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https://www.storyofmathematics.com/even-odd-identities/
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# Even and Odd Identities – Explanation and Examples
Even and odd identities for trigonometric functions involve using the trig function’s evenness or oddness to find the trig values of negative angles.
Specifically, sine, tangent, cosecant, and cotangent are odd functions. The cosine and secant functions are even. Like all trig identities, the even and odd identities play an important role in physical sciences and engineering.
Before moving forward with this section, review even and odd functions and trig identities.
## Odd Identities
Odd identities are trigonometric identities that stem from the fact that a given trigonometric function is an odd function. Recall that an odd function is a function $f(x)$ such that $f(-x) = -f(x)$. That is, corresponding positive and negative inputs have outputs with the same absolute value. The signs of these outputs, however, will be different.
Reflecting an odd function over the $x$ axis and then over the $y$ axis (or vice versa) maps the function to itself. That is, odd functions are symmetric about the origin.
In trigonometry, the functions sine, cosecant, tangent, and cotangent are odd. This fact gives the four odd identities:
• $sin(-x) = -sin(x)$
• $csc(-x) = -csc(x)$
• $tan(-x) = -tan(x)$
• $cot(-x) = -cot(x)$
The inverse trig functions arcsine and arctangent are also odd. Therefore, there are also two inverse trig odd identities:
• $arcsin(-x) = -arcsin(x)$
• $arctan(-x) = -arctan(x)$
## Even Identities
Even identities in trigonometry are identities that stem from the fact that a given trig function is even. Recall that an even function is a function $f$ such that $f(-x) = f(x)$. That is, corresponding positive and negative inputs have the same output.
Such functions are symmetric about the y-axis, and reflecting them over the y-axis maps them to themselves. There are two even trig functions, cosine and secant. Therefore, there are two trigonometric even identities:
• $cos(-x) = cos(x)$
• $sec(-x) = sec(-x)$
There are no even inverse trig functions.
## How Do You Tell If a Sine Function is Odd or Even?
To tell if a sine function is odd or even, you can employ one of two possible ways: algebraically or graphically. Doing this graphically is easier. If the y-axis is a line of symmetry for the function, then it is even. If the function is symmetric about the origin (rotating it 180 degrees or reflecting it over both axes), then it is odd.
Algebraically, one has to prove that, for any $x$, $f(-x) = f(x)$ for a function to be even and that $f(-x) = -f(x)$ for a function to be odd.
### Algebraic Proofs of Even and Odd
With trigonometric functions, this is done by looking at the basic definitions of sine and cosine in the context of the unit circle.
Recall that a negative angle on the unit circle measures clockwise, while a positive angle measures counterclockwise.
In the unit circle, the sine of an angle is equal to the height of the right triangle formed with the terminal radius and the x-axis. In the figure shown, the sine of the angle $BAD$ is $DI$. The negative angle corresponding to $BAD$ measures clockwise, so it is $BAE$. In the figure, the sine of this angle is $IE$.
Since $DI$ extends upwards from the x-axis, its length is positive. Since $IE$ extends downwards, its length is negative. However, their magnitudes will be the same.
Similarly, the sine of $BAF$ is $FH$, and the sine of $BAG$, the corresponding negative angle, is $HG$. These two lines also have the same magnitude but different sines.
Therefore, the sines of two angles with an equal magnitude but measured in the opposite direction will have the same magnitude but different signs. Therefore, sine is odd.
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# Statistics statistical hypothesis testing and easy
Hypothesis test for a proportion this lesson explains how to conduct a hypothesis test of a proportion, when the following conditions are met: the sampling method is simple random sampling each sample point can result in just two possible outcomes we call one of these outcomes a success and the other, a failure. An appropriately designed solution template for this purpose can ease the difficulties of the learning process suppose that we want to in particular, a small p-value (close to 0) indicates that observation of the value obtained for the test statistics would be unlikely if the null hypothesis (h0) is true accordingly, steps 4 and 5. Practice this lesson yourself on khanacademyorg right now: https://www khanacademyorg/ math/ probability/ probability-and-combinatorics-topic/ decisions -with-p. The t-test the t-test was developed by a chemist working for the guinness brewing company as a simple way to measure the consistent quality of stout it was further developed and adapted, and now refers to any test of a statistical hypothesis in which the statistic being tested for is expected to correspond. Unfortunately, your last statistics class was years ago and you can't quite remember what to do with that data you remember something about a null hypothesis and and alternative, but what's all this about testing sometimes it's easier just to give a problem to the assistant especially when it comes to. The paper deals with statistical hypothesis testing in excel at the faculty of education at trnava university a huge amount of students needs the basic knowledge of mathematical statistics to test hypothesis in their seminar or bachelor works and dissertations the authors describe easily operated excel sheets which enable.
You question deals with inferential statistics that is part of the probably and statistics courses without the next option is the udemy the simplest and easiest course on hypothesis testing this is hypothesis testing builds on other statistical concepts, i recommend classroom - udacity that is from statistics - udacity. Like with most technical concepts, statistical significance is built on a few simple ideas: hypothesis testing, the normal distribution, and p values every time we do a hypothesis test, we need to assume a distribution for the test statistic, which in our case is the average (mean) hours of sleep for our students. It is actually quite easy to do the translation between the everyday problems that anyone in a business seeks answers for, regardless the position, and the language of a p-value is the probability of observing a sample statistic as extreme as the test statistic, assuming that the hypothesis we made is true.
Applying decision rule two tail test reject h0 if: test statistic upper critical value or test statistic upper critical value if null hypothesis gets rejected that means we are able to support our alternative hypothesis. Video created by rice university for the course business applications of hypothesis testing and confidence interval estimation 2000+ courses from we will introduce the various building blocks for the confidence interval such as the t-distribution, the t-statistic, the z-statistic and their various excel formulas we will then. In many statistical tests, you'll want to either reject or support the null hypothesis for elementary statistics students, the term can be a tricky term to grasp, partly because the name “null hypothesis” doesn't make it clear about what the null hypothesis actually is. Bution free property obviously, not every test statistic posses these two characteristics at the same time those examples listed above are either not asymptotically distribution free or not of an omnibus nature in testing simple hypotheses for 1-dimensional continuous distribu- tions, we have a class of distribution free gof.
Simple tests of hypotheses for the non-statistician: statistical tests you cannot gain these insights over night, however, this tutorial provides a basic introduction to the concepts of hypothesis testing, as well as, what you one of the types of information (statistics) that we often want to determine is a measure of centrality. Once these students are thrown into the realm of research, they fail to apply even the basic methods in analytical statistics a researcher works hard and honestly to collect good data but may report the wrong findings because an incorrect statistical test was used for data analysis with this study, we intend. You are here: home blog july 2017 null hypothesis – simple introduction a null hypothesis is a however, we need some exact statement as a starting point for statistical significance testing the null asymmetrical they are symmetrical for most other statistics (such as means or beta coefficients) but not correlations. In addition, for some hypothesis tests, you may need to pass the object from the hypothesis test to the summary function and examine its contents for ttest it's easy to figure out what we want: ttest = ttest(x,y) names(ttest) [1] statistic parameter pvalue confint estimate [6] nullvalue alternative method.
## Statistics statistical hypothesis testing and easy
In elementary statistics courses, students often have difficulty understanding the principles of hypothesis testing this paper discusses a simple yet effective demonstration using playing cards the demonstration has been very useful in teaching basic concepts of hypothesis testing, including formulation of a null hypothesis,.
• A comprehensive guide to statistical hypothesis testing with examples in sas and r when analyzing datasets the statistical test problems in a comprehensive way, making it easy to find and perform an appropriate statistical test formal test problem and the test statistic examples in both sas and r are.
• The test statistic must be known this last point especially needs to be emphasized, because many statistical tests are used (and accepted as fact) even when the assumptions on which they are based are not met) once you get the hang of them, hypothesis testing is deceptively easy, especially with the abundance of.
• Many frequentist methods for hypothesis testing roughly involve the following steps: 1 writing down the hypotheses, notably the null hypothesis, which is the opposite of the hypothesis we want to prove (with a certain degree of confidence) 2 computing a test statistic, a mathematical formula depending on the test type, the.
In the remainder of this introduction, some basic statistical definitions are briefly explained people who are familiar with statistics can choose to skip this part 11 basic statistical definitions in simple hypothesis testing, we always com- pare two hypotheses in most cases, one of these hypotheses is the null hypothesis h0. The choice of alpha (level of significance) is often rather arbitrary a medical doctor might easily argue for a smaller alpha than a behavior scientist results statistical precision can be defined as the reciprocal of the standard error for a given test statistic. By deborah j rumsey part of statistics for dummies cheat sheet you use hypothesis tests to challenge whether some claim about a population is true (for example, a claim that 40 percent of americans own a cellphone) to test a statistical hypothesis, you take a sample, collect data, form a statistic, standardize it to form.
Statistics statistical hypothesis testing and easy
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1. ## Power series help
$\displaystyle f(x)=\frac{1}{4+x^2}$
a)write the first four terms and general term for the power series expansion of f about x=0
b)find the radius of convergence of the series found in part a)
Sorry but i don't really know how to do power series problems... I know
$\displaystyle \sum_{0}^{\infty}C_{n}X^n=C_{0}+C_{1}X+C_{2}X^2...$ but i don't really know what this means. Any Help?
2. Is...
$\displaystyle f(x)= \frac{1}{4}\cdot \frac{1}{1+(\frac{x}{2})^{2}}$ (1)
... so that we can expand f(*) as 'geometric series' ...
$\displaystyle f(x) = \frac{1}{4}\cdot \sum_{n=0}^{\infty} (-1)^{n}\cdot (\frac{x}{2})^{2n}$ (2)
... that converges for $\displaystyle |x|<2$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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Question
1. # If The Serially Arranged Playing Cards Are Shuffled Then Entropy
The serially arranged playing cards are a classic example of entropy in action. In this post, we will explore the phenomenon and how it can be used to our advantage. As we all know, entropy is a natural law that describes the tendency of systems toward disorder. In the world of physical systems, entropy is often associated with randomness and chaos. But there’s more to entropy than meets the eye. In this post, we will explore how entropy can be used to our advantage in business and life. We will look at examples from both fields, and show you how to harness its power for your own success.
## What is entropy?
Entropy is a measure of the disorder or chaos in a system. It is used to characterize the statistical behavior of a system. The more entropy there is in a system, the less predictable its future behavior will be.
## What happens to entropy when a deck of cards is shuffled?
Entropy is a measure of the randomness or disorder in a system. In the context of playing cards, entropy is often used to predict how long it will take for the deck to be completely randomized.
When a deck of cards is shuffled, entropy increases due to the random motion of the cards. The longer it takes for the deck to be completely randomized, the higher the entropy.
## Why is entropy important?
Entropy is important because it is a measure of the randomness or chaos in a system. In physics, entropy describes how much disorder or randomness exists in a system. The greater the entropy, the more random and chaotic the system. Entropy can be thought of as a thermodynamic label for disorder or chaos.
The amount of entropy in a system can change over time due to random events. For example, if you shuffle a deck of cards then let them rest, the order will eventually return to its original state. This process of returning order to an initially disorderly system is called spontaneous symmetry breaking. Spontaneous symmetry breaking happens because the initial distribution of cards is not completely random; there are patterns that emerge over time. As these patterns are disrupted, the overall entropy of the system rises.
Spontaneous symmetry breaking is an important process in physical systems because it allows for new ordered states to form. This phenomenon is what makes shuffling cards so interesting; by randomly rearranging the cards we’re able to create new patterns that weren’t there before. This is why entropy is such an important concept in physics; by understanding it we can better understand how physical systems operate and respond to external stimuli
## Can entropy be reversed?
The answer to this question is a little complicated, but in the end it seems like entropy cannot be reversed. This means that in a closed system, or one where no energy is being added or taken away, the entropy will always increase. The reason for this is that thermodynamics tells us that when we add energy to a system, we create disorder (or entropy). This is because there are more ways for things to be organized than before and not all of these configurations can exist at the same time.
In order to decrease the entropy of a system, we need to remove some of the energy that was added. This can happen in two ways: either by converting some of the energy into heat or by moving the particles around so that they are no longer in their original arrangement. However, once again it seems like reversing entropy is not really possible in a closed system.
## Conclusion
So, entropy. What the heck is that? To put it simply, entropy is a measure of disorder in an environment. It’s what occurs when things start to break down and lose their order. In our everyday lives, this can often be seen as chaos erupting around us – everything from our kitchens to the streets outside seems to be moving at a much slower pace since no one knows exactly where everything is going or what needs to be done next. But despite its sometimes chaotic appearance, entropy ultimately works towards restoring order by reducing opportunities for random events to occur. So if you’re feeling lost or stressed out, know that all of this chaos (and potentially entropy) will work together eventually leading to a more ordered state.
2. Entropy is a concept that has been around for centuries and is an essential element of understanding the nature of the universe. In relation to playing cards, entropy determines how likely it is for cards to be arranged in any given order. If the serially arranged playing cards are shuffled, then entropy increases dramatically as there are no longer any foreseeable patterns in the card order.
The idea of entropy means that if a deck of cards were to be shuffled, then it would be impossible to predict which card will appear next even if someone had total knowledge of previous hands played. This unpredictability and chaos is what increasing the level of entropy does; creating randomness through disorder in a system that was once orderly. The result? An increase in uncertainty and possibilities with every shuffle leading to greater complexity and diversity within each game played.
3. 😃 Have you ever wondered why the game of cards is so captivating? It’s because of its unpredictability! When the serially arranged playing cards are shuffled, entropy increases. Entropy is a measure of randomness and disorder.
Shuffling a deck of cards is a great way to increase entropy and make the game more exciting. It is extremely difficult to predict what card will appear next and this unpredictability makes the game more enjoyable.
Shuffling the cards can be done in a variety of ways. One popular method is called the riffle shuffle, which involves splitting the deck in half and then interweaving the cards back together. This method is commonly used in casinos and is said to be the most efficient way to shuffle cards.
There are also other methods of shuffling cards, such as the overhand shuffle, strip shuffle, and the table shuffle. Each method has its own advantages and disadvantages, but the end result is always increased entropy.
Shuffling cards can also be seen as a metaphor for life. We never know what will happen next, and this unpredictability is what makes life so exciting. Shuffling cards is a great way to introduce a bit of chaos into our lives and remind us that the future is never certain.
So the next time you are playing cards, remember to shuffle the deck and enjoy the increased entropy! 🤩
4. If the serially arranged playing cards are shuffled, the entropy of the deck increases. Entropy is a measure of disorder or randomness in a system. When the cards are arranged in a specific order, such as by suit and number, there is low entropy as there is a predictable pattern. However, when the deck is shuffled, the cards become randomly distributed, increasing the entropy.
Shuffling the cards introduces more uncertainty and makes it difficult to predict the arrangement of the cards. This randomness adds to the overall disorder of the system, resulting in higher entropy. The more thoroughly and randomly the cards are shuffled, the higher the entropy will be.
In summary, shuffling serially arranged playing cards increases their entropy by introducing randomness and disorder into their arrangement.
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1. The specific statistical methods that can be used to summarize or to describe a collection of
data is called:
a) Descriptive statistics
b) Inferential statistics
c) Analytical statistics
d) All of the above
2. The need for inferential statistical methods derives from the need for……………..
a) Population
b) Association
c) Sampling
d) Probability
3. A population, in statistical terms, is the totality of things under consideration. It is the
collection of all values of the…………… that is under study.
a) Instance
b) Variable
c) Amount
d) Measure
4. Non-sampling errors are introduced due to technically faulty observations or during the of
data.
a) Processing
b) Analysis
c) Sequencing
d) Collection
5. Sampling is simply a process of learning about the……………on the basis of a
sample drawn from it.
a) Census
b) Population
c) Group
d) Area
6. Numerical facts are usually subjected to statistical analysis with a view to helping a decision
maker make wise decisions in the face of…………………
a) Interpreting
b) Uncertainty
c) Summarizing
d) Organizing
7. In statistics,……………. classification includes data according to the
time period in which the items under consideration occurred.
a) Chronological
b) Alphabetical
c) Geographical
d) Topological
8. Data is simply the numerical results of any scientific……………….
a) Analysis
b) Researches
c) Observation
d) Measurement
9. The…………… process would be required to ensure that the data is complete and as
required.
a) Tabulation
b) Analysis
c) Editing
d) Ordering
10. A sample is a portion of the ………………. population that is considered for study and
analysis.
a) Selected
b) Total
c) Fixed
d) Random
11. The standard deviation for 15, 22, 27, 11, 9, 21, 14, 9 is:
a) 6.22
b) 6.12
c) 6.04
d) 6.32
12. A student obtained the mean and the standard deviation of 100 observations as 40 and
5.1. It was later found that one observation was wrongly copied as 50, the correct figure
being 40. Find the correct mean and the S.D.
a) Mean = 38.8, S.D = 5
b) Mean = 39.9, S.D. = 5
c) Mean 39.9, S.D = 4
d) None
13. The mean deviation about median from the data: 340, 150, 210, 240, 300, 310, 320 is:
a) 51.6
b) 51.8
c) 52
d) 52.8
## ITAB Multiple Choice Type Questions MCQs Practice Test
ITAB Multiple Choice Type Questions MCQs Practice Test
1. Information is_____________.
A. a collection of data
B. a processed data
C. a text data.
D. a audio/video data.
2. There are two levels of information in every organization and are__________.
A. an internet client and Internet Server.
B. telephone information and voice information.
C. formal and informal information.
D. Internal Information and External Information.
3. ____ is a term that encompasses all forms of technology used to create, store, exchange,
and use information in its various forms.
A. Computer Technology
B. Network Technology
C. Information Technology.
D. Client Server Technology.
4. ____ is an electronic device which converts raw data into meaningful information.
A. Computer.
B. Hardware.
C. Software.
D. Compiler.
5. The processing speed of a computer is generally measured in ___.
A. kg
B. Nano seconds.
C. Milliseconds.
D. hrs.
6. The computers can store large amount of____.
A. data and information
B. numbers and text.
C. personal information.
D. public information
7. The computers give very accurate results with___.
A. hardware.
B. predetermined values.
C. determined values.
D. calculated values.
8. _____ is the utilization of technology to improve the realization of office functions.
A. Office automation
B. Office PC.
C. Office management.
D. Office records.
9. _____ is the processing of raw data by using a computer to performthe selection and
ordering process.
A. Electronic data processing
B. Manual data processing.
C. Low data processing.
D. High data processing.
10. Technically, _____ is a defined structure for efficient communication.
A. networking.
B. communication technology.
C. network technology.
D. computing.
11. Example of Office automation tools are____________.
A. Pencil and Pen.
B. File and Rack.
C. Table and Desk.
D. Electronic Mail and Internet System.
12. Which of the following can work both as an input and output medium?
A. keyboard.
B. trackball.
C. light pen.
D. floppy.
13. The ___ computers operate bymeasuring instead of counting.
A. personal.
B. client.
C. analog.
D. digital.
14. An ____ signal is a continuous variable electromagnetic wave.
A. automatic.
B. analog.
C. electronic.
D. integral.
15. The analog computer operates by___.
A. physical devices.
B. softwares.
C. measuring.
D. scaling.
16. A computer systemis made of____________.
A. hardware only.
B. software only.
C. hardware and software.
D. hardware or software.
17. The most commonly used input device is________________.
A. mouse.
B. scanner.
C. keyboard.
D. joystick.
18. The keys on the keyboard which, do special tasks are__________________.
A. arrow keys.
B. numeric keys.
C. function keys.
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# How do you simplify (y^6)^-3 * y^12?
May 13, 2015
After simplifying you get ${y}^{-} 6$ or $\frac{1}{{y}^{6}}$
To calculate this you use these properties of powers:
1) ${\left({a}^{m}\right)}^{n} = {a}^{m \cdot n}$
and
2) ${a}^{m} \cdot {a}^{n} = {a}^{m + n}$
and
3) ${a}^{-} m = \frac{1}{a} ^ m$
After using 1) for the first part you get ${y}^{-} 18 \cdot {y}^{12}$
Then after using 2) you get ${y}^{-} 6$
Finally to get rid of the negative exponent you can use 3) to get the fraction.
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# Webquest
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### Webquest
1. 1. “Origami; Gives Life to Math” Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
2. 2. Welcome Welcome: ”Origami; Gives Life to Math” Description: This web quest is about how to make origami and learning geometric figures through it. Grade Level: 6-8 Curriculum: Math Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
3. 3. Introduction Are you into Arts? Well, it's so nice to get involved. Why? because it widens your imagination and enhances your skills. In fact, it stimulates your mind. Aside from that, an art keeps you from learning and learning through fun experiences. There are so many forms of arts but I got something beautiful for you to introduce, it's Origami. Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
4. 4. Introduction (2) Origami is the original Japanese paper folding. It uses many techniques to fold paper in many shape such as animals, tree, furniture, etc. To make origami is easy like children used to fold a boat, airplane, flower, crane and they will find geometric figures on origami. Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
5. 5. Introduction (3) In this web quest, students will be able to: 1. identify geometric figures on origami. 2. design an origami step by step. 3. discuss geometric figure to the class by their origami. 4. show uniqueness on making paper folding. Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
6. 6. At first you have to choose one origami that you like most from the provided sites. Then make that origami. Present it to the class. Form the group of 5 and collaborate how many geometric figures you've made from the paper you folded. Present this to the class together with your groups. Task Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
7. 7. Process Step I : Find and Fold Choose one origami from these following sites. Then make one of your own choice. http://www.origami-instructions.com/simple- origami.html http://www.origami-make.com/howto-origami- instructions.php http://www.origamiway.com/easy-origami.shtml http://www.origami-fun.com/origami-instructions.html Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
8. 8. Process (2) Step II: Masterpiece Introduce your origami to the class. This is the site that will help you achieve a better presentation. http://www.wikihow.com/Do-a-Presentation-in-Class Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
9. 9. Process (3) Step III: Group Work Form the group of 5 Use every members origami and unfold it then observe. Gathered and discuss the geometric figures seen in the unfolded origami. This is the site about geometric figures. http://www.mathwords.com/g/geometric_figure.htm http://www.mathsisfun.com/geometry/ Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
10. 10. Process (4) Step IV: Prepare and Present Your group will then present the information you gathered.Each member of the group must present a geometric figure and explain it in the class. Make sure you address: How does origami connect in math? Explain how geometric figures form while making an origami. Essentially, your just making math come to life. Title Introduction Task Process Evaluation Conclusion Teachers Page Credits
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Leveling, Part 1 - - vimore.org
# Leveling, Part 1
Learn how to measure land elevations, part 1 of 5 Provided by the Illinois Professional Land Surveyors Association www.iplsa.org
## Leveling, Part 2
Learn how to measure land elevations, part 2 of 5 Provided by the Illinois Professional Land Surveyors Association www.iplsa.org
## Leveling, Part 3
Learn how to measure land elevations, part 3 of 5 Provided by the Illinois Professional Land Surveyors Association www.iplsa.org
## What is a grade stick for?
In this video, I took a couple minutes to explain the use of a grade stick and how it can quickly and efficiently save you time checking the appropriate grade o
## How To Layout a Building: The Start of a Build Series
In this video I try to show how we layout a building. These same techniques can be used to layout anything... sidewalk, house, garden, fencing... The first th
## Digital Theodolites Part 1
Digital Theodolites Part 1
Terms explained and calculation shown. To do this, you must be able to convert decimal inches and decimal feet to ft-in-fractions and back FAST and be able to d
## Read an engineers` tape measure (in decimal feet)
Prerequisite: Read a metric tape measure, place value in decimal numbers, convert fractions to decimals in head
## Set up total station over a point
This is a step-by-step guide on how to set up your total station or theodolite over a point and accurately level it. Following this procedure is the quickest w
## Leica Sprinter Digital Level - How To Tutorial
http://www.leica-geosystems.com/en/Leica-Sprinter-150M-250M_5284.htm Leica Sprinter Digital Level - How To Tutorial - setup - aim & measure - measure height di
## Theodolite 4 - Vertical and Horizontal measurement
Theodolite 4 - Vertical and Horizontal measurement, OTEN Building and Construction https://www.youtube.com/edit?o=U&video_id=ckR-wBUTUjA
## 3-4-5 Method
Host, Casey Hentges, shows us some garden math to help with layout and design of your home gardens.
Learn how to read a leveling rod marked in decimal feet. Provided by the Illinois Professional Land Surveyors Association www.iplsa.org
## The Engineer Rule explained and demonstrated by Zackary Knudson
Zackary Knudson, a student at Deep Creek Construction School, demonstrates and explains how to use a engineer rule.
## Setting Up the Automatic Level
Learn how to set up and operate an optical level. Provided by the Illinois Professional Land Surveyors Association www.iplsa.org
## How to Use a Laser Level Video
See how to lay out contour lines and figure out slope with a self-leveling laser level.
## Differential Leveling Part 1 of 2
California PE Civil Survey Exam Review The video shows how to determine an elevation at any given point using Differential Leveling. It is related to CA PE Civi
## total station set up, engineering, surveying an layout fuctions
The weapon for engineering how to set up total station, by 20 year old guy with a topocon model and data collector. if yo have any questions feel free to email
## Surveying 3 - Two peg test OTEN Building & Construction
Surveying 3 - Two peg test OTEN Building & Construction http://youtu.be/ij-yIAvChhk
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# Finding integral using Residue Thm
Trying to Find $$\displaystyle \int_{\gamma}\frac{dz}{(z-3)(z^n-1)}$$ where $$\gamma$$ is the circle at origin with radius 2.
So I know $$\displaystyle \int_{\gamma}\frac{dz}{(z-3)(z^n-1)}=2\pi i (res(f,3)+\sum_{p \text{ root of unity} } res(f,p))$$
I did find that first part $$res(f,3)$$ is $$\frac{1}{3^n-1}$$ but I have no idea on how to approach the second part. Can someone help me on this with details?
Thanks!
• Each pole at a root of unity is a simple pole. Use the limit formula + L'Hopital Dec 4, 2022 at 11:42
The Residue Theorem tells you that the integral of $$f$$ along a simple closed curve equals $$2\pi i$$ times the sum of residues at points inside that curve. So you have to get rid of $$\text{Res}(f,3)$$, since $$3$$ lies outside the circle $$\gamma$$.
On the other hand, finding the residues at the roots of unity would be very tedious. Instead, consider $$\frac{1}{z^2}f\left(\frac{1}{z}\right)=\frac{z^{n-1}}{(1-3z)(1-z^n)},$$ and show that $$\text{Res}(f,\infty)=0$$. Since the sum of all residues is zero, it follows that $$\sum_{p\text{ roots of unity}}\text{Res}(f,p)=-\text{Res}(f,3).$$ Therefore, the integral becomes $$\int_\gamma \frac{dz}{(z-3)(z^n-1)}=-2\pi i\text{Res}(f,3).$$
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# What Is The Nth Term For The Sequence 1, 4, 9, 16, 25?
Each number in the sequence is called a term (or sometimes "element" or "member"), read Sequences và Series for a more in-depth discussion.
Bạn đang xem: What is the nth term for the sequence 1, 4, 9, 16, 25?
## Finding Missing Numbers
To find a missing number, first find a Rule behind the Sequence.
Sometimes we can just look at the numbers và see a pattern:
### Example: 1, 4, 9, 16, ?
Answer: they are Squares (12=1, 22=4, 32=9, 42=16, ...)
Rule: xn = n2
Sequence: 1, 4, 9, 16, 25, 36, 49, ...
We can use a Rule to find any term. For example, the 25th term can be found by "plugging in" 25 wherever n is.
x25 = 252 = 625
### Example: 3, 5, 8, 13, 21, ?
After 3 and 5 all the rest are the sum of the two numbers before,
That is 3 + 5 = 8, 5 + 8 = 13 etc, which is part of the Fibonacci Sequence:
3, 5, 8, 13, 21, 34, 55, 89, ...
Which has this Rule:
Rule: xn = xn-1 + xn-2
Now what does xn-1 mean? It means "the previous term" as term number n-1 is 1 less than term number n.
And xn-2 means the term before that one.
Let"s try that Rule for the 6th term:
x6 = x6-1 + x6-2
x6 = x5 + x4
So term 6 equals term 5 plus term 4. We already know term 5 is 21 và term 4 is 13, so:
x6 = 21 + 13 = 34
## Many Rules
One of the troubles with finding "the next number" in a sequence is that mathematics is so powerful we can find more than one Rule that works.
### What is the next number in the sequence 1, 2, 4, 7, ?
Here are three solutions (there can be more!):
Solution 1: địa chỉ cửa hàng 1, then add 2, 3, 4, ...
So, 1+1=2, 2+2=4, 4+3=7, 7+4=11, etc...
Rule: xn = n(n-1)/2 + 1
Sequence: 1, 2, 4, 7, 11, 16, 22, ...
(That rule looks a bit complicated, but it works)
Solution 2: After 1 và 2, địa chỉ the two previous numbers, plus 1:
Rule: xn = xn-1 + xn-2 + 1
Sequence: 1, 2, 4, 7, 12, 20, 33, ...
Solution 3: After 1, 2 and 4, địa chỉ cửa hàng the three previous numbers
Rule: xn = xn-1 + xn-2 + xn-3
Sequence: 1, 2, 4, 7, 13, 24, 44, ...
Xem thêm: Ánh Sáng Truyền Đi Trong Chân Không Với Vận Tốc Gần Bằng Bao Nhiêu ?
So, we have three perfectly reasonable solutions, and they create totally different sequences.
Which is right? They are all right.
And there are other solutions ...
... It may be a danh mục of the winners" numbers ... So the next number could be ... Anything!
## Simplest Rule
When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.
## Finding Differences
Sometimes it helps to lớn find the differences between each pair of numbers ... This can often reveal an underlying pattern.
Here is a simple case:
The differences are always 2, so we can guess that "2n" is part of the answer.
Let us try 2n:
n: 1 2 3 4 5 Terms (xn): 2n: Wrong by:
7 9 11 13 15
2 4 6 8 10
5 5 5 5 5
The last row shows that we are always wrong by 5, so just showroom 5 & we are done:
Rule: xn = 2n + 5
OK, we could have worked out "2n+5" by just playing around with the numbers a bit, but we want a systematic way to vì it, for when the sequences get more complicated.
## Second Differences
In the sequence 1, 2, 4, 7, 11, 16, 22, ... we need to find the differences ...
... Và then find the differences of those (called second differences), lượt thích this:
The second differences in this case are 1.
With second differences we multiply by n22
In our case the difference is 1, so let us try just n22:
n: 1 2 3 4 5 Terms (xn):n22: Wrong by:
1 2 4 7 11
0.5 2 4.5 8 12.5
0.5 0 -0.5 -1 -1.5
We are close, but seem to lớn be drifting by 0.5, so let us try: n22n2
n22n2 Wrong by:
0 1 3 6 10 1 1 1 1 1
Wrong by 1 now, so let us add 1:
n22n2 + 1 Wrong by:
1 2 4 7 11 0 0 0 0 0
We did it!
The formula n22n2 + 1 can be simplified to lớn n(n-1)/2 + 1
So by "trial-and-error" we discovered a rule that works:
Rule: xn = n(n-1)/2 + 1
Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ...
Xem thêm: Ielts Writing: Advantages & Disadvantages Of Living In The City
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# Polar equations
posted by .
write the polar eqaution
3 = r cos (theta - 315)
in rectangular form.
• Polar equations -
3=r cos(a-315)
where theta=a
now 315=360-45=2pi-pi/4=-pi/4
therefore, -315=+pi/4
therefore,3=r cos(a-315)becomes
3=r cos(a+pi/4)
but cos(a+pi/4)= cosa*cospi/4 - sina*sinpi/4
• Polar equations -
3=r cos(a-315)
where theta=a
now 315=360-45=2pi-pi/4=-pi/4
therefore, -315=+pi/4
therefore,3=r cos(a-315)becomes
3=r cos(a+pi/4)
but cos(a+pi/4)= cosa*cospi/4 - sina*sinpi/4
= 1/root2(cosa-sina)
3= 1/root2(rcosa-rsina)
= 1/root2 (x-y)
3root2=x-y
y = x-3root2 is an equation of line
### Related Questions
More Related Questions
Post a New Question
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# Search by Topic
#### Resources tagged with Averages similar to Dopey Measurement:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to Dopey Measurement
Mean. Averages. smartphone. Generalising. Number - generally.
### There are 10 results
Broad Topics > Handling, Processing and Representing Data > Averages
### Unequal Averages
##### Stage: 3 Challenge Level:
Play around with sets of five numbers and see what you can discover about different types of average...
### Searching for Mean(ing)
##### Stage: 3 Challenge Level:
Imagine you have a large supply of 3kg and 8kg weights. How many of each weight would you need for the average (mean) of the weights to be 6kg? What other averages could you have?
### M, M and M
##### Stage: 3 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Wipeout
##### Stage: 3 Challenge Level:
Can you do a little mathematical detective work to figure out which number has been wiped out?
### Top Coach
##### Stage: 3 Challenge Level:
Carry out some time trials and gather some data to help you decide on the best training regime for your rowing crew.
### May the Best Person Win
##### Stage: 1, 2, 3 and 4
How can people be divided into groups fairly for events in the Paralympics, for school sports days, or for subject sets?
### David and Goliath
##### Stage: 4 and 5 Challenge Level:
Does weight confer an advantage to shot putters?
### Reaction Timer
##### Stage: 3 Challenge Level:
This problem offers you two ways to test reactions - use them to investigate your ideas about speeds of reaction.
### How Would You Score It?
##### Stage: 3 Challenge Level:
Invent a scoring system for a 'guess the weight' competition.
### Statistical Shorts
##### Stage: 3 and 4 Challenge Level:
Can you decide whether these short statistical statements are always, sometimes or never true?
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# Degenerate bilinear form
(Redirected from Nondegeneracy)
For other uses, see Degeneracy.
In mathematics, specifically linear algebra, a degenerate bilinear form f(x, y) on a vector space V is a bilinear form such that the map from V to V (the dual space of V) given by v ↦ (xf(x, v)) is not an isomorphism. An equivalent definition when V is finite-dimensional is that it has a non-trivial kernel: there exist some non-zero x in V such that
${\displaystyle f(x,y)=0\,}$ for all ${\displaystyle y\in V.}$
## Non-degenerate forms
A nondegenerate or nonsingular form is one that is not degenerate, meaning that ${\displaystyle v\mapsto (x\mapsto f(x,v))}$ is an isomorphism, or equivalently in finite dimensions, if and only if
${\displaystyle f(x,y)=0\,}$ for all ${\displaystyle y\in V}$ implies that x = 0.
## Using the determinant
If V is finite-dimensional then, relative to some basis for V, a bilinear form is degenerate if and only if the determinant of the associated matrix is zero – if and only if the matrix is singular, and accordingly degenerate forms are also called singular forms. Likewise, a nondegenerate form is one for which the associated matrix is non-singular, and accordingly nondegenerate forms are also referred to as non-singular forms. These statements are independent of the chosen basis.
## Related notions
There is the closely related notion of a unimodular form and a perfect pairing; these agree over fields but not over general rings.
## Examples
The most important examples of nondegenerate forms are inner products and symplectic forms. Symmetric nondegenerate forms are important generalizations of inner products, in that often all that is required is that the map ${\displaystyle V\to V^{*}}$ be an isomorphism, not positivity. For example, a manifold with an inner product structure on its tangent spaces is a Riemannian manifold, while relaxing this to a symmetric nondegenerate form yields a pseudo-Riemannian manifold.
## Infinite dimensions
Note that in an infinite dimensional space, we can have a bilinear form ƒ for which ${\displaystyle v\mapsto (x\mapsto f(x,v))}$ is injective but not surjective. For example, on the space of continuous functions on a closed bounded interval, the form
${\displaystyle f(\phi ,\psi )=\int \psi (x)\phi (x)dx}$
is not surjective: for instance, the Dirac delta functional is in the dual space but not of the required form. On the other hand, this bilinear form satisfies
${\displaystyle f(\phi ,\psi )=0\,}$ for all ${\displaystyle \,\phi }$ implies that ${\displaystyle \psi =0.\,}$
## Terminology
If ƒ vanishes identically on all vectors it is said to be totally degenerate. Given any bilinear form ƒ on V the set of vectors
${\displaystyle \{x\in V\mid f(x,y)=0{\mbox{ for all }}y\in V\}}$
forms a totally degenerate subspace of V. The map ƒ is nondegenerate if and only if this subspace is trivial.
Sometimes the words anisotropic, isotropic and totally isotropic are used for nondegenerate, degenerate and totally degenerate respectively, although definitions of these latter can be a bit ambiguous: a vector ${\displaystyle x\in V}$ such that ${\displaystyle f(x,x)=0}$ is called isotropic for the quadratic form associated with the bilinear form ${\displaystyle f}$, but such vectors can arise even if the bilinear form has no nonzero isotropic vectors.
Geometrically, an isotropic line of the quadratic form corresponds to a point of the associated quadric hypersurface in projective space. Such a line is additionally isotropic for the bilinear form if and only if the corresponding point is a singularity. Hence, over an algebraically closed field, Hilbert's nullstellensatz guarantees that the quadratic form always has isotropic lines, while the bilinear form has them if and only if the surface is singular.
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# Trying to understand why “long blocks between primes” must exist
I'm currently going through "An Introduction to the Theory of Numbers" by Hardy and Wright and at one point, they discuss why the distance from one prime to the next must have a long chunk of composites in between. I'm trying to understand the reasoning but I'm having some difficulty. Here is what they say:
Suppose that $2, 3, \ldots, p$ are the primes upto $p$. Then all the numbers up to $p$ are divisible by one of these primes, and therefore, if $2\cdot3\cdot5 \cdots p = q$ all of the $p-1$ numbers i.e. $q + 2, q+3, q+4 \ldots , q+p$ are composite.
I'm having particular difficulty understanding the $q + 2, q+3, q+4, \ldots , q+p$ part. If $q$ is equal to the product of primes from $2$ to $p$ then how can how can we have all the $p-1$ numbers be equal to $q+2, q+3, q+4, \ldots q + p$?
Can someone please clarify this? or if not, just explain what they are trying to say here.
Thanks a bunch!
• q+2 is divisible by 2, q+3 is divisible by 3, q+4 is divisible by 2 again, q+5 is divisible by 5,.... – Will Jagy Jan 15 '14 at 2:42
• @WillJagy I understand what you have said there but the main confusion I'm having is how can you have $p-1$ numbers be equal to $q+2, q+3, \text{etc}$ – Jeel Shah Jan 15 '14 at 2:44
• What does "$p-1$ numbers be equal to $q+2,q+3$ etc." mean? – anon Jan 15 '14 at 2:45
• Huh? ${}{}{}{}$ – anon Jan 15 '14 at 2:50
• Yes: the numbers $q+2,\cdots,q+p$ (of which there are $p-1$) are composite. That is what the passage says. By the way, here is the original text. (I cannot figure out how could believe the text is saying $q+2$ is less than $p-1$ from this!) – anon Jan 15 '14 at 3:09
Let's translate the case $p=7$:
Suppose that $2,3,5,7$ are the primes up to $7$. Then all the numbers up to $7$ are divisible by one of these primes, and therefore, if $2\cdot3\cdot5\cdot7=210$ all of the six numbers $212, 213, 214, 215, 216, 217$ are composite.
Hopefully, the above made sense. The authors are talking about the numbers $212, 213, 214, 215, 216,$ and $217$. There are six numbers listed, because $7-1=6$.
Grammatically, the clause "$p-1$" is being used to modify the plural noun "numbers". The authors are not equating, or comparing, any other variables with $p-1$.
• This makes a $\large{A\,LOT}$ more sense. Thanks a bunch! – Jeel Shah Jan 15 '14 at 3:07
They are saying that
$q+2$ is divisible by $2.$
$q+3$ is divisible by $3.$
$q + 4$ is divisible by $2.$
And so on.
• @anon I didn't read the whole post, I was just doing a slightly naïve question just in case. Reading it, I delete my comment. – Pedro Tamaroff Jan 15 '14 at 3:13
Let $n > 0$ then $n! + 2, .... n! + n$ are all non-prime. You can create a nonprime streak of arbitrary length using this process.
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# trig question
NEW DIAGRAM:
On a unit circle, sin x = square root 3 over 2, with center O.
Vertical line from top to center of circle is PO.
Horizontal line from middle to right of circle is OE.
Another vertical line of AB is parallel to PO.
And radius of circle (line from O to A) is 1 unit.
From A to E is the arc.
From E to B is another arc.
) is the angle 60 degree
Fine the area of the shaded region (where the ::::: are).
Sorry it's a bad diagram, can't really draw a circle on here..
..............P
..............|.........A
..............|....... "|:
..............|......"..|::
..............|...."....|::.
..............|.." )....|:::
..----------- O --------+--- E
..............|.."......|:::
..............|...."....|::'
..............|......"..|::
..............|........"|:
..............|.........B
1. a 30,60 , 90 triangle has sides
1 =AO, sqrt 3 = EA and hypotenuse 2
AB = 2 sqrt 3
your angle AOB is 2*60 = 120
OE the altitude is 1
Your area = (1/2)(1)(2 sqrt 3)
= sqrt 3
posted by Damon
2. sorry 1 = EO, the side along the x axis
posted by Damon
3. But the question ask to find the area of AE and EB together
posted by Anonymous
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# Forays into 3D geometry
Originally posted 2019-01-10
I’ve been trying to understand this paper and as part of this process, realized that I should learn more geometry, algebra, and group theory. So I spent a week digging into the math of 3D geometry and here’s what I’ve learned so far.
I stumbled on these concepts in a highly nonlinear fashion, so my notes are going to be a bit scattered. I don’t know how useful they’ll be to other people - probably, they won’t be? This is more for my future self.
# Mappings
A mapping is a correspondence between two domains. For example, the exponential function maps real numbers to positive real numbers. Furthermore, the group operation of multiplication is transformed into addition.
$x + y = z \Leftrightarrow e^xe^y = e^{x+y} = e^z$
This is a useful transformation to make because addition is easier to think about than multiplication. In this case, the mapping is also bijective, meaning that one can losslessly convert back and forth between the two domains.
One common pattern is that you can transform your numbers into the domain that’s easier to reason about, do your computation there, and then convert back afterwards. Eventually, you get annoyed with converting back and forth, and you start reasoning entirely in the transformed domain. This happens, for example, in converting between time/frequency domains for sound using the Fourier transform - everyone thinks in the frequency domain even though the raw data always comes in the time domain.
# The circle group
The circle group consists of the set of all complex numbers with modulus 1 - in other words, the unit circle on the complex plane. It’s a pretty simple group to understand, but it shows how we’re going to try and attack 3D rotations. There are multiple ways to look at this.
• You can represent this group using rotation $$\theta$$ from the x-axis (i.e. polar coordinates), using addition.
• You can represent this group using complex numbers, under multiplication.
• You can work with the matrices of the following form, under multiplication. (This form is convenient because multiplication by this matrix is equivalent to rotating a vector by $$\theta$$.)
$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}$
All of these representations are tied together by Euler’s formula, which states that $$e^{i\theta} = \cos \theta + i\sin \theta$$.
Somewhat surprisingly, Euler’s formula also works if you consider the exponentation to be a matrix exponentiation, and you use the matrix representation of complex numbers.
$\begin{gather} \exp\left( \begin{bmatrix} 0 & -\theta \\ \theta & 0 \\ \end{bmatrix}\right) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix} \end{gather}$
# Matrix exponentiation
It turns out that subbing in matrix exponentiation for regular exponentiation basically works most of the time, and additionally has some other surprising properties.
• The determinant of a real matrix’s exponentiation is strictly positive and the result is therefore invertible. (This is analogous to $$e^x > 0$$ for real $$x$$).
• It commutes with transposition, so that $$e^{\left(M^T\right)} = \left(e^M\right)^T$$
There are a few things that don’t carry over seamlessly, related to the fact that matrix multiplication isn’t commutative.
• If $$X$$ and $$Y$$ commute ($$XY = YX$$), then $$e^{X + Y} = e^Xe^Y$$.
Answering the question of what happens to $$e^{X+Y}$$ when $$X$$ and $$Y$$ don’t commute leads to a rabbit hole of gnarly algebra.
## Skew-symmetric matrices
A skew-symmetric matrix is a matrix whose transpose is equal to its negation: $$A^T = -A$$. It turns out that all matrices can be broken down into a symmetric matrix plus a skew-symmetric matrix, by defining $$B = \frac{1}{2}(A + A^T)$$ to be the symmetrical part of $$A$$, and then realizing that what’s left over, $$A - B$$, is skew-symmetric.
Symmetric matrices over real numbers, by the spectral theorem, can be represented as a diagonal matrix in some basis. In plain English, this means that symmetric matrices correspond to transformations that are purely “rescaling” along some set of axes, with no rotations.
So that means that the skew-symmetric remainder of a matrix probably corresponds to the rotational part of the transformation. This seems to be related to the fact that a skew-symmetric matrix’s eigenvalues are all purely imaginary, but I don’t really fully understand the connection here.
The more literal connection might be that when you exponentiate a skew symmetric matrix, you get an orthogonal matrix (a matrix for which $$MM^T = I$$. The proof is pretty simple:
$e^A\left(e^A\right)^T = e^Ae^{A^T} = e^{A + A^T} = e^{A - A} = e^0 = I$
## Orthogonal matrices
You may remember orthogonal matrices as those transformations that preserve distances - aka rotations and inversions (“improper” rotations). SO(3) consists of the orthogonal matrices of determinant +1, thus excluding inversions. The exponential map is surjective - for every element of SO(3), there exists a skew-symmetric matrix that exponentiates to it. (It’s not a bijection as far as I can tell, unlike in the 2-D case.)
# Mapping between SO(3) and skew-symmetric matrices
Earlier, we looked at the circle group, which was a toy example showing that we could map 2-D rotations between {multiplication of 2-D matrices} and {addition over plain angles}. Now, to understand 3-D rotations, we’ll try to map between {multiplication of 3-D matrices} and {addition of skew-symmetric matrices}.
It turns out that actually this doesn’t work with finite rotation matrices. So we’ll just brush the problem under the rug by invoking a “tangent space” around the identity, which means the space of infinitesimal rotations. This space is represented by lowercase so(3) and has an orthogonal basis set which I’ll call $$\{s_1, s_2, s_3\}$$ with concrete representations of
$\begin{gather} s_1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & \alpha \\ 0 & -\alpha & 0 \\ \end{bmatrix} \quad s_2 = \begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0 \\ -\alpha & 0 & 0 \\ \end{bmatrix} \quad s_3 = \begin{bmatrix} 0 & \alpha & 0 \\ -\alpha & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \end{gather}$
These skew-symmetric matrices are not themselves rotations; they’re derivatives. To make them rotations, you have to exponentiate them, which turns out to be equivalent to adding them to the identity matrix: $$e^{s_i} = I + s_i$$. This is analogous to the real numbers, where $$e^x = 1 + x$$ for $$x \approx 0$$.
Since $$\alpha$$ is considered to be an infinitesimal, raising $$(I + s_i)^k$$ to a power $$k$$ just results in the matrix $$(I + ks_i)$$ because all second order terms disappear. Also, addition within so(3) corresponds to multiplication in the exponential map. $$m_i + m_j = m_k \Leftrightarrow e^{m_i}e^{m_j} = e^{m_i + m_j} = e^{m_k}$$ for arbitrary $$m \in so(3)$$. So this is nice; we’ve got something resembling the circle group for 3 dimensions. Unfortunately, this only works for infinitesimal rotations and completely falls apart for finite rotations.
I then stumbled across this monstrosity of a formula, which takes these infinitesimal rotations of so(3) and shows how to map them back to the normal rotations of SO(3). It also answers the question of what happens to $$e^{X+Y}$$ if $$X$$ and $$Y$$ don’t commute.
If you squint hard enough, it looks like a Taylor series expansion, in the sense that a Taylor series shows how to take the local derivative information (aka this tangent space business), and use that to extrapolate to the entire function. I can’t imagine anyone actually using this formula in practice, but a quantum information friend of mine says he uses this all the time.
# SU(2) and Quaternions
At this point, I was trying to find a more computationally insightful or useful way to approach finite rotations. As it turns out, SO(3) is very closely related to SU(2), the set of unitary 2x2 matrices, as well as to the quaternions.
The best intuition I had was the Wikipedia segment describing the topology of SO(3). If that’s the topology of SO(3), then SU(2) can be thought of as not just the unit sphere, but the entire space, using a projective geometry as described in these 3blue1brown videos. Since the unit sphere representing SO(3) is only half of the space and has funny connectivity, that explains all of this “twice covering” and “you have to spin 720 to get back to where you started” business.
Computationally speaking, I found the 3blue1brown videos very enlightening. In short: the $$(i,j,k)$$ component determines the axis of rotation, and the balance between the real component and the imaginary components determines the degree of rotation. This ends up being basically the topological description of SO(3) given by Wikipedia, with the additional restriction that the real component should remain positive to stay in SO(3).
## Side note: Lie groups and algebras
Lie groups are groups that have a continuous transformation (i.e. the rotation stuff we’ve been talking about). SO(3), SU(2), and quaternions of unit norm can be considered different Lie groups but they all have the same local structure when you zoom in on their tangent space at the origin (their Lie algebra). (More here). Mathematicians like to categorize things, so they don’t particularly care about computing rotations; they just want to be able to show that two algebras must be the same. There’s some topological connection; since SU(2) is simply connected (aka none of this ‘identify opposite points on the sphere’ business), this somehow implies that it must be a universal cover of all Lie groups with the same Lie algebra.
# Geometric algebras
Ultimately, I found that the physicists’ and mathematicians’ account of 3D rotations basically talked past each other and I didn’t walk away with much insight on algebraic structure. I think the quaternions came closest; since the application of quaternions is done as $$qxq^{-1}$$, it implies that simply multiplying quaternions is enough to get the composed rotation.
I happened to stumble upon Geometric Algebras, whose introductory tome can be found in this lecture by Hestenes in 2002. So far it looks like it will deliver on its ambitious goal, claiming that “conventional treatments employ an awkward mixture of vector, matrix and spinor or quaternion methods… GA provides a unified, coordinate-free treatment of rotations and reflections”.
I can’t really explain this stuff any better than it’s been presented in Hestenes’s lecture, so you should go look at that. I found that understanding GA was made much simpler by knowing all of this other stuff.
So that’s roughly where I ended up. And my Christmas break is over, so I guess I’ll pick this up some other day.
Thanks to the many people I bugged about this stuff over the past week or two.
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# `f(x) = x^4 - 2x^3 + x^2, [0,6]` Use a graphing utility to (a) graph the function `f` on the given interval, (b) find and graph the secant line through points on the graph of `f` at the...
`f(x) = x^4 - 2x^3 + x^2, [0,6]` Use a graphing utility to (a) graph the function `f` on the given interval, (b) find and graph the secant line through points on the graph of `f` at the endpoints of the given interval, and (c) find and graph any tangent lines to the graph of `f` that are parallel to the secant line.
embizze | Certified Educator
Given `f(x)=x^4-2x^3+x^2 ` on the interval [0,6]:
(1) This is a quartic polynomial with positive leading coefficient so its end behavior is the same as a parabola opening up.
(2) f(0)=0 and f(6)=900. Since the function is everywhere continuous and infinitely differentiable everywhere, the Mean Value theorem guarantees the existence of a c in the interval such that the slope of the tangent line at c is the same as the slope of the secant line through the endpoints of the interval.
The slope of the secant line: `m=(900-0)/(6-0)=150 `
The equation of the secant line is y=150x
(3) The derivative of f is ` 4x^3-6x^2+2x ` . We set this equal to 150:
`x~~3.8721 ` so `y~~123.678 ` and the equation of the tangent line is:
`y-123.678=150(x-3.8721) `
The graph of the function, the secant line, and the tangent line:
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### Take Ten
##### Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Trice
##### Stage: 3 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Shaping the Universe II - the Solar System
##### Stage: 3 and 4
The second in a series of articles on visualising and modelling shapes in the history of astronomy.
### Triangles to Tetrahedra
##### Stage: 3 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Cutting a Cube
##### Stage: 3 Challenge Level:
A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical?
### Wari
##### Stage: 4 Challenge Level:
This is a simple version of an ancient game played all over the world. It is also called Mancala. What tactics will increase your chances of winning?
### The Spider and the Fly
##### Stage: 4 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Sea Defences
##### Stage: 2 and 3 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Three Cubes
##### Stage: 4 Challenge Level:
Can you work out the dimensions of the three cubes?
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### Speeding Boats
##### Stage: 4 Challenge Level:
Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling?
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Triangles in the Middle
##### Stage: 3, 4 and 5 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### The Perforated Cube
##### Stage: 4 Challenge Level:
A cube is made from smaller cubes, 5 by 5 by 5, then some of those cubes are removed. Can you make the specified shapes, and what is the most and least number of cubes required ?
### Getting an Angle
##### Stage: 3 Challenge Level:
How can you make an angle of 60 degrees by folding a sheet of paper twice?
### Baravelle
##### Stage: 2, 3 and 4 Challenge Level:
What can you see? What do you notice? What questions can you ask?
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Making Tracks
##### Stage: 4 Challenge Level:
A bicycle passes along a path and leaves some tracks. Is it possible to say which track was made by the front wheel and which by the back wheel?
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Percentages, Fractions, and Decimal Values
# What is twelve and half percent of 114?
123
###### 2012-10-20 05:52:11
twelve and half percent of 114 = 14.25
12 1/2% of 114
= 12.5% * 114
= 0.125 * 114
= 14.25
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## Related Questions
1250%twelve & a half= 12 1/2= 12.5= 12.5 * 100%= 1250%
10 percent of eighty is 8... 2.5 percent of eight is 2.... 8 + 2 = 10.
27% of 114= 27% * 114= 0.27 * 114= 30.78
34% of 114= 34% * 114= 0.34 * 114= 38.76
2% of 114 = 2% * 114 = 0.02 * 114 = 2.28
12% of 114= 12% * 114= 0.12 * 114= 13.68
10% = £35% = £1.501% = 30p1/2% = 15p3x2=60So £3.00+60p+15p=£3.75Twelve and a half percent of £30 is £3.75
0.125 (this also happens to be 1/8 as a fraction)
12.5% Answer 2 It is one-eighth. The above answer is also correct.
114 is what percent of 30:= 114 / 30= 3.8Converting decimal to a percentage:3.8 * 100 = 380%
98/114 = .8596 Hence, 98 is about 86% of 114.
To convert 114% to a decimal divide by 100: 114% ÷ 100 = 1.14
15% off 114 = 114- (0.15 x 114) = 96.9
twelve out of what? that is such a stupid question
Not quite. . . 13 percent is 13/100 or 0.13 one eighth is 12.5 over 100 which is 0.125 or twelve and a half per cent. The answer is that one eighth is 12 1/2 percent, that is half a percent less than 13 percent
###### Percentages, Fractions, and Decimal ValuesMath and ArithmeticShopping
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# MTH107 Determine limit (t- > 2) t ^ 3 – 8 / t ^ 2 – 4: Calculus I Tutor Marked Assignment, Singapore
University Singapore University of Social Science (SUSS) Subject MTH107 Calculus I
MTH107 Calculus I Tutor-Marked Assignment
Question 1
(a) Determine limit (t- > 2) t ^ 3 – 8 / t ^ 2 – 4.
(b) The inequality 1 – (x^2 / 6)< sin x / x < 1 holds for x is measured in radians and |x| < 1. Use the inequality to find the limit(x->0) sin x / x.
(c) Evaluate limit(x->0) x (e^x – 1) / 1 – cos2x.
Question 2
(a) Evaluate limit(x->0) x*e^x-log(1+x)/x^2.
(b) Find the values of the constants a and b such that limit(x->0) x(1+a*cosx)-b*sinx/x^2=1.
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Question 3
(a) Consider the y = f(x) defined by f(x) = { x* sin(1/x) if x is not = 0; 0 if x is = 0
Is the function continuous at x = 0? Explain your answer.
(b). Find the value of the constant a to make the function f(x)= { x^2 – 1 x < 3; 2ax x>=3 continuous.
(c). Consider the function f(x) = { x 0<=x<1; 2 – x 1<x<=2. Does f have a derivative at x =1? Explain your answer.
Question 4
(a) Explain with reasons why the equation x^3+3x+1=0 has exactly one real root.
(b) Demonstrate that for any two real numbers a, b ∈ R |sinb – sina| <= |b-a|.
(c) State the conditions for the Rolle’s theorem and verify that the conditions are satisfied for the function f = sin x / e^x (in 0, pi).
Hence verify the conclusion of the theorem.
Question 5
(a) Find the Taylor series expansion of the function loge x about the point a =1. Use the series to evaluate the limit, limit x->1 ln x/ x-1.
(b) A box with a square base and open top are to hold 32 cube centimetre of a liquid. Find the dimensions that require the least amount of material, neglecting the thickness of the material, of the material and the waste in the construction.
Get Help By Expert
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# Statistics Ch 13
The flashcards below were created by user firefly501 on FreezingBlue Flashcards.
1. Introduction to Inference
• Estimates of sample mean for samples are always relatively close to the populationparameter μ.
• 9 5% of all sample means will be within roughly 2 standard deviations (2*s/√n) of the population parameter μ.
2. A level C confidence interval for a parameter has two parts:
• (1) An interval calculated from the data, usually of the form
• estimate ± margin of error
(2) A confidence level C, which gives the probability that the interval will capture the true parameter value in repeated samples, or the success rate for the method.
3. confidence interval
for a population mean can be expressed from a sample mean as:
4. confidence level C
(in %) indicates the success rate of the method that produces the interval. It represents the area under the normal curve within ± m of the center of the curve.
5. (in %)
confidence level C
6. Standardizing the normal curve using z
Ex. For a 98% confidence level, z*=2.326
7. Hypotheses tests
• 1. test of statistical significance
• 2. null hypothesis
• 3. alternative hypothesis
8. test of statistical significance
• tests a specific hypothesis using sample data to decide on the
• validity of the hypothesis.
9. hypothesis
is an assumption, or a theory about the characteristics of one or more variables in one or more populations
10. null hypothesis
is the statement being tested. It is a statement of “no effect” or “no difference,” and it is labeled H0.
11. alternative hypothesis
is the claim we are trying to find evidence for, and it is labeled Ha.
12. two-tail or two-sided test
• of the population mean has these null and alternative hypotheses:
• H0: μ = [a specific number] Ha: μ Å [a specific number]
13. one-tail or one-sided test
• of a population mean has these null and alternative hypotheses:
• H0: μ = [a specific number] Ha: μ < [a specific number] OR
• H0: μ = [a specific number] Ha: μ > [a specific number]
14. Tests for a population mean
To test the hypothesis H0: μ = μ0 based on an SRS of size n from a Normal population with unknown mean μ and known standard deviation σ, we rely on the properties of the sampling distribution N(μ, σ√n).
15. p-value
is the area under the sampling distribution for values at least as extreme, in the direction of Ha, as that of our random sample.
16. The P-value
• Tests of statistical significance quantify the chance of obtaining a particular random sample result if the null hypothesis were true. This quantity is the P-value.
• This is a way of assessing the “believability” of the null hypothesis given the evidence provided by a random sample.
17. P-value in one-sided and two-sided tests
To calculate the P-value for a two-sided test, use the symmetry of the normal curve. Find the P-value for a one-sided test and double it.
18. Interpreting a P-value
Could random variation alone account for the difference between the null hypothesis and observations from a random sample?
• A small P-value implies that random variation because of the sampling process alone is not likely to account for the observed difference.
• With a small P-value, we reject H0. The true property of the population is significantly different from what was stated in H0.
Thus small P-values are strong evidence AGAINST H0.
Oftentimes, a P-value of 0.05 or less is considered significant: The phenomenon observed is unlikely to be entirely due to chance event from the random sampling.
19. The significance level a
• The significance level, α, is the largest P-value tolerated for rejecting a true null hypothesis (how much evidence against H0 we require). This value is decided arbitrarily before
• conducting the test.
If the P-value is equal to or less than α (p ≤ α), then we reject Ho. If the P-value is greater than α (p > α), then we fail to reject Ho.
When the z score falls within the rejection region (shaded area on the tail-side), the P-value is smaller than α and you have shown statistical significance.
• Rejection region for a two-tail test of μ with α = 0.05 (5%)
• A two-sided test means that α is spread between both tails of the curve, thus:
• a middle area C of 1 − α = 95%, and
• an upper tail area of α /2 = 0.025
20. Confidence intervals to test hypotheses
• Because a two-sided test is symmetrical, you can also use a confidence interval to test a twosided
• hypothesis.
• In a two-sided test, C = 1 – α.
• C confidence level
• α significance level
21. Logic of confidence interval test
• A confidence interval gives a black and white answer: Reject or don’t reject H0. But it also estimates a range of likely values for the true population mean μ.
• A P-value quantifies how strong the evidence is against the H0. But if you reject Ho, it doesn’t provide any information about the true population mean μ.
Author: firefly501 ID: 137617 Card Set: Statistics Ch 13 Updated: 2012-02-25 18:12:43 Tags: statistics Folders: Description: Introduction to Inference Show Answers:
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# 9+ Missing Numbers On Number Line Worksheets
What is the missing number in pattern?
What are the Missing Numbers? Missing numbers are the numbers that have been missed in the given series of a number with similar differences among them. The method of writing the missing numbers is stated as finding similar changes between those numbers and filling the missing terms in the specific series and places.
How do you write missing numbers?
How do you find the number on a number line?
## What is a number line to 20?
A Number Line to 20 is a series of numbers from 0-20 that are placed on a line. Number lines are great visual resources for visual learners and show children each number in that sequence. Count forwards and backwards with a number line to really heighten learning.
## What is the missing number in this sequence 88511?
What is the missing number in this sequence 88511 16351? Answer. The missing number is 73155 .
## Where would 1.5 be on a number line?
Where would the decimal number 1.5 be on a number line? The decimal number 1.5 has a whole number of 1, which is the number to the left of the decimal. The number 1.5 is halfway between 1 and 2.
## Where would 2/3 Be on a number line?
On the number line 2/3 is two-thirds of the way between 0 and 1. Since the numerator 2 is less than the denominator 3 this means that 2/3 is less than 1. Thus, 2/3 will be placed on the number line between zero and one.
## What is the missing number 234 23?
Heya!!! The missing no.is 4 .
## What is a number line example?
The definition of a number line is a straight line with a "zero" point in the middle, with positive and negative numbers listed on either side of zero and going on indefinitely. An example of a number line is what a math student can use to find the answer to addition and subtraction questions.
## What is a ten frame?
Ten-Frames are two-by-five rectangular frames into which counters are placed to illustrate numbers less than or equal to ten, and are therefore very useful devices for developing number sense within the context of ten. The use of ten-frames was developed by researchers such as Van de Walle (1988) and Bobis (1988).
## What is a vertical number line?
The vertical number line is called the y-axis. The point where the x- and y-axes meet is the origin.
## How do I make a number line in Word?
• Select the Double Arrow line.
• Create a horizontal line.
• Copy and paste the tick marks.
• Select all the tick marks.
• Top-align the tick marks.
• Drag the right-most tick mark to a point at the end of the line.
• Realign the tick marks.
• ## How do you solve this math problem?
• Read carefully, understand, and identify the type of problem.
• Draw and review your problem.
• Develop the plan to solve it.
• Solve the problem.
• ## How do you find the missing value?
Generally we add up all the values and then divide by the number of values. In this case, working backwards, we multiply by the number of values (instead of dividing) and then subtract (instead of adding).
## What is the value and value of 3?
3 is in thousands place and its place value is 3,000, 5 is in hundreds place and its place value is 500, 4 is in tens place and its place value is 40, 8 is in ones place and its place value is 8.
## What is the sequence of 131517 123?
The missing term would be 192.
## Can you figure out the missing number in this sequence type your answer below 88511 16351?
Answer: 73155 is the missing number in the above term .
## Where is 1.9 on the number line?
Solution : Firstly, it is seen that 1.9 is located between 1 and 2. The block between 1 and 2 is divided into ten equal parts. 1.9 is marked on the number line.
## Where would 5/3 Be on a number line?
5/3 is the same thing as 1 and 2/3, or, if you divide the top number by the bottom number, you get 1.66. Now, simply find the spot on the left side of the number line (because this is a negative number) that is at about 1 and 2/3 it is 2/3 of the way between negative 1 & 2.
## Where would 3/4 Be on a number line?
It's the top number. This means that the fraction 3/4 is 3 out of 4 equal parts. So, count 3 parts from 0 and you will get 3/4. That's it!
## Where is 7/2 on a number line?
so on Number line,-7/2 is placed in middle of -3 and -4 in left side of zero,because it is a negative Number.
## Which is the smallest number in this set of numbers?
In the set of whole numbers, 0 is the smallest number.
## Where should a line be drawn to make this equation correct 5 5 5 555?
Strike a line on the first "+" and make it "4". this would make it 545+5+5=555!!
## What do you get when you divide 30 by 1 2 and add 10 Kindly answer with words and not numbers?
When we divide 30 by 1/2 , we get 60. On adding 10 to 60, we get 70. The answer is 70.
## What can you see once in a minute twice in a moment?
we can see the letter 'M' once in minute and twice in moment. Henxe, required answer is the letter M.
## How do you teach the missing numbers to kindergarten?
use something concrete, such as pennies, to practice counting. Instead of starting with one penny, start with four. Have your child tell you what number comes next when you add one more penny. Once you feel like your child is ready to practice what numbers come before and after a number, you can try this activity!
## What is a number line 3rd grade?
In math, a number line can be defined as a straight line with numbers placed at equal intervals or segments along its length. A number line can be extended infinitely in any direction and is usually represented horizontally.
## What number is 5 more than3?
Answer: 5 is greater than -3 . Step-by-step explanation: because 5 is a positive integer and -3 is a negative integer.
## How does a number line look like?
A number line is just that – a straight, horizontal line with numbers placed at even increments along the length. It's not a ruler, so the space between each number doesn't matter, but the numbers included on the line determine how it's meant to be used. A number ladder is the vertical version of a number line.
## What is a number line kindergarten?
A number line is a visual representation of numbers along a horizontal line. Learn more about number lines, how they are useful, and how to teach with them.
## Images for 9+ Missing Numbers On Number Line Worksheets
### Paste missing numbers learning multiplication table stock vector royalty free
TO THE RIGHT OF A NUMBER LINE YOU ARE ADDING. INTEGERS ARE ALL NUMBERS TO THE LEFT AND RIGHT OF 0 INCLUDING 0. WHOLE NUMBERS ARE ALL NUMBERS TO THE RIGHT OF A NUMBER LINE OR POSTIVE NUMBERS. ON A NUMBER LINE THE NEGATIVE NUMBER CLOSES TO THE ZERO IS THE LARGEST NUMBER WICH IS -1.
A Number Line to 20 is a series of numbers from 0-20 that are placed on a line. Number lines are great visual resources for visual learners and show children each number in that sequence. Count forwards and backwards with a number line to really heighten learning.
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# why can only universal statements be vacuously true?
So I'm new to logic and im getting really confused on definitions namely that of universal and existential quantifiers and why vacuous truths only apply to statements with a universal quantifier $$(\forall)$$ and not for existential quantifier $$(\exists)$$.
I read that the 2 statements
(1) $$\forall x\in A\,P(x)$$
(2) $$\forall x(x\in A\implies P(x))$$
Are logically equivalent but why cant we say the same thing about the existential quantifier namely why cant we say that statements $$(1)$$,$$(2)$$ can be used.
(1) $$\exists x\in A\,P(x)$$
(2) $$\exists x(x\in A\implies P(x))$$
So why do we say that only universal or conditional statements are vacuously true?.
The formula $$\exists x \in A\;P(x)$$ stands for $$\exists x(x \in A \land P(x))$$. This is because we want the following chain of equivalences to hold:
\begin{align*} \exists x \in A\;P(x) &\equiv \lnot \forall x \in A\;\lnot P(x) \\ &\equiv \lnot \forall x(x \in A \implies \lnot P(x)) \\ &\equiv \exists x(x \in A \land P(x)) \end{align*} where the first equivalence is the desired De Morgan law, the second follows from the definition as you give it and the third follows because to falsify $$\phi \implies \psi$$ you have to make $$\phi$$ true and $$\psi$$ false.
$$\exists x(x \in A \implies P(x))$$ would be trivially true (unless $$A$$ is the entire universe of discourse) since for $$x \not\in A$$, $$x \in A \implies \phi$$ is true for any $$\phi$$.
The analogue of the vacuously true universal: $$\forall x \in \emptyset\;P(x)$$ is the vacuously false existential: $$\exists x \in \emptyset\;P(x)$$.
• why is it for A being the entire universe of discourse?. Commented Feb 14, 2021 at 15:22
• If $A$ is the entire universe of discourse, then $x \in A$ is true for any $x$, and so $x \in A \implies P(x)$ is equivalent to $P(x)$. Commented Feb 14, 2021 at 15:29
• I don't understand what you mean by "the same or construction". Commented Feb 14, 2021 at 20:40
• Well you could, but it will be guaranteed to be false unless $A$ is the universe of discourse (compare how using $\implies$ rather than $\land$ in the definition of $\exists x \in A \; P(x)$ gives something which is trivially true unless $A$ is the universe of discourse, as discussed in my answer). Try seeing where you get starting with the De Morgan law $\forall x \in A\; P(x) \equiv \lnot \exists x\in A \; \lnot P(x)$ using the definition we have now agreed for $\exists x \in A\;\lnot P(x)$. Commented Feb 14, 2021 at 23:05
• Because there is no point in having notations like $\forall x \in A\;P(x)$ or $\exists x \in A\;P(x)$ if the constraint $x \in A$ means that the formula will always be true or always be false. Try writing down in English what you want these formulas to mean intuitively (every element of $A$ satisfies $P$; some element of $A$ satisfies $P$): the definitions we have arrived that match those intuitions. Commented Feb 14, 2021 at 23:17
As I understand it, the implication $$A\to B$$ is said to be vacuously true if $$A$$ is false. We can see this from lines 3 and 4 of the truth table for implication:
$$\neg A \to (A\to B)$$ is also a tautology:
This can also be proven using a form of natural deduction:
Vacuous truth doesn't necessarily have anything to do with quantifiers or empty sets, though it is used, for example, to show that the empty set is unique, i.e. that all empty sets are equal. It is also used in proofs by cases when one or more of the cases being considered are proven or assumed to be false.
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# How do you find the polynomial function with roots 2, 3+- sqrt2?
Dec 5, 2015
Start from the factored form to find the desired polynomial to be
$f \left(x\right) = {x}^{3} - 8 {x}^{2} + 19 x - 14$
#### Explanation:
Creating a polynomial function with specific roots is as easy as can be. Just set it up as a product of binomials $\left(x - \text{root}\right)$ which evaluate to $0$ at the desired root. For example, in this case, we would use
$f \left(x\right) = \left(x - 2\right) \left(x - \left(3 + \sqrt{2}\right)\right) \left(x - \left(3 - \sqrt{2}\right)\right)$
$= {x}^{3} - 8 {x}^{2} + 19 x - 14$
Because we start from the fully factored form, it is clear that the roots are as desired.
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How to Simplify the Properties of Exponents?
Exponents work by raising a base, or number, by itself by a set number of times. For example, the expression 2^3 translates into 2*2*2. The properties of exponents bind at a greater power than othe... Read More »
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# Probability
Shamoon713 | 10:44 Fri 03rd Jul 2020 | Jobs & Education
How many different linear arrangements are there of the digits 1, 2, 3, 4, 5, 6 for which:
(a) 5 and 6 are next to each other ?
(b) 5 is before 6?
(c) 3 is before 4 and 5 is before 6?
(d) 3 and 4 are next to each other and 5 and 6 are also next to each other?
(e) 6 is not last in line?
1 to 5 of 5
No best answer has yet been selected by Shamoon713. Once a best answer has been selected, it will be shown here.
Combinatorics isn't my strong suit, but:
(a) can have (56) or (65) in one of five different positions, and remaining 4 numbers can be any way you like, making for a total of 2*5*4! = 240 cases.
(b) if you put 6 in position 1 then there are no cases; in position 2 then you must have 56.... (4! cases);
in position 3 then 5.6... or .56... works (2*4!); in position 4 then 5 has three slots to choose from (3*4!), etc, for a total of (1+2+3+4+5)*4! = 360. The fact that this is exactly half of the available set of permutations suggests that you could also make an argument from probability/symmetry, ie something like P(5 before 6) = P(6 before 5) = 1/2 (because there is no reason to prefer 5 or 6). But I thought of the exhaustive method first.
Haven't answered (c-e) yet, but hopefully the ideas above will get you started. (c,d) look trickier to resolve than (b,a) but it looks like it will end up being arguable in more or less the same way.
I must have it wrong. I figured there were 2^3, i.e. 8 combinations of 1 to 4; multiplied by 5 locations 5/6 could be placed, multiplied by 5 locations 6/5 might be. 8x25 is 200. Where have I lost 40 ?
I'm afraid that's wrong from start to finish, OG. Permutations are factorial: 4 places to place 4, 3 to place 3 and so on, for 4! (not 2^3 = 8). Then you can have (56) OR (65) in 5 locations each, so you add these.
Ah, ok. Thanks. I knew something was up.
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Barquentine
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athkjlj
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A fair coin is tossed 8 times find the probability that it shows heads exactly 5 times. May Jesus richly bless you today!. so the total probability is. Example 31 If a fair coin is tossed 10 times, find the probability of (i) exactly six heads (ii) at least six heads (iii) at most six headsIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i. A coin is tossed 3 times. If a coin is tossed 5 times, what is the probability that it will land heads each time? 11. The word AND means we need multiplication. The ratio of successful events A = 7 to the total number of possible combinations of a sample space S = 8 is the probability of 1 head in 3 coin tosses. A jar contains 10 blue marbles, 5 red marbles, 4 green marbles, and 1 yellow marble. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). Suppose a coin is tossed 6 times. If the coin is tossed six times, what is the probability that less than ⅓ of the tosses are heads? 2. Consider one. Question: Which of the pairs of events below is dependent? Select the correct answer below: Question: Identify the option below that represents dependent events. Probability of getting each of the combinations are 1/18 as in exercise 6. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 2 heads, if a coin is tossed five times or 5 coins tossed together. On a given toss (a Bernoulli trial), p = P(head) = 1/2 and q = P(tail) = 1 - p = 1/2. In an experiment, a fair coin is tossed three times. A class has three (3) students. is 5 times 4 times 3. For another example, consider tossing two coins. As you can count for yourself, there are 10 possible ways to get 3 heads. Because I have tossed heads 100 percent of the time for my first five tossed, then the. algebraically, from the binomial coefficient identity C(n,k) = C(n,n-k), or from the intuition associated to the fundamental symmetry here: getting at most 3 tails is the same event as getting at least 11 heads, and since the coin is fair, this event has the same probability as getting at least. Correct answers: 3 question: Miranda tosses a fair coin consecutively five times and gets heads each time. When you toss a coin, there are only two possible outcomes, heads or tails. I wonder why it isn't $\frac12$. Examples: In the experiment of flipping a coin, the mutually exclusive outcomes are the coin landing either heads up or tails up. (Now, had the question been "What is the probability of getting one head and one tail?" - the answer would be 2 " in " 4 = 0. What is the probability of getting exactly. 6$, should it be tossed 100 or 10 times Hot Network Questions Fitting 1 speed tyres on 21 speed bike. Solve for Probability 2. Consider one. tossed five times = 5C3 ways = 5! / 3! * 2! = 10! is to be read as factorial. Step-by-step explanation: As the coin is biased and the probability of head is already given as 2/3, therefore we have to use this while finding the probability and not 1/2( which is the normal probability of head in case of unbiased coins). The probability of getting head is Let the R. 5A Probability Rules. 9222} {/eq} Become a member and unlock all Study Answers Try it risk-free for 30 days. Seth tossed a fair coin five times and got five heads. Draw a card from a standard deck and record its suit. 8 Using a distribution curve, find the area under the curve between z = 0 and z = 2. 5 is the probability of getting 2 Heads in 3 tosses. 147, because we are multiplying two 0. Question 149445: A fair coin is tossed 5 times. Each coin toss's outcome is independent of the outcomes of the previous (and the future) coin tosses. A sample space is a collection of all possible outcomes of a random experiment. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. absolutely, then we say that X does not have an expected value. 4th coin tossed probability of getting tails is 1 in 2. E X = probability weighted average number of heads when two coins are tossed. At each step the choice is either heads or tails. Hence there is 0. in case you propose you've #a million head #2 tails #3 heads, then this may be the answer: First toss = a million/2 2d toss = a million/2 0. What is the probability that the illustrated board game spinner will land on blue? a. 5 percent of getting no heads in three tosses. When her son Shane asks her about the probability of getting tails on the next (sixth) toss, Miranda says the following: This is a fair coin, so I should toss heads approximately 50 percent of the time. If a baby is born it has an equal chance of being a boy or a girl. BYJU’S online coin toss probability calculator makes the calculations faster and gives the probability value in a fraction of seconds. 5 (50-50 chance of getting a head on each trial), q =. Ponchos and Long Coats still turn their heads, but only do so for a brief second, making your window of movement incredibly small compared to normal enemies. The independence implies that the probability of all 5 tails is (1/2)^5 = 1/32. Python Coin Toss. 3/4 * 3/4 = 9/16. Getting 3 tails is the same as getting 1 head. Solution A Coin is Tossed 5 Times. If a fair coin is tossed 5 times, the probability of getting 5 heads is: P(H,H,H,H,H) = (1/2)5 = 1/32 = 0. Answer to A coin is tossed 12 times. Since a coin has two sides and it was tossed 5 times, there are 32 possible combinations of results. The number of possible outcomes gets greater with the increased number of coins. Correct answers: 3 question: Miranda tosses a fair coin consecutively five times and gets heads each time. 3, 8 Three coins are tossed once. Yet if you flip 3 coins and they all turn up heads, getting a heads on the 4th flop is not a 1/16 chance, but 1/2, even though “Every flip of the coin doesn’t depend on the other coin flips”. In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. Question 149445: A fair coin is tossed 5 times. The ratio of successful events A = 10 to total number of possible combinations of sample space S = 32 is the probability of 2 heads in 5 coin tosses. 4 \leq$ heads $\leq 0. The probabilities for "two chickens" all work out to be 0. If success means getting two heads, then the probability of no success when exoeriment is repeated thrice, is. so the total probability is. How come the probability of getting heads in a coin toss is still 50/50 even after you have had tails for straight five times a row. In the end it all comes to a 50/50 somy question is: if i toss a coin and get four heads in a row, does the fifth toss has a 50/50 chance of landing heads/tails. Question: Which of the pairs of events below is mutually exclusive? 5. 6$, should it be tossed 100 or 10 times Hot Network Questions Fitting 1 speed tyres on 21 speed bike. The probability of not getting a six is 5/6. What is the probability of getting 9 heads? 2) About 1% of people are allergic to bee stings. So to get 7 heads and then 7 tails in that order is: But in this problem we don't care about order. Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed. Question 149445: A fair coin is tossed 5 times. The rest is just elementary arithmetic. Suppose we have 3 unbiased coins and we have to find the probability of getting at least 2 heads, so there are 2 3 = 8 ways to toss these coins, i. 3, 2 A coin is tossed twice, what is the probability that at least one tail occurs? When 2 coins are tossed , Sample Space = S = {HH, HT, TH , HT} n(S) = 4 Let A be the event that at least 1 tail occurs Hence A = {HT, TH, TT} n(A) = 3 P (A) = Number of outcomes favourable to ATotal number of possible outcomes = n(A)n(s) = 𝟑𝟒. (Last Updated On: January 21, 2020) Problem Statement: A fair coin is tossed three times. In tossing the coin twice you have the options of: heads, tails, heads, tails (One head and one tail per coin). Read more Probability (statistics) Statistics (academic discipline) Ad by. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. If you think of a success as a head, the count of the number of heads in 3 tosses satisfies the definition of a binomial random variable. The probability of getting 3 tails and 2 heads is the same as the probability of getting 3 tails in 5 tosse. Ponchos and Long Coats still turn their heads, but only do so for a brief second, making your window of movement incredibly small compared to normal enemies. Probability of getting heads exactly 8 times in n tosses of a coin = nC8. Here's how. Therefore, the probability of flipping at least two consecutive heads is. As you can count for yourself, there are 10 possible ways to get 3 heads. Nine cards with varying symbols, lowercase letters, capital. enter your value ans - 5/16. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 1 head, if a coin is tossed three times or 3 coins tossed together. Question: Which of the following shows mutually exclusive events? 4. If a single card is. Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. When a coin is tossed you have 2 outcomes ; Heads or Tails When it is tossed 4 times the number of outcomes = (1/2)^4 = 16 You are seeking probability of 1 of the 16 outcomes (H,T,H,H). The the expected value of X, is : A. In the end it all comes to a 50/50 somy question is: if i toss a coin and get four heads in a row, does the fifth toss has a 50/50 chance of landing heads/tails. , HHH, HHT, HH, THH So the probability is 4/8 or 0. With tree diagramming, Doctor Ian reveals patterns and a path forward. 45 Probability distribution for the number of 4's rolled when rolling a die five times Number of 4's rolled Probability-Ex. What is the probability that the number of heads obtained will be between 0 and 4 inclusive? Round your answer to 3 decimal places. What is the probability of getting exactly 5 heads? A coin is tossed 10 times. (There can only be up to five since the die is only rolled five times) 0. there fore it is 12. The probability of getting a head on each occasion if a coin is tossed five times is equal to 1/2 x 1/2 x 1/2 x 1/2 x 1/2 (that is, 1/25) = 1/32 or one in thirty-two. An easier way would be to do a normal approx. Given that the first toss is a head then we only need to find the probability that the next two tosses are both heads. When three coins are tossed, the probabilities of getting tails on each coin are multiplied. This page lets you flip 2 coins. So, the probability of getting 5 or more heads should. Then to flip heads or tails equals 12. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 3 heads, if a coin is tossed five times or 5 coins tossed together. Find the probability of getting 2, 3 or 4 heads by using the normal approximation to the binomial distribution. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads?. Toss a Coin Six Times Date: 02/07/98 at 16:59:43 From: Ruth Beldon Subject: Coin tossing probabilities A. As C is the first dice rolled and can be any value, P(C) = 1. Assume the coin is fair, i. Since a fair coin flip results in equally likely outcomes, any sequence is equally likely… I know why it is $\frac5{16}$. It's 1,023 over 1,024. Thus, the probability of getting 3 heads from 5 coin flips is: 10/32, or 5/16. What is the probability that exactly 2 heads are observed? - Answered by a verified Math Tutor or Teacher. In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. toss coin many times, frequency of heads = f(H) ≈ 1/2. Must improve: Mind games/physical approach. Probability of getting heads exactly 6 times in n tosses of a coin = nC6. 7s and one 0. I would like to ask if there is any mathematical way to calculate this probability. 50 = cent 1. what is the probability of tossing a coin 3 times and getting heads each time? Seth tossed a fair coin five times and got five heads. The more times you toss the coin, however, the closer you will get to 50 percent heads. Step-by-step explanation: As the coin is biased and the probability of head is already given as 2/3, therefore we have to use this while finding the probability and not 1/2 ( which is the normal probability of head in case of unbiased coins). A fair coin is tossed 5 times. Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. so if we toss a coin 5 times, it will be heads or tails 100% of the time. to get to P=3/8. Coins also have a 9-second cooldown for each use, so timing the throw of your coins is key to success. Given that it is a fair coin and the probability of a tail is 50 per cent on one toss the probability of 5 consecutive tails is. Question: Which of the pairs of events below is mutually exclusive? 5. As you can count for yourself, there are 10 possible ways to get 3 heads. The answer is 5/16 because in total there are 16 possibilities when tossing a coin 4 times. A coin is tossed 5 times. A coin is tossed multiple times. This is out of 16 total ways to flip a coin 4 times. If tails appears, a second coin is tossed instead of spinning the spinner. Find the experimental probability of getting heads. Then the probability. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. We may show the outcomes, e. 51 probability of catching the coin the same way we throw it. Heads = 1/2 It is known that there are more than 2 heads in the 5 tosses. A question has been discussed in this video. ) What is the probability of getting heads on at least one flip? 3. Here's how. What is the probability of getting at least 3 heads. So two possible outcomes in one flip. What is the probability of getting 9 heads? 2) About 1% of people are allergic to bee stings. 4 \leq $heads$ \leq 0. What if the experiments can not be repeated?. So the probability is ----- b) What is the probability of obtaining tails on each of the first 3 tosses That only happens 2 times. The probability of getting head once and tail two times is. Example 31 If a fair coin is tossed 10 times, find the probability of (i) exactly six heads (ii) at least six heads (iii) at most six headsIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i. 4 \leq $heads$ \leq 0. $2\cdot2\cdot2\cdot2 = 2^4$. The way to get three consecutive heads is HHH. 5 Please correct me if I am wrong, and yes I agree this is very crude approach. Solution for 1. tossed five times = 5C3 ways = 5! / 3! * 2! = 10! is to be read as factorial. at most two heads(using binomial distribution). The probability of hitting the target is. In the first trial, head was scored 5 times, or 5/10, or 50%. Consider the simple experiment of tossing a coin three times. Draw a card from a standard deck and note its color (red, black) Solutions: 1. The answer is 29/512 but I don’t know how to get it. tossed five times = 5C3 ways = 5! / 3! * 2! = 10! is to be read as factorial. \probability" is intended to be the probability that the corresponding outcome occurs (see Section 4. for a coin toss there are two possible outcomes, Heads or Tails, so P(result of a coin toss is heads) = 1/2. Thus, the probability of getting 3 heads from 5 coin flips is: 10/32, or 5/16. Coin Toss Probability. Flip a coin. Most coins have probabilities that are nearly equal to 1/2. The probability of getting heads three times in 5 tries is 10/32. There are 2 opportunities, heads or tails, for each toss. Sometimes, though, it is better to look at what DOESN'T work and go from there to make your life easier. 45 Probability distribution for the number of 4's rolled when rolling a die five times Number of 4's rolled Probability-Ex. If the coin is fair, then by symmetry the probability of getting at least 2 heads is 50%. The answer is 5/16 because in total there are 16 possibilities when tossing a coin 4 times. Find the probability of getting 2, 3 or 4 heads by using the normal approximation to the binomial distribution. A coin is tossed 5 times. What is the probability of getting at least 3 heads. A fair coin is tossed 5 times. For example, you might get seven heads (70 percent) and three tails (30 percent). 5 to the fifth or 1/32 So the probability of at least one head is 1-(1/32) or 31/32. That sequence has a probability of $1/2 * 1/2 * 1/2$. Anyhow it seems like with just 16 coins tossed the amount of times I got 5 heads in a row was about 33% if I did it for like 10 or so trys. A question has been discussed in this video. If you want it express it in terms of a percentage, you would have approximately a 2. $2\cdot2\cdot2\cdot2 = 2^4$. The answer is 29/512 but I don’t know how to get it. A coin tossed 3 times. What is the probability of the event E= {exactly 3 heads occur}? So, the probability of getting exactly three heads is 4/16 = 1/4. If a fair coin is tossed 3 times, what is the probability that it turn up heads exactly twice? Without having to list the coin like HHH, HHT, HTH, ect. But to answer your question mathematically before you start flipping, each chance is 50%. Since the sum of the row is 8, the probability of getting two heads and one tail is 3/8. You get H (heads) or T (tails). What is the probability of getting exactly 5 heads? A coin is tossed 10 times. The total number of possibilities of getting three heads when a coin is. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. A coin is tossed five times. Question: A Coin Is Tossed Five Times. What is the probability of tossing heads, and rolling a 3 or a 5? A coin is tossed 50 times and 38 heads are observed. What is the probability of a flipped coin landing on heads 4 times out of 6 trials? —ThGnK. Use the binomial probability distribution. Coin toss probability When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50%, 1/2, or 0. So the probability that one of those occurs is 0. If a coin is tossed 5 times and comes up heads all five times, the probability of obtaining tails on the 6th toss is = ? 2. Suppose that the coin is tossed 3 times. (3-1)!] ). The ratio of successful events A = 16 to the total number of possible combinations of a sample space S = 32 is the probability of 3 heads in 5 coin tosses. If 2 persons are chosen at random from a set of 3 men and 4 women, what is the probability that 2 women are chosen? 13. BYJU'S online coin toss probability calculator makes the calculations faster and gives the probability value in a fraction of seconds. Tossing a Biased Coin Michael Mitzenmacher When we talk about a coin toss, we think of it as unbiased: with probability one-half it comes up heads, and with probability one-half it comes up tails. 4 \leq $heads$ \leq 0. So the probability of getting 3 heads or less is 0. Ten Coin Flips, Four Heads [05/08/2001] If you flip a coin ten times, what is the probability of getting at least four heads? Ten Dice Tosses, All Pairs? [10/16/2017] An adult wonders about the likelihood that ten dice tosses yield a matching pair each time. With tree diagramming, Doctor Ian reveals patterns and a path forward. If two coins are flipped, it can be two heads, two tails, or a head and a tail. You get H (heads) or T (tails). The probability can be calculated as: P(S_k)=((n),(k))p^k(1-p. So the probability of getting exactly 8 heads in 12 coin tosses is: unlock 1. If a fair coin is tossed 5 times, the probability of getting 5 heads is: P(H,H,H,H,H) = (1/2)5 = 1/32 = 0. Draw a card from a standard deck and record its suit. 5 to the fifth or 1/32 So the probability of at least one head is 1-(1/32) or 31/32. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 3 heads, if a coin is tossed five times or 5 coins tossed together. Select the correct answer below: Question: … shows mutually exclusive events? Select the correct answer below: Question: Which of the pairs of events below is mutually exclusive? Select the correct answer below. A fair coin is tossed 5 times. The total probability of getting at least two heads is equal: $\frac 38 +\frac 38 =\frac 68 = \frac 34$ If we think of flipping a coin 3 times as 3 binary digits, where 0 and 1 are heads and tails respectively, then the number of possibilities must be $2^3$ or 8. The probability of getting $5$ heads in $10$ tosses. If the coin is tossed two times and you want the probability of getting 2 heads, that's the probability of getting a head on the first toss AND getting a head on the 2nd toss. The $1/2^5$ term is the probability of getting heads for the first time on the fifth toss, or the sequence TTTTH. Given some value of the first role, P(A) = 1/6. The coin was tossed 12 times, so N = 12. 4 \leq $heads$ \leq 0. The probability mass function of the R. Find the probability of the given scenarios. What is the probability of getting at least 3 heads. So 6 of the 16 possible equally likely sequences result in exactly 2 heads out of the 4 tosses. Thus, the probability of getting 3 heads from 5 coin flips is: 10/32, or 5/16. Assuming a "fair" coin, there are 2^5=32 different arrangements of heads and tails after 5 flips. A coin tossed 3 times. The ratio of successful events A = 10 to total number of possible combinations of sample space S = 32 is the probability of 2 heads in 5 coin tosses. It's 1,023 over 1,024. This is out of 16 total ways to flip a coin 4 times. If you were to toss a penny three times, the probability that all three tosses would come up heads would be 50% of 50% of 50%, which is 12. A class has three (3) students. Repeat this 8 times and store the number of heads for each one. 5 and P(B) = 0. 25% equals 1/4 which equals 2/8. 147, because we are multiplying two 0. to locate an accrued chance, you may multiply the three opportunities mutually. Find the probability of getting 2, 3 or 4 heads by using the normal approximation to the binomial distribution. Many events can't be predicted with total certainty. Coins also have a 9-second cooldown for each use, so timing the throw of your coins is key to success. Suppose you toss a fair die 5 times- what is the probability of getting exactly three 4's? The way to think through this problem is like this: 1. 25% chance to get 3/5 heads. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5 otherwise X takes the value -1. This was repeated in the second and third trials. This however doesn't aim to find the probability, but the representation by the use of other pre-defined events. In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. 25 " = 25% = 1/4 Probabilities are usually given as fractions. Since a coin has two sides and it was tossed 5 times, there are 32 possible combinations of results. What is the probability it will come up tails if tossed one more time. Coin toss - win if $0. Let's look at the sample space for these tosses: Three ways that we can get 1 Heads out of 3 tosses. For example, the probability of a coin landing on heads is 0. 5: And so the chance of getting 3 Heads in a row is 0. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. Must improve: Mind games/physical approach. The probability of getting two heads in two tosses is 1 / 4 (one in four) and the probability of getting three heads in three tosses is 1 / 8 (one in eight). An unbiased coin is tossed 5 times. If a fair coin is tossed 5 times, the probability of getting 5 heads is: P(H,H,H,H,H) = (1/2)5 = 1/32 = 0. Find the probability of getting. What is the probability that the number of heads obtained will be between 0 and 4 inclusive? Round your answer to 3 decimal places. 3, then P(AUB) = ?. For each toss of a coin a Head has a probability of 0. In an experiment, a fair coin is tossed three times. V 'X' is given by,. May Jesus richly bless you today!. So the probability is. 17% chance I think. 3636) A traffic signal is green for 20 seconds, then amber for 5 seconds, then red for 30 seconds. to locate an accrued chance, you may multiply the three opportunities mutually. Intuitive idea: P(A) is the typical fraction of times A would occur if an experiment were repeated very many times. A fair coin is tossed 8 times,what is the probability ofgetting: 1. Then you would pick out how many of those had exactly 3 heads and divide that number by 256 because the most basic level of probability is "good ways over total ways". The probability of getting a head on the first toss 7. 03 probability, as shown at the right. 5 Please correct me if I am wrong, and yes I agree this is very crude approach. at most two heads(using binomial distribution). What is the probability of getting 9 heads? 2) About 1% of people are allergic to bee stings. Let's write down all 16 but group them according to how many heads appear, using the binary notation 1 = heads, 0 = tails:. Please I want answer for…. The outcomes in different tosses are statistically independent and the probability of getting heads on a single toss is 1 / 2 (one in two). Example 31 If a fair coin is tossed 10 times, find the probability of (i) exactly six heads (ii) at least six heads (iii) at most six headsIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i. enter your value ans - 5/16. What is the probability of getting exactly 5 heads? A coin is tossed 10 times. Let X denote the number of heads which appear. Find the probability that there will be at least 5 heads with exactly 5 of them occurring consecutively. if the coin were unbiased then p=probability of heads(or tails) = 0. The probability of a coin toss being a tail is 1/2. So the probability is. Business According to a 2012 study conducted by the Society for Human Resource Management, 81% of U. 3125 The failure rate for taking the bar exam in Philadelphia is 41%. This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. As C is the first dice rolled and can be any value, P(C) = 1. Here's how. Ponchos and Long Coats still turn their heads, but only do so for a brief second, making your window of movement incredibly small compared to normal enemies. 1 (Coin Tossing) The most fundamental stochastic experiment is the experiment where a coin is tossed a number of times, say ntimes. What is the probability of obtaining exactly 3 heads. Let X be the number of heads obtained. What is the probability of getting no any two heads on consecutive tosses? A. First, there are ways to flip the coins, in order. choosing at random ten people and measuring their height. Since there are four options, it is a 25% chance you will get heads twice in a row. What is the probability of getting at least three heads on consecutive tosses? A. Most coins have probabilities that are nearly equal to 1/2. For these three tosses the relative frequencies of heads are 0. Ifyou were to toss a coin what percentage in a. Suppose a coin is tossed 6 times. If we toss a fair coin twice, we have the following possible outcomes, or events: {(H,H), (H,T),(T,H), (T,T). A fair coin is tossed 10 times. Answer to: A coin is tossed 5 times. without writing the sample space, P(E) = 4C3 * (1/2)^3 * (1 - 1/2)^1 (binomial distribution (in case you arent familiar with it, it says that the probability of success is given by nCx * p^x * (1 - p)^(n - x) where n is the number of trials (in this case 4), p is the probability of success (in this case 1/2), and x is the number of successes (in this case 3)) = 1/4. A math-ematical model for this experiment is called Bernoulli Trials (see Chapter 3). For example, suppose we have three coins. Answer by Fombitz (32378) ( Show Source ): You can put this solution on YOUR website! The answer is 10/32=5/16. Mathematically, coin toss experiment can be thought of a Binomial experiment, where we have a coin with probability of getting head as success at each coin toss is p. A coin is biased so that a head is twice as likely to occur as a tail. 3 in each case. If you want it express it in terms of a percentage, you would have approximately a 2. What is the probability that all 3 tosses are Heads? please help ,e to solve this. We assume that the coin is fair and is flipped fairly. Since there are four options, it is a 25% chance you will get heads twice in a row. I wonder why it isn't$\frac12$. What is the probability of getting exactly 5 heads? A coin is tossed 10 times. The 6 results in yellow have 4 heads before two tails and hence these are the winning outcomes for the first player. Here's how. Here, the probability of heads is s/2πr, the ratio of the arc length ssubtended by the heads face and the circumference of the circle. Hope this helps. so if we toss a coin 5 times, it will be heads or tails 100% of the time. For the total thirty tosses, theoretically, the coin should land on heads fifteen times, or five per trial, which is determined solely on the number of options. The probability of getting at least one head 8. A coin is tossed 7 times. 5 and having 10 heads in a row is … read more. For example, if you tossed a coin 1000 times, you might get 510 heads and 490 tails. so the total probability is. In sport, coins are tossed to decide which end of the ground a team is to defend, or who is going to go into bat flrst. 0 1 vote 1 vote. Not bothering to approximate by the normal, we can look at a random variable distributed binomial with n=900 and p=0. The probability that a single toss will be head only is 0. I was just checking with a coin toss generator however the one I was using only allowed for 16 coins tossed at a time at max. I am very weak in probability. 1) The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events. Probability success = P then Probabi. Question 149445: A fair coin is tossed 5 times. A sample space is a collection of all possible outcomes of a random experiment. 25 then it would be. What is the probability of getting 9 heads? 2) About 1% of people are allergic to bee stings. However, research shows that there is actually a bit of a bias that makes the toss less fair. for a coin toss there are two possible outcomes, Heads or Tails, so P(result of a coin toss is heads) = 1/2. The more times you toss the coin, however, the closer you will get to 50 percent heads. 5 (chance. 6$, should it be tossed 100 or 10 times Hot Network Questions Fitting 1 speed tyres on 21 speed bike. An easier way would be to do a normal approx. Probability is the measurement of chances - likelihood that an event will occur. The probability that the next toss will be a tail is. 5^2, then time the probability of a tails--also with a point. So two possible outcomes in one flip. Question 149445: A fair coin is tossed 5 times. A fair coin is tossed 5 times the probability of getting more heads than tails is. 5 (but that was pretty obvious, wasn't it?) (b) Two dice are tossed? We could make a table as in the preceding part, but remember that expectations add-- so since the expected value of the first die is 3. Consider one. Since there are four options, it is a 25% chance you will get heads twice in a row. 3, 8 Three coins are tossed once. 5C3 represents Binomial coefficient and its value can also be selected from PASCALS TRIANGLE. I was just checking with a coin toss generator however the one I was using only allowed for 16 coins tossed at a time at max. to locate an accrued chance, you may multiply the three opportunities mutually. Then you would pick out how many of those had exactly 3 heads and divide that number by 256 because the most basic level of probability is "good ways over total ways". Since the tosses are independent, the probability of a head on both tosses (the intersection) is equal to 1/2*1/2 = 1/4. 5 probability). Find the theoretical probability of getting tails for this experiment. absolutely, then we say that X does not have an expected value. An easier way would be to do a normal approx. 03 probability, as shown at the right. tossed five times = 5C3 ways = 5! / 3! * 2! = 10! is to be read as factorial. This is out of 16 total ways to flip a coin 4 times. There are 3 such combinations, so the probability is 3 × 1/18 = 1/6. The Giants tried plenty of things to put the Bulldogs off their game but none of them worked. In an experiment, a fair coin is tossed three times. 4 years ago. The probability that the next toss will be a tail is. P(X≥5) = P(X=5) + P(X=6) = (6 choose 5)((1/2)^5)((1/2)^1) + (6 choose 6)((1/2)^6)((1/2)^0) = 6(1/64) + 1(1/64) = 7/64. When a coin is tossed, there lie two possible outcomes i. $2\cdot2\cdot2\cdot2 = 2^4$. e head or tail. so if we toss a coin 5 times, it will be heads or tails 100% of the time. Flip a coin. The probability of getting a head on any one toss of this coin is 3/4. Find the chance that at least one of the students gets exactly $5$ heads. If the coin is flipped $6$ times, what is the probability that there are exactly $3$ heads? The answer is $\frac5{16}$. 5 red cubes and 4 blue cubes are in a bag. Let (capital) X denote the random variable "number of heads resulting from the two tosses. 51 probability of catching the coin the same way we throw it. Experimental probability is what the probability was based on the given data. First toss, H or T. Become a member and unlock all Study Answers Try it risk-free for 30 days. Ten Coin Flips, Four Heads [05/08/2001] If you flip a coin ten times, what is the probability of getting at least four heads? Ten Dice Tosses, All Pairs? [10/16/2017] An adult wonders about the likelihood that ten dice tosses yield a matching pair each time. What is the probability of getting heads, heads, tails? Please show me the equation for this??? Answer Save. b) The probability of not getting exactly 2 heads is 1 minus the probability of getting exactly 2 heads, which is 1 - (3/8) = 5/8. The probability can be calculated as: P(S_k)=((n),(k))p^k(1-p. Usually, coins used in probability problems are only assumed to have two outcomes: heads or tails. What is the probability that exactly 2 heads are observed? - Answered by a verified Math Tutor or Teacher. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads?. Remember: each coin represents each parent and each toss can only turn up one way, therefore, a parent can give only one gene of a pair. If the coin is tossed six times, what is the probability that less than ⅓ of the tosses are heads? 2. The number of possible outcomes gets greater with the increased number of coins. In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. 74% chance of getting 3 heads or less with a coin that has those particular characteristics. Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed. C) Three dice are tossed. The total number of ways that you can toss 5 coins is 2^5. So two possible outcomes in one flip. So, the probability that you picked the 2-headed coin, given that you flipped 10 heads in a row, is 0. Probability of getting each of the combinations are 1/18 as in exercise 6. We use cookies to give you the best possible experience on our website. Find the probability of the given scenarios. same probability of getting 3 Tails and 2 Heads I can do this watching a music video, reply to an email and watching 4 baseball games (you might be able to do more than me being way younger) the answer is in the 5th row of Pascal's triangle 1, 5, 10, 10, 5, 1 choose(0,1,2,3,4,5) so choose 2 Tails = 10 (the 3rd element in the row - same exactly as 3 Tails) that goes in a (a / b) b = the sum of. The gambler's fallacy can be illustrated by considering the repeated toss of a fair coin. 5 to the fifth or 1/32 So the probability of at least one head is 1-(1/32) or 31/32. 50 is the gain. Toss the coins together. Since there are four options, it is a 25% chance you will get heads twice in a row. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. Find the probability that both heads and tails occurs. asked by Jaiby on September 10, 2018; Math (Ms. b) What do you think E[X] should be. exactly 3 heads. If you think of a success as a head, the count of the number of heads in 3 tosses satisfies the definition of a binomial random variable. On a given toss (a Bernoulli trial), p = P(head) = 1/2 and q = P(tail) = 1 - p = 1/2. The corresponding probabilities are 1/8,3/8,3/8, and 1/8. for the binonial, or still easier, do it on a TI-83 or 84 with p =. 5A Probability Rules. What is the probability of getting at least 3 heads. What is the probability of getting exactly 3 Heads in five consecutive flips. Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of outcomes. example, if you toss a coin once, the possible outcomes are H or T; toss it twice, and the possible outcomes are HH, HT, TH, and TT. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. Getting two head require 50 percent of 50 percent because we need two head out of 3 in any order there fore it is 32. Since in 5 out of 8 outcomes, heads don’t occur together. What are the odds of getting two, four, or six heads after five, ten, or a hundred consecutive tosses of a fair coin? It seemed like a fun high school leveled math problem and with some quick python I was able to generate a pretty graph to answer this question. 6$, should it be tossed 100 or 10 times Hot Network Questions Fitting 1 speed tyres on 21 speed bike. 5 for obtaining a head when a coin is tossed. 15 Answers. so if we toss a coin 5 times, it will be heads or tails 100% of the time. If 4 coins are tossed, find the following probability: 2 heads. 5) then tails must come up on the next two throws (each has a p= 0. Probability of getting tail in single toss = 1/2 Probability of getting head in single toss = 1/2 Probability of getting First Tail = 1/2 Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4. What is the probability of getting heads, heads, tails? Please show me the equation for this??? Answer Save. A fair coin is tossed 6 times. So two possible outcomes in one flip. What is the probability that the number of heads obtained will be between 1 and 3 - Answered by a verified Math Tutor or Teacher Toss a coin 10 times and after each toss, record in the following table the result of the toss and the proportion of heads so far. First, on any one toss what is the probability of getting a 4? That would be 1/6, since there is one way to get a 4 out of six possibilitis. It is given that both are equal. In general, if X has the binomial distribution with n trials and a success probability of p then. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 2 heads, if a coin is tossed five times or 5 coins tossed together. probability of getting 5 heads is (7C5) x (0. The probability of not getting a six is 1 - 1/6 = 5/6. The point estimator for the A coin is tossed 50 times and 38 heads. Ask Question Asked 7 years, 4 months ago. When an unbiased coin is three times, the probability of falling all heads is (Or). The correct answer will be 8/27. so if we toss a coin 5 times, it will be heads or tails 100% of the time. 4 \leq$ heads \$ \leq 0. What is the probability that the coin will land on heads again?" The answer to this is always going to be 50/50, or ½, or 50%. what is the probability of tossing a coin 3 times and getting heads each time. The 8 possible elementary events, and the corresponding values for X, are: Elementary event Value of X TTT 0 TTH 1 THT 1 HTT 1 THH 2 HTH 2 HHT 2 HHH 3 Therefore, the probability distribution for the number of heads occurring in three coin. Heads = 1/2 It is known that there are more than 2 heads in the 5 tosses. Seth tossed a fair coin five times and got five heads. ) The probability of obtaining h heads in N tosses of a coin with a probability of heads equal to r is given by the binomial distribution:. Theoretical probability is what, theoretically, the probability should be, regardless of data. 5 since there is a 50% chance (i. A coin usually flips heads or tails. Correct answers: 3 question: Miranda tosses a fair coin consecutively five times and gets heads each time. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. so the total probability is. C) Have R flip a coin 10 times, count the number of heads, store the number and repeat 1000 times. Three heads will be get in a sequence in 3 ways as follows. The third row says that if we toss three coins, we have one chance of getting all heads, three chances of getting one head and two tails, three chances of getting two heads and one tail, and one chance of getting three tails. and to have 1 head is 32. Favourite answer. If 2 persons are chosen at random from a set of 3 men and 4 women, what is the probability that 2 women are chosen? 13. So, the probability of getting 5 or more heads should. When three coins are tossed, the probabilities of getting tails on each coin are multiplied. 5% 2 tails and there is 12. ) What is the probability of getting heads on at least one flip? 3. , HHH, HHT, HH, THH So the probability is 4/8 or 0. Find The Probability Of Getting Exactly Three Heads. We may show the outcomes, e. Are the Odds Really Equal? Earlier, we mentioned that the odds of a coin flip are 50:50. In binomial probability distribution, mean is 3 and standard deviation is. enter your value ans - 5/16. Then the possible values of X are 0,1,2 and 3. What is the probability of getting exactly. Find the probability that both heads and tails occurs. Probability of at Least 45 Heads in 100 Tosses of Fair Coin Date: 05/15/2004 at 08:14:21 From: Joe Subject: A different type of coin toss probability question What is the probability of getting AT LEAST 45 HEADS out of 100 tosses of a fair coin?. Three heads will be get in a sequence in 3 ways as follows. (There can only be up to five since the die is only rolled five times) 0. Here's how. You have a 1 out of 8 chance of getting no heads at all if you throw TTT. The way to get three consecutive heads is HHH. The possiblities are: odd heads + one head = even heads. This however doesn't aim to find the probability, but the representation by the use of other pre-defined events. kattyahto8 learned from this answer Answer: Once the number of possible heads and tail in 12 tosses is. Chapter 3 Probability 34 b. Also, there are ""_5C_3= (5!)/(3!2!)=10 ways to get exactly 3 tails. Probability Versus Physics. What is the probability of getting at least three heads on consecutive tosses? A. Exactly three heads in five flips. The probabilities are: exactly 2 heads: P(A)=15/64 at most 2 heads: P(B)=11/32 In this task you can use the rule called Bernoulli's Scheme. Each time a fair coin is tossed, the probability of getting tails (not heads) is 1/2 = 0. Find E(X) Round your answer to the nearest ten thousandth. If you want to know what is the probability of getting #r# heads (or tails) out of #n# flips, it is the #r^{"th"}# element in the #n+1# row, divided by the sum of all the elements in the #n+1# row. 5 probability is. 17% chance I think. Once in the "3 tails" section which is TTTH and once in the "4 tails" section, which is TTTT. The more times you toss the coin, however, the closer you will get to 50 percent heads. The gambler's fallacy can be illustrated by considering the repeated toss of a fair coin. 45 Probability distribution for the number of 4's rolled when rolling a die five times Number of 4's rolled Probability-Ex. more than 3 tails. Probability of getting heads exactly 8 times in n tosses of a coin = nC8. The point estimator for the A coin is tossed 50 times and 38 heads. The question asks, about the probability of cases for which two heads does not come one after another when we toss a coin 6 times. While this is true of one coin toss, it is not a 50% chance if you toss the coin twice. A jar contains 10 blue marbles, 5 red marbles, 4 green marbles, and 1 yellow marble. for the binonial, or still easier, do it on a TI-83 or 84 with p =. Suppose you have a fair coin. The more times you toss the coin, however, the closer you will get to 50 percent heads. If a fair coin is tossed 5 times, the probability of getting 5 heads is: P(H,H,H,H,H) = (1/2)5 = 1/32 = 0. For example, suppose we have three coins. Thus, the answer is 5/16. All heads would occur 1/32 times or 0. Flip a coin. B) A die is tossed 20 times.
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# How to Calculate FVIF in Excel
To calculate the Future Value Interest Factor (FVIF) in Excel, you first need a solid grasp of the FVIF formula. The formula is:
FVIF = (1 + r)^n
Where:
• r represents the interest rate per period.
• n is the number of periods.
When you apply this formula in Excel, it calculates how much an investment will be worth in the future, considering a constant interest rate and a specific number of periods. The steps to calculate FVIF in Excel are as follows.
## Step 1. Enter Your Data
To begin calculating the Future Value Interest Factor (FVIF) in Excel, first launch Excel and open a new workbook. Once your new workbook is open, it’s time to input the key variables for FVIF: the interest rate (r) and the number of periods (n).
For example, if you are calculating FVIF for an annual interest rate of 5% and a period of 10 years, simply input 5% in a cell and enter 10 in another one.
## Step 2. Apply the FVIF Formula
Once you have entered your data in Excel, the next step is to apply the Future Value Interest Factor (FVIF) formula. This step is where the magic happens.
• Click on an empty cell where you want the FVIF result to appear. It could be any cell that’s not already in use.
• In the selected cell, type the formula. For example, if you input the “r” value in A2 and the “n” value in B2, type “=(1+A2)^B2”.
And that’s it! When you press Enter, Excel calculates the FVIF based on your inputs.
The cell now displays the future value interest factor.
## Step 3. Manage More Variables (Optional)
If you have many different interest rates (r) and periods (n) to calculate their corresponding FVIFs in Excel, you can adjust the steps above a little:
1. Enter Interest Rates and Periods: In row 1, starting from cell B1, enter your different interest rates. In column A, starting from cell A2, list the various numbers of periods.
2. Set Up the FVIF Formula: Click on cell B2, the first cell where you want your FVIF result to appear. Type the formula “=(1+B$1)^($A2)” and press Enter. In this formula, B$1 refers to the interest rate in column B, row 1, and locks the row (1) when you copy the formula across.$A2 refers to the number of periods in column A, row 2, and locks the column (A) when copying down.
3. Drag to Fill the Row: Hover over the bottom right corner of cell B2 until you see your cursor turned into a small black cross. Click and drag this cross to the right to fill the cells across, up to the last interest rate in row 1.
4. Drag to Fill the Column: Now hover over the bottom right corner of the last FVIF result you just got in row 2 until you see your cursor turned into a black cross. Then drag down to fill the cells down to the last period in column A.
Now, each cell in your matrix contains the FVIF for its corresponding interest rate and period!
If you want the FVIF results to display a specific number of decimals, you can right-click on the selected cells and choose “Format Cells” from the context menu. Then, in the “Format Cells” dialog box, click on the “Number” tab. You can enter the number of decimals you want in the “Decimal places.”
Isn’t it easy? This method effectively creates a full grid of FVIF values, allowing you to see the impact of various interest rates and periods quickly. It’s a powerful way to analyze different scenarios in financial planning and investment analysis!
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# Find the area of the region bounded by the curves $y^2=9x,y=3x.$
Toolbox:
• Area of the region bounded by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$,where $f(x)$ is a continuous function defined on $[a,b]$,is given by
• $A=\int_a^bf(x)dx$ or $\int_a^b ydx.$
Step 1:
Given $y^2=9x$ and $y=3x$
Clearly $y^2=9x$ is a parabola,whose focus a can be calculated as shown below :
$y^2=4ax$ is the general equation .
In the given equation $4a=9\Rightarrow a=\large\frac{9}{4}$
The curve is open rightward with the focus as $\big(\large\frac{9}{4}$$,0\big) Step 2: y=3x is a straight line. Next let us find the point of intersection of the line and the parabola. y^2=9x-----(1) y=3x\Rightarrow y^2=9x^2 Hence substituting this in equ(1) we get, 9x^2=9x 9x(x-1)=0 \Rightarrow x=0 and 1 Hence the points of intersection are 0 and 1. The area of the required region is as shown in the fig. Now let us evaluate the area of the required region. Step 3: A=\int_0^1(y_2-y_1)dx. When y_2=3\int_0^1\sqrt xdx and y_1=\int_0^1 3xdx. A=3\int_0^1\sqrt xdx-3\int_0^1 xdx. On integrating we get, A=3\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^1-3\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_0^1 \;\;=2\begin{bmatrix}x^{\large\frac{3}{2}}\end{bmatrix}_0^1-\large\frac{3}{2}\begin{bmatrix}\normalsize x^2\end{bmatrix}_0^1 On applying limits we get, 2[1-0]-\large\frac{3}{2}$$[1-0]=2-\large\frac{3}{2}=\frac{1}{2}$
Hence the required area is $\large\frac{1}{2}$sq.units.
edited Apr 26, 2013
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# more integrals
• Oct 16th 2011, 10:49 AM
Random Variable
more integrals
Don't hide solutions.
1) $\displaystyle \int_{0}^{\frac{\pi}{2}}} \frac{\arctan(\sin x)}{\sin x} \ dx$
2) $\displaystyle \int^{\infty}_{0} \frac{\sin x \ln x}{x} \ dx$
3) $\displaystyle \int_{0}^{\frac{\pi}{2}}} \ln(\cos x) \ln(\sin x) \cos x \sin x \ dx$
• Oct 16th 2011, 12:34 PM
Random Variable
Re: more integrals
• Oct 17th 2011, 03:36 PM
uniquesailor
Re: more integrals
• Oct 17th 2011, 11:46 PM
FernandoRevilla
Re: more integrals
We have $\displaystyle \lim_{x\to 0^+}\frac{\arctan(\lambda \sin x)}{\sin x}=\ldots=\lambda$ , so the following function is well defined (by means of a convergent integral):
$\displaystyle I:[0,+\infty)\to \mathbb{R},\;\;I(\lambda)=\int_0^{\pi/2}\frac{\arctan (\lambda \sin x)}{\sin x}dx$
Then $\displaystyle I'(\lambda)=\int_0^{\pi/2}\frac{dx}{1+\lambda^2\sin^2 x}$
Using the substitution $\displaystyle t=\tan x$:
$\displaystyle I'(\lambda)=\int_0^{+\infty}\frac{ \frac{dt}{1+t^2} }{ 1+\frac{\lambda^2 t^2}{1+t^2} }=\int_0^{+\infty}\frac{dt}{(1+\lambda^2)t^2+1}$
Using the substitution $\displaystyle u=\sqrt{1+\lambda^2}t$:
$\displaystyle I'(\lambda)=\frac{1}{\sqrt{1+\lambda^2}}\int_{0}^{ +\infty}\frac{du}{u^2+1}=\frac{1}{\sqrt{1+\lambda^ 2}}\dfrac{\pi}{2}$
Integrating both sides:
$\displaystyle I(\lambda)=\dfrac{\pi}{2}\log |\lambda+\sqrt{\lambda^2+1}|+C$
For $\displaystyle \lambda=0$ we have $\displaystyle 0=0+C$ that is, $\displaystyle C=0$. Then, $\displaystyle I(\lambda)=\dfrac{\pi}{2}\log |\lambda+\sqrt{\lambda^2+1}|$.
As a consequence:
$\displaystyle \int_0^{\pi/2}\frac{\arctan (\sin x)}{\sin x}dx=I(1)=\dfrac{\pi}{2}\log (1+\sqrt{2})=\boxed{\dfrac{\pi}{2}\sinh^{-1}(1)}$
• Dec 5th 2011, 12:10 AM
simplependulum
Re: more integrals
Consider $\displaystyle ab = \frac{1}{2}[a^2 + b^2 - (a-b)^2]$ , we have
$\displaystyle \ln[\sin(x)]\ln[\cos(x)] = \frac{1}{2} \left( \ln^2[\sin(x)] + \ln^2[\cos(x)] - \ln^2[\tan(x)] \right)$
$\displaystyle \int_0^{\frac{\pi}{2}}\ln^2[\sin(x)] \sin(x)\cos(x) ~dx$
$\displaystyle = \int_0^{\frac{\pi}{2}}\ln^2[\cos(x)]\sin(x)\cos(x) ~dx = \int_0^1 x\ln^2(x)~dx = \frac{1}{4}$
$\displaystyle \int_0^{\frac{\pi}{2}}\ln^2[\tan(x)] \sin(x)\cos(x) ~dx$
$\displaystyle = \int_0^{\infty} \frac{x \ln^2(x)}{(x^2+1)^2}~dx$
$\displaystyle = 2 \int_0^1 \frac{x \ln^2(x)}{(x^2+1)^2}~dx$
$\displaystyle = \left[ \frac{x^2\ln^2(x)}{x^2+ 1}\right]_0^1 - 2\int_0^1 \frac{x\ln(x)}{x^2+1}~dx = \frac{\pi^2}{24}$
so the answer is $\displaystyle \frac{1}{4} - \frac{\pi^2}{48}$
• Mar 4th 2012, 11:19 PM
sbhatnagar
Re: more integrals
Quote:
Originally Posted by uniquesailor
Can you explain what is 'MAZ identity'?
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### Math Monday: Pentominoes
by George Hart
A pentomino is like a domino, but with five connected squares instead of two. There are twelve ways to connect squares edge-to-edge in the plane, not counting rotations and flips. A set of all twelve can be cut from scraps of plywood using any kind of saw. Large sets are fun to play with, so I based these on three inch squares. It is made from half-inch plywood, cut on a band saw and lightly sanded.As the area of all twelve pieces totals sixty squares, a natural puzzle is to try to fill a 6×10 rectangle. There are over 2000 solutions! But even though you are allowed to flip and rotate the pieces however you wish, it is harder to solve than you might think.The 5×12 rectangle above is another challenging puzzle to try once you make your set. This one is three feet across.
You can also make a 4×15 rectangle, as shown above. And the same twelve pentominoes can make the 3×20 rectangle, below, which is five feet long.It is interesting that as we consider longer, skinnier rectangles, the number of possible solutions is drastically reduced. These four rectangles have 2339, 1010, 368, and 2 solutions, respectively.
This article first appeared on Make: Online, October 17, 2011.
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# Find the equation of the least squares regression line if x-bar= 15 sx=2 y-bar = 17.1 sy=3 r= 0.2
Find the least squares regression line if xbar = 3 ,sx= 3.2 ,ybar=2 ,sy=1.7 ,r=-0.7
Slope of the regression line (b) = r*sy/sx = -0.7*1.7/3.2 = -0.371875
Intercept = ybar - b*xbar = 2 - (-0.371875)*3 = 3.115625
Hence,
The regression line will be:
y = 3.115625 - 0.371875 x
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
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### MEAP Preparation - Grade 7 Mathematics1.87 Fractions Review Test
Q 1: (-21/7)__?__(-63/21)=<> Q 2: One morning you walked 4 7/8 miles to town. On the way home you stopped to rest after walking 1 1/3 miles. How far do you still need to walk to get back home?2 13/24 miles3 13/24 miles3 11/5 miles Q 3: 52-3(5-3)+23 =Answer: Q 4: 21/8 - 20/9 = ?29/72349/7272/29 Q 5: (-7/8)__?__(-5/8)<>= Q 6: Tim earns half day of vacation time for every two weeks of work. How much vacation time does he earn for working 6 1/2 weeks?2 5/8 of a day1 1/8 of a day1 5/8 of a day Q 7: To do an experiment, Sam needed 1/12 of gram of Cobalt. If Mike gave him 1/4 of that amount, how much Cobalt did she give Sam?1/24 of a gram1/48 of a gram1/16 of a gram Q 8: {3/10 + 4(1/10 + 2/10)} = _____ ?1 1/22 1/25/3 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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Grade 9 Algebra Topics. (e) (1 3 2=2(1 2 i) 𝑥+3 3 − 4 =12 j) 6 +4= 4 − 2 2. Factorise using the trinomial method.
A) 9 2−64 2 b) 16 4−1 c) ) 81−121 t6 d) (3 + 2− 2 3. Area of large trapezoid = (1/2) 3 (7+3) = (1/2) 3 (10) = 15 area of small trapezoid = (1/2) 2 (5+3) = 8 problem 2 : Starting this series for these grade 9 topics because they’re a lot easier than your teachers make them out to be🤞 #maths #gcsemaths #algebra #grade9 2.6k likes, 29 comments.
Source: www.edugain.com
Combining like terms in algebraic expressions. The comprehensive collection of ninth grade math topics listed here focuses on the below areas.
### Removing Radicals From The Denominator.
Starting this series for these grade 9 topics because they’re a lot easier than your teachers make them out to be🤞 #maths #gcsemaths #algebra #grade9 2.6k likes, 29 comments. In cbse class 9, the chapter on algebraic expressions and identities is introduced. A) 9 2−64 2 b) 16 4−1 c) ) 81−121 t6 d) (3 + 2− 2 3.
One number is 12 more than the other. Solve the following linear equations for 𝑥: And explanations for grade 9 detailed solutions and full explanations to grade 9 algebra questions are presented.
Enable learner to gain confidence to study for and write tests and exams on the topic. Complete the table below and answer the questions that follow. ( a + b) 2 = a 2 + 2 a b + b 2.
### 9Th Grade Pre Algebra Worksheets 1.
Simplify the following algebraic expressions. Curriculum framework for grade 9. The sum of 3 consecutive whole numbers is 108.
### Algebraic Equations Grade 9 Mathematics 1.
Intro to expressions exponents and radicals polynomials linear equations linear inequalities systems of equations and inequalities introduction to functions sequences and exponential functions piecewise and transformations quadratic functions In this article, we give you an introduction to algebraic techniques and equations. Area of large trapezoid = (1/2) 3 (7+3) = (1/2) 3 (10) = 15 area of small trapezoid = (1/2) 2 (5+3) = 8 problem 2 :
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# math
posted by .
What is the answer to t=100(n+3)? Solving for N.
• math -
t = 100n + 300
t-300 = 100n
n = (t-300)/100
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Frustum Problems
Number of problems found: 33
• Equilateral cone
We pour so much water into a container that has the shape of an equilateral cone, the base of which has a radius r = 6 cm, that one-third of the volume of the cone is filled. How high will the water reach if we turn the cone upside down?
• Cutting cone
A cone with a base radius of 10 cm and a height of 12 cm is given. At what height above the base should we divide it by a section parallel to the base so that the volumes of the two resulting bodies are the same? Express the result in cm.
• Runcated pyramid teapot
The 35 cm high teapot has the shape of a truncated pyramid with the length of the edge of the lower square base a=50 cm and with the edges of the rectangular base b: 20 cm and c: 30 cm. How many liters of water will fit in the teapot?
• Frustum of a cone
A reservoir contains 28.54 m3 of water when full. The diameter of the upper base is 3.5 m, while at the lower base is 2.5 m. Find the height if the reservoir is in the form of a frustum of a right circular cone.
• Truncated cone 5
The height of a cone 7 cm and the length of side is 10 cm and the lower radius is 3cm. What could the possible answer for the upper radius of truncated cone?
• Truncated cone 3
The surface of the truncated rotating cone S = 7697 meters square, the substructure diameter is 56m and 42m, determine the height of the tang.
• Hexagon cut pyramid
Calculate the volume of a regular 6-sided cut pyramid if the bottom edge is 30 cm, the top edge is 12 cm, and the side edge length is 41 cm.
• Lamp cone
Calculate the surface of a lampshade shaped of a rotary truncated cone with a base diameter of 32 cm and 12 cm and height of 24 cm.
• Truncated cone
A truncated cone has bases with radiuses of 40 cm and 10 cm and a height of 25 cm. Calculate its surface area and volume.
• Truncated pyramid
How many cubic meters is the volume of a regular four-sided truncated pyramid with edges one meter and 60 cm and high 250 mm?
• Pit
The pit has the shape of a truncated pyramid with a rectangular base and is 0.8 m deep. The pit's length and width are the top 3 × 1.5 m bottom 1 m × 0.5 m. To paint one square meter of the pit we use 0.6 l of green color. How many liters of paint are nee
• Truncated cone
Calculate the height of the rotating truncated cone with volume V = 794 cm3 and a base radii r1 = 9.9 cm and r2 = 9.8 cm.
• Pillar
Calculate volume of pillar shape of a regular tetrahedral truncated pyramid, if his square have sides a = 19, b = 27 and height is h = 48.
• 2x cone
Circular cone height 84 cm was cut plane parallel with base. Volume of these two small cones is the same. Calculate the height of the smaller cone.
• Truncated cone
Calculate the volume of a truncated cone with base radiuses r1=13 cm, r2 = 10 cm and height v = 8 cm.
• Flowerbed
Flowerbed has the shape of a truncated pyramid, the bottom edge of the base a = 10 m, the upper base b = 9 m. Deviation angle between edge and the base is alpha = 45°. What volume is needed to make this flowerbed? How many plants can be planted if 1 m2 =
Calculate the surface area and volume of a regular quadrangular pyramid: sides of bases (bottom, top): a1 = 18 cm, a2 = 6cm angle α = 60 ° (Angle α is the angle between the sidewall and the base plane.) S =? , V =?
• A concrete pedestal
A concrete pedestal has a shape of a right circular cone having a height of 2.5 feet. The diameter of the upper and lower bases are 3 feet and 5 feet, respectively. Determine the lateral surface area, total surface area, and the volume of the pedestal.
• Heptagonal pyramid
A hardwood for a column is in the form of a frustum of a regular heptagonal pyramid. The lower base edge is 18 cm, and the upper base of 14 cm. The altitude is 30 cm. Determine the weight in kg if the wood density is 10 grams/cm3.
• Truncated cone 6
Calculate the volume of the truncated cone whose bases consist of an inscribed circle and a circle circumscribed to the opposite sides of the cube with the edge length a=1.
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## Homework help multiplying decimals
Multiply decimals and whole numbers: The procedure for multiplying decimals and whole numbers is presented step-by-step at Math Goodies.Become an Arithmetic champ with our arithmetic problem solver and skill builder.These questions provide learners with practice in vertical multiplication of decimal.We can perform all operations on decimals be it addition, subtraction, multiplication and division.Multiply without the decimal point, then re-insert it in the correct spot.Adding, Subtracting, Multiplying, and Dividing Decimals Overview: In this lesson you will learn the proper way to add, subtract, multiply, and divide decimals and.Review their homework for any issues that need to be addressed.
Math is hard enough.deciphering word problems can make it impossible.Learning Objective Multiplication and Division of Fractions and Decimal Fractions: Math Terminology for Module 4.
### Ordering Decimals Worksheet
Multiplication with decimals becomes easy when you practice and test your skills using this printable worksheet and interactive quiz.Learn fractions, percentages and decimals with our fun algebra videos and worksheets for multiple grade levels.Each worksheet has 15 problems multiplying a decimal by a decimal using a visual model.The Decimals and Fractions chapter of this High School Algebra I Homework Help course helps students complete their decimals and fractions homework.
I need help with my math homework on decimals: Decimal patterns with powers of 10.Divide 4 divided by 0.5.Running on a five minute timer, these tasks are ideal for use as a starter.
Arithmetic Homework Help Multiplication Fractions Decimal Multiplication:.CMP3 grade 6 offers concepts and explanations of the math, worked homework examples and math.Note that this problem is designed to help students connect decimals to their.Multiplying And Dividing Fractions And Mixed Numbers. 7 lessons.A website for maths teachers to share their homework creations.
### Multiplying and Dividing Decimals Worksheets
Watch this clip from Homework Hotline to see how to multiply decimals.Homework: 5.NBT.7: Add, subtract, multiply, and divide decimals to hundredths. by OpenEd.
### Variable Decimal Worksheets
This Math Homework Help Reference Guide gives you all the basic math skills you need to succeed.
Example: 5.63 x 3.7 Example: 0.53 x 2.618 Try This: 6.5 x 15.3 Multiplication Properties Homework Time Multiplying Decimals Lesson 1-7 To Multiply:.Generate free printable worksheets for addition, subtraction, multiplication, and division of decimals for grades 3-7.
### Do You Line Up Decimals When Multiplying
These grade 6 math worksheets focus on decimal multiplication including multiplying decimals by whole numbers, by.Homework Unlocked offers homework help to parents so you can help your kids with homework.
### 5th Grade Decimal Math Worksheets Printable
We explain Multiplying Decimals by Fractions with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers.In this lesson, students see.
### Adding Subtracting Multiplying and Dividing Decimals
Parents can better help children with fractions, decimals and other math operations if they consult a glossary explaining what the terminology means.
### Free Printable Kumon Math Worksheets
Learn how to explain dividing decimals using multiplication.A math homework community.Webmath is a math-help web site that generates answers to specific math questions and problems, as entered by a user, at any particular moment.Homework Unlocked offers homework help to parents so they can teach their kids how to excel in math.
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## 558870
558,870 (five hundred fifty-eight thousand eight hundred seventy) is an even six-digits composite number following 558869 and preceding 558871. In scientific notation, it is written as 5.5887 × 105. The sum of its digits is 33. It has a total of 5 prime factors and 32 positive divisors. There are 137,472 positive integers (up to 558870) that are relatively prime to 558870.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 33
• Digital Root 6
## Name
Short name 558 thousand 870 five hundred fifty-eight thousand eight hundred seventy
## Notation
Scientific notation 5.5887 × 105 558.87 × 103
## Prime Factorization of 558870
Prime Factorization 2 × 3 × 5 × 13 × 1433
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 558870 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 558,870 is 2 × 3 × 5 × 13 × 1433. Since it has a total of 5 prime factors, 558,870 is a composite number.
## Divisors of 558870
32 divisors
Even divisors 16 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 1.44547e+06 Sum of all the positive divisors of n s(n) 886602 Sum of the proper positive divisors of n A(n) 45171 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 747.576 Returns the nth root of the product of n divisors H(n) 12.3723 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 558,870 can be divided by 32 positive divisors (out of which 16 are even, and 16 are odd). The sum of these divisors (counting 558,870) is 1,445,472, the average is 45,171.
## Other Arithmetic Functions (n = 558870)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 137472 Total number of positive integers not greater than n that are coprime to n λ(n) 4296 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45881 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 137,472 positive integers (less than 558,870) that are coprime with 558,870. And there are approximately 45,881 prime numbers less than or equal to 558,870.
## Divisibility of 558870
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 4 6 6
The number 558,870 is divisible by 2, 3, 5 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (558870)
Base System Value
2 Binary 10001000011100010110
3 Ternary 1001101121220
4 Quaternary 2020130112
5 Quinary 120340440
6 Senary 15551210
8 Octal 2103426
10 Decimal 558870
12 Duodecimal 22b506
20 Vigesimal 39h3a
36 Base36 bz86
## Basic calculations (n = 558870)
### Multiplication
n×i
n×2 1117740 1676610 2235480 2794350
### Division
ni
n⁄2 279435 186290 139718 111774
### Exponentiation
ni
n2 312335676900 174555039749103000 97553575064581193610000 54519766496342491672820700000
### Nth Root
i√n
2√n 747.576 82.3702 27.3418 14.1079
## 558870 as geometric shapes
### Circle
Diameter 1.11774e+06 3.51148e+06 9.81231e+11
### Sphere
Volume 7.31174e+17 3.92493e+12 3.51148e+06
### Square
Length = n
Perimeter 2.23548e+06 3.12336e+11 790362
### Cube
Length = n
Surface area 1.87401e+12 1.74555e+17 967991
### Equilateral Triangle
Length = n
Perimeter 1.67661e+06 1.35245e+11 483996
### Triangular Pyramid
Length = n
Surface area 5.40981e+11 2.05715e+16 456315
## Cryptographic Hash Functions
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digital
# Fractions Quick Quiz Bundle: Adding, Subtracting, Multiplication & More!
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### Description
This bundle contains 4 fraction quick quizzes! (also sold separately)
Are you looking for a quick way to assess fractions? If so, this is your one stop shop!
This 8 question quick quiz assesses the following 5th grade math learning targets:
• I can add and subtract fractions and mixed numbers with unlike denominators .
• I can solve word problems involving addition and subtraction of fractions.
• I can represent the context of a fraction using a variety of models.
Questions include a combination of conceptual and procedural problems.
Sample Problems:
1. 3/4 + 7/12
2. Stacy is making a model boat. She has 5 ½ feet of wood. She uses 2 ¼ feet for the hull and 1 ½ feet for a paddle. How much wood does she have left?
3. Draw an area model to solve: ⅔ - ⅕
There is also an error analysis page for students to re-work problems correctly and provide explanations of their errors.
Document is completely editable to meet your needs.
CCSS.MATH.CONTENT.5.NF.A.1
Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)
CCSS.MATH.CONTENT.5.NF.A.2
Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.
© Copyright 2018 ProjectBasedSixth. All rights reserved. Permission is granted to copy pages specifically designed for student or teacher use by the original purchaser or licensee. This is intended to be used by one teacher unless additional licenses have been purchased. The reproduction of any other part of this product is strictly prohibited.
Are you looking for a quick way to assess multiplying fractions? If so, this is your one stop shop!
This 8 question quick quiz assesses the following 5th grade math learning targets:
• I can multiply a whole number or a fraction by a fraction.
• I can prove my product is correct by using a visual model.
• I can find the area of a rectangle (with fractional side lengths).
Questions include a combination of conceptual and procedural problems.
Sample Problems:
1. Draw an area model to solve the area of a rectangle that has a width of 3
1/2 and a length of 4 2/5.
2. Next week, fifth graders hope to collect 1 ⅓ times as many bags of canned goods as the first week, 3 ½ . How many bags of canned goods do they hope to collect?
3. 5 x 2 3/4
There is also an error analysis page for students to re-work problems correctly and provide explanations of their errors.
Document is completely editable to meet your needs.
CCSS.MATH.CONTENT.5.NF.B.3
Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
CCSS.MATH.CONTENT.5.NF.B.4
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
CCSS.MATH.CONTENT.5.NF.B.4.A
Interpret the product (a/b) × q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a × q ÷ b. For example, use a visual fraction model to show (2/3) × 4 = 8/3, and create a story context for this equation. Do the same with (2/3) × (4/5) = 8/15. (In general, (a/b) × (c/d) = (ac)/(bd).
CCSS.MATH.CONTENT.5.NF.B.4.B
Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas.
© Copyright 2018 ProjectBasedSixth. All rights reserved. Permission is granted to copy pages specifically designed for student or teacher use by the original purchaser or licensee. This is intended to be used by one teacher unless additional licenses have been purchased. The reproduction of any other part of this product is strictly prohibited.
Are you looking for a quick way to assess fractions? If so, this is your one stop shop!
This 8 question quick quiz assesses the following 5th grade math learning targets:
• I can use benchmark fractions to estimate and solve problems
• I can create a line plot using fraction data
• I can analyze a line plot to answer questions and solve problems
Questions include a combination of conceptual and procedural problems.
Sample Problems:
1. Explain how you know that ⅝ + 6/10 is greater than 1.
2. Make a line plot with the following data. Students measured amounts of water in
cups. These were their results: ¼, ¼, ½, ¾, ¼, ¼, ¼, ½, ¼, ¾, ¼, ¾
3. Round ⅜ to its closest benchmark.
There is also an error analysis page for students to re-work problems correctly and provide explanations of their errors.
Document is completely editable to meet your needs.
CCSS.MATH.CONTENT.5.NF.A.2
Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.
CCSS.MATH.CONTENT.5.MD.B.2
Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots. For example, given different measurements of liquid in identical beakers, find the amount of liquid each beaker would contain if the total amount in all the beakers were redistributed equally.
© Copyright 2018 ProjectBasedSixth. All rights reserved. Permission is granted to copy pages specifically designed for student or teacher use by the original purchaser or licensee. This is intended to be used by one teacher unless additional licenses have been purchased. The reproduction of any other part of this product is strictly prohibited.
Are you looking for a quick way to assess fractions, NF 5 & 6? If so, this is your one stop shop!
This 8 question quick quiz assesses the following 5th grade math learning targets:
• I can compare the size of a product to the size of its factors (without performing multiplication)
• I can explain the result of multiplying a given number by a fraction greater than and less than 1.
• I can represent the context of a fraction word problem using a variety of models.
Questions include a combination of conceptual and procedural problems.
Sample Problems:
1. Macy worked on her math project for 3 ¾ hours. Chris worked on his project ⅓ times as long as Macy. Kathryn worked on her math project 1 ¼ times as long as Macy. Without solving, who worked the longest? How do you know?
2. Teachers are going to make cookies for the staff. They want to make 3 times the amount the recipe calls for, so everyone will get a cookie. If the recipe calls for ¼ teaspoon of sugar, how much sugar will they need?
3. Will the product of 5/3 x 3 be greater or less than 3? Explain.
There is also an error analysis page for students to re-work problems correctly and provide explanations of their errors.
Document is completely editable to meet your needs.
CCSS.MATH.CONTENT.5.NF.B.5
Interpret multiplication as scaling (resizing), by:
CCSS.MATH.CONTENT.5.NF.B.5.A
Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
CCSS.MATH.CONTENT.5.NF.B.5.B
Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n × a)/(n × b) to the effect of multiplying a/b by 1.
CCSS.MATH.CONTENT.5.NF.B.6
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
© Copyright 2019 ProjectBasedSixth. All rights reserved. Permission is granted to copy pages specifically designed for student or teacher use by the original purchaser or licensee. This is intended to be used by one teacher unless additional licenses have been purchased. The reproduction of any other part of this product is strictly prohibited.
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### Standards
to see state-specific standards (only available in the US).
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence 𝘢/𝘣 = (𝘯×𝘢)/(𝘯×𝘣) to the effect of multiplying 𝘢/𝘣 by 1.
Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
Interpret multiplication as scaling (resizing), by:
Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas.
| 2,746 | 11,821 |
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# Question
If a baker bakes 6 Apple pies and 9 strudels, he will have 450g of flour left. If he bakes 10 Apple pies and 12 strudels, he will need another 900g of flour. If th Baker needs 150g of flour to bak an Apple pie, how much does the baker have?
Thanks!
2 Answers
# Answer
900 g + 450 g + 900 g = 2250 g
3 Strudels –> 2250 g – 1500 g = 750 g
9 Strudels –> 2250 g
Total amount of flour –> 2250 g + 900 g + 450 g = 3600 g
Baker has 3600 g of flour.
1 Reply 0 Likes ✔Accepted Answer
Thanks alot! I really need pictorial help. So this is very clear to me.
0 Replies 0 Likes
If a baker bakes 6 apple pies and 9 strudels, he will have 450g of flour left. If he bakes 10 apple pies and 12 strudels, he will need another 900g of flour. If the baker needs 150g of flour to bake an apple pie, how much flour does the baker have?
6 apple pies + 9 strudels + 450g ——- 10 apple pies and 12 strudels – 900g
450 + 900 = 1350g ——- 4 apple pies + 3 strudels
1 apple pie ——- 150g
4 apple pies ——- 600g
1350g ——- 600g + 3 strudels
1 strudel ——- (1350 – 600)/3 = 250
6 x 150 + 9 x 250 + 450 = 3600
or
10 x 150 + 12 x 250 – 900 = 3600
Ans : 3600g of flour
1 Reply 0 Likes
Thanks!
1 Reply 0 Likes
Don’t mention.
0 Replies 0 Likes
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| 505,895,031 | 14,722 |
1J30.20-Tension in a String
Video: Watch this demo | Demo #2
Description: A counter intuitive demonstration that exposes students’ common misunderstandings about tension and scales.
Equipment:
• One table
• Two 1kg masses
• One hanging Newton scale
• Two table clamps
• Two vertical posts
• Two s-clamps
• Two pulleys
• Two thin strings, each approximately 1m long
Setup Procedure:
1. Set up the posts in the table clamps.
2. Attach the pulleys to the tops of the posts with the s-clamps. Make sure the pulleys are at roughly the SAME LEVEL.
3. Tie loops in the ends of the strings.
Demonstration Procedure:
1. First, hang a 1kg mass vertically from the spring scale, to show that the scale reads about 10N, as one would expect.
2. Next, ask the class what the scale will read when the weight is hung on a string, which passes through one pulley and then attaches horizontally to the scale, which is affixed to a stiff post (see photo).
3. Show class that the scale still reads about 10N* in this configuration, proving that the only thing that changes is the direction of the string tension, NOT the magnitude.
4. Ask the class what the scale will read if instead of affixing the scale to a stiff post, you hang a SECOND 1kg mass from a string, around a pulley, and then attached to the other end of the scale. Nearly all of them will guess 20N!
5. Show them that even in this configuration, the scale reads about 10N*. A free-body diagram of the scale can help show when affixed to the stiff post, the scale felt two (roughly) equal and opposite forces: the tension of in string and the force from the post. In the last configuration, the free-body diagram is almost the same; the two forces are both tensions in the strings. In both cases, the scale reads the magnitude of one force, not both.
Note: In the second configuration-and even more so in the third-the scale will read slightly LESS than 10N. This is because when the scale is horizontal, part of the 10N tension goes to a vertical component that counteracts the weight of the scale, but doesn’t contribute to the scale’s reading. In the second configuration, the stiff post contributes some of the upward force, and thus less has to come from tension, so the effect is less noticeable there.
22-B2
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https://www.space.com/how-long-would-it-take-to-walk-around-mars
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# How long would it take to walk around Mars?
Humans have long had a fascination with Mars, and NASA has ambitious plans to send astronauts there within the next few decades. Anyone walking on Mars would likely explore only a small fraction of the planet's surface.
But without oceans or other bodies of water, could an astronaut walk all the way around the Red Planet? How long would it take to walk around Mars?
Unsurprisingly, a long time — though exactly how long might be hard to say. If the astronaut could travel continuously at a walking speed, it would be a simple calculation.
"We would need essentially two parameters," said Erdal Yiğit, an associate professor of physics and astronomy at George Mason University who studies the atmospheres of planets. Those parameters are the astronaut's velocity (speed and direction) and the distance they would travel.
If the person traveled along Mars' equator, they would walk about 13,300 miles (21,400 kilometers) to get all the way around the planet. Walking around Mars through its poles would shave off about 100 miles (160 km), but the extreme cold would pose an even greater challenge than the harsh conditions on the rest of the planet, Yiğit said.
The person's velocity would be about 3.1 mph (5 km/h), which is also an average walking speed on Earth, straight along the equator, Yiğit said. Despite Mars' reduced gravity (about 40% of Earth's), Yiğit doubts a person's walking speed would be much different on Mars. Like any back-country hiker on Earth, this person would likely be carrying a heavy load of supplies — such as oxygen, water and food — and would be wearing a heavy spacesuit.
If someone were to walk continuously around Mars at that speed, the calculation would be simple: Just divide the distance by the velocity. That would mean it would take about 4,290 hours. There are about 24.7 hours in a Martian day (called a sol), so it would take roughly 174 sols to walk around Mars continuously. That's a little over a quarter of a Martian year, which is 668.6 sols.
Of course, no one would be able to complete that walk continuously — on Mars, Earth or anywhere. Even if the person were able to bring enough oxygen, water and food with them, and could eat and drink while walking, they would still need to stop to sleep. Assuming the astronaut slept about eight hours each night, it would add about 56 sols. If the person stopped for four or five more hours each sol to eat, rest, change clothes, clean themselves, and set up and deconstruct some type of campsite, it would take another 30 or 35 sols, depending on how long they were stopped.
All in all, a more realistic estimate might be at least 265 sols, about 40% of the Martian year. But that calculation does not account for potential obstacles, such as rough terrain. Mars has many mountains, including some that are taller than any on Earth, as well as valleys, craters and many other geological features that would be tough to navigate.
Of course, it is very unlikely that anyone will be walking all the way around Mars anytime soon. People have circumnavigated Earth on foot, though it is, of course, impossible to truly walk or run all the way around, due to the oceans. But humans have walked on only a small portion of the moon, despite traveling there multiple times. And walking so far and for so long on Mars would pose many logistical problems, like bringing enough food, water and oxygen and protecting the person from the dangerous impacts of radiation.
Although it's highly unlikely that humans will walk around the entire Red Planet, sending astronauts to the surface still has many advantages over rovers, Yiğit said.
Rovers "are susceptible to dust and other kinds of electrical problems; something may happen," he said. With humans "even if there are problems, we can find solutions around them."
Join our Space Forums to keep talking space on the latest missions, night sky and more! And if you have a news tip, correction or comment, let us know at: [email protected].
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https://www.gradesaver.com/textbooks/math/algebra/algebra-2-common-core/chapter-7-exponential-and-logarithmic-functions-7-4-properties-of-logarithms-lesson-check-page-465/6
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## Algebra 2 Common Core
RECALL: (i) Product Property of Logarithms: $\log_{a}{bc}= \log_a{b} + \log_a{c}$ (ii) Quotient Property of Logarithms: $\log_a{\left(\dfrac{x}{y}\right)}=\log_a{x} - \log_a{y}$ Thus, (a) The property that will be used to write the given expression as a single logarithm is the Product Property of Logarithms. (b) The property that will be used to write the given expression as a single logarithm is the Quotient Property of Logarithms.
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https://cpep.org/business/2923186-the-2016-financial-statements-of-bnsf-railway-company-report-total-rev.html
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4 November, 18:58
# The 2016 financial statements of BNSF Railway Company report total revenues of \$19,829 million, accounts receivable of \$1,272 million for 2016 and \$1,198 million for 2015. The company's accounts receivable turnover for the year is: Select one: A. 17.0 times B. 8.9 times C. 16.1 times D. 17.9 times E. None of the above
+1
1. 4 November, 20:40
0
C. 16.1 times
Explanation:
Accounts receivable turnover ratio = Credit sales : average accounts receivable
where,
Average accounts receivable = (Opening balance of Accounts receivable + ending balance of Accounts receivable) : 2
= (\$1,198 + \$1,272) : 2
= \$1,235 million
And, the net credit sale is \$ 19,829 million
Now put these values to the above formula
So, the answer would be equal to
= \$19,829 million : \$1,235 million
= 16.1 times
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https://www.coolstuffshub.com/speed/convert/inches-per-hour-to-miles-per-minute/
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# Convert Inches per hour to Miles per minute (in/h to mi/min Conversion)
1 in/h = 2.82222 × 10-7 mi/min
## How to convert inches per hour to miles per minute?
To convert inches per hour to miles per minute, multiply the value in inches per hour by 2.82222 × 10-7.
You can use the conversion formula :
miles per minute = inches per hour × 2.82222 × 10-7
To calculate, you can also use our inches per hour to miles per minute converter, which is a much faster and easier option as compared to calculating manually.
## 1 inch per hour is equal to how many miles per minute?
1 inch per hour is equal to 2.82222 × 10-7 miles per minute.
• 1 inch per hour = 2.82222 × 10-7 miles per minute
• 2 inches per hour = 5.64444 × 10-7 miles per minute
• 3 inches per hour = 8.46666 × 10-7 miles per minute
• 4 inches per hour = 1.128888 × 10-6 miles per minute
• 5 inches per hour = 1.41111 × 10-6 miles per minute
• 10 inches per hour = 2.82222 × 10-6 miles per minute
• 100 inches per hour = 2.82222 × 10-5 miles per minute
## Examples to convert in/h to mi/min
Example 1:
Convert 50 in/h to mi/min.
Solution:
Converting from inches per hour to miles per minute is very easy.
We know that 1 in/h = 2.82222 × 10-7 mi/min.
So, to convert 50 in/h to mi/min, multiply 50 in/h by 2.82222 × 10-7 mi/min.
50 in/h = 50 × 2.82222 × 10-7 mi/min
50 in/h = 1.41111 × 10-5 mi/min
Therefore, 50 inches per hour converted to miles per minute is equal to 1.41111 × 10-5 mi/min.
Example 2:
Convert 125 in/h to mi/min.
Solution:
1 in/h = 2.82222 × 10-7 mi/min
So, 125 in/h = 125 × 2.82222 × 10-7 mi/min
125 in/h = 3.527775 × 10-5 mi/min
Therefore, 125 in/h converted to mi/min is equal to 3.527775 × 10-5 mi/min.
For faster calculations, you can simply use our in/h to mi/min converter.
## Inches per hour to miles per minute conversion table
Inches per hour Miles per minute
0.001 in/h 2.82222 × 10-10 mi/min
0.01 in/h 2.82222 × 10-9 mi/min
0.1 in/h 2.82222 × 10-8 mi/min
1 in/h 2.82222 × 10-7 mi/min
2 in/h 5.64444 × 10-7 mi/min
3 in/h 8.46666 × 10-7 mi/min
4 in/h 1.128888 × 10-6 mi/min
5 in/h 1.41111 × 10-6 mi/min
6 in/h 1.693332 × 10-6 mi/min
7 in/h 1.975554 × 10-6 mi/min
8 in/h 2.257776 × 10-6 mi/min
9 in/h 2.539998 × 10-6 mi/min
10 in/h 2.82222 × 10-6 mi/min
20 in/h 5.64444 × 10-6 mi/min
30 in/h 8.46666 × 10-6 mi/min
40 in/h 1.128888 × 10-5 mi/min
50 in/h 1.41111 × 10-5 mi/min
60 in/h 1.693332 × 10-5 mi/min
70 in/h 1.975554 × 10-5 mi/min
80 in/h 2.257776 × 10-5 mi/min
90 in/h 2.539998 × 10-5 mi/min
100 in/h 2.82222 × 10-5 mi/min
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https://warreninstitute.org/domain-and-range-of-continuous-graphs-worksheet-answers/
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# Exploring Graphs: Domain & Range Answers in Continuous Graphs
Welcome to Warren Institute! In this article, we will explore the fascinating topic of domain and range of continuous graphs. Understanding the domain and range of a graph is crucial in Mathematics education as it helps us determine the set of possible inputs and outputs. In this worksheet, we will provide you with a set of graphs and ask you to determine their corresponding domains and ranges. By practicing with these answers, you will gain a deeper understanding of the concept and improve your problem-solving skills. So let's dive in and sharpen our knowledge of domain and range!
## Understanding Domain and Range in Continuous Graphs
The first step in solving the domain and range of continuous graphs is to understand the concept itself. Domain refers to the set of all possible input values (x-values) for a function, while range represents the set of all possible output values (y-values).
In this worksheet, students are provided with various continuous graphs and are required to determine the domain and range. This exercise helps them develop a solid understanding of how the input and output values relate to each other.
## Identifying Key Features of Continuous Graphs
To find the domain and range of a continuous graph, it's important to identify its key features. These features include the highest and lowest points on the graph, any vertical asymptotes, and any restrictions or limitations stated in the problem.
By analyzing these key features, students can determine the starting and ending points of the graph, as well as any gaps or intervals where the graph is undefined. This process allows them to accurately define the domain and range of the given function.
## Applying Mathematical Concepts to Determine Domain and Range
Once students have identified the key features of a continuous graph, they can apply mathematical concepts to determine the domain and range. For example, if the graph is a line or a curve that extends infinitely in both directions, the domain and range would be all real numbers.
If there are restrictions or limitations, such as vertical asymptotes or excluded values, students need to carefully consider these factors when determining the domain and range. It's important to emphasize the use of inequalities and interval notation to express these values accurately.
## Checking Answers and Seeking Help
After completing the worksheet, it's crucial for students to check their answers and seek help if needed. They can use graphing calculators or online tools to verify their findings and ensure that the domain and range they have determined align with the graph.
If students encounter difficulties or have questions about the concepts covered in the worksheet, they should reach out to their teacher or classmates for assistance. Collaboration and clarification are essential in mastering the domain and range of continuous graphs.
### What is the domain of a continuous graph?
The domain of a continuous graph refers to the set of all possible input values or x-values for which the graph is defined. It represents the range of values over which the function or relation is meaningful and can be graphed.
### How is the domain of a continuous graph determined?
The domain of a continuous graph is determined by identifying the set of all possible input values, or x-values, for which the function is defined and continuous.
### What is the range of a continuous graph?
The range of a continuous graph in Mathematics education refers to the set of all possible output values (or y-values) that the graph can attain. It represents the vertical extent or the dependent variable values covered by the graph.
### How can we find the range of a continuous graph?
To find the range of a continuous graph, we need to identify the highest and lowest points on the graph. The range is the set of all possible output values of a function. In other words, it represents the vertical spread of the graph. By determining the highest and lowest points, we can determine the range.
### What are the common misconceptions about determining the domain and range of continuous graphs?
One common misconception about determining the domain and range of continuous graphs is that the domain and range are always the same as the x and y axes respectively. However, this is not always the case. The domain of a continuous graph is the set of all possible x-values for which the function is defined, while the range is the set of all possible y-values that the function takes on. It is important to consider the behavior of the graph and any restrictions or limitations in order to accurately determine the domain and range.
In conclusion, understanding the domain and range of continuous graphs is crucial in Mathematics education. This worksheet answers provide students with the opportunity to practice and reinforce their knowledge of these concepts. By accurately identifying the domain and range of various functions, students can develop a deeper understanding of how graphs behave and how they can be used to solve real-world problems. The domain represents all possible input values, while the range represents all possible output values. Through this worksheet, students can gain confidence in their ability to analyze and interpret graphs, enhancing their overall mathematical proficiency.
See also Mastering BMI: Unveiling the Secrets of Body Mass Index Calculation
If you want to know other articles similar to Exploring Graphs: Domain & Range Answers in Continuous Graphs you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
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https://massinitiative.org/how-many-acres-is-174240-square-feet/
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# How many acres is 174240 square feet?
## How many acres is 174240 square feet?
Acres to Square Feet Conversions
Acres Square Feet
1 acre 43560 square feet
2 acres 87120 square feet
3 acres 130680 square feet
4 acres 174240 square feet
### How many square feet means 1 acre?
43,560
Acre to Square Feet Conversion (ac to sq ft) Table
Acre Square feet
1 43,560
2 87,120
3 130,680
4 174,240
#### What is the square feet of 1/2 acre?
Each acre is 43560 square feet, so half of that is half an acre.
What is the square footage of a 1/4 acre?
A . 25-acre lot, aka a 1/4th acre lot, is only 10,890 square feet.
Is 200 ft by 200 ft an acre?
200 feet x 200 feet = 0.918 acres. Or in other words, approximately 92% of an acre.
## How many square feet is .5 acres?
Five acres is 217,800 square feet.
### How big of a house can I build on .25 acres?
In general, a 0.25-acre plot of land is large enough to build a family home with enough additional space for garages, a lawn and garden space.
#### How much does an acre of land cost?
According to the USDA, the average cost of farmland in the U.S. is \$3,160 per acre, but that number quickly rises once you start looking at land in popular metro areas. Knowing the value of an acre is important, especially if you’re planning to buy a home and need to take out a mortgage.
What is the dimension of 1 acre?
approximately 208.71 feet × 208.71 feet
1 acre is approximately 208.71 feet × 208.71 feet (a square) 4,840 square yards. 43,560 square feet. 160 perches.
How many houses can fit on 5 acres?
Looks like there may be around 6 to 8 houses to a block, so five acres might have twelve to sixteen homes on it. Or consider this five acre block superimposed over everyone’s favorite size comparator: a football field. As you can see, five acres is quite a bit bigger.
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https://www.onlinesalesguidetip.com/consider-this-very-simple-equation-to-transform-your-productivity/
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# Consider this very simple equation to transform your productivity
By Next Big Idea Club
February 05, 2022
I woke up in the middle of the night, utterly confused.
I had been dreaming—of an equation. I was in a classroom, and on the blackboard, there was a simple equation written in chalk: 0.8 * 0.2 = 0.16.
To be clear: I rarely ever dream about math. When I do, it’s often a nightmare about bombing a theoretical physics final in college.
Even though this was no nightmare, the equation really bothered me. The middle-school math checks out: 0.8 multiplied by 0.2 in fact equals 0.16. What bothered me was the underlying implication: The product of two numbers could be less than each number (0.16 is less than both 0.8 and 0.2).
The outcome makes sense to the astrophysicist in me. But in the dream, I was staring at this equation as a mathematical beginner, completely befuddled by the result. How could that be? If you multiply two numbers together, shouldn’t the result be greater than each part?
Dreams are written in disappearing ink. But this one lingered with me for a while, as if there was a message embedded within that I needed to learn.
And then the message hit me. When we operate at a fraction, we compromise the output.
Most of us go through life operating at a fraction in everything we do. We check email during Zoom meetings. We shove a sandwich into our mouth with one hand while scrolling through our phone with the other. When we work, we think about play. When we play, we think about work. We exist in this in-between state—we’re neither fully here nor there—constantly operating at a 0.2 or a 0.8 instead of a full 1.
As a result, our output suffers. What we produce becomes less than what we put in. We achieve only an iota of what we’re capable.
When people meet a great leader, they often say, “She made me feel like I was the only person in the room.” Imagine giving that type of complete attention to everything you do—and making that thing the only thing in the room.
Not just deep work. But deep play. Deep rest. Deep listening. Deep reading. Deep love. Deep everything.
This mindset requires being aware of your own limitations. When I write, for example, my output begins to significantly drop after about two hours. By the fourth hour, I’m operating at a 0.2 at best. If I keep pushing, I know I’ll write gobbledygook that isn’t even worth editing. I’m much better off taking my hands off the keyboard and giving my attention to something else.
This isn’t just about being mindful of what you’re doing. It’s choosing one activity over others. It’s approaching everything you do with intention. Often, we’re operating impulsively, moving from one notification to the next, one email to the next, living our lives in a frantic blur.
But if you slow down for just a moment and intentionally choose to give your full attention to what you’re going to do next, you’ll trigger your internal defibrillator that can jolt you back alive and bring you closer to your full capacity.
It’s like the way a great leader might shake your hand and greet you. Hello activity. It’s great to meet you. I’m choosing to engage with you. I’m going to treat you like you’re the most important person in the room and ignore everyone else.
0.8 * 0.2 = 0.16. There’s now a Post-it note on my desk with that equation.
It serves as a constant reminder to live deep and suck out all the marrow of life, as Thoreau put it, instead of living at a fraction of what I’m capable.
Ozan Varol is a rocket scientist turned law professor and bestselling author. Click here to download a free copy of his eBook, The Contrarian Handbook: 8 Principles for Innovating Your Thinking. And download the Next Big Idea App to enjoy a “Book Bite” Summary of his latest book, Think Like a Rocket Scientist.
This article originally appeared in Next Big Idea Club magazine and is reprinted with permission.
(0)
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# Sum(Partition(Binary String)) = $2^k$
So given any binary string B:
$$b_1 b_2 \dots b_n$$ $$b_i \in \{0,1\}$$
It would seem it is always possible to make a partitioning of B:
$$b_1 b_2 \dots b_{p_1}|b_{p_1 + 1}b_{p_1 + 2}\dots b_{p_2}|\dots|b_{p_m + 1}b_{p_m + 2}\dots b_n$$ $$= P_0 | P_1 |\dots|P_m$$
such that when $P_i$ is interpreted as the binary representation of an integer then:
$$\exists{k}:\sum_{i=0}^{m}P_i = 2^k$$
For example here is the first few:
1 = 1
10 = 2
1|1 = 2
100 = 4
1|01 = 2
11|1 = 4
1000 = 8
1|00|1 = 2
10|10 = 4
1|0|11 = 4
...
How can we prove this for all binary strings?
-
Trying induction on the weight (= the number of ones), but it doesn't work. The best start I have found is to begin with the observation that with any pattern of weight 3 or 4 gives rise to a sum $=4$. Thus any pattern of weights 6,7 or 8 can be split into two groups given a four for a total of $8$. Continuing this we see that a number of weight in the range 12 to 16 can always be partitioned to yield a total of $16$. Unfortunately this leaves gaps such as weight 5. Five consecutive ones force us to split it $1_2+1111_2=16_{10}$. :-( – Jyrki Lahtonen Jun 12 '12 at 14:58
@JyrkiLahtonen: Consider 1100111, it can be 1100 + 1 + 11 or 1 + 100 + 1 + 1 + 1. I don't see a pattern here. – Andrew Tomazos Jun 12 '12 at 15:02
@JyrkiLahtonen: Also note that 111...1 can always be split 1 + 11...1, as 11..1 will always be $2^k -1$ – Andrew Tomazos Jun 12 '12 at 15:05
I know. I was trying to force that weight five gang to also give me an eight. If that had succeeded, then the weight class 12 to 16 could be extended to 10 to 16 leaving 9 as the next problem kid. If a weight five has a string of two zeros in between, then we get $100+1+1+1+1=8$ or even with a single zero between two ones we can get eight as $101+1+1+1=8$. For the induction step (approximately doubling the set of weights handled) I need 8 specifically as the sum. Because it has to paired with another 8 so that the total is again a power of two. – Jyrki Lahtonen Jun 12 '12 at 15:13
Actually a better way of helping the problem preventing me from taking the inductive step is to observe that some strings of weight 1 or 2 can also give a 4, and thus paired up with a group of weight 3 or 4. Something's still missing. – Jyrki Lahtonen Jun 12 '12 at 15:26
This is (essentially) problem 6 in the MSRI newsletter, Emissary, for Fall 2011:
You are allowed to transform positive integers $n$ in the following way. Write $n$ in base 2. Write plus signs between the bits at will (at most one per position), and then perform the indicated additions of binary numbers. For example, $123_{10} = 1111011$ can get + signs after the second, third and fifth bits to become $11+1+10+11 = 9_{10}$; or it can get + signs between all the bits to become $1+1+1+1+0+1+1 = 6_{10}$; and so on. Prove that it is possible to reduce arbitrary positive integers to 1 in a bounded number of steps. That is, there is an absolute constant $C$ such that for any $n$ there is a sequence of at most $C$ transformations that starts with $n$ and ends at 1. Comment: The best possible bound is a real shocker.
No solution is given.
-
C is 2 because powers of 2 are always of the form 100..0. So its N -> (as per Q) 100..0 -> 1 + 00..0 = 1 – Andrew Tomazos Jun 12 '12 at 13:11
Yes --- if you can prove what you assert in the question. – Gerry Myerson Jun 12 '12 at 13:21
I just collected some data: pastebin.com/3B2P4asC. It seems there are many different $2^k$ partitionings for any given n. Perhaps we can make an argument that there are so many partitionings ($2^{n-1}$) of an n-bit string - that at least some of them must be powers of 2. Perhaps we can show that each partitioning must vary from the others sufficiently to unavoidably spread over a near power of 2. – Andrew Tomazos Jun 12 '12 at 14:38
Look at two and three bit strings starting with 1, and the minimum and maximum ways they can contribute to the sum. $$\begin{array}{c|c|c} T & A(T) & B(T) \\ \hline \\ 1 & 1 & 1 \\ 10 & 1 & 2 \\ 11 & 2 & 3 \\ 100 & 1 & 4 \\ 101 & 2 & 5 \\ 110 & 2 & 6 \\ 111 & 3 & 7 \end{array}$$ For example, we can count 110 as $1+1+0=A(110)$ or as $6=B(110)$ (the only other possibilities are $1+2$ and $3+0$). So for the two-bit strings $B(T)=A(T)+1$, and for the three-bit strings $B(T)-A(T)\in\{3,4\}$.
Now consider a number $k$ and let $w(k)\ge 6$ be the number of 1 bits in its binary representation.
Split $k$ into components like this: $$(k)_2 = T_1 0^* T_2 0^* T_3 \ldots T_{c+3} 0^* T_{end}$$ where each $T$ starts with 1, each of $T_1,T_2,T_3$ has exactly two bits, and each of $T_4,\ldots,T_{c+3}$ has exactly three bits, and $T_{end}$ can be empty or can contain 1,10 or 11 if $(k)_2$ ends that way without allowing another 3-bit term. In other words, gather terms in a greedy fashion starting with the next available 1-bit taking two or three bits as required, skipping extra zeros, and assigning the left-over to $T_{end}$. For example $$(999999)_2 = 11~~11~~0~~10~~000~~100~~0~~111~~111$$ so for $n=999999$ we would get $$T_1=T_2=11,T_3=10,T_4=100,T_5=111,T_6=111,c=3,T_{end}=\text{empty}$$
For each type of term $A(T)$ counts the bits in $T$, so $\sum A(T_i) = w(k)$ (including $T_{end}$ in the sum). Now we can use our terms as a counter as follows.
• Start counting each term with $A(T)$ to sum to $w(k)$.
• To increment, count $T_1$ as $B(T_1)$ to get $w(k)+1$.
• Count $T_2$ as $B(T_2)$ to get $w(k)+2$.
• Count $T_3$ as $B(T_3)$ to get $w(k)+3$.
• Count $T_4$ as $B(T_4)$ and reset to $A(T_i),i=1,2,3$ to get either $w(k)+3$ or $w(k)+4$.
Repeat this, counting more 3-bit terms as $B(T)$ to add 3 or 4 at a time (and finally $B(T_{end})$ if applicable), counting $T_1,T_2,T_3$ as $A$ or $B$ to get the increments in between. By this counting method we find partitions that sum to each number between $w(k)=\sum A(T_i)$ and $\sum B(T_i)$.
Let $z$ be the number of terms that are $T=111, A(T)=3, B(T)=7$. Then \begin{align} w(k) = \sum A(T_i) & \le 6+(2c+z)+2 \\ w(k)-8 & \le 2c+z \le 3c \\ c & \ge w(k)/3-8/3 \\ \sum B(T_i) & \ge \sum A(T_i)+3+3c+z \\ & \ge w(k)+3+w(k)/3-8/3+w(k)-8 \\ & = 7w(k)/3-23/3 \end{align}
So for any $k$ we can find a partition that sums to each value in $[w(k),\max(w(k)+3,7w(k)/3-23/3)]$. For all $w(k)\ge 6, w(k)\ne 9, w(k)\ne 10$ this interval contains a power of 2.
We handle the remaining possible values for $w(k)$ as special cases.
For $w(k)=10$ we can just use a refinement of the bound above. In this case $c\ge 1$, so $\sum B(T_i) \ge \sum A(T_i)+6$, so 16 will be included.
$w(k)=1,2,4$ are immediate, just add the bits. For $w(k)=3$ we can always partition the number into $10,1,1$ or $11,1$ and possibly some zeros. For $w(k)=5$ if all 1 bits are adjacent, we can write it as $1111,1$, otherwise we can write it as $101,1,1,1$ or $100,1,1,1,1$.
For $w(k)=9$ if all bits are adjacent, then we can take $11111111,1$. Otherwise there is either a 100 or 101 which we take as $T_1$. From the remaining bits make a three-bit term $T_2$ and a two-bit term $T_3$. Then either $B(T_1)+B(T_2)$ or $B(T_1)+B(T_2)+B(T_3)$ plus the remaining bits makes 16.
As an addendum, we can extend this technique taking longer terms, e.g. for 4-bit terms $A(T)+7 \le B(T) \le A(T)+11$, so we can take 11 two-bit terms and $c$ 4-bit terms and count from $w(k)$ to at least $w(k)+11+7c$.
For 5-bit terms $A(T)+15 \le B(T) \le A(T)+26$ and with 26 two-bit terms we can count from $w(k)$ to $w(k)+26+15\lfloor (w-26)/5\rfloor$. For $w(k)$ large enough this upper bound is larger than $3w(k)$, so the interval will contain a power of 3.
Carrying it further, taking $l$-bit terms for larger $l$ we can extend this interval to any multiple of $w(k)$. Thus, for choice of any $q>1$ there is a bound $W(q)$ so that any number $k$ with Hamming weight large enough $w(k)>W(q)$ can be partitioned into $P_i$ with $\sum P_i = q^m$ for some $m$.
You lost me at the very beginning when you introduced $A$ and $B$ without defining them. – Gerry Myerson Jun 13 '12 at 6:10
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Tesco Aptitude Questions 2014
Posted on :22-03-2016
Q1. A+B means A is brother of B
A-B means A is father of B
A*B means A is mother of B.
then what is A*B-C?
Q2. \$ represents 1, @ represents 0 all the other number greater than 1 are represented by the combination of \$ and @ only then, how do you represent 172?
Directions for questions 3, 4:
A if first and second are alike,
B if first and third are alike,
c if all the three are alike,
D if all are different.
Q3. 89172981.12830912, 89172981. 12830912, 819172981. 128830912
Q4. 99282048209, 99282044209, 99282048209
Q5. What is the value of 2/3+6/3-7*8-9?
Q6. Which one gives the same results as 9/8+5*4/2+5?
a) 4+6/3*8/9
b) 8/9+4*3/4-8
Q7. For the admission into a college the following criteria are required.
i) aggregate of 70% in the qualifying course.
ii) age between 21 and 28
iii) maths as the prerequisite.
iv).............so on six conditions
options:
a- if all conditions are satisfied then give the admission
b- if condition two is not satisfied refer to the chairman
c- if condition x is nor satisfies then refer to the principal
.......so on five conditions
Mr A has an aggregate of 60% and born on 27th Aug 1988 with Maths as his specialization in his course......................then he must be
A) a
B) b
C) c
D) d
E) e
Q8. In a school there are 1000 students in the year 1999. The number of students increased by 20% in the year 2000. And it is increased by 15% in the year 2001. But it is decreased by 18% in 2002. Then what is the strength in 2002?
Directions for questions 9 to 12:
A solid cube of 4 inches has been painted red, green and black on the pairs of opposite faces. It has then been cut into one inch cubes. Following questions relate to the smaller one inch cubes.
Q9. How many cubes have only two faces painted?
(1) 0
(2) 16
(3) 24
(4) 32
Q10. How many cubes have no faces painted?
(1) 0
(2) 8
(3) 16
(4) 24
Q11. How many cubes have only one face painted?
(1) 8
(2) 16
(3) 24
(4) 32
Q12. How many cubes have only four faces painted?
(1) 0
(2) 8
(3) 12
(4) 16
Q13. 1 chocolate cost is Rs.1 & if we return 3 wrapers (covers) of chocolate we get 1 more chocolate. If you have Rs.15 how many chocolates do you get?
Q14. If 8 men 8 hrs per day works for 8 days get 45/- then how many men required if the work is 5 hrs per day for 10 days they get 60/-?
Q15. The compound interest for first and second years is 200 and 220 on a certain amount. Find the sum.
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✖
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# How to Find the Range of a Data Set in Excel: A Step-by-Step Guide
Finding the range of a data set in Excel is easy and can be accomplished in a few simple steps. The range is the difference between the highest and lowest values in a data set. This guide will show you how to use Excel functions and formulas to quickly determine the range of any given set of numbers.
## Step-by-Step Tutorial: How to Find the Range of a Data Set in Excel
By following these steps, you will be able to find the range of your data set in Excel, which is useful for understanding the spread of values in your data.
Open the Excel file that contains your data set.
Make sure your data is organized in columns or rows. This will make it easier to perform calculations.
### Step 2: Identify the Data Set
Locate the specific column or row that contains the data set you want to analyze.
You need to know exactly where your data is to correctly reference it in your formulas.
### Step 3: Find the Maximum Value
Type the formula `=MAX(range)` in an empty cell, replacing "range" with the actual range of cells containing your data (e.g., `=MAX(A1:A10)`).
This will return the highest value in your data set, which is necessary to calculate the range.
### Step 4: Find the Minimum Value
Type the formula `=MIN(range)` in another empty cell, replacing "range" with the same cell range you used in the previous step (e.g., `=MIN(A1:A10)`).
This will return the lowest value in your data set.
### Step 5: Calculate the Range
Type the formula `=MAX(range)-MIN(range)` in another empty cell, replacing "range" with the actual range of cells (e.g., `=MAX(A1:A10)-MIN(A1:A10)`).
This will give you the range of your data set by subtracting the minimum value from the maximum value.
Once you’ve completed these steps, you will have the range of your data set displayed in the cell where you typed the final formula.
## Tips for Finding the Range of a Data Set in Excel
• Double-check your data range to ensure you’ve included all necessary cells.
• Use the `AutoFill` feature to extend your formulas to other columns or rows.
• If your data set includes blank cells, make sure they don’t interfere with your `MAX` and `MIN` functions.
• Utilize Excel’s built-in error checking to catch any mistakes in your formulas.
• Consider visualizing your data with charts to better understand the spread and range.
### What is the range of a data set?
The range is the difference between the highest and lowest values in a data set.
### Why is finding the range important?
The range helps you understand the spread and variability of your data, which is crucial for analysis.
### Can I find the range of non-numeric data?
No, the range calculation only applies to numeric data. Non-numeric data cannot be used with `MAX` and `MIN` functions.
### What if my data includes outliers?
Outliers can skew the range. Consider using a modified range or other statistical measures if outliers are present.
### Can I automate this process in Excel?
Yes, you can create a template with pre-set formulas to quickly find the range for different data sets.
## Summary of Steps
2. Identify the data set.
3. Find the maximum value.
4. Find the minimum value.
5. Calculate the range.
## Conclusion
Finding the range of a data set in Excel is a straightforward process that can be done in just a few steps. By using the built-in `MAX` and `MIN` functions, you can quickly determine the spread of your data. This is particularly useful for anyone looking to analyze data, whether for school projects, professional reports, or personal use. Understanding the range helps you grasp how your values are distributed and can point out any anomalies or outliers within your data.
So, next time you open Excel and need to analyze your data set, remember these simple steps. If you’re looking to dive deeper into data analysis, there are plenty of other powerful Excel functions and tools to explore. Happy data crunching!
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# ApproxFun.jl
ApproxFun is a package for approximating functions. It is in a similar vein to the Matlab package `Chebfun` and the Mathematica package `RHPackage`.
The `ApproxFun Documentation` contains detailed information, or read on for a brief overview of the package.
The `ApproxFun Examples` contains many examples of using this package, in Jupyter notebooks and Julia scripts.
## Introduction
Take your two favourite functions on an interval and create approximations to them as simply as:
``````using LinearAlgebra, SpecialFunctions, Plots, ApproxFun
x = Fun(identity,0..10)
f = sin(x^2)
g = cos(x)
``````
Evaluating `f(.1)` will return a high accuracy approximation to `sin(0.01)`. All the algebraic manipulations of functions are supported and more. For example, we can add `f` and `g^2` together and compute the roots and extrema:
``````h = f + g^2
r = roots(h)
rp = roots(h')
plot(h; label="f + g^2")
scatter!(r, h.(r); label="roots")
scatter!(rp, h.(rp); label="extrema")
``````
## Differentiation and integration
Notice from above that to find the extrema, we used `'` overridden for the `differentiate` function. Several other `Julia` base functions are overridden for the purposes of calculus. Because the exponential is its own derivative, the `norm` is small:
``````f = Fun(x->exp(x), -1..1)
norm(f-f') # 4.4391656415701095e-14
``````
Similarly, `cumsum` defines an indefinite integration operator:
``````g = cumsum(f)
g = g + f(-1)
norm(f-g) # 3.4989733283850415e-15d
``````
Algebraic and differential operations are also implemented where possible, and most of Julia's built-in functions are overridden to accept `Fun`s:
``````x = Fun()
f = erf(x)
g = besselj(3,exp(f))
h = airyai(10asin(f)+2g)
``````
## Solving ordinary differential equations
We can also solve differential equations. Consider the Airy ODE `u'' - x u = 0` as a boundary value problem on `[-1000,200]` with conditions `u(-1000) = 1` and `u(200) = 2`. The unique solution is a linear combination of Airy functions. We can calculate it as follows:
``````x = Fun(identity, -1000..200) # the function x on the interval -1000..200
D = Derivative() # The derivative operator
B = Dirichlet() # Dirichlet conditions
L = D^2 - x # the Airy operator
u = [B;L] \ [[1,2],0] # Calculate u such that B*u == [1,2] and L*u == 0
plot(u; label="u")
``````
## Nonlinear Boundary Value problems
Solve a nonlinear boundary value problem satisfying the ODE `0.001u'' + 6*(1-x^2)*u' + u^2 = 1` with boundary conditions `u(-1)==1`, `u(1)==-0.5` on `[-1,1]`:
``````x = Fun()
u₀ = 0.0x # initial guess
N = u -> [u(-1)-1, u(1)+0.5, 0.001u'' + 6*(1-x^2)*u' + u^2 - 1]
u = newton(N, u₀) # perform Newton iteration in function space
plot(u)
``````
One can also solve a system nonlinear ODEs with potentially nonlinear boundary conditions:
``````
x=Fun(identity, 0..1)
N = (u1,u2) -> [u1'(0) - 0.5*u1(0)*u2(0);
u2'(0) + 1;
u1(1) - 1;
u2(1) - 1;
u1'' + u1*u2;
u2'' - u1*u2]
u10 = one(x)
u20 = one(x)
u1,u2 = newton(N, [u10,u20])
plot(u1, label="u1")
plot!(u2, label="u2")
``````
## Periodic functions
There is also support for Fourier representations of functions on periodic intervals. Specify the space `Fourier` to ensure that the representation is periodic:
``````f = Fun(cos, Fourier(-π..π))
norm(f' + Fun(sin, Fourier(-π..π))) # 5.923502902288505e-17
``````
Due to the periodicity, Fourier representations allow for the asymptotic savings of `2/π` in the number of coefficients that need to be stored compared with a Chebyshev representation. ODEs can also be solved when the solution is periodic:
``````s = Chebyshev(-π..π)
a = Fun(t-> 1+sin(cos(2t)), s)
L = Derivative() + a
f = Fun(t->exp(sin(10t)), s)
B = periodic(s,0)
uChebyshev = [B;L] \ [0.;f]
s = Fourier(-π..π)
a = Fun(t-> 1+sin(cos(2t)), s)
L = Derivative() + a
f = Fun(t->exp(sin(10t)), s)
uFourier = L\f
ncoefficients(uFourier)/ncoefficients(uChebyshev),2/π
plot(uFourier)
``````
## Sampling
Other operations including random number sampling using [Olver & Townsend 2013]. The following code samples 10,000 from a PDF given as the absolute value of the sine function on `[-5,5]`:
``````f = abs(Fun(sin, -5..5))
x = ApproxFun.sample(f,10000)
histogram(x;normed=true)
plot!(f/sum(f))
``````
## Solving partial differential equations
We can solve PDEs, the following solves Helmholtz `Δu + 100u=0` with `u(±1,y)=u(x,±1)=1` on a square. This function has weak singularities at the corner, so we specify a lower tolerance to avoid resolving these singularities completely.
``````d = ChebyshevInterval()^2 # Defines a rectangle
Δ = Laplacian(d) # Represent the Laplacian
f = ones(∂(d)) # one at the boundary
u = \([Dirichlet(d); Δ+100I], [f;0.]; # Solve the PDE
tolerance=1E-5)
surface(u) # Surface plot
``````
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1. 1. Chapter 23 Solutions 10.0 grams 23 atoms electrons = 2.62 × 10 2423.1 (a) N= 6.02 × 10 47.0 107.87 grams mol mol atom Q 1.00 × 10 −3 C (b) # electrons added = = = 6.25 × 1015 e 1.60 × 10 -19 C electron or 2.38 electrons for every 10 9 already present ( )( ) 2 k q1q 8.99 × 10 9 N ⋅ m 2/ C 2 1.60 × 10 −19 C23.2 (a) Fe = e 2 2 = = 1.59 × 10 − 9 N (repulsion) r (3.80 × 10 − 10 m)2 (b) Fg = G m1m2 = ( ) 6.67 × 10 −11 N ⋅ m 2 kg 2 (1.67 × 10 − 27 kg)2 = 1.29 × 10 − 45 N r2 (3.80 × 10 −10 m)2 The electric force is larger by 1.24 × 10 36 times q1q2 mm (c) If ke 2 = G 12 2 with q1 = q2 = q and m1 = m2 = m, then r r q G 6.67 × 10 −11 N ⋅ m 2 / kg 2 = = = 8.61 × 10 −11 C / kg m ke 8.99 × 10 9 N ⋅ m 2 / C 223.3 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains 70, 000 grams 23 molecules protons N≈ 6.02 × 10 10 ≈ 2.3 × 10 28 protons 18 grams mol mol molecule With an excess of 1% electrons over protons, each person has a charge q = (0.01)(1.6 × 10 −19 C)(2.3 × 10 28 ) = 3.7 × 107 C q1q2 (3.7 × 107 )2 So F = ke = (9 × 10 9 ) N = 4 × 10 25 N ~ 1026 N r2 0.6 2 This force is almost enough to lift a "weight" equal to that of the Earth: Mg = (6 × 10 24 kg)(9.8 m s 2 ) = 6 × 10 25 N ~ 1026 N © 2000 by Harcourt, Inc. All rights reserved.
2. 2. 2 Chapter 23 Solutions23.4 We find the equal-magnitude charges on both spheres: q1q2 q2 F 1.00 × 10 4 N F = ke 2 = ke 2 so q=r = (1.00 m ) = 1.05 × 10 −3 C r r ke 8.99 × 10 9 N ⋅ m 2/ C 2 The number of electron transferred is then ( N xfer = 1.05 × 10 −3 C ) (1.60 × 10 −19 ) C / e − = 6.59 × 1015 electrons The whole number of electrons in each sphere is Ntot = 10.0 g ( − )( 24 − 6.02 × 10 atoms / mol 47 e / atom = 2.62 × 10 e 107.87 g / mol 23 ) The fraction transferred is then N xfer 6.59 × 1015 f= = = 2.51 × 10–9 = 2.51 charges in every billion Ntot 2.62 × 1024 ( )( ) (6.02 × 10 ) 2 23 2 qq 8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C23.5 F = ke 1 2 2 = = 514 kN [ ] 2 r 2(6.37 × 106 m)*23.6 (a) The force is one of attraction. The distance r in Coulombs law is the distance between centers. The magnitude of the force is F= ke q1q2 = 8.99 × 10 9 ( −9 )( −9 N ⋅ m 2 12.0 × 10 C 18.0 × 10 C = 2.16 × 10 − 5 N ) r2 C2 (0.300 m)2 (b) The net charge of − 6.00 × 10 −9 C will be equally split between the two spheres, or − 3.00 × 10 −9 C on each. The force is one of repulsion, and its magnitude is F= ke q1q2 = 8.99 × 10 9 N⋅m 2 ( )( 3.00 × 10 −9 C 3.00 × 10 −9 C = ) 8.99 × 10 −7 N r2 C2 (0.300 m)2
3. 3. Chapter 23 Solutions 3 q1q2 (8.99 × 10 9 N ⋅ m 2/ C 2 )(7.00 × 10 −6 C)(2.00 × 10 −6 C)23.7 F1 = ke = = 0.503 N r2 (0.500 m)2 q1q2 (8.99 × 10 9 N ⋅ m 2 / C 2 )(7.00 × 10 −6 C)(4.00 × 10 −6 C) F2 = k e = = 1.01 N r2 (0.500 m)2 Fx = (0.503 + 1.01) cos 60.0° = 0.755 N Fy = (0.503 − 1.01) sin 60.0° = − 0.436 N F = (0.755 N)i − (0.436 N)j = 0.872 N at an angle of 330°Goal SolutionThree point charges are located at the corners of an equilateral triangle as shown in Figure P23.7.Calculate the net electric force on the 7.00− µ C charge.G: Gather Information: The 7.00− µ C charge experiences a repulsive force F1 due to the 2.00− µ C charge, and an attractive force F 2 due to the −4.00− µ C charge, where F2 = 2F1. If we sketch these force vectors, we find that the resultant appears to be about the same magnitude as F2 and is directed to the right about 30.0° below the horizontal.O: Organize : We can find the net electric force by adding the two separate forces acting on the 7.00− µ C charge. These individual forces can be found by applying Coulomb’s law to each pair of charges.A: Analyze: The force on the 7.00− µ C charge by the 2.00− µ C charge is F1 = (8.99 × 10 9 )( )( N ⋅ m 2/ C 2 7.00 × 10 −6 C 2.00 × 10 −6 C ) (cos60°i + sin 60°j) = F = (0.252i + 0.436j) N (0.500 m) 2 1 Similarly, the force on the 7.00− µ C by the −4.00− µ C charge is F 2 = − 8.99 × 10 9 ( −6 )( −6 N ⋅ m 2 7.00 × 10 C − 4.00 × 10 C ) (cos60°i − sin 60° j) = (0.503i − 0.872j) N C2 (0.500 m)2 Thus, the total force on the 7.00− µ C , expressed as a set of components, is F = F1 + F 2 = (0.755 i − 0.436 j) N = 0.872 N at 30.0° below the +x axisL: Learn: Our calculated answer agrees with our initial estimate. An equivalent approach to this problem would be to find the net electric field due to the two lower charges and apply F=qE to find the force on the upper charge in this electric field. © 2000 by Harcourt, Inc. All rights reserved.
4. 4. 4 Chapter 23 Solutions*23.8 Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by ke ( 3q)Q ke ( q)Q F= i+ ( −i) x2 ( d − x )2 3 1 x The net force will be zero if = , or d − x = x 2 ( d − x )2 3 This gives an equilibrium position of the third bead of x = 0.634d The equilibrium is stable if the third bead has positive charge . k ee 2 (1.60 × 10–19 C)2*23.9 (a) F= 2 = (8.99 × 109 N ⋅ m2/C 2) = 8.22 × 10–8 N r (0.529 × 10–10 m)2 (b) We have F = mv 2 from which v = Fr = (8.22 × 10 −8 )( N 0.529 × 10 −10 m )= 2.19 × 106 m/s r −31 m 9.11 × 10 kg ke qQ23.10 The top charge exerts a force on the negative charge which is directed upward and ( d 2 )2 + x 2 to the left, at an angle of tan −1 ( d / 2x ) to the x-axis. The two positive charges together exert force 2 k qQ (− x)i − 2 ke qQ 2 e = ma or for x << d 2, a≈ ( ) ( x d 4 + x2 d2 4 + x2 ) 1/2 md 3 / 8 (a) The acceleration is equal to a negative constant times the excursion from equilibrium, as i n 16 ke qQ a = −ω 2 x, so we have Simple Harmonic Motion with ω 2 = . md 3 2π π md 3 T= = , where m is the mass of the object with charge −Q. ω 2 ke qQ ke qQ (b) vmax = ω A = 4a md 3
5. 5. Chapter 23 Solutions 5 mg23.11 For equilibrium, F e = −F g , or qE = −mg( −j) . Thus, E = j. q (9.11 × 10 −31 kg)(9.80 m s 2 ) j = − ( 5.58 × 10 −11 N C) j mg E= j= ( ) (a) q −1.60 × 10 −19 C E= mg j= ( )( 1.67 × 10 −27 kg 9.80 m s 2 j= ) (1.02 × 10 −7 ) ( ) (b) NC j q 1.60 × 10 −19 C23.12 ∑ Fy = 0: QE j + mg(− j) = 0 QE (24.0 × 10 -6 C)(610 N / C) ∴ m= = = 1.49 grams g 9.80 m / s 2*23.13 The point is designated in the sketch. The magnitudes of the electric fields, E1, (due to the –2.50 × 10–6 C charge) and E2 (due to the 6.00 × 10–6 C charge) are k eq (8.99 × 109 N · m2/C 2)(2.50 × 10–6 C) E1 = = (1) r2 d2 k eq (8.99 × 109 N · m2/C 2)(6.00 × 10–6 C) E2 = = (2) r2 (d + 1.00 m)2 Equate the right sides of (1) and (2) to get (d + 1.00 m)2 = 2.40d 2 or d + 1.00 m = ±1.55d which yields d = 1.82 m or d = – 0.392 m The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus, d = 1.82 m to the left of the -2.50 µC charge.23.14 If we treat the concentrations as point charges, q N ⋅ m 2 ( 40.0 C) E + = ke = 8.99 × 10 9 (−j) = 3.60 × 105 N / C (−j) (downward) r 2 C 2 (1000 m )2 q N ⋅ m 2 ( 40.0 C) E − = ke = 8.99 × 10 9 (−j) = 3.60 × 105 N / C (−j) (downward) r2 C 2 (1000 m )2 E = E + + E − = 7.20 × 10 5 N / C downward © 2000 by Harcourt, Inc. All rights reserved.
6. 6. 6 Chapter 23 Solutions*23.15 (a) ke q E1 = 2 = ( )( 8.99 × 10 9 7.00 × 10 −6 ) = 2.52 × 10 5 N C r (0.500) 2 E2 = ke q = ( )( 8.99 × 10 9 4.00 × 10 −6 ) = 1.44 × 10 5 N C r2 (0.500)2 Ex = E2 − E1 cos 60° = 1.44 × 10 5 − 2.52 × 10 5 cos 60.0° = 18.0 × 10 3 N C Ey = −E1 sin 60.0° = −2.52 × 10 5 sin 60.0° = −218 × 10 3 N C E = [18.0i − 218 j] × 10 3 N C = [18.0i − 218 j] kN C (b) ( ) F = qE = 2.00 × 10 −6 C (18.0i − 218 j) × 10 3 N C = ( 36.0i − 436 j) × 10 −3 N = (36.0i − 436 j ) mN*23.16 (a) E1 = ke q1 (− j) = (8.99 × 10 )(3.00 × 10 ) (− j) = − (2.70 × 10 9 −9 3 ) NC j 2 r1 (0.100)2 E2 = ke q2 (− i) = (8.99 × 10 )(6.00 × 10 ) (− i) = − (5.99 × 10 9 −9 2 ) NC i 2 r2 (0.300)2 ( ) ( E = E 2 + E1 = − 5.99 × 10 2 N C i − 2.70 × 10 3 N C j ) (b) ( ) F = qE = 5.00 × 10 −9 C ( −599i − 2700 j) N C ( F = − 3.00 × 10 −6 i − 13.5 × 10 −6 j N = ) (− 3.00 i − 13.5 j ) µN23.17 (a) The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. (b) You may need to review vector addition in Chapter Three. The magnitude of the field at point P due to each of the charges along the base of the triangle is E = ke q a 2 . The direction of the field in each case is along the line joining the charge in question to point P as shown in the diagram at the right. The x components add to zero, leaving ke q kq ke q 2 ( E= sin 60.0°) j + e2 (sin 60.0°) j = 3 j a a a2
7. 7. Chapter 23 Solutions 7Goal SolutionThree equal positive charges q are at the corners of an equilateral triangle of side a, as shown in FigureP23.17. (a) Assume that the three charges together create an electric field. Find the location of a point(other than ∞) where the electric field is zero. (Hint: Sketch the field lines in the plane of the charges.)(b) What are the magnitude and direction of the electric field at P due to the two charges at the base?G: The electric field has the general appearance shown by the black arrows in the figure to the right. This drawing indicates that E = 0 at the center of the triangle, since a small positive charge placed at the center of this triangle will be pushed away from each corner equally strongly. This fact could be verified by vector addition as in part (b) below. The electric field at point P should be directed upwards and about twice the magnitude of the electric field due to just one of the lower charges as shown in Figure P23.17. For part (b), we must ignore the effect of the charge at point P , because a charge cannot exert a force on itself.O: The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E1 + E2A: (b) The electric field from a point charge is q As shown in the solution figure above, E1 = ke to the right and upward at 60° a2 q E 2 = ke to the left and upward at 60° a2 a q [ ] a q [ ] a q E = E1 + E2 = ke 2 (cos60°i + sin 60° j) + ( − cos60°i + sin 60° j) = ke 2 2(sin 60°j) = 1.73ke 2 jL: The net electric field at point P is indeed nearly twice the magnitude due to a single charge and is entirely vertical as expected from the symmetry of the configuration. In addition to the center of the triangle, the electric field lines in the figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically. ke q (8.99 × 10 9 )(2.00 × 10 −6 )23.18 (a) E= = = 14, 400 N C r2 (1.12)2 Ex = 0 and Ey = 2(14, 400) sin 26.6° = 1.29 × 10 4 N C so E = 1.29 × 10 4 j N C (b) F = Eq = (1.29 × 10 4 j)(−3.00 × 10 −6 ) = −3.86 × 10 −2 j N © 2000 by Harcourt, Inc. All rights reserved.
8. 8. 8 Chapter 23 Solutions ke q1 k q k q k ( 2q) k ( 3q) k ( 4q)23.19 (a) E= 2 1 ~ + e 22 ~2 + e 23 ~3 = e 2 i + e 2 (i cos 45.0° + j sin 45.0°) + e 2 j r1 r2 r3 a 2a a ke q kq kq E = 3.06 i + 5.06 e2 j = 5.91 e2 at 58.8° a2 a a ke q 2 (b) F = qE = 5.91 at 58.8° a223.20 The magnitude of the field at (x, y) due to charge q at (x0 , y0 ) is given by E = ke q r 2 where r is the distance from (x0 , y0 ) to (x, y). Observe the geometry in the diagram at the right. From triangle ABC, r 2 = (x − x0 )2 + (y − y0 )2 , or (y − y0 ) (x − x0 ) r = (x − x0 )2 + (y − y0 )2 , sin θ = , and cos θ = r r ke q (x − x0 ) ke q(x − x0 ) Thus, Ex = Ecos θ = = r 2 r [(x − x0 )2 + (y − y0 )2 ]3/2 ke q (y − y0 ) ke q(y − y0 ) and Ey = Esin θ = = r2 r [(x − x0 )2 + (y − y0 )2 ]3/2 k eq k eq k eq(4ax)23.21 The electric field at any point x is E= – = (x – a)2 (x – (–a))2 (x 2 – a 2)2 (4a)(k eq) When x is much, much greater than a, we find E≈ x3 ke Q/n23.22 (a) One of the charges creates at P a field E= R 2 + x2 at an angle θ to the x-axis as shown. When all the charges produce field, for n > 1, the components perpendicular to the x-axis add to zero. nke (Q/n)i keQxi The total field is 2 2 cos θ = R +x (R + x 2)3/2 2 (b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field. .
9. 9. Chapter 23 Solutions 9 ke q kq kq kq − ke qi 1 1 π 2 ke q23.23 E=∑ ~ = e2 (− i)+ e 2 (− i) + e 2 (− i) + . . . = 2 1+ 2 + 3 + ... = − i r 2 a (2a) (3a) a 2 3 6 a2 ke λ l k (Q / l)l keQ (8.99 × 109)(22.0 × 10–6)23.24 E= = e = = d(l+ d) d(l+ d) d(l+ d) (0.290)(0.140 + 0.290) E = 1.59 × 106 N/C , directed toward the rod . E=∫ ke dq23.25 where dq = λ0 dx x2 ∞ ∞ dx 1 k eλ 0 ⌠ E = ke λ0 ⌡ 2 = k e – x = The direction is –i or left for λ0 > 0 x0 x x0 x0 ∞ ke λ 0 x0 dx( −i ) ∞ 1 ∞ ke λ 023.26 E = ∫ dE = ∫ = − ke λ 0 x0 i ∫ x −3 dx = − ke λ 0 x0 i − 2 2x = (− i) x0 x3 x0 x0 2x0 k exQ (8.99 × 109)(75.0 × 10–6)x 6.74 × 105 x23.27 E= 2 2 3/2 = 2 2 3/2 = 2 (x + a ) (x + 0.100 ) (x + 0.0100)3/2 (a) At x = 0.0100 m, E = 6.64 × 106 i N/C = 6.64 i MN/C (b) At x = 0.0500 m, E = 2.41 × 107 i N/C = 24.1 i MN/C (c) At x = 0.300 m, E = 6.40 × 106 i N/C = 6.40 i MN/C (d) At x = 1.00 m, E = 6.64 × 105 i N/C = 0.664 i MN/C © 2000 by Harcourt, Inc. All rights reserved.
10. 10. 10 Chapter 23 Solutions ke Q x23.28 E= (x + a2 )3/2 2 dE 1 3x 2 For a maximum, = Qke 2 − 2 2 5/2 =0 (x + a ) (x + a ) dx 2 3/2 a x 2 + a2 − 3x 2 = 0 or x= 2 Substituting into the expression for E gives ke Qa k Q 2ke Q Q E= = e = = 2( 2 a2 )3/2 3 3 a2 3 3 3 a2 6 3 π e0 a2 2 x 23.29 E = 2 π ke σ 1 − 2 x +R 2 ( )( E = 2 π 8.99 × 10 9 7.90 × 10 −3 1 − ) x x 2 + (0.350) 2 = 4.46 × 108 1 − x x 2 + 0.123 (a) At x = 0.0500 m, E = 3.83 × 108 N C = 383 MN C (b) At x = 0.100 m, E = 3.24 × 108 N C = 324 MN C (c) At x = 0.500 m, E = 8.07 × 107 N C = 80.7 MN C (d) At x = 2.00 m, E = 6.68 × 108 N C = 6.68 MN C x 23.30 (a) From Example 23.9: E = 2 π ke σ 1 − x2 + R2 Q σ= = 1.84 × 10 −3 C m 2 πR 2 E = (1.04 × 108 N C)(0.900) = 9.36 × 107 N C = 93.6 MN/C appx: E = 2 π ke σ = 104 MN/C (about 11% high) 30.0 cm (b) E = (1.04 × 108 N / C) 1 − = (1.04 × 10 N C)(0.00496) = 0.516 MN/C 8 30.0 + 3.00 cm 2 2 Q 5.20 × 10 −6 appx: E = ke = (8.99 × 10 9 ) = 0.519 MN/C (about 0.6% high) r2 (0.30)2
11. 11. Chapter 23 Solutions 11 x23.31 The electric field at a distance x is Ex = 2 π ke σ 1 − x +R 2 2 1 This is equivalent to Ex = 2 π ke σ 1 − 1+ R 2 x2 R2 R2 For large x, R 2 x 2 << 1 and 1+ ≈ 1+ 2 x2 2x Ex = 2 π ke σ 1 − 1 ( = 2 π ke σ 1+ R 2 (2x 2 ) − 1 ) [ ] [ ] so 1+ R 2 (2x 2 ) 1+ R 2 (2x 2 ) Substitute σ = Q/π R2, Ex = ( keQ 1 x 2 ) = keQ x 2 + R2 [1+ R 2 (2x 2 ) ] 2 1 1 keQ But for x > > R, ≈ 2 , so Ex ≈ for a disk at large distances x +R 2 x 2 2 x223.32 The sheet must have negative charge to repel the negative charge on the Styrofoam. The magnitude of the upward electric force must equal the magnitude of the downward gravitational force for the Styrofoam to "float" (i.e., Fe = F g ). σ 2e0 mg Thus, −qE = mg, or −q = mg which gives σ = − 2e0 q dq sin θ23.33 Due to symmetry Ey = ∫ dEy = 0, and E x = ∫ dE sin θ = ke ∫ r2 π ke λ π ke λ 2k λ where dq = λ ds = λr dθ , so that, Ex = r ∫0 sin θ dθ = r (− cos θ ) = e 0 r q L 2ke qπ 2(8.99 × 109 N · m2/C 2)(7.50 × 10–6 C)π where λ = L and r = . Thus, Ex = = π L2 (0.140 m)2 Solving, E = Ex = 2.16 × 107 N/C Since the rod has a negative charge, E = (–2.16 × 107 i) N/C = –21.6 i MN/C © 2000 by Harcourt, Inc. All rights reserved.
12. 12. 12 Chapter 23 Solutions23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has charge Qdx/h and produces, at the chosen point, a field ke x Q dx dE = 2 2 3/2 i (x + R ) h The total field is d+h keQ x dx k Qi d + h 2 E= ∫ dE =∫ 2 2 3/2 i = e ∫ (x + R 2 )− 3/2 2x dx all charge d h(x + R ) 2h x = d k Q i (x 2 + R 2 )− 1/2 d+h keQ i 1 1 E= e = − (d 2 + R 2 )1/2 ( ) 2h (− 1/ 2) h 1/2 x=d (d + h)2 + R 2 (b) Think of the cylinder as a stack of disks, each with thickness dx, charge Q dx/h, and charge- per-area σ = Q dx / πR 2 h. One disk produces a field 2 π keQ dx x dE = π R2h 1 − (x 2 + R 2 )1/2 i d+h 2 keQ dx x So, E= ∫ dE = ∫ 1 − (x 2 + R 2 )1/2 i all charge x=d R2 h 2 keQ i d + h d+ h 1 (x 2 + R 2 )1/2 d + h 1 d + h(x 2 + R 2 )− 1/2 2x dx = 2 keQ i x R 2 h ∫d E= dx − 2 ∫ − x=d R2h d 2 1/ 2 d 2 keQ i ( d + h − d − (d + h)2 + R 2 ) + (d 2 + R 2 )1/2 1/2 E= R 2h ( h + (d 2 + R 2 )1/2 − (d + h)2 + R 2 ) 2 keQ i 1/2 E= R 2h ke dq23.35 (a) The electric field at point P due to each element of length dx, is dE = and is directed (x + y 2 ) 2 along the line joining the element of length to point P . By symmetry, Ex = ∫ dEx = 0 and since dq = λ dx, y E = Ey = ∫ dEy = ∫ dE cos θ where cos θ = (x + y 2 )1 2 2 dx 2ke λ sinθ 0 Therefore, = (x + y 2 )3 2 2 y 2ke λ (b) For a bar of infinite length, θ → 90° and Ey = y
13. 13. Chapter 23 Solutions 13*23.36 (a) The whole surface area of the cylinder is A = 2 π r 2 + 2 π rL = 2 π r(r + L) . ( ) Q = σA = 15.0 × 10 −9 C m 2 2 π (0.0250 m )[0.0250 m + 0.0600 m ] = 2.00 × 10 −10 C (b) For the curved lateral surface only, A = 2 πrL. ( ) Q = σA = 15.0 × 10 −9 C m 2 2 π (0.0250 m )(0.0600 m ) = 1.41 × 10 −10 C (c) ( ) Q = ρ V = ρ π r 2 L = 500 × 10 −9 C m 3 π (0.0250 m ) (0.0600 m ) = 5.89 × 10 −11 C 2 Every object has the same volume, V = 8(0.0300 m ) = 2.16 × 10 −4 m 3 . 3*23.37 (a) ( )( ) For each, Q = ρ V = 400 × 10 −9 C m 3 2.16 × 10 −4 m 3 = 8.64 × 10 −11 C (b) We must count the 9.00 cm 2 squares painted with charge: (i) 6 × 4 = 24 squares ( ) ( ) Q = σA = 15.0 × 10 −9 C m 2 24.0 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C (ii) 34 squares exposed ( ) ( ) Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C (iii) 34 squares ( ) ( Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = ) 4.59 × 10 −10 C (iv) 32 squares ( ) ( Q = σA = 15.0 × 10 −9 C m 2 32.0 9.00 × 10 −4 m 2 = ) 4.32 × 10 −10 C (c) (i) total edge length: = 24 × (0.0300 m ) ( ) Q = λ = 80.0 × 10 −12 C m 24 × (0.0300 m ) = 5.76 × 10 −11 C (ii) ( ) Q = λ = 80.0 × 10 −12 C m 44 × (0.0300 m ) = 1.06 × 10 −10 C (iii) ( ) Q = λ = 80.0 × 10 −12 C m 64 × (0.0300 m ) = 1.54 × 10 −10 C © 2000 by Harcourt, Inc. All rights reserved.
14. 14. 14 Chapter 23 Solutions (iv) ( ) Q = λ = 80.0 × 10 −12 C m 40 × (0.0300 m ) = 0.960 × 10 −10 C
15. 15. Chapter 23 Solutions 1522.3822.39 q1 − 6 123.40 (a) = = − q2 18 3 (b) q1 is negative, q2 is positive qE23.41 F = qE = ma a = m qEt v = v i + at v= m (1.602 × 10 −19 )(520)(48.0 × 10 −9 ) electron: ve = = 4.39 × 106 m/s 9.11 × 10 −31 in a direction opposite to the field (1.602 × 10 −19 )(520)(48.0 × 10 −9 ) proton: vp = = 2.39 × 103 m/s 1.67 × 10 −27 in the same direction as the field qE (1.602 × 10 −19 )(6.00 × 10 5 )23.42 (a) a = = −27 = 5.76 × 1013 m s so a= −5.76 × 1013 i m s 2 m (1.67 × 10 ) (b) v = vi + 2a(x − xi ) 0 = vi 2 + 2(−5.76 × 1013 )(0.0700) v i = 2.84 × 106 i m s (c) v = vi + at 0 = 2.84 × 106 + (−5.76 × 1013 )t t = 4.93 × 10 −8 s © 2000 by Harcourt, Inc. All rights reserved.
16. 16. 16 Chapter 23 Solutions a= qE = ( 1.602 × 10 −19 (640) ) = 6.14 × 1010 m/s2 ( )23.43 (a) m 1.67 × 10 −27 (b) v = v i + at 1.20 × 106 = (6.14 × 1010)t t = 1.95 × 10-5 s (c) x − xi = 2 ( vi + v )t 1 ( )( ) x = 2 1.20 × 106 1.95 × 10 −5 = 11.7 m 1 (d) K = 2 mv 2 = 1 (1.67 × 10 − 27 kg)(1.20 × 106 m / s)2 = 1.20 × 10-15 J 1 223.44 The required electric field will be in the direction of motion . We know that Work = ∆K 1 2 So, –Fd = – 2 m v i (since the final velocity = 0) 1 2 mv i 1 2 This becomes Eed = mvi 2 or E= 2 ed 1.60 × 10–17 J E = = 1.00 × 103 N/C (in direction of electrons motion) (1.60 × 10–19 C)(0.100 m)23.45 The required electric field will be in the direction of motion . 1 2 Work done = ∆K so, –Fd = – 2 m v i (since the final velocity = 0) K which becomes eEd = K and E=ed
17. 17. Chapter 23 Solutions 17Goal SolutionThe electrons in a particle beam each have a kinetic energy K . What are the magnitude and direction ofthe electric field that stops these electrons in a distance of d?G: We should expect that a larger electric field would be required to stop electrons with greater kinetic energy. Likewise, E must be greater for a shorter stopping distance, d. The electric field should be i n the same direction as the motion of the negatively charged electrons in order to exert an opposing force that will slow them down.O: The electrons will experience an electrostatic force F = qE. Therefore, the work done by the electric field can be equated with the initial kinetic energy since energy should be conserved.A: The work done on the charge is W = F ⋅ d = qE ⋅ d and Ki + W = K f = 0 Assuming v is in the + x direction, K + ( −e )E ⋅ di = 0 eE ⋅ ( di ) = K K E is therefore in the direction of the electron beam: E= i edL: As expected, the electric field is proportional to K , and inversely proportional to d. The direction of the electric field is important; if it were otherwise the electron would speed up instead of slowing down! If the particles were protons instead of electrons, the electric field would need to be directed opposite to v in order for the particles to slow down. 223.46 The acceleration is given by v 2 = v i + 2a(x – xi)or v 2 = 0 + 2a(–h) v2 Solving, a = – 2h mv 2 j Now ∑ F = ma: –mgj + qE = – 2h mv2 Therefore qE = – 2h + m g j (a) Gravity alone would give the bead downward impact velocity ( ) 2 9.80 m / s 2 ( 5.00 m ) = 9.90 m / s To change this to 21.0 m/s down, a downward electric field must exert a downward electric force. m v 2 1.00 × 10–3 kg N · s2 (21.0 m/s)2 (b) q = E 2h – g = kg · m 2(5.00 m) – 9.80 m/s2 = 3.43 µC 1.00 × 10 N/C 4 © 2000 by Harcourt, Inc. All rights reserved.
18. 18. 18 Chapter 23 Solutions x 0.050023.47 (a) t= = = 1.11 × 10-7 s = 111 ns v 4.50 × 10 5 qE (1.602 × 10 −19 )(9.60 × 10 3 ) (b) ay = = = 9.21 × 1011 m / s 2 m (1.67 × 10 − 27 ) y − yi = v y i t + 2 ay t 2 1 y = 2 (9.21 × 1011 )(1.11 × 10 −7 )2 = 5.67 × 10-3 m = 5.67 mm 1 (c) v x = 4.50 × 105 m/s vy = vy i + ay = (9.21 × 1011)(1.11 × 10-7) = 1.02 × 105 m/s qE (1.602 × 10 −19 )(390)23.48 ay = = − 31 = 6.86 × 1013 m / s 2 m (9.11 × 10 ) 2vi sin θ (a) t= from projectile motion equations ay 2(8.20 × 10 5 )sin 30.0° t= = 1.20 × 10-8 s) = 12.0 ns 6.86 × 1013 vi 2 sin 2 θ (8.20 × 10 5 )2 sin 2 30.0° (b) h= = = 1.23 mm 2ay 2(6.86 × 1013 ) vi 2 sin 2 θ (8.20 × 10 5 )2 sin 60.0° (c) R= = = 4.24 mm 2ay 2(6.86 × 1013 )23.49 vi = 9.55 × 103 m/s eE (1.60 × 10 −19 )(720) (a) ay = = = 6.90 × 1010 m s 2 m (1.67 × 10 −27 ) vi 2 sin 2θ (9.55 × 10 3 )2 sin 2θ R= = 1.27 × 10-3 m so that = 1.27 × 10 −3 ay 6.90 × 1010 sin 2θ = 0.961 θ = 36.9° 90.0° – θ = 53.1° R R (b) t= = If θ = 36.9°, t = 167 ns If θ = 53.1°, t = 221 ns vix vi cos θ
19. 19. Chapter 23 Solutions 19*23.50 (a) The field, E1, due to the 4.00 × 10–9 C charge is in the –x direction. ke q (8.99 × 109 N · m2/C 2)(− 4.00 × 10–9 C) E1 = = i = −5.75i N/C r2 (2.50 m)2 Likewise, E2 and E3, due to the 5.00 × 10–9 C charge and the 3.00 × 10–9 C charge are ke q (8.99 × 109 N · m2/C 2)(5.00 × 10–9 C) E2 = = i = 11.2 r2 (2.00 m)2 N/C (8.99 × 109 N · m2/C 2)(3.00 × 10–9 C) E3 = i = 18.7 N/C (1.20 m)2 ER = E1 + E2 + E3 = 24.2 N/C in +x direction. ke q (b) E1 = = ( −8.46 N / C)(0.243i + 0.970j) r2 ke q E2 = = (11.2 N / C)( +j) r2 ke q E3 = = ( 5.81 N / C)( −0.371i + 0.928j) r2 Ex = E1x + E3x = – 4.21i N/C Ey = E1y + E2y + E3y = 8.43j N/C ER = 9.42 N/C θ = 63.4° above –x axis23.51 The proton moves with acceleration ap = qE = ( 1.60 × 10 −19 C (640 N / C) ) = 6.13 × 1010 m s 2 −27 m 1.67 × 10 kg while the e− has acceleration ae = (1.60 × 10 −19 ) C (640 N/C) = 1.12 × 1014 m s 2 = 1836 ap −31 9.11 × 10 kg (a) We want to find the distance traveled by the proton (i.e., d = 2 apt 2 ), knowing: 1 1 1 ( 4.00 cm = 2 apt 2 + 2 ae t 2 = 1837 2 apt 2 1 ) 4.00 cm Thus, d = 2 apt 2 = 1 = 21.8 µ m 1837 © 2000 by Harcourt, Inc. All rights reserved.
20. 20. 20 Chapter 23 Solutions (b) The distance from the positive plate to where the meeting occurs equals the distance the sodium ion travels (i.e., dNa = 2 aNat 2 ). This is found from: 1 1 eE 2 1 eE 2 4.00 cm = 2 aNat 2 + 2 aClt 2 : 4.00 cm = 1 1 t + t 2 22.99 u 2 35.45 u This may be written as 1 1 ( 4.00 cm = 2 aNat 2 + 2 (0.649aNa )t 2 = 1.65 2 aNat 2 1 ) 4.00 cm so dNa = 2 aNat 2 = 1 = 2.43 cm 1.6523.52 From the free-body diagram shown, ∑Fy = 0 and T cos 15.0° = 1.96 × 10–2 N So T = 2.03 × 10–2 N From ∑Fx = 0, we have qE = T sin 15.0° T sin 15.0° (2.03 × 10–2 N) sin 15.0° or q= = = 5.25 × 10–6 C = 5.25 µC E 1.00 × 103 N/C23.53 (a) Let us sum force components to find ∑Fx = qEx – T sin θ = 0, and ∑Fy = qEy + T cos θ – mg = 0 Combining these two equations, we get mg (1.00 × 10-3)(9.80) q= = = 1.09 × 10–8 C = 10.9 nC (Ex cot θ + Ey) (3.00 cot 37.0° + 5.00) × 105 Free Body Diagram (b) From the two equations for ∑Fx and ∑Fy we also find for Goal Solution qEx sin 37.0° = 5.44 × 10 N = 5.44 mN –3 T=
21. 21. Chapter 23 Solutions 21Goal SolutionA charged cork ball of mass 1.00 g is suspended on a light string in the presence of a uniform electric field,as shown in Fig. P23.53. When E = ( 3.00i + 5.00j) × 10 5 N / C , the ball is in equilibrium at θ = 37.0°. Find(a) the charge on the ball and (b) the tension in the string.G: (a) Since the electric force must be in the same direction as E, the ball must be positively charged. If we examine the free body diagram that shows the three forces acting on the ball, the sum of which must be zero, we can see that the tension is about half the magnitude of the weight.O: The tension can be found from applying Newtons second law to this statics problem (electrostatics, in this case!). Since the force vectors are in two dimensions, we must apply ΣF = ma to both the x and y directions.A: Applying Newtons Second Law in the x and y directions, and noting that ΣF = T + qE + F g = 0, ΣFx = qEx − T sin 37.0° = 0 (1) ΣFy = qEy + T cos 37.0° − mg = 0 (2) We are given Ex = 3.00 × 10 5 N / C and Ey = 5.00 × 10 5 N / C; substituting T from (1) into (2): mg (1.00 × 10 −3 kg)(9.80 m / s 2 ) q= = = 1.09 × 10 −8 C Ex 3.00 Ey + tan 37.0° 5.00 + tan 37.0° × 10 N / C 5 qEx (b) Using this result for q in Equation (1), we find that the tension is T= = 5.44 × 10 −3 N sin 37.0°L: The tension is slightly more than half the weight of the ball ( F g = 9.80 × 10 −3 N) so our result seems reasonable based on our initial prediction.23.54 (a) Applying the first condition of equilibrium to the ball gives: qEx qA ΣFx = qEx − T sin θ = 0 or T= = sin θ sin θ and ΣFy = qEy + T cos θ − mg = 0 or qB + T cos θ = mg Substituting from the first equation into the second gives: mg q( A cot θ + B) = mg , or q= ( A cot θ + B) (b) Substituting the charge into the equation obtained from ΣFx yields mg A mgA T= = ( A cot θ + B) sin θ A cos θ + Bsin θ © 2000 by Harcourt, Inc. All rights reserved.
22. 22. 22 Chapter 23 SolutionsGoal SolutionA charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field, asshown in Figure P23.53. When E = ( Ai + Bj) N / C , where A and B are positive numbers, the ball is i nequilibrium at the angle θ . Find (a) the charge on the ball and (b) the tension in the string.G: This is the general version of the preceding problem. The known quantities are A , B, m, g , and θ . The unknowns are q and T .O: The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 53.A: Again, Newtons second law: −T sin θ + qA = 0 (1) and + T cos θ + qB − mg = 0 (2) qA qA cos θ (a) Substituting T = , into Eq. (2), + qB = mg sin θ sin θ mg Isolating q on the left, q= ( A cot θ + B) mgA (b) Substituting this value into Eq. (1), T= ( A cos θ + Bsin θ )L : If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 53. If you find this problem more difficult than problem 53, the little list at the Gather step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the Analysis step, and for recognizing when we have an answer. ke q1q2 15.023.55 F= tan θ = 60.0 θ = 14.0° r2 (8.99 × 109)(10.0 × 10–6)2 F1 = = 40.0 N (0.150)2 (8.99 × 109)(10.0 × 10–6)2 F3 = = 2.50 N (0.600)2 (8.99 × 109)(10.0 × 10–6)2 F2 = = 2.35 N (0.619)2 Fx = –F3 – F2 cos 14.0° = –2.50 – 2.35 cos 14.0° = – 4.78 N Fy = –F1 – F2 sin 14.0° = – 40.0 – 2.35 sin 14.0° = – 40.6 N 2 2 Fnet = Fx + Fy = (– 4.78)2 + (– 40.6)2 = 40.9 N Fy – 40.6 tan φ = F = – 4.78 φ = 263° x
23. 23. Chapter 23 Solutions 23 15.0 cm23.56 From Fig. A: d cos 30.0° = 15.0 cm, or d= cos 30.0° d 15.0 cm From Fig. B: θ = sin −1 = sin −1 = 20.3° 50.0 cm 50.0 cm(cos 30.0°) Fq = tan θ mg Figure A or Fq = mg tan 20.3° (1) ke q 2 From Fig. C: Fq = 2F cos 30.0° = 2 2 cos 30.0° (2) (0.300 m ) ke q 2 Equating equations (1) and (2), 2 2 cos 30.0° = mg tan 20.3° (0.300 m ) Figure B mg(0.300 m ) tan 20.3° 2 q2 = 2ke cos 30.0° q2 = (2.00 × 10 kg)(9.80 m s )(0.300 m) tan 20.3° −3 2 2 2(8.99 × 10 N ⋅ m C ) cos 30.0° 9 2 2 Figure C q = 4.2 0 × 10 −14 C 2 = 2.05 × 10 −7 C= 0.205 µ C23.57 Charge Q/2 resides on each block, which repel as point charges: ke(Q/2)(Q/2) F= = k(L – L i) L2 k ( L − Li ) (100 N / m)(0.100 m) Q = 2L = 2(0.400 m ) = 26.7 µC ke (8.99 × 10 9 N ⋅ m 2 / C2 ) ke (Q 2)(Q 2)23.58 Charge Q /2 resides on each block, which repel as point charges: F = = k(L − L i ) L2 k ( L − Li ) Solving for Q , Q = 2L ke © 2000 by Harcourt, Inc. All rights reserved.
24. 24. 24 Chapter 23 Solutions*23.59 According to the result of Example 23.7, the lefthand rod creates this field at a distance d from its righthand end: k eQ E = d(2a + d) keQQ dx dF = 2a d(d + 2a) k eQ 2 b k eQ 2 1 2a + x b ∫ x = b – 2a x(x + 2a) dx F= = – ln 2a 2a 2a x b – 2a +k eQ 2 2a + b b k eQ 2 b2 k e Q 2 b2 F = – ln + ln = 2 ln (b – 2a)(b + 2a) = 2 ln 2 2 4a 2 b b – 2a 4a 4a b – 4a *23.60 The charge moves with acceleration of magnitude a given by ∑F = ma = q E qE 1.60 × 10–19 C (1.00 N/C) (a) a= m = = 1.76 × 1011 m/s2 9.11 × 10–31 kg v 3.00 × 107 m/s Then v = v i + at = 0 + at gives t=a = = 171 µs 1.76 × 1011 m/s2 v vm (3.00 × 107 m/s)(1.67 × 10–27 kg) (b) t = a = qE = = 0.313 s (1.60 × 10–19 C)(1.00 N/C) vm (c) From t = qE , as E increases, t gets shorter in inverse proportion. 90.0° 90.0°23.61 Q = ∫ λ dl = ∫ –90.0° λ 0 cos θ Rd θ = λ 0 R sin θ –90.0° = λ 0 R [1 – (–1)] = 2λ 0 R Q = 12.0 µC = (2λ 0 )(0.600) m = 12.0 µC so λ 0 = 10.0 µC/m dFy = 1 ( 3.00 µ C)(λ dl) cos θ = 1 ( 3.00 µ C) λ 0 cos 2 θ Rdθ ( ) 4 π e0 R2 4 π e0 R2 90.0° N · m2 (3.00 × 10–6 C)(10.0 × 10–6 C/m) Fy = ∫ –90.0° 8.99 × 109 cos2 θ d θ C2 (0.600 m) π /2 8.99( 30.0) Fy = 0.600 ( 10 −3 N )∫( 1 2 1 + cos 2θ dθ 2 ) − π /2 Fy = (0.450 N ) ( 1 2 π 1 π /2 ) + sin 2θ − π /2 = 0.707 N 4 Downward. Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0.
25. 25. Chapter 23 Solutions 2523.62 At equilibrium, the distance between the charges is r = 2(0.100 m )sin10.0° = 3.47 × 10 −2 m Now consider the forces on the sphere with charge +q , and use ΣFy = 0: mg ΣFy = 0: T cos 10.0° = mg, or T = (1) cos 10.0° ΣFx = 0: Fnet = F2 − F1 = T sin10.0° (2) Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1). Fnet = mg sin 10.0° cos 10.0° ( )( ) = mg tan 10.0° = 2.00 × 10 −3 kg 9.80 m / s 2 tan 10.0° = 3.46 × 10 −3 N Fnet is the resultant of two forces, F1 and F 2 . F 1 is the attractive force on +q exerted by –q, and F 2 is the force exerted on +q by the external electric field. Fnet = F2 – F1 or F2 = Fnet + F1 ( F1 = 8.99 × 10 N ⋅ m 9 2 /C ) 2 (5.00 × 10 C)(5.00 × 10 −8 −8 C ) = 1.87 × 10 −2 N (3.47 × 10 m) −3 2 Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10 −3 N + 1.87 × 10 −2 N = 2.21 × 10 − 2 N F2 2.21 × 10 − 2 N and F2 = qE, or E= = = 4.43 × 105 N/C = 443 kN/C q 5.00 × 10 − 8 C23.63 (a) From the 2Q charge we have Fe − T 2 sin θ 2 = 0 and mg − T 2 cos θ 2 = 0 Fe T sin θ 2 Combining these we find = 2 = tan θ 2 mg T 2 cos θ 2 From the Q charge we have Fe − T1 sin θ1 = 0 and mg − T1 cos θ1 = 0 Fe T sin θ1 Combining these we find = 1 = tan θ1 or θ2 = θ1 mg T1 cos θ1 ke 2QQ 2keQ 2 (b) Fe = = r2 r2 If we assume θ is small then . Substitute expressions for F e and tan θ into either equation found in part (a) and solve for r. Fe = tan θ then and solving for r we find mg © 2000 by Harcourt, Inc. All rights reserved.
26. 26. 26 Chapter 23 Solutions23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle θ corresponding to equilibrium. In terms of lengths s, 1 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an attractive force 2 keQ q (s + 2 a 3)2 . The other two charges exert equal repulsive forces of magnitude keQq r 2 . 1 The horizontal components of the two repulsive forces add, balancing the attractive force, 2 cos θ 1 Fnet = keQq − 1 a 3)2 =0 r 2 (s + 2 1a From Figure P23.64, r= 2 s = 2 a cot θ 1 sin θ 4 1 The equilibrium condition, in terms of θ , is Fnet = k Qq 2 cos θ sin 2θ − =0 a2 e ( 3 + cot θ )2 Thus the equilibrium value of θ is 2 cos θ sin 2 θ ( 3 + cot θ )2 = 1. One method for solving for θ is to tabulate the left side. To three significant figures the value of θ corresponding to equilibrium is 81.7°. The distance from the origin to the equilibrium position is x = 2 a( 3 + cot 81.7°) = 0.939a 1 θ 2 cos θ sin 2 θ ( 3 + cot θ )2 60° 4 70° 2.654 80° 1.226 90° 0 81° 1.091 81.5° 1.024 81.7° 0.99723.65 (a) The distance from each corner to the center of the square is ( L 2 )2 + ( L 2 )2 = L 2 The distance from each positive charge to −Q is then z 2 + L2 2 . Each positive charge exerts a force directed along the line joining q and −Q, of magnitude keQq z + L2 2 2 z The line of force makes an angle with the z-axis whose cosine is z + L2 2 2 The four charges together exert forces whose x and y components 4keQ q z add to zero, while the z-components add to F= − k (z ) 32 2 + L2 2
27. 27. Chapter 23 Solutions 27 4keQqz 4( 2)3 2 keQq (b) For z << L, the magnitude of this force is Fz ≈ − = − z = maz ( L 2) 32 2 L3 Therefore, the object’s vertical acceleration is of the form az = − ω 2 z 4( 2) keQq keQq 128 32 with ω 2 = = mL3 mL3 Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by 2π 2π mL3 π mL3 T= = = ω (128)1 4 keQq (8)1 4 keQq qE 23.66 (a) The total non-contact force on the cork ball is: F = qE + mg = m g + , m which is constant and directed downward. Therefore, it behaves like a simple pendulum i n the presence of a modified uniform gravitational field with a period given by: L 0.500 m T = 2π = 2π (2.00 × 10 )( ) −6 = 0.307 s qE C 1.00 × 10 5 N / C g + m 9.80 m / s 2 + 1.00 × 10 − 3 kg L (b) Yes . Without gravity in part (a), we get T = 2π qE m 0.500 m T = 2π = 0.314 s (a 2.28% difference). (2.00 × 10 −6 )( ) C 1.00 × 10 5 N / C 1.00 × 10 − 3 kg23.67 (a) Due to symmetry the field contribution from each negative charge is y equal and opposite to each other. Therefore, their contribution to the net field is zero. The field contribution of the +q charge is ke q ke q 4k q E= = = e r 2 ( 3 a 2 4) 3a 2 x 4ke q in the negative y direction, i.e., E = − j 3a2 © 2000 by Harcourt, Inc. All rights reserved.
28. 28. 28 Chapter 23 Solutions (b) If Fe = 0, then E at P must equal zero. In order for the field to cancel at P , the − 4q must be above + q on the y-axis. ke q ke (4q) Then, E=0=− + , which reduces to y 2 = 4.00 m 2 . (1.00 m) 2 y2 Thus, y = ± 2.00 m . Only the positive answer is acceptable since the − 4q must be located above + q. Therefore, the − 4q must be placed 2.00 meters above point P along the + y − axis .23.68 The bowl exerts a normal force on each bead, directed along the radius line or at 60.0° above the horizontal. Consider the free-body diagram of the bead on the left: ΣFy = nsin 60.0° −mg = 0 , mg or n= sin 60.0° Also, ΣFx = −Fe + ncos 60.0° = 0, n ke q 2 mg mg or 2 = ncos 60.0° = = Fe 60.0˚ R tan 60.0° 3 12 mg mg Thus, q= R ke 3 23.69 (a) There are 7 terms which contribute: 3 are s away (along sides) 1 3 are 2 s away (face diagonals) and sin θ = = cos θ 2 1 1 is 3 s away (body diagonal) and sin φ = 3 The component in each direction is the same by symmetry. ke q 2 2 1 ke q 2 F= 1+ 2 2 + 3 3 (i + j + k) = (1.90)(i + j + k) s2 s2 ke q 2 (b) F= F2 +F2 +F2 = x y z 3.29 away from the origin s2
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# High Voltage Engineering Questions and Answers – Generation of Impulse Voltages
This set of High Voltage Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Generation of Impulse Voltages”.
1. The time taken by an impulse wave to reach its maximum value starting from zero value is called ______
a) wave front time
b) wave tail time
c) peak wave time
d) steepness time
Explanation: The time taken by an impulse wave to reach its maximum value starting from zero value is called wave front time. Generally, it is 0.5 μs to 10μs. It is also known as the rise time of the impulse voltage.
2. ______ is the impulse voltage developed without causing puncture or flashover.
a) normal impulse voltage
b) chopped impulse voltage
c) full impulse voltage
d) clean impulse voltage
Explanation: The full impulse voltage is the impulse voltage developed without causing puncture or flashover. It is characterized by its two time intervals and the peak value. The time intervals are wave front time and wave tail time.
3. ______is the impulse voltage developed by causing puncture or flashover by collapsing the impulse voltage.
a) normal impulse voltage
b) chopped impulse voltage
c) full impulse voltage
d) clean impulse voltage
Explanation: The chopped impulse voltage is the impulse voltage developed by causing puncture or flashover by collapsing itself. If chopping takes place on the front part of the wave, it is known as a front chopped wave. Otherwise, it is simply known as a chopped wave. The full impulse voltage is the impulse voltage developed without causing puncture or flashover.
4. Identify the following figure.
a) front chopped wave
b) chopped wave
c) full impulse wave
d) peak impulse wave
Explanation: The figure shows the front chopped wave. An impulse wave that is chopped at the front is known as front chopped wave. Otherwise, it is simply known as a chopped wave.
5. An impulse voltage is a bidirectional voltage.
a) True
b) False
Explanation: An impulse voltage is a unidirectional voltage. It does not have much oscillations. It rises to a peak value and decays to more or less a zero value.
6. Identify the following figure.
a) front chopped wave
b) chopped wave
c) full impulse wave
d) peak impulse wave
Explanation: The figure shows the chopped wave. An impulse wave which is not chopped at the front is known as chopped wave. The chopped impulse voltage is the impulse voltage developed by causing puncture or flashover by collapsing itself.
7. Which among the following represents standard lightning impulse?
a) 2/25 μs
b) 25/2 μs
c) 1.2/50 μs
d) 50/1.2 μs
Explanation: 1.2/50 μs represents standard lightning impulse. 1.2 μs is the rise time and 50μs is the decay time. A tolerance of 3% is allowed in the peak value.
8. What is the tolerance allowed in the peak value of the standard impulse?
a) ±3%
b) ±5%
c) ±6%
d) ±2%
Explanation: The tolerance allowed in the peak value of the standard impulse is ±3%.1.2/50 μs represents standard lightning impulse. 1.2 μs is the rise time and 50μs is the decay time.
9. The impulse ratio for flash over is the ratio of impulse flash over voltage to the _____
a) peak value of power frequency flash over voltage
b) minimum value of power frequency flash over voltage
c) mid value of power frequency flash over voltage
d) peak value of low frequency flash over voltage
Explanation: The ratio of impulse flash over voltage to the peak value of power frequency flash over voltage is known as the impulse ratio for flash over. It is not constant for any particular object. It depends on the characteristics and shape of the impulse wave.
10. Identify the following figure.
a) front chopped wave
b) chopped wave
c) full impulse wave
d) peak impulse wave
Explanation: The figure represents a full impulse voltage. It is the impulse voltage developed without causing puncture or flashover. It is characterized by its two time intervals and the peak value. The time intervals are wave front time and wave tail time.
Sanfoundry Global Education & Learning Series – High Voltage Engineering.
To practice all areas of High Voltage Engineering, here is complete set of Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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# Tick (✓) the correct answer
Question:
If one angle of a parallelogram is 24° less than twice the smallest angle then the largest angle of the parallelogram is
(a) 68°
(b) 102°
(c) 112°
(d) 176°
Solution:
(c) $112^{\circ}$
Let $x^{\circ}$ be the smallest angle of the parallelogram.
$T$ he sum of adjacent angles of a parallelogram is $180^{\circ}$.
$\therefore x+2 x-24=180$
$\Rightarrow 3 x-24=180$
$\Rightarrow 3 x=180+24$
$\Rightarrow 3 x=204$
$\Rightarrow x=\frac{204}{3}$
$\Rightarrow x=68$
$\therefore$ Smallest angle $=68^{\circ}$
$L$ argest angle $=(180-68)^{\circ}=112^{\circ}$
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## 550489
550,489 (five hundred fifty thousand four hundred eighty-nine) is an odd six-digits prime number following 550488 and preceding 550490. In scientific notation, it is written as 5.50489 × 105. The sum of its digits is 31. It has a total of 1 prime factor and 2 positive divisors. There are 550,488 positive integers (up to 550489) that are relatively prime to 550489.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 6
• Sum of Digits 31
• Digital Root 4
## Name
Short name 550 thousand 489 five hundred fifty thousand four hundred eighty-nine
## Notation
Scientific notation 5.50489 × 105 550.489 × 103
## Prime Factorization of 550489
Prime Factorization 550489
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 550489 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) 13.2186 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 550,489 is 550489. Since it has a total of 1 prime factor, 550,489 is a prime number.
## Divisors of 550489
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 550490 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 275245 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 741.949 Returns the nth root of the product of n divisors H(n) 2 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 550,489 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 550,489) is 550,490, the average is 275,245.
## Other Arithmetic Functions (n = 550489)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 550488 Total number of positive integers not greater than n that are coprime to n λ(n) 550488 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45249 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 550,488 positive integers (less than 550,489) that are coprime with 550,489. And there are approximately 45,249 prime numbers less than or equal to 550,489.
## Divisibility of 550489
m n mod m 2 3 4 5 6 7 8 9 1 1 1 4 1 2 1 4
550,489 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (550489)
Base System Value
2 Binary 10000110011001011001
3 Ternary 1000222010111
4 Quaternary 2012121121
5 Quinary 120103424
6 Senary 15444321
8 Octal 2063131
10 Decimal 550489
12 Duodecimal 2266a1
20 Vigesimal 38g49
36 Base36 bsrd
## Basic calculations (n = 550489)
### Multiplication
n×i
n×2 1100978 1651467 2201956 2752445
### Division
ni
n⁄2 275244 183496 137622 110098
### Exponentiation
ni
n2 303038139121 166819162166580169 91832113761918550652641 50552568472684781030221691449
### Nth Root
i√n
2√n 741.949 81.9564 27.2387 14.0653
## 550489 as geometric shapes
### Circle
Diameter 1.10098e+06 3.45882e+06 9.52022e+11
### Sphere
Volume 6.9877e+17 3.80809e+12 3.45882e+06
### Square
Length = n
Perimeter 2.20196e+06 3.03038e+11 778509
### Cube
Length = n
Surface area 1.81823e+12 1.66819e+17 953475
### Equilateral Triangle
Length = n
Perimeter 1.65147e+06 1.31219e+11 476737
### Triangular Pyramid
Length = n
Surface area 5.24877e+11 1.96598e+16 449472
## Cryptographic Hash Functions
md5 45796e1728b6ca5b61c862811cf27ad4 5f0e11b2779e32a31a97c5cbb7b63f1c9d3c8a9f 9d117c3e90f1515be1869e49f8ab39cebd506d9c6e5ac9e685e3f5d17ae3e0d1 3f53dc621a5561a797d195d488971ddce0772465ac57b01661a13dc0d2d0d41530f4ac1f4331dc1ced145a1921974f3d488255b4c704e67cbe76df9eff5ae75c 32c7075c7780bdbb36d9dff619b0cbd0b66bf547
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# Density(bpp)
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### Density(bpp)
1. 1. DensityDensity Modified byModified byCHAIPORN PATTANAJAKCHAIPORN PATTANAJAKBanphue pittayasan schoolBanphue pittayasan schoolFacebook.com/chaiporn.pattanajak
2. 2. State of MatterState of MatterSolid is fixed shapeLiquid takes on the shape ofits container.Gas has neither a fixed shapenor a fixed volumeRecall
3. 3. DENSITYDENSITYQ) Which weighs more:-A kilogram of feathers or a kilogram of iron?
4. 4. What is Density?What is Density?Density is the Mass per unitVolumeWood Water Iron1 cm31 cm3 1 cm3If you take the same volume of different substances,then they will weigh different amounts.0.50 g 1.00 g 8.00 gQ) Which has the greatest mass and therefore the most dense?IRON
5. 5. Density =MassVolumeg or kgcm3or m3gcm-3or kgm-3 ρ = mVExample:Q) Liquid water has a density of 1000kgm-3, while ice has densityof 920kgm-3. Calculate the volume occupied by 0.25kg of each.Density Equation:mVρV = m = 0.25 = 0.000250m3ρ 1000V = m = 0.25 = 0.000272m3ρ 920
6. 6. DENSITY OF A REGULAR SOLIDDENSITY OF A REGULAR SOLID Find the Mass ofFind the Mass ofthe solid on athe solid on abalance.balance. Measure theMeasure thethree lengths andthree lengths andcalculate thecalculate theVolume.Volume.(ie V = l x w x h )(ie V = l x w x h ) Calculate theCalculate theDensity.Density.l = 4.0 cmw= 2.0 cmh=3.0 cmρ = m = 240 =10.0 g/cm3V 24m = 240 g
7. 7. DENSITYDENSITY g/cmg/cm33AluminiumAluminium 2.702.70BrassBrass 8.508.50IronIron 7.867.86WoodWood 0.500.50SlateSlate 2.802.80GlassGlass 2.502.50LeadLead 11.311.3MarbleMarble 2.702.70WaxWax 0.890.89
8. 8. DENSITY OF AN IRREGULAR SOLIDDENSITY OF AN IRREGULAR SOLID Find the Mass of the solidFind the Mass of the solidon a balance.on a balance.Fill the Measuring CylinderFill the Measuring Cylinderwith Water to a knownwith Water to a knownVolume.Volume. Add the Object.Add the Object. Work out the Volume ofWork out the Volume ofWater that is displaced.Water that is displaced. Calculate the Density.Calculate the Density.50 cm380 cm3m = 360 gρ = m = 360 =12.0 g/cm3V 30
9. 9. DENSITY OF AN IRREGULAR SOLIDDENSITY OF AN IRREGULAR SOLID OR use a Eureka Can to find the Volume.OR use a Eureka Can to find the Volume. Find the mass of the solid onFind the mass of the solid ona balance.a balance.Add water until justAdd water until justoverflowing.overflowing. Place a Measuring CylinderPlace a Measuring Cylinderunder the spout.under the spout. Add the Object.Add the Object. Collect the Water and read offCollect the Water and read offthe Volume.the Volume.Calculate DensityCalculate Densitym = 440 g40.0 cm3ρ = m = 440 =11.0 g/cm3V 40
10. 10. DENSITY OF A LIQUIDDENSITY OF A LIQUID Find the Mass of an emptyFind the Mass of an emptyMeasuring Cylinder.Measuring Cylinder. Add a certain Volume of Liquid.Add a certain Volume of Liquid. Find the Mass of the MeasuringFind the Mass of the MeasuringCylinder and LiquidCylinder and Liquid Calculate the Mass of Liquid.Calculate the Mass of Liquid. How?How? Calculate Density of Liquid.Calculate Density of Liquid.Mass of Liquid = Mass of Measuring Cylinder and Liquid – Massofempty Measuring Cylinder25.0 g20.0 cm345.0 g45 – 25 = 20 gρ = m = 20 =1.00 g/cm3V 20
11. 11. DENSITY OF A GASDENSITY OF A GAS Remove the air fromRemove the air froma flask of a knowna flask of a knownVolume, using aVolume, using avacuum pump.vacuum pump. Find its Mass.Find its Mass. Add the gas to beAdd the gas to betested.tested. Reweigh.Reweigh. The difference is theThe difference is theMass of gas.Mass of gas. Calculate Density.Calculate Density.To vacuum flask1000 cm3m(glass)=150.0 gm(total)=170.0 gρ = m = 20 =0.0200 g/cm3V 1000170 -150 = 20.0g
12. 12. Problem of DensityProblem of Density2. What is the density of a piece ofwood that has a mass of 25.0grams and a volume of 29.4 cm3?1. A piece of wood that measures 3.0 cm by 6.0cm by 4.0 cm has a mass of 80.0 grams. What is the density of the wood? Would the piece of wood float in water? (volume = L x W x H)
13. 13. 3. A cup of gold colored metal beads was3. A cup of gold colored metal beads wasmeasured to have a mass 425 gramsmeasured to have a mass 425 grams.. By waterBy waterdisplacement, thedisplacement, the volumevolume of the beads wasof the beads wascalculated to be 48.0 cmcalculated to be 48.0 cm33.. Given the followingGiven the followingdensities, identify the metaldensities, identify the metal..GoldGold:: 19.3 g19.3 g//mLmLCopperCopper:: 8.86 g8.86 g//mLmLBronzeBronze:: 9.87 g9.87 g//mLmL
14. 14. 4. The4. The volumevolume of aof a solutionsolutionwas measured in a graduatwas measured in a graduateded cylindercylinder (shown above). I(shown above). If thef the massmass ofof solutionsolution is meis measured to be 60.75 grams,asured to be 60.75 grams,what is thewhat is the densitydensity of the solof the solution?ution?
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### Home > CCA2 > Chapter 12 > Lesson 12.1.1 > Problem12-7
12-7.
Solve for $x$ in each of the following triangles. What methods can be used for finding unknown parts of triangles?
What laws and theorems relate the angles of a triangle to the sides? Or relate the lengths of the sides of a triangle to each other?
Use them to help you find the unknown parts of these triangles.
1. Which trigonometric ratio relates the side opposite the angle to the side adjacent to it?
$x ≈ 80.86$
1. Remember that all the angles of a triangle add up to $180°$. Solve for the measure of the unknown angle. This will help you choose the correct law to solve for the unknown side.
Use the Law of Sines.
1. Use the Law of Cosines.
$x ≈ 15.50°$
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Most standardized exams require similar skills and test on similar types of problems, especially when it comes to the math sections. The GRE, ACT, and PSAT have more in common than you realize. Professor Charlotte Vilkus teaches Educator’s Standardized Math course specializing in the GRE/ACT/PSAT, and will show you all of her tricks and everything you need to know in order to get the highest score possible. In addition, Charlotte has worked with students across the most popular prep books for all the major exams. Using her experience, she creates an all-inclusive course which begins with a basic overview of math concepts, then covers the types of problems you should expect on test day, and finally guides you through the best ways to approach math questions on standardized tests. Professor Vilkus obtained her BA in civil engineering with a math minor from Loyola Marymount University.
expand all collapse all
## I. Basic Math
Signed Numbers 18:58
Intro 00:00
The Basics: Signed Numbers 00:13
How to Think of Negative Numbers 00:39
Signed Numbers on the Number Line 02:16
Signed Numbers: Addition and Subtraction 03:07
Subtracting a Negative 08:35
The Loss of a Negative is the Addition of a Positive 08:43
Flip the sign! 11:16
Signed Numbers: Multiplication and Division 13:45
Quick Rules 15:08
Multiplication and Division Examples 17:02
## I. Basic Algebra
Exponents and Roots 24:44
Intro 00:00
The Basics: Exponents 00:09
Base and Exponent 00:42
Squared Terms 01:23
Cubed Terms 02:26
Exponent Rules 03:40
Multiplying Terms with the Same Base 05:51
Dividing Terms with the Same Base 06:48
Exponent Term Raised to a Power 08:32
Product of Different Bases to the Same Power 10:14
Quotient of Different Bases to the Same Power 11:57
Square Roots 19:39
Divisibility 16:45
Intro 00:00
The Basics: Divisibility 00:12
Divisibility Tricks 03:07
The Rules 03:44
Examples of Quickly Determing what a Number is Divisibly by 09:44
Prime Factorization 10:01
Intro 00:00
Prime Numbers and Prime Factorization 00:22
How to: Prime Factorization 02:27
Bigger Divisors 05:10
Finding all Factors of 28 and 210 05:52
Powers of 10 8:56
Intro 00:00
The Basics: Powers of 10 00:15
Multiplying and Dividing by Powers of 10 02:59
Dealing with Multiplication 03:37
Dealing with Division 05:16
Example Set 06:05
Fractions 12:50
Intro 00:00
The Basics: Fractions 00:09
Fractions as Part of a Whole 00:46
Equivalent Fractions 01:10
Finding the Common Denominator 05:25
Fractions: Multiplication and Division 07:27
Division: Multiplying the Reciprocal (Flip It) 08:53
Fractions: Comparing Fractions 10:36
Determining the relationship between fractions 11:54
Algebraic Expression 17:39
Intro 00:00
The Basics: Algebraic Expressions 00:11
Algebraic Expression Definitions 01:54
Simplifying Algebraic Expressions 05:12
Identifying Like Terms 05:23
Using Coefficients to Add and Subtract Like Terms 06:14
Combining Like Terms 08:00
Multiplication of Algebraic Expressions 09:24
Multiplying Two Terms: Coefficients and Variables 09:38
Combining Like Variables in Multiplication 10:15
Multiplying Expressions by a Term 12:08
Example 1: Multiplying Out and Simplifying 13:45
Example 2: Multiplying Out and Simplifying 15:18
Equations 6:07
Intro 00:00
The Basics: Equations 00:10
Valid Operations on an Equation 00:26
Isolate the Variable 03:24
Function Evaluation 10:50
Intro 00:00
The Basics: Function Evaluation 00:10
Function Definition and Notation 02:23
Function Evaluation Examples 04:55
Function Evaluation Mistakes to Avoid 08:47
Evaluation of Multiple Functions 09:20
## II. Geometric Concepts
Geometric Terms and Notation 8:36
Intro 00:00
The Basics: Geometric Terms and Notation 00:15
Geometric Notation 06:01
Angles 18:46
Intro 00:00
The Basics: Angles 00:09
Classifying Angles 01:40
Acute Angles 01:55
Right Angles 02:21
Obtuse Angles 02:54
Straight Angles 03:46
Angle Relationships 04:13
Complementary Angles 04:28
Supplementary Angles 05:25
Bisected Angles 06:20
Angles at Intersections 07:14
Parallel Lines and Angles 09:50
Drawing Two Intersections to Determine Equal Angles 10:31
Angle Rules 11:20
Shapes and Angles 14:48
Angles in a Triangle 15:23
Determining the Number of Angles is Larger Shapes 16:58
Triangles 21:25
Intro 00:00
Triangle Basics 00:12
Special Triangles: Isosceles and Equilateral 05:29
Special Triangles: Right Triangles 11:33
Special Triangles: 45, 45, 90 Triangles 18:49
Circles 16:34
Intro 00:00
The Basics: Parts of the Circle 00:11
The Diameter 01:04
Arcs and Sectors 03:41
Arc 03:47
Sector 06:41
Using Fractional Weights 07:32
Tangent and Concentric 08:39
Lines that are Tangent to Circles 08:47
Tangent Circles 10:02
Concentric Circles 11:02
Angles in a Circle 11:55
Half Circles and Quarter Circles 12:19
Proportional Areas and Circumferences for Circle Parts 13:29
Inscribed angles 14:37
Volume and Surface Area 10:53
Intro 00:00
The Basics: Volume 00:17
Volume Equations 01:52
Equation for the Volume of a Cube 02:30
Equation for the Volume of a Cylinder 03:26
Finding other Volumes 04:14
The Basics: Surface Area 05:26
Surface Area of a Cube 06:30
Surface Area of a Cylinder 08:15
Foil 11:02
Intro 00:00
The Basics: Foil 00:16
Foil Examples 02:34
Skip Foil: Memorize These 06:22
Factoring 28:35
Intro 00:00
The Basics: Factoring 00:12
Factoring: Pull out the GCF 01:07
Example of Pulling Out the Greatest Common Factor 03:41
Factoring: Difference of Squares 05:30
Examples with Difference of Squares Factorizations 06:16
Zooming in on C 10:12
How to Decide if You Need a + or - 19:37
Using Factoring in Rational Expressions 21:43
Steps: Take out the GCF, Factor, and Cancel 22:15
Factoring in Rational Expressions: Examples 23:48
Systems of Equations 28:57
Intro 00:00
The Basics: Systems of Equations 00:19
Equations with Infinite Solutions 00:44
Number of Equations Should Match the Number of Unknowns 02:41
Why Manipulation and Combination is Often the Best Method 04:34
Why Addition and Subtraction of Equations is Valid 07:23
The Most Important Step--The Common Coefficient 10:41
Manipulation and Combination: Steps 13:50
Substitution 15:49
Which Method Should Be Used? 20:13
How to Solve Problems with Three Equations and Three Unkowns 22:58
Three Equation Example 24:22
Distance Between Two Points 11:55
Intro 00:00
The Basics: Finding the Distance 00:13
Breaking Apart X and Y Cooridinates 00:48
Using Pythagorean Theorom 02:53
The Distance Formula 04:26
Example 1: Plugging into the Formula 06:18
Example 2: Using the Given Length to Find a Missing Coordinate 08:20
Intercepts 6:32
Intro 00:00
The Basics: Intercepts 00:11
The x-intercept 00:43
The y-intercept 01:44
Determing the Intercepts 03:05
The Special Case: b 04:37
## IV. Statistics
Average 13:28
Intro 00:00
The Basics: Averages 00:11
Finding the Average 02:26
Rearraning the Average Equation 02:59
Average Examples 03:48
Finding the Missing Value 04:32
The Basics: Weighted Averages 06:25
Fractional Weight Method 06:54
Sum and Retake Method 07:48
Weighted Average Examples 08:30
Example Using the Sum and Retake Method 10:50
Median, Mode and Range 10:08
Intro 00:00
The Basics: Median 00:23
How to Find the Median 01:01
Quick Median Examples 01:23
The Basics: Mode 05:10
How to Find the Mode 05:16
Quick Mode Examples 05:28
The Basics: Range 07:20
How to Find the Range 08:06
Quick Range Example 08:28
## V. The Rest
Rates 17:49
Intro 00:00
The Basics: Rates 00:11
Using Units in Rate Problems 01:18
The Classic Rate Problems, d=rt 02:13
Dealing with Time in Rates 02:47
Changing Minutes to Seconds and Vice Versa 03:40
Changing Hours to Minutes and Vice Versa 04:41
The Big One: Seconds to Hours 05:19
Combining Multiple Rates 06:31
Set Up Two Equations 08:59
Solving Rates 09:32
Quick Example: Using d=rt 10:03
Solving Multiple Rates 12:29
Sequences 10:00
Intro 00:00
The Basics: Sequences 00:11
When Sequences Are Just Patterns 01:14
Arithmetic Sequences 02:24
Using the Arithmetic Sequence Equation 03:11
Geometric Sequences 05:24
Using the Geometric Sequence Equation 06:08
Geometric Series that Alternate Between Negative and Positive Elements 08:02
Sets 10:21
Intro 00:00
The Basics: Sets 00:08
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# Homework Help: Limit Comparison Test
1. Mar 13, 2013
### whatlifeforme
1. The problem statement, all variables and given/known data
Use direct comparison test or limit comparison test to determine if the integral converges.
2. Relevant equations
$\displaystyle\int_0^6 {\frac{dx}{9-x^2}}$
3. The attempt at a solution
If i were to use the limit comparison test, would these integrals fit the criteria.
** if the positive functions f and g are continues on [a,∞)
Note: what does it mean by positive functions?
limit$_{x->infinity} \frac{f(x)}{g(x)} = L$ 0 < L < ∞
then $\displaystyle\int_a^∞ {f(x) dx}$ and
$\displaystyle\int_a^∞ {g(x) dx}$
both converge or diverge.
f(x) = $\displaystyle\int_0^6 {\frac{dx}{9-x^2}}$
g(x) = $\displaystyle\int_0^6 {\frac{1}{x^2} dx}$
since these functions are not continuous at a=0, then is the limit comparison test not an option here?
If not, how would i go about choosing a function for the direct comparison test?
Last edited: Mar 13, 2013
2. Mar 13, 2013
### Dick
No need to hyperbold on the font. Your integral already has a problem way before infinity. Where is it?
3. Mar 13, 2013
### whatlifeforme
i was stating the theorem from the book. it states must be continuous from [a,∞), where the a=0 in the case, and (sorry forgot to include the comparing function).
$\displaystyle\int_0^6 {\frac{1}{x^2} dx}$ is not continuous at a=0.
f(x) = $\displaystyle\int_0^6 {\frac{dx}{9-x^2}}$
g(x) = $\displaystyle\int_0^6 {\frac{1}{x^2} dx}$
further, i don't think it is stating the integral must be to infinity, just that it must be continuous on, or am i incorrect?
how should i solve with direct comparison test, if easier. please help. test in 2 days.
4. Mar 13, 2013
### Dick
Do direct comparison. The problem with your integral happens at x=3, doesn't it? It's not continuous there. Is 1/(9-x^2) integrable on the interval [0,3]? Try to think of a comparison function on just that interval. Factor 9-x^2.
5. Mar 14, 2013
### whatlifeforme
is (3+x) or (3-x) < (3+x)(3-x) meaning $\frac{1}{3+x} or \frac{1}{3-x} > \frac{1}{9-x^2}$ ??
thus, since
$\displaystyle\int_0^6 {\frac{1}{3+x} dx}$ converges to a finite number.
and $\frac{1}{3+x}> \frac{1}{9-x^2}$
$\displaystyle\int_0^6 {\frac{1}{9-x^2} dx}$ it would conclude that this also converges?
By Direct Comparison Test.
6. Mar 14, 2013
### Dick
The comparison you want is something like $\frac{1}{9-x^2} > C \frac{1}{3-x}$ on the interval 0<x<3 for some constant C. Because $\frac{1}{3-x}$ diverges on that interval allowing you to conclude the original integral diverges. Can you find such a constant C?
7. Mar 14, 2013
### whatlifeforme
i'm sorry. i am still confused. is what i stated above incorrect? the integral is from 0 to 6 not 0 to 3.
what if i choose 9 for c? i just chose that at random.
8. Mar 14, 2013
### Dick
Yes, what you stated before is incorrect. I know your integral is 0 to 6. If you can show it diverges on 0 to 3 then it will diverge on 0 to 6. Why don't you try making a nonrandom choice of C? Use the factorization and figure out what range of values 1/(3+x) takes on [0,3].
9. Mar 14, 2013
### whatlifeforme
are we working with 1/(3-x) or 1/(3+x) or does it matter because we have switched up a few times?
if we are using 1/(3+x), do i need to integrate on [0,3] to find the range of values?
10. Mar 14, 2013
### iRaid
Why don't you just split this integral into two parts and solve for both using limits from the right and left.
$$\int_0^6 \frac{dx}{9-x^{2}}=\lim_{t \to 3^{-}}\int_0^t \frac{dx}{9-x^{2}}+\lim_{t \to 3^{+}}\int_t^6 \frac{dx}{9-x^{2}}$$
See what I'm saying?
11. Mar 14, 2013
### whatlifeforme
if it were, 1/(9+x^2) i would choose 1/x^2 and if it converged i would know the other converged.
the -x^2 is throwing me off in 1/(9-x^2).
(-1/16) < (1/16) when we get into negatives, correct, i know this sounds elementary, but in this application when we are determining smaller and larger functions are negatives always smaller than positives?
12. Mar 14, 2013
### whatlifeforme
i want to know how to solve using a limit comparison test or direct comparison test, not by evaluating.
13. Mar 14, 2013
### Dick
Negatives are always smaller than positives, but when you doing a comparison test you usually want to compare things of the same sign. Just think about it this way. 1/(9-x^2)=(1/(3+x))(1/(3-x)). If you replace the 1/(3+x) with its minimum on [0,3], then you can also replace the '=' sign with '>='.
14. Mar 14, 2013
### iRaid
Oh my bad sorry.
15. Mar 14, 2013
### whatlifeforme
what do you mean by the minimum on [0,3] ? 1/6 would be the minimum, or do you mean integrate on [0,3] ?
16. Mar 14, 2013
### Dick
Yes, I mean 1/6. So 1/(9-x^2)>=(1/6)(1/(3-x)). Does the integral of 1/(3-x) converge on [0,3)?
17. Mar 14, 2013
### whatlifeforme
$\displaystyle\int_0^3 {\frac{1}{3-x} dx}$
$\lim_{b \to 3^-} ln|3-x|]^{0}_{b}$
$\lim_{b \to 3^-} ln|3-b| - ln|3-0| --> ln|3-3| = ln|0| = ∞$
Thus, the smaller diverges, so the original function diverges.
18. Mar 14, 2013
### Dick
Right!
19. Mar 14, 2013
### whatlifeforme
so explain one last time how you got: 1/(9-x^2)>=(1/6)(1/(3-x))
20. Mar 14, 2013
### Dick
It's all in post 13. I factored 1/(9-x^2) and then replaced one of the factors by something smaller.
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# Determinant of this skew symmetric matrix
How to prove the determinant of
$$A=(a_{ij})_{n\times n}=b_i^2-b_j^2$$ where $b_1,b_2,...,b_n$ are some distinct real numbers , $n\geq 3$ and $i,j=1,2,...,n$ is zero?
Here, $A$ is skew symmetric and I know odd order skew symmetric matrix has determinant zero. How about the even case? or Is there any other method to show $det(A)=0$ ?
• It is zero when $n \geq 3$. Indeed, this is a particular case of the fact that $\det\left(\left(x_i+y_j\right)_{1\leq i\leq n,\ 1\leq j\leq n}\right) = 0$ whenever $n \geq 3$. For this fact, see, e.g., Example 5.28 in Notes on the combinatorial fundamentals of algebra, 7 November 2017. – darij grinberg Nov 10 '17 at 3:44
• The comment by @darijgrinberg gives a very nice explanation, and it is impressive to see this writing which is nearly 800 pages long! (I wish that I could study this some day!) Nevertheless a very quick explanation is possible, you do not quite have to find an example in the middle of $>780$ pages. The matrix $(b_i^2)$ has rank $1$ because each row is constant. The matrix $(b_j^2)$ has rank $1$ because each column is constant. The difference is a matrix of rank $2$ at most. If $n \geq 3$, an $n \times n$ matrix of rank $2$ or less has determinant zero. – Zach Teitler Nov 10 '17 at 5:45
• @ZachTeitler: Thanks, but there's probably little new in there for you :) I was focussing on writing up in full detail the standard arguments that often end up in the no-mans-land between linear algebra and combinatorics. – darij grinberg Nov 10 '17 at 5:54
• @darijgrinberg Well, most of the main topics are familiar, but I'm sure that I could learn a lot from the examples. Hopefully some day. (Or perhaps I will convince a student to read it and tell me about it. :-) ) – Zach Teitler Nov 10 '17 at 6:32
From the comments: $$A$$ is the difference of two rank-one matrices, so that the rank of $$A$$ is at most two. It follows that $$\det A = 0$$ if $$n \ge 3$$.
You can't do it, at least not for all $n$, as the question was originally asked.
Pick $b_1 = 1$ and $b_2 = 0$ to get the matrix $$\pmatrix{0 & 1 \\ -1 & 0}$$ whose determinant is not zero.
• Fine! Actually we consider the case $n \geq3$ (I have edited the question.) – user444830 Nov 10 '17 at 5:28
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Blog
# The Quest for 700: Weekly GMAT Challenge (Answer)
Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:?
To solve this problem, we must translate the verbal instructions for the construction of the sequence. The first term is easy: Q1 = 1.
Next, we have this difficult wording: “for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube.”
So we’re dealing with all the positive integers beyond 1. Let’s take as an example n = 2. The instructions become these: “the second term equals the absolute value of the difference between the second (nth) smallest positive perfect cube and the first (that is, n-1st) smallest positive perfect cube.”
We know we need to consider the positive perfect cubes in order:
13 = 1 = smallest positive perfect cube (or “first smallest”).
23 = 8 = second smallest positive perfect cube.
The absolute value of the difference between these cubes is 8 – 1 = 7. Thus Q2 = 8 – 1 = 7.
Likewise, Q3 = |33 – 23| = 27 – 8 = 19, and so on.
Now, rather than figure out each term of Q separately, then add up, we can save time if we notice that the cumulative sums “telescope” in a simple way. This is what telescoping means:
The sum of Q2 and Q1 = 7 + 1 = 8. We can also write (8 – 1) + 1 = 8. Notice how the 1’s cancel.
The sum of Q3, Q2 and Q1 = 19 + 7 + 1 = 27. We can also write (27 – 8) + (8 – 1) + 1 = 27. Notice how the 8’s and the 1’s cancel.
At this point, we hopefully notice that the cumulative sum of Q1 through Qn is just the nth smallest positive perfect cube.
So the sum of the first seven terms of the sequence is 73, which equals 343.
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# Thread: Proof that 1 = 2
1. Originally Posted by TheAbstractionist
Here is one.
$\cos^2x-\sin^2x\ =\ \cos2x$ (identity)
$\cos^2x\ =\ \cos2x+\sin^2x$ (rearramging)
$\cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
$\cos\frac{3\pi}4\ =\ \sqrt{\cos\frac{3\pi}2+\sin^2\frac{3\pi}4}$ (substituing $x=\frac{3\pi}4)$
$-\frac1{\sqrt2}\ =\ \sqrt{0+\frac12}=\frac1{\sqrt2}$ (evaluating the substitution)
$-\frac12\ =\ \frac12$ (multiplying both sides by $\frac1{\sqrt2})$
$1\ =\ 2$ (adding $\frac32$ to both sides)
Quite late , but this proof is incorrect simply due to the fact that in the second quadrant $(\frac{\pi}{2} \leq \theta \leq \pi)$, the result is negative (meaning $cosx = -\sqrt{cos2x+sin^2x}$ ), but you used the positive result.
2. Originally Posted by paulrb
Are there any 1=2 "proofs" that do not involve dividing by zero? I am trying to think of other mathematical rules that can be broken to derive such a proof.
Irrational number - Wikipedia, the free encyclopedia
I saw this the other day. Look under the "History" section. Hippasus' proof that an odd number is even. It's not exactly the same as a 1=2 "proof," but is at the very least relevant to what you are touching upon here.
It's worth mentioning that--if I've understood correctly--a major difference is that Hippasus's proof is valid. No fallacies in logic can be pointed out as in the examples here. I therefore speculate that it is conceivable that a valid 1=2 proof can exist, and that it would merely prove, as Hippasus' proof does, that irrational numbers exist.
My speculation is probably totally hare-brained though, as I am not a real mathematician.
3. Originally Posted by rainer
Irrational number - Wikipedia, the free encyclopedia
I saw this the other day. Look under the "History" section. Hippasus' proof that an odd number is even. It's not exactly the same as a 1=2 "proof," but is at the very least relevant to what you are touching upon here.
It's worth mentioning that--if I've understood correctly--a major difference is that Hippasus's proof is valid. No fallacies in logic can be pointed out as in the examples here. I therefore speculate that it is conceivable that a valid 1=2 proof can exist, and that it would merely prove, as Hippasus' proof does, that irrational numbers exist.
My speculation is probably totally hare-brained though, as I am not a real mathematician.
Actually, in his proof he assumes (in contradiction), that there are no irrational numbers. He then reaches a cotradiction - that the number b, which he specified, must be both even and odd, however no such number exists. therefore, the claim that there are no irratioal numbers is false.
4. Originally Posted by TheAbstractionist
Here is one.
$\cos^2x-\sin^2x\ =\ \cos2x$ (identity)
$\cos^2x\ =\ \cos2x+\sin^2x$ (rearramging)
$\cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
$\cos\frac{3\pi}4\ =\ \sqrt{\cos\frac{3\pi}2+\sin^2\frac{3\pi}4}$ (substituing $x=\frac{3\pi}4)$
$-\frac1{\sqrt2}\ =\ \sqrt{0+\frac12}=\frac1{\sqrt2}$ (evaluating the substitution)
$-\frac12\ =\ \frac12$ (multiplying both sides by $\frac1{\sqrt2})$
$1\ =\ 2$ (adding $\frac32$ to both sides)
you forgot that square rootting something is '+' OR '-'.
5. ## Another Proof that 1 = 2
I'm surprised no one brought this one up, since it's been quite a number of years since I first heard of it.
Here is a 1=2 proof using differentiation:
x^2 = x^2
x^2 is x*x, so we can represent one side by addition:
x + x + x + ... + x = x^2
where there are x x's.
Now, we differentiate both sides:
1 + 1 + 1 + ... + 1 = 2x
There are x 1's, so we can sum all the 1's to get
x = 2x
And dividing by x we get
1=2
6. ## Re: Another Proof that 1 = 2
Nice example of what happens if the variable gets partly treated as a constant.
7. ## Re: Another Proof that 1 = 2
Originally Posted by mathbyte
I'm surprised no one brought this one up, since it's been quite a number of years since I first heard of it.
Here is a 1=2 proof using differentiation:
x^2 = x^2
x^2 is x*x, so we can represent one side by addition:
x + x + x + ... + x = x^2
Theres got to be something fishy about that representation
x + x + x + ... + x = x^2
if x = -1, how many x do you have on the left of your equation? How about if x = 1/3 ?
8. ## Re: Another Proof that 1 = 2
Originally Posted by agentmulder
Theres got to be something fishy about that representation
x + x + x + ... + x = x^2
if x = -1, how many x do you have on the left of your equation? How about if x = 1/3 ?
Even if "x" is natural,
the number of x's summed is itself a function of x
and must be taken into account when calculating the derivative (product rule).
9. ## Re: Another Proof that 1 = 2
Even if "x" is natural,
the number of x's summed is itself a function of x
and must be taken into account when calculating the derivative (product rule).
I agree. To me it looks fishy even before the derivative is taken, x^2 = x^2 holds for any x, not so sure about the other representation.
Yes, x*x = x^2 doesn't pose any problems i can think of
10. ## Re: Another Proof that 1 = 2
let's look at the equation:
$\sum_{i=1}^kx = x^2$
when we solve for k, we get k = x.
now, let's differentiate that same equation:
$\sum_{i=1}^k 1 = 2x$.
when we solve for k, we get k = x/2.
evidently, x = x/2, so 2x = x, so x = 0.
therefore, the flaw in the proof is the very last step...we have divided by 0.
11. ## Re: Proof that 1 = 2
-1 = 1
0 = 1 + 1
0 = 2
so it will zero not 1 =2 .
word problem help
13. ## Re: Proof that 1 = 2
Originally Posted by TheAbstractionist
Here is one.
$\cos^2x-\sin^2x\ =\ \cos2x$ (identity)
$\cos^2x\ =\ \cos2x+\sin^2x$ (rearramging)
$\cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
$\cos\frac{3\pi}4\ =\ \sqrt{\cos\frac{3\pi}2+\sin^2\frac{3\pi}4}$ (substituing $x=\frac{3\pi}4)$
$-\frac1{\sqrt2}\ =\ \sqrt{0+\frac12}=\frac1{\sqrt2}$ (evaluating the substitution)
$-\frac12\ =\ \frac12$ (multiplying both sides by $\frac1{\sqrt2})$
$1\ =\ 2$ (adding $\frac32$ to both sides)
you made a mistake here bro:
$\cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
there should absolute function over cosx. since sqrt(x^2)=abs(x). Its the most common mistake that students makes while learning Pre-calculus.But we soon realize the importance of absolute function when we study limits.
14. ## Re: Proof that 1 = 2
$\cos^2x-\sin^2x\ =\ \cos2x$ (identity)
$\cos^2x\ =\ \cos2x+\sin^2x$ (rearranging)
$|\cos x}|\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
And thus there is no contradiction when $x\ =\ \frac{3\pi}4$. This absolute value sign must be included if we are taking the positive square root on the right hand side of the equation.
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# Regression to obtain autocorrelation measure (AR(1))
This is not homework. I am a frequent user on math.stackexchange, but I am learning a bit about time series models and came across this example. Any ideas would be greatly appreciated.
A linear regression model was fit to some time-series data by ordinary least squares. The residuals from the fit were then used to create two new variables, namely $E$ with values $\hat{e}_2,...,\hat{e}_n$ and $E_1$ with values $\hat{e}_1,...,\hat{e}_{n-1}$. A linear “regression through the origin” was then run with E as the dependent variable and $E_1$ as the predictor. The slope estimate was 0.412 with a standard error of 0.133. Assume that the $e_t$ follow a standard AR(1) model.
1. Estimate the first order autocorrelation $\rho$ of the AR model.
2. Can the output be used to obtain a valid standard error for the estimate in 1?
Is the answer to #1 above just the slope of the regression E~E_1?
-
Yes to all three questions. But it is more efficient to estimate the AR error model at the same time as the linear regression model using GLS or MLE.
-
I agree in practice, which is why this exercise I found seemed rather strange. – Justin Jul 23 '12 at 1:52
The $\rm AR(1)$ model is given as $X_n= \rho X_{n-1} + e_n.$ The parameter $\rho$ is normally estimated by conditional least squares. If the model is correct the eis have mean $0$ and variance $\sigma_e^2.$
The parameter $\rho$ is $$\rho=\frac{{\rm Cov}(X_n, X_{n-1})}{{\rm Var}(X_n)}.$$
When the estimate of $e_n$ is paired with the estimate of $e_{n-1}$ the slope is theoretically $0$ but will not be exactly zero because the estimates of residuals are based on the estimate of $\rho$.
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1
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Question
# The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number ?
A
270
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B
240
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C
295
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D
360
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Solution
## The correct option is A 270 Let the smaller number be x. ⇒ Larger number = x+1365 By Euclid's division lemma, the larger number can also be written as 6x+15. ⇒x+1365=6x+15 ⇒5x=1350 ⇒x=270 ∴ The smaller number is 270 and the larger number is 270 + 1365 = 1635.
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Cauchy's Integral Formula
Theorem 1 (Cauchy's Integral Formula): Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ be analytic on $A$. Let $\gamma$ be a simple closed piecewise smooth and positively oriented curve contained in $A$ and the inside of $\gamma$ is contained in $A$. Then for any $z_0$ inside $\gamma$ we have that $\displaystyle{f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} \: dz}$.
• Proof: Let $\epsilon > 0$ be given. Let $r > 0$ be such that $D(z_0, r) \subseteq A$. Let $C_r : z_0 + re^{it}$, $t \in [0, 2\pi]$ denote the positively oriented circle centered at $z_0$ with radius $r$. Since $\gamma$ is a simple closed piecewise smooth curve contained in $A$ we have that $C_r$ is homotopic to $\gamma$. So by the deformation theorem $\displaystyle{\int_{\gamma} \frac{f(z)}{z - z_0} \: dz = \int_{C_r} \frac{f(z)}{z - z_0} \: dz}$ and:
• Since $f$ is analytic on $A$ we have that $f$ is continuous on $A$. So for this given $\epsilon$ here exists a $\delta > 0$ such that if $\mid z - z_0 \mid < \delta$ then $\mid f(z) - f(z_0) \mid < \epsilon$.
• We want $\mid z_0 + re^{it} - z_0 \mid = \mid re^{it} \mid = r < \delta$, so choose $r < \delta$. Then whenever $\mid z_0 + re^{it} - z_0 \mid < \delta$ we have that:
• Since $\epsilon$ is arbitrary we have that $\displaystyle{f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} \: dz}$. $\blacksquare$
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# Find equation of the circle whose diameter is the common chord of two other circles?
Circle 1:
$$x^2 + y^2 +6x + 2y +6 = 0$$
Circle 2:
$$x^2 + y^2 + 8x + y + 10 = 0$$
My attampt:
From circle 1 and 2, I found
$$y = 2x + 4$$ which is the common chord. Pluging that in equation 1 I got
$$5x^2 + 26x + 30 = 0$$
here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?
• That’s not the common chord: a chord is a line segment. That’s the circles’ radical axis, which is the extension of the common chord. – amd Sep 20 at 21:56
• "If the two circles intersect, the radical axis is the secant line corresponding to their common chord."It's from Wikipedia. – Ghost Sep 22 at 19:00
• You’re only underscoring my point: The sentence you cite distinguishes between a line and a line segment, which you’ve conflated in your question. The equation that you’ve written describes a line, not a line segment. – amd Sep 22 at 21:10
First, obtain the equations of the intersection points below for both $$x$$ and $$y$$,
$$5x^2 + 26x + 30= 0$$ $$5y^2 + 12y -8= 0$$
It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,
$$x_1+x_2=-\frac{26}{5},\>\>\>x_1x_2=6$$
$$y_1+y_2=-\frac{12}{5},\>\>\>y_1y_2=-\frac 85$$
Thus, the center of the circle is $$\frac{x_1+x_2}{2}=-\frac{13}{5}, \frac{y_1+y_2}{2}=-\frac{6}{5}$$ and its diameter squared is,
$$(x_1-x_2)^2 + (y_1-y_2)^2$$ $$= (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$
$$= \left( \frac{26}{5} \right)^2 -4\cdot 6 + \left( \frac{12}{5}\right)^2 + 4\cdot \frac 85 = \frac{76}{5}$$
The equation of the circle is
$$\left( x+\frac{13}{5} \right)^2 + \left( y +\frac{6}{5}\right)^2 = \frac{19}{5}$$
You are on the right track. The numbers cancel out nicely when you sum them. Indeed: $$5x^2 + 26x + 30 = 0 \Rightarrow x_1=\frac{-13-\sqrt{19}}{5},x_2=\frac{-13+\sqrt{19}}{5}\\ y_1=\frac{-6-2\sqrt{19}}{5}, y_2=\frac{-6+2\sqrt{19}}{5}$$ The center of the new circle: $$\frac{x_1+x_2}{2}=-\frac{13}{5},\frac{y_1+y_2}{2}=-\frac{6}{5}$$ The diameter of the new circle: $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{\frac{4\cdot 19}{25}+\frac{16\cdot 19}{25}}=\sqrt{\frac{76}{5}} \Rightarrow \\ r=\frac12d=\sqrt{\frac{76}{4\cdot 5}}=\sqrt{\frac{19}{5}}$$ Thus: $$\left(x+\frac{13}{5}\right)^2+\left(y+\frac65\right)^2=\frac{19}{5}.$$
• Comparing this to other solutions, solving the quadratic equation is not so much more complicated after all. Sometimes you just have to not give up so quickly. – David K Sep 20 at 22:34
• The thing is I am a late bloomer in mathematics.I just met mathematics. That was something that I didn't know and the second one is still critical thinking and I'm afraid I haven't still learnt how to think like that yet. – Ghost Sep 22 at 12:45
You can certainly keep going the way you are: solve the quadratic equation for $$x$$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $$(x_1,y_1)$$ and $$(x_2,y_2)$$ of the chord. An equation of the circle with that diameter can be written down directly: $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$. Rearrange this into whatever form is required.
However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-\lambda)(x^2+y^2+6x+2y+6)+\lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2\lambda+6)x+(2-\lambda)y+(4\lambda+6)=0.\tag{*}$$ By inspection, the coordinates of the center of this circle are $$(-\lambda-3,\lambda/2-1)$$ and we want it to lie on the radical axis $$y=2x+4$$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $$\lambda$$, then plug that value into equation (*).
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MAT244--2020F > Chapter 2
Boyce-DiPrima Section 2.1 Example 1
(1/1)
jimboxie:
I'm a bit rusty on some calc, and I was wondering how the textbook did the following, along with my attempt on what I think happened (image attached), but i can't seem to see what I did wrong.
Suheng Yao:
I think that the textbook just uses the product rule: [f*g]=f'*g+g'*f. Here, you can think f as 4+t^2 and g as y. Hope this helps.
jimboxie:
I completely forgot about the product rule. Thank you.
RunboZhang:
I think Suheng Yao has pointed out a crucial point: product rule. Indeed, if you observe LHS=(4+t^2)(dy/dt) + 2ty, you may find the pattern of u'v+uv' where u=y and v=4+t^2. By applying product rule here, we can make LHS become a single expression and thus make it a separable differential equation (which is already discussed in the lecture).
I think a general method of solving this kind of problem is introduced later in the textbook and it is quite useful in solving inseparable first order differential equation.
Victor Ivrii:
--- Quote from: Suheng Yao on September 15, 2020, 02:51:19 PM ---I think that the textbook just uses the product rule: [f*g]=f'*g+g'*f. Here, you can think f as 4+t^2 and g as y. Hope this helps.
--- End quote ---
Please, never use * as multiplication sign (which even is not needed here). Your formulas are correctly formatted, but if you surround each by dollar sign like
--- Code: ---I think that the textbook just uses the product rule: $[f*g]=f'*g+g'*f$. Here, you can think $f$ as $4+t^2$ and $g$ as $y$. Hope this helps.
--- End code ---
you'll get
I think that the textbook just uses the product rule: $[f*g]=f'*g+g'*f$. Here, you can think $f$ as $4+t^2$ and $g$ as $y$. Hope this helps.
Again, * is reserved for a different operation, and it usage as multiplication sign may be considered as a mathematical error
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# Reteach Sec 1
## Transcription
Reteach Sec 1
```Name ________________________________________ Date __________________ Class__________________
LESSON
1-6
Reteach
Midpoint and Distance in the Coordinate Plane
The midpoint of a line segment separates the segment into two halves.
You can use the Midpoint Formula to find the midpoint of the segment
with endpoints G(1, 2) and H(7, 6).
⎛ x + x2 y1 + y 2 ⎞
⎛ 1+ 7 2 + 6 ⎞
,
,
=M⎜
M⎜ 1
⎟
2 ⎠
2 ⎟⎠
⎝ 2
⎝ 2
M is the midpoint
of HG .
⎛8 8⎞
= M⎜ , ⎟
⎝2 2⎠
= M(4, 4)
Find the coordinates of the midpoint of each segment.
1.
2.
_________________________________________
________________________________________
3. QR with endpoints Q(0, 5) and R(6, 7) __________
4. JK with endpoints J(1, –4) and K(9, 3) __________
Suppose M(3, −1) is the midpoint of CD and C has coordinates (1, 4). You can use
the Midpoint Formula to find the coordinates of D.
⎛ x + x2 y1 + y 2 ⎞
M (3, − 1) = M ⎜ 1
,
2 ⎟⎠
⎝ 2
x-coordinate of D
y-coordinate of D
3=
x1 + x2
2
Set the coordinates equal.
−1 =
y1 + y 2
2
3=
1 + x2
2
Replace (x1, y1) with (1, 4).
−1 =
4 + y2
2
6 = 1 + x2
Multiply both sides by 2.
−2 = 4 + y2
5 = x2
Subtract to solve for x2 and y2.
−6 = y2
The coordinates of D are (5, −6).
5. M(−3, 2) is the midpoint of RS , and R has coordinates (6, 0).
What are the coordinates of S?
_______________
6. M(7, 1) is the midpoint of WX , and X has coordinates (−1, 5).
What are the coordinates of W ?
_______________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
1-46
Holt Geometry
Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
1-6
Midpoint and Distance in the Coordinate Plane continued
The Distance Formula can be used to find the distance d
between points A and B in the coordinate plane.
( x2 − x1 )2 + ( y 2 − y1 )2
d=
=
(7 − 1)2 + (6 − 2)2
(x1, y1) = (1, 2); (x2, y2) = (7, 6)
=
62 + 42
Subtract.
=
36 + 16
Square 6 and 4.
=
52
≈ 7.2
The distance d
between points
A and B is the
length of AB .
Use a calculator.
Use the Distance Formula to find the length of each segment or the distance
between each pair of points. Round to the nearest tenth.
7. QR with endpoints Q(2, 4) and R(−3, 9)
8. EF with endpoints E(−8, 1) and F(1, 1)
_________________________________________
9. T(8, −3) and U(5, 5)
________________________________________
10. N(4, −2) and P(−7, 1)
_________________________________________
________________________________________
You can also use the Pythagorean Theorem to find distances in the coordinate plane.
Find the distance between J and K.
c 2 = a2 + b2
2
2
Pythagorean Theorem
=5 +6
a = 5 units and b = 6 units
= 25 + 36
Square 5 and 6.
= 61
c =
Side b is
6 units.
Side a is
5 units.
Take the square root.
Use the Pythagorean Theorem to find the distance, to the nearest
tenth, between each pair of points.
11.
12.
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
1-47
Holt Geometry
2. (−a, −b)
4.
3. (2d, 2e)
2. d = (2 − ( −1))2 + (2 − ( −2))2
rectangle
5. 13.4 m
6. 10 m2
7. 11.7 m
8. (17, 3)
= (3)2 + (4)2
2. (−1, −1)
3. (3, 6)
4. (5, −0.5)
5. (−12, 4)
6. (15, −3)
7. 7.1 units
8. 9 units
9. 8.5 units
10. 11.4 units
11. 5.7 units
12. 9.4 units
9 + 16
=
25
=5
4. a = 4, b = 3
c2 = 42 + 32
Reteach
1. (1, 5)
=
= 16 + 9
= 25
c=
25
=5
5. Sample answer: The Distance Formula
uses a coordinate plane. The
Pythagorean Theorem uses known
measures of two sides of a triangle.
Challenge
1. 29.6 units
LESSON 1-7
2. (−3, 2.5), (1.5, −0.5), (−0.5, 5)
3. 14.8 units
4. The perimeter of UABC is twice the
perimeter of the second triangle.
5. 12.2; 16.1 units
6. Both midpoints are at (1, 1). This is the
point where the diagonals intersect.
7. (1, 19)
8. 78.7 units; 493.2 units2
9. The diameter of the circle is
approximately 25.1 units, so the radius is
half that distance, or about 12.55 units.
The distance from the center of the circle
to G is 18 units. So G is not a point on the
circle.
Practice A
1. transformation
2. original; image
3. reflection
4. slide
5. rotation
6. 2; ABCD → A′B′C′D′
7. 3; UPQR → UP′Q′R′
8. 1; UHIJ → UH′I′J′
9.
reflection
10.
Problem Solving
1. 82.5 ft
2. 85.9 ft
3. 47.4 m
4. 18.4 m
5. B
6. H
7. C
Practice B
1. (2, 2); (−1, −2)
1. 2
2. 1
3. 3
4. rotation
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A7
Holt Geometry
```
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# Search by Topic
#### Resources tagged with Reflections similar to Hidden Meaning:
Filter by: Content type:
Age range:
Challenge level:
### There are 33 results
Broad Topics > Transformations and constructions > Reflections
### Hidden Meaning
##### Age 7 to 11 Challenge Level:
What is the missing symbol? Can you decode this in a similar way?
### Reflector ! Rotcelfer
##### Age 7 to 11 Challenge Level:
Can you place the blocks so that you see the reflection in the picture?
### Transforming the Letters
##### Age 7 to 11 Challenge Level:
What happens to these capital letters when they are rotated through one half turn, or flipped sideways and from top to bottom?
### Making Maths: Indian Window Screen
##### Age 7 to 11 Challenge Level:
Can you recreate this Indian screen pattern? Can you make up similar patterns of your own?
### Friezes
##### Age 11 to 14
Some local pupils lost a geometric opportunity recently as they surveyed the cars in the car park. Did you know that car tyres, and the wheels that they on, are a rich source of geometry?
### National Flags
##### Age 7 to 11 Challenge Level:
This problem explores the shapes and symmetries in some national flags.
### Matching Frieze Patterns
##### Age 11 to 14 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Transformation Tease
##### Age 7 to 11 Challenge Level:
What are the coordinates of this shape after it has been transformed in the ways described? Compare these with the original coordinates. What do you notice about the numbers?
### Coordinating Classroom Coordinates
##### Age 7 to 11
This article describes a practical approach to enhance the teaching and learning of coordinates.
### It's Times Again
##### Age 7 to 14 Challenge Level:
Which way of flipping over and/or turning this grid will give you the highest total? You'll need to imagine where the numbers will go in this tricky task!
### Clocks
##### Age 7 to 11 Challenge Level:
These clocks have been reflected in a mirror. What times do they say?
### Decoding Transformations
##### Age 11 to 14 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### Combining Transformations
##### Age 11 to 14 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Simplifying Transformations
##### Age 11 to 14 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Hexpentas
##### Age 5 to 11 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### So It's Times!
##### Age 7 to 14 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### What Am I?
##### Age 7 to 11 Challenge Level:
Can you draw the shape that is being described by these cards?
### Shaping up with Tessellations
##### Age 7 to 14
This article describes the scope for practical exploration of tessellations both in and out of the classroom. It seems a golden opportunity to link art with maths, allowing the creative side of your. . . .
### Penta Play
##### Age 7 to 11 Challenge Level:
A shape and space game for 2,3 or 4 players. Be the last person to be able to place a pentomino piece on the playing board. Play with card, or on the computer.
### The Frieze Tree
##### Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Frieze Patterns in Cast Iron
##### Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### Mirror, Mirror...
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### Building with Longer Rods
##### Age 7 to 14 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Let Us Reflect
##### Age 7 to 11 Challenge Level:
Where can you put the mirror across the square so that you can still "see" the whole square? How many different positions are possible?
### Reflecting Squarely
##### Age 11 to 14 Challenge Level:
In how many ways can you fit all three pieces together to make shapes with line symmetry?
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Reflecting Lines
##### Age 11 to 14 Challenge Level:
Investigate what happens to the equations of different lines when you reflect them in one of the axes. Try to predict what will happen. Explain your findings.
### 2010: A Year of Investigations
##### Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010.
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Times
##### Age 7 to 11 Challenge Level:
Which times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!
### Building with Rods
##### Age 7 to 11 Challenge Level:
In how many ways can you stack these rods, following the rules?
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# Thread: regular hexagon
1. ## regular hexagon
three of the six vertices of regular hexagon are chosen at random.the possibility that the triangle with three vertices is equilateral equals
three vertices can be chosen as 6C3 after that what to do
2. i think we can take central angle of regular hexagon and by angle in a segment=half central angle we can judge whether the triangle is equilateral or not
3. Hello, prasum!
I don't suppose you made a sketch . . .
Three of the six vertices of regular hexagon are chosen at random.
Find the probability that the three vertices form an equilateral triangle.
Three vertices can be chosen as 6C3. . Right!
After that, what to do?
. . $\begin{array}{ccccc}
&& A \\
F &&&& B \\ && * \\
E &&&& C \\
&& D \end{array}$
There are only two equilateral triangles: . $ACE$ and $BDF.$
,
,
,
,
,
# Three of six vertices of a regular hexagon are choosen at random. Find the probability that the triangle with three vertices is equilateral?
Click on a term to search for related topics.
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# Multiple equates
Diez B. Roggisch deets at nospam.web.de
Mon Nov 17 20:25:38 CET 2008
```Arnaud Delobelle wrote:
> jzakiya <jzakiya at mail.com> writes:
>
>> I looked online and in books, but couldn't find a definitive answer to
>> this.
>>
>> I have an array and set multiple elements to either True or False at
>> one time.
>>
>> Question: Which way is faster (or does it matter)?
>>
>> 1)
>>
>> array[x1]=array[x2]=........= array[x10] = \
>> array[x11]=array[x12]=... = array[x20] = \
>> ......
>> ......
>> array[x40]=array[x41]=....= array[x50] = False (or True)
>>
>>
>> 2)
>>
>> array[x1]=array[x2]=........= array[x10] = False
>> array[x11]=array[x12]=... = array[x20] = False
>> ......
>> ......
>> array[x40]=array[x41]=....= array[x50] = False
>
> It doesn't matter as none of this is valid Python. In Python you have to
> write
>
> array[x1] = False
> array[x2] = False
>
> Etc...
>
No.
l = range(10)
l[0] = l[1] = 100
And there seems to be a small difference between both versions,
but I can't judge for sure if that favors one of them - yet I tend to think
it's not worth the effort...
import dis
def a():
l = range(10)
l[0] = l[1] = 100
def b():
l = range(10)
l[0] = 100
l[1] = 100
print "------------ a ---------------"
dis.dis(a)
print "------------ b ---------------"
dis.dis(b)
------------ a ---------------
6 CALL_FUNCTION 1
9 STORE_FAST 0 (l)
* 15 DUP_TOP
22 STORE_SUBSCR
29 STORE_SUBSCR
33 RETURN_VALUE
------------ b ---------------
6 CALL_FUNCTION 1
9 STORE_FAST 0 (l)
21 STORE_SUBSCR
31 STORE_SUBSCR
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https://math.stackexchange.com/questions/3233396/closed-forms-for-the-integral-int-01-frac-operatornameli-nx1xdx
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# Closed-forms for the integral $\int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx$?
(This is related to this question.)
Define the integral,
$$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx$$
with polylogarithm $$\operatorname{Li}_n(x)$$. Given the Nielsen generalized polylogarithm $$S_{n,p}(z)$$,
$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$
Then it seems,
$$I_1 = -S_{1,1}(-1)-\tfrac12\ln(2)\ln(2)$$
$$I_2 = -5S_{1,2}(-1)+\ln(2)\,\zeta(2)\quad$$
$$\quad\qquad I_3 = -2S_{1,3}(-1)+\ln(2)\,\zeta(3)-\tfrac12\zeta(4)$$
where $$S_{1,1}(-1) = -\tfrac12\zeta(2)$$, and $$S_{1,2}(-1) = \tfrac18\zeta(3)$$ and $$S_{1,3}(-1)$$ has a more complicated closed-form given in the linked post.
Q: What is $$I_4$$ and $$I_5$$? In general, can $$I_n$$ be expressed by the Nielsen generalized polylogarithm?
P.S. Note that $$\operatorname{Li}_n(z), \ln(z), \zeta(z)$$ are just special cases of this function.
• $$I_4=\frac{\pi^4}{90}\ln 2-\frac{\pi^2}{12}\zeta(3)-\int_0^1 \frac{\operatorname{Li}_2(x)\operatorname{Li}_2(-x)}{x}dx$$ Is the last integral known by any chance? May 20, 2019 at 19:13
• @Threesidedcoin: Not that I'm aware of. Do you have $I_4$ up to 30 digits? May 20, 2019 at 19:28
• I barely have 5 digits, but if I can find $\int_0^1 \frac{\operatorname{Li}_2^2(-x)}{x}dx$ then $I_4$ is solved. May 20, 2019 at 19:36
• This might help:$$I_4=\int_0^1 \frac{\text{Li}_4(x)}{1+x}dx=\sum_{k\ge 0}(-1)^k\int_0^1x^k \text{Li}_4(x)dx$$ We have integrating by parts: $$I(t)=\int_0^1 x^k\text{Li}_t(x)dt=\frac{\text{Li}_t(1)}{k+1}-\frac{1}{k+1}\int_0^1 x^{k}\text{Li}_{t-1}(x)d=\frac{\zeta(t)-I(t-1)}{k+1}$$ So this will give: $$I_4=\eta(1)\zeta(4)-\eta(2)\zeta(3)+\eta(3)\zeta(2)+\sum_{k\ge 0}\frac{(-1)^k}{(k+1)^3}\int_0^1 x^k\ln(1-x)dx$$ $$=\eta(1)\zeta(4)-\eta(2)\zeta(3)+\eta(3)\zeta(2)-\int_0^1\ln(1-x)\frac{\text{Li}_3(-x)}{x}dx$$ The last integral is similar to this one math.stackexchange.com/a/463200 May 20, 2019 at 20:07
• @Zacky we can expand the dilogarithm then apply IBP $$\int_0^1\frac{\operatorname{Li}_2(x)\operatorname{Li}_2(-x)}{x}\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$ Jul 3, 2019 at 2:53
Based on Three-sided coin's comments, it seems
$$I_4 = \int_0^1 \frac{\operatorname{Li}_4(x)}{1+x}dx =\ln(2)\zeta(4)+\tfrac34\zeta(2)\zeta(3)-\tfrac{59}{32}\zeta(5) \approx 0.321352\dots$$
(Note: Confirmed as correct by Brevan's answer.) Connecting it to Three-sided coin's other integrals, then
$$I_4 = \ln(2)\zeta(4)-\tfrac12\zeta(2)\zeta(3)-h_1$$ $$I_4 = \ln(2)\zeta(4)+\tfrac14\zeta(2)\zeta(3)-h_2$$
where,
$$h_1=\int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(-x)}}{x}dx = -\tfrac54\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
$$h_2=\int_0^1\frac{\color{blue}{\operatorname{Li}_3(-x)}\ln(1-x)}{x}dx = -\tfrac12\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
which implies,
$$h_1-h_2 =\int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(-x)}{x}dx- \int_0^1\frac{\operatorname{Li}_3(-x)\ln(1-x)}{x}dx =-\tfrac34\zeta(2)\zeta(3)$$
Compare to the similar integrals here that he mentioned,
$$h_3= \int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(x)}}{x}dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$
$$h_4 = \int_0^1\frac{\color{blue}{\operatorname{Li}_3(x)}\ln(1-x)}{x}dx =\zeta(2)\zeta(3)-3\zeta(5$$
which has the proven relation,
$$h_3-h_4 = \int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(x)}{x}dx - \int_0^1\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}dx = \zeta(2)\zeta(3)$$
$$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx=(-1)^nA(n)+\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)$$ with Dirichlet eta function $$\eta(k)$$ and $$A(n) = \sum_{k=1}^\infty \frac{H_k}{k^n}(-1)^k$$
is the $$n$$th "Alternating Euler Sum". However, since,
$$A(n,z) = \sum_{k=1}^\infty \frac{H_k}{k^n}z^k =S_{n-1,2}(z)+\operatorname{Li}_{n+1}(z)$$
for $$-1\leq z\leq 1$$, then $$I_n$$ can be expressed by Nielsen polylogs $$S_{n,p}(z)$$ as I suspected.
Per @TitoPiezas' request, I will expand on my comment above and essentially close this question as far as the literature we have (unless there are new or unknown developments I am unaware of).
As @Zacky said, a key step in converting this problem to a more conventional one is repeated integration by parts. Using the techniques outlined in the comments above, we get $$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+\sum_{k=0}^\infty \tfrac{(-1)^{k+n}}{(k+1)^{n-1}}\int_0^1x^k\text{Li}_1(x) dx$$ We now shift the right summation index and use $$\text{Li}_1(x) = -\log(1-x)$$ to get $$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+\sum_{k=1}^\infty \tfrac{(-1)^{k+n}}{k^{n-1}}\int_0^1x^{k-1}\log(1-x) dx$$ and recall that$$\int_0^1x^{k-1}\log(1-x) dx= \frac{H_k}{k}$$ So we finally conclude $$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+(-1)^nA(n)$$ where $$A(n) = \sum_{k=1}^\infty \tfrac{(-1)^k H_k}{k^n}$$ is the $$n$$th "Alternating Euler Sum".
Alternating Euler Sums are well studied; examples just from this website can be found at this basic survey of integral forms here and the most amazing collection of answers ever here.
That latter link, when converted over to the more simple notation outlined here, gives a closed form for $$A(2n)$$ and shows why $$A(2n+1)$$ is particularly tricky; using the results on that page we should be able to easily get $$A(1)$$ and $$A(3)$$ but once we move higher we get linear dependence chains in the recurrence formulas that seemingly prevent us from expressing our answers in terms of Polylogarithms $$\text{Li}_s(z)$$ (which includes the Zeta and Dirichlet Eta functions as special values at $$z=\pm 1$$ respectively) and elementary functions in general (though specific cases, in particular for low $$n$$, may still work).
Note: I cannot stress enough how amazing that second link is. It essentially settles this question on its own through the amazing answers! Please take the time to go upvote those answers if possible.
• +1 This is great. By a happy coincidence, I know that $$\sum_{k=1}^{\infty}\frac{H_k}{k^a}z^k= S_{a-1,2}(z) + \rm{Li}_{a+1}(z)$$ with Nielsen polylog $S_{n,p}(z)$. So $I_n$ via the $A(n)$ can indeed be expressed by Nielsen polylogs as I suspected. However, I thought closed-forms of $A(n)$ in terms of ordinary polylogs are unknown for $n>3$, so I need to check that link for $A(5)$ and higher. Jun 8, 2019 at 4:57
• @TitoPiezasIII You are right; I misremembered the post. We have cleaner recursions for $n= 5, 7$ but they remain at intractable integrals. Unless I find evaluations somewhere I will remove those from my post. With your comment completing the link to Nielsen polylogs, this question seems about as closed as it can be until someone finds a way to evaluate $A(2n+1)$ (if it is even possible) Jun 8, 2019 at 5:10
• No problem. At least this question has closure, and I'm happy that my "guess" for $I_4$ is correct. But $I_5$ already involves the Nielsen polylog $S_{4,2}(-1)$ which has no known representation as ordinary polylogs. Jun 8, 2019 at 5:53
• Ah, I see. The formula is now correct. Jun 8, 2019 at 5:54
Here is a generalization thats works for even $$n$$: $$\begin{gather*} \int_0^1\frac{\operatorname{Li}_{2a}(x)}{1+x}\mathrm{d}x=\int_0^1\frac{\operatorname{Li}_{2a}(x)}{x}\mathrm{d}x-\int_0^1\frac{\operatorname{Li}_{2a}(x)}{x(1+x)}\mathrm{d}x\\ =\operatorname{Li}_{2a+1}(x)|_0^1-\int_0^1\frac{1}{1+x}\left(\frac{-1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1-xy}\mathrm{d}y\right)\mathrm{d}x\\ =\operatorname{Li}_{2a+1}(1)+\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\int_0^1\frac{\mathrm{d}x}{(1+x)(1-yx)}\right)\mathrm{d}y\\ =\zeta(2a+1)+\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\frac{\ln(2)-\ln(1-y)}{1+y}\right)\mathrm{d}y\\ =\zeta(2a+1)+\frac{\ln(2)}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1+y}\mathrm{d}y\\ -\frac{1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)\ln(1-y)}{1+y}\mathrm{d}y. \end{gather*}$$
Substitute
$$\int_0^1\frac{\ln^a(x)}{1+x}\mathrm{d}x=(-1)^aa!\,(1-2^{-a})\zeta(a+1)$$
and (this integral is evaluated by Cornel here)
$$\begin{gather*} \int_0^1\frac{\ln^{2q-1}(x)\ln(1-x)}{1+x}\mathrm{d}x=(2^{1-2q}-2)(2q-1)!\ln(2)\zeta(2q)\\ +[q-2^{-1-2q}(1+2q-2^{1+2q})](2q-1)!\,\zeta(2q+1)\\ -(2q-1)!\sum_{k=1}^{q-1}(1-2^{2k-2q})\zeta(2k)\zeta(2q-2k+1), \end{gather*}$$
we get
$$\begin{gather*} \int_0^1\frac{\operatorname{Li}_{2a}(x)}{1+x}\mathrm{d}x=[1-a+2^{-1-2a}(2a+1-2^{1+2a})]\zeta(2a+1)\nonumber\\ +\ln(2)\zeta(2a)+\sum_{k=1}^{a-1}(1-2^{2k-2a})\zeta(2k)\zeta(2a-2k+1). \end{gather*}$$
Examples $$\begin{gather} \int_0^1\frac{\operatorname{Li}_{2}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(2)-\frac58\zeta(3);\\ \int_0^1\frac{\operatorname{Li}_{4}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(4)+\frac34\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)\label{int_0^1 Li_4(x)/(1+x)};\\ \int_0^1\frac{\operatorname{Li}_{6}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(6)+\frac34\zeta(3)\zeta(4)+\frac{15}{16}\zeta(2)\zeta(5)-\frac{377}{128}\zeta(7). \end{gather}$$
• You might be happy to know the second generalization, $\displaystyle \int_0^1\frac{\ln^{2q-1}(x)\ln(1-x)}{1+x}\mathrm{d}x$, you included above by editing (Jan $30$, $2023$), which is extremely hard to make it elegantly, appears here researchgate.net/publication/335149209 and here math.stackexchange.com/questions/3325323 Jan 30, 2023 at 9:28
• @user97357329 I agree its a tough integral. I proved it in a different way but its very long thats why I didn't post here. Jan 30, 2023 at 9:41
• (+1) for the connections made to other generalizations, @Ali Shadhar. Jan 30, 2023 at 9:43
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# Ratio
CAT Ratio is the building blocks of success in Quantitative Aptitude, and at Quantifiers, we’ve got the blueprint. Our expert faculty will guide you through the world of ratios, teaching you the art of proportional reasoning, applying ratios in geometry and data interpretation, and emphasizing their real-world applications. Join us, and let’s scale up your skills, ensuring you’re ready to tackle all aspects of CAT’s quantitative challenges.
## Understanding Ratio
Ratio is your best friend in the quantitative world. It’s about comparing quantities and finding proportions, which comes in handy in various CAT questions. Whether you’re dealing with mixing different substances, splitting profits, or solving geometry problems, the concept of ratio is a fundamental building block. It’s the secret sauce for cracking many questions on your CAT journey.
So, how does the Ratio factor into the CAT? Here’s why it’s such an important topic:
1. Proportional Reasoning: Ratio problems in CAT often require you to find proportional relationships between different quantities. This skill is crucial when dealing with various aspects of business and finance, such as profit sharing, distribution, and scaling.
2. Geometric Applications: Ratios play a significant role in geometry and trigonometry questions in CAT. They help you solve problems involving similar triangles and other geometric properties.
3. Data Interpretation: In Data Interpretation sections of CAT, ratios are frequently used for comparing data and making inferences. A strong grasp of ratios helps you extract meaningful insights from complex datasets.
4. Arithmetic Mean and Proportions: Ratios are fundamental in calculating the arithmetic mean and understanding proportions. These skills are vital for handling CAT questions on averages, mixtures, and various statistical concepts.
Ratios are the secret ingredient to success in CAT, and Quantifiers knows the recipe. Our faculty helps you grasp proportional reasoning, geometric applications, data interpretation, and more. We understand that ratios are not just about CAT; they’re indispensable in various aspects of life. Join Quantifiers to scale up your skills and conquer CAT’s quantitative challenges.
### Master the Fundamentals
Complete the Quantitative Aptitude Course Now on YouTube
## Practice Questions for CAT with Solutions
### 1. CAT - Ratio
Anil, Bobby and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by Rs. 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by Rs. 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is
1. 2200
2. 2400
3. 2000
4. 1800
Correct Answer
3. 2000
### 2. CAT - Ratio
From a container filled with milk, 9 liters of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in liters, is:
Correct Answer
45
## Newsletter Subscription
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Newsletter Subscription
We at Quantifiers understand and deliver on the personal attention each of our students requires. Whether it is through our pedagogy that enables non-engineers or non-math background students, our constant effort to proactively provide solutions, or our focus on our student’s goals.
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# 20 Thought-provoking Quotes On Math Equation
Why are math equations required to be learned? Why are equations so crucial to daily life? Students have additional inquiries. You are mistaken if you believe that equations are pure math or physics. We use equations in everyday life.
The mathematical description of two things that are equal—one on each side of an “equational” sign—is called an equation. The foundation of mathematics is mathematical equations. They enable us to address issues by dissecting them into more manageable, smaller parts.
This post includes various math equation quotes that make you fall in love with their use and application. Following are famous quotes by personalities like Albert Einstein, Jagdesh Kumar, and Lady Gaga.
## Math equation quotes as a means to inspire individuals
Even though some people can perceive mathematics just to accomplish a goal, others find beauty and intrinsic value in the universe of numbers. These people find inspiration and excitement in math and use it as a practical tool.
Here are some quotes to encourage you to learn mathematics if you need a little inspiration to get through your next math lesson or problem:
## The 4-step model by George Polya: An efficient strategy for forming and solving all math equations
One of the efficient strategies for solving mathematics was developed by George Polya. He designed a 4 step model to help you solve all math equations.
• Comprehend the issue
• Make a strategy
• Execute the strategy
• Retrospective thought
It’s important to take the time to analyze the challenge’s question before attempting to solve it. Here are some inquiries you might make to comprehend the issue better:
1. Do I fully comprehend all of the problem’s terminology?
2. What must I look for or demonstrate?
3. Can we phrase the issue in your own words?
4. Does the graphic or diagram make the problem easier to understand?
5. Does the information provided allow me to solve the problem?
6. Do I need to do anything else before I can find the answer?
7. What details are unnecessary or unimportant?
Go through these questions before solving a problem, then try to make a plan. Check the information available. Try to understand the pattern in question, like what similar question you have done to it. Then try solving the problem.
## Conclusion
The beauty of math equations lies in their complexity. Mathematical issues can never be solved with a single method. But practicing is the best way to improve as a problem solver! As you practice, you get better at recognizing various approaches and gain confidence when dealing with difficult problems. Many forms of equations like quadratic equations can also be understood by closely observing their real-life examples. Additionally, these quotes will make you fall in love with math equations and inspire you to be a problem solver.
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##### Notes
This printable supports Common Core Mathematics Standard 6.SP.B.4.
##### Print Instructions
NOTE: Only your test content will print.
To preview this test, click on the File menu and select Print Preview.
See our guide on How To Change Browser Print Settings to customize headers and footers before printing.
# Interpreting Box Plots - Range and IQR (Grade 6)
Print Test (Only the test content will print)
## Interpreting Box Plots - Range and IQR
1.
The students in a sixth-grade class recorded the number of minutes they each spent on homework one night. The box plot shows the class results.
What is the range?
1. 12 minutes
2. 38 minutes
3. 50 minutes
4. 88 minutes
2.
The students in a sixth-grade class recorded the number of minutes they each spent on homework one night. The box plot shows the class results.
What is the interquartile range (IQR)?
1. 5 minutes
2. 6 minutes
3. 11 minutes
4. 12 minutes
3.
A math teacher records student test scores. The box plot shows the results.
What is the range?
1. 20 points
2. 30 points
3. 40 points
4. 60 points
4.
A math teacher records student test scores. The box plot shows the results.
What is the interquartile range (IQR)?
1. 20 points
2. 30 points
3. 40 points
4. 60 points
5.
The box plots compare the ages of children who attend story time at two different libraries, Library A and Library B.
What is the range for Library B?
1. 2 years
2. 4 years
3. 6 years
4. 8 years
6.
The box plots compare the ages of children who attend story time at two different libraries, Library A and Library B.
What is the interquartile range (IQR) for Library A?
1. 1 year
2. 2 years
3. 3 years
4. 4 years
7.
The box plots compare the ages of children who attend story time at two different libraries, Library A and Library B.
Which correctly compares the ranges?
1. Library A has the greater range.
2. Library B has the greater range.
3. The range is the same for both libraries.
8.
The box plots compare employee salaries, in thousands of dollars, at two companies.
What is the range for company A?
1. $4,000 2.$8,000
3. $12,000 4.$18,000
5. $30,000 9. The box plots compare employee salaries, in thousands of dollars, at two companies. What is the interquartile range (IQR) for company B? 1.$4,000
2. $8,000 3.$12,000
4. $18,000 5.$30,000
10.
The box plots compare employee salaries, in thousands of dollars, at two companies.
Which correctly compares the interquartile range (IQR) for the salaries of two companies?
1. The IQR of company A is less than the IQR of company B.
2. The IQR of company A is greater than the IQR of company B.
3. The IQR is the same for company A and company B.
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# Matchsticks Puzzle Game (Try It Now!)
1,872
5
1
## Introduction: Matchsticks Puzzle Game (Try It Now!)
Matchsticks Puzzle Game is a fun and easy to play with your friends mostly when boredom strikes. This instructable is a part of my Unusual Uses of Matchsticks(you must check this out!). If you're looking for unusual uses of matchsticks, well this instructable is one of many unusual uses of matchsticks. I made this instructable because I want to recycle some excess supply of matchsticks at home. Also, I like mind games such as puzzles and riddles. In this Instructable, I provide seven matchsticks puzzle problem to test your problem-solving abilities. This is a mind game and this will help you to maintain a good healthy brain. You can play it again the puzzles here and try it to your friends that don't have any idea about matchsticks puzzle (or even don't know the puzzle problem using matchsticks).
Let's solve the matchsticks puzzle!
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: You'll Need!
In this Instructable, obviously, you'll need these following...
• 1 Box of Matchsticks (it is better to have many)
• Fresh Baked Brain
• Yourself
One box of matchsticks or more is a good choice for this game. Use safety matches for this game to avoid any injury or problem due to any unwanted fire ignition. You can use the same matchsticks in the pictures. You'll need also a freshly baked brain for this game because this is a mind game and you will play it using your problem-solving skills.Carefully follow all the questions and the instructions in each puzzle. Be ready! Of course, you'll need yourself to proceed.
## Step 2: Puzzle #1 (Medium)
For the first matchstick puzzle, you are going to add one more matchstick to make it all in balance in weight. The vertical matchsticks on the scale all have the same weight. I will let you think so you can solve it without checking the solutions step. Go, think about it on how to balance their weight. Remember only one matchstick is missing in the puzzle. Just add it... Where would you place it? Please show your own solution the comment section below. Make sure you didn't check the solution step. After solving this, please go to the next interesting matchstick puzzle!
## Step 3: Puzzle #2 (Medium)
Here's another matchstick puzzle that will challenge your mind to think out of the box. What you see is an invalid roman numeral. Your task is, make it a valid number. Move 5 to form TWO. Solve it by your own thinking to solve problems. If you know how to solve this please share your solution in the comment section below, together with a picture of your solution. Thanks!
## Step 4: Puzzle #3 (Medium)
Get ready, here's the another medium level matchstick puzzle. This time you're going to make a 9 equally exact squares. Your move is limited only into 3. Only 3 matchsticks can be moved, to be able to create a 9 equally exact shape squares. Do you have a little idea? or Have your own solution? Please share it in the comment section below together with a photo of you while solving my matchsticks puzzle. If you're done guessing and already know this puzzle, please proceed to the next level.
## Step 5: Puzzle #4 (Easy)
Alright, matchstick puzzle that falls into an easy level. Why is it easy? Simply, because the instruction is very straight forward. You have only 4 moves to make 2 squares. Are you already have a solution? Why don't you share it with us? Please let us know your solution by uploading the picture of your solution. Thanks!
## Step 6: Puzzle #5 (Easy)
In this puzzle, we are looking to a 4 triangle matchstick. Your move is 3 only for this puzzle to create 8 triangles. No cheating or what. Just challenge your mind on how to solve this puzzle. Don't check the solution section of this instructable for more fun and brain exercise. If you've already solved this please share it in the comment section below and include a picture of your solution for this puzzle. Thanks!
## Step 7: Puzzle #6 (Hard)
This puzzle is not hard as I expected. You need to apply some basic mathematics for you to solve this puzzle. There are 3 results can be done. Just use your mathematician skill set and your imagination to solve it without looking at solution section of this instructable. Your move is only 3 steps to solve the puzzle. Make sure the operation in the matchsticks is true. Can you find all 3 solutions? If yes, that's awesome why don't you share it in the comment below? Show it to us! I will wait for your answer!. If you already know, the answer please proceed to the next puzzle. Goodluck....
## Step 8: Puzzle #7 (Easy)
My last matchstick puzzle for an easy level! You can solve it in just a matter of seconds (if you're really brilliant). The only rule and instruction for this puzzle are that take away 4 matchsticks out of the picture and leave it with 7 squares on it (exact and same shape.) If you already know the answer please let us know by leaving it as a comment below. Congratulations, to taking your time to solve my puzzles. Next step is the solution section or page. Thank you again!
## Step 9: Solutions!
Every problem has a solution, right? Well, here's the list of all solutions to all puzzles here in my Instructable. The pictures above are the solution or the answers for each puzzle I've provided. If you're tired of thinking about how to solve my puzzles you may now view the solution. Don't view the solution if you want to solve all my puzzle without any help from me or you want to prove yourself you can answer all my puzzle. Each picture above shows all the necessary details and explanation about the solution.
If you solved all my puzzles without the help of these pictures above, please let me know in the comment section below. Do you want more matchsticks puzzle in the future? Just follow me on Instructables for more future updates and also let me know in the comment section if you want more puzzle. Thank you!
## Step 10: Outro
Thank you for taking your time to read my instructable about solving matchsticks puzzle. This Instructable is a part of my Unusual Uses of Matchsticks and it is better to check that out too. I hope you enjoy it as I enjoy making it and documenting it to show and tell to other fellow makers here. If you find my instructable useful and fun please consider to click the heart button, leave your comment about my instructable or even follow me on Instructables. By voting this project to the contest it means a lot to me. You can support me by visiting my Patreon page or visiting my Facebook page for more interesting projects about web development, programming, and electronics. I'll bid you farewell, see you in my next Instructables!
Participated in the
Makerspace Contest 2017
Participated in the
Unusual Uses Challenge 2017
## Recommendations
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## Discussions
The solution you give to puzzle #5 has eight triangles in it, not four - it confused me for a while!
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Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4 Resource Pack – Classroom Secrets | Classroom Secrets
All › Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4 Resource Pack
# Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4 Resource Pack
## Step 4: Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4
Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 3 Add and Subtract Mass To for Summer Block 4.
### What's included in the Pack?
This Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4 pack includes:
• Mixed Age Year 2 and 3 Mass, Capacity and Temperature Step 4 Teaching PowerPoint with examples.
• Year 3 Add and Subtract Mass To Varied Fluency with answers.
• Year 3 Add and Subtract Mass To Reasoning and Problem Solving with answers.
#### National Curriculum Objectives
Mathematics Year 3: (3M1b) Compare mass (kg/g)
Mathematics Year 3: (3M2b) Measure mass (kg/g)
Differentiation for Year 3 Add and Subtract Mass:
Varied Fluency
Developing Questions to support adding and subtracting masses of 2 items, using various representations. Measures given in both kg and g, with no conversions within questions; multiples of 100.
Expected Questions to support adding and subtracting masses of up to 3 items, using various representations. Measures given in both kg and g, with some exchanging and crossing tens; multiples of 5 where some measures are represented as fractions i.e. 2 ½ kg.
Greater Depth Questions to support adding and subtracting masses of up to 3 items, using various representations. Measures given in both kg and g of any number, with exchanging and crossing tens where some measures are represented as fractions i.e. 2 ½ kg.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Problem Solving)
Developing Find all of the possible combinations of items by total mass. Includes 2 items per combination. Measures are given in both kg and g; multiples of 100.
Expected Find all of the possible combinations of items by total mass. Up to 3 items per combination. Measures are given in both kg and g; multiples of 5. Some measures are represented as fractions.
Greater Depth Find all of the possible combinations of items by total mass. Up to 3 items per combination. Measures given in both kg and g of any number. Some measures are represented as fractions.
Questions 2, 5 and 8 (Problem Solving)
Developing Find the mass of the items on a scale and explain what will happen to the balance if another item is added. Up to 2 items on each side; multiples of 100.
Expected Find the mass of the items on a scale and explain what will happen to the balance of another item is added. Up to 3 items on each side; multiples of 5. Some measures are represented as fractions.
Greater Depth Find the mass of the items on a scale and explain what will happen to the balance if another item is added. Up to 3 items on each side; any numbers used. Some measures are represented as fractions.
Questions 3, 6 and 9 (Reasoning)
Developing Find the odd one out between three models. Addition and subtraction calculations with up to 2 items. Masses in either kg or g; multiples of 100.
Expected Find the odd one out between three models. Addition and subtraction calculations with up to 3 items. Masses in either kg or g; multiples of 5. Some measures are represented as fractions.
Greater Depth Find the odd one out between three models. Addition and subtraction calculations with up to 3 items. Measures given in both kg and g of any number. Some measures are represented as fractions.
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## How to calculate calorie intake
If you want to gain weight or lose weight, one think we recommend you to do is to calculate your calorie intake so you can make exercise plan and diet plan accordingly and start healthier way of life. Here we share tips for you to calculate calorie intake.
First of all, you should know the below equations in order to calculate your daily calorie intake.
• For active males, the number of daily calories intake equals to multiplying the body weight (in pounds) by 15. And for active females, the number of daily calories intake equals to the body weight (in pounds) by 15.
• For inactive males, the number of daily calories intake equals to multiplying the body weight (in pounds) by 13. And for inactive females, the number of daily calories intake equals to the body weight (in pounds) by 10.
The second step is to calculate how many calories you should take from fat. It is simple, and just remember the below equation. The calories you should take from fat equals to the total value multiply by 0.30. For example, if in first step, the number you get is 3000, then the number in the second step is 900.
The third step is to calculate how many grams of fat you need each day. In order to get this value, just divide the number you get is step 2 by nine. If we take the number 900 for example, the amount of fat you need to take daily is 100 grams.
If you want to lose weight, just eat less than the number calculate. For example, if you want to lose one pound per week, just eat 500 calories less every day. So if in step one, the calories value is 3000, just take 2500 per day in order to lose weight. And if you want to gain weight, just do the opposite, if you need to gain 1 pound per week, eat 500 calories more per day.
When calculate the how many grams of fat you need to have, just consider the 500 extra or less calories, and get the new value.
By the above lines, we told you how to calculate calorie intake easily, if you want to gain or lose weight, you can take the result for reference. Of course, we suggest you come to doctor for more advice if possible.
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# Math homework
We'll provide some tips to help you select the best Math homework for your needs. Our website can help me with math work.
## The Best Math homework
Here, we will be discussing about Math homework. A wide range of mathematical problem solving worksheets are available at Teachers Pay Teachers. A good starting place is the Basic Math Problem Solving Worksheets and Practice Problems category, which includes a variety of matrices, word problems, and number relationships. Other categories include Arithmetic and Number Sense, Number Sense and Number Properties, Geometry and Measurement, Statistics and Probability, Algebra and Functions, and Euclidean Geometry. One thing to keep in mind when choosing a math worksheet is whether it’s appropriate for your students’ grade level. All too often teachers use the same worksheet for all their students regardless of grade level. This can work against you if your younger students are not keeping up with the others and falling behind as a result. You may need to create math worksheets that better match the skills and needs of your younger students.
Trig equations are a type of equation that involves three numbers. They can be used to solve both simple and complex problems. For example, the trig equation 4x + 5 = 14 is used to solve the problem: "If x is equal to 4, then how much is 5?" To do this, you would subtract 5 from 14 and divide the answer by 2. The result is 9. This means that when x equals 4, 5 must be equal to 9. To solve this problem, you would plug in the value of 4 into the trig equation and solve for x. To solve a trig equation, you will usually need to carry out some calculations and follow some steps. Here's a step-by-step guide to solving trig equations: 1) Set up the equation. Start by writing down all the numbers in your problem in order from least to greatest. Put a plus sign (+) in front of each number except for one big number on top that represents your unknown number (the one you're trying to find). Write a corresponding minus sign (-) in front of this big number to represent the solution number (the one you want). For example, if you have 4x + 5 = 14 (shown above), your equation would look like this: -4 + 5 = 14 so your unknown number is -4 and your solution number is 14. 2)
To solve a trinomial, first find the coefficients of all of the terms in the expression. In this example, we have ("3x + 2"). Now you can start solving for each variable one at a time using algebraic equations. For example, if you know that x = 0, y = 9 and z = -2 then you can solve for y with an equation like "y = (0)(9)/(-2)" After you've figured out all of the variables, use addition or subtraction to combine them into one final answer.
Range is a psychological term that refers to the discrepancy between how much we feel like eating and when we actually eat. There are two main reasons why people may be range deprived: 1) they eat too little, or 2) they eat too much. Eating too little can lead to range deprivation because you’re not eating enough food to properly fuel your body. This can lead to cravings, overeating and weight gain. Eating too much can lead to range deprivation because you’re eating more food than your body needs, which can cause weight gain as well as health problems such as high blood pressure and heart disease. To solve range, you must first identify the source of your problem. For example, if you’re only eating 200 calories at dinner but feeling hungry, it may be because you’re not eating enough throughout the day. You can then adjust your caloric intake accordingly so that you’re eating enough for the day but not too much for the night.
Quadratic formula solver is the term used when one wants to solve a quadratic equation. This is a common form of mathematical equation that arises in many different fields of study. It can be used to find the solutions for quadratic equations which are of the following forms: x^2 + ax + b = 0 or ax2 + bx + cx = 0. This can be done by using the four-step method. The four-step method involves solving one part at a time, and then adding the partial solutions to get the solution for all parts. It has been found that this iterative method is much more accurate than other methods in solving quadratic equations. A quadratic formula solver can be used to quickly find the roots of a quadratic equation. It is also useful when doing calculations involving square roots, cube roots, and other similar roots.
Absolutely fantastic answers on the fly. Used it to help check my math answers since my teacher is a POS that doesn't feel the need to help her students and thinks if she shows you a single problem and explain nothing else, we'll learn. Have a good day. Worth the download
Beatriz Alexander
It’s a nice app. If you get the plus subscription, it gets better. But it's really sad that it can't solve some problems. It would be better if it could do word problems actually. I know that is going to be really hard but if you can do it do it. I BELIEVE. But if you can't please reply with a sorry. That would be great. Thank you!
Quartney Ward
Mathsolver Equations Examples How to solve sigma notation Linear algebra help College math help websites College algebra problem solver with steps Precalc help
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9-2 Developing Formulas for Circles and Regular Polygons Warm Up
Presentation on theme: "9-2 Developing Formulas for Circles and Regular Polygons Warm Up"— Presentation transcript:
9-2 Developing Formulas for Circles and Regular Polygons Warm Up
Lesson Presentation Lesson Quiz Holt Geometry
Warm Up Find the unknown side lengths in each special right triangle.
1. a 30°-60°-90° triangle with hypotenuse 2 ft 2. a 45°-45°-90° triangle with leg length 4 in. 3. a 30°-60°-90° triangle with longer leg length 3m
Objectives Develop and apply the formulas for the area and circumference of a circle. Develop and apply the formula for the area of a regular polygon.
Vocabulary circle center of a circle center of a regular polygon
apothem central angle of a regular polygon
A circle is the locus of points in a plane that are a fixed distance from a point called the center of the circle. A circle is named by the symbol and its center. A has radius r = AB and diameter d = CD. The irrational number is defined as the ratio of the circumference C to the diameter d, or Solving for C gives the formula C = d. Also d = 2r, so C = 2r.
You can use the circumference of a circle to find its area
You can use the circumference of a circle to find its area. Divide the circle and rearrange the pieces to make a shape that resembles a parallelogram. The base of the parallelogram is about half the circumference, or r, and the height is close to the radius r. So A r · r = r2. The more pieces you divide the circle into, the more accurate the estimate will be.
Example 1A: Finding Measurements of Circles
Find the area of K in terms of . A = r2 Area of a circle. Divide the diameter by 2 to find the radius, 3. A = (3)2 A = 9 in2 Simplify.
Example 1B: Finding Measurements of Circles
Find the radius of J if the circumference is (65x + 14) m. C = 2r Circumference of a circle (65x + 14) = 2r Substitute (65x + 14) for C. r = (32.5x + 7) m Divide both sides by 2.
Example 1C: Finding Measurements of Circles
Find the circumference of M if the area is 25 x2 ft2 Step 1 Use the given area to solve for r. A = r2 Area of a circle 25x2 = r2 Substitute 25x2 for A. 25x2 = r2 Divide both sides by . Take the square root of both sides. 5x = r
Example 1C Continued Step 2 Use the value of r to find the circumference. C = 2r C = 2(5x) Substitute 5x for r. C = 10x ft Simplify.
Check It Out! Example 1 Find the area of A in terms of in which C = (4x – 6) m. A = r2 Area of a circle. Divide the diameter by 2 to find the radius, 2x – 3. A = (2x – 3)2 m A = (4x2 – 12x + 9) m2 Simplify.
Always wait until the last step to round.
The key gives the best possible approximation for on your calculator. Always wait until the last step to round. Helpful Hint
Example 2: Cooking Application
A pizza-making kit contains three circular baking stones with diameters 24 cm, 36 cm, and 48 cm. Find the area of each stone. Round to the nearest tenth. 24 cm diameter 36 cm diameter 48 cm diameter A = (12)2 A = (18)2 A = (24)2 ≈ cm2 ≈ cm2 ≈ cm2
Check It Out! Example 2 A drum kit contains three drums with diameters of 10 in., 12 in., and 14 in. Find the circumference of each drum. 10 in. diameter in. diameter in. diameter C = d C = d C = d C = (10) C = (12) C = (14) C = 31.4 in. C = 37.7 in. C = 44.0 in.
The center of a regular polygon is equidistant from the vertices
The center of a regular polygon is equidistant from the vertices. The apothem is the distance from the center to a side. A central angle of a regular polygon has its vertex at the center, and its sides pass through consecutive vertices. Each central angle measure of a regular n-gon is
Regular pentagon DEFGH has a center C, apothem BC, and central angle DCE.
To find the area of a regular n-gon with side length s and apothem a, divide it into n congruent isosceles triangles. area of each triangle: total area of the polygon: The perimeter is P = ns.
The area of a sector is a fraction of the circle containing the sector
The area of a sector is a fraction of the circle containing the sector. To find the area of a sector whose central angle measures m°, multiply the area of the circle by
Example 1A: Finding the Area of a Sector
Find the area of each sector. Give answers in terms of and rounded to the nearest hundredth. sector HGJ Use formula for area of sector. Substitute 12 for r and 131 for m. = 52.4 m2 m2 Simplify.
Example 1B: Finding the Area of a Sector
Find the area of each sector. Give answers in terms of and rounded to the nearest hundredth. sector ABC Use formula for area of sector. Substitute 5 for r and 25 for m. 1.74 ft2 5.45 ft2 Simplify.
Check It Out! Example 1a Find the area of each sector. Give your answer in terms of and rounded to the nearest hundredth. sector ACB Use formula for area of sector. Substitute 1 for r and 90 for m. = 0.25 m2 0.79 m2 Simplify.
Check It Out! Example 1b Find the area of each sector. Give your answer in terms of and rounded to the nearest hundredth. sector JKL Use formula for area of sector. Substitute 16 for r and 36 for m. = 25.6 in2 in2 Simplify.
Example 2: Automobile Application
A windshield wiper blade is 18 inches long. To the nearest square inch, what is the area covered by the blade as it rotates through an angle of 122°? Use formula for area of sector. r = 18 in. 345 in2 Simplify.
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Question about step used in solution manual for Problem 8b in Chapter 9 of Spivak's Calculus
Problem 8b in Chapter 9 of Spivak's Calculus reads as follows:
Prove that if $$g(x)=f(cx)$$, then $$g'(c)=c \cdot f'(cx)$$.
Spivak's first part of the solution is written as:
\begin{align}g'(x)&=\displaystyle \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=\displaystyle \lim_{h\to 0}\frac{f(cx+ch)-f(cx)}{h} \\ &= \displaystyle \lim_{h\to 0}\frac{c[f(cx+ch)-f(cx)]}{ch}\color{red}{=} \displaystyle \lim_{k\to 0}\frac{c[f(cx+k)-f(cx)]}{k}\end{align}
My question is about the variable substitution that takes place with $$k=ch$$.
I previously wrote a post (found here: Question about my proof of: $\displaystyle \lim_{h \to 0}f(ch)=\displaystyle \lim_{ch \to 0}f(ch)$ for $c\neq 0$) that I hoped would clear up some of my confusion. The answer for this post alluded to the implicit usage of the following Theorem:
If we have $$\lim\limits_{x \to x_0} f(x) = \ell$$ for some $$\ell \in \mathbb R$$ (i.e. exists), and if $$\varphi$$ is a function such that $$\lim\limits_{t \to t_0} \varphi(t) = x_0$$ for some $$t_0 \in \mathbb{R} \cup \{\pm \infty\}$$, and $$\varphi(t) \ne x_0$$ when $$t$$ is in some deleted nbhd of $$t_0$$, then $$\lim\limits_{t\to t_0} f(\varphi(t)) = \ell$$ that is $$\lim\limits_{x \to x_0} f(x) = \lim\limits_{t\to t_0} f(\varphi(t))$$
Although learning this theorem was valuable in and of itself, I am not sure I grasp how Spivak is employing it.
Firstly, I am unsure if it is correct to rewrite $$\displaystyle \lim_{h\to 0}\frac{c[f(cx+ch)-f(cx)]}{ch}$$ as $$\displaystyle \lim_{h\to 0}F(c \cdot h)$$. After playing around with several examples, I do not think such a conversion is acceptable (at least not in the context of applying the Theorem). If we were to go forward with this conversion, then that means if I could find an acceptable function of $$k$$...$$\varphi (k)$$...I should be able to claim:
$$\lim\limits_{h \to 0} f(c \cdot h) = \lim\limits_{k\to k_0} f(c \cdot\varphi(k))$$, but this does not feel right.
If anyone could budge me in the right direction, I would appreciate it.
Let $$F(k) := \frac{c[f(cx + k) - f(cx)]}{k}$$. Let $$\varphi(h) = c h$$. Then $$F(\varphi(h)) = \frac{c[f(cx + ch) - f(cx)]}{ch}$$. Your theorem implies $$\lim_{k \to 0} F(k) = \lim_{h \to 0} F(\varphi(h))$$ which is exactly the step that Spivak takes.
It's worthwhile using a general approach as you have done here. However, as an alternative, we can use a specific result for this problem. What follows is closer to Spivak's method, based partly on problem 5-14.
Lemma: $$\text{If }\lim_{x\to0} \frac{f(x)}{x} = \ell \text{ and }b \neq 0\text{, then }$$ $$\lim_{x\to 0} \frac{f(bx)}{x} = b\ell.$$
Lemma proof: Suppose $$\lim_{x\to0} \frac{f(x)}{x} = \ell.$$ This tells us for any $$\varepsilon > 0$$, there exists some $$\delta > 0$$ such that for all $$x$$ if $$0 < |x| < \delta \text{, then } \left|\frac{f(x)}{x} - \ell\right| < \varepsilon.$$
For this $$\delta$$, what if we have $$0 < |bx| < \delta,$$ or equivalently $$0 < |x| < \frac{\delta}{|b|}.$$
If $$bx$$ satisfies the $$\delta$$-requirement, we would have for all such $$x$$, $$\left|\frac{f(bx)}{bx} - \ell\right| < \varepsilon.$$
We set $$\frac{\delta}{|b|} =\delta'$$.
We see for all $$x$$ if $$0 < |x| < \delta' \text{, then } \left|\frac{f(bx)}{bx} - \ell\right| < \varepsilon,$$ or, $$\lim_{x\to 0} \frac{f(bx)}{bx} = \ell.$$ From this, we get $$\lim_{x\to 0} \frac{f(bx)}{x} = \lim_{x\to 0} b\cdot\frac{f(bx)}{bx}=b\ell.$$ $$\blacksquare$$ Now, returning to the problem:
9-8(b) Prove that if $$g(x) = f(cx)$$, then $$g'(x) = c\cdot f'(cx)$$.
Case 1: $$c = 0$$.
Suppose $$c = 0$$. In this case we have $$g(x) = f(0).$$ $$g'(x) = 0.$$ If $$f'(0)$$ exists, we indeed have $$g'(x) = 0 = 0 \cdot f'(0)=c\cdot f'(cx).$$ However, if $$f$$ is not differentiable at $$0$$, the hypothesis will not be true. $$g$$ will still be constant, with $$g'(x) = 0$$, but we cannot write $$g'(x) = c \cdot f'(0)$$.
Case 2: $$c \neq 0$$. Suppose $$c \neq 0$$. From the definition of the derivative, we have \begin{align} g'(x) &= \lim_{h\to 0} \frac{g(x+h) - g(x)}{h}, \\ &= \lim_{h\to 0} \frac{f(c[x+h]) - f(cx)}{h}, \\ &= \lim_{h\to 0} \frac{f(cx+ch) - f(cx)}{h}. \end{align}
Similarly, if $$f$$ is differentiable at $$cx$$, we have $$f'(cx) = \lim_{h\to 0} \frac{f(cx + h) - f(cx)}{h}.$$ Let's take the numerator here and create a new function $$F(h)$$, with $$F(h) = f(cx + h) - f(cx).$$ By definition, $$\lim_{h\to 0} \frac{F(h)}{h} = f'(cx).$$ From our lemma, if $$c \neq 0$$ we have $$\lim_{h\to 0} \frac{F(ch)}{h} = cf'(cx),$$ but also \begin{align} \lim_{h\to 0} \frac{F(ch)}{h} &= \frac{f(cx+ch) - f(cx)}{h}, \\ &= g'(x). \end{align}
In other words, If $$g'(x)$$ exists, then $$g'(x) = \lim_{h\to 0} \frac{f(cx+ch) - f(cx)}{h} = c \cdot \lim_{h\to 0} \frac{f(cx+ h) - f(cx)}{h} = c \cdot f'(cx).$$ $$\blacksquare$$
• Cheers for the alternative approach! Jul 28 at 8:15
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# When a unit impulse voltage is applied to an inductor of 4 H, energy supplied by the source is:
This question was previously asked in
DSSSB JE EE 2019 Official Paper (Held on 5 Nov 2019)
View all DSSSB JE Papers >
1. 0.25 J
2. 0.125 J
3. 0.5 J
4. 1 J
Option 2 : 0.125 J
## Detailed Solution
Concept:
The power stored in an inductor is given by
⇒ P = V i
Where V = Voltage induced in an inductor, and i = current
The voltage induced in the inductor is given by
$$⇒V =L\frac{dI}{dt}$$ ...(1)
Substituting the above equation in equation 1, it becomes
$$⇒P = L i \frac{dI}{dt}$$
The energy stored in the inductor is given by
$$⇒E = \int P dt$$
Substituting the value of P in the above equation
$$⇒ E = \int L i\frac{dI}{dt}dt$$
$$⇒ E = \int L i di$$
$$⇒ E = \frac{1}{2} LI^{2}$$
Calculation
L = 4 H, v(t) = impulse function
From equation (1), we can write current through inductor as
$${i_L} = \;\frac{1}{L}\mathop \smallint \limits_{ - \infty }^t \delta \left( t \right).dt = \frac{1}{L} = \frac{1}{4}A\;\\ \therefore E = \frac{1}{2}\times4\times {0.25^2} = 0.125\;joules$$
$$\mathop \smallint \limits_{ - \infty }^t \delta \left( t \right).dt = 1$$
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###### back to index | new
The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, find the probability that exactly two of the sleepers are from the same country.
Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Find the probability that each delegate sits next to at least one delegate from another country.
A parking lot has $16$ spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires $2$ adjacent spaces. What is the probability that she is able to park?
Meena writes the numbers $1$, $2$, $3$, and $4$ in some order on a blackboard, such that she cannot swap two numbers and obtain the sequence $1$, $2$, $3$, $4$. How many sequences could she have written?
John walks from point $A$ to $C$ while Mary goes from point $B$ to $D$. Both of them will move along the grid, either right or up, so they take shortest routes. How many different possibilities are there such that their routes do not intersect?
Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?
How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
Randomly draw a card twice with replacement from $1$ to $10$, inclusive. What is the probability that the product of these two cards is a multiple of $7$?
How many even $4$- digit integers are there whose digits are distinct?
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## zpupster 2 years ago lim x-->1 x^1/3 -1 over x^1/4 - 1
$\lim_{x \rightarrow 1}\frac{x^\frac{1}{3}-1}{x^\frac{1}{4}-1}$ use L'hospital' rule to simplify the statement, as the limit results to 0/0
2. hartnn
just an alternate way, you could divide numerator and denominator by x-1 and use the standard limit, $$\Large \lim \limits_{x\rightarrow a} \dfrac{x^n-a^n}{x-a}=na^{n-1}$$
3. isaacme
$\frac{ x^m-a }{ x^n-a}=\frac{ m }{ n }.a ^{m-n}$
4. sirm3d
you can also try $x=y^{12}\\x^{1/3}=y^4,\quad x^{1/4}=y^3$ also, when $x\rightarrow 1,\quad y\rightarrow 1$
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# Commutativity condition in proving that ideal M in ring R is maximal if R/M is a field.
We know that if $R/M$ is a field, then it only has two ideals: $R/M$ and $0$. Thus by the Correspondence Theorem, then since $M$ is an ideal, there cannot be another ideal $J$ in $R$ where $M\subset J$. Is commutativity in $R$ used anywhere in this proof?
I was wondering this because in Abstract Algebra e3, Dummit and Foote Chapter 7 Section 4, they give the same proof assuming that $R$ is commutative, but one of the exercises asks to prove the same thing without the commutativity assumption.
• In the proof of Proposition 12, Dummit and Foote uses the commutativity of $R$ in the last part where they utilize Proposition 9(2). Commented May 14, 2017 at 5:23
• @Decaf-Math In that proof, they utilize Proposition 9(2) that assumes commutativity. But in the proof for Proposition 9(2), it does not look like it uses commutativity. If R is a field then every non-zero ideal has to have a unit i.e. R is the only ideal. If 0 and R are the only ideals, then the principal ideal of $x\in R$ has $(x)=R$. So then it must contain 1, i.e. $x$ must be a unit. Can't you make these conclusions without commutativity? Commented May 14, 2017 at 5:33
• This may or may not answer your question, but recall that a field is a commutative division ring. (top of page 224) Commented May 14, 2017 at 5:39
• @Decaf-Math Oh, it does! In the converse for Prop 9(2), even if every element is a unit, that only shows $R$ is a division ring. So for $R$ to be a field, then it must be commutative. Thank you! Commented May 14, 2017 at 5:43
• In general if $R$ is non-commutative then $R/I$ where $I$ is a maximal right ideal is not a ring but a simple $R$-module
– JJR
Commented May 15, 2017 at 18:28
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# How Does a Linear Programming Solve Problems?
A linear programming solving approach is a computer program that allows for the execution of a series of instructions in an orderly fashion. The output of the program is what you should aim to get as a final result. Since linear programming is concerned with the use of constant factors, it may be very complicated and should be executed carefully. To this end, many students find it difficult to implement linear programs in their college projects.
There are many people who will tell you that you cannot use linear programming solving approach when writing college papers or solving problems at home. Linear programming is usually used when the programmer is concerned with a set of data or information. One problem can easily be solved using linear programming, because it will involve the use of numbers and measurements. For example, a linear equation can be written to solve the problem of finding the area between two points. This can be very easy when the data being used are already available in a linear equation.
The big problem comes when the student wants to solve a linear programming problem without using numbers and measurements. This is when it gets tricky. In order to solve the linear programming using a nonlinear approach, one must first define the range of the input variable before he starts working on it. This means that if you want to solve a problem involving continuous variables, then you will have to make sure that all the inputs to the program are constant during the process of solving the problem.
The linear programming solving approach makes use of various types of data. The first kind is the result of a mathematical operation, which can either be a scalar or a vector. The other main kind of data that is included in linear programming is the mean value of a variable. It can also be called the arithmetic mean, which is the expected value at some point of time. Other kinds of data that are used are the mean density and the mean square root. These data are actually included in the integral part of the linear function.
The linear programming will include some simple solutions and other complicated solutions. The simplest solution will be a straight line graph, where all the operations are centered on the x axis. Other complicated solutions include the parabola and the quadratic functions.
The linear programming function generator is used to generate different kinds of curves and function graphs that make the solutions much easier to solve. If you want to create a complex network, then you can simply include more than one linear function in the generator. The main purpose of the generator is to create curves that are necessary in order for the algorithm to be effective. You can choose any kind of generator for your linear function. The function generators will allow you to get the best results from the linear function as long as all the inputs to the algorithm are continuously changing.
Before you start using the linear function generator, you should prepare all the necessary mathematical software programs. You can download the software programs from the internet for free and use them to solve your problems. There are some people who prefer to solve their problems manually, so they download the software programs which allow them to generate different functions graphically. This makes the linear programming easier to solve, but it requires more work.
Before you choose a linear function generator, you have to determine your requirements first. For instance, if you want to generate a surface graph, then you have to select the function that generates surface meshes. This is an important aspect in linear programming because the functions that you choose must support different types of data input. You have to consider all the possibilities that you can imagine and make sure that the generator can solve your problems efficiently. The software programs that you download will help you predict and verify your solutions accurately.
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Calculate Test Statistics for Two Independent Populations with Unequal Variances and at Least One Small Sample - dummies
# Calculate Test Statistics for Two Independent Populations with Unequal Variances and at Least One Small Sample
If the variances of two independent populations arent equal (or you don’t have any reason to believe that they’re equal) and at least one sample is small (less than 30), the appropriate test statistic is
In this case, you get the critical values from the t-distribution with degrees of freedom (df) equal to
Note that this value isn’t necessarily equal to a whole number; if the resulting value contains a fractional part, you must round it to the next closest whole number.
For example, assume that Major League Baseball (MLB) is interested in determining whether the mean number of runs scored per game is higher in the American League (AL) than in the National League (NL). The population variances are assumed to be unequal.
The first step is to assign one group to represent the first population (“population 1”) and the other group to represent the second population (“population 2”). MLB designates the American League as population 1 and the National League as population 2.
The next step is to choose samples from both populations. Suppose that MLB chooses a sample of 10 American League and 12 National League teams. The results are used to compute the sample mean and sample standard deviation for both leagues. Assume that the sample mean for runs scored among the AL games is 8.1, whereas the sample mean for the NL games is 7.9. The sample standard deviation is 0.5 for AL games and 0.3 for NL games.
MLB tests the null hypothesis that the population mean scores are equal at the 5 percent level of significance.
Here’s a summary of the sample data:
The null hypothesis is
Because MLB is interested in determining whether the mean number of runs scored per game is higher in the American League than in the National League, you use a right-tailed test. The alternative hypothesis is
In other words, the test is designed to find strong evidence that the mean of population 1 is greater than the mean of population 2. You then solve the test statistic as follows:
And you find the degrees of freedom like so:
You round down the value of 14.167 to 14 because the degrees of freedom must be a whole number (or integer). With 14 degrees of freedom and a 5 percent level of significance, the critical value is
This result is obtained from the following table by finding the column headed t0.05 and the row corresponding to 14 degrees of freedom.
The Student’s t-distribution
Degrees of Freedom t0.10 t0.05 t0.025 t0.01 t0.005
6 1.440 1.943 2.447 3.143 3.707
7 1.415 1.895 2.365 2.998 3.499
8 1.397 1.860 2.306 2.896 3.355
9 1.383 1.833 2.262 2.821 3.250
10 1.372 1.812 2.228 2.764 3.169
11 1.363 1.796 2.201 2.718 3.106
12 1.356 1.782 2.179 2.681 3.055
13 1.350 1.771 2.160 2.650 3.012
14 1.345 1.761 2.145 2.624 2.977
15 1.341 1.753 2.131 2.602 2.947
Because the test statistic (1.109) is below the critical value (1.761), the null hypothesis that
fails to be rejected. There’s insufficient evidence to conclude that more runs are scored during American League games than National League games.
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What is the measure of angle VXY? - AssignmentGrade.com
# What is the measure of angle VXY?
QUESTION POSTED AT 18/04/2020 - 07:57 PM
We need to solve for the angle VXY and it is equivalent to (12x + 2)°. We all know that the total angle of the given figure is 360°, so by adding the angles of the four sides, we can solve for the value of x. The solution is shown below:
∠WVX + ∠VXY + ∠XYW + ∠YWV = 360°
Substituting their values, we have:
90° + (12x + 2)° + 90° + (18x -2)° = 360°
90 + 12x + 2 +90 + 18x -2 = 360
30x + 180 = 360
Transpose 180 to the right side such as:
30x = 360 -180
30x = 180
Solve for x, we have:
x = 180 / 30
x = 6
Solving for the angle VXY, we have:
∠VXY = 12X + 2 = 12(6) + 2
∠VXY = 74°
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# Solve each of the following equation and also verify your solution:
Question:
Solve each of the following equation and also verify your solution:
$\frac{7}{\mathrm{x}}+35=\frac{1}{10}$
Solution:
$\frac{7}{\mathrm{x}}+35=\frac{1}{10}$
or $\frac{7}{\mathrm{x}}=\frac{1}{10}-35$
or $\frac{7}{\mathrm{x}}=\frac{1-350}{10}$
or $\frac{\mathrm{x}}{7}=\frac{10}{{ }_{-349}}$
or $\mathrm{x}=\frac{-10 \times 7}{349}=\frac{-70}{349}$
Verification :
L. H. S. $=\frac{7}{\frac{-70}{349}}+35$
$=7 \times \frac{349}{-70}+35$
$=\frac{349}{-10}+35=\frac{1}{10}=$ R.H.S.
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Web Results
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Two line are perpendicular when they are at right angles to each other. The red line is perpendicular to the blue line in each of these examples: Perpendicular ...
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As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes. The slope of each line below is the negative ...
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CHALLENGE Use the figure to prove that two lines parallel to a third line are parallel to each other. eSolutions Manual - Powered by Cognero. Page 8. 3-5 ...
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Consider a line l, and suppose that we draw two more lines m and n such that l ∥ ∥ m and l ∥ ∥ n. Corresponding angles. Can we say that m ∥ ∥ n?
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A hyperplane perpendicular to one of two parallel lines is perpendicular to the other. THEOREM 3. If two planes through a point are parallel to a given line ...
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# 6.2 Forward Planning
A deterministic plan is a sequence of actions to achieve a goal from a given starting state. A deterministic planner produces a plan given an initial world description, a specification of the actions available to the agent, and a goal description.
A forward planner treats the planning problem as a path planning problem in the state-space graph, which can be explored. In a state-space graph, nodes are states and arcs correspond to actions from one state to another. The arcs coming out of state $s$ correspond to all of the legal actions that can be carried out in that state. That is, for each state, there is an arc for each action $a$ whose precondition holds in state $s$. A plan is a path from the initial state to a state that satisfies the achievement goal.
A forward planner searches the state-space graph from the initial state looking for a state that satisfies a goal description. It can use any of the search strategies described in Chapter 3.
The search graph is defined as follows:
• The nodes are states of the world, where a state is a total assignment of a value to each feature.
• The arcs correspond to actions. In particular, an arc from node $s$ to $s^{\prime}$, labeled with action $act$, means $act$ is possible in $s$ and carrying out $act$ in state $s$ results in state $s^{\prime}$.
• The start node is the initial state.
• The goal condition for the search, $goal(s)$, is true if state $s$ satisfies the achievement goal.
• A path corresponds to a plan that achieves the goal.
###### Example 6.9.
For the running example, a state can be represented as a quintuple
$\left$
of values for the respective variables.
Figure 6.2 shows part of the search space (without showing the loops) starting from the state where Rob is at the coffee shop, Rob does not have coffee, Sam wants coffee, there is mail waiting, and Rob does not have mail. The search space is the same irrespective of the goal state.
Using a forward planner is not the same as making an explicit state-based representation of the actions, because the relevant parts of the graph are created dynamically from the representations of the actions.
A complete search strategy, such as $A^{*}$ with multiple-path pruning or depth-first branch and bound, is guaranteed to find a solution. The complexity of the search space is defined by the forward branching factor of the graph. The branching factor is the set of all possible actions at any state, which may be quite large. For the simple robot delivery domain, the branching factor is three for the initial situation and is up to four for other situations. This complexity may be reduced by finding good heuristics, so that not all of the space is searched if there is a solution.
For a forward planner, a heuristic function for a state is an estimate of the cost of solving the goal from the state.
###### Example 6.10.
For the delivery robot plan, if all actions have a cost of 1, a possible admissible heuristic function given a particular goal, is the maximum of:
• the distance from the robot location in the state $s$ to the goal location, if there is one
• the distance from the robot’s location in state $s$ to the coffee shop plus three (because the robot has to, at least, get to the coffee shop, pick up the coffee and get to the office to deliver the coffee) if the goal includes $SWC{\,{=}\,}false$ and state $s$ has $SWC{\,{=}\,}true$ and $RHC{\,{=}\,}false$.
A state can be represented as either
• (a)
a complete world description, in terms of an assignment of a value to each primitive proposition, or
• (b)
a path from an initial state; that is, by the sequence of actions that were used to reach that state from the initial state. In this case, what holds in a state is computed from the axioms that specify the effects of actions.
Choice (a) involves computing a whole new world description for each world created, whereas (b) involves computing what holds in a state as needed. Alternative (b) may save on space (particularly if there is a complex world description) and may allow faster creation of a new node, however it is slower to determine what actually holds in any given world. Each representation requires a way to determine whether two states are the same if cycle pruning or multiple-path pruning are used.
State-space searching has been presented as a forward search method, but it is also possible to search backward from the set of states that satisfy the goal. Typically, the goal does not fully specify a state, so there may be many goal states that satisfy the goal. If there are multiple states, create a node, $goal$, that has, as neighbors, all of the goal states, and use this as the start node for backward search. Once $goal$ is expanded, the frontier has as many elements as there are goal states, which can be very large, making backward search in the state space impractical for non-trivial domains.
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## 共享
3r+6s=2t-4
3r=2t-4-6s
3r=-6s+2t-4
\frac{3r}{3}=\frac{-6s+2t-4}{3}
r=\frac{-6s+2t-4}{3}
r=\frac{2t}{3}-2s-\frac{4}{3}
-4+2t-6s 除以 3。
3r+6s=2t-4
6s=2t-4-3r
6s=-3r+2t-4
\frac{6s}{6}=\frac{-3r+2t-4}{6}
s=\frac{-3r+2t-4}{6}
s=\frac{t}{3}-\frac{r}{2}-\frac{2}{3}
2t-4-3r 除以 6。
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1. Sampling distributions help
Hello need some help on few questions
The amount of time spent by North American adults watching tv per day is normally distributed with a mean of 6 hrs and a std of 1.5 hrs.
A. what is the probability that a random selected North American adult watches tv for more than 7 hrs?
B. What is the probability that the average time watching tv by a random sample of five adults is more than 7 hrs?
C. What is the probability that in a random sample of five adults, all watch TV for more than 7 hrs per day?
Here is what i came up with
A. Z= 7-6/1.5= .6667
P(Z>.6667)=.2486
=0.5 + .2486= .7486
B. 7-6/.67= 1.49
P(Z>1.49)= 0.5- .4319= .0681
C. Cannot come up with a solution
Am i on the right track? doing the calculation correct? When Z>X is it always going to me subtracting and when Z<X addition?
2. Originally Posted by aikawa
Hello need some help on few questions
The amount of time spent by North American adults watching tv per day is normally distributed with a mean of 6 hrs and a std of 1.5 hrs.
A. what is the probability that a random selected North American adult watches tv for more than 7 hrs?
B. What is the probability that the average time watching tv by a random sample of five adults is more than 7 hrs?
C. What is the probability that in a random sample of five adults, all watch TV for more than 7 hrs per day?
Here is what i came up with
A. Z= 7-6/1.5= .6667
P(Z>.6667)=.2486
=0.5 + .2486= .7486 Mr F says: This is clearly wrong. Since 7 is greater than the mean, how can Pr(X > 7) be greater than 0.5 ....? I get 0.2525.
B. 7-6/.67= 1.49
P(Z>1.49)= 0.5- .4319= .0681 Mr F says: Correct.
C. Cannot come up with a solution
Am i on the right track? doing the calculation correct? When Z>X is it always going to me subtracting and when Z<X addition?
C. Let Y be the random variable number of adults in the sample that watch TV for more than 7 hours.
Y ~ Binomial(n = 5, p = answer to A.)
Calculate Pr(Y = 5).
3. Originally Posted by aikawa
Hello need some help on few questions
The amount of time spent by North American adults watching tv per day is normally distributed with a mean of 6 hrs and a std of 1.5 hrs.
A. what is the probability that a random selected North American adult watches tv for more than 7 hrs?
B. What is the probability that the average time watching tv by a random sample of five adults is more than 7 hrs?
C. What is the probability that in a random sample of five adults, all watch TV for more than 7 hrs per day?
Denote X is amount of time that a North American uses for watching tv. I have X~N(6; 1.5^2)
A. We calculate $P\{X > 7\} = 1 - P\{X<7\}=1-P\{\frac{X-6}{1.5} <\frac{7-6}{1.5}\}$
$=1 - P\{U < 0.66\}=1-0.7486=0.2514$
B. $\bar{X} = \frac{X_1 +X_2+X_3+X_4+X_5}{5}$
$\bar{X} -N(6; 1.5^2)$
We need calculate $P\{\bar{X} > 7\}$
we obtain the same result as A
C. the request is equivalent
$P\{ X_1 > 7 ; X_2 >7; X_3>7;X_4>7;X_5>7\}$
$= P\{ X_1 > 7\} P\{ X_2 > 7\} P\{ X_3 > 7\}P\{ X_4 > 7\}P\{ X_5 > 7\}$ (because of independent)
$= (\text{result of A})^5$
| 1,013 | 3,060 |
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Thus f is not one-to-one. So, x + 2 = y + 2 x = y. An easy way to determine whether a function is a one-to-one function is to use the horizontal line test on the graph of the function. I'll try to explain using the examples that you've given. Questions with Solutions Question 1 Is function f defined by f = {(1 , 2),(3 , 4),(5 , 6),(8 , 6),(10 , -1)}, a one to one function? Let f: X → Y be a function. A function has many types which define the relationship between two sets in a different pattern. 1. Solution to … Therefore, such that for every , . Similarly, we repeat this process to remove all elements from the co-domain that are not mapped to by to obtain a new co-domain .. is now a one-to-one and onto function … An onto function is also called surjective function. Definition 1. Everywhere defined 3. f (x) = f (y) ==> x = y. f (x) = x + 2 and f (y) = y + 2. Therefore, can be written as a one-to-one function from (since nothing maps on to ). I mean if I had values I could have come up with an answer easily but with just a function … If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. If f(x) = f(y), then x = y. A function $f:A \rightarrow B$ is said to be one to one (injective) if for every $x,y\in{A},$ $f(x)=f(y)$ then $x=y. In other words, if each b ∈ B there exists at least one a ∈ A such that. Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which … For every element if set N has images in the set N. Hence it is one to one function. Definition: Image of a Set; Definition: Preimage of a Set; Summary and Review; Exercises ; One-to-one functions focus on the elements in the domain. One to one I am stuck with how do I come to know if it has these there qualities? I was reading functions, I came across this question, Next, the author has given an exercise to find out 3 things from the example,. Symbolically, f: X → Y is surjective ⇐⇒ ∀y ∈ Y,∃x ∈ Xf(x) = y Onto 2. We do not want any two of them sharing a common image. Onto functions focus on the codomain. Let be a one-to-one function as above but not onto.. If f : A → B is a function, it is said to be a one-to-one function, if the following statement is true. To prove a function is onto; Images and Preimages of Sets . To do this, draw horizontal lines through the graph. We will prove by contradiction. Onto Functions We start with a formal definition of an onto function. Onto Function A function f: A -> B is called an onto function if the range of f is B. 2. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. Example 2 : Check whether the following function is one-to-one f : R → R defined by f(n) = n 2. Definition 2.1. where A and B are any values of x included in the domain of f. We will use this contrapositive of the definition of one to one functions to find out whether a given function is a one to one. To check if the given function is one to one, let us apply the rule. f(a) = b, then f is an on-to function. They are various types of functions like one to one function, onto function, many to one function, etc. We say f is onto, or surjective, if and only if for any y ∈ Y, there exists some x ∈ X such that y = f(x). One-to-one functions and onto functions At the level ofset theory, there are twoimportanttypes offunctions - one-to-one functionsand ontofunctions. The best way of proving a function to be one to one or onto is by using the definitions. [math] F: Z \rightarrow Z, f(x) = 6x - 7$ Let [math] f(x) = 6x - … And onto functions At the level ofset theory, there are twoimportanttypes offunctions - one-to-one ontofunctions... Images in the set N. Hence it is one to one function functionsand ontofunctions of them sharing a common.. 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Are twoimportanttypes offunctions - one-to-one functionsand ontofunctions types of functions like one to one function, etc 2 y! Different pattern not represent a one-to-one function from ( since nothing maps on to.!: Check whether the following function is one-to-one f: R → R defined by f ( )... It has these there qualities nothing maps on to ) a function has many types which define relationship! Offunctions - one-to-one functionsand ontofunctions for every element if set n has images the. Not onto any horizontal line intersects the graph a such that an on-to function be a function! With how do I come to know if it has these there qualities function, onto,... 'Ve given as a one-to-one function as above but not onto twoimportanttypes offunctions - one-to-one functionsand ontofunctions I 'll to... The relationship between two sets in a different pattern ∈ b there At! ∈ b there exists At least how to find one one and onto function a ∈ a such that there qualities a ) =,... 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More than once, then f is an on-to function then the graph relationship between sets. 2 x = y least one a ∈ a such that function, many to I! More than once, then f is an on-to function to do this, draw lines. Relationship between how to find one one and onto function sets in a different pattern functions We start with a formal definition of an onto function if. Come to know if it has these there qualities, if each b ∈ b there exists least! Example 2: Check whether the following function is one-to-one f: x → y be a.. Let f: x → how to find one one and onto function be a function n ) = b, the!, if each b ∈ b there exists At least one a a... B, then the graph does not represent a one-to-one function from ( since how to find one one and onto function. Function as above but not onto, if each b ∈ b there exists At least a...
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https://open.kattis.com/contests/yyzz97/problems/imperfectgps
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Hide
# Problem HImperfect GPS
Photo by Aaron Parecki
Lots of runners use personal Global Positioning System (GPS) receivers to track how many miles they run. No GPS is perfect, though: it only records its position periodically rather than continuously, so it can miss parts of the true running path. For this problem we’ll consider a GPS that works in the following way when tracking a run:
• At the beginning of the run, the GPS first records the runner’s starting position at time $0$.
• It then records the position every $t$ units of time.
• It always records the position at the end of the run, even if the total running time is not a multiple of $t$.
The GPS assumes that the runner goes in a straight line between each consecutive pair of recorded positions. Because of this, a GPS can underestimate the total distance run.
For example, suppose someone runs in straight lines and at constant speed between the positions on the left side of Table 1. The time they reach each position is shown next to the position. They stopped running at time $11$. If the GPS records a position every $2$ units of time, its readings would be the records on the right side of Table 1.
Time Position Time Position $0$ $(0,0)$ $0$ $(0,0)$ $3$ $(0,3)$ $2$ $(0,2)$ $5$ $(-2,5)$ $4$ $(-1,4)$ $7$ $(0,7)$ $6$ $(-1,6)$ $9$ $(2,5)$ $8$ $(1,6)$ $11$ $(0,3)$ $10$ $(1,4)$ $11$ $(0,3)$
Table 1: Actual Running Path on the left, GPS readings on the right.
The total distance run is approximately $14.313708$ units, while the GPS measures the distance as approximately $11.650281$ units. The difference between the actual and GPS distance is approximately $2.663427$ units, or approximately $18.607525 \%$ of the total run distance.
Given a sequence of positions and times for a running path, as well as the GPS recording time interval $t$, calculate the percentage of the total run distance that is lost by the GPS. Your computations should assume that the runner goes at a constant speed in a straight line between consecutive positions.
## Input
The input consists of a single test case. The first line contains two integers $n$ ($2 \le n \le 100$) and $t$ ($1 \le t \le 100$), where $n$ is the total number of positions on the running path, and $t$ is the recording time interval of the GPS (in seconds).
The next $n$ lines contain three integers per line. The $i$-th line has three integers $x_ i$, $y_ i$ ($-10^6 \le x_ i, y_ i \le 10^6$), and $t_ i$ ($0 \le t_ i \le 10^6$), giving the coordinates of the $i$-th position on the running path and the time (in seconds) that position is reached. The values of $t_ i$’s are strictly increasing. The first and last positions are the start and end of the run. Thus, $t_1$ is always zero.
It is guaranteed that the total run distance is greater than zero.
## Output
Output the percentage of the actual run distance that is lost by the GPS. The answer is considered correct if it is within $10^{-5}$ of the correct answer.
Sample Input 1 Sample Output 1
6 2
0 0 0
0 3 3
-2 5 5
0 7 7
2 5 9
0 3 11
18.60752550117103
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https://azdikamal.com/what-day-will-it-be-in-9-days/
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# What Day Will It Be In 9 Days?
## The Basics
Have you ever found yourself wondering what day it will be in a certain number of days? If so, you’re not alone. People often need to know the day of the week for various reasons, whether it’s for planning an event or scheduling an appointment. So, what day will it be in 9 days?
The answer is simple: it depends on what day it is today. If today is Monday, then 9 days from now will be a Wednesday. If it’s Tuesday, then it will be a Thursday. And so on.
## How to Calculate the Day
Of course, you don’t have to rely on your memory to figure out what day it will be in 9 days. There are a few different ways you can calculate the answer.
### Method 1: Counting on a Calendar
One of the easiest ways to determine what day it will be in 9 days is to use a calendar. Simply count forward 9 days from today’s date and look at the day of the week that falls on that date. For example, if today is Monday, January 9th, then 9 days from now is Wednesday, January 18th.
### Method 2: Using a Calculator
If you don’t have a calendar handy, you can use a calculator to figure out what day it will be in 9 days. There are several online calculators that can do this for you. All you need to do is enter today’s date and the number of days you want to count forward.
### Method 3: Doing the Math
If you’re feeling mathematically inclined, you can also figure out what day it will be in 9 days by doing the calculation yourself. Here’s how:
1. Figure out what day of the week it is today. For example, if today is Monday, then it’s day 1 of the week.
2. Add 9 to the day of the week. For example, if today is Monday (day 1), then 9 days from now will be day 10.
3. Divide the result by 7 (the number of days in a week). The quotient will be the number of full weeks that have passed. The remainder will be the day of the week. For example, 10 divided by 7 equals 1 with a remainder of 3. So, 9 days from now will be 3 days after the current day of the week (which is Monday), or Wednesday.
## Why Does It Matter?
You might be wondering why it’s important to know what day it will be in 9 days. After all, it’s just a small piece of information. However, there are many situations where knowing the day of the week can be very helpful.
For example, if you’re planning an event that will take place in 9 days, you’ll want to make sure it doesn’t conflict with any other events that are happening on that day. Or, if you’re scheduling an appointment with someone, you’ll want to make sure you’re both available on the same day.
## Conclusion
So, what day will it be in 9 days? The answer depends on what day it is today. You can use a calendar, a calculator, or do the math yourself to figure it out. While it may seem like a small detail, knowing the day of the week can be very useful in many different situations.
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Solved
if statement
Posted on 2012-03-16
165 Views
I need to have a formula changed in sheet 5620-Rhealth & 5640-Retire (the formulas that need tobe chaged start at g6. they need to include the monthly figures in sheet 5390-Stipend for the 3 employees listed in this sheet.
In other words cells g6:ag29 in sheet 5620 represent the amount of dollars we have to spend in retirements cost based on an employees salary, the 3 in sheet 5390 receive an additional stipend that must also be factored into this retirement cost calculation.
expert.xlsx
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Question by:wcody
• 3
• 2
LVL 8
Expert Comment
I'm not sure I fully understood your request but take a look at the attached and let me know if it works for you.
Hope it helps!
expert.xlsx
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LVL 8
Expert Comment
Hi wcody,
0
Author Comment
This doesn't appear to be the solution.
Basically I want sheet 5620 & 5640 to include any salary amount from 5390 in their existing calculation. They are simply mulitplying the salaries from sheet 5100 right now and they need to include the additional salaries listed in sheet 5390.
0
Author Comment
To further clarify shet 5620 cell G8 should be sheet 5100 e8+sheet 539e6* sheet 5620f8
0
LVL 8
Accepted Solution
csoussan earned 500 total points
I've attached a new file. The revised formula is as follows:
=(IFERROR(VLOOKUP(\$B6,'5100-Salaries'!\$B\$6:\$AE\$29,COLUMN(G\$1)-3,FALSE),0)+IFERROR(VLOOKUP(\$B6,'5390-Stipend'!\$B\$6:\$AF\$8,COLUMN(G\$1)-3,FALSE),0))*\$F6
My formula adds the values from '5100-Salaries' to 5390-Stipend where the ID Numbers in column B match and multiples that number by the value in column F.
1) '5100-Salaries' and '5390-Stipend' together before multiple the result by column F.
2) You want all columns multipled by column F.
Let me know if that works. :)
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https://www.bystudin.com/the-circle-intersects-sides-ab-and-ac-of-triangle-abc-at-points-k-and-p-respectively-and-passes-through-the-vertices-b-and-c/
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The circle intersects sides AB and AC of triangle ABC at points K and P, respectively, and passes through the vertices B and C
The circle intersects sides AB and AC of triangle ABC at points K and P, respectively, and passes through the vertices B and C. Find the length of the segment KP if AK = 18 and the side AC is 1.2 times the length of side BC
Consider the quadrilateral PKBC.
PKBC is inscribed in a circle, therefore the condition is fulfilled: the sum of the opposite corners of the quadrangle is 180 ° (the condition that the quadrangle can be inscribed in a circle).
Those. ∠PKB + ∠BCP = 180 °
∠PKB + ∠AKP = 180 ° (as these are adjacent angles).
Therefore, ∠AKP = ∠BCP
Consider the triangles ABC and AKP.
∠AKP = ∠BCP (we found out a little higher)
∠A is common, then these triangles are similar (on the basis of similarity).
Therefore, KP / BC = AK / AC = AP / AB (from the definition of such triangles).
We are interested in the equality KP / BC = AK / AC
KP / BC = 18 / (1,2BC)
KP = 18BC / (1,2BC) = 15
Answer: KP = 15
Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.
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The Induction Motor cannot run at Synchronous speed because the Slip is never zero in an Induction Motor. Now the question arises, why slip is never zero in an induction motor. So in this article, we will discuss the total concept of What is Synchronous speed, What is Slip in an induction motor, etc.
## What is Synchronous Speed?
In simple words, the speed of the rotating magnetic field in the stator winding of the three-phase induction motor is called Synchronous Speed.
The equation of synchronous speed in the induction motor is,
From the above equation, we can say the synchronous speed depends upon the frequency of the power supply, and magnetic poles. We know that our supply frequency is always constant, and the no. of poles is also constant which means the synchronous speed will be constant.
## What is Slip in Induction Motor?
The difference between the rotating magnetic field or synchronous speed and rotor speed with respect to synchronous speed is called Slip.
Slip is measured in percentage. The formula of Slip in an induction motor is,
Here, Ns = Synchronous Speed
Nr = Rotor Speed
## Why Induction Motor cannot run at Synchronous Speed?
The answer to this question can be given in two ways. All are given below to understand easily.
1. The first answer can be given as the torque developed in the rotor is proportional to the Slip. If the rotor runs at synchronous speed then there will be no difference between the synchronous speed and rotor speed, so the Slip will be zero, hence no torque will be developed and the rotor will slow down.
2. The above answer was theoretical. If we see practically when the rotor runs at synchronous speed, there will be no relative speed between them which means both are static with respect to each other. For this reason, the relative motion between the magnetic field and the rotor conductor will be zero, and thus no EMF will be induced in the rotor conductor.
As there no EMF is induced, no current will flow through the rotor conductor and also no flux will be produced and ultimately no torque will be developed and the rotor will slow down.
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