20211130201559-00584.warc.gz | 458,742,087 | 7,369 | # 214886 (number)
214,886 (two hundred fourteen thousand eight hundred eighty-six) is an even six-digits composite number following 214885 and preceding 214887. In scientific notation, it is written as 2.14886 × 105. The sum of its digits is 29. It has a total of 3 prime factors and 8 positive divisors. There are 92,088 positive integers (up to 214886) that are relatively prime to 214886.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 29
• Digital Root 2
## Name
Short name 214 thousand 886 two hundred fourteen thousand eight hundred eighty-six
## Notation
Scientific notation 2.14886 × 105 214.886 × 103
## Prime Factorization of 214886
Prime Factorization 2 × 7 × 15349
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) 214886 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 214,886 is 2 × 7 × 15349. Since it has a total of 3 prime factors, 214,886 is a composite number.
## Divisors of 214886
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) 368400 Sum of all the positive divisors of n s(n) 153514 Sum of the proper positive divisors of n A(n) 46050 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 463.558 Returns the nth root of the product of n divisors H(n) 4.66636 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 214,886 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 214,886) is 368,400, the average is 46,050.
## Other Arithmetic Functions (n = 214886)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 92088 Total number of positive integers not greater than n that are coprime to n λ(n) 15348 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 19147 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 92,088 positive integers (less than 214,886) that are coprime with 214,886. And there are approximately 19,147 prime numbers less than or equal to 214,886.
## Divisibility of 214886
m n mod m 2 3 4 5 6 7 8 9 0 2 2 1 2 0 6 2
The number 214,886 is divisible by 2 and 7.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (214886)
Base System Value
2 Binary 110100011101100110
3 Ternary 101220202202
4 Quaternary 310131212
5 Quinary 23334021
6 Senary 4334502
8 Octal 643546
10 Decimal 214886
12 Duodecimal a4432
20 Vigesimal 16h46
36 Base36 4lt2
## Basic calculations (n = 214886)
### Multiplication
n×y
n×2 429772 644658 859544 1074430
### Division
n÷y
n÷2 107443 71628.7 53721.5 42977.2
### Exponentiation
ny
n2 46175992996 9922574430938456 2132222329166641056016 458184727425302829963054176
### Nth Root
y√n
2√n 463.558 59.8967 21.5304 11.6531
## 214886 as geometric shapes
### Circle
Diameter 429772 1.35017e+06 1.45066e+11
### Sphere
Volume 4.15636e+16 5.80265e+11 1.35017e+06
### Square
Length = n
Perimeter 859544 4.6176e+10 303895
### Cube
Length = n
Surface area 2.77056e+11 9.92257e+15 372193
### Equilateral Triangle
Length = n
Perimeter 644658 1.99948e+10 186097
### Triangular Pyramid
Length = n
Surface area 7.99792e+10 1.16939e+15 175454
## Cryptographic Hash Functions
md5 5b090fe4e9b020ea591a641c7ef17962 c5764a380252d9e863467785c1ee3229b48e091d abbb86e6715249b29b62c1938941dc50d6f264f8891787bf910a9dc71dcbcac4 7e48d131d33a5966ea62e7b185092ad169ce7042b263f57178a1d198fd03bbbc0ababc15bf309e427e3f900d7f194969dc0252b45559237a6f8645ab83bb39e0 aefc2270c3ec68f89baabc6b4bd2d6061661ba16 | 1,435 | 4,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.80632 |
alt3.joachim-breitner.de | 1,719,075,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00546.warc.gz | 79,603,483 | 1,982 | She's the 1
zur Melodie von "She's the one" von Robbie Williams
```"I" was her back in Rome
But she's 1 here at home
And if there's a number
Times n is n,
She's the 1.
If there's a number
Times n is n,
She's the 1.
She and oh's all you need
to work in binary
And if there's a number
Times n is n,
She's the 1.
If there's a number
Times n is n,
She's the 1.
When you take some to what you got
And want to know what's in the set
When you write her left on the row
And then one thousand times "zero"
That'll be to hight, that's mind-boggling
Facutly might be strong
But 1 ! that's just 1
And if there's a number
Times n is n,
She's the 1.
If there's a number
Times n is n,
She's the 1.
When you take some to what you got
And want to know what's in the set
When you write her left on the row
And then one thousand times 0
That'll be to hight, that's mind-boggling
"I" was her back in Rome
But she's 1 here at home
And if there's a number
Times n is n,
She's the 1.
If there's a number
Times n is n,
She's the 1.
If there's a number
Times n is n,
She's the 1.
Yeah she's the 1
If there's a number
Times n is n,
She's the 1.
She's the 1.
If there's a number
Times n is n,
She's the 1.
Singhinweise:
!: "bang"
```
Go up | 410 | 1,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.956744 |
https://www.programming-idioms.org/idiom/18/depth-first-traversing-of-a-tree/2084/python | 1,521,310,134,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645280.4/warc/CC-MAIN-20180317174935-20180317194935-00793.warc.gz | 884,455,369 | 6,208 | This language bar is your friend. Select your favorite languages!
# Idiom #18 Depth-first traversing of a tree
Call a function f on every node of a tree, in depth-first prefix order
```def DFS(f, root):
f(root)
for child in root:
DFS(f, child)```
```void prefixOrderTraversal(alias f)(ref Tree tree)
{
f(tree);
foreach (child; tree.children)
prefixOrderTraversal!f(child);
}```
```traverse(Tree node, f(value)) {
f(node.value);
for (var child in node.children) {
traverse(child, f);
}
}```
```func (t *Tree) Dfs(f func(*Tree)) {
if t == nil {
return
}
f(t)
for _, child := range t.Children {
child.Dfs(f)
}
}```
```preordered (Node pivot left right) =
pivot : preordered left ++ preordered right
preordered Ø = []
f <\$> (preordered tree)
```
```sub depth_first_traversal {
my (\$f, \$treenode) = @_;
\$f->(\$treenode);
depth_first_traversal(\$f, \$_) for @{\$treenode->{children}};
}
```
```def dfs(f, node)
f.(node)
node.children.each do |child|
dfs(f, child)
end
end```
```pub struct Tree<V> {
children: Vec<Tree<V>>,
value: V
}
impl<V> Tree<V> {
pub fn dfs<F: Fn(&V)>(&self, f: F) {
self.dfs_helper(&f);
}
fn dfs_helper<F: Fn(&V)>(&self, f: &F) {
(f)(&self.value);
for child in &self.children {
child.dfs_helper(f)
}
}
// ...
}```
#### Idiom created by
programming-idioms.org | 390 | 1,285 | {"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} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.463926 |
https://www.livescience.com/19048-happy-pi-day-history.html | 1,721,371,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00317.warc.gz | 760,554,863 | 118,299 | Happy Pi Day! Why Geeks Celebrate 3.14...
If you're celebrating Pi Day today (March 14), then you're a certified math geek or physics geek or maybe even a tech geek. If you're just an outside observer, we thought you might like to know why all the hubbub over 3.1415926535 ... well, that could go on forever, so …
On Pi Day, pi enthusiasts wear clothing adorned with the pi symbol, eat pie, and even throw pi-related parties.
March 14 is chosen as the day to celebrate pi, because the numerical date, 3/14, represents the first 3 digits of pi. Hardcore Pi Day celebrants are planning special events for 9:26:53 a.m. on March 14, 2015, as the numerical date 3/14/15 9:26:53 represents the first 7 digits of pi, 3.141592653. [Real Pie Chart: America's Favorite Pies]
The concept of pi is important to mathematics because of its relationship to the circle; it is a constant representing the ratio of a circle's circumference to its diameter. Since pi is found in so many different equations in math, physics and other sciences, it is considered one of the most important mathematical constants.
Pi is an irrational transcendental number, meaning that its decimal places will continue to infinity. It cannot be represented using decimal notation or a rational fraction. As such, 3.14 is not pi, but simply an easy notation for the first 3 places. Even the common use of 22/7 for pi is not exact. To date, pi has been calculated out to more than 1 trillion decimal places, and mathematicians continue to calculate further digits.
Pi Day was started at the Exploratorium, a San Francisco-based science museum known for its interactive exhibits, by staff physicist Larry Shaw in 1988. Staff and visitors celebrated the day by holding a circular parade and then eating fruit pies. The Exploratorium continues to hold an annual Pi Day Celebration, which has gotten larger each year. In 2012, the celebration expanded to the Internet, with both a webcast and a Second Life-based event.
It was in 2009 that Pi Day became a national event, with official recognition from the House of Representatives through Resolution 224. The hope is that official recognition of Pi Day will help to increase interest in math and science among the American public. Schools are urged to use the day to teach their students about the importance of pi and other mathematical concepts.
Fun celebrations for Pi Day have the somewhat pie-in-the-sky goal of showing students that learning about math and science doesn't have to be boring. Interestingly, however, some mathematicians want to say goodbye to pi. | 572 | 2,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.952795 |
http://www.jiskha.com/display.cgi?id=1207613813 | 1,498,574,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321426.45/warc/CC-MAIN-20170627134151-20170627154151-00114.warc.gz | 543,073,377 | 4,442 | # physics
posted by .
1)A 5.0 cm real object is placed at a distance of 30.0 cm from a concave mirror of focal length 10.0 cm. Find the location of the image.
(30)(10)/5.0
15
or (30)(5.0)/10
60
I don't know which one
2)What must be the minimum height of a plane mirror so that a boy of height 162 cm can view himself from head to toe?
162 cm
3)A 5.0 cm real object is placed at a distance of 5.0 cm from a concave mirror of focal length 10 cm. Find the size of the image.
20 cm
• physics -
3)(10)(5.0)/5.0-10
answer: 10 cm (I get -10 from the math but that's not one of the choices)
• physics -
3) I skipped a step but 10 is still the answer.
• physics -
1/do + 1/di = 1/f = 1/10
1/30 + 1/di = 1/10
1/di = 2/30
One of your answers is correct, but you used the wrong formula to get it.
di = 15 cm is the distance to the image. It makes no difference what the object height is.
2) wrong
3) wrong. You are just taking wild guesses. That won't do.
1/5 + 1/di = 1/10
1/di = 1/10 - 2/10 = -1/10
di = -10
Magnification = |di/do| = ?
Image size = (object size) * (magnification)= ?
• physics-it may look like it but im not -
it may look like it but im not
• physics -
So I was right originally on 3, with 20 cm. For 2, 162/2 = 81 cm
• physics -
mag = di/do
mag = -10/5.0
mag = 2
im size = o-size * mag
im size = 5.0 * 2
im size = 10
• physics -
15 | 474 | 1,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-26 | latest | en | 0.868883 |
https://republicofsouthossetia.org/question/you-might-need-calculator-yasemin-deposited-1000-into-a-savings-account-the-relationship-between-16229496-74/ | 1,638,305,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00136.warc.gz | 544,300,731 | 13,805 | ## You might need: Calculator Yasemin deposited $1000 into a savings account. The relationship between the time, t, in years, Question You might need: Calculator Yasemin deposited$1000 into a savings account.
The relationship between the time, t, in years, since the account was first opened, and Yasemin’s account
balance, B(t), in dollars, is modeled by the following function.
B(t) = 1000. 20.03
How many years will it take for Yasemin’s account balance to reach $1500? Round your answer, if necessary, to the nearest hundredth. in progress 0 2 months 2021-10-15T14:01:56+00:00 1 Answer 0 views 0 ## Answers ( ) 1. Answer: The number of years it will take for Yasemin’s account balance to reach$1500 is 13.52 years.
Step-by-step explanation:
Where the model of the relationship between the time, t, in years, since the account was first opened and the and the balance in Yasemin’s account is presented as follows;
To find find out how many years it will take for Yasemin’s account balance to reach $1500, we substitute B(t) =$1500 since we are told that after the years his account balance became $1500 as follows; We now solve for t The number of years it will take for Yasemin’s account balance to reach$1500 = 13.52 years. | 314 | 1,231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.960529 |
https://www.sciforums.com/threads/where-is-most-gravity-inside-or-out.155498/page-5 | 1,718,565,400,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00897.warc.gz | 871,507,336 | 20,784 | # Where is most "gravity", inside or out?
Of course not, but it is correct in this specific case which is what the question was about.
The question was about whether there is more [accumulated] gravity toward the interior or the exterior of a symmetrical spherical mass, ideally ignoring for simplicity sakes local (black hole) peculiarities, and: is it not true, that gravity is mainly on the outside? Even in an evaporating black hole, the interior gravity has to fall to zero at the center, theoretically. or? That mass of dark matter too is inert gravitationally toward the center. imho.
so: if you think that the higher outer orbital velocity have to be the result of dark matter in the peripheral neighbourhood, ?? re-examine that please! perhaps it is really because all the gravity is there , compliments of the old fashioned matter and the shell theorem?
PS: Even outside the Schwarzschild radius, the same mighty gravity of the Black hole mass that falls quickly to the near center, will only slowly peter out approaching infinity. Adding mass will not change the fact that all the gravity acts toward the outside. Sphagettification is reversed once you pass the event horizon, or max gravity.
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Take the case of the singularity, with that small interior dimension, all the mass is there, but no gravity, all the force or projected tension is to the outside, to eternity, but, the force reverses direction through the center, so there must be null gravity in the center and with diminishing gravity inward, there has to be a reverse tidal effect,or?
Take the case of the singularity, with that small interior dimension, all the mass is there, but no gravity, all the force or projected tension is to the outside, to eternity, but, the force reverses direction through the center, so there must be null gravity in the center and with diminishing gravity inward, there has to be a reverse tidal effect,or?
There is a concept of central pressure for a spherical object. This central pressure is non zero.
There is a concept of central pressure for a spherical object. This central pressure is non zero.
Pressure is the result of gravity acting from the outside --(because that is where it is)--inward. While pressure , as kinetic energy can be a contributing source of gravity, it does not cause gravity, a pulling condition/force to go to maximum at the center. or?
PS: A more pressure-compressed mass would not reduce the amount of gravity (intensity x distance) on the outside would it, but would confine its gravity to a smaller volume. so: could we say that the more dense the body the greater the outside to the inside gravity ratio?
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I think this graph will clearly answer your question:
If you compressed the mass with size R into a smaller and smaller volume, The peak of the surface gravity would rise with decreasing radius, but the original curve, from R to 2R to 3R would remain. The interior gravity pictured by the area of the 0Rgs triangle, would have a sum of the same value, but an additional outside gravity potential would be created, defined by area inclosed by the newly added steeper section of curve . so:
can it be said that there is more gravity outside than inside? more so as the central mass becomes denser?
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Gravity is everywhere.
Gravity is everywhere.
except at the centre of entities, or hollow spheres.
except at the centre of entities, or hollow spheres.
Entities?
Like creatures? Why would they follow different laws?
Like creatures? Why would they follow different laws
entities like galactic superclusters, the solar system, the Earth. there is no gravity at their center of gravity, most gravity is outside their :surface, the denser they are, the more gravity is at the outside, stretching to infinity, do not be surprised if a lot is happening outside. faster than we might think. whether they are directly created, creatures, is another, non-technical matter.
There is related material in a thread in the Alternate Theory forum:
"ALMA," lookback time to oldest galaxies. .
In there is proposed a model of the universe as seen having moved away from the BB in all directions through time, to now exist in an expanding sphere membrane.
In such a membrane, all the gravity or other force fields would be on the outside, tapering off into infinity the future, that is outside the universe, none would be left in the empty past, even the BB location , the point in timespace where it all started.
startling?
Why the big bang was not the beginning
First hints are emerging of a universe that existed before our own: an alien world of chaos where time, space and geometry were yet to form --quote from New Scientist mag issue 14 March 2018.
perhaps this condition "world of chaos" outside, and before the present universe, before this universe in timespace was susceptible to gravity?
There was even more gravity outside then inside then?
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There is related material in a thread in the Alternate Theory forum:
"ALMA," lookback time to oldest galaxies. .
In there is proposed a model of the universe as seen having moved away from the BB in all directions through time, to now exist in an expanding sphere membrane.
In such a membrane, all the gravity or other force fields would be on the outside, tapering off into infinity the future, that is outside the universe, none would be left in the empty past, even the BB location , the point in timespace where it all started.
startling?
Please don't drag pseudoscience from the Fringe section into the Science section.
Please don't drag pseudoscience from the Fringe section into the Science section.
I thought an item in New Scientist would be worthy enough to be considered here, even though on a topic that has so far escaped heading south.
If you compressed the mass with size R into a smaller and smaller volume, The peak of the surface gravity would rise with decreasing radius, but the original curve, from R to 2R to 3R would remain. The interior gravity pictured by the area of the 0Rgs triangle, would have a sum of the same value, but an additional outside gravity potential would be created, defined by area inclosed by the newly added steeper section of curve . so:
can it be said that there is more gravity outside than inside? more so as the central mass becomes denser?
I doubt that very much. Gravity itself is caused by mass, regardless of size.
What you are talking about is the size of the gravity field becomes larger with greater mass. But the gravity field itself does not add to the mass of the object, it is caused by the object. It is a result of mass.
But i'd like to ask a counter question: If mass causes a warping of spacetime, would it follow that the smaller the size of the massive object, the greater the warping of spacetime. IOW the gravitational well becomes deeper instead of more spread out.
If I visualize a very small neutron star with the same mass as say, Sirius, which is a very large star, what happens to their gravitational fields? Will the neutron star create a smaller, but deeper well , than Sirius, which would create a larger but shallower warping of spacetime?
Either way the gravitational fields might be equal in overall volume, but differ in dimensional shape?
I thought an item in New Scientist would be worthy enough to be considered here, even though on a topic that has so far escaped heading south.
I am talking about your own stuff in there.
I doubt that very much. Gravity itself is caused by mass, regardless of size.
What you are talking about is the size of the gravity field becomes larger with greater mass. But the gravity field itself does not add to the mass of the object, it is caused by the object. It is a result of mass.
But i'd like to ask a counter question: If mass causes a warping of spacetime, would it follow that the smaller the size of the massive object, the greater the warping of spacetime. IOW the gravitational well becomes deeper instead of more spread out.
If I visualize a very small neutron star with the same mass as say, Sirius, which is a very large star, what happens to their gravitational fields? Will the neutron star create a smaller, but deeper well , than Sirius, which would create a larger but shallower warping of spacetime?
Either way the gravitational fields might be equal in overall volume, but differ in dimensional shape?
Here's a the cross sections of the gravity wells of Sirius, compared to the a equal mass neutron star. It is plotted as gravitational potential along the vertical and distance from the center along the horizontal. The black vertical line marks the center for both bodies and the green lines where the surface of Sirius would be.
Note that outside of the green lines, the wells are identical. It is inside the lines where they differ. The well for Sirius begins to level out again as you pass beneath its surface. The well for the neutron star gets steeper and plunges a lot deeper (once below its surface, it will also start to flatten out.)
The slope of the lines are representative of the local field strength (strength of gravity) at that point. Horizontal =0, Vertical would be infinite. For Sirius, it increases as you get closer to the star, but then decreases after you pass below the surface and becomes zero at the center. With the smaller neutron star, gravity continues to increase after you pass the distance equal to the radius of Sirius up until you reach its surface.
Either way the gravitational fields might be equal in overall volume, but differ in dimensional shape?
good points. The OP question was very general. as you said so much depends on the shape of the object associated with the gravitational field, the strength measured everywhere and plotted.
Take the case, discussed somewhere else, of a universe sized model having expanded into an empty outer shell. No gravity inside at all, only on the outside all the forces reaching to infinity.
Even the various graphs on paper do not give the total strength/volume picture, because even at equal value at a distance from the center, the 3 dimensional inner versus outer volume projection shows that gravity is a phenomenon of the perimeter on, and not of the interior.
Here's a the cross sections of the gravity wells of Sirius, compared to the a equal mass neutron star. It is plotted as gravitational potential along the vertical and distance from the center along the horizontal. The black vertical line marks the center for both bodies and the green lines where the surface of Sirius would be.
Thank you for all the graphs.
with due respect, all the curves shown should go back up to the zero level at the top of the intersection of the horizontal line and dotted and vertical black line. Is that not so?, because you said correctly:
but then decreases after you pass below the surface and becomes zero at the center.
bsw, a similar graph was shown as what timespace would look like if a dense subject stopped moving through time, and the rest of the universe kept moving out from the beginning.
Here's a the cross sections of the gravity wells of Sirius, compared to the a equal mass neutron star. It is plotted as gravitational potential along the vertical and distance from the center along the horizontal. The black vertical line marks the center for both bodies and the green lines where the surface of Sirius would be.
View attachment 1885
Note that outside of the green lines, the wells are identical. It is inside the lines where they differ. The well for Sirius begins to level out again as you pass beneath its surface. The well for the neutron star gets steeper and plunges a lot deeper (once below its surface, it will also start to flatten out.)
The slope of the lines are representative of the local field strength (strength of gravity) at that point. Horizontal =0, Vertical would be infinite. For Sirius, it increases as you get closer to the star, but then decreases after you pass below the surface and becomes zero at the center. With the smaller neutron star, gravity continues to increase after you pass the distance equal to the radius of Sirius up until you reach its surface.
Thanks for that clear illustration of identical masses creating the same gravitational force outside of the wells.
But now the logical next question would be what happens to the spacetime fabric inside the wells. From the illustration it would appear that spacetime becomes much more stretched in the neutron star's well, than for Sirius' well.
So the question would be if the volume of each gravitational well is also equal, due to the much larger circumference of Sirius' gravitational well even as the spacetime fabric seems to be less warped than for the neutron star?
Can this be measured? I guess this is basically the same question as *nebel* posed in #97.
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Thank you for all the graphs.
with due respect, all the curves shown should go back up to the zero level at the top of the intersection of the horizontal line and dotted and vertical black line.
No. Gravity has zero net force at the centre, but you are still deep in a gravitational well. If you could drop a probe to the centre of Sirius to take measurements, you would find that time dilation applies. | 2,731 | 13,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.941263 |
http://web.mat.bham.ac.uk/R.W.Kaye/seqser/completeness2.html | 1,582,568,419,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145966.48/warc/CC-MAIN-20200224163216-20200224193216-00187.warc.gz | 142,139,880 | 2,592 | # The Bolzano-Weierstrass Theorem
## 1. Introduction
We have seen the completeness axiom for the reals in the form of the Monotone Convergence Theorem. This web page discusses an application of this for bounded sequences and proves a beautiful result: that every bounded sequence has a convergent subsequence, the Bolzano-Weierstrass theorem.
## 2. The Bolzano-Weierstrass theorem
We have seen that every convergent sequence is bounded, but that not every bounded sequence is convergent. For example, (-1) defines a non-convergent bounded sequence. Thus there is no full converse to the statement that every convergent sequence is bounded. The Bolzano-Weierstrass theorem, to be presented next, gives a partial converse.
Bolzano-Weierstrass Theorem.
Suppose that ( ) is a bounded sequence of real numbers. Then ( ) has a convergent subsequence.
Proof.
Suppose that ( ) is bounded. For the purposes of this proof, say that the th term of the sequence is dominant if for all . Our proof now splits into two cases.
Case 1. Suppose that ( ) has infinitely many dominant terms
1 , 2 , 3 , 4 ,
where 1 2 3 . Then by the definition of dominant we have 1 2 3 and this gives a bounded monotonic nonincreasing subsequence which converges by the monotone convergence theorem.
Case 2. If not, then the sequence ( ) has only finitely many dominant terms. Choose 1 beyond the last dominant term. (So, for example, we may take the last dominant term , and define 1=+1 .) As 1 is not dominant there is 2 1 such that 1 2 , and as 2 is not dominant there is 3 2 such that 2 3 , and so on. Continuing in this way we obtain a monotonic bounded increasing subsequence 1 2 3 which once again converges by the monotone convergence theorem.
The Bolzano-Weierstrass theorem is an important and powerful result related to the so-called compactness of intervals , in the real numbers, and you may well see it discussed further in a course on metric spaces or topological spaces. Note that the completeness of the reals (in the form of the monotone convergence theorem) is an essential ingredient of the proof. We shall apply it in another section of these notes to give an alternative view of the completeness of . | 534 | 2,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-10 | latest | en | 0.914255 |
https://help.precisely.com/r/Data360-Analyze/3.12/en-US/Data360-Analyze-Server-Help/Reference/Data360-Analyze-Script-help/Conditional-operators | 1,701,199,034,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00581.warc.gz | 339,304,417 | 10,378 | # Conditional operators - 3.12
## Data360 Analyze Server Help
Product
Data360 Analyze
Version
3.12
Language
English
Portfolio
Verify
Product family
Data360
Product name
Data360 Analyze
Title
Data360 Analyze Server Help
2023
First publish date
2016
CAUTION:
This topic relates to Data360 Analyze Script which is the language that is used in some superseded nodes. If you are looking for help configuring the Python-based nodes, please see Python scripting.
Conditional operators can be used to identify whether a specified condition is true or false.
See if.
## if
If the predicate evaluates to true, then execute the consequent statement or statements. Otherwise, execute the alternate statement or statements. Null evaluates to false, so if the predicate is null, then the alternate statement or statements will be executed.
Used in the following format, where `predicate` must be a Boolean, `consequent` and `alternate` may be any expression and `consequent statements` and `alternate statements` may be any set of statements:
```if predicate then consequent [else alternate] ```
```if predicate then { consequent statements } [else { alternate statements }]```
```if predicate then { consequent statements } [else alternate] ```
Note: You do not have to specify `then` in `if` statements, it is optional.
Examples
`if false then “hi” else 3.7 # value: 3.7`
`if true then 1 else null # value: 1`
```if ((2 > 5) or (42 >= 21)) then{ “hi” } else 3.7 # value: “hi”```
```if ((42 < 21) and (42 > 84)) then { someVariable = 12 “hi”} else { someVariable = 24 3.7 } # value: 3.7, and someVariable is 24.```
## switch
Compares the test value to each case value in order. If the test value equals a case value, then the value of the result is the value of the switch expression. If no case value matches, the default code block is used. Uses the same comparison criteria as the equals operator.
Used in the following format, where `test`, `case result` and `default` may be any expression:
``` switch test {case result}* default ```
The return value type is the same as a `result` parameter or `default`.
Examples
``````
# value: 3.5
test="hello"
test.switch (
"hi", 1, # case="hi", result=1
2.7, "bye", # case=2.7, result="bye"
"hello", 3.5, # case="hello", result=3.5
null) # default=null``````
See if. | 621 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.640774 |
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## Comment Re:At what speed? (Score 1)722
Depends on where you are driving and how well timed the lights are. Living in a smaller town, beating the timed lights or the delay after a waiting car signals a change can be very beneficial. The worst case is that you wind up moving forward as soon as everyone else. Best case you get an extra green light.
## Comment Re:At what speed? (Score 1)722
As usual a purely physics based approach assumes a lot of variables. It takes me little thought to imagine many problems with this, I live in Oklahoma and this situation isn't rare. The described deceleration is assuming the cubic object is traveling on a friction-less surface that hits a static object at a known rate in an inelastic collision. So many more variables. Vehicle flips, vehicle spins and takes out the 2-3 vehicles behind it, vehicle is bumped into a guard rail or bridge support and has a very very rapid rate of change in velocity. Or a truck is jarred and the shotgun in the back seat shoots too many holes in your assumptions.
Yes, control systems can react much faster than humans; however, the statement "The car behind will apply maximum braking force the very moment a single cycle of it's control loop happens (probably 1/1000 or a second or so)." is almost patently absurd and shows a best a limited understanding of tuning a control loop. Yes, a reaction will happen within a single cycle, and yes 1kHz is reasonable, but assuming maximum braking force is assuming it is only a proportional control that is tuned to be highly reactive. I would challenge anyone to ride in a car that has a purely proportional control that would react as described. You'd sue me for whiplash.
## Comment Re:Nope (Score 1)341
A few things. First, rifles will be used to shoot down a drone; a shotgun has no chance of hitting it. Second, while unsettling, it isn't all that uncommon for hunters to get 'rained' on while bird hunting with falling shot. The pellets lose too much energy when shot up into the air to be lethal.
## Comment Re:No more time travel! (Score 5, Informative)735
"Even though Abrams' last known direct contribution to Lost was the script to the season 3 premiere, "A Tale of Two Cities" (which he co-wrote with Damon Lindelof), and he had stopped being the main driving force behind the direction of the show as early as season 1, instead leaving Lindelof and Carlton Cuse as the showrunners, a considerable part of the (casual) audience still considers Abrams to be the man in charge of the show." http://lostpedia.wikia.com/wiki/J.J._Abrams
## Comment Re:Another idiot buying into the bitcoin scam. (Score 1, Flamebait)347
The US has a vast supply of gold and land backing its currency, at least theoretically it has something backing it. I realize what a fiat currency is, i am just pointing out that it could pay creditors. | 706 | 3,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.95296 |
https://www.chegg.com/homework-help/college-algebra-4th-edition-chapter-5.3-solutions-9780321639394 | 1,556,259,799,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578760477.95/warc/CC-MAIN-20190426053538-20190426075538-00005.warc.gz | 618,681,218 | 29,659 | # College Algebra plus MyMathLab with Pearson eText -- Access Card Package (4th Edition) Edit edition Solutions for Chapter 5.3
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 6
Consider
• Step 2 of 6
Now, find the ordered pairs that are the solutions of the equationby choosing various values forand calculate the corresponding values as follows:
For;
For;
• Step 3 of 6
For;
For;
• Step 4 of 6
For;
For;
• Step 5 of 6
The above calculated values of are shown in the below table:
0 1 1 3 2 9
• Step 6 of 6
Now plot above points and connect them with a smooth curve.
This gives the graph of the equation as shown below:
Corresponding Textbook
College Algebra plus MyMathLab with Pearson eText -- Access Card Package | 4th Edition
9780321639394ISBN-13: 0321639391ISBN:
Alternate ISBN: 9780321830760 | 238 | 873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-18 | latest | en | 0.769771 |
https://www.hackmath.net/en/example/640 | 1,560,692,730,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998238.28/warc/CC-MAIN-20190616122738-20190616144738-00295.warc.gz | 776,813,502 | 7,240 | # Garden
Area of a square garden is 6/4 of triangle garden with sides 56 m, 35 m, and 35 m.
How many meters of fencing need to fence a square garden?
Result
x = 119 m
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Pythagorean theorem is the base for the right triangle calculator. Do you want to convert length units? See also our trigonometric triangle calculator.
## Next similar examples:
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From the right triangle with legs 12 cm and 20 cm we built a square with the same content as the triangle. How long will be side of the square?
2. Square area
Complete the table and then draw each square. Provide exact lengths. Describe any problems you have. Side Length Area .
3. RT 10
Area of right triangle is 84 cm2 and one of its cathethus is a=10 cm. Calculate perimeter of the triangle ABC.
4. Right triangle ABC
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Calculate the length of the arm r of isosceles triangle ABC, with base |AB| = 18 cm and a height v=17 cm.
7. OPT
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8. Base
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9. Square
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Ladder 10 meters long is staying against the wall so that its bottom edge is 6 meters away from the wall. What height reaches ladder?
11. Oil rig
Oil drilling rig is 23 meters height and fix the ropes which ends are 7 meters away from the foot of the tower. How long are these ropes? | 524 | 2,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-26 | latest | en | 0.896681 |
https://www.r-bloggers.com/working-with-factors-in-r-tutorial-forcats-package/ | 1,597,365,330,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739104.67/warc/CC-MAIN-20200813220643-20200814010643-00111.warc.gz | 616,979,376 | 24,784 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Introduction to factors in R
In R Language, factors represent categorical variables. Conceptually, categorical variables take a limited number of different values but can be represented by either character or integer values. Understanding of factors in R language is critical to developing statistical modeling because character variables are treated differently in statistical models than continuous variables. By the end of this tutorial on forcats package for working with factors in R, you will be able to inspect levels, change the order of levels, change the values of levels, combine levels, and add/drop levels more efficiently.
But before that, let us learn a bit more about factors in R.
The first and foremost thing to remember is that a factor variable in R is represented, or you can say stored as a vector of integer values. Here, each integer represents a character value used to display the levels of character values. You can check that by `str()` function. When you check the structure of the data frame, you will realize that all the factor variables are denoted by 1, 2, 1 after the colon(:). Let’s take a look for a better understanding.
```# Checking structure
str(mtcars)
```
```# Output
'data.frame': 32 obs. of 11 variables:
\$ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
\$ cyl : Factor w/ 3 levels "4","6","8": 2 2 1 2 3 2 3 1 1 2 ...
\$ disp: num 160 160 108 258 360 ...
\$ hp : num 110 110 93 110 175 105 245 62 95 123 ...
\$ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
\$ wt : num 2.62 2.88 2.32 3.21 3.44 ...
\$ qsec: num 16.5 17 18.6 19.4 17 ...
\$ vs : num 0 0 1 1 0 1 0 1 1 1 ...
\$ am : num 1 1 1 0 0 0 0 0 0 0 ...
\$ gear: num 4 4 4 3 3 3 3 4 4 4 ...
\$ carb: num 4 4 1 1 2 1 4 2 2 4 ...
```
Second, both numeric and character variables can be converted to factor variable using `as.factor()` or `factor()` function from the forcats package in R.
Third, the levels of factors are always stored as character values. You can check the levels of factor variables using `levels()` function in R.
Fourth, factors R can be either ordered or unordered. Please do not ignore this point; in some analysis or statistical models, the order of the levels may matter.
Now that you are aware of the factors in R. Let’s learn how to execute some of the most frequently used tasks in R involving factors. All the functions mentioned in this tutorial come from `forcats` package in R. The best part of using `forcats` package is that it returns tibble, and that means consistency.
Convert and check levels of factor variables
As mentioned earlier, here, we will use `factor()` function to covert `cyl` variable from `mtcars` data to factor. We will then check the levels of the variable using `levels()` function. Finally, we check the class of the variable, which will validate our third point that levels are represented as characters.
1. Converting an integer variable to factor variable.
```library(forcats)
mtcars\$cyl <- factor(mtcars\$cyl)
class(mtcars\$cyl)
```
```# Output
[1] "factor"
```
1. Check levels of a factor variable
```levels(mtcars\$cyl)
# Output
[1] "4" "6" "8"
```
You can see the output values are represented using inverted quotes confirming that levels are stored as character values.
Inspecting levels of factor variables
Here we will see how to get the count of each level within a factor using `fct_count()`. While we do so, you will also learn how to sort the levels by count using `sort=` argument. We will then learn how to get the unique values, removing duplicates using `fct_unique()` function.
1. Count the number of values
```fct_count(mtcars\$cyl, sort = TRUE)
```
```# Output
# A tibble: 3 x 2
f n
1 8 14
2 4 11
3 6 7
```
1. Remove duplicates to get unique values
```fct_unique(mtcars\$cyl)
```
```# Output
[1] 4 6 8
Levels: 4 6 8
```
Changing the order of levels for a factor variable
There could be multiple reasons for which you would want to change the order of levels in factor variables. As this tutorial is only about `R programming language` and `forcats` package. The why and when do we need to order the levels of factor variables is out of scope. However, we will still discuss the different logical approaches one can take to reorder the factor variable levels.
1. Manually ordering levels of a factor variable
Here the choice is your that is how you wish to reorder the levels. Let’s say you want to reorder levels of `cyl` variable; then, you can use `fct_relevel()` function as illustrated below.
```fct_relevel(mtcars\$cyl, c("8", "4", "6")
```
1. Reorder factor levels based on the appearance in data
The `fct_inorder()` will reorder the levels of a factor variable in R based on the order in which they appear in the data. Below you will notice that 6 appears, then 4 and lastly 8 and so does are factor levels are arranged.
```fct_inorder(mtcars\$cyl)
# Output
[1] 6 6 4 6 8 6 8 4 4 6 6 8 8 8 8 8 8 4 4 4 4 8 8 8 8 4 4 4 8 6
[31] 8 4
Levels: 6 4 8
```
1. Order factor levels based on the frequency
The `fct_infreq()` function from the `forcats` package arranges the levels of a factor based on each level’s frequency. The level with the highest frequency takes the first place, followed by lesser frequent levels. It seems most cars in the dataset have 8 cylinders followed by 4 and 6 cylinders.
```fct_infreq(mtcars\$cyl)
> fct_infreq(mtcars\$cyl)
[1] 6 6 4 6 8 6 8 4 4 6 6 8 8 8 8 8 8 4 4 4 4 8 8 8 8 4 4 4 8 6
[31] 8 4
Levels: 8 4 6
```
1. Reversing the order of levels
If you are interested in reversing the order of the levels of the factors, you can use the `fct_rev()` function. You can see we end up with exact reverse order. If you wish, you can check the original order using `levels()` function.
```fct_rev(mtcars\$cyl)
[1] 6 6 4 6 8 6 8 4 4 6 6 8 8 8 8 8 8 4 4 4 4 8 8 8 8 4 4 4 8 6
[31] 8 4
Levels: 8 6 4
```
1. Reorder factor levels based on the relationship with other variables
I have not used this function so far, but while I was reading through the documentation of the `forcats` package, I found this of interest and thus sharing it with you. The example here is borrowed from the project documentation itself. The function is mostly useful for display purposes.
In the example, we order the levels of the color variable based upon its relationship with the `a` variables. Specifically, we look for minimum values in column `a`, and you will find that the level `blue` has the minimum value of 1 and then `red` has 2 and so on.
The package provides another function called `fct_reorder2()`; the function takes into account the relationship with two variables instead of just one.
```df <- tibble::tribble(
~color, ~a, ~b,
"blue", 1, 2,
"green", 6, 2,
"purple", 3, 3,
"red", 2, 3,
"yellow", 5, 1
)
df\$color <- factor(df\$color)
fct_reorder(df\$color, df\$a, min)
```
Add or drop factor levels in R
The three functions which are important to know from the addition and deletion perspective are
1. fct_expand() – use it to add new level
2. fct_explicit_na() – use it, if you wish to assign NA as one of the levels. This way, when you plot charts, NA’s will also appear.
3. fct_drop() – use it drop a particular level
Below we have code snippets with examples for better understanding.
```# Adding factor level
fct_expand(mtcars\$cyl,"7")
# Converting NA to factor level
f1 <- factor(c(1, 1, NA, NA,2, 2, NA,2, 1, 2, 2))
f2 <- fct_explicit_na(f1, na_level = "(Unknown)")
# Drop factor level
fac1 <- factor(c("aa","bb"),c("aa","bb","cc"))
fac2 <- fct_drop(fac1)
fac2
```
Changing values of factor levels in R
The task of changing the levels of variables can be done in multiple ways. One, you may be interested in manual recording. Two, You may be interested in collapsing the levels into lesser groups. Three, You may be interested in clubbing the least/most common levels into a single level. Fourth, You may just want to keep/drop some levels and rename everything as others.
Below is an illustration of how to achieve the above tasks using the functions from the `forcats` package in R.
1. Use `fct_collapse()` to manually combine levels into defined groups.
Below we collapse 4 and 6 to form another group called others.
```fct_collapse(mtcars\$cyl, Other = c("4", "6"))
```
1. Use `fct_other()` to replace levels that you don’t want to keep to others. You can also mention levels that you want to drop; here, the level mentioned in the `drop=` argument will be named others. The below code produces the exact same results, as mentioned above.
```# Example showing keep as argument
fct_other(mtcars\$cyl, keep = c("8"))
# Example showing drop as argument
fct_other(mtcars\$cyl, drop = c("4", "6"))
```
1. Use `fct_lump()` to group most/least common levels into a single level. The function is very powerful can provides other statistics to be considered as a measure to combine levels. I encourage you to read more about the function using help(fct_lump). Below we reserve the most common n values. This again results in the exact same output as mentioned above.
```fct_lump(mtcars\$cyl, n = 1)
```
We also have different variants of the above function.
• `fct_lump_min()`: lumps levels that appear fewer than min times.
• `fct_lump_prop()`: lumps levels that appear in fewer prop * n times.
• `fct_lump_n()` lumps all levels except for the n most frequent (or least frequent if n < 0)
• `fct_lump_lowfreq()` lumps together the least frequent levels, ensuring that “other” is the smallest.
1. use `fct_recode()` if you wish to replace the values of the levels manually. The other function which you can use to achieve the same task is `fct_relable()`.
Here we kind of rename the levels of the cyl factor variable to cyl4, cyl6, and cyl8.
```fct_recode(mtcars\$cyl, cyl4 = "4", cyl6 = "6", cly8 = "8")
```
The same task can also be achieved using `fct_relable()`. The syntax of the function obeys purrr::map() syntax. The purrr package is an amazing package, and if you have not explored that package yet, I insist that you must. You can find the detailed tutorial on purrr package here.
Combining factors with different levels
Often, we get data from different sources, which can also potentially lead to some information available in one source and some in another. For categorical variables, that means that we may now have to patch together factors from these sources because these should have the same levels and not different. The below image illustrates what we mean by factors coming from different sources. This should help you digest the concept.
1. To combine different levels you can use `fct_c()` function.
```# Creating two factors with different levels
fac1 <- factor("aa")
fac2 <- factor("bb")
fct_c(fac1, fac2)
```
1. To standardize the factor levels across different sources use `fct_unify()`. This is another approach that you can use. Here both lists will have all the levels irrespective of if the value is present or not in the dataset.
```# Creating two factors with different levels
fac1 <- factor("aa")
fac2 <- factor("bb")
fct_unify(list(fac1, fac2))
```
```# Output
[[1]]
[1] aa
Levels: aa bb
[[2]]
[1] bb
Levels: aa bb
```
Notice how we have both the levels in both the vectors mentioned as part of levels.
With this, we come to an end of the tutorial on the forcats package. I hope you find this tutorial of help and start incorporating some of the functionality discussed here.
Happy Learning! | 3,241 | 11,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-34 | latest | en | 0.771154 |
https://www.vedantu.com/question-answer/the-figure-below-is-the-net-of-a-prism-made-up-class-10-maths-cbse-5ee9f60c52e6816718648e6e | 1,620,303,135,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00553.warc.gz | 1,085,123,211 | 83,395 | Question
# The figure below is the net of a prism made up of identical triangles, what is the total area of the prism if the side of the square is 6 cm.[a] 75 sq-cm[b] 84 sq-cm[c] 95 sq-cm[d] 56 sq-cm
Hint: Find the height of the triangles and use the fact that the area of a triangle $=\dfrac{1}{2}bh$ where b is the base and h is the height and the area of a square $={{a}^{2}}$. There are four triangles and one square. So, the net area will be the sum of 4 times the area of the triangle and area of the square.
Since the length of a side of square = 6cm.
Also since the triangles are identical CG = HF
Since the total height of the diagram = 14.
i.e. CG+HF+6 = 14
i.e. 2CG+6 = 14
Subtracting 6 from both sides we get
2CG+6-6 = 14-6
i.e. 2CG = 8
Dividing both sides by 2 we get
CG = 4cm
Since CG = HF, we have
HF = 4 cm.
Now we know that the area of a triangle $=\dfrac{1}{2}bh$
Using we get the area of triangle ABC $=\dfrac{1}{2}\times AB\times CG=\dfrac{1}{2}\times 6\times 4=\dfrac{24}{2}=12$
Also, we know that the area of a square $={{a}^{2}}$
Using, we get the area of square ABED $=A{{D}^{2}}={{6}^{2}}=36$.
Hence the total area of the prism = 4 times the area of the triangle ABC + area of square ABED
$=4\times 12+36=48+36=84$
Hence the total area of faces of prism = 84 sq-cm.
Hence option [b] is correct.
Note: Although the figure above is referred to be of a prism, it is a pyramid. A pyramid has triangular faces, whereas a prism has rectangular faces. The diagram above is of a pyramid with a square base.
Further pyramids have only one base, whereas prisms have two.
A cone is an example of a pyramid with a circular base. | 524 | 1,646 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-21 | longest | en | 0.874514 |
https://www.popsci.com/entertainment-amp-gaming/article/2009-04/new-spin-bowling/ | 1,726,307,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00602.warc.gz | 870,804,514 | 33,199 | # A New Spin on Bowling
The Breakdown takes on the science of strikes
## Share
Enter the two-handed bowler. Increasingly, we are seeing this novel technique cropping up in bowling alleys across the country. Notice the formidable hook you can generate with this type of delivery — it looks like the ball is headed straight for the gutter, but then, seemingly at the last second, it cuts back into the pocket for another strike. It’s this superior hooking ability that makes two-handed bowling a force to be reckoned with. In order to get some insight into the issue, let’s examine some of the physics involved in tossing a 12- to 16-pound sphere down a lane of polished oily wood.
In order to get a strike you probably already know that the ball needs to strike the pins in one of the “pockets”, which are the regions halfway between the head pin and the pins on either side of the head pin. But why do we need to hook the ball at all? Why not just throw it straight up the alley and directly into the pocket? The answer has to do with conservation of momentum.
When the ball impacts the head pin in a glancing collision, it imparts some of its momentum to the pin. Say the ball knocks the head pin forward and to the left. In order for momentum to be conserved, the ball has to rebound (forward) and to the right. If the ball hits the pocket in a straight roll, the rebound tends to knock the ball away from the main concentration of pins, often leaving you with a disconcertingly difficult spare to contend with.
However, if the ball hooks into the pocket at a steep angle, there is less rebound, and the ball finds itself right in the middle of the action. The result: a strike and that uniquely characteristic and aesthetically pleasing sound of bowling pins flying in all directions.
In order to hook a bowling ball, you need to give it spin. Now obviously, simply rolling the ball down the alley will make it end up spinning. However, the spin is around an axis perpendicular to the alley. In order to curve the ball, you need it to also spin around an axis parallel to the alley. You need to give it a side spin.
Now, notice how the ball always seems to hook very late in its roll. This is because bowling lanes are oiled, but only partway up the lane. The lane is dry the last 10 feet or so in front of the pins. The result is that, due to the low friction from the oil, a ball will tend to slide until it gets to the drier part of the lane. At that point there is sufficient friction for rolling to commence. Without side spin, the ball just rolls forward, but if you give the ball a side spin — let’s say you’re a right-handed bowler so you give the ball a spin to the left — as the ball spins it pushes against the lane towards the right. According to Newton’s third law the lane reacts by pushing back on the ball with an equal amount of force to the left. It’s this reaction force that causes the ball to then accelerate in that direction. The faster you can get that ball spinning, the harder it pushes into the lane, and the more hook you’re going to get. And so … enter the two-handed bowler!
Nowadays, they actually make high-performance bowling balls so that they will naturally tend to hook. They do this by incorporating an asymmetrical core into the ball. The old symmetrical plastic balls don’t curve nearly as easily, which makes those hooks in the video doubly impressive, if, as they claim, he’s using a plastic ball.
Adam Weiner is the author of Don’t Try This at Home! The Physics of Hollywood Movies. | 773 | 3,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.937538 |
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